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6900 | https://www.youtube.com/watch?v=8eEoWQ_KznM | Projectile Motion - How to Find the Maximum Height and Range - Physics
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Posted: 16 Sep 2024
This physics video tutorial explains how to find the maximum height and range quickly using direct formulas.
Projectile Motion - Free Formula Sheet:
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Transcript:
in today's lesson we're going to work on projectile motion problems that will help us to calculate the maximum height and the range of a projectile so let's begin with this problem A ball rolls off a cliff at 15 m/ second and hits the ground 6 seconds later what is the height of the cliff and how far from the base of the cliff will the ball hit the ground now for each of these problems I recommend that you pause the video Try it yourself and see if you can get the right answer and then play it when you want to check your answer now the first thing we needs to do is draw a picture so here we have a ball it's rolling horizontally off a cliff so it's going to follow this particular trajectory now we know the initial speed of the ball it's 15 m/ second and we know the time it takes to hit the ground it's 6 seconds later how can we determine the height of the cliff the height of the cliff is going to be 1 12 GT2 for this particular trajectory you could find this formula in the formula sheet that I have Below in the description section of this video I recommending uh printing that formula because it's going to be very helpful for the problems to come in this video so the height is going to be 1/2 G is going to be 9.8 and T T is 6 6^ 2 is 36 half of that is 18 18 9.8 is [Music] so we already know the height what is the range the range is equal to vxt and for this problem initially the ball is moving in a horizontal Direction in the X Direction so for this particular trajectory V initial is equal to VX and VX doesn't change it's constant throughout this trajectory VX is also equal to V initial cosine Theta however when you're moving in a horizontal Direction the angle is zero and cosine of 0° is 1 so this becomes just the initial T So it's 15 m/s time 6 seconds which will give you 90 M so that's the range for this particular projectile you simply multiply the initial velocity by the time and you'll get the answer now let's move on to the next question a ball is kicked from the ground at 40 m/s at an angle of 30° above the horizontal so what is the maximum height of the ball so let's begin by drawing a picture so here's the ball it's kick from the ground it's going to go up and then it's going to go down so we know the angle it's 30° and we know the initial velocity of the ball it's 40 m/s how can we determine the maximum height with this this information once again if you've printed the formula sheet that I have in the description section below you can find this equation the maximum height is going to be V initial squar sin^2 Theta over 2G so V initial is 40 this is going to be 402 and then s of 30 and then we need to square that / 2 time 9.8 2 9.8 is 19.6 so I'm just going to put 19.6 on the bottom 40 s is600 sin 30 is a half so 1600 a half is 800 times another sin3 that's going to take it to 400 divided by 19.6 you should get 20.4 M for the maximum height attained by the um by the projectile so this is going to give you this answer that's 20.4 m now Part B determine the range of the projectile so the range is going to be this horizontal distance from where it lifted off from the ground to where it lands now you could use the previous formula that we had in the other problem range is equal to V initial cosine Theta T the only problem is we don't know the time it takes to go from point A to point C so we don't want to use this equation because we need to solve for T in order to use it there is however another equation that we could use in the formula sheet down below and here it is the range is equal to V initial squar time sin 2 Theta over G so V initial is 40 Theta is 30 so 2 Theta is 60 and G is 9.8 402 is600 s of 60 that's 13856 divide that by 9.8 you're going to get 141.5 M so that's the range of the projectile for this problem now let's move on to this one a ball is kicked from the top of a 300 M building at an angle of 40° above the horizontal with an initial speed of 50 let's draw a picture so here's the ball it's going to go up and then it's going to go down so it's launched at a 40° angle the initial speed is 50 m/s let's call this point a the top point B and point C now we know the height of the building is 300 M so you can call that H or Y initial part A determine the maximum height attained by the ball relative to the ground in my formula sheet down below this is the maximum height for this trajectory lowercase H would represent the height of the building which is the same as y initial so the formula that you need to calculate the maximum height attained for this specific trajectory here it is it's a y initial or the height of the building plus v initial squ sin 2 Theta over 2 G that's how you can calculate the maximum height so it's going to be 300 plus v initial squared so 50 squ and then sin 40° but we got to square that and 2 g 2 9.8 is 19.6 so 50 s is 2500 if you multiply that by sin 40 2 times that'll give you about 132.0 divide that by 19.6 you get 52.7 add 300 to that that will give you the maximum height which is 352 approximately 7 m so that's the answer to the first part part of the problem again you could find that formula in the formula sheet now Part B how long would it take the ball to hit the ground so we want to find the time it takes to go from point A to point C now there are two formulas that in the formula sheet that will help you to find this here's the first one it's V sin Theta over G plus the sare < TK of 2 the maximum height attain / G so this formula is useful if you already know the maximum height I'll show you another formula that you can use if you don't know the maximum height so this is V initial by the way V initial is 50 sin Theta V initial sin Theta is equivalent to V VY or Vy initial so it's 50 sin Theta over 9.8 plus the sare < TK of 2 35 2.7 over 9.8 so when you plug that in the time is going to be 11. 76 36 seconds so that's how long it's going to take the ball to hit the ground for this particular trajectory okay let's confirm that answer with another formula so you can also get the time using this it's V sin theta plus V initial SAR sin 2 Theta which is basically v y initial squar plus 2G y initial anytime you see V initial sin Theta that's v y initial this equation comes from the quadratic formula the quadratic formula gives you two answers but this simplifies it where it gives you this answer and not the answer that's over here so V initial is going to be 50 time sin 40 and then on the inside it's going to be 50 SAR sin 40 2ar + 2 9.8 and then the initial height y initial which is 300 and that's all within a square root all of this is divided by 9.8 go ahead and plug that into your calculator when you do you should get the same answer 11.76 36 so those are the two ways in which you can calculate the time it takes hit the ground as you could see you don't need the maximum height what you do need is the initial speed the launch angle and the height of the cliff with those three things you can find uh the time it takes to hit the ground the other formula you don't need the height of the cliff but you do need the maximum height attained from the ground so now that we have that let's move on to part C calculate the range of the ball so we need to calculate this distance here the horizontal distance traveled before it hits the ground so for this one we could just use range is equal to V initial cine Theta T we know V initial it's 50 m/s we got the launch angle it's 40° and we now have the time it takes to go from point A to point C that's 11.76 36 seconds let's go ahead and plug this in 50 cosine 40 that'll give us VX which is 38.3 11.76 36 so the range is 450. 57 M so that's how you could determine the range for this type of trajectory now Part D what is the final velocity of the ball just before it hits the ground so we're going to do this two ways if you have the formula sheet you can go straight to this formula which will basically be a straight shot to the answer the final velocity is V initial squ minus 2 GT V initial sin Theta or Theta is is the initial angle plus GT ^2 and we'll confirm this answer using some other formulas all right let's see if we can fit everything in V initial is 50 - 2 9.8 and the time is 11 7636 the initial is 50 s of the angle the angle is 4 40 and then plus GT ^2 I'm just going to put that here plus 9.8 11.76 36 now of course this is still under the square root so don't put this outside the square root this is still inside the square root I just ran out of space so if you plug this in < TK 50 2 - 2 9.8 11.76 36 50 sin 40 plus parentheses 9.8 11 76 36 close parentheses squar and that's all inside the square root so this should give you a final veloc well the final speed of 91.5 42 m/s the reason why I say this is the final speed is because we don't have the direction of this Vector we need to find it our goal is to find the velocity so we need the magnitude which we already have now and we need the direction the angle so there's multiple ways in which we can get the angle but right now instead of using AR tan our cosine is going to be easier because we don't know VY so for this this particular problem given what we already have another way to find the angle is to use R cosine of VX over VF adjacent over hypotenuse now VX is V initial cosine Theta where Theta is the initial angle and this is V final so it's easy to use this formula because we already have everything we need V initial is 50 and then times cosine 40 over V final which we now have 91. 542 so if you plug it in correctly you should get this angle this is the reference angle by the way the angle within a triangle it's 65. 265 I'm going to round that to three so 65.3 de now let's talk about what it means so here's the vector that's VF as you can see here it's going in uh the quadrant four Direction This is VX this is VY and here is the angle Theta so this is the 65.3 de it represents this angle here so it's 65.3 de below the xaxis if you need to find the angle uh measured counterclockwise from the positive x-axis instead of below the x-axis you could do 360 - 65.3 de so it's also 200 94.7 de measured counterclockwise from the positive x-axis so you have multiple ways in which to describe the angle but I'm just going to say 65.3 de below the xaxis so we have the magnitude and the direction below the x-axis now let's confirm our answer so VX we know it's V initial cosine Theta so that's 50 cosine 40 so VX is 38.3 m/s VY it's perhaps you've seen this equation V final is equal to V initial plus a t so you can say v y final is equal to VY initial now a is going to be 9.8 but if you put minus g g is postive 9.8 the effect is overall this should be negative because gravity brings things down now VY initial is basically V initial sin Theta and then we need to minus GT because that velocity that velocity component is under gravitational acceleration so every second the velocity in the y direction will decrease by 9.8 so that's why we have negative GT so this is going to be 50 sin 40° - 9.8 over the times the total time of 11. 7636 so 50 sin 40 That's 3213 - 9.8 11. 7636 we're going to get a negative answer and the reason for this is because it's it's going down so this is 8314 m/s so let's draw a picture if we focus on this Vector here it's still moving to the right that's VX so that's why VX is positive it's going down v y it's negative and V final is going in this general direction now notice that this forms a right triangle so to find V final we could use the Pythagorean theorem C is equal to the square < TK of A2 + b^2 so V final is equal to the square root of vx2 + V y^2 so vx2 that's 38.3 v y once we Square it will be positive so 83.1 4^ squared when you plug this in it will confirm this answer now I got 91.5 796 so that's going to round to 91.6 instead of 91.5 but keep in mind I'm using rounding answers here so that's why it's a little bit off from what we have here this is more accurate because we plugged everything in directly we didn't use any intermediate answers so this answer is more correct but nevertheless 91.6 or 91.5 is still close to 9154 so as you can see we still get about the same answer now let's get the angle Theta so if you know VX and VY you could use the arct tangent formula before we had VY we had VX and V final so we could use AR cosine instead of AR tangent but for this one it's going to be ar tangent VY over VX now for VY plug in positive 83 3.14 don't worry about the negative sign otherwise you'll get a different answer so AR tangent 83.1 14 / 38.3 will give us the same answer 65.3 de so all the formulas that we've used in number three you can see they have been confirmed and again if you want to find those formulas that will lead you to the answer directly uh feel free to check out the links in the description section below I kind of slurred my words a bit but you'll find that formula sheet in the description section below for those of you who want to print it out I think it's going to be very helpful when you're solving difficult project. motion problems it took a while to create that formula sheet but I know some of you have been asking for it so I decided to uh put it on my patreon page and it's free to download uh once you join that patreon account thanks again for watching |
6901 | https://byjus.com/chemistry/kohlrausch-law/ | Kohlrausch Law refers to an electrolyte’s limiting molar conductivity to its constituent ions. It indicates that an electrolyte’s limiting molar conductivity is equal to the sum of the individual limiting molar conductivities of the cations and anions that make up the electrolyte.
The Kohlrausch Law of Independent Migration is another name for this law. The Kohlrausch law and its applications are crucial in the study of dilute liquids as well as electrochemical cells. Among other essential uses, this law is utilised to determine the limiting conductivity of a weak electrolyte.
Table of Contents
What is Kohlrausch’s Law?
Uses of Kohlrausch’s Law
Frequently Asked Questions – FAQs
What is Kohlrausch’s Law?
Kohlrausch’s law states that the equivalent conductivity of an electrolyte at infinite dilution is equal to the sum of the conductances of the anions and cations.
The molar conductivity of a solution at a given concentration is the conductance of the volume of solution containing one mole of electrolyte kept between two electrodes with the unit area of cross-section and distance of unit length. The molar conductivity of a solution increases with the decrease in concentration. This increase in molar conductivity is because of the increase in the total volume containing one mole of the electrolyte. When the concentration of the electrolyte approaches zero, the molar conductivity is known as limiting molar conductivity, Ëm°.
Kohlrausch observed certain regularities while comparing the values of limiting molar conductivities of some strong electrolytes. On the basis of his observations, Kohlrausch proposed “limiting molar conductivity of an electrolyte can be represented as the sum of the individual contributions of the anions and cations of the electrolyte”. This law is popularly known as Kohlrausch law of independent migration of ions. For example, limiting molar conductivity, Λ of sodium chloride can be determined with the knowledge of limiting molar conductivities of sodium ion and chloride ion.
Some important applications of Kohlrausch law of independent migration of ions are:
Kohlrausch law helps us in the determination of limiting molar conductivities for any electrolyte. Weak electrolytes have lower molar conductivities and lower degree of dissociation at higher concentrations. The graph plotted between molar conductivity and c1/2 (where c is the concentration) is not a straight line for weak electrolytes. The molar conductivity of weak electrolyte increases steeply at lower concentrations. Therefore, limiting molar conductivity, Λ cannot be obtained by extrapolation of molar conductivity to zero concentration. Hence, we use the Kohlrausch law of independent migration of ions for the determination of limiting molar conductivity, Λ for weak electrolytes.
Kohlrausch law also helps us in determining the value of dissociation constant from the value of molar conductivity and limiting molar conductivity for a weak electrolyte at a given concentration.
Ë
α = dissociation constant
Λ = molar conductivity
Ëm° = limiting molar conductivity
Uses of Kohlrausch’s law
Calculation of Degree of dissociation
Calculation of solubility of sparingly soluble salt
Calculation of Dissociation Constant for weak electrolytes
Calculation of Molar Conductivity for weak electrolytes at infinite dilution
Recommended Videos
Kohlrausch’s Law and Conductometric Titrations
Electrolysis in Molten State
Frequently Asked Questions – FAQs
Q1
Who discovered the law of independent migration of ions?
This law was discovered by Friedrich Kohlrausch after observing experimental data on conductivities of various electrolytes.
Q2
What is kohlrausch law and its applications?
It’s used to determine an electrolyte’s dissociation constant. It’s used to find out what a weak electrolyte’s limiting molar conductivity is. This law can also be used to determine the degrees of dissociation of weak electrolytes.
Q3
What is Kohlrausch law of independent migration?
According to Kohlrausch’s law of independent ion movement, the limiting molar conductivity of an electrolyte can be described as the sum individual contributions of its cations and anions. Since fewer ions are present for conduction, a solution’s conductivity decreases with dilution.
Q4
Why do we need Kohlrausch law?
The Kohlrausch law can be used to calculate the limiting molar conductivities of any electrolyte. At larger concentrations, weak electrolytes have lower molar conductivities and a lower degree of dissociation.
Q5
What is infinite dilution in electrochemistry?
Infinite dilution is a state of dilution in which the concentration does not change when more solvent is added. In chemistry, the concept of infinite dilution is used to investigate how compounds dissolve in different solvents.
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6902 | https://sjtrem.biomedcentral.com/articles/10.1186/1757-7241-19-69 | Skip to main content
Scandinavian Journal of Trauma, Resuscitation and Emergency Medicine
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Published:
Burn Resuscitation
Frederick W Endorf1 &
David J Dries2
Scandinavian Journal of Trauma, Resuscitation and Emergency Medicine
volume 19, Article number: 69 (2011)
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Abstract
Fluid resuscitation following burn injury must support organ perfusion with the least amount of fluid necessary and the least physiological cost. Under resuscitation may lead to organ failure and death. With adoption of weight and injury size-based formulas for resuscitation, multiple organ dysfunction and inadequate resuscitation have become uncommon. Instead, administration of fluid volumes well in excess of historic guidelines has been reported. A number of strategies including greater use of colloids and vasoactive drugs are now under investigation to optimize preservation of end organ function while avoiding complications which can include respiratory failure and compartment syndromes. Adjuncts to resuscitation, such as antioxidants, are also being investigated along with parameters beyond urine output and vital signs to identify endpoints of therapy. Here we briefly review the state-of-the-art and provide a sample of protocols now under investigation in North American burn centers.
Introduction
One of the most challenging aspects of caring for burned patients is the acute resuscitation. The profound inflammatory response generated by a burn far surpasses that seen in trauma or sepsis, and the resultant fluid needs can be extreme. There is a large and ever-increasing body of research devoted to refining strategies for acute burn resuscitation, and this article attempts to summarize some the most important recent findings in the field.
After treating victims of the infamous Coconut Grove fire in 1942, Cope and Moore first postulated that burn resuscitation needs may have contributions from both the patient's body weight and the size of their burn . Baxter and Shires later built on this knowledge, using canine and human data, to specifically measure fluid requirements by weight and total body surface area (%TBSA). Their formula of 3.5 to 4.5 ml of lactated Ringers per %TBSA per kilogram became known as the Parkland formula after the Dallas medical complex in which their experiments took place . Although the Parkland formula is still the most commonly employed resuscitation formula worldwide, it is far from a perfect solution.
Ongoing research focuses on refining existing formulas to prevent complications of over-resuscitation. This includes devising novel means for titrating resuscitation, such as nurse-driven or computer-driven protocols. The composition of the fluids used in resuscitation has generated significant interest, with a particular focus on colloids and hypertonic saline. Pharmaceutical therapies that attempt to down regulate the inflammatory response such as vitamin C may have a role in acute resuscitation. Likewise, investigators are proposing the use of adjuncts such as plasmapheresis to remove inflammatory mediators from the bloodstream during resuscitation. The final topic inviting significant scrutiny is outcomes of resuscitation, both choosing the most appropriate outcomes to use and how to best measure these outcomes in clinical practice.
Resuscitation Volumes
The Parkland formula is simple to calculate and has a long record of widespread use in the burn community. This has not stopped researchers from frequently examining the efficacy of this formula, and also determining whether practitioners of this resuscitation formula are using it correctly.
The most consistent criticism of the Parkland formula is that patients tend to receive more fluid than the formula would have predicted based on the patient's weight and %TBSA. Whether this is an inherent flaw of the formula or user error by practitioners is debated, but it is clear that fluid volumes used in resuscitation are higher than historical controls. This phenomenon was described by Pruitt as "fluid creep ." Engrav et al. in a multi-center study in 2000 reviewed resuscitations in 50 patients and confirmed that 58% of patients required more fluid than predicted by the Parkland formula, compared to 12% exceeding predicted values in Baxter's original work . Friedrich et al. continued this investigation in 2004, comparing recent patients at their center to those from the 1970's, and found that fluid requirements during resuscitation had doubled during that period . In a related study, they correlated this change with a marked increase in narcotic pain medication usage during the resuscitation period. They theorized that the vasodilatory effects of opioid medications may cause relative hypotension, in turn necessitating increased fluid administration . This finding was later replicated by Wibbenmeyer et al. who showed a strong correlation between opioid equivalents received in the first 24 hours post-burn and the fluid volumes during that same period . Even at Parkland medical center, Blumetti et al. found that 48% of patients were receiving higher than predicted fluid volumes . Despite growing awareness of this "fluid creep," Cartotto et al. found that recent patients at their center were continuing to receive an average of 6.3 ml/kg/%TBSA during resuscitation, with 76% getting more than the 4.3 ml/kg/%TBSA in the original Baxter work .
Resuscitation Protocols
One factor playing into the trend of higher resuscitation volumes may be inadequate titration of fluids by physicians. Cancio et al. reviewed their experience with resuscitations using the modified Brooke formula, which predicts 2 ml/kg/%TBSA fluid volumes. An important finding was that clinicians directing the resuscitation were significantly less likely to reduce the rate of fluid infusion when urine output was high than they were to increase fluid rates when urine output was low . In an effort to reduce the dependence on clinical decision-making, several centers have experimented with standardized algorithms derived from hourly urine output. Nurse-driven resuscitation protocols may ameliorate differences in resuscitation due to practitioner experience. Established algorithms may have better reinforcement of downward titration of fluid volumes when urine output is high, and may even allow for reductions in infusion rates when urine output is adequate. Jenabzadeh et al. used a nurse-driven protocol and showed a significant decrease in fluid volumes during resuscitation and a dramatic decrease in the incidence of abdominal compartment syndrome .
To further remove the element of human error, some centers have been incorporating computer-based resuscitation algorithms. Salinas et al. described a computer model for burn resuscitation that was utilized in the resuscitation of 32 burn patients that were compared to historical controls. Their protocol resulted in less 24-hour and 48-hour crystalloid volumes, and less total crystalloid volumes while in the intensive care unit. Volumes by body weight and by burn area were also significantly less for the computer resuscitation group. The computer protocol also helped patients to more effectively meet hourly goals for urine output . In a follow-up study, Salinas et al. analyzed how practitioners used the computer program's recommendations and found that they followed the computer recommendations 83.2% of the time. Reasons given for refusing the computer recommendations were when clinicians felt the recommended fluid increase was either excessive or inadequate, or if a patient was hypotensive .
Colloid
Historically, prevailing opinion had been that using colloid in the first 24 hours of resuscitation was contraindicated. It was thought that colloid would pass through the "leaky" capillaries in burn shock and exert an osmotic pull, drawing even more fluid into the interstitial space and worsening burn edema. However, investigators more recently are advocating the use of colloid in burn resuscitation, even in the first 24 hours.
Lawrence et al. performed a retrospective review of 52 burn patients with greater than 20% TBSA burns. Twenty-six of these patients received albumin during their resuscitation, and 26 only crystalloid. As part of an institutional resuscitation algorithm, patients requiring more fluid volumes than predicted by the Parkland formula changed to an arm of the algorithm in which they got a third of their hourly fluid volumes as 5% albumin, with the other two-thirds given as lactated Ringer's (LR) solution. After colloid infusion began, patients quickly returned to predicted fluid rates and stayed at those lower volumes for the remainder of their resuscitation. Neither the colloid nor crystalloid group had any patients with abdominal compartment syndrome, though the colloid patients had more extremity escharotomies, likely related to their larger average burn size . In a previous study from the same center, Cochran et al. performed a case-control analysis of large burns (>20% TBSA) that either did or did not receive albumin during their resuscitation. Not only was albumin not harmful, it actually conferred a mortality benefit in their study that confirmed on multivariate analysis . This phenomenon seems to hold in pediatric patients as well. The same group examined 53 pediatric patients with greater than 15% TBSA burns and found that patients with higher than predicted fluid volumes "normalized" with albumin administration. Again, there were no cases of abdominal compartment syndrome, and in the pediatric population there was no difference in the incidence of extremity or torso escharotomy. The albumin group did have a longer length of stay, again likely related to larger burn size and a higher rate of inhalation injury . The use of artificial colloids in burn resuscitation has also generated significant interest. Vlachou et al. randomized 26 adult patients to either a purely crystalloid resuscitation or one substituting 6% hydroxyethylstarch (HES) for one third of the predicted crystalloid volume. They found that patients in the HES arm required less overall fluid volumes in the first 24 hours and subsequently had less increase in body weight. They also measured C-reactive protein as a putative marker for inflammation and found lower values in the HES group . However, caution should be used with higher concentrations of HES. Bechir et al. studied a 10% HES solution versus crystalloid in 30 burned patients and found a trend toward higher rates of renal failure and higher mortality, though neither reached statistical significance .
Hypertonic Fluids
In an attempt to prevent overresuscitation, some investigators have also started using hypertonic saline, alone or in combination with colloid. Belba et al. performed a prospective, randomized study of 110 burned patients, 55 of whom were resuscitated with LR according to the Parkland formula for adults and the Shriner's formula for children. The other 55 patients received a hypertonic saline lactate solution containing sodium (250 mEq/L) and lactate (120 mEq/L). The hypertonic group needed higher fluid rates initially, but both groups decreased to less than predicted by the Parkland formula over the first 24 hours. The hypertonic group used less fluid overall than the isotonic fluid group, but this difference was not statistically significant . Using a hypertonic solution during resuscitation may also lower the risk of abdominal compartment syndrome. Oda et al. reviewed 36 patients with greater than 40% TBSA burns, 14 of whom were resuscitated using a hypertonic lactated saline solution and 22 with LR solution. The hypertonic solution was given in a staggered fashion, starting with a solution of sodium (300 mEq/L), chloride (88 mEq/L), and lactate (212 mEq/L). This was tapered down gradually, ending 48 hours after the burn with a solution of sodium (150 mEq/L), chloride (102 mEq/L), and lactate (48 mEq/L). They found that two of 14 patients in the hypertonic group developed abdominal compartment syndrome versus 11 of 22 in the LR group .
Antioxidants
The extensive inflammation seen in burn injury causes free oxygen radical release, which worsens vascular permeability and subsequently causes significant peripheral edema. The loss of fluid into the interstitium results in higher fluid needs during resuscitation. It is though that the use of antioxidants during resuscitation may help scavenge these free radicals and attenuate vascular permeability.
Tanaka et al. compared two groups of patients, 18 that were resuscitated with LR alone, and the other 19 patients receiving LR plus high dose ascorbic acid (vitamin C, 66 mg/kg/hr). They found average fluid needs of 3 ml/kg/%TBSA in the vitamin C group versus 5.5 ml/kg/%TBSA in the group resuscitated with LR alone. In addition, the vitamin C group had fewer ventilator days . Kahn et al. performed a retrospective review of 33 patients, 17 of whom had high-dose (66 mg/kg/hr) vitamin C plus LR, and 16 who had LR alone. They also found lower average fluid volumes in the vitamin C plus LR group (5.3 ml/kg/%TBSA) compared to the LR group (7.1 ml/kg/%TBSA). There was no difference in outcomes in their study, but also no difference in complications. They conclude that vitamin C is a safe adjunct for decreasing fluid volumes in the first 24 hours of burn resuscitation .
In a murine model, Constantini et al. experimented with the use of pentoxifylline (PTX) after burns. After a 30% steam burn, PTX in saline alone was injected intraperitoneally into one group of mice and saline alone into another group. The group with PTX had decreased intestinal permeability and inflammation. In a secondary finding, they also noted a decreased incidence of acute lung injury in the PTX group. Although there are no existing human trials with PTX in burn resuscitation, it may have promise as an antioxidant immune modulator during acute resuscitation .
Plasmapheresis
In addition to using antioxidants, some centers have looked at using mechanical removal of inflammatory mediators from the bloodstream. Klein et al. reviewed the use of plasma exchange at their institution over a 5-year period, in which 37 burn patients underwent plasma exchange during their acute resuscitation, seven of whom received two treatments for a total of 44 plasma exchanges. These were severe burns with a mean %TBSA of 48.6%, and 73% of the patients had associated inhalation injury. There was no protocol for initiation of plasma exchange, but it was often prompted by reaching twice the predicted Parkland formula resuscitation volume. Average time to initiation of plasma exchange was 17 hours, and the average duration of therapy was 2.4 hours. Albumin (5%) was primarily used as replacement fluid, unless the patient had low fibrinogen or abnormal clotting factors, in which case fresh frozen plasma (FFP) was used. They found that plasma exchange decreased crystalloid administration by 28.3%. When adjusted for patient weight and %TBSA, the average post-exchange fluid resuscitation volumes dropped by 40%. After plasma exchange, the hourly fluid administration rates never returned to preexchange levels in any patient .
Neff et al. performed a retrospective case-control study of 40 patients over a two-year period, all of whom had greater than 20% TBSA burns. Twenty-one of these patients underwent plasma exchange as part of their resuscitation, and they were matched with 19 contemporaneous controls. Plasma exchange was triggered by fluid volumes of 1.2 times that predicted by the Parkland formula, or by continued low urine output or hypotension in the face of escalating fluid rates. They found several physiologic benefits with plasma exchange, including a 24% increase in mean arterial pressure (MAP), a 400% increase in urine output, and a 25% reduction in intravenous fluid resuscitation rates need to maintain vital signs and urine output goals. Lactate levels also decreased, and they noted that an elevated admission lactate independently predicted the need for eventual plasma exchange .
Outcomes of resuscitation and monitoring
The second challenge in management of burn resuscitation is determination of optimal clinical endpoints. Traditionally, urine output has been used as the primary gauge of tissue perfusion during acute resuscitation. Greenhalgh recently published findings from a survey of American Burn Association (ABA) and International Society for Burn Injuries (ISBI) members regarding various topics in resuscitation. Respondents (94.9%) used urine output as a major index of successful resuscitation, with 22.7% using other monitors . Despite its widespread use, urine output is not generally viewed as a perfect measure of overall tissue perfusion. Concern has also been raised about effectiveness of other conventional parameters including blood pressure, heart rate, and central venous pressure. There are several techniques being examined that aim to more accurately measure peripheral perfusion, and allow better titration of fluid volumes on a real-time basis.
Parameters derived from transcardiopulmonary thermodilution using the PiCCO system have shown good correlation with values from a conventional pulmonary artery catheter in burned patients . The use of this system has also confirmed the hyperdynamic physiologic response in pediatric patients with large burns . However, there exist no studies showing any influence of this system on outcomes during acute burn resuscitation.
Jeng et al. studied four patients with severe (average 58% TBSA) burns and shock, using a multisensory probe with three transducers. One transducer was placed in the subcutaneous tissue in a representative second-degree burn, another transducer in the stomach via a tonometric gastric tube, and a third into the bloodstream through a single-lumen femoral line. They then used these transducers to measure tissue pH, CO2, and PaO2. They also simultaneously charted urine output, mean arterial pressure (MAP), and serum lactate. Third, these investigators measured burn wound perfusion with laser Doppler imaging and correlated the previous variables with observed changes in burn wound perfusion. Although changes in all variables were associated with changes in Doppler perfusion, they found that tissue pH and CO2, as well as gastric CO2, had the closest temporal relationship to changes in peripheral perfusion. Urine output, MAP, and lactate did change over time but tended to lag behind changes in peripheral tissue perfusion measured with laser Doppler imaging .
Though lactate may not give "real-time" information about resuscitation success, it does predict morbidity and mortality in burned patients. Cochran et al. reviewed 128 patients with an average of 41.7% TBSA and measured base deficit and lactate levels at 6-hour intervals. They found that the non-survivors had higher lactates at admission, 12, 18, and 24 hours than the group of survivors. Elevation of lactate in the first 48 hours was an independent predictor of mortality, but they were unable to demonstrate a specific threshold for clinical use. The authors caution that treatment should not be withheld based on any individual laboratory value .
Conclusion
Burn resuscitation continues to be a complex and challenging phase of care for burn patients. The long-running trend of increases in crystalloid fluid volumes is now recognized by practitioners, and efforts are being made to reduce excess fluid administration when possible. Refinement in resuscitation protocols, as well as multiple adjunctive therapies, may help reduce excess crystalloid administration. Finding a more accurate measure of resuscitation success could allow better and faster responses to physiologic changes.
The best consensus statement available comes from the American Burn Association in 2008. While there has been significant work examining alternatives to standard resuscitation practices, the ultimate consensus report retains emphasis on a crystalloid-based resuscitation utilizing 2-4 mL/kg/body weight/%TBSA during the first 24 hours. Fluid should be isotonic and titrated to maintain urine output of 0.5 to 1.0 mL/kg/hr in adults and 1.0-1.5 mL/kg/hr in children. Children may require additional fluid due to maintenance requirements. Incremental volume administration may be necessary in patients with significant full-thickness injury, delayed resuscitation or smoke inhalation (Additional File 1, Box 1) .
While these basic consensus guidelines may not reflect the variety of research carried on in burn resuscitation in recent years, this does reflect a safe starting point, particularly for practitioners in centers without extensive experience in burn resuscitation attempting to stabilize a thermal injury victim prior to transfer to a burn center.
With other burn centers in the United States, we are studying approaches to limit crystalloid administration with protocol-based use of vasoactive drugs and colloids in patients failing to respond to the initial resuscitation prescription. Crystalloid administration is capped at 100 mL of fluid/kg with second and third degree injury in our practice. When this physical limit is reached, a transition is made to colloids regardless of the time since injury. While this approach is designed and originally intended for adult patients, we now use it in all age groups (Additional File 2, Figure S1 and Additional File 3, Figure S2).
Traditional practice in the resuscitation of surgical patients included an automatic increase in fluid administration for hypotension with strict avoidance of vasoactive drugs, particularly norepinephrine, and diuretic administration. While we continue to emphasize careful evaluation of urine output and vital signs, vasopressin, norepinephrine or dobutamine may be employed after initial response to fluid administration is evaluated. The hypotensive patient with acceptable central venous pressure may receive vasopressin or norepinephrine. The patient with elevated blood pressure and central venous pressure may receive furosemide and dobutamine in addition to decreased resuscitation fluids. This approach incorporating more resuscitation options should be done in consultation with a center having expertise in burn management.
Please refer to Additional File 4, Box 2 for a summary of key points.
Author Information
Frederick W. Endorf, MD is Staff Surgeon at Regions Hospital, the Level I Trauma and Burn Center, in St. Paul, Minnesota, USA. He is also Clinical Assistant Professor of Surgery at the University of Minnesota.
David J. Dries, MSE, MD, FACS, FCCM, FCCP is the Assistant Medical Director of Surgical Care for HealthPartners Medical Group and Division Head for Surgery at Regions Hospital, the Level I Trauma and Burn Center, in St. Paul, Minnesota, USA. He is also Professor of Surgery, Professor of Anesthesiology and Clinical Adjunct Professor of Emergency Medicine at the University of Minnesota. Dr. Dries also holds the John F. Perry, Jr. Chair of Trauma Surgery at the University of Minnesota.
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Author information
Authors and Affiliations
The Burn Center, Regions Hospital, 640 Jackson Street, St. Paul, MN, 55101, USA
Frederick W Endorf
2. Department of Surgery, Regions Hospital, 640 Jackson Street, St. Paul, MN, 55101, USA
David J Dries
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Frederick W Endorf
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The authors declare that they have no competing interests.
Authors' contributions
FE performed the literature review and wrote the initial draft of the manuscript. DD conceived of the study and edited and rewrote portions of the manuscript. All authors read and approved the final manuscript.
Electronic supplementary material
13049_2011_319_MOESM1_ESM.DOC
Additional file 1: Box 1. American Burn Association Consensus Guidelines- 2008. Basic consensus guidelines in burn resuscitation. (DOC 29 KB)
13049_2011_319_MOESM2_ESM.DOC
Additional file 2: Figure S1. Nurse Driven Resuscitation Protocol. This detailed resuscitation approach incorporates vasoactive drugs, fresh frozen plasma, and albumin, and is designed for adults with greater than 20% Body Surface Area Burn (BASB) or young adults as defined. The starting point for fluid resuscitation is the standard Parkland formula utilizing lactated Ringer's at 2 to 4 mL/kg/% burn. Subsequent therapy is titrated based on urine output and vital signs. In large burns, our crystalloid limit of 100 mL/kg means that transition to colloids or vasoactive drugs occurs well before the conclusion of 24 hours of resuscitation. Note that this protocol is designed for utilization by the bedside nurse. However, if vasoactive drugs or colloids are considered, burn unit faculty are immediately engaged. (DOC 48 KB)
13049_2011_319_MOESM3_ESM.DOC
Additional file 3: Figure S2. Colloid Protocol and Pressor Protocol. Where patients are not successfully resuscitated using a simple crystalloid protocol, options include administration of fresh frozen plasma or albumin, which can be very valuable in children as dilution of serum albumin with crystalloids can rapidly occur. Vasoactive drugs are utilized in conjunction with a central venous catheter and measurement of central venous pressure. Bladder pressures are also monitored via the urinary catheter to identify intraabdominal hypertension and minimize the risk of intraabdominal compartment syndrome in patients receiving large resuscitation volume [32–34]. Finally, our unit continues to use fresh frozen plasma (FFP) as a part of our resuscitation strategy. This is given when crystalloid volumes exceed 100 mL/kg (see section of Additional File 3, Figure S2 marked "Colloid Protocol"). FFP is administered at 0.5 mL/kg/%TBSA, transfused over 8 hours. We recognize that this is controversial and also use albumin in selected patients. (DOC 54 KB)
Additional file 4: Box 2. Key Points. Resuscitation options in the burned patient. (DOC 28 KB)
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Endorf, F.W., Dries, D.J. Burn Resuscitation.
Scand J Trauma Resusc Emerg Med 19, 69 (2011).
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Keywords
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6903 | http://www.nhc.ed.ac.uk/index.php?page=493.470.483.482 | Natural History Collections: SUBCLASS ACTINOPTERYGII: MODERN TELEOSTS
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VERTEBRATE ANIMALS
THE FISH
CLASS OSTEICHTHYES
SUBCLASS ACTINOPTERYGII: MODERN TELEOSTS
SUBCLASS ACTINOPTERYGII: MODERN TELEOSTS
By far the vast majority of species of modern fish are teleosts classified together in the division Teleostei within the infraclass Neopterygii. Teleost fish first appeared in the fossil record in the Jurassic. It has been suggested that the teleosts derived from a chondrostean lineage of fish such as Elonichthys. From a fairly simple structural plan, the main line of fish evolution developed into an immense diversity of forms, resulting in fish as different as eels and flying fish, sea horses and flatfish. The key advance was the swim bladder which allows teleosts to achieve weightlessness - neutral buoyancy - and so frees the fish from having to generate lift.
TELEOST SKELETON
The bony skeleton of a carp (above center) represents the teleost skeleton. The name tele-ostei (end-bone) refers to the specialised bones at the end of the vertebral column that support the symmetrical, homocercal caudal fin. Teleost skeletons typically comprise bony vertebrae and two pairs of ribs enclosing the viscera. There are pectoral, pelvic and anal fins. The fin rays are composed of bony segmented rays (lepidotrichia). The upper part of the cranium protects the brain. The orbit for the eyes is large and easily visible. The gills are covered by an operculum. A series of spines protrude above and below the vertebrae behind the ribs. The ctenoid bony scales are usually light and thin or even absent as in catfish.
EVOLUTION OF MODERN TELEOSTS AND THE DISPLAY
The modern view is that the species belonging to thirty six of the forty two teleost orders represent one main line of bony fish evolution and they are therefore classified together within the infradivision Euteleostei. There appear to be at least six major groups, classified as superorders, within this main line. The other six orders represent three specialised or relict offshoots of the main line, each of which is classified as an infradivision within the division Teleostei.
Our display contains a few species from each of the three specialised or relict offshoots of the main line and from each of the six superorders within the infradivision Euteleostei. The fish displayed here show how flexible the basic teleost plan has been.
SPECIALISED OR RELICT OFFSHOOTS OF MAIN TELEOST LINE
Infradivision Osteoglossomorpha � elephant fish.
Infradivision Elopomorpha � eels.
Infradivision Clupeomorpha - herring.
MAIN LINE OF EVOLUTION: EUTELEOSTS
Superorder Ostariophysi � carp, loaches, roach, catfish.
Superorder Protacanthopterygii � trout, powan.
Superorder Stenopterygii � bristlemouth, anglemouth, dragonfish, hatchetfish.
Superorder Scopelomorpha � spiderfish, lanternfish.
Superorder Paracanthopterygii � Atlantic cod, grenadier, anglerfish.
Superorder Acanthopterygii � flying fish, rock bass, angelfish, mudskipper, eelpout, seahorses, remora, weever fish, porcupine fish and cowfish, sunfish, flatfish.
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6904 | https://math.stackexchange.com/questions/204001/find-an-equation-of-the-tangent-line-to-the-parabola-passing-through-x-0-y-0 | calculus - Find an equation of the tangent line to the parabola passing through $(x_0,y_0)$. - Mathematics Stack Exchange
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Find an equation of the tangent line to the parabola passing through $(x_0,y_0)$.
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Let $(x_0,y_0)$, where $y_o \neq 0$, be a point on the parabola $y^2=2px$. Find an equation of the tangent line to the parabola passing through $(x_0,y_0)$.
I did my work for this, but I cannot get the answer, which my tutor swears on his job (Joking, but I'm pretty sure its correct), that is correct.
Given Answer: $y_0y=p(x+x_0)$
Here's my workings.
Differentiate with respect to $x$, we have $2y\frac{\mathrm{dy} }{\mathrm{d} x}=2p$ Therefore, $\frac{\mathrm{dy} }{\mathrm{d} x}=\frac {p} {y}$
At Point $(x_0,y_0), \frac{\mathrm{dy} }{\mathrm{d} x}=\frac {p} {y_0}$
Using the equation for straight lines, we have $(y-y_0)=\frac{p}{y_0}(x-x_0)$ $$yy_0-y_0^2=px-px_0$$
By now, it should be clear I will not get the answer. Did I make any mistake?
I reverse-engineered my tutor's answer and I realised he used $(y-y_0)=\frac{p}{y}(x-x_0)$ instead. $y^2-yy_0=px-px_0$
$yy_0=px_0-px+y^2$
$yy_0=px_0-px+2px$
$yy_0=px_0+px=p(x+x_0)$
Why is it that when I use $\frac {p}{y}$ versus $\frac {p}{y_0}$, both gives me the same answer? Which should I be using?
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edited Aug 8, 2014 at 17:46
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asked Sep 28, 2012 at 14:08
Yellow SkiesYellow Skies
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1 $\begingroup$The answer given is quadratic in $x$. This cannot be a line and seems to be incorrect.$\endgroup$ivan –ivan 2012-09-28 14:11:54 +00:00 Commented Sep 28, 2012 at 14:11
$\begingroup$Whoops. Edited. there shouldn't be an x$\endgroup$Yellow Skies –Yellow Skies 2012-09-28 14:16:03 +00:00 Commented Sep 28, 2012 at 14:16
$\begingroup$Just use the fact that $y^2=2px$ for $y_0$ and plug back in. You are already there.$\endgroup$ivan –ivan 2012-09-28 14:18:48 +00:00 Commented Sep 28, 2012 at 14:18
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Since $y_0^2=2px_0$, your solution is equivalent to $yy_0=p(x+x_0)$.
Your "reverse-engineered" solution works only by accident; it has two errors that happen to cancel each other out exactly:
Using $p/y$ as the slope is wrong; that amounts to saying: "to find out whether some point is on the tangent line, find out what the slope of the parabola at that point would have been (if a woodchuck could chuck wood), and check whether the connecting line between the test point and $(x_0,y_0)$ is right". That has no right to work -- the slope you want to compare to is the slope at $(x_0,y_0)$, not the slope at $(x,y)$, no matter whether $(x,y)$ is even on the parabola.
You're rewriting $y^2$ to $2px$, but that is only true when $(x,y)$ is actually on the parabola. This is true by assumption about $(x_0,y_0)$, but the rewriting is supposed to preserve the truth or falsity of the original equation for all $(x,y)$, not just those on the parabola.
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edited Sep 28, 2012 at 14:38
answered Sep 28, 2012 at 14:17
hmakholm left over Monicahmakholm left over Monica
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$\begingroup$Ok thanks. Didn't see that lol.$\endgroup$Yellow Skies –Yellow Skies 2012-09-28 14:23:14 +00:00 Commented Sep 28, 2012 at 14:23
$\begingroup$Oh yeah just wondering, I added a couple of steps to my answer. Why is it that when I use $\frac {p}{y}$ versus $\frac {p}{y_0}$, both gives me the same answer? Which should I be using?$\endgroup$Yellow Skies –Yellow Skies 2012-09-28 14:24:31 +00:00 Commented Sep 28, 2012 at 14:24
$\begingroup$See updated answer.$\endgroup$hmakholm left over Monica –hmakholm left over Monica 2012-09-28 14:38:33 +00:00 Commented Sep 28, 2012 at 14:38
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6905 | https://www.ti.com/lit/sboa174 | Application Brief
Current Sensing in an H-Bridge
Arjun Prakash and Mubina Toa Current Sensing Products
The semiconductor industry has always looked for improvements in creating technology that has the ability to enable higher power density systems. One such circuit is an H-bridge. As shown in Figure 1
an H-bridge is a simple circuit consisting of 4 FET transistors connected between the load. An H-bridge is often used when the direction of the current is required to be controlled and managed from the supply to the load. If the load is highly inductive, the energy stored in the load can also be discharged safely to ground by controlling the H-bridge. H-bridge circuits are commonly used in motor control, DC-DC converters, audio sub systems and LED lighting control making systems more safe and reliable. H-bridge consisting of silicon FET transistors often achieve > 95% efficiency, while GaN FET transistors can allow for efficiencies beyond 99%. A higher efficiency H-bridge combined with current sense amplifiers to monitor, manage, and control the load currents leading to improvement in safety, reliable, and overall improvements in the power efficiency of an end equipment. Load
VDD
Q1 Q2 Q3 Q4
High Side
Low Side
Figure 1. H-Bridge Circuit
Full H-Bridge Circuit Configuration and Control
The H-bridge can be controlled by turning ON and OFF the FETs. A pulse-width modulation (PWM) scheme is an effective method used in creating different waveforms to control the flow of current. By controlling the duty cycle of the PWM waveform
the current flowing to the load can be effectively controlled. Figure 2 depicts the PWM waveform with different duty-cycles. By modulating the duty cycle of the PWM generator, the output current to the load can be precisely controlled. Duty Cycle = 10% Duty Cycle = 50% Duty Cycle = 90%
Figure 2. Pulse Width Modulation Scheme for H-
Bridge
While controlling the H-bridge using PWM waveforms careful consideration must be taken into account to make sure there is no direct short from battery to ground. For example, in Figure 1 , do not turn Q1 and Q2 on simultaneously. Such a scenario creates a high current shoot through which can damage the corresponding electronic drive circuitry. Table 1
describes the possible states of a full H-bridge control.
Table 1. Operating States of an H-Bridge
Q1 Q2 Q3 Q4 State of Load
ON OFF OFF ON Current flows from H-bridge to the load
OFF ON ON OFF Direction of the current to the load is reversed
OFF ON OFF ON Provide safe path for the load to discharge to ground
ON OFF ON OFF Recirculation current stored in the load
OFF ON OFF ON Recirculation current stored in the load
ON ON OFF OFF Short circuit from battery to ground
OFF OFF ON ON Short circuit from battery to ground
ON ON ON ON Short circuit from battery to ground
Current Measurement in an H-Bridge for Motor
Control
Bidirectional current sensing in a full H-bridge motor control is critical for safety and reliability to monitor and control a system. An accurate current measurement in an H-bridge can control the torque of the motor precisely or precisely set the position in a stepper motor.
www.ti.com TI Tech Note
SBOA174D – DECEMBER 2016 – REVISED APRIL 2023
Submit Document Feedback Current Sensing in an H-Bridge 1Copyright © 2023 Texas Instruments Incorporated
Figure 3 describes common locations to measure current in an H-bridge: high-side, in-line, and low-side. As motors are highly inductive, the PWM output tends to overshoot during low to high transitions and undershoot during high to low transition. The characteristics of overshoot and undershoot of an amplifier are important in selecting a correct component. A current sense amplifier that can sustain overshoot and undershoot conditions with a fast response time, and is able survive harsh requirements of an inductive system is critical. By providing valuable current sensing data to the system, this helps to detect anomalies in the motor or other inductive system features, which can lead to premature failures.
Table 2 describes the advantages and disadvantages of measuring currents in an H-bridge at multiple locations. VDD
Q1 Q2 Q3 Q4
ON ON
OFF OFF
High-Side
In-Line
Low-Side
VDD
Q1 Q2 Q3 Q4
ON ON
OFF OFF
High-Side
In-Line
Low-Side
Figure 3. Current Sensing Locations in an H-
Bridge Control
Table 2. Current Sensing in an H-Bridge
Current Measurement Pros Cons
High-Side Detect shorted load from battery for diagnostics
High voltage common-mode amplifier
In-Line
Direct motor current measurement, low-bandwidth amplifier
High dv/dt signals. PWM settling time
Low-Side Low-cost, low common-mode voltage
Unable to detect shorted load
The INA240 current sense amplifier can operate from a common-mode voltage ranging from –4 V to 80 V. In an H-bridge application, the INA240 can be used regardless of whether the measurement location is high-side, in-line, or low-side. A low offset of (25 μV) and low voltage offset drift (0.25 μV/°C) combined with a low gain error (0.2%) and gain drift (2.5 ppm/°C) makes it applicable for precise measurements regardless of system temperature. In addition to high performance DC specifications, the INA240 is also designed to operate and reject dv/dt transients enabling real time load current measurements at the in-line measurement location. The system level benefits of in-line sensing enables
higher power density by lowering the processing power requirements for closed loop control system.
Alternate Device Recommendations
The INA241 is an ultra-precise analog current sense amplifier. The INA241 can be used in high-voltage bidirectional applications paired with 1-MHz bandwidth to offer fast response time with precise operation for in-line control within H-bridge applications. The INA241 can measure currents at common-mode voltages of –5 V to 110 V and survive voltages between –20 V to 120 V.
The INA253 or INA254 devices are ultra-precise current sense amplifiers with integrated low-inductive, precision 2-mΩ or 400-μΩ shunts with an accuracy of 0.1% or 0.5%, respectively, with a temperature drift of < 15 ppm/°C. The INA253 is limited to applications that need < ±15 A of continuous current at T A = 85°C, and the INA254 is limited to applications < ±50 A of continuous current at T A = 85°C. The INA253 and
INA254 integrated shunt is internally Kelvin-connected to the INA240 amplifier. The INA253 and INA254
devices provide the performance benefits of the
INA240 amplifier with the inclusion of a precision shunt providing a total uncalibrated system gain accuracy of < 0.2%.
The INA281 can be used in high-voltage applications such as high-side current sensing in a motor. The INA281 can measure currents at common-mode voltages of –4 V to 110 V and survive voltages between –20 V to 120 V, making this device versatile for a variety of applications where voltage can swing negative.
An option for low-side sensing is the INA381 which is a cost-optimized current sense amplifier with an integrated comparator which serves to reduce PCB footprint size and simplifies design.
Table 3. Alternate Device Recommendations
Device Optimized Parameter Performance Trade-Off
INA241 Vcm range: –5- to 110-V bidirectional IQis slightly greater
INA281 Vcm range: –4 V to 110 V Unidirectional
INA381 Integrated comparator Vcm limited to 26 V
INA253 Integrated shunt 2 mΩ, V CM
range: –4 V to 80 V
±15-A maximum continuous current
INA254 Integrated shunt 400 μΩ, V CM
range: –4 V to 80 V
±50-A maximum continuous current
Table 4. Related TI Application Briefs
Document Title
SBOA160 Low-Drift, Precision, In-Line Motor Current Measurements With PWM Rejection
SBOA176 Switching Power Supply Current Measurements
SBOA163 High-Side Current Overcurrent Protection Monitoring
SBOA187 Current Mode Control in Switching Power Supplies
TI Tech Note www.ti.com
2Current Sensing in an H-Bridge SBOA174D – DECEMBER 2016 – REVISED APRIL 2023
Submit Document Feedback
Copyright © 2023 Texas Instruments Incorporated
IMPORTANT NOTICE AND DISCLAIMER
TI PROVIDES TECHNICAL AND RELIABILITY DATA (INCLUDING DATA SHEETS), DESIGN RESOURCES (INCLUDING REFERENCE DESIGNS), APPLICATION OR OTHER DESIGN ADVICE, WEB TOOLS, SAFETY INFORMATION, AND OTHER RESOURCES “AS IS” AND WITH ALL FAULTS, AND DISCLAIMS ALL WARRANTIES, EXPRESS AND IMPLIED, INCLUDING WITHOUT LIMITATION ANY IMPLIED WARRANTIES OF MERCHANTABILITY, FITNESS FOR A PARTICULAR PURPOSE OR NON-INFRINGEMENT OF THIRD PARTY INTELLECTUAL PROPERTY RIGHTS. These resources are intended for skilled developers designing with TI products. You are solely responsible for (1) selecting the appropriate TI products for your application, (2) designing, validating and testing your application, and (3) ensuring your application meets applicable standards, and any other safety, security, regulatory or other requirements. These resources are subject to change without notice. TI grants you permission to use these resources only for development of an application that uses the TI products described in the resource. Other reproduction and display of these resources is prohibited. No license is granted to any other TI intellectual property right or to any third party intellectual property right. TI disclaims responsibility for, and you will fully indemnify TI and its representatives against, any claims, damages, costs, losses, and liabilities arising out of your use of these resources. TI’s products are provided subject to TI’s Terms of Sale or other applicable terms available either on ti.com or provided in conjunction with such TI products. TI’s provision of these resources does not expand or otherwise alter TI’s applicable warranties or warranty disclaimers for TI products. TI objects to and rejects any additional or different terms you may have proposed. IMPORTANT NOTICE Mailing Address: Texas Instruments, Post Office Box 655303, Dallas, Texas 75265 Copyright © 202 3, Texas Instruments Incorporated |
6906 | https://www.khanacademy.org/math/6th-grade-illustrative-math/unit-8-data-sets-and-distribution/lesson-14-comparing-mean-and-median/a/comparing-mean-and-median | Comparing mean and median (article) | Khan Academy
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6th grade math (Illustrative Math-aligned)
Course: 6th grade math (Illustrative Math-aligned)>Unit 8
Lesson 14: Comparing mean and median
Comparing mean and median
Math>
6th grade math (Illustrative Math-aligned)>
Data sets and distribution>
Comparing mean and median
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Comparing mean and median
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New content on this topic will come during the summer of 2026.
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In this lesson, you will compare the mean and median of data sets. Check back in the summer of 2026 to see what new content we have to offer!
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6907 | https://math.libretexts.org/Bookshelves/Precalculus/Precalculus_1e_(OpenStax)/08%3A_Further_Applications_of_Trigonometry/8.02%3A_Non-right_Triangles_-_Law_of_Cosines | 8.2.1
8.2.1
8.2.2
8.2.2
8.2.3A
8.2.3B
8.2.4
8.2.3
8.2.5
8.2.4
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8.2: Non-right Triangles - Law of Cosines
Last updated
: Jan 2, 2021
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8.1: Non-right Triangles - Law of Sines
8.3: Polar Coordinates
Page ID
: 1376
OpenStax
OpenStax
( \newcommand{\kernel}{\mathrm{null}\,})
Learning Objectives
In this section, you will:
Use the Law of Cosines to solve oblique triangles.
Solve applied problems using the Law of Cosines.
Use Heron’s formula to find the area of a triangle.
Suppose a boat leaves port, travels 1010 miles, turns 2020 degrees, and travels another 8 miles as shown in Figure 8.2.18.2.1 How far from port is the boat?
Unfortunately, while the Law of Sines enables us to address many non-right triangle cases, it does not help us with triangles where the known angle is between two known sides, a SAS (side-angle-side) triangle, or when all three sides are known, but no angles are known, a SSS (side-side-side) triangle. In this section, we will investigate another tool for solving oblique triangles described by these last two cases.
Using the Law of Cosines to Solve Oblique Triangles
The tool we need to solve the problem of the boat’s distance from the port is the Law of Cosines, which defines the relationship among angle measurements and side lengths in oblique triangles. Three formulas make up the Law of Cosines. At first glance, the formulas may appear complicated because they include many variables. However, once the pattern is understood, the Law of Cosines is easier to work with than most formulas at this mathematical level.
Understanding how the Law of Cosines is derived will be helpful in using the formulas. The derivation begins with the Generalized Pythagorean Theorem, which is an extension of the Pythagorean Theorem to non-right triangles. Here is how it works: An arbitrary non-right triangle ABCABC is placed in the coordinate plane with
vertex
AA at the origin, side cc drawn along the x-axis, and
vertex
CC located at some point (x,y)(x,y) in the plane, as illustrated in Figure 8.2.28.2.2. Generally, triangles exist anywhere in the plane, but for this explanation we will place the triangle as noted.
We can drop a perpendicular from CC to the x-axis (this is the altitude or height). Recalling the basic trigonometric identities, we know that
cosθ=x(adjacent)b(hypotenuse)cosθ=x(adjacent)b(hypotenuse) and sinθ=y(opposite)b(hypotenuse)sinθ=y(opposite)b(hypotenuse)
In terms of θθ, x=bcosθx=bcosθ and y=bsinθy=bsinθ. The (x,y)(x,y) point located at CC has coordinates (bcosθ,bsinθ)(bcosθ,bsinθ). Using the side (x−c)(x−c) as one leg of a right triangle and yy as the second leg, we can find the length of hypotenuse aa using the Pythagorean Theorem. Thus,
a2=(x−c)2+y2=(bcosθ−c)2+(bsinθ)2Substitute (bcosθ) for x and (bsinθ) for y=(b2cos2θ−2bccosθ+c2)+b2sin2θExpand the perfect square.=b2cos2θ+b2sin2θ+c2−2bccosθGroup terms noting that cos2θ+sin2θ=1=b2(cos2θ+sin2θ)+c2−2bccosθFactor out b2a2=b2+c2−2bccosθa2=(x−c)2+y2=(bcosθ−c)2+(bsinθ)2=(b2cos2θ−2bccosθ+c2)+b2sin2θ=b2cos2θ+b2sin2θ+c2−2bccosθ=b2(cos2θ+sin2θ)+c2−2bccosθa2=b2+c2−2bccosθSubstitute (bcosθ) for x and (bsinθ) for yExpand the perfect square.Group terms noting that cos2θ+sin2θ=1Factor out b2
The formula derived is one of the three equations of the Law of Cosines. The other equations are found in a similar fashion.
Keep in mind that it is always helpful to sketch the triangle when solving for angles or sides. In a real-world scenario, try to draw a diagram of the situation. As more information emerges, the diagram may have to be altered. Make those alterations to the diagram and, in the end, the problem will be easier to solve.
The LAW OF COSINES
The Law of Cosines states that the square of any side of a triangle is equal to the sum of the squares of the other two sides minus twice the product of the other two sides and the cosine of the included angle.
For triangles labeled as in Figure 8.2.38.2.3, with angles αα, ββ and γγ, and opposite corresponding sides aa, bb, and cc, respectively, the Law of Cosines is given as three equations.
a2=b2+c2−2bccosα
a2=b2+c2−2bccosα(8.2.1)
b2=a2+c2−2accosβ
b2=a2+c2−2accosβ(8.2.2)
c2=a2+b2−2abcosγ
c2=a2+b2−2abcosγ(8.2.3)
To solve for a missing side measurement, the corresponding opposite angle measure is needed.
When solving for an angle, the corresponding opposite side measure is needed. We can use another version of the Law of Cosines to solve for an angle.
cosα=b2+c2−a22bc
cosα=b2+c2−a22bc(8.2.4)
cosβ=a2+c2−b22ac
cosβ=a2+c2−b22ac(8.2.5)
cosγ=a2+b2−c22ab
cosγ=a2+b2−c22ab(8.2.6)
How to: Given two sides and the angle between them (SAS), find the measures of the remaining side and angles of a triangle
Sketch the triangle. Identify the measures of the known sides and angles. Use variables to represent the measures of the unknown sides and angles.
Apply the Law of Cosines to find the length of the unknown side or angle.
Apply the Law of Sines or Cosines to find the measure of a second angle.
Compute the measure of the remaining angle.
Example 8.2.18.2.1: Finding the Unknown Side and Angles of a SAS Triangle
Find the unknown side and angles of the triangle in Figure 8.2.48.2.4.
Solution
First, make note of what is given: two sides and the angle between them. This arrangement is classified as SAS and supplies the data needed to apply the Law of Cosines.
Each one of the three laws of cosines begins with the square of an unknown side opposite a known angle. For this example, the first side to solve for is side bb, as we know the measurement of the opposite angle ββ.
b2=a2+c2−2accosβb2=102+122−2(10)(12)cos(30°)Substitute the measurements for the known quantities.b2=100+144−240(√32)Evaluate the cosine and begin to simplify.b2=244−120√3b=√244−120√3Use the square root property.b≈6.013b2=a2+c2−2accosβb2=102+122−2(10)(12)cos(30°)b2=100+144−240(3–√2)b2=244−1203–√b=244−1203–√−−−−−−−−−−√b≈6.013Substitute the measurements for the known quantities.Evaluate the cosine and begin to simplify.Use the square root property.
Because we are solving for a length, we use only the positive square root. Now that we know the length bb, we can use the Law of Sines to fill in the remaining angles of the triangle. Solving for angle αα, we have
sinαa=sinβbsinα10=sin(30°)6.013sinα=10sin(30°)6.013Multiply both sides of the equation by 10.α=sin−1(10sin(30°)6.013)Find the inverse sine of 10sin(30°)6.013.α≈56.3°sinαa=sinβbsinα10=sin(30°)6.013sinα=10sin(30°)6.013α=sin−1(10sin(30°)6.013)α≈56.3°Multiply both sides of the equation by 10.Find the inverse sine of 10sin(30°)6.013.
The other possibility for αα would be α=180°−56.3°≈123.7°α=180°−56.3°≈123.7°. In the original diagram,αα is adjacent to the longest side, so αα is an acute angle and, therefore, 123.7°123.7° does not make sense. Notice that if we choose to apply the Law of Cosines, we arrive at a unique answer. We do not have to consider the other possibilities, as cosine is unique for angles between 0°0° and 180°180°. Proceeding with α≈56.3°α≈56.3°, we can then find the third angle of the triangle.
γ=180∘−30∘−56.3∘≈93.7∘
γ=180∘−30∘−56.3∘≈93.7∘
The complete set of angles and sides is
α≈56.3° a=10
β=30° b≈6.013
γ≈93.7° c=12
Exercise 8.2.1
Find the missing side and angles of the given triangle: α=30°, b=12, c=24.
Answer
: a≈14.9, β≈23.8°, γ≈126.2°.
Example 8.2.2: Solving for an Angle of a SSS Triangle
Find the angle α for the given triangle if side a=20, side b=25, and side c=18.
Solution
For this example, we have no angles. We can solve for any angle using the Law of Cosines. To solve for angle α, we have
a2=b2+c2−2bccosα202=252+182−2(25)(18)cosαSubstitute the appropriate measurements.400=625+324−900cosα Simplify in each step.400=949−900cosα−549=−900cosαIsolate cosα.0.61≈cosαFind the inverse cosine.α≈52.4°
See Figure 8.2.5.
Analysis
Because the inverse cosine can return any angle between 0 and 180 degrees, there will not be any ambiguous cases using this method.
Exercise 8.2.2
Given a=5, b=7, and c=10, find the missing angles.
Answer
: α≈27.7°, β≈40.5°, γ≈111.8°
Solving Applied Problems Using the Law of Cosines
Just as the Law of Sines provided the appropriate equations to solve a number of applications, the Law of Cosines is applicable to situations in which the given data fits the cosine models. We may see these in the fields of navigation, surveying, astronomy, and geometry, just to name a few.
Example 8.2.3A: Using the Law of Cosines to Solve a Communication Problem
On many cell phones with GPS, an approximate location can be given before the GPS signal is received. This is accomplished through a process called triangulation, which works by using the distances from two known points. Suppose there are two cell phone towers within
range
of a cell phone. The two towers are located 6000 feet apart along a straight highway, running east to west, and the cell phone is north of the highway. Based on the signal delay, it can be determined that the signal is 5050 feet from the first tower and 2420 feet from the second tower. Determine the position of the cell phone north and east of the first tower, and determine how far it is from the highway.
Solution
For simplicity, we start by drawing a diagram similar to Figure 8.2.6 and labeling our given information.
Using the Law of Cosines, we can solve for the angle θ. Remember that the Law of Cosines uses the square of one side to find the cosine of the opposite angle. For this example, let a=2420, b=5050, and c=6000. Thus, θ corresponds to the opposite side a=2420.
a2=b2+c2−2bccosθ(2420)2=(5050)2+(6000)2−2(5050)(6000)cosθcosθ≈0.9183θ≈cos−1(0.9183)θ≈23.3°
To answer the questions about the phone’s position north and east of the tower, and the distance to the highway, drop a perpendicular from the position of the cell phone, as in Figure 8.2.7. This forms two right triangles, although we only need the right triangle that includes the first tower for this problem.
Using the angle θ=23.3° and the basic trigonometric identities, we can find the solutions. Thus
cos(23.3°)=x5050x=5050cos(23.3°)x≈4638.15feetsin(23.3°)=y5050y=5050sin(23.3°)y≈1997.5feet
The cell phone is approximately 4638 feet east and 1998 feet north of the first tower, and 1998 feet from the highway.
Example 8.2.3B: Calculating Distance Traveled Using a SAS Triangle
Returning to our problem at the beginning of this section, suppose a boat leaves port, travels 10 miles, turns 20 degrees, and travels another 8 miles. How far from port is the boat? The diagram is repeated here in Figure 8.2.8.
Solution
The boat turned 20 degrees, so the obtuse angle of the non-right triangle is the supplemental angle, 180°−20°=160°. With this, we can utilize the Law of Cosines to find the missing side of the obtuse triangle—the distance of the boat to the port.
x2=82+102−2(8)(10)cos(160°)x2=314.35x=√314.35x≈17.7miles
The boat is about 17.7 miles from port.
Using Heron’s Formula to Find the Area of a Triangle
We already learned how to find the area of an oblique triangle when we know two sides and an angle. We also know the formula to find the area of a triangle using the base and the height. When we know the three sides, however, we can use Heron’s formula instead of finding the height. Heron of Alexandria was a geometer who lived during the first century A.D. He discovered a formula for finding the area of oblique triangles when three sides are known.
HERON’S FORMULA
Heron’s formula finds the area of oblique triangles in which sides a, b,and c are known.
Area=√s(s−a)(s−b)(s−c)
where s=(a+b+c)2 is one half of the perimeter of the triangle, sometimes called the semi-perimeter.
Example 8.2.4: Using Heron’s Formula to Find the Area of a Given Triangle
Find the area of the triangle in Figure 8.2.9 using Heron’s formula.
Solution
First, we calculate s.
s=(a+b+c)2s=(10+15+7)2=16
Then we apply the formula.
Area=√s(s−a)(s−b)(s−c)Area=√16(16−10)(16−15)(16−7)Area≈29.4
The area is approximately 29.4 square units.
Exercise 8.2.3
Use Heron’s formula to find the area of a triangle with sides of lengths a=29.7 ft, b=42.3 ft, and c=38.4 ft.
Answer
: Area = 552 square feet
Example 8.2.5: Applying Heron’s Formula to a Real-World Problem
A Chicago city developer wants to construct a building consisting of artist’s lofts on a triangular lot bordered by Rush Street, Wabash Avenue, and Pearson Street. The frontage along Rush Street is approximately 62.4 meters, along Wabash Avenue it is approximately 43.5 meters, and along Pearson Street it is approximately 34.1 meters. How many square meters are available to the developer? See Figure 8.2.10 for a view of the city property.
Solution
Find the measurement for s, which is one-half of the perimeter.
s=(62.4+43.5+34.1)2s=70mApply Heron's formula.Area=√70(70−62.4)(70−43.5)(70−34.1)Area=√506,118.2Area≈711.4
The developer has about 711.4 square meters.
Exercise 8.2.4
Find the area of a triangle given a=4.38 ft , b=3.79 ft, and c=5.22 ft.
Answer
: about 8.15 square feet
Media
Access these online resources for additional instruction and practice with the Law of Cosines.
Law of Cosines
Law of Cosines: Applications
Law of Cosines: Applications 2
Key Equations
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| Law of Cosines | a2=b2+c2−2bccosα b2=a2+c2−2accosβ c2=a2+b2−2abcosγ |
| Heron’s formula | Area=√s(s−a)(s−b)(s−c) where s=(a+b+c)2 |
Key Concepts
The Law of Cosines defines the relationship among angle measurements and lengths of sides in oblique triangles.
The Generalized Pythagorean Theorem is the Law of Cosines for two cases of oblique triangles: SAS and SSS. Dropping an imaginary perpendicular splits the oblique triangle into two right triangles or forms one right triangle, which allows sides to be related and measurements to be calculated. See Example 8.2.1 and Example 8.2.2.
The Law of Cosines is useful for many types of applied problems. The first step in solving such problems is generally to draw a sketch of the problem presented. If the information given fits one of the three models (the three equations), then apply the Law of Cosines to find a solution. See Example 8.2.3 and Example 8.2.4.
Heron’s formula allows the calculation of area in oblique triangles. All three sides must be known to apply Heron’s formula. See Example 8.2.5 and See Example 8.2.6.
Contributors and Attributions
Jay Abramson (Arizona State University) with contributing authors. Textbook content produced by OpenStax College is licensed under a Creative Commons Attribution License 4.0 license. Download for free at
8.1: Non-right Triangles - Law of Sines
8.3: Polar Coordinates |
6908 | https://shs.hal.science/halshs-00859318/file/2009ProblemsHM_PC.pdf | K. Chemla. p. 1 « On mathematical problems as historically determined artifacts. Reflections inspired by sources from ancient China », Historia Mathematica, 36, 3, 2009, p. 213-246. For Menso Folkerts On the occasion of his 65th birthday As an expression of friendship and appreciation Introduction The main goal of this paper is methodological.
1 It derives from the general observation that historians have often worked under the assumption that the main components of scientific texts —problems, algorithms and so on— are essentially ahistorical objects, which can be approached as some present-day counterparts. To be clear, by this diagnosis, I do not mean that when problems reflect tax systems or civil works, historians have failed to recognize that their statements adhere to a given historical context. Rather, I claim that they have mainly viewed problems in the way we commonly use them today —or as we commonly think we do— that is, as formulations of questions that require something to be determined and call for the execution of this task. In contrast, by focusing on mathematical problems found in ancient Chinese sources, this paper aims at establishing that we cannot take for granted that we know a priori what, in the contexts in which our sources were produced or used, a problem was. Moreover, besides providing evidence to support this claim, it suggests a method that can be used to describe the nature of mathematical problems in a given historical tradition. My conclusion is that developing such descriptions should be a prerequisite to setting out to read sources of the past. To substantiate these statements and illustrate this method, I shall concentrate on the earliest extant mathematical documents composed in China. These sources are of various types, a point that will prove essential for my argument. The book that factually is the oldest mathematical writing that has survived from ancient China, the Book of Mathematical Procedures (Suanshushu 算數書), was recently discovered in a tomb sealed ca. 186 B.C.E.
2 In contrast to this document, which was not handed down, the earliest book that has come down to us through the written tradition, The Nine Chapters on Mathematical Procedures (Jiuzhang suanshu 九章算術), to be abbreviated as The Nine Chapters, was probably 1 A first version of this paper was presented at the conference organized by Roger Hart and Bob Richards, The disunity of Chinese science, which was held in Chicago on May 10-12, 2002. It is my pleasure to take the opportunity to thank the organizers of this meeting and the audience for their comments. A revised version was prepared at the invitation of Della Fenster, for the conference "Exploring the History of Mathematics: How do we know what questions to ask?" that was held at the University of Richmond, Richmond, VA (USA), on May 12-15, 2004. May she be thanked for her generosity and encouragement. Markus Giaquinto, the referees and the editors of Historia Mathematica spent much time trying to help me formulate the argument of this paper more clearly. I owe them a huge debt of gratitude. 2 The first critical edition was published in [PENG Hao 彭浩, 2001]. Since then, several papers suggested philological improvements, such as Guo Shirong郭世榮, ; Guo Shuchun郭書春, . Two translations into English are forthcoming: [Cullen, 2004], the first version of which was published on the internet, [Dauben, 2008]. A new critical edition and translation into Japanese and Chinese were published recently: [張家山漢簡『算數書』研究会編 Chôka san kankan Sansûsho kenkyûkai. Research group on the Han bamboo slips from Zhangjiashan Book of Mathematical Procedures, 2006]. Recently, Cullen presented his rewriting of the mathematics of ancient China based on the discovery of the manuscript. K. Chemla. p. 2 completed in the first century C.E. and considered a “Canon (jing 經)” soon thereafter.
3 Both books are for the most part composed of particular problems and algorithms solving them. What kind of texts were they? How should we read them? These are the main questions I have in view here. The outlines of the problems often echo the concrete problems that bureaucrats or merchants of that time might have confronted in their daily practice. It has hence often been assumed that mathematics in ancient China was merely practical, oriented, as it seems, toward solving concrete problems. In fact, such a hasty conclusion conceals two implicit assumptions regarding problems. The first assumption, specifically attached to sources such as the Chinese ones, is that the situation used to set a problem should be taken at face value: learning how to handle the concrete task presented by the problem would precisely be what motivated its inclusion in a text. The second assumption, much more generally held, is that the sole aim of a problem is to present a mathematical task to be executed and to provide means to do so. In this paper, I shall cast doubt on the second assumption, showing that it distorts our view of the way in which problems were actually used in ancient China. As a side result, the first assumption will also be undermined. This conclusion implies that we should be careful in the ways in which we use the evidence provided by the particular situations described in the problems inserted in mathematical documents. We cannot conclude merely from their outer appearance that they were written only for practical purposes. How can we devise a method for determining what a mathematical problem was in a given historical context? The suggestion I develop in this paper is to look for readers of the past who left evidence regarding how they read, or reacted to, our sources. This is why the Chinese case is most helpful: in the form of commentaries, it yields source material documenting early reading of some mathematical sources. This is also where the difference in type between the two Chinese documents mentioned above is essential. In relation to its status as a Canon, The Nine Chapters was handed down and several commentaries were composed on it. Among these commentaries, two were selected by the written tradition to be handed down together with the text of the Canon: the one completed by Liu Hui 劉徽 in 263 and the one presented to the throne by Li Chunfeng 李淳風 in 656. As we have already stressed above, The Nine Chapters is for the most part composed of problems and algorithms solving them. The commentaries bear on the algorithms and, less frequently, on the problems. They are placed at the end of the piece of text on which they comment or between its statements. In this paper, I shall gather evidence from Liu Hui’s commentary with respect to how he used problems and also how he read those included in The Nine Chapters. This will provide us with source material to reconsider in a critical way the nature of a mathematical problem in ancient China and to establish that problems were submitted to a mathematical practice differing, on many points, from the one we spontaneously attach to them. What does such evidence tell us? To start with, it shows us why the questions we raise on problems are essential and cannot be dismissed if we are to read our sources in a rational way. We shall select two examples of the difficulties which face the historian and which are 3 I argue in favor of this dating in [Chemla and Guo Shuchun, 2004, 475-481]. In chapter B ([Chemla and Guo Shuchun, 2004, 43-56]), Guo Shuchun presents the various views on the date of completion of the book held by scholars in the past and argues that the composition of The Nine Chapters was completed in the first century B.C.E. In what follows, I shall regularly rely on the glossary of mathematical terms I composed to back up the translation into French of The Nine Chapters and the commentaries (see below) [Chemla and Guo Shuchun, 2004, 897-1035]. In it, the entry jing 經, “Classic, Canon,” provides evidence countering the commonly held view that only in the seventh century were such books as The Nine Chapters considered to be “Canons.” In [Chemla, 2001 (forthcoming); Chemla, 2003b], on the basis of the extant evidence regarding The Nine Chapters, I discuss the kind of scripture a Canon constituted in ancient China. In addition to separate contributions, [Chemla and Guo Shuchun, 2004] contains joint work, such as the critical edition of The Nine Chapters and the commentaries on which this paper relies. K. Chemla. p. 3 illustrated by the commentary. First, The Nine Chapters describes a procedure for multiplying fractions after the following problem: (1.19) Suppose one has a field which is 4/7 bu wide and 3/5 bu long. One asks how much the field makes. 今 有 田 廣 七 分 步 之 四 ，從 五 分 步 之 三 ，問 爲 田 幾 何 。 4 [Chemla and Guo Shuchun, 2004, 170-171] Why is it that, in the middle of his commentary on the procedure for multiplying fractions, the third century commentator Liu Hui introduced another problem that can be formulated as follows: 1 horse is worth 3/5 jin of gold. If a person sells 4/7 horse, how much does the person get?
5 Seen from our point of view, the two problems are identical: their solution requires multiplying the two fractions by one another. Obviously, from Liu Hui’s perspective, they differ. Otherwise, he would not need to change one for the other. Inquiring into their difference as perceived by Liu Hui should hence disclose one respect in which the practice of problems in ancient China differs from the one we would be spontaneously tempted to assume. This will be our first puzzle. In the example above, the numerical values remain the same. The first difficulty hence regards the interpretation of the situation chosen to set a problem. In addition, the commentator also regularly changes the numerical values with which a problem is stated, without changing the situation, before he comments on the procedure attached to it. For instance, Problem 5.15 in The Nine Chapters, which requires determining the volume of a pyramid, reads as follows: (5.15) Suppose one has a yangma, which is 5 chi wide, 7 chi long and 8 chi high. One asks how much the volume is. 今有陽馬，廣五尺，袤七尺，高八尺。問積幾何。 6 [Chemla and Guo Shuchun, 2004, 428-429] 4 The shape of the field is designated by the name of its dimensions: a field with only a length (north-south direction) and a width (east-west direction) is rectangular. I discuss the names of geometrical figures and their dimensions in [Chemla and Guo Shuchun, 2004, chapter D, 100-104]. “1.19” indicates the 19th problem of Chapter 1 of The Nine Chapters. The same convention is used for designating other problems in this paper. Note that at the time when The Nine Chapters was composed, the bu was between 1.38 and 1.44 m. The values of the length and width do not seem concrete. 5 Section 3 of this paper describes the text in greater detail. I refer the reader to this section to see precisely how the commentator formulates this problem. Note that procedures for multiplying fractions similar to the one in The Nine Chapters occur in several contexts in the Book of Mathematical Procedures. In some cases, the procedure is given apparently without the context of any problem (bamboo slips 6, 7 [PENG Hao 彭浩, 2001, 38, 40].) In another case, its statement follows a problem of the type Liu Hui introduces in his commentary (selling a fraction of an arrow and getting cash in return, bamboo slips 57-58 [PENG Hao 彭浩, 2001, 65]). In yet other cases, the procedure is formulated in relation to computing the area of a rectangular field and showing how it inverts procedures for dividing (bamboo slips 160-163 [PENG Hao 彭浩, 2001, 114]). 6 The yangma designates a specific pyramid, with a rectangular base. Its shape is defined by the fact that its apex is above a vertex of the rectangular base. See section 4. I opt for not translating the Chinese term to avoid a long expression that would express the yangma as a kind of pyramid, whereas the term in Chinese does not link the shape to that of other geometrical solids. We do not know exactly by means of which object the Chinese term designates the pyramid. Nor do we know whether, at the time when The Nine Chapters was composed, the term had acquired a technical meaning or was only designating a specific object with the geometrical shape required. Note also that, at that time, 1 chi was between 0.23 and 0.24 m. K. Chemla. p. 4 At the beginning of his commentary, which, as always, follows the procedure, Liu Hui states: Suppose 7 the width and the length are each 1 chi, and the height is 1 chi 假令廣袤各一尺，高一尺，… Again, the same question forces itself upon us: for which purpose does the commentator need to change the numerical values here, as he also regularly does when he discusses other surfaces and solids? This will be our second puzzle. My claim is that it is only when we are in a position to argue for an interpretation of the differences between, on the one hand, the situations and, on the other hand, the numerical values that we may think we have devised a non-naive reading of mathematical problems as used in ancient China. This paper develops an argument for solving the two puzzles. Here are the main steps of the argument: First, I shall show that we must discard the obvious explanation that one could be tempted to put forward, that is, that a problem in ancient China only stood for itself. We can put forward evidence showing that the commentators read a particular problem as standing for a class (lei) of problems. Moreover, they determined the extension of the class on the basis of an analysis of the procedure for solving the particular problem rather than simply by a variation of its numerical values. An even stronger statement can be substantiated: one can prove that these readers expected that an algorithm given in The Nine Chapters after a particular problem should allow solving as large a class of problems as possible [Chemla, 2003a]. In making these points, we shall thereby establish the first elements of a description of the use of problems in mathematics in ancient China. Section 1 thus proves that our puzzles cannot be easily solved and require explanations of another kind. In section 2, I introduce some basic information concerning the practice of proving the correctness of algorithms as carried out by commentators like Liu Hui, since this will turn out to be necessary for our argumentation. On this basis, Section 3 concentrates on the situations used to set problems. It argues that, in fact, far from being reduced to questions to be solved, the statements of problems, and more specifically the situations with which they are formulated, were an essential component in the practice of proof as exemplified in the commentaries. Viewing the situations from this angle allows accounting for why the problem with cash was substituted for that about the area of a field. Section 4 focuses on the numerical values given in the problems and suggests that, if we set aside cases in which a change of values aims at exposing the lack of generality of a procedure, it is only within the framework of geometry that Liu Hui changed the values in the statements of problems. In Section 4, I argue that the reason for this is that the commentators introduce material visual tools to support their proofs and that the numerical values given in the problems refer to these tools. This leads me to suggest a parallel between the role played by visual tools in the commentaries and the part devoted to the situations described by problems. Sections 1 to 4 concentrate on how the evidence provided by the commentary allows solving the two puzzles put forward. To be sure, these puzzles could also be grasped only thanks to the commentaries. As a result, my argument establishes how the practice with problems attested to by the commentators differs from our own and how this description 7 The expression for “suppose” is not the same as in The Nine Chapters itself; i.e., jinyou has been replaced by “suppose (jialing),” which in the commentaries as well as in some later treatises is more frequently used to introduce a problem. K. Chemla. p. 5 accounts for the difficulties presented. Even if this is to be considered as the only outcome of the paper, we would have fulfilled our aim of providing an example of a practice with problems that does not conform to our expectations. Now, the question is: can we go one step further and transfer the results established with respect to the use of problems by the commentators to The Nine Chapters itself, even though the earliest extant commentary on The Nine Chapters was composed probably some two or three centuries after the completion of the Canon? This question is addressed somewhat briefly in Section 5. The reason why we must proceed in this indirect way relates to the fact that commentators wrote in a style radically different from that of the Canons. More precisely, commentaries express expectations, motivations and second-order remarks, all these elements being absent from The Nine Chapters, which is mainly composed of problems and algorithms. Commentaries hence allow us to grasp features of mathematical practice that are difficult to approach on the basis of the Canon itself, at least when one demands that a reading of ancient sources be based on arguments. One point must be emphasized: Unless we find new sources, there is no way to reach full certainty about whether what was established on the basis of third century sources holds true with respect to writings composed some centuries earlier. However, this having been said, two remarks can be made. First, relying on commentaries composed more than two centuries after the Canon to interpret the latter appears to be a less inadequate method than relying on one’s personal experience of a mathematical problem. It seems to me to be more plausible that the practice of problems contemporary with the compilation of The Nine Chapters is related to Liu Hui’s practice than that it is related to ours. However, here we are in the realm of hypothesis rather than certainty. Second, once we restore the practice of problems to which the commentaries bear witness, we can find many hints indicating that some conclusions probably hold true for The Nine Chapters itself and even for the Book of Mathematical Procedures. Section 5 is devoted to discussing such hints. With these warnings in mind, let us turn to examining our evidence. K. Chemla. p. 6 1. How does a problem stand for a class of problems? 1.1. A first description of the statement of problems Problem 1.19 quoted above illustrates what, in general, a problem in The Nine Chapters looks like. It is particular in two respects. The statement of the problem refers to a particular and most often apparently concrete situation, such as, in this case, computing the area of a field. Moreover, it mentions a particular numerical value for each of the data involved—in this case, 3/5 bu and 4/7 bu for the data “length” and “width,” respectively.
8 However, some problems are only particular in this latter respect. An example of this is Problem 1.7, one of three that precede the procedure for the addition of fractions: (1. 7) Suppose one has 1/3 (one of three parts), 2/5 (two of five parts). One asks how much one obtains if one gathers them. 今有三分之一，五分之二，問合之得幾何。 Although numerical values are given, the fractions to be added, or in other terms, the units out of which parts are taken, are abstract.
9 All problems in The Nine Chapters are one of the two types exemplified by 1.19 or 1.7. The procedures associated with the problems in The Nine Chapters also show some variation. Problem 1.19 is followed by a procedure for the multiplication of parts, which amounts to . It is expressed in general, abstract terms, since it makes no reference to the situation in the problem, the area of a field: Multiplying parts Procedure: the denominators being multiplied by one another make the divisor. The numerators being multiplied by one another make the dividend. One divides the dividend by the divisor. 乘分 術曰：母相乘爲法，子相乘爲實，實如法而一。 [Chemla and Guo Shuchun, 2004, 170-171]. However, in other cases, the procedure given in The Nine Chapters is expressed with respect to the concrete situation and values described by the problem. An example of this is the following problem from Chapter 9, “Basis and height (gougu),” which is devoted to the right-angled triangle (cf. Figure 1): (9. 9) Suppose one has a log with a circular section, stuck into a wall, with dimensions 8 The same remark holds true for the problems in the context of which the procedure amounting to the “Pythagorean theorem” is discussed. Further, in fact, there is evidence showing that at the latest in the third century the term “field” came to designate a geometrical shape in general. See [Chemla and Guo Shuchun, 2004, Glossary, 992-993]. It is hence difficult to determine whether the statement of the problem in The Nine Chapters still uses the term with a concrete sense or already with a technical meaning. The same problem was raised above regarding the interpretation of yangma. More generally, the qualification of the statement of a problem as “concrete” should be manipulated with care. [Vogel, 1968, 124-127] describes the general format of a problem in The Nine Chapters. He suggests that the problems can be divided into two groups: those dealing with problems of daily life and those that can be considered as recreational problems. In my view, this opposition is anachronistic and does not fit the evidence we have from ancient China (see the following discussion). Moreover, Vogel fails to point out that there are problems formulated in abstract terms. 9 For a general discussion on the form of these problems, see [Chemla, 1997a]. Compare [Cullen, 2007, 17; Guo Shuchun郭書春, 2002, 514-517] for a description of the form of problems in the Book on mathematical procedures. K. Chemla. p. 7 unknown. If one saws it with a saw at a depth of 1 cun (CD), the path of the saw (AB) is 1 chi long. One asks how much the diameter is. Answer: the diameter of the log is 2 chi 6 cun. Procedure: half the path of the saw being multiplied by itself, one divides by the depth of 1 cun, and increases this (the result of the previous operation) by the depth of 1 cun, which gives the diameter of the log. 今有圓材埋在壁中，不知大小。以鐻鐻之，深一寸，鐻道長一尺。問徑幾何。 荅曰：材徑二尺六寸。 術曰：半鐻道自乘，如深寸而一，以深寸增之，即材徑。 10 Figure 1 In modern terms, the procedure amounts to the formula 11 . Since both the problem and the procedure are formulated in the same concrete terms, we might, for such cases at least, be tempted to assume that they are to be read as standing only for themselves. Interestingly, as we will see below, we can find evidence allowing us to determine how Liu Hui read this problem. It clearly shows that even in such cases this assumption must be discarded. Moreover, it also reveals how Liu Hui used this problem as a general statement, and not as a particular one. 1.2. The commentator Liu Hui’s reading and use of problems The piece of evidence on which we can rely to approach the commentator’s reading of the latter problem comes from Liu Hui’s commentary on Problems 1.35/1.36 and the procedure included in The Nine Chapters for the determination of the area of a circular segment. The key point is that, in this piece of commentary, Liu Hui refers to the problem of the log stuck in a wall from the Canon. The Problems 1.35 and 1.36 are similar. The second one reads as follows (cf. Figure 2): 12 Figure 2 (1.36) Suppose again 13 that one has a field in the form of a circular segment, whose chord (AB) is 78 bu 1/2 bu, and whose arrow (CD) is 13 bu 7/9 bu. One asks how much the field makes. 又有弧田，弦七十八步二分步之一，矢十三步九分步之七。問爲田幾何。 [Chemla and Guo Shuchun, 2004, 190-191] The Nine Chapters then provides an algorithm to compute the area of this field, which 10 [Chemla and Guo Shuchun, 2004, 714-715]. In the translation, for the sake of my argument, I have inserted references to a geometrical figure drawn by myself (figure 1). Needless to say, neither the figure nor the references are to be found in The Nine Chapters. More generally, the Canon does not refer to any visual tool. 11 In the right-angled triangle OAC, the difference of the hypotenuse and the side OC is equal to CD. Dividing AC2 by CD yields the sum of the hypotenuse and the side OC; hence the result. 12 Again, I have drawn the figure for the sake of commenting. No such figure is to be found in the original sources. Moreover, I have added to the translation of the original text references to the figure between brackets. 13 This is the usual beginning for problems after the first one, when a mathematical question is dealt with through a sequence of problems. K. Chemla. p. 8 amounts to the formula (AB.CD+CD2)/2. In his analysis of this procedure, Liu Hui first shows that when the circular segment is half of the circle, in fact the algorithm computes the area of the half-dodecagon inscribed in the circle. Moreover, he stresses that the imprecision increases when the circular segment is smaller than the half-circle. This imprecision motivates him to establish a new procedure, which derives from tiling the circular segment with triangles and computing its area as the sum of their areas.
14 It is within this context that Liu Hui first needs to compute the diameter of the circle containing the circular segment from the two data of Problem 1.36, the chord (AB) and the arrow (CD). For this, he refers to Problem 9.9 on the log stuck in a wall as follows: (…) It is appropriate then to rely on the procedure of the (problem) where one saws a log with a circular section in the (Chapter) "Basis (gou) and height (gu)" and to look for the diameter of the corresponding (circle) by taking the chord of the circular segment as the length of the path of the saw, and the arrow as the depth of the piece sawn. Once one knows the diameter of the circle, then one can cut the circular segments in pieces. (...)宜依句股鋸圓材之術,以弧弦爲鋸道長,以矢爲鋸深,而求其徑。既知圓徑,則 弧可割分也。 (My emphasis). [Chemla and Guo Shuchun, 2004, 192-193] This piece of evidence shows that Liu Hui does not read the problem of the log stuck in a wall and the procedure attached to it as merely standing for themselves, but as expressing something more general. It thus reveals that, even in a case like Problem 9.9, where the procedure is expressed with reference to the concrete situation of a log stuck in a wall and particular values, the third-century commentator reads its meaning as exceeding this particular case. This holds true in the entire commentary, in which one can find other pieces of evidence confirming this conclusion, as we will see below. 1.3. The procedure defines the extension of the class meant by the problem Another interesting example of the generality the commentator attaches to a problem and the corresponding procedure is the following one: (6.18) Suppose that 5 persons share 5 coins (units of cash) in such a way that what the two superior persons obtain is equal to what the three inferior persons obtain. One asks how much each obtains. 今有五人分五錢，令上二人所得與下三人等，問各得幾何。 [Chemla and Guo Shuchun, 2004, 526-529] The procedure following the answers to the problem is also expressed with reference to the particular situation and values mentioned in the statement. However, in this case, through the analysis of the procedure that he then develops, Liu Hui brings to light that the procedure is not general. To be more precise, the procedure adequately solves the problem, but cannot solve all similar problems, because it uses features specific to the situation 14 This reasoning implies covering an area with an infinite number of tiles and it has been the topic of much discussion in the literature. It falls outside the topic of this paper. For a concise exposition, compare [Li Yan and Du Shiran, 1987, 68-69]. 15 As the commentator makes clear, it is assumed that the five persons have unequal ranks defined by the integers 5 to 1 and that the share they obtain depends on their rank. We may feel that the statement of the problem is incomplete, but this would mean that we project our own expectations of how a problem should be formulated. Actual description of how problems recorded in historical documents were formulated should replace this anachronistic approach. However, dealing with this topic would exceed the scope of this paper. K. Chemla. p. 9 described in Problem 6.18. Note that this is the only case when this happens for a procedure given in The Nine Chapters and that the commentator immediately exposes the lack of generality. This reaction betrays his expectation that a problem stands not only for itself, but also for a class (lei). Liu Hui reacts to this situation in several steps. First, his analysis determines criteria that enable him to know to which problems the procedure of The Nine Chapters can be successfully applied.
16 In a sense, Liu Hui inquires into the class of problems for which the particular problem stands, and he does so through an examination of the procedure. Second, he formulates another problem similar to that of the Canon, as follows: Suppose 7 persons share 7 coins and they want to do this in such a way that (what) the two superior persons (obtain) is equal to (what) the five inferior persons (obtain). 假令七人分七錢，欲令上二人與下五人等。 [Chemla and Guo Shuchun, 2004, 528-529] The criteria previously put forward immediately show why the procedure of The Nine Chapters does not apply to this problem.
18 Despite the appearances, on the basis of the procedure stated, the two problems do not belong to the same class. In other terms, the category of problems for which a problem stands is not determined by a variation of its numerical values, but rather by the procedure provided to solve it. Third, Liu Hui suggests modifying the procedure in such a way that it solves all similar problems. Seen from another angle, Liu Hui aims at stating a procedure for which all similar problems belong to the same class. We see that the commentary on a procedure analyzes it in such a way as to inquire into the extension of its validity and modifies it to extend the class of problems that can be solved by it (and for which a particular problem stands). As a result, on the basis of the sections of Liu Hui’s commentary examined so far, and in fact of others, we can thus state that, in his view, a problem stands for a class (lei) of problems that is determined on the basis of the procedure described after it. It is not so much the similarity of structure between the situations described by different problems that allow considering them as sharing the same category, but, most importantly, the fact that they are solved by the same procedure.
