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6400	https://tasks.illustrativemathematics.org/content-standards/4/NF/A/1	Illustrative Mathematics Typesetting math: 100% Engage your students with effective distance learning resources. ACCESS RESOURCES>> Grade 4 Domain Number and Operations—Fractions Cluster Extend understanding of fraction equivalence and ordering. Standard Explain why a fraction a/b is equivalent to a fraction (n×a)/(n×b) by using visual fraction models, with attention to how the number and size of the parts differ even though the two fractions themselves are the same size. Use this principle to recognize and generate equivalent fractions. 4.NF.A.1 Explain why a fraction a/b is equivalent to a fraction (n×a)/(n×b) by using visual fraction models, with attention to how the number and size of the parts differ even though the two fractions themselves are the same size. Use this principle to recognize and generate equivalent fractions. View all 4.NF.A.1 TasksDownload all tasks for this grade Tasks Explaining Fraction Equivalence with Pictures View Details Fractions and Rectangles View Details Typeset May 4, 2016 at 18:58:52. Licensed by Illustrative Mathematics under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License.
6401	https://www.sciencedirect.com/topics/medicine-and-dentistry/uremic-pericarditis	Skip to Main content My account Sign in Uremic Pericarditis In subject area:Medicine and Dentistry Uremic pericarditis is defined as an inflammation of the pericardium that may occur in patients with elevated blood urea nitrogen levels, characterized by chest pain, pericardial friction rub, and potential ECG changes. It is typically associated with renal failure and often prompts the initiation of dialysis if the patient is not already receiving renal replacement therapy. AI generated definition based on: Emergency Medicine (Second Edition), 2013 How useful is this definition? Add to Mendeley Discover other topics Chapters and Articles You might find these chapters and articles relevant to this topic. Chapter Renal Failure 2013, Emergency Medicine (Second Edition)William J. Brady, Amita Sudhir Uremic Pericarditis Uremic pericarditis may occur in patients with a high blood urea nitrogen level. Classic findings of chest pain, pericardial friction rub, worsening of pain when supine, and ECG changes may or may not be present. The presence of any of these signs should trigger a differential diagnosis that includes uremic pericarditis. If no signs of pericardial tamponade are present, pain control and hospital admission are appropriate. Uremic pericarditis traditionally prompts the initiation of dialysis in patients not already receiving renal replacement therapy. Dialysis-associated pericarditis can also occur. It is not well understood, but large effusions may be present and require either increased dialysis or surgical correction.7 View chapterExplore book Read full chapter URL: Book2013, Emergency Medicine (Second Edition)William J. Brady, Amita Sudhir Review article The pathology of pericarditis 2022, Diagnostic HistopathologySakda Sathirareuangchai, Susan Armstrong Miscellaneous pericarditis Uremic pericarditis: the pericardium can be involved in end-stage renal disease (ESRD) in several ways. ESRD patients tend to have pleural and pericardial effusions due to volume overload. Uremic pericarditis is a common entity described in ESRD patients. Constrictive pericarditis rarely occurs following uremic pericarditis, but it has been reported. While the true incidence of uremic pericarditis is unknown, the reported incidence in the literature varies between 2 and 21%.37 With better dialysis techniques and equipment, the global incidence has declined to 5% in patients starting dialysis.25 The pathophysiology of uremic pericarditis is unclear, but accumulation of toxic metabolites seems to be the responsible mechanism. Other explanations include increased endothelial permeability and oxidative stress from high levels of free radicals.37 The pathology in uremic pericarditis is that of fibrinous pericarditis. Sarcoid pericarditis: sarcoidosis is a systemic granulomatous inflammatory disease that commonly affects the myocardium. Pericardial involvement in sarcoidosis is rare, with the prevalence of 3% of sarcoidosis patients at autopsy.13 Pericardial effusion is the most common manifestation, and often asymptomatic. Constrictive pericarditis from sarcoidosis is rare, and only few cases have been reported.38 Pericardiectomy specimen rarely shows the characteristic non-necrotizing granulomatous inflammation. View article Read full article URL: Journal2022, Diagnostic HistopathologySakda Sathirareuangchai, Susan Armstrong Chapter The pericardium and its diseases 2022, Cardiovascular Pathology (Fifth Edition)Susan M. Armstrong, ... Jagdish Butany Uremic pericarditis This is seen in 6%–10% of patients with acute or chronic renal failure, but its incidence has decreased markedly since the widespread use of dialysis (Fig. 15.5). The pathophysiology of uremic pericarditis is not clear. However, it is correlated with increased levels of blood urea nitrogen and creatinine, accumulation of toxic metabolites (methlyguanidine, guanidinoacetate, parathyroid hormone, beta2-microglobulin, and uric acid), hypercalcemia, hyperuricemia, hemorrhagic, viral, and autoimmune mechanisms [7,61]. Grossly, the acute and subacute phases are characterized by shaggy, hemorrhagic, fibrinous exudates on both parietal and visceral surfaces and fibrinous (bread and butter) pericarditis with minimal inflammatory reaction. Subsequently, the effusion gets organized, the pericardium thickens due to fibrosis and dense adhesions are formed between the pericardial layers leading to constriction. Large, gradually accumulating effusions are typically seen in up to 20% of cases . Occasionally, a small amount of asymptomatic pericardial effusion or hemorrhagic effusion may be seen due to fluid overload and platelet dysfunction, respectively. Microscopically, a fibrinous exudate, inflammatory cells, and reactive mesothelial cells are seen. View chapterExplore book Read full chapter URL: Book2022, Cardiovascular Pathology (Fifth Edition)Susan M. Armstrong, ... Jagdish Butany Chapter Surgery in End-Stage Renal Disease Patients 2008, Handbook of Dialysis Therapy (Fourth Edition)Michael J. Moritz MD, Vincent T. Armenti MD, PhD Uremic Pericarditis Uremic pericarditis can occur in dialysis patients presumably because of inadequate removal of uremic toxins. Acutely ill catabolic patients may be relatively underdialyzed and at increased risk for pericarditis. Patients with pericarditis may have chest pain, fever, or a pericardial friction rub—and will usually have a pericardial effusion. The standard treatment is intensive (usually daily) dialysis. Other causes of pericardial effusion include viral and tubercular infections, lupus, drugs such as minoxidil, and malignant pericardial involvement. If the effusion enlarges rapidly or acutely, pericardial tamponade can develop. Signs that a patient may be progressing toward tamponade include loss of a friction rub (due to increasing size of the effusion) and hypotension with tachycardia in the absence of hypovolemia. This presentation can occur during dialysis from fluid removal (decreasing the high preload needed to maintain acceptable hemodynamics), but with the patient still above their dry weight. A high index of suspicion and prompt echocardiography are vital to make a timely diagnosis. Indications for surgical treatment (pericardial window) include tamponade (i.e., hemodynamic compromise) or failure of a large effusion to improve with inten-sive dialysis. Any patient with uremic pericarditis is at risk for bleeding into the pericardium. Therefore, anticoagulants should be avoided. View chapterExplore book Read full chapter URL: Book2008, Handbook of Dialysis Therapy (Fourth Edition)Michael J. Moritz MD, Vincent T. Armenti MD, PhD Chapter Cardiovascular Disease in Chronic Kidney Disease 2019, Chronic Kidney Disease, Dialysis, and Transplantation (Fourth Edition)Mark J. Sarnak MD, MS, Daniel E. Weiner MD, MS Pericardial Disease Pericardial disease in CKD is generally associated with stage 5 CKD. It most commonly manifests as acute uremic or dialysis-associated pericarditis, although chronic constrictive pericarditis may also be seen. Uremic pericarditis describes patients who develop clinical manifestations of pericarditis before or within 8 weeks of initiation of kidney replacement therapy, whereas dialysis-associated pericarditis by definition occurs after a patient is stabilized on dialysis. The precise etiology of uremic pericarditis and dialysis-associated pericarditis is unknown but may be related to inadequate dialysis and volume overload. With high-flux dialysis, uremic pericarditis is exceedingly rare among dialysis patients after initiation.256 Pericarditis may be accompanied by nonspecific symptoms including chest pain, fever, chills, malaise, dyspnea, and cough. Physical examination may reveal a pericardial friction rub. When hemodynamically significant, pericardial disease accompanied by an effusion may be characterized by hypotension, particularly during the hemodialysis procedure.257 Although other expected signs of pericardial effusion may be present, dialysis-related pericarditis often does not manifest with the classic ECG finding of diffuse ST segment elevation because there may be only minimal inflammation of the epicardium.258 Echocardiography is helpful to diagnose pericarditis in dialysis patients; however, effusions may be absent in patients who have adhesive, noneffusive pericarditis. Treatment is dependent on symptoms and effusion size. Small, asymptomatic pericardial effusions are fairly common in dialysis patients and require no acute intervention, whereas larger effusions present a risk for tamponade. Initiation of dialysis is the treatment of choice for uremic pericarditis, whereas intensification of hemodialysis is the mainstay of therapy for dialysis-associated pericarditis but is only effective approximately 50% of the time.256 Traditionally, heparin has been avoided during dialysis out of concern for hemorrhagic tamponade, although this has not been rigorously studied. Adjuvant medical therapies, including oral and intravenous glucocorticoids and nonsteroidal antiinflammatory medications, have generally not been effective. For patients with hemodynamic instability, treatment consists of emergent drainage of the pericardial effusion. This is generally accomplished by pericardiocentesis or pericardiotomy with or without pericardiostomy for the instillation of long-acting, nonabsorbable glucocorticoids.259 View chapterExplore book Read full chapter URL: Book2019, Chronic Kidney Disease, Dialysis, and Transplantation (Fourth Edition)Mark J. Sarnak MD, MS, Daniel E. Weiner MD, MS Review article Renal: Part 2 of 3 2011, MedicineVictoria J. Ingham, Lui G. Forni Uraemic pericarditis Uraemic pericarditis is observed in 6–10% of patients with advanced renal failure, resulting from inflammation of both the visceral and parietal membranes of the pericardial sac. Pericarditis in AKI presents with fever and pleuritic chest pain, characteristically worse in the recumbent position, and a pericardial rub may be heard on auscultation, although the ECG does not show the typical diffuse ST and T wave elevations observed with other causes of acute pericarditis. The development of pericarditis in a patient with AKI is an indication to institute RRT, unless there are signs of cardiac tamponade due to a pericardial effusion. Under such conditions, heparin-free haemodialysis or haemofiltration should be used because of the risk of increased bleeding into the pericardial sac with anticoagulation. View article Read full article URL: Journal2011, MedicineVictoria J. Ingham, Lui G. Forni Chapter Acute Pericarditis 2009, Pericardial DiseasesStuart J. Hutchison MD, FRCPC, FACC, FAHA, FASE, FSCMR, FSCCT, ... Stuart J. Hutchison Metabolic Pericarditis Uremic pericarditis is encountered in a third of uremic patients. The associated mortality is high. Dialysis-related pericarditis is common and occurs often without typical features of acute pericarditis. Associated effusions with some cases of tamponade and late cases of constriction are seen. Myxedema is associated in about a third of cases with pericarditis but more often with effusions alone. Myxedema-related effusions may have cholesterol particles. Effusions regress when a euthyroid state is re-established. Amyloidosis is usually associated with small pericardial effusions, with a low likelihood of tamponade. Late constriction has been described. View chapterExplore book Read full chapter URL: Book2009, Pericardial DiseasesStuart J. Hutchison MD, FRCPC, FACC, FAHA, FASE, FSCMR, FSCCT, ... Stuart J. Hutchison Chapter Cardiovascular Disease in Chronic Kidney Disease 2019, Chronic Kidney Disease, Dialysis, and Transplantation (Fourth Edition)Mark J. Sarnak MD, MS, Daniel E. Weiner MD, MS Structural Disease: Percardial and Valvular Conditions Pericardial Disease Pericardial disease in CKD is generally associated with stage 5 CKD. It most commonly manifests as acute uremic or dialysis-associated pericarditis, although chronic constrictive pericarditis may also be seen. Uremic pericarditis describes patients who develop clinical manifestations of pericarditis before or within 8 weeks of initiation of kidney replacement therapy, whereas dialysis-associated pericarditis by definition occurs after a patient is stabilized on dialysis. The precise etiology of uremic pericarditis and dialysis-associated pericarditis is unknown but may be related to inadequate dialysis and volume overload. With high-flux dialysis, uremic pericarditis is exceedingly rare among dialysis patients after initiation.256 Pericarditis may be accompanied by nonspecific symptoms including chest pain, fever, chills, malaise, dyspnea, and cough. Physical examination may reveal a pericardial friction rub. When hemodynamically significant, pericardial disease accompanied by an effusion may be characterized by hypotension, particularly during the hemodialysis procedure.257 Although other expected signs of pericardial effusion may be present, dialysis-related pericarditis often does not manifest with the classic ECG finding of diffuse ST segment elevation because there may be only minimal inflammation of the epicardium.258 Echocardiography is helpful to diagnose pericarditis in dialysis patients; however, effusions may be absent in patients who have adhesive, noneffusive pericarditis. Treatment is dependent on symptoms and effusion size. Small, asymptomatic pericardial effusions are fairly common in dialysis patients and require no acute intervention, whereas larger effusions present a risk for tamponade. Initiation of dialysis is the treatment of choice for uremic pericarditis, whereas intensification of hemodialysis is the mainstay of therapy for dialysis-associated pericarditis but is only effective approximately 50% of the time.256 Traditionally, heparin has been avoided during dialysis out of concern for hemorrhagic tamponade, although this has not been rigorously studied. Adjuvant medical therapies, including oral and intravenous glucocorticoids and nonsteroidal antiinflammatory medications, have generally not been effective. For patients with hemodynamic instability, treatment consists of emergent drainage of the pericardial effusion. This is generally accomplished by pericardiocentesis or pericardiotomy with or without pericardiostomy for the instillation of long-acting, nonabsorbable glucocorticoids.259 Endocarditis Infective endocarditis is a relatively common complication of hemodialysis,260-263 which reflects the relatively high incidence of bacteremia, chronic use of dialysis catheters, and the high prevalence of preexisting valvular abnormalities.264-266 The majority of endocarditis in hemodialysis patients is secondary to gram-positive organisms, with Staphylococcus aureus predominating.263,267-269 Dialysis patients with endocarditis usually have fever; murmurs, leukocytosis, and septic emboli may also be common. The mitral and aortic valves are most often affected.263 Diagnosis is chiefly dependent on positive blood cultures and clinical suspicion; clinical suspicion should be high in settings where bacteremia is persistent and in individuals with prior history of endocarditis. Transthoracic and transesophageal echocardiography are important in establishing the diagnosis. Treatment of endocarditis begins with appropriate antibiotic therapy; even with appropriate therapy, however, survival is often poor, with case series showing 30% mortality during the initial hospitalization and 1-year mortality over 50%.261,263 Surgical intervention may also be appropriate, and indications for surgery are the same as in the general population: progressive valvular destruction, progressive heart failure, recurrent systemic emboli, and failure to respond to appropriate antibiotic therapy. Factors associated with mortality include hypoalbuminemia, involvement of multiple valves, and severe valvular insufficiency. In one study, 30-day survival among patients who had surgery was 80%, whereas it was only 47% among those managed medically.268 Although current data are observational, an important inference to be made is that hemodialysis patients with endocarditis should be considered surgical candidates if they have indications. Mitral Annular Calcification Mitral annular calcification may occur in 30% to 50% of patients on dialysis and is also common in patients during the earlier stages of CKD.270,271 It is recognized on echocardiography as a uniform, echodense, rigid band located near the base of the posterior mitral leaflet and may progressively involve the posterior leaflet. The pathogenesis of mitral calcification may be linked to altered mineral metabolism.271,272 Serious complications of mitral annular calcification include conduction abnormalities, embolic phenomena, mitral valve disease, and an increased risk of endocarditis.273 Aortic Calcification and Stenosis Aortic valve calcification is common in dialysis patients, occurring in 28% to 55% of patients. Although the overall prevalence is similar to that seen in the general population, dialysis patients experience aortic valve calcification 10 to 20 years earlier than the general population.274 Age is the most significant risk factor for aortic valve calcification,273 and abnormal mineral metabolism may also play a role.275 The most significant hazard associated with aortic valve calcification is the potential for the development of progressive immobilization of the aortic leaflets, which eventually restricts flow. Aortic stenosis occurs when the valve leaflets thicken to the extent that commissural fusion can no longer occur and a pressure gradient develops across the aortic valve. In one study of dialysis patients the estimated incidence of symptomatic aortic stenosis was 3.3% per year.275 Progression of aortic valve calcification to aortic stenosis in dialysis patients appears more rapid than in the general population.276 Very little evidence exists in the nondialysis CKD population as to the prevalence and progression of valvular abnormalities. Angina, heart failure, and syncope are the cardinal symptoms of critical aortic stenosis. Clinical evidence of aortic stenosis may be more readily evident in dialysis patients as they may have more frequent episodes of intradialytic hypotension, particularly as ultrafiltration can rapidly reduce preload. Treatment of aortic stenosis is multifaceted, encompassing prevention of progression, eventual repair of the valve, and prevention of complications, including endocarditis consistent with general population guidelines for those with prior infective endocarditis or prosthetic valves.277 Management of mineral metabolism abnormalities could theoretically slow progression of aortic stenosis, although this has not been proven. Valve replacement is the therapy of choice for critical aortic stenosis, and the timing of surgery is dependent on individual patient characteristics with the caveat that surgery should be performed before left ventricular contractility becomes diminished. Currently there is no consensus about a benefit of prosthetic versus bioprosthetic valves in dialysis patients.278 Dialysis patients undergoing valve replacement have a high mortality rate: 17% operative mortality for aortic valve replacement in dialysis patients, 23% for mitral valve replacement, 25% for aortic valve replacement and CABG, and 37% for mitral valve replacement and CABG.279 However, in most cases the prognosis is worse if clinically indicated surgery is not performed or if emergent, rather than elective, surgery is performed.270 Transcatheter aortic valve replacement (TAVR), increasingly being used in advanced CKD and dialysis patients, likely has lower risks than surgery, although one series did report 1-year mortality of 30% after TAVR in a dialysis population, whereas other studies reported 1-year mortality of 24% to 35% in CKD stage 4. Of note, evaluation for TAVR requires intravenous contrast administration.280-282 View chapterExplore book Read full chapter URL: Book2019, Chronic Kidney Disease, Dialysis, and Transplantation (Fourth Edition)Mark J. Sarnak MD, MS, Daniel E. Weiner MD, MS Chapter Management of Ischemic Heart Disease, Heart Failure, and Pericarditis in Patients Undergoing Long-Term Dialysis 2023, Handbook of Dialysis Therapy (Sixth Edition)Ian E. McCoy MD, MS Uremic Pericarditis Because of the earlier initiation of dialysis and higher delivered dialysis doses, clinically significant uremic pericarditis is now a relatively rare event. When it does occur, the symptoms may be sudden and have severe consequences if not recognized and treated promptly. A high index of suspicion is therefore required. Between 50% and 70% of patients with uremic pericarditis will respond to initiation or intensification of dialysis, suggesting uremia itself is responsible in some cases. Cases that do not improve with dialysis may be due to other etiologies, such as viral pericarditis. Clinical Presentation and Diagnosis The typical clinical presentation of pericarditis may include precordial pain, dyspnea, cough, or fever. The chest pain is not related to exertion, may be pleuritic in nature, and is often relieved by leaning forward. Most patients will have a pericardial friction rub. Those patients with a significant pericardial effusion may have jugular venous distention or a paradoxical pulse. Any of these symptoms in the presence of new-onset hypotension makes pericardial tamponade a likely diagnosis. Electrocardiograph findings (classically diffuse ST changes in nonuremic pericarditis) are not sensitive or specific in uremic pericarditis. Echocardiography should be performed in any patient suspected of having a pericardial effusion to estimate its size and guide further management. Management For patients who are hemodynamically stable and without impending tamponade, intensification of dialysis and analgesia is usually all that is required for management. Fluid removal with dialysis should be limited and cautious given preload dependence and the potential for hypotension and cardiogenic shock. Pericardiocentesis can be considered if not improving after 7–14 days. Nonsteroidal anti-inflammatory drugs, used cautiously, are useful for reducing pain but do not appear to hasten the resolution of pericarditis. Prednisone is not recommended and may actually increase morbidity. Heparin should be avoided during dialysis therapy if at all possible to prevent hemorrhagic complications. Patients should be carefully monitored for clinical evidence of tamponade until all symptoms have resolved. Hemodynamically unstable patients who have tamponade require immediate pericardiocentesis. Blind pericardiocentesis is associated with considerable risks and is therefore indicated only in the most urgent cases. Patients with a large pericardial effusion or those with unstable hemodynamics and/or echocardiographic evidence of cardiac-chamber compromise (even with a moderate pericardial effusion) should be referred for pericardiostomy, pericardial window, or pericardiectomy. The selection of the surgical procedure should be based on clinical and hemodynamic status, comorbid conditions, and the experience of the physician. Pericardial window is often the procedure of choice. View chapterExplore book Read full chapter URL: Book2023, Handbook of Dialysis Therapy (Sixth Edition)Ian E. McCoy MD, MS Related terms: Uremia Pericarditis Acute Kidney Injury Chronic Kidney Disease Tumor Lysis Syndrome Renal Replacement Therapy Pericardial Fluid Effusion Hemodialysis End Stage Renal Disease View all Topics
6402	https://mikesmathpage.wordpress.com/2014/11/29/problem-solving-part-2-an-old-aime-problem/	Problem Solving part 2 – an old AIME problem – Mike's Math Page Skip to content Mike's Math Page mikesmathpage Problem Solving part 2 – an old AIME problem mjlawlerUncategorizedNovember 29, 2014 November 29, 2014 4 Minutes [note: I’m up in Boston today and decided to take a trip to the MIT library to check out a book that Jordan Ellenberg mentions in his book “How not to be Wrong.” I didn’t know the MIT libraries were closed until 11:00 am today (they used to be open 24 hours!!), so I wrote this post fairly quickly in the hour I had to kill waiting for the library to open!. Sorry if it seems rushed – it sort of was!] I few months ago I wrote about Tim Gowers live blogging his solution to one of the IMO problems: Problem Solving and Tim Gowers’s live blogging an IMO problem I thought that live blogging problem solving was a good idea because I think that kids (and everyone) needs to see that most solutions you get to problems aren’t the super perfect “official solutions” and don’t come to mind immediately. In that post I talked through my solution to a number theory problem that David Radcliffe had put on Twitter. Today’s blog is my second attempt at live blogging a problem. I saw today’s problem in Art of Problem Solving’s Precalculus book in the chapter 8 challenge problems. It also happens to be problem 15 from the 1991 AIME – see the bottom of the page here: The 1991 AIME Problems hosted by Art of Problem Solving Here is the problem: For a positive integer define to be the minimum value of the sum: where are positive real numbers whose sum is 17. There is a unique positive integer for which is also an integer, find this The “live blogging” of my solution is below: Background: I had a nice (and short) discussion on twitter with Patrick Honner about Art of Problem Solving’s Precalculus book. I personally like this book because of chapter 8 – the geometry of complex numbers. This chapter, and in particular section 8.5, has content that I’ve not seen anywhere else. Advertisement Following that discussion, I grabbed my copy of the book to take a fresh look at chapter 8. I ended up back in the challenge problems and the problem I’m writing about today caught my eye. My first reaction, frankly, was fear. Back in high school I’d not really understood analysis all that well, and did not have a good grasp of theorems that helped with this type of summation problem – the Cauchy Schwarz inequality, for example: The Cauchy-Schwarz Inequality. An “analysis” approach to this problem probably requires an even more general inequality – Hölder’s inequality – though I would not have known that in high school. Whenever I see this type of problem, even today, I remember the fear of the analysis ideas that I had in high school. Funny how those old high school fears never seem to go away! But wait – why would the Art of Problem Solving folks put an analysis problem in the chapter about the geometry of complex numbers? There must be a geometric solution here somewhere? Where is it? So, let’s look at a few terms: (1) The “sum” is just one term . Certainly not an integer, but also certainly reminiscent of the Pythagorean theorem and a right triangle with legs of length 1 and 17. (2) The sum is now . Now there are two right triangles – one with legs 1 and , and another with legs 3 and . The sum of the integer length legs is 4, and the sum of the other legs is 17. Oh, wait, I see what’s going on. The general case is going to have a bunch of right triangles whose “heights” always adds up to 17 and whose “lengths” will sum up to See here for the sum of the odd numbers: Advertisement How cool! The minimum value of the sum we are being asked to consider is going to come when we can make all of the hypotenuses of these triangles form a straight line, and that length will be from a “large” right triangle with length and height 17. Sweet – per the statement of the problem, there must only be one right triangle with integer side lengths where one of the side lengths is 17. That shouldn’t be too hard to find. We know there’s a 3,4,5 triangle, and a 5, 12, 13 triangle, and a 7, 24, 25 triangle – which helps see the pattern fairly quickly. For an odd number, you’ll get a right triangle with integer sides by finding two consecutive numbers that add up to the square of the odd number. From the three examples above: 4 + 5 = 33, 12 + 13 = 55, and 24 + 25 = 77. So . . . 144 + 145 = 289 = 1717, so there must be a 17, 144, 145 triangle. By luck 144 is a perfect square and we indeed have a right triangle with integer sides having one leg equal to 17 and the other leg equal to a perfect square. That means the minimum value of the sum we were looking for is 145, and the value for that particular triangle is 12. Fun and super clever problem. Love the connection between geometry and algebra here – especially because the algebra side of this problem still makes me nervous Most of the thoughts above were in my head – my “work” for the problem is in the picture below. I’m glad I saw the geometric solution before trying to dive into all of the analysis which, I assume, is way more grungy and way less of a fun solution. Share this: Click to share on X (Opens in new window)X Click to share on Facebook (Opens in new window)Facebook Like Loading... Tagged Art of Problem Solving contest math geometry Published by mjlawler View all posts by mjlawler PublishedNovember 29, 2014 November 29, 2014 Post navigation Previous Post The last 4 digits of Graham’s number Next Post Angry Birds and Snap Cubes: Using Bryna Kra’s MoMath public lecture to talk math with kids Leave a comment Cancel reply Δ Search by Date November 2014\| M \| T \| W \| T \| F \| S \| S \| --- --- --- \| \| 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 \| « OctDec » Search by Name Search for: Search By Tag 3d geometry3D Printing4th DimensionalgebraarithmeticArt of Problem SolvingbasesbinarycalculusCatriona Shearercomplex numberscomputersconstructioncontest mathcontinued fractionscountingCut The KnotDan AndersonDavid RadcliffedecimalsEd FrenkelEvelyn LambexponentialsFawn NguyenFibonacci NumbersFive TrianglesFractalsfractionsgamesgeometryGeometry RevisitedGrant Sandersonimaginary numbersinfinite seriesinfinityJames TantonJim ProppJordan EllenbergKate NowakKeith DevlinKelsey Houston-EdwardsLaura Taalmanlearning mathlogarithmsmodelingMoMathMr. WatermanNumberphilenumber sensenumber theoryPascal's trianglePatrick Honnerpattern recognitionpatternsPhysicsplace valueprimesprobabilityproblem solvingproofquadratic formulasnap cubesstatisticsSteven StrogatzsymmetryTerry TaotopologyTracy Johnston Zagertrianglestrigonometryultimate frisbeeunsolved problemsVi Hartworkzometool Create a free website or blog at WordPress.com. 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6403	https://www.atlantapublicschools.us/cms/lib/GA01000924/Centricity/ModuleInstance/9012/Pythagorean_Theorem_Examples.pdf	Extra Practice 12 PYTHAGOREAN THEOREM #5 #5 a b c Any triangle that has a right angle is called a RIGHT TRIANGLE. The two sides that form the right angle, a and b, are called LEGS, and the side opposite (that is, across the triangle from) the right angle, c, is called the HYPOTENUSE. For any right triangle, the sum of the squares of the legs of the triangle is equal to the square of the hypotenuse, that is, a2 + b2 = c2. This relationship is known as the PYTHAGOREAN THEOREM. In words, the theorem states that: (leg)2 + (leg)2 = (hypotenuse)2. Example Draw a diagram, then use the Pythagorean Theorem to write an equation to solve each problem. a) Solve for the missing side. 13 c 17 c2 +132 = 172 c2 +169 = 289 c2 = 120 c = 120 c = 2 30 c ! 10.95 b) Find x to the nearest tenth: 5x x 20 (5x)2 + x2 = 202 25x2 + x2 = 400 26x2 = 400 x2 ! 15.4 x ! 15.4 x ! 3.9 c) One end of a ten foot ladder is four feet from the base of a wall. How high on the wall does the top of the ladder touch? x 4 10 4 x2 + 42 = 102 x2 + 16 = 100 x2 = 84 x ≈ 9.2 The ladder touches the wall about 9.2 feet above the ground. d) Could 3, 6 and 8 represent the lengths of the sides of a right triangle? Explain. 32 + 62 = ? 82 9 + 36 = ? 64 45!64 Since the Pythagorean Theorem relationship is not true for these lengths, they cannot be the side lengths of a right triangle. GEOMETRY Connections 13 Use the Pythagorean Theorem to find the value of x. Round answers to the nearest tenth. 1. 2. 3. 4. 5. 37 22 x x 96 20 x 42 16 46 83 x x 65 72 6. 7. 8. 9. 10. 22 16 x 32 15 x 38 16 x 75 105 x 125 30 x Solve the following problems. 11. A 12 foot ladder is six feet from a wall. How high on the wall does the ladder touch? 12. A 15 foot ladder is five feet from a wall. How high on the wall does the ladder touch? 13. A 9 foot ladder is three feet from a wall. How high on the wall does the ladder touch? 14. A 12 foot ladder is three and a half feet from a wall. How high on the wall does the ladder touch? 15. A 6 foot ladder is one and a half feet from a wall. How high on the wall does the ladder touch? 16. Could 2, 3, and 6 represent the lengths of sides of a right angle triangle? Justify your answer. 17. Could 8, 12, and 13 represent the lengths of sides of a right triangle? Justify your answer. 18. Could 5, 12, and 13 represent the lengths of sides of a right triangle? Justify your answer. 19. Could 9, 12, and 15 represent the lengths of sides of a right triangle? Justify your answer. 20. Could 10, 15, and 20 represent the lengths of sides of a right triangle? Justify your answer. Answers 1. 29.7 2. 93.9 3. 44.9 4. 69.1 5. 31.0 6. 15.1 7. 35.3 8. 34.5 9. 73.5 10. 121.3 11. 10.4 ft 12. 14.1 ft 13. 8.5 ft 14. 11.5 ft 15. 5.8 ft 16. no 17. no 18. yes 19. yes 20. no
6404	https://www.scribd.com/document/450835596/Ch-8-Answer-Key-CK-12-MSM-Concepts-Grade-6-PDF	CH 8 Answer Key CK-12 MSM Concepts - Grade 6 PDF \| PDF \| Ratio \| Fraction (Mathematics) Opens in a new window Opens an external website Opens an external website in a new window This website utilizes technologies such as cookies to enable essential site functionality, as well as for analytics, personalization, and targeted advertising. To learn more, view the following link: Privacy Policy Open navigation menu Close suggestions Search Search en Change Language Upload Sign in Sign in Download free for 30 days 0 ratings 0% found this document useful (0 votes) 326 views 8 pages CH 8 Answer Key CK-12 MSM Concepts - Grade 6 PDF This document contains the answers to problems in Chapter 8 on ratios, proportions, and percents from a math textbook for 6th grade middle school students. The chapter covers topics like equ… Full description Uploaded by Karthi Keyan AI-enhanced title and description Go to previous items Go to next items Download Save Save Ch 8 Answer Key CK-12 MSM Concepts - Grade 6 (PDF) For Later Share 0%0% found this document useful, undefined 0%, undefined Print Embed Ask AI Report Download Save Ch 8 Answer Key CK-12 MSM Concepts - Grade 6 (PDF) For Later You are on page 1/ 8 Search Fullscreen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">>!," "/>!-" "./% !%" "!> !"#; I,46,-.0 0 746.(>-0 8-0:,40 !" %-'!88 " !!'!88 ," .'!88 -" !'!88 %" >0'!88 ." >,'!88 /" !0'!88 >" -'!88 0" 0'!88 !8" !>'!88 !!" >0'!88 !" !88'!88 adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free Share this document Share on Facebook, opens a new window Share on LinkedIn, opens a new window Share with Email, opens mail client Copy link Millions of documents at your fingertips, ad-free Subscribe with a free trial You might also like PHY104 Electricity Lectures 2024RevisedFinal No ratings yet PHY104 Electricity Lectures 2024RevisedFinal 156 pages Coding For Kids Python A Playful Way For - Mark B Bennet 100% (1) Coding For Kids Python A Playful Way For - Mark B Bennet 143 pages Aristotle - History of Animals No ratings yet Aristotle - History of Animals 183 pages Professional Education-Curriculum Development (Let Reviewer) 100% (11) Professional Education-Curriculum Development (Let Reviewer) 7 pages Generator Spare Parts Budget-2020 No ratings yet Generator Spare Parts Budget-2020 106 pages BBMA Flow Diagram No ratings yet BBMA Flow Diagram 212 pages Geometric Series No ratings yet Geometric Series 16 pages Introduction To Well Planning, GTO and Drilling Terms No ratings yet Introduction To Well Planning, GTO and Drilling Terms 73 pages 6.RP.A.3 Ratio and Rates Word Problems No ratings yet 6.RP.A.3 Ratio and Rates Word Problems 17 pages Ratio Proportion Percent Worksheets 100% (2) Ratio Proportion Percent Worksheets 47 pages Landform Development Theories No ratings yet Landform Development Theories 25 pages BS 5493 1977 Amd 2 Code of Practice For Protective Coating o PDF No ratings yet BS 5493 1977 Amd 2 Code of Practice For Protective Coating o PDF 118 pages (L6) - (JEE 2.0) - 3D Geometry - 28th Nov No ratings yet (L6) - (JEE 2.0) - 3D Geometry - 28th Nov 44 pages Cashoutflow No ratings yet Cashoutflow 36 pages Decimals Worksheets No ratings yet Decimals Worksheets 50 pages Catalogo Juntas Rotativas DEUBLIN 100% (1) Catalogo Juntas Rotativas DEUBLIN 32 pages Outstanding Digest No ratings yet Outstanding Digest 10 pages The Effect of Macrocelebrity and Microin Uencer Endorsements On Consumer-Brand Engagement in Instagram No ratings yet The Effect of Macrocelebrity and Microin Uencer Endorsements On Consumer-Brand Engagement in Instagram 21 pages Compitators No ratings yet Compitators 32 pages Mechanics of Structure No ratings yet Mechanics of Structure 16 pages Law Firm Questions No ratings yet Law Firm Questions 5 pages OB - Product Design - Eng No ratings yet OB - Product Design - Eng 29 pages Wharton 2010 No ratings yet Wharton 2010 81 pages Witsoc Reviewer No ratings yet Witsoc Reviewer 11 pages Rizal Paris To Berlin No ratings yet Rizal Paris To Berlin 14 pages Chapter 12 - Fractions, Decimals, Ratios, and Percents - Nelson Math 6 0% (1) Chapter 12 - Fractions, Decimals, Ratios, and Percents - Nelson Math 6 12 pages 123 624 1 PB No ratings yet 123 624 1 PB 14 pages An Optimized Grounded Base Oscillator Design For VHF/UHF No ratings yet An Optimized Grounded Base Oscillator Design For VHF/UHF 12 pages Rex C. 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6405	https://qmplus.qmul.ac.uk/mod/resource/view.php?id=2772599	Notice \| MyQMUL Skip to main content Side panel Home Calendar All Modules HelpRaise a TicketUser Guides More Sitewide search Search Close Perform search Toggle search input You are currently using guest access Log in This course is currently unavailable to students Continue You are currently using guest access (Log in) Get the mobile app Powered by Moodle Student Life Student email My QMUL Queen Mary Students' Union Student Enquiry Centre QMplus for students Careers Library Academic Skills Centre Library Landing Page Library Website Cite Them Right Archives Archive 2024/25 2023/24 2022/23 2021/22 2020/21 ©Queen Mary University of London Site policies and guidelines\|Manual login Loading... All Modules HelpRaise a TicketUser Guides Home Calendar
6406	https://www.youtube.com/watch?v=6RymMUj-9ZE	RL High pass Filter, One zero, One pole, Magnitude, Phase and 3dB Frequency eeKnowHow 4850 subscribers 48 likes Description 8880 views Posted: 14 Oct 2016 RL HIgh pass filter. Calculation of Magnitude, Phase and 3dB frequency from the transfer function. Plotting of approximate Bode plot. 2 comments Transcript: welcome to Anam in this video we will look at a RL high-pass filter and will calculate the magnitude phase and the 3 DB frequency for this so the RL high-pass filter looks like this you'll have R and L in series and we will take the output across the inductor and the R is at the input so this is V out here and VN is here so now if you write the transfer function for this so V out over VN it can be written as a cell which is the reactance of the impedance of the inductor and then divided by R plus SL because the output is dropped across the inductor the input is dropped across the series combination of the R and I so this can be written as s we will divide both the numerator and the denominator width by r so it will be s L over R divided by 1 plus s L over R so this is in the form of we have seen earlier yes tau over 1 plus s tau so we're towers the time constant so you have a 0 at 0 and a pole at 1 over tau so the pole s 1 over L over R that's where the pole is so that is the that is how it looks and now we will kind of simplify it further what we will do is we will forgetting the magnitude or I will divide the numerator and the denominator with SL over R so you would get it as 1 over 1 over s L over R plus so which is nothing but one over one over J Omega L over R plus one which is further simplified to one over the J in the denominator can go up and become minus J so let's let me put the one first one sorry this has to be a minus here so 1 minus J over Omega L so now from this equation we can get both the magnitude in the face so the magnitude in DB let me write it in DB would be 20 log to the base 10 1 over square root of 1 plus 1 by Omega square L square over R yeah we take an example in calculate you'll see a nice you know you could do it in Excel and you can get this magnitude how it varies with frequency so this is how the magnitude of the RL high pass filter varies with frequency now if you write phase now phase yes - why do you put minuses because we are dealing with terms in the denominator minus tan inverse and then you have the imaginary part of the real part the imaginary part is 1 over Omega L over R divided that's minus actually and then over 1 so tan inverse minus X is minus tan X so the minus comes out and get cancelled with the other minus so it would be tan inverse 1 our Omega you know oh so this is the face face with respect to frequency how it varies at the output now we got both of these and now we want to calculate the 3 DB frequency here the 3 DB frequency even said when the magnitude is the real magnet the Norden DB but the actual magnitude becomes 1 over root 2 or in the in the DB it will be minus 3 DB that happens when Omega square L square R square is 1 or that happens when Omega L over R is equal to 1 so 2 pi F 3 DB 2 pi F is Omega L over R is equal to 1 so we can say Omega naught is 1 over L o and then F 3 DB will be equal to 1 over 2 pi L over R so this is the Omega naught in radiance and the F 3 DB in degrees so F 3 DB is here so this is similar to an RC circuit where RC is the time constant so the RC came into the picture here L over R is the time constant so the L over R came into the picture for calculating the 3 DB frequency now if you look at the the phase the magnitude the plot it would look similar and at Omega naught you would get a minus 3 dB and then from then on it increases goes to zero DB and from here on it is 20 DB per decade slope and it keeps falling so 20 DB per decade this is the magnitude plot this is the magnitude in DB with respect to frequency yeah and this is a log scale when we do bode plots it's always log scale on the frequency now if you look at the the phase the phase would look similar to a high-pass filter where initially it starts with zero and that at Omega naught it is 45 degrees and 10 Omega naught we can approximate it to 90 and 0.1 Omega naught we can approximate it to zero so it goes like this this is 90 degrees but in reality the curve we saw that it's going to look like it goes passes through the 45 it'll look like this so if you actually do the calculation for the face using the tan inverse 1 over Omega L over R so we can see the actual curve we can plot that choker so this is the RL high-pass filter
6407	https://www.doubtnut.com/qna/1460705	Show that f(x)=1x is a decreasing function on (0, ∞) . More from this Exercise To show that the function f(x)=1x is a decreasing function on the interval (0,∞), we can follow these steps: Step 1: Find the derivative of the function We start by differentiating the function f(x): f′(x)=ddx(1x) Using the power rule, we can rewrite 1x as x−1: f′(x)=ddx(x−1)=−1⋅x−2=−1x2 Step 2: Analyze the sign of the derivative Next, we need to analyze the sign of f′(x) in the interval (0,∞): f′(x)=−1x2 Since x2>0 for all x>0, it follows that 1x2>0. Therefore, we have: f′(x)<0for all x∈(0,∞) Step 3: Conclusion Since the derivative f′(x) is negative for all x in the interval (0,∞), we conclude that the function f(x)=1x is a decreasing function on the interval (0,∞). Summary Thus, we have shown that f(x)=1x is a decreasing function on (0,∞). --- To show that the function f(x)=1x is a decreasing function on the interval (0,∞), we can follow these steps: Step 1: Find the derivative of the function We start by differentiating the function f(x): f′(x)=ddx(1x) Using the power rule, we can rewrite 1x as x−1: f′(x)=ddx(x−1)=−1⋅x−2=−1x2 Step 2: Analyze the sign of the derivative Next, we need to analyze the sign of f′(x) in the interval (0,∞): f′(x)=−1x2 Since x2>0 for all x>0, it follows that 1x2>0. Therefore, we have: f′(x)<0for all x∈(0,∞) Step 3: Conclusion Since the derivative f′(x) is negative for all x in the interval (0,∞), we conclude that the function f(x)=1x is a decreasing function on the interval (0,∞). Summary Thus, we have shown that f(x)=1x is a decreasing function on (0,∞). Topper's Solved these Questions Explore 176 Videos Explore 1401 Videos Similar Questions Show that f(x)=1x is decreasing function on (0,∞). Show that f(x)=cosx is a decreasing function on (0, π) , increasing in (π, 0) and neither increasing nor decreasing in (π, π) . Show that f(x)=tan−1x−x is decreasing function on R. Show that f(x)=cosx is decreasing function on (0,π), increasing in (−π,0) and neither increasing nor decreasing in (−π,π). Show that f(x)=e1/x, x≠0 is a decreasing function for all x≠0 . Show that f(x)=e−x is a strictly decreasing function on Rgt Show that f(x)=e1x,x≠0 is decreasing function for all x≠0. Show that f(x)=e1/x is a strictly decreasing function for all x>0 Show that the function f(x)=x2 is a strictly increasing function on (0,∞). Show that the function f(x)=x2 is a strictly decreasing function on (−∞, 0] . RD SHARMA-INCREASING AND DECREASING FUNCTION -Solved Examples And Exercises Prove that f(x)=a x+b , where a ,\ b are constants and a >0 is an incr... Prove that f(x)=a x+b , where a ,\ b are constants and a<0 is a decrea... Show that f(x)=1/x is a decreasing function on (0,\ oo) . 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Find the intervals for which f(x)=x^4-2x^2 is increasing or decreasing... Determine the values of x for which f(x)=(x-2)/(x+2),\ \ x!=-1 is incr... Find the intervals in which f(x)=x/2+2/x ,\ \ x!=0 is increasing or de... Find the intervals in which the function f given by f(x)=\ x^3+1/(... Exams Free Textbook Solutions Free Ncert Solutions English Medium Free Ncert Solutions Hindi Medium Boards Resources Doubtnut is No.1 Study App and Learning App with Instant Video Solutions for NCERT Class 6, Class 7, Class 8, Class 9, Class 10, Class 11 and Class 12, IIT JEE prep, NEET preparation and CBSE, UP Board, Bihar Board, Rajasthan Board, MP Board, Telangana Board etc NCERT solutions for CBSE and other state boards is a key requirement for students. Doubtnut helps with homework, doubts and solutions to all the questions. It has helped students get under AIR 100 in NEET & IIT JEE. 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6408	https://en.wikipedia.org/wiki/Ammonia_(data_page)	Published Time: 2005-07-17T17:11:05Z Ammonia (data page) - Wikipedia Jump to content [x] Main menu Main menu move to sidebar hide Navigation Main page Contents Current events Random article About Wikipedia Contact us Contribute Help Learn to edit Community portal Recent changes Upload file Special pages Search Search [x] Appearance Donate Create account Log in [x] Personal tools Donate Create account Log in Pages for logged out editors learn more Contributions Talk Contents move to sidebar hide (Top) 1 Structure and properties 2 Thermodynamic properties 3 Vapor–liquid equilibrium data 4 Heat capacity of liquid and vapor 5 Spectral data 6 Regulatory data 7 Safety data sheet 8 References 9 External links [x] Toggle the table of contents Ammonia (data page) [x] 1 language Magyar Edit links Article Talk [x] English Read Edit View history [x] Tools Tools move to sidebar hide Actions Read Edit View history General What links here Related changes Upload file Permanent link Page information Cite this page Get shortened URL Download QR code Print/export Download as PDF Printable version In other projects Wikidata item Appearance move to sidebar hide From Wikipedia, the free encyclopedia Chemical data page This page provides supplementary chemical data on ammonia. Structure and properties [edit] \| Molecular structure \| \| Point group \| C 3v \| \| Bond length \| 101.2 pm (N–H) \| \| Bond angle \| 106.7° (H–N–H) \| \| Bond strength \| 435 kJ/mol (H–NH 2) \| \| Crystal data \| \| Crystal structure \| ? \| \| Properties \| \| Dipole moment \| 1.46 D \| \| Dielectric constant \| 22 ε 0 at 239 K \| \| Magnetic susceptibility \| diamagnetic \| \| Acidity of NH 4+ (p K a) \| 9.25 \| Thermodynamic properties [edit] Phase diagram and crystalline states of ammonia I. cubic, II. hcp, III. fcc, IV. orthorhombic \| Phase behavior \| \| Triple point \| 195.4 K (−77.75°C), 6.060 kPa \| \| Critical point \| 405.5 K (132.3°C), 11.300 MPa \| \| Std enthalpy change of fusion, Δ fus H~~o~~ \| +5.653 kJ/mol \| \| Std entropy change of fusion, Δ fus S~~o~~ \| +28.93 J/(mol·K) \| \| Std enthalpy change of vaporization, Δ vap H~~o~~ \| +23.35 kJ/mol at BP of −33.4°C \| \| Std entropy change of vaporization, Δ vap S~~o~~ \| +97.41 J/(mol·K) at BP of −33.4°C \| \| Solid properties \| \| Std enthalpy change of formation, Δ f H~~o~~solid \| ? kJ/mol \| \| Standard molar entropy, S~~o~~solid \| ? J/(mol K) \| \| Heat capacity, c p \| ? J/(mol K) \| \| Liquid properties \| \| Std enthalpy change of formation, Δ f H~~o~~liquid \| −80.882 ± 0.053 kJ/mol \| \| Standard molar entropy, S~~o~~liquid \| ? J/(mol K) \| \| Heat capacity, c p \| 80.80 J/(mol K) \| \| Gas properties \| \| Std enthalpy change of formation, Δ f H~~o~~gas \| −45.556 ± 0.029 kJ/mol \| \| Std Gibbs free energy change of formation, Δ f G~~o~~gas \| −16.6 kJ/mol \| \| Standard molar entropy, S~~o~~gas \| 192.77 J/(mol K) \| \| Heat capacity, c p \| 35.06 J/(mol K) \| \| Heat capacity ratio, γ at 15°C \| 1.310 \| \| van der Waals' constants \| a = 422.5 L2 kPa/mol2 b = 0.03707 L/mol \| Vapor–liquid equilibrium data [edit] P in mm Hg1 10 40 100 400 760 1520 3800 7600 15600 30400 45600 T in °C−109.1(s)−91.9(s)−79.2(s)−68.4−45.4−33.6−18.7 4.7 25.7 50.1 78.9 98.3 Table data (above) obtained from CRC Handbook of Chemistry and Physics 44th ed. The (s) notation indicates equilibrium temperature of vapor over solid. Otherwise temperature is equilibrium of vapor over liquid. log 10 of anydrous ammonia vapor pressure. Uses formula shown below. Vapor-pressure formula for ammonia: log 10 P = A – B / (T − C), where P is pressure in kPa, and T is temperature in kelvins; A = 6.67956, B = 1002.711, C = 25.215 for T = 190 K through 333 K. \| Vapor over anhydrous ammonia \| \| Temp. \| Pressure \| ρ of liquid \| ρ of vapor \| Δ vap H \| \| −78°C \| 5.90 kPa \| \| \| \| −75°C \| 7.93 kPa \| 0.73094 g/cm 3 \| 7.8241×10−5 g/cm 3 \| \| \| −70°C \| 10.92 kPa \| 0.72527 g/cm 3 \| 1.1141×10−4 g/cm 3 \| \| \| −65°C \| 15.61 kPa \| 0.71953 g/cm 3 \| 1.5552×10−4 g/cm 3 \| \| \| −60°C \| 21.90 kPa \| 0.71378 g/cm 3 \| 2.1321×10−4 g/cm 3 \| \| \| −55°C \| 30.16 kPa \| 0.70791 g/cm 3 \| 2.8596×10−4 g/cm 3 \| \| \| −50°C \| 40.87 kPa \| 0.70200 g/cm 3 \| 3.8158×10−4 g/cm 3 \| 1417 J/g \| \| −45°C \| 54.54 kPa \| 0.69604 g/cm 3 \| 4.9940×10−4 g/cm 3 \| 1404 J/g \| \| −40°C \| 71.77 kPa \| 0.68999 g/cm 3 \| 6.4508×10−4 g/cm 3 \| 1390 J/g \| \| −35°C \| 93.19 kPa \| 0.68385 g/cm 3 \| 8.2318×10−4 g/cm 3 \| 1375 J/g \| \| −30°C \| 119.6 kPa \| 0.67764 g/cm 3 \| 1.0386×10−3 g/cm 3 \| 1361 J/g \| \| −25°C \| 151.6 kPa \| 0.67137 g/cm 3 \| 1.2969×10−3 g/cm 3 \| 1345 J/g \| \| −20°C \| 190.2 kPa \| 0.66503 g/cm 3 \| 1.6039×10−3 g/cm 3 \| 1330 J/g \| \| −15°C \| 236.3 kPa \| 0.65854 g/cm 3 \| 1.9659×10−3 g/cm 3 \| 1314 J/g \| \| −10°C \| 290.8 kPa \| 0.65198 g/cm 3 \| 2.3874×10−3 g/cm 3 \| 1297 J/g \| \| −5°C \| 354.8 kPa \| 0.64533 g/cm 3 \| 2.8827×10−3 g/cm 3 \| 1280 J/g \| \| 0°C \| 429.4 kPa \| 0.63857 g/cm 3 \| 3.4528×10−3 g/cm 3 \| 1263 J/g \| \| 5°C \| 515.7 kPa \| 0.63167 g/cm 3 \| 4.1086×10−3 g/cm 3 \| 1245 J/g \| \| 10°C \| 614.9 kPa \| 0.62469 g/cm 3 \| 4.8593×10−3 g/cm 3 \| 1226 J/g \| \| 15°C \| 728.3 kPa \| 0.61755 g/cm 3 \| 5.7153×10−3 g/cm 3 \| 1207 J/g \| \| 20°C \| 857.1 kPa \| 0.61028 g/cm 3 \| 6.6876×10−3 g/cm 3 \| 1187 J/g \| \| 25°C \| 1003 kPa \| 0.60285 g/cm 3 \| 7.7882×10−3 g/cm 3 \| 1167 J/g \| \| 30°C \| 1166 kPa \| 0.59524 g/cm 3 \| 9.0310×10−3 g/cm 3 \| 1146 J/g \| \| 35°C \| 1350 kPa \| 0.58816 g/cm 3 \| 1.0431×10−2 g/cm 3 \| 1124 J/g \| \| 40°C \| 1554 kPa \| 0.57948 g/cm 3 \| 1.2006×10−2 g/cm 3 \| 1101 J/g \| \| 45°C \| 1781 kPa \| 0.57130 g/cm 3 \| 1.3775×10−2 g/cm 3 \| 1083 J/g \| \| 50°C \| 2032 kPa \| 0.56287 g/cm 3 \| 1.5761×10−2 g/cm 3 \| 1052 J/g \| \| 55°C \| 2310 kPa \| 0.55420 g/cm 3 \| \| \| \| 60°C \| 2613 kPa \| 0.54523 g/cm 3 \| 2.05×10−2 g/cm 3 \| \| \| 65°C \| 2947 kPa \| 0.53596 g/cm 3 \| \| \| \| 70°C \| 3312 kPa \| 0.52632 g/cm 3 \| 2.65×10−2 g/cm 3 \| \| \| 75°C \| 3711 kPa \| 0.51626 g/cm 3 \| \| \| \| 80°C \| 4144 kPa \| 0.50571 g/cm 3 \| 3.41×10−2 g/cm 3 \| \| \| 85°C \| 4614 kPa \| 0.49463 g/cm 3 \| \| \| \| 90°C \| 5123 kPa \| 0.48290 g/cm 3 \| 4.39×10−2 g/cm 3 \| \| \| 95°C \| 5672 kPa \| 0.47041 g/cm 3 \| \| \| \| 100°C \| 6264 kPa \| 0.45693 g/cm 3 \| 5.68×10−2 g/cm 3 \| \| \| Temp. \| Pressure \| ρ of liquid \| ρ of vapor \| Δ vap H \| \| The table above gives properties of the vapor–liquid equilibrium of anhydrous ammonia at various temperatures. The second column is vapor pressure in kPa. The third column is the density of the liquid phase. The fourth column is the density of the vapor. The fifth column is the heat of vaporization needed to convert one gram of liquid to vapor. \| Freezing curve of ammonia-water system. Three eutectic points I. II. and III. are shown. Left of the I. point the frozen component is ice. Right of the III. point the frozen component is ammonia. \| Vapor over aqueous ammonia solution \| \| Temp. \| %wt NH 3 \| Partial pressure NH 3 \| Partial pressure H 2 O \| \| 0°C \| 4.72 \| 1.52 kPa \| 0.68 kPa \| \| 9.15 \| 3.31 kPa \| 0.71 kPa \| \| 14.73 \| 6.84 kPa \| 0.55 kPa \| \| 19.62 \| 11.0 kPa \| 0.40 kPa \| \| 22.90 \| 14.9 kPa \| 0.37 kPa \| \| 10°C \| 4.16 \| 2.20 kPa \| 1.21 kPa \| \| 8.26 \| 4.96 kPa \| 1.17 kPa \| \| 12.32 \| 8.56 kPa \| 1.01 kPa \| \| 15.88 \| 12.68 kPa \| 0.93 kPa \| \| 20.54 \| 19.89 kPa \| 0.83 kPa \| \| 21.83 \| 22.64 kPa \| 0.73 kPa \| \| 19.9°C \| 4.18 \| 3.65 kPa \| 2.19 kPa \| \| 6.50 \| 6.11 kPa \| 2.15 kPa \| \| 6.55 \| 6.13 kPa \| 2.13 kPa \| \| 7.72 \| 7.49 kPa \| 2.08 kPa \| \| 10.15 \| 10.75 kPa \| 2.01 kPa \| \| 10.75 \| 11.51 kPa \| 1.96 kPa \| \| 16.64 \| 22.14 kPa \| 1.72 kPa \| \| 19.40 \| 28.74 kPa \| 1.64 kPa \| \| 23.37 \| 40.32 kPa \| 1.37 kPa \| \| 30.09°C \| 3.93 \| 5.49 kPa \| 4.15 kPa \| \| 7.43 \| 11.51 kPa \| 3.89 kPa \| \| 9.75 \| 16.00 kPa \| 3.80 kPa \| \| 12.77 \| 23.33 kPa \| 3.55 kPa \| \| 17.76 \| 38.69 kPa \| 3.31 kPa \| \| 17.84 \| 38.81 kPa \| 3.24 kPa \| \| 21.47 \| 53.94 kPa \| 2.95 kPa \| \| 40°C \| 3.79 \| 8.15 kPa \| 7.13 kPa \| \| 7.36 \| 17.73 kPa \| 6.76 kPa \| \| 11.06 \| 29.13 kPa \| 6.55 kPa \| \| 15.55 \| 47.14 kPa \| 5.52 kPa \| \| 17.33 \| 57.02 kPa \| \| \| 20.85 \| 76.81 kPa \| 5.04 kPa \| \| 50°C \| 3.29 \| 10.54 kPa \| 11.95 kPa \| \| 5.90 \| 20.17 kPa \| 11.61 kPa \| \| 8.91 \| 32.88 kPa \| 11.07 kPa \| \| 11.57 \| 45.56 kPa \| 10.75 kPa \| \| 14.15 \| 60.18 kPa \| 10.27 kPa \| \| 14.94 \| 64.94 kPa \| 10.03 kPa \| \| 60°C \| 3.86 \| 18.25 kPa \| 19.21 kPa \| \| 5.77 \| 28.78 kPa \| \| \| 7.78 \| 40.05 kPa \| 18.47 kPa \| \| 9.37 \| 50.09 kPa \| 18.07 kPa \| \| 9.37 \| 63.43 kPa \| 17.39 kPa \| \| Temp. \| %wt NH 3 \| Partial Pressure NH 3 \| Partial Pressure H 2 O \| Heat capacity of liquid and vapor [edit] Heat capacity, c p, of anhydrous ammonia gas. Uses polynomial obtained from CHERIC. Heat capacity of anhydrous liquid ammonia. Uses polynomial obtained from CHERIC. Spectral data [edit] \| UV-Vis \| \| λ max \| None nm \| \| Extinction coefficient, ε \| None \| \| IR \| \| Major absorption bands \| 3444, 3337, 1627, 950 cm−1 \| \| NMR \| \| Proton NMR \| \| \| Carbon-13 NMR \| None – no carbons \| \| Other NMR data \| \| \| MS \| \| Masses of main fragments \| 17 (100%) 16(80%) 15(9%) \| Regulatory data [edit] \| Regulatory data \| \| EINECS number \| 231-635-3 (gas) 215-647-6 (soln.) \| \| EU index number \| 007-001-00-5 (gas) 007-001-01-2 (soln.) \| \| PEL-TWA (OSHA) \| 50 ppm (35 mg/m 3) \| \| IDLH (NIOSH) \| 300 ppm \| \| Flash point \| 11°C \| \| Autoignition temperature \| 651°C \| \| Explosive limits \| 15–28% \| \| RTECS # \| BO0875000 \| Safety data sheet [edit] The handling of this chemical may incur notable safety precautions... It is highly recommend that you seek the Safety Data Sheet (SDS) for this chemical from a reliable source and follow its directions. SIRI Science Stuff (Ammonia Solution) References [edit] Linstrom, Peter J.; Mallard, William G. (eds.); NIST Chemistry WebBook, NIST Standard Reference Database Number 69, National Institute of Standards and Technology, Gaithersburg (MD) ^ abCRC Handbook of Chemistry and Physics, 94th ed. 24 July 2017 at the Wayback Machine. Page 9-26. Retrieved 18 June 2013. ^"Ammonia - NH3 (aq, undissoc)" Active Thermochemical Tables v1.130. ^"Ammonia - NH3(g)" Active Thermochemical Tables v1.130. ^Lange's Handbook of Chemistry, 10th ed. page 1436. ^Lange's Handbook of Chemistry, 10th ed. page 1451 and 1468. ^Friedrich Merkel, Franjo Bošnjaković (1929). Diagramme und Tabellen zur Berechnung der Absorptions-Kältemachienen. Berlin: Julius Springer. p.46. ^Perman, Jour. Chem. Soc. 83 1168 (1903). ^ ab"Pure Components Properties"(Queriable database). Chemical Engineering Research Information Center. Archived from the original on 3 June 2007. Retrieved 1 June 2007. This box: view edit Except where noted otherwise, data relate to Standard temperature and pressure. Reliability of data general note. External links [edit] Phase diagram for ammonia IR spectrum (from NIST) Retrieved from " Category: Chemical data pages Hidden categories: Webarchive template wayback links Articles with short description Short description matches Wikidata Use dmy dates from December 2023 Chemical data pages cleanup This page was last edited on 13 May 2024, at 17:00(UTC). Text is available under the Creative Commons Attribution-ShareAlike 4.0 License; additional terms may apply. By using this site, you agree to the Terms of Use and Privacy Policy. Wikipedia® is a registered trademark of the Wikimedia Foundation, Inc., a non-profit organization. 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6409	https://www.hilarispublisher.com/open-access/the-power-of-pancreatic-amylase-understanding-its-function-in-carbohydrate-digestion-99853.html	GET THE APP The Power of Pancreatic Amylase: Understanding its Function in Carbohydrate Digestion Hepatology and Pancreatic Science ISSN: 2573-4563 Open Access Powered by Translate Submit Manuscript arrow_forward arrow_forward Commentary - (2023) Volume 7, Issue 3 The Power of Pancreatic Amylase: Understanding its Function in Carbohydrate Digestion Felicia Fend Correspondence: Felicia Fend, Department of Gastrointestinal Science, German Cancer Research Center (DKFZ), Im Neuenheimer Feld 280, DE-69120 Heidelberg, Germany, Email: Author information Department of Gastrointestinal Science, German Cancer Research Center (DKFZ), Im Neuenheimer Feld 280, DE-69120 Heidelberg, Germany Received: 01-May-2023, Manuscript No. hps-23-105844; Editor assigned: 03-May-2023, Pre QC No. P-105844; Reviewed: 15-May-2023, QC No. Q-105844; Revised: 20-May-2023, Manuscript No. R-105844; Published: 27-May-2023 , DOI: 10.37421/2573-4563.2023.7.223 Citation: Fend, Felicia. “The Power of Pancreatic Amylase: Understanding its Function in Carbohydrate Digestion.” J Hepatol Pancreat Sci 7 (2023): 223. Copyright: © 2023 Fend F. This is an open-access article distributed under the terms of the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original author and source are credited. Description Carbohydrates are organic compounds made up of carbon, hydrogen, and oxygen atoms, and they exist in various forms, ranging from simple sugars to complex polysaccharides. Common sources of carbohydrates in the diet include fruits, vegetables, grains, legumes, and dairy products. Complex carbohydrates, such as starches and glycogen, consist of long chains of glucose molecules linked together. These chains can be branched or unbranched, and their structural complexity determines the rate at which they can be digested and absorbed by the body. In contrast, simple carbohydrates, such as Monosaccharaides (e.g., glucose, fructose, and galactose), and Disaccharides (e.g., sucrose, lactose, and maltose), consist of one or two sugar molecules and are more readily absorbed. Pancreatic amylase is a key enzyme involved in the digestion of complex carbohydrates. It is produced by the acinar cells of the pancreas and secreted into the small intestine, specifically the duodenum, where carbohydrate digestion primarily takes place . The secretion of pancreatic amylase is regulated by hormonal signals triggered by the presence of carbohydrates in the digestive system. When carbohydrates reach the small intestine, the cells lining the intestinal walls release a hormone called Cholecystokinin (CCK). CCK stimulates the pancreas to release pancreatic amylase into the small intestine, where it can act on the complex carbohydrates in the food. Pancreatic amylase works by hydrolysing the bonds between the glucose molecules in complex carbohydrates, ultimately breaking them down into smaller units. It specifically targets the alpha-1, 4 glycosidic linkages present in starches and glycogen. The hydrolysis process begins when pancreatic amylase binds to the polysaccharide substrate. It then cleaves the alpha-1, 4 glycosidic bonds, releasing shorter carbohydrate chains called dextrin. These dextrins undergo further hydrolysis by pancreatic amylase, resulting in the formation of maltose, a disaccharide consisting of two glucose molecules. Maltose, along with other disaccharides present in the small intestine, is further broken down into monosaccharaides by brush border enzymes, such as maltase, sucrase, and lactase, located on the surface of the intestinal cells . These monosaccharides, primarily glucose, fructose, and galactose, are then absorbed into the bloodstream and transported to various tissues for energy production or storage. Pancreatic amylase is highly effective at breaking down complex carbohydrates. Its ability to hydrolyse the alpha-1,4 glycosidic bonds in starches and glycogen enables the efficient release of glucose units, providing a readily available energy source. Pancreatic amylase complements the action of other carbohydrate-digesting enzymes, such as salivary amylase. Salivary amylase initiates carbohydrate digestion in the mouth, breaking down complex carbohydrates into smaller fragments. However, salivary amylase activity is limited due to the brief exposure to saliva during chewing. Pancreatic amylase takes over in the small intestine, where it further digests complex carbohydrates and ensures complete breakdown. The breakdown of complex carbohydrates into simpler forms by pancreatic amylase and other digestive enzymes facilitates optimal nutrient absorption. Smaller carbohydrate units, such as monosaccharides, are more efficiently absorbed through the intestinal wall into the bloodstream, ensuring that the body can readily utilize the energy they provide. Hormonal release of pancreatic amylase is stimulated by the hormone Cholecystokinin (CCK). CCK is released by specialized cells in the lining of the small intestine in response to the presence of proteins and fats. As these food components enter the small intestine, CCK signals the pancreas to release pancreatic amylase, along with other digestive enzymes, to facilitate digestion . Neural signals, particularly from the parasympathetic nervous system, also play a role in the regulation of pancreatic enzyme secretion. Parasympathetic stimulation, often occurring during the rest and digest state, enhances the secretion of pancreatic enzymes, including amylase. Pancreatic amylase is a powerful enzyme that plays a vital role in the digestion of complex carbohydrates. It breaks down starches and glycogen into smaller units, facilitating the release of glucose molecules that can be readily absorbed by the body. Understanding the function and importance of pancreatic amylase in carbohydrate digestion provides insights into the intricate process of nutrient breakdown and absorption . Disruptions in pancreatic amylase secretion or activity can lead to carbohydrate malabsorption and related health issues. Further research and advancements in our understanding of pancreatic amylase and carbohydrate digestion will continue to shed light on the optimal management and treatment of digestive disorders, as well as inform dietary recommendations for individuals with specific carbohydrate-related conditions . Acknowledgment None. Conflict of Interest There are no conflicts of interest by author. References Sung, Hyuna, Jacques Ferlay, Rebecca L. Siegel and Mathieu Laversanne, et al. "Global cancer statistics 2020: GLOBOCAN estimates of incidence and mortality worldwide for 36 cancers in 185 countries." CA Cancer J Clin 71 (2021): 209-249. Google Scholar, Crossref, Indexed at White, Rebekah R. and Andrew M. Lowy. "Clinical management: Resectable disease." J Cancer 23 (2017): 343-349. Google Scholar, Crossref, Indexed at Van Roessel, Stijn, Gyulnara G. Kasumova, Joanne Verheij and Robert M. Najarian, et al. "International validation of the eighth edition of the American Joint Committee on Cancer (AJCC) TNM staging system in patients with resected pancreatic cancer." JAMA surgery 153 (2018): e183617-e183617. Google Scholar, Crossref, Indexed at Layer, Peter, Alan R. Zinsmeister and Eugene P. DiMagno. "Effects of decreasing intraluminal amylase activity on starch digestion and postprandial gastrointestinal function in humans." Gastroenterology 91 (1986): 41-48. Google Scholar, Crossref, Indexed at Boivin, Michel, Bernard Flourie, Robert A. Rizza and Vay Liang W. Go, et al. "Gastrointestinal and metabolic effects of amylase inhibition in diabetics." Gastroenterology 94 (1988): 387-394. Google Scholar, Crossref, Indexed at Awards & Nominations 50+ Million Readerbase Journal Highlights Google Scholar citation report Citations: 34 Hepatology and Pancreatic Science received 34 citations as per Google Scholar report Hepatology and Pancreatic Science peer review process verified at publons Indexed In Related Links Tweets by Advancemedica4 Open Access Journals arrow_upward arrow_upward Original text Rate this translation Your feedback will be used to help improve Google Translate
6410	https://www.heldermann-verlag.de/jgg/jgg01_05/jgg0404.pdf	Journal for Geometry and Graphics Volume 4 (2000), No. 1, 55–69. Projection from 4D to 3D Svatopluk Zachari´ aˇ s1, Daniela Velichov´ a2 1Faculty of Applied Sciences, West Bohemian University Univerzitn´ ı 22, CZ 306 14 Plzeˇ n, Czech Republic 2Dept. of Mathematics, Mechanical Eng’g Faculty, Slovak Technical University N´ am. slobody 17, SK 812 31 Bratislava, Slovak Republic email: velichov@sjf.stuba.sk Abstract. The aim of this paper is to give a survey on analytic representations of central and orthographic projections from R4 to R3 or R2. There are discussed various aspects of these projections, whereby some special relations were revealed, e.g., the fact that homogeneous coordinates or barycentric coordinates in R3 can be obtained by applying particular projections on a point with given cartesian coordinates in R4. We would also like to demonstrate that by projecting curves or 2-surfaces of R4 interesting shapes in R3 and R2 can be obtained. Key Words: geometry in 4D, projections, quaternions MSC 1994: 51N20, 51N05. 1. Introduction Geometric objects in the space R4 can be projected ﬁrst into the space R3 and then into the plane R2. We prefer orthographic projections against other parallel projections, as they are an approximation of the central projection with large distance. The advantage of the orthographic projection is a rather good realism in visualization of unknown geometric objects. The basic aspect of a realistic view of smooth surfaces after a projection R3 →R2 is to ﬁnd the outline curve; an algorithm is described in . In the projection P : R4 →R3 those 3-dimensional objects are visible, that are in the case of the central projection close to the centre of projection (we restrict our consideration on the projection of only one of the open semi-spaces determined by the hyperplane parallel to the projection plane and incident to the centre of projection). In the case of a parallel projection visible objects are in larger distance from the 3-dimensional projection plane, if these distances are oriented opposite to the rays of sight. Visibility deﬁned in this way will be denoted by W4. The visibility W4 in the projection P is diﬀerent from the visibility W3 in the space R3 that is applied on any projection R3 →R2. E.g., in the projection of a simplex S4 ⊂R4 ISSN 1433-8157/$ 2.50 c ⃝2000 Heldermann Verlag 56 S. Zachari´ aˇ s, D. Velichov´ a: Projection from 4D to 3D we suppose that the centre of projection is an exterior point (for parallel projections this condition is always satisﬁed). A simplex S4 with vertices A1, . . . , A5 is projected onto the convex hull of the ﬁve image points A∗ 1, . . . , A∗ 5. If one of these points A∗ i is an interior point of the tetrahedron formed by the remaining four image points A∗ k, then the four edges passing through A∗ i are non-visible in the visibility W3, but these points are visible in the visibility W4. If any point A∗ i , i = 1, . . . , 5 is an exterior point of the tetrahedron determined by the remaining four points, then one and only one edge is non-visible in both visibilities W 4 and W3. If one point A∗ i is located on one edge of the tetrahedron of the remaining A∗ k, then all edges are visible in the visibility W3. Generally, there is no chance to deﬁne a visibility when any R4-object is projected into R2. The reason is that the ”rays of sight“ are planes, and for any two points in a plane one cannot deﬁne that one point ”hides“ the other. Such a remark can be found also in . Visibility W4 is suitable for enlightening the space R4. In the ﬁgures included in the paper the visibility W3 was applied, as it is easier to realize in the projection plane R2. 2. Central projections, modelling curves and surfaces 2.1. Central projections Let V be a curve or a 2-surface in the space Rn+1, n > 1. There is a central projection of Rn+1 from the origin O of the coordinate system onto any hyperplane R. Under this projection any point B = (x1, . . . , xn+1) of the ﬁgure V \ {O} can be connected with O by the line b = OB intersecting the hyperplane R in the image (x∗ 1, . . . , x∗ n) of B. When the equation of the hyperplane R is in the form xn+1 = 1, the we get the same relation as between homogeneous coordinates (x1, . . . , xn+1) and cartesian coordinates µ x1 xn+1 , . . . xn xn+1 ¶ , xn+1 ̸= 0 of points of the projective extension Pn of the Euclidean space Rn. When the equation of the hyperplane R is in the form x1 + x2 + . . . + xn + xn+1 = 1, we speak about barycentric coordinates in Rn xb i := xi x1 + . . . + xn+1 , i = 1, . . . , n, and we even have to assume that no point of the ﬁgure V is located in the hyperplane x1 + . . . + xn+1 = 0 parallel to R. In this situation we do not speak of the projective space Pn, but of barycentric coordinates in Rn. 2.2. Modelling curves and surfaces Any parabola in Rn+1, n > 1, can be easily determined by parametric equations using quadratic polynomials like xi(t) := ai,2t2 + ai,1t + ai,0 , −∞< t < ∞. S. Zachari´ aˇ s, D. Velichov´ a: Projection from 4D to 3D 57 It is clear that generally a conic section in the projective extension of Rn can be determined by the ratio of quadratic polynomials xi(t) in the form x∗ i (t) = xi(t) xn+1(t), i = 1, . . . , n. In addition to the presented polynomial representation of a parabola, any ellipse in Rn+1 can be represented by the equations xi(t) = ai cos t + bi sin t + ci, 0 ≤t < 2π, ai, bi, ci ∈R for i = 1, . . . , n + 1, any hyperbola by xi(t) = ±ai cosh t + bi sinh t + ci, −∞< t < ∞, for i = 1, . . . , n + 1 or an ellipse in the exponential form xi(t) = ai exp(it) + ai exp(−it) + ci, −∞< t < ∞, ai ∈C, ci ∈R for i = 1, . . . , n + 1, or a hyperbola in the exponential form xi(t) = ±ai exp t + bi exp(−t) + ci, −∞< t < ∞, ai, bi, ci ∈R for i = 1, . . . , n + 1. Generally, a conic section in Rn+1 can be represented by a linear vector combination of diﬀerent basic functions {1, t, t2}, {1, sin t, cos t}, {1, exp t, exp(−t)}, and so on. In the space Rn we get the corresponding ”rational” functions. Quite a wide variety of curves in R3 that are useful for technical applications can be determined with the basis {1, t, t2, t3}. These are curves generated from cubic curves in R4, while the vector coeﬃcients can be four linearly independent vectors in R4. Any aﬃne transformation Rn →Rn or parallel projection Rn →Rn−1 transforms the control polygon (or net) to the control polygon (or net). An aﬃne transformation Rn+1 → Rn+1 or the central projection from the origin O, Rn+1 →Pn (that is the extension of Rn by the hyperplane xn+1 = 0), transforms the control polygon {Qj \| j = 1, . . . , k} ⊂Rn+1 onto the polygon {Q∗ j \| j = 1, . . . , k} ⊂Pn. If all vertices Q∗ j of the polygon are real points of the projective extension of the space Rn, then the function coeﬃcient at the vertex Q∗ j will be of the form f ∗ j = fj P j fj Qj,n+1 , provided the function coeﬃcients fj, j = 1, . . . , k, are linear combinations of polynomial functions in the basis {1, t, t2, t3}, and the coordinates of the control point Qj are denoted as Qj = (Qj,1, . . . , Qj,n+1). The situation is a bit more complicated at the transition to barycentric coordinates: The representation of the point B ∈Rn+1 in terms of the control polygon {Qj \| j = 1, . . . , k} B = k X j=1 fjQj will be replaced by B∗ i = P j fjQj,i P j,m fjQj,m , i = 1, . . . , n + 1. 58 S. Zachari´ aˇ s, D. Velichov´ a: Projection from 4D to 3D Let the control polygon {Q1, . . . , Qk} be a simplex in Rn+1, i.e., k = n+1, then the vertex Q1 can be associated to the (n + 1)-tuple (1, 0, . . . , 0), and vertex Qn+1 to the (n + 1)-tuple (0, . . . , 0, 1). For k > n + 1 the vertices Qi ∈Rn+1, i = 1, . . . , k , can be regarded as the parallel views of the vertices of any simplex in Rk−1. The practical advantage of the determination of barycentric coordinates for the control polygons is that all barycentric coordinates of any point B are positive numbers, if and only if the point is located inside the simplex. In connection to the parallel projection of the simplex, the following statement is valid: If all coeﬃcients determining the point B with respect to the control polygon are positive and the sum of them equals 1, then B is located inside the convex hull of the control polygon. 2-surfaces in R4 can sometimes be modelled as the graph of any complex function f(z) in one complex variable z = x + iy. This gives for f(z) = u(x, y) + iv(x, y) x1 = Re z, x2 = Im z, x3 = u = Re f(z), x4 = v = Im f(z). Figure 1: Central views of 2-surfaces deﬁned by complex functions In Fig. 1 central views of graphs of the functions f(z) = 4 + i + z2 (on the left) and f(z) = 4 + z2 (on the right) under the central projection from the origin (0, 0, 0, 0) onto the hyperplane x4 = 1 are displayed. 2-surfaces in R4 can also be determined by basic functions {1, sin u, cos u, sin v, cos v}, 0 ≤u, v ≤2π. When x4(u, v) is suﬃciently far from zero, we receive in R3 a closed torus-like surface. Some examples are shown in the Figures 2 and 3. The orthographic view of the surface deﬁned by the parametric equations w = 8 + cos u + 2 sin u + cos v + sin v S. Zachari´ aˇ s, D. Velichov´ a: Projection from 4D to 3D 59 Figure 2: Central views of torus-like surfaces Figure 3: Central views of 2-surfaces deﬁned by linear rational functions x = 1 w, y = (2 + sin u) w , z = 0.5 cos v w , 0 ≤u, v < 2π is presented in Fig. 4. The outline of the orthographic view is shown on the left, the net of isoparametric curves is displayed on the right. In the space R4 with the coordinates x1, x2, x3, x4 two tori can share no more than two meridian circles. Let one of them be located in the hyperplane x4 = 0. Both tori in R4 can be projected from the point (0, 0, 0, 0) to the hyperplane x4 = 0 (see Fig. 5). 60 S. Zachari´ aˇ s, D. Velichov´ a: Projection from 4D to 3D Figure 4: Orthographic view of a 2-surface deﬁned parametrically Figure 5: Central views of two tori sharing two meridian circles 3. Spherical coordinates and orthographic projections In R4 the norm of a vector is the l2-norm ∥(x, y, z, w)∥:= p x2 + y2 + z2 + w2. The hypersphere S3 with the implicit equation x2 + y2 + z2 + w2 = r2 can be parametrized in many ways, from which that one will be chosen that is the extension of the geographic spherical coordinates from R3 to R4: x(r, t, u) = r cos t cos u y(r, t, u) = r sin t cos u z(r, u) = r sin u 7→ x(r, t, u, v) = r cos t cos u cos v y(r, t, u, v) = r sin t cos u cos v z(r, u, v) = r sin u cos v w(r, v) = r sin v S. Zachari´ aˇ s, D. Velichov´ a: Projection from 4D to 3D 61 where r = 1 is the radius, t the longitude obeying 0 ≤t < 2π; u is the classical latitude with −π/2 ≤u ≤π/2, v the additional new latitude with −π/2 ≤v ≤π/2. The parametrization of the sphere S2 ∈R2 has a singular subsphere S0, i.e., the two poles v = ±π/2. The parametrization of the hypersphere S3 has a singular subsphere S1, i.e., the circle κ : x = y = 0, z2 + w2 = 1. Excluding this singularity of our parametrization, we receive: v = arcsin w , u = arcsin z √ 1 −w2 , t = arg(x + iy). A spherical motion O in R4 keeping invariant the origin O = (0, 0, 0, 0) is represented by an orthogonal matrix Q with det Q = 1. This matrix Q is an element of the group O+(4) (see ). Similarly, the group of revolutions in R3 is represented by the group of orthogonal matrices of degree 3 with determinant 1 and denoted by O+(3). An orthogonal matrix Q can be obtained from the Jacobian matrix of the transformation (r, t, u, v) 7→(x, y, z, w) in R4, i.e., from the partial derivatives of the vector [r cos t cos u cos v, r sin t cos u cos v, r sin u cos v, r sin v] by normalizing: Q =     cos t cos u cos v sin t cos u cos v sin u cos v sin v −sin t cos t 0 0 −cos t sin u −sin t sin u cos u 0 −cos t cos u sin v −sin t cos u sin v −sin u sin v cos v    . It is easy to prove, that Q T = Q−1 and Q ∈O+(4). At parallel projections from R3 to R2 usually the image of the last coordinate axis is speciﬁed as a vertical line in the projection plane. This can be assured by excluding the row with the partial derivatives with respect to r in the matrix of the projection. After erasing the ﬁrst row in Q the matrix P =   −sin t cos t 0 0 −cos t sin u −sin t sin u cos u 0 −cos t cos u sin v −sin t cos u sin v −sin u sin v cos v   of an orthographic projection R4 →R3 onto the hyperplane passing through the origin O is obtained. The rows of the matrix P are basic vectors of the tangent space of the hypersphere S3 at an arbitrary point determined by parameters (t, u, v). The image hyperplane R is in general position, parallel to this tangent hyperplane of S3. Coordinates of the orthographic views will be denoted by upper asterixes, [x∗, y∗, z∗] T = P · [x, y, z, w] T . The sum of squared norms of the columns in the matrix Q equals 4. The columns in the matrix P determine the orthographic views of four unit vectors along the coordinate axes +x, +y, +z, +w. The sum of squared norms of the columns of P is 3. This is a well-known property when the unit points of a cartesian frame are mapped under an orthogonal projection; this result can be found in [8, 5, 7]. Using the matrix Q instead of P, the ﬁrst additional coordinate will determine the oriented distance of the projected point in R4 to the image hyperplane R passing through O. 62 S. Zachari´ aˇ s, D. Velichov´ a: Projection from 4D to 3D Omitting the visibility, the parametrized hypersphere S3 is projected orthogonally to the ball in R3 with two-parametric nets of space curves. Iso-parametric t-curves on S3 are circles or points (for u = ±π/2). These points are located in the plane (z, w) on the unit circle κ with the centre in the origin. Iso-parametric u-curves are semi-circles or the points S, J (for v = ±π/2), the North and South Pole located on the axis w. The planes of the arcs are perpendicular to the axis w. Iso-parametric v-curves are semi-circles with the end points in S and J. The 2-surface v = const., a sphere S2, is projected to the ellipsoid, that can be also con-tracted to the point S∗or J∗. The 2-surface u = const. is projected to the “garlic pod” (closed surface) with point S∗, J∗that can appear in the form of an ellipsoid (u = 0) transformed contractively up to a semi-ellipse (u = ±π/2). The 2-surface t = const., a semi-sphere, is projected to a semi-ellipsoid. Figure 6: Orthographic views of parametric surfaces on the hypersphere S3 In Fig. 6 orthographic views of 2-surfaces ⊂S3 deﬁned by the constant parameters v = 0 and v = 1.2, t = 0 and u = 0, u = 0.7854, u = 1.2 are displayed. Orthographic views of a 2-surface patch determined by parametric equations in the form x = cos u cos v, y = sin u cos v, z = sin v, w = v, −π/2 ≤u, v ≤π/2, are presented in Fig. 7. Diﬀerent orthographic projections were derived from the spherical motion determined by omitting one row in the randomly deﬁned matrix Q ∈O+(4), Q =     −0.699441 −0.232487 −0.657456 0.156472 −0.628437 0.029543 0.520789 −0.577038 −0.317978 0.707540 0.228128 0.588417 −0.121403 −0.666681 0.494458 0.544343    . S. Zachari´ aˇ s, D. Velichov´ a: Projection from 4D to 3D 63 Figure 7: Diﬀerent orthographic views of the same surface patch The views in Fig. 7 from left to right are obtained from Q by cancelling the ﬁrst, the second, the third, or ﬁnally the last row, respectively. Any helix in R3 is a non-self-intersecting curve, but its orthographic view to the plane is a prolonged cycloidal curve which can be self-intersecting. Similarly, a 2-surface in R4 which is not self-intersecting can have a self-intersecting orthogonal view in R3. Let us inscribe into the hypersphere S3 a regular cross-polytope with vertices (±1, 0, 0, 0), . . . , (0, 0, 0, ±1), the dual to the hypercube. There are four triples of axes, and any triple generates a regular octahedron that is inscribed into a sphere S2 i ⊂S3, i = 1, . . . , 4. In this way we obtain on the hypersphere S3 four congruent concentric spheres S2 i , i = 1, . . . , 4 , any two of them intersect orthogonally in a concentric circle with the same radius. The spheres S2 i intersect the hypersphere S3 in 16 curve-like tetrahedra, precisely in 16 homeomorphic images of a tetrahedron that compose the hypersphere S3. Four ”principal” spheres S2 i are projected by the orthographic projection P : R4 →R3 onto four ellipsoids that are inscribed into the ”outline” (i.e. a sphere S2) in R3, and then projected by a perspective, as it can be seen in Fig. 8. Conjugate diameters of these ellipsoids are three lines from four diagonals of the regular cross-polytope. The central projection from the origin O onto the hyperplane x1 + x2 + x3 + x4 = 1 results in the determination of well-known barycentric coordinates in R3. This hyperplane with the normal vector n = (1, 1, 1, 1) can be regarded as the image plane for an orthogonal axonometry. The unit normal vector is n1 = (1/2, 1/2, 1/2, 1/2), while sin v = 1 2, cos v = r 3 2. The unit vector of the orthographic view of n1 in the plane (x1, x2, x3) is n11 = (1/ √ 3, 1/ √ 3, 1/ √ 3), and then sin u = 1/ √ 3, cos u = 2/ √ 3. The unit vector of the orthographic view of n11 in the plane (x1, x2) is n111 = (1/ √ 2, 1/ √ 2), and then sin t = 1/ √ 2, cos t = √ 3/2. 64 S. Zachari´ aˇ s, D. Velichov´ a: Projection from 4D to 3D Figure 8: Four concentric spheres pairwise intersecting in circles The matrix of the orthogonal isometry R4 →R3 is therefore in the form P =    −1/ √ 2 1/ √ 2 0 0 −1/ √ 6 −1/ √ 6 p 2/3 0 −1/ √ 12 −1/ √ 12 −1/ √ 12 √ 3/2   . Isometric views of the n-space have also been studied in . Another parametrization of the hypersphere S3 of radius r, x = r cos u cos t, y = r cos u sin t, z = r sin u cos v, w = r sin u sin v for 0 ≤t < 2π, 0 ≤u ≤π/2, 0 ≤v < 2π produces a matrix Q =     cos u cos t cos u sin t sin u cos v sin u sin v −sin t cos t 0 0 −sin u cos t −sin u sin t cos u cos v cos u sin v 0 0 −sin v cos v    . 4. Quaternions The space R4 = C 2 can be regarded as the set H of quaternions. Quaternions were deﬁned by W. Hamilton in the year 1843. They are a generalization of complex numbers R2 = C, where the real (scalar) part Re H remains in R and a new imaginary part (vector part, pure quaternion) Im H in {R3 \ O} with three axes {i, j, k} is introduced, i.e. for a1, . . . , a4 ∈R a := a1 + a2i + a3j + a4k ∈H, Re a = a1, Im a = a2i + a3j + a4k. S. Zachari´ aˇ s, D. Velichov´ a: Projection from 4D to 3D 65 The sum of two quaternions a + b can be regarded as the standard vector sum. The multi-plication of quaternions is associative and satisﬁes the distributive laws and the relations ii = jj = kk = −1, ij = −ji = k, jk = −kj = i, ki = −ik = j. Hence ab = (a1 + a2i + a3j + a4k)(b1 + b2i + b3j + b4k) = (a1b1 −a2b2 −a3b3 −a4b4) + (a2b1 + a1b2 −a4b3 + a3b4) i+ +(a3b1 + a4b2 + a1b3 −a2b4) j + (a4b1 −a3b2 + a2b3 + a1b4) k. It follows from these properties that the product of two pure quaternions from Im H, i.e. a1 = b1 = 0, can be expressed in the cartesian coordinates as the diﬀerence of the vector product a × b minus the scalar product a · b: (a2i + a3j + a4k) (b2i + b3j + b4k) = = −a2b2 −a3b3 −a4b4 + (a3b4 −a4b3)i −(a2b4 −a4b2)j + (a2b3 −a3b2)k = = (a × b) −(a · b). If c ∈Im H, then c2 < 0. In the terms of algebra, the quaternions form an associative non-commutative ﬁeld. With respect to the Frobenius theorem (1877), quaternions form the unique associative non-commutative ﬁnitely dimensional algebra with a unit element and without zero divisors. It is interesting to point out that both (scalar and vector) products a · b and a × b were born historically in the theory of quaternions. Conjugate quaternions and their norms satisfy the following formulae: a := a1 −a2i −a3j −a4k = Re a −Im a, a b = b a, ∥a∥= √ a a, a−1 = a ∥a∥2 for a ̸= 0. The product ab of quaternions a, b can be expressed in several matrix forms: ab = ¡ 1 i j k ¢     a1 −a2 −a3 −a4 a2 a1 −a4 a3 a3 a4 a1 −a2 a4 −a3 a2 a1         b1 b2 b3 b4    = = ¡ 1 i j k ¢ A▷     b1 b2 b3 b4    = = ¡ a1 a2 a3 a4 ¢     b1 b2 b3 b4 −b2 b1 −b4 b3 −b3 b4 b1 −b2 −b4 −b3 b2 b1         1 i j k    = = ¡ a1 a2 a3 a4 ¢ ◁B     1 i j k    . 66 S. Zachari´ aˇ s, D. Velichov´ a: Projection from 4D to 3D For quaternions from Im H the matrices A▷and ◁B are antisymmetric. For unit quaternions the matrices A▷and ◁B are orthogonal and their determinant is equal 1; they are elements of the group O+(4). In analogy to the fact that the complex number b1 + b2i corresponds homomorphically to the real matrix µ b1 b2 −b1 b1 ¶ , Cayley showed 1858 that the quaternion b1 + b2i + b3j + b4k corresponds to the complex matrix (see ) µ b1 + b2i b3 + b4i −b3 + b4i b1 −b2i ¶ . In this sense the imaginary unit i∈C corresponds to the unit j ∈H. 1 ↔ µ 1 0 0 1 ¶ , i ↔ µ i 0 0 −i ¶ , j ↔ µ 0 1 −1 0 ¶ , k ↔ µ 0 i i 0 ¶ . In this model the quaternion b clearly refers to the real matrix ◁B. J.W. Gibbs is the founder of the vector analysis without quaternions, which is today widely used. 5. Spherical motions and orthographic projections A spherical motion O ﬁxing the origin O ∈R4 can be expressed by an orthogonal matrix M with det M = 1, which is regarded as an element of the group O+(4). Compared with the group of revolutions O+(3) that is simple, the group O+(4) has a non-trivial normal subgroup (see ). Any pair of unit quaternions (r, s) deﬁnes an element O of the group O+(4) according to O: H →H, q 7→r q s . The converse statement is also valid (see ) which means, that for any O ∈O+(4) there exist two unit quaternions r, s generating this spherical motion. The pairs (r, s) = (1, 1) and (−1, −1) correspond to the identity IdR4 in O+(4). The multiplication in the group O+(4) can be expressed by means of the quaternion product: Let two elements from O+(4) correspond to two pairs (r, s) and (r′, s′) of unit quaternions, respectively: The product of these pairs is the pair (r s, r′s′), because the following holds: q 7→rr′q ss′ = r(r′q s′)s. Comparing quaternion and matrix expression of the elements of O+(4) we have r(q s) = ¡ 1 i j k ¢ R▷◁S T     q1 q2 q3 q4    = ¡ 1 i j k ¢ M▷     q1 q2 q3 q4    , M▷= R▷◁S T = R▷◁S = S. Zachari´ aˇ s, D. Velichov´ a: Projection from 4D to 3D 67 =     r1 −r2 −r3 −r4 r2 r1 −r4 r3 r3 r4 r1 −r2 r4 −r3 r2 r1         s1 s2 s3 s4 −s2 s1 −s4 s3 −s3 s4 s1 −s2 −s4 −s3 s2 s1    . For r = s we receive the classical revolution in Im H: M▷= R▷◁R =     1 0 0 0 0 1 −2(r2 3 + r2 4) 2(r2r3 −r1r4) 2(r1r3 + r2r4) 0 2(r2r3 + r1r4) 1 −2(r2 2 + r2 4) 2(r3r4 −r1r2) 0 2(r2r4 −r1r3) 2(r1r2 + r3r4) 1 −2(r2 2 + r2 3)    ∈O+(3). It is easy to show, that for r1 = q1 = 0 and all λ ∈R holds r (λr) r = λr, Re (r q r) = 0, r · q = 0 = ⇒r · (r q r) = 0 . From this follows that the matrix M▷which is the matrix of a revolution (about the plane determined by the real axis and Im r) acts only on Im H. The matrix (−M▷) operates in Im H as the symmetry with respect to the plane passing through the origin and perpendicular to Im r (proof in ). All unit quaternions form a hypersphere S3 ⊂R4, and any point r located on this hypersphere corresponds to an orthogonal matrix R▷with det R▷= 1. The quaternion r will be related to the oriented revolution in Im H about the axis determined by Im r according to q 7→r q r . Instead of the unit quaternion r it is possible to choose a non-zero quaternion d. Any such revolution is regarded as the element of the group O+(3). The angle θ of revolution, 0 ≤θ ≤π, can be expressed as (see or ) θ = 2 arctan µ∥Im r∥ \|Re r\| ¶ = 2 arctan µ∥Im d∥ \|Re d\| ¶ . Re r = r1 = 0 implies θ = π, and we have an axial symmetry in Im H with respect to the line λr. To determine the angle θ in the easier way, we can express, while choosing the unit quaternion as c ∈Im H, the unit quaternion r of the revolution q 7→r q r in the form presented in : r = cos θ 2 + c sin θ 2 . The group O+(3) has three free parameters, the same as its model S3 ⊂H. Limits for the angle θ in the interval [0, π] are not essential, because the complementary interval [π, 2π] can be achieved by a revolution about the axis determined by the vector (−c) for the angles also in the interval [0, π]. This topological problem in R3 will be not a problem in the projective space P3, because this is homeomorphic to O+(3). The converse statement is also valid (see in ), to any revolution from O+(3) there can be related a non-zero quaternion generating this revolution in Im H. Modelling a spherical motion in R4 enables us to construct diﬀerent views of 4-dimensional objects in such a way, that the object is ﬁrstly “revolved” and then projected into the 3-dimensional space deﬁned by the ﬁrst three coordinate axes (similarly as in the projection R3 →R2). For an orthographic view it is suﬃcient to exclude one coordinate. There exists a de-composition of the matrix Q ∈O+(4) (see Section 1), Q▷= R▷◁S , anyhow the construction 68 S. Zachari´ aˇ s, D. Velichov´ a: Projection from 4D to 3D Figure 9: Images of a hypersphere S3 under projections R4 →R3 of matrices R▷, ◁S is a diﬀerent problem. On the other hand, with the help of the random generator of quaternions r, s it is easy to create a random matrix in O+(4) as the product (R▷◁S). Excluding one of the rows the random matrix P of a projection from R4 to R3 can be obtained. This projection includes also the revolution of the orthographic view in R3, when the orthographic views of the axis k are not necessarily coincident with the z-axis. Several examples are illustrated in Fig. 9, where the spherical motion is determined by the quaternion a′ = r a s, while r = cos u + i sin u, s = j cos v + k sin v, a = 0.5 + 0.5i + 0.5j + 0.5k , and intervals for parameters u, v are given in the ﬁgure. The corresponding projection is applied to the hypersphere S3, and its image is then projected by a linear perspective from R3 to R2. The other possible access is to compose revolutions about six pairs of coordinate axes. Six corresponding angles are in the interval [0, 2π]. The number of free parameters of the pair (r, s) with \|r\| = \|s\| = 1 is six. The product of these two revolutions needs not be a classical revolution, but it is a spherical motion.     cos α −sin α 0 0 sin α cos α 0 0 0 0 1 0 0 0 0 1         1 0 0 0 0 1 0 0 0 0 cos β −sin β 0 0 sin β cos β    = S. Zachari´ aˇ s, D. Velichov´ a: Projection from 4D to 3D 69 =     cos α −sin α 0 0 sin α cos α 0 0 0 0 cos β −sin β 0 0 sin β cos β    . In general the last matrix has not two real eigenvectors that would determine a plane, about which the classical revolution could be performed. References M. Berger: G´ eom´ etrie. CEDIC and Fernand Nathan, Paris 1977. B.A. Dubrovin, S.P. Novikov, A.T. Fomenko: Sovremennaya geometriya, metody i priloˇ zeniya. Nauka, Moskva 1986. J.C. Hart, G.K. Francis, L.H. Kauffman: Visualizing Quaternion Rotation. ACM Transactions on Graphics 13 (3), 256–276 (1994). G.C. Laghi: About the Contour Shape of a Hypercube in Isometric Axonometry. Pro-ceedings of the 3rd ICECGDG, Vienna 1988, vol. 1, 300–304. H. Stachel: Mehrdimensionale Axonometrie. Proceedings Congress of Geometry, Thes-saloniki 1987, 159–168. H. Stachel: The Right-Angle-Theorem in Four Dimensions. Journal of Theoretical Graphics and Computing 3, 4–13 (1990). H. Stachel: Parallel Projections in Multidimensional Space. Proceedings Compugraph-ics ‘91, Sesimbra (Portugal) 1991, vol. I, 119–128. W. Wunderlich: Zur normalen Axonometrie des vierdimensionalen Raumes. Monatsh. Math. 80, 231–240 (1975). S. Zachari´ aˇ s: Projection in barycentric coordinates. Proceedings WSCG 96, Plzeˇ n (CZ), 1996. S. Zachari´ aˇ s: Projekce hladk´ ych ploch. Proceedings Seminars on Computational Ge-ometry SCG’97, Koˇ covce (SR), 1997, 189–193. Received February 14, 2000; ﬁnal form June 12, 2000
6411	https://www.quora.com/What-are-the-properties-of-mean	What are the properties of mean? - Quora Something went wrong. Wait a moment and try again. Try again Skip to content Skip to search Sign In Mathematics Central Tendency Data and Analysis Mean (statistics) Properties (mathematics) Summary Statistics Statistical Analysis Statistical Properties Quantitative Data Analysi... + 5 What are the properties of mean? All related (38) Sort Recommended Assistant Bot · 1y The mean, often referred to as the average, has several important properties in statistics. Here are the key properties: 1. Definition: The mean of a set of numbers is calculated by summing all the values and dividing by the count of those values. For a dataset X={x 1,x 2,…,x n}X={x 1,x 2,…,x n}, the mean μ μ is given by: μ=1 n∑n i=1 x i μ=1 n∑i=1 n x i 2. Sensitivity to Outliers: The mean is sensitive to extreme values (outliers). A single very high or very low value can significantly affect the mean. 3. Uniqueness: There is only one mean for a given set of numbers. It is a single value that summarize Continue Reading The mean, often referred to as the average, has several important properties in statistics. Here are the key properties: 1. Definition: The mean of a set of numbers is calculated by summing all the values and dividing by the count of those values. For a dataset X={x 1,x 2,…,x n}X={x 1,x 2,…,x n}, the mean μ μ is given by: μ=1 n∑n i=1 x i μ=1 n∑i=1 n x i 2. Sensitivity to Outliers: The mean is sensitive to extreme values (outliers). A single very high or very low value can significantly affect the mean. 3. Uniqueness: There is only one mean for a given set of numbers. It is a single value that summarizes the central tendency of the dataset. 4. Additivity: If you add a constant to each value in the dataset, the mean increases by that constant. For example, if each value in a dataset is increased by c c, the new mean will be μ+c μ+c. 5. Scaling: If you multiply each value in the dataset by a constant, the mean is also multiplied by that constant. For instance, if each value is multiplied by k k, the new mean will be k×μ k×μ. 6. Linear Combination: The mean of a linear combination of two datasets can be expressed in terms of the means of the individual datasets. For two datasets A A and B B with means μ A μ A and μ B μ B, the mean of the combined dataset is: μ A+B=n A μ A+n B μ B n A+n B μ A+B=n A μ A+n B μ B n A+n B where n A n A and n B n B are the number of observations in datasets A A and B B respectively. 7. Central Tendency: The mean is a measure of central tendency, representing the "center" of a dataset. It is particularly useful for normally distributed data. 8. Existence: The mean exists for any finite set of numbers, provided that the sum of the numbers is finite. These properties make the mean a widely used and valuable statistic in data analysis and interpretation. However, due to its sensitivity to outliers, it's often recommended to consider other measures of central tendency (like the median) alongside the mean, especially in skewed distributions. Upvote · Liban Ali Mohamud Office (2015–present) · Author has 112 answers and 394.9K answer views ·6y Originally Answered: What are the Properties of mean? · The Properties of mean are :- 1. The mean is found by using all the values of the data. 2. The mean varies less than the median or mode when samples are taken from the same population and all three measures are computed for these samples. 3. The mean is used in computing other statistics, such as the variance. 4. The mean for the data set is unique and not necessarily one of the data values. 5. The mean cannot be computed for the data in a frequency distribution that has an open-ended class. 6. The mean is affected by extremely high or low values, called outliers, and may not be the appropriate aver Continue Reading The Properties of mean are :- 1. The mean is found by using all the values of the data. 2. The mean varies less than the median or mode when samples are taken from the same population and all three measures are computed for these samples. 3. The mean is used in computing other statistics, such as the variance. 4. The mean for the data set is unique and not necessarily one of the data values. 5. The mean cannot be computed for the data in a frequency distribution that has an open-ended class. 6. The mean is affected by extremely high or low values, called outliers, and may not be the appropriate average to use in these situations. Upvote · 99 11 Sponsored by Grammarly Is your writing working as hard as your ideas? Grammarly’s AI brings research, clarity, and structure—so your writing gets sharper with every step. Learn More 99 18 Rod Lacey Owner, Principal Consultant at Pacific North Consulting (2014–present) ·Updated 4y “Mean” is the value arithmetically centered within the dataset; it is not necessarily one of the values within the dataset. For example, consider the following dataset: [ 0, 1, 1, 1, 2, 2, 2, 3 ] The total number of values is 8 (n=8). The sum of all values is 12 (sum n1 thru n8 = 12). The mean value of the dataset is 12 divided by 8, or 1.5. Note that none of the values even are 1.5!! Depending on the meaning of the values in the dataset, the mean may be, or might not be, a useful attribute. For example, if the dataset describes the time off per week that employees at a business had, then the mean i Continue Reading “Mean” is the value arithmetically centered within the dataset; it is not necessarily one of the values within the dataset. For example, consider the following dataset: [ 0, 1, 1, 1, 2, 2, 2, 3 ] The total number of values is 8 (n=8). The sum of all values is 12 (sum n1 thru n8 = 12). The mean value of the dataset is 12 divided by 8, or 1.5. Note that none of the values even are 1.5!! Depending on the meaning of the values in the dataset, the mean may be, or might not be, a useful attribute. For example, if the dataset describes the time off per week that employees at a business had, then the mean is sensible (“Of the 8 employees at Stinko's Carwash, they averaged a day and a half of time off each week.“) However, if the set described the number of children that couples in your neighborhood had, the mean is NOT particularly useful (“Since folks around here average 1.5 kids per couple, I think Wilma and I will have a normal first kid, then probably just a half of a kid. We're scared to death!) Remember to consider what your data describes, and how it will be used, before simply calculating and attempting to utilize the various attributes of the dataset. For Wilma and her partner, the Median value would probably more helpful (and less horrible to imagine!) If it doesn't make sense, something might be off. Never lose track of the big picture! Upvote · 9 1 Related questions More answers below What are some characteristics of mean median and mode? What are the six properties? Which are the properties of the average? What is the formula of the mean? What are the properties of and how do you calculate? The Trevor Former Computer Technician at Nat West (1970–2012) · Author has 213 answers and 54.8K answer views ·Feb 13 Originally Answered: What is the meaning of properties? · One meaning of properties is the characteristics of something. One property of water is that if it is frozen to below zero degrees, it turns into ice. Another property of water, if it’s clean, is that it is transparent. The properties help to describe a thing, different things have different properties. Another meaning of property is an area of land, or a building or a thing that is owned by a person, or a business. A land-owner might construct a building on his or her piece of land. A car owner might park his or her car on the property. An extra complexity is that a property has properties. The Continue Reading One meaning of properties is the characteristics of something. One property of water is that if it is frozen to below zero degrees, it turns into ice. Another property of water, if it’s clean, is that it is transparent. The properties help to describe a thing, different things have different properties. Another meaning of property is an area of land, or a building or a thing that is owned by a person, or a business. A land-owner might construct a building on his or her piece of land. A car owner might park his or her car on the property. An extra complexity is that a property has properties. The land might have trees, or a hill or a river running through it. A building might have a red roof, two bedrooms and six windows. The car might have a six-cylinder engine and is painted blue. Upvote · Muhammad.Munawar Quality Adutior at US Appear (2016–present) · Author has 145 answers and 218.5K answer views ·5y The mean can be used for both continuous and discrete numeric data. Limitations of the mean: The mean cannot be calculated for categorical data, as the values cannot be summed. As the mean includes every value in the distribution the mean is influenced by outliers and skewed distributions. Upvote · Sponsored by Amazon Business Solutions and supplies to support learning. Save on essentials and reinvest in students and staff. Sign Up 99 39 Jim Hastings Studied at California State University Los Angeles · Author has 6.9K answers and 2.5M answer views ·Feb 12 Originally Answered: What is the meaning of properties? · Generally “Property” refers to land you own or rent but can also mean any objects (like furniture) you may own. Upvote · Related questions More answers below What are the properties of Arithmetic mean? What are the main avantages and properties of mean, median and mode? What exactly is a property in mathematics? How can we determine property lines? What are the three properties of the mean and how each of these properties is unique to the mean and not to the median or the mode? Peter Banos Eclectic reader; political news-hound · Author has 10.7K answers and 1.5M answer views ·Feb 12 Originally Answered: What is the meaning of properties? · Either the qualities of some object, the sort of things you would say to describe and identify it, or (in a different context) items of real estate (mostly buildings and plots of land). Upvote · Sponsored by CDW Corporation How can you strengthen security against malicious attacks? With CDW, discover how Trellix’s GenAI-powered security platform empowers SecOps with broad protection. Learn More 999 172 Doug Reiss B.S. in Physics&Mathematics, The Cooper Union for the Advancement of Science and Art (Graduated 1971) · Author has 922 answers and 1.3M answer views ·5y Progressively, unkind, foul, angry, horrible and sadistic, unless you happens to “mean” average!! (Or something entirely different, such as the “mean” in mathematics). Please specify. Upvote · Muthoka Jacob PhD candidate in economics, policy analyst at University of Nairobi (2017–present) · Author has 68 answers and 142.2K answer views ·6y Related What is the importance of mean? You will be surprised that the mean is perhaps the most important statistic in data because it forms the basis of conducting and understanding all other complex statistics. The mean is the “center of gravity” of your data, and is meant to carry a piece of information from every member of the sample. It is the most basic statistic that carries something from every respondent, and suggests the middle ground or the generally acceptable response. Regression analyses, validity issues, representativeness of sample, variance, etc, are pegged around the mean. It may be simplistic, but it founds the co Continue Reading You will be surprised that the mean is perhaps the most important statistic in data because it forms the basis of conducting and understanding all other complex statistics. The mean is the “center of gravity” of your data, and is meant to carry a piece of information from every member of the sample. It is the most basic statistic that carries something from every respondent, and suggests the middle ground or the generally acceptable response. Regression analyses, validity issues, representativeness of sample, variance, etc, are pegged around the mean. It may be simplistic, but it founds the complicated stuff you see in statistics all over. Upvote · 99 17 Sponsored by Grommet What are the smartest inventions of 2025? Here are some of the smartest (and most useful) gadgets we've seen this year. Take a look! Learn More 9 2 Art Cole Professional graphic designer since 1997. ·7y Related What is the importance of mean? It is a meeting point that is democratic and plays equally into the hand of all parties. The mean is more than just calculating the average. It’s our fastest path to the highest shared values. The mean levels the field and let’s everyone play a role and have a portion of the voice. If you’re “in the game” and struck out, you still contributed. You made an attempt and caused the other team to throw at least three pitches, which had a positive effect in that your “at-bat” depleated the opponent’s pitch-count during your hitting attempt. The mean can only be true when all contributors were account Continue Reading It is a meeting point that is democratic and plays equally into the hand of all parties. The mean is more than just calculating the average. It’s our fastest path to the highest shared values. The mean levels the field and let’s everyone play a role and have a portion of the voice. If you’re “in the game” and struck out, you still contributed. You made an attempt and caused the other team to throw at least three pitches, which had a positive effect in that your “at-bat” depleated the opponent’s pitch-count during your hitting attempt. The mean can only be true when all contributors were accounted for. Upvote · 9 4 Andrew Rasanen Studied Ancient Greek History (Graduated 1989) · Author has 864 answers and 423.6K answer views ·7y Related What is meant by mean? As an adjective: lacking distinction or eminence (humble); lacking in mental discrimination (dull); of inferior quality or status (as in “mean streets”); lacking dignity or honor (base); contemptible or worthy of little regard (often used negatively, as in “no mean feat”); excellent, effective (“plays a mean trumpet”); characterized by petty selfishness or malice (“mean girls”). For the verb, consult a dictionary. Upvote · Don DePasquale Former HS Math and Physics Teacher (Ret) at Albany High School (1968–2005) · Author has 626 answers and 176.6K answer views ·2y Related What are the three properties of the mean and how each of these properties is unique to the mean and not to the median or the mode? The mean is the average of a certain set of numbers. The mean also allows you to get the sum of the numbers. And finally if you know the sum of the numbers and the mean you can predict the number of elements in the sample. This comes from the sum of the numbers divided by the number of elements is the mean. Sum / n= mean. Sum = n x mean and n = sum / mean. The median is the middle number of the sample or the average of two middle numbers. The mode is the most frequent number in the set of numbers. A set of numbers can have more than one mode. Upvote · Dr Ulhas Chandekar Senior Consultant (2019–present) · Author has 1.5K answers and 759.2K answer views ·1y Related What is the meaning of mean? The word has many meanings. The word ‘Mean’ is used when one wants to either understand or explain something. For example, “What do you mean by a chemical reaction?” or “Square of a number means a multiplication of that number with the same number; 7 X 7 means 7 square.” The word ‘Mean’ is the same as average. For example, in a class of 40 students, the mean of marks obtained (out of 100) was 65. The word ‘Mean’ is used to describe a person’s nature/character. For example, the lady next door is quite mean and does not allow kids to enter her premises to collect a cricket ball. Another example: he Continue Reading The word has many meanings. The word ‘Mean’ is used when one wants to either understand or explain something. For example, “What do you mean by a chemical reaction?” or “Square of a number means a multiplication of that number with the same number; 7 X 7 means 7 square.” The word ‘Mean’ is the same as average. For example, in a class of 40 students, the mean of marks obtained (out of 100) was 65. The word ‘Mean’ is used to describe a person’s nature/character. For example, the lady next door is quite mean and does not allow kids to enter her premises to collect a cricket ball. Another example: he is so mean, that he finishes all the rice from the bowl, without any consideration to other guests. I hope my answer makes sense. Upvote · Related questions What are some characteristics of mean median and mode? What are the six properties? Which are the properties of the average? What is the formula of the mean? What are the properties of and how do you calculate? What are the properties of Arithmetic mean? What are the main avantages and properties of mean, median and mode? What exactly is a property in mathematics? How can we determine property lines? What are the three properties of the mean and how each of these properties is unique to the mean and not to the median or the mode? What are the different kinds of properties? What are the properties of a function? What is the meaning of properties? What are the properties of a function definition? What is allodial property? Related questions What are some characteristics of mean median and mode? What are the six properties? Which are the properties of the average? What is the formula of the mean? What are the properties of and how do you calculate? What are the properties of Arithmetic mean? 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6412	https://www.math.purdue.edu/~peterson/Teaching/IMPALectureNotes.pdf	Lecture Notes on Random Walks in Random Environments Jonathon Peterson ∗ Purdue University February 21, 2013 This lecture notes arose out of a mini-course I taught in January 2013 at Instituto Nacional de Matem´ atica pura e Aplicada (IMPA) in Rio de Janeiro, Brazil. In these lecture notes I do not always give all the details of the proofs, nor do I prove all the results in their greatest generality. A more detailed treatment of most of these topics can be found in Zeitouni’s lecture notes on RWRE [Zei04]. ∗e-mail: peterson@math.purdue.edu 1 1 Introduction to RWRE We begin with a very brief introduction into the model of RWRE. For simplicity, we will begin by describing the model of nearest-neighbor RWRE on Z. Once that model is understood it is easy for the reader to understand how to deﬁne RWRE on other graphs such as multi-dimensional integer lattices, trees, and other random graphs. The case of one-dimensional RWRE is the simplest to describe since in that case an environment is an elment ω = {ωx}x∈Z ∈[0, 1]Z. For any environment ω and any x ∈Z, we can construct a Markov chain Xn on Z with distribution given by P x ω deﬁned by P x ω(X0 = x) = 1 and P x ω(Xn+1 = z \| Xn = y) =      ωy z = y + 1 1 −ωy z = y −1 0 otherwise. Since we will often be concerned with RWRE starting at x = 0 we will use the notation Pω for P 0 ω. Considering a random walk in an arbitrary environment is obviously too general, and so we wish to give some additional structure to the environment by assuming that the environment ω is an Ω-valued random variable with distribution P on the space of environments Ω. Then, since for any ﬁxed event G for the random walk, P x ω(G) is a [0, 1]-valued random variable since ω is random. Thus, we can deﬁne another probability measure Px on Xn by Px(·) = EP [P x ω(·)]. Again, for simplicity we will use the notation P for P0. In general, distirbution on environments is assumed to be such that the sequence {ωx}x∈Z is stationary and ergodic. However, as an introduc-tion to the model it is often best to consider the simplest example where the environment {ωx} is an i.i.d. sequence. Since there are two diﬀerent sources of randomness in the model of RWRE (the environment and the walk), there are two diﬀerent types of probabilistic questions that can be asked. • Quenched The distribution Pω of the RWRE for a ﬁxed environment is called the quenched law of the RWRE. Under the quenched law Xn is a Markov chain, and so all the tools of Markov chains are available. However, the challenge is typically to prove a result that is true under the quenched law Pω for P-a.e. environment ω. • Averaged/Annealed The distribution P is called the averaged law for the RWRE (some prefer the term “annealed” over “averaged,” but we will use averaged in these notes). Under the averaged law the RWRE is no longer a Makov chain since the past history gives information about the environment. For instance, note that P(X1 = 1) = EP [ω0] but P(X3 = 1 \| X1 = 1, X2 = 0) = P(X1 = 1, X2 = 0, X3 = 1) P(X1 = 1, X2 = 0) = EP [ω2 0(1 −ω1)] EP [ω0(1 −ω1)]. On the other hand, due to the averaging over all environments the averaged law has ho-mogeneity that the quenched law is lacking. For instance, due to the stationarity of the environment ω it is true that P(Xn = X0) = Px(Xn = X0) for any starting location x ∈Z. 2 To make sure one understands the model of RWRE, it is helpful to consider a speciﬁc example. Example 1.1. Suppose that the environment ω = {ωx}x∈Z is i.i.d. with distribution P(ω0 = 3/4) = p, P(ω0 = 1/3) = 1 −p, for some p ∈[0, 1]. An example of part of such an environment is shown in Figure 1.1 where sites with ωx = 3/4 are colored red and sites with ωx = 1/3 are colored blue. −5 −4 −3 −2 −1 0 1 2 3 4 5 Figure 1: An example of an environment from Example 1.1. Sites colored red are such that ωx = 3/4 and blue sites are such that ωx = 1/3. Thus far we have explained the model of RWRE only in the nearest-neighbor case on Z. How-ever, it is easy to see that the model can be expanded to other graphs besides Z and that the distribution on environments does not need to be i.i.d. We now give some examples, leaving the details of making the model precise to the reader. Example 1.2 (Random walk among random conductances). For any graph G (common examples would be Z or Zd), assign a conductance cxy = cyx to every edge (x, y) of the graph. Given these conductances, the random walk then chooses an adjacent edge to move along with probability proportional to the conductance of the edge. That is, Pω(Xn+1 = y \| Xn = x) = cxy P z∼x cxz . Typically the conductances are chosen to be i.i.d., but this does not make the environment i.i.d. in the sense that ωx and ωy are dependent if x and y are connected by an edge. Example 1.3 (Random walk on Galton-Watson trees). In this example, part of the randomness of the environment is the choice of the graph on which the process evolves. That is, we ﬁrst choose a random Galton-Watson tree. Then we can assign transition probabilities ωx to every vertex x of the tree in some deterministic or random manner. For instance, possible choices are • Simple random walk - choose one of the neighboring vertices with equal porobability. • Biased random walk - Fix a parameter β > 0. If the vertex x has k “descendants” then move to a descendant of x with probability β/(1 + βk) and to the ancestor of x with probability 1/(1 + βk). • Choose transition probabilities randomly in some way. For instance do a biased random walk but with a diﬀerent bias factor βx > 0 at each vertex, where the βx are i.i.d. Example 1.4 (Random walk on super-critical percolation clusters). Let p > pc(d) be ﬁxed, where pc(d) is the critical value for edge percolation on Zd. Choose an instance of p-edge percolation on Zd, conditioned on 0 being in the unique inﬁnite component. Then perform a simple random walk on the remaining edges. Note that this is a special case of the random conductance example where the conductances on the edges of Zd are Bernoulli(p). 3 2 One-dimensional RWRE - First Order Asymptotics Having introduced the model of RWRE, we now turn our study to one-dimensional nearest neighbor RWRE. Recall that for a RWRE on Z, the environment ω = {ωx} ∈[0, 1]Z. To avoid certain degeneracy complications, and to make the proofs easier we will make the following assumptions. Assumption 1. There exists a c > 0 such that P(ω0 ∈[c, 1 −c]) = 1. Assumption 2. The distribution P is such that {ωx}x is an i.i.d. sequence. In this section, we will study the ﬁrst order asymptotics of the behavior of the RWRE: criterion for recurrence/transience and a law of large numbers. 2.1 Recurrence/Transience In Solomon’s seminal paper on RWRE [Sol75], Solomon gave an explicit criterion for recurrence or transience. While a naive guess might be that the RWRE is transient to +∞if and only if P(X1 = 1) = EP [ω0] > 1/2 this is not the case. In fact, the recurrence or transience of the RWRE is determined by the quantity EP [log ρ0], where ρx = 1 −ωx ωx , x ∈Z. (1) Theorem 2.1. Let Assumptions 1 and 2 hold. Then, EP [log ρ0] < 0 = ⇒ lim n→∞Xn = +∞, P −a.s. EP [log ρ0] > 0 = ⇒ lim n→∞Xn = −∞, P −a.s. EP [log ρ0] = 0 = ⇒ lim inf n→∞Xn = −∞, lim sup n→∞Xn = +∞, P −a.s. Remark 2.2. Note that the statement of Theorem 2.1 is under the averaged measure P, but that it also holds quenched. For instance, if EP [log ρ0] < 0 then 1 = P( lim n→∞Xn = ∞) = EP [Pω( lim n→∞Xn = ∞)], and so we can conclude that Pω(limn→∞Xn = ∞) = 1 for P-a.e. environment ω. Example 2.3. If the distribution on environments is as in Example 1.1 then Xn is transient to +∞if and only if p > log(2)/ log(6) ≈0.3869. Note that EP [ω0] > 1/2 if and only if p > 0.4, which demonstrates the gap between the true criterion for transience and the naive guess. Proof. The key to the proof of Theorem 2.1 is an explicit formula for hitting probabilities. To this end, we introduce some notation. For a ﬁxed environment ω, we deﬁne the potential V of the environment by V (k) =      Pk−1 i=0 log ρi k ≥1 0 k = 0 −P−1 i=k log ρi k ≤−1. (2) 4 Also, for any x ∈Z deﬁne the hitting time Tx by Tx = inf{n ≥0 : Xn = x}. (3) Then, since under the quenched law Pω the random walk is simply a birth-death Markov chain, for any ﬁxed a ≤x ≤b we have the following formula for hitting probabilities. P x ω(Ta < Tb) = Pb i=x+1 eV (i) Px i=a+1 eV (i) + Pb i=x+1 eV (i). (4) To see this, it is enough to note that if we denote the right side by h(x) then h(a) = 1, h(b) = 0 and h(x) = ωxh(x + 1) + (1 −ωx)h(x −1), a < x < b. We will prove that the RWRE is transient to +∞when EP [log ρ0] < 0 and leave the remaining cases to the reader. First, note that if EP [log ρ0] < 0 then since the environment is i.i.d. it follows that V (i) ∼EP [log ρ0]i as i →±∞. In particular this implies that P∞ i=1 eV (i) < ∞and P0 i=−∞eV (i) = ∞. Therefore, from the hitting probability formula in (4) we obtain that Pω(Tn < ∞) = lim a→−∞Pω(Tn < Ta) = lim a→∞ P0 i=a+1 eV (i) P0 i=a+1 eV (i) + Pn i=1 eV (i) = 1, and lim a→−∞Pω(Ta < ∞) = lim a→−∞lim b→∞Pω(Ta < Tb) = lim a→−∞lim b→∞ Pb i=1 eV (i) P0 i=a+1 eV (i) + Pb i=1 eV (i) = lim a→−∞ P∞ i=1 eV (i) P0 i=a+1 eV (i) + P∞ i=1 eV (i) = 0. The ﬁrst of these implies that lim supn→∞Xn = ∞, Pω-a.s. The second can be used to show that lim infn→∞Xn = ∞, Pω-a.s. as well. Indeed otherwise the random walk would return inﬁnitely often to some vertex, and by uniform ellipticity each time there would be a positive probability of reaching site a before returning to x. Thus, if any site is visited inﬁnitely often then Ta < ∞for all a. 2.2 Law of Large Numbers Having established a criterion for recurrence/transience we now turn toward a law of large numbers. That is, we wish to show that the limit limn→∞Xn/n exists and doesn’t depend on the environment ω. Theorem 2.4 ([Sol75]). If Assumptions 1 and 2 hold, then lim n→∞ Xn n =        1−EP [ρ0] 1+EP [ρ0] EP [ρ0] < 1 0 EP [ρ0] ≥1 and EP [ρ−1 0 ] ≥1 −1−EP [ρ−1 0 ] 1+EP [ρ−1 0 ] EP [ρ−1 0 ] < 1, P-a.s. 5 Remark 2.5. Jensen’s inequality implies that 1/EP [ρ−1 0 ] ≤EP [ρ0], and thus it cannot happen that EP [ρ0] < 1 and EP [ρ−1 0 ] < 1. Also, Jensen’s inequality implies that it is possible to have EP [log ρ0] < 0 and EP [ρ0] ≥1 (see the example below) so that the RWRE can be transient but with asymptotically zero speed. Example 2.6. Again, if the distribution on environments is as in Example 1.1 then the speed is pos-itive if p > 0.6. Thus, the RWRE is transient with asymptotically zero speed if p ∈(0.3689..., 0.6]. We will give the proof of Theorem 2.4 when EP [log ρ0] ≤0 (that is, when the random walk is recurrent or transient to the right). The formula for the limiting speed when the walk is transient to the left is obtained by symmetry. The starting point for the proof of Theorem 2.4 is the following lemma. Lemma 2.7. Suppose that lim supn→∞Xn = ∞and limn→∞Tn/n = c ∈[1, ∞]. Then, lim n→∞ Xn n = ( 1 c if c < ∞ 0 if c = ∞. Proof. Let X∗ n = maxk≤n Xk denote the maximum distance to the right that the random walk has reached by time n. Then, TX∗ n ≤n < TX∗ n+1 so that TX∗ n X∗ n ≤n X∗ n ≤TX∗ n+1 X∗ n + 1 X∗ n + 1 X∗ n . Since X∗ n →∞, the fact that Tk/k →c implies that lim n→∞ X∗ n n = ( 1 c if c < ∞ 0 if c = ∞. It remains to show that Xn/n has the same limit as X∗ n/n. Since Xn ≤X∗ n this is trivial when c = ∞(that is, when X∗ n/n →0), and so it is enough to show that limn→∞(X∗ n −Xn)/n = 0 when c < ∞. Since the step sizes are at most 1 we have that X∗ n −Xn ≤n −TX∗ n, and thus lim sup n→∞ X∗ n −Xn n ≤lim n→∞1 −lim n→∞ TX∗ n X∗ n X∗ n n = 1 −c 1 c = 0. Next, we introduce some notation. For any k ≥1 let τk := Tk −Tk−1. (Recall that we are assuming the random walk is recurrent or transient to the right so that τk < ∞for all k ≥1.) Lemma 2.8. Under the averaged measure P, the sequence {τk}k≥1 is ergodic. Proof. Let {ξk,j}k∈Z, j≥0 be a i.i.d. collection of U(0, 1) random variables that is independent of ω. Then, given an environment ω we can use the random variables ξk,j to construct the random walk. If X∗ n = k and n −TX∗ n = j then Xn+1 = 1{ξk,j<ωXn} −1{ξk,j≥ωXn} if k = X∗ n and j = n −TX∗ n. 6 It is clear that the random walk constructed this way has the same distribution as the averaged law for the RWRE. Note that to construct the path of the RWRE up until time T1, only the random variables {ξ0,j}j≥0 are needed. Similarly, the path of the random walk on the time interval [Tk, Tk+1] only depends on {ξk,j}j≥1. Now, denote Ξk = {ξk,j}j≥0 and let θ be the left shift operator on environments so that (θkω)n = ωk+n. Then it is clear from the above construction of the random walk that there is a deterministic function f such that τk = f(θk−1ω, Ξk−1). Since the environment is i.i.d. and the sequence {Ξk}k∈Z is independent of ω, it follows that {(θkω, Ξk)}k∈Z is ergodic and therefore τk = f(θk−1ω, Ξk−1) is ergodic as well. The ﬁnal ingredient we need before giving the proof of Theorem 2.4 is a formula for the quenched mean of T1. Lemma 2.9. If EP [log ρ0] < 0, then for P-a.e. environment ω Eω[τ1] = 1 ω0 + ∞ X k=1 1 ω−k ρ−k+1ρ−k+2 · · · ρ0 = 1 + 2 ∞ X k=0 ρ−kρ−k+1 · · · ρ0. (5) Proof. First we give the idea of the proof. By conditioning on the ﬁrst step of the random walk we obtain that Eω[τ1] = ω0 + (1 −ω0)E−1 ω [1 + T1] = 1 + (1 −ω0)E−1 ω [T1] = 1 + (1 −ω0) (Eθ−1ω[τ1] + Eω[τ1]) . Then, solving for Eω[τ1] we obtain that Eω[τ1] = 1 ω0 + ρ0Eθ−1ω[τ1]. (6) Iterating this formula we obtain that for any m < ∞ Eω[τ1] = 1 ω0 + m X k=1 1 ω−k ρ−k+1ρ−k+2 · · · ρ0 + ρ−mρ−m+1 · · · ρ0Eθ−m−1ω[τ1]. (7) Finally, taking m →∞we obtain the ﬁrst equality in (5). There are two diﬃculties in the above argument. First of all, in order to solve for Eω[τ1] as in (6) we need Eω[τ1] < ∞, and to iterate this we need Eθ−kω[τ1] < ∞for any k ≥1 as well. Secondly, even if all these quenched expectations are ﬁnite we need to prove that the last term in (7) vanishes as m →0. Both of these diﬃculties can be handled by truncating the hitting times. For a ﬁxed M < ∞it is easy to see that Eω[τ1 ∧M] = 1 + (1 −ω0)E−1 ω [(1 + T1) ∧M] ≤1 + (1 −ω0) (Eθ−1ω[τ1 ∧M] + Eω[τ1 ∧M]) , 7 and since now all expectations are ﬁnite we obtain that Eω[τ1 ∧M] ≤1 ω0 + ρ0Eθ−1ω[τ1 ∧M]. Iterating this gives Eω[τ1 ∧M] = 1 ω0 + m X k=1 1 ω−k ρ−k+1ρ−k+2 · · · ρ0 + ρ−mρ−m+1 · · · ρ0Eθ−m−1ω[τ1 ∧M] The assumption that EP [log ρ0] < ∞implies that ρ−mρ−m+1 · · · ρ0 →0 as m →∞and last quenched expectation is bounded above by M. Thus, can take m →∞to obtain that Eω[τ1 ∧M] ≤1 ω0 + ∞ X k=1 1 ω−k ρ−k+1ρ−k+2 · · · ρ0 . Taking M →∞, the monotone convergence theorem then gives Eω[τ1] ≤1 ω0 + ∞ X k=1 1 ω−k ρ−k+1ρ−k+2 · · · ρ0 . (8) To obtain the corresponding lower bound to (5), note that the sum on the right side of (5) is ﬁnite P-a.s. since EP [log ρ0] < 1. Therefore, Eω[τ1] < ∞for almost every environment ω, and since the environment ω = {ωx}x∈Z is stationary it follows that Eθ−kω[τ1] < ∞for all k ∈Z for almost every environment ω. Thus, the argument leading to (7) is valid and by omitting the last term we obtain that Eω[τ1] ≥1 ω0 + m X k=1 1 ω−k ρ−k+1ρ−k+2 · · · ρ0 . Finally, taking m →∞proves a matching lower bound to (8). We have thus proved the ﬁrst equality in (5). The second equality follows easily from the fact that 1/ωx = 1 + ρx. We are now ready to give the proof of Theorem 2.4. Proof. Since the sequence {τk}k≥1 is ergodic under P, Birkhoﬀ’s ergodic theorem implies that lim n→∞ Tn n = lim n→∞ 1 n n X k=1 τk = E[τ1]. Using the second formula for Eω[τ1] in (5) and the fact that the environment is i.i.d., we obtain 8 that E[τ1] = EP [Eω[τ1]] = 1 + 2 ∞ X k=0 EP [ρ−kρ−k+1 · · · ρ0] = 1 + 2 ∞ X k=0 EP [ρ0]k+1 = ( 1+EP [ρ0] 1−EP [ρ0] if EP [ρ0] < 1 ∞ if EP [ρ0] ≥1. This gives a formula for limn→∞Tn/n. The formula for limn→∞Xn/n follows from Lemma 2.7. 2.3 Notes The results in this section are true under much weaker assumptions. • Theorem 2.1 holds as long as the environment is ergodic and EP [log ρ0] exists (including +∞ or −∞). The only part of the proof that is more diﬃcult without the i.i.d. assumption is proving recurrence when EP [log ρ0] = 0. For this what is needed is that Pn−1 j=0 log ρj changes sign inﬁnitely many times as n →∞. Zeitouni uses a Lemma of Kesten to show that this is indeed the case [Zei04]. • The law of large numbers also holds under the weaker assumptions of ergodic environments and EP [log ρ0] being well deﬁned. However, if the environment is not i.i.d. then the formula for the speed vP does not simplify as much. Instead, the best we can do is vP = 1 + 2 ∞ X k=0 EP [ρ−kρ−k+1 · · · ρ0] !−1 . 9 3 Limiting Distributions - Central Limit Theorems Having given a characterization of recurrence/transience and a formula for the limiting velocity, the next natural step is to consider ﬂuctuations from the deterministic velocity - that is, limiting distributions. In this section we focus on the case when the limiting distributions are Gaussian. We will see in the next section that this is certainly not always the case. 3.1 Limiting Distributions for Hitting Times As was the case with the proof of the law of large numbers, we will deduce limiting distributions for Xn by ﬁrst proving limiting distributions for Tn. We begin with the following quenched CLT for the hitting times. Theorem 3.1. If Assumptions 1 and 2 hold and EP [ρ2 0] < 1 then lim n→∞Pω Tn −EωTn σ1 √n ≤t = Z t −∞ 1 √ 2πe−z2/2 dz =: Φ(t), ∀t ∈R, P-a.s., (9) where σ2 1 = EP [Varω(T1)] < ∞. Remark 3.2. As stated, the convergence in (9) is true for P-a.e. environment and any ﬁxed t. However, since Φ(t) is a continuous function and both sides are monotone in t it follows that the convergence is uniform in t. That is, lim n→∞sup t∈R Pω Tn −EωTn σ1 √n ≤t −Φ(t) = 0, P-a.s. A key element in the proof of Theorem 3.1 will be the following Lemma. Lemma 3.3. If EP [ρ2 0] < 1 then E[τ 2 1 ] < ∞. Proof. we ﬁrst derive a formula for Eω[τ 2 1 ] in a similar manner to the derivation of the formula for Eω[τ1] in Lemma 2.9. By conditioning on the ﬁrst step of the random walk, Eω[τ 2 1 ] = ω0 + (1 −ω0)E−1 ω [(1 + T1)2] = 1 + (1 −ω0) 2Eθ−1ω[τ1] + 2Eω[τ1] + 2(Eθ−1ω[τ1])(Eω[τ1]) + Eθ−1ω[τ 2 1 ] + Eω[τ 2 1 ] . Then we can solve for Eω[τ 2 1 ] to obtain Eω[τ 2 1 ] = 1 ω0 + ρ0 2Eθ−1ω[τ1] + 2Eω[τ1] + 2(Eθ−1ω[τ1])(Eω[τ1]) + Eθ−1ω[τ 2 1 ] . At this point, we can simplify things by noting that ρ0Eθ−1ω[τ1] = Eω[τ1] − 1 ω0 . Combining this with the above formula for Eω[τ 2 1 ] and doing a little bit of algebra one obtains that Eω[τ 2 1 ] = 2(Eω[τ1])2 −1 ω0 + ρ0Eθ−1ω[τ 2 1 ]. 10 Iterating this m times and then taking m →∞we can arrive at the following formula for Eω[τ 2 1 ]. Eω[τ 2 1 ] = 2(Eω[τ1])2 + 2 ∞ X n=1 (ρ−n+1ρ−n+2 · · · ρ0) (Eθ−nω[τ1])2 −Eω[τ1]. (10) We remark that the argument leading to (10) we have ignored some technical diﬃculties that arise. However, as in the proof of Lemma 2.9 the formula in (10) can be justiﬁed by repeating the above argument for the truncated second moment Eω[(τ1 ∧M)2]. We leave the details to the interested reader. Having proved the formula (10), we now note that since the environment is i.i.d. that E[τ 2 1 ] = 2EP [(Eωτ1)2] ∞ X n=0 EP [ρ0]n −E[τ1]. (Note that here we have used that Eθ−nω[τ1] depends only on ωx for x ≤n.) Since EP [ρ2 0] < 1 implies that EP [ρ0] < 1 as well, it will follow that E[τ 2 1 ] < ∞if we can show that EP [(Eωτ1)2] < ∞. To this end, from the second formula for Eω[τ1] in (5) it follows that EP [(Eω[τ1])2] ≤4EP   ∞ X k=0 ρ−kρ−k+1 · · · ρ0 !2  = 4EP  X k≥0 (ρ−kρ−k+1 · · · ρ0)2 + 2 X 0≤k<n (ρ−n · · · ρ−k−1)(ρ−kρ−k+1 · · · ρ0)2   = 4    X k≥0 (EP [ρ2 0])k+1 + X 0≤k<n (EP [ρ0])n−k(EP [ρ2 0])k+1   , (11) and these last sums are ﬁnite when EP [ρ2 0] < 1. Remark 3.4. Since σ2 1 = EP [Eω[τ 2 1 ] −(Eω[τ1])2] = E[τ 2 1 ] −EP [(Eω[τ1])2], it follows from Lemma 3.3 that σ2 1 < ∞if EP [ρ2 0] < 1. In fact, by being more careful with the argument in the proof of Lemma 3.3 one can derive the following formula for σ2 1 in terms of EP [ρ0] and EP [ρ2 0]. σ2 1 = 4(1 + EP [ρ0])(EP [ρ0] + EP [ρ2 0]) (1 −EP [ρ0])2(1 −EP [ρ2 0]) . Proof of Theorem 3.1. Under the quenched measure, Tn −EωTn = Pn k=1(τk −Eω[τk]) is the sum of n independent zero mean random variables (note that the random variables are not identically distributed). The main idea is to use the Lindberg-Feller criterion to prove a central limit theorem. That is, the statement of the theorem will follow if we can check that lim n→∞ 1 n n X k=1 Eω[(τk −Eω[τk])2] = σ2 1, P-a.s., (12) and lim n→∞ 1 n n X k=1 Eω h (τk −Eωτk)21{\|τk−Eω[τk]\|≥ε√n} i = 0, ∀ε > 0, P-a.s. (13) 11 To prove (12), note that lim n→∞ 1 n n X k=1 Eω[(τk −Eω[τk])2] = lim n→∞ 1 n n X k=1 Varθk−1ωτ1 = EP [Varωτ1], where the last equality follows from Birkohﬀ’s ergodic Theorem. The proof of (13) is similar. Fix ε > 0 and M < ∞. Then, lim sup n→∞ 1 n n X k=1 Eω h (τk −Eωτk)21{\|τk−Eω[τk]\|≥ε√n} i ≤lim n→∞ 1 n n X k=1 Eω (τk −Eωτk)21{\|τk−Eω[τk]\|≥M} = EP Eω (τk −Eωτk)21{\|τk−Eω[τk]\|≥M} , where again the last equality follows from Birkhoﬀ’s ergodic Theorem. Since σ2 1 = EP [Eω[(τ1 − Eω[τ1])2] < ∞, it follows that the right side can be made arbitrarily small by taking M →∞. This ﬁnishes the proofs of (12) and (13), and thus also the proof of the theorem. Having proved the quenched central limit theorem for hitting times, we next give an limiting distribution under the averaged measure. Theorem 3.5. If Assumptions 1 and 2 hold and EP [ρ2 0] < 1 then lim n→∞Pω Tn −n/vP σ√n ≤t = Φ(t), ∀t ∈R, where σ2 = σ2 1 + σ2 2, with σ2 1 deﬁned as in Theorem 3.1 and σ2 2 = Var(Eω[τ1]) + 2 n−1 X k=1 Cov(Eω[τ1], Eθkω[τ1]) < ∞. Remark 3.6. Using the second formula in (5) for Eω[τ1], it is not too diﬃcult to compute Var(Eω[τ1]) and Cov(Eω[τ1], Eθkω[τ1]). By doing this, one can derive the following formula for σ2 2. σ2 2 = 4(1 + EP [ρ0])VarP (ρ0) (1 −EP [ρ0])3(1 −EP [ρ2 0]). A ﬁrst step in proving the averaged CLT is to prove the following CLT for the quenched mean of the hitting times. Theorem 3.7. If Assumptions 1 and 2 hold and EP [ρ2 0] < 1 then lim n→∞P Eω[Tn] −n/vP σ2 √n ≤t = Φ(t), ∀t ∈R, where σ2 2 < ∞is deﬁned as in Theorem 3.5. Proof. Note that Eω[Tn] −n/vP = Pn k=1(Eω[τk] −1/vP ) = Pn−1 k=0(Eθkω[τ1] −1/vP ) is the sum of an ergodic, zero-mean sequence. Then the proof of the CLT for Eω[Tn] will follow if we can check 12 the condition for the CLT for sums of ergodic sequences in [Dur96, p. 417]. That is, we need to show that ∞ X n=0 r EP h (EP [Eω[τ1] −1/vP \| F−n])2i < ∞, where F−n = σ(ωx : x ≤−n). (14) However, it is clear from the second formula for Eω[τ1] in (5) that EP [Eω[τ1] −1/vP \| F−n] = 1 + 2 n X k=1 EP [ρ0]k + 2EP [ρ0]n X k≥n ρ−kρ−k+1 · · · ρ−n −1 vP = EP [ρ0]n   − 1 1 −EP [ρ0] + 2 X k≥n ρ−kρ−k+1 · · · ρ−n   , where the second inequality follows from the fact that 1/vP = E[τ1] = (1 + EP [ρ0])/(1 −EP [ρ0]) and a little bit of algebra. Therefore, ∞ X n=0 r EP h (EP [Eω[τ1] −1/vP \| F−n])2i = ∞ X n=0 EP [ρ0]n v u u u tEP    − 1 1 −EP [ρ0] + 2 X k≥n ρ−kρ−k+1 · · · ρ−n   2  = v u u u tEP    − 1 1 −EP [ρ0] + 2 X k≥0 ρ−kρ−k+1 · · · ρ0   2  ∞ X n=0 EP [ρ0]n where the last equality follows from the fact that the environment is a stationary sequence. Fi-nally, the computation in (11) shows that the expectation under the square rootis ﬁnite, and since EP [ρ0] < 1 the sum is ﬁnite as well. This completes the proof of (14) and thus also of the theo-rem. Proof of Theorem 3.5. The proof of the averaged CLT for hitting times follows easily from Theo-rems 3.1 and 3.7. The idea is that Tn −n/vP √n = Tn −Eω[Tn] √n + Eω[Tn] −n/vP √n , and Theorems 3.1 and 3.7 imply that the terms on the right side are asymptotically zero mean Gaussian random variables with variance σ2 1 and σ2 2 respectively. Moreover, the second term on the right depends only on the environment, while the ﬁrst term is asymptotically independent of the environment (since the limiting distribution in Theorem 3.1 doesn’t depend on ω). Therefore, we expect that right side should be asymptotically the sum of two independent mean zero Gaussians with varianes σ2 1 and σ2 2. To make the proof precise, we ﬁrst write P Tn −n/vP σ√n ≤t = P Tn −Eω[Tn] σ√n ≤t −Eω[Tn] −n/vP σ√n = EP Pω Tn −Eω[Tn] σ1 √n ≤σt σ1 −σ2 σ1 Eω[Tn] −n/vP σ2 √n . 13 Since, as noted in Remark 3.2, the convergence in the quenched CLT is uniform in t it follows that lim n→∞P Tn −n/vP σ√n ≤t = lim n→∞EP Φ σt σ1 −σ2 σ1 Eω[Tn] −n/vP σ2 √n = E Φ σt σ1 −σ2 σ1 Z , with Z ∼N(0, 1). (15) Note that the last equality above follows from Theorem 3.7. Finally, note that Φ σt σ1 −σ2 σ1 Z = P Z′ ≤σt σ1 −σ2 σ1 Z = P σ1 σ Z′ + σ2 σ Z ≤t , where Z′ is a N(0, 1) random variable that is independent of Z. Since σ1 σ Z′ + σ2 σ Z ∼N(0, 1) (recall that σ2 = σ2 1 + σ2 2) it follows that the last line of (15) is equal to Φ(t). 3.2 Limiting Distributions for the Position of the RWRE We now show how to deduce quenched and averaged CLTs for Xn from the corresponding CLTs for the hitting times Tn. Theorem 3.8. If Assumptions 1 and 2 hold and EP [ρ2 0] < 1, then lim n→∞P Xn −nvP v3/2 P σ√n < t ! = Φ(t), ∀t ∈R, where as in Theorem 3.5 σ2 = σ2 1 + σ2 2 < ∞. Proof. Recall the deﬁnition of X∗ n = maxk≤n Xk. We will ﬁrst prove the averaged CLT for X∗ n in place of Xn and then show that Xn is close enough to X∗ n for the same limiting distribution to hold. Note that {X∗ n < k} = {Tk > n}. Then, for any t ∈R and n ≥1 let x(n) := ⌈nvP + v3/2 P σ√nt⌉so that P X∗ n −nvP v3/2 P σ√n < t ! = P X∗ n < nvP + v3/2 P σ√nt = P Tx(n,t) > n = P Tx(n,t) −x(n, t)/vP σ p x(n, t) > n −x(n, t)/vP σ p x(n, t) ! . It follows from the above deﬁnition of x(n, t) that lim n→∞ n −x(n, t)/vP σ p x(n, t) = −t. Thus, we can conclude from Theorem 3.5 that lim n→∞P X∗ n −nvP v3/2 P σ√n < t ! = lim n→∞P Tx(n,t) −x(n, t)/vP σ p x(n, t) > −t ! = 1 −Φ(−t) = Φ(t). It remains to show that Xn is close enough to X∗ n to have the same limiting distribution. To this end, the following Lemma is more than enough to ﬁnish the proof of the CLT for Xn. 14 Lemma 3.9. If Assumptions 1 and 2 hold and EP [log ρ0] < ∞, then lim n→∞ X∗ n −Xn (log n)2 = 0, P-a.s. Proof. By the Borel-Cantelli Lemma, it is enough to show that X n≥1 P(X∗ n −Xn ≥δ(log n)2) < ∞, ∀δ > 0. (16) To this end, note that the event {X∗ n −Xn ≥δ(log n)2} implies that after ﬁrst hitting some k ≤n the random walk then backtracks to k −⌈δ(log n)2⌉. Thus, by the strong Markov property (using the quenched law) Pω(X∗ n −Xn ≥δ(log n)2) ≤ n X k=1 P k ω(Tk−⌈δ(log n)2⌉< ∞). Taking expectations with respect to P we obtain P(X∗ n −Xn ≥δ(log n)2) ≤ n X k=1 EP h P k ω(Tk−⌈δ(log n)2⌉< ∞) i = nP(T−⌈δ(log n)2⌉< ∞), where in the last equality we used the stationarity of the distribution on environments. Then, the proof of (16) will be completed if we can show that there exist constants C1, C2 > 0 such that P(T−k < ∞) ≤C1e−C2k, ∀k ≥1. (17) To this end, note that from the formula for hitting probabilities (4) we can see that Pω(T−k < ∞) = P j≥1 eV (j) P j≥−k+1 eV (j) ≤ X j≥1 eV (j)−V (−k+1). Typically, V (j)−V (−k+1) is close to (j+k−1)EP [log ρ0] and so for k large we expect Pω(T−k < ∞) to be exponentially small. To this end, ﬁx c > 0 and note that P(T−k < ∞) = EP [Pω(T−k < ∞)] ≤ e−kc 1 −e−c + P   ∞ X j=1 eV (j)−V (−k+1) > e−kc 1 −e−c   ≤ e−kc 1 −e−c + X j≥1 P eV (j)−V (−k+1) > e−c(j+k−1) = e−kc 1 −e−c + X j≥1 P (V (j) −V (−k + 1) > −c(j + k −1)) = e−kc 1 −e−c + X j≥k P (V (j) > −cj) , (18) where the last equality follows from the fact that V (j) −V (−k + 1) has the same distribution as V (j + k −1) since the environment ω is stationary. Now since V (j) = Pj−1 i=0 log ρi is the sum of i.i.d. bounded random variables, it follows from Cramer’s Theorem [DZ98, Theorem 2.2.3] that 15 P(V (j) > −cj) decays exponentially in j if c < −EP [log ρ0]. That is, for 0 < c < −EP [log ρ0] there exists a δ > 0 (depending on c) such that P(V (j) > −cj) ≤e−δj for all j suﬃciently large. Applying this to (18) we obtain that P(T−k < ∞) ≤ e−kc 1 −e−c + e−δk 1 −ε−δ for all k suﬃciently large. This proves (17) and thus also the lemma. We can also prove a quenched CLT for Xn. However, since the centering is random (depending on the environment) instead of deterministic in the the quenched CLT for Tn, determining the proper centering for Xn is more diﬃcult. Theorem 3.10. If Assumptions 1 and 2 hold and EP [ρ2 0] < 1, then lim n→∞Pω Xn −nvP + Zn(ω) v3/2 P σ1 √n < t ! = Φ(t), ∀t ∈R, (19) where σ2 1 is deﬁned as in Theorem 3.1 and Zn(ω) = vP P⌊nvP ⌋ k=1 (Eω[τk] −1/vP ). Sketch of the proof. The idea of the proof is essentially the same as the proof of Theorem 3.8. As mentioned above, the main diﬃculty is determining a proper quenched centering. Let cn(ω) be some possible environment-dependent centering scheme. Then, denoting y(n, t, ω) = ⌈cn(ω)+v3/2 P σ1 √nt⌉ Pω X∗ n −cn(ω) v3/2 P σ1 √n < t ! = Pω X∗ n < cn(ω) + v3/2 P σ1 √nt = Pω Ty(n,t,ω) > n = Pω Ty(n,t,ω) −Eω Ty(n,t,ω) σ1 p y(n, t, ω) > n −Eω Ty(n,t,ω) σ1 p y(n, t, ω) ! . We wish to choose the centering scheme cn(ω) so that lim n→∞ y(n, t, ω) n = vP , and lim n→∞ n −Eω Ty(n,t,ω) σ1√nvP = −t, ∀t, P-a.s., (20) in which case it would follow from Theorem 3.1 that lim n→∞Pω X∗ n −cn(ω) v3/2 P σ1 √n < t ! = lim n→∞Pω Ty(n,t,ω) −Eω Ty(n,t,ω) σ1 p y(n, t, ω) > −t ! = 1 −Φ(−t) = Φ(t). It remains to check that the conditions in (20) are satisﬁed for cn(ω) = nvP −Zn(ω). We will not give a completely rigorous proof of these facts, but instead explain why they indeed hold and leave the details to the reader. It will be crucial below to note that Theorem 3.7 and the deﬁnition of Zn(ω) imply that Zn(ω)/√n converges in distribution to a zero-mean Gaussian random variable. 16 Informally, this implies that Zn(ω) is typically of size O(√n). The ﬁrst condition in (20) is easily checked since lim n→∞ y(n, t, ω) n = lim n→∞ ⌈nvP −Zn(ω) + v3/2 P σ1 √nt⌉ n = vP −lim n→∞ Zn(ω) n = vP , P-a.s. Checking the second condition in (20) is more diﬃcult. First note that n −Eω[Ty(n,t,ω)] = n −Eω[T⌊nvP −Zn(ω)⌋] − ⌈nvP −Zn(ω)+v3/2 P σ1 √nt⌉ X k=⌊nvP −Zn(ω)⌋+1 Eω[τk]. (21) Since the last sum on the right is the sum of v3/2 P σ1 √nt ergodic random variables with mean 1/vP it should be true that lim n→∞ 1 σ1√nvP ⌈nvP −Zn(ω)+v3/2 P σ1 √nt⌉ X k=⌊nvP −Zn(ω)⌋+1 Eω[τk] = t, P-a.s. (22) Next, note that Eω[T⌊nvP −Zn(ω)⌋] −n = ⌊nvP −Zn(ω)⌋ X k=1 Eω[τk] −1 vP + ⌊nvP −Zn(ω)⌋ vP −n = ⌊nvP −Zn(ω)⌋ X k=1 Eω[τk] −1 vP −Zn(ω) vP + δn(ω) = − ⌊nvP ⌋ X k=⌊nvP −Zn(ω)⌋+1 Eω[τk] −1 vP + δn(ω) where \|δn(ω)\| < 1/vP is an error term coming from the integer eﬀects of the ﬂoor function, and the last equality follows from the deﬁnition of Zn(ω). Since this last sum is the sum of Zn(ω) zero-mean ergodic terms and Zn(ω)/√n converges in distribution, it should be the case that lim n→∞ n −Eω[T⌊nvP −Zn(ω)⌋] √n = lim n→∞ 1 √n ⌊nvP ⌋ X k=⌊nvP −Zn(ω)⌋+1 Eω[τk] −1 vP = 0, P-a.s. (23) Combining (21), (22) and (23) veriﬁes the second condition in (20). Remark 3.11. As mentioned above, the above justiﬁcation of the centering scheme cn(ω) = nvP − Zn(ω) skips some technical details (in particular we are trying to apply Birkhoﬀ’s ergodic theorem with ω-dependent endpoints of the summands). For more details and other centering schemes that can be used see [Gol07]. 3.3 Notes The above proofs of the quenched and averaged CLTs diﬀer from those given at other places in the literature. 17 • Kesten, Kozlov, and Spitzer [KKS75] also deduce limiting distributions for Xn from corre-sponding limiting distributions for Tn. However, since they are primarily interested in the non-Gaussian limits when EP [ρ2 0] < 1 (see the next section) they prove much more than is needed to obtain a CLT in the case when EP [ρ2 0] < 1. Also, in [KKS75] only averaged limiting distributions are proved for Xn and Tn. • Zeitouni [Zei04] proves an averaged CLT for Xn using a method known as the environment viewed from the point of view of the particle. With this method he is able to prove a CLT for certain ergodic, non-i.i.d. laws on environments. As a byproduct he comes very close to proving a quenched CLT, but with the limit (19) holding only in P-probability instead of P-a.s. • The idea of using the Lindberg-Feller criterion for proving a quenched CLT for Tn ﬁrst used by Alili [Ali99]. However, it wasn’t until later that Goldsheid [Gol07] and independently Peterson [Pet08] showed how to obtain a quenched CLT for Xn by choosing an appropriate environment-dependent centering scheme. Goldsheid is able to prove the quenched CLT for certain uniformly ergodic environment. Peterson’s proof on the other hand proves a functional CLT (convergence to Brownian motion) for both Tn and Xn and is valid for environments satifying a certain technical mixing condition. 18 4 Limiting Distributions - The non-Gaussian Case In the previous section we proved quenched and averaged central limit theorems under the assump-tion that EP [ρ2 0] < 1. In this section we will examine the (quenched and averaged) limiting distri-butions when this assumption is removed. We will, however, continue to assume that EP [ρ0] < 0 so that the RWRE is transient to the right. The recurrent case is very diﬀerent, but at the end of the section we will make some remarks about the limiting distributions in the recurrent case. The reader should be warned that this section begins a change in the notes where we will omit the proofs of certain technical arguments. Many of the remaining results are quite technical, and to aid the reader we will instead try to give a hueristic understanding of the technical parts and only give the full arguments for the less technical sections. Throughout this section, we will always be assuming Assumptions 1 and 2 and that EP [log ρ0] < 0. If in addition we have P(ω0 ≥1/2) = 1, then P(ρ0 ≤1) = 1 and P(ρ0 < 1) > 0. In this case EP [ρ2 0] < 1 and so the central limit theorems from the previous section apply. Thus, we will assume instead that P(ω0 < 1/2) > 0 and we claim that in this case there exists a unique κ = κ(P) > 0 such that EP [ρκ 0] = 1. (24) To see this, note that φ(γ) = EP [ργ 0] = EP [eγ log ρ0] is the moment generating function for log ρ0. Therefore, φ(γ) is a convex function in γ with slope φ′(0) = EP [log ρ0] < 0 at the origin. Therefore, φ(γ) < r(0) = 1 for some γ > 0. On the other hand, since P(ω0 < 1/2) = P(ρ0 > 1) > 0 then it follows that φ(γ) →∞as γ →∞(note that Assumption 1 implies that φ(γ) < ∞for all γ ∈R). Since φ(γ) is convex there is thus a unique κ > 0 satisfying (24). φ(γ) = EP [ργ 0] κ 1 γ Figure 2: A visual depiction of the parameter κ = κ(P). Note that the derivative of the curve at the origin is EP [log ρ0] < 0. Also, it is clear from the picture that EP [ργ 0] < 1 ⇐ ⇒γ ∈(0, κ). Note that some of the results in the previous sections can be stated in terms of κ. • The random walk is transient with zero speed if κ ∈(0, 1] and with positive speed if κ > 1. • The central limit theorems and the moment bounds on τ1 in Section 3 all hold if and only if κ > 2. 19 The main results in this section will also need the following technical assumption. Assumption 3. The distribution of log ρ0 is non-arithmetic. That is, the support of log ρ0 is not contained in a + bZ for any a, b ∈R. The key place that this assumption is used is in the following Lemma. Lemma 4.1. Let Assumptions 1, 2 and 3 hold and let κ > 0 be deﬁned as in (24). Then, there exists a constant C > 0 such that lim t→∞P(Eω[τ1] > t) ∼Ct−κ, as t →∞. (25) Proof. This is essentially a direct application of [Kes73, Theorem 5]. Remark 4.2. The proof of Lemma 4.1 is rather technical and so we will content ourselves with only giving a reference to the paper [Kes73]. However, to give some intuition of the result note that Lemma 4.1 implies that EP [(Eω[τ1])γ] < ∞if and only if γ < κ. Since Eω[τ1] = 1 + 2 P∞ k=0(ρ−k · · · ρ0) it is reasonable to expect that EP [(Eω[τ1])γ] < ∞if and only if EP [ργ 0] < 1, but the deﬁnition of the parameter κ implies that EP [ργ 0] < 1 if and only if γ ∈(0, κ). 4.1 Background - Stable Distributions and Poisson Point Processes Before discussing the limiting distributions when κ ∈(0, 2) we need to review some facts about stable distributions and Poisson point processes. 4.1.1 Stable Distribution Recall that a (non-degenerate) distribution F is a stable distribution if for any n ≥2 there exist constants cn ∈R and an > 0 such that if X1, X2, . . . Xn are i.i.d. with common distribution F, then (X1 + X2 + · · · + Xn −cn)/an also has distribution F. The stable distributions are characterized ﬁrst of all by their index α ∈(0, 2]. We will refer to a stable distribution with index α as an α-stable distribution. The 2-stable distributions are the two-parameter family of Normal/Gaussian distributions N(µ, σ2). The family of α-stable distributions with α ∈(0, 2) are a characterized by three parameters: centering µ ∈R, scaling b > 0, and skewness γ ∈[−1, 1]. The centering and scaling parameters have the same roles as the mean and variance of the normal family of distributions. However, note that α-stable random variables have inﬁnite variance when α < 2 and inﬁnite mean when α < 1. Unlike the Normal distributions, the α-stable distributions are symmetric only when the skewness parameter γ = 0. When the skewness parameter γ = 1 or γ = −1, the distribution is said to be totally skewed to the right or left, respectively. In general, there are not explicit formulas for the stable distributions. However, there are a few special cases where the densities are known. 20 • The standard Cauchy distribution has density f(x) = 1 π(1+x2). This is a 1-stable distribution with µ = 0, b = 1, and γ = 0. • The L´ evy distribution has density f(x) = (2π)−1/2x−3/2e−1 2x 1{x>0}. This is a 1 2-stable dis-tribution with µ = 0, b = 1, and γ = 1. Aside from the normal distributions and the above two examples, the stable distributions are generally deﬁned by their characteristic function. Example 4.3. The stable distributions that we will be interested in with regard to RWRE are the α-stable distributions that are totally skewed to the right. We will denote by Lα,b the α-stable distribution with scaling parameter b, centering µ = 0, and skew γ = 1. These are the distributions with characteristic functions ˆ Lα,b(u) = Z R eixu Lα,b(dx) = exp −b\|u\|α 1 −i u \|u\|φα(u) where φα(u) := ( tan απ 2 α ̸= 1 2 π log \|u\| α = 1. Stable distributions arise naturally as limiting distributions of sums of i.i.d. random variables. If the i.i.d. random variables have ﬁnite mean, then the central limit theorem implies that (after centering and scaling properly) the limiting distribution is Gaussian. On the other hand, stable distributions with α < 2 arise as limits of sums of i.i.d. random variables with inﬁnite variance. However, while the central limit theorem is robust in the sense that only a ﬁnite moment is needed, to obtain α-stable limiting distributions more precise information on the tail asymptotics is needed. Example 4.4. Let ξ1, ξ2, ξ3, . . . be a sequence of non-negative i.i.d. random variables, and suppose that there exists some b > 0 and α ∈(0, 2) such that P(ξ1 > t) ∼bt−α, as t →∞. (26) Recall that Lα,b are the distribution functions for the totally skewed to the right α-stable distribu-tions. Then, lim n→∞P Pn i=1 ξi −Cα(n) n1/α ≤x = Lα,b(x), where the centering term Cα(n) =      0 α ∈(0, 1) nE[ξ11{ξ1≤n}] α = 1 nE[ξ1] α ∈(1, 2). Note that when α = 1 the tail asymptotics (26) imply that C1(n) ∼bn log n, but that in general we cannot replace the centering term by bn log n in this case since it may be that lim sup n→∞ \|C1(n) −bn log n\| n = lim sup n→∞ E[ξ11{ξ1≤n}] −log n > 0. 21 Example 4.5. Again, let ξ1, ξ2, ξ3, . . . be i.i.d. random variables, but now assume that P(ξ1 > t) ∼bt−2 as t →∞. Note that this tail decay implies that Var(ξ1) = ∞so that the central limit theorem does not apply. Nevertheless, if we take a scaling that is slightly larger than √n we can still obtain a Gaussian limiting distribution. That is, there exists an a > 0 such that lim n→∞P Pn i=1 ξi −nE[ξ1] a√n log n ≤x = Φ(x). 4.1.2 Stable Distributions and Poisson Point Processes Next, we brieﬂy recall the relationship between Poisson point process and α-stable distributions when α < 2. Recall that a point proccess N = P i≥1 δxi is a measure valued random variable. The xi are called the atoms of the point process N (note that the ordering of the atoms does not matter), and for any (Borel-measurable) A ⊂R, N(A) is the number of atoms contained in A. Deﬁnition 4.1. N is a non-homogeneous Poisson point process with intensity λ(x) if (i) N ∼Poisson R A λ(x) dx for all A ⊂R. (ii) {N(A1), N(A2), . . . , N(Ak)} are independent if the sets A1, A2, . . . , Ak are disjoint. Example 4.6. Let M = P i≥1 δti be a homogeneous rate 1 Poisson point process on (0, ∞) (that is the intensity 1x>0). Fix a constant λ > 0 and α > 0 and let Nλ,α be the transformed point process Nλ,α = X i≥1 δλ1/αx−1/α i . Then Nλ,α is a Poisson point process with intensity λαx−α−1. This is a standard exercise in transformed Poisson point process, but as a review we will note that since λ1/αx−1/α ∈[a, b] if and only if x ∈[λb−α, λa−α] it follows that Nλ,α([a, b]) = M([λb−α, λa−α]) ∼Poisson λ(a−α −b−α) = Poisson Z b a λαx−α−1 dx . We now show how the Poisson point processes from Example 4.6 are related to the totally skewed to the right α-stable distributions from Example 4.3. Example 4.7. Let Nλ,α be a Poisson point process with intensity λαx−α−1. If α ∈(0, 1] the random variable Z = Z x Nλ,α(dx). is almost surely well deﬁned and has distribution Lα,λ as deﬁned in Example 4.3. To see that Z is well deﬁned let Nλ,α = P i≥1 δzi so that Z = P i≥1 zi. Recall from Example 4.6 that we can represent the atoms zi of Nλ,α by zi = λ1/αx−1/α i , where the xi are the atoms of a homogeneous Poisson process with rate 1. Since we know that xi ∼i as i →∞it follows that zi ∼λ1/αi−1/α as i →∞. Since P i≥1 i−1/α < ∞when α < 1 this shows that Z is almost surely well deﬁned. 22 The fact that Z has distribution Lα,λ can be veriﬁed by directly computing the characteristic function. This is a somewhat involved computation, but more simply one can easily check that Z has a stable distribution. Let N1, N2, . . . , Nn be n independent point processes, all with the same distribution as Nλ,α. Then if Zi = R x Ni(dx), the random variables Z1, . . . , Zn are i.i.d. and all with the same distribution as Z. It follows from the superposition of Poisson point processes that n X i=1 Zi = Z x N(dx), where N is a Poisson point process with intensity nλαx−α−1. Also, if Nλ,α = P i≥1 δxi then P i≥1 δn1/αxi is a Poisson point process with intensity nλαx−α−1 as well. From this it is clear that Z1 + Z2 + · · · Zn n1/α Law = Z. Example 4.8. Let Nλ,α be a Poisson point process with intensity λαx−α−1. If α ∈(1, 2) then the random variable Z = lim δ→0 Z ∞ δ x Nλ,α(dx) −λαδ1−α α −1 , (27) is almost surely well deﬁned and has distribution Lα,λ as deﬁned in Example 4.3. For convenience of notation, let Zδ = R ∞ δ x Nλ,α(dx) −λαδ1−α α−1 . To see that the limit limδ→0 Zδ exists, ﬁrst note that E Z ∞ δ x Nλ,α(dx) = Z ∞ δ λαx−α dx = λαδ1−α α −1 , so that E[Zδ] = 0 for all δ > 0. Secondly, note that for any 0 < ε < δ Var(Zδ −Zε) = Var Z δ ε x Nλ,α(dx) = Z δ ε λαx1−α dx = λα 2 −α δ2−α −ε2−α ≤λαδ2−α 2 −α . Since α < 2 this vanishes as δ →0. This shows that Zδ is Cauchy in probability and thus converges in probability. In fact, it can be shown that the limit converges almost surely, but we will content ourselves with the above argument for now. As in the previous example it can be checked using the superposition of Poisson processes that n−1/α(Z1 + · · · Zn) has the same distribution as Z for any n. Finally, it can be checked by direct computation that limδ→0 E[eiuZδ] = ˆ Lα,λ(u) so that Z does have distribution Lα,λ. 4.2 Averaged Limiting Distributions - κ ≤2 Having reviewed the necessary information on stable distributions, we are now ready to begin the study of the limiting distributions for RWRE when κ ∈(0, 2). In contrast to the previous section we will discuss the averaged limiting distributions ﬁrst since they are much easier. However, as in the previous section we ﬁll ﬁrst study the limiting distributions for hitting times and then deduce the corresponding limiting distributions for Xn. Theorem 4.9. Let Assumptions 1, 2, and 3 hold. If κ is deﬁned as in (24) then 23 (i) If κ ∈(0, 1), then there exists a b > 0 such that lim n→∞P Tn n1/κ ≤t = Lκ,b(t), ∀t. (ii) If κ = 1, then there exists a constant b > 0 and a sequence D(n) ∼b log n such that lim n→∞P Tn −nD(n) n ≤t = L1,b(t), ∀t. (iii) If κ ∈(1, 2), then there exists a b > 0 such that lim n→∞P Tn −n/vP n1/κ ≤t = Lκ,b(t), ∀t. (iv) If κ = 2, then there exists a b > 0 such that lim n→∞P Tn −n/vP b√n log n ≤t = Φ(t), ∀t. Note the similarity in the above limiting distributions to those for sums of i.i.d. non-negative heavy tailed random variables in Examples 4.4 and 4.5. On the one hand this is not surprising since Tn = Pn k=1 τk, but under the averaged measure P the random variables τk are neither independent nor identically distributed. However, as we will see below the main idea of the proof is that if we group certain of the τk together the sums of the τk within the groups become essentially independent and identically distributed with heavy tails. Recall the deﬁnition of the potential of the environment V in (2). For a given environment ω we will deﬁne a sequence of “ladder locations” of the environment as follows. ν0 = 0, and νk = inf{i > νk−1 : V (i) < V (νk−1)} for k ≥1. (28) The idea of the proof of Theorem 4.9 is to show that the times to cross between ladder locations Uk := Tνk −Tνk−1 (29) have slowly varying tails and are approximately i.i.d. First of all, we note that the crossing times Uk are “almost” a stationary sequence. The reason for this is that the distribution on the environment near ν0 = 0 is diﬀerent from the distribution of the environment near νk for k ≥1. In particular, while the deﬁnition of the ladder locations implies that V (j) > V (νk) for all j ∈[0, νk) it is possible that V (j) ≤V (0) = 0 for some j ≤0. To rectify this problem we deﬁne a new measure Q on environments by Q(·) = P(· \| V (j) > 0, ∀j ≤−1). (30) Note that the measure Q is well deﬁned since EP [log ρ0] < 0 implies that P(V (j) > 0, ∀j ≤−1) > 0. The environment is no longer i.i.d. under the measure Q, but it does have the following useful properties. Lemma 4.10. If the measure Q is deﬁned as in (30) then 24 (i) Under the measure Q on environments the environment ω is stationary under shifts of the ladder locations in the sense that {θνkω}k≥0 is a stationary sequence. (ii) P and Q can be coupled so that there exists a distribution on pairs of environments (ω, ω′) such that ω ∼P, ω′ ∼Q, and ωx = ω′ x for all x ≥0. (iii) The “blocks” of the environment between ladder locations are i.i.d. That is, if Bk = (ωνk−1, ωνk−1+1, . . . , ωνk−1), then the sequence {Bk}k≥1 is i.i.d. Proof. Stationarity under shifts of the ladder locations follows easily from the deﬁnition of Q. The coupling of P and Q is easy to construct since the conditioning event in the deﬁnition of Q only depends on the environment to the left of the origin. It is easy to see that the blocks between ladder locations Bk are i.i.d. under the measure P since the environment is i.i.d. under P and νk −νk−1 only depends on the environment to the right of νk−1. Finally, since the Bk only depend on the environment to the right of the origin they have the same distribution under Q as under P. We will use the notation Q to denote the averaged distribution of the RWRE when the envi-ronment has distribution Q. That is Q(·) = EQ[Pω(·)]. Part (1) of Lemma 4.10 can easily be seen to imply the following Corollary. Corollary 4.11. Under the measure Q, the crossing times of ladder locations Uk = Tνk −Tνk−1 are a stationary sequence. Remark 4.12. The proof of Corollary 4.11 is essentially the same as that of Lemma 2.8 and is therefore ommitted. Also, it can be shown in fact that under Q the environment is ergodic under the shifts of the ladder locations and thus that Uk is ergodic under Q. We still need to show that the sequence Uk has (nicely behaved) heavy tails and is “fast-mixing” so that it is almost i.i.d. To accomplish this, it will be helpful to seperate the randomness in Uk due to the environment and the random walk, respectively. Deﬁne for any k ≥1, βk = βk(ω) = Eω[Uk] = Eνk−1 ω [Tνk]. (31) Also, by possibly expanding the probability space Pω, let {ηk}k≥1 be a sequence of i.i.d. Exp(1) random variables. The main idea of the proof is to show that P k Uk has approximately the same distribution as P k βkηk. Lemma 4.13. Under the measure Q, the sequence {βk}k≥1 is stationary. Moreover, there exists a constant C0 > 0 such that Q(β1 > t) ∼C0t−κ, as t →∞. Proof. The stationarity of the βk under Q follows easily from Theorem 4.10 part (1). The proof of the tail asymptotics of β1 is quite technical and therefore ommitted (the details can be found in [PZ09]). However, note that the nice polynomial tail decay is not surprising in light of the similar tail decay of Eω[τ1] as stated in Lemma 4.1. 25 Corollary 4.14. If C0 is the constant from Lemma 4.13, then Q(β1η1 > t) ∼Γ(κ + 1)C0t−κ, as t →∞. Proof. We need to show that limt→∞tκQ(β1η1 > t) = Γ(κ+1)C0. By conditioning on η1 we obtain tκQ(β1η1 > t) = Z ∞ 0 tκQ(β1 > t/y)e−y dy. The tail asymptotics of β1 from Lemma 4.13 imply that there is a constant K < ∞such that tκQ(β1 > t) ≤K for all t ≥0. Thus, tκQ(β1 > t/y) = yκ(t/y)κQ(β1 > t/y) ≤yκK and so we may apply the dominated convergence theorem and Lemma 4.13 to obtain lim t→∞Q(β1η1 > t) = Z ∞ 0 lim t→∞tκQ(β1 > t/y)e−y dy = Z ∞ 0 C0yκe−y dy = C0Γ(κ + 1). As noted above, the βk are stationary but not independent. However, the following Lemma shows that they the dependence is rather weak. Lemma 4.15. For each n ≥1, there exists a stationary sequence {β(n) k }k≥1 such that (i) If I ⊂N is such that \|k −j\| > √n for all k, j ∈I with k ̸= j, then {βk}k∈I is an independent family of random variables. (ii) There exist constants C, C′ > 0 such that Q \|βk −β(n) k \| > e−n1/4 ≤Ce−C√n. Sketch of proof. For any k, n ≥1, let ω(k,n) be then environment ω modiﬁed by ω(k,n) x = ( 1 if x = νk−1 −⌊√n⌋ ωx if x ̸= νk−1 −⌊√n⌋. That is, we modify the environment by putting a reﬂecting barrier to the right at a distance √n to the left of the ladder location νk−1. We then deﬁne β(n) k = Eνk−1 ω(n,k)[Tνk]. The claimed independence properties of the sequence {β(n) k }k≥1 is then obvious. Since backtracking of the random walk is exponentially unlikely (recall Lemma 4.13), it seems reasonable that modifying the environment a distance √n to the left of the starting location won’t change the expected crossing time by much. In fact, by using the exact formulas for quenched expectations of hitting times in (5) it can be shown that βk −β(n) k = 2   νk−1 X j=νk−1 j Y i=νk−1 ρi     X i≤νk−1−⌊√n⌋ νk−1−1 Y j=i ρj  . Note that in the second term in parenthesis on the right, each product in the sum has at least √n terms, and as was shown in the proof of Lemma 4.13 with high probability these products are exponentially small in the number of terms in the product. This can be used to show the second claimed property of the sequence {β(n) k }. 26 We are now ready to give a (sketch) of the proof of Theorem 4.9 Sketch of proof of Theorem 4.9. First of all, since our assumptions imply that the random walk is transient, the walk spends only ﬁnitely many steps to the left of the origin. Since the measures P and Q can be coupled so that they only diﬀer to the left of the origin it can be shown that if the limiting distributions hold under Q then they also hold under P. Secondly, note that the gaps between ladder locations νk −νk−1 are i.i.d. and thus lim n→∞ νn n = lim n→∞ 1 n n X k=1 νk −νk−1 = EP [ν1] In fact, it is easy to see that ν1 has exponential tails so that the deviations of νn/n are exponentially unlikely. From this, it can be shown that if ¯ α = 1/EP [ν1] then n−1/κ(Tν¯ αn −Tn) →0 in Q-probability. We have thus reduced the problem to proving limiting distributions for Tνn = Pn k=1 Uk under the measure Q. As mentioned above, the key will be to be able to approximate the crossing times between ladder locations Uk = Tνk −Tνk−1 by βkηk. To this end, we will create a coupling of the random variables Uk and βkηk. For simplicity we will describe this coupling when k = 1 only. The crossing time U1 = Tν1 can be thought of as a series of excursions away from the origin. There will be a random number G of excursions that return to the origin before reaching ν1 (we will call these excursions “failures”) followed by an excursions that goes from 0 to ν1 without ﬁrst returning to 0 (we will call this a “success” excursion). That is, if we time of the i-th failure excursion by Fi and the time of the success excursion by S then we can represent Tν1 = G X i=1 Fi + S. It is easy to see that the number of failure excursions in this decomposition is geometric with distribution Pω(G = k) = (1 −pω)kpω, where pω = ω0P 1 ω(Tν1 < T0). We will couple Tν1 with β1η1 by coupling the exponential random variable η1 with the geometric random variable G. This is accomplished by letting G = ⌊cωη1⌋, where cω = −1 log(1 −pω). It can be shown that this coupling is good enough enough so that lim n→∞n−2/κV arω Tνn − n X k=1 βkηk ! = 0, in Q-probability. (32) 27 Now, for any ε, δ > 0 Q Tνn − n X k=1 βkηk > δn1/κ ! = EQ " Pω Tνn − n X k=1 βkηk > δn1/κ !# ≤ε + Q Pω Tνn − n X k=1 βkηk > δn1/κ ! ε ! ≤ε + Q n−2/κVarω Tνn − n X k=1 βkηk ! ≥εδ ! . Since (32) implies that this last probability vanishes as n →∞and since ε > 0 was arbitrary we can conclude that lim n→∞n−1/κ Tνn − n X k=1 βkηk ! = 0, in Q-probability. Finally, we are down to proving a limiting distribution for Pn k=1 βkηk under the measure Q. However, Lemma 4.14 shows that the random variables βkηk have well behaved polynomial tails, and Lemma 4.15 shows that they are close enough to i.i.d. to have limiting distributions of the same form as in Examples 4.4 and 4.5. We now state the corresponding averaged limiting distributions for Xn when κ ∈(0, 2]. Theorem 4.16. Let Assumptions 1, 2, and 3 hold. If κ is deﬁned as in (24) then (i) If κ ∈(0, 1), then there exists a b > 0 such that lim n→∞P Tn n1/κ ≤t = 1 −Lκ,b(t−1/κ), ∀t. (ii) If κ = 1, then there exists a constant b > 0 and a sequence δ(n) ∼n/(b log n) such that lim n→∞P Xn −δ(n) n/(log n)2 ≤t = 1 −L1,b(−b2t), ∀t. (iii) If κ ∈(1, 2), then there exists a b > 0 such that lim n→∞P Tn −n/vP n1/κ ≤t = 1 −Lκ,b(tv−1−1/κ P ), ∀t. (iv) If κ = 2, then there exists a b > 0 such that lim n→∞P Tn −n/vP v3/2 P b√n log n ≤t ! = Φ(t), ∀t. 28 Remark 4.17. Note that we have stated Theorem 4.16 so that the constants b in each case are the same scaling parameters appearing in the limiting distributions of the hitting times in Theorem 4.9. Note that when κ ∈[1, 2) one can simplify the limits on the right hand side by using the fact that Lκ,b(ct) = Lκ,bc−κ(t) for any c > 0. In this way the right hand side can be written as 1 −Lκ,¯ b(−t), where ¯ b = ( 1 b κ = 1 bv1+κ P κ ∈(1, 2). From this it is clear that when κ ∈[1, 2) the limiting distribution is a totally skewed to the left κ-stable distribution. In contrast, when κ ∈(0, 1) the limiting distribution is 1 −Lκ,b(t−1/κ) which is not a κ-stable distribution but is instead a transformation of a κ-stable distribution. This particular transformation of κ-stable distributions is sometimes referred to as a Mittag-Leﬄer distribution. Proof. The proof of Theorem 4.16 follows from Theorem 4.9 in essentially the same way that Theorem 3.8 followed from Theorem 3.5. For example, when κ ∈(0, 1) we have that P X∗ n nκ < t = P(X∗ n < tnκ) = P(T⌈tnκ⌉> n) = P T⌈tnκ⌉ ⌈tnκ⌉1/κ > n ⌈tnκ⌉1/κ . Since n ⌈tnκ⌉1/κ →t−1/κ as n →∞it follows from Theorem 4.9 that lim n→∞P X∗ n nκ < t = lim n→∞P T⌈tnκ⌉ ⌈tnκ⌉1/κ > t−1/κ = 1 −Lκ,b(t−1/κ). The proofs of the cases when κ ∈(1, 2) or κ = 2 are similar and therefore ommitted. The proof of the case when κ = 1 is slightly more diﬃcult due to the somewhat strange centering term nD(n) in the limiting distribution for Tn. While Theorem 4.9 states that D(n) ∼b log n, one actually better control of the function D(n) to prove the limiting distribution for Xn in this case. In fact, it turns out that the proof of Theorem 4.9 gives D(n) = (1/¯ ν)EQ[β11{β1≤n}] (note the similarity to the centering term in Example 4.4 when α = 1). From this explicit form for D(n) and the tail asymptotics of β1 in Lemma 4.13, it follows that there exists a function δ(x) such that δ(x)D(δ(x)) = x + o(1), as x →∞. If the centering term for Xn is chosen in this way, then one can prove the claimed limiting distribu-tion for Xn in the case κ = 1. (The details of this argument in the case when κ = 1 can be found in [KKS75, pp. 167-8].) 4.3 Weak Quenched Limiting Distributions - κ < 2 We now turn our attention the study of the asymptotics of the quenched distribution of hitting times when κ < 2. Much of the work that we did in the proof of the averaged limiting distributions in Theorem 4.9 was done with this in mind. Recall that in the case κ > 2 we proved a quenched central limiting distribution. We will refer to this as a strong quenched limiting distribution since the convergence holds for P-a.e. environment ω. The main result in this subsection shows that there is no such strong quenched limiting distribution for the hitting times when κ < 2. Instead, we will prove a what we will call a weak quenched limiting distribution. 29 Let M1(R) denote the space of probability measures on (R, B(R)) where B(R) is the Borel σ-ﬁeld. Recall that prohorov metric ρ on M1(R) is deﬁned by ρ(µ, π) = inf{ε : µ(A) ≤π(A(ε)) + ε, ∀A ∈B(R)}, where A(ε) = {x : dist(x, A) < ε}. The Prohorov metric ρ induces the topology of weak convergence (i.e., convergence in distribution) on the space M1(R), and the metric space (M1(R), ρ) is a Polish space. We will be interested in studying random probability measures - that is M1(R)-valued random variables. If πn is a sequence of M1(R)-valued random variables and π is another M1(R) valued random variable we will use the notation µn = ⇒µ to denote convergence in distribution of M1(R)-valued random variables. Remark 4.18. Note that the notation = ⇒for convergence in distribution of random probability measures should not be confused with the standard convergence of measures in the space M1(R). The notation µn = ⇒µ means that lim n→∞E[φ(µn)] = E[φ(µ)], for all bounded continuous φ : M1(R) →R. On the other hand, pointwise convergence in the space M1(R), which we would denote µn →µ, is equivalent to lim n→∞ Z φ(x) dµn = Z φ(x) dµ, for all bounded continuous φ : R →R. Now, for any environment ω ∈Ωand any n ≥1 deﬁne µn,ω,κ ∈M1(R) by µn,ω,κ(·) = Pω Tn −Eω[Tn] n1/κ ∈· . Since the environment ω is itself random, then we can view µn,ω,κ as a M1(R)-valued random variable (or a random probability measure). In order to deﬁne the limiting random probability measures that will arise we need to introduce some notation. Let Mp denote the space of Radon point processes on (0, ∞] - i.e., point processes with ﬁnitely many points on [x, ∞] for any x > 0. We will equip Mp with the standard topology of vague convergence (see [Res08] for more information on point processes and the deﬁnition of vague convergence). Let F ⊂Mp denote the subset of point processes N = P i≥1 δxi such that Z x2 N(dx) = X i≥1 x2 i < ∞. Then, deﬁne the function H : Mp →M1(R) by H(N) = ( Pη P k≥1 xk(ηk −1) ∈· if N = P k≥1 δxk ∈F, δ0 if N / ∈F, (33) where {ηk}k≥1 is an i.i.d. sequence of Exp(1) random variables with distribution Pη. Remark 4.19. Note that the condition N ∈F guarantees that the random sum P k≥1 xkηk is ﬁnite Pη-a.s. The deﬁnition of H(N) when N / ∈F is arbitrary and will not matter since we will only be considering point processes that are almost surely in F. 30 Theorem 4.20. If Assumptions 1, 2, and 3 hold and the parameter κ < 2, then there exists a λ > 0 such that µn,ω,κ = ⇒H(Nλ,κ), where Nλ,κ is a non-homogeneous Poisson point process with intensity λκx−κ−1. Sketch of proof. As in the proof of the averaged stable limit laws for Tn, the proof of Theorem 4.20 is accomplished by the following reductions. First we show that µn,ω,κ has approximately the same distribution on M1(R) when ω ∼P and when ω ∼Q. Secondly, we show that it is enough to prove a similar weak quenched limiting distribution for the quenched distribution of Tνn instead of Tn, and ﬁnally we show that we can couple Tνn with a sum of exponential random variables so that it is enough to study the quenched distribution of Pn k=1 βkηk, where βk = βk(ω) is as deﬁned in (31) and the ηk are i.i.d. Exp(1) random variables that are independent of the βk. That is, letting σn,ω,κ ∈M1(R) be the random probability measure deﬁned by σn,ω,κ(·) = Pη 1 n1/κ n X k=1 βk(ηk −1) ∈· ! , (34) it is enough to show that there exists a λ′ > 0 such that σn,ω,κ = ⇒H(Nλ′,κ) as n →∞when ω has distribution Q. Note that in the deﬁnition of σn,ω,κ in (34), the distribution is entirely determined by the coeﬃcients βk. Thus, the key to understanding the random probability distribution σn,ω,κ is un-derstanding the joint distribution of the coeﬃcients βk. To this end, let Nn,ω,κ be the point process Nn,ω,κ = n X k=1 δβk/n1/κ. (35) Then, it can be shown that Nn,ω,κ converges in distribution under Q to a non-homogeneous Poisson point process Nλ′,κ. Recall that in the proof of Theorem 4.9 we remarked that under the distribution Q the βk have heavy tails and are fast-mixing enough to be close to i.i.d. If the βk were i.i.d. then the convergence of Nn,ω,κ to the Poisson process Nλ′,κ would be standard, but since the βk are not quite i.i.d. it takes a little extra work. Recall the deﬁnition of the function H : Mp →M1(R) from (33). Then, the deﬁnitions of the point process Nn,ω,κ and the random measure σn,ω,κ imply that σn,ω,κ = H(Nn,ω,κ). Since we know the point processes Nn,ω,κ converge in distribution, this suggests that σn,ω,κ should converge in distribution to H(Nλ′,κ). Unfortunately the function H is not continuous, and so we need to do a modiﬁcation. The details of this truncation and the rest of the full proof of Theorem 4.20 can be found in [PS10]. 31 5 RWRE on Zd - d ≥2 We now turn to discussion of multi-dimesional RWRE. For nearest-neighbor RWRE on Z an en-vironment can be encoded by a single number ωx ∈[0, 1] at every site x ∈Z. However, for multi-dimensional RWRE we need a probability vector at every site. To simplify things we will only consider the case of nearest-neighbor RWRE, but obviously one can consider RWRE on Zd with bounded jumps as well (although less is known in the more general bounded jumps case). In this case an environment ω = {ωx}x∈Zd where ωx is a probability distribution on E = {z ∈Zd : \|z\| = 1} = {±ei, i = 1, 2, . . . d} in the sense that ωx = (ωx(z))z∈E ∈[0, 1]E with X z∈E ωx(z) = 1. Given an environment ω = {ωx}x = {ωx(z)}x,z, the quenched transition probabilities for the random walk are given by Pω(Xn+1 = x + z \| Xn = x) = ( ωx(z) if z ∈E 0 otherwise. As with our coverage of one-dimensional RWRE we will restrict ourselves to i.i.d. uniformly elliptic environments. Assumption 4. The environment ω = {ωx}x∈Zd is i.i.d. under the distribution P on environments. Assumption 5. There exists a constant c > 0 such that P(ωx(z) ≥c, ∀x ∈Zd, z ∈E) = 1. 5.1 Directional transience/recurrence The ﬁrst natural thing to study for multi-dimensional RWRE is the question of recurrence or transience. Unfortunately, as we will see below, in contrast to the one-dimensional case where there is a nice explicit criterion for recurrence/transience (see Theorem 2.1) even the question of directional transience/recurrence is not yet settled. Let Sd−1 = {z ∈Rd : \|z\| = 1} be the d −1-dimensional sphere. We will refer to a ﬁxed ℓ∈Sd−1 as a direction in Rd. For any such ﬁxed direction ℓwe will deﬁne the event of transience in direction ℓby Aℓ= n lim n→∞Xn · ℓ= +∞ o . (36) The following lemma was proved by Kalikow in [Kal81]. Lemma 5.1 (Kalikow’s 0-1 Law). If the distribution on environments P satisﬁes Assumptions 4 and 5, then P(Aℓ∪A−ℓ) ∈{0, 1}, ∀ℓ∈Sd−1. Proof. First, we claim that P(Aℓ∪A−ℓ∪Oℓ) = 1, where Oℓis the event Oℓ= {Xn · ℓchanges sign inﬁnitely many times}. 32 If none of the events Aℓ, A−ℓor Oℓare satisﬁed, then either 0 ≤lim inf n→∞Xn · ℓ< ∞ or −∞< lim sup n→∞Xn · ℓ≤0. (37) Indeed, in the ﬁrst case in (37) there must be some x > 0 such that \|Xn · ℓ−x\| ≤1 inﬁnitely many times. However, by uniform ellipticity the probability of the random walk visiting {\|z · ℓ−x\| ≤1} inﬁnitely many times without ever reaching the half-space {z · ℓ< 0} is zero. Next, for ℓ∈Sd−1 let Dℓ:= inf{n ≥0 : Xn · ℓ< X0 · ℓ} be the ﬁrst time the random walk “backtracks” in direction ℓfrom its initial location (note that Dℓis a stopping time and that we have stated the deﬁnition to account for starting locations other than X0 = 0). If P(Dℓ= ∞) = 0 then P x ω(Dℓ< ∞) = 1 for all x ∈Zd and P-a.e. environment ω. By the strong Markov property this implies that P(lim infn→∞Xn · ℓ< 0) = 1 and so P(Aℓ) = 0. Taking the contrapositive of this we obtain that P(Aℓ) > 0 = ⇒ P(Dℓ= ∞) > 0. To complete the proof of the lemma, we may assume that either P(Aℓ) > 0 or P(A−ℓ) > 0 since otherwise the conclusion of the lemma is obvious. Without loss of generality we will assume that P(Aℓ) > 0. Since we showed above that P(Aℓ∪A−ℓ∪Oℓ) = 1, it will be enough to show that P(Oℓ) = 0 whenever P(Aℓ) > 0. To this end, ﬁrst note that P(Oℓ∩{sup n≥0 Xn · ℓ< ∞}) = 0, for by uniform ellipticity every time Xn · ℓswitches from negitive the probability that the random walk reaches the halfspace {z ·ℓ> x} before {z ·ℓ< 0} is uniformly bounded below. Next, we claim that P(Oℓ∩{supn≥0 Xn · ℓ= ∞}) = 0 as well. To see this, we introduce a sequence of stopping times B1 ≤F1 ≤B2 ≤F2 ≤. . . deﬁned as follows. B1 = Dℓ, Fk = inf{n > Bk : Xn · ℓ> max i Fk : Xn · ℓ< 0}. (Note that if Bk = ∞for some k then Fj = Bj = ∞also for all j ≥k.) The times Bk are certain “backtracking” times where the random walk enters the halfspace to the left of the origin, and the Fk are the ﬁrst “fresh times” where the random walk reaches a new portion of the environment farther to the right than it had previously reached. On the event Oℓ∩{supn≥0 Xn · ℓ< ∞} it is clear that Bk < ∞for all k < ∞. However, when Bk+1 < ∞by decomposing according to the location of the random walk at time Fk we obtain P(Bk+1 < ∞) = X z P(Fk < ∞, XFk = z, Bk+1 < ∞) = X z EP Pω(Fk < ∞, XFk = z)P z ω(inf n≥0 Xn · ℓ< 0) ≤ X z EP [Pω(Fk < ∞, XFk = z)P z ω(Dℓ< ∞)] Note that the quenched probabilities inside the last expectation are independent since Pω(Fk < ∞, XFk = z) is σ(ωx : x · ℓ< z · ℓ)-measureable and P z ω(Dℓ< ∞) is σ(ωx : x · ℓ≥z · ℓ)-measurable. 33 Thus, we obtain that P(Bk+1 < ∞) ≤ X z P(Fk < ∞, XFk = z)Pz(Dℓ< ∞) = P(Fk < ∞)P(Dℓ< ∞) ≤P(Bk < ∞)P(Dℓ< ∞). Since P(B1 < ∞) = P(Dℓ< ∞) we obtain by induction that P(Bk < ∞) ≤P(Dℓ< ∞)k for any k ≥1. Since P(Dℓ< ∞) < 1 whenever P(Aℓ) > 0 we have that P(Oℓ∩{sup n≥0 Xn · ℓ= ∞}) ≤P(Bk < ∞, ∀k ≥1) ≤lim k→∞P(Dℓ< ∞)k = 0. Thus we have shown that P(Oℓ) = 0 whenever P(Aℓ) > 0 and so P(Aℓ∪A−ℓ) = 1 whenever P(Aℓ) > 0. MORE YET TO BE ADDED.... 34 References [Ali99] S. Alili. Asymptotic behaviour for random walks in random environments. J. Appl. Probab., 36(2):334–349, 1999. [Dur96] Richard Durrett. Probability: theory and examples. Duxbury Press, Belmont, CA, second edition, 1996. [DZ98] Amir Dembo and Ofer Zeitouni. Large deviations techniques and applications, volume 38 of Applications of Mathematics (New York). Springer-Verlag, New York, second edition, 1998. [Gol07] Ilya Ya. Goldsheid. Simple transient random walks in one-dimensional random environ-ment: the central limit theorem. Probab. Theory Related Fields, 139(1-2):41–64, 2007. [Kal81] Steven A. Kalikow. Generalized random walk in a random environment. Ann. Probab., 9(5):753–768, 1981. [Kes73] Harry Kesten. Random diﬀerence equations and renewal theory for products of random matrices. Acta Math., 131:207–248, 1973. [KKS75] H. Kesten, M. V. Kozlov, and F. Spitzer. A limit law for random walk in a random environment. Compositio Math., 30:145–168, 1975. [Pet08] Jonathon Peterson. Limiting distributions and large deviations for random walks in random environments. PhD thesis, University of Minnesota, 2008. Available at arXiv:0810.0257v1. [PS10] Jonathon Peterson and Gennady Samorodnitsky. Weak quenched limiting distributions for transient one-dimensional random walk in a random environment, 2010. To appear in Ann. Inst. Henri. Poincar´ e Probab. Stat. Preprint available at [PZ09] Jonathon Peterson and Ofer Zeitouni. Quenched limits for transient, zero speed one-dimensional random walk in random environment. Ann. Probab., 37(1):143–188, 2009. [Res08] Sidney I. Resnick. Extreme values, regular variation and point processes. Springer Series in Operations Research and Financial Engineering. Springer, New York, 2008. Reprint of the 1987 original. [Sol75] Fred Solomon. Random walks in a random environment. Ann. Probability, 3:1–31, 1975. [Zei04] Ofer Zeitouni. Random walks in random environment. In Lectures on probability theory and statistics, volume 1837 of Lecture Notes in Math., pages 189–312. Springer, Berlin, 2004. 35
6413	https://brainly.com/question/35255986	[FREE] Simplify. Express the answer using positive exponents. \frac{-2a b^2 b^4}{4a b^{-8}}, \quad a \neq 0, \ b - brainly.com 3 Search Learning Mode Cancel Log in / Join for free Browser ExtensionTest PrepBrainly App Brainly TutorFor StudentsFor TeachersFor ParentsHonor CodeTextbook Solutions Log in Join for free Tutoring Session +60k Smart guidance, rooted in what you’re studying Get Guidance Test Prep +21,6k Ace exams faster, with practice that adapts to you Practice Worksheets +6,8k Guided help for every grade, topic or textbook Complete See more / Mathematics Expert-Verified Expert-Verified Simplify. Express the answer using positive exponents. 4 a b−8−2 a b 2 b 4,a=0,b=0 1 See answer Explain with Learning Companion NEW Asked by kmmjones7941 • 08/04/2023 0:00 / 0:15 Read More Community by Students Brainly by Experts ChatGPT by OpenAI Gemini Google AI Community Answer This answer helped 20843704 people 20M 4.5 0 Upload your school material for a more relevant answer The simplified expression for -2ab² b⁴ / 4ab⁻⁸, where a ≠ 0 and b ≠ 0, is -1/2 b¹⁴. Let's simplify the expression -2ab² b⁴ / 4ab⁻⁸, where a ≠ 0 and b ≠ 0. We'll simplify step by step while ensuring all exponents are positive: Given expression: -2ab² b⁴ / 4ab⁻⁸ Combine the like terms in the numerator: -2 a b⁴ / 4ab⁻⁸ Simplify the coefficient -2/4, which is -1/2: (1/2) a b⁶ / ab⁻⁸ Simplify the fraction by canceling out common terms in the numerator and denominator: (1/2) b⁶ / b⁻⁸ When dividing with the same base and different exponents, subtract the exponents: (1/2) b(⁶ ⁻ ⁻⁸) (1/2) b¹⁴ So, the simplified expression is -1/2 b¹⁴, where a ≠ 0 and b ≠ 0. The expression is now in its simplest form with positive exponents. For more such information on: expression brainly.com/question/1859113 SPJ11 Answered by robinkumar9613 •22.4K answers•20.8M people helped Thanks 0 4.5 (2 votes) Expert-Verified⬈(opens in a new tab) This answer helped 20843704 people 20M 4.5 0 Upload your school material for a more relevant answer The simplified expression for 4 a b−8−2 a b 2 b 4 is −2 1b 14. This simplification combines terms, reduces coefficients, and ensures all exponents remain positive. The final answer maintains the conditions that a=0 and b=0. Explanation To simplify the expression 4 a b−8−2 a b 2 b 4, we will follow these steps while ensuring that we only use positive exponents: Combine the terms in the numerator: The terms in the numerator are −2 a b 2 and b 4. We can combine them as follows: −2 a b 2 b 4=−2 a(b 2+4)=−2 a b 6 Now, rewrite the entire expression: 4 a b−8−2 a b 6 Simplify the coefficients: We can simplify the coefficient −2/4: 4−2=−2 1 Now, the expression looks like: −2 1⋅a b−8 a b 6 Cancel out common terms in the numerator and denominator: We observe that a a=1 since a=0. So we simplify the b terms: b−8 b 6=b 6−(−8)=b 6+8=b 14 Combine the results: Now we have: −2 1b 14 Thus, the simplified expression is: −2 1b 14, where a=0 and b=0. All exponents in this expression are positive. Examples & Evidence For example, if a=1 and b=2, substituting these values into −2 1b 14 yields −2 1⋅2 14=−16384, demonstrating how the simplified expression evaluates with specific values. The steps shown clearly illustrate the application of exponent rules and fraction simplification, which are fundamental principles in algebra, ensuring the methodology is accurate and reliable. Thanks 0 4.5 (2 votes) Advertisement kmmjones7941 has a question! Can you help? Add your answer See Expert-Verified Answer ### Free Mathematics solutions and answers Community Answer Simplify the expression. Express the answer so that all exponents are positive. Whenever an exponent is 0 or negative, we assume that the base is not 0 . ((-8x^(4)y^(-4))/(5z^(4)))^(-2) Community Answer Simplify. ((2b^(2))^(3))/((3b^(4))^(2)) Write your answer using only positive exponents. Community Answer 16 Simplify each expression. 3a^2b^-4/12a^-2b^-2, a=0, b=0 Community Answer Simplify. ((a^(3)b^(5))/(ab^(3)))^(4) Write your answer using only positive exponents. Community Answer 4.6 12 Jonathan and his sister Jennifer have a combined age of 48. If Jonathan is twice as old as his sister, how old is Jennifer Community Answer 11 What is the present value of a cash inflow of 1250 four years from now if the required rate of return is 8% (Rounded to 2 decimal places)? Community Answer 13 Where can you find your state-specific Lottery information to sell Lottery tickets and redeem winning Lottery tickets? (Select all that apply.) 1. Barcode and Quick Reference Guide 2. Lottery Terminal Handbook 3. Lottery vending machine 4. OneWalmart using Handheld/BYOD Community Answer 4.1 17 How many positive integers between 100 and 999 inclusive are divisible by three or four? Community Answer 4.0 9 N a bike race: julie came in ahead of roger. julie finished after james. david beat james but finished after sarah. in what place did david finish? Community Answer 4.1 8 Carly, sandi, cyrus and pedro have multiple pets. carly and sandi have dogs, while the other two have cats. sandi and pedro have chickens. everyone except carly has a rabbit. who only has a cat and a rabbit? New questions in Mathematics What is the equation of the line that is parallel to the given line and passes through the point (−2,2)? A. y=5 1x+4 B. y=5 1x+5 12 C. y=−5 x+4 D. y=−5 x+5 12 Factor. w 2+12 w+36 Divide. 5×1 0−1 2×1 0 1 Which of the binomials below is a factor of this trinomial? x 2−5 x−36 A. x−4 B. x 2+7 C. x−7 D. x+4 Identify all of the correct coefficients of the expression below. $x^2-3 x+6 A. 1,−3,6 B. 3 C. 1,−3 D. x 2,−3 x,6 Previous questionNext question Learn Practice Test Open in Learning Companion Company Copyright Policy Privacy Policy Cookie Preferences Insights: The Brainly Blog Advertise with us Careers Homework Questions & Answers Help Terms of Use Help Center Safety Center Responsible Disclosure Agreement Connect with us (opens in a new tab)(opens in a new tab)(opens in a new tab)(opens in a new tab)(opens in a new tab) Brainly.com Dismiss Materials from your teacher, like lecture notes or study guides, help Brainly adjust this answer to fit your needs. Dismiss
6414	https://www.quora.com/How-do-you-convert-1-joule-into-erg-step-by-step	How to convert 1 joule into erg step by step - Quora Something went wrong. Wait a moment and try again. Try again Skip to content Skip to search Sign In Physics Conversions Erg (SI unit) Energy (power) Units of Measure Joule (SI unit) Unit Converter Conversion (math) Physical Energy 5 How do you convert 1 joule into erg step by step? All related (33) Sort Recommended Greg Lehey Fascinated observer of measurements · Author has 13.6K answers and 16.3M answer views ·6y You convert it to base units kg.m².s⁻². You convert the mass component. 1 kg = 10³ g, so you have 10³ g.m².s⁻². You convert the length (area) component. 1 m² = 10⁴ cm², so you have 10⁷ g.cm².s⁻² Upvote · 9 1 Sponsored by Hudson Financial Partners How do I invest in shares in Australia? Buy direct ASX shares or pool your money with other investors in managed funds. Call us for more. Contact Us 9 3 John Steele Author has 10.2K answers and 11.3M answer views ·7y A joule is a unit of energy equal to a newton-meter An erg is a CGS unit of energy equal to a dyne-centimeter, and all named derived units of the CGS are deprecated by the SI Brochure, so I would never do the conversion. However, the SI Brochure defines the dyne as 10^-5 N (so 1 N is 10^5 dynes_ and of course 1 m is 100 cm. So 1 J must be 10^5 dynes times 10^2 cm or 10^7 ergs. Note that the SI Brochure specifies that if these deprecated units are used, they should be defined before use in any publication. Erghhh. Upvote · Satya Parkash Sud Former Professor at Himachal Pradesh University Shimla (1986–2002) · Author has 8.1K answers and 27.4M answer views ·Updated 5y Related 1 joule = how many ergs? Both Joule and erg are units of work/energy. The erg is the cgs unit of work/energy and Joule is MKS/ SI unit of the same. By definition, 1 erg is the work done when a force of one dyne produces a displacement of 1 cm in an object in the direction of the force. And one dyne is that force which produces an acceleration of 1 cm/s² in an object of 1 gm. So one erg = 1 gm × 1 cm/s² × 1 cm. By definition, one Joule is the work done by a force of 1 Newton in producing a displacement of 1 metre in an object in the direction of the force. One Newton in turn is that force which produces in an object of 1 kg Continue Reading Both Joule and erg are units of work/energy. The erg is the cgs unit of work/energy and Joule is MKS/ SI unit of the same. By definition, 1 erg is the work done when a force of one dyne produces a displacement of 1 cm in an object in the direction of the force. And one dyne is that force which produces an acceleration of 1 cm/s² in an object of 1 gm. So one erg = 1 gm × 1 cm/s² × 1 cm. By definition, one Joule is the work done by a force of 1 Newton in producing a displacement of 1 metre in an object in the direction of the force. One Newton in turn is that force which produces in an object of 1 kg an acceleration of 1 m/s². So 1 Newton = 1 kg × 1 m/s² = 1000 gm × 100 cm/s² = 10⁵ dyne. 1 Joule = 1 Newton × 1 m displacement = 10⁵ dyne × 100 cm displacement = 10⁷ ergs. 1 Joule = 10⁷ ergs. Upvote · 99 80 99 13 Related questions More answers below What is the relationship between a joule and an erg? What is the value of one joule in erg? How do I convert 1 joule to ERG by using dimensional analysis? 1 joule = how many ergs? What's the value of 1 erg in a joule? Assistant Bot · 1y To convert joules to ergs, you can use the conversion factor between these two units of energy. The conversion is as follows: 1 joule = 10 7 10 7 ergs. Here’s a step-by-step breakdown of the conversion: Step 1: Understand the Conversion Factor 1 joule is equivalent to 10 7 10 7 ergs. This means that for every joule, there are ten million ergs. Step 2: Set Up the Conversion To convert joules to ergs, you multiply the number of joules by the conversion factor: Energy in ergs=Energy in joules×10 7 Energy in ergs=Energy in joules×10 7 Step 3: Perform the Calculation If you want to convert 1 joule to ergs: 1 joule=1\time 1 joule=1\time Continue Reading To convert joules to ergs, you can use the conversion factor between these two units of energy. The conversion is as follows: 1 joule = 10 7 10 7 ergs. Here’s a step-by-step breakdown of the conversion: Step 1: Understand the Conversion Factor 1 joule is equivalent to 10 7 10 7 ergs. This means that for every joule, there are ten million ergs. Step 2: Set Up the Conversion To convert joules to ergs, you multiply the number of joules by the conversion factor: Energy in ergs=Energy in joules×10 7 Energy in ergs=Energy in joules×10 7 Step 3: Perform the Calculation If you want to convert 1 joule to ergs: 1 joule=1×10 7 ergs=10,000,000 ergs 1 joule=1×10 7 ergs=10,000,000 ergs Conclusion Thus, 1 joule is equal to 10,000,000 ergs. Upvote · Umar M Physics Mentor \| Spiritual Seeker · Author has 97 answers and 129.6K answer views ·2y Related How do you convert one erg into joules? One erg is a unit of energy and mechanical work equal to 10^-7 joules 123. To convert ergs to joules, you can multiply the number of ergs by 10^-7. For example, 1000 ergs is equal to 0.0001 joules 3. Upvote · Sponsored by Grammarly 92% of professionals who use Grammarly say it has saved them time Work faster with AI, while ensuring your writing always makes the right impression. Download 999 209 Arch Stanton Undergrad chemistry student ·Updated 9y Related How do I convert joules to hertz? Energy and frequency are different physical quantities but for photons they’re related in a simple way by E = hν So while a frequency is not an energy, when talking about photons they’re interchangeable since one can go from one another simply by multiplying or dividing by a constant. Also, note that since in SI units Planck’s constant h is given in J⋅s and frequency in 1/s, both members of the equation are energies expressed in joules. Now say you want to know how many Hz it takes to reach 1 J (many…), how do you get the value? You write down the equation 1[J] = h[J⋅s]x[1/s] where the unknown x is Continue Reading Energy and frequency are different physical quantities but for photons they’re related in a simple way by E = hν So while a frequency is not an energy, when talking about photons they’re interchangeable since one can go from one another simply by multiplying or dividing by a constant. Also, note that since in SI units Planck’s constant h is given in J⋅s and frequency in 1/s, both members of the equation are energies expressed in joules. Now say you want to know how many Hz it takes to reach 1 J (many…), how do you get the value? You write down the equation 1[J] = h[J⋅s]x[1/s] where the unknown x is the number of Hz you’re looking for. Bear in mind that all the numbers are unitless since the units are considered separately; nonetheless, you have to choose the right values based on your units (you’ll have to take a different value for h if you’re considering electron volts for example). Now you just solve for x: x = 1/h All the units simplified to 1 and you get the numerical value of the frequency in Hz of a photon of energy 1 J. Your response is private Was this worth your time? This helps us sort answers on the page. Absolutely not Definitely yes Upvote · 99 14 9 1 Related questions More answers below What is the relationship between joules, ergs, calories, and electro volts? How do you convert one erg into joules? How do you convert 1 erg into a joule by using the dimensional method? The SI and the C.G.S. unit are joule and erg. How many ergs are equal to 1J? How do you convert 50 erg in joules by dimensional analysis? Lucy Thompson Lives in Jefferson City, MO · Author has 2.9K answers and 2.8M answer views ·3y Related How do you convert energy into joules? First, let us look at the definition of the Joule. joule [jo͞ol] NOUN the SI unit of work or energy, equal to the work done by a force of one newton when its point of application moves one meter in the direction of action of the force, equivalent to one 3600th of a watt-hour. A Joule IS energy. It happens to be the SI unit of energy. Below is a cut from the Whackapedia page on the Joule: A joule is 10^7 ergs. Or 9.4782×10^-4 BTU Continue Reading First, let us look at the definition of the Joule. joule [jo͞ol] NOUN the SI unit of work or energy, equal to the work done by a force of one newton when its point of application moves one meter in the direction of action of the force, equivalent to one 3600th of a watt-hour. A Joule IS energy. It happens to be the SI unit of energy. Below is a cut from the Whackapedia page on the Joule: A joule is 10^7 ergs. Or 9.4782×10^-4 BTU Upvote · 9 1 Sponsored by RedHat Know what your AI knows, with open source models. Your AI should keep records, not secrets. Learn More 99 36 Aditya Agarwal 5y Related How do I convert 1 joule to ERG by using dimensional analysis? Joule is the unit for energy Dimension Formula for Joule is [M¹L²T⁻²] Let n1(joule)=n2(erg) n1(M₁¹L₁²T₁⁻²]=n2[M₂¹L₂²T₂⁻²] n2=n1(M₁¹L₁²T₁⁻²] /M₂¹L₂²T₂⁻²] =n1x(M1/M2)¹x(L1/L2)²x(T1/T2)⁻² =1x(1kg/gm)¹x(1m/1cm)²x(1s/1s)⁻² =1x(1000g/gm)x(100cm/1cm)²x(1s/1s)⁻² =10⁷ Therefore n1(joule)=n2(erg) ∴1 Joule=10⁷erg Upvote · 9 1 Chauhan Studied at University of Mumbai ·7y Related How do I convert joule to eV? 1 Joule = Energy required to move a charge of 1 Coulomb through a potential difference of 1 volt. Thus 1 Joule = 1 Coulomb x 1 Volt 1 eV = Energy required to move an electron (Charge of an electron = 1.6 x 10^-19 Coulomb) through a potential difference of 1 Volt Thus 1 eV = 1.6 x 10^-19 Coulomb x 1 Volt = 1.6 x 10^-19 Joule Thus 1 eV/1.6 x 10^-19 = 1 Joule 0.625 x 10^19 eV = 1 Joule Upvote · 99 70 9 3 Sponsored by All Out Kill Dengue, Malaria and Chikungunya with New 30% Faster All Out. Chance Mat Lo, Naya All Out Lo - Recommended by Indian Medical Association. Shop Now 999 624 Devasish Reddy B.E in Mechanical Engineering&Manufacturing, Anna University, Tamil Nadu, India (Graduated 2019) ·7y Related What is the value of one joule in erg? Both Joule and erg are units of work/energy. The erg is the cgs unit of work/energy Joule is MKS/ SI unit of the same. By definition, one erg = 1 gm × 1 cm/s² × 1 cm. By definition, one Newton = 1 kg × 1 m/s² = 1000 gm × 100 cm/s² = 10^5 dyne. 1 Joule = 1 Newton × 1 m displacement = 10 ^5 dyne × 100 cm displacement = 10^7 ergs. 1 Joule = 10^7 ergs. Upvote · 99 10 9 2 Shyam Active in innovative experiments in PHYSICS · Author has 692 answers and 1.1M answer views ·2y Related How do I convert 1.5 bar m³ to joules? Bar is unit of pressure and pressure Pis given as P=Force/area Or, pressure in (N/m^2)=F in Newton/A in sq meter 1 atmospheric pressure = 1bar And 1 atmospheric pressure= 1 bar=pressure of 76 cm high mercury column(h d g) =76×13.6 × 980 dyne/cm^2=10^6 dyne/cm^2; But 1 kg wi= 1g N = 1 kg 9.8 m/s^2=9.8 N Also 1kg wt= 1000 gm ×980 cm/s^=9.8×10^5 dyne 9.8 N=9.8 ×10^5 dyne Or 1N=10^5 dyne Upvote · 9 2 Jim.Moore Physics, Math, Systems Engineer, Educator · Author has 22.3K answers and 17.5M answer views ·2y Related How do I convert 1 joules to Newton meters and 1 kgm2s-2? I see that Joule is defined as both kg m² s-2 and as Nm So it appears they are all the same thing. Here is Table 4 from the BIPM SI (metric) manual Continue Reading I see that Joule is defined as both kg m² s-2 and as Nm So it appears they are all the same thing. Here is Table 4 from the BIPM SI (metric) manual Upvote · 9 1 Archan Desai Student ·7y Related 1 joule = how many ergs? Unit of energy is joule = kgm²/s² Joule is a MKS unit and erg is a CGS unit MKS = CGS 1 m = 100 cm 1 s = 1 s 1 kg = 1000 gm From above : 1 joule = 1 kgm²/s² = 1000 10000 / 1 cmm²/s²= 10^7 gmcm²/s² = 10^7 erg So, Answer : 1 joule = 10^7 erg Upvote · 99 47 9 1 9 1 Related questions What is the relationship between a joule and an erg? What is the value of one joule in erg? How do I convert 1 joule to ERG by using dimensional analysis? 1 joule = how many ergs? What's the value of 1 erg in a joule? What is the relationship between joules, ergs, calories, and electro volts? How do you convert one erg into joules? How do you convert 1 erg into a joule by using the dimensional method? The SI and the C.G.S. unit are joule and erg. How many ergs are equal to 1J? How do you convert 50 erg in joules by dimensional analysis? How many joules are present in 100 ergs? How do I convert 120 kg m to joule? How do I convert a joule into a foot pound? How do I convert 1 joules to Newton meters and 1 kgm2s-2? Can you convert joules into horsepower? Related questions What is the relationship between a joule and an erg? What is the value of one joule in erg? How do I convert 1 joule to ERG by using dimensional analysis? 1 joule = how many ergs? What's the value of 1 erg in a joule? What is the relationship between joules, ergs, calories, and electro volts? Advertisement About · Careers · Privacy · Terms · Contact · Languages · Your Ad Choices · Press · © Quora, Inc. 2025
6415	https://openstax.org/books/statistics/pages/8-3-a-population-proportion	Skip to ContentGo to accessibility pageKeyboard shortcuts menu Statistics 8.3 A Population Proportion Statistics8.3 A Population Proportion Search for key terms or text. During an election year, we see articles in the newspaper that state confidence intervals in terms of proportions or percentages. For example, a poll for a particular candidate running for president might show that the candidate has 40 percent of the vote within 3 percentage points (if the sample is large enough). Often, election polls are calculated with 95 percent confidence, so the pollsters would be 95 percent confident that the true proportion of voters who favored the candidate would be between 0.37 and 0.43 (0.40 – 0.03, 0.40 + 0.03). Investors in the stock market are interested in the true proportion of stocks that go up and down each week. Businesses that sell personal computers are interested in the proportion of households in the United States that own personal computers. Confidence intervals can be calculated for the true proportion of stocks that go up or down each week and for the true proportion of households in the United States that own personal computers. The procedure to find the confidence interval, the sample size, the error bound for a population (EBP), and the confidence level for a proportion is similar to that for the population mean, but the formulas are different. How do you know you are dealing with a proportion problem? First, the data that you are collecting is categorical, consisting of two categories: Success or Failure, Yes or No. Examples of situations where you are the following trying to estimate the true population proportion are the following: What proportion of the population smoke? What proportion of the population will vote for candidate A? What proportion of the population has a college-level education? The distribution of the sample proportions (based on samples of size n) is denoted by P′ (read “P prime”). The central limit theorem for proportions asserts that the sample proportion distribution P′ follows a normal distribution with mean value p, and standard deviation , where p is the population proportion and q = 1 -– p. The confidence interval has the form (p′ – EBP, p′ + EBP). EBP is error bound for the proportion. p′ = the estimated proportion of successes (p′ is a point estimate for p, the true proportion.) x = the number of successes n = the size of the sample The error bound for a proportion is where q′ = 1 – p′. This formula is similar to the error bound formula for a mean, except that the "appropriate standard deviation" is different. For a mean, when the population standard deviation is known, the appropriate standard deviation that we use is . For a proportion, the appropriate standard deviation is . However, in the error bound formula, we use as the standard deviation, instead of . In the error bound formula, the sample proportions p′ and q′, are estimates of the unknown population proportions p and q. The estimated proportions p′ and q′ are used because p and q are not known. The sample proportions p′ and q′ are calculated from the data: p′ is the estimated proportion of successes, and q′ is the estimated proportion of failures. The confidence interval can be used only if the number of successes np′ and the number of failures nq′ are both greater than five. That is, in order to use the formula for confidence intervals for proportions, you need to verify that both and . Example 8.10 Problem Suppose that a market research firm is hired to estimate the percentage of adults living in a large city who have cell phones. Five hundred randomly selected adult residents in this city are surveyed to determine whether they have cell phones. Of the 500 people surveyed, 421 responded yes, they own cell phones. Using a 95 percent confidence level, compute a confidence interval estimate for the true proportion of adult residents of this city who have cell phones. Solution The first solution is step-by-step (Solution A). The second solution uses a function of the TI-83, 83+, or 84 calculators (Solution B). Let X = the number of people in the sample who have cell phones. X is binomial. . To calculate the confidence interval, you must find p′, q′, and EBP. n = 500 x = the number of successes = 421 p′ = 0.842 is the sample proportion; this is the point estimate of the population proportion. Because CL = 0.95, then α = 1 – CL = 1 – 0.95 = 0.05 = 0.025. Then, Use the TI-83, 83+, or 84+ calculator command invNorm(0.975,0,1) to find z0.025. Remember that the area to the right of z0.025is 0.025, and the area to the left of z0.025is 0.975. This can also be found using appropriate commands on other calculators, using a computer, or using a standard normal probability table. The confidence interval for the true binomial population proportion is (p′ – EBP, p′ + EBP) = (0.810, 0.874). Interpretation We estimate with 95 percent confidence that between 81 percent and 87.4 percent of all adult residents of this city have cell phones. Explanation of 95 percent Confidence Level Ninety-five percent of the confidence intervals constructed in this way would contain the true value for the population proportion of all adult residents of this city who have cell phones. Solution Using the TI-83, 83+, 84, 84+ Calculator Press STAT and arrow over to TESTS. Arrow down to A:1-PropZint. Press ENTER. Arrow down to x and enter 421. Arrow down to n and enter 500. Arrow down to C-Level and enter .95. Arrow down to Calculate and press ENTER. The confidence interval is (0.81003, 0.87397). Try It 8.10 Suppose 250 randomly selected people are surveyed to determine whether they own tablets. Of the 250 surveyed, 98 reported owning tablets. Using a 95 percent confidence level, compute a confidence interval estimate for the true proportion of people who own tablets. Example 8.11 Problem For a class project, a political science student at a large university wants to estimate the percentage of students who are registered voters. He surveys 500 students and finds that 300 are registered voters. Compute a 90 percent confidence interval for the true percentage of students who are registered voters, and interpret the confidence interval. Solution The first solution is step-by-step (Solution A). The second solution uses a function of the TI-83, 83+, or 84 calculators (Solution B). Solution A Because CL = 0.90, then α = 1 – CL = 1 – 0.90 = 0.10 = 0.05. Use the TI-83, 83+, or 84+ calculator command invNorm(0.95,0,1) to find z0.05. Remember that the area to the right of z0.05 is 0.05, and the area to the left of z0.05 is 0.95. This can also be found using appropriate commands on other calculators, using a computer, or using a standard normal probability table. The confidence interval for the true binomial population proportion is (p′ – EBP , p′ + EBP) = (0.564, 0.636). Interpretation We estimate with 90 percent confidence that the true percentage of all students who are registered voters is between 56.4 percent and 63.6 percent. Alternate wording: We estimate with 90 percent confidence that between 56.4 percent and 63.6 percent of all students are registered voters. Explanation of 90 percent Confidence Level Ninety percent of all confidence intervals constructed in this way contain the true value for the population percentage of students who are registered voters. Solution Solution B Using the TI-83, 83+, 84, 84+ Calculator Press STAT and arrow over to TESTS. Arrow down to A:1-PropZint. Press ENTER. Arrow down to x and enter 300. Arrow down to n and enter 500. Arrow down to C-Level and enter 0.90. Arrow down to Calculate and press ENTER. The confidence interval is (0.564, 0.636). Try It 8.11 A student polls her school to determine whether students in the school district are for or against the new legislation regarding school uniforms. She surveys 600 students and finds that 480 are against the new legislation. a. Compute a 90 percent confidence interval for the true percentage of students who are against the new legislation, and interpret the confidence interval. b. In a sample of 300 students, 68 percent said they own an iPod and a smartphone. Compute a 97 percent confidence interval for the true percentage of students who own an iPod and a smartphone. Plus-Four Confidence Interval for p There is a certain amount of error introduced into the process of calculating a confidence interval for a proportion. Because we do not know the true proportion for the population, we are forced to use point estimates to calculate the appropriate standard deviation of the sampling distribution. Studies have shown that the resulting estimation of the standard deviation can be flawed. Fortunately, there is a simple adjustment that allows us to produce more accurate confidence intervals: We simply pretend that we have four additional observations. Two of these observations are successes, and two are failures. The new sample size, then, is n + 4, and the new count of successes is x + 2. Computer studies have demonstrated the effectiveness of the plus-four confidence interval for p method. It should be used when the confidence level desired is at least 90 percent and the sample size is at least ten. Example 8.12 Problem A random sample of 25 statistics students was asked: “Have you used a product in the past week?” Six students reported using the product within the past week. Use the plus-four method to find a 95 percent confidence interval for the true proportion of statistics students who use the product weekly. Solution Solution A Six students out of 25 reported using a product within the past week, so x = 6 and n = 25. Because we are using the plus-four method, we will use x = 6 + 2 = 8, and n = 25 + 4 = 29. Because CL = 0.95, we know α = 1 – 0.95 = 0.05, and = 0.025. We are 95 percent confident that the true proportion of all statistics students who use the product is between 0.113 and 0.439. Solution Using the TI-83, 83+, 84, 84+ Calculator Press STAT and arrow over to TESTS. Arrow down to A:1-PropZint. Press ENTER. Arrow down to x and enter 8. Arrow down to n and enter 29. Arrow down to C-Level and enter 0.95. Arrow down to Calculate and press ENTER. The confidence interval is (0.113, 0.439). Reminder Remember that the plus-four method assumes an additional four trials: two successes and two failures. You do not need to change the process for calculating the confidence interval; simply update the values of x and n to reflect these additional trials. Try It 8.12 Out of a random sample of 65 freshmen at State University, 31 students have declared their majors. Use the plus-four method to find a 96 percent confidence interval for the true proportion of freshmen at State University who have declared their majors. Example 8.13 Problem A group of researchers recently conducted a study analyzing the privacy management habits of teen internet users. In a group of 50 teens, 13 reported having more than 500 friends on a social media site. Use the plus four method to find a 90 percent confidence interval for the true proportion of teens who would report having more than 500 online friends. Solution Using plus-four, we have x = 13 + 2 = 15, and n = 50 + 4 = 54. 8.5 Because CL = 0.90, we know α = 1 – 0.90 = 0.10, and = 0.05. We are 90 percent confident that between 17.8 percent and 37.8 percent of all teens would report having more than 500 friends on a social media site. Solution Using the TI-83, 83+, 84, 84+ Calculator Press STAT and arrow over to TESTS. Arrow down to A:1-PropZint. Press ENTER. Arrow down to x and enter 15. Arrow down to n and enter 54. Arrow down to C-Level and enter 0.90. Arrow down to Calculate and press ENTER. The confidence interval is (0.178, 0.378). Try It 8.13 The research group referenced in Example 8.13 talked to teens in smaller focus groups but also interviewed additional teens over the phone. When the study was complete, 588 teens had answered the question about their social media site friends, with 159 saying that they have more than 500 friends. Use the plus-four method to find a 90 percent confidence interval for the true proportion of teens who would report having more than 500 online friends based on this larger sample. Compare the results to those in Example 8.13. Calculating the Sample Size n If researchers desire a specific margin of error, then they can use the error bound formula to calculate the required sample size. The margin of error formula for a population proportion is , where p′ is the sample proportion, q′ = 1 – p′, and n is the sample size. Solving for n gives you an equation for the sample size. . This formula tells us that we can compute the sample size n required for a confidence level of by taking the square of the critical value , multiplying by the point estimate p′, and by q′ = 1 – p′ and finally dividing the result by the square of the margin of error. Always remember to round up the value of n. Example 8.14 Problem Suppose a mobile phone company wants to determine the current percentage of customers ages 50+ who use text messaging on their cell phones. How many customers ages 50+ should the company survey in order to be 90 percent confident that the estimated (sample) proportion is within 3 percentage points of the true population proportion of customers ages 50+ who use text messaging on their cell phones? Assume that p′ = 0.5. Solution From the problem, we know that EBP = 0.03 (3 percent=0.03), and z0.05 = 1.645 because the confidence level is 90 percent. To calculate the sample size n, use the formula and make the substitutions. Round the answer to the next higher value. The sample size should be 752 cell phone customers ages 50+ in order to be 90 percent confident that the estimated (sample) proportion is within 3 percentage points of the true population proportion of all customers ages 50+ who use text messaging on their cell phones. Try It 8.14 An internet marketing company wants to determine the current percentage of customers who click on ads on their smartphones. How many customers should the company survey in order to be 90 percent confident that the estimated proportion is within 5 percentage points of the true population proportion of customers who click on ads on their smartphones? Assume that the sample proportion p′ is 0.50. PreviousNext Order a print copy Citation/Attribution This book may not be used in the training of large language models or otherwise be ingested into large language models or generative AI offerings without OpenStax's permission. Want to cite, share, or modify this book? This book uses the Creative Commons Attribution License and you must attribute Texas Education Agency (TEA). The original material is available at: Changes were made to the original material, including updates to art, structure, and other content updates. Attribution information If you are redistributing all or part of this book in a print format, then you must include on every physical page the following attribution: Access for free at If you are redistributing all or part of this book in a digital format, then you must include on every digital page view the following attribution: Access for free at Citation information Use the information below to generate a citation. We recommend using a citation tool such as this one. Authors: Barbara Illowsky, Susan Dean Publisher/website: OpenStax Book title: Statistics Publication date: Mar 27, 2020 Location: Houston, Texas Book URL: Section URL: © Oct 10, 2024 Texas Education Agency (TEA). The OpenStax name, OpenStax logo, OpenStax book covers, OpenStax CNX name, and OpenStax CNX logo are not subject to the Creative Commons license and may not be reproduced without the prior and express written consent of Rice University.