19 We hence reach the conclusion that far from being only the 16 These criteria are as fallows: the number of inferior persons must exceed the number of superior ones by only 1; moreover, the sum of the coefficients attached to the superiors (5 and 4) must be greater than that attached to the inferiors (3, 2, 1). I do not enter into any detail here, referring the interested reader to [Chemla, 2003a]. 17 Here too, the expression jinyou for “suppose” in The Nine Chapters has been replaced by jialing. In this paper, I do not discuss the numerical values chosen to set a problem. However, clearly they call for comment. The figures used in Problem 6.18 in The Nine Chapters are the simplest possible with which the mathematical question can be formulated. In his commentary, Liu Hui introduces values that are the simplest possible to make his point. Also, the figures occurring in Problem 1.19 (3, 4, 5, 7; see above) are probably chosen on purpose. Compare my introduction to Chapter 9 in [Chemla and Guo Shuchun, 2004, 663-665, 684-689]. Further research is needed in this respect. 18 There are three more persons among the inferiors than among the superiors. Moreover, the sum of the coefficients attached to the superiors (7+6) is smaller than the sum of those attached to the inferiors (5+4+3+2+1). 19 On this point, I refer the reader to [Chemla, 1997a], where I analyze how Liu Hui uses the term lei “class, category” with respect to problems. As rightly stressed by C. Cullen, in the earliest known theoretical discussion on the modes and methods of inquiry in mathematics and cosmography, i.e., in the opening sections of The Gnomon of the Zhou (Zhou bi), which he dates to the beginning of the common era, the concept and practice of “categories” in mathematics are central [Cullen, 1996, 74-75, 177]. In my glossary of mathematical terms, I discuss more generally the various uses of the term lei in the commentaries on The Nine Chapters and in philosophical texts of antiquity [Chemla and Guo Shuchun, 2004, 948-949]. It is interesting that the Book of K. Chemla. p. 10 sequence of operations allowing a given problem to be solved, the procedure is read beyond the specific context within which it is formulated, and, further, it even determines the scope of generality of a given problem. We shall come back to this issue in Section 3. In the fourth step, the one we are most interested in here, Liu Hui suggests an entirely different, more general procedure for dealing with Problem 6.18. In fact, the commentator does this simply by suggesting “to imitate the procedure” given in the Canon for the next problem, 6.19. This problem reads as follows: (6.19) Suppose that a bamboo has 9 internodes 20 and that the 3 inferior internodes have a capacity of 4 sheng whereas the 4 superior internodes have a capacity of 3 sheng. One asks, if one wants that between two (neighboring) inner internodes capacities be uniformly distributed, 21 how much they each contain. 今有竹九節，下三節容四升，上四節容三升。問中間二節欲均容，各多少。 Again, the procedure described after the statement of Problem 6.19 refers to the particular situation and values displayed in the statement. However, despite the differences on both counts between Problems 6.18 and 6.19, Liu Hui directly imitates the procedure solving 6.19 (that is, transfers it step by step) to solve Problem 6.18, and, beyond, the problems that now belong to the same class. As a result, the new procedure shapes Problem 6.18 as standing for a much larger class. In this case, as in the case of the commentary following Problem 1.36 quoted above, the same phenomenon recurs: the procedure circulates from one context to the other, disregarding the change in situation and in numerical values.
22 What is particularly noteworthy, however, is how Liu Hui does so in both contexts. Let us explain on the example of the commentary on the circular segment (Problem 1.36) what we mean by the “circulation” of a problem. Liu Hui does not feel the need to express a more abstract statement or procedure that would capture the “essence” of Problem 9.9 and could be applied to similar cases such as Problem 1.36. On the contrary, he directly makes use of the procedure given after 9.9, with its own terms, in the context of 1.36, by establishing a term-to-term correspondence, “taking the chord of the circular segment as the length of the path of the saw, and the arrow as the depth of the piece sawn.” This seems to indicate that the situation described in Problem 9.9 can be directly put into play in other concretely different situations. The particular appears to be used to state the general in the most straightforward way possible. Further, to describe a more general procedure for problems of the same class as 6.18, Liu Hui imitates the procedure for 6.19 within the context of the most singular of all problems (6.18) and not that of the more “generic” one, which he introduced as a counterexample. More importantly, for all problems such as 6.18, he could simplify the procedure given for 6.19 to make it fit certain specific features that these problems all share —for problems like 6.18, in contrast to the bamboo problem and its middle internodes, there are no persons who do not belong to either the group of inferiors or that of superiors. Instead, Liu Hui prefers to keep the procedure with the higher generality that characterizes it. The conclusion of the previous paragraph can be stated in a stronger way: the most particular of all paradigms is used to Mathematical Procedures also attests to the use of the term lei in the description of mathematical procedures as early as the second century B.C.E. (see bamboo slip 21 [PENG Hao 彭浩, 2001, 45]. 20 The trunks of bamboos have nodes. The Chinese term refers here to the space between two nodes and considers the “capacity” of the volume thus formed. The problem hence deals with nine terms of an arithmetic progression —the capacities of the successive spaces between the nodes. The sums of the first three terms and of the last four ones, respectively, are given and it is asked to determine all the terms. 21 Like in the previous problem, the capacities of the cavities form an arithmetic progression. 22 Compare the analysis of transfer between situations in [Volkov, 1992, Volkov, 1994]. K. Chemla. p. 11 formulate the most general of all algorithms and consequently it now stands for a much wider class. A remark concerning the use of problems in the context of commentaries is here in order and will prove useful below. Liu Hui uses the procedure solving Problem 9.9 in a commentary in which he describes a new procedure for computing the area of the circular segment. At the same time as he shapes the procedure, he shows why it is correct. Such a concern for the correctness of algorithms drives the greatest part of the commentaries, which systematically establish that the procedures in The Nine Chapters are correct. Problems play a key role in achieving this goal. In the context examined, the use of Problem 9.9 is signaled by the verb “to look for qiu 求.” 23 More generally, this term signals the use of a problem in a proof. The commentary on the area of a circular segment illustrates the following use of a problem in a proof: The task of establishing the new procedure is divided into subtasks, which are identified with problems known —to start with, Problem 9.9. On the one hand, using the procedure that The Nine Chapters gives to solve Problem 9.9 yields the first segment of the procedure now sought for. On the other hand, since the correctness of the procedure solving 9.9 has already been established, the commentator can rely on the fact that it yields the magnitude needed at this point of the reasoning, that is, the diameter of the circle. In the terms Liu Hui uses to speak of the proof of the correctness of a procedure, the “meaning yi 意” of the result of the first segment of the procedure has been ascertained. More generally, let us stress the fact that, within the proof of the correctness of an algorithm he is shaping, Liu Hui uses problems and procedures solving them to determine step by step the “meaning” of the whole sequence of operations, that is, to determine that the procedure he establishes yields the area of a circular segment. 1.4. A similar way of reading and using problems in later sources In ancient China, this way of using procedures stated in one context directly in another, illustrated above for the use of the procedure solving Problem 9.9 for dealing with the area of a circular segment, was not specific to Liu Hui. In fact, one can find a similar piece of evidence four centuries later, in a seventh-century commentary on the Mathematical Canon Continuing the Ancients (Qigu suanjing), written by Wang Xiaotong in the first half of the seventh century.
24 The commentary relates the first problem of the book, which is devoted to astronomical matters, to a problem dealing with a dog pursuing a rabbit, indicating that the latter problem is included in The Nine Chapters.
25 In this case, too, the latter problem is not reformulated in astronomical terms, nor is a third and abstract description of it introduced as a middle term, to allow the result concerning the dog and the rabbit to be applied in astronomy. In exactly the same way as described above, although The Nine Chapters presents the problem within a particular concrete context, the first reader that we can observe, namely the seventh-century commentator, reads it as exemplifying a set of problems sharing a similar 23 Note that the verb occurs in the statement of problems and tasks in the Book of Mathematical Procedures, see bamboo slip 160-163 [PENG Hao 彭浩, 2001, 114]. 24 Volume 2 of [Qian Baocong, 1963] contains a critical edition of Wang Xiaotong’s Qigu suanjing. The problem and commentary mentioned are to be found in [Qian Baocong, 1963, 2, 495-496]. [Eberhard (Bréard), 1997] discusses this example. Part of her discussion and her translation are published in [Bréard, 1999, 41-43, 333-336]. For a further study concerning transmissions of problems of this type, see [Bréard, 2002]. 25 Although The Nine Chapters contains similar problems, the extant editions do not contain precisely the one quoted. Since we are only interested here in how a problem dealing with a given situation is used in another context, this textual problem can be left aside. K. Chemla. p. 12 structure and solved by the same algorithm. Furthermore, as in the previous example, the commentator feels free to make the problem and procedure, which apparently do not relate to astronomy, "circulate" as such into a different, astronomical context. This seems to indicate that there was an ongoing tradition in ancient China that did not mind discussing general mathematical procedures in the particular terms of the problems in which they had been formulated, although the questions discussed exceeded the case illustrated by the particular situation. The above evidence from the third and the seventh centuries leads to the same conclusions. This indicates that it would not be farfetched to assume that this was also the way in which the authors of The Nine Chapters conceived of the problems that they included in the Canon. This seems all the more reasonable because, as we have indicated above, except for one case (6.18), all procedures following the statement of problems are general.
26 We shall come back to this issue in greater detail in Section 5. Even though it is perhaps less striking in comparison with our own uses of problems, let us stress that, in fact, the same pieces of evidence show that these conclusions hold true with respect to the numerical values. Although Liu Hui regularly comments on a given problem and procedure on the basis of particular numerical values, he understands the meaning of his discussion as extending beyond this particular set and as, in fact, general. Again, in this respect, the commentator thus proves to discuss the general in terms of a particular [Chemla, 1997a]. The problems of logs stuck in walls and dogs pursuing rabbits that can be found in The Nine Chapters may be perceived as recreational by some readers of today, because of the terms in which they are cast. The evidence examined proves that things are not so simple. The historian is thus warned against the assumption that the category of "mathematical problem" remained invariant in time. Such a historical reconstruction guards us against mistaking a problem as merely particular or practical, when Chinese scholars read it as general and meaningful beyond its own context, or mistaking it as merely recreational when it was put to use in concrete situations. Now that we have seen that, in Liu Hui’s practice, a problem did practically stand for a category of problems, and how it did so, we have discarded the simple solution that could have accounted for our puzzles. We are hence left with the question: Why is it that, within the context of his commentary on the procedure for “multiplying parts,” or on that for the volume of the yangma, Liu Hui feels it necessary to substitute one situation for another, or one set of values for another, although both the original and the substituted problem seem to us to share the same category? One may even say that after examining the evidence presented so far, our puzzles look even more intriguing. Elaborating a solution for these puzzles will compel us to enter more deeply into the practice of mathematical problems as exemplified in Liu Hui’s commentary. 26 This fact is only one feature among the many hints indicating that generality was a key epistemological value that inspired the composition of The Nine Chapters. Another hint is provided by how the “chapters zhang 章” are composed. They each embody a part of mathematics that derives from a unique procedure, and in correlation with this fact, their text is organized around the generality of the procedure placed at the beginning, which by derivation commands the whole chapter. Alexei Volkov stresses this fact in his translation of the title of the Canon as Computational Procedures for Nine Categories [of Mathematical Problems] [Volkov, 1986; 2001]. Compare the translations of the title as Nine Chapters on the Mathematical Art [Needham and Wang Ling, 1959, 19], Nine Books on Arithmetical Techniques [Vogel, 1968], Nine Chapters of Calculation [Wang Ling, 1956, 15], Mathematical Methods in a Nine-Fold Categorization [Cullen, 2002, 784], and Mathematical Procedures under Nine Headings [Cullen, 2004, 1]. Among the meanings that may be intended by the title, the term zhang designates a division in a writing or a stage in a process of development, as well as, more generally, a distinction. Whatever the interpretation of the title may be, the division of the book into nine chapters manifests the influence of the value of generality. K. Chemla. p. 13 2. Proving the correctness of procedures in order to find out general formal strategies in mathematics The interpretation of Liu Hui’s commentary on the procedure for “multiplying parts” or computing the volume of the yangma requires that we recall some basic information regarding the mathematical practices linked to the exegesis of such a Canon as The Nine Chapters. As has been recalled above, after virtually every procedure given by The Nine Chapters for solving a problem or a set of problems, the commentators systematically establish its correctness. However, the way in which they deal with the issue of correctness manifests a specific practice of proof, which can be linked to the context of exegesis within which it develops.
27 I shall sketch its main characteristics, since it will prove useful for solving our puzzles. To this end, I shall first evoke Liu Hui’s commentary on the algorithm given by The Nine Chapters for adding up fractions, which appears to be a pivotal section in his text.
28 2.1. Proving the correctness of the procedure for the addition of fractions In Chapter 1, where the arithmetic of fractions is dealt with, the Canon presents three problems similar to Problem 1.7, after which it offers the following general and abstract algorithm to add up fractions, equivalent to ( ): Gathering partsProcedure: The denominators multiply the numerators that do not correspond to them; one adds up and takes this as the dividend (shi). The denominators being multiplied by one another make the divisor (fa). One divides the dividend by the divisor. (…) 合分 術曰：母互乘子，幷以爲實。母相乘爲法。 實如法而一。不滿法者，以法命之。 [Chemla, Guo, 2004, pp. 156-161]. Let us outline how, in his commentary on this section of the Canon, Liu Hui establishes the correctness of this procedure. The Canon and the commentaries approach the object “fraction,” or in Chinese terms: “parts,” in two ways. The expression for m/n used in The Nine Chapters, “m of n parts” (n fen zhi m n 分之 m), gives the fraction as being composed of “parts.” This dimension is the one emphasized in what I call the “material” approach to fractions. The expression also displays a numerator and a denominator (“m of n parts”), which are the basis for what I designate as the “numerical” approach to fractions. These two approaches to fractions appear to have been combined by the mathematicians of ancient China. On the one hand, the problem asking us to add up fractions requires gathering various disparate parts together to form a single quantity, which must hence be evaluated ( ). On the other hand, the algorithm 27 The question of the relationship between the exegesis of a Canon, on the one hand, and Liu Hui’s or Li Chunfeng’s specific practice of proof, on the other hand, is addressed in [Chemla, 2001 (forthcoming)]. In Chapter A of [Chemla and Guo Shuchun, 2004, 27-39], I discuss the fundamental operations that the commentaries put into play when proving the correctness of procedures. 28 I discuss in detail this annotation of Liu Hui’s commentary in [Chemla, 1997b]. For further argumentation regarding what is stated in this section, the reader is referred to this paper. K. Chemla. p. 14 prescribes computations on numerators and denominators —the numerical dimension of the fractions involved— to yield a value as the result of a division ( —where ad+cb is the dividend and bd the divisor). Establishing the correctness of the algorithm requires proof that the value obtained ( ) measures the quantity formed by bringing together the various parts ( ). In his commentary on the simplification of fractions, a topic dealt with immediately before the addition of fractions, Liu Hui had approached the fractions as entities to be manipulated by the procedure concerned, i.e. as a pair consisting of a numerator and a denominator, and he had stressed the potential variability of their expression: one can multiply, or divide, the numerator and the denominator by the same number, he had stated, without changing the quantity meant. In this context, to divide (e.g., 2/4 becoming 1/2) is to “simplify yue 約” the fraction. The opposed operation (e.g., 2/4 becoming 4/8), which Liu Hui had then introduced and called by opposition “to complicate fan 繁,” is needed only for the sake of proving the correctness of algorithms dealing with fractions. At the beginning of his commentary on the procedure for “gathering parts,” Liu Hui, then, considers the counterpart of these operations with respect to the fractions regarded as parts: “simplified” fractions correspond to “coarser parts” (halves instead of fourths), “complicated” ones to “finer” parts (eighths instead of fourths). At this level, Liu Hui again stresses the invariability of the quantity, beyond possible variations in the way of composing it (using halves, fourths or eighths). Now to prove that the commentator shows that the algorithm carries out the following program: by multiplication of both the numerator and the denominator, it refines the disparate parts to make them have the same size (a/b becoming ad/bd and c/d becoming cb/bd, so that all parts are bdths). Quoting the Canon, Liu Hui expounds the actual “meaning” of each step, in terms of both parts and numerators/denominators, making clear how they combine to fulfill the program outlined.
29 When “the denominators are multiplied by one another” —an operation that, in the course of the proof, Liu Hui names “to equalize”—, this computes the denominator common to all fractions (bd) and defines the size that the different parts can have in common: when parts have a common size, the fractions can be added. Moreover, when “the denominators multiply the numerators that do not correspond to them,” to yield ad and bc, the numerators, he says, are made homogeneous with the denominators to which they correspond. The overall operation is a “complication.” It has been previously shown to be valid and it ensures that the “original quantities are not lost.” Here, too, in passing, Liu Hui confers a name to this set of operations: “to homogenize.” “Equalizing” the denominators and “homogenizing” the numerators, the algorithm thus actually yields a correct measure of the quantity formed by gathering the various fractions. 29 As above, in the discussion of the commentary on Problem 9.9, the term “meaning” refers to what the operations carry out, with respect to the situation in which they are applied. One term in Liu Hui’s terminology can be interpreted to correspond to this concept: “meaning, intention yi 意,” see my glossary in [Chemla and Guo Shuchun, 2004, 1018-1022]. K. Chemla. p. 15 2.2. The correctness of algorithms and the search for fundamental operations We, contemporary readers, may read Liu Hui’s commentary on the procedure for “gathering parts” as establishing the correctness of an algorithm. But what were the aims pursued by the commentator when writing it? They are highlighted by the following part of this commentary, which continues with highly abstract and philosophical considerations, concluded by a key declaration: “Multiply to disaggregate them, simplify to assemble them, homogenize and equalize to make them communicate, how could those not be the key points of computations/mathematics (suan)?” [Chemla and Guo Shuchun, 2004, 158-159].
30 Opaque as it may seem, this declaration is essential. It clearly shows that something else is at stake in the previous proof besides establishing the correctness of a procedure. The proof had exhibited operations at play in the algorithm: multiplying numerators and denominators to disaggregate the parts they represent, dividing them to assemble the parts into coarser parts, equalizing denominators, homogenizing numerators. These operations constitute the topic of the subsequent considerations: exhibiting them appears to be one motivation for carrying out the proof. Moreover, all these operations were introduced in relation to fractions, for which they referred to precise operations on numerators and denominators. However —and this is the point difficult to understand—, Liu Hui’s concluding declaration indicates that their relevance far exceeds this limited context, since they are now listed among the “key points of mathematics.” How can we interpret this claim? In fact, when we read Liu Hui’s commentary as a whole, we observe that these operations recur in several other proofs that the commentator formulates to establish the correctness of other procedures described in the Canon [Chemla, 1997b]. Let us allude to an example.
31 Chapter 8 of the Canon is devoted to solving systems of simultaneous linear equations. If we represent a system by the equations: the fundamental algorithm given in The Nine Chapters amounts to transforming them into Then, by subtraction of the two equations, one eliminates x and determines the value of y, after which the value of x is easily obtained. When Liu Hui accounts for the correctness of the algorithm, he brings to light that one can multiply and divide the coefficients of an 30 In the English translation of The Nine Chapters by Shen Kangshen et al., various critical parts of the work are not translated accurately and their importance is hence overlooked. For the passage considered, compare the translation by Shen Kangshen et al.: “Multiplying [the denominators] means fine division and reducing means rough division; the rules of homogenizing and uniformizing are used to get a common denominator. Are they not the key rules of arithmetic?” [Shen Kangshen, et al., 1999, 72]. In addition to inaccuracies (such as, for instance, translating “assemble” by “rough division”), the theoretical import and generality of the statement, which is one of the most important of the commentary, is completely missed. The consequences should appear clearly with what is explained below, in this section and in the following one. 31 For the purpose of clarity the example has been simplified in its detail: cf. [Chemla, 2000]. In Section 3, I shall describe with greater detail how “equalizing” and “homogenizing” occur in another context. K. Chemla. p. 16 equation by the same number without altering the relationship it expresses between the unknowns. Moreover, he points out that the algorithm “equalizes” the values of the coefficients of x, whereas it “homogenizes” the values of the other coefficients. This is how all the operations included in the key declaration quoted above occur again in the proof of the correctness of another algorithm. The same holds true for other cases, as we shall shortly see. This fact explains why Liu Hui’s declaration can be so general and why he makes a statement, the validity of which goes beyond the context of fractions. However, if we compare the two situations alluded to, in which Liu Hui identifies “equalization,” clearly in the context of equations we do not have an “equalization of denominators,” since what is “equalized” is the coefficients of x. In other words, what is common between the two contexts is not the concrete meaning that “equalization” takes, although in each context the specific concrete meaning of “equalization” is what is required for the proof to work. The declaration invites us to find something else that is common to the various contexts in which the operations identified occur and which would justify its validity. This leads us to note that in each proof in which they occur, the terms designating the operations have in fact two meanings. Let us explain this point for the term “equalization.” In the context of adding up fractions, equalization was interpreted as the operation equalizing denominators. In that of equations, it was interpreted as the operation that made the coefficients of x equal. Similarly, in all the other contexts in which “equalizing” occurs, it can be interpreted in terms of its effect with respect to the particular situation to which it is being applied. This material effect constitutes the first meaning of “equalization,” one that changes according to the context. However, the fact that the operation recurs in different contexts reveals that the term takes a second meaning, a formal one that is common to all contexts: the term “equalizing” points to how the algorithms work. All algorithms for which the proof of correctness highlights that “equalizing” and “homogenizing” are at play proceed by “equalizing” some quantities while “homogenizing” others. The second meaning of the two terms captures and expresses the strategy followed by the procedure. And the parallel between the proofs discloses that, in fact, the algorithms follow the same formal strategy of equalizing and homogenizing. Even though the concrete meanings of equalizing and homogenizing vary according to the contexts in which they are at play, formally, they operate in the same way. These remarks reveal a key feature shared by the proofs: They bring to light that the same fundamental algorithm underlies various procedures. It is on the basis of the actual reasons accounting for the correctness of the algorithms that, through the proofs, a concealed formal connection between them is unveiled. This conclusion reveals that while proving the correctness of an algorithm, the commentator concentrates on formal dimensions in the procedure.
32 This concern relates to one of the reasons for Liu Hui to carry out proofs, i.e., bringing to light such fundamental formal strategies common to the various procedures provided by The Nine Chapters.
33 Such key algorithms, such as “equalizing/homogenizing,” allow a reduction of the variety of procedures of the Canon, uncovering a small number of strategies systematically used in designing all its procedures. Multiplying or dividing all numbers in an adequate set as well as equalizing and homogenizing appear to underlie many of the algorithms of the Canon in domains that for us belong to arithmetic or algebra. This is why, when they first occur, in the context of the procedure for “gathering parts,” Liu Hui 32 We could capture this point in the way in which the commentary unfolds. Several hints indicate that the authors of The Nine Chapters also considered procedures from a similar perspective. However, dealing with this issue would exceed the scope of this article. 33 Such a motivation appears to be driving later commentators as well [Chemla, 2001 (forthcoming), Chemla, 2003b]. K. Chemla. p. 17 immediately stresses their importance. Thus his declaration appears to gather together the most fundamental algorithms underlying the procedures of The Nine Chapters, those procedures being brought to light by the proofs contained in his commentary. 2.3. A fundamental operation in geometry The same motivation of disclosing fundamental operations common to various algorithms appears to permeate the proofs that Liu Hui develops in the context of geometry. This can be deduced from the fact that the proofs establishing the correctness of the most important algorithms related to geometrical shapes all bring to light that these algorithms use the same formal strategy, which Liu Hui captures with the expression “one uses the excess to fill up the void.” 34 Let us illustrate this point with the example of the trapezoid prism (see Figure 3 —again no figure is to be found for this problem either in the commentary or in The Nine Chapters). This solid is the first one considered in Chapter 5, in which most problems of that type are gathered. Figure 3 The procedure given in The Nine Chapters to compute the volume of solids of this shape reads as follows: Procedure: one adds the upper and lower widths and halves this (the result). One multiplies this (the result) by the height or the depth. Again, one multiplies this (the result) by the length, hence the chi of the volume 術曰：并上下廣而半之。以高若深乘之，又以袤乘之，即積尺。[Chemla and Guo Shuchun, 2004, 410-413] To establish its correctness Liu Hui writes: In this procedure, the reason why “one adds the upper and lower widths and halves this” is that if one uses the excess to fill up the void, this yields the average width. “Multiplying this (the result) by the height or the depth” yields the erected surface of a front. The reason why “again, one multiplies this (the result) by the length” is that it yields the volume corresponding to the solid, this is why this makes “the chi of the volume” 此術并上下廣而半之者，以盈補虛，得中平之廣。以高若深乘之，得一頭之立冪 。又以袤乘之者，得立實之積，故爲積尺”。(My emphasis) [Chemla and Guo Shuchun, 2004, 412-413] Given the position in which I have drawn the trapezoid prism, what Liu Hui calls a front is represented in the upper and lower planes. The first steps of the procedure are interpreted as computing the area of a face, by means of its transformation into a rectangle. With the expression “one uses the excess to fill up the void,” Liu Hui indicates the concrete 34 This point was first discussed in great detail in [WU Wenjun吳文俊, 1982]. See further developments in [Volkov, 1994]. 35 For reasons that will be presented in Section 4, I have drawn the solid in a position different from the one the procedure refers to. What corresponds to the “lower” and “upper width” are shown on the figure as the front and rear width. The following argumentation is not affected by this rotation. K. Chemla. p. 18 transformation of the solid that is at the basis of the proof (it is illustrated in Figure 3) and that allows him to interpret the “meaning” of the successive steps of the procedure. In addition, most importantly for us here, the commentator refers to this transformation by the same sentence that he uses to designate other different concrete transformations that make the proof work in other geometrical contexts. In each context, the actual transformations differ. However, the recurrence of the same formula to refer to them reveals that viewed from a certain angle, they are formally the same. This conclusion confirms what we have seen with “multiplying,” “dividing,” “equalizing,” and “homogenizing”: the proof again appears as a means for bringing to light formal patterns that are common to various algorithms despite the apparent difference between them that derives from the fact that they prescribe different computations.
36 How does the proof fulfill the function of revealing such formal patterns? The example of the procedure for “multiplying parts,” which we will analyze in the following section, highlights that the problems play a key part in enabling the proof to fulfill this function and, in this case, disclose the hidden action of “equalizing” and “homogenizing” in its process. It is from this perspective that we can now go back to our first puzzle and offer a solution for it. 3. The situation in the statement of a problem as a condition for exhibiting formal strategies As has already been mentioned, Problem 1.19, after which The Nine Chapters states the procedure for “multiplying parts,” requires computing the area of a rectangular field, 3/5 bu long and 4/7 bu wide. However, the procedure itself is formulated without reference to any concrete situation. Liu Hui’s commentary on the procedure provides key evidence for understanding the part played by problems for proofs to fulfill the function brought to light in the previous section. Let us analyze it in greater detail. 3.1. The first proof of the correctness of the procedure for multiplying fractions In the first part of his commentary on this procedure, Liu Hui develops abstract reasoning to account for its correctness. This argument shows one way in which multiplication and division are at play in the design of the procedure. Its opening section can be translated as follows: In each of the cases when a dividend does not fill up a divisor, they hence have the name of numerator and denominator. If there are parts (i.e. a fraction), and if, when 36 For the sake of clarity, we opposed the first set of general operations, presented in Liu Hui’s key declaration, to the transformation “one uses the excess to fill up the void,” which occurs only in relation to geometrical shapes. Most probably, this type of transformation was conceived of as one of the general patterns with which the operations identified in the key declaration could take shape. 37 This technical expression refers to the case when the dividend is smaller than the divisor. Note that “dividend” designates the content of a position on the calculating surface, and not a determined number —such a way of employing terms corresponds to the assignment of variables, whose use for the description of algorithms is characteristic in ancient China, in contrast to other ancient traditions. As a result, in what follows, the word “dividend” will designate different values, depending on the operations that have been applied to the value put in the position at each step of the computation. I have respected this technical use of terms in the translation. 38 In such cases, the result of the division is the fraction whose numerator and denominator are respectively the dividend and the divisor. “Numerator” and “denominator” refer to the numbers as constituting a fraction; “dividend” and “divisor” refer to them as the terms of the operation yielding the fraction. The K. Chemla. p. 19 expanding the corresponding dividend by multiplication, then, correlatively, it (i.e., the dividend produced by the multiplication) fills up the divisor, the (division) hence only yields an integer. If, furthermore, one multiplies something by the numerator, the denominator must consequently divide (the product) in return (baochu). “Dividing in return,” this is dividing the dividend by the divisor. 凡實不滿法者而有母、子之名。若有分，以乘其實而長之，則亦滿法，乃爲全耳 。又以子有所乘，故母當報除。報除者，實如法而一也。(…)。 [Chemla and Guo Shuchun, 2004, 170-171, 768, footnote 176] Before translating the end of the argument, let us explain the meaning of some of the technical expressions. The expression of “dividing in return” (baochu) is particularly important to note. The commentator introduces it here for the first time in the whole text (Canon and commentaries). Seen as an operation, as Liu Hui makes it clear, it consists in a division. However, the qualification “in return” adds something to the prescription of a division: it makes explicit the reason for dividing. Using “dividing in return” means that it was needed, for some reason, earlier in the procedure, to carry out a multiplication, which was superfluous with regard to the sought-for result: this division compensates for the earlier multiplication, deleting its effect. This general idea clearly makes sense in the passage of Liu Hui’s commentary translated above. The use of the technical term indicates that multiplying by the numerator is to be interpreted as follows: instead of multiplying “something” by a fraction a/b, one multiplies the “something” by its numerator a. Multiplying the fraction a/b by b, one obtains the numerator a. Having multiplied the “something” by the numerator a, instead of a/b, one has multiplied by a value that was b times what was desired. Consequently, one has to “divide in return” by that with which one had superfluously multiplied, that is, by b. Multiplying something by a/b is hence shown to be the same as multiplying by a alone, and dividing the result by b. Most importantly for our purpose, “dividing in return” is one of the qualifications of division that one can find in The Nine Chapters itself too. This fact calls for two remarks. First, the expression clearly adheres to the sphere of justification, since the very prescription of the division indicates a reason for carrying it out. In other words, such hints prove that The Nine Chapters does refer to arguments supporting the correctness of algorithms in some specific ways. Second, we can find the expression both in The Nine Chapters, and in the commentaries. This fact points to continuities between the two, which will be useful in our argumentation below. The continuation of Liu Hui’s first proof of the correctness of the procedure for multiplying fractions interprets the meaning of the procedure as follows: Now, “the numerators are multiplied by one another,” 40 the denominators must hence each divide in return. 今子相乘則母各當報除。 [Chemla and Guo Shuchun, 2004, 170-171] For each of the numerators of the fractions to be multiplied, the argument developed above applies. If one multiplies the numerators instead of multiplying the fractions, one must divide the product by each of the denominators. The last sentence of Liu Hui’s first proof transforms the sequence of operations just established to carry out the multiplication of fractions (multiplication of the numerators, division by a denominator, division by the other commentary alternately uses the two sets of terms, with the greatest precision. 39 If the division of a by b yields a/b, then, if a is multiplied by kb, the new dividend akb divided by b yields the integer ak. 40 This is a quotation from the procedure of the Canon. K. Chemla. p. 20 denominator) into the procedure given in the Canon, as follows: Consequently, one makes the “denominators multiply one another” and one divides altogether (lianchu) (by their product). 因令分母相乘而連除也。 [Chemla and Guo Shuchun, 2004, 170-171] 41 Transforming a sequence of two divisions into a unique division by the product of the divisors is a valid transformation because the results of division are exact. Liu Hui proves to be aware of the link between the two facts, that is, between the validity of the transformation and the exactness of the results of division [Chemla, 1997/1998]. As above, the formulation “to divide altogether” prescribes a division in a way that indicates the reason why the division prescribed is to be carried out in this particular way. This expression recurs regularly in Liu Hui’s commentary. The qualification adheres to the sphere of justification. In contrast to the “division in return,” however, the expression lianchu never occurs in The Nine Chapters itself. This sentence concludes Liu Hui’s first proof of the correctness of the procedure. One can see how the proof discloses that multiplication and division, two of the fundamental operations listed in Liu Hui’s declaration, are put into play, as opposed operations, for designing the procedure as it stands. For our purpose, it is interesting to note that this proof of the correctness of the algorithm develops independently of the framework of the problems in the context of which the “procedure for multiplying parts” was formulated, that is, the problems about rectangular fields. The arguments only make use of general properties of dividend and divisor, numerator and denominator. Moreover, they bring into play the properties of multiplication and division with respect to each other. However, Liu Hui does not end here his commentary on this procedure. He goes on to develop a second proof, which highlights how, seen from another angle, this procedure also puts into play “equalizing” and “homogenizing.” This is where we go back to the first puzzle that we have presented. 3.2. The second proof of the correctness of the procedure for multiplying fractions The sentence linking the two parts of the commentary is important for our purpose. Liu Hui states: If here one makes use of the formulation of a field with length and width, it is difficult to have (the procedure) be understood with a greater generality nan yi guang yu 此田有廣從，難以廣諭。 [Chemla and Guo Shuchun, 2004, 170-171] 41 Shen Kangshen et al. [1999, 82] translate the whole passage as follows: “When the dividend is smaller than the divisor, then [one] gets a [proper] fraction. When the numerator is multiplied by the dividend, the product may be larger than the denominator (divisor), thus yielding an integer. If the numerator is multiplied, the product should be divided by the denominator. Since the product of [all the] numerators is taken as dividend, it should be divided by the product of [all the] denominators, i.e., take the continued product of the denominators as divisor.” I do not see what, in the second sentence, “multiplying the numerator by the dividend” may mean. Moreover, there is no word for “numerator” in the text at this point. Later, “if the numerator is multiplied” does not conform to the syntax of the Chinese and the translation expresses a meaning unclear to me, leaving aside (as below) the key word “divide in return.” In correlation with this, the following sentence is not translated. The transformation of two divisions into a division by the product is hidden by the addition, in many places, of the term “product,” which is not in the original text. The mathematical explanation given in footnote 1 (p. 83) does not seem to me to fit with the meaning of the text. It takes the commentary as distinguishing various cases in the multiplication and fails to read the argument that Liu Hui makes in it. K. Chemla. p. 21 In fact, this sentence justifies why the commentator discards Problem 1.19 in the context of which the procedure for “multiplying parts” is described in The Nine Chapters and why he introduces, immediately thereafter, another problem equivalent to the original one: 1 horse is worth 3/5 jin of gold. If a person sells 4/7 horse, how much does the person get? Let us analyze more closely the assertion introducing the alternative proof that the commentator develops in the sequel. Since it is a key assertion for my argument, I shall insist on it and distinguish all the facts that it reveals. First, Liu Hui’s above statement establishes a link between the context of a problem — here the problem provided by The Nine Chapters, i.e. that of a rectangular field— and the aim of understanding the procedure. It appears here that a given problem can prevent one from gaining a wider understanding of the procedure. Second, this seems to indicate that Liu Hui also perceives the preceding passage as contributing to an understanding of the procedure. This link between establishing the correctness of a procedure and understanding it is not fortuitous. It will recur again in the passage of the commentary following the above statement. Now, the change of problem is justified by the attempt to gain a more general understanding of the procedure. Strikingly enough, if we bring all these facts together, the statement indicates a possible link between the context of a problem and the proof of the correctness of a procedure. Again, this link will be confirmed in what follows. In fact, Liu Hui introduces next a sequence of three new problems, all formulated with respect to a single kind of situation, which make it possible to develop the second proof of the correctness of the procedure conforming to the practice of demonstration sketched in Section 2. Let us stress right at the outset the essential consequence that can be derived from both the above statement and the analysis developed below: problems appear here not to be reduced to questions that require a solution —as we would readily assume—, but they also play a part in proofs. Observing Liu Hui’s next development should hence allow us to progress on two fronts. It should provide evidence showing how problems intervene in proving the correctness of a procedure. And this is where the difference between the problem with the field and those with the persons, the horses and the gold should become apparent. Moreover, it should help us grasp how Liu Hui conceives of “understanding a procedure.” Let us first indicate, in modern terms, the main idea of the second proof of the correctness of the procedure. It can be represented by the following sequence of expressions: The second proof brings to light that the computations of ac and bd, with which the procedure for “multiplying parts” begins, have in fact the meaning of “homogenizations.” This meaning can be seized if the “equalizing” of bc, which is essential for disclosing the pattern, is revealed. These are the key points of the second argument. We shall examine below how Liu Hui puts them into play in providing his second proof. In this, an essential part will be played by an operation the importance of which was already stressed with respect to proofs: interpreting the “meaning” (yi) of the computations prescribed by an algorithm, that is, expressing the “meaning” of their results —their “semantics”— in terms of the situation described by the problem. This remark enables us to infer why the first problem cannot be used here. The situation of the field with length and width is unsuitable from a semantic point of view: it does not offer possibilities of interpretation rich enough for the “meaning” of the computation of bc to be expressed in a natural way. As a consequence, the problem of computing the area of a rectangular field does not allow bringing to light the “equalization” that underlies the procedure for “multiplying parts.” The scheme of “equalizing” and “homogenizing” cannot be unfolded in this context. K. Chemla. p. 22 In contrast, the situation in which the sequence of three new problems is formulated, with the three components constituted by the quantities of gold, persons, and horses, offers richer possibilities for interpreting the effects of operations. It is semantically rich enough for disclosing that the scheme of “homogenizing” and “equalizing” underlies the procedure for “multiplying parts.” Consequently, the new situation allows revealing in another way how the procedure for “multiplying parts” relates to the fundamental operations identified by Liu Hui and discussed in Section 2. This is what is at stake in the second proof. The key point here is that the interest in bringing to light how equalization is at play in “multiplying parts” belongs only to the sphere of proving. The equalization plays no role in the actual computation of the result. This is why the problem in the context of which the algorithm is described differs from the problems in relation to which the proof is carried out. A link is thereby established between bringing to light a formal strategy accounting for the correctness of the procedure and interpreting the “intentions,” the “meanings” (yi) of the operations in the field of interpretation offered by the situation of a problem. The example of the multiplication of fractions, in which the description of the algorithm and the development of its proof require a sequence of different problems, reveals the essential part played by problems in establishing the correctness of algorithms. Note, however, that the numerical values are common to all problems: Liu Hui introduced the new problem mentioned above in such a way that the task to be carried out is still to multiply 3/5 by 4/7. As a consequence, the computations leading to a solution of the problem included in The Nine Chapters (3 times 4, 5 times 7, dividing the former result by the latter) are an actual subset of the operations involved in proving the correctness of the procedure for “multiplying parts” (which furthermore includes computing 4 times 5). With these observations in mind, let us examine in detail how Liu Hui uses problems to conduct his second proof. 3.3. The part played by mathematical problems in a proof Liu Hui’s second proof of the correctness of the procedure for “multiplying parts” consists in articulating, in an adequate way, a sequence of equivalent problems that are transformed one into the other. What would be done in modern mathematics by formal computations is here carried out by interpreting the results of operations with respect to a succession of problems. In the alternative proof, Liu Hui’s first step consists in formulating a first problem, the solution of which requires use of the last operation of the procedure for “multiplying parts,” i.e., dividing 12 (ac) by 35 (bd). It reads as follows: Suppose that one asks: 20 (bc) horses are worth 12 (ac) jin of gold. If one sells 20 (bc) horses and if 35 (bd) persons share this [the gain], how much does a person get? Answer: 12/35 jin. To solve it, one must follow the procedure for “directly sharing (jingfen 經分)” and take 12 jin of gold as dividend and 35 persons as divisor. 設有問者曰，馬二十匹，直金十二斤。今賣馬二十匹，三十五人分之，人得幾何 。荅曰：三十五分斤之十二。其爲之也，當如經分術，以十二斤金爲實，三十五 人爲法。[Chemla and Guo Shuchun, 2004, 170-171] 42 This procedure, which is described in The Nine Chapters just before the procedure for “multiplying parts,” covers all possible cases of division involving integers and fractions [Chemla, 1992a]. Note that division is dealt with before multiplication and that, in correlation with this fact, the exegete makes use of the procedure for division in his commentary on multiplication. To make my argument clearer, I have inserted in the translation modern symbolic expressions between parentheses. K. Chemla. p. 23 The fact that the procedure given to solve this problem is correct was established by Liu Hui in his commentary on the preceding section of The Nine Chapters, devoted to “sharing parts.” This problem is immediately followed by another problem, presented as a transformation of the former one: Suppose that, modifying (the problem), one says: 5 (b) horses are worth 3 (a) jin of gold. If one sells 4 (c) horses and if 7 (d) persons share this (the gain), how much does a person get? Answer: 12/35 jin. 設更言馬五匹，直金三斤。今賣四匹，七人分之，人得幾何。荅曰：人得三十五 分斤之十二。 [Chemla and Guo Shuchun, 2004, 170-171] The key point here is that the procedure for solving this second problem is first described in terms of “homogenizing”: To do it, one has to homogenize these quantities of gold (ac) and persons (bd). They then all conform to the first problem and this is solved by the (procedure for) “directly sharing.” 其爲之也，當齊其金、人之數，皆合初問入於經分矣。[Chemla and Guo Shuchun, 2004, 170-171] The use of the term “homogenize” implies that, in parallel, an equalization is carried out. Only the operation of equalization can confer the meaning of “homogenization” to the other operations. Liu Hui will make this point explicitly in one of the subsequent sentences of his proof. “Homogenizing” quantities of gold and persons goes along with “equalizing” quantities of horses. This is why, as Liu Hui states, the homogenizations prescribed lead to the “first problem”: indeed, they yield the two following statements, in which we recognize the first problem: 20 (bc) horses are worth 12 (ac) jin of gold. If one sells 20 (bc) horses and if 35(bd) persons share [the gain], how much does a person get? This transformation of the second problem into the first one involves computing ac and bd, which amount to the first part of the procedure for “multiplying parts.” The transformation is clearly correct: it does not alter the meaning of the relationship between the values considered. The procedure solving the first problem can then conclude the solution of the second one. Now, if one considers the whole procedure that correctly solves the second problem, through transforming it into the first one, one realizes that the procedure for “multiplying parts” is embedded in it. What makes the difference between the two is that the procedure for “multiplying parts” does not prescribe the computation of bc. However, if we pay a closer attention to the way in which Liu Hui formulates the transformation of the second problem into the first one, we observe that there, too, he evokes the computation of bc only in an allusive way, by referring to the computation of ac and bd as “homogenizations,” and by stating that the second problem is reduced to the first one. In fact, the sequence of operations that is concretely given to solve the second problem corresponds exactly to the procedure given by The Nine Chapters for “multiplying parts.” To recapitulate, Liu Hui describes the procedure that correctly solves the second problem as being the list of operations that constitute the procedure for “multiplying parts.” While stressing in his following statements 43 Shen Kangshen et al. translate this sentence as follows: “By the Homogenization and Uniformization Rule one can get the same answer as by the rule of division.” [Shen Kangshen, et al., 1999, 83], which does not fit with the Chinese. Again, the argument developed by Liu Hui cannot be understood from the translation. K. Chemla. p. 24 the identity between the two procedures step by step, most importantly, Liu Hui transfers the interpretation in terms of homogenization and equalization into the procedure for “multiplying parts.” He writes: If this is so, “multiplying the numerators by one another to make the dividend (ac)” is like homogenizing the corresponding gold. “Multiplying the denominators by one another to make the divisor (bd)” is like homogenizing the corresponding persons. Equalizing the corresponding denominators makes 20 (bc). But the fact that the horses be equalized plays no role. One only wants to find the homogenized (quantities) and this is all. 然則分子相乘爲實者，猶齊其金也；母相乘爲法者，猶齊其人也。同其母爲二十 ，馬無事於同，但欲求齊而已。 [Chemla and Guo Shuchun, 2004, 170-171] At this point, it is established that, on the one hand, the procedure for “multiplying parts” solves the second problem correctly and, on the other hand, the procedure involves only homogenizations. Liu Hui devotes the following statement to considering the ambiguous status of “equalization”. The computation of bc as the equal quantity in the statement of the first problem is essential to ascertain that ac and bd correctly correspond to each other in a pattern similar to that of the first problem. In other words, it is essential for the proof. But this computation is useless in obtaining the result: once one knows the reason why ac and bd correspond to each other, it suffices to divide one by the other to yield the solution. This explains why Liu Hui emphasizes that the “equalization” plays no part in the procedure itself, except for allowing an interpretation of its first steps as “homogenizations.” In turn, the interpretation of these steps as “homogenizations” is what lies at the basis of the correctness of the procedure. The key point to note here is that exhibiting the homogenizations and equalizations can only be done within the framework of the new situation with persons, gold and horses, in which the equalization can be interpreted and thereby brought to light, and not within the framework of computing the area of a rectangular field. Liu Hui’s last step is to show that, as far as the question raised is concerned, the second problem is equivalent to a third problem, itself strictly identical to Problem 1.19. The idea is the following: we now know that the procedure for “multiplying parts” correctly solves the second problem. We want to show that this procedure correctly solves 1.19. The step to be taken is to show that the two problems amount to the same. Liu Hui establishes the equivalence between the two by showing the equivalence between the second problem and one strictly identical to Problem 1.19, but formulated in terms of horses, gold and persons. He states: Moreover, that 5 horses are worth 3 jin of gold, these are lü in integers. If one expresses them in parts, then this makes 1 horse worth 3/5 jin of gold. That 7 persons sell 4 horses is that 1 person sells 4/7 horse. 又，馬五匹，直金三斤，完全之率；分而言之，则爲一匹直金五分斤之三。七人 賣四馬，一人賣七分馬之四。 [Chemla and Guo Shuchun, 2004, 170-171] Qualifying as lü the data in each of the two statements of the quote designates the ability of the pairs to be possibly multiplied or divided by a common number, without altering the meaning of the relationship [Guo Shuchun郭書春, 1984b]. If we use this property, as Liu Hui suggests, to transform the outline of the second problem into an equivalent problem, we obtain the third problem, the one with which we have formulated our first puzzle: 1 horse is worth 3/5 (a/b) jin of gold. If one sells 4/7 (c/d) horse, how much does the K. Chemla. p. 25 person get? This quotation brings to a close the passage of Liu Hui’s commentary on “multiplying parts” that we needed to analyze to solve our first puzzle and to account for why Liu Hui substituted the third problem for Problem 1.19. We now return to the two questions we have raised in our analysis above. First, how can we qualify the understanding of the procedure for “multiplying parts” yielded by Liu Hui’s commentary here? In fact, as was the case for the procedure for “gathering parts,” Liu Hui proves the correctness of the procedure by bringing to light how the “procedure of homogenizing and equalizing” underlies it. As we have emphasized above, the “meanings” of “equalizing” and “homogenizing” differ in the two contexts. In the procedure for “gathering parts,” “equalizing” meant equalizing the denominators. Here, “equalizing” refers to the fact that a denominator and a numerator are made equal so that the procedure can fulfill its task. Even though the concrete meanings differ, the procedures use the same strategy formally: in both contexts, it is by making some quantities equal and then making other quantities homogeneous that one can obtain the solution. This is what the proofs disclose. This remark indicates yet again the sense in which, in Lu Hui’s eyes, the fundamental operations he listed can be deemed fundamental for mathematics. Each proof sheds a different light on the procedure. Through highlighting how the first operations prescribed by the procedure for “multiplying parts” amount to homogenizing some quantities, the second proof discloses a new meaning for the algorithm, that is, another way of conceiving its formal strategy.
44 The new problems introduced appear to be an essential condition for formulating the new understanding, based on the pattern of homogenizing and equalizing. This is because they allow the disclosure of the part played by “equalization” by providing the means to interpret the meaning of the operation. It is in this way that we can interpret Liu Hui’s introductory statement, where he claims that he needs to discard the problem of the area of the rectangular field in order to “have (the procedure) be understood with a greater generality.” The second proof requires introducing a procedure —the procedure of homogenization and equalization— in which the procedure for “multiplying parts” is embedded. This brings us back to the remark I made in Section 1, on the commentary Liu Hui devoted to Problem 1.36 and the procedure for the area of the circular segment: proofs of the correctness of procedures regularly include establishing, or formulating, new procedures. For multiplying fractions, the interpretation of the operations of a procedure developed for the sake of the proof required the introduction of a new problem. In the case of the commentary on the topic of Problem 1.36, in order to formulate his new procedure, Liu Hui introduces new elements in the figure of the circular segment: the circle from which it derives, and then triangles tiling its surface. These elements are in fact the topics of other problems in The Nine Chapters and are thus associated with procedures that the commentator uses to establish his new procedure for the area of the circular segment. In particular, Liu Hui employs these new elements to express the “meaning” of the steps of the new procedure. We hence see here a parallel emerging between the uses of, on the one hand, geometrical figures and, on the other hand, problems. In what follows, we shall analyze further this parallel between problems and geometrical elements in a figure. The distinction between the procedure required by the proof and the procedure for 44 The commentators use another term to designate this second type of “meaning.” To avoid confusion with the first one, I transcribe it as yi’ 義. See the corresponding entry in my glossary in [Chemla and Guo Shuchun, 2004, 1022-1023]. K. Chemla. p. 26 solving a problem is important for explaining the substitution of one problem for the other in the case of the multiplication of fractions. The difference between the new problem and the one described in The Nine Chapters now appears precisely to lie in the fact that the procedure developed within the context of the proof can be “interpreted” only with respect to the new problem. The key point is that the operation of “equalization,” which is part of the procedure needed by the proof, is not needed by the procedure that computes the result. In correlation with this, the new problem, and not the one in The Nine Chapters, allows an interpretation of the “meaning yi” of the procedure. These remarks highlight how problems play a key part in Liu Hui’s practice of proof. They account for the fact that problems that appear to be the same to us are in fact different for him. This is how I suggest that our first puzzle can be solved. Its solution reveals that, in the ancient Chinese mathematical practice to which these documents bear witness, problems did not boil down to being statements requiring a solution, but were also used as providing a situation in which the semantics of the operations used by an algorithm could be formulated in order to establish its correctness. This second function becomes apparent when the problem fulfilling the first function cannot fulfill the second. This aspect appears to be an essential component of the practice of proof as carried out by the commentators. What is most interesting, furthermore, as the case discussed here shows, is that the interpretation of the operations with respect to the situation described to formulate a problem is used to carry out the proof and thereby disclose the latter’s most formal dimensions. Second, the previous development allows us to go back to the question of determining the class of problems for which a particular problem stands. In Section 1, we have shown that a problem stands for a class of problems sharing the same category. But this conclusion seemed to be contradicted by the need to replace the problem about the area of a rectangular field by a problem with horses, gold and persons, both problems sharing exactly the same particular values: the two problems looked identical to each other. In fact, there is no contradiction if we recall that the category (lei) associated to a problem is defined by examining the procedure attached to it. With respect to the procedure for “multiplying parts,” these problems belong to the same category, but, with respect to the procedure developed for the sake of the proof, they can no longer be substituted one for the other. However, what is essential is that all problems similar to the one with horses, gold, and persons, in which the procedure of the proof, i.e., equalization and homogenization, can be unfolded share the same category. Consequently, even though the proof is discussed in terms of horse, gold and person, with particular values, it is meant to be read as general.
45 This conclusion holds true for a procedure as well as for a proof: here as above, when the problem of the piece sawn was used within the context of the field with the shape of a circular segment, the general is discussed in terms of the particular. This conclusion cannot be underestimated. Proofs developed by the commentators were regularly written down in ways that conflict with our expectations, since they seem to be formulated only with respect to particular cases. As a result, they were interpreted as lacking generality. Such an interpretation is in my view anachronistic, because it reads our sources in ways that project our modes of interpretation onto documents that require other readings. I think that the previous discussion establishes why they must be read as expressing general arguments. 45 In his unpublished dissertation, [Wang Ling, 1956, 179-181] discusses the commentary on the multiplication of fractions in relation to the problem with horses, people and cash. He mentioned neither the change of problem, nor the pattern of “homogenizing”/“equalizing.” Moreover, his entire analysis develops in terms of “proportion,” which does not fit with the concepts in the text. However, his conclusion of the use of the context of a problem is worth quoting (p. 181): “Evidently, he [Liu Hui] chose these specific numbers, 3, 4, 5, 7 and problems about horses, money and persons in a representative sense. Thus we have here an example demonstrating all the logical steps to prove a rule.” K. Chemla. p. 27 The same conclusion can be drawn not only for the situation in which a problem is formulated, but also for the particular values it involves. Proofs relying on problems with particular values are also meant to be general in this sense: the example of the pyramid with a square basis (yangma) provides a clear illustration. We shall now use this example to establish this point. 4. The values in the statement of a problem: a parallel between problems and visual tools To analyze the relevance of the numerical values used in the statement of a problem, we shall use the same method as we did in the previous paragraph, in which we focused on changes in the situations used to formulate a problem. Here, we shall examine cases in which the commentator modifies the numerical values of a problem presented in The Nine Chapters, before he begins commenting on a given topic. In our discussion of Problem 6.18 in Section 1, we have seen that Liu Hui changed the values of the problem to expose the lack of generality of the procedure given by The Nine Chapters. This use employs problems as counterexamples, in a fashion quite common today. The change of numerical values in Problem 5.15, which deals with the volume of the pyramid yangma, already mentioned in the Introduction, is more difficult to account for. We will now examine the context in which it is carried out and the way it is used. 4.1. The proofs for the singular and the general cases Chapter 5 of The Nine Chapters deals with the determination of the volume of various kinds of solids. The type of pyramid called yangma 陽馬, in Problem 5.15, is represented on Figure 4. Also in this case, there is no diagram in the sources. However, Liu Hui’s commentary does refer to a visual tool of a different kind, namely “blocks qi 棊.” This is the only type of visual tool the commentators use for space geometry, whereas for plane geometry they use “figures tu 圖.” In the diagrams below, I have tried to picture only what the text says about blocks without adding more modern ways of dealing with visual aids. In particular, I have not added letters by which one could refer to the vertices of a block. Moreover, I have drawn some of the diagrams in such a way as to give a sense that the actual visual tools used were composed of solid pieces assembled in space. Figure 4: The pyramid called Yangma The procedure stated by The Nine Chapters for solving Problem 5.15 prescribes multiplying all three dimensions (“width guang 廣,” “length zong 從” and “height gao 高”) by each other and then dividing the result by 3 to yield the volume. In his commentary Liu Hui sets out to establish the correctness of this very algorithm.
46 The beginning of his proof is 46 This commentary was the topic of quite a few publications, among which: [Li Yan, 1958, 53-54] (I could not consult the first edition of this book, but the second sketches the essential points of the proof), [Wagner, 1979], and [Guo Shuchun郭書春, 1984a, 51-53, Li Jimin李繼閔, 1990, 295-303]. See also [Chemla, 1992b], [Chemla and Guo Shuchun, 2004, 396-398, 428-433, 820-824], in which a French translation is provided. Note that [Wagner, 1979] provides a full translation of the commentary into English. I refer the reader K. Chemla. p. 28 the most important part for us here. To present his proof, as we have seen in the Introduction, Liu Hui first suggests modifying the values in the problem by considering a yangma with dimensions (width, length, height) equal to 1 chi. He then remarks that three such yangma form a cube, as is shown in Figure 5. Figure 5: Three yangma 陽馬 with equal dimensions form a cube It follows immediately that the volume of the yangma is one third of the volume of the cube with the same dimensions, which is obtained by multiplying its length, width and height by one another. However, Liu Hui notes —and this is the key point for our purpose— that this reasoning does not extend to the general case, when the dimensions are not all equal. The reason for this is that when the dimensions are different, the three pyramids into which the parallelepiped with the same dimensions can be decomposed are not identical, as was the case for the cube. Thus from this decomposition, one cannot conclude that the volumes of the three pyramids are equal before the procedure under study has been proved correct. As a result, Liu Hui discards the first reasoning and starts a second one, general and much more complex. So far, one may think that the change of the three numerical values of Problem 5.15 to 1 chi was intended to focus the reader’s attention on a specific preliminary case. Surprisingly, in the general reasoning, instead of making use of the values of Problem 5.15, or of no values at all, Liu Hui again introduces three other values for the length, the width and the height, this time making each equal to 2 chi. Why does he change the values of the problem at all? Why does he change the values he has just introduced to new values that are also equal to each other? We shall sketch the first and main part of Liu Hui’s general proof —the only one important for the points we want to make— to find answers to these questions. For the second reasoning, Liu Hui considers simultaneously a yangma and a specific tetrahedron named bienao 鼈臑, which together form a half-parallelepiped called qiandu 壍堵 (see Figure 6). Both solids, the qiandu and the bienao, are the topics of problems of The Nine Chapters itself (5.14 and 5.16, respectively). On the basis of the qiandu thus formed, Liu Hui sets himself a new goal: establishing that the yangma occupies twice as much volume in the qiandu than the bienao. From this property, the correctness of the procedure for computing the volume of the pyramid can be derived immediately. Figure 6: A yangma and a bienao form a qiandu (half-parallelepiped) However, we do not need to examine this part of the reasoning here, since to make our points, it suffices to evoke the beginning of Liu Hui’s reasoning whereby he establishes the proposition that is his new target. His first step consists in forming the bienao and the yangma with dimensions all equal to 2 chi from blocks of the following types: parallelepiped, qiandu, yangma, bienao, all with dimensions of 1 chi. 47 Let us describe the composition of the yangma and the bienao to these publications for greater detail, concentrating here only on the features that are important for my argument. 47 The bienao is formed with vermillion blocks whereas black blocks are used for the yangma. But I will not mention the colors any further here, because they do not matter for the point I want to make. The reader interested is referred to [Chemla, 1994]. More generally, the reader is referred to the published translations of Liu Hui’s commentary on the yangma to get a more precise idea of the original text (see footnote 46). K. Chemla. p. 29 successively. The yangma is decomposed into several kinds of pieces, the dimensions of which are all half of the dimensions of the original body (see Figure 7). On top, and in front on the right hand side, we see two small yangma, similar to the one we started with. In the back, on the right hand side, and in front on the left, we see two small qiandu, whereas in the back, below and on the left hand side, we have a small parallelepiped. This is how, for space geometry, Liu Hui uses blocks to compose solids. The reasoning that follows is typical of Liu Hui’s reasonings of that kind. Figure 7: Halving the dimensions of the yangma pyramid decomposes the solid into unit blocks As for the bienao, the original body is composed of two kinds of blocks, the dimensions of which are all half of the original dimensions (see Figure 8). On top in the rear, and in front on the right hand side, we see two small bienao similar to the one we started with. In front, on the left hand side, on top of each other, we see two small qiandu. Figure 8: Halving the dimensions of the bienao tetrahedron decomposes the solid into unit blocks If we bring together the solids thus composed, we obtain a global decomposition of the original qiandu, as shown on Figure 9. I have drawn it with the blocks set apart to make easier the enumeration of the components (see Figure 10). However, it must be kept in mind that Liu Hui’s commentary refers throughout to the qiandu as a whole. Liu Hui considers separately two zones in the qiandu. The first zone is the space in it that is occupied by pieces similar to the original bienao and yangma, but the dimensions of which are all half of the original ones. We know that there are two such pieces of each kind. Within the qiandu, they form, two by two, smaller qiandu, one situated in the lower and front part, on the right, of Figure 9 (or 10, indifferently) and the other one situated on the upper and rear part of Figure 9 (or 10), on the left. Figure 9: The half-parallelepiped decomposed The second zone, in the space within the qiandu, is composed of smaller qiandu and a small parallelepiped. All these pieces may be oriented in different ways, but their three dimensions are uniformly half the length, half the width and half the height of the original qiandu. Liu Hui has already established that half a parallelepiped has a volume that is half the volume of the corresponding parallelepiped. The volume of these pieces is hence easy to compute. However, this is not the relevant feature in the situation —and Liu Hui does not mention it. The only useful information is that the volumes of all these half parallelepipeds are the same. But there are two other features that are essential. The first key feature derives from evaluating the relative occupation, in the second zone, of blocks coming from the yangma and blocks coming from the bienao. It turns out that in the second zone, there is twice as much space occupied by pieces coming from the yangma as there is space occupied by pieces coming from the bienao. The property sought for (i.e., the yangma occupies twice as much volume in the qiandu than the bienao) is hence established in the second zone. K. Chemla. p. 30 As for the first zone, it replicates, on a different scale, the situation we started with. The second key feature in the situation concerns the evaluation of the respective proportion of the space in which the situation is known and the space where it still needs to be clarified. Figure 10: The half parallelepiped decomposed, with the pieces set apart To determine this proportion, by making two cuts in the qiandu, Liu Hui removes the upper half-parallelepiped in the front and brings it closer to the similar half-parallelepiped in the lower rear part of the figure. In addition, he removes the rear upper half-parallelepiped, which is composed of a smaller bienao and a smaller yangma, and moves it near the similar one in the lower front part of the figure (see Figure 11). This rearrangement, which moves only two pieces —two smaller qiandu—, transforms the original qiandu into a parallelepiped, composed of four smaller parallelepipeds, the dimensions of which are all half of the original ones. Seen from this perspective, the second zone considered above —the one in which the property sought for is established— clearly appears to occupy 3/4 of the original body, whereas the first zone occupies only ¼ of it. These relative proportions ensure that when one repeats the same reasoning, 3/4 of the remaining ¼ of the original qiandu can be shown to hold the property, and so on. Figure 11: The original qiandu rearranged into four smaller qiandu, which in fact compose a parallelepiped 4.2. The proof is general, but it makes use of specific values To make the points I have in mind about the meaning of the numerical values appearing in the statements of problems, we need not discuss the final part of Liu Hui’s general proof, in which he establishes the proportion of 2 to 1 that he chose as his new goal to prove the correctness of the algorithm computing the volume of the yangma. We have seen enough of this proof to note how different it is from the proof for the special case. Yet both proofs are constructed using particular values, for which all the dimensions of the solids involved are equal. If Liu Hui was only aiming at making a proof valid for the special case — precisely the one he introduces by his change of values for the dimensions of the yangma—, he would use a simpler argument: this argument is simply the first argument he makes and then discards as not general. Two conclusions can be derived from these remarks. To begin with, when Liu Hui first changes the values of the dimensions of the yangma, his goal is not to introduce a simpler case for which a straightforward argument can be made to establish the correctness of the algorithm. Second, the second proof, conducted and presented within the framework of the simplest situation possible —that with dimensions all equal— is meant to be general. We are hence led to the same conclusion as we have drawn at the end of the previous section: with respect to the proof, the commentator discusses the general in terms of the particular, and even the most particular possible. The same conclusion holds true for the various problems and procedures, whether they are those from The Nine Chapters itself or those used within the K. Chemla. p. 31 context of Liu Hui’s proof.
48 In the case with gold, persons and horses, discussed in Section 3, as well as in the case of the yangma, the extension of the validity of the reasoning is determined on the basis of the operations put into play. We meet again with a parallel between problems and visual tools, based on the fact that they are used in similar ways. We shall come back to this remark shortly. In Section 2, we have shown how the proof of the correctness of the procedure for the trapezoidal prism brought to light that the geometrical transformation needed could be subsumed under the general formal operation of “using the excess to fill up the void.” In my view, the way of conducting the general proof for the yangma also reveals how it brings into play the same general operation.
49 To explain this point, we need to sketch how, after Problem 5.14 and the procedure for it, Liu Hui comments on the volume of the qiandu. His commentary on the algorithm computing the volume of this solid offers two ways of accounting for the correctness of the procedure. The first one relies on the fact that two identical qiandu make a parallelepiped. However, there is a second argument, which is not easy to understand. I have suggested that it refers to the qiandu as a particular case of the trapezoidal prism [Chemla, 1992b]. As a consequence, the procedure given for the qiandu appears simply to be an application of that for the prism and the proof of its correctness derives from the general proof for the prism, applied to the specific case of the qiandu (see Figure 12). Figure 12 (a) & (b): The qiandu and the geometrical transformation underlying the proof of the correctness of the procedure computing its volume In turn, the proof for the correctness of the procedures for the yangma and the bienao, which depends on that for the qiandu, fits within this framework, except that the piece to be moved needs to be cut in the middle and the pieces obtained have to be exchanged before they are moved. In this way, again, the operation of “using the excess to fill up the void” appears to be efficient and in fact to underlie the transformations needed to conclude the proof. From the previous remarks, we can conclude that in the case of the yangma, as in the other cases examined above, through proving, the commentator also here seems to aim at identifying general and fundamental transformations underlying a number of different algorithms. Yet, in contrast with the multiplication of fractions examined in Section 3, for which the change of the situation of the problem was linked to the goal of exhibiting the general formal operations underlying the procedure, for the yangma the change of the values is not linked to this question. If we are to understand the use of the particular values provided by problems, we must therefore answer the following question: Why does Liu Hui twice change the values of the problem, when clearly any particular values could do? The answer is clear when we observe the two proofs above: the values of the dimensions used by Liu Hui are determined by the visual tools he used, that is, the blocks, which were concrete objects, all dimensions of which were equal to 1 chi. In fact, the first reasoning brings together three identical blocks. In relation to the physical operation carried out, the values needed are all equal to 1 chi. The second reasoning illustrates the composition 48 Indeed, as in Liu Hui’s commentary on the area of the circular segment, it is clear that in order to unfold, the proof needs the original bodies to be cut into other bodies. These bodies are associated with procedures that compute their volumes, and these procedures are employed in the proof. 49 I make this point in greater detail in [Chemla, 1992b]. K. Chemla. p. 32 of a yangma and a bienao with blocks. The same blocks being used, the values needed to refer to the composition of the solids are 2 chi. As for the iteration of the decomposition of the yangma and the bienao in the second step of this proof, its analysis is based on the same configuration as in the first step. Blocks with the simplest dimensions have the property of being polyvalent for all these uses. Using the same blocks, Liu Hui can develop a proof that is particular (the first one) or a proof that is general (the second one). It is thus the physical features of the objects with which the proof is conducted that dictate the change of values. This conclusion implies that the commentaries were written down by reference to objects used in the course of proving. Note that there is no hint that blocks were used at the time when The Nine Chapters itself was compiled.
50 One may assume that these blocks —or some of their uses— were introduced for the sake of exegesis. This hypothesis would explain why the commentator had to change the values to refer to the concrete use of objects in relation to his reasonings, whereas the problems in The Nine Chapters would not mention values relating to material objects. In any event, the conclusion reached accounts for the fact that, except for the introduction of new values for a problem employed as a counterexample, the only systematic cases in which the commentators change the numerical values of problems occur in the context of geometry and in relation to the use of material visual tools. The new values are all determined by the objects used as visual tools to compose the body under consideration. The generality of the proofs developed is an issue that is completely dissociated from the fact that the texts could be written as referring to specific material objects. The next point will highlight this conclusion from another angle. 4.3. Visual devices and problems as tools to express the “meaning yi” of operations Although the particular and the general proofs are different on various accounts, they both use blocks to express the “meaning yi” of operations in the same way. In the former proof, gathering the three blocks yields a solid, the volume of which expresses the “meaning” of the multiplication together of length, width and height. The coefficient 3 is interpreted as related to the composition of the cube with three yangma. In the latter proof, like in the commentary following Problem 1.36 and bearing on the area of the circular segment, the commentator introduces a procedure that computes iteratively the extension of the space, within the qiandu, in which the volumes of the pieces coming, respectively, from the yangma and the bienao are in the proportion of 2 to 1. In order for its various steps to receive a 50 Several points should be stressed about the solids used in the earliest surviving documents and the hints about the early use of blocks. First, in contrast to The Nine Chapters, the Book of Mathematical Procedures does not mention the yangma or the bienao, that is, pieces among the blocks essential to establish the correctness of the procedures for other solids. However, the book treats other solids also found in The Nine Chapters in ways that require knowing the volumes of the yangma and the bienao. Several algorithms for volumes in both The Nine Chapters and the Book of Mathematical Procedures point to a geometrical interpretation, which provides reasons for the correctness of the procedures. This is how the commentators read them in the case of The Nine Chapters: they interpret the procedures step by step with the help of blocks [Chemla, 1990]. In such commentaries, Liu Hui also regularly changes the values of the problems and uses blocks in the first way described above, that is, in a way that is not geometrically general. Does this indicate that the commentators knew that objects of the type of blocks were used at the time when our earliest extant sources were composed, but used only in a certain way? Or did they introduce blocks as a tool for exegesis? It is impossible to give a certain answer to this question. However, the generic terms for “figure” or “block” do not occur in the earliest sources. The first known occurrences of these terms are found in the commentaries. Moreover, we may assume that if blocks did not exist, geometrical practice may have been similar to what some commentaries betray within the context of some problems dealing with geometrical topics: instead of referring to specific visual aids, the commentators interpret the operations directly in terms of the physical features of the situation of the problem. See my introduction to Chapter 9 in [Chemla and Guo Shuchun, 2004, 673-684]. K. Chemla. p. 33 “meaning,” the interpretation of the procedure requires that the volume of the yangma be divided. Interestingly, this question of interpretation brings us back to the parallel between problems and visual tools alluded to above. In the case analyzed in Section 3, a problem was changed into another problem, the situation of which was richer in possibilities of interpretation, since this feature was required for making explicit the “meaning” of the specific operations required by a proof. The same phenomenon recurs here in relation to the yangma: the change of values in the problem of The Nine Chapters is correlated to the introduction of an inner decomposition of the solid, which creates further possibilities for the interpretation of the steps needed by the proof. Not only does this conclusion offer a similar explanation for the change of a situation and for a change of numerical values. It also leads to an interesting observation. Seen from this angle, the use of a problem and the use of a visual tool are similar: 51 On the one hand, they “illustrate” a situation. On the other hand, they are put into play to express the “meaning” of operations, and when they are not rich enough to support the needs of interpretation required by a proof, they are replaced. In this context, it is worth recalling how the commentator introduces blocks when he first resorts to them: “ ‘Speech cannot exhaust the “meaning yi” ’ (yan bu jin yi) 52, hence to dissect/analyze (jie) this (volume), one must use blocks; this is the only way to get to understanding (the procedure). 言不盡意，解此要當以棊，乃得明耳。” We find here again, in relation to visual devices, the combination of terms (meaning, understanding) which in Section 3 was used in relation to problems. The previous discussion highlighted why in the context of the commentary on the yangma the values of the problem were changed into other values, which refer to the dimensions of a block: In fact, the continuity between the two kinds of item —problem and visual tool— is thereby manifested by, and inscribed in, the text. All that has been established for blocks in fact also accounts for the specificities of what is known regarding figures. This is why, even though all the evidence examined here bears on blocks, we have formulated our conclusions for visual devices in general. The perspective developed further suggests an interpretation of Liu Hui’s own description of his activity, when, in his preface to the book, he writes: “The internal constitutions (理 li) are analyzed with statements (辭 ci) and the bodies are dissected with figures (圖 tu).
53 所析理以辭，解體用圖。” [Chemla and Guo Shuchun, 2004, 126-127]. This concludes what I wanted to establish about Liu Hui’s practice with problems. Until this point in the paper, I have mainly concentrated on the commentaries, since they provide evidence to support conclusions about the practice with problems. To which extent can these conclusions on the practice with problems obtained thanks to the evidence found in the commentary be transferred to the mathematical activity at the time of the composition of The Nine Chapters itself, or even the Book of Mathematical Procedures? This question will be examined in Section 5. With the example of visual tools we have already seen reasons to believe that not all mathematical practices were the same. What about problems? This point is 51 Even though, in his unpublished dissertation, Wang Ling did not analyze in detail how problems and figures were used in the course of proving, it is interesting that he used the same term of “model” to refer to both problems (he speaks of “‘model’ problems” or “variant models”) and figures. Moreover, his conclusion regularly stresses that in the mathematics of ancient China, the particular was used to deal with the general [Wang Ling, 1956, 211, 282, 287, 295]. However, he did not provide evidence to support his claims, which is my main purpose in this paper. 52 The commentator quotes here the “Great commentary” (Xici dazhuan) to the Book of changes (first chapter, paragraph 12). Compare [Chemla and Guo Shuchun, 2004, 374-375]. 53 Here, probably, the generic term of “figure” stands for all the visual tools used in the commentary. On the terms occurring in the statement, see my glossary. K. Chemla. p. 34 all the more important since in the earliest extant writings problems play a prominent part. 5. Back to The Nine Chapters : Connecting the evidence from the commentaries and the Canon How can one determine whether the editors of the Canon also used problems in a way similar to that of Liu Hui described above? Let us repeat that, unless new sources are found, we will not be able to answer this question with full certainty. The method I will suggest here is to gather hints in The Nine Chapters indicating continuities with respect to the practice evidenced by Liu Hui’s more prolific writings. I shall allude to some of these hints below. However, another paper would be needed to deal with the question more systematically. One essential point should be first stressed. Much of what I have said bears on the use of problems within the context of proof. If we believe, as is often stated, that in sheer contrast to the commentaries The Nine Chapters contains nothing relating to proof and betrays no interest in this dimension of mathematical activity, this would deny from the outset the possibility of a real continuity. However, I have argued several times that, even though The Nine Chapters includes no fully developed proof, various facts indicate that the authors had an interest in understanding why their procedures were correct. We have already indicated some of them. For example, we have noticed that the qualification of division as “dividing in return (baochu)” adheres to the sphere of justifying procedures. It is hence quite striking that this expression occurs several times in The Nine Chapters itself. Why should one prescribe to “divide in return” instead of “divide,” unless to indicate the reason for carrying out the operation? In addition, we have mentioned that Liu Hui interpreted some algorithms for computing the volume of solids as indicating a proof of their correctness through the description of the procedure (see footnote 50). A further hint in support of the claim that the authors of The Nine Chapters had an interest in the correctness of algorithms is provided by how Liu Hui appears to interpret the fact that they described procedures in the context of problems. Not only does this piece of evidence indicate that he reads a concern for proof in the Canon, but it also reveals that he links the problems in the Canon to this concern. Let us therefore examine this evidence in greater detail. In fact, there are cases where procedures in the Canon are described outside of the context of problems. Such is the case, for instance, for the rule of three (jinyoushu 今有術) [Chemla and Guo Shuchun, 2004, 222-225]. This observation indicates that a problem is not an indispensable component in presenting a procedure. It is interesting, incidentally, that, to establish the correctness of the rule of three, Liu Hui’s commentary introduces a problem allowing the interpretation of the “meaning” (yi) of the operations. This fact confirms from yet another angle the part played by problems in proofs in relation to making explicit the “meaning” of the operations of an algorithm. What is most interesting in this case, however, lies elsewhere. The commentator states about this procedure for the rule of three: “This is a universal procedure (此都術也 ci doushu ye).” There is only one other passage in which Liu Hui repeats this statement. There it applies to the procedure for solving systems of simultaneous linear equations (方程術 fangchengshu), mentioned in Section 2 above [Chemla and Guo Shuchun, 2004, 616-617]. However, in contrast with the rule of three, in this case the “universal procedure” is described by The Nine Chapters within the context of a problem about different types of millet. It reads as follows: (8.1) Suppose that 3 bing of high-quality millet, 2 bing of medium-quality millet and 1 K. Chemla. p. 35 bing of low-quality millet produce (shi) 39 dou; 2 bing of high-quality millet, 3 bing of medium-quality millet and 1 bing of low-quality millet produce 34 dou; 1 bing of high-quality millet, 2 bing of medium-quality millet and 3 bing of low-quality millet produce 26 dou. One asks how much is produced respectively by one bing of high-, medium- and low-quality millet. 今有上禾三秉，中禾二秉，下禾一秉，實三十九斗；上禾二秉，中禾三秉，下禾 一秉，實三十四斗；上禾一秉，中禾二秉，下禾三秉，實二十六斗。問上、中、 下禾實一秉各幾何。 [Chemla and Guo Shuchun, 2004, 616-617]. The piece of evidence in which I am interested is found immediately after the sentence in which Liu Hui qualifies the procedure as being “universal.” The statement he adds to this can be interpreted as accounting for why, here, in contrast to the case of the rule of three, the Canon uses the context of problems to present the procedure. The commentator writes: It would be difficult to understand (the procedure) with abstract expressions (kongyan), this is why one deliberately linked it to (a problem of) millet to eliminate the obstacle. 以空言難曉，故特繫之禾以决之。 This statement reveals that, in Liu Hui’s perspective, the purpose of the Canon for presenting a procedure in the context of a problem was related to the aim of having the procedure be understood. We have already shown the relationship between “understanding” a procedure and establishing its correctness. In the text that follows, Liu Hui uses the context of the problem on millets to interpret the operations of the procedure and thereby bring to light that the pattern of “equalizing” and “homogenizing” underlies it, thus proving its correctness (see Section 2). The piece of evidence that the above statement constitutes may hence indicate that Liu Hui reads the Canon as providing problems to be put into play to prove algorithms. Several additional features of the problems used in the same chapter are quite interesting to support our argument that the practice of problems in The Nine Chapters presents continuities with Liu Hui’s commentary [Chemla, 2000]. First, in fact, the statement of Problem 8.1, regardless of whether one interprets it in terms of production of millet or of capacities (cf. footnote 54), can be interpreted in a still completely different way: (8.1, alternative interpretation) Suppose that 3 bing of high-position millet, 2 bing of medium-position millet and 1 bing of low-position millet correspond to the dividend (shi) 39 dou; 2 bing of high-position millet, 3 bing of medium-position millet and 1 bing 54 Bing秉 designates a unit of capacity from a system of units different from that of the dou. I am grateful to Michel Teboul who suggested another interpretation for this problem: the statement may be understood as referring to distinct units of capacity all named bing and the value of which would depend on the grain measured. I have shown that in The Nine Chapters, the Canon gives evidence of a similar phenomenon with respect to the unit of capacity hu 斛, which had different values for different grains [Chemla and Guo Shuchun, 2004, introduction to Chapter 2, 201-205]. The key term shi in Problem 8.1 could thus be interpreted either as “fill up” or as “capacity” (the basic sentence would then read: “Suppose that 3 bing of high-quality grain, 2 bing of medium-quality grain and 1 bing of low-quality grain (correspond to) a capacity (shi) of 39 dou”). The meaning of shi as “capacity” is attested in a passage from the Records on the scrutiny of the crafts (Kaogongji 考工記, dated to the third century B.C.E.), which deals with standard vessels and is quoted in Liu Hui’s commentary on Problem 5.25 [Chemla and Guo Shuchun, 2004, 450-453]. However, the actual values in the problem still make me prefer the first interpretation, which I have therefore inserted here. 55 On the interpretation of kongyan, see [Chemla, 1997a] and the entry for kong “abstract,” in my glossary, [Chemla and Guo Shuchun, 2004, 947]. [Chemla, 2000] develops a reading of Chapter 8 of The Nine Chapters along these lines, that is, by assuming that the millets were intended to enable an interpretation of the operations in the procedures. K. Chemla. p. 36 of low-position millet correspond to the dividend 34 dou; 1 bing of high-position millet, 2 bing of medium-position millet and 3 bing of low-position millet correspond to the dividend 26 dou. One asks how much is the dividend corresponding respectively to one bing of high-, medium- and low-position millet. In addition to interpreting “high,” “middle,” and “low” as positions on the calculating surface, this second reading interprets the term 實 shi (“produce”/capacity) with its technical meaning in mathematics: “dividend.” From the first century C.E. till the fourteenth century at least, in ancient China an algebraic equation was conceptualized as an opposition between divisors (the various coefficients of the unknown) and a dividend (the constant term). In fact, it can be shown that Liu Hui also reads the statement of Problem 8.1 in this alternative way and understands a linear equation as opposing a dividend (its constant term) to divisors. This first problem of Chapter 8 is followed by five others similar to it, all relating to grain. The full procedure for solving systems of linear equations, with positive and negative numbers, is then unfolded in relation to them. If Problem 8.6 were interpreted along the same lines as the previous ones, it would correspond to a system of equations with two negative constant terms. However, all ancient sources give the two “productions/dividends” as positive, a fact that the commentators do not stress as erroneous. In other terms, it seems that the interpretation of the “dividends” in terms of “production” or “capacities” imposes that the shi remains positive. If this is the case, this constraint, which derives from the interpretation of the operations in the terms of the problem, poses limits to the full presentation of the mathematical topic. What follows in The Nine Chapters supports this hypothesis: Problems 8.7 and 8.8 turn to a new situation —buying animals— to discuss new types of systems in which dividends —in this context no longer “productions” or “capacities”— are either positive, negative or zero. This seems to indicate that in The Nine Chapters the situation with millet producing grain (or contained in units of capacities) was used to deal with systems of equations as long as the interpretation allowed by the situation did not conflict with the mathematical requirement. As soon as a divergence arose between the interpretation in the terms of the problems and the mathematical meaning, the situation for discussing the topic was changed to allow further discussion. If our interpretation is correct, this implies that, in agreement with Liu Hui’s explanation, the use of the situation of a problem for interpreting the operations of an algorithm, in relation to understanding and proving it, dates to the time of The Nine Chapters. Such a conclusion suggests that there may have existed a mathematical culture of situations selected for their usefulness in presenting procedures. Another element appears to confirm this hypothesis: many mathematical questions are discussed within the framework of the same kind of situation throughout the centuries. One way of accounting for this would be that these situations proved particularly suitable in relation to interpreting the operations of the procedures. Inquiring further into this question exceeds the framework of this paper. Let me simply claim for now that this is quite plausible. 6. Conclusion We have seen that not only does the commentator Liu Hui attest to a practice of problems, described in Sections 1 to 4, that is peculiar and differs from the one most common today, but he also seems to assume that his way of using problems was in continuity with former practices, especially those contemporary with the making of the Canon. In fact, we have found hints in The Nine Chapters that appear to support his belief. In this tradition, whether problems were practical in form or abstract, they were read as general statements, the K. Chemla. p. 37 extension of which was determined on the basis of the procedure relating to them. This fleshes out what the opening sections of The Gnomon of the Zhou intended when depicting intellectual activity in mathematics or astronomy as aiming at widening “classes (lei)” of problems. Instead of interpreting these statements on the basis of our own experience, it seems to me more appropriate to rely on the evidence we have about mathematical activity in ancient China to interpret this early theoretical description of what mathematics was about. Furthermore, the results obtained in this paper give us elements for establishing a method to interpret our earliest Chinese sources. As we have seen, the procedure given for a problem can either solve it or be used for establishing the correctness of another algorithm. This is why problems can be either questions to be solved or statements describing a situation in order to interpret the meaning of operations. Clearly, ignoring such facts would be quite detrimental when reading mathematical sources from ancient China. This is one of the main errors responsible for the mistaken idea that these texts can be adequately interpreted as merely practical. However, the benefit of promoting a new method of reading is not limited to this aspect. Instead of assuming that mathematical practice has been uniform in space and time, such a way of approaching texts, attempting to establish how they should be read before one sets out to read them, contributes to restoring the diversity of mathematical practice. I hope the above arguments will inspire further research that by gathering specific evidence will enable us to restore various practices of mathematical problems. Once we have assembled several similar case studies, we shall be in a position to outline a research program that may create new conditions for interpreting our sources. Bibliography: Bréard, A., 1999. Re-Kreation eines mathematischen Konzeptes im chinesischen Diskurs : "Reihen" vom 1. bis zum 19. Jahrhundert. Franz Steiner Verlag, Stuttgart. Bréard, A., 2002. Problems of pursuit: Recreational mathematics or astronomy. In: Dold-Samplonius, Y., Dauben, J.W., Folkerts, M., and Dalen, B.van (Eds.), From China to Paris: 2000 Years of Mathematical Transmission. Proceedings of a Conference held in Bellagio, Italy, May 8-12, 2000. Franz Steiner Verlag, Stuttgart, pp. 57-86. Chemla, K., 1990. De l'algorithme comme liste d'opérations. In: Jullien, F. (Ed.), L'art de la liste. Extrême Orient, Extrême Occident. 12. Presses Universitaires de Vincennes, Saint-Denis, pp. 79-94. Chemla, K., 1992a. Les fractions comme modèle formel en Chine ancienne. In: Benoit, P. Chemla, K., Ritter, J. (Ed.), Histoire de fractions fractions d'histoire. Birkhäuser, Basel, pp. 189—207, 405, 410. Chemla, K., 1992b. Résonances entre démonstration et procédure: Remarques sur le commentaire de Liu Hui (IIIe siècle) aux Neuf chapitres sur les procédures mathématiques (1er siècle). In: Chemla, K. (Ed.), Regards obliques sur l'argumentation en Chine. Extrême-Orient, Extrême-Occident. 14. Presses Universitaires de Vincennes, Saint-Denis, pp. 91—129. Chemla, K., 1994. De la signification mathématique de marqueurs de couleurs dans le commentaire de Liu Hui. In: Peyraube, A., Tamba, I., and Lucas, A. (Eds.), Linguistique et Asie Orientale. Mélanges en hommage à Alexis Rygaloff, pp. 61—76. Chemla, K., 1997a. Qu'est-ce qu'un problème dans la tradition mathématique de la Chine ancienne? Quelques indices glanés dans les commentaires rédigés entre le IIIe et le VIIe siècles au classique Han Les neuf chapitres sur les procédures mathématiques. In: Chemla, K. (Ed.), La valeur de l'exemple. Extrême-Orient, Extrême-Occident. 19. K. Chemla. p. 38 Presses Universitaires de Vincennes, Saint-Denis, pp. 91—126. Chemla, K., 1997b. What is at stake in mathematical proofs from third-century China? Science in context 10, 227-251. Chemla, K., 1997/1998. Fractions and irrationals between algorithm and proof in ancient China. Studies in History of Medicine and Science. New Series. 15, 31—54. Chemla, K., 2000. Les problèmes comme champ d'interprétation des algorithmes dans les Neuf chapitres sur les procédures mathématiques et leurs commentaires. De la résolution des systèmes d'équations linéaires. Oriens Occidens. Sciences Mathématiques et Philosophie de l'Antiquité à l'Age Classique 3, 189-234. Chemla, K., Classic and commentary: An outlook based on mathematical sources. Critical Problems in the History of East Asian Science, Dibner Institute, Cambridge (Mass.), 2001 (forthcoming). See a version at: Chemla, K., 2003a. Generality above abstraction: the general expressed in terms of the paradigmatic in mathematics in ancient China. Science in Context. 16, 413--458. Chemla, K., 2003b. Les catégories textuelles de "Classique" et de "Commentaire" dans leur mise en oeuvre mathématique en Chine ancienne. In: Berthelot, J.-M. (Ed.), Figures du texte scientifique. Presses Universitaires de France, Paris, pp. 55-79. Chemla, K., and Guo Shuchun, 2004. Les neuf chapitres. Le Classique mathématique de la Chine ancienne et ses commentaires. Dunod, Paris. Cullen, C., 1996. Astronomy and mathematics in ancient China : the Zhou bi suan jing. In: Needham Research Institute studies ; 1, vol. Cambridge University Press, Cambridge [England] ; New York. Cullen, C., 2002. Learning from Liu Hui? A Different Way to Do Mathematics. Notices of the American Mathematical Society. 49, 783-790. Cullen, C., 2004. The Suan shu shu 筭數書 ‘Writings on reckoning’: A translation of a Chinese mathematical collection of the second century BC, with explanatory commentary. In: Needham Research Institute Working Papers vol. Needham Research Institute, Cambridge. Cullen, C., 2007. The Suan shu shu 筭數書 ‘Writings on reckoning’: Rewriting the history of early Chinese mathematics in the light of an excavated manuscript. Historia Mathematica. 34, 10-44. Dauben, J. W., 2008. 算數書. Suan Shu Shu (A Book on Numbers and Computations). English Translation with Commentary. Archive for history of exact sciences. 62, 91-178. Eberhard (Bréard), A., 1997. Re-création d'un concept mathématique dans le discours chinois: Les "séries" du Ier au XIXe siècle, (Ph.D.), University Paris 7 & Technische Universitaet Berlin. Guo Shirong郭世榮, 2001. Suanshushu kan wu算術書勘誤 (Errors in editing the Book of Mathematical Procedures). Neimenggu shida xuebao ziran kexue (Han wen) ban 內蒙古師大學報自然科學 漢文)版 (Journal of the Normal University of Inner Mongolia—Science (part in Chinese)). 30, 276-285. Guo Shuchun郭書春, 1984a. Liu Hui de Tiji Lilun 劉徽的體積理論 (Liu Hui's Volume Theory). Kexueshi Jikan (Collected Papers in History of Science)科學史集刊. 11, 47-62. Guo Shuchun郭書春, 1984b. Jiuzhang suanshu he Liu Hui zhu zhong de lü gainian ji qi yingyong shixi《九章算術》和劉徽注中之率概念及其應用試析 (Analysis of the concept of lü and its uses in The Nine Chapters on Mathematical Procedures and Liu Hui’s commentary) (in Chinese). Kejishi Jikan 科技史集刊 (Journal for the history of science and technology). 11, 21-36. K. Chemla. p. 39 Guo Shuchun郭書春, 2001. Suanshushu jiaokan 算數書校勘(Critical edition of the Book of Mathematical Procedures). Zhongguo keji shiliao 中國科技史料 (China Historical Materials of Science and Technology). 22, 202-219. Guo Shuchun郭書春, 2002. Shilun Suanshushu de lilun gongxian yu bianzuan 試論算數書的理論貢獻與編纂(On the theoretical achievements and the compilation of the Book of Mathematical Procedures). Faguo hanxue 法國漢學 Sinologie française. 6, 505-537. Li Jimin李繼閔, 1990. Dongfang shuxue dianji Jiuzhang suanshu ji qi Liu Hui zhu yanjiu 東方數學典籍覧《九章算術》及其劉徽注研究 (Research on the Oriental mathematical Classic The Nine Chapters on Mathematical Procedures and on its Commentary by Liu Hui). Shaanxi renmin jiaoyu chubanshe, Xi'an. Li Yan, 1958. Zhongguo shuxue dagang. Xiuding ben (Outline of the history of mathematics in China. Revised edition) Kexue chubanshe, Beijing. Li Yan, and Du Shiran, 1987. Chinese mathematics : a concise history. Oxford Science Publications. Clarendon Press, Oxford. Needham, J., and Wang Ling, 1959. section 19: Mathematics, Science and civilisation in China. Cambridge University Press, Cambridge (UK), pp. 1-168. Peng Hao 彭浩, 2001. Zhangjiashan hanjian «Suanshushu» zhushi 張家山漢簡《算數書》注釋 (Commentary on the Book of Mathematical Procedures, a writing on bamboo slips dating from the Han and discovered at Zhangjiashan). Science Press (Kexue chubanshe), Beijing. Qian Baocong, 1963. Suanjing shishu (Qian Baocong jiaodian) (Critical punctuated edition of The Ten Classics of Mathematics). Zhonghua shuju, Beijing. Shen Kangshen, Crossley, John N., and Lun, Anthony W.-C., 1999. The Nine Chapters on the Mathematical Art. Companion and commentary. Oxford University Press & Science Press, Oxford & Beijing. Vogel, K., 1968. Neun Bücher arithmetischer Technik. Friedr. Vieweg & Sohn, Braunschweig. Volkov, A., 1986. O nazvanii odnogo driévniékitaïskogo matiématitchiéskogo traktata (On the title of an ancient Chinese mathematical treatise). In: Volkov, S. (Ed.), Istoriia i koul'toura Vostotchnoï i Iougo-Vostotchnoï Azii (History and Culture of East and Southeast Asia). Nauka Press, Moscow, pp. 193-199 Volkov, A., 1992. Analogical reasoning in ancient China: Some examples. In: Chemla, K. (Ed.), Regards obliques sur l'argumentation en Chine. Extrême-Orient, Extrême-Occident. 14. Presses Universitaires de Vincennes, Saint-Denis, pp. 15-48. Volkov, A., 1994. Transformations of geometrical objects in Chinese mathematics and their evolution. In: Alleton, V., and Volkov, A. (Eds.), Notions et perceptions du changement en Chine. Collège de France, Institut des Hautes Etudes Chinoises, Paris, pp. 133-148. Volkov, A., 2001. Capitolo XII: La matematica. 1. Le bacchette. In: Karine Chemla, with the collobaration of Francesca Bray, Fu Daiwie, Huang Yilong, and Métailié, G. (Eds.), Vol. II. In : S. Petruccioli (Gen.Ed.) La scienza in Cina. Enciclopedia Italiana, Roma, pp. 125-133. Wagner, D. B., 1979. An early Chinese derivation of the volume of a pyramid: Liu Hui, 3rd century A.D. Historia Mathematica. 6, 164-188. See a version at: Wang Ling, 1956. The 'Chiu Chang Suan Shu' and the History of Chinese Mathematics during the Han Dynasty, (Dissertation for the Degree of Doctor of Philosophy), Trinity College, Cambridge. Wu Wenjun吳文俊, 1982. Churu xiangbu yuanli 出入相補原理(The principle "What comes K. Chemla. p. 40 in and what goes out compensate each other"). In: Wu Wenjun吳文俊 (Ed.), Jiuzhang suanshu yu Liu Hui 九章算數與劉徽 [The Nine Chapters on Mathematical Procedures and Liu Hui]. Beijing Shifan Daxue Chubanshe北京師範大學出版社, Beijing北京, pp. 58-75. 張家山漢簡『算數書』研究会編 Chôka san kankan Sansûsho kenkyûkai. Research group on the Han bamboo slips from Zhangjiashan Book of Mathematical Procedures, 2006. 漢簡『算數書』Kankan Sansûsho. The Han bamboo slips from Zhangjiashan Book of Mathematical Procedures. 朋友書店Hôyû shoten, 京都 Kyoto. |
6909 | https://zhuanlan.zhihu.com/p/503538267 | 0-1 分布、二项分布（期望与方差） - 知乎
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他叫小胖子呐：常见分布的数学期望和方差及相关证明 497 赞同 · 14 评论 文章
0-1 分布
1.定义：
0-1分布又称两点分布或伯努利（ Bernoulli）分布.
设随机变量X的分布律为
则称X服从参数为 p(0<p<1)p(0<p<1) 的0-1分布.