6416	https://www.isorb.cn/sarvalue/	大连东信微波吸收材料有限公司-吸波材料,微波暗室,微波吸收材料,EMC电磁兼容暗室,电波暗室,天线暗室,吸波片,吸波体,海绵吸波材料,泡沫吸波材料,橡胶吸波片,吸波涂料,附着力促进剂 – 什么是SAR值首页产品中心暗室吸波材料电波微波暗室设备吸波材料测试服务测试系统技术交流新闻动态关于我们联系我们 English 首页产品中心暗室吸波材料电波微波暗室设备吸波材料测试服务测试系统技术交流新闻动态关于我们联系我们 English 什么是SAR值首页/技术交流/ 什么是SAR值技术交流什么是SAR值首先，什么是手机幅射？当人们使用手机时，手机会向发射基站传送无线电波，而任何一种无线电波或多或少会被人体吸收，这种无线电波就称为手机辐射。手机辐射可分为两种：手机机身辐射和基站辐射。手机辐射的测试标准就是SAR。SAR是Specific Absorbtion Ratio的缩写，即“吸收比率”。通俗地讲，就是测量手机辐射对人体的影响是否符合标准。以任意6分钟记时平均，每公斤人体组织吸收的电磁辐射能量（瓦）。以手机辐射为例，SAR指的是辐射被头部的软组织吸收的比率，SAR值越低，辐射被脑部吸收的量越少，但是，这并不表示SAR等级与手机用户的健康直接有关。手机辐射标准(SAR限值)：根据国际电信联盟和世界卫生组织推荐的衡量手机辐射的技术标准SAR值的要求，国际电信联盟标准限值为2.0瓦特/千克，美国联邦通信委员会标准限值为1.6瓦特/千克。 SAR值即为生物体（包括人体）每单位（kg）允许吸收的辐射值，也就是代表辐射对人体的影响，是最直接的测试值，SAR有针对全身的、局部的、四肢的数据。SAR值越低，辐射被吸收的量越少。国际上通用SAR值来表示辐射对人体的影响，值越高对人体影响就越大。不同的科研测试论证的结果不同，有的科学家认为SAR值低于1.3W/kg，对人体基本没有伤害，有的科学家则认为，SAR值只要低于2.0W/kg才对人体无害，目前没有定论。目前通用的标准为SAR值不高于2.0W/kg。因为我国目前还没有相关国家标准，所以国内手机检测基本上按照上述国际标准。我国也曾酝酿过自己的标准，2001年，国家标准化管理委员会发布了一个《电磁辐射暴露限值和测量方法》国家强制性标准的征求意见稿，其中规定的手机SAR值比欧美更低，为1.0W/kg，但考虑到这个标准“太严”，可能使得大部分手机不达标，因此至今这个标准也没出台。下表是世界主要国家和地区对SAR的限值规定和测试方法澳大利亚新西兰 USA 欧洲国际组织测量方法 ACA（旧）EN50361（新）ANSI C95.3；IEEE1528 EN50361 IEC62209（草案）限值 AS2772.1；ARPANSA NZS2772.1 ANSI C95.1 EN50360 ICNIRP 人体整体 0.08 W/kg 0.08 W/kg 0.08 W/kg 0.08 W/kg 0.08 W/kg 空间峰值 2 W/kg 2 W/kg 1.6 W/kg 2 W/kg 2 W/kg 平均时间 6 min 6 min 30 min 6 min 6 min 平均物质 10g 10g 1g 10g 10g 大连东信微波吸收材料有限公司™ 辽ICP备11005778号辽公网安备21021102001179号 design by jeraystudio
6417	https://www.isinj.com/mt-usamo/Inequalities%20A%20Mathematical%20Olympiad%20Approach%20-%20Manfrino,%20Ortega,%20and%20Delgado%20(Birkhauser,%202009).pdf	Inequalities A Mathematical Olympiad Approach Radmila Bulajich Manfrino José Antonio Gómez Ortega Rogelio Valdez Delgado Birkhäuser Basel · Boston · Berlin 2000 Mathematical Subject Classiﬁcation 00A07; 26Dxx, 51M16 Library of Congress Control Number: 2009929571 Bibliograﬁsche Information der Deutschen Bibliothek Die Deutsche Bibliothek verzeichnet diese Publikation in der Deutschen National-bibliograﬁe; detaillierte bibliograﬁsche Daten sind im Internet über abrufbar. ISBN 978-3-0346-0049-1 Birkhäuser Verlag, Basel – Boston – Berlin This work is subject to copyright. All rights are reserved, whether the whole or part of the material is concerned, speciﬁcally the rights of translation, reprinting, re-use of illustra-tions, recitation, broadcasting, reproduction on microﬁlms or in other ways, and storage in data banks. For any kind of use permission of the copyright owner must be obtained. © 2009 Birkhäuser Verlag AG Basel · Boston · Berlin Postfach 133, CH-4010 Basel, Schweiz Ein Unternehmen von Springer Science+Business Media Gedruckt auf säurefreiem Papier, hergestellt aus chlorfrei gebleichtem Zellstoff. TCF ∞ Printed in Germany ISBN 978-3-0346-0049-1 e-ISBN 978-3-0346-0050-7 9 8 7 6 5 4 3 2 1 www.birkhauser.ch Autors: Radmila Bulajich Manfrino Rogelio Valdez Delgado Facultad de Ciencias Universidad Autónoma Estado de Morelos Av. Universidad 1001 Col. Chamilpa 62209 Cuernavaca, Morelos México e-mail: bulajich@uaem.mx valdez@uaem.mx José Antonio Gómez Ortega Departamento de Matemàticas Facultad de Ciencias, UNAM Universidad Nacional Autónoma de México Ciudad Universitaria 04510 México, D.F. México e-mail: jago@fciencias.unam.mx Introduction This book is intended for the Mathematical Olympiad students who wish to pre-pare for the study of inequalities, a topic now of frequent use at various levels of mathematical competitions. In this volume we present both classic inequalities and the more useful inequalities for confronting and solving optimization prob-lems. An important part of this book deals with geometric inequalities and this fact makes a big diﬀerence with respect to most of the books that deal with this topic in the mathematical olympiad. The book has been organized in four chapters which have each of them a diﬀerent character. Chapter 1 is dedicated to present basic inequalities. Most of them are numerical inequalities generally lacking any geometric meaning. How-ever, where it is possible to provide a geometric interpretation, we include it as we go along. We emphasize the importance of some of these inequalities, such as the inequality between the arithmetic mean and the geometric mean, the Cauchy-Schwarz inequality, the rearrangement inequality, the Jensen inequality, the Muir-head theorem, among others. For all these, besides giving the proof, we present several examples that show how to use them in mathematical olympiad prob-lems. We also emphasize how the substitution strategy is used to deduce several inequalities. The main topic in Chapter 2 is the use of geometric inequalities. There we ap-ply basic numerical inequalities, as described in Chapter 1, to geometric problems to provide examples of how they are used. We also work out inequalities which have a strong geometric content, starting with basic facts, such as the triangle inequality and the Euler inequality. We introduce examples where the symmetri-cal properties of the variables help to solve some problems. Among these, we pay special attention to the Ravi transformation and the correspondence between an inequality in terms of the side lengths of a triangle a, b, c and the inequalities that correspond to the terms s, r and R, the semiperimeter, the inradius and the circumradius of a triangle, respectively. We also include several classic geometric problems, indicating the methods used to solve them. In Chapter 3 we present one hundred and twenty inequality problems that have appeared in recent events, covering all levels, from the national and up to the regional and international olympiad competitions. vi Introduction In Chapter 4 we provide solutions to each of the two hundred and ten exer-cises in Chapters 1 and 2, and to the problems presented in Chapter 3. Most of the solutions to exercises or problems that have appeared in international math-ematical competitions were taken from the oﬃcial solutions provided at the time of the competitions. This is why we do not give individual credits for them. Some of the exercises and problems concerning inequalities can be solved us-ing diﬀerent techniques, therefore you will ﬁnd some exercises repeated in diﬀerent sections. This indicates that the technique outlined in the corresponding section can be used as a tool for solving the particular exercise. The material presented in this book has been accumulated over the last ﬁf-teen years mainly during work sessions with the students that won the national contest of the Mexican Mathematical Olympiad. These students were develop-ing their skills and mathematical knowledge in preparation for the international competitions in which Mexico participates. We would like to thank Rafael Mart´ ınez Enr´ ıquez, Leonardo Ignacio Mart´ ınez Sandoval, David Mireles Morales, Jes´ us Rodr´ ıguez Viorato and Pablo Sober´ on Bravo for their careful revision of the text and helpful comments for the improve-ment of the writing and the mathematical content. Contents Introduction vii 1 Numerical Inequalities 1 1.1 Order in the real numbers . . . . . . . . . . . . . . . . . . . . . . . 1 1.2 The quadratic function ax2 + 2bx + c . . . . . . . . . . . . . . . . . 4 1.3 A fundamental inequality, arithmetic mean-geometric mean . . . . . . . . . . . . . . . . . . . 7 1.4 A wonderful inequality: The rearrangement inequality . . . . . . . . . . . . . . . . . . . . . 13 1.5 Convex functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20 1.6 A helpful inequality . . . . . . . . . . . . . . . . . . . . . . . . . . 33 1.7 The substitution strategy . . . . . . . . . . . . . . . . . . . . . . . 39 1.8 Muirhead’s theorem . . . . . . . . . . . . . . . . . . . . . . . . . . 43 2 Geometric Inequalities 51 2.1 Two basic inequalities . . . . . . . . . . . . . . . . . . . . . . . . . 51 2.2 Inequalities between the sides of a triangle . . . . . . . . . . . . . . 54 2.3 The use of inequalities in the geometry of the triangle . . . . . . . 59 2.4 Euler’s inequality and some applications . . . . . . . . . . . . . . . 66 2.5 Symmetric functions of a, b and c . . . . . . . . . . . . . . . . . . . 70 2.6 Inequalities with areas and perimeters . . . . . . . . . . . . . . . . 75 2.7 Erd˝ os-Mordell Theorem . . . . . . . . . . . . . . . . . . . . . . . . 80 2.8 Optimization problems . . . . . . . . . . . . . . . . . . . . . . . . . 88 3 Recent Inequality Problems 101 4 Solutions to Exercises and Problems 117 4.1 Solutions to the exercises in Chapter 1 . . . . . . . . . . . . . . . . 117 4.2 Solutions to the exercises in Chapter 2 . . . . . . . . . . . . . . . . 140 4.3 Solutions to the problems in Chapter 3 . . . . . . . . . . . . . . . . 162 Notation 205 viii Contents Bibliography 207 Index 209 Chapter 1 Numerical Inequalities 1.1 Order in the real numbers A very important property of the real numbers is that they have an order. The order of the real numbers enables us to compare two numbers and to decide which one of them is greater or whether they are equal. Let us assume that the real numbers system contains a set P, which we will call the set of positive numbers, and we will express in symbols x > 0 if x belongs to P. We will also assume the following three properties. Property 1.1.1. Every real number x has one and only one of the following prop-erties: (i) x = 0, (ii) x ∈P (that is, x > 0), (iii) −x ∈P (that is, −x > 0). Property 1.1.2. If x, y ∈P, then x+y ∈P (in symbols x > 0, y > 0 ⇒x+y > 0). Property 1.1.3. If x, y ∈P, then xy ∈P (in symbols x > 0, y > 0 ⇒xy > 0). If we take the “real line” as the geometric representation of the real numbers, by this we mean a directed line where the number “0”has been located and serves to divide the real line into two parts, the positive numbers being on the side containing the number one “1”. In general the number one is set on the right hand side of 0. The number 1 is positive, because if it were negative, since it has the property that 1 · x = x for every x, we would have that any number x ̸= 0 would satisfy x ∈P and −x ∈P, which contradicts property 1.1.1. Now we can deﬁne the relation a is greater than b if a −b ∈P (in symbols a > b). Similarly, a is smaller than b if b −a ∈P (in symbols a < b). Observe that 2 Numerical Inequalities a < b is equivalent to b > a. We can also deﬁne that a is smaller than or equal to b if a < b or a = b (using symbols a ≤b). We will denote by R the set of real numbers and by R+ the set P of positive real numbers. Example 1.1.4. (i) If a < b and c is any number, then a + c < b + c. (ii) If a < b and c > 0, then ac < bc. In fact, to prove (i) we see that a + c < b + c ⇔(b + c) −(a + c) > 0 ⇔ b −a > 0 ⇔a < b. To prove (ii), we proceed as follows: a < b ⇒b −a > 0 and since c > 0, then (b −a)c > 0, therefore bc −ac > 0 and then ac < bc. Exercise 1.1. Given two numbers a and b, exactly one of the following assertions is satisﬁed, a = b, a > b or a < b. Exercise 1.2. Prove the following assertions. (i) a < 0, b < 0 ⇒ab > 0. (ii) a < 0, b > 0 ⇒ab < 0. (iii) a < b, b < c ⇒a < c. (iv) a < b, c < d ⇒a + c < b + d. (v) a < b ⇒−b < −a. (vi) a > 0 ⇒1 a > 0. (vii) a < 0 ⇒1 a < 0. (viii) a > 0, b > 0 ⇒a b > 0. (ix) 0 < a < b, 0 < c < d ⇒ac < bd. (x) a > 1 ⇒a2 > a. (xi) 0 < a < 1 ⇒a2 < a. Exercise 1.3. (i) If a > 0, b > 0 and a2 < b2, then a < b. (ii) If b > 0, we have that a b > 1 if and only if a > b. The absolute value of a real number x, which is denoted by \|x\|, is deﬁned as \|x\| = x if x ≥0, −x if x < 0. Geometrically, \|x\| is the distance of the number x (on the real line) from the origin 0. Also, \|a −b\| is the distance between the real numbers a and b on the real line. 1.1 Order in the real numbers 3 Exercise 1.4. For any real numbers x, a and b, the following hold. (i) \|x\| ≥0, and is equal to zero only when x = 0. (ii) \|−x\| = \|x\|. (iii) \|x\|2 = x2. (iv) \|ab\| = \|a\| \|b\|. (v) a b = \|a\| \|b\| , with b ̸= 0. Proposition 1.1.5 (Triangle inequality). The triangle inequality states that for any pair of real numbers a and b, \|a + b\| ≤\|a\| + \|b\| . Moreover, the equality holds if and only if ab ≥0. Proof. Both sides of the inequality are positive; then using Exercise 1.3 it is suﬃ-cient to verify that \|a + b\|2 ≤(\|a\| + \|b\|)2: \|a + b\|2 = (a + b)2 = a2 + 2ab + b2 = \|a\|2 + 2ab + \|b\|2 ≤\|a\|2 + 2 \|ab\| + \|b\|2 = \|a\|2 + 2 \|a\| \|b\| + \|b\|2 = (\|a\| + \|b\|)2 . In the previous relations we observe only one inequality, which is obvious since ab ≤\|ab\|. Note that, when ab ≥0, we can deduce that ab = \|ab\| = \|a\| \|b\|, and then the equality holds. □ The general form of the triangle inequality for real numbers x1, x2, . . . , xn, is \|x1 + x2 + · · · + xn\| ≤\|x1\| + \|x2\| + · · · + \|xn\|. The equality holds when all xi’s have the same sign. This can be proved in a similar way or by the use of induction. Another version of the last inequality, which is used very often, is the following: \|±x1 ± x2 ± · · · ± xn\| ≤\|x1\| + \|x2\| + · · · + \|xn\|. Exercise 1.5. Let x, y, a, b be real numbers, prove that (i) \|x\| ≤b ⇔−b ≤x ≤b, (ii) \|\|a\| −\|b\|\| ≤\|a −b\|, (iii) x2 + xy + y2 ≥0, (iv) x > 0, y > 0 ⇒x2 −xy + y2 > 0. Exercise 1.6. For real numbers a, b, c, prove that \|a\| + \|b\| + \|c\| −\|a + b\| −\|b + c\| −\|c + a\| + \|a + b + c\| ≥0. 4 Numerical Inequalities Exercise 1.7. Let a, b be real numbers such that 0 ≤a ≤b ≤1. Prove that (i) 0 ≤b −a 1 −ab ≤1, (ii) 0 ≤ a 1 + b + b 1 + a ≤1, (iii) 0 ≤ab2 −ba2 ≤1 4. Exercise 1.8. Prove that if n, m are positive integers, then m n < √ 2 if and only if √ 2 < m+2n m+n . Exercise 1.9. If a ≥b, x ≥y, then ax + by ≥ay + bx. Exercise 1.10. If x, y > 0, then x2 y + y2 x ≥√x + √y. Exercise 1.11. (Czech and Slovak Republics, 2004) Let a, b, c, d be real numbers with a + d = b + c, prove that (a −b)(c −d) + (a −c)(b −d) + (d −a)(b −c) ≥0. Exercise 1.12. Let f(a, b, c, d) = (a −b)2 + (b −c)2 + (c −d)2 + (d −a)2. For a < b < c < d, prove that f(a, c, b, d) > f(a, b, c, d) > f(a, b, d, c). Exercise 1.13. (IMO, 1960) For which real values of x the following inequality holds: 4x2 (1 −√1 + 2x)2 < 2x + 9? Exercise 1.14. Prove that for any positive integer n, the fractional part of √ 4n2 + n is smaller than 1 4. Exercise 1.15. (Short list IMO, 1996) Let a, b, c be positive real numbers such that abc = 1. Prove that ab a5 + b5 + ab + bc b5 + c5 + bc + ca c5 + a5 + ca ≤1. 1.2 The quadratic function ax2 + 2bx + c One very useful inequality for the real numbers is x2 ≥0, which is valid for any real number x (it is suﬃcient to consider properties 1.1.1, 1.1.3 and Exercise 1.2 of the previous section). The use of this inequality leads to deducing many other inequalities. In particular, we can use it to ﬁnd the maximum or minimum of a quadratic function ax2 + 2bx + c. These quadratic functions appear frequently in optimization problems or in inequalities. 1.2 The quadratic function ax2 + 2bx + c 5 One common example consists in proving that if a > 0, the quadratic function ax2 + 2bx + c will have its minimum at x = −b a and the minimum value is c −b2 a . In fact, ax2 + 2bx + c = a x2 + 2 b ax + b2 a2 + c −b2 a = a x + b a 2 + c −b2 a . Since x + b a 2 ≥0 and the minimum value of this expression, zero, is attained when x = −b a, we conclude that the minimum value of the quadratic function is c −b2 a . If a < 0, the quadratic function ax2+2bx+c will have a maximum at x = −b a and its value at this point is c−b2 a . In fact, since ax2+2bx+c = a x + b a 2+c−b2 a and since a x + b a 2 ≤0 (because a < 0), the greatest value of this last expression is zero, thus the quadratic function is always less than or equal to c −b2 a and assumes this value at the point x = −b a. Example 1.2.1. If x, y are positive numbers with x + y = 2a, then the product xy is maximal when x = y = a. If x + y = 2a, then y = 2a −x. Hence, xy = x(2a −x) = −x2 + 2ax = −(x −a)2 + a2 has a maximum value when x = a, and then y = x = a. This can be interpreted geometrically as “of all the rectangles with ﬁxed perimeter, the one with the greatest area is the square”. In fact, if x, y are the lengths of the sides of the rectangle, the perimeter is 2(x + y) = 4a, and its area is xy, which is maximized when x = y = a. Example 1.2.2. If x, y are positive numbers with xy = 1, the sum x+y is minimal when x = y = 1. If xy = 1, then y = 1 x. It follows that x + y = x + 1 x = √x − 1 √x 2 + 2, and then x + y is minimal when √x − 1 √x = 0, that is, when x = 1. Therefore, x = y = 1. This can also be interpreted geometrically in the following way, “of all the rectangles with area 1, the square has the smallest perimeter”. In fact, if x, y are the lengths of the sides of the rectangle, its area is xy = 1 and its perimeter is 2(x + y) = 2 x + 1 x = 2 √x − 1 √x 2 + 2 ≥4. Moreover, the perimeter is 4 if and only if √x − 1 √x = 0, that is, when x = y = 1. Example 1.2.3. For any positive number x, we have x + 1 x ≥2. 6 Numerical Inequalities Observe that x + 1 x = √x − 1 √x 2 + 2 ≥2. Moreover, the equality holds if and only if √x − 1 √x = 0, that is, when x = 1. Example 1.2.4. If a, b > 0, then a b + b a ≥2, and the equality holds if and only if a = b. It is enough to consider the previous example with x = a b . Example 1.2.5. Given a, b, c > 0, it is possible to construct a triangle with sides of length a, b, c if and only if pa2 + qb2 > pqc2 for any p, q with p + q = 1. Remember that a, b and c are the lengths of the sides of a triangle if and only if a + b > c, a + c > b and b + c > a. Let Q = pa2 + qb2 −pqc2 = pa2 + (1 −p)b2 −p(1 −p)c2 = c2p2 + (a2 −b2 −c2)p + b2, therefore Q is a quadratic function1 in p and Q > 0 ⇔ △= a2 −b2 −c22 −4b2c2 < 0 ⇔ a2 −b2 −c2 −2bc a2 −b2 −c2 + 2bc < 0 ⇔ a2 −(b + c)2 a2 −(b −c)2 < 0 ⇔ [a + b + c] [a −b −c] [a −b + c] [a + b −c] < 0 ⇔ [b + c −a][c + a −b][a + b −c] > 0. Now, [b + c −a][c + a −b][a + b −c] > 0 if the three factors are positive or if one of them is positive and the other two are negative. However, the latter is impossible, because if [b + c −a] < 0 and [c + a −b] < 0, we would have, adding these two inequalities, that c < 0, which is false. Therefore the three factors are necessarily positive. Exercise 1.16. Suppose the polynomial ax2 + bx + c satisﬁes the following: a > 0, a + b + c ≥0, a −b + c ≥0, a −c ≥0 and b2 −4ac ≥0. Prove that the roots are real and that they belong to the interval −1 ≤x ≤1. Exercise 1.17. If a, b, c are positive numbers, prove that it is not possible for the inequalities a(1 −b) > 1 4, b(1 −c) > 1 4, c(1 −a) > 1 4 to hold at the same time. 1A quadratic function ax2 + bx + c with a > 0 is positive when its discriminant Δ = b2 −4ac is negative, in fact, this follows from ax2 + bx + c = a(x + b 2a )2 + 4ac−b2 4a . Remember that the roots are −b±√ b2−4ac 2a , and they are real when Δ ≥0, otherwise they are not real roots, and then ax2 + bx + c will have the same sign; this expression will be positive if a > 0. 1.3 Arithmetic mean-geometric mean 7 1.3 A fundamental inequality, arithmetic mean-geometric mean The ﬁrst inequality that we consider, fundamental in optimization problems, is the inequality between the arithmetic mean and the geometric mean of two non-negative numbers a and b, which is expressed as a + b 2 ≥ √ ab, (AM-GM). Moreover, the equality holds if and only if a = b. The numbers a+b 2 and √ ab are known as the arithmetic mean and the ge-ometric mean of a and b, respectively. To prove the inequality we only need to observe that a + b 2 − √ ab = a + b −2 √ ab 2 = 1 2 √a − √ b 2 ≥0. And the equality holds if and only if √a = √ b, that is, when a = b. Exercise 1.18. For x ≥0, prove that 1 + x ≥2√x. Exercise 1.19. For x > 0, prove that x + 1 x ≥2. Exercise 1.20. For x, y ∈R+, prove that x2 + y2 ≥2xy. Exercise 1.21. For x, y ∈R+, prove that 2(x2 + y2) ≥(x + y)2. Exercise 1.22. For x, y ∈R+, prove that 1 x + 1 y ≥ 4 x+y. Exercise 1.23. For a, b, x ∈R+, prove that ax + b x ≥2 √ ab. Exercise 1.24. If a, b > 0, then a b + b a ≥2. Exercise 1.25. If 0 < b ≤a, then 1 8 (a−b)2 a ≤a+b 2 − √ ab ≤1 8 (a−b)2 b . Now, we will present a geometric and a visual proof of the following inequal-ities, for x, y > 0, 2 1 x + 1 y ≤√xy ≤x + y 2 . (1.1) x y h g A B C D E O 8 Numerical Inequalities Let x = BD, y = DC and let us construct a semicircle of diameter BC = x + y. Let A be the point where the perpendicular to BC in D intersects the semicircle and let E be the perpendicular projection from D to the radius AO. Let us write AD = h and AE = g. Since ABD and CAD are similar right triangles, we deduce that h y = x h, then h = √xy. Also, since AOD and ADE are similar right triangles, we have g √xy = √xy x+y 2 , then g = 2xy x + y = 2 1 x + 1 y . Finally, the geometry tells us that in a right triangle, the length of one leg is always smaller than the length of the hypotenuse. Hence, g ≤h ≤x+y 2 , which can be written as 2 1 x + 1 y ≤√xy ≤x + y 2 . The number 2 1 x + 1 y is known as the harmonic mean of x and y, and the left inequality in (1.1) is known as the inequality between the harmonic mean and the geometric mean. Some inequalities can be proved through the multiple application of a simple inequality and the use of a good idea to separate the problem into parts that are easier to deal with, a method which is often used to solve the following exercises. Exercise 1.26. For x, y, z ∈R+, (x + y)(y + z)(z + x) ≥8xyz. Exercise 1.27. For x, y, z ∈R, x2 + y2 + z2 ≥xy + yz + zx. Exercise 1.28. For x, y, z ∈R+, xy + yz + zx ≥x√yz + y√zx + z√xy. Exercise 1.29. For x, y ∈R, x2 + y2 + 1 ≥xy + x + y. Exercise 1.30. For x, y, z ∈R+, 1 x + 1 y + 1 z ≥ 1 √xy + 1 √yz + 1 √zx. Exercise 1.31. For x, y, z ∈R+, xy z + yz x + zx y ≥x + y + z. Exercise 1.32. For x, y, z ∈R, x2 + y2 + z2 ≥x y2 + z2 + y √ x2 + z2. The inequality between the arithmetic mean and the geometric mean can be extended to more numbers. For instance, we can prove the following inequa-lity between the arithmetic mean and the geometric mean of four non-negative numbers a, b, c, d, expressed as a+b+c+d 4 ≥ 4 √ abcd, in the following way: a + b + c + d 4 = 1 2 a + b 2 + c + d 2 ≥1 2 √ ab + √ cd ≥ √ ab √ cd = 4 √ abcd. 1.3 Arithmetic mean-geometric mean 9 Observe that we have used the AM-GM inequality three times for two numbers in each case: with a and b, with c and d, and with √ ab and √ cd. Moreover, the equality holds if and only if a = b, c = d and ab = cd, that is, when the numbers satisfy a = b = c = d. Exercise 1.33. For x, y ∈R, x4 + y4 + 8 ≥8xy. Exercise 1.34. For a, b, c, d ∈R+, (a + b + c + d) 1 a + 1 b + 1 c + 1 d ≥16. Exercise 1.35. For a, b, c, d ∈R+, a b + b c + c d + d a ≥4. A useful trick also exists for checking that the inequality a+b+c 3 ≥ 3 √ abc is true for any three non-negative numbers a, b and c. Consider the following four numbers a, b, c and d = 3 √ abc. Since the AM-GM inequality holds for four numbers, we have a+b+c+d 4 ≥ 4 √ abcd = 4 √ d3d = d. Then a+b+c 4 ≥d −1 4d = 3 4d. Hence, a+b+c 3 ≥d = 3 √ abc. These ideas can be used to justify the general version of the inequality for n non-negative numbers. If a1, a2, . . . , an are n non-negative numbers, we take the numbers A and G as A = a1 + a2 + · · · + an n and G = n √a1a2 · · · an. These numbers are known as the arithmetic mean and the geometric mean of the numbers a1, a2, . . . , an, respectively. Theorem 1.3.1 (The AM-GM inequality). a1 + a2 + · · · + an n ≥ n √a1a2 · · · an. First proof (Cauchy). Let Pn be the statement G ≤A, for n numbers. We will proceed by mathematical induction on n, but this is an induction of the following type. (1) We prove that the statement is true for 2 numbers, that is, P2 is true. (2) We prove that Pn ⇒Pn−1. (3) We prove that Pn ⇒P2n. When (1), (2) and (3) are veriﬁed, all the assertions Pn with n ≥2 are shown to be true. Now, we will prove these statements. (1) This has already been done in the ﬁrst part of the section. (2) Let a1, . . . , an−1 be non-negative numbers and let g = n−1 √a1 · · · an−1. Using this number and the numbers we already have, i.e., a1, . . . , an−1, we get n numbers to which we apply Pn, a1 + · · · + an−1 + g n ≥ n √a1a2 · · · an−1g = n gn−1 · g = g. 10 Numerical Inequalities We deduce that a1+· · ·+an−1+g ≥ng, and then it follows that a1+···+an−1 n−1 ≥ g, therefore Pn−1 is true. (3) Let a1, a2, . . . , a2n be non-negative numbers, then a1 + a2 + · · · + a2n = (a1 + a2) + (a3 + a4) + · · · + (a2n−1 + a2n) ≥2 √a1a2 + √a3a4 + · · · + √a2n−1a2n ≥2n √a1a2 √a3a4 · · · √a2n−1a2n 1 n = 2n (a1a2 · · · a2n) 1 2n . We have applied the statement P2 several times, and we have also applied the statement Pn to the numbers √a1a2, √a3a4, . . . , √a2n−1a2n. □ Second proof. Let A = a1+···+an n . We take two numbers ai, one smaller than A and the other greater than A (if they exist), say a1 = A −h and a2 = A + k, with h, k > 0. We exchange a1 and a2 for two numbers that increase the product and ﬁx the sum, deﬁned as a′ 1 = A, a′ 2 = A + k −h. Since a′ 1 + a′ 2 = A + A + k −h = A −h + A + k = a1 + a2, clearly a′ 1 + a′ 2 + a3 + · · · + an = a1 + a2 + a3 + · · · + an, but a′ 1a′ 2 = A(A + k −h) = A2 + A(k −h) and a1a2 = (A+ k)(A−h) = A2 +A(k −h)−hk, then a′ 1a′ 2 > a1a2 and thus it follows that a′ 1a′ 2a3 · · · an > a1a2a3 · · · an. If A = a′ 1 = a′ 2 = a3 = · · · = an, there is nothing left to prove (the equality holds), otherwise two elements will exist, one greater than A and the other one smaller than A and the argument is repeated. Since every time we perform this operation we create a number equal to A, this process can not be used more than n times. □ Example 1.3.2. Find the maximum value of x(1 −x3) for 0 ≤x ≤1. The idea of the proof is to exchange the product for another one in such a way that the sum of the elements involved in the new product is constant. If y = x(1 −x3), it is clear that the right side of 3y3 = 3x3(1 −x3)(1 −x3)(1 −x3), expressed as the product of four numbers 3x3, (1 −x3), (1 −x3) and (1 −x3), has a constant sum equal to 3. The AM-GM inequality for four numbers tells us that 3y3 ≤ 3x3 + 3(1 −x3) 4 4 = 3 4 4 . Thus y ≤ 3 4 3 √ 4. Moreover, the maximum value is reached using 3x3 = 1 −x3, that is, if x = 1 3 √ 4. 1.3 Arithmetic mean-geometric mean 11 Exercise 1.36. Let xi > 0, i = 1, . . . , n. Prove that (x1 + x2 + · · · + xn) 1 x1 + 1 x2 + · · · + 1 xn ≥n2. Exercise 1.37. If {a1, . . . , an} is a permutation of {b1, . . . , bn} ⊂R+, then a1 b1 + a2 b2 + · · · + an bn ≥n and b1 a1 + b2 a2 + · · · + bn an ≥n. Exercise 1.38. If a > 1, then an −1 > n a n+1 2 −a n−1 2 . Exercise 1.39. If a, b, c > 0 and (1 + a)(1 + b)(1 + c) = 8, then abc ≤1. Exercise 1.40. If a, b, c > 0, then a3 b + b3 c + c3 a ≥ab + bc + ca. Exercise 1.41. For non-negative real numbers a, b, c, prove that a2b2 + b2c2 + c2a2 ≥abc(a + b + c). Exercise 1.42. If a, b, c > 0, then a2b + b2c + c2a ab2 + bc2 + ca2 ≥9a2b2c2. Exercise 1.43. If a, b, c > 0 satisfy that abc = 1, prove that 1 + ab 1 + a + 1 + bc 1 + b + 1 + ac 1 + c ≥3. Exercise 1.44. If a, b, c > 0, prove that 1 a + 1 b + 1 c ≥2 1 a + b + 1 b + c + 1 c + a ≥ 9 a + b + c. Exercise 1.45. If Hn = 1 + 1 2 + · · · + 1 n, prove that n(n + 1) 1 n < n + Hn for n ≥2. Exercise 1.46. Let x1, x2, . . . , xn > 0 such that 1 1+x1 +· · ·+ 1 1+xn = 1. Prove that x1x2 · · · xn ≥(n −1)n . Exercise 1.47. (Short list IMO, 1998) Let a1, a2, . . . , an be positive numbers with a1 + a2 + · · · + an < 1, prove that a1a2 · · · an [1 −(a1 + a2 + · · · + an)] (a1 + a2 + · · · + an) (1 −a1) (1 −a2) · · · (1 −an) ≤ 1 nn+1 . 12 Numerical Inequalities Exercise 1.48. Let a1, a2, . . . , an be positive numbers such that 1 1+a1 +· · ·+ 1 1+an = 1. Prove that √a1 + · · · + √an ≥(n −1) 1 √a1 + · · · + 1 √an . Exercise 1.49. (APMO, 1991) Let a1, a2, . . . , an, b1, b2, . . . , bn be positive numbers with a1 + a2 + · · · + an = b1 + b2 + · · · + bn. Prove that a2 1 a1 + b1 + · · · + a2 n an + bn ≥1 2(a1 + · · · + an). Exercise 1.50. Let a, b, c be positive numbers, prove that 1 a3 + b3 + abc + 1 b3 + c3 + abc + 1 c3 + a3 + abc ≤ 1 abc. Exercise 1.51. Let a, b, c be positive numbers with a + b + c = 1, prove that 1 a + 1 1 b + 1 1 c + 1 ≥64. Exercise 1.52. Let a, b, c be positive numbers with a + b + c = 1, prove that 1 a −1 1 b −1 1 c −1 ≥8. Exercise 1.53. (Czech and Slovak Republics, 2005) Let a, b, c be positive numbers that satisfy abc = 1, prove that a (a + 1)(b + 1) + b (b + 1)(c + 1) + c (c + 1)(a + 1) ≥3 4. Exercise 1.54. Let a, b, c be positive numbers for which 1 1+a + 1 1+b + 1 1+c = 1. Prove that abc ≥8. Exercise 1.55. Let a, b, c be positive numbers, prove that 2ab a + b + 2bc b + c + 2ca c + a ≤a + b + c. Exercise 1.56. Let a1, a2, . . . , an, b1, b2, . . . , bn be positive numbers, prove that n i=1 1 aibi n i=1 (ai + bi)2 ≥4n2. Exercise 1.57. (Russia, 1991) For all non-negative real numbers x, y, z, prove that (x + y + z)2 3 ≥x√yz + y√zx + z√xy. 1.4 A wonderful inequality 13 Exercise 1.58. (Russia, 1992) For all positive real numbers x, y, z, prove that x4 + y4 + z2 ≥ √ 8xyz. Exercise 1.59. (Russia, 1992) For any real numbers x, y > 1, prove that x2 y −1 + y2 x −1 ≥8. 1.4 A wonderful inequality: The rearrangement inequality Consider two collections of real numbers in increasing order, a1 ≤a2 ≤· · · ≤an and b1 ≤b2 ≤· · · ≤bn. For any permutation (a′ 1, a′ 2, . . . , a′ n) of (a1, a2, . . . , an), it happens that a1b1 + a2b2 + · · · + anbn ≥a′ 1b1 + a′ 2b2 + · · · + a′ nbn (1.2) ≥anb1 + an−1b2 + · · · + a1bn. (1.3) Moreover, the equality in (1.2) holds if and only if (a′ 1, a′ 2, . . . , a′ n)=(a1, a2, . . . , an). And the equality in (1.3) holds if and only if (a′ 1, a′ 2, . . . , a′ n) = (an, an−1, . . . , a1). Inequality (1.2) is known as the rearrangement inequality. Corollary 1.4.1. For any permutation (a′ 1, a′ 2, . . . , a′ n) of (a1, a2, . . . , an), it follows that a2 1 + a2 2 + · · · + a2 n ≥a1a′ 1 + a2a′ 2 + · · · + ana′ n. Corollary 1.4.2. For any permutation (a′ 1, a′ 2, . . . , a′ n) of (a1, a2, . . . , an), it follows that a′ 1 a1 + a′ 2 a2 + · · · + a′ n an ≥n. Proof (of the rearrangement inequality). Suppose that b1 ≤b2 ≤· · · ≤bn. Let S = a1b1 + a2b2 + · · · + arbr + · · · + asbs + · · · + anbn, S′ = a1b1 + a2b2 + · · · + asbr + · · · + arbs + · · · + anbn. The diﬀerence between S and S′ is that the coeﬃcients of br and bs, where r < s, are switched. Hence S −S′ = arbr + asbs −asbr −arbs = (bs −br)(as −ar). Thus, we have that S ≥S′ if and only if as ≥ar. Repeating this process we get the result that the sum S is maximal when a1 ≤a2 ≤· · · ≤an. □ 14 Numerical Inequalities Example 1.4.3. (IMO, 1975) Consider two collections of numbers x1 ≤x2 ≤· · · ≤ xn and y1 ≤y2 ≤· · · ≤yn, and one permutation (z1, z2, . . . , zn) of (y1, y2, . . . , yn). Prove that (x1 −y1)2 + · · · + (xn −yn)2 ≤(x1 −z1)2 + · · · + (xn −zn)2 . By squaring and rearranging this last inequality, we ﬁnd that it is equivalent to n i=1 x2 i −2 n i=1 xiyi + n i=1 y2 i ≤ n i=1 x2 i −2 n i=1 xizi + n i=1 z2 i , but since n i=1 y2 i = n i=1 z2 i , then the inequality we have to prove turns to be equivalent to n i=1 xizi ≤ n i=1 xiyi, which in turn is inequality (1.2). Example 1.4.4. (IMO, 1978) Let x1, x2, . . . , xn be distinct positive integers, prove that x1 12 + x2 22 + · · · + xn n2 ≥1 1 + 1 2 + · · · + 1 n. Let (a1, a2, . . . , an) be a permutation of (x1, x2, . . . , xn) with a1 ≤a2 ≤ · · · ≤an and let (b1, b2, . . . , bn) = 1 n2 , 1 (n−1)2 , . . . , 1 12 ; that is, bi = 1 (n+1−i)2 for i = 1, . . . , n. Consider the permutation (a′ 1, a′ 2, . . . , a′ n) of (a1, a2, . . . , an) deﬁned by a′ i = xn+1−i, for i = 1, . . . , n. Using inequality (1.3) we can argue that x1 12 + x2 22 + · · · + xn n2 = a′ 1b1 + a′ 2b2 + · · · + a′ nbn ≥anb1 + an−1b2 + · · · + a1bn = a1bn + a2bn−1 + · · · + anb1 = a1 12 + a2 22 + · · · + an n2 . Since 1 ≤a1, 2 ≤a2, . . . , n ≤an, we have that x1 12 + x2 22 + · · ·+ xn n2 ≥a1 12 + a2 22 +· · ·+ an n2 ≥1 12 + 2 22 +· · · + n n2 = 1 1 + 1 2 +· · · + 1 n. Example 1.4.5. (IMO, 1964) Suppose that a, b, c are the lengths of the sides of a triangle. Prove that a2 (b + c −a) + b2 (a + c −b) + c2 (a + b −c) ≤3abc. 1.4 A wonderful inequality 15 Since the expression is a symmetric function of a, b and c, we can as-sume, without loss of generality, that c ≤b ≤a. In this case, a (b + c −a) ≤ b (a + c −b) ≤c (a + b −c) . For instance, the ﬁrst inequality is proved in the following way: a (b + c −a) ≤b (a + c −b) ⇔ ab + ac −a2 ≤ab + bc −b2 ⇔ (a −b) c ≤(a + b) (a −b) ⇔ (a −b) (a + b −c) ≥0. By (1.3) of the rearrangement inequality, we have a2(b+c−a)+b2(c+a−b)+c2(a+b−c) ≤ba(b+c−a)+cb(c+a−b)+ac(a+b−c), a2(b+c−a)+b2(c+a−b)+c2(a+b−c) ≤ca(b+c−a)+ab(c+a−b)+bc(a+b−c). Therefore, 2 a2(b + c −a) + b2(c + a −b) + c2(a + b −c) ≤6abc. Example 1.4.6. (IMO, 1983) Let a, b and c be the lengths of the sides of a triangle. Prove that a2b(a −b) + b2c(b −c) + c2a (c −a) ≥0. Consider the case c ≤b ≤a (the other cases are similar). As in the previous example, we have that a(b+c−a) ≤b(a+c−b) ≤c(a+b−c) and since 1 a ≤1 b ≤1 c, using Inequality (1.2) leads us to 1 aa(b + c −a) + 1 bb(c + a −b) + 1 c c(a + b −c) ≥1 ca(b + c −a) + 1 ab(c + a −b) + 1 b c(a + b −c). Therefore, a + b + c ≥a (b −a) c + b(c −b) a + c (a −c) b + a + b + c. It follows that a(b−a) c + b(c−b) a + c(a−c) b ≤0. Multiplying by abc, we obtain a2b (a −b) + b2c(b −c) + c2a(c −a) ≥0. Example 1.4.7 (Cauchy-Schwarz inequality). For real numbers x1, . . . , xn, y1, . . . , yn, the following inequality holds: n i=1 xiyi 2 ≤ n i=1 x2 i n i=1 y2 i . The equality holds if and only if there exists some λ ∈R with xi = λyi for all i = 1, 2, . . ., n. 16 Numerical Inequalities If x1 = x2 = · · · = xn = 0 or y1 = y2 = · · · = yn = 0, the result is evident. Otherwise, let S = n i=1 x2 i and T = n i=1 y2 i , where it is clear that S, T ̸= 0. Take ai = xi S and an+i = yi T for i = 1, 2, . . ., n. Using Corollary 1.4.1, 2 = n i=1 x2 i S2 + n i=1 y2 i T 2 = 2n i=1 a2 i ≥a1an+1 + a2an+2 + · · · + ana2n + an+1a1 + · · · + a2nan = 2x1y1 + x2y2 + · · · + xnyn ST . The equality holds if and only if ai = an+i for i = 1, 2, . . . , n, or equivalently, if and only if xi = S T yi for i = 1, 2, . . . , n. Another proof of the Cauchy-Schwarz inequality can be established using Lagrange’s identity n i=1 xiyi 2 = n i=1 x2 i n i=1 y2 i −1 2 n i=1 n j=1 (xiyj −xjyi)2. The importance of the Cauchy-Schwarz inequality will be felt throughout the remaining part of this book, as we will use it as a tool to solve many exercises and problems proposed here. Example 1.4.8 (Nesbitt’s inequality). For a, b, c ∈R+, we have a b + c + b c + a + c a + b ≥3 2. Without loss of generality, we can assume that a ≤b ≤c, and then it follows that a + b ≤c + a ≤b + c and 1 b+c ≤ 1 c+a ≤ 1 a+b. Using the rearrangement inequality (1.2) twice, we obtain a b + c + b c + a + c a + b ≥ b b + c + c c + a + a a + b, a b + c + b c + a + c a + b ≥ c b + c + a c + a + b a + b. Hence, 2 a b + c + b c + a + c a + b ≥ b + c b + c + c + a c + a + a + b a + b = 3. Another way to prove the inequality is using Inequality (1.3) twice, c + a b + c + a + b c + a + b + c a + b ≥3, a + b b + c + b + c c + a + c + a a + b ≥3. 1.4 A wonderful inequality 17 Then, after adding the two expressions, we get 2a+b+c b+c + 2b+c+a c+a + 2c+a+b a+b ≥6, therefore 2a b + c + 2b c + a + 2c a + b ≥3. Example 1.4.9. (IMO, 1995) Let a, b, c be positive real numbers with abc = 1. Prove that 1 a3(b + c) + 1 b3(c + a) + 1 c3(a + b) ≥3 2. Without loss of generality, we can assume that c ≤b ≤a. Let x = 1 a, y = 1 b and z = 1 c, thus S = 1 a3(b + c) + 1 b3(c + a) + 1 c3(a + b) = x3 1 y + 1 z + y3 1 z + 1 x + z3 1 x + 1 y = x2 y + z + y2 z + x + z2 x + y . Since x ≤y ≤z, we can deduce that x + y ≤z + x ≤y + z and also that x y+z ≤ y z+x ≤ z x+y. Using the rearrangement inequality (1.2), we show that x2 y + z + y2 z + x + z2 x + y ≥ xy y + z + yz z + x + zx x + y , x2 y + z + y2 z + x + z2 x + y ≥ xz y + z + yx z + x + zy x + y , which in turn leads to 2S ≥x + y + z ≥3 3 √xyz = 3. Therefore, S ≥3 2. Example 1.4.10. (APMO, 1998) Let a, b, c ∈R+, prove that 1 + a b 1 + b c 1 + c a ≥2 1 + a + b + c 3 √ abc . Observe that 1 + a b 1 + b c 1 + c a ≥2 1 + a + b + c 3 √ abc ⇔ 1 + a b + b c + c a + a c + c b + b a + abc abc ≥2 1 + a + b + c 3 √ abc ⇔ a b + b c + c a + a c + c b + b a ≥2(a + b + c) 3 √ abc . 18 Numerical Inequalities Now we set a = x3, b = y3, c = z3. We need to prove that x3 y3 + y3 z3 + z3 x3 + x3 z3 + z3 y3 + y3 x3 ≥2 x3 + y3 + z3 xyz . But, if we consider (a1, a2, a3, a4, a5, a6) = x y , y z , z x, x z , z y , y x , (a′ 1, a′ 2, a′ 3, a′ 4, a′ 5, a′ 6) = y z , z x, x y , z y, y x, x z , (b1, b2, b3, b4, b5, b6) = x2 y2 , y2 z2 , z2 x2 , x2 z2 , z2 y2 , y2 x2 , we are led to the following result: x3 y3 + y3 z3 + z3 x3 + x3 z3 + z3 y3 + y3 x3 ≥x2 y2 y z + y2 z2 z x + z2 x2 x y + x2 z2 z y + z2 y2 y x + y2 x2 x z = x2 yz + y2 zx + z2 xy + x2 zy + z2 yx + y2 xz = 2 x3 + y3 + z3 xyz . Example 1.4.11 (Tchebyshev’s inequality). Let a1 ≤a2 ≤· · · ≤an and b1 ≤b2 ≤ · · · ≤bn, then a1b1 + a2b2 + · · · + anbn n ≥a1 + a2 + · · · + an n · b1 + b2 + · · · + bn n . Applying the rearrangement inequality several times, we get a1b1 + · · · + anbn = a1b1 + a2b2 + · · · + anbn, a1b1 + · · · + anbn ≥a1b2 + a2b3 + · · · + anb1, a1b1 + · · · + anbn ≥a1b3 + a2b4 + · · · + anb2, . . . . . . . . . a1b1 + · · · + anbn ≥a1bn + a2b1 + · · · + anbn−1, and adding together all the expressions, we obtain n (a1b1 + · · · + anbn) ≥(a1 + · · · + an) (b1 + · · · + bn) . The equality holds when a1 = a2 = · · · = an or b1 = b2 = · · · = bn. Exercise 1.60. Any three positive real numbers a, b and c satisfy the following inequality: a3 + b3 + c3 ≥a2b + b2c + c2a. 1.4 A wonderful inequality 19 Exercise 1.61. Any three positive real numbers a, b and c, with abc = 1, satisfy a3 + b3 + c3 + (ab)3 + (bc)3 + (ca)3 ≥2(a2b + b2c + c2a). Exercise 1.62. Any three positive real numbers a, b and c satisfy a2 b2 + b2 c2 + c2 a2 ≥b a + c b + a c . Exercise 1.63. Any three positive real numbers a, b and c satisfy 1 a2 + 1 b2 + 1 c2 ≥a + b + c abc . Exercise 1.64. If a, b and c are the lengths of the sides of a triangle, prove that a b + c −a + b c + a −b + c a + b −c ≥3. Exercise 1.65. If a1, a2, . . . , an ∈R+ and s = a1 + a2 + · · · + an, then a1 s −a1 + a2 s −a2 + · · · + an s −an ≥ n n −1. Exercise 1.66. If a1, a2, . . . , an ∈R+ and s = a1 + a2 + · · · + an, then s s −a1 + s s −a2 + · · · + s s −an ≥ n2 n −1. Exercise 1.67. If a1, a2, . . . , an ∈R+ and a1 + a2 + · · · + an = 1, then a1 2 −a1 + a2 2 −a2 + · · · + an 2 −an ≥ n 2n −1. Exercise 1.68. (Quadratic mean-arithmetic mean inequality) Let x1, . . . , xn ∈ R+, then x2 1 + x2 2 + · · · + x2 n n ≥x1 + x2 + · · · + xn n . Exercise 1.69. For positive real numbers a, b, c such that a + b + c = 1, prove that ab + bc + ca ≤1 3. Exercise 1.70. (Harmonic, geometric and arithmetic mean) Let x1, . . . , xn ∈R+, prove that n 1 x1 + 1 x2 + · · · + 1 xn ≤ n √x1x2 · · · xn ≤x1 + x2 + · · · + xn n . And the equalities hold if and only if x1 = x2 = · · · = xn. 20 Numerical Inequalities Exercise 1.71. Let a1, a2, . . . , an be positive numbers with a1a2 · · · an = 1. Prove that an−1 1 + an−1 2 + · · · + an−1 n ≥1 a1 + 1 a2 + · · · + 1 an . Exercise 1.72. (China, 1989) Let a1, a2, . . . , an be positive numbers such that a1 + a2 + · · · + an = 1. Prove that a1 √1 −a1 + · · · + an √1 −an ≥ 1 √n −1(√a1 + · · · + √an). Exercise 1.73. Let a, b and c be positive numbers such that a + b + c = 1. Prove that (i) √4a + 1 + √ 4b + 1 + √4c + 1 < 5, (ii) √4a + 1 + √ 4b + 1 + √4c + 1 ≤ √ 21. Exercise 1.74. Let a, b, c, d ∈R+ with ab + bc + cd + da = 1, prove that a3 b + c + d + b3 a + c + d + c3 a + b + d + d3 a + b + c ≥1 3. Exercise 1.75. Let a, b, c be positive numbers with abc = 1, prove that a b + b c + c a ≥a + b + c. Exercise 1.76. Let x1, x2, . . . , xn (n > 2) be real numbers such that the sum of any n−1 of them is greater than the element left out of the sum. Set s = n k=1 xk. Prove that n k=1 x2 k s −2xk ≥ s n −2. 1.5 Convex functions A function f : [a, b] →R is called convex in the interval I = [a, b] if for any t ∈[0, 1] and for all a ≤x < y ≤b, the following inequality holds: f(ty + (1 −t)x) ≤tf(y) + (1 −t)f(x). (1.4) Geometrically, the inequality in the deﬁnition means that the graph of f between x and y is below the segment which joins the points (x, f(x)) and (y, f(y)). 1.5 Convex functions 21 x y (x, f(x)) (y, f(y)) In fact, the equation of the line joining the points (x, f(x)) and (y, f(y)) is expressed as L(s) = f(x) + f(y) −f(x) y −x (s −x). Then, evaluating at the point s = ty + (1 −t)x, we get L(ty + (1 −t)x) = f(x) + f(y) −f(x) y −x (t(y −x)) = f(x) + t(f(y) −f(x)) = tf(y) + (1 −t)f(x). Hence, Inequality (1.4) is equivalent to f(ty + (1 −t)x) ≤L(ty + (1 −t)x). Proposition 1.5.1. (1) If f is convex in the interval [a, b], then it is convex in any subinterval [x, y] ⊂[a, b] . (2) If f is convex in [a, b], then for any x, y ∈[a, b], we have that f x + y 2 ≤1 2(f(x) + f(y)). (1.5) (3) (Jensen’s inequality) If f is convex in [a, b], then for any t1, . . . , tn ∈[0, 1], with n i=1 ti = 1, and for x1, . . . , xn ∈[a, b], we can deduce that f(t1x1 + · · · + tnxn) ≤t1f(x1) + · · · + tnf(xn). (4) In particular, for x1, . . . , xn ∈[a, b], we can establish that f x1 + · · · + xn n ≤1 n (f(x1) + · · · + f(xn)) . 22 Numerical Inequalities Proof. (1) We leave the proof as an exercise for the reader. (2) It is suﬃcient to choose t = 1 2 in (1.4). (3) We have f (t1x1 + · · · + tnxn) = f((1 −tn) ( t1 1 −tn x1 + · · · + tn−1 1 −tn xn−1) + tnxn) ≤(1 −tn) f t1 1 −tn x1 + · · · + tn−1 1 −tn xn−1 + tnf(xn), by convexity ≤(1 −tn) t1 1 −tn f(x1) + · · · + tn−1 1 −tn f(xn−1) + tnf(xn), by induction = t1f(x1) + · · · + tnf(xn). (4) We only need to apply (3) using t1 = t2 = · · · = tn = 1 n. □ Observations 1.5.2. (i) We can see that (4) holds true only under the assumption that f satisﬁes the relation f x+y 2 ≤f(x)+f(y) 2 for any x, y ∈[a, b]. (ii) We can observe that (3) is true for t1, . . . , tn ∈[0, 1] rational numbers, only under the condition that f satisﬁes the relation f x+y 2 ≤f(x)+f(y) 2 for any x, y ∈[a, b]. We will prove (i) using induction. Let us call Pn the assertion f x1 + · · · + xn n ≤1 n (f(x1) + · · · + f(xn)) for x1, . . . , xn ∈[a, b]. It is clear that P1 and P2 are true. Now, we will show that Pn ⇒Pn−1. Let x1, . . . , xn ∈[a, b] and let y = x1+···+xn−1 n−1 . Since Pn is true, we can establish that f x1 + · · · + xn−1 + y n ≤1 nf (x1) + · · · + 1 nf(xn−1) + 1 nf(y). But the left side is f(y), therefore n · f(y) ≤f (x1) + · · · + f(xn−1) + f(y), and f(y) ≤ 1 n −1 (f (x1) + · · · + f(xn−1)) . Finally, we can observe that Pn ⇒P2n. Let D = f x1+···+xn+xn+1+···+x2n 2n = f u+v 2 , where u = x1+···+xn n and v = xn+1+···+x2n n . Since f u+v 2 ≤1 2(f(u) + f(v)), we have that D ≤1 2(f(u) + f(v)) = 1 2 f x1 + · · · + xn n + f xn+1 + · · · + x2n n ≤1 2n (f (x1) + · · · + f(xn) + f(xn+1) + · · · + f(x2n)) , 1.5 Convex functions 23 where we have used twice the statement that Pn is true. To prove (ii), our starting point will be the assertion that f x1+···+xn n ≤ 1 n(f(x1) + · · · + f(xn)) for x1, . . . , xn ∈[a, b] and n ∈N. Let t1 = r1 s1 , . . . , tn = rn sn be rational numbers in [0, 1] with n i=1 ti = 1. If m is the least common multiple of the si’s, then ti = pi m with pi ∈N and n i=1 pi = m, hence f(t1x1 + · · · + tnxn) = f p1 m x1 + · · · + pn m xn = f ⎛ ⎜ ⎝1 m ⎡ ⎢ ⎣(x1 + · · · + x1) p1−terms + · · · + (xn + · · · + xn) pn−terms ⎤ ⎥ ⎦ ⎞ ⎟ ⎠ ≤1 m ⎡ ⎢ ⎣(f(x1) + · · · + f(x1)) p1−terms + · · · + (f(xn) + · · · + f(xn)) pn−terms ⎤ ⎥ ⎦ = p1 m f(x1) + · · · + pn m f(xn) = t1f(x1) + · · · + tnf(xn). Observation 1.5.3. If f : [a, b] →R is a continuous2 function on [a, b] and satisﬁes hypothesis (2) of the proposition, then f is convex. We have seen that if f satisﬁes (2), then f(qx + (1 −q)y) ≤qf(x) + (1 −q)f(y) for any x, y ∈[a, b] and q ∈[0, 1] rational number. Since any real number t can be approximated by a sequence of rational numbers qn, and if these qn belong to [0, 1], we can deduce that f(qnx + (1 −qn)y) ≤qnf(x) + (1 −qn)f(y). Now, by using the continuity of f and taking the limit, we get f(tx + (1 −t)y) ≤tf(x) + (1 −t)f(y). We say that a function f : [a, b] →R is concave if −f is convex. 2A function f : [a, b] →R is continuous at a point c ∈[a, b] if lim x→c f(x) = f(c), and f is continuous on [a, b] if it is continous in every point of the interval. Equivalently, f is continuous at c if for every sequence of points {cn} that converges to c, the sequence {f(cn)} converges to f(c). 24 Numerical Inequalities Observation 1.5.4. A function f : [a, b] →R is concave if and only if f(ty + (1 −t)x) ≥tf(y) + (1 −t)f(x) for 0 ≤t ≤1 and a ≤x < y ≤b. Now, we will consider some criteria to decide whether a function is convex. Criterion 1.5.5. A function f : [a, b] →R is convex if and only if the set {(x, y)\| a ≤x ≤b, f(x) ≤y} is convex.3 Proof. Suppose that f is convex and let A = (x1, y1) and B = (x2, y2) be two points in the set U = {(x, y) \| a ≤x ≤b, f(x) ≤y}. To prove that tB + (1 −t)A = (tx2 + (1 −t)x1, ty2 + (1 −t)y1) belongs to U, it is suﬃcient to demonstrate that a ≤tx2 + (1−t)x1 ≤b and f(tx2 + (1−t)x1) ≤ty2 + (1−t)y1. The ﬁrst condition follows immediately since x1 and x2 belong to [a, b]. As for the second condition, since f is convex, it follows that f(tx2 + (1 −t)x1) ≤tf(x2) + (1 −t)f(x1). Moreover, since f(x2) ≤y2 and f(x1) ≤y1, we can deduce that f(tx2 + (1 −t)x1) ≤ty2 + (1 −t)y1. Conversely, we will observe that f is convex if U is convex. Let x1, x2 ∈[a, b] and let us consider A = (x1, f(x1)) and B = (x2, f(x2)). Clearly A and B belong to U, and since U is convex, the segment that joins them belongs to U, that is, the points of the form tB + (1 −t)A for t ∈[0, 1]. Thus, (tx2 + (1 −t)x1, tf(x2) + (1 −t)f(x1)) ∈U, but this implies that f(tx2 + (1 −t)x1) ≤tf(x2) + (1 −t)f(x1). Hence f is convex. □ Criterion 1.5.6. A function f : [a, b] →R is convex if and only if, for each x0 ∈ [a, b], the function P(x) = f(x)−f(x0) x−x0 is non-decreasing for x ̸= x0. Proof. Suppose that f is convex. To prove that P(x) is non-decreasing, we take x < y and then we show that P(x) ≤P(y). One of the following three situations can arise: x0 < x < y, x < x0 < y or x < y < x0. Let us consider the ﬁrst of these 3A subset C of the plane is convex if for any pair of points A, B in C, the segment determined by these points belongs entirely to C. Since the segment between A and B is the set of points of the form tB + (1 −t)A, with 0 ≤t ≤1, the condition is that any point described by this expression belongs to C. 1.5 Convex functions 25 cases and then the other two can be proved in a similar way. First note that P(x) ≤P(y) ⇔ f(x) −f(x0) x −x0 ≤f(y) −f(x0) y −x0 ⇔ (f(x) −f(x0))(y −x0) ≤(f(y) −f(x0))(x −x0) ⇔ f(x)(y −x0) ≤f(y)(x −x0) + f(x0)(y −x) ⇔ f(x) ≤f(y)x −x0 y −x0 + f(x0) y −x y −x0 ⇔ f x −x0 y −x0 y + y −x y −x0 x0 ≤f(y)x −x0 y −x0 + f(x0) y −x y −x0 . The result follows immediately. □ Criterion 1.5.7. If the function f : [a, b] →R is diﬀerentiable4 with a non-decreasing derivative, then f is convex. In particular, if f is twice diﬀerentiable and f ′′(x) ≥0, then the function is convex. Proof. It is clear that f ′′(x) ≥0, for x ∈[a, b], implies that f ′(x) is non-decreasing. We see that if f ′(x) is non-decreasing, the function is convex. Let x = tb + (1 −t)a be a point on [a, b]. Recalling the mean value theorem,5 we know there exist c ∈(a, x) and d ∈(x, b) such that f(x) −f(a) = (x −a)f ′(c) = t(b −a)f ′(c), f(b) −f(x) = (b −x)f ′(d) = (1 −t)(b −a)f ′(d). Then, since f ′(x) is non-decreasing, we can deduce that (1 −t) (f(x) −f(a)) = t(1 −t)(b −a)f ′(c) ≤t(1 −t)(b −a)f ′(d) = t(f(b) −f(x)). After rearranging terms we get f(x) ≤tf(b) + (1 −t)f(a). □ Let us present one geometric interpretation of convexity (and concavity). Let x, y, z be points in the interval [a, b] with x < y < z. If the vertices of the triangle XY Z have coordinates X = (x, f(x)), Y = (y, f(y)), Z = (z, f(z)), then the area of the triangle is given by Δ = 1 2 det A, where A = ⎛ ⎝ 1 x f(x) 1 y f(y) 1 z f(z) ⎞ ⎠. 4A function f : [a, b] →R is diﬀerentiable in a point c ∈[a, b] if the function f′(c) = lim x→c f(x)−f(c) x−c exists and f is diﬀerentiable in A ⊂[a, b] if it is diﬀerentiable in every point of A. 5Mean value theorem. For a continuous function f : [a, b] →R, which is diﬀerentiable in (a, b), there exists a number x ∈(a, b) such that f ′(x)(b −a) = f(b) −f(a). See [21, page 169]. 26 Numerical Inequalities The area can be positive or negative, this will depend on whether the triangle XY Z is positively oriented (anticlockwise oriented) or negatively oriented. For a convex function, we have that Δ > 0 and for a concave function, Δ < 0, as shown in the following graphs. x y z (x, f(x)) (y, f(y)) (z, f(z)) x y z (x, f(x)) (y, f(y)) (z, f(z)) In fact, Δ > 0 ⇔ det A > 0 ⇔ (z −y)f(x) −(z −x)f(y) + (y −x)f(z) > 0 ⇔ f(y) < z −y z −xf(x) + y −x z −xf(z). If we take t = y−x z−x, we have 0 < t < 1, 1 −t = z−y z−x, y = tz + (1 −t)x and f(tz + (1 −t)x) < tf(z) + (1 −t)f(x). Now, let us introduce several examples where convex functions are used to establish inequalities. Example 1.5.8. The function f(x) = xn, n ≥1, is convex in R+ and the function f(x) = xn, with n even, is also convex in R. This follows from the fact that f ′′(x) = n(n −1)xn−2 ≥0 in each case. As an application of this we get the following. (i) Since a+b 2 2 ≤ a2+b2 2 , we can deduce that a+b 2 ≤ a2+b2 2 , which is the inequality between the arithmetic mean and the quadratic mean. (ii) Since a+b 2 n ≤an+bn 2 , we can deduce that an + bn ≥ 1 2n−1 , for a and b positive numbers such that a + b = 1. (iii) If a and b are positive numbers, 1 + a b n + 1 + b a n ≥2n+1. This follows 1.5 Convex functions 27 from 2n = f(2) ≤f a+b a + a+b b 2 ≤1 2 ' f 1 + a b + f 1 + b a ( = 1 2 ' 1 + a b n + 1 + b a n( . Example 1.5.9. The exponential function f(x) = ex is convex in R, since f ′′(x) = ex > 0, for every x ∈R. Let us observe several ways in which this property can be used. (i) (Weighted AM-GM inequality) If x1, . . . , xn, t1, . . . , tn are positive numbers and n i=1 ti = 1, then xt1 1 · · · xtn n ≤t1x1 + · · · + tnxn. In fact, since xti i = eti log xi and ex is convex, we can deduce that xt1 1 · · · xtn n = et1 log x1 · · · etn log xn = et1 log x1+···+tn log xn ≤t1elog x1 + · · · + tnelog xn = t1x1 + · · · + tnxn. In particular, if we take ti = 1 n, for 1 ≤i ≤n, we can produce another proof of the inequality between the arithmetic mean and the geometric mean for n numbers. (ii) (Young’s inequality) Let x, y be positive real numbers. If a, b > 0 satisfy the condition 1 a + 1 b = 1, then xy ≤1 axa + 1 byb. We only need to apply part (i) as follows: xy = (xa) 1 a yb 1 b ≤1 axa + 1 byb. (iii) (H¨ older’s inequality) Let x1, x2, . . . , xn, y1, y2, . . . , yn be positive numbers and a, b > 0 such that 1 a + 1 b = 1, then n i=1 xiyi ≤ n i=1 xa i 1/a n i=1 yb i 1/b . Let us ﬁrst assume that n i=1 xa i = n i=1 yb i = 1. Using part (ii), xiyi ≤1 axa i + 1 b yb i , then n i=1 xiyi ≤1 a n i=1 xa i + 1 b n i=1 yb i = 1 a + 1 b = 1. 28 Numerical Inequalities Now, suppose that n i=1 xa i = A and n i=1 yb i = B. Let us take x′ i = xi A1/a and y′ i = yi B1/b . Since n i=1 (x′ i)a = n i=1 xa i A = 1 and n i=1 (y′ i)b = n i=1 yb i B = 1, we can deduce that 1 ≥ n i=1 x′ iy′ i = n i=1 xiyi A1/aB1/b = 1 A1/aB1/b n i=1 xiyi. Therefore, n i=1 xiyi ≤A1/aB1/b. If we choose a = b = 2, we get the Cauchy-Schwarz inequality. Let us introduce a consequence of H¨ older’s inequality, which is a generaliza-tion of the triangle inequality. Example 1.5.10 (Minkowski’s inequality). Let a1, a2, . . . , an, b1, b2, . . . , bn be positive numbers and p > 1, then n k=1 (ak + bk)p 1 p ≤ n k=1 (ak)p 1 p + n k=1 (bk)p 1 p . We note that (ak + bk)p = ak(ak + bk)p−1 + bk(ak + bk)p−1, so that n k=1 (ak + bk)p = n k=1 ak(ak + bk)p−1 + n k=1 bk(ak + bk)p−1. (1.6) We apply H¨ older’s inequality to each term of the sum on the right-hand side of (1.6), with q such that 1 p + 1 q = 1, to get n k=1 ak(ak + bk)p−1 ≤ n k=1 (ak)p 1 p n k=1 (ak + bk)q(p−1) 1 q , n k=1 bk(ak + bk)p−1 ≤ n k=1 (bk)p 1 p n k=1 (ak + bk)q(p−1) 1 q . Putting these inequalities into (1.6), and noting that q(p −1) = p, yields the required inequality. Note that Minkowski’s inequality is an equality if we allow p = 1. For 0 < p < 1, the inequality is reversed. 1.5 Convex functions 29 Example 1.5.11. (Short list IMO, 1998) If r1, . . . , rn are real numbers greater than 1, prove that 1 1 + r1 + · · · + 1 1 + rn ≥ n n √r1 · · · rn + 1. First note that the function f(x) = 1 1+ex is convex for R+, since f ′(x) = −ex (1+ex)2 and f ′′(x) = ex(ex−1) (ex+1)3 ≥0 for x > 0. Now, if ri > 1, then ri = exi for some xi > 0. Since f(x) = 1 1+ex is convex, we can establish that 1 e( x1+···+xn n ) + 1 ≤1 n 1 1 + ex1 + · · · + 1 1 + exn , hence n n √r1 · · · rn + 1 ≤ 1 1 + r1 + · · · + 1 1 + rn . Example 1.5.12. (China, 1989) Prove that for any n real positive numbers x1, . . . , xn such that n i=1 xi = 1, we have n i=1 xi √1 −xi ≥ n i=1 √xi √n −1 . We will use the fact that the function f(x) = x √1−x is convex in (0, 1), since f ′′(x) > 0, 1 n n i=1 xi √1 −xi = 1 n n i=1 f(xi) ≥f n i=1 1 n xi = f 1 n = 1 √n√n −1, hence n i=1 xi √1 −xi ≥ √n √n −1. It is left to prove that n i=1 √xi ≤√n, but this follows from the Cauchy-Schwarz inequality, n i=1 √xi ≤ n i=1 xi n i=1 1 = √n. Example 1.5.13. (Hungary–Israel, 1999) Let k and l be two given positive integers, and let aij, 1 ≤i ≤k and 1 ≤j ≤l, be kl given positive numbers. Prove that if q ≥p > 0, then ⎛ ⎝ l j=1 k i=1 ap ij q p ⎞ ⎠ 1 q ≤ ⎛ ⎜ ⎝ k i=1 ⎛ ⎝ l j=1 aq ij ⎞ ⎠ p q ⎞ ⎟ ⎠ 1 p . 30 Numerical Inequalities Deﬁne bj = k i=1 ap ij for j = 1, 2, . . ., l, and denote the left-hand side of the required inequality by L and the right-hand side by R. Then Lq = l j=1 b q p j = l j=1 b q−p p j k i=1 ap ij = k i=1 ⎛ ⎝ l j=1 b q−p p j ap ij ⎞ ⎠. Using H¨ older’s inequality we obtain Lq ≤ k i=1 ⎡ ⎢ ⎣ ⎛ ⎝ l j=1 b q−p p j q q−p ⎞ ⎠ q−p q ⎛ ⎝ l j=1 (ap ij) q p ⎞ ⎠ p q ⎤ ⎥ ⎦ = k i=1 ⎡ ⎢ ⎣ ⎛ ⎝ l j=1 b q p j ⎞ ⎠ q−p q ⎛ ⎝ l j=1 aq ij ⎞ ⎠ p q ⎤ ⎥ ⎦ = ⎛ ⎝ l j=1 b q p j ⎞ ⎠ q−p q · ⎡ ⎢ ⎣ k i=1 ⎛ ⎝ l j=1 aq ij ⎞ ⎠ p q ⎤ ⎥ ⎦= Lq−pRp. The inequality L ≤R follows by dividing both sides of Lq ≤Lq−pRp by Lq−p and taking the p-th root. Exercise 1.77. (i) For a, b ∈R+, with a + b = 1, prove that a + 1 a 2 + b + 1 b 2 ≥25 2 . (ii) For a, b, c ∈R+, with a + b + c = 1, prove that a + 1 a 2 + b + 1 b 2 + c + 1 c 2 ≥100 3 . Exercise 1.78. For 0 ≤a, b, c ≤1, prove that a b + c + 1 + b c + a + 1 + c a + b + 1 + (1 −a)(1 −b)(1 −c) ≤1. 1.5 Convex functions 31 Exercise 1.79. (Russia, 2000) For real numbers x, y such that 0 ≤x, y ≤1, prove that 1 √ 1 + x2 + 1 1 + y2 ≤ 2 √1 + xy . Exercise 1.80. Prove that the function f(x) = sin x is concave in the interval [0, π]. Use this to verify that the angles A, B, C of a triangle satisfy sin A+sin B+sin C ≤ 3 2 √ 3. Exercise 1.81. If A, B, C, D are angles belonging to the interval [0, π], then (i) sin A sin B ≤sin2 A+B 2 and the equality holds if and only if A = B, (ii) sin A sin B sin C sin D ≤sin4 A+B+C+D 4 , (iii) sin A sin B sin C ≤sin3 A+B+C 3 , Moreover, if A, B, C are the internal angles of a triangle, then (iv) sin A sin B sin C ≤3 8 √ 3, (v) sin A 2 sin B 2 sin C 2 ≤1 8, (vi) sin A + sin B + sin C = 4 cos A 2 cos B 2 cos C 2 . Exercise 1.82. (Bernoulli’s inequality) (i) For any real number x > −1 and for every positive integer n, we have (1 + x)n ≥1 + nx. (ii) Use this inequality to provide another proof of the AM-GM inequality. Exercise 1.83. (Sch¨ ur’s inequality) If x, y, z are positive real numbers and n is a positive integer, we have xn(x −y)(x −z) + yn(y −z)(y −x) + zn(z −x)(z −y) ≥0. For the case n = 1, the inequality can take one of the following forms: (a) x3 + y3 + z3 + 3xyz ≥xy(x + y) + yz(y + z) + zx(z + x). (b) xyz ≥(x + y −z)(y + z −x)(z + x −y). (c) If x + y + z = 1, 9xyz + 1 ≥4(xy + yz + zx). Exercise 1.84. (Canada, 1992) For any three non-negative real numbers x, y and z we have x(x −z)2 + y(y −z)2 ≥(x −z)(y −z)(x + y −z). Exercise 1.85. If a, b, c are positive real numbers, prove that a (b + c)2 + b (c + a)2 + c (a + b)2 ≥ 9 4(a + b + c). 32 Numerical Inequalities Exercise 1.86. Let a, b and c be positive real numbers, prove that 1 + 3 ab + bc + ca ≥ 6 a + b + c. Moreover, if abc = 1, prove that 1 + 3 a + b + c ≥ 6 ab + bc + ca. Exercise 1.87. (Power mean inequality) Let x1, x2, . . . , xn be positive real numbers and let t1, t2, . . . , tn be positive real numbers adding up to 1. Let r and s be two nonzero real numbers such that r > s. Prove that (t1xr 1 + · · · + tnxr n) 1 r ≥(t1xs 1 + · · · + tnxs n) 1 s with equality if and only if x1 = x2 = · · · = xn. Exercise 1.88. (Two extensions of H¨ older’s inequality) Let x1, x2, . . . , xn, y1, y2, . . . , yn, z1, z2, . . . , zn be positive real numbers. (i) If a, b, c are positive real numbers such that 1 a + 1 b = 1 c, then n i=1 (xiyi)c ) 1 c ≤ n i=1 xia ) 1 a n i=1 yib ) 1 b . (ii) If a, b, c are positive real numbers such that 1 a + 1 b + 1 c = 1, then n i=1 xiyizi ≤ n i=1 xia ) 1 a n i=1 yib ) 1 b n i=1 zic ) 1 c . Exercise 1.89. (Popoviciu’s inequality) If I is an interval and f : I →R is a convex function, then for a, b, c ∈I the following inequality holds: 2 3 ' f a + b 2 + f b + c 2 + f c + a 2 ( ≤f(a) + f(b) + f(c) 3 + f a + b + c 3 . Exercise 1.90. Let a, b, c be non-negative real numbers. Prove that (i) a2 + b2 + c2 + 3 3 √ a2b2c2 ≥2(ab + bc + ca), (ii) a2 + b2 + c2 + 2abc + 1 ≥2(ab + bc + ca). Exercise 1.91. Let a, b, c be positive real numbers. Prove that b + c a + c + a b + a + b c ≥4 a b + c + b c + a + c a + b . 1.6 A helpful inequality 33 1.6 A helpful inequality First, let us study two very useful algebraic identities that are deduced by consid-ering a special factor of a3 + b3 + c3 −3abc. Let P denote the cubic polynomial P(x) = x3 −(a + b + c)x2 + (ab + bc + ca)x −abc, which has a, b and c as its roots. By substituting a, b, c in the polynomial, we obtain a3 −(a + b + c)a2 + (ab + bc + ca)a −abc = 0, b3 −(a + b + c)b2 + (ab + bc + ca)b −abc = 0, c3 −(a + b + c)c2 + (ab + bc + ca)c −abc = 0. Adding up these three equations yields a3 + b3 + c3 −3abc = (a + b + c)(a2 + b2 + c2 −ab −bc −ca). (1.7) It immediately follows that if a + b + c = 0, then a3 + b3 + c3 = 3abc. Note also that the expression a2 + b2 + c2 −ab −bc −ca can also be written as a2 + b2 + c2 −ab −bc −ca = 1 2[(a −b)2 + (b −c)2 + (c −a)2]. (1.8) In this way, we obtain another version of identity (1.7), a3 + b3 + c3 −3abc = 1 2(a + b + c)[(a −b)2 + (b −c)2 + (c −a)2]. (1.9) This presentation of the identity leads to a short proof of the AM-GM inequality for three variables. From (1.9) it is clear that if a, b, c are positive numbers, then a3 + b3 + c3 ≥3abc. Now, if x, y, z are positive numbers, taking a = 3 √x, b = 3 √y and c = 3 √z will lead us to x + y + z 3 ≥ 3 √xyz with equality if and only if x = y = z. Note that identity (1.8) provides another proof of Exercise 1.27. Exercise 1.92. For real numbers x, y, z, prove that x2 + y2 + z2 ≥\|xy + yz + zx\|. 34 Numerical Inequalities Exercise 1.93. For positive real numbers a, b, c, prove that a2 + b2 + c2 abc ≥1 a + 1 b + 1 c . Exercise 1.94. If x, y, z are real numbers such that x < y < z, prove that (x −y)3 + (y −z)3 + (z −x)3 > 0. Exercise 1.95. Let a, b, c be the side lengths of a triangle. Prove that 3 a3 + b3 + c3 + 3abc 2 ≥max{a, b, c}. Exercise 1.96. (Romania, 2007) For non-negative real numbers x, y, z, prove that x3 + y3 + z3 3 ≥xyz + 3 4\|(x −y)(y −z)(z −x)\|. Exercise 1.97. (UK, 2008) Find the minimum of x2 + y2 + z2, where x, y, z are real numbers such that x3 + y3 + z3 −3xyz = 1. A very simple inequality which may be helpful for proving a large number of algebraic inequalities is the following. Theorem 1.6.1 (A helpful inequality). If a, b, x, y are real numbers and x, y > 0, then the following inequality holds: a2 x + b2 y ≥(a + b)2 x + y . (1.10) Proof. The proof is quite simple. Clearing out denominators, we can express the inequality as a2y(x + y) + b2x(x + y) ≥(a + b)2xy, which simpliﬁes to become the obvious (ay −bx)2 ≥0. We see that the equality holds if and only if ay = bx, that is, if and only if a x = b y. Another form to prove the inequality is using the Cauchy-Schwarz inequality in the following way: (a + b)2 = a √x √x + b √y √y 2 ≤ a2 x + b2 y (x + y). □ Using the above theorem twice, we can extend the inequality to three pairs of numbers a2 x + b2 y + c2 z ≥(a + b)2 x + y + c2 z ≥(a + b + c)2 x + y + z , 1.6 A helpful inequality 35 and a simple inductive argument shows that a2 1 x1 + a2 2 x2 + · · · + a2 n xn ≥(a1 + a2 + · · · + an)2 x1 + x2 + · · · + xn (1.11) for all real numbers a1, a2, . . . , an and x1, x2, . . . , xn > 0, with equality if and only if a1 x1 = a2 x2 = · · · = an xn . Inequality (1.11) is also called the Cauchy-Schwarz inequality in Engel form or Arthur Engel’s Minima Principle. As a ﬁrst application of this inequality, we will present another proof of the Cauchy-Schwarz inequality. Let us write a2 1 + a2 2 + · · · + a2 n = a2 1b2 1 b2 1 + a2 2b2 2 b2 2 + · · · + a2 nb2 n b2 n , then a2 1b2 1 b2 1 + a2 2b2 2 b2 2 + · · · + a2 nb2 n b2 n ≥(a1b1 + a2b2 + · · · + anbn)2 b2 1 + b2 2 + · · · + b2 n . Thus, we conclude that (a2 1 + a2 2 + · · · + a2 n)(b2 1 + b2 2 + · · · + b2 n) ≥(a1b1 + a2b2 + · · · + anbn)2 and the equality holds if and only if a1 b1 = a2 b2 = · · · = an bn . It is worth to mention that there are other forms of the Cauchy-Schwarz inequality in Engel form. Example 1.6.2. Let a1, . . . , an, b1, . . . , bn be positive real numbers. Prove that (i) a1 b1 + · · · + an bn ≥(a1 + · · · + an)2 a1b1 + · · · + anbn , (ii) a1 b2 1 + · · · + an b2 n ≥ 1 a1 + · · · + an a1 b1 + · · · + an bn 2 . Both inequalities are direct consequence of inequality (1.11), as we can see as follows. (i) a1 b1 + · · · + an bn = a2 1 a1b1 + · · · + a2 n anbn ≥(a1 + · · · + an)2 a1b1 + · · · + anbn , (ii) a1 b2 1 + · · · + an b2 n = a2 1 b2 1 a1 + · · · + a2 n b2 n an ≥ 1 a1 + · · · + an a1 b1 + · · · + an bn 2 . 36 Numerical Inequalities Example 1.6.3. (APMO, 1991) Let a1, . . . , an, b1, . . . , bn be positive real numbers such that a1 + a2 + · · · + an = b1 + b2 + · · · + bn. Prove that a2 1 a1 + b1 + · · · + a2 n an + bn ≥1 2(a1 + · · · + an). Observe that (1.11) implies that a2 1 a1 + b1 + · · · + a2 n an + bn ≥ (a1 + a2 + · · · + an)2 a1 + a2 + · · · + an + b1 + b2 + · · · + bn = (a1 + a2 + · · · + an)2 2(a1 + a2 + · · · + an) = 1 2(a1 + a2 + · · · + an). The following example consists of a proof of the quadratic mean-arithmetic mean inequality. Example 1.6.4 (Quadratic mean-arithmetic mean inequality). For positive real numbers x1, . . . , xn, we have x2 1 + x2 2 + · · · + x2 n n ≥x1 + x2 + · · · + xn n . Observe that using (1.11) leads us to x2 1 + x2 2 + · · · + x2 n n ≥(x1 + x2 + · · · + xn)2 n2 , which implies the above inequality. In some cases the numerators are not squares, but a simple trick allows us to write them as squares, so that we can use the inequality. Our next application shows this trick and oﬀers a shorter proof for Example 1.4.9. Example 1.6.5. (IMO, 1995) Let a, b, c be positive real numbers such that abc = 1. Prove that 1 a3(b + c) + 1 b3(a + c) + 1 c3(a + b) ≥3 2. Observe that 1 a3(b + c) + 1 b3(c + a) + 1 c3(a + b) = 1 a2 a(b + c) + 1 b2 b(c + a) + 1 c2 c(a + b) ≥ ( 1 a + 1 b + 1 c)2 2(ab + bc + ca) = ab + bc + ca 2(abc) ≥3 3 (abc)2 2 = 3 2, 1.6 A helpful inequality 37 where the ﬁrst inequality follows from (1.11) and the second is a consequence of the AM-GM inequality. As a further example of the use of inequality (1.11), we provide a simple proof of Nesbitt’s inequality. Example 1.6.6 (Nesbitt’s inequality). For a, b, c ∈R+, we have a b + c + b c + a + c a + b ≥3 2. We multiply the three terms on the left-hand side of the inequality by a a, b b, c c, respectively, and then we use inequality (1.11) to produce a2 a(b + c) + b2 b(c + a) + c2 c(a + b) ≥ (a + b + c)2 2(ab + bc + ca). From Equation (1.8) we know that a2 + b2 + c2 −ab −bc −ca ≥0, that is, (a + b + c)2 ≥3(ab + bc + ca). Therefore a b + c + b c + a + c a + b ≥ (a + b + c)2 2(ab + bc + ca) ≥3 2. Example 1.6.7. (Czech and Slovak Republics, 1999) For a, b and c positive real numbers, prove the inequality a b + 2c + b c + 2a + c a + 2b ≥1. Observe that a b + 2c + b c + 2a + c a + 2b = a2 ab + 2ca + b2 bc + 2ab + c2 ca + 2bc. Then using (1.11) yields a2 ab + 2ca + b2 bc + 2ab + c2 ca + 2bc ≥ (a + b + c)2 3(ab + bc + ca) ≥1, where the last inequality follows in the same way as in the previous example. Exercise 1.98. (South Africa, 1995) For a, b, c, d positive real numbers, prove that 1 a + 1 b + 4 c + 16 d ≥ 64 a + b + c + d. Exercise 1.99. Let a and b be positive real numbers. Prove that 8(a4 + b4) ≥(a + b)4. 38 Numerical Inequalities Exercise 1.100. Let x, y, z be positive real numbers. Prove that 2 x + y + 2 y + z + 2 z + x ≥ 9 x + y + z . Exercise 1.101. Let a, b, x, y, z be positive real numbers. Prove that x ay + bz + y az + bx + z ax + by ≥ 3 a + b. Exercise 1.102. Let a, b, c be positive real numbers. Prove that a2 + b2 a + b + b2 + c2 b + c + c2 + a2 c + a ≥a + b + c. Exercise 1.103. (i) Let x, y, z be positive real numbers. Prove that x x + 2y + 3z + y y + 2z + 3x + z z + 2x + 3y ≥1 2. (ii) (Moldova, 2007) Let w, x, y, z be positive real numbers. Prove that w x + 2y + 3z + x y + 2z + 3w + y z + 2w + 3x + z w + 2x + 3y ≥2 3. Exercise 1.104. (Croatia, 2004) Let x, y, z be positive real numbers. Prove that x2 (x + y)(x + z) + y2 (y + z)(y + x) + z2 (z + x)(z + y) ≥3 4. Exercise 1.105. For a, b, c, d positive real numbers, prove that a b + c + b c + d + c d + a + d a + b ≥2. Exercise 1.106. Let a, b, c, d, e be positive real numbers. Prove that a b + c + b c + d + c d + e + d e + a + e a + b ≥5 2. Exercise 1.107. (i) Prove that, for all positive real numbers a, b, c, x, y, z with a ≥b ≥c and z ≥y ≥x, the following inequality holds: a3 x + b3 y + c3 z ≥(a + b + c)3 3(x + y + z). (ii) (Belarus, 2000) Prove that, for all positive real numbers a, b, c, x, y, z, the following inequality holds: a3 x + b3 y + c3 z ≥(a + b + c)3 3(x + y + z). 1.7 The substitution strategy 39 Exercise 1.108. (Greece, 2008) For x1, x2, . . . , xn positive integers, prove that x2 1 + x2 2 + · · · + x2 n x1 + x2 + · · · + xn kn t ≥x1 · x2 · · · · · xn, where k = max {x1, x2, . . . , xn} and t = min {x1, x2, . . . , xn}. Under which con-dition the equality holds? 1.7 The substitution strategy Substitution is a useful strategy to solve inequality problems. Making an adequate substitution we can, for instance, change the diﬃcult terms of the inequality a little, we can simplify expressions or we can reduce terms. In this section we give some ideas of what can be done with this strategy. As always, the best way to do that is through some examples. One useful suggestion for problems that contain in the hypothesis an extra condition, is to use that condition to simplify the problem. In the next example we apply this technique to eliminate the denominators in order to make the problem easier to solve. Example 1.7.1. If a, b, c are positive real numbers less than 1, with a + b + c = 2, then a 1 −a b 1 −b c 1 −c ≥8. After performing the substitution x = 1 −a, y = 1 −b, z = 1 −c, we obtain that x + y + z = 3 −(a + b + c) = 1, a = 1 −x = y + z, b = z + x, c = x + y. Hence the inequality is equivalent to y + z x z + x y x + y z ≥8, and in turn, this is equivalent to (x + y)(y + z)(z + x) ≥8xyz. This last inequality is quite easy to prove. It is enough to apply three times the AM-GM inequality under the form (x + y) ≥2√xy (see Exercise 1.26). It may be possible that the extra condition is used only as part of the solution, as in the following two examples. Example 1.7.2. (Mexico, 2007) If a, b, c are positive real numbers that satisfy a + b + c = 1, prove that √ a + bc + √ b + ca + √ c + ab ≤2. 40 Numerical Inequalities Using the condition a + b + c = 1, we have that a + bc = a(a + b + c) + bc = (a + b)(a + c), then, by the AM-GM inequality it follows that √ a + bc = (a + b)(a + c) ≤2a + b + c 2 . Similarly, √ b + ca ≤2b + c + a 2 and √ c + ab ≤2c + a + b 2 . Thus, after adding the three inequalities we obtain √ a + bc + √ b + ca + √ c + ab ≤2a + b + c 2 + 2b + c + a 2 + 2c + a + b 2 = 4a + 4b + 4c 2 = 2. The equality holds when a + b = a + c, b + c = b + a and c + a = c + b, that is, when a = b = c = 1 3. Example 1.7.3. If a, b, c are positive real numbers with ab + bc + ca = 1, prove that a √ a2 + 1 + b √ b2 + 1 + c √ c2 + 1 ≤3 2. Note that (a2 + 1) = a2 + ab + bc + ca = (a + b)(a + c). Similarly, b2 + 1 = (b + c)(b + a) and c2 + 1 = (c + a)(c + b). Now, the inequality under consideration is equivalent to a (a + b)(a + c) + b (b + c)(b + a) + c (c + a)(c + b) ≤3 2. Using the AM-GM inequality, applied to every element of the sum on the left-hand side, we obtain a (a + b)(a + c) + b (b + c)(b + a) + c (c + a)(c + b) ≤1 2 a a + b + a a + c + 1 2 b b + c + b b + a + 1 2 c c + a + c c + b = 3 2. Many inequality problems suggest which substitution should be made. In the following example the substitution allows us to make at least one of the terms in the inequality look simpler. Example 1.7.4. (India, 2002) If a, b, c are positive real numbers, prove that a b + b c + c a ≥c + a c + b + a + b a + c + b + c b + a. 1.7 The substitution strategy 41 Making the substitution x = a b , y = b c, z = c a the left-hand side of the inequality is now more simple, x + y + z. Let us see how the right-hand side changes. The ﬁrst element of the sum is modiﬁed as follows: c + a c + b = 1 + a c 1 + b c = 1 + a b b c 1 + b c = 1 + xy 1 + y = x + 1 −x 1 + y . Similarly, a + b a + c = y + 1 −y 1 + z and b + c b + a = z + 1 −z 1 + x. Now, the inequality is equivalent to x −1 1 + y + y −1 1 + z + z −1 1 + x ≥0 with the extra condition xyz = 1. The last inequality can be rewritten as (x2 −1)(z + 1) + (y2 −1)(x + 1) + (z2 −1)(y + 1) ≥0, which in turn is equivalent to x2z + y2x + z2y + x2 + y2 + z2 ≥x + y + z + 3. But, from the AM-GM inequality, we have x2z+y2x+z2y ≥3 3 x3y3z3 = 3. Also, x2 + y2 + z2 ≥1 3(x + y + z)2 = x+y+z 3 (x + y + z) ≥ 3 √xyz(x + y + z) = x + y + z, where the ﬁrst inequality follows from inequality (1.11). In order to make a substitution, sometimes it is necessary to work a little bit beforehand, as we can see in the following example. This example also helps us to point out that we may need to make more than one substitution in the same problem. Example 1.7.5. Let a, b, c be positive real numbers, prove that (a + b)(a + c) ≥2 abc(a + b + c). Dividing both sides of the given inequality by a2 and setting x = b a, y = c a, the inequality becomes (1 + x)(1 + y) ≥2 xy(1 + x + y). Now, dividing both sides by xy and making the substitution r = 1 + 1 x, s = 1 + 1 y, the inequality we need to prove becomes rs ≥2 √ rs −1. This last inequality is equivalent to (rs −2)2 ≥0, which become evident after squaring both sides and doing some algebra. It is a common situation for inequality problems to have several solutions and also to accept several substitutions that help to solve the problem. We will see an instance of this in the next example. 42 Numerical Inequalities Example 1.7.6. (Korea, 1998) If a, b, c are positive real numbers such that a + b + c = abc, prove that 1 √ 1 + a2 + 1 √ 1 + b2 + 1 √ 1 + c2 ≤3 2. Under the substitution x = 1 a, y = 1 b, z = 1 c, condition a + b + c = abc becomes xy + yz + zx = 1 and the inequality becomes equivalent to x √ x2 + 1 + y y2 + 1 + z √ z2 + 1 ≤3 2. This is the third example in this section. Another solution is to make the substitution a = tan A, b = tan B, c = tan C. Since tan A + tan B + tan C = tan A tan B tan C, then A + B + C = π (or a multiple of π). Now, since 1 + tan2 A = (cos A)−2, the inequality is equivalent to cosA + cos B + cos C ≤3 2, which is a valid result as will be shown in Example 2.5.2. Note that the Jensen inequality cannot be applied in this case because the function f(x) = 1 √ 1+x2 is not concave in R+. We note that not all substitutions are algebraic, since there are trigonometric substitutions that can be useful, as is shown in the last example and as we will see next. Also, as will be shown in Sections 2.2 and 2.5 of the next chapter, there are some geometric substitutions that can be used for the same purposes. Example 1.7.7. (Romania, 2002) If a, b, c are real numbers in the interval (0, 1), prove that √ abc + (1 −a)(1 −b)(1 −c) < 1. Making the substitution a = cos2 A, b = cos2 B, c = cos2 C, with A, B, C in the interval (0, π 2 ), we obtain that √1 −a = √ 1 −cos2 A = sin A, √ 1 −b = sin B and √1 −c = sin C. Therefore the inequality is equivalent to cos A cos B cos C + sin A sin B sin C < 1. But observe that cos A cos B cos C + sin A sin B sin C < cos A cos B + sin A sin B = cos(A −B) ≤1. Exercise 1.109. Let x, y, z be positive real numbers. Prove that x3 x3 + 2y3 + y3 y3 + 2z3 + z3 z3 + 2x3 ≥1. Exercise 1.110. (Kazakhstan, 2008) Let x, y, z be positive real numbers such that xyz = 1. Prove that 1 yz + z + 1 zx + x + 1 xy + y ≥3 2. 1.8 Muirhead’s theorem 43 Exercise 1.111. (Russia, 2004) If n > 3 and x1, x2, . . . , xn are positive real numbers with x1x2 · · · xn = 1, prove that 1 1 + x1 + x1x2 + 1 1 + x2 + x2x3 + · · · + 1 1 + xn + xnx1 > 1. Exercise 1.112. (Poland, 2006) Let a, b, c be positive real numbers such that ab + bc + ca = abc. Prove that a4 + b4 ab(a3 + b3) + b4 + c4 bc(b3 + c3) + c4 + a4 ca(c3 + a3) ≥1. Exercise 1.113. (Ireland, 2007) Let a, b, c be positive real numbers, prove that 1 3 bc a + ca b + ca b ≥ a2 + b2 + c2 3 ≥a + b + c 3 . Exercise 1.114. (Romania, 2008) Let a, b, c be positive real numbers with abc = 8. Prove that a −2 a + 1 + b −2 b + 1 + c −2 c + 1 ≤0. 1.8 Muirhead’s theorem In 1903, R.F. Muirhead published a paper containing the study of some algebraic methods applicable to identities and inequalities of symmetric algebraic functions of n variables. While considering algebraic expressions of the form xa1 1 xa2 2 · · · xan n , he an-alyzed symmetric polynomials containing these expressions in order to create a “certain order” in the space of n-tuples (a1, a2, . . . , an) satisfying the condition a1 ≥a2 ≥· · · ≥an. We will assume that xi > 0 for all 1 ≤i ≤n. We will denote by ! F(x1, . . . , xn) the sum of the n! terms obtained from evaluating F in all possible permutations of (x1, . . . , xn). We will consider only the particular case F(x1, . . . , xn) = xa1 1 xa2 2 · · · xan n with xi > 0, ai ≥0. We write [a] = [a1, a2, . . . , an] = 1 n! ! xa1 1 xa2 2 · · · xan n . For instance, for the vari-ables x, y, z > 0 we have that [1, 1] = xy, [1, 1, 1] = xyz, [2, 1, 0] = 1 3![x2(y + z) + y2(x + z) + z2(x + y)]. 44 Numerical Inequalities It is clear that [a] is invariant under any permutation of the (a1, a2, . . . , an) and therefore two sets of a are the same if they only diﬀer in arrangement. We will say that a mean value of the type [a] is a symmetrical mean. In particular, [1, 0, . . ., 0] = (n−1)! n! (x1 + x2 + · · · + xn) = 1 n n i=1 xi is the arithmetic mean and [ 1 n, 1 n, . . . , 1 n] = n! n!(x 1 n 1 · x 1 n 2 · · · x 1 n n ) = n √x1x2 · · · xn is the geometric mean. When a1 +a2 +· · ·+an = 1, [a] is a common generalization of both the arithmetic mean and the geometric mean. If a1 ≥a2 ≥· · · ≥an and b1 ≥b2 ≥· · · ≥bn, usually [b] is not comparable to [a], in the sense that there is an inequality between their associated expressions valid for all n-tuples of non-negative real numbers x1, x2, . . . , xn. Muirhead wanted to compare the values of the symmetric polynomials [a] and [b] for any set of non-negative values of the variables occurring in both polynomials. From now on we denote (a) = (a1, a2, . . . , an). Deﬁnition 1.8.1. We will say that (b) ≺(a) ((b) is majorized by (a)) when (a) and (b) can be rearranged to satisfy the following two conditions: (1) n i=1 bi = n i=1 ai ; (2) ν i=1 bi ≤ ν i=1 ai for all 1 ≤ν < n. It is clear that (a) ≺(a) and that (b) ≺(a) and (c) ≺(b) implies (c) ≺(a). Theorem 1.8.2 (Muirhead’s theorem). [b] ≤[a] for any n-tuple of non-negative numbers (x1, x2, . . . , xn) if and only if (b) ≺(a). Equality takes place only when (b) and (a) are identical or when all the xis are equal. Before going through the proof, which is quite diﬃcult, let us look at some examples. First, it is clear that [2, 0, 0] cannot be compared with [1, 1, 1] because the ﬁrst condition in Deﬁnition 1.8.1 is not satisﬁed, but we can see that [2, 0, 0] ≥ [1, 1, 0], which is equivalent to x2 + y2 + z2 ≥xy + yz + zx. In the same way, we can see that 1. x2 + y2 ≥2xy ⇔[2, 0] ≥[1, 1], 2. x3 + y3 + z3 ≥3xyz ⇔[3, 0, 0] ≥[1, 1, 1], 3. x5 + y5 ≥x3y2 + x2y3 ⇔[5, 0] ≥[3, 2], 4. x2y2 + y2z2 + z2x2 ≥x2yz + y2xz + z2xy ⇔[2, 2, 0] ≥[2, 1, 1], 1.8 Muirhead’s theorem 45 and all these inequalities are satisﬁed if we take for granted Muirhead’s theorem. Proof of Muirhead’s theorem. Suppose that [b] ≤[a] for any n positive numbers x1, x2, . . . , xn. Taking xi = x, for all i, we obtain x bi = [b] ≤[a] = x ai. This can only be true for all x if bi = ai. Next, take x1 = x2 = · · · = xν = x, xν+1 = · · · = xn = 1 and x very large. Since (b) and (a) are in decreasing order, the index of the highest powers of x in [b] and [a] are b1 + b2 + · · · + bν, a1 + a2 + · · · + aν, respectively. Thus, it is clear that the ﬁrst sum can not be greater than the second and this proves (2) in Deﬁnition 1.8.1. The proof in the other direction is more diﬃcult to establish, and we will need a new deﬁnition and two more lemmas. We deﬁne a special type of linear transformation T of the a’s, as follows. Suppose that ak > al, then let us write ak = ρ + τ, al = ρ −τ (0 < τ ≤ρ). If now 0 ≤σ < τ ≤ρ, then a T -transformation is deﬁned by T (ak) = bk = ρ + σ = τ + σ 2τ ak + τ −σ 2τ al, T (al) = bl = ρ −σ = τ −σ 2τ ak + τ + σ 2τ al, T (aν) = aν (ν ̸= k, ν ̸= l). If (b) arises from (a) by a T -transformation, we write b = Ta. The deﬁnition does not necessarily imply that either the (a) or the (b) are in decreasing order. The suﬃciency of our comparability condition will be established if we can prove the following two lemmas. Lemma 1.8.3. If b = Ta, then [b] ≤[a] with equality taking place only when all the xi’s are equal. Proof. We may rearrange (a) and (b) so that k = 1, l = 2. Thus [a] −[b] = [ρ + τ, ρ −τ, a3, . . .] −[ρ + σ, ρ −σ, a3, . . .] = 1 2n! ! xa3 3 · · · xan n (xρ+τ 1 xρ−τ 2 + xρ−τ 1 xρ+τ 2 ) −1 2n! ! xa3 3 · · · xan n (xρ+σ 1 xρ−σ 2 + xρ−σ 1 xρ+σ 2 ) = 1 2n! !(x1x2)ρ−τxa3 3 · · · xan n (xτ+σ 1 −xτ+σ 2 )(xτ−σ 1 −xτ−σ 2 ) ≥0 with equality being the case only when all the xi’s are equal. □ 46 Numerical Inequalities Lemma 1.8.4. If (b) ≺(a), but (b) is not identical to (a), then (b) can be derived from (a) using the successive application of a ﬁnite number of T -transformations. Proof. We call the number of diﬀerences aν−bν which are not zero, the discrepancy between (a) and (b). If the discrepancy is zero, the sets are identical. We will prove the lemma by induction, assuming it to be true when the discrepancy is less than r and proving that it is also true when the discrepancy is r. Suppose then that (b) ≺(a) and that the discrepancy is r > 0. Since n i=1 ai = n i=1 bi, and (aν −bν) = 0, and not all of these diﬀerences are zero, there must be positive and negative diﬀerences, and the ﬁrst which is not zero must be positive because of the second condition of (b) ≺(a). We can therefore ﬁnd k and l such that bk < ak, bk+1 = ak+1, . . . , bl−1 = al−1, bl > al; (1.12) that is, al −bl is the ﬁrst negative diﬀerence and ak −bk is the last positive diﬀerence which precedes it. We take ak = ρ + τ, al = ρ −τ, and deﬁne σ by σ = max(\|bk −ρ\| , \|bl −ρ\|). Then 0 < τ ≤ρ, since ak > al. Also, one (possible both) of bl −ρ = −σ or bk −ρ = σ is true, since bk ≥bl, and σ < τ, since bk < ak and bl > al. Hence 0 ≤σ < τ ≤ρ. We now write a′ k = ρ + σ, a′ l = ρ −σ, a′ ν = aν (ν ̸= k, ν ̸= l). If bk −ρ = σ, a′ k = bk, and if bl −ρ = −σ, then a′ l = bl. Since the pairs ak, bk and al, bl each contributes one unit to the discrepancy r between (b) and (a), the discrepancy between (b) and (a′) is smaller, being equal to r −1 or r −2. Next, comparing the deﬁnition of (a′) with the deﬁnition of the T -transfor-mation, and observing that 0 ≤σ < τ ≤ρ, we can infer that (a′) arises from (a) by a T -transformation. Finally, let us prove that (b) ≺(a′). In order to do that, we must verify that the two conditions of ≺are satisﬁed and that the order of (a′) is non-increasing. For the ﬁrst one, we have a′ k + a′ l = 2ρ = ak + al, n i=1 bi = n i=1 ai = n i=1 a′ i. For the second one, we must prove that b1 + b2 + · · · + bν ≤α′ 1 + α′ 2 + · · · + α′ ν. Now, this is true if ν < k or ν ≥l, as can be established by using the deﬁnition of (a′) and also the second condition of (b) ≺(a). It is true for ν = k, because it is true for ν = k −1 and bk ≤a′ k, and it is true for k < ν < l because it is valid for ν = k and the intervening b and a′ are identical. 1.8 Muirhead’s theorem 47 Finally, we observe that bk ≤ρ + \|bk −ρ\| ≤ρ + σ = a′ k, bl ≥ρ −\|bl −ρ\| ≥ρ −σ = a′ l, and then, using (1.12), a′ k−1 = ak−1 ≥ak = ρ + τ > ρ + σ = a′ k ≥bk ≥bk+1 = ak+1 = a′ k+1, a′ l−1 = al−1 = bl−1 ≥bl ≥a′ l = ρ −σ > ρ −τ = al ≥al+1 = a′ l+1. The inequalities involving a′ are as required. We have thus proved that (b) ≺(a′), a set arising from (a) using a transfor-mation T and having a discrepancy from (b) of less than r. This proves the lemma and completes the proof of Muirhead’s theorem. □ The proof of Muirhead’s theorem demonstrates to us how the diﬀerence be-tween two comparable means can be decomposed as a sum of obviously positive terms by repeated application of the T -transformation. We can produce from this result a new proof for the AM-GM inequality. Example 1.8.5 (The AM-GM inequality). For real positive numbers y1, y2, . . . , yn, y1 + y2 + · · · + yn n ≥ n √y1y2 · · · yn. Note that the AM-GM inequality is equivalent to 1 n n i=1 xn i ≥x1x2 · · · xn, where xi = n √yi. Now, we observe that 1 n n i=1 xn i = [n, 0, 0, . . ., 0] and x1x2 · · · xn = [1, 1, . . . , 1]. By Muirhead’s theorem we can show that [n, 0, 0, . . ., 0] ≥[1, 1, . . . , 1]. Next, we provide another proof for the AM-GM inequality, something we shall do by following the ideas inherent in the proof of Muirhead’s theorem in 48 Numerical Inequalities order to illustrate how it works. 1 n n i=1 xn i −(x1x2 · · · xn) = [n, 0, 0, . . . , 0] −[1, 1, . . ., 1] = ([n, 0, 0, . . . , 0] −[n −1, 1, 0, . . ., 0]) + ([n −1, 1, 0, . . ., 0] −[n −2, 1, 1, 0, . . ., 0]) + ([n −2, 1, 1, 0, . . ., 0] −[n −3, 1, 1, 1, 0, . . ., 0]) + · · · + ([2, 1, 1, . . ., 1] −[1, 1, . . .1]) = 1 2n! !(xn−1 1 −xn−1 2 )(x1 −x2) + !(xn−2 1 −xn−2 2 )(x1 −x2)x3 + !(xn−3 1 −xn−3 2 )(x1 −x2)x3x4 + · · · . Since (xν r −xν s)(xr −xs) > 0, unless xr = xs, the inequality follows. Example 1.8.6. If a, b are positive real numbers, then a2 b + b2 a ≥√a + √ b. Setting x = √a, y = √ b and simplifying, we have to prove x3 + y3 ≥xy(x + y). Using Muirhead’s theorem, we get [3, 0] = 1 2(x3 + y3) ≥1 2xy(x + y) = [2, 1], and thus the result follows. Example 1.8.7. If a, b, c are non-negative real numbers, prove that a3 + b3 + c3 + abc ≥1 7(a + b + c)3. It is not diﬃcult to see that (a + b + c)3 = 3[3, 0, 0] + 18[2, 1, 0] + 36[1, 1, 1]. Then we need to prove that 3[3, 0, 0] + 6[1, 1, 1] ≥1 7(3[3, 0, 0] + 18[2, 1, 0] + 36[1, 1, 1]), that is, 18 7 [3, 0, 0] + 6 −36 7 [1, 1, 1] ≥18 7 [2, 1, 0] 1.8 Muirhead’s theorem 49 or 18 7 ([3, 0, 0] −[2, 1, 0]) + 6 −36 7 [1, 1, 1] ≥0. This follows using the inequalities [3, 0, 0] ≥[2, 1, 0] and [1, 1, 1] ≥0. Example 1.8.8. If a, b, c are non-negative real numbers, prove that a + b + c ≤a2 + b2 2c + b2 + c2 2a + c2 + a2 2b ≤a3 bc + b3 ca + c3 ab. The inequalities are equivalent to the following: 2(a2bc + ab2c + abc2) ≤ab(a2 + b2) + bc(b2 + c2) + ca(c2 + a2) ≤2(a4 + b4 + c4), which is in turn equivalent to [2, 1, 1] ≤[3, 1, 0] ≤[4, 0, 0]. Using Muirhead’s theo-rem we arrive at the result. Exercise 1.115. Prove that any three positive real numbers a, b and c satisfy a5 + b5 + c5 ≥a3bc + b3ca + c3ab. Exercise 1.116. (IMO, 1961) Let a, b, c be the lengths of the sides of a triangle, and let (ABC) denote its area. Prove that 4 √ 3(ABC) ≤a2 + b2 + c2. Exercise 1.117. Let a, b, c be positive real numbers. Prove that a (a + b)(a + c) + b (b + c)(b + a) + c (c + a)(c + b) ≤ 9 4(a + b + c). Exercise 1.118. (IMO, 1964) Let a, b, c be positive real numbers. Prove that a3 + b3 + c3 + 3abc ≥ab(a + b) + bc(b + c) + ca(c + a). Exercise 1.119. (Short list Iberoamerican, 2003) Let a, b, c be positive real num-bers. Prove that a3 b2 −bc + c2 + b3 c2 −ca + a2 + c3 a2 −ab + b2 ≥a + b + c. Exercise 1.120. (Short list IMO, 1998) Let a, b, c be positive real numbers such that abc = 1. Prove that a3 (1 + b)(1 + c) + b3 (1 + c)(1 + a) + c3 (1 + a)(1 + b) ≥3 4. Chapter 2 Geometric Inequalities 2.1 Two basic inequalities The two basic geometric inequalities we will be refering to in this section involve triangles. One of them is the triangle inequality and we will refer to it as D1; the second one is not really an inequality, but it represents an important observation concerning the geometry of triangles which points out that if we know the greatest angle of a triangle, then we know which is the longest side of the triangle; this observation will be denoted as D2. D1. If A, B and C are points on the plane, then AB + BC ≥AC. Moreover, the equality holds if and only if B lies on the line segment AC. D2. In a triangle, the longest side is opposite to the greatest angle and vice versa. Hence, if in the triangle ABC we have ∠A > ∠B, then BC > CA. Exercise 2.1. (i) If a, b, c are positive numbers with a < b + c, b < c + a and c < a + b, then a triangle exists with side lengths a, b and c. (ii) To be able to construct a triangle with side lengths a ≤b ≤c, it is suﬃcient that c < a + b. (iii) It is possible to construct a triangle with sides of length a, b and c if and only if there are positive numbers x, y, z such that a = x + y, b = y + z and c = z + x. Exercise 2.2. (i) If it is possible to construct a triangle with side-lengths a < b < c, then it is possible to construct a triangle with side-lengths √a < √ b < √c. (ii) The converse of (i) is false. 52 Geometric Inequalities (iii) If it is possible to construct a triangle with side-lengths a < b < c, then it is possible to construct a triangle with side-lengths 1 a+b, 1 b+c and 1 c+a. Exercise 2.3. Let a, b, c, d and e be the lengths of ﬁve segments such that it is possible to construct a triangle using any three of them. Prove that there are three of them that form an acute triangle. Sometimes the key to solve a problem lies in the ability to identify certain quantities that can be related to geometric measurements, as in the following example. Example 2.1.1. If a, b, c are positive numbers with a2 + b2 −ab = c2, prove that (a −b)(b −c) ≤0. Since c2 = a2 + b2 −ab = a2 + b2 −2ab cos 60◦, we can think that a, b, c are the lengths of the sides of a triangle such that the measure of the angle opposed to the side of length c is 60◦. The angles of the triangle ABC satisfy ∠A ≤60◦ and ∠B ≥60◦, or ∠A ≥60◦and ∠B ≤60◦; hence, using property D2 we can deduce that a ≤c ≤b or a ≥c ≥b. In any case it follows that (a −b)(b −c) ≤0. Observation 2.1.2. We can also solve the example above without the identiﬁcation of a, b and c with the lengths of the sides of a triangle. First suppose that a ≤b, then the fact that a2 + b2 −ab = c2 implies that a(a −b) = c2 −b2 = (c −b)(c + b), hence c −b ≤0 and therefore (a −b)(b −c) ≤0. Similarly, a ≥b implies c −b ≥0, and hence (a −b)(b −c) ≤0. Another situation where it is not obvious that we can identify the elements with a geometric inequality, or that the use of geometry may be helpful, is shown in the following example. Example 2.1.3. If a, b, c are positive numbers, then a2 + ac + c2 ≤ a2 −ab + b2 + b2 −bc + c2. The radicals suggest using the cosine law with angles of 120◦and of 60◦as follows: a2 + ac + c2 = a2 + c2 −2ac cos 120◦, a2 −ab + b2 = a2 + b2 −2ab cos 60◦ and b2 −bc + c2 = b2 + c2 −2bc cos 60◦. A B C D b a c 60◦ 60◦ 2.1 Two basic inequalities 53 Then, if we consider a quadrilateral ABCD, with ∠ADB = ∠BDC = 60◦and ∠ADC = 120◦, such that AD = a, BD = b and CD = c, we can deduce that AB = √ a2 −ab + b2, BC = √ b2 −bc + c2 and CA = √ a2 + ac + c2. The inequality we have to prove becomes the triangle inequality for the triangle ABC. Exercise 2.4. Let ABC be a triangle with ∠A > ∠B, prove that BC > 1 2AB. Exercise 2.5. Let ABCD be a convex quadrilateral, prove that (i) if AB + BD < AC + CD, then AB < AC, (ii) if ∠A > ∠C and ∠D > ∠B, then BC > 1 2AD. Exercise 2.6. If a1, a2, a3, a4 and a5 are the lengths of the sides of a convex pentagon and if d1, d2, d3, d4 and d5 are the lengths of its diagonals, prove that 1 2 < a1 + a2 + a3 + a4 + a5 d1 + d2 + d3 + d4 + d5 < 1. Exercise 2.7. The length ma of the median AA′ of a triangle ABC satisﬁes ma > b+c−a 2 . Exercise 2.8. If the length ma of the median AA′ of a triangle ABC satisﬁes ma > 1 2a, prove that ∠BAC < 90◦. Exercise 2.9. If AA′ is the median of the triangle ABC and if AB < AC, then ∠BAA′ > ∠A′AC. Exercise 2.10. If ma, mb and mc are the lengths of the medians of a triangle with side-lengths a, b and c, respectively, prove that it is possible to construct a triangle with side-lengths ma, mb and mc, and that 3 4 (a + b + c) < ma + mb + mc < a + b + c. Exercise 2.11. (Ptolemy’s inequality) If ABCD is a convex quadrilateral, then AC · BD ≤AB · CD + BC · DA. The equality holds if and only if ABCD is a cyclic quadrilateral. Exercise 2.12. Let ABCD be a cyclic quadrilateral. Prove that AC > BD if and only if (AD −BC)(AB −DC) > 0. Exercise 2.13. (Pompeiu’s problem) Let ABC be an equilateral triangle and let P be a point that does not belong to the circumcircle of ABC. Prove that PA, PB and PC are the lengths of the sides of a triangle. Exercise 2.14. If ABCD is a paralelogram, prove that \|AB2 −BC2\| < AC · BD. 54 Geometric Inequalities Exercise 2.15. If a, b and c are the lengths of the sides of a triangle, ma, mb and mc represent the lengths of the medians and R is the circumradius, prove that (i) a2 + b2 mc + b2 + c2 ma + c2 + a2 mb ≤12R, (ii) ma(bc −a2) + mb(ca −b2) + mc(ab −c2) ≥0. Exercise 2.16. Let ABC be a triangle whose sides have lengths a, b and c. Suppose that c > b, prove that 1 2(c −b) < mb −mc < 3 2(c −b), where mb and mc are the lengths of the medians. Exercise 2.17. (Iran, 2005) Let ABC be a triangle with ∠A = 90◦. Let D be the intersection of the internal angle bisector of ∠A with the side BC and let Ia be the center of the excircle of the triangle ABC opposite to the vertex A. Prove that AD DIa ≤ √ 2 −1. 2.2 Inequalities between the sides of a triangle Inequalities involving the lengths of the sides of a triangle appear frequently in mathematical competitions. One sort of problems consists of those where you are asked to prove some inequality that is satisﬁed by the lengths of the sides of a triangle without any other geometric elements being involved, as in the following example. Example 2.2.1. The lengths a, b and c of the sides of a triangle satisfy a (b + c −a) < 2bc. Since the inequality is symmetric in b and c, we can assume, without loss of generality, that c ≤b. We will prove the inequality in the following cases. Case 1. a ≤b. Since they are the lengths of the sides of a triangle, we have that b < a + c; then b + c −a = b −a + c < c + c = 2c ≤2bc a . Case 2. a ≥b. In this case b −a ≤0, and since a < b + c ≤2b, we can deduce that b + c −a = c + b −a ≤c < 2bc a . Another type of problem involving the lengths of the sides of a triangle is when we are asked to prove that a certain relationship between the numbers a, b and c is suﬃcient to construct a triangle with sides of the same length. 2.2 Inequalities between the sides of a triangle 55 Example 2.2.2. (i) If a, b, c are positive numbers and satisfy, a2 + b2 + c22 > 2 a4 + b4 + c4 , then a, b and c are the lengths of the sides of a triangle. (ii) If a, b, c, d are positive numbers and satisfy a2 + b2 + c2 + d22 > 3 a4 + b4 + c4 + d4 , then,using any three of them we can construct a triangle. For part (i), it is suﬃcient to observe that a2 + b2 + c22 −2 a4 + b4 + c4 = (a+b+c)(a+b−c)(a−b+c)(−a+b+c) > 0, and then note that none of these factors is negative. Compare this with Example 1.2.5. For part (ii), we can deduce that 3 a4 + b4 + c4 + d4 < a2 + b2 + c2 + d22 = a2 + b2 + c2 2 + a2 + b2 + c2 2 + d2 2 ≤ a2 + b2 + c2 2 2 + a2 + b2 + c2 2 2 + d4 ) √ 3 2 . The second inequality follows from the Cauchy-Schwarz inequality; hence, a4 + b4 + c4 < 2(a2+b2+c2) 4 2 . Using the ﬁrst part we can deduce that a, b and c can be used to construct a triangle. Since the argument we used is symmetric in a, b, c and d, we obtain the result. There is a technique that helps to transform one inequality between the lengths of the sides of a triangle into an inequality between positive numbers (of course related to the sides).This is called the Ravi transformation. If the incircle (I, r) of the triangle ABC is tangent to the sides BC, CA and AB at the points X, Y and Z, respectively, we have that x = AZ = Y A, y = ZB = BX and z = XC = CY . B C A I X Y Z x x y y z z It is easily seen that a = y + z, b = z + x, c = x + y, x = s −a, y = s −b and z = s −c, where s = a+b+c 2 . Let us now see how to use the Ravi transformation in the following example. 56 Geometric Inequalities Example 2.2.3. The lengths of the sides a, b and c of a triangle satisfy (b + c −a)(c + a −b)(a + b −c) ≤abc. First, we have that (b + c −a)(c + a −b)(a + b −c) = 8(s −a)(s −b)(s −c) = 8xyz, on the other hand abc = (x + y)(y + z)(z + x). Thus, the inequality is equivalent to 8xyz ≤(x + y)(y + z)(z + x). (2.1) The last inequality follows from Exercise 1.26. Example 2.2.4. (APMO, 1996) Let a, b, c be the lengths of the sides of a triangle, prove that √ a + b −c + √ b + c −a + √ c + a −b ≤√a + √ b + √c. If we set a = y + z, b = z + x, c = x + y, we can deduce that a + b −c = 2z, b + c −a = 2x, c + a −b = 2y. Hence, the inequality is equivalent to √ 2x + 2y + √ 2z ≤√x + y + √y + z + √ z + x. Now applying the inequality between the arithmetic mean and the quadratic mean (see Exercise 1.68), we get √ 2x + 2y + √ 2z = √ 2x + √2y 2 + √2y + √ 2z 2 + √ 2z + √ 2x 2 ≤ 2x + 2y 2 + 2y + 2z 2 + 2z + 2x 2 = √x + y + √y + z + √ z + x. Moreover, the equality holds if and only if x = y = z, that is, if and only if a = b = c. Also, it is possible to express the area of a triangle ABC, its inradius, its circumradius and its semiperimeter in terms of x, y, z. Since a = x + y, b = y + z and c = z + x, we ﬁrst obtain that s = a+b+c 2 = x + y + z. Using Heron’s formula for the area of a triangle, we get (ABC) = s(s −a)(s −b)(s −c) = (x + y + z)xyz. (2.2) The formula (ABC) = sr leads us to r = (ABC) s = (x + y + z)xyz x + y + z = xyz x + y + z . 2.2 Inequalities between the sides of a triangle 57 Finally, from (ABC) = abc 4R we get R = (x + y)(y + z)(z + x) 4 (x + y + z)xyz . Example 2.2.5. (India, 2003) Let a, b, c be the side lengths of a triangle ABC. If we construct a triangle A′B′C′ with side lengths a + b 2, b + c 2, c + a 2, prove that (A′B′C′) ≥9 4(ABC). Since a = y + z, b = z + x and c = x + y, the side lengths of the triangle A′B′C′ are a′ = x+2y+3z 2 , b′ = 3x+y+2z 2 , c′ = 2x+3y+z 2 . Using Heron’s formula for the area of a triangle, we get (A′B′C′) = 3(x + y + z)(2x + y)(2y + z)(2z + x) 16 . Applying the AM-GM inequality to show that 2x+y ≥3 3 x2y, 2y +z ≥3 3 y2z, 2z + x ≥3 3 √ z2x, will help to reach the inequality (A′B′C′) ≥ 3(x + y + z)27(xyz) 16 = 9 4(ABC). Equation (2.2) establishes the last equality. Exercise 2.18. Let a, b and c be the lengths of the sides of a triangle, prove that 3(ab + bc + ca) ≤(a + b + c)2 ≤4(ab + bc + ca). Exercise 2.19. Let a, b and c be the lengths of the sides of a triangle, prove that ab + bc + ca ≤a2 + b2 + c2 ≤2(ab + bc + ca). Exercise 2.20. Let a, b and c be the lengths of the sides of a triangle, prove that 2 a2 + b2 + c2 ≤(a + b + c)2. Exercise 2.21. Let a, b and c be the lengths of the sides of a triangle, prove that 3 2 ≤ a b + c + b c + a + c a + b < 2. Exercise 2.22. (IMO, 1964) Let a, b and c be the lengths of the sides of a triangle, prove that a2 (b + c −a) + b2(c + a −b) + c2(a + b −c) ≤3abc. 58 Geometric Inequalities Exercise 2.23. Let a, b and c be the lengths of the sides of a triangle, prove that a b2 + c2 −a2 + b c2 + a2 −b2 + c a2 + b2 −c2 ≤3abc. Exercise 2.24. (IMO, 1983) Let a, b and c be the lengths of the sides of a triangle, prove that a2b(a −b) + b2c(b −c) + c2a(c −a) ≥0. Exercise 2.25. Let a, b and c be the lengths of the sides of a triangle, prove that a −b a + b + b −c b + c + c −a c + a < 1 8. Exercise 2.26. The lengths a, b and c of the sides of a triangle satisfy ab+bc+ca = 3. Prove that 3 ≤a + b + c ≤2 √ 3. Exercise 2.27. Let a, b, c be the lengths of the sides of a triangle, and let r be the inradius of the triangle. Prove that 1 a + 1 b + 1 c ≤ √ 3 2r . Exercise 2.28. Let a, b, c be the lengths of the sides of a triangle, and let s be the semiperimeter of the triangle. Prove that (i) (s −a)(s −b) < ab, (ii) (s −a)(s −b) + (s −b)(s −c) + (s −c)(s −a) ≤ab + bc + ca 4 . Exercise 2.29. If a, b, c are the lengths of the sides of an acute triangle, prove that cyclic a2 + b2 −c2 a2 −b2 + c2 ≤a2 + b2 + c2, where cyclic stands for the sum over all cyclic permutations of (a, b, c). Exercise 2.30. If a, b, c are the lengths of the sides of an acute triangle, prove that cyclic a2 + b2 −c2 a2 −b2 + c2 ≤ab + bc + ca, where cyclic represents the sum over all cyclic permutations of (a, b, c). 2.3 The use of inequalities in the geometry of the triangle 59 2.3 The use of inequalities in the geometry of the triangle A problem which shows the use of inequalities in the geometry of the triangle was introduced in the International Mathematical Olympiad in 1961; for this problem there are several proofs and its applications are very broad, as will be seen later on. Meanwhile, we present it here as an example. Example 2.3.1. If a, b and c are the lengths of the sides of a triangle with area (ABC), then 4 √ 3(ABC) ≤a2 + b2 + c2. Since an equilateral triangle of side-length a has area equal to √ 3 4 a2, the equality in the example holds for this case; hence we will try to compare what happens in any triangle with what happens in an equilateral triangle of side length a. B C A D d e b c h Let BC = a. If AD is the altitude of the triangle at A, its length h can be expressed as h = √ 3 2 a + y, where y measures its diﬀerence in comparison with the length of the altitude of the equilateral triangle. We also set d = a 2 −x and e = a 2 + x, where x can be interpreted as the diﬀerence that the projection of A on BC has with respect to the projection of A on BC in an equilateral triangle, which in this case is the midpoint of BC. We obtain a2 + b2 + c2 −4 √ 3(ABC) = a2 + h2 + a 2 + x 2 + h2 + a 2 −x 2 −4 √ 3ah 2 = 3 2a2 + 2h2 + 2x2 −2 √ 3 a √ 3 2 a + y = 3 2a2 + 2 √ 3 2 a + y 2 + 2x2 −3a2 −2 √ 3ay = 3 2a2 + 3 2a2 + 2 √ 3ay + 2y2 + 2x2 −3a2 −2 √ 3ay = 2(x2 + y2) ≥0. Moreover, the equality holds if and only if x = y = 0, that is, when the triangle is equilateral. 60 Geometric Inequalities Let us give another proof for the previous example. Let ABC be a triangle, with side-lengths a ≥b ≥c, and let A′ be a point such that A′BC is an equilateral triangle with side-length a. If we take d = AA′, then d measures, in a manner, how far is ABC from being an equilateral triangle. A A′ B C a b c a d Using the cosine law we can deduce that d2 = a2 + c2 −2ac cos(B −60◦) = a2 + c2 −2ac(cos B cos 60◦+ sin B sin 60◦) = a2 + c2 −ac cos B −2 √ 3ac sin B 2 = a2 + c2 −ac a2 + c2 −b2 2ac −2 √ 3(ABC) = a2 + b2 + c2 2 −2 √ 3(ABC). But d2 ≥0, hence we can deduce that 4 √ 3(ABC) ≤a2 + b2 + c2, which is what we wanted to prove. Moreover, the equality holds if d = 0, that is, if A′ = A or, equivalently, if ABC is equilateral. It is quite common to ﬁnd inequalities that involve elements of the triangle among mathematical olympiad problems. Some of them are based on the following inequality, which is valid for positive numbers a, b, c (see Exercise 1.36 of Section 1.3): (a + b + c) 1 a + 1 b + 1 c ≥9. (2.3) Moreover, we recall that the equality holds if and only if a = b = c. Another inequality, which has been very helpful to solve geometric-related problems, is Nesbitt’s inequality (see Example 1.4.8 of Section 1.4). It states that for a, b, c positive numbers, we always have a b + c + b c + a + c a + b ≥3 2. (2.4) 2.3 The use of inequalities in the geometry of the triangle 61 The previous inequality can be proved using inequality (2.3) as follows: a b + c + b c + a + c a + b = a + b + c b + c + a + b + c c + a + a + b + c a + b −3 = (a + b + c) 1 b + c + 1 c + a + 1 a + b −3 = 1 2 [(a + b) + (b + c) + (c + a)] · 1 b + c + 1 c + a + 1 a + b −3 ≥9 2 −3 = 3 2. The equality holds if and only if a + b = b + c = c + a, or equivalently, if a=b=c. Let us now observe some examples of geometric inequalities where such re-lationships are employed. Example 2.3.2. Let ABC be an equilateral triangle of side length a, let M be a point inside ABC and let D, E, F be the projections of M on the sides BC, CA and AB, respectively. Prove that (i) 1 MD + 1 ME + 1 MF ≥6 √ 3 a , (ii) 1 MD + ME + 1 ME + MF + 1 MF + MD ≥3 √ 3 a . A B C E F D M Let x = MD, y = ME and z = MF. Remember that we denote the area of the triangle ABC as (ABC), then (ABC) = (BCM) + (CAM) + (ABM), hence ah = ax + ay + az, where h = √ 3 2 a represents the length of the altitude of ABC. Therefore, h = x + y + z. (This result is known as Viviani’s lemma; see Section 2.8). Using inequality (2.3) we can deduce that h 1 x + 1 y + 1 z ≥9 and, after solving, that 1 x + 1 y + 1 z ≥9 h = 6 √ 3 a . 62 Geometric Inequalities To prove the second part, using inequality (2.3), we can establish that (x + y + y + z + z + x) 1 x + y + 1 y + z + 1 z + x ≥9. Therefore, 1 x+y + 1 y+z + 1 z+x ≥ 9 2h = 3 √ 3 a . Example 2.3.3. If ha, hb and hc are the lengths of the altitudes of the triangle ABC, whose incircle has center I and radius r, we have (i) r ha + r hb + r hc = 1, (ii) ha + hb + hc ≥9r. In order to prove the ﬁrst equation, observe that r ha = r·a ha·a = (IBC) (ABC). Simi-larly, r hb = (ICA) (ABC), r hc = (IAB) (ABC). Adding the three equations, we have that r ha + r hb + r hc = (IBC) (ABC) + (ICA) (ABC) + (IAB) (ABC) = (IBC) + (ICA) + (IAB) (ABC) = 1. A B C I r ha The desired inequality is a straightforward consequence of inequality (2.3), since (ha + hb + hc) 1 ha + 1 hb + 1 hc · r ≥9r. Example 2.3.4. Let ABC be a triangle with altitudes AD, BE, CF and let H be its orthocenter. Prove that (i) AD HD + BE HE + CF HF ≥9, (ii) HD HA + HE HB + HF HC ≥3 2. 2.3 The use of inequalities in the geometry of the triangle 63 A B C H E F D To prove part (i), consider S = (ABC), S1 = (HBC), S2 = (HCA), S3 = (HAB). Since triangles ABC and HBC share the same base, their area ratio is equal to their altitude ratio, that is, S1 S = HD AD . Similarly, S2 S = HE BE and S3 S = HF CF . Then, HD AD + HE BE + HF CF = 1. Using inequality (2.3) we can state that AD HD + BE HE + CF HF HD AD + HE BE + HF CF ≥9. If we substitute the equality previously calculated, we get (i). Moreover, the equality holds if and only if HD AD = HE BE = HF CF = 1 3, that is, if S1 = S2 = S3 = 1 3S. To prove the second part observe that HD HA = HD AD−HD = S1 S−S1 = S1 S2+S3 , and similarly, HE HB = S2 S3+S1 and HF HC = S3 S1+S2 , then using Nesbitt’s inequality leads to HD HA + HE HB + HF HC ≥3 2. Example 2.3.5. (Korea, 1995) Let ABC be a triangle and let L, M, N be points on BC, CA and AB, respectively. Let P, Q and R be the intersection points of the lines AL, BM and CN with the circumcircle of ABC, respectively. Prove that AL LP + BM MQ + CN NR ≥9. Let A′ be the midpoint of BC, let P ′ be the midpoint of the arc BC, let D and D′ be the projections of A and P on BC, respectively. It is clear that AL LP = AD PD′ ≥ AD P ′A′ . Thus, the minimum value of AL LP + BM MQ + CN NR is attained when P, Q and R are the midpoints of the arcs BC, CA and AB. This happens when AL, BM and CN are the internal angle bisectors of the triangle ABC. Hence, without loss of generality, we will assume that AL, BM and CN are the internal angle bisectors of ABC. Since AL is an internal angle bisector, we have6 BL = ca b + c, LC = ba b + c and AL2 = bc 1 − a b + c 2 . 6See [6, pages 74 and 105] or [9, pages 10,11]. 64 Geometric Inequalities A B C A′ L P D P ′ D′ Moreover, AL LP = AL2 AL · LP = AL2 BL · LC = (bc) 1 − a b+c 2 a2bc (b+c)2 = (b + c)2 −a2 a2 . Similarly, for the internal angle bisectors BM and CN, we have BM MQ = (c + a)2 −b2 b2 and CN NR = (a + b)2 −c2 c2 . Therefore, AL LP + BM MQ + CN NR = b + c a 2 + c + a b 2 + a + b c 2 −3 ≥1 3 b + c a + c + a b + a + b c 2 −3 ≥1 3 (6)2 −3 = 9. The ﬁrst inequality follows from the convexity of the function f(x) = x2 and the second inequality from relations in the form a b + b a ≥2. Observe that equality holds if and only if a = b = c. Another way to ﬁnish the problem is the following: b + c a 2 + c + a b 2 + a + b c 2 −3 = a2 b2 + b2 a2 + b2 c2 + c2 b2 + c2 a2 + a2 c2 + 2 ab c2 + bc a2 + ca b2 −3 ≥2 · 3 + 2 · 3 −3 = 9. 2.3 The use of inequalities in the geometry of the triangle 65 Here we made use of the fact that a2 b2 + b2 a2 ≥2 and that ab c2 + bc a2 + ca b2 ≥ 3 3 (ab)(bc)(ca) a2b2c2 = 3. Example 2.3.6. (Shortlist IMO, 1997) The lengths of the sides of the hexagon ABCDEF satisfy AB = BC, CD = DE and EF = FA. Prove that BC BE + DE DA + FA FC ≥3 2. A B C F E D a b c Set a = AC, b = CE and c = EA. Ptolemy’s inequality (see Exercise 2.11), applied to the quadrilateral ACEF, guarantees that AE ·FC ≤FA·CE +AC ·EF. Since EF = FA, we have that c · FC ≤FA · b + FA · a. Therefore, FA FC ≥ c a + b. Similarly, we can deduce the inequalities BC BE ≥ a b + c and DE DA ≥ b c + a. Hence, BC BE + DE DA + F A F C ≥ a b+c + b c+a + c a+b ≥3 2; the last inequality is Nesbitt’s inequality. Exercise 2.31. Let a, b, c be the lengths of the sides of a triangle, prove that: (i) a b + c −a + b c + a −b + c a + b −c ≥3, (ii) b + c −a a + c + a −b b + a + b −c c ≥3. Exercise 2.32. Let AD, BE, CF be the altitudes of the triangle ABC and let PQ, PR, PS be the distances from a point P to the sides BC, CA, AB, respectively. Prove that AD PQ + BE PR + CF PS ≥9. 66 Geometric Inequalities Exercise 2.33. Through a point O inside a triangle of area S three lines are drawn in such a way that every side of the triangle intersects two of them. These lines divide the triangle into three triangles with common vertex O and areas S1, S2 and S3, and three quadrilaterals. Prove that (i) 1 S1 + 1 S2 + 1 S3 ≥9 S , (ii) 1 S1 + 1 S2 + 1 S3 ≥18 S . Exercise 2.34. The cevians AL, BM and CN of the triangle ABC concur in P. Prove that AP PL + BP PM + CP PN = 6 if and only if P is the centroid of the triangle. Exercise 2.35. The altitudes AD, BE, CF intersect the circumcircle of the triangle ABC in D′, E′ and F ′, respectively. Prove that (i) AD DD′ + BE EE′ + CF FF ′ ≥9, (ii) AD AD′ + BE BE′ + CE CF ′ ≥9 4. Exercise 2.36. In the triangle ABC, let la, lb, lc be the lengths of the internal bisectors of the angles of the triangle, and let s and r be the semiperimeter and the inradius of ABC. Prove that (i) lalblc ≤rs2, (ii) lalb + lblc + lcla ≤s2, (iii) l2 a + l2 b + l2 c ≤s2. Exercise 2.37. Let ABC be a triangle and let M, N, P be arbitrary points on the line segments BC, CA, AB, respectively. Denote the lengths of the sides of the triangle by a, b, c and the circumradius by R. Prove that bc AM + ca BN + ab CP ≤6R. Exercise 2.38. Let ABC be a triangle with side-lengths a, b, c. Let ma, mb and mc be the lengths of the medians from A, B and C, respectively. Prove that max {a ma, b mb, c mc} ≤sR, where R is the radius of the circumcircle and s is the semiperimeter. 2.4 Euler’s inequality and some applications Theorem 2.4.1 (Euler’s theorem). Given the triangle ABC, where O is the cir-cumcenter, I the incenter, R the circumradius and r the inradius, then OI2 = R2 −2Rr. 2.4 Euler’s inequality and some applications 67 Proof. Let us give a proof7 that depends only on Pythagoras theorem and the fact that the circumcircle of the triangle BCI has center D, the midpoint of the arc8 BC. For the proof we will use directed segments. A B C D M O Q I Let M be the midpoint of BC and let Q be the orthogonal projection of I on the radius OD. Then OB2 −OI2 = OB2 −DB2 + DI2 −OI2 = OM 2 −MD2 + DQ2 −QO2 = (MO + DM) (MO −DM) + (DQ + QO)(DQ −QO) = DO(MO + MD + DQ + OQ) = R(2MQ) = 2Rr. Therefore OI2 = R2 −2Rr. □ As a consequence of the last theorem we obtain the following inequality. Theorem 2.4.2 (Euler’s inequality). R ≥2r. Moreover, R = 2r if and only if the triangle is equilateral.9 7Another proof can be found in [6, page 122] or [9, page 29]. 8The proof can be found in [6, observation 3.2.7, page 123] or [1, page 76]. 9There are direct proofs for the inequality (that is, without having to use Euler’s formula). One of them is the following: the nine-point circle of a triangle is the circumcircle of the medial triangle A′B′C′. Because this triangle is similar to ABC with ratio 2:1, we can deduce that the radius of the nine-point circle is R 2 . Clearly, a circle that intersects the three sides of a triangle must have a greater radius than the radius of the incircle, therefore R 2 ≥r. 68 Geometric Inequalities Theorem 2.4.3. In a triangle ABC, with circumradius R, inradius r and semipe-rimeter s, it happens that r ≤ s 3 √ 3 ≤R 2 . Proof. We will use the fact that10 (ABC) = abc 4R = sr. Using the AM-GM inequality, we can deduce that 2s = a + b + c ≥3 3 √ abc = 3 3 √ 4Rrs. Thus, 8s3 ≥27(4Rrs) ≥27(8r2s), since R ≥2r. Therefore, s ≥3 √ 3r. The second inequality, s 3 √ 3 ≤R 2 , is equivalent to a+b+c ≤3 √ 3R. But using the sine law, this is equivalent to sin A + sin B + sin C ≤3 √ 3 2 . Observe that the last inequality holds because the function f(x) = sin x is concave on [0, π], thus sin A+sin B+sin C 3 ≤sin A+B+C 3 = sin 60◦= √ 3 2 . □ Exercise 2.39. Let a, b and c be the lengths of the sides of a triangle, prove that (a + b −c)(b + c −a)(c + a −b) ≤abc. Exercise 2.40. Let a, b and c be the lengths of the sides of a triangle, prove that 1 ab + 1 bc + 1 ca ≥1 R2 , where R denotes the circumradius. Exercise 2.41. Let A, B and C be the measurements of the angles in each of the vertices of the triangle ABC, prove that 1 sin A sin B + 1 sin B sin C + 1 sin C sin A ≥4. Exercise 2.42. Let A, B and C be the measurements of the angles in each of the vertices of the triangle ABC, prove that sinA 2 sinB 2 sinC 2 ≤1 8. Exercise 2.43. Let ABC be a triangle. Call A, B and C the angles in the vertices A, B and C, respectively. Let a, b and c be the lengths of the sides of the triangle and let R be the radius of the circumcircle. Prove that 2A π 1 a 2B π 1 b 2C π 1 c ≤ 2 3 √ 3 R . Theorem 2.4.4 (Leibniz’s theorem). In a triangle ABC with sides of length a, b and c, and with circumcenter O, centroid G and circumradius R, the following holds: OG2 = R2 −1 9 a2 + b2 + c2 . 10See [6, page 97] or [9, page 13]. 2.4 Euler’s inequality and some applications 69 Proof. Let us use Stewart’s theorem which states11 that if L is a point on the side BC of a triangle ABC and if AL = l, BL = m, LC = n, then a l2 + mn = b2m + c2n. A B C A′ G O R Applying Stewart’s theorem to the triangle OAA′ to ﬁnd the length of OG, where A′ is the midpoint of BC, we get AA′ OG2 + AG · GA′ = A′O2 · AG + AO2 · GA′. Since AO = R, AG = 2 3AA′ and GA′ = 1 3AA′, substituting we get OG2 + 2 9(A′A)2 = A′O2 · 2 3 + R2 · 1 3. On the other hand12, since (A′A)2 = 2(b2+c2)−a2 4 and A′O2 = R2 −a2 4 , we can deduce that OG2 = R2 −a2 4 2 3 + 1 3R2 −2 9 2 b2 + c2 −a2 4 = R2 −a2 6 −2 b2 + c2 −a2 18 = R2 −a2 + b2 + c2 9 . □ One consequence of the last theorem is the following inequality. Theorem 2.4.5 (Leibniz’s nequality). In a triangle ABC with side-lengths a, b and c, with circumradius R, the following holds: 9R2 ≥a2 + b2 + c2. 11For a proof see [6, page 96] or [9, page 6]. 12See [6, page 83] or [9, page 10]. 70 Geometric Inequalities Moreover, equality holds if and only if O = G, that is, when the triangle is equi-lateral. Example 2.4.6. In a triangle ABC with sides of length a, b and c, it follows that 4 √ 3(ABC) ≤ 9abc a + b + c. Using that 4R (ABC) = abc, we have the following equivalences: 9R2 ≥a2 + b2 + c2 ⇔ a2b2c2 16(ABC)2 ≥a2 + b2 + c2 9 ⇔4(ABC) ≤ 3abc √ a2 + b2 + c2 . Cauchy-Schwarz inequality says that a + b + c ≤ √ 3 √ a2 + b2 + c2, hence 4 √ 3(ABC) ≤ 9abc a + b + c. Exercise 2.44. Let A, B and C be the measurements of the angles in each of the vertices of the triangle ABC, prove that sin2A + sin2B + sin2C ≤9 4. Exercise 2.45. Let a, b and c be the lengths of the sides of a triangle, prove that 4 √ 3(ABC) ≤3 3 √ a2b2c2. Exercise 2.46. Suppose that the incircle of ABC is tangent to the sides BC, CA, AB, at D, E, F, respectively. Prove that EF 2 + FD2 + DE2 ≤s2 3 , where s is the semiperimeter of ABC. Exercise 2.47. Let a, b, c be the lenghts of the sides of a triangle ABC and let ha, hb, hc be the lenghts of the altitudes over BC, CA, AB, respectively. Prove that a2 hbhc + b2 hcha + c2 hahb ≥4. 2.5 Symmetric functions of a, b and c The lengths of the sides a, b and c of a triangle have a very close relationship with s, r and R, the semiperimeter, the inradius and the circumradius of the triangle, respectively. The relationships that are most commonly used are a + b + c = 2s, (2.5) ab + bc + ca = s2 + r2 + 4rR, (2.6) abc = 4Rrs. (2.7) 2.5 Symmetric functions of a, b and c 71 The ﬁrst is the deﬁnition of s and the third follows from the fact that the area of the triangle is abc 4R = rs. Using Heron’s formula for the area of a triangle, we have the relationship s(s −a)(s −b)(s −c) = r2s2, hence s3 −(a + b + c)s2 + (ab + bc + ca)s −abc = r2s. If we substitute (2.5) and (2.7) in this equality, after simplifying we get that ab + bc + ca = s2 + r2 + 4Rr. Now, since any symmetric polynomial in a, b and c can be expressed as a polyno-mial in terms of (a + b + c), (ab + bc + ca) and (abc), it can also be expressed as a polynomial in s, r and R. For instance, a2 + b2 + c2 = (a + b + c)2 −2(ab + bc + ca) = 2 s2 −r2 −4Rr , a3 + b3 + c3 = (a + b + c)3 −3(a + b + c)(ab + bc + ca) + 3abc = 2 s3 −3r2s −6Rrs . These transformations help to solve diﬀerent problems, as will be seen later on. Lemma 2.5.1. If A, B and C are the measurements of the angles within each of the vertices of the triangle ABC, we have that cos A + cos B + cos C = r R + 1. Proof. cos A + cos B + cos C = b2 + c2 −a2 2bc + c2 + a2 −b2 2ca + a2 + b2 −c2 2ab = a b2 + c2 + b c2 + a2 + c a2 + b2 − a3 + b3 + c3 2abc = (a + b + c) (a2 + b2 + c2) −2(a3 + b3 + c3) 2abc = 4s s2 −r2 −4Rr −4 s3 −3r2s −6Rrs 8Rrs = s2 −r2 −4Rr −(s2 −3r2 −6Rr) 2Rr = 2r2 + 2Rr 2Rr = r R + 1. □ Example 2.5.2. If A, B and C are the measurements of the angles in each of the vertices of the triangle ABC, we have that cos A + cos B + cos C ≤3 2. Lemma 2.5.1 guarantees that cos A+cos B+cos C = r R +1, and using Euler’s inequality, R ≥2r, we get the result. 72 Geometric Inequalities We can give another direct proof. Observe that, a(b2+c2−a2)+b(c2+a2−b2)+c(a2+b2−c2) = (b+c−a)(c+a−b)(a+b−c)+2abc. Then, cos A + cos B + cos C = b2 + c2 −a2 2bc + c2 + a2 −b2 2ca + a2 + b2 −c2 2ab = (b + c −a)(c + a −b)(a + b −c) 2abc + 1, and since (b + c −a)(c + a −b)(a + b −c) ≤abc, we have the result. Example 2.5.3. (IMO, 1991) Let ABC be a triangle, let I be its incenter and let L, M, N be the intersections of the internal angle bisectors of A, B, C with BC, CA, AB, respectively. Prove that 1 4 < AI AL BI BM CI CN ≤ 8 27. B C A I L M Using the angle bisector theorem BL LC = AB CA = c b and the fact that BL + LC = a, we can deduce that BL = ac b+c and LC = ab b+c. Again, the angle bisector theorem applied to the internal angle bisector BI of the angle ∠ABL gives us IL AI = BL AB = ac (b+c)c = a b+c. Hence, AL AI = AI + IL AI = 1 + IL AI = 1 + a b + c = a + b + c b + c . Then, AI AL = b+c a+b+c.13 Similarly, BI BM = c+a a+b+c and CI CN = a+b a+b+c. Therefore, the inequality that we have to prove in terms of a, b and c is 1 4 < (b + c)(c + a)(a + b) (a + b + c)3 ≤8 27. The AM-GM inequality guarantees that (b + c)(c + a)(a + b) ≤ (b + c) + (c + a) + (a + b) 3 3 = 8 27(a + b + c)3, 13Another way to prove the identity is as follows. Consider α = (ABI), β = (BCI) and γ = (CAI). It is clear that AI AL = α+γ α+β+γ = r(c+b) r(a+c+b) = c+b a+c+b. 2.5 Symmetric functions of a, b and c 73 hence the inequality on the right-hand side is now evident. To prove the inequality on the left-hand side, ﬁrst note that (b + c)(c + a)(a + b) (a + b + c)3 = (a + b + c)(ab + bc + ca) −abc (a + b + c)3 . Substitute above, using equations (2.5), (2.6) and (2.7), to get (b + c)(c + a)(a + b) (a + b + c)3 = 2s(s2 + r2 + 4Rr) −4Rrs 8s3 = 2s3 + 2sr2 + 4Rrs 8s3 = 1 4 + 2r2 + 4Rr 8s2 > 1 4. We can also use the Ravi transformation a = y + z, b = z + x, c = x + y, to reach the ﬁnal result in the following way: (b + c)(c + a)(a + b) (a + b + c)3 = (x + y + z + x)(x + y + z + y)(x + y + z + z) 8(x + y + z)3 = 1 8 1 + x x + y + z 1 + y x + y + z 1 + z x + y + z = 1 8 1 + x + y + z x + y + z + xy + yz + zx x + y + z + xyz x + y + z > 1 4. Exercise 2.48. Let A, B and C be the values of the angles in each one of the vertices of the triangle ABC, prove that sin2 A 2 + sin2 B 2 + sin2 C 2 ≥3 4. Exercise 2.49. Let a, b and c be the lengths of the sides of a triangle. Using the tools we have studied in this section, prove that 4 √ 3(ABC) ≤ 9abc a + b + c. Exercise 2.50. Let a, b and c be the lengths of the sides of a triangle. Using the tools presented in this section, prove that 4 √ 3(ABC) ≤3 3 √ a2b2c2. Exercise 2.51. (IMO, 1961) Let a, b and c be the lengths of the sides of a triangle, prove that 4 √ 3(ABC) ≤a2 + b2 + c2. Exercise 2.52. Let a, b and c be the lengths of the sides of a triangle, prove that 4 √ 3(ABC) ≤a2 + b2 + c2 −(a −b)2 −(b −c)2 −(c −a)2 . 74 Geometric Inequalities Exercise 2.53. Let a, b and c be the lengths of the sides of a triangle, prove that 4 √ 3(ABC) ≤ab + bc + ca. Exercise 2.54. Let a, b and c be the lengths of the sides of a triangle, prove that 4 √ 3(ABC) ≤3(a + b + c)abc ab + bc + ca . Exercise 2.55. Let a, b and c be the lengths of the sides of a triangle. If a+b+c = 1, prove that a2 + b2 + c2 + 4abc < 1 2. Exercise 2.56. Let a, b and c be the lengths of the sides of a triangle, let R and r be the circumradius and the inradius, respectively, prove that (b + c −a)(c + a −b)(a + b −c) abc = 2r R . Exercise 2.57. Let a, b and c be the lengths of the sides of a triangle and let R be the circumradius, prove that 3 √ 3R ≤ a2 b + c −a + b2 c + a −b + c2 a + b −c. Exercise 2.58. Let a, b and c be the lengths of the sides of a triangle. Set x = b+c−a 2 , y = c+a−b 2 and z = a+b−c 2 . If τ1 = x + y + z, τ2 = xy + yz + zx and τ3 = xyz, verify the following relationships. (1) (a −b)2 + (b −c)2 + (c −a)2 = (x −y)2 + (y −z)2 + (z −x)2 = 2(τ2 1 −3τ2). (2) a + b + c = 2τ1. (3) a2 + b2 + c2 = 2τ2 1 −2τ2. (4) ab + bc + ca = τ2 1 + τ2. (5) abc = τ1τ2 −τ3. (6) 16(ABC)2 = 2(a2b2 + b2c2 + c2a2) −(a4 + b4 + c4) = 16r2s2 = 16τ1τ3. (7) R = τ1τ2 −τ3 4√τ1τ3 . (8) r = τ3 τ1 . (9) τ1 = s, τ2 = r(4R + r), τ3 = r2s. 2.6 Inequalities with areas and perimeters 75 2.6 Inequalities with areas and perimeters We begin this section with the following example. Example 2.6.1. (Austria–Poland, 1985) If ABCD is a convex quadrilateral of area 1, then AB + BC + CD + DA + AC + BD ≥4 + √ 8. Set a = AB, b = BC, c = CD, d = DA, e = AC and f = BD. The area of the quadrilateral ABCD is (ABCD) = ef sin θ 2 , where θ is the angle between the diagonals, which makes it clear that 1 = ef sin θ 2 ≤ef 2 . Since (ABC) = ab sin B 2 ≤ab 2 and (CDA) = cd sin D 2 ≤cd 2 , we can deduce that 1 = (ABCD) ≤ab+cd 2 . Similarly, 1 = (ABCD) ≤bc+da 2 . These two inequalities imply that ab + bc + cd + da ≥4. Finally, since (e + f)2 = 4ef + (e −f)2 ≥4ef ≥8 and (a + b + c + d)2 = 4(a + c)(b + d) + ((a + c) −(b + d))2 ≥4(a + c)(b + d) = 4(ab + bc + cd + da) ≥16, we can deduce that a + b + c + d + e + f ≥4 + √ 8. Example 2.6.2. (Iberoamerican, 1992) Using the triangle ABC, construct a hexa-gon H with vertices A1, A2, B1, B2, C1, C2 as shown in the ﬁgure. Show that the area of the hexagon H is at least thirteen times the area of the triangle ABC. B C A A1 A2 B1 B2 C1 C2 a a a b b b c c c It is clear, using the area formula (ABC) = ab sin C 2 , that (A1A2B1B2C1C2) =(A1BC2) + (A2CB1) + (B2AC1) + (AA1A2) + (BB1B2) + (CC1C2) −2(ABC) =(c + a)2sin B 2 + (a + b)2sin C 2 + (b + c)2sin A 2 + a2sin A 2 + b2sin B 2 + c2sin C 2 −2(ABC) 76 Geometric Inequalities = (a2 + b2 + c2)(sin A + sin B + sin C) 2 + ca sin B + ab sin C + b c sin A −2(ABC) = (a2 + b2 + c2)(sin A + sin B + sin C) 2 + 4(ABC). Therefore, (A1A2B1B2C1C2) ≥13(ABC) if and only if (a2 + b2 + c2)(sinA + sinB + sinC) 2 ≥9(ABC) = 9abc 4R . Using the sine law, sinA a = 1 2R, we can prove that the inequality is true if and only if (a2+b2+c2)(a+b+c) 4R ≥9abc 4R , that is, (a2 + b2 + c2)(a + b + c) ≥9abc. The last inequality can be deduced from the AM-GM inequality, from the re-arrangement inequality or by using Tchebychev’s inequality. Moreover, the equal-ity holds only in the case a = b = c. Example 2.6.3. (China, 1988 and 1993) Consider two concentric circles of radii R and R1 (R1 > R) and a convex quadrilateral ABCD inscribed in the small circle. The extensions of AB, BC, CD and DA intersect the large circle at C1, D1, A1 and B1, respectively. Show that (i) perimeter of A1B1C1D1 perimeter of ABCD ≥R1 R ; (ii) (A1B1C1D1) (ABCD) ≥ R1 R 2 . A1 B1 C1 D1 A B C D O 2.6 Inequalities with areas and perimeters 77 To prove (i), we use Ptolemy’s inequality (see Exercise 2.11) applied to the quadrilaterals OAB1C1, OBC1D1, OCD1A1 and ODA1B1, which implies that AC1 · R1 ≤B1C1 · R + AB1 · R1, BD1 · R1 ≤C1D1 · R + BC1 · R1, (2.8) CA1 · R1 ≤D1A1 · R + CD1 · R1, DB1 · R1 ≤A1B1 · R + DA1 · R1. Then, when we add these inequalities together and write AC1, BD1, CA1 and DB1, and express them as AB + BC1, BC + CD1, CD + DA1 and DA + AB1, respectively, we get R1· perimeter (ABCD) + R1(BC1 + CD1 + DA1 + AB1) ≤R · perimeter (A1B1C1D1) + R1(AB1 + BC1 + CD1 + DA1). Therefore, perimeter (A1B1C1D1) perimeter (ABCD) ≥R1 R . To prove (ii), we use the fact that (ABCD) = ad sin A+bc sin A 2 = sin A 2 (ad+bc) and also that (ABCD) = ab sin B+cd sin B 2 = sin B 2 (ab + cd), where A = ∠DAB and B = ∠ABC. A1 B1 C1 D1 A B C D O a b c d x y z w Since (AB1C1) = x(a+y)sin (180◦−A) 2 = x(a+y)sin A 2 , we can produce the identity (AB1C1) (ABCD) = x(a+y) ad+bc . Similarly, (BC1D1) (ABCD) = y(b+z) ab+cd , (CD1A1) (ABCD) = z(c+w) ad+bc , (DA1B1) (ABCD) = w(d+x) ab+cd . Then, (A1B1C1D1) (ABCD) = 1 + x(a + y) + z(w + c) ad + bc + y(b + z) + w(d + x) ab + cd . The power of a point in the larger circle with respect to the small circle is equal to R2 1 −R2. In particular, the power of A1, B1, C1 and D1 is the same. On the other hand, we know that these powers are w(w + c), x(x + d), y(y + a) and z(z + b), respectively. 78 Geometric Inequalities Substituting this in the previous equation implies that the area ratio is (A1B1C1D1) (ABCD) =1+(R2 1−R2) ' x y(ad + bc) + z w(ad + bc) + y z(ab + cd) + w x(ab + cd) ( . Using the AM-GM inequality allows us to deduce that (A1B1C1D1) (ABCD) ≥1 + 4(R2 1 −R2) (ad + bc)(ab + cd) . Since 2 (ad + bc)(ab + cd) ≤ad+bc+ab+cd = (a+c)(b+d) ≤1 4(a+b+c+d)2 ≤ (4 √ 2R)2 4 = 8R2, the ﬁrst two inequalities follow from the AM-GM inequality, and the last one follows from the fact that, of all the quadrilaterals inscribed in a circle, the square has the largest perimeter. Thus (A1B1C1D1) (ABCD) ≥1 + 4(R2 1 −R2) 4R2 = R1 R 2 . Moreover, the equalities hold when ABCD is a square and only in this case. Since in order to reduce inequalities (2.8) to identities, it must be the case that the four quadrilaterals OAB1C1, OBC1D1, OCD1A1 and ODA1B1 are cyclic. Thus, OA is an internal angle bisector of the angle BAD, and the same happens for OB, OC and OD. There are problems that, even when they are not presented in a geomet-ric form, they invite us to search for geometric relationships, as in the following example. Example 2.6.4. If a, b, c are positive numbers with c < a and c < b, we can deduce that c(a −c) + c(b −c) ≤ √ ab. Consider the isosceles triangles ABC and ACD, both sharing the common side AC of length 2√c; we take the ﬁrst triangle as having equal sides AB = BC of length √a and the second one satisfying CD = DA with length √ b. The area of the quadrilateral ABCD is, on the one hand, (ABCD) = (ABC) + (ACD) = c(a −c) + b(b −c); and on the other hand, (ABCD) = 2(ABD) = 2 √ ab sin ∠BAD 2 . This last procedure for calculating the area clearly proves that (ABCD) ≤ √ ab, and thus the result is obtained. 2.6 Inequalities with areas and perimeters 79 A B C D E √a √a √ b √ b √c √c Another solution is as follows. Since AC and BD are perpendiculars, Py-thagoras theorem implies that DE = √ b −c and EB = √a −c. By Ptolemy’s inequality (see Exercise 2.11), ( √b −c + √a −c) · (2√c) ≤√a √ b + √a √ b and then the result. Exercise 2.59. On every side of a square with sides measuring 1, choose one point. The four points will form a quadrilateral of sides of length a, b, c and d, prove that (i) 2 ≤a2 + b2 + c2 + d2 ≤4, (ii) 2 √ 2 ≤a + b + c + d ≤4. Exercise 2.60. On each side of a regular hexagon with sides measuring 1, we choose one point. The six points form a hexagon of perimeter h. Prove that 3 √ 3 ≤h ≤6. Exercise 2.61. Consider the three lines tangent to the incircle of a triangle ABC which are parallel to the sides of the triangle; these, together with the sides of the triangle, form a hexagon T . Prove that the perimeter of T ≤2 3 the perimeter of (ABC). Exercise 2.62. Find the radius of the circle of maximum area that can be covered using three circles with radius 1. Exercise 2.63. Find the radius of the circle of maximum area that can be covered using three circles with radii r1, r2 and r3. Exercise 2.64. Two disjoint squares are located inside a square of side 1. If the lengths of the sides of the two squares are a and b, prove that a + b ≤1. Exercise 2.65. A convex quadrilateral is inscribed in a circumference of radius 1, in such a way that one of its sides is a diameter and the other sides have lengths a, b and c. Prove that abc ≤1. 80 Geometric Inequalities Exercise 2.66. Let ABCDE be a convex pentagon such that the areas of the triangles ABC, BCD, CDE, DEA and EAB are equal. Prove that (i) (ABCDE) 4 < (ABC) < (ABCDE) 3 , (ii) (ABCDE) = 5 + √ 5 2 (ABC). Exercise 2.67. If AD, BE and CF are the altitudes of the triangle ABC, prove that perimeter (DEF) ≤s, where s is the semiperimeter. Exercise 2.68. The lengths of the internal angle bisectors of a triangle are at most 1, show that the area of such a triangle is at most √ 3 3 . Exercise 2.69. If a, b, c, d are the lengths of the sides of a convex quadrilateral, show that (i) (ABCD) ≤ab + cd 2 , (ii) (ABCD) ≤ac + bd 2 and (iii) (ABCD) ≤ a + c 2 b + d 2 . 2.7 Erd˝ os-Mordell Theorem Theorem 2.7.1 (Pappus’s theorem). Let ABC be a triangle, AA′B′B and CC′A′′A two parallelograms constructed on AC and AB such that both either are inside or outside the triangle. Let P be the intersection of B′A′ with C′A′′. Construct another parallelogram BP ′P ′′C on BC such that BP ′ is parallel to AP and of the same length. Thus, we will have the following relationships between the areas: (BP ′P ′′C) = (AA′B ′B) + (CC′A ′′A). Proof. See the picture on the next page. □ 2.7 Erd˝ os-Mordell Theorem 81 P A A′ A′′ B B′ C C′ P ′ P ′′ Theorem 2.7.2 (Erd˝ os-Mordell theorem). Let P be an arbitrary point inside or on the boundary of the triangle ABC. If pa, pb, pc are the distances from P to the sides of ABC, of lenghts a, b, c, respectively, then PA + PB + PC ≥2 (pa + pb + pc) . Moreover, the equality holds if and only if the triangle ABC is equilateral and P is the circumcenter. Proof (Kazarinoﬀ). Let us reﬂect the triangle ABC on the internal bisector BL of angle B. Let A′ and C′ be the reﬂections of A and C. The point P is not reﬂected. Now, let us consider the parallelograms determined by B, P and A′, and by B, P and C′. P A C B C′ A′ L The sum of the areas of these parallelograms is cpa + apc and this is equal to the area of the parallelogram A′P ′P ′′C′, where A′P ′ is parallel to BP and of the same length. The area of A′P ′P ′′C′ is at most b · PB. Moreover, the areas are equal if BP is perpendicular to A′C′ and this happens if and only if P is on BO, where O is the circumcenter of ABC.14 Then, cpa + apc ≤bPB. 14BP is perpendicular to A′C′ if and only if ∠P BA′ = 90◦−∠A′, but ∠A′ = ∠A and OBC = 90◦−∠A, then P should be on BO. 82 Geometric Inequalities P B C′ A′ c pa a pc P C′ P ′′ B P ′ A′ b Therefore, PB ≥c bpa + a b pc. Similarly, PA ≥b apc + c apb and PC ≥b cpa + a c pb. If we add together these inequalities, we have PA + PB + PC ≥ b c + c b pa + c a + a c pb + a b + b a pc ≥2 (pa + pb + pc) , since b c + c b ≥2. Moreover, the equality holds if and only if a = b = c and P is on AO, BO and CO, that is, if the triangle is equilateral and P = O. □ Example 2.7.3. Using the notation of the Erd˝ os-Mordell theorem, prove that aPA + bPB + cPC ≥4(ABC). Consider the two parallelograms that are determined by B, C, P and B, A, P as shown in the ﬁgure, and the parallelogram that is constructed following Pappus’s theorem. It is clear that bPB ≥apa + cpc. P B A C b c a pa pc 2.7 Erd˝ os-Mordell Theorem 83 Similarly, it follows that aPA ≥bpb + cpc, cPC ≥apa + bpb. Hence, aPA + bPB + cPC ≥2(apa + bpb + cpc) = 4(ABC). Example 2.7.4. Using the notation of the Erd˝ os-Mordell theorem, prove that paPA + pbPB + pcPC ≥2 (papb + pbpc + pcpa) . As in the previous example, we have that aPA ≥bpb + cpc. Hence, paPA ≥b apapb + c apcpa. Similarly, we can deduce that pbPB ≥a b papb + c bpbpc, pcPC ≥a c pcpa + b cpbpc. If we add together these three inequalities, we get paPA + pbPB + pcPC ≥ a b + b a papb + b c + c b pbpc + c a + a c pcpa ≥2 (papb + pbpc + pcpa) . Example 2.7.5. Using the notation of the Erd˝ os-Mordell theorem, prove that 2 1 PA + 1 PB + 1 PC ≤1 pa + 1 pb + 1 pc . A B C P A′ B′ C′ A1 B′ 1 C1 A′ 1 C′ 1 Let us apply inversion to the circle of center P and radius d = pb. If A′, B′, C′ are the inverse points of A, B, C, respectively, and A′ 1, B′ 1, C′ 1 are the inverse points 84 Geometric Inequalities of A1, B1, C1, we can deduce that PA · PA′ = PB · PB′ = PC · PC′ = d2, PA1 · PA′ 1 = PB1 · PB′ 1 = PC1 · PC′ 1 = d2. Moreover, A′, B′ and C′ are on B′ 1C′ 1, C′ 1A′ 1 and A′ 1B′ 1, respectively, and the segments PA′, PB′ and PC′ are perpendicular to B′ 1C′ 1, C′ 1A′ 1 and A′ 1B′ 1, respec-tively. An application of the Erd˝ os-Mordell theorem to the triangle A′ 1B′ 1C′ 1 shows that PA′ 1 + PB′ 1 + PC′ 1 ≥2 (PA′ + PB′ + PC′). Since PA′ 1 = d2 PA1 , PB′ 1 = d2 PB1 , PC′ 1 = d2 PC1 , PC′ = d2 PC , PB′ = d2 PB , PA′ = d2 PA, then d2 1 PA1 + 1 PB1 + 1 PC1 ≥2d2 1 PA + 1 PB + 1 PC , that is, 2 1 PA + 1 PB + 1 PC ≤ 1 pa + 1 pb + 1 pc . Example 2.7.6. Using the notation of the Erd˝ os-Mordell theorem, prove that PA · PB · PC ≥R 2r (pa + pb) (pb + pc) (pc + pa) . P A C1 C B b c c pa pc Let C1 be a point on BC such that BC1 = AB. Then AC1 = 2c sin B 2 , and Pappus’s theorem implies that PB 2c sin B 2 ≥c pa + c pc. Therefore, PB ≥pa + pc 2 sin B 2 . 2.7 Erd˝ os-Mordell Theorem 85 Similarly, PA ≥pb + pc 2 sin A 2 and PC ≥pa + pb 2 sin C 2 . Then, after multiplication, we get PA · PB · PC ≥1 8 1 sin A 2 sin B 2 sin C 2 (pa + pb) (pb + pc) (pc + pa) . The solution of Exercise 2.42 helps us to prove that sin A 2 sin B 2 sin C 2 = r 4R, then the result follows. Example 2.7.7. (IMO, 1991) Let P be a point inside the triangle ABC. Prove that at least one of the angles ∠PAB, ∠PBC, ∠PCA is less than or equal to 30◦. Draw A1, B1 and C1, the projections of P on sides BC, CA and AB, re-spectively. Using the Erd˝ os-Mordell theorem we get PA + PB + PC ≥2PA1 + 2PB1 + 2PC1. A B C A1 B1 C1 P Thus, one of the following inequalities will be satisﬁed: PA ≥2PC1, PB ≥2PA1 or PC ≥2PB1. If, for instance, PA ≥2PC1, we can deduce that 1 2 ≥PC1 PA = sin ∠PAB, then ∠PAB ≤30◦or ∠PAB ≥150◦. But, if ∠PAB ≥150◦, then it must be the case that ∠PBC < 30◦and thus in both cases the result follows. Example 2.7.8. (IMO, 1996) Let ABCDEF be a convex hexagon such that AB is parallel to DE, BC is parallel to EF and CD is parallel to FA. Let RA, RC, RE denote the circumradii of triangles FAB, BCD, DEF, respectively, and let P denote the perimeter of the hexagon. Prove that RA + RC + RE ≥P 2 . Let M, N and P be points inside the hexagon in such a way that MDEF, NFAB and PBCD are parallelograms. Let XY Z be the triangle formed by the 86 Geometric Inequalities lines through B, D, F and perpendicular to FA, BC, DE, respectively, where B is on Y Z, D on ZX and F on XY . Observe that MNP and XY Z are similar triangles. X Y Z A B C D E F P M N Since the triangles DEF and DMF are congruent, they have the same circumra-dius; moreover, since XM is the diameter of the circumcircle of triangle DMF, then XM = 2RE. Similarly, Y N = 2RA and ZP = 2RC. Thus, the inequality that needs to be proven can be written as XM + Y N + ZP ≥BN + BP + DP + DM + FM + FN. The case M = N = P is the Erd˝ os-Mordell inequality, on which the rest of the proof is based. Let Y ′, Z′ denote the reﬂections of Y and Z on the internal angle bisector of X. Let G, H denote the feet of the perpendiculars of M and X on Y ′Z′, respectively. X Z′ Y ′ D H F G M Since (XY ′Z′) = (XMZ′) + (Z′MY ′) + (Y ′MX), we obtain XH · Y ′Z′ = MF · XZ′ + MG · Y ′Z′ + MD · Y ′X. 2.7 Erd˝ os-Mordell Theorem 87 If we set x = Y ′Z′, y = ZX′, z = XY ′, the above equality becomes xXH = xMG + zDM + yFM. Since ∠XHG = 90◦, then XH = XG sin ∠XGH ≤XG. Moreover, using the triangle inequality, XG ≤XM + MG, we can deduce that XM ≥XH −MG = z xDM + y xFM. Similarly, Y N ≥x y FN + z yBN, ZP ≥y z BP + x y DP. After adding together these three inequalities, we get XM + Y N + ZP ≥z xDM + y xFM + x y FN + z y BN + y z BP + x z DP. (2.9) Observe that y z BP + z yBN = y z + z y BP + BN 2 + y z −z y BP −BN 2 . Since the triangles XY Z and MNP are similar, we can deﬁne r as r = FM −FN XY = BN −BP Y Z = DP −DM ZX . If we apply the inequality y z + z y ≥2, we get y z BP + z yBN = y z + z y BP + BN 2 −r 2 yx z −zx y ≥BP + BN −r 2 yx z −zx y . Similar inequalities hold for x y FN + y xFM ≥FN + FM −r 2 xz y −yz x , z xDM + x z DP ≥DM + DP −r 2 zy x −xy z . If we add the inequalities and substitute them in (2.9), we have XM + Y N + ZP ≥BN + BP + DP + DM + FM + FN, which completes the proof. 88 Geometric Inequalities Exercise 2.70. Using the notation of the Erd˝ os-Mordell theorem, prove that PA · PB · PC ≥4R r papbpc. Exercise 2.71. Using the notation of the Erd˝ os-Mordell theorem, prove that (i) PA2 pbpc + PB2 pcpa + PC2 papb ≥12, (ii) PA pb + pc + PB pc + pa + PC pa + pb ≥3, (iii) PA √pbpc + PB √pcpa + PC √papb ≥6, (iv) PA · PB + PB · PC + PC · PA ≥4(papb + pbpc + pcpa). Exercise 2.72. Let ABC be a triangle, P be an arbitrary point in the plane and let pa, pb y pc be the distances from P to the sides of a triangle of lengths a, b and c, respectively. If, for example, P and A are on diﬀerent sides of the segment BC, then pa is negative, and we have a similar situation for the other cases. Prove that PA + PB + PC ≥ b c + c b pa + c a + a c pb + a b + b a pc. 2.8 Optimization problems In this section we present two classical examples known as the Fermat-Steiner problem and the Fagnano problem. The Fermat-Steiner problem. This problem seeks to ﬁnd a point in the interior or on the sides of a triangle such that the sum of the distances from the point to the vertices of the triangle is minimum. We will present three solutions and point out the methods used to solve the problem. Torricelli’s solution. It takes as its starting point the following two lemmas. Lemma 2.8.1 (Viviani’s lemma). The sum of the distances from an interior point to the sides of an equilateral triangle is equal to the altitude of the triangle. Proof. Let P be a point in the interior of the triangle ABC. Draw the triangle A′B′C′ with sides parallel to the sides of ABC, with P on C′A′ and B′C′ on the line through B and C. 2.8 Optimization problems 89 A B C A′ B′ C′ L′ M N L P P ′ P ′′ M ′ If L, M and N are the feet of the perpendiculars of P on the sides, it is clear that PM = NM ′, where M ′ is the intersection of PN with A′B′. Moreover, PM ′ is the altitude of the equilateral triangle AP ′P. If A′P ′′ is the altitude of the triangle AP ′P from A′, it is clear that PM ′ = A′P ′′. Let L′ be a point on B′C′ such that A′L′ is the altitude of the triangle A′B′C′ from A′. Thus, PL+PM +PN = PL+PN +NM ′ = PL+A′P ′′ = A′P ′′ +P ′′L′ = A′L′. □ Next, we present another two proofs of Viviani’s lemma for the sake of com-pleteness. Observation 2.8.2. (i) The following is another proof of Viviani’s lemma which is based on the use of areas. We have that (ABC) = (ABP) + (BCP) + (CAP). Then, if a is the length of the side of the triangle and h is the length of its altitude, we have that ah = aPN + aPL + aPM, that is, h = PN + PL + PM. (ii) Another proof of Viviani’s lemma can be deduced from the following diagram. A B C M N L P M ′ 90 Geometric Inequalities Lemma 2.8.3. If ABC is a triangle with all angles less than or equal to 120◦, there is a unique point P such that ∠APB = ∠BPC = ∠CPA = 120◦. The point P is known as the Fermat point of the triangle. Proof. First, we will proof the existence of P. On the sides AB and CA we con-struct equilateral triangles ABC′ and CAB′. Their circumcircles intersect at A and at another point that we denote as P. B C A C′ B′ P Since APCB′ is cyclic, we have that ∠CPA = 180◦−∠B′ = 120◦. Similarly, since APBC′ is cyclic, ∠APB = 120◦. Finally, ∠BPC = 360◦−∠APB −∠CPA = 360◦−120◦−120◦= 120◦. To prove the uniqueness, suppose that Q satisﬁes ∠AQB = ∠BQC = ∠CQA = 120◦. Since ∠AQB = 120◦, the point Q should be on the circumcircle of ABC′. Similarly, it should be on the circumcircle of CAB′, hence Q = P. □ We will now study Torricelli’s solution to the Fermat-Steiner problem. Given the triangle ABC with angles less than or equal to 120◦, construct the Fermat point P, which satisﬁes ∠APB = ∠BPC = ∠CPA = 120◦. Now, through A, B and C we draw perpendiculars to AP, BP and CP, respectively. These perpendiculars determine a triangle DEF which is equilateral. This is so because the quadrilateral PBDC is cyclic, having angles of 90◦in B and C. Now, since ∠BPC = 120◦, we can deduce that ∠BDC = 60◦. This argument can be repeated for each angle. Therefore DEF is indeed equilateral. We know that the distance from P to the vertices of the triangle ABC is equal to the length of the altitude of the equilateral triangle DEF. Observe that any other point Q inside the triangle ABC satisﬁes AQ ≥A′Q, where A′Q is the distance from Q to the side EF, similarly BQ ≥B′Q and CQ ≥C′Q. Therefore AQ + BQ + CQ is greater than or equal to the altitude of DEF which is AP + BP + CP, which in turn is equal to A′Q + B′Q + C′Q as can be seen by using Viviani’s lemma. 2.8 Optimization problems 91 B C A P D E F B′ C′ A′ Q Hofmann-Gallai’s solution. This way of solving the problem uses the ingenious idea of rotating the ﬁgure to place the three segments that we need next to each other, in order to form a polygonal line and then add them together. Thus, when we join the two extreme points with a segment of line, since this segment of line represents the shortest path between them, it is then necessary to ﬁnd the conditions under which the polygonal line lies over such segment. This proof was provided by J. Hofmann in 1929, but the method for proving had already been discovered and should be attributed to the Hungarian Tibor Gallai. Let us recall this solution. Consider the triangle ABC with a point P inside it; draw AP, BP and CP. Next, rotate the ﬁgure with its center in B and through an angle of 60◦, in a positive direction. B C A C′ P ′ P 60◦ We should point out several things. If C′ is the image of A and P ′ is the image of P, the triangles BPP ′ and BAC′ are equilateral. Moreover, if AP = P ′C′ and BP = P ′B = P ′P, then AP + BP + CP = P ′C′ + P ′P + CP. The path CP + PP ′ + P ′C is minimum when C, P, P ′ and C′ are collinear, which in turn requires that ∠C′P ′B = 120◦and ∠BPC = 120◦; but since ∠C′P ′B = ∠APB, 92 Geometric Inequalities the point P should satisfy ∠APB = ∠BPC = 120◦(and then also ∠CPA = 120◦). An advantage of this solution is that it provides another description of the Fermat point and another way of ﬁnding it. If we review the proof, we can see that the point P is on the segment CC′, where C′ is the third vertex of the equilateral triangle with side AB. But if instead of the rotation with center in B, we rotate it with its center in C, we obtain another equilateral triangle AB′C and we can conclude that P is on BB′. Hence we can ﬁnd P as the intersection of BB′ and CC′. Steiner’s solution. When we solve maximum and minimum problems we are prin-cipally faced with three questions, (i) is there a solution?, (ii) is there a unique solution? (iii) what properties characterize the solution(s)? Torricelli’s solution demonstrates that among all the points in the triangle, this particular point P, from which the three sides of the triangle are observed as having an angle of 120◦, provides the minimum value of PA + PB + PC. In this sense, this point answers the three questions we proposed and does so in an elegant way. However, the solu-tion does not give us any clue as to why Torricelli chose this point, or what made him choose that point; probably this will never be known. But in the following we can consider a sequence of ideas that bring us to discover that the Fermat point is the optimal point. These ideas belong to the Swiss geometer Jacob Steiner. Let us ﬁrst provide the following two lemmas. Lemma 2.8.4 (Heron’s problem). Given two points A and B on the same side of a line d, ﬁnd the shortest path that begins at A, touches the line d and ﬁnishes at B. B P A d The shortest path between A and B, touching the line d, can be found re-ﬂecting B on d to get a point B′; the segment AB′ intersects d at a point P ∗that makes AP ∗+ P ∗B represent the minimum between the numbers AP + PB, with P on d. To convince ourselves it is suﬃcient to observe that AP ∗+ P ∗B = AP ∗+ P ∗B′ = AB′ ≤AP + PB′ = AP + PB. This point satisﬁes the following reﬂection principle: The incident angle is equal to the reﬂection angle. It is evident that the point which has this property is the minimum. 2.8 Optimization problems 93 B B′ P P ∗ A d α α Lemma 2.8.5 (Heron’s problem using a circle). Given two points A and B outside the circle C, ﬁnd the shortest path that starts at A, touches the circle and ﬁnishes at B. A C B We will only give a sketch of the solution. Let D be a point on C, then we have that the set {P : PA+PB = DA+DB} is an ellipse ED with foci points A and B, and that the point D belongs to ED. In general Ed = {P : PA + PB = d}, where d is a positive number, is an ellipse with foci A and B (if d > AB). Moreover, these ellipses have the property that Ed is a subset of the interior of Ed′ if and only if d < d′. We would like to ﬁnd a point Q on C such that the sum QA+QB is minimum. The optimal point Q will belong to an ellipse, precisely to EQ. Such an ellipse EQ does not intersect C in other point; in fact, if C′ is another common point of EQ and C, then every point C′′ of the circumference arc between Q and C′ of C would be in the interior of EQ, therefore C′′A + C′′B < QA + QB and so Q is not the optimal point, that is, a contradiction. Thus, the point Q that minimize AQ+QB should satisfy that the ellipse EQ is tangent to C. The common tangent line to EQ and C happens to be perpendicular to the radius CQ, where C is the center of C and, because of the reﬂection property of the ellipse (the incidence angle is equal to the reﬂection angle), it follows that 94 Geometric Inequalities the line CQ is the internal bisector of the angle ∠AQB, that is, ∠BQC = ∠CQA. A B C Q α β Now let us go back to Steiner’s solution of the Fermat-Steiner problem. A point P that makes the sum PA+PB +PC a minimum can be one of the vertices A, B, C or a point of the triangle diﬀerent from the vertices. In the ﬁrst case, if P is one of the vertices, then one term of the sum PA + PB + PC is zero and the other two are the lengths of the sides of the triangle ABC that have in common the chosen vertex. Hence, the sum will be minimum when the chosen vertex is opposite to the longest side of the triangle. In order to analize the second case, Steiner follows the next idea (very useful in optimization problems and one which can be taken to belong to the strategy of “divide and conquer”), which is to keep ﬁxed some of the variables and optimize the rest. This procedure would provide conditions in the variables not ﬁxed. Such restrictions will act as restrains in the solution space until we reach the optimal solution. Speciﬁcally, we proceed as follows. Suppose that PA is ﬁxed; that is, P belongs to the circle of center A and radius PA, where we need to ﬁnd the point P that makes the sum PB +PC minimum. Note that B should be located outside of such circle, otherwise PA ≥AB and, using the triangle inequality, PB + PC > BC. From this, it follows that PA+PB = PC > AB+BC, which means B would be a more suitable point (instead of P). Similarly, C should be outside of such circle. Now, since B and C are points outside the circle C = (A, PA), the optimal point for the problem of minimizing PB + PC with the condition that P is on the circle C is, by Lemma 2.8.5, a point Q on the circle C, such that this circle is tangent to the ellipse with foci B and C in Q, and the point Q is such that the angles ∠AQB and ∠CQA are equal. Since the role of A, B, C can be exchanged, if now we ﬁx B (and PB), then the optimal point Q will satisfy the condition ∠AQB = ∠BQC and therefore ∠AQB = ∠BQC = ∠CQA = 120◦. This means Q should be the Fermat point. All the above work in the second case is to assure that Q is inside of ABC, if the angles of the triangle are not greater than 120◦. The Fagnano problem. The problem is to ﬁnd an inscribed triangle of minimum perimeter inside an acute triangle. We present two classical solutions, where the reﬂection on lines play a central role. One is due to H. Schwarz and the other to L. Fejer. 2.8 Optimization problems 95 Schwarz’s solution. The German mathematician Hermann Schwarz provided the following solution to this problem for which he took as starting point two ob-servations that we present as lemmas. These lemmas will demonstrate that the inscribed triangle with the minimum perimeter is the triangle formed using the feet of the altitudes of the triangle. Such a triangle is known as the ortic triangle. Lemma 2.8.6. Let ABC be a triangle, and let D, E and F be the feet of the altitudes on BC, CA and AB as they fall from the vertices A, B and C, respectively. Then the triangles ABC, AEF, DBF and DEC are similar. Proof. It is suﬃcient to see that the ﬁrst two triangles are similar, since the other cases are proved in a similar way. B C A E F H D Since these two triangles have a common angle at A, it is suﬃcient to see that ∠AEF = ∠ABC. But, since we know that ∠AEF +∠FEC = 180◦and ∠ABC + ∠FEC = 180◦because the quadrilateral BCEF is cyclic, then ∠AEF = ∠ABC. □ Lemma 2.8.7. Using the notation of the previous lemma, we can deduce that the reﬂection of D on AB is collinear with E and F, and the reﬂection of D on CA is collinear with E and F. Proof. It follows directly from the previous lemma. □ B C A E F D′ D′′ D 96 Geometric Inequalities Using these elements we can now continue with the solution proposed by H. Schwarz for the Fagnano problem. We will now prove that the triangle with minimum perimeter is the ortic tri-angle. Denote this triangle as DEF and consider another triangle LMN inscribed in ABC. B C A E F D B A C A B F ′ N N ′ M L Reﬂect the complete ﬁgure on the side BC, so that the resultant triangle is re-ﬂected on CA, then on AB, on BC and ﬁnally on CA. We have in total six congruent triangles and within each of them we have the ortic triangle and the inscribed triangle LMN. The side AB of the last triangle is parallel to the side AB of the ﬁrst, since as a result of the ﬁrst reﬂection, the side AB is rotated in a negative direction through an angle of 2B, and then in a negative direction through an angle of 2A, the third reﬂection is invariant and the fourth is rotated through an angle of 2B in a positive direction and in the ﬁfth it is also rotated in a positive direction through an angle of 2A. Thus the total angle of rotation of AB is zero. The segment FF ′ is twice the perimeter of the ortic triangle, since FF ′ is composed of six pieces where each side of the ortic triangle is taken twice. Also, the broken line NN ′ is twice the perimeter of LMN. Moreover, NN ′ is parallel to the line FF ′ and of the same length, then since the length of the broken line NN ′ is greater than the length of the segment NN ′, we can deduce that the perimeter of DEF is less than the perimeter of LMN. The Fejer’s solution. The solution due to the Hungarian mathematician L. Fejer also uses reﬂections. Let LMN be a triangle inscribed on ABC. Take both the reﬂection L′ of the point L on the side CA, and L′′ the reﬂection of L on the side AB, and draw the segments ML′ and NL′′. It is clear that LM = ML′ and L′′N = NL. Hence the perimeter of LMN satisﬁes LM + MN + NL = L′′N + NM + ML′ ≥L′L′′. 2.8 Optimization problems 97 B C A M N L′ L′′ L Thus, we can conclude that if we ﬁx the point L, the points M and N that make the minimum perimeter LMN are the intersections of L′L′′ with CA and AB, respectively. Now, let us see which is the best option for the point L. We already know that the perimeter of LMN is L′L′′, thus L should make this quantity a minimum. B C A M N L′ L′′ L It is evident that AL = AL′ = AL′′ and that AC and AB are internal angle bisectors of the angles LAL′ and L′′AL, respectively. Thus ∠L′′AL′ = 2∠BAC = 2α which is a ﬁxed angle. The cosine law applied to the triangle AL′′L′ guarantees that (L′L′′)2 = (AL′)2 + (AL′′)2 −2AL′ · AL′′ cos 2α = 2AL2(1 −cos 2α). Then, L′L′′ is minimum when AL is minimum, which will be the case when AL is the altitude.15 A similar analysis using the points B and C will demonstrate that 15This would be enough to ﬁnish Fejer’s proof for the Fagnano’s problem. This is the case because if AL is the altitude and L′L′′ intersects sides CA and AB in E and F , respectively, then BE and CF are altitudes. Let us see why. The triangle AL′′L′ is isosceles with ∠L′′AL′ = 2∠A, then ∠AL′L′′ = 90◦−∠A and by symmetry ∠ELA = 90◦−∠A. Therefore ∠CLE = ∠A. Then AELB is a cyclic quadrilateral, therefore ∠AEB = ∠ALB = 90◦, which implies that BE is an altitude. Similarly, it follows that CE is an altitude. 98 Geometric Inequalities BM and CN should also be altitudes. Thus, the triangle LMN with minimum perimeter is the ortic triangle. Exercise 2.73. Let ABCD be a convex cyclic quadrilateral. If O is the intersection of the diagonals AC and BD, and P, Q, R, S are the feet of the perpendiculars of O on the sides AB, BC, CD, DA, respectively, prove that PQRS is the quadrilateral of minimum perimeter inscribed in ABCD. Exercise 2.74. Let P be a point inside the triangle ABC. Let D, E and F be the points of intersection of AP, BP and CP with the sides BC, CA and AB, respectively. Determine P such that the area of the triangle DEF is maximum. Exercise 2.75. (IMO, 1981) Let P be a point inside the triangle ABC. Let D, E, F be the feet of the perpendiculars from P to the lines BC, CA, AB, respectively. Find the point P that minimizes BC PD + CA PE + AB PF . Exercise 2.76. Let P, D, E and F be as in Exercise 2.75. For which point P is the sum of BD2 + CE2 + AF 2 minimum? Exercise 2.77. Let P, D, E and F be as in Exercise 2.75. For which point P is the product of PD · PE · PF maximum? Exercise 2.78. Let P be a point inside the triangle ABC. For which point P is the sum of PA2 + PB2 + PC2 minimum? Exercise 2.79. For every point P on the circumcircle of a triangle ABC, we draw the perpendiculars PM and PN to the sides AB and CA, respectively. Determine for which point P the length MN is maximum and ﬁnd that length. Exercise 2.80. (Turkey, 2000) Let ABC be an acute triangle with circumradius R; let ha, hb and hc be the lengths of the altitudes AD, BE and CF, respectively. Let ta, tb and tc be the lengths of the tangents from A, B and C, respectively, to the circumcircle DEF. Prove that t2 a ha + t2 b hb + t2 c hc ≤3 2R. Exercise 2.81. Let ha, hb, hc be the lengths of the altitudes of a triangle ABC, and let pa, pb, pc be the distances from a point P to the sides BC, CA, AB, respectively, where P is a point inside the triangle ABC. Prove that (i) ha pa + hb pb + hc pc ≥9, (ii) hahbhc ≥27papbpc, (iii) (ha −pa)(hb −pb)(hc −pc) ≥8papbpc. Exercise 2.82. If h is the length of the largest altitude of an acute triangle, then r + R ≤h. 2.8 Optimization problems 99 Exercise 2.83. Of all triangles with a common base and the same perimeter, the isosceles triangle has the largest area. Exercise 2.84. Of all triangles with a given perimeter, the one with largest area is the equilateral triangle. Exercise 2.85. Of all inscribed triangles on a given circle, the one with largest perimeter is the equilateral triangle. Exercise 2.86. If P is a point inside the triangle ABC, l = PA, m = PB and n = PC, prove that (lm + mn + nl)(l + m + n) ≥a2l + b2m + c2n. Exercise 2.87. (IMO, 1961) Let a, b and c be the lengths of the sides of a triangle and let (ABC) be the area of that triangle, prove that 4 √ 3(ABC) ≤a2 + b2 + c2. Exercise 2.88. Let (ABC) be the area of a triangle ABC and let F be the Fermat point of the triangle. Prove that 4 √ 3(ABC) ≤(AF + BF + CF)2. Exercise 2.89. Let P be a point inside the triangle ABC, prove that PA + PB + PC ≥6r. Exercise 2.90. (The area of the pedal triangle). For a triangle ABC and a point P on the plane, we deﬁne the “pedal triangle” of P with respect to ABC as the triangle A1B1C1 where A1, B1, C1 are the feet of the perpendiculars from P to BC, CA, AB, respectively. Prove that (A1B1C1) = (R2 −OP 2)(ABC) 4R2 , where O is the circumcenter. We can thus conclude that the pedal triangle of maximum area is the medial triangle. Chapter 3 Recent Inequality Problems Problem 3.1. (Bulgaria, 1995) Let SA, SB and SC denote the areas of the reg-ular heptagons A1A2A3A4A5A6A7, B1B2B3B4B5B6B7 and C1C2C3C4C5C6C7, respectively. Suppose that A1A2 = B1B3 = C1C4, prove that 1 2 < SB + SC SA < 2 − √ 2. Problem 3.2. (Czech and Slovak Republics, 1995) Let ABCD be a tetrahedron such that ∠BAC + ∠CAD + ∠DAB = ∠ABC + ∠CBD + ∠DBA = 180◦. Prove that CD ≥AB. Problem 3.3. (Estonia, 1995) Let a, b, c be the lengths of the sides of a triangle and let α, β, γ be the angles opposite to the sides. Prove that if the inradius of the triangle is r, then a sin α + b sin β + c sin γ ≥9r. Problem 3.4. (France, 1995) Three circles with the same radius pass through a common point. Let S be the set of points which are interior to at least two of the circles. How should the three circles be placed so that the area of S is minimized? Problem 3.5. (Germany, 1995) Let ABC be a triangle and let D and E be points on BC and CA, respectively, such that DE passes through the incenter of ABC. If S = area(CDE) and r is the inradius of ABC, prove that S ≥2r2. Problem 3.6. (Ireland, 1995) Let A, X, D be points on a line with X between A and D. Let B be a point such that ∠ABX = 120◦and let C be a point between B and X. Prove that 2AD ≥ √ 3 (AB + BC + CD). 102 Recent Inequality Problems Problem 3.7. (Korea, 1995) A ﬁnite number of points on the plane have the prop-erty that any three of them form a triangle with area at most 1. Prove that all these points lie within the interior or on the sides of a triangle with area less than or equal to 4. Problem 3.8. (Poland, 1995) For a ﬁxed positive integer n, ﬁnd the minimum value of the sum x1 + x2 2 2 + x3 3 3 + · · · + xn n n , given that x1, x2, . . . , xn are positive numbers satisfying that the sum of their reciprocals is n. Problem 3.9. (IMO, 1995) Let ABCDEF be a convex hexagon with AB = BC = CD and DE = EF = FA such that ∠BCD = ∠EFA = π 3 . Let G and H be points in the interior of the hexagon such that ∠AGB = ∠DHE = 2π 3 . Prove that AG + GB + GH + DH + HE ≥CF. Problem 3.10. (Balkan, 1996) Let O be the circumcenter and G the centroid of the triangle ABC. Let R and r be the circumradius and the inradius of the triangle. Prove that OG ≤ R(R −2r). Problem 3.11. (China, 1996) Suppose that x0 = 0, xi > 0 for i = 1, 2, . . . , n, and n i=1 xi = 1. Prove that 1 ≤ n i=1 xi √1 + x0 + · · · + xi−1 √xi + · · · + xn < π 2 . Problem 3.12. (Poland, 1996) Let n ≥2 and a1, a2, . . . , an ∈R+ with n i=1 ai = 1. Prove that for x1, x2, . . . , xn ∈R+, with n i=1 xi = 1, we have 2 i<j xixj ≤n −2 n −1 + n i=1 aix2 i 1 −ai . Problem 3.13. (Romania, 1996) Let x1, x2, . . . , xn, xn+1 be positive real numbers with x1 + x2 + · · · + xn = xn+1. Prove that n i=1 xi(xn+1 −xi) ≤ + + , n i=1 xn+1(xn+1 −xi). Problem 3.14. (St. Petersburg, 1996) Let M be the intersection of the diagonals of a cyclic quadrilateral, let N be the intersection of the segments that join the opposite midpoints and let O be the circumcenter. Prove that OM ≥ON. Recent Inequality Problems 103 Problem 3.15. (Austria–Poland, 1996) If w, x, y and z are real numbers satisfying w + x + y + z = 0 and w2 + x2 + y2 + z2 = 1, prove that −1 ≤wx + xy + yz + zw ≤0. Problem 3.16. (Taiwan, 1997) Let a1, . . . , an be positive numbers such that ai−1+ai+1 ai is an integer for all i = 1, . . . , n, a0 = an, an+1 = a1 and n ≥3. Prove that 2n ≤an + a2 a1 + a1 + a3 a2 + a2 + a4 a3 + · · · + an−1 + a1 an ≤3n. Problem 3.17. (Taiwan, 1997) Let ABC be an acute triangle with circumcenter O and circumradius R. Prove that if AO intersects the circumcircle of OBC at D, BO intersects the circumcircle of OCA at E and CO intersects the circumcircle of OAB at F, then OD · OE · OF ≥8R3. Problem 3.18. (APMO, 1997) Let ABC be a triangle. The internal bisector of the angle in A meets the segment BC at X and the circumcircle at Y . Let la = AX AY . Deﬁne lb and lc in the same way. Prove that la sin2 A + lb sin2 B + lc sin2 C ≥3 with equality if and only if the triangle is equilateral. Problem 3.19. (IMO, 1997) Let x1, . . . , xn be real numbers satisfying \|x1 + · · · + xn\| = 1 and \|xi\| ≤n+1 2 for all i = 1, . . . , n. Prove that there exists a permutation y1, . . . , yn of x1, . . . , xn such that \|y1 + 2y2 + · · · + nyn\| ≤n + 1 2 . Problem 3.20. (Czech and Slovak Republics, 1998) Let a, b, c be positive real numbers. A triangle exists with sides of lengths a, b and c if and only if there exist numbers x, y and z such that y z + z y = a x, z x + x z = b y , x y + y x = c z . Problem 3.21. (Hungary, 1998) Let ABCDEF be a centrally symmetric hexagon and let P, Q, R be points on the sides AB, CD, EF, respectively. Prove that the area of the triangle PQR is at most one-half of the area of the hexagon. Problem 3.22. (Iran, 1998) Let x1, x2, x3 and x4 be positive real numbers such that x1x2x3x4 = 1. Prove that x3 1 + x3 2 + x3 3 + x3 4 ≥max x1 + x2 + x3 + x4, 1 x1 + 1 x2 + 1 x3 + 1 x4 . 