其分布律又可写成
P{X=k}=p k(1−p)1−k,k=0,1 P{X=k}=p^{k}(1-p)^{1-k}, \quad k=0,1
常用它来表示两个状态的问题（即随机试验的结果只有两个，称为伯努利试验）
2.数学期望与方差
X ~ B(1, p) 二项分布的数学期望：p
X ~ B(n, p) 二项分布的方差：p(1-p)
二项分布
1.定义：
将上述伯努利试验独立地做n次，称为n重伯努利试验，
P{X=k}=\left(\begin{array}{l} n \ k \end{array}\right) p^{k}(1-p)^{n-k}, k=0,1, \cdots, n
称具有上述分布律的随机变量为服从参数为n, p的二项分布, 记为： X~B(n, p)
特别地, 0-1分布即为B(1, p). 【0-1分布就是进行一次伯努利实验】
注意：二项分布理论上只适用于有放回抽样, 但当n很大时, 也可近似用于无放回抽样
因为要是没有放回，就会改变整体的概率了。如果量很大的时候，拿走一个并不会对概率有多大的变化，所以可以近似。
2.数学期望与方差
X ~ B(n, p) 二项分布的数学期望：np
证明：二项分布的分布律为
P{X=k}=\mathrm{C}_{n}^{k} p^{k} q^{n-k} \quad(k=0,1,2, \cdots, n, p+q=1)
由公式 k \cdot \mathrm{C}{n}^{k}=n \cdot \mathrm{C}{n-1}^{k-1} 有
\begin{aligned} E(X) &=\sum_{k=0}^{n} k \cdot P{X=k} \ &=\sum_{k=0}^{n} k \cdot \mathrm{C}{n}^{k} p^{k} q^{n-k} \ &=n p \sum{k=1}^{n} \mathrm{C}_{n-1}^{k-1} p^{k-1} q^{n-k} \ &=n p(p+q)^{n-1} \ &=n p \end{aligned}
X ~ B(n, p) 二项分布的方差：np(1-p)
（证明如下）
3.例题：
1.(电子元件)一大批电子元件有10%已损坏, 若从这批元件中随机选取20只来组成一个线路, 问这线路能正常工作的概率是多少？
解：
因为元件的数量很大，所以取20只元件可看作是有放回抽样, 记X表示20只元件中好品的数量，则
X ~ B(20,0.9)
例题：
2.(人寿保险)若一年中某类保险者里面每个人死亡的概率等于0.005， 现有10000个人参加这类人寿保险， 试求在未来一年中在这些保险者里面，
⑴ 有40个人死亡的概率; ⑵ 死亡人数不超过70个的概率 。
解：
记X 为未来一年中在这些人中死亡的人数，则X ~ B(10000,0.005)
小概率事件原理
在利用概率模型进行推断问题中, 常认为小概率事件是不发生的, 如果发生了, 则说明原概率模型不真实
编辑于 2022-04-29 01:52
二项分布
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Wishing
零一分布中的方差符号打错了吧，应该是x～B（1，p）=p（1-p）不是x～B（n，p）=p（1-p）
2022-11-23
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袁国为的giegie
请问一下二项分布的期望与方差正好是01分布的n倍，这是不是因为二项分布相当于n次独立的01分布，所以期望与方差是n倍呀
2024-05-20
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lhaiy
。。。。。
2024-04-13
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随遇而安
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hualong
应该是打错了
2023-04-09
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6910 | https://www.youtube.com/watch?v=Xq9Q2IgGHTA | reciprocals (KristaKingMath)
Krista King
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Posted: 2 Dec 2014
► My Pre-Algebra course:
In this video we learn about reciprocals: what they are, and how to find the reciprocal of fractions, whole numbers, and negative numbers.
A reciprocal is just the number that, when multiplied by the original value, makes the product 1. You can find the reciprocal of a fraction by switching the numerator and denominator.
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● ● ● CONNECT WITH KRISTA ● ● ●
Hi, I’m Krista! I make math courses to keep you from banging your head against the wall. ;)
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in this video we're going to be talking about reciprocals and the only thing you need to know about reciprocals is that the reciprocal of something is whatever you can multiply this thing by in order to make it equal to one the reciprocal is also called the multiplicative inverse but of course we never use that term when we could just use the term reciprocal because it's a lot easier to say so what do we have to multiply by 3 over4 in order to get one well in fact it's just as easy as flipping this fraction upside down so the denominator is going to become the numerator and the numerator is going to become the denominator so we just flipped the place of the numerator and the denominator we flipped the Straight Upside Down flip their places and now when we multiply straight across 3 4 gives me 12 4 3 gives me 12 and anything over itself we know is 1 so 12 over 12 is equal to 1 so 4/3 was what I had to multiply by 3/4s in order to get one and that's always going to be the case except when we have zero and we'll talk about that in in a second but in order to find the reciprocal which is whatever you have to multiply by in order to get one you just flip this upside down so if we have 5 over 2 we just flip it upside down and we can say that the reciprocal is 2 fths so our reciprocal is going to be 2 fths just like our reciprocal was 4/3 and we can prove that 2 fths is the reciprocal by multiplying straight across 5 2 is 10 2 5 is 10 10 over 10 is equal to 1 so when we multiply by 2 fths we get 1 so we know that 2 fths is the reciprocal now what happens when we try to find the reciprocal of three well before when we were dealing with a fraction it was really easy because we just swapped the places of the numerator and the denominator what happens when we just have a whole number well in that case we remember that this is still a fraction three is actually 3 over 1 and now we have a fraction we flip the fraction upside down and we get 1 over three so the numerator becomes the denominator and the denominator becomes the numerator and we say 3 1 is 3 and 1 3 is 3 and 3 3 over 3 is 1 so we can say that 1/3 is our reciprocal so when you have a whole number when you just have three you make sure that you say 3 over 1 then you flip it upside down and you get the reciprocal now how does that apply when we have zero well you might think I can say this is 0 over one because 0 over one is still zero now I have a fraction I just multiply by the reciprocal which is what I get when I flip it upside down so I do 0 over 1 I get 0 1 is 0 I get 1 0 is 0 and maybe 0 over Z is one because anything over itself is one well there's so many things wrong with this it's hard to even know where to start first of all we can never have zero in the denominator so we can't have one divided by Z because that's undefined so this doesn't work here 0 over 0 first of all we can't have zero in the denominator and 0 over 0 would be zero if anything but either way this is going to be undefined also so what we say is that the only number that doesn't have a reciprocal is zero everything else we're going to be able to find a reciprocal for but there is no reciprocal for zero what about negative numbers here we have -1 over 4 so we just flip the numerator and the denominator and we say 4 over 11 If I multiply straight across I'm going to get 11 4 which is 44 I'm going to get 4 11 which is 44 but I have this negative sign out in front so I would put a negative sign out in front and I would get negative 1 but remember the reciprocal is what makes this value here positive 1 not negative 1 so in order to find the reciprocal of a negative number not only do we have to flip the numerator and the denominator but we also have to include a negative if I include a negative here with my reciprocal then I get a negative time a negative gives me a positive and I end up with positive one so then my reciprocal is -4 over 11 exactly what I get when I flip the numerator and denominator and keep the negative sign so that's the only difference here we keep the negative sign that's important what about here negx you can do this with more than just fractions or whole numbers here we have to remember that this is X over 1 we're going to flip the numerator and the denominator and get 1 /x but like we learned in this example we need to make sure we keep the negative sign then to check to make sure I did this right I can say x 1 gives me X I can say 1 X gives me X I know that negative a negative gives me a positive they cancel and I just get X overx which I know is 1 so I can say that my reciprocal is -1x |
6911 | https://www.youtube.com/watch?v=bGmbyztdCHE | Percent Composition By Mass
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This chemistry video tutorial explains how to find the percent composition by mass of each element in a compound using a simple formula. You need to find the molar mass of the entire compound in order to calculate the mass percent of each element. This video contains plenty of examples and practice problems.
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Introduction in this video we're going to talk about how to calculate the percent composition of an element so let's go over the formula the mass percent is equal to the mass of the element divided by the total mass times 100 now you can calculate the mass percent if you're given the grams or if you're given the molar mass in terms of grams per mole the answer will still be the same in this example we're going to use the atomic weights or the molar mass that can be found in the periodic table so let's begin Example C12H22O11 let's start with c12h22o11 sucrose let's calculate the percent by mass of carbon now the first thing i would recommend doing is finding the molar mass of the entire compound first so let's do that we have 12 carbon atoms 22 hydrogen atoms and 11 oxygen atoms carbon has an atomic mass of 12.01 and hydrogen is 1.008 oxygen 16. so if you type this in exactly the way you see it in your calculator you should get 342.296 grams per mole so that's the molar mass of the entire thing now if you want to find the percent by carbon you need to take the molar mass of all 12 carbon atoms and divide it by the total and then multiply by 100 so each carbon is 12.01 that's the atomic mass the total molar mass is this number that we have 342 0.296 and then we're going to multiply by 100 so 12 times 12.01 that's 144.12 divide that by 342.296 and then take that result multiply it by 100 and you should get 42.1 so that's the percentage by mass of the carbon atoms in sucrose now i'm just going to rewrite the answer on the top so we can reference it later so it's 42.1 percent now let's do the same thing with hydrogen so we need to take the mass of all the hydrogen atoms there's 22 of them each has an atomic mass of 1.008 and divided by the same total mass for the entire compound which is 342.296 and then multiply everything by a hundred percent so 22 times 1.008 that's 22.176 and then divide that by 342.296 and then multiply by 100 this will give you about six point five percent if you round it now let's do the same thing with oxygen so in sucrose there are 11 oxygen atoms each with an atomic mass of 16 then divided by the total molar mass 11 times 16 that's 176 and then divided by the molar mass and multiply by 100 you should get 51.4 so that's the percent by oxygen or the percent by mass of oxygen now if you add up 51.4 42.1 and 6.5 this will give you 100 that's how you can tell if you did everything correctly so now let's move on to that next example Example C5H9O2 let's calculate the percent composition of every element in c5 h9 and o2 so first let's find the mole mass of the entire compound so we have five carbon atoms nine hydrogen atoms one nitrogen atom and two oxygen atoms now let's replace carbon with its atomic mass of 12.01 h is 1.008 nitrogen 14.01 and oxygen it's 16. now 5 times 12.01 is 60.05 9 times 1.008 that's this number and then 2 times 16 is 32 so let's add everything so the total molar mass is 115.132 grams per mole so now that we have that we can calculate the mass percent of every l excuse me every element so let's start with the percent of carbon so it's going to be the five carbon atoms which we could see that it has a mass of 60.05 divided by the total molar mass of 115.132 times 100 percent so it's 52.16 percent by carbon now let's move on to the percent by hydrogen or the percent of hydrogen rather so it's going to be the nine hydrogen atoms which has a mass of 9.072 divided by the same total molar mass of 115.132 times 100 percent and so that's going to be about 7.88 percent now let's do the same for the remaining two elements so let's move on to nitrogen so we have the mass of nitrogen which is uh 14.01 divided by the total times 100 so once you do this a few times it becomes second nature you'll automatically get the hang of it so it's 12.1 percent for nitrogen now let's move on to oxygen the last one there are two oxygen atoms and so the molar mass of those two oxygen atoms is 32 and just as we've been doing before we're going to divide it by the total so nothing has changed except the top number and i got 27 for the value of oxygen so that's the mass percent of oxygen in that compound so now you know how to find the percent composition of every element within a molecular compound and that's it for this problem number two which of the following compounds contain the greatest percentage of carbon by mass is it going to be a b c d or e well the most simplest way to find the answer is to determine the percentage of carbon for each compound let's start with ch4 so the percent of carbon is going to be the mass of carbon there's only one carbon atom and we know the atomic mass is 12.01 divided by the total which is the mass of carbon plus the four hydrogen atoms and then we're going to multiply by 100 percent now let's find the molar mass of ch4 let's start with that first so 12.01 plus 4 times 1.008 that's 16.042 and let's multiply everything by 100 so this is 74.87 percent mass by carbon so let's look at the second example let's see if there's anything that can top that so for b there are two carbon atoms so it's two times 12.01 divided by the total which will be the two carbon atoms and as a total of seven hydrogen atoms five plus two is seven and we have a nitrogen atom as well so let's calculate the molar mass first so for the molal mass i got 45.086 2 times 12 that's going to be 24. so the percentage is 53.3 percent so we could eliminate answer choice b so far a is the highest now let's move on to part c so this time we have five carbon atoms it's gonna be five times twelve point zero one divided by those five carbon atoms and there's a 12 hydrogen atoms 911 and we have an oxygen atom so let's multiply that by 100 percent so five times twelve point zero one that's going to be sixty point zero five and the total molar mass of c5h11oh is 88.146 if i did that correctly so the mass percent of carbon in this sample is 68.1 percent which is still less than 74. so we can get rid of that now let's move on to part d so we have six carbon atoms this time divided by everything else there's 12 hydrogen atoms and six oxygen atoms six times 12.01 that's going to be 72.06 and the molar mass of c6h12o6 that's 180.156 so the percentage is about 40 carbon now the last thing we need to check is carbon dioxide which is probably not going to be more than 50 probably a lot less so we have one carbon which is 12.01 divided by the total the molar mass of co2 is 44.01 so you should get 27.3 percent which is the lowest so therefore the answer is a number three a molecule contains a single atom of nitrogen if the mass percent of nitrogen in this compound is 13.32 percent what is the molar mass of the molecule so how can we figure this out well let's start with the formula the percent of nitrogen is going to equal the mass of nitrogen divided by the total molar mass multiplied by 100 so we have the percentage of nitrogen in this compound it's given to us it's 13.32 now we know that there's a single nitrogen atom and we need to multiply that by the atomic mass of nitrogen which is 14.01 we don't know the molar mass so let's call it some variable x and then we got to multiply everything by 100 so at this point it's basically algebra so what i would prefer doing is dividing both sides by a hundred so this disappears 13.32 divided by 100 is 0.1332 so that's equal to 14.01 divided by x now let's put this over one and let's cross multiply so we have one times 14.01 and that's equal to 0.1332 times x so now to find the value of x let's divide both sides by 0.1332 so the molar mass is 105 point 18 grams per mole so that's the molar mass of the unknown compound so if you have the mass percent of a single element and if you know how many atoms of that element are in the molecule you can calculate the molar mass of the unknown molecule you |
6912 | https://www.youtube.com/watch?v=v9rLdEVA544 | 6.28 | Part of riding a bicycle involves leaning at the correct angle when making a turn, as seen in
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Part of riding a bicycle involves leaning at the correct angle when making a turn, as seen in Figure 6.36. To be stable, the force exerted by the ground must be on a line going through the center of gravity. The force on the bicycle wheel can be resolved into two perpendicular components—friction parallel to the road (this must supply the centripetal force), and the vertical normal force (which must equal the system’s weight).
(a) Show that θ (as defined in the figure) is related to the speed v and radius of curvature r of the turn in the same way as for an ideally banked roadway—that is,θ = tan^–1 v^2 / rg
(b) Calculate θ for a 12.0 m/s turn of radius 30.0 m (as in a race).
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number 28 part of riding a bicycle involves leaning at the correct angle when making a turn that's seen in Figure 6.36 to be stable the force exerted by the ground must be on a line going through the center of gravity the force of the bicycle wheel can be resolved into two perpendicular components friction parallel to the road this must apply the centripetal force and the vertical normal force which must equal the system's weight letter a show that theta as defined in the figure is related to the speed V and radius of curvature R for the turn in the same way as for an ideally banked runway that is this equation alright so here I created a picture that basically is the same thing as representing up here another thing I added is just this top parallel line to the frictional force just for illustrative purposes so they gave me this angle in here theta but I'm also saying that this angle over here is the same now that should make sense right let's just say that this is 30 if this angle here is 30 and you know that this is a right angle what is this angle up here well that's going to be 60 right remember all angles add up to 120 in a triangle and if this is 60 knowing that this creates a right angle here what is this missing angle it's also 30 right so they do match so okay so now we got that out of the way so those Thetas are the same now they told me in the problem here they said that the centripetal force is is is applied by the friction right between the tires and the road so the frictional force here is the same as the centripetal force so I'm gonna make them equal okay now let's start off with the formula let's start off with the centripetal force formula over here on the right hand side okay so let's write centripetal force is equal to mass times acceleration and that's centripetal acceleration so what I need now need to do is somehow try to get well first theta involved into this equation how can I do so well I see probably the easiest way to do so is by relating theta to the the centripetal force here right by using some trig and I'm going to incorporate the hypotenuse of this okay so if I'm incorporating the hypotenuse of my triangle here right in this angle and the side opposite what am i doing I'm doing sine so sine of theta is equal to the opposite side of the hypotenuse so the sine of that angle should be equal to the centripetal force divided by F solve this for centripetal force are going to do is cross multiply here right so centripetal force will equal F whatever this force is multiplied by sine of theta okay now this centripetal force is that literally the same thing as this so what that means is I can take this and plug it on in for it in the formula so here we go so this is f now sine of theta is equal to MA see great now let's see somehow right now I'm thinking about it somehow I'm looking at my equation here and I realized well there's no mass in it but I have mass here so somehow I got it I got to get it to go bye-bye all right I got to cancel it so how am I gonna do so well I'm looking at my picture and I'm like well I know the weight right is equal to mg and if somehow I can get the weight in here for F I can cancel those masses that may help so I'm thinking now well how how do I get weight into the formula via F well how are they related how are these two things related well the related via this angle right it's a triangle again so I'm thinking about maybe using some cosine here right because hypotenuse angle this is the side adjacent so let's do that so cosine of theta is equal to the adjacent side over the hypotenuse cosine of that angle will equal the weight divided by F now solve this thing for F okay remember this is really over one I mean I drew it down a little low but you can basically just switch this numerator come on this numerator with this denominator okay so when you solve that on out you should get let me just backtrack you should get now F is equal to double you over cosine of theta now this is great because literally now I know this F here right and this F are the same thing I mean I used the same F in both of these equations so what that means is that now I can substitute this right on in to my equation so instead of writing f here I'm gonna write now W cosine of theta so let's write that sewed up well not W cosine W divided by cosine of theta right then times the sine of theta will be equal to Ma sea all right now I see one thing right away I got sine divided by cosine essentially right now that has to ring some alarm bells in a good way right sine of a cosine is what tangent oh look at that oh I see tangent in here all right we must be getting warmer so now W times the tangent of theta will equal MA see let's expand this w to be mg right because that is the case so we can now see that we're going to cancel the M on that side right the M on the other side that's what my plan was so we got them to cancel so now we got G right times the tangent of theta is equal to centripetal acceleration okay now let me take a step back I see I have centripetal acceleration in my formula down here but I look back to here and I don't have it at all but I'm noticing some some things I recognize right I'm seeing what I got a V squared I got an R I know I have a G in here right but I notice I'm missing a V squared in R so I'm thinking to myself well how is there any way that centripetal acceleration is related to those things oh wait a minute look on the right-hand side guys there it is there's the relationship okay so I know that now I can plug in V squared over R for the centripetal acceleration so let's do that so this is G times the tangent of theta is equal to V squared over R we are very very close now okay so now basically I got to get rid of the G on this side so you can divide you know divide this side by G or multiply it by one over you might see that a little better multiply them on the other side by one over G the G's cancel here we're left with follow me on the top here now it's tangent of theta will be equal to V squared divided by R times G we are right there all we have to now do is get rid of the tangent and we do that by taking the inverse tan of both sides okay so it's basically right tan to the minus 1 of this thing and then tan to the minus 1 of that whole side so then that gets us this cancels the tangent over here so we get now theta equaling inverse tangent of V squared over R G and voila there it is ladies and gentlemen we found it all right so not too bad that was letter A let's take now let's take a look now at a letter B okay so let's take a look at letter B calculate theta for a 12 meter per second turn of radius 330 meters and then erase all right so it looks like we're just plugging it in right thank God all right so theta here is going to equal the inverse tangent right of the velocity which they gave to us as 12 right 12 point 0 that's gonna be squared divide that by then the radius which was 30 multiplied by G 9.8 oh all right then I love these type of problems where they just plug in right I mean do you agree with me I agree so inverse tangent of 12 squared divided by 4 at the C 30 times 9.8 close those parentheses and we had a value of 26.1 all right so 26 point one degrees that is the angle okay so excellent thank you guys so much for tuning in I hope this helped it's not that I it's not that bad right it's it's just a couple of substitutions it's like the hardest thing I always thought when I was doing a class like this was how know where to start now that's always sometimes the hardest part now sometimes it takes a little foresight and little planning okay and it takes a little experience too and as you do more and more of these questions you will begin to see some of the patterns it's tough to see at the start don't get frustrated keep going all right I know you can get there so thanks guys for tuning in please remember to subscribe and look forward to helping you in the next question take care |
6913 | https://cs.uwaterloo.ca/journals/JIS/VOL12/Kitaev/kitaev4.pdf | 23 11 Article 09.5.1 Journal of Integer Sequences, Vol. 12 (2009), 2 3 6 1 47 Equidistribution of Descents, Adjacent Pairs, and Place-Value Pairs on Permutations Emeric Deutsch Department of Mathematics Polytechnic Institute of NYU Brooklyn, NY 11201 USA deutsch@duke.poly.edu Sergey Kitaev1 The Mathematics Institute School of Computer Science Reykjav´ ık University IS-103 Reykjav´ ık Iceland sergey@ru.is Jeffrey Remmel2 Department of Mathematics University of California, San Diego La Jolla, CA 92093-0112 USA remmel@math.ucsd.edu 1The work presented here was supported by grant no. 090038011 from the Icelandic Research Fund.
2Partially supported by NSF grant DMS 0654060.
1 Abstract An (X, Y )-descent in a permutation is a pair of adjacent elements such that the first element is from X, the second element is from Y , and the first element is greater than the second one.
An (X, Y )-adjacency in a permutation is a pair of adjacent elements such that the first one is from X and the second one is from Y . An (X, Y )-place-value pair in a permutation is an element y in position x, such that y is in Y and x is in X. It turns out, that for certain choices of X and Y some of the three statistics above become equidistributed. Moreover, it is easy to derive the distribution formula for (X, Y )-place-value pairs thus providing distribution for other statistics under consideration too. This generalizes some results in the literature. As a result of our considerations, we get combinatorial proofs of several remarkable identities. We also conjecture existence of a bijection between two objects in question preserving a certain statistic.
1 Introduction Let Sn denote the set of permutations of [n] = {1, . . . , n} and N = {1, 2, . . .}. Also, E and O denote the set of even and odd numbers, respectively. For σ = σ1 · · · σn ∈Sn and X, Y ⊆N define the following permutation statistics desX,Y (σ) = |{i : σi > σi+1, & σi ∈X & σi+1 ∈Y }|, adjX,Y (σ) = |{i : σi ∈X & σi+1 ∈Y }|, valX,Y (σ) = |{i : i ∈X & σi ∈Y }|, excX,Y (σ) = |{i : σi > i & i ∈X & σi ∈Y }|, and the following corresponding polynomials DX,Y n (x) = X σ∈Sn xdesX,Y (σ) = n−1 X s=0 DX,Y n,s xs, AX,Y n (x) = X σ∈Sn xadjX,Y (σ) = n−1 X s=0 AX,Y n,s xs, V X,Y n (x) = X σ∈Sn xvalX,Y (σ) = n−1 X s=0 V X,Y n,s xs, EX,Y n (x) = X σ∈Sn xexcX,Y (σ) = n−1 X s=0 EX,Y n,s xs.
Objects counted by desX,Y are called (X, Y )-descents in . Similarly, we can talk of (X, Y )-adjacencies, (X, Y )-place-value pairs, and (X, Y )-excedances.
Foata’s first transformation exchanging excedances and descents (to be used in the paper) can most easily be explained with an example. The permutation w = 61437258 has three excedances: 6, 4, and 7 in positions 1, 3, and 5, respectively. We write w in cycle form: (162)(34)(57)(8).
Next, write each cycle with largest element last, and order the cycles by increasing largest element: (34)(216)(57)(8). Finally, reverse each cycle and erase the parentheses to get the outcome permutation 43612758 with the descents 43, 61, and 75.
Remark 1. Using Foata’s first transformation, one obtains that DX,Y n (x) = EY,X n (x). Thus, we do not need to provide any arguments for the polynomial EX,Y n (x) and its coefficients, instead studying the other three polynomials.
In this paper, we use the following notation for any X ⊆N and integer n ≥1: Xn = [n] ∩X, xn = |Xn|, Xc n = [n] −X, and xc n = |Xc n|.
Collecting some data on the polynomials, we noticed several equidistributions among the statistics, and nice formulas associated with them, for particular choices of sets X and Y .
We collect those observations in Table 1, where an,k denotes the number of permutations in Sn with k occurrences of the corresponding statistic.
Many of formulas listed in Table 1 are known (see, e.g., ). Others are new but quite easy to prove. Our idea to establish the equidistribution results is to prove general recurrence relations for the statistics for arbitrary choice of sets X and Y .
Then we will get the equidistributions in Table 1 as a simple corollary to the fact that the recurrences for the statistics in a given block are the same for a particular choice of X and Y . For example, we will show that whenever X and Y are disjoint subsets of N, then AX,Y n,s = V X,Y n,s for all n and s. Indeed, the recursions that we develop will allow us to give a bijective proof of this fact.
Other equidistribution results follow from simple bijections. For example, it is easy to see that for any X and Y , AX,Y n,s = AY,X n,s since if σiσi+1 is an (X, Y )-adjacency in σ = σ1 · · · σn, then σi+1σi is a (Y, X)-adjacency in the reverse of σ, σr = σnσn−1 · · · σ1.
Several of our formulas are quite easy to prove for one of our three statistics. For example, it is always easy to compute V X,Y n,s .
Theorem 2. For any X, Y ⊆N, n ≥1, and 0 ≤s ≤n, V X,Y n,s = s!(xn −s)!(xc n)!
xn s yn s yc n xn −s .
(1) Proof. To count the number of permutations of length n with s occurrences of valX,Y , we can first pick s positions from Xn in xn s ways for the places where we will have values of Y occurring in the places corresponding to Xn. Then we pick s values from Yn in yn s ways, and permute the values in s! ways to arrange the s occurrences of values in Yn in the places in Xn. In the remaining xn −s places in Xn, we must choose values from Y c n. We thus have yc n xn−s ways to choose those values and (xn −s)! ways to rearrange them. Finally we have xc n! ways to arrange the elements in places outside of Xn.
Similarly, it is easy to count AX,X n,s for any set X ⊆N. That is, we have the following theorem.
3 Stat.
Description, related polynomial, and enumeration S1 # of even descent-tops (DE,N n,k ). E.g., S1(215436) = 2.
S2 # of even excedance values (EN,E n,k ). E.g., S2(215436) = 1.
S3 # of even entries in even positions (V E,E n,k ). E.g., S3(215436) = 2.
a2n,k = n!
n k 2; a2n+1,k = n!(n + 1)!
n k n+1 k+1 .
S4 # of odd descent-bottoms (DN,O n,k ). E.g., S4(215436) = 2.
S5 # of odd excedance positions (EO,N n,k ). E.g., S5(215436) = 2.
S6 # of even entries in odd positions (V O,E n,k ). E.g., S6(215436) = 1.
S7 # of odd entries in even positions (V E,O n,k ). E.g., S7(215436) = 1.
S8 # of (odd,even) pairs (AO,E n,k ). E.g., S8(2154 36) = 2.
S9 # of (even, odd) pairs (AE,O n,k ). E.g., S9(215436) = 2.
a2n,k = n!
n k 2; a2n+1,k = n!(n + 1)!
n k n+1 k .
S10 # of odd descent-tops (DO,N n,k ). E.g., S10(215436) = 1.
S11 # of odd excedance values (EN,O n,k ). E.g., S11(215436) = 1.
S12 # of (odd,odd) pairs (AO,O n,k ). E.g., S12(215436) = 1.
a2n,k = (n!)2 n−1 k n+1 k+1 ; a2n+1,k = n!(n + 1)!
n k n+1 k .
S13 # of even descent-bottoms (DN,E n,k ). E.g., S13(215436) = 1.
S14 # of even excedance positions (EE,N n,k ). E.g., S14(215436) = 0.
a2n,k = (n!)2 n−1 k n+1 k+1 ; a2n+1,k = n!(n + 1)!
n k n+1 k+1 .
S15 # of odd entries in odd positions (V O,O n,k ). E.g., S15(215436) = 2.
a2n,k = n!
n k 2; a2n+1,k = n!(n + 1)!
n k−1 n+1 k .
S16 # of (even,even) pairs (AE,E n,k). E.g., S16(215436) = 0.
a2n,k = (n!)2 n−1 k n+1 k+1 ; a2n+1,k = n!(n + 1)!
n−1 k n+2 k+2 .
Table 1: 16 statistics under consideration classified into 6 statistic groups.
4 Theorem 3. For any X ⊆N, n ≥1, and 0 ≤s ≤n −1, AX,X n,s = (xn)!(xc n)!
xn −1 s xc n + 1 xn −s .
(2) Proof. Fix n ≥1. First we pick a permutation σ of X ∩[n] and a permutation τ of [n] −X.
Clearly, we have (xn)!(xc n)! ways to pick σ and τ. We are now interested in finding the number of permutations of γ of Sn such that γ restricted to the elements in X ∩[n] yields the permutation σ, γ restricted to the elements in [n] −X yields the permutation τ, and adjX,X(γ) = s. Next in σ1σ2 · · · σxn, we think of choosing s spaces from the xn −1 spaces between the elements of σ to create the adjacencies that will appear in such a γ. For example, if n = 12, s = 2, X = E, σ = 4 2 10 8 6 12, and we pick spaces 2 and 5, then our choice partitions σ into four blocks, 4, 2 −10, 8 and 6 −12. Our idea is to insert these blocks into the spaces that either lie immediately before an element of τ or immediately after the last element of τ. We label these spaces from left to right. For example, suppose τ = 5 1 7 9 3 11 and we pick spaces 2, 4, 5, and 7. Then we would insert the block 4 immediately before 1, the block 2 −10 immediately before 9, the block 8 immediately before 3, 6−12 immediately after 11 to obtain the permutation 5 4 1 7 2 10 9 8 3 11 6 12.
Clearly there are xn−1 s ways to choose the spaces to obtain our s adjacencies. This will leave us with xn −s blocks. Then there are xc n+1 xn−s to choose the spaces for τ where we insert the blocks.
Hall and Remmel gave direct combinatorial proofs of a pair of formulas for DX,Y n,s which combined with our equidistribution results, gives formulas for the other polynomials under consideration. We state these results here together with an example of using them.
Theorem 4. For any X, Y ⊆N, n ≥1, and 0 ≤s ≤n −1, DX,Y n,s = |Xc n|!
s X r=0 (−1)s−r |Xc n| + r r n + 1 s −r Y x∈Xn (1 + r + αX,n,x + βY,n,x), (3) Theorem 5. For any X, Y ⊆N, n ≥1, and 0 ≤s ≤n −1, DX,Y n,s = |Xc n|!
|Xn|−s X r=0 (−1)|Xn|−s−r |Xc n| + r r n + 1 |Xn| −s −r Y x∈Xn (r + βX,n,x −βY,n,x), (4) where for any set A and any j, 1 ≤j ≤n, we define αA,n,j = |Ac ∩{j + 1, j + 2, . . . , n}| = |{x : j < x ≤n & x / ∈A}|, and βA,n,j = |Ac ∩{1, 2, . . . , j −1}| = |{x : 1 ≤x < j & x / ∈A}|.
Example 1. Suppose X = {2, 3, 4, 6, 7, 9}, Y = {1, 4, 8}, and n = 6. Thus X6 = {2, 3, 4, 6}, Xc 6 = {1, 5}, Y6 = {1, 4}, Y c 6 = {2, 3, 5, 6}, and we have the following table of values of αX,6,x, βY,6,x, and βX,6,x.
5 x 2 3 4 6 αX,6,x 1 1 1 0 βY,6,x 0 1 2 3 βX,6,x 1 1 1 2 Equation (3) gives DX,Y 6,2 = 2!
2 X r=0 (−1)2−r 2 + r r 7 2 −r (2 + r)(3 + r)(4 + r)(4 + r) = 2 (1 · 21 · 2 · 3 · 4 · 4 −3 · 7 · 3 · 4 · 5 · 5 + 6 · 1 · 4 · 5 · 6 · 6) = 2(2016 −6300 + 4320) = 72, while (4) gives DX,Y 6,2 = 2!
2 X r=0 (−1)2−r 2 + r r 7 2 −r (1 + r)(0 + r)(−1 + r)(−1 + r) = 2 (1 · 21 · 1 · 0 · (−1) · (−1) −3 · 7 · 2 · 1 · 0 · 0 + 6 · 1 · 3 · 2 · 1 · 1) = 2(0 −0 + 36) = 72.
The paper is organized as follows. In Section 2 we find general recurrence relations for DX,Y n,k , AX,Y n,k , and V X,Y n,k , and use them to explain the facts in Table 1.
In Section 3 we generalize several of the results that appear in Table 1, and use this to obtain combinatorial proofs of several remarkable identities. Finally, in Section 4, we discuss some directions for further research.
2 Recurrence relations for DX,Y n,k , AX,Y n,k , and V X,Y n,k In this section, we derive recurrence relations for DX,Y n,k , AX,Y n,k , and V X,Y n,k . We notice that the recurrences we get for AX,Y n,k and V X,Y n,k are almost identical, except for the case when the element n+1 ∈X ∩Y — the recurrences differ by “1+.” However, assuming X ∩Y = ∅, we do not have this case, leading, in particular, to the explanation of all of the equidistributions in Table 1, and to many more results for other choices of X and Y , X ∩Y = ∅.
Another thing to observe is that in the case of the same recurrence relations, we naturally get bijective proofs for the corresponding equidistributed statistics. Indeed, one can label positions in a permutation, say from left to right, in which we insert the largest element, n + 1, or do the other insertion procedure (see Subsection 2.3); then, it is enough to match insertions in the positions having the same labels. However, such straightforward approach is not necessarily the best one, as labeling positions differently, rather than just from left to right, one may preserve extra statistics in bijections (see Section 4 for conjectures, which should be possible to prove using our approach with different labeling).
6 2.1 Recurrences for DX,Y n,k A recursion for DX,Y n,k is derived in : DX,Y n+1,k = ( (k + 1)DX,Y n,k+1 + (n + 1 −k)DX,Y n,k , if n + 1 ̸∈X; (yn −(k −1))DX,Y n,k−1 + (n + 1 −(yn −k))DX,Y n,k , if n + 1 ∈X.
An argument for deriving the recursion is as follows. We are thinking of inserting the element n + 1 in a permutation σ = σ1 · · · σn, and we consider which of the obtained permutations are counted by DX,Y n,k . If n + 1 ̸∈X then one never increases the number of (X, Y )-descents by inserting n + 1. More precisely, the number of (X, Y )-descents is either unchanged, or it is decreased by 1, when n + 1 is inserted between σi ∈X and σi+1 ∈Y where σi > σi+1.
The corresponding recursion case follows.
For the second case, notice that if n + 1 ∈X, then the number of (X, Y )-descents is unchanged if n + 1 is inserted at the end of the permutation, in front of σj ̸∈Y , or between σi ∈X and σi+1 ∈Y where σi > σi+1, and it is increased by 1 in other cases (that is, when n + 1 is inserted in front of σj ∈Y not involved in an (X, Y )-descent). The second recursion case follows.
We use a similar approach to derive recurrence relations for AX,Y n,k . Our derivations for V X,Y n,k use a different insertion procedure.
2.2 Recurrences for AX,Y n,k We consider 4 cases.
Case 1. n + 1 ̸∈X ∪Y . The number of (X, Y )-adjacent pairs is decreased by 1 when n + 1 is inserted between σi ∈X and σi+1 ∈Y and it is unchanged otherwise. Thus, in this case AX,Y n+1,k = (k + 1)AX,Y n,k+1 + (n + 1 −k)AX,Y n,k .
Case 2. n + 1 ∈X ∩Y . Adding n + 1 after a σi ∈X or before a σj ∈Y increases adjX,Y by 1, while it keeps adjX,Y (σ) unchanged otherwise. However, we note that the place between σi ∈X and σi+1 ∈Y is after a σi ∈X and before a σi+1 ∈Y . Thus, in this case AX,Y n+1,k = (xn + yn −(k −1))AX,Y n,k−1 + (n + 1 −(xn + yn −k))AX,Y n,k .
Case 3. n+1 ∈X −Y . Inserting n+1 to the left of a σi ̸∈Y does not change adjX,Y (σ), which is also the case if n + 1 is inserted between σi ∈X and σi+1 ∈Y , or n + 1 is inserted at the very end. On the other hand, if n + 1 is inserted between σi ̸∈X and σi+1 ∈Y , the number of (X, Y )-adjacent pairs is increased by 1. Thus, in this case AX,Y n+1,k = (yn −(k −1))AX,Y n,k−1 + (n + 1 −(yn −k))AX,Y n,k .
Case 4. n + 1 ∈Y −X. Inserting n + 1 to the right of a σi ̸∈X does not change adjX,Y (σ), which is also the case if n + 1 is inserted between σi ∈X and σi+1 ∈Y , or n + 1 is inserted at the very beginning. On the other hand, if n + 1 is inserted between σi ∈X and σi+1 ̸∈Y , the number of (X, Y )-adjacent pairs is increased by 1. Thus, in this case AX,Y n+1,k = (xn −(k −1))AX,Y n,k−1 + (n + 1 −(xn −k))AX,Y n,k .
7 2.3 Recurrences for V X,Y n,k Instead of inserting the largest element, n+1, in all possible places, we use another insertion procedure I(i) n (σ) that generates Sn+1 from Sn. For σ = σ1 · · · σn, let I(n+1) n+1 (σ) = σ(n + 1) = σ1 · · · σn(n + 1), and for 1 ≤i ≤n, let I(i) n+1(σ) = σ1 · · · σi−1(n + 1)σi+1 · · · σnσi (that is, in the last case we replace σi in σ by n + 1 and move σi to the very end).
We now consider 4 cases.
Case 1. n + 1 ̸∈X ∪Y . In this case, one can only decrease the number of (X, Y )-place-value pairs. This happens when n + 1 occupies position i ∈X in I(i) n+1(σ) for some σ, such that σi ∈Y (σi is in position n + 1 in I(i) n+1(σ)). Thus, in this case V X,Y n+1,k = (k + 1)V X,Y n,k+1 + (n + 1 −k)V X,Y n,k .
Case 2. n+1 ∈X ∩Y . It is straightforward to see that the number of (X, Y )-place-value pairs is unchanged if i ̸∈X and σi ̸∈Y , and it increases by 1 in each of the following three cases: i ∈X and σi ∈Y , i ∈X and σi ̸∈Y , and i ̸∈X and σi ∈Y . Note that we add 1 for each i ∈X and 1 for each σi ∈Y , so we count i ∈X and σi ∈Y twice. Moreover, having n + 1 in position n + 1 gives one more (X, Y )-place-value pair. Thus, in this case V X,Y n+1,k = (1 + xn + yn −(k −1))V X,Y n,k−1 + (n + 1 −(1 + (xn + yn −k)))V X,Y n,k .
Case 3. n + 1 ∈X −Y . One can check that in this case, the number of (X, Y )-place-value pairs increases by 1 if i ̸∈X and σi ∈Y , and it is unchanged otherwise. Thus, in this case V X,Y n+1,k = (yn −(k −1))V X,Y n,k−1 + (n + 1 −(yn −k))V X,Y n,k .
Case 4. n + 1 ∈Y −X. One can check that in this case, the number of (X, Y )-place-value pairs increases by 1 if i ∈X and σi ̸∈Y , and it is unchanged otherwise. Thus, in this case V X,Y n+1,k = (xn −(k −1))V X,Y n,k−1 + (n + 1 −(xn −k))V X,Y n,k .
There are a number of cases where the recursions for V A,B n,k , AC,D n,k , and DE,F n,k coincide so that we immediately have equality between the various pairs of statistics. For example, comparing the recursions for AX,Y n,k and V X,Y n,k , we immediately have the following theorem.
Theorem 6. For all X, Y ⊆N such that X ∩Y = ∅, n ≥1, and 0 ≤k ≤n, V X,Y n,k = AX,Y n,k .
In fact, it is easy to see that our proofs of the recursions can be used to give an inductive proof that there exists a bijection from Sn onto Sn for all n that will witness this equality.
That is, our proofs of the recursions immediately allow us to construct inductively bijections Θn : Sn →Sn for all n such that for all σ ∈Sn, adjX,Y (σ) = valX,Y (Θn(σ)).
For example, suppose that we have constructed Θn and n + 1 ̸∈X ∪Y . First consider our insertion procedure to prove the recursions for AX,Y n,s . If σ ∈Sn, then we consider the places 8 where we can insert n + 1 to σ. We first label the spaces between the elements σi ∈X and σi+1 ∈Y from left to right with 1, . . . , adjX,Y (σ) and then label the rest of the spaces from left to right with adjX,Y (σ)+1, . . . , n+1. For example, if X = E, Y = O, and σ = 1 4 3 2 5, the spaces would be labeled by − 31− 44− 13− 52− 25− 6.
We then let σ(i) be the permutation that results by inserting n + 1 into the space labeled i.
For example, in our example, σ(4) = 1 6 4 3 2 5. Next we consider our insertion procedure for proving the recursions for V X,Y n,s . Now if τ ∈Sn, then we label the positions of τ by first labeling the positions i such that i ∈X and τi ∈Y from left to right with 1, . . . , valX,Y (τ) and then label the remaining positions from left to right with valX,Y (τ) + 1, . . . , n. For example, if X = O and Y = E and τ = 1 4 2 5 3, then we would label the positions 1 2 4 3 2 1 5 4 3 5 where we have indicated the labels in boldface.
If label j is in position i, then we let τ (j) = I(i) n (τ) and we let τ (n+1) = I(n+1) n (τ). For example, in our case, τ (2) = 6 4 2 5 3 1.
Then for any σ ∈Sn and i ∈{1, . . . , n + 1}, we can define Θn+1(σ(i)) = Θn(σ)(i).
We can extend Θn to Θn+1 in the other cases of the recursions in a similar manner.
Similarly, comparing the recursions for the V A,B n,k , AC,D n,k , and DE,F n,k , we can also derive bijective proofs of the following theorems.
Theorem 7. If X and Y are subsets of N, A = X ∪Y and there exists a B ⊆N such that bn = |B ∩[n]| satisfies bn = xn + yn = |X ∩[n]| + |Y ∩[n]|, if n + 1 ∈X ∩Y ; yn = |Y ∩[n]|, if n + 1 ∈X −Y ; xn = |X ∩[n]|, if n + 1 ∈Y −X, then DA,B n,k = AX,Y n,k .
Theorem 8. If X and Y are subsets of N, A = X ∪Y and there exists a B ⊆N such that bn = |B ∩[n]| satisfies bn = 1 + xn + yn = 1 + |X ∩[n]| + |Y ∩[n]|, if n + 1 ∈X ∩Y ; yn = |Y ∩[n]|, if n + 1 ∈X −Y ; xn = |X ∩[n]|, if n + 1 ∈Y −X, then DA,B n,k = V X,Y n,k .
9 2.4 Explanation of Table 1 using our general results 1. The first group of statistics. DE,N n,k = EN,E n,k by Foata’s first transformation. Also, DE,N n,k = V E,E n,k by Theorem 8. Indeed, in this case A = X = Y = E and B = N leading to A = X ∪Y , X −Y = Y −X = ∅, and bn = n = 1 + 2|E ∩[n]| if n + 1 ∈E. As for the formulas, we can apply Theorem 2 with X = Y = E: a2n,k = V X,Y 2n,k = k!(n −k)!n!
n k n k n n −k = n!
n k 2 ; a2n+1,k = V X,Y 2n+1,k = k!(n −k)!(n + 1)!
n k n k n + 1 n −k = n!(n + 1)!
n k n + 1 k + 1 .
2. The second group. DN,O n,k = EO,N n,k by Foata’s first transformation. Applying the reverse operation to each permutation, one sees that AO,E n,k = AE,O n,k . Applying the inverse operation to each permutation, one gets V O,E n,k = V E,O n,k . By Theorem 6, V O,E n,k = AO,E n,k as O ∩E = ∅. Finally, by Theorem 7, DN,O n,k = AO,E n,k . Indeed, in this case A = N, B = X = O, and Y = E leading to A = X ∪Y , and bn = # of odd numbers in [n] = E ∩[n], if n + 1 ̸∈O; O ∩[n], if n + 1 ∈E.
As for the formulas, we can apply Theorem 2 with X = E and Y = O: a2n,k = V O,E 2n,k = k!(n −k)!n!
n k n k n n −k = n!
n k 2 ; a2n+1,k = V O,E 2n+1,k = k!(n −k)!(n + 1)!
n k n + 1 k n n −k = n!(n + 1)!
n k n + 1 k .
3. The third group. Again, DO,N n,k = EN,O n,k by Foata’s first transformation. Moreover, by Theorem 7, DO,N n,k = AO,O n,k . Indeed, in this case A = X = Y = O and B = N leading to A = X ∪Y , X −Y = Y −X = ∅, and bn = n = 2|O ∩[n]| if n + 1 ∈O. As for the formulas, we can apply Theorem 3 with X = O: a2n,k = AO,O 2n,k = (n!)2 n −1 k n + 1 k + 1 ; a2n+1,k = AO,O 2n+1,k = n!(n + 1)!
n k n + 1 k .
4. The fourth group. DN,E n,k = EE,N n,k by Foata’s first transformation. The formulas for DN,E n,k are proved in [4, Section 4].
10 5. The fifth group. We use Theorem 2 with X = Y = O, to get a2n,k = V O,O 2n,k = k!n!n!
n k n k n n −k = n!
n k 2 ; a2n+1,k = V O,O 2n+1,k = k!(n+1−k)!n!
n + 1 k n + 1 k n k −1 = n!(n+1)!
n k −1 n + 1 k .
6. The sixth group. We use Theorem 3 with X = E, to get a2n,k = AE,E 2n,k = (n!)2 n −1 k n + 1 k + 1 ; a2n+1,k = AE,E 2n+1,k = n!(n + 1)!
n −1 k n + 2 n −k .
3 Applications In this section, we shall generalize several of the results that appear in Table 1. That is, in Table 1, we consider the parity of the elements in a descent, adjacency, or place-value pair. We shall show that we can get similar formulas when we consider the equivalence class modulo k of the elements in a descent, adjacency, or place-value pair. See for related research on descents generalizing results of . For any k ≥2 and 0 ≤i ≤k −1, we let i + kN = {i + kn : n ≥0}.
First, we shall consider V X,Y n,k and AX,Y n,k where X = i + kN and Y = j + kN and 0 ≤i < j ≤k −1.
It follows from Theorem 6 that V X,Y n,k = AX,Y n,k in this case.
Sup-pose that A = i + kN ∪j + kN and B = i + kN. Note that when m + 1 = kn + i ∈X −Y , then ym = n = bm = |B ∩[m]| and when m + 1 = kn + j ∈Y −X, then xm = n + 1 = bm.
Thus it follows from Theorems 7 and 8 that DA,B n,k = V X,Y n,s = AX,Y n,s for all n and s. We then have three cases.
Case 1. m = kn + t where 0 ≤t < i. In this case, xm = |X ∩[m]| = ym = |Y ∩[m]| = n and xc m = |[m] −X| = yc m = |[m] −Y | = (k −1)n + t. Thus it follows from Theorem 2 that V X,Y m,s = n s 2(k −1)n + t n −s s!(n −s)!((k −1)n + t)!.
(5) On the other hand, it follows from Theorem 4 that DA,B m,s = |Ac m|!
s X r=0 (−1)s−r |Ac m| + r r m + 1 s −r Y x∈Am (1 + r + αA,m,x + βB,m,x).
(6) In this case |Ac m| = kn + t −2n = (k −2)n + t. For any x, it is easy to see that αA,m,x + βB,m,x = kn + t −1 −|A ∩[x + 1, kn + t] −|B ∩[x −1]| 11 where [x + 1, kn + t] = {r : x + 1 ≤r ≤kn + t}. Thus for any 0 ≤a ≤n −1, αA,m,ak+i + βB,m,ak+i = kn + t −1 −(2n −(2a + 1)) −a = (k −2)n + t + a (7) and αA,m,ak+j + βB,m,ak+j = kn + t −1 −(2n −(2a + 2)) −(a + 1) = (k −2)n + t + a.
(8) Thus Y x∈Am (1 + r + αA,m,x + βB,m,x) = n−1 Y a=0 (1 + r + αA,m,ak+i + βB,m,ak+i) ! n−1 Y a=0 (1 + r + αA,m,ak+j + βB,m,ak+j) !
= n−1 Y a=0 (1 + r + (k −2)n + t + a) !2 = (1 + r + (k −2)n + t)n(1 + r + (k −2)n + t)n where we define (a)n by (a)0 = 1 and (a)n = a(a+1) · · · (a+n−1) for n ≥1. Since we have a combinatorial proof of the fact that V X,Y m,s = DA,B m,s in this case and the proof of Theorem 4 is also completely combinatorial, it follows that we have a combinatorial proof of the following identity: n s 2(k −1)n + t n −s s!(n −s)!((k −1)n + t)! = (9) ((k −2)n + t)!
s X r=0 (−1)s−r (k −2)n + t + r r kn + t + 1 s −r × (1 + r + (k −2)n + t)n(1 + r + (k −2)n + t)n.
Case 2.
m = kn + t where i ≤t < j.
In this case, xm = |X ∩[m]| = n + 1 and ym = |Y ∩[m]| = n and xc m = |[m]−X| = (k −1)n+t−1 and yc m = |[m]−Y | = (k −1)n+t.
Thus it follows from Theorem 2 that V X,Y m,s = n + 1 s n s (k −1)n + t n + 1 −s s!(n + 1 −s)!((k −1)n + t −1)!.
(10) On the other hand, we can obtain a formula for V X,Y m,s = DA,B m,s from equation (6). In this case |Ac m| = kn + t −(2n + 1) = (k −2)n + t −1. For any 0 ≤a ≤n, αA,m,ak+i + βB,m,ak+i = kn + t −1 −(2n + 1 −(2a + 1)) −a = (k −2)n + t −1 + a (11) 12 and, for any 0 ≤a ≤n −1 αA,m,ak+j + βB,m,ak+j = kn + t −1 −(2n + 1 −(2a + 2)) −(a + 1) = (k −2)n + t −1 + a.
(12) Thus Y x∈Am (1 + r + αA,m,x + βB,m,x) = n Y a=0 (1 + r + αA,m,ak+i + βB,m,ak+i) ! n−1 Y a=0 (1 + r + αA,m,ak+j + βB,m,ak+j) !
= (1 + r + (k −2)n + t −1)n+1(1 + r + (k −2)n + t −1)n = (r + (k −2)n + t)n+1(r + (k −2)n + t)n.
As in Case 1, it follows that we have a combinatorial proof of the following identity: n + 1 s n s (k −1)n + t n + 1 −s s!(n −s)!((k −1)n + t −1)! = (13) ((k −2)n + t −1)!
s X r=0 (−1)s−r (k −2)n + t −1 + r r kn + t + 1 s −r × (r + (k −2)n + t)n+1(r + (k −2)n + t)n.
Case 3. m = kn+t where j ≤t ≤k−1. In this case, xm = |X∩[m]| = ym = |Y ∩[m]| = n+1 and xc m = |[m] −X| = yc m = |[m] −Y | = (k −1)n + t −1. Thus it follows from Theorem 2 that V X,Y m,s = n + 1 s 2(k −1)n + t −1 n + 1 −s s!(n + 1 −s)!((k −1)n + t −1)!.
(14) On the other hand, we can obtain a formula for V X,Y m,s = DA,B m,s from equation (6). In this case |Ac m| = kn + t −(2n + 2) = (k −2)n + t −2. For any 0 ≤a ≤n, αA,m,ak+i + βB,m,ak+i = kn + t −1 −(2n + 2 −(2a + 1)) −a = (k −2)n + t −2 + a (15) and, for any 0 ≤a ≤n αA,m,ak+j + βB,m,ak+j = kn + t −1 −(2n + 2 −(2a + 2)) −(a + 1) = (k −2)n + t −2 + a.
(16) Thus Y x∈Am (1 + r + αA,m,x + βB,m,x) = n Y a=0 (1 + r + αA,m,ak+i + βB,m,ak+i) ! n Y a=0 (1 + r + αA,m,ak+j + βB,m,ak+j) !
= (1 + r + (k −2)n + t −2)n+1(1 + r + (k −2)n + t −2)n+1 = (r + (k −2)n + t −1)n+1(r + (k −2)n + t −1)n+1.
13 It follows that we have a combinatorial proof of the following identity: n + 1 s 2(k −1)n + t −1 n + 1 −s s!(n + 1 −s)!((k −1)n + t −1)! = (17) ((k −2)n + t −1)!
s X r=0 (−1)s−r (k −2)n + t −1 + r r kn + t + 1 s −r × (r + (k −2)n + t −1)n+1(r + (k −2)n + t −1)n+1.
Next we shall consider V X,Y n,k and AX,Y n,k where X = Y = i + kN for 0 ≤i ≤k −1 and k ≥2. In this case, it is no longer the case that AX,X m,s = V X,X m,s so we will handle the cases of AX,X m,s and V X,X m,s separately.
First we shall consider V X,Y n,s . Note that if A = i+kN and B = i+kN∪i+1+kN, then for m+1 = kn+i ∈X ∩Y , then xn = |X ∩[m]| = n = ym = |Y ∩[m]| and bm = |B ∩[m]| = 2n.
Thus it follows from Theorem 8 that V X,X n,s = DA,B n,s for all n and s. We then have two cases.
Case I. m = kn + t where 0 ≤t < i.
In this case, xm = |X ∩[m]| = n and xc n = |[m] −X| = (k −1)n + t. Then it follows from Theorem 2 that V X,X m,s = n s 2(k −1)n + t n −s s!(n −s)!((k −1)n + t)!.
(18) On the other hand, we can obtain a formula for V X,X m,s = DA,B m,s from equation (6). In this case |Ac m| = kn + t −n = (k −1)n + t. For any 0 ≤a ≤n, αA,m,ak+i + βB,m,ak+i = kn + t −1 −(n −(a + 1)) −2a = (k −1)n + t −a.
(19) Thus Y x∈Am (1 + r + αA,m,x + βB,m,x) = n−1 Y a=0 (1 + r + αA,m,ak+i + βB,m,ak+i) = n−1 Y a=0 (1 + r + (k −1)n + t −a) = (1 + r + (k −1)n + t) ↓n where (a) ↓n is defined by (a) ↓0= 1 and (a) ↓n= a(a −1) · · · (a −n + 1) for n ≥1. Thus it follows that n s 2(k −1)n + t n −s s!(n −s)!((k −1)n + t)! = (20) ((k −1)n + t)!
s X r=0 (−1)s−r (k −1)n + t + r r kn + t + 1 s −r (1 + r + (k −1)n + t) ↓n .
14 Case II. m = kn + t where i ≤t ≤k −1. In this case, xm = |X ∩[m]| = n + 1 and xc n = |[m] −X| = (k −1)n + t −1. Then it follows from Theorem 2 that V X,X m,s = n + 1 s 2(k −1)n + t −1 n + 1 −s s!(n + 1 −s)!((k −1)n + t −1)!.
(21) On the other hand, we can obtain a formula for V X,X m,s = DA,B m,s from equation (6). In this case |Ac m| = kn + t −(n + 1) = (k −1)n + t −1. For any 0 ≤a ≤n, αA,m,ak+i + βB,m,ak+i = kn + t −1 −(n + 1 −(a + 1)) −(2a + 1) = (k −1)n + t −1 −a.
(22) Thus Y x∈Am (1 + r + αA,m,x + βB,m,x) = n Y a=0 (1 + r + αA,m,ak+i + βB,m,ak+i) = n−1 Y a=0 (1 + r + (k −1)n + t −1 −a) = (r + (k −1)n + t) ↓n+1 .
Thus it follows that n + 1 s 2(k −1)n + t −1 n + 1 −s s!(n + 1 −s)!((k −1)n + t −1)! = (23) ((k −1)n + t −1)!
s X r=0 (−1)s−r (k −1)n + t −1 + r r kn + t + 1 s −r (r + (k −1)n + t) ↓n+1 .
Next we consider the case of computing AX,Y n,s where X = Y = i + kN where k ≥2 and 0 ≤i ≤k −1. Let A = i + kN and B = i −1 + kN. Then it is easy to see that if m + 1 = kn + i ∈X ∩Y = X, then xm = ym = n and bm = 2n + 1. Thus it follows from Theorem 7 that AX,Y n,k = DA,B n,s for all n and s in this case. We then have two cases.
Case A. m = kn + t where 0 ≤t < i −1. In this case, xm = n and xc m = (k −1)n + t. Thus if follows from Theorem 3 that AX,X m,s = n!((k −1)n + t)!
n −1 s (k −1)n + t + 1 n −s .
(24) On the other hand, we can obtain a formula for AX,X m,s = DA,B m,s from equation (6). In this case |Ac m| = (k −1)n + t. For any 0 ≤a ≤n, αA,m,ak+i + βB,m,ak+i = kn + t −1 −(n −(a + 1)) −(2a + 1) = (k −1)n + t −1 −a.
(25) 15 Thus Y x∈Am (1 + r + αA,m,x + βB,m,x) = n−1 Y a=0 (1 + r + αA,m,ak+i + βB,m,ak+i) = n−1 Y a=0 (1 + r + (k −1)n + t −1 −a) = (r + (k −1)n + t) ↓n .
Thus it follows that n!((k −1)n + t)!
n −1 s (k −1)n + t + 1 n −s = (26) ((k −1)n + t)!
s X r=0 (−1)s−r (k −1)n + t + r r kn + t + 1 s −r (r + (k −1)n + t) ↓n .
Case B. m = kn+t where i ≤t < k −1. In this case, xm = n+1 and xc m = (k −1)n+t−1.
Thus if follows from Theorem 3 that AX,X m,s = (n + 1)!((k −1)n + t −1)!
n s (k −1)n + t n + 1 −s .
(27) On the other hand, we can obtain a formula for AX,X m,s = DA,B m,s from equation (6). In this case |Ac m| = (k −1)n + t −1. For any 0 ≤a ≤n, αA,m,ak+i + βB,m,ak+i = kn + t −1 −(n + 1 −(a + 1)) −(2a + 1) = (k −1)n + t −2 −a.
(28) Thus Y x∈Am (1 + r + αA,m,x + βB,m,x) = n+1 Y a=0 (1 + r + αA,m,ak+i + βB,m,ak+i) = n+1 Y a=0 (1 + r + (k −1)n + t −2 −a) = (r + (k −1)n + t −1) ↓n+1 .
Thus it follows that n!((k −1)n + t −1)!
n + 1 s (k −1)n + t n + 1 −s = (29) ((k −1)n + t −1)!
s X r=0 (−1)s−r (k −1)n + t −1 + r r kn + t + 1 s −r (r + (k −1)n + t −1) ↓n+1 .
16 4 Direction for future research In this section, we shall describe some problems for further research that naturally arise from the work in this paper.
There are other statistics which are closely related to the statistics that we consider in this paper. For example, suppose that X, Y ⊆N and define γX,Y (σ) = |{i ∈X : σi ∈X} ∪{i ∈Y : σi ∈Y }|.
Let ΓX,Y n,s = |{σ ∈Sn : γX,Y (σ) = s}|. Then we have the following theorem.
Theorem 9. For any X and Y such that X ∪Y = N and X ∩Y = ∅, we have ΓX,Y n,s = 0 unless s = 2k + yn −xn for some k, in which case ΓX,Y n,2k+yn−xn = (xn)!(yn)!
xn k yn xn −k .
Proof. Suppose we pick k positions in Xn to contain elements in Xn in xn k ways and we pick xn −k positions Yn to put the other elements of Xn in yn xn−k ways. The remaining positions in the permutation must be filled with elements of Yn. Next arrange elements in Xn in (xn)! ways and we arrange elements in Yn in (yn)!. Clearly the number of permutations σ that can be constructed in this way is (xn)!(yn)!
xn k yn xn−k . Note that our construction forces yn −(xn −k) elements of Yn to be in positions in Yn so that for any σ constructed in this way γX,Y (σ) = 2k + yn −xn.
Note that in the special case where X = E and Y = O, we have that ΓE,O 2s,2n = (n!)2 n s 2 and ΓE,O 2s+1,2n+1 = n!(n + 1)!
n s n+1 s+1 which agrees with other formulas in our table in these special cases. However, for general X and Y , we get quite different recursions. For example, suppose that n + 1 ∈X ∩Y , σ ∈Sn, i ∈X −Y , and σi ∈Y −X. Then it is easy to see that γX,Y (I(i) n+1(σ)) = γX,Y (σ) + 2 so that the value of γX,Y can jump by 2 with a single insertion. This type of phenomenon does not happen with any of the other statistics studied in this paper. Thus it would be interesting to further study ΓX,Y n,s for arbitrary X, Y ⊆N to see if one can prove explicit formulas for ΓX,Y n,s which are similar to the formulas for DX,Y n,s given in Theorems 4 and 5.
Even though we found solutions to all the bijective questions related to the objects in our table, in some cases, one should be able to modify our bijections (find new ones) to preserve more than one statistic.
Recall that the statistic S10 is the number of odd descent-tops, and S12 is the number of (odd,odd) pairs. In this section, we will use the following statistics as well.
• S17 — the length of the maximal subsequence of the form 12 · · · i in a permutation.
E.g., S17(34152) = 2 while the increasing permutation of length n gives the maximum value of S17 in Sn. A modification of this statistic was studied by Zeilberger in connection with 2-stack sortable permutations.
17 • T1 = S10 but not S12.
• T2 = S12 but not S10.
• T3 = S10 and S12.
Our first conjecture is the following joint equidistribution: Conjecture 10. The following should be true: (S10, S12, S17) ∼(S12, S10, S17).
Notice, that Conjecture 10 suggests existence of an involution turning S10 to S12 and vice versa. This conjecture can be refined as follows.
Conjecture 11. (T1, T2, T3, S17) ∼(T2, T1, T3, S17). That is, if the involution mentioned above exists, it is likely to leave pairs that are both S10 and S12 untouched.
Observe that to preserve statistic S17 in Conjectures 10 and 11, we need to require the increasing n-permutation to go to itself, and this is the only thing we need to worry about in our recursive construction of the bijection regarding S17 as otherwise it is not changed and thus preserved by induction no matter where we stick the largest element. So, it seems like we should be able to have the increasing permutation as a fixed point.
Here is how a proof of Conjecture 11 could be arranged for even n assuming the rest is constructed by induction. For odd n’s things seem to be much more complicated.
For n = 1, 1 is mapped to 1. Suppose we have constructed a bijection from Sn−1 to Sn−1 (n is even) such that it sends k (resp. ℓ, s) occurrences of T1 (resp. T2, T3) to k (resp.
ℓ, s) occurrences of T2 (resp. T1, T3). Inserting n in T1 (resp. T2, T3) pair decreases the number of T1 (resp. T2, T3) by 1 keeping all other statistics unchanged. Clearly, we can manage the corresponding insertion on the other side that would decrease by 1 the number of occurrences of the corresponding statistic. Inserting n at any other position does not change a thing in either side and can be matched to each other. In particular, inserting n at the end corresponds to inserting n at the end and this guarantees that the statistic S17 is preserved (either it is unchanged in both cases, or assuming we deal with the increasing permutation 12 · · · (n −1) going to itself, S17 is increased by 1 in both cases).
References D. C. Foata and M. P. Sch¨ utzenberger, Th´ eorie G´ eom´ etrique des Polynˆ omes Euleriens, Lecture Notes in Math. 138, Springer-Verlag, Berlin, 1970.
J. Hall and J. Remmel, Counting descent pairs with prescribed tops and bottoms, J.
Combin. Theory Ser. A 115 (2008), 693–725.
S. Kitaev and J. Remmel, Classifying descents according to equivalence mod k, Electron.
J. Combin. 13 (2006), #R64.
S. Kitaev and J. Remmel, Classifying descents according to parity, Ann. Comb. 11 (2007), 173–193.
18 D. Zeilberger, A proof of Julian West’s conjecture that the number of two-stack-sortable permutations of length n is 2(3n)!/((n+1)!(2n+1)!), Discrete Math. 102 (1992), 85–93.
2000 Mathematics Subject Classification: Primary 05A15, 05A19 Secondary 05A05.
Keywords: descent, adjacent pair, place-value-pair, equidistribution, binomial identity.
Received March 14 2009; revised version received June 21 2009. Published in Journal of Integer Sequences, June 21 2009.
Return to Journal of Integer Sequences home page.
19 |
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Gonorrhea
Gonorrhea
Gonorrhea, caused by Neisseria gonorrhoeae, is the second most common bacterial STI and results in substantial morbidity and economic cost worldwide. Gonorrhea is spread by having unprotected vaginal, anal, or oral sex with someone who has gonorrhea, or from mother-to-child during childbirth. Correct and consistent use of condoms significantly decreases the risk of sexual transmission.
Gonorrhea is often asymptomatic in women. If untreated, gonorrhea infection may lead to serious complications.
Of all the STIs, gonorrhea is the most antibiotic-resistant. Increased resistance to most antibiotics used to treat gonococcal infections has been reported worldwide, raising concerns about the eventual development of untreatable gonococcal infections with serious sexual and reproductive health consequences.
To the extent possible, countries should update their national guidelines for the treatment of gonococcal infection based on recent antimicrobial resistance (AMR) surveillance. If local ARM surveillance in not yet functional, PAHO urges countries to adopt the latest WHO treatment guidelines for N. gonorrhoeae.
Key facts
According to the Latin American AMR Surveillance Network (ReLAVRA), ciprofloxacin resistance has steadily grown, with isolates increasing from 35% in 2009 to 62% in 2015. Moreover, reduced sensitivity to broad spectrum cephalosporins and macrolides is beginning to emerge in Latin America and the Caribbean (LAC).
While there are documented increases in gonococcal resistance to antimicrobial drugs, only 36% of the countries in the Americas systematically monitor this resistance to support treatment decisions
In 2017, only 8% of the countries in LAC reported the use of ceftriaxone plus azithromycin, as recommended in the WHO treatment guidelines.
Fact sheet
Gonorrhea is the second most common sexually transmitted infection (STI) caused by bacteria in the world. In women is mostly asymptomatic and it can cause serious health problems if not treated. Of all the STIs, gonorrhea is the most resistant to antibiotics. Because of widespread resistance, older and cheaper antibiotics have lost their effectiveness in treatment of the infection, and gonorrhea infection may become untreatable.
You can get gonorrhea by having unprotected vaginal, anal, or oral sex with someone who has gonorrhea. You can still get gonorrhea even if you had it in the past and got treatment for it. It can also be transmitted from mother-to-child during childbirth.
Correct and consistent use of condoms significantly decreases the risk of infection during anal, vaginal or oral sexual contact. For all neonates, the WHO STI guideline recommends topical ocular prophylaxis for the prevention of gonococcal and chlamydial ophthalmia neonatorum (eye infection).
Many people with gonorrhea do not show any symptoms, with less than half of infected women expressing non-specific symptoms. If you do have symptoms, they can appear in different forms. Women might notice irregular vaginal discharge, pain when urinating, discomfort in the lower part of the stomach, or bleeding from the vagina after sex or between periods. Men might notice pain when urinating, pus-like discharge from the penis, or pain or swelling in one testicle.
In both men and women rectal infection often do not cause symptoms, but in some cases, you may have rectal and anal pain or discharge. Throat infections can also occur and, in most cases, are asymptomatic, but mild sore throat or inflammation in the back of the throat could happen.
Gonorrhea is often asymptomatic in women, and the lack of discernible symptoms results in unrecognized and untreated infections. If left untreated, gonorrhea usually resolves naturally, but can also cause serious health problems. For women this could include pelvic inflammatory disease (PID), which can lead to ectopic pregnancy (pregnancy outside the womb) and inability to have a baby. In men, health problems could include epididymitis (inflammation of the tube that stores and carries sperm at the back of the testicle), urethral stricture (narrowing of the urine canal) and inability to have a baby. The risk of complications increases with repeated infection.
Infants of mothers with gonococcal infection can be infected at delivery, resulting in neonatal conjunctivitis (eye infection) manifesting as purulent ocular discharge and swollen eyelids. Untreated conjunctivitis may lead to scarring and blindness.
Diagnosis is made in a laboratory using different samples, including urine, and vulvovaginal, cervical, and urethral swabs. However, since laboratory diagnostic tests are not available in many countries, a qualified care provider may treat you based on the presence of symptoms such as vaginal and urethral discharge.
There have been major developments in the diagnosis of gonorrhea including using rapid molecular diagnostic tests, some combined with tests for chlamydia.
The WHO guidelines for the treatment of gonorrhea recommends that local antimicrobial resistance (AMR) data should guide national treatment guidelines. In settings where local AMR data are not available, WHO suggests dual therapy over single therapy for people with genital or anorectal gonorrhea (conditional recommendation, low quality evidence)
Dual therapy (one of the following):
ceftriaxone 250 mg intramuscular (IM) as a single dose PLUS azithromycin 1 g orally as a single dose.
cefixime 400 mg orally as a single dose PLUS azithromycin 1 g orally as a single dose.
With the right treatment, gonorrhea infection can be cured completely. However, it is becoming harder to treat some forms of gonorrhea infection, because the bacteria have developed resistance to nearly all the antibiotics used for its treatment.
However, since laboratory diagnostic tests are not available in many countries, a qualified care provider may treat you based on the presence of symptoms such as vaginal and urethral discharge.
Antimicrobial resistance happens when microorganisms (such as bacteria, fungi, viruses, and parasites) change when they are exposed to antimicrobial drugs (such as antibiotics, antifungals, antivirals, antimalarials, and anthelmintics). Microorganisms that develop antimicrobial resistance are sometimes referred to as "superbugs". As a result, the medicines become ineffective and infections persist in the body, increasing the risk of spread to others. Antimicrobial resistance is an increasingly serious threat to global public health that requires action across all government sectors and society.
PAHO Response
The global response to STI is currently guided by the Global health sector strategy on sexually transmitted infections, 2016 - 2021
The Plan of Action for the Prevention and Control of HIV and Sexually Transmitted Infections (2016-2021) provides lines of action for Member States and PAHO to enhance and expand the prevention and control of HIV and STIs in the Americas
The Plan of Action for Antimicrobial Resistance (2015-2020) provides committed and coordinated support to the countries’ efforts to contain antimicrobial resistance
PAHO’s Latin American Surveillance Network of Antimicrobial Resistance (ReLAVRA for its Spanish Acronym) aims at reporting the magnitude and trends of antimicrobial resistance in the Region, using routine data of microbiology laboratories
WHO’s Gonococcal Antimicrobial Surveillance Programme (GASP) tackles this public health problem through a worldwide laboratory network that is coordinated by regional coordinating centers, including ReLAVRA
In 2016, WHO developed the Guidelines for the Treatment of Neisseria gonorrhoeae providing updated recommendations for the treatment of gonorrhea based on the most recent evidence
WHO and PAHO released, in 2018, a “Standard protocol to assess prevalence of gonorrhoea and chlamydia among pregnant women in antenatal care clinics” to support national and regional estimates of the burden of N. gonorrhoeae infections.
And in 2020, PAHO launched a new guidance to support countries implementing and/or improving their antimicrobial resistance (AMR) surveillance of N. gonorrhoeae, integrated with the surveillance of gonorrhea infections
PAHO facilitates and foster technical cooperation among countries, key partners and civil society organizations and provides direct technical assistance to countries to support the achievement of the elimination of priority sexually transmitted infections (STIs) as a public health problem by 2030
This is one of the 30 conditions that the Elimination Initiative has proposed to eliminate in the Region of the Americas by the year 2030.
Documents
All
Final Statement. Advisory Meeting: Integrating and strengthening responses to mpox, syphilis, and other STIs on the road to elimination. São Paulo, 3–4 April 2024
Prevention and control of sexually transmitted infections (STIs) in the era of oral pre-exposure prophylaxis (PrEP) for HIV
Mandates and Strategies |
6915 | https://www.quora.com/In-a-right-angled-triangle-with-sides-6m-and-8m-on-the-shorter-sides-What-is-the-radius-of-the-largest-circle-that-would-fit-inside | In a right angled triangle with sides 6m and 8m on the shorter sides. What is the radius of the largest circle that would fit inside? - Quora
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In a right angled triangle with sides 6m and 8m on the shorter sides. What is the radius of the largest circle that would fit inside?
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Haresh Sagar
Studied Science&Mathematics (Graduated 1988) · Author has 6.2K answers and 7M answer views
·4y
It's a right angled triangle and shorter sides are 6 6 and 8 8 metres so it's hypotenuse is √6 2+8 2=√100=10 6 2+8 2=100=10 metres.
It's perimeter is (6+8+10)(6+8+10)24 24 metres and semi perimeter is 12 12 metres.
Area of triangle is (0.5×6×8)(0.5×6×8)24 24 m 2 m 2.
Radius of largest circle, that fits inside it is given by
R=A r e a S e m i p e r i m e t e r R=A r e a S e m i p e r i m e t e r⟹⟹
R=24 12 R=24 12==2 2 metres.
Continue Reading
It's a right angled triangle and shorter sides are 6 6 and 8 8 metres so it's hypotenuse is √6 2+8 2=√100=10 6 2+8 2=100=10 metres.
It's perimeter is (6+8+10)(6+8+10)24 24 metres and semi perimeter is 12 12 metres.
Area of triangle is (0.5×6×8)(0.5×6×8)24 24 m 2 m 2.
Radius of largest circle, that fits inside it is given by
R=A r e a S e m i p e r i m e t e r R=A r e a S e m i p e r i m e t e r⟹⟹
R=24 12 R=24 12==2 2 metres.
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To find the radius of the largest circle that can fit inside a right-angled triangle (also known as the incircle), we can use the formula for the radius r r of the incircle:
r=A s r=A s
where:
A A is the area of the triangle,
s s is the semi-perimeter of the triangle.
Step 1: Calculate the Area A A
For a right-angled triangle, the area can be calculated using the formula:
A=1 2×base×height A=1 2×base×height
In this case, the base and height are the two shorter sides (6 m and 8 m):
A=1 2×6×8=1 2×48=24 m 2 A=1 2×6×8=1 2×48=24 m 2
Step 2: Calculate the Sem
Continue Reading
To find the radius of the largest circle that can fit inside a right-angled triangle (also known as the incircle), we can use the formula for the radius r r of the incircle:
r=A s r=A s
where:
A A is the area of the triangle,
s s is the semi-perimeter of the triangle.
Step 1: Calculate the Area A A
For a right-angled triangle, the area can be calculated using the formula:
A=1 2×base×height A=1 2×base×height
In this case, the base and height are the two shorter sides (6 m and 8 m):
A=1 2×6×8=1 2×48=24 m 2 A=1 2×6×8=1 2×48=24 m 2
Step 2: Calculate the Semi-perimeter s s
The semi-perimeter s s is calculated as follows:
s=a+b+c 2 s=a+b+c 2
where a a and b b are the lengths of the two shorter sides, and c c is the length of the hypotenuse. First, we need to find the hypotenuse c c:
Using the Pythagorean theorem:
c=√a 2+b 2=√6 2+8 2=√36+64=√100=10 m c=a 2+b 2=6 2+8 2=36+64=100=10 m
Now we can calculate the semi-perimeter:
s=6+8+10 2=24 2=12 m s=6+8+10 2=24 2=12 m
Step 3: Calculate the Radius r r
Now we can substitute A A and s s into the formula for the radius:
r=A s=24 12=2 m r=A s=24 12=2 m
Conclusion
The radius of the largest circle that would fit inside the triangle is 2 meters.
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Edwin Koh
Author has 1.1K answers and 2M answer views
·4y
(a - r) + (b - r) = 2R. (a + b) - 2r = 2R. (6 + 8) - 2r = 2(5). 2r = 14 - 10 = 4. r = 2 m.
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(a - r) + (b - r) = 2R. (a + b) - 2r = 2R. (6 + 8) - 2r = 2(5). 2r = 14 - 10 = 4. r = 2 m.
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Ansh Keer
Former Owner · Author has 6K answers and 1.7M answer views
·4y
Sides are 6,8,10
Area = 1/268 = 24 m^2
s = 6+8+10 /2 = 12 m
Radius of incircle = A/s = 24/12 = 2 m
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Artificial Intelligence (AI) CHATGPT openAi
Answered by
Richard Scutt
·
·Apr 13, 2021
Umm…the question should be “What is the radius of the smallest circle that the triangle would fit inside?” Currently the answer would be an infinite radius since the circle could be infinitely big and the triangle would still fit inside.
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Pramodkumar Tandon
Former Hed of Mechanical Engg. Department at Institute of Engineering and Rural Technology (1965–2002) · Author has 2.1K answers and 1.2M answer views
·Updated 1y
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What is the radius of a circle circumscribing a triangle of sides 6m, 8m, and 12m respectively?
In our case let —
a = 6m, b = 8m and c = 12 m
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In our case let —
a = 6m, b = 8m and c = 12 m
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Haresh Sagar
Studied Science&Mathematics (Graduated 1988) · Upvoted by
Lalit Narayan Vyas
, MSc. , Mathematics, BU Bhopal (2001) · Author has 6.2K answers and 7M answer views
·Updated 3y
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What is the radius of a circle circumscribing a triangle of sides 6m, 8m, and 12m respectively?
Area of triangle ⟹⟹
[A B C]=√13×7×5×1=√455[A B C]=13×7×5×1=455
C r=6×8×12 4√455=144√455≈6.75 C r=6×8×12 4 455=144 455≈6.75
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Area of triangle ⟹⟹
[A B C]=√13×7×5×1=√455[A B C]=13×7×5×1=455
C r=6×8×12 4√455=144√455≈6.75 C r=6×8×12 4 455=144 455≈6.75
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Dean Rubine
I am obsessed with triangles, especially their area. · Author has 10.6K answers and 23.7M answer views
·3y
Related
What is the radius of a circle circumscribing a triangle of sides 6m, 8m, and 12m respectively?
The radius of the circumcenter is the product of the sides divided by four times the area; it’s usually more useful squared:
R 2=a 2 b 2 c 2 16 Δ 2 R 2=a 2 b 2 c 2 16 Δ 2
There are lots of expressions of 16 Δ 2,16 Δ 2,
[math]R^2= \dfrac{a^2b^2c^2}{4a^2b^2 -(a^2+b^2-c^2)^2 }[/math]
[math]R^2= \dfrac{6^2 8^2 12^2}{4 \ 6^2 8^2 - (6^2+8^2 - 12^2)^2 } = \dfrac{ 20736}{455}[/math]
[math]R = \dfrac{144}{\sqrt{455}}[/math]
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9 7
Aravind Kumar
Worked at CUMMINS India Limited (2012–2012) · Author has 8.4K answers and 3.3M answer views
·3y
Related
What is the radius of a circle circumscribing a triangle of sides 6m, 8m, and 12m respectively?
circum circle radius R= ahc/4area
area of the triangle with sides a,b ,c is
(s(s-a)(s-b)(s-c)))^0.5
where s=(a+b+c)/2)
hence s=(6+8+12)/2=13
hence area is (13×7×5×1)=(455)^0.5
hence radius is 12×6×8/4(455)^0.5
=576/4(455)^0.5=144/(455)^0.5
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9 3
Pramodkumar Tandon
Former Hed of Mechanical Engg. Department at Institute of Engineering and Rural Technology (1965–2002) · Author has 2.1K answers and 1.2M answer views
·3y
Related
What is the radius of a circle inscribed in a right-angled triangle whose base and altitude are 6 cm and 8 cm respectively?
D,E and F are points where the circle touches the sides of the right triangle.
If Radius of the circle is = r, then
OD = OF = r
From the figure we find that —
FC = CE and AD = AE
FC = BC - r = 6 - r
=> CE = 6 - r
=> AE = AC - CE = 10 - ( 6 - r ) = 4 + r
Again AD = 8 - r
=> AD = AE = 4 + r
=> 4 + r = 8 - r
=> 2 r = 4
=>r = 2 cm ……. Answer
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D,E and F are points where the circle touches the sides of the right triangle.
If Radius of the circle is = r, then
OD = OF = r
From the figure we find that —
FC = CE and AD = AE
FC = BC - r = 6 - r
=> CE = 6 - r
=> AE = AC - CE = 10 - ( 6 - r ) = 4 + r
Again AD = 8 - r
=> AD = AE = 4 + r
=> 4 + r = 8 - r
=> 2 r = 4
=>r = 2 cm ……. Answer
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9 3
Padmini Kalyanaraman
School Teacher · Author has 4K answers and 4.6M answer views
·6y
Related
How can I find the radius of circle inscribed in a right angled triangle?
Radius =(a+b-c)/2
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Radius =(a+b-c)/2
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Pradeep Hebbar
Many years of Structural Engineering & Math enthusiasm · Author has 9.3K answers and 6.2M answer views
·Feb 10
Related
If sides of a right angled triangle are 21, 28, and 35, then what will be the radius of the circle inscribed in that triangle?
Let [math]a[/math] and [math]b[/math] be the lengths of catheti and [math]c[/math] be the length of hypotenuse.
Let [math]r[/math] and [math]s[/math] be the inradius and semi perimeter of triangle respectively.
Given: [math]a=21[/math], [math]b=28[/math], [math]c=35[/math]
[math]s= \dfrac{a+b+c}{2}= \dfrac{21+28+35}{2} = 42[/math]
[math]r= s-c = 42-35 = 7[/math]
Alternatively,
[math]r=\dfrac{ab}{a+b+c}[/math]
[math]r=\dfrac{21 \times 28}{21+28+35} = 7[/math]
Unconventional method
Reciprocal of inradius is equal to sum of reciprocals of altitudes.
In a right triangle, [math]2[/math] catheti are altitudes. Third altitude is the one drawn on hypotenuse. Let [math]f[/math] be its length.
[math]f= \dfrac{ab}{c} = \dfrac{21 \times 28}{35} = \dfrac{84}{5}[/math]
[math]\dfrac{1}{r}=\dfrac{1}{a}+\dfrac{1}{b}+\d[/math]
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Let [math]a[/math] and [math]b[/math] be the lengths of catheti and [math]c[/math] be the length of hypotenuse.
Let [math]r[/math] and [math]s[/math] be the inradius and semi perimeter of triangle respectively.
Given: [math]a=21[/math], [math]b=28[/math], [math]c=35[/math]
[math]s= \dfrac{a+b+c}{2}= \dfrac{21+28+35}{2} = 42[/math]
[math]r= s-c = 42-35 = 7[/math]
Alternatively,
[math]r=\dfrac{ab}{a+b+c}[/math]
[math]r=\dfrac{21 \times 28}{21+28+35} = 7[/math]
Unconventional method
Reciprocal of inradius is equal to sum of reciprocals of altitudes.
In a right triangle, [math]2[/math] catheti are altitudes. Third altitude is the one drawn on hypotenuse. Let [math]f[/math] be its length.
[math]f= \dfrac{ab}{c} = \dfrac{21 \times 28}{35} = \dfrac{84}{5}[/math]
[math]\dfrac{1}{r}=\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{f}[/math]
[math]\dfrac{1}{r}=\dfrac{1}{21}+\dfrac{1}{28}+\dfrac{5}{84}[/math]
[math]r=7[/math]
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Paul Grimshaw
IT Architect in the Computer Industry (1983–present) · Author has 3.5K answers and 14.9M answer views
·Feb 10
Related
How can I find the radius of circle inscribed in a right angled triangle?
This is quite an interesting result, particularly when the right triangle has integer sides which is when their lengths are known as a Pythagorean Triple. But let’s say generally that the side lengths are [math]a[/math], [math]b[/math], and [math]c=\sqrt{a^2+b^2}[/math].
There are a number of ways of deriving the inscribed circle radius [math]r[/math] and I prefer an algebraic rather than a geometric approach. First note that the circle with centre math[/math] must have [math]p=q=r[/math] as the circle is touching both the [math]x[/math] and [math]y[/math] axes and therefore the centre is distance [math]r[/math] from each axis.
The line corresponding with the hypotenuse passes through math[/math] and math[/math] an
Continue Reading
This is quite an interesting result, particularly when the right triangle has integer sides which is when their lengths are known as a Pythagorean Triple. But let’s say generally that the side lengths are [math]a[/math], [math]b[/math], and [math]c=\sqrt{a^2+b^2}[/math].
There are a number of ways of deriving the inscribed circle radius [math]r[/math] and I prefer an algebraic rather than a geometric approach. First note that the circle with centre math[/math] must have [math]p=q=r[/math] as the circle is touching both the [math]x[/math] and [math]y[/math] axes and therefore the centre is distance [math]r[/math] from each axis.
The line corresponding with the hypotenuse passes through math[/math] and math[/math] and therefore the line equation can be written [math]bx+ay-ab=0[/math]. The circle also touches this line and so the distance of the line from the centre math[/math] is also [math]r[/math].
The standard equation for the distance of a line [math]fx+gy+h=0[/math] from a point math[/math] is:
[math]\qquad d=\dfrac{|fu+gv+h|}{\sqrt{f^2+g^2}}[/math]
and so in this case we have:
[math]\qquad r=\dfrac{|(a+b)r-ab|}{\sqrt{a^2+b^2}}[/math]
for some unknown [math]r[/math] which we’re looking to find. To solve this equation for [math]r[/math] then let’s first square both sides and multiply by the denominator giving:
[math]\qquad (a^2+b^2)r^2=((a+b)r-ab)^2=(a^2+b^2+2ab)r^2-2ab(a+b)r+a^2b^2[/math]
Cancelling common terms and dividing what’s left by [math]ab[/math] (as neither [math]a[/math] nor [math]b[/math] are zero) we find we have a quadratic in [math]r[/math]:
[math]\qquad 2r^2-2(a+b)r+ab=0[/math]
The quadratic roots equation tells us:
[math]\qquad r=\frac{1}{2}\left(a+b\pm\sqrt{(a+b)^2-2ab}\right)[/math]
which using [math]c=\sqrt{a^2+b^2}[/math] simplifies to:
[math]\qquad = \frac{1}{2}(a+b\pm c)[/math]
So we have two solutions, for example:
Clearly the inscribed circle is the smaller of the two, so [math]r=½(a+b-c)[/math].
In the case of a Pythagorean triple then [math]a,b[/math] and [math]c[/math] are all integers and [math]a+b-c[/math] is an even number (if [math]c[/math] is even then [math]a[/math] and [math]b[/math] are either both odd or both even, whereas if [math]c[/math] is odd then [math]a[/math] and [math]b[/math] must be odd and even). And so we find that [math]r[/math] is always an integer.
In the example above with the triple [math]5[/math], [math]12[/math] and [math]13[/math] we have [math]r=2[/math].
As an aside, a well known way of forming some of the Pythagorean triples is to choose any integer [math]n>2[/math] and then for odd [math]n[/math] the triple is [math]n, ½(n^2-1)[/math] and [math]½(n^2+1)[/math], and for even [math]n[/math] we pick [math]n,¼n^2-1[/math] and [math]¼n^2+1[/math]. So for odd [math]n[/math] we find that [math]a+b-c=n-1[/math] and for even [math]n[/math] it’s [math]a+b-c=n-2[/math]. This means that we can choose any positive integer [math]r[/math] and use both [math]n=2r+1[/math] and [math]n=2r+2[/math] to build Pythagorean triple right triangles having a radius [math]r[/math] inscribed circle.
So for example to find right triangles with inscribed circle [math]r=3[/math] we can use [math]n=7[/math] giving [math]a=7[/math], [math]b=24[/math] and [math]c=25[/math], and [math]n=8[/math] giving [math]a=8[/math], [math]b=15[/math] and [math]c=17[/math].
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6916 | https://www.youtube.com/watch?v=skErSV1AAoc | Integrating A Radical Function | #Calculus
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11 comments
Transcript:
hello everyone in this video we're going to be integrating a radical function we have the square root of x + S root of x we're going to integrate this with respect to X so I'm going to use substitution when I saw this problem first I thought about either calling square root of x something or the whole thing something now when you go ahead and set the whole thing equal to something let's say U then you're going to square both sides and then of course you know isolate the radical and then Square both sides again but this kind of uh brings some difficulties as far as I can see and the solution does not look very straightforward this didn't really get anywhere so instead I tried a different approach and I think that's going to work and that approach is just calling the square root of XT why T because we already used U so U is done okay square root of x set it equal to T that's going to be my new variable and then from here everything unrolls let's go ahead and see how that plays out I I'll square both sides and then once I get X I should be evaluating DX which means differentiate and then multiply by the D of the variable which is DT in this case so the derivative of t^2 is 2 2T and then you can multiply now another way to look at this D thing is if you find DX over DT derivative of x with respect to T is just going to be 2T right 2T or not 2T and then cross multiply you're going to get the same answer make sense all right that's where the D is come from now we pretty much have everything we need so let's go ahead and start substituting so the square root of X+ otk of X DX can be written as now the square root of x which is t^2 Plus T and then I'm I'm going to replace DX with 2D 2T DT which is this one 2T DT make sense that is DX awesome now this is our new integral and it's do you think this is easier than the first one the original one yes because at least we don't have to worry about denesting this because denesting would probably be really difficult for this one especially anyways I haven't tried it by the way but this is pretty straightforward because we have a quadratic under the radical we're going to take care of it there's a couple different ways to approach it which I'm going to show you but not only a radical it's also multiply by 2T we can take out the two and just focus on T but actually I'd like to keep the 2T because if you differentiate T ^2 + T you'll get 2T + one so 2T is actually a good thing to use here so I'm going to keep it inside but I want to do the following let's go ahead and take this expression without the integral symbols and write it as follows I'm going to go ahead and since I do need a one for the derivative of t^2 + T I could probably just do the following replace or write the 2T as 2T plus one minus one does that make sense I basically I take 2T and then add one and then subtract one which is essentially zero so why are we adding zero because we're going to separate this part that's actually going to be very important because it's the derivative of the inside therefore we can use the chain rule but in the opposite direction so we can kind of unchain things right okay that's how U substitution Works chain Rule and then reverse it that's going to be U substitution make sense okay great let's go ahead and plug this in to the integration sign symbol whatever you want to call it and this is what it's going to look like > t^2 + T is going to be multiplied by this so it's going to be the sare otk of T ^2 + T 2 t + 1 DT since there's a minus sign here after I distribute I'm going to go ahead and split this up and write it as a difference of two integrals and the second integral is actually looks fairly simple but the first one is easier because we can use use substitution for the second one we're going to do something else so let's take one at a time let's call this the first integral and let's call this the second integral I'm going to go ahead and handle the first one first and it's fairly easy let's go ahead and call this something how about U if you call that U this becomes U Prime which means If U is equal to T ^2 + T then du becomes 2T + 1 DT which is what I exactly have so this integral turns into the square root of U duu let's go ah and handle one number one first and then we'll do number two and then we'll subtract them make sense okay how do you integrate square root of U you probably know the derivative of squ root of U which is 1/ 2 < tk2 1/ 2un U right but this is the integral so we must use the power rule I mean we don't have to but I'd like to write it that way and if you use the power rule you're supposed to add one and divide by whatever results from adding one and here we could add a constant but we don't need that right now because at the end we're going to add it so please don't take points off because I didn't add the constant okay thank you now what is you what is that so we can write this as 2 over 3 U to the^ 3 halfes or 2 over 3 the square otk of U cubed but U is T ^2 + T so I can back substitute and write this as 2 over 3 theare < TK of t^2 + T cubed a lot of times people are going to write this with the power three Hales which is perfectly fine and I think your professor should also be okay hopefully you never know right uh with the three halves okay you just got to be careful about the domains but this is an indefinite integral so we should be good but this is not the final answer because our integral or integrant or the result whatever needs to be in terms of X what is T what is X let's go back to the basics and T is square root of x awesome so we're going to go ahead and replace t with square root of x and that's going to give us 2 over 3 and then T = root of x so t^ s is going to be X so this will be x +un x cubed under the radical make sense that's the first piece pce which is kind of the easier piece now comes the hard part a little harder not too bad so let's go ahead and take a look at the second piece what is the second piece the second piece number two remember is the integral of the square root of T ^2 + T so they kind of looks like a nice expression right well there's a couple ways to go about it I think there's one thing called Oilers substitution formula something I can't remember what it is but you said it equal to T minus something squared but I'm not going to use I'm going to use trigonometric approach anyways T ^2 + T can be written as t^2 + t plus 1/4 it's like where nor 1/4 comes from well we are trying to complete the square here and if I add 1/4 to this it becomes t + 12 2ar minus 12 squ or just 1/4 notice that 12 squ is 1/4 which is nice now here's what we do we're going to substitute something for t + 12 and notice that this is in the form should I say X or Y I guess I could use y here y^2 minus a^ 2 and whenever you have something especially under the radical it's going to work well right as well you should replace y with a secant Theta why because of Y here's Y a^ 2 secant 2 Theta minus a 2 becomes a sare secant 2 Theta - 1 and secant 2 Theta - 1 is tangent s because secant sare is 1 plus tangent Square I hope you know that if you didn't now you know it right okay there's no excuses now so notice that if you square root it you're going to get a tan Theta don't worry about the plus minus signs at this point just take it out make sense so that's why this is a good substitution in our case this is what it's going to look like I'm going to replace t with or t + one half it actually the whole thing inside the radical with 12 secant Theta and everything is going to proceed the same way let's go ahead and erase this area clean it up and then we'll our work so now this is what I'm supposed to do from here obviously you do need DT and DT is going to be if you just differentiate T it's going to be one and if you differentiate 1 12 sec Theta with respect to Theta you're going to get 12 secant Theta tangent Theta D Theta don't forget the D Theta because it's needed for the integral notice that the derivative of secant is just secant times tangent so it's kind of weird it's derivative includes itself okay and how do you prove that you can write the secant as one over cosine use the reciprocal function and so on and so forth all right now this is what we have and what happens if you replace t+ 1 half it that 12 secant Theta 2 - 1/4 is going to give you 1/4 SEC - 1 which is tangent s Theta when you square root it it's going to become 12 tangent Theta this is going to become 1/ 12 tangent Theta does that make sense let me go ahead and write it that way 1 12 tangent Theta and then I'm going to replace DT with what one 12 secant Theta tangent Theta D Theta and another one 12 secant Theta tangent Theta don't forget the secant tangent because that's the derivative remember and then the D Theta at the end and integrate it so that's the conversion now let's go and simplify this up a little bit let me go ahead and clean up this area I hope you don't mind because I just want you to be able to see the original problems and where this comes from great hopefully this made sense to you this is the trigonometric substitution so we did the use the U substitution and now we're going to use the trigonometric substitution okay 1/2 1/2 is going to be 1/4 take it out tangent tangent is tangent squar and then we end up with see can Theta tangent s Theta D Theta uhoh this is not a very good integral right is it well we're going to do the following I'm going to try to shorten the process so since tangent squ + 1 is secant squar I can replace C tangent squ with secant 2qu minus one notice that these formulas are commonly used back and forth back and forth because our goal is to turn this into a secant and tangent integral to a secant only integral and I'm going to share with you like you can go ahead and check out if you're are not familiar with powers of secant because I'm going to use a formula here you can go ahead and check out this video make sense okay now let's go ahead and by the way let me tell you something if you had tangent Theta secant s Theta D Theta this would be a much easier integral because you could use U substitution okay but here you can't too bad now this is going to give you secant Cube Theta and then of course you're going to subtract secant Theta D Theta for for those I'm going to go ahead and use a formula but again check out those videos if you're new to this so the integral of secant cubed is going to be the following you're going to get 1/4 right and that's basically going to give me the integral of secant cube is going to be 12 secant Theta tangent theta plus 12 Ln the absolute value of secant plus tangent which is kind of weird because the integral of secant Cube actually contains the integral of um half of the integral of secant because of uh you know some reason and then you're going to subtract from this the integral of secant which is Ln secant plus tangent which is kind of like an interesting integral right and then plus C you can add at the end but still we have to add another piece so let's leave it like that and then you can go and simplify this but uh you know we got to do some conversion so what am I going to do for tangent Theta right what is tangent Theta what is secant Theta let's go ahead and see where we use the secant Theta first and uh then we're going to go ahead and back substitute so we said okay I'm going to call wait a minute where does this come from okay yeah I did use uh this as I think it's where I used it before did I okay where did I use the secant first oh yes I replaced T I probably didn't write it I'm sorry about that but basically what I used here was tal2 I don't know if I wrote it anywhere I just can't seem to find it but I did replace t with 1 12 uh secant Theta oh yeah I erased it because we already talked about it remember okay that's why I can't find it so now we have to use that expression so T is equal to 12 secant Theta and from here secant Theta becomes 2T or not 2T just keeps popping up and I'm going to draw a right triangle with Theta as my angle in a right triangle secant is one over cosine so it's going to look like this this over that uh oh where does that come from so secant is one over cosine cosine is this over that so this is going to be 2T and this is going to be 1 and this is supposed to be the square root of 4 t^ 2us 1 make sense awesome now let's go ahead and back substitute everything and then we're going to come up with our integral in T and now we're going to go to x but this is what it's going to look like at the end because we also integrated the other piece remember and turned it into X did I did we turn it into X anyways I don't remember but what happens here is if you subtract one Ln something from 1 12 Ln something you're going to get negative 1/ 12 Ln something does that make sense and then of course I do need to distribute to 1/4 so that's going to bring in a one e that's where it comes from so it's going to be 1/8 secant Theta tangent Theta minus 1/8 Ln hopefully if I didn't make any mistakes which I probably did plus tangent Theta this is the Theta part right and then we're going to go ahead and we need to replace it what with uh the we we need to go back to the X integral from here but let's go ahead and replace uh to switch to T values it's going to be 1 1/ 18 secant is 2T remember and tangent is going to be the < TK of 4 T ^2 - 1 - 1/8 Ln of 2T + the < TK of 4 T ^ 2us 1 this is part of the integral now I'm going to go ahead and put it all together so the squ of x + of X DX it's probably not going to fit here but I'm going to try my best 2/3 and then we're going to write the square root of x +qu otk of X cubed plus 1/4 < TK of x the < TK of 4x - 1 this is T by the way okay and then minus 1/8 Ln 2 < tkx + the square < TK of 4X -1 + C all right that should be the whole thing hopefully I didn't make any mistakes because I'm going to show you go ahead and take a look but I'm going to show you the result from W from alpha as well so you can compare okay let's go now the result from W Alpa is pretty different right obviously there's a sign hyperbolic so on and so forth why did it come out differently I have no idea I haven't checked it but anyways this brings us to the end of this video thank for watching I hope you enjoyed it please let me know don't forget to comment like And subscribe I'll see you next time with another video Until then be safe take care I apologize this was a long video anyways it took a while and bye-bye |
6917 | https://blog.lewolang.com/en/posts/88/15-phrasal-verbs-related-to-crime | 15 Phrasal verbs related to crime. | Lewolang
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15 Phrasal verbs related to crime.
1020Published by Nuri at 10/10/2019
You know that here at Lewolang we want your English to improve and therefore, nothing better than to give you a batch of phrasal verbs. You know, those verbs that if you add a preposition or an adverb to them, they mutate, they become something else, as if they were gremlins, when they get wet they turn into monsters.
This is what phrasal verbs are like, apparently normal verbs that, when combined with those hateful particles called prepositions, become a nightmare for Spanish speakers who are trying to learn English.
Having said this, and since we have no choice but to learn them if we want to show off our mastery of English, we want to give you a sample. To prevent you from having an anxiety attack, we will give them to you in small doses, so that they will be better assimilated.
On this occasion the phrasal verbs are related to crime. Why do we need to know them if we have nothing to do with this subject? For example, to understand films or crime series in English. Maybe you'll hear them on the TV news or read them in the newspapers. So we'll leave you with the following examples:
1.Break into. To enter a building in order to steal something.
Our neighbour's house was broken into last night.
2.Break out (of). To escape from prison.
This is his third attempt at breaking out of prison.
3. Tip off. To give secret information about illegal activities.
Who tipped off the police?
4. Stake out. To wait and watch outside a building because something illegal is taking place inside.
The police have been staking out the house on the corner for days.
5. Bring in. To take someone to the police station, usually to be questioned about a crime.
The suspect has been brought in for questioning.
6. Lock up. To imprison.
I hope he's locked up for what he did.
7. Hold up. To rob using a gun.
The whereabouts of the robbers who held up the bank are still unknown.
8. Get away. To escape, especially when someone is chasing you.
How did the thief manage to get away?
9. Track down. To find someone or something by looking for information.
We'll use all our resources to track down the murderer.
10. Own up to. Admit or confess.
Did you get the suspect to own up to the crime?
11. Grass somebody up / grass on somebody. Inform the police about someone who has carried out a crime.
Apparently, someone grassed him up to the police.
12. Turn in. To take someone or yourself to the police.
The robber finally turned himself in to the police .
13. Do away with.To kill someone.
I think our neighbour has done away with his wife.
14. Get away with. To manage not to be punished for something bad you've done.
He thought he could defraud the taxman and get away with it.
15. Set someone up.To make someone to be blamed for something they haven't done.
He claims he didn't steal the money, that he was set up.
Learn these phrasal verbs and you will become experts in criminal matters. Next time you see a crime series or film, prick up your ears and see if you catch any. Don't hesitate to put other phrasal verbs that you want to share with us about this topic in your comments.
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Misdiagnosis of chronic kidney disease and parathyroid hormone testing during the past 16 years
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Published: 22 September 2023
Misdiagnosis of chronic kidney disease and parathyroid hormone testing during the past 16 years
Haojie Liu1,
Huan Zhao2,
Danna Zheng2,
Wenfang He3,
Yueming Liu2,
Juan Jin3,
Qiang He3&
…
Bo Lin2,4
Show authors
Scientific Reportsvolume 13, Article number:15838 (2023) Cite this article
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2 Citations
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Abstract
Chronic kidney disease (CKD) is a prevalent pathological condition worldwide. Parathyroid hormone (PTH) is an important index related to bone metabolism in CKD patients and has not received enough attention. This study was performed to investigate the incidence and diagnostic rate of CKDin hospital as well as PTH testing and treatment for secondary hyperparathyroidism (SHPT) in patients with stage 3 to 5 CKD. The data of patients who visited Zhejiang Provincial People's Hospital from February 2006 to April 2022 were retrieved from the hospital database. All data were divided into three subgroups using PTH testing and SHPT treatment as major comparative indicators for analysis. The data were then analyzed for overall PTH testing, CKD incidence, and diagnostic rate. Among 5,301,391 patients, the incidence of CKD was 13.14%. The missed diagnosis rate for CKD was 65.76%. The total PTH testing rate was 1.22%, of which 15.37% of PTH testing was performed in patients with stage 3 to 5 CKD. The overall diagnosis rate of SHPT in patients with stage 3 to 5 CKD was 31.0%. The prophylactic medication rate was 7.4%, and the rate of post-diagnostic drug therapy was 22.2% in patients who underwent SHPT treatment. The high misdiagnosis rate and low PTH testing rate of CKD requires prompt attention from clinicians. SHPT treatment should be considered especially in patients with stage 3 to 5 CKD.
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Introduction
The incidence of chronic kidney disease (CKD) has been increasing worldwide and is regarded as a public health concern requiring greater attention1. The global incidence of stage 3 to 5 CKD was 10.6% in 20162. Moreover, a recent cross-sectional study in China showed the prevalence of CKD in Chinese adults was 8.2%3. Although CKD patients who have visited the hospital undergo extensive examinations for kidney disease, the diagnosis of CKD is still missed in many patients.
Parathyroid hormone (PTH) is a peptide-derived hormone secreted from the chief cells in the parathyroid gland. It is a major modulator of bone and mineral metabolism and a key regulator of calcium homeostasis4. PTH plays a vital role in regulating the balance of calcium and phosphorus metabolism by promoting calcium reabsorption in the renal tubules, calcium absorption in the gastrointestinal tract, bone calcium mobilization, and phosphate excretion in the kidney5. The synthesis and secretion of PTH are mainly regulated by extracellular calcium, which binds and activates calcium-sensitive receptors on parathyroid cells, thereby reducing PTH release6.
Secondary hyperparathyroidism (SHPT) is a type of hypermetabolism caused by underlying primary diseases, such as chronic renal failure and post-hemodialysis. The low blood calcium caused by these diseases induces over-secretion of PTH7. SHPT is a consequence and critical mediator of CKD–mineral bone disorder (CKD-MBD)8. The pathogenesis of SHPT is complex and driven by many factors, including deficiency of 1,25-dihydroxyvitamin D, hypocalcemia, and hyperphosphatemia9,10; CKD is regarded as the primary etiology. The imbalance in the fibroblast growth factor (FGF) –calcium–phosphate–vitamin D–PTH axis translates into vascular calcification and bone dystrophy in patients with stage 3 CKD, resulting in a low estimated glomerular filtration rate (eGFR)11,12. However, recent studies have shown that low PTH levels (less than twice the upper limit of normal) in patients with end-stage kidney disease are also associated with higher mortality risk13,14.
The exploration of CKD-MBD, which is associated with hyperphosphatemia, hypocalcemia, low serum levels of vitamin D, and increased PTH secretion, has gradually increased in recent years15. SHPT is one of the most important clinical manifestations of CKD-MBD, and PTH is also a crucial indicator of prognosis in CKD8,11,12. However, there are some indications that clinicians do not seem to pay as much attention to the diagnosis of CKD and the detection of PTH. We aim to amplify this phenomenon and discuss it by analyzing a large sample of data.
The present study was performed to investigate the incidence and diagnostic rate of CKD, the rate of PTH testing, and the status of SHPT treatment in patients with stage 3 to 5 CKD in a single hospital in China over 16 years.
Methods
Patients and data collection
The data of all patients who visited Zhejiang Provincial People's Hospital from February 2006 to April 2022 with recorded general and medical data were retrospectively retrieved from the hospital records, and the data were scrutinized for the patients’ general information, laboratory data, and drug treatment. The inclusion criteria were abnormal renal function lasting > 3 months, abnormal kidney imaging examination findings lasting > 3 months and an eGFR of ≤ 60 for > 3 months. Patients who had undergone kidney transplantation were also eligible for inclusion. Patients who had been diagnosed with acute renal insufficiency were excluded from the study. 199,072 patients were selected and included in the CKD group. The study flow chart is shown in Fig.1.
Figure 1
Flow of participants.
Full size image
This study was conducted according to the Declaration of Helsinki principles. This is a retrospective analysis using Zhejiang Provincial People’s Hospital record database. The study protocol was approved by the Institution Ethics Committees of Zhejiang Provincial People’s Hospital and granted a waiver of written informed consent.
Definitions and explanations
The eGFR was calculated using the CKD Epidemiology Collaboration formula, and the diagnosis of stage 3 to 5 CKD was made using the 2021 Kidney Disease: Improving Global Outcomes (KDIGO) guidelines as a standard (eGFR of < 60 mL/min/1.73 m 2)16. The KDIGO 2021 guidelines were further used for the diagnosis of CKD16. Similarly, the diagnosis of hyperparathyroidism was based on the 2021 KDIGO guidelines: PTH of > 70 pg/mL was the diagnostic criterion for stage 3 CKD, PTH of > 110 pg/mL for stage 4, and PTH of > 300 pg/mL for stage 516.
Research methods
The data were analyzed for the overall incidence of CKD and the rate of PTH testing in the total cohort, and patients who had undergone kidney-related examinations were screened. Eligible patients who met the inclusion criteria were selected. The incidence and diagnosis rates of CKD were calculated. After analysis of the overall and department-specific PTH testing rates, all selected patients with CKD were divided based on the eGFR, and the data of patients with stage 3 to 5 CKD were obtained for PTH-related studies. And the annual increase in the number of PTH tests and patients with stage 3 to 5 CKD was calculated. All selected patients were divided into three groups: those with stage 3 to 5 CKD (Group A), those with stage 3 to 5 CKD who had undergone PTH testing (Group B), and those with stage 3 to 5 CKD who had been diagnosed with SHPT (Group C). All patients' general and clinical data were collected and analyzed for age, sex, visit type, CKD stage, and the use of drugs for SHPT treatment, including calcium supplements, active vitamin D3, lanthanum, sevelamer and cinacalcet.
Statistical analysis
All data were statistically analyzed using SPSS version 25 (IBM Corp., Armonk, NY, USA), and figures were drawn using GraphPad Prism version 9 (GraphPad Software, San Diego, CA, USA). Measurement data with a normal distribution are expressed as mean ± standard deviation. Histograms and line graphs were developed for the visual expression of the data.
Results
Incidence and diagnosis of CKD
During the selected study period, 5,301,391 patients visited Zhejiang Provincial People’s Hospital, among whom 1,514,914 were examined for kidney-related disorders. Among these patients, 199,072 satisfied the definition of CKD. The incidence of CKD was 13.14%, among whom 68,163 patients were diagnosed with CKD (diagnostic rate of 34.24%). The data regarding the incidence and diagnosis of CKD are shown in Fig.2.
Figure 2
Incidence and diagnostic rate of CKD.
Full size image
Overall and department-specific PTH testing rates
Among all selected patients, 64,727 patients underwent PTH testing (testing rate of only 1.22%). The PTH testing situation at the departmental level was then analyzed. The number of PTH tests from each department during the study period was obtained, resulting in 59,107 patient records in the database. Among the screened departments, the top five were Thyroid and Breast Surgery, Endocrinology, Orthopedics, Health Management Center, and Nephrology. Head and Neck Surgery, which ranked sixth, was not far behind the top five. All the other departments had a far lower number of PTH tests than the top six. These results are shown in Fig.3.
Figure 3
Testing rate and Department composition of PTH.
Full size image
General information of patients in the three study groups
As shown in Table 1, Group A had a significantly higher proportion of men (59.0%) than women (41.0%), whereas no significant difference in sex was observed in Group B or C. The mean age of the patients in Groups A, B, and C was 74.2 ± 15.7, 69.7 ± 17.2, and 66.2 ± 18.0 years, respectively. Regarding the types of visits (outpatient vs. inpatient), only the first PTH test was counted in Group B, which resulted in the detection of 6,652 (92.5%) patients during hospitalization. In comparison, only 541 (7.5%) patients were tested in the outpatient and emergency departments.
Table 1 Basic information of patients in groups A, B, and C.
Full size table
In Group A, stage 3, 4, and 5 CKD was present in 29,321 (62.7%), 8922 (19.1%), and 8555 (18.3%) patients, respectively. In Group B, stage 3, 4, and 5 CKD was present in 2961 (41.2%), 1148 (16.0%), and 3084 (42.9%) patients, respectively. In Group C, stage 3, 4, and 5 CKD was present in 694 (31.1%), 428 (19.2%), and 1108 (49.7%) patients, respectively. The proportion of patients with stage 5 CKD was significantly higher among patients who had undergone PTH testing.
PTH testing in patients with stage 3 to 5 CKD
Combined with the primary data, the annual rate of new PTH testing in patients with stage 3 to 5 CKD was calculated and plotted as a function of time (Fig.4). Overall, the curve showed an upward trend, and the overall testing rate of patients with stage 3 to 5 CKD was 15.4%. Furthermore, the independent analysis of each stage of CKD demonstrated a 10.1% testing rate for stage 3, 12.9% for stage 4, and 36.0% for stage 5.
Figure 4
Annual trend of the newly increased number of PTH testing and CKD 3–5 stage.
Full size image
Medications used and diagnosis of SHPT
The drugs used and diagnosis of SHPT were analyzed, and the data are presented in Table 1. We used the time of the first diagnosis of SHPT to divide the drug treatments into two groups: prophylactic and post-diagnostic. The medications used before the diagnosis of SHPT were considered prophylactic, with a prophylaxis rate of 7.4%, while the post-diagnosis drug treatment rate was 22.2%. We made detail statistics on the medications used of prophylactic and post-diagnosis in Table 2. As can be seen in Table 2, calcium and active vitamin D were most used drugs, and the use of all medications increased significantly after diagnosis. The diagnosis rate of SHPT in patients with stage 3 to 5 CKD was also calculated, and the results showed that the overall diagnosis rate was 31.0%. The diagnostic rate of SHPT in patients with stage 3, 4, and 5 CKD was 23.4%, 37.3%, and 36.0%, respectively.
Table 2 Prophylactic and post-diagnostic medication used.
Full size table
Discussion
In this study, the incidence and diagnostic rate of CKD were investigated in patients who visited Zhejiang Provincial People's Hospital from February 2006 to April 2022 by retrospectively retrieving the patients’ data from the hospital database. The data showed that the incidence of CKD was significantly higher than the data reported in 2016 and 20232,3.
The patients who visited the hospital had poorer health conditions with a higher risk of CKD than those in the regional cohort. Surprisingly, the proportion of patients with a missed diagnosis of CKD exceeded 65%. A possible reason for this high rate is a lack of familiarity with CKD diagnosis in other departments17. Another possible reason is the acute reversible kidney injuries (e.g., recurrent kidney stones) as the cause of the examination abnormalities.
The focus of this study differs from that of previous cross-sectional studies in different provinces and cities in China18,19 in the respect that we investigated and analyzed the overall testing rate of PTH. Our results showed that this rate was only 1.22% during the study period (Fig.2), indicating that PTH has been neglected as an indicator of CKD. Our data showed that the Thyroid and Breast Surgery Department was the most common department conducting PTH testing during the past 16 years. PTH is a hormone with systemic action; hence, patients with hormonal disturbances usually prefer to visit the Endocrinology Department as their first choice. Furthermore, PTH testing is also performed during orthopedic examinations because PTH is closely associated with osteoblast and osteoclast activities, which determine growth and bone development20. The PTH testing rate in the Health Management Center unexpectedly ranked fourth, reflecting the importance of PTH measurement as a physical examination tool. In the Nephrology Department, PTH is often measured in patients with CKD because these patients often have CKD-MBD21. China has a well-developed health insurance system. PTH testing as a basic treatment item is allowed to be fully reimbursed. There is no financial burden on the patients. Thus, we don’t think this is one of the reasons for the low PTH testing.
This study involved patients with stage 3 to 5 CKD. A Japanese study showed that a reduction in the eGFR to < 60 mL/min/1.73 m 2 with deficient vitamin D is a significant risk factor for an increased serum PTH concentration22. To highlight a more typical effect, we selected more representative sample data.
We found no significant sex difference among patients with stage 3 to 5 CKD who had SHPT. Concerning age, the patients with stage 3 to 5 CKD who underwent PTH testing were relatively younger than those who did not undergo testing, probably because younger patients paid more attention to PTH testing. Analysis of the types of visits among patients first tested for PTH showed that many patients were tested during hospitalization. PTH testing takes a long time to perform. If the test results cannot be made available to the attending physician on the same day of the outpatient visit, the complexity of outpatient treatment is significantly increased. Furthermore, outpatient testing programs are not comprehensive, and only the necessary tests are often selected for the primary assessment of the condition. During hospitalization, however, PTH is used for a comprehensive assessment of the patient’s condition.
Our data showed that the overall PTH testing rate in patients with stage 3 to 5 CKD was only 15.4%. Patients with stage 3 to 5 CKD develop severe parathyroid gland function insufficiency, resulting in loss of the metabolic regulation ability of calcium, phosphorus, and active vitamin D3. Moreover, numerous clinical trials have confirmed that high PTH is a critical risk factor for the prognosis of patients with CKD, directly leading to an increased risk of cardiovascular disease23,24. The testing rate of 15.4% indicates that the parathyroid function of patients with CKD demands more attention.
In the present study, the PTH testing rates of patients at different stages of CKD were grouped, and the results demonstrated that patients with stage 3, 4, and 5 CKD had a testing rate of 10.1%, 12.9%, and 36.0%, respectively. These low PTH testing rates for patients with stage 3 and 4 CKD indicate that PTH testing in patients with early CKD still requires more attention. A retrospective study in the United States showed that only 46% of patients with stage 4 CKD and 41% of those with stage 5 underwent PTH testing25. These data suggest that the PTH testing rates in patients with CKD worldwide are concerning. The statistical analysis of our data for patients with stage 3 to 5 CKD from February 2006 to April 2022 showed that the annual trend of newly detected patients generally increased yearly (Fig.4). This shows that as in-depth research on CKD increased, clinicians began to pay more attention to PTH.
Further statistical analysis revealed that the diagnosis rate for SHPT in patients with stage 3 to 5 CKD was 31.0%. The prevalence of SHPT in CKD has been well described, with estimates ranging from 20 to 80% depending on the severity of CKD26. A recent study conducted in Italy showed a 38% incidence of SHPT in patients with stage 3 to 5 CKD27. By contrast, in our study, the diagnosis rate of SHPT was 23.4%, 37.3%, and 36.0% in patients with stage 3, 4, and 5 CKD, respectively (Table 1). The meager testing rate of PTH indicates that many patients with CKD who possibly have SHPT have yet to be detected and diagnosed promptly, which significantly impacts the prognosis of patients with CKD.
Based on the pathophysiology of SHPT, a variety of preventive and therapeutic approaches can be used, including reducing phosphorus intake, using phosphate binders, and using active vitamin D or calcimimetics28. In recent years, the use of active vitamin D and cinacalcet has significantly increased and the performance of parathyroidectomy has decreased29,30,31. As shown in Table 1, 7.35% of patients who had stage 3 to 5 CKD with SHPT had received prophylaxis before the diagnosis of SHPT, whereas the proportion of patients who received medications increased to 22.24% following the diagnosis of SHPT. This could be due to a lack of attention to calcium and phosphorus metabolism disorders in patients with CKD.
Furthermore, although the risk of SHPT is higher in patients with stage 3 to 5 CKD, some patients’ serum calcium and phosphorus levels remain within the normal range, discouraging therapy use32. The KDIGO guidelines also do not recommend routine use of calcitriol or its analogs in patients with stage 3 to 5 CKD because of an increased risk of hypercalcemia9. Moreover, dialysis can partially regulate electrolyte disturbances in the body, which may have some influence on the choice of drug therapy. An additional reason for low drug treatment could be the availability of a wide variety of calcium supplements on the market that is convenient for patients to purchase themselves, resulting in hospitals not prescribing drugs. In detail use of medications of SHTP, the calcium and active vitamin D were used most frequently. The reasons might be cheaper prices and more accessible.
This study had several limitations. First, the data in this study were obtained from a single center. Although the center had a strong regional representation, the data source was still a single institution, and there may have been some sampling errors. Second, some data from the data platform were incomplete, which may have caused some errors. Finally, drug therapy for SHPT only came from the single center involved in the study, and the records of some drugs used outside the hospital were incomplete.
Conclusion
The clinical situation regarding the diagnosis of CKD, testing of PTH, and treatment of SHPT is not optimistic. The rate of missed diagnosis of CKD is relatively high, requiring clinicians to pay more attention to this disorder. Our data suggest that clinicians should pay more attention to parathyroid function in patients with stage 3 to 5 CKD.
Data availability
The datasets used and/or analysed during the current study available from the corresponding author on reasonable request.
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Funding
This study was supported by grant from Zhejiang Medical and Health Science and Technology Plan (2023KY052, 2023KY546).
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Haojie Liu
Urology and Nephrology Center, Department of Nephrology, Zhejiang Provincial People’s Hospital (Affiliated People’s Hospital, Hangzhou Medical College), Hangzhou, China
Huan Zhao,Danna Zheng,Yueming Liu&Bo Lin
Department of Nephrology, the First Affiliated Hospital of Zhejiang Chinese Medical University (Zhejiang Provincial Hospital of Traditional Chinese Medicine), Hangzhou, 310000, Zhejiang, China
Wenfang He,Juan Jin&Qiang He
Department of Nephrology, Zhejiang Provincial People’s Hospital Bijie Hospital, Guizhou, 551700, China
Bo Lin
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B.L. and Q.H. conceived and designed the study. H.L., H.Z., D.Z., W.H. were responsible for data acquisition. Y.L. and J.J. analyzed the data. H.L., H.Z., D.Z. contributed to data interpretation. H.Z., D.Z., W.H. search related literature. H.L. wrote the manuscript. B.L. reviewed the manuscript. All authors read and approved the final version of the manuscript.
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6919 | https://math.stackexchange.com/questions/1531800/how-to-compute-sum-of-sum-linear-analysis | sequences and series - How to compute sum of sum (linear analysis) - Mathematics Stack Exchange
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How to compute sum of sum (linear analysis)
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I'm not sure if I should call this "the sum of a sum" but I can't think of another phrase for it.
Anyhow, I'm confused about excercises such as this one
∑j=2∞(∑k=2∞k−j)∑j=2∞(∑k=2∞k−j)
where I want to compute the sum. I'm thinking maybe I could rewrite it as
∑k=2∞1 k 2+∑k=2∞1 k 3+...∑k=2∞1 k 2+∑k=2∞1 k 3+...
but I don't know where to go from there or if that is even correct thinking. (I just began my first course in linear analysis.)
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asked Nov 16, 2015 at 12:15
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Hint: you can try and rearrange the summands - all of them are positive. In particular, you can try to sum by j j first
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answered Nov 16, 2015 at 12:21
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Let S S denote your sum. First question: is it true that S=∑k=2∞∑j=2∞k−j S=∑k=2∞∑j=2∞k−j? Second question: what is the sum ∑j=2∞k−j∑j=2∞k−j for k≥2 k≥2?Jytug –Jytug 2015-11-16 12:41:14 +00:00 Commented Nov 16, 2015 at 12:41
Thank you for commenting further. I've been trying now to solve ∑∞j=2 k−j∑j=2∞k−j but the indice beginning at 2 messes things up for me. If it said from j=0 j=0 to infinity I would say that for each k k it would be 1/(1−k)1/(1−k)... Then I thought maybe I could use that same formula but subtract 1 1 and k k (since ∑∞j=0 k−j=1+k+k 2+...=1/(1−k))∑j=0∞k−j=1+k+k 2+...=1/(1−k)) but that didn't help me.m.bing –m.bing 2015-11-16 13:08:31 +00:00 Commented Nov 16, 2015 at 13:08
Oh, and to answer the first question, yes, I would say that it is true.m.bing –m.bing 2015-11-16 13:11:44 +00:00 Commented Nov 16, 2015 at 13:11
It did help! Now I solved it. Thank you and sorry for spamming the comments. :)m.bing –m.bing 2015-11-16 13:23:59 +00:00 Commented Nov 16, 2015 at 13:23
To be fair, your previous comment was wrong: ∑j=0∞k−j=1/(1−(1/k))∑j=0∞k−j=1/(1−(1/k))Jytug –Jytug 2015-11-16 13:39:43 +00:00 Commented Nov 16, 2015 at 13:39
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6920 | https://math.stackexchange.com/questions/2747827/probability-of-certain-permutations-in-a-set | Skip to main content
Probability of certain permutations in a set
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1. Consider a uniformly random permutation of the set {1,2,...,77}. Define the event A= “in the permutation, both 8 and 4 are to the left of
3”.
What is Pr(A)?
Answer: 1/3
I am trying to understand why this the answer and the question somewhat. I can understand that the possibilities for the permutation would be
8−4−3 and,
4−8−3.
Since there are 2 such permutations, we have Pr(A)=23!=1/3.
Why are we dividing by 3! in 23! ?
The set has 77 elements, so shouldn't the denominator be 77!? When I read this question, there would be 1 spot in the random 77-element permutation where there is a 3, then everything to the left of that would contain an 8 and 3, but they don't necessairly have to be next to eachother as long as they are to the left of 3. Since it's a random permutation of 77 elements, what if 3 lands as the first element? 8 and 4 would have no where to go.
2. Let n≥2 be an integer. Consider permutations ai,a2,...,an of the set {1,2,...,n} for which a1<a2.
How many such permutations are there?
Answer: n!2
Here is my intuition. I would like to know how to arrive to this answer.
The first two elements can be
(1,k),2≤k≤6⟹ 5 ways, or
(2,k),3≤k≤6⟹ 4 ways, or
(3,k),4≤k≤6⟹ 3 ways, or
(4,k),5≤k≤6⟹ 2 ways, or
(5,6)⟹ 1 way.
This comes out to 5!, but 5! also counts the pairs the other way around, so we divide by 2 to get
5!2=(n−1)2
Why is it n!2?
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edited Apr 21, 2018 at 20:38
udpcon
asked Apr 21, 2018 at 20:19
udpconudpcon
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For the first question, you could indeed choose to have the denominator be 77! at which point your numerator should count how many permutations of all 77 elements satisfy your desired condition. By first picking the three positions occupied by 8,4,3 and then picking where everything else is, and then picking which order 8,4,3 appear in, we arrive at an answer of (773)⋅2⋅74!77! which simplifies quite nicely to 13, agreeing with our earlier method.
– JMoravitz
Commented
Apr 21, 2018 at 20:25
1
The main point to take away from this is... although you can choose to have a very descriptive sample space, you do not need to have an overly descriptive sample space. So long as whatever collection of outcomes you have in your sample space are equally likely to occur and the event you wish to find the probability of can be expressed as a set of outcomes from the sample space, then you may use a more efficient smaller sample space & counting methods to proceed and the fewer things in the sample space the easier the arithmetic becomes (though it may become conceptually more difficult).
– JMoravitz
Commented
Apr 21, 2018 at 20:30
1
The same idea can be used for the second problem. Letting our sample space consist of information only about ai and a2, we should find that there are n⋅(n−1) possible equally likely outcomes, (n2) of which have ai less than a2, giving us that it occurs with probability 12, implying that half of the permutations have this property, i.e. n!2 such permutations have this property.
– JMoravitz
Commented
Apr 21, 2018 at 20:35
1
Sometimes problems are designed specifically so that numbers get very large but simplify nicely precisely to encourage pen-and-paper approaches. Other times they may be designed to have very large numbers because seeing it in an unsimplified form gives insight as to the thought process behind the answer and the unsimplified form is far more useful of an answer than a simplified form. A good question in my opinion is always answerable without a calculator.
– JMoravitz
Commented
Apr 21, 2018 at 20:43
2
Choose the value for a1. You have n choices. Choose the value for a2. You have (n−1) choices. Apply multiplication principle of counting to see there are n(n−1) possible. Compare this to having chosen both of the values for a1 and a2 simultaneously and remembering that there is only one way to order the cards in ascending order to get there are (n2) ways to select the values for a1,a2 where a1<a2. Taking the ratio simplifies to 12.
– JMoravitz
Commented
Apr 21, 2018 at 21:18
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For question 1, you uniformly draw a permutation of the numbers {1,2,…,77}. As you write, the order of the numbers 3, 4, and 8 in this permutation will be one of the following 3! options.
3,4,8
3,8,4
4,3,8
4,8,3
8,3,4
8,4,3
As our choice is uniform, each possibility is equally likely, and since exactly two of them has both 8 and 4 to the left of 3, you get P(A)=23!=13.
For question 2, it looks like you are not counting the permutations correctly. For instance, if n=6, and if a1 occupies the 5th position, then a2 has to occupy the 6th position - but there is not just one permutation that fulfills this, there are 4!, since the first four positions can be permuted at will.
The answer is n!2, because in any of the n! permutations of {1,2,…,n}, we have either a1<a2 or a2<a1, and because of uniformity, exactly half of them will have a1<a2.
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answered Apr 21, 2018 at 21:13
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Yes, there are 77 numbers in the set so 77! permutations- but that is not at all relevant! Instead look at "3", "4", and "8". There are 3!= 6 permutations of those 3 numbers only. We can list them: 348, 384, 438, 483, 834, and 843. In two of them, 483 and 843, a proportion of 1/3, have "4 and 8 before 3". In all 77! of the permutations, all possible permutations of "348" also appear with "3 and 4 appear before 3" in 1/3 of them.
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answered Apr 21, 2018 at 20:27
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This makes much more sense now. Thank you for clearing up the first question!
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6921 | https://pmc.ncbi.nlm.nih.gov/articles/PMC7224427/ | Malignant Otitis Externa: Causes for Various Treatment Responses - PMC
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J Int Adv Otol
. 2020 Mar 23;16(1):98–103. doi: 10.5152/iao.2020.7709
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Malignant Otitis Externa: Causes for Various Treatment Responses
Nenad Arsovic
Nenad Arsovic
1 Clinic of Otorhinolaryngology and Maxillofacial Surgery, Clinical Center of Serbia, Belgrade, Serbia (NA, NR, SJ, LC, ZD); Clinic of ENT, Clinical and Hospital Centre Zvezdara, Belgrade, Serbia (SB)
Find articles by Nenad Arsovic
1, Nemanja Radivojevic
Nemanja Radivojevic
1 Clinic of Otorhinolaryngology and Maxillofacial Surgery, Clinical Center of Serbia, Belgrade, Serbia (NA, NR, SJ, LC, ZD); Clinic of ENT, Clinical and Hospital Centre Zvezdara, Belgrade, Serbia (SB)
Find articles by Nemanja Radivojevic
1,✉, Snezana Jesic
Snezana Jesic
1 Clinic of Otorhinolaryngology and Maxillofacial Surgery, Clinical Center of Serbia, Belgrade, Serbia (NA, NR, SJ, LC, ZD); Clinic of ENT, Clinical and Hospital Centre Zvezdara, Belgrade, Serbia (SB)
Find articles by Snezana Jesic
1, Snezana Babac
Snezana Babac
1 Clinic of Otorhinolaryngology and Maxillofacial Surgery, Clinical Center of Serbia, Belgrade, Serbia (NA, NR, SJ, LC, ZD); Clinic of ENT, Clinical and Hospital Centre Zvezdara, Belgrade, Serbia (SB)
Find articles by Snezana Babac
1, Ljiljana Cvorovic
Ljiljana Cvorovic
1 Clinic of Otorhinolaryngology and Maxillofacial Surgery, Clinical Center of Serbia, Belgrade, Serbia (NA, NR, SJ, LC, ZD); Clinic of ENT, Clinical and Hospital Centre Zvezdara, Belgrade, Serbia (SB)
Find articles by Ljiljana Cvorovic
1, Zoran Dudvarski
Zoran Dudvarski
1 Clinic of Otorhinolaryngology and Maxillofacial Surgery, Clinical Center of Serbia, Belgrade, Serbia (NA, NR, SJ, LC, ZD); Clinic of ENT, Clinical and Hospital Centre Zvezdara, Belgrade, Serbia (SB)
Find articles by Zoran Dudvarski
1
Author information
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1 Clinic of Otorhinolaryngology and Maxillofacial Surgery, Clinical Center of Serbia, Belgrade, Serbia (NA, NR, SJ, LC, ZD); Clinic of ENT, Clinical and Hospital Centre Zvezdara, Belgrade, Serbia (SB)
✉
Corresponding Address: Nemanja Radivojevic E-mail: nemanjardv@gmail.com
✉
Corresponding author.
Received 2019 Sep 13; Revised 2019 Dec 6; Accepted 2019 Dec 13; Issue date 2020 Apr.
© Copyright 2020 The European Academy of Otology and Neurotology
Content of this journal is licensed under a Creative Commons Attribution-NonCommercial 4.0 International License.
PMC Copyright notice
PMCID: PMC7224427 PMID: 32209516
Abstract
OBJECTIVES
Malignant (necrotizing) otitis externa (MOE) is an aggressive form of skin inflammation of the external ear with a tendency to spread infection to the temporal bone. The study aimed to evaluate a causal relationship between treatment responses and clinical features in patients with MOE.
MATERIALS and METHODS
In a retrospective, descriptive section study, the database was analyzed between January 2008 and December 2018 in our department, all patients with diagnosed MOE were identified.
RESULTS
A total of 30 patients were evaluated, of which 27 men and 3 women. The youngest patient was 52 years old while he was eldest 88 years, (mean-71 years old). As the most common comorbidity, diabetes mellitus was found in 23 (76%) subjects. Median duration of symptoms was about 3 months. The most common isolated pathogen was Pseudomonas aeruginosa (47%). Patients with facial nerve palsy and erosion of temporal bone find on computerized tomography affect prolonged stationary treatment (Mean, SD 29.2±8.5 and 26,7±11.6 days), while 80% of patients with facial nerve palsy had recurrence of disease (p=0.005) with mean duration of clinical remission of 60±17.3 days. Overall length of treatment is also increased in the presence of comorbidities as well as in patients with cranial nerve involvement.
CONCLUSION
Patients with cranial nerve involvement, erosion of temporal bone and presence of comorbidities affect prolonged treatment and adverse prognosis. Early diagnosis and initiation of aggressive therapy are essential for stopping the further spread of the disease and prevention of serious complications.
Keywords: Malignant otitis externa, Pseudomonas aeruginosa, facial nerve palsy, diabetes mellitus
INTRODUCTION
Malignant (necrotizing) otitis externa (MOE) is an aggressive form of skin inflammation of the external ear with a tendency to spread the infection to the temporal cortical bone, leading to potential skull base osteomyelitis . The condition was first reported by Toulmouche in 1838 and is mentioned as the first case report of this disease. The most common causative pathogen is Pseudomonas aeruginosa, especially in immunocompromised patients with diabetes, HIV, leukemia, granulocytopenia, anemia, on immunosuppressive therapy. In addition, some species of fungi have been described as a causative agent, such as Aspergillus and Candida species. Several cases of methicillin-resistant Staphylococcus aureus (MRSA) have been described, and there is an ongoing increase in the MOE causative agents, such as Klebsiella and Proteus mirabilis.
The mechanism of tissue damage involves coagulation tissue necrosis because of microangiopathy of small blood vessels. The most common clinical findings are severe otalgia, otorrhea, impaired hearing, and granulation polyps. The facial nerve is the most commonly involved cranial nerve, but glossopharyngeal, vagus, accessory, or hypoglossal nerves could also be affected.
Diagnosis was based on anamnesis; clinical examination; audiological assessment; microbiological analysis of ear swab; and CT (computed tomography) scan of the temporal bone, skull base, and endocranium . Notably, the diagnostic criteria have changed over time. A recent systematic review described 27 different sets of diagnostic criteria for this condition .
The primary treatment of MOE is long-term antimicrobial therapy. Other treatment strategies include close follow-up of blood glucose levels and inflammation markers and repeated local debridement of necrotic tissue. Lately, increased use of the hyperbaric oxygen chamber has been noted as one of the therapeutic modalities. Notably, surgery has a limited role in the treatment of MOE. Despite MOE being recognized for decades, the literature reveals a low level of evidence supporting its management . Therefore, in the absence of strong evidence to guide decision-making, clinicians depend on their individual experience and judgment .
Our study aimed to evaluate the causal relationship between treatment responses and clinical features of this disease.
MATERIALS AND METHODS
The database was searched for all patients with MOE who were treated between 2008 and 2018 in our department. In this series, a diagnosis of MOE was made if all the following main criteria listed in Cohen and Friedman were met: clinical findings of severe otalgia, edema, and exudate in the external auditory canal (EAC) or granulation tissue in EAC, positive finding on CT scan of the temporal bone, failure of local treatment after more than 1 week, histological exclusion of other causes of external otitis like tumors and cholesteatomas, and microabscesses on pathological samples obtained during surgery. Anamnestic data were used from medical history, laboratory findings, CT scans, and magnetic resonance imaging. Clinical findings were recorded along with comorbidities for each patient. Predictors of treatment response were defined as the duration of inpatient admission, length of antibiotic treatment, and disease-related hospital readmission.
Patients with incomplete medical documentation were excluded. The study was approved by the Institutional Ethics Committee. Informed consent was obtained, where appropriate.
Statistical Analysis
All the results were analyzed using the Statistical Packages for the Social Sciences (SPSS) Statistics for Windows, version 21.0 (IBM Corp., Armonk, NY, USA). Absolute numbers and percentages described the data. Numerically continuous data were expressed using mean±standard deviation. Comparison of categorical variables was performed using the chi-square test and the Fisher’s test. A P value of less than 0.05 was considered statistically significant. Analysis of variance was used to check if the means of two or more independent samples were significantly different from each other. The Mann–Whitney U test was used for non-normally distributed variables.
RESULTS
Overall, 30 patients were analyzed, including 27 men and 3 women. The youngest patient was 52 years old, and the oldest was 88 years old (mean age 71±8.8 years). Notably, 24 patients had two or more symptoms. The most common symptoms were otorrhea and otalgia (Table 1).
Table 1.
Signs and symptoms of patients with malignant otitis externa
| Clinical characteristics | All patients (30) | % |
:---:
| Otalgia | 23 | 76 |
| Otorrhea | 24 | 79 |
| Edema and granulation in EAC | 10 | 31 |
| Edema in EAC | 14 | 45 |
| Chronic suppurative otitis media | 4 | 14 |
| Polypus in EAC | 5 | 17 |
| Edema of the preauricular and retroauricular regions | 3 | 12 |
| Peripheral paralysis of the facial nerve | 5 | 17 |
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EAC: External auditory canal
The mean duration of symptoms was about 3 months. The minimum and maximum duration of stay in our department were 5 and 44 days, respectively (mean: 17.8±11.7 days).
In almost 50% of patients, urea was elevated. Regarding inflammatory markers, leukocytes and CRP (C-reactive protein) were within normal limits in most patients. Erythrocyte sedimentation rate (ESR) was elevated in 30% (n=9), with values ranging between 40–105 mm/h. No relationship was observed between ESR and the degree of disease spread on the CT scan. Blood glucose level was elevated in 46% (n=14), with the mean glycemic value of 10.1 mmol/L. The most common isolated pathogen was P. aeruginosa. Fungi were isolated in five patients (Table 2). Among the five swab findings, several pathogens were isolated, but no relationship with other variables was observed.
Table 2.
Types of microbiological findings
| Swab finding | No of patients | % |
:---:
| Pseudomonas aeruginosa | 14 | 47 |
| Staphylococcus aureus | 3 | 10 |
| Candida | 5 | 17 |
| Enteroccocus | 1 | 3 |
| Escherichia coli | 2 | 7 |
| Proteus mirabilis | 1 | 3 |
| Streptococcus pyogenes | 1 | 3 |
| Normal findings | 8 | 27 |
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Mixed hearing loss and conductive hearing loss were observed in 14 and 15 patients, respectively. A biopsy was obtained in 15 patients. Pathohistological findings described non-specific granulation tissue and inflammation. Conservative therapy was initiated in 20% of patients, and 80% of patients underwent surgery in combination with antibiotic therapy. Notably, resultant otologic complications were observed in 15 patients.
Involvement of the Cranial Nerves
Paralysis of the facial nerve, one of the most common otologic complications, was documented in five patients, three infected with Pseudomonas and two non-Pseudomonas-infected patients. Erosion of temporal bone on CT scan was described in 12 patients, with 50% of them having cranial nerve involvement and Pseudomonas positive swab.
Bilateral laryngeal paralysis and paralysis of the soft palate was observed in one patient because of the probable expansion of the infective process to the skull base, affecting the ninth and tenth nerves (Figure 1). In addition, these nerves, along with the seventh cranial nerve, were affected in another patient who had infection spreading in the area of the infratemporal and pterygopalatine fossa, clivus, sphenoidal sinuses with infiltration of adjacent meninges. Moreover, the hypoglossal nerve was affected in one patient with advanced disease.
Figure 1. a–c.
Open in a new tab
Coronal MRI images of patient who had expansion of the process in soft tissue. a) spread of process in the left tympanic cavity and soft tissue swelling in the infratemporal space. b) left pharyngeal wall swelling. c) The spread of infection in left parapharyngeal space and affected reccurent nerve.
Comorbidities
Diabetes mellitus type 2 was observed in 23 patients (76%), 13 of them on therapy with insulin, and was the most common comorbidity, as observed in 36% infected with Pseudomonas, 23% non-Pseudomonas infected patients, and 17% with normal swab findings. Complications with diabetes were documented in 23% of patients, including polyneuropathies, renal insufficiency, and peripheral occlusive arterial disease.
Regarding other comorbidities, arterial hypertension was documented in 13 patients, stroke in 3 patients, and myocardial infarction in 1 patient. Three patients were defined as immunocompromised, one with bladder cancer and on chemotherapy and the other two diagnosed with secondary anemia.
MOE Management
The typical duration of antimicrobial treatment is at least 4–6 weeks. Depending on the antibiogram findings, all patients were treated with aggressive antibiotic therapy, most commonly systemic quinolones, such as ciprofloxacin, ceftriaxone, meropenem, or clindamycin. Fluconazole was the first treatment option in the case of fungal infection. In 21 patients, different types of surgical treatment were performed, including tympanomastoidectomy in 38%. Only one patient required radical temporal bone trepanation because of skull base osteomyelitis. When clinical findings demonstrated recovery, the treatment was stopped, and a CT scan was performed a few days later if needed. Negative findings on the CT scan confirmed complete recovery. Sometimes several CT scans were required to be performed in some of our patients until disease remission was confirmed.
Predictive Values
The contribution of patient clinical characteristics on readmission to our department is presented in Table 3. Facial nerve paralysis with the involvement of other cranial nerves and erosion of temporal bone were identified as factors that could influence the length of stay (LOS). We observed a statistically significant correlation between the presence of these factors and LOS (Table 4). The mean value of LOS for patients with facial nerve paralysis was 29.2 days, and that for temporal bone erosion was 26.7 days.
Table 3.
Hospital readmission in relation to different variables associated variables
| | Hospital readmission | p |
:---:
| yes | no |
| Facial nerve paralysis | 4 | 1 | 0.005 |
| Erosion of temporal bone | 5 | 7 | 0.067 |
| Pseudomonas aeruginosa | 4 | 9 | 0.405 |
| Cranial nerve involvement | 5 | 4 | 0.083 |
| Diabetes | 5 | 18 | 0.532 |
| Surgical treatment | 6 | 18 | 0.567 |
| Comorbidities | 5 | 12 | 0.326 |
Open in a new tab
Table 4.
LOT and LOS in relation to different variables
| associated variables | LOT (weeks) Mean, SD | LOS (days) Mean, SD |
:---:
| Facial nerve paralysis | p=0.024 31.6±29.7 | p=0.012 29.2±8.5 |
| Erosion of temporal bone | p=0.074 22.3±23.8 | p<0.001 26.7±11.6 |
| Pseudomonas aeruginosa | p=0.199 20±23.5 | p=0.819 18±13.2 |
| Diabetes mellitus | p=0.980 15.5±18.9 | p=0.484 18.3±12 |
| Cranial nerve involvement | p=0.012 24.0±23.2 | p<0.001 24.6±10.7 |
| Comorbidities | p=0.048 19.82±20.5 | p=0.502 19.12±11.0 |
Open in a new tab
LOT: Length of treatment; LOS: length of hospital stay; SD: Standard Deviation
In some cases, the disease was persistent and aggressive despite prolonged treatment. In this subgroup, two patients died from complications of the disease. We documented disease recurrence rate of 23% after complete remission, involving 71% of patients having P. aeruginosa infection and temporal bone erosion and 60% with facial nerve paralysis. The mean duration of clinical remission was 55.7±27.7 days. However, in patients with facial nerve palsy and temporal bone erosion, the duration of clinical remission was 60±17.3 and 52±20.9 days, respectively.
Regarding the length of treatment (LOT), some patients were treated in the outpatient setting as well, so we have identified several variables that affect slower therapy response. The mean duration of overall treatment was 15±18.4 weeks. Nevertheless, patients were treated for a significantly longer time if they had comorbidities and cranial nerve involvement (Figure 2).
Figure 2.
Open in a new tab
Influence of different variables on the length of treatment.
DISCUSSION
Our series of cases over the last 10 years recorded successful management of patients with MOE, with an observed mortality rate of 7% directly attributed to MOE that is concordant with values from some previous studies, 17.8% and 21% [6, 8]. The recurrence rate of MOE after treatment completion was 23% compared with the 10% reported in the literature .
The link with diabetes was well established, with recent studies reporting the prevalence of diabetes in MOE to be between 65% and 85% [10,11]. We documented 23 patients with diabetes mellitus type 2 (76%). As reported by Joshua et al. MOE was associated with diabetes and a history of diabetic complications, which led to significantly longer treatment and shorter survival periods. In contrast, some authors reported that the presence of diabetes was not a significant prognostic factor in itself [10, 13].
The most frequent symptoms were otorrhea, otalgia, edema, and granulations in the EAC. Other symptoms were polypus in the EAC and edema of the preauricular area. In one study, all patients had severe otalgia and chronic otorrhea, and 56% of them had granulations in the EAC . Laboratory findings did not reveal elevated values of inflammatory markers, such as leukocytes and CRP. ESR levels were elevated in nine patients. Most publications have revealed elevated ESR and considered it a non-specific inflammatory marker for the diagnosis of disease [15,16]. Bhat et al. reported elevated ESR in all patients except one. Moreover, they observed the total leukocyte count was within normal limits in 14 patients, and only 1 patient had an elevated total leukocyte count of 12,700/mm 3. The mean value of ESR level in the study of Lee et al. was 96.1±30.8 mm/hr for the uncontrolled group of patients who either remained alive with the disease or died from the progression of the disease.
In our series, 14 patients (46%) had isolated P. aeruginosa, which was the most commonly isolated microbiological agent. In a study conducted by Shavit et al. P. aeruginosa was isolated in 39 (44.3%) of 88 patients and S. aureus in 7 (8%) patients. In our study, the most common fungal pathogen was Candida albicans. Nevertheless, an increasing number of normal swab findings could be observed, which makes it imperative to consider MOE in non-diabetic patients with normal swab findings.
Overall, 12 patients had erosion of temporal bone. The facial nerve was the most commonly affected cranial nerve. Nevertheless, only one patient had nerve function recovery after therapy completion. Some authors compared patients with facial paralysis and without facial paralysis who were diagnosed with MOE and observed no significant difference in survival [20, 21]. In a study conducted by Franco-Vidal et al. 20% of 46 patients had facial nerve involvement. According to Stevens et al. facial nerve palsy defines a severe form of MOE and correlated with increased mortality. Facial nerve involvement (palsy or paresis) was documented in 15.5% of patients and was associated with a significantly longer mean LOS of 12.9±19.6 days, compared with the unaffected patients.
Despite biopsy being considered the only form of surgical approach by some authors in the available medical literature, we demonstrated different results. Surgery was performed in 70% of all cases. Failure of local treatment and poor antibiotic response were probably responsible for that high percentage of surgical treatment. Notably, recent studies have observed an increased resistance for fluoroquinolones, with resistance rates of 20%–54% [22–24]. Furthermore, with the increasing frequency of non-pseudomonal MOE, ciprofloxacin may not always be an effective treatment because it has poor gram-positive coverage and is ineffective against MRSA . The role of surgery is to debride the necrotic material in cases with extensive disease. Mastoidectomy, as part of debridement, may be needed to clear the disease in severe cases . However, a growing trend toward surgical intervention is observed in severe cases. Glikson et al. performed surgery on 68% of patients (n=17), with five patients undergoing mastoidectomy under general anesthesia. In another study, surgical interventions were performed on 50% of patients who responded poorly to intravenous antibiotic treatment .
Factors like facial nerve paralysis and temporal bone erosion significantly contributed to the increase in the number of days spent in our department and readmission. In a systematic review by Hatch et al. patients with weight loss, diabetes with chronic complications, CHF, coagulopathy, and liver disease were noted to have a significantly prolonged hospitalization compared with the other patients. Loh et al. reported that the presence of major findings on initial CT scans was not predictive of outcome (33.3% in the poor outcome group and 36.4% in the good outcome group had major findings; p=1.00). As reported by Soudry et al. facial nerve involvement indicated progression of MOE, but did not, by itself, worsen the prognosis. Some other studies determined that facial nerve palsy affected the overall patient mortality risk . According to the analysis by Kaya et al. uncontrolled diabetes mellitus and older age are associated with the LOS. Moreover, the age of 70 years and above, diabetes mellitus, facial nerve palsy, and positive findings on the CT scan at diagnosis were significant factors, according to Shavit et al. .
Notably, surgical debridement and parenteral or oral antibiotic therapy for 4–6 weeks is the recommended treatment for osteomyelitis of peripheral bones . Park et al. suggested that 4–6 weeks may be inadequate for patients at high risk of recurrent infections, such as those with infection because of MRSA and end-stage renal disease. Nonetheless, the optimal length of therapy for osteomyelitis in MOE remains unclear, warranting further research in this field and a more accurate therapy protocol.
CONCLUSION
Patients with cranial nerve involvement, erosion of temporal bone, and the presence of comorbidities require prolonged treatment and have an adverse prognosis.
Early diagnosis and initiation of aggressive therapy based on microbiological cultures can prevent severe complications and disease spread, thereby resulting in better outcomes and reduced mortality. Moreover, it may be suggested that the duration of therapy be not less than 6 weeks, especially in the sensitive group of patients at high risk of recurrent infections.
In addition, regular check-ups are recommended in preventing the recurrence of the disease. Nevertheless, the treatment regimen should be individualized, with multidisciplinary cooperation among ENT specialists, infectious diseases, and others.
Nonetheless, this study had some limitations. The study was a descriptive, retrospective, observational study, and the available records limited the study data. Therefore, further analysis of a larger number of respondents will provide a better understanding of this unpredictable illness and perhaps compensate for the lack of clarity, especially in terms of etiopathogenesis and diagnosis.
MAIN POINTS.
Malignant (necrotizing) otitis externa (MOE) is an aggressive form of skin inflammation of the external ear especially in immunocompromised patients.
Despite MOE being recognized for decades, the literature reveals a low level of evidence supporting its management
There is a growing trend toward surgical intervention as part of comprehensive MOE treatment.
Duration of antibiothic therapy should not be less than 6 weeks, especially in the sensitive group of patients at high risk of recurrent infections.
Patients with cranial nerve involvement, erosion of temporal bone, and the presence of comorbidities require prolonged treatment and have an adverse prognosis.
Footnotes
Ethics Committee Approval: Ethics committee approval was received for this study from the Ethics Committee of Clinical Center of Serbia.
Informed Consent: Informed consent was obtained, where appropriate.
Peer-review: Externally peer-reviewed.
Author Contributions: Concept – N.R., A.R.; Design N.R., Z.D.; Supervision N.A., S.B.; Resource S.J., L.C.; Materials N.R., Z.D.; Data Collection and/or Processing N.R., N.A., Z.D.; Analysis and/or Interpretation L.C.; Literature Search N.R., L.C.; Writing N.R., N.A.; Critical Reviews J.S., S.B.;
Conflict of Interest: The authors have no conflict of interest to declare.
Financial Disclosure: The authors declared that this study has received no financial support.
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6922 | https://www.gauthmath.com/solution/1985512816119300/Work-1-Show-that-the-function-fx-x3-4x2-3-is-strictly-increasing-at-the-point-wh | Solved: Work 1 Show that the function f(x)=x^3-4x^2+3 is strictly increasing at the point where . [Math]
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Work 1 Show that the function f(x)=x^3-4x^2+3 is strictly increasing at the point where . x=3.
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Answer
The answer is $$f'(3) = 3 > 0$$f′(3)=3>0, therefore the function is strictly increasing at $$x = 3$$x=3
Explanation
Find the derivative of the function $$f(x)$$f(x)
To determine if a function is strictly increasing at a point, we need to find its derivative, $$f'(x)$$f′(x), and evaluate it at that point. If $$f'(x) > 0$$f′(x)>0 at $$x = 3$$x=3, then the function is strictly increasing at that point.
Given $$f(x) = x^{3} - 4x^{2} + 3$$f(x)=x 3−4 x 2+3, we find its derivative:
$$f'(x) = 3x^{2} - 8x$$f′(x)=3 x 2−8 x
Evaluate the derivative at $$x = 3$$x=3
Now, we substitute $$x = 3$$x=3 into the derivative $$f'(x)$$f′(x):
$$f'(3) = 3(3)^{2} - 8(3) = 3(9) - 24 = 27 - 24 = 3$$f′(3)=3(3)2−8(3)=3(9)−24=27−24=3
Check the sign of $$f'(3)$$f′(3)
Since $$f'(3) = 3 > 0$$f′(3)=3>0, the function $$f(x)$$f(x) is strictly increasing at $$x = 3$$x=3
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6923 | https://teachingcalculus.com/2014/08/18/darbouxs-theorem/ | Skip to primary content
Teaching Calculus
The pleasure lies not in discovering the truth, but in searching for it.
Darboux’s Theorem
Posted on by Lin McMullin
Jean Gaston Darboux1842 – 1917
Jean Gaston Darboux was a French mathematician who lived from 1842 to 1917. Of his several important theorems the one we will consider says that the derivative of a function has the Intermediate Value Theorem property – that is, the derivative takes on all the values between the values of the derivative at the endpoints of the interval under consideration.
Darboux’s Theorem is easy to understand and prove but is not usually included in a first-year calculus course (and is not included on the AP exams). Its use is in the more detailed study of functions in a real analysis course.
You may want to use this as an enrichment topic in your calculus course, or a topic for a little deeper investigation. The ideas here are certainly within the range of what first-year calculus students should be able to follow. They relate closely to the Mean Value Theorem (MVT). I will suggest some ideas (in blue) to consider along the way.
More precisely Darboux’s theorem says that
If f is differentiable on the closed interval [a, b] and r is any number between f ’(a) and f ’(b), then there exists a number c in the open interval (a, b) such that f‘ (c) = r.
Differentiable on a closed interval?
Most theorems in beginning calculus require only that the function be differentiable on an open interval. Here, obviously, we need a closed interval so that there will be values of the derivative for r to be between.
The limit definition of derivative requires a regular two-sided limit to exist; at the endpoint of an interval there is only one side. For most theorems this is enough. Here the definition of derivative must be extended to allow one-sided limits as x approaches the endpoint values from inside the interval. Also note that if a function is differentiable on (a, b), then it is differentiable on any closed sub-interval of (a, b) that does not include a or b.
Geometric proof
Consider the diagram below, which shows a function in blue. At each endpoint draw a line with the slope of r. Notice that these two lines have a slope less than that of the function at the left end and greater than the slope at the right end. At least one of these lines must intersect the function at an interior point of the interval. Before reading on, see if you and your students can complete the proof from here. (Hint: What theorem does the top half of the figure remind you of?)
On the interval between the intersection point and the end point we can apply the Mean Value Theorem and determine the value of c where the tangent line will be parallel to the line through the endpoint. At this point f‘(c) = r. Q.E.D.
Analytic Proof
Consider the function . Since f(x) is differentiable, it is continuous; is also continuous and differentiable. Therefore, h(x) is continuous and differentiable on [a, b]. By the Extreme Value Theorem, there must be a point, x = c, in the open interval (a, b) where h(x) has an extreme value. At this point h’(c) = 0.
Before reading on see if you can complete the proof from here.
Q.E.D.
Exercise: Compare and contrast the two proofs.
In the geometric proof, what does represent? Where does it show up in the diagram?
How do both proofs relate to the Mean Value Theorem (or Rolle’s Theorem).
The function represents the vertical distance from f(x) to . In the diagram, this is a vertical segment connecting f(x) to .This expression may be positive, negative, or zero. In the diagram, at the point(s) where the line through the right endpoint intersects the curve and at the endpoint h(x) = 0. Therefore, h(x) meets the hypotheses of Rolle’s Theorem (and the MVT), and the result follows.
The line through the right endpoint will have equation the This makes . When differentiated and the result will be the same expression as in the analytic proof.
Also, you may move this line upwards parallel to its original position and eventually it will be tangent to the graph of the function. (See my posts on MVT 1 and especially MVT 2).
Exercise:
Consider the function f(x) = sin(x)
On the interval [1,3] what values of the derivative of f are guaranteed by Darboux’s Theorem? .
Does Darboux’s theorem guarantee any value on the interval ? Why or why not?
Answers:
f ‘(x) = cos(x). f ‘ (1) = 0.54030 and f ‘ (3) = -0.98999. So the guaranteed values are from -0.98999 to 0.54030.
No. f ‘ (x) = 1 at both endpoints, so there are no values between one and one.
Another interesting aspect of Darboux’s Theorem is that there is no requirement that the derivative f‘(x) be continuous!
A common example of such a function is
With .
This function (which has appeared on the AP exams) is differentiable (and therefore continuous).There is an oscillating discontinuity at the origin. The derivative is not continuous at the origin. Yet, every interval containing the origin as an interior point meets the conditions of Darboux’s Theorem, so the derivative while not continuous has the intermediate value property.
AP exam question 1999 AB3/BC3 part c:
Finally, what inspired this post was a recent discussion on the AP Calculus Community bulletin board about the AP exam question 1999 AB3/BC3 part c. This question gave a table of values for the rate, R, at which water was flowing out of a pipe as a differentiable function of time t. The question asked if there was a time when R’(t) = 0. It was expected that students would use Rolle’s Theorem or the MVT. There was a discussion about using Darboux’s theorem or saying something like the derivative increased (or was positive), then decreased (was negative) so somewhere the derivative must be zero (implying that derivative had the intermediate value property). Luckily, no one tried this approach, so it was a moot point.
Take a look at the problem with your students and see if you can use Darboux’s theorem. Be sure the hypotheses are met.
Answer (try it yourself before reading on):
The function is not differentiable at the endpoints. But consider an interval like [0,3]. Using the given values in the table, by the MVT there is a time t = c where R‘(c) = 0.8/3 > 0, and there is a time t= d on the interval [21, 24] where R‘(d) = -0.6/3 < 0. The function is differentiable on the closed interval [c, d] so by Darboux’s Theorem there must exist a time when R’(t) = 0. Admittedly, this is a bit of overkill.
References:
After Nitecki, Zbigniew H. Calculus Deconstructed A Second Course in First-Year Calculus, ©2009, The Mathematical Association of America, ISBN 978-0-883835-756-4, p. 221-222.
After Dunham, William The Calculus Gallery Masterpieces from Newton to Lebesque, © 2005, Princeton University Press, ISBN 978-0-691-09565-3, p. 156.
Both these book are good reference books.
Updated: August 20, 2014, and October 4, 2017
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This is a good theorem
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6924 | https://www.quora.com/The-point-P-3-2-5-lies-on-the-curve-for-which-dy-dx-2e-3-2x-The-point-Q-1-k-also-lies-on-the-curve-The-normals-to-the-curve-at-the-points-P-and-Q-intersect-at-R-What-are-the-coordinates-of-R | The point P (3 / 2 , 5) lies on the curve for which dy / dx = 2e ^ (3 - 2x). The point Q (1 , k) also lies on the curve. The normals to the curve at the points P and Q intersect at R. What are the coordinates of R? - Quora
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The point P (3 / 2 , 5) lies on the curve for which dy / dx = 2e ^ (3 - 2x). The point Q (1 , k) also lies on the curve. The normals to the curve at the points P and Q intersect at R. What are the coordinates of R?
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Ak Murthy
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The curve is y = —e^(3–2x) + c, using P, c —1 = 5 or c= 6 So the curve is
y = —e^(3–2x) + 6, When x= 1 y = 6—e= k
Slope of normal at any point is —dx/dy = ( —1/2)e^(2x—3)
N1 has slope = —1/2 and it's equation is. y + x/2 = 23/4…..(1)
The slope of the second = —1/2e. and Q = ( 1, 6—e) .SO normal ar Q is
y +x/2e =6—e +1/2e ……………………………….(2)
Solving (1) and (2) (e—1)x/2e = e—1/2e—1/4 So ( e—1)x = 2e²—1—e/2
x = (4e²— e—2)/2(e—1) and
y = 23/4 — (4e²— e—2)/4(e—1)= (25e—21–4e²)/4(e—1)
R= ( (4e²— e—2)/2(e—1), (25e—21–4e²)/4(e—1))
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dy/dx = 2.e^(3–2x).
or, dy = 2.e^(3–2x).dx.
Integrating both sides.
or, y = 2.e^(3–2x)/(-2) + C.
or, y = - e^(3–2x) + C. , point P(3/2,5) lies on the curve therefore :-
5 = - e^0 + C. =>. C = 5+1 = 6.
Thus , the equation of the curve is :-
y = - e^(3–2x). + 6………………..(1) , the point Q(1,k) lies on (1). therefore:-
k = - e + 6. , thus coordinates of point Q are (1, 6-e ).
dy/dx at point P(3/2,5) = 2.e^(3–3). = 2×1 = 2.
Equation of the normal at P(3/2,5) is :-
y - 5 = - 1/2.(x-3/2).
or, y - 5 = -x/2 + 3/4.
or, 4y - 20 = -2x + 3.
or, 2x + 4y = 23. …………………(2).
dy/dx at point Q(1, 6–e) = 2.e.
Equation of the normal
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dy/dx = 2.e^(3–2x).
or, dy = 2.e^(3–2x).dx.
Integrating both sides.
or, y = 2.e^(3–2x)/(-2) + C.
or, y = - e^(3–2x) + C. , point P(3/2,5) lies on the curve therefore :-
5 = - e^0 + C. =>. C = 5+1 = 6.
Thus , the equation of the curve is :-
y = - e^(3–2x). + 6………………..(1) , the point Q(1,k) lies on (1). therefore:-
k = - e + 6. , thus coordinates of point Q are (1, 6-e ).
dy/dx at point P(3/2,5) = 2.e^(3–3). = 2×1 = 2.
Equation of the normal at P(3/2,5) is :-
y - 5 = - 1/2.(x-3/2).
or, y - 5 = -x/2 + 3/4.
or, 4y - 20 = -2x + 3.
or, 2x + 4y = 23. …………………(2).
dy/dx at point Q(1, 6–e) = 2.e.
Equation of the normal at Q(1, 6-e) is :-
y - 6+e = - 1/2e.(x -1).
or, 2e.y -12e +2e^2 = -x +1.
or, x + 2e.y = 1+12e - 2e^2. ………………(3).
Multiplying eqn. (3) by 2 and subtracting the eqn.(2).
4e.y - 4.y = -21+ 24e-4e^2.
y.(4e-1) = -(4e^2–24e+21). => y = -(4e^2–24e+21)/(4e-1).
From eqn. (2). x = (23 -4y)/2 = {23 + (16e^2-96e+84)/(4e-1)}.
or, x = (92e-23+16e^2–96e+84)/(4e-1).= (16e^2–4e+61)/(4e-1).
Thus , the coordinates of the point R are :-
((16e^2–4e+61)/(4e-1), - (4e^2–24e+21)/(4e-1)). Answer.
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To find the coordinates of point R R, we first need to determine the equation of the curve by integrating d y d x d y d x and then find the normals at points P P and Q Q. Finally, we will find the intersection of the normals.
Step 1: Find the equation of the curve
Given:
d y d x=2 e 3−2 x d y d x=2 e 3−2 x
We integrate this with respect to x x:
y=∫2 e 3−2 x d x y=∫2 e 3−2 x d x
Using substitution, let u=3−2 x u=3−2 x, then d u=−2 d x d u=−2 d x or d x=−d u 2 d x=−d u 2:
y=∫2 e u(−d u 2)=−∫e u d u=−e u+C=−e 3−2 x+C y=∫2 e u(−d u 2)=−∫e u d u=−e u+C=−e 3−2 x+C
Thus, the equation of the curve is:
y=−e 3−2 x+C y=−e 3−2 x+C
Step 2: Find the co
Continue Reading
To find the coordinates of point R R, we first need to determine the equation of the curve by integrating d y d x d y d x and then find the normals at points P P and Q Q. Finally, we will find the intersection of the normals.
Step 1: Find the equation of the curve
Given:
d y d x=2 e 3−2 x d y d x=2 e 3−2 x
We integrate this with respect to x x:
y=∫2 e 3−2 x d x y=∫2 e 3−2 x d x
Using substitution, let u=3−2 x u=3−2 x, then d u=−2 d x d u=−2 d x or d x=−d u 2 d x=−d u 2:
y=∫2 e u(−d u 2)=−∫e u d u=−e u+C=−e 3−2 x+C y=∫2 e u(−d u 2)=−∫e u d u=−e u+C=−e 3−2 x+C
Thus, the equation of the curve is:
y=−e 3−2 x+C y=−e 3−2 x+C
Step 2: Find the constant C C
Using point P(3 2,5)P(3 2,5):
5=−e 3−2⋅3 2+C 5=−e 3−2⋅3 2+C
5=−e 3−3+C⟹5=−1+C⟹C=6 5=−e 3−3+C⟹5=−1+C⟹C=6
So the equation of the curve is:
y=−e 3−2 x+6 y=−e 3−2 x+6
Step 3: Find the coordinates of point Q Q
Point Q Q is given as (1,k)(1,k). We can find k k by substituting x=1 x=1 in the curve equation:
y=−e 3−2⋅1+6 y=−e 3−2⋅1+6
y=−e 1+6⟹k=−e+6 y=−e 1+6⟹k=−e+6
Step 4: Find the slopes at points P P and Q Q
At point P P:
d y d x∣∣∣P=2 e 3−2⋅3 2=2 e 0=2 d y d x|P=2 e 3−2⋅3 2=2 e 0=2
The slope of the normal at P P is:
m n P=−1 2 m n P=−1 2
At point Q Q:
d y d x∣∣∣Q=2 e 3−2⋅1=2 e 1=2 e d y d x|Q=2 e 3−2⋅1=2 e 1=2 e
The slope of the normal at Q Q is:
m n Q=−1 2 e m n Q=−1 2 e
Step 5: Find the equations of the normals
Normal at P P:
Using point-slope form:
y−5=−1 2(x−3 2)y−5=−1 2(x−3 2)
Simplifying:
y−5=−1 2 x+3 4 y−5=−1 2 x+3 4
y=−1 2 x+23 4 y=−1 2 x+23 4
Normal at Q Q:
Using point-slope form:
y−k=−1 2 e(x−1)y−k=−1 2 e(x−1)
Substituting k=−e+6 k=−e+6:
y−(−e+6)=−1 2 e(x−1)y−(−e+6)=−1 2 e(x−1)
y+e−6=−1 2 e x+1 2 e y+e−6=−1 2 e x+1 2 e
Rearranging:
y=−1 2 e x+1 2 e+6−e y=−1 2 e x+1 2 e+6−e
y=−1 2 e x+1+12 e−12 e 2 2 e y=−1 2 e x+1+12 e−12 e 2 2 e
Step 6: Find the intersection point R R
Setting the two equations equal:
−1 2 x+23 4=−1 2 e x+1+12 e−12 e 2 2 e−1 2 x+23 4=−1 2 e x+1+12 e−12 e 2 2 e
Multiplying through by 4 e 4 e to eliminate fractions:
−2 e x+23 e=−2 x+2+48 e−48 e 2−2 e x+23 e=−2 x+2+48 e−48 e 2
Rearranging gives us:
(2−2 e)x=48 e 2−25 e+2(2−2 e)x=48 e 2−25 e+2
Thus,
x=48 e 2−25 e+2 2−2 e x=48 e 2−25 e+2 2−2 e
Substituting this x x back into one of the normal equations to find y y.
Final Coordinates of R R
The coordinates of R R depend on the value of e e and can be calculated as shown. The exact values can be computed numerically.
If you would like, I can help with the numerical approximation of these coordinates!
Your response is private
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Related questions
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Aside Points P (3/2,5) and Q(1,k) lies on the curve dy/dx=2e^ (3-2x). The normals to the curve at the point P and Q intersect at the points R. What are the coordinates of R?
The normal to the curve y = x3 - 2x + l at the point (2, 5) intersects they-axis at point P. What are the coordinates of P?
The point P (3 / 2 , 5) lies on the curve for which dy / dx = 2e ^ (3 - 2x). The point Q (1 , 6 - e) also lies on the curve. The normals to the curve at the points P and Q intersect at R. What are the coordinates of R in term of e?
A curve has dy/dx = 2 (kx - 1) ^5. The curve passes through points (0,1) and (1,8). What is the equation of the curve?
The translation (-3 2) maps the point P to the point with coordinates (-1, -5). What will be the coordinates of the point P?
Robert Paxson
BSME in Mechanical Engineering, Lehigh University (Graduated 1983) · Author has 3.9K answers and 4M answer views
·1y
Originally Answered: The points P (3/2, 5) and Q (1, k) lie on a curve f(x), the derivative of which, f'(x), is 2e^ (3 - 2x). The normals to the curve at P and Q intersect at a point R. What are the coordinates of R in terms of e? ·
y′=2 e(3−2 x)y′=2 e(3−2 x)
y=∫(2 e(3−2 x))d x=−e(3−2 x)+c y=∫(2 e(3−2 x))d x=−e(3−2 x)+c
At (3/2,5)(3/2,5), we find:
5=−e(3−2[3 2])+c 5=−e(3−2[3 2])+c
c=6 c=6
or
y=6−e(3−2 x)y=6−e(3−2 x)
such that:
k=−e(3−2)+6 k=−e(3−2)+6
k=6−e k=6−e
and (1,6−e)(1,6−e) is Point Q.
The slope of a tangent is:
m=y′=2 e(3−2 x)m=y′=2 e(3−2 x)
such that the slope of a normal is:
n=−1 m=−1 2 e(3−2 x)n=−1 m=−1 2 e(3−2 x)
At Point P, the normal is:
y−5=−1 2(x−3 2)y−5=−1 2(x−3 2)
y=−1 2 x+23 4 y=−1 2 x+23 4
At Point Q, the normal is:
y−(6−e)=−1 2 e(x−1)y−(6−e)=−1 2 e(x−1)
y=−1 2 e x+1 2 e−e+6 y=−1 2 e x+1 2 e−e+6
The normals intersect at:
y=−1 2 e x+1 2 e−e+6=−1 2 x+23 4 y=−1 2 e x+1 2 e−e+6=−1 2 x+23 4
(1 2−1 2 e)x=23 4−1 2 e+e−6(1 2−1 2 e)x=23 4−1 2 e+e−6
\f\f
Continue Reading
y′=2 e(3−2 x)y′=2 e(3−2 x)
y=∫(2 e(3−2 x))d x=−e(3−2 x)+c y=∫(2 e(3−2 x))d x=−e(3−2 x)+c
At (3/2,5)(3/2,5), we find:
5=−e(3−2[3 2])+c 5=−e(3−2[3 2])+c
c=6 c=6
or
y=6−e(3−2 x)y=6−e(3−2 x)
such that:
k=−e(3−2)+6 k=−e(3−2)+6
k=6−e k=6−e
and (1,6−e)(1,6−e) is Point Q.
The slope of a tangent is:
m=y′=2 e(3−2 x)m=y′=2 e(3−2 x)
such that the slope of a normal is:
n=−1 m=−1 2 e(3−2 x)n=−1 m=−1 2 e(3−2 x)
At Point P, the normal is:
y−5=−1 2(x−3 2)y−5=−1 2(x−3 2)
y=−1 2 x+23 4 y=−1 2 x+23 4
At Point Q, the normal is:
y−(6−e)=−1 2 e(x−1)y−(6−e)=−1 2 e(x−1)
y=−1 2 e x+1 2 e−e+6 y=−1 2 e x+1 2 e−e+6
The normals intersect at:
y=−1 2 e x+1 2 e−e+6=−1 2 x+23 4 y=−1 2 e x+1 2 e−e+6=−1 2 x+23 4
(1 2−1 2 e)x=23 4−1 2 e+e−6(1 2−1 2 e)x=23 4−1 2 e+e−6
e−1 2 e x=2 e 2−e 2−1 2 e e−1 2 e x=2 e 2−e 2−1 2 e
x=2 e 2−e 2−1 e−1 x=2 e 2−e 2−1 e−1
y=−1 2(2 e 2−e 2−1 e−1)+23 4 y=−1 2(2 e 2−e 2−1 e−1)+23 4
y=e+2−4 e 2 4(e−1)+23(e−1)4(e−1)y=e+2−4 e 2 4(e−1)+23(e−1)4(e−1)
y=24 e−21−4 e 2 4(e−1)=6 e−21 4−e 2 e−1 y=24 e−21−4 e 2 4(e−1)=6 e−21 4−e 2 e−1
or
(2 e 2−e 2−1 e−1,6 e−21 4−e 2 e−1)≈(7.227552,2.136224)(2 e 2−e 2−1 e−1,6 e−21 4−e 2 e−1)≈(7.227552,2.136224)
A plot looks like this:
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9 2
Gordon M. Brown
Math Tutor at San Diego City College (2018-Present) · Author has 6.2K answers and 4.3M answer views
·1y
Originally Answered: The points P (3/2, 5) and Q (1, k) lie on a curve f(x), the derivative of which, f'(x), is 2e^ (3 - 2x). The normals to the curve at P and Q intersect at a point R. What are the coordinates of R in terms of e? ·
This question is a slight variant of a question posted only the day before, by another Quora user. The original poster had initially given the coordinates of Q as (1, k), presumably leaving it to readers to solve for k in the course of their replies. But then, rather mysteriously, the poster had reposted the coordinates of Q as (1, 6 - e). This version of the question retains the original coordinates of Q, and provides the solution method for k, as this makes for the far more interesting problem.
Continue Reading
This question is a slight variant of a question posted only the day before, by another Quora user. The original poster had initially given the coordinates of Q as (1, k), presumably leaving it to readers to solve for k in the course of their replies. But then, rather mysteriously, the poster had reposted the coordinates of Q as (1, 6 - e). This version of the question retains the original coordinates of Q, and provides the solution method for k, as this makes for the far more interesting problem.
Upvote ·
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L Viswanathan
Loves Math, Physics. Enjoys Sanskrit · Author has 3K answers and 1.4M answer views
·4y
step 1.Find the equation of the curve. integrating, y = 2e^(3–2x)/(-2) + K. Since P lies on curve, K = 5 +e^0 = 5+ 1 = 6. Curve equation is y - 6 = -e^(3–2x).
Calculate the slope dy/dx at P = 2e^(0) = 2 .The slope of the normal = -1/2. The equation of the normal at P is y = (-1/2)(x -3/2) + 6 → (a)
Since Q lies on the curve k - 6 = -e^(3 -2). ==> k = 6 -e
The slope of the tangent at Q is 2e. The slope of the normal at Q is -1/(2e).
The equation of the normal at Q is y = (-1/2e)(x -1) + 6-e → (b).
R is the intersection of equations (a) and (b): -e(x - 3/2) = -(x - 1) -2e^2
==> (e -1)x = (2e^2 + 3e/
Continue Reading
step 1.Find the equation of the curve. integrating, y = 2e^(3–2x)/(-2) + K. Since P lies on curve, K = 5 +e^0 = 5+ 1 = 6. Curve equation is y - 6 = -e^(3–2x).
Calculate the slope dy/dx at P = 2e^(0) = 2 .The slope of the normal = -1/2. The equation of the normal at P is y = (-1/2)(x -3/2) + 6 → (a)
Since Q lies on the curve k - 6 = -e^(3 -2). ==> k = 6 -e
The slope of the tangent at Q is 2e. The slope of the normal at Q is -1/(2e).
The equation of the normal at Q is y = (-1/2e)(x -1) + 6-e → (b).
R is the intersection of equations (a) and (b): -e(x - 3/2) = -(x - 1) -2e^2
==> (e -1)x = (2e^2 + 3e/2 - 1). ==>x = (4e^2 + 3e -2)/(2(e-1))
Plug in the value of x in eq. (a) or (b) to find the value of y.
(There may be algebraic and/or arithmetic errors!)
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Related questions
More answers below
The points P (3/2, 5) and Q (1, k) lie on a curve f(x), the derivative of which, f'(x), is 2e^ (3 - 2x). The normals to the curve at P and Q intersect at a point R. What are the coordinates of R in terms of e?
The point p (3,5) lies on the curve y= (1/x-1)-(9/x–5). What is the x-coordinate of the point where the normal to the curve at p intersects the x–axis?
How do you find the integral ∫2(2 x−3)(1/5)−1(2 x−3)(2 x−3)(1/5)+(2 x−3)(1/5)d x∫2(2 x−3)(1/5)−1(2 x−3)(2 x−3)(1/5)+(2 x−3)(1/5)d x?
A point P moves along the curve y=2x^3-12x^2 +x+9. What are the coordinates of P such that the normal to the curve at P will have a gradient that is either a maximum or a minimum?
At what points does the normal line to the curve, f(x) = x^2 - 3x + 5 at the point (3,5), intersect the curve?
Jörg Straube
M.Sc. in Computer Science, ETH Zurich (Graduated 1987) · Author has 6.2K answers and 1.7M answer views
·1y
Originally Answered: The points P (3/2, 5) and Q (1, k) lie on a curve f(x), the derivative of which, f'(x), is 2e^ (3 - 2x). The normals to the curve at P and Q intersect at a point R. What are the coordinates of R in terms of e? ·
f’(x) = 2e^(3 - 2x) → f(x) = -e^(3 - 2x) + C
P: f(3/2) = -e^0 + C = 5 → C = 6
Q: f(1) = -e^1 + 6 = k → k = 6 - e
normal at P: y - 5 = -1/2(x - 3/2)
normal at Q: y - k = -1/(2e)(x - 1)
R: -1/2(x - 3/2) + 5 = -1/(2e)(x - 1) + k ; multiply by -2e
e(x - 3/2) - 10e = x - 1 - 2ke → ex - x = 10e + 3e/2 - 1 - 2ke
x = (23e/2 - 1 - 2ke) / (e - 1) = (2e^2 - e/2 - 1) / (e - 1) ≈ 7.23
Continue Reading
f’(x) = 2e^(3 - 2x) → f(x) = -e^(3 - 2x) + C
P: f(3/2) = -e^0 + C = 5 → C = 6
Q: f(1) = -e^1 + 6 = k → k = 6 - e
normal at P: y - 5 = -1/2(x - 3/2)
normal at Q: y - k = -1/(2e)(x - 1)
R: -1/2(x - 3/2) + 5 = -1/(2e)(x - 1) + k ; multiply by -2e
e(x - 3/2) - 10e = x - 1 - 2ke → ex - x = 10e + 3e/2 - 1 - 2ke
x = (23e/2 - 1 - 2ke) / (e - 1) = (2e^2 - e/2 - 1) / (e - 1) ≈ 7.23
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Anthony Madden
Writer for Betterbuck
·Updated Aug 15
What are the weirdest mistakes people make on the internet right now?
Here are a couple of the worst mistakes I’ve seen people make:
Not using an ad blocker
If you aren’t using an ad blocker yet, you definitely should be.
A good ad blocking app will eliminate virtually all of the ads you’d see on the internet before they load.
No more YouTube ads, no more banner ads, no more pop-up ads, etc.
Most people I know use Total Adblock (link here) - it’s about £2/month, but there are plenty of solid options.
Ads also typically take a while to load, so using an ad blocker reduces loading times (typically by 50% or more). They also block ad tracking pixels to protect your pr
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Here are a couple of the worst mistakes I’ve seen people make:
Not using an ad blocker
If you aren’t using an ad blocker yet, you definitely should be.
A good ad blocking app will eliminate virtually all of the ads you’d see on the internet before they load.
No more YouTube ads, no more banner ads, no more pop-up ads, etc.
Most people I know use Total Adblock (link here) - it’s about £2/month, but there are plenty of solid options.
Ads also typically take a while to load, so using an ad blocker reduces loading times (typically by 50% or more). They also block ad tracking pixels to protect your privacy, which is nice.
More often than not, it saves even more than 50% on load times - here’s a test I ran:
Using an ad blocker saved a whopping 6.5+ seconds of load time.
Here’s a link to Total Adblock, if you’re interested.
Not getting paid for your screentime
Apps like Freecash will pay you to test new games on your phone.
Some testers get paid as much as £270/game. Here are a few examples right now (from Freecash's website):
You don't need any kind of prior experience or degree or anything: all you need is a smartphone (Android or IOS).
If you're scrolling on your phone anyway, why not get paid for it?
I've used Freecash in the past - it’s solid. (They also gave me a £3 bonus instantly when I installed my first game, which was cool).
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Elizabeth Jean Stapel
tutor (1989 to present), instructor (1991+), author (1998+) · Author has 5.9K answers and 3.3M answer views
·1y
Originally Answered: The point P (3 / 2 , 5) lies on the curve for which dy / dx = 2e ^ (3 - 2x). The point Q (1 , 6 - e) also lies on the curve. The normals to the curve at the points P and Q intersect at R. What are the coordinates of R in term of e? ·
The normals are perpendicular to the tangents, so first find the tangent lines at each of the points.
The slopes of the tangent lines are the values of the derivative at the given points:
y′(3 2)=2 e 3−2(3/2)=2 e 0=2=m P t y′(3 2)=2 e 3−2(3/2)=2 e 0=2=m P t
y′(1)=2 e 3−2(1)=2 e 1=2 e=m Q t y′(1)=2 e 3−2(1)=2 e 1=2 e=m Q t
The normals have negative reciprocal slopes, so the normals’ slopes are:
m P n=−1 2 m P n=−1 2
m Q n=−1 2 e m Q n=−1 2 e
Plug these slopes, along with their points, into the point-slope formula for straight-line equations:
y P−5=−1 2(x−3 2)y P−5=−1 2(x−3 2)
Continue Reading
The normals are perpendicular to the tangents, so first find the tangent lines at each of the points.
The slopes of the tangent lines are the values of the derivative at the given points:
y′(3 2)=2 e 3−2(3/2)=2 e 0=2=m P t y′(3 2)=2 e 3−2(3/2)=2 e 0=2=m P t
y′(1)=2 e 3−2(1)=2 e 1=2 e=m Q t y′(1)=2 e 3−2(1)=2 e 1=2 e=m Q t
The normals have negative reciprocal slopes, so the normals’ slopes are:
m P n=−1 2 m P n=−1 2
m Q n=−1 2 e m Q n=−1 2 e
Plug these slopes, along with their points, into the point-slope formula for straight-line equations:
y P−5=−1 2(x−3 2)y P−5=−1 2(x−3 2)
y P=−x 2+3 4+5 y P=−x 2+3 4+5
=−x 2+23 4=−x 2+23 4
=23−2 x 4=23−2 x 4
…and:
y Q−(6−e)=−1 2 e(x−1)y Q−(6−e)=−1 2 e(x−1)
y Q−6+e=−x 2 e+1 2 e y Q−6+e=−x 2 e+1 2 e
=1−x 2 e=1−x 2 e
y Q=1−x 2 e+6−e y Q=1−x 2 e+6−e
=1−x 2 e+12 e 2 e−2 e 2 2 e=1−x 2 e+12 e 2 e−2 e 2 2 e
=1−x−2 e 2+12 e 2 e=1−x−2 e 2+12 e 2 e
The intersection point is where the two lines cross; that is, where their equations are equal. So set them equal:
23−2 x 4=1−x−2 e 2+12 e 2 e 23−2 x 4=1−x−2 e 2+12 e 2 e
Solve for the value of x x:
23−2 x 4=1−x−2 e 2+12 e 2 e 23−2 x 4=1−x−2 e 2+12 e 2 e
(23−2 x)(e)=(1−x−2 e 2+12 e)(2)(23−2 x)(e)=(1−x−2 e 2+12 e)(2)
23 e−2 e x=2−2 x−4 e 2+24 e 23 e−2 e x=2−2 x−4 e 2+24 e
2 x−2 e x=2−4 e 2+24 e−23 e 2 x−2 e x=2−4 e 2+24 e−23 e
x(2−2 e)=2+e−4 e 2 x(2−2 e)=2+e−4 e 2
x=2+e−4 e 2 2−2 e x=2+e−4 e 2 2−2 e
Plug this into either of the normals, and simplify to find the corresponding value of y y. I plugged into the normal at P P and got:
y=21−24 e+4 e 2 4−4 e y=21−24 e+4 e 2 4−4 e
Note: Simple integration shows that the original curve was:
y=6−e 3−2 x y=6−e 3−2 x
This allows us to confirm the above results by looking at the graph:
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Pramodkumar Tandon
Retired as Prof. & Head at Institute of Engineering and Rural Technology (1965–present) · Author has 2.1K answers and 1.2M answer views
·May 25
Related
The curve y= (1+4x) ^1/2 and point P(6,5) lies on the curve. The line PQ intersects the x-axis at Q(8,0). How can you show that PQ is normal to the curve?
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Rajeev Raj
B.sc-Hons ♡-->Mathematics from Magadh University (Graduated 2021) · Author has 8.4K answers and 2M answer views
·1y
Originally Answered: The point P (3 / 2 , 5) lies on the curve for which dy / dx = 2e ^ (3 - 2x). The point Q (1 , 6 - e) also lies on the curve. The normals to the curve at the points P and Q intersect at R. What are the coordinates of R in term of e? ·
Hint:→ Given curve dy/dx =2e^(3–2x)
At point P(3/2 , 5)
dy/dx =2e^(3 - 2×3/2) =2e^(0) =2 is slope of tangent
then slope of normal =-1/2
Equ. of normal
y - 5 =-1/2 (x - 3/2)
y - 5 =-x/2 + 3/2
y =-x/2 + 3/2 + 5
2y =-x + 3 + 10
x +2y =13
At point Q(1, 6-e)
dy/dx =2e^(3 -2×1) =2e is slope of tangen
then slope of normal =-1/2e
Equ. of normal
y - 6 + e =-1/2e (x - 1)
…………solve
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Sums & Series (Maths)
Answered by
Bhalchandra Khare
· Author has 4.3K answers and 1.4M answer views
·Jun 9, 2021
The slope of normal at P =- 0.5e^—(3-23/2)=0.5e^-0=-0.5.
Equation of the normal is y-5=-0.5(x-1.5)
4y-20=-2x+3 or 2x+4y-23=0
Slope of normal at (1,k) is -0.5e^—(3-2)=-0.5/e
Equation of this normal is y-k=-(0.5/e)(x-1)
2y-2k+(1/e)x-(0.5/e)=0 e=2.718 1/e=0.37approx.
0.37x+2y-2k-0.185=0
0.74x+4y-4k-0.37=0
2x-0.74x-23+4k+0.37=0
1.26x=22.63-4k
x=17.96–3.17k
4y=23-2x y=5.75-0.5(17.96-3.17k)
y=5.75-8.98+1.585k=-3.23+1.585k.
Coordinates of R=(17.96–3.17k,
-3.23+1.585k)
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Andrew Chee
MBA in Business Studies&Master of Business Administration Degrees, The University of Hong Kong (Graduated 1985) · Author has 3.6K answers and 1.6M answer views
·4y
Slope of tangent at P = 2e^(3-3) = 2
Slope of tangent at Q = 2e^(3-2) = 2e
Slope of normal at P = - 1/2
Equation of normal at P is
y = - 1/2 x + c or 2y +x = c
Passing (3/2, 5)
10 + 3/2 = c
Equation of normal at P is 2y + x = 11 1/2____(1)
Slope of normal at Q = - 1/2e
Equation of normal at Q is
y = - 1/2e x + c or 2ey + x = c
Passing (1,k)
2ek +1 = c
Equation of normal at Q is
2ey + x = 2ek + 1______(2)
Solving equations (1)&(2)
2ey-2y = 2ek-10 1/2
y(e-1) = ek - 5 1/4
y = (ek - 5 1/4)/(e-1) = (2.72k-5.25)/1.72
= 1.58k - 3.05
x = 11.5- 2(1.58k-3.05) = 17.6 - 3.16k
Therefore coordinates of intersection point R are
1
Continue Reading
Slope of tangent at P = 2e^(3-3) = 2
Slope of tangent at Q = 2e^(3-2) = 2e
Slope of normal at P = - 1/2
Equation of normal at P is
y = - 1/2 x + c or 2y +x = c
Passing (3/2, 5)
10 + 3/2 = c
Equation of normal at P is 2y + x = 11 1/2____(1)
Slope of normal at Q = - 1/2e
Equation of normal at Q is
y = - 1/2e x + c or 2ey + x = c
Passing (1,k)
2ek +1 = c
Equation of normal at Q is
2ey + x = 2ek + 1______(2)
Solving equations (1)&(2)
2ey-2y = 2ek-10 1/2
y(e-1) = ek - 5 1/4
y = (ek - 5 1/4)/(e-1) = (2.72k-5.25)/1.72
= 1.58k - 3.05
x = 11.5- 2(1.58k-3.05) = 17.6 - 3.16k
Therefore coordinates of intersection point R are
17.6-3.16k and 1.58k-3.05
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Pradeep Hebbar
Many years of Structural Engineering & Math enthusiasm · Author has 9.3K answers and 6.2M answer views
·Aug 24
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Points P Q and R in that order are dividing a line segment joining A(1 6) and B (5 -2) in four equal parts. What are the coordinates of P Q and R?