104 Recent Inequality Problems Problem 3.23. (Iran, 1998) Let x, y, z be numbers greater than 1 and such that 1 x + 1 y + 1 z = 2. Prove that √x + y + z ≥ √ x −1 + y −1 + √ z −1. Problem 3.24. (Mediterranean, 1998) Let ABCD be a square inscribed in a circle. If M is a point on the arc AB, prove that MC · MD ≥3 √ 3MA · MB. Problem 3.25. (Nordic, 1998) Let P be a point inside an equilateral triangle ABC of length side a. If the lines AP, BP and CP intersect the sides BC, CA and AB of the triangle at L, M and N, respectively, prove that PL + PM + PN < a. Problem 3.26. (Spain, 1998) A line that contains the centroid G of the triangle ABC intersects the side AB at P and the side CA at Q. Prove that PB PA · QC QA ≤1 4. Problem 3.27. (Armenia, 1999) Let O be the center of the circumcircle of the acute triangle ABC. The lines CO, AO and BO intersect the circumcircles of the triangles AOB, BOC and AOC, for the second time, at C1, A1 and B1, respectively. Prove that AA1 OA1 + BB1 OB1 + CC1 OC1 ≥9 2. Problem 3.28. (Balkan, 1999) Let ABC be an acute triangle and let L, M, N be the feet of the perpendiculars from the centroid G of ABC to the sides BC, CA, AB, respectively. Prove that 4 27 < (LMN) (ABC) ≤1 4. Problem 3.29. (Belarus, 1999) Let a, b, c be positive real numbers such that a2 + b2 + c2 = 3. Prove that 1 1 + ab + 1 1 + bc + 1 1 + ca ≥3 2. Problem 3.30. (Czech and Slovak Republics, 1999) For arbitrary positive numbers a, b and c, prove that a b + 2c + b c + 2a + c a + 2b ≥1. Recent Inequality Problems 105 Problem 3.31. (Ireland, 1999) Let a, b, c, d be positive real numbers with a + b + c + d = 1. Prove that a2 a + b + b2 b + c + c2 c + d + d2 d + a ≥1 2. Problem 3.32. (Italy, 1999) Let D and E be given points on the sides AB and CA of the triangle ABC such that DE is parallel to BC and DE is tangent to the incircle of ABC. Prove that DE ≤AB + BC + CA 8 . Problem 3.33. (Poland, 1999) Let D be a point on the side BC of the triangle ABC such that AD > BC. The point E on CA is deﬁned by the equation AE EC = BD AD−BC . Prove that AD > BE. Problem 3.34. (Romania, 1999) Let a, b, c be positive real numbers such that ab + bc + ca ≤3abc. Prove that a + b + c ≤a3 + b3 + c3. Problem 3.35. (Romania, 1999) Let x1, x2, . . . , xn be positive real numbers such that x1x2 · · · xn = 1. Prove that 1 n −1 + x1 + 1 n −1 + x2 + · · · + 1 n −1 + xn ≤1. Problem 3.36. (Romania, 1999) Let n ≥2 be a positive integer and x1, y1, x2, y2, . . . , xn, yn be positive real numbers such that x1 + x2 + · · · + xn ≥x1y1 + x2y2 + · · · + xnyn. Prove that x1 + x2 + · · · + xn ≤x1 y1 + x2 y2 + · · · + xn yn . Problem 3.37. (Russia, 1999) Let a, b and c be positive real numbers with abc = 1. Prove that if a + b + c ≤1 a + 1 b + 1 c, then an + bn + cn ≤ 1 an + 1 bn + 1 cn for every positive integer n. Problem 3.38. (Russia, 1999) Let {x} = x −[x] denote the fractional part of x. Prove that for every natural number n, n2 j=1 - j . ≤n2 −1 2 . Problem 3.39. (Russia, 1999) The positive real numbers x and y satisfy x2 + y3 ≥ x3 + y4. Prove that x3 + y3 ≤2. 106 Recent Inequality Problems Problem 3.40. (St. Petersburg, 1999) Let x0 > x1 > · · · > xn be real numbers. Prove that x0 + 1 x0 −x1 + 1 x1 −x2 + · · · + 1 xn−1 −xn ≥xn + 2n. Problem 3.41. (Turkey, 1999) Prove that (a + 3b)(b + 4c)(c + 2a) ≥60abc for all real numbers 0 ≤a ≤b ≤c. Problem 3.42. (United Kingdom, 1999) Three non-negative real numbers a, b and c satisfy a + b + c = 1. Prove that 7(ab + bc + ca) ≤2 + 9abc. Problem 3.43. (USA, 1999) Let ABCD be a convex cyclic quadrilateral. Prove that \|AB −CD\| + \|AD −BC\| ≥2 \|AC −BD\| . Problem 3.44. (APMO, 1999) Let {an} be a sequence of real numbers satisfying ai+j ≤ai + aj for all i, j = 1, 2, . . .. Prove that a1 + a2 2 + · · · + an n ≥an for all n ∈N. Problem 3.45. (IMO, 1999) Let n ≥2 be a ﬁxed integer. (a) Determine the smallest constant C such that 1≤i 0 and x ≥y ≥z > 0. Prove that a2x2 (by + cz)(bz + cy) + b2y2 (cz + ax)(cx + az) + c2z2 (ax + by)(ay + bx) ≥3 4. Problem 3.48. (Mediterranean, 2000) Let P, Q, R, S be the midpoints of the sides BC, CD, DA, AB, respectively, of the convex quadrilateral ABCD. Prove that 4(AP 2 + BQ2 + CR2 + DS2) ≤5(AB2 + BC2 + CD2 + DA2). Recent Inequality Problems 107 Problem 3.49. (Austria–Poland, 2000) Let x, y, z be non-negative real numbers such that x + y + z = 1. Prove that 2 ≤(1 −x2)2 + (1 −y2)2 + (1 −z2)2 ≤(1 + x)(1 + y)(1 + z). Problem 3.50. (IMO, 2000) Let a, b, c be positive real numbers with abc = 1. Prove that a −1 + 1 b b −1 + 1 c c −1 + 1 a ≤1. Problem 3.51. (Balkan, 2001) Let a, b, c be positive real numbers such that abc ≤ a + b + c. Prove that a2 + b2 + c2 ≥ √ 3 abc. Problem 3.52. (Brazil, 2001) Prove that (a + b)(a + c) ≥2 abc(a + b + c), for all positive real numbers a, b, c. Problem 3.53. (Poland, 2001) Prove that the inequality n i=1 ixi ≤ n 2 + n i=1 xi i holds for every integer n≥2 and for all non-negative real numbers x1, x2, . . . , xn. Problem 3.54. (Austria–Poland, 2001) Prove that 2 < a + b c + b + c a + c + a b −a3 + b3 + c3 abc ≤3, where a, b, c are the lengths of the sides of a triangle. Problem 3.55. (IMO, 2001) Prove that for a, b and c positive real numbers we have a √ a2 + 8bc + b √ b2 + 8ca + c √ c2 + 8ab ≥1. Problem 3.56. (Short list IMO, 2001) Let x1, x2, . . ., xn be real numbers, prove that x1 1 + x2 1 + x2 1 + x2 1 + x2 2 + · · · + xn 1 + x2 1 + · · · + x2 n < √n. Problem 3.57. (Austria, 2002) Let a, b, c be real numbers such that there exist α, β, γ ∈{−1, 1} with αa + βb + γc = 0. Determine the smallest positive value of a3+b3+c3 abc 2 . Problem 3.58. (Balkan, 2002) Prove that 2 b(a + b) + 2 c(b + c) + 2 a(c + a) ≥ 27 (a + b + c)2 for positive real numbers a, b, c. 108 Recent Inequality Problems Problem 3.59. (Canada, 2002) Prove that for all positive real numbers a, b, c, a3 bc + b3 ca + c3 ab ≥a + b + c, and determine when equality occurs. Problem 3.60. (Ireland, 2002) Prove that x 1 −x + y 1 −y + z 1 −z ≥ 3 3 √xyz 1 − 3 √xyz for positive real numbers x, y, z less than 1. Problem 3.61. (Rioplatense, 2002) Let a, b, c be positive real numbers. Prove that a b + c + 1 2 b c + a + 1 2 c a + b + 1 2 ≥1. Problem 3.62. (Rioplatense, 2002) Let a, b, c be positive real numbers. Prove that a + b c2 + b + c a2 + c + a b2 ≥ 9 a + b + c + 1 a + 1 b + 1 c . Problem 3.63. (Russia, 2002) Prove that √x + √y + √z ≥xy + yz + zx for x, y, z positive real numbers such that x + y + z = 3. Problem 3.64. (APMO, 2002) Let a, b, c be positive real numbers satisfying 1 a + 1 b + 1 c = 1. Prove that √ a + bc + √ b + ca + √ c + ab ≥ √ abc + √a + √ b + √c. Problem 3.65. (Ireland, 2003) The lengths a, b, c of the sides of a triangle are such that a + b + c = 2. Prove that 1 ≤ab + bc + ca −abc ≤1 + 1 27. Problem 3.66. (Romania, 2003) Prove that in any triangle ABC the following inequality holds: 1 mbmc + 1 mcma + 1 mamb ≤ √ 3 S , where S is the area of the triangle and ma, mb, mc are the lengths of the medians. Problem 3.67. (Romania, 2003) Let a, b, c, d be positive real numbers with abcd = 1. Prove that 1 + ab 1 + a + 1 + bc 1 + b + 1 + cd 1 + c + 1 + da 1 + d ≥4. Recent Inequality Problems 109 Problem 3.68. (Romania, 2003) In a triangle ABC, let la, lb, lc be the lengths of the internal angle bisectors, and let s be the semiperimeter. Prove that la + lb + lc ≤ √ 3s. Problem 3.69. (Russia, 2003) Let a, b, c be positive real numbers with a + b + c = 1. Prove that 1 1 −a + 1 1 −b + 1 1 −c ≥ 2 1 + a + 2 1 + b + 2 1 + c. Problem 3.70. (APMO, 2003) Prove that (an + bn) 1 n + (bn + cn) 1 n + (cn + an) 1 n < 1 + 2 1 n 2 , where n > 1 is an integer and a, b, c are the side-lengths of a triangle with unit perimeter. Problem 3.71. (IMO, 2003) Given n > 2 and real numbers x1 ≤x2 ≤· · · ≤xn, prove that ⎛ ⎝ i,j \|xi −xj\| ⎞ ⎠ 2 ≤2 3(n2 −1) i,j (xi −xj)2, where equality holds if and only if x1, x2, . . . , xn form an arithmetic progression. Problem 3.72. (Short list Iberoamerican, 2004) If the positive numbers x1, x2, . . . , xn satisfy x1 + x2 + · · · + xn = 1, prove that x1 x2(x1 + x2 + x3) + x2 x3(x2 + x3 + x4) + · · · + xn x1(xn + x1 + x2) ≥n2 3 . Problem 3.73. (Czech and Slovak Republics, 2004) Let P(x) = ax2 + bx + c be a quadratic polynomial with non-negative real coeﬃcients. Prove that for any positive number x, P(x)P 1 x ≥(P(1))2. Problem 3.74. (Croatia, 2004) Prove that the inequality a2 (a + b)(a + c) + b2 (b + c)(b + a) + c2 (c + a)(c + b) ≥3 4 holds for all positive real numbers a, b, c. Problem 3.75. (Estonia, 2004) Let a, b, c be positive real numbers such that a2 + b2 + c2 = 3. Prove that 1 1 + 2ab + 1 1 + 2bc + 1 1 + 2ca ≥1. 110 Recent Inequality Problems Problem 3.76. (Iran, 2004) Let x, y, z be real numbers such that xyz = −1, prove that x4 + y4 + z4 + 3(x + y + z) ≥x2 y + x2 z + y2 x + y2 z + z2 x + z2 y . Problem 3.77. (Korea, 2004) Let R and r be the circumradius and the inradius of the acute triangle ABC, respectively. Suppose that ∠A is the largest angle of ABC. Let M be the midpoint of BC and let X be the intersection of the tangents to the circumcircle of ABC at B and C. Prove that r R ≥AM AX . Problem 3.78. (Moldova, 2004) Prove that for all real numbers a, b, c ≥0, the following inequality holds: a3 + b3 + c3 ≥a2√ bc + b2√ca + c2√ ab. Problem 3.79. (Ukraine, 2004) Let x, y, z be positive real numbers with x+y+z = 1. Prove that √xy + z + √yz + x + √zx + y ≥1 + √xy + √yz + √zx. Problem 3.80. (Ukraine, 2004) Let a, b, c be positive real numbers such that abc ≥1. Prove that a3 + b3 + c3 ≥ab + bc + ca. Problem 3.81. (Romania, 2004) Find all positive real numbers a, b, c which satisfy the inequalities 4(ab + bc + ca) −1 ≥a2 + b2 + c2 ≥3(a3 + b3 + c3). Problem 3.82. (Romania, 2004) The real numbers a, b, c satisfy a2 + b2 + c2 = 3. Prove the inequality \|a\| + \|b\| + \|c\| −abc ≤4. Problem 3.83. (Romania, 2004) Consider the triangle ABC and let O be a point in the interior of ABC. The straight lines OA, OB, OC meet the sides of the triangle at A1, B1, C1, respectively. Let R1, R2, R3 be the radii of the circumcircles of the triangles OBC, OCA, OAB, respectively, and let R be the radius of the circumcircle of the triangle ABC. Prove that OA1 AA1 R1 + OB1 BB1 R2 + OC1 CC1 R3 ≥R. Problem 3.84. (Romania, 2004) Let n ≥2 be an integer and let a1, a2, . . . , an be real numbers. Prove that for any non-empty subset S ⊂{1, 2, . . ., n}, the following inequality holds: i∈S ai 2 ≤ 1≤i≤j≤n (ai + · · · + aj)2. Recent Inequality Problems 111 Problem 3.85. (APMO, 2004) For any positive real numbers a, b, c, prove that (a2 + 2)(b2 + 2)(c2 + 2) ≥9(ab + bc + ca). Problem 3.86. (Short list IMO, 2004) Let a, b and c be positive real numbers such that ab + bc + ca = 1. Prove that 3 3 1 abc + 6(a + b + c) ≤ 3 √ 3 abc. Problem 3.87. (IMO, 2004) Let n ≥3 be an integer. Let t1, t2, . . . , tn be positive real numbers such that n2 + 1 > (t1 + t2 + · · · + tn) 1 t1 + 1 t2 + · · · + 1 tn . Prove that ti, tj, tk are the side-lengths of a triangle for all i, j, k with 1 ≤i < j < k ≤n. Problem 3.88. (Japan, 2005) Let a, b and c be positive real numbers such that a + b + c = 1. Prove that a 3 √ 1 + b −c + b 3 √ 1 + c −a + a 3 √ 1 + a −b ≤1. Problem 3.89. (Russia, 2005) Let x1, x2, . . . , x6 be real numbers such that x2 1 + x2 2 + · · · + x2 6 = 6 and x1 + x2 + · · · + x6 = 0. Prove that x1x2 · · · x6 ≤1 2. Problem 3.90. (United Kingdom, 2005) Let a, b, c be positive real numbers. Prove that a b + b c + c a 2 ≥(a + b + c) 1 a + 1 b + 1 c . Problem 3.91. (APMO, 2005) Let a, b and c be positive real numbers such that abc = 8. Prove that a2 (1 + a3)(1 + b3) + b2 (1 + b3)(1 + c3) + c2 (1 + c3)(1 + a3) ≥4 3. Problem 3.92. (IMO, 2005) Let x, y, z be positive real numbers such that xyz ≥1. Prove that x5 −x2 x5 + y2 + z2 + y5 −y2 y5 + z2 + x2 + z5 −z2 z5 + x2 + y2 ≥0. Problem 3.93. (Balkan, 2006) Let a, b, c be positive real numbers, prove that 1 a(b + 1) + 1 b(c + 1) + 1 c(a + 1) ≥ 3 1 + abc. 112 Recent Inequality Problems Problem 3.94. (Estonia, 2006) Let O be the circumcenter of the acute triangle ABC and let A′, B′ and C′ be the circumcenter of the triangles BCO, CAO and ABO, respectively. Prove that the area of the triangle ABC is less than or equal to the area of the triangle A′B′C′. Problem 3.95. (Lithuania, 2006) Let a, b, c be positive real numbers, prove that 1 a2 + bc + 1 b2 + ca + 1 c2 + ab ≤1 2 1 ab + 1 bc + 1 ca . Problem 3.96. (Turkey, 2006) Let a1, a2, . . . , an be positive real numbers such that a1 + a2 + · · · + an = a2 1 + a2 2 + · · · + a2 n = A. Prove that i̸=j ai aj ≥(n −1)2A A −1 . Problem 3.97. (Iberoamerican, 2006) Consider n real numbers a1, a2, . . . , an, not necessarily distinct. Let d be the diﬀerence between the maximum and the minimum value of the numbers and let s = i 1. Problem 3.104. (Mediterranean, 2007) Let x, y, z be real numbers such that xy + yz + zx = 1. Prove that xz < 1 2. Is it possible to improve the bound 1 2? Problem 3.105. (Mediterranean, 2007) Let x > 1 be a real number which is not an integer. Prove that x + {x} [x] − [x] x + {x} + x + [x] {x} − {x} x + [x] > 9 2, where [x] and {x} represent the integer part and the fractional part of x, respec-tively. Problem 3.106. (Peru, 2007) Let a, b, c be positive real numbers such that a+b+ c ≥1 a + 1 b + 1 c. Prove that a + b + c ≥ 3 a + b + c + 2 abc. Problem 3.107. (Romania, 2007) Let a, b, c be positive real numbers such that 1 a + b + 1 + 1 b + c + 1 + 1 c + a + 1 ≥1. Prove that a + b + c ≥ab + bc + ca. Problem 3.108. (Romania, 2007) Let ABC be an acute triangle with AB = AC. For every interior point P of ABC, consider the circle with center A and radius AP; let M and N be the intersections of the sides AB and AC with the circle, respectively. Determine the position of P in such a way that MN + BP + CP is minimum. Problem 3.109. (Romania, 2007) The points M, N, P on the sides BC, CA, AB, respectively, are such that the triangle MNP is acute. Let x be the length of the shortest altitude in the triangle ABC and let X be the length of the largest altitude in the triangle MNP. Prove that x ≤2X. 114 Recent Inequality Problems Problem 3.110. (APMO, 2007) Let x, y, z be positive real numbers such that √x + √y + √z = 1. Prove that x2 + yz 2x2(y + z) + y2 + zx 2y2(z + x) + z2 + xy 2z2(x + y) ≥1. Problem 3.111. (Baltic, 2008) If the positive real numbers a, b, c satisfy a2 + b2 + c2 = 3, prove that a2 2 + b + c2 + b2 2 + c + a2 + c2 2 + a + b2 ≥(a + b + c)2 12 . Under which circumstances the equality holds? Problem 3.112. (Canada, 2008) Let a, b, c be positive real numbers for which a + b + c = 1. Prove that a −bc a + bc + b −ca b + ca + c −ab c + ab ≤3 2. Problem 3.113. (Iran, 2008) Find the least real number K such that for any positive real numbers x, y, z, the following inequality holds: x√y + y√z + z√x ≤K (x + y)(y + z)(z + x). Problem 3.114. (Ireland, 2008) If the positive real numbers a, b, c, d satisfy a2 + b2 + c2 + d2 = 1, prove that a2b2cd + ab2c2d + abc2d2 + a2bcd2 + a2bc2d + ab2cd2 ≤3 32. Problem 3.115. (Ireland, 2008) Let x, y, z be positive real numbers such that xyz ≥1. Prove that (a) 27 ≤(1 + x + y)2 + (1 + y + z)2 + (1 + z + x)2, (b) (1 + x + y)2 + (1 + y + z)2 + (1 + z + x)2 ≤3(x + y + z)2. The equalities hold if and only if x = y = z = 1. Problem 3.116. (Romania, 2008) If a, b, c are positive real numbers with ab+bc+ ca = 3, prove that 1 1 + a2(b + c) + 1 1 + b2(c + a) + 1 1 + c2(a + b) ≤ 1 abc. Problem 3.117. (Romania, 2008) Determine the maximum value for the real num-ber k if (a + b + c) 1 a + b + 1 b + c + 1 c + a −k ≥k for all real numbers a, b, c ≥0 and with a + b + c = ab + bc + ca. Recent Inequality Problems 115 Problem 3.118. (Serbia, 2008) Let a, b, c be positive real numbers such that a + b + c = 1. Prove that a2 + b2 + c2 + 3abc ≥4 9. Problem 3.119. (Vietnam, 2008) Let x, y, z be distinct non-negative real numbers. Prove that 1 (x −y)2 + 1 (y −z)2 + 1 (z −x)2 ≥ 4 xy + yz + zx. When is the case that the equality holds? Problem 3.120. (IMO, 2008) (i) If x, y, z are three real numbers diﬀerent from 1 and such that xyz = 1, prove that x2 (x −1)2 + y2 (y −1)2 + z2 (z −1)2 ≥1. (ii) Prove that the equality holds for an inﬁnite number of x, y, z, all of them being rational numbers. Chapter 4 Solutions to Exercises and Problems In this chapter we present solutions or hints to the exercises and problems that appear in this book. In Sections 1 and 2 we provide the solutions to the exercises in Chapters 1 and 2, respectively, and in Section 3 the solutions to the problems in Chapter 3. We recommend that the reader should consult this chapter only after having tried to solve the exercises or the problems by himself. 4.1 Solutions to the exercises in Chapter 1 Solution 1.1. It follows from the deﬁnition of a < b and Property 1.1.1 for the number a −b. Solution 1.2. (i) If a < 0, then −a > 0. Also use (−a)(−b) = ab. (ii) (−a)b > 0. (iii) a < b ⇔b −a > 0, now use property 1.1.2. (iv) Use property 1.1.2. (v) If a < 0, then −a > 0. (vi) a 1 a = 1 > 0. (vii) If a < 0, then −a > 0. (viii) Use (vi) and property 1.1.3. (ix) Prove that ac < bc and that bc < bd. (x) Use property 1.1.3 with a −1 > 0 and a > 0. (xi) Use property 1.1.3 with 1 −a > 0 and a > 0. Solution 1.3. (i) a2 < b2 ⇔b2 −a2 = (b + a)(b −a) > 0. (ii) If b > 0, then 1 b > 0, now use Example 1.1.4. Solution 1.4. For (i), (ii) and (iii) use the deﬁnition, and for (iv) and (v) remember that \|a\|2 = a2. 118 Solutions to Exercises and Problems Solution 1.5. (i) x ≤\|x\| and −x ≤\|x\| . (ii) Consider \|a\| = \|a −b + b\| and \|b\| = \|b −a + a\|, and apply the triangle inequal-ity. (iii) (x2 + xy + y2)(x −y) = x3 −y3. (iv) (x2 −xy + y2)(x + y) = x3 + y3. Solution 1.6. If a, b or c is zero, the equality follows. Then, we can assume \|a\| ≥ \|b\| ≥\|c\| > 0. Dividing by \|a\|, the inequality is equivalent to 1 + b a + c a − 1 + b a − b a + c a − 1 + c a + 1 + b a + c a ≥0. Since b a ≤1 and c a ≤1, we can deduce that 1 + b a = 1+ b a and 1 + c a = 1+ c a. Thus, it is suﬃcient to prove that b a + c a − b a + c a − 1 + b a + c a + 1 + b a + c a ≥0. Now, use the triangle inequality and Exercise 1.5. Solution 1.7. (i) Use that 0 ≤b ≤1 and 1 + a > 0 in order to see that 0 ≤b(1 + a) ≤1 + a ⇒0 ≤b −a ≤1 −ab ⇒0 ≤b −a 1 −ab ≤1. (ii) The inequality on the left-hand side is clear. Since 1 + a ≤1 + b, it follows that 1 1+b ≤ 1 1+a, and then prove that a 1 + b + b 1 + a ≤ a 1 + a + b 1 + a = a + b 1 + a ≤1. (iii) For the inequality on the left-hand side, use that ab2 −ba2 = ab(b −a) is the product of non-negative real numbers. For the inequality on the right-hand side, note that b ≤1 ⇒b2 ≤b ⇒−b ≤−b2, and then ab2 −ba2 ≤ab2 −b2a2 = b2(a −a2) ≤a −a2 = 1 4 −(1 2 −a)2 ≤1 4. Solution 1.8. Prove in general that x < √ 2 ⇒1 + 1 1+x > √ 2 and that x > √ 2 ⇒ 1 + 1 1+x < √ 2. Solution 1.9. ax + by ≥ay + bx ⇔(a −b)(x −y) ≥0. Solution 1.10. We can assume that x ≥y. Then, use the previous exercise substi-tuting with √ x2, y2, 1 √y and 1 √x. Solution 1.11. Observe that (a −b)(c −d) + (a −c)(b −d) + (d −a)(b −c) = 2(a −b)(c −d) = 2(a −b)2 ≥0. 4.1 Solutions to the exercises in Chapter 1 119 Solution 1.12. It follows from f(a, c, b, d) −f(a, b, c, d) = (a −c)2 −(a −b)2 + (b −d)2 −(c −d)2 = (b −c)(2a −b −c) + (b −c)(b + c −2d) = 2(b −c)(a −d) > 0, f(a, b, c, d) −f(a, b, d, c) = (b −c)2 −(b −d)2 + (d −a)2 −(c −a)2 = (d −c)(2b −c −d) + (d −c)(c + d −2a) = 2(d −c)(b −a) > 0. Solution 1.13. In order for the expressions in the inequality to be well deﬁned, it is necessary that x ≥−1 2 and x ̸= 0. Multiply the numerator and the denominator by (1 + √1 + 2x)2. Perform some simpliﬁcations and show that 2√2x + 1 < 7; then solve for x. Solution 1.14. Since 4n2 < 4n2 + n < 4n2 + 4n + 1, we can deduce that 2n < √ 4n2 + n < 2n + 1. Hence, its integer part is 2n and then we have to prove that √ 4n2 + n < 2n + 1 4, this follows immediately after squaring both sides of the inequality. Solution 1.15. Since (a3 −b3)(a2 −b2) ≥0, we have that a5 + b5 ≥a2b2(a + b), then ab a5 + b5 + ab ≤ ab a2b2(a + b) + ab = abc2 a2b2c2(a + b) + abc2 = c a + b + c. Similarly, bc b5+c5+bc ≤ a a+b+c and ca c5+a5+ca ≤ b a+b+c. Hence, ab a5 + b5 + ab + bc b5 + c5 + bc + ca c5 + a5 + ca ≤ c a + b + c + a a + b + c + b a + b + c, but c a+b+c + a a+b+c + b a+b+c = c+a+b a+b+c = 1. Solution 1.16. Consider p(x) = ax2 +bx+c, using the hypothesis, p(1) = a+b+c and p(−1) = a −b + c are not negative. Since a > 0, the minimum value of p is attained at −b 2a and its value is 4ac−b2 4a < 0. If x1, x2 are the roots of p, we can deduce that b a = −(x1 + x2) and c a = x1x2, therefore a+b+c a = (1 −x1)(1 −x2), a−b+c a = (1 + x1)(1 + x2) and a−c a = 1 −x1x2. Observe that, (1 −x1)(1 −x2) ≥0, (1 + x1)(1 + x2) ≥0 and 1 −x1x2 ≥0 imply that −1 ≤x1,x2 ≤1. Solution 1.17. If the inequalities are true, then a, b and c are less than 1, and a(1 −b)b(1 −c)c(1 −a) > 1 64. On the other hand, since x(1 −x) ≤1 4 for 0 ≤x ≤1, then a(1 −b)b(1 −c)c(1 −a) ≤ 1 64. Solution 1.18. Use the AM-GM inequality with a = 1, b = x. 120 Solutions to Exercises and Problems Solution 1.19. Use the AM-GM inequality with a = x, b = 1 x. Solution 1.20. Use the AM-GM inequality with a = x2, b = y2. Solution 1.21. In the previous exercise add x2 + y2 to both sides. Solution 1.22. Use the AM-GM inequality with a = x+y x , b = x+y y and also use the AM-GM inequality for x and y. Or reduce this to Exercise 1.20. Solution 1.23. Use the AM-GM inequality with ax and b x. Solution 1.24. Use the AM-GM inequality with a b and b a. Solution 1.25. a+b 2 − √ ab = ( √a− √ b) 2 2 , simplify and ﬁnd the bounds using 0 < b ≤ a. Solution 1.26. x + y ≥2√xy. Solution 1.27. x2 + y2 ≥2xy. Solution 1.28. xy + zx ≥2x√yz. Solution 1.29. See Exercise 1.27. Solution 1.30. 1 x + 1 y ≥ 2 √xy . Solution 1.31. xy z + yz x ≥2 xy2z zx = 2y. Solution 1.32. x2+(y2+z2) 2 ≥x y2 + z2. Solution 1.33. x4 + y4 + 8 = x4 + y4 + 4 + 4 ≥4 4 x4y416 = 8xy. Solution 1.34. (a + b + c + d) ≥4 4 √ abcd, 1 a + 1 b + 1 c + 1 d ≥4 4 1 abcd. Solution 1.35. a b + b c + c d + d a ≥4 4 a b b c c d d a = 4. Solution 1.36. (x1 + · · · + xn) ≥n n √x1 · · · xn, 1 x1 + · · · + 1 xn ≥n n 1 x1···xn . Solution 1.37. a1 b1 + a2 b2 + · · · + an bn ≥n n a1···an b1···bn = n. Solution 1.38. an −1 > n a n+1 2 −a n−1 2 ⇔(a −1) an−1 + · · · + 1 > na n−1 2 (a− 1) ⇔an−1+···+a+1 n > a n−1 2 , but 1+a+···+an−1 n > n a (n−1)n 2 = a n−1 2 . Solution 1.39. 1 = 1+a 2 1+b 2 1+c 2 ≥√a √ b√c = √ abc. Solution 1.40. Using the AM-GM inequality, we obtain a3 b + b3 c + bc ≥3 3 a3 b · b3 c · bc = 3ab. 4.1 Solutions to the exercises in Chapter 1 121 Similarly, b3 c + c3 a + ca ≥3bc and c3 a + a3 b + ab ≥3ca. Therefore, 2( a3 b + b3 c + c3 a ) + (ab + bc + ca) ≥3(ab + bc + ca). Second solution. The inequality can also be proved using Exercise 1.107. Solution 1.41. If abc = 0, the result is clear. If abc > 0, then we have ab c + bc a + ca b = 1 2 a b c + c b + b c a + a c + c a b + b a ≥1 2(2a + 2b + 2c), and the result is evident. Solution 1.42. Apply the AM-GM inequality twice over, a2b + b2c + c2a ≥3abc, ab2 + bc2 + ca2 ≥3abc. Solution 1.43. 1+ab 1+a = abc+ab 1+a = ab 1+c 1+a , 1 + ab 1 + a + 1 + bc 1 + b + 1 + ca 1 + c = ab 1 + c 1 + a + bc 1 + a 1 + b + ca 1 + b 1 + c ≥3 3 (abc)2 = 3. Solution 1.44. 1 a+b + 1 b+c + 1 c+a (a + b + c) ≥9 2 is equivalent to 1 a + b + 1 b + c + 1 c + a (a + b + b + c + c + a) ≥9, which follows from Exercise 1.36. For the other inequality use 1 a + 1 b ≥ 4 a+b. See Exercise 1.22. Solution 1.45. Note that n + Hn n = (1 + 1) + (1 + 1 2) + · · · + (1 + 1 n) n . Now, apply the AM-GM inequality. Solution 1.46. Setting yi = 1 1+xi , then xi = 1 yi −1 = 1−yi yi . Observe that y1 +· · ·+ yn = 1 implies that 1 −yi = j̸=i yi, then j̸=i yi ≥(n −1) 0 j̸=i yj 1 n−1 and 1 i xi = 1 i 1 −yi yi = 0 i j̸=i yj 0 i yi ≥ (n −1)n 0 i 0 j̸=i yj 1 n−1 0 i yi = (n −1)n. 122 Solutions to Exercises and Problems Solution 1.47. Deﬁne an+1 = 1−(a1 +· · ·+an) and xi = 1−ai ai for i = 1, . . . , n+1. Apply Exercise 1.46 directly. Solution 1.48. n i=1 1 1+ai = 1 ⇒n i=1 ai 1+ai = n −1. Observe that n i=1 √ai −(n −1) n i=1 1 √ai = n i=1 1 1 + ai n i=1 √ai − n i=1 ai 1 + ai n i=1 1 √ai = i,j ai −aj (1 + aj)√ai = i>j (√ai√aj −1)(√ai −√aj)2(√ai + √aj) (1 + ai)(1 + aj)√ai√aj . Since 1 ≥ 1 1+ai + 1 1+aj = 2+ai+aj 1+ai+aj+aiaj , we can deduce that aiaj ≥1. Hence the terms of the last sum are positive. Solution 1.49. Let Sa = n i=1 a2 i ai+bi and Sb = n i=1 b2 i ai+bi . Then Sa −Sb = n i=1 a2 i −b2 i ai + bi = n i=1 ai − n i=1 bi = 0, thus Sa = Sb = S. Hence, we have 2S = n i=1 a2 i + b2 i ai + bi ≥1 2 n i=1 (ai + bi)2 ai + bi = n i=1 ai, where the inequality follows after using Exercise 1.21. Solution 1.50. Since the inequality is homogeneous16 we can assume that abc = 1. Setting x = a3, y = b3 and z = c3, the inequality is equivalent to 1 x + y + 1 + 1 y + z + 1 + 1 z + x + 1 ≤1. Let A = x + y + 1, B = y + z + 1 and C = z + x + 1, then 1 A + 1 B + 1 C ≤1 ⇔(A −1)(B −1)(C −1) −(A + B + C) + 1 ≥0 ⇔(x + y)(y + z)(z + x) −2(x + y + z) ≥2 ⇔(x + y + z)(xy + yz + zx −2) ≥3. Now, use that x + y + z 3 ≥(xzy) 1 3 and xy + yz + zx 3 ≥(xyz) 2 3 . 16A function f(a, b, . . .) is homogeneous if f(ta, tb, . . .) = tf(a, b, . . .) for each t ∈R. Then, an inequality of the form f(a, b, . . .) ≥0, in the case of a homogeneous function, is equivalent to f(ta, tb, . . .) ≥0 for any t > 0. 4.1 Solutions to the exercises in Chapter 1 123 Second solution. Follow the ideas used in the solution of Exercise 1.15. Start with the inequality (a2 −b2)(a −b) ≥0 to guarantee that a3 + b3 + abc ≥ab(a + b + c), then 1 a3 + b3 + abc ≤ c abc(a + b + c). Solution 1.51. Note that abc ≤ a+b+c 3 3 = 1 27. 1 a + 1 1 b + 1 1 c + 1 = 1 + 1 a + 1 b + 1 c + 1 ab + 1 bc + 1 ca + 1 abc ≥1 + 3 3 √ abc + 3 3 (abc)2 + 1 abc = 1 + 1 3 √ abc 3 ≥43. Solution 1.52. The inequality is equivalent to b+c a a+c b a+b c ≥8. Now, we use the AM-GM inequality for each term of the product and the inequality follows immediately. Solution 1.53. Notice that a (a + 1)(b + 1) + b (b + 1)(c + 1) + c (c + 1)(a + 1) = (a + 1)(b + 1)(c + 1) −2 (a + 1)(b + 1)(c + 1) = 1 − 2 (a + 1)(b + 1)(c + 1) ≥3 4 if and only if (a+ 1)(b+ 1)(c+ 1) ≥8, and this last inequality follows immediately from the inequality a+1 2 b+1 2 c+1 2 ≥√a √ b√c = 1. Solution 1.54. Observe that this exercise is similar to Exercise 1.52. Solution 1.55. Apply the inequality between the arithmetic mean and the harmonic mean to get 2ab a + b = 2 1 a + 1 b ≤a + b 2 . We can conclude that equality holds when a = b = c. Solution 1.56. First use the fact that (a + b)2 ≥4ab, and then take into account that n i=1 1 aibi ≥4 n i=1 1 (ai + bi)2 . Now, use Exercise 1.36 to prove that n i=1 (ai + bi)2 n i=1 1 (ai + bi)2 ≥n2. 124 Solutions to Exercises and Problems Solution 1.57. Using the AM-GM inequality leads to xy + yz ≥2y√xz. Adding similar results we get 2(xy + yz + zx) ≥2(x√yz + y√zx + z√xy). Once again, using AM-GM inequality, we get x2 +x2+y2+z2 ≥4x√yz. Adding similar results once more, we obtain x2 + y2 + z2 ≥x√yz + y√zx + z√xy. Now adding both results, we reach the conclusion (x+y+z)2 3 ≥x√yz + y√zx + z√xy. Solution 1.58. Using the AM-GM inequality takes us to x4+y4 ≥2x2y2. Applying AM-GM inequality once again shows that 2x2y2 + z2 ≥ √ 8xyz. Or, directly we have that x4 + y4 + z2 2 + z2 2 ≥4 4 x4y4z4 4 = √ 8xyz. Solution 1.59. Use the AM-GM inequality to obtain x2 y −1 + y2 x −1 ≥2 xy (x −1)(y −1) ≥8. The last inequality follows from x √x−1 ≥2, since (x −2)2 ≥0. Second solution. Let a = x −1 and b = y −1, which are positive numbers, then the inequality we need to prove is equivalent to (a+1)2 b + (b+1)2 a ≥8. Now, by the AM-GM inequality we have (a + 1)2 ≥4a and (b + 1)2 ≥4b. Then, (a+1)2 b + (b+1)2 a ≥4 a b + b a ≥8. The last inequality follows from Exercise 1.24. Solution 1.60. Observe that (a, b, c) and (a2, b2, c2) have the same order, then use inequality (1.2). Solution 1.61. By the previous exercise a3 + b3 + c3 ≥a2b + b2c + c2a. Observe that ( 1 a, 1 b, 1 c) and ( 1 a2 , 1 b2 , 1 c2 ) can be ordered in the same way. Then, use inequality (1.2) to get (ab)3 + (bc)3 + (ca)3 = 1 a3 + 1 b3 + 1 c3 ≥1 a2 1 c + 1 b2 1 a + 1 c2 1 b = b a + c b + a c = a2b + b2c + c2a. Adding these two inequalities leads to the result. Solution 1.62. Use inequality (1.2) with (a1, a2, a3) = (b1, b2, b3) = a b , b c, c a and (a′ 1, a′ 2, a′ 3) = b c, c a, a b . 4.1 Solutions to the exercises in Chapter 1 125 Solution 1.63. Use inequality (1.2) with (a1, a2, a3) = (b1, b2, b3) = 1 a, 1 b, 1 c and (a′ 1, a′ 2, a′ 3) = 1 b, 1 c, 1 a . Solution 1.64. Assume that a ≤b ≤c, and consider (a1, a2, a3) = (a, b, c), then use the rearrangement inequality (1.2) twice over with (a′ 1, a′ 2, a′ 3) = (b, c, a) and (c, a, b), respectively. Note that we are also using (b1, b2, b3) = 1 b + c −a, 1 c + a −b, 1 a + b −c . Solution 1.65. Use the same idea as in the previous exercise, but with n variables. Solution 1.66. Turn to the previous exercise and the fact that s s−a1 = 1 + a1 s−a1 . Solution 1.67. Apply Exercise 1.65 to the sequence a1, . . . , an, a1, . . . , an. Solution 1.68. Apply Example 1.4.11. Solution 1.69. Note that 1 = (a2 + b2 + c2) + 2(ab + bc + ca), and use the previous exercise as follows: 1 3 = a + b + c 3 ≤ a2 + b2 + c2 3 . Therefore 1 3 ≤a2 + b2 + c2. Hence, 2(ab + bc + ca) ≤2 3, and the result is evident. Second solution. The inequality is equivalent to 3(ab + bc + ca) ≤(a + b + c)2, which can be simpliﬁed to ab + bc + ca ≤a2 + b2 + c2. Solution 1.70. Let G = n √x1x2 · · · xn be the geometric mean of the given numbers and (a1, a2, . . . , an) = x1 G , x1x2 G2 , . . . , x1x2···xn Gn . Using Corollary 1.4.2, we can establish that n ≤a1 a2 + a2 a3 + · · · + an−1 an + an a1 = G x2 + G x3 + · · · + G xn + G x1 , thus n 1 x1 + · · · + 1 xn ≤G. Also, using Corollary 1.4.2, n ≤a1 an + a2 a1 + · · · + an an−1 = x1 G + x2 G + · · · + xn G , then G ≤x1 + x2 + · · · + xn n . The equalities hold if and only if a1 = a2 = · · · = an, that is, if and only if x1 = x2 = · · · = xn. 126 Solutions to Exercises and Problems Solution 1.71. The inequality is equivalent to an−1 1 + an−1 2 + · · · + an−1 n ≥a1 · · · an a1 + a1 · · · an a2 + · · · + a1 · · · an an , which can be veriﬁed using the rearrangement inequality several times over. Solution 1.72. First note that n i=1 ai √1−ai = n i=1 1 √1−ai −n i=1 √1 −ai. Use the AM-GM inequality to obtain 1 n n i=1 1 √1 −ai ≥ n + + , n 1 i=1 1 √1 −ai = / 1 n 0n i=1(1 −ai) ≥ / 1 1 n n i=1(1 −ai) = n n −1. Moreover,the Cauchy-Schwarz inequality serves to show that n i=1 √ 1 −ai ≤ + + , n i=1 (1 −ai)√n = n(n −1) and n i=1 √ai ≤√n. Solution 1.73. (i) √4a + 1 < 4a+1+1 2 = 2a + 1. (ii) Use the Cauchy-Schwarz inequality with u = (√4a + 1, √ 4b + 1, √4c + 1) and v = (1, 1, 1). Solution 1.74. Suppose that a ≥b ≥c ≥d (the other cases are similar). Then, if A = b + c + d, B = a + c + d, C = a + b + d and D = a + b + c, we can deduce that 1 A ≥1 B ≥1 C ≥1 D. Apply the Tchebyshev inequality twice over to show that a3 A + b3 B + c3 C + d3 D ≥1 4(a3 + b3 + c3 + d3) 1 A + 1 B + 1 C + 1 D ≥1 16(a2 + b2 + c2 + d2)(a + b + c + d) 1 A + 1 B + 1 C + 1 D = 1 16(a2 + b2 + c2 + d2) A + B + C + D 3 1 A + 1 B + 1 C + 1 D . Now, use the Cauchy-Schwarz inequality to derive the result a2 + b2 + c2 + d2 ≥ab + bc + cd + da = 1 and the inequality (A + B + C + D)( 1 A + 1 B + 1 C + 1 D) ≥16. Solution 1.75. Apply the rearrangement inequality to (a1, a2, a3) = 3 a b , 3 b c, 3 c a , (b1, b2, b3) = ⎛ ⎝3 a b 2 , 3 /b c 2 , 3 c a 2 ⎞ ⎠ 4.1 Solutions to the exercises in Chapter 1 127 and the permutation (a′ 1, a′ 2, a′ 3) = 3 b c, 3 c a, 3 a b to derive a b + b c + c a ≥ 3 a2 bc + 3 b2 ca + 3 c2 ab. Finally, use the fact that abc = 1. Second solution. The AM-GM inequality and the fact that abc = 1 imply that 1 3 a b + a b + b c ≥ 3 a b a b b c = 3 a2 bc = 3 a3 abc = a. Similarly, 1 3 b c + b c + c a ≥b and 1 3 c a + c a + a b ≥c, and the result follows. Solution 1.76. Using the hypothesis, for all k, leads to s −2xk > 0. Turn to the Cauchy-Schwarz inequality to show that n k=1 x2 k s −2xk n k=1 (s −2xk) ≥ n k=1 xk 2 = s2. But 0 < n k=1(s −2xk) = ns −2s, therefore n k=1 x2 k s −2xk ≥ s n −2. Solution 1.77. The function f(x) = x + 1 x 2 is convex in R+. Solution 1.78. The function f(a, b, c) = a b + c + 1 + b a + c + 1 + c a + b + 1 + (1 −a)(1 −b)(1 −c) is convex in each variable, therefore its maximum is attained at the endpoints. Solution 1.79. If x = 0, then the inequality reduces to 1 + 1 √ 1+y2 ≤2, which is true because y ≥0. By symmetry, the inequality holds for y = 0. Now, suppose that 0 < x ≤1 and 0 < y ≤1. Let u ≥0 and v ≥0 such that x = e−u and y = e−v, then the inequality becomes 1 √ 1 + e−2u + 1 √ 1 + e−2v ≤ 2 √ 1 + e−(u+v) , 128 Solutions to Exercises and Problems that is, f(u) + f(v) 2 ≤f u + v 2 , where f(x) = 1 √ 1+e−2x . Since f ′′(x) = 1−2e2x (1+e−2x)5/2e4x , the function is concave in the interval [0, ∞). Thus the previous inequality holds. Solution 1.80. Find f′′(x). Solution 1.81. Use log(sin x) or the fact that sin A sin B = sin A + B 2 + A −B 2 sin A + B 2 −A −B 2 . Solution 1.82. (i) If 1+nx ≤0, the inequality is evident since (1+x)n ≥0. Suppose that (1 + nx) > 0. Apply AM-GM inequality to the numbers (1, 1, . . . , 1, 1 + nx) with (n −1) ones. (ii) Let a1,. . ., an be positive numbers and deﬁne, for each j = 1, . . . n, σj = a1+···+aj j . Apply Bernoulli’s inequality to show that σj σj−1 j ≥j σj σj−1 −(j −1), which implies σj j ≥σj j−1 j σj σj−1 −(j −1) = σj−1 j−1(jσj −(j −1)σj−1) = ajσj−1 j−1. Then, σn n ≥anσn−1 n−1 ≥anan−1σn−2 n−2 ≥· · · ≥anan−1 · · · a1. Solution 1.83. If x ≥y ≥z, we have xn(x −y)(x −z) ≥yn(x −y)(y −z) and zn(z −x)(z −y) ≥0. Solution 1.84. Notice that x(x−z)2 +y(y−z)2−(x−z)(y−z)(x+y−z) ≥0 if and only if x(x −z)(x −y) + y(y −z)(y −x) + z(x −z)(y −z) ≥0. The inequality now follows from Sch¨ ur’s inequality. Alternatively, we can see that the last expression is symmetric in x, y and z, then we can assume x ≥z ≥y, and if we return to the original inequality, it becomes clear that x(x −z)2 + y(y −z)2 ≥0 ≥(x −z)(y −z)(x + y −z). Solution 1.85. The inequality is homogeneous, therefore we can assume that a + b + c = 1. Now, the terms on the left-hand side are of the form x (1−x)2 and the function f(x) = x (1−x)2 is convex, since f ′′(x) = 4+2x (1−x)4 > 0. By Jensen’s inequality it follows that a (1−a)2 + b (1−b)2 + c (1−c)2 ≥3f a+b+c 3 = 3f 1 3 = 3 2 2. Solution 1.86. Since (a+b+c)2 ≥3(ab+bc+ca), we can deduce that 1+ 3 ab+bc+ca ≥ 1 + 9 (a+b+c)2 . Thus, the inequality will hold if 1 + 9 (a + b + c)2 ≥ 6 (a + b + c). 4.1 Solutions to the exercises in Chapter 1 129 But this last inequality follows from 1 − 3 a+b+c 2 ≥0. Now, if abc = 1, consider x = 1 a, y = 1 b and z = 1 c; it follows immediately that xyz = 1. Thus, the inequality is equivalent to 1 + 3 xy + yz + zx ≥ 6 x + y + z which is the ﬁrst part of this exercise. Solution 1.87. We will use the convexity of the function f(x) = xr for r ≥1 (its second derivative is r(r −1)xr−2). First suppose that r > s > 0. Then Jensen’s inequality for the convex function f(x) = x r s applied to xs 1, . . . , xs n gives t1xr 1 + · · · + tnxr n ≥(t1xs 1 + · · · + tnxs n) r s and taking the 1 r -th power of both sides gives the desired inequality. Now suppose 0 > r > s. Then f(x) = x r s is concave, so Jensen’s inequality is reversed; however, taking 1 r-th powers reverses the inequality again. Finally, in the case r > 0 > s, f(x) = x r s is again convex, and taking 1 r-th powers preserves the inequality. Solution 1.88. (i) Apply H¨ older’s inequality to the numbers xc 1, . . . , xc n, yc 1, . . . , yc n with a′ = a c and b′ = b c. (ii) Proceed as in Example 1.5.9. The only extra fact that we need to prove is xiyizi ≤1 axa i + 1 byb i + 1 czc i , but this follows from part (i) of that example. Solution 1.89. By the symmetry of the variables in the inequality we can assume that a ≤b ≤c. We have two cases, (i) b ≤a+b+c 3 and (ii) b ≥a+b+c 3 . Case (i): b ≤a+b+c 3 . It happens that a+b+c 3 ≤a+c 2 ≤c, and it is true that a+b+c 3 ≤b+c 2 ≤c. Then, there exist λ, μ ∈[0, 1] such that c + a 2 = λc + (1 −λ) a + b + c 3 and b + c 2 = μc + (1 −μ) a + b + c 3 . Adding these equalities, we obtain a + b + 2c 2 = (λ + μ)c + (2 −λ −μ) a + b + c 3 = (2 −λ −μ) a + b −2c 3 + 2c. Hence, a + b −2c 2 = (2 −λ −μ) a + b −2c 3 , therefore 2 −(λ + μ) = 3 2 and (λ + μ) = 1 2. 130 Solutions to Exercises and Problems Now, since f is a convex function, we have f a + b 2 ≤1 2 (f(a) + f(b)) f b + c 2 ≤μf(c) + (1 −μ)f a + b + c 3 f c + a 2 ≤λf(c) + (1 −λ)f a + b + c 3 thus, adding these inequalities we get f a + b 2 + f b + c 2 + f c + a 2 ≤1 2 (f(a) + f(b) + f(c)) + 3 2f a + b + c 3 . Case (ii): b ≥a+b+c 3 . It is similar to case (i), using the fact that a ≤a+c 2 ≤a+b+c 3 and a ≤a+b 2 ≤a+b+c 3 . Solution 1.90. If any of a, b or c is zero, the inequality is evident. Applying Popovi-ciu’s inequality (see the previous exercise) to the function f : R →R+ deﬁned by f(x) = exp(2x), which is convex since f ′′(x) = 4 exp(2x) > 0, we obtain exp(2x) + exp(2y) + exp(2z) + 3 exp 2(x + y + z) 3 ≥2 [exp(x + y) + exp(y + z) + exp(z + x)] = 2 [exp(x) exp(y) + exp(y) exp(z) + exp(z) exp(x)] . Setting a = exp(x), b = exp(y), c = exp(z), the previous inequality can be rewrit-ten as a2 + b2 + c2 + 3 3 √ a2b2c2 ≥2(ab + bc + ca). For the second part apply the AM-GM inequality in the following way: 2abc + 1 = abc + abc + 1 ≥3 3 √ a2b2c2. Solution 1.91. Apply Popoviciu’s inequality to the convex function f(x) = x + 1 x. We will get the inequality 1 a + 1 b + 1 c + 9 a+b+c ≥ 4 b+c + 4 c+a + 4 a+b. Then multiply both sides by (a + b + c) to ﬁnish the proof. Solution 1.92. Observe that by using (1.8), we obtain x2 + y2 + z2 −\|x\|\|y\| −\|y\|\|z\| −\|z\|\|x\| = 1 2(\|x\| −\|y\|)2 + 1 2(\|y\| −\|z\|)2 + 1 2(\|z\| −\|x\|)2, which is clearly greater than or equal to zero. Hence \|xy + yz + zx\| ≤\|x\|\|y\| + \|y\|\|z\| + \|z\|\|x\| ≤x2 + y2 + z2. Second solution. Apply Cauchy-Schwarz inequality to (x, y, z) and (y, z, x). 4.1 Solutions to the exercises in Chapter 1 131 Solution 1.93. The inequality is equivalent to ab + bc + ca ≤a2 + b2 + c2, which we know is true. See Exercise 1.27. Solution 1.94. Observe that if a + b + c = 0, then it follows from (1.7) that a3 + b3 + c3 = 3abc. Since (x −y) + (y −z) + (z −x) = 0, we can derive the following factorization: (x −y)3 + (y −z)3 + (z −x)3 = 3(x −y)(y −z)(z −x). Solution 1.95. Assume, without loss of generality, that a ≥b ≥c. We need to prove that −a3 + b3 + c3 + 3abc ≥0. Since −a3 + b3 + c3 + 3abc = (−a)3 + b3 + c3 −3(−a)bc, the latter expression factors into 1 2(−a + b + c)((a + b)2 + (a + c)2 + (b −c)2). The conclusion now follows from the triangle inequality, b + c > a. Solution 1.96. Let p = \|(x −y)(y −z)(z −x)\|. Using AM-GM inequality on the right-hand side of identity (1.8), we get x2 + y2 + z2 −xy −yz −zx ≥3 2 3 p2. (4.1) Now, since \|x −y\| ≤x + y, \|y −z\| ≤y + z, \|z −x\| ≤z + x, it follows that 2(x + y + z) ≥\|x −y\| + \|y −z\| + \|z −x\|. (4.2) Applying again the AM-GM inequality leads to 2(x + y + z) ≥3 3 √p, and the result follows from inequalities (4.1) and (4.2). Solution 1.97. Using identity (1.7), the condition x3 + y3 + z3 −3xyz = 1 can be factorized as (x + y + z)(x2 + y2 + z2 −xy −yz −zx) = 1. (4.3) Let A = x2 + y2 + z2 and B = x + y + z. Notice that B2 −A = 2(xy + yz + zx). By identity (1.8), we have that B > 0. Equation (4.3) now becomes B A −B2 −A 2 = 1, therefore 3A = B2 + 2 B . Since B > 0, we may apply the AM-GM inequality to obtain 3A = B2 + 2 B = B2 + 1 B + 1 B ≥3, that is, A ≥1. For instance, the minimum A = 1 is attained when (x, y, z) = (1, 0, 0). 132 Solutions to Exercises and Problems Solution 1.98. Inequality (1.11) helps to establish 1 a + 1 b + 4 c + 16 d ≥(1 + 1 + 2 + 4)2 a + b + c + d = 64 a + b + c + d. Solution 1.99. Apply inequality (1.11) twice over to get a4 + b4 = a4 1 + b4 1 ≥(a2 + b2)2 2 ≥( (a+b)2 2 )2 2 = (a + b)4 8 . Solution 1.100. Express the left-hand side as ( √ 2)2 x + y + ( √ 2)2 y + z + ( √ 2)2 z + x and use inequality (1.11). Solution 1.101. Express the left-hand side as x2 axy + bzx + y2 ayz + bxy + z2 azx + byz , and then use inequality (1.11) to get x2 axy + bzx + y2 ayz + bxy + z2 azx + byz ≥ (x + y + z)2 (a + b)(xy + yz + zx) ≥ 3 a + b, where the last inequality follows from (1.8). Solution 1.102. Rewrite the left-hand side as a2 a + b + b2 b + c + c2 a + c + b2 a + b + c2 b + c + a2 a + c, and then apply inequality (1.11). Solution 1.103. (i) Express the left-hand side as x2 x2 + 2xy + 3zx + y2 y2 + 2yz + 3xy + z2 z2 + 2zx + 3yz and apply inequality (1.11) to get x x + 2y + 3z + y y + 2z + 3x + z z + 2x + 3y ≥ (x + y + z)2 x2 + y2 + z2 + 5(xy + yz + zx). Now it suﬃces to prove that (x + y + z)2 x2 + y2 + z2 + 5(xy + yz + zx) ≥1 2, 4.1 Solutions to the exercises in Chapter 1 133 but this is equivalent to x2 + y2 + z2 ≥xy + yz + zx. (ii) Proceed as in part (i), expressing the left-hand side as w2 xw + 2yw + 3zw + x2 xy + 2xz + 3xw + y2 yz + 2yw + 3xy + z2 zw + 2xz + 3yz , then use inequality (1.11) to get w x + 2y + 3z + x y + 2z + 3w + y z + 2w + 3x + z w + 2x + 3y ≥ (w + x + y + z)2 4(wx + xy + yz + zw + wy + xz). Then, the inequality we have to prove becomes (w + x + y + z)2 4(wx + xy + yz + zw + wy + xz) ≥2 3, which is equivalent to 3(w2 +x2 +y2 +z2) ≥2(wx+xy +yz+zw +wy +xz). This follows by using the AM-GM inequality six times under the form x2 + y2 ≥2xy. Solution 1.104. We again apply inequality (1.11) to get x2 (x + y)(x + z) + y2 (y + z)(y + x) + z2 (z + x)(z + y) ≥ (x + y + z)2 x2 + y2 + z2 + 3(xy + yz + zx). Also, the inequality (x + y + z)2 x2 + y2 + z2 + 3(xy + yz + zx) ≥3 4 is equivalent to x2 + y2 + z2 ≥xy + yz + zx. Solution 1.105. We express the left-hand side as a2 a(b + c) + b2 b(c + d) + c2 c(d + a) + d2 d(a + b) and apply inequality (1.11) to get a2 a(b + c) + b2 b(c + d) + c2 c(d + a) + d2 d(a + b) ≥ (a + b + c + d)2 a(b + 2c + d) + b(c + d) + d(b + c). 134 Solutions to Exercises and Problems On the other hand, observe that (a + b + c + d)2 (ac + bd) + (ab + ac + ad + bc + bd + cd) = a2 + b2 + c2 + d2 + 2ab + 2ac + 2ad + 2bc + 2bd + 2cd (ac + bd) + (ab + ac + ad + bc + bd + cd) . To prove that this last expression is greater than 2 is equivalent to showing that a2 + c2 ≥2ac and b2 + d2 ≥2bd, which can be done using the AM-GM inequality. Solution 1.106. We express the left-hand side as a2 ab + ac + b2 bc + bd + c2 cd + ce + d2 de + ad + e2 ae + be and apply inequality (1.11) to get a2 ab + ac + b2 bc + bd + c2 cd + ce + d2 de + ad + e2 ae + be ≥(a + b + c + d + e)2 ab . Since (a + b + c + d + e)2 = a2 + 2 ab, we have to prove that 2 a2 + 4 ab ≥5 ab, which is equivalent to 2 a2 ≥ ab. The last inequality follows from a2 ≥ ab. Solution 1.107. (i) Using Tchebyshev’s inequality with the collections (a ≥b ≥c) and ( a2 x ≥b2 y ≥c2 z ), we obtain 1 3 a3 x + b3 y + c3 z ≥ a2 x + b2 y + c2 z 3 · a + b + c 3 , then by (1.11), we can deduce that a2 x + b2 y + c2 z ≥(a + b + c)2 x + y + z . Therefore a3 x + b3 y + c3 z ≥(a + b + c)2 x + y + z · a + b + c 3 . 4.1 Solutions to the exercises in Chapter 1 135 (ii) By Exercise 1.88, we have a3 x + b3 y + c3 z 1 3 (1 + 1 + 1) 1 3 (x + y + z) 1 3 ≥a + b + c. Raising to the cubic power both sides and then dividing both sides by 3(x+y +z) we obtain the result. Solution 1.108. Using inequality (1.11), we obtain x2 1 + x2 2 + · · · + x2 n x1 + x2 + · · · + xn = x2 1 x1 + x2 + · · · + xn + x2 2 x1 + x2 + · · · + xn + · · · + x2 n x1 + x2 + · · · + xn ≥(x1 + x2 + · · · + xn)2 n(x1 + x2 + · · · + xn) = x1 + x2 + · · · + xn n . Thus, it is enough to prove that x1 + x2 + · · · + xn n kn t ≥x1 · x2 · · · · · xn. Since k = max {x1, x2, . . . , xn} ≥min {x1, x2, . . . , xn} = t, we have that kn t ≥n and since x1+x2+···+xn n ≥1, because all the xi are positive integers, it is enough to prove that x1 + x2 + · · · + xn n n ≥x1 · x2 · · · · · xn, which is equivalent to the AM-GM inequality. Because all the intermediate inequalities are valid as equalities when x1 = x2 = · · · = xn, we conclude that equality happens when x1 = x2 = · · · = xn. Solution 1.109. Using the substitution a = x y, b = y z and c = z x, the inequality takes the form a3 a3 + 2 + b3 b3 + 2 + c3 c3 + 2 ≥1, and with the extra condition, abc = 1. In order to prove this last inequality the extra condition is used as follows: a3 a3 + 2 + b3 b3 + 2 + c3 c3 + 2 = a3 a3 + 2abc + b3 b3 + 2abc + c3 c3 + 2abc = a2 a2 + 2bc + b2 b2 + 2ca + c2 c2 + 2ab ≥ (a + b + c)2 a2 + b2 + c2 + 2bc + 2ca + 2ab = 1. The inequality above follows from inequality (1.11). 136 Solutions to Exercises and Problems Solution 1.110. With the substitution x = a b , y = b c, z = c a, the inequality takes the form a b + c + b c + a + c a + b ≥3 2, which is Nesbitt’s inequality (Example 1.4.8). Solution 1.111. Use the substitution x1 = a2 a1 , x2 = a3 a2 , . . . , xn = a1 an . Since 1 1+x1+x1x2 = 1 1+ a2 a1 + a2 a1 a3 a2 = a1 a1+a2+a3 and similarly for the other terms on the left-hand side of the inequality, the inequality we have to prove becomes a1 a1 + a2 + a3 + a2 a2 + a3 + a4 + · · · + an an + a1 + a2 > 1. But this inequality is easy to prove. It is enough to observe that for all i = 1, . . . , n we have ai + ai+1 + ai+2 < a1 + a2 + · · · + an. Solution 1.112. Using the substitution x = 1 a, y = 1 b, z = 1 c, the condition ab + bc + ca = abc becomes x + y + z = 1 and the inequality is equivalent to x4 + y4 x3 + y3 + y4 + z4 y3 + z3 + z4 + x4 z3 + x3 ≥1 = x + y + z. Tchebyshev’s inequality can be used to prove that x4 + y4 2 ≥x3 + y3 2 x + y 2 , thus x4 + y4 x3 + y3 + y4 + z4 y3 + z3 + z4 + x4 z3 + x3 ≥x + y 2 + y + z 2 + z + x 2 . Solution 1.113. The inequality on the right-hand side follows from inequality (1.11). For the inequality on the left-hand side, the substitution x = bc a , y = ca b , z = ab c transforms the inequality into x + y + z 3 ≥ yz + zx + xy 3 . Squaring both sides, we obtain 3(xy + yz + zx) ≤(x + y + z)2, which is valid if and only if (xy + yz + zx) ≤x2 + y2 + z2, something we already know. Solution 1.114. Note that a −2 a + 1 + b −2 b + 1 + c −2 c + 1 ≤0 ⇔3 −3 1 a + 1 + 1 b + 1 + 1 c + 1 ≤0 ⇔1 ≤ 1 a + 1 + 1 b + 1 + 1 c + 1. 4.1 Solutions to the exercises in Chapter 1 137 Using the substitution a = 2x y , b = 2y z , c = 2z x , we get 1 a + 1 + 1 b + 1 + 1 c + 1 = 1 2x y + 1 + 1 2y z + 1 + 1 2z x + 1 = y 2x + y + z 2y + z + x 2z + x = y2 2xy + y2 + z2 2yz + z2 + x2 2zx + x2 ≥ (x + y + z)2 2xy + y2 + 2yz + z2 + 2zx + x2 = 1. The only inequality in the expression follows from inequality (1.11). Solution 1.115. Observe that [5, 0, 0] = 2 6(a5 + b5 + c5) ≥2 6(a3bc + b3ca + c3ab) = [3, 1, 1], where Muirhead’s theorem has been used. Solution 1.116. Using Heron’s formula for the area of a triangle, we can rewrite the inequality as a2 + b2 + c2 ≥4 √ 3 (a + b + c) 2 (a + b −c) 2 (a + c −b) 2 (b + c −a) 2 . This is equivalent to (a2 + b2 + c2)2 ≥3[((a + b)2 −c2)(c2 −(b −a)2)] = 3(2c2a2 + 2c2b2 + 2a2b2 −(a4 + b4 + c4)), that is, a4 + b4 + c4 ≥a2b2 + b2c2 + c2a2, which, in terms of Muirhead’s theorem, is equivalent to proving [4, 0, 0] ≥[2, 2, 0]. Second solution. Using the substitution x = a + b −c, y = a −b + c, z = −a + b + c, we obtain x + y + z = a + b + c; then, using Heron’s formula we get 4(ABC) = (a + b + c)(xyz) ≤ (a + b + c)(x + y + z)3 27 = (a + b + c)2 3 √ 3 . Now we only need to prove that (a + b + c)2 ≤3(a2 + b2 + c2). This last inequality follows from Muirhead’s theorem, since [1, 1, 0] ≤[2, 0, 0]. 138 Solutions to Exercises and Problems Solution 1.117. Notice that a (a + b)(a + c) + b (b + c)(b + a) + c (c + a)(c + b) ≤ 9 4(a + b + c) ⇔8(ab + bc + ca)(a + b + c) ≤9(a + b)(b + c)(c + a) ⇔24abc + 8 (a2b + ab2) ≤9 (a2b + ab2) + 18abc ⇔6abc ≤a2b + ab2 + b2c + bc2 + c2a + ca2 ⇔[1, 1, 1] ≤[2, 1, 0]. Solution 1.118. The inequality is equivalent to a3 + b3 + c3 ≥ab(a + b −c) + bc(b + c −a) + ca(c + a −b). Setting x = a + b −c, y = b + c −a, z = a + c −b, we get a = z+x 2 , b = x+y 2 , c = y+z 2 . Then, the inequality we have to prove is 1 8((z+x)3+(x+y)3+(y+z)3) ≥1 4((z+x)(x+y)x+(x+y)(y+z)y+(y+z)(z+x)z), which is again equivalent to 3(x2y + y2x + · · · + z2x) ≥2(x2y + · · · ) + 6xyz or x2y + y2x + y2z + z2y + z2x + x2z ≥6xyz, and applying Muirhead’s theorem we obtain the result when x, y, z are non-negative. If one of them is negative (and it cannot be more than one at a time), we will get x2(y + z) + y2(z + x) + z2(x + y) = x22c + y22a + z22b ≥0 but 6xyz is negative, which ends the proof. Solution 1.119. Observe that a3 b2 −bc + c2 + b3 c2 −ca + a2 + c3 a2 −ab + b2 ≥a + b + c is equivalent to the inequality a3(b + c) b3 + c3 + b3(c + a) c3 + a3 + c3(a + b) a3 + b3 ≥a + b + c, which in turn is equivalent to a3(b + c)(a3 + c3)(a3 + b3) + b3(c + a)(b3 + c3)(a3 + b3) + c3(a + b)(a3 + c3)(b3 + c3) ≥(a + b + c)(a3 + b3)(b3 + c3)(c3 + a3). 4.1 Solutions to the exercises in Chapter 1 139 The last inequality can be written in the terminology of Muirhead’s theorem as [9, 1, 0] + [6, 4, 0] + [6, 3, 1] + [4, 3, 3] ≥ 1 2[1, 0, 0] [6, 3, 0] + 1 3[3, 3, 3] = [7, 3, 0] + [6, 4, 0] + [6, 3, 1] + [4, 3, 3] ⇔[9, 1, 0] ≥[7, 3, 0], a direct result of Muirhead’s theorem. Solution 1.120. Suppose that a ≤b ≤c, then 1 (1 + b) (1 + c) ≤ 1 (1 + c) (1 + a) ≤ 1 (1 + a) (1 + b). Use Tchebyshev’s inequality to prove that a3 (1 + b) (1 + c) + b3 (1 + c) (1 + a) + c3 (1 + a) (1 + b) ≥1 3(a3 + b3 + c3) 1 (1 + b)(1 + c) + 1 (1 + a)(1 + c) + 1 (1 + a)(1 + b) = 1 3(a3 + b3 + c3) 3 + (a + b + c) (1 + a)(1 + b)(1 + c). Finally, use the facts that 1 3(a3 + b3 + c3) ≥( a+b+c 3 )3, a+b+c 3 ≥1 and (1 + a)(1 + b)(1 + c) ≤ 3+a+b+c 3 3 to see that 1 3(a3 + b3 + c3) 3 + (a + b + a) (1 + a)(1 + b)(1 + c) ≥ a + b + c 3 3 6 (1 + a+b+c 3 )3 ≥6 8. For the last inequality, notice that a+b+c 3 1 + a+b+c 3 ≥1 2. Second solution. Multiplying by the common denominator and expanding both sides, the desired inequality becomes 4(a4 + b4 + c4 + a3 + b3 + c3) ≥3(1 + a + b + c + ab + bc + ca + abc). Since 4(a4 + b4 + c4 + a3 + b3 + c3) = 4(3[4, 0, 0] + 3[3, 0, 0]) and 3(1 + a + b + c + ab + bc + ca + abc) = 3([0, 0, 0] + 3[1, 0, 0] + 3[1, 1, 0] + [1, 1, 1]), the inequality is equivalent to 4[4, 0, 0] + 4[3, 0, 0] ≥[0, 0, 0] + 3[1, 0, 0] + 3[1, 1, 0] + [1, 1, 1]. Now, note that [4, 0, 0] ≥ '4 3, 4 3, 4 3 ( = a 4 3 b 4 3 c 4 3 = 1 = [0, 0, 0], 140 Solutions to Exercises and Problems where it has been used that abc = 1. Also, 3[4, 0, 0] ≥3[2, 1, 1] = 31 3(a2bc + b2ca + c2ab) = 31 3(a + b + c) = 3[1, 0, 0] and 3[3, 0, 0] ≥3 '4 3, 4 3, 1 3 ( = 31 3 a 4 3 b 4 3 c 1 3 + b 4 3 c 4 3 a 1 3 + c 4 3 a 4 3 b 1 3 = 31 3(ab + bc + ca) = 3[1, 0, 0]. Finally, [3, 0, 0] ≥[1, 1, 1]. Adding these results, we get the desired inequality. 4.2 Solutions to the exercises in Chapter 2 Solution 2.1. (i) Draw a segment BC of length a, a circle with radius c and center in B, and a circle with radius b and center in C, under what circumstances do they intersect? (ii) It follows from (i). (iii) a = x + y, b = y + z, c = z + x ⇔x = a+c−b 2 , y = a+b−c 2 , z = b+c−a 2 . Solution 2.2. (i) c < a + b ⇒c < a + b + 2 √ ab = (√a + √ b)2 ⇒√c < √a + √ b. (ii) With 2, 3 and 4 it is possible to construct a triangle but with 4, 9 and 16 it is not possible to do so. (iii) a < b < c ⇒a + b < a + c < b + c ⇒ 1 b+c < 1 c+a < 1 a+b, then it is suﬃcient to see that 1 a+b < 1 b+c + 1 c+a, and it will be even easier to see that 1 c < 1 b+c + 1 c+a. Solution 2.3. Use the fact that if a, b, c are the lengths of the sides of a triangle, the angle that is opposed to the side c is either 90◦or acute or obtuse if c2 is equal, less or greater than a2 + b2, respectively. Now, suppose that a ≤b ≤c ≤d ≤e and that the segments (a, b, c) and (c, d, e) do not form an acute triangle; since c2 ≥a2 + b2 and e2 ≥c2 + d2, we deduce that e2 ≥a2 + b2 + d2 ≥a2 + b2 + c2 ≥ a2 + b2 + a2 + b2 = (a + b)2 + (a −b)2 ≥(a + b)2, hence a + b ≤e, which is a contradiction. Solution 2.4. Since ∠A > ∠B then BC > CA. Using the triangle inequality we obtain AB < BC + CA, and by the previous statement, AB < 2BC. Solution 2.5. (i) Let O be the intersection point of the diagonals AC and BD. Apply the triangle inequality to the triangles ABO and CDO. Adding the inequal-ities, we get AB + CD < AC + BD. On the other hand, by hypothesis we have that AB + BD < AC + CD. Adding these last two inequalities we get AB < AC. (ii) Let DE be parallel to BC, then ∠EDA < ∠BCD < ∠A; therefore DE > 1 2AD and hence 1 2AD < DE < BC. Refer to the previous exercise. 4.2 Solutions to the exercises in Chapter 2 141 Solution 2.6. Each di is less than the sum of the lengths of two sides. Also, use the fact that in a convex quadrilateral the sum of the lengths of two opposite sides is less than the sum of the lengths of the diagonals. Solution 2.7. Use the triangle inequality in the triangles ABA′ and AA′C to prove that c < ma + 1 2a and b < ma + 1 2a. Solution 2.8. If α, β, γ are the angles of a triangle in A, B and C, respectively, and if α1 = ∠BAA′ and α2 = ∠A′AC, then, using D2, β > α1 and γ > α2. Therefore, 180◦= α + β + γ > α1 + α2 + α = 2α. Or, if we draw a circle with diameter BC, A should lie outside the circle and then ∠BAC < 90◦. Solution 2.9. Construct a parallelogram ABDC, with one diagonal BC and the other AD which is equal to two times the length of AA′ and use D2 on the triangle ABD. Solution 2.10. Complete a parallelogram as in the previous solution to prove that ma < b+c 2 . Similarly, mb < a+c 2 and mc < a+b 2 . To prove the left hand side inequality, let A′, B′ and C′ be the midpoints of the sides BC, CA and AB, respectively. B C A A′ C′ A′′ B′ Extend the segment C′B′ to a point A′′ such that C′A′′ = BC. Apply the previous result to the triangle AA′A′′ with side-lengths ma, mb and mc. Solution 2.11. Consider the quadrilateral ABCD and let O be a point on the exterior of the quadrilateral so that AOB is similar to ACD, and thus OAC and BAD are also similar. If O, B and C are collinear, we have an equality, otherwise we have an inequality.17 Solution 2.12. Set a = AB, b = BC, c = CD, d = DA, m = AC and n = BD. Let R be the radius of the circumcircle of ABCD. Thus we have18 (ABCD) = (ABC) + (CDA) = m(ab + cd) 4R , (ABCD) = (BCD) + (DAB) = n(bc + ad) 4R . 17See [6, page 136] or [1, page 128]. 18See [6, page 97] or [9, page 13]. 142 Solutions to Exercises and Problems Therefore m n = bc + ad ab + cd > 1 ⇔bc + ad > ab + cd ⇔(d −b)(a −c) > 0. Solution 2.13. Apply to the triangle ABP a rotation of 60◦with center at A. Under the rotation the point B goes to the point C, and let P ′ be the image of P. The triangle PP ′C has as sides PP ′ = PA, P ′C = PB and PC, and then the result. A B C P P ′ A B C P P ′ Second solution. Apply Ptolemy’s inequality (see Exercise 2.11) to the quadrilat-erals ABCP, ABPC and APBC; after cancellation of common terms we obtain that PB < PC + PA, PA < PC + PB and PC < PA + PB, respectively, which establish the existence of the triangle. Third solution. For the case when P is inside ABC. Let P ′ be the point where AP intersects the side BC. Next, use that AP < AP ′ < AB = BC < PB + PC. In a similar way, the other inequalities PB < PC + PA and PC < PA + PB hold. Solution 2.14. Set a = AB, b = BC, x = AC, y = BD. Remember that in a paralelogram we have 2(a2 + b2) = x2 + y2. We can suppose, without loss of generality, that a ≤b. It is clear that 2b < x + y, therefore (2b)2 < (x + y)2 = x2 + y2 + 2xy = 2(a2 + b2) + 2xy. Simplifying, we get 2(b2 −a2) < 2xy. Solution 2.15. (i) Extend the medians AA′, BB′ and CC′ until they intersect the circumcircle at A1, B1 and C1, respectively. Use the power of A′ to establish that A′A1 = a2 4ma . Also, use the facts that ma + A′A1 ≤2R and that the length of the median satisﬁes m2 a = 2(b2+c2)−a2 4 , that is, 4m2 a + a2 = 2(b2 + c2). We have analogous expressions for mb and mc. (ii) Use Ptolemy’s inequality in the quadrilaterals AC′GB′, BA′GC′ and CB′GA′, where G denotes the centroid. For instance, from the ﬁrst quadrilateral we get 2 3ma a 2 ≤b 2 mc 3 + c 2 mb 3 , then 2maa2 ≤abmc + camb. 4.2 Solutions to the exercises in Chapter 2 143 Solution 2.16. Using the formula 4m2 b +b2 = 2(c2+a2), we observe that m2 b −m2 c = 3 4(c2−b2). Now, using the triangle inequality, prove that mb+mc < 3 2(b+c). From this you can deduce the left-hand side inequality. The right-hand side inequality can be obtained from the ﬁrst when applied to the triangle of sides19 with lengths ma, mb and mc. Solution 2.17. Let a, b, c be the lengths of the sides of ABC. If E and F are the projections of Ia on the sides AB and CA, respectively, it is clear that if ra is the radius of the excircle, we have that ra = IaE = EA = AF = FIa = s, where s is the semiperimeter of ABC. Also, if ha is the altitude of the triangle ABC from vertex A, then AD DIa = ha ra . Since aha = bc, we have that AD DIa = ha ra = bc as = abc 4R 4Rr a2 1 rs = 4Rr a2 , where r and R are the inradius and the circumradius of ABC, respectively. Since 2R = a and 2r = b+c−a, therefore AD DIa = b+c−a a = b+c a −1. Then, it is enough to prove that b+c a ≤ √ 2 or, equivalently, that 2bc ≤a2, but bc = √ b2c2 ≤b2+c2 2 = a2 2 . Solution 2.18. Simplifying, the ﬁrst inequality is equivalent to ab + bc + ca ≤ a2+b2+c2, which follows from Exercise 1.27. For the second one, expand (a+b+c)2 and use the triangle inequality to obtain a2 < a(b + c). Solution 2.19. Use the previous suggestion. Solution 2.20. Expand and you will get the previous exercise. Solution 2.21. The ﬁrst inequality is the Nesbitt’s inequality, Example 1.4.8. For the second inequality use the fact that a + b > a+b+c 2 , then c a+b < 2c a+b+c. Solution 2.22. Observe that a2 (b + c −a)+b2 (c + a −b)+c2 (a + b −c)−2abc = (b + c −a) (c + a −b) (a + b −c), now see Example 2.2.3. Solution 2.23. Observe that a b2 + c2 −a2 +b c2 + a2 −b2 + c a2 + b2 −c2 = a2 (b + c −a) + b2 (c + a −b) + c2 (a + b −c) , now see Exercise 2.22. Solution 2.24. Use Ravi’s transformation with a = y + z, b = z + x, c = x + y to see ﬁrst that a2b(a −b) + b2c(b −c) + c2a(c −a) = 2(xy3 + yz3 + zx3) −2(xy2z + x2yz + xyz2). Then, the inequality is equivalent to x2 y + y2 z + z2 x ≥x+y+z. Apply then inequality (1.11). 19See the solution of Exercise 2.10. 144 Solutions to Exercises and Problems Solution 2.25. a −b a + b + b −c b + c + c −a c + a = a −b a + b · b −c b + c · c −a c + a < cab (a + b)(b + c)(c + a) ≤1 8. For the last inequality, see the solution of Example 2.2.3. Solution 2.26. By Exercise 2.18, 3(ab + bc + ca) ≤(a + b + c)2 ≤4(ab + bc + ca). Then, since ab + bc + ca = 3, it follows that 9 ≤(a + b + c)2 ≤12, and then the result. Solution 2.27. Use Ravi’s transformation, a = y + z, b = z + x and c = x + y. The AM-GM inequality and the Cauchy-Schwarz inequality imply 1 a + 1 b + 1 c = 1 y + z + 1 z + x + 1 x + y ≤1 2 1 √yz + 1 √zx + 1 √xy = √x + √y + √z 2√xyz ≤ √ 3√x + y + z 2√xyz = √ 3 2 x + y + z xyz = √ 3 2r . For the last identity, see the end of the proof of Example 2.2.4. Solution 2.28. The part (i) follows from the following equivalences: (s −a)(s −b) < ab ⇔s2 −s(a + b) < 0 ⇔a + b + c < 2(a + b) ⇔c < a + b. For (ii), use Ravi’s transformation, a = y + z, b = z + x, c = x + y, in order to see that the inequality is equivalent to 4(xy + yz + zx) ≤(y + z)(z + x) + (z + x)(x + y) + (x + y)(y + z). In turn, the last inequality follows from the inequality xy +yz +zx ≤x2 +y2 +z2, which is Exercise 1.27. 4.2 Solutions to the exercises in Chapter 2 145 Another way to obtain (ii) is the following: the given inequality is equivalent to 3s2 −2s(a + b + c) + (ab + bc + ca) ≤ab+bc+ca 4 , which in turn is equivalent to 3(ab + bc + ca) ≤4s2. The last inequality can be rewritten as 3(ab + bc + ca) ≤ (a + b + c)2. Solution 2.29. Applying the cosine law, we can see that a2 + b2 −c2 a2 −b2 + c2 = √ 2ab cos C √ 2ac cos B = 2a (b cos C)(c cos B) ≤2ab cos C + c cos B 2 = a2. Solution 2.30. Using the Cauchy-Schwarz inequality, for any x, y, z, w ≥0, we have that √xy + √zw ≤ (x + z)(y + w). Therefore cyclic a2 + b2 −c2 a2 −b2 + c2 = 1 2 cyclic a2 + b2 −c2 a2 −b2 + c2 + c2 + a2 −b2 c2 −a2 + b2 ≤1 2 cyclic (2a2)(2c2) = cyclic ac. Solution 2.31. Consider positive numbers x, y, z with a = y + z, b = z + x and c = x + y. The inequalities are equivalent to proving that y + z 2x + z + x 2y + x + y 2z ≥3 and 2x y + z + 2y z + x + 2z x + y ≥3. For the ﬁrst inequality use the fact that y x + x y ≥2 and for the second inequality use Nesbitt’s inequality. Solution 2.32. Since in triangles with the same base, the ratio between its altitudes is equal to the ratio of theirs areas, we have that PQ AD + PR BE + PS CF = (PBC) (ABC) + (PCA) (ABC) + (PAB) (ABC) = (ABC) (ABC) = 1. Use inequality (2.3) of Section 2.3. Solution 2.33. (i) Recall that (S1 + S2 + S3)( 1 S1 + 1 S2 + 1 S3 ) ≥9. (ii) The non-common vertices of the triangles form a hexagon which is divided into 6 triangles S1, S2, S3, T1, T2, T3, where Si and Ti have one common angle. 146 Solutions to Exercises and Problems Using the formula for the area that is related to the sine of the angle, prove that S1S2S3 = T1T2T3. After this, use the AM-GM inequality as follows: S 1 S1 + 1 S2 + 1 S3 ≥(S1 + S2 + S3 + T1 + T2 + T3) 1 S1 + 1 S2 + 1 S3 ≥18 6 √S1S2S3T1T2T3 3 √S1S2S3 = 18. The equality holds when the point O is the centroid of the triangle and the lines through O are the medians of the triangle; in this case S1 = S2 = S3 = T1 = T2 = T3 = 1 6S. Solution 2.34. If P = G is the centroid, the equality is evident since AG GL = BG GM = CG GN = 2. On the other hand, if AP PL + BP PM + CP PN = 6, we have AL PL + BM PM + CN PN = 9. It is not diﬃcult to see that PL AL = (PBC) (ABC), PM BM = (PCA) (ABC) and P N CN = (P AB) (ABC), therefore PL AL + PM BM + PN CN = 1. This implies that AL PL + BM PM + CN PN PL AL + PM BM + PN CN = 9. By inequality (2.3), the equality above holds only in the case when AL PL = BM PM = CN PN = 3, which implies that P is the centroid. Solution 2.35. (i) It is known that HD = DD′, HE = EE′ and HF = FF ′, where H is the orthocenter.20 Thus, the solution follows from part (i) of Example 2.3.4. (ii) Since AD′ AD = AD+DD′ AD = 1 + HD AD , we also have, after looking at the solution to Example 2.3.4, that AD′ AD + BE′ BE + CF ′ CF = 1 + HD AD + 1 + HE BE + 1 + HF CF = 4. Since AD AD′ + BE BE′ + CF CF ′ AD′ AD + BE′ BE + CF ′ CF ≥9, we have the result. Solution 2.36. As it has been mentioned in the proof of Example 2.3.5, the length of the internal bisector of angle A satisﬁes l2 a = bc 1 − a b + c 2 = 4bc (b + c)2 (s(s −a)). Since 4bc ≤(b + c)2, it follows that l2 a ≤s(s −a) and lalb ≤s (s −a)(s −b) ≤ s (s−a)+(s−b) 2 = s c 2. Therefore, lalblc ≤s s(s −a)(s −b)(s −c) = s(sr), lalb + lblc + lcla ≤ s a+b+c 2 = s2 and l2 a + l2 b + l2 c ≤s(s −a) + s(s −b) + s(s −c) = s2. 20See [6, page 85] or [9, page 37]. 4.2 Solutions to the exercises in Chapter 2 147 Solution 2.37. Let α = ∠AMB, β = ∠BNA, γ = ∠APC, and let (ABC) be the area of ABC. We have (ABC) = 1 2a · AM sin α = abc 4R . Hence, bc AM = 2R sin α. Similarly, ca BN = 2R sin β and ab CP = 2R sin γ. Thus, bc AM + ca BN + ab CP = 2R(sin α + sin β + sin γ) ≤6R. Equality is attained if M, N and P are the feet of the altitudes. Solution 2.38. Let A1, B1, C1 be the midpoints of the sides BC, CA, AB, re-spectively, and let B2, C2 be the reﬂections of A1 with respect to AB and CA, respectively. Also, consider D as the intersection of AB with A1B2 and E the intersection of CA with A1C2. Then, 2DE = B2C2 ≤C2B1 + B1C1 + C1B2 = A1B1 + B1C1 + C1A1 = s. Use the fact that the quadrilateral A1DAE is inscribed on a circle of diameter AA1 and the sine law on ADE, to deduce that DE = AA1 sin A = ma sin A. Then, s ≥2DE = 2ma sin A = 2ma a 2R = ama R , that is, ama ≤sR. Similarly, we have that bmb ≤sR and cmc ≤sR. Solution 2.39. The inequality is equivalent to 8(s −a)(s −b)(s −c) ≤abc, where s is the semiperimeter. Since (ABC) = sr = abc 4R = s(s −a)(s −b)(s −c), where r and R denote the inradius and the circumradius of ABC, respectively; we only have to prove that 8sr2 ≤abc, that is, 8sr2 ≤4Rrs, which is equivalent to 2r ≤R. Solution 2.40. The area of a triangle ABC satisﬁes the equalities (ABC) = abc 4R = (a+b+c)r 2 , therefore 1 ab+ 1 bc+ 1 ca = 1 2Rr ≥ 1 R2 , where R and r denote the circumradius and the inradius, respectively. Solution 2.41. Use Exercise 2.40 and the sine law. Solution 2.42. Use that21 sin A 2 = (s−b)(s−c) bc , where s denotes the semiperimeter of the triangle ABC, and similar expressions for sin B 2 and sin C 2 , to see that sin A 2 sin B 2 sin C 2 = (s −a)(s −b)(s −c) abc = sr2 abc = r 4R ≤1 8, where R and r are the circumradius and the inradius of ABC, respectively. 21Notice that sin2 A 2 = 1 −cos A 2 = 1 −b2+c2−a2 2bc 2 = a2 −(b −c)2 4bc = (s −b)(s −c) bc . 148 Solutions to Exercises and Problems Solution 2.43. From inequality (2.3), we know that (a + b + c) 1 a + 1 b + 1 c ≥9. Since a + b + c ≤3 √ 3R, we have 1 a + 1 b + 1 c ≥ √ 3 R . (4.4) Applying, once more, inequality (2.3), we get 1 3 π 2A + π 2B + π 2C ≥ 3 2 π(A + B + C) = 3 2. (4.5) Let f(x) = log π 2x, since f ′′(x) = 1 x2 > 0, f is convex. Using Jensen’s inequality, we get 1 3 log π 2A + log π 2B + log π 2C ≥log '1 3 π 2A + π 2B + π 2C ( . Applying (4.5) and the fact that log x is a strictly increasing function, we obtain 1 3 log π 2A + log π 2B + log π 2C ≥log 3 2. (4.6) We can suppose that a ≤b ≤c, which implies A ≤B ≤C. Therefore 1 a ≥1 b ≥1 c and log π 2A ≥log π 2B ≥log π 2C . Using Tchebyshev’s inequality, 1 a log π 2A + 1 b log π 2B + 1 c log π 2C ≥ 1 a + 1 b + 1 c log π 2A + log π 2B + log π 2C 3 . Therefore, using (4.4) and (4.6) leads us to 1 a log π 2A + 1 b log π 2B + 1 c log π 2C ≥ √ 3 R log 3 2. Now, raising the expresions to the appropriate powers and taking the reciprocals, we obtain the desired inequality. In all the above inequalities, the equality holds if and only if a = b = c (this means, equality is obtained if and only if the triangle is equilateral). Solution 2.44. By the sine law, it follows that sin A a = sin B b = sin C c = 1 2R, 4.2 Solutions to the exercises in Chapter 2 149 where a, b, c are the lengths of the sides of the triangle and R is the circumradius of the triangle. Thus, sin2A + sin2B + sin2C = a2 4R2 + b2 4R2 + c2 4R2 = 1 4R2 (a2 + b2 + c2) ≤ 1 4R2 · 9R2 = 9 4, where the inequality follows from Leibniz’s inequality. Solution 2.45. Use Leibniz’s inequality and the fact that the area of a triangle is given by (ABC) = abc 4R . Solution 2.46. We note that the incircle of ABC is the circumcircle of DEF. Applying Leibniz’s inequality to DEF, we get EF 2 + FD2 + DE2 ≤9r2, where r is the inradius of ABC. On the other hand, using Theorem 2.4.3 we obtain s2 ≥27r2, hence EF 2 + FD2 + DE2 ≤s2 3 . Solution 2.47. a2 hbhc + b2 hcha + c2 hahb = a2bc + b2ca + c2ab 4(ABC)2 = abc(a + b + c) 4(ABC)2 = abc(a + b + c) 4 abc 4R (a+b+c)r 2 = 2R r ≥4. Solution 2.48. Remember that sin2 A 2 = 1−cos A 2 and use that cos A + cos B + cosC ≤3 2 (see Example 2.5.2). Solution 2.49. Observe that 4 √ 3(ABC) ≤ 9abc a + b + c ⇔4 √ 3rs ≤9 · 4 Rrs 2s ⇔2 √ 3s ≤9 R ⇔ 2s 3 √ 3 ≤R. The last inequality was proved in Theorem 2.4.3. Solution 2.50. Use the previous exercise and the inequality between the harmonic mean and the geometric mean, 3 1 ab + 1 bc + 1 ca ≤ 3 √ a2b2c2. 150 Solutions to Exercises and Problems Solution 2.51. Use the previous exercise and the AM-GM inequality, 3 √ a2b2c2 ≤a2 + b2 + c2 3 . Solution 2.52. First, observe that if s = a+b+c 2 , then a2 + b2 + c2−(a −b)2 −(b −c)2 −(c −a)2 = = a2 −(b −c)2 + b2 −(c −a)2 + c2 −(a −b)2 = 4{(s −b)(s −c) + (s −c)(s −a) + (s −a)(s −b)}. Hence, if x = s −a, y = s −b, z = s −c, then the inequality is equivalent to √ 3 xyz(x + y + z) ≤xy + yz + zx. Squaring and simplifying the last inequality, we get xyz(x + y + z) ≤x2y2 + y2z2 + z2x2. This inequality can be deduced using Cauchy-Schwarz’s inequality with (xy, yz, zx) and (zx, xy, yz). Solution 2.53. Use Exercise 2.50 and the inequality 3 3 (ab)(bc)(ca) ≤ab+bc+ca. Solution 2.54. Note that 3(a + b + c)abc ab + bc + ca ≥ 9abc a + b + c ⇔ (a + b + c)2 ≥3(ab + bc + ca) ⇔ a2 + b2 + c2 ≥ab + bc + ca, now, use Exercise 2.49. Solution 2.55. Using (2.5), (2.6) and (2.7) we can observe that a2+b2+c2+4abc = 1 2 −2r2. Solution 2.56. Observe the relationships used in the proof of Exercise 2.39, (b + c −a)(c + a −b)(a + b −c) abc = 8(s −a)(s −b)(s −c) abc = 8s(s −a)(s −b)(s −c) 4Rs( abc 4R ) = 8(rs)2 4Rs(rs) = 2r R . 4.2 Solutions to the exercises in Chapter 2 151 Solution 2.57. Observe that a2 b + c −a + b2 c + a −b + c2 a + b −c = 1 2 a2 s −a + b2 s −b + c2 s −c = 1 2 sa s −a −a + sb s −b −b + sc s −c −c = s 2 a s −a + b s −b + c s −c −s = s 2 '(a + b + c)s2 −2(ab + bc + ca)s + 3abc (s −a)(s −b)(s −c) ( −s = s 2 '2s3 −2s(s2 + r2 + 4rR) + 3(4Rrs) r2s ( −s = 2s(R −r) r ≥2s(R −R 2 ) r ≥3 √ 3rR r = 3 √ 3R, the last two inequalities follow from the fact that R ≥2r (which implies that −r ≥−R 2 ) and from s ≥3 √ 3r, respectively. Solution 2.58. Start on the side of the equations which expresses the relationship between the τ’s and perform the operations. Solution 2.59. If x1, 1 −x1, x2, 1 −x2, . . . are the lengths into which each side is divided for the corresponding point, we can deduce that a2 + b2 + c2 + d2 = (x2 i + (1 −xi)2). Prove that 1 2 ≤2(xi −1 2)2 + 1 2 = x2 i + (1 −xi)2 ≤1. For part (ii), the inequality on the right-hand side follows from the triangle inequality. For the one on the left-hand side, use reﬂections on the sides, as you can see in the ﬁgure. b a c d Solution 2.60. This is similar to part (ii) of the previous problem. 152 Solutions to Exercises and Problems Solution 2.61. If ABC is the triangle and DEFGHI is the hexagon with DE, FG, HI parallel to BC, AB, CA, respectively, we have that the perimeter of the hexagon is 2(DE + FG + HI). Let X, Y , Z be the tangency points of the incircle with the sides BC, CA, AB, respectively, and let p = a+b+c be the perimeter of the triangle ABC. Set x = AZ = AY , y = BZ = BX and z = CX = CY , then we have the relations DE a = AE + ED + DA p = 2x p . Similarly, we have the other relations FG c = 2z p , HI b = 2y p . Therefore, p(DEFGHI) = 4(xa + yb + zc) p = 4(a(s −a) + b(s −b) + c(s −c)) 2s = 4((a + b + c)s −(a2 + b2 + c2)) 2s = 2(a + b + c) −4(a2 + b2 + c2) (a + b + c) , but a2 + b2 + c2 ≥ 1 3(a + b + c)(a + b + c) by Tchebyshev’s inequality. Thus, p(DEFGHI) ≤2(a + b + c) −4 3(a + b + c) = 2 3(a + b + c). Solution 2.62. Take the circumcircle of the equilateral triangle with side length 2. The circles with centers the midpoints of the sides of the triangle and radii 1 cover a circle of radius 2. If a circle of radius greater than 2 √ 3 3 is covered by three circles of radius 1, then one of the three circles covers a chord of length greater than 2. Solution 2.63. Take the acute triangle with sides of lengths 2r1, 2r2 and 2r3, if it exists. Its circumradius is the solution. If the triangle does not exist, the maximum radius between r1, r2 and r3 is the answer. Solution 2.64. We need two lemmas. Lemma 1. If a square of side-length a lies inside a rectangle of sides c and d, then a ≤min {c, d}. c d a 4.2 Solutions to the exercises in Chapter 2 153 Through the vertices of the square draw parallel lines to the sides of the rectangle in such a way that those lines enclose the square as in the ﬁgure. Since the parallel lines form a square inside the rectangle and such a square contains the original square, we have the result. Lemma 2. The diagonal of a square inscribed in a right triangle is less than or equal to the length of the internal bisector of the right angle. Let ABC be the right triangle with hypotenuse CA and let PQRS be the inscribed square. It can be assumed that the vertices P and Q belong to the legs of the right triangle (otherwise, translate the square) and let O be the intersection point of the diagonals PR and QS. A B C R S T P Q O O′ V Since BQOP is cyclic (∠B = ∠O = 90◦), it follows that ∠QBO = ∠QPO = 45◦, then O belongs to the internal bisector of ∠B. Let T be the intersection of BO with RS, then ∠QBT = ∠QST = 45◦, therefore BQTS is cyclic and the center O′ of the circumcircle of BQTS is the intersection of the perpendicular bisectors of SQ and BT. But the perpendicular bisector of SQ is PR, hence the point O′ belongs to PR, and if V is the midpoint of BT, we have that V OO′ is a right triangle. Since O′O > O′V , then the chords SQ and BT satisfy SQ < BT, and the lemma follows. Let us ﬁnish now the proof of the problem. Let ABCD be the square of side 1 and let l be a line that separates the two squares. If l is parallel to one of the sides of the square ABCD, then Lemma 1 applies. Otherwise, l intersects every line that determines a side of the square ABCD. Suppose that A is the farthest vertex from l. 154 Solutions to Exercises and Problems a b A B C D G H E F If l intersects the sides of ABCD in E, F, G, H as in the ﬁgure, we have, by Lemma 2, that the sum of the lengths of the diagonals of the small squares is less than or equal to AC, that is, √ 2(a + b) ≤ √ 2, then the result follows. Solution 2.65. If α, β, γ are the central angles which open the chords of length a, b, c, respectively, we have that a = 2 sin α 2 , b = 2 sin β 2 and c = 2 sin γ 2 . Therefore, abc = 8 sin α 2 sin β 2 sin γ 2 ≤8 sin3 α + β + γ 6 = 8 sin3(30◦) = 1, where the inequality follows from Exercise 1.81. Solution 2.66. The ﬁrst observation that we should make is to check that the diagonals are parallel to the sides. Let X be the point of intersection between the diagonals AD and CE. Now, the pentagon can be divided into (ABCDE) = (ABC) + (ACX) + (CDE) + (EAX). Since ABCX is a parallelogram, we have (ABC) = (CXA) = (CDE). Let a = (CDX) = (EAX) and b = (DEX), then we get a b = AX XD = (CXA) (CDX) = a+b a , that is, a b = 1+ √ 5 2 . Now we have all the elements to ﬁnd (ABCDE). Solution 2.67. Prove that sr = s1R = (ABC), where s1 is the semiperimeter of the triangle DEF. To deduce this equality, it is suﬃcient to observe that the radii OA, OB and OC are perpendicular to EF, FD and DE, respectively. Use also that R ≥2r. Solution 2.68. Suppose that the maximum angle is A and that it satisﬁes 60◦≤ A ≤90◦, then the lengths of the altitudes hb and hc are also less than 1. Now, use the fact that (ABC) = hbhc 2 sin A and that √ 3 2 ≤sin A ≤1. The obtuse triangle case is easier. 4.2 Solutions to the exercises in Chapter 2 155 Solution 2.69. Let ABCD be the quadrilateral with sides of length a = AB, b = BC, c = CD and d = DA. (i) (ABCD) = (ABC) + (CDA) = ab sin B 2 + cd sin D 2 ≤ab+cd 2 . (ii) If ABCD is the quadrilateral mentioned with sides of length a, b, c and d, consider the triangle BC′D which results from the reﬂection of DCB with respect to the perpendicular bisector of side BD. The quadrilaterals ABCD and ABC′D have the same area but the second one has sides of length a, c, b and d, in this order. Now use (i). (iii) (ABC) ≤ab 2 , (BCD) ≤bc 2 , (CDA) ≤cd 2 and (DAB) ≤da 2 . Solution 2.70. In Example 2.7.6 we proved that PA · PB · PC ≥R 2r(pa + pb)(pb + pc)(pc + pa). Use the AM-GM inequality. Solution 2.71. (i) PA2 pbpc + PB2 pcpa + PC2 papb ≥3 3 PA2 pbpc P B2 pcpa P C2 papb ≥3 3 4R r 2 ≥12. (ii) PA pb+pc + P B pc+pa + PC pa+pb ≥3 3 PA pb+pc PB pc+pa PC pa+pb ≥3 3 R 2r ≥3. (iii) PA √pbpc + PB √pcpa + PC √papb ≥3 3 PA √pbpc PB √pcpa PC √papb ≥3 3 4R r ≥6. For the last inequalities in (i) and (iii), we have used Exercise 2.70. For the last inequality in (ii), we have resorted to Example 2.7.6. (iv) Proceed as in Example 2.7.5, that is, apply inversion in a circle with center P and radius d (arbitrary, for instance d = pb). Let A′, B′, C′ be the inverses of A, B, C, respectively. Let p′ a, p′ b, p′ c be the distances from P to the sides B′C′, C′A′, A′B′, respectively. Let us prove that p′ a = paPB′·PC′ d2 . We have p′ aB′C′ = 2(PB′C′) = PB′ · PC′ · B′C′ PA′ 1 = paPB′ · PC′ · B′C′ d2 , where A′ 1 is the inverse of A1, the projection of P on BC. Similarly, p′ b = pbPC′·P A′ d2 and p′ c = pcPA′·PB′ d2 . The Erd˝ os-Mordell inequality, applied to the triangle A′B′C′, guarantees us that PA′ + PB′ + PC′ ≥2(p′ a + p′ b + p′ c). Now, since PA · PA′ = PB · PB′ = PC · PC′ = d2, after substitution we get 1 PA + 1 PB + 1 PC ≥2 pa PB · PC + pb PC · PA + pc PC · PA and this inequality is equivalent to PB · PC + PC · PA + PA · PB ≥2(paPA + pbPB + pcPC). Finally, to conclude use example 2.7.4. 156 Solutions to Exercises and Problems Solution 2.72. If P is an interior point or a point on the perimeter of the triangle ABC, see the proof of Theorem 2.7.2. If ha is the length of the altitude from vertex A, we have that the area of the triangle ABC satisﬁes 2(ABC) = aha = apa + bpb + cpc. Since ha ≤PA + pa (even if pa ≤0, that is, if P is a point on the outside of the triangle, on a diﬀerent side of BC than A), and because the equality holds if P is exactly on the segment of the altitude from the vertex A, therefore aPA+apa ≥ aha = apa + bpb + cpc, hence aPA ≥bpb + cpc. This inequality can be applied to triangle AB′C′ symmetric to ABC with respect to the internal angle bisector of A, where aPA ≥cpb + bpc, with equality when AP passes through the point O. Similarly, bPB ≥apc + cpa and cPC ≥apb + bpa, therefore PA + PB + PC ≥ b c + c b pa + c a + a c pb + a b + b a pc. We have the equality when P is the circumcenter O. Second solution. Let L, M and N be the feet of the perpendicular from point P to the sides BC, CA and AB, respectively. Let H and G be the orthogonal projections of B and C, respectively, over the segment MN. Then BC ≥HG = HN + NM + MG. Since ∠BNH = ∠ANM = ∠APM, the right triangles BNH and APM are similar, therefore HN = PM PA BN. In an analogous way we get MG = P N PA CM. Applying Ptolemy’s theorem to AMPN, we obtain PA · MN = AN · PM + AM · PN, hence MN = AN · PM + AM · PN PA , from there we get BC ≥PM PA BN + AN · PM + AM · PN PA + PN PA CM. Therefore, BC · PA ≥PM · AB + PN · CA. Then, PA ≥pb c a + pc b a. Similarly for the other two inequalities. Solution 2.73. Take a sequence of reﬂections of the quadrilateral ABCD, as shown in the ﬁgure. 4.2 Solutions to the exercises in Chapter 2 157 A B C D A′ A′′ B′ B′′ D′ C′ S P Q R R′ P ′′ S′′ R′′ Note that the perimeter of PQRS is the sum of the lengths of the piecewise line PQR′S′′P ′′. Note also that A′′B′′ is parallel to AB and that the shortest distance is AA′′ as can be seen if we project O on the sides of the quadrilateral. Solution 2.74. First note that (DEF) = (ABC) −(AFE) −(FBD) −(EDC). If x = BD, y = CE, z = AF, a −x = DC, b −y = EA and c −z = FB, we have (AFE) (ABC) = z(b −y) cb , (FBD) (ABC) = x(c −z) ac and (EDC) (ABC) = y(a −x) ba . Therefore, (DEF) (ABC) = 1 −z c 1 −y b −x a 1 −z c −y b 1 −x a = 1 −x a 1 −y b 1 −z c + x a · y b · z c = 2x a · y b · z c . The last equality follows from the fact that x a−x · y b−y · z c−z = 1 which is guaranteed because the cevians occur. Now, the last product is maximum when x a = y b = z c , and since the segments concur the common value is 1 2. Thus P must be the centroid. Solution 2.75. If x = PD, y = PE and z = PF, we can deduce that 2(ABC) = ax + by + cz. Using the Cauchy-Schwarz inequality, (a + b + c)2 ≤ a x + b y + c z (ax + by + cz) . Then a x + b y + c z ≥(a+b+c)2 2(ABC) and the equality holds when x = y = z, that is, when P is the incenter. 158 Solutions to Exercises and Problems Solution 2.76. First, observe that BD2 +CE2 +AF 2 = DC2 +EA2 +FB2, where BD2 −DC2 = PB2 −PC2 and similar relations have been used. Now, (BD + DC)2 = a2, hence BD2 + DC2 = a2 −2BD · DC. Similarly for the other two sides. Thus, BD2 + DC2 + CE2 + AE2 + AF 2 + FB2 = a2 + b2 + c2 −2(BD · DC + CE · AE + AF · FB). In this way, the sum is minimum when (BD · DC + CE · AE + AF · FB) is maximum. But BD·DC ≤ BD+DC 2 2 = a 2 2 and the maximum is attained when BD = DC. Similarly, CE = EA and AF = FB, therefore P is the circumcenter. Solution 2.77. Since 3 (aPD)(bPE)(cPF) ≤aPD+bP E+cP F 3 = 2(ABC) 3 , we can deduce that PD · PE · PF ≤ 8 27 (ABC)3 abc . Moreover, the equality holds if and only if aPD = bPE = cPF. But c · PF = b · PE ⇔(ABP) = (CAP) ⇔P is on the median AA′. Similarly, we can see that P is on the other medians, thus P is the centroid. Solution 2.78. Using the technique for proving Leibniz’s theorem, verify that 3PG2 = PA2 + PB2 + PC2 −1 3(a2 + b2 + c2), where G is the centroid of ABC. Therefore, the optimal point must be P = G. Solution 2.79. The quadrilateral APMN is cyclic and it is inscribed in the circle of diameter AP. The chord MN always opens the angle A (or 180◦−∠A), therefore the length of MN will depend proportionally on the radius of the circumscribed circle to APMN. The biggest circle will be attained when the diameter AP is the biggest possible. This happens when P is diametrally opposed to A. In this case M N A P C B M and N coincide with B and C, respectively. Therefore the maximum chord MN is BC. Solution 2.80. The circumcircle of DEF is the nine-point circle of ABC, therefore it intersects also the midpoints of the sides of ABC and goes through L, M, N, the midpoints of AH, BH, CH, respectively. 4.2 Solutions to the exercises in Chapter 2 159 C O H M F A′ D A N B ∠A E L Note that t2 a = AL · AD, then t2 a ha = AL · AD AD = AL = OA′ = R cos A ≤3R cos A + B + C 3 = 3R cos 60◦= 3 2R. Observe that we can prove a stronger result t2 a ha = R + r, using the fact that cosA + cos B + cos C = r R + 1. See Lemma 2.5.2. Solution 2.81. (i) Notice that pa ha + pb hb + pc hc = apa aha + bpb bhb + cpc chc = 2(PBC) + 2(PCA) + 2(PAB) 2(ABC) = 1. Now use the fact that pa ha + pb hb + pc hc ha pa + hb pb + hc pc ≥9. (ii) Using the AM-GM inequality, we have 27 pa ha pb hb pc hc ≤ pa ha + pb hb + pc hc 3 = 1, where the last equality follows from (i). (iii) Let x = (PBC), y = (PCA) and z = (PAB). Observe that a(ha −pa) = aha −apa = 2(y + z) ≥4√yz. Similarly, we have that b(hb −pb) ≥4√zx y c(hc −pc) ≥4√xy. Then, a(ha −pa)b(hb −pb)c(hc −pc) ≥64xyz = 8(apabpbcpc). Therefore, (ha −pa)(hb −pb)(hc −pc) ≥8papbpc. 160 Solutions to Exercises and Problems Solution 2.82. Assume that a < b < c, then of all the altitudes of ABC, AD is the longest. If E is the projection of I on AD, it is enough to prove that AE ≥AO = R. Remember that the internal bisector of ∠A is also the internal bisector of ∠EAO. If I is projected on E′ in the diameter AA′, then AE = AE′. Now prove that AE′ ≥AO, by proving that I is inside the acute triangle COF, where F is the intersection of AA′ with BC. To see that COF is an acute triangle, use that the angles of ABC satisfy ∠A < ∠B < ∠C, so that 1 2∠B < 90◦−∠A, 1 2∠C < 90◦−∠A. Use also that ∠COF = ∠A + ∠C −∠B < 90◦. Solution 2.83. Let ABC be a triangle with sides of lengths a, b and c. Using Heron’s formula to calculate the area of the triangle, we have that (ABC) = s(s −a)(s −b)(s −c), where s = a + b + c 2 . (4.7) If s and c are ﬁxed, then s −c is also ﬁxed. Then the product 16(ABC)2 is maximum when (s −a)(s −b) is maximum, that is, if s −a = s −b, which is equivalent to a = b. Therefore the triangle is isosceles. Solution 2.84. Let ABC be a triangle with sides of length a, b and c. Since the perimeter is ﬁxed, the semi-perimeter is also ﬁxed. Using (4.7), we have that 16(ABC)2 is maximum when (s −a)(s −b)(s −c) is maximum. The product of these three numbers is maximum when (s −a) = (s −b) = (s −c), that is, when a = b = c. Therefore, the triangle is equilateral. Solution 2.85. If a, b, c are the lengths of the sides of the triangle, observe that a+b+c = 2R(sin ∠A+sin ∠B+sin ∠C) ≤6R sin ∠A+∠B+∠C 3 , since the function sin x is concave. Moreover, equality holds when sin ∠A = sin ∠B = sin ∠C. Solution 2.86. The inequality (lm + mn + nl)(l + m + n) ≥a2l + b2m + c2n is equivalent to l2 + m2 −c2 lm + m2 + n2 −b2 mn + n2 + l2 −a2 nl + 3 ≥0 ⇔cos ∠APB + cos ∠BPC + cos ∠CPA + 3 2 ≥0. Now use the fact that cos α + cos β + cos γ + 3 2 ≥0 is equivalent to (2 cos α+β 2 + cos α−β 2 )2 + sin2( α−β 2 ) ≥0. Solution 2.87. Consider the Fermat point F and let p1 = FA, p2 = FB and p3 = FC, then observe ﬁrst that (ABC) = 1 2(p1p2 + p2p3 + p3p1) sin 120◦= √ 3 4 (p1p2 + p2p3 + p3p1). Also, a2 + b2 + c2 = 2p2 1 + 2p2 2 + 2p2 3 −2p1p2 cos 120◦−2p2p3 cos 120◦−2p3p1 cos 120◦ = 2(p2 1 + p2 2 + p2 3) + p1p2 + p2p3 + p3p1. 4.2 Solutions to the exercises in Chapter 2 161 Now, using the fact that x2 + y2 ≥2xy, we can deduce that a2 + b2 + c2 ≥ 3(p1p2 + p2p3 + p3p1) = 3 4 3 √ 3(ABC) . Then, a2 + b2 + c2 ≥4 √ 3(ABC). Moreover, the equality a2 + b2 + c2 = 4 √ 3(ABC) holds when p2 1 + p2 2 + p2 3 = p1p2 + p2p3 + p3p2, that is, when p1 = p2 = p3 or, equivalently, when the triangle is equilateral. Solution 2.88. Let a, b, c be the lengths of the sides of the triangle ABC. In the same manner as we proceeded in the previous exercise, deﬁne p1 = FA, p2 = FB and p3 = FC. From the solution of the previous exercise we know that 4 √ 3(ABC) = 3(p1p2 + p2p3 + p3p1). Thus, we only need to prove that 3(p1p2 + p2p3 + p3p1) ≤(p1 + p2 + p3)2, but this is equivalent to p1p2 + p2p3 + p3p1 ≤p2 1 + p2 2 + p2 3, which is Exercise 1.27. Solution 2.89. As in the Fermat problem there are two cases, when in ABC all angles are less than 120◦or when there is an angle greater than 120◦. In the ﬁrst case the minimum of PA + PB + PC is CC′, where C′ is the image of A when we rotate the ﬁgure in a positive direction through an angle of 60◦having B as the center. Using the cosine law, we obtain (CC′)2 = b2 + c2 −2bc cos(A + 60◦) = b2 + c2 −bc cosA + bc √ 3 sin A = 1 2(a2 + b2 + c2) + 2 √ 3(ABC). Now, use the fact that a2 + b2 + c2 ≥4 √ 3(ABC) to obtain (CC′)2 ≥4 √ 3(ABC). Applying Theorem 2.4.3 we have that (ABC) ≥3 √ 3r2, therefore (CC′)2 ≥36 r2. When ∠A ≥120◦, the point that solves Fermat-Steiner problem is the point A, then PA + PB + PC ≥AB + AC = b + c. It suﬃces to prove that b + c ≥6r. Moreover, we can use the fact that b = x + z, c = x + y and r = xyz x+y+z. Second solution. It is clear that PA + pa ≥ha, where pa is the distance from P to BC and ha is the length of the altitude from A. Then ha + hb + hc ≤ (PA + PB + PC) + (pa + pb + pc) ≤3 2(PA + PB + PC), where the last inequality follows from Erd˝ os-Mordell’s theorem. Now using Exercise 1.36 we have that 9 ≤(ha + hb + hc)( 1 ha + 1 hb + 1 hc ) = (ha + hb + hc)( 1 r). Therefore, 9r ≤ha + hb + hc ≤3 2(PA + PB + PC) and the result follows. Solution 2.90. First, we note that (A1B1C1) = 1 2A1B1 ·A1C1 ·sin ∠B1A1C1. Since PB1CA1 is a cyclic quadrilateral with diameter PC, applying the sine law leads us to A1B1 = PC sin C. Similarly, A1C1 = PB sin B. 162 Solutions to Exercises and Problems Call Q the intersection of BP with the circumcircle of triangle ABC, then ∠B1A1C1 = ∠QCP. In fact, since PB1CA1 is a cyclic quadrilateral we have ∠B1CP = ∠B1A1P. Similarly, ∠C1BP = ∠C1A1P. Then ∠B1A1C1 = ∠B1A1P + ∠C1A1P = ∠B1CP + ∠C1BP, but ∠C1BP = ∠ABQ = ∠ACQ. Therefore, ∠B1A1C1 = ∠B1CP + ∠ACQ = ∠QCP. Once again, the sine law guarantees that sin ∠QCP sin ∠BQC = PQ PC . (A1B1C1) = 1 2A1B1 · A1C1sin ∠B1A1C1 = 1 2PB · PCsin B sin C sin ∠QCP = 1 2PB · PC · sin B sin C PQ PC sin ∠BQC = 1 2PB · PQ · sin A sin B sin C = (R2 −OP 2)(ABC) 4R2 . The last equality holds true because the power of the point P with respect to the circumcircle of ABC is PB · PQ = R2 −OP 2, and because (ABC) = 2R2sin A sin B sin C. The area of A1B1C1 is maximum when P = O, that is, when A1B1C1 is the medial triangle. B C A B1 A1 P C1 O Q 4.3 Solutions to the problems in Chapter 3 Solution 3.1. Let a = A1A2, b = A1A3 and c = A1A4. Using Ptolemy’s theorem in the quadrilateral A1A3A4A5, we can deduce that ab + ac = bc or, equivalently, a b + a c = 1. 4.3 Solutions to the problems in Chapter 3 163 Since the triangles A1A2A3 and B1B2B3 are similar, B1B2 B1B3 = A1A2 A1A3 = a b and from there we obtain B1B2 = a2 b . Similarly C1C2 = a2 c . Therefore SB+SC SA = a2 b2 + a2 c2 = a2c2+a2b2 b2c2 = b2+c2 (b+c)2 > (b+c)2 2(b+c)2 = 1 2. The third equality follows from ab + ac = bc and the inequality follows from inequality (1.11). The inequality is strict since b ̸= c. Note that a2 b2 + a2 c2 = a b + a c 2 −2 a2 bc = 1 −2 a2 bc . The sine law applied to the triangle A1A3A4 leads us to a2 bc = sin2 π 7 sin 2π 7 sin 4π 7 = sin2 π 7 2 sin 2π 7 sin 2π 7 cos 2π 7 = sin2 π 7 2(1 −cos2 2π 7 ) cos 2π 7 = sin2 π 7 2 cos 2π 7 (1 + cos 2π 7 )(1 −cos 2π 7 ) = sin2 π 7 4 cos 2π 7 (1 + cos 2π 7 ) sin2 π 7 = 1 4 cos 2π 7 (1 + cos 2π 7 ) > 1 4 cos π 4 (1 + cos π 4 ) = 1 4 √ 2 2 (1 + √ 2 2 ) = √ 2 −1 2 . Thus a2 b2 + a2 c2 = 1 −2 a2 bc < 1 −( √ 2 −1) = 2 − √ 2. Solution 3.2. Cut the tetrahedron along the edges AD, BD, CD and place it on the plane of the triangle ABC. The faces ABD, BCD and CAD will have as their image the triangles ABD1, BCD2 and CAD3. Observe that D3, A and D1 are collinear, as are D1, B and D2. Moreover, A is the midpoint of D1D3 (since both D1A and D3A are equal in length to DA), and similarly B is the midpoint of D1D2. Then AB = 1 2D2D3 and by the triangle inequality, D2D3 ≤CD3 + CD2 = 2CD. Hence AB ≤CD, as desired. Solution 3.3. Letting S be the area of the triangle, we have the formulae sin α = 2S bc , sin β = 2S ca , sin γ = 2S ab and r = S s = 2S a+b+c. Using these formulae we ﬁnd that the inequality to be proved is equivalent to a bc + b ca + c ab (a + b + c) ≥9, which can be proved by applying the AM-GM inequality to each factor on the left side. Solution 3.4. Suppose that the circles have radii 1. Let P be the common point of the circles and let A, B, C be the second intersection points of each pair of circles. We have to minimize the common area between any pair of circles, which will be minimum if the point P is in the interior of the triangle ABC (otherwise, rotate one circle by 180◦around P, and this will reduce the common area). The area of the common parts is equal to π −(sin α + sin β + sin γ), where α, β, γ are the central angles of the common arcs of the circles. It is clear that 164 Solutions to Exercises and Problems γ B β α A γ P α C β α+β+γ = 180◦. Since the function sin x is concave, the minimum is reached when α = β = γ = π 3 , which implies that the centers of the circles form an equilateral triangle. Solution 3.5. Let I be the incenter of ABC, and draw the line through I perpendic-ular to IC. Let D′, E′ be the intersections of this line with BC, CA, respectively. First prove that (CDE) ≥(CD′E′) by showing that the area of D′DI is greater than or equal to the area of EE′I; to see this, observe that one of the triangles DD′I, EE′I lies in the opposite side to C with respect to the line D′E′, if for instance, it is DD′I, then this triangle will have a greater area than the area of EE′I, then the claim. Now, prove that the area (CD′E′) is 2r2 sin C ; to see this, note that CI = r sin C 2 and that D′I = r cos C 2 , then (CD′E′) = 1 2D′E′ · CI = D′I · CI = 2r2 2 sin C 2 cos C 2 = 2r2 sin C ≥2r2. Solution 3.6. The key is to note that 2AX ≥ √ 3(AB+BX), which can be deduced by applying Ptolemy’s inequality (Exercise 2.11) to the cyclic quadrilateral ABXO that is formed when we glue the triangle ABX to the equilateral triangle AXO of side AX, and then observing that the diameter of the circumcircle of the equilateral triangle is 2 √ 3AX, that is, AX(AB + BX) = AX · BO ≤AX · 2 √ 3AX. Hence 2AD = 2(AX + XD) ≥ √ 3(AB + BX) + 2XD ≥ √ 3(AB + BC + CX) + √ 3XD ≥ √ 3(AB + BC + CD). Solution 3.7. Take the triangle A′B′C′ of maximum area between all triangles that can be formed with three points of the given set of points; then its area satisﬁes (A′B′C′) ≤1. Construct another triangle ABC that has A′B′C′ as its 4.3 Solutions to the problems in Chapter 3 165 medial triangle; this has an area (ABC) = 4(A′B′C′) ≤4. In ABC we can ﬁnd all the points. Indeed, if some point Q is outside of ABC, it will be in one of the half-planes determined by the sides of the triangle and opposite to the half-plane where the third vertex lies. For instance, if Q is in the half-plane determined by BC, opposite to where A lies, the triangle QB′C′ has greater area than A′B′C′, a contradiction. Solution 3.8. Let M = 1+ 1 2 +· · ·+ 1 n. Let us prove that M is the desired minimum value, which is achieved by setting x1 = x2 = · · · = xn = 1. Using the AM-GM inequality, we get xk k + (k −1) ≥kxk for all k. Therefore x1+ x2 2 2 + x3 3 3 +· · ·+ xn n n ≥x1+x2−1 2 +· · ·+xn−n −1 n = x1+x2+· · ·+xn−n+M. On the other hand, the arithmetic-harmonic inequality leads us to x1 + x2 + · · · + xn n ≥ n 1 x1 + 1 x1 + · · · + 1 xn = 1. We conclude that the given expression is at least n −n + M = M. Since M is achieved, it is the desired minimum. Second solution. Apply the weighted AM-GM inequality to the numbers {xj j} with weights -tj = 1 j 1 j . , to get xj j j ≥ 1 j (x1x2 · · · xn) 1 1 j ≥ 1 j . The last inequality follows from n n 1 x1 · · · 1 xn ≤ 1 xj = n. Solution 3.9. Note that AFE and BDC are equilateral triangles. Let C′ and F ′ be points outside the hexagon and such that ABC′ and DEF ′ are also equilateral triangles. Since BE is the perpendicular bisector of AD, it follows that C′ and F ′ are the reﬂections of C and F on the line BE. Now use the fact that AC′BG and EF ′DH are cyclic in order to conclude that AG + GB = GC′ and DH + HE = HF ′. Solution 3.10. Leibniz’s theorem implies OG2 = R2 −1 9(a2 + b2 + c2). Since rs = abc 4R , we can deduce that 2rR = abc a+b+c. Then we have to prove that abc ≤ (a+b+c) 3 (a2+b2+c2) 3 , for which we can use the AM-GM inequality. Solution 3.11. The left-hand side of the inequality follows from 1 + x0 + x1 + · · · + xi−1 √xi + · · · + xn ≤1 2 (1 + x0 + · · · + xn) = 1. 166 Solutions to Exercises and Problems For the right-hand side consider θi = arcsin (x0 + · · · + xi) for i = 0, . . . , n. Note that 1 + x0 + · · · + xi−1 √xi + · · · + xn = 1 + sin θi−1 1 −sin θi−1 = cos θi−1. It is left to prove that sin θi−sin θi−1 cos θi−1 < π 2 . But sin θi −sin θi−1 = 2 cos θi + θi−1 2 sin θi −θi−1 2 < (cos θi−1)(θi −θi−1). To show the inequality, use the facts that cos θ is a decreasing function and that sin θ ≤θ for 0 ≤θ ≤π 2 . Then sin θi −sin θi−1 cos θi−1 < θi −θi−1 = θn −θ0 = π 2 . Solution 3.12. If n i=1 xi = 1, then 1 = (n i=1 xi)2 = n i=1 x2 i + 2 i<j xixj. Therefore the inequality that we need to prove is equivalent to 1 n −1 ≤ n i=1 x2 i 1 −ai . Use the Cauchy-Schwarz inequality to prove that n i=1 xi 2 ≤ n i=1 x2 i 1 −ai n i=1 (1 −ai) . Solution 3.13. First prove that n i=1 xn+1(xn+1−xi) = (n−1)x2 n+1. The inequality that we need to prove is reduced to n i=1 xi(xn+1 −xi) ≤ √ n −1xn+1. Now use the Cauchy-Schwarz inequality with the following two n-sets of real num-bers: (√x1, . . . , √xn) and (√xn+1 −x1, . . . , √xn+1 −xn). Solution 3.14. First, recall that N is also the midpoint of the segment that joins the midpoints X and Y of the diagonals AC and BD. The circle of diameter OM goes through X and Y since OX and OY are perpendiculars to the corresponding diagonals, and ON is a median of the triangle OXY . Solution 3.15. The inequality on the right-hand side follows from wx + xy + yz + zw = (w + y)(x + z) = −(w + y)2 ≤0. 4.3 Solutions to the problems in Chapter 3 167 For the inequality on the left-hand side, note that \|wx + xy + yz + zw\| = \|(w + y)(x + z)\| ≤1 2 (w + y)2 + (x + z)2 ≤w2 + x2 + y2 + z2 = 1. We can again use the Cauchy-Schwarz inequality to obtain \|wx + xy + yz + zw\|2 ≤(w2 + x2 + y2 + z2)(x2 + y2 + z2 + w2) = 1. Solution 3.16. For the inequality on the left-hand side, rearrange as follows: an + a2 a1 + a1 + a3 a2 + · · · + an−1 + a1 an = a1 a2 + a2 a1 + a3 a2 + a2 a3 + · · · + a1 an + an a1 , now, use the fact that x y + y x ≥2. Set Sn = an+a2 a1 + a1+a3 a2 + a2+a4 a3 +· · ·+ an−1+a1 an . Using induction, prove that Sn ≤3n. First, for n = 3, we need to see that b+c a + c+a b + a+b c ≤9. If a = b = c, then b+c a + c+a b + a+b c = 6 and the inequality is true. Suppose that a ≤b ≤c and that not all numbers are equal, then we have three cases: a = b < c, a < b = c, a < b < c. In all of them, we have a ≤b and a < c. Hence 2c = c + c > a + b and a+b c < 2, and since a+b c is a positive integer we have c = a + b. Thus, b+c a + c+a b + a+b c = a+2b a + 2a+b b + 1 = 3 + 2 b a + 2 a b . Since 2 b a and 2 a b are positive integers, and since 2 b a 2 a b = 4, we have that either both numbers are equal to 2 or one number is 1 and the other is 4. This means the sum is at most 8, which is less than 9, then the result. We continue with the induction. Suppose that Sn−1 ≤3(n −1). Consider {a1, . . . , an}, if all are equal, then Sn = 2n and the inequality is true. Suppose instead that there are at least two diﬀerents ai’s. Take the maximum of the ai’s; its neighbors (ai−1, ai+1) can be equal to this maximum value, but since there are two diﬀerent numbers between the ai’s for some maximum ai, we have that one of its neighbors is less than ai. We can then assume, without loss of generality, that an is maximum and that one of its neighbors, an−1 or a1, is less than an. Then, since 2an > an−1 + a1, we have that an−1+a1 an < 2 and then an−1+a1 an = 1, for which an = an−1 + a1. When we substitute this value of an in Sn, we get Sn = an−1 + a1 + a2 a1 + a2 + a3 a2 + · · · + an−2 + an−1 + a1 an−1 + an−1 + a1 an−1 + a1 = 1 + an−1 + a2 a1 + a2 + a3 a2 + · · · + an−2 + a1 an−1 + 1 + 1. Since Sn−1 ≤3(n −1), this implies that Sn ≤3n. 168 Solutions to Exercises and Problems Solution 3.17. Since the quadrilateral OBDC is cyclic, use Ptolemy’s theorem to prove that OD = R BD BC + DC BC , where R is the circumradius of ABC. On the other hand, since the triangles BCE and DCA are similar, as well as the triangles ABD and FBC, it happens that R BD BC + DC BC = R AD F C + AD EB . We can ﬁnd similar equalities for OE and OF, OE = R BE AD + BE CF and OF = R CF BE + CF AD . Multiplying these equalities and applying the AM-GM inequality, the result is attained. Another way to prove this is using inversion. Let D′, E′ and F ′ be the intersection points of AO, BO and CO with the sides BC, CA and AB, re-spectively. Invert the sides BC, CA and AB with respect to (O, R), obtain-ing the circumcircles of the triangles OBC, OCA and OAB, respectively. Then, OD · OD′ = OE · OE′ = OF · OF ′ = R2. If x = (ABO), y = (BCO) and z = (CAO), we can deduce that AO OD′ = z + x y , BO OE′ = x + y z and CO OF ′ = y + z x . This implies, using the AM-GM inequality, that R3 OD′·OE′·OF ′ ≥8; therefore, OD · OE · OF ≥8R3. Solution 3.18. First, observe that AY ≤2R and that ha ≤AX, where ha is the length of the altitude on BC. Then we can deduce that la sin2 A = AX AY sin2 A ≥ ha 2R sin2 A = ha a sin A since sin A a = 1 2R ≥3 3 ha a sin A hb b sin B hc c sin C = 3 since ha = b sin C, hb = c sin A, hc = a sin B. Solution 3.19. Without loss of generality, x1 ≤x2 ≤· · · ≤xn. Since 1 < 2 < · · · < n, we have, using the rearrangement inequality (1.2), that A = x1 + 2x2 + · · · + nxn ≥nx1 + (n −1)x2 + · · · + xn = B. Then, \|A + B\| = \|(n + 1) (x1 + · · · + xn)\| = n + 1, hence A + B = ±(n + 1). Now, if A + B = (n + 1) it follows that B ≤n+1 2 ≤A, and if A + B = −(n + 1), it is the case that B ≤−n+1 2 ≤A If we now assume that n+1 2 or −n+1 2 is between B and A, otherwise A or B would be in the interval −n+1 2 , n+1 2 , then either \|A\| or \|B\| is less than or equal to n+1 2 and we can solve the problem. 4.3 Solutions to the problems in Chapter 3 169 Suppose therefore that B ≤−n+1 2 < n+1 2 ≤A. Let y1, . . ., yn be a permutation of x1, . . ., xn such that 1y1+2y2+· · ·+nyn = C takes the maximum value with C ≤−n+1 2 . Take i such that y1 ≤y2 ≤· · · ≤yi and yi > yi+1 and consider D = y1 + 2y2 + · · · + iyi+1 + (i + 1)yi + (i + 2)yi+2 + · · · + nyn D −C = iyi+1 + (i + 1)yi −(iyi + (i + 1)yi+1) = yi −yi+1 > 0. Since \|yi\|, \|yi+1\| ≤n+1 2 , we can deduce that D −C = yi −yi+1 ≤n + 1; hence D ≤C + n + 1 and therefore C < D ≤C + n + 1 ≤n+1 2 . On the other hand, D ≥−n+1 2 , since C is the maximum sum which is less than −n+1 2 . Thus −n+1 2 ≤D ≤n+1 2 and then \|D\| ≤n+1 2 . Solution 3.20. Among the numbers x, y, z two have the same sign (say x and y), since c = z x y + y x is positive, we can deduce that z is positive. Note that a + b −c = 2xy z , b + c −a = 2yz x , c + a −b = 2zx y are positive. Conversely, if u = a + b −c, v = b + c −a and w = c + a −b are positive, taking u = 2xy z , v = 2yz x , w = 2zx y , we can obtain a = u+w 2 = x y z + z y , and so on. Solution 3.21. First, prove that a centrally symmetric hexagon ABCDEF has opposite parallel sides. Thus, (ACE) = (BDF) = (ABCDEF ) 2 . Now, if we reﬂect the triangle PQR with respect to the symmetry center of the hexagon, we get the points P ′, Q′, R′ which form the centrally symmetric hexagon PR′QP ′RQ′, inscribed in ABCDEF with area 2(PQR). Solution 3.22. Let X = 4 i=1 x3 i , Xi = X −x3 i ; it is then evident that X = 1 3 4 i=1 Xi. Using the AM-GM inequality leads to 1 3X1 ≥ 3 x3 2x3 3x3 4 = 1 x1 ; similar inequalities hold for the other indexes and this implies that X ≥4 i=1 1 xi . Using Tchebyshev’s inequality we obtain x3 1 + x3 2 + x3 3 + x3 4 4 ≥x2 1 + x2 2 + x2 3 + x2 4 4 · x1 + x2 + x3 + x4 4 . Thanks to the AM-GM inequality we get x2 1+x2 2+x2 3+x2 4 4 ≥ 4 (x1x2x3x4)2 = 1, and therefore X ≥ i=1 xi. Solution 3.23. Use the Cauchy-Schwarz inequality with u = √x−1 √x , √y−1 √y , √z−1 √z and v = √x, √y, √z . Solution 3.24. If α = ∠ACM and β = ∠BDM, then MA·MB MC·MD = tan α tan β and α+β = π 4 . Now use the fact that tan α tan β tan γ ≤tan3 α+β+γ 3 , where γ = π 4 . Another method uses the fact that the inequality is equivalent to (MCD) ≥ 3 √ 3(MAB) which is equivalent to h+l h ≥3 √ 3, where l is the length of the side of the square and h is the length of the altitude from M to AB. Find the maximum h. 170 Solutions to Exercises and Problems Solution 3.25. First note that PL AL + PM BM + PN CN = 1. Now, use the fact that AL, BM and CN are less than a. Solution 3.26. Since PB PA · QC QA ≤1 4 PB PA + QC QA 2 , it is suﬃcient to see that P B PA + QC QA = 1. A B C G A′ B′ C′ P Q Draw BB′, CC′ parallel to the median AA′ in such a way that B′ and C′ are on PQ. The triangles APG and BPB′ are similar, as well as AQG and CQC′, thus P B P A = BB′ AG and QC QA = CC′ AG . Use this together with the fact that AG = 2GA′ = BB′ + CC′. Solution 3.27. Let Γ be the circumcircle of ABC, and let R be its radius. Consider the inversion in Γ. For any point P other than O, let P ′ be its inverse. The inverse of the circumcircle of OBC is the line BC, then A′ 1, the inverse of A1, is the intersection point between the ray OA1 and BC. Since22 P ′Q′ = R2 · PQ OP · OQ for two points P, Q (distinct from O) with inverses P ′, Q′, we have AA1 OA1 = R2 · A′A′ 1 OA′ · OA′ 1 · OA1 = AA′ 1 OA = x + y + z y + z , where x, y, z denote the areas of the triangles OBC, OCA, OAB, respectively. Similarly, we have that BB1 OB1 = x + y + z z + x and CC1 OC1 = x + y + z x + y . Thus AA1 OA1 + BB1 OB1 + CC1 OC1 = (x + y + z) 1 y + z + 1 z + x + 1 x + y ≥9 2. For the last inequality, see Exercise 1.44. 22See [5, page 132] or [9, page 112]. 4.3 Solutions to the problems in Chapter 3 171 Solution 3.28. The area of the triangle GBC is (GBC) = (ABC) 3 = a·GL 2 . Therefore GL = 2(ABC) 3a . Similarly, GN = 2(ABC) 3c . In consequence, (GNL) = GL · GN sin B 2 = 4(ABC)2 sin B 18ac = 4(ABC)2b2 (18abc)(2R) = (ABC)2 b2 (9R abc 4R )(4R) = (ABC) b2 9 · 4R2 . Similarly, (GLM) = (ABC) c2 9·4R2 and (GMN) = (ABC) a2 9·4R2 . Therefore, (LMN) (ABC) = 1 9 a2 + b2 + c2 4R2 = R2 −OG2 4R2 . The inequality in the right follows easily. For the other inequality, note that OG = 1 3OH. Since the triangle is acute, H is inside the triangle and HO ≤R. Therefore, (LMN) (ABC) = R2 −1 9OH2 4R2 ≥R2 −1 9R2 4R2 = 2 9 > 4 27. Solution 3.29. The function f(x) = 1 1+x is convex for x > 0. Thus, f(ab) + f(bc) + f(ca) 3 ≥f ab + bc + ca 3 = 3 3 + ab + bc + ca ≥ 3 3 + a2 + b2 + c2 = 1 2, the last inequality follows from ab + bc + ca ≤a2 + b2 + c2. We can also begin with 1 1 + ab + 1 1 + bc + 1 1 + ca ≥ 9 3 + ab + bc + ca ≥ 9 3 + a2 + b2 + c2 = 3 2. The ﬁrst inequality follows from inequality (1.11) and the second from Exercise 1.27. Solution 3.30. Set x = b + 2c, y = c + 2a, z = a + 2b. The desired inequality becomes x y + y x + y z + z y + z x + x z + 3 y x + z y + x z ≥15, 172 Solutions to Exercises and Problems which can be proved using the AM-GM inequality. Another way of doing it is the following: a b + 2c + b c + 2a + c a + 2b = a2 ab + 2ca + b2 bc + 2ab + c2 ca + 2bc ≥ (a + b + c)2 3(ab + bc + ca). The inequality follows from inequality (1.11). It remains to prove the inequality (a + b + c)2 ≥3(ab + bc + ca), which is a consequence of the Cauchy-Schwarz inequality. Solution 3.31. Use the inequality (1.11) or use the Cauchy-Schwarz inequality with a √ a + b, b √ b + c, c √ c + d, d √ d + a and ( √ a + b, √ b + c, √ c + d, √ d + a). Solution 3.32. Let x = b + c −a, y = c + a −b and z = a + b −c. The similarity between the triangles ADE and ABC gives us DE a = perimeter of ADE perimeter of ABC = 2x a + b + c. Thus, DE = x(y+z) x+y+z; that is, the inequality is equivalent to x(y+z) x+y+z ≤x+y+z 4 . Now use the AM-GM inequality. Solution 3.33. Take F on AD with AF = BC and deﬁne E′ as the intersection of BF and AC. Using the sine law in the triangles AE′F, BCE′ and BDF, we obtain AE′ E′C = AF sin F sin E′ · sin E′ BC sin B = sin F sin B = BD FD = AE EC , therefore E′ = E. Subsequently, consider G on BD with BG = AD and H the intersection point of GE with the parallel to BC passing through A. Use the fact that the triangles ECG and EAH are similar and also use Menelaus’s theorem for the triangle CAD with transversal EFB to conclude that AH = DB. Hence, BDAH is a parallelogram, BH = AD and BHG is isosceles with BH = BG = AD > BE. Solution 3.34. Note that ab + bc + ca ≤3abc if and only if 1 a + 1 b + 1 c ≤3. Since (a + b + c) 1 a + 1 b + 1 c ≥9, we should have that (a + b + c) ≥3. Then 3(a + b + c) ≤(a + b + c)2 = a3/2a−1/2 + b3/2b−1/2 + c3/2c−1/2 2 ≤ a3 + b3 + c3 1 a + 1 b + 1 c ≤3 a3 + b3 + c3 . 4.3 Solutions to the problems in Chapter 3 173 Solution 3.35. Take yi = xi n−1 for all i = 1, 2, . . ., n and suppose that the inequality is false, that is, 1 1 + y1 + 1 1 + y2 + · · · + 1 1 + yn > n −1. Then 1 1 + yi > j̸=i 1 − 1 1 + yj = j̸=i yj 1 + yj ≥(n −1) n−1 / y1 · · · ˆ yi · · · yn (1 + y1) · · · (1 + yi) · · · (1 + yn) , where y1 · · · ˆ yi · · · yn is the product of the y’s except yi. Then n 1 i=1 1 1 + yi > (n −1)n y1 · · · yn (1 + y1) · · · (1 + yn), and this implies 1 > x1 · · · xn, a contradiction. Solution 3.36. Use the Cauchy-Schwarz inequality with x1 y1 , . . . , xn yn and √x1y1, . . . , √xnyn to get (x1 + · · · + xn)2 = x1 y1 √x1y1 + · · · + xn yn √xnyn 2 ≤ x1 y1 + · · · + xn yn (x1y1 + · · · + xnyn) . Now use the hypothesis xiyi ≤ xi. Solution 3.37. Since abc = 1, we have (a −1)(b −1)(c −1) = a + b + c − 1 a + 1 b + 1 c and similarly (an −1)(bn −1)(cn −1) = an + bn + cn − 1 an + 1 bn + 1 cn . The proof follows from the fact that the left sides of the equalities have the same sign. Solution 3.38. We prove the claim using induction on n. The case n = 1 is clear. Now assuming the claim is true for n, we can prove it is true for n + 1. Since n < √ n2 + i < n + 1, for i = 1, 2, . . ., 2n, we have - n2 + i . = n2 + i −n < / n2 + i + i 2n 2 −n = i 2n. 174 Solutions to Exercises and Problems Thus (n+1)2 j=1 - j . = n2 j=1 - j . + (n+1)2 j=n2+1 - j . ≤n2 −1 2 + 1 2n 2n i=1 i = (n + 1)2 −1 2 . Solution 3.39. Let us prove that the converse aﬃrmation, that is, x3 + y3 > 2, implies that x2 + y3 < x3 + y4. The power mean inequality x2+y2 2 ≤ 3 x3+y3 2 implies that x2 + y2 ≤(x3 + y3)2/3 3 √ 2 < (x3 + y3)2/3(x3 + y3)1/3 = x3 + y3. Then x2 −x3 < y3 −y2 ≤y4 −y3. The last inequality follows from the fact that y2(y −1)2 ≥0. Second solution. Since (y −1)2 ≥0, we have that 2y ≤y2 + 1, then 2y3 ≤y4 + y2. Thus, x3 + y3 ≤x3 + y4 + y2 −y3 ≤x2 + y2, since x3 + y4 ≤x2 + y3. Solution 3.40. The inequality is equivalent to (x0 −x1) + 1 (x0 −x1) + (x1 −x2) + · · · + (xn−1 −xn) + 1 (xn−1 −xn) ≥2n. Solution 3.41. Since a+3b 4 ≥ 4 √ ab3, b+4c 5 ≥ 5 √ bc4 and c+2a 3 ≥ 3 √ ca2, we can deduce that (a + 3b)(b + 4c)(c + 2a) ≥60a 11 12 b 19 20 c 17 15 . Now prove that c 2 15 ≥a 1 12 b 1 20 or, equivalently, that c8 ≥a5b3. Solution 3.42. We have an equivalence between the following inequalities: 7(ab + bc + ca) ≤2 + 9 abc ⇔ 7(ab + bc + ca)(a + b + c) ≤2(a + b + c)3 + 9 abc ⇔ a2b + a b2 + b2c + b c2 + c2a + c a2 ≤2(a3 + b3 + c3.) For the last one use the rearrangement inequality or Tchebyshev’s inequality. Solution 3.43. Let E be the intersection of AC and BD. Then the triangles ABE and DCE are similar, which implies \|AB −CD\| \|AC −BD\| = \|AB\| \|AE −EB\|. Using the triangle inequality in ABE, we have \|AB\| \|AE−EB\| ≥1 and we therefore conclude that \|AB −CD\| ≥\|AC −BD\|. Similarly, \|AD −BC\| ≥\|AC −BD\|. 4.3 Solutions to the problems in Chapter 3 175 Solution 3.44. First of all, show that a1 + · · · + aj ≥j(j+1) 2n an, for j ≤n, in the following way. First, prove that the inequality is valid for j = n, that is, a1 + · · ·+ an ≥n+1 2 an; use the fact that 2(a1+· · ·+an) = (a1+an−1)+· · ·+(an−1+a1)+2an. Next, prove that if bj = a1+···+aj 1+···+j , then b1 ≥b2 ≥· · · ≥bn ≥an n (to prove by induction that bj ≥bj+1, we need to show that bj ≥aj+1 j+1 which, on the other hand, follows from the ﬁrst part for n = j + 1). We provide another proof of a1 + · · · + aj ≥ j(j+1) 2n an, once again using induction. It is clear that a1 ≥a1, a1 + a2 2 = a1 2 + a1 2 + a2 2 ≥a2 2 + a2 2 = a2. Now, let us suppose that the aﬃrmation is valid for n = 1, . . . , j, that is, a1 ≥a1 a1 + a2 2 ≥a2 . . . a1 + a2 2 + · · · + aj j ≥aj. Adding all the above inequalities, we obtain ja1 + (j −1)a2 2 + · · · + aj j ≥a1 + · · · + aj. Adding on both sides the identity a1 + 2a2 2 + · · · + j aj j = aj + · · · + a1, we obtain (j + 1) a1 + a2 2 + · · · + aj j ≥(a1 + aj) + (a2 + aj−1) + · · · + (aj + a1) ≥jaj+1. Hence a1 + a2 2 + · · · + aj j ≥ j j + 1aj+1. Finally, adding aj+1 j+1 on both sides of the inequality provides the ﬁnal step in the induction proof. Now, a1 + a2 2 + · · · + an n = 1 n(a1 + · · · + an) + n−1 j=1 1 j − 1 j + 1 (a1 + · · · + aj) ≥1 n n(n + 1) 2n an + n−1 j=1 1 j(j + 1) j(j + 1) 2n an = an. 176 Solutions to Exercises and Problems Solution 3.45. ⎛ ⎝ 1≤i≤n xi ⎞ ⎠ 4 = ⎛ ⎝ 1≤i≤n x2 i + 2 1≤i<j≤n xixj ⎞ ⎠ 2 ≥4 ⎛ ⎝ 1≤i≤n x2 i ⎞ ⎠ ⎛ ⎝2 1≤i<j≤n xixj ⎞ ⎠ = 8 ⎛ ⎝ 1≤i≤n x2 i ⎞ ⎠ ⎛ ⎝ 1≤i<j≤n xixj ⎞ ⎠ = 8 1≤i 0. Using the AM-GM inequality we have 3x4y2 ≤x6 + x3y3 + x3y3 and 3x2y4 ≤y6 + x3y3 + x3y3, with equality if and only if x6 = x3y3 = y6 or, equivalently, if and only if x = y. Adding together these two inequalities and adding x6 + y6 to both sides, we get x6 + y6 + 3x2y2(x2 + y2) ≤2(x6 + y6 + 2x3y3). Equality occurs when x = y, that is, when a = b. Solution 3.47. Denote the left-hand side of the inequality as S. Since a ≥b ≥c and x ≥y ≥z, using the rearrangement inequality we have bz +cy ≤by +cz, then (by + cz)(bz + cy) ≤(by + cz)2 ≤2((by)2 + (cz)2). Setting α = (ax)2, β = (by)2, γ = (cz)2, we obtain a2x2 (by + cz)(bz + cy) ≥ a2x2 2((by)2 + (cz)2) = α 2(β + γ). 4.3 Solutions to the problems in Chapter 3 177 Adding together the other two similar inequalities, we get S ≥1 2 α β + γ + β γ + α + γ α + β . Use Nesbitt’s inequality to conclude the proof. Solution 3.48. If XM is a median in the triangle XY Z, then XM 2 = 1 2XY 2 + 1 2XZ2 −1 4Y Z2, a result of using Stewart’s theorem. If we take (X, Y, Z, M) to be equal to (A, B, C, P), (B, C, D, Q), (C, D, A, R) and (D, A, B, S), and then substitute them in the formula, we then add together the four resulting equations to get a ﬁfth equation. Multiplying both sides of the ﬁfth equation by 4, we ﬁnd that the left-hand side of the desired inequality equals AB2+BC2 +CD2+DA2+ 4(AC2 + BD2). Thus, it is suﬃcient to prove that AC2 + BD2 ≤AB2 + BC2 + CD2 + DA2. This inequality is known as the “parallelogram inequality”. To prove it, let O be an arbitrary point on the plane, and for each point X let x denote the vector from O to X. We expand each term in AB2 + BC2 + CD2 + DA2 −AC2 − BD2, writing for instance AB2 = \|a −b\|2 = \|a\|2 −2a · b + \|b\|2 and then ﬁnding that the expression equals \|a\|2 + \|b\|2 + \|c\|2 + \|d\|2 −2(a · b + b · c + c · d + d · a −a · c −b · d) = \|a + c −b −d\|2 ≥0, with equality if and only if a + c = b + d, that is, only if the quadrilateral ABCD is a parallelogram. Solution 3.49. Put A = x2 + y2 + z2, B = xy + yz + zx, C = x2y2 + y2z2 + z2x2, D = xyz. Then 1 = A+2B, B2 = C +2xyz(x+y+z) = C +2D and x4 +y4+z4 = A2 −2C = 4B2 −4B + 1 −2C = 2C −4B + 8D + 1. Then, the expression in the middle is equal to 3 −2A + (2C −4B + 8D + 1) = 2 + 2C + 8D ≥2, with equality if and only if two out of the x, y, z are zero. Now, the right-hand expression is equal to 2 +B +D. Thus we have to prove that 2C +8D ≤B+D or B−2B2−3D ≥0. Using the Cauchy-Schwarz inequality, we get A ≥B, so that B(1 −2B) = BA ≥B2. Thus it is suﬃcient to prove that B2 −3D = C −D ≥0. But C ≥xyyz + yzzx + zxxy = D as can be deduced from the Cauchy-Schwarz inequality. Solution 3.50. Suppose that a = x y , b = y z , c = z x. The inequality is equivalent to x y −1 + z y y z −1 + x z z x −1 + y x ≤1 178 Solutions to Exercises and Problems and can be rewritten as (x+z −y)(x+y−z)(y+z −x) ≤xyz. This last inequality is valid if x, y, z are the lengths of the sides of a triangle. See Example 2.2.3. A case remains when some out of the u = x+z−y, v = x+y−z, w = y+z−x are negative. If one or three of them are negative, then the left side is negative and the inequality is clear. If two of the values u, v, w are negative, for instance u and v, then u + v = 2x is also negative; but x > 0, so that this last situation is not possible. Solution 3.51. First note that abc ≤a + b + c implies (abc)2 ≤(a + b + c)2 ≤ 3(a2 + b2 + c2), where the last inequality follows from inequality (1.11). By the AM-GM inequality, a2 + b2 + c2 ≥3 3 (abc)2, then (a2 + b2 + c2)3 ≥ 33(abc)2. Therefore (a2 + b2 + c2)4 ≥32(abc)4. Solution 3.52. Using the AM-GM inequality, (a + b)(a + c) = a(a + b + c) + bc ≥2 abc(a + b + c). Second solution. Setting x = a + b, y = a + c, z = b + c, and since a, b, c are positive, we can deduce that x, y, z are the side lengths of a triangle XY Z. Thus, the inequality is equivalent to xy 2 ≥(XY Z) as can be seen using the formula for the area of a triangle in Section 2.2. Now, recall that the area of a triangle with side lengths x, y, z is less than or equal to xy 2 . Solution 3.53. Since xi ≥0, then xi −1 ≥−1. Next, we can use Bernoulli’s inequality for all i to get (1 + (xi −1))i ≥1 + i(xi −1). Adding these inequalities together for 1 ≤i ≤n, gives us the result. Solution 3.54. Subtracting 2, we ﬁnd that the inequalities are equivalent to 0 < (a + b −c)(a −b + c)(−a + b + c) abc ≤1. The left-hand side inequality is now obvious. The right-hand side inequality is Example 2.2.3. Solution 3.55. If we prove that a √ a2+8bc ≥ a4/3 a4/3+b4/3+c4/3 , it will be clear how to get the result. The last inequality is equivalent to a4/3 + b 4/3 + c 4/3 2 ≥a2/3(a2 + 8bc). Apply the AM-GM inequality to each factor of a4/3 + b 4/3 + c 4/3 2 − a4/3 2 = b 4/3 + c 4/3 a4/3 + a4/3 + b 4/3 + c 4/3 . 4.3 Solutions to the problems in Chapter 3 179 Another method for solving this exercise is to consider the function f(x) = 1 √x, this function is convex for x > 0 (f ′′(x) = 3 4 √ x5 > 0). For 0 < a, b, c < 1, with a + b + c = 1, we can deduce that a √x + b √y + c √z ≥ 1 √ax+by+cz. Applying this to x = a2 + 8bc, y = b2 + 8ca and z = c2 + 8ab (previously multiplying by an appropriate factor to have the condition a + b + c = 1), we get a √ a2 + 8bc + b √ b2 + 8ca + c √ c2 + 8ab ≥ 1 √ a3 + b3 + c3 + 24abc . Also use the fact that (a + b + c)3 = a3 + b3 + c3 + 3(a2b + a2c + b2a + b2c + c2a + c2b) + 6abc ≥a3 + b3 + c3 + 24abc. Solution 3.56. Using the Cauchy-Schwarz inequality aibi ≤ a2 i b2 i with ai = 1, bi = xi 1+x2 1+x2 2+···+x2 i , we can deduce that x1 1 + x2 1 + x2 1 + x2 1 + x2 2 + · · · + xn 1 + x2 1 + · · · + x2 n ≤√n b2 i . Then, it suﬃces to show that b2 i < 1. Observe that for i ≥2, b2 i = xi 1 + x2 1 + · · · + x2 i 2 = x2 i (1 + x2 1 + · · · + x2 i )2 ≤ x2 i (1 + x2 1 + · · · + x2 i−1)(1 + x2 1 + · · · + x2 i ) = 1 (1 + x2 1 + · · · + x2 i−1) − 1 (1 + x2 1 + · · · + x2 i ). For i = 1, use the fact that b2 1 ≤ x2 1 1+x2 1 = 1 − 1 1+x2 1 . Adding together these inequal-ities, the right-hand side telescopes to yield b2 i = n i=1 xi 1 + x2 1 + · · · + x2 i 2 ≤1 − 1 1 + x2 1 + · · · + x2 n < 1. Solution 3.57. Since there are only two possible values for α, β, γ, the three must either all be equal, or else two are equal and one is diﬀerent from these two. Therefore, we have two cases to consider. (1) α = β = γ. In this case we have a + b + c = 0, and therefore a3 + b3 + c3 abc 2 = a3 + b3 −(a + b)3 −ab(a + b) 2 = (a + b)2 −a2 + ab −b2 ab 2 = 3ab ab 2 = 9. 180 Solutions to Exercises and Problems (2) Without loss of generality, we assume that α = β, γ ̸= α, then c = a + b and a3 + b3 + c3 abc = a3 + b3 + (a + b)3 ab(a + b) = (a + b)2 + a2 −ab + b2 ab = 2a2 + 2b2 + ab ab = 2 a b + b a + 1. If a and b have the same sign, we see that this expression is not less than 5, and its square is therefore no less than 25. If the signs of a and b are not the same, we have a b + b a ≤−2, therefore 2 a b + b a + 1 ≤−3 and 2 a b + b a + 1 2 ≥9. Thus, the smallest possible value is 9. Solution 3.58. Using the AM-GM inequality, 1 b(a+b) + 1 c(b+c) + 1 a(c+a) ≥ 3 XY , where X = 3 √ abc, Y = 3 (a + b)(b + c)(c + a). Using AM-GM inequality again gives X ≤a+b+c 3 and Y ≤2 a+b+c 3 , then 3 XY ≥ 27 2 1 (a+b+c)2 . Solution 3.59. The inequality is equivalent to a4 + b4 + c4 ≥a2bc + b2ca + c2ab, which follows using Muirhead’s theorem since [4, 0, 0] ≥[2, 1, 1]. Second solution. a3 bc + b3 ca + c3 ab = a4 abc + b4 abc + c4 abc ≥(a2 + b2 + c2)2 3abc ≥(a + b + c)4 27abc = a + b + c 3 3 (a + b + c) abc ≥(abc) a + b + c abc = a + b + c. In the ﬁrst two inequalities we applied inequality (1.11), and in the last inequality we used the AM-GM inequality. Solution 3.60. Take f(x) as f(x) = x 1−x. Since f ′′(x) = 2 (1−x)3 > 0, f(x) is convex. Using Jensen’s inequality we get f(x) + f(y) + f(z) ≥3f( x+y+z 3 ). But since f is increasing for x < 1, and because the AM-GM inequality helps us to establish that x+y+z 3 ≥ 3 √xyz, then we can deduce that f( x+y+z 3 ) ≥f( 3 √xyz). Solution 3.61. a b + c + 1 2 b c + a + 1 2 c a + b + 1 2 ≥1 is equivalent to (2a + b + c)(2b + c + a)(2c + a + b) ≥8(b + c)(c + a)(a + b). Now, observe that (2a + b + c) = (a + b + a + c) ≥2 (a + b)(c + a). 4.3 Solutions to the problems in Chapter 3 181 Solution 3.62. The inequality of the problem is equivalent to the following inequal-ity: (a + b −c)(a + b + c) c2 + (b + c −a)(b + c + a) a2 + (c + a −b)(c + a + b) b2 ≥9, which in turn is equivalent to (a+b)2 c2 + (b+c)2 a2 + (c+a)2 b2 ≥12. Since (a + b)2 ≥4ab, (b + c)2 ≥4bc and (c + a)2 ≥4ca, we can deduce that (a + b)2 c2 + (b + c)2 a2 + (c + a)2 b2 ≥4ab c2 + 4bc a2 + 4ca b2 ≥12 3 (ab)(bc)(ca) c2a2b2 = 12. Solution 3.63. By the AM-GM inequality, x2 + √x + √x ≥3x. Adding similar inequalities for y, z, we get x2 + y2 + z2 + 2(√x + √y + √z) ≥3(x + y + z) = (x + y + z)2 = x2 + y2 + z2 + 2(xy + yz + zx). Solution 3.64. If we multiply the equality 1 = 1 a + 1 b + 1 c by √ abc, we get √ abc = ab c + bc a + ca b . Then, it is suﬃcient to prove that √ c + ab ≥√c + ab c . Squaring shows that this is equivalent to c + ab ≥c + ab c + 2 √ ab, c + ab ≥ c + ab(1 −1 a −1 b ) + 2 √ ab or a + b ≥2 √ ab. Solution 3.65. Since (1 −a)(1 −b)(1 −c) = 1 −(a + b + c) + ab + bc + ca −abc and since a + b + c = 2, the inequality is equivalent to 0 ≤(1 −a)(1 −b)(1 −c) ≤1 27. But a < b + c = 2 −a implies that a < 1 and, similarly, b < 1 and c < 1, therefore the left inequality is true. The other one follows from the AM-GM inequality. Solution 3.66. It is possible to construct another triangle AA1M with sides AA1, A1M, MA of lengths equal to the lengths of the medians ma, mb, mc. A C C1 A1 B1 M B Moreover, (AA1M) = 3 4(ABC). Then the inequality we have to prove is 1 mamb + 1 mbmc + 1 mcma ≤3 4 √ 3 (AA1M). 182 Solutions to Exercises and Problems Now, the last inequality will be true if the triangle with side-lengths a, b, c and area S satisﬁes the following inequality: 1 ab + 1 bc + 1 ca ≤3 √ 3 4S . Or equivalently, 4 √ 3 S ≤ 9abc a+b+c, which is Example 2.4.6. Solution 3.67. Substitute cd = 1 ab and da = 1 bc, so that the left-hand side (LHS) inequality becomes 1 + ab 1 + a + 1 + ab ab + abc + 1 + bc 1 + b + 1 + bc bc + bcd (4.8) = (1 + ab) 1 1 + a + 1 ab + abc + (1 + bc) 1 1 + b + 1 bc + bcd . Now, using the inequality 1 x + 1 y ≥ 4 x+y, we get (LHS) ≥(1 + ab) 4 1 + a + ab + abc + (1 + bc) 4 1 + b + bc + bcd = 4 1 + ab 1 + a + ab + abc + 1 + bc 1 + b + bc + bcd = 4 1 + ab 1 + a + ab + abc + a + abc a + ab + abc + abcd = 4. Solution 3.68. Using Stewart’s theorem we can deduce that l2 a = bc 1 − a b + c 2 = bc (b + c)2 ((b + c)2 −a2) ≤1 4((b + c)2 −a2). Using the Cauchy-Schwarz inequality leads us to (la + lb + lc)2 ≤3(l2 a + l2 b + l2 c) ≤3 4((a + b)2 + (b + c)2 + (c + a)2 −a2 −b2 −c2) ≤3 4(a + b + c)2. Solution 3.69. Since 1 1−a = 1 b+c, the inequality is equivalent to 1 b + c + 1 c + a + 1 a + b ≥ 2 2a + b + c + 2 2b + a + c + 2 2c + a + b. Now, using the fact that 1 x + 1 y ≥ 4 x+y, we have 2 1 b + c + 1 c + a + 1 a + b ≥ 4 a + b + 2c + 4 b + c + 2a + 4 c + a + 2b which proves the inequality. 4.3 Solutions to the problems in Chapter 3 183 Solution 3.70. We may take a ≤b ≤c. Then c < a + b and n √ 2 2 = n √ 2 2 (a + b + c) > n √ 2 2 (2c) = n √ 2cn ≥ n √ bn + cn. Since a ≤b, we can deduce that b + a 2 n = bn + nbn−1 a 2 + other positive terms > bn + n 2 abn−1 ≥bn + an. Similarly, since a ≤c, we have (c + a 2)n > cn + an, therefore (an + bn) 1 n + (bn + cn) 1 n + (cn + an) 1 n < b + a 2 + n √ 2 2 + c + a 2 = a + b + c + n √ 2 2 = 1 + n √ 2 2 . Second solution. Remember that a, b, c are the lengths of the sides of a triangle if and only if there exist positive numbers x, y, z with a = y+z, b = z +x, c = x+y. Since a + b + c = 1, we can deduce that x + y + z = 1 2. Now, we use Minkowski’s inequality n i=1 (xi + yi)m 1 m ≤ n i=1 xm i 1 m + n i=1 ym i 1 m to get (an + bn) 1 n = ((y + z)n + (z + x)n) 1 n ≤(xn + yn) 1 n + (2zn) 1 n < c + n √ 2z. Similarly, (bn + cn) 1 n < a + n √ 2x and (cn + an) 1 n < b + n √ 2y. Therefore (an + bn) 1 n + (bn + cn) 1 n + (cn + an) 1 n < a + b + c + n √ 2(x + y + z) = 1 + n √ 2 2 . Solution 3.71. First notice that if we restrict the sums to i < j, then they are halved. The left-hand side sum is squared while the right-hand side sum is not, so that the desired inequality with sums restricted to i < j has (1/3), instead of (2/3), on the right-hand side. Consider the sum of all the \|xi −xj\| with i < j. The number x1 appears in (n −1) terms with negative sign, x2 appears in one term with positive sign and (n −2) terms with negative sign, and so on. Thus, we get −(n −1)x1 −(n −3)x2 −(n −5)x3 −· · · + (n −1)xn = (2i −1 −n)xi. 184 Solutions to Exercises and Problems We can now apply the Cauchy-Schwarz inequality to show that the square of this sum is less than x2 i (2i −1 −n)2. Looking at the sum at the other side of the desired inequality, we immediately see that it is n x2 i −( xi)2. We would like to get rid of the second term, which is easy because if we add h to every xi the sums in the desired inequality are unaﬀected (since they only involve diﬀerences of the xi), so we can choose an h to make xi zero. Thus, we can ﬁnish if we can prove that (2i−1−n)2 = n(n2−1) 3 , (2i −1 −n)2 = 4 i2 −4(n + 1) i + n(n + 1)2 = 2 3n(n + 1)(2n + 1) −2n(n + 1)2 + n(n + 1)2 = 1 3n(n + 1)(2(2n + 1) −6(n + 1) + 3(n + 1)) = 1 3n(n2 −1). This establishes the required inequality. Second solution. The inequality is of the Cauchy-Schwarz type, and since the prob-lem asks us to prove that equality holds when x1, x2, . . . , xn form an arithmetic progression, that is, when xi−xj = r(i−j) with r > 0, then consider the following inequality which is true, as can be inferred from the Cauchy-Schwarz inequality, ⎛ ⎝ i,j \|i −j\|\|xi −xj\| ⎞ ⎠ 2 ≤ i,j (i −j)2 i,j (xi −xj)2. Here, we already know that equality holds if and only if (xi −xj) = r(i −j), with r > 0. Since i,j (i −j)2 = (2n −2) · 12 + (2n −4) · 22 + · · · + 2 · (n −1)2 = n2(n2−1) 6 , we need to prove that i,j \|i −j\| \|xi −xj\| = n 2 i,j \|xj −xj\|. To see that it happens compare the coeﬃcient of xi in each side. On the left-hand side the coeﬃcient is (i −1) + (i −2) + · · · + (i −(i −1)) −((i + 1) −i) + ((i + 2) −i) + · · · + (n −i)) = (i −1)i 2 −(n −i)(n −i + 1) 2 = n(2i −n −1) 2 . The coeﬃcient of xi on the right-hand side is n 2 ⎛ ⎝ ii −1 ⎞ ⎠= n 2 ((i −1) −(n −i)) = n(2i −n −1) 2 . Since they are equal we have ﬁnished the proof. 4.3 Solutions to the problems in Chapter 3 185 Solution 3.72. Let xn+1 = x1 and xn+2 = x2. Deﬁne ai = xi xi+1 and bi = xi + xi+1 + xi+2, i ∈{1, . . . , n}. It is evident that n 1 i=1 ai = 1, n i=1 bi = 3 n i=1 xi = 3. The inequality is equivalent to n i=1 ai bi ≥n2 3 . Using the AM-GM inequality, we can deduce that 1 n n i=1 bi ≥ n b1 · · · bn ⇔3 n ≥ n b1 · · · bn ⇔ 1 n √b1 · · · bn ≥n 3 . On the other hand and using again the AM-GM inequality, we get n i=1 ai bi ≥n n a1 b1 · · · an bn = n n √a1 · · · an n √b1 · · · bn = n n √b1 · · · bn ≥n2 3 . Solution 3.73. For any a positive real number, a + 1 a ≥2, with equality occurring if and only if a = 1. Since the numbers ab, bc and ca are non-negative, we have P(x)P 1 x = (ax2 + bx + c) a 1 x2 + b 1 x + c = a2 + b2 + c2 + ab x + 1 x + bc x + 1 x + ca x2 + 1 x2 ≥a2 + b2 + c2 + 2ab + 2bc + 2ca = (a + b + c)2 = P(1)2. Equality takes place if and only if either x = 1 or ab = bc = ca = 0, which in view of the condition a > 0 means that b = c = 0. Consequently, for any positive real x we have P(x)P 1 x ≥(P(1))2 with equality if and only if either x = 1 or b = c = 0. 186 Solutions to Exercises and Problems Second solution. Using the Cauchy-Schwarz inequality we get P(x)P 1 x = (ax2 + bx + c) a 1 x2 + b 1 x + c = (√ax)2 + ( √ bx)2 + (√c)2 ⎛ ⎝ √a x 2 + √ b √x 2 + (√c)2 ⎞ ⎠ ≥ √ax √a x + √ bx √ b √x + √c√c 2 = (a + b + c)2 = (P(1))2. Solution 3.74. a2(b + c) + b2(c + a) + c2(a + b) (a + b)(b + c)(c + a) ≥3 4 ⇔ a2b + a2c + b2c + b2a + c2a + c2b 2abc + a2b + a2c + b2c + b2a + c2a + c2b ≥3 4 ⇔ a2b + a2c + b2c + b2a + c2a + c2b −6abc ≥0 ⇔ [2, 1, 0] ≥[1, 1, 1]. The last inequality follows after using Muirhead’s theorem. Second solution. Use inequality (1.11) and the Cauchy-Schwarz inequality. Solution 3.75. Applying the AM-GM inequality to each denominator, one obtains 1 1 + 2ab + 1 1 + 2bc + 1 1 + 2ca ≥ 1 1 + a2 + b2 + 1 1 + b2 + c2 + 1 1 + c2 + a2 . Now, using inequality (1.11) leads us to 1 1 + a2 + b2 + 1 1 + b2 + c2 + 1 1 + c2 + a2 ≥ (1 + 1 + 1)2 3 + 2(a2 + b2 + c2) = 9 3 + 2 · 3 = 1. Solution 3.76. The inequality is equivalent to each of the following ones: x4 + y4 + z4 + 3(x + y + z) ≥−(x3z + x3y + y3x + y3z + z3y + z3x), x3(x + y + z) + y3(x + y + z) + z3(x + y + z) + 3(x + y + z) ≥0, (x + y + z)(x3 + y3 + z3 −3xyz) ≥0. Identity (1.9) shows us that the last inequality is equivalent to 1 2(x + y + z)2((x −y)2 + (y −z)2 + (z −x)2) ≥0. 4.3 Solutions to the problems in Chapter 3 187 Solution 3.77. Let O and I be the circumcenter and the incenter of the acute triangle ABC, respectively. The points O, M, X are collinear and OCX and OMC are similar right triangles. Hence we have OC OX = OM OC . Since OC = R = OA, we have OA OM = OX OA . Hence OAM and OXA are similar, so we have AM AX = OM R . It now suﬃces to show that OM ≤r. Let us compare the angles ∠OBM and ∠IBM. Since ABC is an acute triangle, O and I lie inside ABC. Now we have ∠OBM = π 2 −∠A = 1 2(∠A+∠B+∠C)−∠A = 1 2(∠B+∠C−∠A) ≤∠B 2 = ∠IBM. Similarly, we have ∠OCM ≤∠ICM. Thus the point O lies inside IBC, so we get OM ≤r. Solution 3.78. Setting a = x2, b = y2, c = z2, the inequality is equivalent to x6 + y6 + z6 ≥x4yz + y4zx + z4xy. This follows from Muirhead’s theorem since [6, 0, 0] ≥[4, 1, 1]. Solution 3.79. Use the Cauchy-Schwarz inequality to see that √xy + z = √x√y + √z√z ≤√x + z√y + z = xy + z(x + y + z) = √xy + z. Similarly, √yz + x ≤ √yz + x and √zx + y ≤√zx + y. Therefore, √xy + z + √yz + x + √zx + y ≥√xy + √yz + √zx + x + y + z. Solution 3.80. Using Example 1.4.11, we have a3 + b3 + c3 ≥(a + b + c)(a2 + b2 + c2) 3 . Now, a3 + b3 + c3 ≥a + b + c 3 (a2 + b2 + c2) ≥ 3 √ abc(ab + bc + ca) ≥ab + bc + ca, where we have used the AM-GM and the Cauchy-Schwarz inequalities. Solution 3.81. Using Example 1.4.11, we get (a+b+c)(a2+b2+c2) ≤3(a3+b3+c3), but by hypothesis a2 + b2 + c2 ≥3(a3 + b3 + c3), hence a + b + c ≤1. On the other hand, 4(ab + bc + ca) −1 ≥a2 + b2 + c2 ≥ab + bc + ca, therefore 3(ab + bc + ca) ≥1. As 1 ≤3(ab + bc + ca) ≤(a + b + c)2 ≤1, we obtain a+b+c = 1. Consequently, a+b+c = 1 and 3(ab+bc+ca) = (a+b+c)2, which implies a = b = c = 1 3. 188 Solutions to Exercises and Problems Solution 3.82. The Cauchy-Schwarz inequality yields (\|a\| + \|b\| + \|c\|)2 ≤3(a2 + b2 + c2) = 9. Hence \|a\| + \|b\| + \|c\| ≤3. From the AM-GM inequality it follows that a2 + b2 + c2 ≥3 3 (abc)2 or \|abc\| ≤1, which implies −abc ≤1. The requested inequality is then obtained by summation. Solution 3.83. Notice that OA1 AA1 = (OBC) (ABC) = OB · OC · BC 4R1 · 4R AB · AC · BC . Now, we have to prove that OB · OC · BC + OA · OB · AB + OA · OC · AC ≥AB · AC · BC. We consider the complex coordinates O(0), A(a), B(b), C(c) and obtain \|b\| · \|c\| · \|b −c\| + \|a\| · \|b\| · \|a −b\| + \|a\| · \|c\| · \|c −a\| ≥\|a −b\| · \|b −c\| · \|c −a\|. That is, \|b2c −c2b\| + \|a2b −b2a\| + \|c2a −a2c\| ≥\|ab2 + bc2 + ca2 −a2b −b2c −c2a\|, which is obvious by the triangle inequality. Solution 3.84. Let S = {i1, i1 + 1, . . . , j1, i2, i2 + 1, . . . , j2, . . . , ip, . . . , jp} be the ordering of S, where jk < ik+1 for k = 1, 2, . . ., p−1. Take Sp = a1 +a2 +· · ·+ap, S0 = 0. Then i∈S ai = Sjp −Sip−1 + Sjp−1 −Sip−1−1 + · · · + Sj1 −Si1−1 and 1≤i≤j≤n (ai + · · · + aj)2 = 0≤i≤j≤n (Si −Sj)2. It suﬃces to prove an inequality with the following form: (x1 −x2 + · · · + (−1)p+1xp)2 ≤ 1≤i<j≤p (xj −xi)2 + p i=1 x2 i , (4.9) because this means neglecting the same non-negative terms on the right-hand side of the inequality. Thus inequality (4.9) reduces to 4 1≤i≤j≤p j−i even xixj ≤(p −1) p i=1 x2 i . 4.3 Solutions to the problems in Chapter 3 189 This can be obtained adding together inequalities with the form 4xixj ≤2(x2 i +x2 j), i < j, j −i =even (for odd i, xi appears in such inequality p−1 2 times, and for even i, xi appears in such inequality p 2 −1 times). Solution 3.85. Let x = a+b+c, y = ab+bc+ca, z = abc. Then a2+b2+c2 = x2−2y, a2b2+b2c2+c2a2 = y2−2xz, a2b2c2 = z2, and the inequality to be proved becomes z2 + 2(y2 −2xz) + 4(x2 −2y) + 8 ≥9y or z2 + 2y2 −4xz + 4x2 −17y + 8 ≥0. Now, from a2 + b2 + c2 ≥ab + bc + ca = y we obtain x2 = a2 + b2 + c2 + 2y ≥3y. Also, a2b2 + b2c2 + c2a2 = (ab)2 + (bc)2 + (ca)2 ≥ab · ac + bc · ab + ac · bc = (a + b + c)abc = xz, and thus y2 = a2b2 + b2c2 + c2a2 + 2xz ≥3xz. Hence, z2 + 2y2 −4xz + 4x2 −17y + 8 = z −x 3 2 + 8 9(y −3)2 + 10 9 (y2 −3xz) + 35 9 (x2 −3y) ≥0, as required. Second solution. Expanding the left-hand side of the inequality we obtain the equivalent inequality (abc)2 + 2(a2b2 + b2c2 + c2a2) + 4(a2 + b2 + c2) + 8 ≥9(ab + bc + ca). Since 3(a2+b2+c2) ≥3(ab+bc+ca) and 2(a2b2+b2c2+c2a2)+6 ≥4(ab+bc+ca) (for instance, 2a2b2 + 2 ≥4 √ a2b2 = 4ab), it is enough to prove that (abc)2 + a2 + b2 + c2 + 2 ≥2(ab + bc + ca). Part (i) of Exercise 1.90 tells us that it is enough to prove that (abc)2 + 2 ≥ 3 3 √ a2b2c2, but this follows from the AM-GM inequality. Solution 3.86. Let us write 3 3 √ 3 3 1 abc + 6(a + b + c) = 3 3 √ 3 3 1 + 6a2bc + 6b2ac + 6c2ab abc = 3 3 √ 3 3 1 + 3ab(ac + bc) + 3bc(ba + ca) + 3ca(ab + bc) abc , and consider the condition ab + bc + ca = 1 to obtain 3 3 √ 3 3 1 + 3ab −3(ab)2 + 3bc −3(bc)2 + 3ca −3(ca)2 abc = 3 3 √ 3 3 4 −3((ab)2 + (bc)2 + (ca)2) abc . 190 Solutions to Exercises and Problems It is easy to see that 3((ab)2 + (bc)2 + (ac)2) ≥(ab + bc + ac)2 (use the Cauchy-Schwarz inequality). Then, it is enough to prove that 3 3 √ 3 3 3 abc ≤ 1 abc, which is equivalent to (abc)2 ≤ 1 27. But this last inequality follows from the AM-GM inequality, (abc)2 = (ab)(bc)(ca) ≤ ab + bc + ca 3 3 = 1 27. The equality holds if and only if a = b = c = 1 √ 3. Solution 3.87. Using symmetry, it suﬃces to prove that t1 < t2 + t3. We have n i=1 ti n i=1 1 ti = n + 1≤i 0 and using the hypothesis, we arrive at n2 + 1 > n i=1 ti n i=1 1 ti ≥n + 2 t1 √t2t3 + 2 √t2t3 t1 + 2 'n2 −n 2 −2 ( = 2a + 2 a + n2 −4. Hence 2a + 2 a −5 < 0, which implies 1/2 < a = t1/√t2t3 < 2. So t1 < 2√t2t3, and one more application of the AM-GM inequality yields t1 < 2√t2t3 ≤t2 + t3, as needed. Solution 3.88. Note that 1 + b −c = a + b + c + b −c = a + 2b ≥0. Then a 3 √ 1 + b −c ≤a 1 + 1 + (1 + b −c) 3 = a + ab −ac 3 . Similarly, b 3 √ 1 + c −a ≤b + bc −ba 3 c 3 √ 1 + a −b ≤c + ca −cb 3 . Adding these three inequalities, we get a 3 √ 1 + b −c + b 3 √ 1 + c −a + c 3 √ 1 + a −b ≤a + b + c = 1. 4.3 Solutions to the problems in Chapter 3 191 Solution 3.89. If any of the numbers is zero or if an odd number of them are negative, then x1x2 · · · x6 ≤0 and the inequality follows. Therefore, it can only be 2 or 4 negative numbers between the numbers in the inequality. Suppose that neither of them are zero and that there are 2 negative numbers (in the other case, change the signs of all numbers). If yi = \|xi\|, then it is clear that y2 1 + y2 2 + · · · + y2 6 = 6, y1 + y2 = y3 + · · · + y6 and that x1x2 · · · x6 = y1y2 · · · y6. From the AM-GM inequality we get y1y2 ≤ y1 + y2 2 2 = A2. Also, the AM-GM inequality yields y3y4y5y6 ≤ y3 + y4 + y5 + y6 4 4 = y1 + y2 4 4 = 1 24 A4. Therefore, y1y2 · · · y6 ≤ 1 24 A6. On the other hand, the Cauchy-Schwarz inequality implies that 2(y2 1 + y2 2) ≥(y1 + y2)2 = 4A2 4(y2 3 + y2 4 + y2 5 + y2 6) ≥(y3 + y4 + y5 + y6)2 = 4A2. Thus, 6 = y2 1 + y2 2 + · · · + y2 6 ≥2A2 + A2 = 3A2 and then y1y2 · · · y6 ≤ 1 24 A6 ≤ 23 24 = 1 2. Solution 3.90. Use the Cauchy-Schwarz inequality with (1, 1, 1) and ( a b , b c, c a) to obtain (12 + 12 + 12) a2 b2 + b2 c2 + c2 a2 ≥ a b + b c + c a 2 . The AM-GM inequality leads us to a b + b c + c a ≥3 3 abc bca = 3, then a2 b2 + b2 c2 + c2 a2 ≥ a b + b c + c a . Similarly, a c + b a + c b ≥3 3 abc bca = 3. Therefore, a2 b2 + b2 c2 + c2 a2 + a c + b a + c b ≥3 + a b + b c + c a. Adding a c + b a + c b to both sides yields the result. 192 Solutions to Exercises and Problems Solution 3.91. Note that a2 + 2 2 = (a2 −a + 1) + (a + 1) 2 ≥ (a2 −a + 1)(a + 1) = 1 + a3. After substituting in the given inequality, we need to prove that a2 (a2 + 2)(b2 + 2) + b2 (b2 + 2)(c2 + 2) + c2 (c2 + 2)(a2 + 2) ≥1 3. Set x = a2, y = b2, z = c2, then xyz = 64 and x (x + 2)(y + 2) + y (y + 2)(z + 2) + z (z + 2)(x + 2) ≥1 3 if and only if 3[x(z + 2) + y(x + 2) + z(y + 2)] ≥(x + 2)(y + 2)(z + 2). Now, 3(xy + yz + zx) + 6(x + y + z) ≥xyz + 2(xy + yz + zx) + 4(x + y + z) + 8 if and only if xy + yz + zx + 2(x + y + z) ≥xyz + 8 = 72, but using the AM-GM inequality leads to x + y + z ≥12 and xy + yz + zx ≥48, which ﬁnishes the proof. Solution 3.92. Observe that x5 −x2 x5 + y2 + z2 − x5 −x2 x3(x2 + y2 + z2) = x2(y2 + z2)(x3 −1)2 x3(x5 + y2 + z2)(x2 + y2 + z2) ≥0. Then x5 −x2 x5 + y2 + z2 ≥ x5 −x2 x3(x2 + y2 + z2) = 1 x2 + y2 + z2 x2 −1 x ≥ 1 x2 + y2 + z2 (x2 −yz) ≥0. The second inequality follows from the fact that xyz ≥1, that is, 1 x ≤yz. The last inequality follows from (1.8). Second solution. First, note that x5 −x2 x5 + y2 + z2 = x5 + y2 + z2 −(x2 + y2 + z2) x5 + y2 + z2 = 1 −x2 + y2 + z2 x5 + y2 + z2 . Now we need to prove that 1 x5 + y2 + z2 + 1 x5 + z2 + x2 + 1 x5 + x2 + y2 ≤ 3 x2 + y2 + z2 . 4.3 Solutions to the problems in Chapter 3 193 Using the Cauchy-Schwarz inequality we get (x2 + y2 + z2)2 ≤(x2 · x3 + y2 + z2)(x2 · 1 x3 + y2 + z2) and since xyz ≥1, then x2 · 1 x3 = 1 x ≤yz, and we have that (x2 + y2 + z2)2 ≤(x5 + y2 + z2)(yz + y2 + z2) therefore 1 x5 + y2 + z2 ≤ yz + y2 + z2 (x2 + y2 + z2)2 ≤ y2+z2 2 + y2 + z2 (x2 + y2 + z2)2 = 3 x2 + y2 + z2 . Solution 3.93. Notice that (1 + abc) 1 a(b + 1) + 1 b(c + 1) + 1 c(a + 1) + 3 = 1 + abc + ab + a a(b + 1) + 1 + abc + bc + b b(c + 1) + 1 + abc + ca + c c(a + 1) = 1 + a a(b + 1) + b(c + 1) (b + 1) + 1 + b b(c + 1) + c(a + 1) (c + 1) + 1 + c c(a + 1) + a(b + 1) (a + 1) ≥6. The last inequality follows after using the AM-GM inequality for six numbers. Solution 3.94. Let R be the circumradius of the triangle ABC. Since ∠BOC = 2∠A, ∠COA = 2∠B and ∠AOB = 2∠C, we have that (ABC) = (BOC) + (COA) + (AOB) = R2 2 (sin 2A + sin 2B + sin 2C) ≤R2 2 3 sin 2A + 2B + 2C 3 = R2 2 3 sin 2π 3 = 3 √ 3R2 4 . The inequality follows since the function sin x is concave in [0, π]. On the other hand, since BOC is isosceles, the perpendicular bisector OA′ of BC is also the internal bisector of the angle ∠BOC, so that ∠BOA′ = ∠COA′ = ∠A; similarly ∠COB′ = ∠AOB′ = ∠B and ∠AOC′ = ∠BOC′ = ∠C. In the triangle B′OC′ the altitude on the side B′C′ is R 2 and B′C′ = R 2 (tan B + tan C). 194 Solutions to Exercises and Problems Therefore, the area of the triangle B′OC′ is (B′OC′) = R2 8 (tan B + tan C). Simi-larly, (C′OA′) = R2 8 (tan C + tan A) and (A′OB′) = R2 8 (tan A + tan B). Then, (A′B′C′) = (B′OC′) + (C′OA′) + (A′OB′) = R2 4 (tan A + tan B + tan C) ≥R2 4 3 tan A + B + C 3 = R2 4 3 tan π 3 = 3 √ 3R2 4 . The inequality follows since the function tan x is convex in [0, π 2 ]. Hence, (A′B′C′) ≥3 √ 3R2 4 ≥(ABC). Solution 3.95. First, note that a2+bc ≥2 √ a2bc = 2 √ ab√ca and similarly b2+ca ≥ 2 √ bc √ ab, c2 + ab ≥2√ca √ bc; then it follows that 1 a2 + bc + 1 b2 + ca + 1 c2 + ab ≤1 2 1 √ ab√ca + 1 √ bc √ ab + 1 √ca √ bc . Now, using the Cauchy-Schwarz inequality in the following way 1 √ ab√ca + 1 √ bc √ ab + 1 √ca √ bc 2 ≤ 1 ab + 1 bc + 1 ca 1 ca + 1 ab + 1 bc , the result follows. Solution 3.96. From the Cauchy-Schwarz inequality we get i̸=j ai aj i̸=j aiaj ≥ ⎛ ⎝ i̸=j ai ⎞ ⎠ 2 = (n −1) n i=1 ai 2 = (n −1)2A2. On the other hand, i̸=j aiaj = n i=1 ai 2 − n i=1 a2 i = A2 −A. Solution 3.97. Without loss of generality, take a1 ≤· · · ≤an. Let dk = ak+1 −ak for k = 1, . . . , n. Then d = d1 + · · · + dn−1. For i < j we have that \|ai −aj\| = aj −ai = di + · · · + dj−1. Then, 4.3 Solutions to the problems in Chapter 3 195 s = i 1. Solution 3.104. It will be enough to consider the case x ≤y ≤z. Then x = y −a, z = y + b with a, b ≥0. On the one hand, we have xz = 1 −xy −yz = 1 −(y −a)y −y(y + b) = 1 −2y2 + ay −by and on the other, xz = (y −a)(y + b) = y2 −ay + by −ab. Adding both identities, we get 2xz = 1 −y2 −ab, so that 2xz −1 = −y2 −ab ≤0. If 2xz = 1, then y = 0 and xz = 1, a contradiction, therefore xz < 1 2. The numbers x = y = 1 n and z = 1 2(n−1 n) satisfy x ≤y ≤z and xy+yz+zx = 1. However, xz = 1 2n(n −1 n) = 1 2 − 1 2n2 can be as close as we wish to 1 2, therefore, the value 1 2 cannot be improved. Solution 3.105. Suppose that a = [x] and that r = {x}. Then, the inequality is equivalent to a + 2r a − a a + 2r + 2a + r r − r 2a + r > 9 2. This inequality reduces to 2 r a + a r − a a + 2r + r 2a + r > 5 2. But since r a + a r ≥2, it is enough to prove that a a + 2r + r 2a + r < 3 2. But a + 2r ≥a + r and 2a + r ≥a + r; moreover, the two equalities cannot hold at the same time (otherwise a = r = 0), therefore a a + 2r + r 2a + r < a a + r + r a + r = 1 < 3 2. 198 Solutions to Exercises and Problems Solution 3.106. Inequality (1.11) shows that a + b + c ≥1 a + 1 b + 1 c ≥ 32 a + b + c, so that a+b+c 3 ≥ 3 a+b+c. Thus, it is enough to prove that a + b + c ≥ 3 abc. Since (x + y + z)2 ≥3(xy + yz + zx), we have (a + b + c)2 ≥ 1 a + 1 b + 1 c 2 ≥3 1 ab + 1 bc + 1 ca = 3 abc(a + b + c), and from here it is easy to conclude the proof. Solution 3.107. By means of the Cauchy-Schwarz inequality we get (a + b + 1)(a + b + c2) ≥(a + b + c)2. Then a + b + c2 (a + b + c)2 + a2 + b + c (a + b + c)2 + a + b2 + c (a + b + c)2 ≥ 1 a + b + 1 + 1 b + c + 1 + 1 c + a + 1 ≥1. Therefore, 2(a + b + c) + (a2 + b2 + c2) ≥(a + b + c)2 = a2 + b2 + c2 + 2(ab + bc + ca), and the result follows. Solution 3.108. For an interior point P of ABC, consider the point Q on the perpendicular bisector of BC satisfying AQ = AP. Let S be the intersection of BP with the tangent to the circle at Q. Then, SP + PC ≥SC, therefore BP + PC = BS + SP + PC ≥BS + SC. On the other hand, BS + SC ≥BQ + QC, then BP + PC is minimum if P = Q. Let T be the midpoint of MN. Since the triangle AMQ is isosceles and MT is one of its altitudes, then MT = ZQ where Z is the foot of the altitude of Q over AB. Then MN + BQ + QC = 2(MT + QC) = 2(ZQ + QC) is minimum when Z, Q, C are collinear and this means CZ is the altitude. By symmetry, BQ should be also an altitude and then P is the orthocenter. Solution 3.109. Let H be the orthocenter of the triangle MNP, and let A′, B′, C′ be the projections of H on BC, CA, AB, respectively. Since the triangle MNP is acute, H belongs to the interior of the triangle MNP; hence, it belongs to the interior of the triangle ABC too, and therefore x ≤HA′ + HB′ + HC′ ≤HM + HN + HP ≤2X. 4.3 Solutions to the problems in Chapter 3 199 The second inequality is evident, the other two will be presented as the following two lemmas. Lemma 1. If H is an interior point or belongs to the sides of a triangle ABC, and if A′, B′, C′ are its projections on BC, CA, AB, respectively, then x ≤ HA′ + HB′ + HC′, where x is the length of the shortest altitude of ABC. Proof. HA′ + HB′ + HC′ x ≥HA′ ha + HB′ hb + HC′ hc = (BHC) (ABC) + (CHA) (ABC) + (AHB) (ABC) = 1. □ Lemma 2. If MNP is an acute triangle and H is its orthocenter, then HM +HN + HP ≤2X, where X is the length of the largest altitude of the triangle MNP. Proof. Suppose that ∠M ≤∠N ≤∠P, then NP ≤PM ≤MN and so it happens that X is equal to the altitude MM ′. We need to prove that HM + HN + HP ≤ 2MM ′ = 2(HM + HM ′) or, equivalently, that HN + HP ≤HM + 2HM ′. □ Let H′ be the symmetric point of H with respect to NP; since MNH′P is a cyclic quadrilateral, Ptolemy’s theorem tells us that H′M · NP = H′N · MP + H′P · MN ≥H′N · NP + H′P · NP, and then we get H′N + H′P ≤H′M = HM + 2HM ′. Solution 3.110. Without loss of generality, we can suppose that x ≤y ≤z. Then x + y ≤z + x ≤y + z, xy ≤zx ≤yz, 2z2(x + y) ≥2y2(z + x) ≥2x2(y + z), 1 √ 2z2(x+y) ≤ 1 √ 2y2(z+x) ≤ 1 √ 2x2(y+z). If we resort to the rearrangement inequality and apply it twice, we have 2yz 2x2(y + z) ≥ xy + zx 2x2(y + z) . Now, adding 2x2 √ 2x2(y+z) to both sides of the last inequality, we obtain 2x2 + 2yz 2x2(y + z) ≥ 2x2 + xy + zx 2x2(y + z) = 2x2 + x(y + z) 2x2(y + z) ≥ 2 2x3(y + z) 2x2(y + z) = 2(√x + √y + √z) = 2. 200 Solutions to Exercises and Problems Second solution. First, note that x2 + yz 2x2(y + z) = x2 −x(y + z) + yz 2x2(y + z) + x(y + z) 2x2(y + z) = (x −y)(x −z) 2x2(y + z) + y + z 2 ≥(x −y)(x −z) 2x2(y + z) + √y + √z 2 . Similarly for the other two elements of the sum; then x2 + yz 2x2(y + z) ≥ (x −y)(x −z) 2x2(y + z) + √x + √y + √z. Then, it is enough to prove that (x −y)(x −z) 2x2(y + z) + (y −z)(y −x) 2y2(z + x) + (z −x)(z −y) 2z2(x + y) ≥0. Without loss of generality, suppose that x ≥y ≥z. Then (x−y)(x−z) √ 2x2(y+z) ≥0, and (y −z)(y −x) 2y2(z + x) + (z −x)(z −y) 2z2(x + y) = (x −z)(y −z) 2z2(x + y) −(y −z)(x −y) 2y2(z + x) ≥(x −y)(y −z) 2z2(x + y) −(y −z)(x −y) 2y2(z + x) = (y −z)(x −y) 1 2z2(x + y) − 1 2y2(z + x) ≥0. The last inequality is a consequence of having y2(z+x) = y2z+y2x ≥yz2+z2x = z2(x + y). Solution 3.111. Inequality (1.11) leads to a2 2 + b + c2 + b2 2 + c + a2 + c2 2 + a + b2 ≥ (a + b + c)2 6 + a + b + c + a2 + b2 + c2 . Then, we need to prove that 6+a+b+c+a2+b2+c2 ≤12, but since a2+b2+c2 = 3, it is enough to prove that a + b + c ≤3. But we also have (a + b + c)2 = a2 + b2 + c2 + 2(ab + bc + ca) ≤3(a2 + b2 + c2) = 9. The equality holds if and only if a = b = c = 1. Solution 3.112. First, note that 1 −a −bc a + bc = 2bc 1 −b −c + bc = 2bc (1 −b)(1 −c) = 2bc (c + a)(a + b). 4.3 Solutions to the problems in Chapter 3 201 Then, the inequality is equivalent to 2bc (c + a)(a + b) + 2ca (a + b)(b + c) + 2ab (b + c)(c + a) ≥3 2. This last inequality can be simpliﬁed to 4 [bc(b + c) + ca(c + a) + ab(a + b)] ≥3(a + b)(b + c)(c + a), which in turn is equivalent to the inequality ab + bc + ca ≥9abc. But this inequality follows from (a + b + c)( 1 a + 1 b + 1 c) ≥9. Solution 3.113. Notice that (x√y + y√z + z√x)2 = x2y + y2z + z2x + 2(xy√yz + yz√zx + zx√xy). The AM-GM inequality implies that xy√yz = √xyz xy2 ≤xyz + xy2 2 , then (x√y + y√z + z√x)2 ≤x2y + y2z + z2x + xy2 + yz2 + zx2 + 3xyz. Since (x + y)(y + z)(z + x) = x2y + y2z + z2x + xy2 + yz2 + zx2 + 2xyz, we obtain (x√y + y√z + z√x)2 ≤(x + y)(y + z)(z + x) + xyz ≤(x + y)(y + z)(z + x) + 1 8(x + y)(y + z)(z + x) = 9 8(x + y)(y + z)(z + x). Therefore K2 ≥9 8, and then K ≥ 3 2 √ 2. When x = y = z, the equality holds with K = 3 2 √ 2, hence this is the minimum value. Second solution. Apply the Cauchy-Schwarz inequality in the following way: x√y + y√z + z√x = √x√xy + √y√yz + √z√zx ≤ (x + y + z)(xy + yz + zx). After that, use the AM-GM inequality several times to produce (x + y + z) 3 (xy + yz + zx) 3 ≤ 3 √xyz 3 x2y2z2 = xyz ≤(x + y) 2 (y + z) 2 (z + x) 2 . 202 Solutions to Exercises and Problems Solution 3.114. The left-hand side of the inequality can be written as a2b2cd+ab2c2d+abc2d2+a2bcd2+a2bc2d+ab2cd2 = abcd(ab+bc+cd+ac+ad+bd). The AM-GM inequality implies that a2b2c2d2 ≤( a2+b2+c2+d2 4 )4 = 1 4 4, hence abcd ≤ 1 16. To see that the factor (ab + bc + cd + ac + ad + bd) is less than 3 2 we can proceed in two forms. The ﬁrst way is to apply the Cauchy-Schwarz inequality to obtain (ab + bc + cd + ac + ad + bd + ba + cb + dc + ca + da + db) ≤(a2 + b2 + c2 + d2 + a2 + b2 + c2 + d2 + a2 + b2 + c2 + d2) = 3. The second way consists in applying the AM-GM inequality as follows: (ab + bc + cd + ac + ad + bd) ≤a2 + b2 2 + b2 + c2 2 + c2 + d2 2 +a2 + c2 2 + a2 + d2 2 + b2 + d2 2 = 3 2. Solution 3.115. (a) After some algebraic manipulation and some simpliﬁcations we obtain (1 + x + y)2 + (1 + y + z)2 + (1 + z + x)2 = 3 + 4(x + y + z) + 2(xy + yz + zx) + 2(x2 + y2 + z2). Now, the AM-GM inequality implies that (x + y + z) ≥3 3 √xyz ≥3, (xy + yz + zx) ≥3 3 x2y2z2 ≥3, (x2 + y2 + z2) ≥3 3 x2y2z2 ≥3. Then, (1 + x + y)2 + (1 + y + z)2 + (1 + z + x)2 ≥3 + 4 · 3 + 2 · 3 + 2 · 3 = 27. The equality holds when x = y = z = 1. (b) Again, after simpliﬁcation, the inequality is equivalent to 3 + 4(x + y + z) + 2(xy + yz + zx) + 2(x2 + y2 + z2) ≤3(x2 + y2 + z2) + 6(xy + yz + zx) and also to 3 + 4u ≤u2 + 2v, where u = x + y + z ≥3 and v = xy + yz + zx ≥3. But u ≥3 implies that (u −2)2 ≥1, then (u −2)2 + 2v ≥1 + 6 = 7. The equality holds when u = 3 and v = 3, that is, when x = y = z = 1. Solution 3.116. Notice that 1 1 + a2(b + c) = 1 1 + a(ab + ac) = 1 1 + a(3 −bc) = 1 3a + 1 −abc. 4.3 Solutions to the problems in Chapter 3 203 The AM-GM inequality implies that 1 = ab+bc+ca 3 ≥ 3 √ a2b2c2, then abc ≤1. Thus 1 1 + a2(b + c) = 1 3a + 1 −abc ≤1 3a. Similarly, 1 1+b2(c+a) ≤ 1 3b and 1 1+c2(a+b) ≤ 1 3c. Therefore, 1 1 + a2(b + c) + 1 1 + b2(c + a) + 1 1 + c2(a + b) ≤1 3a + 1 3b + 1 3c = bc + ca + ab 3abc = 1 abc. Solution 3.117. The inequality is equivalent to (a + b + c) 1 a + b + 1 b + c + 1 c + a ≥k + (a + b + c)k = (a + b + c + 1)k. On the other hand, using the condition a + b + c = ab + bc + ca, we have 1 a + b + 1 b + c + 1 c + a = a2 + b2 + c2 + 3(ab + bc + ca) (a + b)(b + c)(c + a) = a2 + b2 + c2 + 2(ab + bc + ca) + (ab + bc + ca) (a + b)(b + c)(c + a) = (a + b + c)(a + b + c + 1) (a + b + c)2 −abc . Hence (a + b + c) (a + b + c + 1) 1 a + b + 1 b + c + 1 c + a = (a + b + c)2 (a + b + c)2 −abc ≥1, and since the equality holds if and only if abc = 0, we can conclude that k = 1 is the maximum value. Solution 3.118. Multiplying both sides of the inequality by the factor (a + b + c), we get the equivalent inequality 9(a + b + c)(a2 + b2 + c2) + 27abc ≥4(a + b + c)3, which in turn is equivalent to the inequality 5(a3 + b3 + c3) + 3abc ≥3(ab(a + b) + ac(a + c) + bc(b + c)). By the Sch¨ ur inequality with n = 1, Exercise 1.83, it follows that a3 + b3 + c3 + 3abc ≥ab(a + b) + bc(b + c) + ca(c + a), and the Muirhead’s inequality tells us that 2[3, 0, 0] ≥2[2, 1, 0], which is equivalent to 4(a3 + b3 + c3) ≥2(ab(a + b) + ac(a + c) + bc(b + c)). Adding these last inequalities, we get the result. 204 Solutions to Exercises and Problems Solution 3.119. Lemma. If a, b > 0, then 1 (a−b)2 + 1 a2 + 1 b2 ≥ 4 ab. Proof. In order to prove the lemma notice that 1 (a−b)2 + 1 a2 + 1 b2 −4 ab = (a2+b2−3ab)2 a2b2(a−b)2 . □ Without loss of generality, z = min{x, y, z}; now apply the lemma with a = (x −z) and b = (y −z), to obtain 1 (x −y)2 + 1 (y −z)2 + 1 (z −x)2 ≥ 4 (x −z)(y −z). Now, it is left to prove that xy + yz + zx ≥(x −z)(y −z); but this is equivalent to 2z(y + x) ≥z2, which is evident. Solution 3.120. In the case of part (i), there are several ways to prove it. First form. We can prove that x2 (x −1)2 + y2 (y −1)2 + z2 (z −1)2 −1 = (yz + zx + xy −3)2 (x −1)2(y −1)2(z −1)2 . Second form. With the substitution a = x x−1, b = y y−1, c = z z−1, the inequality is equivalent to a2 + b2 + c2 ≥1, and the condition xyz = 1 is equivalent to abc = (a −1)(b −1)(c −1) or (ab + bc + ca) + 1 = a + b + c. With the previous identities we can obtain a2 + b2 + c2 = (a + b + c)2 −2(ab + bc + ca) = (a + b + c)2 −2(a + b + c −1) = (a + b + c −1)2 + 1, therefore a2 + b2 + c2 = (a + b + c −1)2 + 1. Part (ii) can be proved depending on how we prove part (i). For instance, if we used the second form, the equality holds when a2+b2+c2 = 1 and a+b+c = 1. (In the ﬁrst form, the equality holds when xyz = 1 and xy + yz + zx = 3). From the equations we can cancel out one variable, for instance c (and since c = 1−a−b, if we ﬁnd that a and b are rational numbers, then c will be a rational number too), to obtain a2 + b2 + ab −a −b = 0, an identity that we can think of as a quadratic equation in the variable b with roots b = 1−a±√ (1−a)(1+3a) 2 , which will be rational numbers if (1 −a) and (1 + 3a) are squares of rational numbers. If a = k m, then m −k and m + 3k are squares of integers, for instance, if m = (k −1)2 + k, then m −k = (k −1)2 and m + 3k = (k + 1)2. Thus, the rational numbers a = k m, b = m−k+k2−1 2m and c = 1−a−b, when k varies in the integer numbers, are rational numbers where the equality holds. There are some exceptions, that is, when k = 0, 1, since the values a = 0 or 1 are not allowed. Notation We use the following standard notation: N the positive integers (natural numbers) R the real numbers R+ the positive real numbers ⇔ iﬀ, if and only if ⇒ implies a ∈A the element a belongs to the set A A ⊂B A is a subset of B \|x\| the absolute value of the real number x {x} the fractional part of the real number x [x] the integer part of the real number x [a, b] the set of real numbers x such that a ≤x ≤b (a, b) the set of real numbers x such that a < x < b f : [a, b] →R the function f deﬁned in [a, b] with values in R f′(x) the derivative of the function f(x) f ′′(x) the second derivative of the function f(x) det A the determinant of the matrix A n i=1 ai the sum a1 + a2 + · · · + an 0n i=1 ai the product a1 · a2 · · · an 0 i̸=j ai the product of all a1, a2, . . . , an except aj max{a, b, . . .} the maximum value between a, b, . . . min{a, b, . . .} the minimum value between a, b, . . . √x the square root of the positive real number x n √x the n-th root of the real number x exp x = ex the exponential function cyclic f(a, b, . . . ) represents the sum of the function f evaluated in all cyclic permutations of the variables a, b, . . . 206 Solutions to Exercises and Problems We use the following notation for the section of Muirhead’s theorem: ! F(x1, . . . , xn) the sum of the n! terms obtained from evaluating F in all possible permutations of (x1, . . . , xn) (b) ≺(a) (b) is majorized by (a) [b] ≤[a] 1 n! ! xb1 1 xb2 2 · · · xbn n ≤1 n! ! xa1 1 xa2 2 · · · xan n . We use the following geometric notation: A, B, C the vertices of the triangle ABC a, b, c the lengths of the sides of the triangle ABC A′, B′, C′ the midpoints of the sides BC, CA and AB ∠ABC the angle ABC ∠A the angle in the vertex A or the measure of the angle A (ABC) the area of the triangle ABC (ABCD...) the area of the polygon ABCD... ma, mb, mc the lengths of the medians of the triangle ABC ha, hb, hc the lengths of the altitudes of the triangle ABC la, lb, lc the lengths of the internal bisectors of the triangle ABC s the semiperimeter of the triangle ABC r the inradius of the triangle ABC, the radius of the incircle R the circumradius of the triangle ABC, the radius of the circumcircle I, O, H, G the incenter, circumcenter, orthocenter and centroid of the triangle ABC Ia, Ib, Ic the centers of the excircles of the triangle ABC. We use the following notation for reference of problems: IMO International Mathematical Olympiad APMO Asian Paciﬁc Mathematical Olympiad (country, year) problem corresponding to the mathematical olympiad celebrated in that country, in that year, in some stage. Bibliography Altshiller, N., College Geometry: An Introduction to Modern Geometry of the Triangle and the Circle. Barnes and Noble, 1962. Andreescu, T., Feng, Z., Problems and Solutions from Around the World. Mathematical Olympiads 1999-2000. MAA, 2002. Andreescu, T., Feng, Z., Lee, G., Problems and Solutions from Around the World. Mathematical Olympiads 2000-2001. MAA, 2003. Andreescu, T., Enescu, B., Mathematical Olympiad Treasures. Birkh¨ auser, 2004. Barbeau, E.J., Shawyer, B.L.R., Inequalities. A Taste of Mathematics, vol. 4, 2000. Bulajich, R., G´ omez Ortega, J.A., Geometr´ ıa. Cuadernos de Olimpiadas de Matem´ aticas, Instituto de Matem´ aticas, UNAM, 2002. Bulajich, R., G´ omez Ortega, J.A., Geometr´ ıa. Ejercicios y Problemas. Cuadernos de Olimpiadas de Matem´ aticas, Instituto de Matem´ aticas, UNAM, 2002. Courant, R., Robbins, H., ¿Qu´ e son las Matem´ aticas? Fondo de Cultura Econ´ omica, 2002. Coxeter, H., Greitzer, S., Geometry Revisited. New Math. Library, MAA, 1967. Dorrie, H., 100 Great Problems of Elementary Mathematics. Dover, 1965. Engel, A.,Problem-Solving Strategies. Springer-Verlag, 1998. Fomin, D., Genkin, S., Itenberg, I., Mathematical Circles. Mathematical World, Vol. 7. American Mathematical Society, 1996. Hardy, G.H., Littlewood, J.E., P` olya, G., Inequalities. Cambridge at the University Press, 1967. Honsberger, R., Episodes in Nineteenth and Twentieth Century Euclidean Geometry. New Math. Library, MAA, 1995. 208 Bibliography Kazarinoﬀ, N., Geometric Inequalities. New Math. Library, MAA. 1961. Larson, L., Problem-Solving Through Problems. Springer-Verlag, 1990. Mitrinovic, D., Elementary Inequalities. NoordhoﬀLtd., Groningen, 1964. Niven, I., Maxima and Minima Without Calculus. The Dolciani Math. Expositions, MAA, 1981. Shariguin, I., Problemas de Geometr´ ıa. Editorial Mir, 1989. Soulami, T., Les Olympiades de Math´ ematiques. Ellipses, 1999. Spivak, M., Calculus. Editorial Benjamin, 1967. Index Absolute value, 2 Concavity Geometric interpretation, 25 Convexity Geometric interpretation, 25 discrepancy, 46 Erd˝ os-Mordell theorem, 81–84, 88 Euler theorem, 66 Fermat point, 90, 92 Function concave, 23 convex, 20 quadratic, 4 Greater than, 1 Inequality arithmetic mean–geometric mean, 9, 47 weighted, 27 Bernoulli, 31 Cauchy-Schwarz, 15, 35 Engel form, 35 Euler, 67 H¨ older, 27 generalized, 32 harmonic mean–geometric mean, 8 helpful, 34 Jensen, 21 Leibniz, 69 Minkowski, 28 Nesbitt, 16, 37, 65 Popoviciu, 32 power mean, 32 Ptolemy, 53 quadratic mean–arithmetic mean, 19, 36 rearrangement, 13 Schur, 31 Tchebyshev, 18 triangle, 3 general form, 3 Young, 27 Leibniz theorem, 68 Mean arithmetic, 7, 9, 19, 31 geometric, 7, 9, 19, 31 harmonic, 8, 19 power, 32 quadratic, 19 Muirhead theorem, 43, 44 Ortic triangle, 95, 98 Pappus theorem, 80 Pedal triangle, 99 210 Index Problem Fagnano, 88, 94 Fermat-Steiner, 88 Heron, 92 with a circle, 93 Pompeiu, 53 Real line, 1 Smaller than, 1 Smaller than or equal to, 2 Solution Fagnano problem Fej´ er L., 96 Schwarz H., 96 Fermat-Steiner problem Hofmann-Gallai, 91 Steiner, 92, 94 Torricelli, 88, 90 Transformation Ravi, 55, 73 Viviani lemma, 88, 90
6418	https://artofproblemsolving.com/wiki/index.php/Factoring_Quadratics?srsltid=AfmBOorcdzGfW3wskxTrGLqLVOql03nHjspzl4lrJYAUZVIRM8MEsFPU	Art of Problem Solving Factoring Quadratics - AoPS Wiki Art of Problem Solving AoPS Online Math texts, online classes, and more for students in grades 5-12. Visit AoPS Online ‚ Books for Grades 5-12Online Courses Beast Academy Engaging math books and online learning for students ages 6-13. Visit Beast Academy ‚ Books for Ages 6-13Beast Academy Online AoPS Academy Small live classes for advanced math and language arts learners in grades 2-12. Visit AoPS Academy ‚ Find a Physical CampusVisit the Virtual Campus Sign In Register online school Class ScheduleRecommendationsOlympiad CoursesFree Sessions books tore AoPS CurriculumBeast AcademyOnline BooksRecommendationsOther Books & GearAll ProductsGift Certificates community ForumsContestsSearchHelp resources math training & toolsAlcumusVideosFor the Win!MATHCOUNTS TrainerAoPS Practice ContestsAoPS WikiLaTeX TeXeRMIT PRIMES/CrowdMathKeep LearningAll Ten contests on aopsPractice Math ContestsUSABO newsAoPS BlogWebinars view all 0 Sign In Register AoPS Wiki ResourcesAops Wiki Factoring Quadratics Page ArticleDiscussionView sourceHistory Toolbox Recent changesRandom pageHelpWhat links hereSpecial pages Search Factoring Quadratics The purpose of factoring a quadratic is to turn the quadratic into a product of 2 binomials. Contents 1 Method 1 1.1 Example 1.2 Limitations 2 Method 2 2.1 Example 2.2 Limitations 3 Also See Method 1 Method 1 starts with factoring the product of the roots. Let the quadratic we are factoring be . When factored, it will be in the form of where and are the roots of the quadratic, and where and . Example Since the coefficient on the term is , we know are quadratic factors in the form of . We know that the factor pairs of 12 are and We can find that only and satisfy our equations and , so the factored form of is . Limitations This method cannot be used to factor quadratics with complex or irrational roots. Method 2 Method 2 starts by using the sum. Let the quadratic we are factoring be . When factored, it will be in the form of where and are the roots of the quadratic, and where and . Example We know that , so we can set and . Then, we get that , giving us that , or . Because we have both and as our roots, it doesn't matter which one is plugged in, giving us that the factored form of is . Limitations None currently known. Also See Quadratic equation Factoring Vieta's Formulas Retrieved from " Categories: Algebra Quadratic equations Art of Problem Solving is an ACS WASC Accredited School aops programs AoPS Online Beast Academy AoPS Academy About About AoPS Our Team Our History Jobs AoPS Blog Site Info Terms Privacy Contact Us follow us Subscribe for news and updates © 2025 AoPS Incorporated © 2025 Art of Problem Solving About Us•Contact Us•Terms•Privacy Copyright © 2025 Art of Problem Solving Something appears to not have loaded correctly. Click to refresh.
6419	https://mathsmartinthomas.wordpress.com/2015/08/18/images-of-loci-of-complex-numbers/	A minimum of blind calculation "All limits, especially national ones, are contrary to the nature of mathematics… Mathematics knows no races… For mathematics the whole cultural world is a single country" – David Hilbert. "Face problems with a minimum of blind calculation, a maximum of seeing thought" – Hermann Minkowski Complex loci: circle-plus ↔ circle-plus Linear transformations in two dimensions are defined by transforming lines into lines. If we look at things more in a polar-coordinates, complex-numbers way, we find another set of transformations which transform shapes of a particular simple sort into shapes of the same sort. Circles are in some ways for polar coordinates what lines are for cartesian coordinates. The new set of transformations is represented in complex numbers by equations like: All transformations like that transform circles into circles – so long as we interpret lines as special “circles” with centre at infinity and infinite radius. In the same sort of way, some linear transformations in two dimensions transform lines into special “lines”, i.e. points. If the locus of z is a circle (or line), then the locus of w=(az+b)/(cz+d) is a circle (or line). A diameter of the locus of z passing through the point z=d/c transforms into a diameter of the locus of w. Proof: If the locus of w+k is a circle (or line), then the locus of w is a circle (or line), just shifted by k. If the locus of mw is a circle (or line), then the locus of w is a circle (or line), just shifted and, if it’s a circle, made smaller or larger. So it’s enough to consider the transformation w=1/z. If that transforms circles (or lines) into circles (or lines), then all transformations of form w=(az+b)/(cz+d) do that too. We go by way of looking at the transformation 1/z. If z = r cis θ, then 1/z = (1/r) cis θ. 1/z is in the same direction as z, but (modulus of z)×(modulus of (1/z))=1 First take the case where the origin is inside the circle. (This is called the “Intersecting Chords Theorem”). Let origin = P in the diagram above. Geometry tells us that in the diagram above, AP·PD = BP·PC, in other words AP.PD is constant whatever point A we choose on the circumference if D is the other end of the chord through P. Thus for some constant k, as z moves round the circle, the reflection in the imaginary axis of k/z moves round the same circle; so k/z moves round a circle; so 1/z moves round a circle; and so 1/z does too. And a diameter going through P transforms into a diameter. The geometry: triangles ABP and CDP are similar because: ∠BAD = ∠BCD, as inscribed angles subtended by the same chord BD, ∠ABC = ∠ADC, as inscribed angles subtended by the same chord AC, ∠APB = ∠CPD, as a pair of vertical angles. So AP/BP = DP/CP, and thus AP·PD = BP·PC ▇ If the origin D (which is the pole, or centre of inversion) is outside the circle, as above, then there is a similar result, the “Intersecting Secants Theorem”. CLAIM: if z moves on a circle, the tangent from the origin to the circle has length k, and m=2, then w=m/z moves on the same circle, and the diameter of the z-circle ⟼ diameter of the w-circle. PROOF: A is any point on the circle; E is the centre; F is the midpoint of CA, so angle CFE is a right angle. D is the origin. Because DFE and CFE are both right-angled triangles Pythagoras ⇒ DE2−DF2=EF2=CE2−CF2 ⇒ DE2−CE2=DF2−CF2 [] But DC.DA = (DF−CF).(DF+FA) = (DF−CF).(DF+CF) = DF2−CF2 and DG.DH = (DE−GE).(DE+EH) = (DE−CE).(DE+CE) = DE2−CE2 since GE, CE, and EH are all radii of the circle So equation [] ⇒ DC.DA = DG.DH. Since this is true wherever A is on the circle, it is also true when A is at B, and so DC.DA = DG.DH = DB2 = k2 So DC is in the same direction as DA, but of length (or magnitude, or modulus) equal to k2/\|DA\|. If A is represented by the complex number z, C is represented by the complex number k2/z Since the whole transformation is symmetrical around the line DH, the diameter GH of the z-circle ⟼ the diameter HG of the w-circle (though in general other diameters of the z-circle do not ⟼ diameters of the w-circle). ▇ The final case is when the origin (pole) is on the circle, i.e. neither inside the circle nor outside it. When the pole is very close to the original circle, then the image circle is very big. So we’d expect that when the origin (pole) is actually on the original circle, the image will be a line. So it is. We have one of the diagrams below, where D is the origin and A is such that DA is the diameter, with DA.DA’=DB.DB’ ∠ADB is common to the triangles ΔABD and B’A’D, and DA/DB=DB’/DA’. Therefore the triangles are similar. Therefore ∠B’A’D = ∠ABD = π/2, since ∠ABD is an angle in a semicircle. Therefore the locus of B’, i.e. of 1/z, if B represents z, is a straight line perpendicular to the diameter of the circle through D and distant 1/d from D where d=diameter. Conversely, if B’ represents z, moving on a line with perpendicular distance a from D, this reasoning shows us that the locus of 1/z is a circle going through D, with diameter perpendicular to the locus of z, and diameter 1/a. The two ends of the diameter are the images under the transformation of infinity and of the point on the line closest to D. The final case is when z moves along a line passing through the origin D. 1/z has the same argument as z, and its modulus takes all real values (including 0; z→∞ ⟼ 1/z→0), so the locus of 1/z is the same line. In short: If z moves on a circle not passing through the origin, m/z moves on the same circle, for some positive constant m. If z moves on a circle passing through the origin of diameter d, 1/z moves on a line perpendicular to the diameter through the origin. If z moves on a line not passing through the origin, 1/z moves on a circle passing through the origin. If z moves on a line passing through the origin, 1/z moves on the same line. A transformation like w=(az+b)/(z+d) (I’ve divided through by a common factor to make c=1) can be broken down into successive transformations: z ⟼z+d (translation) z+d ⟼ 1/(z+d) (inversion+conjugate) 1/(z+d) ⟼ 1/(z+d) (inversion) 1/(z+d) ⟼ (b−ad) [1/(z+d)] (enlargement and rotation) (b−ad) [1/(z+d)] ⟼ a + (b−ad) [1/(z+d)] = (az+ad+b−ad) / (z+d) = (az+b) / (z+d) (translation) We can calculate the image by drawing each transformation in turn. But there’s an easier rule of thumb: You identify whether the locus of w is a circle or a line by seeing whether z takes a value which will make w go to infinity (so its locus is a line), or doesn’t. Then you identify the w-circle by finding a diameter, given by the w-values got from z-values at two ends of a diameter along the line connecting the centre of the z-circle from the centre of inversion (z=−d/c). Or, if z moves on a line, a diameter of the w-circle is given by the w-values got from z at infinity and from z at the closest point on the line to the centre of inversion. You identify the w-line by using any two convenient z-values to calculate two w-values. ——– Alternatively: if the locus of z is \|z\|=k, to identify either two ends of a diameter of the w-circle, or the two points from which the w-line is equidistant, do the following. Rearrange the equation w=(az+b)/(cz+d) and we get z=(−b+dw)/(a−cw), or: the distance of w from b/d is k\|c\|/\|b\| times the distance of w from a/c. Let λ=k\|c\|/\|b\|. Look at the line connecting b/d and a/c. Unless λ=1, there will be two points on that line which have (distance from b/d)=λ(distance of w from a/c), one (call it M) between b/d and a/c and one (call it L) outside. The circle must be symmetrical around LM, because if a point on one side fits the equation (distance from b/d)=λ(distance of w from a/c) then the reflection of that point in LM also fits the equation. Therefore LM is the diameter of the circle. See here for another, more traditional, proof of this result, which is called the “Circle of Apollonius”. Share this: Click to share on X (Opens in new window) X Click to share on Facebook (Opens in new window) Facebook Like Loading... Related Möbius transformations: the geometric approach, and worked answers to all textbook questions and all recent FP2 exam questionsIn "Geometry" Contents page for FP2 transforming loci in the complex planeIn "Geometry" Algebraic method for calculating Möbius transformationsIn "Mobius transforms" Reblog Subscribe Subscribed A minimum of blind calculation Already have a WordPress.com account? Log in now. 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6420	https://blog.csdn.net/qq_45513182/article/details/106088496	阿拉伯数字转换成中文读法的C语言程序_c 数字转中文-CSDN博客博客下载学习社区 GitCode InsCodeAI 会议搜索 AI 搜索登录登录后您可以：复制代码和一键运行与博主大V深度互动解锁海量精选资源获取前沿技术资讯立即登录会员·新人礼包消息历史创作中心创作阿拉伯数字转换成中文读法的C语言程序最新推荐文章于 2025-07-14 10:06:07 发布 BigBlackOX于 2020-05-13 00:09:53 发布阅读量8k收藏 40 点赞数 16 CC 4.0 BY-SA版权文章标签：编程语言c++ 版权声明：本文为博主原创文章，遵循CC 4.0 BY-SA版权协议，转载请附上原文出处链接和本声明。本文链接：本文介绍了一道关于将阿拉伯数字转化为中文读法的编程题，回顾了中文数字读法与英文的不同，并提供了C语言实现的算法思路，包括如何处理中间的“零”和不同数量级的单位转换。记一道评论区的题目导语 : 作为一个西电的菜鸡 , 每天只能和不超过两天的ddl较劲 , 啥也不懂 , 啥也不会. 因为家里硬盘不太够了所以把作业的代码传到博客里保存. 可以预见到的 , 遭到了真正大佬的嘲笑. 其中 , 二哥在讨论区里留下一道题目 , 我觉得还挺有意思的. 所以试着实现了一下 , 作为博客的第二篇拙作. 1.题目分析一下 , 是想实现一个小写数字到中文大写数字的转化. 那么 , 如果我们想要实现这个功能 , 首先我们要重新回顾一下中文数字的读法和英文读法的区别. 2.回顾背后的数学知识我们在小学二年级的时候就学过数字的读法 , 在初中的时候基本上了解到了英文的数字的读法. 比如对于31980 , 中文读作"三万一千九百八十" , 英文读作"tirty-one thousand and nine hundred eighty". 那么中文和英文对于数字的读法来说有什么不同呢? 首先 , 我们要了解一下中文和英文当中对于数字的单位. 在中文中 , 有个 , 十 , 百 , 千 , 万 , 十万 , 百万 , 千万 , 亿 , 十亿… 在英文中 , 有1-20的不同的读法 , x0的读法 , hundred , thousand , million , billion… 通过比较我们不难发现其规律. 中文的数字单位是以四个为一循环 , 比如四位引出一个万 , 再四位引出一个亿. 而英语中是以三位为一个计数段 , 引出一个定义的单位. 1,000,000,000 1,000,000,000 1,0 0 0,0 0 0,0 0 0 英语国家的人看数字的方式. 10,0000,0000 10,0000,0000 1 0,0 0 0 0,0 0 0 0 中国人看数字的方式. 两者相似的 , 对于比较小的位数划分的比较详细一些 , 而对于比较大的位数 , 中国人采取与小位数的单位相结合同时适度引入新单位来记录数字 , 而英语采用每三位定义一个单位的方式记录数字. 对于中文来说 , 比较特殊的一点是 , 如果在数的中间有0的话习惯中间读一个"零"表示区分层次. 如 : 1023读作"一千零二十三" , 但是如果从某一位开始之后全是0的情况下则不读这个0. 而这个"零"的读法便是这个小demo最有意思的地方. 3.程序设计和算法思路通过上面对问题的回顾 , 我们大致可以想到一个思路. 将一个数字按4位为一组分隔开. 将0-9999的数实现向中文的转换 , 然后在后面加上单位即可. 比如12345 , 可以看作1,2345 , 那么第一组2345就读作"二千三百四十五" , 这里为了叙述方便 , 将中文大写数字"壹"写作了中文小写数字"一". 这一组因为最低位是个位 , 所以单位为空. 而第二组只有一个1 , 读作"一" , 它处于万位上. 所以后面加上单位"万". 所以读作"一万二千三百四十五". 下面主要讨论如何将0-9999内的数字实现. 很自然的 , 我们想到模拟人来读数字的方法. 首先,我们将所读数的千位 , 百位 , 十位 , 个位存在一个整型数组里. 下面列出需要讨论的几种情况:12341,2,3,410,0,0,110011,0,0,110001,0,0,011101,1,1,000,0,0,0 下面列出需要讨论的几种情况:\ 1234\ 1,2,3,4\ 1\ 0,0,0,1\ 1001\ 1,0,0,1\ 1000\ 1,0,0,0\ 1110\ 1,1,1,0\ 0\ 0,0,0,0 下面列出需要讨论的几种情况:1 2 3 4 1,2,3,4 1 0,0,0,1 1 0 0 1 1,0,0,1 1 0 0 0 1,0,0,0 1 1 1 0 1,1,1,0 0 0,0,0,0 首先来看第一种. 第一种读作"一千二百三十四" , 可以看到 , 我们开始读是从第0位开始读的 , 而且每一位都读到了. 这里我们可以得出对应规则 : 第0位对应单位为"千",如果不为0应将其读了数字名后加"千" , 之后类推. 再看第二种 , 显然之前的三位都没有读到 , 是从第四位开始读的. 这里我们可以体会到 : 应该设置一个变量 , 我们设置为start. 我们通过start的状态确定读数的开始和停止. 第三种读作"一千零一". 这里我们引出了中间读"零"的问题. 我们看到中间虽然有两个0 , 但是只读一次. 我们可以认为只读第一次出现的0. 所以这里我们引出变量firstzero , 用来确定是否为第一个0 , 只有第一个0会读. 第四种读作"一千". 这里我们可以发现一个概念. 若从某位数开始全是0 , 则这个第一个0不读. 所以引入alwayszero变量 , 用来确定是否之后全是0 , 若是则不读 , 不是则读第一个0. 最低0.47元/天解锁文章 200万优质内容无限畅学确定要放弃本次机会？福利倒计时 : : 立减 ¥ 普通VIP年卡可用立即使用 BigBlackOX 关注关注 16点赞踩 40 收藏觉得还不错? 一键收藏 6评论分享复制链接分享到 QQ 分享到新浪微博扫一扫举报举报 C# 实现阿拉伯数字转换为中文数字 3种实现思路（递归拼接、循环拼接、if拼接）旭东怪的博客 09-08 6155 1、数字 1-9 转换为中文数字 //数字 1-9 转换为中文数字 public string OneBitNumberToChinese(int num){ string result = ""; switch(num){ case 1:result = "一";break; case 2:result = "二";break; case 3:result = "三";break; case 4:result = "四";break; case 5:resul. 怎样将c 语言的字体变中文,请问，在c 语言中如何将阿拉伯数字转换成汉字，求代码... weixin_36357218的博客 05-16 1968 该楼层疑似违规已被系统折叠隐藏此楼查看此楼#includeint main(){int a,b,c,d,e,n,f;while (scanf("%d", &a) != EOF){if (a >= 0 && a <= 9999){if (a >= 1000 && a < 10000){switch (n = a / 1000){cas... 6 条评论您还未登录，请先登录后发表或查看评论实现阿拉伯数字与中文数字转换的C 程序代码 08-09 实现阿拉伯数字与中文数字转换的C 程序代码 c 语言数字转汉语读法 11-15 可以将任意长度的数字转换为汉语口语的读法、对于支持语音模块的单片机可以直接实现口语报数例如： 12345 一万两千三百四十五 10001 一万零一 9020012 九百零二万零一十二 C 语言实现汉语与阿拉伯数字串互转工具 weixin_36369848的博客 07-14 360 在 C 语言中，字符串通常被定义为字符数组，且以空字符（’\0’）结尾。字符串的初始化可以通过字符字面量（char literals）或者字符数组的方式进行。// 字符串定义示例// 使用字符数组定义并初始化字符串// 指针指向字符串常量初始化字符串时，C 语言编译器会自动在字符串的末尾添加一个空字符（’\0’），用于标识字符串的结束。因此，一个包含 n 个字符的字符串实际上需要 n+1 个字节的空间。在头文件中，我们定义了必要的宏，并声明了需要对外公开的函数接口。// 定义字符常量。 c 语言编程阿拉伯数字转中文,阿拉伯数字转中文大(小)写的函数 weixin_29655587的博客 05-19 891 阿拉伯数字转中文大(小)写的函数 erquan 发表于 2002-7-10 17:43 PHP编程 //chinese_money.php //Changing Arab Money Num to Chinese Money Num /====================== Functions List: Chinese_Money_Max() 小数点前此函数可以单独拿出用于将... C 语言实现阿拉伯数字与中文数字互转程序最新发布 08-07 C 语言实现阿拉伯数字与中文数字互转程序是一项涉及数字系统理解与程序设计的技能，它是计算机科学与应用数学的交叉产物，主要应用于需要将阿拉伯数字转换为中文表述，或者反之的场景。由于中文数字系统与阿拉伯数字... C 语言实战项目案例：阿拉伯数字转汉字描述中提到的“阿拉伯数字转换汉字”实质上是一种数字格式转换，即将数字表示的数值转换为对应的中文汉字表示。在中文中，数字的汉字表示通常有一套完整的体系，包括个位、十位、百位、千位等的特定汉字（一、十、百... C 语言中数字转化为中文读法 03-20 在C 语言中，可以通过编写自定义逻辑来实现将阿拉伯数字转换为其对应的中文读法。这种功能通常涉及以下几个方面： 1. 映射表构建：创建一个数组或结构体，用来存储数字与其对应汉字之间的关系。 2. 分组处理... 3.中文系统的开发具有重要意义，如何将数字转化为中文读法很重要。例如面对用户输入的是501，输出五百零一；输入00810，输出八百一十。请给出代码处理用户输入不大于五位数（包括五位数）的阿拉伯数字，输出中文读法。用c 语言 03-15 好的，我需要解决如何将最多五位数的阿拉伯数字转换为中文读法的C 语言程序。首先，我得理解中文数字的读法规则，然后考虑如何处理各种情况，比如中间的零、末尾的零，以及不同位数的处理。首先，中文数字的单位有... c 语言编程阿拉伯数字转中文,阿拉伯数字到中文大写数字的转换 weixin_32808853的博客 05-19 3420 1 #include 2 #include 34 struct output //存放中文大写单位或数字的结构体类型5 {6 char ch;7 struct output next;8 };9 struct output head1, psnew1, p1;10 typedef structoutput output1;1112 void Num(output1 ... c++ 将 _数字转换为汉字 04-22 #include"iostream.h" #include"string" int main() { int nNumber; cout<>nNumber; char str[]={"零"}; switch(nNumber) { case 0: { strcpy(str,"零"); break; } case 1: { strcpy(str,"一"); break; } case 2: { strcpy(str,"二"); break; } case 3: { strcpy(str,"三"); break; } case 4: { strcpy(str,"四"); break; } case 5: { strcpy(str,"五"); break; } case 6: { strcpy(str,"六"); break; } case 7: { strcpy(str,"七"); break; } case 8: { strcpy(str,"八"); break; } case 9: { strcpy(str,"九"); break; } default: { cout<<"输入错误 ! "; return ; } } cout<<nNumber<<" 对应的大写汉字是 "<<str<<"\n"; } C#阿拉伯数字转为中文数字 01-21 这里一个阿拉伯数字转为中文数字的C#写的类！@ 阿拉伯数字转中文 07-10 自己写的一个阿拉伯数字转中文的工具方法，主要考虑了0的重叠和所在位置不同的翻译。网上找的好多没考虑0的问题。欢迎大家测试 C实现数字变换成汉字 l8947943的博客 03-07 6484 问题:在一个文件里面写入“我有1支铅笔，但是仓库里面有353628支”，读出刚才的文件内容，将1变为一，以此类推，写入新的文件。新文件内容应为“我有一支铅笔，但是仓库里面有三五三六二八支”#include<stdio.h> #include<stdlib.h> #include<string.h> #include<math... c 语言如何输入汉字_ 阿拉伯数字转汉字 weixin_29002209的博客 11-29 2506 最近看到一个笔试题，阿拉伯数字和汉字数字的互相转化，网上比较多的是python和java的实现，这里对 C++ 的实现做了整理和改动。要实现的功能是将正整数转成汉字数字，忽略小数部分(小数部分就是一对一的映射，也比较容易实现)。例如这几个测试用例：18->十八180->一百八十20930->两万零九百三十129809->十二万九千八百零九200182190->二... C 语言将数字转成大写中文数字热门推荐 xyphf的博客 07-20 1万+ #include #include //内置函数头文件int main() { //转换大小写 //printf("大写：%c", toupper('a')); //将小写字母转大写字母 //思考题：如何将用户输入的小写数字转换成中文大写 //如：1234 - 壹仟贰佰叁拾肆元整 int money, count = 0; c 语言，数字转汉字并返回盗墓者 07-05 4382 //盗墓者是个丑奴儿，原 #include <stdio.h> #include <stdlib.h> #include <string.h> #include <time.h> //准备汉字数组 char (s[])={"零","一","二","三","四","五",&q 大写汉字转阿拉伯数字 c 语言,阿拉伯数字转中文数字方法详解（C++ 实现） weixin_30651345的博客 05-22 2247 阿拉伯数字与中文数字没有一一对应关系，不存在直接转换的公式化算法，因此需要根据两种数字体系的特点精心构造转换算法。中文计数有一个特点，就是“零”的使用变化多端。阿拉伯数字中数字的权位依靠数字在整个数字长度中的偏移位置确定，因此数字中间出现的0用于标记数字的偏移位置，即便是连续出现的0也不能省略。中文计数方式中每个数字的权位都直接跟在数字后面，因此代表连续出现的若干个0。尽管如此，也不是所有的情况都... 关于我们招贤纳士商务合作寻求报道 400-660-0108 kefu@csdn.net 在线客服工作时间 8:30-22:00 公安备案号11010502030143 京ICP备19004658号京网文〔2020〕1039-165号经营性网站备案信息北京互联网违法和不良信息举报中心家长监护网络110报警服务中国互联网举报中心 Chrome商店下载账号管理规范版权与免责声明版权申诉出版物许可证营业执照 ©1999-2025北京创新乐知网络技术有限公司 BigBlackOX 博客等级码龄6年 7 原创29 点赞 54 收藏 5 粉丝关注私信 🚀 支持40+常用图表，20+主流数据源，拖拉拽制作数据大屏。开源免费，支持二开。广告 TA的精选新现代图片选择器（PHPicker）在 SwiftUI 应用 3016 阅读新记录tesseract-OCR配置的一些问题 362 阅读热在Visual Studio Code中配置python环境 1947 阅读热华为云EAI应用（1）定制语言合成 858 阅读热小记~随笔[记录一个Matlab的中的一个错误] 709 阅读查看更多 2021年 1篇 2020年 6篇大家在看上下文工程驱动智能体向规则引擎与神经网络共生 537 上下文工程与智能体认知进化 135 CoRL 2025最佳论文深度解析上下文工程驱动智能体向量子计算融合推理 204 上下文工程驱动智能体向价值观对齐机制 84 上一篇：关于进制转换和高精度加减法的算法,希望帮到做作业的同学下一篇：小记~随笔[记录一个Matlab的中的一个错误] 目录记一道评论区的题目 1.题目 2.回顾背后的数学知识 3.程序设计和算法思路展开全部收起目录记一道评论区的题目 1.题目 2.回顾背后的数学知识 3.程序设计和算法思路展开全部收起 🚀 支持40+常用图表，20+主流数据源，拖拉拽制作数据大屏。开源免费，支持二开。广告上一篇：关于进制转换和高精度加减法的算法,希望帮到做作业的同学下一篇：小记~随笔[记录一个Matlab的中的一个错误] 登录后您可以享受以下权益：免费复制代码和博主大V互动下载海量资源发动态/写文章/加入社区 ×立即登录评论 6 被折叠的条评论为什么被折叠?到【灌水乐园】发言查看更多评论添加红包祝福语请填写红包祝福语或标题红包数量个红包个数最小为10个红包总金额元红包金额最低5元余额支付当前余额 3.43 元前往充值 > 需支付：10.00 元取消确定成就一亿技术人! 领取后你会自动成为博主和红包主的粉丝规则 hope_wisdom 发出的红包实付元使用余额支付点击重新获取扫码支付钱包余额 0 抵扣说明： 1.余额是钱包充值的虚拟货币，按照1:1的比例进行支付金额的抵扣。 2.余额无法直接购买下载，可以购买VIP、付费专栏及课程。余额充值确定取消举报选择你想要举报的内容（必选）内容涉黄政治相关内容抄袭涉嫌广告内容侵权侮辱谩骂样式问题其他原文链接（必填）请选择具体原因（必选）包含不实信息涉及个人隐私请选择具体原因（必选）侮辱谩骂诽谤请选择具体原因（必选）搬家样式博文样式补充说明（选填）取消确定 AI助手 AI 搜索智能体 AI 编程 AI 作业助手下载APP 程序员都在用的中文IT技术交流社区公众号专业的中文 IT 技术社区，与千万技术人共成长视频号关注【CSDN】视频号，行业资讯、技术分享精彩不断，直播好礼送不停！客服返回顶部
6421	https://teachablemath.com/bar-model-example-before-and-after/	Skip to content TeachableMath Singapore math support and resources Start Free TrialStart Free Trial Bar Model example – Before After 1 Comment / Other Topics / By Tze-Ping Low Here is a bar model example for “before and after” type of word problems. Tom had $190 and Jane had $60. After each of them received an equal amount of money from their father, Tom had twice as much money as Jane. How much did their father give each of them? See more 3rd grade worked solutions videos Related Resources For more related resources, please refer to our Bar Models page. 1 thought on “Bar Model example – Before After” Des Duthie Thank you – I am very excited about teaching the use of bar modelling while learning about its exceptional power of problem representation! Your teachable math emails have been/ are gold – thank you Kar-Hwee and Tze-Ping. Reply Leave a Comment Cancel Reply
6422	https://www.geeksforgeeks.org/maths/harmonic-functions/	Harmonic Function Harmonic functions are one of the most important functions in complex analysis, as the study of any function for singularity as well residue we must check the harmonic nature of the function. For any function to be Harmonic, it should satisfy the lapalacian equation i.e., ∇2u = 0. In this article, we have provided a basic understanding of the concept of Harmonic Function including its definition, examples, as well as properties. Other than this, we will also learn about the steps to identify any harmonic function. What is Harmonic Function? A harmonic function is a function which meets two criteria. First, it needs to be smooth, meaning it can be continuously and easily differentiated twice. Second, it must follow a specific rule called Laplace's equation, expressed as: \frac{\partial^2u}{\partial x^2} + \frac{\partial^2u}{\partial y^2} = 0 Does it refer In simpler terms, for a function [u(x, y)] to be harmonic, the sum of its second partial derivatives with respect to x and y must be zero. Harmonic Function Definition Any smooth function u(x,y) is said to be harmonic if ∇2u = 0. Where, ∇2 is the laplacian operator i.e., ∂2/∂x2 + ∂2/∂y2. In simple words, if any smooth function u(x, y) satisfy the equation uxx + uyy = 0, then this function u is harmonic fucntion. Where uxx and uyy represent second order partial derivative with respect to x and y respectively. Examples of Harmonic Function Some of the common examples of harmonic function are: Some other examples with three variable are: Where r = x2 + y2 + z2. What are Conjugate Harmonic Functions? In situations where you have an analytic function ω(z)=u+iv, you can think of "v" as the conjugate harmonic function of "u" and vice versa. In other words: If you have an analytic function ω1(z)=u+iv, then ω2(z)=−v+iu is also an analytic function. In this context, u and v are considered harmonic conjugates. This means that these functions are connected in a special way, and when you swap the real and imaginary parts, the resulting function remains analytic. Properties of Harmonic Functions Some of the common properties of harmonic functions are: How to Identify Harmonic Function? To identify a harmonic function, you can follow these steps. Step 1: Understand the Basics Step 2: Examine the Function Consider a function u(x, y), for example, u(x, y) = x2 - y2 Step 3: Check Continuity and Differentiability Ensure that u(x, y) is smooth, meaning it is continuous and has continuous first and second derivatives. In our example, u(x, y) = x2 - y2 is a polynomial, so it's smooth everywhere. Step 4: Verify Laplace's Equation Apply Laplace's equation: (\frac{\partial^2u}{\partial x^2} + \frac{\partial^2u}{\partial y^2} = 0). Since the sum is zero, the function u(x, y) = x2 - y2 satisfies Laplace's equation, indicating that it is a harmonic function. Read More, Solved Examples on Harmonic Function Example 1: Determine if the function u(x,y) = ln(x2 +y2) is harmonic. Solution: Calculate the partial derivatives of u. ⇒ \frac{\partial^2u}{\partial x^2} = \frac{2y^2 - x^2}{(x^2 + y^2)^2} ⇒ \frac{\partial^2u}{\partial y^2} = \frac{2x^2 - y^2}{(x^2 + y^2)^2} Sum the second partial derivatives. ⇒ \frac{\partial^2u}{\partial x^2} + \frac{\partial^2u}{\partial y^2} = 0 Since the sum is zero, u(x, y) = In(x2 + y2) is harmonic. Example 2: Check the harmonic nature of u(x, y) = cos(x) cosh(y). Solution: Compute the second partial derivatives of u. \frac{\partial^2u}{\partial x^2} = -cos(x) cosh(y) \frac{\partial^2u}{\partial y^2} = \os(x) cosh(y) Sum the second partial derivatives:. \frac{\partial^2u}{\partial x^2} + \frac{\partial^2u}{\partial y^2} = 0 The sum is zero, indicating that u(x, y) = cos(x) cosh(y) is a harmonic function. Practice Problems on Harmonic Functions Problem 1: Investigate whether the function u(x,y)=x3−3xy2+3x2−3y2+1 is harmonic. Problem 2: Show that u(x,y)=2x(1−y) is a harmonic function. Determine its harmonic conjugate v(x,y). Problem 3: Prove that the function u(x,y)=ex2−y2 cos(2xy) is harmonic. Find the harmonic conjugate v(x,y) of u, considering the ambiguity of a constant. S Explore Maths Basic Arithmetic What are Numbers? Arithmetic Operations Fractions - Definition, Types and Examples What are Decimals? 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6423	https://www.sciencedirect.com/science/article/abs/pii/S0096300310003309	On the midpoint method for solving equations - ScienceDirect Typesetting math: 100% Skip to main contentSkip to article Journals & Books Access throughyour organization Purchase PDF Search ScienceDirect Article preview Abstract Introduction Section snippets References (18) Cited by (8) Applied Mathematics and Computation Volume 216, Issue 8, 15 June 2010, Pages 2321-2332 On the midpoint method for solving equations Author links open overlay panel Ioannis K.Argyros a, Yeol Je Cho b, Saïd Hilout c Show more Add to Mendeley Share Cite rights and content Abstract Using our idea of recurrent functions, we provide a new semilocal convergence analysis for the midpoint method (MPM) introduced by Argyros and Chen. We show that this way the error estimates are tighter, and the sufficient convergence conditions can be weaker. Moreover, we also show that the Newton-type method (NTM) introduced by Wu and Zhao (using the same information as (MPM)) can always be replaced by the (MPM). Numerical results where our results apply to solve nonlinear equations, but others cannot are also provided in this study. Introduction In this study, we are concerned with the problem of approximating a locally unique solution x★ of the equation F(x)=0,where F is a twice Fréchet-differentiable operator defined on a convex subset D of a Banach space X with values in a Banach space Y. A large number of problems in applied mathematics and also in engineering are solved by finding the solutions of certain equations. For example, dynamic systems are mathematically modeled by difference or differential equations and their solutions usually represent the states of the systems. For the sake of simplicity, assume that a time-invariant system is driven by the equation x˙=Q(x), for some suitable operator Q, where x is the state. Then the equilibrium states are determined by solving the Eq. (1.1). Similar equations are used in the case of discrete systems. The unknowns of engineering equations can be functions (difference, differential, and integral equations), vectors (systems of linear or nonlinear algebraic equations), or real or complex numbers (single algebraic equations with single unknowns). Except in special cases, the most commonly used solution methods are iterative – when starting from one or several initial approximations a sequence is constructed that converges to a solution of the equation. Iteration methods are also applied for solving optimization problems. In such cases, the iteration sequences converge to an optimal solution of the problem at hand. Since all of these methods have the same recursive structure, they can be introduced and discussed in a general framework. We use the midpoint method (MPM):y n=x n-F′(x n)-1 F(x n),z n=x n+y n 2,x n+1=x n-F′(z n)-1 F(x n),n⩾0,x 0∈D,to generate a sequence approximating x★. A non-affine convergence analysis was provided in , , , , where the cubical convergence of (MPM) was established. (MPM) uses two inverses and one function evaluation at every step. A survey on iterative methods and related topics can be found in (see, also , , , , , , , , , , , , , , , ). Here, we revisit (MPM). In Section 2, we provide more precise majorizing sequences for (MPM) under the same hypotheses and computational cost as before. In particular, we show that the Newton-type method (NTM) extending the Frontini and Sormani ideas , in a Banach space setting, which was introduced by Wu and Zhao :y n=x n-F′(x n)-1 F(x n),x n+1=x n-2 F′(x n)+F′(y n)-1 F(x n),n⩾0,x 0∈D,can be replaced by the (MPM). This is due to the fact that, under the same computational cost and weaker hypotheses, the (MPM) converges faster than the (NTM) (see Remark 2.3). Moreover, in Section 3, we provide new sufficient affine-convergence conditions, which can be weaker than before, using our new idea of recurrent functions. This idea is different from the technique of recurrent sequences inaugurated by Rall , and essentially used in Section 2. A numerical example is also provided to show that our results apply to solve equations, but earlier ones cannot. Access through your organization Check access to the full text by signing in through your organization. Access through your organization Section snippets Semilocal convergence analysis I for (MPM) We shall show that (MPM) can always replace (NTM). By simply replacing F by F′(x 0)-1 F in Theorem 4.1 in , , we obtain the following semilocal convergence theorem for the (MPM): Theorem 2.1 Let F:D⊆X→Y be a twice Fréchet-differentiable operator. Assume that there exist x 0∈D , constants M⩾0,N⩾0,K>0 and η⩾0 such that, for all x,y∈D , F′(x 0)-1∈L(Y,X),‖F′(x 0)-1 F(x 0)‖⩽η,‖F′(x 0)-1 F″(x)‖⩽M,‖F′(x 0)-1(F″(x)-F″(y))‖⩽N‖x-y‖,M 1+7 N 6 M 2⩽K,M≠0,h=K η⩽1 2,and U¯y 0,t★-η=y∈X:‖y-y 0‖⩽t★-η⊆D. Then the sequences{x n}and{y n}(n⩾0) Semilocal convergence analysis II of (MPM) We introduce our new idea of recurrent functions in this section. It is convenient for us to define some constants, sequences, and functions. Let M 0⩾0,M⩾0,M 1⩾0,N 0⩾0,N⩾0 and η⩾0 be given constants. Define constants δ 0,δ 1,δ 2 and w∞ by:δ 0=M 1 η 1-M 0 2 η<2,M 0 η≠2,δ 1=2 M M 0+M+(M 0+M)2+12 M 0 M<1,M 0≠0 or M≠0,δ 2=2(12 M+7 N η)7 N η+24 M 0+(7 N η+24 M 0)2+48(M+2 M 0)(12 M+7 N η)<1,w∞=1-M 0 η 1+M 0 η,the sequences {t n},{s n}(n⩾0) by t 0=0,s 0=η,t n+1=s n+M 2(s n-t n)2 2 1-M 0 2(s n+t n),s n+1=t n+1+12 M(t n+1-s n)2+7 N 1(s n-t n)3+6 M 2(s n-t n)3 1-M 0 2(s n+t n)24(1-M 0 t n+ Conclusion In order to approximate a locally unique solution of a nonlinear equation in a Banach space, involving a twice differentiable operator, we provided a semilocal convergence analysis for midpoint method. We also established a comparison with Newton-type method, and proved that the order of convergence is three. Using our new idea of recurrent functions, we provided a semilocal convergence analysis with the following advantages over the work in : weaker sufficient convergence conditions, and Acknowledgement This work was supported by the Korea Research Foundation Grant funded by the Korean Government (KRF-2008-313-C00050). Recommended articles References (18) I.K. Argyros An improved error analysis for Newton-like methods under generalized conditions J. Comput. Appl. Math. (2003) I.K. Argyros On the comparison of a weak variant of the Newton–Kantorovich and Miranda theorems J. Comput. Appl. Math. (2004) I.K. Argyros et al. The midpoint method in Banach spaces and the Pták error estimates Appl. Math. Comput. (1994) J.A. Ezquerro et al. Remark on the convergence of the midpoint method under mild differentiability conditions J. Comput. Appl. Math. (1998) M. Frontini et al. Modified Newton’s method with third-order convergence and multiple roots J. Comput. Appl. Math. (2003) M. Frontini et al. Third-order methods from quadrature formulae for solving systems of nonlinear equations Appl. Math. Comput. (2004) J.M. Gutiérrez et al. An acceleration of Newton’s method: super-Halley method Appl. Math. Comput. (2001) M.A. Hernández A modification of the classical Kantorovich conditions for Newton’s method J. Comput. Appl. Math. (2001) M.A. Hernández et al. Modification of the Kantorovich assumptions for semilocal convergence of the Chebyshev method J. Comput. Appl. Math. (2000) There are more references available in the full text version of this article. Cited by (8) Two-step Newton methods 2014, Journal of Complexity Show abstract We present sufficient convergence conditions for two-step Newton methods in order to approximate a locally unique solution of a nonlinear equation in a Banach space setting. The advantages of our approach over other studies such as Argyros et al. (2010) , Chen et al. (2010) , Ezquerro et al. (2000) , Ezquerro et al. (2009) , Hernández and Romero (2005) , Kantorovich and Akilov (1982) , Parida and Gupta (2007) , Potra (1982) , Proinov (2010) , Traub (1964) for the semilocal convergence case are: weaker sufficient convergence conditions, more precise error bounds on the distances involved and at least as precise information on the location of the solution. In the local convergence case more precise error estimates are presented. These advantages are obtained under the same computational cost as in the earlier stated studies. Numerical examples involving Hammerstein nonlinear integral equations where the older convergence conditions are not satisfied but the new conditions are satisfied are also presented in this study for the semilocal convergence case. In the local case, numerical examples and a larger convergence ball are obtained. ### EXTENDED CONVERGENCE OF TWO-STEP ITERATIVE METHODS FOR SOLVING EQUATIONS WITH APPLICATIONS 2024, Journal of Numerical Analysis and Approximation Theory ### An improved semilocal convergence analysis for the Halley's method 2018, Advances in Nonlinear Variational Inequalities ### A contemporary study of iterative methods: Convergence, dynamics and applications 2018, A Contemporary Study of Iterative Methods Convergence Dynamics and Applications ### Newton-Kantorovich Convergence Theorem of a Modified Newton's Method Under the Gamma-Condition in a Banach Space 2013, Journal of Optimization Theory and Applications ### Semilocal convergence for a fifth-order Newton's method using recurrence relations in Banach spaces 2011, Journal of Applied Mathematics View all citing articles on Scopus View full text Copyright © 2010 Elsevier Inc. All rights reserved. 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6424	https://www3.nd.edu/~jhauenst/preprints/ghRealSolving.pdf	Real solutions to systems of polynomial equations and parameter continuation Zachary A. Griﬃn∗ Jonathan D. Hauenstein† May 7, 2013 Abstract Given a parameterized family of polynomial equations, a fundamental question is to determine upper and lower bounds on the number of real solutions a member of this family can have and, if possible, compute where the bounds are sharp. A computational approach to this problem was developed by Dietmaier in 1998 which used a local linearization procedure to move in the parameter space to change the number of real solutions. He used this approach to show that there exists a Stewart-Gough platform that attains the maximum of forty real assembly modes. Due to the necessary ill-conditioning near the discriminant locus, we propose replacing the local linearization near the discriminant locus with a homotopy-based method derived from the method of gradient descent arising in optimization. This new hybrid approach is then used to develop a new result in real enumerative geometry. Keywords. real solutions, parameter space, discriminant, numerical algebraic ge-ometry, polynomial system, homotopy continuation, enumerative geometry AMS Subject Classiﬁcation. 65H10, 68W30, 14Q99 Introduction Parameterized families of systems of real polynomial equations naturally arise in many areas including economics, engineering, enumerative geometry, and physics where the real solutions for each member are often the solutions of interest. A fundamental and diﬃcult question is to compute upper and lower bounds on the number of real solutions and, if possible, compute parameter values where such bounds are attained. For systems arising from geometry, some bounds on the number of real solutions are known and ∗Harold Vance Department of Petroleum Engineering, Mailstop 3116, Texas A&M University, College Station, TX 77843 (zacgriﬃn21@neo.tamu.edu, people.tamu.edu/~zacgriffin21). This author was partially supported by NSF grant DMS-1114336. †Department of Mathematics, North Carolina State University, Raleigh, NC 27695-8205 (hauen-stein@ncsu.edu, www.math.ncsu.edu/~jdhauens). This author was partially supported by NSF grants DMS-0915211 and DMS-1262428. 1 summarized in . Two notable examples are the real analog of Steiner’s problem of conics and the number of real assembly modes of Stewart-Gough platforms [8, 24]. Steiner’s problem of conics is to count the number of plane conics tangent to ﬁve given plane conics in general position, with the answer being 3264. The solving of Steiner’s problem lead to the development of intersection theory and ultimately to many techniques used in the ﬁeld of numerical algebraic geometry (see for a general introduction to numerical algebraic geometry). The real analog of Steiner’s problem, proposed by Fulton , is to determine if there exists ﬁve real plane conics such that there are 3264 real plane conics tangent to the given ﬁve conics. This was answered in the aﬃrmative later by Fulton, a result he did not publish, and independently in . A general Stewart-Gough platform has 40 assembly modes. Dietmaier consid-ered, and answered in the aﬃrmative, the existence of a Stewart-Gough platform with 40 real assembly modes. This platform was found using a computational approach that is summarized in Section 1.4. In short, the approach moves through the real param-eter space based on solving a linear optimization problem obtained by constructing a local linearization of the system of equations. The shortcoming of this approach is the linearization near the discriminant locus is necessarily ill-conditioned. This shortcoming of Dietmaier’s method is addressed in Section 3 by introducing a homotopy-based approach for moving to and through the discriminant locus. This new approach uses homotopies that we call gradient descent homotopies, described in Section 2, based on their relationship to the method of gradient descent used for solving optimization problems. Gradient descent homotopies arise from the computation of critical points of the distance function between a given point and the set of real solutions of a system of polynomial equations. Such critical points have been used in several other algorithms in real algebraic geometry including [1, 11, 18, 19]. Section 4 demonstrates using this hybrid approach together with an a posteriori real certiﬁcation technique developed in arising from Smale’s α-theory [4, 20] to prove the following real enumerative geometric theorem. Theorem 1. There exists eight lines in R3 met by 92 real plane conics. Since eight general lines in C3 are met by 92 plane conics, Theorem 1 shows that the real analog of this enumerative geometry problem is answered in the aﬃrmative. 1 Background 1.1 Algebraic sets For a polynomial system g : CN →Cn, let V(g) = {x ∈CN \| g(x) = 0} and VR(g) = V(g) ∩RN = {x ∈RN \| g(x) = 0} be the set of solutions and the set of real solutions of g(x) = 0, respectively. A set A ⊂CN is called an algebraic set if A = V(g) for some polynomial system g. An 2 algebraic set A ⊂CN is reducible if there exists algebraic sets B, C ⊂CN such that B, C ⊊A and A = B ∪C. If an algebraic set is not reducible, it is said to be irreducible. Let Jg(x) denote the n × N Jacobian matrix of g evaluated at x. When N = n, the system g is said to be a square system. In this case, a point x ∈V(g) is nonsingular if det Jg(x) ̸= 0 and singular if det Jg(x) = 0. 1.2 Discriminant locus Let P ⊂CP be an irreducible algebraic set and PR = P ∩RP . We will focus on parameterized polynomial systems f : CN × P →CN having real coeﬃcients such that f(x, p) = 0 has ﬁnitely many isolated solutions in CN, all of which are nonsingular, for general parameter values p ∈P. Thus, there exists an integer MP ≥0 and an algebraic set Q ⊊P such that, for all p ∈P \ Q, f(x, p) = 0 has exactly MP solutions, all of which are nonsingular. (1) The discriminant locus of f, denoted ∆(f), is the set of p ∈P such that (1) does not hold. The real discriminant locus of f is ∆R(f) = ∆(f) ∩RP . If C is a connected component of PR\∆R(f), the number of real solutions is constant on C. That is, there exist an integer 0 ≤MC ≤MP, with MP −MC even, such that f(x, p) = 0 has exactly MC real solutions, all of which are nonsingular, for every p ∈C. Example 2. For P = C2 and f(x, p) = x2 + p1x + p2, we have MP = 2 and ∆(f) = V(p2 1 −4p2). In particular, the equation f(x, p) = 0 has two distinct real solutions for p ∈R2 such that p2 1 > 4p2, one real solution of multiplicity 2 when p2 1 = 4p2, and no real solutions when p2 1 < 4p2. Example 3. Let P = V(p2 3 −p1p3 + p2) ⊂C3 and f : C2 × P →C2 deﬁned by f(x, p) = x2 1 + p1x1 + p2 x1x2 + p3x2 −1 . We have MP = 1 and ∆(f) = P ∩V(2p3 −p1). In particular, the system of equations f(x, p) = 0 has one real solution for p ∈PR such that 2p3 ̸= p1 and no solutions in C2 when p ∈∆(f). We note that MC3 = 2. 1.3 Trackable paths Homotopy methods rely upon the construction of solution paths which are trackable. The following, from , deﬁnes a trackable path starting at a nonsingular solution. Deﬁnition 4. Let H(x, t) : CN × C →CN be polynomial in x and complex analytic in t. If y ∈CN is a nonsingular solution of H(x, 1) = 0, then y is said to be trackable for t ∈(0, 1] from t = 1 to t = 0 using H(x, t) if there is a smooth map ξy : (0, 1] →CN such that ξy(1) = y and, for t ∈(0, 1], ξy(t) is a nonsingular solution of H(x, t) = 0. The solution path starting at y is said to converge if limt→0+ ξy(t) ∈CN, where limt→0+ ξy(t) is called the endpoint (or limit point) of the path. 3 A trackable path may converge to a nonsingular solution, converge to a singular solution, or diverge. Numerical path tracking algorithms (see for a general overview) use endgames and projective space to handle the singular and divergent cases. 1.4 Dietmaier’s approach The following is a summary of the approach used by Dietmaier to compute a Stewart-Gough platform [8, 24] that achieves the maximum of 40 real assembly modes. In this section, we assume that P = CP and consider polynomial systems of the form f : CN ×CP →CN having real coeﬃcients such that f(x, p) = 0 has MCP ≥2 solutions, all of which are nonsingular, for general parameter values p ∈CP . The basic idea is to move in the real parameter space PR = RP so that two complex conjugate solutions ﬁrst merge and then become two distinct real solutions. Clearly, the parameter value where the two solutions coincide is a point on the real discriminant locus ∆R(f). Suppose that p∗∈RP \ ∆R(f) and x∗∈CN \ RN such that f(x∗, p∗) = 0. If ∆x∗∈CN and ∆p∗∈RP so that f(x∗+∆x∗, p∗+∆p∗) = 0, linearizing at (x∗, p∗) yields ∆x∗≈−Jxf(x∗, p∗)−1Jpf(x∗, p∗)∆p∗ where Jxf(x∗, p∗) and Jpf(x∗, p∗) are the Jacobian matrices of f with respect to x and p evaluated at (x∗, p∗), respectively. Since p∗/ ∈∆R(f), the matrix Jxf(x∗, p∗) is indeed invertible. The value of ∆p∗is chosen to minimize the distance, using some appropriate norm, between x∗+ ∆x∗and RN subject to the following conditions: 1. Since this linearization is only acceptable on a small neighborhood, ∆p∗must be constrained to avoid too large of a step. 2. To maintain the same number of distinct real solutions, for every distinct pair y1, y2 ∈RN such that f(yi, p∗) = 0, the distance between the approximations of the corresponding solutions of f(x, p∗+ ∆p∗) = 0, namely yi + ∆yi, must remain larger than a given bound. 3. To maintain the same number of ﬁnite solutions, for every z ∈CN such that f(z, p∗) = 0, the norm of the approximation of the corresponding solution of f(x, p∗+ ∆p∗) = 0, namely z + ∆z, must remain below a given bound. For a properly selected ∆p∗, the solutions to f(x, p∗+ ∆p∗) = 0 are computed with one natural approach being a parameter homotopy (see [2, 21]). If no acceptable ∆p∗can be found, the process is restarted using a diﬀerent nonreal solution of f(x, p∗) = 0. The process repeats until some complex conjugate pair merges (up to numerical tolerance). If no acceptable value of ∆p∗can be found for all nonreal solutions, the process fails. If a complex conjugate pair has successfully merged, the parameter value needs to be moved so that the merged pair becomes distinct real solutions. We note that does not directly address how to move through the discriminant locus to initially yield two distinct real solutions, but one approach is to simply continue to move in the direction 4 of the last ∆p∗computed. Updating p∗to be the new parameter value with the newly created solutions x1, x2 ∈RN of f(x, p∗) = 0, the approach uses the same constraints above with the objective of maximizing the distance between x1 + ∆x1 and x2 + ∆x2. As before, once ∆p∗is computed, the solutions to f(x, p∗+∆p∗) = 0 are computed and this process is repeated until the two new real solutions are suﬃciently far apart. 2 Gradient descent homotopies Suppose that f : CN →Cn is a polynomial system with real coeﬃcients and y ∈RN. Since we will be using deformations of f, we impose two assumptions on f. First, we assume that N ≥n so that f is not overconstrained. Second, we assume that Jf(x) has rank n for general x ∈CN. That is, there exists an algebraic set Q ⊊CN such that rank Jf(x) = n for all x ∈CN \ Q. We note that these assumptions can always be satisﬁed by replacing f with sums of squares. For x ∈RN, deﬁne dy(x) = ∥x −y∥2 = (x −y)T (x −y) and consider the polynomial optimization problem (P) min {dy(x) \| x ∈VR(f)}. The basic idea is to construct a homotopy and deﬁnes a solution path emanating from the given point y. The aim is to compute a real critical point for problem (P), which is a point x ∈VR(f) such that rank x −y ∇f1(x)T · · · ∇fn(x)T ≤n where ∇fi(x) is the gradient vector of fi evaluated at x. In particular, the set of real crtical points for (P) is the set π(V(G))∩RN where π(x, λ) = x and G : CN×Pn →CN+n is the polynomial system deﬁned by G(x, λ) = f(x) λ0(x −y) + λ1∇f1(x)T + · · · λn∇fn(x)T . (2) Consider the homotopy H : CN × Pn × C →CN+n deﬁned by H(x, λ, t) = f(x) −tf(y) λ0(x −y) + λ1∇f1(x)T + · · · λn∇fn(x)T . (3) Since we are interested in one solution path, namely the path starting at (y, 1, 0, . . . , 0) when t = 1, we will also consider an aﬃne version of H which performs computations on an aﬃne patch in Pn. This homotopy Ha : CN × Cn+1 × C →CN+n+1 is deﬁned by Ha(x, λ, t) =   f(x) −tf(y) λ0(x −y) + λ1∇f1(x)T + · · · λn∇fn(x)T λ0 + α1λ1 + · · · + αnλn −α0   (4) where αi ∈R \ {0}. For Ha, we will consider the start point (y, α0, 0, . . . , 0) at t = 1. Due to Proposition 6 below, we call H and Ha gradient descent homotopies. 5 Remark 5. We note that even though we have constructed homotopies H and Ha for a polynomial system f, one could consider these homotopies for a more general class of systems, e.g., analytic systems or twice diﬀerentiable systems. 2.1 Gradient descent Consider the homotopy Ha deﬁned by (4) when f : CN →C is a polynomial with real coeﬃcients, y ∈RN such that f(y) ̸= 0 and ∇f(y) ̸= 0, and α0, α1 ∈R \ {0}. It is easy to see that the Jacobian matrix of Ha with respect to x and λ is JHa x,λ(x, λ0, λ1, t) =   ∇f(x) 0 0 λ0IN + λ1Hf(x) x −y ∇f(x)T 0 1 α1   where IN is the N × N identity matrix and Hf(x) is the Hessian of f. In particular, JHa x,λ(y, α0, 0, 1) =   ∇f(y) 0 0 α0IN 0 ∇f(y)T 0 1 α1   is full rank since \| det JHa x,λ(y, α0, 0, 1)\| = \|α0\|N−1∥∇f(y)∥2 ̸= 0. (5) The implicit function theorem yields that the solution path starting at (y, α0, 0) at t = 1 exists, and is smooth, locally near t = 1. The following proposition shows that this path, as t decreases from 1, moves in the direction of gradient descent if f(y) > 0 and in the direction of gradient ascent if f(y) < 0. Proposition 6. Let f : CN →C be a polynomial with real coeﬃcients and y ∈RN such that f(y) ̸= 0 and ∇f(y) ̸= 0. Let α0, α1 ∈R \ {0} and Ha be the homotopy deﬁned by (4). If (x(t), λ(t)) is the solution path starting with (y, α0, 0) at t = 1, then dx dt t=1 = f(y) ∥∇f(y)∥2 ∇f(y)T . Proof. Deﬁne γ = f(y)/∥∇f(y)∥2. The result immediately follows from (5) and JHa x,λ(y, α0, 0)   γ∇f(y)T γα0α1 −γα0  =   f(y) 0 0  = −JHa t (y, α0, 0) where JHa t (y, α0, 0) is the vector corresponding to the Jacobian matrix of Ha with respect to t evaluated at (y, α0, 0). 6 Figure 1: Plot of the x coordinates of the solution path for the Rosenbrock polynomial on (a) the graph and (b) a contour plot. (a) (b) Since the path need not move in the gradient descent or ascent directions for t ̸= 1, this solution path avoids one of the drawbacks of the method of gradient descent, also known as the method of steepest descent (see [15, §3.2-3.3]), as shown in the following classical example. Example 7. Let f(x1, x2) = 100(x2−x2 1)2+(1−x1)2 be the Rosenbrock polynomial . Clearly, VR(f) = {(1, 1)} and f > 0 on R2 \ VR(f). For this polynomial, the iterative method of gradient descent has poor convergence due to this method using orthogonal steps to move through the “curved valley,” which surrounds the parabola x2 = x2 1 and contains the unique real root. Following , we took y = (−1.2, 1) so that the x coordinates of the solution path descend into the “curved valley” and then follow the valley around to the point (1, 1). Figure 1 plots the x coordinates of the solution path on the graph and on a contour plot. 2.2 Theory The following provides some theoretical results for gradient descent homotopies. Proposition 8. Let N ≥n, f : CN →Cn be a polynomial system with real coeﬃcients, y ∈RN such that f(y) ̸= 0, α ∈(R \ {0})n+1, and Ha be the homotopy deﬁned by (4). If the solution path deﬁned by Ha starting at (y, α0, 0, . . . , 0) ∈RN × Rn+1 is trackable on (0, 1] and converges as t →0 with endpoint (x∗, λ∗), then x∗∈VR(f) is a real critical point for (P). Proof. Since the homotopy Ha has real coeﬃcients and the start point is real, trackabil-ity and convergence immediately imply that every point on the path for t ∈[0, 1] is real. 7 Figure 2: Plot of the x1 coordinate of the solution path on the graph of f(x1, 0). Since α0 ̸= 0 and Ha(x∗, λ∗, 0) = 0, we know λ∗̸= 0. Therefore, if we consider λ∗∈Pn, this yields G(x∗, λ∗) = 0 where G is deﬁned in (2). Thus, by deﬁnition, x∗∈VR(f) is a real critical point for (P). Even though the endpoint is a real critical point for (P), the following example shows that it need not be the global minimizer of the distance measured from y. Example 9. Consider the polynomial f(x1, x2) = x2 2 + x2 1(x1 −1)(x1 −2) with y = (0.6, 0). It is easy to verify that the Euclidean distance between y and VR(f) is 0.4 which is attained at the point (1, 0) ∈VR(f). For the homotopy Ha deﬁned in (4) with α = (4, 2), the endpoint of the path starting at (0.6, 0, 4, 0) is (0, 0, 0, 2). That is, this path ended at the real critical point x∗= (0, 0) which is not the global minimizer of (P). Since every point on the path has an x2 coordinate of zero, Figure 2 plots the x1 coordinate of the solution path on the graph of f(x1, 0). The local maximum of f(x1, 0) between 0 and 1 occurs at β = (9 − √ 17)/8 ≈0.6096 which yields that the gradient descent path for y = (z, 0) will yield (0, 0) when 0 < z < β and (1, 0) when β < z < 1. Proposition 8 immediately yields the following. Corollary 10. Let N ≥n and f : CN →Cn be a polynomial system with real coef-ﬁcients such that VR(f) = ∅. For any y ∈RN and any α ∈Rn+1, the solution path deﬁned by the homotopy Ha from (4) starting at (y, α0, 0, . . . , 0) is either not trackable on (0, 1] or does not converge in CN × Cn+1 as t →0. The following examples demonstrate the two cases of Corollary 10. 8 Example 11. Consider the univariate polynomial f(x) = x2 + 1 for which VR(f) = ∅ with y = 1 and α = (3, 2). For the homotopy Ha deﬁned by (4), namely Ha(x, λ, t) =   x2 + 1 −2t λ0(x −1) + 2λ1x λ0 + 2λ1 −3  , it is easy to verify that the solution path starting at (1, 3, 0) is not trackable on (0, 1] due to a singularity at t = 1/2. Example 12. Consider the polynomial f(x1, x2) = x2 1 +(x1x2 −1)2 from for which VR(f) = ∅with y = (1, 1) and α = (3, 2). Clearly, f > 0 on R2, but 0 is the inﬁmum of f on R2 since lims→∞f(1/s, s) = 0. For the homotopy Ha deﬁned by (4), it is easy to verify that the solution path starting at (1, 1, 3, 0) is trackable on (0, 1], but does not converge in C2 × C2 as t →0. 2.3 Application to discriminants With only a few changes, the gradient descent homotopies can be applied to param-eterized polynomial systems for computing points on the real discriminant locus. For simplicity, we assume that r : CP →Cu is a polynomial system with real coeﬃcients such that P = V(r) is an irreducible algebraic set of dimension P −u. Let f : CN × P →CN be a polynomial system with real coeﬃcients such that (1) holds generically on P. That is, there is an algebraic set A ⊊P such that (1) holds for all p ∈P \ A. Given (y, q) ∈RN × RP such that q ∈RP \ ∆R(f), we will use gradient descent homotopies with the aim of computing a point on ∆R(f) of minimal distance to q. We note that q is not assumed to be in PR. Let Jxf(x, p) be the Jacobian matrix of f with respect to x evaluated at (x, p). Con-sider the polynomial system g : CN × CP →CN+u+1 having real coeﬃcients deﬁned by g(x, p) =   f(x, p) r(p) det Jxf(x, p)  . (6) One may attempt to use a gradient descent homotopy for g starting with (y, q). If successful, the limit point (x∗, p∗) ∈VR(g) is a critical point of the distance function deﬁned over RN × RP and x∗is a singular solution of f(x, p∗) = 0. However, since it is more natural to consider the critical points of the distance function deﬁned over RP , i.e., remove the dependence on x, we simply replace λ0 x −y p −q with λ0 0 p −q in the gradient descent homotopies. 9 Figure 3: Plot for (a) f(x) = x2 2 + x2 1(x1 −1)(x1 −2) and (b) g(x) = x2 2 −x2 1(x1 + 1). (a) (b) Example 13. Consider P = C3 with polynomial f(x, p) = p1x2+p2x+p3, y = 0.5, and q = (1, −3, 1). A gradient descent homotopy for g deﬁned in (6), where r(p) is simply removed since P = C3, starting with (y, q) yields, to four decimal places, x∗= 0.9172 and p∗= (1.4478, −2.6559, 1.2180). One can verify that (x∗, p∗) solves min{∥(x, p) −(y, q)∥2 \| f(x, p) = 0, p ∈∆R(f)} where ∥z∥2 = zT z for z ∈R4. After removing the dependency upon x, the gradient descent homotopy yields x∗= 1 and p∗= (4/3, −8/3, 4/3). One can verify that p∗solves min{∥p −q∥2 \| p ∈∆R(f)} where ∥w∥2 = wT w for w ∈R3. 2.4 Illustrative examples We conclude our discussion of gradient descent homotopies with illustrative examples. Example 14. Let f be as in Example 9 and g(x1, x2) = x2 2 −x2 1(x1 + 1). Clearly, the curve VR(f) is compact while VR(g) is unbounded. Figure 3 displays VR(f) and VR(g) along with 500 random points and their corresponding endpoint on the real curves connected by a straight line computed using a gradient descent homotopy. Example 15. For the Griewank-Osborne polynomial system f(x1, x2) = 29/16x3 1 −2x1x2 x2 −x2 1 , 10 Figure 4: Plot of ∆R(f) for (a) P = VC(p3 −1) and (b) P = V(p2 1 + p2 2/2 −p2 3 −1). (a) (b) Newton’s method diverges starting from every point in C2 \ {(0, 0)}. Using a graident descent homotopy starting at the point (1, 1), the path converges to (0, 0). We note that the path starting with (1, −1) is not trackable on (0, 1]. Example 16. Example 13 demonstrates computing points on the discriminant locus of a quadratic polynomial with P = C3. We used gradient descent homotopies starting at 1000 random points to compute points on the real discriminant locus using the plane P = V(p3 −1) ⊂C3 and the hyperboloid of one sheet P = V(p2 1 + p2 2/2 −p2 3 −1). The points obtained are displayed in Figure 4. Example 17. Let P = C2 and consider the polynomial system F : C2 × P →C2 from deﬁned by F(x, p) = x6 1 + p1x3 2 −x2 x6 2 + p2x3 1 −x1 . The discriminant locus ∆(F) is an algebraic curve of degree 90, which was studied in via A-discriminants. Figure 5 plots the points on ∆R(F) computed by using gradient descent homotopies starting at 2300 random points. 3 Changing the number of real solutions Dietmaier’s approach , which is summarized in Section 1.4, uses a local linearization approach to move through a real parameter space changing the number of real solutions. Due to the necessary ill-conditioning near the discriminant locus, we propose incorpo-rating a modiﬁed gradient descent homotopy into Dietmaier’s approach. The modiﬁed 11 Figure 5: Plot of ∆R(F). homotopies are presented in Section 3.1 and used in Sections 3.2 and 3.3 to increase and decrease, respectively, the number of real solutions. Throughout this section, we assume that P = CP so that PR = RP and f : CN × CP →CN is a polynomial system with real coeﬃcients such that f(x, p) = 0 has MCP ≥2 solutions, all of which are nonsingular, for general parameter values p ∈CP . 3.1 Modiﬁed gradient descent homotopies Let y ∈CN \ RN and q ∈RP \ ∆R(f) such that f(y, q) = 0. For the ﬁrst modiﬁed gradient descent homotopy, we aim to move q to the real discriminant locus so that the solution corresponding to y becomes a real singular solution. Let y = yr + iyi where yr, yi ∈RN and i = √−1. Consider g, h : CN × CN × CP →CN deﬁned by g(a, b, p) = f(a + ib, p) + f(a −ib, p) 2 and h(a, b, p) = f(a + ib, p) −f(a −ib, p) 2i . which are the real and imaginary parts of f(a + ib, p), respectively, with g(a, −b, p) = g(a, b, p), g(a, 0, p) = f(a, p), h(a, −b, p) = −h(a, b, p), h(a, 0, p) = 0. Let α ∈R2N+1 and consider Hc : CN × CP × C2N+1 × C →C3N+P+1 deﬁned by Hc(x, p, λ, t) =    g(x, t · yi, p) h(x, t · yi, p) λ0 0 p −q + PN j=1 λj ∇agj(x, t · yi, p)T ∇pgj(x, t · yi, p)T + PN j=1 λN+j ∇ahj(x, t · yi, p)T ∇phj(x, t · yi, p)T λ0 + α1λ1 + · · · + α2Nλ2N −α0    (7) 12 where ∇aF(a, b, p)T and ∇pF(a, b, p)T are the gradient vectors of a polynomial F with respect to a and p evaluated at (a, b, p), respectively, with start point (yr, q, α0, 0, . . . , 0). Suppose that this solution path is trackable and converges to (x∗, p∗, λ∗). Since the start point and the coeﬃcients of Hc are real, the endpoint is real, i.e., x∗∈RN and p∗∈RP . Since x∗arises as the limit of complex solutions to a family of real polynomial systems, it immediately follows that x∗is a real singular solution of f(x, p∗) = 0. Similarly, let y1, y2 ∈RN and q ∈RP \ ∆R(f) such that f(yj, q) = 0. For the second modiﬁed gradient descent homotopy, we aim to move q to the real discriminant locus so that the solutions corresponding to y1 and y2 coincide. Let α ∈R2N+1 and consider the homotopy Hr : CN × CP × C2N+1 × C →C3N+P+1 deﬁned by Hr(x, p, λ, t) =    f(x, p) f(x + t(y2 −y1), p) λ0 0 p −q + PN j=1 λj ∇xfj(x, p)T ∇pfj(x, p)T + PN j=1 λN+j ∇xfj(x + t(y2 −y1), p)T ∇pfj(x + t(y2 −y1), p)T λ0 + α1λ1 + · · · + α2Nλ2N −α0    (8) with start point (y1, q, α0, 0, . . . , 0). Suppose that this solution path is trackable and converges to (x∗, p∗, λ∗). As in the Hc case, since the start point and the coeﬃcients of Hr are real, the limit point is real, i.e., x∗∈RN and p∗∈RP . Since x∗arises as the limit of two real solutions to a family of real polynomial systems, it immediately follows that x∗is a real singular solution of f(x, p∗) = 0. 3.2 Increase the number of real solutions Given y ∈CN \ RN and q ∈RP \ ∆R(f) such that f(y, q) = 0, the ﬁrst part of our two-part approach for attempting to increase the number of real solutions is outlined in the following summary. 1. Attempt to use Hc deﬁned by (7) to yield a new parameter value such that the corresponding polynomial system has a real singular solution. (a) If successful, compute all solutions for the new parameter value. (b) If the real solutions have persisted, update the parameter values and solu-tions, and terminate the process. 2. If using Hc was not successful or some real solutions have disappeared, compute ∆q using the approach of Dietmaier attempting to move y closer to RN. (a) If ∆q can not be computed, the process is terminated. Otherwise, compute all solutions for the new parameter value. (b) If the real solutions have persisted, update the parameter values and the solutions, and return to Item 1. Otherwise, return to Item 2 computing a shorter ∆q. 13 If this process terminates, it has either computed a parameter p∗∈∆R(f) and a point x∗∈RN that is a singular solution of f(x, p∗) = 0, or has failed. If it has failed, the approach is repeated starting with a diﬀerent nonreal solution to f(x, q) = 0. If all nonreal solutions of f(x, q) = 0 fail, the procedure has failed. One could then restart the procedure after picking another value of q and a nonreal solution of f(x, q) = 0. If the ﬁrst part has succeeded, the second part attempts to produce new distinct real solutions and is outlined in the following summary. 1. Use points along the solution path tracked using Hc to compute a unit vector v ∈RN which is approximately tangent to the projection of path into the param-eter space at t = 0 and points in the direction of the path as t decreases. 2. Compute ∆p∗to be a nonzero vector in the direction of v. 3. Compute all solutions of f(x, p∗+ ∆p∗) = 0. (a) If the number of real solutions has increased, use the approach of Dietmaier to attempt to increase the distance between the real solutions and then ter-minate the process. Otherwise, return to Item 2 and compute a shorter ∆p∗. If successful, the two-part process has passed through the discriminant locus in such a way to increase the number of real solutions. 3.3 Decrease the number of real solutions The process for attempting to decrease the number of real solutions follows the same basic setup as the process for attempting to increase the number of real solutions in Section 3.2, with only a few minor changes. The ﬁrst part of the two-part process starts with two real solutions, y1 and y2, and uses the homotopy Hr deﬁned in (8). The persistence of the real solutions in Items 1b and 2b is replaced with the persistence of nonreal solutions. Also, the local linearization in Item 2 is setup to minimize the distance between y1 and y2. Upon failure, this part is repeated using a new pair of real solutions. For the second part, we replace Hc with Hr in Item 1. The only other change occurs in Item 3a where, if ∆p∗has been computed such that the number of nonreal solutions has increased, we use the approach of Dietmaier to attempt to increase the distance between the nonreal solutions and RN. 3.4 Illustrative example As an illustration, reconsider the polynomial f : C × C2 →C from Example 2, namely f(x, p) = x2 + p1x + p2. 14 We ﬁrst consider increasing the number real solutions starting with y = −1 + 2i and q = (2, 5). For α = (2, −5, 7), the homotopy Hc deﬁned by (7) yields x∗= −2 and p∗= (4, 4). The ﬁrst part was successful since x∗is a real singular solution of f(x, p∗) = 0. In the second part, we took, to three decimal places, v = (0.970, −0.243), which is the unit vector pointing in the direction of the vector between the parameter values at t = 10−5 and p∗. Taking ∆p∗= 10−4v, the equation f(x, p∗+ ∆p∗) = 0 has two real solutions separated by a distance of 0.030. We now consider decreasing the number of real solutions starting with y1 = −1, y2 = 3, and q = (−2, −3). For α = (2, −5, 7), the homotopy Hr deﬁned by (8) yields, to three decimal places, x∗= −0.388 and p∗= (−0.777, 0.151). The ﬁrst part was successful since x∗is a real singular solution of f(x, p∗) = 0. In the second part, we took, to three decimal places, v = (0.149, 0.989), which is the unit vector pointing in the direction of the vector between the parameter values at t = 10−5 and p∗. Taking ∆p∗= 10−4v, the equation f(x, p∗+∆p∗) = 0 has two complex solutions separated by a distance of 0.021. 4 Real enumerative geometry: points, lines, and conics The approach presented in Section 3 can be used to develop new results in real enu-merative geometry. To demonstrate, consider the geometric problem of computing the plane conics in C3 which meet k given points and 8 −2k given lines for k = 0, 1, 2. Table 1 shows the number of plane conics Nk when the k points and 8 −2k lines are in general position. Table 1: Number of plane conics k 2 1 0 Nk 4 18 92 By solving random instances of these problems, data is presented in showing that every possible number of real solutions can be achieved except • 18 nonreal solutions for k = 1, and • 92 real solutions for k = 0. The following sections summarize using the approach presented in Section 3 to reproduce some of these results as well as complete the k = 0 case. That is, this approach has indeed computed eight lines in R3 with 92 real plane conics meeting them. The following computations used Bertini to perform the the path tracking and updating of solutions, Matlab to solve the linear programs, and alphaCertified to certify the number of real solutions using rational arithmetic. We note that all numbers provided in the following sections are exact. The website of the last author contains additional details regarding the computations described below. 15 4.1 Two points and four lines We started with two randomly selected points (−0.506, 1.57, −6.01) and (0.725, 0.604, 2.84), and four randomly selected lines Li = {pi + tvi \| t ∈C} where p1 = (−0.297, 0.164, −0.846) v1 = (1.52, 1.69, −0.767) p2 = (−0.972, −1.32, 2.34) v2 = (−1.88, 69.9, 1.03) p3 = (0.475, −0.368, −0.863) v3 = (−0.696, 0.0421, 1.35) p4 = (1.1, 0.67, 0.902) v4 = (54.3, 0.0202, −3.24). There are four nonreal plane conics passing through these two points and meeting these four lines. The following summarizes using the approach of Section 3 to systematically increase the number of real solutions. Since there are four nonreal plane conics, the ﬁrst goal was to ﬁnd two points and four lines which describe two real and two nonreal plane conics. This was accomplished using the approach of Section 3 which yielded the two points (−0.469, 1.53, −6.05) and (0.738, 0.594, 2.8), and the four lines Li deﬁned by p1 = (−0.334, 0.201, −0.809) v1 = (1.48, 1.73, −0.793) p2 = (−1.01, −1.36, 2.3) v2 = (−1.85, 69.9, 1.03) p3 = (0.482, −0.405, −0.853) v3 = (−0.722, 0.00478, 1.32) p4 = (1.14, 0.644, 0.939) v4 = (54.3, 0.0462, −3.28). The next goal was to force the two remaining nonreal solutions to merge and then be-come two distinct real solutions. The two points (−0.416, 1.48, −6.1) and (0.756, 0.54, 2.76), and the four lines Li deﬁned by p1 = (−0.362, 0.251, −0.756) v1 = (1.47, 1.74, −0.772) p2 = (−0.963, −1.31, 2.33) v2 = (−1.86, 69.9, 1.02) p3 = (0.429, −0.459, −0.906) v3 = (−0.776, −0.049, 1.27) p4 = (1.17, 0.641, 0.979) v4 = (54.2, 0.5006, −3.32) were computed by the approach of Section 3. It is easy to verify that there are four real plane conics passing through these two points and meeting these four lines. 4.2 One point and six lines We started with the randomly selected point (−1.01, −0.011, 0.01) and six randomly selected lines Li = {pi + tvi \| t ∈C} where p1 = (1.6, 0.136, −4.43) v1 = (2.99, −2.16, 2.05) p2 = (−2.59, 4.38, 4.43) v2 = (−3.63, −0.01, −4.24) p3 = (1.1, 2.67, 0.9) v3 = (−1.38, 2.47, −2.13) p4 = (−4.77, 1.52, −1.31) v4 = (1.51, 1.1, 1.42) p5 = (0.204, 0.85, 1.44) v5 = (−1.69, −1.37, 0.987) p6 = (−4.6, −1.01, 0.841) v6 = (−2.01, 0.755, −0.436). 16 There are 4 real and 14 nonreal plane conics passing through this point and meeting these six lines. The approach of Section 3 was able to decrease the number of real solutions down to 2, but failed when trying to remove the remaining two real solutions. This computation is consistent with the results presented in Table 5 of . In particular, for the point (−1.01, 0.001, −0.001) and the six lines Li deﬁned by p1 = (1.59, 0.126, −4.44) v1 = (3.001, −2.16, 2.04) p2 = (−2.57, 4.39, 4.41) v2 = (−3.61, 0.0092, −4.26) p3 = (1.12, 2.65, 0.883) v3 = (−1.4, 2.49, −2.11) p4 = (−4.78, 1.51, −1.3) v4 = (1.49, 1.09, 1.44) p5 = (0.215, 0.839, 1.43) v5 = (−1.68, −1.36, 0.996) p6 = (−4.38, −1.01, 0.851) v6 = (−1.99, 0.745, −0.446), there are 2 real and 16 nonreal plane conics passing through this point and meeting these lines. 4.3 Eight lines We started with the eight randomly selected lines Li = {pi + tvi \| t ∈C} where p1 = (0.4096, −3.903, −2.287) v1 = (3.222, 1.433, −0.3969) p2 = (3.027, 0.5909, 3.208) v2 = (0.3143, 1.228, 4.478) p3 = (0.2573, 0.9133, 0.6372) v3 = (−3.261, −1.43, 1.502) p4 = (−4.276, 3.802, −1.097) v4 = (−0.926, 3.681, −2.706) p5 = (1.505, −0.7109, −4.401) v5 = (−3.741, 4.067, −2.589) p6 = (0.6752, −4.367, −2.557) v6 = (1.009, 1.772, −1.654) p7 = (−3.625, 3.66, 1.698) v7 = (−4.53, 1.966, 1.868) p8 = (1.444, 3.607, 0.5243) v8 = (−3.045, −2.643, −0.7563) for which there are 82 real and 10 nonreal plane conics meeting these eight lines. The approach of Section 3 systematically increased the number of real solutions up to 92. The following theorem is a restatement of Theorem 1 which includes the eight lines. Theorem 18. There are 92 real plane conics meeting the 8 lines Li = {pi + tvi \| t ∈C} deﬁned by p1 = (0.46978, −3.988, −2.3527) v1 = (2.9137, 1.546, −0.27448) p2 = (3.19, 0.5752, 3.0953) v2 = (0.56569, 1.108, 4.3629) p3 = (0.40308, 0.78659, 0.9053) v3 = (−3.0656, −1.4638, 1.4096) p4 = (−4.3743, 4.0046, −1.0243) v4 = (−0.9163, 3.6495, −2.6528) p5 = (1.5198, −0.86125, −4.5963) v5 = (−3.8418, 3.9541, −2.5494) p6 = (0.46801, −4.0308, −2.4411) v6 = (1.0225, 1.6422, −1.5925) p7 = (−3.3382, 3.8432, 1.693) v7 = (−4.4657, 1.9618, 1.6865) p8 = (1.3536, 3.6311, 0.42864) v8 = (−3.1442, −2.4915, −0.63586). 17 Proof. For i = 1, . . . , 8, let xi(t), yi(t), and zi(t) be the ﬁrst, second, and third coordi-nates of pi+tvi, respectively. Consider the polynomial system F : C16 →C16 deﬁned by F(t1, . . . , t8, c1, . . . , c5, a1, a2, a3) = a1xi(ti) + a2yi(ti) + a3 −zi(ti), i = 1, . . . , 8 C(xi(ti), yi(ti), c1, . . . , c5), i = 1, . . . , 8 where C(x, y, c1, . . . , c5) = c1x2 + c2xy + c3y2 + c4x + c5y −1. In particular, if (t, c, a) ∈ V(F) where t = (t1, . . . , t8), c = (c1, . . . , c5), and a = (a1, a2, a3), then the plane conic Ca,c = {(x, y, a1x + a2y + a3) ∈C3 \| c1x2 + c2xy + c3y2 + c4x + c5y = 1} passes through the point pi + tivi ∈Li ⊂C3 for i = 1, . . . , 8. Suppose that (t, c, a), (˜ t, ˜ c, ˜ a) ∈V(F) such that Ca,c = C˜ a,˜ c, and deﬁne Pa = V(a1x + a2y + a3 −z) and P˜ a = V(˜ a1x + ˜ a2y + ˜ a3 −z). If Pa ̸= P˜ a, then Ca,c is contained in the line Pa ∩P˜ a. However, since one can easily verify that no line meets the eight lines Li, we must have Pa = P˜ a which immediately implies that a = ˜ a. This shows that if (t, c, a), (˜ t, ˜ c, ˜ a) ∈V(F) such that a ̸= ˜ a, then Ca,c and C˜ a,˜ c are distinct plane conics meeting the eight lines Li. We used Bertini to compute a set of 92 points X ⊂C16, represented using ﬂoating point, such that each point in X is heuristically within 10−50 of a point in V(F). Since F is a square system with rational coeﬃcients, after approximating the coordinates of the points in X using rational numbers, the results of based on α-theory [4, 20] implemented in alphaCertified using exact rational arithmetic proved the following statements. • Newton’s method with respect to F starting at each point in X quadratically converges to a point in V(F). • If Z ⊂V(F) is the set of points which arise as the limit of Newton’s method with respect to F starting at some point in X, then Z consists of 92 distinct real points. This computation also proved that the maximum distance from each point in X to the corresponding point in Z ⊂V(F) is bounded above by 3 · 10−53. Since, for distinct (t, c, a), (˜ t, ˜ c, ˜ a) ∈X, we have ∥a −˜ a∥≥0.01, the triangle inequality with these two bounds show that the 92 points in Z indeed correspond to 92 distinct real conics passing through the eight given lines Li. Theorem 18 together with Table 6 of shows that, for ℓ= 0, 2, 4, . . . , 92, there exists 8 real lines such that there are ℓreal and 92−ℓnonreal plane conics meeting them. Acknowledgments Both authors would like to thank Frank Sottile for his help and support. The second author would also like to thank Chris Aholt and Rehka Thomas for related discussions. 18 References P. Aubry, F. Rouillier, and M. Safey El Din. Real solving for positive dimensional systems. J. Symbolic Comput., 34 (6), 543–560, 2002. D.J. Bates, D.A. Brake, and M.E. Niemerg. 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6425	https://www.statpearls.com/physician/cme/activity/89672/	CME Activity \| Mandible Osteoradionecrosis \| MDs & PAs PhysicianPhysician Board ReviewsPhysician Associate Board ReviewsCMELifetime CMEFree CMEMATE and DEA ComplianceCPD StudentUSMLE® Step 1USMLE® Step 2USMLE® Step 3COMLEX-USA® Level 1COMLEX-USA® Level 2COMLEX-USA® Level 396 Medical School ExamsStudent Resource Center NCLEX - RNNCLEX - LPN/LVN/PN24 Nursing Exams Nurse PractitionerAPRN/NP Board ReviewsCNS Certification ReviewsCE - Nurse PractitionerFREE CE NurseRN Certification ReviewsCE - NurseFREE CE PharmacistPharmacy Board Exam PrepCE - Pharmacist AlliedAllied Health Exam PrepDentist ExamsCE - Social WorkerCE - Dentist Point of Care Institutional Sales Student Resources Search Sign-upLogin PhysicianPhysician Board ReviewsPhysician Associate Board ReviewsCMELifetime CMEFree CMEMATE and DEA ComplianceCPD StudentUSMLE® Step 1USMLE® Step 2USMLE® Step 3COMLEX-USA® Level 1COMLEX-USA® Level 2COMLEX-USA® Level 396 Medical School ExamsStudent Resource Center NCLEX - RNNCLEX - LPN/LVN/PN24 Nursing Exams Nurse PractitionerAPRN/NP Board ReviewsCNS Certification ReviewsCE - Nurse PractitionerFREE CE NurseRN Certification ReviewsCE - NurseFREE CE PharmacistPharmacy Board Exam PrepCE - Pharmacist AlliedAllied Health Exam PrepDentist ExamsCE - Social WorkerCE - Dentist Point of Care Free CME/CE Mandible Osteoradionecrosis Home Physician MD/DO CME Mandible Osteoradionecrosis Overview 5 out of 5 (7 Reviews) Credits 1.00 Post-Assessment Questions 11 Start Date 1 Sep 2023 Last Review Date 1 Sep 2023 Expiration Date 31 Aug 2026 Estimated Time To Finish 60 Minutes Start This Activity Unlimited Physician CME Stay up to date on the latest medical knowledge with 6632 CME activities. In these online self-assessment activities, read our reference articles and test your knowledge with more than 7895.5 hours of CME. Learn About Lifetime CME Single Activity Take this single activity $49 1 activity Buy Now 6 Month Unlimited Physician CME Access to all the Unlimited Physician CME activities in all specialties. $329 per half year per user Buy Now 1 Year Unlimited Physician CME Access to all the Unlimited Physician CME activities in all specialties. $499 per 1 year per user Buy Now Need Help?If you have a system or content concerns, please contactsupport@statpearls.com Activity Description Osteoradionecrosis (ORN) of the mandible is a severe iatrogenic disease of devitalized bone caused by radiation therapy of oral and oropharyngeal cancers. It is a state of injured bone tissue with inadequate healing or remodeling response of at least three to six months. The wound can result from radiotherapy combined with mechanical insult or radiotherapy exposure alone. This activity reviews the evaluation, treatment, and complications of mandibular osteoradionecrosis and underscores the importance of an interprofessional team approach to its management. Target Audience This activity has been designed to meet the educational needs of physicians, physician associates, nurses, pharmacists, nurse practitioners, and dentists. Learning Objectives At the conclusion of this activity, the learner will be better able to: Describe the etiology and pathophysiology of mandibular osteoradionecrosis. Review the examination process for evaluation of a patient with mandible osteoradionecrosis, including indicated diagnostic imaging. Summarize the preventative strategies for and treatment of mandibular osteoradionecrosis. Explain modalities to improve care coordination among interprofessional team members in order to improve outcomes for patients affected by osteoradionecrosis of the mandible. 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6426	https://www.vedantu.com/maths/conversion-of-temperature	Courses for Kids Free study material Offline Centres Talk to our experts Maths Temperature Conversion Explained: Celsius, Fahrenheit & More Temperature Conversion Explained: Celsius, Fahrenheit & More Reviewed by: Rama Sharma Download PDF NCERT Solutions NCERT Solutions for Class 12 NCERT Solutions for Class 11 NCERT Solutions for Class 10 NCERT Solutions for class 9 NCERT Solutions for class 8 NCERT Solutions for class 7 NCERT Solutions for class 6 NCERT Solutions for class 5 NCERT Solutions for class 4 NCERT Solutions for Class 3 NCERT Solutions for Class 2 NCERT Solutions for Class 1 CBSE CBSE class 3 CBSE class 4 CBSE class 5 CBSE class 6 CBSE class 7 CBSE class 8 CBSE class 9 CBSE class 10 CBSE class 11 CBSE class 12 NCERT CBSE Study Material CBSE Sample Papers CBSE Syllabus CBSE Previous Year Question Paper CBSE Important Questions Marking Scheme Textbook Solutions RD Sharma Solutions Lakhmir Singh Solutions HC Verma Solutions TS Grewal Solutions DK Goel Solutions NCERT Exemplar Solutions CBSE Notes CBSE Notes for class 12 CBSE Notes for class 11 CBSE Notes for class 10 CBSE Notes for class 9 CBSE Notes for class 8 CBSE Notes for class 7 CBSE Notes for class 6 Step-by-Step Guide: Convert Between Celsius, Fahrenheit, and Kelvin Conversion of temperature is the process of changing the value of temperature from one unit to another. There are many methods of conversion of temperature, among them Kelvin, Celsius and Fahrenheit are the most frequently used methods. According to the Kelvin scale, the freezing point of water is 273.15K and the boiling point is 373.15K and according to the Fahrenheit scale, the freezing point of water is 32°F and the boiling point is 212°F. In the same way, according to the Celsius temperature scale, the freezing point of water is 0°C and the boiling point is 100°C. Celsius Scale The Celsius scale is also called the centigrade scale, and it is based on 0° C for the freezing point and 100° C for the boiling point of water. Initially, On the Celsius scale, 0° is used to denote the boiling point of water and 100° C to denote the freezing point of water. Further, these values were inverted 0° for the freezing point and 100° for the boiling point of water. This form of scale gained widespread use all over the world. Fahrenheit Scale This scale is based on 32° for the freezing point of the water and 212° for the boiling point of the water, the interval between the freezing point and the boiling point in being divided into 180 equal parts. The Fahrenheit temperature scale was first introduced in the 18th century by a German physicist named Daniel Gabriel Fahrenheit. Initially, he selected zero of his scale, as the temperature of the ice salt mixture. And then selected the values of 30° and 90° for the freezing point of water and normal body temperature, Later they were revised to 32° and 96°. But, the final scale requires an adjustment to 98.6° for the latter value. The Fahrenheit scale is used in the US and the Celsius or centigrade scale is used in most of the other countries, for scientific purposes worldwide. The conversion formula for a temperature that is expressed on the Celsius (°C) scale to its Fahrenheit (°F) formula is given below: °F = (9/5 × °C) + 32. Interactive Thermometer [Image will be Uploaded Soon] An interactive thermometer consists of the accompanying measures to illustrate the temperature as shown in the above figure. The scale is shown in both Celsius and Fahrenheit readings on either side. This ranges from 100°C to -40°C and is marked in intervals of ten degrees. A slider is dragged in the interactive thermometer to alter the measure. And note that the red ‘mercury’ level does not raise or lower on the interactive. Conversion of Temperature Temperature Conversion Formulas There are mainly three types of temperature conversion given below: Conversion Between Celsius and Kelvin Conversion Between Fahrenheit and Kelvin Conversion Between Celsius and Fahrenheit The Formula Used for Temperature Conversion of Celsius and Kelvin is: The formula used to convert temperature from Celsius to Kelvin is K = C + 273.15. The formula used to convert temperature from Kelvin to Celcius is C = K − 273.15. The Formula Used for Temperature Conversion of Fahrenheit and Celsius is: The conversion of the temperature from Fahrenheit to Celsius formula is C = (F − 32) × 5 ⁄ 9. conversion of the temperature from Celsius to Fahrenheit formula is F = C(9 ⁄ 5) + 32. The Formula Used for Temperature Conversion of Fahrenheit and Kelvin is: The formula used to convert temperature from Fahrenheit to Kelvin is, K = (F − 32) × 5 ⁄ 9 + 273.15. The formula used to convert temperature from Kelvin to Fahrenheit is, F = (K – 273.15) × 9 ⁄ 5 + 32. Fahrenheit to Celsius Chart Table \| \| \| \| --- \| Fahrenheit \| Celcius \| Description \| \| -459.67 °F \| -273.15 °C \| absolute zero temperature \| \| 50 °F \| 10.00 °C \| \| 32 °F \| 0 °C \| freezing/melting point of water \| \| 70 °F \| 21.11 °C \| room temperature \| \| 98.6 °F \| 37 °C \| average body temperature \| \| 212 °F \| 100 °C \| The boiling point of water \| \| 300 °F \| 148.89 °C \| Temperature Conversion Formula Table \| \| \| \| \| --- --- \| \| Unit \| To Celsius \| To Fahrenheit \| To Kelvin \| \| Celsius (C) \| C (°) \| C(9 ⁄ 5) + 32 \| C + 273.15 \| \| Fahrenheit \| (F − 32) × 5 ⁄ 9 \| F \| (F − 32) × 5 ⁄ 9 + 273.15 \| \| Kelvin \| K – 273.15 \| (K – 273.15) 9 / 5 + 32 \| K \| Solved Example Question Question 1: What is 30°C in Kelvin? Solution: The formula used to convert temperature from Celsius to Kelvin is, K = C + 273.15 K = 30 + 273.15 K = 303.15 Question 2: Convert 50°C to °F. Solution: Celsius to Fahrenheit conversion formula is given by: °F = °C(9 / 5) + 32 = 50 (9 / 5) + 32 = 90 + 32 = 122 Therefore, 50°C = 122°F Question 3: Convert 113°F to Kelvin. Solution: Fahrenheit to Kelvin conversion formula is given by: K = (F − 32) × 5 / 9 + 273.15 = (113 – 32) × (5 / 9) + 273.15 = 81 × (5 / 9) + 273.15 = 45 + 273.15 = 318.15 Therefore, 113°F = 318.15 K Question 4: Convert 225 K to Celsius. Solution: Kelvin to Celsius conversion formula is given by: C = K – 273.15 = 225 – 273.15 = -48.15 Therefore, 225 K = -48.15°C Recently Updated Pages Master Maths Concepts & Formulas \| Fast Learning GuideKnot Theory in Maths: Concepts, Applications & ExamplesMobius Strip Explained: Definition, Properties & UsesFermat’s Theorem: Meaning, Proofs, Examples & ApplicationsInfinitesimal in Maths: Concepts, Examples & ApplicationsFuzzy Logic in Maths: Concepts, Uses & Examples Explained Master Maths Concepts & Formulas \| Fast Learning GuideKnot Theory in Maths: Concepts, Applications & ExamplesMobius Strip Explained: Definition, Properties & Uses Fermat’s Theorem: Meaning, Proofs, Examples & ApplicationsInfinitesimal in Maths: Concepts, Examples & ApplicationsFuzzy Logic in Maths: Concepts, Uses & Examples Explained Trending topics 1 Billion in Rupees: Explained with Examples & ConversionComposite Numbers from 1 to 100: Full List ExplainedMetric Length Made Simple: SI Units & ConversionsHow to Find Percentage of Marks EasilyMaths Quiz Questions with Answers: Practice for Class 6 to 10 ExamsTemperature Conversion Explained: Celsius, Fahrenheit & More 1 Billion in Rupees: Explained with Examples & ConversionComposite Numbers from 1 to 100: Full List ExplainedMetric Length Made Simple: SI Units & Conversions How to Find Percentage of Marks EasilyMaths Quiz Questions with Answers: Practice for Class 6 to 10 ExamsTemperature Conversion Explained: Celsius, Fahrenheit & More Other Pages Ganesh Chaturthi 2025: History, Rituals, and How to CelebrateTeachers Day Speech 2025 in English for School CelebrationsOnam 2025: History, Significance, and How Kerala CelebratesImportant Days and Dates in AugustIndependence Day Speech in English for Students 2025Independence Day Essay for Students 2025 - Key Facts Explained Ganesh Chaturthi 2025: History, Rituals, and How to CelebrateTeachers Day Speech 2025 in English for School CelebrationsOnam 2025: History, Significance, and How Kerala Celebrates Important Days and Dates in AugustIndependence Day Speech in English for Students 2025Independence Day Essay for Students 2025 - Key Facts Explained
6427	http://hyperphysics.phy-astr.gsu.edu/hbase/Class/p3012.html	\| \| \| \| \| --- --- \| \| Chapter 4: Causes of Motion \| \| \| --- \| \| Describe Newton's Laws and their applications to motion. Given F=ma, calculate the motion of an object If a 5 kg object is acted upon by a force of 20 N, what is its acceleration? If it starts from rest, what will be its speed after 3 seconds? What is the difference between mass and weight? If and object has a mass of 2 kg, what is its weight under normal Earth gravity? If a crate weighs 120 pounds, what is its mass? Describe static and kinetic friction and their applications If the coefficient of static friction of a surface is 0.5 and the coefficient of kinetic friction is 0.4, then how much force will be required to set in motion a 40 kg crate? How much force will be required to keep it going at a constant velocity a after starting? If a large truck and a Volkswagen collide, which is acted upon by the larger force? If more than one force acts on an object, how do you predict its motion from Newton's second law? If a 10 kg object falls in the Earth's gravity but is acted upon by a 40 N air resistance force, what will be its acceleration? \| Motion Concepts Newton's Laws F = ma Mass and Weight Elevator Problem Application of F=ma Motion calculation Weight calculation Friction Force to overcome friction Newton's 3rd Law Combinations of forces \| \| Index \| \| HyperPhysicsPhysics 2010/3010 \| Go Back \| \| \| \| \| \| --- --- \| \| \| \| \| --- \| \| Chapter 5: Circular Motion Does motion in a circle at a constant speed involve an acceleration? What is the difference between centrifugal force and centripetal force? Which can you measure with a scale? Calculate the centripetal force on an object moving in a curved path, given the mass, velocity, and radius of curvature of the path. If a Ferris wheel ride has a diameter of 20 meters and a speed of 8 m/s, what is the centripetal acceleration of a rider? What is the centripetal force on a 70 kg rider? Why do you feel lighter at the top of the ride and heavier at the bottom? How does the force of gravity change with distance? If an object weighs 600 N at the Earth's surface, what will it weigh twice as far out from the Earth's center? ...three times as far out? What causes the tides? Why are there two per day instead of just one? \| Centripetal acceleration Vector velocities Centripetal force calculation Law of gravity Inverse square law Circular orbit Gravity in orbit Tides Sun Tide Jupiter effect Syncom satellites \| \| Index \| \| HyperPhysicsPhysics 2010/3010 \| Go Back \| \| \| \| \| \| --- --- \| \| \| \| \| --- \| \| Chapter 8: Rotational Motion What is torque? How can you increase the torque on a wrench to turn a stubborn bolt? What is required for equilibrium of a large object? Why do you pick up a long board at it's center? Why is it difficult to lift it by one end alone? Why does a skater spin faster if they pull their legs and arms in? What is the difference between centrifugal force and centripetal force? Which can you measure with a scale? \| Torque Equilibrium Force combination Lifting example Conservation of angular momentum Centripetal force Centrifugal force Coriolis force \| \| Index \| \| HyperPhysicsPhysics 2010/3010 \| Go Back \| \| \| \| \| \| --- --- \| \| Chapter 9: Fluids \| \| \| --- \| \| Define pressure - how does it differ from force? If the same force is applied to twice the area, what happens to the pressure? If the pressure of the atmosphere is about 15 pounds per square inch, and the top of your car is about 50 inches by 50 inches, what total force is exerted on the top of your car by the atmosphere? Why doesn't it crush the car top in? How do you find the pressure at a depth in a fluid? If you dive to a depth of 10 meters, how much additional pressure is exerted upon you? What is Pascal's Principle? What is a hydraulic press and what is it good for? How is Pascal's Principle used in the hydraulic press? Can a hydraulic press multiply force? Can it cause something to move further than the force which is applied to the press? Can it do both at the same time? What is the Bernoulli Principle? Is there really such a thing as a curveball pitch in baseball, or is it an optical illusion? How can the Bernoulli Principle cause a ball to curve? What is Archimedes' Principle? What was the basis for Archimedes' conclusion that the king's crown was not made out of solid gold? How can you make use of the buoyant force on an object to determine its density? If your heart were not pumping enough blood through an artery to supply your body's needs, what could be changed to increase the flowrate? If you could increase the blood pressure by 10%, or increase the radius of the artery by 10%, which would be more effective in increasing blood flow? If an artery started to expand because its wall was too weak to withstand the internal blood pressure, would that expansion relieve the stress or tension of the artery wall? What is the origin of the force which pushes the mercury up in a barometer? What force pushes a liquid up a drinking straw? What force moves air into your lungs when you breathe? \| Pressure definition Atmospheric pressure fluid pressure Pascal's principle hydraulic press hydraulic brakes auto lift Bernoulli equation Baseball curve Archimedes' principle buoyancy Poiseuille's law Blood pressure LaPlace's law barometer Drinking through straw Breathing pressure Ideal gas law \| \| Index \| \| HyperPhysics Physics 2010/3010 \| Go Back \|
6428	https://brainly.com/question/32761869	[FREE] What is the sample space? You toss a fair coin five times. a. What is the sample space if you record the - brainly.com Search Learning Mode Cancel Log in / Join for free Browser ExtensionTest PrepBrainly App Brainly TutorFor StudentsFor TeachersFor ParentsHonor CodeTextbook Solutions Log in Join for free Tutoring Session +37,8k Smart guidance, rooted in what you’re studying Get Guidance Test Prep +38,6k Ace exams faster, with practice that adapts to you Practice Worksheets +8k Guided help for every grade, topic or textbook Complete See more / Mathematics Textbook & Expert-Verified Textbook & Expert-Verified What is the sample space? You toss a fair coin five times. a. What is the sample space if you record the result of each toss (H or T)? b. What is the sample space if you record the number of heads? 1 See answer Explain with Learning Companion NEW Asked by lizzbugg3912 • 05/29/2023 0:01 / 0:15 Read More Community by Students Brainly by Experts ChatGPT by OpenAI Gemini Google AI Community Answer This answer helped 8683679 people 8M 0.0 0 Upload your school material for a more relevant answer Sample space can be defined as the set of all possible outcomes of an experiment. When you toss a fair coin five times, the sample space can be calculated as follows: a) Sample space if you record the result of each toss (H or T):The sample space is calculated by the formula 2^n, where n is the number of tosses. Here, the coin is tossed 5 times, so the sample space will be: 2^5 = 32. The 32 possible outcomes of the experiment are:HHHHH, HHHHT, HHHTH, HHHTT, HHTHH, HHTHT, HHTTH, HHTTT, HTHHH, HTHHT, HTHTH, HTHTT, HTTHH, HTTHT, HTTTH, HTTTT, THHHH, THHHT, THHTH, THHTT, THTHH, THTHT, THTTH, THTTT, TTHHH, TTHTH, TTHTT, TTTHH, TTTHT, TTTTH, TTTTT. b) Sample space if you record the number of heads:The sample space is calculated by the formula n + 1, where n is the maximum number of heads possible. Here, the coin is tossed 5 times, so the maximum number of heads is 5. Therefore, the sample space will be 5 + 1 = 6. The 6 possible outcomes of the experiment are:0 heads, 1 head, 2 heads, 3 heads, 4 heads, 5 heads. To know more about probability visit: brainly.com/question/11234923 SPJ11 Answered by LianaWarner •9.7K answers•8.7M people helped Thanks 0 0.0 (0 votes) Textbook &Expert-Verified⬈(opens in a new tab) This answer helped 8683679 people 8M 0.0 0 Quantitative Methods for Plant Breeding - Walter Suza Statistics - Barbara Illowsky, Susan Dean Introduction to Physics - Openstax Upload your school material for a more relevant answer The sample space for tossing a coin five times and recording each toss is 32 different sequences of H and T. When recording only the number of heads obtained, the sample space includes six outcomes: 0, 1, 2, 3, 4, and 5 heads. Thus, both scenarios provide a clear understanding of possible results from the experiment. Explanation To understand sample space, it's important to know it refers to the set of all possible outcomes of an experiment. Let’s break it down step by step for the case of tossing a fair coin five times: a. Sample space if you record the result of each toss (H or T): When you toss a coin once, it can land in one of two ways: Heads (H) or Tails (T). Since each toss is independent, when you toss the coin five times, the total number of outcomes can be calculated using the formula for the sample space of independent events, which is 2 n where n is the number of tosses. Here, n=5, so: 2 5=32 This means there are a total of 32 outcomes recorded as different sequences of H and T. Some examples of these sequences include: HHHHH HHHHT HTHTT TTTTT All these sequences represent different combinations of heads and tails across five tosses. b. Sample space if you record the number of heads: In this case, we are interested in how many heads appear regardless of the order. The maximum number of heads you could get in five tosses is 5 (if all tosses are heads) and the minimum is 0 (if all tosses are tails). Therefore, the possible outcomes representing the number of heads are: 0 heads 1 head 2 heads 3 heads 4 heads 5 heads This gives us a total of six different possible outcomes when recording just the number of heads. Examples & Evidence For example, if the results of the five coin tosses were HHTHT, it would contribute to the sample space of distinct sequences, while counting those results yields 3 heads, falling under the outcome '3 heads' in the second scenario. The calculations of sample space are based on principles of probability, where the total number of outcomes for independent events is determined by the formula 2 n, demonstrably accurate for sequences of binary results (heads and tails) from a fair coin. Thanks 0 0.0 (0 votes) Advertisement lizzbugg3912 has a question! Can you help? Add your answer See Expert-Verified Answer ### Free Mathematics solutions and answers Community Answer A random experiment consists of tossing a fair coin five times and recording the number of tails. What is the sample space? Community Answer 4.6 12 Jonathan and his sister Jennifer have a combined age of 48. If Jonathan is twice as old as his sister, how old is Jennifer Community Answer 11 What is the present value of a cash inflow of 1250 four years from now if the required rate of return is 8% (Rounded to 2 decimal places)? Community Answer 13 Where can you find your state-specific Lottery information to sell Lottery tickets and redeem winning Lottery tickets? (Select all that apply.) 1. Barcode and Quick Reference Guide 2. Lottery Terminal Handbook 3. Lottery vending machine 4. OneWalmart using Handheld/BYOD Community Answer 4.1 17 How many positive integers between 100 and 999 inclusive are divisible by three or four? Community Answer 4.0 9 N a bike race: julie came in ahead of roger. julie finished after james. david beat james but finished after sarah. in what place did david finish? Community Answer 4.1 8 Carly, sandi, cyrus and pedro have multiple pets. carly and sandi have dogs, while the other two have cats. sandi and pedro have chickens. everyone except carly has a rabbit. who only has a cat and a rabbit? Community Answer 4.1 14 richard bought 3 slices of cheese pizza and 2 sodas for $8.75. Jordan bought 2 slices of cheese pizza and 4 sodas for $8.50. How much would an order of 1 slice of cheese pizza and 3 sodas cost? A. $3.25 B. $5.25 C. $7.75 D. $7.25 Community Answer 4.3 192 Which statements are true regarding undefinable terms in geometry? Select two options. A point's location on the coordinate plane is indicated by an ordered pair, (x, y). A point has one dimension, length. A line has length and width. A distance along a line must have no beginning or end. A plane consists of an infinite set of points. New questions in Mathematics Hasmeet walks once round a circle with diameter 80 metres. There are 8 points equally spaced on the circumference of the circle. Find the distance Hasmeet walks between one point and the next point. f(x)=sec x Show that f′′(x)=2 sec 3 x−sec x Consider the following incomplete deposit slip: \| Description \| Amount ($) \| \| ---: \| Cash, including coins \| 150 \| 75 \| \| Check 1 \| 564 \| 81 \| \| Check 2 \| 2192 \| 43 \| \| Check 3 \| 4864 \| 01 \| \| Subtotal \| ???? \| ?? \| \| Less cash received \| ???? \| ?? \| \| Total \| 7050 \| 50 \| \| \| \| \| How much cash did the person who filled out this deposit slip receive? Which sequence is generated by the function f(n+1)=f(n)−2 for f(1)=10? A. −2,8,18,28,38,… B. 10,8,6,4,2,… C. 8,18,28,38,48,… D. −10,−12,−14,−16,−18,… i) Write down the expansion of (1+x)3. ii) Find the first four terms in the expansion (1−x)−4 in ascending powers of x. For what values is the expansion valid? iii) When the expansion is valid, find the values of a and b in the equation below. (1−x)4(1+x)3=1+7 x+a x 2+b x 3+…… Previous questionNext question Learn Practice Test Open in Learning Companion Company Copyright Policy Privacy Policy Cookie Preferences Insights: The Brainly Blog Advertise with us Careers Homework Questions & Answers Help Terms of Use Help Center Safety Center Responsible Disclosure Agreement Connect with us (opens in a new tab)(opens in a new tab)(opens in a new tab)(opens in a new tab)(opens in a new tab) Brainly.com Dismiss Materials from your teacher, like lecture notes or study guides, help Brainly adjust this answer to fit your needs. Dismiss
6429	https://artofproblemsolving.com/wiki/index.php/1988_USAMO_Problems/Problem_4?srsltid=AfmBOor9COz0C55m--kljJejcuu0uPPn5hkLb1H1Bmyg9Rz9uLrF4dQA	Art of Problem Solving 1988 USAMO Problems/Problem 4 - AoPS Wiki Art of Problem Solving AoPS Online Math texts, online classes, and more for students in grades 5-12. Visit AoPS Online ‚ Books for Grades 5-12Online Courses Beast Academy Engaging math books and online learning for students ages 6-13. Visit Beast Academy ‚ Books for Ages 6-13Beast Academy Online AoPS Academy Small live classes for advanced math and language arts learners in grades 2-12. Visit AoPS Academy ‚ Find a Physical CampusVisit the Virtual Campus Sign In Register online school Class ScheduleRecommendationsOlympiad CoursesFree Sessions books tore AoPS CurriculumBeast AcademyOnline BooksRecommendationsOther Books & GearAll ProductsGift Certificates community ForumsContestsSearchHelp resources math training & toolsAlcumusVideosFor the Win!MATHCOUNTS TrainerAoPS Practice ContestsAoPS WikiLaTeX TeXeRMIT PRIMES/CrowdMathKeep LearningAll Ten contests on aopsPractice Math ContestsUSABO newsAoPS BlogWebinars view all 0 Sign In Register AoPS Wiki ResourcesAops Wiki 1988 USAMO Problems/Problem 4 Page ArticleDiscussionView sourceHistory Toolbox Recent changesRandom pageHelpWhat links hereSpecial pages Search 1988 USAMO Problems/Problem 4 Contents 1 Problem 2 Solution 2.1 Solution 1 2.2 Solution 2 2.3 Solution 3 3 Solution 4 4 See Also Problem is a triangle with incenter . Show that the circumcenters of , , and lie on a circle whose center is the circumcenter of . Solution Solution 1 Let the circumcenters of , , and be , , and , respectively. It then suffices to show that , , , , , and are concyclic. We shall prove that quadrilateral is cyclic first. Let , , and . Then and . Therefore minor arc in the circumcircle of has a degree measure of . This shows that , implying that . Therefore quadrilateral is cyclic. This shows that point is on the circumcircle of . Analagous proofs show that and are also on the circumcircle of , which completes the proof. Solution 2 Let denote the midpoint of arc . It is well known that is equidistant from , , and (to check, prove ), so that is the circumcenter of . Similar results hold for and , and hence , , and all lie on the circumcircle of . Solution 3 Extend to point on . By The Incenter-Excenter Lemma, B, I, A are all concyclic. Thus, L is the circumcenter of triangle . In other words, , so is on . Similarly, we can show that and are on , and thus, are all concyclic. It follows that the circumcenters are equal. Solution 4 Let the centers be . We want to show that and have the same circumcircle. By Fact 5 we know that lie on a circle and similarly with the others. Thus the two triangles have the same circumcircle. ~coolmath_2018 See Also 1988 USAMO (Problems • Resources) Preceded by Problem 3Followed by Problem 5 1•2•3•4•5 All USAMO Problems and Solutions These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Retrieved from " Category: Olympiad Geometry Problems Art of Problem Solving is an ACS WASC Accredited School aops programs AoPS Online Beast Academy AoPS Academy About About AoPS Our Team Our History Jobs AoPS Blog Site Info Terms Privacy Contact Us follow us Subscribe for news and updates © 2025 AoPS Incorporated © 2025 Art of Problem Solving About Us•Contact Us•Terms•Privacy Copyright © 2025 Art of Problem Solving Something appears to not have loaded correctly. Click to refresh.
6430	https://www.unitconverters.net/numbers/base-6-to-decimal.htm	Convert Base-6 to Decimal Home / Numbers Conversion / Convert Base-6 to Decimal Convert Base-6 to Decimal Please provide values below to convert base-6 to decimal, or vice versa. From:base-6 To:decimal Popular Numbers Unit Conversions binary to decimal decimal to binary decimal to hex hex to decimal binary to hex hex to binary Convert Base-6 to Other Numbers Units Base-6 to Binary Base-6 to Octal Base-6 to Hexadecimal Base-6 to Break Base-6 to Base-2 Base-6 to Base-3 Base-6 to Base-4 Base-6 to Base-5 Base-6 to Base-7 Base-6 to Base-8 Base-6 to Base-9 Base-6 to Base-10 Base-6 to Base-11 Base-6 to Base-12 Base-6 to Base-13 Base-6 to Base-14 Base-6 to Base-15 Base-6 to Base-16 Base-6 to Base-17 Base-6 to Base-18 Base-6 to Base-19 Base-6 to Base-20 Base-6 to Base-21 Base-6 to Base-22 Base-6 to Base-23 Base-6 to Base-24 Base-6 to Base-25 Base-6 to Base-26 Base-6 to Base-27 Base-6 to Base-28 Base-6 to Base-29 Base-6 to Base-30 Base-6 to Base-31 Base-6 to Base-32 Base-6 to Base-33 Base-6 to Base-34 Base-6 to Base-35 Base-6 to Base-36 All Converters Common Converters LengthWeight and MassVolumeTemperatureAreaPressureEnergyPowerForceTimeSpeedAngleFuel ConsumptionNumbersData StorageVolume - DryCurrencyCase Engineering ConvertersHeat ConvertersFluids ConvertersLight ConvertersElectricity ConvertersMagnetism ConvertersRadiology ConvertersCommon Unit Systems about us \| terms of use \| privacy policy \| sitemap © 2008 - 2025 unitconverters.net
6431	https://math.stackexchange.com/questions/1456304/understanding-why-x-log-by-y-log-bx	logarithms - Understanding why $x^{\log_b(y)} = y^{\log_b(x)}$ - Mathematics Stack Exchange Join Mathematics By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy. Sign up with Google OR Email Password Sign up Already have an account? Log in Skip to main content Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. 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Upvoting indicates when questions and answers are useful. What's reputation and how do I get it? Instead, you can save this post to reference later. Save this post for later Not now Thanks for your vote! You now have 5 free votes weekly. Free votes count toward the total vote score does not give reputation to the author Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation. Got it!Go to help center to learn more Understanding why x log b(y)=y log b(x)x log b⁡(y)=y log b⁡(x) Ask Question Asked 10 years ago Modified5 years, 11 months ago Viewed 1k times This question shows research effort; it is useful and clear 2 Save this question. Show activity on this post. According to wikipedia, we have that x log b(y)=y log b(x)x log b⁡(y)=y log b⁡(x) because x log b(y)=b log b(x)log b(y)=b log b(y)log b(x)=y log b(x)x log b⁡(y)=b log b⁡(x)log b⁡(y)=b log b⁡(y)log b⁡(x)=y log b⁡(x) But what justifies that first leap? x log b(y)=b log b(x)log b(y)x log b⁡(y)=b log b⁡(x)log b⁡(y) logarithms Share Share a link to this question Copy linkCC BY-SA 3.0 Cite Follow Follow this question to receive notifications asked Sep 29, 2015 at 11:36 PP1211PP1211 31 2 2 bronze badges 1 1 Take l o g b l o g b of both sides of the original equation.lulu –lulu 2015-09-29 11:39:18 +00:00 Commented Sep 29, 2015 at 11:39 Add a comment\| 5 Answers 5 Sorted by: Reset to default This answer is useful 4 Save this answer. Show activity on this post. It starts from the fact that x x is equal to b log b(x)b log b⁡(x). To elaborate, the log to the base b b of x x is defined as the number which, when b b is raised to this number, gives back x x. Now, since x=b log b(x)x=b log b⁡(x), we can always replace any occurrence of x x in any equation whatsoever with b log b(x)b log b⁡(x). So, breaking that first leap into two steps, we first have: x l o g b(y)=(b log b(x))l o g b(y)x l o g b(y)=(b log b⁡(x))l o g b(y) From here, we can apply a known rule of exponents which is that (s r)t=s r t(s r)t=s r t. Applying that rule we get: (b log b(x))l o g b(y)=b log b(x)l o g b(y)(b log b⁡(x))l o g b(y)=b log b⁡(x)l o g b(y) Which is the first step in the above. Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Follow Follow this answer to receive notifications edited Oct 25, 2019 at 14:27 answered Sep 29, 2015 at 11:38 Colm BhandalColm Bhandal 4,847 16 16 silver badges 40 40 bronze badges Add a comment\| This answer is useful 3 Save this answer. Show activity on this post. Notice that a b=e b ln(a)a b=e b ln⁡(a) and log c(d)=ln(d)ln(c)log c⁡(d)=ln⁡(d)ln⁡(c). Therefore: x log b(y)=e log b(y)ln(x)=e ln(y)ln(b)ln(x)=e ln(y)ln(x)ln(b)=e ln(y)log b(x)=y log b(x)x log b⁡(y)=e log b⁡(y)ln⁡(x)=e ln⁡(y)ln⁡(b)ln⁡(x)=e ln⁡(y)ln⁡(x)ln⁡(b)=e ln⁡(y)log b⁡(x)=y log b⁡(x) Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications edited Apr 21, 2017 at 15:12 user21820 61.1k 9 9 gold badges 109 109 silver badges 282 282 bronze badges answered Sep 29, 2015 at 11:39 SurbSurb 57.4k 11 11 gold badges 68 68 silver badges 119 119 bronze badges 1 1 Your original had a "x x" instead of a "y y", probably because it was too tiny too be noticed. I fixed that and made the fractions bigger. =)user21820 –user21820 2017-04-21 15:13:26 +00:00 Commented Apr 21, 2017 at 15:13 Add a comment\| This answer is useful 1 Save this answer. Show activity on this post. Their proof is kind of crappy. Just take the log of both sides of the equation, and you'll get log(x)log(y)=log(y)log(x) Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications answered Sep 29, 2015 at 11:40 Jerry GuernJerry Guern 2,782 1 1 gold badge 17 17 silver badges 31 31 bronze badges Add a comment\| This answer is useful 1 Save this answer. Show activity on this post. For an intuitive (purposefully non-rigorous) reason, consider writing logarithms in an infix notation, as a log–––b a log _ b rather than log b a log b⁡a. Then the relation is x^(y log–––b)=y^(x log–––b)x^(y log _ b)=y^(x log _ b) which is just one arithmetic step up from a relation you probably already know: x×(y÷b)=y×(x÷b)x×(y÷b)=y×(x÷b) which itself is also just one arithmetic step up from: x+(y−b)=y+(x−b)x+(y−b)=y+(x−b) It makes a nice pattern. Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications answered Sep 29, 2015 at 12:07 DanielVDanielV 24.6k 5 5 gold badges 41 41 silver badges 70 70 bronze badges 1 Hello! I strongly urge you to either delete this answer or to explain in full the analogy. The reason is that (as you definitely know) one cannot do mathematics by following intuitive patterns. Worse still, in this case it isn't at all obvious what the pattern is, and exponentiation is not even commutative. There is a deep algebraic reason (group actions and their inverses), but maybe not one that should be gotten into at this level. Agree? =)user21820 –user21820 2017-04-21 15:28:18 +00:00 Commented Apr 21, 2017 at 15:28 Add a comment\| This answer is useful -1 Save this answer. Show activity on this post. Method-1: Notice, we know that 1 log m(n)=log n(m)1 log m⁡(n)=log n⁡(m)&m log m(n)=n m log m⁡(n)=n Now, we have x log b(y)=y log b(x)x log b⁡(y)=y log b⁡(x) ⟺x 1 log b(x)=y 1 log b(y)⟺x 1 log b⁡(x)=y 1 log b⁡(y) ⟺x log x(b)=y log y(b)⟺x log x⁡(b)=y log y⁡(b) ⟺b=b⟺b=b Method-2: we have x log b(y)=y log b(x)x log b⁡(y)=y log b⁡(x) taking log with base x x on both the sides, we get log x(x log b(y))=log x(y log b(x))log x⁡(x log b⁡(y))=log x⁡(y log b⁡(x)) log b(y)log x(x)=log b(x)log x(y)log b⁡(y)log x⁡(x)=log b⁡(x)log x⁡(y) log b(y)log b(x)=log x(y)log b⁡(y)log b⁡(x)=log x⁡(y) log b(y)⋅log x(b)=log x(y)log b⁡(y)⋅log x⁡(b)=log x⁡(y) log x(y)=log x(y)log x⁡(y)=log x⁡(y) Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications answered Sep 29, 2015 at 12:21 Harish Chandra RajpootHarish Chandra Rajpoot 38.6k 135 135 gold badges 85 85 silver badges 120 120 bronze badges 1 Your answer is simply wrong when x=1 x=1.user21820 –user21820 2017-04-21 15:16:15 +00:00 Commented Apr 21, 2017 at 15:16 Add a comment\| You must log in to answer this question. 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Featured on Meta Introducing a new proactive anti-spam measure Spevacus has joined us as a Community Manager stackoverflow.ai - rebuilt for attribution Community Asks Sprint Announcement - September 2025 Report this ad Related 5Why is log(b,n)=⌊log b(n)⌋log⁡(b,n)=⌊log b⁡(n)⌋ primitive recursive? 2What is the value of log b m log b n log b⁡m log b⁡n? 1Proving that log b(r s)=s log b(r)log b⁡(r s)=s log b⁡(r) 3trouble understanding when to use b log b(x)=x b log b⁡(x)=x 0If a>1 a>1, b>1 b>1 and p=log b(log b a)log b a p=log b⁡(log b⁡a)log b⁡a, then what's a p a p? 1x log b a=2 x log b⁡a=2 ; find x x. 0Proving log a x log c x=log a x−log b x log b x−log c x log a⁡x log c⁡x=log a⁡x−log b⁡x log b⁡x−log c⁡x, when b 2=a c b 2=a c 0How Can Solving (2 n)(log b 2)=(5 n)(log b 5)(2 n)(log b⁡2)=(5 n)(log b⁡5) be Generalized for any Base of log b n log b⁡n? 1Finding the Taylor Series of log b(x)log b⁡(x) Hot Network Questions How to rsync a large file by comparing earlier versions on the sending end? 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6432	https://www.youtube.com/watch?v=4xHPDxVrJqc	Photosynthesis: Part 5: Light Reactions \| HHMI BioInteractive Video biointeractive 263000 subscribers 1960 likes Description 187164 views Posted: 23 Jan 2019 This animation provides a detailed look at the chemical processes in the light reactions of photosynthesis. It is appropriate for advanced high school biology through college-level students. The light reactions of photosynthesis transform light energy into chemical energy stored in molecules of ATP and NADPH. Chlorophyll molecules absorb photons of light in two photosystems, which are embedded in the thylakoid membrane of the chloroplast. The photons energize electrons in the chlorophyll molecules, which drives the electrons through a series of oxidation and reduction reactions in photosystem II. Meanwhile, more electrons are released by the splitting of water molecules, which produces oxygen as a byproduct. The energy from the light-energized electrons generates a proton (hydrogen ion) gradient. As protons flow through ATP synthase, ATP is formed. Photons energize electrons again at photosystem I, eventually forming the NADPH molecule. Both ATP and NADPH provide energy for making sugars in the Calvin cycle. To view the rest of this series on photosynthesis, go to the full playlist at To view a specific part of the series, use the following links: Part 1: Overview Part 2: Chemical Process Part 3: Leaf Structure Part 4: Chloroplasts Part 5: Light Reactions Part 6: Calvin Cycle Part 7: Biosynthesis To view the entire series as a full-length animation, use the following link: To download these animations and discover related material, go to 42 comments Transcript: The thylakoid membranes contain specialized molecules that work together to perform the light reactions. Light is absorbed by protein-pigment complexes called photosystems. There are two photosystems: photosystem I and photosystem II. The photosystems transform light energy to chemical energy by exciting and then shuttling electrons from molecule to molecule in a chainlike fashion on the thylakoid membrane. This process is called an electron transport chain. Let’s take a closer look to see how this process works. First, photons of light hit chlorophyll, a light-absorbing pigment in photosystem II. Electrons in the chlorophyll are excited to a higher energy level. The excited electrons are passed to an electron carrier. Meanwhile, water splits and releases electrons. These electrons replace those lost at photosystem II. The byproduct of this reaction is oxygen, which is eventually released into the air. The other products are protons, or hydrogen ions, which are released into the inside of the thylakoid, or lumen. The excited electrons move to the cytochrome complex. Some of the energy from the electrons is used by the cytochrome complex to transport additional protons into the lumen. The second electron carrier, a protein inside the lumen, receives the electrons and passes them to photosystem I. These electrons have now lost most of the energy they gained from light in photosystem II. Photons of light hit chlorophyll in photosystem I and excite the electrons again. The electrons are then passed to the third electron carrier. Finally, these electrons are either recycled or they interact with an enzyme and NADP+, the final electron acceptor of the light reactions, to form NADPH. Some of the energy from light is now stored in the reduced molecule NADPH. Some of the energy released from the transfer of electrons established a proton gradient across the thylakoid membrane. Protons that accumulated in the lumen diffuse into the stroma through an enzyme called ATP synthase. ATP synthase uses the potential energy of the proton gradient to combine ADP with inorganic phosphate to form ATP. In this way, the potential energy is transformed into chemical energy stored as ATP. ATP and NADPH now have stored energy from the light reactions. This energy can be used in the Calvin cycle. This light driven electron transport chain is usually continuous in the presence of sunlight. It encompasses a series of chemical reactions that involve light absorption, energy conversion and electron transfer carried out by the photosystems and other enzymes on the membrane of the thylakoids.
6433	https://app.jove.com/science-education/v/11147/ideal-gas-law-derivation-assumptions-and-the-dumas-method-concept	Video: Ideal Gas Law - Concept We value your privacy We use cookies to enhance your browsing experience, serve personalised ads or content, and analyse our traffic. By clicking "Accept All", you consent to our use of cookies.Cookie Policy Customise Accept All Customise Consent Preferences We use cookies to help you navigate efficiently and perform certain functions. You will find detailed information about all cookies under each consent category below. The cookies that are categorised as "Necessary" are stored on your browser as they are essential for enabling the basic functionalities of the site. ...Show more Necessary Always Active Necessary cookies are required to enable the basic features of this site, such as providing secure log-in or adjusting your consent preferences. These cookies do not store any personally identifiable data. 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Save My Preferences Accept All Research AllBehaviorBiochemistryBioengineeringBiologyCancer ResearchChemistryDevelopmental BiologyEngineeringEnvironmentGeneticsImmunology and InfectionMedicineNeuroscience JoVE JournalJoVE Encyclopedia of ExperimentsJoVE Visualize Education AllBiologyChemistryClinicalEngineeringEnvironmental SciencesPharmacologyPhysicsPsychologyStatistics JoVE CoreJoVE Science EducationJoVE Lab Manual JoVE QuizJoVE Playlists Business AllFinanceMarketingMicroeconomics All All Research Education en EN - English CN - 中文 DE - Deutsch ES - Español KR - 한국어 IT - Italiano FR - Français PT - Português TR - Türkçe JA - 日本語 PL - Polski RU - Русский HE - עִברִית AR - العربية Sign In Start Free Trial JoVE Lab Manual> Chemistry> Ideal Gas Law ConceptsInstructor PrepStudent Protocol JoVE Lab Manual Lab: Chemistry A subscription to JoVE is required to view this content. Sign in to start your free trial. live 00:00 00:00 1x Speed × 0.75x 1x 1.5x 2x CC Subtitles × MEDIA_ELEMENT_ERROR: Format error Ideal Gas Law Embed Video Add to Playlist en EN - EnglishCN - 中文DE - DeutschES - EspañolKR - 한국어IT - ItalianoFR - FrançaisPT - PortuguêsTR - TürkçeJA - 日本語PL - PolskiRU - РусскийHE - עִברִיתAR - العربية Text Related Derivation of the Ideal Gas Law Gases are a fundamental state of matter. A gas is a collection of molecules that have a significant distance between their molecules. Due to this distance, colorless gases are invisible to the human eye and are studied using four measurable parameters: pressure (P), volume (V), number of moles (n), and temperature (T). The ideal gas law is a mathematical equation that relates all of these parameters. It is a combination of several different laws that describe the behavior of gases. In 1662, Robert Boyle confirmed a prior discovery relating the pressure of a gas to its volume. Boyle’s law states that the pressure of a gas is inversely proportional to its volume if the temperature and number of moles of the gas are held constant. Boyle’s law can be extended to calculate the new pressure or volume of a gas if the initial pressure and volume are known. In the 1780s, the unpublished work of French scientist Jacques Charles was credited by French scientist Joseph Louis Gay-Lussac for describing the direct relationship between the volume and temperature of a gas. Charles’s law allows us to calculate the new volume or temperature of a gas if the initial volume and temperature are known, and the pressure and number of moles are constant. Joseph Louis Gay-Lussac provided an extension to Charles’s law by relating pressure and temperature. Gay-Lussac’s law establishes that the pressure of an enclosed gas is directly proportional to its temperature. Therefore, if a change is applied to a gas at a constant volume and number of moles, the new pressure or temperature can be calculated if the initial pressure and temperature are known. Finally, in 1811, Amedeo Avogadro proposed the direct proportionality between the volume of a gas and the number of moles present. The law describes how equal volumes of two gases, with the same temperature and pressure, contain an equal number of molecules. All of these relationships combine to form the ideal gas law, first proposed by Emile Clapeyron in 1834, as a way to combine these laws of physical chemistry. The ideal gas law accounts for pressure (P), volume (V), moles of gas (n), and temperature (T), with an added proportionality constant, the ideal gas constant (R). The universal gas constant, R, is equal to 8.314 J·K-1 mol-1. Assumptions of the Ideal Gas Law The ideal gas law assumes that gases behave ideally, meaning they adhere to the following characteristics: (1) the collisions occurring between molecules are elastic and their motion is frictionless, meaning that the molecules do not lose energy; (2) the total volume of the individual molecules is magnitudes smaller than the volume that the gas occupies; (3) there are no intermolecular forces acting between the molecules or their surroundings; (4) the molecules are constantly in motion, and the distance between two molecules is significantly larger than the size of an individual molecule. As a result of all these assumptions, an ideal gas would not form a liquid at room temperature. However, as we know, many gases become liquids at room temperature and therefore deviate from ideal behavior. In 1873, Johannes D. Van der Waals modified the ideal gas law to account for the molecular size, intermolecular forces, and volume that define real gases. In the Van der Waals equation, parameters a and b are constants that can be determined experimentally and differ from one gas to another. Parameter a will experience larger values for gases with strong intermolecular forces (i.e., water) and smaller values for gases that have weak intermolecular forces (i.e., inert gases). Parameter b represents the volume that 1 mole of gas molecules occupies; thus, when b decreases, the pressure increases as a result. The Dumas Method Invented by Jean Baptiste Andre Dumas, the Dumas method utilizes the ideal gas law to study gas samples. The ideal gas law includes Avogadro’s law, where the number of moles of two gas samples occupying the same volume is the same at a constant pressure and temperature. This relationship allows the Dumas method to calculate the molar mass of an unknown gas sample. To accomplish this, a Dumas tube is used. A Dumas tube is an elongated glass bulb with a long capillary neck. Prior to the experiment, the volume and mass of the tube are measured. Then, a small amount of a volatile compound is placed in the Dumas tube. Volatile compounds have a high vapor pressure at room temperature and are vaporized at low temperatures. Thus, when the Dumas tube containing the volatile liquid is placed in boiling water, the liquid vaporizes and forces the air out of the tube, and the tube is solely filled with vapor. When the tube is removed from the water bath and left at room temperature, the vapor condenses back to a liquid. Since mass is conserved, the mass of the liquid in the tube is equal to the mass of the gas in the tube. Using the known mass and volume of the gas, along with the known water bath temperature and room pressure, the moles and therefore molecular weight of the gas can be calculated using the ideal gas law. Here, three assumptions are made: (1) the vapor is acting ideally, (2) the volume of the tube does not vary between the room temperature and the working temperature, and (3) the gas and the water bath are at thermal equilibrium. References Kotz, J.C., Treichel Jr, P.M., Townsend, J.R. (2012) Chemistry and Chemical Reactivity. Belmont, CA: Brooks/Cole, Cengage Learning. Gay-Lussac, J.L. (1809). Memoir on the Combination of Gaseous Substances with Each Other. Mémoires de la Société d'Arcueil, Vol. 2, 207. Van der Waals, J.D. (1967). The equation of state for gases and liquids. Nobel Lectures, Physics. Elsevier: Amsterdam, pp. 254-265. Silderberg, M.S. (2009). Chemistry: The Molecular Nature of Matter and Change. Boston, MA: McGraw Hill. Transcript A gas is simply a dispersed sample of matter that is fluid and expands freely to occupy available space. However, a certain number of gas molecules occupy a specific volume under a defined temperature and pressure. We can describe the behavior of a gas under these parameters using the ideal gas law, which uses the universal gas constant, R, to relate all of these variables. The universal gas constant is equal to 8.314 joules per mole Kelvin. This equation enables us to understand state relationships in a gaseous system. For example, in a system of constant temperature and pressure, we know that the addition of more moles of gas results in an increase in volume. Similarly, we can look at a system of constant temperature and moles and see that a decrease in volume results in an increase in pressure. One challenge is that the ideal gas law describes gases behaving ideally. So what do we mean by that? Ideal behavior assumes that first, the molecules themselves are infinitesimally small and essentially have no volume and that the distance between the molecules is significantly larger than the size of the individual molecule. Second, we assume that the molecules are constantly in motion. Any collisions occurring between the molecules are elastic, and their motion is frictionless, meaning that the molecules do not lose energy. Finally, we assume that there are no intermolecular forces acting between the molecules and their surroundings. Unfortunately, most gases do not behave ideally. At very low temperature or high pressure, molecules are very close together and slow-moving, so intermolecular interactions are significant. Similarly, gases with a high molecular weight experience increased interactions due to their large size and mass. However, the ideal gas relationship serves as a good approximation in general. So how do we use the ideal gas law to study the behavior of a gas in the laboratory? Pressure, volume, and temperature are generally more easily measured, but how about moles, and by extension, mass? One of the simplest ways to measure the mass of a gas is by the Dumas method. To perform this test, a small amount of a volatile compound in its liquid phase is placed inside a Dumas tube, and the tube is then placed in boiling water. A volatile compound has a high vapor pressure at room temperature. The vapor pressure is the pressure exerted by a vapor in equilibrium with its liquid phase. Thus, a volatile compound with high vapor pressure transitions from liquid to gas rapidly. When this happens, the newly formed gas forces the air out of the Dumas tube so that it is solely filled with gas. Once the tube is removed from the water bath and left at room temperature, the gas condenses to form a liquid again. Since mass is conserved, we know that the mass of the condensed liquid is equal to the mass of the gas that filled the known volume of the Dumas tube. In this lab, you'll explore the ideal gas law by using the Dumas Method to determine the molar mass of an unknown volatile substance. You'll then measure the temperature, pressure, and volume of the system and see how much this gas deviates from ideality. Tags JoVE Lab ChemLab 37 Concept Research BehaviorBiochemistryBioengineeringBiologyCancer ResearchChemistryDevelopmental BiologyEngineeringEnvironmentGeneticsImmunology and InfectionMedicineNeuroscience JoVE JournalJoVE Encyclopedia of Experiments JoVE VisualizeMethods CollectionsArchive Education JoVE CoreJoVE Science EducationJoVE Lab Manual Faculty Site Business FinanceMarketingMicroeconomics JoVE Business Highschools Authors OverviewPublishing ProcessEditorial BoardScope and PoliciesPeer ReviewFAQSubmit Librarians OverviewTestimonialsSubscriptionsAccessResourcesLibrary Advisory BoardFAQ About JoVE OverviewLeadershipBlogJoVE Help CenterJoVE Sitemap Contact UsRecommend to libraryJoVE Newsletters Privacy policyTerms of UsePolicies Copyright © 2025 MyJoVE Corporation. All rights reserved
6434	https://open.oregonstate.education/animalnutrition/chapter/chapter-18/	Skip to content XVIII. Water in Animal Nutrition This chapter discusses the role and importance of water in food-producing animals. New Terms Metabolic water Polioencephalomalacia Total dissolved solids Water quality Chapter Objective To introduce and discuss the physiological role, requirement, and quality of water for maintaining animal health and productivity Water is essential for sustaining life and ranks second to oxygen in importance. Water is needed in greater quantity than any other orally ingested substance and is classified as a macronutrient. Sources of water include drinking water, metabolic water (produced during catabolism of carbohydrates, fats, and proteins to carbon dioxide and water), and the water that presents as moisture in different feed ingredients. Metabolic water serves as the sole source of water in desert and hibernating animals, and feed water is the major water source for marine animals. Where Animals Obtain Water Feed Drinking water Metabolic water Why Is Water Important for Livestock? Water is important for all organisms. Water makes up one-half to two-thirds of the body mass of adult animals and more than 90% of the body mass of newborn animals. Water is an essential constituent of almost all secretions of the body. Within the body, water is a universal solvent that facilitates cellular biochemical reactions involving digestion, absorption, and transportation of nutrients. The aqueous medium of water helps different digestives juices and food components interact, enhancing digestion, and helps in the excretion of waste products in the form of urine, feces, and perspiration, sweat, from the animal body. Because of the high specific heat of water, it helps in regulating body temperature, by absorbing the heat generated through different metabolic reactions. Water also regulates body temperature through evaporation as sweat or transports heat away from organs through blood. Water provides shape to body cells. Water helps in maintaining the acid-base balance of the body. Water acts as a cushion for tissue cells and the nervous system and protects the various vital organs against shocks and injuries. Water is the single most important nutrient in the animal body. It is essential for all metabolic processes, chemical reactions, temperature regulation, eliminating waste from the body, and ultimately, health and survival. Water Sources and Losses The animal body derives water from different sources. These include drinking water, water present as part of feeds (moisture), or those liberated during several metabolic reactions. The importance of these different sources varies among species, habitat, and diet. For example, hibernating animals and desert rodents depend on metabolic water to keep them alive, whereas marine animals depend on their diet to derive their water requirements. Metabolic water also depends on the type of nutrient catabolized. Oxidation of fat produces the greatest amount of metabolic water. However, overall contribution of metabolic water to daily water needs is less than 5% to 10% in most animals. The water content of feed consumed by ruminant and nonruminant animals vary highly. Forages consumed by ruminant animals vary from 5% to 7% for mature plant products, such as hay, to more than 90% for young lush green vegetation. Animals such as sheep depend a lot on water derived from such green forages for their need. Most commercial diets fed to nonruminant animals such as pigs and poultry may contain 7% to 10% moisture. Some of the canned foods fed to pet animals such as dogs and cats may contain more than 75% moisture. All animals experience daily water loss through different venues such as urine, feces, sweat, saliva, evaporation from the lungs through respiration, and milk in lactating animals. Among these, urinary loss accounts for the major loss. Water lost through urine serves as a tool to dispose of the toxic products of metabolism. Some animals such as birds are capable of concentrating urine and excreting it as uric acid instead of urea and thus conserving water. Urinary water loss depends on weather and on the type of food consumed. Consumption of excess water during heat stress can increase urinary volume. Animals that consume high-fibrous diets excrete more water in their feces. Fecal water excretion is higher in cows (30%–32%) compared to sheep (13%–24%) that void pellet-type dry feces to minimize water loss. Sweating is a means to dissipate body heat. In animals such as horses, loss of water through sweating is high. Animals such as chickens and dogs have very poorly developed sweat glands and compensate for heat loss by panting and increasing water intake. Daily clean water consumption is needed to make up for all the losses and is extremely important during periods of heat stress, especially in animals such as poultry. \| \| \| \| --- \| Animal \| L/day \| Gal/day \| \| Beef cattle \| 26-66 \| 6.5-17 \| \| Dairy cattle \| 38-110 \| 9.5-28 \| \| Horses \| 30-45 \| 7-12 \| \| Sheep and goats \| 4-15 \| 1-4 \| Water Requirements An animal’s water requirement depends on several factors such as ambient temperature, diet (energy level, fiber content, salt), physiological state (age, growth, pregnancy, ability to conserve water), level of exercise, and health. Environmental temperature (and associated humidity) is a major factor contributing to water intake. Water consumption, when expressed by unit of body weight for non-heat-stressed, nonlactating cattle, is around 5% to 6% of the body weight per day (or 2–5 kg of water for every kg of dry feed consumed) and can go up to 12% or more under heat stress. Water intake increases with higher environmental temperatures and increasing physical activity because of water lost through evaporative loss. Dietary dry matter intake and feed water content are highly correlated with water intake at moderate temperatures. High-energy, high-fat, and high-protein diets increase water intake because of increases in metabolic waste and urinary excretion of urea as well as increases in heat produced by metabolism. The salt content of a diet increases water consumption. Diets high in fiber (bran, dry forages) increase water intake as well. Young animals have higher water requirements per body size as compared to large animals. Similarly, animals that conserve water, such as sheep and poultry, need lower levels than cattle. Pregnancy and milk production increases water intake too. Dairy cattle may require 38–110 L/d compared to beef cattle at 22–66 L/d. Ambient temperature is the major factor affecting an animal’s water intake. Other factors include age, type of diet, level of exercise, stage of growth, or pregnancy. Water Restriction and Toxicity Water shortage affects both domestic and wild animals. To compensate for the losses and to maintain all related physiological functions, animals should have access to a clean supply of water. Lack of enough water could lead to a reduction in feed intake and productivity. Dehydration of the body leads to a reduction in body weight and the consequences are worse in high environmental temperatures. Dehydration is accompanied by a loss of electrolytes, an increase in body temperature, and an increase in respiratory rate. Animals become highly irritable, and prostration and death follow after severe water deprivation. Water intoxication may occur as a result of a sudden ingestion of large amounts of water after a short period of deprivation and is due to the slow adaptation of the kidneys to the high water load. Water restriction reduces feed consumption and is very stressful for animals. Water Quality Water per se is almost nontoxic, but problems with water arise from contamination with microbes, parasites, minerals, and various other toxic substances, such as pesticides. Water quality affects consumption, productivity, and animal health. Substances such as salts, pathogenic organisms, algae, and pesticides pollute water supplies and can affect palatability. Mineral salts include carbonate and bicarbonates, sulfates, and chlorides of Ca, Mg, Na, and K. Other toxic substances in water include nitrate, iron salts, and hydrocarbons. Contamination with nitrate is common in farming intensive areas. Concentrations above 1,500 ppm may cause toxicity causing death from anoxia. Iron salts in groundwater cause rust deposits on pipes and may cause bacterial contamination by iron-utilizing bacteria. Pesticides such as malathion and organophosphates may get into water systems and can be toxic. Certain blue-green algae (cyanogenic) in lakes can produce toxic substances. Toxicity in livestock causing vomiting, frothing, muscle tremors, liver damage, and death are reported due to blue-green algae toxicosis. Water quality can influence the development of polioencephalomalacia, a noninfectious disease affecting the brain in feedlot cattle. Most affected animals show aimless wandering, disorientation, blindness, recumbency, star-gazing posture, and edema in the brain, causing a “softness” in the brain. Water high in sulfates promotes polioencephalomalacia, apparently via a complex interaction with other minerals and B vitamins. Most domestic animals can tolerate a total dissolved solid concentration of 15,000 to 17,000 mg/L. Water containing less than 1,000 mg/L of total soluble salt is safe for all classes of livestock. At higher (> 5,000–7,000 mg/L) levels, it may cause mild diarrhea and an increase in mortality in poultry, but it could be acceptable to other livestock. A guideline for the interpretation of total dissolved solids in water is shown in table 18.1. Table 18.1. Guidelines for interpretation of total dissolved solids content in water \| Total dissolved solids, mg/L \| Interpretation \| \| <1000 \| Suitable for all classes of livestock \| \| 1000-1999 \| Satisfactory for all classes of livestock; may produce transient diarrhea in animals \| \| 2000-4999 \| Temporary water refusal and diarrhea may be seen when animals are exposed to such water sources. May reduce productivity in dairy cattle. \| \| 5000-6999 \| Likely to reduce productivity in dairy cattle. May reduce growth rates. May result in water refusal and diarrhea. Avoid if possible. \| \| 7000-10000 \| Unfit for swine, very risky in all other species. Avoid. \| \| > 10000 \| Dangerous, avoid. \| Key Points Water is one of the most important nutrients, yet it is almost always neglected. Water serves as the fluid matrix of the animal body. Water gives form and structure and provides protection from environmental stress. The high solvent power of water permits the formation of solutions within which metabolic reactions occur. There are different sources of water (e.g., diet, drinking, metabolic). Metabolic water is produced during catabolism of carbohydrates, fats, and proteins to carbon dioxide and water and is important for hibernating animals. Species difference, type of metabolism, and digestive tract type affect water requirements. For example, birds and fish have low requirements, while ruminants need a large quantity of water to suspend ingesta in the rumen. An animal’s water requirement is highly influenced by environmental temperature, humidity, diet (energy level, fiber content, salt), physiological state (age, growth, pregnancy), and level of exercise and health. Water restriction affects animal health, growth, and productivity and is very stressful for animals. Water containing less than 1,000 mg/L of total soluble salt is safe for all classes of livestock. Review Questions Water is one of the most important essential nutrients. Functions of water include: Can act as a diluent Carrier of waste from the body Transport of nutrients All of the above List the sources of water in the animal body. Major loss of water in a beef cattle is through which medium? Water quality can affect cattle health. Name a non infectious disease in feedlot cattle associated with water quality.
6435	https://chem.libretexts.org/Bookshelves/General_Chemistry/ChemPRIME_(Moore_et_al.)	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Password Sign in Sign in Sign in Forgot password Contents 1. Home 2. Bookshelves 3. General Chemistry 4. ChemPRIME (Moore et al.) Expand/collapse global location ChemPRIME (Moore et al.) Front Matter 1: Introduction - The Ambit of Chemistry 2: Atoms, Molecules, and Chemical Reactions 3: Using Chemical Equations in Calculations 4: The Structure of Atoms 5: The Electronic Structure of Atoms 6: Chemical Bonding - Electron Pairs and Octets 7: Further Aspects of Covalent Bonding 8: Properties of Organic Compounds 9: Gases 10: Solids, Liquids and Solutions 11: Reactions in Aqueous Solutions 12: Chemistry of the Representative Elements 13: Chemical Equilibrium 14: Ionic Equilibria in Aqueous Solutions 15: Thermodynamics- Atoms, Molecules and Energy 16: Entropy and Spontaneous Reactions 17: Electrochemical Cells 18: Chemical Kinetics 19: Nuclear Chemistry 20: Molecules in Living Systems 21: Spectra and Structure of Atoms and Molecules 22: Metals Back Matter ChemPRIME (Moore et al.) Last updated Dec 24, 2024 Save as PDF Detailed Licensing Front Matter picture_as_pdf Full Book Downloads Full PDF Import into LMS Individual ZIP Buy Print Copy Print Book Files Buy Print CopyReview / Adopt Submit Adoption Report Submit a Peer Review View on CommonsDonate Page ID 49241 Ed Vitz, John W. Moore, Justin Shorb, Xavier Prat-Resina, Tim Wendorff, & Adam Hahn Chemical Education Digital Library (ChemEd DL) ( \newcommand{\kernel}{\mathrm{null}\,}) Table of contents No headers ChemPRIME (Chemical Principles through Integrated Multiple Exemplars) is designed so that chemistry concepts can be presented in an order that reflects the conceptual structure of the discipline and students are able to learn chemistry within a broad range of contexts that relate chemistry to other disciplines and everyday life. ChemPRIME is a collaborative project in which concepts and contexts can be contributed by anyone and used free of charge by anyone for non-commercial purposes. Front Matter 1: Introduction - The Ambit of Chemistry 2: Atoms, Molecules, and Chemical Reactions 3: Using Chemical Equations in Calculations 4: The Structure of Atoms 5: The Electronic Structure of Atoms 6: Chemical Bonding - Electron Pairs and Octets 7: Further Aspects of Covalent Bonding 8: Properties of Organic Compounds 9: Gases 10: Solids, Liquids and Solutions 11: Reactions in Aqueous Solutions 12: Chemistry of the Representative Elements 13: Chemical Equilibrium 14: Ionic Equilibria in Aqueous Solutions 15: Thermodynamics- Atoms, Molecules and Energy 16: Entropy and Spontaneous Reactions 17: Electrochemical Cells 18: Chemical Kinetics 19: Nuclear Chemistry 20: Molecules in Living Systems 21: Spectra and Structure of Atoms and Molecules 22: Metals Back Matter Thumbnail: Lead Iodide (CC BY 2.0 Generic;Paige PowersviaFlickr) This page titled ChemPRIME (Moore et al.) is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Ed Vitz, John W. Moore, Justin Shorb, Xavier Prat-Resina, Tim Wendorff, & Adam Hahn. Back to top Detailed Licensing Front Matter Was this article helpful? Yes No Recommended articles Front Matter Back Matter 9: Gases 1: Introduction - The Ambit of ChemistryThe science of chemistry is concerned with the composition, properties, and structure of matter and with the ways in which substances can change from ... 5: The Electronic Structure of AtomsThe electronic structures of atoms developed during the first half of the twentieth century. The periodic repetition of chemical properties discovered... Article typeBook or UnitAuthorChemPRIMECover PageSet Cover Page/Add to Download CenterLicenseCC BY-NC-SALicense Version4.0Show Page TOCno on page Tags lulu@ChemPRIME@John Moore et al.@University of Wisconsin-Madison@ChemPRIME © Copyright 2025 Chemistry LibreTexts Powered by CXone Expert ® ? 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6436	https://www.youtube.com/watch?v=XWEC4rfTjuU	Fraunhoffer diffraction at a circular aperture. Learning Science 5690 subscribers 6 likes Description 441 views Posted: 30 Nov 2021 Fraunhoffer diffraction at a circular aperture Discussion 1) The radius of the central maximum or the Airy's disc decreases as the radius of the circle ( circular aperture b/2) increases. 2) The radius of the central maximum depends upon wavelength. So, it is greater for red light than for blue light. 3) With monochromatic light the diffraction pattern consists of alternate bright and dark rings. But, with white light, central maximum is white and it is surrounded by coloured rings. learningscience fraunhofferdiffractionatcircularaperture circularaperture 2 comments Transcript: hello everyone welcome to learning science atopic hair from hofer diffraction at a circular aperture archangel have frowned off a diffraction at a circular aperture suppose a plane wavefront is incident on a circular aperture of diameter b as shown in figure first hum suppose plane wavefront hair incident the secondary waves originating from the primary wavefront and travelling parallel to the incident light come to focus at p or secondary waves yogi primary where france originates on the diameter of the circular aperture there are two points on either side of o judi meter has circular the secondary waves originating from these points and going parallel to incident beam after passing through the lens reach b without a path difference the secondary waves is points parallel hoga incident beam so point p is the position of principle or central maximum to your point p have a position hoga principle yes central maximum if we consider the secondary waves diffracted at angle theta as shown in figure first agar hum secondary difference between the secondary waves originating from a and b to b and have a path different soga secondary b sine theta k the point p 1 will be minimum if b n is integral multiple of lambda that is when b sine theta p is equal to p lambda point p 1 j or your point p have a maximum tab difference to p plus 1 lambda upon 2 that is b sine theta p is equal to 2 p plus 1 lambda upon 2 i if p p1 is the position of minimum intensity agar p1 minimum intensity position hair then all points at a distance y from the point p will be of minimum intensity due to a circular aperture consists of central bright disc called disc to joe diffraction pattern has circular apertures the intensity of bright rings goes on decreasing as we move away from p or bright rings intensity if the screen is far away from the lens that the lens is of large focal length or agar ham consider where double slit also for the first secondary minimum the equation first can be written as first secondary minimum condition order of sine theta y upon f sine theta lambda upon b sine theta y upon f or lambda upon b equal hyga or y hamada f lambda upon b here y is the radius of erie disc origin radius but in actual practice it has been found by erie himself that y is equal to 1.22 f lambda upon a b equation fourth formula 1.22 f lambda upon b the radius of the central maximum or the airy disc so it is greater for red light than for blue light or yeah wavelengths a directly proportional hair red light wavelength greater hoga red light [Music] with monochromatic light the diffraction pattern consists of alternate bright and dark rings monochromatic light used but with white light central maximum is white and it is surrounded by colored rings thank you so much for watching
6437	https://ocw.mit.edu/courses/18-01sc-single-variable-calculus-fall-2010/15f8e38786ae3b738375debb112ee660_MIT18_01SCF10_ex80sol.pdf	Exploring a Parametric Curve a) Describe the curve traced out by the parametrization: x = t cos t y = t sin t, where 0 ≤ t ≤ 4π. b) Set up and simplify, but do not integrate, an expression for the arc length Z 4π ds dt of this curve. dt 0 Solution a) Describe the curve traced out by the parametrization x = t cos t y = t sin t, where 0 ≤ t ≤ 4π. We know that the equations: x = a cos t y = a sin t describe a circle of radius a, where a is a constant. In the problem we’ve been given, the multiple a is variable: x = a(t) cos t y = a(t) sin t, where a(t) = t. The curve still moves counter-clockwise around the origin as t increases; however, its distance from the origin also increases as t increases. We could compute the coordinates of several points on the curve to get a better idea of its behavior. The table below gives a few of these coordinates. ( π, 0) − π π π 0 π 4 π t (x, y) (0, 0) π √ 2 π √ 2 8 , 8 0, 2 2 π 3 2 0, −3 2 We see that the curve starts at (0, 0) and proceeds counter-clockwise, moving away from the origin. It wraps twice around the origin as t increases from 0 to 4π, tracing out a ﬁgure known as an Archimedean spiral. 1 p s 0 b) Set up and simplify, but do not integrate, an expression for the arc length Z 4π ds dt of this curve. dt We know that ds = dx2 + dy2, so: 2 2 ds dx dy = + . dt dt dt We compute that: dx dt = −t sin t + cos t dy = t cos t + sin t dt dx dt 2 = t2 sin2 t − 2t sin t cos t + cos2 t dy 2 = t2 cos 2 t + 2t sin t cos t + sin2 t dt dx dt 2 + dy dt 2 = t2(sin2 t + cos2 t) − 2t sin t cos t + 2t sin t cos t + (cos2 t + sin2 t) dx 2 + dy 2 = t2 + 1. dt dt Thus we have: Z 4π ds Arc length = dt dt 0 2 Z 4π p = t2 + 1 dt. 0 If we wished we could complete this calculation using the inverse substitution t = tan θ and then integrating sec3 θ. We could then check our work by comparing our ﬁnal result to the circumference of an appropriate circle. 3 MIT OpenCourseWare 18.01SC Single Variable Calculus Fall 2010 For information about citing these materials or our Terms of Use, visit:
6438	https://mathcentral.uregina.ca/QQ/database/QQ.09.08/h/fred1.html	Twelve golfers - Math Central ← BACKPRINT+ TEXT SIZE –SEARCHHOME Math Central Quandaries & Queries Hello- There are twelve golfers in 3 foursomes for three days. Is there a combination of four man foursomes that allow for no duplicates? In other words, is there a formula to insure the greatest amount of variety in the foursomes. Looking at it math like you have 1,2,3,4,5,6,7,8,9,10,11,12 grouped into four numbers each for three days. So if your have the first day 1,2,3,4 in one group, 5,6,7,8 in the second and 9,10,11,12 in the third what can you do for the next two days to have the fewest duplicates or have the greatest variety in the foursomes? Thanks Fred, I would suggest something like this: Day 1: (1,2,3,4), (5,6,7,8), (9,10,11,12) Day 2: (1,2,7,11), (3,5,6,12), (4,8,9,10) Day 3: (1,6,7,10), (2,3,8,9), (4,5,11,12) Then the pairs of people who play together twice are: (1,2), (5,6), (9,10), (Day 1 and Day 2) (2,3), (6,7),(11,12) (Day 1 and Day 3) (1,7), (5,12), (8,9) (Day 2 and Day 3) On any pair of days you will have to have at least 3 pairs of people who play together both days. This is simply because if you split 4 people into 3 teams at least one team will have 2 players (this idea is important enough to have a name, it is called the "pigeon hole principle") This schedule has the minimal number of pairs of people who play together twice. Karen Twelve is a hard number for golf schedules. It isn't possible to have no duplicates, even if only on two days. Six duplicate pairs is a theoretical minimum. The reason is that on the second day, each group of 4 must contain two players who were together on day 1. After selecting one from each of the foursomes from the first day -- the only way to get no duplicates -- the fourth player in the group came from the same foursome as one of the first three. The same reasoning holds for the third day. Let's suppose first there are 9 players in three threesomes per day. Later a fourth player will be added to each threesome. A schedule for 9 players with no duplicates is Day 1: 123, 456, 789 Day 2: 147, 258, 369 Day 3: 159, 267, 348 Now add three more players A, B and C, one to each group on each day, however you think works best. It looks like six duplicate pairs is easy to achieve in this way. Best of luck, and have fun. Victoria Math Central is supported by the University of Regina and The Pacific Institute for the Mathematical Sciences.
6439	https://arxiv.org/pdf/1310.7268	arXiv:1310.7268v1 [math.HO] 27 Oct 2013 Parallel Weighings Tanya Khovanova MIT October 29, 2013 Abstract I introduce, solve and generalize a new coin puzzle that involves parallel weighings. 1 Ancient Coin Problems I heard my first coin problem when I was very young: Given 9 coins, one of them fake and lighter, find the fake coin in two weighings on a balance scale. I believed that this problem was thousands of years old, that Pythagoras could have invented it. But surprisingly its first publication was by E. D. Schell in the January 1945 issue of the American Mathematical Monthly . This does not prove that Pythagoras did not invent it, but it makes the said event highly unlikely. I will not be surprised if everyone who reads this paper has heard this problem before and knows how to solve it. But I still need to discuss it to establish the methods that are used later in my featured problem. First, the problem implies that all real coins weigh the same, and that we need to find a strategy that guarantees finding the fake coin in two weighings. By the way, two weighings is the smallest number of weighings that guar-antee finding the fake coin. In addition, 9 is the largest number of coins such that the fake coin can be found in two weighings. Rather than discussing the solution for the 9-coin problem, I would like to generalize it to any number of coins: 1Given N coins, one of them fake and lighter, find the minimum number of weighings that will guarantee finding the fake coin. Here is a way to think about it. In one weighing divide all coins into three piles: the coins that go onto the left pan, the coins that go onto the right pan, and the coins that do not go on the scale. Clearly, we need to put the same number of coins on the pans, otherwise, we do not get any meaningful information. If the scale balances then the fake coin is in the leftover pile. If the scale does not balance, then the fake coin is in the pile that is lighter. Either way, the coin is in one of three piles and we have to start all over with this smaller pile. So to minimize the number of weighings, we need to divide all the coins into three piles so that the largest pile is the smallest. Oops, that didn’t sound right. I mean, we need to minimize the size of the largest pile. Dividing the coins into three piles as evenly as possible allows us with n weighings to find the fake coin among up to 3 n coins. In particular, we can find the fake coin among 9 coins in two weighings. Another famous coin puzzle appeared almost at the same time as the previous puzzle . There are 12 coins; one of them is fake. All real coins weigh the same. The fake coin is either lighter or heavier than the real coins. Find the fake coin and figure out whether it is heavier or lighter in 3 weighings on a balance scale. The solution is well known and quite beautiful. Unsurprisingly it gen-erated more publications than the 9-coin problem , , , . Readers who do not know the solution should try it. What is the minimum number of weighings in this puzzle’s setting for any number of coins? I just want to point out that we need to assume that the number of coins is more than 2, otherwise we cannot solve it at all. If there are N coins, then there are 2 N possible answers to this puzzle. We need to pinpoint the fake coin and say whether it is heavier or lighter. Each weighing divides information into three parts, so in n weighings we can give 3n different answers. Thus, the expected number of weighings should be of the order log 3 2N. The exact answer can be calculated using this additional constraint of having the same number of coins on each pan in each weighing. The exact answer is (3 n − 3) /2, see , . In the following important variation of the latter puzzle we need to find the fake coin, but do not need to tell whether is it heavier or lighter . 2There are N coins; one of them is fake. All real coins weigh the same. The fake coin is either lighter or heavier than the real coins. What is the maximum number of coins for which you can guarantee finding the fake coin with n weighings on a balance scale? This problem is very similar to the previous one. Let me call the previous problem the find-and-label problem, as opposed to this problem that I will call the just-find problem. The answer to the just-find problem is (3 n − 1) /2, see . In particular, 13 coins is the best we can do in 3 weighings. Notice that for every strategy for the find-and-label problem that resolves n coins, we can produce a strategy for the just-find problem that resolves n+1 coins by adding a coin that is never on the scale. Indeed, in the find-and-label strategy by the last weighing at least one of the weighings needs to be unbalanced to label the fake coin. Thus, if all the weighings balance at the end, the fake coin is the extra coin. 2 The Original Parallel Weighings Puzzle We have all been hearing about parallel computing, and now it has turned up in a coin-weighing puzzle invented by Konstantin Knop. The puzzle appeared at 2012 Russia-Ukraine Puzzle Tournament and in Konstantin Knop’s blog . We have N indistinguishable coins. One of them is fake, and it is not known whether it is heavier or lighter than the genuine coins, which all weigh the same. There are two balance scales that can be used in parallel. Each weighing lasts one minute. What is the largest number of coins N for which it is possible to find the fake coin in five minutes? 3 The Road Map Section 4 describes the similarity of the original puzzle with a multiple-pans problem: a coin weighing puzzle involving balance scales with not two, but any number of pans. The notion of a coin’s potential—a useful technical 3tool in solving coin weighing puzzles—is defined in Section 5. How many coins with known potential can be processed in n minutes is discussed there too. Section 6 provides a solution to the parallel weighing problem in case we have an unlimited supply of real coins. It is followed by a solution to the original puzzle and its generalization for any number of minutes in Section 7. Section 8 generalizes these results to the use of more than two scales in parallel. The find-and-label variation of this problem for any number of minutes is discussed in Section 9. The last Section 10 compares the find-and-label problem with the just-find problem. 4 Warm-up: Multiple Pans Problem Knop’s puzzle reminds me of another coin-weighing problem, where in a similar situation you need to find a fake coin by using five weighings on one scale with four pans. The answer in this variation would be 5 5 = 3125. Divide coins in five groups with the same number of coins and put four groups on the four pans of the scale. If one of the pans is different (heavier or lighter), then this pan contains the fake coin. As it is one out of four pans, then after the weighing we will know the deviation of the fake coin. Otherwise, the leftover group contains the fake coin. The strategy is to divide the coins into five piles as evenly as possible. This way each weighing reduces the pile with the fake coin by a factor of five. Thus, it is possible to resolve 5 n coins in n weighings. I leave it to the reader to check that, excluding the case of two coins, any number of coins greater than 5 n−1 and not greater than 5 n can be optimally resolved in n weighings. One scale with four pans gives you more information than two scales with two pans used in parallel. We can expect that Knop’s puzzle requires at least the same number of weighings as the four-pan puzzle for the same number of coins. So the answer to Knop’s puzzle should not not exceed 3125. But what will it be? If you know Russian, you can read the author’s solution to the original puzzle at Knop’s blog , otherwise, bear with me and you will get the answer to this puzzle, as well as the answers to this puzzle’s generalizations. 45 Coins Potential While weighing coins, we may be able to determine some incomplete infor-mation about a coin’s reality. For instance, we may be able to rule out the possibility that a given coin is fake-and-heavy, without being able to tell whether that coin is real or fake-and-light. Let us call such a coin potentially light ; and conversely, let us say a coin is potentially heavy if it could be real or fake-and-heavy but cannot be fake-and-light. How many coins with known potential can be processed in n minutes? If all the coins are potentially light then we can find the fake coin out of 5 n coins in n minutes. Indeed, in this case using two scales or one scale with four pans (see Section 4) gives us the same information. The pan that is lighter contain the fake coin. If everything balances, then the fake coin is not on the scales. What if there is a mixture of potentials? Can we expect the same answer? How much more complicated could it be? Suppose there are five coins: two of them are potentially light and three are potentially heavy. Then on the first scale we compare one potentially light coin with the other such coin. On the other scale we compare one potentially heavy coin against another potentially heavy coin. The fake coin can be determined in one minute. Our intuition suggests that it is a bad idea to compare a potentially heavy coin on one pan with a potentially light coin on the other pan. Such a weighing, if unbalanced, will not produce any new information. On the other hand, if we compare a potentially heavy coin with a potentially heavy coin, then we will get new information. If the scale balances, then both coins are real. If the scale does not balance, then the fake coin is the heavier coin out of the two that are potentially heavy. Does this mean that we should only put coins with the same potential on the same scale? Actually, we can mix the coins. For example, suppose we put 3 potentially light coins and 5 potentially heavy coins on each pan of the same scale. If the left pan is lighter, then the potentially heavy coins on the left pan and potentially light coins on the right pan must be genuine. The fake coin must be either one of the three potentially light coins on the left pan or one of the five potentially heavy coins on the right pan. In general, after each minute, the best hope is to have the number of coins that are not determined to be real to be reduced by a factor of 5. If one of the weighings on one scale is unbalanced, then the potentially light coins on the lighter pan, plus the potentially heavy coins on the heavier pan 5would contain the fake coin. We do not want this number to be bigger than one-fifth of the total number of coins being processed. So, divide coins in pairs with the same potential and from each pair put the coins on different pans of the same scale. In one minute we can divide the group into five equal, or almost equal, groups. If there is an odd number of coins with the same potential, then the extra coin does not go on the scales. The only thing left to check is what happens if the number of coins is small. Namely, we need to check what happens when the number of potentially light coins is odd and the number of potentially heavy coins is odd, and the total number of coins is not more than five. In this case the algorithm requires us to put aside two coins: one potentially heavy and one potentially light, but the put-aside pile cannot have more than one coin. After checking small cases, we see that we cannot resolve the problem in one minute when there are 2 coins of different potential, or when the 4 coins are distributed as 1 and 3. On the other hand, if there are extra coins that are known to be real, then the above cases can be resolved. This means that the small cases are only a problem if they happen in the first minute. Hence, Lemma 1. Any number of coins N > 4 with known potential can be resolved in ⌈log 5 N⌉ minutes. 6 Unlimited Supply of Real Coins We say that the coin that is potentially light or potentially heavy has known potential . The notion of known potential is important in solving Knop’s puzzle and many other coin weighing puzzles due to the following theorem: Theorem 2. In a coin-weighing puzzle, where only one coin is fake, any coin that visited the scales is either genuine or its potential is known. Proof. If the scale ever balanced, all coins that were on it on any such occa-sion are real. Any coin that appeared on both a heavier pan and a lighter pan is also real. Otherwise, the coins that only visited lighter pans are potentially light and the coins that only visited heavier pans are potentially heavy. Now let us go back to the original problem, in which we do not know the coins’ potential at the start. Let us temporarily add an additional assumption to the original problem. Suppose there is an unlimited supply of coins that 6we know to be real. Let u(n) be the maximum number of coins we can process in n minutes if we do not know their potential and have an unlimited supply of real coins. Lemma 3. u(n) = 2 · 5n−1 + u(n − 1) . Proof. What information do we get after the first minute? Both scales might be balanced, meaning that the fake coin is in the leftover pile of coins with unknown potential. So we have to leave out not more than u(n − 1) coins. On the other hand, exactly one scale might be unbalanced. In this case, all the coins on this scale will have their potential revealed. The number of these coins cannot be more than 5 n−1, so u(n) ≤ 2 · 5n−1 + u(n − 1). Can we achieve this bound? Yes. On each scale, put 5 n−1 unknown coins on one pan, and 5 n−1 real coins from the supply on the other. Thus, u(n) = 2 · 5n−1 + u(n − 1). We also can see that u(1) = 3. Indeed, on each scale put one coin against one real coin and have one coin in the leftover pile. Thus, the corollary: Corollary 4. u(n) = (5 n + 1) /2. Thus, the answer to the puzzle problem with the additional resource of an unlimited supply of real coins is (5 n + 1) /2. Clearly the answer without the additional resource cannot be larger. But what is it? We assumed that there is an unlimited supply of real coins. But how many extra coins do we really need? The extra coins are needed for the first minute only, because after the first minute at least one of the scales will balance and many coins will be determined to be real. In the first minute, we need to put 5 n−1 coins from the unknown pile on each scale. The coins do not have to be on the same pan. The only problem is that the number of coins is odd, so we need one extra real coin to make this number even. So our unlimited supply need not be unlimited—we just need two extra coins, one for each scale. 7 The Puzzle Solution Now let us go back and remember that the formula for u(n) assumes an unlimited supply of real coins. The unlimited supply need not be more than two real coins. 7So, how can we solve the original problem? We already know that the only adjustment that is needed is in the first minute. In the first minute we put unknown coins against unknown coins, not more than 5 n−1 on each scale, and, since the number on each scale must be even, the best we can do is put 5n−1 − 1 coins on each scale. Thus, the answer to the puzzle is (5 n − 3) /2. Do not forget that we cannot find the fake coin out of 2 coins, ever. Theorem 5. Given two scales in parallel, the number of coins N that can be optimally resolved in exactly n minutes is: (5 n−1 − 3) /2 ≤ N < (5 n − 3) /2,with one exception: N = 2 , for which the fake coin cannot be identified. Going back to the original puzzle: the largest number of coins that can be resolved in 5 minutes is 1561. 8 More Scales It is straightforward to generalize the just-find problem to any number of scales used in parallel. Suppose the number of scales is k. The following problems can be solved in n minutes: Known Potential. If all the coins have known potential, then any number of coins up to (2 k + 1) n can be resolved. Unlimited Supply of Real Coins. If we do not know the potential of any coin and there is an unlimited supply of real coins, the maxi-mum number of coins that can be solved is defined by a recursion: uk(n) = k · (2 k + 1) n−1 + uk(n − 1) and uk(1) = k + 1. Any number of coins up to ((2 k + 1) n + 1) /2 can be resolved. General Case. If we do not know the potential of any coin and there are no extra real coins, then any number of coins between 3 and uk(n) − k =((2 k + 1) n + 1) /2 − k can be resolved. Let me draw your attention to the fact that if k = 1, then the general case is the classic problem of just finding the fake coin. So plugging in k = 1 into the formula ((2 k + 1) n + 1) /2 − k above should give the answer given in Section 1: (3 n + 1) /2 − 1. In particular, for n = 3, it should be 13. And it is. 89 Find and Label The methods described above can be used to answer another common ques-tion in the same setting: Find the fake coin and say whether it is heavier or lighter. If all coins have known potential, then the just-find problem is equivalent to the find-and-label problem. The find-and-label problem can be solved by similar methods to the just-find problem. Namely, let us denote by Uk(n) the number of coins that can be resolved in n minutes in parallel on k scales when there is an unlimited supply of extra real coins. Then the recursion is the same as for the just-find problem: Uk(n) = k · (2 k + 1) n−1 + Uk(n − 1). The difference is in the starting point: and Uk(1) = k.Similarly, if we do not have an unlimited supply of real coins, then the bound is described by the following theorem: Theorem 6. There are N coins one of which is fake, and it is not known whether it is heavier or lighter. There are also k balance scales that can be used in parallel, one weighing per one minute. The maximum number of coins that requires n minutes to find and label the fake coin is ((2 k+1) n+1) /2−k−1.If N = 2 , then the problem cannot be resolved. Again, if k = 1, then this is the classic problem of just funding and labeling the fake coin. So plugging in k = 1 into the formula ((2 k + 1) n +1) /2 − k − 1 above should give the answer from Section 1: (3 n + 1) /2 − 2. In particular, for n = 3, it should be 12. And it is. 10 Lazy Coin You might have noticed that the answer for the just-find and for the find-and-label problems differ by one: Lemma 7. In the parallel weighing problem with one fake coin, the maximum number of coins that can be optimally resolved in n weighings for the just-find problem is one more than the maximum number of coins that can be optimally resolved in the find-and-label problem in the same number of weighings. We already proved the lemma by explicitly calculating the answer. It would be nice if there was a simple argument to prove it without calculations. And such an argument exists if we restrict ourselves to static strategies. In a 9static or non-adaptive strategy you decide beforehand what your weighings are. Then, seeing the results of the weighings you can find the fake coin and label it if needed. In a static strategy for the find-and-label problem every coin has to visit the scales at some point, or, otherwise, if the coin does not visit the scales and it happens to be fake, it can not be labeled. In a static strategy for the just-find problem if every coin visits the scale, then all the coins can be labeled. Suppose we add an extra coin that is never on the scales. If all the weighings balance, then the extra coin is the fake one. We can not have two such coins. Indeed, if one of them is fake we can not differentiate between them. If all the coins visit the scales in the just-find problem, then all the coins can be labeled at the end. Let us call a strategy that resolves the maximum number of coins in a given number of weighings a maximal strategy. We just showed that a maximal static strategy for the just-find problem has to have a coin that does not go on the scales. Therefore, there is a bijection between maximal static strategies for the just-find and the find-and-label problems. The strategies differ by an extra coin that sits lazily outside the scales all the time. In dynamic or adaptive strategies the next weighing depends on the re-sults of the previous weighings. With dynamic strategies the story is more complicated. There is no a bijection any more. On one hand it is possible to add a lazy coin to a strategy in the find-and-label problem to get a strategy in the just-find problem. But there exist maximal strategies in the just-find problem where all the coins can end up on the scale. For example, consider the following strategy to just-find the fake coin out of 4 coins in 2 weighings on one scale. In the first weighing we balance the first coin against the second. If the weighing unbalances, we know that one of the participating coins is fake, and we know the potential of every participating coin. Then in the second weighing we balance the first and the second coins against the third and the fourth. In this example, all coins might visit the scale and 4 is the maximum number of coins that can be processed in 2 weighings. Alas! There is no simple argument, but at least it is easy to remember, that the maximal strategies for these two problems differ by one coin. 10 11 Acknowledgements I am grateful to Daniel Klain and Alexey Radul for helpful discussions. References B. Descartes, The twelve coin problem, Eureka, 13 (1950) 7, 20. F. J. Dyson, Note 1931—The problem of the pennies, Math. Gaz., 30 (1946) 231–234. D. Eves, Problem E712—The extended coin problem, Amer. Math. Monthly, 53 (1946) 156. N. J. Fine, Problem 4203—The generalized coin problem, Amer. Math. Monthly, 53 (1946) 278. Solution, 54 (1947) 489–491. R. L. Goodstein, Note 1845—Find the penny, Math. Gaz., 29 (1945) 227–229. [Erroneous solution] H. D. Grossman, The twelve-coin problem, Scripta Math., 11 (1945) 360– 361. K. Knop, Weighings on two scales, available at 2013/04/blog-post_11.html (2013) (in Russian) J. G. Mauldon, Strong solutions for the counterfeit coin problem, IBM Research Report RC 7476 (#31437) (1978) E. D. Schell, Problem E651—Weighed and found wanting, Amer. Math. Monthly, 52 (1945) 42. 2012 Ukraine-Russian Puzzle Tournament, available at: tvpark.ua/_ARC/2012/KG_12_38_12.PDF (in Russian) Lothrop Withington, Another solution of the 12-coin problem, Scripta Math., 11 (1945) 361–363. 11
6440	https://www.youtube.com/watch?v=5NHAHlltACc	Piercing Points and Plane Intersections \| Basic #2 Descriptive Geometry Pro 707 subscribers 47 likes Description 3899 views Posted: 18 Apr 2022 📐Edge-View Method Assuming that the straight line is neither parallel to nor in the plane, it will intersect the plane at a point common to both the line and the plane. The limits of the line and plane as given may have to be extended to determine this "pierce point". Since an edge-view of the plane contains all points in the plane, the view showing the plane as an edge will also show the point where the line pierces the plane. ❗️Another method would involve showing the line as a point. This point of view of the line will contain the pierce point common to both the line and the plane. 📋Example 1: (Given Plane Appears as an Edge) The plane ABCD and the line EF are given in both the plan and front elevation views. The plane appears as an edge in the plan view. The intersection of the line and the plane is determined by point M which is common to both. Using careful visualization, we notice that the FM portion of the line lies wholly in front of the plane and therefore will appear visible in the front view. The plan view shows that the EM Jportion of the line lies wholly behind the plane and, therefore, will be hidden in the front view. 📋Example 2: (Oblique Plane) The plane ABC and the line EF are given in both the plan and front elevation views. Draw an auxiliary elevation view showing the plane as an edge. The pierce point M in this view is the point of intersection which can now be projected back to the plan and front elevation views. Careful visualization reveals the visible and hidden portions of the line. ❗️The pierce point could also have been determined by showing the plane as an edge in an inclined view projected from the front view. 📐Two-View Cutting Plane Method The intersection of a line and an oblique plane can be determined by using a vertical cutting plane that contains the given line. The line of intersection of the cutting plane with the oblique plane and the given line must intersect or be parallel because they both lie in the vertical cutting plane. Since the cutting plane appears as an edge in the plan view, the relationship between the line of intersection and the given line is not apparent in the plan view. The related view, however, reveals this relationship. Should the two lines intersect in the related view, it is evident that the point of intersection is common to both the given plane and the given line and therefore determines the pierce point of the given line and given plane. 📋Example The oblique plane ABC and the line EF are given in both the plan and front elevation views. A vertical cutting plane, coincidental with and containing the given line EF, appears as an edge in the plan view. The intersection of the given plane ABC and the vertical cutting plane containing EF is line 12. The lines EF and 12 both lie in the vertical cutting plane and intersect each other at point M in the front view. Since point M is on line 12, it is also on plane ABC because line 12 is on plane ABC. Therefore point M is the required point, being common to both the given line EF and the given plane ABC. It can now be projected to the related view. Use careful visualization to determine what portion of the line should be visible in each view. ❗️If line 12 had appeared parallel to EF in the front view, it would have indicated that the line EF was parallel to plane ABC and therefore it would have no point of intersection with the given plane. descriptivegeometry #point #intersection 1 comments Transcript:
6441	https://www.sciencedirect.com/science/article/abs/pii/S0196070922002952	Robotic cochlear implantation in post-meningitis ossified cochlea - ScienceDirect Skip to main contentSkip to article Journals & Books Access throughyour organization Purchase PDF Patient Access Other access options Search ScienceDirect Article preview Abstract Introduction Section snippets References (12) Cited by (4) American Journal of Otolaryngology Volume 44, Issue 1, January–February 2023, 103668 Robotic cochlear implantation in post-meningitis ossified cochlea☆ Author links open overlay panel Mohammad Al Saadi a, Emilie Heuninck a, Leo De Raeve b, Paul Van de Heyning c, Vedat Topsakal a Show more Add to Mendeley Share Cite rights and content Abstract Aim To report the experience of an image-guided and navigation-based robot arm as an assistive surgical tool for cochlear implantation in a case with a labyrinthitis ossificans. Patient A 55-years-old man with a history of childhood meningitis whose hearing deteriorated progressively to bilateral profound sensorineural hearing loss. Intervention Robotic Assisted Cochlear Implant Surgery (RACIS) with a straight flexible lateral wall electrode. Primary outcome measures Electrode cochlear insertion depth with RACIS with facial recess approach and autonomous inner ear access with full electrode insertion of a flexible straight cochlear implant array. Conclusions Intra cochlear ossifications pose a challenge for entering the cochlea and full-length insertion of a cochlear implant. RACIS has shown that computations of radiological images combined with navigation-assisted robot arm drilling can provide efficient access to the inner ear. Introduction Labyrinthitis ossificans (LO) is the formation of fibrous tissue and new bone in the membranous labyrinth. It often occurs after a severe inflammatory process to the inner ear, such as infections, far advanced otosclerosis, and immune-mediated inner ear diseases . Meningitis is the most common cause of LO, mainly caused by Streptococcus Pneumonia . Five percent of streptococcus meningitis may be associated with profound sensorineural hearing loss (SNHL). LO may occur as early as 3 to 21 days after meningitis . Therefore, the hearing should be evaluated as soon as possible in meningitis patients, and a long follow-up period is advisory for late-onset SNHL. The ossification is best evaluated in the early stage with magnetic resonance imaging (MRI). A loss of liquid density on an MRI may hint toward early fibrosis and already challenging surgical Cochlear implant (CI) placement . Bony ossification is also visible on CT scans. Some consensus and protocols suggest the best timing of the surgery is during the first month to avoid the risk of incomplete electrode insertion because of fibrosis or ossification, but it remains very challanging . For years, cochlear ossification was considered a contraindication for cochlear implant (CI) surgery, not only because of the difficulty of inserting the electrode through the ossified cochlea but also because it had been thought that surviving spiral ganglion cells would be affected for adequate stimulation . Years ago, surgical techniques were developed to insert the array even in patients with totally cochlear ossification. However, it remains challenging for otolaryngologists and audiologists to gain auditory benefits . In the surgical context, technological innovations have always been of interest to surgeons to overcome challenges of access to the inner ear. Here we report an autonomous robotic system for inner ear access of 1.0 mm diameter by a trajectory passing through the facial recess with a keyhole tunnel of 1.8 mm diameter . Pre-operative planning was performed with dedicated software that can segment and reconstruct CT imaging of the temporal bone to obtain an accurate cochlear view, calculate the estimation of cochlear duct length (CDL) to predict the electrode insertion depth, and find the best keyhole trajectory. The most optimal trajectory to access the cochlea simulated for robotic keyhole surgery is the best alienation with the basal turn of the cochlea . We have planned this trajectory with dedicated software (Otoplan® CAScination, Bern, Switzerland) and have performed a robotic surgical procedure in a challenging case with a post-meningitis partial cochlear ossification . We report intracochlear findings during this Robotic-Assisted Cochlear Implant Surgery (RACIS) that was proven safe and efficient for the surgical placement of a flexible lateral straight electrode. Access through your organization Check access to the full text by signing in through your organization. Access through your organization Section snippets Case presentation An otherwise healthy 55-year-old male with profound bilateral SNHL was referred for cochlear implant candidacy. His medical history involved childhood meningitis. He became hearing aid dependent. Work-up for Cochlear Implantation was performed when his hearing thresholds deteriorated over time. It included audiological tests, imaging, and psychological evaluation. The pure-tone audiometry and speech audiometry in quit showed profound bilateral deafness. A high-resolution temporal bone CT scan Discussion The primary goal of CI surgery is a full and atraumatic insertion of the electrode array. Cochlear ossification is a challenge for otologists where atraumatic is less of a focus than full insertion. The insertion of the array is extremely complex due to the ossification and fibrosis, and damage to remaining neurosensory cells should be reduced. Several approaches have been reported to increase the insertion part of the electrode: drilling out the ossified part of scala tympani , inserting CI Conclusion Only a very experienced otologist can manage challenges such as cochlear ossification. The robotically assisted surgery applied in the present case provides a novel and efficient approach with high safety and accuracy during the insertion of the array that is much more based on image guidance rather than surgical experience. Cochlear ossification is no longer considered a surgical contraindication for CI. Despite unpredictable audiological outcomes due to the challenge of signal transduction, Recommended articles References (12) J.B. Hinderink et al. Development and application of a health-related quality-of-life instrument for adults with cochlear implants: the Nijmegen Cochlear implant questionnaire Otolaryngol Head Neck Surg (2000) J.D. Green et al. Labyrinthitis ossificans: histopathologic consideration for Cochlear implantation Otolaryngol Head Neck Surg (1990) S.P. Tinling et al. Location and timing of initial osteoid deposition in postmeningitic labyrinthitis ossificans determined by multiple fluorescent labels Laryngoscope (2004) P. Merkus Dutch Cochlear implant group (CI-ON) consensus protocol on postmeningitis hearing evaluation and treatment Otol Neurotol (2010) T. Balkany Surgical technique for implantation of the totally ossified cochlea Laryngoscope (1998) V. Topsakal First study in men evaluating a surgical robotic tool providing autonomous inner ear access for Cochlear implantation Front Neurol (2022) There are more references available in the full text version of this article. Cited by (4) Artificial Intelligence and Pediatric Otolaryngology 2024, Otolaryngologic Clinics of North America Citation Excerpt : With pediatric CI, Bom Braga and colleagues demonstrated the feasibility of planning and performing implantation in phantom models of pediatric subjects.29 Robotic-assisted surgical planning and navigation also have the potential to improve outcomes in complex clinical scenarios such as CI in cases of postmeningitic labyrinthitis ossificans.30 Beyond utilization for CI, robotic assistance for the optimization of transcanal and endoscopic approaches to the middle ear—for example, for congenital cholesteatoma resection—holds promise but remains in very early stages.31 ### Training and validation of a deep learning U-net architecture general model for automated segmentation of inner ear from CT 2024, European Radiology Experimental Show abstract The intricate three-dimensional anatomy of the inner ear presents significant challenges in diagnostic procedures and critical surgical interventions. Recent advancements in deep learning (DL), particularly convolutional neural networks (CNN), have shown promise for segmenting specific structures in medical imaging. This study aimed to train and externally validate an open-source U-net DL general model for automated segmentation of the inner ear from computed tomography (CT) scans, using quantitative and qualitative assessments. In this multicenter study, we retrospectively collected a dataset of 271 CT scans to train an open-source U-net CNN model. An external set of 70 CT scans was used to evaluate the performance of the trained model. The model’s efficacy was quantitatively assessed using the Dice similarity coefficient (DSC) and qualitatively assessed using a 4-level Likert score. For comparative analysis, manual segmentation served as the reference standard, with assessments made on both training and validation datasets, as well as stratified analysis of normal and pathological subgroups. The optimized model yielded a mean DSC of 0.83 and achieved a Likert score of 1 in 42% of the cases, in conjunction with a significantly reduced processing time. Nevertheless, 27% of the patients received an indeterminate Likert score of 4. Overall, the mean DSCs were notably higher in the validation dataset than in the training dataset. This study supports the external validation of an open-source U-net model for the automated segmentation of the inner ear from CT scans. This study optimized and assessed an open-source general deep learning model for automated segmentation of the inner ear using temporal CT scans, offering perspectives for application in clinical routine. The model weights, study datasets, and baseline model are worldwide accessible. A general open-source deep learning model was trained for CT automated inner ear segmentation. The Dice similarity coefficient was 0.83 and a Likert score of 1 was attributed to 42% of automated segmentations. The influence of scanning protocols on the model performances remains to be assessed. Meta-Analysis of Robotic Cochlear Implantation 2025, Laryngoscope ### Defining the ideal trajectory into the inner ear in image-guided cochlear implant surgery 2024, Scientific Reports ☆ Conflict of interest: The authors declare no conflict of interest. Funding: This study did not receive any specific grant from the public, commercial, or not-for-profit funding agencies. View full text © 2022 Elsevier Inc. All rights reserved. Recommended articles Peri-tumoral infiltrate in OSCC: “The simpler, the better” temptation American Journal of Otolaryngology, Volume 44, Issue 1, 2023, Article 103666 Robin Baudouin, …, Stéphane Hans ### Endoscopic stapler-assisted Zenker's diverticulotomy: The surgical technique with video American Journal of Otolaryngology, Volume 44, Issue 1, 2023, Article 103686 Guglielmo Ronzani, …, Roberto Saetti ### Concordant palliative care delivery in advanced head and neck cancer American Journal of Otolaryngology, Volume 44, Issue 1, 2023, Article 103675 Sydney T.Cannon, …, Buddy Marterre ### Leveraging the virtual landscape to promote diversity, equity, and inclusion in Otolaryngology-Head & Neck Surgery American Journal of Otolaryngology, Volume 44, Issue 1, 2023, Article 103673 Carlos A.Ortega, …, Brandon I.Esianor ### Structured histopathology and laboratory evidence in nasal polyposis with different pathogenesis American Journal of Otolaryngology, Volume 44, Issue 1, 2023, Article 103649 Giuseppe Brescia, …, Gino Marioni ### Role of mastoid pneumatization in predicting tympanoplasty results: Will it have the same importance in different age groups? American Journal of Otolaryngology, Volume 44, Issue 1, 2023, Article 103680 Joana Raquel Costa, …, Luís Meireles Show 3 more articles About ScienceDirect Remote access Contact and support Terms and conditions Privacy policy Cookies are used by this site.Cookie settings All content on this site: Copyright © 2025 Elsevier B.V., its licensors, and contributors. All rights are reserved, including those for text and data mining, AI training, and similar technologies. For all open access content, the relevant licensing terms apply. We use cookies that are necessary to make our site work. We may also use additional cookies to analyze, improve, and personalize our content and your digital experience. You can manage your cookie preferences using the “Cookie Settings” link. For more information, see ourCookie Policy Cookie Settings Accept all cookies Cookie Preference Center We use cookies which are necessary to make our site work. 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6442	https://documents1.worldbank.org/curated/en/858341468161667176/pdf/371480RUSSIAN01rilJune200601PUBLIC1.pdf	Beyond Transition Экономический вестник о вопросах переходной экономики Апрель — июнь 2006 • Номер 10 (2006) Тема номера: Энергетика Интервью с Кеннетом Рогоффом: “В ближайшие 5 — 7 лет [...] нефть будет стоить 20 долларов за баррель” 3 Кредитноденежная политика и мировые рынки сырой нефти Нуреддин Кричин 5 Нефтяной потенциал стран СНГ Рудигер Аренд и Уильям Томпсон 7 Вставка: "Ресурсное проклятие" и свобода СМИ Георгий Егоров, Сергей Гуриев, Константин Сонин 8 Управление энергоресурсами в Китае Жианг Кежун 9 Каспийская нефть: изменение расстановки сил Ядвига Семиколенова 11 Интервью с Владимиром Миловым: "Государство должно уйти из энергетического сектора" 12 Человеческий капитал и "ресурсное проклятие" Наталья Волчкова, Елена Суслова 14 Новое в экономической науке Энергетическая бедность в Македонии и Чехии Стефан Бузар 15 Либерализация сферы услуг и производительность в Чехии Йенс Арнольд, Беата Яворчик и Аадитья Матту 17 Влияние СМИ на корпоративное управление в России Александр Дайк, Наталья Волчкова, Луиджи Зингалес 19 Существует ли в Чехии "стеклянный потолок"? Степан Джурайда и Теодора Палигорова 21 Иностранные предприятия и эффективность производства Валентин Зеленюк 23 Россия и ВТО: бремя "неприсоединившегося" Богдан Лиссоволик и Ярослав Лиссоволик 24 Новости Всемирного банка 25 Новые публикации 27 Календарь событий 30 www.cefir.ru Рост мирового спроса на нефть тыс. баррелей в день Европа Страны бывшего СССР Ближний Восток Северная Америка Латинская Америка Африка Азия 809 88 197 212 ?4 6 170 47 47 402 324 332 1279 414 470 275 133 117 98 83 68 Источник: Международное энергетическое агенство, www.iea.org 2004/2005/2006 37148 Public Disclosure Authorized Public Disclosure Authorized Public Disclosure Authorized Public Disclosure Authorized Public Disclosure Authorized Public Disclosure Authorized Public Disclosure Authorized Public Disclosure Authorized Beyond Transition • Апрель — июнь 2006 2 · Уважаемые читатели, Высокие и крайне неустойчивые цены на нефть последних нескольких лет создают различные по своему характеру проблемы для импортеров и экспортеров энергоресурсов. Хотя на первый взгляд кажется, что страны"импортеры находятся в гораздо менее выигрышном положении, Кеннет Рогофф в своем интервью BT утверждает, что экспортеры сталкиваются с куда более серьезными проблемами. Рогофф считает — и в статье Нуреддина Кричина приводятся данные в поддержку такого мнения, — что причиной экономи" ческого спада в 1970"е годы являлась не столько нефть, сколько неэффективная кредитно"денежная по" литика. За последние три десятка лет страны"импортеры смогли укрепить финансовую систему и улуч" шить кредитно"денежную политику, благодаря чему они в состоянии выдержать текущее повышение цен и избежать кризиса производства. К тому же не стоит забывать, что сегодняшний высокий уровень цен, как это всегда случается, не будет длиться вечно. По прогнозу Рогоффа, в течение следующих 5—7 лет мы станем свидетелями падения цен на нефть, что, несомненно, вызовет экономическую нестабильность в странах"экспортерах, финансовая система и макроэкономическая политика которых недостаточно ус" тойчивы. Повышение роли частного сектора в нефтегазовой отрасли могло бы помочь странам"экспортерам более безболезненно пережить па" дение цен, считает Рогофф. В действительности же мы наблюдаем усиление контроля государства над нефтегазовым сектором во мно" гих добывающих странах. Богатые энергоресурсами страны СНГ, такие как Россия и Казахстан, также движутся по пути усиления конт" роля государства, несмотря на достаточно высокую в прошлом степень участия частных и иностранных собственников в нефтяном сек" торе. Как наглядно демонстрируют Аренд и Томпсон, прямым результатом такой политики стал резкий спад темпов роста добычи и ин" вестиций. Страны могут столкнуться с еще более серьезными проблемами в будущем, ведь многочисленные примеры уже доказали, что государственные компании менее эффективны в разработке новых месторождений по сравнению с частными. Например, российский газовый монополист "Газпром" не сумел разработать ни одного крупного газового месторождения за последние 15 лет (Милов). Однако даже у государственных компаний могут появиться стимулы повышать свою производительность, если они поставлены в усло" вия жесткой конкуренции. Недавно вступившие в строй трубопроводы в каспийском регионе, идущие в обход России, несомненно, уве" личат конкуренцию между нефтегазовыми компаниями стран СНГ. Однако Азербайджан, чьи нефтяные запасы оказались существенно переоцененными, возможно, не сыграет большой роли в конкурентной борьбе (Семиколенова). В отличие от развитых энергопотребляющих стран, которые, по мнению Рогоффа, достаточно подготовлены к росту цен на нефть, мно" гие развивающиеся страны столкнулись с серьезными проблемами. Китай, чья быстрорастущая экономика, по распространенному мне" нию, стала не последним фактором в стимулировании роста цен, оказался в особенно серьезном положении. Чтобы поддерживать вы" сокие и устойчивые темпы роста, стране необходимо выработать комплексную стратегию управления энергоресурсами. Жианг Кежун рассматривает различные сценарии будущего спроса и предложения энергоресурсов в Китае и рекомендует систему мер, направленных на снижение текущих темпов роста спроса. Энергетическая безопасность — один из основных вопросов повестки дня на саммите "Большой восьмерки" в Санкт"Петербурге. В статьях этого номера показано, что наиболее эффективным способом достижения энергетической безопасности как в странах"экс" портерах, так и странах"импортерах является совершенствование институциональной среды. Первым шагом в этом направлении может быть обеспечение свободы средств массовой информации, что будет способствовать усилению контроля над чиновниками в ходе реали" зации реформ (Егоров, Гуриев, Сонин) и улучшению корпоративного управления в частных компаниях (Волчкова, Дайк, Зингалес). По" вышение эффективности сектора услуг, в частности, финансового сектора, также способствует продвижению в этом направлении. Как утверждают Арнольд, Яворчик и Матту, либерализация сектора услуг и открытие его для прямых иностранных инвестиций позволят наиболее быстро добиться действенных результатов. Ксения Юдаева, Управляющий редактор От редактора: BT: Нефть сегодня стоит более 70 дол ларов за баррель, и скорее всего цены ос танутся на высоком уровне в ближайшем будущем. Не грозит ли миру нефтяной кризис, подобный кризису 1970х? Нет, я не думаю, что это кризисная си" туация, в том смысле, что надвигается гло" бальный экономический спад. Высокие це" ны на нефть — серьезная проблема для мировой экономики, прежде всего потому что они приносят огромные прибыли экс" портерам нефти и столь же огромные из" держки потребителям. Однако за послед" ние несколько лет рост цен в основном был обусловлен значительным ростом мировой экономики, а не фактическими или ожида" емыми перебоями с поставками нефти. На самом деле, экономисты уже давно сомневаются в том, что именно нефть явля" лась основной причиной низких темпов роста производительности и высокой инф" ляции в развитых странах в 1970"е годы. Исследования последних десяти лет пока" зали, что намного более значимую роль сыграла неэффективная кредитно"денеж" ная политика. В 1971—1972 гг. в попытке увеличить спрос и способствовать переизб" ранию американского президента Никсона Федеральная резервная система увеличила денежную массу более чем на 30%. Даже без нефтяного кризиса Федеральная ре" зервная система была бы вынуждена час" тично снизить огромный избыток денег, что" бы избежать гигантского взлета инфляции. Так что вероятность экономического спада после выборов была предопределена задол" го до того, как ОПЕК решила освободиться от западного контроля цен на нефть. Рост цен на нефть способствовал углублению спада, но не был его основным фактором. BT: Насколько лучше приспособлена сегодняшняя мировая экономика к высо ким ценам на энергоносители по срвнению с 1970ми гг.? Наиболее важное изменение заключа" ется в том, что теперь центральные банки могут более спокойно, чем раньше, отно" ситься к повышению цен на нефть. Пос" ледние два десятилетия экономисты назы" вают эпохой "великой умеренности". Не" стабильность снизилась, финансовые рын" ки окрепли, а рынки товаров и труда стали более гибкими. Денежно"кредитная поли" тика явно улучшилась, благодаря глобали" зации и большей независимости централь" ных банков, уделяющих серьезное внима" ние проблемам инфляции. Поскольку те" перь инфляционные ожидания лучше обос" новываются, центробанкам не нужно по" вышать процентные ставки так быстро, как раньше, когда инфляция росла из"за повышения цен на энергоносители. Более того, с 1970"х гг. доля энергии в производстве промышленно развитых стран снизилась почти на 50%, и нефть те" перь намного больше используется в транспортной сфере, чем в производстве. Поэтому рост цен на нее не вызывает цеп" ную реакцию в экономике, как раньше. Однако мы не должны спокойно отно" ситься к высоким ценам на нефть. Одна из серьезных проблем — их влияние на ог" ромный дефицит платежного баланса Сое" диненных Штатов, составляющий на сегод" няшний день 6% национального дохода. По оценкам МВФ, половина роста этого дефи" цита за последние два года была обусловле" на импортом нефти. Принимая во внима" ние, что неупорядоченная корректировка платежного баланса США подвергнет ми" ровую экономику существенным рискам и что эта проблема осложняется повышени" ем цен на нефть, текущий ценовой цикл мо" жет не иметь "счастливого конца". BT: Какие страны наиболее уязвимы перед лицом нефтяного кризиса? На самом деле, экспортеры нефти вы" зывают у меня гораздо больше беспокой" ства, чем страны"импортеры. Высокие це" ны выгодны экспортерам, но рост цен в определенной мере— часть цикла, и вопрос в том, как экспортеры будут реагировать на падающие цены. В большинстве экспорти" рующих стран очень слабо развиты финан" совые системы, поэтому странам трудно справляться с макроэкономической вола" тильностью. Высокие цены на нефть также сдерживают проведение важных структур" ных реформ по диверсификации экономики. Из"за этого большинство стран"экспорте" ров нефти не ощутили благ "великой уме" ренности" в той мере, в какой это сделали промышленно развитые страны. Россия, в частности, конечно, все еще чрезвычайно чувствительна к волатильнос" ти цен на нефть. В финансовой системе по" ка преобладают государственные банки, при том что другие формы финансирования, к сожалению, недостаточно развиты. Для развития финансовых рынков нужны боль" шая прозрачность, более эффективные инс" титуты и система управления. Центральный банк России менее независим, чем было бы желательно для долгосрочного сдержива" ния инфляции и развития экономики. К то" му же Россия пока не вступила в ВТО. Та" ким образом, по всем этим показателям · 3 Тема номера: Энергетика Всемирный банк и ЦЭФИР Кеннет Рогофф: "В ближайшие пятьсемь лет […] нефть будет стоить 20 долларов за баррель" Кеннет Рогофф — профессор экономики и государственной политики в Гарвардском универси? тете. Он — автор множества научных работ по международным финансам, в частности по воп? росам валютных курсов, структуре центрального банка и финансовой глобализации. Сейчас проф. Рогофф работает над книгой "Нефть и мировая экономика". Во время недавнего визита в Москву проф. Рогофф поделился с BT своими взглядами на рынки энергоносителей. Россия более уязвима чем, скажем, Канада — еще один экспортер энергоноси" телей и член "Большой восьмерки". То есть, несмотря на рост цен на нефть, благодаря спросу в Индии и Китае ни одна страна не может рассчитывать на то, что цена в 70 долларов за баррель будет веч" ной. В ближайшие пять"семь лет непре" менно наступит, по крайней мере, короткий период, когда нефть будет стоить 20 долла" ров за баррель. Когда это произойдет, та" кие страны, как Венесуэла, где производ" ство снизилось на 40% и где чрезвычайно возросли бюджетные расходы, столкнутся с огромными проблемами. BT: Какие меры могли бы способство вать снижению макроэкономических рисков, связанных с волатильностью цен на нефть? У стран — производителей нефти масса проблем. В первую очередь, им необходимы более развитые финансовые рынки, кото" рые помогут справиться со всеми видами волатильности, включая цены на нефть. Способность диверсифицировать некото" рые из рисков на международных рынках также была бы полезна. К сожалению, на" ционализация нефтяных компаний очень усложняет диверсификацию рисков; около 70% мировых запасов нефти находится в руках национальных нефтяных компаний. Государственные нефтяные компании до некоторой степени могут осуществлять ди" версификацию через финансовые контрак" ты, но они не могут сделать многого без фактического отказа от своей собственнос" ти. Наличие более гибкого валютного курса могло бы помочь, тем не менее диверсифи" кация остается очень серьезной проблемой. Между прочим, некоторые считают, что нефтяное богатство — это "ресурсное проклятие". Этот термин придумал гар" вардский историк экономической мысли Дэвид Ландес, который утверждал, что, когда государство располагает большими запасами природных ресурсов, оно в мень" шей степени заинтересовано в развитии среднего класса для стимулирования эко" номического роста, так как экономика мо" жет расти и благодаря природным ресур" сам. Но я думаю, что эта идея была слиш" ком раздута. Между прочим, в США боль" шие запасы природных ресурсов, так же как в Канаде, Австралии и Новой Зелан" дии. Так что если бы меня спросили, хотел бы я жить в стране с большими запасами полезных ископаемых, я бы рискнул отве" тить "да". Главное — иметь при этом эф" фективные институты. BT: А что Вы можете сказать о стра нах — потребителях нефти? Большинство этих стран хорошо усвои" ли уроки [прошлого кризиса]: у них более развитые финансовые рынки, хорошая нормативно"правовая база и эффективная валютная политика. Если бы страны"экс" портеры смогли достичь того же уровня, они, конечно, испытывали бы гораздо меньше трудностей. BT: Новые члены ЕС тоже "усвоили уроки"? Нет, Вы правы, у этих стран все еще много нерешенных вопросов, и высокие цены на нефть создают им много проблем. Новые члены ЕС по"прежнему используют много нефти в производстве, что, как я уже отмечал, может повлечь "цепную реакцию" [в экономике], нежели когда нефть пре" имущественно используется в транспорте. BT: Возвращаясь к вопросу национа лизации, как Вы объясняете недавнюю серию национализаций в нефтегазовой отрасли? На самом деле это давняя тенденция, и ситуация отличается по регионам. Ближне" восточные страны богаты в одних облас" тях, но слабо развиты во многих других. Они пока остаются развивающимися стра" нами, работающими над созданием инсти" тутов и децентрализацией экономики. Ког" да несколько десятилетий назад эти страны начали самостоятельно управлять своими нефтяными богатствами, перестав зави" сеть от иностранных нефтяных компаний, большинство их граждан, конечно, выигра" ло от этого. Но времена прошли, и теперь эти страны должны думать о расширении присутствия частного сектора. Россия, очевидно, иной случай. Беспо" рядочная приватизация в России в 1990"х гг. закончилась передачей многих богатств уз" кому кругу лиц. Это фактическое разграбле" ние национальных ресурсов привело ко все" возможным проблемам в области управле" ния и законности. Стремление нынешнего правительства страны исправить некоторые последствия очень понятно, хотя возвраще" ние к государственному контролю вряд ли поможет. Я согласен с г"ном Илларионовым [бывшим советником президента по эконо" мический политике], который многократно указывал, что страны с преобладанием част" ной собственности в нефтяной отрасли бо" лее экономически благополучны. BT: Как Вы прокомментируете недав ние события в Венесуэле и Боливии? Трудно найти какие"либо положитель" ные стороны недавних национализаций в Латинской Америке. В Венесуэле добыча нефти сейчас упала до 60% от прежнего уровня, после того как Чавес взял нефтя" ную отрасль под контроль. Вместо нацио" нализации нефтяных компаний было бы лучше усовершенствовать перераспреде" ление доходов и применение налогового за" конодательства. Ситуация в Боливии так" же проблематична. Там произошла класси" ческая конфискация иностранных активов, несущая краткосрочные выгоды, но потен" циально огромные долгосрочные издержки. К сожалению, недавние национализации энергоресурсов в Латинской Америке — гигантский шаг назад, последствия которо" го, вероятно, будут отражаться на регионе еще не одно десятилетие. BT: Сейчас в России много спорят о том, что делать со стабилизационным фондом, в котором уже накопилось более 60 млрд долларов США. Как, по Вашему мнению, можно наилучшим образом ис пользовать этот фонд? Вообще, аккумулировать часть непред" виденных доходов от временно высоких цен на нефть — хорошая идея. Я надеюсь, что в итоге правительство найдет способ напра" вить часть высоких доходов на развитие не" которых наиболее бедных регионов, в кото" рых слабо развиты инфраструктура, здра" воохранение и образование. К сожалению, как это часто бывает в российской истории, большая часть денег оседает в Москве и немного достается Санкт"Петербургу. BT: Один из главных аргументов про тив использования стабфонда —инфля ционные риски… На самом деле, накопление денег в нефтяном фонде ослабляет инфляционное давление, так как выводит денежные ре" сурсы из экономики. Тем не менее ускоре" ние укрепления рубля было бы достаточно для снижения уровня инфляции в России ниже двузначного показателя (одного из худших в мире). Многие в России считают, что валютные интервенции помогли под" держанию низкого реального обменного курса, но я сомневаюсь, что совокупное воздействие на реальный валютный курс составило более 10 — 15%. Вместо этого инфляции пришлось нести основное бремя корректировки реального валютного кур" са. Это нежелательный компромисс. Сле" дуя нынешней политике низкого номиналь" ного укрепления рубля, Россия добивается только высокой инфляции. Разрешив по" вышение номинального валютного курса, можно было бы достичь таких же показате" лей экономического роста. Интервью брала Ольга Мосина BT · 4 Beyond Transition • Апрель — июнь 2006 Тема номера: Энергетика · 5 Всемирный банк и ЦЭФИР Цены на сырую нефть резко выросли примерно с 30 долл. США за баррель в 2004 г. до почти 70 долл./барр. в сентябре 2005 г., примерно на 133%. Несмотря на столь стремительное повышение цен, объем поставок сырой нефти почти не изменился, оставшись на уровне 84—85 млн барр. в день, что свидетельствует о серьезных ресу" рсных ограничениях на добычу сырой нефти. В течение всего периода были значи" тельными и колебания цен. Между февра" лем и сентябрем 2005 г. подразумеваемая неустойчивость цены опциона на покупку сырой нефти в среднем составила около 30%, что означало значительную неопреде" ленность ожиданий рынка в отношении движения цен (Рис. 1). Волатильность впоследствии возросла до 40%, что свиде" тельствует о большой чувствительности рынка даже к небольшим потрясениям и новой информации. Например, в сентябре 2005 г. после урагана Катрина, нанесшего урон американским нефтеперерабатываю" щим заводам в Мексиканском заливе, цены на нефть подскочили выше 70 долл./барр. Неэластичный спрос и жесткое предложение Каковы основные свойства нефтяных рынков, вызывающие существенные коле" бания цен на нефть? Модель на основе системы уравнений на основе ежеквартальных (с 1"го квартала 1984 г. по 2"й квартал 2005 г.) и ежегодных (1970—2005 гг.) данных подтверждает ги" потезу низкой эластичности кратковремен" ного спроса по цене (от "0,02 до "0,03). Это означает, что изменения цен на нефть ока" зывают лишь незначительное частичное влияние на спрос. Таким образом, неустой" чивость нефтяных рынков и их уязвимость к небольшим потрясениям объясняется тем, что в краткосрочном плане потребление энергии определяется наличием оборудова" ния и преобладающими технологиями. Это позволяет лишь в ограниченной степени реагировать на изменения цен. Долговре" менный спрос на сырую нефть также не" эластичен по ценам, хотя и выше кратков" ременного и составляет от "0,03 до "0,08. Краткосрочная эластичность по доходу находится в пределах 0,12 — 0,19. Это яс" но показывает, что спрос на нефть реаги" рует на изменения экономической актив" ности: ее повышение приводит к росту спроса на нефть. И краткосрочный, и долгосрочный спрос на сырую нефть находится в отрица" тельной зависимости от номинального эф" фективного курса доллара. Эластичность обменного курса колеблется в пределах от "0,03 до "0,09, т. е. повышение курса долла" ра вызывает повышение цен на нефть и сни" жение спроса на нее. Процентная ставка от" рицательно связана со спросом на сырую нефть: повышение процентной ставки сни" жает спрос, и наоборот. Изменения про" центных ставок не сразу отражаются на эко" номической активности и ценах — как изве" стно, этот фактор действует с задержкой. Предложение сырой нефти в краткос" рочной перспективе неэластично по цене. При повышении цены не происходит увели" чения добычи в краткосрочной перспективе из"за ограниченных производственных мощностей, фиксированных квот или жела" ния сохранить высокий уровень цен. Точно так же производители не сокращают добы" чу в связи со значительным снижением цен. В некоторых случаях при низких ценах на нефть производители могут продавать ее сверх квоты, чтобы получить крайне необ" ходимые доходы в бюджет. На краткосроч" ное предложение сырой нефти значительно влияет добыча природного газа. Краткос" рочная ценовая эластичность находится в пределах от 0,11 до 0,26, поэтому рост до" бычи природного газа может сопровож" даться увеличением добычи сырой нефти. Долгосрочная ценовая эластичность ос" тается на низком уровне 0,08 — это озна" чает, что предложение нефти на рынок оп" ределяется технологическими факторами и открытиями новых месторождений и не так чувствительно к ценам. Природный газ и в долгосрочной перспективе играет важную роль для поставок сырой нефти: в течение 1970 — 2005 гг. между нефтью и газом прослеживалась линейная корреляция. Таким образом, основным свойством нефтяных рынков является сочетание низ" кой цены, высокой эластичности по дохо" дам и неэластичности предложения. Это объясняет высокую и постоянную неустой" чивость рынков, а также рыночную силу производителей. Двусторонняя зависи мость Период с 1970 по 2005 гг. можно разде" лить на три этапа: • 1970 —1986 гг. — шок предло" жения, максимальная цена на нефть в 1980 г. достигла 41 долл./барр. В результа" те мировая инфляция выросла до двузнач" ного показателя, составив в период с 1974 по 1981 гг. в среднем 10,1%. Чтобы спра" виться с нефтяным шоком и вызванным им инфляционным давлением, были необхо" димы меры денежно"кредитной политики. Процентные ставки росли вслед за ценами на нефть и дошли до максимума, только после того как цены на нефть достигли пи" ка. Так, в 1981 г. процентная ставка по фе" Кредитноденежная политика и мировые рынки сырой нефти Нуреддин Кричин Скачок спроса на нефть, вызванный рекордно низкими процентными ставками, привел в 2004 — 2005 гг. к чрезмерному росту нефтяных цен Рис 1. Подразумеваемая волатильность цен на сырую нефть, февраль — август 2005 г. 45 40 35 30 25 20 15 10 5 0 Подразумеваемая волатильность 10.02.2005 24.02.2005 10.03.2005 24.03.2005 07.04.2005 21.04.2005 05.05.2005 19.05.2005 02.06.2005 16.06.2005 30.06.2005 14.07.2005 28.07.2005 деральным фондам достигла максимально" го значения 19,08%. Из"за высоких про" центных ставок значительно вырос номи" нальный эффективный курс доллара, и нефть стала еще дороже. Следствием этого стало резкое замедление мирового эконо" мического роста в период 1980 — 1982 гг. (в среднем до 0,8%), вызвавшее продол" жительное снижение как цен на сырую нефть, так и темпов мировой инфляции в 1981 — 1986 гг. • 1986 —1999 гг. — период отно" сительной стабильности цен на нефть и ус" тойчивости номинального эффективного курса доллара. Однако процентные ставки были крайне подвижными, что означало проведение активной денежно"кредитной политики, которая в основном была направлена на поддержание экономичес" кого роста и стабильности цен. • 1999 — 2005 гг. — период ре" кордно низких процентных ставок и сниже" ния номинального эффективного курса доллара. Процентные ставки очевидно оп" ределяли цены на сырую нефть. Процент" ная ставка по федеральным фондам поддер" живалась на уровне 1% в течение 2003 — 2004 гг. Цены на сырую нефть начали быстро повышаться, превысив отметку в 70 долл./барр. в сентябре 2005 г. Связь между ценами на сырую нефть, процентными ставками и номинальным эффективным курсом доллара характери" зуется двусторонней зависимостью, в зави" симости от типа шока. Во время шока предложения цены на нефть влияют на процентные ставки, а во время шока спро" са, наоборот, процентные ставки воздей" ствуют на цены на сырую нефть. Жесткая кредитноде нежная политика и стаби лизация нефтяных рынков В 1974 — 1981 гг. во время шока пред" ложения при сохранении устойчивого спроса на нефть процентные ставки опре" делялись ценами на нефть. Кредитно"де" нежная политика была направлена на сни" жение цен через снижение спроса на нефть. Это осуществлялось через сущест" венное повышение реальных процентных ставок. Процентные ставки увеличивались на шесть процентных пунктов на каждые 100% повышения цен на нефть. Так, дан" ные показывают, что в этот период цены на сырую нефть повысились с 11,17 долл./барр. до 40,97 долл./барр., в то вре" мя как процентная ставка по федеральным фондам повысилась с 4,61% до 19,1%. В период снижения цен на нефть в 1982 — 1986 гг. кредитно"денежная политика пос" тепенно смягчалась. В период шока предложения необходи" мо было предпринимать решительные действия в отношении процентных ставок, чтобы сдержать повышение цен на нефть, а это повлекло резкое снижение темпов ми" рового экономического роста. Эти действия подкреплялись и другими мерами по сдер" живанию роста цен на нефть: высокие на" логи на потребление нефти, энергозамеще" ние и энергосбережение, технологические усовершенствования, направленные на бо" лее рациональное использования энергии. В период скачка спроса на нефть в пос" ледние годы, притом что предложение ос" тавалось стабильным, цены на нефть опре" делялись процентными ставками. Низкие процентные ставки вызвали повышение спроса на сырую нефть, что привело к рос" ту цен. Следовательно, существенное уве" личение реальных процентных ставок, как это было в 1974 — 1981 гг., могло бы при" вести спрос в соответствие с предложени" ем и сдержать инфляцию, вызванную вы" сокими ценами на нефть. Однако эти действия могут вызвать временный спад роста мировой экономики. Наша модель предсказывает, что рост доходов на 1% приводит к повышению до" бычи сырой нефти на 0,49%, а ее цены — на 2,8%. Таким образом, рост мировой эко" номики мог в значительной мере способ" ствовать повышению цен на нефть. Рост добычи природного газа на 1% приводит к увеличению производства сырой нефти на 0,30% и снижению ее цены на 1,47%. По" вышение процентных ставок ведет в дол" госрочной перспективе к снижению как спроса на сырую нефть, так и цены на нее. Таким образом, кредитно"денежная по" литика, использующая изменения про" центных ставок и денежной массы, имеет существенное и долгосрочное влияние на совокупный спрос на товары и услуги, а также на цены активов, таких как обмен" ные курсы, цены на жилье и акции. Посто" янное подталкивание вверх цены на нефть в 2004 — 2005 гг. можно объяснять кре" дитно"денежной политикой, направленной на стимулирование экономического роста и приведшей к падению процентных ставок на едином международном рынке капитала до рекордно низкого уровня. Стимулируе" мый низким уровнем процентных ставок и обесцениванием доллара США спрос на нефть растет быстрее, чем предложение. Принимая во внимание краткосрочную це" новую неэластичность как спроса на нефть, так и ее предложения, равновесие достигается посредством значительного повышения цен на нефть. Главный вывод заключается в том, что для стабилизации нефтяных рынков необ" ходимо ужесточение кредитно"денежной политики и повышение ожидаемых реаль" ных процентных ставок. Основываясь на данных за 1970 — 1986 гг., можно утверж" дать, что для сдерживания роста цен на нефть и соответствующих инфляционных последствий может потребоваться сущест" венное ужесточение кредитно"денежной политики. При этом, возможно, придется выбирать между инфляцией и выпуском. А данные за 1986 — 2000 гг. показывают, что устойчивый неинфляционный рост ми" ровой экономики требует определенной стабильности на нефтяных рынках. Нуреддин Кричин (Noureddine Krichene) — экономист МВФ. Полный текст работы дос! тупен по адресу: /external/pubs/cat/longres.cfm?sk=18890.0. Взгляды автора, изложенные в данной рабо! те, могут не совпадать с точкой зрения и по! литикой МВФ. BT · 6 Beyond Transition • Апрель июнь 2006 Ежеквартальный мировой спрос на нефть Ежеквартальные поставки нефти на мировой рынок 87 86 85 84 83 82 81 80 79 78 млн. бар. в день 1Q2004 2Q2004 3Q2004 4Q2004 1Q2005 2Q2005 3Q2005 4Q2005 1Q2006 2Q2006 3Q2006 4Q2006 86 85 84 83 82 81 80 1Q2004 2Q2004 3Q2004 4Q2004 1Q2005 2Q2005 3Q2005 4Q2005 1Q2006 81,8 82,4 83,3 84,2 83,7 84,4 84 84,2 84,8 Источник: Международное энергетическое агентство, 82,6 81,3 82,2 84,2 84,6 82,5 83,3 84,2 85 85,5 87,7 86,4 Тема номера: Энергетика млн. бар. в день · 7 Всемирный банк и ЦЭФИР Нефтяной потенциал стран СНГ Рудигер Аренд и Уильям Томпсон В период с 1998 по 2004 гг. 60% роста мировых поставок нефти приходилось на страны СНГ. По данным Международного энергетического агентства (МЭА), до 2010 г. ожидается рост доли СНГ в миро" вых поставках нефти — в основном за счет Казахстана и Азербайджана. Вероятно, СНГ останется наиболее крупным нефтя" ным регионом после Ближнего Востока, хотя и не станет реальным конкурентом или альтернативой ОПЕК, доля которой на рынке возрастет примерно с 40% в 2002 г. до 53% в 2030 г. Плоды рыночной страте гии В 1990"х гг. политика в отношении нефтяной промышленности трех крупней" ших нефтедобывающих стран СНГ — Рос" сии, Казахстана и Азербайджана — была почти полностью рыночно ориентирован" ной и опиралась на частный сектор: • В России была практически пол" ностью приватизирована нефтяная про" мышленность, в результате чего собствен" ность оказалась в руках инсайдеров отрас" ли или российских финансовых групп. В собственности государства осталась толь" ко инфраструктура и небольшая часть до" бычных и нефтеперерабатывающих предп" риятий. Несмотря на барьеры для иност" ранного капитала, сомнительный характер многих приватизационных сделок и конф" ликты между новыми владельцами и влас" тями, сектор развивался очень динамично. После завершения приватизации инвести" ции, производство и экспорт начали расти быстрыми темпами, что стало возможным благодаря использованию новых техноло" гий на уже разрабатываемых месторожде" ниях и во многих случаях привлечения за" падных сервисных компаний (см. Рис 1). • Казахстан и Азербайджан, остро нуждавшиеся после распада Советского Союза в иностранном капитале и техноло" гиях, пошли по пути значительно большего привлечения в отрасль зарубежных инвес" тиций. В Казахстане были приватизирова" ны предприятия нефтяного сектора, а раз" работка основных нефтяных месторожде" ний велась на основе концессий и соглаше" ний о долевом разделе продукции (СРП) с иностранными инвесторами. Азербайджан также выбрал СРП для привлечения иностранных нефтяных компаний, но при этом азербайджанская государственная нефтяная компания стала партнером во всех проектах. Благодаря разработке но" вых крупных месторождений иностранны" ми консорциумами добыча и экспорт кас" пийской нефти резко возросли. • Второстепенные производители нефти в СНГ — Узбекистан и Туркменистан —избрали третий путь, при котором госуда" рственная собственность сочеталась с нез" начительными иностранными инвестициями или вовсе без них. Наличие альтернативных источников дохода от экспорта (продажи хлопка) позволило им не уделять особого внимания развитию нефтегазового сектора. Обе страны добывают относительно немно" го нефти, и на сегодняшний день не наблю" дается признаков возможного роста нефте" добычи в ближайшие годы, хотя Туркменис" тан может обладать намного большими за" пасами нефти, чем принято считать. Необходимость разработ ки новых месторождений Тенденция к усилению государственно" го контроля над нефтяным и другими "стра" тегическими секторами" в России заставля" ет сомневаться в возможности разработки крайне необходимых новых месторожде" ний. Атака на ЮКОС вновь вызвала неуве" ренность в гарантиях прав собственности, к тому же власти повысили налоговую наг" рузку на нефтяной сектор. Государственный произвол очевиден как в сфере налогообло" жения, так и в сфере выдачи лицензий. Ко" нечно, некоторое повышение налогов в нефтяном секторе было оправданным, од" нако недавние изменения привели к росту искажений, вызванных системой налогооб" ложения, в которой ставка налога на добы" чу полезных ископаемых не зависит от при" были. Это создает существенные препят" ствия как для роста нефтедобычи, так и для инвестиций в разведку и разработку новых месторождений. В то же время в отноше" нии государства к нефтяной инфраструкту" ре превалируют геополитические, а не ком" мерческие соображения. Это в определен" ной степени привело к транспортным огра" ничениям экспорта, при том что изменения в налоговом законодательстве уже и так сделали экспорт менее выгодным. В последние годы в Казахстане также заметно усилились противоречия между го" сударством и частными (иностранными) ин" весторами в основном в результате все бо" лее агрессивного подхода государства к концессионным договорам и СРП. Срок действия самого крупного СРП страны был законодательно аннулирован, несмотря на предыдущие обязательства по обеспечению стабильности контрактов, имели место не" однократные угрозы пересмотра основных соглашений. Новые налоговая и норматив" ная базы, введенные в 2004 — 2005 гг., яв" ляются главным препятствием для новых проектов, так как предусматривают 50% долю государственной нефтегазовой ком" пании КазМунайГаз в каждом из них. Политические изменения по отноше" нию к нефтяному сектору уже вызвали за" медление его роста. В 2004 г. инвестиции в российский нефтяной сектор резко сокра" тились, а вскоре произошло падение тем" пов роста добычи и экспорта. Влияние но" вых подходов на рост нефтяного сектора в Казахстане было менее выраженным, так как многие из законодательных нововведе" ний применяются только к проектам, запу" В 1998 — 2004 гг. страны СНГ обеспечили 60% мирового роста добычи нефти 550 450 350 250 1992 1993 1994 1995 1996 1997 1998 1999 2000 2001 2002 2003 2004 2005 Источник: МЭА, Oil Information 2005 Database. Данные за 2005 г. предварительные Россия Азербайджан Узбекистан Казахстан Туркменистан Бывший СССР Рис 1. Добыча сырой нефти и сжиженного природного газа (млн.тонн) щенным с 1 января 2004 г. Даже при этом ужесточившаяся нормативная политика явно сдерживает рост нефтедобычи. Из трех стран, возможно, наиболее благопри" ятный климат для инвесторов в нефтяном секторе в Азербайджане; после открытия трубопровода Баку—Тбилиси— Джейхан добыча азербайджанской нефти возросла. Но в результате закрытия многих проектов после неудачных попыток обнаружить про" мышленные запасе нефти стабильность ситуации в долгосрочной перспективе вы" зывает сомнения. Взгляд в будущее Растущая открытость Узбекистана и Туркменистана для иностранных инвести" ций в нефтегазовом секторе свидетельству" ет об их понимании роли иностранных ин" вестиций и технологий для реализации сво" его нефтяного потенциала. В настоящее время сочетание внешнеполитических со" ображений с недостаточно благоприятным инвестиционным климатом в этих странах позволяет предположить, что роль инвес" торов все в большей степени будут брать на себя государственные российские компа" нии и национальные нефтяные компании быстроразвивающихся стран"потребите" лей, таких как Китай и Индия. В нефтяном секторе России и Казахста" на прослеживаются тенденции роста нало" говой нагрузки, увеличения доли государ" ственной собственности и все более агрес" сивной и авторитарной политики в отноше" нии лицензирования и регулирования от" расли. Сейчас Россия пытается исправить искажения, вызванные не зависящей от прибыли системой налогообложения неф" тяного сектора, но долгожданная и крайне необходимая реформа законодательства по НДПИ застопорилась. Скорее всего, воз" растание роли государства приведет к еще большей путанице и проволочкам, когда де" ло дойдет до главных решений, таких как разведка и инвестиции в новые месторож" дения. Обе страны отличаются слабостью административного, нормативного и пра" воприменительного потенциала, неудов" летворенностью властей своей неспособ" ностью извлечь большие доходы в условиях высоких цен на нефть. Однако высокая до" ля государственной собственности в нефтя" ном секторе совсем не обязательно обеспе" чит государству возможность получить эти доходы. Может произойти их распыление из"за неэффективной деятельности и извлечения ренты инсайдерами. В конце 2003 г. на контролируемые го" сударством нефтяные компании России приходилось около 17% добычи сырой нефти. С тех пор эта цифра выросла более чем в два раза и может достичь 45% к мо" менту продажи активов ЮКОСа. Неэф" фективность большинства российских го" сударственных компаний в отношении контроля за издержками, производитель" ности, корпоративного управления и инно" ваций не сулит ничего хорошего дальней" шему развитию нефтяной отрасли. Увели" чение доли государственной собственности в нефтедобывающем секторе также может отрицательно повлиять на стимулы остав" шихся частных нефтяных компаний, вызывая недобросовестную конкуренцию с крупными государственными нефтепроиз" водителями. Учитывая объем средств, выделенных на существующие проекты, у Казахстана все еще остаются очень хорошие долго" срочные перспективы развития нефтяной отрасли. Но при существующих налоговом и нормативном режимах запуск новых крупных проектов маловероятен, и, скорее всего, стране не удастся утроить производ" ство нефти к 2015 г., как это запланирова" но, не в последнюю очередь из"за конф" ликта с инвесторами в сфере развития экс" портной инфраструктуры. В России иная ситуация: сейчас ее доля составляет почти 80% добычи нефти в СНГ, и существует значительный потенциал для дальнейшего постепенного развития отрасли. Однако, если в России не начнут своевременно раз" рабатываться новые месторождения, за" медление или даже падение добычи в сред" несрочной перспективе намного более ве" роятны, чем в Казахстане. Рудигер Аренд (Rudiger Ahrend) — эконо! мист, Уильям Томпсон (William Tompson ) — старший экономист департамента экономи! ки ОЭСР. Полный текст работы: o/ECO!WKP(2006)12. Взгляды, изложенные в статье, принадлежат авторам и не обяза! тельно отражают точку зрения ОЭСР или государств!