Given line joins A(1,6)A(1,6) and B(5,−2)B(5,−2) .Points P P, Q Q and R R divide the line into 4 4 parts
Let us write the parametric equation of A B A B
x(t)=1+(5−1)t=1+4 t x(t)=1+(5−1)t=1+4 t
y(t)=6+(−2−6)t=6−8 t y(t)=6+(−2−6)t=6−8 t
We have 3 3 points that divide A B A B into 4 4 segments
Point P P
t=1 4 t=1 4
x(t)=1+4⋅1 4=2 x(t)=1+4⋅1 4=2
y(t)=6−8⋅1 4=4 y(t)=6−8⋅1 4=4
Point Q Q
t=2 4 t=2 4
x(t)=1+4⋅2 4=3 x(t)=1+4⋅2 4=3
y(t)=6−8⋅2 4=2 y(t)=6−8⋅2 4=2
Point R R
x(t)=1+4⋅3 4=4 x(t)=1+4⋅3 4=4
y(t)=6−8⋅3 4=0 y(t)=6−8⋅3 4=0
Now we can write 3 3 points
P(2,4)P(2,4), Q(3,2)Q(3,2) and R(4,0)R(4,0)
Aliter
Equation of line A B A B:
y−6 x−1=−2−6 5−1⟹2 x+y=8 y−6 x−1=−2−6 5−1⟹2 x+y=8
Let us first locate C(5,6)C(5,6)
Continue Reading
Given line joins A(1,6)A(1,6) and B(5,−2)B(5,−2) .Points P P, Q Q and R R divide the line into 4 4 parts
Let us write the parametric equation of A B A B
x(t)=1+(5−1)t=1+4 t x(t)=1+(5−1)t=1+4 t
y(t)=6+(−2−6)t=6−8 t y(t)=6+(−2−6)t=6−8 t
We have 3 3 points that divide A B A B into 4 4 segments
Point P P
t=1 4 t=1 4
x(t)=1+4⋅1 4=2 x(t)=1+4⋅1 4=2
y(t)=6−8⋅1 4=4 y(t)=6−8⋅1 4=4
Point Q Q
t=2 4 t=2 4
x(t)=1+4⋅2 4=3 x(t)=1+4⋅2 4=3
y(t)=6−8⋅2 4=2 y(t)=6−8⋅2 4=2
Point R R
x(t)=1+4⋅3 4=4 x(t)=1+4⋅3 4=4
y(t)=6−8⋅3 4=0 y(t)=6−8⋅3 4=0
Now we can write 3 3 points
P(2,4)P(2,4), Q(3,2)Q(3,2) and R(4,0)R(4,0)
Aliter
Equation of line A B A B:
y−6 x−1=−2−6 5−1⟹2 x+y=8 y−6 x−1=−2−6 5−1⟹2 x+y=8
Let us first locate C(5,6)C(5,6) such that A C⊥B C A C⊥B C
A C=5−1=4 A C=5−1=4
We divide A C A C into 4 4 parts of 1 1 unit each and draw 3 3 lines parallel to y y-axis.
We have their equations, x=2 x=2, x=3 x=3 and x=4 x=4. These lines respectively intersect the line A B A B at P P, Q Q, R R
x=2⟹2(2)+y=8⟹y=4 x=2⟹2(2)+y=8⟹y=4
x=3⟹2(3)+y=8⟹y=2 x=3⟹2(3)+y=8⟹y=2
x=4⟹2(4)+y=8⟹y=0 x=4⟹2(4)+y=8⟹y=0
Therefore, we have 3 3 points
P(2,4)P(2,4), Q(3,2)Q(3,2) and R(4,0)R(4,0)
Upvote ·
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EP.73 | Simple proof of Stewart's theorem with the law of cosine
This theorem describes the relationship between the sides and a cevian in any arbitrary triangle. A cevian is a line segment that joins a vertex to a point on the opposite side of the triangle (in this example, line 𝑑 is a cevian).
#math #mathematics #mathisfun #algebra #geometry #triangle #mathtok #maths #foryoupage #fyp #studytok
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but my question is in physics why we take
R²=A²+B²+2ABcos(μ)
R²=A²+B²-2ABcos(μ)
dependensing on what please🙏🙏
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Aquí solo los que se quedaron por la canción 🛸🗿
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Watching this video like I understand it 😭
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I searched simple geometry😔
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💟Monkeydiva_rite(◍•ᴗ•◍)❤
what is happening?
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math vs science
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수학이 원래이렇게 간지낫나
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اقسم بالله يخوان الرياضيات متعة اذا فمهتها
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Love? F#k.
Репост за всех 😂
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ERM what
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KENAPA ASYIK LALU FYP NI.. MEMANG NK SURUH AKU STUDY KEE 😭
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Biệt đội chống phản quốc 1945
Khi trong não của một đứa giỏi toán khi thi 🗿
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nhìn vẫn dễ hiểu hơn trên lớp
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☆:.｡. o(≧▽≦)o .｡.:☆
this is so easy 😲
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Baru paham sin cos tan, tapi gapapa lah. Keep spirit 🗣️‼️🔥
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действительно круто
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я в 8 классе, математичка уже это нам преподает и я ничего не понимаю
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Largest number of sides and diagonals
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I have asked to solve the following: For given an integer number n≥3 n≥3, find the largest positive number k n k n for which: for every convex n−n−polygon (with n n sides), we can find k n k n segments, each segment is a side or a diagonal of this n−n−polygon, such that for any two segments, they always has a common point.
For example: n=3 n=3: Triangle A B C A B C we can choose the set {A B,B C,C A}{A B,B C,C A} that is k 3=3 k 3=3.
n=4 n=4: Quadrilateral A B C D A B C D then we can choose {A B,B C,A C,B D}{A B,B C,A C,B D} that is k 4=4 k 4=4.
I have no idea to solve this? Could anyone help me?
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edited Jan 1, 2018 at 13:52
Jie FanJie Fan
asked Jan 1, 2018 at 13:47
Jie FanJie Fan
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Does A B A B and C D C D have any common point or am I misunderstanding the question? Because from that set, "any" two segments does not have a common point.ArsenBerk –ArsenBerk 2018-01-01 13:49:50 +00:00 Commented Jan 1, 2018 at 13:49
My mistake, thanks Jie Fan –Jie Fan 2018-01-01 13:51:36 +00:00 Commented Jan 1, 2018 at 13:51
A C A C and B D B D still does not satisfy the condition.ArsenBerk –ArsenBerk 2018-01-01 13:53:49 +00:00 Commented Jan 1, 2018 at 13:53
no AC,BD are diagonals so they must has a common point Jie Fan –Jie Fan 2018-01-01 13:55:32 +00:00 Commented Jan 1, 2018 at 13:55
Oh, that was my bad, I did not see the part "side or diagonal", my apologies.ArsenBerk –ArsenBerk 2018-01-01 13:56:14 +00:00 Commented Jan 1, 2018 at 13:56
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I think k n=n k n=n. My reasoning is that we can choose the segments systematically in order to see this result:
First, let's choose a point from the n n-gon A 1 A 2...A n A 1 A 2...A n. Without loss of generality, we can assume it is regular and choose A 1 A 1 by using the symmetry. Then every segment that has an end point A 1 A 1 is included to our count (because A 1 A 1 will be the common vertex) so there are (n−1)(n−1) such segments. Now, for the sides A 1 A 2 A 1 A 2 and A 1 A n A 1 A n, the diagonal A 2 A n A 2 A n also intersects all of other (n−1)(n−1) segments. And this the maximum we can get because except that diagonal, we cannot find any other diagonal such that it has common points with A 1 A 2 A 1 A 2 and A 1 A n A 1 A n both. Therefore, the answer is k n=n k n=n.
Now, if we don't have a regular n n-gon, this argument still holds because A 1 A 2 A 1 A 2 and A 1 A n A 1 A n are two sides of the triangle A 1 A 2 A n A 1 A 2 A n and all other vertices A i A i (with i≠1,2,n i≠1,2,n) of n n-gon cannot be placed such that ∠A 1 A 2 A i>180∘∠A 1 A 2 A i>180∘ or ∠A 1 A n A i>180∘∠A 1 A n A i>180∘ because n n-gon is convex (Also notice that even the angle was 180∘180∘, since in this case A 1 A 2∩A 1 A i=A 1 A 2 A 1 A 2∩A 1 A i=A 1 A 2 there would have been more than one mutual points). Therefore for all A i A i, A 1 A i A 1 A i intersects A n A 2 A n A 2. And the same reasoning for k n=n k n=n to be maximum is still valid.
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edited Jan 1, 2018 at 14:44
answered Jan 1, 2018 at 14:11
ArsenBerkArsenBerk
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Haha nice. We have the same answer, but you added a diagram QuIcKmAtHs –QuIcKmAtHs 2018-01-01 14:23:04 +00:00 Commented Jan 1, 2018 at 14:23
And the OP specified otherwise and also s/he gave an example for n=4 n=4 as his/her specification. So I took the intersection point as a common point. I did not see a question in this context before so had no chance to know what does that mean unless OP specifies.ArsenBerk –ArsenBerk 2018-01-01 14:29:58 +00:00 Commented Jan 1, 2018 at 14:29
Then this answer is wrong, for k 6 k 6 we can get 7 7 Stella Biderman –Stella Biderman 2018-01-01 14:31:07 +00:00 Commented Jan 1, 2018 at 14:31
"We cannot find any other diagonal such that it has common points with A 1 A 2 A 1 A 2 and A 1 A n A 1 A n both". Can we?ArsenBerk –ArsenBerk 2018-01-01 14:33:25 +00:00 Commented Jan 1, 2018 at 14:33
In the 9 9-got you’ve drawn, A 3 A n A 3 A n intersects every other line Stella Biderman –Stella Biderman 2018-01-01 14:34:08 +00:00 Commented Jan 1, 2018 at 14:34
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ArsenBerk provided an algorithm that shows choosing n n such segments is possible, but not that choosing more is impossible. Here is a proof of that.
Suppose that for some convex n n-gon, n+1 n+1 segments have been chosen meeting the conditions. We will obtain a contradiction.
By pigeonhole principle, there exists a vertex which is the endpoint of at least two chosen segments. Label that vertex A 0 A 0, and all the others, in clockwise direction, starting and ending with its neighbours, as A 1,A 2,…,A n−1 A 1,A 2,…,A n−1. Let the first of these which is connected to A 0 A 0 (here and henceforth, "connected" means "connected by a chosen segment") be A r A r and the last one be A n−s A n−s, with 1≤r<n−s≤n−1 1≤r<n−s≤n−1. The maximum number of chosen segments with A 0 A 0 as endpoint is then (n−s)−(r−1)=n−s−r+1(n−s)−(r−1)=n−s−r+1.
Any chosen segments not having A 0 A 0 as an endpoint must be of the form A i A j A i A j with 1≤ir i>r, then A 0 A r∩A i A j=∅A 0 A r∩A i A j=∅ (because they are non-adjacent sides in the convex quadrilateral A 0 A r A i A j A 0 A r A i A j), so we must have i≤r i≤r. Similarly, if j<n−s j<n−s, A i A j∩A n−s A 0=∅A i A j∩A n−s A 0=∅, therefore j≥n−s j≥n−s.
Let us relabel the points A j A j, for which j≥n−s j≥n−s (there are s s of those) as B 1,B 2,…,B s B 1,B 2,…,B s, keeping the same clockwise-is-ascending order: A n−s A n−s is now B 1 B 1, and so on. A useful visualization could be that we are painting these points blue; if we also imagine A 1,…,A r A 1,…,A r as red, then we have shown that all chosen segments not attached to A 0 A 0 must connect a red point with a blue point.
Lemmma: In the convex polygon A 1 A 2…A r B 1 B 2…B s A 1 A 2…A r B 1 B 2…B s, with r,s≥1 r,s≥1, the maximum number of chosen segments of the type A i B j A i B j so that any two have a point in common, is r+s−1 r+s−1.
Proof: By induction. If r+s=2 r+s=2, there is altogether only one segment, so that is the maximum that can be chosen. Assume the claim is true for every such polygon with r+s=t−1 r+s=t−1 for some t>2 t>2. We prove it for r+s=t r+s=t.
If r=1 r=1, then every chosen segment connects the same red point with one of s=t−1 s=t−1 blue points, which makes at most t−1 t−1 segments. Similarly for the case s=1 s=1.
Now suppose r,s>1 r,s>1, and at least t t segments have been chosen. Consider the vertices A 1 A 1 and B 1 B 1. If both of these are endpoints of more than one chosen segment, then each must be connected to at least one other point. Let A 1 A 1 be connected to B j B j (j>1 j>1) and B 1 B 1 be connected to A i A i (i>1 i>1). But then A 1 A i B 1 B j A 1 A i B 1 B j is a convex quadrilateral with two non-adjacent sides chosen, a contradiction. So at least one of A 1 A 1 or B 1 B 1 is an endpoint of at most one chosen segment. If we remove it, in the resulting (still convex) t−1 t−1-gon we have lost at most one chosen segment, so at least t−1 t−1 chosen segments remain, contradicting the inductive hypothesis. Therefore at most t−1=r+s−1 t−1=r+s−1 segments can be chosen.
Going back to the original polygon, it has at most n−s−r+1 n−s−r+1 chosen segments attached to A 0 A 0, and from the lemma, at most r+s−1 r+s−1 segments not attached to A 0 A 0, or in total at most n n chosen segments.
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answered Jan 2, 2018 at 21:44
Nick PavlovNick Pavlov
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6927 | https://teachy.ai/en/project/high-school-en-US/us-9th-grade/math/exploring-circles-arc-lengths-and-sector-areas | Log In
Project: Exploring Circles: Arc Lengths and Sector Areas
Lara from Teachy
SubjectMath
Math
SourceTeachy Original
Teachy Original
TopicCircles: Arc Lengths and Areas of Sectors
Circles: Arc Lengths and Areas of Sectors
Contextualization
Introduction to Circles, Arc Lengths and Areas of Sectors
Circles are fascinating geometrical figures with multiple intriguing properties. The circle itself is a simple shape, but the mathematics involved is deep and far-reaching. Within a circle, we can find arcs and sectors that play a significant role in various mathematical applications. One of these concepts is the Arc Length - this is the distance along the curve of a section (arc) of the circle's circumference. Another is the Area of the Sector, which refers to the region enclosed by two radii of the circle and their intercepted arc. These two measures depend on a simple but important value in a circle: the central angle.
The concept of the central angle is tied intimately to both arc length and sector area. Using this angle, we can form a ratio that allows us to calculate the arc length and the area of a sector using the circle's radius.
Relevance of Circles, Arc Lengths and Areas of Sectors
This mathematics plays a crucial role in many areas of life and science. For instance, in Astronomy, determining the arc length between two stars from an observer's point of view allows astronomers to plot constellations. Meanwhile, in the field of Urban Planning, the area of a sector is commonly used to design roundabouts and circular parks. Even in sports like Football or Track & Field, understanding these concepts allows us to design the fields and tracks.
Mastering these concepts not only provides you with the knowledge necessary for more advanced mathematics, but it also gives you practical skills applicable in everyday life, from understanding a pie chart to determining the shortest path between two points on a map, which is a part of a circle's arc.
Resources
To delve deeper into these topics and ensure your understanding, here are some reliable resources:
Khan Academy: Arc Length & Sector Area
BBC Bitesize: Arc Lengths and Areas of Sectors
Math is Fun: Sectors, Segments, Arcs and Radians
Book: Geometry, Grades 9-12: Mcdougal Littell High School Math (2000)
Video: Math Antics - Circles, What Is PI?
Why not start by browsing these resources, and see where your curiosity takes you? It's time to dive into the marvelous world of circles!
Practical Activity
Activity Title: "Creating Arcs and Discovering Sectors"
Objective of the project: The main objective of this project is to help students understand the mathematics behind calculating the arc length and the area of a sector in a circle. It aims to cultivate a hands-on experience for students to learn these concepts by doing and to appreciate their practical relevance.
Detailed Description of the Project: In groups of 4, students will create models of various circles and determine the lengths of different arcs and areas of various sectors within those circles. They will use assigned central angles for these calculations and compare their findings. Consequently, they will analyze and discuss their results in a report.
Materials Needed:
Large sheets of cardboard
Protractor
Scissors
Compass
Measuring tape or ruler
Marker
Calculator
Notebook
Step-by-step to carry out the activity:
Each group first needs to draw multiple circles of different radii on the cardboard using a compass. Make sure to mark the center point of the circle.
Using a protractor, draw several central angles (of different degrees) from the center of the circle to the edge.
Using the scissors, cut along the lines of the drawn angles, creating multiple sectors in each circle.
Now, for each sector, measure the radius, and using a measuring tape or ruler, measure the length of the arc (the circumference portion of the sector).
For each sector, calculate the arc length and the sector area using the formulas:
Arc Length = (Central Angle/360) 2π radius
Area of Sector = (Central Angle/360) π radius²
Record the calculated values and the measured arc length in your notebook.
Compare your calculated arc lengths with the measured lengths. Discuss why there might be differences.
Repeat steps 4-7 for each circle and their respective sectors.
Project Deliverables:
The end product will be a written document detailing the process and discoveries made during this activity. The document should ideally contain photographs of the work process, i.e., drawings, measurements, and calculations. The report should follow these guidelines:
Introduction: The student must contextualize the theme, its relevance, and real-world applications, along with the objective of this project.
Development: The student must detail the theory behind the arc length and area of a sector within circles, explaining the activity in detail, indicating the methodology used, and presenting and discussing the obtained results.
Conclusion: The student must conclude the work by revisiting its main points, explicitly stating the learnings obtained and the conclusions drawn about the project.
Bibliography: The student must indicate the sources they relied upon to work on the project, such as books, web pages, videos, etc.
The goal of this report is not merely to demonstrate understanding but also to communicate findings in a clear and organized manner. This fosters the development of communication skills and the ability to collaborate and work as a team. Remember, mathematics is not just about finding the right answer, but also about the journey taken to get there, understanding the process, and learning from any mistakes made along the way.
Need materials to present the project topic in class?
On the Teachy platform, you can find a variety of ready-to-use materials on this topic! Games, slides, activities, videos, lesson plans, and much more...
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6928 | https://www.khanacademy.org/math/grade-6-virginia/x99d65df986ffa9b5:integers/x99d65df986ffa9b5:intro-to-absolute-value/a/absolute-value-review | Absolute value review (article) | Integers | Khan Academy
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6th grade math (VA SOL)
Course: 6th grade math (VA SOL)>Unit 1
Lesson 4: Absolute value
Absolute value examples
Intro to absolute value
Meaning of absolute value
Meaning of absolute value
Finding absolute values
Absolute value review
Math>
6th grade math (VA SOL)>
Integers>
Absolute value
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Absolute value review
GA.Math: 6.NR.3.5, 6.NR.3.6
Google Classroom
Microsoft Teams
Review absolute value, and try some practice problems.
What is absolute value?
The absolute value of a number is its distance from 0.
Example: positive number
The absolute value of 4 is 4:
−5−4−3−2−10123454
Example: negative number
The absolute value of −4 is also 4:
−5−4−3−2−10123454
4 and −4 are the same distance from 0, so they have the same absolute value of 4.
Want to learn more about absolute value? Check out this video.
The absolute value symbol
The symbol for absolute value is a bar | on each side of the number.
Example:
"The absolute value of −3.2" can be written |−3.2|.
Practice
Problem 1
Current
What is |5|?
Check Explain
Want to try more problems like this? Check out these exercises:Finding absolute value
Comparing absolute values
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Avani Mohindra 8 years ago Posted 8 years ago. Direct link to Avani Mohindra's post “Is it possible for a deci...” more Is it possible for a decimal number to have absolute value?
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Kim Seidel 8 years ago Posted 8 years ago. Direct link to Kim Seidel's post “Yes... you can find the a...” more Yes... you can find the absolute value of all real numbers.
For example:
| -2.6 | = 2.6
| -8/9 | = 8/9
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~𝕋𝕪𝕝𝕖𝕣 ℍ𝕚𝕝𝕝~ 7 years ago Posted 7 years ago. Direct link to ~𝕋𝕪𝕝𝕖𝕣 ℍ𝕚𝕝𝕝~'s post “What are absolute values ...” more What are absolute values in the real world😎😎😎
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Gray 4 years ago Posted 4 years ago. Direct link to Gray's post “One of the videos talked ...” more One of the videos talked about how a table leg should be 150 milliliters wide with and absolute deviation of say 2.5. Then, you would use absolute value
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Jonathan Vargas 7 years ago Posted 7 years ago. Direct link to Jonathan Vargas's post “Is absolute value the opp...” more Is absolute value the opposite of a number. What I mean by this is that if a number was positive, it would be a negative? I am very puzzled by this.
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Nicholas 7 years ago Posted 7 years ago. Direct link to Nicholas's post “No. Absolute value is _al...” more No. Absolute value is always positive. Since it's the distance a number is from 0, it would always be positive. So, the absolute value of positive 5, would be positive 5. You can apply this to physics too. In one dimensional motion, to the left would be negative displacement, and to the right would be positive displacement, but simply moving without specifying a direction would merely be distance, which is always positive.
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AteMyName4Breakfast 2 years ago Posted 2 years ago. Direct link to AteMyName4Breakfast's post “The ll things are the sig...” more The ll things are the signs for the distance from 0 to whatever number. For instance, l-4l is 4 because there are the two lines. Why’s it’s not -4 is because the distance between 0 & -4 is four. You can use a number line to help solve this. Another trick is to just look at the number and don’t care about the - if there is one and that will tell you the distance from 0 to that number, whether positive or negative.
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Nicole Hidalgo 2 years ago Posted 2 years ago. Direct link to Nicole Hidalgo's post “who elses parents make th...” more who elses parents make them do this 5 hours a day[eye roll]
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PROJ.IO 2 years ago Posted 2 years ago. Direct link to PROJ.IO's post “I do it for 30 [sigh]” more I do it for 30 [sigh]
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#khanWight 7 years ago Posted 7 years ago. Direct link to #khanWight's post “If there's a opposite for...” more If there's a opposite for every number, is there a opposite for 0?
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Aiena 7 years ago Posted 7 years ago. Direct link to Aiena's post “Hello, The opposite of z...” more Hello,
The opposite of zero will be zero itself.
So, we take zero as the focal point, and all numbers to its left are positive and numbers to its left are negative.
Opposite of 2 will be (-2). In this case, you jump two places to the left to reach zero and then jump two places again to reach (-2).
Is there any place to jump from zero to reach zero? No, isn't it?
Hence, the opposite of zero is zero.
I hope this helped.
Aiena.
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Kai Davids 5 years ago Posted 5 years ago. Direct link to Kai Davids's post “So do all numbers have a...” more So do all numbers have an absolute value no matter what number it is?
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27cooka 5 years ago Posted 5 years ago. Direct link to 27cooka's post “is there such thing as -0” more is there such thing as -0
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Oliver 5 years ago Posted 5 years ago. Direct link to Oliver's post “0 is in the middle of pos...” more 0 is in the middle of positive and negative, which means that it is not technically either. You could add a negative to 0, or you could add a positive to a 0, but the answer will still be 0 as the number 0 means nothing. If you added a negative to nothing, you'd still get nothing, and that goes the same with positives.
Hope this answered your question.
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AteMyName4Breakfast 2 years ago Posted 2 years ago. Direct link to AteMyName4Breakfast's post “The ll things are the sig...” more The ll things are the signs for the distance from 0 to whatever number. For instance, l-4l is 4 because there are the two lines. Why’s it’s not -4 is because the distance between 0 & -4 is four. You can use a number line to help solve this. Another trick is to just look at the number and don’t care about the - if there is one and that will tell you the distance from 0 to that number, whether positive or negative.
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6929 | https://planetmath.org/leftidentityandrightidentity | left identity and right identity
Let be a groupoid. An element is called a left identity element if for all . Similarly, is a right identity element if for all .
An element which is both a left and a right identity is an identity element.
A groupoid may have more than one left identify element: in fact the operation defined by for all defines a groupoid (in fact, a semigroup) on any set , and every element is a left identity.
But as soon as a groupoid has both a left and a right identity, they are necessarily unique and equal. For if is a left identity and is a right identity, then .
| | |
--- |
| Title | left identity and right identity |
| Canonical name | LeftIdentityAndRightIdentity |
| Date of creation | 2013-03-22 13:02:05 |
| Last modified on | 2013-03-22 13:02:05 |
| Owner | mclase (549) |
| Last modified by | mclase (549) |
| Numerical id | 5 |
| Author | mclase (549) |
| Entry type | Definition |
| Classification | msc 20N02 |
| Classification | msc 20M99 |
| Related topic | IdentityElement |
| Related topic | Unity |
| Defines | left identity |
| Defines | right identity | |
6930 | https://courses.lumenlearning.com/atd-herkimer-westerncivilization/chapter/feudalism/ | Feudalism
Learning Objective
Recall the structure of the feudal state and the responsibilities and obligations of each level of society
Key Points
Feudalism flourished in Europe between the 9th and 15th centuries.
Feudalism in England determined the structure of society around relationships derived from the holding and leasing of land, or fiefs.
In England, the feudal pyramid was made up of the king at the top with the nobles, knights, and vassals below him.
Before a lord could grant land to a tenant he would have to make him a vassal at a formal ceremony. This ceremony bound the lord and vassal in a contract.
While modern writers such as Marx point out the negative qualities of feudalism, such as the exploitation and lack of social mobility for the peasants, the French historian Marc Bloch contends that peasants were part of the feudal relationship; while the vassals performed military service in exchange for the fief, the peasants performed physical labour in return for protection, thereby gaining some benefit despite their limited freedom.
The 11th century in France saw what has been called by historians a “feudal revolution” or “mutation” and a “fragmentation of powers” that increased localized power and autonomy.
Terms
homage
In the Middle Ages this was the ceremony in which a feudal tenant or vassal pledged reverence and submission to his feudal lord, receiving in exchange the symbolic title to his new position.
fealty
An oath, from the Latin fidelitas (faithfulness); a pledge of allegiance of one person to another.
vassals
Persons who entered into a mutual obligation to a lord or monarch in the context of the feudal system in medieval Europe.
fiefs
Heritable property or rights granted by an overlord to a vassal.
mesne tenant
A lord in the feudal system who had vassals who held land from him, but who was himself the vassal of a higher lord.
Overview
Feudalism was a set of legal and military customs in medieval Europe that flourished between the 9th and 15th centuries. It can be broadly defined as a system for structuring society around relationships derived from the holding of land, known as a fiefdom or fief, in exchange for service or labour.
The classic version of feudalism describes a set of reciprocal legal and military obligations among the warrior nobility, revolving around the three key concepts of lords, vassals, and fiefs. A lord was in broad terms a noble who held land, a vassal was a person who was granted possession of the land by the lord, and a fief was what the land was known as. In exchange for the use of the fief and the protection of the lord, the vassal would provide some sort of service to the lord. There were many varieties of feudal land tenure, consisting of military and non-military service. The obligations and corresponding rights between lord and vassal concerning the fief formed the basis of the feudal relationship.
Feudalism, in its various forms, usually emerged as a result of the decentralization of an empire, especially in the Carolingian empires, which lacked the bureaucratic infrastructure necessary to support cavalry without the ability to allocate land to these mounted troops. Mounted soldiers began to secure a system of hereditary rule over their allocated land, and their power over the territory came to encompass the social, political, judicial, and economic spheres.
Many societies in the Middle Ages were characterized by feudal organizations, including England, which was the most structured feudal society, France, Italy, Germany, the Holy Roman Empire, and Portugal. Each of these territories developed feudalism in unique ways, and the way we understand feudalism as a unified concept today is in large part due to critiques after its dissolution. Karl Marx theorized feudalism as a pre-capitalist society, characterized by the power of the ruling class (the aristocracy) in their control of arable land, leading to a class society based upon the exploitation of the peasants who farm these lands, typically under serfdom and principally by means of labour, produce, and money rents.
While modern writers such as Marx point out the negative qualities of feudalism, the French historian Marc Bloch contends that peasants were an integral part of the feudal relationship: while the vassals performed military service in exchange for the fief, the peasants performed physical labour in return for protection, thereby gaining some benefit despite their limited freedom. Feudalism was thus a complex social and economic system defined by inherited ranks, each of which possessed inherent social and economic privileges and obligations. Feudalism allowed societies in the Middle Ages to retain a relatively stable political structure even as the centralized power of empires and kingdoms began to dissolve.
Structure of the Feudal State in England
Feudalism in 12th-century England was among the better structured and established systems in Europe at the time. The king was the absolute “owner” of land in the feudal system, and all nobles, knights, and other tenants, termed vassals, merely “held” land from the king, who was thus at the top of the feudal pyramid.
Below the king in the feudal pyramid was a tenant-in-chief (generally in the form of a baron or knight), who was a vassal of the king. Holding from the tenant-in-chief was a mesne tenant—generally a knight or baron who was sometimes a tenant-in-chief in their capacity as holder of other fiefs. Below the mesne tenant, further mesne tenants could hold from each other in series.
Vassalage
Before a lord could grant land (a fief) to someone, he had to make that person a vassal. This was done at a formal and symbolic ceremony called a commendation ceremony, which was composed of the two-part act of homage and oath of fealty. During homage, the lord and vassal entered into a contract in which the vassal promised to fight for the lord at his command, while the lord agreed to protect the vassal from external forces.
Once the commendation ceremony was complete, the lord and vassal were in a feudal relationship with agreed obligations to one another. The vassal’s principal obligation to the lord was “aid,” or military service. Using whatever equipment the vassal could obtain by virtue of the revenues from the fief, he was responsible for answering calls to military service on behalf of the lord. This security of military help was the primary reason the lord entered into the feudal relationship. In addition, the vassal could have other obligations to his lord, such as attendance at his court, whether manorial or baronial, or at the king’s court.
The vassal’s obligations could also involve providing “counsel,” so that if the lord faced a major decision he would summon all his vassals and hold a council. At the level of the manor this might be a fairly mundane matter of agricultural policy, but could also include sentencing by the lord for criminal offenses, including capital punishment in some cases. In the king’s feudal court, such deliberation could include the question of declaring war. These are only examples; depending on the period of time and location in Europe, feudal customs and practices varied.
Feudalism in France
In its origin, the feudal grant of land had been seen in terms of a personal bond between lord and vassal, but with time and the transformation of fiefs into hereditary holdings, the nature of the system came to be seen as a form of “politics of land.” The 11th century in France saw what has been called by historians a “feudal revolution” or “mutation” and a “fragmentation of powers” that was unlike the development of feudalism in England, Italy, or Germany in the same period or later. In France, counties and duchies began to break down into smaller holdings as castellans and lesser seigneurs took control of local lands, and (as comital families had done before them) lesser lords usurped/privatized a wide range of prerogatives and rights of the state—most importantly the highly profitable rights of justice, but also travel dues, market dues, fees for using woodlands, obligations to use the lord’s mill, etc. Power in this period became more personal and decentralized.
Sources
Candela Citations
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Boundless World History. Authored by: Boundless. Located at: License: CC BY-SA: Attribution-ShareAlike
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6931 | https://math.stackexchange.com/questions/213711/the-intersection-of-a-line-with-a-circle?rq=1 | analytic geometry - The intersection of a line with a circle - Mathematics Stack Exchange
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The intersection of a line with a circle
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Get the intersections of the line y=x+2 y=x+2 with the circle x 2+y 2=10 x 2+y 2=10
What I did:
y 2=10−x 2 y 2=10−x 2
y=10−x 2−−−−−−√y=10−x 2 or y=−10−x 2−−−−−−√y=−10−x 2
x+2=y=10−x 2−−−−−−√x+2=y=10−x 2
If you continue, x=−3 x=−3 or x=1 x=1 , so you get 2 points (1,3)(1,3), (−3,−1)(−3,−1)
But then, and here is where the problems come:
x+2=−10−x 2−−−−−−√x+2=−10−x 2
I then, after a while get the point (−3 1 2,−1 1 2(−3 1 2,−1 1 2) but this doesn't seem to be correct. What have I done wrong at the end?
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asked Oct 14, 2012 at 16:50
JohnPhtevenJohnPhteven
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Personally, I'd not take the square root, but substitute for y y from the linear equation into the quadratic, which gives you a quadratic in x x and two solutions.Mark Bennet –Mark Bennet 2012-10-14 17:02:06 +00:00 Commented Oct 14, 2012 at 17:02
Yes, I know, that is true, but what is wrong about my way?JohnPhteven –JohnPhteven 2012-10-14 17:02:31 +00:00 Commented Oct 14, 2012 at 17:02
It involves more manipulation and therefore more chance to make a mistake.Mark Bennet –Mark Bennet 2012-10-14 17:04:06 +00:00 Commented Oct 14, 2012 at 17:04
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If you do everything correct, the solution of x+2=−10−x 2−−−−−−√x+2=−10−x 2 would be almost the same - (1,−3)(1,−3), (−3,1)(−3,1). It's refer to intersaction of y=−(x+2)y=−(x+2) and original circle.
By the way, your answer is correct.
I have no idea how you'd get the point (−3 1 2,−1 1 2)(−3 1 2,−1 1 2).
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edited Aug 8, 2014 at 22:00
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answered Oct 14, 2012 at 16:54
m0nhawkm0nhawk
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x+2=−10−x 2−−−−−−√x+2=−10−x 2. x 2+4 x+4=−10+x 2 x 2+4 x+4=−10+x 2. 4 x=−14 4 x=−14. x=3,5 x=3,5.JohnPhteven –JohnPhteven 2012-10-14 16:56:19 +00:00 Commented Oct 14, 2012 at 16:56
1 @ZafarS - the square of −10−x 2−−−−−−√−10−x 2 is 10−x 2 10−x 2 Mark Bennet –Mark Bennet 2012-10-14 17:03:26 +00:00 Commented Oct 14, 2012 at 17:03
ahh, that's what I wanted, thanks.JohnPhteven –JohnPhteven 2012-10-14 17:18:14 +00:00 Commented Oct 14, 2012 at 17:18
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Let the intersection be (a,b)(a,b), so it must satisfy both the given eqaution.
So, a=b+2 a=b+2 also a 2+b 2=10 a 2+b 2=10
Putting b=a+2 b=a+2 in the given circle a 2+(a+2)2=10 a 2+(a+2)2=10
2 a 2+4 a+4=10⟹a=1 2 a 2+4 a+4=10⟹a=1 or −3−3
If a=1,b=a+2=3 a=1,b=a+2=3
If a=−3,b=−3+2=−1 a=−3,b=−3+2=−1
So, the intersections are (−3,−1)(−3,−1) and (1,3)(1,3)
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answered Oct 14, 2012 at 16:52
lab bhattacharjeelab bhattacharjee
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I am aware it could be done that way, but what is wrong mathematically with my method? The correction model gives your method too, but I want to know why this is incorrect.JohnPhteven –JohnPhteven 2012-10-14 16:54:45 +00:00 Commented Oct 14, 2012 at 16:54
x+2=−10−x 2−−−−−−√x+2=−10−x 2 actually refers to the intersection of y=−(x+2)y=−(x+2) and x 2+y 2=10 x 2+y 2=10, the points of intersection being (−3,1),(1,−3)(−3,1),(1,−3)lab bhattacharjee –lab bhattacharjee 2012-10-14 16:57:31 +00:00 Commented Oct 14, 2012 at 16:57
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6933 | https://piyuscs.com/context-sensitive-grammar-for-net-set-computer-science/ | Context Sensitive Grammar for NET SET Computer Science - Piyu's CS
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Context Sensitive Grammar for NET SET Computer Science
Context-Sensitive Grammars (CSGs)
1. Definition of a Context-Sensitive Grammar
A Context-Sensitive Grammar (CSG) is a formal grammar where production rules depend on the surrounding context of a non-terminal. This means a non-terminal can be replaced by a string only if it appears within specific surrounding symbols.
CSGs are more powerful than context-free grammars (CFGs) but less powerful than unrestricted (Type-0) grammars.
Formal Definition:
A CSG is defined as a quadruple:
G = (V, T, P, S)
Where:
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V = A finite set of variables (non-terminals)
T = A finite set of terminals
P = A finite set of production rules of the form:
αAβ→αγβ
Where:
A∈V (a non-terminal)
α,β∈(V∪T) (any strings of terminals and/or non-terminals)
γ≠ε (γ is a non-empty string of terminals and/or non-terminals)
S∈V = Start symbol
2. Key Characteristics of Context-Sensitive Grammars
Context Dependence
Replacement of a non-terminal depends on its surrounding context (α andβ). The rule applies only if the non-terminal appears with the specified neighbors.
Non-Shortening Rules
Production rules do not reduce the length of the string:
|αAβ| ≤ |αγβ|
Exception: Some definitions allow S→ε only if the empty string belongs to the language.
Non-Terminal Replacement in Context
The grammar allows replacing non-terminals only under specific surrounding conditions.
3. Examples of Context-Sensitive Grammars
Example 1: Grammar for aⁿbⁿcⁿ, where n ≥ 1
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This language cannot be generated by any context-free grammar, but can be generated by a context-sensitive grammar:
G = ({S, B, C}, {a, b, c}, P, S)
Production Rules (P):
S→a B C
S→a S B C
C B→B C
a B→a b
b B→b b
b C→b c
c C→c c
This grammar generates strings like:
aⁿbⁿcⁿ, for n ≥ 1
Example 2: Context Dependence Illustration
Productions:
S→a A B C
A B→B B
In this case, A is replaced by B only when it is directly followed by B, showing clear context dependence.
4. Key Concepts Related to CSGs
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Derivation
Sequence of production rules applied to derive a string, considering the surrounding context.
Parse Tree
A graphical representation of the derivation. The structure is more complex than CFG parse trees due to context-sensitive rules.
Linear Bounded Automata (LBA)
CSGs are equivalent to LBAs, which are Turing machines with restricted memory (the tape is limited to the length of the input).
5. Relationship to Linear Bounded Automata
Every Context-Sensitive Grammar corresponds to a Linear Bounded Automaton that accepts the same language.
LBAs are restricted Turing machines with memory proportional to the input size.
Context-sensitive languages lie between context-free and unrestricted languages in the Chomsky hierarchy.
6. Importance of Context-Sensitive Grammars
CSGs are essential in formal language theory.
Applications include:
Natural language processing
Compiler design
Modelingcomplex language structures
CSGs offer a balance between expressiveness and computability, extending beyond the limitations of context-free grammars while maintaining practical constraints.
Identifying Context-Sensitive Grammars (CSGs)
Context-sensitive grammars (CSGs) are more complex to identify than context-free grammars. Follow these guidelines to identify a CSG:
1. Check the Form of Production Rules
CSGs have rules of the form:
αAβ→αγβ
Where:
A is a non-terminal
α and β are strings of terminals and/or non-terminals (the context)
γ is a non-empty string of terminals and/or non-terminals
The key idea: The replacement of A depends on its surrounding context (α and β)
Length condition:
|αAβ| ≤ |αγβ|
Meaning, no production can shorten the string, except possibly the rule S→ε if the empty string belongs to the language
2. Look for Context Dependence
Check if the replacement of a non-terminal depends on neighboring symbols
If a production applies differently based on what symbols appear around a non-terminal, the grammar is likely context-sensitive
3. Check for Non-Shortening Rules
Ensure no rule reduces the overall length of the string (except possibly S→ε)
Most CSGs prevent shortening to maintain structure
4. Consider the Language Generated
Some languages are known to be context-sensitive but not context-free
Example: L = { aⁿbⁿcⁿ| n ≥ 1 } is a classic context-sensitive language
5. Examples
Example 1: Context-Sensitive Grammar
Production: a A b→a B b
Here, A is replaced by B only when surrounded by a and b
Context-dependent → This is context-sensitive
Example 2: Context-Sensitive Grammar for L = { aⁿbⁿcⁿ| n ≥ 1 }
Productions:
S→a B C
S→a S B C
C B→B C
a B→a b
b B→b b
b C→b c
c C→c c
Explanation:
The rule C B→B C applies only when C is followed by B, showing context sensitivity
Example 3: Not Context-Sensitive (Context-Free)
Productions:
A→a A
A→b
Explanation:
A is replaced independently of surrounding symbols → This is context-free
Example 4: Not Context-Sensitive (Unrestricted Grammar)
Production:
A B→ε
Explanation:
This rule shortens the string, which is not allowed in CSGs (except for S→ε)
This may belong to a Type-0 (unrestricted) grammar
6. Use the Pumping Lemma
If you suspect a language is context-sensitive:
Apply the Pumping Lemma for context-free languages
If the language fails the pumping lemma for CFGs, it might be context-sensitive (but not guaranteed)
7. Consider the Recognizing Automaton
Context-sensitive languages are recognized by Linear Bounded Automata (LBA)
If an LBA can be constructed for the language, it indicates context sensitivity
8. Challenges in Identifying CSGs
Context dependencies can be subtle and hard to detect
Unlike CFGs, no universal parsing algorithm exists for CSGs
9. Summary of Context-Sensitive Grammars
CSG Requirement: Replacement depends on the surrounding context
Production Form:
αAβ→αγβ
A is replaced based on its neighborsα and β
Non-Shortening Condition: Length does not decrease (except S→ε)
Not a CSG if:
Rules apply independently of context
Rules shorten the string improperly
Table of Contents
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6934 | https://tochno.st/materials/neft-i-gaz-obespecivaiut-do-poloviny-doxodov-federalnogo-biudzeta-rossii-no-tolko-v-dvux-bogatyx-resursami-regionax-uroven-zizni-silno-vyse-srednego-po-strane | Нефть и газ обеспечивают до половины доходов бюджета — но только в двух богатых ресурсами регионах уровень жизни сильно выше среднего по стране
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Нефть и газ обеспечивают до половины доходов бюджета — но только в двух богатых ресурсами регионах уровень жизни сильно выше среднего по стране
29 августа 2024
Нефть и газ обеспечивают до половины доходов бюджета — но только в двух богатых ресурсами регионах уровень жизни сильно выше среднего по стране
Разбираемся, почему природные ресурсы далеко не всегда приводят к процветанию
Нейросеть Midjourney
Основной источник богатства России — полезные ископаемые, в первую очередь, нефть и газ. За последние 10 лет больше трети доходов федеральный бюджет получил от добычи и экспорта углеводородов. Почти вся добыча приходится на 17 регионов — но это совсем не обязательно делает богаче местных жителей. Мы проанализировали уровень доходов и неравенства, структуру бюджетных расходов и демографию в ресурсных регионах и во всех остальных. Рассказываем, каким регионам удается извлечь выгоду из своих богатств и почему нефть и газ не гарантируют благосостояния жителям добывающих территорий.
В восьми регионах РФ почти половина экономики — это добыча нефти и газа
У понятия «ресурсный регион» нет общепринятого определения — мы будем называть так регионы, в которых в 2019 году (в большинстве случаев для расчетов мы берем этот год как последний доковидный и докризисный) было добыто не меньше 5 млн тонн нефти и газа (в пересчете на нефтяной эквивалент). Их в России 17. Выделяются Ханты-Мансийский и Ямало-Ненецкий автономные округа: в 2019 году на ХМАО пришлось 41% всей российской нефти, на ЯНАО — 84% всего газа.
Среди ресурсных регионов можно выделить сугубо сырьевые — такие, где в 2010–2019 годах доля добывающей отрасли в валовом региональном продукте (ВРП)Стоимость всех произведенных в регионе товаров и услуг за вычетом стоимости товаров и услуг, потребленных в процессе производства. превышала 50%. Таких регионов четыре: три автономных округа — Ханты-Мансийский, Ямало-Ненецкий, Ненецкий — и Сахалинская область. Чуть меньше половины ВРП добывающий сектор занимает еще в четырех регионах — Коми, Якутии, Оренбургской и в Астраханской областях.
В ХМАО, ЯНАО и НАО доля добывающей промышленности составляет 70-75% ВРП и не сильно меняется с начала нулевых. В отличие от Сахалина, где в конце 2000-х началось активное освоение газовых месторождений на шельфе острова и доля нефтегаза выросла почти в четыре раза — с 17% в 2004 до 64% в 2019 году. В Охотском море добывают 4% российского газа — больше, чем в любом другом регионе, кроме ЯНАО.
В большинстве нефтегазовых регионов активная добыча идет еще с советских времен, однако в пяти из них активное развитие отрасли развернулось лишь в последние 10-20 лет. В Красноярском крае, Иркутской и Тюменской областях, на которые в 2019 году пришлась десятая часть всей добытой в России нефти, в 2000 году ее еще практически не добывали. Якутия за 20 лет нарастила добычу нефти с 0,5 до 14,2 млн тонн.
Правда, слишком полагаться на данные о ВРП нельзя
Во-первых, данные о ВРП доступны только по очень крупным классам, таким как добыча полезных ископаемых, обрабатывающие производства, оптовая и розничная торговля и так далее. Чем считать первичную переработку нефти, при которой от нее отделяются газ, вода и механические примеси — частью добывающего сектора или уже обрабатывающего?
Во-вторых, добыча полезных ископаемых не обязательно предполагает именно нефть или газ, например, в случае Якутии это еще и алмазы.
Наконец, структура нефтегазовых холдингов устроена таким образом, что основной вид деятельности многих входящих в них компаний связан не с добычей, а с торговлей, транспортом и разными сопутствующими услугами.
В результате, комментирует директор Института энергетики и финансов Марсель Салихов, в отраслевой структуре ВВП России на добывающую промышленность приходится всего 10% произведенной в России добавленной стоимости, хотя очевидно, что влияние нефтегазового сектора гораздо выше. Искажение той же природы действует и на региональном уровне.
Производство растет не везде, запасы не бесконечны. В целом по России уровень добычи углеводородов за последние 10 лет вырос не сильно. В старых добывающих регионах добыча падает или стагнирует, а, например, в Тюмени и Якутии — быстро растет. Ресурсные регионы обеспечивают всю страну: за последние 10 лет больше трети доходов федеральный бюджет получил от добычи и экспорта углеводородов. Но как при этом живут сами жители нефтегазовых территорий? Чтобы узнать, как добыча полезных ископаемых влияет на уровень жизни в регионе, мы сравнили уровень доходов и неравенства, структуру бюджетных расходов и демографию в ресурсных регионах и во всех остальных.
Как мы считали
Информацию о бюджетных расходах взяли с сайта Федерального казначейства (при сравнении расходов делали поправку на значения индекса бюджетных расходов — с сайта Министерства финансов). Данные о налоговых поступлениях в разрезе по регионам и типам деятельности организаций есть на сайте ФНС. Значения всех остальных использованных показателей можно найти на Росстате.
Чтобы получить представление об уровне жизни в добывающих регионах, мы сравнили их по нескольким показателям с остальной Россией. Среди остальных регионов выделили еще две группы, представляющие крайние случаи — два столичных города-региона с одной стороны и дотационные регионы с другой. Дотационными считали регионы, бюджетные доходы которых в 2020 году более чем на 20% состояли из дотаций из федерального бюджета.
Только в шести добывающих регионах доходы значительно выше медианы по стране
В 2000 году среднедушевые доходы населения по добывающим регионам — с поправкой на местные цены — в среднем были на треть выше, чем в нересурсных (без учета Москвы и Санкт-Петербурга). Однако спустя двадцать лет разница сократилась до 5%. В нересурсных регионах в нулевые годы доходы росли быстрее, а в период 2010–2019 оставались примерно на том же уровне, в то время как в ресурсных — даже немного сократились.
В последние годы доходы ресурсных регионов находятся близко к медианному уровню по стране. Отчасти это может объясняться тем, что многие сотрудники добывающих предприятий работают вахтовым методом и приезжают из других регионов, поэтому их доходы не идут в учет в регионе добычи.
Исключение составляют только Татарстан и Якутия, где они значительно выше, а также сугубо сырьевые регионы (Сахалин, ХМАО, ЯНАО, НАО) — в них доходы выше среднего по стране за счет того, что большая часть населения занята в одной отрасли.
За последние 20 лет разрыв между ресурсными и нересурсными регионами сократился. В 2000 году среднедушевые доходы населения по добывающим регионам — с поправкой на местные цены — в среднем были на треть выше, чем в нересурсных (без учета Москвы и Санкт-Петербурга). Спустя двадцать лет разница сократилась до 5%. В нересурсных регионах в нулевые годы доходы росли быстрее, а в период 2010–2019 оставались примерно на том же уровне, в то время как в ресурсных — даже немного сократились.
Раньше доходы в добывающих регионах были выше среднего — почему это изменилось?
Как объясняет Наталья Зубаревич в статье «Мифы и реалии пространственного неравенства», у сокращения этого разрыва есть две основных причины. Во-первых, государство несколько раз поднимало зарплаты работников бюджетных организаций, которые составляют большинство занятых в слаборазвитых регионах. Во-вторых, «политика бизнеса по снижению издержек, раньше других начатая крупными сырьевыми компаниями, привела к сокращению занятости и более медленному росту уже достаточно высокой заработной платы в ресурсодобывающих отраслях».
В нефтегазовых регионах более высокое неравенство
Высокий средний доход — важный показатель, но нужно помнить, что сам по себе он еще не достаточно говорит об уровне жизни. Как минимум потому, что где высокие доходы, там обычно и высокое неравенство. Так, в Якутии индекс Джини, показывающий неравномерность распределения доходов, — на уровне Санкт-Петербурга (при этом доходы в Якутии гораздо ниже). В Ненецком и Ямало-Ненецком автономных округах индекс Джини выше, чем в Москве. В некоторых ресурсных субъектах (Красноярский край, Тюменская область, Башкортостан) доходы примерно на уровне среднего региона — а неравенство выше.
Правда, с учетом неравенства, как и с учетом доходов, все очень сложно. Оценку искажает и нехватка данных о самых богатых (например, Росстат не учитывает доходы от акций, торгующихся на западных биржах) и неформальная занятость, скрытая от статистического наблюдения. По оценке доцента кафедры теории и социально-экономической статистики РЭУ им. Плеханова Ольги Лебединской, Росстат «не видит» около 30% всего фонда оплаты труда.
Но если про самых богатых официальные данные сообщают нам мало, то про самых бедных мы все же что-то знаем. По данным выборочного обследования Росстата, в 2022 году в Якутии, Красноярском крае и Коми доля населения с доходом ниже прожиточного минимума составляла 15-16%. Это примерно на треть выше медианы.
Ресурсные регионы тратят больше, но в основном на ЖКХ
Еще одно измерение качества жизни в регионе — это бюджетные траты на душу населения. В 2019 году больше всего бюджетных денег в пересчете на душу населения потратила Сахалинская область — немного меньше Москвы (на 16%) и больше Петербурга (на 24%).
Однако это не значит, что на Сахалине очень высокая обеспеченность социальными услугами. Почти пятая часть всех бюджетных расходов в Сахалинской области приходится на ЖКХ и благоустройство. Здесь снова играет роль небольшое население региона, а также большое число населенных пунктов, в которых живет по несколько десятков человек. Это повышает затраты на содержание инфраструктуры: чем больше население и чем компактнее проживание, тем дешевле обходится содержание инфраструктуры в пересчете на душу.
Похожая картина в ЯНАО и НАО, которые вошли в пятерку лидеров по тратам: как поясняет экономист Наталья Зубаревич, высокая доля коммунальной статьи в бюджетных расходах этих регионов связана с тем, что государство должно поддерживать жилищно-коммунальные организации, иначе их услуги были бы запредельно дорогими.
В результате на ЖКХ в ЯНАО и НАО тратится столько же или больше, чем на здравоохранение или на социальную политику. В нересурсных регионах на ЖКХ обычно тратят значительно меньше: так, в Санкт-Петербурге на эту статью идет 9% бюджета, в Свердловской области — 5%. Зато Москва по доле затрат на ЖКХ и благоустройство (17-18%) выглядит как северный нефтедобывающий регион.
За исключением перечисленных регионов, только в Тюменской области, Татарстане и Иркутской области душевые расходы бюджета сильно выше, чем в среднем по нересурсным регионам. В остальных — все примерно на том же уровне.
Нефть и газ не мешают оттоку населения
Динамика численности населения обычно говорит о привлекательности региона для жизни. Стабильные доходы и уверенность в завтрашнем дне положительно влияют на рождаемость. Кроме того, комфортные условия привлекают мигрантов из других регионов.
В целом большинство нефтегазовых регионов движется в общероссийском демографическом тренде — их население сокращается. В среднем темп депопуляции в ресурсных регионах немного ниже, чем в нересурсных Здесь и далее мы называем так остальные регионы России за вычетом Москвы и Петербурга, с одной стороны, и дотационных регионов, с другой., однако некоторые пустеют особенно быстро.
В первую очередь это республика Коми, население которой с 2004 по 2022 год сократилось на 20% — против 8% в среднем по нересурсными регионам и 2% по ресурсным. Опережающим темпом снижается численность населения Оренбургской области и Сахалина (-13% в обоих случаях).
Почти не сокращается население Башкортостана и Астраханской области. Только в пяти добывающих регионах численность населения растет — это Томская область (по сравнению с 2004 годом численность населения в 2022 году выросла на 2% ), Татарстан (+5%), Якутия (+6%) и особенно ХМАО (+18%) и Тюменская область (без АО; +22%). В двух последних регионах темп прироста примерно как в Санкт-Петербурге и Москве.
Одно из объяснений депопуляции некоторых нефтегазовых регионах состоит в том, что в СССР производства часто были организованы по географическому принципу (рядом с крупными месторождениями) — и после распада структуры экономики люди стали уезжать из «мертвеющих» мест. Как отмечают авторы книги The Siberian Curse, постсоветская Россия характеризуется миграцией с севера на юг и с востока на запад.
Еще один демографический признак, по которому можно косвенно судить о качестве жизни в регионе — ее ожидаемая продолжительность, то есть число лет, которое в среднем предстоит прожить человеку при условии, что факторы смертности останутся такими же, как в год его рождения Это условие фактически невыполнимо, поэтому показатель нельзя воспринимать буквально, но он позволяет сравнивать регионы по качеству жизни..
В большинстве ресурсных регионов ожидаемая продолжительность жизни (ОПЖ) — на уровне среднего российского региона. В негативном смысле выделяются только Иркутская область и Коми, где ОПЖ и мужчин и женщин на 1-2 года ниже медианы. В богатых добывающих регионах — ХМАО, ЯНАО, Тюменской области и Татарстане, — а также в Башкортостане ожидаемая продолжительность жизни, особенно мужская, даже выше медианных значений. Правда, до столичных значений им пока все равно далеко: к ним приближаются только ХМАО и Татарстан и только по женской ОПЖ.
Кроме того, в части ресурсных регионов высокая смертность от причин, которые принято связывать с отчаянием (подробнее об этом читайте в нашем недавнем тексте), — к ним относятся алкоголизм, наркотики и суициды.
Так, в Удмуртии и Иркутской области смертность, связанная с алкоголем, выше медианы по стране на 87% и 77% соответственно. Выделяются также Коми (63%), Оренбургская область (54%), Пермский край (38%) и Якутия (32%). В Красноярском крае и Томской области высокая смертность от наркотиков — в 6 и в 4,5 раза выше, чем в среднем регионе; много таких смертей регистрируют также в Пермском крае (в 3,4 раза больше медианы) и Тюменской области (в 2,6 раза больше). Ненецкий автономный округ один из антилидеров по суицидам: судя по данным Росстата, жители региона заканчивают жизнь самоубийством в 4 раза чаще, чем в среднем регионе; много таких смертей в Удмуртии (в 2,2 раза больше медианы), Пермском крае (тоже в 2,2 раза), Коми (2,1 раза), Оренбургской области (1,8 раза) и Якутии (1,5 раза).
Самые благоприятные для жизни — Тюменская область и Татарстан
Так хорошо ли жить в регионе, богатом нефтью или газом? На этот вопрос не получается дать однозначный ответ. Наш анализ показывает, что обеспеченность ресурсами не означает автоматически высоких доходов населения; не гарантирует богатства и отсутствия социальных проблем; не обязательно делает регион привлекательным для жизни.
Однако ресурсные регионы очень разные — и у них довольно мало общего в том, что касается уровня жизни. Даже сугубо сырьевые регионы — Ханты-Мансийский, Ямало-Ненецкий, Ненецкий автономные округа и Сахалин — могут сильно отличаться друг от друга. Они близки по структуре экономики и сильно опережают большинство других субъектов по уровню доходов (как, впрочем, и по уровню неравенства), но при этом население ХМАО быстро растет, а Сахалина — сокращается; ожидаемая продолжительность жизни в ХМАО и ЯНАО сильно выше медианы по стране, в НАО и на Сахалине — в районе среднего региона. Остальные ресурсные регионы сильно отличаются друг от друга по множеству параметров, не связанных с добычей, — не удивительно, что и по уровню жизни разброс среди них шире.
Некоторые регионы извлекают выгоду из своих ресурсных богатств, но это не всегда работает так. Самыми благоприятными для жизни выглядят Татарстан и Тюменская область: ожидаемая продолжительность жизни там сильно выше среднего, смертность «от отчаяния» низкая, население прибывает, доходы сравнительно высокие, а неравенство не сильно выше среднего. Хуже всего выглядит Коми: средний доход в республике на уровне Тюменской области, при этом она быстро теряет население, ОПЖ ниже среднего, смертность от алкоголя — одна из самых высоких в стране.
Большая часть налогов добывающих предприятий уходит в федеральный бюджет
Ключевой фактор в экономике российских регионов — это система перераспределения доходов от добычи углеводородов. Основной налог, который платят нефтегазовые компании — это налог на добычу полезных ископаемых (НДПИ). Он был введен в 2002 году, в ходе реформы налогового законодательства: тогда 60% этого налога шло в региональный бюджет, 40% — в федеральный.
Уже на следующий год долю регионов сократили до 20%. В 2004 году федеральный центр забрал себе весь НДПИ на природный газ и оставил регионам 14,4% НДПИ на другие полезные ископаемые. Министр финансов Алексей Кудрин объяснял это тем, что добывающие регионы уже прошли фазу слабого развития, на которой государство было вынуждено оставлять в их бюджетах значительную природной ренты, а федеральному бюджету нужны деньги на прорывные направления и фундаментальные научные исследования. В 2005 году регионами оставили 5% НДПИ на нефть. В 2009 году в федеральный бюджет забрали и их.
Региону же остаются большая часть налога на прибыль (17-18% из 20%), налог на имущество компаний, транспортный налог, НДФЛ и другие, менее значительные платежи. Для сравнения: в 2019 году НДПИ, оплаченный добывающими компаниями из ресурсных регионов, был в среднем в 12 раз больше, чем все их налоговые отчисления в региональные бюджеты.
Андрей Пискунов
управляющий директор группы рейтингов устойчивого развития рейтингового агентства НКР
Основу налоговых и неналоговых доходов региональных бюджетов составляют региональная доля налога на прибыль — 17% из 20%, НДФЛ — 100% остается в регионе, и налог на имущество юридических лиц — тоже 100% в регионе. Добывающие предприятия, как правило, платят более высокие зарплаты, обладают обширным имущественным комплексом и, если центр прибыли не «уводится» в другой регион, платят налог на прибыль по месту добычи. Нужно смотреть на то, где находится центр прибыли. К примеру, вхождение предприятия «Карельский окатыш» в «Северсталь» с последующим переносом центра прибыли в Череповец оставило в бюджете Республики Карелия дыру в размере около 300 млн рублей. Уровень добычи при этом не менялся.
Более того, региональные налоги на нефтегазовую прибыль не обязательно попадают в бюджет того субъекта, на территории которого ведется добыча: добывающая компания может быть зарегистрирована в другом регионе и в таком случае платить налоги по месту регистрации. Так, крупнейшие российские налогоплательщики — «Газпром» и «Роснефть» — платят свои налоги соответственно в основном в бюджеты Санкт-Петербурга и Москвы.
Это не значит, что все их отчисления попадают в столицы, поясняет директор Института энергетики и финансов Марсель Салихов. У крупных компаний сложная структура, принадлежащие им юрлица обычно зарегистрированы по месту добычи — и там же платят налог на прибыль. Но часть прибыли они передают в головную компанию, так что в итоге значительная доля региональных налогов попадает туда, где находится центральный офис.
Например, по данным ФНС за 2019 год, в бюджеты Москвы и Ленинградской области поступило больше налогов на прибыль от добычи, чем в бюджеты Коми, Томской, Астраханской и Сахалинской областей. По сумме НДФЛ с зарплат сотрудников нефтегазовых компаний Москва уступила только ХМАО и ЯНАО.
Смена прописки нефтегазовой компании может привести к образованию огромной дыры в бюджете региона. Так было, например, в Омской области: в 2005 году компания «Сибнефть» — после покупки «Газпромом» и переименования в «Газпром нефть» — перерегистрировалась в Санкт-Петербурге. В результате из бюджета региона выпали около 14 млрд рублей — для сравнения, годом ранее все доходы бюджета Омской области составили 33,4 млрд.
Из-за этой специфики перераспределения нефтегазовых доходов местные бюджеты могут оказываться без собственных средств для развития социальной, транспортной и производственной инфраструктуры. Это, в свою очередь, ведет к оттоку людей в столичные регионы, где аккумулируются не только финансы, но и трудовой и инновационный потенциал, а региональное неравенство усиливается.
Как правительство пыталось решить проблему
Для более справедливого распределение доходов от прибыли больших холдингов в 2012 году был запущен институт консолидированных групп налогоплательщиков: их участники получали возможность более выгодным для себя образом формировать налоговую базу, а доходы от их прибыли распределялись между регионами пропорционально численности работающих в них сотрудников компаний. Однако, по оценке Минфина, это решение себя не оправдало — эффектом особого налогового режима стало снижение поступлений в региональные бюджеты. В итоге было принято решение о ликвидации института КГН начиная с 2023 года. Чтобы регионы могли адаптироваться к изменениям, в первые три года после отмены доходы бывших участников консолидированных групп продолжат распределяться между регионами согласно установленным пропорциям.
В результате регионы, богатые природными ресурсами и обеспечивающие федеральный бюджет существенной частью доходов, нередко оказываются в отстающем положении. Все это не значит, что правильным было бы просто оставить все доходы от ресурсов добывающим регионам: перераспределение необходимо для поддержки отстающих субъектов федерации. Другой вопрос — как сделать эту систему справедливой и на что тратить деньги от продажи ресурсов.
Медленное развитие богатых ресурсами территорий — не исключительно российская проблема. Экономисты активно изучают ее с конца 90-х — начала 00-х, когда обратили внимание на парадокс: в 1960–1990-х годах страны богатые ресурсами развивались медленнее тех, которые не обладали таким богатством. Этот феномен получил название «ресурсное проклятие». Но с существованием «проклятия» согласны далеко не все. В любом случае, российские сырьевые регионы очень разные и на их положение влиет слишком много факторов, чтобы можно было объяснить ситуацию в каждом из них «ресурсным проклятием».
Сильнее всего на развитие территорий в России влияют география и налоговая система, считают опрошенные нами эксперты.
Марсель Салихов
президент Института энергетики и финансов
Нефть, конечно, дает буст экономике региона. Но главное — география и особенности перераспределения нефтегазовых доходов. Российская налоговая система так устроена, что большая часть налогов уходит в федеральный центр. Какие-то плюсы для региона от добычи есть, но они не очень большие.
В этих условиях различия в развитии больше связаны с географической спецификой, чем с решениями местных властей. Я с трудом представляю себе развитие кластера обрабатывающей промышленности на территории, условно, Ямало-Ненецкого автономного округа. Просто потому что далеко и холодно.
Есть пример другого развития — Татарстан. Но он просто находится в центре Поволжья, среди других регионов, где более-менее нормальные климатические условия и развитая промышленность была еще до того, как в 1950-е годы начали добывать нефть на Ромашкинском месторождении. Но нельзя сказать, что есть «хорошие» и «плохие» пути развития нефтегазовых регионов.
Автор:Сергей Бондарьков
Инфографика:Татьяна Цвирова, Екатерина Буркова
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6935 | https://journals.sagepub.com/doi/10.1177/2632010X221126987 | Histological Spectrum of Post Covid Debridement Tissues: Salient Histomorphological Features With Respect to Identification Fungal Elements - Preeti Agarwal, Devanshi Brajesh Dubey, Madhu Kumar, Pratima Verma, Menka Mishra, Shalini Rawat, Damini Singh, Virendra Verma, Ravindra Kumar Garg, 2022
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First published online September 30, 2022
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Histological Spectrum of Post Covid Debridement Tissues: Salient Histomorphological Features With Respect to Identification Fungal Elements
Preeti Agarwalpreavn@gmail.com, Devanshi Brajesh Dubey, […], Madhu Kumar, Pratima Verma, Menka Mishra, Shalini Rawat, Damini Singh, Virendra Verma, and Ravindra Kumar Garg+6-6View all authors and affiliations
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Abstract
Background:
Secondary bacterial and fungal infections in COVID patients have been documented during current pandemic. The present study provides detailed account of histomorphology of debridement tissue received for suspected fungal infections. The primary objective was to determine the morphological characteristics that must be recognized for the identification of fungal hyphae.
Methods:
The detailed histological examination of debridement tissue was performed. Demographic and clinical findings with treatment provided was recorded. Presence or absence of necrosis and lecocytoclasis was noted.
Results:
A total of 110 cases of debrided tissues were included in the study. Eosinophilic granular necrosis with lecocytoclasis was observed in 103cases; fungal elements were identified in 89.3% (92/103) of these. Eleven cases where necrosis was observed, strong suspicion of fungus was reported, 6 of them displayed fungus on KOH preparation, 3 on repeat biopsy. However, in 2 of these cases, neither KOH nor repeat biopsies identified the fungus. Mucor with aspergillus was observed in 7 cases and actinomyces in 3. In all these 10 cases dense fungal colonies were evident. In 7 cases careful observation revealed fruiting bodies of aspergillus. Cotton ball appearance of actinomyces was evident. Mucor infection in current disease was so rampant that aseptate ribbon like branching mucor hyphae were evident on H&E sections. Diabetes was significantly associated with fungal infection (97.2%; 70/72; P< .005). 90% [19/21] of the patients who were on room air and diagnosed with fungal infection were diabetic.
Conclusions:
Eosinophilic granular necrosis with the presence of neutrophilic debris in a case of suspected fungal disease suggests the presence of fungal elements. This warrants processing of the entire tissue deposited for examination, careful observation, application of fungal stains, and repeat biopsy if clinical suspicion is strong. Moreover, uncontrolled diabetes is more frequently associated with secondary fungal infection in COVID patients as compared to oxygen therapy.
Introduction
One of the major reverberations of the second wave of COVID 19 in the Indian subcontinent included fungal infections, leading to death and disability in many.1 The major etiological agent seen was mucor mycosis, followed by aspergillus and candida species.2 These were commonly referred to as black, white, and yellow fungi. Multiple factors have been implicated to the post covid fungal infections namely injudicious use of steroids, uncontrolled diabetes, use and reuse of oxygen masks, industrial oxygen, innate delta variant nature and others. Some of these have been justified by adequate evidence, while others have been circulated in the media unwisely.3
The spike protein of SARS-CoV-2 is the key viral protein responsible for binding of the virus to angiotensin-converting enzyme- 2 (ACE-2) receptor in humans. It confers both infectivity and antigenicity. Multiple versions of SARS-CoV-2 have emerged since its initial emergence in Wuhan province of China. The delta variant was the prominent variant which caused the second and the most devastating wave in India. Mutations in the terminal domain of the Spike protein N, single Spike variations in SARS-CoV-2 have been reported.4 This led to increased infectivity, transmission, and antibody resistance. Complimented with injudicious use of steroids along with multiple clinical and logistic factors, secondary bacterial and fungal infections conferred poor outcome with delta variant infection.5
Fungal infections, especially mucormycosis, are an uncommon histological finding. Mucor mold is ubiquitous. It may be found in soil, manure, rotting produce, even in nose and mucus of healthy individuals. They often are seen as opportunistic infections in immunocompromised and debilitated patients. In our setup, we saw occasional cases in prepandemic times.6 Mucor mycosis is an angioinvasive disease caused by mold fungi of the genus Rhizopus, Mucor, Rhizomucor, Cunninghamella, and Absidia of order Mucorales, class Zygomycetes. Rhizopus Oryzae is the most common type and responsible for nearly 60% of Mucor mycosis cases in humans. It also accounts for 90% of the Rhinoorbital Cerebral (ROCM) form.3,7,8 The mode of contamination occurs by inhalation of fungal spores. Infection with this invasive fungus may be life threatening. In sections stained with hematoxylin and eosin (H&E) they may be visualized as broad aseptate branching fungal hyphae. Angioinvasion and thrombosis are also evident in mucor-infected tissues. Periodic acid shiff (PAS) and Grocott methenamine silver (GMS) are used to stain and identify these fungal elements. It has been postulated that SARS-CoV-2 infection may alter and effect cell mediated immunity, especially CD4+ and CD8+ T cells. T cells are involved in the production of cytokines that damage fungal hyphae, namely interleukin 4, 10, 17 and interferon gamma. This superadded with altered cell-mediated immunity seen in diabetics may contribute in invasive fungal infections seen in delta wave.9,10
Surgical debridement tissue samples from patients with suspected fungal infections were received in large numbers during the first months of 2021. In this study, we present a detailed account of the histology of these samples and discuss the salient morphological features for their identification. Highlighting the features that must be recognized while suspecting or diagnosing these fungal infections.
Material and Methods
Ours was a covid-19 L3 facility. The patients were referred from all towns, community centers and even district hospitals. Later we were converted to a fungal management center when secondary fungal infections in patients with COVID started to increase. A robust surgical team consisting of ENT, ophthalmic, oral, and maxillofacial surgeons was formed to complete the procedure.
Standard operating protocol for processing of debridement tissue: To ensure that deadly fungal elements are not missed, standard operating procedures were formed at the laboratory level and it was decided that all tissue would be processed for histology and submitted for fungal identification.11 Periodic acid shiff (PAS) and Grocott methenamine silver (GMS) stains were also performed as and when necessary. The definitive report was dispatched only after 2 consultant pathologists had interpreted the sections.
In this retrospective observational study, sections obtained from debridement tissues received in histopathology laboratory were recovered from records and reviewed. Collected patient data was de-identified. All the investigators were blinded to the patient identifying details.
The slides were first scanned by third-year resident pathologists (PV, MM, DBD), then by senior residents (SR and DS) followed by consultant pathologists (PA and MK). Objective identification of morphological parameters was performed, namely (1) presence and absence of fungal elements, (2) presence and absence of necrosis, (3) type of necrosis, and (4) vascular invasion and any other morphological characteristics.
After careful review of the histology sections, few elements of patient clinical details were retrieved form records namely history and medication for diabetes mellitus, intervention in form of nebulization or oxygen support and management in terms of intravenous antifungals.
An MS EXCEL spreadsheet was used for data entry; SPSS version 16.0 was used for analysis. The Chi-Square test /Fisher’s exact test were used to compare the study variables. For this study, the significance level was set at .05.
Results
Demographic variables
A total of 110 debridement tissue samples were studied. Mean age was 50.16 years which included patients from 25 to 75 years of age. Male: female ratio was 1.97:1. The most common site of fungal infection was the nasal cavity, followed by the maxilla, eyeball, and palate (Table 1). 65.4% (72/110) of the patients were diabetic. Fungal elements were not found in 14 cases. Ninety-four patients received antifungals, mainly intravenous amphotericin and Posaconazole. Diabetes was significantly associated with fungal infection (97.2%; 70/72; P< .005). 90% [19/21] of the patients who were on room air and diagnosed with fungal infection were diabetic. In 96 cases, morphological fungal elements were identified. Mucor mycosis was most frequent, dual infection with both aspergillus and mucor was seen in 7 cases and 2 showed actinomyces with mucor hyphae. 66% (6/9) of these patients with dual infection were diabetic and 77% (7/9) received nebulization. No fungal elements that were morphologically consistent with candida were observed in any of the debridement samples studied.
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Table 1. Demographic and clinical variables of the studied debridement tissues: n = 110.
| S. no. | Patient characteristic | Value |
---
| 1 | Age | Mean 50.16 year (25-75) |
| 2 | Male:Female ratio | 1.97:1 (73:37) |
| 3 | Site: |
| 1. Nasal Cavity | 73 |
| 2. Maxilla | 17 |
| 3. Eyeball | 8 |
| 4. Palate | 7 |
| 5. Maxilla + Palate | 6 |
| 6. Brain | 1 |
| 7. Endotracheal | 1 |
| 8. Endobronchial | 1 |
| 4 | Room Air | 21 (19.09%) |
| 5 | Nebulization plus oxygen | 88 (80%) |
| 6 | Diabetes Mellitus (n = 77) |
| 1. Present | 72 |
| 2. Absent | 14 |
| 7 | No Fungal Elements | 14 |
| 8 | Fingal elements identified | 96 |
| 1. Only mucor | 83 |
| 2. Mucor on repeat biopsy | 3 |
| 3. Mucor + Aspergillus | 7 |
| 4. Mucor + Actinomyces | 3 |
| 9 | Treatment |
| 1. Death/Lama | 14 |
| 2. Anti-fungal given | 94 |
| 3. Conservative | 2 |
Morphological variables
Necrosis
Significantly large to small focus of necrosis was evident in 96 cases. The necrosis was very peculiar morphologically. It was dense eosinophilic granular enclosing fragmented polymorphs and debris (Figure 1a and b). Associated bone necrosis and pigment deposition were also extensively observed in cases (Figure 1c and d). This necrosis was observed in almost all cases in which fungal elements were present (Figure 1e). In 11 cases where necrosis was seen; and no fungal elements were identified, histological suspicion of fungal infection was reported. Subsequently repeat debridement tissue in 3 out them displayed mucor hyphae. In 6 cases, the KOH smear was positive, which was submitted in microbiology (Table 2). There were only 2 of them that were negative for fungal elements. They were managed conservatively. Associated granulomas with entrapped fungal hyphal structures were also evident, although granulomas were morphologically observed in 5 cases only (Figure 2a and b).
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Table 2. Histological presence of Necrosis and lecocytoclasis in association with fungal element identification.
| Fungus on HPE | Necrosis and lecocytoclasis |
--- |
| Present | Absent |
| Present | Mucor = 92 (89.3%) | 0 |
| Absent | N = 11 (11.7%) Repeat biopsy positive = 3 (27.27%) KOH positive = 6 (54.54%) No further Fungus identified = 2 (18.18%) | 7 (7.4%) |
| | 103 | 7 |
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Figure 1. The section shows eosinophilic granular necrosis with an amorphous texture similar to “magic sand” with enclosed neutrophilic debris in (a) (H&E x200) and (b) (H&E x100). Bone necrosis is also observed in (c) and brown pigment is also observed in (d) (H&E x200). Fungal elements entrapped in necrosis are seen to be intermixed with fragmented neutrophils ((e); H&E x400).
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Figure 2. The section shows a well-formed granuloma with central necrosis ((a); H&Ex40) along with giant cells, in higher power view in (b) (H&E x400) trapped fragmented fungal elements are seen. (c-e) (H&E x400) show pauciseptate ribbon-shaped fungal hyphae showing various thicknesses of the wall and branching at the right angle as well. Small blob-like ends are also seen. The fungal elements in this section show background mucus in them.
Mucor mycosis
Of the 96 fungal infections, 86 (89.58%) were Mucor mycosis alone. In 7 cases, co-infection with aspergillus and in 3 with actinomyces was evident. To our surprise, the fungal elements were readily identifiable in H&E-stained sections as a result of the presence of large numbers of fungal hyphae. Furthermore, PAS and GMS stains merely served to confirm. Large numbers of fungal elements were seen trapped in mucus (Figures 2c to e and 3a).
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Figure 3. Section (a) shows a dense fungal colony with a large amount of hemosiderin pigment (H&E x200), fungal elements morphologically consistent with aspergillus were observed at higher magnification ((b); H&E x200). The fungal hyphae are acute angle branched with readily appreciable septations. Beautiful fruiting bodies can be seen in (c and d) (H&E x400) with a central core and hyphal arranged in a crown-like manner, and multiple red blood cells are seen in close apposition. Long slender bacteria forming cotton balls are seen in (e) (H&E x400).
Morphologically, the fungal hyphae of the mucor were thick, irregular, and branching. They were pauciseptate and right-angle branched. Few broken forms to large hyphae were seen (Figure 2c to e). No septations were evident in many. Fungal hyphae were also seen infiltrating the vascular wall (Figure 4a and b). Fine focus examination could reveal the pale and dark areas within the wall of the fungal structure on H&E. Interestingly, all stages of the fungal life cycle were seen in the same case, and hemosiderin pigment due to red blood cell lysis was evident in many.
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Figure 4. Mucor hyphae are seen infiltrating the vascular walls in (a and b) (H&E x200).
Aspergillus
Aspergillosis, when present with mucor in co-infections, had acute angled septate branching hyphal structures (Figure 3b), with intervening blob- or flower-like fruiting bodies (Figure 3c and d). Aspergillus hyphae were thin compared to mucor. They were present in large groups or colonies (Figure 3a).
Actinomyces
Superadded actinomyces with Mucor mycosis were also observed. Cotton ball like structures composed of slender bacteria were seen (Figure 3e), which were well stained with hematoxylin.
Discussion
The prevalence of Mucor mycosis varies from 0.005 to 1.7 per million people worldwide. Its prevalence is approximately 80 times higher (0.14 per 1000) in India when compared to western countries, according to a recent estimate of the year 2019-2020.7,8 Diabetes mellites remains the leading risk factor associated with Mucor mycosis worldwide, with a global mortality of 46%.12
During the second wave in India, March 2021to July 2021; this number increased exponentially. SARS-CoV-2 infection with comorbidities, superimposed with unopposed use of steroids and other immunomodulators, increased the incidence of fungal infections in these patients. It was also seen that use of invasive procedures like intubation, central venous catheters for management predisposed to opportunistic fungal infections.13,14
The virologist played an important role in the diagnosis of Covid-19 and the mycologist in the initial identification of fungal hyphae in KOH preparations. However, histology and morphological typing of fungal elements in the debridement tissue was significant. The patients received additional intravenous antifungal therapy based on morphological diagnosis or culture reports.15,16
The present study highlights a detailed morphological account of debridement tissue received from suspected fungal infections with the objective of identifying the characteristic features that must be recognized while interpreting these tissues.
Necrosis reflects severe injury to cells, characterized by denaturation of proteins and cellular content leakage. The main cause of necrosis is cellular ischemia. Mucor mycosis is basically an Angio invasive fungus17 that leads to ischemic necrosis accompanied by acute inflammation. Commonly the ischemic necrosis morphologically shows increased eosinophilia, with glassy homogenous appearance of the necrosed cells.18 However, necrosis observed in tissues infected with mucor infection was eosinophilic, amorphous, displaying leucocytoclasis. Its texture was similar to “magic sand.” The defense of the host against mucor is especially through macrophages and neutrophils. Neutrophils initiate oxidative burst which cause death of proliferating hyphae and macrophages inhibit germination of fungal spores in diseased area.19 Presence of both these inflammatory cells was seen in abundance in debrided tissue.
Interestingly mucor was also seen flourishing in the entrapped mucus of the tissue along with surrounding hemorrhagic areas. Mucor is known to thrive where iron is present and tend to invade the blood vessels. Iron metabolism related genes have been identified to be involved in iron uptake by them. They regulate reductive system via ferric reductase, copper oxidases and high affinity permease. Siderophore permeases and uptake from heme via heme oxygenase are also seen.3,20
Morphological identification of fungal elements is vital for any histopathologist who reports on surgical pathology samples. Patients often present with symptoms such as black discharge from the nose, mouth, or gums. It may be accompanied by facial weakness or squinting due to the involvement of cranial nerves. In our study, the patients presented with swelling, black discoloration, pain or discharge from these areas. Usually, these fungus are resistant to multidrug therapy and cause serious life-threatening systemic infection if left untreated. The newer triazoles, posaconazole (POSA) and isavuconazole (ISAV) have been found to be effective therapy in their management.21
Another important pertinent morphological feature apart from hyphae being pauci septate and right-angle branching was that fine focus revealed differential density of the thick fungal hyphae which was not seen in either aspergillus or candida. They have been reported to vary in width and may also appear crinkled or folded.19,22
In cases where dual infection was observed, the fungal colonies were dense with seemingly intermingling of fungal elements. Clinically and in KOH preparations dual infections are difficult to identify. Histology is diagnostic in this scenario.23 Morphological identification of the 2 separate fungal elements on basis of the described features helped to individualize both. The presence of fruiting bodies on histology was quite diagnostic of aspergillus infection. There are few case reports in literature which report histological features of aspergillus fruiting bodies.24,25 Fruiting bodies were observed in 5/7 cases with a central core and fungal elements growing out of it (Figure 3c and d). Careful observation and scanning of the sections are needed to find these readily identifiable formations.
Diabetes was seen to be significantly associated with mucor infection in patients who underwent debridement in our study, as compared to patients who underwent intervention or were given oxygen therapy. Patel et al, in their large metacentric study in patients with COVID 19, also reported that accompanying mucor mycosis was seen more frequently in patients with uncontrolled diabetes.26 Similarly, Jeong et al in their meta-analysis collated the above observations. The plausible reason for this observation is defective innate immunity, cell-mediated immunity, chemotaxis along with the function of polymorphs and macrophages.27,28
The strength of the present study is that it provides a detailed account of the morphology of the fungal debridement tissues and supports diabetes being the major cofactor of opportunistic fungal infections in Covid 19 patients. The limitation is that it is a selective study of patients with fungal infections in SARS-CoV-2 patients. The observations cannot be generalized until further investigated. A study that evaluates the morphology of fungal infections in patients who are not infected with SARS-CoV-2 may establish and support our findings.
Conclusion
To conclude, mucormycosis was the most common opportunistic fungal infection seen in cases of Covid-19 in our set-up followed by aspergillus. Eosinophilic granular necrosis with the presence of neutrophilic debris was an important characteristic feature associated with fungal infection. So much so that the presence of this type of necrosis warrants careful observation, embedding of the entire sampled tissue, additional special stains, and repeat biopsy if required. We also describe the salient morphological characteristics for the identification of mucor myosis and aspergillus, along with the histomorphology of the fruiting bodies of aspergillus. Finally, we recommend that while interpreting a case with suspected fungal infection, the morphological identification of fungal elements must be performed with caution. Fine features must be kept in mind, and the entire tissue must be thoroughly examined to avoid misdiagnosis and most importantly “missed diagnoses.”
Acknowledgments
We are thankful to King George’s Medical University for providing the infrastructure to perform our work. We are also grateful to the surgical team and all KGMU doctors and staff for their unrelenting efforts and dedication to patients. We lost many of our staff and doctors in this fight against COVID and dedicate this article to them.
Consent for Publication
The manuscript has been reviewed by all authors and they are of the view that this information will be useful to clinicians and pathologists alike. We hereby transfer, assign, or otherwise convey all copyright ownership, including all rights incidental thereto, exclusively to the journal, if such work is published by the journal.
Statement of Ethics
The work was approved by Institutional Ethics Committee, King George’s Medical University. An ethical waiver has been granted, all the information provided in this manuscript has been de-identified and the patients cannot be traced directly. All the investigators were blinded to the patient identifying details.
Declaration of conflicting interests:
The author(s) declared no potential conflicts of interest with respect to the research, authorship, and/or publication of this article.
Funding:
The author(s) received no financial support for the research, authorship, and/or publication of this article.
Data availability statement
All data generated or analyzed during this study are included in this article. Further enquiries can be directed to the corresponding author.
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Clinical Pathology
Volume 15
Article first published online: September 30, 2022
Issue published: January-December 2022
Keywords
Mucor mycosis
debridement tissue
fungal infection
histology
Covid-19
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Authors
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Preeti Agarwal
Department of Pathology, King George’s Medical University, Lucknow, Uttar Pradesh, India
preavn@gmail.com
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Devanshi Brajesh Dubey
Department of Pathology, King George’s Medical University, Lucknow, Uttar Pradesh, India
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Madhu Kumar
Department of Pathology, King George’s Medical University, Lucknow, Uttar Pradesh, India
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Pratima Verma
Department of Pathology, King George’s Medical University, Lucknow, Uttar Pradesh, India
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Menka Mishra
Department of Pathology, King George’s Medical University, Lucknow, Uttar Pradesh, India
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Shalini Rawat
Department of Pathology, King George’s Medical University, Lucknow, Uttar Pradesh, India
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Damini Singh
Department of Pathology, King George’s Medical University, Lucknow, Uttar Pradesh, India
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Virendra Verma
Department of ENT, King George’s Medical University, Lucknow, Uttar Pradesh, India
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Ravindra Kumar Garg
Department of Neurology, King George’s Medical University, Lucknow, Uttar Pradesh, India
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Notes
Preeti Agarwal, Department of Pathology, KGMU, Shah Mina Road, Lucknow, Uttar Pradesh 226003, India. Email: preavn@gmail.com
Author Contributions
Preeti Agarwal: Conceptualization, Methodology, Software, Validation, Formal analysis, Data Curation, Writing—Original Draft, Writing—Review & Editing, Visualization, Project administration. Devanshi B Dubey: Software, Writing—Original Draft. Madhu Kumar: Validation. Pratima verma: Software. Menka Mishra: Software. Shalini Rawat: Software. Damini Singh: Software. Veerendra Verma: Investigation, Resources, Supervision. Ravindra Kumar Garg: Investigation, Resources, Supervision.
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Figures
Figure 1. The section shows eosinophilic granular necrosis with an amorphous texture similar to “magic sand” with enclosed neutrophilic debris in (a) (H&E x200) and (b) (H&E x100). Bone necrosis is also observed in (c) and brown pigment is also observed in (d) (H&E x200). Fungal elements entrapped in necrosis are seen to be intermixed with fragmented neutrophils ((e); H&E x400).
Go to FigureOpen in Viewer
Figure 2. The section shows a well-formed granuloma with central necrosis ((a); H&Ex40) along with giant cells, in higher power view in (b) (H&E x400) trapped fragmented fungal elements are seen. (c-e) (H&E x400) show pauciseptate ribbon-shaped fungal hyphae showing various thicknesses of the wall and branching at the right angle as well. Small blob-like ends are also seen. The fungal elements in this section show background mucus in them.
Go to FigureOpen in Viewer
Figure 3. Section (a) shows a dense fungal colony with a large amount of hemosiderin pigment (H&E x200), fungal elements morphologically consistent with aspergillus were observed at higher magnification ((b); H&E x200). The fungal hyphae are acute angle branched with readily appreciable septations. Beautiful fruiting bodies can be seen in (c and d) (H&E x400) with a central core and hyphal arranged in a crown-like manner, and multiple red blood cells are seen in close apposition. Long slender bacteria forming cotton balls are seen in (e) (H&E x400).
Go to FigureOpen in Viewer
Figure 4. Mucor hyphae are seen infiltrating the vascular walls in (a and b) (H&E x200).
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Tables
Table 1. Demographic and clinical variables of the studied debridement tissues: n = 110.
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Table 2. Histological presence of Necrosis and lecocytoclasis in association with fungal element identification.
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6936 | https://thirdspacelearning.com/us/math-resources/topic-guides/number-and-quantity/decimal-number-line/ | High Impact Tutoring Built By Math Experts
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What is a decimal number line?
Common Core State Standards
How to fill in values on a decimal number line
Decimal number line examples
Example 1: fill in the scale Example 2: fill in the scale Example 3: fill in the scale Example 4: finding a missing value Example 5: finding a missing value Example 6: finding a missing value
Teaching tips for decimal number lines
Easy mistakes to make
Related lessons on decimals
Practice decimal number line questions
Decimal number line FAQs
Next lessons
Still stuck?
Decimal number line
Here you will learn about decimal number lines, including what decimal number lines are and how to use them.
Students will first learn about decimal number lines as part of numbers and operations in base 10 in 5th grade. They will build on this knowledge as part of the number system in 6th grade.
What is a decimal number line?
A decimal number line is a number line that includes decimal increments and can help us to visualize decimal numbers.
Decimals are numbers that have parts that are not whole. They represent values in between whole numbers such as tenths, hundredths, thousandths, and so on.
For example, this number line goes up from 0 to 1 by tenths (0.1).
Below, the number 0.638 has been marked on a decimal number line. Its position has been estimated between 0.6 and 0.7, but closer to 0.6.
If you zoom in between 0.6 and 0.7 to a number line that goes up by hundredths (0.01), you can position the decimal 0.638 more accurately. This time the position has been estimated between 0.63 and 0.64, but closer to 0.64.
If you zoom in once again between 0.63 and 0.64 to a number line that goes up by thousandths (0.001), then you can place the decimal 0.638 in its exact position.
Tenths, hundredths and thousandths are common decimal scales, but they are not the only ones you can use. Decimal number lines can use any decimal scale as long as the intervals are split into equal parts.
For example, this decimal number line goes up from 0 to 1 by two-tenths (0.2).
For example, this decimal number line goes up from 0 to 5 by five-tenths (0.5).
For example, this decimal number line goes up from 0 to 2 by twenty-five hundredths (0.25).
Decimal number lines can also be used to represent negative numbers.
For example, this decimal number line goes up from -3 to 2 by five-tenths (0.5).
What is a decimal number line?
Common Core State Standards
How does this relate to 5th grade math and 6th grade math?
Grade 5 – Numbers and Operations in Base 10 (5.NBT.A.3)Read, write, and compare decimals to thousandths.
Grade 6 – The Number System (6.NS.C.6.3)
Find and position integers and other rational numbers on a horizontal or vertical number line diagram.
How to fill in values on a decimal number line
In order to fill in values on a decimal number line:
Consider the top and bottom of the scale and the number of intervals.
Calculate the length of each interval by finding the difference in the top and bottom of the scale and dividing by the number of intervals.
Fill in the missing values.
[FREE] Decimal Number Line Worksheet (Grade 5 and 6)
Use this worksheet to check your 5th grade and 6th grade students’ understanding of decimal number line. 15 questions with answers to identify areas of strength and support!
DOWNLOAD FREE
x
[FREE] Decimal Number Line Worksheet (Grade 5 and 6)
Use this worksheet to check your 5th grade and 6th grade students’ understanding of decimal number line. 15 questions with answers to identify areas of strength and support!
DOWNLOAD FREE
Decimal number line examples
Example 1: fill in the scale
Fill in the scale.
Consider the bottom and top of the scale and the number of intervals.
The bottom of the scale is 7, the top of the scale is 8 and there are 5 intervals.
2Calculate the length of each interval by finding the difference in the top and bottom of the scale and dividing by the number of intervals.
To calculate the length of an interval step you can use the formula.
(\text{Top of the scale } - \text{ Bottom of the scale}) \; \div \text{ Number of intervals}
This gives,
(8-7)\div 5=0.2
So the scale goes up by two-tenths (0.2).
3Fill in the missing values.
Start from 7 and go up the scale, adding 0.2 each time.
Example 2: fill in the scale
Fill in the scale.
Consider the bottom and top of the scale and the number of intervals.
The bottom of the scale is 5, the top of the scale is 9 and there are 8 intervals.
Calculate the length of each interval by finding the difference in the top and bottom of the scale and dividing by the number of intervals.
To calculate the length of an interval step you can use the formula.
(\text{Top of the scale } - \text{ Bottom of the scale}) \; \div \text{ Number of intervals}
This gives,
(9-5)\div 8=0.5
So the scale goes up by five-tenths (0.5).
Fill in the missing values.
Start from 5 and go up the scale, adding 0.5 each time.
Example 3: fill in the scale
Fill in the scale.
Consider the bottom and top of the scale and the number of intervals.
The bottom of the scale is 12, the top of the scale is 14 and there are 5 intervals.
Calculate the length of each interval by finding the difference in the top and bottom of the scale and dividing by the number of intervals.
To calculate the length of an interval step you can use the formula.
(\text{Top of the scale } - \text{ Bottom of the scale}) \; \div \text{ Number of intervals}
This gives,
(14-12)\div 5=0.4
So the scale goes up by four-tenths (0.4).
Fill in the missing values.
Start from 12 and go up the scale, adding 0.4 each time.
Example 4: finding a missing value
Find the missing value.
Consider the bottom and top of the scale and the number of intervals.
The bottom of the scale is 5.2, the top of the scale is 6.0 and there are 8 intervals.
Calculate the length of each interval by finding the difference in the top and bottom of the scale and dividing by the number of intervals.
To calculate the length of an interval step you can use the formula.
(\text{Top of the scale } - \text{ Bottom of the scale}) \; \div \text{ Number of intervals}
This gives,
(6.0-5.2)\div 8=0.1
So the scale goes up by one-tenth (0.1).
Fill in the missing values.
Start from 5.2 and go up the scale, adding 0.1 each time.
The missing value is 5.6.
Example 5: finding a missing value
Find the missing value.
Consider the bottom and top of the scale and the number of intervals.
The bottom of the scale is 14, the top of the scale is 17 and there are 6 intervals.
Calculate the length of each interval by finding the difference in the top and bottom of the scale and dividing by the number of intervals.
To calculate the length of an interval step you can use the formula.
(\text{Top of the scale } - \text{ Bottom of the scale}) \; \div \text{ Number of intervals}
This gives,
(17-14)\div 6=0.5
So the scale goes up by five-tenths (0.5).
Fill in the missing values.
Start from 14 and go up the scale, adding 0.5 each time.
The missing value is 16.5.
Example 6: finding a missing value
Find the missing value.
Consider the bottom and top of the scale and the number of intervals.
The bottom of the scale is 7.1, the top of the scale is 7.5 and there are 8 intervals.
Calculate the length of each interval by finding the difference in the top and bottom of the scale and dividing by the number of intervals.
To calculate the length of an interval step you can use the formula.
(\text{Top of the scale } - \text{ Bottom of the scale}) \; \div \text{ Number of intervals}
This gives,
(7.5-7.1)\div 8=0.05
So the scale goes up by five-hundredths (0.05).
Fill in the missing values.
Start from 7.1 and go up the scale, adding 0.05 each time.
The missing value is 7.45.
Teaching tips for decimal number lines
Let students spend some time working with blank number lines to create their own representations from scratch. This gives them opportunities to compare and contrast strategies with others and also develop a better understanding of how decimal number lines are created. However, ensure that students understand the importance of creating equal intervals. In fact, it may be beneficial to compare two number lines (one with equal intervals and one without) and have students discuss why decimal number lines are only valid if they have equal intervals.