членов этой организации. BT · 8 Beyond Transition • Апрель июнь 2006 "Ресурсное проклятие" и свобода СМИ В 1985 г. перед Михаилом Горбачевым возникла сложная ди" лемма. Без свободы слова реформирование крайне неэффектив" ного бюрократического аппарата и командной экономики каза" лось невозможным. Но свободный обмен информацией поставил бы под угрозу основы господства Коммунистической партии. Эта дилемма характерна для любого авторитарного государ" ственного руководителя. В недемократических обществах глава страны нуждается в независимых источниках информации о ре" зультатах его деятельности (таких, как свободные СМИ). Иначе ему будет сложно предоставить стимулы бюрократии, что может привести к неэффективности экономики и, в итоге, потере им сво" его поста. Глава государства может пойти по пути разрешения сво" боды СМИ, однако независимые и конкурирующие средства ин" формации скорее всего будут предоставлять гражданам в том числе и негативную информацию, что может повысить риск свер" жения лидера страны. Либо глава государства может создать сек" ретную службу, докладывающую о деятельности бюрократии не" посредственно ему. В этом случае появляется риск сговора между контролирующей организацией и бюрократами. Компромисс между необходимостью влиять на бюрократию и потребностью "разделять и властвовать" посредством блокирова" ния информационных потоков особенно заметен в богатых природ" ными ресурсами развивающихся странах. В среднем экономика та" ких стран менее эффективна, чем экономика бедных ресурсами го" сударств. Ученые все больше сходятся во мнении, что главная при" чина замедления экономического роста в богатых природными ре" сурсами странах носит институциональный характер. Наша теорети" ческая модель предсказывает отрицательную связь между изобили" ем природных ресурсов и свободой средств информации; эта связь должна быть особенно ярко выраженной в странах со слаборазви" той демократией. Богатство природных ресурсов снижает необходи" мость обеспечивать надлежащие стимулы бюрократии, что также снижает стремление главы государства иметь независимые СМИ. На основе данных Freedom House о степени свободы СМИ, показателей Polity IV по уровню демократии и автократии и дан" ных BP по запасам нефти мы делаем вывод, что, учитывая уровень экономического развития и демократии, средства информации ме" нее независимы в богатых нефтью странах. Отрицательная связь между наличием природных ресурсов и свободой СМИ особенно ярко выражена в странах с менее развитой демократией, тогда как по"настоящему демократические государства относительно защи" щены от неблагоприятного влияния нефтяных богатств. Сергей Гуриев — профессор корпоративных финансов и ректор Российской экономической школы (РЭШ), Москва. Константин Со! нин — профессор экономики РЭШ. Георгий Егоров — докторант Гар! вардского университета. Статья основана на работе, доступной в препринтах из серии "Научные труды": www.cefir.ru (№ 63). BT Тема номера: Энергетика · 9 Всемирный банк и ЦЭФИР Управление энергоресурсами в Китае Жианг Кежун Благодаря быстрому росту экономики Китая общее потребление первичных энер" гетических ресурсов в стране возросло с 400 млн тонн нефтяного эквивалента (МТНЭ) в 1978 г. до почти 1,32 млрд тонн в 2004 г. при среднем ежегодном темпе роста в 4,7% (см. Рис. 1). Основная причина рез" кого повышения потребности в энергоноси" телях — быстрое расширение энергоемко" го производства. Например, производство стали возросло с 131 млн тонн в 2000 г. до 297 млн тонн в 2004 г. Приблизительно в такой же пропорции увеличилось производ" ство другой энергоемкой продукции. Самый крупный произ водитель угля в мире • Уголь. Китай — самый крупный производитель и потребитель угля. С 1980 по 2004 гг. совокупная добыча рядового уг" ля возросла с 620 млн тонн до более 1900 млн тонн при среднем ежегодном темпе рос" та в 4,8%. Значительная зависимость Китая от угля приводит к серьезным экологичес" ким проблемам и является тяжелым бреме" нем для транспортной системы страны. • Электроэнергия. Совокупный объем производимой электроэнергии вы" рос с 66 Гигаватт (ГВт) в 1980 до 440 ГВт в 2004 г., причем доля гидроэнергетики сос" тавила 23%. Летом 2004 г., по сообщени" ям, 24 из 31 провинций страны испытыва" ли нехватку электроэнергии. • Нефть. С 1980 по 2004 гг. сово" купная добыча сырой нефти в Китае воз" росла с 106 млн тонн до 175 млн тонн. Доля страны в мировой добыче сырой нефти сос" тавляет 4,7%. В связи с быстрым ростом потребности Китая в нефти импорт стано" вится существенным фактором для страны, что в значительной мере влияет на между" народные нефтяные рынки и будущие стра" тегии китайских нефтяных компаний. Рост потребности в энергоносителях привел к их нехватке и экологическим проблемам. Понимая серьезность ситуа" ции, китайское правительство предложило ряд мер и законодательных инициатив в этой области. В стратегическом развитии энергетики первоочередное значение при" дается эффективному использованию энергии и ее экономии. Правительство Ки" тая поставило цель сократить энергопот" ребление на 20% к 2010 г. по сравнению с 2005 г. В 2020 году потребности возрастут на 4080% Для исследования потребности Китая в энергоносителях, энергопроизводства и энергоснабжения в будущем с учетом дос" тупности энергоресурсов и возможного возникновения экологических проблем мы разработали три сценария на основе моде" лей Канадской ассоциации независимых нефтепромышленников (IPAC"Emission и IPAC"AIM). Основные предположения при построении моделей включают вероятно" стный рост численности населения до 1,53 млрд человек к 2030 г. и рост ВВП на 8,2% до 2010 г., на 7% с 2010 по 2020 гг. и на 5,6% с 2020 по 2030 гг. • Базовый сценарий предполагает расширение международной торговли и дальнейшую глобализацию китайской эко" номики и, следовательно, импорт энерго" ресурсов. Результаты базового сценария показы" вают, что к 2020 г. потребность в первичной энергии может достичь 2,1 млрд ТНЭ и 2,7 млрд ТНЭ в 2030 г. Уголь останется главным источником энергии и сможет пок" рыть 58% совокупного спроса на энергоно" сители, доля природного газа в удовлетво" рении энергетических потребностей быстро возрастет с 4% в 2000 г. до 17, 3% в 2030 г. Потребность в электроэнергии возрастет почти в четыре раза и составит 451 МТНЭ к 2030 г. Потребность в нефтепродуктах, используемых в транспортном секторе, уве" личится в 5,5 раз — до 410 МТНЭ к 2030 г., в связи с быстрым увеличением ко" личества транспортных средств. • Сценарий, прогнозирующий высокий спрос, предполагает увеличение потребности Китая в энергоносителях, поскольку после вступления в ВТО страна превратится в глобальный производствен" ный центр. В итоге в Китае возрастет про" изводство в энергоемких отраслях, таких как сталелитейная, цветная металлургия и изготовление строительных материалов. Этот сценарий предполагает привнесение в страну новых технологий и методов энер" госбережения. По этому сценарию потребность в пер" вичной энергии достигнет 2,9 млрд ТНЭ к 2030 г., что на 250 млн тонн больше, чем по К 2030 г. импорт Китаем ископаемого топлива может достичь 680 млн тонн нефтяного эквивалента Рис 1. Производство и потребеление энергоносителей 1800 1600 1400 1200 1000 800 600 400 200 0 1950 1970 1990 2005 Потребление Производство МТНЭ Год Рис 2. Импорт по сценарию с высоким спросом 2000 2010 2020 2030 800 700 600 500 400 300 200 100 0 Пр. газ Нефть Уголь МТНЭ Год базовому сценарию. Уголь обеспечит 59,1% совокупной потребности в первич" ной энергии, нефть — 16,1%, природный газ — 17,8%, атомная энергия — 1,2%. Поскольку этот сценарий предполагает бо" лее глубокую интеграцию в международ" ные рынки, Китай будет в большей степени опираться на импорт природного газа и нефти. • Третий сценарий предполагает осуществление различных энергетичес! ких и экологических программ, снижение потребления энергии в целях экономии энергоресурсов и охраны окружающей сре" ды. Меры энергетической политики могут включать: продвижение энергосбережения конечными потребителями; введение стан" дартов эффективного энергопотребления для зданий; налоги на транспортные сред" ства и энергоносители; разработка возоб" новляемых источников энергии; повыше" ние доли общественного транспорта в горо" дах; повышение эффективности транспорта и использование транспортных средств с низким потреблением топлива; повышение эффективности электростанций, работаю" щих на угле; рост поставок природного газа; развитие атомной энергетики. По этой модели к 2030 г. потребность в энергоресурсах будет ниже почти на 280 МТНЭ по сравнению с базовым сценари" ем. Для этого необходимо своевременно осуществить вышеупомянутые меры. Богатые альтернативные источники энергии Ожидается, что уголь будет и в даль" нейшем играть ключевую роль в обеспече" нии энергетической стабильности страны, но, вероятно, его доля в совокупных ресур" сах ископаемого топлива снизится по срав" нению с нынешними 96%. Природный газ всегда играл в Китае небольшую роль, но за последние два года на материке были обнаружены три очень крупных месторож" дения природного газа. Что касается гидро" электроэнергии, то приоритетной задачей является передача электроэнергии с бога" того водными ресурсами юго"запада Китая в восточную часть, а также развитие малой гидроэнергетики. У Китая также хорошие предпосылки для развития атомной энерге" тики: производства энергии, поставки ядерного топлива, утилизации отходов, а также развития необходимых технологий. Альтернативные источники энергии — биомасса, получаемая в сельском хозяй" стве, лесоводстве и лесной промышлен" ности, а также городские отходы — обла" дают достаточным потенциалом, для того чтобы играть определяющую роль в энер" госнабжении Китая. Например, из отходов сельского хозяйства можно произвести почти 80 млрд кубометров биогаза. Об" ширная территория и протяженная берего" вая линия позволяют Китаю довольно ус" пешно развивать ветроэнергетику. По оценкам научно"исследовательского инс" титута метеорологии Китая, наземные и океанские промышленные ветряные ре" сурсы позволяют получить около 1000 ГВт электроэнергии. Китай уже занимает деся" тое место в мире по совокупному потенци" алу своих 40 ветроэнергетических станций. Сценарии энергоснабже ния Модель будущего производства энер" гии в Китае позволяет сделать прогнозы: • К 2030 г. производство угля мо" жет достичь 1,48 млрд ТНЭ; таким обра" зом, потребность в угле может превысить внутреннюю добычу угля. • Добыча нефти достигнет 190 млн тонн к 2020 г. и 175 млн тонн к 2030 г. • Добыча газа достигнет к 2030 г. 312 млрд кубометров. • Выработка электроэнергии на АЭС существенно возрастет — до 344 млрд кВт/ч к 2030 г., по сравнению с 16,7 млрд кВт/ч в 2000 г., — но из"за высоких издержек составит лишь небольшую часть совокупного производства энергии. • Выработка электроэнергии на гидростанциях увеличится с 224 млрд кВт/ч в 2000 г. до 722 млрд кВт/ч в 2030 г. Следовательно, по базовому сценарию ежегодная потребность Китая в импорте ископаемого топлива составит 375 МТНЭ к 2020 г. и 562 МТНЭ — к 2030 г. Для сравнения: США в 2000 г. импортировали 870 МТНЭ. Нефть возглавит список им" портируемых энергоресурсов, но после 2020 г. Китаю придется импортировать да" же уголь при ежегодной потребности в 129 млн тонн этого топлива. Согласно сценарию с повышенным спросом импорт энергоносителей будет значительно больше (см. Рис. 2). Совокуп" ный импорт ископаемого топлива составит 445 МТНЭ к 2020 г. и 680 МТНЭ — к 2030 г. По этому сценарию предполагается еще больший импорт угля. Чтобы облегчить напряженную ситуа" цию с энергоснабжением, Китаю необхо" димо разработать эффективную энергети" ческую стратегию. Действенными мерами могут оказаться: разработка технологий нового поколения, введение налогов на энергию и использование природных ре" сурсов, введение экспортных пошлин на энергоемкую продукцию для стимулирова" ния энергосбережения, развитие возоб" новляемых источников энергии и создание диверсифицированной системы энергос" набжения. Крайне необходимы также тех" нологии по уменьшению загрязнения окру" жающей среды в результате сгорания угля. Благодаря низким затратам на производ" ство Китай, скорее всего, станет мировым производственным центром энерго" и ре" сурсоемкой продукции. Эта тенденция должна находиться под постоянным конт" ролем, и, кроме того, необходимо следить за включением издержек на охрану окру" жающей среды в производственные расхо" ды. Жианг Кежун (Jiang Kejun) — профессор Научно!исследовательского энергетическо! го института. Полный текст работы см. по адресу: INTDECABCTOK2006/Resources/Kejun_Energy _China.pdf BT · 10 Согласованность действий международного сообщества игра" ет важную роль в обеспечении энергетической безопасности. По" вестку дня формируют следующие ключевые вопросы: • Активная пропаганда эффективного использования энергии во всех странах посредством разработки программ, опре" деляющих широкий спектр задач в международном контексте и способствующих обмену информацией и технологиями; • Осуществление доступа к иностранным инвестициям и обеспечение безопасности международной транспортировки энергоресурсов, чтобы добиться диверсификации компаний"пос" тавщиков энергии на международных и национальных рынках; • Расширение обязательства, принятого на саммите "Большой восьмерки" в Глениглз в отношении снижения выбро" сов в атмосферу двуокиси углерода, и адаптации к условиям изме" нения климата, путем включения этого обязательства в повестку дня по энергетической безопасности; • Инициативы международных организаций по оказанию поддержки беднейшим странам в их борьбе с кратковременными ценовыми шоками, вызванными колебаниями на энергетическом рынке, а в долгосрочной перспективе ведущими к доступности экологически безопасных энергоресурсов для всех граждан; • Обеспечение прозрачности всей последовательности производства и использования энергоресурсов, в особенности рынка нефти, путем поддержки и расширения необходимых ас" пектов действующих инициатив. Источник: Всемирный банк, /INTRUSSIANFEDERATION/Resources/Energy_Security_eng.pdf BT Beyond Transition • Апрель — июнь 2006 Тема номера: Энергетика · 11 Всемирный банк и ЦЭФИР Каспийская нефть: изменение расстановки сил Ядвига Семиколенова После распада советской империи Азер" байджану и Казахстану удалось вызвать ог" ромный интерес к первичной разведке своих малоизученных месторождений нефти и га" за. Несмотря на низкие цены на нефть, не" достоверные данные о ее запасах и отсут" ствие соответствующей законодательной базы, в 1997—1998 гг. в каспийский регион хлынул мощный поток иностранных инвес" тиций. К 1999 г. только в Азербайджане бы" ло подписано более 20 контрактов на раз" ведку нефти, что составило 30 млрд долл. США долгосрочных капиталовложений и 2,5 млрд подтвержденных инвестиций. В Казахстане в период с 1991 по 1996 гг. прямые иностранные инвестиции в нефте" газовую отрасль составили 2 млрд долл. США. Все ожидали, что малоисследован" ные, но потенциально крупные запасы кас" пийской нефти изменят конфигурацию гло" бального энергетического рынка и создадут альтернативу ближневосточной нефти. Ожидания, однако, были сильно пре" увеличены, и вскоре шумиха вокруг кас" пийской нефти спала. Большинство проек" тов было закрыто вследствие недостаточ" ности запасов для коммерческой разработ" ки месторождений. После 1999 г. большин" ство контрактов было аннулировано или заморожено; а две нефтяные компании, Arco и Conoco, решили уйти из региона. Бум как результат "стад ного инстинкта"? Нефтяная лихорадка на Каспии в сере" дине 1990"х с трудом поддается объяснению: • Объем доказанных запасов кас" пийской нефти был недостаточно велик, чтобы привлечь такой мощный поток ин" вестиций за столь короткий срок. Доказан" ные запасы нефти в Азербайджане и Ка" захстане составляли 3,6 и 10 млрд барр., соответственно, что намного меньше, чем, например, 56 млрд барр. в России. • Информация о потенциальном объеме запасов каспийской нефти была весьма недостоверна. Оценки верхней гра" ницы вероятных запасов нефти в Азер" байджане колебались от 13,2 млрд барр. по данным Министерства внутренних дел США до 27 млрд барр. по данным Государ" ственного департамента США. • Когда большинство нефтяных компаний в 1997—1998 гг. начали развед" ку нефти на Каспии, мировые цены на нефть падали. Когда же к 2000 г. цены на нефть вернулись к прежнему уровню, кас" пийская нефтяная лихорадка закончилась. Динамику каспийской нефтяной лихо" радки можно объяснить "стадным инстинк" том" под влиянием информационной кампа" нии. Возможно, многие инвесторы появи" лись в Каспийском регионе, привлеченные информацией в СМИ, — в частности, о первых инвестиционных проектах иност" ранных компаний и результатах бурения первых разведочных скважин. На инвесто" ров также могли повлиять аналитические доклады уважаемых учреждений. Большин" ство компаний появились в регионе в 1997—1998 гг., после обнародования Г ос" депом США высоких оценок потенциальных запасов каспийской нефти и выхода первого проекта на стадию добычи. Лихорадка за" кончилась после появления сообщений о пустых скважинах и закрытия некоторых проектов в конце 1998 — начале 1999 гг. БТД: новый энергетичес кий порядок? Однако все изменилось весной 2006 г., когда вступил в строй трубопровод Баку — Тбилиси — Джейхан (БТД), позволивший перекачивать каспийскую нефть в турец" кий порт Джейхан. Несколькими днями раньше по только что построенному Юж" но"Кавказскому трубопроводу (ЮКТ) на грузинский и турецкий рынки начал посту" пать природный газ. Переговоры о транзи" те нефти в обход России начались еще в 1994 г. По одному из пунктов соглашения о разделе продукции, заключенного Азер" байджаном и BP в 1994 г., иностранные инвесторы должны были проложить экс" портный трубопровод для азербайджанс" кой нефти. Среди нескольких рассматри" вавшихся маршрутов были из Баку в Ново" российск через российскую территорию, из Баку к грузинскому черноморскому порту Супса и БТД через Г рузию в Турцию. До 2003 г. ни один из трубопроводов в обход России не казался достаточно серьез" ной угрозой российской региональной моно" полии на транспортировку нефти. По мне" нию специалистов, пропускную способность трубопроводов Баку — Супса и Баку — Новороссийск можно было увеличить толь" ко до 450 000 барр. каспийской нефти в день, что было бы недостаточно для пере" качки азербайджанской сырой нефти, когда ее добыча на месторождениях Азери, Чираг и глубоководном Гюнешли (проект Азер" байджанской международной операцион" ной компании (АМОК) под руководством BP) вышла бы в 2007 — 2010 гг. на полную мощность в 700 000 — 800 000 барр. в день. До 2003 г. казалось, что после выхода АМОК на полную мощность российские нефтепроводы останутся единственным пу" тем экспорта азербайджанской нефти. Однако в конце 2003 г. вероятность транспортировки нефти из Азербайджана в обход России приобрела реальные очерта" ния. Благодаря повышению цен на нефть, созданная в 2002 г. и возглавляемая BP компания БТД привлекла достаточное ко" личество инвесторов, чтобы продолжить осуществление проекта. В следующем году представители Азербайджана и Казахстана начали переговоры о транспортировке нефти из Казахстана по трубопроводу БТД. Осенью 2005 г. строительство нефтепро" вода было завершено, и в июне 2006 г. пер" вая нефть поступила по нему на терминал в Джейхане. Появление нефтепровода Баку — Тби" лиси — Джейхан и Южно"Кавказского га" зопровода — серьезный этап, знаменую" щий конец российского господства на ев" ропейских энергетических рынках. Это повлияет на потребление энергоресурсов и геополитические расклады в регионе. Во" первых, будет ослаблена почти полная мо" нополия России на транзит энергоресурсов из прикаспийских государств на мировые рынки. Во"вторых, России придется вклю" чаться в международную конкуренцию, но каким образом это произойдет, сказать трудно. В"третьих, у европейских потреби" телей появляется доступ к каспийской нефти в обход России. Наконец, положе" ние Турции на перекрестке энергопотоков позволит ей играть более важную роль на европейском энергетическом рынке. Ядвига Семиколенова (Yadviga Semikolenova) — профессор факультета экономики и бизнеса горного университета штата Колорадо, США. Статья основана на работах, которые можно получить у ав! тора по запросу по эл. почте: yvsst2@pitt.edu BT Появление новых трубопроводов знаменует конец российской монополии на европейских рынках · 12 Beyond Transition • Апрель июнь 2006 Владимир Милов: "Государство долж но уйти из энергетического сектора" BT: Энергетическая безопасность — одна из основных тем саммита "восьмер ки" в СанктПетербурге в этом году. Что под этим понимают Россия в целом и "Г азпром" в частности? В связи с крайне неравномерным гео" графическим распределением энергети" ческих ресурсов справедливо говорить о том, что понимание энергетической безо" пасности варьируется по странам в зависи" мости от обеспеченности ресурсами или возможности доступа к ресурсам в других государствах. Сам термин появился после арабского нефтяного эмбарго в 1970"е гг., но за прошедшее время консенсусного по" нимания энергетической безопасности так и не сложилось. Несмотря на то что эта те" ма была предложена Россией для саммита "восьмерки" год назад, я с сожалением вы" нужден констатировать, что за прошедший год мир не приблизился к общему понима" нию того, что такое энергетическая безо" пасность. Подготовка саммита была про" валена интеллектуально, и Россия, к сожа" лению, смогла предложить миру только набор штампов, который не соотносится с реальным положением дел. Более того, своими практическими действиями мы только подогрели неопределенность в этой сфере. BT: Почему страны могут быть заин тересованы в достижении консенсуса? Президент Путин правильно написал в своей статье в "The Wall Street Journal" в этом году, что энергетический эгоизм не окупается. Примерно три четверти миро" вых энергетических ресурсов сосредоточе" ны в десятке стран, которые производят чуть больше 5% мирового ВВП по парите" ту покупательной способности. Страны, производящие более 70% мирового ВВП, владеют только 10% нефтегазовых ресур" сов, которые постепенно истощаются. Та" кой глобальный дисбаланс несет серьез" ный конфликтный потенциал. Если не удастся создать устойчивую систему обес" печения человечества энергоресурсами из нескольких ресурсно"богатых стран, то нас ждут ресурсные войны. Думаю, в первую очередь это будет зависеть от таких стран, как Россия, Саудовская Аравия и Иран. Роль России в нефтяной сфере не нас" только глобальна, как у Саудовской Ара" вии, но Россия контролирует больше трети мировых доказанных запасов газа, а с уче" том неразведанных запасов — почти поло" вину. Газ будет пользоваться все возраста" ющим спросом на мировом рынке — за последние 15 лет почти во всех регионах мира потребление газа росло гораздо бо" лее быстрыми темпами, чем потребление других видов топлива. В настоящий момент, особенно с появ" лением у власти контроля над важнейшими ресурсами, в первую очередь "Г азпромом", Россия ведет себя довольно эгоистично и требует специальных условий на поставку энергоресурсов на международные рынки. Тем самым нарушается некая неформаль" ная договоренность с Западом времен Со" ветского Союза — мы вам поставляем энергию, вы нам за это платите очень хоро" шие деньги, — создавшая стране репута" цию надежного поставщика. Россия теперь хочет большего. Сегодня российские власти говорят, что всего лишь хотят получить дос" туп к распределительным сетям в европейс" ких странах взамен на предоставление дос" тупа к своим ресурсам, но что они потребу" ют завтра? Если все ресурсно"богатые страны начнут вести себя подобным обра" зом, может сложиться серьезная ситуация, и признаки подобного поведения, например, со стороны Ирана и Венесуэлы, уже налицо. Другой подход к пониманию энергети" ческой безопасности заключается в нахож" дении основ для глобального консенсуса и создания долгосрочных гарантий для ре" сурсно"богатых стран в том, что они смогут осуществлять инвестиции в энергетику с минимальным для себя риском. BT: Какие механизмы гарантирова ния Вы имеете в виду — долгосрочные контракты? Я бы не хотел говорить о конкретных механизмах, их много, долгосрочные конт" ракты — самый простой из них. Важно признание того факта, что страны"произ" водители нуждаются в некоторых коммер" чески привлекательных условиях разра" ботки своих ресурсов, а страны"импорте" ры — в гарантиях получения достаточных ресурсов на рыночных условиях. При этом ни одна страна не будет пытаться узурпи" ровать ресурсы и требовать получения специальных политических привилегий на международной арене. BT: У многих экспертов и инвесторов вызывает беспокойство падение добычи на основных месторождениях "Г азпрома" в Западной Сибири. Но предправления Миллер недавно оптимистично заявил на мировом газовом конгрессе, что "в 2005 году "Г азпром" продемонстрировал ре кордный за период с 1993 г. прирост запа сов газа". Насколько оправданы опасения о возможном дефиците газа и какими мо гут быть последствия для потребителей? Проблема не в том, что у России нет за" пасов в земле — обеспеченность текущей добычи газа его запасами превышает 80 лет. Есть огромная проблема с экономической системой и стимулами, которые обеспечи" вали бы своевременное извлечение запасов из"под земли и их поставку на рынок. В 1992 г., когда создавался "Г азпром", об" суждалась идея создать на его основе нес" колько независимых частных газодобываю" щих компаний, конкурирующих между со" бой. Победили сторонники создания цент" рализованной системы, главным аргумен" том которых было то, что только мощная Владимир Милов — президент Института энергетической политики в Москве, в прошлом замми? нистра энергетики Российской Федерации. В своем интервью BT он поделился своими взгляда? ми на энергетическую безопасность и проблемы российской газовой отрасли. Тема номера: Энергетика вертикально интегрированная компания сможет разрабатывать сверхкрупные мес" торождения. В России есть очень крупные месторождения, например, только два из них на Ямальском полуострове содержат более 10% всех российских доказанных за" пасов газа. Проблема в том, что за 15 лет с момента создания "Газпром" разработал только проект технико"экономического обоснования освоения Бованенковского месторождения, который был отклонен в прошлом году из"за низкого качества под" готовки материалов. На мой взгляд, это приговор "Газпрому". Сверхкрупные месторождения, введен" ные еще в советское время и дающие сегод" ня основную часть российской добычи, быстро истощаются. В ближайшее время "Г азпром" сможет только ввести сателлит" ные месторождения в действующих регионах добычи, чтобы избежать падения добычи. Но этот потенциал будет быстро исчерпан. Кумулятивные инвестиции "Газпрома" в развитие газодобычи за последние 7 лет составят всего 12,5 млрд долл. в сегодняш" них ценах. Для сравнения: инвестиции в до" бычу нефти с 1999 по 2004 гг. составили, по официальным данным, 37 млрд долл., и рост нефтедобычи в последние годы — это действительно инвестиционный рост. Газ" пром же предпочитает инвестировать в другие проекты. В предыдущие годы основ" ным приоритетом в структуре капитальных вложений "Газпрома" были инвестиции в развитие новых трубопроводов, в первую очередь экспортных, и приобретение акти" вов в газовой отрасли, нефтехимии, нефти, электроэнергетике и т. д. Дефицит газа не "предстоит", он уже есть. Так как спрос на газ носит ярко выра" женный сезонный характер и достигает пи" ка зимой, дефицит ощущается только в пи" ковый период. Во время холодов прошлой зимой пострадали все потребители — и российские, и внешние. Подача газа на электростанции (электроэнергетика — ос" новной потребитель газа в России) в энер" гозонах Центра, Волги и Северо"Запада в некоторые дни ограничивалась до уровня в 15% по сравнению с согласованными объ" емами. В то же время "Газпром" не мог обеспечить достаточными объемами газа украинских и европейских потребителей. Украинцы, первыми имеющие доступ к "трубе", могли отбирать газ, а европейцы недополучили существенные объемы газа. BT: Верно ли, что реальные интересы “Г азпрома” связаны не с добычей газа, а с его экспортом, переработкой и продажей? "Газпром" предпочитает инвестировать в увеличение своей монопольной силы. Компания работает всего лишь с годичным бюджетом и не занимается долгосрочным финансовым планированием. В таких усло" виях осуществлять долгосрочные рискован" ные проекты противоречит экономической логике существования компании. "Газпром" рассматривает заимствования не как источ" ник финансирования новых проектов, а как источник покрытия кассовых разрывов бюджета. Причем понятны трудности "Газпрома" с балансированием своего бюд" жета во времена неплатежей со стороны российских потребителей и низких евро" пейских цен на газ, но и сегодня компания умудряется продолжать еле сводить концы с концами в условиях космических цен в Ев" ропе и роста внутренних цен в реальном выражении в 3 раза за последние 6 лет. На мой взгляд, это доказывает, что экономи" ческая система "Газпрома" не является са" мостоятельным генератором экономичес" кой и инвестиционной логики, а всего лишь игрушкой в руках внешних и внутренних лоббистов. Это приводит к тому, что "Газ" пром" вкладывает средства в те проекты, которые могут принести доход инсайдерам либо могут увеличить его монопольную си" лу и обеспечить более короткий возврат этих средств по сравнению с долгими ин" вестициями в новую газодобычу. BT: Как Вы относитесь к заявлениям "Г азпрома" об обмене активами с евро пейскими компаниями? Это естественный путь интеграции рос" сийского бизнеса с международным. Рос" сийские нефтегазовые компании, сформи" рованные в результате приватизации в на" чале 1990"х гг., неконкурентоспособны на международном рынке. Им необходимо ин" тегрироваться с крупными международны" ми структурами, чтобы иметь возможность влиять на деятельность международных корпораций, получать новые технологии и учиться умению инвестировать в сложные проекты на длительный срок. Интеграция цепочки создания стоимости, в которой присутствуют перекрестные интересы раз" ных наций, закладывает основы для непо" литического эффективного и стабильного взаимодействия между ресурсно"богатыми и ресурсно"бедными странами. Интеграция с европейскими компаниями была бы воз" можна, если бы наши руководители не вме" шивались со своими личными интересами и попытками все контролировать. BT: Так стоит ли европейцам бояться инвестиций "Г азпрома" в свои газорасп ределительные сети? "Газпром" воспринимается как не са" мый удачный агент для интеграции. Во"пер" вых, у компании много внутренних проблем и большая внутренняя неэффективность, что затрудняет создание альянсов по ком" мерческим соображениям. Во"вторых, "Газпром" — государственная компания, и его побаиваются по политическим причи" нам. Но Россия и сама ведет себя подобным образом. Например, государственная ки" тайская компания CNPC не была допущена на аукцион по "Славнефти" в 2002 г. "Г азпром" хвастается своей договорен" ностью с BASF/Wintershall о продаже доли в Южно"Русском месторождении, но это лишь единичный случай и, скорее, исключе" ние из правил из"за того, что Wintershall в хороших отношениях с некоторыми инсайде" рами “Г азпрома”. Других примеров успеш" ного партнерства у "Г азпрома" нет. Главное, из"за чего не получаются партнерства, — политическая составляющая "Г азпрома". BT: Может ли обмен активами при вести к разделению "Г азпрома"? И какое будущее может ожидать компанию? Я думаю, "Газпром" ожидает худшее бу" дущее, чем Советский Союз в конце 1980"х гг. Я не хотел бы быть мрачным пророком, но думаю, нас ждет полная де" зинтеграция отрасли с очень серьезными последствиями для рынка газа. У "Газпрома" есть лицензии на огром" ные месторождения, которые не разраба" тываются. Продав лицензии сторонним ин" весторам или отдав им долю акционерного капитала в компаниях, эксплуатирующих месторождения, можно быстро решить проблему накопленного долга "Газпрома" и перестроить весь его экономический меха" низм. В то же время сторонние инвесторы получили бы возможность быстрой разра" ботки месторождений. Пока же "Газпром" везде настаивает на сохранении контрольного пакета, что при нынешней ситуации с финансами, систе" мой принятия решений и общей эффектив" ности работы компании означает невоз" можность акционерного финансирования крупных проектов. На примере разработки нефтяных мес" торождений частными (иностранными) ин" весторами, например проектов Сахалин"1 и Сахалин"2, видно, что активы, попадаю" щие в частные руки, уже дают отдачу. Эти проекты обещают быть успешными, в от" личие от других крупных сахалинских про" ектов, где государство само захотело пору" ководить и где до сих пор ничего не проис" ходит. Государство должно уйти из энерге" тического сектора и открыть его для иност" ранных компаний, иначе не будет никакого движения вперед. Интервью брала Ольга Мосина BT · 13 Всемирный банк и ЦЭФИР · 14 Человеческий капитал и "ресурсное проклятие" Наталья Волчкова, Елена Суслова Один из основных вопросов для рос" сийской экономики — ускорение экономи" ческого роста и создание инфраструктуры, способной обеспечить устойчивость этого роста. Существенным препятствием на этом пути может стать так называемое "ре" сурсное проклятие" — эмпирическая зако" номерность, заключающаяся в том, что экономики богатых природными ресурсами стран в среднем растут медленнее, чем бед" ных ресурсами стран. Три канала распростра нения "проклятия" В экономической науке есть три основ" ные группы гипотез относительно каналов распространения "проклятия". • Макроэкономический канал связан с нестабильностью цен на ресурсы на мировых рынках и возникающей в этой связи волатильностью ВВП и доходов гос" бюджета богатых ресурсами стран, что мо" жет служить существенным препятствием для достижения долгосрочного стабильно" го роста экономики. • Наиболее часто обсуждаемым микроэкономическим каналом является так называемая "голландская болезнь", ме" ханизм, состоящий в перетоке факторов производства из обрабатывающей промыш" ленности в сектора экономики, производя" щие товары и услуги исключительно для внутреннего рынка, и добывающие сектора экономики в ответ на рост доходов в добы" вающих отраслях. Однако эмпирические ис" следования по многим ресурсозависимым странам не поддерживают гипотезы о суще" ствовании отрицательной зависимости тем" пов долгосрочного роста и волатильности условий торговли или об отрицательном влиянии размера добывающего сектора промышленности на размер обрабатываю" щего. Исследования российской экономики в течение 15 лет также не позволяют гово" рить о “голландской болезни”в России. • Все больше исследователей склонны считать, что основным препят" ствием для дальнейшего развития ресурсо" зависимых стран являются недостаточ! но развитые институты — как рыноч" ные, так и государственные. Однако в сырьевых экономиках, в силу высокого объема экономической ренты в добываю" щих отраслях, институциональная отста" лость может приводить к тому, что меры экономической политики ведут к противо" положным результатам и усиливают сырь" евую зависимость страны. Сочетание значительной ресурсной рен" ты, плохой защиты прав собственности, не" достаточно развитых и несовершенных рын" ков и слабой правовой системы может иметь деструктивные последствия для эко" номики. Большая доля гражданских конф" ликтов XX века была инициирована именно попытками захватить контроль над рентой, в менее выраженных случаях борьба за рен" ту влечет коррупцию в бизнесе и государ" стве, что существенно искажает эффектив" ность размещения ресурсов в экономике. Наличие обширных запасов ресурсов и генерируемый ими доход не способствует созданию стимулов у государства для про" ведения экономических реформ, повыше" нию эффективности бюрократического уп" равления, улучшению качества институтов или, говоря другими словами, обширные запасы ресурсного капитала вытесняют ка" питал социальный. Человеческий капитал в ресурсозависимых странах Анализ 44 стран и 11 отраслей за период 1980—1990 гг. позволяет сделать выводы относительно развития человеческого капи" тала в ресурсозависимых экономиках и его влияния на темпы роста экономики в целом. Этот канал является частью институци" онального, и гипотеза о его существовании базируется на двух предпосылках. Во"пер" вых, значительное число исследований указывает на то, что в богатых природными ресурсами экономиках возможно недоин" вестирование в человеческий капитал, поскольку ресурсоинтенсивные сектора экономики привлекают основную часть ин" вестиций в экономике, но не создают высо" коквалифицированных рабочих мест. Это снижает стимулы как частного, так и госу" дарственного секторов вкладывать сред" ства в образование. Во"вторых, как классическая, так и но" вая теория роста подчеркивает важность накопления человеческого капитала для генерации долгосрочного экономического роста. Так что можно ожидать, что страны, богатые природными ресурсами, проигры" вают в темпах роста странам без ресурсов в силу недостаточного развития в первых странах человеческого капитала. Наш подход к исследованию проблемы “ресурсного проклятия” основан на методе “разницы разниц”, что позволяет отслежи" вать разницу в темпах роста разных секторов промышленности в разных странах, учиты" вая страновые и отраслевые эффекты. От" расли промышленности ранжированы сог" ласно спросу этой отрасли на человеческий капитал определенного уровня, а страны — по обеспеченности природными ресурсами. Результаты анализа показали, что раз" ница в темпах роста отраслей, нуждающих" ся в большем количестве работников с вы" соким уровнем человеческого капитала, и отраслей, менее нуждающихся в таких ра" ботниках, меньше в странах с высокой до" лей первичного экспорта. Иначе говоря, в богатых ресурсами странах отрасли, ис" пользующие работников с высоким уров" нем человеческого капитала, а это — неф" техимическая промышленность, машино" строение (кроме электроники) и др., нахо" дятся в менее выгодном положении по от" ношению, например, к пищевой промыш" ленности, чем в бедных ресурсами страна" ми. Этот результат был также подтвержден в отношении еще одного критерия ресурс" ного богатства — производства нефти и углеводородов. Суммируя результаты, получаем, что: • интенсивное использование при" родных ресурсов подавляет рост в отраслях с высоким уровнем человеческого капитала; • чем выше зависимость отрасли от человеческого капитала, тем больше она проигрывает от разработки природных ре" сурсов. Данные результаты подчеркивают роль инвестиций в образование как механизма преодоления "ресурсного проклятия". Наталья Волчкова — старший эконо! мист ЦЭФИР, Москва. Елена Суслова — вы! пускница магистерской программы РЭШ. Статья будет опубликована в серии “Науч! ные труды” на www.cefir.ru BT Интенсивное использование природных ресурсов ведет к подавлению роста в отраслях, в которых заняты работники с высоким уровнем человеческого капитала Beyond Transition • Апрель июнь 2006 Тема номера: Энергетика Энергетическая бедность в Македонии и Чехии Понятие “энергетическая бедность” оз" начает, что либо средняя дневная темпера" тура в жилых помещениях ниже 21°C, что является необходимым для поддержания здоровья уровнем, либо отопление в домах ниже субъективного минимума, позволяю" щего человеку вести нормальный повсе" дневный образ жизни. Многие страны Центральной и Восточ" ной Европы, а также бывшие республики Советского Союза в последние годы суще" ственно повысили цены на энергоснабже" ние, с целью ликвидировать унаследован" ные от социализма ценовые перекосы. Пос" кольку правительствам большинства госу" дарств не удалось создать при этом необхо" димую систему социальной защиты наибо" лее уязвимых домохозяйств, то существует риск, что энергетическая бедность может стать реальностью для миллионов людей в регионе, не оставляя им другого выхода, кроме как сократить энергопотребление. Энергетическая бедность может при" вести к возникновению порочного круга между инвестициями, энергетической по" литикой и социальным неблагополучием. Поскольку конечный уровень полезного тепла в доме зависит от энергосберегаю" щих свойств стройматериалов, системы отопления и бытовой техники, характер и степень энергетической бедности напря" мую взаимосвязаны с уровнем инвестиций в основные фонды и их обслуживание. В странах, где реформы энергетики прод" вигаются медленно, одной из причин пос" тоянного перекрестного субсидирования остается опасение, что повышение цен на энергоснабжение вынудит значительное количество домохозяйств сократить энер" гопотребление, что в результате может вы" литься в социальные и политические вол" нения. Но сохранение в жилом секторе цен на энергию ниже ее себестоимости не спо" собствует инвестициям в строительство энергоэффективных зданий и поощряет расточительное отношение к энергии. Наше исследование институциональ" ных, географических и социальных факто" ров энергетической бедности в Македонии и Чехии опирается на интервью с политика" ми, экспертами и домохозяйствами в обеих странах, а также на анализ моделей доходов и расходов, субъективного восприятия бла" госостояния и оценок качества жилья. Македония: энергетиче ская бедность среди бедных и среднего класса Одной из главных целей недавних эко" номических преобразований в Македонии было реструктурирование государственного предприятия Electric Power Company of Macedonia (ESM), которое являлось моно" полистом в производстве, передаче и рас" пределении электроэнергии. В 1990"х гг. при подготовке ESM к приватизации тари" фы на потребление электроэнергии населе" нием и уровень отключений были повыше" ны более чем вдвое. Однако стране не уда" лось создать комплексную программу ин" вестиций в эффективное энергопотребле" ние в жилом секторе. Сегодня в Македонии нет соответствующих законодательных и организационных основ для разработки и осуществления политики эффективного энергопотребления, отсутствуют и действенные механизмы влияния на энер" госберегающие свойства строящегося жилья. И это несмотря на тот факт, что поч" ти все жилье в Македонии является част" ным и не сдаваемым в аренду. Проблема энергетической бедности возникла на фоне быстрого роста общей бедности. В настоящее время в Македонии почти 30% населения живет ниже уровня бедности — это многократный рост по сравнению с 4% в 1991 г. Но последова" тельной политики преодоления энергети" ческой бедности в Македонии пока нет. Единственный механизм — мягкая поли" тика отключений, скрыто проводимая предприятиями энергоснабжения, которые часто позволяют населению пользоваться электричеством или отоплением, несмотря на многомесячные задолженности по опла" те услуг. Все это привело к увеличению до" ли использования дровяного отопления, которое в настоящее время используется около 70% населения, особенно в сельских районах. Централизованное отопление су" ществует только в столице страны Скопье. Домохозяйства в небольших городах без централизованного отопления пользуются для обогрева электричеством — число тако" вых выросло приблизительно до 30% Численность населения, живущего в состоянии энергетической бедности, неиз" вестна, поскольку непосредственных иссле" дований этого вопроса не проводилось. Од" нако масштаб проблемы можно оценить с помощью метода "компенсирующих изме" нений", примененного к национальным оп" росам по расходам домохозяйств. Этот ме" тод позволяет определить, как в процентном выражении должны были бы измениться доходы домохозяйств в 2004 г., чтобы опла" та ими потребления электроэнергии соста" вила ту же долю, что и ранее. При этом в ка" честве точки отсчета для сравнения исполь" зовался средний показатель по стране в 1995 г., когда цены на энергию были отно" сительно низкими. Выяснилось, что 60% домохозяйств с самыми низкими доходами должны были бы получить дотации в размере от 1% до 27% от общего дохода. В то же время доходы сле" довало бы "отобрать" у 30% самых богатых домохозяйств. Это означает, что относи" тельные расходы на энергоснабжение сос" тоятельных домашних хозяйств возросли по сравнению с уровнем 1995 г., в то время как менее успешные 60% домохозяйств были вынуждены снизить свои затраты на энерго" потребление (см. Таблицу). Доля в 60% соответствует данным, по" лученным в результате опросов субъектив" ной оценки благосостояния, согласно кото" рым только 38% всех домохозяйств в 2003 г. считали, что могут нормально отапливать свое жилье, хотя всего тремя годами раньше доля таковых составляла 46%. Таким обра" зом, энергетическая бедность намного бо" лее распространена среди населения, чем Энергетическая бедность — результат неудовлетворительной координации мер энергетической, социальной и жилищной политики Стефан Бузар · 15 Новое в экономической науке Всемирный банк и ЦЭФИР 60% беднейших домохозяйств в Македонии потребовался бы дополнительный доход для сохранения потребления электро? энергии на уровне 1995 г. бедность по доходам, отраженная в статис" тике. Результаты анализа также свидетель" ствуют о том, что улучшения энергоэффек" тивности пока произошли только среди обеспеченной части населения При нор" мальных условиях их расходы на энергопот" ребление снизились бы в результате ис" пользования более эффективного оборудо" вания и/или перехода на другое топливо. Таким образом, основываясь на анализе по методу "компенсирующих изменений" и двух небольших опросах, проведенных в репрезентативных городских районах, мож" но сделать вывод, что от повышения цен на энергию страдают: • Домохозяйства с низкими дохода" ми в общепринятом смысле: иждивенцы, домохозяйства с безработными взрослыми, домохозяйства с несколькими детьми и семьи, доход которых полностью зависит от сельского хозяйства. • Семьи, относящиеся к группе рис" ка в силу своих жилищных условий — в ос" новном пенсионеры и семьи с маленькими детьми. В их случае энергетическая бед" ность может быть частично отнесена за счет недостаточной эффективности использова" ния энергии в домах и ежедневных потреб" ностях в энергии выше среднего уровня. Таким образом, поскольку энергетичес" кая бедность зависит от целого ряда жилищ" ных и социальных условий (помимо просто низких доходов), она затрагивает как бед" ных, так и средний класс в Македонии. Чехия: тесная связь с де мографией В Чешской Республике осуществление энергетических реформ закончилось офици" альной ликвидацией монополии на электро" энергию. Г осударственная компания Czech Electricity Company (CEZ) производит 10 ТВт электроэнергии в основном на атомных и угольных электростанциях. Высоковольт" ная сеть полностью контролируется находя" щимся в собственности CEZ филиалом, а распределительная сеть поделена между во" семью региональными электрическими ком" паниями. Типы собственности на сеть цент" рального отопления варьируются от муни" ципальной до полностью частной; нацио" нальная система газопроводов полностью находится в собственности транснациональ" ной энергетической компании и эксплуати" руется ею; эта компания обладает также контрольными пакетами акций в большин" стве предприятий газоснабжения. Страна лидирует в эффективности ис" пользования энергии среди других переход" ных стран и осуществила широкий спектр программ капиталовложений. Однако по сравнению с другими областями разработки в области повышения энергоэффективности получали недостаточное финансирование, а управление ими было рассредоточено по нескольким ведомствам. В жилом секторе в качестве всесторон" него механизма социальный защиты госу" дарство попыталось использовать регулиро" вание оплаты жилья. Этот подход привел к неприемлемым уровням оплаты жилья ни" же рыночных и деформации отношений между собственниками и арендаторами, что негативно повлияло на обслуживание жило" го сектора и, следовательно, эффективность энергопользования, а также на географи" ческую мобильность домохозяйств. Повышение цен на энергию в Чешской Республике было гораздо менее резким, чем в Македонии. Кроме того, страна распола" гает более разнообразной структурой топ" ливного баланса для энергоснабжения жи" лого сектора по сравнению с Македонией. В общем объеме энергоснабжения газ сос" тавляет около 40%, а оставшаяся доля поч" ти поровну распределяется между электро" и теплоснабжением. Анализ "компенсирующих изменений" в период с 1995 по 2004 гг. (см. Таблицу) по" казывает, что чешские домохозяйства реа" гировали тремя разными способами на по" вышение цен на энергию. • Нижний дециль снизил расход энергии на 6% по денежному доходу, что яв" ляется признаком энергетической бедности. Опросы по благосостоянию домохозяйств показали, что 8,2% из них не удовлетворены уровнем отопления своих домов. • Децили 2 — 6 отметили относи" тельное повышение своих расходов на энер" гопотребление — до 49% в третьем дециле. И хотя возможно, что это увеличение связа" но с быстрым ростом цен на энергоснабже" ние с 1995 г., скорее всего, домашние хозяй" ства были вынуждены понести дополнитель" ные расходы на оплату энергопотребления. • Расходы на энергопотребление че" тырех верхних децилей фактически снизи" лись с 1995 г. — вероятно, благодаря более дешевым и/или более эффективным видам топлива, при одновременном улучшении технических характеристик жилого фонда. Очевидно, что в Чешской Республике энергетическая бедность более тесно связа" на с демографическими характеристиками и затрагивает до 10% населения. Наиболее подвержены энергетической бедности не" полные семьи, домохозяйства с несколькими детьми и пенсионеры. На основе имеющихся данных мы заключаем, что в Чешской Рес" публике доход играет более важную роль в энергетической бедности, чем в Македонии, хотя состояние жилого сектора также имеет значение, особенно в случае пенсионеров. Выводы Значительное число домохозяйств в Ма" кедонии и немного меньше в Чешской Рес" публике находится в состоянии энергети" ческой бедности, форме "энергетического голодания", появившейся в бывших социа" листических странах. Проблема имеет об" ширные социально"экономические послед" ствия для всего региона, поскольку распо" ложенные в зоне холодного климата страны с переходной экономикой прошли через пе" риод потрясений как в отношении доходов населения, так и цен на энергию и распад инфраструктуры. Кроме того, они отлича" ются неэффективным использованием энергии и отсутствием последовательной энергетической политики. Полученные данные свидетельствуют о связи энергетической бедности с неэффек" тивной координацией между соответствую" щими ведомствами в энергетической, соци" альной и жилищной сферах. Одна из глав" ных проблем заключается в отсутствии ин" тегрированного подхода руководства стран к социальным преобразованиям, эффектив" ному использованию энергии, ее доступнос" ти и энергетической бедности. Рост энерге" тической бедности связан и с отсутствием комплексной системы развития эффектив" ного энергопотребления в обеих странах. Стефан Бузар (Stefan Buzar) — стар! ший научный сотрудник географического факультета Оксфордского университета и приглашенный профессор факультета эко! номической географии Гданьского универси! тета, Польша. BT · 16 Новое в экономической науке Децили I II III IV V VI VII VIII IX X Македония 6,1% ?2,7% ?48,6% ?44,4% ?13,9% ?10,4% 18,0% 25,3% 14,8% 2,3% Чешская Республика 26,8% 13,9% 9,3% 6,4% 3,8% 1,1% ?0,1% ?2,5% ?0,7% ?6,0% Примечание: расчеты автора произведены на основе данных опроса по расходам домохозяйств (Македония) и опроса по семейному бюджету (Чешская Республика). Расходы домохозяйств на энергию, по децилям Beyond Transition • Апрель июнь 2006 · 17 Всемирный банк и ЦЭФИР Йенс Арнольд, Беата Яворчик и Аадитья Матту Если по результатам либерализации торговли товарами существуют обширные эмпирические исследования, то послед" ствия либерализации сферы услуг пока мало изучены. Предоставление телеком" муникационных, финансовых, консалтин" говых, транспортных и страховых услуг — неотъемлемая часть широкого ряда про" мышленных отраслей и сферы обслужи" вания. Повышение предложения и каче" ства услуг в результате либерализации этой сферы может принести большую вы" году. Учитывая ограниченные возможнос" ти использования международной торгов" ли в предоставлении услуг, можно было бы ожидать, что эффективность отраслей будет больше зависеть от качества и ас" сортимента услуг, предоставляемых мест" ными поставщиками, чем это происходит с промежуточными материальными ресур" сами. Результаты нашего исследования, основанные на подробных данных по предприятиям, свидетельствуют о поло" жительной зависимости между глубиной реформ в сфере услуг и производитель" ностью промышленности, использующих эти услуги. Значительные реформы индустрии ус" луг в Чешской Республике в том числе отк" рыли доступ в этот сектор иностранным инвесторам. В 1998 — 2003 гг. приток прямых иностранных инвестиций (ПИИ) в сектор услуг Чехии был стабильно более высоким, чем в производство (см. Рис. 1). Через стимулирование выхода на ры" нок услуг новых местных или иностранных компаний либерализация должна способ" ствовать расширению выбора поставщи" ков для конечных потребителей. Это поз" воляет повысить качество услуг, оказывае" мых предприятиям. Например, инвестиции в инфраструктуру со стороны новых мест" ных или иностранных акционеров могут по" высить надежность международной теле" фонной связи или обеспечение электро" энергией, или увеличить скорость приня" тия решений по кредитам в условиях расту" щей конкуренции банков. Это в свою оче" редь позволяет уменьшить количество сбо" ев производства и снизить текущие расхо" ды в промышленности. Кроме того, новые услуги могут поя" виться с внедрением передового междуна" родного опыта. В качестве примеров мож" но привести инновационные финансовые продукты и инструменты управления пото" ками наличных денег, интермодальные транспортные или цифровые телекоммуни" кационные услуги. Наличие подобных ус" луг позволяет производителям вводить но" вации, способствующие росту эффектив" ности производства, например, получение заказов по Интернету или онлайновые аук" ционы для поставщиков. Очень большое значение может иметь выход на внутренний рынок иностранных поставщиков услуг. Иностранные компа" нии могут принести с собой новые техноло" гии и знания, а также международный пе" редовой опыт. Устанавливая более высокие стандарты обслуживания и предоставляя новые услуги, они могут также стимулиро" вать местных поставщиков проводить по" добные улучшения. Например, крупный австрийский банк первым предложил уда" ленные банковские услуги по телефону или Интернету, и 70 000 его клиентов в Чехии воспользовались новыми услугами в тече" ние первых двух месяцев. Сейчас этими ус" лугами пользуются около 800 000 человек, и другие банки тоже начали предлагать по" добные возможности. По использованию мобильных телефо" нов в 1998 г. Чешская Республика отстава" ла от всех своих западноевропейских сосе" дей. Однако уже в 2003 г. по числу мобиль" ных телефонов на 100 человек она уступа" ла только Италии и Швеции. Либерализация повыси ла качество, доступность и спектр услуг Потенциальный положительный эф" фект от либерализации сферы услуг и иностранного присутствия отражен в ре" зультатах опроса фирм, проведенного Все" мирным банком в Чешской Республике в 2004 г. Большинство из 350 опрошенных предприятий Чехии полагают, что либера" Ограничения на присутствие иностранных фирм в сфере услуг могут привести к снижению про? изводительности в промышленности Либерализация сферы услуг и производительность в Чехии Рис 1. Приток ПИИ в Чешскую Республику в 1998 2002 гг. (млн. евро) Источник: Чешский национальный банк Сфера производства Сфера услуг 1998 1999 2000 2001 2002 8000 7000 6000 5000 4000 3000 2000 1000 0 Приток прямых иностранных инвестиций в чешский сектор услуг в 1998 — 2003 гг.был стабильно выше, чем приток инвестиций в производство лизация индустрии услуг способствовала повышению качества, ассортимента и спектра услуг в стране. На Рис. 2 показано восприятие руководителями чешских фирм либерализации сектора телекомму" никаций. Подобные же результаты (не приведенные здесь) были получены в от" ношении банковской сферы, финансовой отчетности, страхования и транспортных услуг. Более тщательное изучение сферы ус" луг в Чешской Республике показывает, что существуют значительные различия между местными и иностранными поставщиками. Зарубежные компании отличаются более высокой производительностью труда и бо" лее склонны к капиталовложениям. Дан" ные свидетельствуют также о повышении производительности и инвестиций чешских фирм после того, как они были приобрете" ны иностранными владельцами. Кроме то" го, после приобретения иностранцами эти компании увеличивают долю на рынке, что говорит о явном предпочтении потребите" лей услуг компаниям, принадлежащим иностранным владельцам. Чтобы изучить связь между реформа" ми сектора услуг и производительностью потребителей услуг, мы проверяем зависи" мость каждой отрасли промышленности от каждого сектора услуг, которая оценивает" ся на основе национальной матрицы меж" отраслевого баланса. Таким образом мы можем оценить влияние реформ в сфере услуг на промышленные сектора. Затем мы связываем совокупную производитель" ность факторов производства промышлен" ных предприятий с уровнем либерализа" ции в секторах услуг. Анализ основан на данных коммерческой базы данных Amadeus, включающей финансовые отче" ты и информацию о владельцах приблизи" тельно 10 000 чешских ком" паний за период 1998 — 2003 гг. Чтобы отразить степень либерализации в секторах услуг, в нашем исследова" нии используется несколь" ко показателей. Первый — индексы реформ Европейс" кого банка реконструкции и развития, отражающие сте" пень реформирования дан" ного сектора услуг. Показа" тели с динамикой измене" ния во времени имеются для банковского сектора, телекоммуникаций, электроэнергетики, железнодорожного транспорта, автотра" нспорта и водоснабжения. Другие показа" тели охватывают конкретные аспекты ли" берализации: (а) уровень присутствия иностранных инвесторов, равный доле ус" луг, предоставляемой компаниями с иностранными владельцами; (б) прогресс приватизации, равный доле услуг, произ" веденных частными компаниями; (в) уро" вень конкуренции, измеренный долей ус" луг на рынке четырех самых крупных пос" тавщиков. В дополнение к этим показателям, в эмпирическом анализе мы также учитыва" ем целый ряд каналов, через которые воз" росшая открытость экономики может по" влиять на производительность компаний. В частности, мы учитываем доступность материальных ресурсов, произведенных компаниями с иностранным капиталом в первичных отраслях промышленности, а также тарифы на импортные промежуточ" ные материальные ресурсы. Чтобы оце" нить уровень конкуренции на рынке конеч" ной продукции, мы учитываем иностранное присутствие и тарифный протекционизм в этом секторе. Иностранное присут ствие в секторе услуг влияет на производительность в промышленности Результаты нашего исследования де" монстрируют положительную зависимость между либерализацией в секторе услуг и производительностью промышленных предприятий, пользующихся этими услу" гами. Положительная и статистически значимая связь установлена также для по" казателей степени реформирования, при" сутствия иностранных поставщиков и сте" пени приватизации в секторе услуг. Наи" более отчетлива связь между присутстви" ем иностранных поставщиков в секторе услуг и производительностью компаний, пользующихся этими услугами. Из этого следует, что открытость сектора услуг для иностранных компаний может являться ключевым механизмом, через который ре" формирование этого сектора влияет на производительность в промышленности. Полученные нами результаты четко ука" зывают на связь между либерализацией сектора услуг, и в особенности притоком ПИИ в этот сектор, ростом предложения, ассортимента и качества услуг, что в свою очередь способствует повышению произ" водительности предприятий, пользующих" ся данными услугами. Как указывалось в недавних дебатах в Европейском союзе и Всемирной торговой организации, либерализация сектора услуг — очень острый вопрос. Большин" ство барьеров на пути зарубежных инвес" тиций на сегодняшний день находится не в области торговли товарами, а в сфере пре" доставления услуг. Это отражает нежела" ние правительств, особенно в развиваю" щихся странах, разрешить неограниченное иностранное присутствие в секторах эко" номики, которые они считают "стратеги" ческими". Наш анализ показывает, что ог" раничения для иностранного присутствия в сфере обслуживания производства могут серьезно подорвать рост производитель" ности в промышленности. Йенс Арнольд (Jens Arnold) — консуль! тант, Беата Яворчик (Beata Javorcik) — старший экономист и Аадитья Матту (Aaditya Mattoo) — ведущий экономист Г руппы исследований экономики развиваю! щихся стран Всемирного банка, 1818 H Street, NW; MSN MC3!303; Washington, DC, 20433. Статья основана на совместной ра! боте авторов, которая готовится к публи! кации в серии "Научные исследования" Все! мирного банка. BT · 18 Новое в экономической науке Характер реформ, уровень присутствия иностранных фирм и степень приватизации компаний в секторе услуг тесно взаимосвязаны друг с другом Рис 2. Восприятие влияния либерализации сферы услуг на телекоммуникационную отрасль предприятиями в Чешской Республике Источник: Всемирный банк Цены Качество Ассортимент услуг Предложение услуг Положительно Отрицательно Никакого эффекта Beyond Transition • Апрель июнь 2006 Положительно Положительно Положительно Отрицательно Отрицательно Отрицательно Никакого эффекта Никакого эффекта Никакого эффекта · 19 Всемирный банк и ЦЭФИР Влияние СМИ на корпоративное управление в России Александр Дайк, Наталья Волчкова, Луиджи Зингалес Освещение нарушений норм корпоративного управления в британских и американских СМИ влияет на вероятность изменения компанией своего поведения В двух российских компаниях — неф" тегазовой "Сиданко" и телекоммуникаци" онной МГТС — крупные акционеры попы" тались размыть долю других акционеров, выпустив акции ниже их рыночной стои" мости для продажи инсайдерам. Действия нефтяной компании вызвали значительный резонанс в международной деловой прессе, и Федеральная комиссия по рынку ценных бумаг аннулировала эмиссию этих акций. А телекоммуникационная компания прив" лекла значительно меньше внимания СМИ, и акции были выпущены. Так действительно ли СМИ могут повлиять на действия компаний? СМИ могут сыграть важную роль в от" ношении нарушений норм корпоративного управления через воздействие на репута" цию вовлеченных в конфликт сторон. Но это влияние эффективно, только если ин" формация доводится до сведения целевой аудитории, мнение которой значимо для компании. Однако очень часто такая ауди" тория не занимается целенаправленным сбором сведений о действиях руководства корпорации и/или политических деятелей, если только сбор информации не требует больших усилий или материал не подан ин" тересно журналистами. Однако когда фак" ты злоупотреблений привлекают внимание значимой аудитории, для корпоративного руководства важно позаботиться о созда" нии себе репутации лиц, действующих в интересах акционеров. То же самое верно и для чиновников, контролирующих соблю" дение норм корпоративного управления: они скорее будут выполнять свои обязан" ности, если будут знать, что за их действи" ями следит широкая общественность. На данных по России очень легко прос" ледить влияние средств массовой инфор" мации на корпоративное управление. Во" первых, в конце 1990"х нарушения норм корпоративного управления в России были повсеместными, заметными и очень серь" езными. Во"вторых, в России стандартные механизмы возмещения ущерба при зло" употреблениях либо не существовали, ли" бо были совершенно неэффективны (нап" ример, суды были коррумпированы). Это позволяет определить, действительно ли средства массовой информации влияли на ситуацию. В"третьих, и это самое важное, в России существует инвестиционный фонд Hermitage Capital, целенаправленно ис" пользующий стратегию работы со СМИ, благодаря чему мы можем установить на" личие причинных связей между освещени" ем в СМИ и исходом случая корпоративно" го нарушения. Основанный в 1996 г. как универсаль" ный инвестиционный фонд для работы в России, Hermitage Capital оказался вовле" ченным в войны в сфере корпоративного управления. Крупнейший портфельный иностранный инвестор не мог не реагиро" вать на крупные нарушения в корпоратив" ном управлении. Не располагая достаточно сильными правовыми рычагами для защи" ты своих инвестиций, фонд выбрал страте" гию активной дискредитации компаний в российской и международной прессе, наде" ясь нанести ущерб их репутации и репута" ции тех государственных чиновников, кото" рые занимались регулированием в соотве" тствующей отрасли. Безусловно, Hermitage Capital заинтересован действо" вать таким образом только в отношении тех компаний, где у него есть доля акций. Ситуация с двумя компаниями, упомя" нутыми выше, очень показательна. В обоих случаях крупные акционеры пытались раз" мыть доли других акционеров, однако Hermitage Capital владел значительной до" лей в "Сиданко" и не владел акциями МГТС. Мы установили, что действия руко" водителей "Сиданко" были освещены в 23 новостных материалах, 14 из которых были опубликованы в солидных международных изданиях (9 в газете Financial Times, 4 в Wall Street Journal и 1 в The Economist). А о действиях руководства МГТС в извест" ных международных СМИ были опублико" ваны только 3 статьи (все в газете Financial Times). В "Сиданко" размывание стоимос" ти компании было полностью остановлено, а в МГТС быстро завершилось. Исправлены 17 из 57 на рушений Начальная выборка из 250 потенциаль" ных злоупотреблений в сфере корпоратив" ного управления в период с 1998 по 2002 г. была составлена нами на основе новостей, опубликованных крупным российским ин" вестиционным банком "Тройка Диалог" в своем еженедельном обзоре. Поскольку мы рассматривали только те случаи, которые могли нанести значительный ущерб инте" ресам миноритарных акционеров, и когда предлагаемая ответная реакция, в принци" пе, могла ограничиваться действиями ми" норитарных акционеров и их союзников, в выборке осталось 57 случаев. В 24 случаях, где мы располагали надежными данными по котировкам, реакция цен на акции подтвер" дила правильность нашей оценки. Если на" рушение в сфере корпоративного управле" ния было неожиданным для рынка, оно должно было негативно повлиять на курс акций. Действительно, курс акций реагиро" вал резко отрицательно, причем среднее снижение совокупного избыточного дохода составило 15,3%. Что касается фактичес" ких результатов, то положительный эффект в виде значительного или частичного устра" нения нарушения имел место в 17 из 57 рассмотренных случаев. Эффективность иност ранной прессы Влияет ли каким"либо образом осве" щение в прессе на вероятность того, что нарушение норм корпоративного управле" ния будет частично или полностью компен" сировано? В стране, где средства правовой защиты неэффективны, единственным гарантом не" допущения подобных нарушений остается международная репутация компании, кото" рую мы оцениваем по трем параметрам: • доля иностранной собственности в фирме; Инвестиционный фонд the Hermitage Fund, действующий в России, активно применяет медиа?стратегию • наличие Европейского банка ре" конструкции и развития (ЕБРР) в списке кредиторов компании; • количество совместных предпри" ятий компании с иностранными партнера" ми. Из вышеперечисленных факторов только наличие ЕБРР в числе кредиторов компании значительно влияет на результат и повышает вероятность полного возмеще" ния ущерба на 18 процентных пунктов. Добавление к основным параметрам показателя освещаемости в иностранной прессе (количество статей, опубликован" ных в газете Financial Times и Wall Street Journal в течение двух месяцев после фак" та злоупотребления) показывает, что осве" щение в прессе дает положительный и ста" тистически значимый эффект. Увеличение числа статей, опубликованных в иностран" ных газетах, на одно стандартное отклоне" ние повышает вероятность полной компен" сации на 10 процентных пунктов. Публикация в Wall Street Journal, похо" же, влияет сильнее: увеличение числа ста" тей в Wall Street Journal на одно стандарт" ное отклонение повышает вероятность по" ложительного результата на 10 процентных пунктов, а в Financial Times — только на 1 процентный пункт . Российские газеты — даже весьма авторитетные — похоже, не имеют значимого влияния. Таким образом, мы приходим к выводу, что главное сред" ство воздействия — доступ к международ" ной аудитории. Наличие фонда Hermitage Capital сре" ди акционеров повышает среднее число публикаций более чем втрое и подчерки" вает наличие прямой зависимости между средствами массовой информации и соб" людением норм корпоративного управле" ния. Освещение в СМИ вы нуждает принимать меры Каковы главные механизмы влияния прессы? Мы сгруппировали случаи с поло" жительным исходом событий в соответ" ствии с причиной, по которой нарушение было пресечено: • Около 30% случаев заканчива" ются (по крайней мере, частично) положи" тельно в результате вмешательства регуля" тивного органа. Какое отношение имеет пресса к решению регулятивного органа о вмешательстве в ситуацию? Пресса ин" формирует большое количество людей о соответствующих фактах, повышая таким образом угрозу дискредитации репутации регулятивного органа в случае бездей" ствия. • В 18% случаев дело решается благодаря политическому вмешательству. В типичной демократической стране поли" тики вынуждены вмешиваться в ситуации, получающие широкую огласку, чтобы не рисковать своей политической репутацией. В России важный фактор — репутация в отношениях с иностранными, особенно англо"американскими, инвесторами. • Еще в 24% случаев положитель" ный результат имеет место благодаря усиле" нию уже существующей оппозиции после ос" вещения в прессе. Например, в случае с за" водом "КамАЗ" ЕБРР боролся с дроблением акций, одобренным компанией. Освещение событий в печати укрепило позиции ЕБРР , так как способствовало информированию инвесторов, что повысило угрозу репутации компании в случае продолжения нарушений. • В остальных 24% случаев, похоже, компания добровольно изменила свое пове" дение, и здесь сложнее выявить роль СМИ. Таким образом, основной механизм, пос" редством которого освещение фактов злоу" потреблений в печати оказывает влияние на ситуацию, — это угроза дискредитации ре" путации компании в глазах соответствую" щей аудитории. Очевидно, что успех страте" гии сильно зависит от значения, которое ключевые игроки придают своей репутации в глазах этой аудитории. Полученные нами результаты могут объясняться особым пери" одом в России вскоре после дефолта, когда страна была заинтересована в восстановле" нии международного доверия. Результат исследования может иметь важные политические последствия. Очевидно, что развитые страны, взаимо" действуя с развивающимися, могут оказы" вать позитивное влияние не только на госу" дарственное, но и на корпоративное управ" ление. Поскольку политики и бизнесмены стремятся "хорошо выглядеть" в англо" американском общественном мнении, та" кой механизм влияния может стимулиро" вать их улучшить стандарты управления. Александр Дайк (Alexander Dyck) — про! фессор университета Торонто; Наталья Волчкова — старший научный сотрудник ЦЭФИР; Луиджи Зингалес (Luigi Zingales) — профессор бизнес!школы Чикагского универ! ситета. Статья основана на работе "The corporate governance role of the media: evi! dence from Russia", представленной на кон! ференции Европейской финансовой ассоциа! ции, в Москве в 2005 г. BT · 20 Новое в экономической науке В экономической науке выделяют несколько каналов влияния: • Увеличение доступа к источникам внешнего финансирования у фирм, что ведет к увеличению инвестиций, быстрому экономическому росту, созданию новых рабочих мест. Финансовые рынки и рынки капитала лучше развиты в странах с эффективными механизмами защиты прав собственности. • Более низкая стоимость капитала и более высокая оценка стоимости фирм, что привлекательнее для инвесторов и ведет к экономическому росту и повышению занятости. Аутсайдеры менее склонны инвестировать и устанавливают более высокие процентные ставки, если они не уверены, что коэффициент окупаемости капитала будет адекватным • Улучшение организационной деятельности через более эффективное управление и распределение ресурсов, улучшение трудовой политики и других усовершенствований, ведущих к сбережению финансовых средств. • Уменьшение риска финансовых кризисов является особенно важным фактором, поскольку они вызывают большие эономические и социальные издержки. Во время Восточноазиатского финансового кризиса совокупный доход от акционерного капитала тех фирм, в которых управляющие менеджеры обладали высокой степенью контроля, но не имели непосредственной доли в собственности, был на 10 — 20 процентных пунктов ниже, чем в других фирмах. • Улучшение отношений со всеми заинтересованными сторонами, включая банки, держателей облигаций, работников, местные и национальные правительства. Каждая из этих групп контролирует, мотивирует, оказывает влияние на управление в фирме самыми различными способами, улучшая социальные и трудовые отношения и такие сферы, как, например, защита окружающей среды. Источник: Stijn Claessens, "Корпоративное управление и развитие", s BT Корпоративное управление, рост и развитие Публикации в The Wall Street Journal о нарушениях в области корпоративного управления приносят наибольший эффект Beyond Transition • Апрель июнь 2006 · 21 Всемирный банк и ЦЭФИР Существует ли в Чехии "стеклян ный потолок"? Степан Джурайда и Теодора Палигорова Все больше фактов свидетельствует о том, что в Чешской Республике женщины упираются в "стеклянный потолок" — прег" раду на пути к карьерному росту, которая не позволяет им получать высокооплачивае" мые должности и равную с мужчинами зара" ботную плату, особенно на высших долж" ностях. Анализ гендерных разрывов в зара" ботной плате менеджеров особенно важен для бывших соцстран Центральной Европы, где стратегии фирм в отношении персонала и корпоративное управление начинают приближаться к западным стандартам. Наше исследование в отношении ме" неджеров компаний в Чехии охватывает не только высших руководителей фирм, но также менеджеров среднего звена и рядо" вых сотрудников, позволяя таким образом обобщить относительное положение жен" щин на всех уровнях корпоративной иерар" хии. Мы классифицируем высших руково" дителей компаний и директоров как топ" менеджеров, а руководителей производ" ства, специалистов и супервайзеров (от" ветственных работников низшего звена) как менеджеров нижнего уровня. Мы ис" пользуем данные национального опроса ра" ботодателей, Информационной системы по средним доходам, с 2000 по 2004 гг. и работаем только со статистикой по зарплате, поскольку данные по полному компенсационному пакету отсутствуют. Используемые нами данные не охваты" вают государственный сектор (образо" вание, здравоохранение и госуправле" ние), где размер заработной платы уста" навливается в соответствии с бюджетом и основан на тарифных сетках. В нашем исследовании мы рассматриваем только крупные фирмы с количеством работни" ков более 250. Полная выборка предс" тавляет собой данные по заработной плате 1692 топ"менеджеров и более 36000 менеджеров нижнего уровня. В госкомпаниях самая маленькая разница в оп лате По нашим данным, топ"менеджеры зарабатывают в час в среднем в 2,69 ра" за больше, чем менеджеры нижнего уровня, зарплата которых, в свою очередь, в 2,46 раза превышает зарплату рядовых сотрудников. Какова доля женщин в этих высокооплачиваемых группах? Если счи" тать, что она такая же, как в общей структу" ре занятости в стране, то руководители" женщины должны составить около 46% от общего числа руководителей. Но наши дан" ные показывают, что женщины составляют только 7% от общего числа топ"менедже" ров и 32% — от общего числа менеджеров нижнего уровня. Женщины"руководители высшего звена зарабатывают в час в сред" нем на 41% меньше, чем их коллеги"муж" чины, а гендерный разрыв в заработной плате рядовых работников составляет 22%. Женщины составляют относительно высокую долю от общего числа молодых и особенно менее образованных топ"менед" жеров — но эти группы также находятся в конце списка оплаты труда. Г ендерный раз" рыв в заработной плате высокообразован" ных менеджеров нижнего звена невелик, но в этой группе также относительно нем" ного женщин. Доля женщин и гендерный разрыв в заработной плате наиболее сба" лансированы в группе рядовых сотрудников. Рассматривая структуру занятости и уровень заработной платы в компаниях различных размеров, типов собственности и отраслей промышленности, мы установи" ли, что наименьший гендерный разрыв в заработной плате наблюдается в госкомпа" ниях, так же как и большее, по сравнению с другими, количество женщин на вершине иерархии. Деление фирм на четыре квар" тиля по размеру (общему количеству ра" ботников) показывает, что женщины зани" мают должности топ"менеджеров всего в нескольких очень крупных предприятиях. Наконец, больше всего женщин среди ру" ководства компаний в относительно низко" оплачиваемых секторах — розничной тор" говле, транспорте и связи, где работает 38% от общего числа женщин"руководи" телей. Гендерный разрыв в заработной плате топ"менеджеров варьируется по от" раслям экономики и особенно велик в роз" ничной торговле. Если рассматривать менеджеров ниж" него звена, то, как и в случае с топ"менед" жерами, доля женщин здесь больше всего в государственных и очень крупных компа" ниях. Г ендерный разрыв в заработной пла" те этой группы мало зависит от типа компании, здесь также больше всего женщин в розничной торговле и транс" порте. Несомненно, 7% женщин"руково" дителей высшего звена — это очень низкий показатель, и между группами топ"менеджеров и менеджеров нижнего звена имеет место явное гендерное раз" личие. Тот факт, что разрыв в заработ" ной плате мужчин и женщин растет с уровнем корпоративной иерархии, сви" детельствует о все более неблагоприят" ных карьерных возможностях для жен" щин по мере их продвижения по служеб" ной лестнице. Что является причиной гендерного разрыва в зарплатах? В какой степени гендерный разрыв в заработной плате объясняется стан" дартными схемами найма, учитываю" щими гендерный фактор, и различиями Среди руководителей чешских компаний только 7% женщин, и они зарабатывают на 20% меньше своих коллег?мужчин Межстрановое сравнение относительно" го гендерного состава руководителей крупных фирм (профессиональная группа 12 по клас" сификации ISCO"88) с использованием дан" ных опросов домохозяйств показывает, что доля женщин"руководителей крупных фирм сильно различается по странам: от 17% в Бельгии до 43% в США. Разрыв в почасовой оплате, определяемый как отношение зара" ботной платы женщин к заработной плате мужчин минус один, самый высокий в России, Испании и США и наименьший — в Ирлан" дии и Словении. В Чешской Республике жен" щины составляют лишь около 23% руководи" телей корпораций (группа 12 по ISCO), и со" ответствующий гендерный разрыв в заработ" ной плате в 24% близок к среднему разрыву для этой выборки стран. Источник: Stepan Jurajda, Teodora Paligorova, "Female managers and their wages in Central Europe". в демографических характеристиках ме" неджеров? Используя традиционный метод Оакса" ки"Блайндера, мы установили, что около трети гендерного разрыва в заработной плате в группе топ"менеджеров и менед" жеров нижнего звена можно объяснить гендерными различиями в возрасте и обра" зовании. По другому обстоит дело с рядо" выми сотрудниками, среди которых жен" щины имеют сравнительно лучшие демо" графические характеристики. Мы также установили, что женщины" руководители как высшего, так и нижнего звена чаще встречаются в менее "щед" рых" фирмах. Однако наш анализ показы" вает, что существует значительный "не" объяснимый" компонент гендерного раз" рыва в заработной плате руководителей. Используя метод сопоставительной де" композиции, мы приходим к выводу, что этот "необъяснимый" разрыв в заработ" ной плате (для мужчин и женщин сравни" мых по демографическим характеристи" кам и типу работодателя) составляет приблизительно 20% для каждой группы работников; это относится как к руково" дителям обоих уровней, так и рядовым сотрудникам. Связав относительное положение жен" щин на всех трех уровнях корпоративной иерархии, мы приходим к выводу, что суще" ствует статистически значимая положи" тельная зависимость между долей женщин на всех трех уровнях иерархии фирм и от" рицательная — между гендерным разры" вом в заработной плате на данном уровне иерархии и долей женщин на других уров" нях. Таким образом, чешские компании (в рамках одной отрасли и класса) системати" ческим образом отличаются по своему от" ношению к работникам"женщинам. Чешский "стеклянный потолок" в сравнении Чтобы сравнить гендерный разрыв в заработной плате в Чехии с аналогичной ситуацией в США, мы рассмотрели по пять самых высокооплачиваемых руководите" лей каждой фирмы из нашей выборки. Доля женщин среди пяти самых высо" кооплачиваемых руководителей чешских фирм в нашей выборке составила 9% — это более высокий показатель, по сравне" нию с 6% в США. В Чехии отношение за" работной платы женщин"руководителей к заработной плате мужчин"руководителей составляет 74%, что сопоставимо с соот" ветствующим показателем по США — 73%. Однако гендерный разрыв в США, основанный не на заработной плате, а на оценке общего компенсационного пакета, оказался больше — 67%. Если сравнивать объединенную группу руководителей выс" шего и нижнего звена в Чехии и США, то разрыв в Чехии по чистой заработной пла" те оказывается меньше примерно на треть. Какова структура гендерного разрыва в заработной плате самых высокооплачивае" мых руководителей компаний в Чехии и США? Около пятой части гендерного раз" рыва в заработной плате в Чехии можно объяснить низким присутствием женщин в фирмах с высокой оплатой труда, в США этим объясняется только треть. В Чешской Республике 74% пятерки самых высокооп" лачиваемых женщин"руководителей рабо" тает в самых мелких фирмах. Значительная часть различия в оплате труда в обеих стра" нах объясняется профессиональным соста" вом женщин"руководителей. К сожалению, данные по Чехии не позволяют нам полу" чить полную картину по генеральным ди" ректорам компаний, что, возможно, являет" ся одной из причин, по которой "необъясни" мый" разрыв в Чехии больше, чем в США. В целом мы приходим к выводу, что от" носительное положение женщин в руко" водстве американских и чешских фирм весьма схоже. Мы также считаем, что в Чехии величина гендерного разрыва в за" работной плате, который не может быть связан со значимыми различиями между мужчинами и женщинами или компаниями, где они заняты, довольно одинаков во всей корпоративной иерархии. Если толковать этот обусловленный разрыв как дискрими" нацию в оплате труда по гендерному приз" наку, то можно считать, что к женщинам относятся одинаково как на верхних ступе" нях служебной лестницы, так и на нижних. Большая часть среднего гендерного разры" ва в заработной плате топ"менеджеров связана с различием в типах фирм: женщи" ны обычно не руководят компаниями с вы" сокой оплатой труда. Рекомендация, следу" ющая из результатов нашего исследова" ния, заключается в том, что меры, направ" ленные на установление гендерного балан" са среди руководителей, скорее всего, бу" дут эффективными в уравнивании заработ" ной платы топ"менеджеров — мужчин и женщин, — если эти меры сосредоточатся на политике продвижения женщин в наи" более престижных компаниях. Степан Джурайда ( Stepan Jurajda) — заместитель директора по исследованиям, профессор Центра экономических исследова! ний и образования Карлова университета и Института экономики Академии наук Чеш! ской Республики (CERGE!EI); Теодора Пали! горова (Teodora Paligorova) — младший на! учный сотрудник, докторант CERGE!EI. Полный текст работы: ei.cz/pdf/wp/Wp296.pdf. Исследование явля! ется частью проекта ЕС по равенству "Пятьдесят на пятьдесят: равные возмож! ности для женщин и мужчин" и, в частнос! ти, финансировалось Европейским социаль! ным фондом ЕС и государственным бюдже! том Чешской Республики BT · 22 Новое в экономической науке Три четверти самых высокооплачиваемых топ?менеджеров? женщин работает в мелких фирмах Женщины в законодательстве, высшем управляющем звене (% от общего числа) Страна % Страна % Страна % Филиппины 58 Германия 36 Болгария 30 Фиджи 51 Венгрия 34 Грузия 28 Танзания 49 Польша 34 Чешская Республика 26 Соединенные Штаты 46 Великобритания 33 Гонконг, Китай 26 Латвия 40 Словения 33 Япония 10 Молдова 40 Румыния 31 Турция 6 Российская Федерация 39 Монголия 30 Пакистан 2 Примечание: информация приведена за последний из доступных годов в период 1992 — 2003 гг. Источник: рассчитано на основе данных по занятости Международной организации труда 2005. Beyond Transition • Апрель июнь 2006 · 23 Всемирный банк и ЦЭФИР Иностранные предприятия и эффективность производства Валентин Зеленюк Экономисты уже утвердились во мне" нии, что в большинстве отраслей промыш" ленности частные компании в среднем опе" режают государственные по эффективнос" ти. Но если сравнивать по этому критерию иностранные и местные компании, то все не столь очевидно. Определяется ли преи" мущество одних над другими характером отрасли промышленности? Уровень адап" тации иностранных фирм может отличать" ся в разных отраслях промышленности (например, в связи местными особеннос" тями). Однако прогрессивные технологии, привнесенные иностранцами, также могут в разной степени осваиваться и распрост" раняться местными фирмами. Мы анализируем эффективность предприятий в Китае и на Украине по двухступенчатой методологии: оцениваем показатели эффективности для каждой от" расли промышленности в каждой провин" ции/области, а затем анализируем показа" тели по отдельности. Украинские компании эффективнее иностранных Исследование по Украине с использова" нием данных на микроуровне показало, что местные частные компании оказались более эффективными, чем иностранные фирмы. Недостаточность выборки по каждой отрас" ли не позволила рассмотреть, было ли это различие в эффективности характерным для всех отраслей или только для некоторых. Изучая эффективность китайских предприятий, мы использовали последние агрегированные данные по типу их собственности для 29 провинций. Полу" ченные результаты подтвердили, что част" ные предприятия превосходят государ" ственные по эффективности. Мы также установили, что в легкой промышленнос" ти, в отличие от тяжелой, имеет место яв" ная территориальная концентрация про" изводства. Кроме того, предприятия с иностранным капиталом оказались гораз" до эффективнее, чем предприятия с дру" гим типом собственности, но только в тя" желой промышленности. Полученные по Китаю результаты показывают, что предприятия легкой промышленности с иностранными владельцами оказались в среднем менее эффективными, чем част" ные китайские предприятия. Дополни" тельные данные по Украине подтвержда" ют эту тенденцию — в легкой промыш" ленности местные компании в среднем превосходят по эффективности своих иностранных конкурентов. Каковы причины этих несколько нео" жиданных результатов? Дальнейшее об" суждение результатов исследования со специалистами"практиками помогли их объяснить и подтвердить теорию распрост" ранения и освоения технологий. Главным является не тип промыш" ленности — "легкая" или "тяже" лая", — а степень ее капиталоем" кости. Тяжелая промышленность в среднем является более капитало" емкой, и, следовательно, покупка новых технологий и их внедрение требуют больших объемов капита" ловложений. Иностранные инвес" торы в таких отраслях (обычно ги" гантские международные корпора" ции), скорее всего, имеют преиму" щество над местными фирмами в отношении капиталовложений и дорогостоящих технологий, что позволяет им работать эффективнее своих местных конкурентов. С другой стороны, даже когда иност" ранцы имеют изначальное технологичес" кое и финансовое преимущество, местные частные фирмы в легкой промышленности могут легче и быстрее осваивать, внедрять и адаптировать технологии в соответствии с местными условиями. А поскольку лег" кая промышленность, как правило, более трудоемка по сравнению с тяжелой, ее эф" фективность, вероятно, больше зависит от местной культуры, традиций, привычек и т. д. Таким образом местные фирме при одинаковом с иностранными компаниями уровне использования технологий могут сделать последние более эффективными. Поэтому, возможно, что в Китае и на Украине, там, где иностранные фирмы про" игрывают местным, приток иностранных инвестиций в легкую промышленность мо" жет замедлиться или уйти в другие отрасли. В то же время продолжится поступление новых иностранных инвестиций в более ка" питалоемкие отрасли, где иностранцы еще долго смогут пользоваться своим техноло" гическим и финансовым преимуществом. Поэтому странам с переходной экономикой можно рекомендовать уделять больше вни" мания образованию, направленному в пер" вую очередь на развитие человеческого ка" питала для применения в капиталоемких отраслях, особенно высокотехнологичных. Это будет способствовать ускорению осво" ения и дальнейшего распространения пе" редовых западных технологий. Валентин Зеленюк — старший научный сотрудник Киевского института экономики и директор UPEG в Консорциуме экономичес! ких исследований и образования (EERC) Го! сударственного университета "Киево!Мо! гилянска академия", Киев, Украина, адрес эл. почты: zelenyuk@stat.ucl.ac.be. Статья обобщает результаты исследований по Ки! таю и Украине автора, приведенных в Биб! лиографии. BT В Китае и на Украине частные компании в некоторых отраслях промышленности догоняют иностранных конкурентов и даже опережают их по эффективности Преимущество местных компаний в более трудоемкой легкой промышленности объясняется их знанием местных условий · 24 Новое в экономической теории Россия и ВТО: бремя "неприсоединившегося" Богдан Лиссоволик и Ярослав Лиссоволик Вступление России в ВТО считается ключевым этапом для дальнейшего углуб" ления рыночных преобразований в стране. С одной стороны, вступление России в ВТО могло бы способствовать гармониза" ции ее законодательства и правопримене" ния с законами ее основных торговых парт" неров. С другой — это устранило бы боль" шую часть сохраняющихся барьеров на пу" ти российского экспорта в страны — члены ВТО, существенно повысив выигрыш Рос" сии от торговли. Наличие нереализованно" го потенциала дальнейшей торговой перео" риентации России подтверждается тем фак" том, что российский экспорт в страны — члены ВТО составлял "всего" около 80% от общего объема экспорта в 2002 г., при" том что доля членов ВТО в мировой тор" говле составляет 95%. Более того, доля российского экспорта в страны, ставшие членами ВТО в 1995 г., почти не измени" лась в период с 1995 по 2002 гг., и рост до" ли экспорта во все страны ВТО был в зна" чительной степени обусловлен увеличени" ем количества членов этой организации. Однако несмотря на эти аргументы и провозглашение вступления в ВТО одним из приоритетов российской политики, де" баты вокруг выгод от членства России в торговой организации в последнее время становятся все более неоднозначными, за" медляя и без того затянувшийся процесс. Российские противники вступления в ВТО утверждают, что выгоды от членства будут невелики, так как многие члены ВТО уже предоставили России режим наибольшего благоприятствования, а некоторые разви" тые страны предоставили льготные усло" вия в соответствии с Общей системой пре" ференций. Анализ таких утверждений тре" бует тщательной оценки выгод от вступле" ния России в ВТО для ее торговли, и осо" бенно экспорта. Для лучшего понимания факторов, влияющих на показатели российского экс" порта, мы используем гравитационную мо" дель, которая позволяет выяснить, влияет ли статус "неприсоединившегося" на структуру экспорта. Используемые данные за период с 1995 по 2002 гг. охватывают 171 страну, большинство из которых — члены ВТО, в т. ч. недавно вступившие. Главный результат нашего исследова" ния в том, что структура торговли России остается в некотором роде "неоптималь" ной" или "отличающейся" от эталонных показателей гравитационной модели в том смысле, что Россия "слишком мало" торгу" ет с членами ВТО и/или "слишком много" с не"членами ВТО. Этот результат нельзя назвать неожиданным, принимая во вни" мание, что многие страны, не являющиеся членами ВТО, — бывшие соцстраны, име" ющие глубокие исторические и системные связи с Россией. Однако, с другой стороны, итоги анализа несколько удивляют, по" скольку мы учитываем эти особенности посредством введения различных регио" нальных переменных. Почему структура российской торговли отличается от "эталонной", согласно моде" ли? Очевидно, что особая структура тор" говли России обусловлена различными факторами — от статистических и истори" ческих до более значимых экономических и политических. Из нашего эмпирического анализа следует, что наиболее существен" ными являются два фактора. Во"первых, российский экспорт может быть осложнен ограничениями со стороны торговых парт" неров России в ВТО — либо потому что они не реализуют все выгоды от либерали" зации торговли, либо вследствие офици" альных или неофициальных барьеров, вве" денных отдельными членами этой органи" зации. Членам ВТО проще вводить подоб" ные ограничения для стран, не состоящих в организации, чем для других членов, пос" кольку первые не могут принять ответные меры, например, обратиться в органы ВТО по урегулированию споров или предпри" нять другие шаги. Во"вторых, возможно, на структуру российского экспорта влияют собственные внутренние ограничения. В то время как большинство экспортных пошлин в России постепенно снизилось к середине 1996 г., в 1999—2000 гг. были введены некоторые новые пошлины или восстановлены ста" рые. Во всяком случае некоторые данные указывают на роль внутренних ограничений экспорта: значимость переменной ВТО в модели существенно снижается в 1997 г. и резко падает в 1998 г., когда внутренние ог" раничения экспорта были минимальными. Другие вероятные факторы, такие как исключительно выгодная позиция России на переговорах с некоторыми странами — не"членами ВТО, нельзя назвать достаточ" но существенными. Числовые коэффициенты в наших рег" рессионных уравнениях показывают, что, в итоге, российский экспорт в страны — чле" ны ВТО может возрасти примерно на 50%. Главный вопрос, произойдут ли изменения в виде общего увеличения экспорта или его переориентации на другие страны. Наибо" лее вероятно, будут иметь место как эффект расширения, так и замещения. В частности, если полученные результаты экономически обусловлены какими"либо торговыми огра" ничениями, то маловероятно, что баланс между экспортными и неторгуемыми това" рами останется неизменным после сниже" ния экспортных барьеров. Таким образом, вступление в ВТО, по" хоже, самый разумный способ решить проблему диспропорции в структуре рос" сийской торговли. Что касается текущей торговой политики, то в результате вступ" ления в ВТО Россия должна выиграть от концентрации усилий на создании единооб" разных и равных конкурентных условий в региональных торговых структурах. Огром" ный размер и неоднородность страны гово" рят в пользу торговой интеграции через членство в ВТО, а не посредством регио" нальных соглашений. Кроме того, вступле" ние России в ВТО придаст организации практически глобальный характер и, таким образом, возможно, станет мощным им" пульсом для развития мировой торговли. Богдан Лиссоволик (Bogdan Lissovolik) — старший экономист европейского департа! мента Международного валютного фонда в Вашингтоне. Ярослав Лиссоволик — глав! ный экономист Deutsche UFG в Москве. Пол! ный текст работы находится по адресу: 2006/01/pdf/lissovol.pdf (IMF Staff Papers, Vol. 53, №.1). BT После вступления России в ВТО российский экспорт в страны — члены ВТО может вырасти на 50% В настоящее время Россия “слишком мало” торгует со странами — членами ВТО и “слишком много” с не?членами Beyond Transition • Апрель июнь 2006 Третья международная конференция по адресным трансфертам 26 — 30 июня в Стамбуле состоялась Третья международная конференция по адресным трансфертам, которую, как и предыду" щие, спонсировали Всемирный банк и правительство Турции при поддержке ряда доноров. Цель конференции — обмен международ" ным опытом и знаниями по вопросам эффективности осуществляе" мых программ как с политической, так и с технической точки зре" ния. Программы адресных трансфертов предоставляют денежные средства бедным домохозяйствам на условиях их вложения в чело" веческий капитал, например, обучение детей в школе или прохож" дение ими регулярного медосмотра и программ вакцинации. Подоб" ные программы начали реализовываться относительно недавно, но они уже работают во многих странах. Оценка результатов пилотных программ показывает, что адресные трансферты способствуют эф" фективному развитию человеческого капитала в бедных домохозяй" ствах. Очевидное подтверждение успеха программ — увеличение численности учащихся школ, улучшения в области профилактичес" кого здравоохранения и повышение потребительского уровня домо" хозяйств. Однако существуют и проблемы, требующие разрешения. Всемирный банк готов к реализации Стратегий партнерства с будущими члена ми ЕС 13 июня Совет директоров Всемирного банка обсудил Страте" гии партнерства Болгарии и Румынии. Стратегии разработаны на период с 2006 по 2009 гг., в течение которого оба государства пла" нируют вступить в Европейский cоюз. Стратегии партнерства для различных стран закладывают основу для работы Всемирного бан" ка в этих странах. В Болгарии и Румынии Банк оказывает содей" ствие в сближении законов, норм и стандартов этих стран с ЕС и освоении фондов Европейского союза. В Болгарии Стратегии нап" равлены на повышение производительности труда и занятости, стабилизацию бюджета, а также освоение средств из фондов ЕС и социальную интеграцию незащищенных групп. Стратегии для Ру" мынии должны способствовать решению трех основных проблем: ускорения структурных и институциональных реформ, направлен" ных на экономический рост, повышения стабильности бюджета и модернизации государственного сектора экономики, сокращения бедности и поддержки социальной интеграции незащищенных групп населения. Дополнительную информацию о работе Всемир" ного банка в этих странах можно получить на сайтах Обсуждение реформ рынка труда на Форуме в Стамбуле 12 — 13 июня в Стамбуле состоялся Форум по вопросам заня" тости и созданию рабочих мест в Восточной Европе и Централь" ной Азии. Мероприятие было организовано Департаментом Все" мирного банка по развитию человеческого потенциала в Европе и Центральной Азии совместно с Организацией государственного планирования Турции. В конференции приняли участие специа" листы со всего региона, которые обсудили результаты недавних реформ рынка труда в своих странах и практические выводы не" давних исследований, в т.ч. Всемирного банка. Участники конфе" ренции также попытались определить, какие реформы необходи" мы для повышения занятости и улучшения качества рабочих мест в регионе и каким образом Всемирный банк и другие доноры мог" ли бы практически помочь странам"клиентам в проведении необ" ходимых реформ. Новое исследование по рынку труда Турции дос" тупно по адресу: NAL/COUNTRIES/ECAEXT/TURKEYEXTN/0,,contentMDK:20873 325~pagePK:141137~piPK:141127~theSitePK:361712,00.html Комиссар ЕС обещает поддержку евро пейским цыганам Европейская комиссия обещает предоставить финансирова" ние (структурные фонды для новых членов ЕС и фонды для подго" товки стран к вступлению в ЕС) для финансирования программ поддержки цыганских меньшинств в Центральной и Юго"Восточ" ной Европе. Комиссар ЕС Владимир Шпидла заявил об этом на международной встрече в Брюсселе, организованной совместно с вице"премьером Румынии Марко Бела. На встрече 12 июня соб" рались высокопоставленные представители восьми стран, кото" рые в 2005 г. начали реализацию инициативы "Десятилетие интег" рации цыган". В мероприятии приняли участие лидеры цыганских общин и представители международных организаций, включая Институт открытого общества и Всемирный банк. 1 июля болгарс" кое правительство принимает у Румынии переходящий пост пред" седателя Международного руководящего комитета декады. Оно планирует в течение будущего года укрепить сотрудничество меж" ду национальными правительствами и ЕС. Г енеральный директор ЕС по занятости, социальным вопросам и равным возможностям Николаус Ван Дер Пас в заключение встречи подчеркнул важ" ность максимального использования потенциала цыганской моло" дежи. "Повышение присутствия цыган на рынке труда принесет су" щественные экономические выгоды Европе, переживающей демог" рафический кризис из"за сокращения ее молодого населения", — заявил он. Дополнительную информацию можно получить по ад" ресу: Россия и Всемирный банк объявили о новой программе сотрудничества в под держку глобального развития 9 июня президент Всемирного банка Пол Вулфовиц и Ми" нистр финансов России Алексей Кудрин договорились о сотруд" ничестве с целью оказания помощи бедным странам в области экономического развития, снижения долгового бремени, борьбы с инфекционными заболеваниями и развития энергетики. Заявле" ние о соглашении было сделано в рамках встречи министров фи" нансов Большой восьмерки. П. Вулфовиц и А. Кудрин договори" лись о сотрудничестве в области программы снижения долгового бремени для стимулирования развития, предусматривающей нап" равление 250 млн дол. США на первоочередные нужды стран Аф" рики, расположенных к югу от Сахары. Всемирный банк и Россия объединят усилия по борьбе с инфекционными заболеваниями в странах Африки и Центральной Азии. Всемирный банк и Россия также договорились о расширении доступа к современным энер" гетическим услугам для наименее развитых стран, при этом особо выделив Африку. Всемирный банк обеспечит техническую по" мощь российским властям в создании национальной системы го" сударственной помощи развивающимся странам. Пол Вулфовиц и Алексей Кудрин также договорились об оценке хода реализации программ сотрудничества два раза в год во время весенних и еже" годных совместных конференций Всемирного банка и МВФ. Бли" жайшая из них состоится 19 — 20 сентября текущего года в Син" гапуре. · 25 Новости Всемирного банка Всемирный банк и ЦЭФИР Финансовая помощь странам Европы и Центральной Азии (ЕЦА) для борьбы с птичьим гриппом Армения и Г рузия вошли в список стран ЕЦА, реализующих программы борьбы с птичьим гриппом, в результате решения Со" вета директоров Всемирного банка от 1 июня о финансировании программ повышения информированности населения и предотв" ращения распространения птичьего гриппа. Несмотря на то что в Армении не отмечено вспышек птичьего гриппа, распространение вируса в соседних странах, в частности в Г рузии, вызвало необхо" димость предпринять профилактические мероприятия. Проекты, направленные на борьбу с распространением птичьего гриппа, состоят из трех частей: защита животных от инфицирования, про" филактические мероприятия для населения и кампания по распро" странению информации о птичьем гриппе. Если первые две сос" тавные части направлены на повышение готовности государствен" ных учреждений и системы здравоохранения к эффективным действиям в случае возникновения вспышек заболевания, то ин" формационная кампания предполагает повышение информиро" ванности гражданского общества и населения в целом о рисках и возможных последствиях заболевания. Дополнительная информа" ция о деятельности Всемирного банка в Армении и Г рузии доступ" на на сайтах: http: // www.worldbank.org.am http: // www.worldbank.org/ge Региональный фонд по борьбе со СПИДом объявляет о первых грантах в Центральной Азии В конце мая Региональный фонд по борьбе со СПИДом объя" вил о начале программы предоставления грантов странам, участ" вующим в Центрально"Азиатском проекте по борьбе со СПИДом при финансировании Всемирного банка и Министерства между" народного развития Великобритании (DFID). В проекте участву" ют Казахстан, Кыргызстан, Таджикистан и Узбекистан. Г ранты предоставляются на конкурсной основе для осуществления наци" ональных и региональных инициатив неправительственным орга" низациям, государственным, общественным и частным учрежде" ниям и компаниям. Критерии предоставления грантов опублико" ваны на сайте Проекта: Региональный фонд в первую очередь стремится финансировать работу в таких областях, как профилактика ВИЧ/СПИД; повышение количества добровольного тестирования на ВИЧ/СПИД и обращений за кон" сультациями; расширение доступности лечения и ухода для лиц с ВИЧ/СПИД; эпидемиологический контроль; воспитание толе" рантности в обществе к лицам с ВИЧ/СПИД и борьба с дискри" минацией; укрепление потенциала организаций, объединяющих лиц с ВИЧ/СПИД, и профилактика наркомании, лечение и реаби" литация наркозависимости. Заявки на гранты необходимо отпра" вить по электронному адресу: caap_kz@yahoo.com. Полная ин" формация на русском языке доступна по адресу: Ежеквартальный доклад об экономике восьми стран ЕС в основном оптимистичен Свежий ежеквартальный доклад по экономике 8 стран ЕС, представленный в Варшаве и Братиславе 30 мая, констатирует стабильность и устойчивость экономического роста к внутриполи" тическим колебаниям и непростым внешним условиям в 2005 г. Несмотря на недавние признаки снижения оптимизма по отноше" нию к странам с переходной экономикой, краткосрочные экономи" ческие перспективы достаточно благоприятны. Специальная тема доклада рассматривает взаимосвязь между государственным бюд" жетом и экономическим ростом в регионе. Анализ показывает, что госбюджеты стали больше ориентироваться на экономический рост, однако далеко не все возможности в этой области использо" ваны. В докладе также рассматриваются экономические послед" ствия птичьего гриппа в регионе, планирование использования структурных фондов ЕС в течение бюджетного периода 2007— 2013 гг., прогресс в достижении целей Лиссабонской программы и роль стимулов в привлечении прямых иностранных инвестиций. Доклад доступен по адресу Президент Всемирного банка выступил с докладом по Турции На Второй ежегодной конференции им. Сакипа Сабанчи, про" веденной 23 мая университетом Сабанчи и институтом Брукингса в Вашингтоне, президент Всемирного банка Пол Вулфовиц выс" тупил с докладом: "Турция: у слияния Востока и Запада". П. Вул" фовиц отметил экономические и социальные достижения Турции, подчеркнув их значение как для соседей этой страны, так и для всего мира. Однако несмотря на серьезные экономические преоб" разования, уровень безработицы в Турции остается слишком вы" соким. В отношении вступления Турции в Европейский союз Вул" фовиц отметил, что оно принесет ЕС определенные выгоды. Он также добавил, что Турции необходимо обратить внимание на сис" тему образования и рынок труда. П. Вулфовиц также подчеркнул светский характер государства и уважение к религии в Турции. Полный текст речи президента Всемирного банка доступен по ад" ресу TRIES/ECAEXT/TURKEYEXTN/0,,contentMDK:20933010~menu PK:361718~pagePK:141137~piPK:141127~theSitePK:361712,0 0.html Награды за инновации на западнобал канском Конкурсе идей по экономическо му развитию2006 17 — 18 мая в Белграде состоялся первый на Западных Бал" канах Конкурс идей по экономическому развитию. 55 финалистов из Албании, Боснии и Г ерцеговины, Македонии, Сербии и Черно" гории, а также Косова, победившие в национальных конкурсах, представили инновационные идеи по созданию рабочих мест в официальном секторе занятости и боролись за получение грантов в размере 35 000 долларов США на проект. Международное жю" ри под председательством представителя Всемирного банка Орса" лии Каланцопулос выбрало 21 проект, получившие в общей слож" ности 712 000 долл., предоставленных донорами. 17 мая в рамках конкурса прошел Форум знаний, посвященный занятости молоде" жи. Форум был организован при поддержке известной местной неправительственной организации "Г ражданские инициативы". В мероприятии приняли участие более 100 молодых представителей неправительственных организаций и политиков из стран — участ" ниц конкурса. Они обменялись опытом по вопросам расширения возможностей молодежи и включения ее в рынок труда. Информация предоставлена Мерел Такк и Кристиной Лакатос, Отдел внешних связей по Европе и Центральной Азии, Всемирный банк, Вашингтон BT · 26 Новости Всемирного банка Beyond Transition • Апрель июнь 2006 · 27 Новые публикации Всемирный банк и ЦЭФИР Всемирный банк Neil Parison, Yelena Dobrolyubova, Gord Evans, Nick Manning и Yulia Shirokova Повышение эффективности управления: подходы к админи стративной реформе в Российской Федерации (Increasing Government Effectiveness: Approaches to Administrative Reform in the Russian Federation) Report 36142, май 2006 г. В условиях России административная реформа означает осу" ществление преобразований, связанных со структурой и функци" ями государственного аппарата. Кроме того, она должна открыть новые подходы к повышению эффективности работы государства (власти) (стратегическому планированию, контролю эффектив" ности деятельности, внутренней подотчетности), а также повыше" нию уровня исполнения и ответственности (прозрачности, качест" ва работы исполнительных органов и внешней подотчетности). Данная работа рассматривает вопросы, значимые для следующего этапа российских административных реформ. Работа содержит: а) обзор подходов к административным реформам и методов их осу" ществления в других странах (применительно к условиям и конк" ретным задачам России), включая изучение отдельных ситуаций и накопленный опыт; б) перечень интернет"ресурсов со ссылками как на конкретный инструментарий, используемый чиновниками и специалистами"практиками ряда стран, так и на результаты его применения; в) оценку значения международного опыта для даль" нейшей разработки и осуществления административных реформ в Российской Федерации. Roberta Gatti и Inessa Love Влияет ли доступность кредитов на повышение производи тельности? Данные по болгарским компаниям (Does Access to Credit Improve Productivity? Evidence from Bulgarian Firms) WPS3921, май 2006 г. Несмотря на распространенное мнение, что развитие финан" совой сферы способствует экономическому росту, данных о ка" налах влияния кредитования на экономический рост на микроу" ровне пока недостаточно. Опираясь на структурные данные по болгарским компаниям, авторы работы оценивают влияние дос" тупности кредитов (измеренное возможностью фирм получить кредит или возможностью овердрафта) на производительность. Авторы исследуют данные о предыдущем развитии фирм и их кредитовании, заключая, что кредитование является безуслов" ным и мощным фактором повышения уровня производительнос" ти компаний. Anqing Shi Миграция в города Китая, история трех провинций: о чем говорят предварительные данные переписи населения 2000 г. (Migration in Towns in China, a Tale of Three Provinces: Evidence from the 2000 Census Preliminary Tabulations) WPS3890, апрель 2006 г. В Китае возникла проблема остановки роста городов, вслед" ствие чего затормозилось создание несельскохозяйственных ра" бочих мест в аграрных регионах. Для исследования этой пробле" мы автор использует данные переписи населения 2000 г., рас" сматривая уровень образования, первоначальное место прожива" ния и профессиональный состав мигрантов в городах трех китай" ских провинций — Чжэцзян, Хэнань и Сычуань. Автор устанав" ливает, что в целом мигранты обладают более высоким уровнем образования, чем местное городское население, особенно в ме" нее развитых западных и центральных провинциях. Приток чело" веческого капитала может способствовать развитию городов. Также подтверждается, что по мере возрастания экономических возможностей в городах, как это, например, происходит в относи" тельно благополучных прибрежных районах провинции Чжэцзян, более образованное население в сельских районах, вероятно, бу" дет перемещаться из сельского хозяйства в несельскохозяйствен" ный сектор ближайших городов, не покидая своего места житель" ства. Это может снизить остроту проблемы миграции в крупные города. Наконец, рынок труда в городах менее развитых западных и центральных регионов более гибок по отношению к мигрантам, по сравнению, например, с экономически развитой провинцией Чжэцзян, где наблюдается разделение на рынке труда между им" мигрантами и местным населением. Paloma Lopez"Garcia Бизнессреда и рынок труда в Европе и странах Централь ной Азии (Business Environment and Labor Market Outcomes in Europe and Central Asia Countries) WPS3885, апрель 2006 г. Данная работа помогает расширить знания о бизнес"среде в странах Европы и Центральной Азии (ЕЦА). Она изучает, какие из формирующих бизнес"среду институтов наиболее важны для эффективного функционирования рынка труда. Автор группирует институты по их влиянию на выход фирм на рынок и на выживание и расширение бизнеса и выводит совокупные показатели. Затем анализируется влияние институтов на рост занятости в частном секторе, измеряемый уровнем занятости в сфере обслуживания. Автор делает вывод, что доступ к финансированию — наиболее важный фактор роста занятости во всех странах ЕЦА. С другой стороны, ограниченные возможности получения финансирования в Болгарии, Хорватии и особенно Румынии являются главной при" чиной недостаточного развития частного сектора этих стран. Су" щественное влияние на занятость оказывают регулирование рын" ка, затраты на создание бизнеса и налоговое бремя. Peter Huber Развитие региональных рынков труда в переходный период (Regional Labor Market Developments in Transition) WPS3896, апрель 2006 Автор работы приходит к заключению, что значительные и ус" тойчивые различия между региональными рынками труда появи" лись практически во всех странах с переходной экономикой, чему послужили различия в стартовых условиях и выходе на рынок в переходный период. Структура заработной платы на региональ" ных рынках труда лишь немногим менее жесткая, чем во многих странах Европейского союза. Кроме того, низок уровень межре" гиональной миграции, а капитал охотнее движется в города с вы" сокой заработной платой и занятостью, чем в отсталые регионы. Необходима долгосрочная политика, направленная на преодоле" ние региональных различий, устранение миграционных барьеров, преобразование институтов региональной политики и координа" цию программ региональной политики и политики в отношении рынка труда. Paolo Verme Препятствия для экономического роста и создания рабочих мест в странах СНГ с низким доходом (Constraints to Growth and Job Creation in LowIncome CIS Countries) WPS3893, апрель 2006 г. · 28 Новые публикации Несмотря на устойчивый рост производства с 1997 г. в странах Содружества Независимых Государств (СНГ"7) с низким доходом, он не повлиял на уровень занятости. Это явление, наблюдаемое также в других странах с переходной экономикой, называется "экономическим ростом без роста занятости". Автор рассматрива" ет его причины в СНГ"7. Он доказывает, что количество рабочих мест не увеличивается из"за сочетания структурных факторов, среди которых — капиталоемкий экономический рост, большой потенциал повышения производительности труда у занятых работ" ников и сосуществование различных экономических структур, наиболее ярко проявляющееся в двойной структуре рынка труда. Если правительства стран и международное сообщество не начнут инвестировать средства в сельское хозяйство и промышленные программы, направленные на возрождение обрабатывающей про" мышленности, экономический рост без создания новых рабочих мест, скорее всего, продолжится. Jan Rutkowski Развитие рынка труда в экономике переходного периода (Labor Market Developments during Economic Transition) WPS3894, апрель 2006 г. В работе рассматривается развитие рынка труда в переходных экономиках стран Европы и Центральной Азии. Автор доказывает, что дефицит рабочих мест и возрастающая сегментация рынка тру" да — две главные проблемы в области занятости, стоящие перед странами с переходной экономикой. В европейских странах с пере" ходной экономикой нехватка рабочих мест вызвала устойчивую открытую безработицу. В СНГ это привело к скрытой безработице. Безработица в европейских странах с переходной экономикой соче" тается с развитой системой социальной защиты. В СНГ большин" ство рабочих либо остается на своих старых неэффективных рабо" чих местах на нереструктурированных предприятиях, либо работа" ет в неофициальном секторе, либо переходит на натуральное хо" зяйство. Сегментация рынка труда была вызвана резким увеличе" нием дифференциации доходов, сопутствующим ростом количества низкооплачиваемых рабочих мест, значительными различиями ус" ловий на рынке труда в зависимости от региона и, наконец, расши" рением неофициального сектора. Сегментация рынка труда и соп" ровождающее ее неравенство сильнее проявляются в странах СНГ , чем в европейских государствах с переходной экономикой. John S. Earle and David J. Brown Микроэкономика создания производительных рабочих мест: обобщение исследований по предприятиям в странах пе реходного периода (The Microeconomics of Creating Productive Jobs: A Synthesis of FirmLevel Studies in Transition Economies) WPS3886, апрель 2006 г. Серьезной проблемой политики занятости в экономиках пере" ходного периода стала необходимость создания производительных рабочих мест для компенсации резкого спада занятости. Авторы показывают охват и эффективность перераспределения рабочих мест и работников. Они рассматривают результаты приватиза" ции, либерализации товарного рынка и рынка труда и барьеры, препятствующие росту нового частного сектора и его эффектив" ности в Венгрии, Румынии, России и Украине. Авторы приходят к выводу, что рыночная реформа привела к значительному росту темпов перераспределения рабочих мест, особенно между секто" рами экономики, а также в результате движения фирм на рынке, что способствовало росту совокупной производительности. Влия" ние приватизации на производительность компаний отличается по странам и не всегда положительно. Рост производительности компаний в результате их приватизации обычно происходил не за счет работников, а скорее был связан с повышением заработной платы и занятости. Felix Eschenbach и Bernard Hoekman Стратегии сектора услуг в экономиках переходного периода: Европейский союз и ВТО как механизмы принятия обяза тельств (Services Policies in Transition Economies: on the European Union and the WTO as Commitment Mechanisms) WPS3951, июнь 2006 г. Авторы анализируют, в какой степени 15 стран ЕС и 16 стран с переходной экономикой использовали Г енеральное соглашение о торговле услугами (ГАТС) ВТО для проведения реформ в секторе услуг. Они сравнивают обязательства по ГАТС с реальными мера" ми политики в сфере услуг за прошедшие годы и приходят к выво" ду, что между объемом обязательств по ГАТС и "качеством" фак" тической политики в сфере услуг по оценке частного сектора су" ществует отрицательная связь. Частично это объясняется тем, что перспектива вступления в ЕС снижает эффективность ГАТС в ка" честве механизма принятия обязательств. С другой стороны, для многих стран, не вступающих в ЕС, вступление в ВТО представля" ется слабым стимулом для выполнения обязательств. Еще одна причина заключается в том, что небольшой размер рынка услуг не является достаточным внешним стимулом для выполнения обяза" тельств по ГАТС. Чтобы расширить выгоды от вступления в ВТО для небольших стран, авторы предлагают увеличить общие инвес" тиции членов ВТО в мониторинг и повышение прозрачности. Институт переходной экономики Банка Финляндии (BOFIT) Olena Havrylchyk и Emilia Jurzyk Доходность иностранных банков в Центральной и Восточной Европе: имеет ли значение способ выхода на рынок? (Profitability of Foreign Banks in Central and Eastern Europe: Does the Entry Mode Matter?) BOFIT DP № 5/2006 В работе анализируется доходность местных и иностранных банков в странах Центральной и Восточной Европы (ЦВЕ) на ос" нове данных по 265 банкам за 1995 — 2003 гг. Авторы показыва" ют, что иностранные банки, особенно новые на этом рынке, полу" чают более высокую прибыль, чем местные. Это свойство скорее приобретенное, чем унаследованное, поскольку иностранные бан" ки обычно приобретают менее прибыльные местные финансовые учреждения. Прибыль филиалов иностранных банков в ЦВЕ пре" вышает прибыль "материнских" банков. Авторы изучают издерж" ки и выгоды иностранной собственности, анализируя факторы до" ходности местных, приобретенных иностранцами и созданных "с нуля" иностранных банков на данном рынке. На прибыль иност" ранных банков меньше влияют макроэкономические условия в странах их происхождения. Однако созданные "с нуля" банки бо" лее чувствительны к ситуации в головных офисах. Местные банки получают высокую прибыль на более концентрированных рынках, а банки, приобретенные иностранцами, страдают от отрицатель" ного эффекта масштаба, вследствие того, что обычно это крупные структуры. Laura Solanko Проблема недостающей социальной инфраструктуры: ана лиз российских промышленных предприятий (Coping with Beyond Transition • Апрель июнь 2006 Missing Public Infrastructure: An Analysis of Russian Industrial Enterprises) BOFIT DP № 2/2006 г. На основе данных по 404 крупным и средним промышленным предприятиям в 40 регионах России в работе рассматривается обеспечение функционирования общественной инфраструктуры российскими промышленными предприятиями. Автор приходит к выводу, что, во"первых, в значительной степени обеспечение инфраструктурой — измеренное как обеспечение централизо" ванным отоплением — является наследием советских времен. Во"вторых, предприятия, обеспечивающие отоплением не только район своего местонахождения, но и другие районы, обычно име" ют более тесные взаимоотношения с местными органами власти и в других сферах. Центр экономических и финан совых исследований и разработок (ЦЭФИР) Tuuli Juurikkala и Ольга Лазарева Лоббирование на местном уровне: социальные активы рос сийских компаний (Lobbying at the Local Level: Social Assets in Russian Firms) WP 61, Январь 2006 г. Во времена плановой экономики предприятия отвечали за обеспечение своих работников социальными благами. В России в переходный период 1990"х социальная инфраструктура должна была перейти из ведения предприятий в государственный сектор. В течение нескольких лет права собственности на часть социаль" ных активов, особенно жилье, не были должным образом опреде" лены, поскольку решения по передаче объектов на баланс госуда" рства в основном принимали игроки местного уровня. Авторы ис" пользуют данные недавнего опроса 404 средних и крупных про" мышленных предприятий в 40 российских регионах и применяют данные по их реорганизации для установления фактора выбора времени отчуждения общественных активов. Результаты исследо" вания показывают, что в муниципалитетах с высокой долей собственных доходов в местном бюджете и, таким образом, мень" шими налоговыми льготами, предприятия использовали социаль" ные активы в качестве рычага получения бюджетной помощи и других льгот от местных властей. Авторы приходят к заключению, что менее конкурентоспособные фирмы использовали социаль" ные активы для защиты от рыночной конкуренции. Центр социальных и экономичес ких исследований (CASE) www.case.com.pl Алексей Кисенков, Петр Козаржевский, Ирина Лукашова, Мария Лукашова, Юлия Миронова Система корпоративного управления в Кыргызстане (The System of Corporate Governance in Kyrgyzstan) Studies and Analysis № 326, июнь 2006 г. В работе рассматривается формирование современной систе" мы корпоративного управления в Республике Кыргызстан. Ис" следуются основные факторы, влияющие на этот процесс, напри" мер, законодательная база и практика приватизации, корпоратив" ное и антимонопольное законодательство, финансовые рынки, действия акционеров и т. д. Авторы приходят к выводу, что в сфе" ре корпоративного управления в Кыргызстане произошли суще" ственные положительные изменения. В стране, у которой отсут" ствовал опыт частной собственности и рыночных отношений, бы" ли сформированы институты корпоративного управления, начал" ся процесс приобретения владельцами и менеджерами компаний знаний по управлению бизнесом на основе имеющейся законода" тельной и нормативной базы. Но этот процесс далеко не закончен, поскольку существующие корпоративные отношения все еще очень искажены. По мнению авторов, улучшение корпоративного управления в стране требует комплексного подхода: совершен" ствование законодательства должно сопровождаться активными мерами по улучшению ситуации во всех сферах, влияющих на ка" чество корпоративного управления. Основная задача — создать благоприятный нормативно"правовой климат, который способ" ствовал бы повышению качества корпоративного управления и привлечению иностранных инвестиций. Jens Holscher, Mariusz Jarmuzek, Roman Matousek, и Eva Katalin Polgar Прозрачность бюджета в экономиках переходного периода (Fiscal Transparency in Transition Economies) Studies and Analysis № 328, июнь 2006 г. Прозрачность бюджета стала темой общественных и научных дискуссий после мексиканского и азиатского финансовых кризи" сов. Однако понятие прозрачности бюджета в значительной мере качественное, и поэтому измерить эту характеристику непросто В данной работе предлагается показатель прозрачности бюджета по различным аспектам разработки бюджетной политики. Индекс рассчитан для двадцати семи стран с переходной экономикой и ос" нован на анализе открытой информации. Анализ индекса прозрач" ности бюджета выявляет четкую структуру: страны Центральной и Восточной Европы отличаются по всем показателям от Юго" Восточной Европы (ЮВЕ) и СНГ. Вероятно, это результат вклю" чения новых членов ЕС в структуры и процедуры Европейского cоюза. Страны ЮВЕ отстают от первой группы и сталкиваются с серьезными затруднениями в бюджетном процессе и соблюдении стандартов отчетности. Страны СНГ все еще далеки от междуна" родных стандартов в этой области и должны улучшить методы ра" боты с бюджетом и процедуры отчетности. Прием рецензий Научное издание "St. Antony's International Review" объявляет о приеме рецензий на книги для публикации в своем следующем номере, посвященном демократизации. Заинтересованные в теме ученые, аспиранты, специалисты могут предложить книги, рецен" зии на которые они хотели бы предоставить, редактору рубрики Калину Иванову (Kalin.ivanov@politics.ox.ac.uk). Крайний срок подачи предложений — 1 сентября 2006 г. Общие требования, предъявляемые к рецензиям, изложены в "Руководстве для авто" ров": www.sant.ox.ac.uk/jcr/STAIR/files/ReviewerGuidelines.pdf "St. Antony's International Review" — научный журнал, осно" ванный научными сотрудниками Колледжа Св. Антония в Оксфо" рдском университете и рецензируемый членами научного сообще" ства. Журнал издается дважды в год и способствует развитию междисциплинарного диалога о глобальных вопросах современ" ности. BT · 29 Всемирный банк и ЦЭФИР · 30 Календарь событий 21й Ежегодный конгресс Европейской экономи ческой ассоциации и европейские встречи Эконо метрического общества (EEA ESEM) 24 — 28 августа 2006, Вена, Австрия Европейская экономическая ассоциация и Эконометрическое общество — международные научные организации, целью кото" рых является продвижение и распространение результатов эконо" мических исследований в эмпирической и теоретической областях. 21"й Ежегодный конгресс будет проведен совместно с Институтом повышения квалификации и Университетом Вены. Дополнительная информация: esem2006.org/ Летний европейский симпозиум по экономике труда 2006 (ESSLE) 14 сентября 2006, Аммерси, Г ермания Цель летнего симпозиума — пригласить к участию экономис" тов, занимающихся исследованиями в области труда, из Европы и других регионов. Участники конференции получат возможность обсудить результаты своей работы со старшими коллегами, уста" новить долгосрочное сотрудничество. Дополнительная информация: ence_files/essle2006/call_for_papers Ежегодная встреча Международного валютного фонда и Группы Всемирного банка 16 — 18 сентября, 2006, Сингапур Тема встречи: "Роль Азии в глобальном пространстве, миро" вые тенденции в Азии". В программу включены более 30 рабочих сессий, на которых будут обсуждаться вопросы международных финансов и развития, оказывающие влияние на жизнь людей и функционирование бизнеса в регионе и в мире: • Рынки капитала и финансовый сектор в странах с разви" вающейся экономикой • Коррупция, управление и экономический рост • Инфраструктура в 21"м веке • Инвестиционный климат; торговля и инвестиции • Энергетика и безопасность • Экономический рост и социальное равенство • Региональная экономическая и финансовая интеграция • Инновации и технологии. Дополнительная информация: /external/am/2006/schedule.htm Устанавливая равенство в области здравоохра нения 29 — 30 сентября 2006, Хельсинки, Финляндия В конференции примут участие исследователи и специалисты" практики из развитых и развивающихся стран, которые проанали" зируют причины неравного доступа к услугам здравоохранения и обсудят последствия неравенства в этой области для жителей раз" ных стран. Они рассмотрят тенденции развития ситуации и опре" делят альтернативные методы решения проблемы и направления будущих исследований. Дополнительная информация: edu/conference/conferences.htm Вторая международная конференция "Роль го сударства в экономике в XXI веке" 13 — 14 октября, 2006, Москва, Россия Организаторы: Global Institute, BP , ЦЭФИР в РЭШ (Москва), Центр стратегических разработок (Москва), Центр "Китай в ми" ровой экономике" (Пекин), Центр экономических исследований КНР (Пекин). Темы панельных сессий включают: • Стратегии развития в XXI веке: баланс между государ" ственным регулированием и рыночными отношениями • Стимулирование чиновников госаппарата • Роль государства в макроэкономике: таргетирование ре" ального валютного курса или инфляции? • Государство в глобальной экономике • Цель 2030: государство и долгосрочный рост • Государство и экономика знаний • Энергетические вызовы • Неравенство и социальная стабильность. Дополнительная информация: /index.php?l=rus&id=156 Курс обучения им. Марии Кюри по проблемам кооперации в странах с переходной экономикой 9 — 13 октября 2006, София, Болгария Курс обучения будет организован Институтом изучения коопе" рации Университета им. Гумбольдта, Берлин, и Институтом эконо" мики сельского хозяйства, София. Это первый курс из серии ме" роприятий, запланированных в рамках Проекта им. Марии Кюри "Современное сельское хозяйство в странах Центральной и Вос" точной Европы: инструменты анализа и управление изменения" ми". В рамках курса участники ознакомятся с исследованиями в области кооперации и кооперативов как экономическим и соци" альным способом организации деятельности, а также рассмотрят эмпирическую методологию, применяемую при анализе. Послед" ний срок подачи заявки на участие и краткого описания работы — 31 августа 2006 г. Подробная информация: www.mace!events.org. Контакты: Marlis Werner, Institute for Cooperative Studies, Humboldt University Berlin, Unter den Linden 6,10099 Berlin, Germany,Tел:+49 30 2093 6500,Факс: +49 30 2093 6501, email: ifg@agrar.hu!berlin.de Седьмая ежегодная исследовательская конфе реция МВФ им. Жака Полака 9 — 10 ноября 2006, Вашингтон, США Тема конференции: "Потоки капитала". На конференции будет обсуждаться спектр вопросов: • Роль потоков капитала в экономическом развитии • Подъемы и спады в потоках капитала в странах с разви" вающейся рыночной экономикой и меры, смягчающие эти эффек" ты • Глобальные финансовые диспропорции и пути их устра" нения • Влияние движения капитала на внутреннюю политику стран и их институты • Факторы, определяющие направление, размер и структу" ру международных потоков капитала. Дополнительная информация: nal/np/res/seminars/2006/arc/index.htm BT Beyond Transition • Апрель июнь 2006 · 31 Библиография Всемирный банк и ЦЭФИР "Кредитноденежная политика и мировые рынки сырой нефти" Нуреддин Кричин Bernanke, B.S., and I. Mihov, 1998, "Measuring Monetary Policy", Quarterly Journal of Economics, Vol. 113, No. 3, pp. 869"902. Bernanke, B.S., M. Gertler, and M. Watson, 1997, "Systematic Monetary Policy and the Effects of Oil Shocks", Brookings Papers on Economic Activity, Vol. 1997, No. 1, pp. 91"157. 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Helpman, E. and Rangel, A., 1999, "Adjusting to a New Technology: Experience and Training", Journal of Economic Growth, 4, pp. 359"383. Simar, L., Wilson, P .W ., 2006, "Estimation and Inference in Two"Stage, Semi"Parametric Models of Production Processes", Journal of Econometrics, forthcoming. Simar, L., Zelenyuk, V ., 2006a, "Statistical Inference For Aggregates Of Farrell"Type Efficiencies", Journal of Applied Econometrics, forthcoming. Simar, L., Zelenyuk, V ., 2006b, "On Testing Equality of Two Distribution Functions of Efficiency Scores Estimated from DEA", Econometric Reviews, forthcoming. Shiu, A., Zelenyuk, V , "Production Efficiency vs. Ownership: The Case of China", mimeo, Kyiv Economics Institute, Kyiv, Ukraine. Zelenyuk, V ., Zheka, V ., 2006, "Corporate Governance and Firm's Efficiency: The Case of a Transitional Country", Ukraine. Journal of Productivity Analysis 25, pp.143"168. "Россия и ВТО: бремя "неприсоединившегося" Богдан Лиссоволик и Ярослав Лиссоволик Elborgh"Woytek, K., 2003, "Of Openness and Distance: Trade Developments in the Commonwealth of Independent States, 1993"2002", IMF Working Paper 03/207. Washington: International Monetary Fund. Lissovolik, Y., and N, Liventsev, 2002, Actualnyye Pmblemy Prisoyedmeniya Rossii к VTO, Moscow: Ekonomika. Rose, A.K., 2002, "Do WE Really Know That the WTO Increases Trade?" NBER Working Paper No. 9273, Cambridge, Massachusetts: National Bureau of Economic Research. Subramanian, A., and S.J. Wei, 2003, "The WTO Promotes Trade, Strongly but Unevenly", IMF Working Paper 03/185, Washington: International Monetary Fund. · 32 Подписаться на печатную версию Подписаться на электронную версию Если Вы или Ваша организация хотите бесплатно получать вестник "Beyond Transition", пожалуйста, заполните форму и отправьте ее нам по факсу, почте или на адрес электронной почты: editor@cefir.ru. 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6443	https://gmcancer.org.uk/wp-content/uploads/2021/10/nice-melanoma-assessment-and-management-guideline.pdf	Melanoma: assessment and management Melanoma: assessment and management NICE guideline Published: 29 July 2015 nice.org.uk/guidance/ng14 © NICE 2015. All rights reserved. Y Your responsibility our responsibility The recommendations in this guideline represent the view of NICE, arrived at after careful consideration of the evidence available. When exercising their judgement, professionals are expected to take this guideline fully into account, alongside the individual needs, preferences and values of their patients or service users. The application of the recommendations in this guideline are not mandatory and the guideline does not override the responsibility of healthcare professionals to make decisions appropriate to the circumstances of the individual patient, in consultation with the patient and/or their carer or guardian. Local commissioners and/or providers have a responsibility to enable the guideline to be applied when individual health professionals and their patients or service users wish to use it. They should do so in the context of local and national priorities for funding and developing services, and in light of their duties to have due regard to the need to eliminate unlawful discrimination, to advance equality of opportunity and to reduce health inequalities. Nothing in this guideline should be interpreted in a way that would be inconsistent with compliance with those duties. Commissioners and providers have a responsibility to promote an environmentally sustainable health and care system and should assess and reduce the environmental impact of implementing NICE recommendations wherever possible. Melanoma: assessment and management (NG14) © NICE 2015. All rights reserved. Page 2 of 61 Contents Contents Overview ................................................................................................................................................................................ 5 Who is it for? ...................................................................................................................................................................................... 5 Introduction .......................................................................................................................................................................... 6 Safeguarding children .................................................................................................................................................................... 6 Medicines............................................................................................................................................................................................ 7 Patient-centred care.......................................................................................................................................................... 8 Key priorities for implementation................................................................................................................................ 9 Communication and support....................................................................................................................................................... 9 Assessing melanoma....................................................................................................................................................................... 10 Managing suboptimal vitamin D levels.................................................................................................................................... 10 Staging investigations .................................................................................................................................................................... 10 Managing stage III melanoma ..................................................................................................................................................... 11 Follow-up after treatment for melanoma .............................................................................................................................. 12 Stages of melanoma ...........................................................................................................................................................14 1 Recommendations ..........................................................................................................................................................15 1.1 Communication and support ............................................................................................................................................... 15 1.2 Assessing melanoma ............................................................................................................................................................... 16 1.3 Managing suboptimal vitamin D levels............................................................................................................................ 18 1.4 Managing concurrent drug treatment ............................................................................................................................. 18 1.5 Staging investigations............................................................................................................................................................. 18 1.6 Managing stages 0–II melanoma........................................................................................................................................ 20 1.7 Managing stage III melanoma.............................................................................................................................................. 20 1.8 Managing stage IV melanoma.............................................................................................................................................. 22 1.9 Follow-up after treatment for melanoma....................................................................................................................... 24 2 Research recommendations .......................................................................................................................................28 2.1 Techniques for confirming a diagnosis in people with suspected atypical spitzoid melanocytic lesions .................................................................................................................................................................................................. 28 Melanoma: assessment and management (NG14) © NICE 2015. All rights reserved. Page 3 of 61 2.2 Surgical excision for people with lentigo maligna ....................................................................................................... 28 2.3 Follow-up surveillance imaging........................................................................................................................................... 29 2.4 Vitamin D supplementation ................................................................................................................................................. 29 2.5 The effect of drug therapy for concurrent conditions on melanoma survival.................................................. 30 3 Other information...........................................................................................................................................................31 3.1 Scope and how this guideline was developed................................................................................................................ 31 3.2 Related NICE guidance........................................................................................................................................................... 31 4 The Guideline Development Group, National Collaborating Centre and NICE project team, and declarations of interests...................................................................................................................................................34 4.1 Guideline Development Group........................................................................................................................................... 34 4.2 National Collaborating Centre for Cancer..................................................................................................................... 35 4.3 NICE project team.................................................................................................................................................................... 36 4.4 Declarations of interests ....................................................................................................................................................... 37 Implementation: getting started...................................................................................................................................53 Challenge 1 – Using dermoscopy (dematoscopy) to assess pigmented lesions...................................................... 53 Challenge 2 – Measuring vitamin D levels and advising on supplementation......................................................... 54 Challenge 3 – Considering sentinel lymph node biopsy and completion lymphadenectomy ........................... 55 Further resources ............................................................................................................................................................................ 57 Changes after publication................................................................................................................................................58 About this guideline ...........................................................................................................................................................59 Strength of recommendations.................................................................................................................................................... 59 Other versions of this guideline................................................................................................................................................. 60 Implementation ................................................................................................................................................................................ 60 Your responsibility........................................................................................................................................................................... 60 Melanoma: assessment and management (NG14) © NICE 2015. All rights reserved. Page 4 of 61 This guideline is the basis of QS130. Ov Overview erview This guideline covers the assessment and management of melanoma (a type of skin cancer) in children, young people and adults. It aims to reduce variation in practice and improve survival. Who is it for? Healthcare professionals working in primary, secondary and tertiary care People with melanoma and their families and carers Melanoma: assessment and management (NG14) © NICE 2015. All rights reserved. Page 5 of 61 Introduction Introduction Melanoma is the third most common skin cancer in the UK. It accounts for more cancer deaths than all other skin cancers combined. In 2011 there were 13,348 new cases of melanoma and 2209 deaths from melanoma. Although melanoma is more often diagnosed in older people, it is increasingly affecting younger people. More than 900 adults aged under 35 are now diagnosed with melanoma annually in the UK, and it is the second most common cancer in adults aged between 25 and 49. Melanoma therefore leads to more years of life lost overall than many more common cancers. Most melanomas occur in people with pale skin. The risk factors are skin that tends to burn in the sun, having many moles, intermittent sun exposure and sunburn. This guideline addresses areas where there is uncertainty or variation in practice. It contains recommendations on: assessing and staging melanoma, including the use of sentinel lymph node biopsy treating stages 0–IV melanoma, including adjuvant chemotherapy and immunotherapy treating in-transit melanoma metastases treating metastatic melanoma follow-up after treatment for melanoma. The guideline also includes advice on managing vitamin D levels and drug therapy for intercurrent conditions in people diagnosed with melanoma. The guideline covers suspected or newly diagnosed cutaneous melanoma (including vulval and penile melanoma) in children, young people and adults. However, there was insufficient high-quality evidence on which to make specific recommendations for vulval and penile melanoma. It does not cover primary ocular melanoma or melanoma arising in mucosal sites. Safeguarding children Remember that child maltreatment: Melanoma: assessment and management (NG14) © NICE 2015. All rights reserved. Page 6 of 61 is common can present anywhere may co-exist with other health problems, including melanoma. See the NICE guideline on child maltreatment for clinical features that may be associated with maltreatment. Medicines The guideline will assume that prescribers will use a medicine's summary of product characteristics to inform decisions made with individual patients. This guideline recommends some medicines for indications for which they do not have a UK marketing authorisation at the date of publication, if there is good evidence to support that use. The prescriber should follow relevant professional guidance, taking full responsibility for the decision. The patient (or those with authority to give consent on their behalf) should provide informed consent, which should be documented. See the General Medical Council's Prescribing guidance: prescribing unlicensed medicines for further information. Where recommendations have been made for the use of medicines outside their licensed indications ('off-label use'), these medicines are marked with a footnote in the recommendations. Melanoma: assessment and management (NG14) © NICE 2015. All rights reserved. Page 7 of 61 P Patient-centred care atient-centred care This guideline offers best practice advice on the care of children, young people and adults with suspected or diagnosed melanoma. Patients and healthcare professionals have rights and responsibilities as set out in the NHS Constitution for England – all NICE guidance is written to reflect these. Treatment and care should take into account individual needs and preferences. Patients should have the opportunity to make informed decisions about their care and treatment, in partnership with their healthcare professionals. If the patient is under 16, their family or carers should also be given information and support to help the child or young person to make decisions about their treatment. Healthcare professionals should follow the Department of Health's advice on consent. If someone does not have capacity to make decisions, healthcare professionals should follow the code of practice that accompanies the Mental Capacity Act and the supplementary code of practice on deprivation of liberty safeguards. NICE has produced guidance on the components of good patient experience in adult NHS services. All healthcare professionals should follow the recommendations in patient experience in adult NHS services. If a young person is moving between paediatric and adult services, care should be planned and managed according to the best practice guidance described in the Department of Health's Transition: getting it right for young people. Adult and paediatric healthcare teams should work jointly to provide assessment and services to young people with suspected or diagnosed melanoma. Diagnosis and management should be reviewed throughout the transition process, and there should be clarity about who is the lead clinician to ensure continuity of care. Melanoma: assessment and management (NG14) © NICE 2015. All rights reserved. Page 8 of 61 K Ke ey priorities for implementation y priorities for implementation The following recommendations have been identified as priorities for implementation. The full list of recommendations is in section 1. See implementation: getting started for information about putting the recommendations on dermoscopy, managing suboptimal vitamin D levels, sentinel lymph node biopsy and completion lymphadenectomy into practice. Communication and support To help people make decisions about their care, follow the recommendations on communication, information provision and support in NICE's guideline on improving outcomes for people with skin tumours including melanoma, in particular the following 5 recommendations: 'Improved, preferably nationally standardised, written information should be made available to all patients. Information should be appropriate to the patients' needs at that point in their diagnosis and treatment, and should be repeated over time. The information given must be specific to the histopathological type of lesion, type of treatment, local services and any choice within them, and should cover both physical and psychosocial issues.' 'Those who are directly involved in treating patients should receive specific training in communication and breaking bad news.' 'Patients should be invited to bring a companion with them to consultations.' 'Each LSMDT [local hospital skin cancer multidisciplinary team] and SSMDT [specialist skin cancer multidisciplinary team] should have at least one skin cancer clinical nurse specialist (CNS) who will play a leading role in supporting patients and carers. There should be equity of access to information and support regardless of where the care is delivered.' 'All LSMDTs and SSMDTs should have access to psychological support services for skin cancer patients.' Melanoma: assessment and management (NG14) © NICE 2015. All rights reserved. Page 9 of 61 Assessing melanoma Dermoscop Dermoscopy and other visualisation techniques y and other visualisation techniques Assess all pigmented skin lesions that are either referred for assessment or identified during follow-up in secondary or tertiary care, using dermoscopy carried out by healthcare professionals trained in this technique. Photogr Photograph aphy y For a clinically atypical melanocytic lesion that does not need excision at first presentation in secondary or tertiary care: use baseline photography (preferably dermoscopic) and and review the clinical appearance of the lesion, and compare it with the baseline photographic images, 3 months after first presentation to identify early signs of melanoma. T Taking tumour samples for genetic testing aking tumour samples for genetic testing If targeted systemic therapy is a treatment option, offer genetic testing using: a secondary melanoma tissue sample if there is adequate cellularity or or a primary melanoma tissue sample if a secondary sample is not available or is of inadequate cellularity. Managing suboptimal vitamin D levels Measure vitamin D levels at diagnosis in secondary care in all people with melanoma. Staging investigations Sentinel lymph node biopsy Sentinel lymph node biopsy Consider sentinel lymph node biopsy as a staging rather than a therapeutic procedure for people with stage IB–IIC melanoma with a Breslow thickness of more than 1 mm, and give them detailed verbal and written information about the possible advantages and disadvantages, using the table below. Melanoma: assessment and management (NG14) © NICE 2015. All rights reserved. Page 10 of 61 P Possible advantages of sentinel lymph node ossible advantages of sentinel lymph node biopsy biopsy P Possible disadvantages of sentinel lymph node ossible disadvantages of sentinel lymph node biopsy biopsy The operation helps to find out whether the cancer has spread to the lymph nodes. It is better than ultrasound scans at finding very small cancers in the lymph nodes. The purpose of the operation is not to cure the cancer. There is no good evidence that people who have the operation live longer than people who do not have it. The operation can help predict what might happen in the future. For example, in people with a primary melanoma that is between 1 and 4 mm thick: around 1 out of 10 die within 10 years if the sentinel lymph node biopsy is negative around 3 out of 10 die within 10 years if the sentinel lymph node biopsy is positive. The result needs to be interpreted with caution. Of every 100 people who have a negative sentinel lymph node biopsy, around 3 will subsequently develop a recurrence in the same group of lymph nodes. People who have had the operation may be able to take part in clinical trials of new treatments for melanoma. These trials often cannot accept people who haven't had this operation. A general anaesthetic is needed for the operation. The operation results in complications in between 4 and 10 out of every 100 people who have it. Managing stage III melanoma Completion lymphadenectom Completion lymphadenectomy y Consider completion lymphadenectomy for people whose sentinel lymph node biopsy shows micro-metastases and give them detailed verbal and written information about the possible advantages and disadvantages, using the table below. P Possible advantages of completion ossible advantages of completion lymphadenectom lymphadenectomy y P Possible disadvantages of completion ossible disadvantages of completion lymphadenectom lymphadenectomy y Melanoma: assessment and management (NG14) © NICE 2015. All rights reserved. Page 11 of 61 Removing the rest of the lymph nodes before cancer develops in them reduces the chance of the cancer returning in the same part of the body. Lymphoedema (long-term swelling) may develop, and is most likely if the operation is in the groin and least likely in the head and neck. The operation is less complicated and safer than waiting until cancer develops in the remaining lymph nodes and then removing them. In 4 out of 5 people, cancer will not develop in the remaining lymph nodes, so there is a chance that the operation will have been done unnecessarily. People who have had the operation may be able to take part in clinical trials of new treatments to prevent future melanoma. These trials often cannot accept people who have not had this operation. There is no evidence that people who have this operation live longer than people who do not have it. Having any operation can cause complications. Adjuvant r Adjuvant radiother adiotherap apy y Do not offer adjuvant radiotherapy to people with stage IIIB or IIIC melanoma unless a reduction in the risk of local recurrence is estimated to outweigh the risk of significant adverse effects. Follow-up after treatment for melanoma F Follow-up for all people who ha ollow-up for all people who hav ve had melanoma e had melanoma Consider personalised follow-up for people who are at increased risk of further primary melanomas (for example people with atypical mole syndrome, previous melanoma, or a history of melanoma in first-degree relatives or other relevant familial cancer syndromes). F Follow-up after stage ollow-up after stage IIC melanoma with no sentinel lymph node biopsy or stage IIC melanoma with no sentinel lymph node biopsy or stage III III melanoma melanoma Consider surveillance imaging as part of follow-up for people who have had stage IIC melanoma with no sentinel lymph node biopsy or stage III melanoma and who would become eligible for systemic therapy as a result of early detection of metastatic disease if: there is a clinical trial of the value of regular imaging or or Melanoma: assessment and management (NG14) © NICE 2015. All rights reserved. Page 12 of 61 the specialist skin cancer multidisciplinary team agrees to a local policy and specific funding for imaging 6-monthly for 3 years is identified. Take into account the possible advantages and disadvantages of surveillance imaging and discuss these with the person, using the table below. P Possible advantages of surv ossible advantages of surveillance imaging (ha eillance imaging (having regular ving regular scans) scans) P Possible disadvantages of ossible disadvantages of surv surveillance imaging (ha eillance imaging (having ving regular scans) regular scans) If the melanoma comes back (recurrent melanoma), it is more likely to be detected sooner. It is possible that this could lead to a better outcome by allowing treatment with drugs (such as immunotherapy drugs) to start earlier. Although early drug treatment of recurrent melanoma might improve survival, there is currently no evidence showing this. Some people find it reassuring to have regular scans. Some people find that having regular scans increases their anxiety. Scans expose the body to radiation, which can increase the risk of cancer in the future. Scans of the brain and neck increase the risk of developing cataracts. Scans of the chest cause a very small increase in the risk of thyroid cancer. Scans may show abnormalities that are later found to be harmless, causing unnecessary investigations and anxiety. Melanoma: assessment and management (NG14) © NICE 2015. All rights reserved. Page 13 of 61 Stages of melanoma Stages of melanoma The stages of melanoma referred to in this guideline are from the American Joint Committee on Cancer's Melanoma of the skin staging (7th edition). Staging of primary melanoma can be carried out in 2 steps. The initial staging is based on the histopathological features reported by the pathologist looking at the microscopic sections of the tumour. The melanoma is staged as 0–IIC, based on factors such as the thickness of the tumour and the presence or absence of ulceration. In many hospitals in the UK, this first step is followed by the option of a second, which is a sampling of the lymph nodes most likely to contain secondary melanoma cells (sentinel lymph node biopsy). If a sentinel lymph node biopsy is performed and microscopic disease is detected, the melanoma becomes stage III. If no microscopic disease is detected then the initial stage is used. Melanoma: assessment and management (NG14) © NICE 2015. All rights reserved. Page 14 of 61 1 1 Recommendations Recommendations The following guidance is based on the best available evidence. The full guideline gives details of the methods and the evidence used to develop the guidance. The wording used in the recommendations in this guideline (for example, words such as 'offer' and 'consider') denotes the certainty with which the recommendation is made (the strength of the recommendation). See about this guideline for details. These recommendations cover suspected and diagnosed melanoma. All recommendations relate to children, young people and adults unless specified otherwise. 1.1 Communication and support 1.1.1 To help people make decisions about their care, follow the recommendations on communication, information provision and support in NICE's guideline on improving outcomes for people with skin tumours including melanoma, in particular the following 5 recommendations: 'Improved, preferably nationally standardised, written information should be made available to all patients. Information should be appropriate to the patients' needs at that point in their diagnosis and treatment, and should be repeated over time. The information given must be specific to the histopathological type of lesion, type of treatment, local services and any choice within them, and should cover both physical and psychosocial issues.' 'Those who are directly involved in treating patients should receive specific training in communication and breaking bad news.' 'Patients should be invited to bring a companion with them to consultations.' 'Each LSMDT [local hospital skin cancer multidisciplinary team] and SSMDT [specialist skin cancer multidisciplinary team] should have at least one skin cancer clinical nurse specialist (CNS) who will play a leading role in supporting patients and carers. There should be equity of access to information and support regardless of where the care is delivered.' 'All LSMDTs and SSMDTs should have access to psychological support services for skin cancer patients.' Melanoma: assessment and management (NG14) © NICE 2015. All rights reserved. Page 15 of 61 1.1.2 Follow the recommendations on follow-up in NICE's guideline on improving outcomes for people with skin tumours including melanoma, in particular the following 2 recommendations: 'All patients should be given written instruction on how to obtain quick and easy access back to see a member of the LSMDT/SSMDT when necessary.' 'All patients should be given both oral and written information about the different types of skin cancer and instruction about self-surveillance.' 1.1.3 Give people with melanoma and their families or carers advice about protecting against skin damage caused by exposure to the sun while avoiding vitamin D depletion. 1.1.4 Carry out a holistic needs assessment to identify the psychosocial needs of people with melanoma and their needs for support and education about the likelihood of recurrence, metastatic spread, new primary lesions and the risk of melanoma in their family members. 1.1.5 Follow the recommendations on communication and patient-centred care in NICE's guideline on patient experience in adult NHS services. 1.2 Assessing melanoma Dermoscop Dermoscopy and other visualisation techniques y and other visualisation techniques See implementation: getting started for information about putting recommendation 1.2.1 into practice. 1.2.1 Assess all pigmented skin lesions that are either referred for assessment or identified during follow-up in secondary or tertiary care, using dermoscopy carried out by healthcare professionals trained in this technique. 1.2.2 Do not routinely use confocal microscopy or computer-assisted diagnostic tools to assess pigmented skin lesions. Melanoma: assessment and management (NG14) © NICE 2015. All rights reserved. Page 16 of 61 Photogr Photograph aphy y 1.2.3 For a clinically atypical melanocytic lesion that does not need excision at first presentation in secondary or tertiary care: use baseline photography (preferably dermoscopic) and and review the clinical appearance of the lesion, and compare it with the baseline photographic images, 3 months after first presentation to identify early signs of melanoma. Assessing and managing atypical spitzoid lesions Assessing and managing atypical spitzoid lesions 1.2.4 Discuss all suspected atypical spitzoid lesions at the specialist skin cancer multidisciplinary team meeting. 1.2.5 Make the diagnosis of a spitzoid lesion of uncertain malignant potential on the basis of the histology, clinical features and behaviour. 1.2.6 Manage a spitzoid lesion of uncertain malignant potential as melanoma. T Taking tumour samples for genetic testing aking tumour samples for genetic testing 1.2.7 If targeted systemic therapy is a treatment option, offer genetic testing using: a secondary melanoma tissue sample if there is adequate cellularity or or a primary melanoma tissue sample if a secondary sample is not available or is of inadequate cellularity. Genetic testing in early-stage melanoma Genetic testing in early-stage melanoma 1.2.8 Do not offer genetic testing of stage IA–IIB primary melanoma at presentation except as part of a clinical trial. 1.2.9 Consider genetic testing of stage IIC primary melanoma or the nodal deposits or in-transit metastases for people with stage III melanoma. Melanoma: assessment and management (NG14) © NICE 2015. All rights reserved. Page 17 of 61 1.2.10 If insufficient tissue is available from nodal deposits or in-transit metastases, consider genetic testing of the primary tumour for people with stage III melanoma. 1.3 Managing suboptimal vitamin D levels See implementation: getting started for information about putting recommendations 1.3.1 and 1.3.2 into practice. 1.3.1 Measure vitamin D levels at diagnosis in secondary care in all people with melanoma. 1.3.2 Give people whose vitamin D levels are thought to be suboptimal advice on vitamin D supplementation and monitoring in line with local policies and NICE's guideline on vitamin D. 1.4 Managing concurrent drug treatment 1.4.1 Do not withhold or change drug treatment for other conditions, except immunosuppressants, on the basis of a diagnosis of melanoma. 1.4.2 Consider minimising or avoiding immunosuppressants for people with melanoma. 1.5 Staging investigations Sentinel lymph node biopsy Sentinel lymph node biopsy See implementation: getting started for information about putting recommendation 1.5.2 into practice. 1.5.1 Do not offer imaging or sentinel lymph node biopsy to people who have stage IA melanoma or those who have stage IB melanoma with a Breslow thickness of 1 mm or less. 1.5.2 Consider sentinel lymph node biopsy as a staging rather than a therapeutic procedure for people with stage IB–IIC melanoma with a Breslow thickness of Melanoma: assessment and management (NG14) © NICE 2015. All rights reserved. Page 18 of 61 more than 1 mm, and give them detailed verbal and written information about the possible advantages and disadvantages, using the table below. P Possible advantages of sentinel lymph ossible advantages of sentinel lymph node biopsy node biopsy P Possible disadvantages of sentinel lymph ossible disadvantages of sentinel lymph node biopsy node biopsy The operation helps to find out whether the cancer has spread to the lymph nodes. It is better than ultrasound scans at finding very small cancers in the lymph nodes. The purpose of the operation is not to cure the cancer. There is no good evidence that people who have the operation live longer than people who do not have it. The operation can help predict what might happen in the future. For example, in people with a primary melanoma that is between 1 and 4 mm thick: around 1 out of 10 die within 10 years if the sentinel lymph node biopsy is negative around 3 out of 10 die within 10 years if the sentinel lymph node biopsy is positive. The result needs to be interpreted with caution. Of every 100 people who have a negative sentinel lymph node biopsy, around 3 will subsequently develop a recurrence in the same group of lymph nodes. People who have had the operation may be able to take part in clinical trials of new treatments for melanoma. These trials often cannot accept people who haven't had this operation. A general anaesthetic is needed for the operation. The operation results in complications in between 4 and 10 out of every 100 people who have it. Imaging Imaging 1.5.3 Offer CT staging to people with stage IIC melanoma who have not had sentinel lymph node biopsy, and to people with stage III or suspected stage IV melanoma. Melanoma: assessment and management (NG14) © NICE 2015. All rights reserved. Page 19 of 61 1.5.4 Include the brain as part of imaging for people with suspected stage IV melanoma. 1.5.5 Consider whole-body MRI for children and young people (from birth to 24 years) with stage III or suspected stage IV melanoma. 1.6 Managing stages 0–II melanoma Ex Excision cision 1.6.1 Consider a clinical margin of at least 0.5 cm when excising stage 0 melanoma. 1.6.2 If excision for stage 0 melanoma does not achieve an adequate histological margin, discuss further management with the multidisciplinary team. 1.6.3 Offer excision with a clinical margin of at least 1 cm to people with stage I melanoma. 1.6.4 Offer excision with a clinical margin of at least 2 cm to people with stage II melanoma. Imiquimod for stage Imiquimod for stage 0 melanoma 0 melanoma 1.6.5 Consider topical imiquimod to treat stage 0 melanoma in adults if surgery to remove the entire lesion with a 0.5 cm clinical margin would lead to unacceptable disfigurement or morbidity. 1.6.6 Consider a repeat skin biopsy for histopathological assessment after treatment with topical imiquimod for stage 0 melanoma, to check whether it has been effective. 1.7 Managing stage III melanoma Completion lymphadenectom Completion lymphadenectomy y See implementation: getting started for information about putting recommendation 1.7.1 into practice. Melanoma: assessment and management (NG14) © NICE 2015. All rights reserved. Page 20 of 61 1.7.1 Consider completion lymphadenectomy for people whose sentinel lymph node biopsy shows micro-metastases and give them detailed verbal and written information about the possible advantages and disadvantages, using the table below. P Possible advantages of completion ossible advantages of completion lymphadenectom lymphadenectomy y P Possible disadvantages of ossible disadvantages of completion lymphadenectom completion lymphadenectomy y Removing the rest of the lymph nodes before cancer develops in them reduces the chance of the cancer returning in the same part of the body. Lymphoedema (long-term swelling) may develop, and is most likely if the operation is in the groin and least likely in the head and neck. The operation is less complicated and safer than waiting until cancer develops in the remaining lymph nodes and then removing them. In 4 out of 5 people, cancer will not develop in the remaining lymph nodes, so there is a chance that the operation will have been done unnecessarily. People who have had the operation may be able to take part in clinical trials of new treatments to prevent future melanoma. These trials often cannot accept people who have not had this operation. There is no evidence that people who have this operation live longer than people who do not have it. Having any operation can cause complications. L Lymph node dissection ymph node dissection 1.7.2 Offer therapeutic lymph node dissection to people with palpable stage IIIB–IIIC melanoma or nodal disease detected by imaging. Adjuvant r Adjuvant radiother adiotherap apy y 1.7.3 Do not offer adjuvant radiotherapy to people with stage IIIA melanoma. 1.7.4 Do not offer adjuvant radiotherapy to people with stage IIIB or IIIC melanoma unless a reduction in the risk of local recurrence is estimated to outweigh the risk of significant adverse effects. Melanoma: assessment and management (NG14) © NICE 2015. All rights reserved. Page 21 of 61 P Palliativ alliative treatment for in-tr e treatment for in-transit metastases ansit metastases 1.7.5 Refer the care of all people with newly diagnosed or progressive in-transit metastases to the specialist skin cancer multidisciplinary team (SSMDT). 1.7.6 If palliative treatment for in-transit metastases is needed, offer palliative surgery as a first option if surgery is feasible. 1.7.7 If palliative surgery is not feasible for people with in-transit metastases, consider the following options: systemic therapy (for more information see recommendations 1.8.5–1.8.9) isolated limb infusion isolated limb perfusion radiotherapy electrochemotherapy in line with NICE's interventional procedure guidance on electrochemotherapy for metastases in the skin from tumours of non-skin origin and melanoma CO2 laser a topical agent such as imiquimod . P Palliativ alliative treatment for superficial skin metastases e treatment for superficial skin metastases 1.7.8 Consider topical imiquimod to palliate superficial melanoma skin metastases. 1.8 Managing stage IV melanoma Management of oligometastatic stage Management of oligometastatic stage IV melanoma IV melanoma 1.8.1 Refer the care of people who appear to have oligometastatic melanoma to the specialist skin cancer multidisciplinary team (SSMDT) for recommendations about staging and management. 1.8.2 Consider surgery or other ablative treatments (including stereotactic radiotherapy or radioembolisation) to prevent and control symptoms of Melanoma: assessment and management (NG14) © NICE 2015. All rights reserved. Page 22 of 61 oligometastatic stage IVmelanomain consultation with site-specific MDTs (such as an MDT for the brain or for bones). Br Brain metastases ain metastases 1.8.3 Discuss the care of people with melanoma and brain metastases with the SSMDT. 1.8.4 Refer people with melanoma and brain metastases that might be suitable for surgery or stereotactic radiotherapy to the brain and other central nervous system tumours MDT for a recommendation about treatment. Systemic anticancer treatment Systemic anticancer treatment T Targeted tr argeted treatments eatments 1.8.5 For adults, see NICE's technology appraisal guidance on dabrafenibfor treating unresectable or metastatic BRAF V600 mutation-positive melanoma . 1.8.6 For adults, 'Vemurafenib is recommended as an option for treating BRAF V600 mutation-positive unresectable or metastatic melanoma only if the manufacturer provides vemurafenib with the discount agreed in the patient access scheme' . [This recommendation is from NICE's technology appraisal guidance on vemurafenib for treating locally advanced or metastatic BRAF V600 mutation-positive malignant melanoma.] Immunother Immunotherapy apy 1.8.7 For adults, see NICE's technology appraisal guidance on ipilimumab for previously treated advanced (unresectable or metastatic) melanoma and ipilimumab for previously untreated advanced (unresectable or metastatic) melanoma . Cytoto Cytotoxic chemother xic chemotherapy apy 1.8.8 Consider dacarbazine for people with stage IV metastatic melanoma if immunotherapy or targeted therapy are not suitable . Melanoma: assessment and management (NG14) © NICE 2015. All rights reserved. Page 23 of 61 1.8.9 Do not routinely offer further cytotoxic chemotherapy for stage IV metastatic melanoma to people previously treated with dacarbazine except in the context of a clinical trial. 1.9 Follow-up after treatment for melanoma F Follow-up for all people who ha ollow-up for all people who hav ve had melanoma e had melanoma 1.9.1 Perform a full examination of the skin and regional lymph nodes at all follow-up appointments. 1.9.2 Consider personalised follow-up for people who are at increased risk of further primary melanomas (for example people with atypical mole syndrome, previous melanoma, or a history of melanoma in first-degree relatives or other relevant familial cancer syndromes). 1.9.3 Consider including the brain for people having imaging as part of follow-up after treatment for melanoma. 1.9.4 Consider imaging the brain if metastatic disease outside the central nervous system is suspected. 1.9.5 Consider CT rather than MRI of the brain for adults having imaging as part of follow-up or if metastatic disease is suspected. 1.9.6 Consider MRI rather than CT of the brain for children and young people (from birth to 24 years) having imaging as part of follow-up or if metastatic disease is suspected. 1.9.7 Provide psychosocial support for the person with melanoma and their family or carers at all follow-up appointments. 1.9.8 All local follow-up policies should include reinforcing advice about self-examination (in line with recommendation 1.1.2), and health promotion for people with melanoma and their families, including sun awareness, avoiding vitamin D depletion (in line with recommendation 1.1.3), and NICE guidance on smoking cessation. Melanoma: assessment and management (NG14) © NICE 2015. All rights reserved. Page 24 of 61 1.9.9 Continue to manage drug treatment for other conditions in line with recommendations 1.4.1 and 1.4.2 after treatment for melanoma. F Follow-up after stage ollow-up after stage 0 melanoma 0 melanoma 1.9.10 Discharge people who have had stage 0 melanoma after completion of treatment and provide advice in line with recommendation 1.9.8. F Follow-up after stage ollow-up after stage IA melanoma IA melanoma 1.9.11 For people who have had stage IA melanoma, consider follow-up 2–4 times during the first year after completion of treatment and discharging them at the end of that year. 1.9.12 Do not routinely offer screening investigations (including imaging and blood tests) as part of follow-up to people who have had stage IA melanoma. F Follow-up after stages ollow-up after stages IB–IIB melanoma or stage IB–IIB melanoma or stage IIC melanoma (fully staged using IIC melanoma (fully staged using sentinel lymph node biopsy) sentinel lymph node biopsy) 1.9.13 For people who have had stages IB–IIB melanoma or stage IIC melanoma with a negative sentinel lymph node biopsy, consider follow-up every 3 months for the first 3 years after completion of treatment, then every 6 months for the next 2 years, and discharging them at the end of 5 years. 1.9.14 Do not routinely offer screening investigations (including imaging and blood tests) as part of follow-up to people who have had stages IB–IIB melanoma or stage IIC melanoma with a negative sentinel lymph node biopsy. F Follow-up after stage ollow-up after stage IIC melanoma with no sentinel lymph node biopsy or stage IIC melanoma with no sentinel lymph node biopsy or stage III III melanoma melanoma 1.9.15 For people who have had stage IIC melanoma with no sentinel lymph node biopsy, or stage III melanoma, consider follow-up every 3 months for the first 3 years after completion of treatment, then every 6 months for the next 2 years, and discharging them at the end of 5 years. 1.9.16 Consider surveillance imaging as part of follow-up for people who have had stage IIC melanoma with no sentinel lymph node biopsy or stage III melanoma Melanoma: assessment and management (NG14) © NICE 2015. All rights reserved. Page 25 of 61 and who would become eligible for systemic therapy as a result of early detection of metastatic disease if: there is a clinical trial of the value of regular imaging or or the specialist skin cancer multidisciplinary team agrees to a local policy and specific funding for imaging 6-monthly for 3 years is identified. Take into account the possible advantages and disadvantages of surveillance imaging and discuss these with the person, using the table below. P Possible advantages of surv ossible advantages of surveillance imaging (ha eillance imaging (having regular ving regular scans) scans) P Possible disadvantages of ossible disadvantages of surv surveillance imaging (ha eillance imaging (having ving regular scans) regular scans) If the melanoma comes back (recurrent melanoma), it is more likely to be detected sooner. It is possible that this could lead to a better outcome by allowing treatment with drugs (such as immunotherapy drugs) to start earlier. Although early drug treatment of recurrent melanoma might improve survival, there is currently no evidence showing this. Some people find it reassuring to have regular scans. Some people find that having regular scans increases their anxiety. Scans expose the body to radiation, which can increase the risk of cancer in the future. Scans of the brain and neck increase the risk of developing cataracts. Scans of the chest cause a very small increase in the risk of thyroid cancer. Scans may show abnormalities that are later found to be harmless, causing unnecessary investigations and anxiety. Melanoma: assessment and management (NG14) © NICE 2015. All rights reserved. Page 26 of 61 F Follow-up after stage ollow-up after stage IV melanoma IV melanoma 1.9.17 Offer personalised follow-up to people who have had stage IV melanoma. At the time of publication (July 2015) topical imiquimod did not have a UK marketing authorisation for this indication. The prescriber should follow relevant professional guidance, taking full responsibility for the decision. Informed consent should be obtained and documented. See the General Medical Council's Prescribing guidance: prescribing unlicensed medicines for further information. At the time of publication (July 2015) topical imiquimod did not have a UK marketing authorisation for this indication or for use in children and young people. The prescriber should follow relevant professional guidance, taking full responsibility for the decision. Informed consent should be obtained and documented. See the General Medical Council's Prescribing guidance: prescribing unlicensed medicines for further information. Dabrafenib has a marketing authorisation in the UK in monotherapy for the treatment of adult patients with unresectable or metastatic melanoma with a BRAF V600 mutation. Vemurafenib has a UK marketing authorisation for 'the treatment of adult patients with BRAF V600 mutation-positive unresectable or metastatic melanoma'. Ipilimumab has a UK marketing authorisation for 'the treatment of advanced (unresectable or metastatic) melanoma in adults'. Although this use is common in UK clinical practice, at the time of publication (July 2015), dacarbazine did not have a UK marketing authorisation for this indication or for use in children and young people. The prescriber should follow relevant professional guidance, taking full responsibility for the decision. Informed consent should be obtained and documented. See the General Medical Council's Prescribing guidance: prescribing unlicensed medicines for further information. Melanoma: assessment and management (NG14) © NICE 2015. All rights reserved. Page 27 of 61 2 2 Research recommendations Research recommendations The Guideline Development Group has made the following recommendations for research, based on its review of evidence, to improve NICE guidance and patient care in the future. 2.1 Techniques for confirming a diagnosis in people with suspected atypical spitzoid melanocytic lesions In people with reported atypical spitzoid lesions, how effective are fluorescence in-situ hybridization (FISH), comparative genomic hybridization (CGH) and tests to detect driver mutations compared with histopathological examination alone in predicting disease-specific survival? This should be investigated in a prospective diagnostic study. Secondary outcomes should include sensitivity, specificity, accuracy, positive predictive value, disease-specific survival and progression-free survival. Wh Why this is important y this is important Atypical spitzoid lesions continue to be diagnostically challenging. There are no reliably reproducible histological, immunohistochemistry or molecular features that allow exact typing and prognostic assessment of these lesions. The current 'gold standard' is histological examination with expert review, but it is not always possible to distinguish spitzoid melanoma from benign spitzoid melanocytic lesions. Current molecular technologies such as FISH and CGH provide some help, but the results are difficult to interpret and may not be conclusive. Understanding and mapping changes in molecular pathways could predict outcome and inform individual treatment planning. 2.2 Surgical excision for people with lentigo maligna For people with lentigo maligna (stage 0 in sun-damaged skin, usually on the face) how effective is Mohs micrographic surgery, compared with excision with a 0.5 cm clinical margin, in preventing biopsy-proven local recurrence at 5 years? This should be investigated in a randomised controlled trial. Secondary outcomes should include cosmetic and functional outcomes. Melanoma: assessment and management (NG14) © NICE 2015. All rights reserved. Page 28 of 61 Wh Why this is important y this is important Mohs micrographic surgery is a microscopically controlled surgical technique designed to allow complete excision of the tumour with minimal tissue loss. The technique can be useful for people with lentigo maligna because their lesions can be very large and located in a cosmetically sensitive site where surgery may cause significant scarring. However, the histological detection of small numbers of melanocytes at the edge of a sample is difficult, and can lead to false negative results. In addition, lentigo maligna may occur in an area of field change with a risk of skip lesions at the edge. Therefore, although Mohs micrographic surgery may ensure complete excision of lentigo maligna, it can be accompanied by the recurrence of a similar lesion in adjacent skin. 2.3 Follow-up surveillance imaging In people treated for high-risk stage II and III melanoma, does regular surveillance imaging improve melanoma-specific survival compared with routine clinical follow-up alone? This should be investigated in a randomised controlled trial. Secondary outcomes should include time to recurrence, site of recurrence, proportion of people receiving active therapy at recurrence, cost effectiveness and quality of life. Wh Why this is important y this is important Until recently there have been no effective therapies for metastatic melanoma and no strong rationale for early detection of relapse through surveillance imaging. However, new, effective targeted treatments and immunotherapy agents are now available and further treatments are likely to become available in the near future. In particular, immunotherapy can offer long-term disease-free survival but takes a number of months to take effect. In this situation, early detection of relapse may identify people likely to be fit enough to receive the treatment for long enough to benefit. Although early detection of relapse through surveillance imaging might appear likely to improve outcomes, there is no evidence to confirm this. In addition, routine imaging has resource implications and involves more hospital visits and increased radiation exposure for the person. 2.4 Vitamin D supplementation In people with stage I–III melanoma does vitamin D supplementation improve overall survival? Melanoma: assessment and management (NG14) © NICE 2015. All rights reserved. Page 29 of 61 This should be investigated in a placebo-controlled randomised trial. Secondary outcomes should include disease-specific survival and toxicity, including the development of renal stones and hypercalcaemia. Wh Why this is important y this is important It has been reported that suboptimal levels of vitamin D at diagnosis are common in people with melanoma from the north of England and that higher levels are associated with lower melanoma-related mortality. However, vitamin D levels are higher in leaner, fitter people and the nature of the relationship between vitamin D levels and melanoma survival is unclear. There are 2 adjuvant trials of vitamin D supplementation listed as active currently, 1 in Italy and 1 in Australia. However, there are many uncertainties about the design of vitamin D trials, which might become clearer in the next few years. These include the dose of vitamin D, use of concurrent aspirin therapy and the baseline level at which vitamin D supplementation would be started. 2.5 The effect of drug therapy for concurrent conditions on melanoma survival In people diagnosed with melanoma what is the effect of drug therapy to treat concurrent conditions on disease-specific survival? This should be investigated in a national prospective cohort study. Secondary outcomes should include overall survival and quality of life. Wh Why this is important y this is important Drugs such as immunosuppressants and those used to treat conditions such as diabetes have effects that may affect survival in people with melanoma. For example metformin, the most frequently prescribed drug for type 2 diabetes, is thought to reduce overall cancer rates in people with diabetes but to increase mortality from melanoma in the approximately 40% of these people who have a somatic BRAF mutation. There is a need to balance the risk of melanoma deaths with the benefits from the most effective treatment of the concurrent conditions. But there is currently no evidence to inform this decision. Melanoma: assessment and management (NG14) © NICE 2015. All rights reserved. Page 30 of 61 3 3 Other information Other information 3.1 Scope and how this guideline was developed NICE guidelines are developed in accordance with a scope that defines what the guideline will and will not cover. How this guideline was de How this guideline was dev veloped eloped NICE commissioned the National Collaborating Centre for Cancer to develop this guideline. The Centre established a Guideline Development Group (see section 4), which reviewed the evidence and developed the recommendations. The methods and processes for developing NICE clinical guidelines are described in the guidelines manual. 3.2 Related NICE guidance Further information is available on the NICE website. Published Published Gener General al Vitamin D (2014) NICE guideline PH56 Neutropenic sepsis (2012) NICE guideline CG151 Opioids in palliative care (2012) NICE guideline CG140 Patient experience in adult NHS services (2012) NICE guideline CG138 MIST therapy system for the promotion of wound healing in chronic and acute wounds (2011) NICE medical technology guidance 5 Medicines adherence (2009) NICE guideline CG76 Surgical site infection (2008) NICE guideline CG74 Smoking cessation (2008) NICE guideline PH10 Melanoma: assessment and management (NG14) © NICE 2015. All rights reserved. Page 31 of 61 Condition-specific Condition-specific Suspected cancer: recognition and referral (2015) NICE guideline NG12 Dabrafenib for treating unresectable or metastatic BRAF V600 mutation-positive melanoma (2014) NICE technology appraisal guidance 321 Ipilimumab for previously untreated advanced (unresectable or metastatic) melanoma (2014) NICE technology appraisal guidance 319 Electrochemotherapy for metastases in the skin from tumours of non-skin origin and melanoma (2013) NICE interventional procedure guidance 446 Vemurafenib for treating locally advanced or metastatic BRAF V600 mutation-positive malignant melanoma (2012) NICE technology appraisal guidance 269 Ipilimumab for previously treated advanced (unresectable or metastatic) melanoma (2012) NICE technology appraisal guidance 268 Endoscopic radical inguinal lymphadenectomy (2011) NICE interventional procedure guidance 398 Skin cancer prevention (2011) NICE guideline PH32 Improving outcomes for people with skin tumours including melanoma (2010) NICE guideline CSGSTIM Improving outcomes in children and young people with cancer (2005) NICE guideline CSGCYP Improving supportive and palliative care for adults with cancer (2004) NICE guideline CSGSP Under de Under dev velopment elopment NICE is developing the following guidance: Sunlight exposure: communicating the benefits and risks to the general public. NICE guideline. Publication expected September 2015. Skin cancer: the VivaScope 1500 and 3000 systems for detecting and monitoring skin lesions. NICE diagnostics guidance. Publication expected November 2015. Pembrolizumab for treating unresectable, metastatic melanoma after progression with ipilimumab. NICE technology appraisal guidance. Publication expected December 2015. Melanoma: assessment and management (NG14) © NICE 2015. All rights reserved. Page 32 of 61 Dabrafenib and trametinib for treating advanced unresectable or metastatic BRAFV600 mutation-positive melanoma. NICE technology appraisal guidance. Publication expected August 2016. Ipilimumab for the adjuvant treatment of completely resected high risk stage III or IV melanoma. NICE technology appraisal guidance. Publication date to be confirmed. Melanoma: assessment and management (NG14) © NICE 2015. All rights reserved. Page 33 of 61 4 4 The Guideline De The Guideline Dev velopment Group, National Collabor elopment Group, National Collaborating Centre ating Centre and NICE project team, and declar and NICE project team, and declarations of interests ations of interests 4.1 Guideline Development Group Gill Godsell Gill Godsell Nurse Consultant (Skin Cancer), Nottingham NHS Treatment Centre Laszlo LaszloIgali Igali Consultant Histopathologist, Norfolk and Norwich University Hospital NHS Foundation Trust Richard Jackson Richard Jackson Patient and carer member Charles K Charles Kelly elly Consultant Clinical Oncologist, Northern Centre for Cancer Care, Freeman Hospital, Newcastle Stephen K Stephen Keohane eohane Consultant Dermatologist, Portsmouth Hospitals NHS Trust, Portsmouth Dermatology Centre F Fergus Macbeth ( ergus Macbeth (Chair) Chair) Clinical adviser, Wales Cancer Trials Unit, Cardiff University Julia Newton-Bishop ( Julia Newton-Bishop (Clinical L Clinical Lead) ead) Professor of Dermatology, University of Leeds Christine P Christine Parkinson arkinson Consultant in Medical Oncology, Addenbrooke's Hospital, Cambridge Barry P Barry Powell owell Consultant Plastic Surgeon, St George's Hospital, London Saskia Reek Saskia Reeken en Clinical Nurse Specialist, Skin Cancer and Dermatology, Kingston Hospital NHS Foundation Trust, Surrey Melanoma: assessment and management (NG14) © NICE 2015. All rights reserved. Page 34 of 61 Rachael Robinson Rachael Robinson GP, Stockwell Road Surgery, Knarlesborough, GPwSI Dermatology, Harrogate District Foundation Trust, GPwSI Durnford Dermatology, Middleton, Manchester Simon Rodwell Simon Rodwell (from April 2013 until October 2013) Patient and carer member John Rouse John Rouse (from November 2013) Patient and carer member Julia Schofield Julia Schofield Principal Lecturer, University of Herefordshire, Consultant Dermatologist, United Lincolnshire Hospitals NHS Trust Jonathan Smith Jonathan Smith Consultant Radiologist, Leeds Teaching Hospital Trust Sar Sara Stoneham a Stoneham Paediatric and Adolescent Oncology Consultant, University College Hospital, London Martin T Martin Telfer elfer Consultant Maxillofacial Surgeon, York Teaching Hospital NHS Foundation Trust 4.2 National Collaborating Centre for Cancer Stephanie Arnold Stephanie Arnold Information Specialist Nathan Bromham Nathan Bromham Senior Researcher Laur Laura Bunting a Bunting Researcher Andrew Champion Andrew Champion Centre Manager Melanoma: assessment and management (NG14) © NICE 2015. All rights reserved. Page 35 of 61 John Gr John Graham aham Director Lianne Gwillim Lianne Gwillim Project Manager James Ha James Hawkins wkins Health Economist Cor Coral McCarth al McCarthy y Project Manager (from May 2014 until February 2015) Anghar Angharad Morgan ad Morgan Researcher Delyth Morris Delyth Morris Information Specialist (until April 2014) Susan O Susan O' 'Connell Connell Researcher Matthew Prettyjohns Matthew Prettyjohns Senior Health Economist 4.3 NICE project team Christine Carson Christine Carson Guideline Lead Mark Bak Mark Baker er Clinical Adviser Katie P Katie Perryman F erryman Ford ord Guideline Commissioning Manager Jennifer W Jennifer Watson-Henry atson-Henry Guideline Coordinator (until November 2014) Melanoma: assessment and management (NG14) © NICE 2015. All rights reserved. Page 36 of 61 Thomas F Thomas Feist eist Guideline Coordinator (from November 2014) Nichole T Nichole Task aske e Technical Lead Bhash Naidoo Bhash Naidoo Health Economist Judy McBride Judy McBride Editor 4.4 Declarations of interests The following members of the Guideline Development Group made declarations of interests. All other members of the Group stated that they had no interests to declare. The conflicts of interest policy (2007) was followed until September 2014, when an updated policy was published. Member Member Interest declared Interest declared T Type of ype of interest interest Decision tak Decision taken en Barry Powell Received a fee from Roche for chairing an advisory board on BRAF inhibitors in malignant melanoma. Donate fee to charity. Personal pecuniary; specific Declare and withdraw from discussions on all topics regarding BRAF inhibitors until July 2013 Barry Powell Novartis have offered a fee to take part in a future advisory board on MEK inhibitors in melanoma. Not yet accepted. Personal pecuniary; specific If accepted, declare and withdraw from discussions on all topics regarding the MEK inhibitors until 12 months after date of advisory board Melanoma: assessment and management (NG14) © NICE 2015. All rights reserved. Page 37 of 61 Barry Powell Enrols patients into the EORTC 18091 trial. No fee received for doing this and no involvement past enrolling of patients. Personal non-pecuniary; specific Declare and participate Barry Powell Principal investigator for the UK for the EORTC MINITUB study. Study not yet started. Funded by individual trusts. Personal non-pecuniary; specific Declare and participate Barry Powell Chair of the Pathway Group for Skin Cancer for the London Cancer Alliance (working group on provision of skin cancer care in London). Personal non-pecuniary Declare and participate Barry Powell Wrote an editorial for Surgery journal giving opinions on the management of malignant melanoma. Personal non-pecuniary Declare and participate Barry Powell Received reimbursement of travelling expenses and subsistence from IGEA for attending a meeting regarding data collection for electrochemotherapy. Personal pecuniary; specific Declare and participate Christine Parkinson Received a fee from Boehringer Ingelheim for attending an advisory board and giving advice on a trial for their ovarian cancer drug BIBF1120. Fee was donated to charity. Personal pecuniary non-specific Declare and participate Christine Parkinson Received reimbursement of registration fee and accommodation from Boehringer Ingelheim for attending the International Gynaecological Cancer Society conference. Personal pecuniary interest, non-specific Declare and participate Christine Parkinson Co-investigator on the COMBI-V study. Funded by GSK. Non-personal pecuniary; specific Declare and participate Christine Parkinson Co-investigator on PACMEL. Sponsored by University of Oxford. Funded by GSK. Non-personal pecuniary; specific Declare and participate Melanoma: assessment and management (NG14) © NICE 2015. All rights reserved. Page 38 of 61 Christine Parkinson Co-investigator on the Phase 1, Open Label, Dose Finding Study to Assess the Safety and Tolerability of IMCgp100, a Monoclonal T Cell Receptor Anti-CD3 scFv Fusion Protein in Patients With Advanced Malignant Melanoma. Sponsored and funded by Immunocore Ltd. Non-personal pecuniary; specific Declare and participate Christine Parkinson Co-investigator on NICAM. Sponsored by Royal Marsden Foundation Trust and Institute of Cancer Research. Funded by CTAAC. Non-personal pecuniary; specific Declare and participate Christine Parkinson Co-investigator on the IMAGE study. Funded by Bristol Myers Squibb. Non-personal pecuniary; specific Declare and participate Christine Parkinson Co-investigator on the SUAVE study. Sponsor is Clatterbridge Centre for Oncology NHS Trust. Funded by Pfizer Limited and CTAAC. Non-personal pecuniary; specific Declare and participate Christine Parkinson Co-investigator on the MelResist study. Funded by Cambridge University Hospitals NHS Foundation Trust. Non-personal pecuniary; specific Declare and participate Christine Parkinson Principle investigator on the PARAGON trial. Sponsored by NHS Greater Glasgow & Clyde. Funded by CRUK. Non-personal pecuniary; non-specific Declare and participate Christine Parkinson Received reimbursement of travel and subsistence expenses from CLOVIS for attending an investigator meeting for the ARIEL2 and ARIEL3 trials for ovarian cancer. Personal pecuniary; non-specific Declare and participate Fergus Macbeth Chief investigator of a CRUK-funded trial supported by Pfizer with free drug and unrestricted educational grant. Non-personal pecuniary; non-specific Declare and participate Melanoma: assessment and management (NG14) © NICE 2015. All rights reserved. Page 39 of 61 Fergus Macbeth Received reimbursement of travel and subsistence expenses for attending the World Lung Cancer conference. Personal pecuniary, non-specific Declare and participate Gill Godsell Received reimbursement of travel and subsistence expenses from Almirall (manufacturers of topical treatments for pre-cancerous lesions) for attending a European Academy of Dermatology and Venerology meeting. Personal pecuniary; specific Declare and participate Gill Godsell Vice Chair of the Karen Clifford Skin Cancer Charity, until November 2013. Gave advice on clinical aspects of skin cancer – not specific treatments. Personal non-pecuniary Declare and participate Jonathan Smith Reviewed a systematic review on PET-CT in stage III melanoma for publication in the Journal of Surgical Oncology. Personal non-pecuniary; specific Declare and participate Jonathan Smith Received reimbursement of subsistence and course fee from Nucletron for attending the annual UK prostate brachytherapy course. Personal pecuniary; non-specific Declare and participate Jonathan Smith Received travel and accommodation from the Royal College of Radiologists to give a lecture on 'how to run a radiology discrepancy' at the Royal College of Radiology autumn scientific meeting. Personal pecuniary; non-specific Declare and participate Jonathan Smith Reports CT studies in the STAR trial, which is an RCT multi-centre trial in drug therapy for metastatic renal cell cancer. Non-specific Declare and participate Julia Newton-Bishop Received an honorarium from Roche for giving advice on cutaneous toxicity from vemurafenib. Personal pecuniary; specific Declare and participate Julia Newton-Bishop Department research fund received payment from Roche for giving advice on cutaneous toxicity from vemurafenib. Non-personal pecuniary Declare and participate Melanoma: assessment and management (NG14) © NICE 2015. All rights reserved. Page 40 of 61 Julia Newton-Bishop Received reimbursement of travelling expenses from Irish Association of Dermatologists for giving a talk on vitamin D and melanoma. Personal pecuniary; specific Declare and participate Julia Newton-Bishop Received an honorarium from Irish Association of Dermatologists for giving a talk on vitamin D and melanoma. Personal pecuniary; specific Declare and participate Julia Newton-Bishop Received reimbursement of travelling expenses from the Melanoma Study Group for giving a talk at the Focus on Melanoma conference on the levels of vitamin D in melanoma patients. Personal pecuniary; specific Declare and participate Julia Newton-Bishop Received reimbursement of travelling expenses from Beatson Institute for attending a seminar and giving a talk on the genetics of susceptibility and survival of melanoma. Personal pecuniary Declare and participate Julia Newton-Bishop Received reimbursement of travelling expenses from London Strategic Health Authority for attending an ECRIC Cancer Registry meeting to discuss NCIN work designed to understand cancer registration. Personal pecuniary; non-specific Declare and participate Julia Newton-Bishop Received reimbursement of travelling expenses from Public Health England for chairing an NCIN Chair's meeting regarding national data collection on skin cancer. Personal pecuniary; non-specific Declare and participate Julia Newton-Bishop Received reimbursement of travelling expenses from Public Health England for chairing the skin SSCRG group covering national data collection on skin cancer. Personal pecuniary; non-specific Declare and participate Melanoma: assessment and management (NG14) © NICE 2015. All rights reserved. Page 41 of 61 Julia Newton-Bishop Received reimbursement of travelling expenses from conference organisers giving a talk on the genetics of melanoma survival at the 8th World Congress of Melanoma. Personal pecuniary; specific Declare and participate Julia Newton-Bishop Received reimbursement of travelling expenses from Public Health England for chairing an NCIN workshop on national data collection on skin cancer. Personal pecuniary; non-specific Declare and participate Julia Newton-Bishop Received reimbursement of travelling expenses from Roche for attending a meeting and giving a talk on the biology of melanoma. Personal pecuniary; specific Declare and participate Julia Newton-Bishop Department received payment from Roche for giving an introductory talk on the biology of melanoma. Non-personal pecuniary; specific Declare and participate Julia Newton-Bishop Received an honorarium from Roche for attending an advisory board meeting on the management of skin toxicity (April 2011). Personal pecuniary; specific Declare and participate Julia Newton-Bishop Received an honorarium from Roche for attending an advisory board meeting on the management of skin toxicity (July 2011). Personal pecuniary; specific Declare and participate Julia Newton-Bishop Department received payment from Roche for attending an advisory board meeting on the management of skin toxicity. Personal pecuniary; specific Declare and participate Julia Newton-Bishop Department received payment from Roche for making a training video on the management of skin toxicity. Non-personal pecuniary; specific Declare and participate Melanoma: assessment and management (NG14) © NICE 2015. All rights reserved. Page 42 of 61 Julia Newton-Bishop Department received payment from Roche for giving a talk on 'why do people get melanoma and what determines whether or not they survive' at the annual British Association of Dermatologists conference. Non-personal pecuniary; specific Declare and participate Julia Newton-Bishop Co-author on paper published in 2013 regarding the toxicity of vemurafenib. Personal non-pecuniary Declare and participate Julia Schofield Received a fee from Basilea for giving advice on their product toctino (treatment for hand eczema) into the marketplace. Personal pecuniary; non-specific Declare and participate Julia Schofield Received a fee and reimbursement of travel expenses from Leo Pharmaceuticals for giving a lecture on GPs with a special interest. Personal pecuniary Declare and participate Julia Schofield Received a fee and reimbursement of travel expenses from the British Dermatology Nursing Group for giving a lecture on dermoscopy and teledermatology in relation to skin cancer (including melanoma). Personal pecuniary; specific Declare and participate Julia Schofield Received a fee and reimbursement of travel expenses from the Dowling Club (national dermatology educational society) to present at a meeting for dermatology trainees on delivering dermatology services. Personal pecuniary; non-specific Declare and participate Julia Schofield Received a fee and reimbursement of travel expenses from the Primary Care Dermatology Society for presenting at a meeting on the management of pre-cancerous lesions in primary care. Personal pecuniary; non-specific Declare and participate Melanoma: assessment and management (NG14) © NICE 2015. All rights reserved. Page 43 of 61 Julia Schofield Received a fee and reimbursement of travel expenses from the Irish Primary Care Dermatology Society for presenting at a meeting on recognising skin lesions and paediatric dermatology problems. Personal pecuniary;, non-specific Declare and participate Julia Schofield During 2012, acted as an advisor to Buckinghamshire NHS Trust on redesigning their dermatology services. Personal pecuniary; non-specific Declare and participate Julia Schofield External advisor to All Party Parliamentary Group on Skin. Personal non-pecuniary Declare and participate Julia Schofield Trustee of the Psoriasis Association. Personal non-pecuniary Declare and participate Julia Schofield Received a fee and reimbursement of travel expenses from Leo Pharmaceuticals for giving a lecture on GPs with a special interest. Personal pecuniary interest Declare and participate Julia Schofield Received travel and subsistence from Conference Plus for giving a lecture to GPs in Namibia on non-melanoma skin cancer, eczema and topical dermatology. Personal pecuniary Declare and participate Julia Schofield Received a fee and reimbursement of travel expenses from the Irish Primary care Dermatology Society for giving talks on hyperhidrosis, skin lesion recognition and optimising primary/secondary care. Personal pecuniary interest Declare and participate Laszlo Igali Received a fee from St James' University Hospital, Leeds for speaking at a symposium on alopecia and immunohistochemistry in dermatopathology. Personal pecuniary, non-specific Declare and participate Laszlo Igali Received reimbursement of travelling expenses from the Royal College of Pathologists for attending a council meeting. Personal pecuniary Declare and participate Melanoma: assessment and management (NG14) © NICE 2015. All rights reserved. Page 44 of 61 Laszlo Igali Involved in the EUR-GAST II study (investigating environmental factors, H. pylori infection and genetic susceptibility in gastric cancer risk in the European population). Was the pathologist responsible for coordinating specimen collection and evaluation from the UK. No commercial funding. Non-personal pecuniary; non-specific Declare and participate Laszlo Igali Involved in the EPIC study. Did selective pathology data collection and evaluation. No commercial funding. Non-personal pecuniary Declare and participate Laszlo Igali Supervised an MSc student investigating optimal fixation of metastatic melanoma for tissue banking. Non-personal pecuniary; specific Declare and participate Laszlo Igali Involved in a new prospective study looking at BRAF immunostaining in metastatic melanoma to stratify patients for future treatment. Role is to do the immunohistochemistry and report on the BRAF status. Research funded by employer. Non-personal pecuniary; specific Declare and participate Laszlo Igali Ran a workshop on teledermatopathology as part of the American Society of Dermatopathology annual congress. No fee received for this activity. Personal non-pecuniary Declare and participate Laszlo Igali Holds the post of Editor of the Bulletin of the Royal College of Pathology. Personal non-pecuniary Declare and participate Laszlo Igali Provides ad-hoc advice to EZDerm on developing an integrated dermatology/ electronic record system. No fee received for this activity. Personal non-pecuniary Declare and participate Melanoma: assessment and management (NG14) © NICE 2015. All rights reserved. Page 45 of 61 Laszlo Igali Member of the Interim Body to the Professional Records Standard Body. Provides IT advice on how their electronic records should be set up. Personal non-pecuniary Declare and participate Laszlo Igali Received travelling expenses and accommodation from the British Association of Plastic, Reconstructive and Aesthetic Surgeons (BAPRAS) for giving a lecture at the Skin Cancer course on Basal cell carcinoma and squamous cell carcinoma, conventional and Mohs histology. Personal pecuniary; non-specific Declare and participate Laszlo Igali Treasurer for the professional record standard body (PRSB) for patient data standards. Personal non-pecuniary Declare and participate Martin Telfer Gave a presentation on 'Anatomical restrictions in the surgical excision of Scalp Sq CCa: does this effect local recurrence and regional nodal metastasis?' to the British Association of Oral and Maxillofacial Surgeons. No fee received. Personal non-pecuniary Declare and participate Martin Telfer Presented at the Yorkshire & Humber Regional Clinical Effectiveness Meeting on 'Facial skin cancer surgery: patient satisfaction'. No fee received. Personal non-pecuniary Declare and participate Rachael Robinson Received a fee from the RCGP for taking part in a panel reviewing a musculoskeletal e-learning package. Personal pecuniary; non-specific Declare and participate Rachael Robinson Received a fee from Galderma for chairing an educational meeting of the Leeds Skin Club on the treatment of acne and the red face. Personal pecuniary; non-specific Declare and participate Melanoma: assessment and management (NG14) © NICE 2015. All rights reserved. Page 46 of 61 Rachael Robinson Received reimbursement of travel expenses from the Yorkshire Deanery for attending a meeting to talk about the new curriculum for GP registrars. Personal pecuniary; non-specific Declare and participate Rachael Robinson Practice recruits patients into the 3C – cough complications cohort study, organised by Oxford University. Practice receives an income for this activity which is shared amongst the GPs. Non-personal pecuniary; non-specific Declare and participate Rachael Robinson Practice recruits patients into the early arthritis study, organised by Leeds University. Practice receives an income for this activity which is shared amongst the GPs. Non-personal pecuniary; non-specific Declare and participate Rachael Robinson Practice recruits patients into a study on transdermal patches for the treatment of chronic pain, organised by IMS Health. Practice receives an income for this activity which is shared amongst the GPs. Non-personal pecuniary; non-specific Declare and participate Rachael Robinson Currently involved in reviewing an acne decision aid tool for the BMJ patient decision aid group. No fee is being received. Personal non-pecuniary non-specific Declare and participate Sara Stoneham Received a fee from the Royal Marsden for giving a lecture on renal tumours in paediatric oncology as part of their MSc in Oncology. Personal pecuniary; non-specific Declare and participate Sara Stoneham Principal investigator for the CNS 9204 trial (Neuropsychological, academic and functional outcomes in survivors of infant ependymoma (UKCCSG CNS 9204)). Funded by CRUK. Not involved in designing the trial protocol. Non-personal pecuniary non-specific Declare and participate Melanoma: assessment and management (NG14) © NICE 2015. All rights reserved. Page 47 of 61 Sara Stoneham Was principal investigator for the GC 2005 04 (GC-3) trial (Protocol for the treatment of Extracranial Germ Cell Tumours in children and adolescents). Trial closed in 2009, 1 patient still in follow up. Sponsored by University Hospitals of Leicester NHS Trust. Funded by Children's Cancer and Leukaemia Group (CCLG). Non-personal pecuniary; non-specific Declare and participate Sara Stoneham Co-investigator in the HERBY trial (study of high grade paediatric glioma). Funded by Roche. Non-personal pecuniary; non-specific Declare and participate Saskia Reeken Received an honorarium from Leo Pharmaceuticals for attending an advisory board on dermatology (their psoriasis treatments and new products – none relating to melanoma). Personal pecuniary, non-specific Declare and participate. Saskia Reeken Received an honorarium from the British Dermatology Nursing Group for giving a lecture on topical treatments for dermatology (specifically steroid creams). Personal pecuniary; non-specific Declare and participate Saskia Reeken Received reimbursement of travel expenses (from the organiser) for attending the British Association of Dermatology Nursing annual conference. Personal pecuniary; non-specific Declare and participate Saskia Reeken Received a fee from Janssen for giving a lecture to dermatology nurses on the recognition of skin cancer lesions (including melanoma) in patients with psoriasis and the practical skills for lymph node examination. Personal pecuniary; specific Declare and withdraw from discussions on all topics regarding the recognition of melanoma until May 2013 Melanoma: assessment and management (NG14) © NICE 2015. All rights reserved. Page 48 of 61 Saskia Reeken Received reimbursement of travel and subsistence expenses from the Danish Embassy in Copenhagen for attending a meeting on sun radiation and the effect on the environment. Personal pecuniary; non-specific Declare and participate Saskia Reeken Member of the CRUK Sun Smart Advisory Board – looks at strategies for sun awareness and health promotion. Personal non-pecuniary Declare and participate Saskia Reeken Member of the Melanoma Task Force – interested in improving the care of patients with melanoma. Personal non-pecuniary Declare and participate Saskia Reeken Nurse representative on the British Association of Dermatology skin cancer committee. Personal non-pecuniary Declare and participate Saskia Reeken Nurse representative on Skin Cancer UK – provides advice on skin cancer issues. Personal non-pecuniary Declare and participate Saskia Reeken Received sponsorship from LEO pharmaceuticals and Dermal Laboratories Limited for attending a study day on Maximising Capacity and Productivity in your Dermatology Service. Personal pecuniary; non-specific Declare and participate Saskia Reeken Received a practice development award of £900 from the British Dermatology Nursing Group. The award is to be used for professional development and will be put towards an MSc module of child health. Personal pecuniary; non-specific Declare and participate Stephen Keohane Received a fee from Meda for attending an advisory board on their new treatment for actinic keratosis (Zyclara). Personal pecuniary; non-specific Declare and participate Stephen Keohane Received a fee from Almirall for giving a lecture on new advances in non-melanoma skin cancer. Personal pecuniary; non-specific Declare and participate Melanoma: assessment and management (NG14) © NICE 2015. All rights reserved. Page 49 of 61 Stephen Keohane Received a fee from Leo Pharmaceuticals for attending an advisory board on their new treatment for actinic keratosis (Picato). Personal pecuniary, non-specific Declare and participate Stephen Keohane Received a fee from Roche for attending an advisory board on their treatment for advanced basal cell carcinoma (Erivedge). Personal pecuniary; non-specific Declare and participate Stephen Keohane Received reimbursement of expenses (travel, accommodation, subsistence and conference fee) from Leo Pharmaceuticals for attending the American Academy of Dermatology conference. Personal pecuniary; non-specific Declare and participate Stephen Keohane Local principal investigator for a trial on Ingenol (treatment of facial and scalp actinic keratoses). Trial is funded by Leo Pharmaceuticals. Responsible for administrating the trial locally. Not involved in designing the trial protocol. Non-personal pecuniary; non-specific Declare and participate Stephen Keohane Chaired a meeting on advanced melanoma management (content of the meeting was investigation and management and covered new therapeutic treatments including ipilimumab, vemurafenib, MEK inhibitors and DNA vaccines. The event was sponsored by Bristol Myers Squibb. Did not receive a fee or organise the meeting. Personal non-pecuniary Declare and participate Stephen Keohane Member of the National Cancer Intelligence Network Skin Reference Group – look at changing trends in skin cancer and how these impact on service provision. Personal non-pecuniary Declare and participate Melanoma: assessment and management (NG14) © NICE 2015. All rights reserved. Page 50 of 61 Stephen Keohane Chair of the British Association of Dermatologists Skin Cancer Committee – look at service provision and ensuring the quality of skin cancer care provided by dermatologists is equitable across the UK. Personal non-pecuniary Declare and participate Stephen Keohane Chair of the Skin Cancer Site Specific Group of the Central South Coast Cancer Network – look at local service provision and coordinate regional audits etc. Personal non-pecuniary Declare and participate John Rouse Member of the NCRI/AstraZeneca patient reference panel. Personal non-pecuniary Declare and participate John Rouse Received travelling expenses, subsistence allowance and overnight accommodation for a NCRI/AstraZeneca patient reference meeting at Alderley Park on 26 September 2013. Personal pecuniary; non-specific Declare and participate John Rouse Received travelling expenses, subsistence allowance and overnight accommodation from ESO and M-icab for attending a conference on Patient Participation in Melanoma Clinical Research. Personal pecuniary; specific Declare and participate John Rouse Received a bursary from the NCRN to attend the NCRI conference in Liverpool Personal pecuniary Declare and participate John Rouse Received travelling expenses, overnight accommodation and subsistence allowance paid for by CRUK for attending the NCRN/ECMC Combinations Alliance AZ Workshop. Personal pecuniary Declare and participate John Rouse Received travelling expenses costs from Macmillan Cancer support and accommodation costs from the meeting organisers for attending the Britain Against Cancer conference and Quality in Care awards. Personal pecuniary Declare and participate Melanoma: assessment and management (NG14) © NICE 2015. All rights reserved. Page 51 of 61 Richard Jackson Interviewed for the Daily Mail on the effectiveness of ipilimumab for metastatic melanoma. Personal non-pecuniary Declare and participate Richard Jackson Interviewed for the BBC on medical breakthroughs and the use of Ipilimumab received during treatment. Personal non-pecuniary Declare and participate Richard Jackson Photographed and filmed by Bristol-Myers Squibb Pharmaceuticals Ltd. Discussed his experience of metastatic melanoma. Film is to be used by BMS to make colleagues more aware of patients' unmet medical needs. No payment received. Personal non-pecuniary Declare and participate. Julia Schofield Received travel and accommodation costs from Conference Plus for giving a lecture in an educational programme for GPs. The lecture includes a session on skin lesion diagnosis. Non-personal pecuniary; non-specific Declare and participate John Rouse Received a bursary from the NCRN to attend the NCRI conference in Liverpool. Personal pecuniary; non-specific Declare and participate John Rouse Received travelling expenses, overnight accommodation and subsistence allowance paid for by CRUK for attending the NCRN/ECMC Combinations Alliance AZ Workshop. Personal pecuniary; non-specific Declare and participate John Rouse Received travelling expenses costs from Macmillan Cancer support and accommodation costs from the meeting organisers for attending the Britain Against Cancer conference and Quality in Care awards. Personal pecuniary; non-specific Declare and participate Melanoma: assessment and management (NG14) © NICE 2015. All rights reserved. Page 52 of 61 Implementation: getting started Implementation: getting started While developing this guideline, the Guideline Development Group identified 10 recommendations in 6 areas as key priorities for implementation. This section highlights 3 of those areas that could have a significant impact on practice and be challenging to implement. They have been identified with the help of stakeholders and members of the Guideline Development Group, using the criteria outlined in developing NICE guidelines: the manual, section 9.4. See 'Further resources' below for details of where to get help to address these challenges. Challenge 1 – Using dermoscopy (dematoscopy) to assess pigmented lesions See recommendation 1.2.1. P Potential benefits of implementation otential benefits of implementation Dermoscopy performed in secondary care by suitably trained specialists is both more sensitive and more specific in classifying skin lesions than clinical examination with the naked eye alone. It lessens the chance of missing a diagnosis of melanoma and reduces the number of unnecessary surgical procedures to remove benign lesions. Challenges for implementation Challenges for implementation For healthcare professionals in secondary care skin cancer clinics: Using dermoscopy routinely. Dermoscopy is integral to most dermatology services but is thought to be less commonly used in some clinics, for example clinics staffed by plastic and reconstructive surgeons. For healthcare professionals who assess pigmented lesions: Developing competencies in assessment. Gaining experience in dermoscopy through regular practice. Including formal training in dermoscopy in their continuing professional development and revalidation work. Gaining access to new equipment (in some areas). For relevant royal colleges and speciality training organisations: Melanoma: assessment and management (NG14) © NICE 2015. All rights reserved. Page 53 of 61 Including dermoscopy in speciality training curricula for healthcare professionals who assess pigmented lesions. Making the changes happen Making the changes happen Commissioners of services could: Include provision of dermoscopy in local service specifications. Providers of secondary skin cancer clinics could: Arrange for healthcare professionals who assess pigmented lesions to have formal training in dermoscopy. There are a range of academic institutions that deliver national and local courses. Routinely provide experiential training for staff in specialist clinics. This could include competency-based assessment. Include reference to ongoing experience and competency in the appraisals of healthcare professionals who perform dermoscopy. The relevant royal colleges, supported by the speciality training organisations, could: Include dermoscopy where relevant in their specialty training curricula, and look at the specialty training curriculum for dermatology from the Joint Royal Colleges of Physicians Training Board as an example. Consider developing work-based assessments for measuring competency in the use of dermoscopy. Challenge 2 – Measuring vitamin D levels and advising on supplementation See recommendations 1.3.1 and 1.3.2. P Potential benefits of implementation otential benefits of implementation Measuring vitamin D levels at diagnosis allows healthcare professionals to identify people with melanoma whose vitamin D levels are low and who might benefit from supplementation in line with national policies, as well as people with high vitamin D levels who do not need supplementation and in whom supplementation might be harmful. Knowing a person's vitamin D level will also improve the accuracy of the advice given to them about the risks and benefits of sunlight exposure. Melanoma: assessment and management (NG14) © NICE 2015. All rights reserved. Page 54 of 61 Challenges for implementation Challenges for implementation For dermatologists (and possibly oncologists) in skin cancer multidisciplinary teams: Measuring vitamin D levels routinely at diagnosis of melanoma. Developing expertise in interpreting vitamin D levels. Providing advice about vitamin D supplementation if needed. Making the changes happen Making the changes happen Dermatologists (and possibly oncologists) in skin cancer multidisciplinary teams could: Refer to the Scientific Advisory Committee on Nutrition for information on vitamin D supplementation. Refer to NICE's pathway on vitamin D: increasing supplement use among at-risk groups, which covers vitamin D supplementation in people with low or no exposure to the sun. Listen to the podcast produced by NICE to explain the evidence behind the recommendations on vitamin D and its supplementation. Use the Advice for melanoma health care teams – vitamin D and melanoma produced by GenoMEL (the Melanoma Genetics Consortium). Challenge 3 – Considering sentinel lymph node biopsy and completion lymphadenectomy See recommendations 1.5.2 and 1.7.1. P Potential benefits of implementation otential benefits of implementation Considering sentinel lymph node biopsy (SLNB) for people who have stage IB–IIC melanoma with a Breslow thickness of more than 1 mm, and discussing the possible advantages and disadvantages with them, will enable people with these melanomas to make an informed decision about whether or not to have this procedure. Those who choose to have SLNB may benefit from more accurate staging, giving a better indication of outcome (including survival and risk of relapse). SLNB is more sensitive than ultrasound, so lymphatic spread may be diagnosed earlier. In addition, people who have SLNB may be able to participate in clinical trials of new treatments. Melanoma: assessment and management (NG14) © NICE 2015. All rights reserved. Page 55 of 61 Similarly, discussing the possible advantages and disadvantages of completion lymphadenectomy with people who have a positive SLNB will enable them to make an informed decision about whether or not to have this procedure after SLNB. Completion lymphadenectomy can reduce the chance of the melanoma returning and may enable the person to participate in clinical trials of new treatments. Challenges for implementation Challenges for implementation For clinicians in skin cancer multidisciplinary teams: Explaining the value of SLNB as a staging tool to people with melanoma, because there are no clear survival benefits from it. Providing comprehensive information about the possible advantages and disadvantages of having the procedure. Explaining the benefits of proceeding to completion lymphadenectomy to people with a positive SLNB result. For commissioners: Providing SLNB in services. Making the changes happen Making the changes happen Clinicians in skin cancer multidisciplinary teams could: Listen to the explanation of the evidence behind the SLNB recommendations on the NICE podcast. Use the option grids that NICE has produced to help discuss the possible risks and benefits of having SLNB and, for people with a positive SLNB, proceeding to completion lymphadenectomy. Commissioners could: Ensure that service specifications include provision of SLNB. This may not be delivered locally. Visit the NICE local practice collection to see examples of SLNB services. Melanoma: assessment and management (NG14) © NICE 2015. All rights reserved. Page 56 of 61 Further resources Further resources are available from NICE that may help to support implementation. NICE produces indicators annually for use in the Quality and Outcomes Framework (QOF) for the UK. The process for this and the NICE menu can be found here. Uptake data about guideline recommendations and quality standard measures are available on the NICE website. Melanoma: assessment and management (NG14) © NICE 2015. All rights reserved. Page 57 of 61 Changes after publication Changes after publication No Nov vember 2015: ember 2015: Corrected a link to the Joint Royal Colleges of Physicians Training Board in the implementation section. Melanoma: assessment and management (NG14) © NICE 2015. All rights reserved. Page 58 of 61 About this guideline About this guideline NICE clinical guidelines are recommendations about the treatment and care of people with specific diseases and conditions. NICE guidelines are developed in accordance with a scope that defines what the guideline will and will not cover. This guideline was developed by the National Collaborating Centre for Cancer, which is based at the Velindre NHS Trust in Cardiff. The Collaborating Centre worked with a Guideline Development Group, comprising healthcare professionals (including consultants, GPs and nurses), patients and carers, and technical staff, which reviewed the evidence and drafted the recommendations. The recommendations were finalised after public consultation. The methods and processes for developing NICE clinical guidelines are described in the guidelines manual. NICE produces guidance, standards and information on commissioning and providing high-quality healthcare, social care, and public health services. We have agreements to provide certain NICE services to Wales, Scotland and Northern Ireland. Decisions on how NICE guidance and other products apply in those countries are made by ministers in the Welsh government, Scottish government, and Northern Ireland Executive. NICE guidance or other products may include references to organisations or people responsible for commissioning or providing care that may be relevant only to England. Strength of recommendations Some recommendations can be made with more certainty than others. The Guideline Development Group makes a recommendation based on the trade-off between the benefits and harms of an intervention, taking into account the quality of the underpinning evidence. For some interventions, the Guideline Development Group is confident that, given the information it has looked at, most patients would choose the intervention. The wording used in the recommendations in this guideline denotes the certainty with which the recommendation is made (the strength of the recommendation). For all recommendations, NICE expects that there is discussion with the patient about the risks and benefits of the interventions, and their values and preferences. This discussion aims to help them to reach a fully informed decision (see also patient-centred care). Melanoma: assessment and management (NG14) © NICE 2015. All rights reserved. Page 59 of 61 Interv Interventions that must ( entions that must (or must not) be used or must not) be used We usually use 'must' or 'must not' only if there is a legal duty to apply the recommendation. Occasionally we use 'must' (or 'must not') if the consequences of not following the recommendation could be extremely serious or potentially life threatening. Interv Interventions that should ( entions that should (or should not) be used – a 'strong' recommendation or should not) be used – a 'strong' recommendation We use 'offer' (and similar words such as 'refer' or 'advise') when we are confident that, for the vast majority of patients, an intervention will do more good than harm, and be cost effective. We use similar forms of words (for example, 'Do not offer…') when we are confident that an intervention will not be of benefit for most patients. Interv Interventions that could be used entions that could be used We use 'consider' when we are confident that an intervention will do more good than harm for most patients, and be cost effective, but other options may be similarly cost effective. The choice of intervention, and whether or not to have the intervention at all, is more likely to depend on the patient's values and preferences than for a strong recommendation, and so the healthcare professional should spend more time considering and discussing the options with the patient. Other versions of this guideline The full guideline, melanoma: assessment and management of melanoma contains details of the methods and evidence used to develop the guideline. It is published by the National Collaborating Centre for Cancer. The recommendations from this guideline have been incorporated into a NICE pathway. We have produced information for the public about this guideline. Implementation Implementation tools and resources to help you put the guideline into practice are also available. Your responsibility This guidance represents the view of NICE, which was arrived at after careful consideration of the evidence available. Healthcare professionals are expected to take it fully into account when Melanoma: assessment and management (NG14) © NICE 2015. All rights reserved. Page 60 of 61 exercising their clinical judgement. However, the guidance does not override the individual responsibility of healthcare professionals to make decisions appropriate to the circumstances of the individual patient, in consultation with the patient and/or guardian or carer, and informed by the summaries of product characteristics of any drugs. Implementation of this guidance is the responsibility of local commissioners and/or providers. Commissioners and providers are reminded that it is their responsibility to implement the guidance, in their local context, in light of their duties to have due regard to the need to eliminate unlawful discrimination, advance equality of opportunity and foster good relations. Nothing in this guidance should be interpreted in a way that would be inconsistent with compliance with those duties. ISBN: 978-1-4731-1322-0 Accreditation Melanoma: assessment and management (NG14) © NICE 2015. All rights reserved. Page 61 of 61
6444	https://math.stackexchange.com/questions/846101/expected-value-equals-sum-of-probabilities	probability - Expected value equals sum of probabilities - Mathematics Stack Exchange Join Mathematics By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy. Sign up with Google OR Email Password Sign up Already have an account? Log in Skip to main content Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Visit Stack Exchange Loading… Tour Start here for a quick overview of the site Help Center Detailed answers to any questions you might have Meta Discuss the workings and policies of this site About Us Learn more about Stack Overflow the company, and our products current community Mathematics helpchat Mathematics Meta your communities Sign up or log in to customize your list. more stack exchange communities company blog Log in Sign up Home Questions Unanswered AI Assist Labs Tags Chat Users Teams Ask questions, find answers and collaborate at work with Stack Overflow for Teams. Try Teams for freeExplore Teams 3. Teams 4. Ask questions, find answers and collaborate at work with Stack Overflow for Teams. Explore Teams Teams Q&A for work Connect and share knowledge within a single location that is structured and easy to search. Learn more about Teams Hang on, you can't upvote just yet. You'll need to complete a few actions and gain 15 reputation points before being able to upvote. Upvoting indicates when questions and answers are useful. What's reputation and how do I get it? Instead, you can save this post to reference later. Save this post for later Not now Thanks for your vote! You now have 5 free votes weekly. Free votes count toward the total vote score does not give reputation to the author Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation. Got it!Go to help center to learn more Expected value equals sum of probabilities [duplicate] Ask Question Asked 11 years, 3 months ago Modified11 years, 3 months ago Viewed 10k times This question shows research effort; it is useful and clear 8 Save this question. Show activity on this post. This question already has answers here: Find the Mean for Non-Negative Integer-Valued Random Variable (3 answers) Closed 11 years ago. Let X X be a random variable that takes non-negative integer values. Show that, E[X]=∑k=1∞P(X≥k)E[X]=∑k=1∞P(X≥k) I'm having trouble following the solution. Could someone help clarify some steps? Thanks. By definition, P(X≥k)=∑i=k∞p X(i)P(X≥k)=∑i=k∞p X(i) Therefore, we substitute to get ∑k=1∞P(X≥k)=∑k=1∞∑i=k∞p X(i)∑k=1∞P(X≥k)=∑k=1∞∑i=k∞p X(i) Now here is where I'm confused. ∑k=1∞∑i=k∞p X(i)=∑i=1∞∑k=1 i p X(i)=∑i=1∞i p X(i)∑k=1∞∑i=k∞p X(i)=∑i=1∞∑k=1 i p X(i)=∑i=1∞i p X(i) I don't understand how we are manipulating the summations in the first equality and how we derive i p X(i)i p X(i) in the second equality. probability Share Share a link to this question Copy linkCC BY-SA 3.0 Cite Follow Follow this question to receive notifications edited Jun 24, 2014 at 17:03 user940 asked Jun 24, 2014 at 17:00 ConvergiiConvergii 233 3 3 silver badges 12 12 bronze badges 3 1 Put P(X≥k)=P(X=k)+P(X=k+1)+⋯P(X≥k)=P(X=k)+P(X=k+1)+⋯ as the k k'th row of an infinite matrix, padded at the start with k−1 k−1 zeroes. Now sum over the columns.David Mitra –David Mitra 2014-06-24 17:03:52 +00:00 Commented Jun 24, 2014 at 17:03 As in the first answer here.David Mitra –David Mitra 2014-06-24 17:10:36 +00:00 Commented Jun 24, 2014 at 17:10 1 This is a recurring question here, but it's hard to find it because the titles vary so much.Michael Hardy –Michael Hardy 2014-06-24 17:18:37 +00:00 Commented Jun 24, 2014 at 17:18 Add a comment\| 2 Answers 2 Sorted by: Reset to default This answer is useful 6 Save this answer. Show activity on this post. =0 P(X=0)+1 P(X=1)P(X=1)+++2 P(X=2)P(X=2)P(X=2)++++3 P(X=3)P(X=3)P(X=3)P(X=3)+++++⋯⋯⋯⋯⋯0 P(X=0)+1 P(X=1)+2 P(X=2)+3 P(X=3)+⋯=P(X=1)+P(X=2)+P(X=3)+⋯+P(X=2)+P(X=3)+⋯+P(X=3)+⋯+⋯ The sum in the first row is P(X>0)P(X>0); that in the second row is P(X>1)P(X>1); that in the third row is P(X>2)P(X>2), and so on. The equality of ∑k=1∞∑i=k∞⋯=∑i=1∞∑k=1 i⋯∑k=1∞∑i=k∞⋯=∑i=1∞∑k=1 i⋯ can be seen by observing that both are the sum over the set {(k,i):1≤k≤i}.{(k,i):1≤k≤i}. The next equality involves a sum of the form ∑k=1 i a∑k=1 i a where a a does not change as k k goes from 1 1 to i i, so it is a+a+a+⋯+ai terms=a i.a+a+a+⋯+a⏟i terms=a i. Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications edited Jun 24, 2014 at 17:30 answered Jun 24, 2014 at 17:17 Michael HardyMichael Hardy 1 Add a comment\| This answer is useful 3 Save this answer. Show activity on this post. Perhaps you can show the pointwise identity X=∑k=1 X 1=∑k=1∞1 X⩾k X=∑k=1 X 1=∑k=1∞1 X⩾k and use linearity of expectation. Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications edited Jun 24, 2014 at 17:33 answered Jun 24, 2014 at 17:09 DidDid 285k 27 27 gold badges 334 334 silver badges 613 613 bronze badges Add a comment\| Start asking to get answers Find the answer to your question by asking. 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6445	https://blogs.ifas.ufl.edu/polkco/2022/07/22/overwatering-can-be-as-problematic-as-underwatering/	Overwatering Can Be as Problematic as Underwatering Skip Navigation or Skip to Content Blogs About Home Blog Directory Subscribe Agriculture Agribusiness Crops Farm Management Horticulture Livestock Pests & Disease Turf Natural Resources Coasts & Marine Conservation Forests Invasive Species Recreation Water Wildlife Home Landscapes Fruits & Vegetables Florida-Friendly Landscaping Lawn Pests & Disease Work & Life Disaster Preparation Food Safety Health & Nutrition Home Management Money Matters Relationships & Family Community Volunteers 4-H & Youth Camp Curriculum Clubs & Volunteers What's New Agriculture Agriculture Agribusiness Crops Farm Management Horticulture Livestock Pests & Disease Turf Natural Resources Natural Resources Coasts & Marine Conservation Forests Invasive Species Recreation Water Wildlife Home Landscapes Home Landscapes Fruits & Vegetables Florida-Friendly Landscaping Lawn Pests & Disease Work & Life Work & Life Disaster Preparation Food Safety Health & Nutrition Home Management Money Matters Relationships & Family Community Volunteers 4-H & Youth 4-H & Youth Camp Curriculum Clubs & Volunteers Subscribe More What's New About Home Blog Directory Home Home Blog Directory Blog Directory Blogs.IFAS Blogs.IFAS Home » UF/IFAS Extension Polk County » Overwatering Can Be As Problematic As Underwatering Overwatering Can Be as Problematic as Underwatering Email 0Facebook 0Twitter 0Reddit 0 XLinkedin 0Stumbleupon 0 It’s that time of year again when we can expect a thunderstorm to roll in mid-afternoon and pour for half an hour before dissipating as quickly as it came. These rains provide relief to thirsty lawns that spend the rest of the day baking in the sun. However, that is only half the story. Unfortunately, these same rains can provide too much water if irrigation times are not adjusted to compensate, especially if previously raised for the high water needs of late spring/early summer. According to a study done by Romero and Dukes (2011), the May average irrigation requirement for the Tampa area is just under 6 inches, which comes out to 1.5 inches per week. In contrast, in July, the irrigation requirement for the same area is just under 2 inches, which is only half an inch per week. Although plants need more water now because they are actively growing, our typical precipitation mostly covers it. For this reason, weeds, fungi, and diseases take advantage of plant stress and can become widespread once the summer rains begin. So, what can you do to ensure your lawn isn’t overwatered? A contractor installs a rain sensor at a home in Davenport, FL. First, m ake sure your rain sensor is in the correct location and working. Sensors should be out in the open to collect rain, not underneath overhangs or trees. Testing is easier with two people since one likely needs to climb a ladder to get to the sensor. Each model works slightly differently, but the most common type has a button on top for testing. Once you have manually turned on the sprinklers, push and hold the button. If your sprinklers turn off, you know your sensor is working. If not, you may need a new sensor. Contact your water utility to see if they are offering free replacement sensors or you can visit SavePolkWater.org to see what sort of incentives are available in your area. Remember, a working rain sensor is required by law in Florida for all properties with automatic irrigation systems. Second, you can look for physical signs that your lawn is getting too much water. For instance, does water run off your lawn when you turn on your sprinklers? Do you have water-loving plants like dollar weed infiltrating your grass? Other indicators include fungus, root rot, yellowing grass, thatch, and the death of patches of grass. If you have these problems, you should consider making changes to your timer. Third, make sure your irrigation controller is only set for the days prescribed in your watering restrictions. Through most of Florida, water restrictions are year-round as prescribed by the water management district, unless local municipalities have more stringent rules. Be careful you don’t accidentally have more than one program running. You can find out by pressing the “program” button while setting up the run times. Alternatively, you can run your sprinklers manually only when there is no rain forecasted within the next day, nor has it rained within the last couple of days. You are still only allowed to water on your authorized days, but you can set a reminder on your phone to check your lawn for water stress and decide if it needs a drink. Symptoms of water stress include turning an unusual bluish color, the leaf blades folding over, or footprints not disappearing. Be sure to only water before 10 AM or after 4 PM to minimize water lost to evaporation. For more information, contact UF/IFAS Extension Polk County at (863) 519-1041 or visit us online at The Plant Clinic is open Monday-Friday, 9:00 am-4:00 pm to answer your gardening and landscaping questions. Visit us in person, give us a call, or email us at polkmg@ifas.ufl.edu. An Equal Opportunity Institution. 0 by Anne Yasalonis and Beth Robertson Posted: July 22, 2022 Category: More From Blogs.IFAS Monarchs and Milkweed: An Important Update Beware of Volcano Mulching! Composting for Beginners Hydroponic Production Methods- Part 2 Author Profile Anne Yasalonis Anne is the Residential Horticulture Agent and Master Gardener Volunteer Coordinator at UF/IFAS Extension Polk County. She offers education to residents on landscaping and gardening in central Florida. ... Beth Robertson Beth is the water conservation specialist for UF/IFAS Polk Extension. She educates the public on water saving strategies and programs available to help in these endeavors. Links Blogs.IFAS Solutions for Your Life UF/IFAS Naturally Florida Podcast Subscribe to UF/IFAS Extension Polk County Blog Search By Date Explore by Category Subscribe Get a daily or weekly email digest. 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6446	https://www.cut-the-knot.org/m/Geometry/HomothetyInThreeTangentCircles.shtml	Homothety in Three Tangent Circles Typesetting math: 100% Site... What's new Content page Front page Index page About Privacy policy Help with math Subjects... Arithmetic Algebra Geometry Probability Trigonometry Visual illusions Articles... Cut the knot! What is what? Inventor's paradox Math as language Problem solving Collections... Outline mathematics Book reviews Interactive activities Did you know? Eye opener Analogue gadgets Proofs in mathematics Things impossible Index/Glossary Simple math... Fast Arithmetic Tips Stories for young Word problems Games and puzzles Our logo Make an identity Elementary geometry Homothety in Three Tangent Circles What Is This About? 11 March 2017, Created with GeoGebra Problem Solution 1 Join D and F to A. Due to the tangency of the circles, B∈A D and C∈A F. Circle (A) is obtained from (B) by a homothety with center D such that B E∥A G. Similarly E C∥H A. However, since circles (B) and (C) are tangent at E,E∈B C. In other words, B,E C are collinear, implying that so are H,A, and G. Solution 2 The problem is a particular case of a more general one, treated on a separate page: Given three circles (A),(B),(C), such that (A) and (B) are tangent internally at D while (A) and (C) are tangent internally at F. Assume (B) and (C) intersect at E and E′ as shown below. F E′, extend, intersects (A) in H;D E′, extended, intersects (A) in G. Prove that ∠B E C=∠H A G. Obviously, (B) and (C) are tangent at E, if and only if ∠B E C=180∘, which reduces to the previous problem. Acknowledgment I learned of this problem from a tweet by Tim Brzezinski, with a dynamic illustration of the problem posted by Antonio Gutierrez. \|Contact\|\|Up\|\|Front page\|\|Contents\|\|Geometry\| Copyright © 1996-2018 Alexander Bogomolny 73255727
6447	https://www.geeksforgeeks.org/dsa/find-the-largest-prime-number-in-string/	Find the largest prime number in String Last Updated : 23 Oct, 2024 Suggest changes 1 Like Given a string S, representing a large integer, the task is to return the largest-valued prime integer (as a string) that is a substring of the given string S. Note: A substring is a contiguous sequence of characters within a string. A null string (“”) is also a substring. Examples: Input: S = "504" Output: "5" Explanation: The only substring 5 is the largest prime number. Input: S = "1323" Output: "23" Explanation: The only substring 23 is the largest prime number. Input: S = "24617" Output: "617" Explanation: The only substring "617" is the largest prime number. Input: S = "4444" Output: "" Explanation: There is no substring of a prime number. Approach 1: (Bruteforce approach ) To solve the problem follow the below idea: The idea is to check each and every substring by running two loops and checking whether the substring is prime or not and returning the largest prime number substring. Below are the steps involved in the implementation of the code: Call the Function Find the largest prime by passing the string s as a parameter. Now iterate through each and every substring of the string and check whether it is prime or not. if is it prime check for the largest prime previous by taking an extra variable. Return the integer value of the largest substring in to_string format. Below is the implementation for the above approach: C++ ```` // C++ code for the above approach: include include include using namespace std; // Function to check whether substring // is prime or not bool is_prime(int n) { if (n <= 1) return false; for (int i = 2; i <= sqrt(n); i++) { if (n % i == 0) return false; } return true; } // Function to find largest value prime // substring string find_largest_prime(string s) { int largest_prime = 0; // Start Iterating for (int i = 0; i < s.length(); i++) { for (int j = i + 1; j <= s.length(); j++) { string substring = s.substr(i, j - i); if (substring.find_first_not_of("0123456789") == string::npos && is_prime(stoi(substring))) { largest_prime = max(largest_prime, stoi(substring)); } } } if (largest_prime < 0) return " "; else return to_string(largest_prime); } // Driver Code int main() { string s; s = "132346"; // Function call string largest_prime = find_largest_prime(s); cout << largest_prime << endl; return 0; } ```` // C++ code for the above approach: // C++ code for the above approach: ``` include #include ``` ``` include #include ``` ``` include #include ``` using namespace std; using namespace std // Function to check whether substring // Function to check whether substring // is prime or not // is prime or not bool is_prime(int n) bool is_prime int n { if (n <= 1) if n<= 1 return false; return false for (int i = 2; i <= sqrt(n); i++) {for int i = 2 i<= sqrt n i ++ if (n % i == 0) if n% i == 0 return false; return false } return true; return true } // Function to find largest value prime // Function to find largest value prime // substring // substring string find_largest_prime(string s) string find_largest_prime string s { int largest_prime = 0; int largest_prime = 0 // Start Iterating // Start Iterating for (int i = 0; i < s.length(); i++) {for int i = 0 i< s length i ++ for (int j = i + 1; j <= s.length(); j++) {for int j = i + 1 j<= s length j ++ string substring = s.substr(i, j - i); string substring = s substr i j - i if (substring.find_first_not_of("0123456789") if substring find_first_not_of "0123456789" == string::npos ==string::npos && is_prime(stoi(substring))) {&& is_prime stoi substring largest_prime largest_prime = max(largest_prime, stoi(substring)); = max largest_prime stoi substring } } } if (largest_prime < 0) if largest_prime< 0 return " "; return " " else else return to_string(largest_prime); return to_string largest_prime } // Driver Code // Driver Code int main() int main { string s; string s s = "132346"; s = "132346" // Function call // Function call string largest_prime = find_largest_prime(s); string largest_prime = find_largest_prime s cout << largest_prime << endl; cout<< largest_prime<< endl return 0; return 0 } Java ```` // Java code for the above approach import java.util.; public class GFG { // Function to check whether substring is prime or not static boolean isPrime(int n) { if (n <= 1) return false; for (int i = 2; i <= Math.sqrt(n); i++) { if (n % i == 0) return false; } return true; } // Function to find largest value prime substring static String findLargestPrime(String s) { int largestPrime = 0; // Start iterating for (int i = 0; i < s.length(); i++) { for (int j = i + 1; j <= s.length(); j++) { String substring = s.substring(i, j); if (substring.matches("[0-9]+") && isPrime( Integer.parseInt(substring))) { largestPrime = Math.max( largestPrime, Integer.parseInt(substring)); } } } if (largestPrime < 0) return " "; else return String.valueOf(largestPrime); } // Driver code public static void main(String[] args) { String s = "132346"; // Function call String largestPrime = findLargestPrime(s); System.out.println(largestPrime); } } // This code is contributed by Susobhan Akhuli ```` Python ```` Python code for the above approach: import math Function to check whether substring is prime or not def is_prime(n): if n <= 1: return False for i in range(2, int(math.sqrt(n)) + 1): if n % i == 0: return False return True Function to find largest value prime substring def find_largest_prime(s): largest_prime = 0 # Start Iterating for i in range(len(s)): for j in range(i + 1, len(s) + 1): substring = s[i:j] if substring.isdigit() and is_prime(int(substring)): largest_prime = max(largest_prime, int(substring)) if largest_prime < 0: return " " else: return str(largest_prime) Driver Code if name == "main": s = "132346" # Function call largest_prime = find_largest_prime(s) print(largest_prime) ```` C# ```` // C# code for the above approach using System; using System.Linq; public class GFG { // Function to check whether substring // is prime or not static bool IsPrime(int n) { if (n <= 1) return false; for (int i = 2; i <= Math.Sqrt(n); i++) { if (n % i == 0) return false; } return true; } // Function to find the largest value prime // substring static string FindLargestPrime(string s) { int largestPrime = 0; // Start Iterating for (int i = 0; i < s.Length; i++) { for (int j = i + 1; j <= s.Length; j++) { string substring = s.Substring(i, j - i); if (substring.All(char.IsDigit) && IsPrime(int.Parse(substring))) { largestPrime = Math.Max( largestPrime, int.Parse(substring)); } } } if (largestPrime < 0) return " "; else return largestPrime.ToString(); } // Driver Code public static void Main() { string s = "132346"; // Function call string largestPrime = FindLargestPrime(s); Console.WriteLine(largestPrime); } } // This code is contributed by Susobhan Akhuli ```` JavaScript ```` // JavaScript code for the above approach // Function to check whether a number is prime or not function isPrime(n) { if (n <= 1) { return false; } for (let i = 2; i <= Math.sqrt(n); i++) { if (n % i === 0) { return false; } } return true; } // Function to find the largest value prime substring function findLargestPrime(s) { let largestPrime = 0; // Start iterating for (let i = 0; i < s.length; i++) { for (let j = i + 1; j <= s.length; j++) { let substring = s.substring(i, j); if (/^\d+$/.test(substring) && isPrime(parseInt(substring))) { largestPrime = Math.max(largestPrime, parseInt(substring)); } } } if (largestPrime < 0) { return " "; } else { return largestPrime.toString(); } } // Driver Code let s = "132346"; // Function call let largestPrime = findLargestPrime(s); console.log(largestPrime); // This code is contributed by Susobhan Akhuli ```` Output 23 Time Complexity: O(N^3) O(log (N)), where N is the length of the string. Auxiliary Space: O(1). Approach 2: To solve the problem follow the below idea: The idea is to genrate all prime numbers upto the maximum substring number present in string. Below is the implementation for the above approach: C++ ```` // C++ code for the above approach: include using namespace std; // Function to genrate all prime number void genratePrimes(bool isPrime[], int n) { for (int p = 2; p p <= n; p++) { // If prime[p] is not changed, then it is a prime if (isPrime[p] == true) { // Update all multiples of p greater than or // equal to the square of it numbers which are // multiple of p and are less than p^2 are // already been marked. for (int i = p p; i <= n; i += p) isPrime[i] = false; } } } // Function to find largest value prime // substring string find_largest_prime(string s, bool isPrime[]) { int largest_prime = 0; // Start Iterating for (int i = 0; i < s.length(); i++) { // using temp variable to store substring int temp=0; for (int j = i; j < s.length(); j++) { temp = temp 10 + (s[j] - '0'); // check wheather temp is prime or not if (isPrime[temp]) largest_prime = max(largest_prime, temp); } } if (largest_prime < 0) return " "; else return to_string(largest_prime); } // Driver Code int main() { string s; s = "132346"; int n = stoi(s); // Create a boolean array "isPrime[0..n]" and initialize // all entries it as true. A value in isPrime[i] will // finally be false if i is Not a prime, else true. bool isPrime[n + 1]; memset(isPrime, true, sizeof(isPrime)); // Genrate all primes numbers upto max substring number // in string s genratePrimes(isPrime, n); // Function call for largest prime string largest_prime = find_largest_prime(s, isPrime); cout << largest_prime << endl; return 0; } ```` // C++ code for the above approach: // C++ code for the above approach: ``` include #include ``` using namespace std; using namespace std // Function to genrate all prime number // Function to genrate all prime number void genratePrimes(bool isPrime[], int n) void genratePrimes bool isPrime int n { for (int p = 2; p p <= n; p++) {for int p = 2 p p<= n p ++ // If prime[p] is not changed, then it is a prime // If prime[p] is not changed, then it is a prime if (isPrime[p] == true) {if isPrime p == true // Update all multiples of p greater than or // Update all multiples of p greater than or // equal to the square of it numbers which are // equal to the square of it numbers which are // multiple of p and are less than p^2 are // multiple of p and are less than p^2 are // already been marked. // already been marked. for (int i = p p; i <= n; i += p) for int i = p p i<= n i += p isPrime[i] = false; isPrime i = false } } } // Function to find largest value prime // Function to find largest value prime // substring // substring string find_largest_prime(string s, bool isPrime[]) string find_largest_prime string s bool isPrime { int largest_prime = 0; int largest_prime = 0 // Start Iterating // Start Iterating for (int i = 0; i < s.length(); i++) {for int i = 0 i< s length i ++ ``` ``` // using temp variable to store substring // using temp variable to store substring int temp=0; int temp = 0 for (int j = i; j < s.length(); j++) {for int j = i j< s length j ++ temp = temp 10 + (s[j] - '0'); temp = temp 10 + s j - '0' ``` ``` // check wheather temp is prime or not // check wheather temp is prime or not if (isPrime[temp]) if isPrime temp largest_prime = max(largest_prime, temp); largest_prime = max largest_prime temp } } if (largest_prime < 0) if largest_prime< 0 return " "; return " " else else return to_string(largest_prime); return to_string largest_prime } // Driver Code // Driver Code int main() int main { string s; string s s = "132346"; s = "132346" int n = stoi(s); int n = stoi s // Create a boolean array "isPrime[0..n]" and initialize // Create a boolean array "isPrime[0..n]" and initialize // all entries it as true. A value in isPrime[i] will // all entries it as true. A value in isPrime[i] will // finally be false if i is Not a prime, else true. // finally be false if i is Not a prime, else true. bool isPrime[n + 1]; bool isPrime n + 1 memset(isPrime, true, sizeof(isPrime)); memset isPrime true sizeof isPrime // Genrate all primes numbers upto max substring number // Genrate all primes numbers upto max substring number // in string s // in string s genratePrimes(isPrime, n); genratePrimes isPrime n // Function call for largest prime // Function call for largest prime string largest_prime = find_largest_prime(s, isPrime); string largest_prime = find_largest_prime s isPrime cout << largest_prime << endl; cout<< largest_prime<< endl return 0; return 0 } Java ```` // Java code for the above approach import java.util.Arrays; public class GFG { // Function to generate all prime numbers static void generatePrimes(boolean isPrime[], int n) { for (int p = 2; p p <= n; p++) { // If isPrime[p] is not changed, then it is a // prime if (isPrime[p]) { // Update all multiples of p greater than or // equal to the square of it; numbers which // are multiples of p and are less than p^2 // are already marked. for (int i = p p; i <= n; i += p) isPrime[i] = false; } } } // Function to find the largest value prime // substring static String findLargestPrime(String s, boolean isPrime[]) { int largestPrime = 0; // Start Iterating for (int i = 0; i < s.length(); i++) { // Using temp variable to store substring int temp = 0; for (int j = i; j < s.length(); j++) { temp = temp 10 + (s.charAt(j) - '0'); // Check whether temp is prime or not if (isPrime[temp]) largestPrime = Math.max(largestPrime, temp); } } if (largestPrime < 0) return " "; else return Integer.toString(largestPrime); } // Driver Code public static void main(String[] args) { String s = "132346"; int n = Integer.parseInt(s); // Create a boolean array "isPrime[0..n]" and // initialize all entries as true. A value in // isPrime[i] will finally be false if i is not a // prime, else true. boolean[] isPrime = new boolean[n + 1]; Arrays.fill(isPrime, true); // Generate all prime numbers up to max substring // number in string s generatePrimes(isPrime, n); // Function call for the largest prime String largestPrime = findLargestPrime(s, isPrime); System.out.println(largestPrime); } } // This code is contributed by Susobhan Akhuli ```` Python ```` Python code for the above approach Function to generate all prime numbers up to n def generate_primes(n): is_prime = [True] (n + 1) p = 2 while p p <= n: if is_prime[p]: for i in range(p p, n + 1, p): is_prime[i] = False p += 1 return is_prime Function to find the largest prime substring def find_largest_prime(s, is_prime): largest_prime = 0 # Start iterating through the string for i in range(len(s)): temp = 0 for j in range(i, len(s)): temp = temp 10 + int(s[j]) # Check whether temp is prime or not if is_prime[temp]: largest_prime = max(largest_prime, temp) if largest_prime < 0: return " " else: return str(largest_prime) Driver Code if name == "main": s = "132346" n = int(s) # Create a boolean array "is_prime[0..n]" and initialize # all entries it as true. A value in is_prime[i] will # finally be False if i is not a prime, else True. is_prime = generate_primes(n) # Function call for finding the largest prime substring largest_prime = find_largest_prime(s, is_prime) print(largest_prime) This code is contributed by Susobhan Akhuli ```` C# ```` // C# code for the above approach using System; public class GFG { // Function to genrate all prime number static void genratePrimes(bool[] isPrime, int n) { for (int p = 2; p p <= n; p++) { // If prime[p] is not changed, then it is a // prime if (isPrime[p] == true) { // Update all multiples of p greater than or // equal to the square of it numbers which // are multiple of p and are less than p^2 // are already been marked. for (int i = p p; i <= n; i += p) { isPrime[i] = false; } } } } // Function to find largest value prime substring static string find_largest_prime(string s, bool[] isPrime) { int largest_prime = 0; // Start Iterating for (int i = 0; i < s.Length; i++) { // using temp variable to store substring int temp = 0; for (int j = i; j < s.Length; j++) { temp = temp 10 + (s[j] - '0'); // check wheather temp is prime or not if (isPrime[temp]) { largest_prime = Math.Max(largest_prime, temp); } } } if (largest_prime < 0) { return " "; } else { return largest_prime.ToString(); } } // Driver Code static void Main() { string s; s = "132346"; int n = int.Parse(s); // Create a boolean array "isPrime[0..n]" and // initialize all entries it as true. A value in // isPrime[i] will finally be false if i is Not a // prime, else true. bool[] isPrime = new bool[n + 1]; for (int i = 0; i <= n; i++) { isPrime[i] = true; } // Genrate all primes numbers upto max substring // number in string s genratePrimes(isPrime, n); // Function call for largest prime string largest_prime = find_largest_prime(s, isPrime); Console.WriteLine(largest_prime); } } // This code is contributed by Susobhan Akhuli ```` JavaScript ```` // JavaScript code for the above approach // Function to generate all prime numbers function generatePrimes(isPrime, n) { for (let p = 2; p p <= n; p++) { // If isPrime[p] is true, then it is a prime if (isPrime[p] === true) { // Update all multiples of p greater than or equal to p^2 // Numbers which are multiple of p and are less than p^2 are already marked for (let i = p p; i <= n; i += p) { isPrime[i] = false; } } } } // Function to find the largest value prime substring function findLargestPrime(s, isPrime) { let largestPrime = 0; // Start iterating for (let i = 0; i < s.length; i++) { // Using temp variable to store substring let temp = 0; for (let j = i; j < s.length; j++) { temp = temp 10 + parseInt(s[j]); // Check whether temp is prime or not if (isPrime[temp]) { largestPrime = Math.max(largestPrime, temp); } } } if (largestPrime < 0) { return " "; } else { return largestPrime.toString(); } } let s = "132346"; let n = parseInt(s); // Create a boolean array "isPrime[0..n]" and initialize all entries as true // A value in isPrime[i] will finally be false if i is not a prime, else true let isPrime = new Array(n + 1).fill(true); // Generate all prime numbers up to the maximum substring number in string s generatePrimes(isPrime, n); // Function call for the largest prime let largestPrime = findLargestPrime(s, isPrime); document.write(largestPrime); // This code is contributed by Susobhan Akhuli ```` Output 23 Time Complexity: O(N^2) + O(Nlog(log(N))) ≈ O(N^2), Nlog(log(N)) is used for generating all prime numbers to where N is the length of the string. Auxiliary Space: O(N) M mdjabir786 Improve Article Tags : Strings DSA Prime Number substring cpp-strings Practice Tags : cpp-strings Prime Number Strings Similar Reads Basics & Prerequisites Logic Building Problems Logic building is about creating clear, step-by-step methods to solve problems using simple rules and principles. It€™s the heart of coding, enabling programmers to think, reason, and arrive at smart solutions just like we do.Here are some tips for improving your programming logic: Understand the pro 2 min readAnalysis of Algorithms Analysis of Algorithms is a fundamental aspect of computer science that involves evaluating performance of algorithms and programs. 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6448	https://en.wikipedia.org/wiki/Saxitoxin	Jump to content Saxitoxin العربية تۆرکجه Čeština Dansk Deutsch Español فارسی Français 한국어 Italiano Nederlands 日本語 Norsk bokmål Norsk nynorsk Português Русский Slovenščina Српски / srpski Srpskohrvatski / српскохрватски Suomi ไทย Українська 中文 Edit links From Wikipedia, the free encyclopedia Paralytic shellfish toxin Saxitoxin \| \| \| \| \| \| \| --- \| \| \| \| \| \| \| Names \| \| \| IUPAC name [(3aS,4R,10aS)-10,10-dihydroxy-2,6-diiminooctahydro-1H,8H-pyrrolo[1,2-c]purin-4-yl]methyl carbamate \| \| \| Identifiers \| \| \| CAS Number \| 35523-89-8Y \| \| 3D model (JSmol) \| Interactive image \| \| ChEBI \| CHEBI:34970N \| \| ChEMBL \| ChEMBL501134N \| \| ChemSpider \| 34106N \| \| ECHA InfoCard \| 100.160.395 \| \| IUPHAR/BPS \| 2625 \| \| KEGG \| C13757Y \| \| PubChem CID \| 37165 \| \| UNII \| Q0638E899BY \| \| CompTox Dashboard (EPA) \| DTXSID3074313 \| \| InChI InChI=1S/C10H17N7O4/c11-6-15-5-4(3-21-8(13)18)14-7(12)17-2-1-9(19,20)10(5,17)16-6/h4-5,19-20H,1-3H2,(H2,12,14)(H2,13,18)(H3,11,15,16)/t4-,5-,10-/m0/s1N Key: RPQXVSUAYFXFJA-HGRQIUPRSA-NN InChI=1/C10H17N7O4/c11-6-15-5-4(3-21-8(13)18)14-7(12)17-2-1-9(19,20)10(5,17)16-6/h4-5,19-20H,1-3H2,(H2,12,14)(H2,13,18)(H3,11,15,16)/t4-,5-,10-/m0/s1 Key: RPQXVSUAYFXFJA-HGRQIUPRBO \| \| \| SMILES O=C(OC[C@@H]2/N=C(/N)N3[C@]1(/N=C(\N[C@H]12)N)C(O)(O)CC3)N \| \| \| Properties \| \| \| Chemical formula \| C10H17N7O4 \| \| Molar mass \| 299.291 g·mol−1 \| \| Except where otherwise noted, data are given for materials in their standard state (at 25 °C [77 °F], 100 kPa). N verify (what is YN ?) Infobox references \| \| Chemical compound Saxitoxin (STX) is a potent neurotoxin and the best-known paralytic shellfish toxin. Ingestion of saxitoxin by humans, usually by consumption of shellfish contaminated by toxic algal blooms, is responsible for the illness known as paralytic shellfish poisoning (PSP). The term saxitoxin originates from the genus name of the butter clam (Saxidomus) from which it was first isolated. But the term saxitoxin can also refer to the entire suite of more than 50 structurally related neurotoxins (known collectively as "saxitoxins") produced by protists, algae and cyanobacteria which includes saxitoxin itself (STX), neosaxitoxin (NSTX), gonyautoxins (GTX) and decarbamoylsaxitoxin (dcSTX). Saxitoxin has a large environmental and economic impact, as its presence in bivalve shellfish such as mussels, clams, oysters and scallops frequently leads to bans on commercial and recreational shellfish harvesting in many temperate coastal waters around the world including the Northeastern and Western United States, Western Europe, East Asia, Australia, New Zealand, and South Africa. In the United States, paralytic shellfish poisoning has occurred in California, Oregon, Washington, Alaska, and New England. Source in nature [edit] Saxitoxin is a neurotoxin naturally produced by certain species of marine dinoflagellates (Alexandrium sp., Gymnodinium sp., Pyrodinium sp.) and freshwater cyanobacteria (Dolichospermum cicinale sp., some Aphanizomenon spp., Cylindrospermopsis sp., Lyngbya sp., Planktothrix sp.) Saxitoxin accumulates in "planktivorous invertebrates, including mollusks (bivalves and gastropods), crustaceans, and echinoderms". Saxitoxin has also been found in at least twelve marine puffer fish species in Asia and one freshwater fish tilapia in Brazil. The ultimate source of STX is often still uncertain. The dinoflagellate Pyrodinium bahamense is the source of STX found in Florida. Recent research shows the detection of STX in the skin, muscle, viscera, and gonads of "Indian River Lagoon" southern puffer fish, with the highest concentration (22,104 μg STX eq/100 g tissue) measured in the ovaries. Even after a year of captivity, Landsberg et al. found the skin mucus remained highly toxic. The concentrations in puffer fish from the United States are similar to those found in the Philippines, Thailand, Japan, and South American countries. Puffer fish also accumulate a structurally distinct toxin, tetrodotoxin. Structure and synthesis [edit] Saxitoxin dihydrochloride is an amorphous hygroscopic solid, but X-ray crystallography of crystalline derivatives enabled the structure of saxitoxin to be determined. Oxidation of saxitoxin generates a highly fluorescent purine derivative which has been utilized to detect its presence. Several total syntheses of saxitoxin have been accomplished. Mechanism of action [edit] Main article: Sodium channel blocker Saxitoxin is a neurotoxin that acts as a selective, reversible, voltage-gated sodium channel blocker. One of the most potent known natural toxins, it acts on the voltage-gated sodium channels of neurons, preventing normal cellular function and leading to paralysis. The voltage-gated sodium channel is essential for normal neuronal functioning. It exists as integral membrane proteins interspersed along the axon of a neuron and possessing four domains that span the cell membrane. Opening of the voltage-gated sodium channel occurs when there is a change in voltage or some ligand binds in the right way. It is of foremost importance for these sodium channels to function properly, as they are essential for the propagation of an action potential. Without this ability, the nerve cell becomes unable to transmit signals and the region of the body that it enervates is cut off from the nervous system. This may lead to paralysis of the affected region, as in the case of saxitoxin. Saxitoxin binds reversibly to the sodium channel. It binds directly in the pore of the channel protein, occluding the opening, and preventing the flow of sodium ions through the membrane. This leads to the nervous shutdown described above. Biosynthesis [edit] Although the biosynthesis of saxitoxin seems complex, organisms from two different kingdoms, indeed two different domains, species of marine dinoflagellates and freshwater cyanobacteria are capable of producing these toxins. While the prevailing theory of production in dinoflagellates was through symbiotic mutualism with cyanobacteria, evidence has emerged suggesting that dinoflagellates themselves also possess the genes required for saxitoxin synthesis. Saxitoxin biosynthesis is the first non-terpene alkaloid pathway described for bacteria, though the exact mechanism of saxitoxin biosynthesis is still essentially a theoretical model. The precise mechanism of how substrates bind to enzymes is still unknown, and genes involved in the biosynthesis of saxitoxin are either putative or have only recently been identified. Two biosyntheses have been proposed in the past. Earlier versions differ from a more recent proposal by Kellmann, et al. based on both biosynthetic considerations as well as genetic evidence not available at the time of the first proposal. The more recent model describes a STX gene cluster (Sxt) used to obtain a more favorable reaction. The most recent reaction sequence of Sxt in cyanobacteria is as follows. Refer to the diagram for a detailed biosynthesis and intermediate structures. It begins with the loading of the acyl carrier protein (ACP) with acetate from acetyl-CoA, yielding intermediate 1. This is followed by SxtA-catalyzed methylation of acetyl-ACP, which is then converted to propionyl-ACP, yielding intermediate 2. Later, another SxtA performs a Claisen condensation reaction between propionyl-ACP and arginine producing intermediate 4 and intermediate 3. SxtG transfers an amidino group from an arginine to the α-amino group of intermediate 4 producing intermediate 5. Intermediate 5 then undergoes retroaldol-like condensation by SxtBC, producing intermediate 6. SxtD adds a double bond between C-1 and C-5 of intermediate 6, which gives rise to the 1,2-H shift between C-5 and C-6 in intermediate 7. SxtS performs an epoxidation of the double bond yielding intermediate 8, and then an opening of the epoxide to an aldehyde, forming intermediate 9. SxtU reduces the terminal aldehyde group of the STX intermediate 9, thus forming intermediate 10. SxtIJK catalyzes the transfer of a carbamoyl group to the free hydroxyl group on intermediate 10, forming intermediate 11. SxtH and SxtT, in conjunction with SxtV and the SxtW gene cluster, perform a similar function which is the consecutive hydroxylation of C-12, thus producing saxitoxin and terminating the STX biosynthetic pathway. Illness and poisoning [edit] Toxicology [edit] Saxitoxin is highly toxic to guinea pigs, fatal at only 5 μg/kg when injected intramuscularly. The median lethal dose (LD50) for mice is very similar with varying administration routes: i.v. is 3.4 μg/kg, i.p. is 10 μg/kg and p.o. is 263 μg/kg. The oral LD50 for humans is 5.7 μg/kg, therefore approximately 0.57 mg of saxitoxin is lethal if ingested and the lethal dose by injection is about one-tenth of that (approximately 0.6 μg/kg). The human inhalation toxicity of aerosolized saxitoxin is estimated to be 5 mg·min/m3. Saxitoxin can enter the body via open wounds and a lethal dose of 50 μg/person by this route has been suggested. Illness in humans [edit] The human illness associated with ingestion of harmful levels of saxitoxin is known as paralytic shellfish poisoning, or PSP, and saxitoxin and its derivatives are often referred to as "PSP toxins". The medical and environmental importance of saxitoxin derives from the consumption of contaminated shellfish and certain finfish which can concentrate the toxin from dinoflagellates or cyanobacteria. The blocking of neuronal sodium channels which occurs in PSP produces a flaccid paralysis that leaves its victim calm and conscious through the progression of symptoms. Death often occurs from respiratory failure. PSP toxins have been implicated in various marine animal mortalities involving trophic transfer of the toxin from its algal source up the food chain to higher predators. Studies in animals have shown that the lethal effects of saxitoxin can be reversed with 4-aminopyridine, but there are no studies on human subjects. As with any paralytic agent, where the acute concern is respiratory failure, mouth-to-mouth resuscitation or artificial ventilation of any means will keep a poisoned victim alive until antidote is administered or the poison wears off. Military interest [edit] See also: United States biological weapons program Saxitoxin, by virtue of its extremely low LD50, readily lends itself to weaponization. In the past, it was considered for military use by the United States and was developed as a chemical weapon by the US military. It is known that saxitoxin was developed for both overt military use as well as for covert purposes by the CIA. Among weapons stockpiles were M1 munitions that contained either saxitoxin, botulinum toxin or a mixture of both. On the other hand, the CIA is known to have issued a small dose of saxitoxin to U-2 spy plane pilot Francis Gary Powers in the form of a small injection hidden within a silver dollar, for use in the event of his capture and detainment. After the 1969 ban on biological warfare by President Nixon, the US stockpiles of saxitoxin were destroyed, and development of saxitoxin as a military weapon ceased. venom against Nixon's orders which was then destroyed or distributed to researchers. It is listed in schedule 1 of the Chemical Weapons Convention. The United States military isolated saxitoxin and assigned it the chemical weapon designation TZ. See also [edit] Action potential – Neuron communication by electric impulses Anabaena circinalis – Species of bacterium Alexandrium tamarense – Species of single-celled organism Canadian Reference Materials Ciguatoxin – Group of chemical compounds Domoic acid Harmful algal bloom – Population explosion of organisms that can kill marine life Okadaic acid Paralytic shellfish poisoning – Syndrome of shellfish poisoning Tetrodotoxin – NeurotoxinPages displaying short descriptions with no spaces References [edit] ^ Jump up to: a b Clark R. F., Williams S. R., Nordt S. P., Manoguerra A. S. (1999). "A review of selected seafood poisonings". Undersea Hyperb Med. 26 (3): 175–84. PMID 10485519. Archived from the original on October 7, 2008. Retrieved 2008-08-12. ^ Landsberg JH (2002). "The Effects of Harmful Algal Blooms on Aquatic Organisms". Reviews in Fisheries Science. 10 (2): 113–390. Bibcode:2002RvFS...10..113L. doi:10.1080/20026491051695. S2CID 86185142. ^ Jump up to: a b c d e "Saxitoxin". Retrieved April 10, 2022. ^ Galvão JA, Oetterer M, Bittencourt-Oliveira Mdo MD, Gouvêa-Barros S, Hiller S, Erler K, Luckas B, Pinto E, Kujbida P (2009). "Saxitoxins accumulation by freshwater tilapia (Oreochromis niloticus) for human consumption". Toxicon. 54 (6): 891–894. Bibcode:2009Txcn...54..891G. doi:10.1016/j.toxicon.2009.06.021. PMID 19560484. ^ Smith EA, Grant F, Ferguson CM, Gallacher S (2001). "Biotransformations of Paralytic Shellfish Toxins by Bacteria Isolated from Bivalve Molluscs". Applied and Environmental Microbiology. 67 (5): 2345–2353. Bibcode:2001ApEnM..67.2345S. doi:10.1128/AEM.67.5.2345-2353.2001. PMC 92876. PMID 11319121. ^ Jump up to: a b c Sato S, Kodama M, Ogata T, Saitanu K, Furuya M, Hirayama K, Kakinuma K (1997). "Saxitoxin as a toxic principle of a freshwater puffer, Tetraodon fangi, in Thailand". Toxicon. 35 (1): 137–140. Bibcode:1997Txcn...35..137S. doi:10.1016/S0041-0101(96)00003-7. PMID 9028016. ^ Landsberg JH, Hall S, Johannessen JN, White KD, Conrad SM, Abbott JP, Flewelling LJ, Richardson RW, Dickey RW, Jester EL, Etheridge SM, Deeds JR, Van Dolah FM, Leighfield TA, Zou Y, Beaudry CG, Benner RA, Rogers PL, Scott PS, Kawabata K, Wolny JL, Steidinger KA (2006). "Saxitoxin Puffer Fish Poisoning in the United States, with the First Report of Pyrodinium bahamense as the Putative Toxin Source". Environmental Health Perspectives. 114 (10): 1502–1507. Bibcode:2006EnvHP.114.1502L. doi:10.1289/ehp.8998. PMC 1626430. PMID 17035133. ^ Deeds JR, Landsberg JH, Etheridge SM, Pitcher GC, Longan SW (2008). "Non-Traditional Vectors for Paralytic Shellfish Poisoning". Marine Drugs. 6 (2): 308–348. doi:10.3390/md6020308. PMC 2525492. PMID 18728730. ^ Lagos NS, Onodera H, Zagatto PA, Andrinolo D́, Azevedo SM, Oshima Y (1999). "The first evidence of paralytic shellfish toxins in the freshwater cyanobacterium Cylindrospermopsis raciborskii, isolated from Brazil". Toxicon. 37 (10): 1359–1373. Bibcode:1999Txcn...37.1359L. doi:10.1016/S0041-0101(99)00080-X. PMID 10414862. ^ For a more comprehensive discussion of TTX-producing bacterial species associated with metazoans from which the toxin has been isolated or toxicity observed, and biosynthesis, see Chau R, Kalaitzis JA, Neilan BA (Jul 2011). "On the origins and biosynthesis of tetrodotoxin" (PDF). Aquatic Toxicology. 104 (1–2): 61–72. Bibcode:2011AqTox.104...61C. doi:10.1016/j.aquatox.2011.04.001. PMID 21543051. Archived from the original (PDF) on 2016-03-05. Retrieved 2022-04-10. ^ Bordner J., Thiessen W. E., Bates H. A., Rapoport H. (1975). "The structure of a crystalline derivative of saxitoxin. The structure of saxitoxin". Journal of the American Chemical Society. 97 (21): 6008–12. Bibcode:1975JAChS..97.6008B. doi:10.1021/ja00854a009. PMID 1176726. ^ Schantz E. J., Ghazarossian V. E., Schnoes H. K., Strong F. M., Springer J. P., Pezzanite J. O., Clardy J. (1975). "The structure of saxitoxin". Journal of the American Chemical Society. 97 (5): 1238–1239. Bibcode:1975JAChS..97.1238S. doi:10.1021/ja00838a045. PMID 1133383. ^ Bates H. A., Kostriken R., Rapoport H. (1978). "A chemical assay for saxitoxin. Improvements and modifications". Journal of Agricultural and Food Chemistry. 26 (1): 252–4. Bibcode:1978JAFC...26..252B. doi:10.1021/jf60215a060. PMID 621331. ^ Tanino H., Nakata T., Kaneko T., Kishi Y. (1997). "A stereospecific total synthesis of d,l-saxitoxin". Journal of the American Chemical Society. 99 (8): 2818–9. Bibcode:1977JAChS..99.2818T. doi:10.1021/ja00450a079. PMID 850038. ^ Bhonde V. R., Looper R. E. (2011). "A stereocontrolled synthesis of (+)-saxitoxin". Journal of the American Chemical Society. 133 (50): 20172–4. Bibcode:2011JAChS.13320172B. doi:10.1021/ja2098063. PMC 3320040. PMID 22098556. ^ Fleming J. J., McReynolds M. D., Du Bois J. (2007). "(+)-Saxitoxin: a first and second generation stereoselective synthesis". Journal of the American Chemical Society. 129 (32): 9964–75. Bibcode:2007JAChS.129.9964F. doi:10.1021/ja071501o. PMID 17658800. ^ Handbook of toxicology of chemical warfare agents. Gupta, Ramesh C. (Ramesh Chandra), 1949- (Second ed.). London: Academic Press. 21 January 2015. p. 426. ISBN 978-0-12-800494-4. OCLC 903965588.{{cite book}}: CS1 maint: others (link) ^ Huot RI, Armstrong DL, Chanh TC (June 1989). "Protection against nerve toxicity by monoclonal antibodies to the sodium channel blocker tetrodotoxin". Journal of Clinical Investigation. 83 (6): 1821–1826. doi:10.1172/JCI114087. PMC 303901. PMID 2542373. ^ Jump up to: a b Stüken A, Orr R, Kellmann R, Murray S, Neilan B, Jakobsen K (18 May 2011). "Discovery of Nuclear-Encoded Genes for the Neurotoxin Saxitoxin in Dinoflagellates". PLOS ONE. 6 (5): e20096. Bibcode:2011PLoSO...620096S. doi:10.1371/journal.pone.0020096. PMC 3097229. PMID 21625593.{{cite journal}}: CS1 maint: article number as page number (link) ^ Jump up to: a b Kellmann R, Mihali TK, Jeon YJ, Pickford R, Pomati F, Neilan BA (2008). "Biosynthetic Intermediate Analysis and Functional Homology Reveal a Saxitoxin Gene Cluster in Cyanobacteria". Applied and Environmental Microbiology. 74 (13): 4044–4053. Bibcode:2008ApEnM..74.4044K. doi:10.1128/AEM.00353-08. PMC 2446512. PMID 18487408. ^ Patocka J, Stredav L (April 23, 2002). Price R (ed.). "Brief Review of Natural Nonprotein Neurotoxins". ASA Newsletter. 02–2 (89): 16–23. ISSN 1057-9419. Retrieved 26 May 2012. ^ Benton BJ, Keller SA, Spriggs DL, Capacio BR, Chang FC (1998). "Recovery from the lethal effects of saxitoxin: A therapeutic window for 4-aminopyridine (4-AP)". Toxicon. 36 (4): 571–588. Bibcode:1998Txcn...36..571B. doi:10.1016/s0041-0101(97)00158-x. PMID 9643470. ^ Chang FC, Spriggs DL, Benton BJ, Keller SA, Capacio BR (1997). "4-Aminopyridine reverses saxitoxin (STX)- and tetrodotoxin (TTX)-induced cardiorespiratory depression in chronically instrumented guinea pigs". Fundamental and Applied Toxicology. 38 (1): 75–88. doi:10.1006/faat.1997.2328. PMID 9268607. S2CID 17185707. ^ Chen H, Lin C, Wang T (1996). "Effects of 4-Aminopyridine on Saxitoxin Intoxication". Toxicology and Applied Pharmacology. 141 (1): 44–48. doi:10.1006/taap.1996.0258. PMID 8917674. ^ "Paralytic shellfish poisoning (PSP)" (PDF). Fish Dept. Sabah Malaysia. Archived from the original (PDF) on October 25, 2021. Retrieved April 10, 2022. ^ Stewart CE (2006). Weapons of Mass Casualties and Terrorism Response Handbook. Jones & Bartlett Learning. p. 175. ISBN 978-0-7637-2425-2. Retrieved 4 May 2015. ^ Jump up to: a b c Unauthorized Storage of Toxic Agents. Church Committee Reports. Vol. 1. The Assassination Archives and Research Center (AARC). 1975–1976. p. 7. ^ Jump up to: a b Wheelis M, Rozsa L, Dando M (2006). Deadly Cultures: Biological Weapons since 1945. President and Fellows of Harvard College. p. 39. ISBN 978-0-674-01699-6. Retrieved 4 May 2015. ^ Mauroni AJ (2000). America's Struggle with Chemical-biological Warfare. 88 Post Road West, Westport, CT 06881: Praeger Publishers. p. 50. ISBN 978-0-275-96756-7. Retrieved 4 May 2015.{{cite book}}: CS1 maint: location (link) ^ "Saxitoxin fact sheet" (PDF). Organisation for the Prohibition of Chemical Weapons (OPCW). June 2014. External links [edit] Paralytic Shellfish Poisoning Neil Edwards. The Chemical Laboratories. School of Chemistry, Physics & Environmental Science. University of Sussex at Brighton. Saxitoxin - from food poisoning to chemical warfare Toxic cyanobacteria in water: A guide to their public health consequences, monitoring and management. Edited by Ingrid Chorus and Jamie Bartram, 1999. Published by World Health Organization. ISBN 0-419-23930-8 \| v t e Cyanotoxins \| \| --- \| \| Neurotoxins \| Anatoxin-a Guanitoxin Antillatoxin BMAA Kalkitoxin Saxitoxin \| \| Hepatotoxins \| Cylindrospermopsin Microcystins (e.g. Microcystin-LR) Nodularin \| \| Other \| Aplysiatoxin Apratoxin A Caldoramide Coibamide A Cyanobacterin Cyanopeptolin Cyclamide Debromoaplysiatoxin Lyngbyatoxin-a Microviridin Aetokthonotoxin \| \| Category \| \| \| v t e Agents used in chemical warfare incapacitation riot control \| \| --- \| \| Blood agents \| Cyanogen Cyanogen bromide Cyanogen chloride (CK) Hydrogen cyanide (AC) Arsine Cacodyl cyanide Cacodyl oxide Hydrogen sulfide Phosphine Carbon monoxide Phosphorus trifluoride Methyl cyanoformate Iron pentacarbonyl Nickel tetracarbonyl 2,3,7,8-Tetrachlorodibenzodioxin Glycolonitrile Lactonitrile Acetone cyanohydrin Stibine Chloral cyanohydrin \| \| Blister agents \| \| \| \| --- \| \| Arsenicals \| - Ethyldichloroarsine (ED) - Methyldichloroarsine (MD) - Phenyldichloroarsine (PD) - Lewisite (L) - Lewisite 2 (L2) - Lewisite 3 (L3) \| \| Sulfur mustards \| Levinstein mustard (EA-229) T Q CEES \| \| Nitrogen mustards \| HN1 HN2 HN3 TL-301 \| \| Nettle agents \| Phosgene oxime (CX) \| \| Other \| KB-16 Dibutylchloromethyltin chloride Selenium oxychloride Phenacyl bromide Dimethyl acetylenedicarboxylate Dimethyl fumarate Chalcone hydrazone Methylnitronitrosoguanidine Norcantharidin Cantharidin T-2 mycotoxin \| \| \| Nerve agents \| \| \| \| --- \| \| G-agents \| Tabun (GA) Sarin (GB) Chlorosarin (ClGB) Thiosarin (SGB) Soman (GD) Chlorosoman (ClGD) Ethylsarin (GE) GH Cyclosarin (GF) GP Fluorotabun EA-1356 EA-4352 Crotylsarin \| \| V-agents \| EA-2192 EA-3148 VE VG VM VP VR VS VX EA-1763 Chinese VX V-sub x (GD-7) \| \| GV agents \| GV (EA-5365) \| \| Novichok agents \| A-208 A-232 A-234 A-242 A-262 C01-A035 C01-A039 C01-A042 \| \| Carbamates \| Dimethylcarbamoyl fluoride EA-3887 EA-3887A EA-3966 EA-3990 EA-4056 T-1123 T-1152 T-1194 Octamethylene-bis(5-dimethylcarbamoxyisoquinolinium bromide) TL-599 TL-1238 TL-1299 TL-1317 Miotine (AR-28/T-1843) 3152 CT 4-686-293-01 (Agent 1-10) \| \| Other \| Diisopropyl fluorophosphate Dicyclohexyl phosphorofluoridate EA-2012 EA-2054 EA-2098 EA-2613 2-Ethoxycarbonyl-1-methylvinyl cyclohexyl methylphosphonate Neopentylene fluorophosphate Selenophos Phospholine R-16661 Ro 3-0422 Methanesulfonyl fluoride Dimefox (TL-792) MSPI \| \| Precursors \| Acetonitrile AT ATO AlP A.P.C. complex Chlorosarin Chlorosoman Cyclohexanol 1,8-Dibromooctane N,N-Diisopropylaminoethanol (KB) EA-1250 DIHP ZS DEHP EA-1224 Dimethylamidophosphoric dichloride Dimethylamidophosphoric dicyanide DMHP Ethylphosphonoselenoic dichloride Formaldoxime Nital 4-Hydroxycoumarin Isopropyl alcohol (TB) Methyldichlorophosphine (SW) Methylphosphonyl difluoride (difluoro) (DF) Methylphosphonyl dichloride (dichloro) Nitromethane OPA mixture Phosphoryl chloride Phosphorus pentachloride Phosphorus trichloride (TH) Pinacolone Pinacolyl alcohol Phenacyl chloride QL QL sulfide 2,4,5-Trichlorophenol 3,3,5-Trimethylcyclohexanol Triethyl phosphite Trimethyl phosphite TC TG \| \| \| Neurotoxins \| Anatoxin-a Saxitoxin (TZ) Coniine Bungarotoxin Botulinum toxin (BTX) Tetanospasmin (TeNT) Ryanodine Ciguatoxin (CTX) Guanitoxin (GTX) Chlorophenylsilatrane Palytoxin (PTX) Maitotoxin (MTX) Tetrodotoxin Veratridine Aconitine Brevetoxin (PbTX) Strychnine Antillatoxin (ATX) Tetraethyllead Dimethylmercury HN1 hydrochloride HN2 hydrochloride HN3 hydrochloride A-8564 Picrotoxin Gelsemine Sulfuryl fluoride Tremorine Oxotremorine Batrachotoxin Tetramethylenedisulfotetramine (TETS) Bicyclic phosphates + IPTBO + TBPO + TBPS Cloflubicyne Trimethylolpropane phosphite Domoic acid \| \| Pulmonary/ choking agents \| Chlorine Bromine Phosgene (CG) Fluorine Perfluoroisobutene Chloropicrin (PS) Dimethyl(trifluoromethylthio)arsine Diphosgene (DP) Disulfur decafluoride (Z) Acrolein Ethyl bromoacetate Perchloromethyl mercaptan Phenylcarbylamine chloride Tetranitromethane Tetrachlorodinitroethane Chlorine trifluoride Perchloryl fluoride Cadmium oxide Cadmium chloride Mercuric chloride Selenium dioxide Selenoyl fluoride Trifluoronitrosomethane Trichloronitrosomethane Nitric oxide Nitrogen dioxide Dinitrogen tetroxide Sulfur dioxide Phosphorus trichloride Methyl isocyanate Ethenone Methyl vinyl ketone Trifluoroacetyl chloride Salcomine Fluomine Uranium hexafluoride Diborane Green Cross \| \| Vomiting agents \| Adamsite (DM) Chloropicrin Litharge-glycerine Diphenylchlorarsine Diphenylcyanoarsine Cacodyl cyanide o-Dianisidine Antimony potassium tartrate Lobeline \| \| Incapacitating agents \| BZ (CS-4030) Apomorphine Butyrophenone EA-4941 (CS-4640) Etorphine EA-2092 CS-4297 Etoxadrol Dimethylheptylpyran (DMHP) Canbisol EA-2148 EA-3167 Nicotine EA-3443 Methaqualone Pethidine EA-3580 Ibogaine EA-3834 Kolokol-1 LSD-25 PAVA spray Psilocybin Sleeping gas Carfentanil JB-318 JB-336 CS-27349 CAR-226,086 CAR-301,060 CAR-302,196 CAR-302,282 CAR-302,668 Benperidol Desflurane Enflurane Bufotenin Isoflurane Halothane Sevoflurane Pentazocine Procarbazine Fluphenazine Chlorpromazine \| \| Lachrymatory agents \| Xylyl bromide Pepper spray (OC) Mace (spray) Phorbol esters Ingenol mebutate RTX CN CS CR CNS Benzyl chloride Benzyl bromide Benzyl iodide Bromobenzyl cyanide Thiophosgene Chloroacetone Bromoacetone Bromomethyl ethyl ketone Acrolein Phenacyl bromide Chloroacetophenone oxime Ethyl bromoacetate Ethyl iodoacetate Iodoacetone Allyl isothiocyanate Hexamethylene diisocyanate Crotonaldehyde Croton oil DRC-5593 \| \| Malodorant agents \| Thioacetone Allicin Skatole Cadaverine Putrescine \| \| Cornea-clouding agents \| Lewisite CX KB-16 Methyl cyanoacrylate N-Methylmorpholine Allyl alcohol Osmium tetroxide Acrolein Cacodyl cyanide Diglycolic acid esters \| \| Biological toxins \| Abrin Aconitine Cyclopiazonic acid Histrionicotoxins Aflatoxins Anatoxin-a Batrachotoxin Botulinum toxin Brevetoxin Ciguatoxin Domoic acid Enterotoxin type B Grayanotoxin Guanitoxin Maitotoxin Modeccin Palytoxin Ricin Saxitoxin Shiga toxin T-2 mycotoxin Tetanospasmin Tetrodotoxin Volkensin Veratridine \| \| Tumor promoting agents \| Phorbol esters 12-O-Tetradecanoylphorbol-13-acetate Croton oil \| \| Other \| Methyl fluoroacetate Napalm (variants and mixtures) Fluoroethyl fluoroacetate Depleted uranium + post-combustion uranium oxides Plutonium and its compounds Polonium White phosphorus \| \| List of chemical warfare agents CB military symbol \| \| \| v t e Neurotoxins \| \| --- \| \| Animal toxins \| Batrachotoxin Bestoxin Birtoxin Bungarotoxin Charybdotoxin Conotoxin Fasciculin Huwentoxin Poneratoxin Saxitoxin Tetrodotoxin Vanillotoxin Spooky toxin (SsTx) Epibatidine Zetekitoxin AB Dendrotoxin \| \| Bacterial \| Botulinum toxin Tetanospasmin \| \| Cyanotoxins \| Anatoxin-a Guanitoxin BMAA Saxitoxin \| \| Plant toxins \| Aconitine Bicuculline Penitrem A Picrotoxin Strychnine Tutin Rotenone Ginkgotoxin Cicutoxin Oenanthotoxin Thujone Volkensin Veratridine \| \| Mycotoxins \| Ibotenic acid Muscarine Muscimol \| \| Pesticides \| Fenpropathrin Tetramethylenedisulfotetramine Bromethalin Crimidine Methamidophos Endosulfan Fipronil Phenylsilatrane Chlorophenylsilatrane Sulfuryl fluoride Mipafox Schradan Dimefox \| \| Nerve agents \| Cyclosarin EA-3148 Novichok agent Sarin Soman Tabun VE VG VM VP VR VX GV EA-3990 EA-4056 T-1123 Octamethylene-bis(5-dimethylcarbamoxyisoquinolinium bromide) Fluorotabun Chinese VX EA-2192 \| \| Bicyclic phosphates \| TBPS TBPO IPTBO \| \| Cholinergic neurotoxins \| Acetylcholine mustard Catecholine Choline mustard Ethylcholine mustard Hemicholinium mustard \| \| Psychoactive drugs \| Alcohol Nitrous oxide \| \| Other \| Dimethylcadmium Dimethylmercury Toxopyrimidine IDPN Tetraethyllead \| \| v t e Plankton \| \| \| --- \| About plankton \| Algal bloom CLAW hypothesis High lipid content microalgae Holoplankton Marine microorganisms Meroplankton Mycoplankton Milky seas effect Paradox of the plankton Planktivore Planktology Red tide Spring bloom Thin layers More... \| \| \| By size \| Eukaryotic picoplankton Heterotrophic picoplankton Marine microplankton Microphyte (microalgae) Nanophytoplankton Photosynthetic picoplankton Picobiliphyte Picoeukaryote Picoplankton \| \| Bacterioplankton \| Aeromonas salmonicida Cyanobacteria Cyanobiont Cyanotoxin Enteric redmouth disease Flavobacterium Flavobacterium columnare Pelagibacter ubique Marine bacteriophage SAR11 clade Streptococcus iniae \| \| Phytoplankton \| \| \| \| --- \| \| Auxospore Axodine Bacteriastrum Chaetoceros Chaetocerotaceae Coccolithophore Emiliania huxleyi Eustigmatophyte Frustule Stramenopile Nannochloropsis Navicula Prasinophyceae Raphidophyte Thalassiosira pseudonana \| \| \| Diatom orders \| Centrales Pennales + Classes: Coscinodiscophyceae + Fragilariophyceae + Bacillariophyceae \| \| \| Flagellates \| Brevetoxin Choanoflagellates Dinoflagellates Flagellum Pfiesteria piscicida Saxitoxin Symbiodinium Velvet (fish disease) \| \| Zooplankton \| \| \| \| --- \| \| Chaetognatha Ciguatera Ctenophora Gelatinous zooplankton Ichthyoplankton Jellyfish Marine larvae Crustacean larvae Salmon louse Sea louse \| \| \| Copepod orders \| Calanoida Canuelloida Cyclopoida Gelyelloida Harpacticoida Misophrioida Monstrilloida Mormonilloida Platycopioida Siphonostomatoida More... \| \| \| Related topics \| Aeroplankton Algaculture Algal mat Algal nutrient solutions Artificial seawater Autotrophs Biological pump Diel vertical migration Dimethylsulfoniopropionate f-ratio Fish diseases and parasites Heterotroph HNLC Macroalgae Manta trawl Marine mucilage Microbial mat Ocean acidification Marine microorganisms Marine primary production Pseudoplankton Stromatolite Tychoplankton Zoid C-MORE CPR AusCPR MOCNESS SCAR \| \| v t e Fish diseases and parasites \| \| --- \| \| Pathogens \| Aeromonas salmonicida Nervous necrosis virus Columnaris Enteric redmouth Fin rot Fish dropsy Flavobacterium Hematopoietic necrosis Heterosigma akashiwo Hole in the head Hypodermal and hematopoietic necrosis Infectious pancreatic necrosis Koi herpes virus Mycobacterium marinum (Aquarium granuloma) Novirhabdovirus Pfiesteria piscicida Photobacterium damselae ssp piscicida Salmon anemia Streptococcus iniae Spring viraemia of carp Taura syndrome UDN VHS White spot Yellowhead \| \| Parasites \| Abergasilus Amoebic gill disease Anisakis Carp lice Ceratomyxa shasta Clinostomum marginatum Dactylogyrus vastator Diphyllobothrium Cymothoa exigua Eustrongylidosis Epizootic ulcerative syndrome Flukes Glugea Gyrodactylus salaris Henneguya zschokkei Ich (freshwater) Ich (marine) Kudoa thyrsites Lernaeocera branchialis Microsporidia Monogenea Myxobolus cerebralis Myxosporea Nanophyetus salmincola Pseudorhabdosynochus spp. Salmon lice Saprolegnia Schistocephalus solidus Sea louse Sphaerothecum destruens Swim bladder disease Tetracapsuloides bryosalmonae Velvet Xenoma \| \| Fish groups \| Diseases and parasites in cod Diseases and parasites in salmon Disease in ornamental fish List of aquarium diseases \| \| Related topics \| Amnesic shellfish poisoning Brevetoxin Ciguatera Diarrheal shellfish poisoning Fish kill Marine viruses Neurotoxic shellfish poisoning Paralytic shellfish poisoning Saxitoxin Scombroid food poisoning \| \| v t e Toxins + cardiotoxin + cytotoxin + enterotoxin + hemotoxin + hepatotoxin + neurotoxin + phototoxin \| \| --- \| \| Bacterial toxins \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| --- --- --- --- --- --- --- --- --- --- --- \| \| Exotoxin \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| --- --- --- --- --- --- --- --- \| \| Gram positive \| \| \| \| \| \| \| \| --- --- --- \| \| Bacilli \| \| \| \| --- \| \| Clostridium: \| tetani + Tetanospasmin + Tetanolysin perfringens + Alpha toxin + Enterotoxin difficile + A + B botulinum + Botox \| \| Other: \| Anthrax toxin Listeriolysin O \| \| \| Cocci \| \| \| \| --- \| \| Streptolysin Leukocidin + Panton–Valentine leukocidin \| \| \| Staphylococcus \| Staphylococcus aureus alpha/beta/delta Exfoliatin Toxic shock syndrome toxin Staphylococcal Enterotoxin B (SEB) \| \| \| Actinomycetota \| Cord factor Diphtheria toxin \| \| \| Gram negative \| Shiga toxin/Verotoxin E. coli heat-stable enterotoxin Cholera toxin/Heat-labile enterotoxin Pertussis toxin Pseudomonas exotoxin Extracellular adenylate cyclase \| \| Mechanisms \| type I + Superantigen type II + Pore-forming toxin type III + AB toxin/AB5 \| \| \| Endotoxin \| Lipopolysaccharide + Lipid A Bacillus thuringiensis delta endotoxin + Cry1Ac + Cry3Bb1 Other B. thuringiensis toxins + Cry6Aa + Cry34Ab1 \| \| Virulence factor \| Clumping factor A Fibronectin binding protein A \| \| \| Mycotoxins \| Aflatoxin Amatoxin (alpha-amanitin, beta-amanitin, gamma-amanitin, epsilon-amanitin, Amanullin, Amanullinic acid, Amaninamide, Amanin, Proamanullin) beta-Nitropropionic acid Citrinin Cytochalasin Ergotamine Fumonisin (Fumonisin B1, Fumonisin B2, Fumonisin B3, Fumonisin B4) Gliotoxin Ibotenic acid Lolitrem B Muscarine Muscimol Orellanine Ochratoxin Patulin Phallotoxin (Phalloidin) Satratoxin-H Sterigmatocystin T-2 mycotoxin Trichothecene Vomitoxin Zeranol Zearalenone \| \| Plant toxins \| Amygdalin Anisatin Antiarin Brucine Chaconine Cicutoxin Coniine Daphnin Delphinine Divicine Djenkolic acid Falcarinol Gossypol Helenalin Ledol Linamarin Lotaustralin Mimosine Oenanthotoxin Oleandrin Persin Protoanemonin Pseudaconitine Retronecine Resiniferatoxin Scopolamine Solamargine Solanidine Solanine Solasodamine Solasodine Solasonine Solauricidine Solauricine Strychnine Swainsonine Tagetitoxin Tinyatoxin Tomatine Toxalbumin + Abrin + Ricin Tutin \| \| Invertebrate toxins \| \| \| \| --- \| \| Scorpion: \| Androctonus australis hector insect toxin Charybdotoxin Maurotoxin Agitoxin Margatoxin Slotoxin Scyllatoxin Hefutoxin HgeTx1 HsTx1 Kaliotoxin Lq2 Birtoxin Bestoxin BmKAEP Phaiodotoxin Imperatoxin Pi3 \| \| Spider: \| Latrotoxin + Alpha-latrotoxin CSTX Cupiennins PhTx3 Stromatoxin Vanillotoxin Huwentoxin \| \| Mollusca: \| Conotoxin Eledoisin Onchidal Saxitoxin Tetrodotoxin \| \| \| Vertebrate toxins \| \| \| \| --- \| \| Fish: \| Ciguatoxin Tetrodotoxin \| \| Amphibian: \| (+)-Allopumiliotoxin 267A Batrachotoxin Bufotoxins + Arenobufagin + Bufotalin + Bufotenin + Cinobufagin + Marinobufagin Epibatidine Histrionicotoxin Pumiliotoxin 251D Samandarin Samandaridine Tarichatoxin Zetekitoxin AB \| \| Reptile/ Snake venom: \| Bungarotoxin + α-Bungarotoxin + β-Bungarotoxin Calciseptine Taicatoxin Calcicludine Cardiotoxin III \| \| \| note: some toxins are produced by lower species and pass through intermediate species \| \| \| Category \| \| \| v t e Ion channel modulators \| \| --- \| \| Calcium \| \| \| \| \| \| \| \| --- --- --- \| \| VDCCsTooltip Voltage-dependent calcium channels \| \| \| \| --- \| \| Blockers \| L-type-selective: Dihydropyridines: Amlodipine (+telmisartan and indapamide) Aranidipine Azelnidipine Barnidipine Clevidipine Cronidipine Darodipine Dexniguldipine Elgodipine Elnadipine Felodipine Flordipine Furnidipine Iganidipine Isradipine Lacidipine Lemildipine Lercanidipine Levamlodipine Levniguldipine Manidipine Mepirodipine Mesudipine Nicardipine Nifedipine Niguldipine Niludipine Nilvadipine Nimodipine Nisoldipine Nitrendipine Olradipine Oxodipine Palonidipine Pranidipine Ryodipine (riodipine) Sagandipine Sornidipine Teludipine Tiamdipine Trombodipine Vatanidipine; Diltiazem derivatives: Clentiazem Diltiazem Iprotiazem Nictiazem Siratiazem; Phenylalkylamines: Anipamil Dagapamil Devapamil Dexverapamil Emopamil Etripamil Falipamil Gallopamil Levemopamil Nexopamil Norverapamil Ronipamil Tiapamil Verapamil; Others: AH-1058 Brinazarone Budiodarone Celivarone Cyproheptadine Dronedarone Fantofarone SR-33805 Tetrahydropalmatine N-type-selective: ω-Conotoxins ω-Conotoxin GVIA Caroverine Huwentoxin XVI Leconotide (ω-conotoxin CVID) PD-173212 Ralfinamide Safinamide Z160 Ziconotide (ω-conotoxin MVIIA) P-type-selective: ω-Agatoxin IVA ω-Agatoxin IVB R-type-selective: SNX-482 T-type-selective: ABT-639 ML-218 Niflumic acid NNC 55-0396 ProTx I Z944 Zonisamide Non-selective: ω-Agatoxin TK ω-Conotoxin MVIIC Benidipine Bepridil Cilnidipine Cinnarizine Dotarizine Efonidipine Flunarizine Lamotrigine Levetiracetam Lomerizine Loperamide Mibefradil NP078585 Ruthenium red TROX-1 α2δ subunit-selective (gabapentinoids): 4-Methylpregabalin Arbaclofen Arbaclofen placarbil Atagabalin Baclofen Gabapentin Gabapentin enacarbil Imagabalin Mirogabalin PD-200,347 PD-217,014 PD-299,685 Phenibut Pregabalin Others/unsorted: Bencyclane Berbamine Bevantolol Canadine Carboxyamidotriazole Cycleanine Dauricine Dimeditiapramine Diproteverine Enpiperate Eperisone Elpetrigine Ethadione Ethanol (alcohol) Ethosuximide Fasudil Fendiline Fostedil Imepitoin JTV-519 Lidoflazine Magnesium Manoalide Mesuximide Monatepil Naftopidil Ochratoxin A Osthol Otilonium bromide Paramethadione Phensuximide Pinaverium bromide Prenylamine Rhynchophylline Sesamodil Silperisone Sipatrigine Terodiline Tetrandrine Tolperisone Trimethadione Valperinol \| \| Activators \| L-type-selective: Bay K8644 \| \| \| \| Potassium \| \| \| \| \| \| \| \| --- --- --- \| \| VGKCsTooltip Voltage-gated potassium channels \| \| \| \| --- \| \| Blockers \| 3,4-Diaminopyridine (amifampridine) 4-Aminopyridine (fampridine/dalfampridine) Adekalant Almokalant Amiodarone Azimilide Bretylium Bunaftine Charybdotoxin Clamikalant Conotoxins Dalazatide Dendrotoxin Dofetilide Dronedarone E-4031 Hanatoxin HgeTx1 HsTx1 Ibutilide Inakalant Kaliotoxin Linopirdine Lolitrem B Maurotoxin Nifekalant Notoxin Paxilline Pinokalant Quinidine ShK-186 Sotalol Tedisamil Terikalant Tetraethylammonium Vernakalant hERG (KCNH2, Kv11.1)-specific: Ajmaline Amiodarone AmmTX3 Astemizole Azaspiracid AZD1305 Azimilide Bedaquiline BeKm-1 BmTx3 BRL-32872 Chlorpromazine Cisapride Clarithromycin Darifenacin Dextropropoxyphene Diallyl trisulfide Domperidone E-4031 Ergtoxins Erythromycin Gigactonine Haloperidol Ketoconazole Norpropoxyphene Orphenadrine Pimozide PNU-282,987 Promethazine Quinidine Ranolazine Roxithromycin Sertindole Solifenacin Tamulotoxin Terodiline Terfenadine Thioridazine Tolterodine Vanoxerine Vernakalant KCNQ (Kv7)-specific: Linopirdine XE-991 Spooky toxin (SsTx) \| \| Activators \| KCNQ (Kv7)-specific: Flupirtine Retigabine \| \| \| IRKsTooltip Inwardly rectifying potassium channel \| \| \| \| --- \| \| Blockers \| KATPTooltip ATP-sensitive potassium channel-specific: Acetohexamide Carbutamide Chlorpropamide Glibenclamide (glyburide) Glibornuride Glicaramide Gliclazide Glimepiride Glipizide Gliquidone Glisoxepide Glyclopyramide Glycyclamide Metahexamide Mitiglinide Nateglinide Repaglinide Tolazamide Tolbutamide GIRKTooltip G protein-coupled inwardly rectifying potassium channel-specific: Barium Caramiphen Cloperastine Clozapine Dextromethorphan Ethosuximide Ifenprodil Tertiapin Tipepidine \| \| Activators \| KATPTooltip ATP-sensitive potassium channel-specific: Aprikalim Bimakalim Cromakalim Diazoxide Emakalim Levcromakalim Mazokalim Minoxidil Minoxidil sulfate Naminidil Nicorandil Pinacidil Rilmakalim Sarakalim GIRKTooltip G protein-coupled inwardly rectifying potassium channel-specific: ML-297 (VU0456810) \| \| \| KCaTooltip Calcium-activated potassium channel \| \| \| \| --- \| \| Blockers \| BKCa-specific: Ethanol (alcohol) GAL-021 \| \| Activators \| BKCa-specific: Flufenamic acid Meclofenamic acid Niflumic acid Nimesulide Rottlerin (mallotoxin) Tolfenamic acid \| \| \| K2PsTooltip Tandem pore domain potassium channel \| \| \| \| --- \| \| Blockers \| 12-O-Tetradecanoylphorbol-13-acetate Arachidonic acid Fluoxetine Norfluoxetine \| \| Activators \| \| \| \| \| Sodium \| \| \| \| \| \| \| \| --- --- --- \| \| VGSCsTooltip Voltage-gated sodium channels \| \| \| \| --- \| \| Blockers \| Antianginals: Ranolazine Antiarrhythmics (class I): Ajmaline Aprindine Disopyramide Dronedarone Encainide Flecainide Lidocaine Lorajmine Lorcainide Mexiletine Moricizine Pilsicainide Prajmaline Procainamide Propafenone Quinidine Sparteine Tocainide Anticonvulsants: Acetylpheneturide Carbamazepine Cenobamate Chlorphenacemide Elpetrigine Eslicarbazepine acetate Ethotoin Fosphenytoin Lamotrigine Lacosamide Licarbazepine Mephenytoin Oxcarbazepine Oxitriptyline Phenacemide Pheneturide Phenytoin Rufinamide Sipatrigine Topiramate Sodium valproate Valnoctamide Valproate pivoxil Valproate semisodium Valproic acid Valpromide Zonisamide Local anesthetics: pFBT Amylocaine Articaine Benzocaine Bupivacaine (Levobupivacaine, Ropivacaine) Butacaine Butamben Chloroprocaine Cinchocaine Cocaine Cyclomethycaine Dimethocaine Diphenhydramine Etidocaine Hexylcaine Iontocaine Lidocaine Mepivacaine Meprylcaine Metabutoxycaine Orthocaine Piperocaine Prilocaine Procaine Propoxycaine Proxymetacaine Risocaine Tetracaine Trimecaine Analgesics: AZD-3161 DSP-2230 Funapide GDC-0276 NKTR-171 PF-04531083 PF-05089771 Ralfinamide Raxatrigine RG7893 (GDC-0287) Suzetrigine Toxins: Conotoxins Neosaxitoxin Saxitoxin Tetrodotoxin Zetekitoxin AB Others: Buprenorphine Evenamide Menthol (mint) Safinamide Tricyclic antidepressants \| \| Activators \| Aconitine Atracotoxins (ω-Atracotoxin, Robustoxin, Versutoxin) Batrachotoxin Ciguatoxins Grayanotoxins Poneratoxin \| \| \| ENaCTooltip Epithelial sodium channel \| \| \| \| --- \| \| Blockers \| Amiloride Benzamil Triamterene \| \| Activators \| \| \| \| ASICsTooltip Acid-sensing ion channel \| \| \| \| --- \| \| Blockers \| A-317567 Amiloride Aspirin Ibuprofen PcTX1 \| \| \| \| Chloride \| \| \| \| \| \| \| \| --- --- --- \| \| CaCCsTooltip Calcium-activated chloride channel \| \| \| \| --- \| \| Blockers \| Crofelemer DIDS Ethacrynic acid Flufenamic acid Fluoxetine Furosemide Glibenclamide Mefloquine Mibefradil Niflumic acid \| \| Activators \| \| \| \| CFTRTooltip Cystic fibrosis transmembrane conductance regulator \| \| \| \| --- \| \| Blockers \| Glibenclamide Lonidamine Piretanide \| \| Activators \| 1,7-Phenanthroline 1,10-Phenanthroline 4,7-Phenanthroline 7,8-Benzoquinoline Ivacaftor Phenanthridine \| \| \| Unsorted \| \| \| \| --- \| \| Blockers \| Bumetanide Flufenamic acid Meclofenamic acid Mefenamic acid Mepacrine Niflumic acid Talniflumate Tolfenamic acid Trifluoperazine \| \| \| \| Others \| \| \| \| --- \| \| TRPsTooltip Transient receptor potential channels \| See here instead. \| \| LGICsTooltip Ligand gated ion channels \| See here instead. \| \| \| See also: Receptor/signaling modulators • Transient receptor potential channel modulators \| \| \| \| \| --- \| \| Authority control databases: National \| \| Retrieved from " Categories: Marine neurotoxins Carbamates Guanidine alkaloids Phycotoxins Cyanotoxins Sodium channel blockers Bacterial alkaloids Voltage-gated sodium channel blockers Non-protein ion channel toxins Geminal diols Heterocyclic compounds with 3 rings Nitrogen heterocycles Pyrrolopurines Hidden categories: CS1: unfit URL CS1 maint: others CS1 maint: article number as page number CS1 maint: location Articles with short description Short description is different from Wikidata Articles with changed EBI identifier Articles with changed ChemSpider identifier ECHA InfoCard ID from Wikidata Articles with changed InChI identifier Articles containing unverified chemical infoboxes Short description matches Wikidata Pages displaying short descriptions with no spaces via Module:Annotated link
6449	https://www.theexpertta.com/book-files/OpenStaxUniversityPhysicsVol1/UP1_Ch10.%20Fixed-Axis%20Rotation.pdf	10 \| FIXED-AXIS ROTATION Figure 10.1 Brazos wind farm in west Texas. As of 2012, wind farms in the US had a power output of 60 gigawatts, enough capacity to power 15 million homes for a year. (credit: modification of work by U.S. Department of Energy) Chapter Outline 10.1 Rotational Variables 10.2 Rotation with Constant Angular Acceleration 10.3 Relating Angular and Translational Quantities 10.4 Moment of Inertia and Rotational Kinetic Energy 10.5 Calculating Moments of Inertia 10.6 Torque 10.7 Newton’s Second Law for Rotation 10.8 Work and Power for Rotational Motion Introduction In previous chapters, we described motion (kinematics) and how to change motion (dynamics), and we defined important concepts such as energy for objects that can be considered as point masses. Point masses, by definition, have no shape and so can only undergo translational motion. However, we know from everyday life that rotational motion is also very important and that many objects that move have both translation and rotation. The wind turbines in our chapter opening image are a prime example of how rotational motion impacts our daily lives, as the market for clean energy sources continues to grow. We begin to address rotational motion in this chapter, starting with fixed-axis rotation. Fixed-axis rotation describes the rotation around a fixed axis of a rigid body; that is, an object that does not deform as it moves. We will show how to apply all the ideas we’ve developed up to this point about translational motion to an object rotating around a fixed axis. In the next chapter, we extend these ideas to more complex rotational motion, including objects that both rotate and translate, and objects that do not have a fixed rotational axis. Chapter 10 \| Fixed-Axis Rotation 471 10.1 \| Rotational Variables Learning Objectives By the end of this section, you will be able to: • Describe the physical meaning of rotational variables as applied to fixed-axis rotation • Explain how angular velocity is related to tangential speed • Calculate the instantaneous angular velocity given the angular position function • Find the angular velocity and angular acceleration in a rotating system • Calculate the average angular acceleration when the angular velocity is changing • Calculate the instantaneous angular acceleration given the angular velocity function So far in this text, we have mainly studied translational motion, including the variables that describe it: displacement, velocity, and acceleration. Now we expand our description of motion to rotation—specifically, rotational motion about a fixed axis. We will find that rotational motion is described by a set of related variables similar to those we used in translational motion. Angular Velocity Uniform circular motion (discussed previously in Motion in Two and Three Dimensions) is motion in a circle at constant speed. Although this is the simplest case of rotational motion, it is very useful for many situations, and we use it here to introduce rotational variables. In Figure 10.2, we show a particle moving in a circle. The coordinate system is fixed and serves as a frame of reference to define the particle’s position. Its position vector from the origin of the circle to the particle sweeps out the angle θ , which increases in the counterclockwise direction as the particle moves along its circular path. The angle θ is called the angular position of the particle. As the particle moves in its circular path, it also traces an arc length s. Figure 10.2 A particle follows a circular path. As it moves counterclockwise, it sweeps out a positive angle θ with respect to the x-axis and traces out an arc length s. The angle is related to the radius of the circle and the arc length by (10.1) θ = s r. 472 Chapter 10 \| Fixed-Axis Rotation This OpenStax book is available for free at The angle θ , the angular position of the particle along its path, has units of radians (rad). There are 2π radians in 360°. Note that the radian measure is a ratio of length measurements, and therefore is a dimensionless quantity. As the particle moves along its circular path, its angular position changes and it undergoes angular displacements Δθ. We can assign vectors to the quantities in Equation 10.1. The angle θ →is a vector out of the page in Figure 10.2. The angular position vector r → and the arc length s → both lie in the plane of the page. These three vectors are related to each other by (10.2) s →= θ →× r →. That is, the arc length is the cross product of the angle vector and the position vector, as shown in Figure 10.3. Figure 10.3 The angle vector points along the z-axis and the position vector and arc length vector both lie in the xy-plane. We see that s →= θ →× r →. All three vectors are perpendicular to each other. The magnitude of the angular velocity, denoted by ω , is the time rate of change of the angle θ as the particle moves in its circular path. The instantaneous angular velocity is defined as the limit in which Δt →0 in the average angular velocity ω – = Δθ Δt : (10.3) ω = lim Δt →0 Δθ Δt = dθ dt , where θ is the angle of rotation (Figure 10.2). The units of angular velocity are radians per second (rad/s). Angular velocity can also be referred to as the rotation rate in radians per second. In many situations, we are given the rotation rate in revolutions/s or cycles/s. To find the angular velocity, we must multiply revolutions/s by 2π , since there are 2π radians in one complete revolution. Since the direction of a positive angle in a circle is counterclockwise, we take counterclockwise rotations as being positive and clockwise rotations as negative. We can see how angular velocity is related to the tangential speed of the particle by differentiating Equation 10.1 with respect to time. We rewrite Equation 10.1 as s = rθ. Taking the derivative with respect to time and noting that the radius r is a constant, we have ds dt = d dt(rθ) = θdr dt + rdθ dt = rdθ dt where θdr dt = 0 . Here ds dt is just the tangential speed vt of the particle in Figure 10.2. Thus, by using Equation 10.3, we arrive at Chapter 10 \| Fixed-Axis Rotation 473 (10.4) vt = rω. That is, the tangential speed of the particle is its angular velocity times the radius of the circle. From Equation 10.4, we see that the tangential speed of the particle increases with its distance from the axis of rotation for a constant angular velocity. This effect is shown in Figure 10.4. Two particles are placed at different radii on a rotating disk with a constant angular velocity. As the disk rotates, the tangential speed increases linearly with the radius from the axis of rotation. In Figure 10.4, we see that v1 = r1 ω1 and v2 = r2 ω2 . But the disk has a constant angular velocity, so ω1 = ω2 . This means v1 r1 = v2 r2 or v2 = ⎛ ⎝ r2 r1 ⎞ ⎠v1 . Thus, since r2 > r1 , v2 > v1 . Figure 10.4 Two particles on a rotating disk have different tangential speeds, depending on their distance to the axis of rotation. Up until now, we have discussed the magnitude of the angular velocity ω = dθ/dt, which is a scalar quantity—the change in angular position with respect to time. The vector ω →is the vector associated with the angular velocity and points along the axis of rotation. This is useful because when a rigid body is rotating, we want to know both the axis of rotation and the direction that the body is rotating about the axis, clockwise or counterclockwise. The angular velocity ω → gives us this information. The angular velocity ω →has a direction determined by what is called the right-hand rule. The right-hand rule is such that if the fingers of your right hand wrap counterclockwise from the x-axis (the direction in which θ increases) toward the y-axis, your thumb points in the direction of the positive z-axis (Figure 10.5). An angular velocity ω → that points along the positive z-axis therefore corresponds to a counterclockwise rotation, whereas an angular velocity ω →that points along the negative z-axis corresponds to a clockwise rotation. 474 Chapter 10 \| Fixed-Axis Rotation This OpenStax book is available for free at Figure 10.5 For counterclockwise rotation in the coordinate system shown, the angular velocity points in the positive z-direction by the right-hand-rule. Similar to Equation 10.2, one can state a cross product relation to the vector of the tangential velocity as stated in Equation 10.4. Therefore, we have (10.5) v →= ω →× r →. That is, the tangential velocity is the cross product of the angular velocity and the position vector, as shown in Figure 10.6. From part (a) of this figure, we see that with the angular velocity in the positive z-direction, the rotation in the xy-plane is counterclockwise. In part (b), the angular velocity is in the negative z-direction, giving a clockwise rotation in the xy-plane. Figure 10.6 The vectors shown are the angular velocity, position, and tangential velocity. (a) The angular velocity points in the positive z-direction, giving a counterclockwise rotation in the xy-plane. (b) The angular velocity points in the negative z-direction, giving a clockwise rotation. Chapter 10 \| Fixed-Axis Rotation 475 Example 10.1 Rotation of a Flywheel A flywheel rotates such that it sweeps out an angle at the rate of θ = ωt = (45.0 rad/s)t radians. The wheel rotates counterclockwise when viewed in the plane of the page. (a) What is the angular velocity of the flywheel? (b) What direction is the angular velocity? (c) How many radians does the flywheel rotate through in 30 s? (d) What is the tangential speed of a point on the flywheel 10 cm from the axis of rotation? Strategy The functional form of the angular position of the flywheel is given in the problem as θ(t) = ωt , so by taking the derivative with respect to time, we can find the angular velocity. We use the right-hand rule to find the angular velocity. To find the angular displacement of the flywheel during 30 s, we seek the angular displacement Δθ , where the change in angular position is between 0 and 30 s. To find the tangential speed of a point at a distance from the axis of rotation, we multiply its distance times the angular velocity of the flywheel. Solution a. ω = dθ dt = 45 rad/s . We see that the angular velocity is a constant. b. By the right-hand rule, we curl the fingers in the direction of rotation, which is counterclockwise in the plane of the page, and the thumb points in the direction of the angular velocity, which is out of the page. c. Δθ = θ(30 s) −θ(0 s) = 45.0(30.0 s) −45.0(0 s) = 1350.0 rad . d. vt = rω = (0.1 m)(45.0 rad/s) = 4.5 m/s . Significance In 30 s, the flywheel has rotated through quite a number of revolutions, about 215 if we divide the angular displacement by 2π . A massive flywheel can be used to store energy in this way, if the losses due to friction are minimal. Recent research has considered superconducting bearings on which the flywheel rests, with zero energy loss due to friction. Angular Acceleration We have just discussed angular velocity for uniform circular motion, but not all motion is uniform. Envision an ice skater spinning with his arms outstretched—when he pulls his arms inward, his angular velocity increases. Or think about a computer’s hard disk slowing to a halt as the angular velocity decreases. We will explore these situations later, but we can already see a need to define an angular acceleration for describing situations where ω changes. The faster the change in ω , the greater the angular acceleration. We define the instantaneous angular acceleration α as the derivative of angular velocity with respect to time: (10.6) α = lim Δt →0 Δω Δt = dω dt = d2 θ dt2 , where we have taken the limit of the average angular acceleration, α – = Δω Δt as Δt →0 . The units of angular acceleration are (rad/s)/s, or rad/s2 . In the same way as we defined the vector associated with angular velocity ω →, we can define α →, the vector associated with angular acceleration (Figure 10.7). If the angular velocity is along the positive z-axis, as in Figure 10.5, and dω dt is positive, then the angular acceleration α →is positive and points along the +z- axis. Similarly, if the angular velocity ω →is along the positive z-axis and dω dt is negative, then the angular acceleration is negative and points along the +z -476 Chapter 10 \| Fixed-Axis Rotation This OpenStax book is available for free at axis. Figure 10.7 The rotation is counterclockwise in both (a) and (b) with the angular velocity in the same direction. (a) The angular acceleration is in the same direction as the angular velocity, which increases the rotation rate. (b) The angular acceleration is in the opposite direction to the angular velocity, which decreases the rotation rate. We can express the tangential acceleration vector as a cross product of the angular acceleration and the position vector. This expression can be found by taking the time derivative of v →= ω →× r →and is left as an exercise: (10.7) a →= α →× r →. The vector relationships for the angular acceleration and tangential acceleration are shown in Figure 10.8. Figure 10.8 (a) The angular acceleration is the positive z-direction and produces a tangential acceleration in a counterclockwise sense. (b) The angular acceleration is in the negative z-direction and produces a tangential acceleration in the clockwise sense. We can relate the tangential acceleration of a point on a rotating body at a distance from the axis of rotation in the same way that we related the tangential speed to the angular velocity. If we differentiate Equation 10.4 with respect to time, noting that the radius r is constant, we obtain (10.8) at = rα. Thus, the tangential acceleration at is the radius times the angular acceleration. Equation 10.4 and Equation 10.8 are important for the discussion of rolling motion (see Angular Momentum). Let’s apply these ideas to the analysis of a few simple fixed-axis rotation scenarios. Before doing so, we present a problem-Chapter 10 \| Fixed-Axis Rotation 477 solving strategy that can be applied to rotational kinematics: the description of rotational motion. Problem-Solving Strategy: Rotational Kinematics 1. Examine the situation to determine that rotational kinematics (rotational motion) is involved. 2. Identify exactly what needs to be determined in the problem (identify the unknowns). A sketch of the situation is useful. 3. Make a complete list of what is given or can be inferred from the problem as stated (identify the knowns). 4. Solve the appropriate equation or equations for the quantity to be determined (the unknown). It can be useful to think in terms of a translational analog, because by now you are familiar with the equations of translational motion. 5. Substitute the known values along with their units into the appropriate equation and obtain numerical solutions complete with units. Be sure to use units of radians for angles. 6. Check your answer to see if it is reasonable: Does your answer make sense? Now let’s apply this problem-solving strategy to a few specific examples. Example 10.2 A Spinning Bicycle Wheel A bicycle mechanic mounts a bicycle on the repair stand and starts the rear wheel spinning from rest to a final angular velocity of 250 rpm in 5.00 s. (a) Calculate the average angular acceleration in rad/s2 . (b) If she now hits the brakes, causing an angular acceleration of −87.3 rad/s2 , how long does it take the wheel to stop? Strategy The average angular acceleration can be found directly from its definition α – = Δω Δt because the final angular velocity and time are given. We see that Δω = ωfina −ωinitial = 250 rev/min and Δt is 5.00 s. For part (b), we know the angular acceleration and the initial angular velocity. We can find the stopping time by using the definition of average angular acceleration and solving for Δt , yielding Δt = Δω α . Solution a. Entering known information into the definition of angular acceleration, we get α – = Δω Δt = 250 rpm 5.00 s . Because Δω is in revolutions per minute (rpm) and we want the standard units of rad/s2 for angular acceleration, we need to convert from rpm to rad/s: Δω = 250 rev min · 2π rad rev · 1 min 60 s = 26.2rad s . Entering this quantity into the expression for α , we get α = Δω Δt = 26.2 rad/s 5.00 s = 5.24 rad/s2. b. Here the angular velocity decreases from 26.2 rad/s (250 rpm) to zero, so that Δω is −26.2 rad/s, and α is given to be –87.3 rad/s2 . Thus, 478 Chapter 10 \| Fixed-Axis Rotation This OpenStax book is available for free at 10.1 Δt = −26.2 rad/s −87.3 rad/s2 = 0.300 s. Significance Note that the angular acceleration as the mechanic spins the wheel is small and positive; it takes 5 s to produce an appreciable angular velocity. When she hits the brake, the angular acceleration is large and negative. The angular velocity quickly goes to zero. Check Your Understanding The fan blades on a turbofan jet engine (shown below) accelerate from rest up to a rotation rate of 40.0 rev/s in 20 s. The increase in angular velocity of the fan is constant in time. (The GE90-110B1 turbofan engine mounted on a Boeing 777, as shown, is currently the largest turbofan engine in the world, capable of thrusts of 330–510 kN.) (a) What is the average angular acceleration? (b) What is the instantaneous angular acceleration at any time during the first 20 s? Figure 10.9 (credit: “Bubinator”/ Wikimedia Commons) Example 10.3 Wind Turbine A wind turbine (Figure 10.10) in a wind farm is being shut down for maintenance. It takes 30 s for the turbine to go from its operating angular velocity to a complete stop in which the angular velocity function is ω(t) = [(ts−1−30.0)2/100.0]rad/s . If the turbine is rotating counterclockwise looking into the page, (a) what are the directions of the angular velocity and acceleration vectors? (b) What is the average angular acceleration? (c) What is the instantaneous angular acceleration at t = 0.0, 15.0, 30.0 s? Chapter 10 \| Fixed-Axis Rotation 479 Figure 10.10 A wind turbine that is rotating counterclockwise, as seen head on. Strategy a. We are given the rotational sense of the turbine, which is counterclockwise in the plane of the page. Using the right hand rule (Figure 10.5), we can establish the directions of the angular velocity and acceleration vectors. b. We calculate the initial and final angular velocities to get the average angular acceleration. We establish the sign of the angular acceleration from the results in (a). c. We are given the functional form of the angular velocity, so we can find the functional form of the angular acceleration function by taking its derivative with respect to time. Solution a. Since the turbine is rotating counterclockwise, angular velocity ω →points out of the page. But since the angular velocity is decreasing, the angular acceleration α →points into the page, in the opposite sense to the angular velocity. b. The initial angular velocity of the turbine, setting t = 0, is ω = 9.0 rad/s . The final angular velocity is zero, so the average angular acceleration is α – = Δω Δt = ω −ω0 t −t0 = 0 −9.0 rad/s 30.0 −0 s = −0.3 rad/s2. c. Taking the derivative of the angular velocity with respect to time gives α = dω dt = (t −30.0)/50.0 rad/s2 α(0.0 s) = −0.6 rad/s2, α(15.0 s) = −0.3 rad/s2, and α(30.0 s) = 0 rad/s. Significance We found from the calculations in (a) and (b) that the angular acceleration α and the average angular acceleration α – are negative. The turbine has an angular acceleration in the opposite sense to its angular velocity. We now have a basic vocabulary for discussing fixed-axis rotational kinematics and relationships between rotational variables. We discuss more definitions and connections in the next section. 480 Chapter 10 \| Fixed-Axis Rotation This OpenStax book is available for free at 10.2 \| Rotation with Constant Angular Acceleration Learning Objectives By the end of this section, you will be able to: • Derive the kinematic equations for rotational motion with constant angular acceleration • Select from the kinematic equations for rotational motion with constant angular acceleration the appropriate equations to solve for unknowns in the analysis of systems undergoing fixed-axis rotation • Use solutions found with the kinematic equations to verify the graphical analysis of fixed-axis rotation with constant angular acceleration In the preceding section, we defined the rotational variables of angular displacement, angular velocity, and angular acceleration. In this section, we work with these definitions to derive relationships among these variables and use these relationships to analyze rotational motion for a rigid body about a fixed axis under a constant angular acceleration. This analysis forms the basis for rotational kinematics. If the angular acceleration is constant, the equations of rotational kinematics simplify, similar to the equations of linear kinematics discussed in Motion along a Straight Line and Motion in Two and Three Dimensions. We can then use this simplified set of equations to describe many applications in physics and engineering where the angular acceleration of the system is constant. Rotational kinematics is also a prerequisite to the discussion of rotational dynamics later in this chapter. Kinematics of Rotational Motion Using our intuition, we can begin to see how the rotational quantities θ, ω, α , and t are related to one another. For example, we saw in the preceding section that if a flywheel has an angular acceleration in the same direction as its angular velocity vector, its angular velocity increases with time and its angular displacement also increases. On the contrary, if the angular acceleration is opposite to the angular velocity vector, its angular velocity decreases with time. We can describe these physical situations and many others with a consistent set of rotational kinematic equations under a constant angular acceleration. The method to investigate rotational motion in this way is called kinematics of rotational motion. To begin, we note that if the system is rotating under a constant acceleration, then the average angular velocity follows a simple relation because the angular velocity is increasing linearly with time. The average angular velocity is just half the sum of the initial and final values: (10.9) ω – = ω0 + ωf 2 . From the definition of the average angular velocity, we can find an equation that relates the angular position, average angular velocity, and time: ω – = Δθ Δt . Solving for θ , we have (10.10) θf = θ0 + ω – t, where we have set t0 = 0 . This equation can be very useful if we know the average angular velocity of the system. Then we could find the angular displacement over a given time period. Next, we find an equation relating ω , α , and t. To determine this equation, we start with the definition of angular acceleration: α = dω dt . Chapter 10 \| Fixed-Axis Rotation 481 We rearrange this to get αdt = dω and then we integrate both sides of this equation from initial values to final values, that is, from t0 to t and ω0 to ωf . In uniform rotational motion, the angular acceleration is constant so it can be pulled out of the integral, yielding two definite integrals: α∫ t0 t dt′ = ∫ ω0 ωf dω. Setting t0 = 0 , we have αt = ω f −ω0. We rearrange this to obtain (10.11) ωf = ω0 + αt, where ω0 is the initial angular velocity. Equation 10.11 is the rotational counterpart to the linear kinematics equation vf = v0 + at . With Equation 10.11, we can find the angular velocity of an object at any specified time t given the initial angular velocity and the angular acceleration. Let’s now do a similar treatment starting with the equation ω = dθ dt . We rearrange it to obtain ωdt = dθ and integrate both sides from initial to final values again, noting that the angular acceleration is constant and does not have a time dependence. However, this time, the angular velocity is not constant (in general), so we substitute in what we derived above: ⌠ ⌡ t0 t f (ω0 + αt′)dt′ = ∫ θ0 θf dθ; ⌠ ⌡ t0 t ω0 dt + ∫ t0 t αtdt = ∫ θ0 θf dθ = ⎡ ⎣ω0 t′ + α⎛ ⎝ (t′)2 2 ⎞ ⎠ ⎤ ⎦t0 t = ω0 t + α⎛ ⎝ t2 2 ⎞ ⎠= θf −θ0, where we have set t0 = 0 . Now we rearrange to obtain (10.12) θf = θ0 + ω0 t + 1 2αt2. Equation 10.12 is the rotational counterpart to the linear kinematics equation found in Motion Along a Straight Line for position as a function of time. This equation gives us the angular position of a rotating rigid body at any time t given the initial conditions (initial angular position and initial angular velocity) and the angular acceleration. We can find an equation that is independent of time by solving for t in Equation 10.11 and substituting into Equation 10.12. Equation 10.12 becomes 482 Chapter 10 \| Fixed-Axis Rotation This OpenStax book is available for free at θf = θ0 + ω0 ⎛ ⎝ ωf −ω0 α ⎞ ⎠+ 1 2α⎛ ⎝ ωf −ω0 α ⎞ ⎠ 2 = θ0 + ω0 ωf α −ω0 2 α + 1 2 ωf 2 α −ω0 ωf α + 1 2 ω0 2 α = θ0 + 1 2 ωf 2 α −1 2 ω0 2 α , θf −θ0 = ωf 2 −ω0 2 2α or (10.13) ωf 2 = ω0 2 + 2α(Δθ). Equation 10.10 through Equation 10.13 describe fixed-axis rotation for constant acceleration and are summarized in Table 10.1. Angular displacement from average angular velocity θf = θ0 + ω – t Angular velocity from angular acceleration ωf = ω0 + αt Angular displacement from angular velocity and angular acceleration θf = θ0 + ω0 t + 1 2αt2 Angular velocity from angular displacement and angular acceleration ωf 2 = ω0 2 + 2α(Δθ) Table 10.1 Kinematic Equations Applying the Equations for Rotational Motion Now we can apply the key kinematic relations for rotational motion to some simple examples to get a feel for how the equations can be applied to everyday situations. Example 10.4 Calculating the Acceleration of a Fishing Reel A deep-sea fisherman hooks a big fish that swims away from the boat, pulling the fishing line from his fishing reel. The whole system is initially at rest, and the fishing line unwinds from the reel at a radius of 4.50 cm from its axis of rotation. The reel is given an angular acceleration of 110 rad/s2 for 2.00 s (Figure 10.11). (a) What is the final angular velocity of the reel after 2 s? (b) How many revolutions does the reel make? Chapter 10 \| Fixed-Axis Rotation 483 Figure 10.11 Fishing line coming off a rotating reel moves linearly. Strategy Identify the knowns and compare with the kinematic equations for constant acceleration. Look for the appropriate equation that can be solved for the unknown, using the knowns given in the problem description. Solution a. We are given α and t and want to determine ω . The most straightforward equation to use is ωf = ω0 + αt , since all terms are known besides the unknown variable we are looking for. We are given that ω0 = 0 (it starts from rest), so ωf = 0 + (110 rad/s2)(2.00 s) = 220 rad/s. b. We are asked to find the number of revolutions. Because 1 rev = 2π rad , we can find the number of revolutions by finding θ in radians. We are given α and t, and we know ω0 is zero, so we can obtain θ by using θf = θi + ωit + 1 2αt2 = 0 + 0 + (0.500)⎛ ⎝110 rad/s2⎞ ⎠(2.00 s)2 = 220 rad. Converting radians to revolutions gives Number of rev = (220 rad) 1 rev 2π rad = 35.0 rev. Significance This example illustrates that relationships among rotational quantities are highly analogous to those among linear quantities. The answers to the questions are realistic. After unwinding for two seconds, the reel is found to spin at 220 rad/s, which is 2100 rpm. (No wonder reels sometimes make high-pitched sounds.) In the preceding example, we considered a fishing reel with a positive angular acceleration. Now let us consider what happens with a negative angular acceleration. Example 10.5 Calculating the Duration When the Fishing Reel Slows Down and Stops Now the fisherman applies a brake to the spinning reel, achieving an angular acceleration of −300 rad/s2 . How 484 Chapter 10 \| Fixed-Axis Rotation This OpenStax book is available for free at 10.2 long does it take the reel to come to a stop? Strategy We are asked to find the time t for the reel to come to a stop. The initial and final conditions are different from those in the previous problem, which involved the same fishing reel. Now we see that the initial angular velocity is ω0 = 220 rad/s and the final angular velocity ω is zero. The angular acceleration is given as α = −300 rad/s2. Examining the available equations, we see all quantities but t are known in ωf = ω0 + αt , making it easiest to use this equation. Solution The equation states ωf = ω0 + αt. We solve the equation algebraically for t and then substitute the known values as usual, yielding t = ωf −ω0 α = 0 −220.0 rad/s −300.0 rad/s2 = 0.733 s. Significance Note that care must be taken with the signs that indicate the directions of various quantities. Also, note that the time to stop the reel is fairly small because the acceleration is rather large. Fishing lines sometimes snap because of the accelerations involved, and fishermen often let the fish swim for a while before applying brakes on the reel. A tired fish is slower, requiring a smaller acceleration. Check Your Understanding A centrifuge used in DNA extraction spins at a maximum rate of 7000 rpm, producing a “g-force” on the sample that is 6000 times the force of gravity. If the centrifuge takes 10 seconds to come to rest from the maximum spin rate: (a) What is the angular acceleration of the centrifuge? (b) What is the angular displacement of the centrifuge during this time? Example 10.6 Angular Acceleration of a Propeller Figure 10.12 shows a graph of the angular velocity of a propeller on an aircraft as a function of time. Its angular velocity starts at 30 rad/s and drops linearly to 0 rad/s over the course of 5 seconds. (a) Find the angular acceleration of the object and verify the result using the kinematic equations. (b) Find the angle through which the propeller rotates during these 5 seconds and verify your result using the kinematic equations. Chapter 10 \| Fixed-Axis Rotation 485 Figure 10.12 A graph of the angular velocity of a propeller versus time. Strategy a. Since the angular velocity varies linearly with time, we know that the angular acceleration is constant and does not depend on the time variable. The angular acceleration is the slope of the angular velocity vs. time graph, α = dω dt . To calculate the slope, we read directly from Figure 10.12, and see that ω0 = 30 rad/s at t = 0 s and ωf = 0 rad/s at t = 5 s . Then, we can verify the result using ω = ω0 + αt . b. We use the equation ω = dθ dt ; since the time derivative of the angle is the angular velocity, we can find the angular displacement by integrating the angular velocity, which from the figure means taking the area under the angular velocity graph. In other words: ∫ θ0 θf dθ = θf −θ0 = ∫ t0 tf ω(t)dt. Then we use the kinematic equations for constant acceleration to verify the result. Solution a. Calculating the slope, we get α = ω −ω0 t −t0 = (0 −30.0) rad/s (5.0 −0) s = −6.0 rad/s2. We see that this is exactly Equation 10.11 with a little rearranging of terms. b. We can find the area under the curve by calculating the area of the right triangle, as shown in Figure 10.13. 486 Chapter 10 \| Fixed-Axis Rotation This OpenStax book is available for free at Figure 10.13 The area under the curve is the area of the right triangle. Δθ = area ⎛ ⎝triangle ⎞ ⎠; Δθ = 1 2(30 rad/s)(5 s) = 75 rad. We verify the solution using Equation 10.12: θf = θ0 + ω0 t + 1 2αt2. Setting θ0 = 0 , we have θ0 = (30.0 rad/s)(5.0 s) + 1 2(−6.0 rad/s2)(5.0 rad/s)2 = 150.0 −75.0 = 75.0 rad. This verifies the solution found from finding the area under the curve. Significance We see from part (b) that there are alternative approaches to analyzing fixed-axis rotation with constant acceleration. We started with a graphical approach and verified the solution using the rotational kinematic equations. Since α = dω dt , we could do the same graphical analysis on an angular acceleration-vs.-time curve. The area under an α-vs.-t curve gives us the change in angular velocity. Since the angular acceleration is constant in this section, this is a straightforward exercise. 10.3 \| Relating Angular and Translational Quantities Learning Objectives By the end of this section, you will be able to: • Given the linear kinematic equation, write the corresponding rotational kinematic equation • Calculate the linear distances, velocities, and accelerations of points on a rotating system given the angular velocities and accelerations Chapter 10 \| Fixed-Axis Rotation 487 In this section, we relate each of the rotational variables to the translational variables defined in Motion Along a Straight Line and Motion in Two and Three Dimensions. This will complete our ability to describe rigid-body rotations. Angular vs. Linear Variables In Rotational Variables, we introduced angular variables. If we compare the rotational definitions with the definitions of linear kinematic variables from Motion Along a Straight Line and Motion in Two and Three Dimensions, we find that there is a mapping of the linear variables to the rotational ones. Linear position, velocity, and acceleration have their rotational counterparts, as we can see when we write them side by side: Linear Rotational Position x θ Velocity v = dx dt ω = dθ dt Acceleration a = dv dt α = dω dt Let’s compare the linear and rotational variables individually. The linear variable of position has physical units of meters, whereas the angular position variable has dimensionless units of radians, as can be seen from the definition of θ = s r , which is the ratio of two lengths. The linear velocity has units of m/s, and its counterpart, the angular velocity, has units of rad/s. In Rotational Variables, we saw in the case of circular motion that the linear tangential speed of a particle at a radius r from the axis of rotation is related to the angular velocity by the relation vt = rω . This could also apply to points on a rigid body rotating about a fixed axis. Here, we consider only circular motion. In circular motion, both uniform and nonuniform, there exists a centripetal acceleration (Motion in Two and Three Dimensions). The centripetal acceleration vector points inward from the particle executing circular motion toward the axis of rotation. The derivation of the magnitude of the centripetal acceleration is given in Motion in Two and Three Dimensions. From that derivation, the magnitude of the centripetal acceleration was found to be (10.14) ac = vt 2 r , where r is the radius of the circle. Thus, in uniform circular motion when the angular velocity is constant and the angular acceleration is zero, we have a linear acceleration—that is, centripetal acceleration—since the tangential speed in Equation 10.14 is a constant. If nonuniform circular motion is present, the rotating system has an angular acceleration, and we have both a linear centripetal acceleration that is changing (because vt is changing) as well as a linear tangential acceleration. These relationships are shown in Figure 10.14, where we show the centripetal and tangential accelerations for uniform and nonuniform circular motion. 488 Chapter 10 \| Fixed-Axis Rotation This OpenStax book is available for free at Figure 10.14 (a) Uniform circular motion: The centripetal acceleration ac has its vector inward toward the axis of rotation. There is no tangential acceleration. (b) Nonuniform circular motion: An angular acceleration produces an inward centripetal acceleration that is changing in magnitude, plus a tangential acceleration at . The centripetal acceleration is due to the change in the direction of tangential velocity, whereas the tangential acceleration is due to any change in the magnitude of the tangential velocity. The tangential and centripetal acceleration vectors a → t and a → c are always perpendicular to each other, as seen in Figure 10.14. To complete this description, we can assign a total linear acceleration vector to a point on a rotating rigid body or a particle executing circular motion at a radius r from a fixed axis. The total linear acceleration vector a →is the vector sum of the centripetal and tangential accelerations, (10.15) a →= a → c + a → t. The total linear acceleration vector in the case of nonuniform circular motion points at an angle between the centripetal and tangential acceleration vectors, as shown in Figure 10.15. Since a → c ⊥a → t , the magnitude of the total linear acceleration is \| a →\| = ac 2 + at 2. Note that if the angular acceleration is zero, the total linear acceleration is equal to the centripetal acceleration. Chapter 10 \| Fixed-Axis Rotation 489 Figure 10.15 A particle is executing circular motion and has an angular acceleration. The total linear acceleration of the particle is the vector sum of the centripetal acceleration and tangential acceleration vectors. The total linear acceleration vector is at an angle in between the centripetal and tangential accelerations. Relationships between Rotational and Translational Motion We can look at two relationships between rotational and translational motion. 1. Generally speaking, the linear kinematic equations have their rotational counterparts. Table 10.2 lists the four linear kinematic equations and the corresponding rotational counterpart. The two sets of equations look similar to each other, but describe two different physical situations, that is, rotation and translation. Rotational Translational θf = θ0 + ω – t x = x0 + v – t ωf = ω0 + αt vf = v0 + at θf = θ0 + ω0 t + 1 2αt2 xf = x0 + v0 t + 1 2at2 ωf 2 = ω0 2 + 2α(Δθ) vf 2 = v0 2 + 2a(Δx) Table 10.2 Rotational and Translational Kinematic Equations 2. The second correspondence has to do with relating linear and rotational variables in the special case of circular motion. This is shown in Table 10.3, where in the third column, we have listed the connecting equation that relates the linear variable to the rotational variable. The rotational variables of angular velocity and acceleration have subscripts that indicate their definition in circular motion. Rotational Translational Relationship ( r = radius) θ s θ = s r ω vt ω = vt r α at α = at r Table 10.3 Rotational and Translational Quantities: Circular Motion 490 Chapter 10 \| Fixed-Axis Rotation This OpenStax book is available for free at Rotational Translational Relationship ( r = radius) ac ac = vt 2 r Table 10.3 Rotational and Translational Quantities: Circular Motion Example 10.7 Linear Acceleration of a Centrifuge A centrifuge has a radius of 20 cm and accelerates from a maximum rotation rate of 10,000 rpm to rest in 30 seconds under a constant angular acceleration. It is rotating counterclockwise. What is the magnitude of the total acceleration of a point at the tip of the centrifuge at t = 29.0s? What is the direction of the total acceleration vector? Strategy With the information given, we can calculate the angular acceleration, which then will allow us to find the tangential acceleration. We can find the centripetal acceleration at t = 0 by calculating the tangential speed at this time. With the magnitudes of the accelerations, we can calculate the total linear acceleration. From the description of the rotation in the problem, we can sketch the direction of the total acceleration vector. Solution The angular acceleration is α = ω −ω0 t = 0 −(1.0 × 104)2π/60.0 s(rad/s) 30.0 s = −34.9 rad/s2. Therefore, the tangential acceleration is at = rα = 0.2 m(−34.9 rad/s2) = −7.0 m/s2. The angular velocity at t = 29.0 s is ω = ω0 + αt = 1.0 × 104 ⎛ ⎝2π 60.0 s ⎞ ⎠+ ⎛ ⎝−34.9 rad/s2⎞ ⎠(29.0 s) = 1047.2 rad/s −1012.71 = 35.1 rad/s. Thus, the tangential speed at t = 29.0 s is vt = rω = 0.2 m(35.1 rad/s) = 7.0 m/s. We can now calculate the centripetal acceleration at t = 29.0 s : ac = v2 r = (7.0 m/s)2 0.2 m = 245.0 m/s2. Since the two acceleration vectors are perpendicular to each other, the magnitude of the total linear acceleration is \| a →\| = ac 2 + at 2 = (245.0)2 + (−7.0)2 = 245.1 m/s2. Since the centrifuge has a negative angular acceleration, it is slowing down. The total acceleration vector is as shown in Figure 10.16. The angle with respect to the centripetal acceleration vector is θ = tan−1 −7.0 245.0 = −1.6°. The negative sign means that the total acceleration vector is angled toward the clockwise direction. Chapter 10 \| Fixed-Axis Rotation 491 10.3 Figure 10.16 The centripetal, tangential, and total acceleration vectors. The centrifuge is slowing down, so the tangential acceleration is clockwise, opposite the direction of rotation (counterclockwise). Significance From Figure 10.16, we see that the tangential acceleration vector is opposite the direction of rotation. The magnitude of the tangential acceleration is much smaller than the centripetal acceleration, so the total linear acceleration vector will make a very small angle with respect to the centripetal acceleration vector. Check Your Understanding A boy jumps on a merry-go-round with a radius of 5 m that is at rest. It starts accelerating at a constant rate up to an angular velocity of 5 rad/s in 20 seconds. What is the distance travelled by the boy? Check out this PhET simulation ( to change the parameters of a rotating disk (the initial angle, angular velocity, and angular acceleration), and place bugs at different radial distances from the axis. The simulation then lets you explore how circular motion relates to the bugs’ xy-position, velocity, and acceleration using vectors or graphs. 10.4 \| Moment of Inertia and Rotational Kinetic Energy Learning Objectives By the end of this section, you will be able to: • Describe the differences between rotational and translational kinetic energy • Define the physical concept of moment of inertia in terms of the mass distribution from the rotational axis • Explain how the moment of inertia of rigid bodies affects their rotational kinetic energy • Use conservation of mechanical energy to analyze systems undergoing both rotation and translation • Calculate the angular velocity of a rotating system when there are energy losses due to nonconservative forces So far in this chapter, we have been working with rotational kinematics: the description of motion for a rotating rigid body 492 Chapter 10 \| Fixed-Axis Rotation This OpenStax book is available for free at with a fixed axis of rotation. In this section, we define two new quantities that are helpful for analyzing properties of rotating objects: moment of inertia and rotational kinetic energy. With these properties defined, we will have two important tools we need for analyzing rotational dynamics. Rotational Kinetic Energy Any moving object has kinetic energy. We know how to calculate this for a body undergoing translational motion, but how about for a rigid body undergoing rotation? This might seem complicated because each point on the rigid body has a different velocity. However, we can make use of angular velocity—which is the same for the entire rigid body—to express the kinetic energy for a rotating object. Figure 10.17 shows an example of a very energetic rotating body: an electric grindstone propelled by a motor. Sparks are flying, and noise and vibration are generated as the grindstone does its work. This system has considerable energy, some of it in the form of heat, light, sound, and vibration. However, most of this energy is in the form of rotational kinetic energy. Figure 10.17 The rotational kinetic energy of the grindstone is converted to heat, light, sound, and vibration. (credit: Zachary David Bell, US Navy) Energy in rotational motion is not a new form of energy; rather, it is the energy associated with rotational motion, the same as kinetic energy in translational motion. However, because kinetic energy is given by K = 1 2mv2 , and velocity is a quantity that is different for every point on a rotating body about an axis, it makes sense to find a way to write kinetic energy in terms of the variable ω , which is the same for all points on a rigid rotating body. For a single particle rotating around a fixed axis, this is straightforward to calculate. We can relate the angular velocity to the magnitude of the translational velocity using the relation vt = ωr , where r is the distance of the particle from the axis of rotation and vt is its tangential speed. Substituting into the equation for kinetic energy, we find K = 1 2mvt 2 = 1 2m(ωr)2 = 1 2(mr2)ω2. In the case of a rigid rotating body, we can divide up any body into a large number of smaller masses, each with a mass m j and distance to the axis of rotation r j , such that the total mass of the body is equal to the sum of the individual masses: M = ∑ j m j . Each smaller mass has tangential speed v j , where we have dropped the subscript t for the moment. The total kinetic energy of the rigid rotating body is K = ∑ j 1 2m j v j 2 = ∑ j 1 2m j (r j ω j)2 Chapter 10 \| Fixed-Axis Rotation 493 and since ω j = ω for all masses, (10.16) K = 1 2 ⎛ ⎝ ⎜∑ j m j r j 2⎞ ⎠ ⎟ω2. The units of Equation 10.16 are joules (J). The equation in this form is complete, but awkward; we need to find a way to generalize it. Moment of Inertia If we compare Equation 10.16 to the way we wrote kinetic energy in Work and Kinetic Energy, ⎛ ⎝1 2mv2⎞ ⎠, this suggests we have a new rotational variable to add to our list of our relations between rotational and translational variables. The quantity ∑ j m j r j 2 is the counterpart for mass in the equation for rotational kinetic energy. This is an important new term for rotational motion. This quantity is called the moment of inertia I, with units of kg · m2 : (10.17) I = ∑ j m j r j 2. For now, we leave the expression in summation form, representing the moment of inertia of a system of point particles rotating about a fixed axis. We note that the moment of inertia of a single point particle about a fixed axis is simply mr2 , with r being the distance from the point particle to the axis of rotation. In the next section, we explore the integral form of this equation, which can be used to calculate the moment of inertia of some regular-shaped rigid bodies. The moment of inertia is the quantitative measure of rotational inertia, just as in translational motion, and mass is the quantitative measure of linear inertia—that is, the more massive an object is, the more inertia it has, and the greater is its resistance to change in linear velocity. Similarly, the greater the moment of inertia of a rigid body or system of particles, the greater is its resistance to change in angular velocity about a fixed axis of rotation. It is interesting to see how the moment of inertia varies with r, the distance to the axis of rotation of the mass particles in Equation 10.17. Rigid bodies and systems of particles with more mass concentrated at a greater distance from the axis of rotation have greater moments of inertia than bodies and systems of the same mass, but concentrated near the axis of rotation. In this way, we can see that a hollow cylinder has more rotational inertia than a solid cylinder of the same mass when rotating about an axis through the center. Substituting Equation 10.17 into Equation 10.16, the expression for the kinetic energy of a rotating rigid body becomes (10.18) K = 1 2Iω2. We see from this equation that the kinetic energy of a rotating rigid body is directly proportional to the moment of inertia and the square of the angular velocity. This is exploited in flywheel energy-storage devices, which are designed to store large amounts of rotational kinetic energy. Many carmakers are now testing flywheel energy storage devices in their automobiles, such as the flywheel, or kinetic energy recovery system, shown in Figure 10.18. 494 Chapter 10 \| Fixed-Axis Rotation This OpenStax book is available for free at Figure 10.18 A KERS (kinetic energy recovery system) flywheel used in cars. (credit: “cmonville”/Flickr) The rotational and translational quantities for kinetic energy and inertia are summarized in Table 10.4. The relationship column is not included because a constant doesn’t exist by which we could multiply the rotational quantity to get the translational quantity, as can be done for the variables in Table 10.3. Rotational Translational I = ∑ j m j r j 2 m K = 1 2Iω2 K = 1 2mv2 Table 10.4 Rotational and Translational Kinetic Energies and Inertia Example 10.8 Moment of Inertia of a System of Particles Six small washers are spaced 10 cm apart on a rod of negligible mass and 0.5 m in length. The mass of each washer is 20 g. The rod rotates about an axis located at 25 cm, as shown in Figure 10.19. (a) What is the moment of inertia of the system? (b) If the two washers closest to the axis are removed, what is the moment of inertia of the remaining four washers? (c) If the system with six washers rotates at 5 rev/s, what is its rotational kinetic energy? Chapter 10 \| Fixed-Axis Rotation 495 Figure 10.19 Six washers are spaced 10 cm apart on a rod of negligible mass and rotating about a vertical axis. Strategy a. We use the definition for moment of inertia for a system of particles and perform the summation to evaluate this quantity. The masses are all the same so we can pull that quantity in front of the summation symbol. b. We do a similar calculation. c. We insert the result from (a) into the expression for rotational kinetic energy. Solution a. I = ∑ j m j r j 2 = (0.02 kg)(2 × (0.25 m)2 + 2 × (0.15 m)2 + 2 × (0.05 m)2) = 0.0035 kg · m2 . b. I = ∑ j m j r j 2 = (0.02 kg)(2 × (0.25 m)2 + 2 × (0.15 m)2) = 0.0034 kg · m2 . c. K = 1 2Iω2 = 1 2(0.0035 kg · m2)(5.0 × 2π rad/s)2 = 1.73 J . Significance We can see the individual contributions to the moment of inertia. The masses close to the axis of rotation have a very small contribution. When we removed them, it had a very small effect on the moment of inertia. In the next section, we generalize the summation equation for point particles and develop a method to calculate moments of inertia for rigid bodies. For now, though, Figure 10.20 gives values of rotational inertia for common object shapes around specified axes. 496 Chapter 10 \| Fixed-Axis Rotation This OpenStax book is available for free at Figure 10.20 Values of rotational inertia for common shapes of objects. Applying Rotational Kinetic Energy Now let’s apply the ideas of rotational kinetic energy and the moment of inertia table to get a feeling for the energy associated with a few rotating objects. The following examples will also help get you comfortable using these equations. First, let’s look at a general problem-solving strategy for rotational energy. Problem-Solving Strategy: Rotational Energy 1. Determine that energy or work is involved in the rotation. 2. Determine the system of interest. A sketch usually helps. 3. Analyze the situation to determine the types of work and energy involved. 4. If there are no losses of energy due to friction and other nonconservative forces, mechanical energy is conserved, that is, Ki + Ui = Kf + Uf . 5. If nonconservative forces are present, mechanical energy is not conserved, and other forms of energy, such as heat and light, may enter or leave the system. Determine what they are and calculate them as necessary. 6. Eliminate terms wherever possible to simplify the algebra. 7. Evaluate the numerical solution to see if it makes sense in the physical situation presented in the wording of the problem. Chapter 10 \| Fixed-Axis Rotation 497 Example 10.9 Calculating Helicopter Energies A typical small rescue helicopter has four blades: Each is 4.00 m long and has a mass of 50.0 kg (Figure 10.21). The blades can be approximated as thin rods that rotate about one end of an axis perpendicular to their length. The helicopter has a total loaded mass of 1000 kg. (a) Calculate the rotational kinetic energy in the blades when they rotate at 300 rpm. (b) Calculate the translational kinetic energy of the helicopter when it flies at 20.0 m/s, and compare it with the rotational energy in the blades. Figure 10.21 (a) Sketch of a four-blade helicopter. (b) A water rescue operation featuring a helicopter from the Auckland Westpac Rescue Helicopter Service. (credit b: modification of work by “111 Emergency”/Flickr) Strategy Rotational and translational kinetic energies can be calculated from their definitions. The wording of the problem gives all the necessary constants to evaluate the expressions for the rotational and translational kinetic energies. Solution a. The rotational kinetic energy is K = 1 2Iω2. We must convert the angular velocity to radians per second and calculate the moment of inertia before we can find K. The angular velocity ω is ω = 300 rev 1.00 min 2π rad 1 rev 1.00 min 60.0 s = 31.4 rad s . The moment of inertia of one blade is that of a thin rod rotated about its end, listed in Figure 10.20. The total I is four times this moment of inertia because there are four blades. Thus, I = 4Ml2 3 = 4 × (50.0 kg)(4.00 m)2 3 = 1067.0 kg · m2. Entering ω and I into the expression for rotational kinetic energy gives K = 0.5(1067 kg · m2)(31.4 rad/s)2 = 5.26 × 105 J. b. Entering the given values into the equation for translational kinetic energy, we obtain K = 1 2mv2 = (0.5)(1000.0 kg)(20.0 m/s)2 = 2.00 × 105 J. 498 Chapter 10 \| Fixed-Axis Rotation This OpenStax book is available for free at To compare kinetic energies, we take the ratio of translational kinetic energy to rotational kinetic energy. This ratio is 2.00 × 105 J 5.26 × 105 J = 0.380. Significance The ratio of translational energy to rotational kinetic energy is only 0.380. This ratio tells us that most of the kinetic energy of the helicopter is in its spinning blades. Example 10.10 Energy in a Boomerang A person hurls a boomerang into the air with a velocity of 30.0 m/s at an angle of 40.0° with respect to the horizontal (Figure 10.22). It has a mass of 1.0 kg and is rotating at 10.0 rev/s. The moment of inertia of the boomerang is given as I = 1 12mL2 where L = 0.7 m . (a) What is the total energy of the boomerang when it leaves the hand? (b) How high does the boomerang go from the elevation of the hand, neglecting air resistance? Figure 10.22 A boomerang is hurled into the air at an initial angle of 40° . Strategy We use the definitions of rotational and linear kinetic energy to find the total energy of the system. The problem states to neglect air resistance, so we don’t have to worry about energy loss. In part (b), we use conservation of mechanical energy to find the maximum height of the boomerang. Solution a. Moment of inertia: I = 1 12mL2 = 1 12(1.0 kg)(0.7m)2 = 0.041 kg · m2 . Angular velocity: ω = (10.0 rev/s)(2π) = 62.83 rad/s . The rotational kinetic energy is therefore KR = 1 2(0.041 kg · m2)(62.83 rad/s)2 = 80.93 J. The translational kinetic energy is KT = 1 2mv2 = 1 2(1.0 kg)(30.0 m/s)2 = 450.0 J. Chapter 10 \| Fixed-Axis Rotation 499 10.4 Thus, the total energy in the boomerang is KTotal = KR + KT = 80.93 + 450.0 = 530.93 J. b. We use conservation of mechanical energy. Since the boomerang is launched at an angle, we need to write the total energies of the system in terms of its linear kinetic energies using the velocity in the x- and y-directions. The total energy when the boomerang leaves the hand is EBefore = 1 2mvx 2 + 1 2mvy 2 + 1 2Iω2. The total energy at maximum height is EFinal = 1 2mvx 2 + 1 2Iω2 + mgh. By conservation of mechanical energy, EBefore = EFinal so we have, after canceling like terms, 1 2mvy 2 = mgh. Since vy = 30.0 m/s(sin 40°) = 19.28 m/s , we find h = (19.28 m/s)2 2(9.8 m/s2) = 18.97 m. Significance In part (b), the solution demonstrates how energy conservation is an alternative method to solve a problem that normally would be solved using kinematics. In the absence of air resistance, the rotational kinetic energy was not a factor in the solution for the maximum height. Check Your Understanding A nuclear submarine propeller has a moment of inertia of 800.0 kg · m2 . If the submerged propeller has a rotation rate of 4.0 rev/s when the engine is cut, what is the rotation rate of the propeller after 5.0 s when water resistance has taken 50,000 J out of the system? 10.5 \| Calculating Moments of Inertia Learning Objectives By the end of this section, you will be able to: • Calculate the moment of inertia for uniformly shaped, rigid bodies • Apply the parallel axis theorem to find the moment of inertia about any axis parallel to one already known • Calculate the moment of inertia for compound objects In the preceding section, we defined the moment of inertia but did not show how to calculate it. In this section, we show how to calculate the moment of inertia for several standard types of objects, as well as how to use known moments of inertia to find the moment of inertia for a shifted axis or for a compound object. This section is very useful for seeing how to apply a general equation to complex objects (a skill that is critical for more advanced physics and engineering courses). Moment of Inertia We defined the moment of inertia I of an object to be I = ∑ i miri 2 for all the point masses that make up the object. 500 Chapter 10 \| Fixed-Axis Rotation This OpenStax book is available for free at Because r is the distance to the axis of rotation from each piece of mass that makes up the object, the moment of inertia for any object depends on the chosen axis. To see this, let’s take a simple example of two masses at the end of a massless (negligibly small mass) rod (Figure 10.23) and calculate the moment of inertia about two different axes. In this case, the summation over the masses is simple because the two masses at the end of the barbell can be approximated as point masses, and the sum therefore has only two terms. In the case with the axis in the center of the barbell, each of the two masses m is a distance R away from the axis, giving a moment of inertia of I1 = mR2 + mR2 = 2mR2. In the case with the axis at the end of the barbell—passing through one of the masses—the moment of inertia is I2 = m(0)2 + m(2R)2 = 4mR2. From this result, we can conclude that it is twice as hard to rotate the barbell about the end than about its center. Figure 10.23 (a) A barbell with an axis of rotation through its center; (b) a barbell with an axis of rotation through one end. In this example, we had two point masses and the sum was simple to calculate. However, to deal with objects that are not point-like, we need to think carefully about each of the terms in the equation. The equation asks us to sum over each ‘piece of mass’ a certain distance from the axis of rotation. But what exactly does each ‘piece of mass’ mean? Recall that in our derivation of this equation, each piece of mass had the same magnitude of velocity, which means the whole piece had to have a single distance r to the axis of rotation. However, this is not possible unless we take an infinitesimally small piece of mass dm, as shown in Figure 10.24. Figure 10.24 Using an infinitesimally small piece of mass to calculate the contribution to the total moment of inertia. The need to use an infinitesimally small piece of mass dm suggests that we can write the moment of inertia by evaluating Chapter 10 \| Fixed-Axis Rotation 501 an integral over infinitesimal masses rather than doing a discrete sum over finite masses: (10.19) I = ∑ i miri 2 becomes I = ∫r2 dm. This, in fact, is the form we need to generalize the equation for complex shapes. It is best to work out specific examples in detail to get a feel for how to calculate the moment of inertia for specific shapes. This is the focus of most of the rest of this section. A uniform thin rod with an axis through the center Consider a uniform (density and shape) thin rod of mass M and length L as shown in Figure 10.25. We want a thin rod so that we can assume the cross-sectional area of the rod is small and the rod can be thought of as a string of masses along a one-dimensional straight line. In this example, the axis of rotation is perpendicular to the rod and passes through the midpoint for simplicity. Our task is to calculate the moment of inertia about this axis. We orient the axes so that the z-axis is the axis of rotation and the x-axis passes through the length of the rod, as shown in the figure. This is a convenient choice because we can then integrate along the x-axis. Figure 10.25 Calculation of the moment of inertia I for a uniform thin rod about an axis through the center of the rod. We define dm to be a small element of mass making up the rod. The moment of inertia integral is an integral over the mass distribution. However, we know how to integrate over space, not over mass. We therefore need to find a way to relate mass to spatial variables. We do this using the linear mass density λ of the object, which is the mass per unit length. Since the mass density of this object is uniform, we can write λ = m l or m = λl. If we take the differential of each side of this equation, we find dm = d(λl) = λ(dl) since λ is constant. We chose to orient the rod along the x-axis for convenience—this is where that choice becomes very helpful. Note that a piece of the rod dl lies completely along the x-axis and has a length dx; in fact, dl = dx in this situation. We can therefore write dm = λ(dx) , giving us an integration variable that we know how to deal with. The distance of each piece of mass dm from the axis is given by the variable x, as shown in the figure. Putting this all together, we obtain I = ∫r2 dm = ∫x2 dm = ∫x2 λdx. The last step is to be careful about our limits of integration. The rod extends from x = −L/2 to x = L/2 , since the axis is in the middle of the rod at x = 0 . This gives us I = ∫ −L/2 L/2 x2 λdx = λx3 3\|−L/2 L/2 = λ⎛ ⎝1 3 ⎞ ⎠ ⎡ ⎣ ⎢⎛ ⎝L 2 ⎞ ⎠ 3 −⎛ ⎝−L 2 ⎞ ⎠ 3⎤ ⎦ ⎥ = λ⎛ ⎝1 3 ⎞ ⎠L3 8 (2) = M L ⎛ ⎝1 3 ⎞ ⎠L3 8 (2) = 1 12ML2. Next, we calculate the moment of inertia for the same uniform thin rod but with a different axis choice so we can compare the results. We would expect the moment of inertia to be smaller about an axis through the center of mass than the endpoint 502 Chapter 10 \| Fixed-Axis Rotation This OpenStax book is available for free at axis, just as it was for the barbell example at the start of this section. This happens because more mass is distributed farther from the axis of rotation. A uniform thin rod with axis at the end Now consider the same uniform thin rod of mass M and length L, but this time we move the axis of rotation to the end of the rod. We wish to ﬁnd the moment of inertia about this new axis (Figure 10.26). The quantity dm is again defined to be a small element of mass making up the rod. Just as before, we obtain I = ∫r2 dm = ∫x2 dm = ∫x2 λdx. However, this time we have different limits of integration. The rod extends from x = 0 to x = L , since the axis is at the end of the rod at x = 0 . Therefore we find I = ∫ 0 L x2 λdx = λx3 3\|0 L = λ⎛ ⎝1 3 ⎞ ⎠[(L)3 −(0)3] = λ⎛ ⎝1 3 ⎞ ⎠L3 = M L ⎛ ⎝1 3 ⎞ ⎠L3 = 1 3ML2. Figure 10.26 Calculation of the moment of inertia I for a uniform thin rod about an axis through the end of the rod. Note the rotational inertia of the rod about its endpoint is larger than the rotational inertia about its center (consistent with the barbell example) by a factor of four. The Parallel-Axis Theorem The similarity between the process of finding the moment of inertia of a rod about an axis through its middle and about an axis through its end is striking, and suggests that there might be a simpler method for determining the moment of inertia for a rod about any axis parallel to the axis through the center of mass. Such an axis is called a parallel axis. There is a theorem for this, called the parallel-axis theorem, which we state here but do not derive in this text. Parallel-Axis Theorem Let m be the mass of an object and let d be the distance from an axis through the object’s center of mass to a new axis. Then we have (10.20) Iparallel-axis = Icenter of mass + md2. Let’s apply this to the rod examples solved above: Iend = Icenter of mass + md2 = 1 12mL2 + m⎛ ⎝L 2 ⎞ ⎠ 2 = ⎛ ⎝1 12 + 1 4 ⎞ ⎠mL2 = 1 3mL2. This result agrees with our more lengthy calculation from above. This is a useful equation that we apply in some of the examples and problems. Chapter 10 \| Fixed-Axis Rotation 503 10.5 Check Your Understanding What is the moment of inertia of a cylinder of radius R and mass m about an axis through a point on the surface, as shown below? A uniform thin disk about an axis through the center Integrating to find the moment of inertia of a two-dimensional object is a little bit trickier, but one shape is commonly done at this level of study—a uniform thin disk about an axis through its center (Figure 10.27). Figure 10.27 Calculating the moment of inertia for a thin disk about an axis through its center. Since the disk is thin, we can take the mass as distributed entirely in the xy-plane. We again start with the relationship for the surface mass density, which is the mass per unit surface area. Since it is uniform, the surface mass density σ is constant: σ = m A or σA = m, so dm = σ(dA). Now we use a simplification for the area. The area can be thought of as made up of a series of thin rings, where each ring is a mass increment dm of radius r equidistanct from the axis, as shown in part (b) of the figure. The infinitesimal area of each ring dA is therefore given by the length of each ring ( 2πr ) times the infinitesimmal width of each ring dr: A = πr2, dA = d(πr2) = πdr2 = 2πrdr. The full area of the disk is then made up from adding all the thin rings with a radius range from 0 to R. This radius range then becomes our limits of integration for dr, that is, we integrate from r = 0 to r = R . Putting this all together, we have I = ⌠ ⌡ 0 R r2 σ(2πr)dr = 2πσ∫ 0 R r3dr = 2πσr4 4\|0 R = 2πσ⎛ ⎝ R4 4 −0⎞ ⎠ = 2πm A ⎛ ⎝ R4 4 ⎞ ⎠= 2π m πR2 ⎛ ⎝ R4 4 ⎞ ⎠= 1 2mR2. Note that this agrees with the value given in Figure 10.20. 504 Chapter 10 \| Fixed-Axis Rotation This OpenStax book is available for free at Calculating the moment of inertia for compound objects Now consider a compound object such as that in Figure 10.28, which depicts a thin disk at the end of a thin rod. This cannot be easily integrated to find the moment of inertia because it is not a uniformly shaped object. However, if we go back to the initial definition of moment of inertia as a summation, we can reason that a compound object’s moment of inertia can be found from the sum of each part of the object: (10.21) Itotal = ∑ i Ii. It is important to note that the moments of inertia of the objects in Equation 10.21 are about a common axis. In the case of this object, that would be a rod of length L rotating about its end, and a thin disk of radius R rotating about an axis shifted off of the center by a distance L + R , where R is the radius of the disk. Let’s define the mass of the rod to be mr and the mass of the disk to be md. Figure 10.28 Compound object consisting of a disk at the end of a rod. The axis of rotation is located at A. The moment of inertia of the rod is simply 1 3mr L2 , but we have to use the parallel-axis theorem to find the moment of inertia of the disk about the axis shown. The moment of inertia of the disk about its center is 1 2md R2 and we apply the parallel-axis theorem Iparallel-axis = Icenter of mass + md2 to find Iparallel-axis = 1 2md R2 + md (L + R)2. Adding the moment of inertia of the rod plus the moment of inertia of the disk with a shifted axis of rotation, we find the moment of inertia for the compound object to be Itotal = 1 3mr L2 + 1 2md R2 + md (L + R)2. Applying moment of inertia calculations to solve problems Now let’s examine some practical applications of moment of inertia calculations. Example 10.11 Person on a Merry-Go-Round A 25-kg child stands at a distance r = 1.0 m from the axis of a rotating merry-go-round (Figure 10.29). The merry-go-round can be approximated as a uniform solid disk with a mass of 500 kg and a radius of 2.0 m. Find the moment of inertia of this system. Chapter 10 \| Fixed-Axis Rotation 505 Figure 10.29 Calculating the moment of inertia for a child on a merry-go-round. Strategy This problem involves the calculation of a moment of inertia. We are given the mass and distance to the axis of rotation of the child as well as the mass and radius of the merry-go-round. Since the mass and size of the child are much smaller than the merry-go-round, we can approximate the child as a point mass. The notation we use is mc = 25 kg, rc = 1.0 m, mm = 500 kg, rm = 2.0 m . Our goal is to find Itotal = ∑ i Ii . Solution For the child, Ic = mcr2 , and for the merry-go-round, Im = 1 2mmr2 . Therefore Itotal = 25(1)2 + 1 2(500)(2)2 = 25 + 1000 = 1025 kg · m2. Significance The value should be close to the moment of inertia of the merry-go-round by itself because it has much more mass distributed away from the axis than the child does. Example 10.12 Rod and Solid Sphere Find the moment of inertia of the rod and solid sphere combination about the two axes as shown below. The rod has length 0.5 m and mass 2.0 kg. The radius of the sphere is 20.0 cm and has mass 1.0 kg. Strategy Since we have a compound object in both cases, we can use the parallel-axis theorem to find the moment of inertia 506 Chapter 10 \| Fixed-Axis Rotation This OpenStax book is available for free at about each axis. In (a), the center of mass of the sphere is located at a distance L + R from the axis of rotation. In (b), the center of mass of the sphere is located a distance R from the axis of rotation. In both cases, the moment of inertia of the rod is about an axis at one end. Refer to Table 10.4 for the moments of inertia for the individual objects. a. Itotal = ∑ i Ii = IRod + ISphere ; ISphere = Icenter of mass + mSphere(L + R)2 = 2 5mSphere R2 + mSphere(L + R)2 ; Itotal = IRod + ISphere = 1 3mRod L2 + 2 5mSphere R2 + mSphere(L + R)2; Itotal = 1 3(2.0 kg)(0.5 m)2 + 2 5(1.0 kg)(0.2 m)2 + (1.0 kg)(0.5 m + 0.2 m)2; Itotal = (0.167 + 0.016 + 0.490) kg · m2 = 0.673 kg · m2. b. ISphere = 2 5mSphere R2 + mSphere R2 ; Itotal = IRod + ISphere = 1 3mRod L2 + 2 5mSphere R2 + mSphere R2 ; Itotal = 1 3(2.0 kg)(0.5 m)2 + 2 5(1.0 kg)(0.2 m)2 + (1.0 kg)(0.2 m)2 ; Itotal = (0.167 + 0.016 + 0.04) kg · m2 = 0.223 kg · m2. Significance Using the parallel-axis theorem eases the computation of the moment of inertia of compound objects. We see that the moment of inertia is greater in (a) than (b). This is because the axis of rotation is closer to the center of mass of the system in (b). The simple analogy is that of a rod. The moment of inertia about one end is 1 3mL2 , but the moment of inertia through the center of mass along its length is 1 12mL2 . Example 10.13 Angular Velocity of a Pendulum A pendulum in the shape of a rod (Figure 10.30) is released from rest at an angle of 30° . It has a length 30 cm and mass 300 g. What is its angular velocity at its lowest point? Figure 10.30 A pendulum in the form of a rod is released from rest at an angle of 30°. Chapter 10 \| Fixed-Axis Rotation 507 Strategy Use conservation of energy to solve the problem. At the point of release, the pendulum has gravitational potential energy, which is determined from the height of the center of mass above its lowest point in the swing. At the bottom of the swing, all of the gravitational potential energy is converted into rotational kinetic energy. Solution The change in potential energy is equal to the change in rotational kinetic energy, ΔU + ΔK = 0 . At the top of the swing: U = mghcm = mgL 2(cos θ) . At the bottom of the swing, U = mgL 2. At the top of the swing, the rotational kinetic energy is K = 0 . At the bottom of the swing, K = 1 2Iω2 . Therefore: ΔU + ΔK = 0 ⇒(mgL 2(1 −cos θ) −0) + (0 −1 2Iω2) = 0 or 1 2Iω2 = mgL 2(1 −cos θ). Solving for ω , we have ω = mgL I (1 −cos θ) = mg L 1/3mL2(1 −cos θ) = g3 L(1 −cos θ). Inserting numerical values, we have ω = 9.8 m/s2 3 0.3 m(1 −cos 30) = 3.6 rad/s. Significance Note that the angular velocity of the pendulum does not depend on its mass. 10.6 \| Torque Learning Objectives By the end of this section, you will be able to: • Describe how the magnitude of a torque depends on the magnitude of the lever arm and the angle the force vector makes with the lever arm • Determine the sign (positive or negative) of a torque using the right-hand rule • Calculate individual torques about a common axis and sum them to find the net torque An important quantity for describing the dynamics of a rotating rigid body is torque. We see the application of torque in many ways in our world. We all have an intuition about torque, as when we use a large wrench to unscrew a stubborn bolt. Torque is at work in unseen ways, as when we press on the accelerator in a car, causing the engine to put additional torque on the drive train. Or every time we move our bodies from a standing position, we apply a torque to our limbs. In this section, we define torque and make an argument for the equation for calculating torque for a rigid body with fixed-axis rotation. Defining Torque So far we have defined many variables that are rotational equivalents to their translational counterparts. Let’s consider what the counterpart to force must be. Since forces change the translational motion of objects, the rotational counterpart must be related to changing the rotational motion of an object about an axis. We call this rotational counterpart torque. 508 Chapter 10 \| Fixed-Axis Rotation This OpenStax book is available for free at In everyday life, we rotate objects about an axis all the time, so intuitively we already know much about torque. Consider, for example, how we rotate a door to open it. First, we know that a door opens slowly if we push too close to its hinges; it is more efficient to rotate a door open if we push far from the hinges. Second, we know that we should push perpendicular to the plane of the door; if we push parallel to the plane of the door, we are not able to rotate it. Third, the larger the force, the more effective it is in opening the door; the harder you push, the more rapidly the door opens. The first point implies that the farther the force is applied from the axis of rotation, the greater the angular acceleration; the second implies that the effectiveness depends on the angle at which the force is applied; the third implies that the magnitude of the force must also be part of the equation. Note that for rotation in a plane, torque has two possible directions. Torque is either clockwise or counterclockwise relative to the chosen pivot point. Figure 10.31 shows counterclockwise rotations. Figure 10.31 Torque is the turning or twisting effectiveness of a force, illustrated here for door rotation on its hinges (as viewed from overhead). Torque has both magnitude and direction. (a) A counterclockwise torque is produced by a force F →acting at a distance r from the hinges (the pivot point). (b) A smaller counterclockwise torque is produced when a smaller force F →′ acts at the same distance r from the hinges. (c) The same force as in (a) produces a smaller counterclockwise torque when applied at a smaller distance from the hinges. (d) A smaller counterclockwise torque is produced by the same magnitude force as (a) acting at the same distance as (a) but at an angle θ that is less than 90° . Now let’s consider how to define torques in the general three-dimensional case. Torque When a force F →is applied to a point P whose position is r →relative to O (Figure 10.32), the torque τ →around O is (10.22) τ →= r →× F →. Chapter 10 \| Fixed-Axis Rotation 509 Figure 10.32 The torque is perpendicular to the plane defined by r →and F →and its direction is determined by the right-hand rule. From the definition of the cross product, the torque τ → is perpendicular to the plane containing r →and F → and has magnitude \| τ →\| = \| r →× F →\| = rFsin θ, where θ is the angle between the vectors r →and F →. The SI unit of torque is newtons times meters, usually written as N · m . The quantity r⊥= rsin θ is the perpendicular distance from O to the line determined by the vector F → and is called the lever arm. Note that the greater the lever arm, the greater the magnitude of the torque. In terms of the lever arm, the magnitude of the torque is (10.23) \| τ →\| = r⊥F. The cross product r →× F →also tells us the sign of the torque. In Figure 10.32, the cross product r →× F →is along the positive z-axis, which by convention is a positive torque. If r →× F → is along the negative z-axis, this produces a negative torque. If we consider a disk that is free to rotate about an axis through the center, as shown in Figure 10.33, we can see how the angle between the radius r →and the force F →affects the magnitude of the torque. If the angle is zero, the torque is zero; if the angle is 90° , the torque is maximum. The torque in Figure 10.33 is positive because the direction of the torque by the right-hand rule is out of the page along the positive z-axis. The disk rotates counterclockwise due to the torque, in the same direction as a positive angular acceleration. 510 Chapter 10 \| Fixed-Axis Rotation This OpenStax book is available for free at Figure 10.33 A disk is free to rotate about its axis through the center. The magnitude of the torque on the disk is rFsin θ .When θ = 0° , the torque is zero and the disk does not rotate. When θ = 90° , the torque is maximum and the disk rotates with maximum angular acceleration. Any number of torques can be calculated about a given axis. The individual torques add to produce a net torque about the axis. When the appropriate sign (positive or negative) is assigned to the magnitudes of individual torques about a specified axis, the net torque about the axis is the sum of the individual torques: (10.24) τ → net = ∑ i \| τ → i\|. Calculating Net Torque for Rigid Bodies on a Fixed Axis In the following examples, we calculate the torque both abstractly and as applied to a rigid body. We first introduce a problem-solving strategy. Problem-Solving Strategy: Finding Net Torque 1. Choose a coordinate system with the pivot point or axis of rotation as the origin of the selected coordinate system. 2. Determine the angle between the lever arm r →and the force vector. 3. Take the cross product of r →and F →to determine if the torque is positive or negative about the pivot point or axis. 4. Evaluate the magnitude of the torque using r⊥F . 5. Assign the appropriate sign, positive or negative, to the magnitude. 6. Sum the torques to find the net torque. Example 10.14 Calculating Torque Four forces are shown in Figure 10.34 at particular locations and orientations with respect to a given xy-coordinate system. Find the torque due to each force about the origin, then use your results to find the net Chapter 10 \| Fixed-Axis Rotation 511 torque about the origin. Figure 10.34 Four forces producing torques. Strategy This problem requires calculating torque. All known quantities––forces with directions and lever arms––are given in the figure. The goal is to find each individual torque and the net torque by summing the individual torques. Be careful to assign the correct sign to each torque by using the cross product of r →and the force vector F →. Solution Use \| τ →\| = r⊥F = rFsin θ to find the magnitude and τ →= r →× F →to determine the sign of the torque. The torque from force 40 N in the first quadrant is given by (4)(40)sin 90° = 160 N · m . The cross product of r →and F →is out of the page, positive. The torque from force 20 N in the third quadrant is given by −(3)(20)sin 90° = −60 N · m . The cross product of r →and F →is into the page, so it is negative. The torque from force 30 N in the third quadrant is given by (5)(30)sin 53° = 120 N · m . The cross product of r →and F →is out of the page, positive. The torque from force 20 N in the second quadrant is given by (1)(20)sin 30° = 10 N · m . The cross product of r →and F →is out of the page. The net torque is therefore τnet = ∑ i \|τi\| = 160 −60 + 120 + 10 = 230 N · m. Significance Note that each force that acts in the counterclockwise direction has a positive torque, whereas each force that acts in the clockwise direction has a negative torque. The torque is greater when the distance, force, or perpendicular components are greater. 512 Chapter 10 \| Fixed-Axis Rotation This OpenStax book is available for free at Example 10.15 Calculating Torque on a rigid body Figure 10.35 shows several forces acting at different locations and angles on a flywheel. We have \| F → 1\| = 20 N, \| F → 2\| = 30 N , \| F → 3\| = 30 N , and r = 0.5 m . Find the net torque on the flywheel about an axis through the center. Figure 10.35 Three forces acting on a flywheel. Strategy We calculate each torque individually, using the cross product, and determine the sign of the torque. Then we sum the torques to find the net torque. Solution We start with F → 1 . If we look at Figure 10.35, we see that F → 1 makes an angle of 90° + 60° with the radius vector r →. Taking the cross product, we see that it is out of the page and so is positive. We also see this from calculating its magnitude: \| τ → 1\| = rF1 sin 150° = 0.5 m(20 N)(0.5) = 5.0 N · m. Next we look at F → 2 . The angle between F → 2 and r →is 90° and the cross product is into the page so the torque is negative. Its value is \| τ → 2\| = −rF2 sin 90° = −0.5 m(30 N) = −15.0 N · m. When we evaluate the torque due to F → 3 , we see that the angle it makes with r →is zero so r →× F → 3 = 0. Therefore, F → 3 does not produce any torque on the flywheel. We evaluate the sum of the torques: τnet = ∑ i \|τi\| = 5 −15 = −10 N · m. Chapter 10 \| Fixed-Axis Rotation 513 10.6 Significance The axis of rotation is at the center of mass of the flywheel. Since the flywheel is on a fixed axis, it is not free to translate. If it were on a frictionless surface and not fixed in place, F → 3 would cause the flywheel to translate, as well as F → 1 . Its motion would be a combination of translation and rotation. Check Your Understanding A large ocean-going ship runs aground near the coastline, similar to the fate of the Costa Concordia, and lies at an angle as shown below. Salvage crews must apply a torque to right the ship in order to float the vessel for transport. A force of 5.0 × 105 N acting at point A must be applied to right the ship. What is the torque about the point of contact of the ship with the ground (Figure 10.36)? Figure 10.36 A ship runs aground and tilts, requiring torque to be applied to return the vessel to an upright position. 10.7 \| Newton’s Second Law for Rotation Learning Objectives By the end of this section, you will be able to: • Calculate the torques on rotating systems about a fixed axis to find the angular acceleration • Explain how changes in the moment of inertia of a rotating system affect angular acceleration with a fixed applied torque In this section, we put together all the pieces learned so far in this chapter to analyze the dynamics of rotating rigid bodies. We have analyzed motion with kinematics and rotational kinetic energy but have not yet connected these ideas with force and/or torque. In this section, we introduce the rotational equivalent to Newton’s second law of motion and apply it to rigid bodies with fixed-axis rotation. Newton’s Second Law for Rotation We have thus far found many counterparts to the translational terms used throughout this text, most recently, torque, the rotational analog to force. This raises the question: Is there an analogous equation to Newton’s second law, Σ F →= m a →, which involves torque and rotational motion? To investigate this, we start with Newton’s second law for 514 Chapter 10 \| Fixed-Axis Rotation This OpenStax book is available for free at a single particle rotating around an axis and executing circular motion. Let’s exert a force F →on a point mass m that is at a distance r from a pivot point (Figure 10.37). The particle is constrained to move in a circular path with fixed radius and the force is tangent to the circle. We apply Newton’s second law to determine the magnitude of the acceleration a = F/m in the direction of F →. Recall that the magnitude of the tangential acceleration is proportional to the magnitude of the angular acceleration by a = rα . Substituting this expression into Newton’s second law, we obtain F = mrα. Figure 10.37 An object is supported by a horizontal frictionless table and is attached to a pivot point by a cord that supplies centripetal force. A force F →is applied to the object perpendicular to the radius r, causing it to accelerate about the pivot point. The force is perpendicular to r. Multiply both sides of this equation by r, rF = mr2 α. Note that the left side of this equation is the torque about the axis of rotation, where r is the lever arm and F is the force, perpendicular to r. Recall that the moment of inertia for a point particle is I = mr2 . The torque applied perpendicularly to the point mass in Figure 10.37 is therefore τ = Iα. The torque on the particle is equal to the moment of inertia about the rotation axis times the angular acceleration. We can generalize this equation to a rigid body rotating about a fixed axis. Newton’s Second Law for Rotation If more than one torque acts on a rigid body about a fixed axis, then the sum of the torques equals the moment of inertia times the angular acceleration: (10.25) ∑ i τi = Iα. The term Iα is a scalar quantity and can be positive or negative (counterclockwise or clockwise) depending upon the sign of the net torque. Remember the convention that counterclockwise angular acceleration is positive. Thus, if a rigid body is rotating clockwise and experiences a positive torque (counterclockwise), the angular acceleration is positive. Equation 10.25 is Newton’s second law for rotation and tells us how to relate torque, moment of inertia, and rotational kinematics. This is called the equation for rotational dynamics. With this equation, we can solve a whole class of problems involving force and rotation. It makes sense that the relationship for how much force it takes to rotate a body would include the moment of inertia, since that is the quantity that tells us how easy or hard it is to change the rotational motion of an object. Chapter 10 \| Fixed-Axis Rotation 515 Deriving Newton’s Second Law for Rotation in Vector Form As before, when we found the angular acceleration, we may also find the torque vector. The second law Σ F →= m a → tells us the relationship between net force and how to change the translational motion of an object. We have a vector rotational equivalent of this equation, which can be found by using Equation 10.7 and Figure 10.8. Equation 10.7 relates the angular acceleration to the position and tangential acceleration vectors: a →= α →× r →. We form the cross product of this equation with r →and use a cross product identity (note that r →· α →= 0 ): r →× a →= r →× ( α →× r →) = α →( r →· r →) −r →( r →· α →) = α →( r →· r →) = α →r2. We now form the cross product of Newton’s second law with the position vector r →, Σ( r →× F →) = r →× (m a →) = m r →× a →= mr2 α →. Identifying the first term on the left as the sum of the torques, and mr2 as the moment of inertia, we arrive at Newton’s second law of rotation in vector form: (10.26) Σ τ →= I α →. This equation is exactly Equation 10.25 but with the torque and angular acceleration as vectors. An important point is that the torque vector is in the same direction as the angular acceleration. Applying the Rotational Dynamics Equation Before we apply the rotational dynamics equation to some everyday situations, let’s review a general problem-solving strategy for use with this category of problems. Problem-Solving Strategy: Rotational Dynamics 1. Examine the situation to determine that torque and mass are involved in the rotation. Draw a careful sketch of the situation. 2. Determine the system of interest. 3. Draw a free-body diagram. That is, draw and label all external forces acting on the system of interest. 4. Identify the pivot point. If the object is in equilibrium, it must be in equilibrium for all possible pivot points––chose the one that simplifies your work the most. 5. Apply ∑ i τi = Iα , the rotational equivalent of Newton’s second law, to solve the problem. Care must be taken to use the correct moment of inertia and to consider the torque about the point of rotation. 6. As always, check the solution to see if it is reasonable. Example 10.16 Calculating the Effect of Mass Distribution on a Merry-Go-Round Consider the father pushing a playground merry-go-round in Figure 10.38. He exerts a force of 250 N at the edge of the 50.0-kg merry-go-round, which has a 1.50-m radius. Calculate the angular acceleration produced (a) when no one is on the merry-go-round and (b) when an 18.0-kg child sits 1.25 m away from the center. Consider the merry-go-round itself to be a uniform disk with negligible friction. 516 Chapter 10 \| Fixed-Axis Rotation This OpenStax book is available for free at Figure 10.38 A father pushes a playground merry-go-round at its edge and perpendicular to its radius to achieve maximum torque. Strategy The net torque is given directly by the expression ∑ i τi = Iα , To solve for α , we must first calculate the net torque τ (which is the same in both cases) and moment of inertia I (which is greater in the second case). Solution a. The moment of inertia of a solid disk about this axis is given in Figure 10.20 to be 1 2MR2. We have M = 50.0 kg and R = 1.50 m , so I = (0.500)(50.0 kg)(1.50 m)2 = 56.25 kg-m2. To find the net torque, we note that the applied force is perpendicular to the radius and friction is negligible, so that τ = rFsin θ = (1.50 m)(250.0 N) = 375.0 N-m. Now, after we substitute the known values, we find the angular acceleration to be α = τ I = 375.0 N-m 56.25 kg-m2 = 6.67rad s2 . b. We expect the angular acceleration for the system to be less in this part because the moment of inertia is greater when the child is on the merry-go-round. To find the total moment of inertia I, we first find the child’s moment of inertia Ic by approximating the child as a point mass at a distance of 1.25 m from the axis. Then Ic = mR2 = (18.0 kg)(1.25 m)2 = 28.13 kg-m2. The total moment of inertia is the sum of the moments of inertia of the merry-go-round and the child (about the same axis): I = 28.13 kg-m2 + 56.25 kg-m2 = 84.38 kg-m2. Substituting known values into the equation for α gives Chapter 10 \| Fixed-Axis Rotation 517 10.7 α = τ I = 375.0 N-m 84.38 kg-m2 = 4.44rad s2 . Significance The angular acceleration is less when the child is on the merry-go-round than when the merry-go-round is empty, as expected. The angular accelerations found are quite large, partly due to the fact that friction was considered to be negligible. If, for example, the father kept pushing perpendicularly for 2.00 s, he would give the merry-go-round an angular velocity of 13.3 rad/s when it is empty but only 8.89 rad/s when the child is on it. In terms of revolutions per second, these angular velocities are 2.12 rev/s and 1.41 rev/s, respectively. The father would end up running at about 50 km/h in the first case. Check Your Understanding The fan blades on a jet engine have a moment of inertia 30.0 kg-m2 . In 10 s, they rotate counterclockwise from rest up to a rotation rate of 20 rev/s. (a) What torque must be applied to the blades to achieve this angular acceleration? (b) What is the torque required to bring the fan blades rotating at 20 rev/s to a rest in 20 s? 10.8 \| Work and Power for Rotational Motion Learning Objectives By the end of this section, you will be able to: • Use the work-energy theorem to analyze rotation to find the work done on a system when it is rotated about a fixed axis for a finite angular displacement • Solve for the angular velocity of a rotating rigid body using the work-energy theorem • Find the power delivered to a rotating rigid body given the applied torque and angular velocity • Summarize the rotational variables and equations and relate them to their translational counterparts Thus far in the chapter, we have extensively addressed kinematics and dynamics for rotating rigid bodies around a fixed axis. In this final section, we define work and power within the context of rotation about a fixed axis, which has applications to both physics and engineering. The discussion of work and power makes our treatment of rotational motion almost complete, with the exception of rolling motion and angular momentum, which are discussed in Angular Momentum. We begin this section with a treatment of the work-energy theorem for rotation. Work for Rotational Motion Now that we have determined how to calculate kinetic energy for rotating rigid bodies, we can proceed with a discussion of the work done on a rigid body rotating about a fixed axis. Figure 10.39 shows a rigid body that has rotated through an angle dθ from A to B while under the influence of a force F →. The external force F → is applied to point P, whose position is r →, and the rigid body is constrained to rotate about a fixed axis that is perpendicular to the page and passes through O. The rotational axis is fixed, so the vector r → moves in a circle of radius r, and the vector d s → is perpendicular to r →. 518 Chapter 10 \| Fixed-Axis Rotation This OpenStax book is available for free at Figure 10.39 A rigid body rotates through an angle dθ from A to B by the action of an external force F →applied to point P. From Equation 10.2, we have s →= θ →× r →. Thus, d s →= d( θ →× r →) = d θ →× r →+ d r →× θ →= d θ →× r →. Note that d r →is zero because r →is fixed on the rigid body from the origin O to point P. Using the definition of work, we obtain W = ∫∑F →· d s →= ∫∑F →· (d θ →× r →) = ∫d θ →· ( r →× ∑F →) where we used the identity a →· ( b →× c →) = b →· ( c →× a →) . Noting that ( r →× ∑F →) = ∑τ →, we arrive at the expression for the rotational work done on a rigid body: (10.27) W = ∫∑τ →· d θ →. The total work done on a rigid body is the sum of the torques integrated over the angle through which the body rotates. The incremental work is (10.28) dW = ⎛ ⎝ ⎜∑ i τi ⎞ ⎠ ⎟dθ where we have taken the dot product in Equation 10.27, leaving only torques along the axis of rotation. In a rigid body, all particles rotate through the same angle; thus the work of every external force is equal to the torque times the common incremental angle dθ . The quantity ⎛ ⎝ ⎜∑ i τi ⎞ ⎠ ⎟is the net torque on the body due to external forces. Similarly, we found the kinetic energy of a rigid body rotating around a fixed axis by summing the kinetic energy of each particle that makes up the rigid body. Since the work-energy theorem Wi = ΔKi is valid for each particle, it is valid for the Chapter 10 \| Fixed-Axis Rotation 519 sum of the particles and the entire body. Work-Energy Theorem for Rotation The work-energy theorem for a rigid body rotating around a fixed axis is (10.29) WAB = KB −KA where K = 1 2Iω2 and the rotational work done by a net force rotating a body from point A to point B is (10.30) WAB = ⌠ ⌡ θA θB⎛ ⎝ ⎜∑ i τi ⎞ ⎠ ⎟dθ. We give a strategy for using this equation when analyzing rotational motion. Problem-Solving Strategy: Work-Energy Theorem for Rotational Motion 1. Identify the forces on the body and draw a free-body diagram. Calculate the torque for each force. 2. Calculate the work done during the body’s rotation by every torque. 3. Apply the work-energy theorem by equating the net work done on the body to the change in rotational kinetic energy. Let’s look at two examples and use the work-energy theorem to analyze rotational motion. Example 10.17 Rotational Work and Energy A 12.0 N · m torque is applied to a flywheel that rotates about a fixed axis and has a moment of inertia of 30.0 kg · m2 . If the flywheel is initially at rest, what is its angular velocity after it has turned through eight revolutions? Strategy We apply the work-energy theorem. We know from the problem description what the torque is and the angular displacement of the flywheel. Then we can solve for the final angular velocity. Solution The flywheel turns through eight revolutions, which is 16π radians. The work done by the torque, which is constant and therefore can come outside the integral in Equation 10.30, is WAB = τ(θB −θA). We apply the work-energy theorem: WAB = τ(θB −θA) = 1 2IωB 2 −1 2IωA 2. With τ = 12.0 N · m, θB −θA = 16.0π rad, I = 30.0 kg · m2, and ωA = 0 , we have 12.0 N-m(16.0π rad) = 1 2(30.0 kg · m2)(ωB 2) −0. 520 Chapter 10 \| Fixed-Axis Rotation This OpenStax book is available for free at Therefore, ωB = 6.3 rad/s. This is the angular velocity of the flywheel after eight revolutions. Significance The work-energy theorem provides an efficient way to analyze rotational motion, connecting torque with rotational kinetic energy. Example 10.18 Rotational Work: A Pulley A string wrapped around the pulley in Figure 10.40 is pulled with a constant downward force F →of magnitude 50 N. The radius R and moment of inertia I of the pulley are 0.10 m and 2.5 × 10−3kg-m2 , respectively. If the string does not slip, what is the angular velocity of the pulley after 1.0 m of string has unwound? Assume the pulley starts from rest. Figure 10.40 (a) A string is wrapped around a pulley of radius R. (b) The free-body diagram. Strategy Looking at the free-body diagram, we see that neither B →, the force on the bearings of the pulley, nor M g →, the weight of the pulley, exerts a torque around the rotational axis, and therefore does no work on the pulley. As the pulley rotates through an angle θ, F →acts through a distance d such that d = Rθ. Solution Since the torque due to F →has magnitude τ = RF , we have W = τθ = (FR)θ = Fd. If the force on the string acts through a distance of 1.0 m, we have, from the work-energy theorem, Chapter 10 \| Fixed-Axis Rotation 521 10.8 WAB = KB −KA Fd = 1 2Iω2 −0 (50.0 N)(1.0 m) = 1 2(2.5 × 10−3kg-m2)ω2. Solving for ω , we obtain ω = 200.0 rad/s. Power for Rotational Motion Power always comes up in the discussion of applications in engineering and physics. Power for rotational motion is equally as important as power in linear motion and can be derived in a similar way as in linear motion when the force is a constant. The linear power when the force is a constant is P = F →· v →. If the net torque is constant over the angular displacement, Equation 10.25 simplifies and the net torque can be taken out of the integral. In the following discussion, we assume the net torque is constant. We can apply the definition of power derived in Power to rotational motion. From Work and Kinetic Energy, the instantaneous power (or just power) is defined as the rate of doing work, P = dW dt . If we have a constant net torque, Equation 10.25 becomes W = τθ and the power is P = dW dt = d dt(τθ) = τdθ dt or (10.31) P = τω. Example 10.19 Torque on a Boat Propeller A boat engine operating at 9.0 × 104 W is running at 300 rev/min. What is the torque on the propeller shaft? Strategy We are given the rotation rate in rev/min and the power consumption, so we can easily calculate the torque. Solution 300.0 rev/min = 31.4 rad/s; τ = P ω = 9.0 × 104 N · m/s 31.4 rad/s = 2864.8 N · m. Significance It is important to note the radian is a dimensionless unit because its definition is the ratio of two lengths. It therefore does not appear in the solution. Check Your Understanding A constant torque of 500 kN · m is applied to a wind turbine to keep it rotating at 6 rad/s. What is the power required to keep the turbine rotating? Rotational and Translational Relationships Summarized The rotational quantities and their linear analog are summarized in three tables. Table 10.5 summarizes the rotational variables for circular motion about a fixed axis with their linear analogs and the connecting equation, except for the 522 Chapter 10 \| Fixed-Axis Rotation This OpenStax book is available for free at centripetal acceleration, which stands by itself. Table 10.6 summarizes the rotational and translational kinematic equations. Table 10.7 summarizes the rotational dynamics equations with their linear analogs. Rotational Translational Relationship θ x θ = s r ω vt ω = vt r α at α = at r ac ac = vt 2 r Table 10.5 Rotational and Translational Variables: Summary Rotational Translational θf = θ0 + ω – t x = x0 + v – t ωf = ω0 + αt vf = v0 + at θf = θ0 + ω0 t + 1 2αt2 xf = x0 + v0 t + 1 2at2 ωf 2 = ω2 0 + 2α(Δθ) vf 2 = v2 0 + 2a(Δx) Table 10.6 Rotational and Translational Kinematic Equations: Summary Rotational Translational I = ∑ i miri 2 m K = 1 2Iω2 K = 1 2mv2 ∑ i τi = Iα ∑ i F → i = m a → WAB = ⌠ ⌡ θA θB⎛ ⎝ ⎜∑ i τi ⎞ ⎠ ⎟dθ W = ∫F →· d s → P = τω P = F →· v → Table 10.7 Rotational and Translational Equations: Dynamics Chapter 10 \| Fixed-Axis Rotation 523 angular acceleration angular position angular velocity instantaneous angular acceleration instantaneous angular velocity kinematics of rotational motion lever arm linear mass density moment of inertia Newton’s second law for rotation parallel axis parallel-axis theorem rotational dynamics rotational kinetic energy rotational work surface mass density torque total linear acceleration work-energy theorem for rotation CHAPTER 10 REVIEW KEY TERMS time rate of change of angular velocity angle a body has rotated through in a fixed coordinate system time rate of change of angular position derivative of angular velocity with respect to time derivative of angular position with respect to time describes the relationships among rotation angle, angular velocity, angular acceleration, and time perpendicular distance from the line that the force vector lies on to a given axis the mass per unit length λ of a one dimensional object rotational mass of rigid bodies that relates to how easy or hard it will be to change the angular velocity of the rotating rigid body sum of the torques on a rotating system equals its moment of inertia times its angular acceleration axis of rotation that is parallel to an axis about which the moment of inertia of an object is known if the moment of inertia is known for a given axis, it can be found for any axis parallel to it analysis of rotational motion using the net torque and moment of inertia to find the angular acceleration kinetic energy due to the rotation of an object; this is part of its total kinetic energy work done on a rigid body due to the sum of the torques integrated over the angle through with the body rotates mass per unit area σ of a two dimensional object cross product of a force and a lever arm to a given axis vector sum of the centripetal acceleration vector and the tangential acceleration vector the total rotational work done on a rigid body is equal to the change in rotational kinetic energy of the body KEY EQUATIONS Angular position θ = s r Angular velocity ω = lim Δt →0 Δθ Δt = dθ dt Tangential speed vt = rω Angular acceleration α = lim Δt →0 Δω Δt = dω dt = d2 θ dt2 Tangential acceleration at = rα Average angular velocity ω – = ω0 + ωf 2 Angular displacement θf = θ0 + ω – t 524 Chapter 10 \| Fixed-Axis Rotation This OpenStax book is available for free at Angular velocity from constant angular acceleration ωf = ω0 + αt Angular velocity from displacement and constant angular acceleration θf = θ0 + ω0 t + 1 2αt2 Change in angular velocity ωf 2 = ω0 2 + 2α(Δθ) Total acceleration a →= a → c + a → t Rotational kinetic energy K = 1 2 ⎛ ⎝ ⎜∑ j m j r j 2⎞ ⎠ ⎟ω2 Moment of inertia I = ∑ j m j r j 2 Rotational kinetic energy in terms of the moment of inertia of a rigid body K = 1 2Iω2 Moment of inertia of a continuous object I = ∫r2 dm Parallel-axis theorem Iparallel-axis = Icenter of mass + md2 Moment of inertia of a compound object Itotal = ∑ i Ii Torque vector τ →= r →× F → Magnitude of torque \| τ →\| = r⊥F Total torque τnet = ∑ i \|τi\| Newton’s second law for rotation ∑ i τi = Iα Incremental work done by a torque dW = ⎛ ⎝ ⎜∑ i τi ⎞ ⎠ ⎟dθ Work-energy theorem WAB = KB −KA Rotational work done by net force WAB = ⌠ ⌡ θA θB⎛ ⎝ ⎜∑ i τi ⎞ ⎠ ⎟dθ Rotational power P = τω SUMMARY 10.1 Rotational Variables • The angular position θ of a rotating body is the angle the body has rotated through in a fixed coordinate system, which serves as a frame of reference. • The angular velocity of a rotating body about a fixed axis is defined as ω(rad/s) , the rotational rate of the body in radians per second. The instantaneous angular velocity of a rotating body ω = lim Δt →0 Δω Δt = dθ dt is the derivative Chapter 10 \| Fixed-Axis Rotation 525 with respect to time of the angular position θ , found by taking the limit Δt →0 in the average angular velocity ω – = Δθ Δt . The angular velocity relates vt to the tangential speed of a point on the rotating body through the relation vt = rω , where r is the radius to the point and vt is the tangential speed at the given point. • The angular velocity ω →is found using the right-hand rule. If the fingers curl in the direction of rotation about a fixed axis, the thumb points in the direction of ω →(see Figure 10.5). • If the system’s angular velocity is not constant, then the system has an angular acceleration. The average angular acceleration over a given time interval is the change in angular velocity over this time interval, α – = Δω Δt . The instantaneous angular acceleration is the time derivative of angular velocity, α = lim Δt →0 Δω Δt = dω dt . The angular acceleration α → is found by locating the angular velocity. If a rotation rate of a rotating body is decreasing, the angular acceleration is in the opposite direction to ω →. If the rotation rate is increasing, the angular acceleration is in the same direction as ω →. • The tangential acceleration of a point at a radius from the axis of rotation is the angular acceleration times the radius to the point. 10.2 Rotation with Constant Angular Acceleration • The kinematics of rotational motion describes the relationships among rotation angle (angular position), angular velocity, angular acceleration, and time. • For a constant angular acceleration, the angular velocity varies linearly. Therefore, the average angular velocity is 1/2 the initial plus final angular velocity over a given time period: ω – = ω0 + ωf 2 . • We used a graphical analysis to find solutions to fixed-axis rotation with constant angular acceleration. From the relation ω = dθ dt , we found that the area under an angular velocity-vs.-time curve gives the angular displacement, θf −θ0 = Δθ = ∫ t0 t ω(t)dt . The results of the graphical analysis were verified using the kinematic equations for constant angular acceleration. Similarly, since α = dω dt , the area under an angular acceleration-vs.-time graph gives the change in angular velocity: ω f −ω0 = Δω = ∫ t0 t α(t)dt . 10.3 Relating Angular and Translational Quantities • The linear kinematic equations have their rotational counterparts such that there is a mapping x →θ, v →ω, a →α . • A system undergoing uniform circular motion has a constant angular velocity, but points at a distance r from the rotation axis have a linear centripetal acceleration. • A system undergoing nonuniform circular motion has an angular acceleration and therefore has both a linear centripetal and linear tangential acceleration at a point a distance r from the axis of rotation. • The total linear acceleration is the vector sum of the centripetal acceleration vector and the tangential acceleration vector. Since the centripetal and tangential acceleration vectors are perpendicular to each other for circular motion, the magnitude of the total linear acceleration is \| a →\| = ac 2 + at 2 . 526 Chapter 10 \| Fixed-Axis Rotation This OpenStax book is available for free at 10.4 Moment of Inertia and Rotational Kinetic Energy • The rotational kinetic energy is the kinetic energy of rotation of a rotating rigid body or system of particles, and is given by K = 1 2Iω2 , where I is the moment of inertia, or “rotational mass” of the rigid body or system of particles. • The moment of inertia for a system of point particles rotating about a fixed axis is I = ∑ j m j r j 2 , where m j is the mass of the point particle and r j is the distance of the point particle to the rotation axis. Because of the r2 term, the moment of inertia increases as the square of the distance to the fixed rotational axis. The moment of inertia is the rotational counterpart to the mass in linear motion. • In systems that are both rotating and translating, conservation of mechanical energy can be used if there are no nonconservative forces at work. The total mechanical energy is then conserved and is the sum of the rotational and translational kinetic energies, and the gravitational potential energy. 10.5 Calculating Moments of Inertia • Moments of inertia can be found by summing or integrating over every ‘piece of mass’ that makes up an object, multiplied by the square of the distance of each ‘piece of mass’ to the axis. In integral form the moment of inertia is I = ∫r2 dm . • Moment of inertia is larger when an object’s mass is farther from the axis of rotation. • It is possible to find the moment of inertia of an object about a new axis of rotation once it is known for a parallel axis. This is called the parallel axis theorem given by Iparallel-axis = Icenter of mass + md2 , where d is the distance from the initial axis to the parallel axis. • Moment of inertia for a compound object is simply the sum of the moments of inertia for each individual object that makes up the compound object. 10.6 Torque • The magnitude of a torque about a fixed axis is calculated by finding the lever arm to the point where the force is applied and using the relation \| τ →\| = r⊥F , where r⊥ is the perpendicular distance from the axis to the line upon which the force vector lies. • The sign of the torque is found using the right hand rule. If the page is the plane containing r → and F →, then r →× F →is out of the page for positive torques and into the page for negative torques. • The net torque can be found from summing the individual torques about a given axis. 10.7 Newton’s Second Law for Rotation • Newton’s second law for rotation, ∑ i τi = Iα , says that the sum of the torques on a rotating system about a fixed axis equals the product of the moment of inertia and the angular acceleration. This is the rotational analog to Newton’s second law of linear motion. • In the vector form of Newton’s second law for rotation, the torque vector τ → is in the same direction as the angular acceleration α →. If the angular acceleration of a rotating system is positive, the torque on the system is also positive, and if the angular acceleration is negative, the torque is negative. 10.8 Work and Power for Rotational Motion • The incremental work dW in rotating a rigid body about a fixed axis is the sum of the torques about the axis times the incremental angle dθ . Chapter 10 \| Fixed-Axis Rotation 527 • The total work done to rotate a rigid body through an angle θ about a fixed axis is the sum of the torques integrated over the angular displacement. If the torque is a constant as a function of θ , then WAB = τ(θB −θA) . • The work-energy theorem relates the rotational work done to the change in rotational kinetic energy: WAB = KB −KA where K = 1 2Iω2. • The power delivered to a system that is rotating about a fixed axis is the torque times the angular velocity, P = τω . CONCEPTUAL QUESTIONS 10.1 Rotational Variables 1. A clock is mounted on the wall. As you look at it, what is the direction of the angular velocity vector of the second hand? 2. What is the value of the angular acceleration of the second hand of the clock on the wall? 3. A baseball bat is swung. Do all points on the bat have the same angular velocity? The same tangential speed? 4. The blades of a blender on a counter are rotating clockwise as you look into it from the top. If the blender is put to a greater speed what direction is the angular acceleration of the blades? 10.2 Rotation with Constant Angular Acceleration 5. If a rigid body has a constant angular acceleration, what is the functional form of the angular velocity in terms of the time variable? 6. If a rigid body has a constant angular acceleration, what is the functional form of the angular position? 7. If the angular acceleration of a rigid body is zero, what is the functional form of the angular velocity? 8. A massless tether with a masses tied to both ends rotates about a fixed axis through the center. Can the total acceleration of the tether/mass combination be zero if the angular velocity is constant? 10.3 Relating Angular and Translational Quantities 9. Explain why centripetal acceleration changes the direction of velocity in circular motion but not its magnitude. 10. In circular motion, a tangential acceleration can change the magnitude of the velocity but not its direction. Explain your answer. 11. Suppose a piece of food is on the edge of a rotating microwave oven plate. Does it experience nonzero tangential acceleration, centripetal acceleration, or both when: (a) the plate starts to spin faster? (b) The plate rotates at constant angular velocity? (c) The plate slows to a halt? 10.4 Moment of Inertia and Rotational Kinetic Energy 12. What if another planet the same size as Earth were put into orbit around the Sun along with Earth. Would the moment of inertia of the system increase, decrease, or stay the same? 13. A solid sphere is rotating about an axis through its center at a constant rotation rate. Another hollow sphere of the same mass and radius is rotating about its axis through the center at the same rotation rate. Which sphere has a greater rotational kinetic energy? 10.5 Calculating Moments of Inertia 14. If a child walks toward the center of a merry-go-round, does the moment of inertia increase or decrease? 15. A discus thrower rotates with a discus in his hand before letting it go. (a) How does his moment of inertia change after releasing the discus? (b) What would be a good approximation to use in calculating the moment of inertia of the discus thrower and discus? 16. Does increasing the number of blades on a propeller increase or decrease its moment of inertia, and why? 17. The moment of inertia of a long rod spun around an axis through one end perpendicular to its length is mL2 /3 . Why is this moment of inertia greater than it would be if you spun a point mass m at the location of the center of mass of the rod (at L/2) (that would be mL2 /4 )? 528 Chapter 10 \| Fixed-Axis Rotation This OpenStax book is available for free at 18. Why is the moment of inertia of a hoop that has a mass M and a radius R greater than the moment of inertia of a disk that has the same mass and radius? 10.6 Torque 19. What three factors affect the torque created by a force relative to a specific pivot point? 20. Give an example in which a small force exerts a large torque. Give another example in which a large force exerts a small torque. 21. When reducing the mass of a racing bike, the greatest benefit is realized from reducing the mass of the tires and wheel rims. Why does this allow a racer to achieve greater accelerations than would an identical reduction in the mass of the bicycle’s frame? 22. Can a single force produce a zero torque? 23. Can a set of forces have a net torque that is zero and a net force that is not zero? 24. Can a set of forces have a net force that is zero and a net torque that is not zero? 25. In the expression r →× F → can \| r →\| ever be less than the lever arm? Can it be equal to the lever arm? 10.7 Newton’s Second Law for Rotation 26. If you were to stop a spinning wheel with a constant force, where on the wheel would you apply the force to produce the maximum negative acceleration? 27. A rod is pivoted about one end. Two forces F →and −F → are applied to it. Under what circumstances will the rod not rotate? PROBLEMS 10.1 Rotational Variables 28. Calculate the angular velocity of Earth. 29. A track star runs a 400-m race on a 400-m circular track in 45 s. What is his angular velocity assuming a constant speed? 30. A wheel rotates at a constant rate of 2.0 × 103 rev/min . (a) What is its angular velocity in radians per second? (b) Through what angle does it turn in 10 s? Express the solution in radians and degrees. 31. A particle moves 3.0 m along a circle of radius 1.5 m. (a) Through what angle does it rotate? (b) If the particle makes this trip in 1.0 s at a constant speed, what is its angular velocity? (c) What is its acceleration? 32. A compact disc rotates at 500 rev/min. If the diameter of the disc is 120 mm, (a) what is the tangential speed of a point at the edge of the disc? (b) At a point halfway to the center of the disc? 33. Unreasonable results. The propeller of an aircraft is spinning at 10 rev/s when the pilot shuts off the engine. The propeller reduces its angular velocity at a constant 2.0 rad/s2 for a time period of 40 s. What is the rotation rate of the propeller in 40 s? Is this a reasonable situation? 34. A gyroscope slows from an initial rate of 32.0 rad/s at a rate of 0.700 rad/s2 . How long does it take to come to rest? 35. On takeoff, the propellers on a UAV (unmanned aerial vehicle) increase their angular velocity from rest at a rate of ω = (25.0t) rad/s for 3.0 s. (a) What is the instantaneous angular velocity of the propellers at t = 2.0 s ? (b) What is the angular acceleration? 36. The angular position of a rod varies as 20.0t2 radians from time t = 0 . The rod has two beads on it as shown in the following figure, one at 10 cm from the rotation axis and the other at 20 cm from the rotation axis. (a) What is the instantaneous angular velocity of the rod at t = 5 s? (b) What is the angular acceleration of the rod? (c) What are the tangential speeds of the beads at t = 5 s? (d) What are the tangential accelerations of the beads at t = 5 s? (e) What are the centripetal accelerations of the beads at t = 5 s? Chapter 10 \| Fixed-Axis Rotation 529 10.2 Rotation with Constant Angular Acceleration 37. A wheel has a constant angular acceleration of 5.0 rad/s2 . Starting from rest, it turns through 300 rad. (a) What is its final angular velocity? (b) How much time elapses while it turns through the 300 radians? 38. During a 6.0-s time interval, a flywheel with a constant angular acceleration turns through 500 radians that acquire an angular velocity of 100 rad/s. (a) What is the angular velocity at the beginning of the 6.0 s? (b) What is the angular acceleration of the flywheel? 39. The angular velocity of a rotating rigid body increases from 500 to 1500 rev/min in 120 s. (a) What is the angular acceleration of the body? (b) Through what angle does it turn in this 120 s? 40. A flywheel slows from 600 to 400 rev/min while rotating through 40 revolutions. (a) What is the angular acceleration of the flywheel? (b) How much time elapses during the 40 revolutions? 41. A wheel 1.0 m in diameter rotates with an angular acceleration of 4.0 rad/s2 . (a) If the wheel’s initial angular velocity is 2.0 rad/s, what is its angular velocity after 10 s? (b) Through what angle does it rotate in the 10-s interval? (c) What are the tangential speed and acceleration of a point on the rim of the wheel at the end of the 10-s interval? 42. A vertical wheel with a diameter of 50 cm starts from rest and rotates with a constant angular acceleration of 5.0 rad/s2 around a fixed axis through its center counterclockwise. (a) Where is the point that is initially at the bottom of the wheel at t = 10 s? (b) What is the point’s linear acceleration at this instant? 43. A circular disk of radius 10 cm has a constant angular acceleration of 1.0 rad/s2 ; at t = 0 its angular velocity is 2.0 rad/s. (a) Determine the disk’s angular velocity at t = 5.0 s . (b) What is the angle it has rotated through during this time? (c) What is the tangential acceleration of a point on the disk at t = 5.0 s? 44. The angular velocity vs. time for a fan on a hovercraft is shown below. (a) What is the angle through which the fan blades rotate in the first 8 seconds? (b) Verify your result using the kinematic equations. 45. A rod of length 20 cm has two beads attached to its ends. The rod with beads starts rotating from rest. If the beads are to have a tangential speed of 20 m/s in 7 s, what is the angular acceleration of the rod to achieve this? 10.3 Relating Angular and Translational Quantities 46. At its peak, a tornado is 60.0 m in diameter and carries 500 km/h winds. What is its angular velocity in revolutions per second? 47. A man stands on a merry-go-round that is rotating at 2.5 rad/s. If the coefficient of static friction between the man’s shoes and the merry-go-round is µS = 0.5 , how far from the axis of rotation can he stand without sliding? 48. An ultracentrifuge accelerates from rest to 100,000 rpm in 2.00 min. (a) What is the average angular acceleration in rad/s2 ? (b) What is the tangential acceleration of a point 9.50 cm from the axis of rotation? (c) What is the centripetal acceleration in m/s2 and multiples of g of this point at full rpm? (d) What is the total distance travelled by a point 9.5 cm from the axis of rotation of the ultracentrifuge? 49. A wind turbine is rotating counterclockwise at 0.5 rev/ s and slows to a stop in 10 s. Its blades are 20 m in length. (a) What is the angular acceleration of the turbine? (b) What is the centripetal acceleration of the tip of the blades at t = 0 s? (c) What is the magnitude and direction of the total linear acceleration of the tip of the blades at t = 0 s? 50. What is (a) the angular speed and (b) the linear speed of a point on Earth’s surface at latitude 30° N. Take the radius of the Earth to be 6309 km. (c) At what latitude 530 Chapter 10 \| Fixed-Axis Rotation This OpenStax book is available for free at would your linear speed be 10 m/s? 51. A child with mass 40 kg sits on the edge of a merry-go-round at a distance of 3.0 m from its axis of rotation. The merry-go-round accelerates from rest up to 0.4 rev/s in 10 s. If the coefficient of static friction between the child and the surface of the merry-go-round is 0.6, does the child fall off before 5 s? 52. A bicycle wheel with radius 0.3 m rotates from rest to 3 rev/s in 5 s. What is the magnitude and direction of the total acceleration vector at the edge of the wheel at 1.0 s? 53. The angular velocity of a flywheel with radius 1.0 m varies according to ω(t) = 2.0t . Plot ac(t) and at(t) from t = 0 to 3.0 s for r = 1.0 m . Analyze these results to explain when ac ≫at and when ac ≪at for a point on the flywheel at a radius of 1.0 m. 10.4 Moment of Inertia and Rotational Kinetic Energy 54. A system of point particles is shown in the following figure. Each particle has mass 0.3 kg and they all lie in the same plane. (a) What is the moment of inertia of the system about the given axis? (b) If the system rotates at 5 rev/s, what is its rotational kinetic energy? 55. (a) Calculate the rotational kinetic energy of Earth on its axis. (b) What is the rotational kinetic energy of Earth in its orbit around the Sun? 56. Calculate the rotational kinetic energy of a 12-kg motorcycle wheel if its angular velocity is 120 rad/s and its inner radius is 0.280 m and outer radius 0.330 m. 57. A baseball pitcher throws the ball in a motion where there is rotation of the forearm about the elbow joint as well as other movements. If the linear velocity of the ball relative to the elbow joint is 20.0 m/s at a distance of 0.480 m from the joint and the moment of inertia of the forearm is 0.500 kg-m2 , what is the rotational kinetic energy of the forearm? 58. A diver goes into a somersault during a dive by tucking her limbs. If her rotational kinetic energy is 100 J and her moment of inertia in the tuck is 9.0 kg · m2 , what is her rotational rate during the somersault? 59. An aircraft is coming in for a landing at 300 meters height when the propeller falls off. The aircraft is flying at 40.0 m/s horizontally. The propeller has a rotation rate of 20 rev/s, a moment of inertia of 70.0 kg-m2 , and a mass of 200 kg. Neglect air resistance. (a) With what translational velocity does the propeller hit the ground? (b) What is the rotation rate of the propeller at impact? 60. If air resistance is present in the preceding problem and reduces the propeller’s rotational kinetic energy at impact by 30%, what is the propeller’s rotation rate at impact? 61. A neutron star of mass 2 × 1030 kg and radius 10 km rotates with a period of 0.02 seconds. What is its rotational kinetic energy? 62. An electric sander consisting of a rotating disk of mass 0.7 kg and radius 10 cm rotates at 15 rev/sec. When applied to a rough wooden wall the rotation rate decreases by 20%. (a) What is the final rotational kinetic energy of the rotating disk? (b) How much has its rotational kinetic energy decreased? 63. A system consists of a disk of mass 2.0 kg and radius 50 cm upon which is mounted an annular cylinder of mass 1.0 kg with inner radius 20 cm and outer radius 30 cm (see below). The system rotates about an axis through the center of the disk and annular cylinder at 10 rev/s. (a) What is the moment of inertia of the system? (b) What is its rotational kinetic energy? 10.5 Calculating Moments of Inertia 64. While punting a football, a kicker rotates his leg about the hip joint. The moment of inertia of the leg is 3.75 kg-m2 and its rotational kinetic energy is 175 J. (a) What is the angular velocity of the leg? (b) What is the Chapter 10 \| Fixed-Axis Rotation 531 velocity of tip of the punter’s shoe if it is 1.05 m from the hip joint? 65. Using the parallel axis theorem, what is the moment of inertia of the rod of mass m about the axis shown below? 66. Find the moment of inertia of the rod in the previous problem by direct integration. 67. A uniform rod of mass 1.0 kg and length 2.0 m is free to rotate about one end (see the following figure). If the rod is released from rest at an angle of 60° with respect to the horizontal, what is the speed of the tip of the rod as it passes the horizontal position? 68. A pendulum consists of a rod of mass 2 kg and length 1 m with a solid sphere at one end with mass 0.3 kg and radius 20 cm (see the following figure). If the pendulum is released from rest at an angle of 30° , what is the angular velocity at the lowest point? 69. A solid sphere of radius 10 cm is allowed to rotate freely about an axis. The sphere is given a sharp blow so that its center of mass starts from the position shown in the following figure with speed 15 cm/s. What is the maximum angle that the diameter makes with the vertical? 70. Calculate the moment of inertia by direct integration of a thin rod of mass M and length L about an axis through the rod at L/3, as shown below. Check your answer with the parallel-axis theorem. 10.6 Torque 71. Two flywheels of negligible mass and different radii are bonded together and rotate about a common axis (see below). The smaller flywheel of radius 30 cm has a cord that has a pulling force of 50 N on it. What pulling force needs to be applied to the cord connecting the larger flywheel of radius 50 cm such that the combination does not rotate? 72. The cylindrical head bolts on a car are to be tightened with a torque of 62.0 N · m . If a mechanic uses a wrench of length 20 cm, what perpendicular force must he exert on the end of the wrench to tighten a bolt correctly? 73. (a) When opening a door, you push on it perpendicularly with a force of 55.0 N at a distance of 0.850 m from the hinges. What torque are you exerting relative to the hinges? (b) Does it matter if you push at the same height as the hinges? There is only one pair of hinges. 74. When tightening a bolt, you push perpendicularly on a wrench with a force of 165 N at a distance of 0.140 m from the center of the bolt. How much torque are you exerting in newton-meters (relative to the center of the bolt)? 75. What hanging mass must be placed on the cord to keep the pulley from rotating (see the following figure)? The 532 Chapter 10 \| Fixed-Axis Rotation This OpenStax book is available for free at mass on the frictionless plane is 5.0 kg. The inner radius of the pulley is 20 cm and the outer radius is 30 cm. 76. A simple pendulum consists of a massless tether 50 cm in length connected to a pivot and a small mass of 1.0 kg attached at the other end. What is the torque about the pivot when the pendulum makes an angle of 40° with respect to the vertical? 77. Calculate the torque about the z-axis that is out of the page at the origin in the following figure, given that F1 = 3 N, F2 = 2 N, F3 = 3 N, F4 = 1.8 N . 78. A seesaw has length 10.0 m and uniform mass 10.0 kg and is resting at an angle of 30° with respect to the ground (see the following figure). The pivot is located at 6.0 m. What magnitude of force needs to be applied perpendicular to the seesaw at the raised end so as to allow the seesaw to barely start to rotate? 79. A pendulum consists of a rod of mass 1 kg and length 1 m connected to a pivot with a solid sphere attached at the other end with mass 0.5 kg and radius 30 cm. What is the torque about the pivot when the pendulum makes an angle of 30° with respect to the vertical? 80. A torque of 5.00 × 103 N · m is required to raise a drawbridge (see the following figure). What is the tension necessary to produce this torque? Would it be easier to raise the drawbridge if the angle θ were larger or smaller? 81. A horizontal beam of length 3 m and mass 2.0 kg has a mass of 1.0 kg and width 0.2 m sitting at the end of the beam (see the following figure). What is the torque of the system about the support at the wall? 82. What force must be applied to end of a rod along the x-axis of length 2.0 m in order to produce a torque on the rod about the origin of 8.0k ^ N · m ? 83. What is the torque about the origin of the force Chapter 10 \| Fixed-Axis Rotation 533 (5.0 i ^ −2.0 j ^ + 1.0k ^) N if it is applied at the point whose position is: r →= ⎛ ⎝−2.0 i ^ + 4.0 j ^⎞ ⎠m? 10.7 Newton’s Second Law for Rotation 84. You have a grindstone (a disk) that is 90.0 kg, has a 0.340-m radius, and is turning at 90.0 rpm, and you press a steel axe against it with a radial force of 20.0 N. (a) Assuming the kinetic coefficient of friction between steel and stone is 0.20, calculate the angular acceleration of the grindstone. (b) How many turns will the stone make before coming to rest? 85. Suppose you exert a force of 180 N tangential to a 0.280-m-radius, 75.0-kg grindstone (a solid disk). (a)What torque is exerted? (b) What is the angular acceleration assuming negligible opposing friction? (c) What is the angular acceleration if there is an opposing frictional force of 20.0 N exerted 1.50 cm from the axis? 86. A flywheel ( I = 50 kg-m2 ) starting from rest acquires an angular velocity of 200.0 rad/s while subject to a constant torque from a motor for 5 s. (a) What is the angular acceleration of the flywheel? (b) What is the magnitude of the torque? 87. A constant torque is applied to a rigid body whose moment of inertia is 4.0 kg-m2 around the axis of rotation. If the wheel starts from rest and attains an angular velocity of 20.0 rad/s in 10.0 s, what is the applied torque? 88. A torque of 50.0 N-m is applied to a grinding wheel ( I = 20.0 kg-m2 ) for 20 s. (a) If it starts from rest, what is the angular velocity of the grinding wheel after the torque is removed? (b) Through what angle does the wheel move while the torque is applied? 89. A flywheel ( I = 100.0 kg-m2 ) rotating at 500.0 rev/ min is brought to rest by friction in 2.0 min. What is the frictional torque on the flywheel? 90. A uniform cylindrical grinding wheel of mass 50.0 kg and diameter 1.0 m is turned on by an electric motor. The friction in the bearings is negligible. (a) What torque must be applied to the wheel to bring it from rest to 120 rev/min in 20 revolutions? (b) A tool whose coefficient of kinetic friction with the wheel is 0.60 is pressed perpendicularly against the wheel with a force of 40.0 N. What torque must be supplied by the motor to keep the wheel rotating at a constant angular velocity? 91. Suppose when Earth was created, it was not rotating. However, after the application of a uniform torque after 6 days, it was rotating at 1 rev/day. (a) What was the angular acceleration during the 6 days? (b) What torque was applied to Earth during this period? (c) What force tangent to Earth at its equator would produce this torque? 92. A pulley of moment of inertia 2.0 kg-m2 is mounted on a wall as shown in the following figure. Light strings are wrapped around two circumferences of the pulley and weights are attached. What are (a) the angular acceleration of the pulley and (b) the linear acceleration of the weights? Assume the following data: r1 = 50 cm, r2 = 20 cm, m1 = 1.0 kg, m2 = 2.0 kg . 93. A block of mass 3 kg slides down an inclined plane at an angle of 45° with a massless tether attached to a pulley with mass 1 kg and radius 0.5 m at the top of the incline (see the following figure). The pulley can be approximated as a disk. The coefficient of kinetic friction on the plane is 0.4. What is the acceleration of the block? 94. The cart shown below moves across the table top as the block falls. What is the acceleration of the cart? Neglect friction and assume the following data: 534 Chapter 10 \| Fixed-Axis Rotation This OpenStax book is available for free at m1 = 2.0 kg, m2 = 4.0 kg, I = 0.4 kg-m2, r = 20 cm 95. A uniform rod of mass and length is held vertically by two strings of negligible mass, as shown below. (a) Immediately after the string is cut, what is the linear acceleration of the free end of the stick? (b) Of the middle of the stick? 96. A thin stick of mass 0.2 kg and length L = 0.5 m is attached to the rim of a metal disk of mass M = 2.0 kg and radius R = 0.3 m . The stick is free to rotate around a horizontal axis through its other end (see the following figure). (a) If the combination is released with the stick horizontal, what is the speed of the center of the disk when the stick is vertical? (b) What is the acceleration of the center of the disk at the instant the stick is released? (c) At the instant the stick passes through the vertical? 10.8 Work and Power for Rotational Motion 97. A wind turbine rotates at 20 rev/min. If its power output is 2.0 MW, what is the torque produced on the turbine from the wind? 98. A clay cylinder of radius 20 cm on a potter’s wheel spins at a constant rate of 10 rev/s. The potter applies a force of 10 N to the clay with his hands where the coefficient of friction is 0.1 between his hands and the clay. What is the power that the potter has to deliver to the wheel to keep it rotating at this constant rate? 99. A uniform cylindrical grindstone has a mass of 10 kg and a radius of 12 cm. (a) What is the rotational kinetic energy of the grindstone when it is rotating at 1.5 × 103 rev/min? (b) After the grindstone’s motor is turned off, a knife blade is pressed against the outer edge of the grindstone with a perpendicular force of 5.0 N. The coefficient of kinetic friction between the grindstone and the blade is 0.80. Use the work energy theorem to determine how many turns the grindstone makes before it stops. 100. A uniform disk of mass 500 kg and radius 0.25 m is mounted on frictionless bearings so it can rotate freely around a vertical axis through its center (see the following figure). A cord is wrapped around the rim of the disk and pulled with a force of 10 N. (a) How much work has the force done at the instant the disk has completed three revolutions, starting from rest? (b) Determine the torque due to the force, then calculate the work done by this torque at the instant the disk has completed three revolutions? (c) What is the angular velocity at that instant? (d) What is the power output of the force at that instant? 101. A propeller is accelerated from rest to an angular velocity of 1000 rev/min over a period of 6.0 seconds by a constant torque of 2.0 × 103 N · m . (a) What is the moment of inertia of the propeller? (b) What power is being provided to the propeller 3.0 s after it starts rotating? 102. A sphere of mass 1.0 kg and radius 0.5 m is attached to the end of a massless rod of length 3.0 m. The rod rotates about an axis that is at the opposite end of the sphere (see below). The system rotates horizontally about the axis at a Chapter 10 \| Fixed-Axis Rotation 535 constant 400 rev/min. After rotating at this angular speed in a vacuum, air resistance is introduced and provides a force 0.15 N on the sphere opposite to the direction of motion. What is the power provided by air resistance to the system 100.0 s after air resistance is introduced? 103. A uniform rod of length L and mass M is held vertically with one end resting on the floor as shown below. When the rod is released, it rotates around its lower end until it hits the floor. Assuming the lower end of the rod does not slip, what is the linear velocity of the upper end when it hits the floor? 104. An athlete in a gym applies a constant force of 50 N to the pedals of a bicycle to keep the rotation rate of the wheel at 10 rev/s. The length of the pedal arms is 30 cm. What is the power delivered to the bicycle by the athlete? 105. A 2-kg block on a frictionless inclined plane at 40° has a cord attached to a pulley of mass 1 kg and radius 20 cm (see the following figure). (a) What is the acceleration of the block down the plane? (b) What is the work done by the gravitational force to move the block 50 cm? 106. Small bodies of mass m1 and m2 are attached to opposite ends of a thin rigid rod of length L and mass M. The rod is mounted so that it is free to rotate in a horizontal plane around a vertical axis (see below). What distance d from m1 should the rotational axis be so that a minimum amount of work is required to set the rod rotating at an angular velocity ω? ADDITIONAL PROBLEMS 107. A cyclist is riding such that the wheels of the bicycle have a rotation rate of 3.0 rev/s. If the cyclist brakes such that the rotation rate of the wheels decrease at a rate of 0.3 rev/s2 , how long does it take for the cyclist to come to a complete stop? 108. Calculate the angular velocity of the orbital motion of Earth around the Sun. 109. A phonograph turntable rotating at 33 1/3 rev/min slows down and stops in 1.0 min. (a) What is the turntable’s angular acceleration assuming it is constant? (b) How many revolutions does the turntable make while stopping? 110. With the aid of a string, a gyroscope is accelerated from rest to 32 rad/s in 0.40 s under a constant angular acceleration. (a) What is its angular acceleration in rad/s2 ? (b) How many revolutions does it go through in the process? 111. Suppose a piece of dust has fallen on a CD. If the spin rate of the CD is 500 rpm, and the piece of dust is 4.3 cm from the center, what is the total distance traveled by the dust in 3 minutes? (Ignore accelerations due to getting the CD rotating.) 112. A system of point particles is rotating about a fixed axis at 4 rev/s. The particles are fixed with respect to each other. The masses and distances to the axis of the point 536 Chapter 10 \| Fixed-Axis Rotation This OpenStax book is available for free at particles are m1 = 0.1 kg, r1 = 0.2 m , m2 = 0.05 kg, r2 = 0.4 m , m3 = 0.5 kg, r3 = 0.01 m . (a) What is the moment of inertia of the system? (b) What is the rotational kinetic energy of the system? 113. Calculate the moment of inertia of a skater given the following information. (a) The 60.0-kg skater is approximated as a cylinder that has a 0.110-m radius. (b) The skater with arms extended is approximated by a cylinder that is 52.5 kg, has a 0.110-m radius, and has two 0.900-m-long arms which are 3.75 kg each and extend straight out from the cylinder like rods rotated about their ends. 114. A stick of length 1.0 m and mass 6.0 kg is free to rotate about a horizontal axis through the center. Small bodies of masses 4.0 and 2.0 kg are attached to its two ends (see the following figure). The stick is released from the horizontal position. What is the angular velocity of the stick when it swings through the vertical? 115. A pendulum consists of a rod of length 2 m and mass 3 kg with a solid sphere of mass 1 kg and radius 0.3 m attached at one end. The axis of rotation is as shown below. What is the angular velocity of the pendulum at its lowest point if it is released from rest at an angle of 30°? 116. Calculate the torque of the 40-N force around the axis through O and perpendicular to the plane of the page as shown below. 117. Two children push on opposite sides of a door during play. Both push horizontally and perpendicular to the door. One child pushes with a force of 17.5 N at a distance of 0.600 m from the hinges, and the second child pushes at a distance of 0.450 m. What force must the second child exert to keep the door from moving? Assume friction is negligible. 118. The force of 20 j ^N is applied at r →= (4.0 i ^ −2.0 j ^) m . What is the torque of this force about the origin? 119. An automobile engine can produce 200 N · m of torque. Calculate the angular acceleration produced if 95.0% of this torque is applied to the drive shaft, axle, and rear wheels of a car, given the following information. The car is suspended so that the wheels can turn freely. Each wheel acts like a 15.0-kg disk that has a 0.180-m radius. The walls of each tire act like a 2.00-kg annular ring that has inside radius of 0.180 m and outside radius of 0.320 m. The tread of each tire acts like a 10.0-kg hoop of radius 0.330 m. The 14.0-kg axle acts like a rod that has a 2.00-cm radius. The 30.0-kg drive shaft acts like a rod that has a 3.20-cm radius. 120. A grindstone with a mass of 50 kg and radius 0.8 m maintains a constant rotation rate of 4.0 rev/s by a motor while a knife is pressed against the edge with a force of 5.0 N. The coefficient of kinetic friction between the grindstone and the blade is 0.8. What is the power provided by the motor to keep the grindstone at the constant rotation rate? Chapter 10 \| Fixed-Axis Rotation 537 CHALLENGE PROBLEMS 121. The angular acceleration of a rotating rigid body is given by α = (2.0 −3.0t) rad/s2 . If the body starts rotating from rest at t = 0 , (a) what is the angular velocity? (b) Angular position? (c) What angle does it rotate through in 10 s? (d) Where does the vector perpendicular to the axis of rotation indicating 0° at t = 0 lie at t = 10 s ? 122. Earth’s day has increased by 0.002 s in the last century. If this increase in Earth’s period is constant, how long will it take for Earth to come to rest? 123. A disk of mass m, radius R, and area A has a surface mass density σ = mr AR (see the following figure). What is the moment of inertia of the disk about an axis through the center? 124. Zorch, an archenemy of Rotation Man, decides to slow Earth’s rotation to once per 28.0 h by exerting an opposing force at and parallel to the equator. Rotation Man is not immediately concerned, because he knows Zorch can only exert a force of 4.00 × 107 N (a little greater than a Saturn V rocket’s thrust). How long must Zorch push with this force to accomplish his goal? (This period gives Rotation Man time to devote to other villains.) 125. A cord is wrapped around the rim of a solid cylinder of radius 0.25 m, and a constant force of 40 N is exerted on the cord shown, as shown in the following figure. The cylinder is mounted on frictionless bearings, and its moment of inertia is 6.0 kg · m2 . (a) Use the work energy theorem to calculate the angular velocity of the cylinder after 5.0 m of cord have been removed. (b) If the 40-N force is replaced by a 40-N weight, what is the angular velocity of the cylinder after 5.0 m of cord have unwound? 538 Chapter 10 \| Fixed-Axis Rotation This OpenStax book is available for free at
6450	https://www.algebra-class.com/calculating-slope.html	Algebra Class Making Algebra easier for you! Home Pre-Algebra Topics Algebra Topics Shop Helpful Resources About Contact Algebra Calculator Algebra Cheat Sheet Algebra Practice Test Algebra Readiness Test Algebra Formulas Want to Build Your Own Website? Login Algebra E-Course Login In Sign In / Register Calculating SlopeThe Key to Graphing Equations What do you think of when you hear the term, slope? Do you think of a skier, skiing down a large mountain? Or, maybe you think of the sliding board at the playground. Whatever you are thinking, it's probably something that's on an incline! When we study slope in Algebra we are going to study the incline and other characteristics of a line on a graph. Slope is a very important concept to understand in Algebra. Therefore, I've created three different lessons to help you gain a full understanding. We will start here with defining and calculating slope by analyzing a graph. Then we will move on to graphing slope and finally to using slope intercept form to create your graph. Start here from the beginning, or move onto the concept of slope that you need help with! What is Slope in Algebra? Slope is used very often in Mathematics. It can be used to actually find how steep a particular line is, or it can be used to show how much something has changed over time. We calculate slope by using the following definition. In Algebra, slope is defined as the rise over the run. This is written as a fraction like this: Rules for Calculating the Slope of Line Find two points on the line. Count the rise (How many units do you count up or down to get from one point to the next?) Record this number as your numerator. Count the run (How many units do you count left or right to get to the point?) Record this number as your denominator. Simplify your fraction if possible. Important Note: If you count up or right your number is positive. If you count down or left your number is negative. There are two examples below. You can watch how I calculate slope in the video lesson, or you can skip the video and check out examples 1 and 2 below. Example 1: Calculating Slope Let's take a look at another example where the slope is a fraction. Example 2: Calculating Slope Just Remember: Slope is a fraction: rise/run. Calculate your rise and then your run! If your rise is downhill it is negative. If your run is to the left it is also negative! You can choose any two points on the line to calculate your slope! The slope is constant throughout the whole line. Home Graphing Equations Calculating Slope Comments We would love to hear what you have to say about this page! Need More Help With Your Algebra Studies? Get access to hundreds of video examples and practice problems with your subscription! Click here for more information on our affordable subscription options. Not ready to subscribe? Register for our FREE Pre-Algebra Refresher course. ALGEBRA CLASS E-COURSE MEMBERS Click here for more information on our Algebra Class e-courses. Need Help? Try This Online Calculator! Affiliate Products... On this site, I recommend only one product that I use and love and that is Mathway If you make a purchase on this site, I may receive a small commission at no cost to you. About Me Affiliates Privacy Policy Disclaimer Let Us Know How we are doing! send us a message to give us more detail! \| \| \| --- \| \| Share this page: What’s this? \| FacebookX \| Enjoy this page? Please pay it forward. Here's how... Would you prefer to share this page with others by linking to it? Click on the HTML link code below. Copy and paste it, adding a note of your own, into your blog, a Web page, forums, a blog comment, your Facebook account, or anywhere that someone would find this page valuable. Copyright © 2009-2020 \| Karin Hutchinson \| ALL RIGHTS RESERVED. Cookies help us deliver our services. By using our services, you agree to our use of cookies. Learn more.
6451	https://www.youtube.com/watch?v=PMT7IOuSG4s	From Word Problems to Linear Equations in One Variable - SAT® Math: 1.3 Quizlet 27500 subscribers 4 likes Description 289 views Posted: 23 Dec 2024 Struggling with SAT® word problems? This video will show you how to translate word problems into linear equations in one variable, making these tricky questions manageable. Learn how to identify key phrases, set up equations, and solve efficiently. Algebra is an essential starting point for mastering SAT® Math, and we're excited to offer foundational resources freely available on our site. For further exploration into advanced topics, additional learning opportunities will be available soon. In the meantime, check out this link for more SAT® Math practice examples, expert-verified content, and practice tests! Quizlet SAT® Math: Algebra Playlist: SAT® is a registered trademark of College Board which is not affiliated with and does not endorse Quizlet or the information provided on Quizlet’s website. Transcript: Today, we’ll explore Linear Equations in One Variable, focusing on Verbal Equations In this video, we’ll break down word problems and learn how to translate them into linear equations. Word problems may seem tricky at first, but once you understand how to convert them into equations, they become much more manageable. By the end of this video, you’ll have the skills to solve these problems and turn words into numbers confidently! Here’s what we’ll cover today: First is how to identify key information from a word problem. Next is how to write a linear equation from that information. And finally, we’ll solve some examples, starting with simple problems and gradually moving to more complex ones. Let's dive in! Now that we’ve introduced the topic, let’s break down what we mean by “Verbal Equations” and how we can solve them. A verbal equation is simply a word problem that describes a mathematical relationship. Our job is to translate the words into an equation that we can solve. This process involves identifying the key information from the problem and deciding what the unknown value, our variable, should represent. Here’s the key idea. When you see a word problem, look for clues in the form of numbers, relationships like ”more than” or ”less than”, and keywords like ”total” or ”difference”. These clues will help you form the equation. For example, ”Tom has 3 more than twice the number of apples as Sarah”. You could represent Sarah’s apples as x, and Tom’s apples as 2x + 3. Remember: Once you’ve solved for the variable, always check your solution by plugging it back into the problem! Let's take on our first example. Tom has 3 more than twice the number of apples as Sarah. If Sarah has 5 apples, how many apples does Tom have? Let's break this down step by step. First, we need to identify the key information from the problem. The problem already tells us how many apples Sarah has, which is 5. Now, let's focus on how Tom’s number of apples relates to Sarah’s. The problem says that Tom has 3 more than twice the number of apples as Sarah, so we’ll start by multiplying Sarah’s apples by 2. Then, the problem says Tom has 3 more than that. So we’ll add 3 to 10. Finally, Tom has 13 apples. Let's take a look at another example. A cable company charges a monthly fee of $40, plus an additional $5 for each movie rented. If your total bill for the month was $85, how many movies did you rent? First, we need to identify the key information. The problem tells us that the monthly fee is $40, and each movie rented costs $5. We also know the total bill was $85. What we need to find is how many movies were rented. This will then become our variable. Let's call this value x. This means that the total cost for renting movies would be 5x dollars. But the total bill of $85 wasn't accumulated solely by the number of movies rented , as this number also includes the monthly fee of $40. So the equation for the total bill is as such. Next, we’ll solve this equation step by step. First, subtract 40 from both sides to isolate the movie rental term. Now, to find x, divide both sides of the equation by 5: So, the customer rented 9 movies this month. Let's now discuss a more challenging example. A concert ticket costs $30 each, and there is a one-time service fee of $15. If you paid a total of $165 for your tickets, how many tickets did you buy? First, identify the key information. We know that each concert ticket costs $30, and that there is a one-time service fee of $15. We also know the total amount paid was $165. What we need to figure out is how many tickets were purchased. This will then become our variable. Let's call this value x. This means that the total cost for the tickets would be 30x dollars. But the total bill of $165 wasn't accumulated solely by the number of tickets purchased , as this number also includes the one-time fee of $15. So the equation for the total bill is as such. Next, we’ll solve this equation step by step. First, subtract 15 from both sides to isolate x. Now, to find x, divide both sides by 30: So, the customer bought 5 tickets. Let’s check our answer by substituting x = 5 back into the equation. That checks out, so the correct answer is 5 tickets. Great job working through these verbal equations today! Let’s quickly recap what we’ve learned. Today, we learned how to translate a word problem into a linear equation. This is the key skill for solving verbal equations. This starts with identifying the variable, where we figure out what we're solving for. Next is setting up the equation, where we looked for specific keywords to write relationships between quantities. Last is solving, where we utilized algebra to find the solution. The key formula to remember is the fixed cost plus the product of the variable cost and the number of items equals the total cost. When applied, that looks like this Well done today, and keep up the great work! We’ll see you in the next lesson.
6452	https://artofproblemsolving.com/wiki/index.php/Arithmetic_sequence?srsltid=AfmBOor9DaNTMUt2jlMtXRmyYLUnlprIxqutChONCpSBfs5JSphZzMnk	Art of Problem Solving Arithmetic sequence - AoPS Wiki Art of Problem Solving AoPS Online Math texts, online classes, and more for students in grades 5-12. Visit AoPS Online ‚ Books for Grades 5-12Online Courses Beast Academy Engaging math books and online learning for students ages 6-13. Visit Beast Academy ‚ Books for Ages 6-13Beast Academy Online AoPS Academy Small live classes for advanced math and language arts learners in grades 2-12. Visit AoPS Academy ‚ Find a Physical CampusVisit the Virtual Campus Sign In Register online school Class ScheduleRecommendationsOlympiad CoursesFree Sessions books tore AoPS CurriculumBeast AcademyOnline BooksRecommendationsOther Books & GearAll ProductsGift Certificates community ForumsContestsSearchHelp resources math training & toolsAlcumusVideosFor the Win!MATHCOUNTS TrainerAoPS Practice ContestsAoPS WikiLaTeX TeXeRMIT PRIMES/CrowdMathKeep LearningAll Ten contests on aopsPractice Math ContestsUSABO newsAoPS BlogWebinars view all 0 Sign In Register AoPS Wiki ResourcesAops Wiki Arithmetic sequence Page ArticleDiscussionView sourceHistory Toolbox Recent changesRandom pageHelpWhat links hereSpecial pages Search Arithmetic sequence In algebra, an arithmetic sequence, sometimes called an arithmetic progression, is a sequence of numbers such that the difference between any two consecutive terms is constant. This constant is called the common difference of the sequence. For example, is an arithmetic sequence with common difference and is an arithmetic sequence with common difference ; however, and are not arithmetic sequences, as the difference between consecutive terms varies. More formally, the sequence is an arithmetic progression if and only if . A similar definition holds for infinite arithmetic sequences. It appears most frequently in its three-term form: namely, that constants , , and are in arithmetic progression if and only if . Contents [hide] 1 Properties 2 Sum 3 Problems 3.1 Introductory problems 3.2 Intermediate problems 4 See Also Properties Because each term is a common distance from the one before it, every term of an arithmetic sequence can be expressed as the sum of the first term and a multiple of the common difference. Let be the first term, be the th term, and be the common difference of any arithmetic sequence; then, . A common lemma is that given the th term and th term of an arithmetic sequence, the common difference is equal to . Proof: Let the sequence have first term and common difference . Then using the above result, as desired. Another common lemma is that a sequence is in arithmetic progression if and only if is the arithmetic mean of and for any consecutive terms . In symbols, . This is mostly used to perform substitutions, though it occasionally serves as a definition of arithmetic sequences. Sum An arithmetic series is the sum of all the terms of an arithmetic sequence. All infinite arithmetic series diverge. As for finite series, there are two primary formulas used to compute their value. The first is that if an arithmetic series has first term , last term , and total terms, then its value is equal to . Proof: Let the series be equal to , and let its common difference be . Then, we can write in two ways: Adding these two equations cancels all terms involving ; and so , as required. The second is that if an arithmetic series has first term , common difference , and terms, it has value . Proof: The final term has value . Then by the above formula, the series has value This completes the proof. Problems Here are some problems with solutions that utilize arithmetic sequences and series. Introductory problems 2005 AMC 10A Problem 17 2006 AMC 10A Problem 19 2012 AIME I Problems/Problem 2 2004 AMC 10B Problems/Problem 10 2006 AMC 10A, Problem 9 2006 AMC 12A, Problem 12 Intermediate problems 2003 AIME I, Problem 2 Find the roots of the polynomial , given that the roots form an arithmetic progression. See Also Geometric sequence Harmonic sequence Sequence Series Retrieved from " Categories: Algebra Sequences and series Definition Art of Problem Solving is an ACS WASC Accredited School aops programs AoPS Online Beast Academy AoPS Academy About About AoPS Our Team Our History Jobs AoPS Blog Site Info Terms Privacy Contact Us follow us Subscribe for news and updates © 2025 AoPS Incorporated © 2025 Art of Problem Solving About Us•Contact Us•Terms•Privacy Copyright © 2025 Art of Problem Solving Something appears to not have loaded correctly. Click to refresh.