Use number line worksheets when applicable, but look for worksheets that show intervals other than tenths or hundredths. Often students see these types in so many activities, that they stop paying attention to the given numbers and intervals – leading to simple mistakes.
Use real world distance applications of decimal number lines whenever possible. Perhaps there is a rope in the school gym that needs to be measured and labeled at different lengths. Or maybe students can place a piece of masking tape on the floor to create a number line showing the distance from the classroom door to the wall. Using distance in this way allows students to make connections to measurement conversions and ratios and proportions.
Sixth graders should spend just as much time creating negative decimal number lines as positive decimal number lines.
To challenge students, have them create a decimal number line that also shows the equivalent fractions and mixed numbers it represents.
Our favorite mistakes
Assuming a decimal number line goes up in tenths
It is worth double checking what the intervals in a number line are. It is easy to assume it will be tenths, but often other steps are used.
For example,
Related lessons on decimals
Decimals
Decimal place value
Decimal places
Adding decimals
Multiplying and dividing decimals
Multiplying decimals
Dividing decimals
Adding and subtracting decimals
Subtracting decimals
Comparing decimals
Practice decimal number line questions
Find the missing value.
6.1
6.5
6.2
6.4
The bottom of the scale is 6, the top of the scale is 7 and there are 5 intervals.
To calculate the length of an interval step you can use the formula:
(\text{Top of the scale } – \text{ Bottom of the scale}) \; \div \text{ Number of intervals}
This gives,
(7-6)\div 5=0.2
Start from 6 and go up the scale, adding 0.2 each time.
The missing value is 6.4.
Find the missing value.
2.12
2.3
2.2
2.35
The bottom of the scale is 2.1, the top of the scale is 2.7 and there are 6 intervals.
To calculate the length of an interval step you can use the formula:
(\text{Top of the scale } – \text{ Bottom of the scale}) \; \div \text{ Number of intervals}
This gives,
(2.7-2.1)\div 6=0.1
Start from 2.1 and go up the scale, adding 0.1 each time.
The missing value is 2.3.
Find the missing value.
5.25
5.26
5.5
5.7
The bottom of the scale is 5.2, the top of the scale is 5.3 and there are 10 intervals.
To calculate the length of an interval step you can use the formula:
(\text{Top of the scale } – \text{ Bottom of the scale}) \; \div \text{ Number of intervals}
This gives,
(5.3-5.2)\div 10=0.01
Start from 5.2 and go up the scale, adding 0.01 each time.
The missing value is 5.25.
Find the missing value.
16.6
17.5
17.8
16.8
The bottom of the scale is 16.2, the top of the scale is 18.2 and there are 5 intervals.
To calculate the length of an interval step you can use the formula:
(\text{Top of the scale } – \text{ Bottom of the scale}) \; \div \text{ Number of intervals}
This gives,
(18.2-16.2)\div 5=0.4
Start from 16.2 and go up the scale, adding 0.4 each time.
The missing value is 17.8.
Find the missing value.
7.25
6.8
8.3
7.5
The bottom of the scale is 6.5, the top of the scale is 9.0 and there are 10 intervals.
To calculate the length of an interval step you can use the formula:
(\text{Top of the scale } – \text{ Bottom of the scale}) \; \div \text{ Number of intervals}
This gives,
(9.0-6.5)\div 10=0.25
Start from 6.5 and go up the scale, adding 0.25 each time.
The missing value is 7.25.
Find the missing value.
5.8
7.1
6.4
6.5
The bottom of the scale is 5.5, the top of the scale is 7.3 and there are 6 intervals.
To calculate the length of an interval step you can use the formula:
(\text{Top of the scale } – \text{ Bottom of the scale}) \; \div \text{ Number of intervals}
This gives,
(7.3-5.5)\div 6=0.3
Start from 5.5 and go up the scale, adding 0.3 each time.
The missing value is 6.4.
Decimal number line FAQs
What can decimal number lines be used for?
Decimal number lines can help students understand decimal place value and be used to order and compare decimals. They are also helpful as models for adding, subtracting, multiplying or dividing decimals. Students will also use decimal number lines when creating graphs and plotting data.
Should decimal number lines be created by hand?
There is value in having students explore decimal number lines by drawing them themselves. However, there are drawbacks as it can be more time consuming and ensuring that all the intervals are equal can be difficult. For each activity, use discretion to decide whether providing students with premade number lines or letting them create their own is more appropriate for the learning goal.
The next lessons are
Exponents
Algebraic expression
Math equations
Rounding decimals
Percents
Converting fractions, decimals, and percentages
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Introduction
What is a decimal number line?
Common Core State Standards
How to fill in values on a decimal number line
Decimal number line examples
↓
Example 1: fill in the scale Example 2: fill in the scale Example 3: fill in the scale Example 4: finding a missing value Example 5: finding a missing value Example 6: finding a missing value
Teaching tips for decimal number lines
Easy mistakes to make
Related lessons on decimals
Practice decimal number line questions
Decimal number line FAQs
Next lessons
Still stuck?
x
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6937 | https://pmc.ncbi.nlm.nih.gov/articles/PMC1401766/ | Hexamethonium - PMC
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. 1982 Jan;13(1):7–14. doi: 10.1111/j.1365-2125.1982.tb01331.x
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PMCID: PMC1401766 PMID: 7066156
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6938 | https://math.arizona.edu/~glickenstein/math323s13/test3practicesol.pdf | Some solutions Some things to be sure to know for Test 3 (note: this is not ALL you need for the test, but this is a good start) 1) De…nitions of function, injective/injection, surjective/surjection, bijec-tive/bijection, domain, codomain, range, image, preimage.
2) De…nitions of equinumerous, …nite, in…nite, denumerable, countable, un-countable 3) De…nitions of relation, re‡exive, symmetric, transitive, equivalence rela-tion, equivalence class.
4) If f : A ! B and C; C0 A and D; D0 B; then you should be able to prove basic things like: C f 1 (f (C)) f f 1 (D) D f (C \ C0) f (C) \ f (C0) f (C [ C0) = f (C) [ f (C0) f 1 (D \ D0) = f 1 (D) \ f 1 (D0) f 1 (D \ D0) = f 1 (D) \ f 1 (D0) Recall that if there is an equal, there are two directions to prove. It is extremely important that you know how to start and end these proofs.
Answers: Some of these are on p. 66/72.
5) Basics of compositions of functions: Let f : A ! B and g : B ! C be functions. You should be able to prove: a) If f and g are surjective, then g f is surjective.
Answer: Given c 2 C; since g is surjective, there exists b 2 B sych that g (b) = c: Since f is surjective, there exists a 2 A such that f (a) = b: Hence g (f (a)) = c: b) If g f is surjective, then g is surjective, but f may not be.
Answer: If g is not surjective, then there exists c 2 C such that g (b) 6= c for all b 2 B: But then g (f (a)) 6= c for all a 2 A: Thus we have proven the contrapositive, and we …nd that if g f is surjective then g is surjective. It is possible that f is not. For instance, take A = f1; 2g ; B = f1; 2; 3g ; C = f1g ; f (x) = x and g (y) = 1 for all y: The function f is not surjective, but the composition is.
1 c) If f and g are injective, then g f is injective.
d) If g f is injective, then f is injective, but g may not be.
6) More things you should be able to prove: a) Any …nite set is not equinumerous to N.
b) N is equinumerous to N n f1g.
Answer: The function f : N ! N n f1g given by f (n) = n + 1 is a bijection (inverse is g (n) = n 1).
c) N is equinumerous to Z.
Answer: We can consider the function f : N ! Z given by f (n) = n 2 if n is even n 1 2 if n is odd 7) You should know the de…nition of arbitrary intersection and arbitrary unions. Also, you should be able to prove: a) T n2N 0; 1 n = f0g b) T n2N (0; 1 n] = ?
c) S n2N 0; 1 1 n = [0; 1) d) S x2(0;1) x; 1 x = (0; 1) 2 |
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6940 | https://alipourmiri.github.io/files/ISAAC13.pdf | Faster Rumor Spreading with Multiple Calls Konstantinos Panagiotou1, Ali Pourmiri2, and Thomas Sauerwald2,3 1 The Ludwig Maximilian University of Munich, Germany 2 Max Planck Institute for Informatics, Saarbr¨ ucken, Germany 3 University of Cambridge, United Kingdom Abstract. We consider the random phone call model introduced by Demers et al. , which is a well-studied model for information dissem-ination on networks. One basic protocol in this model is the so-called Push protocol which proceeds in synchronous rounds. Starting with a single node which knows of a rumor, every informed node calls a ran-dom neighbor and informs it of the rumor in each round. The Push-Pull protocol works similarly, but additionally every uninformed node calls a random neighbor and may learn the rumor from that neighbor.
While it is well-known that both protocols need Θ(log n) rounds to spread a rumor on a complete network with n nodes, we are interested by how much we can speed up the spread of the rumor by enabling nodes to make more than one call in each round. We propose a new model where the number of calls of a node u is chosen independently according to a probability distribution R with bounded mean determined at the beginning of the process. We provide both lower and upper bounds on the rumor spreading time depending on statistical properties of R such as the mean or the variance. If R follows a power law distribution with exponent ∈(2, 3), we show that the Push-Pull protocol spreads a rumor in Θ(log log n) rounds.
1 Introduction Rumor spreading is an important primitive for information dissemination in networks. The goal is to spread a piece of information, the so-called rumor, from an arbitrary node to all the other nodes. The random phone call model is based on the simple idea that every node picks a random neighbor and these two nodes are able to exchange information in that round. This paradigm ensures that the protocol is local, scalable and robust against network failures (cf. ). Therefore these protocols have been successfully applied in other contexts such as replicated databases , failure detection , resource discovery , load balancing , data aggregation , and analysis of the spread of computer viruses .
A basic protocol for spreading a rumor in the phone call model is the Push protocol. At the beginning, there is a single node who knows of some rumor. Then in each of the following rounds every informed node calls a random neighbor chosen independently and uniformly at random and informs it of the rumor.
The Pull protocol is symmetric, here every uninformed node calls a random neighbor chosen independently and uniformly at random, and if that neighbor L. Cai, S.-W. Cheng, and T.-W. Lam (Eds.): ISAAC2013, LNCS 8283, pp. 446–456, 2013.
c ⃝Springer-Verlag Berlin Heidelberg 2013 Faster Rumor Spreading with Multiple Calls 447 happens to be informed the node becomes informed. The Push-Pull protocol is simply the combination of both protocols. Most studies in randomized rumor spreading concern the runtime of these protocols which is the number of rounds required until a rumor initiated by a single node reaches all other nodes.
In one of the first papers in this area, Frieze and Grimmett proved that if the underlying graph is a complete graph with n nodes, then the runtime of the Push protocol is log2 n + log n ± o(log n) with high probability1, where log denotes the natural logarithm. This result was later strengthened by Pittel .
For the standard Push-Pull protocol, Karp et al. proved a runtime bound of log3 n + O(log log n). In order to overcome the large number of Θ(n log n) calls, Karp et al. also presented an extension of the Push-Pull protocol together with a termination mechanism that spreads a rumor in O(log n) rounds using only O(n log log n) messages. More recently Doerr and Fouz proposed a new protocol using only Push calls with runtime (1+o(1)) log2 n using only O(n·f(n)) calls (and messages), where f(n) is an arbitrarily slow growing function.
Besides the complete graph, the randomized rumor spreading protocols men-tioned above have been shown to be efficient also on other topologies. In particu-lar, their runtime is at most logarithmic in the number of nodes n for topologies ranging from basic networks, such as random graphs [11,12] and hypercubes , random regular graphs , graphs with constant conductance [20,5,15], constant weak conductance or constant vertex expansion , to more complex struc-tures including preferential attachment graphs modeling social networks [10,13].
In particular, recent studies establishing a sub-logarithmic runtime on certain social network models [10,13] raise the question whether it is possible to achieve a sub-logarithmic runtime also on complete graphs. In addition to analyses on static graphs, there are also studies on mobile geometric graphs, e.g., [7,21], that have deal with strong correlations as nodes are moving according to a random walk.
Since the Push protocol, the Pull protocol and the Push-Pull protocol all re-quire Θ(log n) rounds to spread the rumor on a complete graph, we equip nodes with the possibility of calling more than one node in each round. Specifically, we assume that the power of each node u, denoted by Cu is determined by a probability distribution R on the positive integers which is independent of u. In order to keep the overall communication cost small, we focus on distributions R satisfying u∈V Cu = O(n) with high probability – in particular, R has bounded mean. While being a natural extension from a theoretical perspective, different Cu values could arise due to varying battery capacities, processor speeds or clock synchronizations. Our aim is to understand the impact of the distribution R on the runtime of randomized rumor spreading. In particular, we seek for conditions on R which are necessary (or sufficient) for a sublogarithmic runtime.
Our first result concerns the Push protocol for the case where R has bounded mean and bounded variance. As this is the most basic setting, our runtime bound 1 By with high probability we refer to an event which holds with probability 1 −o(1) as n →∞. For simplicity, we sometimes omit the “with high probability” in the introduction.
448 K. Panagiotou, A. Pourmiri, and T. Sauerwald is even tight up to low-order terms. To this end, let Tn = min{t | Pr [It = n] ⩾ 1 −g(n)} be the first round in which all nodes are informed with probability 1 −g(n), where g(n) is a function tending to zero as n goes to infinity (for simplicity we do not specify g(n)).
Theorem 1.1. Consider the Push protocol and let R be a distribution with E [R]= O(1) and Var [R]=O(1). Then|Tn −(log1+E[R] n + logeE[R] n)|=o(log n).
Note that by putting R ≡1, we retain the classic result by Frieze and Grim-mett for the standard Push protocol. If we drop the assumption on the variance, then the theorem below provides a lower bound of Ω(log n). Although this result is less precise than Theorem 1.1, it demonstrates that it is necessary to consider the Push-Pull protocol in order to achieve a sub-logarithmic runtime.
Theorem 1.2. Assume that R is any distribution with E [R] = O(1). Then with prob. 1 −o(1), the Push protocol needs Ω(log n) rounds to inform all nodes.
We point out that the lower bound in Theorem 1.2 is tight up to constant factors, as the results in [14,22] for the standard Push-Pull protocol already imply an upper bound of O(log n) rounds. Next we consider the Push-Pull protocol and extend the lower bound of Ω(log n) from Theorem 1.1.
Theorem 1.3. Assumethat RisanydistributionwithE [R] = O(1)andVar [R] = O(1). Then for any constant ϵ > 0, with probability 1−ϵ the Push-Pull protocol needs at least Ω(log n) rounds to inform all nodes.
Theorem 1.3 establishes that an unbounded variance is necessary to break the Ω(log n) lower bound. An important distribution with bounded mean but unbounded variance is the power law distribution with exponent β < 3, i.e., there are constants 0 < c1 ⩽c2 such that c1z1−β ⩽Pr [R ⩾z] ⩽c2z1−β for any z ⩾1, and Pr [R ⩾1] = 1. We are especially interested in power law distributions, because they are scale invariant and have been observed in a variety of settings in real life. Our main result below shows that this natural distribution achieves a sublogarithmic runtime.
Theorem 1.4. Assume that R is a power law distribution with 2 < β < 3. Then the Push-Pull protocol informs all nodes in Θ(log log n) rounds with prob. 1−o(1).
Notice that if R is a power law distribution with β > 3, then Theorem 1.3 ap-plies because the variance of R is bounded. Hence our results reveal a dichotomy in terms of the exponent β: if 2 < β < 3, then the Push-Pull protocol finishes in O(log log n) rounds, whereas for β > 3 the Push-Pull protocol finishes in Θ(log n) rounds 2. While a very similar dichotomy was shown in for random graphs with a power law degree distribution, our result here concerns the spread of the rumor from one to all nodes (and not only to a constant fraction as in ).
2 We do not consider the case β ⩽2, since then there exists at least one node with degree Ω(n) and the rumor is spread in constant time (additionally, E [R] is no longer bounded). The analysis of the case β = 3 is an interesting open problem.
Faster Rumor Spreading with Multiple Calls 449 In addition, the distribution of the edges used throughout the execution of the Push-Pull protocol is different from the distribution of the edges in a power law random graph, as the latter is proportional to the product of the two nodes weights. Therefore it seems difficult to apply the previous techniques for power law random graphs used for the average distance and rumor spreading .
Besides the power law distribution, one may also consider a simple two point distribution, where for instance, R = n with probability n−1 and R = 1 other-wise. It is then straightforward to see that with constant probability, the Push-Pull protocol informs all nodes in O(1) rounds. The same result also holds if R = nϵ with probability n−ϵ and R = 1 otherwise. However, the power law distribution is arguably a more natural distribution which occurs in a variety of instances in practice.
Finally, we also show that it is crucial that the Cu’s do not change over time.
Instead, suppose we generate a new variable Ct u according to the distribution R for the number of calls made by node u in each round t. Then one can prove a lower bound of Ω(log n) for the Push-Pull protocol for any distribution R with bounded mean. Based on this lower bound it seems crucial to have a fixed set of powerful nodes (i.e. nodes u with large Cu) in order to obtain a sublogarithmic rumor spreading time.
2 Definitions and Notations We now provide additional definitions and notations (note that the classic Push, Pull and Push-Pull protocols have already been defined before). Here we gener-alize the classic Push, Pull and Push-Pull to the following statistical model on a complete graph with n nodes.
Before the protocol starts, every node u generates a random integer Cu ⩾1 according to a distribution R. Then, the rumor is placed on a randomly chosen node3. Our generalized Push, Pull and Push-Pull protocol proceed like the classic ones except that every (un)informed node u calls Cu node(s) chosen indepen-dently and uniformly at random and sends (request) the rumor.
Let It be the set of all informed nodes in round t (which means after the execution of round t) and Ut be the complement of It, i.e., the set of uninformed nodes. The size of It and Ut are denoted by It and Ut. We indicate the set of newly informed nodes in round t + 1 by Nt and its size is denoted by Nt. Let St be the number of Push calls in round t + 1, so St = u∈It Cu ⩾It. Let us define N Pull t and N Push t to be the set of newly informed nodes by Pull and Push calls in round t + 1, respectively. The size of N Pull t and N Push t are denoted by N Pull t and N Push t . The size of every set divided by n will be denoted by the 3 This is equivalent to saying that the initial node which knows the rumor has to be chosen without knowing the sequence Cu, u ∈V. We make this assumption through-out the paper, as it is frequently needed for lower bounding the runtime, e.g., the lower bound in Theorem 1.2 may not hold if the rumor initiates from the node with the largest Cu.
450 K. Panagiotou, A. Pourmiri, and T. Sauerwald corresponding small letter, so it, nt and st are used to denote It/n, Nt/n, and St/n, respectively. Further, we define the set L(z) := {u ∈V : Cu ⩾z}.
The size of L(z) is denoted by L(z). We define Δ to be maxu∈V Cu.
3 Push Protocol 3.1 Push Protocol with Bounded Variance (Thm. 1.1) In this subsection we assume that the random numbers Cu’s are generated ac-cording to some distribution R with bounded mean and variance. Recall that Tn := min{t | Pr [It = n] ⩾1 −o(1)}, i.e., the first round in which all nodes are informed with probability 1 −o(1). In Theorem 1.1 we show that if R is a distribution with E [R] = O(1) and Var [R] = O(1), then |Tn −(log1+E[R] n + logeE[R] n)| = o(log n).
To prove this result, we study the protocol in three consecutive phases. In the following we give a brief overview of the proof.
– The Preliminary Phase. This phase starts with one informed node and ends when It ⩾log5 n and St ⩽logO(1) n. Similar to the Birthday Paradox we show that in each round every Push call informs a different uninformed node and thus the number of informed nodes increases by St ⩾It. Hence after O(log log n) rounds there are at least log5 n informed nodes. Further, since E [R] = O(1), after O(log log n) rounds we also have St ⩽logO(1) n.
– The Middle Phase. This phase starts when log5 n ⩽It ⩽St ⩽logO(1) n and ends when It ⩾ n log log n. First we show that the number of Push calls St increases by a factor of approximately 1 + E [R] as long as the number of informed nodes is o(n). Then we prove that the number of newly informed nodes in round t + 1 is roughly the same as St. Therefore an inductive argument shows that it takes log1+E[R] n ± o(log n) rounds to reach n log log n informed nodes.
– The Final Phase. This phase starts when It ⩾ n log log n and ends when all nodes are informed with high probability. In this phase, we first prove that after o(log n) rounds the number of uninformed nodes decreases to n log5 n.
Then we show the probability that an arbitrary uninformed node remains uninformed is e−E[R]±o( 1 log n ), so Ut decreases by this probability. Finally, an inductive argument establishes that it takes logeE[R] n±o(log n) rounds until every node is informed.
3.2 Push Protocol with Arbitrary Variance (Thm. 1.2) We prove that if R is any distribution with E [R] = O(1), then with probability 1 −o(1) the Push protocol needs at least Ω(log n) rounds to inform all nodes.
In the Push protocol, in round t + 1, at most St randomly chosen uninformed Faster Rumor Spreading with Multiple Calls 451 nodes are informed. Hence the total contribution of newly informed nodes to E [St+1] is at most E [R] · St. Applying the law of total expectation shows that E [St+1] ⩽(1 + E [R])t E [R] which implies that Ω(log n) rounds are necessary to inform all nodes.
4 Push-Pull Protocol 4.1 Push-Pull Protocol with Bounded Variance (Thm 1.3) In this part we consider the case where R is a distribution with bounded mean and bounded variance. We prove that with probability at least 1 −ϵ, the Push-Pull protocol needs at least Ω(log n) rounds to inform all nodes. One interesting example for a distribution R with bounded mean and bounded variance is a power law distribution with parameter β > 3.
The crucial ingredient of the proof is to bound the Cu’s of the nodes that become informed by using Pull, i.e., the Cu’s of uninformed nodes that call an informed node. Note that the contribution of an uninformed node u ∈Ut to E [St+1] is Cu times the probability that it gets informed, which is at most Cu · (It/n) ⩽Cu · (St/n). Hence the contribution of u ∈Ut is at most C2 u · (St/n). Now using the assumption that R has bounded variance, we have that u∈V C2 u = O(n) which implies that St increases only exponentially in t.
4.2 Push-Pull Protocol with Power Law Distr. 2 < β < 3 (Thm. 1.4) In this section we analyze the Push-Pull protocol where R is a power law distri-bution with 2 < β < 3 and show that it only takes Θ(log log n) rounds to inform all with probability 1 −o(1).
To prove the upper bound of O(log log n), we study the protocol in three consecutive phases and show each phase takes only O(log log n) rounds. The proof of the lower bound is ommitted in this extended abstract.
Proof of the Upper Bound. The following lemmas about Push will be used throughout this section.
Lemma 4.1. Consider the Push protocol and suppose that St ⩽logc n, where c > 0 is any constant. Then with probability 1−O( log2c n n ) we have It+1 = It+St.
Lemma 4.2. Consider the Push protocol.Then with probability 1 −o( 1 log n) we have that st −2s2 t −2 st log log n n ⩽nt ⩽st.
We will also use the following fact about Power law distributions.
Lemma 4.3. Let {Cu : u ∈V} be a set of n independent random variables and assume that each Cu is generated according to a power law distribution with exponent β > 2. Then for every z = O(n 1 β−1 / log n), it holds with probability 1 −o( 1 n) n · c1 · z1−β 2 ⩽L(z) ⩽3 · n · c2 · z1−β 2 .
452 K. Panagiotou, A. Pourmiri, and T. Sauerwald The Preliminary Phase. This phase starts with just one informed node and ends when It ⩾n 1 β−1 /(2 log n). Let T1 be the number of rounds needed so that the number of informed nodes exceeds n 1 β−1 /(2 log n). We will show that with probability 1 −o(1), T1 = O(log log n). At first we prove the following lemma.
Lemma 4.4. Let c > 0 be any constant. Then with probability 1 −o(1), the number of rounds needed to inform logc n nodes is bounded by O(log log n).
Proof. In order to prove our lemma we only consider Push calls and apply Lemma 4.1 which states that as long as St ⩽logc n, with probability 1 −O( log2c n n ), It+1 = It + St ⩾2It.
Thus as long as St ⩽logc n, in each round the number of informed nodes is at least doubled. So we conclude that with probability 1−o(1), O(log log n) rounds are sufficient to inform logc n nodes.
⊓ ⊔ Lemma 4.5. With probability 1 −o(1), T1 = O(log log n).
Proof. Let T0 be the first round when IT0 ⩾log 2 3−β n. Let us define the constant γ := 3−β 2(β−2) > 0. Let T be the first round such that I(1+γ) T −1 ⩽n 1 β−1 / log n < I(1+γ) T .
Now for any T0 ⩽t ⩽T , we can apply Lemma 4.3 and conclude that with probability 1 −o( 1 n), u∈L(I1+γ t ) Cu ⩾L(I1+γ t ) · I1+γ t ⩾n · c1 · I(1+γ)(2−β) t 2 .
So, It n u∈L(I1+γ t ) Cu ⩾c1 · I1+(1+γ)(2−β) t 2 = c1 · I3−β+γ(2−β) t 2 .
We will bound the probability that none of u ∈L(I1+γ t ) gets informed by Pull calls in round t + 1 as follows, u∈L(I1+γ t ) 1 −It n Cu = 1 −It n u∈L(I1+γ t ) Cu ⩽e−c1·I3−β+γ(2−β) t = e−c1·I 3−β 2 t .
Since for any t ⩾T0, It ⩾log 2 3−β n, we have that with probability at least 1−n−c1, at least one node in L(I1+γ t ) gets informed by Pull in round t+1. Hence we have that St+1 ⩾I1+γ t . Let us now consider the Push calls in round t + 2.
Faster Rumor Spreading with Multiple Calls 453 By applying Lemma 4.1 we know that as long as St+1 = o(n) with probability 1 −o( 1 log n), St+1(1 −o(1)) ⩽Nt+1. Thus, It+2 ⩾It+1 + St+1(1 −o(1)) > I1+γ t 2 .
An inductive argument shows that for any integer k ⩾1 as long as I1+γ T0+2k−2 ⩽ n 1 β−1 / log n, with probability 1 −o( k log n) IT0+2k > 1 2 k−1 i=0 (1+γ)i I(1+γ)k T0 = IT0 2γ (1+γ)k · 21/γ > H log 2 3−β n C′ I(1+γ)k , where C′ = 2γ = O(1). So we conclude that after T0 + 2k rounds, where k = o(log1+γ log n), there are two cases: either IT0+2k ⩾n 1 β−1 /(2 log n) which means T1 ⩽T0 + 2k = O(log log n) and we are done, or IT0+2k < n 1 β−1 /(2 log n) < n 1 β−1 / log n < I1+γ T0+2k.
In the latter case, we change the value γ to γ′ which satisfies I1+γ′ T0+2k = n 1 β−1 / log n and a similar argument shows that IT0+2k+2 ⩾n 1 β−1 /(2 log n).
⊓ ⊔ The Middle Phase. This phase starts with at least n 1 β−1 /(2 log n) informed nodes and ends when It ⩾ n log n. Let T2 be the first round in which n log n nodes are in-formed. We will show that T2 −T1 = O(log log n). In contrast to the Preliminary Phase where we focus only on an informed node with maximal Cu, we now con-sider the number of informed nodes u with a Cu above a certain threshold Zt+1 which is inversely proportional to It.
Lemma 4.6. Suppose that It ⩾n 1 β−1 /(2 log n) for some round t. Let Zt+1 := n log log n It . Then with probability 1 −o( 1 n), |L(Zt+1) ∩It+1| ⩾1 4L(Zt+1).
Proof. We consider two cases. If at least 1 4 of the nodes in L(Zt+1) are already informed (before round t+1), then the statement of the lemma is true. Otherwise |L(Zt+1) ∩Ut+1| > 3 4L(Zt+1). In the latter case, we define L′(Zt+1) := L(Zt+1) ∩Ut+1.
Let Xu be an indicator random variable for every u ∈L′(Zt+1) so that Xu = 1 if u gets informed by Pull in round t + 1 and Xu = 0 otherwise.
454 K. Panagiotou, A. Pourmiri, and T. Sauerwald Then we define a random variable X to be X := u∈L′(Zt+1) Xu. Since for every u ∈L′(Zt+1), Cu ⩾Zt+1 = n log log n It , it follows that Pr [Xu = 1] = 1 − 1 −It n Cu ⩾1 − 1 −It n Zt+1 = 1 −e−Ω(log log n) = 1 −o(1).
Thus Pr [Xu = 1] > 3 4 and E [X] = u∈L′(Zt+1) Pr [Xu = 1] > 3 4|L′(Zt+1)|.
Since |L′(Zt+1)| = |L(Zt+1) ∩Ut+1| > 3 4L(Zt+1), E [X] ⩾ 9 16L(Zt+1). We know that It ⩾n 1 β−1 /(2 log n) and also It is a non-decreasing function in t, so Zt+1 = n log log n It ⩽2 · n β−2 β−1 log n log log n < n 1 β−1 /log n, where the last inequality holds because β < 3. Now we can apply Lemma 4.3 (see appendix) to infer that with probability 1 −o( 1 n), L(Zt+1) ⩾n · c1 · Z1−β t+1 2 ⩾c1 · logβ−1 n 2 .
Therefore, E [X] ⩾9·c1·logβ−1 n 32 . Then applying a Chernoffbound results into Pr X < E [X] 2 ⩽Pr |X −E [X] | ⩾E [X] 2 < 2e−E[X] 10 ⩽2e−Ω(logβ−1 n).
So with probability 1 −o( 1 n), we have that |L(Zt+1) ∩It+1| ⩾X ⩾E [X] 2 > 3|L′(Zt+1)| 8 ⩾1 4L(Zt+1), where the last inequality holds because |L′(Zt+1)| > 3 4L(Zt+1).
⊓ ⊔ Lemma 4.7. With probability 1 −o(1), T2 −T1 = O(log log n).
Proof. Since It ⩾n 1 β−1 /(2 log n), Zt+1 = n log log n It < n 1 β−1 / log n, using Lemma 4.6 results into a lower bound for |L(Zt+1) ∩It+1|. So with probability 1 −o( 1 n), St+1 = u∈It+1 Cu ⩾|L(Zt+1 ∩It+1)| · Zt+1 ⩾1 4L(Zt+1) · Zt+1.
By applying Lemma 4.3, we conclude that with probability 1 −o( 1 n), L(Zt+1) ⩾ n·c1·Z1−β t+1 2 . Therefore, with probability 1 −o( 1 n), St+1 ⩾ n·c1·Z2−β t+1 8 . As long as St+1 = o(n), we can apply Lemma 4.2 for the Push protocol to round t + 2 implying that with probability 1 −o( 1 log n), It+2 = It+1 + Nt ⩾It+1 + St+1(1 −o(1)).
Faster Rumor Spreading with Multiple Calls 455 Thus, It+2 > St+1 2 ⩾c1 16n · Z2−β t+1 = c1 16 · n3−β · log log2−β n · Iβ−2 t .
By an inductive argument, we obtain that for any integer k ⩾1 with St+k = o(n), it holds with probability 1 −o( k log n), It+2k > c 16n3−β · log log2−β n k−1 i=0 (β−2)i I(β−2)k t = c 16n3−β · log log2−β n 1−(β−2)k 3−β I(β−2)k t .
Therefore there exists k = O(log 1 β−2 log n) such that It+2k ⩾ c 16n3−β · log log2−β n 1−O(1/ log n) 3−β I1/ log n t = Ω H n1−O(1/ log n) c 16 · log log2−β n 1−O(1/ log n) 3−β I = Ω n log logδ n , where δ = β−2 3−β(1 −O(1/ log n)) > 0. Hence T2 ⩽T1 + 2k = T1 + O(log log n) with probability 1 −o(1).
⊓ ⊔ The Final Phase. This phase starts with at least n log n informed nodes. Since the runtime of our Push-Pull protocol is stochastically smaller than the runtime of the standard Push-Pull protocol (i.e. Cu = 1 for every u ∈V ), we simply use the result by Karp et. al in [18, Theorem 2.1] for the standard Push-Pull protocol which states that once It ⩾ n log n, additional O(log log n) rounds are with probability 1 −o(1) sufficient to inform all n nodes.
References 1. Berenbrink, P., Els¨ asser, R., Friedetzky, T.: Efficient randomised broadcasting in random regular networks with applications in peer-to-peer systems. In: Proc. 27th Symp. Principles of Distributed Computing (PODC), pp. 155–164 (2008) 2. Berger, N., Borgs, C., Chayes, J., Saberi, A.: On the Spread of Viruses on the Internet. In: Proc. 16th Symp. Discrete Algorithms (SODA), pp. 301–310 (2005) 3. Boyd, S., Ghosh, A., Prabhakar, B., Shah, D.: Randomized gossip algorithms. IEEE Transactions on Information Theory 52(6), 2508–2530 (2006) 4. Censor-Hillel, K., Shachnai, H.: Partial information spreading with application to distributed maximum coverage. In: Proceedings of the 29th Annual ACM Sympo-sium on Principles of Distributed Computing, pp. 161–170 (2010) 5. Chierichetti, F., Lattanzi, S., Panconesi, A.: Almost tight bounds for rumour spreading with conductance. In: Proc. 42nd Symp. Theory of Computing (STOC), pp. 399–408 (2010) 6. Chung, F.R.K., Lu, L.: The average distance in a random graph with given expected degrees. Internet Mathematics 1(1), 91–113 (2003) 456 K. Panagiotou, A. Pourmiri, and T. Sauerwald 7. Clementi, A.E.F., Pasquale, F., Silvestri, R.: Opportunistic manets: Mobility can make up for low transmission power. IEEE/ACM Trans. Netw. 21(2), 610–620 (2013) 8. Demers, A., Greene, D., Hauser, C., Irish, W., Larson, J., Shenker, S., Sturgis, H., Swinehart, D., Terry, D.: Epidemic algorithms for replicated database maintenance.
In: Proc. 6th Symp. Principles of Distributed Computing (PODC), pp. 1–12 (1987) 9. Doerr, B., Fouz, M.: Asymptotically optimal randomized rumor spreading. In: Aceto, L., Henzinger, M., Sgall, J. (eds.) ICALP 2011, Part II. LNCS, vol. 6756, pp. 502–513. Springer, Heidelberg (2011) 10. Doerr, B., Fouz, M., Friedrich, T.: Social networks spread rumors in sublogarithmic time. In: Proc. 43rd Symp. Theory of Computing (STOC), pp. 21–30 (2011) 11. Feige, U., Peleg, D., Raghavan, P., Upfal, E.: Randomized broadcast in networks.
Random Structures and Algorithms 1(4), 447–460 (1990) 12. Fountoulakis, N., Huber, A., Panagiotou, K.: Reliable broadcasting in random networks and the effect of density. In: Proc. 29th IEEE Conf. Computer Commu-nications (INFOCOM), pp. 2552–2560 (2010) 13. Fountoulakis, N., Panagiotou, K., Sauerwald, T.: Ultra-fast rumor spreading in social networks. In: Proc. 23rd Symp. Discrete Algorithms (SODA), pp. 1642–1660 (2012) 14. Frieze, A., Grimmett, G.: The shortest-path problem for graphs with random-arc-lengths. Discrete Applied Mathematics 10, 57–77 (1985) 15. Giakkoupis, G.: Tight bounds for rumor spreading in graphs of a given conductance.
In: Proc. 28th Symp. Theoretical Aspects of Computer Science (STACS), pp. 57–68 (2011) 16. Giakkoupis, G.: Tight bounds for rumor spreading with vertex expansion. In: Proc. 25th Symp. Discrete Algorithms, SODA (to appear, 2014) 17. Harchol-Balter, M., Leighton, F.T., Lewin, D.: Resource discovery in distributed networks. In: Proc. 18th Symp. Principles of Distributed Computing (PODC), pp.
229–237 (1999) 18. Karp, R., Schindelhauer, C., Shenker, S., V¨ ocking, B.: Randomized Rumor Spread-ing. In: Proc. 41st Symp. Foundations of Computer Science (FOCS), pp. 565–574 (2000) 19. Kempe, D., Dobra, A., Gehrke, J.: Gossip-based computation of aggregate informa-tion. In: Proc. 44th Symp. Foundations of Computer Science (FOCS), pp. 482–491 (2003) 20. Mosk-Aoyama, D., Shah, D.: Fast distributed algorithms for computing separable functions. IEEE Transactions on Information Theory 54(7), 2997–3007 (2008) 21. Peres, Y., Sinclair, A., Sousi, P., Stauffer, A.: Mobile geometric graphs: Detection, coverage and percolation. In: Proc. 22nd Symp. Discrete Algorithms (SODA), pp.
412–428 (2011) 22. Pittel, B.: On spreading a rumor. SIAM Journal on Applied Mathematics 47(1), 213–223 (1987) 23. van Renesse, R., Minsky, Y., Hayden, M.: A gossip-style failure detection service.
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6941 | https://arxiv.org/pdf/2001.02084 | Counting walks by their last erased self-avoiding polygons using sieves
Pierre-Louis Giscard a
a
Universit´ e du Littoral Cˆ ote d’Opale, EA2597-LMPA Laboratoire de Math´ ematiques Pures et Appliqu´ ees Joseph Liouville, Calais, France.
Abstract
Let G be an infinite, vertex-transitive lattice with degree λ and fix a vertex on it. Consider all cycles of length exactly l from this vertex to itself on
G. Erasing loops chronologically from these cycles, what is the fraction
Fp/λ (p) of cycles of length l whose last erased loop is some chosen self-avoiding polygon p of length(p), when l → ∞ ? We use combinatorial sieves to prove an exact formula for Fp/λ `(p) that we evaluate explicitly. We further prove that for all self-avoiding polygons p, Fp ∈ Q[χ] with χ an irrational number depending on the lattice, e.g. χ = 1 /π on the infinite square lattice. In stark contrast we current methods, we proceed via purely deterministic arguments relying on Viennot’s theory of heaps of pieces seen as a semi-commutative extension of number theory. Our approach also sheds light on the origin of the difference between exponents stemming from loop-erased walk and self-avoiding polygon models, and suggests a natural route to bridge the gap between both.
Keywords: Self-avoiding polygons, Heaps of pieces, Sieves
2000 MSC: 05C38, 05A15, 11N36
?
This work is supported by the French National Research Agency (ANR) throught the ANR-19-CE40-0006 project Al gebraic Co mbinatorics of Hikes On Lattices ( Alcohol ).
Email address: mailto:giscard@univ-littoral.fr (Pierre-Louis Giscard)
URL: (Pierre-Louis Giscard)
Preprint submitted to Elsevier March 3, 2022
arXiv:2001.02084v2 [math.CO] 3 Nov 2020
Context
1.1. Self-avoiding objects and loop-erased walks
The “widely open problem of counting Self-Avoiding Walks (SAWs) and Self-Avoiding Polygons (SAPs) on lattices” (quoting Flajolet & Sedgewick) was first conceived from the study of polymer chemistry, but these objects arise in a wide range of physical and mathematical problems; e.g. as phase boundaries and in percolation clusters [2, 3, 4] or, as B. Mandelbrot ob-served , from the outer frontier of Brownian motion . Because in such models SAWs and SAPs are invariably realised through random processes, the problem of studying and counting them has so far only been attacked with tools from statistical physics and probability theory. Works along these direc-tions have yielded deep insights into renormalisation and conformal mappings such as the relation between self-avoiding curves and the Schramm-Loewner Evolution (SLE) [7, 8], and the result of H. Duminil-Copin and S. Smirnov who proved the value of the connective constant of the honeycomb lattice . One of the most active subfields of this research concerns loop-erased random walks (LERWs) introduced by G. Lawler . Lawler’s original motivation was to produce yet another model of random generation of self-avoiding objects. Indeed, removing loops from an ordinary random walk in chronological order–the so called loop-erasing procedure–yields the self-avoiding ‘skeleton’ of the walk. LERWs are now the object of much research on conformal loop ensembles and loop soups [11, 12, 13]. At the heart of this activity are conformally invariant loop measures, one of which associates to a SAP p the proportion of all cycles–also known as closed walks–whose last erased loop is the SAP p. This measure, studied within the framework of probability theory, remains very difficult to evaluate explicitly. Indeed, this requires a complicated mapping to Abelian sand-pile models followed by arduous calculations which were completed only for the shortest SAPs (length less than 6). But the straightforward combinatorial meaning of the measure as a proportion means that a purely deterministic counting method should be able to evaluate it as well. This is indeed so, using number theoretic sieves and heaps of cycles.
1.2. Heaps of cycles and number theory
The theory of heaps of pieces [14, 15, 16], which describes the combi-natorics of piles of arbitrary pieces, has now found a remarkable number of 2applications throughout mathematics, from combinatorics to group theory. Studying specifically heaps of cycles on graphs, called hikes , P.-L. Giscard and P. Rochet showed that these constitute a very natural semi-commutative extension of number theory . This extension comprises all the fundamen-tal objects of number theory (zeta functions, sieve methods) as well as most of the relations between these. In this context, the hikes themselves ex-tend the natural integers, the walks (a.k.a. paths) extend the integers of the form pk with p prime and SAWs and SAPs extend the primes. This last observation implies that in the semi-commutative framework of the hikes,
the extension of the prime number theorem will give the asymptotic number of SAPs on regular lattices . What is even more remarkable here is that the set of prime factors of a walk as dictated by the extension of number theory precisely coincides with the loops erased from a walk in Lawler’s procedure. The fundamental premise of this work is that the tools offered by number theory are also effective in the study of self-avoiding objects.
1.3. Layout
In this self-contained contribution (definitions in §3), we use purely deter-ministic sieve techniques for evaluating the asymptotic fraction of all closed walks whose last erased loop is any given SAP p on any finite graph ( §4) or in-finite vertex-transitive lattice ( §5). We arrive at an exact, closed-form, easy to evaluate formula for this fraction, immediately presented in the next section. Within the framework of probability-theory, this fraction is a conformally-invariant loop-measure known and studied by Lawler . The sieves also make it clear that the loop-measure comes with seemingly hitherto unnoticed error-terms (Appendix E), which explain combinatorially the difference be-tween LERW and SAP exponents. A route for overcoming this final hurdle using 1990s work is suggested ( §6).
Statement of the main result and illustrations
2.1. Main theorem
The main result, which we prove in the remainder of this work is as follows:
Infinite Sieve Theorem. Let G be an infinite vertex transitive graph of bounded degree λ and let AG be its adjacency matrix. Let p be a self-avoiding polygon on G. Let {GTor
N
} be a the small tori sequence of vertex-transitive
3graphs on N vertices converging to G as N → ∞ (see Appendix A). Let R(z)
be the ordinary generating function of all closed walks on G with fixed initial vertex. Then, on GT or N , the fraction of all hikes, i.e. heaps of cycles, which are closed walks whose last erased loop is p is given asymptotically for N 1
by
αN
NFp
λ`(p) .
In this expression α := lim z→1/λ − exp (∫ 1
z
(R(z) − 1)dz ), α ∈]0 , 1[ , is well defined and Fpλ−`(p) designates the fraction of all closed walks defined up to translation on G whose last erased loop is p. This fraction is explicitly given by Fp
λ`(p) = 1
λ`(p)+1 deg T. adj
(
I + CG
∣∣p.Bp
)
. 1, (1)
where adj( .) designates the adjugate operator, Bp is the adjacency matrix of the graph Gp induced by p and its immediate neighbours on G, 1 des-ignates the vector full of 1 and deg = diag( B2
p
) is the vector of vertex-degrees on Gp. Finally, CG|p is the restriction to Gp of the matrix CG := lim z→1/λ − (I − Pλ)R(z), with Pλ the projector onto the eigenspace associated with the dominant eigenvalue and R(z) := ( I − zAG)−1 is the resolvent of graph G.
Corollary 2.1. Let G be an infinite vertex-transitive lattice of degree λ. Let
p be a self-avoiding polygon on G and let Fp λ−`(p) be the fraction of all closed walks whose last erased loop is the SAP p.- If G is a d-dimensional hypercubic lattice, then Fp ∈ Q[1 /π d−1].- If G is the triangular, hexagonal or Kagom´ e lattice, then Fp ∈ Q[√3/π ].
The error terms generated by the sieve on infinite graphs are given in Appendix E together with a discussion of their relevance for analytical esti-mates of the asymptotic growth of the number of SAPs of length as → ∞ .See also the discussion of §6. From a practical point of view, we observe that the matrix Bp is of size
e(p) × e(p), where e(p) is the number of vertices of G at distance at most 1 from p. Since clearly (p) ≤ e(p) ≤ λ (p), computing the fraction Fp/λ `(p)
costs O(`(p)3) operations. 4The lattice constant α relates the densities of walks and of hikes on the infinite lattice G. On the square lattice, α = 14 e4Cπ ' 0.8025 ... with C Cata-lan’s constant. The matrix CG is easy to obtain on regular graphs because its entries obey the same recursion relations as the graph resolvent. More precisely, let (CG
)
m,n
designate the entry of the matrix corresponding to jumping from vertex m to vertex n. Then λ(CG
)
m,n
= ∑
i∈N (n)
(CG
)
m,i
∑
j∈N (m)
(CG
)
j,n
,where N (n) and N (m) designate the set of vertices that are neighbours to
n and m on G, respectively. On the square lattice this implies that CG has the following explicit expression:
(CG
)
ij
= − 1
π
∫ ∞
0
1
τ
(
1 −
(τ − i
τ + i
)xij −yij (τ − 1
τ + 1
)xij +yij
)
dτ
where i2 = −1, xij and yij are the distance along x and y between vertices i
and j of Gp, respectively. In particular if xij = yij = m then
(CG
)
ij
= − 4
π
m−1
∑
k=1
12k + 1 = − 2
π
(
Hm− 12
log(4)
)
,
with Hm the mth harmonic number. Explicit expressions for CG have already been determined on many more lattices owing to its relation with lattice Green’s functions and the resistor problem [18, 19].
2.2. Illustrations
We may now illustrate the Infinite Sieve Theorem with concrete results on the infinite square lattice.
I The fraction of closed walks whose last erased loop is a given edge e is
Fe
42 = 18 = 0 .125 .
Since a point is connected to 4 edges, this means that an edge is the last erased loop of 1 /2 of all closed walks on the square lattice. In the language of the semi-commutative extension of number theory that holds for walks and hikes on graph, this indicates that 1 /2 of all closed walks on the square 5●●●●● ●● ● ● ● ● ● ● ● ● ●●●●●●●●● ●●●●●●●●●●●●●●●●●●●● ●● ● ●
128 (π - 2)
π 3
5 10 50 100 500 1000 02468Walk length
Exact fraction ⨯ 4 4
Figure 1: Exact fraction of closed walks of length ` on the infinite square lattice whose last erased loop is a 1 × 1 square as a function of the length of these walks. The exact fraction was obtained from the extension of Viennot’s lemma to infinite graphs Proposition 5.1 and Eq. (3). The exact fraction converges to its asymptotic value proportionally with the inverse of the walk length, as dictated by an analysis of the error terms associated with the Infinite Sieve Theorem, see Appendix E.
lattice whose unique right prime divisor is an edge cycle.
I The fraction of closed walks whose last erased loop is a 1 × 1 square is
F1×1
44 = 128( π − 2) 44π3 ' 0.0184 . (2) This result was first obtained via a complicated mapping between Fp and Abelian sand-pile models [20, 21]. According to the authors this process “becomes very tedious” for longer self-avoiding polygons, so that very few explicit values for Fp have been published in the literature so far, and all pertain to self-avoiding polygons of length ` ≤ 6. See Fig. (1) for an illustration of the convergence of the fraction of closed walks on the infinite square lattice whose last erased loop is a 1 × 1 square to the above number. Here the extension of Viennot’s lemma to infinite graphs yields the ordinary generating function of closed walks whose last erased loop 61 23456123456 2 1345621345632145623145632 14564312563412564 31256451236 5 41236Figure 2: Illustration of the 12 walks of length 6 on the square lattice whose last erased loop is right-oriented 1 × 1 square cycle (highlighted in red). The numbers next to the highlighted edges indicate the order in which these edges are traversed.
is an oriented 1 × 1 square, which we denote R1×1(z), as
R1×1(z) = 1256 π4z4
( (16 z2 − 1) K(16 z2) + E(16 z2)
)2
× (3)
( (1 − 16 z2) K(16 z2)2 + 2 K(16 z2) (8πz 2 − E(16 z2)) − 4π2z2 + E(16 z2)2
)
,
= z4 + 12 z6 + 144 z8 + 1804 z10 + 23464 z12 + · · · ,
where K(x) := ∫ π/ 20
(1−x sin 2(θ))−1/2dθ and E(x) := ∫ π/ 20
(1−x sin 2(θ))1/2dθ
are the complete elliptic integrals of the first and second kind, respectively. In Fig. (2) we illustrate the 12 walks of length 6 on the square lattice whose last erased loop is a 1 × 1 oriented square, as correctly counted by the coef-ficient of z6 in R1×1(z), denoted [ z6]R1×1(z) = 12. Eq. (2) establishes that 7asymptotically [z2n]R1×1(z) ∼ 128( π − 2) 44π3
(2nn
)2
, as n → ∞ .
I The fraction of closed walks whose last erased loop is a 1 × 2 rectangle is
F1×2
46 = 32( π − 8)( π − 4)(3 π − 8)(3 π − 4) 46π4 ' 0.002585 .
I The fraction of closed walks whose last erased loop is a 1 × 3 rectangle is
F1×3
48 = 1024(16 − 3π)(64 + 3( π − 12) π)(64 + 27( π − 4) π)(128 + π(27 π − 124)) 48 × 81 π7 ,
' 0.00035499 .
I The fraction of closed walks whose last erased loop is a 2 × 2 square is
F2×2
48 = 32768( π − 8) 2(π − 4)(3 π − 8) 3(9 π − 32) 48 × 81 π7 ' 0.00044623 .
I As an example of longer SAP, consider:
Then the fraction of all closed walks whose last erased loop is this SAP is
Fp
418 = 8388608 418 × 8303765625 π12
(
1721510367131231944781594624
− 6733029120634416611029155840 π + 12001725045126647537146527744 π2
− 12895675745638007921939841024 π3 + 9303982639359984674575220736 π4
− 4748903115679537036020154368 π5 + 1758418560456019196044640256 π6
− 475910723284488375970037760 π7 + 93430267561362281294131200 π8
− 12973459941155225172708000 π9 + 1209211981439562793530000 π10
− 67906363349663583525000 π11 + 1736896666805181140625 π12
)
,
' 7.7644 × 10 −9.
8This result is well beyond what is realistically achievable from conformally invariant measures mapped to Abelian sandpile models. Fractions of walks whose last erased loop is a much longer SAP are easily obtained numerically, costing no more than O(`(p)3) to compute as outlined earlier. For example, the fraction of closed walks whose last erased loop in a 70 × 70 square is F70 ×70
4280 ' 1.5236 × 10 −108 . (4) Analytically speaking these fractions become very involved very quickly as a function of SAP length and there is no reason to believe that there exists a simpler expression for them than that given by Eq. (1) of the Infinite Sieve Theorem. For example, the analytical expression for the fraction of closed walks whose last erased loop is a 6 ×6 square already involves sums and prod-ucts of up to 16-digits prime integers. In fully expanded form this fraction involves a 67-digits prime integer (!). Overall, we have calculated the frac-tions for over 100,700,000 SAPs analytically , and for more than 3,480,000,000 SAPs numerically on the square lattice. These results as well as the accom-panying algorithm will soon be presented in a separate contribution. The hexagonal and triangular lattices will be also be treated.
Definitions
3.1. Hikes and related objects
In the general setting, we consider (weighted di)graphs G = ( V; E) with
N = |V| nodes and M = |E| edges, both of which may be infinite but the de-gree of G must be bounded. The ordinary adjacency matrix of G is denoted
AG or simply A. If G is weighted then the entry Aij is the weight of the edge
eij from i to j if this edge exists, and 0 otherwise. The labelled adjacency matrix of G is denoted W and its entries are formal variables belonging to the Cartier-Foata monoid, Wij = eij .A induced subgraph H of G, also called simply a subgraph of G and de-noted H ≺ G, is a set of vertices VH ⊆ V together with the set of all edges linking these vertices in G, EH = {eij ∈ E : i, j ∈ V H }.A walk w of length (w) from vi to vj on G is a sequence w = eii 1 ei1i2 · · · ei−1j
of ` contiguous edges. The walk w is open if i 6 = j and closed otherwise. A simple cycle , also known in the literature under the names loop , cy-cle , elementary circuit and Self-Avoiding Polygon or SAP , is a closed walk 9a)
b)
c)
d)
p1
p2
p3Figure 3: Illustration of the main objects of the hike family on a graph G:a) a hike comprising two walks; b) a walk w=p2p3p1≡p3p2p1;c) a self-avoiding hike comprising three vertex disjoint simple cycles; d) a simple cycle. The walk whas a unique right prime divisor, here the triangle p1. In the Cartier-Foata monoids, the starting point of this walk can be any of the three vertices of p1and p1will always be the last erased loop of the walk.
w = eii 1 ei1i2 · · · ei−1i which does not cross the same vertex twice, that is, the indices i, i 1, . . . , i−1 are all different. Two simple cycles differing only by orientation are considered distinct (Rule 1), but two simple cycles differing only by their starting point are taken to be identical (Rule 2). The necessity of these choices and of the definition of hikes below can be found in Cartier and Foata’s foundational work . From now on, we employ the letter P to designate sets of simple cycles, in particular PG will be the set of all simple cycles on a graph G.The central objects of the present work are hikes , a hike h being an un-ordered collection of vertex-disjoint closed walks. Hikes can be also be seen as equivalence classes on words W = pi1 pi2 · · · pin over the alphabet of simple cycles pi of a graph. Two words W and W ′ are equivalent if and only if W ′
can be obtained from W through allowed permutations of consecutive simple cycles. In this context, two simple cycles are allowed to commute if and only if they are vertex disjoint V(pi) ∩ V (pj ) = ∅ ⇐⇒ pipj = pj pi.For example, if p1 and p2 commute but neither commute with p3, then
p1p2 and p2p1 represent the same hike, but p1p3p2 and p2p3p1 are distinct hikes. 10 The letters pi1 , · · · , p in found in a hike h are called its prime divisors. This terminology is due to the observation that simple cycles obey the defining property of prime elements in the semi-commutative Cartier-Foata monoid H
of hikes equipped with the concatenation. In the context of Viennot’s theory of heaps of pieces, hikes are heaps of simple cycles modulo Rules 1 and 2. In this work, we use the terminologies “primes” and “SAPs” interchangeably. Two special types of hikes will be important for our purpose here: A self-avoiding hike is a hike all prime factors of which commute with one another. In other terms, it is collection of vertex-disjoint simple cycles. If P
designates a set of simple cycles, we designate Ps.a. the set of all self-avoiding hikes that can be built from cycles of P.In Viennot’s terminology, a hike is a walk if and only if it is a pyramid of cycles, i.e. a heap of cycle with a unique top cycle. Equivalently, this means that a hike is a walk if and only if it has a unique right prime divisor p . In this situation we say that p right divides the walk w, denoted p|rw or that
w is a (left) multiple of p. Remarkably, these notions are also identical with those produced by G. Lawler’s loop erasing procedure : in this framework the unique right prime divisor of a closed walk is the last erased loop of this walk. As an example, let p1, p2 and p3 be three simples cycles such that
p2 and p3 commute with one another, but p1 commutes with neither p2 nor
p3. Then the hike h = p2p3p1 ≡ p3p2p1 is a walk whose unique right prime divisor is p1, i.e. p1 is the last erased loop of h. This is illustrated in Fig. (3).
3.2. Functions on hikes
A function on hikes is a complex-valued function f : H 7 → C. The most important example here will be that of rank function .A rank function, is a function ρ : H 7 → R that is totally additive over the hikes, ρ(hh ′) = ρ(h) + ρ(h′) and which respects the divisibility order, i.e.
h ≤ h′ ⇒ ρ(h) ≤ ρ(h′). The reverse implication does not hold in general. Examples of rank function include the length and the number of self-crossings of a hike. In general, we will denote ρ(h) the rank of h as per the rank func-tion ρ(.) and will denote % any given rank, i.e. value taken by the function ρ.In the case of the length function, `(h) is the length of hike and l denotes a length. The set of hikes with a given rank % is H% := {h ∈ H , ρ (h) = %}. The cardinality of this set is denoted |H %| := card( H%), if the graph is weighted this is understood to mean the total weight carried by hikes of rank %.11 A function f : H 7 → C on hikes is associated with a formal series ∑
h∈H
f (h)h. This series is rarely accessible as such, rather linear alge-bra provides tools to access a related ordinary generating function F (z) := ∑
h∈H
f (h)z`(h). Important examples of functions on hikes and their related series include: The zeta function on hikes is the identity function over the hikes ∀h ∈H, ζ (h) = 1. The associated ordinary generating function will be denoted
ζ(z), it is given by
ζ(z) = ∑
h∈H
z`(h) = 1det (I − zAG
),
where AG is the adjacency matrix of the graph G on which the hikes live. The M¨ obius function on hikes is the convolution inverse of the zeta func-tion. We have
∀h ∈ H , μ (h) =
{
(−1) Ω( h), if h is self-avoiding 0, otherwise Here Ω( h) is the prime factor counting function , its count the number of prime right-divisors of h. The associated ordinary generating function will be denoted μ(z), it is given by μ(z) = det (I − zAG
).The walk von Mangoldt function on hikes Λ is defined as the number Λ( h) of contiguous is the number of possible contiguous rearrangements of the edges in h, obtained without permuting two edges with the same starting point. This is equivalent to
∀h ∈ H , Λ( h) =
{
`(p), if h is a walk with unique right prime divisor p,
0, otherwise .
On a graph with N vertices, the ordinary generating function associated with the walk von Mangoldt function is given by Λ( z) = Tr (I − zAG
) − N .
The asymptotics of hikes and walks on finite graphs
The aim of this section is to develop sieving tools to asymptotically count hikes satisfying certain properties on finite graphs. The main results here will 12 be the Finite Sieve Theorem and its length corollary. Before we state and prove these results, there is an important precedent to be found in Viennot’s work [15, 22], which provides the ordinary generating functions of hikes which are closed walk multiples of any chosen prime p, i.e. whose last erased loop is the SAP p. Since the asymptotic expansion of this result is among the results obtained below, we start by recalling Viennot’s result.
Viennot’s lemma [15, 22]. Let G be a finite graph. Let p be a prime on this graph and let Wp := ∑
w:p|rw
w be the formal series of closed walks whose unique right prime divisor is p. Then
Wp = det (I − WG\p
)
det (I − WG
) p.
where WG\p and WG designate the labelled adjacency matrices of G\p and G,respectively.
Viennot gave a beautiful bijective proof of this result in . There are at least four more proofs, one of which is the spirit of sieves and is provided below. The extension of Viennot’s lemma to infinite graphs is obtained in Section 5.1.
Proof. Let H[p,. ]6 =0 be the set of hikes none of whose connected components commute with p. Clearly, for all h ∈ H [p,. ]6 =0 , hp is a walk multiple of p and we need only determine H[p,. ]6 =0 to obtain the lemma. But this set is the set of all hikes minus the set of hikes such that at least one connected component commutes with p. Such a component must be divisible on the right by prime(s) p′ commuting with p, hence V(p′) ⊆ G\p. Let PG\p be the set of all such primes. The exclusion-inclusion principle then yields ∑
h∈H [p,. ]6=0
h =
∑
d∈P s.a G\p
μ(d)M (d), with the convention that 1 ∈ P s.a G\p and M (d) is the formal series of the left-multiples of d. For any hike this is M (h) = det( I − W)−1h
since all hikes multiplied by h on the right are left multiples of h. Then
∑
h∈H [p,. ]6=0
h =
∑
d∈P s.a G\p
μ(d)d
det( I − W) = det( I − WG\p)det( I − W) .
The series Wp is obtained upon multiplying the above by p on the right. 13 The sieve based proof of Viennot’s lemma suggests a wider family of results to count exactly or asymptotically families of hikes satisfying chosen properties on finite graphs. Concentrating on asymptotic expansions, we have:
Finite Sieve Theorem. Let G be a finite (weighted di)graph with adjacency matrix A. Let H be an induced subgraph of G and let PH be the set of primes on H. Let ρ : H 7 → R be a rank function on hikes such that |H %| = λ%f (%)
with λ a real constant and f (.) a bounded function such that lim %→∞ f (%)
exists. Then the number (weight) 1 S(H%, PH ) of hikes of rank ρ(h) = % which are not multiples of primes on H is asymptotically given by
S(H%, PH )
|H %| ∼ ∑
d∈P s.a H
μ(d) λ−ρ(d), as % → ∞ .
The Finite Sieve Theorem’s most important application here will be with the length rank-function and sieving subgraph H = G\p for p a prime. This provides the asymptotic expansion of Viennot’s lemma:
Length corollary. Let G be a finite (weighted di)graph with adjacency ma-trix A and dominant eigenvalue λ, which we assume to be unique. 2 Let p be a simple cycle or a simple path on G of length `(p) and let S(Hl, PG\p) be defined as in the Finite Sieve Theorem. Then S(Hl, PG\p) is equal to the number (weight) of closed walks of length
`(w) = l on G whose unique right prime divisor is p and is asymptotically given by
S(Hl, PG\p)
|H l| ∼ 1
λ`(p) det
(
I − 1
λAG\p
)
as ` → ∞ .
Let Err( Hl, PG\p) be the difference between the two terms above. Let f (l) := [zl]ζ(z/λ ) be the coefficient of zl in the expansion of ζ(z/λ ). Then f is
1The notation S(Hρ,PH) for this quantity is employed in keeping with conventions from sieve theory.
2The theorem extends if λis not unique upon replacing λ−1by λ−gwith gits multi-plicity.
14 bounded, lim l→∞ f (l) exists, and
Err( Hl, PG\p) = 1
λ`(p)
∞
∑
k≥0
(∇kf
f (l) λk k! − δk, 0
)
det (k)
(
I − 1
λAG\p
)
.
with δk, 0 the Kronecker delta and ∇ is the backward difference operator 3.Here, det (k)(I− 1
λ
AG\p) stands for the kth derivative of det( I−zAG\p) evaluated in z = 1 /λ .Proof of the Finite Sieve Theorem. The proof relies on an inclusion-exclusion principle in the poset of the hikes ordered by right-divisibility, see for an overview of this poset. Let P ( H be a set of primes and Ps.a. the set of all self-avoiding hikes constructible from P. Let S(H%, P) be the num-ber (weight) of hikes in H% which are not right-divisible by any prime of P.The inclusion-exclusion principle–here the extension to hikes of the sieve of Erathostenes-Legendre–yields
S(H%, P) = ∑
d∈P s.a.
μ(d)|M d|,
with |M d| the number of multiples of d in Hρ and μ(d) is the M¨ obius function on hikes. In order to progress, we seek a multiplicative function prob( .) such that
|M d| = prob( d)|H %| + r(d). In this expression, prob( d) approximates the probability that a hike taken uniformly at random in H% is right-divisible by d. If edge-weights are present, the hikes are not all uniformly probable but follow a distribution dependent on these weights. No knowledge of this distribution is required here. Similarly, m(d) = prob( d)|H %| is the expected number of multiples of d in H%. Finally, r(d) is the associated error term, arising from the fact that |M d| is not truly multiplicative. Supposing that we can identify the m(.) function, we would obtain
S(H%, P) = ∑
d∈P s.a.
μ(d)m(d) + ∑
d∈P s.a.
μ(d)r(d).
3This operator acts on a function Fof a variable xas ∇F := F(x)−F(x−1), and
∇kF designates the kth iteration of operator ∇, i.e. ∇kF = ∑kj=0 (−1) j(kj
)F(x−
j)
15 Contrary to number theory, the first term does not admit any simpler form without further assumptions on P. This is because of the possible lack of commutativity between some elements of P. We note however that since
μ(d) is non-zero if and only if d is self-avoiding, and since we have required that m(.) be multiplicative, 4 then it follows that the first term is determined solely from the values of m(.) over the primes of P.We therefore turn to determining m(p) for p prime. The set of left-multiples of p in H is Mp := {hp, h ∈ H} , hence in bijection with the set
H. Thus, the number of left-multiples of p in H%, is exactly |H %−ρ(p)|. Then prob( p) + r(p)
|H %| = |H %−ρ(p)||H %| .
Seeking the best possible probability function prob( ρ), let us suppose that we can choose this function such that the error term of the above equation vanishes in the limit ρ → ∞ . If this is true, then we obtain prob( ρ) = lim
ρ→∞
|H %−ρ(p)||H %| .
In order to progress, we have to make an important assumption regarding the cardinality of the set H%:
Assumption 4.1. There exists a scaling constant λ and bounded function
f : R 7 → R such that lim %→∞ f (%) exists and for % ∈ N∗
|H %| = λ%f (%).
In the case of the length rank function, this assumption is actually a propo-sition:
Proposition 4.2. Let G be a finite (weighted di)graph with dominant eigen-value λ of multiplicity g. Let H:= {h ∈ H :(h) = } be set of all hikes on G of length. Then, there exists a bounded function f : N 7 → R such that
lim →∞ f () exists and for ` ∈ N∗ we have exactly
|H | = λg f (`).
4But not necessarily totally multiplicative.
16 Proof. This follows directly from the ordinary zeta function on hikes ζ(z) = det( I − zA)−1, from which we have
|H | = [ z]
( 1det( I − zA)
)
= ∑
i1,··· , i N``
λi1
1
λi2
2
· · · λiN
N
= λ` ∑
i1,··· , i N``
λi1−`λi2
2
· · · λiN
N
where the sums run over all non-negative values of ij ≥ 0 such that ∑
j
ij = `
and λ ≡ λ1 is the eigenvalue of the graph with the largest absolute value. We assume for the moment that λ is unique and let f (`) := ∑
i1,··· , i N``
λi1−`λi2
2
· · · λiN
N
=[z`]ζ(z/λ ). This function is clearly bounded and lim
`→∞
f (`) = lim
z→1/λ −
(1 − zλ )ζ(z),
exists and is finite. If |λ| is not unique and has multiplicity g, then the scaling constant for the number of hikes becomes λg and then f () = [ z]ζ(z/λ g). Proceeding with Assumption 4.1–or in the case of the length rank function Proposition 4.2–the existence of the limit for f gives prob( p) = lim
%→∞
λ%−ρ(p)f (% − ρ(p))
λ%f (%) = λ−ρ(p).
The prob(.) function is multiplicative over the primes as desired and yields
m(p) = |H %|λ−ρ(p). The associated error term is
r(%) = |H %−ρ(p)| − |H %|λ−ρ(p) = λ%−ρ(p)
(
f (% − ρ(p)) − f (%)
)
.
To establish the validity of these results, we need only verify that they are consistent with our initial supposition concerning the error term, namely that r(p)/|H %| vanishes in the limit % → ∞ . The existence of the limit of f
implies lim %→∞ |f (% − ρ(p)) − f (%)| = 0 and therefore that lim
%→∞
r(p)
|H %| = lim
%→∞
λ−ρ(p)
(
f (% − ρ(p)) − f (%)
)
= 0 ,
as required. We are now ready to proceed with general self-avoiding hikes. Let d =
p1 · · · pΩ( d) be self-avoiding. Since m is multiplicative and the rank function 17 is totally additive over H, m(d) = ∏
i
m(pi) = λ− ∑
iρ(pi)
= λ−ρ(d). The associated error term follows as
r(d) = |H %−ρ(d)| − |H %|λ−ρ(d) = λ%−ρ(d)(f (% − ρ(d)) − f (%)).
Inserting these forms for m(d) and r(d) in the sieve yields
S(H%, P) = |H %| ∑
d∈P s.a.
μ(d)λ−ρ(d) + λ% ∑
d∈P s.a.
μ(d)λ−ρ(d)(f (% − ρ(d)) − f (%)).
(5) We can now progress much further on making an additional assumption concerning the nature of the prime set P. We could consider two possibilities: i) that P is the set of all primes on an induced subgraph H ≺ G; or ii) that
P is a cut-off set, e.g. one disposes of all the primes of length `(p) ≤ Θ. In the situation where all primes commute with one-another–i.e. when Vi-ennot’s theory of heaps of pieces reduces to number theory –then one may have both i) and ii) simultaneously. This is because in the semi-commutative extension of number theory that holds on the monoid of hikes, coprimality extends to being vertex-disjoint . Therefore, requiring all primes to com-mute is equivalent to forcing the graph G to be made of disjoint oriented simple cycles. Consequently, we can choose H to be the induced subgraph of G comprising all simple cycles of length up to some cut-off Θ and both situations i) and ii) are realised. For this reason, number theoretic sieves benefit from the advantages of both situations: ii) guarantees that sieves can be used to obtain estimates on the number of primes, while i) allows these estimates to be computable. On general graphs however, i) and ii) are not compatible and while ii) could be used to obtain direct estimates for the number of primes of any length, a problem of great interest, this actually makes the sieve NP-hard to implement. We therefore focus on the first situ-ation. Let H ≺ G be an induced subgraph of the graph G and let that P ≡ P H
be the set of all primes (here simple cycles) on H. To conclude the proof we need only show that the error term of Eq. (5) is asymptotically dominated by the first term ∑
d∈P s.a.
H
μ(d)λ−ρ(d). To this end, we note that since H is 18 finite 5
λ% ∑
d∈P s.a.
H
μ(d)λ−ρ(d)(f (% − ρ(d)) − f (%)),
is a sum involving finitely many self-avoiding hikes d. In addition, given that lim %→∞ f (%) exists (either by Assumption 4.1 or by Proposition 4.2 for the length rank function), lim %→∞ f (% − ρ(d)) − f (%) = 0 as long as ρ(d) is finite, which is guaranteed by the finiteness of H. We have consequently established that the error term comprises finitely many terms, each of which vanishes in the % → ∞ limit. As a corollary, the first term is asymptotically dominant:
S(H%, PH )
|H %| ∼ ∑
d∈P s.a.
H
μ(d)λ−ρ(d) as % → ∞ ,
where we assume that |H %| 6 = 0. We now turn to establishing the length corollary of the Finite Sieve The-orem. We are specifically looking for the number of closed walks which are multiples of a prime p. To this end, we need only choose H correctly. Let h
be a hike, for w = hp to be a walk of length l, then h must have length l−`(p)and be such that none of its right-prime divisor commutes with p. The sieve must thus eliminate all hikes h which are left-multiples of primes commuting
with p. Observe that all such primes are on H = G\p. Consequently the Finite Sieve Theorem yields, for |H l−`(p)| 6 = 0,
S(Hl, PG\p) = |H l−`(p)| ∑
d∈P s.a.
μ(d)λ−`(d)
λl−`(p) ∑
d∈P s.a.
μ(d)λ−(d)(f (l −(p) − (d)) − f (l −(p)) ),
where λ is now the graph dominant eigenvalue per Proposition 4.2. The asymptotically dominant term is a sum over all the self-avoiding hikes on
G\p, each with coefficient μ(d)λ−(d) and is equal to det( I − λ−1AG\p). Since furthermore |H l−(p)| = |H l|λ−(p)f (l −(p))/f (l), we have asymptotically for
l 1 S(Hl, PG\p)
|H l| ∼ λ−`(p) det
(
I − 1
λAG\p
)
,
5Gis finite and so are all its induced subgraphs.
19 while the error terms is Err( Hl, PG\p) := S(Hl, PG\p)
|H l| − λ−`(p) det
(
I − 1
λAG\p
)
,
= λ−(p)(f (l −(p))/f (l) − 1) det
(
I − 1
λAG\p
)
+
λ−`(p)
f (l)
∑
d∈P s.a.
μ(d)λ−(d)(f (l −(p) − (d)) − f (l −(p)) ).
The last line can be brought in determinantal form as well, since
f (l − (p) −(d)) − f (l − `(p)) =
`(d)
∑
k≥1
∇kf
k!
(`(d))
(k)
,
with ( a)(k) := ∏k−1
i=0
(a − i) the falling factorial and ∇ the backward difference operator. This operator acts on a function F of a variable x as ∇F :=
F (x) − F (x − 1), and ∇kF designates the kth iteration of operator ∇,i.e. ∇kF = ∑kj=0 (−1) j (kj
)F (x − j). Then
λ−`(p)
f (l)
∑
d∈P s.a.
μ(d)λ−(d)(f (l −(p) − (d)) − f (l −(p)) ).
= λ−`(p) ∑
d∈P s.a.
H
μ(d)λ−`(d)
`(d)
∑
k≥1
∇kf
f (l) k!
(`(d))
(k)
.
The upper limit of the inner sum over k can be extended to ∞ since all terms with k > (d) are nul. Noting that ((d))
(k)
z`(d) = zk d k
dz k
z`(d), this allows us to invert the two sums, yielding
λ−`(p)
∞
∑
k≥1
∇kf
f (l)λk k! det (k)
(
I − 1
λAG\p
)
.
Now setting k = 0 in the above recovers λ−(p)(f (l−(p))/f (l)−1) det (I − 1
λ
AG\p
)
with the exception of the −1 in the parenthesis, which can be introduced as
−δk, 0. This establishes the Finite Sieve Theorem and its length corollary. 20 �) �) �)Figure 4: a) A self-avoiding polygon p on the square lattice; b) Polygon p and all the edges that have at least one endpoint on p; c) Graph Gp.
Infinite graphs
5.1. Viennot’s lemma on infinite graphs
As we have seen, when counting closed walk multiples of a prime ac-cording to their length on finite graphs 6, the Finite Sieve Theorem produces the asymptotics of Viennot’s lemma. To put the infinite graphs results in context we thus start by proving that this lemma extends to infinite graphs with bounded degree. Here, we state only the ordinary generating function version of the extension here, that for formal series on hikes is provided in Appendix B.
Proposition 5.1. Let G be an infinite (weighted di)graph with bounded de-gree. Let p be a prime on G, with support V(p) and neighborhood N (p). Let
Gp ≺ G be the induced subgraph of G with vertex set V(Gp) = V(p)∪N (p) (see Fig. 4 for an example) and Bp its adjacency matrix. Let R(z) = (I−zA)−1∣∣Gp
be the restriction of the resolvent of G to Gp.Then the ordinary generating function Rp(z) of closed walks whose last erased loop is p is given by
Rp(z) = z`(p) det (I + zR(z)Bp
). (6) Observe that neither the series det( I − zAG\p) nor det( I − zA) appearing in Viennot’s original lemma are well defined on infinite graphs—e.g. all their
6Recall that this is the same thing as counting closed walk whose last erased loop is a certain simple cycle.
21 finite order coefficients can be infinite. Their ratio evaluated on a sequence of finite graphs converging to G (as defined in Appendix A) nonetheless gives rise to a well defined series in the sense of the Proposition above.
Proof. Let {GTor
N
}N ∈N be the small torus sequence of graphs converging to G
as defined in Appendix A. For any SAP p, define N (p) ∈ N such for all N ≥
N (p), then Gp is an induced subgraph of GN .7 The result follows by using Viennot’s lemma on GTor
N
and grouping all terms into a single determinant. Let p be a prime of finite support, hence finite length `(p) on G. Let N ≥
N (p), RN (z) := ( I − zAGTor
N
)−1 and BN (p) := AGTor
N
− AGTor
N\p
. Then Viennot’s lemma on GTor
N
yields the ordinary generating function RN,p (z) of closed walks whose last erased loop is p on GTor
N
as
RN,p (z) = det( I − zAGTor
N\p
)det( I − zAGTor
N
) = det (RN (z)) det (I − zAGTor
N
zBN,p
),
= det (I + zRN (z)BN,p
).
Given that the sequence of small tori GTor
N
converges to G, lim N →∞ RN,p (z) =
Rp(z) provided we can show that det (I + zRN (z)BN,p
) is well defined under this limit. Now since (BN,p
)
ij
= 0 unless both i, j ∈ V (p) ∪ N (p) and since p is of finite length, then det (I + zRN (z)BN,p
) is equal to the determinant of the finite |V (p) ∪ N (p)| × |V (p) ∪ N (p)| matrix Qij := (I + zRN (z)BN,p
)
ij
, for i, j
in p and its neighborhood. Given that BN,p
∣∣Gp = Bp for all N ≥ N (p) and since the graph has bounded degree lim N →∞ RN (z) = R(z) is well defined. We consequently have
Rp(z) = lim
N→∞
RN,p (z) = lim
N→∞
det
((I + zRN (z)BN,p
)|Gp
)
= det (I + zR(z)Bp
).
We emphasise that the determinant det (I + zR(z)Bp
) is equal to that of the finite matrix Q and no considerations pertaining to the determinants of infinite matrices is needed. We illustrate Viennot’s lemma on infinite graphs with the ordinary gen-erating function Re(z) of closed walks whose last erased loop is an edge cycle
7Existence of N(p) is guaranteed for finite length SAPs as the small torus GTor
N:= n2
contains the ball of radius n/ 2 centred on the starting point of pon G.
22 21 34213431 4214 23142331 241243Figure 5: Illustration of the 7 walks of length 4 on the square lattice whose last erased loop is an edge cycle (highlighted in red). The numbers next to the highlighted edges indicate the order in which these edges are traversed.
e on the square lattice. Direct application of Eq. (6) gives
Re(z) = π
4 − 116 + 116 π2
(64 z2 − 4)K(16 z2)2 + 14π
(
K(16 z2) − π2
)
, (7) = z2 + 7 z4 + 70 z6 + 807 z8 + 10 , 046 z10 + 131 , 206 z12 + · · ·
where K(x) := ∫ π/ 20
(1 − x sin 2(θ))−1/2dθ is the complete elliptic integral of the first kind. In Fig. (5) we illustrate the 7 walks of length 4 on the square lattice whose last erased loop is an edge cycle, as correctly counted by [ z4]Re(z) = 7. Typically, generating functions for walks whose last erased loop is a certain self-avoiding polygon p, such as Re(z) above, are not mero-morphic as functions of z. Consequently, the asymptotic growth of their coefficients cannot be determined with the traditional tools of meromorphic asymptotics . In the next section we develop a generic method as a re-placement that allows us to determine this asymptotic growth for any prime
p.
5.2. Asymptotic expansion of Viennot’s lemma
We now aim at establishing the formula for the fraction of walks whose last erased loop is any chosen self-avoiding polygon on any infinite vertex-transitive graph.
Proof of Corollary 2.1. The first difficulty in extending the Finite Sieve The-orem to infinite graphs comes from the proliferation of hikes on such graphs: 23 there are either exactly 0 or infinitely many hikes of any given length. Fur-thermore, the number of hikes increases uncontrollably with the length as there are also infinitely many more hikes of any length L′ > L than of length
L. These observations continue to be true even when hikes are considered up to translation. To make matters worse, the fraction of all hikes which are walks is exactly 0; and even with edge weights uniformly set to 1 /λ , the total weight carried by all walks whose last erased loop is any SAP p is still divergent. To resolve these serious difficulties requires us to separate the finite sieve results into two contributions, the first of which relates hikes to walks and the second relates closed walks to walks with fixed last erased loop p. This second contribution must itself be dealt with carefully to cure divergences stemming from the non-meromorphic nature of the generating functions pro-duced by the extension of Viennot’s Lemma to infinite graphs. We illustrate every step of the proof with explicit results on the square lattice. Let us consider the dominant term of the asymptotic expansion of Vi-ennot’s lemma on the sequence of finite graphs GTor
N
. Fixing N , consider p
a self-avoiding polygon on GTor
N
. Using the transformation presented in the extension of Viennot’s lemma to infinite graphs, we can express the asymp-totic fraction of hikes of length l that are walks whose last erased loop is p
as l → ∞ as
S(Hl, PGTor
N\p
)
|H l| ∼ λ−`(p) det
(
I − 1
λAGTor
N\p
)
= λ−`(p) lim
z→1/λ −
ζN (z)−1 det ( I + zRN (z)Bp) .
This suggests a strategy consisting of proving separate convergence in z →
1/λ − of the two terms in the limit above. This naive strategy ultimately fails, but the procedure that works is best understood once the nature of this failure is made apparent and several results we will obtain along the way are necessary to implement the correct proof strategy. In this spirit, we pretend to follow the naive approach and thus first examine the behaviour of the limit lim z→λ−1 ζN (z)−1 asymptotically in N :
Lemma 5.2. Let {GTor
N
}N ∈N be the small tori sequence of vertex-transitive graphs converging to the infinite bounded-degree vertex-transitive graph G
with maximum eigenvalue λ. Let ζN (z) be the zeta function of hikes on GTor
N
24 and let R(z) = R(z)ii be the ordinary generating function of closed walks on
G. Then,
lim
N→∞
ζN (z)1/N = ˜ζ(z) = exp
(∫ 1
z
(R(z) − 1)dz
)
. (8)
Furthermore α := lim z→1/λ − ˜ζ−1(z) is well defined. All the coefficients [zn]˜ζ(z)
are positive integers, while the coefficients [zn]˜ζ−1(z) are integers. Proof. Since the log-derivative of the hike zeta function is the trace of the resolvent, the log derivative of ζ(z)1/N is, on vertex-transitive graphs, a single diagonal entry of the resolvent, yielding Eq. (8). Assuming that the dominant eigenvalue is unique, R(z) diverges at worse as 1 /(1 − zλ ) around z ∼ 1/λ
so that necessarily ∫ 1
z
(R(z) − 1)dz converges in 1 /λ .To understand the coefficients of [ zn]˜ζ(z), let h be a hike and let w1, · · · wn
be vertex-disjoint walks making up hike h, i.e. h = w1w2 · · · wn modulo the fact that all these walks commute with one another. A rooted hike hroot is the object obtained from h on translating all walks wi so that the origin lies on their unique right prime divisor but retaining the fact that they commute. Then [ zn]˜ζ(z) counts the number of rooted hikes of length n, ˜ζ(z) is the zeta function of rooted hikes, while ˜ μ(z) := ˜ζ−1(z) is the M¨ obius function on rooted hikes.
Corollary 5.3. On the square lattice, the generating function of rooted hikes is
˜ζ(z) = exp
(∑
n≥1
(2nn
)2 z2n
2n
)
,
= 1 + 2 z2 + 11 z4 + 86 z6 + 805 z8 + 8402 z10 + 94306 z12 + . . . ,
This is illustrated on Fig. (6). The corresponding M¨ obius function is
˜μ(z) := 1 /˜ζ(z),
= 1 − 2z2 − 7z4 − 50 z6 − 456 z8 − 4728 z10 − 53095 z12 + . . . ,
and α := ˜ζ−1(1 /λ ) = 14 e4Cπ ' 0.8025 ... with C Catalan’s constant.
25 Figure 6: The [ z4]˜ζ(z) = 11 objects of length 4 giving rise to distinct rooted hikes of length 4 on the infinite square lattice. Rigorously speaking, rooted hikes are obtained on considering the above objets up to translation of each walk.
Proof. While this result is an immediate corollary of the precedent lemma, we can prove it directly by considering the sequence of small n × n square lattices GSq
n2
converging to G as described in Appendix A. The eigenvalues of GSq
N:= n2
are λ(i, k ) = 2 cos
(
π jn+1
)
2 cos
(
π kn+1
)
, where j and k are two integers between 1 and n. Product-integration yields lim
N→∞
ζN (z)1/N = exp
(∫
S
log (1 − 2z cos( πx ) − 2z cos( πy ))dxdy
)
.
with S the unit square [0 , 1] × [0 , 1]. We now obtain the relation to R(z)directly. Observe that
∫
S
log (1 − 2z cos( πx ) − 2z cos( πy ))dxdy =
− ∑
n≥1
(2 z)2n
2n
∫
S
( cos( πx ) + cos( πy ))2ndxdy,
where we used the fact that odd powers of the sum of the cosines must have 0 integral over S since both cosines are symmetric functions on S. Then we 26 have
∫
S
( cos( πx ) + cos( πy ))2ndxdy =
n
∑
h=0
(2n
2h
) ∫ 10
cos( πx )2n−2hdx
∫ 10
cos( πy )2hdy,
=
n
∑
h=0
(2n
2h
)Γ (n − h + 12
)
√π(n − h)! Γ (h + 12
)
√πh ! ,
= 4nΓ (n + 12
)2
π(n!) 2 =
(2nn
)2
4−n.
It follows that ∫
S
log (1 − 2z cos( πx ) − 2z cos( πy ))dxdy = − ∑∞
n=1
(2nn
)2 z2n
2n
=
− ∫ 1
z
(R(z) − 1)dz.
Lemma 5.2 implies that asymptotically, when N 1, lim z→1/λ − ζN (z)−1 ∼
αN is well defined. We will see later on that αN relates the density of hikes to that of walks on an infinite graphs. More precisely, it can be interpreted as follows: every time a vertex is added to a graph GTor
N
or GSq
N
, N 1, the fraction of hikes which are closed walks is multiplied by α.Returning to the fraction of hikes on GTor
N
which are walk whose last erased loop is p, Lemma 5.2 and our naive strategy suggest that we express this fraction asymptotically for N 1 as det
(
I − 1
λAGTor
N\p
) ?
= αN lim
z→λ−1
det (I + zRN (z)Bp
). (9) In fact the right hand side is divergent under the limit z → 1/λ −, but we will see below that its divergence is generic: it is the same for any SAP p as well as for all closed walks. In other terms, the right hand side diverges only because the series of all walks diverges at the point z → 1/λ −, hinting that the fraction of walks whose last erased loop is p with respect to all walks might itself be well defined. To put this observation on firm foundations, we need to show that lim z→1/λ − det (I+zRN (z)Bp
)/R (z) is well defined, and that the fraction of all walks with respect to all hikes is well defined as well, in spite of the generic divergence. These two results rely on different techniques. We start with proving of the second fact without which the first would be useless. 27 Lemma 5.4. Let G be an infinite vertex-transitive graph of bounded degree. Let {GTor
N
}N ∈N be the small tori sequence of vertex-transitive graphs converg-ing to G. Let • be any vertex of G. Then the fraction F• of hikes which are closed walks from • to itself is asymptotically given by
F• ∼ αN
N , N → ∞ .
In this expression α is the constant defined in Lemma 5.2.
The 1 /N factor in F• originates from that we have fixed the vertex •.If instead we consider translation invariant quantities, i.e. we consider all closed walks irrespectively of their starting point, then the fraction of hikes which are closed walks is asymptotically αN , N 1.
Remark 5.1. Let |W |N () and |H|N () be the total number of walks and of hikes up to length ` ≥ 0 on GTor
N
, respectively. Then, Lemma (5.4) is equivalent to lim
N→∞
lim
`→∞
1
N log (|W |N ()/|H|N ()) = log( α),
that is the density of the logarithm of the fraction of hikes which are closed walks is well defined on G.
Proof. On the finite graphs GTor
N
, the fraction of all closed walks off • with respect to hikes is given by F• = det( I − 1
λ
AGTor
N\•
). Let BN, • be the N × N
identically zero matrix except on edges adjacent to • where its value is 1, i.e. so that AGTor
N\•
= AGTor
N
− BN, •. Then F• = det( I − 1
λ
AGTor
N
1
λ
BN, •) and we determine this determinant by expansion around AGTor
N
, hereafter denoted
AN in order to alleviate the equations. This gives
F• = det
(
I − 1
λAN
)
1
λ Tr
(
adj
(
I − 1
λAN
)
BN, •
)
− 12λ2 lim
z→λ−1
det ( I − zAN )
[
Tr 2(RN (z)BN, •
) − Tr (RN (z)BN, •
)2]
.
All orders of degree higher than two are exactly zero. In this expression
RN (z) := (I − zAN
)−1, Tr( M) designates the trace of a matrix M and adj( M)its adjugate. 28 Order zero of the expansion is null since λ is an eigenvalue of GTor
N
. By the same token, the adjugate matrix of the first order is proportional to the projector PN,λ onto the eigenvector associated with λ on GTor
N
adj
(
I − 1
λN
AN
)
= αN PN,λ ,
where αN := ∏λi∈Sp( AN )
λi6=λN
(1 − λi/λ N ). Since GTor
N
is regular PλN = 1
N
J, where
Jij = 1. Then 1
λ Tr
(
adj
(
I − 1
λAN
)
BN, •
)
= αN
λ × 2λN = 2 αN
N ,
since BN, • has exactly 2 λ non-zero entries. The second order of the expansion is always well defined, as is readily seen from the equivalent form
− 12λ2 lim
z→λ−1
det ( I − zAN )
[
Tr 2(RN (z)BN, •
) − Tr (RN (z)BN, •
)2]
= αN
λ2
[Tr (PN,λ BN, •
)Tr (CN BN, •
) − Tr (PλN BN, •CN BN, •
)] . (10) where
CN := lim
z→1/λ −
(I − PN,λ
)RN (z), (11) and
αN := lim
z→1/λ −
(1 − zλ )−1 det ( I − zAN ) = ∏
λi∈Sp( AN)
λi6=λN
(1 − λi/λ N ),
which, by Lemma 5.2, is asymptotically given by αN for N 1 and α =˜ζ(1 /λ )−1. The second line of Eq. (10) stems directly from the observation that PN,λ is a rank one projector. Indeed for any such projector Q and matrix
M of bounded norm we have Tr ((QM )2) = Tr 2(QM ).Each term of the second line of Eq. (10) can be evaluated generically. For convenience, let c0 be the diagonal entry of CG, c1 the entry relating first neighbours on G and c2,i , all the distinct entries of CG relating second neighbours on G. Then Tr (PN,λ BN, •
) = 2 λ/N , Tr (CN BN, •
) = 2 λc 1 and
N Tr (PλN BN, •CN BN, •
) = ∑
ordered vertex pairs vi, v j
deg( vi)deg( vj ) Cvivj ,
= ( λ2 + λ)c0 + 2 λ2c1 + ∑
i
n2,i c2,i ,
29 �)�)�)Figure 7: In thick red edges, corolla graphs G• on: a) the square lattice; b) the triangular lattice; c) the hexagonal lattice. The same construction and proof given in this section applies to any vertex transitive graph, in particular it is not limited to planar lattices and corollas can be considered in any dimension.
where deg( vi) is the degree of vi on the corolla G• (see Fig. 7 for an illustration of corolla graphs) and n2,i is the number of times entry c2,i appears. Recursion relations between the entries of CG presented in Appendix C give c0 = 0,
c1 = −1, ∑bλ/ 2c
i=1
n2,i c2,i = −λ2. The second order then generically evaluates to αN
N λ 2
(−4λ2 − (−2λ2 − λ2)) = −αN
N .
We can finally put the zeroth, first and second orders of the determinant expansion together, yielding det
(
I − 1
λAN
)
= 0 + 2 αN
N − αN
N = αN
N .
Asymptotically, for N 1, αN ∼ αN as per Lemma 5.2, which gives the result. The method employed in the proof of Lemma 5.4 to calculate the fraction of hikes which are closed walks extends to any self-avoiding polygon. In this more general situation, the expansion of the determinant always terminates at a finite order which grows with the polygon’s length. The formulas to be evaluated become very involved however, to the point of being effectively im-practical even for short SAPs. We nonetheless provide the general expression of the formal expansion at all orders for any SAP in Appendix D. To further exemplify this determinant expansion it is worth considering the calculation of the fraction Fe of hikes which are walk whose last erased loop is an edge e on the square lattice. In this case, the expansion of Fe
around det( I − 1
λ
AG) has exactly four non-zero orders each of which involves 30 complicated sums of entries of the graph resolvent R(z) For example, the second order is asymptotically equal to lim
z→1/λ −
αN
λ2N
(
112 R10 (z) − 38 R00 (z) − 40 R11 (z)
− 20 R20 (z) − 12 R21 (z) − 2R30 (z)
)
= αN
λ2N × − 10 ,
for N 1. Although λ = 4 on the square lattice, we left it unevaluated to help see that this is the second order of the expansion. Here Rij (z) designates the entry of the resolvent relating two vertices with distance i along x and j
along y. Taken together, the four orders of the determinant expansion give asymptotically for N 1
Fe = 1
λ2
(
14 αN
λN − 10 αN
λ2N − 80 αN
λ3N + 96 αN
λ4N
)
= 2αN
N λ 2 = αN
8N ,
where the overall 1 /λ 2 factor in front arises from the fact that the edge has length 2 and would simply be 1 /λ (p) for a general SAP. Lemma 5.4 indicates that αN /N relates hikes to closed walks, while the remaining 1 /8 factor is thus the asymptotic fraction of closed walks which are walk whose last erased loop is an edge cycle. This means in particular that, whenever [ z]R(z) 6 = 0, lim
`→∞
[z]Re(z)[z]R(z) = 18,
where Re(z) is the generating function given in Eq. (7). It might struck the reader that a remarkable number of simplifications must have taken place to yield such a simple answer as 1 /8 from the compli-cated orders of the expansion. This hints at the existence of a much simpler calculation procedure, and this is precisely the procedure in the spirit of Eq. (9). Now however, we will be able to remove the generic divergence of the naive approach, having separately obtained an asymptotic expression for the fraction of walks with respect to hikes by Lemma 5.4.
Lemma 5.5. Let G be an infinite vertex-transitive graph of bounded-degree and let λ be the supremum of its spectrum. Let {GTor
N
}N ∈N be the small tori sequence of vertex-transitive graphs converging to G. Then the asymptotic
31 fraction of closed walks which are walk whose last erased loop is p is well defined and given by
Fp
λ`(p) = lim
N→∞
lim
z→1/λ −
z`(p) det (I + zRN (z)Bp
)
det (I + zRN (z)B•
),
= 1
λ`(p)+1 deg T. adj
(
I + CG
∣∣p.Bp
)
. 1,
where CG
∣∣p is the restriction to Gp of CG := lim z→1/λ − (I − Pλ)R(z).Proof. According to the Finite Sieve Theorem, the asymptotic fraction of hikes which are closed walks whose last erased loop is p on GTor
N
is lim z→1/λ − z`(p) det( I−
zAGTor
N\p
). Since all the limits taken here are finite and well defined (as ev-erything takes place on GTor
N
), the asymptotic fraction of closed walks whose last erased loop is p on these finite graphs is lim z→1/λ − z`(p) det( I − zAGTor
N\p
)lim z→1/λ − z0 det( I − zAGTor
N\•
) = lim
z→1/λ −
z`(p) ζN (z) det( I − zAGTor
N\p
)
ζN (z) det( I − zAGTor
N\•
),
= lim
z→1/λ −
z`(p) det (I + zRN (z)Bp
)
det (I + zRN (z)B•
).