6453	https://english.stackexchange.com/questions/620533/being-in-advantage-as-used-in-the-video-gaming-world-of-fighting-games	phrases - "Being in advantage", as used in the video gaming world of fighting games - English Language & Usage Stack Exchange Join English Language & Usage By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy. Sign up with Google OR Email Password Sign up Already have an account? Log in Skip to main content Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. 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You'll need to complete a few actions and gain 15 reputation points before being able to upvote. Upvoting indicates when questions and answers are useful. What's reputation and how do I get it? Instead, you can save this post to reference later. Save this post for later Not now Thanks for your vote! You now have 5 free votes weekly. Free votes count toward the total vote score does not give reputation to the author Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation. Got it!Go to help center to learn more "Being in advantage", as used in the video gaming world of fighting games Ask Question Asked 1 year, 6 months ago Modified1 year, 6 months ago Viewed 74 times This question shows research effort; it is useful and clear 1 Save this question. Show activity on this post. In fighting games such as Street Fighter, it is common to say that you're "in advantage" to say that you're "in an advantageous state" as opposed to your opponent. Is it grammatically correct to use "advantage" as a noun and to say that you are in advantage like you would say that you are "in difficulty"? phrases expressions nouns gaming Share Share a link to this question Copy linkCC BY-SA 4.0 Improve this question Follow Follow this question to receive notifications asked Mar 13, 2024 at 12:00 Stefan SchoutenStefan Schouten 113 3 3 bronze badges 4 2 'Advantage' is a noun. Grammar doesn't enter into this. Whether it colligates (the grammatical equivalent of 'collocates') with 'in' is the question you're asking.Edwin Ashworth –Edwin Ashworth 2024-03-13 12:21:19 +00:00 Commented Mar 13, 2024 at 12:21 It's probably a bad video game translation. I wouldn't trust Japanese video games to learn English from. (Also, in old video games they probably can't display a long text.)Stuart F –Stuart F 2024-03-13 12:41:02 +00:00 Commented Mar 13, 2024 at 12:41 "to have the advantage" in a gameLambie –Lambie 2024-03-13 15:23:49 +00:00 Commented Mar 13, 2024 at 15:23 1 This sounds like a game-specific idiom. So grammaticality may not enter into it. Consider how "advantage" is used in tennis, for instance.Barmar –Barmar 2024-03-13 19:43:19 +00:00 Commented Mar 13, 2024 at 19:43 Add a comment\| 2 Answers 2 Sorted by: Reset to default This answer is useful 1 Save this answer. Show activity on this post. There is, as you point out, a modelling after the valid expressions 'in trouble', 'in difficulty', 'in clover' ... but is 'in advantage' acceptable? There are some reasonable-looking examples of this candidate colligation to be found: Mr. Kerry tried to emphasize the positive aspects of relations with Brazil by focusing on subjects like the government's plans here to send tens of thousands of Brazilian students to American universities and vigorous bilateral trade, tilted in advantage of the United States in the form of an ample surplus. [The New York Times] The simulation results show that the proposed method can handle unpredicted disturbance and data uncertainties very well in advantage of the effectiveness of observation and control. [Applied Mathematical Modelling] The question of building production plants centralized offshore or decentralized onshore is rather in advantage of decentralized production because grid-connected systems are easier to operate, hydrogen transportation will be unnecessary and building on land is less expensive than offshore. [International Journal of Energy and Environmental Engineering] For sex matching, differences between investigated groups were no more than 10%% in advantage of either gender. [Acta Neuropathologica] On the contrary, we detected a non-significant difference in means of 2.3 in SOFA score in advantage of placebo, a trend not explained by the bilirubin changes. [Critical Care] [All quotes courtesy of Ludwig.guru] These show that educated writers consider the expression to be acceptable. But note that all of them include the transitivising preposition/particle 'of': '10% in advantage of women' say. A Google search for << "are in advantage" -"in advantage of" -built >> yields very few relevant and non-substandard results that don't apply to games blogs. One (seeming to mean '[are] advantageous') is That's where particle beams, in our case protons, are in advantage. Due to their finite range and Bragg-peak energy deposition ... [S Lautenschlaeger;NIH: National Library of Medicine; 2019] Also, the fact that many dictionaries (OALD, Cambridge Dictionary, Collins, Britannica Dictionary, Merriam-Webster, AHD, ) fail to give similar examples (while Cambridge even has a separate entry for say be in difficulty) points to the fact that it is rather rare. But labelling it as incorrect is overprescriptive. Not that I'd choose 'John is in advantage' over 'John is at an advantage' / 'John has an/the advantage [over Jack]'. Though I'd probably say 'John is winning.' Share Share a link to this answer Copy linkCC BY-SA 4.0 Improve this answer Follow Follow this answer to receive notifications edited Mar 13, 2024 at 13:05 answered Mar 13, 2024 at 12:35 Edwin AshworthEdwin Ashworth 91.1k 14 14 gold badges 161 161 silver badges 288 288 bronze badges 1 Thanks for the in-depth answer! I had the same results when looking around on Google, but it did not sound completely incorrect. This helps a lot!Stefan Schouten –Stefan Schouten 2024-03-14 13:42:55 +00:00 Commented Mar 14, 2024 at 13:42 Add a comment\| This answer is useful 0 Save this answer. Show activity on this post. According to the Cambridge dictionary, the correct expression is to be at an advantage. If we are talking about disadvantages, according to same source, the correct expresion would be to be at a disadvantage. Share Share a link to this answer Copy linkCC BY-SA 4.0 Improve this answer Follow Follow this answer to receive notifications answered Mar 13, 2024 at 12:15 fonema Jimenafonema Jimena 42 5 5 bronze badges Add a comment\| You must log in to answer this question. Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions phrases expressions nouns gaming See similar questions with these tags. 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6454	https://anthroholic.com/half-life?srsltid=AfmBOopQbgUt_uh4p0Xnu0bQECOYuFtM8gihKr6m-NUHOs3B-jnClp7i	Concept of Half-Life in Archaeology \| Anthroholic Skip to content Main Menu Anthropology Courses Anthropology Self Study Video Course Anthropology Optional Foundation Course Anthropology Daily Answer Writing Series Super 50 Anthropology Test Series 2026 Free Resources Anthropology PYQs UPSC Anthropology Syllabus YouTube Anthroholic Blog Blogs & Current Affairs Books/Study Materials Anthropology Courses Anthropology Self Study Video Course Anthropology Optional Foundation Course Anthropology Daily Answer Writing Series Super 50 Anthropology Test Series 2026 Free Resources Anthropology PYQs UPSC Anthropology Syllabus YouTube Anthroholic Blog Blogs & Current Affairs Books/Study Materials Contact Us to Know More Login Sign Up Username or Email Address Password [x] Remember Me Forgot Password? Log In Email A link to set a new password will be sent to your email address. Your personal data will be used to support your experience throughout this website, to manage access to your account, and for other purposes described in our privacy policy. Register Username or Email Address Get New Password ← Back to login +91-7303290503, +91-9557169661 \| MON to SUN 10:00 AM - 6:00 PM Anthropology Courses Anthropology Self Study Video Course Anthropology Optional Foundation Course Anthropology Daily Answer Writing Series Super 50 Anthropology Test Series 2026 Free Resources Anthropology PYQs UPSC Anthropology Syllabus YouTube Anthroholic Blog Blogs & Current Affairs Books/Study Materials Login Menu Login Menu HomeAnthropologyArchaeology Half-Life Last Updated:Sep 8, 2025 Half-life, in nuclear physics and chemistry, refers to the time required for half the atoms in a sample of a radioactive substance to disintegrate or achieve a transformation that alters their fundamental properties . This concept is essential in various fields, including archaeology, geology, and medicine, and it’s a crucial part of understanding nuclear chemistry and physics. Definition and Principles of Half-Life What is Half-Life? As the term suggests, half-life represents a “half-life period,” during which half of a radioactive element’s atoms decay or transform. Once the half-life period has elapsed, the quantity of radioactive atoms and the rate of decay both reduce by half. Hence, after two half-lives, only a quarter of the original radioactive atoms will remain. This characteristic exponential decay continues until virtually no radioactive atoms are left . Principles Governing Half-Life Three key principles govern the concept of half-life: Randomness: Decay events are random, implying that it is impossible to predict precisely when a specific atom will decay. However, for a large number of atoms, the average rate of decay becomes predictable. Exponential Decay: Radioactive decay follows an exponential decay pattern. Hence, after each half-life, the number of undecayed atoms is halved. Independence of Physical Conditions: The half-life of a radioactive substance does not depend on physical conditions like temperature, pressure, or the compound’s chemical state . The Half-Life Formula Half-life is mathematically expressed using the equation: T½ = (ln(2) / λ) Here, ‘T½’ represents the half-life, ‘ln(2)’ is the natural logarithm of 2 (approximately equal to 0.693), and ‘λ’ is the decay constant, which characterizes the rate of decay of the radioactive substance. The decay constant ‘λ’ can further be calculated from the total number of atoms ‘N’ and the number of atoms that disintegrate per unit time ‘dN/dt’, expressed as λ = - (1/N) (dN/dt). Real-world Applications of Half-Life The concept of half-life has practical implications in multiple sectors: Radiocarbon Dating: In archaeology, the half-life of Carbon-14 (approximately 5730 years) is used to date ancient organic materials. This method, known as radiocarbon dating, allows archaeologists to estimate the age of artifacts and skeletal remains. Medicine: In healthcare, radioactive isotopes are used in the diagnosis and treatment of various diseases. The chosen isotope’s half-life should be compatible with the time frame of the desired medical procedure. Nuclear Power: Understanding half-life is crucial for managing nuclear waste. Some waste products have half-lives of thousands of years, necessitating long-term storage strategies . \| Application \| Isotope \| Half-life \| --- \| Radiocarbon Dating \| Carbon-14 \| 5730 years \| \| Medical Diagnostics \| Technetium-99m \| 6 hours \| \| Nuclear Waste \| Plutonium-239 \| 24,000 years \| Table: Some common isotopes and their half-lives in various applications Understanding Different Types of Decay Radioactive decay isn’t a uniform process, and different substances can undergo various forms of decay. Understanding these various types of decay not only helps grasp the concept of half-life better but also sheds light on the wide-ranging applications of nuclear physics and chemistry . Alpha Decay Alpha decay involves the emission of an alpha particle, comprising two protons and two neutrons (essentially a helium nucleus). This type of decay decreases the atomic number by two and the mass number by four. Alpha particles, due to their relatively large mass and positive charge, are the least penetrating but can cause significant biological damage if ingested . Beta Decay In beta decay, a neutron in the nucleus converts into a proton, with the emission of an electron (beta particle) and an electron antineutrino. This decay increases the atomic number by one but leaves the mass number unchanged. Beta particles are more penetrating than alpha particles but less damaging per event . Gamma Decay Gamma decay involves the emission of high-energy photons (gamma rays), often following alpha or beta decay. The atomic number and mass number remain unchanged, but the atom loses energy. Gamma rays are the most penetrating of the three and require heavy shielding to protect against . \| Type of Decay \| Particle Emitted \| Change in Atomic Number \| Change in Mass Number \| Penetration Power \| --- --- \| Alpha Decay \| Alpha Particle (2 protons, 2 neutrons) \| -2 \| -4 \| Least \| \| Beta Decay \| Electron and Electron Antineutrino \| +1 \| No change \| Medium \| \| Gamma Decay \| Gamma Rays \| No change \| No change \| High \| Table: A comparison of different types of radioactive decay. Half-Life and Radiometric Dating Radiometric dating, an offshoot of the half-life concept, is a method used to determine the age of rocks and other materials. This method is based on the half-life of the radioactive isotopes contained within these materials . Principles of Radiometric Dating Parent and Daughter Nuclides: The radioactive isotope (parent) decays to produce a stable isotope (daughter). For instance, Uranium-238 decays to form Lead-206. Rate of Decay: The rate of decay is constant and is expressed in terms of the half-life of the radioactive isotope. Age Determination: The ratio of parent to daughter nuclides can help determine the age of a sample, assuming no initial daughter product was present and the system has remained closed . Conclusion The concept of half-life is an integral part of understanding our world at both macro and micro levels. While the mathematics behind it may seem daunting, the principle itself is straightforward: over a defined period, half of the radioactive atoms in a sample will decay. This consistent, predictable decay underpins numerous applications, from medical diagnostics to nuclear power. As we advance technologically and scientifically, our understanding and application of these fundamental principles, such as half-life, will continue to evolve and shape our future. References “Half-life.” Encyclopædia Britannica, Encyclopædia Britannica, Inc Sublette, Carey. “Nuclear Weapons Frequently Asked Questions.” Nuclear Weapon Archive, www.nuclearweaponarchive.org/Nwfaq/Nfaq12.html. Clark, Jim. “Half-Life.” Chemguide, www.chemguide.co.uk/physical/radioact/half-life.html. The Editors of Encyclopaedia Britannica. “Decay constant.” Encyclopædia Britannica, Encyclopædia Britannica, Inc., www.britannica.com/science/decay-constant. Radioactive Decay. (n.d.). In Wikipedia. Retrieved July 23, 2023, from “Radioactive decay.” Physics LibreTexts, “Alpha Decay.” HyperPhysics “Beta Decay.” HyperPhysics, “Gamma Decay.” HyperPhysics, “Radiometric Dating.” National Geographic Society Tags # Anthropology# Archaeology# UPSC Share your love Aman Yadav Aman Yadav is the founder of Anthroholic, a platform dedicated to making Anthropology accessible and engaging for learners worldwide. Driven by a deep passion for the subject, he created Anthroholic to share knowledge and spark curiosity about human societies, cultures, and evolution. Alongside his work in Anthropology education, Aman is also an experienced marketing strategist, having worked with global organizations across various sectors. He hails from India and is committed to bridging the gap between academic insight and practical learning. Articles:164 Previous Post EthnopoeticsNext Post Humans or Homo-Sapiens Newsletter Updates Enter your email address below and subscribe to our newsletter Subscribe [x] I accept the Privacy Policy Related Posts Matrilineal Descent Media Anthropology Individualism Anthropology Course for UPSC: The Ultimate Guide to Choosing the Best Coaching for Success Leave a ReplyCancel Reply You must be logged in to post a comment. Anthroholic Anthroholic helps the world learn Anthropology for Free. 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6455	https://www.fengshuilondon.net/kaplan-kaplan-preference-model-offers-model-designing-engaging-environments/	Kaplan and Kaplan Preference Model offers a Model of Designing Engaging Environments - Feng Shui London UK • The Capital Feng Shui Consultant We value your privacy We use cookies to enhance your browsing experience, serve personalized ads or content, and analyze our traffic. By clicking "Accept All", you consent to our use of cookies. Customize Reject All Accept All Customize Consent Preferences We use cookies to help you navigate efficiently and perform certain functions. You will find detailed information about all cookies under each consent category below. 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Reject All Save My Preferences Accept All Call / text now 07956 288574 (+44 7956 288574) Home Services Feng Shui For Homes Feng Shui For Business Feng Shui For Buying & Selling Properties Remote Feng Shui Consultations Corporate Feng Shui Feng Shui For Branding & Logo Design Feng Shui for Architects, Interior Designers, Developers and Estate Agents Environmental Psychology Feng Shui Courses Vastu Shastra Feng Shui for Gardens Spiritual Feng Shui Feng Shui Mentoring EMF & EMR Surveys Tips Videos Shop Blog Testimonials FAQs About Contact Menu Kaplan and Kaplan Preference Model offers a Model of Designing Engaging Environments Posted on 04/02/2010 by Jan Cisek – Feng Shui Consultant Kaplan and Kaplan Preference Modeloffers a model of designing engaging environments which are nourishing and foster creativity and ingenuity Environmental Psychology theory Kaplan and Kaplan’s preference model (1989) looks at four preference factors coherence, legibility, complexity, and mysterywhich are key to engaging, stimulating and creative environments. People, whether at home or workplace, thrive in environments that comply with the Kaplan and Kaplan model. Kaplan and Kaplan Preference Model Complexity and fractality Complexity can be accomplished with rich, nourishing environments. Nature is complex. Fractals are complex. Our bodies are mostly fractal. Fractality is a measure of complexity.A fractal is a mathematical set that usually displays self-similar patterns (Benoît Mandelbrot, 1975). The concept of fractality extends beyond self-similarity and can include detailed patterns repeating themselves. Examples of fractal patterns can be seen in: cauliflowers, your DNA, Pollock’s paintings, snowflakes, roses, and some cities such as Prague which was designed on the pattern of a rose (see below – bottom right) and your lungs and waves. Examples of fractals Fractals Fractal vs non fractal Less fractal vs fractal Which house is more fractal and more engaging? For more info and examples read:Environmental Psychology for Design, Second Edition by Dak Kopec Posted in Feng shui Environmental Psychology and tagged Fractals, Kaplan and Kaplan Preference Model. Post navigation ← Astrocartography for Feng Shui: Understanding… The Power of Monstera: The… → Book a feng shui consultation for your home, business, work or personal life Call Jan on07956 288 574(+44 7956 288 574 from abroad) to discuss how feng shui can help to improve your home,personal life, work or business.Or email himusing the contact form Feng Shui Blog – The latest Feng Shui Tips Subscribe to the feng shui blog for the latest news on feng shui and Environmental Psychology Feng Shui Store Feng shui fish: koi carps 3D photo – for prosperity and luck Start improving the feng shui of your home/workplace by neutralising geopathic stress & electrosmog. GetHelios3 – geopathic stress and electro-smog harmoniser and more... 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6456	https://www.scribd.com/document/435664570/Damped-Harmonic-Oscillator	Damped Harmonic Oscillator \| PDF \| Differential Calculus \| Physical Quantities Opens in a new window Opens an external website Opens an external website in a new window This website utilizes technologies such as cookies to enable essential site functionality, as well as for analytics, personalization, and targeted advertising. To learn more, view the following link: Privacy Policy Open navigation menu Close suggestions Search Search en Change Language Upload Sign in Sign in Download free for 30 days 0 ratings 0% found this document useful (0 votes) 1K views 3 pages Damped Harmonic Oscillator The document discusses the damped harmonic oscillator. It begins by presenting the Newton's second law motion equation for a damped harmonic oscillator. It then describes how this equation c… Full description Uploaded by Md Chand Asique AI-enhanced description Go to previous items Go to next items Download Save Save Damped Harmonic Oscillator For Later Share 0%0% found this document useful, undefined 0%, undefined Print Embed Ask AI Report Download Save Damped Harmonic Oscillator For Later You are on page 1/ 3 Search Fullscreen Damped Harmonic Oscillator Damping coefficient Undamped oscillator Driven oscillator The Newton's 2nd Law motion equation is This is in the form of a homogeneous second order differential equation and has a solution of the form Substituting this form gives an auxiliary equation for λ The roots of the quadratic auxiliary equation areThe three resulting cases for the damped oscillator areIndex Periodic motion concepts HyperPhysics Mechanics R Nave Go Back Damping Coefficient When a damped oscillator is subject to a damping force which is linearly dependent upon the velocity, such as viscous da mping, the oscillation will Index Periodic adDownload to read ad-free have exponential decay terms which depend upon a damping coefficient. If the damping force is of the form then the damping coefficient is given by This will seem logical when you note that the damping force is proportional to c, but its influence inversely proportional to the mass of the oscillator.motion concepts HyperPhysics Mechanics R Nave Go Back Underdamped Oscillator For any value of the damping coefficient γ less than the critical damping factor the mass will overshoot the zero point and oscillate about x=0. The behavior is shown for one-half and one-tenth of the critical damping factor.Index Periodic motion concepts Reference Barger &Olsson adDownload to read ad-free Also shown is an example of the overdamped case with twice the critical damping factor.Note that these examples are for the same specific initial conditions, i.e., a release from rest at a position x 0 . For other initial conditions, the curves would look different, but the the behavior with time would still decay according to the damping factor.Damped Oscill ator Furthe r detail s for un derdam ped cas e HyperPhysics Mechanics R Nave Go Back Underdamped Oscillator When a damped oscillator is underdamped, it approaches zero faster than in the case of critical damping, but oscillates about that zero.The equation is that of an exponentially decaying sinusoid.The damping coefficient is less than the undamped resonant frequency. The sinusoid frequency is given by but the motion is not strictly periodic.Index Periodic motion concepts HyperPhysics Mechanics R Nave Go Back Share this document Share on Facebook, opens a new window Share on LinkedIn, opens a new window Share with Email, opens mail client Copy link Millions of documents at your fingertips, ad-free Subscribe with a free trial You might also like Engineering Utilities Module 1 74% (19) Engineering Utilities Module 1 42 pages Linear Algebra and Its Applications - D. C. Lay No ratings yet Linear Algebra and Its Applications - D. C. 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6457	https://galileo.ai/blog/best-practices-for-ai-model-validation-in-machine-learning	Book a Demo Get Started for Free Oct 27, 2024 · 10 min read Best Practices for AI Model Validation in Machine Learning Conor Bronsdon Head of Developer Awareness In machine learning, AI model validation checks how well a model performs on unseen data, ensuring accurate predictions before deployment. A data-centric approach focuses on improving the quality and utility of data used in model validation. However, the complexity of modern models and datasets introduces significant challenges in effectively validating these models. We recently explored this topic on our Chain of Thought podcast, where industry experts shared practical insights and real-world implementation strategies. Importance of Model Validation Validating models confirms they generalize beyond training data. But why is model validation so crucial? It helps to: Detect overfitting and underfitting. Align model performance with business goals. Build confidence in the model's reliability. Identify issues early for correction. Moreover, the growing reliance on AI models in business decisions has led to significant consequences when models are inaccurate. A McKinsey report indicates that 44% of organizations have reported negative outcomes due to AI inaccuracies. This highlights the essential role of AI model validation in mitigating risks such as data drift and LLM hallucinations. In practical terms, the importance of model validation is underscored by the rise in synthetic data usage. According to Gartner, synthetic data is projected to be used in 75% of AI projects by 2026. Synthetic data provides a viable alternative when real data is unavailable or costly to obtain, enabling organizations to develop and train AI models without compromising privacy or security. However, synthetic data may not capture all the complexities of real-world scenarios. Therefore, rigorous model validation is essential to ensure that models trained on synthetic data perform effectively in actual operational conditions. This helps bridge the gap between synthetic training environments and real-world applications, preventing potential errors and ensuring reliability. Recognizing these challenges shows the need for strong validation tools that make the process easier and provide useful information. Key Concepts and Terminology Important terms include: Training Data: Data used to train the model. Validation Data: Data used to evaluate the model during development. Test Data: Data used to assess the final model's performance after training is complete. Overfitting: When a model is too closely tailored to the training data, causing poor performance on new data. Underfitting: When a model is too simple to capture the underlying patterns in the data, leading to poor performance. Cross-Validation: A method to estimate model performance on unseen data by partitioning the dataset into multiple training and validation sets. Performance Metrics: Quantitative measures used to evaluate model performance, such as accuracy, precision, recall, and F1 score. Types of Validation Techniques Validating your AI model with appropriate techniques ensures it performs well on new, unseen data. Cross-Validation Cross-validation splits your dataset into subsets to assess how the model generalizes to independent data. Common approaches include K-Fold Cross-Validation, dividing data into K parts and using each part as a validation set in turns, and Stratified K-Fold Cross-Validation, ensuring each fold represents the class distribution. Leave-One-Out Cross-Validation (LOOCV) uses each data point as its own validation set, offering detailed insights but can be computationally intensive. Holdout Validation In holdout validation, you reserve a portion of your dataset exclusively for testing. You split data into training and holdout sets, providing an unbiased evaluation of the model's performance on unseen data. Bootstrap Methods Bootstrap methods involve resampling your dataset with replacement to create multiple training samples. By measuring performance variance across different subsets, bootstrap methods assess model stability, which makes them useful when data is limited. Domain-Specific Validation Techniques As AI models become increasingly tailored to specific industries and use cases, domain-specific validation techniques are gaining importance. According to Gartner, by 2027, 50% of AI models will be domain-specific, requiring specialized validation processes for industry-specific applications. This trend necessitates validation strategies that account for the unique characteristics and requirements of each domain. In such cases, it's crucial to evaluate LLMs for RAG using methods tailored to their specific applications. In industry-specific contexts, traditional validation methods may not suffice due to specialized data types, regulatory considerations, and unique performance metrics. For example, in healthcare, AI models must comply with stringent privacy laws and clinical accuracy standards, requiring validation processes that address these concerns. Similarly, in finance, models must be validated for compliance with financial regulations and risk management practices. Domain-specific validation techniques might include the involvement of subject matter experts, customized performance metrics aligned with industry standards, and validation datasets that reflect the particularities of the domain. Incorporating these specialized validation processes ensures that AI models are not only technically sound but also practically effective and compliant within their specific industry contexts. Performance Metrics for Model Validation Selecting the right performance metrics is essential to determine how well your model will perform on new data. Accuracy, Precision, and Recall Accuracy measures the proportion of correct predictions. For more insights, consider: Precision: The ratio of true positive predictions to total predicted positives. Recall: The ratio of true positive predictions to all actual positives. Using both helps you understand trade-offs between detecting positive instances and avoiding false alarms, following a metrics-first approach. F1 Score and ROC-AUC The F1 score combines precision and recall into a single metric. The ROC-AUC evaluates the model's ability to distinguish between classes across thresholds. An AUC close to 1 indicates excellent ability, while near 0.5 suggests random performance. Applying these metrics effectively can help improve RAG performance. Data Preparation for Model Validation Proper data preparation ensures accurate model performance. Handling Missing Data Address missing values by: Cleaning and preprocessing data: Identify and decide whether to remove or fill them in to fix ML data errors. Handling outliers: Manage outliers to prevent skewed predictions. Ensuring high-quality data: Confirm data represents the scenarios your model will encounter. Data Normalization and Standardization Standardizing data helps the model interpret features, especially on different scales. It involves: Processing data into the final format: Transform raw data for modeling. Applying model-specific transformations: Adjust data to meet algorithm needs. Feature Selection and Engineering Choosing the right features enhances performance and interpretability by: Evaluating input strength: Assess features that influence predictions. Generating new features: Create features for learning patterns effectively. Avoiding bias: Ensure steps don't introduce bias into the model. Overfitting and Underfitting: Detection and Solutions Finding the right balance between complexity and generalization is crucial. Identifying Overfitting and Underfitting Overfitting occurs when a model captures noise as patterns, which leads to poor performance on new data. Indications include high training set accuracy but low validation accuracy. Underfitting happens when a model is too simple to capture the data structure, resulting in low accuracy across datasets. Techniques to Mitigate Overfitting To address overfitting: Cross-Validation: Use techniques like K-Fold Cross-Validation. Simplifying the Model: Reduce complexity by limiting features. Regularization: Apply penalties to discourage fitting noise. Data Augmentation: Increase your training dataset size for more general patterns. Hyperparameter Tuning: Fine-tune hyperparameters to optimize model performance. This is especially important when optimizing LLM performance. Recent industry reports indicate that proper hyperparameter optimization can improve model performance by up to 20%, highlighting its significant impact on validation results. Balancing Model Complexity Achieve optimal performance by balancing complexity. Use feature selection and hyperparameter tuning, guided by cross-validation insights. Model Validation in Practice Validating AI models effectively ensures they perform accurately and reliably. Step-by-Step Model Validation with Galileo Using advanced tools like Galileo can simplify the validation process. Here's how to validate your model using Galileo: Import Your Model and Data: Upload your trained model and validation dataset into our platform. Evaluate Model Performance: Use Galileo to assess your model's performance with relevant metrics like accuracy, precision, and recall. Visualize Results: Utilize our visualization tools to interpret results, including ROC curves and confusion matrices. Identify Weaknesses: Pinpoint areas where the model underperforms, such as specific classes or feature interactions. Iterate and Improve: Make data or model adjustments guided by Galileo's insights, and re-validate to measure improvements. By following these steps, you can efficiently validate your model, ensuring it meets the necessary standards before deployment. Real-World Example: Investment and Accounting Solution Improves Efficiency with Galileo A leading investment and accounting solution achieved significant efficiency gains and reduced mean-time-to-detect from days to minutes using our monitoring and validation tools. For more insights, check out Galileo case studies. Comparing Galileo with Competitors While tools like Langsmith offer basic validation features, they may lack scalability and advanced capabilities needed for comprehensive model validation. On the other hand, Scikit-learn and TensorFlow provide built-in validation functions, but these are often limited to model evaluation metrics and may not offer extensive monitoring or error analysis tools. We offer advanced features for monitoring and managing post-deployment model performance. These include detailed error analysis, continuous monitoring for model drift detection, and tools to maintain model freshness in production. The platform provides an intuitive dashboard for easy navigation and interpretation of validation results, supports large datasets and complex models, and facilitates collaboration with integrated documentation and sharing capabilities. For more details, you can explore our blog post on building high-quality models using high-quality data at scale. By choosing Galileo over competitors like Langsmith, AI engineers gain access to a comprehensive tool that enhances model validation processes and supports the long-term success of AI initiatives. Common Challenges and Mistakes Data Leakage: Don't include test data in training. Overfitting to Validation Data: Avoid excessive tuning to validation results. Ignoring Data Quality Issues: Address missing values or outliers to avoid ML data blindspots. Neglecting Real-World Conditions: Validate under realistic scenarios to address GenAI evaluation challenges. Bias and Fairness Oversight: Check for biases across groups. Misinterpreting Metrics: Don't rely on a single metric. In sensitive fields like healthcare, validation challenges such as data leakage and overfitting to validation data can pose significant security and privacy risks. These issues not only compromise the integrity of the model but also potentially expose confidential data. Validation tools must account for these risks to ensure models meet compliance standards, especially under evolving regulations like the EU AI Act. Failing to address these concerns can lead to legal repercussions and loss of trust among stakeholders. Best Practices Use Multiple Evaluation Metrics: Gain a comprehensive view. Align with Business Objectives: Reflect specific goals in metrics. Address Bias and Fairness: Mitigate potential biases. Simulate Real-World Conditions: Test under likely scenarios. Implement Continuous Monitoring: Track performance over time. Document the Validation Process: Maintain transparency and reproducibility. Involve Domain Experts: Collaborate to interpret results. Tools and Frameworks for Model Validation Leading Libraries and Platforms Improve your model validation with these tools: Galileo: Offers an end-to-end solution for model validation with advanced analytics, visualization, and collaboration features. Simplifies the validation process with automated insights and easy integration into workflows. Scikit-learn: Functions for cross-validation and scoring metrics. TensorFlow: Model evaluation APIs and TensorFlow Model Analysis (TFMA). PyTorch: Utilities for handling validation sets and testing. Langsmith: Provides basic validation tools but may lack advanced features for comprehensive analysis. Integration with Machine Learning Pipelines Integrating validation steps into pipelines ensures model reliability: Automate Validation Processes: Use tools like Galileo to simplify cross-validation and evaluation. Continuously Monitor Model Performance: Track performance for drift detection with platforms that support ongoing monitoring. Collaborate Effectively: Utilize shared spaces and documentation features to facilitate teamwork. Conclusion and Future Directions Summary of Best Practices Key best practices include: Define Clear Validation Criteria: Align metrics with business goals. Use Advanced Tools: Utilize platforms like Galileo for efficient validation. Use Diverse Test Datasets: Reflect real-world scenarios. Implement Automated Validation Pipelines: Ensure consistency and efficiency. Involve Cross-Functional Teams: Improve validation processes through collaboration. Document Validation Processes: Maintain transparency and compliance. Continuous Monitoring and Updating: Monitor performance and update practices. Emerging Trends in Model Validation Emerging trends include: Focus on Security and Safety Testing: Automated tools for vulnerability identification. Integration with Development Workflows: CI/CD pipeline inclusion. Compliance with AI Regulations: Meet standards from organizations like NIST. Emphasis on Explainability: Address the "black box" problem. Ongoing Model Governance: Manage AI model risks with frameworks. Improving Your AI Model Validation By embracing these best practices and using advanced tools like Galileo, you can ensure your AI models are both reliable and effective in real-world applications. Our GenAI Studio simplifies AI agent evaluation. 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6458	https://solveordie.com/riddles-about-household-items/	4 riddles about household items - SOLVE or DIE Selected riddles all Difficulty for adults - all Length all Riddles about household items Choose popular riddles or all riddles Clean Clever Crazy Detective Dirty Einstein’s Funny Interview Logic Love Math Mystery Poems Punny Scary Simple Story Stupid Tricky What am I Who am I Riddles for adults All Easy Medium Hard Difficult Riddles for teens All Easy Medium Hard Difficult Riddles for kids All Easy Medium Hard Difficult Riddles by lenght All riddles Short riddles Long riddles With simple answers Let someone else guess riddles about Do not use for your own guessing, just for friends or family. Because topics might suggest you the answer. This filter reset others. 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The clerk says "One will cost you 12 cents, ten will cost your 24 cents, and one hundred will cost you 36 cents." What is the man buying? The man is buying physical numbers to nail to the front of his house. Each number costs 12 cents, and so "1" will cost 12 cents, "10" will cost 24 cents, and "100" will cost 36 cents. Show me the answer 74.40% 98 votes Share riddle logictricky Hardware store in Boston Fred went to a hardware store in Boston with Alex, Ben, and George. He noted that a hammer cost ten times as much as a screwdriver and a power saw cost ten times as much as a hammer. The storekeeper said that Ben could buy a power saw, George could buy a screwdriver and Alex could buy a hammer. Based on this what would the storekeeper let Fred buy? Alex's full name is Alexander and Ben's full name is Benjamin. George was Alex's boss and good friend. Fred could buy all three (the power saw, hammer and screw driver) since he had $111 with him (a $1 bill - George Washington, a $10 Alexander Hamilton, and a $100 bill - Ben Franklin). Boston is in the USA and therefore uses the US currency I just described. Show me the answer 72.80% 79 votes Share riddle cleansimplelogic Blind in a hardware store A blind man walks into a hardware store to buy a hammer. There are hammers hanging behind the front desk, but obviously the blind man isn't able to see them. And yet a few minutes later, he happily walks out of the store, having just purchased a new hammer. How did he do it? He walks up the the front desk where the clerk is working and says "I'd like to buy a hammer." Show me the answer 70.83% 57 votes Share riddle cleversimpletricky Store clerk Elmer Johnson went to the hardware store to make a purchase for his house. He asked the store clerk, "How much will one cost?" The clerk thought for a moment and said, "Three dollars." Elmer Johnson, who looked a little puzzled said, "Well then, how much will twelve cost?" "Six dollars," replied the clerk. Elmer Johnson scratched his head and said, "If I were to purchase two hundred, how much would that cost?" "That," said the clerk, "will cost you nine dollars." What was Elmer Johnson buying? He was buying house numbers. Show me the answer 69.95% 71 votes Share riddle 1 CHOOSE ANOTHER > Page 1 of 1. Popular riddles cleanclevercrazydetectivedirtyEinstein’sfunnyinterviewlogiclovemathmysterypoemspunnyscarysimplestorystupidtrickywhat am Iwho am I Favourite easy riddles More noise than a dog in your house Wordplay Mt. Everest People can’t survive without it You bury me when I am alive Do you know a perfect riddle? Add your own riddlewith answer. Try hardest riddles Sphinx riddle Unusual paragraph Runs forever, but never moves at all Gray, eats fish, and lives in Washington D.C. The largest living ant in the world You solved? 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6459	https://mathquestions.quora.com/How-many-integer-solutions-are-possible-to-the-equation-math-x-4y-y-1-math	How many integer solutions are possible to the equation [math]x!=4y(y+1)[/math]? - Math Questions - Quora Something went wrong. Wait a moment and try again. Try again Skip to content Skip to search Sign In Math Questions PROBABLY THE MOST ACTIVE MATH COMMUNITY ON THE WEB!! UP/DOWN VOTE GOOD/BAD Qs/As Follow · 379.1K 379.1K How many integer solutions are possible to the equation [math]x!=4y(y+1)[/math]? Submission accepted by Abu Khan · See parent question Answer Request Follow · 9 1 3 Answers Sort Recommended Joaquim Nadal Vidal Former Professor at Generalitat De Catalunya (1977–2013) ·1y 4! = 24 = = 4 · 2 · 3 ===> x = 4 and y = 2 5! = 120 = 4 · 5 · 6 ===> x = 5 and y = 5 9 1 Stephen Davey Studied Engineering&Electronics at Christchurch Polytechnic Institute of Technology ·Updated 1y y = X! = gamma(x + 1) [blue line], does not intercept x = (4 y (y + 1), [red line]. To plot: x = 4 y ( y + 1), I used a parametric function. To plot: y = x! I used the equivalent function y = gamma( x + 1 ), to get a continuous curve. Using the substitute parametric variable ‘t’... The graph uses [x = 4 t ( t + 1)] on the shared x-axis. Gamma(t + 1) [green line], and y(t) = t [red line], are plotted on the y-axis. Where these lines intersect gives the x-y vaules which are common to both functions. Continue Reading y = X! = gamma(x + 1) [blue line], does not intercept x = (4 y (y + 1), [red line]. To plot: x = 4 y ( y + 1), I used a parametric function. To plot: y = x! I used the equivalent function y = gamma( x + 1 ), to get a continuous curve. Using the substitute parametric variable ‘t’... The graph uses [x = 4 t ( t + 1)] on the shared x-axis. Gamma(t + 1) [green line], and y(t) = t [red line], are plotted on the y-axis. Where these lines intersect gives the x-y vaules which are common to both functions. 9 2 Nimat Ullah M.sc physics in Physics&Mathematics, Science (Graduated 2016) ·1y The equation x! = 4y(y+1) has only two integer solutions: x = 4, y = 1 x = 5, y = 1 Here's why: For x > 5, x! is divisible by 6, but 4y(y+1) is not. For x = 3, 3! < 4y(y+1) for all y. For x = 2, 2! < 4y(y+1) for all y. For x = 1, 1! < 4y(y+1) for all y. For x = 0, 0! = 1, but 4y(y+1) ≥ 4 for all y. So, there are only two integer solutions: (4,1) and (5,1). View 1 other answer on parent question Related questions What is 5000x 454 x67 x90? If 5^2(x-1) × 5^x+1=0.04, what is the value of x? Can 41 books be placed on more than one shelf, provided that each shelf has the same number of books? Do you write 0.85 as a fraction? When we look at the equation 2=2, there is a 2 on the left side of the equal sign and a 2 on the right side of the equal sign, so how does the equal sign represent complete equality? Can we guess the correct prime number divisor of 90861 in our head? Does Euclidean Space only work because Gödel's first order completeness? Is Euclidean space an objective feature of the universe, or is it a mental abstraction produced by human perception? If Euclidean space is an abstraction derived from reality, can it ever capture the “completeness” of reality itself? And is this something set theory should have resolved? If Euclidean space is a cognitive encoding of perceptual deltas rather than an ontological substrate, does this imply that orthodox mathematics and physics are structurally flawed in their foundational assumptions? Did Gödel’s incompleteness theorems prove that Descartes’ Euclidean sandbox cannot prove or disprove reality, and that Plato’s assertions about eternal mathematical forms are therefore false? What is the explicit form of [math]u_{n-1}=3a^n - 2a^{n-1}u_1+(2a+u_1)u_{n-1}[/math]? How do I figure it out the value of [math]a[/math] that the function [math]f(x) = \begin{cases}ax + 2; & x \leq 2 \ \dfrac{x^{3} - 2x^{2} + 2x - 4}{x - 2}, & x > 2 \end{cases}[/math] will be continous? For [math]a = 1[/math] find the inverse function of [math]f[/math] and show that is one-to-one. The real sequence is defined recursively by [math]b_{n + 1} = \dfrac{1}{2}(b_{n} + \dfrac{3}{b_{n}})[/math] with [math]b_{1} = 2[/math]. How do I show that this sequence is convergent and how to find its limit? How do I find the values of limits [math]\displaystyle \lim_{x \to 1}\dfrac{\sqrt{x + 3} - 2}{x - 1}[/math], [math]\displaystyle \lim_{x \to 0}\dfrac{1 - \cos(x)}{x\tan(3x)}[/math] and [math]\displaystyle \lim_{x \to \infty}(\sqrt{x^{2} + 5x} - x)[/math]? About · Careers · Privacy · Terms · Contact · Languages · Your Ad Choices · Press · © Quora, Inc. 2025
6460	https://www.sciencedirect.com/topics/chemistry/azeotropic-mixture	Skip to Main content Sign in Chapters and Articles You might find these chapters and articles relevant to this topic. Solvent Selection II Using Azeotropes to Select Solvents An azeotrope is a constant-boiling mixture with a constant mole fraction composition of components. Azeotropes consist of two, three, or more components, and can be homogeneous or heterogeneous (more than one phase) . Abbott researchers have detailed using the water contents and solvent concentrations using azeotropes in chasing H2O and i-PrOAc with i-PrOH . Organic chemists probably first experience azeotropes by removing water from a refluxing mixture using a Dean–Stark trap, as with the heterogeneous azeotrope of toluene–water. Most of the azeotropes important to process chemists are minimum-boiling azeotropes, i.e., the boiling point of the mixture is less than the boiling point of any component. (A familiar exception is concentrated aqueous HCl, which is formed as a positive azeotrope.) Extensive data are available from Gmeling's Dortmund Database . All heterogeneous azeotropes are minimum-boiling . Unlike liquids may form azeotropes if the boiling points are relatively similar. For example, toluene azeotropes with H2O, EtOH, ethylenediamine, and AcOH, but toluene and hexanoic acid do not form an azeotrope. Many organic solvents form azeotropes with water (Table 5.5), and azeotroping out H2O can conveniently dry product-rich extracts and equipment. For process scientists the value of azeotropes lies primarily in the ability to efficiently remove volatile components of reaction mixtures. Azeotropic removal of volatile components can drive reactions, as shown in Figure 5.8. Azeotropic removal of water formed an imine key to a domino sequence that led to an oseltamivir precursor . Azeotropic removal of water led to the protection of an alcohol as the trityl ether . Azeotropes may facilitate separations and workups. For instance, excess ethylenediamine was removed by azeotropic distillation with toluene in the hydrazinolysis of S-serine methyl ester (Figure 5.8); in this case separating the polar hydrazine from the polar, water-soluble product by extraction could be difficult. Hexamethyldisiloxane, a byproduct from acid-catalyzed deprotection of trimethylsilyl compounds, forms an azeotrope with esters, alcohols, and trimethylsilanol . For the workup of the TMS-protected product in Figure 12.2 hexamethyldisiloxane was removed as an azeotrope with the solvent acetonitrile . Even if an azeotrope is not rigorously characterized removal of components may be facilitated by a lower boiling point. An azeotrope may also be an economical solvent if the solvent can be recovered and reused, as is the case for the agricultural intermediate in Figure 5.8 . Expect azeotropes to be present in processing. Exploit them in developing processes. Some useful binary azeotropes of solvents are shown in Table 5.9. When the composition of a binary azeotrope is closer to 1:1 (as is the case for i-PrOH–i-PrOAc, for instance) solvent chasing from either solvent to the other will require less solvent. Some ternary azeotropes (solvent–solvent–water) that may be encountered are shown in Table 5.10 . Some ambiguities exist in the literature around azeotropic data. TABLE 5.9. Some Binary Azeotropes of Note (1,2) \| More-polar Solvent \| Heptane (bp 98 °C) \| \| Toluene (bp 111 °C) \| \| Other Solvent \| \| \| --- --- --- --- \| \| Azeotrope bp \| % of Polar Solvent \| Azeotrope bp \| % of Polar Solvent \| Second Solvent \| Azeotrope bp \| % of More-polar Solvent \| \| MeOH (bp 65 °C) \| 59.1 °C \| 51.5 wt% \| 63.8 °C \| 69 wt% \| MTBE \| 15 wt% \| 52 °C \| \| EtOH (bp 78 °C) \| 70.9 °C \| 49 wt% \| 76.7 °C \| 68 wt% \| Cyclohexane \| 31 wt% \| 65 °C \| \| i-PrOH (bp 82 °C) \| 76.4 °C \| 50 wt% \| 80.6 °C \| 69 wt% \| i-PrOAc \| 52 wt% \| 80 °C \| \| 2-BuOH (bp 98 °C) \| 88 °C \| 37 wt% \| 95 °C \| 55 wt% \| H2O \| 73 wt% \| 87 °C \| \| t-BuOH (bp 83 °C) \| 78 °C \| 62 wt% \| None \| None \| H2O \| 88 wt% \| 80 °C \| \| t-AmOH (bp 102 °C) \| 92.2 \| 26.5 wt% \| 100.5 °C \| 56 wt% \| H2O \| 72 wt% \| 87 °C \| \| acetone (bp 56 °C) \| 56 °C \| 89.5 wt% \| None \| None \| \| \| \| \| CH3CN (bp 81 °C) \| 69.4 °C \| 44 wt% \| 81.1 °C \| 78 wt% \| EtOH \| 44 wt% \| 73 °C \| \| EtOAc (bp 77 °C) \| 77 °C (2) \| 6 wt% (2) \| None \| None \| EtOH \| 74 wt% \| 72 °C \| \| i-PrOAc (bp 89 °C) \| 88 °C \| 67 wt% \| None \| None \| MeOH \| 20 wt% \| 65 °C \| \| CH3CO2H (bp 118 °C) \| 95 °C \| 17 wt% \| 104 °C \| 32 wt% \| \| \| \| \| HCO2H (bp 100 °C) \| 78.2 °C \| 56.5 wt% \| 86 °C \| 50 wt% \| \| \| \| (1) : Unless otherwise indicated, data are from McConville, F. X. The Pilot Plant Real Book; 2nd ed.; FXM Engineering & Design: Worcester MA, 2007; Section 6. (2) : Lowenthal, H. J. E. A Guide for the Perplexed Organic Experimentalist; 2nd ed.; Wiley: New York, 1990; p 171. TABLE 5.10. Some Ternary (Solvent–Solvent–Water) Azeotropes that May be Encountered (1) \| Solvent A \| Solvent B \| Azeotrope bp (Atmospheric Pressure) \| Azeotrope Composition A:B:water (wt%) \| --- --- \| \| EtOH (bp 78 °C) \| EtOAc (bp 77 °C) \| 70 °C \| 8:83:9 \| \| i-PrOH (bp 82 °C) \| i-PrOAc (bp 111 °C) \| 76 °C \| 13:76:11 \| \| i-PrOH (bp 82 °C) \| PhCH3 (bp 111 °C) \| 76 °C \| 38:49:13 \| \| 2-BuOH (bp 88 °C) \| Heptane (bp 98 °C) \| 76 °C \| 22:67:11 \| (1) : McConville, F. X. The Pilot Plant Real Book; 2nd ed.; FXM Engineering & Design: Worcester MA, 2007; Section 6. Often conducting a distillation at a reduced pressure will reduce the fraction of the minor component in the distillate, as shown in Table 5.11 for the EtOAc–H2O azeotrope; this is termed “breaking an azeotrope.” This behavior is not found for the i-PrOH–H2O azeotrope. McConville has graphically shown the effects of reduced pressure on the composition of some azeotropes . TABLE 5.11. Effect of Reducing Distillation Pressure on EtOAc–H2O and iPrOH–H2O Azeotropes \| EtOAc–H2O Azeotrope \| \| \| iPrOH–H2O Azeotrope \| \| \| --- --- --- \| \| Pressure (mm) \| bp (°C) \| Water in Azeotrope (wt%) \| Pressure (mm) \| bp (°C) \| Water in Azeotrope (wt%) \| \| 760 \| 70.4 \| 8.5 \| 760 \| \| 12 \| \| 250 \| 42.6 \| 6.3 \| 319 \| \| 12 \| \| 25 \| 1.9 \| 3.6 \| 153 \| \| 12 \| \| \| \| \| 67 \| \| 12 \| Sources: Horsley, L. H. Azeotropic Data – III; Advances in Chemistry Series, 116; ACS: Washington, D.C., 1973. Gmehling, J.; Menke, J.; Fischer, K.; Krafczyk, J. Azeotropic Data, Parts I and II; VCH: Weinheim, 1994. Reichardt, C. Solvents and Solvent Effects in Organic Chemistry; 2nd ed.; VCH: Weinheim, 1990. 4 CRC Handbook of Chemistry and Physics; 78th ed., Lide, D. R., Ed., CRC Press: Boca Raton, FL, pp 97–98. Industrial Solvents Handbook; 4th ed., Flick, E. W., Ed., Noyes Data Corporation: Park Ridge, NJ, 1990. View chapterExplore book Read full chapter URL: Book2012, Practical Process Research and Development (Second Edition)Neal G. Anderson Chapter Solvent Azeotropes 2014, Cleaning with SolventsJohn B Durkee II 8.5.7 Replacement Binary Azeotropes for HCFC-141b HCFC-141b is also a Class II ODC. Its manufacture and use are banned by the Clean Air Act implementing the Montreal Protocol in the US. Nonetheless, its use and manufacture is current in other countries (more details are given in Chapter 9, Footnote 23). There are at least 105 azeotropes in the database of binary azeotropes which have HSP values similar to those of HCFC-141b, as indicated by calculated RA values less than 8 MPa½ using Equation 2.8. Eight of them meet the additional requirements for replacement—chiefly higher values of exposure limits and low boiling point. They are listed in Table 8.7. Table 8.7. Binary Azeotropes Suitable to Replace HCFC-141b for Solvent Cleaning Operations \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| --- --- --- --- --- --- --- \| Empty Cell \| Azeotrope Composition \| \| wt. % A \| T Boiling °C \| Density g/cc \| Surface Tension dyne/cm \| Patent Coverage \| Flash Point? \| Exposure Limits, ppm \| \| HSP, MPa1/2 \| \| \| \| \| Empty Cell \| A \| B \| Empty Cell \| Empty Cell \| Empty Cell \| Empty Cell \| Empty Cell \| Empty Cell \| A \| B \| RA \| δD \| δP \| δH \| \| CAS \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| 1717-00-6 \| HCFC-141b \| \| \| 32 \| 1.23 \| 21 \| \| \| \| 500 \| 0 \| 15.7 \| 4 \| 1 \| \| Azeo \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| 651 \| Acetone \| Cyclopentane \| 36.00 \| 41 \| 0.761 \| 22.4 \| \| \| 750 \| 600 \| 2.7 \| 16.1 \| 3.4 \| 3.5 \| \| 652 \| Acetone \| Pentane \| 20.00 \| 36 \| 0.653 \| 16.5 \| \| \| 750 \| 1000 \| 3.32 \| 14.6 \| 1.4 \| 1 \| \| 1083 \| HFC-365mfc \| Pentane \| 66.00 \| 27.4 \| 0.941 \| 15.3 \| USP 6,303,668 \| \| 1000 \| 1000 \| 4.12 \| 15.7 \| 0 \| 0 \| \| 1285 \| HFE-7100 \| Ethanol \| 93.00 \| 52 \| 1.428 \| 15.3 \| WO 96/36689 & USP 6,426,327 \| NOT? \| 750 \| 1000 \| 4.16 \| 13.8 \| 2.6 \| 2 \| \| 349 \| Ethanol \| Cyclopentane \| 7.50 \| 44.7 \| 0.749 \| 21.9 \| \| \| 1000 \| 600 \| 4.16 \| 16.4 \| 0.4 \| 2.6 \| \| 1082 \| HFC-365mfc \| Water \| 98.00 \| 38 \| 1.263 \| 15.7 \| USP 6,365,566 \| \| 1000 \| Not Listed \| 4.22 \| 16.4 \| 0.1 \| 0 \| \| 1036 \| HFC-365mfc \| Ethanol \| 98.40 \| 39.2 \| 1.258 \| 15.3 \| USP 5,445,757 \| \| 1000 \| 1000 \| 4.25 \| 16.4 \| 0.1 \| 0.2 \| \| 1027 \| HFE-7100 \| HFC-365mfc \| 1.00 \| 40 \| 1.272 \| 15 \| USP 6,951,835 \| No Flash \| 750 \| 1000 \| 4.33 \| 16.4 \| 0 \| 0 \| Binary azeotrope 1082 (HFC-356mfc and water) appears to offer potential for replacing HCFC-141b because both components are VOC exempt in the US, and the exposure limits of both components raise no concern. This composition is claimed in USP 6,635,56633, although use as a cleaning solvent is not claimed. Other potentially useful azeotropes are 651 (acetone and cyclopentane), 652 (acetone and pentane), and 349 (ethanol and cyclopentane). An additional two azeotropes, however, offer a potential advantage over HCFC141b—in addition to their SHE advantages. That is reduced surface tension. They are 1083 (HFC-365mfc and pentane), 1285 (HFE-7100 and ethanol), and 1036 (HFC-365mfc and ethanol). All are claimed compositions in US patents. Dissimilarity of these replacements and HCFC-141b are shown in Figure 8.11. All of these seven azeotropes will raise concerns about flammability. View chapterExplore book Read full chapter URL: Book2014, Cleaning with SolventsJohn B Durkee II Chapter Basic Laboratory Operations 2016, Experimental Organic ChemistryJoaquín Isac-García, ... Henar Martínez-García 4.11.5 Azeotropic distillation (Dean-Stark) An azeotrope is a mixture of two or more liquid components under constant boiling, and distillation processes are performed as if they were a pure compound (see Table 4.6). Azeotropic distillation is a useful procedure for removing a liquid from a crude reaction by a co-distillation with an immiscible organic solvent. This technique is often used in equilibrium reactions where water is formed as byproducts of the reaction. Water removal shifts the equilibrium of the reaction toward the product side. If the reaction is carried out, for example, with toluene, which is less dense than water, the steam in the reflux condenser will consist of an azeotropic mixture of toluene and water. When this mixture is condensed, it falls into the so-called Dean-Stark, forming two layers: the top layer will consist of toluene and the bottom layer water. When the liquid level in the Dean-Stark trap reaches the top of the side arm, the toluene flows back into the reaction flask. The water can be removed through a stopcock in the bottom of the Dean-Stark trap (see Figure 4.29). Table 4.6. Different azeotropes formed by the most common solvents. \| Component A \| \| Component B \| \| Azeotrope \| --- --- \| B.p. (°C) \| % (weigh) \| B.p. (°C) \| % (weigh) \| B.p. (°C) \| \| H2O (100) \| 1.3 \| Diethyl ether (34.5) \| 98.7 \| 34.2 \| \| H2O (100) \| 1.4 \| Pentane (36.1) \| 98.6 \| 34.6 \| \| MeOH (64.7) \| 12.1 \| Ketone (56.1) \| 87.9 \| 55.5 \| \| MeOH (64.7) \| 72.5 \| Toluene (110.7) \| 27.5 \| 63.5 \| \| EtOH (78.3) \| 68.0 \| Toluene (110.7) \| 32.0 \| 76.7 \| \| H2O (100) \| 13.5 \| Toluene (110.7) \| 86.5 \| 84.1 \| View chapterExplore book Read full chapter URL: Book2016, Experimental Organic ChemistryJoaquín Isac-García, ... Henar Martínez-García Chapter Solution Thermodynamics—Use of the Second and Third Derivatives of 2017, Solution Thermodynamics and its Application to Aqueous Solutions (Second Edition)Yoshikata Koga [II-9] Azeotrope The azeotrope occurs when the liquid phase composition and the gas phase composition are identical in a binary mixture B-W at the equilibrium, or the total vapor pressure as a function of xB shows an extremum, as depicted in Fig. II-9, i.e., with . It follows then that, (II-85) Taking into account the Gibbs-Duhem relation. Eq. (II-25), and the expression for the chemical potential using the partial pressure, Eq. (II-9), Eq. (II-85) is rewritten as, (II-86) Recall that at a critical point (dμB/dxB) = 0 and hence (dpB/dxB) = 0. Thus, unless the system is at a critical point, the following relation holds at an azeotrope, (II-87) This occurs quite commonly in aqueous alcohols. As will be discussed in Chapter III, this situation is problematic when analyzing the vapor pressure data to obtain chemical potentials by the Boissonnas method (Boissonnas, 1939). We will circumvent this difficulty as detailed in Section [III-7]. View chapterExplore book Read full chapter URL: Book2017, Solution Thermodynamics and its Application to Aqueous Solutions (Second Edition)Yoshikata Koga Chapter 27 European Symposium on Computer Aided Process Engineering 2017, Computer Aided Chemical EngineeringYeong-Gak Yoon, ... Chul-Jin Lee 3.1 Experiment Setup The azeotrope distillation column on the pilot-scale, as shown in Figure 2, is a packed distillation column with an internal diameter of 53.5 mm and a height of 4.77 m. The packing material is Dixone 3 mm, and the HETP (height equivalent to a theoretical plate) of the packing is 0.055 m. The packing height is 3.1 m. The theoretical number of trays is 56. The feed is a mixture of 2-MPA (≧ 99 %, Sigma Aldrich) and MEK (≧ 99.5 %, Sigma Aldrich), and water was introduced to 27th tray by a pump (P101). A hot-oil-based thermosiphon reboiler (E102) was used to supply the heat to the column. The column-bottom level was adjusted by a pump (P104) with a level transmitter (LT), and the bottom stream was cooled by a cooler (E103). A total condenser at the top of the column was used to condense the entire overhead vapor stream that was gathered in a decanter (D101), and the condensed vapor was separated to two liquid phases. The organic phase was returned to the top tray of the column by a reflux pump (P102), and the stream was adjusted by a flow controller (FC). The aqueous phase, which includes most of the water from the azeotrope distillation column process, was discharged as an top product, and the D101 level was controlled by a solenoid valve (SV). The temperatures throughout the system were recorded by thermocouple sensors (T1 ~ T9) with a resolution of 0.1° C. The cyclohexane for the make-up was fed to the reflux line by a pump (P103). View chapterExplore book Read full chapter URL: Book series2017, Computer Aided Chemical EngineeringYeong-Gak Yoon, ... Chul-Jin Lee Chapter Common Physical Techniques Used In Purification 2022, Purification of Laboratory Chemicals (Ninth Edition)Wilfred L.F. Armarego Types of condensers: Air condenser. A glass tube such as the inner part of a Liebig condenser. Used for liquids with boiling points above 90°. Can be of any length. Allihn condenser. The inner tube of a Liebig condenser is modified by having a series of bulbs to increase the condensing surface. Further modifications of the bubble shapes give the Julian and Allihn-Kronbitter (glass-jacketed) condensers. Bailey-Walker condenser. A type of all-metal condenser fitting into the neck of extraction apparatus and being supported by the rim. Used for high-boiling liquids. Coil condenser. An open tube, into which is sealed a glass coil or spiral through which water circulates. The tube is sometimes also surrounded by an outer cooling jacket. Double surface condenser. A tube in which the vapour is condensed between an outer and inner water-cooled jacket after impinging on the latter. Very useful for liquids boiling below 40°. Friedrichs condenser. A ‘cold-finger’ type of condenser sealed into a glass jacket open at the bottom and near the top. The cold finger is formed into glass screw threads. Graham condenser. A type of coil condenser. Hopkins condenser. A cold-finger type of condenser resembling that of Friedrichs. Liebig condenser. An inner glass tube surrounded by a glass jacket through which water is circulated. Othmer condenser. A large-capacity condenser which has two coils of relatively large bore glass tubing inside it, through which the water flows. The two coils join at their top and bottom. West condenser. A Liebig condenser with a light-walled inner tube and a heavy-walled outer tube, with only a narrow space between them. Wiley condenser. A condenser resembling the Bailey-Walker type. [For more information see: Vacuum distillation. This expression is commonly used to denote a distillation under reduced pressure lower than that of the normal atmosphere. As the boiling point of a substance depends on the pressure, it is often possible to distil materials at a temperature low enough to avoid partial or complete decomposition by lowering the pressure, even if they are unstable when boiled at atmospheric pressure. Sensitive or high-boiling liquids should invariably be distilled or fractionally distilled under reduced pressure. The apparatus is essentially as described for distillation except that ground joints connecting the different parts of the apparatus should be air-tight by using grease, or better Teflon sleeves. For low, moderately high, and very high temperatures Apiezon L, M and T greases, respectively, are very satisfactory. Alternatively, it is often preferable to avoid grease and to use thin Teflon sleeves in the joints. The distillation flask, must be supplied with a capillary bleed (which allows a fine stream of air, nitrogen or argon into the flask), and the receiver should be of the fraction collector type. When distilling under vacuum it is very important to place a loose packing of glass wool above the liquid to buffer sudden boiling of the liquid. The flask should not be more than two-thirds full of liquid. The vacuum must have attained a steady state, i.e. the liquid has been completely degassed, before the heating source is applied, and the temperature of the heat source must be raised very slowly until boiling is achieved. If the pump is a filter pump off a high-pressure water supply, its performance will be limited by the temperature of the water because the vapour pressure of water at 10°, 15°, 20° and 25° is 9.2, 12.8, 17.5 and 23.8 mm Hg, respectively. The pressure can be measured with an ordinary manometer. For vacuums in the range of 10-2 mm Hg to 10 mm Hg, rotary mechanical pumps (oil pumps) are used and the pressure can be measured with a Vacustat McLeod-type gauge. If still higher vacuums are required, for example for high vacuum sublimations, a mercury diffusion pump is suitable. Such a pump can provide a vacuum up to 10-6 mm Hg. For better efficiencies, the diffusion pump can be backed up by a mechanical pump. In all cases, the mercury pump is connected to the distillation apparatus through several traps to remove mercury vapours. These traps may operate by chemical action, for example the use of sodium hydroxide pellets to react with acid vapours, or by condensation, in which case empty tubes cooled in solid carbon dioxide-ethanol or liquid nitrogen (contained in wide-mouthed Dewar flasks) are used. Special oil or mercury traps are available commercially, and a liquid-nitrogen (b -209.9°C) trap is the most satisfactory one to use between these and the apparatus. It has an advantage over liquid air or oxygen in that it is non-explosive if it becomes contaminated with organic matter. Air should not be sucked through the apparatus before starting a distillation because this will cause liquid oxygen (b -183°C) to condense in the liquid nitrogen trap, and this is potentially explosive (especially in mixtures with organic materials). Due to the potential lethal consequences of liquid oxygen/organic material mixtures, care must be exercised when handling liquid nitrogen. Hence, it is advisable to degas the system for a short period before the trap is immersed into the liquid nitrogen (which is kept in a Dewar flask). Spinning-band distillation. Factors which limit the performance of distillation columns include the tendency to flood (which occurs when the returning liquid blocks the pathway taken by the vapour through the column) and the increased hold-up (which decreases the attainable efficiency) in the column that should, theoretically, be highly efficient. To overcome these difficulties, especially for distillation under high vacuum of heat sensitive or high-boiling highly viscous fluids, spinning band columns are commercially available. In such units, the distillation columns contain a rapidly rotating, motor-driven, spiral band, which may be of polymer-coated metal, stainless steel or platinum. The rapid rotation of the band in contact with the walls of the still gives intimate mixing of descending liquid with ascending vapour while the screw-like motion of the band drives the liquid towards the still-pot, helping to reduce hold-up. There is very little pressure drop in such a system, and very high throughputs are possible, with high efficiency. For example, a 765-mm long 10-mm diameter commercial spinning-band column is reported to have an efficiency of 28 plates and a pressure drop of 0.2 mm Hg for a throughput of 330 mL/hour. The columns may be either vacuum jacketed or heated externally. The stills can be operated down to 10-5 mm Hg. The principle, which was first used commercially in the Podbielniak Centrifugal Superfractionator, has also been embodied in descending-film molecular distillation apparatus. Steam distillation. When two immmiscible liquids distil, the sum of their (independent) partial pressures is equal to the atmospheric pressure. Hence in steam distillation, the distillate has the composition where the P’s are vapour pressures (in mm Hg) in the boiling mixture. The customary technique consists of heating the substance and water in a flask (to boiling), usually with the passage of steam, followed by condensation and separation of the aqueous and non-aqueous phases in the distillate. Its advantages are those of selectivity (because only some water-insoluble substances, such as naphthalene, nitrobenzene, phenol and aniline are volatile in steam) and of ability to distil certain high-boiling substances well below their boiling point. It also facilitates the recovery of a non-steam-volatile solid at a relatively low temperature from a high-boiling solvent such as nitrobenzene. The efficiency of steam distillation is increased if superheated steam is used (because the vapour pressure of the organic component is increased relative to water). In this case the flask containing the material is heated (without water) in an oil bath and the steam passing through it is superheated by prior passage through a suitable heating device (such as a copper coil heated electrically or an oil bath). Azeotropic distillation. In some cases two or more liquids form constant-boiling mixtures, or azeotropes. Azeotropic mixtures are most likely to be found with components which readily form hydrogen bonds or are otherwise highly associated, especially when the components are dissimilar, for example an alcohol and an aromatic hydrocarbon, but have similar boiling points. Examples where the boiling point of the distillate is a minimum (less than either pure component) include: Water with ethanol, n-propanol and isopropanol, tert-butanol, propionic acid, butyric acid, pyridine, methanol with methyl iodide, methyl acetate, chloroform, ethanol with ethyl iodide, ethyl acetate, chloroform, benzene, toluene, methyl ethyl ketone, benzene with cyclohexane, acetic acid with toluene. Although less common, azeotropic mixtures are known which have higher boiling points than their components. These include water with most of the mineral acids (hydrofluoric, hydrochloric, hydrobromic, perchloric, nitric and sulfuric) and formic acid. Other examples are acetic acid-pyridine, acetone-chloroform, aniline-phenol, and chloroform-methyl acetate. The following azeotropes are important commercially for drying ethanol: \| \| \| --- \| \| ethanol 95.5% (by weight) - water 4.5% \| b 78.1° \| \| ethanol 32.4% - benzene 67.6% \| b 68.2° \| \| ethanol 18.5% - benzene 74.1% - water 7.4% \| b 64.9° \| Materials are sometimes added to form an azeotropic mixture with the substance to be purified. Because the azeotrope boils at a different temperature, this facilitates separation from substances distilling in the same range as the pure material. (Conversely, the impurity might form the azeotrope and be removed in this way.) This method is often convenient, especially where the impurities are isomers or are otherwise closely related to the desired substance. Formation of low-boiling azeotropes also facilitates distillation. One or more of the following methods can generally be used for separating the components of an azeotropic mixture: 1. : By using a chemical method to remove most of one species prior to distillation. (For example, water can be removed by suitable drying agents; aromatic and unsaturated hydrocarbons can be removed by sulfonation). 2. : By redistillation with an additional substance which can form a ternary azeotropic mixture (as in the ethanol-water-benzene example given above). 3. : By selective adsorption of one of the components. (For example, of water on to silica gel or molecular sieves, or of unsaturated hydrocarbons onto alumina). 4. : By fractional crystallisation of the mixture, either by direct freezing or by dissolving in a suitable solvent. Kügelrohr distillation. The Aldrich Kugelrohr Distillation Apparatus (see Aldrich-Sigma Labware catalogue) is made up of small glass bulbs (ca 4-5 cm diameter) that are joined together via Quickfit joints at each pole of the bulbs. The liquid (or low melting solid) to be purified is placed in the first bulb of a series of bulbs joined end to end, and the system can be evacuated. The first bulb is heated in a furnace (e.g. Buchi Kugelrohr micro distillation oven from Sigma-Aldrich Labware catalogue) at a high temperature whereby most of the material distils into the second bulb (which is outside of the furnace). The second bulb is then moved into the furnace and the furnace temperature is reduced by ca 5° whereby the liquid in the second bulb distils into the third bulb (at this stage the first bulb is now out at the back of the furnace, and the third and subsequent bulbs are outside the front of the furnace). The furnace temperature is lowered by a further ca 5°, and the third bulb is moved into the furnace. The lower boiling material will distil into the fourth bulb. The process is continued until no more material distils into the subsequent bulb. The vacuum (if applied) and the furnace are removed, the bulbs are separated and the various fractions of distillates are collected from the individual bulbs. For volatile liquids, it may be necessary to cool the receiving bulb with solid CO2 held in a suitable container (a Kugelrohr distillation apparatus with an integrated cooling system is available). This procedure is used for preliminary purification and the distillates are then redistilled or recrystallised. Isopiestic, isobaric or isothermal distillation. This technique can be useful for the preparation of metal-free solutions of volatile acids and bases for use in trace metal studies. The procedure involves placing two beakers, one of distilled water and the other of a solution of the material to be purified, in a desiccator. The desiccator is sealed and left to stand at room temperature for several days. The volatile components distribute themselves between the two beakers whereas the non-volatile contaminants remain in the original beaker. This technique has afforded metal-free pure solutions of ammonia, hydrochloric acid and hydrogen fluoride. The procedure can also be used to crystallise and purify inorganic compounds. For example, if a concentrated aqueous solution of an inorganic compound, e.g. AsO3, is placed in an open beaker or Petri dish in a large container that can be sealed, together with a second dish containing a pure solution of the anion e.g. HCl, and set aside undisturbed for a day or more, then pure AsCl3 can be obtained. [For ‘Room-temperature isopiestic distillation of in situ generated arsenious chloride and its application for the determination of trace level impurities in arsenious oxide’, see Chaurassia, Sahayam and Mishra Anal Chem 74(23) 6102 2002, DOI: 10.1021/ac020136a; and ‘An improved isopiestic method for measurement of water activities in aqueous polymer and salt solutions’ see Lin, Mei, Zhu and Han Fluid PhaseEquilibria 118(2) 241 1996, View chapterExplore book Read full chapter URL: Book2022, Purification of Laboratory Chemicals (Ninth Edition)Wilfred L.F. Armarego Chapter 21 European Symposium on Computer Aided Process Engineering 2011, Computer Aided Chemical EngineeringLaszlo Hegely, Peter Lang Abstract In multicomponent azeotropic mixtures (e.g. waste solvents), the products obtainable by batch distillation and their maximal amount ishighly dependent on the charge composition. A method is presented for the determination of product sequences for any number of components based only on the boiling points of pure components and azeotropes, and azeotropic compositions, without the knowledge of further VLE data. The method is suitable for heteroazeotropes as well. The stability of fixed points are determined with the assumption that the Serafimov-type of the mixture occurs in Reshetov's statistics. On the basis of the stabilities, we enumerate all feasible product sequences using the algorithm of Ahmad et al. (1998). Finally, the amount of products are determined assuming maximal separation for the given charge composition. The results are presented for the system acetone-chloroform-methanol-ethanol-benzene. View chapterExplore book Read full chapter URL: Book series2011, Computer Aided Chemical EngineeringLaszlo Hegely, Peter Lang Chapter 22 European Symposium on Computer Aided Process Engineering 2012, Computer Aided Chemical EngineeringJ. Cortez-González, ... A. Hernández-Aguirre Abstract The separation of azeotropic mixtures with help of distillation is an important unit operation in chemical and pharmaceutical industry. The literature focuses on different alternatives for the separation of azeotropic mixtures, such as extractive distillation, azeotropic distillation, heterogeneous azeotropic distillation, vacuum distillation and the Pressure Swing Distillation (PSD). This work describes an approach for PSD in continuous flow for binary azeotropic mixture separation sensible to the pressure changes. The main advantages of this process compared to the others are: i) no additional substances (entrainer) have to be used, and ii) for the continuous flow operation heat integration is possible and it can save energy. The PSD has been optimized with a novel stochastic algorithm called Boltzmann Univariate Marginal Distribution Algorithm (BUMDA). The performance of BUMDA is robust and highly efficient as shown by the experiments conducted (although it does not guarantee optimality). This work makes two specific contributions: 1) the optimization of the PSD in continuous for a binary azeotropic mixture; 2) the application of the BUMDA stochastic algorithm, and the constraint handling technique. View chapterExplore book Read full chapter URL: Book series2012, Computer Aided Chemical EngineeringJ. Cortez-González, ... A. Hernández-Aguirre Chapter Vapor–Liquid Equilibrium and Physical Properties for Distillation 2014, DistillationJürgen Gmehling, Michael Kleiber 2.7 Conditions for the occurrence of azeotropic behavior The greatest separation problem for distillation processes is the occurrence of binary, ternary, and quaternary azeotropic points . At the azeotropic point, for homogeneous systems the mole fractions of all components in the liquid phase are identical with the mole fractions in the vapor phase. This leads to the fact that all K-factors (relative volatilities) show values of unity at the azeotropic point and that the system cannot be separated by ordinary distillation. Therefore a reliable knowledge of all azeotropic points for the system to be separated is of great importance for the synthesis and design of the separation step. For a binary system, the following relations can be derived for homogeneous systems at the azeotropic point using the simplified Eqn (2.8) of the γ–ϕ approach: (2.23) Using the ϕ–ϕ approach the following relation is obtained for the azeotropic point starting from Eqn (2.7): (2.24) It can be seen that starting from Eqn (2.23) azeotropic behavior occurs if for a given composition the ratio of the activity coefficients γ2/γ1 is identical to the ratio of the pure component vapor pressures . The typical curvature of the γ2/γ1-ratio is shown in Figure 2.14 in logarithmic form for an azeotropic system with a positive (left-hand side) or negative (right-hand side) deviation from Raoult's law at constant temperature. The azeotropic composition can directly be obtained from the intersection of the straight line for the vapor pressure ratio and the curve for the ratio of the activity coefficients γ2/γ1. Since the ratio of the vapor pressures (component 1 low boiler) is always greater than in the case of positive deviation from Raoult's law and always greater than in the case of negative deviation, azeotropic behavior always occurs when: (2.25) From Eqn (2.23) it can be concluded that binary azeotropes easily occur if the vapor pressures of the two components are very similar, since in this case already very small deviations from Raoult's law are sufficient to fulfill Eqn (2.25) and to produce an azeotropic point either with positive or negative deviations from Raoult's law. At the Bancroft point, where the vapor pressure curves intersect, even ideal systems such as water–heavy water show azeotropic behavior . From Figure 2.14 and Eqn (2.21), it can be concluded that the occurrence of azeotropic points as a function of temperature can be calculated if in addition to the activity coefficients at infinite dilution the ratio of the vapor pressures is known for the temperature range covered. The vapor pressures and the activity coefficients at infinite dilution depend on temperature following the Clausius–Clapeyron or Gibbs–Helmholtz equations . The result can be that azeotropic behavior occurs or disappears with temperature (pressure). For the ethanol–1,4-dioxane and acetone–water systems, these values are shown in Figure 2.15 together with the calculated results using the NRTL equation. While for the ethanol–1,4-dioxane system the azeotropic behavior disappears at higher temperature; the opposite is true for the acetone–water system, which shows azeotropic behavior at temperatures above 343 K. The experimental and calculated azeotropic points are shown in Figure 2.16. As can be seen, the occurrence and disappearance of the azeotropic behavior is described reliably with the NRTL model. The procedure for the determination of ternary and quaternary homogeneous and heterogeneous azeotropes using gE models or equations of state is explained in detail in Ref. . Azeotropic behavior is not rare; approximately 45% of the published 55,500 azeotropic and zeotropic data for binary and higher systems stored in the DDB and published in a data compilation show azeotropic behavior. View chapterExplore book Read full chapter URL: Book2014, DistillationJürgen Gmehling, Michael Kleiber Chapter Design and Operation of Batch Distillation 2014, DistillationEva Sorensen 5.6.2 Extractive batch distillation Extractive distillation is an important process in chemical industries for the separation of azeotropes and close boiling mixtures, and continuous extractive distillation is a well-known and widespread technology. In contrast, extractive distillation in batch mode is a relatively new process in the literature, with the first studies published by Bernot et al. [42,43], and so far the process has had limited instances (or at least reports) of industrial implementation. As the process offers the advantages of both batch and extractive distillation, it has been extensively studied in the academic literature over the past decade. The choice of entrainer is clearly of great importance and should ideally be considered as part of the overall synthesis and design problem (although this is not considered here). The focus in the literature has been on two strands: either focusing on graphical methods with the primary objective of assessing the feasibility of different entrainers and column configurations for given mixtures, or considering the optimal design and/or operation for given entrainers and configurations. The most common azeotropes encountered industrially are minimum boiling azeotropes, that is, the azeotrope is the lightest fraction, and these are normally separated using a low volatile entrainer, that is, the entrainer is the heaviest fraction. The easiest and most commonly used mode of operation to separate these mixtures in a regular batch still is to add the entrainer directly to the still at the start; this is also called solvent enhanced batch distillation , as illustrated in Alternative A in Figure 5.13. It is also possible to continuously add the entrainer to the still during the operation (Alternative B). In the academic literature, the process is commonly operated using a four-step operating policy for binary minimum azeotropic mixtures in a regular column based on Alternative C in Figure 5.13: 1. : Total reflux operation without entrainer feeding. This establishes the azeotropic composition at the top. 2. : Total reflux with entrainer feeding near the top of the column. This breaks the azeotrope and achieves a high purity of the light component at the top. 3. : Finite reflux operation with entrainer feeding to withdraw the light component. 4. : Finite reflux operation without entrainer feeding. The heavy component from the original feed mixture is withdrawn over the top and the highly pure entrainer is left in the reboiler. Offcut fractions may also be taken off between each product. The entrainer is normally added either at the top of the column section, in which case the whole column section becomes an extractive zone, or to a point in the column section, separating the column into a rectifying section above the entrainer feed and an extractive section below the entrainer feed. Some authors have considered using intermediate boiling entrainers to separate minimum boiling azeotropes (e.g. ), or have studied the separation of maximum boiling azeotropes (e.g. ). Extractive batch distillation in middle vessel configurations has also been considered, with different alternatives for addition of entrainer, as illustrated in Figure 5.14; that is, either feed to the middle vessel, to a location toward the top of the rectification section, or to a location toward the top of the stripping section (e.g. [9,47]). The feed can be added either to the middle vessel or the reboiler or can be distributed between the two. The intermediate component can be withdrawn from the middle vessel, although most authors have not considered this option, only allowing removal as distillate or as bottoms. For the separation of binary minimum azeotropic mixtures in a middle vessel column using a low boiling entrainer, the following operations have been considered : 1. : Total reflux and total reboil operation without entrainer feeding. 2. : Total reflux and reboil operation with entrainer feeding toward the top of the rectification section. 3. : Entrainer feeding toward the top of the rectification section with distillate removal of the lightest component but without bottom removal, that is, total reboil operation. 4. : Entrainer feeding toward the top of the rectification section with both bottom and distillate removal (the bottom stream contains mainly the entrainer, which is therefore recycled back to the top of the column section). View chapterExplore book Read full chapter URL: Book2014, DistillationEva Sorensen Related terms: Distillation Columns Dimethyl Carbonate Pervaporation Ionic Liquid Methanol Purity Phase Composition Boiling Acetic Acid View all Topics
6461	https://www.princetonreview.com/grad-school-advice/gre-practice-questions	This site uses various technologies, as described in our Privacy Policy, for personalization, measuring website use/performance, and targeted advertising, which may include storing and sharing information about your site visit with third parties. By continuing to use this website you consent to our Privacy Policy and Terms of Use. I understand Menu Account Account Call Cart < Back to All Articles GRE Practice Questions Are you ready for the GRE? Try these GRE sample questions and get a quick preview of the question types you'll face on test day. Check your answers against our in-depth explanations to see how you did. We pulled these GRE practice questions from our book Cracking the GRE and from our GRE prep course materials. For even more practice, take a free GRE practice test with us held under realistic testing conditions. GRE Math Practice GRE Verbal Practice GRE Analytical Writing Practice Note: On the GRE you'll be given an on-screen calculator with the five basic operations: addition, subtraction, multiplication, division and square root, plus a decimal function and positive/negative feature. Don't use anything fancier when you tackle this GRE math practice! GRE Quantitative Comparison Practice Quantitative comparison questions ask you to compare Quantity A to Quantity B. Your job is to compare the two quantities and decide which of the following describes the relationship: Quantity A is greater. Quantity B is greater. The two quantities are equal. The relationship cannot be determined from the information given. 1. The average (arithmetic mean) high temperature for x days is 70 degrees. The addition of one day with a high temperature of 75 degrees increases the average to 71 degrees. \| Quantity A \| Quantity B \| \| x \| 5 \| (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. Answer: (B) Quantity B is greater. If the average high temperature for x days is 70 degrees, then the sum of those x high temperatures is 70x. The sum of the high temperatures, including the additional day that has a temperature of 75 degrees is, therefore, 70x + 75. Next, use the average formula to find the value of x: In this formula, 71 is the average, 70x + 75 is the total, and there are x + 1 days. Substituting this information into the formula gives: To solve, cross-multiply to get 71x + 71 = 70x + 75. Next, simplify to find that x = 4. Therefore, Quantity B is greater. The correct answer is (B). [+] See the Answer 2. a and b are integers. a 2 = b 3 \| Quantity A \| Quantity B \| \| a \| b \| (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. Answer: (D) The relationship cannot be determined from the information given. Try different integers for a and b that satisfy the equation a 2 = b 3 such as a = 8 and b = 4. These numbers satisfy the equation as 8 2 = 4 3 = 64. In this case, Quantity A is greater. Because Quantity B is not always greater nor are the two quantities always equal, choices (B) and (C) can be eliminated. Next, try some different numbers. When choosing a second set of numbers, try something less common such as making a = b = 1. Again, these numbers satisfy the equation provided in the problem. In this case, however, the quantities are equal. Because Quantity A is not always greater, choice (A) can now be eliminated. The correct answer is (D). [+] See the Answer GRE Multiple-Choice Practice For question 3, select one answer from the list of five answer choices. 3. A certain pet store sells only dogs and cats. In March, the store sold twice as many dogs as cats. In April, the store sold twice the number of dogs that it sold in March, and three times the number of cats that it sold in March. If the total number of pets the store sold in March and April combined was 500, how many dogs did the store sell in March? (A) 80 (B) 100 (C) 120 (D) 160 (E) 180 Answer: (B) 100 Plug In the Answers, starting with the middle choice. If 120 dogs were sold in March, then 60 cats were sold that month. In April, 240 dogs were sold, along with 180 cats. The total number of dogs and cats sold during those two months is 600, which is too large, so eliminate (C), (D), and (E). Try (B). If there were 100 dogs sold in March, then 50 cats were sold; in April, 200 dogs were sold along with 150 cats. The correct answer is (B) because 100 + 50 + 200 + 150 = 500. [+] See the Answer 4. µ ABC has an area of 108 cm 2 . If both x and y are integers, which of the following could be the value of x ? Indicate all such values. (A) 4 (B) 5 (C) 6 (D) 8 (E) 9 Answer: (A), (C), (D) and (E) Plug the information given into the formula for the area of a triangle to learn more about the relationship between x and y: The product of x and y is 216, so x needs to be a factor of 216. The only number in the answer choices that is not a factor of 216 is 5. The remaining choices are possible values of x. [+] See the Answer GRE Numeric Entry Practice Some questions on the GRE won't have answer choices, and you'll have to generate your own answer. 5. Each month, Renaldo earns a commission of 10.5% of his total sales for the month, plus a salary of $2,500. If Renaldo earns $3,025 in a certain month, what were his total sales? Answer: 5,000 If Renaldo earned $3,025, then his earnings from the commission on his sales are $3,025 - $2,500 = $525. So, $525 is 10.5% of his sales. Set up an equation to find the total sales: Solving this equation, x = 5,000. [+] See the Answer 6. At a recent dog show, there were 5 finalists. One of the finalists was awarded "Best in Show" and another finalist was awarded "Honorable Mention." In how many different ways could the two awards be given out? Answer: 20 In this problem order matters. Any of the 5 finalists could be awarded "Best in Show." There are 4 choices left for "Honorable Mention," because a different dog must be chosen. Therefore, the total number of possibilities is 5 x 4, or 20. [+] See the Answer Text Completion Practice Text Completion questions include a passage composed of one to five sentences with one to three blanks. There are three answer choices per blank, or five answer choices if there is a single blank. There is a single correct answer, consisting of one choice for each blank. You receive no credit for partially correct answers. 1. Upon visiting the Middle East in 1850, Gustave Flaubert was so belly dancing that he wrote, in a letter to his mother, that the dancers alone made his trip worthwhile. \| \| \| (A) overwhelmed by \| \| (B) enamored by \| \| (C) taken aback by \| \| (D) beseeched by \| \| (E) flustered by \| Answer: (B) enamored by Choose carefully here. The clue is "the dancers alone made his trip worthwhile." Thus, Flaubert was impressed by them. Enamored by is the only choice that captures such a feeling. Overwhelmed by is extreme, and implies that Flaubert got more than he could handle. Taken aback by, in contrast, merely suggests that our traveler was surprised by the dancers; we cannot be sure that his surprise was pleasant. Meanwhile, beseeched by does not indicate how Flaubert felt, whereas if he were flustered by the performers, he would not likely have found his encounter with them worthwhile. [+] See the Answer 2. Increasingly, the boundaries of congressional seats are drawn in order to protect incumbents, as legislators engineer the demographics of each district such that those already in office can coast to (i) victory. Of course, there is always the possibility that the incumbent will face a challenge from within his or her own party. Nevertheless, once the primary is over, the general election is (ii) . \| Blank (i) \| Blank (ii) \| --- \| \| (A) an ineluctable \| (B) seldom nugatory \| \| (C) an invidious \| (D) remarkably contentious \| \| (E) a plangent \| (F) merely denouement \| Answer: (A) an ineluctable and (F) merely denouement If district boundaries are designed to protect incumbents —that is, those already in office—then victory for those incumbents should be close to assured or inevitable. Ineluctable is synonymous with these words. Invidious means "causing envy" and plangent means "full of lamentation," neither of which is as well supported as the credited response. The second blank comes after a couple of transition words. The first is Of course, which might sound like the passage is continuing in the same direction, but here indicates a change of direction: The author is conceding that sometimes incumbents face challenges. The second, Nevertheless, also changes direction, meaning that the passage has returned to where it started, arguing that elections are essentially decided before they begin. That is what merely denouement means. Seldom nugatory means rarely inconsequential, which is the opposite of what the passage calls for; remarkably contentious is wrong for the same reason, as that phrase would indicate that the general election is fiercely contested. [+] See the Answer Sentence Equivalence Practice Sentence Equivalence questions consist of one sentence with six answer choices. Your job is to choose the two answer choices that logically complete the sentence. 3. Possessed of an insatiable sweet tooth, Jim enjoyed all kinds of candy, but he had a special for gumdrops, his absolute favorite. (A) container (B) affinity (C) odium (D) nature (E) disregard (F) predilection Answer: (B) affinity and (F) predilection The word in the blank is used to describe Jim's feelings for gumdrops. The clues "enjoyed all kinds of candy" and "his absolute favorite" dictates that the blank means liking. Both affinity and predilection mean liking. Odium and disregard go in the wrong direction. Container might sound right, but it is not related to the clue. Nature does not mean liking. [+] See the Answer 4. The twins' heredity and upbringing were identical in nearly every respect, yet one child remained unfailingly sanguine even in times of stress while her sister was prone to angry outbursts that indicated an exceptionally choleric . (A) genotype (B) environment (C) physiognomy (D) incarnation (E) incarnation (E) temperament (F) humor Answer: (E) temperament and (F) humor The main clues are that one twin is described as sanguine, the other choleric; even if you don't know these words, the phrases "even in times of stress" and "angry outbursts" suggest that words are used to describe personality. Temperament is a good synonym for personality. While it is frequently used to mean comedy, humor can also mean personality, especially in conjunction with the words such as sanguine and choleric, which derive from the ancient belief that temperament was shaped by the levels of different fluids or humors, in a person's body. The remaining choices don't fit. Environment means one's surroundings while the other three words are concerned with the physical rather than the mental. [+] See the Answer Reading Comprehension Practice Questions 5-6 are based on the following reading passage. Called by some the “island that time forgot,” Madagascar is home to a vast array of unique, exotic creatures. One such animal is the aye-aye. First described by western science in 1782, it was initially categorized as a member of the order Rodentia. Further research then revealed that it was more closely related to the lemur, a member of the primate order. Since the aye-aye is so different from its fellow primates, however, it was given its own family: Daubentoniidae. The aye-aye has been listed as an endangered species and, as a result, the government of Madagascar has designated an island off the northeastern coast of Madagascar as a protected reserve for aye-ayes and other wildlife. Long before Western science became enthralled with this nocturnal denizen of Madagascar’s jungles, the aye-aye had its own reputation with the local people. The aye-aye is perhaps best known for its large, round eyes and long, extremely thin middle finger. These adaptations are quite sensible, allowing the aye-aye to see well at night and retrieve grubs, which are one of its primary food sources, from deep within hollow branches. However, the aye-aye’s striking appearance may end up causing its extinction. The people of Madagascar believe that the aye-aye is a type of spirit animal, and that its appearance is an omen of death. Whenever one is sighted, it is immediately killed. When combined with the loss of large swaths of jungle habitat, this practice may result in the loss of a superb example of life’s variety. 5. Based on the information given in the passage, the intended audience would most likely be (A) visitors to a natural science museum (B) professors of evolutionary science (C) a third-grade science class (D) students of comparative religions (E) attendees at a world cultural symposium Answer: (A) The passage contains a mixture of information about the aye-aye, both from a scientific and cultural background. it gives an overview of the animal without giving a lot of detail in any one area. Choice (B) is incorrect because the passage mentions evolution only briefly, at the end. This choice is too narrow. Choice (C) is incorrect because the style of the passage is too advanced for young students. Choice (D) is incorrect because the passage mentions religion only as it relates to the fate of the aye-aye. Choice (E) is incorrect because the information given is focuses more on the aye-aye itself than on the culture of Madagascar. [+] See the Answer 6. Consider each of the choices separately and select all that apply. Which of the following statements can be logically inferred from the passage? (A) Taxonomic classifications are not always absolute. (B) The traditional religion of Madagascar involves augury. (C) There are no longer enough resources on the main island to support the aye-aye population. Answer: (A) and (B) Choices (A) and (B) can both be inferred from the passage. Choice (A) is supported by the first paragraph. The classification of the aye-aye changes, which demonstrates that such classifications are not absolute. Choice (B) is supported by the part of the passage dealing with the future of the aye-aye. It states that aye-aye is seen as an omen of death in the traditional religion of the Madagascar. Augury refers to the use of omens, so this statement must be true. Choice (C), however, is not supported. Although the passage states that the aye-aye is in danger, it does not directly discuss whether this is due to limited resources on the main island. [+] See the Answer The Issue Essay The Issue Essay of the GRE requires you to present your opinion on the provided topic. Issue Topic You will be given a brief quotation that states or implies an issue of general interest and specific instructions on how to respond to that issue. You will have 30 minutes to plan and compose a response in which you develop a position on the issue according the specific instructions. A response to any other issue will receive a score of zero. \| \| \| "True beauty is found not in the exceptional but in the commonplace." Write an essay in which you take a position on the statement above. In developing and supporting your essay, consider instances in which the statement does and does not hold true. \| A high-scoring Issue essay accomplishes four key tasks: (1) considers the complexities of the issue; (2) supports the position with relevant examples; (3) is clear and well organized; (4) demonstrates superior facility with the conventions of standard written English. Make sure that you respond to the specific instructions and support your position on the issue with reasons and examples drawn from such areas as your reading, experience, observations, and/or academic studies. [+] See the Answer The Argument Essay The Argument Essay of the GRE asks you to examine and critique the logic of an argument. Argument Topic You will be given a short passage that presents an argument, or an argument to be completed, and specific instructions on how to respond to that passage. You will have 30 minutes to plan and compose a response in which you analyze the passage according to specific instructions. A response to any other issue will receive a score of zero. \| \| \| The director of the International Health Foundation recently released this announcement: "A new medical test that allows the early detection of a particular disease will prevent the deaths of people all over the world who would otherwise die from the disease. The test has been extremely effective in allowing doctors to diagnose the disease six months to a year before it would have been spotted by conventional means. As soon as we can institute this test as a routine procedure in hospitals around the world, the death rate from this disease will plummet." Write a response in which you examine the stated and/or unstated assumptions of the argument. Be sure to explain how the argument depends on the assumptions and what the implications are if the assumptions prove unwarranted. \| A high-scoring Argument essay accomplishes these tasks: (1) clearly identifies and insightfully analyzes important features of the argument; (2) develops ideas clearly and logically with smooth transitions; (3) effectively supports the main points of the critique; (4) demonstrates superior facility with the conventions of standard written English. Note that you are NOT being asked to present your own views on the subject. Make sure you that you respond to the specific instructions and support your analysis with relevant reasons and/or examples. [+] See the Answer How will you score on the GRE? Take a GRE exam sample test with us under the same conditions as the real thing. You'll get a personalized score report highlighting your strengths and areas of improvement. 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6462	https://handwiki.org/wiki/Cauchy%E2%80%93Binet_formula	Cauchy–Binet formula - HandWiki Anonymous Not logged in Create account Log in Hand W iki Search Cauchy–Binet formula From HandWiki bookmark [x] Namespaces Page Discussion More More Page actions Read View source History ZWI Export In mathematics, specifically linear algebra, the Cauchy–Binet formula, named after Augustin-Louis Cauchy and Jacques Philippe Marie Binet, is an identity for the determinant of the product of two rectangular matrices of transpose shapes (so that the product is well-defined and square). It generalizes the statement that the determinant of a product of square matrices is equal to the product of their determinants. The formula is valid for matrices with the entries from any commutative ring. [x] Contents 1 Statement 2 Special cases 2.1 In the case n=3 3 A simple proof 4 Proof 5 Relation to the generalized Kronecker delta 6 Geometric interpretations 7 Generalization 8 Continuous version 9 References 10 External links Statement Let A be an m×n matrix and B an n×m matrix. Write [n] for the set {1,...,n}, and ([n]m) for the set of m-combinations of [n] (i.e., subsets of [n] of size m; there are (n m) of them). For S∈([n]m), write A[m],S for the m×m matrix whose columns are the columns of A at indices from S, and B S,[m] for the m×m matrix whose rows are the rows of B at indices from S. The Cauchy–Binet formula then states det(A B)=∑S∈([n]m)det(A[m],S)det(B S,[m]). Example: Taking m=2 and n=3, and matrices A=(1 1 2 3 1−1) and B=(1 1 3 1 0 2), the Cauchy–Binet formula gives the determinant det(A B)=\|1 1 3 1\|⋅\|1 1 3 1\|+\|1 2 1−1\|⋅\|3 1 0 2\|+\|1 2 3−1\|⋅\|1 1 0 2\|. Indeed A B=(4 6 6 2), and its determinant is −28 which equals −2×−2+−3×6+−7×2 from the right hand side of the formula. Special cases If n<m then ([n]m) is the empty set, and the formula says that det(AB)=0 (its right hand side is an empty sum); indeed in this case the rank of the m×m matrix AB is at most n, which implies that its determinant is zero. If n = m, the case where A and B are square matrices, ([n]m)={[n]} (a singleton set), so the sum only involves S=[n], and the formula states that det(AB)=det(A)det(B). For m=0, A and B are empty matrices (but of different shapes if n>0), as is their product AB; the summation involves a single term S=Ø, and the formula states 1=1, with both sides given by the determinant of the 0×0 matrix. For m=1, the summation ranges over the collection ([n]1) of the n different singletons taken from [n], and both sides of the formula give ∑j=1 n A 1,j B j,1, the dot product of the pair of vectors represented by the matrices. The smallest value of m for which the formula states a non-trivial equality is m=2; it is discussed in the article on the Binet–Cauchy identity. In the case n=3 Let a,b,c,d,x,y,z,w be three-dimensional vectors. 1=1(m=0)a⋅x=a 1 x 1+a 2 x 2+a 3 x 3(m=1)\|a⋅x a⋅y b⋅x b⋅y\|=\|a 2 a 3 b 2 b 3\|\|x 2 y 2 x 3 y 3\|+\|a 3 a 1 b 3 b 1\|\|x 3 y 3 x 1 y 1\|+\|a 1 a 2 b 1 b 2\|\|x 1 y 1 x 2 y 2\|=(a×b)⋅(x×y)(m=2)\|a⋅x a⋅y a⋅z b⋅x b⋅y b⋅z c⋅x c⋅y c⋅z\|=\|a 1 a 2 a 3 b 1 b 2 b 3 c 1 c 2 c 3\|\|x 1 y 1 z 1 x 2 y 2 z 2 x 3 y 3 z 3\|=[a⋅(b×c)]x⋅(y×z)\|a⋅x a⋅y a⋅z a⋅w b⋅x b⋅y b⋅z b⋅w c⋅x c⋅y c⋅z c⋅w d⋅x d⋅y d⋅z d⋅w\|=0(m=4) In the case m>3, the right-hand side always equals 0. A simple proof The following simple proof relies on two facts that can be proven in several different ways: For any 1≤k≤n the coefficient of z n−k in the polynomial det(z I n+X) is the sum of the k×k principal minors of X. If m≤n and A is an m×n matrix and B an n×m matrix, then det(z I n+B A)=z n−m det(z I m+A B). Now, if we compare the coefficient of z n−m in the equation det(z I n+B A)=z n−m det(z I m+A B), the left hand side will give the sum of the principal minors of B A while the right hand side will give the constant term of det(z I m+A B), which is simply det(A B), which is what the Cauchy–Binet formula states, i.e. det(A B)=∑S∈([n]m)det((B A)S,S)=∑S∈([n]m)det(B S,[m])det(A[m],S)=∑S∈([n]m)det(A[m],S)det(B S,[m]). Proof There are various kinds of proofs that can be given for the Cauchy−Binet formula. The proof below is based on formal manipulations only, and avoids using any particular interpretation of determinants, which may be taken to be defined by the Leibniz formula. Only their multilinearity with respect to rows and columns, and their alternating property (vanishing in the presence of equal rows or columns) are used; in particular the multiplicative property of determinants for square matrices is not used, but is rather established (the case n=m). The proof is valid for arbitrary commutative coefficient rings. The formula can be proved in two steps: use the fact that both sides are multilinear (more precisely 2 m-linear) in the rows of A and the columns of B, to reduce to the case that each row of A and each column of B has only one non-zero entry, which is 1. handle that case using the functions [m]→[n] that map respectively the row numbers of A to the column number of their nonzero entry, and the column numbers of B to the row number of their nonzero entry. For step 1, observe that for each row of A or column of B, and for each m-combination S, the values of det(AB) and det(A[m],S)det(B S,[m]) indeed depend linearly on the row or column. For the latter this is immediate from the multilinear property of the determinant; for the former one must in addition check that taking a linear combination for the row of A or column of B while leaving the rest unchanged only affects the corresponding row or column of the product AB, and by the same linear combination. Thus one can work out both sides of the Cauchy−Binet formula by linearity for every row of A and then also every column of B, writing each of the rows and columns as a linear combination of standard basis vectors. The resulting multiple summations are huge, but they have the same form for both sides: corresponding terms involve the same scalar factor (each is a product of entries of A and of B), and these terms only differ by involving two different expressions in terms of constant matrices of the kind described above, which expressions should be equal according to the Cauchy−Binet formula. This achieves the reduction of the first step. Concretely, the multiple summations can be grouped into two summations, one over all functions f:[m]→[n] that for each row index of A gives a corresponding column index, and one over all functions g:[m]→[n] that for each column index of B gives a corresponding row index. The matrices associated to f and g are L f=((δ f(i),j)i∈[m],j∈[n])and R g=((δ j,g(k))j∈[n],k∈[m]) where "δ" is the Kronecker delta, and the Cauchy−Binet formula to prove has been rewritten as ∑f:[m]→[n]∑g:[m]→[n]p(f,g)det(L f R g)=∑f:[m]→[n]∑g:[m]→[n]p(f,g)∑S∈([n]m)det((L f)[m],S)det((R g)S,[m]), where p(f,g) denotes the scalar factor (∏i=1 m A i,f(i))(∏k=1 m B g(k),k). It remains to prove the Cauchy−Binet formula for A=L f and B=R g, for all f,g:[m]→[n]. For this step 2, if f fails to be injective then L f and L f R g both have two identical rows, and if g fails to be injective then R g and L f R g both have two identical columns; in either case both sides of the identity are zero. Supposing now that both f and g are injective maps [m]→[n], the factor det((L f)[m],S) on the right is zero unless S=f([m]), while the factor det((R g)S,[m]) is zero unless S=g([m]). So if the images of f and g are different, the right hand side has only null terms, and the left hand side is zero as well since L f R g has a null row (for i with f(i)∉g([m])). In the remaining case where the images of f and g are the same, say f([m])=S=g([m]), we need to prove that det(L f R g)=det((L f)[m],S)det((R g)S,[m]). Let h be the unique increasing bijection [m]→S, and π,σ the permutations of [m] such that f=h∘π−1 and g=h∘σ; then (L f)[m],S is the permutation matrix for π, (R g)S,[m] is the permutation matrix for σ, and L f R g is the permutation matrix for π∘σ, and since the determinant of a permutation matrix equals the signature of the permutation, the identity follows from the fact that signatures are multiplicative. Using multi-linearity with respect to both the rows of A and the columns of B in the proof is not necessary; one could use just one of them, say the former, and use that a matrix product L f B either consists of a permutation of the rows of B f([m]),[m] (if f is injective), or has at least two equal rows. Relation to the generalized Kronecker delta As we have seen, the Cauchy–Binet formula is equivalent to the following: det(L f R g)=∑S∈([n]m)det((L f)[m],S)det((R g)S,[m]), where L f=((δ f(i),j)i∈[m],j∈[n])and R g=((δ j,g(k))j∈[n],k∈[m]). In terms of generalized Kronecker delta, we can derive the formula equivalent to the Cauchy–Binet formula: δ g(1)…g(m)f(1)…f(m)=∑k:[m]→[n]k(1)<⋯<k(m)δ k(1)…k(m)f(1)…f(m)δ g(1)…g(m)k(1)…k(m). Geometric interpretations If A is a real m×n matrix, then det(A A T) is equal to the square of the m-dimensional volume of the parallelotope spanned in Rn by the m rows of A. Binet's formula states that this is equal to the sum of the squares of the volumes that arise if the parallelepiped is orthogonally projected onto the m-dimensional coordinate planes (of which there are (n m)). In the case m=1 the parallelotope is reduced to a single vector and its volume is its length. The above statement then states that the square of the length of a vector is the sum of the squares of its coordinates; this is indeed the case by the definition of that length, which is based on the Pythagorean theorem. In tensor algebra, given an inner product spaceV of dimension n, the Cauchy–Binet formula defines an induced inner product on the exterior algebra∧m V, namely: ⟨v 1∧⋯∧v m,w 1∧⋯∧w m⟩=det(⟨v i,w j⟩)i,j=1 m. Generalization The Cauchy–Binet formula can be extended in a straightforward way to a general formula for the minors of the product of two matrices. Context for the formula is given in the article on minors, but the idea is that both the formula for ordinary matrix multiplication and the Cauchy–Binet formula for the determinant of the product of two matrices are special cases of the following general statement about the minors of a product of two matrices. Suppose that A is an m × n matrix, B is an n × p matrix, I is a subset of {1,...,m} with k elements and J is a subset of {1,...,p} with k elements. Then [AB]I,J=∑K[A]I,K[B]K,J where the sum extends over all subsets K of {1,...,n} with k elements. Continuous version A continuous version of the Cauchy–Binet formula, known as the Andréief-Heine identity or Andréief identity appears commonly in random matrix theory. It is stated as follows: let {f j(x)}j=1 N and {g j(x)}j=1 N be two sequences of integrable functions, supported on I. Then ∫I⋯∫I det[f j(x k)]j,k=1 N det[g j(x k)]j,k=1 N d x 1⋯d x N=N!det[∫I f j(x)g k(x)d x]j,k=1 N. Proof Let S N be the permutation group of order N, \|s\| be the sign of a permutation, ⟨f,g⟩=∫I f(x)g(x)d x be the "inner product".left side=∑s,s′∈S N(−1)\|s\|+\|s′\|∫I N∏j f s(j)(x j)∏k g s′(k)(x k)=∑s,s′∈S N(−1)\|s\|+\|s′\|∫I N∏j f s(j)(x j)g s′(j)(x j)=∑s,s′∈S N(−1)\|s\|+\|s′\|∏j∫I f s(j)(x j)g s′(j)(x j)d x j=∑s,s′∈S N(−1)\|s\|+\|s′\|∏j⟨f s(j),g s′(j)⟩=∑s′∈S N(−1)\|s′\|+\|s′\|∑s∈S N(−1)\|s\|+\|s′−1\|∏j⟨f(s∘s′−1)(j),g j⟩=∑s′∈S N∑s∈S N(−1)\|s∘s′−1\|∏j⟨f(s∘s′−1)(j),g j⟩=right side Forrester describes how to recover the usual Cauchy–Binet formula as a discretisation of the above identity. Proof Pick t 1<⋯<t m in [0,1], pick f 1,…,g n, such that f j(t k)=A j,k and the same holds for g and B. Now plugging in f j(x k)∑l δ(x k−t l) and g j(x k) into the Andreev identity, and simplifying both sides, we get: ∑l 1,…,l n∈[1:m]det[f j(t l k)]det[g j(t l k)]=n!det[∑l f j(t l)g k(t l)] The right side is n!det(A B), and the left side is n!∑S⊂[1:m],\|S\|=n det(A[1:m],S)det(B S,[1:m]). References ↑Tao, Terence (2012). Topics in random matrix theory. Graduate Studies in Mathematics. 132. Providence, RI: American Mathematical Society. p.253. doi:10.1090/gsm/132. ISBN978-0-8218-7430-1. ↑C. Andréief, Mem. de la Soc. Sci. de Bordeaux 2, 1 (1883) ↑Mehta, M.L. (2004). Random Matrices (3rd ed.). Amsterdam: Elsevier/Academic Press. ISBN0-12-088409-7. ↑Forrester, Peter J. (2018). "Meet Andréief, Bordeaux 1886, and Andreev, Kharkov 1882–83". arXiv:1806.10411 [math-ph]. Joel G. Broida & S. Gill Williamson (1989) A Comprehensive Introduction to Linear Algebra, §4.6 Cauchy-Binet theorem, pp 208–14, Addison-Wesley ISBN:0-201-50065-5. Jin Ho Kwak & Sungpyo Hong (2004) Linear Algebra 2nd edition, Example 2.15 Binet-Cauchy formula, pp 66,7, Birkhäuser ISBN:0-8176-4294-3. I. R. Shafarevich& A. O. Remizov (2012) Linear Algebra and Geometry, §2.9 (p.68) & §10.5 (p.377), SpringerISBN:978-3-642-30993-9. M.L. Mehta (2004) Random matrices, 3rd ed., ElsevierISBN:9780120884094. External links Aaron Lauve (2004) A short combinatoric proof of Cauchy–Binet formula[\|permanent dead link\|dead link}}] from Université du Québec à Montréal. 0.00 (0 votes) Original source: formula. 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6463	https://math.stackexchange.com/questions/90782/the-minimum-of-two-independent-geometric-random-variables	probability - The minimum of two independent geometric random variables - Mathematics Stack Exchange Join Mathematics By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy. Sign up with Google OR Email Password Sign up Already have an account? Log in Skip to main content Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. 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Upvoting indicates when questions and answers are useful. What's reputation and how do I get it? Instead, you can save this post to reference later. Save this post for later Not now Thanks for your vote! You now have 5 free votes weekly. Free votes count toward the total vote score does not give reputation to the author Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation. Got it!Go to help center to learn more The minimum of two independent geometric random variables Ask Question Asked 13 years, 9 months ago Modified3 years, 10 months ago Viewed 31k times This question shows research effort; it is useful and clear 12 Save this question. Show activity on this post. here's a question I got for homework (sorry if my translation is a bit unclear): Let X∼‬G(p 1)X∼‬G(p 1), Y∼‬G(p 2)Y∼‬G(p 2), X X and Y Y are independent. Prove that the minimum is also geometric, meaning: min(X,Y)∼G(1−(1−p 1)(1−p 2))min(X,Y)∼G(1−(1−p 1)(1−p 2)). Instructions: first calculate the probability P(min(X,Y)>k)P(min(X,Y)>k) and compare it to the parallel probability in (of?) a geometric random variable. I have no idea where to start, even with the great clue that they've supplied. Any hints? probability Share Share a link to this question Copy linkCC BY-SA 4.0 Cite Follow Follow this question to receive notifications edited Jun 12, 2020 at 10:38 CommunityBot 1 asked Dec 12, 2011 at 12:48 yotamooyotamoo 2,871 9 9 gold badges 36 36 silver badges 48 48 bronze badges 5 Well, just follow the clue. Calculate P(min(X,Y) > k), for a general k. Did you try that? what did you get?Prateek –Prateek 2011-12-12 13:05:07 +00:00 Commented Dec 12, 2011 at 13:05 2 note that P(min(X,Y)>k)P(min(X,Y)>k) is the same as P(X>k,Y>k)P(X>k,Y>k).deinst –deinst 2011-12-12 13:17:53 +00:00 Commented Dec 12, 2011 at 13:17 Ok I did that and it was quite easy. P[min(X,Y)>k] = ((1-p1)^K)((1-p2)^K) - is that enough to prove what I need to prove?yotamoo –yotamoo 2011-12-12 15:29:58 +00:00 Commented Dec 12, 2011 at 15:29 @yotamoo Is this the same as the probability that a G(1−(1−p 1)(1−p 2))G(1−(1−p 1)(1−p 2)) geometric variable is greater than k k?deinst –deinst 2011-12-12 15:42:48 +00:00 Commented Dec 12, 2011 at 15:42 1 I found several later posts of exactly the same question. Judging by the content, it's hard to decide which one should be deemed the original and the others duplicate. In chronological order: 845706, 1040620, 1056296, 1169142, and 1207241.Lee David Chung Lin –Lee David Chung Lin 2018-03-21 01:10:05 +00:00 Commented Mar 21, 2018 at 1:10 Add a comment\| 1 Answer 1 Sorted by: Reset to default This answer is useful 15 Save this answer. Show activity on this post. Let X X and Y Y be independent random variables having geometric distributions with probability parameters p 1 p 1 and p 2 p 2 respectively. Then if Z Z is the random variable min(X,Y)min(X,Y) then Z Z has a geometric distribution with probability parameter 1−(1−p 1)(1−p 2)1−(1−p 1)(1−p 2). There are essentially two ways to see this: First, the method outlined by the hint in your homework - Note that the cdf of X X is 1−(1−p 1)k 1−(1−p 1)k and the cdf of Y Y is 1−(1−p 2)k 1−(1−p 2)k, so the probability that X>k X>k is (1−p 1)k(1−p 1)k and the probability that Y>k Y>k is (1−p 2)k(1−p 2)k and so the probability that both are greater than k k is [(1−p 1)(1−p 2)]k[(1−p 1)(1−p 2)]k. But the probability that both are greater than k k is the same as the probability that the minimum of the two is greater than k k. From this we can get the cdf of Z Z as 1−[(1−p 1)(1−p 2)]k 1−[(1−p 1)(1−p 2)]k, and we can note that this is the cdf of a geometric random variable with probability parameter 1−(1−p 1)(1−p 2)1−(1−p 1)(1−p 2). Second, and more intuitively to me, we can go back to the definition of a geometric random variable with probability parameter p p : the number of Bernoulli trials with probability p p needed to get one success. So min(X,Y)min(X,Y) is the number of trials of simultaneously running a Bernoulli experiment with probability p 1 p 1 and one with probability p 2 p 2 before one or the other experiments succeeds. The probability of one of the two experiments succeeding at any step is just 1−(1−p 1)(1−p 2)1−(1−p 1)(1−p 2), so Z Z is a geometric random variable with probability parameter 1−(1−p 1)(1−p 2)1−(1−p 1)(1−p 2). Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Follow Follow this answer to receive notifications edited Nov 16, 2021 at 12:10 Falrach 4,272 16 16 silver badges 40 40 bronze badges answered Dec 13, 2011 at 16:10 deinstdeinst 5,716 29 29 silver badges 33 33 bronze badges 2 As an addendum and a slightly different way of calculating the results, the probability of a "success" on a compound trial is that either of the two separate sub-trials results in a success. This has probability p 1+p 2−p 1 p 2 p 1+p 2−p 1 p 2 (using P(A∪B)=P(A)+P(B)−P(A∩B)=P(A)+P(B)−P(A)P(B)P(A∪B)=P(A)+P(B)−P(A∩B)=P(A)+P(B)−P(A)P(B) by independence of individual sub-trials) and so we have a geometric random variable with parameter p 1+p 2−p 1 p 2=1−(1−p 1)(1−p 2)p 1+p 2−p 1 p 2=1−(1−p 1)(1−p 2) just as you found.Dilip Sarwate –Dilip Sarwate 2011-12-13 16:46:38 +00:00 Commented Dec 13, 2011 at 16:46 A minor correction to my comment: in the first line, for "the probability of", please read "the event"Dilip Sarwate –Dilip Sarwate 2011-12-13 18:56:07 +00:00 Commented Dec 13, 2011 at 18:56 Add a comment\| You must log in to answer this question. Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions probability See similar questions with these tags. Featured on Meta Introducing a new proactive anti-spam measure Spevacus has joined us as a Community Manager stackoverflow.ai - rebuilt for attribution Community Asks Sprint Announcement - September 2025 Report this ad Linked 2Minimum of two geometric random variables is geometric 2The distribution of the minimum of two independent geometric random variables 2Show that min(X,Y)min(X,Y) has geometric distribution 1Variance of alternate flipping rounds 1Expected hitting time of 2 independent Markov chains 1Minimum of n n geometric random variables 1If X,Y X,Y are independent and geometric, then Z=min(X,Y)Z=min(X,Y) is also geometric 0Minimum of two geometric distributions Related 2What is this idea of "Minimum Correlation"? 4Two series of independent Bernoulli trials. 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6464	http://faculty.etsu.edu/gardnerr/1910/notes-11E/c3s5.pdf	3.5 The Chain Rule and Parametric Equations 1 Chapter 3. Diﬀerentiation 3.5 The Chain Rule and Parametric Equations Theorem 3. The Chain Rule. If f(u) is diﬀerentiable at the point u = g(x) and g(x) is diﬀerentiable at x, then the composite function (f ◦g)(x) = f(g(x)) is diﬀerentiable at x, and (f ◦g)′(x) = f ′(g(x))[g′(x)]. In Leibniz’s notation, if y = f(u) and u = g(x), then dy dx = dy du · du dx, where dy/du is evaluated at u = g(x). Note. The proof of the Chain Rule is rather complicated — see Section 3.8. Note. If f(u) = un where n is an integer, then d dx[f(g(x))] = d dx[(g(x))n] = ng(x)n−1[g′(x)]. 3.5 The Chain Rule and Parametric Equations 2 Examples. Page 199 numbers 8, 43, and 54, page 200 number 64. Deﬁnition. A curve is given parametrically if its graph is determined by graphing (x(t), y(t)) for t (the parameter) ranging over some interval. Example. x(t) = cos t, y(t) = sin t for t ∈[0, 2π) parametrically determine the unit circle x2 + y2 = 1. Note. Parametric Formula for dy/dx and d2y/dx2. If x = x(t) and y = y(t) where x(t) and y(t) are diﬀerentiable with x′(t) ̸= 0 then dy dx = dy/dt dx/dt and d2y dx2 = dy′/dt dx/dt . Examples. Pages 201 number 108, page 202 number 114.
6465	https://www.iue.tuwien.ac.at/phd/schwaha/wholese9.html	\| \| \| \| --- \| \| \| \| 4.2 Algebraic Structures Beside the bare structures described so far, the structures resulting from operations deﬁned on top of the sets as a combined union gives rise to algebraic structures. They are of importance as they provide rules, how elements interacts and create new elements. In the setting of digital computers types are models or extensions of algebraic structures. Therefore, several algebraic structures are introduced in the following with an increasing number of requirements imposed on elements and/or operations. Deﬁnition 10 (Monoid) A set of elements on which a binary operation is deﬁned is called a monoid , if the binary operation satisﬁes the following conditions: Closed: The result of the binary operation is again an element of the initial set : Associativity: The order in which consecutive operations are applied does not change the result: An identity (neutral) element exists. such that: Further requirements on Deﬁnition 10 result in Deﬁnition 11 (Group) A monoid , where the binary operation additionally fulﬁls the condition that for every element in an inverse element exists, which produces the identity element under the binary operation is called a group and as a further qualiﬁcation Deﬁnition 12 (Abelian group) When the order of the operands of the binary operation of a group does not change its result, the group is called Abelian or commutative. Groups are basic building blocks in the exploration of further structures, which can be deﬁned by demanding additional operations on the basic sets. Deﬁnition 13 (Ring) An Abelian group (Deﬁnition 12), which is equipped with an additional binary operation under which it is a monoid (Deﬁnition 10), is called a ring , if the two binary relations are distributive: The structure of the described ring is, however, insuﬃcient to describe the basic notion of real numbers or the complex numbers . To this end the following deﬁnition is required: Deﬁnition 14 (Field) If the multiplication operation of a ring is invertible , it is called a ﬁeld. The next deﬁnition describes the algebraic structure of entities (Deﬁnition 61), which have proven to be immensely useful. Deﬁnition 15 (Module) A Module over a ring (Deﬁnition 13) is an Abelian group (Deﬁnition 12) with respect to the operation of the addition of two elements , while additionally being a ring with respect to the operation of multiplying elements by elements , which are called scalars. A related deﬁnition with somewhat stronger requirements yields a structure, which is essential for the construction of simple geometric settings. Deﬁnition 16 (Vector space) A vector space over a ﬁeld is an Abelian group with respect to the operation of the addition of two elements , while additionally being a ring with respect to the operation of multiplying elements by elements . Elements of a vector space are called vectors. While the elements of vector spaces, the vectors, constitute a powerful concept, they are insuﬃcient to describe all the entities required in modelling scientiﬁc processes. Additional entities are therefore required. It is not limited to entities introduced later and therefore provided here to clearly distinguish the algebraic structure from the elements. Deﬁnition 17 (Algebra) An algebra over a ﬁeld is a vector space equipped with an additional binary relation (Deﬁnition 4) A further qualiﬁcation of the just deﬁned structure may be possible. The availability of the following term allows a more precise classiﬁcation as found in Deﬁnition 56 and in conjunction with Deﬁnition 60. Deﬁnition 18 (Graded algebra) In case the algebra admits the decomposition into additive groups of the form where additionally the multiplication operation results in it is called a graded algebra. Among the most versatile and useful, almost ubiquitous, algebraic entities are: Deﬁnition 19 (Polynomials) A formal prescription of the form with coeﬃcients is called a polynomial over the ring (Deﬁnition 13) in the variable . The variable in the purely algebraic deﬁnition is a formal symbol and need not be an element of a ﬁeld (Deﬁnition 14), such as or , as in the case of polynomial codes. The algebraic considerations, however, assert that polynomials deﬁned in this fashion can be added (subtracted) and multiplied, thus forming a ring (Deﬁnition 13). The case that the variables are either from or is of particular usefulness in many ﬁelds of mathematics, with the ﬁeld of interpolation as well integration among them. Then the expression is simply the th power of a variable and values can be derived by simple multiplication within their respective ﬁelds. \| \| \| \| --- \| \| \| \|
6466	https://en.bab.la/sentences/english/far-from-being	far from being example sentences - Use far from being in a sentence To support our work, we invite you to accept cookies or to subscribe. You have chosen not to accept cookies when visiting our site. The content available on our site is the result of the daily efforts of our editors. They all work towards a single goal: to provide you with rich, high-quality content. All this is possible thanks to the income generated by advertising and subscriptions. By giving your consent or subscribing, you are supporting the work of our editorial team and ensuring the long-term future of our site. 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English This is farfrombeing the case.volume_upmore_vert open_in_new Link to source warning Request revision English This is farfrombeing the case.volume_upmore_vert open_in_new Link to source warning Request revision English They are farfrombeing solved.volume_upmore_vert open_in_new Link to source warning Request revision English The process is farfrombeing completed.volume_upmore_vert open_in_new Link to source warning Request revision English But that is farfrombeing a perfect solution.volume_upmore_vert open_in_new Link to source warning Request revision CULTURE & TRAVEL Video Player is loading. Play Video Play Next Unmute Current Time / Duration 0:00/2:53 Loaded: 2.30% 0:00 Stream Type LIVE Seek to live, currently behind live LIVE 1x Playback Rate Chapters Chapters Descriptions descriptions off, selected Captions captions settings, opens captions settings dialog captions off, selected Quality 240p 360p 480p 720p 1080p Auto, selected Audio Track original, selected Fullscreen This is a modal window. Beginning of dialog window. Escape will cancel and close the window. Text Color Transparency Background Color Transparency Window Color Transparency Font Size Text Edge Style Font Family Reset restore all settings to the default values Done Close Modal Dialog End of dialog window. Budget-friendly hangouts in Brussels Budget-friendly hangouts in Brussels Why not plan your next visit around the biennial flower carpet? Source: “Dining Traveler’s Best of Brussels”.Rahul Venkitunder CC BY 4.0,2016-09-14. English Their markets are farfrombeing saturated.volume_upmore_vert open_in_new Link to source warning Request revision English We are emphatically farfrombeing so.volume_upmore_vert open_in_new Link to source warning Request revision English This contribution is, therefore, farfrombeing exclusive.volume_upmore_vert open_in_new Link to source warning Request revision English The challenge of nation building is farfrombeing won.volume_upmore_vert open_in_new Link to source warning Request revision English Egypt is still farfrombeing a democracy.volume_upmore_vert open_in_new Link to source warning Request revision English Michelin is unfortunately farfrombeing the exception in this area.volume_upmore_vert open_in_new Link to source warning Request revision English The report submitted to us is farfrombeing neutral.volume_upmore_vert open_in_new Link to source warning Request revision English This is farfrombeing the case with animal products.volume_upmore_vert open_in_new Link to source warning Request revision English The political climate is certainly farfrombeing stabilised.volume_upmore_vert open_in_new Link to source warning Request revision English Mr President, ladies and gentlemen, not only is the draft Constitution farfrombeing ideal, it is also farfrombeing the compromise it could have been.volume_upmore_vert open_in_new Link to source warning Request revision English We should not bury our heads in the sand and ignore the problems that are farfrombeing completely resolved and public opinion is farfrombeing ready.volume_upmore_vert open_in_new Link to source warning Request revision English This is farfrombeing only an animal protection issue; it is about preventing diseases.volume_upmore_vert open_in_new Link to source warning Request revision English In the case of the American energy market, liberalisation was farfrombeing a success.volume_upmore_vert open_in_new Link to source warning Request revision English Peace is very farfrombeing restored in Timor and Aceh.volume_upmore_vert open_in_new Link to source warning Request revision English The consolidation process is still farfrombeing complete, however.volume_upmore_vert open_in_new Link to source warning Request revision English An overabundance of information, farfrombeing a handicap, can be very rewarding.volume_upmore_vert open_in_new Link to source warning Request revision English It is very important that we recognise that those impacts are farfrombeing insignificant.volume_upmore_vert open_in_new Link to source warning Request revision English You may be the largest group, but you are farfrombeing the majority.volume_upmore_vert open_in_new Link to source warning Request revision English It also underlines the fact that Parliament is farfrombeing a spent political force.volume_upmore_vert open_in_new Link to source warning Request revision English Current events, I have to say, show that this is farfrombeing the case.volume_upmore_vert open_in_new Link to source warning Request revision English At the time the majority of the waste was biodegradable which is farfrombeing the case today.volume_upmore_vert open_in_new Link to source warning Request revision EnglishFarfrombeing opposed, these two aims mutually strengthen each other.volume_upmore_vert open_in_new Link to source warning Request revision English Its studies show that the BSE epidemic is farfrombeing under control.volume_upmore_vert open_in_new Link to source warning Request revision English Their economies, I remind you, are still farfrombeing in line.volume_upmore_vert open_in_new Link to source warning Request revision English Parliament, farfrombeing a spending Parliament, has proved to be responsible.volume_upmore_vert open_in_new Link to source warning Request revision English I regret to say that this is farfrombeing the case at the moment, particularly in the agriculture sector.volume_upmore_vert open_in_new Link to source warning Request revision English The new text, which is still farfrombeing final, was forwarded by the Council to Parliament for information.volume_upmore_vert open_in_new Link to source warning Request revision English The discussion about how the European policy of economic and social cohesion should be shaped after 2006 is farfrombeing completed.volume_upmore_vert open_in_new Link to source warning Request revision English The EU is no longer just a form of West European cooperation, but not farfrombeing pan-European.volume_upmore_vert open_in_new Link to source warning Request revision English Despite the fall of the Taliban regime and the installation of Hamid Karzai as the country's president, the battle for a stable Afghanistan is farfrombeing won.volume_upmore_vert open_in_new Link to source warning Request revision English I wish to thank the interpreters for their excellent translation of what was farfrombeing a straightforward text.volume_upmore_vert open_in_new Link to source warning Request revision English I firmly believe that the old fifteen members, and not us, were farfrombeing prepared for enlargement.volume_upmore_vert open_in_new Link to source warning Request revision English For the sake of future generations, we should not forget that this is very farfrombeing the case.volume_upmore_vert open_in_new Link to source warning Request revision English It follows that the claim that this policy is close to the public’ s concerns is still very farfrombeing backed up.volume_upmore_vert open_in_new Link to source warning Request revision English The directive is not as forceful as we might have hoped; the compromise is farfrombeing perfect and there are no binding targets, only indicative ones.volume_upmore_vert open_in_new Link to source warning Request revision English We will need to be ready to fight with regard to the dossier on geographical origins, which is a battle farfrombeing won.volume_upmore_vert open_in_new Link to source warning Request revision EnglishFarfrombeing a key player, Mr Solana, the consequences of Europe’ s failure are plain for all to see.volume_upmore_vert open_in_new Link to source warning Request revision English This research appears harmless but may quite well be developed and used in areas which are farfrombeing in our interests.volume_upmore_vert open_in_new Link to source warning Request revision English The mad cow business has not enhanced the image of the Union, although the Union is farfrombeing the party primarily responsible for that disaster.volume_upmore_vert open_in_new Link to source warning Request revision English One thing is sure: the situation in Zaire is farfrombeing secure and therefore it will be extremely difficult to ask NGOs to go.volume_upmore_vert open_in_new Link to source warning Request revision EnglishFarfrombeing satisfactory, I see it merely as a first step, although it does, in some respects, go in the right direction.volume_upmore_vert open_in_new Link to source warning Request revision English For others, including me, this is farfrombeing quite so obvious and, above all, quite so automatic.volume_upmore_vert open_in_new Link to source warning Request revision English Real prices of calls from mobiles continue to be high and the transparency of costs and prices is farfrombeing guaranteed.volume_upmore_vert open_in_new Link to source warning Request revision English As Mr Blokland mentioned, incineration is, of course, farfrombeing the ideal way of disposing of waste.volume_upmore_vert open_in_new Link to source warning Request revision English Above all, that good primary education is still farfrombeing granted the priority status that it deserves.volume_upmore_vert open_in_new Link to source warning Request revision English However, this joint situation should be in line with the country that is most favourable to refugees, which is farfrombeing the case.volume_upmore_vert open_in_new Link to source warning Request revision English A functioning internal market in services, farfrombeing a reality, remains an objective to be achieved.volume_upmore_vert open_in_new Link to source warning Request revision English Europe needs to provide enough funds to support its policies, and that is farfrombeing the case for the Caucasian regions.volume_upmore_vert open_in_new Link to source warning Request revision EnglishFarfrombeing innovative, this programme restricts itself mainly to the setting up of an epidemiological monitoring network.volume_upmore_vert open_in_new Link to source warning Request revision English Some might think that gives us some satisfaction, but for my part that is farfrombeing the case.volume_upmore_vert open_in_new Link to source warning Request revision English The consent or non-consent of the victim is not always relevant - in fact, it is farfrombeing relevant.volume_upmore_vert open_in_new Link to source warning Request revision English It was a project which was farfrombeing able to be realised when I arrived here.volume_upmore_vert open_in_new Link to source warning Request revision English Mr President, let us be clear from the outset: when it comes to cooperation, coherence is farfrombeing the watchword.volume_upmore_vert open_in_new Link to source warning Request revision English As far as non-conventional medicines are concerned, i. e. therapeutic approaches not recognized by the medical order, this is farfrombeing so.volume_upmore_vert open_in_new Link to source warning Request revision English I am farfrombeing the sort of person who declaims against genetic engineering - quite the reverse.volume_upmore_vert open_in_new Link to source warning Request revision English This is an encouraging sign that shows that flexibility and the defence of workers ' legitimate aspirations are farfrombeing mutually exclusive.volume_upmore_vert open_in_new Link to source warning Request revision English We now have directives that are beginning to form a consistent and coherent whole, but they are farfrombeing enforced uniformly throughout the Community.volume_upmore_vert open_in_new Link to source warning Request revision EnglishFarfrombeing a harmless treaty, the Treaty of Amsterdam is considerably accentuating the process of federalization and centralization of the European Union.volume_upmore_vert open_in_new Link to source warning Request revision English(Laughter) Another finding is that the unconscious, farfrombeing dumb and sexualized, is actually quite smart.volume_upmore_vert open_in_new Link to source warning Request revision English And farfrombeing a coldly materialistic view of nature, it's a new humanism, it's a new enchantment.volume_upmore_vert open_in_new Link to source warning Request revision English This suggestion is farfrombeing premature as Mauritania has been participating in the process since it first began.volume_upmore_vert open_in_new Link to source warning Request revision English Mr Metten, farfrombeing a mere secretariat of the Council, the Commission this year put forward recommendations for each country.volume_upmore_vert open_in_new Link to source warning Request revision EnglishFarfrombeing a safeguard against violence, the presence of the Indonesian army has continued to be a cause of the problem.volume_upmore_vert open_in_new Link to source warning Request revision English However, the political aim, which caused us to resort to force, is still farfrombeing achieved.volume_upmore_vert open_in_new Link to source warning Request revision English These problems are farfrombeing resolved and they are still affecting European citizens within the euro area.volume_upmore_vert open_in_new Link to source warning Request revision English The report speaks of common principles, of a European electoral system, but it is farfrombeing complete.volume_upmore_vert open_in_new Link to source warning Request revision English Unfortunately, they are spreading like gangrene at all levels of our society, and although the fight is not lost, it is farfrombeing won.volume_upmore_vert open_in_new Link to source warning Request revision English He ordered them to be beaten, but Justus and Pastor, farfrombeing afraid rejoiced and showed themselves willing to die for Christ volume_upmore_vert open_in_new Link to source warning Request revision English The latest report on human rights in the European Union shows that we are farfrombeing beyond reproach in this area.volume_upmore_vert open_in_new Link to source warning Request revision English It is farfrombeing a futile gesture for us to put forward a European code of conduct.volume_upmore_vert open_in_new Link to source warning Request revision EnglishFarfrombeing restricted, freedom of research finds its fulfilment under the law.volume_upmore_vert open_in_new Link to source warning Request revision English The last fifteen years are farfrombeing a success story as the postponement of reforms wasted several valuable years.volume_upmore_vert open_in_new Link to source warning Request revision English Europe imports 70 % of the raw tobacco it needs, and so the European Union is farfrombeing the world's biggest importer.volume_upmore_vert open_in_new Link to source warning Request revision English Subsequent developments have been alarming and very farfrombeing in accordance with the principles of a constitutional state.volume_upmore_vert open_in_new Link to source warning Request revision English Some of those who had originally voted for this law have since said in the media that this was farfrombeing what they wanted.volume_upmore_vert open_in_new Link to source warning Request revision English There have already been some amendments, but they were, to say the very least, farfrombeing enough.volume_upmore_vert open_in_new Link to source warning Request revision English I refer, Commissioner, to the recycling of food waste, an issue which I see as being farfrombeing resolved once and for all.volume_upmore_vert open_in_new Link to source warning Request revision EnglishFarfrombeing something that the Commission can sort out with the Council behind closed doors, that must be done in public view, here in the directly-elected Parliament.volume_upmore_vert open_in_new Link to source warning Request revision English We nonetheless feel a certain hesitation in doing so because Lebanon is farfrombeing a democracy, and there are flagrant breaches of human rights.volume_upmore_vert open_in_new Link to source warning Request revision English The relevant summary included approximately 400 practical – although not especially practical – points requiring cooperation, but these are still farfrombeing resolved.volume_upmore_vert open_in_new Link to source warning Request revision English It follows that our own countries, farfrombeing able to rest upon the laurels they won in the past, must go down roads that are new and innovative for Europe.volume_upmore_vert open_in_new Link to source warning Request revision English On domestic support, the US offer – which I strongly welcomed at the time and continue to do so – is farfrombeing as ambitious as it seemed.volume_upmore_vert open_in_new Link to source warning Request revision English The fight against Aids and, I must stress, many other diseases afflicting mainly the poorest populations is farfrombeing won.volume_upmore_vert open_in_new Link to source warning Request revision English However, in many countries worldwide the situation is very farfrombeing satisfactory and the question put to us today is to know how the Union can do more.volume_upmore_vert open_in_new Link to source warning Request revision English And all this against the rather mediaeval backdrop of territorial dispute between two Member States, a dispute which is farfrombeing settled.volume_upmore_vert open_in_new Link to source warning Request revision English Finally, could I make the observation that farfrombeing the Commission's work programme for 2001, we might rename it the work programme of the Commission, Parliament and Council.volume_upmore_vert open_in_new Link to source warning Request revision English We must also be honest with ourselves: our own delivery of our high data protection standards is farfrombeing ideal.volume_upmore_vert open_in_new Link to source warning Request revision English These criteria, farfrombeing devised for Turkey, apply generally and in the same way to all the candidates there have been to date, and to all that there will be.volume_upmore_vert open_in_new Link to source warning Request revision English This task too – the setting of uniform financial management standards – was farfrombeing a minor undertaking; indeed, it was akin to a cultural revolution in our international organisation.volume_upmore_vert open_in_new Link to source warning Request revision English Then there is the authorities ' continuing arbitrary treatment of the Kurds, who, even after formal changes to the law, are farfrombeing treated as having equal rights.volume_upmore_vert open_in_new Link to source warning Request revision English It is very farfrombeing that simple, rather we are going to have to make concentrated efforts, during talks with both communities, to encourage them to reach an understanding.volume_upmore_vert open_in_new Link to source warning Request revision English The fine 'polluter pays ' principle, so dear to many Member States, including my own, is farfrombeing complied with!volume_upmore_vert open_in_new Link to source warning Request revision English To the members of the Committee on Industry, External Trade, Research and Energy I want to say that the UN resolution in accordance with the Gothenburg Protocol is farfrombeing adequate.volume_upmore_vert open_in_new Link to source warning Request revision English But we would like to draw attention to the fact that the problem of the poor safety of nuclear power stations is farfrombeing solved by closing the power station at Chernobyl.volume_upmore_vert open_in_new Link to source warning Request revision English It will, of course, be farfrombeing a question of luxury, but it does involve a substantial sum, and it will be noticeable in connection with all the activities under the programme.volume_upmore_vert open_in_new Link to source warning Request revision Search by language German English Indonesian Danish Greek Spanish French Italian Hungarian Dutch Norwegian Polish Portuguese Romanian Russian Finnish Swedish Turkish comment Request revision Close Game recommendation Guess the English word ma_i_ Can you guess the word before your phone dies? 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6467	https://www.chegg.com/homework-help/questions-and-answers/2-x-sim-n-left-mu-sigma-2-right-find-b-p-left-b-q102658908	Your solution’s ready to go! Our expert help has broken down your problem into an easy-to-learn solution you can count on. Question: 2. If X∼N(μ,σ2), find b so that P(−b<σX−μ Given information: Not the question you’re looking for? Post any question and get expert help quickly. Chegg Products & Services CompanyCompany Company Chegg NetworkChegg Network Chegg Network Customer ServiceCustomer Service Customer Service EducatorsEducators Educators
6468	https://www.cuemath.com/calculus/derivative-of-e-2x/	Derivative of e^2x Before going to find the derivative of e2x, let us recall a few facts about the exponential functions. In math, exponential functions are of the form f(x) = ax, where 'a' is a constant and 'x' is a variable. Here, the constant 'a' should be greater than 0 for f(x) to be an exponential function. Some other forms of exponential functions are abx, abkx, ex, pekx, etc. Thus, e2x is also an exponential function and the derivative of e2x is 2e2x. We are going to find the derivative of e2x in different methods and we will also solve a few examples using the same. \| \| \| --- \| \| 1. \| What is the Derivative of e^2x? \| \| 2. \| Derivative of e^2x Proof by First Principle \| \| 3. \| Derivative of e^2x Proof by Chain Rule \| \| 4. \| Derivative of e^2x Proof by Logarithmic Differentiation \| \| 5. \| n^th Derivative of e^2x \| \| 6. \| FAQs on Derivative of e^2x \| What is the Derivative of e^2x? The derivative of e2x with respect to x is 2e2x. We write this mathematically as d/dx (e2x) = 2e2x (or) (e2x)' = 2e2x. Here, f(x) = e2x is an exponential function as the base is 'e' is a constant (which is known as Euler's number and its value is approximately 2.718) and the limit formula of 'e' is lim ₙ→∞ (1 + (1/n))n. We can do the differentiation of e2x in different methods such as: Using the first principle Using the chain rule Using logarithmic differentiation Derivative of e^2x Formula The derivative of e2x is 2e2x. It can be written as d/dx (e2x) = 2e2x(or) (e2x)' = 2e2x Let us prove this in different methods as mentioned above. Derivative of e^2x Proof by First Principle Here is the differentiation of e2x by the first principle. For this, let us assume that f(x) = e2x. Then f(x + h) = e2(x + h) = e2x + 2h. Substituting these values in the formula of the derivative using first principle (which is also known as the limit definition of the derivative), f'(x) = limₕ→₀ [f(x + h) - f(x)] / h f'(x) = limₕ→₀ [e2x + 2h - e2x] / h = limₕ→₀ [e2x e2h - e2x] / h = limₕ→₀ [e2x (e2h - 1) ] / h = e2x limₕ→₀ (e2h - 1) / h Assume that 2h = t. Then as h → 0, 2h → 0. i.e., t → 0 as well. Then the above limit becomes, = e2x limₜ→₀ (et - 1) / (t / 2) = 2e2x limₜ→₀ (et - 1) / t Using limit formulas, we have limₜ→₀ (et - 1) / t = 1. So f'(x) = 2e2x (1) = 2e2x Thus, the derivative of e2x is found by the first principle. Derivative of e^2x Proof by Chain Rule We can do the differentiation of e2x using the chain rule because e2x can be expressed as a composite function. i.e., we can write e2x = f(g(x)) where f(x) = ex and g(x) = 2x (one can easily verify that f(g(x)) = e2x). Then f'(x) = ex and g'(x) = 2. By chain rule, the derivative of f(g(x)) is f'(g(x)) · g'(x). Using this, d/dx (e2x) = f'(g(x)) · g'(x) = f'(2x) · (2) = e2x (2) = 2e2x Thus, the derivative of e2x is found by using the chain rule. Derivative of e^2x Proof by Logarithmic Differentiation We know that the logarithmic differentiation is used to differentiate an exponential function and hence it can be used to find the derivative of e2x. For this, let us assume that y = e2x. As a process of logarithmic differentiation, we take the natural logarithm (ln) on both sides of the above equation. Then we get ln y = ln e2x One of the properties of logarithms is ln am = m ln a. Using this, ln y = 2x ln e We know that ln e = 1. So ln y = 2x Differentiating both sides with respect to x, (1/y) (dy/dx) = 2(1) dy/dx = 2y Substituting y = e2x back here, d/dx(e2x) = 2e2x Thus, we have found the derivative of e2x by using logarithmic differentiation. n^th Derivative of e^2x nth derivative of e2x is the derivative of e2x that is obtained by differentiating e2x repeatedly for n times. To find the nth derivative of e2xx, let us find the first derivative, second derivative, ... up to a few times to understand the trend. 1st derivative of e2x is 2 e2x 2nd derivative of e2x is 4 e2x 3rd derivative of e2x is 8 e2x 4th derivative of e2x is 16 e2xand so on. Thus, the nth derivative of e2x is: dn/(dxn) (e2x) = 2n e2x Important Notes on Derivative of e2x: The derivative of e2x is NOT just e2x, but it is 2e2x. In general, the derivative of eax is aeax. For example, the derivative of e-2x is -2e-2x, the derivative of e5x is 5e5x, etc. ☛ Related Topics: Derivative of log x Derivative of ln x Derivative Calculator Calculus Calculator Read More Download FREE Study Materials SHEETS Derivative of e^2x Worksheet Calculus Worksheet Examples Using Derivative of e^2x Example 1: Find the derivative of e2x + 1. Solution: Let f(x) = e2x + 1 Using the chain rule, f'(x) = e2x + 1 d/dx (2x + 1) = e2x + 1 (2) = 2e2x + 1 Answer: The derivative of e2x + 1 is 2e2x + 1. 2. Example 2: What is the derivative of e2x + e-2x? Solution: Let f(x) = e2x + e-2x Using the chain rule, f'(x) = e2x · d/dx (2x) + e-2x · d/dx (-2x) = e2x (2) + e-2x (-2) = 2 (e2x - e-2x) Answer: The derivative of e2x + e-2x is 2 (e2x - e-2x). 3. Example 3: What is the derivative of e2x sin x. Solution: Let f(x) = e2x sin x. Using the product rule, f'(x) = e2x d/dx (sin x) + sin x d/dx (e2x) = e2x (cos x) + sin x (2e2x) = e2x (cos x + 2 sin x) Answer: The derivative of e2x sin x is e2x (cos x + 2 sin x). View Answer > Have questions on basic mathematical concepts? Become a problem-solving champ using logic, not rules. Learn the why behind math with our certified experts Book a Free Trial Class Practice Questions on Derivative of e^2x Check Answer > FAQs on Derivative of e^2x What is the Formula of Derivative of e2x? The derivative of e2x is 2e2x. Mathematically, it is written as d/dx(e2x) = 2e2x (or) (e2x)' = 2e2x. How to Differentiate e to the Power of 2x? Let f(x) = e2x. By applying chain rule, the derivative of e2x is, e2x d/dx (2x) = e2x (2) = 2 e2x. Thus, the derivative of e to the power of 2x is 2e2x. What is the Derivative of e3x? Let f(x) = e3x. By applying chain rule, the derivative of e3x is, e3x d/dx (3x) = e3x (2) = 3 e3x. Thus, the derivative of e3x is 3e3x. How to Find the Derivative of e2x + 3? Let us assume that f(x) = e2x + 3. Using the chain rule, f'(x) = e2x + 3 d/dx (2x + 1) = e2x + 3 (2) = 2e2x + 3. Thus, the derivative of e2x + 3 is 2e2x + 3. Is the Derivative of e2x Same as the Integral of e2x? No, the derivative of e2x is NOT the same as the integral of e2x. The derivative of e2x is 2e2x. The integral of e2x is e2x / 2. What is the Derivative of e2x²? Let f(x) = e2x². By the application of chain rule, f'(x) = e2x² d/dx (2x2) = e2x² (4x) = 4x e2x². Thus, the derivative of e2x² is 4x e2x². How to Find the Derivative of e2x by First Principle? Let f(x) = e2x. By first principle, f'(x) = limₕ→₀ [f(x + h) - f(x)] / h = limₕ→₀ [e2x + 2h - e2x] / h = limₕ→₀ [e2x e2h - e2x] / h = limₕ→₀ [e2x (e2h - 1) ] / h = 2e2x (1) = 2e2x. Thus, the derivative of e2x by first principle is 2e2x. What is the Derivative of e2x sin 3x? Let f(x) = e2x sin 3x. By product rule, f'(x) = e2x d/dx (sin 3x) + sin 3x d/dx (e2x) = e2x (cos 3x) d/dx (3x) + sin 3x (2e2x) = e2x (3 cos 3x + 2 sin 3x). Thus, the derivative of e2x sin 3x is e2x (3 cos 3x + 2 sin 3x). 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6469	https://researcharchive.calacademy.org/research/scipubs/pdfs/v56/proccas_v56_n18_SuppI.pdf	A Unified Concept of Species and Its Consequences for the Future of Taxonomy Kevin de Queiroz Department of Zoology, National Museum of Natural History, Smithsonian Institution, Washington, DC 20560, Email: dequeiroz.kevin@nmnh.si.edu Contemporary species concepts are diverse. Nonetheless, all share the fundamental idea that species are segments of lineages at the population level of biological organ-ization. They differ in the secondary properties (e.g., intrinsic reproductive isolation, monophyly, diagnosability) that are treated as necessary for considering lineages to be species. A unified species concept can be achieved by interpreting the common fundamental idea of being a separately evolving lineage segment as the only neces-sary property of species and viewing the various secondary properties either as lines of evidence relevant to assessing lineage separation or as properties that define dif-ferent subcategories of the species category (e.g., reproductively isolated species, monophyletic species, diagnosable species). This unified species concept has a num-ber of consequences for taxonomy, including the need to acknowledge that undiffer-entiated and undiagnosable lineages are species, that species can fuse, that species can be nested within other species, that the species category is not a taxonomic rank, and that new taxonomic practices and conventions are needed to accommodate these conclusions. Although acceptance of a unified species concept has some radical con-sequences for taxonomy, it also reflects a change in the general conceptualization of the species category that has been underway for more than a half-century — a shift from viewing the species category as one member of the hierarchy of taxonomic ranks to viewing it as a natural kind whose members are the units at one of the lev-els of biological organization. This change is related to a more general shift in the pri-mary concern of the discipline of systematics (including taxonomy), from the utili-tarian activity of classifying organisms to the scientific activity of testing hypotheses about lineage boundaries and phylogenetic relationships. The unified species concept is a natural outcome of this conceptual shift and represents the more complete acceptance of the idea that species are one of the fundamental units of biology. As such, the unified species concept is central to the future of taxonomy. It is widely held that species are one of the fundamental units of biology (e.g., Mayr 1982; Ereshefsky 1992; Claridge et al. 1997). Any time biologists compare different organisms, they con-sider it critical whether those organisms come from the same or from different species. In fact, they often consider their studies to be comparative only if those studies involve multiple species. Moreover, species are used as units of comparison in virtually all fields of biology — from anato-my, to behavior, development, ecology, evolution, genetics, molecular biology, paleontology, phys-iology, and systematics (including taxonomy). Species are considered so important that biologists have developed a formal system of rules for naming them, which they use in an attempt to give each and every species its own unique name (e.g., ICZN 1999; IBC 2000). PROCEEDINGS OF THE CALIFORNIA ACADEMY OF SCIENCES Volume 56, Supplement I, No. 18, pp. 196–215, 3 figs, 1 table. June 3, 2005 196 According to some authors, the concept of species is not only one of the central concepts of biology but also one of that discipline’s oldest and most vexing problems (e.g., Dobzhansky 1976). The problem is that biologists have been unable to reach a general agreement about the nature of species and thus about the definition of the species category. Many papers have been written about this topic, and many definitions (i.e., descriptions of species concepts) have been proposed, but despite all the attention that species concepts have received, no single definition (or its correspon-ding concept) has proved optimal for all of the different uses to which biologists put the term. As a consequence, although one definition or concept has often come to predominate for a certain peri-od of time, or among a certain subgroup of biologists, no single definition or concept has become universal within biology as a whole. This lack of agreement about the concept of species has come to be known as “the species problem” (e.g., Mayr 1957; Dobzhansky 1976). In this paper, I will review a proposed solution to the species problem that unifies diverse con-temporary views on the nature of species (de Queiroz 1998, 1999). The solution is based on iden-tifying a common element in the diverse contemporary views about the nature of species, which not only clarifies the nature of the species problem but also suggests a straightforward solution, the result of which is a unified concept of species. After describing this unified species concept, I will consider some of its consequences, arguing that several have been foreshadowed by recent devel-opments in the study of species. Finally, I will discuss how the unified concept of species repre-sents the more complete acceptance of a historical shift in the conceptualization of the species cat-egory that is already widely held among biologists. Because of the theoretical importance of species and the unresolved nature of the species prob-lem, a unified concept of species is critical to the future of taxonomy. I hope that my proposal will contribute to ending the long-standing debate about the nature of species (see also O’Hara 1993, 1994; Pigliucci 2003) so that biologists in general, and systematists in particular, can focus their attention on methods for determining the boundaries of species (e.g., Sites and Marshall 2003), the processes responsible for the diversification of species (e.g., TRENDS in Ecology and Evolution, July 2001), and the enormous task of inventorying the species of the world (as discussed in the papers from the Biodiversity symposium included in this volume). THE DIVERSITY OF CONTEMPORARY SPECIES CONCEPTS Most systematic and evolutionary biologists are familiar with the existence of alternative species concepts. Many readers may be surprised, however, by the number of different concepts that have been proposed. Mayden (1997, 1999), for example, listed 24 named species concepts. As a point of departure, I will adopt a taxonomy that recognizes 13 major categories of species con-cepts and their corresponding definitions (some of which are subsets of others), based on proper-ties that distinguish the different concepts from one another (Table 1). Space prohibits me from describing these alternative species concepts in detail, so I refer readers to Mayden (1997) and de Queiroz (1998) for reviews. I hope that readers are familiar with at least a couple of the different concepts, though such familiarity is not necessary to follow my arguments. What is important is to know two general things. First, the different species concepts and their corresponding definitions are based, in part, on different biological properties. For example, the biological species concept is based (in part) on reproductive isolation, the ecological species concept is based on the occupation of a distinct niche or adaptive zone, one version of the phylogenetic species concept is based on diagnosability, and another version is based on monophyly. The second important thing to realize is that many of the different species concepts are incompatible with one another in that they lead to the recognition of different species taxa — that is, to different species boundaries and, thus, to different numbers of DE QUEIROZ: A UNIFIED CONCEPT OF SPECIES 197 recognized species. For example, it is commonly the case that adopting the diagnosable version of the phylogenetic species concept leads to the recognition of many more species taxa than would be recognized under the biological species concept (e.g., Cracraft 1983; Zink 1996). This situation creates a problem given that species are used as basic units of comparison in diverse types of stud-ies. On the one hand, if a researcher were to use the species taxa recognized by several different authors specializing on different taxonomic groups, those species taxa likely would not be equiva-lent to one another. On the other hand, if a researcher were to use species taxa based on a single species concept, that person might obtain a very different result than if he or she were to use species taxa based on a different species concept. 198 PROCEEDINGS OF THE CALIFORNIA ACADEMY OF SCIENCES Volume 56, Supplement I, No. 18 Species Concept (Traditional Name) Distinctive Properties (Species Criteria) Advocates Proposing Explicit Species Definitions Biological Potential interbreeding/Intrinsic reproductive isolation Wright (1940); Mayr (1942, 1963); Dobzhansky (1950) Isolation Isolating mechanisms Mayr (1942, 1963); Dobzhansky (1970) Recognition Compatible mate recognition and fertilization systems Paterson (1978, 1985) Evolutionary Unitary evolutionary role, tendencies, and fate Simpson (1951, 1961); Wiley (1978, 1981) Ecological Distinct adaptive zone (niche) Van Valen (1976) Cohesion Intrinsic cohesion mechanisms Templeton (1989) Phylogenetic Association with Phylogenetic Systematics (Cladistics) See below Hennigian Species bounded at both ends by clado-genetic (lineage splitting) 1 Hennig (1966); Ridley (1989) Monophyletic (Apomorphic) Monophyly (as evidenced by apo-morphies = derived character states) Rosen (1979); Mishler (1985) Diagnosable Diagnosability (possession of fixed character state differences) Cracraft (1983); Nixon and Wheeler (1990) Genealogical2 Exclusive coalescence of alleles for multiple loci Baum and Shaw (1995) Phenetic Phenetic cluster (group of similar organisms separated by gaps from other such groups) Michener (1970); Sneath and Sokal (1973) Genotypic Cluster Deficit of genotypic intermediates (heterozygotes) at multiple loci Mallet (1995) TABLE 1. Alternative species concepts and the properties that distinguish them (after de Queiroz 1998). Indented concepts are subsets (not necessarily mutually exclusive) of the non-indented concept immediately preceding them. 1 Species can also be bounded at one end by extinction. 2 The genealogical species concept could be considered an example of the monophyletic species concept (rather than a mutually exclusive concept) in that exclusive coalescence is equivalent to one interpretation of monophyly (see de Queiroz and Donoghue 1990). On the other hand, the monophyletic version of the phylogenetic species concept is usually concerned with monophyly as it relates to component populations or organisms rather than alleles. The existence of alternative and incompatible species concepts reflects a basic disagreement about the nature of species (though, as I will argue below, there is also considerable agreement). This situation may not be particularly troubling to an individual researcher who is convinced that one of the concepts is superior to the others. The problem is that other researchers exhibit equal conviction in their commitments to different species concepts. In addition, the situation is getting worse rather than better, which is to say that the number of alternative species concepts is increas-ing, rather than decreasing. Of the 24 concepts listed by Mayden (1997), a full one-third were pro-posed in the preceding ten years. Moreover the biological species concept, which was once the dominant concept and is still perhaps the most widely adopted, seems to be less popular now than it was 30 years ago. The existence of diverse species concepts makes a certain amount of sense, because the differ-ent concepts are based on properties that are of greatest interest to different subgroups of biologists. For example, biologists who study hybrid zones tend to emphasize reproductive barriers, system-atists tend to emphasize diagnosability and monophyly, and ecologists tend to emphasize niche dif-ferences. Paleontologists and museum taxonomists tend to emphasize morphological differences, and population geneticists and molecular systematists tend to emphasize genetic ones. In addition, the biological properties that are most important in determining the limits of species likely differ among taxonomic groups (e.g., birds versus cyanobacteria), and this situation likely influences the properties emphasized by biologists who specialize on different groups. Nevertheless, for those researchers who are able to step back from their own personal investments and research interests, all of the concepts seem to have some merits in that they are all based on important biological prop-erties (Table 1). RECONCILIATION The reconciliation of these diverse views has two basic components (de Queiroz 1998, 1999). The first is identifying a common fundamental element shared by all modern species concepts. The second is re-evaluating the differences among alternative species concepts in the context of this common element. Before I describe this solution to the species problem, I want to say that regard-less of whether one accepts my proposal, a solution is unlikely to come from the general approach that people have been taking for the last 50 years. I refer to the approach of identifying a particu-lar biological property — whether reproductive isolation, ecological distinctiveness, monophyly, diagnosablity, or anything else — as the basis of a species concept, and then advocating that con-cept because of its supposed theoretical and/or operational superiority over rival concepts. That approach is unlikely to succeed, and it certainly has not been successful so far. Rather than solving the species problem, it has caused (and later aggravated) the problem. Rather than leading to agree-ment on a single species concept, it has lead to a proliferation of alternative concepts and more dis-agreement than ever. For this reason, I have taken a completely different approach. Instead of pro-posing yet another species concept based on yet another biological property, I have proposed a way to unify the existing species concepts. The Common Element Previous attempts to solve the species problem have tended to obscure the solution by empha-sizing the differences, rather than the similarities, among alternative species concepts. As it turns out, all contemporary species concepts share a common element, and more importantly, that shared element is fundamental to the way in which species are conceptualized. Virtually all contemporary species concepts equate species with populations or population lineages — or more accurately, with segments of population level lineages. DE QUEIROZ: A UNIFIED CONCEPT OF SPECIES 199 LINEAGES: Because the concept of a lineage is central to my proposal, I need to clarify some things about lineages. When I use the term lineage, I am not talking about a clade or a monophylet-ic group (see de Queiroz 1998, 1999), and thus, I am not advocating a version of the phylogenetic species concept. A lineage, in the sense that I am using the term (see also Simpson 1951; Hull 1980), is a line of direct ancestry and descent (Fig. 1). Such lineages commonly are not monophyletic in that their later members or parts share more recent common ancestors with recently diverged side branches (which are parts of different lineages) than they do with earlier members of the same lineage. Lineages are formed by biological entities at several dif-ferent levels of organization. For example, every person can trace his or her ancestry back along an organism lineage that passes through a series of ancestral organisms — a parent, a grandparent, a great grandparent, and so forth. Similarly, each species can trace its ancestry back along a population level lineage that pass-es through a series of ancestral species. I also want to point out that lineages, not only population level lineages but also those at other levels of biological organization, are the entities that actually evolve (Hull 1980). In fact, I have argued (de Queiroz 1999:82) that the common claim that populations, rather than organisms, are the entities that evolve (e.g., Futuyma 1986:7), which is reflected in the common definition of evolution as changes in gene frequencies in populations (see Mayr 1982:400), is attributable to the temporal extendedness, rather than the organizational level, of populations. Even if organisms them-selves do not evolve, organism lineages do evolve, and this conclusion suggests that evo-lution can be defined generally as heritable changes in lineages. (This definition is conceptually similar to Darwin’s descent with modification but incorporates the requirement that the modifications must be heritable. It includes gene frequen-cy changes in populations as a special case.) Thus, the concept of a lineage is fundamental to the concept of evolution itself, and it also turns out to be common to all species concepts formulated in the context of an evolutionary worldview. Because lineages at the species/population level are made up of several species, species them-selves are segments of such lineages. The diagrams in Figure 2 illustrate this point in the context of three general models of speciation (Foote 1996). In these diagrams, the vertical lines represent species, and the horizontal ones represent speciation events. In the bifurcation model, where ances-tral species become extinct at speciation events, species correspond more or less precisely with the lineage segments between those events. In the budding model, where ancestral species persist 200 PROCEEDINGS OF THE CALIFORNIA ACADEMY OF SCIENCES Volume 56, Supplement I, No. 18 FIGURE 1. Representation of a species (population) level lineage and its component species within a larger branching tree (modified from de Queiroz 1999). The lineage of inter-est, represented by the darker lines, is made up of a series of ancestral and descendant species, labeled with the letters a through d. Note that the lineage is not monophyletic in that some of its later members (e.g., species ‘d’) share more recent common ancestors with recently diverged side branch-es (gray lines) than with earlier members of the lineage (e.g., species ‘a’). through speciation events, the lineage segments that correspond with species originate, but don’t necessarily terminate, in those events. Finally, in the phyletic transformation model, where speciation occurs within an unbranched lineage, species once again correspond more or less precisely with the lineage seg-ments between speciation events, though what counts as a specia-tion event differs from the other two models. In the bifurcation and budding models, speciation corresponds with lineage split-ting (cladogenesis), while in the phyletic transformation model, speciation corresponds with change within an unbranched lin-eage (anagenesis). With the exception of speciation via hybridization (which might be considered a variant of the bud-ding or bifurcation models, depending on how it occurs), these three general models cover the range of possibilities, and all contemporary species concepts are consistent with (and sometimes imply) one or more of them. Notice that in all three models, species correspond not with entire lineages but instead with lineage segments. As I pointed out earlier, biological entities at various organizational levels form lineages — from genes, to organelles, cells, organisms, and species. The lineages at each level are made up of lower level lineages. Thus, each population level lineage is made up of several organism lineages. In the case of sexual or biparental reproduction, the process of reproduction itself unites organism lineages to form a higher (population) level lineage, because the organism lineages come together at each reproductive event to form an anastomosing nexus. In the case of asexual or uniparental reproduction, the organism level lineages are not bound together in this manner. Therefore, if uni-fication of asexual organism lineages occurs, it must result from other processes than reproduction. Whether asexual organisms do in fact form such higher level lineages is controversial, but the answer is not important to my argument. What is important is that species definitions that are intended to apply to asexual organisms assume that they do. With these clarifications in mind, let me reiterate that all contemporary species concepts are variations on the general theme that species are segments of population-level lineages. Here, I am using the term population in a very general sense that refers to a level of organization above that of the organism, and which applies — at least potentially — to both sexual and asexual beings. I have previously referred to this common theme as the general lineage concept of species to empha-size that the concept of the population level lineage is general in the sense of being common to all contemporary species concepts (de Queiroz 1998, 1999). It is important to understand that this gen-DE QUEIROZ: A UNIFIED CONCEPT OF SPECIES 201 FIGURE 2. Three general models of speciation (modified from Foote 1996). A. Bifurcation model, in which speciation corresponds with lineage splitting and ancestral species terminate upon giving rise to two descendants. B. Budding model, in which speciation corresponds with lineage splitting and ancestral species persist after giving rise to one or more descendants. C. Phyletic transformation model, in which speciation corresponds with change in an unbranched lineage and ancestral species terminate after transforming into descendants. The budding and bifurcation models can be classified as cladogenetic models, in which speciation corresponds with lineage splitting; the phyletic transformation model can be classified as an anagenetic model, in which speciation corresponds with change in an unbranched lineage. Vertical lines represent species; horizontal lines represent speciation events. That which constitutes a speciation event is deliberately left undefined to accommo-date diverse species concepts. eral lineage concept is not an alternative to the various contemporary species concepts; instead, it is a more general concept that subsumes all of them. An early example of a species concept con-forming to the general lineage concept can be found in Darwin’s (1859) Origin of Species, where species are described and illustrated as “lines of descent.” More importantly, the general lineage concept underlies virtually every species concept described during the last half century. The Conformity of Diverse Species Definitions to the General Lineage Concept I have previously presented evidence that diverse modern views on the nature of species all conform to the general concept of species as segments of population-level lineages (de Queiroz 1998, 1999). The following list summarizes this evidence for a diversity of papers proposing explicit species definitions (those used as the source of quoted species definitions in de Queiroz 1998). It is organized in terms of the nature of the evidence, which can be divided into five cate-gories: 1) Papers that explicitly equate species with lineages in their proposed species definitions. Examples are the evolutionary definitions of Simpson (1951, 1961) and Wiley (1978, 1981) and the ecological definition of Van Valen (1976), all of which begin with some variant of the phrase “a species is a lineage.” 2) Papers that explicitly equate species with lineages in their extended discussions, as opposed to their concise definitions. Examples are Mishler’s (1985) and Nixon and Wheeler’s (1990) papers describing different versions (monophyletic and diagnosable) of the phylogenetic species concept, Ridley’s (1989, 1990) on the cladistic (Hennigian) species concept, Templeton’s (1989, 1998) on the cohesion species concept, and Baum and Shaw’s (1995) on the genealogical species concept. 3) Papers that represent species as lineages using diagrams. In these diagrams, species are repre-sented either as single lines (e.g., Darwin 1859, figure 1) or trunks (e.g., Hennig 1966, figures 14, 15), and their component organisms (if they are also illustrated) are represented by dots, which may be connected by lines representing relationships of descent and thus illustrating organism-level lin-eages (e.g., Hennig 1966, figures 3, 4, 6). Examples of such diagrams can be found in numerous papers, including (among those presenting explicit species definitions) those by Simpson (1951, 1961), Hennig (1966), Wiley (1981), Ridley (1989), Nixon and Wheeler (1990), and Baum and Shaw (1995). 4) Papers that implicitly equate species with lineages by equating them with populations in their proposed species definitions. As Simpson (1951) pointed out, a lineage is a population extended through time, and conversely, a population is a segment, in some cases an instantaneous cross sec-tion, of a lineage (see Simpson 1951, figure 3). Thus, definitions that equate species with popula-tions and those that equate species with lineages simply represent time-limited and time-extended versions of the same general species concept. Examples include Wright’s (1940), Mayr’s (1942, 1963, 1982), and Dobzhansky’s (1950, 1970) definitions of the biological species concept, Rosen’s (1979) apomorphic version of the phylogenetic species concept, and Paterson’s (1985) recognition species concept. 5) Papers that implicitly equate species with lineages by equating them with populations in their extended discussions. Examples include Cracraft’s paper on the diagnosable version of the phylo-genetic species concept (1983), Michener’s (1970) and Sneath and Sokal’s (1973) writings on the phenetic species concept, and Mallet’s (1995) paper proposing the genotypic cluster species defini-tion. Even those modern species definitions that seem to diverge most drastically from the rest are at least consistent with — if not actually based on — the general lineage concept of species. For example, phenetic species definitions describe species as phenetic clusters (e.g., Michener 1970; Sneath and Sokal 1973) rather than populations or lineages. These definitions do not, however, 202 PROCEEDINGS OF THE CALIFORNIA ACADEMY OF SCIENCES Volume 56, Supplement I, No. 18 contradict the equation of species with populations or lineages; instead, they simply emphasize the evidence and procedures that are used to recognize species in practice (e.g., Rogers and Appan 1969; Michener 1970; Sokal and Crovello 1970; Sneath and Sokal 1973; Doyen and Slobodchikoff 1974). Similarly, species definitions that emphasize the property of monophyly (e.g., Rosen 1979; Donoghue 1985; Mishler 1985) seem to deny that species differ in any important respect from higher taxa or clades. The views underlying these definitions are also consistent with the equation of species with lineages. Advocates of the definitions in question stress the importance of mono-phyly; however, contrary to the way their views are sometimes portrayed, they do not require all species taxa to be monophyletic. They acknowledge that the members of a single species may not always be mutually most closely related in terms of their common ancestry relationships. The authors in question still refer to these lineages using species names; however, to call attention to their non-monophyletic status, they designate such species paraspecies (Ackery and Vane-Wright 1984), if the evidence suggests that they are paraphyletic, or metaspecies (Donoghue 1985; see also Graybeal 1995) if the evidence is equivocal. DIFFERENCES AMONG ALTERNATIVE SPECIES CONCEPTS Once we realize that all contemporary species concepts share the common view that species are segments of population-level lineages, the next problem is to explain how it is that so much dis-agreement about species concepts can exist in spite of this general agreement. The answer to this question becomes apparent when we consider the differences among alternative species concepts in the context of the common element. The answer is as follows: if we consider the common ele-ment — existence as a separate lineage — as the primary defining property of species (primary species criterion), then the diversity of species concepts can be accounted for by recognizing that each alternative species concept adopts a different property of lineages as a secondary defining property of species (secondary species criterion). In other words, under all species concepts, a species is a population lineage, but under the biological species concept, for example, the lineage also has to be reproductively isolated from other lineages. Under the ecological species concept, the lineage also has to occupy a different niche or adaptive zone. Under the phenetic species con-cept, it also has to form a phenetic cluster. Under the diagnosable version of the phylogenetic species concept, it also has to have a unique combination of character states. Other concepts adopt still other secondary properties. SECONDARY PROPERTIES AND LINEAGE DIVERGENCE. The reason that these different secondary properties — these secondary species criteria — lead to incompatible species concepts is that they commonly arise at different times during the process of lineage divergence. Lineage divergence can be conceptualized in terms of a few general evolutionary processes — processes such as mutation, migration (or the reduction thereof), natural selection, and genetic drift. In contrast, the characters affected by those processes are highly diverse. They can be genetic or phenotypic; qualitative or quantitative; selectively advantageous, disadvantageous, or neutral. Moreover, they involve many different aspects of organismal biology — including genetics, development, morphology, physiol-ogy, and behavior. With regard to the species problem, the important point is that changes in these characters lead to the acquisition of a number of different properties by diverging lineages. Thus, as two lineages diverge, they become phenetically (quantitatively) distinguishable. They become diagnosable in terms of fixed character states. Their genitalia, gametes, and developmental systems become incompatible. Their mate recognition systems diverge to the point where their component organ-DE QUEIROZ: A UNIFIED CONCEPT OF SPECIES 203 isms no longer recognize one another as potential mates. They evolve distinctive ecologies. And they pass through polyphyletic, paraphyletic, and monophyletic stages in terms of their component genes and organisms (Neigel and Avise 1986). These different properties are not all expected to evolve at the same time, nor are they necessarily expected to evolve in a regular order (de Queiroz 1998). The problem is that each different species concept adopts a different one of these properties as a defining (necessary) property of species. This situation is what causes the different species concepts — despite their general conceptual unity — to result in conflicting conclusions concern-ing which lineages deserve to be recognized as species. In short, although all contemporary species concepts equate species with segments of population lineages, different concepts treat different events in the process of lineage divergence as marking the beginnings of those segments. Figure 3 is a highly simplified diagram representing the process of lineage divergence. The shades of gray represent the daughter lineages becoming more and more different from one anoth-er through time, and the numbered lines (1–8) represent the times at which they acquire different properties relative to each other — for exam-ple, when they become phenetically distin-guishable, diagnosable, reciprocally mono-phyletic, reproductively incompatible, ecologi-cally distinct, and so forth. This set of proper-ties forms a broad gray zone within which alter-native species concepts come into conflict. On either side of the gray zone, there will be unan-imous agreement about the number of species. Before the acquisition of the first property, everyone will agree that there is one species, and after the acquisition of the last property, everyone will agree that there are two. But in between, there will be disagreement. Some people will draw the cutoff where loss or fixa-tion of a character in one of the lineages makes them diagnosable. Others will draw the cutoff where the lineages develop an intrinsic repro-ductive barrier. Still others will draw the cutoff where both lineages form exclusive groups in terms of multiple gene trees. Moreover, dis-agreements will be exacerbated if further split-ting and divergence (acquisition of earlier properties) occurs before some of the later properties are acquired. This is cause of the species problem. This is the reason that there are so many incompatible definitions of the species category despite widespread agreement about the general nature of species. A UNIFIED SPECIES CONCEPT On the other hand, the situation I have just described suggests a very simple solution to the 204 PROCEEDINGS OF THE CALIFORNIA ACADEMY OF SCIENCES Volume 56, Supplement I, No. 18 FIGURE 3. Lineage divergence and alternative species cri-teria (modified from de Queiroz 1998). The diagram repre-sents the process of lineage divergence through a cladogenet-ic (lineage splitting) event. Progressive darkening and light-ening of the daughter lineages represents their progressive divergence through time. The numbered horizontal lines (1-8) represent the times at which the daughter lineages acquire different properties relative to each other (e.g., when they become phenetically distinguishable, diagnosable, recipro-cally monophyletic, reproductively incompatible, ecological-ly distinct, and so forth). The species problem results from disagreements about which of these properties are necessary (defining) properties of the species category (species crite-ria). The entire set of properties defines a zone in which there will be disagreement about the number of species among authors adopting different properties as their species criteria. species problem. This solution involves a minor yet fundamental shift in the way we conceptual-ize species. It retains the element that is common to all contemporary species concepts, and it elim-inates the conflicts between rival concepts without denying the importance of the properties that underlie their obvious differences. In short, it represents a unified species concept. The solution has two components. First, we retain the common element — the general concept of species as separately evolving segments of population level lineages. In other words, we retain the primary species criterion. Second, we interpret this property as the only necessary property of species. In other words, we reinterpret all the other properties that have previously been treated as necessary properties of species — the properties that created the incompatibilities among alterna-tive species concepts — as no longer being defining properties of the species category. They can be thought of instead as contingent properties: properties that species may or may not acquire dur-ing the course of their existence. In the context of this proposal, there are no secondary species cri-teria. Lineages do not have to be phenetically distinguishable, or diagnosable, or monophyletic, or reproductively isolated, or ecologically divergent, or anything else, to be species. They only have to be evolving separately from other lineages. This unified species concept is related, but not identical, to the general lineage species con-cept. As noted above, the general lineage concept is the element that is common to all contempo-rary species concepts, which represent variations on this general theme. In addition, the general lin-eage concept is agnostic with regard to the differences among its variants, the alternative species concepts — that is, with regard to interpreting one or another secondary property of lineages as a necessary property of species. This agnosticism is necessary for the concept to be general — for it to subsume, rather than being an alternative to, the other contemporary species concepts. The uni-fied species concept is based on the common element represented by the general lineage concept; however, in contrast with the general lineage concept and its variants, the unified concept treats the common element as the only necessary property of species. In other words, the unified species con-cept is not agnostic with respect to interpreting one or another secondary property of lineages as a necessary property of species; it rejects those interpretations. Nonetheless, the unified concept truly represents a unification in that it does not reject the diverse secondary properties themselves, rec-ognizing that all of those properties continue to play important roles in the study of species. Roles of Secondary Properties I stated above that this proposed solution to the species problem eliminates the conflicts among rival species concepts without denying the importance of the properties that underlie their differ-ences. Under a unified concept of species, the secondary properties — the former secondary species criteria — remain important in two ways. First, they continue to serve as important lines of evidence relevant to assessing the separation of lineages. These properties — properties such as phenetic distinguishability, reciprocal monophyly, pre- and post-zygotic reproductive barriers, eco-logical differences, and so forth — are, after all, some of the best available lines of evidence regarding lineage separation. However, in contrast with the focus on one or a few of these proper-ties under one or another of the alternative species concepts, under the unified species concept, all of the properties are important. That is, the more lines of evidence that can be brought to bear on the question of lineage separation, the better. Second, the secondary properties can be used to define subcategories of the species category — that is, to recognize different classes of species based on the properties that they possess. To use an organism level analogy, biologists commonly recognize different subcategories of the general category organism based on properties possessed by organisms. For example, they recognize sex-DE QUEIROZ: A UNIFIED CONCEPT OF SPECIES 205 ually mature organisms, fully grown organisms, socially dominant organisms, and so forth. Similarly, under a unified species concept, biologists would recognize different subcategories of the general category species based on properties possessed by species. For example, they might recognize reproductively isolated species, ecologically distinct species, monophyletic species, and so forth. (Incidentally, names based on the relevant properties, as in the preceding examples, describe these categories more clearly than do overly general terms such as biological species, eco-logical species, phylogenetic species, etc.) Thus, a unified species concept would not deny the importance of any of the properties that have been considered important by previous authors. It just would not treat those properties as necessary properties of species. CONSEQUENCES OF A UNIFIED SPECIES CONCEPT The solution to the species problem just outlined is very simple — so simple that one wonders if the reason it has been elusive has to do with an assumption that people have not thought to ques-tion. I will suggest that that assumption is related to the historical treatment of the species catego-ry as part of the hierarchy of taxonomic ranks, which has hindered biologists from fully accepting an important shift in the way they conceptualize the species category. I will return to this idea in the next section of this paper (A Shift in the Conceptualization of Species). In this section, I would like to address some consequences of a unified species concept. I anticipate that some people are going to have difficulty accepting some of the consequences that I will describe, at least initially. The reason is that certain consequences of a unified species concept go against long-standing traditions — traditions that are related to the taxonomic assump-tion that I just mentioned. I would argue, however, that it is counterproductive to reject theoretical proposals simply because they conflict with taxonomic conventions. Given that the purpose of tax-onomies is to convey (theoretically significant) information, it is more important for taxonomic conventions to be consistent with systematic theory rather than for systematic theory to be consis-tent with taxonomic conventions. Therefore, I ask readers to bear with me while I describe some consequences of a unified species concept. After I have finished describing those consequences, I will try to explain why biologists should accept them. Nevertheless, I will also describe how sev-eral of the consequences in question have been anticipated by recent trends in the way that biolo-gists treat species. All Lineages are Species One consequence of adopting a unified species concept is that all separately evolving popula-tion level lineages are species. This conclusion follows directly from adopting the unified concept, which treats only existence as a separately evolving lineage, and not any of the contingent proper-ties of lineages, as a necessary property of species. Thus, not only reproductively isolated lineages are species, nor only ecologically differentiated ones, nor only diagnosable ones, nor only phenet-ically distinguishable ones. Even undifferentiated and undiagnosable lineages are species. As long as a lineage exists, which is to say as long as it is (or was, in the case of an extinct lineage) evolv-ing separately from other lineages, it is a species. And lineages can be separated by many different factors, including extrinsic (e.g., geographic) barriers. A corollary of this consequence is that there are many more species on Earth than biologists have been prepared to accept under traditional views. In addition to those species for which no organisms have yet been discovered, many of the species taxa that have been recognized under traditional species concepts are likely made up of multiple species. This consequence of the unified species concept has been foreshadowed by a couple of recent 206 PROCEEDINGS OF THE CALIFORNIA ACADEMY OF SCIENCES Volume 56, Supplement I, No. 18 trends. One of these trends is related to the development of several of the alternative species con-cepts, according to which allopatric, diagnosable taxa formerly ranked as subspecies are regarded as species (e.g., Cracraft 1983; Frost and Hillis 1990). Given that the taxa in question are allopatric and diagnosable relative to other populations formerly considered conspecific with them, they pre-sumably represent separately evolving lineages. If so, then their recognition as species is in agree-ment with the unified concept. This proposition should not, however, be misinterpreted as justifi-cation for treating all diagnosable units as species. For example, many recent studies based on mitochondrial DNA recognize groups of individuals or populations that are geographically con-tiguous and monophyletic (as well as diagnosable and phenetically distinguishable) in terms of their mtDNA haplotypes as species (reviewed by Avise 2000). Although several such groups with-in a previously recognized species may indeed correspond with separately evolving lineages, because mtDNA is maternally inherited, it is important to examine paternally or autosomally inher-ited genes to rule out the alternative hypothesis that the phylogeographic pattern results from female philopatry within a single lineage (Avise 2000). Another recent trend foreshadowing the unified species concept is the recognition of “evolu-tionarily significant units” or “ESUs” within traditional species. Originally proposed in the context of conservation biology (Ryder 1986), an ESU is a population or set of populations that is morpho-logically and genetically, or evolutionarily, distinct from other populations. Several of the criteria proposed for ESU recognition, such as reciprocal monophyly for mtDNA alleles and significant divergence of allele frequencies at nuclear loci (e.g., Moritz 1994), correspond with secondary species criteria associated with the some of the alternative species concepts (e.g., Baum and Shaw 1995; Highton 2000). Indeed, Vogler and DeSalle (1994) have explicitly proposed using the species criteria associated with the one of the versions of the phylogenetic species concept (diagnosability criterion) for identifying ESUs. Given that these same criteria are lines of evidence used to infer the separation of lineages, many ESUs would likely be considered species under the unified species concept. This situation should be beneficial to conservation, given that many of the relevant regu-lations (such as the U.S. Endangered Species Act) emphasize species. Species Fusion Another consequence of a unified species concept is that species can fuse. Traditionally, it has been common to think of species as permanently or irreversibly separated lineages (e.g., Mayr 1982:296; Bush 1995). However, if all separately evolving lineages are species, then the separation of many species from other species may be temporary or reversible. This situation seems obvious for species that are separated only by extrinsic (i.e., geographic) barriers — at least ephemeral ones. However, it would also seem to hold for at least some cases in which separation is intrinsic (see below). As a consequence, collections of organisms or populations that form two species at a given time may fuse to form a single species at a later time. This consequence is not unique to the unified species concept. It also holds under several of the alternative species concepts — in particular, those based on secondary criteria thought not to be indicative of permanent separation, such as distinguishability, diagnosability, and monophyly. Differences related to these properties commonly evolve between populations (lineages) separated by extrinsic barriers, leading to their recognition as separate species under the alternative species concepts in question. However, if those differences do not involve traits influencing reproductive compatibility (crossability of Mayr 1942), and if the extrinsic barriers subsequently disappear, then there is nothing to prevent the lineages from fusing. Species fusion can also occur under secondary species criteria that are commonly thought to DE QUEIROZ: A UNIFIED CONCEPT OF SPECIES 207 be indicative of permanent separation — namely, intrinsic reproductive isolation (e.g., Mayr 1982). In the case of certain kinds of premating barriers, potential breakdown or reversal seems uncontro-versial. For example, premating barriers based on ecological differences can break down if habi-tats change, which would seem likely to be a relatively common occurrence in the face of current large-scale, human-induced habitat changes. Moreover, in the case of postmating reproductive incompatibilities, it is at least theoretically possible for natural selection to eliminate the elements responsible for such an incompatibility, or for factors that reduce the deleterious effects of previ-ously incompatible elements to evolve (e.g., Ritchie and Hewitt 1995). The realization that lineag-es exhibiting intrinsic reproductive isolation can fuse has been acknowledged by at least some advocates of a species criterion based on this property. Thus, Turner (2002), an advocate of the bio-logical species concept (potential interbreeding criterion), explicitly acknowledged the possibility of species fusion, which he termed despeciation. Species within Species Another consequence of a unified species concept is that species can be nested within other species. Taxonomic tradition treats the species category as one rank or level in the hierarchy of tax-onomic categories. In this context, taxa assigned to the same category (rank) are considered mutu-ally exclusive (i.e., to have no members in common). As a consequence, taxa composed of more than one species must be assigned to higher ranks (such as subgenus or genus), and taxa within a species must be assigned to lower ranks (such as subspecies or variety). In other words, a species cannot be nested within another species. This convention, however, is inadequate for dealing with many real biological situations involving species. In particular, it has problems with situations involving incomplete or partial lin-eage separation, as exemplified by cases of introgressive hybridization. These situations cause end-less taxonomic problems under the traditional assumption that all taxa ranked as species are mutu-ally exclusive and therefore cannot contain, or be contained within, other species. In such cases, taxonomies commonly vacillate between treating the partially separated lineages as the same species and treating them as different species. Some classic examples are found among North American birds, such as Bullock’s (Icterus bullockii) and Baltimore Orioles (Icterus galbula) — Northern Orioles (Icterus galbula) when considered a single species — and Red-shafted (Colaptes cafer) and Yellow-shafted Flickers (Colaptes auratus) — Common Flickers (Colaptes auratus) when considered a single species (AOU 1998). This problem can be remedied by allowing species taxa to be nested, a taxonomic innovation that is implied by the unified species concept. Contrary to traditional practice, the question of whether particular organisms belong to the same species cannot always be answered with a simple “yes” or “no.” Sometimes lineages are only partially separated, which implies that their component organisms are simultaneously parts of both the same and different species. In other words, some species are nested within larger species. Moreover, such incompletely separated species do not have to be sister species to be parts of a single more inclusive species (see Omland, et al. 1999, for an example involving orioles). The Species Category is Not a Rank Another consequence of the unified species concept, which is related to several of the previ-ous ones, is that the species category is not a taxonomic rank. Traditionally (i.e., under any of the alternative species concepts), only those separately evolving lineages that have evolved a particu-lar secondary property (e.g., reproductive isolation, a distinct ecological niche, a unique combina-208 PROCEEDINGS OF THE CALIFORNIA ACADEMY OF SCIENCES Volume 56, Supplement I, No. 18 tion of character states) are considered to merit taxonomic recognition as species. Other separate-ly evolving lineages either are not granted formal taxonomic recognition at all, or they are assigned to a taxonomic category of lower rank, such as subspecies. This practice effectively treats both the species and subspecies categories as taxonomic ranks, which is in keeping with the taxonomic tra-dition in which the species category is a rank in the hierarchy of taxonomic categories (the rank below subgenus and above subspecies). It also has the undesirable consequence of ignoring or downplaying the significance of many separately evolving lineages. Under a unified species concept, the species category is not a rank in the hierarchy of taxo-nomic categories but a class or kind made up of the entities or units at one of the fundamental lev-els of biological organization. Species are the entities that from lineages at the population level of biological organization just as organisms are the entities that form lineages at the organism level of biological organization (de Queiroz 1999). And just as all such entities at the organism level of organization are organisms (i.e., not only those that are postnatal, sexually mature, fully grown, etc.), similarly, all such entities at the population level of organization are species (i.e., not only those that are diagnosable, reproductively isolated, ecologically distinct, etc.). Because all sepa-rately evolving population level lineages are species, any taxa traditionally assigned to lower tax-onomic ranks, such as subspecies and varieties, either are species, or they represent entirely differ-ent phenomena, such as morphs or artificial divisions in continuous patterns of geographic varia-tion. This consequence of the unified species concept has been foreshadowed by recent proposals, justified in the context of the evolutionary species concept (e.g., Simpson 1951, 1961; Wiley 1978, 1981), either to recognize former subspecies taxa as species (commonly subject to a secondary cri-terion of diagnosability) or to treat them as artificially defined parts of species (commonly as class-es of organisms sharing one or more necessary and sufficient character states). For example, Grismer (2002), dealing with species of amphibians and reptiles inhabiting Baja California, recog-nized no subspecies whatsoever; instead, he treated all previously recognized subspecies either as species or as artificial “pattern classes” (see also Cracraft 1983; Frost 1995). Another manifesta-tion of treating the species category as something other than a taxonomic rank is the view, to which I will return shortly, that the species category is fundamentally different from the other traditional taxonomic categories (e.g., Simpson 1961; Mayr 1969; see also de Queiroz 1997). Current Taxonomic Conventions are Inadequate All of the above conclusions suggest that traditional taxonomic practices are inadequate to accommodate a unified concept of species. Under such a concept, taxonomists will need to recog-nize many more species than are recognized in traditional taxonomies. They will need to recognize as species lineages that are separated now but may not be separated in the future. They will need to allow some species to be nested within other species (even if the former are not sister species). And they will need to stop treating the species category as a taxonomic rank. In short, taxonomists need new taxonomic conventions for representing the relationships among species, and they will also need new nomenclatural rules for governing the names of species. Equally importantly, biologists need to be able to distinguish clearly and consistently between species that possess different properties — including both those that have previously been adopted as secondary species criteria (such as quantitative and qualitative differences, exclusivity, mono-phyly, ecology, and various kinds of reproductive barriers, from ecological and behavioral differ-ences to incompatible genitalia and developmental systems) and those that have not (such as pop-ulation size, type of population structure, amount or nature of genetic and phenotypic variation, and DE QUEIROZ: A UNIFIED CONCEPT OF SPECIES 209 others). Biologists need to come to terms with the fact that no single property of species is suffi-cient to address all of the diverse questions that species are used to answer. This means that biolo-gists need to be able to identify — for any given study — those species possessing the property or properties that are relevant to answering the particular questions addressed in that study. Just as cer-tain questions about organisms (e.g., those related to mating behavior) can only be answered using particular kinds of organisms (e.g., sexually mature ones), similarly, certain questions about species (e.g., those related to the phenomenon of reinforcement) can only be answered using particular kinds of species (e.g., those exhibiting postmating reproductive barriers). A species taxonomy — or even a traditional species database that includes information on geographic distribution and organismal traits — simply is inadequate for identifying the relevant species to use in a particular study. What is required is a species database that includes information on the diverse properties of species. A SHIFT IN THE CONCEPTUALIZATION OF SPECIES Some people are likely to have difficulty accepting at least some of the consequences of a uni-fied species concept. Rather than trying to anticipate specific objections and presenting counter-arguments, I will instead present a general perspective explaining how the unified species concept that I have described in this paper represents the more complete acceptance of an idea that is already widely accepted. The idea in question is a manifestation of a shift in the way that biologists conceptualize the species category that was well underway at least a half-century ago and has con-tinued to gain ground, as indicated by the trends described in the previous section. Thus, even though some of the consequences of the unified species concept are at odds with taxonomic tradi-tions, the concept itself is not particularly radical. It simply represents the next stage in an ongoing historical process. In this section, I will describe this shift in the conceptualization of the species category and how it relates to my proposed solution to the species problem. Traditionally, the species category was one of the ranks in the hierarchy of taxonomic cate-gories — the familiar kingdom, phylum/division, class, order, family, genus, and species — devel-oped by Linnaeus (e.g., 1753, 1758) and other early naturalists. These categories were ranks that conveyed the relative inclusiveness of taxonomic groups (taxa): species were included within gen-era, genera within families, families within orders, and so forth. The taxa themselves, regardless of their rank, were all considered to be more or less the same kind of entities — groups of organisms sharing particular characters — some were just more or less inclusive than others. In other words, taxa assigned to the species category were not considered to differ fundamentally from those assigned to higher taxonomic categories; they were just smaller groups separated by smaller degrees of difference. Darwin (1859) held the view of the species category that I have just described, though he pro-vided an explanation both for the existence of the groups and for the differences among them. For Darwin, the species category was just another categorical rank — one that applied to groups of organisms differing more than varieties but less than genera. The following quotations from the Origin of Species illustrate Darwin’s (1859) views on the species category and its relationships to the other taxonomic categories. I look at the term species, as one arbitrarily given for the sake of convenience to a set of individ-uals closely resembling each other, . . . it does not essentially differ from the term variety, which is given to less distinct and more fluctuating forms (p. 52). the natural system . . . is genealogical in its attempted arrangement, with the grades of acquired difference marked by the terms varieties, species, genera, families, orders, and classes (p. 456). 210 PROCEEDINGS OF THE CALIFORNIA ACADEMY OF SCIENCES Volume 56, Supplement I, No. 18 Some time after Darwin, a fundamental change occurred in how biologists viewed the species category. This change came to the forefront during the period of the Modern or Evolutionary Synthesis (e.g., Huxley 1942; Mayr and Provine 1980) in the middle of the 20th Century and formed the basis of what was then called the New Systematics (e.g., Huxley 1940). During this time, a new general concept of species emerged that resulted in a decoupling of the species cate-gory from the rest of the taxonomic hierarchy (de Queiroz 1997). Under this new view, species were conceptualized as inclusive populations (e.g., Wright 1940; Mayr 1942; Dobzhansky 1950), or as ancestor-descendant lineages of such populations (e.g., Simpson 1951, 1961). As a conse-quence, the species category came to be viewed as differing fundamentally from the higher taxo-nomic categories. The species category was no longer viewed simply as a taxonomic rank applied to entities of the same basic kind as genera and families; instead, the species category came to des-ignate a particular kind of biological entity — the inclusive population or population lineage. In contrast, the higher taxonomic categories continued to be treated as ranks, which were now applied to more and less inclusive groups of species. The following quotations, from two of the great sys-tematic biologists of the Synthesis Era, give evidence of this new view of species: there are units in nature that have a special evolutionary status not fully shared with taxa either above or below them in the hierarchy . . . Many of them . . . recognized before Darwin had been called species, and it was inevitable that the term should be transferred to the evolutionary units (Simpson 1961). The unique position of species in the hierarchy of taxonomic categories has been pointed out by many authors . . . It is the only taxonomic category for which the boundaries between taxa at that level are defined objectively (Mayr 1969). This new view of species is perhaps epitomized in the statement by Mayr (1982:297) that “the species is as important a unit of biology as is the cell at a lower level of integration.” The unified species concept described in this paper represents the more complete acceptance of the general conceptual shift just described — the shift from viewing the species category as a rank in the hierarchy of taxonomic categories to viewing the species category as a natural kind rep-resenting the units at one of the fundamental levels of biological organization. Conversely, this newer view of species reinforces the solution to the species problem represented by the unified species concept. My point is that if biologists are going to accept Mayr’s proposition about species — if they are going to claim that the species is a fundamental category of biological organization, comparable to the categories cell and organism — then, to be consistent, they must adopt the uni-fied concept of species. More specifically, they must discontinue the practice of treating certain secondary properties of lineages as necessary properties of species. Requiring population lineages to be diagnosable, or monophyletic, or reproductively isolated before those lineages are considered species is, to use an organism level analogy, like requiring living beings to be born, or sexually mature, or fully grown before they are considered organisms. Such a view not only prevents biol-ogists from achieving a generally accepted definition of the species category, thus perpetuating the species problem, it also denies the species category the status of a fundamental category of biolog-ical organization and thus also of a truly central concept in biology. CONCLUSION The unified concept of species and the shift in the conceptualization of the species category that it represents bear on the history and the future of taxonomy. In one sense, taxonomy is among the oldest scientific disciplines. That is, taxonomy was among the earliest branches of human DE QUEIROZ: A UNIFIED CONCEPT OF SPECIES 211 knowledge to adopt explicit methods — to be approached systematically. In another sense, howev-er, taxonomy has only recently become a science. Although the discipline of taxonomy has exist-ed for a very long time, it has only recently experienced a shift from being primarily concerned with the utilitarian exercise of classifying to being primarily concerned with the scientific endeav-or of testing hypotheses. Historically, taxonomists have been concerned with classifying organisms into groups based on shared traits, and then further classifying those groups into the categories of the taxonomic hierarchy, from kingdom to species. In contrast, modern systematic biologists, despite the fact that they still use data taking the same basic form of similarities and differences among organisms, are increasingly devoting their efforts to testing hypotheses about lineage boundaries and phylogenetic relationships. An important manifestation of this shift is the increas-ing realization that the categories of greatest importance to taxonomists are not kingdom, phylum/division, class, order, family, genus, and species (the last term being used here in the older sense of a taxonomic rank) — the important categories are clade and species (the second term now used in the newer sense of a category of biological organization). To the extent that the unified species concept represents the more complete acceptance of this newer view of species, it repre-sents a central component in the future of taxonomy. ACKNOWLEDGMENTS This paper is based on my Past-President’s Address for the Society of Systematic Biologists (Champaign-Urbana, Illinois 2002), which was subsequently modified for the symposium “The Future of Taxonomy” celebrating the 150th anniversary of the California Academy of Sciences (San Francisco, California 2003). I would like to thank Stan Blum and Nina Jablonski for inviting me to speak in the 150th Anniversary Symposium and to contribute this paper to the symposium proceedings. Michael Ghiselin coordinated the reviews and Benoit Dayrat and an anonymous reviewer provided valuable comments on an earlier version of the paper. I also wish to express my appreciation for the California Academy of Sciences as an institution. I held a Tilton Postdoctoral Fellowship at the Academy from 1989 to 1991 in the Department of Herpetology. My time at the Academy was both a thoroughly enjoyable experience and an important part of my professional development. 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6470	https://web.math.princeton.edu/~nalon/PDFS/kriv4.pdf	Constructive bounds for a Ramsey-type problem Noga Alon ∗ Michael Krivelevich † Abstract For every ﬁxed integers r, s satisfying 2 ≤r < s there exists some ϵ = ϵ(r, s) > 0 for which we construct explicitly an inﬁnite family of graphs Hr,s,n, where Hr,s,n has n vertices, contains no clique on s vertices and every subset of at least n1−ϵ of its vertices contains a clique of size r. The constructions are based on spectral and geometric techniques, some properties of Finite Geometries and certain isoperimetric inequalities. 1 Introduction The Ramsey number R(s, t) is the smallest integer n such that every graph on n vertices contains either a clique Ks of size s or an independent set of size t. The problem of determining or estimating the function R(s, t) received a considerable amount of attention, see, e.g., and some of its references. A more general function was ﬁrst considered (for a special case) by Erd˝ os and Gallai in . Suppose 2 ≤r < s ≤n are integers, and let G be a Ks-free graph on n vertices. Let fr(G) denote the maximum cardinality of a subset of vertices of G that contains no copy of Kr, and deﬁne, following , : fr,s(n) = min fr(G), where the minimum is taken over all Ks-free graphs G on n vertices. It is easy to see that for r = 2, we have f2,s(n) < t if and only if the Ramsey number R(s, t) satisﬁes R(s, t) > n, showing that the problem of determining the function fr,s(n) extends that of determining R(s, t). ∗Department of Mathematics, Raymond and Beverly Sackler Faculty of Exact Sciences, Tel Aviv Uni-versity, Tel Aviv, Israel. Email: noga@math.tau.ac.il. Supported in part by the Fund for Basic Research administered by the Israel Academy of Sciences. †Department of Mathematics, Raymond and Beverly Sackler Faculty of Exact Sciences, Tel-Aviv Uni-versity, Tel-Aviv, Israel. Email: krivelev@math.tau.ac.il. Supported in part by a Charles Clore Fellowship. 1 Erd˝ os and Rogers combined a geometric idea with probabilistic arguments and showed that fs−1,s(n) ≤O(n1−1/O(s4 log s)). This bound has been improved in several subsequent papers , , and the best known bounds, proved in , , are c1n 1 s−r+1 (log log n)1− 1 s−r+1 ≤fr,s(n) ≤c2n r s+1 (log n) 1 r−1 , where c1, c2 are positive constants depending only on r and s. Note that to place an upper bound on fr,s(n) one has to prove the existence of a graph with certain properties. As is the case with the problem of bounding the usual Ramsey numbers, the existence of these graphs is usually proved by probabilistic arguments. In fact there is no known explicit construction that provides any non-trivial upper bound for fr,s(n) for any value of r other than 2. By explicit we mean here a construction that supplies a deterministic algorithm to construct a graph with the desired properties in time polynomial in the size of the graph. It is worth noting that for the case r = 2, corresponding to the usual Ramsey numbers, there are several known explicit constructions; see , , , , , . Despite a considerable amount of eﬀort, all these constructions supply bounds that are inferior to those proved by applying probabilistic arguments. The problem of ﬁnding explicit constructions matching the best known bounds is of great interest, and may have algorithmic applications as well. In the present note we describe two diﬀerent explicit constructions providing nontrivial upper bounds for the function fr,s(n) in the case r > 2. The ﬁrst one is based on a spectral technique together with some of the properties of ﬁnite geometries and implies that for every ﬁxed r, s we have: fr,s(n) = O n 1 2 + 2r−3 2s−4 . The second construction is based on a geometric idea and certain isoperimetric inequal-ities, and shows that for every s ≥2, fs,s+1(n) ≤n1−ϵ(s), where ϵ(s) = (1 + o(1)) 2 s2(s + 1)2 ln s and the o(1) term tends to 0 as n tends to inﬁnity. Both constructions are explicit according to all common deﬁnitions of this notion and, in particular, provide a linear time deterministic algorithm to construct the appropriate graph as well as an algorithm that determines if two given vertices are connected using a constant number of arithmetic or bit operations on words of length O(log n), where n is the number of vertices. In the rest of this note we describe these two constructions and prove their properties. 2 2 The ﬁrst construction The ﬁrst construction we present applies ﬁnite geometries and the proof of its properties is based on the spectral technique used in for a similar purpose, together with some additional ideas. Graphs considered in this section may have loops. Each loop contributes one to the degree of a vertex incident to it and contributes 1/2 when we count the number of edges spanned by a set of vertices. We need the following lemma. Lemma 2.1 Let G be a d-regular graph on n vertices with at most one loop at each vertex and suppose that the absolute value of any eigenvalue of G but the ﬁrst is at most λ. For every integer r ≥2 denote sr = (λ + 1)n d 1 + n d + · · · + n d r−2! . Then every set of more than sr vertices of G contains a copy of Kr. Proof. The proof relies on the following simple statement proved (in a slightly stronger form) in (see also , Chapter 9, Corollary 2.6.) Proposition 2.2 Let G = (V, E) be a d-regular graph on n vertices (with loops allowed) and suppose that the absolute value of each of its eigenvalues but the ﬁrst is at most λ. Let B be an arbitrary subset of bn vertices of G and let e(B) denote the number of edges in the induced subgraph of G on B. Then \|e(B) −1 2b2dn\| ≤1 2λbn . To deduce the lemma from the above proposition we apply induction on r. Note that if S is a subset of vertices of size k, then, by the proposition above, e(S) ≥1 2 k2d n −1 2λk. If r = 2 and k > (λ + 1)n/d then e(S) ≥1 2k kd n −λ > 1 2k, and therefore S contains at least one non-loop edge, as needed. Assuming the assertion of the lemma holds for all integers between 2 and r we prove it for r +1 ( ≥3). Since e(S) ≥1 2 k2d n −1 2λk, there exists a vertex v ∈S which is incident with at least kd n −λ edges in S, implying that v has at least kd n −λ −1 neighbours in S other than itself. Let N denote the set of all these neighbours. By the induction hypothesis if kd n −λ −1 > sr, then N contains a copy of Kr, that together with v forms a copy of Kr+1. Hence, any set of more than (sr + λ + 1)n d = (λ + 1)n d 1 + n d + · · · + n d r−1! = sr+1 3 vertices of G contains a copy of Kr+1, completing the proof. 2 For any integer t ≥2 and for any power q = 2g of 2 let PG(t, q) denote the ﬁnite geometry of dimension t over the ﬁeld GF(q). The interesting case for our purposes here is that of ﬁxed t and large q. It is well known (see, e.g., ) that the points and hyperplanes of PG(t, q) can be described as follows. Let Bt denote the set of all nonzero vectors ¯ x = (x0, . . . , xt) of length t + 1 over GF(q) and deﬁne an equivalence relation on Bt by calling two vectors equivalent if one is a multiple of the other by an element of the ﬁeld. The points of PG(t, q) as well as the hyperplanes can be represented by the equivalence classes of Bt with respect to this relation, where a point ¯ x = (x0, . . . , xt) lies in the hyperplane ¯ y = (y0, . . . , yt) if and only if their inner product ⟨x, y⟩= x0y0+. . .+xtyt over GF(q) is zero. Let G(t, q) denote the graph whose vertices are the points of PG(t, q), where two (not necessarily distinct) vertices ¯ x and ¯ y as above are connected by an edge if and only if ⟨x, y⟩= x0y0 + . . . + xtyt = 0, that is, the point represented by ¯ x lies on the hyperplane represented by ¯ y. The graphs G(t, q) have been considered by several authors - see, e.g., , . It is easy to see that the number of vertices of G(t, q) is nt,q = (qt+1 −1)/(q −1) = qt(1 + o(1)) and that it is dt,q = (qt −1)/(q −1) = qt−1(1 + o(1))– regular, where here and in what follows the o(1) term tends to zero as q tends to inﬁnity. It is also easy to see that the number of vertices of G(t, q) with loops is precisely dt,q = (qt−1)/(q−1), since the equation x2 0+. . .+x2 t = 0 over GF(q) is equivalent to the linear equation x0+. . .+xt = 0, which has exactly qt−1 nonzero solutions. This is the only place we use the fact that the ﬁeld GF(q) is of characteristic 2. A similar construction exists for any prime power q, but the computation in this case is (slightly) more complicated. We claim that each copy of Kt+2 in G(q, t) contains at least one vertex represented by a vector ¯ x with ⟨¯ x, ¯ x⟩= 0. Indeed, let ¯ x1, . . . , ¯ xt+2 be vectors of PG(t, q) such that ⟨¯ xi, ¯ xj⟩= 0 for every pair 1 ≤i ̸= j ≤t + 2. Since these are t + 2 vectors in a vector space of dimension t + 1 there are ν1, . . . , νt+2 ∈GF(q) which are not all zero, such that ν1¯ x1 + . . . νt+2¯ xt+2 = ¯ 0. Multiplying this equality by ¯ xi for every 1 ≤i ≤t + 2, we conclude that νi⟨¯ xi, ¯ xi⟩= 0, and hence there exists at least one index i such that ⟨¯ xi, ¯ xi⟩= 0. The eigenvalues of G(t, q) are known and easy to compute. To see this, let A be the adjacency matrix of G(t, q). By the properties of PG(t, q), A2 = AAT = µJ + (dt,q −µ)I, where µ = (qt−1 −1)/(q −1), J is the nt,q × nt,q all 1-s matrix and I is the nt,q × nt,q identity matrix. Therefore the largest eigenvalue of of A2 is d2 t,q and all other eigenvalues are dt,q −µ. It follows that the largest eigenvalue of A is dt,q and the absolute value of all other eigenvalues is (dt,q −µ)1/2. Put λ = (dt,q −µ)1/2 = q(t−1)/2. By Lemma 2.1 every set of λ(nt,q/dt,q)r−1(1 + o(1)) = n 1 2 + 2r−3 2t t,q (1 + o(1)) vertices spans a copy of Kr. Let H(t, q) denote the graph obtained from G(t, q) by deleting all vertices represented by vectors ¯ x with ⟨¯ x, ¯ x⟩= 0. This is a Kt+2-free graph on n = qt vertices, for which the 4 following holds: Theorem 2.3 For every ﬁxed t and large enough q the graph H(t, q) on n = qt vertices has the following properties: 1. Kt+2 ̸⊂G. 2. For every r ≥2 every set of n0(r, t) vertices spans a copy of Kr, where n0(r, t) = n 1 2+ 2r−3 2t (1 + o(1)). To get a result for a general n, we can, for example, start with H(t, q), where q is the minimal integer of the form q = 2g for which qt ≥n. Then 2tn ≥qt. Now delete from H(t, q) an arbitrary subset of qt −n vertices, thus obtaining a graph on n vertices with the desired properties. This gives Corollary 2.4 By an explicit construction for every ﬁxed r, t satisfying 2 ≤r < (t + 1)/2, and for every n, fr,t(n) = O n 1 2+ 2r−3 2t−4 . 3 The second construction The second construction we present borrows the core idea from the construction of Erd˝ os and Rogers . Their argument is non-constructive and relies on the concentration of measure phenomenon in the high-dimensional sphere. Our construction is explicit and somewhat simpler to handle. We begin with some notation. Let s ≥2 and k be positive integers (where s is assumed to be ﬁxed while k tends to inﬁnity). Denote V = [s]k. Thus, the elements of V are vectors ¯ x of length k. We endow V with the normalized counting measure P, that is, P(A) = \|A\|/\|V \| for every subset A ⊆V . For every two vectors ¯ x, ¯ y ∈V denote by d(¯ x, ¯ y) the Hamming distance between ¯ x and ¯ y, that is, d(¯ x, ¯ y) = \|{1 ≤i ≤k : xi ̸= yi}\| . Also, for a set ∅̸= U ⊆V and a vector ¯ x ∈V let d(¯ x, U) = min{d(¯ x, ¯ y) : ¯ y ∈U} denote the distance between ¯ x and U. For every integer δ > 0 deﬁne the δ-neighbourhood U(δ) of a nonempty subset U ⊆V as U(δ) = {¯ x ∈V : d(¯ x, U) ≤δ} ; 5 thus U(0) = U. Deﬁne a graph G = G(s, k) as follows. The vertex set of G is V , and two vectors ¯ x, ¯ y ∈V are connected by an edge in G if and only if d(¯ x, ¯ y) > k 1 − s+1 2 −1 . Let us investigate the properties of this graph. Proposition 3.1 The graph G does not contain a copy of Ks+1. Proof. Suppose indirectly that ¯ x1, . . . , ¯ xs+1 are the vertices of Ks+1 ⊆G, then according to the deﬁnition of G we have d(¯ xi1, ¯ xi2) > k(1− s+1 2 −1) for every pair 1 ≤i1 ̸= i2 ≤s+1. For every 1 ≤j ≤k, there exists at least one pair of vertices of Ks+1 having the same value in the j-th coordinate. Therefore, summing over all k coordinates and averaging, we obtain that there exists at least one pair of vectors ¯ xi1, ¯ xi2, that agree on at least k s+1 2 −1 coordinates. Thus d(¯ xi1, ¯ xi2) ≤k −k s+1 2 −1, supplying the desired contradiction. 2 We next prove that every suﬃciently large subset of V spans a copy of Ks. Deﬁne an s-simplex S to be a set of s vectors ¯ x1, . . . , ¯ xs ∈V with d(¯ xi1, ¯ xi2) = k for every pair 1 ≤i1 ̸= i2 ≤s. Proposition 3.2 If V0 ⊆V and P[V0] > (s−1)/s, then V0 contains a copy of an s-simplex. Proof. Clearly, each vector of V lies in the same number of s-simplices. Choose randomly and uniformly a simplex S among all s-simplices in V . Then P[S ̸⊂V0] = P[ at least one vertex of S does not belong to V0 ] ≤ s\|V \| −\|V0\| \|V \| < s 1 −s −1 s = 1. Hence there exists at least one s-simplex S with all vertices in V0. 2 The next step is to obtain a good isoperimetric inequality for the ﬁnite metric space (V, d). We use martingales as in, e.g., , , . Lemma 3.3 For c > 0 denote δ(c) = ⌈( p ln s/2 + c) √ k⌉. If A is a subset of V with P[A] ≥1/s, then P[A(δ)] > 1 −e−2c2. Proof. Deﬁne a function f : V →R by f(¯ x) = d(¯ x, A). This function is clearly Lipschitz with constant 1, that is, \|f(¯ x) −f(¯ y)\| ≤d(¯ x, ¯ y) for all ¯ x, ¯ y ∈V . Also, f(¯ x) = 0 for every ¯ x ∈A. Let Ef = X0, X1, . . . , Xk = f be the coordinate exposure martingale with respect 6 to f, that is, Xi(¯ x) = E[f(¯ y) : ¯ y ∈V : ¯ yj = ¯ xj ∀j ≤i]. Hence Hoeﬀding’s inequality (see, e.g. Lemma 1.2 of ) implies: P[Xk −X0 < −c √ k] < e−2c2, (1) P[Xk −X0 > c √ k] < e−2c2 (2) for all c > 0. In particular, substituting c = p ln s/2 in (1) and recalling that P[A] ≥1/s, we see that there exists at least one point ¯ x ∈A, for which Xk(¯ x) −X0(¯ x) = f(¯ x) −X0 ≥− q ln s/2 √ k . However, since ¯ x ∈A, one has Xk(¯ x) = 0, and therefore X0 ≤ p ln s/2 √ k. Thus (2) implies that P[f > ( q ln s/2 + c) √ k] < e−2c2 . The left-hand side of the above inequality is at least P[V \ A(δ)] with δ = δ(c) as in the formulation of the lemma, and hence P[A(δ)] > 1 −e−2c2 . 2 Remark. McDiarmid gives in an isoperimetric inequality for graph product spaces (see , Prop. 7.12), implying directly our Lemma 3.3. We chose however to present its proof here for the sake of completeness. Moreover, it is possible to improve the inequality and obtain an asymptotically tight isoperimetric inequality. This and related results will appear in . For our purpose here the present estimate suﬃces. The result of the lemma can be reformulated in the following more convenient way: for every c > 0, if U ⊆V and P[U] ≥e−2c2, then P[U(δ)] > (s −1)/s. Indeed, assuming P[U(δ)] ≤(s−1)/s, denote W = V \U(δ), then P[W] ≥1/s and therefore P[W(δ)] > 1−e−2c2, contradicting the fact that W(δ) ∩U = ∅. Deﬁne c = √ k 2 s+1 2 − q ln s/2 −1 = √ k s(s + 1)(1 + o(1)) , (3) where the o(1) tends to 0 as k tends to inﬁnity. Then if P[U] ≥e−2c2, then P[U(δ)] > (s −1)/s, where δ = δ(c). Therefore, by Proposition 3.2, the set U(δ) contains an s-simplex S. Let ¯ x1, . . . , ¯ xs denote its vertices. Let ¯ yi ∈U satisfy d(¯ xi, ¯ yi) ≤δ(c), where 1 ≤i ≤s. By the triangle inequality d(¯ yi1, ¯ yi2) ≥ d(¯ xi1, ¯ xi2) −2δ(c) 7 > k − k s+1 2 + 2 √ k −2 ≥ k 1 − 1 s+1 2 ! . This means that the vertices ¯ y1, . . . , ¯ ys ∈U form a copy of Ks. Recalling (3), we see that every subset U ⊆V of size \|U\| = \|V \| e−2c2 = \|V \| e−(1+o(1))2k/(s2(s+1)2) spans a copy of Ks. Summing up the above, we obtain the following Theorem 3.4 For every ﬁxed s and large k the graph G = G(s, k) described above is a graph on n = sk vertices, having the following properties: 1. Ks+1 ̸⊂G; 2. Every set of at least n1−ϵ(s) vertices spans a copy of Ks, where ϵ(s) = (1+o(1))2/(s2(s+ 1)2 ln s) and the o(1) term tends to 0 as k tends to inﬁnity. Corollary 3.5 By an explicit construction, for every ﬁxed s ≥2, fs,s+1(n) ≤n1−ϵ(s) , where ϵ(s) = (1 + o(1))2/(s2(s + 1)2 ln s). Remarks. 1. For the case s = 2 the above construction coincides with the construction of Erd˝ os , proving a lower bound for the Ramsey number R(3, t). In this case instead of applying our approach based on isoperimetric inequalities one can calculate exactly the maximum size of a subset of vertices of G not containing an edge K2, using the well-known result of Kleitman . 2. For general s, our estimate on ϵ(s) is better by a factor of 4 than the one of Erd˝ os and Rogers. 3. A similar construction, requiring somewhat more complicated analysis, yields an im-provement of the expression for ϵ(s) by a logarithmic factor. Below we give an outline of the argument, leaving all technical details to the reader. For a given s, deﬁne t = t(s) = min{2 ≤i ≤s : s(mod i) < i/2} . It can be shown that t(s) is at most polylogarithmic in s. Put s = tq + r. It follows from the deﬁnition of t that r < t/2. Let V = [t]k. Deﬁne α1(s) = 1 −(r + 1) q+1 2 + (t −r −1) q 2 s+1 2 , α2(s) = 1 −r q+1 2 + (t −r) q 2 s 2 . 8 It is easy to see that α1(s)k is an upper bound for the minimum Hamming distance between a pair in any family of s + 1 vectors in V . Similarly, α2(s)k is an upper bound for the minimum Hamming distance between a pair in any family of s vectors in V . Note that since r < t/2 we have α2(s) −α1(s) = 2q(t −r −1) (s −1)s(s + 1) ≥ 1 2s(s + 1) . Now deﬁne a graph G with vertex set V by joining two vectors ¯ x, ¯ y ∈V by an edge if their Hamming distance exceeds α1(s)k. Then G is clearly Ks+1-free. Deﬁne an s-simplex to be a family of s vectors in G, whose mutual Hamming distances are all equal to α2(s)k. Assuming ”nice” divisibility properties of k, we can claim that such an s-simplex indeed exists and that every s−1 s \|V \| vertices of G span an s-simplex. Now the same argument as in the proof of Theorem 3.4 shows that every \|V \|e−ck/s4 = n1−c/s4 ln t = n1−c′/s4 ln ln s vertices of G span a copy of Ks, where n = \|V \| and c, c′ are some absolute constants. This gives the bound fs,s+1(n) ≤n1− c0 s4 ln ln s , where c0 is an absolute constant. 4. The idea applied in the construction of Theorem 3.4 can be used also for obtaining constructive upper bounds for the function fr,s(n) for values of r other than s −1. The bounds obtained (as well as the above bound for fs−1,s(n)) are considerably weaker that the ones proved in by probabilistic arguments. It would be interesting to ﬁnd explicit examples providing bounds closer to the last ones. Acknowledgement. We would like to thank Benny Sudakov and an anonymous referee for many helpful comments. References N. Alon, Eigenvalues, geometric expanders, sorting in rounds, and Ramsey theory, Combinatorica 6 (1986), 207–219. N. Alon, Tough Ramsey graphs without short cycles, J. Algebraic Combinatorics 4 (1995), 189–195. N. Alon, Explicit Ramsey graphs and orthonormal labelings, The Electronic J. Combi-natorics 1 (1994), R12, 8pp. N. Alon, R. B. Boppana and J. H. Spencer, An asymptotic isoperimetric inequality, in preparation. 9 N. Alon and F. R. K. Chung, Explicit construction of linear sized tolerant networks, Discrete Math. 72 (1988), 15-19. N. Alon and J. H. Spencer, The Probabilistic Method, Wiley, New York, 1992. B. Bollob´ as, Random Graphs, Academic Press, London, 1985. B. Bollob´ as and H. R. Hind, Graphs without large triangle-free subgraphs, Discrete Math. 87 (1991), 119–131. F. R. K. Chung, R. Cleve and P. Dagum, A note on constructive lower bounds for the Ramsey numbers R(3, t), J. Comb. Th. Ser. B 57 (1993), 150–155. P. Erd˝ os, On the construction of certain graphs, J. Comb. Th. 1 (1966), 149–153. P. Erd˝ os and T. Gallai, On the minimal number of vertices representing the edges of a graph, Publ. Math. Inst. Hungar. Acad. Sci 6 (1961), 181–203. P. Erd˝ os and C. A. Rogers, The construction of certain graphs, Canad. J. Math. 14 (1962), 702–707. P. Frankl and R.M. Wilson, Intersection theorems with geometric consequences, Com-binatorica 1 (1981), 357–368. R.L. Graham, B.L. Rothschild and J. H. Spencer, Ramsey Theory (Second Edition), Wiley, New York, 1990. M. Hall, Combinatorial Theory (Second Edition), Wiley, New York, 1986. D. J. Kleitman, On a combinatorial problem of Erd˝ os, J. Comb. Th. 1 (1966), 209–214. M. Krivelevich, Ks-free graphs without large Kr-free subgraphs, Comb., Prob. and Computing 3 (1994), 349–354. M. Krivelevich, Bounding Ramsey numbers through large deviation inequalities, Ran-dom Str. Alg. 7 (1995), 145–155. C. J. H. McDiarmid, On the method of bounded diﬀerences, in Surveys in Combina-torics 1989, London Math. Society Lecture Notes Series 141 (Siemons J., ed.), Cam-bridge Univ. Press (1989), 148–188. V. D. Milman and G. Schechtman, Asymptotic Theory of Finite Dimensional Normed Spaces, Lecture Notes in Mathematics 1200, Springer Verlag, Berlin and New York, 1986. 10 M. Talagrand, Concentration of measure and isoperimetric inequalities in product spaces, Inst. Hautes ´ Etudes Sci. Publ. Math. 81 (1995), 73–205. 11