We now turn to studying the behaviour of the right hand side as N → ∞ .To this end, we expand I + zRN (z)Bp around 1 /λ with z < 1/λ .We need to distinguish behaviours based on the dimensionality d > 1 of the lattice under study. We ignore the trivial 1D case (for which the only SAP is the edge, and the fraction of closed walks whose last erased loop is the left or right edge attached to any vertex is 1/2). On d > 1 dimensional lattices we have,
I + zRN (z)Bp = CN Bp − 1
λπ PN,λ Bp Li d/ 2(1 − zλ ) + O(1 − zλ ), (12) with CN as defined in Eq. (11) and Li a(x) := ∑
n> 1
xn
na
is the a-polylogarithm function. Combinatorially, it arises here from summations over closed walks weighted by λ−`, which leaves a residual total weight asymptotically given by
−d/ 2 for all closed walks of length 1. The generic nature of the behaviour exhibited by I + zRN (z)Bp is now readily apparent: 1) divergence occurs only on 2D lattices, where it is logarithmic; 2) it is the same for all SAPs; and 3) it is also the same for all closed walks (which are readily recovered upon taking 32 p to be length 0, i.e. Bp ≡ B• is a corolla). Thanks to these observations, the determinant expansion at 1 /λ − is det (I + zRN (z)Bp
) = det
(
CN Bp − 1
λπ PN,λ Bp Li d/ 2(1 − zλ ) + O(1 − zλ )
)
,
= − 1
πN Li d/ 2(1 − zλ ) 1
λ deg T. adj ( I + CN .Bp) . 1
o(Li d/ 2(1 − zλ )),
where we used the matrix-determinant lemma and the QR decomposition
− 1
λπ PN,λ Bp = − 1
πN × 1
λ × 1.deg T.
This decomposition relies on the observation that PN,λ is the projector onto
1, i.e. that all GTor
N
are regular. 8 We recall that in the above expression,
deg = Bp.1 = diag( B2
p
) is the degree of vertices on Gp. Similarly, at 1 /λ −,det (I + zRN (z)B•
) = − 1
πN Li d/ 2(1 − zλ ) + o(Li d/ 2(1 − zλ )),
and finally
z`(p) det (I + zRN (z)Bp
)
det (I + zRN (z)B•
) = 1
λ`(p)+1 deg T. adj ( I + CN .Bp) . 1 + o(1) ,
which yields the result after taking the limits z → 1/λ − and N → ∞ now both clearly well-defined, even when d = 2. Combinatorially, the diver-gence curing on 2D lattices effected here comes from relating walk whose last erased loop is a fixed SAP to all closed walks rather than directly to the hikes. The relation between closed walks and hikes is performed separately through Lemma 5.4.
Remark 5.2. Recall that R(z) and Rp(z) are the ordinary generating func-tions of all closed walks on G and of closed walk whose last erased loop is p on
G, respectively. The finite sieve here indicated that the limit in z → 1/λ − of
8This step can be adapted should we consider lattices which are not vertex-transitive but for which G/ Aut( G) has finitely many vertices. This is beyond the scope of this work.
33 the ratio of power series Rp(z)/R (z) yields the asymptotic behaviour of the term-by-term ratio [ z]Rp(z)/[z]R(z) as ` → ∞ , which rather corresponds to an Hadamard division of Rp(z) by R(z), i.e. lim
z→1/λ −
Rp(z)
R(z) = lim
`→∞
[z]Rp(z)[z]R(z) .
This is a corollary of the fact that the leading divergence of both Rp(z) and
R(z) located at z = 1 /λ . For a general discussion on the relation between Hadamard products and singularity analysis, we refer to . This concludes the proof of the Infinite Sieve Theorem.
5.3. Probabilistic interpretations
The results of the Infinite Sieve Theorem have a probabilistic interpre-tation which motivated the study of the fraction Fp/λ (p) in the literature ultimately leading to SLE 2, albeit without the relation to hikes and the ex-plicit form of Eq. (1). Here the interpretation of Fp/λ(p) as a probability distribution over the SAPs is first manifested in the following:
∑
p: SAP
Fp
λ`(p) = 1 . (13) Combinatorially, this trivially states that any walk has a unique right prime factor and consequently the total fraction of all closed walks whose last erased loop is any SAP is exactly 1.
Remark 5.3. All simple cycles passing through some chosen vertex of G
are present in the sum of Eq. (13). Following the rules in the Cartier-Foata monoid of the hikes, a SAP of length (p) thus appears at least 2(p) times in the sum. Here, the factor of 2 accounts for the orientation; and `(p) reflects all the valid starting points for the SAP as a closed walk. 9 For example, on
9Rigorously, what matters is the size of the equivalence class of all words on edges that define the same hike hin the original Cartier-Foata monoid. As mentioned earlier, this size is the value of the von Mangoldt function Λ( h), which is the length of the unique right factor of a walk if his a walk and 0 otherwise . Seen as a closed walk, a SAP pis its own unique right-prime divisor and thus Λ( p) = `(p).
34 the square lattice, the 1 × 1 square gets a factor of 8 = 2 × 4 × 1 and the 1 × 2 rectangle has a factor of 12 = 2 × 6. Strictly speaking, the sum Eq. (13) therefore runs over simple cycles with fixed starting point and the SAP index is a (harmless) notational abuse. We can go further in the probabilistic interpretation using purely combi-natorial arguments:
Proposition 5.6. Let G be an infinite vertex transitive planar graph and let p be a SAP on it. Let w be a random walk with uniform edge-transition probability 1/λ . Run the walk until it comes back to its starting point. Then the probability P(w 7 → p) that the last erased loop of w be p is equal to the fraction of all closed walks (including those passing an arbitrary number of times through the origin) whose last erased loop is p, P(w 7 → p) = Fp/λ `(p).Proof. Let GN be a family of finite vertex-transitive graphs of degree λ con-verging to G as N → ∞ . Let WN,p (z) be the ordinary generating function of closed walks on GN with right prime divisor p and such that these walks never revisit their starting point except on their final step. By construction, we have P(w 7 → p) = lim z→1/λ − lim N →∞ WN,p (z) provided both limits are well defined. But
WN,p (z) = z`(p) det (I − zAGN \p
)
det (I − zAGN \•
),
= z`(p)
(
ζN (z) det (I − zAGN \p
))(
ζN (z) det (I − zAGN \•
))−1
.
This has a well defined limit when N → ∞ as per the extension of Viennot’s lemma to infinite graphs
Wp(z) := lim
N→∞
WN,p (z) = z`(p) det (I + zR(z)Bp
)
det (I + zR(z)B•
),
and by the Infinite Sieve Theorem the limit z → 1/λ − of the above is also well defined. It is given by lim z→1/λ − Wp(z) = Fp/λ (p) with Fp as per Eq. (1). The logarithm of Fp also has a probabilistic interpretation that has not appeared in the literature so far. It is based on results by Espinasse and Rochet : 35 Proposition 5.7 (Espinasse and Rochet, 2019) . Let G be an infinite vertex-transitive graph and let p be a SAP on it. Let Ew(.) designate the expectation value of a random variable with respect to the closed random walks defined up to translation and weighted with probability λ−(w). Then
log( Fp) = ∑
w:walk
Λp(w)
p(w) λ−(w) = Ew
(Λp(w)
`p(w)
)
.
Here Λp(w) counts the vertices that are both in p and the unique right divisor of w and `p(w) is the number of vertices of p visited by w, counted with multiplicity.
The reason for the notation Λ p(w) is that this function is similar to the von Mangoldt function Λ defined on hikes . Defining Λ p(h) to be 0 when the hike h is not a walk and otherwise Λ p(h) is as in the Proposition above, we have Λ p(h) = Λ( h) when p is the unique right prime divisor of h.The combination of Propositions 5.6 and 5.7 implies the rather uncom-mon result that the entropy of the distribution of Fpλ−`(p) values itself has a probabilistic interpretation. Consequences of this observation will be ex-plored in a separate work.
Discussion
6.1. Extension to SAWs and further lattices
In this contribution, we presented fully deterministic combinatorial ar-guments based on number-theoretic sieves for counting walk multiples of SAPs on any finite or infinite vertex-transitive graphs. This is equivalent to counting all the walks whose last erased loop following Lawler’s loop erasing procedure is some chosen self-avoiding polygon p on such graphs. In fact, all the results presented here are immediately valid for self-avoiding walks (SAWs) as well. In particular, Eq. (1) of the Infinite Sieve Theorem giving the fraction Fp is immediately correct. For a self-avoiding walk p, Fp/λ `(p)
is the fraction of all open walks with the same fixed starting and end points for which p is the self-avoiding skeleton remaining after loop-erasing. Finally, the arguments presented here should extend without fundamental changes to infinite graphs that are not vertex transitive as long as G/ Aut( G)is finite. This nonetheless requires further work and is beyond the scope of the present contribution. 36 6.2. Counting the self-avoiding polygons
We recall that R(z) and Rp(z) are the ordinary generating functions of closed walks and of walks whose last erased loop is a SAP p, respectively. The research presented here suggests a natural strategy to tackle the open problem of asymptotically counting SAWs and SAPs. First, observe that we know the exact number R(z)[ L] of closed walks of length L defined up to translation. Then, if we could determine the exact number Rp(z)[ L]of closed walks of length exactly L whose last erased loop is p, it would be sufficient to sum this over all SAPs of length strictly less than L and subtract the result from R(z)[ L] to determine the number π(L) of SAPs of length exactly L:
π(L) = R(z)[ L] − ∑
p: SAP
`(p)<L
Rp(z)[ L].
While such a precise count is not feasible in practice as L → ∞ , an asymp-totic estimate of the number of walks whose last erased loop is any chosen SAP may seem, at first, to be sufficient to gain an insight into the number of SAPs themselves. Following this idea, we would rather write
π(L)
R(z)[ L] = 1 − ∑
p: SAP
`(p)<L
Rp(z)[ L]
R(z)[ L] . (14) and use Rp(z)[ L]/R (z)[ L] ∼ Fpλ−`(p) for L 1. Thus, we would only need to estimate sums like
S(L) := ∑
p: SAP
`(p)≤L
Fp
λ`(p) ,
for L 1, in order to work out an asymptotic expansion for π(L). Such an estimate can already be determined from R. Kenyon’s seminal results , we find
S(L) = 1 − L−3/5 + O(L−3/5). (15) See also Fig 8 for a numerical illustration. This result of course wildly differs from the ( μ/λ )LL−1/2 expected here from the numerically conjectured scaling for π(L). 10 From the point of view of probability theory, the origin of this
10 The correction term is L−1/2here because we count simple cycles rather than SAPs.
37 ●●● ● ● ● ● ● ● ● ● ● ● ● ● ●
� � �� �� �� �� �� ��� ��� ��� ��� ��� ��� ������� ������� ������ �
�� ���
ℓ(�)≤�
� �
�ℓ (�)Figure 8: In red points and dashed red line, the sum over all self-avoiding polygons with fixed starting point on the square lattice of length at most L of the fraction Fp/4(p),as a function of L. The corresponding data table is presented in Appendix F. The solid blue line is the first term of asymptotic expansion for this quantity, that is 1 − L−3/5.The discrepancy between −3/5 and the −1/2 expected from SAP counting is due to an accumulation of error terms in the Infinite Sieve Theorem, which require us to include terms beyond the dominant Fp/4(p) or use stronger two-sided sieves.
discrepancy is clear: the law governing Lawler’s loop erased random walks essentially converges to SLE 2 rather than the conjectured SLE 8/3 for SAP and SAW models. From the point of view of sieve techniques however, the chasm between these results originates from an uncontrolled accumulation of error terms affecting the estimate Rp(z)[ L]/R (z)[ L] ∼ Fpλ−`(p).11
It is important to recall that Fpλ−(p) is only the first, asymptotically dominant term of the asymptotic expansion of the number of walks whose last erased loop is p. In particular Fpλ−(p) is a good approximation to
Rp(z)[ L]/R (z)[ L] only when L `(p) (see Appendix E). Yet, when we subtract walks with fixed last erased loops from all closed walks of length
This is responsible for a factor of L in front of the L−5/2. Since in addition, R(z)[ L] ∼
λL/(πL ) for L 1, this accounts for another factor of L and finally we get L×L×L−5/2 =
L−1/2.
11
These errors have the same origin as those affecting the Eratosthenes-Legendre sieve in number theory!
38 L, we must consider the walks whose last erased loop p is of length up to
(p) = L − 1. Given the exponential growth in the number of SAPs, this means that most of our estimates are affected by large, uncontrolled errors, and it is impossible to exploit Eq. (14) using solely S(L). This problem has two potential solutions. The first idea is to take into ac-count some error terms Err p(L) in the asymptotic expansion of Rp(z)[ L]/R (z)[ L]so as to determine this quantity more precisely. Since all the error terms are exactly available 12 , it seems possible that an extension to Kenyon’s argu-ments would allow us to estimate sums of such error terms generically; just as Eq. (15) does for the dominant term. This idea suffers from a major drawback: error terms actually grow with L if we consider classes of SAPs for which L −(p) is fixed. Since most SAPs of length up to L − 1 are close in length to L, this means that the overall error term affecting Eq. (14) grows uncontrollably with L. Thus, an increasingly (and unrealistically) detailed knowledge of the errors is needed as L → ∞ , so that this strategy collapses completely with respect to rigorous arguments.
6.3. The path to rigorous progress
The second approach relies on a crucial foundational work by M. Bousquet-M´ elou regarding the enumeration of heaps of pieces satisfying both left and right constraints [27, 28]. This work opens the way for two-sided sieves in the same manner as Viennot’s Lemma relates to the Finite and Infinite Sieve Theorems: they give control over both the left and right prime divisors of a walk. Consequently, the maximum length of the primes to be considered in Eq. (14) is reduced to only L/ 2. The “sieving gap” between L and L/ 2dramatically reduces the importance of the error terms to the extend that, in accordance with Appendix E, we expect them all to vanish under the limit
L → ∞ .13 At the same time, the dominant contribution is not Fp/λ `(p) any-more. The resulting calculations will be presented in a separate work.
12 In fact Rp(z)[ L] itself is in principle exactly available from the extention of Viennot’s Lemma to infinite graphs. In this situation however, it is a precise estimate for the sum over SAPs of Rp(z)[ L] which is utterly lacking.
13 Seeing heaps of pieces as an extension of number theory as in , shows that this
L/ 2 is the extension of the √xgap present in all standard number-theoretic sieves. We can similarly show that the fraction Fpλ−`(p)extends the quantity log( x)/log( p) and all identities given here extend (and hence reduce to) valid number-theoretic identities. Non-trivial (novel) results on partial sums of the M¨ obius function also follow heuristically.
39 Appendix A Converging sequences of graphs
We here recall the notion of a converging sequence of graphs. We follow directly the work of : Let G and G′ be two bounded degree graphs and let r and r′ be vertices called the roots on G and G′, respectively. A topology on the space X of isomorphism classes of rooted connected graphs is induced by the following metric. Let BG(r, n ), n ∈ N∗ be the ball of radius n centred on the root
r in graph G. Let k be the supremum of all n such that ( BG(r, n ), r ) and (BG′ (r′, n ), r ′) be isomorphic as rooted graphs and define the distance d be-tween ( G, r ) and ( G′, r ′) as 2 −k. Then d is a metric on X . Now we say that a sequence ( GN , r N ) of rooted graphs converges to a rooted graph ( G, r ) if and only if lim N →∞ d((GN , r N ), (G, r )) = 2 −∞ := 0. In this work we specifically deal with two sequences converging to infinite vertex-transitive graphs (termed lattices ): i) Small graph sequence: where Gn is a finite cut-out of the lattice G that includes BG(r, n ) but not BG(r, n + 1). In the context of planar lattices we construct a specific small graph sequence as follows. Define GSq
N
the induced subgraph of G which is the cut-out of G
contained within the square of side length n ∈ N∗ centred on r (edges being given a length of 1). Here N stands for the number of vertices of this square cut-out, the precise relation between n and N depending on the underlying lattice G. We can now also define: ii) Small torus sequence: where the small graphs GSq
N
are wrapped around a torus. The resulting graphs GTor
N
are vertex-transitive tori with the same degree as G.Since all tori GTor
N
as well as G are vertex-transitive, the roots rN and r are irrelevant when considering convergence of this sequence. We therefore sim-ply say that the sequence of graphs GTor
N
converges to the lattice G.The small torus sequence generalises to non-planar lattices by wrapping around genus g > 2 tori the small graphs GHyp
N
defined as the induced sub-graph of G which is the cut-out of G contained within the d-hypercube of 40 �) �)Figure 9: Examples of finite graphs part of the sequences converging to the (infinite) square lattice: a) small graph GSq 30 with a self-avoiding polygon on it; b) corresponding small torus GTor 30 with the same self-avoiding polygon.
side length n ∈ N∗ centred on r. In this case, the dimension d > 2 is the smallest dimension of the space Rd such that G can be embedded in Rd in a way that no two edges cross.
Appendix B Viennot’s lemma on infinite graphs with formal se-ries
The proof is entirely similar to that given in the case of ordinary gener-ating functions Proposition 5.1. We have:
Proposition B.1. Let G be an infinite (di)graph with bounded degree with labelled adjacency matrix W. Let p be a prime on G, with support V(p)
and neighborhood N (p). Let Gp ≺ G be the induced subgraph of G with vertex set V(Gp) = V(p) ∪ N (p) and Wp its labelled adjacency matrix. Let
R = (I − W)−1∣∣Gp be the restriction of the formal resolvent of G to Gp.Then the formal series of closed walks whose last erased loop is p is given by ∑
wwalk
p|r w
w = det (I + R W p
)p. (16) 41 Appendix C Recursion relations for C and R(z)
Recursion relations between entries of the resolvent naturally arise from the basic observations that walks of length ` going from vertex vi to vertex
vj were walks of length − 1 from vi to a neighbours of vj . We have a similar relation relating walks of length with those of length ` − 2. In terms of generating functions, these read
λz R(z)1 = R(z) − 1,z2 ∑
i
n2,i R(z)2,i + λz 2R(z) = R(z) − 1where R(z), R(z)1 and R(z)2,i designate a diagonal entry of the resolvent, the entry of the resolvent relating first neighbours, and entries of the resolvent relating second neighbours, respectively. The quantity n2,i is the number of times R(z)2,i appears. Similar recursion relations are thus obeyed by the entries of C
c1 = c0 − 1, 1
λc0 +
bλ/ 2c
∑
i=1
n2,i c2,i = λ2(c0 − 1) .
where c0, c1 and c2,i are defined similarly to R(z), R(z)1 and R(z)2,i respec-tively. These recursion relations simplify further on noting that c0 = 0:
Lemma C.1. Let G be an infinite vertex transitive graph with bounded degree
λ and resolvent R(z). Let Pλ be the projector onto the eigenspace associated with eigenvalue λ and let CG := lim 1/λ −
(I − Pλ
)R(z). Then all diagonal entries c0 of CG are 0. Proof. Since G is regular ( I − Pλ
)
ij
= (1 − 1/N )δi,j + (1 /N )δi6 =j . Then
((I − Pλ
)R(z)
)
kk
=
(
1 − 1
N
)
R(z)k,k + 1
N
∑
j6=k
R(z)j,k = 0 .
42 Appendix D All orders of the determinant expansion
The expansion of determinants such as det( M1 + M2) around det( M1)is well known. Here we report the slight variant valid when M1 is singular. Indeed in the case of interest here M1 ≡ I − zAG which is singular in z = 1 /λ .Taking into consideration that the adjugate matrix of a singular matrix with a kernel of dimension 1 is the projector Pλ onto that kernel times a constant which, for us, is asymptotically αN , N 1, the kth order of the expansion of det( I − λ−1AG\p) for an arbitrary SAP p is: Order (k) = αN
λk
∑
π∈Π( k)
(−1) |π|
k
∑
j=1
{
1(πj − 1)! jπj −1 Tr
[( − zPλBp
)( − zCB p
)j−1]
× Tr
[( − zCG|pBp
)j ]πj −1 k∏
m=1
m6=j
Tr [( − zCG|pBp
)m]πm 1
m! mπm
}
.
In this expression, Π( k) designates the set of partitions π := {πi}1≤i≤k of
k such that ∑
i
iπ i = k and |π| = ∑
i
πi. The matrices CG|p and Bp are as defined in the Infinite Sieve Theorem.
Appendix E Error terms in the Infinite Sieve Theorem
The Infinite Sieve Theorem establishes the dominant term in the asymp-totic expansion of the fraction of closed walks whose unique right prime factor is a given SAP p. This term is dominant only under the limit ` → ∞ , where the closed walks under consideration have diverging length. In this section we evaluate the accompanying error terms which come into consideration for finite length walks. Concretely, these error terms cannot be neglected when this finite length is ‘too close’ to that of the SAP p, in a sense which will be made precise below. The Infinite Sieve Theorem produces the asymptotic expansion of the ex-tension of Viennot’s Lemma to infinite graphs which gives, for the generating function Rp(z) of walks whose last erased loop is the SAP p,
Rp(z) = z`(p) det (I + zR(z)Bp
).
The various divergences encountered in the asymptotic expansion were specif-ically avoided with the following form,
Rp(z) = R(z)
( z`(p)
R(z) det (I + zR(z)Bp
))
. (17) 43 This presentation of Rp(z) is also supported by its combinatorial meaning: with Lemma 5.4 the first term relates to the density of walks amongst the hikes, while the terms in parentheses give rise to the finite asymptotic fraction of walks whose last erased loop is p among all closed walks. As we have seen, we should not expect any of the generating functions appearing here to be meromorphic in general. For this reason, we should only use properties of products of generating functions rather than ratios. One such standard property is recalled below:
Lemma E.1. Let F (z) = ∑
`≥0
λf ()zwith λ real and f () a function such that lim →∞ f () exists and is finite. Then for any function G(z) of z such that G(1 /λ ) is finite,
(F (z)G(z))[L] =
(
G(1 /λ ) + ∑
k> 0
∇kf
λk k! f (L)G(k)(1 /λ )
)
× F (z)[ L],
where G(k)(1 /λ ) stands for the kth derivative of G(z) evaluated in 1/λ .
Returning to Eq. (17), we introduce the walk correction function fw(`) :=
R(z/λ ), so that the number of closed walks of length on G is exactly λfw(`). On the square lattice, we have
fw(2 `) =
(2``
)2
4−2` ∼ 1
π for 1,
while fw(2 + 1) = 0, and lim→∞ fw() = 0. In general, on a d dimensional lattice fw() is on the order of −d/ 2 for 1. We may now use Lemma E.1 on Eq. (17). Together with the Infinite Sieve Theorem, this gives [z`]Rp(z) =
(
Fp
λ`(p) +
∞
∑
k≥1
∇kfw
fw(`)λk k! Frac (k)
p
(1 /λ )
)
[z`]R(z),
with Frac p(z) := Rp(z)/R (z) and Fp as per Eq. (1). Let us denote, Err p(2 `) :=
∞
∑
k≥1
∇kfw
fw(2 `)λk k! Frac (k)
p
(1 /λ ),
and observe that since Frac (k)
p
(1 /λ ) does not depend on , we may now esti-mate the decay of the correction terms with respect to the dominant contri-bution Fp/λ(p) as → ∞ . Taking fw() ∼ `−d/ 2 as a guide on a d dimensional lattice, two situations arise: 44 Length L 2 4 6 8 10 12 14
S(L) 0.5 0.6473 0.7093 0.7493 0.7774 0.7984 0.8149
Length L 16 18 20 22 24 26 28
S(L) 0.8282 0.8392 0.8485 0.8565 0.8635 0.8696 0.8751
Length L 30 32
S(L) 0.8799 0.8843
Table 1: Table of value for the sum S(L) as a function of L. The asterisk ∗indicates that all calculations for this length were analytical.
i) If 1 and − (p) 1, then the error terms decrease proportionally to ( − (p))−1 on all d-dimensional vertex transitive lattices; ii) If 1 but −(p) is on the order of 1, then the error terms actually increase with , asd/ 2.This analysis indicates that [ z]R(z) × Fp/λ(p) is a good approximation to [z]Rp(z) when is much larger than (p). This confirms the discussion of §6: using Fp/λ(p) to obtain an estimate π(L) via Eq. (14) is misguided precisely because most of the SAPs fall in situation ii), where using Fp/λ (p) to estimate [z]Rp(z) is outright wrong.
Appendix F Data table
Table 1 gives all computed values of
S(L) := ∑
p: SAP
`(p)≤L
Fp
4`(p)
as a function of L and evaluated on the square lattice. Although most tabu-lated values of S(L) were computed analytically , we here report only numer-ical results rounded at 10 −4 owing to length concerns. At length 32, S(L)requires computing Fp values for 3 , 484 , 564 , 613 self-avoiding polygons.
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6942 | https://people.math.ethz.ch/~sudakovb/grid-ramsey-talk.pdf | Grid Ramsey problem and related questions Benny Sudakov, ETH joint with D. Conlon, J. Fox and C. Lee Hales-Jewett Theorem Definition: Let [m] = {1, 2, . . . , m}, a ∈[m]n and let S be a non-empty set of coordinates.
A combinatorial line is aS(1), aS(2), . . . , aS(m), where aS(t) is a vector b ∈[m]n such that bi = t, i ∈S and bi = ai, i ̸∈S.
Hales-Jewett Theorem Definition: Let [m] = {1, 2, . . . , m}, a ∈[m]n and let S be a non-empty set of coordinates.
A combinatorial line is aS(1), aS(2), . . . , aS(m), where aS(t) is a vector b ∈[m]n such that bi = t, i ∈S and bi = ai, i ̸∈S.
Example: m = 3, n = 6 1 3 1 2 1 3 2 3 2 2 2 3 3 3 3 2 3 3 Hales-Jewett Theorem Definition: Let [m] = {1, 2, . . . , m}, a ∈[m]n and let S be a non-empty set of coordinates.
A combinatorial line is aS(1), aS(2), . . . , aS(m), where aS(t) is a vector b ∈[m]n such that bi = t, i ∈S and bi = ai, i ̸∈S.
Example: m = 3, n = 6 1 3 1 2 1 3 2 3 2 2 2 3 3 3 3 2 3 3 Theorem: (Hales-Jewett 1963) For every m, r and sufficiently large n, every r-coloring of [m]n contains a monochromatic line.
Hales-Jewett Theorem Theorem: (Hales-Jewett 1963) For every m, r and sufficiently large n, every r-coloring of [m]n contains a monochromatic line.
Informally, if the cells of a n-dimensional m × m × · · · × m cube are colored with r colors, there must be one row, column, or certain diagonal all of whose cells are the same color, i.e., the multi-player tic-tac-toe game cannot end in a draw if the board has high dimesion.
Easy corollary Theorem: (van der Waerden 1927) For every r, m and sufficiently large N, every r-coloring of [N] contains a monochromatic arithmetic progression of length m.
Easy corollary Theorem: (van der Waerden 1927) For every r, m and sufficiently large N, every r-coloring of [N] contains a monochromatic arithmetic progression of length m.
Proof. Consider mapping f from [m]n into [N], f (a1, . . . , an) = n X i=1 aimi−1.
Color every a ∈[m]n by the color of f (a). Then a monochromatic line in this coloring gives a monochromatic arithmetic progression of length m in the original coloring of [N].
□ Easy corollary Definition: A set U ⊂Zd is a homothetic copy of V ⊂Zd iff U = u + λV for some vector u ∈Zd and integer λ.
Easy corollary Definition: A set U ⊂Zd is a homothetic copy of V ⊂Zd iff U = u + λV for some vector u ∈Zd and integer λ.
Theorem: (Gallai-Witt 1943, 1951) For all r and V ⊂Zd, every r-coloring of Zd contains a monochromatic homothetic copy of V .
Easy corollary Definition: A set U ⊂Zd is a homothetic copy of V ⊂Zd iff U = u + λV for some vector u ∈Zd and integer λ.
Theorem: (Gallai-Witt 1943, 1951) For all r and V ⊂Zd, every r-coloring of Zd contains a monochromatic homothetic copy of V .
Proof. Let V = {v1, . . . , vm}. Map [m]n into Zd, f (a1, . . . , an) = n X i=1 vai.
Color every a ∈[m]n by the color of f (a). Then a monochromatic line in this coloring gives a homothetic copy of V in the original coloring of Zd.
□ Hales-Jewett numbers Definition: The Hales-Jewett number HJ(r, m) is the minimum n such that every r-coloring of [m]n contains a monochromatic line.
Hales-Jewett numbers Definition: The Hales-Jewett number HJ(r, m) is the minimum n such that every r-coloring of [m]n contains a monochromatic line.
Theorem: (Shelah 1988) HJ(r, m) is at most iterated tower function, i.e., primitive recursive.
Hales-Jewett numbers Definition: The Hales-Jewett number HJ(r, m) is the minimum n such that every r-coloring of [m]n contains a monochromatic line.
Theorem: (Shelah 1988) HJ(r, m) is at most iterated tower function, i.e., primitive recursive.
Remarks: Greatly improves the original Ackermann type bound.
Main step in the proof is the “Grid-type lemma”, which reduces the size of the alphabet from m to m −1.
Grid graph Definition: A Grid graph Γm,n is a graph on the set of vertices [m] × [n] such that (i, j) is adjacent to (i′, j′) iffi = i′ or j = j′.
Grid graph Definition: A Grid graph Γm,n is a graph on the set of vertices [m] × [n] such that (i, j) is adjacent to (i′, j′) iffi = i′ or j = j′.
We call the ith row the set of vertices {i} × [n] and [m] × {j} is called the jth column.
Rows, columns of Γm,n are complete graphs Kn, Km respectively.
Grid graph A Grid graph Γm,n is a graph on the set of vertices [m] × [n] such that (i, j) is adjacent to (i′, j′) iffi = i′ or j = j′.
Definition: A rectangle in an edge-colored Γm,n is alternating if vertical/horizontal pairs of edges have the same color.
Coloring is alternating-free if it has no such rectangle.
(i, j) (i, j') (i', j) (i', j') Grid graph A Grid graph Γm,n is a graph on the set of vertices [m] × [n] such that (i, j) is adjacent to (i′, j′) iffi = i′ or j = j′.
Definition: A rectangle in an edge-colored Γm,n is alternating if vertical/horizontal pairs of edges have the same color.
Coloring is alternating-free if it has no such rectangle.
(i, j) (i, j') (i', j) (i', j') Grid Ramsey function: G(r) is the minimum integer n such that every r-edge coloring of Γn,n contains an alternating rectangle.
Grid lemma Lemma: (Shelah 1988) G(r) ≤r(r+1 2 ) + 1.
Grid lemma Lemma: (Shelah 1988) G(r) ≤r(r+1 2 ) + 1. j j' i i' Proof. Let n = r(r+1 2 ) +1 and consider an r-edge coloring of Γr+1,n. Recall that every column is a complete graph Kr+1 and thus there at most r(r+1 2 ) ways to r-color its edges.
Grid lemma Lemma: (Shelah 1988) G(r) ≤r(r+1 2 ) + 1. j j' i i' Proof. Let n = r(r+1 2 ) +1 and consider an r-edge coloring of Γr+1,n. Recall that every column is a complete graph Kr+1 and thus there at most r(r+1 2 ) ways to r-color its edges.
Since n > r(r+1 2 ) there are two columns j, j′, whose edges are identically colored.
There are r + 1 edges of the grid graph between vertices in these columns. Since there are only r colors, two of these edges (say in rows i and i′) have the same color. Then {i, i′}×{j, j′} is an alternating rectangle.
□ Grid lemma Lemma: (Shelah 1988) G(r) ≤r(r+1 2 ) + 1.
Grid lemma Lemma: (Shelah 1988) G(r) ≤r(r+1 2 ) + 1.
Remarks: A very similar (more general) grid-type lemma is a key step in Shelah’s proof.
This simple bound is difficult to improve. The only known improvement is by an additive lower-order term (Gy´ arf´ as).
Best lower bound has order r3 (Faudree-Gy´ arf´ as-Sz¨ onyi and Heinrich) Grid lemma Lemma: (Shelah 1988) G(r) ≤r(r+1 2 ) + 1.
Remarks: A very similar (more general) grid-type lemma is a key step in Shelah’s proof.
This simple bound is difficult to improve. The only known improvement is by an additive lower-order term (Gy´ arf´ as).
Best lower bound has order r3 (Faudree-Gy´ arf´ as-Sz¨ onyi and Heinrich) Question: (Graham, Rothschild and Spencer 1990) Is the function G(r) polynomial in r?
Some new results Question: (Graham, Rothschild and Spencer 1990) Is the function G(r) polynomial in r?
Some new results Question: (Graham, Rothschild and Spencer 1990) Is the function G(r) polynomial in r?
Theorem: (Conlon, Fox, Lee and S. 2014+) The grid Ramsey function is not polynomial. Moreover, G(r) ≥rlog3/2−o(1) r.
Some new results Question: (Graham, Rothschild and Spencer 1990) Is the function G(r) polynomial in r?
Theorem: (Conlon, Fox, Lee and S. 2014+) The grid Ramsey function is not polynomial. Moreover, G(r) ≥rlog3/2−o(1) r.
There exist an alternating-free r-edge coloring of Γm,n with m = cr2 and n = r( 1 2 −o(1))r2.
Some new results Question: (Graham, Rothschild and Spencer 1990) Is the function G(r) polynomial in r?
Theorem: (Conlon, Fox, Lee and S. 2014+) The grid Ramsey function is not polynomial. Moreover, G(r) ≥rlog3/2−o(1) r.
There exist an alternating-free r-edge coloring of Γm,n with m = cr2 and n = r( 1 2 −o(1))r2.
Remark: The second result gives some evidence why it is hard to improve the upper bound on G(r).
Key observation Definition: For two edge colorings c1, c2 of Kn, let Gc1,c2 be the subgraph of Kn containing all the edges e with c1(e) = c2(e).
Key observation Definition: For two edge colorings c1, c2 of Kn, let Gc1,c2 be the subgraph of Kn containing all the edges e with c1(e) = c2(e).
Lemma: There is an alternating-free r-edge coloring of Γm,n iffthere are r-edge colorings c1, . . . , cm of the complete graph Kn with χ(Gci,cj) ≤r, for all i ̸= j.
Key observation Lemma: There is an alternating-free r-edge coloring of Γm,n if there are r-edge colorings c1, . . . , cm of the complete graph Kn with χ(Gci,ci′) ≤r, for all i ̸= i′.
Key observation Lemma: There is an alternating-free r-edge coloring of Γm,n if there are r-edge colorings c1, . . . , cm of the complete graph Kn with χ(Gci,ci′) ≤r, for all i ̸= i′.
Proof. Color the edges of the ith row using ci.
Key observation Lemma: There is an alternating-free r-edge coloring of Γm,n if there are r-edge colorings c1, . . . , cm of the complete graph Kn with χ(Gci,ci′) ≤r, for all i ̸= i′.
Proof. Color the edges of the ith row using ci.
Color vertical edge (i, j) ∼(i′, j) between rows i, i′ with the color of the vertex j in the r-coloring of Gci,ci′.
Key observation Lemma: There is an alternating-free r-edge coloring of Γm,n if there are r-edge colorings c1, . . . , cm of the complete graph Kn with χ(Gci,ci′) ≤r, for all i ̸= i′.
Proof. Color the edges of the ith row using ci.
Color vertical edge (i, j) ∼(i′, j) between rows i, i′ with the color of the vertex j in the r-coloring of Gci,ci′.
By definition, if horizontal edges (i, j) ∼(i, j′) and (i′, j) ∼(i′, j′) have the same color, then the vertices j, j′ are adjacent in Gci,ci′ and have distinct colors.
Key observation Lemma: There is an alternating-free r-edge coloring of Γm,n if there are r-edge colorings c1, . . . , cm of the complete graph Kn with χ(Gci,ci′) ≤r, for all i ̸= i′.
Proof. Color the edges of the ith row using ci.
Color vertical edge (i, j) ∼(i′, j) between rows i, i′ with the color of the vertex j in the r-coloring of Gci,ci′.
By definition, if horizontal edges (i, j) ∼(i, j′) and (i′, j) ∼(i′, j′) have the same color, then the vertices j, j′ are adjacent in Gci,ci′ and have distinct colors.
Hence, the vertical edges (i, j) ∼(i′, j) and (i, j′) ∼(i′, j′) have distinct colors as well.
Key observation Lemma: There is an alternating-free r-edge coloring of Γm,n if there are r-edge colorings c1, . . . , cm of the complete graph Kn with χ(Gci,ci′) ≤r, for all i ̸= i′.
Proof. Color the edges of the ith row using ci.
Color vertical edge (i, j) ∼(i′, j) between rows i, i′ with the color of the vertex j in the r-coloring of Gci,ci′.
By definition, if horizontal edges (i, j) ∼(i, j′) and (i′, j) ∼(i′, j′) have the same color, then the vertices j, j′ are adjacent in Gci,ci′ and have distinct colors.
Hence, the vertical edges (i, j) ∼(i′, j) and (i, j′) ∼(i′, j′) have distinct colors as well.
This shows that there are no alternating rectangles.
□ Proof recipe Choose a partition E(Kn) = E1 ∪· · · ∪Et such that any union of “few ” parts has “small ” chromatic number.
Generate ci by assigning to every part Ej randomly one of r colors.
Any two such colorings will agree only on small number of parts, i.e., chromatic number of Gci,ci′ will be small.
An edge partition of Kn Identify the vertex set [n] with binary strings of length t = log n, i.e., x = (x1, . . . , xt) if x −1 = P xi2i−1.
An edge partition of Kn Identify the vertex set [n] with binary strings of length t = log n, i.e., x = (x1, . . . , xt) if x −1 = P xi2i−1.
Definition: An edge partition E1 ∪· · · ∪Et of the complete graph Kn is obtained by taking Ei to be all the edges (x, y), such that i is the minimum index for which xi ̸= yi.
An edge partition of Kn Identify the vertex set [n] with binary strings of length t = log n, i.e., x = (x1, . . . , xt) if x −1 = P xi2i−1.
Definition: An edge partition E1 ∪· · · ∪Et of the complete graph Kn is obtained by taking Ei to be all the edges (x, y), such that i is the minimum index for which xi ̸= yi.
Easy observation: The union of any s parts in this partition have chromatic number at most 2s.
An edge partition of Kn Identify the vertex set [n] with binary strings of length t = log n, i.e., x = (x1, . . . , xt) if x −1 = P xi2i−1.
Definition: An edge partition E1 ∪· · · ∪Et of the complete graph Kn is obtained by taking Ei to be all the edges (x, y), such that i is the minimum index for which xi ̸= yi.
Easy observation: The union of any s parts in this partition have chromatic number at most 2s.
Indeed, every Ei is a bipartite graph with parts containing all x, xi = 0 and all y, yi = 1. Therefore χ(Ei) = 2.
Since χ(H ∪H′) ≤χ(H) · χ(H′) for any pair of graphs on the same vertex set, the claim follows.
Superpolynomial bound Theorem: (Conlon, Fox, Lee and S.) G(r) ≥r0.4 log r Superpolynomial bound Theorem: (Conlon, Fox, Lee and S.) G(r) ≥r0.4 log r Proof. Let n = r0.4 log r and E1 ∪· · · ∪Et be the edge partition of Kn with union of any s parts having chromatic number ≤2s.
Superpolynomial bound Theorem: (Conlon, Fox, Lee and S.) G(r) ≥r0.4 log r Proof. Let n = r0.4 log r and E1 ∪· · · ∪Et be the edge partition of Kn with union of any s parts having chromatic number ≤2s.
Form ci, i = 1, . . . , n by coloring every part Ej randomly and independently in one of r colors.
Superpolynomial bound Theorem: (Conlon, Fox, Lee and S.) G(r) ≥r0.4 log r Proof. Let n = r0.4 log r and E1 ∪· · · ∪Et be the edge partition of Kn with union of any s parts having chromatic number ≤2s.
Form ci, i = 1, . . . , n by coloring every part Ej randomly and independently in one of r colors.
Note, χ(Gci,ci′) > r only if ci, ci′ agree on at least log(r + 1) parts.
Superpolynomial bound Theorem: (Conlon, Fox, Lee and S.) G(r) ≥r0.4 log r Proof. Let n = r0.4 log r and E1 ∪· · · ∪Et be the edge partition of Kn with union of any s parts having chromatic number ≤2s.
Form ci, i = 1, . . . , n by coloring every part Ej randomly and independently in one of r colors.
Note, χ(Gci,ci′) > r only if ci, ci′ agree on at least log(r + 1) parts.
Since t = log n = 0.4 log2 r, this event has probability P ≤ t log(r + 1) r−log(r+1) = r−(1+o(1)) log r ≪1/n2.
Superpolynomial bound Theorem: (Conlon, Fox, Lee and S.) G(r) ≥r0.4 log r Proof. Let n = r0.4 log r and E1 ∪· · · ∪Et be the edge partition of Kn with union of any s parts having chromatic number ≤2s.
Form ci, i = 1, . . . , n by coloring every part Ej randomly and independently in one of r colors.
Note, χ(Gci,ci′) > r only if ci, ci′ agree on at least log(r + 1) parts.
Since t = log n = 0.4 log2 r, this event has probability P ≤ t log(r + 1) r−log(r+1) = r−(1+o(1)) log r ≪1/n2.
Thus, with high probability χ(Gci,ci′) ≤r for all i ̸= i′, which gives alternating-free r-edge coloring of Γn,n.
□ Ramsey-type problem Definition: A (p, q)-coloring of the complete k-uniform hypergraph K (k) n is an edge-coloring in which every copy of K (k) p receives at least q colors.
Ramsey-type problem Definition: A (p, q)-coloring of the complete k-uniform hypergraph K (k) n is an edge-coloring in which every copy of K (k) p receives at least q colors.
Remarks: (p, q)-colorings were introduced by Erd˝ os-Shelah in 1975 and then were systematically studied by Erd˝ os-Gy´ arf´ as in the 90s.
The case q = 2 is a classical Ramsey coloring with no monochromatic p-clique.
Ramsey-type problem Definition: A (p, q)-coloring of the complete k-uniform hypergraph K (k) n is an edge-coloring in which every copy of K (k) p receives at least q colors.
Remarks: (p, q)-colorings were introduced by Erd˝ os-Shelah in 1975 and then were systematically studied by Erd˝ os-Gy´ arf´ as in the 90s.
The case q = 2 is a classical Ramsey coloring with no monochromatic p-clique.
Claim: Every (4, 3)-coloring of K (3) 2n which uses r colors gives an alternating-free r-coloring of Γn,n.
Ramsey-type problem Claim: Every (4, 3)-coloring of K (3) 2n which uses r colors gives an alternating-free r-coloring of Γn,n.
Ramsey-type problem Claim: Every (4, 3)-coloring of K (3) 2n which uses r colors gives an alternating-free r-coloring of Γn,n.
Proof. Let A ∪B be an arbitrary partition of the vertices into two equal parts, which we regard as two copies of [n].
Ramsey-type problem Claim: Every (4, 3)-coloring of K (3) 2n which uses r colors gives an alternating-free r-coloring of Γn,n.
Proof. Let A ∪B be an arbitrary partition of the vertices into two equal parts, which we regard as two copies of [n].
Color the vertical edge (i, j) ∼(i′, j) by the color of {j, i, i′} with j ∈A and i, i′ ∈B, color the horizontal edge (i, j) ∼(i, j′) of the grid by the color of the triple {j, j′, i} with j, j′ ∈A and i ∈B.
A B Ramsey-type problem Claim: Every (4, 3)-coloring of K (3) 2n which uses r colors gives an alternating-free r-coloring of Γn,n.
Proof. Let A ∪B be an arbitrary partition of the vertices into two equal parts, which we regard as two copies of [n].
Color the vertical edge (i, j) ∼(i′, j) by the color of {j, i, i′} with j ∈A and i, i′ ∈B, color the horizontal edge (i, j) ∼(i, j′) of the grid by the color of the triple {j, j′, i} with j, j′ ∈A and i ∈B.
Every alternating rectangle in the grid gives a copy of K (3) 4 with only two colors.
□ A B Ramsey-type problem Claim: Every (4, 3)-coloring of K (3) 2n which uses r colors gives an alternating-free r-coloring of Γn,n.
Remarks: Converse statement is also true. By amplifying “slightly ” the number of colors one can construct from an alternating-free coloring of grid a (4, 3)-coloring of a complete 3-uniform hypergraph.
Ramsey-type problem Definition: F(r, p, q) = min n such that every r-edge-coloring of a complete graph Kn contains a Kp with at most q −1 colors on its edges.
Ramsey-type problem Definition: F(r, p, q) = min n such that every r-edge-coloring of a complete graph Kn contains a Kp with at most q −1 colors on its edges.
Note that F(r, p, 2) is the usual r-colored Ramsey number of Kp. Thus F(r, p, 2) ≥2r is exponential in r.
Ramsey-type problem Definition: F(r, p, q) = min n such that every r-edge-coloring of a complete graph Kn contains a Kp with at most q −1 colors on its edges.
Note that F(r, p, 2) is the usual r-colored Ramsey number of Kp. Thus F(r, p, 2) ≥2r is exponential in r.
Erd˝ os-Gy´ arf´ as noticed that F(r, 9, 34) is connected to the celebrated Ruzsa-Szemer´ edi’s (6, 3)-theorem.
Ramsey-type problem Definition: F(r, p, q) = min n such that every r-edge-coloring of a complete graph Kn contains a Kp with at most q −1 colors on its edges.
Note that F(r, p, 2) is the usual r-colored Ramsey number of Kp. Thus F(r, p, 2) ≥2r is exponential in r.
Erd˝ os-Gy´ arf´ as noticed that F(r, 9, 34) is connected to the celebrated Ruzsa-Szemer´ edi’s (6, 3)-theorem.
If n = r + 2 then at every vertex there are 2 edges of the same color, giving K3 with at most 2 colors. So F(r, 3, 3) ≤r + 2.
Ramsey-type problem Definition: F(r, p, q) = min n such that every r-edge-coloring of a complete graph Kn contains a Kp with at most q −1 colors on its edges.
Note that F(r, p, 2) is the usual r-colored Ramsey number of Kp. Thus F(r, p, 2) ≥2r is exponential in r.
Erd˝ os-Gy´ arf´ as noticed that F(r, 9, 34) is connected to the celebrated Ruzsa-Szemer´ edi’s (6, 3)-theorem.
If n = r + 2 then at every vertex there are 2 edges of the same color, giving K3 with at most 2 colors. So F(r, 3, 3) ≤r + 2.
For p > 3, F(r, p, p 2 ) ≈ √ 2r, since all the edge-colors in such coloring must be distinct.
Ramsey-type problem Definition: F(r, p, q) = min n such that every r-edge-coloring of a complete graph Kn contains a Kp with at most q −1 colors on its edges.
Note that F(r, p, 2) is the usual r-colored Ramsey number of Kp. Thus F(r, p, 2) ≥2r is exponential in r.
If n = r + 2 then at every vertex there are 2 edges of the same color, so F(r, 3, 3) ≤r + 2. For p > 3, F(r, p, p 2 ) ≈ √ 2r, since all the edge-colors in such coloring must be distinct.
Ramsey-type problem Definition: F(r, p, q) = min n such that every r-edge-coloring of a complete graph Kn contains a Kp with at most q −1 colors on its edges.
Note that F(r, p, 2) is the usual r-colored Ramsey number of Kp. Thus F(r, p, 2) ≥2r is exponential in r.
If n = r + 2 then at every vertex there are 2 edges of the same color, so F(r, 3, 3) ≤r + 2. For p > 3, F(r, p, p 2 ) ≈ √ 2r, since all the edge-colors in such coloring must be distinct.
Question: (Erd˝ os-Gy´ arf´ as 90s) As q varies from 2 to p 2 , when F(r, p, q) becomes polynomial?
Polynomial growth Proposition: (Erd˝ os-Gy´ arf´ as) F(r, p, p) = O rp−2 .
Polynomial growth Proposition: (Erd˝ os-Gy´ arf´ as) F(r, p, p) = O rp−2 .
Proof. By induction on p. We already have that F(r, 3, 3) ≤r + 2.
Polynomial growth Proposition: (Erd˝ os-Gy´ arf´ as) F(r, p, p) = O rp−2 .
Proof. By induction on p. We already have that F(r, 3, 3) ≤r + 2.
Suppose we have an r-edge-coloring of Kn such that every p-clique has at least p colors. Fix a vertex v.
Polynomial growth Proposition: (Erd˝ os-Gy´ arf´ as) F(r, p, p) = O rp−2 .
Proof. By induction on p. We already have that F(r, 3, 3) ≤r + 2.
Suppose we have an r-edge-coloring of Kn such that every p-clique has at least p colors. Fix a vertex v.
There is a color that appears on at least n−1 r edges incident to v, so let X be the set of vertices connected to v by these edges.
Polynomial growth Proposition: (Erd˝ os-Gy´ arf´ as) F(r, p, p) = O rp−2 .
Proof. By induction on p. We already have that F(r, 3, 3) ≤r + 2.
Suppose we have an r-edge-coloring of Kn such that every p-clique has at least p colors. Fix a vertex v.
There is a color that appears on at least n−1 r edges incident to v, so let X be the set of vertices connected to v by these edges.
Every (p −1)-set S in X has at least p −1 colors, otherwise S ∪{v} will have fewer than p colors, contradiction.
Polynomial growth Proposition: (Erd˝ os-Gy´ arf´ as) F(r, p, p) = O rp−2 .
Proof. By induction on p. We already have that F(r, 3, 3) ≤r + 2.
Suppose we have an r-edge-coloring of Kn such that every p-clique has at least p colors. Fix a vertex v.
There is a color that appears on at least n−1 r edges incident to v, so let X be the set of vertices connected to v by these edges.
Every (p −1)-set S in X has at least p −1 colors, otherwise S ∪{v} will have fewer than p colors, contradiction.
Hence |X| = O(rp−3) and therefore n = O(r|X|) = O(rp−2).
□ Erd˝ os-Gy´ arf´ as conjecture Question: (Erd˝ os-Gy´ arf´ as 90s) What is the minimum q such that F(r, p, q) becomes polynomial in r? Is it q = p?
Erd˝ os-Gy´ arf´ as conjecture Question: (Erd˝ os-Gy´ arf´ as 90s) What is the minimum q such that F(r, p, q) becomes polynomial in r? Is it q = p?
Remark: We saw that F(r, 3, 2) ≥2r, so this holds for p = 3. It was established also for p = 4, 5 by Mubayi 1998 and Eichhorn-Mubayi 2000.
Erd˝ os-Gy´ arf´ as conjecture Question: (Erd˝ os-Gy´ arf´ as 90s) What is the minimum q such that F(r, p, q) becomes polynomial in r? Is it q = p?
Remark: We saw that F(r, 3, 2) ≥2r, so this holds for p = 3. It was established also for p = 4, 5 by Mubayi 1998 and Eichhorn-Mubayi 2000.
Theorem: (Conlon, Fox, Lee and S. 2014+) For all p ≥4, F(r, p, p −1) ≥rc log 1 p−3 r.
Concluding remarks and open problem Question: (Conlon, Fox, Lee and S.) Is there an r-edge coloring of Kn such that the union of any q colors has chromatic number at most p?
Concluding remarks and open problem Question: (Conlon, Fox, Lee and S.) Is there an r-edge coloring of Kn such that the union of any q colors has chromatic number at most p?
Remark: Since χ(H ∪H′) ≤χ(H) · χ(H′) for any pair of graphs on the same vertex set, one must have pr/q ≥n.
Can this be achieved?
Concluding remarks and open problem Question: (Conlon, Fox, Lee and S.) Is there an r-edge coloring of Kn such that the union of any q colors has chromatic number at most p?
Remark: Since χ(H ∪H′) ≤χ(H) · χ(H′) for any pair of graphs on the same vertex set, one must have pr/q ≥n.
Can this be achieved?
Note that, for r = log n and p = 2q such coloring exists.
Concluding remarks and open problem Question: (Conlon, Fox, Lee and S.) Is there an r-edge coloring of Kn such that the union of any q colors has chromatic number at most p?
Remark: Since χ(H ∪H′) ≤χ(H) · χ(H′) for any pair of graphs on the same vertex set, one must have pr/q ≥n.
Can this be achieved?
Note that, for r = log n and p = 2q such coloring exists.
Question: What happens if we want the union of q colors to have chromatic number ≪2q?
Concluding remarks and open problem Proposition: (Conlon, Fox, Lee and S.) There is an edge-coloring of Kn with r = 23√log n colors in which the union of any q colors has chromatic number at most 23√q log q.
Concluding remarks and open problem Proposition: (Conlon, Fox, Lee and S.) There is an edge-coloring of Kn with r = 23√log n colors in which the union of any q colors has chromatic number at most 23√q log q.
Remark: This result was used to prove that G(r) ≥rlog3/2−o(1) r.
Concluding remarks and open problem Proposition: (Conlon, Fox, Lee and S.) There is an edge-coloring of Kn with r = 23√log n colors in which the union of any q colors has chromatic number at most 23√q log q.
Remark: This result was used to prove that G(r) ≥rlog3/2−o(1) r.
Question: (Conlon, Fox, Lee and S.) What is the maximum n = n(r) such that there is an r-edge coloring of Kn in which union of every 2 colors has chromatic number at most 3? |
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The Method of Coordinates
byGelfand, I. M.
ISBN13: 9780817635336
ISBN10: 0817635335
Format: Paperback
Pub. Date: 1991-02-01
Publisher(s): Birkhauser
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"All through both volumes [Functions & Graphs and The Methods of Coordinates] , one finds a careful description of the step-by-step thinking process that leads up to the correct definition of a concept or to an argument that clinches in the proof of a theorem. We are ... very fortunate that an account of this caliber has finally made it to printed pages... Anyone who has taken this guided tour will never be intimidated by n ever again... High school students (or teachers) reading through these two books would learn an enormous amount of good mathematics. More importantly, they would also get a glimpse of how mathematics is done." -- H. Wu, The Mathematical Intelligencer The need for improved mathematics education at the high school and college levels has never been more apparent than in the 1990's. As early as the 1960's, I.M. Gelfand and his colleagues in the USSR thought hard about this same question and developed a style for presenting basic mathematics in a clear and simple form that engaged the curiosity and intellectual interest of thousands of high school and college students. These same ideas, this development, are available in the following books to any student who is willing to read, to be stimulated, and to learn. The Method of Coordinates is a way of transferring geometric images into formulas, a method for describing pictures by numbers and letters denoting constants and variables. It is fundamental to the study of calculus and other mathematical topics. Teachers of mathematics will find here a fresh understanding of the subject and a valuable path to the training of students in mathematical concepts and skills. Contents Preface Foreword Introduction PART I Chapter 1 The Coordinates of Points on a Line 1. The Number Axis 2. The Absolute Value of Number 3. The Distance Between Two Points Chapter 2 The Coordinates of Points in the Plane 4. The Coordinate Plane 5. Relations Connecting Coordinates 6. The Distance Between Two Points 7. Defining Figures 8. We Begin to Solve Problems 9. Other Systems of Coordinates Chapter 3 The Coordinates of a Point in Space 10. Coordinate Axes and Planes 11. Defining Figures in Space PART II Chapter 1 Introduction 1. Some General Considerations 2. Geometry as an Aid in Calculation 3. The Need for Introducing Four-Dimensional Space 4. The Peculiarities of Four-Dimensional Space 5. Some Physics Chapter 2 Four-Dimensional Space 6. Coordinate Axes and Planes 7. Some Problems Chapter 3 The Four-Dimensional Cube 8. The Definition of the Sphere and the Cube 9. The Structure of the Four-Dimensional Cube 10. Problems on the Cube
Table of Contents
Preface
Foreword
Introduction
Part I
The Coordinates of Points on a Line
The Coordinates of Points in the Plane
The Coordinates of a Point in Space
Part II
Introduction
Four-Dimensional Space
The Four Dimensional Cube
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6944 | https://mathoverflow.net/questions/294919/how-can-a-mathematician-handle-the-pressure-to-discover-something-new | soft question - How can a mathematician handle the pressure to discover something new? - MathOverflow
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How can a mathematician handle the pressure to discover something new?
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Suppose I'm an aspiring mathematician-to-be, who started doing research. Although this is really what I love doing, I found that one disturbing point is that there's always the pressure of discovering something new. When I'm doing mathematical research, there's always the fear in the back of my mind, that maybe, I don't get new results. In the past, I could think freely about mathematics, without the pressure, but now that it's "my job", I have these problems.
How to handle this? Since this site is for mathematicians from the graduate level onwards, maybe somebody has a good suggestion. Note that, although this is a mathematics forum, I think this question is appropriate here, since it perfectly matches with the description of the "soft question"-tag.
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edited Mar 12, 2018 at 12:26
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20 maybe, I don't get new results - It's true that usually you don't, but sometimes you do. And as long as you get figure out something that is new to you, you've accomplished something.Kimball –Kimball 2018-03-11 15:25:28 +00:00 Commented Mar 11, 2018 at 15:25
36 You may want to read Emotional Perils of Mathematics (1965).Rodrigo de Azevedo –Rodrigo de Azevedo 2018-03-11 16:40:30 +00:00 Commented Mar 11, 2018 at 16:40
44 A vague cautionary note from my own life: you should always carefully evaluate the emotional origins of any precived "pressure to achieve," since it's really easy for any latent tendency toward escapism and denial, to masquerade as the perception that one must wholly devote themselves to a particular activity. For example, while I was in graduate school, my father and mother were diagnosed with aggressive forms of cancer; instead trying to cope with this situation in a healthy way, I became obsessed with trying to solve some long standing open problems. ...Not Mike –Not Mike 2018-03-11 18:21:37 +00:00 Commented Mar 11, 2018 at 18:21
41 (for the record, they are both alive and now relatively healthy.)Not Mike –Not Mike 2018-03-11 19:26:15 +00:00 Commented Mar 11, 2018 at 19:26
12 @RodrigodeAzevedo: Nice article! The pdf is available in the following link: math.uh.edu/~tomforde/Articles/Emotional-Perils.pdfSebastien Palcoux –Sebastien Palcoux 2018-03-12 07:45:36 +00:00 Commented Mar 12, 2018 at 7:45
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This is ancient history, and considering my age, I may have told this story here before. I started at Harvard graduate school in 1957, the same year that Hironaka arrived there to work with Zariski. He was already an accomplished mathematician, even if he didn’t yet have a PhD. Early that year, I must have said to him that I couldn’t imagine ever doing research, and he said, in essence, Oh, you learn about some subject, think about it in depth, and before you know it, you’re proving Theorems. I thought, This guy must be in Cloud Cuckoo Land, I’ll never do that. But of course, that’s exactly what happens.
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answered Mar 12, 2018 at 13:38
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Lubin
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24 It's a beautiful story. I first read it here: mathoverflow.net/questions/43690/… :-)Vladimir Dotsenko –Vladimir Dotsenko 2018-03-12 19:29:05 +00:00 Commented Mar 12, 2018 at 19:29
50 @VladimirDotsenko, thanks for pointing out my duplication. We geezers tell the same story over and over!Lubin –Lubin 2018-03-12 21:25:25 +00:00 Commented Mar 12, 2018 at 21:25
18 @Lubin I was very happy to see it again, no complaints whatsoever!Vladimir Dotsenko –Vladimir Dotsenko 2018-03-12 22:06:36 +00:00 Commented Mar 12, 2018 at 22:06
26 The only problem I have with this story is that it is told by someone who happened to succeed.SK19 –SK19 2018-03-19 08:00:22 +00:00 Commented Mar 19, 2018 at 8:00
2 @SK19 Responding to your comment 5 years later, but better late than ... Anyway, Lubin's point is that this seems to happen to most people in graduate school. If someone enters with a love of mathematics and follows Hironaka's advice of learning some subject and thinking about it in depth (which I take to mean, working hard), then new results seems to happen. (Also, talk to lots of people.) Of course, the results are not always what you set out to prove. My primary thesis problem was a conjecture of Lang's that's still open, but I ended up proving some special cases and some weaker versions.Joe Silverman –Joe Silverman 2023-10-21 01:11:03 +00:00 Commented Oct 21, 2023 at 1:11
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The theoretical physicist Richard Feynman was in a similar state of mind, he referred to it as a "burn-out" feeling: Now that he had landed the University professorship he had strived for, he felt the obligation to do something "important", but he had lost the joy of doing science just "for fun". In his autobiography he describes how he recovered:
Then I had another thought: Physics disgusts me a little bit now, but I used to enjoy doing physics. Why did I enjoy it? I used to play with it. I used to do whatever I felt like doing - it didn't have to do with whether it was important for the development of nuclear physics, but whether it was interesting and amusing for me to play with. So I got this new attitude. Now that I am burned out and I'll never accomplish anything, I've got this nice position at the university teaching classes which I rather enjoy, and just like I read the Arabian Nights for pleasure, I'm going to play with physics, whenever I want to, without worrying about any importance whatsoever.
Within a week I was in the cafeteria and some guy, fooling around, throws a plate in the air. As the plate went up in the air I saw it wobble, and I noticed the red medallion of Cornell on the plate going around. It was pretty obvious to me that the medallion went around faster than the wobbling. I had nothing to do, so I start figuring out the motion of the rotating plate. [...] Then I thought about how the electron orbits start to move in relativity. Then there's the Dirac equation in electrodynamics. And then quantum electrodynamics. [...] The whole business that I got the Nobel prize for came from that fiddling around with the wobbling plate.
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answered Mar 11, 2018 at 14:05
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Carlo Beenakker
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15 That's a good comment. I have myself that feeling, that I must do something better. Things that I liked I cannot do, because I have a backlog of unaccapted papers. So I must do something better, and mainstream, and the playing is away. Not because I am a professor, but because I have no job. And that's the pressure. For almost 3 years I am completely weak in maths. Just going to dead ends.hänsel –hänsel 2018-03-11 14:48:42 +00:00 Commented Mar 11, 2018 at 14:48
3 As a graduate student, it is hard to follow this advice as you are inexperienced and you just get to do what your supervisor wants you to do.user35360 –user35360 2019-05-09 01:54:55 +00:00 Commented May 9, 2019 at 1:54
2 This is very good. Unfortunately in the current academic climate, there is a high risk of ruining your career (if you don't have a permanent position yet) or having a negative impact on someone else's (PhD students and postdoc cannot "play" with you)fqq –fqq 2020-06-09 13:02:30 +00:00 Commented Jun 9, 2020 at 13:02
3 This advice is good for an established professor with tenure but perhaps not ideal for a PhD student.Hollis Williams –Hollis Williams 2022-01-31 19:38:32 +00:00 Commented Jan 31, 2022 at 19:38
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A friend of mine once told me quite brightly about mathematics that "when you search, you find". As naive as it sounds, the more the times passes, the more I believe it.
Moreover, mathematics is not only about finding. Part of the job is also made of acquiring a deeper understanding of concepts, relearning your field, sharing your knowledge and interests, and many "new" ideas come from this crossed points of view and interactive community. Take any topic, spend time into it as a whole, and you will get used enough to it to naturally understand what is missing, what could be asked, what could be tried, etc. In this spirit, "failure" (not finding as easily and what expected) is also part of the job: failing to do something is underlining what is missing, being able to share it, and this will provoke ideas, will lead you to ask specific researchers for the missing steps, and will plant seeds that will eventually blossom.
What is not easy to handle is that it takes time, and hence it requires confidence. This is an important psychological aspect of doing mathematics. When you share with other researchers, you realize that they are just as human as you, and not the apparently perfect spirits they seem behind their articles :) I believe it is necessary to mourn the will to be perfect and embrace everything, we are all doing part of the effort, this is essential to understand and believe.
Maybe could I suggest reading Netz's article on "deuteronomic texts". It is a famous work on history of mathematics that shows how much of the advances and efforts in (greek) mathematics have been made after the one who we credit with the discovery, by "digestion": treating examples, spreading the ideas, rewriting, unifying, choosing brighter notations, etc. I always thought it gives a good taste of how a community should work. And mathematicians are a community, it always worked through sharing, attempts and failures (look even at Gauss' or Serre's letters).
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edited Mar 12, 2018 at 18:11
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3 revs, 2 users 88%Desiderius Severus
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Your first sentence suggests that one should not at all worry about whether one gets new results (since as long one is working hard for a long time, one will always discover somthing new). Do I have this right?user103598 –user103598 2018-03-11 15:39:21 +00:00 Commented Mar 11, 2018 at 15:39
2 @user419308 It is a bit provocative on purpose, the sentence is not to be taken too literarily. Doing research is quite challenging (as so many other jobs). Yet, it carries some truth I believe I have given some clues for, and I emphasize also Kimball's comment. Keeping learning, trying, sharing, working is the key. What is true is: you don't have to worry about it. (Some deadlines, applications and people will make you worry, there is some pressure, but you don't have to live under pressure, after it is a personal choice)Desiderius Severus –Desiderius Severus 2018-03-11 16:03:26 +00:00 Commented Mar 11, 2018 at 16:03
64 There may be a survivorship bias here. Plenty of failed mathematicians researched but didn't find. I'm in this camp. Or maybe I just found out I'm no good at research. Your interpretations may vary.Matthew Drury –Matthew Drury 2018-03-11 19:46:31 +00:00 Commented Mar 11, 2018 at 19:46
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I consider my job of mathematician not as intended to produce new mathematics, but to explore and spread mathematics. Then exploring mathematics consists in
hearing/reading mathematics (which you can find great, great but badly written/explained, uninteresting for you, uninteresting at all, etc),
thinking about some questions without really knowing their "status"
rethinking about mathematics that are written in a way you've initially thought as confusing
solve problems, that you've formulated or read, or a mixture
fail to solve problems
[etc...! and the distinction between all this is fuzzy of course; here I put aside from this list the various mathematical extra-activities of professional mathematicians, notably in the evaluation of other's works.]
This is "personal work" and the counterpart is to communicate your work (in the large sense suggested above):
talk math with your colleagues
give talks or lectures about things you've learnt
give talks/lectures and possibly write surveys or books about things you've learnt and reprocessed
write papers with results you consider as new.
ask mathematical questions (in private discussions, in talks, in written papers, in web sites such as here...)
This is not comprehensive: for instance the above list restricts to spreading maths among somewhat specialized mathematicians; it could include spreading towards a non-specialist or non-mathematicians, or trying to develop interactions and applications of pure mathematics... the latter are important activities of a number of mathematicians and these are essential roles; not so far in my personal experience.
Nevertheless, there's a growing tendency (especially in some countries), towards a job exclusively intended in the purpose of publishing "new research" and this is indeed the most visible, and in particular, for a young mathematician, the way considered as most accurate to judge the ability to work as a mathematician. That it tends to remain, all along the carrier, the main/exclusive way a mathematician is now judged, is more of a problem.
Last and not least, the meaning of "new" (new results, new research, new ideas, new maths)... is very subjective!
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answered Mar 11, 2018 at 18:44
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YCor
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7 As someone more interested in consuming mathematics than producing it, better languages and notations have been the most valuable things to me, and have clearly had enormous and enduring impact historically. Our ability to routinely teach children arithmetic, algebra, and calculus, for example is likely primarily due to linguistic and notational improvements that make what traditionally would have been horribly complex problems relatively simple. This happens at more advanced levels with things like topos theory. I don't see this as at all in conflict with discovering new results either.Derek Elkins left SE –Derek Elkins left SE 2018-03-18 04:43:37 +00:00 Commented Mar 18, 2018 at 4:43
16 Conor McBride has an excellent quote that I wish was more representative of mathematics research: "The task of the academic is not to scale great intellectual mountains, but to flatten them." I think this provides a less anxiety-producing perspective (and a more valuable one, at least for everyone else).Derek Elkins left SE –Derek Elkins left SE 2018-03-18 04:44:01 +00:00 Commented Mar 18, 2018 at 4:44
1 In a totally different vein, another area that I see growing in prominence (including commercially) is formalizing mathematics, as in in machine checkable form. As demand for formal methods grows in software engineering, so will demand for people with experience proving things. I also don't see building (mathematical) tools (often in the form of better languages to connect to my previous comment) to deal with the large amounts of complexity involved as having no value to "pure" mathematicians, but maybe that's just the programmer in me speaking.Derek Elkins left SE –Derek Elkins left SE 2018-03-18 04:44:04 +00:00 Commented Mar 18, 2018 at 4:44
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I can share my own experience: similar considerations always disturbed me too. To deal with this, I always considered mathematics research as a hobby. From the very beginning I thought that my main job (from which I make a living) must be teaching. And I do research as a hobby in the time which remains from my teaching duties. When I was a student, I knew that if my research is successful, I will probably qualify for some university teaching job, if not, I will go to high school. In fact, after graduation, I worked for 12 years on a pure research job (without teaching duties) but I always tried to have some connection with a university (did part-time teaching job) in the hope that if my research results are not good enough I can always switch to teaching. Then I switched to a full time teaching, as a professor at a university, and feel comfortable with this job.
Actually few countries have many pure research positions in mathematics. Most mathematicians are affiliated with universities. Of course, in many universities there is a pressure to publish. But in most cases it is not so strong, after you are promoted to a permanent position, and one can write textbooks, for example, survey papers, etc. or do some other useful things, after all:-)
Additional remarks. By "teaching" I mean teaching on all levels. I know some very good mathematicians whose main contribution to mathematics is their graduate students, rather than own research results. Another activity somewhat close to teaching is writing books. There are many very respectable mathematicians whose main contribution to mathematics consists of several good books.
This point of view, that mathematics is really a hobby is shared by many professional mathematicians. For example Eugenio Calabi called math his favorite hobby, and said: “To follow your hobbies as a profession is the extraordinary luck I’ve had in my life.”
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edited Oct 20, 2023 at 16:15
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5 revsAlexandre Eremenko
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8 How can you work for 12 years on a pure research job while regarding teaching your main job from which you make your living?user103598 –user103598 2018-03-11 18:19:10 +00:00 Commented Mar 11, 2018 at 18:19
2 Also, is a job as a professor at a university really a "full time teaching" job? I know at least some countries where this isn't the case, where professors are supposed to do research.user103598 –user103598 2018-03-11 18:22:01 +00:00 Commented Mar 11, 2018 at 18:22
7 @user419308: Our wishes do not always coincide with reality.Alexandre Eremenko –Alexandre Eremenko 2018-03-12 13:25:45 +00:00 Commented Mar 12, 2018 at 13:25
5 Agreed about "to make living". Still I suspect you put more effort into research than into an occasional hike. Also, not being able to solve a problem, most likely, bites you harder than the weather not cooperating with your outdoor plans. I admit that many of us (including myself) have to live with the crispy clear realization that we are no good in what we are doing but still try to do it because we either love it, or can do nothing else, or both, so pretending that we "don't really care much" is a natural way to keep the self-esteem afloat. Still, "hobby" seems to be a wrong word. :-)fedja –fedja 2018-03-18 17:13:29 +00:00 Commented Mar 18, 2018 at 17:13
7 @fedja: as I said the number of pure research positions in this country is VERY small. Even very high class mathematicians have to teach to support themselves. In this sense, their research is a "hobby". It is true that people QUALIFY for the job mostly by their research performance, but once you obtained a permanent position, your research becomes a hobby, even if it gives by some additional income from the grants and salary increases.Alexandre Eremenko –Alexandre Eremenko 2018-03-18 17:36:53 +00:00 Commented Mar 18, 2018 at 17:36
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There’s nothing particular to mathematics with this problem - performers (actors, musicians, comedians) have a much stronger version of it and writers are in much the same position as us.
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answered Mar 11, 2018 at 15:26
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Chris Godsil
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2 Good point for the comparison with writing. You may think you have more freedom as a writer (after all, if mathematical entities just worked the way we want, proofs would get quite easy) but on the other hand in math you can study an already existing field. When you write a novel, you have to create everything from scratch. As time goes by, I find it more difficult to write than to (attempt to) do some math.Sylvain JULIEN –Sylvain JULIEN 2018-03-11 19:30:56 +00:00 Commented Mar 11, 2018 at 19:30
8 One difference is that performers usually do at least produce results that can reach out to reasonably wide audience. Our endeavours are often a bit more solitary.Vladimir Dotsenko –Vladimir Dotsenko 2018-03-12 09:43:27 +00:00 Commented Mar 12, 2018 at 9:43
4 As a composer and guy with some interest in mathematics, I disagree completely. As a composer, what I'm creating has a stamp of personality to it--which is appropriate. All I have to do is be true to myself and I'm very unlikely to become a carbon copy of something existing. Once I knew what I was doing, I could choose what compositional ideas to work on and feel certain, well ahead of time, that the result will be distinct and valuable... and satisfying 90% of the time.composerMike –composerMike 2018-03-13 04:19:14 +00:00 Commented Mar 13, 2018 at 4:19
As a musician, I dont think its a much stronger version of it, music doesn't have any boundaries or rules, so we can be a lot more flexible there, unlike mathematics.Courage –Courage 2018-03-20 06:07:38 +00:00 Commented Mar 20, 2018 at 6:07
1 The boundaries or rules in music are less immediately visible than the axioms of mathematics. I am sure that there are rules / boundaries in any field. Listen to music or read a novel made by a complete dilettante, to see what I mean.santker heboln –santker heboln 2020-05-24 16:06:15 +00:00 Commented May 24, 2020 at 16:06
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I think the only way to handle this is just to carry on. Then, after a while, you may become confident enough in your ability and desire to do mathematics at a level acceptable to you. If that does not happen, you may/should consider other occupations.
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answered Mar 11, 2018 at 13:54
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Iosif Pinelis
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2 Though this answer is short, I like it the best. For someone just starting out, it's encouraging to hear that successful people also felt the same anxieties early on. But we should also acknowledge that some people don't succeed at math research. So it's not entirely honest to say, "Stop worrying and you'll succeed at research." It is important to stop worrying, but there's no guarantee you'll achieve the level of success that others expect of you.Timothy Chow –Timothy Chow 2023-01-29 14:21:00 +00:00 Commented Jan 29, 2023 at 14:21
6 The only way to really stop worrying, IMO, is to recognize that you'll be fine even if you fail at mathematical research. You'll just do something else, and be fine. Once failure no longer frightens you, you will paradoxically have a better chance (though again, not a guarantee) of succeeding.Timothy Chow –Timothy Chow 2023-01-29 14:22:12 +00:00 Commented Jan 29, 2023 at 14:22
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From the way your opening sentence is phrased ("an aspiring mathematician-to-be, who started doing research"), I am guessing that you are at stage of being a graduate student (or not far from that). The answer I am giving is based on this assumption.
Basically, for an "aspiring mathematician-to-be", a matter which is as important as producing new results, or maybe much more important than that, is to identify areas in which you enjoy trying solving problems, and spend some time learning ideas, methods, and interesting questions in those areas. I use the plural form "areas" intentionally: it often helps to (a) have a backup problem to think about when you are stuck and (b) have knowledge of something outside of your primary research area because that's how you sometimes stumble upon unexpected interesting connections.
And, well, if your immediate environment creates a constant pressure to discover something new, do keep in mind that it is a feature of your environment rather than of life of a mathematician in general, and that great things rarely happen under pressure. In fact, from my personal experience, it is pretty much the opposite: one keeps working on a question for hours, days, weeks, months, maybe years, often without tangible progress, and then, on some occasions, breakthrough moments happen and you produce something new. Eventually, one either learns to be comfortable with working as a research mathematician, or finds a profession that has a mathematical component but slightly more instant gratification.
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edited Mar 12, 2018 at 9:40
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2 revsVladimir Dotsenko
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It's fun.
It's like being a hunter, pursuing the prey through the forest, looking for its tracks, the scraps of fur it left behind, piecing together its movements, building up to the moment when you have it in the sights of your gun.
Or like being a miner, digging deep into the mysterious earth, using your axe to get through intervening layers of rock, giving up on one tunnel, trying another one, hoping for that moment when you suddenly happen upon the sparkle of a vein of rare gems. With, along the way, a lot of sweat.
Or like being a creative artist, an artisan, or perhaps even just a house painter (which I was in college). You have a vague idea in your head of what you want the final outcome to be. There are many small details that have to be worked out, that are hard to anticipate, but are fun and interesting to ponder as you work through them. Eventually, you're done, you can stand back and admire your work. Be it a painted house or a theorem.
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edited Mar 11, 2018 at 17:14
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There is a quotation I once heard that perhaps someone could provide a reference to. It may even have been an MO comment...
The gist of it was that as one gets along in one's mathematical career one's research often starts to feel less like tackling some impossible obstacle than preparing a meal for friends.
As I understand this, we mathematicians find a corner of the subject where we want to resolve some problems that are not so flashy as to draw international attention and gain us a Fields Medal or a tenure-track job at Princeton, but are interesting and important for the area of mathematics we love and are aiding in the growth of. The former sort of goal may drive one's early career, the scratching, biting and clawing that seems necessary to secure the time to think with a decent academic job. Later on, post-tenure, one can try to focus a bit more (with equal or greater intensity) on figuring out what is actually going on with a bit of mathematics. Usually, this corner is interesting to a small community of mathematicians who really will care about the work. It really feels that research is a conversation with these people, and that the work is pursued in a spirit of mutual appreciation of the beauty of that subject and a shared, deep desire for increased clarity...although the realities of promotion and pay still drive the occasional flare up to attempt to become famous...
The fewer of these flare-ups, the better, I think. They are driven by economic forces, rather than honest mathematics. The greatest of us probably think of these sorts of economic forces the least, either because they developed their mathematics early enough to head off these forces early on and now have coveted jobs, or simply they are used to living a spartan lifestyle that shields them from typical economic concerns.
This answer, by the way, is a comment mainly intended for the OP, and may be too riddled with opinion to be appropriate. If down voted sufficiently, I will remove it.
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answered Mar 15, 2018 at 14:11
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Although I'm not a professional mathematician, I am a professional scientist who relies heavily on mathematics, and have made a few small research contributions to mathematics (combinatorics, statistics, linear algebra, ...) proper along the way. Perhaps my experiences would be of some help.
To excel or even work in any advanced and valuable field (physics, mathematics, neuroscience, ...), you will be under pressure. There is competition for employment, tenure, grants, recognition, and so on. At base, the only way you'll thrive is if the day-to-day act of working brings you some joy—even if it can (and will) be often frustrating. Taking breaks, especially for vacations and exercise, can reduce stress and indirectly help your research.
Perhaps overlooked is this: Avoid stress elsewhere in your life. Avoid relationships (romantic or otherwise), living situations, financial arrangements (getting into debt), and so on that add to stress and burden. We all have different tolerances for stress, so if you can expend your "stress budget" on mathematics, you'll likely be happier and more productive.
Try to tackle problems at the "right" level of difficulty. As David Hilbert stated, the best problems are not too hard and not too easy. Spend time judging the apparent difficulty of a problem before devoting yourself to it. For instance, I would not recommend approaching the Collatz Conjecture (unless you're a Fields Medalist) because every mathematician who has looked into it recognizes it as extraordinarily difficult. (Erdös said: "Mathematics isn't ready for problems such as that.")
And play with mathematics! John Conway (and others) have brought a sense of play to a number of areas of mathematics... not worrying about whether something deep arises. (And yet, in his hands, it so often did!)
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answered Mar 14, 2018 at 1:19
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Others had said it better than I will, but let me share my own experience.
In the past, when I was working on my PhD, I felt that pressure quite hard. But it was not a “moral” pressure like yours: it was a “material” one. If I couldn’t find something enough interesting to be named a PhD thesis, I wouldn’t be a mathematician at all.
That was an awful feeling: working with that feeling in the back of my mind was the worst thing I could have done. In the end, after some years of suffering, and changing my thesis advisor, the thesis was written. I have to credit my second advisor for it very much. But also, I guess, because without noticing it, I just forgot my fears and began to enjoy the maths I was doing for the sake of it. Because the problems I was facing –and eventually solving, some of them- where great fun!
My career as a math researcher has had its ups and downs after that, but the best moments are always when and I have fun doing it: reading papers from other colleagues, studying books to learn a new subject, writing some papers, making a few speeches now and then… Eventually, reading, studying, writing… you eventually ask yourself questions: “Really? –I don’t understand this.” Or, “I don’t understand this THIS way. Let’s try it to write it the other way around…” And, maybe, sometimes, this try becomes a paper –most of the times it doesn’t, though.
Either way, the most important thing, I insist, I’ve been doing this because I enjoy it: it’s a challenge to understand someone’s other paper, a new technique, a new subject… It it’s not, if it’s boring, if I’m doing it because “I have to” –and not because it’s fun-, then I just quit: let’s forget this s.
Turns out that, now and then, some colleagues found some of my papers deserved to be published in some not too bad journals. Great! –This is good for my status and, in time I get some extra pay because of it: not too bad. Others enjoy some of my speeches. Good: you feel good when others like what you’re doing. I have interesting conversations about maths with colleagues here and there and, since I’ve been very lucky, I’ve found funny people around the world to have some laughing and share beers talking about spectral sequences that don’t want to converge when you need them to do it: silly, isn’t it?
Most of the time, though, research is a lonely business: you, your sheet of paper, and your pen.
Do you like it? Do you keep soiling sheets and sheets of papers, from which you’re going to save just 10% -or less? Do you like it, or do you do it because “you have to”?
Well, I couldn’t survive doing it because “I have to”. But because I like it. Because it’s great fun.
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answered Mar 14, 2018 at 3:40
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A small practical suggestion on how I sometimes handle the pressure you mention.
Something new is really a fuzzy concept: maybe something you think to be known has a yet undiscovered subtle mistake in the original paper (and thus is technically not known to be true), maybe something apparently brand new already appears as a lemma in some article you've never heard of in another field, maybe it is an almost trivial variation on a well known idea and it does not deserve to be called new, maybe every expert in the field kind of knows it but no one ever wrote it down in full detail and rigour. Novelty is really, because of the way we do mathematics in this age and time, a continuum ranging between the "published many times over in different guises" and the "no one has any clue even about where to start", rather than a binary distinction between known and unknown.
(The only distinction where you can precisely draw a line is between what is known and unknown to you. Whence the other answers stressing the importance of learning known facts as a preparation to and a diversion from research.)
So, if it helps, trick yourself into thinking of the open problem you're working on as if it were something whose truth is already known to humanity but whose proof is buried in a very old journal issue which is not to be found anywhere, or which needs a simpler proof than the long and complicated known one, or whose solution already is in the air in the mathematical community and which just needs a keen young researcher to connect the dots.
Sometimes a bit of reductionism can ease the psychological pressure that's behind the novelty part of a new result; do not think of yourself as a daring explorer who ventures in the vast uncharted lands of the unknown, but as a modest worker in a familiar place who is putting together pieces of several familiar puzzles in a moderately new arrangement. (I think someone once said as a hyperbole that not a single original idea has ever been had in mathematics, to emphasize the aspect of combining known facts in new ways and seeing them in new lights.)
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answered Mar 12, 2018 at 0:07
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1 Please note that I also do hold the apparently opposite belief that mathematics is the highest and most worthwhile expression of the human spirit and in some ways transcends what we can conceive. This and my answer really are two faces of the same coin and we get to pick which one suits our needs each time.user41593 –user41593 2018-03-12 00:11:18 +00:00 Commented Mar 12, 2018 at 0:11
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Many other answers make what seem to me important points. Emanuele Tron’s observation that novelty is rather fuzzy and relative, and YCor’s suggestion that good research is not just about “new results”, particularly resonated with me.
Implicit in those answers is the fact that your "job" can be at most to produce work which is new enough to be publishable. This is a highly specific and technical sense of “new”! For example bringing together ingredients from two different published papers, to prove a result which does not obviously follow from either, might absolutely be “new” in this sense.
Of course it is good to be more ambitious than this. In fact it would be nice to come up with some utterly unprecedented new ideas! But that would be extraordinarily rare and even if you did do it, you might not realise that you had. Most new mathematical ideas seem to me to come from understanding existing work more deeply, so as to extend it to new cases or prove stronger results with it.
In practice a lot of the struggle can be to work out what results of a certain type have appeared in, or are well known to follow from, published work. Then you will be able to see the gaps where new results could fit. Because every person has a slightly different perspective, other people may be surprised by the work that results!
If you follow the approach in the last paragraph, there might be a few false starts proving theorems that others do not find exciting or which are not publishable, but this is also part of the learning process. Remember that suggestions for alterations, extensions, alternatives or improvements can all be positive responses which show confidence in your ability.
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edited Mar 15, 2018 at 13:42
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6 revsSimon L Rydin Myerson
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2 My point is that indeed good math is not just about "new research", but also that new research is not just about "new results". Trying to "make the subject one's own" is, in my opinion, an essential research step. Of course it's not the easiest that can leave a written trace.YCor –YCor 2018-03-13 22:04:24 +00:00 Commented Mar 13, 2018 at 22:04
That is better put, if you don't mind I will edit my answer accordingly. Of course feel free to modify it to better reflect what you meant.Simon L Rydin Myerson –Simon L Rydin Myerson 2018-03-14 11:38:04 +00:00 Commented Mar 14, 2018 at 11:38
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Unless you work on an important open problem and someone solves it before you, not much harm is done. Very often your solution provides a different perspective on the problem and you can prove a slightly different or stronger result with your approach. That happened to me a few times. Also, if your solutions is more elegant and easier to read, people will be interested in your paper as it will clarify and popularize the field.
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edited Mar 28, 2018 at 16:25
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2 revs, 2 users 80%Piotr Hajlasz
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I will address the non mathematical aspects of this. Even though I understand it from the perspective of a graduate student of mathematics, I agree with Chris Godsil that it happens in many different areas apart from mathematics.
The key points in your post that strike me are pressure and "my job". Can you reliably identify the pressure and its source? Is your conception of "my job" accurate enough that the pressure it might generate reasonable and appropriate? I recommend using not only your advisor, counseling services, colleagues, friends, and other face-to-face resources to help you consider these issues, but also to reexamine your value system, and see if what you are doing needs to change to fit your values.
I'm going to guess that your position and situation are similar to mine when I started graduate school. I had a teaching assistant fellowship and enough resources to support a four year course of study. I thought of it as a way to occupy myself, with the major pressure being to graduate. This meant looking at many things of interest and writing a dissertation on something "new". There were other duties too, but I took the perspective that this was how I wanted to spend my time, and that the requirements should exert only a reasonable amount of pressure.
One suggestion that might help is to change your idea of "my job" to adjust the pressure. Maybe your job is to try different avenues of investigation and write a daily or weekly report of your findings, whether good or bad or neither. This has the advantage in that you use and develop your perspective, and the report you produce is always new. The impact on others may not be, but out of 52 weekly reports, only one or two really need to have a great impact on others so the others will also consider it new. For example, John Baez does a community service by his blog, even if he never does anything highly original other than present thoughts he sees or hears from other people. He might even get some new research out of it. Would you like that kind of job?
(For another example, my four years turned to seven, the dissertation while somewhat new was minor and had contributions from my advisors, and never got formally submitted. I also found the pressure to graduate greatly reduced right after I presented my results in a conference. So my "my job" got done, but in a different way than I had originally thought. And I understood more about my values in the process.)
Gather advice from many sources, but seek from within to find an answer.
Gerhard "That's What Research Is, Right?" Paseman, 2018.03.11.
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answered Mar 11, 2018 at 19:13
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When someone discovers "new mathematics", they have in essence discovered how to think precisely about a concept genuinely foreign to the human race.
An essential piece of this process is the formalization of your intuitions regarding these new thoughts; the nature of your true goal as a mathematician precludes the possibility of anyone really 'holding your hand' for the final steps in the process.
There is no way better to hone your formalization skills (and your intuition) than to do math all day, every day, regardless of whether it is new or not. This is what keeps me going -- if it ends up being the case that I am too simple to really discover anything of novelty so be it, that is ultimately irrelevant.
Doubt is somewhat intrinsic to the process, and with good reason -- what mathematicians seek to do is not trivial, and is (in my personal opinion) one of the highest callings possible for a sentient being.
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answered Mar 11, 2018 at 16:43
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I think this is a common question in almost all fields and not just math. Hence although I am not a mathematician (not even remotely), I feel the urge to share my thoughts.
As a general advice, you might be interested in reading some materials about Goal-setting theory. In a glance
Goal-setting theory is based on the notion that individuals sometimes have a drive to reach a clearly defined end state. Often, this end state is a reward in itself. A goal's efficiency is affected by three features: proximity, difficulty and specificity. One common goal setting methodology incorporates the SMART criteria, in which goals are: specific, measurable, attainable/achievable, relevant, and time-bound.
An ideal goal should present a situation where the time between the initiation of behavior and the end state is close. With an overly restricting time restraint, the subject could potentially feel overwhelmed, which could deter the subject from achieving the goal because the amount of time provided is not sufficient or rational.
Most people are not optimally motivated, as many want a challenge (which assumes some kind of insecurity of success). At the same time people want to feel that there is a substantial probability that they will succeed. Specificity concerns the description of the goal in their class.
The goal should be objectively defined and intelligible for the individual. Similarly to Maslow's Hierarchy of Needs, a larger end goal is easier to achieve if the subject has smaller, more attainable yet still challenging goals to achieve first in order to advance over a period of time. A classic example of a poorly specified goal is trying to motivate oneself to run a marathon when s/he has not had proper training. A smaller, more attainable goal is to first motivate oneself to take the stairs instead of an elevator or to replace a stagnant activity.
IMHO, while the question is clear, but it points to a surprisingly complex issue.
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edited Jun 15, 2020 at 7:27
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2 If I complain about the downvote I got for this answer, it means that I care, which is pointless since this is CW. Even if it wasn't, I don't think it would have made any difference. But still, it's kinda irritating to think that someone didn't like this answer or found it not suitable, etc. and didn't take the liberty to explain why... Why you people leave such hard puzzles behind?polfosol –polfosol 2018-03-12 12:29:36 +00:00 Commented Mar 12, 2018 at 12:29
2 I did't downvote, but I agree with Lucia's comment: "it would be useful to have some idea of mathematical research before trying to answer it". Especially you're claiming that "this is a common question in almost all fields and not just math". If so this question would be closed as off-topic. The interest and relevance of this question here is it concerns mathematical research and that it has answers that are, at least partly, specific to mathematical research.YCor –YCor 2018-03-14 08:37:13 +00:00 Commented Mar 14, 2018 at 8:37
@YCor where is Lucia's comment? and btw, I disagree with the phrase If so this question would be closed as off-topic. You may ask a question about mathematical research but it might be useful for others as well, and so it might be interesting to know some other insights, or assuming that the big picture is not restricted to math. That wouldn't make it off topic.polfosol –polfosol 2018-03-14 08:51:45 +00:00 Commented Mar 14, 2018 at 8:51
1 Lucia's comment is a comment to the question (sorry I don't know how to link a particular comment). OK I should rather have said "if so the question might have been closed...". At least I would have voted to close it, agreeing that it should better take place on another forum, not specific to math. Then the main argument in favor of not closing it would have probably been that the question addresses issues specific to math!YCor –YCor 2018-03-14 09:37:15 +00:00 Commented Mar 14, 2018 at 9:37
1 @VladimirDotsenko : that's a really strange proposition. And obviously, it's not so clear as you claimed in the beginning. Anyway, let's not start an argument. The case is closed I think.polfosol –polfosol 2018-04-10 04:24:20 +00:00 Commented Apr 10, 2018 at 4:24
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The trite but true answer is to discover new things but how does one do that? What you really should be asking is how to discover what needs to be discovered? The framing of questions or ideas that require inspection or further examination can, sometimes, lead to a discovery. All of the advice you receive on how to discover is not very helpful if you don't know first where to look and then recognize that there is something either missing, incomplete, or plain incorrect that needs to be addressed. This is why so many students end up studying/researching in the field(s) of their mentors and not necessarily being very successful as a result. (OK-it is my opinion that some folks believe that there aren't that many new, good, ideas to go around and so when they find one that is productive they will work it until it is no longer fun)
What has worked the most frequently for me is the "why the hell is that?" or the "well, that makes no damn sense at all!" method of question framing. You'll note that neither of these actually are suited to making a discovery but to identifying where to start exploring. You will quickly learn what other people think that they have found if you are familiar with the area (have read and digested all of the relevant literature, chatted up all of the major and minor players, and know the history of how those discoveries came about). The process of identifying what needs to be done or what remains to be addressed in a particular subject will, frequently, be clear if not immediately obvious. The discoveries that you seek typically end up being the mechanics of the solution used to address those questions. Feynman's and Newton's discoveries hinged on the formation/assembly of the question(s) to be answered. (It is also my opinion that the more practice one gets in making discoveries that the easier it becomes to make new discoveries - that is at least what I have discovered for myself)
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answered Mar 11, 2018 at 22:23
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Being a student is a stage of learning. Being a research student is also a learning stage. The difference is mind-shift. The latter demands your focus on a single thing. You devote most of your time on this focused work. Nobody expects you to find new things. On the other hand, when you focus all your energy on a single thing the chances are more that you might find something new.
I'm not a research student. I had never been. However, I have never stopped learning things. Teaching is my profession. Students ask me questions. I answer them If I know. Otherwise, I study them and then answer them. This process help me to learn new things. Also, it helps me to find something new. Sometime back, a student asked me a question that could have been answered using an existing formula (Derangement formula). Having not come across this formula, I tried my own way of solving the question. In the process, I found two new formulas that are different from derangement formula. Though the approach to the problem is the same, my processes were different. That is how my mind worked out. When I realized that a formula was already existing, it did not discourage me. Instead, I felt the 'aha' feeling.