6471	https://create.kahoot.it/details/understanding-scientific-notation/749e3309-c33f-4ebf-8b1b-ff12455e6874	Understanding Scientific Notation - Details - Kahoot! New page Understanding Scientific Notation - Details - Kahoot! Jump to main content < Back to kahoot.com < kahoot.comLog inSign up FREE Discover Learn Present Make Join Game Sign up FREE MrThompsonBUSD Verified profile Kahoot Understanding Scientific Notation Dive into the world of scientific notation with this engaging quiz! Test your knowledge on converting numbers to and from scientific notation, and discover how this mathematical tool simplifies calcul…Show more 13 plays 121 participants Kahoot session Host live Display on a big screen Host live Host live Assign Send to participants Assign Assign Kahoot self-study Learn Study on your own Learn Learn Flashcards Memorize and master Flashcards Flashcards Play solo Self-paced quiz games Play solo Play solo Updated 1 week ago• Visibility: Public Questions (20) [x] Show answers Show items as list, grid or compact list Show items in a list Show items in a grid Show items in a compact list what is 363000 in scientific notation? which field uses scientific notation often? what is 0.0044 in scientific notation? True or false: Scientific notation simplifies large numbers. what is 0.007 in scientific notation? what is 0.086 in scientific notation? what is 976 in scientific notation? what is 46500 in scientific notation? how does scientific notation help in calculations? what is 8.2 x 10-3 in standard notation? why is scientific notation important? what is 7.04 x 10 0 in standard notation? which number is larger: 1 x 10 6 or 1 x 10 5? what is 3.7 x 10-5 in standard notation? what is 3.7 x 10-2 in standard notation? what is 8.99 x 10 5 in standard notation? 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6472	https://discountexhaustcompany.com/catalytic-converters-key-to-cleaner-air-and-why-you-should-maintain-them/	Skip to content Address:86 Washington St. Unit F1 Plainville MA 02762☎️ 508-316-1982 Free Exhaust System Inspection & Estimate Catalytic Converters: Key to Cleaner Air and Why You Should Maintain Them View Larger Image Catalytic converters are an essential component in modern vehicles, playing a crucial role in reducing harmful emissions and protecting our environment. These devices have been mandated in cars since the 1970s, and their importance has only grown as environmental regulations become stricter. Understanding how catalytic converters work and why they are vital can help everyone appreciate their role in keeping the air we breathe cleaner. What Is a Catalytic Converter? A catalytic converter is a device located in a vehicle’s exhaust system. Its primary function is to convert harmful pollutants produced by the engine into less harmful emissions before they are released into the atmosphere. Catalytic converters contain a honeycomb structure coated with precious metals like platinum, palladium, and rhodium, which act as catalysts to trigger chemical reactions that reduce emissions. The catalytic converter works by converting three main harmful compounds found in a vehicle’s exhaust gases: Carbon Monoxide (CO) – A poisonous gas that can be deadly when inhaled in large quantities. Hydrocarbons (HC) – A contributor to smog, resulting from unburned fuel. Nitrogen Oxides (NOx) – A cause of acid rain and smog, contributing to respiratory problems. The catalytic converter breaks these compounds down into less harmful gases like carbon dioxide (CO₂), nitrogen (N₂), and water vapor (H₂O), significantly reducing the environmental impact of vehicle emissions. Why Are Catalytic Converters Important? Environmental Protection Catalytic converters are essential in the fight against air pollution. Vehicles that lack functional catalytic converters release large amounts of dangerous pollutants into the air, contributing to smog formation and worsening climate change. By converting harmful emissions into less damaging substances, catalytic converters help protect the environment and improve air quality. Compliance with Emission Standards In most parts of the world, vehicles are required to meet strict emission standards, which are enforced to reduce air pollution. Without a catalytic converter, a vehicle would not be able to pass emissions tests or meet legal requirements, making it illegal to drive on public roads. Maintaining a properly functioning catalytic converter ensures that a vehicle stays compliant with these standards, avoiding fines and penalties. Health Benefits Reduced vehicle emissions mean cleaner air, which can have significant health benefits, especially in urban areas where traffic is dense. Air pollution has been linked to respiratory issues such as asthma, bronchitis, and even cardiovascular disease. Catalytic converters help minimize the impact of vehicle emissions on public health by decreasing the amount of harmful gases released into the air. Vehicle Efficiency While catalytic converters are primarily designed to reduce emissions, they can also contribute to the overall efficiency of a vehicle’s exhaust system. A well-maintained catalytic converter helps ensure that your vehicle runs smoothly and that its engine operates at optimal efficiency, potentially improving fuel consumption and engine performance. The Growing Problem of Catalytic Converter Theft Due to the valuable metals used in their construction, catalytic converters have become a target for theft. Thieves often steal catalytic converters to sell the precious metals, causing significant inconvenience and expense for vehicle owners. Replacing a catalytic converter can be costly, and without one, the vehicle becomes non-compliant with emission standards. To combat this growing problem, vehicle owners can take preventive measures such as parking in well-lit areas, installing catalytic converter shields, and using vehicle alarm systems designed to detect tampering. Law enforcement agencies and governments have also increased efforts to crack down on catalytic converter theft by enforcing stricter regulations on the resale of the stolen components. Maintaining Your Catalytic Converter To ensure that your catalytic converter continues to function effectively, regular maintenance is key. If your car begins to exhibit signs of a failing catalytic converter, such as reduced engine performance, the smell of sulfur (rotten eggs), or a check engine light, it’s important to get the issue diagnosed and repaired quickly. Keeping your catalytic converter in good condition will not only help your vehicle run more efficiently but will also ensure that it remains compliant with emission laws. Conclusion Catalytic converters are an indispensable part of modern vehicles, helping to reduce the environmental impact of driving by cutting down on harmful emissions. As we continue to strive for cleaner air and a healthier planet, the role of catalytic converters will remain crucial in the battle against air pollution. Vehicle owners should prioritize the care and maintenance of their catalytic converters, not only to comply with emission regulations but to contribute to a cleaner, healthier environment for everyone. 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6473	https://www.sciencedirect.com/topics/medicine-and-dentistry/lung-adenocarcinoma	Lung Adenocarcinoma - an overview \| ScienceDirect Topics Skip to Main content Journals & Books Lung Adenocarcinoma In subject area:Medicine and Dentistry NSCLC is defined as a category of lung cancer characterized by a complex genomic landscape with numerous genetic and epigenetic alterations involved in carcinogenesis, including somatic mutations, chromosomal rearrangements, and variations in oncogenic protein fusions. It primarily includes the subtypes adenocarcinoma (ADC) and squamous cell carcinoma (SCC), which exhibit distinct molecular alterations despite some overlap. AI generated definition based on:Archives of Medical Research, 2020 How useful is this definition? Press Enter to select rating, 1 out of 3 stars Press Enter to select rating, 2 out of 3 stars Press Enter to select rating, 3 out of 3 stars About this page Add to MendeleySet alert Also in subject area s: Nursing and Health Professions Pharmacology, Toxicology and Pharmaceutical Science Discover other topics 1. On this page On this page Definition Chapters and Articles Related Terms Recommended Publications Featured Authors On this page Definition Chapters and Articles Related Terms Recommended Publications Featured Authors Chapters and Articles You might find these chapters and articles relevant to this topic. Review article Bronchioloalveolar Carcinoma in Small Pulmonary Nodules: Clinical Relevance 2005, Seminars in Thoracic and Cardiovascular SurgeryNasser K. Altorki MD, ... Claudia I. Henschke PhD, MD The WHO Classification defines lung adenocarcinoma as a malignant epithelial neoplasm with glandular differentiation or mucin production. Lung adenocarcinoma is further classified based on histologic pattern of growth into subtypes that include acinar, papillary, bronchioloalveolar, solid with mucin production, or a mixture of these growth patterns.9 The diagnosis of BAC is restricted to cases in which the neoplastic cells grow along preexisting alveolar structures (bronchioloalveolar or lepidic growth pattern) without stromal, vascular, or pleural invasion. View article Read full article URL: Journal2005, Seminars in Thoracic and Cardiovascular SurgeryNasser K. Altorki MD, ... Claudia I. Henschke PhD, MD Review article Lung Cancer 2016, Surgical Oncology Clinics of North AmericaMin Zheng MD, PhD Adenocarcinoma Adenocarcinoma is the most common type of lung cancer, accounting for more than 40% of lung cancers, 60% of the NSCC, and more than 70% of surgically resected cases.3,4 The incidence of adenocarcinoma has risen steadily over the past few decades. Lung adenocarcinoma commonly forms a peripherally located mass with central fibrosis and pleural puckering. It can also have a variety of other gross appearances, including centrally located mass, diffuse lobar consolidation, bilateral multinodular distribution, and pleural thickening. By definition, lung adenocarcinoma is a malignant epithelial neoplasm with glandular differentiation or mucin production. When such morphologic features are recognized, the tumor can be designated as adenocarcinoma, even in small biopsy specimens. Lung adenocarcinoma cells usually express pneumocytic markers. Thyroid transcription factor (TTF-1) and NapsinA are expressed in more than 85% of the lung adenocarcinoma cases and thus can serve as markers of adenocarcinoma or adenocarcinoma differentiation in poorly differentiated tumor and in limited biopsy sampling material (Fig.1).5–7Tumor classification based on ancillary tests such as immunohistochemistry (IHC) is designated as “NSCC, favor adenocarcinoma” in a small biopsy specimen. Resection specimens allow a more detailed subclassification. There has been significant refinement in adenocarcinoma classification in recent years based on close pathologic and clinical correlation.3,8 The major histologic types have been validated to bear prognostic significance delineated by the tumor grade.9–14 Multiple gene alterations can occur in adenocarcinomas, with approved molecular targeted therapy available to improve patient survival (see later disccusion). Sign in to download hi-res image Fig.1. Adenocarcinomas. (A) Lepidic adenocarcinoma (H&E, original magnification ×200). (B) Acinar adenocarcinoma (H&E, original magnification ×200). (C) Papillary adenocarcinoma (H&E, original magnification ×100). (D) Micropapillary adenocarcinoma (H&E, original magnification ×200). (E) Solid adenocarcinoma (H&E, original magnification ×100). (F) Solid adenocarcinoma (TTF-1 stain, original magnification ×100). Preinvasive or Minimally Invasive Adenocarcinoma Adenocarcinoma in situ (AIS) represents relatively small sized tumors (≤3 cm) with neoplastic cells growing along preexisting alveolar structures (lepidic growth pattern) without evidence of stromal, vascular, or pleural invasion. Lepidic is a descriptive term for rind or membranous growth pattern and is now specifically used to describe tumor cells proliferating along the surface of intact alveolar walls.15 Lepidic growth usually correlates with ground glass opacity on radiograms. Most AISs are the nonmucinous type, with mild to moderate pleomorphic cuboidal to columnar tumor cells linearly lining alveolar walls. There is no secondary papillary or micropapillary growth pattern. A minority of such tumors are of mucinous or mixed type. If the tumor contains a small focus (<5 mm) of invasive growth, the tumor is classified microinvasive adenocarcinoma (MIA). Invasion usually induces formation of a desmoplastic stroma. Invasion can also manifest as nonlepidic growth, such as acinar, papillary, micropapillary, or solid patterns. MIA is defined not only by limited size invasive growth but also by a lack of more advanced invasive pattern, such as tumor necrosis, lymphovascular invasion, and pleural invasion. Both types of tumor are low grade and have a nearly 100% 5-year survival rate.3 Invasive Adenocarcinoma Most invasive adenocarcinomas are composed of mixed morphological subtypes; these are classified according to the predominant architectural structures rather than lumped together as mixed subtype. Each tumor is classified according to one predominant growth pattern, including lepidic, acinar, papillary, micropapillary, and solid patterns (see Fig.1). Each additional subpattern is recorded semiquantitatively as estimated percentage in 5% increments.3,8 This architectural-driven classification has prognostic significance, with most favorable prognosis for lepidic-predominant adenocarcinomas, intermediate survival rate for acinar and papillary predominant adenocarcinomas, and poor prognosis for solid and micropapillary predominant adenocarcinomas.11,12,16 Lepidic Adenocarcinoma Lepidic growth is commonly seen in lung adenocarcinoma. The lepidic growth pattern denotes tumor cells spreading along preexisting alveolar structures, although there may be sclerotic thickening of alveolar septa. When it is the predominant growth pattern with additional findings that set it apart from previously described AIS and MIA, it is designated as lepidic adenocarcinoma. These additional findings include any one or more of the following: more than 5 mm of invasion (presence of desmoplastic or myofibroblastic stroma); spread through air spaces; lymphatic or vascular invasion; pleural invasion; tumor necrosis. Although such tumors were previously classified as bronchioloalveolar carcinoma, this term is no longer used because it encompasses a heterogeneous group of adenocarcinomas. This category convenes a significantly better prognosis than other subtypes.3 Acinar Adenocarcinoma Acinar adenocarcinoma is a common type of adenocarcinoma with tumor cells arranged in classic glandular structure on a fibroelastic stroma. It is important to separate demosplastic stroma in this pattern from preexisting alveolar structures with thickened fibroelastic alveolar septa sometimes seen in a lepidic pattern. It is of note that the tumor cells displaying more complex growth patterns, such as a cribriform pattern, likely represent a poor prognostic subtype convening a significant risk of recurrence.17 Papillary Adenocarcinoma The tumor cells form papillary architecture with tumor cells lining the surface of branching fibrovascular cores. The presence of fibrovascular cores separates this tumor type from micropapillary adenocarcinoma. Micropapillary Adenocarcinoma The tumor cells form individual cellular tufts without fibrovascular core. The tumor cells appear as detached small and solid individual cell groups. Psammoma bodies may be seen. This subtype of adenocarcinoma has distinctly poor prognosis in the early stage compared with other subtypes of adenocarcinoma.18–20 Solid Adenocarcinoma The tumor cells form patternless sheets and lack any other recognizable patterns, including poorly differentiated/undifferentiated carcinomas expressing pneumocytic markers (such as TTF-1 and NapsinA), which were formerly grouped in the LCC category.3 It is of note that certain markers commonly associated with squamous cell differentiation, such as p63, and less commonly, p40, can show focal expression in solid adenocarcinoma.3,21 Rare Variants of Invasive Adenocarcinoma Rare variants of invasive adenocarcinoma include invasive mucinous adenocarcinoma, colloid and fetal adenocarcinoma, and enteric adenocarcinoma. Invasive mucinous adenocarcinoma is frequently multicentric and the tumor cells lack expression of TTF-1 commonly seen for lung adenocarcinoma.22 Enteric adenocarcinoma should be distinguished from metastatic colorectal adenocarcinoma.23 Show more View article Read full article URL: Journal2016, Surgical Oncology Clinics of North AmericaMin Zheng MD, PhD Chapter Lung 2009, Modern Surgical Pathology (Second Edition)SAUL SUSTER, CESAR A. MORAN Adenocarcinoma Pulmonary adenocarcinoma is a primary epithelial neoplasm of the lung showing glandular differentiation. The incidence of adenocarcinoma of the lung has continued to rise steadily, and it is now the most common form of lung cancer in the United States.9,34 Most patients with adenocarcinoma have a history of cigarette smoking, although statistically, the association is not as strong as that with squamous and small cell carcinoma. Most adenocarcinomas of the lung are peripheral and generally cause symptoms late in the disease course. Many adenocarcinomas are associated with a scar in the periphery of the lung. In the past, such tumors were designated scar carcinoma and were thought to arise from the scar tissue. More recent observations suggest that the scar may be a host reaction to the tumor and may be unrelated to tumorigenesis.35 Multicentricity is common in adenocarcinoma of the lung. Grossly, most peripheral adenocarcinomas are soft, ill-defined masses with a glistening tan to gray-white cut surface (Fig. 15-9). The overlying pleura is often indented or puckered. Occasionally, the tumors present with diffuse pleural thickening, simulating malignant pleural mesothelioma. Such cases have been designated pseudomesotheliomatous adenocarcinoma, and it may be a formidable challenge to distinguish them from malignant mesothelioma.36,37 On histologic evaluation, the lesions may be indistinguishable from the tubulopapillary variant of malignant mesothelioma. In contrast to malignant mesothelioma, however, the tumor cells are strongly positive for CEA in all cases and also react with Ber-EP4, Leu-M1, B72.3, and TTF-1, four markers that are expressed only rarely in mesothelioma.38 Grading of adenocarcinoma into well differentiated, moderately differentiated, and poorly differentiated is based on the degree and extent of gland formation. Well-formed glands lined by atypical cells that infiltrate the surrounding stroma characterize well-differentiated neoplasms (Fig. 15-10). Poorly differentiated tumors have a few abortive or poorly formed glandular structures or may grow as sheets of tumor cells that show focal intracellular mucin using histochemical stains, such as periodic acid–Schiff (PAS) or mucicarmine (Figs. 15-11 and 15-12). In tumors displaying a predominantly solid growth pattern with scant or inconspicuous glandular structures, demonstration of intracytoplasmic mucin production by the tumor cells may be necessary for diagnosis. Most conventional adenocarcinomas fall into one of three predominant growth patterns: acinar (glandular), papillary, or solid. A variety of other unusual morphologic appearances can be adopted by pulmonary adenocarcinomas (Table 15-4), including spindle cell (sarcomatoid) adenocarcinoma and clear cell adenocarcinoma. The latter may be mistaken for metastases of clear cell carcinomas from other organs, such as the kidney, ovary, or uterus. Another unusual variant that may pose diagnostic difficulties is signet ring cell adenocarcinoma.39,40 Such tumors can easily be mistaken for metastases from occult gastrointestinal or breast primary neoplasms. Mucinous (so-called colloid) carcinoma is another recently described variant of adenocarcinoma of the lung.41 In the past, tumors with identical histologic features were called a variety of terms, including mucinous cystadenoma and mucinous cystic tumor of borderline malignancy.42 These tumors are characterized by the accumulation of pools of mucin that infiltrate and destroy lung parenchyma. The alveolar walls within the lesion may be lined focally by atypical mucinous epithelium (Fig. 15-13). Another distinctive feature is the presence of epithelial tumor cells, either small clusters or single and scattered, floating in pools of mucin. Because of these tumors’ demonstrated ability to behave aggressively and metastasize in distant sites, we prefer to group them collectively under the term mucinous carcinoma.43 The possibility of metastasis from another organ in which such tumors are known to arise (e.g., colon, breast, ovary) must be excluded clinically. Papillary carcinoma is diagnosed when more than 75% of the tumor contains papillary structures supported by fibrovascular cores with secondary and tertiary branches.44 The tumor cells lining the papillae show marked cytologic atypia and areas of necrosis (Fig. 15-14). Papillary carcinoma can mimic papillary tumors arising from other organs, such as the ovary and thyroid. Close clinical correlation is important to arrive at the proper diagnosis. Psammoma bodies may or may not be present and are not required for diagnosis. Bronchioalveolar carcinoma is characterized by a distinctive “lepidic” growth pattern, whereby the tumor cells completely replace the lining of preexisting alveolar walls, with preservation of the underlying architecture.45 Conventional well-differentiated adenocarcinoma often adopts an identical growth pattern, and metastatic adenocarcinoma to the lung originating from the pancreas, thyroid, stomach, colon, kidney, ovary, and breast may show this distinctive growth pattern as well.46,47 Because of the significant overlap between conventional adenocarcinoma and bronchioalveolar carcinoma, and because of the spectrum of cell types encountered in this tumor, the latter diagnosis is often subjective and controversial. Previous studies have shown interobserver variability and little reproducibility in the classification of such tumors. On histologic evaluation, two variants of bronchioalveolar carcinoma are recognized: (1) the mucinous variant, composed of tall columnar cells with abundant mucinous cytoplasm and minimal cytologic atypia (Fig. 15-15), and (2) the nonmucinous variant, composed of cuboid or low columnar cells with scant cytoplasm and hyperchromatic nuclei that often exhibit a “hobnail” configuration (Fig. 15-16). Admixtures of these two patterns can be encountered. The mucinous type shows a high propensity to spread through the airways, and for that reason, these tumors are frequently multifocal. On gross inspection, consolidation of large parts of a lobe or the entire lung is characteristic, displaying a glistening, mucinous cut surface (Fig. 15-17). These tumors bear a striking resemblance to so-called jaagsiekte (pulmonary adenomatosis), a disease of presumed viral origin in sheep, although no viral association has been shown in humans. The nonmucinous variant is generally well circumscribed and single, although multifocal tumors have been described. They more often originate in the periphery of the lung beyond grossly recognizable, cartilage-bearing bronchi. The nuclei of the tumor cells are round to oval, with dense chromatin, inconspicuous nucleoli, and minimal nuclear atypia, and are surrounded by a scant rim of cytoplasm. Ultrastructurally, differentiation toward Clara cells and type II pneumocytes has been shown. Mitotic figures are only rarely encountered, and necrosis of tumor cells is absent. A distinctive feature of those tumors displaying ultrastructural characteristics of type II pneumocytes is the presence of clear intranuclear inclusions that are positive for surfactant apoprotein (Fig. 15-18).48 Bronchioalveolar carcinoma also may show a papillary pattern of growth, with small papillary tufts attached to the walls of the alveoli by short fibrovascular cores. The lining cells, however, should not display marked cytologic atypia or necrosis. Psammoma bodies are encountered occasionally in such cases but are neither distinctive nor pathognomonic.49 The biologic behavior of bronchioalveolar carcinoma is variable and may depend on several factors. The most important consideration is whether the lesion is single or multifocal. The 5-year survival rate for solitary tumors is dramatically better than that for multifocal lesions (approximately 70% versus 0%). The most important criteria for separating bronchioalveolar carcinoma from well-differentiated adenocarcinoma are a well-circumscribed tumor, with preservation of the underlying lung architecture; absence of infiltration of the stroma or pleura; and lack of significant cytologic atypia, mitotic activity, or tumor necrosis. Tumors that exhibit the characteristic lepidic pattern of growth of bronchioalveolar carcinoma but do not display the aforementioned features are best categorized as “adenocarcinoma or papillary carcinoma with bronchioalveolar growth pattern.” Adenocarcinomas of the lung generally express keratins associated with simple epithelia (i.e., cytokeratins 7, 8, 18, and 19). Surfactant apoprotein or Clara cell protein is present in many lung adenocarcinomas and may play a role in the differential diagnosis of metastatic carcinoma to the lung.50 CEA is expressed in more than 75% of lung adenocarcinomas and is a useful marker for distinguishing peripheral lung tumors from malignant mesothelioma. Other monoclonal antibodies of value in this setting include CD15 (Leu M-1), Ber-EP4, B72.3, and MOC-31, which are positive in most adenocarcinomas but usually not expressed in mesothelioma. None of these markers is specific for adenocarcinoma of the lung, however, and may be expressed in cancers from other sites. A valuable marker that seems to be more specific for lung adenocarcinoma is TTF-1.33 Nuclear positivity for TTF-1 may help establish the primary nature of an adenocarcinoma in the lung (Fig. 15-19). Differential expression of CK7 and CK20 may be of value in determining whether adenocarcinoma originates in the lung or is metastatic from the colon. Most lung adenocarcinomas express only CK7, whereas colon adenocarcinomas preferentially express only CK20. Demonstration of nuclear positivity for CDX2 is also helpful in this setting; it is positive in metastases from colorectal cancer and negative in primary lung tumors.51 Show more View chapterExplore book Read full chapter URL: Book 2009, Modern Surgical Pathology (Second Edition)SAUL SUSTER, CESAR A. MORAN Review article Lung Cancer, Part I: Screening, Diagnosis, and Staging 2013, Thoracic Surgery ClinicsMorihito Okada MD, PhD Introduction Primary lung cancer is the leading cause of cancer-related mortality worldwide, and non–small-cell lung cancer accounts for about 85% of all lung cancers, of which approximately 70% are adenocarcinomas. Lung adenocarcinoma is a radiologically, histomorphologically, molecularly, and clinically heterogeneous tumor with a broad range of malignant behavior that cannot be predicted by known prognostic factors such as the TNM staging system at diagnosis or surgery. Because a reliable and easily measured predictor of prognosis is tumor grade based on histomorphologic criteria for several cancer entities, different histopathologic features of adenocarcinoma subtypes might generate further survival evidence. The recent universally accepted criteria for adenocarcinoma subtypes, in particular tumors that were formerly classified as bronchioloalveolar carcinoma (BAC), are of considerable importance. Cumulative data have shown that the nature of lung adenocarcinoma growth is associated with differences in survival.1–11 In 1995, Noguchi and colleagues 1 examined the histologic characteristics of small lung adenocarcinoma, particularly those of BAC, and based an original classification system on tumor growth patterns that were closely related to prognosis. Type C in Noguchi’s classification is a mixed subtype that comprises more than 60% of all small adenocarcinomas. The mixed BAC category was introduced in the 1999 World Health Organization (WHO) classification of lung adenocarcinoma,2 and maintained in the 2004 WHO classification of lung tumors (Box 1).3 The mixed subtype is the most frequent, accounting for more than 80% of all resected small adenocarcinomas 4; this is a major limitation of these WHO classifications. The histologic classification of lung adenocarcinoma must be improved, based on survival outcomes, to overcome this limitation. Thus the new International Multidisciplinary Lung Adenocarcinoma Classification (Box 2, Figs. 1 and 2) was established by the International Association for the Study of Lung Cancer, American Thoracic Society, and European Respiratory Society (IASLC/ATS/ERS),12 containing several key modifications: (1) the term BAC is no longer applied and the type is described as lepidic; (2) adenocarcinoma in situ is suggested for small (≤3 cm) tumors with completely lepidic growth and no invasion; (3) minimally invasive adenocarcinoma refers to small (≤3 cm) predominantly lepidic tumors accompanied by slight invasion (≤0.5 cm); and (4) invasive adenocarcinoma is grouped along with the predominant subtype. Box 1 Summary of 1999 and 2004 classifications of lung adenocarcinoma <1999 Classification> Acinar Papillary BAC: Nonmucinous, Mucinous, Mixed mucinous/nonmucinous Solid with mucin Variants: Colloid, Fetal, Signet-ring, Clear-cell <2004 Classification> Mixed subtype Acinar Papillary BAC: Nonmucinous, Mucinous Solid with mucin production Variants: Fetal, Mucinous (Colloid), Signet-ring, Clear-cell Abbreviation: BAC, bronchioloalveolar carcinoma. Box 2 IASLC/ATS/ERS classification of lung adenocarcinoma (2011) Preinvasive lesions Atypical adenomatous hyperplasia Adenocarcinoma in situ (≤3 cm, formerly BAC) Nonmucinous, Mucinous, Mixed mucinous/nonmucinous Minimally invasive adenocarcinoma (≤3 cm lepidic-predominant tumor with ≤5 mm invasion) Nonmucinous, Mucinous, Mixed mucinous/nonmucinous Invasive adenocarcinoma Lepidic predominant (formerly nonmucinous BAC pattern) Acinar predominant Papillary predominant Micropapillary predominant Solid predominant with mucin production Variants Invasive mucinous adenocarcinoma (formerly mucinous BAC) Colloid Fetal Enteric Abbreviation: BAC, bronchioloalveolar carcinoma. Sign in to download hi-res image Fig.1. Examples of early adenocarcinoma growth patterns. Representative hematoxylin and eosin–stained tumor slides of (A, B) adenocarcinoma in situ and (C, D) minimally invasive adenocarcinoma. A and C × 12.5; B × 100; D× 40. Sign in to download hi-res image Sign in to download hi-res image Fig.2. Examples of invasive adenocarcinoma growth patterns. Representative hematoxylin and eosin–stained tumor slides of (A, B) lepidic-, (C, D) acinar-, (E, F) papillary-, (G, H) micropapillary-, and (I, J) solid-predominant adenocarcinomas. A × 12.5; B, C, D, F, H and J × 100; E, G and I × 40. Show more View article Read full article URL: Journal2013, Thoracic Surgery ClinicsMorihito Okada MD, PhD Chapter Classic Anatomic Pathology and Lung Cancer 2018, IASLC Thoracic Oncology (Second Edition)Ignacio I. Wistuba, ... Masayuki Noguchi Adenocarcinoma Adenocarcinoma is a malignant epithelial tumor with glandular differentiation or mucin production, showing various growth patterns, with expression of either mucin or thyroid transcription factor-1 (TTF-1). A significant change in pathologic classification of lung cancer occurred in 2011 with the publication of the revised classification of lung adenocarcinoma under the sponsorship of the International Association for the Study of Lung Cancer (IASLC), the American Thoracic Society (ATS), and the European Respiratory Society (ERS). This new classification of adenocarcinoma outlined many paradigm shifts that affect clinical diagnosis and management and open new avenues for research.7 A major point in this classification was the concept that personalized medicine for patients with advanced lung cancer is determined by histology and genetics and that strategic tissue management of small biopsy specimens is critical for pathologic and molecular diagnosis. This publication was a multidisciplinary effort rather than one primarily addressed by pathologists; clinicians, radiologists, molecular biologists, and surgeons were involved. This collaboration led to an emphasis on correlations between pathology and clinical, radiologic, and molecular characteristics. In addition, the experts recognized that 70% of patients with lung cancer present with advanced-stage disease, which is usually diagnosed based on small biopsy and cytology specimens. Because the previous (2004) classifications from the WHO focused on lung cancer diagnosis in resection specimens, which are obtained in only 30% of cases, a major new effort was made in this new classification to define terminology and criteria to be used in small biopsies and cytology specimens.8,9 Therefore, this classification is divided into two sections based on the diagnostic modalities of lung cancer (Table 17.2). These changes have been reflected in the recently released 2015 WHO Classification of Lung Cancer.5 Resection specimens apply for patients with early-stage disease who are eligible for surgical resection, and small biopsy and cytology specimens for patients with advanced-stage lung cancer. Several important modifications were made to the 2004 WHO classification concerning specimens in the 2011 IASLC/ATS/ERS and 2015 WHO classifications. The most significant change is the discontinuation of the term bronchiolo–alveolar cell carcinoma (BAC). This term had been used for at least five different entities with disparate clinical and molecular properties, leading to great confusion in routine clinical care and research.7,8 To address two of these entities, the concepts of adenocarcinoma in situ (AIS) and minimally invasive adenocarcinoma (MIA) were proposed for small (no more than 3 cm), solitary adenocarcinomas with a lepidic pattern that either lack invasion (AIS) or only show small foci of invasion measuring no more than 0.5 cm (MIA). AIS and MIA should define patients with either 100% or near 100% 5-year disease-free survival if the tumor is completely resected. The term mixed subtype was discontinued, and invasive adenocarcinomas were classified according to their predominant subtype. Using this approach, the proportions of each of the histologic subtypes should be estimated in a semiquantitative manner and a predominant pattern designated. The term lepidic predominant adenocarcinoma was proposed for nonmucinous tumors classified formerly as mixed subtype where the predominant subtype consists of the former nonmucinous BAC. Micropapillary adenocarcinoma was introduced as a major histologic subtype as multiple studies have shown that patients with such tumors have a poor prognosis.10–13 The tumors formerly classified as mucinous BAC are now reclassified as mucinous AIS or MIA or invasive mucinous adenocarcinoma; these tumors on computed tomography (CT) scan frequently show nodules of consolidation with air bronchograms and a multinodular and multilobar distribution. Finally, clear cell and signet ring adenocarcinomas were discontinued as major subtypes because they represent cytologic features that can occur in multiple histologic patterns of adenocarcinoma; however, now these features can be recorded when any amount is present.7,14 In the new classification, tumors formerly regarded as BAC included a wide spectrum of entities with varied clinical behavior such as AIS, MIA, lepidic predominant adenocarcinoma, overtly invasive adenocarcinoma with a lepidic component, and invasive mucinous adenocarcinoma. AIS should not be equated with tumors previously classified as BAC, particularly in registry databases such as Surveillance, Epidemiology, and End Results Program (SEER).15 Such data could be misleading as AIS is the rarest lung adenocarcinoma, representing only 0.2% to 3% of cases in white populations and up to 5% in a Japanese series.16–18 Most cases previously classified as BAC represent tumors with invasive components. Since the publication of the 2011 IASLC/ETS/ERS classification, a series of studies validated various aspects of the classification in resection specimens. Studies from Australia,17 Europe,19 Asia,18 and North America have demonstrated that the proposed subtyping has prognostic value.16,20 Atypical Adenomatous Hyperplasia Atypical adenomatous hyperplasia (AAH) is considered to be a precursor of adenocarcinoma.21,22AAH is a discrete parenchymal lesion in the alveoli close to terminal and respiratory bronchioles. Because of their size, AAH cells are usually incidental histologic findings, but they may be detected grossly, especially if they are 0.5 cm or larger. The increasing use of high-resolution CT scans for screening purposes has led to an increasing awareness of AAH, which remains one of the most important differential diagnoses of air-filled peripheral lesions (so-called ground-glass opacities). AAH maintains an alveolar structure lined by rounded, cuboidal, or low columnar cells (Fig. 17.1A). The postulated progression of AAH to adenocarcinoma with predominant lepidic growth features, apparent from the increasingly atypical morphology, is supported by the results of morphometric, cytofluorometric, and molecular studies.22,23 The origin of AAH is still unknown, but the differentiation phenotype derived from immunohistochemical and ultrastructural features suggests an origin from the progenitor cells of the peripheral airways, such Clara cells and type II pneumocytes.24,25 Adenocarcinoma in Situ AIS was added to the group of preinvasive lesions along with AAH (Table 17.2).7,8,14 AIS is defined as a localized, small (no more than 3 cm) adenocarcinoma consisting of neoplastic pneumocytes growing along preexisting alveolar structures (lepidic growth), lacking stromal, vascular, or pleural invasion (Fig. 17.1B). No papillary or micropapillary patterns should be present, and intra-alveolar tumor cells are absent. AIS is typically nonmucinous, consisting of type II pneumocytes and/or Clara cells, but rare cases of mucinous AIS do occur. The concept of AIS was proposed with the intention of defining lesions that should correlate with a 100% disease-free survival if completely resected. This proposal was supported by retrospective observational studies in tumors measuring either 2 cm or less or 3 cm or less.7 In the setting of multiple tumors, the criteria for AIS as well as MIA should be applied only if the other tumors are regarded as synchronous primaries rather than intrapulmonary metastases. Minimally Invasive Carcinoma MIA is defined as a small (no more than 3 cm), solitary adenocarcinoma with a predominantly lepidic pattern and invasion of no more than 5 mm in greatest dimension (Fig. 17.1C–D).26,27 MIA is usually nonmucinous, but it may rarely be mucinous.16 Measurement of the invasive component of MIA should include the following: (1) histologic subtypes other than a lepidic pattern (i.e., acinar, papillary, micropapillary, and/or solid) or (2) tumor cells infiltrating myofibroblastic stroma. MIA should not be diagnosed if the tumor invades lymphatics, blood vessels, or pleura or contains tumor necrosis. More details about measuring invasive size are explained elsewhere.7,9 The concept of MIA was introduced to define a population of patients who should have a 100% or near 100% 5-year disease-free survival rate if the lesion is completely resected. Although less evidence was found to support the concept of MIA compared with AIS,26,27 all published cases using these criteria have shown a 5-year disease-free survival rate of 100%.16–18,20 The diagnosis of AIS or MIA requires that the tumor be completely sampled histologically (i.e., the patient has undergone a surgical resection). Both lesions should also have a discrete circumscribed border without miliary spread of small foci of tumor into adjacent lung parenchyma and/or with lobar consolidation. Review of CT scans may be helpful in evaluating pathologic specimens because the extent of ground-glass (usually lepidic) and solid (usually invasive) patterns can guide pathologists in assessing whether the lesion has been properly measured and/or sampled. For lesions suspected to be AIS or MIA larger than 3 cm, the term lepidic predominant adenocarcinoma is best applied, along with a comment that an invasive component cannot be excluded, because the data are insufficient to show that such patients will have 100% 5-year disease-free survival. Invasive Adenocarcinoma Because of the rarity of AIS and MIA, overtly invasive adenocarcinomas represent more than 70% to 90% of surgically resected lung adenocarcinomas. These tumors typically consist of a complex, heterogeneous mixture of histologic patterns, thus explaining the former category of adenocarcinoma, mixed subtype (Fig. 17.2). The major subtypes of invasive adenocarcinoma are now classified according to the predominant component, after performing comprehensive histologic subtyping (Fig. 17.2B–F, and Table 17.2). Comprehensive histologic subtyping is performed by making a semiquantitative estimation of each of the patterns in 5% increments. It is useful to record in diagnostic reports each adenocarcinoma subtype that is present with the respective percentages. This approach may also provide a basis for architectural grading of lung adenocarcinomas.16,28,29 Since the 2011 classification was initially published, a growing number of studies of resected lung adenocarcinomas have demonstrated its utility in identifying significant prognostic subsets and molecular correlations according to the predominant patterns.17–19,28,30–32 Lepidic Predominant Invasive Adenocarcinoma This subtype consists of a proliferation of bland type II or Clara cells growing along the surface of alveolar walls, similar to the morphology defined in the earlier section on AIS and MIA (Fig. 17.2C). Invasive adenocarcinoma is present in at least one focus measuring more than 5 mm in greatest dimension. Invasion is defined as follows: (1) histologic subtypes other than a lepidic pattern (i.e., acinar, papillary, micropapillary, and/or solid); (2) myofibroblastic stroma associated with invasive tumor cells; (3) vascular or pleural invasion; and (4) spread through alveolar spaces (STAS).33–35 Lepid predominant adenocarcinoma is diagnosed rather than MIA if the cancer invades lymphatics, blood vessels, or pleura or contains tumor necrosis. Several recent studies of early-stage adenocarcinomas published since 2011 have demonstrated that lepidic predominant tumors have a favorable prognosis, with 5-year disease-free survival rates of 86% to 90%.17–19 The term adenocarcinoma with lepidic pattern corresponds to some cases previously referred to as adenocarcinoma with BAC features. The term lepidic predominant adenocarcinoma should not be used in the context of invasive mucinous adenocarcinoma with predominant lepidic growth; these tumors should be classified as mucinous adenocarcinomas. Acinar Predominant Invasive Adenocarcinoma This subtype shows a major component of glands that are round to oval with a central luminal space surrounded by tumor cells (Fig. 17.2B).6 The neoplastic cells and/or glandular spaces may contain mucin. Cribriform structures can be found in the acinar pattern, are considered to be high grade, and are associated with poor prognosis.36 Tumor cells may form aggregates of polarized cell without clear lumen, which is still recognized as an acinar pattern. Papillary Predominant Adenocarcinoma This subtype shows a major component of a growth of glandular cells along central fibrovascular cores (Fig. 17.2D).6 If a tumor has lepidic growth, but the alveolar spaces are filled with papillary or micropapillary structures, the tumor is classified as papillary or micropapillary adenocarcinoma, respectively. Micropapillary Predominant Adenocarcinoma This subtype has tumor cells growing in papillary tufts or florets that lack fibrovascular cores (Fig. 17.2E).6 These may appear detached and/or connected to the alveolar walls. The tumor cells are usually small and cuboidal with minimal nuclear atypia. STAS is a newly suggested pattern of invasion, often seen with the micropapillary pattern; it can occur with micropapillary clusters, solid nests, or single cells. STAS probably contributes to the high recurrence rate for patients with small stage I adenocarcinomas who undergo limited surgical resections and the poor prognosis observed by others.12,33 Solid Predominant Invasive Adenocarcinoma The solid subtype with mucin production consists of a major component of polygonal tumor cells forming sheets but without any clear acinar, papillary, micropapillary, or lepidic growth (Fig. 17.2F).6 If the tumor is 100% solid, intracellular mucin should be present in at least five tumor cells in each of two high-power fields, confirmed with histochemical stains for mucin.6,9 Tumors formerly classified as large cell carcinomas that have immunohistochemical expression of TTF-1 and/or napsin A, even if mucin is not identified, are now classified as solid adenocarcinomas. Solid adenocarcinomas must be distinguished from nonkeratinizing squamous cell carcinomas and large cell carcinomas, both of which may show rare cells with intracellular mucin. Neuroendocrine markers, such as neural cell adhesion molecule (NCAM)/CD56, dense core granule associated protein chromogranin A, and synaptic vesicle protein synaptophysin should be performed only if neuroendocrine morphology is present, to allow the diagnosis of large cell neuroendocrine carcinoma (LCNEC). Variants of Invasive Adenocarcinoma Four variants of clinically relevant invasive lung adenocarcinomas are recognized: (1) invasive mucinous adenocarcinoma, with tumor cells that have a goblet or columnar cell morphology with abundant intracytoplasmic mucin (Fig. 17.3A); (2) colloid adenocarcinoma, with abundant mucin pools filling alveolar spaces; (3) fetal adenocarcinoma, resembling fetal lung; and, (4) enteric adenocarcinoma, an adenocarcinoma of the lung resembling enteric adenocarcinoma. Invasive Mucinous Adenocarcinoma Multiple studies have shown major differences in clinical, radiologic, pathologic, and genetic features between invasive mucinous adenocarcinomas and the tumors formerly classified as nonmucinous BAC.37–42 Invasive mucinous adenocarcinomas have tumor cells with a goblet or columnar cell morphology with abundant intracytoplasmic mucin and aligned, basally located nucleoli (Fig. 17.3A). This pathognomonic cellular feature can be recognized in small lung samples. Similar to nonmucinous tumors, invasive mucinous adenocarcinomas may show the same heterogeneous mixture of lepidic, acinar, papillary, micropapillary, and solid growth, which will not be described in detail and quantified for this specific subtype. Although invasive mucinous adenocarcinomas frequently show lepidic predominant growth, extensive sampling usually shows invasive foci. Therefore, reports on biopsy specimens should include a remark such as mucinous adenocarcinoma with lepidic pattern. However, if a resection specimen of the mucinous tumor fulfills the diagnostic criteria of AIS or MIA, the tumor should be diagnosed as mucinous AIS or MIA, respectively, although such tumors are extremely rare. In some cases, mucinous adenocarcinoma appears in CT scans and pathology specimens with a pseudo-pneumonia growth pattern (Fig. 17.3B–C). Colloid Adenocarcinoma This subtype shows abundant extracellular mucin in pools, which distend the alveolar spaces and destroy their walls, showing an overtly invasive growth pattern into the alveolar spaces. Mucin deposits enlarge and dissect the lung parenchyma, creating pools of mucin-rich matrix, while tumor elements consist of foci of tall columnar cells with goblet-like features growing in a lepidic fashion. Tumor glands often float into the mucoid material, becoming poorly recognizable and then requiring extensive tumor sampling. Fetal Adenocarcinoma This subtype consists of complex glandular structures composed of glycogen-rich, nonciliated cells resembling a developing epithelium in the pseudoglandular phase of the fetal lung, with low nuclear atypia and morule formation.43 Primary Pulmonary Adenocarcinoma With Enteric Differentiation This term is used to indicate a primary lung cancer resembling colorectal adenocarcinoma metastatic to the lung.44–46 Its histologic characteristics include eosinophilic tall columnar cells with brush border, vesicular nuclei, central geographic or dotted necrosis, occasional central scar and pleural indentation, and papillo–tubular (or gland-like) structure. The histologic resemblance to colorectal cancer is a hallmark of this tumor. Although some tumors have enteric differentiation with positive immunohistochemical expression of CDX-2 (which encodes for an intestine-specific transcription factor) and cytokeratin (CK) 20, and negative expression of CK7, others have only enteric morphology. Immunohistochemistry of Adenocarcinomas The immunohistochemical expression of lung adenocarcinomas varies somewhat based on the subtype and degree of differentiation. TTF-1 and napsin A are nearly specific markers for lung adenocarcinoma, except that thyroid cancer expresses TTF-1 and renal cell carcinoma expresses napsin A. Approximately 75% of invasive adenocarcinomas are positive for TTF-1 (Fig. 17.4), and none of the squamous cell carcinomas express TTF-1. Among the adenocarcinoma subtypes, most lepidic and papillary predominant tumors are positive for TTF-1, as are the lepidic components of AIS and MIA, whereas positive frequency is lower in cases of solid predominant cancer.32,47 The sensitivity of napsin A is comparable to that of TTF-1, but its specificity is far lower.48 CK7 is another marker for lung adenocarcinoma, and its sensitivity, but not its specificity, is higher than that of TTF-1 and napsin A. Furthermore, tumor protein p63, which has been used as a marker for a squamous cell carcinoma, is also positive in some lung adenocarcinoma (up to 38%),49 and a portion of these adenocarcinomas are positive for anaplastic lymphoma kinase (ALK) gene rearrangement.50–52 In contrast, protein p40, an isoform of p63,50 is never positive in adenocarcinomas except for those contained in adenosquamous carcinomas. Some adenocarcinomas may show a squamous appearance. In these cases, phenotyping with a limited panel of immunomarkers (including p40, and TTF-1) and mucin stain is necessary.7,49,53,54 Histologic and Molecular Correlations Despite the discovery of multiple molecular abnormalities in lung adenocarcinoma, no significant, specific histologic and molecular correlations have been found. A number of driver gene alterations are now known to exist in lung adenocarcinomas including mutations of EGFR, KRAS, BRAF, and ERBB2/HER2, and rearrangements of ALK, RET, ROS1, NTRK1, and NRG1.55–61 Among these mutations, EGFR and ALK are clinically relevant because molecular targeted drugs can be used in patients whose tumors have these molecular abnormalities. Adenocarcinomas with BRAF, HER2, ROS1, and NTRK1 abnormalities share clinicopathologic features with the EGFR-mutant and ALK-rearranged tumors in terms of involvement that is nearly specific to adenocarcinoma in lung cancer, particularly frequent in TTF-1–positive expression, and preferentially present in never-smokers and in women. The frequent finding of KRAS mutation and consistent lack of EGFR mutation in invasive mucinous adenocarcinoma is the strongest histologic and molecular correlation found in lung cancer. Most histologic subtypes of adenocarcinoma harbor EGFR and KRAS mutations, as well as ALK rearrangement. EGFR mutations are encountered most frequently in nonmucinous adenocarcinomas with a lepidic or papillary predominant pattern, whereas KRAS mutations tend to be found in solid predominant adenocarcinomas. ALK rearrangement has mostly been associated with an acinar pattern including a cribriform morphology and with signet-ring cell features, particularly in those tumors with TTF-1 and p63 coexpression.51,62,63 Impact of the New Classification on Tumor, Node, Metastasis Staging The 2011 IASLC/ATS/ERS classification of adenocarcinoma can affect tumor, node, metastasis (TNM) staging in several ways. First, it may help in comparing histologic characteristics of multiple lung adenocarcinomas to determine whether they are intrapulmonary metastases or separate primaries. Using comprehensive histologic subtyping along with other histologic characteristics has been shown to correlate well with molecular analyses and clinical behavior.64,65 Second, it may be more meaningful clinically to measure tumor size in lung adenocarcinomas that have a lepidic component by using the invasive size rather than total size to determine the final size of the tumor for TNM staging. It is possible that in the next edition of the TNM, AIS may be regarded as tumor carcinoma in situ (Tis) and MIA may be regarded as tumor microinvasive (Tmi). Small Biopsy and Cytology Samples In the past, NSCLCs were lumped together without attention to more specific histologic typing (e.g., adenocarcinoma, squamous cell carcinoma, etc.).8 One of the major new proposals in the IASLC/ATS/ERS classification was the development of standardized criteria and terminology for the pathologic diagnosis of lung cancer in small biopsy and cytology specimens (Table 17.3).8,9 In addition to the criteria and terminology, two paradigm shifts have emerged for pathologists in terms of tumor classification and management of specimens. The first is the need to perform immunohistochemistry to further classify tumors formerly diagnosed as NSCLC not otherwise specified (NOS). Because the distinction between histologic types of lung cancer, particularly adenocarcinoma and squamous cell carcinoma, is so important, the new classification recommends that pathologists use special stains to try to further subtype carcinomas that are difficult to classify by light microscopic evaluation of histologic sections alone. For tumors with classic morphologic features the diagnostic terms adenocarcinomas and squamous cell carcinomas can be used (Table 17.3). The morphologic features of these tumors are described in detail elsewhere.6,7,9 If an NSCLC does not show clear glandular or squamous morphology in a small biopsy or cytology specimen, it is classified as NSCLC-NOS.7,66 Tumors with this morphologic appearance should be studied with a limited special stain workup in an attempt to classify them further. It is recommended to use a single adenocarcinoma marker (e.g., TTF-1), a single squamous marker (e.g., p40), and/or mucin stain.66,67 Tumors that are positive for an adenocarcinoma marker or mucin are classified as NSCLC, favor adenocarcinoma. Tumors that are positive for a squamous cell carcinoma immunohistochemical marker with negative adenocarcinoma markers are classified as NSCLC, favor squamous cell carcinoma. Cytology is a powerful diagnostic tool that can accurately subtype NSCLC in most cases,66 and immunohistochemistry is readily available when cell blocks are prepared for the cytology samples.68 Show more View chapterExplore book Read full chapter URL: Book 2018, IASLC Thoracic Oncology (Second Edition)Ignacio I. Wistuba, ... Masayuki Noguchi Review article Solid predominant subtype in lung adenocarcinoma is related to poor prognosis after surgical resection: A systematic review and meta-analysis 2019, European Journal of Surgical OncologyNaofumi Miyahara, ... Konrad Hoetzenecker 1 Introduction Adenocarcinoma is the most common histologic type of lung cancer accounting for almost half of all lung cancers . The International Association for the Study of Lung Cancer, the American Thoracic Society, and the European Respiratory Society (IASLC/ATS/ERS) proposed a new histologic classification of lung adenocarcinoma (ADC) in 2011, recognizing five major histologic subtypes (lepidic, acinar, papillary, solid and micropapillary) . To date, a number of cohort studies including patients after surgical resection for lung adenocarcinoma (LADC) have reported unfavorable prognosis associated with solid predominant (SP) histological subtype [3–9]. Furthermore, it has been shown that the presence of SP subtype represents an additional risk factor for N1/2 lymph node involvement and early recurrence [6,10,11]. However, this reported negative impact of SP subtype of LADC after pulmonary resection has not been validated in larger studies. Thus, it is yet unclear if specific histological subtyping of LADC should be considered as a prognostic factor for overall survival in addition to the current TNM staging system. To give light upon this issue, we therefore comprehensively reviewed available data and conducted a meta-analysis to clarify whether SP subtype represents a significant prognostic factor for LADC. View article Read full article URL: Journal2019, European Journal of Surgical OncologyNaofumi Miyahara, ... Konrad Hoetzenecker Review article Advances in differential diagnosis of pulmonary ground glass opacity on high resolution computed tomography and histopathology 2020, Radiology of Infectious DiseasesYaoyao Zhuo, ... Zhiyong Zhang 2.5 Tumors Lung adenocarcinoma is classified as atypical adenomatous hyperplasia (AAH), adenocarcinoma in situ (AIS), minimally invasive adenocarcinoma (MIA), and invasive mucinous adenocarcinoma in 2015 [28,29]. Bronchioloalveolar carcinoma (BAC) is initially considered to be a subtype of lung adenocarcinoma and has the following characteristics: surrounding the foci, well-differentiated tumor cells, non-invasive alveolar septum, and the trend of airway or lymphatic dissemination. Adenocarcinoma may come in different forms: single nodular, multinodular or diffuse pattern [30–32]. Radiologically, BAC has a variety of manifestations, the most common of which are marginal, lobulated or isolated lung nodules at the edge of the lesion . Invasive mucinous adenocarcinoma is a type of lung adenocarcinoma, which spreads to the other lobes or contralateral lung and shows diffuse increased density opacity, including GGO on CT. View article Read full article URL: Journal2020, Radiology of Infectious DiseasesYaoyao Zhuo, ... Zhiyong Zhang Review article Practical Pediatric Imaging 2017, Radiologic Clinics of North AmericaMatthew A. Zapala MD, PhD, ... Edward Y. Lee MD, MPH Adenocarcinoma Pulmonary adenocarcinomas are malignant epithelial neoplasms much more common in the adult population.47 However, adenocarcinomas have been linked with congenital pulmonary airway malformations in children. Also, a rare subtype of adenocarcinoma called fetal adenocarcinoma of the lung presents in pediatric patients and histologically resembles epithelium of the fetal lung.48 Affected pediatric patients present with postobstructive pneumonias if the masses are located centrally and otherwise may be asymptomatic in peripheral masses that are found incidentally. Adenocarcinomas demonstrate glandular differentiation with mucin production histologically. Imaging features of pediatric adenocarcinomas have been confined to the nonradiology case reports given their rarity. Pulmonary adenocarcinomas in the pediatric population have been described to have a similar varied appearance as in the adult population ranging from solitary pulmonary nodule to consolidative opacity.49,50 CT findings have also been described in similar fashion ranging from ground glass opacities to more solid nodular opacities. There are no standard treatment protocols in pediatric patients for lung adenocarcinomas. Surgical resection is the treatment but metastases are usually present and prognosis remains poor. View article Read full article URL: Journal2017, Radiologic Clinics of North AmericaMatthew A. Zapala MD, PhD, ... Edward Y. Lee MD, MPH Review article Mediastinal Lesions Encountered on Fine Needle Aspiration and Small Biopsy Specimens 2020, Seminars in Diagnostic PathologyAlexander P. Smith, ... Derek B. Allison Lung Adenocarcinoma is the most common histologic subtype and represents about 50% of all lung carcinomas. ADC of the lung is more commonly peripheral and is the most common subtype that occurs in non-smokers.45,46 Because of advances in genomics and personalized medicine, it has become increasingly important to morphologically subtype non-small cell lung carcinomas.23 More precisely, there are approved targeted therapies for lung adenocarcinomas with specific EGFR and BRAF mutations, ALK and ROS1 rearrangements, and NTRK fusions. In addition, there are number of off-label targets and therapies offered on clinical trial only. As a result, it is imperative to 1) prove the histologic subtype as ADC, 2) preserve tissue for mutational profiling and PD-L1 testing. Furthermore, if there is inadequate material on paraffin-embedded tissue, FNA smears can be used for diagnostic molecular testing.47 Due to the fact that 70% of lung cancers present with locally advanced or metastatic disease, most commonly to the mediastinum, FNA and CNB specimens are the most common diagnostic modality utilized for this evaluation.38 Lung ADCs may produce a variety of growth patterns (mucinous, non-mucinous, papillary, micropapillary, and enteric, to name a few) in the primary lesion. In metastatic lesions detected by FNA/CNB, howeverk, the specific ADC growth patterns show little correlation to resection specimens and the architectural distinctions are often lost in metastases.23 In small samples, ADCs are usually diagnosed based on the presence of sheets and three-dimensional clusters that are disorganized with nuclear overlapping, nuclear enlargement, nuclear size and shape variation, clumpy chromatin, and prominent nucleoli (Fig.5). The presence of mucin vacuoles may or may not be readily appreciable or present. These are more readily appreciable on the Diff-Quik stain as magenta vacuoles than on the Pap stain, which may show faint cytoplasmic clear vacuoles. On H&E, mucin vacuoles range from clear to gray/blue depending on the optimization of the stain. Metastases from moderately and poorly differentiated tumors may show no mucin and instead contain large individual cells in a necrotic background. Well-differentiated tumors may have open chromatin with nuclear grooves and pseudoinclusions, which can mimic papillary thyroid carcinoma (PTC) and warrant a full panel of IHC stains incorporating Napsin-A and PAX8 (see Table 5 for IHC pearls and pitfalls). Sign in to download hi-res image Fig. 5. Metastatic adenocarcinomas A. Lung adenocarcinoma showing disordered clusters of large, pleomorphic cells with hyperchromatic nuclei (Pap stain, 60x). B. Lung adenocarcinoma showing the disordered “drunken honeycomb” architecture and nuclear size and shape variation (Diff-Quik stain, 60x). C. Lung adenocarcinoma on a cell block preparation (60x, H&E). D. Lung adenocarcinoma showing nuclear positivity with TTF1 (TTF1 stain, 60x). E. Lung adenocarcinoma showing granular cytoplasmic staining with Napsin-A (Napsin-A stain, 60x). F. Breast adenocarcinoma, ductal type on core needle biopsy (H&E stain, 60x). G. Breast adenocarcinoma showing strong and diffuse nuclear staining with ER (ER stain, 60x). H. Breast adenocarcinoma showing nuclear positivity with GATA3 (GATA3 stain, 60x). I. Clear cell renal cell carcinoma with abundant vacuolated cytoplasm and enlarged nuclei (Diff-Quik stain, 40x). J. Clear cell renal cell carcinoma comprised of vascularized nests of cells with abundant vacuolated cytoplasm (H&E stain, 40x). K. Clear cell renal cell carcinoma showing nuclear positivity with PAX8 (PAX8 stain, 40x). L. Colon adenocarcinoma showing well-formed glands with acute inflammation and necrotic debris; CDX2 staining produces nuclear positivity (H&E stain, 60x; CDX2 stain, 60x). Table 5. Pearls and pitfalls for staining tumors of unknown primary in the mediastinum \| • p16: Positive staining can be seen in many tumors, including carcinomas (a subset of cutaneous and lung SqCCs), sarcomas, and lymphoproliferative neoplasms with no relation to HPV. • p40/p63: Positive staining can be seen in many carcinomas, including squamous cell carcinoma, urothelial carcinoma, myoepithelial/biphasic salivary gland neoplasms, thymoma neoplasms, and NUT carcinoma. • GATA3: Positive staining can be seen in neoplasms from the urinary bladder, kidney, breast, cutaneous and adnexal tumors of the skin, salivary gland, parathyroid gland, and in paraganglioma. A subset of T-lymphocytes is also GATA3 positive. • CDX2 and SATB2: Nuclear staining can be seen in enteric-type tumors from sites other than the colon such as the lung, bladder, and ovary. CDX2 can also be positive in pancreatobiliary adenocarcinomas while SATB2 is typically negative. • PAX8: Positive staining in neoplasms of the thyroid, parathyroid, kidney, upper urothelial tract (15-30%), Mullerian origin, pancreas (NET), thymus, and Merkel cell carcinoma (2/3). • HepPar1: Tumors with a hepatoid appearance from sites other than the liver can be positive. • Napsin-A: Positive staining can be seen in lung adenocarcinoma, clear cell carcinoma of the gynecologic tract, and a subset of renal tumors. • Cytokeratins: Melanoma, lymphoma, and sarcomas can occasionally show varying amounts of cytokeratin staining. • MiT family translocation renal cell carcinomas, PEComas, myoepithelial lesions, and more can stain with melanoma markers. • Prostate adenocarcinoma can be positive for neuroendocrine stains, limiting distinction between high grade prostate cancer and small cell carcinoma. \| Most lung ADCs are CK7 positive and further immunophenotyping with TTF-1 and Napsin-A positivity, both of which are about 80% sensitive, and negative staining with p40 will confirm the majority of these histologic subtypes as primary to the lung.40,48,49 Notable exceptions include mucinous/enteric type lung ADC which may co-express CK7 and CK20 and are often TTF-1 and Napsin-A negative and CDX-2 positive; however, SATB2 may be useful for a more specific marker for a colonic origin than CDX-2.50–52 In addition, there is a newly described, relatively rare variant that may somewhat resemble HCC morphologically and is commonly positive for CK7 and HepPar-1 while negative for TTF-1. These tumors have been classified as SMARCA4-deficient pulmonary ADC based on the loss of SMARCA4/BRG1 staining by IHC.53,54 For further IHC and morphologic pearls and pitfalls, refer to Table 5. Show more View article Read full article URL: Journal2020, Seminars in Diagnostic PathologyAlexander P. Smith, ... Derek B. Allison Related terms: Neoplasm Programmed Cell Death Metastatic Carcinoma Cancer Cell Malignant Neoplasm Breast Cancer Long Untranslated RNA Lung Cancer microRNA Non Small Cell Lung Cancer View all Topics Recommended publications Journal of Thoracic OncologyJournal Lung CancerJournal The Annals of Thoracic SurgeryJournal CHESTJournal Browse books and journals Featured Authors Travis, William DavidMemorial Sloan-Kettering Cancer Center, New York, United States Citations124,209 h-index164 Publications245 Wistuba, Ignacio IvansThe University of Texas MD Anderson Cancer Center, Houston, United States Citations106,399 h-index149 Publications298 Chen, HaiquanFudan University, Shanghai, China Citations16,901 h-index67 Publications150 Suzuki, KenjiJuntendo University School of Medicine, Tokyo, Japan Citations10,304 h-index35 Publications112 About ScienceDirect Remote access Advertise Contact and support Terms and conditions Privacy policy Cookies are used by this site. 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6474	https://math.stackexchange.com/questions/127082/optimization-of-a-rectangular-box	Skip to main content Optimization of a rectangular box Ask Question Asked Modified 5 years, 3 months ago Viewed 15k times This question shows research effort; it is useful and clear 3 Save this question. Show activity on this post. I am suppose to find the volume if 1200 cm^2 of material is available to make a box with a square base and an open top, find the largest possible volume of the box. I think what I need to do is set up the formulas to be 4(lw)+w2=1200 for area lwh=v for volume I know that if the base is a square than the rectangle will have the same dimensions and the only different variable would be the height so I can solve for length like so l=1200−w24w Now that I have that I can put it in my formula lwh=v for volume which I can rewrite as l2∗h=v I then take the derivative of this and I get some ridiculous answer that is wrong. 1200w2−w44w the derivative 300−3w24 Which gives me +−20 which is an incorrect answer. calculus Share CC BY-SA 3.0 Follow this question to receive notifications edited Apr 2, 2012 at 1:51 asked Apr 2, 2012 at 0:39 user138246user138246 Add a comment \| 2 Answers 2 Reset to default This answer is useful 4 Save this answer. Show activity on this post. When you write down a formula, you must write down what the letters stand for. When you write down 4lw+w2=1200, you must add, "where l is the length of the base of the box, and w is the width of the base of the box". If you do that, you might see right away where you've messed up. You've written down a formula for the area which doesn't include the height of the box---that can't possibly be right, can it? In fact, the formula you have written down only makes sense if l is the height of the box, right? So go back and identify the variables explicitly and then write down formulas that make sense. Share CC BY-SA 3.0 Follow this answer to receive notifications answered Apr 2, 2012 at 0:46 Gerry MyersonGerry Myerson 186k1313 gold badges231231 silver badges404404 bronze badges 11 I don't really know, I have trouble keeping track of all these formulas I need to memorize and all these rules to doing these types of problems. There are just so many parts I need to keep track of that I always mess up a few of them. – user138246 Commented Apr 2, 2012 at 0:48 For some reason I keep getting a derivative of 480022−12w44w2 which is ofcourse wrong. – user138246 Commented Apr 2, 2012 at 0:59 1 To add to Gerry's great suggestions, I would recommend drawing a good figure of your box and label it with all of your variables. This is another way to catch your error. – Matthew Conroy Commented Apr 2, 2012 at 1:39 1 If the height of the box is l, Jordan, then what is the h that you have written down? – Gerry Myerson Commented Apr 2, 2012 at 1:56 1 @robjohn, I've interacted with Jordan before, and I know something about the real problems. I despair of doing much about them, but maybe if I can convince Jordan not to write down formulas without also writing down what the letters mean I can be of some help. – Gerry Myerson Commented Apr 2, 2012 at 1:59 \| Show 6 more comments This answer is useful 1 Save this answer. Show activity on this post. The formula V=lwh means "volume = length times width times height." The variable l is length, the variable w is width, and the variable h is height. Using these, the total area is actually 2(l×h)+2(w×h)+w2=1200. We know that l=w (because the base of the box is square), so this is 4wh+w2=1200. This allows us to solve for the height h in terms of width w as h=(1200−w2)/(4w). We have the formula for h in terms of w, and know l=w, so we have the volume function V=lwh=w21200−w24w=300w−14w3. Now can you take the derivative of this, equate it with zero and solve for w? Share CC BY-SA 3.0 Follow this answer to receive notifications answered Apr 2, 2012 at 2:06 anonanon 156k1414 gold badges249249 silver badges421421 bronze badges 2 The zero is 20 but I am not sure what to do with it or what it means anymore.I think I got it now, I just substitute 20 into the other formula and then I find that the height is 10 and put all that into the volume formula which gives me 4,000. – user138246 Commented Apr 2, 2012 at 2:10 @Jordan: As per the derivative test (and a couple other checks), this means the volume is optimized when the width is twenty. Plug w=20 into the formula for V and you will have the optimized volume. – anon Commented Apr 2, 2012 at 2:15 Add a comment \| You must log in to answer this question. Featured on Meta Community help needed to clean up goo.gl links (by August 25) Related 4 Optimization, volume of a box 1 Optimization Problem, (rectangular box) 0 Rectangular Box Optimization Problem 1 Solving Volume with area only given 0 Lagrange Multiplier- Open Rectangular Box 0 Largest volume of an open box. 1 Largest possible volume of this box. Hot Network Questions What would a "big dumb satellite" be like? If Metamagic is used on a spell, then that spell is stopped by Counterspell, are the Sorcery Points lost as well? Should I add arrow indicators to dropdown-style buttons in my mobile app, or is it unnecessary visual clutter? Can I expect flight prices around a public holiday to drop in the 7 months leading up to the holiday? Why did all my surge protectors smoke when my home became ungrounded? Why do Austrian political parties have so many members? Why is the gain of the op amp zero? 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6475	https://www.statology.org/residuals/	A residual is the difference between an observed value and a predicted value in regression analysis. It is calculated as: Residual = Observed value – Predicted value Recall that the goal of linear regression is to quantify the relationship between one or more predictor variables and a response variable. To do this, linear regression finds the line that best “fits” the data, known as the least squares regression line. This line produces a prediction for each observation in the dataset, but it’s unlikely that the prediction made by the regression line will exactly match the observed value. The difference between the prediction and the observed value is the residual. If we plot the observed values and overlay the fitted regression line, the residuals for each observation would be the vertical distance between the observation and the regression line: An observation has a positive residual if its value is greater than the predicted value made by the regression line. Conversely, an observation has a negative residual if its value is less than the predicted value made by the regression line. Some observations will have positive residuals while others will have negative residuals, but all of the residuals will add up to zero. Example of Calculating Residuals Suppose we have the following dataset with 12 total observations: If we use some statistical software (like R, Excel, Python, Stata, etc.) to fit a linear regression line to this dataset, we’ll find that the line of best fit turns out to be: y = 29.63 + 0.7553x Using this line, we can calculate the predicted value for each Y value based on the value of X. For example, the predicted value of the first observation would be: y = 29.63 + 0.7553(8) = 35.67 We can then calculate the residual for this observation as: Residual = Observed value – Predicted value = 41 – 35.67 = 5.33 We can repeat this process to find the residual for every single observation: If we create a scatterplot to visualize the observations along with the fitted regression line, we’ll see that some of the observations lie above the line while some fall below the line: Properties of Residuals Residuals have the following properties: Each observation in a dataset has a corresponding residual. So, if a dataset has 100 total observations then the model will produce 100 predicted values, which results in 100 total residuals. The sum of all residuals adds up to zero. The mean value of the residuals is zero. How Are Residuals Used in Practice? In practice, residuals are used for three different reasons in regression: 1. Assess model fit. Once we produce a fitted regression line, we can calculate the residuals sum of squares (RSS), which is the sum of all of the squared residuals. The lower the RSS, the better the regression model fits the data. 2. Check the assumption of normality. One of the key assumptions of linear regression is that the residuals are normally distributed. To check this assumption, we can create a Q-Q plot, which is a type of plot that we can use to determine whether or not the residuals of a model follow a normal distribution. If the points on the plot roughly form a straight diagonal line, then the normality assumption is met. 3. Check the assumption of homoscedasticity. Another key assumption of linear regression is that the residuals have constant variance at every level of x. This is known as homoscedasticity. When this is not the case, the residuals are said to suffer from heteroscedasticity. To check if this assumption is met, we can create a residual plot, which is a scatterplot that shows the residuals vs. the predicted values of the model. If the residuals are roughly evenly scattered around zero in the plot with no clear pattern, then we typically say the assumption of homoscedasticity is met. Additional Resources Introduction to Simple Linear Regression Introduction to Multiple Linear Regression The Four Assumptions of Linear Regression How to Create a Residual Plot in Excel Zach Bobbitt Hey there. My name is Zach Bobbitt. I have a Masters of Science degree in Applied Statistics and I’ve worked on machine learning algorithms for professional businesses in both healthcare and retail. I’m passionate about statistics, machine learning, and data visualization and I created Statology to be a resource for both students and teachers alike. My goal with this site is to help you learn statistics through using simple terms, plenty of real-world examples, and helpful illustrations. One Reply to “What Are Residuals in Statistics?” Thank you! Reply Leave a Reply Cancel reply ×
6476	https://www.mdarc.org/operating-aids/metric-system/metric-conversions	MDARC - Metric Conversions Search this site Embedded Files Skip to main content Skip to navigation MDARC Home Membership Meetings General Meetings Speakers Speakers 2023 Speakers 2021-2022 Speakers 2019-2020 Speakers 2017-2018 Speakers 2015-2016 Board Meetings Technical Meetings Pacificon Meetings Digital Media Meetings Education and Testing Class Details License Testing Auxiliary Classes On-Line Resources Exam.Tools Public Service Repeater Systems ATV Digital Voice Comparison of DV Modes D-Star D-Star Linking D-Star Registration DMR C4FM (Fusion) M17 EchoLink & IRLP Repeater Rules Repeater Weather Meshtastic About Us COVID Statement Club Announcements Committees Communicating with You Contact Us History of the Club Award Recipients Charter Members History Summary Photos, Videos Members-Special News of the Club Past Officers QSL Cards Rocky Ridge Group Silent Keys W6CX W6LGW Join/Renew/Update Officers & Chairs Official Documents Articles of Incorporation Bylaws Equipment Selling Groups.io Guidelines Policies Privacy Policy Record Retention Policy Terms of Use Zoom Guidelines Store Managing Gmail Managing Groups.io Carrier Newsletter Activities Auction Campout Party Digital Systems DMR NBEMS Winlink DX & Contests Emergency Communications Wilderness Protocol Field Day Holiday Banquet Leap Day Net Nets on MDARC Systems Nets on Other Systems Pacificon Table Vehicles Trailer Van Operating Aids Azimuthal Map Calling Frequencies CTCSS (PL) Greek Alphabet Latitude, Longitude, UTM Lending Library LSB or USB? 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Glossary Links of Interest Other Ham Groups Regulations Amateur Radio Bands Amateur Radio Regulations Emission Designators FCC Monitoring Stations License Copy Acquisition Line A, Line C Obtain an FRN Secondary Status Spectrum Safety Electric Current in the Human Body Technical Coax & Ladder Conductor Properties Constants and Formulas Decibels Peak Envelope Power Periodic Table Schematic Symbols Skin-Effect More Home Membership Meetings General Meetings Speakers Speakers 2023 Speakers 2021-2022 Speakers 2019-2020 Speakers 2017-2018 Speakers 2015-2016 Board Meetings Technical Meetings Pacificon Meetings Digital Media Meetings Education and Testing Class Details License Testing Auxiliary Classes On-Line Resources Exam.Tools Public Service Repeater Systems ATV Digital Voice Comparison of DV Modes D-Star D-Star Linking D-Star Registration DMR C4FM (Fusion) M17 EchoLink & IRLP Repeater Rules Repeater Weather Meshtastic About Us COVID Statement Club Announcements Committees Communicating with You Contact Us History of the Club Award Recipients Charter Members History Summary Photos, Videos Members-Special News of the Club Past Officers QSL Cards Rocky Ridge Group Silent Keys W6CX W6LGW Join/Renew/Update Officers & Chairs Official Documents Articles of Incorporation Bylaws Equipment Selling Groups.io Guidelines Policies Privacy Policy Record Retention Policy Terms of Use Zoom Guidelines Store Managing Gmail Managing Groups.io Carrier Newsletter Activities Auction Campout Party Digital Systems DMR NBEMS Winlink DX & Contests Emergency Communications Wilderness Protocol Field Day Holiday Banquet Leap Day Net Nets on MDARC Systems Nets on Other Systems Pacificon Table Vehicles Trailer Van Operating Aids Azimuthal Map Calling Frequencies CTCSS (PL) Greek Alphabet Latitude, Longitude, UTM Lending Library LSB or USB? Metric System Metric Units Metric Prefixes Metric Conversions Morse Code Operating Procedures Phonetic Alphabet Q-Signals & Prosigns Roman Numerals RST System UTC Time Weather Channels Resources Basics Clubs in NorCal, NV Donate to MDARC Equipment Dipole Length Calcs Dipole Tuning HF Equipment IP Codes Getting Started in Ham Radio 1. License Class Description 2. Getting Your First License 3. Upgrading Your License 4A. Getting On the Air 4B. Getting On the Air-HF 5. Choosing Your First Radio 6. What Do Hams Do 7. Where did “Ham” Come From? Glossary Links of Interest Other Ham Groups Regulations Amateur Radio Bands Amateur Radio Regulations Emission Designators FCC Monitoring Stations License Copy Acquisition Line A, Line C Obtain an FRN Secondary Status Spectrum Safety Electric Current in the Human Body Technical Coax & Ladder Conductor Properties Constants and Formulas Decibels Peak Envelope Power Periodic Table Schematic Symbols Skin-Effect Operating Aids: Azimuthal Map \| Calling Frequencies \| CTCSS (PL) \| Greek Alphabet \| Latitude, Longitude, UTM \| Lending Library \| LSB or USB? \| Metric System \| Morse Code \| Operating Procedures \| Phonetic Alphabet \| Q-Signals & Prosigns \| Roman Numerals \| RST System \| UTC Time \| Weather Channels Metric System:Units \| Prefixes \| Conversions Metric Conversions Below are the factors for converting between SI units and some non-SI units commonly used in the U.S. A large number of other conversion factors exist, far too many to list here. 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6477	https://resources.system-analysis.cadence.com/blog/msa2021-the-relationship-of-the-input-impedance-of-a-transmission-line-to-load-and-characteristic-impedance	Skip to main content The Relationship of the Input Impedance of a Transmission Line to Load and Characteristic Impedance Author Cadence System Analysis Key Takeaways The input impedance of a transmission line is the impedance seen by any signal entering it. It is caused by the physical dimensions of the transmission line and its downstream circuit elements. If a transmission line is ideal, there is no attenuation to the signal amplitudes and the propagation constant turns out to be purely imaginary. When the transmission line length is infinite, the input impedance is equal to the characteristic impedance. Designing electrical circuits involves reading datasheets for component details and specifications While designing a circuit, designers must read through component datasheets to see how much internal resistance is associated with each component. If this is ignored, designers might end up with a circuit drawing a current that is different from its specification. Resistance and impedance are inherent properties of any circuit that resists the flow of current through it. Signals face impedance while in a waveguide, integrated circuit, or transmission line. The input impedance of a transmission line is the impedance seen by any signal entering it. It is caused by the physical dimensions of the transmission line and its downstream circuit elements. It is important for designers to understand input impedance, which is why we’ve put together the following information—read on to learn more. The Input Impedance of a Transmission Line At the entry point of a transmission line, signals encounter input impedance that limits the flow of current through it. The input impedance depends on the complete set of elements present in the circuit. In high-speed and high-frequency circuits, signals can undergo serious degradation due to input impedance. When designing a circuit, input impedance should be considered to ensure signal integrity. There is no way to get rid of input impedance in a circuit. Each of the elements present in the circuit provides some fraction of input impedance to the signal entering the circuit. To determine the input impedance of a circuit with only passive elements such as resistors, capacitors, and inductors, circuit analysis concepts can be used. However, there are non-linear devices, switches, diodes, integrated circuits, cavity resonators, and waveguides in RF and microwave circuits, and, in such circuits, the input impedance is correlated to the voltage level of the signal. In structures such as waveguides, transmission lines, and cavity resonators, the geometry influences the value of input impedance. Considering all these aspects of input impedance, it is safe to say that it plays a significant role in impedance matching networks. If not properly calculated, input impedance will result in poor impedance matching. Calculating the Input Impedance Consider a lossless, high-frequency transmission line where the voltage and currents are given by equations 1 and 2, with the input impedance, characteristic impedance, and load impedance as Zin, Z0, and ZL, respectively. As the transmission line is ideal, there is no attenuation to the signal amplitudes and the propagation constant turns out to be purely imaginary. Let’s define the output terminals with axis point z=0 and input terminals z=-L. Our objective is to find the impedance of the circuit when looking from Z=-L: The input impedance is the ratio of input voltage to the input current and is given by equation 3. By substituting equation 5 into equation 4, we can obtain the input impedance, as given in equation 6: From equation 6, we can conclude that the input impedance of the transmission line depends on the load impedance, characteristic impedance, length of the transmission line, and the phase constant of the signals propagating through it. It is already a known fact that the characteristic impedance Z0 is dependent on the distributed parameters of the transmission line, such as resistance, inductance, capacitance, and conductance (as given by equation 7), which are usually defined per unit length. Whenever any change is made in the circuit, the input impedance changes. The relationship between the characteristic impedance and input impedance can be deduced for certain transmission lines. In the derivation of the input impedance equation, we have considered the finite length of the transmission line. When the transmission line length is infinite, then the input impedance of the transmission line is equal to the characteristic impedance. Whenever the transmission line of finite length is terminated by a load impedance that is equal to the characteristic impedance, there is no reflection of signals (according to equation 7). In this case, the input impedance equals characteristic impedance. The calculation of the input impedance of a transmission line over a range of frequencies is useful to understanding the signal behavior in a circuit. Luckily, Cadence’s software provides tools to calculate impedances and S-parameters at various frequencies. Subscribe to our newsletter for the latest updates. If you’re looking to learn more about how Cadence has the solution for you, talk to us and our team of experts. Contact Us
6478	https://proofwiki.org/wiki/Results_Concerning_Set_Difference_with_Union	Results Concerning Set Difference with Union - ProofWiki Results Concerning Set Difference with Union From ProofWiki Jump to navigationJump to search [x] Contents 1 Theorem 1.1 Set Difference with Union 1.2 Set Difference is Right Distributive over Union 1.3 Set Difference with Set Difference is Union of Set Difference with Intersection 1.4 Set Difference is Subset of Union of Differences Theorem Let: S∖T S∖T denote set differenceS∪T S∪T denote set unionS∩T S∩T denote set intersection. Set Difference with Union R∖(S∪T)=(R∪T)∖(S∪T)=(R∖S)∖T=(R∖T)∖S R∖(S∪T)=(R∪T)∖(S∪T)=(R∖S)∖T=(R∖T)∖S Set Difference is Right Distributive over Union (R∪S)∖T=(R∖T)∪(S∖T)(R∪S)∖T=(R∖T)∪(S∖T) Set Difference with Set Difference is Union of Set Difference with Intersection R∖(S∖T)=(R∖S)∪(R∩T)R∖(S∖T)=(R∖S)∪(R∩T) Set Difference is Subset of Union of Differences R∖S⊆(R∖T)∪(T∖S)R∖S⊆(R∖T)∪(T∖S) Retrieved from " Categories: Set Difference Set Union Navigation menu Personal tools Log in Request account Namespaces Page Discussion [x] English Views Read View source View history [x] More Search Navigation Main Page Community discussion Community portal Recent changes Random proof Help FAQ P r∞f W i k i P r∞f W i k i L A T E X L A T E X commands ProofWiki.org Proof Index Definition Index Symbol Index Axiom Index Mathematicians Books Sandbox All Categories Glossary Jokes To Do Proofread Articles Wanted Proofs More Wanted Proofs Help Needed Research Required Stub Articles Tidy Articles Improvements Invited Refactoring Missing Links Maintenance Tools What links here Related changes Special pages Printable version Permanent link Page information This page was last modified on 4 February 2023, at 13:01 and is 742 bytes Content is available under Creative Commons Attribution-ShareAlike License unless otherwise noted. Privacy policy About ProofWiki Disclaimers
6479	https://www.reddit.com/r/chemhelp/comments/zokbr2/how_to_find_the_ph_value_at_equivalence_point_of/	How to find the pH value at Equivalence Point of a weak acid - strong base titration ? : r/chemhelp Skip to main content How to find the pH value at Equivalence Point of a weak acid - strong base titration ? : r/chemhelp Open menu Open navigation Go to Reddit Home r/chemhelp A chip A close button Get App Get the Reddit app Log In Log in to Reddit Expand user menu Open settings menu Go to chemhelp r/chemhelp r/chemhelp r/chemhelp has made the decision to go dark in light of recently announced reddit API changes. To follow news of the blackout, please check here: 94K Members Online • 2 yr. ago [deleted] How to find the pH value at Equivalence Point of a weak acid - strong base titration ? General/High School Propionic acid (pKa = 4.87) is titrated with a sodium hydroxide solution. All substance concentrations are 0.2 mol/ liter. At exactly what pH value is the equivalence point of the titration of propanoic acid? No titration curve is given, so I don't know how the titration looks like I know the pH value at Equivalence Point of a Strong Acid - Strong Base titration is always 7. But Propionic acid is a weak acid. Is there a formula to find the pH at Equivalence Point of a weak acid - strong base titration by just knowing the pKa and concentration ? After googling a bit, I found this Quora link , which says The pH at the equivalent point of a weak acid and strong base is pH=1/2pKw + 1/2pKa + 1/2log[acid]; [ ] is molar acid concentration. I don't remember having seen it in my lecture, but maybe I didn't pay attention enough. Is this formula accurate ? Read more officialsquarespace • Promoted PSA, Redditors: You don't need a business to have a website. All you need is yourself. And with Squarespace, you can easily create a website that reflects your personal brand, individuality, and identity–all by using its intuitive design, AI, and expressibility tools. Sign Up squarespace.com Sort by: Best Open comment sort options Best Top New Controversial Old Q&A hohmatiy • 2y ago Top 1% Commenter I've never used this formula in your post. No need, as you can just easily calculate it. What is present in the solution at the equivalence point? Which compound(s)? Reply reply [deleted] • 2y ago The conjugate base of the weak acid ? Reply reply 10 more replies 10 more replies More replies More replies FoolishChemist • 2y ago Let's derive it. At the equivalence point, you add enough OH- to neutralize all the HA, so you'll have A- left over A- + H2O <\=> HA + OH- and we can use Kb Kb = Kw/Ka = [HA][OH-]/[A-] Since [HA] = [OH-] from the above equilibrium reaction, we can simplify it as [HA] = sqrt(Kw [A-]/Ka) Now to Henderson-Hasselbach pH = pKa + log [A-]/[HA] = pKa + log (sqrt(Ka [A-]/Kw)) Remember our log rules log xy = log x + log y log xn = n log x -log Ka = pKa pH = pKa + (1/2) log Ka + (1/2) log [A-] - (1/2)log Kw pH = pKa - (1/2) pKa + (1/2) log [A-] + (1/2)pKw pH = (1/2) pKa + (1/2) log [A-] + (1/2)pKw It looks pretty close, the difference being (1/2) log [A-] vs (1/2) log [acid]. However if you just neutralized all your acid, it would all be A- But there is one subtlety that needs to be addressed, the original concentration of the acid is not the same as the concentration of the A- because during the titration, you added a bunch of water so [A-] = [OH-] V_base/V_total where [OH-] is the concentration of NaOH added, V_base is the volume of base added and V_total is the total (base+acid) volume Reply reply [deleted] • 2y ago That makes a lot of sense, thanks a lot ! Reply reply carpetlist • 7mo ago At the equivalence point, you add enough OH- to neutralize all the HA Therefore the concentration of HA is 0? And log(A/HA) is infinite/undefined? Reply reply More replies New to Reddit?Create your account and connect with a world of communities. Continue with GoogleContinue with Google Continue with Email Continue With Phone Number By continuing, you agree to our User Agreement and acknowledge that you understand the Privacy Policy. More posts you may like How to find pH of a weak acid + strong base reaction r/chemhelp• 2 yr. ago r/chemhelp r/chemhelp has made the decision to go dark in light of recently announced reddit API changes. To follow news of the blackout, please check here: 94K Members Online ### How to find pH of a weak acid + strong base reaction 2 upvotes · 7 comments Chemistry professor insists this is correct. Is it? r/chemhelp• 3 mo. ago r/chemhelp r/chemhelp has made the decision to go dark in light of recently announced reddit API changes. To follow news of the blackout, please check here: 94K Members Online ### Chemistry professor insists this is correct. Is it? 31 upvotes · 44 comments Why does phosphorous have an oxidation state of -3 and not +3 in PH3? r/chemhelp• 6 mo. ago r/chemhelp r/chemhelp has made the decision to go dark in light of recently announced reddit API changes. 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To follow news of the blackout, please check here: 94K Members Online ### Can someone help explain this reaction? 34 upvotes · 26 comments Question on pH of a strong acid given M r/chemhelp• 2 yr. ago r/chemhelp r/chemhelp has made the decision to go dark in light of recently announced reddit API changes. To follow news of the blackout, please check here: 94K Members Online ### Question on pH of a strong acid given M 2 upvotes · 2 comments I heard ammonia can't form 4 hydrogen bonds but why doesn't this work? r/chemhelp• 22 days ago r/chemhelp r/chemhelp has made the decision to go dark in light of recently announced reddit API changes. To follow news of the blackout, please check here: 94K Members Online ### I heard ammonia can't form 4 hydrogen bonds but why doesn't this work? 40 upvotes · 15 comments what is this chemical? r/chemhelp• 13 days ago r/chemhelp r/chemhelp has made the decision to go dark in light of recently announced reddit API changes. 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To follow news of the blackout, please check here: 94K Members Online ### Does this compound have chirality? 38 upvotes · 13 comments First ever titration yay! r/chemistry• 3 mo. ago r/chemistry A community for chemists and those who love chemistry 3.8M Members Online ### First ever titration yay! 428 upvotes · 52 comments How would I synthesize this? r/chemhelp• 4 mo. ago r/chemhelp r/chemhelp has made the decision to go dark in light of recently announced reddit API changes. To follow news of the blackout, please check here: 94K Members Online ### How would I synthesize this? 33 upvotes · 29 comments How to determine if molecule dissolves in water or not? (Ignore the pencil marks) r/chemhelp• 2 mo. ago r/chemhelp r/chemhelp has made the decision to go dark in light of recently announced reddit API changes. To follow news of the blackout, please check here: 94K Members Online ### How to determine if molecule dissolves in water or not? 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To follow news of the blackout, please check here: 94K Members Online ### What is the silver ion concentration after adjusting an Ag⁺ solution to: a) 0.010 M potassium iodide b) pH 13 c) 0.0010 M sodium sulphide 1 upvote · 13 comments The distribution coefficient K_d for formic acid for the 1-octanol/water system is 2.59. 2 mmol of formic acid are dissolved in 50 ml of water. a) What quantity of formic acid can be extracted with 20 ml of 1-octanol? b) What is the concentration of formic acid in the water after the extraction? r/chemhelp• 2 yr. ago r/chemhelp r/chemhelp has made the decision to go dark in light of recently announced reddit API changes. To follow news of the blackout, please check here: 94K Members Online ### The distribution coefficient K_d for formic acid for the 1-octanol/water system is 2.59. 2 mmol of formic acid are dissolved in 50 ml of water. a) What quantity of formic acid can be extracted with 20 ml of 1-octanol? b) What is the concentration of formic acid in the water after the extraction? 1 upvote · 6 comments how to choose which pKa value to use for Henderson Hasselbalch equation? r/chemhelp• 2 yr. ago r/chemhelp r/chemhelp has made the decision to go dark in light of recently announced reddit API changes. To follow news of the blackout, please check here: 94K Members Online ### how to choose which pKa value to use for Henderson Hasselbalch equation? 1 upvote · 10 comments [College Chemistry] How to calculate pH value of a Carbonate solution ? r/HomeworkHelp• 2 yr. ago r/HomeworkHelp Need help with homework? We're here for you! The purpose of this subreddit is to help you learn (not complete your last-minute homework), and our rules are designed to reinforce this. 668K Members Online ### [College Chemistry] How to calculate pH value of a Carbonate solution ? 1 upvote · 4 comments Can someone explain this reaction? I'm confused as to how the major product became so small. r/chemhelp• 5 mo. ago r/chemhelp r/chemhelp has made the decision to go dark in light of recently announced reddit API changes. To follow news of the blackout, please check here: 94K Members Online ### Can someone explain this reaction? I'm confused as to how the major product became so small. 90 upvotes · 18 comments Perfect NaOH Titration r/chemistry• 8 mo. ago r/chemistry A community for chemists and those who love chemistry 3.8M Members Online ### Perfect NaOH Titration 60 upvotes · 11 comments I don't understand this!! Why is molecule C correct? r/chemhelp• 5 mo. ago r/chemhelp r/chemhelp has made the decision to go dark in light of recently announced reddit API changes. To follow news of the blackout, please check here: 94K Members Online ### I don't understand this!! Why is molecule C correct? 39 upvotes · 25 comments Public Anyone can view, post, and comment to this community Top Posts Reddit reReddit: Top posts of December 18, 2022 Reddit reReddit: Top posts of December 2022 Reddit reReddit: Top posts of 2022 Reddit Rules Privacy Policy User Agreement Reddit, Inc. © 2025. All rights reserved. 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6480	https://www.cuemath.com/geometry/vertical-angles/	LearnPracticeDownload Vertical Angles Vertical angles are formed when two lines meet each other at a point. They are always equal to each other. In other words, whenever two lines cross or intersect each other, 4 angles are formed. We can observe that two angles that are opposite to each other are equal and they are called vertical angles. They are also referred to as 'Vertically opposite angles' as they lie opposite to each other. \| \| \| --- \| \| 1. \| What are Vertical Angles? \| \| 2. \| Vertical Angles Theorem \| \| 3. \| Vertically Opposite Angles Worksheet \| \| 4. \| FAQs on Vertical Angles \| What are Vertical Angles? When two lines intersect, four angles are formed. There are two pairs of nonadjacent angles. These pairs are called vertical angles. In the image given below, (∠1, ∠3) and (∠2, ∠4) are two vertical angle pairs. Vertical Angles Definition Vertical angles are a pair of non-adjacent angles formed by the intersection of two straight lines. In simple words, vertical angles are located across from one another in the corners of the "X" formed by two straight lines. They are also called vertically opposite angles as they are situated opposite to each other. Vertical Angles Theorem Vertical angles theorem or vertically opposite angles theorem states that two opposite vertical angles formed when two lines intersect each other are always equal (congruent) to each other. Let's learn about the vertical angles theorem and its proof in detail. Statement: Vertical angles (the opposite angles that are formed when two lines intersect each other) are congruent. Vertical Angles Proof The proof is simple and is based on straight angles. We already know that angles on a straight line add up to 180°. So in the above figure, ∠1 + ∠2 = 180° (Since they are a linear pair of angles) --------- (1) ∠1 +∠4 = 180° (Since they are a linear pair of angles) --------- (2) From equations (1) and (2), ∠1 + ∠2 = 180° = ∠1 +∠4. According to transitive property, if a = b and b = c then a = c. Therefore, we can rewrite the statement as ∠1 + ∠2 = ∠1 +∠4. --------(3) By eliminating ∠1 on both sides of the equation (3), we get ∠2 = ∠4. Similarly. we can use the same set of statements to prove that ∠1 = ∠3. Therefore, we conclude that vertically opposite angles are always equal. To find the measure of angles in the figure, we use the straight angle property and vertical angle theorem simultaneously. Let us look at some solved examples to understand this. Vertically Opposite Angles Worksheet The following table is consists of creative vertical angles worksheets. These worksheets are easy and free to download. Try and practice few questions based on vertically opposite angles and enhance the knowledge about the topic. \| \| \| --- \| \| Vertical Angles Worksheet - 1 \| Download PDF \| \| Vertical Angles Worksheet - 2 \| Download PDF \| \| Vertical Angles Worksheet - 3 \| Download PDF \| \| Vertical Angles Worksheet - 4 \| Download PDF \| Important Notes Vertical angles are always equal. Vertical angles can be supplementary as well as complimentary. Vertical angles are always nonadjacent. Topics Related to Vertical Angles Check out some interesting articles related to vertical angles. Angles Alternate Angles Alternate Interior Angles Theorem Complementary Angles Complementary Angle Calculator Supplementary Angles Geometry Read More Download FREE Study Materials Angles Worksheets Vertical Angles Worksheets Worksheets on Angles Vertical Angles Examples Example 1: Find the measure of ∠f from the figure using the vertical angles theorem. Solution: In the image given below, we can observe that AE and DC are two straight lines. Here, ∠DOE and ∠AOC are vertical angles. ∠DOE = ∠AOC 118° = 90° + ∠f ∠f = 118° - 90° ∠f = 28° Therefore, ∠f = 28° 2. Example 2: In the figure shown below ∠f is equal to 79° because vertically opposite angles are equal. Is the statement right? Justify your answer. Solution: Here, 79° and f are located opposite, but they are not vertical angles as the angles are not formed by the intersection of two straight lines. Here, BD is not a straight line. Therefore, ∠f is not equal to 79°. The given statement is false. 3. Example 3: If angle b is three times the size of angle a, find out the values of angles a and b by using the vertical angles theorem. Solution: From the figure, we can observe that 80° and the sum of the angles a and b are vertically opposite. Which means ∠a + ∠b = 80°. It is given that ∠b = 3∠a. Substituting the values in the equation of ∠a + ∠b = 80°, we get, ∠a + 3∠a = 80° 4∠a = 80° ∠a= 80°/4 ∠a= 20° Since ∠b = 3∠a, ∠b = 3 × 20° ∠b = 60° Therefore, ∠a= 20° and ∠b= 60° Show Answer > Have questions on basic mathematical concepts? Become a problem-solving champ using logic, not rules. Learn the why behind math with our Cuemath’s certified experts. Book a Free Trial Class Practice Questions on Vertical Angles View Answer > FAQs on Vertical Angles What are Vertical Angles in Geometry? Vertical angles are formed when two lines intersect each other. Out of the 4 angles that are formed, the angles that are opposite to each other are vertical angles. They are also referred to as 'vertically opposite angles. These angles are always equal. ☛Also Read Pairs of Angles Transversals and Related Angles Interior Angles Are Vertical Angles Congruent? When two straight lines intersect each other vertical angles are formed. Vertical angles are always congruent and equal. Vertical angles are congruent as the two pairs of non-adjacent angles formed by intersecting two lines superimpose on each other. ☛Check out and read Congruent Angles Congruent Congruent Triangles Are Vertical Angles Supplementary? When any two angles sum up to 180°, we call them supplementary angles. If there is a case wherein, the vertical angles are right angles or equal to 90°, then the vertical angles are 90° each. Therefore, the sum of these two angles will be equal to 180°. So in such cases, we can say that vertical angles are supplementary. It is to be noted that this is a special case, wherein the vertical angles are supplementary. Otherwise, in all the other cases where the value of each of the vertical angles is less than or more than 90 degrees, they are not supplementary. ☛Check out the difference between the following: Supplementary Angles Complementary Angles Linear Pair of Angles What is the Vertical Angle Theorem? The vertical angle theorem states that the angles formed by two intersecting lines which are called vertical angles are congruent. The vertical angles are of equal measurements. For example, If ∠a, ∠b, ∠c, ∠d are the 4 angles formed by two intersecting lines and ∠a is vertically opposite to ∠b and ∠c is vertically opposite to ∠d, then ∠a is congruent to ∠b and ∠c is congruent to ∠d. Can Vertical Angles Be Right Angles? Yes, vertical angles can be right angles. When the two opposite vertical angles measure 90° each, then the vertical angles are said to be right angles. This can be observed from the x-axis and y-axis lines of a cartesian graph. ☛Check out Right Angle 90 Degree Angle 180 Degree Angle How to Measure the Value of Vertical Angle? While solving such cases, first we need to observe the given parameters carefully. If the angle next to the vertical angle is given then it is easy to determine the value of vertical angles by subtracting the given value from 180 degrees to As it is proved in geometry that the vertical angle and its adjacent angle are supplementary (180°) to each other. How do you tell if an Angle is an Adjacent or Vertical Angle? Vertical angles are the angles formed when two lines intersect each other. The opposite angles formed by these lines are called vertically opposite angles. Whereas, adjacent angles are two angles that have one common arm and a vertex. Can Vertical Angles Be Adjacent? Vertical angles are opposite from each other whereas, adjacent angles are the ones next to each other. Thus, vertical angles can never be adjacent to each other. Are Vertical Angles Always Congruent? Yes, vertical angles are always congruent. The intersection of two lines makes 4 angles. In this, two pairs of vertical angles are formed. They are equal in measure and are congruent. Q1: What is the value of each angle if two lines intersect in such a way that the four vertical angles are equal? Q2: Two vertical angles are: Q3: What is the measure of the vertically opposite angle whose value is 70°? Q4: A pair of opposite angles formed by intersecting lines are called _____ Q5: If two vertical angles measure (x+10°) and 105°, what is the value of x? 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6481	https://www.youtube.com/watch?v=xp6ibuI8UuQ	Visualizing vectors in 2 dimensions \| Two-dimensional motion \| Physics \| Khan Academy Khan Academy 9090000 subscribers 2580 likes Description 836057 views Posted: 16 Jun 2011 Courses on Khan Academy are always 100% free. Start practicing—and saving your progress—now: Visualizing, adding and breaking down vectors in 2 dimensions. Created by Sal Khan. Watch the next lesson: Missed the previous lesson? Physics on Khan Academy: Physics is the study of the basic principles that govern the physical world around us. We'll start by looking at motion itself. Then, we'll learn about forces, momentum, energy, and other concepts in lots of different physical situations. To get the most out of physics, you'll need a solid understanding of algebra and a basic understanding of trigonometry. About Khan Academy: Khan Academy offers practice exercises, instructional videos, and a personalized learning dashboard that empower learners to study at their own pace in and outside of the classroom. We tackle math, science, computer programming, history, art history, economics, and more. Our math missions guide learners from kindergarten to calculus using state-of-the-art, adaptive technology that identifies strengths and learning gaps. We've also partnered with institutions like NASA, The Museum of Modern Art, The California Academy of Sciences, and MIT to offer specialized content. For free. For everyone. Forever. #YouCanLearnAnything Subscribe to Khan Academy’s Physics channel: Subscribe to Khan Academy: 99 comments Transcript: [Voiceover] All the problems we've been dealing with so far have essentially been happening in one dimension. You could go forward or back. So you could go forward or back. Or right or left. Or you could go up or down. What I wanna start to talk about in this video is what happens when we extend that to two dimensions or we can even just extend what we're doing in this video to three or four, really an arbitrary number of dimensions. Although if you're dealing with classical mechanics you normally don't have to go more than three dimensions. And if you're gonna deal with more than one dimension, especially in two dimensions, we're also gonna be dealing with two-dimensional vectors. And I just wanna make sure, through this video, that we understand at least the basics of two-dimensional vectors. Remember, a vector is something that has both magnitude and direction. So the first thing I wanna do is just give you a visual understanding of how vectors in two dimensions would add. So let's say I have a vector right here. That is vector A. So, once again, its magnitude is specified by the length of this arrow. And its direction is specified by the direction of the arrow. So it's going in that direction. Now let's say I have another vector. Let's call it vector B. Let's call it vector B. It looks like this. Now what I wanna do in this video is think about what happens when I add vector A to vector B. So there's a couple things to think about when you visually depict vectors. The important thing is, for example, for vector A, that you get the length right and you get the direction right. Where you actually draw it doesn't matter. So this could be vector A. This could also be vector A. Notice, it has the same length and it has the same direction. This is also vector A. I could draw vector A up here. It does not matter. I could draw vector A up there. I could draw vector B. I could draw vector B over here. It's still vector B. It still has the same magnitude and direction. Notice, we're not saying that its tail has to start at the same place that vector A's tail starts at. I could draw vector B over here. So I can always have the same vector but I can shift it around. So I can move it up there. As long as it has the same magnitude, the same length, and the same direction. And the whole reason I'm doing that is because the way to visually add vectors... If I wanted to add vector A plus vector B... And I'll show you how to do it more analytically in a future video. I can literally draw vector A. I draw vector A. So that's vector A, right over there. And then I can draw vector B, but I put the tail of vector B to the head of vector A. So I shift vector B over so its tail is right at the head of vector A. And then vector B would look something like this. It would look something like this. And then if you go from the tail of A all the way to the head of B, all the way to the head of B, and you call that vector C, that is the sum of A and B. And it should make sense, if you think about it. Let's say these were displacement vectors. So A shows that you're being displaced this much in this direction. B shows that you're being displaced this much in this direction. So the length of B in that direction. And if I were to say you have a displacement of A, and then you have a displacement of B, what is your total displacement? So you would have had to be, I guess, shifted this far in this direction, and then you would be shifted this far in this direction. So the net amount that you've been shifted is this far in that direction. So that's why this would be the sum of those. Now we can use that same idea to break down any vector in two dimensions into, we could say, into its components. And I'll give you a better sense of what that means in a second. So if I have vector A. Let me pick a new letter. Let's call this vector "vector X." Let's call this "vector X." I can say that vector X is going to be the sum of this vector right here in green and this vector right here in red. Notice, X starts at the tail of the green vector and goes all the way to the head of the magenta vector. And the magenta vector starts at the head of the green vector and then finishes, I guess, well where it finishes is where vector X finishes. And the reason why I do this... And, you know, hopefully from this comparable explanation right here, says, okay, look, the green vector plus the magenta vector gives us this X vector. That should make sense. I put the head of the green vector to the tail of this magenta vector right over here. But the whole reason why I did this is, if I can express X as a sum of these two vectors, it then breaks down X into its vertical component and its horizontal component. So I could call this the horizontal component, or I should say the vertical component. X vertical. And then I could call this over here the X horizontal. Or another way I could draw it, I could shift this X vertical over. Remember, it doesn't matter where I draw it, as long as it has the same magnitude and direction. And I could draw it like this. X vertical. And so what you see is is that you could express this vector X... Let me do it in the same colors. You can express this vector X as the sum of its horizontal and its vertical components. As the sum of its horizontal and its vertical components. Now we're gonna see over and over again that this is super powerful because what it can do is it can turn a two-dimensional problem into two separate one-dimensional problems, one acting in a horizontal direction, one acting in a vertical direction. Now let's do it a little bit more mathematical. I've just been telling you about length and all of that. But let's actually break down... Let me just show you what this means, to break down the components of a vector. So let's say that I have a vector that looks like this. Let me do my best to... Let's say I have a vector that looks like this. It's length is five. So let me call this vector A. So vector A's length is equal to five. And let's say that its direction... We're gonna give its direction by the angle between the direction its pointing in and the positive X axis. So maybe I'll draw an axis over here. So let's say that this right over here is the positive Y axis going in the vertical direction. This right over here is the positive X axis going in the horizontal direction. And to specify this vector's direction I will give this angle right over here. And I'm gonna give a very peculiar angle, but I picked this for a specific reason, just so things work out neatly in the end. And I'm gonna give it in degrees. It's 36.8699 degrees. So I'm picking that particular number for a particular reason. Now what I wanna do is I wanna figure out this vector's horizontal and vertical component. So I wanna break it down into something that's going straight up or down and something that's going straight right or left. So how do I do this? Well, one, I could just draw them, visually, see what they look like. So its vertical component would look like this. It would start... Its vertical component would look like this. And its horizontal component would look like this. Its horizontal component would look like this. The horizontal component, the way I drew it, it would start where vector A starts and go as far in the X direction as vector A's tip, but only in the X direction, and then you need to, to get back to the head of vector A, you need to have its vertical component. And we can sometimes call this, we could call the vertical component over here A sub Y, just so that it's moving in the Y direction. And we can call this horizontal component A sub X. Now what I wanna do is I wanna figure out the magnitude of A sub Y and A sub X. So how do we do that? Well, the way we drew this, I've essentially set up a right triangle for us. This is a right triangle. We know the length of this triangle, or the length of this side, or the length of the hypotenuse. That's going to be the magnitude of vector A. And so the magnitude of vector A is equal to five. We already knew that up here. So how do we figure out the sides? Well, we could use a little bit of basic trigonometry. If we know the angle, and we know the hypotenuse, how do we figure out the opposite side to the angle? So this right here, this right here is the opposite side to the angle. And if we forgot some of our basic trigonometry we can relearn it right now. Soh-cah-toa. Sine is opposite over hypotenuse. Cosine is adjacent over hypotenuse. Tangent is opposite over adjacent. So we have the angle, we want the opposite, and we have the hypotenuse. So we could say that the sine of our angle, the sine of 36.899 degrees, is going to be equal to the opposite over the hypotenuse. The opposite side of the angle is the magnitude of our Y component. ...is going to be equal to the magnitude of our Y component, the magnitude of our Y component, over the magnitude of the hypotenuse, over this length over here, which we know is going to be equal to five. Or if you multiply both sides by five, you get five sine of 36.899 degrees, is equal to the magnitude of the vertical component of our vector A. Now before I take out the calculator and figure out what this is, let me do the same thing for the horizontal component. Over here we know this side is adjacent to the angle. And we know the hypotenuse. And so cosine deals with adjacent and hypotenuse. So we know that the cosine of 36.899 degrees is equal to... Cosine is adjacent over hypotenuse. So it's equal to the magnitude of our X component over the hypotenuse. The hypotenuse here has... Or the magnitude of the hypotenuse, I should say, which has a length of five. Once again, we multiply both sides by five, and we get five times the cosine of 36.899 degrees is equal to the magnitude of our X component. So let's figure out what these are. Let me get the calculator out. Let me get my trusty TI-85 out. I wanna make sure it's in degree mode. So let me check. Yep, we're in degree mode right over there. Don't wanna... Make sure we're not in radian mode. Now let's exit that. And we have the vertical component is equal to five times the sine of 36.899 degrees, which is, if we round it, right at about three. So this is equal to... So the magnitude of our vertical component is equal to three. And then let's do the same thing for our horizontal component. So now we have five times the cosine of 36.899 degrees, is, if once again we round it to, I guess, our hundredths place, we get it to being four. So we get it to being four. So we see here is a situation where we have... This is a classic three-four-five Pythagorean triangle. The magnitude of our horizontal component is four. The magnitude of our vertical component, right over here, is equal to three. And once again, you might say, Sal, why are we going through all of this trouble? And we'll see in the next video that if we say something has a velocity, in this direction, of five meters per second, we could break that down into two component velocities. We could say that that's going in the upwards direction at three meters per second, and it's also going to the right in the horizontal direction at four meters per second. And it allows us to break up the problem into two simpler problems, into two one-dimensional problems, instead of a bigger two-dimensional one.
6482	https://www.aafp.org/pubs/afp/issues/2008/0515/p1415.html	MARK D. ANDREWS, MD, AND MARIANTHE BURNS, MD Am Fam Physician. 2008;77(10):1415-1420 Author disclosure: Nothing to disclose. Article Sections The common dermatophyte genera Trichophyton, Microsporum, and Epidermophyton are major causes of superficial fungal infections in children. These infections (e.g., tinea corporis, pedis, cruris, and unguium) are typically acquired directly from contact with infected humans or animals or indirectly from exposure to contaminated soil or fomites. A diagnosis usually can be made with a focused history, physical examination, and potassium hydroxide microscopy. Occasionally, Wood's lamp examination, fungal culture, or histologic tissue examination is required. Most tinea infections can be managed with topical therapies; oral treatment is reserved for tinea capitis, severe tinea pedis, and tinea unguium. Topical therapy with fungicidal allylamines may have slightly higher cure rates and shorter treatment courses than with fungistatic azoles. Although oral griseofulvin has been the standard treatment for tinea capitis, newer oral antifungal agents such as terbinafine, itraconazole, and fluconazole are effective, safe, and have shorter treatment courses. The common dermatophyte genera Trichophyton, Microsporum, and Epidermophyton are major causes of superficial fungal infections in children. These infections (e.g., tinea corporis, pedis, cruris, and unguium) are typically acquired directly from contact with infected humans or animals or indirectly from exposure to contaminated soil or fomites. A diagnosis usually can be made with a focused history, physical examination, and potassium hydroxide microscopy. Occasionally, Wood's lamp examination, fungal culture, or histologic tissue examination is required. Most tinea infections can be managed with topical therapies; oral treatment is reserved for tinea capitis, severe tinea pedis, and tinea unguium. Topical therapy with fungicidal allylamines may have slightly higher cure rates and shorter treatment courses than with fungistatic azoles. Although oral griseofulvin has been the standard treatment for tinea capitis, newer oral antifungal agents such as terbinafine, itraconazole, and fluconazole are effective, safe, and have shorter treatment courses. Tinea refers to dermatophyte infections, which are generally classified by anatomic location: tinea capitis is located on the scalp, tinea pedis on the feet, tinea corporis on the body, tinea cruris on the groin, and tinea unguium on the nails. Tinea is also called ringworm, especially when located on the body, and is caused by a group of fungi that infect only the outer keratinous layer of skin, hair, and nails. These fungi cannot survive on mucosal surfaces, such as the mouth or vaginal area. Superficial tinea infections are some of the most common dermatologic conditions in children.1 Clinical recommendation \| Evidence rating \| References Direct visualization of hyphae with potassium hydroxide microscopy is the mainstay for the office-based diagnosis of tinea. \| C \| 1,2,4,5 Griseofulvin (Grifulvin) and terbinafine (Lamisil; approved by the U.S. Food and Drug Administration for use in children two years and older) are effective therapies in the treatment of tinea capitis. \| B \| 8–12 Topical fungicidal allylamines such as terbinafine and butenafine (Mentax) are usually sufficient to treat tinea corporis, tinea cruris, and tinea pedis and are superior to the older fungistatic azoles clotrimazole (Lotrimin AF) and miconazole (Micatin). \| A \| 17,18 Oral and topical treatment regimens for tinea unguium have low sustained clinical success rates. \| B \| 24–26 Dermatophytes are aerobic fungi that are divided into three genera (i.e., Trichophyton, Microsporum, and Epidermophyton). Tinea infection is acquired directly from contact with infected humans (anthropophilic organisms) or animals (zoophilic organisms) or indirectly from exposure to contaminated soil or fomites (geophilic organisms). The clinical manifestations of dermatophyte infections varies by the infection site and the patient's immunologic response; genetic susceptibility may play a role in vulnerability to infection.2 Table 1 presents the differential diagnosis of tinea infections,3 and Table 2 summarizes common therapies. Infection \| Common causative species \| Differential diagnosis Tinea capitis \| Trichophyton tonsuransMicrosporum audouiniiZoophilic Microsporum canis \| Alopecia areataImpetigoPediculosisPsoriasisSeborrheic dermatitisTraction alopeciaTrichotillomania Tinea corporis \| Trichophyton rubrumEpidermophyton floccosum \| Cutaneous lupus erythematousDrug eruptionEczemaErythema multiformeGranuloma annulareNummular eczematous dermatitisPityriasis roseaPsoriasisSecondary syphilisTinea (pityriasis) versicolor Tinea cruris \| T. rubrumE. floccosumTrichophyton mentagrophytes \| Candidal intertrigoContact dermatitisErythrasmaPsoriasisSeborrhea Tinea manuum \| T. rubrum \| Same as with tinea pedis Tinea pedis \| T. rubrumT. mentagrophytesE. floccosum \| Bacterial or candidal infectionContact or atopic dermatitisDyshidrosisEczemaPitted keratolysisPsoriasis Tinea unguium \| T. rubrumT. mentagrophytes \| Contact dermatitisLichen planusOnychodystrophyPsoriasis Agent \| Dosage \| Cost (generic) Tinea unguium Terbinafine (Lamisil) \| 250 mg orally every day for 12 weeks†or \| $1,171 ($1,047 to $1,080) 500 mg orally every day during the first week of each month for four months \| $780 ($696 to $720) Itraconazole (Sporanox) \| 200 mg orally every day for 12 weeksor \| $1,838 ($1,241 to $1,713) 400 mg orally every day during the first week of each month for four months‡ \| $1,225 ($828 to $1,142) Tinea capitis Griseofulvin, micronized (Grifulvin) \| 20 mg per kg per day for eight weeks \| $467 ($467 to $578) for a child weighing 30 kg Terbinafine \| 62.5 mg per day for four weeks§ \| $105 (—) for a child older than two years weighing 20 kg Tinea corporis, pedis, cruris Butenafine (Mentax) \| Applied to the lesion and a 2-cm area surrounding the lesion once daily for approximately 14 days \| 30 g: $84 (—) Terbinafine \| Applied to the lesion and a 2-cm area surrounding the lesion twice daily for approximately 14 days \| 15 g: — ($32 to $38) Miconazole (Micatin) \| Applied to the lesion and a 2-cm area surrounding the lesion twice daily for approximately 14 days \| 30 g, 2%: — ($3.20 to $3.30) Clotrimazole (Lotrimin AF) \| Applied to the lesion and a 2-cm area surrounding the lesion twice daily for approximately 14 days \| 24 g: $9.77 ($6.42 to $8.60) Diagnostic Tests MICROSCOPY Potassium hydroxide (KOH) microscopy is essential for the office-based diagnosis of tinea infections.1,2,4,5 This technique directly shows hyphae and confirms infection. The specimen is examined under the microscope after a drop of 10 to 20 percent KOH solution is added to the scraping from the active border of the lesion. KOH microscopy has good sensitivity and is more sensitive than a fungal culture. A positive test result justifies initiation of treatment. KOH microscopy has a 76.5 percent sensitivity and an 81.6 percent negative predictive value for the diagnosis of tinea unguium compared with a 53.2 percent sensitivity and 69.0 percent negative predictive value with culture.4 CULTURE Culture techniques have a limited role in the evaluation and treatment of suspected tinea infection because of the expense and time requirement. However, when the need for long-term oral therapy is anticipated, the infection seems resistant to standard topical therapy, or the diagnosis is unclear, a culture is an appropriate approach for laboratory confirmation. WOOD'S LAMP EXAMINATION Wood's lamp examination for the diagnosis of tinea infection has decreased to near disuse because of the gradually declining number of dermatophytes that fluoresce under ultraviolet light.2,5 Exceptions include tinea capitis caused by zoophilic Microsporum canis and Microsporum audouinii, which fluoresce blue-green; tinea (pityriasis) versicolor caused by Malassezia furfur, which fluoresces pale yellow to white; and erythrasma caused by Corynebacterium minutissimum, which fluoresces bright coral red. Candida infection and tinea cruris do not fluoresce. Tinea Capitis DIAGNOSIS Tinea capitis is the most common fungal infection in children. In the United States, more than 90 percent of tinea capitis cases are caused by Trichophyton tonsurans, and fewer than 5 percent are caused by Microsporum species. Both of these species go beyond superficial infection and invade hair shafts. Patients typically present with scaling of the scalp or circumscribed alopecia with broken hair at the scalp (Figure 1). The differential diagnosis of tinea capitis includes alopecia areata, trichotillomania, traction alopecia, and seborrheic dermatitis. A tinea capitis diagnosis is usually based on physical findings.6 One study showed that children presenting with adenopathy, alopecia, pruritus, and scaling invariably had positive culture results, whereas children who lacked adenopathy and scaling did not have positive culture results (Table 3).6 Sign or symptom \| Positive likelihood ratio \| Sensitivity (%) \| Specificity (%) Adenopathy \| 7.5 \| 94 \| 87 Alopecia \| 3.3 \| 84 \| 75 Pruritus \| 1.4 \| 75 \| 46 Scaling \| 1.1 \| 71 \| 35 Direct KOH microscopy of the skin scraping and hair shafts can confirm infection when positive, but false negatives are common even with experienced examiners. Wood's lamp examination can help identify Microsporum species, but Trichophyton species, which are increasingly common in the United States, do not fluoresce under a Wood's lamp. If necessary, fungal culture can be obtained (using a toothbrush or cotton swab) and then inoculated onto a fungal culture medium. The cotton-swab method is performed by moistening a sterile cotton swab and vigorously rubbing it over the affected areas of the scalp.7 TREATMENT Tinea capitis must be treated with oral agents to penetrate the affected hair shafts. Although griseofulvin (Grifulvin) is often considered the treatment of choice for children, other agents are increasingly more cost-effective and may be more convenient.8 If griseofulvin is used, micronized oral griseofulvin is recommended at a dosage of 20 mg per kg per day for at least eight weeks. The long duration of treatment and the adverse effects of nausea, vomiting, and photosensitivity often negatively affect patient compliance. Four weeks of terbinafine (Lamisil) therapy is as effective as eight weeks of griseofulvin therapy for the treatment of Trichophyton infections.9,10 Terbinafine is not as effective against Microsporum infections, which are much less common.11 A recent meta-analysis showed no significant difference in tolerability or adverse effects between griseofulvin and terbinafine.11 The recommended dosage of terbinafine is based on patient weight (62.5 mg per day in children weighing less than 20 kg [44 lb]; 125 mg per day in children weighing 20 to 40 kg [44 to 88 lb, 3 oz]; and 250 mg per day in children weighing more than 40 kg). Terbinafine is approved for use in children two years or older.12 The cost of a complete regimen of terbinafine is less than that of griseofulvin because of the shorter duration of therapy.13 The shorter regimen may also help increase patient compliance and decrease resistance secondary to early termination of therapy. Adverse effects from terbinafine are uncommon, although there are concerns about liver toxicity. Liver function testing (e.g., aspartate and alanine transaminase) is advised in all patients taking terbinafine at baseline and at follow-up in patients with baseline abnormalities or in those who experience gastrointestinal symptoms.14 Ketoconazole (Nizoral) and itraconazole (Sporanox) are not recommended for treating tinea capitis because of the limited data on their effectiveness and safety. Adjunctive selenium sulfide should be used to decrease the spread of tinea capitis.15 Selenium sulfide 2.5% shampoo is most effective and should be applied at least two to three times per week and should remain on the scalp for at least five minutes per application.16 Tinea Corporis DIAGNOSIS Tinea corporis is most commonly caused by Trichophyton species. Patients typically present with an annular patch or plaque with an advancing, raised, scaling border and central clearing (Figure 2). The differential diagnosis includes other annular skin lesions. Most patients with tinea corporis are diagnosed clinically. KOH microscopy of a skin scraping can determine if hyphae are present. Culture confirmation is usually not required. TREATMENT Topical treatment is often sufficient to cure tinea corporis, although oral medications can be used for patients with severe infection or for infections that do not respond to topical therapy. A topical antifungal agent is applied to the lesion and a 2-cm area surrounding the lesion once or twice daily; therapy is continued for one week after the lesion appears to resolve. Data show that newer butenafine (Mentax) and terbinafine therapies are more effective than older miconazole (Micatin) and clotrimazole therapies, primarily because of the fungicidal properties of the newer agents compared with the fungistatic action of the older agents.17,18 Combination antifungal/corticosteroid preparations are not recommended because of a greater risk of adverse effects, primarily from the higher-potency steroid component.19 Cure rates are lower and the cost is higher with combination therapy than with antifungal creams alone. Tinea Corporis Gladiatorum This special classification of tinea lesions is acquired in wrestlers through skin-to-skin contact. Annular lesions do not always occur; instead, they may appear as erythematous, scaling papules and plaques. They typically occur on the head, neck, and arms, and rarely occur on the legs.20 To prevent transmission, localized lesions should be covered with dressings, and patients with more extensive lesions should not wrestle. Patients can resume participation after one week of topical or systemic therapy. Tinea Cruris (Jock Itch) DIAGNOSIS Tinea cruris is a superficial infection of the groin occurring predominantly in adolescent and young adult men. However, it has become more common in postpubertal females who are overweight or who often wear tight jeans or pantyhose. Because the common causes of tinea pedis (e.g., Trichophyton rubrum, Trichophyton mentagrophytes, Epidermophyton floccosum) are also common causes of tinea cruris, the two infections usually occur together. Although the clinical presentation of tinea cruris varies, the lesion border is usually active to some degree with pustules or vesicles (unless the lesion is chronic). The background rash is red to reddish-brown and is usually a symmetric macule with fairly well-demarcated borders (Figure 3). The rash usually spares the scrotum and is often pruritic; acute rashes also may have a burning quality. TREATMENT The feet should always be evaluated as a possible initial source of tinea cruris. Although the appearance and symptoms of tinea cruris are distinctive, the condition must be distinguished from other common disorders, including candidal intertrigo and erythrasma. Candidal intertrigo is more uniformly red with no central clearing and may have satellite lesions, whereas erythrasma is a bacterial infection that is more uniformly brown with slight scaling and no active border. Topical therapy is usually a sufficient treatment for tinea cruris. Therapies include terbinafine cream or spray applied once daily for one week and butenafine 1% cream applied once daily for two weeks. Tinea Pedis (Athlete's Foot) DIAGNOSIS Tinea pedis is caused by the same dermatophyte species as tinea cruris. It is more common in adolescents but rare in prepubertal children. The most important predisposing factor to acquiring tinea pedis appears to be exposure to a moist environment and maceration of the skin. The infection may be less prevalent in societies that do not commonly wear shoes.21 Tinea pedis typically appears as a white macerated area between the toes, although a more diffuse dry scaling process often caused by T. rubrum (“moccasin type”) may also occur. Another pattern of presentation is characterized by an inflammatory vesiculobullous eruption occurring primarily on the soles of the feet (Figure 4). Conditions that may mimic tinea pedis in children include contact and allergic dermatitis and occasionally atopic dermatitis. Unlike tinea pedis, dermatitis generally spares the intertriginous areas. TREATMENT Topical antifungal agents such as terbinafine applied once daily for one week or butenafine 1% applied once daily for two weeks are effective for most forms of tinea pedis. The therapies can be applied directly to the web spaces and other affected areas. As with treatment of tinea cruris, these therapies are more effective than clotrimazole (Lotrimin AF) and miconazole. Occasionally, systemic terbinafine therapy, 250 mg taken orally every day for two weeks, or fluconazole (Diflucan), 150 mg taken once weekly for three weeks, can be used to treat severe or refractory infections. Tinea Unguium DIAGNOSIS Tinea unguium is an infection of the finger- or toenails (Figure 5) and is a subset of onychomycosis, a broader category of infections that also include yeast and nondermatophyte molds. Tinea unguium is fairly uncommon in prepubertal children. Risk factors in adolescents include associated tinea pedis, improperly fitting shoes, and diabetes. Tinea unguium is a common cause of nail dystrophy; however, it should be differentiated from other acquired and congenital conditions. A nail scraping for KOH microscopy with or without a culture is recommended for confirmation. Although studies have shown that histologic analysis of nail clippings using a periodic acid-Schiff stain is the most accurate test, it is costly.22 A recent study found that a KOH preparation (20 percent KOH, briefly heated) of a subungual debris scraping was 80 percent sensitive and 72 percent specific for the diagnosis of tinea unguium. This is compared with a culture, which was only 59 percent sensitive and 82 percent specific at four weeks.23 TREATMENT The best evidence supports terbinafine for treating adolescents with tinea unguium,24 although griseofulvin is usually used in children. Terbinafine has similar effectiveness and adverse effect profile as itraconazole with a lower cost and a more favorable drug interaction profile. Data show that terbinafine is more effective than itraconazole or fluconazole.24 A recent study found that continuous terbinafine therapy is more effective than pulse (intermittent) terbinafine therapy.25 However, true “success” rates are not high for antifungal therapy, with only 35 to 50 percent of patients who are disease-free at one year.26 Smith SD, Relman DA. Dermatophytes. In: Wilson WR, Sande MA. Current Diagnosis and Treatment in Infectious Diseases. New York, NY: McGraw-Hill Professional; 2001:777–778. Hostetter MK. Fungal infections in normal children. In: Gershon AA, Hotez PJ, Katz SL, Krugman S, eds. Gershon: Krugman's Infectious Diseases of Children. 11th ed. St. Louis, Mo.: Mosby; 2004. Noble SL, Forbes RC, Stamm PL. Diagnosis and management of common tinea infections. Am Fam Physician. 1998;58(1):163-174. Karimzadegan-Nia M, Mir-Amin-Mohammadi A, Bouzari N, Firooz A. Comparison of direct smear, culture and histology for the diagnosis of onychomycosis. Australas J Dermatol. 2007;48(1):18-21. Aly R. Ecology, epidemiology and diagnosis of tinea capitis. Pediatr Infect Dis J. 1999;18(2):180-185. Hubbard TW. The predictive value of symptoms in diagnosing childhood tinea capitis. Arch Pediatr Adolesc Med. 1999;153(11):1150-1153. Friedlander SF, Pickering B, Cunningham BB, Gibbs NF, Eichenfield LF. Use of the cotton swab method in diagnosing tinea capitis. Pediatrics. 1999;104(2 pt 1):276-279. Bennett ML, Fleischer AB, Loveless JW, Feldman SR. Oral griseofulvin remains the treatment of choice for tinea capitis in children. Pediatr Dermatol. 2000;17(4):304-309. Friedlander SF, Aly R, Krafchik B, et al. Terbinafine in the treatment of Trichophyton tinea capitis: a randomized, double-blind, parallel-group, duration-finding study. Pediatrics. 2002;109(4):602-607. Ergin S, Ergin C, Erdogan BS, Kaleli I, Evliyaoglu D. An experience from an outbreak of tinea capitis gladiatorum due to Trichophyton tonsurans. Clin Exp Dermatol. 2006;31(2):212-214. Fleece D, Gaughan JP, Aronoff SC. Griseofulvin versus terbinafine in the treatment of tinea capitis: a meta-analysis of randomized, clinical trials. Pediatrics. 2004;114(5):1312-1315. Pickering LK. Tinea capitis. In: Pickering LK. Red Book: 2006 Report of the Committee on Infectious Diseases. Elk Grove Village, Ill.: American Academy of Pediatrics; 2006:654–656. Cáceres-Ríos H, Rueda M, Ballona R, Bustamante B. Comparison of terbinafine and griseofulvin in the treatment of tinea capitis. J Am Acad Dermatol. 2000;42(1 pt 1):80-84. Lamisil (terbinafine). In: Physician's Desk Reference Red Book 2007. Oxford, U.K.: Blackwell; 2007:2232. Allen HB, Honig PJ, Leyden JJ, McGinley KJ. Selenium sulfide: adjuncitve therapy for tinea capitis. Pediatrics. 1982;69(1):81-83. Givens TG, Murray MM, Baker RC. Comparison of 1% and 2.5% selenium sulfide in the treatment of tinea capitis. Arch Pediatr Adolesc Med. 1995;149(7):808-811. Hart R, Bell-Syer SE, Crawford F, Torgerson DJ, Young P, Russell I. Systematic review of topical treatments for fungal infections of the skin and nails of the feet. BMJ. 1999;319(7202):79-82. Davis R, Balfour JA. Terbinafine. A pharmacoeconomic evaluation of its use in superficial fungal infections. Pharmacoeconomics. 1995;8(3):253-269. Greenberg HL, Shwayder TA, Bieszk N, Fivenson DP. Clotrimazole/beta-methasone diproprionate: a review of costs and complications in the treatment of common cutaneous fungal infections. Pediatr Dermatol. 2002;19(1):78-81. Adams BB. Tinea corporis gladiatorum: a cross-sectional study. J Am Acad Dermatol. 2000;43(6):1039-1041. Habif TP. Superficial fungal infections. In: Clinical Dermatology: A Color Guide to Diagnosis and Therapy. 4th ed. New York, NY: Mosby; 2004. Lilly KK, Koshnick RL, Grill JP, Khalil ZM, Nelson DB, Warshaw EM. Cost-effectiveness of diagnostic tests for toenail onychomycosis: a repeated measure, single-blinded, cross-sectional evaluation of 7 diagnostic tests. J Am Acad Dermatol. 2006;55(4):620-626. Weinberg JM, Koestenblatt EK, Tutrone WD, Tishler HR, Najarian L. Comparison of diagnostic methods in the evaluation of onychomycosis. J Am Acad Dermatol. 2003;49(2):193-197. Crawford F, Young P, Godfrey C, et al. Oral treatments for toenail onychomycosis: a systematic review. Arch Dermatol. 2002;138(6):811-816. Warshaw EM, Fett DD, Bloomfield HE, et al. Pulse versus continuous terbinafine for onychomycosis: a randomized, double-blind, controlled trial. J Am Acad Dermatol. 2005;53(4):578-584. Epstein E. How often does oral treatment of toenail onychomycosis produce a disease-free nail? An analysis of published data. Arch Dermatol. 1998;134(12):1551-1554. Continue Reading More in AFP More in PubMed Article Sections Copyright © 2008 by the American Academy of Family Physicians. 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6483	https://teaching.martahidegkuti.com/shared/lnotes/5Trig/arcs.pdf	Arcs and Sectors in Circles page 1 . Theorem: If a circle has radius r, then its circumference and area can be computed as C = 2πr and A = πr2 Sample Problems 1. The radius of a circle is 8m. a) Find the length of an arc subtended by a central angle of 20◦. b) Find the area of a sector subtended by a central angle of 20◦. 2. Phoenix, AZ and Salt Lake City, UT have approximately the same longitude. The radius of the earth is approximately 3960 miles. The latitude of Phoenix is 33.5◦and that of Salt Lake City is 40. 7◦. Find the distance to the nearest mile between the two cities. 3. Find the radius of a circle if we know that a sector subtended by a central angle of 67◦has an area of 42m2. 4. The minute hand of a clock is 5 cm long. Find the speed of the top of the minute hand. Express your answer in meter per second. 5. A satellite can be seen over the same point on Earth above the equator. It is 200 miles above the surface. Find the speed of the satellite in miles per hour. (The radius of the earth is 3960 mi). 6. Consider the cricles C1 and C2. An arc subtended by a central angle of 40◦in C1 is has the same length as an arc in C2 that is subtended by a central angle of 60◦. a) Find the ratio between the radii of C1 to C2. b) Find the ratio between the areas of C1 to C2. 7. Consider the image shown. There are four semicircles in a square, each with a radius of 1 unit. Find the exact and approximate value of the shaded area. Practice Problems 1. The radius of a circle is 20m. a) Find the length of an arc subtended by a central angle of 58◦. b) Find the area of a sector subtended by a central angle of 58◦. 2. Seattle, WA and San Fracisco, CA have approximately the same longitude. The radius of the earth is approximately 3960 miles. The latitude of Seattle is 47.67◦and that of San Fracisco is 37. 83◦. Find the distance to the nearest mile between the two cities. 3. Find the radius of a circle if we know that a sector subtended by a central angle of 32◦has an area of 5m2. © Hidegkuti, 2022 Last revised: December 27, 2022 Arcs and Sectors in Circles page 2 4. The hour hand of a clock is 4cm long. Find the speed of the top of the hour hand. Express your answer in meter per second. 5. A satellite can be seen over the same point on Earth above the equator. It is 300 miles above the surface. Find the speed of the satellite in miles per hour. (The radius of the earth is 3960 mi). 6. Consider the circles C1 and C2. An arc subtended by a central angle of 80◦in C1 is has the same length as an arc in C2 that is subtended by a central angle of 48◦. a) Find the ratio between the radii of C1 to C2. b) Find the ratio between the areas of C1 to C2. 7. This problem is from the great Catriona Shearer. The picture shows two quarter circles and a semicircle. What is the shaded area? Enrichment 1. Suppose the earth were a perfect sphere with a perfectly ﬁtting belt of 24 000 miles surrounding it along a great circular path. Suppose the belt was cut, and one hundred feet of additional material was added to the belt, with the ”loose ﬁt” evenly distributed around the earth so that the new belt was still circular with its center at the center of the earth. Which of the following best describes the resulting situation? A) We could slip a piece of paper between the belt and the earth. B) We could get our ﬁngers under the belt. C) We could crawl under the belt. D) We could walk upright under the belt. E) We could drive a truck under the belt. Answers Sample Problems 1. a) 2. 793m b) 11. 170m2 2. 498 miles 3. 8. 475 5m 4. 0.0000873m s = 8. 73 × 10−5 m s 5. 1089. 085 45mi h 6. a) r1 r2 = 3 2 b) A1 A2 = 9 4 7. 8 −2π ≈1. 716 8 (unit2) Practice Problems 1. a) 20. 245 82m b) 202. 458 2m2 2.) 680 miles 3.) 4. 231 42m 4. 0.000005817 76m s = 5. 817 76 × 10−6 m s 5. 1115. 265 4mi h 6. a) r1 r2 = 3 5 b) A1 A2 = 9 25 7. 18π © Hidegkuti, 2022 Last revised: December 27, 2022 Arcs and Sectors in Circles page 3 Sample Problems Solutions 1. The radius of a circle is 8m. a) Find the length of an arc subtended by a central angle of 20◦. Solution: The circumference of the circle is C = 2πr = 2π (8m) = 16πm. This is the arc belonging to the central angle of 360◦. If we wanted to ﬁnd the arc length belonging to a central angle of 1◦, we would just need to divide the circumference by 360. To obtain the arc length belonging to a central angle of 20◦, we would need to take the 1◦degree arc twenty times. x = C 360 (20) = C 18 = 16π m 18 = 8 9π m ≈2. 793 m b) Find the area of a sector subtended by a central angle of 20◦. Solution: The area of the circle is A = πr2 = π (8m)2 = 64πm2. This is the area of the sector belonging to the central angle of 360◦. If we wanted to ﬁnd the area of a sector with a central angle of 1◦, we would just need to divide the area of the circle by 360. To obtain the area of a sector with a central angle of 20◦, we would need to take the 1◦degree sector twenty times. x = A 360 (20) = A 18 = 64πm2 18 = 32 9 πm2 ∼ = 11. 170 m2 2. Phoenix, AZ and Salt Lake City, UT have approximately the same longitude. The radius of the earth is approximately 3960 miles. The latitude of Phoenix is 33.5◦and that of Salt Lake City is 40. 7◦. Find the distance to the nearest mile between the two cities. Solution: This is just an arc length problem. Consider the picture shown. Let E be the point on the equator located on the same longitude as Salt Lake City (point S) and Phoenix (point P). It is given that angle SCE = 40. 7◦and angle PCE = 33.5◦. The distance between the cities is the arc length SP, belonging to the central angle 40. 7◦−33.5◦= 7. 2◦. s = C 360 (7.2) = 2πR (7.2) 360 = 2π (3960mi) (7.2) 360 ≈497. 628 276mi We round this result to 498 miles. 3. Find the radius of a circle if we know that a sector subtended by a central angle of 67◦has an area of 42m2. Solution: The ratio between the area of the circle and the sector is the same as the ratio of 360◦to 67◦. Also, recall that the area of a circle with radius r is A = πr2. πr2 42m2 = 360◦ 67◦ solve for r r2 = 360◦42m2 67◦(π) r = r 360 · 42m2 67π = r 15 120 67π m ≈8. 475 5m © Hidegkuti, 2022 Last revised: December 27, 2022 Arcs and Sectors in Circles page 4 4. The minute hand of a clock is 5 cm long. Find the speed of the top of the minute hand. Express your answer in meter per second. Solution: The minute hand travels a full circle in one hour. (Note that 1cm = 0.01m ) speed = distance time = 2πr t = 2π (5cm) 1hr = 10πcm 3600s = 10π (0.01m) 3600s = 0.1π 3600 m s = 8. 727 × 10−5 m s 5. A satellite can be seen over the same point on Earth above the equator. It is 200 miles above the surface. Find the speed of the satellite in miles per hour. (The radius of the earth is 3960 mi). Solution: If the satellite appears to be at the same point, it must travel along with the Earth, covering a full circle in exactly one day. The radius of the circle is the sum of the radius of the Earth and the distance from the surface. speed = distance time = 2π (3960mi + 200mi) 24h = 1040π 3 mi h ≈1089. 085 45mi h 6. Consider the cricles C1 and C2. An arc subtended by a central angle of 40◦in C1 is has the same length as an arc in C2 that is subtended by a central angle of 60◦. a) Find the ratio between the radii of C1 to C2. b) Find the ratio between the areas of C1 to C2. Solution: Let r1 and r2 denote the radii of C1 and C2. We write an equation expressing the arc lengths and simplify the equation. 2πr1 360 (40) = 2πr2 360 (60) divide by 2π, multiply by 360 40r1 = 60r2 divide by 20 2r1 = 3r2 divide by 2r2 r1 r2 = 3 2 Thus the ratio of r1 to r2 is 3 to 2. b) We express the ratio between the ares and hope that we can relate that to the ratio of the radii we found in part a) A1 A2 = πr2 1 πr2 2 = r2 1 r2 2 = r1 r2 2 = 3 2 2 = 9 4 So the ratio between the areas is 9 to 4 . 7. Consider the image shown. There are four semicircles in a square, each with a radius of 1 unit. Find the exact and approximate value of the shaded area. Solution: Square ABCD has an area of 1 unit2. Sector ABD is subtended by a central angle of 90◦. Therefore, its area is exactly 1 4 of the area of a full circle, that is, Asector = 1 4π (1)2 = π 4 (unit2) The area shaded with yellow is the difference of the two: Ayellow = 1 −π 4 (unit2) The next image shows that the area we need to ﬁnd is composed of the same shape, repeated eight times. Therefore, the total shaded area we are looking for is A = 8 1 −π 4 = 8 −2π ≈1. 716 8 (unit2) For more documents like this, visit our page at and click on Lecture Notes. E-mail questions or comments to mhidegkuti@ccc.edu. © Hidegkuti, 2022 Last revised: December 27, 2022
6484	https://courses.grainger.illinois.edu/cs573/fa2010/notes/98-induction.pdf	Algorithms Appendix: Proof by Induction Jeder Genießende meint, dem Baume habe es an der Frucht gelegen; aber ihm lag am Samen. [Everyone who enjoys thinks that the fundamental thing about trees is the fruit, but in fact it is the seed.] — Friedrich Wilhelm Nietzsche, Vermischte Meinungen und Sprüche [Mixed Opinions and Maxims] (1879) In view of all the deadly computer viruses that have been spreading lately, Weekend Update would like to remind you: When you link up to another computer, you’re linking up to every computer that that computer has ever linked up to. — Dennis Miller, “Saturday Night Live”, (c. 1985) Anything that, in happening, causes itself to happen again, happens again. — Douglas Adams (2005) Proof by Induction Induction is a method for proving universally quantiﬁed propositions—statements about all elements of a (usually inﬁnite) set. Induction is also the single most useful tool for reasoning about, developing, and analyzing algorithms. These notes give several examples of inductive proofs, along with a standard boilerplate and some motivation to justify (and help you remember) why induction works. 1 Prime Divisors: Proof by Smallest Counterexample A divisor of a positive integer n is a positive integer p such that the ratio n/p is an integer. The integer 1 is a divisor of every positive integer (because n/1 = n), and every integer is a divisor of itself (because n/n = 1). A proper divisor of n is any divisor of n other than n itself. A positive integer is prime if it has exactly two divisors, which must be 1 and itself; equivalently; a number is prime if and only if 1 is its only proper divisor. A positive integer is composite if it has more than two divisors (or equivalently, more than one proper divisor). The integer 1 is neither prime nor composite, because it has exactly one divisor, namely itself. Let’s prove our ﬁrst theorem: Theorem 1. Every integer greater than 1 has a prime divisor. The very ﬁrst thing that you should notice, after reading just one word of the theorem, is that this theorem is universally quantiﬁed—it’s a statement about all the elements of a set, namely, the set of positive integers larger than 1. If we were forced at gunpoint to write this sentence using fancy logic notation, the ﬁrst character would be the universal quantiﬁer ∀, pronounced ‘for all’. Fortunately, that won’t be necessary. There are only two ways to prove a universally quantiﬁed statement: directly or by contradiction. Let’s say that again, louder: There are only two ways to prove a universally quantiﬁed statement: directly or by contradiction. Here are the standard templates for these two methods, applied to Theorem 1: Direct proof: Let n be an arbitrary integer greater than 1. . . . blah blah blah . . . Thus, n has at least one prime divisor. □ c ⃝Copyright 2010 Jeff Erickson. Released under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 License ( Free distribution is strongly encouraged; commercial distribution is expressly forbidden. See for the most recent revision. 1 Algorithms Appendix: Proof by Induction Proof by contradiction: For the sake of argument, assume there is an integer greater than 1 with no prime divisor. Let n be an arbitrary integer greater than 1 with no prime divisor. . . . blah blah blah . . . But that’s just silly. Our assumption must be incorrect. □ The shaded boxes . . . blah blah blah . . . indicate missing proof details (which you have to ﬁll in). Most people ﬁnd proofs by contradiction easier to discover than direct proofs, so let’s try that ﬁrst. Proof by contradiction: For the sake of argument, assume there is an integer greater than 1 with no prime divisor. Let n be an arbitrary integer greater than 1 with no prime divisor. Since n is a divisor of n, and n has no prime divisors, n cannot be prime. Thus, n must have at least one divisor d such that 1 < d < n. Let d be an arbitrary divisor of n such that 1 < d < n. Since n has no prime divisors, d cannot be prime. Thus, d has at least one divisor d′ such that 1 < d′ < d. Let d′ be an arbitrary divisor of d such that 1 < d′ < d. Because d/d′ is an integer, n/d′ = (n/d) · (d/d′) is also an integer. Thus, d′ is also a divisor of n. Since n has no prime divisors, d′ cannot be prime. Thus, d′ has at least one divisor d′′ such that 1 < d′′ < d′. Let d′′ be an arbitrary divisor of d′ such that 1 < d′′ < d′. Because d′/d′′ is an integer, n/d′′ = (n/d′) · (d′/d′′) is also an integer. Thus, d′′ is also a divisor of n. Since n has no prime divisors, d′′ cannot be prime. . . . blah HELP! blah I’M STUCK IN AN INFINITE LOOP! blah . . . But that’s just silly. Our assumption must be incorrect. □ We seem to be stuck in an inﬁnite loop, looking at smaller and smaller divisors d > d′ > d′′ > ···, none of which are prime. But this loop can’t really be inﬁnite. There are only n −1 positive integers smaller than n, so the proof must end after at most n−1 iterations. But how do we turn this observation into a formal proof? We need a single, self-contained proof for all integers n; we’re not allowed to write longer proofs for bigger integers. The trick is to jump directly to the smallest counterexample. Proof by smallest counterexample: For the sake of argument, assume that there is an integer greater than 1 with no prime divisor. Let n be the smallest integer greater than 1 with no prime divisor. Since n is a divisor of n, and n has no prime divisors, n cannot be prime. Thus, n has a divisor d such that 1 < d < n. Let d be a divisor of n such that 1 < d < n. Because n is the smallest counterexample, d has a prime divisor. Let p be a prime divisor of d. Because d/p is an integer, n/p = (n/d) · (d/p) is also an integer. Thus, p is also a divisor of n. But this contradicts our assumption that n has no prime divisors! So our assumption must be incorrect. □ Hooray, our ﬁrst proof! We’re done! Um. . . well. . . no, we’re deﬁnitely not done. That’s a ﬁrst draft up there, not a ﬁnal polished proof. We don’t write proofs just to convince ourselves; proofs are primarily a tool to convince other people. (In particular, ‘other people’ includes the people grading your homeworks and exams.) And while 2 Algorithms Appendix: Proof by Induction proofs by contradiction are usually easier to write, direct proofs are almost always easier to read. So as a service to our audience (and our grade), let’s transform our minimal-counterexample proof into a direct proof. Let’s ﬁrst rewrite the indirect proof slightly, to make the structure more apparent. First, we break the assumption that n is the smallest counterexample into three simpler assumptions: (1) n is an integer greater than 1; (2) n has no prime divisors; and (3) there are no smaller counterexamples. Second, instead of dismissing the possibility than n is prime out of hand, we include an explicit case analysis. Proof by smallest counterexample: Let n be an arbitrary integer greater than 1. For the sake of argument, suppose n has no prime divisor. Assume that every integer k such that 1 < k < n has a prime divisor. There are two cases to consider: Either n is prime, or n is composite. • Suppose n is prime. Then n is a prime divisor of n. • Suppose n is composite. Then n has a divisor d such that 1 < d < n. Let d be a divisor of n such that 1 < d < n. Because no counterexample is smaller than n, d has a prime divisor. Let p be a prime divisor of d. Because d/p is an integer, n/p = (n/d) · (d/p) is also an integer. Thus, p is a prime divisor of n. In each case, we conclude that n has a prime divisor. But this contradicts our assumption that n has no prime divisors! So our assumption must be incorrect. □ Now let’s look carefully at the structure of this proof. First, we assumed that the statement we want to prove is false. Second, we proved that the statement we want to prove is true. Finally, we concluded from the contradiction that our assumption that the statement we want to prove is false is incorrect, so the statement we want to prove must be true. But that’s just silly. Why do we need the ﬁrst and third steps? After all, the second step is a proof all by itself! Unfortunately, this redundant style of proof by contradiction is extremely common, even in professional papers. Fortunately, it’s also very easy to avoid; just remove the ﬁrst and third steps! Proof by induction: Let n be an arbitrary integer greater than 1. Assume that every integer k such that 1 < k < n has a prime divisor. There are two cases to consider: Either n is prime or n is composite. • First, suppose n is prime. Then n is a prime divisor of n. • Now suppose n is composite. Then n has a divisor d such that 1 < d < n. Let d be a divisor of n such that 1 < d < n. Because no counterexample is smaller than n, d has a prime divisor. Let p be a prime divisor of d. Because d/p is an integer, n/p = (n/d) · (d/p) is also an integer. Thus, p is a prime divisor of n. In both cases, we conclude that n has a prime divisor. □ This style of proof is called induction.1 The assumption that there are no counterexamples smaller than n is called the induction hypothesis. The two cases of the proof have different names. The 1Many authors use the high-falutin’ name the principle of mathematical induction, to distinguish it from inductive reasoning, the informal process by which we conclude that pigs can’t whistle, horses can’t ﬂy, and NP-hard problems cannot be solved in polynomial time. We already know that every proof is mathematical (and arguably, all mathematics is proof), so as a description of a proof technique, the adjective ‘mathematical’ is simply redundant. 3 Algorithms Appendix: Proof by Induction ﬁrst case, which we argue directly, is called the base case. The second case, which actually uses the induction hypothesis, is called the inductive case. You may ﬁnd it helpful to actually label the induction hypothesis, the base case(s), and the inductive case(s) in your proof. The following point cannot be emphasized enough: The only difference between a proof by induc-tion and a proof by smallest counterexample is the way we write down the argument. The essential structure of the proofs are exactly the same. The core of our original indirect argument is a proof of the following implication for all n: n has no prime divisor =⇒some number smaller than n has no prime divisor. The core of our direct proof is the following logically equivalent implication: every number smaller than n has a prime divisor =⇒n has a prime divisor The left side of this implication is just the induction hypothesis. The proofs we’ve been playing with have been very careful and explicit; until you’re comfortable writing your own proofs, you should be equally careful. A more mature proof-writer might express the same proof more succinctly as follows: Proof by induction: Let n be an arbitrary integer greater than 1. Assume that every integer k such that 1 < k < n has a prime divisor. If n is prime, then n is a prime divisor of n. On the other hand, if n is composite, then n has a proper divisor d. The induction hypothesis implies that d has a prime divisor p. The integer p is also a divisor of n. □ A proof in this more succinct form is still worth full credit, provided the induction hypothesis is written explicitly and the case analysis is obviously exhaustive. A professional mathematician would write the proof even more tersely: Proof: Induction. □ And you can write that tersely, too, when you’re a professional mathematician. 2 The Axiom of Induction Why does this work? Well, let’s step back to the original proof by smallest counterexample. How do we know that a smallest counterexample exists? This seems rather obvious, but in fact, it’s impossible to prove without using the following fact: Every non-empty set of positive integers has a smallest element. Every set X of positive integers is the set of counterexamples to some proposition P(n) (speciﬁcally, the proposition n ̸∈X). Thus, the Axiom of Induction can be rewritten as follows: If the proposition P(n) is false for some positive integer n, then the proposition (P(1) ∧P(2) ∧··· ∧P(n −1) ∧¬P(n)) is true for some positive integer n. Equivalently, in English: If some statement about positive integers has a counterexample, then that statement has a smallest counterexample. 4 Algorithms Appendix: Proof by Induction We can write this implication in contrapositive form as follows: If the proposition (P(1) ∧P(2) ∧··· ∧P(n −1) ∧¬P(n)) is false for every positive integer n, then the proposition P(n) is true for every positive integer n. Finally, let’s rewrite the ﬁrst half of this statement in a logically equivalent form, by replacing ¬(p∧¬q) with p →q. If the implication (P(1) ∧P(2) ∧··· ∧P(n −1)) →P(n) is true for every positive integer n, then the proposition P(n) is true for every positive integer n. This formulation is usually called the Axiom of Induction. In a proof by induction that P(n) holds for all n, the conjunction (P(1) ∧P(2) ∧··· ∧P(n −1)) is the inductive hypothesis. A proof by induction for the proposition “P(n) for every positive integer n” is nothing but a di-rect proof of the more complex proposition “(P(1) ∧P(2) ∧··· ∧P(n −1)) →P(n) for every posi-tive integer n”. Because it’s a direct proof, it must start by considering an arbitrary positive integer, which we might as well call n. Then, to prove the implication, we explicitly assume the hypothesis (P(1) ∧P(2) ∧··· ∧P(n −1)) and then prove the conclusion P(n) for that particular value of n. The proof almost always breaks down into two or more cases, each of which may or may not actually use the inductive hypothesis. Here is the boilerplate for every induction proof. Read it. Learn it. Use it. Theorem: P(n) for every positive integer n. Proof by induction: Let n be an arbitrary positive integer. Assume inductively that P(k) is true for every positive integer k < n. There are several cases to consider: • Suppose n is . . . blah blah blah . . . Then P(n) is true. • Suppose n is . . . blah blah blah . . . The inductive hypothesis implies that . . . blah blah blah . . . Thus, P(n) is true. In each case, we conclude that P(n) is true. □ Some textbooks distinguish between several different types of induction: ‘regular’ induction versus ‘strong’ induction versus ‘complete’ induction versus ‘structural’ induction versus ‘transﬁnite’ induction versus ‘Noetherian’ induction. Distinguishing between these different types of induction is pointless hairsplitting; I won’t even deﬁne them. Every ‘different type’ of induction proof is provably equivalent to a proof by smallest counterexample. (Later we will consider inductive proofs of statements about partially ordered sets other than the positive integers, for which ‘smallest’ has a different meaning, but this difference will prove to be inconsequential.) 3 Stamps and Recursion Let’s move on to a completely different example. 5 Algorithms Appendix: Proof by Induction Theorem 2. Given an unlimited supply of 5-cent stamps and 7-cent stamps, we can make any amount of postage larger than 23 cents. We could prove this by contradiction, using a smallest-counterexample argument, but let’s aim for a direct proof by induction this time. We start by writing down the induction boilerplate, using the standard induction hypothesis: There is no counterexample smaller than n. Proof by induction: Let n be an arbitrary integer greater than 23. Assume that for any integer k such that 23 < k < n, we can make k cents in postage. . . . blah blah blah . . . Thus, we can make n cents in postage. □ How do we ﬁll in the details? One approach is to think about what you would actually do if you really had to make n cents in postage. For example, you might start with a 5-cent stamp, and then try to make n−5 cents in postage. The inductive hypothesis says you can make any amount of postage bigger than 23 cents and less than n cents. So if n −5 > 23, then you already know that you can make n −5 cents in postage! (You don’t know how to make n −5 cents in postage, but so what?) Let’s write this observation into our proof as two separate cases: either n ≥30 (where our approach works) and n ≤29 (where we don’t know what to do yet). Proof by induction: Let n be an arbitrary integer greater than 23. Assume that for any integer k such that 23 < k < n, we can make k cents in postage. There are two cases to consider: Either n > 28 or n ≤28. • Suppose n > 28. Then 23 < n −5 < n. Thus, the induction hypothesis implies that we can make n−5 cents in postage. Adding one more 5-cent stamp gives us n cents in postage. • Now suppose n ≤28. . . . blah blah blah . . . In both cases, we can make n cents in postage. □ What do we do in the second case? Fortunately, this case considers only ﬁve integers: 24, 25, 26, 27, and 28. There might be a clever way to solve all ﬁve cases at once, but why bother? They’re small enough that we can ﬁnd a solution by brute force in less than a minute. To make the proof more readable, I’ll unfold the nested cases and list them in increasing order. Proof by induction: Let n be an arbitrary integer greater than 23. Assume that for any integer k such that 23 < k < n, we can make k cents in postage. There are six cases to consider: n = 24, n = 25, n = 26, n = 27, n = 28, and n > 28. • 24 = 7 + 7 + 5 + 5 • 25 = 5 + 5 + 5 + 5 + 5 • 26 = 7 + 7 + 7 + 5 • 27 = 7 + 5 + 5 + 5 + 5 • 28 = 7 + 7 + 7 + 7 • Suppose n > 28. Then 23 < n −5 < n. Thus, the induction hypothesis implies that we can make n−5 cents in postage. Adding one more 5-cent stamp gives us n cents in postage. In all cases, we can make n cents in postage. □ Voilà! An induction proof! More importantly, we now have a recipe for discovering induction proofs. 6 Algorithms Appendix: Proof by Induction 1. Write down the boilerplate. Write down the universal invocation (‘Let n be an arbitrary. . . ’), the induction hypothesis, and the conclusion, with enough blank space for the remaining details. Don’t be clever. Don’t even think. Just write. This is the easy part. To emphasize the common structure, the boilerplate will be indicated in green for the rest of this handout. 2. Think big. Don’t think how to solve the problem all the way down to the ground; you’ll only make yourself dizzy. Don’t think about piddly little numbers like 1 or 5 or 10100. Instead, think about how to reduce the proof about some absfoluckingutely ginormous value of n to a proof about some other number(s) smaller than n. This is the hard part. 3. Look for holes. Look for cases where your inductive argument breaks down. Solve those cases directly. Don’t be clever here; be stupid but thorough. 4. Rewrite everything. Your ﬁrst proof is a rough draft. Rewrite the proof so that your argument is easier for your (unknown?) reader to follow. The cases in an inductive proof always fall into two categories. Any case that uses the inductive hypothesis is called an inductive case. Any case that does not use the inductive hypothesis is called a base case. Typically, but not always, base cases consider a few small values of n, and the inductive cases consider everything else. Induction proofs are usually clearer if we present the base cases ﬁrst, but I ﬁnd it much easier to discover the inductive cases ﬁrst. In other words, I recommend writing induction proofs backwards. Well-written induction proofs very closely resemble well-written recursive programs. We computer scientists use induction primarily to reason about recursion, so maintaining this resemblance is ex-tremely useful—we only have to keep one mental pattern, called ‘induction’ when we’re writing proofs and ‘recursion’ when we’re writing code. Consider the following C and Scheme programs for making n cents in postage: void postage(int n) { assert(n>23); switch ($n$) { case 24: printf("7+7+5+5"); break; case 25: printf("5+5+5+5+5"); break; case 26: printf("7+7+7+5"); break; case 27: printf("7+5+5+5+5"); break; case 28: printf("7+7+7+7"); break; default: postage(n-5); printf("+5"); } } (define (postage n) (cond ((= n 24) (5 5 7 7)) ((= n 25) (5 5 5 5 5)) ((= n 26) (5 7 7 7)) ((= n 27) (5 5 5 5 7)) ((= n 28) (7 7 7 7)) ((> n 28) (cons 5 (postage (- n 5)))))) 7 Algorithms Appendix: Proof by Induction The C program begins by declaring the input parameter (“Let n be an arbitrary integer. . . ") and asserting its range (“. . . greater than 23."). (Scheme programs don’t have type declarations.) In both languages, the code branches into six cases: ﬁve that are solved directly, plus one that is handled by invoking the inductive hypothesis recursively. 4 More on Prime Divisors Before we move on to different examples, let’s prove another fact about prime numbers: Theorem 3. Every positive integer is a product of prime numbers. First, let’s write down the boilerplate. Hey! I saw that! You were thinking, weren’t you? Stop that this instant! Don’t make me turn the car around. First we write down the boilerplate. Proof by induction: Let n be an arbitrary positive integer. Assume that any positive integer k < n is a product of prime numbers. There are some cases to consider: . . . blah blah blah . . . Thus, n is a product of prime numbers. □ Now let’s think about how you would actually factor a positive integer n into primes. There are a couple of different options here. One possibility is to ﬁnd a prime divisor p of n, as guaranteed by Theorem 1, and recursively factor the integer n/p. This argument works as long as n ≥2, but what about n = 1? The answer is simple: 1 is the product of the empty set of primes. What else could it be? Proof by induction: Let n be an arbitrary positive integer. Assume that any positive integer k < n is a product of prime numbers. There are two cases to consider: either n = 1 or n ≥2. • If n = 1, then n is the product of the elements of the empty set, each of which is prime, green, sparkly, vanilla, and hemophagic. • Suppose n > 1. Let p be a prime divisor of n, as guaranteed by Theorem 2. The inductive hypothesis implies that the positive integer n/p is a product of primes, and clearly n = (n/p) · p. In both cases, n is a product of prime numbers. □ But an even simpler method is to factor n into any two proper divisors, and recursively handle them both. This method works as long as n is composite, since otherwise there is no way to factor n into smaller integers. Thus, we need to consider prime numbers separately, as well as the special case 1. Proof by induction: Let n be an arbitrary positive integer. Assume that any positive integer k < n is a product of prime numbers. There are three cases to consider: either n = 1, n is prime, or n is composite. • If n = 1, then n is the product of the elements of the empty set, each of which is prime, red, broody, chocolate, and lycanthropic. • If n is prime, then n is the product of one prime number, namely n. • Suppose n is composite. Let d be any proper divisor of n (guaranteed by the deﬁ-nition of ‘composite’), and let m = n/d. Since both d and m are positive integers smaller than n, the inductive hypothesis implies that d and m are both products of prime numbers. We clearly have n = d · m. In both cases, n is a product of prime numbers. □ 8 Algorithms Appendix: Proof by Induction 5 Summations Here’s an easy one. Theorem 4. n X i=0 3i = 3n+1 −1 2 for every non-negative integer n. First let’s write down the induction boilerplate, which empty space for the details we’ll ﬁll in later. Proof by induction: Let n be an arbitrary non-negative integer. Assume inductively that k X i=0 3i = 3k+1 −1 2 for every non-negative integer k < n. There are some number of cases to consider: . . . blah blah blah . . . We conclude that n X i=0 3i = 3n+1 −1 2 . □ Now imagine you are part of an inﬁnitely long assembly line of mathematical provers, each assigned to a particular non-negative integer. Your task is to prove this theorem for the integer 8675310. The regulations of the Mathematical Provers Union require you not to think about any other integer but your own. The assembly line starts with the Senior Master Prover, who proves the theorem for the case n = 0. Next is the Assistant Senior Master Prover, who proves the theorem for n = 1. After him is the Assistant Assistant Senior Master Prover, who proves the theorem for n = 2. Then the Assistant Assistant Assistant Senior Master Prover proves the theorem for n = 3. As the work proceeds, you start to get more and more bored. You attempt strike up a conversation with Jenny, the prover to your left, but she ignores you, preferring to focus on the proof. Eventually, you fall into a deep, dreamless sleep. An undetermined time later, Jenny wakes you up by shouting, “Hey, doofus! It’s your turn!” As you look around, bleary-eyed, you realize that Jenny and everyone to your left has ﬁnished their proofs, and that everyone is waiting for you to ﬁnish yours. What do you do? What you do, after wiping the drool off your chin, is stop and think for a moment about what you’re trying to prove. What does that P notation actually mean? Intuitively, we can expand the notation as follows: 8675310 X i=0 3i = 30 + 31 + ··· + 38675309 + 38675310. Notice that this expression also contains the summation that Jenny just ﬁnished proving something about: 8675309 X i=0 3i = 30 + 31 + ··· + 38675308 + 38675309. Putting these two expressions together gives us the following identity: 8675310 X i=0 3i = 8675309 X i=0 3i + 38675310 In fact, this recursive identity is the deﬁnition of P . Jenny just proved that the summation on the right is equal to (38675310 −1)/2, so we can plug that into the right side of our equation: 8675310 X i=0 3i = 8675309 X i=0 3i + 38675310 = 38675310 −1 2 + 38675310. 9 Algorithms Appendix: Proof by Induction And it’s all downhill from here. After a little bit of algebra, you simplify the right side of this equation to (38675311 −1)/2, wake up the prover to your right, and start planning your well-earned vacation. Let’s insert this argument into our boilerplate, only using a generic ‘big’ integer n instead of the speciﬁc integer 8675310: Proof by induction: Let n be an arbitrary non-negative integer. Assume inductively that k X i=0 3i = 3k+1 −1 2 for every non-negative integer k < n. There are two cases to consider: Either n is big or n is small . • If n is big , then n X i=0 3i = n−1 X i=0 3i + 3n [deﬁnition of P ] = 3n −1 2 + 3n [induction hypothesis, with k = n −1] = 3n+1 −1 2 [algebra] • On the other hand, if n is small , then . . . blah blah blah . . . In both cases, we conclude that n X i=0 3i = 3n+1 −1 2 . □ Now, how big is ‘big’, and what do we do when n is ‘small’? To answer the ﬁrst question, let’s look at where our existing inductive argument breaks down. In order to apply the induction hypothesis when k = n −1, the integer n −1 must be non-negative; equivalently, n must be positive. But that’s the only assumption we need: The only case we missed is n = 0. Fortunately, this case is easy to handle directly. Proof by induction: Let n be an arbitrary non-negative integer. Assume inductively that k X i=0 3i = 3k+1 −1 2 for every non-negative integer k < n. There are two cases to consider: Either n = 0 or n ≥1. • If n = 0, then n X i=0 3i = 30 = 1, and 3n+1 −1 2 = 31 −1 2 = 1. • On the other hand, if n ≥1, then n X i=0 3i = n−1 X i=0 3i + 3n [deﬁnition of P ] = 3n −1 2 + 3n [induction hypothesis, with k = n −1] = 3n+1 −1 2 [algebra] In both cases, we conclude that n X i=0 3i = 3n+1 −1 2 . □ 10 Algorithms Appendix: Proof by Induction Here is the same proof, written more tersely; the non-standard symbol IH == indicates the use of the induction hypothesis. Proof by induction: Let n be an arbitrary non-negative integer, and assume inductively that Pk i=0 3i = (3k+1 −1)/2 for every non-negative integer k < n. The base case n = 0 is trivial, and for any n ≥1, we have n X i=0 3i = n−1 X i=0 3i + 3n IH == 3n −1 2 + 3n = 3n+1 −1 2 . □ This is not the only way to prove this theorem by induction; here is another: Proof by induction: Let n be an arbitrary non-negative integer, and assume inductively that Pk i=0 3i = (3k+1 −1)/2 for every non-negative integer k < n. The base case n = 0 is trivial, and for any n ≥1, we have n X i=0 3i = 30 + n X i=1 3i = 30 + 3 · n−1 X i=0 3i IH == 30 + 3 · 3n −1 2 = 3n+1 −1 2 . □ In the remainder of these notes, I’ll give several more examples of induction proofs. In some cases, I give multiple proofs for the same theorem. Unlike the earlier examples, I will not describe the thought process that lead to the proof; in each case, I followed the basic outline on page 7. 6 Tiling with Triominos The next theorem is about tiling a square checkerboard with triominos. A triomino is a shape composed of three squares meeting in an L-shape. Our goal is to cover as much of a 2n × 2n grid with triominos as possible, without any two triominos overlapping, and with all triominos inside the square. We can’t cover every square in the grid—the number of squares is 4n, which is not a multiple of 3—but we can cover all but one square. In fact, as the next theorem shows, we can choose any square to be the one we don’t want to cover. Almost tiling a 16 × 16 checkerboard with triominos. Theorem 5. For any non-negative integer n, the 2n × 2n checkerboard with any square removed can be tiled using L-shaped triominos. Here are two different inductive proofs for this theorem, one ‘top down’, the other ‘bottom up’. 11 Algorithms Appendix: Proof by Induction Proof by top-down induction: Let n be an arbitrary non-negative integer. Assume that for any non-negative integer k < n, the 2k ×2k grid with any square removed can be tiled using triominos. There are two cases to consider: Either n = 0 or n ≥1. • The 20×20 grid has a single square, so removing one square leaves nothing, which we can tile with zero triominos. • Suppose n ≥1. In this case, the 2n ×2n grid can be divided into four smaller 2n−1 × 2n−1 grids. Without loss of generality, suppose the deleted square is in the upper right quarter. With a single L-shaped triomino at the center of the board, we can cover one square in each of the other three quadrants. The induction hypothesis implies that we can tile each of the quadrants, minus one square. In both cases, we conclude that the 2n × 2n grid with any square removed can be tiled with triominos. □ Top-down inductive proof of Theorem 4. Proof by bottom-up induction: Let n be an arbitrary non-negative integer. Assume that for any non-negative integer k < n, the 2k ×2k grid with any square removed can be tiled using triominos. There are two cases to consider: Either n = 0 or n ≥1. • The 20×20 grid has a single square, so removing one square leaves nothing, which we can tile with zero triominos. • Suppose n ≥1. Then by clustering the squares into 2 × 2 blocks, we can transform any 2n × 2n grid into a 2n−1 × 2n−1 grid. Suppose square (i, j) has been removed from the 2n×2n grid. The induction hypothesis implies that the 2n−1×2n−1 grid with block (⌊i/2⌋,⌊j/2⌋) removed can be tiled with double-size triominos. Each double-size triomono can be tiled with four smaller triominos, and block (⌊i/2⌋,⌊j/2⌋) with square (i, j) removed is another triomino. In both cases, we conclude that the 2n × 2n grid with any square removed can be tiled with triominos. □ Second proof of Theorem 4. 7 Binary Numbers Exist Theorem 6. Every non-negative integer can be written as the sum of distinct powers of 2. 12 Algorithms Appendix: Proof by Induction Intuitively, this theorem states that every number can be represented in binary. (That’s not a proof, by the way; it’s just a restatement of the theorem.) I’ll present four distinct inductive proofs for this theorem. The ﬁrst two are standard, by-the-book induction proofs. Proof by top-down induction: Let n be an arbitrary non-negative integer. Assume that any non-negative integer less than n can be written as the sum of distinct powers of 2. There are two cases to consider: Either n = 0 or n ≥1. • The base case n = 0 is trivial—the elements of the empty set are distinct and sum to zero. • Suppose n ≥1. Let k be the largest integer such that 2k ≤n, and let m = n −2k. Observe that m < 2k+1 −2k = 2k. Because 0 ≤m < n, the inductive hypothesis implies that m can be written as the sum of distinct powers of 2. Moreover, in the summation for m, each power of 2 is at most m, and therefore less than 2k. Thus, m + 2k is the sum of distinct powers of 2. In either case, we conclude that n can be written as the sum of distinct powers of 2. □ Proof by bottom-up induction: Let n be an arbitrary non-negative integer. Assume that any non-negative integer less than n can be written as the sum of distinct powers of 2. There are two cases to consider: Either n = 0 or n ≥1. • The base case n = 0 is trivial—the elements of the empty set are distinct and sum to zero. • Suppose n ≥1, and let m = ⌊n/2⌋. Because 0 ≤m < n, the inductive hypothesis implies that m can be written as the sum of distinct powers of 2. Thus, 2m can also be written as the sum of distinct powers of 2, each of which is greater than 20. If n is even, then n = 2m and we are done; otherwise, n = 2m + 20 is the the sum of distinct powers of 2. In either case, we conclude that n can be written as the sum of distinct powers of 2. □ The third proof deviates slightly from the induction boilerplate. At the top level, this proof doesn’t actually use induction at all! However, a key step requires its own (straightforward) inductive proof. Proof by algorithm: Let n be an arbitrary non-negative integer. Let S be a multiset containing n copies of 20. Modify S by running the following algorithm: while S has more than one copy of any element 2i Remove two copies of 2i from S Insert one copy of 2i+1 into S Each iteration of this algorithm reduces the cardinality of S by 1, so the algorithm must eventually halt. When the algorithm halts, the elements of S are distinct. We claim that just after each iteration of the while loop, the elements of S sum to n. Proof by induction: Consider an arbitrary iteration of the loop. Assume in-ductively that just after each previous iteration, the elements of S sum to n. Before any iterations of the loop, the elements of S sum to n by deﬁnition. The induction hypothesis implies that just before the current iteration begins, the elements of S sum to n. The loop replaces two copies of some number 2i with their sum 2i+1, leaving the total sum of S unchanged. Thus, when the iteration ends, the elements of S sum to n. □ Thus, when the algorithm halts, the elements of S are distinct powers of 2 that sum to n. We conclude that n can be written as the sum of distinct powers of 2. □ 13 Algorithms Appendix: Proof by Induction The fourth proof uses so-called ‘weak’ induction, where the inductive hypothesis can only be applied at n−1. Not surprisingly, tying all but one hand behind our backs makes the resulting proof longer, more complicated, and harder to read. It doesn’t help that the algorithm used in the proof is overly speciﬁc. Nevertheless, this is the ﬁrst approach that occurs to most students who have not truly accepted the Recursion Fairy into their hearts. Proof by baby-step induction: Let n be an arbitrary non-negative integer. Assume that any non-negative integer less than n can be written as the sum of distinct powers of 2. There are two cases to consider: Either n = 0 or n ≥1. • The base case n = 0 is trivial—the elements of the empty set are distinct and sum to zero. • Suppose n ≥1. The inductive hypothesis implies that n −1 can be written as the sum of distinct powers of 2. Thus, n can be written as the sum of powers of 2, which are distinct except possibly for two copies of 20. Let S be this multiset of powers of 2. Now consider the following algorithm: i ←0 while S has more than one copy of 2i Remove two copies of 2i from S Insert one copy of 2i+1 into S i ←i + 1 Each iteration of this algorithm reduces the cardinality of S by 1, so the algorithm must eventually halt. We claim that for every non-negative integer i, the follow-ing invariants are satisﬁed after the ith iteration of the while loop (or before the algorithm starts if i = 0): – The elements of S sum to n. Proof by induction: Let i be an arbitrary non-negative integer. Assume that for any non-negative integer j ≤i, after the jth iteration of the while loop, the elements of S sum to n. If i = 0, the elements of S sum to n by deﬁnition of S. Otherwise, the induction hypothesis implies that just before the ith iteration, the elements of S sum to n; the ith iteration replaces two copies of 2i with 2i+1, leaving the sum unchanged. □ – The elements in S are distinct, except possibly for two copies of 2i. Proof by induction: Let i be an arbitrary non-negative integer. Assume that for any non-negative integer j ≤i, after the jth iteration of the while loop, the elements of S are distinct except possibly for two copies of 2j. If i = 0, the invariant holds by deﬁnition of S. So suppose i > 0. The induction hypothesis implies that just before the ith iteration, the elements of S are distinct except possibly for two copies of 2i. If there are two copies of 2i, the algorithm replaces them both with 2i+1, and the invariant is established; otherwise, the algorithm halts, and the invariant is again established. □ The second invariant implies that when the algorithm halts, the elements of S are distinct. In either case, we conclude that n can be written as the sum of distinct powers of 2. □ Repeat after me: “Doctor! Doctor! It hurts when I do this!” 14 Algorithms Appendix: Proof by Induction 8 Irrational Numbers Exist Theorem 7. p 2 is irrational. Proof: I will prove that p2 ̸= 2q2 (and thus p/q ̸= p 2) for all positive integers p and q. Let p and q be arbitrary positive integers. Assume that for any positive integers i < p and j < q, we have i2 ̸= 2j2. Let i = ⌊p/2⌋and j = ⌊q/2⌋. There are three cases to consider: • Suppose p is odd. Then p2 = (2i + 1)2 = 4i2 + 4i + 1 is odd, but 2q2 is even. • Suppose p is even and q is odd. Then p2 = 4i2 is divisible by 4, but 2q2 = 2(2j + 1)2 = 4(2j2 + 2j) + 2 is not divisible by 4. • Finally, suppose p and q are both even. The induction hypothesis implies that i2 ̸= 2j2. Thus, p2 = 4i2 ̸= 8j2 = 2q2. In every case, we conclude that p2 ̸= 2q2. □ For some reason, this proof is almost always presented as a proof by inﬁnite descent. Notice that the induction hypothesis assumed that both p and q were as small as possible. Notice also that the ‘base cases’ included every pair of integers p and q where at least one of the integers is odd. 9 Fibonacci Parity The Fibonacci numbers 0,1,1,2,3,5,8,13,21,34,55,89,144,... are recursively deﬁned as follows: Fn =    0 if n = 0 1 if n = 1 Fn−1 + Fn−2 if n ≥2 Theorem 8. For all non-negative integers n, Fn is even if and only if n is divisible by 3. Proof: Let n be an arbitrary non-negative integer. Assume that for all non-negative integers k < n, Fk is even if and only if n is divisible by 3. There are three cases to consider: n = 0, n = 1, and n ≥2. • If n = 0, then n is divisible by 3, and Fn = 0 is even. • If n = 1, then n is not divisible by 3, and Fn = 1 is odd. • If n ≥2, there are two subcases to consider: Either n is divisible by 3, or it isn’t. – Suppose n is divisible by 3. Then neither n −1 nor n −2 is divisible by 3. Thus, the inductive hypothesis implies that both Fn−1 and Fn−2 are odd. So Fn is the sum of two odd numbers, and is therefore even. – Suppose n is not divisible by 3. Then exactly one of the numbers n −1 and n−2 is divisible by 3. Thus, the inductive hypothesis implies that exactly one of the numbers Fn−1 and Fn−2 is even, and the other is odd. So Fn is the sum of an even number and an odd number, and is therefore odd. In all cases, Fn is even if and only if n is divisible by 3. □ 15 Algorithms Appendix: Proof by Induction 10 Recursive Functions Theorem 9. Suppose the function F : N →N is deﬁned recursively by setting F(0) = 0 and F(n) = 1+F(⌊n/2⌋) for every positive integer n. Then for every positive integer n, we have F(n) = 1+⌊log2 n⌋. Proof: Let n be an arbitrary positive integer. Assume that F(k) = 1 + ⌊log2 k⌋for every positive integer k < n. There are two cases to consider: Either n = 1 or n ≥2. • Suppose n = 1. Then F(n) = F(1) = 1+ F(⌊1/2⌋) = 1+ F(0) = 1 and 1+⌊log2 n⌋= 1 + ⌊log2 1⌋= 1 + ⌊0⌋= 1. • Suppose n ≥2. Because 1 ≤⌊n/2⌋< n, the induction hypothesis implies that F(⌊n/2⌋) = 1 + ⌊log2⌊n/2⌋⌋. The deﬁnition of F(n) now implies that F(n) = 1 + F(⌊n/2⌋) = 2 + ⌊log2⌊n/2⌋⌋. Now there are two subcases to consider: n is either even or odd. – If n is even, then ⌊n/2⌋= n/2, which implies F(n) = 2 + ⌊log2⌊n/2⌋⌋ = 2 + ⌊log2(n/2)⌋ = 2 + ⌊(log2 n) −1⌋ = 2 + ⌊log2 n⌋−1 = 1 + ⌊log2 n⌋. – If n is odd, then ⌊n/2⌋= (n −1)/2, which implies F(n) = 2 + ⌊log2⌊n/2⌋⌋ = 2 + ⌊log2((n −1)/2)⌋ = 1 + ⌊log2(n −1)⌋ = 1 + ⌊log2 n⌋ by the algebra in the even case. Because n > 1 and n is odd, n cannot be a power of 2; thus, ⌊log2 n⌋= ⌊log2(n −1)⌋. In all cases, we conclude that F(n) = 1 + ⌊log2 n⌋. □ 11 Trees Recall that a tree is a connected undirected graph with no cycles. A subtree of a tree T is a connected subgraph of T; a proper subtree is any tree except T itself. Theorem 10. In every tree, the number of vertices is one more than the number of edges. This one is actually pretty easy to prove directly from the deﬁnition of ‘tree’: a directed acyclic graph. Proof: Let T be an arbitrary tree. Choose an arbitrary vertex v of T to be the root, and direct every edge of T outward from v. Because T is connected, every node except v has at least one edge directed into it. Because T is acyclic, every node has at most one edge directed into it, and no edge is directed into v. Thus, for every node x ̸= v, there is exactly one edge directed into x. We conclude that the number of edges is one less than the number of nodes. □ 16 Algorithms Appendix: Proof by Induction But we can prove this theorem by induction as well, in several different ways. Each inductive proof is structured around a different recursive deﬁnition of ‘tree’. First, a tree is either a single node, or two trees joined by an edge. Proof: Let T be an arbitrary tree. Assume that in any proper subtree of T, the number of vertices is one more than the number of edges. There are two cases to consider: Either T has one vertex, or T has more than one vertex. • If T has one vertex, then it has no edges. • Suppose T has more than one vertex. Because T is connected, every pair of ver-tices is joined by a path. Thus, T must contain at least one edge. Let e be an arbitrary edge of T, and consider the graph T \ e obtained by deleting e from T. Because T is acyclic, there is no path in T \ e between the endpoints of e. Thus, T has at least two connected components. On the other hand, because T is con-nected, T \ e has at most two connected components. Thus, T \ e has exactly two connected components; call them A and B. Because T is acyclic, subgraphs A and B are also acyclic. Thus, A and B are sub-trees of T, and therefore the induction hypothesis implies that \|E(A)\| = \|V(A)\| −1 and \|E(B)\| = \|V(B)\| −1. Because A and B do not share any vertices or edges, we have \|V(T)\| = \|V(A)\| + \|V(B)\| and \|E(T)\| = \|E(A)\| + \|E(B)\| + 1. Simple algebra now implies that \|E(T)\| = \|V(T)\| −1. In both cases, we conclude that the number of vertices in T is one more than the number of edges in T. □ Second, a tree is a single node connected by edges to a ﬁnite set of trees. Proof: Let T be an arbitrary tree. Assume that in any proper subtree of T, the number of vertices is one more than the number of edges. There are two cases to consider: Either T has one vertex, or T has more than one vertex. • If T has one vertex, then it has no edges. • Suppose T has more than one vertex. Let v be an arbitrary vertex of T, and let d be the degree of v. Delete v and all its incident edges from T to obtain a new graph G. This graph has exactly d connected components; call them G1, G2,..., Gd. Because T is acyclic, every subgraph of T is acyclic. Thus, every subgraph Gi is a proper subtree of G. So the induction hypothesis implies that \|E(Gi)\| = \|V(Gi)\| −1 for each i. We conclude that \|E(T)\| = d + d X i=1 \|E(Gi)\| = d + d X i=1 (\|V(Gi)\| −1) = d X i=1 \|V(Gi)\| = \|V(T)\| −1. In both cases, we conclude that the number of vertices in T is one more than the number of edges in T. □ But you should never attempt to argue like this: Not a Proof: The theorem is clearly true for the 1-node tree. So let T be an arbitrary tree with at least two nodes. Assume inductively that the number of vertices in T is one more than the number of edges in T. Suppose we add one more leaf to T to get a new tree T ′. This new tree has one more vertex than T and one more edge than T. Thus, the number of vertices in T ′ is one more than the number of edges in T ′. □ 17 Algorithms Appendix: Proof by Induction This is not a proof. Every sentence is true, and the connecting logic is correct, but it does not imply the theorem, because it doesn’t explicitly consider all possible trees. Why should the reader believe that their favorite tree can be recursively constructed by adding leaves to a 1-node tree? It’s true, of course, but that argument doesn’t prove it. Remember: There are only two ways to prove any universally quantiﬁed statement: Directly (“Let T be an arbitrary tree. . . ") or by contradiction (“Suppose some tree T doesn’t. . . "). Here is a correct inductive proof using the same underlying idea. In this proof, I don’t have to prove that the proof considers arbitrary trees; it says so right there on the ﬁrst line! As usual, the proof very strongly resembles a recursive algorithm, including a subroutine to ﬁnd a leaf. Proof: Let T be an arbitrary tree. Assume that in any proper subtree of T, the number of vertices is one more than the number of edges. There are two cases to consider: Either T has one vertex, or T has more than one vertex. • If T has one vertex, then it has no edges. • Otherwise, T must have at least one vertex of degree 1, otherwise known as a leaf. Proof: Consider a walk through the graph T that starts at an arbitrary vertex and continues as long as possible without repeating any edge. The walk can never visit the same vertex more than once, because T is acyclic. Whenever the walk visits a vertex of degree at least 2, it can continue further, because that vertex has at least one unvisited edge. But the walk must eventually end, because T is ﬁnite. Thus, the walk must eventually reach a vertex of degree 1. □ Let ℓbe an arbitrary leaf of T, and let T ′ be the tree obtained by deleting ℓfrom T. Then we have the identity \|E(T)\| = \|E(T ′)\| −1 = \|V(T ′)\| −2 = \|V(T)\| −1, where the ﬁrst and third equalities follow from the deﬁnition of T ′, and the second equality follows from the inductive hypothesis. In both cases, we conclude that the number of vertices in T is one more than the number of edges in T. □ 12 Strings Recall that a string is any ﬁnite sequence of symbols. More formally, a string is either empty or a single symbol followed by a string. Let x · y denote the concatenation of strings x and y, and let reverse(x) denote the reversal of string x. For example, now · here = nowhere and reverse(stop) = pots. Theorem 11. For all strings x and y, we have reverse(x · y) = reverse(y) · reverse(x). 18 Algorithms Appendix: Proof by Induction Proof: Let x and y be arbitrary strings. Assume for any proper substring w of x that reverse(w · y) = reverse(y) · reverse(w). There are two cases to consider: Either x is the empty string, or not. • If x is the empty string, then reverse(x) is also the empty string, so reverse(x· y) = reverse(y) = reverse(y) · reverse(x). • Suppose x is not the empty string. Then x = a · w, for some character a and some string w, and therefore reverse(x · y) = reverse((a · w) · y) = reverse(a · (w · y)) = reverse(w · y) · a = reverse(y) · reverse(w) · a [induction hypothesis] = reverse(y) · reverse(x). In both cases, we conclude that reverse(x · y) = reverse(y) · reverse(x). □ 13 Regular Languages Theorem 12. Every regular language is accepted by a non-deterministic ﬁnite automaton. Proof: In fact, we will show something stronger: Every regular language is accepted by an NFA with exactly one accepting state. Let R be an arbitrary regular expression over the ﬁnite alphabet Σ. Assume that for any sub-expression S of R, the corresponding regular language is accepted by an NFA with one accepting state, denoted S . There are six cases to consider—three base cases and three recursive cases—mirroring the recursive deﬁnition of a regular expression. • If R = ∅, then L(R) = ∅is accepted by an NFA with no transitions: . • If R = ϵ, then L(R) = {ϵ} is accepted by the NFA ε . • If R = a for some character a ∈Σ, then L(R) = {a} is accepted by the NFA a . • Suppose R = ST for some regular expressions S and T. The inductive hypothesis implies that S and T are accepted by NFAs S and T , respectively. Then L(T) = {uv \| u ∈L(S), v ∈L(T)} is accepted by the NFA S T . • Suppose R = S+T for some regular expressions S and T. The inductive hypothesis implies that S and T are accepted by NFAs S and T , respectively. Then L(R) = L(S) ∪L(T) is accepted by the NFA S T ε ε ε ε . • Finally, suppose R = S∗for some regular expression S. The inductive hypothesis implies that S is accepted by an NFA S . Then the language L(R) = L(S)∗is accepted by the NFA S ε ε . In every case, L(x) is accepted by an NFA with one accepting state. □ c ⃝Copyright 2010 Jeff Erickson. Released under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 License ( Free distribution is strongly encouraged; commercial distribution is expressly forbidden. See for the most recent revision. 19
6485	https://fiveable.me/key-terms/principles-physics-ii/joule-per-coulomb	Joule per coulomb - (Principles of Physics II) - Vocab, Definition, Explanations \| Fiveable \| Fiveable new!Printable guides for educators Printable guides for educators. Bring Fiveable to your classroom ap study content toolsprintablespricing my subjectsupgrade All Key Terms Principles of Physics II Joule per coulomb 🎢principles of physics ii review key term - Joule per coulomb Citation: MLA Definition Joule per coulomb is the unit of electric potential, which measures the amount of energy in joules that is transferred per unit charge in coulombs. This relationship highlights how electric potential is fundamentally about energy transfer and charge movement, making it crucial for understanding electrical circuits and systems. The concept connects energy and charge, emphasizing how voltage influences the behavior of electric fields and charges in a variety of applications. 5 Must Know Facts For Your Next Test The joule per coulomb (J/C) is equivalent to one volt (V), making it a critical concept in understanding how electrical systems operate. Higher electric potential means that more energy is available to do work on charges moving through an electric field. In circuits, differences in joule per coulomb help determine the flow of current and how components like resistors or capacitors will behave. The conversion from joules to coulombs can help calculate work done by or on charges as they move through an electric potential. Understanding joule per coulomb is essential for analyzing electrical energy consumption and efficiency in various devices. Review Questions How does the concept of joule per coulomb relate to the functioning of electrical circuits? The concept of joule per coulomb directly ties into how electrical circuits function by indicating the energy transfer involved when charges move through a circuit. When a charge moves between two points with different electric potentials, the joules per coulomb indicates how much energy that charge gains or loses. This understanding helps predict current flow and energy consumption in various components within an electrical circuit. Explain how changes in electric potential affect the movement of charges in an electric field, using joule per coulomb as a reference. Changes in electric potential, measured in joule per coulomb, have a direct impact on the movement of charges within an electric field. When a charge moves from a point of higher potential to lower potential, it experiences a decrease in energy represented by joules. Conversely, moving from lower to higher potential requires work done against the field, highlighting how joule per coulomb serves as a measure of energy available for charge movement. Evaluate the importance of understanding joule per coulomb when discussing electrical energy efficiency in modern devices. Understanding joule per coulomb is crucial when evaluating electrical energy efficiency because it provides insights into how much energy is consumed versus how much useful work is achieved. Devices operate optimally when they maintain appropriate voltage levels, ensuring that energy transfer via joules per coulomb aligns with minimal waste. This knowledge aids in designing more efficient circuits and components that maximize performance while reducing unnecessary energy loss. Related terms Voltage: Voltage, often synonymous with electric potential, is the difference in electric potential energy per unit charge between two points in an electric field. Electric Field:An electric field is a region around a charged object where other charges experience a force, and it is directly related to the electric potential. Capacitance: Capacitance is the ability of a system to store electric charge, which is influenced by the voltage across the capacitor and the amount of charge stored. 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6486	https://liucs.net/cs101s14/n1-binary.html	CS101 - Number systems and binary Toggle navigationCS101 Schedule Check-ins Notes Number systems and binary Text encoding Multimedia encoding Boolean logic Algorithms More algorithms Python language Operating systems The web Databases Artificial Intelligence Security & privacy Assignments Assignment 1 – alternative binary representations Assignment 2 – text compression Assignment 3 – hex images Assignment 3 solutions Assignment 4 – circuit diagrams Assignment 5 – sorting Assignment 6 – pseudo-code Assignment 7 – Python Assignment 8 – Linux server Assignment 9 – database Assignment 10 – AI Assignment 11 – Cryptography Exams Quiz 2 solutions Quiz 3 solutions Practice midterm (PDF) Practice midterm solutions (PDF) Quiz 6 solutions (PDF) Practice final (PDF) Practice final solutions (PDF) Number systems and binary Here are some informal notes on number systems and binary numbers. See also sections 3.1–3.2 of the textbook. Positional numbering system Our normal number system is a positional system, where the position (column) of a digit represents its value. Starting from the right, we have the ones column, tens column, hundreds, thousands, and so on. Thus the number 3724 stands for three THOUSAND, seven HUNDRED, two TENS (called twenty), and four ONES. The values of those columns derive from the powers of ten, which is then called the base of the number system. The base ten number system is also called decimal. There is nothing special about base ten, except that it’s what you learned from a young age. A positional numbering system can use any quantity as its base. Let’s take, for example, base five. In base five, the columns represent the quantities (from right to left) one, five, twenty-five, and a hundred twenty-five. We need to use five symbols to indicate quantities from zero up to four. For simplicity, let’s keep the same numerals we know: 0, 1, 2, 3, and 4. The number shown in this figure, 3104 in base five, represents the same quantity that we usually write as 404 in base ten. That’s because it is three × one hundred twenty-five (= 375), plus one × twenty-five (= 25) plus four ones (= 4), so 375 + 25 + 4 = 404. You can count directly in base five; it looks like this: 0, 1, 2, 3, 4, 10, 11, 12, 13, 14, 20, 21, 22, 23, 24, 30, 31, 32, 33, 34, 40, 41, 42, 43, 44, 100. (Those correspond to quantities from zero to twenty-five.) You should try counting and converting between other bases. Below is an interesting video overview of the dozenal (base twelve) number system. Binary numbers Computer systems use binary numbers – that just means they are in base two. Using two as the base is really convenient and flexible, because we need only two ‘symbols’ and there are so many ways we can represent them: zero/one, on/off, up/down, high/low, positive/negative, etc. In binary, the columns are (from right to left) 1, 2, 4, 8, 16, 32, and so on. Using a zero means we exclude that column’s quantity, and a one means we include it. So the binary number 10110 is the quantity 16 + 4 + 2 = 22. Each binary digit (a one or a zero) is called a bit. The largest five-bit binary number, then is 11111 = 16 + 8 + 4 + 2 + 1 = 31. It’s worthwhile to learn to count in binary, at least from zero to fifteen: 0000 = 0 0100 = 4 1000 = 8 1100 = 12 0001 = 1 0101 = 5 1001 = 9 1101 = 13 0010 = 2 0110 = 6 1010 = 10 1110 = 14 0011 = 3 0111 = 7 1011 = 11 1111 = 15 Binary arithmetic It’s relatively easy to add numbers directly in binary. Line up the columns and then proceed from right to left, as usual. There are only four possible cases: If a column has no ones, write a zero below. If a column has one one, write a one below. If a column has two ones, write a zero and carry a one to the next column (to the left). Finally, if a column has three ones (possible due to an incoming carry), write a one and carry a one to the next column. Below is an example of adding 10110 plus 11100. The result is 110010, and you can see the carry bits above the original numbers, in orange. When adding this way, it’s always a good idea to check your work by converting the numbers to decimal and checking the addition. In this case, we’re adding 22 (10110) to 28 (11100) to get 50 (110010). Fixed-size binary numbers We generally arrange numbers along a line, that goes off to infinity. Indeed, in binary we can always continue counting by adding more and more columns that are powers of two. However, in most computer systems and programs we use fixed-size numbers. That is, we decide in advance how many bits will be used to represent the number. For example, a 32-bit computer represents most of its numbers and addresses using 32 bits. The largest such number is 2³²–1 = 4,294,967,295. When your numbers have a fixed number of bits, then there is no number line heading off into infinity. Instead, we arrange the numbers around a circle, like a clock. Below is the number wheel for 3-bit integers. The smallest 3-bit integer is zero, and the largest is seven. Then, if you attempt to keep counting, it just wraps around to zero again. When you perform arithmetic with fixed-size numbers, you throw away any extra carry bit; the result cannot exceed the designated size. For example, see what happens if we try to add 110 + 011 using 3-bit integers: In 3-bit arithmetic, 6 plus 3 is 1. You can make sense of this on the number wheel. Addition corresponds to walking clock-wise around the wheel. So start at 6, and go clockwise by 3. That lands on 1, which is 6+3. Below is a video about fixed-size binary numbers in old video games. Signed magnitude Now we’ll look at signed numbers – that is, numbers that can be positive or negative. There are two techniques for encoding signed numbers. The first one is called signed magnitude. It appears simple at first, but that simplicity hides some awkward properties. Here’s how it works. We use a fixed width, and then the left-most bit represents the sign. So 4-bit signed magnitude looks like this: ``` sign 4 2 1 ``` where having the sign bit set to ‘1’ means the magnitude is interpreted as negative. Thus, 0110 is +6 whereas 1110 is -6. In this system, the largest positive number is 0111 = +7 and the most negative number is 1111 = -7. One of the unfortunate effects of this representation is there are two ways to write zero: 0000 and also 1000. There is no such thing as negative zero, so this doesn’t really make sense. Two’s complement The second way to represent signed quantities is called two’s complement. Although this looks trickier at first, it actually works really well. Below is the interpretation of 4-bit two’s complement. All we need to do compared to normal unsigned numbers is negate the value of the left-most bit. ``` -8 4 2 1 ``` So +6 is 0110 as before, but what about -6? We need to turn on the negative 8, and then add two: 1010. To represent -1, you turn on all the bits: 1111, because that produces -8+4+2+1 = -8+7 = -1. The nice thing about two’s complement is that you can add these numbers and everything just works out. Let’s try adding 7 and -3: ``` 0 1 1 1 = 7 1 1 0 1 = -3 0 1 0 0 = 4 ``` It’s also relatively easy to negate a number – that is, to go from +6 to -6 or from -3 to +3. Here are the steps: First, flip all the bits. That is, all the zeroes become ones and all the ones become zeroes. Next, add one. For example here is how we produce -6 from +6: ``` 0 1 1 0 = +6 1 0 0 1 (flip all the bits) + 1 (add one) 1 0 1 0 = -6 ``` You don’t even have to reverse these steps in order to convert back: ``` 1 0 1 0 = -6 0 1 0 1 (flip all the bits) + 1 (add one) 0 1 1 0 = +6 ``` Here’s a cartoon from XKCD about counting sheep using two’s complement! How many bits is this person using? Octal and hexadecimal Finally, I want to introduce two number systems that are very useful as abbreviations for binary. They work so well because their bases are powers of two. Octal is base eight, so we use the symbols 0–7 and the values of the columns are: ``` 512 64 8 1 8³ 8² 8¹ 8⁰ ``` The real value of octal, however, is that each octal digit maps to exactly three binary digits. So, on octal number like 3714 maps as shown: ``` 3 7 1 4 octal number 0 1 1 1 1 1 0 0 1 1 0 0 binary number (4 2 1 4 2 1 4 2 1 4 2 1) ``` Hexadecimal is base sixteen, so we use the symbols 0–9 and then A to represent ten, B for eleven, and so on up to F for fifteen. The values of the columns are: ``` 4096 256 16 1 16³ 16² 16¹ 16⁰ ``` So a hexadecimal number like 2A5C has the value 2×4096 + 10×256 + 5×16 + 12×1 = 10844 in base ten. In hexadecimal, each digit maps to exactly four bits. So here is that same number in binary: ``` 2 A 5 C 0 0 1 0 1 0 1 0 0 1 0 1 1 1 0 0 (8 4 2 1 8 4 2 1 8 4 2 1 8 4 2 1) ``` Below are two great video overviews of hexadecimal. (Note when you watch these – the Brits often pronounce zero as ‘naught’.) Practice problems Convert the following base ten (decimal) numbers into binary. 6 18 51 63 Convert the following unsigned binary numbers into base ten. 1010 1101 1000 10001 What do all odd numbers have in common, when written in binary? (Hint: try writing the quantities 3, 5, 7, 9, 11 in binary.) Using 7-bit signed (two’s complement) binary numbers, what is the largest positive number? What is the most negative number? Convert the following 5-bit signed (two’s complement) binary numbers into base ten. 01101 01111 10011 11111 Convert the following 16-bit binary number into hexadecimal, and then into octal. 0 1 1 1 1 1 1 1 0 0 1 1 1 0 1 0 7. Convert the following hexadecimal numbers into binary: 1. 9D 2. C4 3. A17E Add and verify the following unsigned binary numbers. 1 0 1 1 1 1 1 1 0 1 1 1 + 1 1 1 0 1 + 1 0 0 1 0 0 —————————————— —————————————— Solutions here Extra: floating-point @jaffathecake on Twitter Contents Positional numbering system Binary numbers Binary arithmetic Fixed-size binary numbers Signed magnitude Two’s complement Octal and hexadecimal Practice problems Extra: floating-point PDF version © 2014 Christopher League · Some rights reserved — CC by-sa
6487	https://revise.im/physics/unit-5/thermal-energy	Thermal Energy - Revise.im Chemistry Physics Books About Thermal Energy Temperature Scales The temperature of an object is a measure of hotness of an object. The hotter an object is, the more internal energy it has. If a warm object is placed in a cold environment, the object loses thermal energy to the surroundings through heat transfer. When the object reaches the same temperature as the environment, they are in thermal equilibrium. A temperature scale is defined in terms of fixed points: The Celsius scale of temperature, °C is defined in terms of the ice point, 0°C, the temperature at which ice melts to water and 100°C, the temperature at which liquid vaporises to steam. The absolute scale of temperature in kelvin (K) is defined in terms of absolute zero, 0 K which is the lowest possible temperature and the triple point of water. Triple Point of Water The temperature and pressure at which the three phases of water; ice, water and water vapour exist in thermal equilibrium. This is at a temperature of 0.01°C and pressure of 0.006 a t m. Internal Energy The internal energy of an object is the sum of the random distributions of the kinetic and potential energies of its molecules. Increasing the internal energy of an object through heat transfer increases the kinetic and potential energies associated with the random motion and positions of its molecules. The internal energy of an object changes if: Heat transfer or energy transfer through radiation occurs with another object. Work is done on or by the object. Specific Heat Capacity The heat capacity of an object is the energy required to raise its temperature by 1 K. The specific heat capacity of a material is the energy needed to raise the temperature of a 1 k g mass of an object by 1 K. It is measured in J k g−1 K−1. Δ Q=m c Δ T Where Δ Q is the energy change (J), m is the mass (k g), c is the specific heat capacity in J k g−1 K−1 and Δ T is the temperature change in kelvin (K) or celsius (C°). Inversion Tube Experiment A method used to determine the specific heat capacity of a substance is the inversion tube experiment. A hollow tube containing lead shot can be inverted multiple times, with the temperature of the lead shot measured after each inversion using a thermometer. As the tube is inverted and the lead shot switches between the ends of the tube, the temperature rises. This is because the gravitational potential energy is converted into internal energy of the lead shot. Assuming no energy is lost to the surroundings or the tube, the loss in gravitational potential energy is equal to the gain in internal energy. m c Δ T c=m g L n=g L n Δ T Where c is the specific heat capacity (J k g−1 K−1), g is the acceleration under gravity (9.81 m s−2), L is the length of the tube (m), n is the number of inversions and Δ T is the temperature change (K). Calorimeter A method of determining the specific heat capacity of a liquid is by using a calorimeter and a heater. With the heater in the liquid, the liquid should be stirred and the temperature measured over time. As the calorimeter is well insulated, it can be assumed no energy is lost to the surroundings. With a known current and voltage, the energy supplied by the heater can be calculated by: energy supplied by heater=I V Δ t Where I is the current (A), V is the voltage (V) and Δ t is the time (s) the heater is in the liquid. As some energy is used to heat the liquid and some to heat the calorimeter: I V Δ t=m 1 c 1 Δ T+m c a l c c a l Δ T Where m 1 and m c a l are the masses of the liquid and the calorimeter and c 1 and c c a l are the specific heat capacities of the liquid and the calorimeter respectively. Δ T is the temperature change (K) measured by the thermometer. A similar approach can be used to determine the specific heat capacity of a solid whereby a heater is inserted into the solid and the temperature difference is measured over time. These methods depend on a known energy input and the assumption of no energy loss to the surroundings. Specific Latent Heat During a phase transition where the state of a substance is changed, energy is required to do work against the attractive forces holding the molecules together. In this process, the energy transferred to the object as heat does not lead to an increase in temperature. The energy required to change the state of a substance with no change in temperature is known as its latent heat, l. The specific latent heat of fusion of a substance is the energy required to change 1 k g of a solid into 1 k g of a liquid, with no change in temperature. The specific latent heat of vaporisation is the energy required to change 1 k g of a liquid into 1 k g of a gas, with no change in temperature. This graph shows the temperature of a solid against the time it is heated for a constant rate. The flat sections of the curve are when the substance is changing state. The energy, Δ Q needed to change the state of a substance of mass m can be calculated with: Δ Q=m l Where Δ Q is the energy required (J), m is the mass of the liquid (m) and l is the specific heat of fusion or vaporisation (J k g−1). Return to Physics<< Nuclear EnergyGas Laws >> © Andrew Deniszczyc, 2025 
6488	https://flexbooks.ck12.org/cbook/ck-12-interactive-geometry-for-ccss/section/9.2/primary/lesson/pyramids-and-cones-geo-ccss/	Skip to content Elementary Math Grade 1 Grade 2 Grade 3 Grade 4 Grade 5 Math 6 Math 7 Math 8 Algebra I Geometry Algebra II Math 6 Math 7 Math 8 Algebra I Geometry Algebra II Probability & Statistics Trigonometry Math Analysis Precalculus Calculus What's the difference? 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Learn. Interact. eXplore. CCSS Math Concepts and FlexBooks aligned to Common Core NGSS Concepts aligned to Next Generation Science Standards Certified Educator Stand out as an educator. Become CK-12 Certified. Webinars Live and archived sessions to learn about CK-12 Other Resources CK-12 Resources Concept Map Testimonials CK-12 Mission Meet the Team CK-12 Helpdesk FlexLets Know the essentials. Pick a Subject Donate Sign Up 9.2 Pyramids and Cones Written by:CK-12 \| Kaitlyn Spong Fact-checked by:The CK-12 Editorial Team Last Modified: Sep 01, 2025 Lesson Recall that a pyramid is a solid with a polygon base and triangular lateral faces that meet in a vertex. Pyramids are named by their base shape. You have seen the formula for the volume of a pyramid before. Where does this formula come from? Recall that to find the volume of a prism or a cylinder, you need to find the area of the base and multiply by the height. The difference between these two formulas is the division by 3. The key to understanding where the 3 comes from is remembering Cavalieri's principle and investigating a square based pyramid. Let's take a look at some problems involving pyramids. The two square pyramids below are each constructed within cubes of the same size. The pyramid on the left has a vertex at the center of a face of the cube. The pyramid on the right has a vertex at one of the vertices of the cube. Is the volume of each pyramid the same? Because each pyramid is constructed within the same cube, the heights of the pyramids are the same and the areas of their bases are the same. When a plane parallel to the base of the pyramid is constructed through the center of each cube, the cross sections of each pyramid are the same. Each cross section is a square. The length of the side of the square is half the length of an edge of the original cube, since the plane was constructed through the middle of the cube. Because cross sections are the same area, heights are the same, and bases are the same, these pyramids must have the same volume due to Cavalieri's principle. How many square based pyramids congruent to the one below would it take to fill the cube? Can you visualize this? It will take exactly 3 congruent pyramids to fill the cube. The image below shows each pyramid being added to the cube. Use the answers from the previous problems to explain why the volume of a pyramid is . The volume of a rectangular prism is . This is because gives the volume of one "layer", and multiplying by the height scales that base volume by the number of "layers" of the prism. Three congruent pyramids fit inside the cube in #2, so the volume of each pyramid must be the volume of the cube. Therefore, the volume of a pyramid is . Remember that pyramids with the same base area and height will have the same volume due to Cavalieri's principle, so both of the pyramids below will have a volume of . Note: This is an informal argument for the formula for the volume of a pyramid. A rigorous derivation of the formula that considers pyramids of any base shape will be developed in calculus. Examples Example 1 The volume of a pyramid is given by . How does this formula help you find the formula for the volume of a cone? A cone is essentially a pyramid with a circular base. The volume of a pyramid is given by . Since the area of the base of a cone is , the formula for the volume of a cone is . Example 2 The base area of the pyramid below is . What is the volume of the pyramid? Example 3 Find the volume of a pyramid with a height of 10 inches and a regular pentagon base with an apothem of 1 inch. . To find the area of the base, divide the pentagon into five congruent triangles. The apothem is the height of each of these triangles. Use trigonometry to find the base length of each of these triangles. Example 4 Find the volume of a cone with a height of 10 inches and a radius of 1 inch. CK-12 PLIX Interactive \| \| \| Summary \| \| A pyramid is a three-dimensional object with a base that is a polygon and triangular faces that meet at one vertex. The volume of a pyramid is A cone is a solid three-dimensional figure with a circular base and one vertex. The volume of a cone is \| Review Explain the connections between a prism and a pyramid. Why do you divide by three when calculating the volume of a pyramid? Explain the connections between a cone and a cylinder. Why do you divide by three when calculating the volume of a cone? Find the volume of each solid based on its description. Keep your answers in terms of when applicable. A cone with a diameter of 4 inches and a height of 12 inches. A pyramid with a height of 15 inches and a regular hexagon base with an apothem of 4 inches. Round your answer to the nearest hundredths. A cone with a radius of 8 centimeters and a height of 15 centimeters. A square based pyramid such that the slant height of the pyramid is 10 inches and each triangular face has a base of 12 inches. An hourglass is created by placing two congruent cones inside of a cylinder with the same base area. The radius is 5 inches and the height of the cylinder is 20 inches. Find the volume of one of the cones. Find the volume of the cylinder. Find the volume of the space between the cones and the cylinder. You want to fill one of the cones with a thick liquid. If one cup of liquid has a volume of approximately , how much liquid will you need to fill one of the cones? Round your answer to the nearest hundredths place. A cone and a square pyramid have the same volume and height. The volume of each solid is . If the radius of the cone is 2.82 centimeters, what is the length of a side of the base of the pyramid? Round your answer to the nearest hundredths place. The ratio of the area of the red circle to the area of the base is 1:9. If the height of the cone is 15 inches, what is the length of ? The height of the cone below is 10 inches. Find the length of . A regular tetrahedron is a triangular pyramid with four congruent equilateral faces. If each of the triangular faces of the tetrahedron has a base length of and a height of what is the volume of the tetrahedron? The length of each side of the triangular faces making up a tetrahedron is . What is the volume of the tetrahedron in terms of ? Imagine a mountain roughly in the shape of a cone. It has a base whose circumference is roughly 78 miles and a height of roughly 8,800 feet. Sketch the scenario. Find the volume in terms of The Great Pyramid of Giza has a height of roughly 756 feet, and a square base with a side length of roughly 455 feet. What is the volume of the Great Pyramid? Review (Answers) Click HERE to see the answer key or go to the Table of Contents and click on the Answer Key under the 'Other Versions' option. Asked by Students Here are the top questions that students are asking Flexi for this concept: Overview A pyramid is a three-dimensional object with a base that is a polygon and triangular faces that meet at one vertex. The volume of a pyramid is @$V=\frac{\text{base}\times \text{height}}{3}@$ A cone is a solid three-dimensional figure with a circular base and one vertex. The volume of a cone is @$V=\frac{\pi r^2\times h}{3}@$ Vocabulary Pyramid prism cylinder Area Height square vertex Face plane center cross section Edge Congruent Rectangular Prism Argument Pentagon trigonometry radius Test Your Knowledge Question 1 Find the volume of a cone with a radius of 6.5 inches and a height of 12 inches. Use 3.14 to represent π. a 163.28 in3 b 1591.98 in3 c 979.68 in3 d 530.66 in3 Let: @$\begin{align}r = 6.5\ \text{in}\end{align}@$ @$\begin{align}h = 12\ \text{in}\end{align}@$ @$\begin{align}\pi =3.14\end{align}@$ The formula for the volume of a cone is @$\begin{align}V=\frac{\pi {r}^{2}h}{3}\end{align}@$. Substitute the given values into the formula below: @$$\begin{align}\eqalign{ V &=\frac{\pi r^2 h}{3} \ V &=\frac{(3.14)(6.5)^2(12)}{3} \ V &=\frac{(3.14)(42.25)(12)}{3} \ V &=\frac{1591.98}{3} \ V &=530.66 }\end{align}@$$ The volume of the cone is @$\begin{align}530.66\ \text{in}^3\end{align}@$. Question 2 Find the volume of the cone with a radius of 10 mm and a height of 12 mm using the formula @$\begin{align}V=\frac{\pi {r}^{2}h}{3}\end{align}@$ and 3.14 for π. a @$\begin{align}3768\ \text{mm}^3\end{align}@$ b @$\begin{align}251.2\ \text{mm}^3\end{align}@$ c @$\begin{align}1205.76\ \text{mm}^3\end{align}@$ d @$\begin{align}1256\ \text{mm}^3\end{align}@$ Let: @$\begin{align}r = 10\ \text{mm}\end{align}@$ @$\begin{align}h = 12\ \text{mm}\end{align}@$ @$\begin{align}\pi =3.14\end{align}@$ The formula for the volume of a cone is @$\begin{align}V=\frac{\pi {r}^{2}h}{3}\end{align}@$. Substitute the given values into the formula below: @$$\begin{align}\eqalign{ V &=\frac{\pi r^2 h}{3} \ V &=\frac{(3.14)(10)^2(12)}{3} \ V &=\frac{(3.14)(100)(12)}{3} \ V &=\frac{3768}{3} \ V &=1256 }\end{align}@$$ The volume of the cone is @$\begin{align}1256\ \text{mm}^3\end{align}@$. Study Guide Go to Study Guide Asked by Students Here are the top questions that students are asking Flexi for this concept: Related Content Pyramids and Cones Study Guide Pyramids Principles - Basic Pyramids Examples - Basic Cones Principles - Basic Cones Examples - Basic Burying a Pharaoh Those Who Live in Glass Pyramids Triangular Living Surface Area and Volume of Pyramids Volume of Cones: Strawberry Ice Cream Surface Area and Volume of Pyramids Pyramid \| Image \| Reference \| Attributions \| --- \| \| \| Credit: CK 12 Source: CK 12 License: CC BY-NC 3.0 \| \| \| \| Credit: CK 12 Source: CK 12 License: CC BY-NC 3.0 \| \| \| \| Credit: CK 12 Source: CK 12 License: CC BY-NC 3.0 \| \| \| \| License: CC BY-NC-SA \| \| \| \| License: CC BY-NC-SA \| \| \| \| License: CC BY-NC-SA \| \| \| \| License: CC BY-NC \| \| \| \| License: CC BY-NC \| \| \| \| Credit: Tom Barrett Source: License: CC BY-NC \| \| \| \| Credit: Steve Snodgrass Source: \| \| \| \| License: CC BY-NC \| \| \| \| Credit: T477ModelA Source: \| \| \| \| License: CC BY-NC \| \| \| \| Credit: Brady Holt Source: License: CC BY-NC \| \| \| \| License: CC BY-NC \| \| \| \| Credit: USFWS Mountain Prarie Source: \| \| \| \| Source: CK-12 Foundation \| \| \| \| Credit: CK-12 Source: CK-12 License: CC BY-NC 3.0 \| \| \| \| License: CC BY-NC \| \| \| \| License: CC BY-NC \| Student Sign Up Are you a teacher? Having issues? Click here By signing up, I confirm that I have read and agree to the Terms of use and Privacy Policy Already have an account? Save this section to your Library in order to add a Practice or Quiz to it. (Edit Title)18/ 100 This lesson has been added to your library. \|Searching in: \| \| \| Looks like this FlexBook 2.0 has changed since you visited it last time. We found the following sections in the book that match the one you are looking for: Go to the Table of Contents No Results Found Your search did not match anything in . Student Sign Up Are you a teacher? Having issues? Click here or By signing up, I confirm that I have read and agree to the Terms of use and Privacy Policy Already have an account?
6489	https://www.ms.uky.edu/~lee/ma111fa18/M111Nov16Graph3.pdf	Intro to Contemporary Math Planar Graphs Nicholas Nguyen nicholas.nguyen@uky.edu Department of Mathematics UK Announcements ▶Your project (all parts) must be uploaded on Canvas by Tuesday, November 20th. ▶ There will be a homework assignment on WebWork It will be due Monday, November 26th. ▶ Mini-Exam 4 is Wednesday, November 28th. Graphs with no Overlaps Denition: A graph is planar if it can be drawn so that its edges do not cross. Faces of a Graph In any planar graph, drawn with no intersections, the edges divide the planes into dierent regions. ▶The regions enclosed by the planar graph are called interior faces of the graph. ▶The region surrounding (outside) the planar graph is called the exterior face of the graph. ▶When we say faces of the graph we mean BOTH the interior AND the exterior faces. We usually denote the number of faces of a planar graph by f . Faces of a Graph In any planar graph, drawn with no intersections, the edges divide the planes into dierent regions. ▶The regions enclosed by the planar graph are called interior faces of the graph. ▶The region surrounding (outside) the planar graph is called the exterior face of the graph. ▶When we say faces of the graph we mean BOTH the interior AND the exterior faces. We usually denote the number of faces of a planar graph by f . Faces of a Graph In any planar graph, drawn with no intersections, the edges divide the planes into dierent regions. ▶The regions enclosed by the planar graph are called interior faces of the graph. ▶The region surrounding (outside) the planar graph is called the exterior face of the graph. ▶When we say faces of the graph we mean BOTH the interior AND the exterior faces. We usually denote the number of faces of a planar graph by f . Face Count Example This graph has a total of three faces: f = 3 ▶Two interior faces (A, B) ▶One exterior face (C) ?(3.2) Face Count Example 2 How many faces does this graph have (what is f )? Type and send a number. Face Count Example 2 This graph has 4 faces: A, B, and C are interior faces D is the exterior face Degree of a Face For a planar graph drawn without edges crossing, the number of edges bordering a particular face is called the degree of the face. The two interior faces A and B have degree 3, while the exterior face C has degree 4. ?(3.3) Degree of a Face Example 2 ▶The interior faces A and B each have degree 3. ▶The degree of face C is what number? Degree of a Face Example 2 ▶The interior faces A and B each have degree 3. ▶The interior face C has degree 4. ▶The exterior face D has degree 4. Sum of Degrees In a planar graph, ▶If you add up the degrees of every vertex and divide by 2, you get the number of edges. ▶If you add up the degrees of every face and divide by 2, you get the number of edges. ▶Conversely, if you take the number of edges and multiply by 2, you get the sum of degrees of the vertices or faces. Sum of Degrees In a planar graph, ▶If you add up the degrees of every vertex and divide by 2, you get the number of edges. ▶If you add up the degrees of every face and divide by 2, you get the number of edges. ▶Conversely, if you take the number of edges and multiply by 2, you get the sum of degrees of the vertices or faces. Sum of Degrees In a planar graph, ▶If you add up the degrees of every vertex and divide by 2, you get the number of edges. ▶If you add up the degrees of every face and divide by 2, you get the number of edges. ▶Conversely, if you take the number of edges and multiply by 2, you get the sum of degrees of the vertices or faces. Sum of Degrees Picture (Vertices) One edge is attached to TWO vertices, and gets counted in degree of left vertex and right vertex. Sum of Degrees Picture (Faces) One edge borders TWO faces, and gets counted in degree of upper face and lower face. One face Another face ?(3.4) Sum of Degrees Example A graph has degree list 2, 2, 3, 3, 4, 4, 5, 5. How many edges does it have? Type and send a number. Sum of Degrees Example A graph has degree list 2, 2, 3, 3, 4, 4, 5, 5. Add up degrees of vertices: 2+2+3+3+4+4+5+5 = 28 then divide by 2 to get the number of edges e: 28 2 = 14 Note: Since the degree list has 8 entries, the graph must have 8 vertices. Euler's Formula There is another relationship between the number of vertices, edges, and faces: ▶For a connected (one-piece) planar graph with v vertices, e edges, and f faces, v −e +f = 2 Euler's Formula Showcase v = 5, e = 7, f = 4: v −e +f = 5−7+4 = 2. ?(3.5) Euler's Formula Example A connected planar graph has 24 vertices and 30 faces. How many edges does the graph have? Type and send a number. Euler's Formula Example v = 24 and f = 30, so in Euler's formula, v −e +f = 2 24−e +30 = 2 −e +54 = 2 −e = 2−54 = −52 e = 52 so there are 52 edges: e = 52. ?(3.6) Euler's Formula Example v −e +f = 2 24−e +30 = 2 −e +54 = 2 −e = 2−54 = −52 e = 52 so there are 52 edges: e = 52. ▶What number would you get if you add up the degrees of the vertices? v = 24 and f = 30, so in Euler's formula, v −e +f = 2 24−e +30 = 2 −e +54 = 2 −e = 2−54 = −52 e = 52 so there are 52 edges: e = 52. ▶Thus, the sum of the degrees of every vertex is 2 times the number of edges: 2e = 2×52 = 104 Next Time ▶Graphs with Labeled Edges
6490	https://artofproblemsolving.com/wiki/index.php/1972_USAMO_Problems/Problem_5?srsltid=AfmBOorCtCiR6f1cS5g2lZ3BqaCNTPuSnQTvR9fotGhLjN5IaRrjhDrm	Art of Problem Solving 1972 USAMO Problems/Problem 5 - AoPS Wiki Art of Problem Solving AoPS Online Math texts, online classes, and more for students in grades 5-12. Visit AoPS Online ‚ Books for Grades 5-12Online Courses Beast Academy Engaging math books and online learning for students ages 6-13. Visit Beast Academy ‚ Books for Ages 6-13Beast Academy Online AoPS Academy Small live classes for advanced math and language arts learners in grades 2-12. Visit AoPS Academy ‚ Find a Physical CampusVisit the Virtual Campus Sign In Register online school Class ScheduleRecommendationsOlympiad CoursesFree Sessions books tore AoPS CurriculumBeast AcademyOnline BooksRecommendationsOther Books & GearAll ProductsGift Certificates community ForumsContestsSearchHelp resources math training & toolsAlcumusVideosFor the Win!MATHCOUNTS TrainerAoPS Practice ContestsAoPS WikiLaTeX TeXeRMIT PRIMES/CrowdMathKeep LearningAll Ten contests on aopsPractice Math ContestsUSABO newsAoPS BlogWebinars view all 0 Sign In Register AoPS Wiki ResourcesAops Wiki 1972 USAMO Problems/Problem 5 Page ArticleDiscussionView sourceHistory Toolbox Recent changesRandom pageHelpWhat links hereSpecial pages Search 1972 USAMO Problems/Problem 5 Problem A given convex pentagon has the property that the area of each of the five triangles , , , , and is unity. Show that all pentagons with the above property have the same area, and calculate that area. Show, furthermore, that there are infinitely many non-congruent pentagons having the above area property. Solution Lemma: Convex pentagon has the property that if and only if for (indices taken mod 5). Proof: For the "only if" direction, since , and are equidistant from , and since the pentagon is convex, . The other four pairs of parallel lines are established by a symmetrical argument, and the proof for the other direction is just this, but reversed. Let be the inner pentagon, labeled so that and are opposite each other, and let be the side lengths of the inner pentagon labeled opposite their corresponding vertices. From the lemma, pentagon is similar to pentagon with and parallelogram (and cyclic) has area 2. Supposing we have Since and are parallelograms, . Triangle is similar to triangle , so , and with the requirement that , Now, we compute that , and similar computation for the other four triangles gives . From the aforementioned pentagon similarity, . Solving for , we have To show that there are infinitely many pentagons with the given property, we start with triangle and construct the original pentagon using the required length ratios. This triangle has three continuous degrees of freedom, and the condition that the equal areas be specifically unity means that there are two continuous degrees of freedom in which the configuration can vary, which implies the desired result. Alternatively, one can shear a regular pentagon, a transformation which preserves areas and the property of a pair of lines being parallel. See Also 1972 USAMO (Problems • Resources) Preceded by Problem 4Followed by Last Question 1•2•3•4•5 All USAMO Problems and Solutions These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Retrieved from " Category: Olympiad Geometry Problems Art of Problem Solving is an ACS WASC Accredited School aops programs AoPS Online Beast Academy AoPS Academy About About AoPS Our Team Our History Jobs AoPS Blog Site Info Terms Privacy Contact Us follow us Subscribe for news and updates © 2025 AoPS Incorporated © 2025 Art of Problem Solving About Us•Contact Us•Terms•Privacy Copyright © 2025 Art of Problem Solving Something appears to not have loaded correctly. Click to refresh.
6491	https://math.stackexchange.com/questions/2496826/minimum-slope-of-a-chord-to-parabola	geometry - Minimum slope of a chord to parabola - Mathematics Stack Exchange Join Mathematics By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy. Sign up with Google OR Email Password Sign up Already have an account? Log in Skip to main content Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. 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Upvoting indicates when questions and answers are useful. What's reputation and how do I get it? Instead, you can save this post to reference later. Save this post for later Not now Thanks for your vote! You now have 5 free votes weekly. Free votes count toward the total vote score does not give reputation to the author Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation. Got it!Go to help center to learn more Minimum slope of a chord to parabola Ask Question Asked 7 years, 11 months ago Modified7 years, 10 months ago Viewed 261 times This question shows research effort; it is useful and clear 1 Save this question. Show activity on this post. A line is drawn from (−2,0)(−2,0) to intersect y 2=4 x y 2=4 x in P, Q within the first quadrant, such that 1 A P+1 A Q<1 4 1 A P+1 A Q<1 4 Find the minimum value of the line slope. A is the origin. I had basically let the coordinates of the parabola in parametric form and then used the given condition, but couldn't find the minimal slope. geometry Share Share a link to this question Copy linkCC BY-SA 3.0 Cite Follow Follow this question to receive notifications edited Nov 4, 2017 at 1:43 Anatoly 17.2k 2 2 gold badges 24 24 silver badges 53 53 bronze badges asked Oct 30, 2017 at 16:40 user481779user481779 757 4 4 silver badges 19 19 bronze badges 3 What is A A, the point (−2,0)(−2,0)? Using what you tried, were you able to make any progress, or did you get stuck somewhere?pjs36 –pjs36 2017-10-30 17:25:07 +00:00 Commented Oct 30, 2017 at 17:25 @pjs36 when I used the condition of sum of reciprocal of distances I couldn't simplify it to get any relationship with the slope user481779 –user481779 2017-10-30 17:42:30 +00:00 Commented Oct 30, 2017 at 17:42 1 I don’t believe that there is such a line. If you parameterize it by its slope m m then for 0<m<1/2–√0<m<1/2 (the upper bound is the slope of the tangent through (−2,0)(−2,0)), 1/6–√<1/A P+1/A Q<1/2 1/6<1/A P+1/A Q<1/2.amd –amd 2017-10-30 18:41:32 +00:00 Commented Oct 30, 2017 at 18:41 Add a comment\| 2 Answers 2 Sorted by: Reset to default This answer is useful 3 Save this answer. +25 This answer has been awarded bounties worth 25 reputation by Community Show activity on this post. As it is stated in the current form, this problem - as shown below - has no real solutions. I suspect that the original formulation might have been slightly different, in particular with A A indicating the point (−2,0)(−2,0) and maybe with a more general equation of the parabola (again see below). In this answer, I will assume that A=(−2,0)A=(−2,0) and that P P and Q Q are distinct points. Also, I will firstly consider the case of the specific parabola described in the OP, and then the case of a more general parabola equation. Lastly, I will add a brief comment on the case where A A is the origin. Since the line crosses the point (−2,0)(−2,0), it has the form y=a x+2 a y=a x+2 a The intersection points of this line with the parabola have x x-coordinates given by 2(1−a 2±1−2 a 2−−−−−−√)a 2 2(1−a 2±1−2 a 2)a 2 whereas the y y-coordinates are 2(1−a 2±1−2 a 2−−−−−−√)a+2 a 2(1−a 2±1−2 a 2)a+2 a Calculating the length of A P¯¯¯¯¯¯¯¯A P¯ using standard formulas yields A P¯¯¯¯¯¯¯¯=⎡⎣(2(1−a 2−1−2 a 2−−−−−−√)a 2)2+(2(1−a 2−1−2 a 2−−−−−−√)a+2 a)2⎤⎦1/2 A P¯=[(2(1−a 2−1−2 a 2)a 2)2+(2(1−a 2−1−2 a 2)a+2 a)2]1/2 which can be simplified as A P¯¯¯¯¯¯¯¯=2 2−2 a 2−a 4−2 1−2 a 2−−−−−−√−−−−−−−−−−−−−−−−−−−−−√a 2 A P¯=2 2−2 a 2−a 4−2 1−2 a 2 a 2 In a similar manner we get A Q¯¯¯¯¯¯¯¯=2 2−2 a 2−a 4+2 1−2 a 2−−−−−−√−−−−−−−−−−−−−−−−−−−−−√a 2 A Q¯=2 2−2 a 2−a 4+2 1−2 a 2 a 2 Note that since we must have 1−2 a 2>0 1−2 a 2>0 to have two separate intersection points, we get the condition −1 2–√<a<1 2–√−1 2<a<1 2 to obtain a different real value for A P¯¯¯¯¯¯¯¯A P¯ and A Q¯¯¯¯¯¯¯¯A Q¯. Substituting the expressions above in 1 A P¯¯¯¯¯¯¯¯+1 A Q¯¯¯¯¯¯¯¯<1 4 1 A P¯+1 A Q¯<1 4 and simplifying, we get 1 2 1+a 2−−−−−√<1 4 1 2 1+a 2<1 4 whose positive solution is a>3–√a>3 Because this solution has no intersection with the range of a a needed to have A P A P and A Q A Q real, we conclude that there are no real values of a a that satisfy the conditions stated in the OP. Notably, after some searching, I found on Google a nearly identical version of this problem with no solution but four possible answers (<3–√<3, >3–√>3, ≥3–√≥3, none of them), with the only difference that the initial parabola has equation y 2=4 k x y 2=4 k x. The point A A is explicitly defined as the point (−2,0)(−2,0). This more general version of the problem is interesting because, after calculations that are similar to those shown above, it leads to the same final inequality 1/(2 1+a 2−−−−−√)<1/4 1/(2 1+a 2)<1/4 with solution a>3–√a>3 (in other words, in the final part of the calculations and simplifications, k k is canceled out). In contrast, k k is not canceled out in the condition necessary for A P A P and A Q A Q to be distinct and real, which becomes −k 2−−√<a<k 2−−√−k 2<a<k 2 allowing to find, for k>6 k>6, an intersection between this range and that stated in the solution, and giving the answer a>3–√a>3 as the inferior bound of the slope. Note that such condition, for k=1 k=1, is the same obtained above considering the problem for the specific parabola y 2=4 x y 2=4 x. Lastly, repeating the same calculations for the case where A A is the origin leads to similar results. The final inequality in this case yields the numerical solution a>1.23414...a>1.23414... (apparently with no closed form). Again, because of the condition needed for A P¯¯¯¯¯¯¯¯A P¯ and A Q¯¯¯¯¯¯¯¯A Q¯ to be real and distinct, there are no real solutions for the case with parabola equation y 2=4 x y 2=4 x. On the other hand, in the general case with parabola equation y 2=4 k x y 2=4 k x, there are real solutions only for k>2⋅1.23414...2=3.04620...k>2⋅1.23414...2=3.04620... and the numerical solution a>1.23414...a>1.23414... is the lower bound of the slope. Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications edited Nov 4, 2017 at 2:17 answered Nov 4, 2017 at 1:38 AnatolyAnatoly 17.2k 2 2 gold badges 24 24 silver badges 53 53 bronze badges Add a comment\| This answer is useful 1 Save this answer. Show activity on this post. Convenient to shift reference to origin y 2=4(x−2)y 2=4(x−2) Convenient to use polar coordinates (x,y)=r(cos θ,sin θ)(x,y)=r(cos⁡θ,sin⁡θ) Plug in r 2 sin 2 θ=4(r cos θ−2)r 2 sin 2⁡θ=4(r cos⁡θ−2) Simplify, let me abbreviate for easier Latex typing 4 r s 2 t−4 r c t+8=0 4 r s t 2−4 r c t+8=0 If roots are r 1,r 2(A P,A Q)r 1,r 2(A P,A Q) for quadratic, sum and product are r 1+r 2=4 c t/s 2 t r 1+r 2=4 c t/s t 2 and r 1⋅r 2=8/s 2 t r 1⋅r 2=8/s t 2 Trick is to divide getting 1 r 1+1 r 2=c t/2 1 r 1+1 r 2=c t/2 ( BTW harmonic mean of the two segments A P,A Q=4 sec θ A P,A Q=4 sec⁡θ ) Now if c t/2<1 4 c t/2<1 4 then for maximum slope t=π/3,m=s l o p e=3–√t=π/3,m=s l o p e=3 Next find slope of tangent onto parabola from origin for maximum possible slope, we find intersections between y 2=4(x−2),y=m x y 2=4(x−2),y=m x Plug in and get quadratic in, say x x. m 2 x 2−4 x+8=0 m 2 x 2−4 x+8=0 Discriminant is zero for tangent condition Δ=16−4⋅m 2⋅8=0 Δ=16−4⋅m 2⋅8=0 of tangent line L2 m=t a n g e n t s l o p e=±1 2–√m=t a n g e n t s l o p e=±1 2 For given slope line L1 3–√>1 2–√3>1 2 So given inequality can never be satisfied with the real numbers. All the above are included together in a single graph below: For line L 1 L 1 neither the segments nor their harmonic mean exist in reals. Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications edited Nov 11, 2017 at 5:21 answered Nov 10, 2017 at 18:00 NarasimhamNarasimham 42.5k 7 7 gold badges 46 46 silver badges 112 112 bronze badges Add a comment\| You must log in to answer this question. Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions geometry See similar questions with these tags. 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Reject All Save My Preferences Accept All Skip to ContentGo to accessibility pageKeyboard shortcuts menu Log in Elementary Algebra 2e Introduction Elementary Algebra 2eIntroduction Contents Contents Highlights Table of contents Preface 1 Foundations 2 Solving Linear Equations and Inequalities Introduction 2.1 Solve Equations Using the Subtraction and Addition Properties of Equality 2.2 Solve Equations using the Division and Multiplication Properties of Equality 2.3 Solve Equations with Variables and Constants on Both Sides 2.4 Use a General Strategy to Solve Linear Equations 2.5 Solve Equations with Fractions or Decimals 2.6 Solve a Formula for a Specific Variable 2.7 Solve Linear Inequalities Chapter Review Exercises 3 Math Models 4 Graphs 5 Systems of Linear Equations 6 Polynomials 7 Factoring 8 Rational Expressions and Equations 9 Roots and Radicals 10 Quadratic Equations Answer Key Index Search for key terms or text. Close Figure 2.1 The rocks in this formation must remain perfectly balanced around the center for the formation to hold its shape. Chapter Outline 2.1 Solve Equations Using the Subtraction and Addition Properties of Equality 2.2 Solve Equations using the Division and Multiplication Properties of Equality 2.3 Solve Equations with Variables and Constants on Both Sides 2.4 Use a General Strategy to Solve Linear Equations 2.5 Solve Equations with Fractions or Decimals 2.6 Solve a Formula for a Specific Variable 2.7 Solve Linear Inequalities If we carefully placed more rocks of equal weight on both sides of this formation, it would still balance. Similarly, the expressions in an equation remain balanced when we add the same quantity to both sides of the equation. In this chapter, we will solve equations, remembering that what we do to one side of the equation, we must also do to the other side. 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Authors: Lynn Marecek, MaryAnne Anthony-Smith, Andrea Honeycutt Mathis Publisher/website: OpenStax Book title: Elementary Algebra 2e Publication date: Apr 22, 2020 Location: Houston, Texas Book URL: Section URL: © Jul 8, 2025 OpenStax. Textbook content produced by OpenStax is licensed under a Creative Commons Attribution License . The OpenStax name, OpenStax logo, OpenStax book covers, OpenStax CNX name, and OpenStax CNX logo are not subject to the Creative Commons license and may not be reproduced without the prior and express written consent of Rice University. Our mission is to improve educational access and learning for everyone. OpenStax is part of Rice University, which is a 501(c)(3) nonprofit. Give today and help us reach more students. Help Contact Us Support Center FAQ OpenStax Press Newsletter Careers Policies Accessibility Statement Terms of Use Licensing Privacy Policy Manage Cookies © 1999-2025, Rice University. 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6494	https://www.medicalnewstoday.com/articles/mcv-levels	Health Conditions Alzheimer's & Dementia Anxiety Asthma & Allergies Atopic Dermatitis Breast Cancer Cancer Cardiovascular Health COVID-19 Diabetes Endometriosis Environment & Sustainability Exercise & Fitness Eye Health Headache & Migraine Health Equity HIV & AIDS Human Biology Leukemia LGBTQIA+ Men's Health Mental Health Multiple Sclerosis (MS) Nutrition Parkinson's Disease Psoriasis Sexual Health Ulcerative Colitis Women's Health Health Products All + Nutrition & Fitness + Vitamins & Supplements + CBD + Sleep + Mental Health + At-Home Testing + Men’s Health + Women’s Health ### News Latest News ### Original Series Medical Myths Honest Nutrition Through My Eyes New Normal Health ### Podcasts All + Does the Mediterranean diet hold the key to longevity? + AMA: Registered dietitian answers 5 key questions about fiber and weight loss + Health misinformation and disinformation: How to avoid it + Brain health, sleep, diet: 3 health resolutions for 2025 + 5 things everyone should know about menopause + 3 ways to slow down type 2 diabetes-related brain aging ### General Health Drugs A-Z Health Hubs Newsletter ### Health Tools Find a Doctor BMI Calculators and Charts Blood Pressure Chart: Ranges and Guide Breast Cancer: Self-Examination Guide Sleep Calculator ### Quizzes RA Myths vs Facts Type 2 Diabetes: Managing Blood Sugar Ankylosing Spondylitis Pain: Fact or Fiction About Medical News Today Who We Are Our Editorial Process Content Integrity Conscious Language ### Find Community Bezzy Breast Cancer Bezzy MS Bezzy Migraine Bezzy Psoriasis ### Follow Us What does a mean corpuscular volume level measure? Medically reviewed by Megan Soliman, MD — Written by Lauren Martin — Updated on December 13, 2024 Mean corpuscular volume, or MCV, measures red blood cell size. A typical adult MCV level is 80 to 100 femtoliters (fl). The above figure comes from a 2024 topic reviewTrusted Source. A doctor usually requests an MCV test as part of a complete blood count, which analyzes many blood components, including white blood cells and platelets. If a doctor suspects that a person has anemia, they will use an MCV test to confirm the type of anemia. Different MCV levels indicate specific types of anemia. However, it is possible to have anemia with a normal MCV level. In this article, we examine what MCV levels measure. We also examine what different MCV levels mean and what may cause these changes. What is an MCV level blood test? Red blood cells have many characteristics that a doctor can measure using specific indicators. MCV indicates the average red blood cell size and volume. Other red blood cell indicators include: Mean corpuscular hemoglobin (MCH): This is the average hemoglobin level within a red blood cell. Mean corpuscular hemoglobin concentration (MCHC): This is the average hemoglobin concentration in red blood cells. Red cell distribution width (RDW): This measures the variation in red blood cells’ size. Doctors use these measurements when diagnosing specific types of anemia, as well as other health conditions. Learn more about MCH levels in blood tests here. What happens during an MCV level blood test? A person does not need to prepare for an MCV blood test. During an MCV blood test, a doctor draws blood from a vein to collect a sample. While the doctor is taking the blood sample, a person may feel a little pain and a stinging sensation. Are there any risks? Drawing blood carries few risks. However, everyone is different, and sometimes blood collection is more straightforward in some people than in others. The side effects of an MCV blood test may include: bruising excessive bleeding feeling lightheaded infection Why do doctors measure MCV levels? A doctor typically measures someone’s MCV levels if they are presenting with: dry, cracked lips cold intolerance fatigue bruising or bleeding easily unexplained weight loss cold hands and feet looking paler than usual jaundice These symptoms indicate conditions that affect the size of red blood cells, making them smaller or larger than usual. When red blood cells are smaller than expected, a person likely has microcytic anemia. If they are larger, a person may have macrocytic anemia. A note about sex and gender Sex and gender exist on spectrums. This article will use the terms “male,” “female,” or both to refer to sex assigned at birth. Learn more. What is a normal MCV level? A typical adult MCV level is 80 to 100 femtoliters (fl)Trusted Source. However, typical ranges vary between age groups and sexes. A 2022 analysisTrusted Source found the following average results. \| Age \| Male \| Female \| --- \| 1-19 \| 81.9 – 87.3 fl \| 82.4 – 87.3 fl \| \| 20+ \| 89.8 – 93.6 fl \| 90 – 92.5 fl \| Children ages 6 to 12 tend to have an MCV of 86 fl. MCV results may differ among labs, so people should not worry if their reading is slightly above or below these ranges. Low MCV level When a person has an MCV level below 80 flTrusted Source, this suggests they have microcytic anemia. Microcytic anemia is a type of anemia in which red blood cells are smaller than usual. Iron deficiency causes microcytic anemia. A person usually develops an iron deficiency due to an underlying health condition or factors such as diet and medications. Common causes Iron deficiency causes include: Blood loss: A person may experience bleeding from the gastrointestinal (GI) tract when they have colon cancer or take nonsteroidal anti-inflammatory drugs (NSAIDs), such as aspirin. People with heavy periods may also lose a lot of iron through menstrual blood. Diet: If someone eats a plant-based diet or an omnivorous diet low in iron, they may need to take iron supplements or focus on eating iron-rich foods. Reduced iron absorption: Ulcerative colitis, Crohn’s disease, weight loss surgery, and Helicobacter pylori infection may all reduce the absorption of iron. Pregnancy: During pregnancy, people may need to supplement their iron intake, as the body needs more iron to support fetal development. Thalassemia Thalassemia is a condition where the body does not make enough normal hemoglobin. It is a genetic condition that a person inherits from their parents. Thalassemia can rangeTrusted Source from mild to severe. If a person has mild thalassemia, they may have mild anemia or not present with any signs or symptoms. If someone has severe thalassemia, they may require regular red blood cell transfusions. High MCV level If someone has a high MCV level, their red blood cells are larger than usual, and they have macrocytic anemia. Macrocytosis occurs in people with an MCV level higher than 100 flTrusted Source. Megaloblastic anemia is a type of macrocytic anemia. Deficiencies in cobalamin (vitamin B12) and folate (vitamin B9) are the most common causes of megaloblastic anemia. Vegan diet Vitamin B12 deficiency may occur when following a vegan diet. This is because vitamin B12 naturally occurs in animal sources. A person can take vitamin B12 supplements or consume vitamin B12-fortified foods to alleviate symptoms. Autoimmune gastritis This is an inflammatory condition that affects the stomach, particularly the parietal cells that make intrinsic factor. If the parietal cells do not make enoughTrusted Source intrinsic factor, vitamin B12 cannot enter the bloodstream via the distal small intestine. Doctors treat the deficiency with vitamin B12 injections and, sometimes, iron infusions. Other causes Other factors that may contribute to vitamin B12 and vitamin B9 deficiencies include: chestfeeding consuming too much alcohol tapeworm Crohn’s disease cancer treatment medications When to contact a doctor about MCV levels If someone is frequently tired and feels cold all the time, they may have anemia. People experiencing symptoms of anemia should contact a doctor. A doctor may ask about a person’s family’s medical history. Some conditions that cause anemia, such as thalassemia and Crohn’s disease, run in families. Summary An MCV test measures the size and volume of red blood cells. A typical MCV range is roughly 80 to 100 flTrusted Source. If someone’s MCV level is below 80 fl, they will likely develop or have microcytic anemia. Alternatively, if their MCV levels are greater than 100 fl, they could experience macrocytic anemia. People with microcytic anemia may feel the cold more and look paler than usual. People with macrocytic anemia may experience jaundice. If someone has symptoms of anemia, they should contact a doctor for advice. Usually, treating the underlying cause of the anemia alleviates symptoms. Read this article in Spanish. Blood / Hematology Medical Devices / Diagnostics Primary Care How we reviewed this article: Medical News Today has strict sourcing guidelines and relies on peer-reviewed studies, academic research institutions, and medical journals and associations. We only use quality, credible sources to ensure content accuracy and integrity. You can learn more about how we ensure our content is accurate and current by reading our editorial policy. Anemia, megaloblastic. (2023). Iron-deficiency anemia. (2022). Kulnigg-Dabsch, S. (2016). Autoimmune gastritis. Lee JY, et al. (2022). Age-related changes in mean corpuscular volumes in patients without anaemia: An analysis of large-volume data from a single institute. Maner BS, et al. (2024). Mean corpuscular volume. Nagao T, et al. (2017). Diagnosis and treatment of macrocytic anemias in adults. What is thalassemia? (2022). Share this article Medically reviewed by Megan Soliman, MD — Written by Lauren Martin — Updated on December 13, 2024 Latest news Coffee drinkers have a lower risk of liver diseases, evidence shows Aspirin may help reduce colorectal cancer recurrence in high-risk people 4 types of foods can boost happiness, well-being in aging adults Green Mediterranean diet, including green tea, may help slow brain aging Even small amounts of alcohol may increase dementia risk, study finds Was this article helpful? Yes No Related Coverage What is a complete blood count? Medically reviewed by Alana Biggers, M.D., MPH A complete blood count (CBC) is a test that measures white blood cells, red blood cells, and platelets. Learn more about a CBC and what the results… READ MORE Hypoglycemia and anemia: Is there a link? Hypoglycemia and anemia are not related conditions, but they can share similar symptoms and some of the same root causes. Learn more here. READ MORE What is vitamin B12 deficiency? Vitamin B12 is crucial for the formation of red blood cells and the healthy functioning and health of nerve tissue. Learn more here about vitamin B12… READ MORE
6495	https://energyeducation.ca/encyclopedia/Entropy	Entropy - Energy Education EnglishFrançaisEspañol Energy Education Navigation menu ENERGY SOURCES Fuels Fossil Fuels Biofuels Nuclear Fuels Flows Hydro Solar Wind Geothermal ENERGY USE Carriers Electricity Gasoline Hydrogen Sectors Transportation Residential Industrial ENERGY IMPACTS Living standard Pollution Acid Rain Smog Pollutants Climate Change Climate Feedback Ocean Acidification Rising Sea Level INDEX Search Entropy Figure 1: With entropy of a closed system naturally increasing, this means that the energy quality will decrease. This is why low quality heat cannot be transferred completely into useful work. Entropy is a measure of the number of ways a thermodynamic system can be arranged, commonly described as the "disorder" of a system. This concept is fundamental to physics and chemistry, and is used in the Second law of thermodynamics, which states that the entropy of a closedsystem (meaning it doesn't exchange matter or energy with its surroundings) may never decrease. This means that the "multiplicity", or number of ways a system can be arranged will never decrease, and that the system will naturally tend to higher disorder. The maximum disorder of a system occurs when it is at thermal equilibrium, therefore, this is what all isolated systems will tend to over time. Entropy can also be described as a system's thermal energy per unit temperature that is unavailable for doing useful work. Therefore entropy can be regarded as a measure of the effectiveness of a specific amount of energy. Shown in Figure 1, this is represented as the "energy quality", which decreases as the entropy of a system increases. It can be seen that heat has a lower energy quality than mechanical energy or electricity, so this can be used to understand why an amount of heat cannot be converted completely into the same amount of these higher quality forms of energy. In a reversible thermodynamic process, such as a Carnot engine, the change in entropy over a full cycle must be equal to zero. This can be explored in more detail on the Hyperphysics website. Order out of Chaos Figure 2: Plants and other complex organisms are able to decrease their entropy due to the fact that they are not isolated systems. There are many examples of complexity in our world around us. Things such as Plants growing from tiny seeds, Single-celled fertilized eggs growing into complex life, Complex molecules and formations, and Vast increases of knowledge and information are extremely complex systems which appear to violate the Entropy Statement of the Second law of thermodynamics. Since entropy is increasing, and the Second law entails that out of this increase comes disorder, randomness and simplicity, how is it that there is so much order and complexity around us? The answer is that this disorder only applies to systems that do not exchange energy with their environment, known as isolated systems. This is a major confusion in understanding entropy, and it is important to distinguish between an isolated (closed) and non-isolated (open) system. Open systems are free to interact with their environment, and therefore energy can be added to or removed from them. Systems that become more ordered as time passes are called self-organizing systems. In order for this decrease in entropy to be possible, they must take in energy from an outside source. However, because the open system's entropy is decreasing, there must be an increase of entropy outside of the system. For example, this is why water can freeze into complex structures. The water forms a highly organized crystal, and the entropy of the water decreases as it forms this structure. This happens because heat energy is transferred from the water to the surrounding air, therefore increasing the entropy of the air. This increase in the air must be more than the decrease in the water, because the whole system's entropy must increase. This is analogous to refrigeration, as work must be input to the refrigerator in order to cool it down, therefore decreasing its entropy. This concept of decreasing a non-isolated system's entropy can be visualized in Figure 3 below. (The arrangement of the bricks is not a literal representation of the brick's entropy, rather it is just to demonstrate the idea of "multiplicity". The actual entropy of the bricks has to do with their internal temperature.) Figure 3: In order to decrease the number of possible arrangements of the bricks, work must be done to the system. The bricks are a non-isolated system because of this. Entropy as the arrow of time Since the entropy of a system tends to more disorder over time, and never in reverse, it is said to give us "time's arrow". If snapshots of a system at two different times shows one state which is more disordered, then it could be implied that this state came later in time. Figure 4: If particles are confined to a closed system as illustrated here, which direction must time be flowing? (Hint: Think in terms of "multiplicity" or "randomness"). For Further Reading Thermodynamics First law of thermodynamics Second law of thermodynamics Thermal equilibrium Thermodynamics Or explore a random page References ↑ Jump up to: 1.01.1R. Wolfson, "Entropy, Heat Engines, and the Second Law of Thermodynamics" in Energy, Environment, and Climate, 2nd ed., New York, NY: W.W. Norton & Company, 2012, ch. 4, sec. 7, pp. 81-84 ↑Hyperphysics, A More General View of Temperature [Online], Available: ↑Encyclopaedia Britannica, Entropy [Online], Available: ↑Wikipedia Commons [Online], Available: ↑ Jump up to: 5.05.1R. D. Knight, "Order Out of Chaos" in Physics for Scientists and Engineers: A Strategic Approach, 3nd ed. San Francisco, U.S.A.: Pearson Addison-Wesley, 2008, ch.18, pp. 557 ↑Hyperphysics, Entropy as Time's Arrow [Online], Available: ↑Hyperphysics, Second Law: Entropy [Online], Available: ↑Adapted from Hyperphysics, Entropy as Time's Arrow [Online], Available: Get Citation Contact usAbout usPrivacy policyTerms of use
6496	https://www.youtube.com/watch?v=kGzXIbauGQk	Plotting complex numbers on the complex plane \| Precalculus \| Khan Academy Khan Academy 9090000 subscribers 482 likes Description 399311 views Posted: 15 Aug 2013 Courses on Khan Academy are always 100% free. Start practicing—and saving your progress—now: Practice this lesson yourself on KhanAcademy.org right now: Watch the next lesson: Missed the previous lesson? Precalculus on Khan Academy: You may think that precalculus is simply the course you take before calculus. You would be right, of course, but that definition doesn't mean anything unless you have some knowledge of what calculus is. Let's keep it simple, shall we? Calculus is a conceptual framework which provides systematic techniques for solving problems. These problems are appropriately applicable to analytic geometry and algebra. Therefore....precalculus gives you the background for the mathematical concepts, problems, issues and techniques that appear in calculus, including trigonometry, functions, complex numbers, vectors, matrices, and others. There you have it ladies and gentlemen....an introduction to precalculus! About Khan Academy: Khan Academy offers practice exercises, instructional videos, and a personalized learning dashboard that empower learners to study at their own pace in and outside of the classroom. We tackle math, science, computer programming, history, art history, economics, and more. Our math missions guide learners from kindergarten to calculus using state-of-the-art, adaptive technology that identifies strengths and learning gaps. We've also partnered with institutions like NASA, The Museum of Modern Art, The California Academy of Sciences, and MIT to offer specialized content. For free. For everyone. Forever. #YouCanLearnAnything Subscribe to Khan Academy’s Precalculus channel: Subscribe to Khan Academy: 18 comments Transcript: Move the orange dot to negative 2 plus 2i. So we have a complex number here. It has a real part, negative 2. It has an imaginary part, you have 2 times i. And what you see here is we're going to plot it on this two-dimensional grid, but it's not our traditional coordinate axes. In our traditional coordinate axis, you're plotting a real x value versus a real y-coordinate. Here on the horizontal axis, that's going to be the real part of our complex number. And our vertical axis is going to be the imaginary part. So in this example, this complex number, our real part is the negative 2 and then our imaginary part is a positive 2. And so that right over there in the complex plane is the point negative 2 plus 2i. Let's do a few more of these. So 5 plus 2i. Once again, real part is 5, imaginary part is 2, and we're done. Let's do two more of these. 1 plus 5i. 1-- that's the real part-- plus 5i right over that Im. All right, let's do one more of these. 4 minus 4i. Real part is 4, imaginary part is negative 4. And we're done.
6497	https://shelleygrayteaching.com/using-bar-models-to-effectively-solve-math-word-problems/	Skip to content Using Bar Models to Effectively Solve Math Word Problems By Shelley Gray No Comments Have you tried Model Drawing in your classroom? I recently read the book 8 Step Model Drawing by Bob Hogan and Char Forsten. I have always been a fan of bar models, but this book was an eye opener as to just how effective they are as a problem solving tool. The authors state that about 80% of word problems that students encounter can be effectively solved using models like these. My mind has been spinning with all of the other potential uses for bar models beyond basic problem-solving. As I continued to read more about model drawing, it became clear to me that they can be connected beautifully to so many different math manipulatives. I think every teacher should know about model drawing, so I decided to put together a 10-part video series with ideas you can take directly back to your classroom. These videos will incorporate both math manipulatives and model drawing for different problems including fractions, multiplication, division, and more! As the videos are posted on Instagram, TikTok, Facebook, and YouTube, I’ll also add links here so you can find them easily. Or sign up here to have the link delivered to your inbox. I hope you enjoy this video series! If you’d like to know even more about model drawing, be sure to check out Step-by-Step Model Drawing as well as 8-Step Model Drawing. Day 1: Using Models and Manipulatives to Solve Fraction Problems Bar models are naturally an incredible visual for fraction problems. Fractions can often be such an abstract concept for our students, and models can transform them to easy-to-understand. View this video below or on these social media outlets: Instagram Facebook Day 2: Using Models and Manipulatives to Solve (More) Fraction Problems I decided to focus on fractions for 2 out of the 10 videos in this series, because models are so effective for visualizing fraction problems. The problem below is one of my favorites because it makes a 2-part, more complex problem so simple to visualize. View this video below or on these social media outlets: Instagram Facebook Day 3: Using Models and Manipulatives to Make Addition and Subtraction Word Problems Visual Models are very well-suited to addition and subtraction problems. They not only make the problem easier to understand, but reinforce the inverse relationships between addition and subtraction. This video shows two different problems and how bar models can be used. Note I neglected to make the parts of the bar different sizes in the second problem. This is my mistake, and I will be re-creating this video soon. In the meantime, please note that the pieces should be more to scale to show the differences in the magnitude of the numbers. View this video below or on these social media outlets: Instagram Day 4: Using Models and Manipulatives to Build Multiplicative Reasoning Models are a must-use tool to develop multiplicative reasoning. They beautifully illustrate the relationship between multiplication and repeated addition, and can take complex multiplication problems from difficult to easy-to-imagine. View this video on these social media outlets: Instagram Facebook Day 5: Using Models to Make Division Problems Visual Bar models for division are pretty straightforward, and we might neglect to use them because they don’t feel necessary. However, remember that modeling simple problems is important so that students feel comfortable with models as a tool for more complex problems. Additionally, I love how division models effectively connect multiplication and division, as well as build a foundation for fraction concepts. View this video below or on the following social media outlets: Instagram Facebook Day 6: Making Ratio Conceptual Using Models and Concrete Manipulatives Let’s take a look at how we can make ratio more conceptual by using models and concrete manipulatives! View this video below or on the following social media outlets: Instagram Facebook Day 7: Solving a Rate Word Problem Using Math Modelling and Manipulatives Rate can be a tough concept for students to grasp, but is much easier to visualize and understand when we incorporate math manipulatives and visual models! View this video below or on the following social media outlets: Instagram TikTok Facebook Day 8: Solving Percent Problems Using Math Manipulatives and Models View this video below or on the following social media outlets: Instagram TikTok Facebook Day 9: Solving a Math Word Problem With Mixed Operations Using Bar Models and Math Manipulatives View this video below or on the following social media outlets: Instagram TikTok Facebook Day 10: Solving Algebra Word Problems Using Math Models and Manipulatives View this video below or on the following social media outlets: Instagram TikTok Facebook PrevpreviousTeaching Mean, Median, and Mode? Try these creative ideas! NextFive Simple Activities to Help Your Students Build Fraction SenseNext Leave a Reply Cancel reply This site uses Akismet to reduce spam. Learn how your comment data is processed. Free Fast Finisher Activities Master the Multiplication Facts! Make math relevant! A fun math game! 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6498	https://definicion.de/movimiento-uniforme/	\| \| \| Escuchar \| Movimiento uniforme En un movimiento uniforme, la velocidad se mantiene constante. Es importante establecer en un primer momento de dónde proceden, etimológicamente hablando, las palabras que le dan forma al término movimiento uniforme. Así, podemos establecer que las dos emanan del latín: • Movimiento es fruto de la suma de dos componentes: el verbo «movere», que puede traducirse como «mudar de un lado a otro», y el sufijo «-miento», que es indicativo de «acción y efecto». • Uniforme, por otro lado, es el resultado de haber unido dos elementos latinos: «unis», que es equivalente a «uno», y «forme», que ejerce como sinónimo de «forma». Qué es el movimiento uniforme La acción y efecto de mover se conoce como movimiento. El verbo mover refiere al estado del cuerpo mientras cambia de posición o de lugar. Mover, por lo tanto, es hacer que un cuerpo pase de un lugar a otro, o agitar una parte de él o de una cosa. />Uniforme es el adjetivo que refiere a las cosas que tienen la misma forma o que resultan iguales o semejantes. Como sustantivo, uniforme es un traje distintivo que utilizan aquellos que pertenecen a un mismo cuerpo o entidad. Con esto en mente, podemos hacer referencia a la noción de movimiento uniforme. Se trata de aquel movimiento cuya velocidad, ya sea de traslación o de rotación, permanece constante. El movimiento uniforme puede ser rectilíneo o circular. Clasificación según el tipo Es posible distinguir entre movimiento rectilíneo uniforme y movimiento circular uniforme. En el primer caso, el cuerpo realiza una trayectoria recta con velocidad constante en el tiempo (es decir, con aceleración nula). En el segundo caso, la trayectoria que realiza el cuerpo es circular, con rapidez constante (no así la velocidad, que es una magnitud vectorial). Para poder llevar a cabo el cálculo del movimiento rectilíneo uniforme se hace necesario contar con una serie de variables o parámetros al respecto. En concreto, nos estamos refiriendo a la posición, a la velocidad y a la aceleración. Mediante dichos valores es como se puede desarrollar, por ejemplo, el cálculo de la rapidez promedio de un atleta que recorre al correr por una pista recta y al realizar dos tramos diferentes de la misma. El movimiento uniformemente acelerado, por su parte, es el movimiento en el que la velocidad se incrementa de manera proporcional al tiempo transcurrido, mientras que el movimiento uniformemente retardado es lo inverso (la velocidad se reduce proporcionalmente al tiempo). Ejemplo de movimiento uniforme Para entender fácilmente qué es el movimiento uniforme, podemos considerar el caso de un automóvil que avanza en línea recta a una velocidad constante de 100 kilómetros por hora. Podrá decirse, en este caso, que el coche realiza un movimiento rectilíneo uniforme. Para poder acometer lo que sería el cálculo de ese mencionado movimiento uniformemente acelerado se hace imprescindible contar con valores tales como la aceleración, que se representa en la fórmula con la letra «a»; la velocidad inicial, que se representa mediante la «Vo», y la velocidad final, la «Vf». Así, por ejemplo, si se desea calcular el tiempo, tendríamos que decir que este es igual a la resta de la Vo a la Vf y luego su resultado dividirlo por la «a». Es decir, t = Vf – Vo / a. Copyright © 2008-2025 - Definicion.de Queda prohibida la reproducción total o parcial de los contenidos de esta web Quiénes somos - Privacidad - Contacto - Arriba ↑
6499	https://semanticsmorphology.weebly.com/inflectional-and-derivational-morphemes.html	HOME MORPHOLOGY CONTENT WORDS AND FUNCTION WORDS MORPHEMES AND THEIR COMPOSITIONS INFLECTIONAL AND DERIVATIONAL MORPHEMES WORD FORMATION THROUGH DERIVATION OTHER WORD FORMATION PROCESSES OTHER WORD FORMATION PROCESSES Morphology inflectional AND DERIVATIONAL MORPHEMES We can make a further distinction within the set of bound morphemes in English. One type of bound morphemes consists of derivational morphemes that are used to create new words or to “make words of a different grammatical class from the stem” (Yule, 2010, p. 69). For example, the addition of the derivational morpheme -ize changes the adjective normal to the verb normalize. Similarly, we can derive the adjectives helpful and helpless by adding the derivational morphemes -ful and less to the noun help. The second type of bound morphemes consists of inflectional morphemes that are used to show some aspects of the grammatical function of a word. We use inflectional morphemes to indicate if a word is singular or plural, whether it is past tense or not, and whether it is a comparative or possessive form. In fact, inflection exists in many languages, but compared to other languages of the world there is relatively little inflection in English. Today there are only eight inflectional morphemes in English, as shown in the image below: Differences between Derivational and Inflectional Morphemes There are some differences between inflectional and derivational morphemes. First, inflectional morphemes never change the grammatical category (part of speech) of a word. For example, tall and taller are both adjectives. The inflectional morpheme -er (comparative marker) simply produces a different version of the adjective tall. However, derivational morphemes often change the part of speech of a word. Thus, the verb read becomes the noun reader when we add the derivational morpheme -er. It is simply that read is a verb, but reader is a noun. However, some derivational morphemes do not change the grammatical category of a word. For example, such derivational prefixes as re- and un- in English generally do not change the category of the word to which they are attached. Thus, both happy and unhappy are adjectives, and both fill and refill are verbs, for example. The derivational suffixes -hood and -dom, as in neighborhood and kingdom, are also the typical examples of derivational morphemes that do not change the grammatical category of a word to which they are attached. Second, when a derivational suffix and an inflectional suffix are added to the same word, they always appear in a certain relative order within the word. That is, inflectional suffixes follow derivational suffixes. Thus, the derivational (-er) is added to read, then the inflectional (-s) is attached to produce readers. Similarly, in organize– organizes the inflectional -s comes after the derivational -ize. When an inflectional suffix is added to a verb, as with organizes, then we cannot add any further derivational suffixes. It is impossible to have a form like organizesable, with inflectional -s after derivational -able because inflectional morphemes occur outside derivational morphemes and attach to the base or stem. For these reasons, Akmajian et al. (2010, p. 46) state that derivational morphemes show the “inner” layer of words, whilst inflectional suffixes mark the “outer” layer of words. A third point worth emphasizing is that certain derivational morphemes serve to create new base forms or new stems to which we can attach other derivational or inflectional affixes. For example, we use the derivational -atic to create adjectives from nouns, as in words like systematic and problematic. Then, we can further add -al to these two words to create systematical and problematical. Similarly, the derivational suffix -ize is often added to create verbs from adjectives, as in modernize, and we can add the inflectional suffix -s (modernizes) to such -ize verbs. To sum up, we can state that certain derivational affixes produce new members for a given class of words, but inflectional affixes are always added to available members of a given class of words. Some Exceptions in English Morphology As in many languages of the world, English also has some irregularities or exceptions in its morphology. So far, we have only exemplified English words in which various inflectional and derivational morphemes can be simply recognized as distinct minimal units of meaning or grammatical function. The plural morpheme -s is attached to boy, and the plural boys is created. Although the vast majority of nouns pluralize in English by adding –s, some nouns form their plurals irregularly. Typical examples of such irregular plurals in English are man-men, woman-women, foot-feet, mouse-mice, tooth-teeth, sheep-sheep and deer-deer. There are also exceptions in the forms of the past tense verbs in English, for example began, broke, went and ran. Both historical influences and the effect of borrowed words play a role in accounting for such irregularities. For pedagogical purposes, most grammar and writing textbooks include long lists of these exceptions in English morphology. \| \| \| --- \| \| \| \|

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