There was a pressure that I had to answer the question. But I took my own time to solve it. Once I put my whole mind on that, It was just a matter of few minutes to get the result.
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answered Mar 14, 2018 at 2:20
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6945 | https://pubmed.ncbi.nlm.nih.gov/40008208/ | An overview of hereditary spherocytosis and the curative effects of splenectomy - PubMed
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. 2025 Feb 11:16:1497588.
doi: 10.3389/fphys.2025.1497588. eCollection 2025.
An overview of hereditary spherocytosis and the curative effects of splenectomy
Kyril Turpaev1,Elizaveta Bovt12,Soslan Shakhidzhanov12,Elena Sinauridze12,Nataliya Smetanina2,Larisa Koleva12,Nikita Kushnir12,Anna Suvorova2,Fazoil Ataullakhanov134
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1 Center for Theoretical Problems of Physicochemical Pharmacology, Russian Academy of Sciences, Moscow, Russia.
2 Dmitry Rogachev National Medical Research Center of Pediatric Hematology, Oncology and Immunology, Moscow, Russia.
3 Moscow Institute of Physics and Technology, National Research University, Dolgoprudny, Russia.
4 Perelman School of Medicine, University of Pennsylvania, Philadelphia, PA, United States.
PMID: 40008208
PMCID: PMC11850534
DOI: 10.3389/fphys.2025.1497588
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An overview of hereditary spherocytosis and the curative effects of splenectomy
Kyril Turpaev et al. Front Physiol.2025.
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. 2025 Feb 11:16:1497588.
doi: 10.3389/fphys.2025.1497588. eCollection 2025.
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Kyril Turpaev1,Elizaveta Bovt12,Soslan Shakhidzhanov12,Elena Sinauridze12,Nataliya Smetanina2,Larisa Koleva12,Nikita Kushnir12,Anna Suvorova2,Fazoil Ataullakhanov134
Affiliations
1 Center for Theoretical Problems of Physicochemical Pharmacology, Russian Academy of Sciences, Moscow, Russia.
2 Dmitry Rogachev National Medical Research Center of Pediatric Hematology, Oncology and Immunology, Moscow, Russia.
3 Moscow Institute of Physics and Technology, National Research University, Dolgoprudny, Russia.
4 Perelman School of Medicine, University of Pennsylvania, Philadelphia, PA, United States.
PMID: 40008208
PMCID: PMC11850534
DOI: 10.3389/fphys.2025.1497588
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Abstract
Hereditary spherocytosis is a common hemolytic anemia with different severity. The causes of hereditary spherocytosis are mutations in genes that encode red blood cell (RBC) membrane and cytoskeletal proteins, including ankyrin-1, Band 3 (or AE1), α spectrin, β spectrin, and protein 4.2. Molecular defects in these proteins decrease membrane integrity, leading to vesiculation, decreased membrane surface area, and reduced deformability of the cells. Eventually, this leads to the trapping the abnormal RBCs (spherocytes) in the spleen. In most severe cases, splenectomy may be necessary to prevent general RBC collapse during the passage of RBCs through the narrow slits of venous sinuses in the spleen. The clinical benefit of splenectomy results from elimination the primary site of RBC damage and destruction. Splenectomy is a curative approach but can cause complications and should be undertaken after examination by various laboratory approaches. Splenectomy does not correct most genetically determined membrane abnormalities in erythrocytes in patients with hereditary spherocytosis. The transformation of biconcave erythrocytes into spherocytes continues, although to a lesser degree than before surgery. Nevertheless, splenectomy increases the lifespan of red cells, significantly reducing the severity of anemia and improving many physiological signs of HS.
Keywords: erythrocyte cytoskeleton; erythrocyte deformability; gene mutations; hereditary spherocytosis; interendothelial slits; microcirculation; spherocytes; splenectomy.
Copyright © 2025 Turpaev, Bovt, Shakhidzhanov, Sinauridze, Smetanina, Koleva, Kushnir, Suvorova and Ataullakhanov.
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Conflict of interest statement
The authors declare that the research was conducted in the absence of any commercial or financial relationships that could be construed as a potential conflict of interest.
Figures
FIGURE 1
A schematic representation of the…
FIGURE 1
A schematic representation of the interaction between the RBC membrane and the underlying…
FIGURE 1
A schematic representation of the interaction between the RBC membrane and the underlying cytoskeleton. Cross-section of the membrane. The Band 3 – Ankyrin-1 complex is centered by a Band 3 tetramer. Among the Band 3 associated proteins, there is ankyrin-1, which also binds to the β-spectrin chain and protein 4.2. Glycophorin A occurs as a dimer. Apparently that the Band 3 complex and transmembrane Rh complex form a higher order macromolecular complex. The Rh complex consists of the rhesus factor proteins, RhAG, CD47 (also known as a thrombospondin receptor), and glycoprotein LW, also known as intercellular adhesion molecule-4 (ICAM-4). CD47 and Rh interact with ankyrin-1 via protein 4.2. The junctional complex. Pivotal protein 4.1R interacts with N-terminal domain of β-spectrin. This region contains short filaments of the F-actin and actin-binding proteins like dematin, tropomyosin, and tropomodulin. The transmembrane part of the junctional complex includes glycophorins C/D (GP C/D), glucose transporter 1 (Glut1), and protein p55. In addition, the complex interacts with a Band 3 population (presumably dimers), which is separate from the Band 3 – Ankyrin complex. The α 2 β 2 tetramers of spectrin form a dense network lining the inner surface of the lipid bilayer. Spectrin tetramer is shown in a straight configuration, although they may be curved in reality (upper drawing). Top view of the membrane. The membrane skeleton is a network formed by quasi hexagonal units centered around actin filaments that are attached to glycophorins C/D. Band 3 and ankyrin interact with C-terminal part of β-spectrin, approximately in the middle of the α 2 β 2 tetramers. The junctional complex is located at the nodes of hexagons. The exact relative positions of the depicted proteins to each other within the complexes may be unknown. The proteins and lipids are not drawn in their actual scale and shape (lower drawing). This illustration is a merged figure adapted from both (Delaunay, 2007; Perrotta et al., 2008; Lux, 2016).
FIGURE 2
The structure of RBC spectrin…
FIGURE 2
The structure of RBC spectrin tetramer The α 2 β 2 tetramers of…
FIGURE 2
The structure of RBC spectrin tetramer The α 2 β 2 tetramers of spectrin form a network lining the inner surface of the lipid bilayer. In spectrin dimers, the α- and β-chains are arranged tandemly. Two dimers connect side-by-side in an antiparallel manner with the N-terminals region of α-chains facing C-terminal regions of β-chains. Each chain contains a series of repetitive domains, each consisting of 106-amino acids, as well as additional specialized functional domains. α-Spectrin has 21 numbered repeats (α1 – α21), plus a partial repeat (α0) at the N-terminus. The repeat α10 comprises the src homology 3 (SH3) subdomain. β-Spectrin has 16 numbered repeats (β1 – β16) and a partial C-terminal repeat (β17). Dimer-tetramer self-association of α- and β-spectrin chains takes place at a nucleation domain (ND) near the edges of spectrin tetramers (repeats β1 and β2 pair with α21 and α20). Actin and protein 4.1R bind to calponin homology (CH) actin-binding domains at the N-end of β-spectrin. Adjacent EF hand domain on α-spectrin binds Ca 2+ ions.
FIGURE 3
Schematic representation of the structure…
FIGURE 3
Schematic representation of the structure of erythroid ankyrin and Band 3 proteins. Ankyrin-1…
FIGURE 3
Schematic representation of the structure of erythroid ankyrin and Band 3 proteins. Ankyrin-1 consists of membrane binding, spectrin binding, and C-terminal domains. The N-terminal membrane binding domain has 4 subdomains (D1-D4). There are two binding sites for the Band 3 protein, one located in the subdomain D2, D3, and D4. The spectrin binding domain contains three subdomains, of which ZU5A contains the binding site for β-spectrin (repeats β14 and β15). The function of presumably regulatory C-terminal domain and neighboring so-called death domain (DD) remains unclear (upper drawing). Band 3 contains two functionally distinct domains: 1) responsible for binding of cytoplasmic proteins domain and 2) a transmembrane domain that forms the anion-exchange channel. In the cytoplasmic domain, the glycolytic enzymes (PFK, aldolase, and GAPDH) bind to amino acid sites located nearby in the folded protein. PK, LDH, and enolase do not interact directly with Band 3, but they form a complex with the three glycolytic enzymes mentioned. These enzymes are displaced by deoxyhemoglobin, which also binds to the N-terminus of the Band 3, or by phosphorylation of tyrosine residues (Y8 and Y21). Ankyrin and protein 4.1R also interact with the cytoplasmic domain of Band 3. The transmembrane domain is composed of 16 subdomains that form ion channels responsible the intake of bicarbonate ions from the blood (lower drawing). Abbreviations in the figure: GAPDH, glyceraldehyde-3-phosphate dehydrogenase; Hb, hemoglobin; Hb-O 2, oxyhemoglobin; LDH, lactate dehydrogenase; PFK, phosphofructokinase; PK, pyruvate kinase; Y, tyrosine phosphorylation site.
FIGURE 4
General representation of the microcirculation…
FIGURE 4
General representation of the microcirculation within the spleen. A map of the main…
FIGURE 4
General representation of the microcirculation within the spleen. A map of the main spleen compartments. Afferent splenic (or trabecular) arteries branch into central arterioles that enter the white-pulp, which consists of an area containing T cells and dendritic cells (DCs) as well as lymphoid follicles that accumulate activated B cells. Arterioles end in white pulp capillaries, from where the blood enters the venous sinuses which gather in the efferent splenic vein, forming the closed blood circulation pathway. The open circulation consists of the flow of RBCs through penicillar arterioles opening into the reticular meshwork of the red pulp followed by filtration through the interendothelial slits (IES) in the venous sinuses. The interior blood pressure in red pulp is higher than that in the sinus lumen that determines the direction of RBC flow (upper drawing). A scheme of the venous sinus. The venous sinuses are formed by spindle shaped endothelial cells supported by ring (or annular) fibers. Stress fibers (not shown) extend under the basal plasma membrane of the endothelial cells, running parallel to the cell axis (lower drawing). This illustration is a merged figure adapted from both (Mebius and Kraal, 2005; Neri et al., 2021).
FIGURE 5
Comparison of RBC deformation during…
FIGURE 5
Comparison of RBC deformation during the passage through capillaries and spleen IES. (A) …
FIGURE 5
Comparison of RBC deformation during the passage through capillaries and spleen IES. (A) A scheme of an RBC transit through a capillary. Depending on the diameter of the capillaries, RBCs acquire either an axisymmetric “parachute-like” (left drawing) or cylindrical “cigar-like” shape (right drawing). The latter occurs when they pass through capillaries with almost minimal diameters (about 3 μm). RBC shapes in capillaries also depends from the stream velocity and shear forces. Capillaries are depicted by the dashed lines. (B) A scheme of an RBC transit through the interendothelial slit (IES) (left drawing). Apparently, in the middle of the passage, the red cells are divided into two equal spheres located on either side of the slit and joined by a thin connection. At this moment red cell is divided equally into two equal spheres on both sides of the slit. This configuration requires a high surface area-to-volume (S/V) ratio, which is inherent for discocytes, but not for spherocytes. In circulating healthy RBCs, the observed S/V ratio is approximately 1.56 which is not more than 15% larger than the minimum value that allows the formation of two equal spheres. Solid red lines indicate the edges of the biconcave disks of RBCs. The concave area of RBCs is depicted as less shaded. Blood flow is from left to right. Right drawing: a scheme of IES.
FIGURE 6
HS and the loss of…
FIGURE 6
HS and the loss of membrane surface by vesiculation. A hallmark of HS…
FIGURE 6
HS and the loss of membrane surface by vesiculation. A hallmark of HS is a decrease of RBC surface area and reduction in the cell surface to volume ratio Two different pathways lead to membrane shrinkage, which occurs during the passage of RBCs through the spleen. 1) Defects in spectrin, ankyrin, or protein 4.2 lead to reduced density of the membrane skeleton and destabilization of the lipid bilayer. This results in the generation of Band-3-containing microvesicles. 2) Defects in Band 3 lead to a weakening of the lipid bilayer connection with cytoskeleton, causing the formation of Band-3-depleted microvesicles. Both pathways lead to a reduction in the surface area of red cells and the formation of spherocytes, which have increased rigidity and decreased deformability. Spherocytes are subsequently retained in the spleen. This illustration is a merged figure adapted from both (Perrotta et al., 2008; Reliene et al., 2002).
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References
Anosa V. O. (1976). Postsplenectomy blood values, marrow cytology, erythrocyte life-span, and sequestration in mice. Am. J. Physiol. 231, 1254–1257. 10.1152/ajplegacy.1976.231.4.1254 - DOI - PubMed
Bain B. J. (2008). Irregularly contracted cells. Am. J. Hematol. 83, 592. 10.1002/ajh.21132 - DOI - PubMed
Baldari L., Boni L., Giuliani B., Cassinotti E. (2023). Porto-spleno-mesenteric venous thrombosis after elective splenectomy: a retrospective cohort study. Front. Immunol. 14, 1216283. 10.3389/fimmu.2023.1216283 - DOI - PMC - PubMed
Baskurt O. K., Hardeman M. R., Uyuklu M., Ulker P., Cengiz M., Nemeth N., et al. (2009). Comparison of three commercially available ektacytometers with different shearing geometries. Biorheology 46, 251–264. 10.3233/BIR-2009-0536 - DOI - PubMed
Bennett V., Healy J. (2008). Organizing the fluid membrane bilayer: diseases linked to spectrin and ankyrin. Trends Mol. Med. 14, 28–36. 10.1016/j.molmed.2007.11.005 - DOI - PubMed
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6946 | https://www.quora.com/How-can-I-find-the-center-and-radius-of-the-circle-x-2-y-2-4x-8y-5-0 | How to find the center and radius of the circle x^2+y^2 - 4x +8y -5=0 - Quora
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How can I find the center and radius of the circle x^2+y^2 - 4x +8y -5=0?
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Gary Ward
MaEd in Education&Mathematics, Austin Peay State University (Graduated 1997) · Author has 4.9K answers and 7.6M answer views
·1y
Originally Answered: Can you solve the center and radius of the circle x^2 + y^2 + 8x - 6y - 15= 0? ·
Can you solve the center and radius of the circle x^2 + y^2 + 8x - 6y - 15= 0?
Yes.
Move constant(s) to the right side of the equal sign (add 15 to both sides of the equation).
x² + y² + 8x - 6y = 15
Group like variables together inside parentheses.
(x² + 8x) + (y² - 6y) = 15
Make each variable a perfect square. Take the constant in front of the unsquared variable, divide it by two and square it. Add it inside the parentheses and also to the right side of the equation.
(x² + 8x + 16) + (y² - 6y + 9) = 15+ 16 + 9 = 40
Take the square root of the three parts presented as squares.
(x + 4)² + (y - 3)² = (2
Continue Reading
Can you solve the center and radius of the circle x^2 + y^2 + 8x - 6y - 15= 0?
Yes.
Move constant(s) to the right side of the equal sign (add 15 to both sides of the equation).
x² + y² + 8x - 6y = 15
Group like variables together inside parentheses.
(x² + 8x) + (y² - 6y) = 15
Make each variable a perfect square. Take the constant in front of the unsquared variable, divide it by two and square it. Add it inside the parentheses and also to the right side of the equation.
(x² + 8x + 16) + (y² - 6y + 9) = 15+ 16 + 9 = 40
Take the square root of the three parts presented as squares.
(x + 4)² + (y - 3)² = (2√10)²
Compare to the form of the standard equation (x - x1)² + (y - y1)² = r²
Center is (-4, 3) with radius 2√10.
Upvote ·
9 5
9 2
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What is the center of the circle with the equation x^2+y^2-6x-8y+2=0?
What is the center of the circle 2x^2 + 2y^2 - 12x + 8y + 16=0?
What is the value of k if the circle x^2 + y^2 + 2x – 8y – k = 0 has a radius of 4 units?
How do I find the center and radius of a circle with the equation x^2+y^2+8x-2y+15=0?
What are the center and radius of the circle whose equation is x^2+y^2+4x=5?
Subramanya R
Former Retired Govt Employee, Interested in All Fields · Author has 2K answers and 1.5M answer views
·Jun 22
Originally Answered: Can you solve the center and radius of the circle x^2 + y^2 + 8x - 6y - 15= 0? ·
Can you solve the center and radius of the circle x^2 + y^2 + 8x - 6y - 15= 0?
[math]\text{General equation of circle is }[/math]
[math]x^2+y^2-2ax-2by+a^2+b^2-R^2=0 \text{ ; here (a, b) is centre.} [/math]
[math]\text{Given equation is }x^2+y^2+8x-6y-15=0[/math]
[math]\text{Comparing general equation with given equation we get:}(a, b)\to(-4, 3)[/math]
[math]\text{Substituting this value in the given equation, and x coordinate of centre -4 } [/math]
[math]\text{as x coordi[/math]
Continue Reading
Can you solve the center and radius of the circle x^2 + y^2 + 8x - 6y - 15= 0?
[math]\text{General equation of circle is }[/math]
[math]x^2+y^2-2ax-2by+a^2+b^2-R^2=0 \text{ ; here (a, b) is centre.} [/math]
[math]\text{Given equation is }x^2+y^2+8x-6y-15=0[/math]
[math]\text{Comparing general equation with given equation we get:}(a, b)\to(-4, 3)[/math]
[math]\text{Substituting this value in the given equation, and x coordinate of centre -4 } [/math]
[math]\text{as x coordinate of the coordinate of the point on circle, we get equation of } [/math]
[math]\text{corresponding ordinates.}[/math]
[math]\text{That is : }(-4)^2+y...
Dave Benson
trying to make maths easy. · Author has 6.1K answers and 2.1M answer views
·11mo
Compare x²+y²-2gx–2fy+c = 0. ¢re (-g-f) = (2,-4) & radius = √(g²+f²-c) = 5 ans
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Compare x²+y²-2gx–2fy+c = 0. ¢re (-g-f) = (2,-4) & radius = √(g²+f²-c) = 5 ans
Upvote ·
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Pradeep Hebbar
Many years of Structural Engineering & Math enthusiasm · Author has 9.3K answers and 6.2M answer views
·1y
Originally Answered: Here is an equation of a circle, x²+y²-8x-6y=0. What is the centre and radius of the circle? ·
Given an equation of circle,
[math]x^2+y^2-8x-6y=0\ldots(1)[/math]
Ler [math]O(h,k)[/math] be the center and [math]r[/math] be the radius
We start with the equation of circle,
math^2+(y-k)^2= r^2[/math]
[math]x^2+h^2-2hx+y^2+k^2-2ky=r^2[/math]
[math]x^2+y^2-2hx-2ky+(h^2+k^2-r^2)=0 \ldots(2)[/math]
Comparing eqn. (1) and (2),
[math]-2h= -8[/math]
[math]-2k=-6[/math]
[math]h^2+k^2-r^2 = 0[/math]
From above, [math]h=4[/math] and [math]k=3[/math]
Further,
[math]4^2+3^2-r^2 = 0[/math]
[math]r^2= 25[/math]
[math]r=5[/math]
Center is at math[/math] and radius is [math]5[/math]
Continue Reading
Given an equation of circle,
[math]x^2+y^2-8x-6y=0\ldots(1)[/math]
Ler [math]O(h,k)[/math] be the center and [math]r[/math] be the radius
We start with the equation of circle,
math^2+(y-k)^2= r^2[/math]
[math]x^2+h^2-2hx+y^2+k^2-2ky=r^2[/math]
[math]x^2+y^2-2hx-2ky+(h^2+k^2-r^2)=0 \ldots(2)[/math]
Comparing eqn. (1) and (2),
[math]-2h= -8[/math]
[math]-2k=-6[/math]
[math]h^2+k^2-r^2 = 0[/math]
From above, [math]h=4[/math] and [math]k=3[/math]
Further,
[math]4^2+3^2-r^2 = 0[/math]
[math]r^2= 25[/math]
[math]r=5[/math]
Center is at math[/math] and radius is [math]5[/math]
Upvote ·
9 4
Related questions
What is the center of the circle with the equation x^2+y^2-6x-8y+2=0?
What is the center of the circle 2x^2 + 2y^2 - 12x + 8y + 16=0?
What is the value of k if the circle x^2 + y^2 + 2x – 8y – k = 0 has a radius of 4 units?
How do I find the center and radius of a circle with the equation x^2+y^2+8x-2y+15=0?
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What is the centre and the radius of the circle of x^2+y^2-6y=0?
2.The general form of a circle is given as x^(2) +y^(2) +4x-12y+4=0. (a) What are the coordinates of the center of the circle? (b) What is the length of the radius of the circle?
How do I find the centre and radius of the circle; x^2 + y^2 - 6x - 8y + 5 = 0?
What is the equation of the circle, x^2+y^2-8y+12y-100= 0? What is the radius of this circle?
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What is the center of the circle with the equation x^2+y^2-6x-8y+2=0?
What is the center of the circle 2x^2 + 2y^2 - 12x + 8y + 16=0?
What is the value of k if the circle x^2 + y^2 + 2x – 8y – k = 0 has a radius of 4 units?
How do I find the center and radius of a circle with the equation x^2+y^2+8x-2y+15=0?
What are the center and radius of the circle whose equation is x^2+y^2+4x=5?
What is the center and radius of the circle, x2+y2-8x+6y+8=0?
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. 2012 Jun;88(1040):335-47.
doi: 10.1136/postgradmedj-2011-130213. Epub 2012 Jan 25.
Lobar collapse demystified: the chest radiograph with CT correlation
Rebecca Mullett1,Abhishek Jain,Sripali Kotugodella,John Curtis
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1 Radiology Department, University Hospital Aintree, Lower Lane, Liverpool L9 7AL, UK. rebeccamullett@doctors.org.uk
PMID: 22282740
DOI: 10.1136/postgradmedj-2011-130213
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Lobar collapse demystified: the chest radiograph with CT correlation
Rebecca Mullett et al. Postgrad Med J.2012 Jun.
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. 2012 Jun;88(1040):335-47.
doi: 10.1136/postgradmedj-2011-130213. Epub 2012 Jan 25.
Authors
Rebecca Mullett1,Abhishek Jain,Sripali Kotugodella,John Curtis
Affiliation
1 Radiology Department, University Hospital Aintree, Lower Lane, Liverpool L9 7AL, UK. rebeccamullett@doctors.org.uk
PMID: 22282740
DOI: 10.1136/postgradmedj-2011-130213
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Abstract
Collapse of a lobe of a lung is an important indicator of a range of conditions, including malignancy. Clinical symptoms and signs may suggest a diagnosis of lobar collapse; however, it is often diagnosed, and always needs to be confirmed, with radiological examination. The radiological signs may be subtle, difficult to interpret and sometimes confusing to both clinicians and radiologists. Although multidetector CT (MDCT) is now widely in use for confirming and diagnosing lobar collapse, the plain chest radiograph is usually the first imaging modality performed and so recognition on the plain film remains of vital importance. The basics of chest radiograph interpretation are reviewed, concentrating on the concepts of radiographic density and the silhouette sign. MDCT images are used to demonstrate the general radiological signs of collapse, and the signs of collapse that are specific to the different lobes of the lung are reviewed.
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6950 | https://www.khanacademy.org/math/mappers/number-and-operations-220-223/x261c2cc7:comparing-negative-numbers/v/comparing-negatives-with-variables | Negative numbers, variables, number line (video) | Khan Academy
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Skip to lesson content
MAP Recommended Practice
Course: MAP Recommended Practice>Unit 35
Lesson 11: Comparing negative numbers
Ordering negative numbers
Ordering negative numbers
Ordering small negative numbers
Ordering rational numbers
Ordering rational numbers
Compare rational numbers
Negative numbers, variables, number line
Compare rational numbers using a number line
Writing numerical inequalities
Math>
MAP Recommended Practice>
Numbers and Operations 218-221>
Comparing negative numbers
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Negative numbers, variables, number line
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Given a, b, and c shown on the number line, Sal determines if statements like -b < c are true.
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shresthsahoo 2 years ago Posted 2 years ago. Direct link to shresthsahoo's post “i do not understand weren...” more i do not understand werent negative no.s to the left of zero?
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Layla M 5 years ago Posted 5 years ago. Direct link to Layla M's post “This could be explained b...” more This could be explained better. It does not address the natural confusion that arises of why letters are numbers and why items to the left of zero are not negative.
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susaa362360 2 years ago Posted 2 years ago. Direct link to susaa362360's post “Normally, to the left of ...” more Normally, to the left of the zero are negative numbers and to the right of the zero are positive numbers in number line, now how a negative number can be to the right of the zero. !
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Joel a year ago Posted a year ago. Direct link to Joel's post “Negative number is never ...” more Negative number is never to the right of zero on the number line. Sal tries to show the opposite variable using the variables as example (a,b,c,x,y,z).
Good observation!
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Rebbehkah 7 years ago Posted 7 years ago. Direct link to Rebbehkah's post “a, b, and c aren't number...” more a, b, and c aren't numbers. Or is he just saying that because they stand for numbers?
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Andrew Celio 7 years ago Posted 7 years ago. Direct link to Andrew Celio's post “Yes, they stand for numbe...” more Yes, they stand for numbers. That's the concept of a variable :)
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Jshill 5 years ago Posted 5 years ago. Direct link to Jshill's post “Why did Sal put the -b on...” more Why did Sal put the -b on the right of the number line?
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EddieWolfy 4 years ago Posted 4 years ago. Direct link to EddieWolfy's post “because a negative variab...” more because a negative variable is the opposite of that number. So if b starts as a negative, then negative b would be a positive because the opposite of a negative number is its positive counterpart.
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Archi 7 years ago Posted 7 years ago. Direct link to Archi's post “I thought that whole numb...” more I thought that whole numbers were bigger than negative numbers. And I know they are. But what confuses me, Is that The number line was a bit hard to understand. I was quite confused on which side the whole numbers were. But I'm still a bit confused with this number line. Can anyone explain this in simpler terms than Sal did?
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Gideon 8 years ago Posted 8 years ago. Direct link to Gideon's post “Why is -b greater than 0?” more Why is -b greater than 0?
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Mohammed Al-Abdali 8 years ago Posted 8 years ago. Direct link to Mohammed Al-Abdali's post “b is on the left side of ...” more b is on the left side of 0 so -b will be on the right side of 0, which means, -b>0
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ArturoA 2 years ago Posted 2 years ago. Direct link to ArturoA's post “when was math invented” more when was math invented
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ArturoA 2 years ago Posted 2 years ago. Direct link to ArturoA's post “this is the best” more this is the best
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#1 vegan 7 years ago Posted 7 years ago. Direct link to #1 vegan's post “wait what comes after a n...” more wait what comes after a negative a double negative or just a negative negative?
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Sara Lynn Younes 6 years ago Posted 6 years ago. Direct link to Sara Lynn Younes's post “It’s either negative, neu...” more It’s either negative, neutral, or positive. No double negative. Hope that explains.
Have a wonderful day!
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Video transcript
[Voiceover] So we have a number line here with zero at the center and then on that number line we've marked off some numbers. So, to the left of the number line we have the number a, and then we have the number b here, a little bit closer to zero, and then on the right side of zero we have the number c. And then after that, we have a bunch of statements dealing with inequalities. Now, what I want you to do is pause the video and think about which of these statements are true, which of these statements are false, and maybe which of these statements you don't have enough information to figure out. So, I'm assuming you've had a go at it. You've tried to figure out which of these are true, which of these are false, and which of these you can't figure out. So, let's do them together. So, this first statement says negative b is less than negative c. So, we do know for a fact that b is less than c. We know that b is less than c. How do we know that? Well, b is to the left of c on the number line. It's that straight forward. So, we know this. This is definitely true. But what about negative b? Is negative b less than negative c? So, let's think about where negative b is on this number line. So, negative b I will do in yellow. So, negative, negative b means the opposite of b. So if b is one hash mark to the left of zero, negative b is going to be one hash mark to the right of zero. So, that right over there, is going to be negative b. And now, where is negative c? Once again, negative c, this literally means the opposite of c. C is one, two, three, four, five hash marks to the right of zero. And so negative c is going to be one, two, three, four, five hash marks to the left of zero. And actually, let me do this in a different color. So, negative c I will do in purple. This right over here is negative c. So, let's compare, is negative b less than negative c? No, negative b is to the right of negative c on the number line. Negative b is greater than negative c. So, this is not true. Negative b is to the right of negative c. Negative b is greater than negative c. And if this is a little confusing, just think about it. Since b is a negative number, negative b is going to be a positive number. And since c is a positive number, negative c is going to be a negative number. So, it makes complete sense that a positive number is going to be greater than a negative number. And you see it here, negative b is to the right of negative c on the number line. So, we can rule this one out. So the next question, is negative b greater than zero? Well, we already plotted negative b, it's going to be one to the right, or one hash mark to the right, we don't know how much each of these hash marks represent, but it's going to be to the right of zero. So, it is greater than zero. This is true. That is true. All right, now is a greater than b? Well, let's look at it. A is to the left of b on the number line. A is more negative than b. So, a is less than b, not greater than b. So, this is not going to be the case. In order for something to be greater than something else, it would have to be to the right of it. For a to the greater than b, it would have to be to the right of b. But we see, a is to the left of b. A is less than b. All right, one more to think about. Negative a is greater than c. So, we know that a isn't greater than c, a is to the left of c. A is a negative number, it's to the left of zero, c is a positive number, it's to the right of zero. But what about negative a? Well, let's draw that. Let me do this in a color that I haven't used yet. Negative a, where would that be? Well a is one, two, three, four, five, six hash marks to the left of zero. And so negative a is going to be six hash marks to the right of zero. So, let's count that. One, two, three, four, five, six. So, negative a is going to be right over there. And notice, negative a is to the right of c, so negative a is greater than c. This is true. And if you get confused, if you say, wait, this looks like a negative, how can it be larger than a positive? Remember, negative a itself is a negative number. And a itself is six hash marks to the left. So if you take the opposite of that, you're going to get a positive number. You're going to get six hash marks to the right. And c, which was already a positive number, is only five hash marks to the right. And so, negative a, this is going to be a positive number, and it's going to be greater than c. It's to the right of c.
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第一单元 生活智慧与时代精神
第一课 美好生活的向导
生活处处有哲学
哲学与生活的关系
关于世界观的学说
哲学是对自然、社会和思维知识的概括和总结
第二课 百舸争流的思想
哲学的基本问题：思维和存在的关系
唯物主义和唯心主义
第三课 时代精神的精华
哲学在社会发展中的作用
马克思主义哲学的地位和作用 产生
马克思主义哲学的基本特征
马克思主义中国化的三大理论成果
综合探究 走进哲学 问辩人生
第二单元 探索世界与追求真理
第四课 探究世界的本质
世界的物质性 自然界、人类社会的物质性
认识运动 把握规律
运动是物质的根本属性
运动是有规律的
第五课 把握思维的奥妙
意识的本质 思维统一于存在
意识的作用 人能够能动地认识世界
人能够能动地改造世界
一切从实际出发 实事求是
第六课 求索真理的历程
人的认识从何而来 实践及其特点
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What's the number of solutions for x + y + z = 7?
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Neville Fogarty
Math Professor at Christopher Newport University · Upvoted by
David Vanderschel
, PhD Mathematics & Physics, Rice (1970) · Author has 404 answers and 2.5M answer views
·9y
Originally Answered: What's the number of solutions for x + y + z = 7 ? ·
Natural numbers is an ambiguous expression here (some consider 0 to be a natural number), so I'll give you two solutions.
First, let's assume that by natural numbers you mean positive integers. An equivalent question to ask is, "Given a row of 7 stars, how many ways can I place 2 bars in the row to partition the stars into 3 groups?" This method is called the stars and bars method. The number of stars in the leftmost group corresponds to x, the number of stars in the middle group corresponds to y, and the number of stars in the rightmost group corresponds to z.
So your row of stars looks like th
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Natural numbers is an ambiguous expression here (some consider 0 to be a natural number), so I'll give you two solutions.
First, let's assume that by natural numbers you mean positive integers. An equivalent question to ask is, "Given a row of 7 stars, how many ways can I place 2 bars in the row to partition the stars into 3 groups?" This method is called the stars and bars method. The number of stars in the leftmost group corresponds to x, the number of stars in the middle group corresponds to y, and the number of stars in the rightmost group corresponds to z.
So your row of stars looks like this:
And one possible way to place the bars is: | |
How many ways can we place the bars? Well, the bars have to go in two of the six slots between stars. There are (6 choose 2) ways to do this. That is:
6!2!(6−2)!=6⋅5⋅4⋅3⋅2⋅1 2⋅1⋅4⋅3⋅2⋅1=6⋅5 2=15 6!2!(6−2)!=6⋅5⋅4⋅3⋅2⋅1 2⋅1⋅4⋅3⋅2⋅1=6⋅5 2=15
So there are 15 positive integer triple solutions to x+y+z=7 x+y+z=7.
Okay, but what if by natural numbers, you meant non-negative integers? I'd rewrite the problem to read "What's the number of positive integer triple solutions for (x+1)+(y+1)+(z+1)=(7+3)(x+1)+(y+1)+(z+1)=(7+3)?" Now this is the same problem as before, as (x+1), (y+1), and (z+1) are all necessarily positive integers. Thus we'd ask the same question as in the first part, but with 10 instead of 7. There are (9 choose 2) ways to partition 10 stars into 3 non-empty groups, so there are 36 positive integer triple solutions to x+y+z=10 x+y+z=10. Thus there are 36 non-negative integer triple solutions to x+y+z=7 x+y+z=7.
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If x,y, z are natural numbers, how many solutions are there to x+y+z=7?
How do I find number of solutions for x+y+z=36 if x,y,z are whole numbers?
If x,y, z ϵ {0,1,2,3,4,5,6,7,8,9} then how many solutions does x+y+z=10 have?
X + y + z + w = 10 how many solutions does it have?
What is the answer if y/x = z, x/y = ?
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To find the number of non-negative integer solutions to the equation x+y+z=7 x+y+z=7, we can use the "stars and bars" theorem. This theorem provides a way to calculate the number of ways to distribute n n indistinguishable objects (stars) into k k distinguishable boxes (variables).
In this case:
The total number of stars (the sum) is n=7 n=7.
The number of variables (boxes) is k=3 k=3 (for x x, y y, and z z).
According to the stars and bars theorem, the number of solutions is given by:
(n+k−1 k−1)(n+k−1 k−1)
Substituting the values:
(7+3−1 3−1)=(9 2)(7+3−1 3−1)=(9 2)
Calculating (9 2)(9 2):
(9 2)=\fr(9 2)=\fr
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To find the number of non-negative integer solutions to the equation x+y+z=7 x+y+z=7, we can use the "stars and bars" theorem. This theorem provides a way to calculate the number of ways to distribute n n indistinguishable objects (stars) into k k distinguishable boxes (variables).
In this case:
The total number of stars (the sum) is n=7 n=7.
The number of variables (boxes) is k=3 k=3 (for x x, y y, and z z).
According to the stars and bars theorem, the number of solutions is given by:
(n+k−1 k−1)(n+k−1 k−1)
Substituting the values:
(7+3−1 3−1)=(9 2)(7+3−1 3−1)=(9 2)
Calculating (9 2)(9 2):
(9 2)=9×8 2×1=72 2=36(9 2)=9×8 2×1=72 2=36
Thus, the number of non-negative integer solutions to the equation x+y+z=7 x+y+z=7 is 36.
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Konstantinos Michailidis
BSc Mathematics ; MSc · Upvoted by
Parag Kalita
, Interested in numbers. · Author has 1K answers and 2.4M answer views
·9y
Originally Answered: What's the number of solutions for x + y + z = 7 ? ·
This is classic stars and bars combinatorics question.
We need to find 3 numbers, i.e. 3-tuples of positive integers such their sum is 7. So the number of solution is:
Refer to Stars and bars (combinatorics)
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This is classic stars and bars combinatorics question.
We need to find 3 numbers, i.e. 3-tuples of positive integers such their sum is 7. So the number of solution is:
Refer to Stars and bars (combinatorics)
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Nancy Mitchell
used to be a teacher. · Author has 3.4K answers and 8M answer views
·Updated 6y
Originally Answered: How many solutions of x+y+z=6 are there if x, y, and z are natural numbers? ·
How many solutions of x+y+z=6 are there if x, y and z are natural numbers?
This problem has two cases.
1.) The equation
x 1+x 2+⋯+x m=r,x 1+x 2+⋯+x m=r,
where
x 1,x 2,…,x m≥0,x 1,x 2,…,x m≥0,
has(m+r−1 m−1)has(m+r−1 m−1)
non-negative integer solutions.
Now let
x,y and z replace x 1,x 2 and x 3,x,y and z replace x 1,x 2 and x 3,
where x,y,z≥0.where x,y,z≥0.
Then
m=3 and r=6 for x+y+z=6.m=3 and r=6 for x+y+z=6.
So, t
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How many solutions of x+y+z=6 are there if x, y and z are natural numbers?
This problem has two cases.
1.) The equation
x 1+x 2+⋯+x m=r,x 1+x 2+⋯+x m=r,
where
x 1,x 2,…,x m≥0,x 1,x 2,…,x m≥0,
has(m+r−1 m−1)has(m+r−1 m−1)
non-negative integer solutions.
Now let
x,y and z replace x 1,x 2 and x 3,x,y and z replace x 1,x 2 and x 3,
where x,y,z≥0.where x,y,z≥0.
Then
m=3 and r=6 for x+y+z=6.m=3 and r=6 for x+y+z=6.
So, the equation has
(3+6−1 3−1)(3+6−1 3−1)
=(8 2)=28=(8 2)=28
non-negative whole-number solutions.
2.) When
x 1≥1,x 2≥1,…,x k≥1 x 1≥1,x 2≥1,…,x k≥1
for a number k. Then there are
(m+r−k−1 m−1)(m+r−k−1 m−1)
positive whole-number solutions to
x 1+x 2+⋯+x m=r.x 1+x 2+⋯+x m=r.
Then for x+y+z=6,k=3 Then for x+y+z=6,k=3
and x,y,z≥1,and x,y,z≥1,
there are
(3+6−3−1 3−1)(3+6−3−1 3−1)
=(5 2)=10=(5 2)=10
positive integer solutions.
For understandable proofs of these formulas, see the article The number of n Ace-King adjacencies in a shuffled deck by John Holte and Mark Holte in the Mathematical Gazette, vol. 77, no. 480, Nov. 1993, pp. 368–370.
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Is (5,7) a solution to 3x+y=22 and x-y=-10?
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What is the number of integral solutions of x+y+z=0 with x>=5,y>=-5 and z>=-5?
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Chiranjeet Saxena
Mathematician · Author has 254 answers and 420.3K answer views
·8y
Originally Answered: What's the number of solutions for x + y + z = 7 ? ·
When X=1 Y=1 Z=5
X=1 Y=2 Z=4
When X=1 Y=3 Z=3
When X=1 Y=4 Z=2
When X=1 Y=5 Z=1
When the value of X is fixed as 1 we get 5 solutions for the equation
For value of X=2 there will be 4 solutions
For X=3 there will be 3 solutions
For X=4 there will be 2 solutions
For X=5 there will be 1 solution
For X=6 there will be No solution(Since one of the variable has to be 0)
Thus the total number of solution to the equation is
5+4+3+2+1=15 solutions
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Nancy Mitchell
used to be a teacher. · Upvoted by
Parag Kalita
, Interested in numbers. · Author has 3.4K answers and 8M answer views
·Updated 7y
Originally Answered: What is the number of the solution for x+y+z=7? ·
What is the number of solution for x+y+z=7?
x+y+z=7 x+y+z=7 is an equation of a plane in Euclidean 3-space. So this equation has infinitely many solutions.
There is an algorithm for finding integer solutions to equations like the one given.
This plane intersects the x-axis at the point (7, 0, 0), the y-axis at (0, 7, 0) and the z-axis at (0, 0, 7).These intercepts give you three solutions of the equation.
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Alon Amit
PhD in Mathematics; Mathcircler. · Upvoted by
Mark Harrison
, BA (Oxon), MA (Oxon), Mathematics and Computation joint honours and
Dan Grubb
, Ph. D. Mathematics, Kansas State University (1986) · Author has 8.8K answers and 173.8M answer views
·Updated 1y
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How do you find the positive integer solutions to x y+z+y z+x+z x+y=4?x y+z+y z+x+z x+y=4?
I was playing around with several cubic representation problems in the style of previous work by Andrew and Richard Guy. The numerical results were fascinating…” (comment on MathOverflow)
This is how Allan MacLeod, a retired mathematician, stumbled upon this equation just a few years ago, and it’s a really fascinating one. Honestly, it’s one of the finest diophantine equations I’ve seen, and I’ve seen a few.
I came across it when it was making the rounds on the web as a nerd-sniping, faux-meme image designed by some cruel soul (Sridhar, was that you?). I had no idea what I was looking at. It loo
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I was playing around with several cubic representation problems in the style of previous work by Andrew and Richard Guy. The numerical results were fascinating…” (comment on MathOverflow)
This is how Allan MacLeod, a retired mathematician, stumbled upon this equation just a few years ago, and it’s a really fascinating one. Honestly, it’s one of the finest diophantine equations I’ve seen, and I’ve seen a few.
I came across it when it was making the rounds on the web as a nerd-sniping, faux-meme image designed by some cruel soul (Sridhar, was that you?). I had no idea what I was looking at. It looked like this:
You may have seen such meme images before. They are always pure click-bait garbage: “95% of MIT graduates couldn’t solve this!”, where “this” is some inane, or ill-defined, or trivial brain teaser.
This one is not. The meme is a clever, or wicked, joke. Roughly 99.999995% of the people don’t stand a chance at solving it, and that includes a good number of mathematicians at leading universities who just don’t happen to be number theorists. It is solvable, yes, but it's really, genuinely hard.
(By the way, it wasn't actually Sridhar, or not exactly. See this comment for the history).
You may think that, if all else fails, we can just throw computers at the problem. It’s really easy to write a computer program that searches for solutions of this simple-looking equation. Surely a computer will find them eventually, if they do exist. Wrong. A brute-force search with a computer is totally useless here, unless you have a few novemdecillion years to spare.
I don’t know how to fit the entire solution in a Quora answer without assuming that everyone already knows everything they need to know about elliptic curves. All I can do here is a brief survey. The main reference is a wonderful, and fairly recent, paper by Bremner and MacLeod called “An unusual cubic representation problem”, published in 2014 in the Annales Mathematicae et Informaticae.
Let’s begin.
We are looking for solutions in positive integers of the equation
a b+c+b a+c+c a+b=4(1)(1)a b+c+b a+c+c a+b=4
(I switched the variable names to match those in the paper).
The first thing to do when you’re looking at any equation is to try and place it in the right context. What sort of thing is this? Well, we are being asked to find solutions in integers, so this is a problem in number theory. As it stands, the equation involves rational functions (polynomials divided by other polynomials), but it’s clear that we can multiply by a common multiple of the denominators to clear things up and get just polynomials, so we have a Diophantine equation. The “positive” requirement is somewhat unusual, and as we shall see it makes things harder.
Now, how many variables do we have here? That seems like a silly question: clearly, we have three, namely a a, b b and c c. Well, not so fast. A trained number theorist immediately notices that the equation is homogenous. What this means is that if (a,b,c)(a,b,c) is any solution of that thing, then so is (7 a,7 b,7 c)(7 a,7 b,7 c). Do you see why? Multiplying each variable by some constant number (7 7 is just an example) doesn’t change anything, because the constant cancels out in each of the fractions.
t a t b+t c=t a t(b+c)=a b+c t a t b+t c=t a t(b+c)=a b+c.
This means that the equation only pretends to be three-dimensional. It is actually two-dimensional. Geometrically, we have a surface (a single equation in three variables generically defines a two-dimensional surface. Generally, k k equations in n n variables define a d d-dimensional manifold, with d=n−k d=n−k). But this surface is actually swept by a line passing through the origin and waving about; the resulting surface can be understood by focusing on how it cuts a single plane. This is a projective curve.
In the most elementary terms, this is what this reduction means: we can classify the solutions, whatever they are, to those where c=0 c=0 and those where c≠0 c≠0. The first class involves just the two variables a a and b b, and for the second class we can just divide through by c c and get a solution with c=1 c=1. So we can really just look for rational solutions in a a and b b for the case c=1 c=1, multiply them by a common denominator, and obtain an integer solution in a,b a,b and c c. Generally speaking, integer solutions of homogenous equations correspond to rational solutions of the dehomogenized version, which is one dimension lower.
Next up: what is the degree of our equation? The degree is the highest power of any term showing up, where “term” is the product of some variables, whose “power” is then the number of terms being multiplied. For example, if you have a 2 b c 4 a 2 b c 4, that’s a term of degree 7=2+1+4 7=2+1+4.
Diophantine equations behave very differently depending on their degree. In broad terms:
Degree 1 1 is easy
Degree 2 2 is fully understood, and can usually be handled by relatively elementary means
Degree 3 3 is a vast ocean of deep theory and a million open problems
Degree 4 4 and up… Really, really hard.
We are in degree 3 3. Why? Well, just multiply through by the denominators:
a(a+b)(a+c)+b(b+a)(b+c)+c(c+a)(c+b)=4(a+b)(a+c)(b+c)a(a+b)(a+c)+b(b+a)(b+c)+c(c+a)(c+b)=4(a+b)(a+c)(b+c)
Even without explicitly expanding everything out, you can see how the degree is 3 3: we’re never multiplying more than three occurrences of any variable. We’re going to have things like a 3 a 3 and b 2 c b 2 c and a b c a b c, but never anything that weighs more than 3 3 units. If you carry out the actual rearrangement, you will find that the equation is
a 3+b 3+c 3−3(a 2 b+a b 2+a 2 c+a c 2+b 2 c+b c 2)−5 a b c=0 a 3+b 3+c 3−3(a 2 b+a b 2+a 2 c+a c 2+b 2 c+b c 2)−5 a b c=0.
You may object that multiplying through by the denominators is illegal if any of them happens to be 0 0. This is true, and indeed our new equation has a few solutions that don’t correspond to anything in the original equation. But that’s actually a good thing. The polynomial version adds a few “patches” to the original and makes it easier to work with; for any particular solution we find, we just need to check that it doesn’t make any of the original denominators vanish.
In fact, the polynomial equation is easily solved with, for instance, a=−1 a=−1, b=1 b=1, c=0 c=0. That’s good: we have a rational solution, or a rational point. This means that our cubic equation (degree-3) is actually an elliptic curve.
When you find that your equation is an elliptic curve you A) rejoice and B) despair, because there’s so much to learn. This equation is a great example of how the powerful theory of elliptic curves can be applied to discover insanely hard to find solutions.
The first thing you do with an elliptic curve is bring it to Weierstrass form. This is an equation that looks like
y 2=x 3+a x+b y 2=x 3+a x+b
or sometimes
y 2=x 3+a x 2+b x+c y 2=x 3+a x 2+b x+c (this is called the long Weierstrass form. It is not strictly needed, but sometimes it’s more convenient).
It’s a basic fact that any elliptic curve can be brought to this form (except if you’re working over fields of small characteristic, which you needn’t worry about here). It’ll take me a whole big answer to explain how to find the right transformation. Just know that it’s a completely mechanical process (it relies, crucially, on the fact that we have at least one rational point, which we do). There are several computer algebra packages which will do this for you, easily.
But even if you don’t know how to find the transformation, verifying it is easy – or, at least, purely mechanical. The required transformation in our case is given by the scary-looking formulas
x=−28(a+b+2 c)6 a+6 b−c x=−28(a+b+2 c)6 a+6 b−c
y=364(a−b)6 a+6 b−c y=364(a−b)6 a+6 b−c
I know this looks like random voodoo, but please believe me that it’s not. Once you have those transformations, a tedious but straightforward algebraic calculation confirms that
y 2=x 3+109 x 2+224 x(2)(2)y 2=x 3+109 x 2+224 x
This equation, though it looks very different, is actually a faithful model of the original. Graphically, it looks like this – a typical elliptic curve with two real components:
The “fish tail” on the right keeps growing to infinity and beyond. The close oval shape on the left is, well, closed, and it will turn out that it’s where the fun is, for us.
Given any solution (x,y)(x,y) of this equation, you can recover the needed a,b,c a,b,c using
a=56−x+y 56−14 x a=56−x+y 56−14 x
b=56−x−y 56−14 x b=56−x−y 56−14 x
c=−28−6 x 28−7 x c=−28−6 x 28−7 x
(Remember that the triplet (a:b:c)(a:b:c) is to be understood projectively – whatever values you get through those equations, you can always multiply them by whatever constant you wish).
The two maps we exhibited, from a,b,c a,b,c to x,y x,y and vice versa, show that these two equations are “the same” from the perspective of number theory: rational solutions of one give you rational solutions of the other. The technical term is birational equivalence, and it’s a very fundamental concept in algebraic geometry. As we noted, there could be some exceptional points that don’t genuinely map correctly; those are the cases where a+b,a+c a+b,a+c or b+c b+c turn out to be 0 0. This is standard fare in birational equivalences, and shouldn’t cause any concern.
Let’s look at an example.
The elliptic curve (2) has a nice rational point on it: x=−100 x=−100, y=260 y=260. It’s not easy to find perhaps, but it’s super easy to verify: Just plug in those values and see that the two sides are the same (I didn’t choose this point at random, but never mind for now). We can simply check which a,b,c a,b,c values it yields. We get
a=2/7,b=−1/14,c=11/14 a=2/7,b=−1/14,c=11/14
and since we can multiply by a common denominator, this can be modified into
a=4,b=−1,c=11 a=4,b=−1,c=11.
And indeed,
4−1+11+−1 4+11+11 4−1=4 4−1+11+−1 4+11+11 4−1=4
as you can readily verify. This is a simple solution to our original equation in integers, but alas – not positive integers. This solution is not easy to see by hand, but it’s also not hard to discover with some patience without all the machinery we are reviewing here. It’s the positive solutions that are the lair of dragons.
Now, as soon as you have a rational point on an elliptic curve, such as P=(−100,260)P=(−100,260) on our curve (2), you can start generating others using the chord and tangent technique, which we covered on a previous Quora episode.
To begin with, we can add our point P P to itself by finding the tangent to the curve at P P, and seeing where it meets the curve again. The result turns out to be a bit scarier:
P+P=2 P=(8836/25,−950716/125)P+P=2 P=(8836/25,−950716/125)
and again, this new point corresponds to values of a,b,c a,b,c which solve our original equation:
(a,b,c)=(9499,−8784,5165)(a,b,c)=(9499,−8784,5165).
This solution is definitely not easy to find by hand, but it’s still manageable with a computer. However, it’s still not positive.
Undeterred, we continue to calculate 3 P=2 P+P 3 P=2 P+P which is done by connecting P P and 2 P 2 P with a straight line and finding its third point of intersection with the curve. And again, we calculate a,b,c a,b,c, and again, the result isn’t positive. And so it is with 4 P 4 P, 5 P 5 P and so on… until we hit 9 P 9 P.
9P=(-66202368404229585264842409883878874707453676645038225/13514400292716288512070907945002943352692578000406921,
58800835157308083307376751727347181330085672850296730351871748713307988700611210/1571068668597978434556364707291896268838086945430031322196754390420280407346469)
definitely not easy to find, but given our machinery, all we had to do was iterate a simple geometric procedure nine times. The corresponding values of a,b,c a,b,c are amazing:
a=154476802108746166441951315019919837485664325669565431700026634898253202035277999,
b=36875131794129999827197811565225474825492979968971970996283137471637224634055579,
c=4373612677928697257861252602371390152816537558161613618621437993378423467772036
Those are 80 80-digit numbers! You cannot find 80 80 digit numbers with a computer using brute force search. But unbelievable as it may seem, those huge numbers, when plugged into the simple expression a/(b+c)+b/(a+c)+c/(a+b)a/(b+c)+b/(a+c)+c/(a+b), yield exactly 4 4.
Those are, in fact, the smallest solutions to the problem. As we continue to add the point P P to itself, the denominators just keep growing. It is not quite easy to prove that, as there's always a possibility of some cancellation, but the theory of heights on an elliptic curve allows us to show that those astronomical numbers are, indeed, the simplest solutions to the question.
Back to some theory. An elliptic curve over the rationals has a rank, which is the number of points we need to get started with the chord and tangent method and know that we will eventually find all rational points on the curve. Our specific elliptic curve (2) has rank 1, which means it has infinitely many rational points but they are all generated from a single one, which is none other than our point P=(−100,260)P=(−100,260). The algorithms for calculating the rank and finding such a generator are far from trivial, but SageMath does it in under a second, using a few short lines of code. You can see my code here. It reconstructs the whole solution from scratch, but of course using Sage’s built-in methods for handling elliptic curves.
In our case, the point P P lies on the oval part of the curve, and so do the points m P m P for any positive odd integer m m. They sort of run around the oval and eventually cover it fairly uniformly. This is lucky, because only a small fraction of that oval yields positive solutions in terms of a,b,c a,b,c: it is the thickened portion in the following diagram, taken from Bremner and MacLeod’s paper.
The points P,2 P P,2 P and so on don’t lie on the black part, but 9 P 9 P does, which is how we got our 80-digit positive solutions.
Bremner and MacLeod looked at what happens when we replace the 4 4 with something else. If you think our solutions are big, wait till you see what happens when you try to represent 178 178 in this way. Instead of 80 digits, you’ll need 398,605,460 digits. Yes, that’s just the number of digits in the solution. If you try 896 896, you’ll be up to trillions of digits. Trillions. For this innocuous equation:
a/(b+c)+b/(a+c)+c/(a+b)=896 a/(b+c)+b/(a+c)+c/(a+b)=896.
This is a striking example of the way diophantine equations with tiny coefficients can have enormous solutions. This isn’t merely awe-inspiring, it is profound. The negative solution of Hilbert’s 10th problem means that the growth of the solutions as the coefficients get larger is an uncomputable function, for if it were computable, we would have had a simple algorithm for solving diophantine equations, and there isn’t one (simple or complex). Here, the correspondence 4→4→ 80-digit numbers, 178→178→ hundreds-of-millions-digit numbers and 896→896→ trillions of digits gives us a glimpse into the first few tiny steps of that monstrous, uncomputable function. Tweak the numbers in your equation, and the solutions promptly exceed anything that fits in our puny little universe.
What a wonderfully sneaky little equation.
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Deb P. Choudhury
Former Professor at University of Allahabad · Author has 10K answers and 8M answer views
·7y
Originally Answered: If x,y, z are natural numbers, how many solutions are there to x+y+z=7? ·
This is a problem of distributing 7 non-distinct objects into 3 distinct boxes with unconstrained repetition. If we were given the condition that each of x, y, z are non-negative, i.e. >/=0, then the number of solutions would have been C(7+3–1,7)=C(9,7).
But we are given that x, y and z are natural numbers, i.e. each is >/=1, hence we put x=x'+1, y=y'+1, z=z'+1, so that x'+y'+z'+3 =7, i.e. we have to count all the non-negative integer solutions of x'+y'+z' =4. Now by the earlier formula this is given by C(4+3–1,4) =C(6,4)=C(6,2) =6!/(2!×4!) = 6×5/(1×2) = 15. The explicit solutions have been lis
Continue Reading
This is a problem of distributing 7 non-distinct objects into 3 distinct boxes with unconstrained repetition. If we were given the condition that each of x, y, z are non-negative, i.e. >/=0, then the number of solutions would have been C(7+3–1,7)=C(9,7).
But we are given that x, y and z are natural numbers, i.e. each is >/=1, hence we put x=x'+1, y=y'+1, z=z'+1, so that x'+y'+z'+3 =7, i.e. we have to count all the non-negative integer solutions of x'+y'+z' =4. Now by the earlier formula this is given by C(4+3–1,4) =C(6,4)=C(6,2) =6!/(2!×4!) = 6×5/(1×2) = 15. The explicit solutions have been listed by Nirmalya.
Remember that number of non-negative integer solutions of the equation:
(x_1)+(x_2)+…..+(x_ r) =n is C(n+r-1,n) which is also the number of ways of distributing n identical objects into r distinct boxes with unconstrained repetition (subject to availability). Note that some boxes may remain empty. The above basic formula may be modified to suit specific requirements as was the case above.
Upvote ·
9 6
Surbhi
Former Math Expert at Photomath (2021–2022) · Author has 330 answers and 761.8K answer views
·6y
Originally Answered: How many solutions of x+y+z=6 are there if x, y, and z are natural numbers? ·
I have a formula , you can say a short trick ,
So x+y+z=k , where k is a natural no.
Case 1: Now,suppose if you need to find the no. of non-negative solutions for the equation.
For the formula:
(i)[constant on the RHS+ no . of variables on the LHS -1]= k+3–1=k+2
(ii) [constant on RHS]- 1 = k-1.
No. of non-negative solutions : is [(k+2)C ( k-1)], where C is for comination.
Case 2: Now if you are asked for the no.of positive solutions ,where x,y,z are natural number
The above formula is useful only if x,y,z starts from zero….
So for new case , we'll assume
x=X+1; y=Y+1; = z=Z+1
Now replace it it the given
Continue Reading
I have a formula , you can say a short trick ,
So x+y+z=k , where k is a natural no.
Case 1: Now,suppose if you need to find the no. of non-negative solutions for the equation.
For the formula:
(i)[constant on the RHS+ no . of variables on the LHS -1]= k+3–1=k+2
(ii) [constant on RHS]- 1 = k-1.
No. of non-negative solutions : is [(k+2)C ( k-1)], where C is for comination.
Case 2: Now if you are asked for the no.of positive solutions ,where x,y,z are natural number
The above formula is useful only if x,y,z starts from zero….
So for new case , we'll assume
x=X+1; y=Y+1; = z=Z+1
Now replace it it the given equation , new eqn becomes :
X+1 + Y +1 + Z+1= k
So , X+Y+Z=k-3….(i)
We'll the non-negative solutions of this equation
This eqn will give us the solutions if x,y,z are natural numbers because X,Y,Z will take the values starting from zero , and then x,y,z will take values from atleast 1 .
So the No .of Solutions =
[K-3+3–1]C[k-3-1]
[K-1]C[k-4]
So the answer of this particular question is given by:
(6–1)C(6–4)
5C2
= 10
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Max Gretinski
Studied Mathematics · Author has 6.6K answers and 2.5M answer views
·2y
Originally Answered: If x,y, z are natural numbers, how many solutions are there to x+y+z=7? ·
Definition: The Natural Numbers consist of 1, 2, 3, 4, ….
7 = 5 + 1 + 1
7 = 4 + 2 + 1
7 = 3 + 3 + 1
7 = 3 + 2 + 2
These are the only partitions of 7 into exactly three addends.
If the order (x, y, z) does not matter to us … that is, if we treat (x, y, z) the same as (x, z, y) … then there are four partitions of 7 into three addends, and we are done.
However, if the order (x, y, z) matters to us, each partition may be divided into different orders.
In the information below, I use “=” to indicate “has the same three addends that sum to 7.”
(5, 1, 1) = (1, 5, 1) = (1, 1, 5).
(4, 2, 1) = (4, 1, 2) = (2, 4,
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Definition: The Natural Numbers consist of 1, 2, 3, 4, ….
7 = 5 + 1 + 1
7 = 4 + 2 + 1
7 = 3 + 3 + 1
7 = 3 + 2 + 2
These are the only partitions of 7 into exactly three addends.
If the order (x, y, z) does not matter to us … that is, if we treat (x, y, z) the same as (x, z, y) … then there are four partitions of 7 into three addends, and we are done.
However, if the order (x, y, z) matters to us, each partition may be divided into different orders.
In the information below, I use “=” to indicate “has the same three addends that sum to 7.”
(5, 1, 1) = (1, 5, 1) = (1, 1, 5).
(4, 2, 1) = (4, 1, 2) = (2, 4, 1) = (2, 1, 4) = (1, 4, 2) = (1, 2, 4)
(3, 3, 1) = (3, 1, 3) = (1, 3, 3).
(3, 2, 2) = (2, 3, 2) = (2, 2, 3).
Accounting for the different orderings of the same partitions, we have fifteen solutions.
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Roddy MacPhee
knows set theory · Author has 483 answers and 556K answer views
·7y
Originally Answered: What is the number of the solution for x+y+z=7? ·
On the positive integers: (1,2,4),(1,3,3),(1,1,5),and (2,2,3), non-negative integers: (0,1,6),(0,2,5),and (0,3,4) get added. After that infinitely many.
Upvote ·
9 1
Thomas Subia
Statistician · Author has 3.2K answers and 3.6M answer views
·4y
Originally Answered: If x,y, z are natural numbers, how many solutions are there to x+y+z=7? ·
Here are two graphs which can illustrate this.
Lets stipulate that x,y and z are integers
Graph 1 represents the values of x, y and z whose sum = 7
Notice here that we’ve defined x and y to be integers whose values are constrained to be between 0 and 3
Hence for integers whose x and y values are confined between 0 and 3, there are 16 x,y,z triples whose sum = 7.
This makes sense since with 4 elements of x and 4 elements of y, there are 16 different pairs of (x,y)
Graph 2: Confine x and y to be between -2 and 2.
Here we see that there are 25 (x,y,z) triples whose sum is equal to 7.
What we observe here
Continue Reading
Here are two graphs which can illustrate this.
Lets stipulate that x,y and z are integers
Graph 1 represents the values of x, y and z whose sum = 7
Notice here that we’ve defined x and y to be integers whose values are constrained to be between 0 and 3
Hence for integers whose x and y values are confined between 0 and 3, there are 16 x,y,z triples whose sum = 7.
This makes sense since with 4 elements of x and 4 elements of y, there are 16 different pairs of (x,y)
Graph 2: Confine x and y to be between -2 and 2.
Here we see that there are 25 (x,y,z) triples whose sum is equal to 7.
What we observe here is that number of triples whose sum = 7 is the square of the number of integers within a given interval. This assumes that both x and y intervals contain equal number of integers within that interval.
Let’s say the the range of x is confined from -2 to 2 and y is confined to y = -2, Then the number of triples (x,y,z) which satisfy our equation is 5. This makes sense since there are 5 possible integers for x and only one integer value for y.
So in general, the number of integer solutions for x + y + z = 7 might be found by computing the product of the number of integers with a range of x and the number of integers within a range of y.
As always, this was not obvious when we first encountered the problem. Only by graphing this could we have come to this conclusion.
These graphs were generated using Minitab. One great feature of Minitab is its ability to create a grid of numbers. Without that feature, generating these graphs would have been more difficult!
Hope this helps!
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9 1
Theresa Knott
Independent State School Coordinator (2018–present) · Upvoted by
Yuri Ginsburg
, M. S. Mathematics & Computer Science (1974) · Author has 90 answers and 123.3K answer views
·7y
Originally Answered: If x,y, z are natural numbers, how many solutions are there to x+y+z=7? ·
Ok I'm going to assume that we don't actually care about the labels x, y and z or the order they are written in. ( If you do then any solutions that I find can be swapped around to give others)
Since x, y, and z are all natural numbers then can only be in the range 1 to 7. So we don't have that much checking to do.
Can one of the numbers be 7? No because the other two would have to sum to 0 and there is no way to do that only using natural numbers.
Can one of the numbers be 6? No because the other two would have to sum to 1 and you can't do that with two natural numbers.
Can one of the numbers be
Continue Reading
Ok I'm going to assume that we don't actually care about the labels x, y and z or the order they are written in. ( If you do then any solutions that I find can be swapped around to give others)
Since x, y, and z are all natural numbers then can only be in the range 1 to 7. So we don't have that much checking to do.
Can one of the numbers be 7? No because the other two would have to sum to 0 and there is no way to do that only using natural numbers.
Can one of the numbers be 6? No because the other two would have to sum to 1 and you can't do that with two natural numbers.
Can one of the numbers be 5? Yes!
5, 1, 1 is a possible solution (So is 1, 5, 1 and 1, 1, 5 but I'm ignoring those as they as they are just swappings around of x, y and z)
4, 1, 2 is another solution
3, 3, 1 and 3, 2, 2 are the last two.
That's it!
Upvote ·
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1: Basics of Geometry
Geometry
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Sat, 01 Oct 2022 18:13:44 GMT
1.1: Geometry Terms
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1.1: Geometry Terms
Last updated Oct 1, 2022
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1: Basics of Geometry
1.2: Shapes
Page ID 2117
( \newcommand{\kernel}{\mathrm{null}\,})
Table of contents
1. Geometric Definitions
2. Postulates
3. Review
4. Review (Answers)
5. Vocabulary
6. Additional Resources
Define and use terms, including points, lines, planes, space, and postulates.
Geometric Definitions
A point is an exact location in space. It describes a location, but has no size. Examples are shown below:
Figure 1.1.1
Table 1.1.1| Label It | Say It |
--- |
| A | point A |
A line is infinitely many points that extend forever in both directions. Lines have direction and location and are _always straight_.
Figure (\PageIndex{2})
Table 1.1.2| _Label It_ | _Say It_ |
--- |
| line g | line g |
| PQ↔ | line PQ |
A plane is a flat surface that contains infinitely many intersecting lines that extend forever in all directions. Think of a plane as a huge sheet of paper with no thickness that goes on forever.
Figure 1.1.3
Table 1.1.3| Label It | Say It |
--- |
| Plane M | Plane M |
| Plane ABC | Plane ABC |
We can use point, line, and plane to define new terms.
Space is the set of all points extending in threedimensions. Think back to the plane. It extended in two dimensions, what we think of as up/down and left/right. If we add a third dimension, one that is perpendicular to the other two, we arrive at three-dimensional space.
Points that lie on the same line are collinear. P,Q,R,S, and T are collinear because they are all on line w. If a point U were located above or below line w, it would be non-collinear.
Figure 1.1.4
Points and/or lines within the same plane are coplanar. Lines h and i, and points A,B,C,D,G, and K are coplanar in Plane J. Line KF↔ and point E are non-coplanar with Plane J.
Figure 1.1.5
An endpoint is a point at the end of a line segment. A line segment is a portion of a line with two endpoints. Or, it is a finite part of a line that stops at both ends. Line segments are labeled by their endpoints. Order does not matter.
Figure 1.1.6
Table 1.1.4| Label It | Say It |
--- |
| AB― | Segment AB |
| BA― | Segment BA |
A ray is a part of a line. It begins with an endpoint and extends forever away from the endpoint in one direction, perfectly straight. A ray is labeled by its endpoint and one other point on the ray. For rays, order does matter. When labeling, put the endpoint under the side WITHOUT the arrow.
Figure 1.1.7
Table 1.1.5| Label It | Say It |
--- |
| CD→ | Ray CD |
| DC← | Ray CD |
An intersection is a point or set of points where lines, planes, segments, or rays overlap.
Figure 1.1.8
Postulates
A postulate is a basic rule of geometry. Postulates are assumed to be true (rather than proven), much like definitions. The following is a list of some basic postulates.
Postulate #1: Given any two distinct points, there is exactly one (straight) line containing those two points.
Figure 1.1.9
Postulate #2: Given any three non-collinear points, there is exactly one plane containing those three points.
Figure 1.1.10
Postulate #3: If a line and a plane share two points, then the entire line lies within the plane.
Figure 1.1.11
Postulate #4: If two distinct lines intersect, the intersection will be one point.
Figure 1.1.12
Lines I and m intersect at point A.
Postulate #5: If two distinct planes intersect, the intersection will be a line.
Figure 1.1.13
When making geometric drawings, be sure to be clear and label all points and lines.
What if you were given a picture of a figure or object, like a map with cities and roads marked on it? How could you explain that picture geometrically?
Example 1.1.1
What best describes San Diego, California on a globe: point, line, or plane?
Solution
A city is usually labeled with a dot, or point, on a globe.
Example 1.1.2
Use the picture below to answer these questions.
Figure (\PageIndex{14})
List another way to label Plane J.
List another way to label line h.
Are K and F collinear?
Are E,B and F coplanar?
Solution
Plane BDG is one possibility. Any combination of three coplanar points that are not collinear would be correct.
AB↔. Any combination of two of the letters A, B, or C would also be correct.
Yes, they both lie on KF↔.
Yes, even though E is not in Plane J, any three points make a distinct plane. Therefore, the three points create Plane EBF.
Example 1.1.3
What best describes a straight road that begins in one city and stops in a second city: ray, line, segment, or plane?
Solution
The straight road connects two cities, which are like endpoints. The best term is segment.
Example 1.1.4
Answer the following questions about the picture.
Figure 1.1.15
Is line l coplanar with Plane V, Plane W, both, or neither?
Are R and Q collinear?
What point belongs to neither Plane V nor Plane W?
List three points in Plane W.
Solution
Neither
Yes
S
Any combination of P, O, T, and Q would work.
Example 1.1.5
Draw and label a figure matching the following description: Line AB↔ and ray CD→ intersect at point (C). Then, redraw so that the figure looks different but is still true to the description.
Solution
Neither the position of A or B on the line, nor the direction that CD→ points matter.
Figure 1.1.16
For the second part, this is one way to draw the diagram differently:
Figure 1.1.17
Review
For questions 1-5, draw and label a figure to fit the descriptions.
CD→ intersecting AB― and Plane P containing AB― but not CD→.
Three collinear points A, B, and C. B is also collinear with points D and E.
XY→, XZ→, and XW→, such that XY→ and XZ→ are coplanar, but XW→ is non-coplanar with both of the other rays.
Two intersecting planes, P and Q, with GH―, where G is in plane P and H is in plane Q.
Four non-collinear points I, J, K,and L, with line segments connecting all points to each other.
Name this line in five ways.
Figure 1.1.18
Name the geometric figure in three different ways.
Figure 1.1.19
Name the geometric figure in two different ways.
Figure 1.1.20
What is the best possible geometric model for a soccer field? Explain your answer.
List two examples of where you see rays in real life.
What type of geometric object is the intersection of a line and a plane? Draw your answer.
What is the difference between a postulate and a theorem?
For 13-16, use geometric notation to explain each picture in as much detail as possible.
Figure 1.1.21
Figure 1.1.22
Figure 1.1.23
Figure 1.1.24
For 17-25, determine if the following statements are true or false.
Any two points are collinear.
Any three points determine a plane.
A line is two rays with a common endpoint.
A line segment is infinitely many points between two endpoints.
A point takes up space.
A line is one-dimensional.
Any four points are coplanar.
AB→ could be read “ray AB” or “ray BA.”
AB↔ could be read “line AB” or “line BA.”
Review (Answers)
To see the Review answers, open this PDF file and look for section 1.1.
Vocabulary
| Term | Definition |
--- |
| collinear | Three or more points are collinear when they lie on the same line. |
| coplanar | Points and/or lines located on the same plane are coplanar. |
| endpoint | An endpoint is a point at one end of a line segment or ray. |
| intersection | A point or set of points where lines, planes, segments, or rays cross. |
| line | Infinitely many points that extend forever in both directions. |
| line segment | A line segment is a part of a line that has two endpoints. |
| plane | A plane is a flat, two-dimensional surface. It can be conceptualized as a sheet of paper of infinite area. |
| Point | A point is a location in space that does not have size or shape. |
| postulate | A postulate is a statement that is accepted as true without proof. |
| ray | A part of a line with one endpoint that extends forever in the direction opposite that point. |
| space | Space is the set of all points extending in three dimensions. |
| Dimensions | Dimensions are the measurements that define the shape and size of a figure. |
| Non-collinear | A non-collinear point is located above or below a line. |
| Non-coplanar | A non-coplanar point is located above or below a plane. |
| theorem | A theorem is a statement that can be proven true using postulates, definitions, and other theorems that have already been proven. |
Additional Resources
Interactive Element
Video: Basic Geometric Definitions Principles - Basic
Activities: Basic Geometric Definitions Discussion Questions
Study Aids: Basics of Geometry Study Guide
Practice: Geometry Terms
Real World: Basic Geometric Definitions
This page titled 1.1: Geometry Terms is shared under a CK-12 license and was authored, remixed, and/or curated by CK-12 Foundation via source content that was edited to the style and standards of the LibreTexts platform.
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6956 | https://access.openupresources.org/curricula/our6-8math/en/grade-6/unit-7/lesson-4/index.html | Grade 6 Mathematics, Unit 7.4 - Open Up Resources
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Practice Problems
Lesson 4 Ordering Rational Numbers
Let’s order rational numbers.
Learning Targets:
I can compare and order rational numbers.
I can use phrases like “greater than,” “less than,” and “opposite” to compare rational numbers.
4.1 How Do They Compare?
Use the symbols >, <, or =to compare each pair of numbers.Be prepared to explain your reasoning.
12 _____ 19
212 _____ 190
15 _____ 1.5
9.02 _____ 9.2
6.050 _____ 6.05
0.4 _____
4.2 Ordering Rational Number Cards
Your teacher will give you a set of number cards. Order them from least to greatest.
Your teacher will give you a second set of number cards. Add these to the correct places in the ordered set.
4.3 Comparing Points on A Line
Use each of the following terms at least once to describe or compare the values of points , , , .
greater than
less than
opposite of (or opposites)
negative number
Tell what the value of each point would be if:
is
is -0.4
is 200
is -15
Are you ready for more?
The list of fractions between 0 and 1 with denominators between 1 and 3 looks like this: We can put them in order like this:
Now let’s expand the list to include fractions with denominators of 4. We won’t include , because is already on the list.
Expand the list again to include fractions that have denominators of 5.
Expand the list you made to include fractions have have denominators of 6.
When you add a new fraction to the list, you put it in between two “neighbors.” Go back and look at your work. Do you see a relationship between a new fraction and its two neighbors?
Lesson 4 Summary
To order rational numbers from least to greatest, we list them in the order they appear on the number line from left to right.For example, we can see that the numbers
-2.7, -1.3, 0.8
are listed from least to greatest because of the order they appear on the number line.
Lesson 4 Practice Problems
Selectall of the numbers that are greater than.
1.3
Order these numbers from least to greatest: , 0, 1, , ,
Here are the boiling points of certain elements in degrees Celsius:
Argon: -185.8
Chlorine: -34
Fluorine: -188.1
Hydrogen: -252.87
Krypton: -153.2
List the elements from least to greatest boiling points.
Explain why zero is considered its own opposite.
Explain how to make these calculations mentally.
Find the quotients.
Over several months, the weight of a baby measured in pounds doubles. Does its weight measured in kilograms also double? Explain.
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6957 | https://www.youtube.com/watch?v=Huaez7lXFcM | How to Find Local Maximum and Minimum Values of a Polynomial Function | Glass of Numbers
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In this video, we use the First Derivative Test to find the local maximum and minimum values of a polynomial function.
How to find the local extreme values? We first need to find critical numbers of the function. Then plot the critical numbers on the number line (x-axis) to construct a sign analysis chart. This chart will tell us on what interval the function is increasing or decreasing. We can next apply the First Derivative Test to find the local maximum and minimum values.
Note that because we are finding the max and min values, we will finding the y-values of the function. Do not confuse this with the location of the extrema, which are the x-values.
See here how to find local extreme values for a rational function:
See here how to deal with a trigonometric function with infinitely many local max and min:
See here how to find the local extreme values for a sum of exponential function:
See here how to find the local extreme values for a polynomial function in factored form:
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hello everyone its Wilson here we want to find the local maximum and minimum values for this function without actually graphing the function so how do we do that we can use calculus to do it the way to do it is that we can use what we call the first derivative test to find the local maximum and minimum points ok so first thing is that we need to do the we need to find the critical numbers and to find a critical numbers we will need to take the derivative of this function first ok so let's start by taking the derivative of this function so we have F prime of X ok that's equal to 12x squared and then plus 6x and then minus 6 okay so that's our derivative and then when we need to find a critical numbers and for the for the critical numbers we need we have two cases either f prime is equal to 0 or F prime does not exist and so for the case when F prime is equal to 0 then let's just let's just start writing down this information here we need to find the critical numbers ok so first case is that F prime is equal to 0 then in this case what are we going to be getting here so that means that 12 X square we can set this derivative the first derivative equal to 0 and then we try to solve this equation and so we can factor out 6 from all the terms then we are going to we are going to get what we are going to get 6 and then we are having 2 X square and then plus X and then minus 1 and that's equal to 0 and then now we can factor this 2x squared plus X minus 1 so to factor that we are going to get 6 then times now we will factor this trinomial as a product of two binomials right so first we need to come up with two turns that will multiply to 2x square and so that would be that would be just X right so let me put down the X right here X and then 2x and then we also need to figure out two numbers when we multiply them we are going to get the 1 so we are going to get the 1 here and then the 1 here so 1 times 1 will give you the 1 and then now we need to worry about the signs because we are having a length of 1 right here so that means one of the ones has a pause this has a positive sign and then the other one will have a negative sign in front of it ok so now let's check if we look at the middle turn the middle turn has a positive 1x and to get positive 1x we actually need to get the this one times the 2x to be a positive positive 2x so that we can get the positive x and so what we are going to do is that we are going to put a plus sign in front of this one right here and then we are going to get a put a negative sign here okay and then we can do a quick trick we have 2x and then that's negative x so that will give us the positive x so now that's good and so if we are solving this equation then we are going to get x equals like the one here and then the other possibility is that X is equal to 1 over 2 all right 1 over 2 when we substitute into this X right here we get 2 times 1 over 2 which is 1 1 minus 1 we get 0 ok so those are the two solutions for this equation and so that means those two are the critical numbers and what about the other case the other case for the critical numbers for finding the critical numbers is that F prime does not exist ok but because this is a polynomial so we cannot really find any values for which F prime does not exist and so that means we are not going to be getting any values right here so there is no value for this sub case here no values okay so we have our two critical numbers so next step is that we are going to make a sign analysis chart so that we can use the we can use that what is that the first derivative test okay so how do we do that we are going to first start by drawing a lumber line here so let's do that so let's draw a number line okay did you draw this lumber line right here and then this is actually our x-axis okay and then we can just do some labelings on there what do we need to label on this lumber line for this lumber line all we'll need to label is that the two critical numbers on there we do not need to label any other the other numbers so we are getting like the one and then the other one is the one over two okay so and those two critical numbers will break the lumber line into three intervals so there is one interval for all the X values less than negative one one interval for all the X values between negative one and one over two and then all the X values that's that are greater than one over two okay so now we are going to pick a lumber so we are going to pick a lumber and then we are going to plug it back into our derivative and do the checking and so what we can do is that we can we can pick we can pick what can we pick here we can pick zero because zero is between a legged lumber and a positive number so that's for the ease of the calculation actually it doesn't matter what lumber is that you pick between the leg of the one in the one half you can pick 0.1 0.2 0.3 as long as it's less than one half and greater than negative one then that will be okay but usually we just pick an easy lumber for the purpose of the making the calculation easier now for this one we are going to pick since this is a polynomial the domain is all real numbers so that means we can pick any numbers that we have here right and so what we can pick here is that we are going to just pick a large number we pick 1000 and then on the left side of this negative one here we are going to pick a leg that 1000 and then you may say why do I pick those big numbers you see in a few seconds why picking those big numbers will be really helpful for checking the signs because all we care are the signs right we don't really care about what the answer is when you plug in the those lumbers into the first derivative so now if you plug in a zero in here we can do the calculation really quickly that's actually just the y-intercept or the first derivative so that would be negative six and so that means we have been put a minus sign right here so that's minus sign okay so that's that and then the next one the next one would be what we plug in the 1000 in there and as you can see here when you plug in the 1000 it becomes really obvious that the the sign will be positive even though we don't know what this number is equal to and so we are going to get positive and then the same thing here when you plug in negative 1000 you are going to square that leg at 1000 so it become it will become a million and then you multiply that by 12 and even though you are going to plug in the leg that 1000 in here which will give you a turn of leggett of 6,000 but that's too small compared to your leading turn here right your dominant turn for this first derivative so this would still be a positive quantity and so you may say why do we care about all the signs it's really because we let's say for this interval the derivative is negative right what does that mean that means the function the original function the F not the F prime is decreasing on this interval so that means F is decreasing from negative 1 to 1 / - so we can now actually write down the result right here the function is decreasing okay and then for the intervals that are that in which the F prime is positive then the function the original function is increasing so now we can actually write down those information right here if we are writing down the information then we are going to we are going to get what we are going to have this so the increasing intervals for the function it's going to be first one is from negative infinity to negative one right so we are going to just put that down right here so what is that that's that's like the infinity to negative one and then what is the other interval the function is also increasing from 1 over 2 to infinity so we are going to get the second interval so 1/2 to infinity okay so that's that and then we can also write down the decreasing interval right here or really happen is that the there is only one interval that F is decreasing which is from negative 1 to 1 over 2 so we can write it down here so that would be negative 1 0 1 over 2 okay now how do we find the local maximum and minimum points if we just try to imagine a little bit what the function looks like and we can actually do it like a really rough sketch on the sign analysis chart to show it how do we show it it's this now we know that the function has been increasing since negative infinity and then it will change to decreasing when it goes past like the 1 right so we know that it's going to be increasing and then once it hits this point it will get a horizontal tangent how do we know that it's the horizontal tangent because that's like the one is the value that will make the when we plug that into the first derivative the first derivative will become zero so we are getting a horizontal tangent here and then be beyond this point then we are going to have a function have the portion of the function that's decreasing so it's going to go down like that okay so now as you can see here we just look at this local area right here then we can see that that's actually a what a local maximum okay now on the other hand for the 1/2 we know that the function has been decreasing so we are going to get the decreasing right here then do you see what's going on here once it passes the 1/2 it will turn into increasing right so the functions that's increasing after the 1/2 so you can see that that will be a local minimum okay so we already have determined the maximum and the minimum and then so now what do we do we can just write down the result here for the local Max value right or the local Max value so how do we get that we are going to plug in the negative 1 into the original function because you are finding the Y value of the function so we have f of with f of what is that that's going to be like the one in there and then what do we get here we are going to have 4 times something cubed + square minus 6 times something and then plus 1 and then all those access all those blanks we are going to fill in the leg of the one and then I'm not going to spend the time doing the calculation right here I'm just going to just write on the answer the answer will be 6 the answer will be 6 for this one than to just make it easier easier for us to read way so I'm going to just underline all those words right here so now it's easier to read and then we also have a local minimum right so let's write it down as well so we have our local minimum value and let me just underline that okay so what is the that value so let's just write down the function first then we just feel that in what is that that's one half that we are going to plug in in there a so we have 1/2 1/2 and 1/2 1/2 and then you do all that calculation right you can use your calculator or you can do it manually by hand it doesn't really matter and eventually we are going to get the answer to be Leggett 3 over 4 and so our local maximum value will be 6 and our local minimum now you'll be Leggett at 3 over 4 okay and then the X the corresponding X values were those two and so if you want to just come up with a point you can actually write it as negative 1 6 and then 1/2 and negative 3 over 4 so yeah so if we won the point right if we won the point then the point for this one so local max point and what is this equal to that's going to be like the 1 and then 6 now if you want a local minimum point for this one right so local minimum point then it would be 1 over 2 and then negative 3 over 4 so that's how we come with the point ok so that's it for this problem if you like this video please subscribe to my channel and then give me some support you know we really appreciate it thank you for watching I will see you next time |
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2.1.4: Calculating Atomic Masses
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2.1.3: The Structure of the Atom and How We Represent It (Problems)
2.1.5: Calculating Atomic Masses (Problems)
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Learning Objectives
Define the atomic mass unit and average atomic mass
Calculate average atomic mass and isotopic abundance
Define the amount unit mole and the related quantity Avogadro’s number
Explain the relation between mass, moles, and numbers of atoms or molecules, and perform calculations deriving these quantities from one another
Video : A review of counting subatomic particles and a preview of isotopes and relative atomic mass.
Isotopes
The symbol for a specific isotope of any element is written by placing the mass number as a superscript to the left of the element symbol (Figure ). The atomic number is sometimes written as a subscript preceding the symbol, but since this number defines the element’s identity, as does its symbol, it is often omitted. For example, magnesium exists as a mixture of three isotopes, each with an atomic number of 12 and with mass numbers of 24, 25, and 26, respectively. These isotopes can be identified as 24Mg, 25Mg, and 26Mg. These isotope symbols are read as “element, mass number” and can be symbolized consistent with this reading. For instance, 24Mg is read as “magnesium 24,” and can be written as “magnesium-24” or “Mg-24.” 25Mg is read as “magnesium 25,” and can be written as “magnesium-25” or “Mg-25.” All magnesium atoms have 12 protons in their nucleus. They differ only because a 24Mg atom has 12 neutrons in its nucleus, a 25Mg atom has 13 neutrons, and a 26Mg has 14 neutrons.
Figure : The symbol for an atom indicates the element via its usual two-letter symbol, the mass number as a left superscript, the atomic number as a left subscript (sometimes omitted), and the charge as a right superscript.
Information about the naturally occurring isotopes of elements with atomic numbers 1 through 10 is given in Table . Note that in addition to standard names and symbols, the isotopes of hydrogen are often referred to using common names and accompanying symbols. Hydrogen-2, symbolized 2H, is also called deuterium and sometimes symbolized D. Hydrogen-3, symbolized 3H, is also called tritium and sometimes symbolized T.
Table : Nuclear Compositions of Atoms of the Very Light Elements
| Element | Symbol | Atomic Number | Number of Protons | Number of Neutrons | Mass (amu) | % Natural Abundance |
| hydrogen | (protium) | 1 | 1 | 0 | 1.0078 | 99.989 |
| (deuterium) | 1 | 1 | 1 | 2.0141 | 0.0115 |
| (tritium) | 1 | 1 | 2 | 3.01605 | — (trace) |
| helium | | 2 | 2 | 1 | 3.01603 | 0.00013 |
| | 2 | 2 | 2 | 4.0026 | 100 |
| lithium | | 3 | 3 | 3 | 6.0151 | 7.59 |
| | 3 | 3 | 4 | 7.0160 | 92.41 |
| beryllium | | 4 | 4 | 5 | 9.0122 | 100 |
| boron | | 5 | 5 | 5 | 10.0129 | 19.9 |
| | 5 | 5 | 6 | 11.0093 | 80.1 |
| carbon | | 6 | 6 | 6 | 12.0000 | 98.89 |
| | 6 | 6 | 7 | 13.0034 | 1.11 |
| | 6 | 6 | 8 | 14.0032 | — (trace) |
| nitrogen | | 7 | 7 | 7 | 14.0031 | 99.63 |
| | 7 | 7 | 8 | 15.0001 | 0.37 |
| oxygen | | 8 | 8 | 8 | 15.9949 | 99.757 |
| | 8 | 8 | 9 | 16.9991 | 0.038 |
| | 8 | 8 | 10 | 17.9992 | 0.205 |
| fluorine | | 9 | 9 | 10 | 18.9984 | 100 |
| neon | | 10 | 10 | 10 | 19.9924 | 90.48 |
| | 10 | 10 | 11 | 20.9938 | 0.27 |
| | 10 | 10 | 12 | 21.9914 | 9.25 |
Use this Build an Atom simulator to build atoms of the first 10 elements, see which isotopes exist, check nuclear stability, and gain experience with isotope symbols.
Atomic Mass
Because each proton and each neutron contribute approximately one amu to the mass of an atom, and each electron contributes far less, the atomic mass of a single atom is approximately equal to its mass number (a whole number). However, the average masses of atoms of most elements are not whole numbers because most elements exist naturally as mixtures of two or more isotopes.
The mass of an element shown in a periodic table or listed in a table of atomic masses is a weighted, average mass of all the isotopes present in a naturally occurring sample of that element. This is equal to the sum of each individual isotope’s mass multiplied by its fractional abundance.
For example, the element boron is composed of two isotopes: About 19.9% of all boron atoms are 10B with a mass of 10.0129 amu, and the remaining 80.1% are 11B with a mass of 11.0093 amu. The average atomic mass for boron is calculated to be:
It is important to understand that no single boron atom weighs exactly 10.8 amu; 10.8 amu is the average mass of all boron atoms, and individual boron atoms weigh either approximately 10 amu or 11 amu.
Example : Calculation of Average Atomic Mass
A meteorite found in central Indiana contains traces of the noble gas neon picked up from the solar wind during the meteorite’s trip through the solar system. Analysis of a sample of the gas showed that it consisted of 91.84% 20Ne (mass 19.9924 amu), 0.47% 21Ne (mass 20.9940 amu), and 7.69% 22Ne (mass 21.9914 amu). What is the average mass of the neon in the solar wind?
Solution
The average mass of a neon atom in the solar wind is 20.15 amu. (The average mass of a terrestrial neon atom is 20.1796 amu. This result demonstrates that we may find slight differences in the natural abundance of isotopes, depending on their origin.)
Exercise
A sample of magnesium is found to contain 78.70% of 24Mg atoms (mass 23.98 amu), 10.13% of 25Mg atoms (mass 24.99 amu), and 11.17% of 26Mg atoms (mass 25.98 amu). Calculate the average mass of a Mg atom.
Answer
: 24.31 amu
We can also do variations of this type of calculation, as shown in the next example.
Example : Calculation of Percent Abundance
Naturally occurring chlorine consists of 35Cl (mass 34.96885 amu) and 37Cl (mass 36.96590 amu), with an average mass of 35.453 amu. What is the percent composition of Cl in terms of these two isotopes?
Solution
The average mass of chlorine is the fraction that is 35Cl times the mass of 35Cl plus the fraction that is 37Cl times the mass of 37Cl.
If we let x represent the fraction that is 35Cl, then the fraction that is 37Cl is represented by 1.00 − x.
(The fraction that is 35Cl + the fraction that is 37Cl must add up to 1, so the fraction of 37Cl must equal 1.00 − the fraction of 35Cl.)
Substituting this into the average mass equation, we have:
So solving yields: x = 0.7576, which means that 1.00 − 0.7576 = 0.2424. Therefore, chlorine consists of 75.76% 35Cl and 24.24% 37Cl.
Exercise
Naturally occurring copper consists of 63Cu (mass 62.9296 amu) and 65Cu (mass 64.9278 amu), with an average mass of 63.546 amu. What is the percent composition of Cu in terms of these two isotopes?
Answer
: 69.15% Cu-63 and 30.85% Cu-65
Use this simulator to make mixtures of the main isotopes of the first 18 elements, gain experience with average atomic mass, and check naturally occurring isotope ratios.
Figure : Analysis of zirconium in a mass spectrometer produces a mass spectrum with peaks showing the different isotopes of Zr.
The occurrence and natural abundances of isotopes can be experimentally determined using an instrument called a mass spectrometer. Mass spectrometry (MS) is widely used in chemistry, forensics, medicine, environmental science, and many other fields to analyze and help identify the substances in a sample of material. In a typical mass spectrometer (Figure ), the sample is vaporized and exposed to a high-energy electron beam that causes the sample’s atoms (or molecules) to become electrically charged, typically by losing one or more electrons. These cations then pass through a (variable) electric or magnetic field that deflects each cation’s path to an extent that depends on both its mass and charge (similar to how the path of a large steel ball bearing rolling past a magnet is deflected to a lesser extent that that of a small steel BB). The ions are detected, and a plot of the relative number of ions generated versus their mass-to-charge ratios (a mass spectrum) is made. The height of each vertical feature or peak in a mass spectrum is proportional to the fraction of cations with the specified mass-to-charge ratio. Since its initial use during the development of modern atomic theory, MS has evolved to become a powerful tool for chemical analysis in a wide range of applications.
Video : Watch this video from the Royal Society for Chemistry for a brief description of the rudiments of mass spectrometry.
The Mole
The identity of a substance is defined not only by the types of atoms or ions it contains, but by the quantity of each type of atom or ion. For example, water, H2O, and hydrogen peroxide, H2O2, are alike in that their respective molecules are composed of hydrogen and oxygen atoms. However, because a hydrogen peroxide molecule contains two oxygen atoms, as opposed to the water molecule, which has only one, the two substances exhibit very different properties. Today, we possess sophisticated instruments that allow the direct measurement of these defining microscopic traits; however, the same traits were originally derived from the measurement of macroscopic properties (the masses and volumes of bulk quantities of matter) using relatively simple tools (balances and volumetric glassware). This experimental approach required the introduction of a new unit for amount of substances, the mole, which remains indispensable in modern chemical science.
The mole is an amount unit similar to familiar units like pair, dozen, gross, etc. It provides a specific measure of the number of atoms or molecules in a bulk sample of matter. A mole is defined as the amount of substance containing the same number of discrete entities (such as atoms, molecules, and ions) as the number of atoms in a sample of pure 12C weighing exactly 12 g. One Latin connotation for the word “mole” is “large mass” or “bulk,” which is consistent with its use as the name for this unit. The mole provides a link between an easily measured macroscopic property, bulk mass, and an extremely important fundamental property, number of atoms, molecules, and so forth.
The number of entities composing a mole has been experimentally determined to be , a fundamental constant named Avogadro’s number (NA) or the Avogadro constant in honor of Italian scientist Amedeo Avogadro. This constant is properly reported with an explicit unit of “per mole,” a conveniently rounded version being .
Video : What is Avogadro's Number?
Consistent with its definition as an amount unit, 1 mole of any element contains the same number of atoms as 1 mole of any other element. The masses of 1 mole of different elements, however, are different, since the masses of the individual atoms are drastically different. The molar mass of an element (or compound) is the mass in grams of 1 mole of that substance, a property expressed in units of grams per mole (g/mol) (Figure ).
Figure : Each sample contains atoms —1.00 mol of atoms. From left to right (top row): 65.4 g zinc, 12.0 g carbon, 24.3 g magnesium, and 63.5 g copper. From left to right (bottom row): 32.1 g sulfur, 28.1 g silicon, 207 g lead, and 118.7 g tin. (credit: modification of work by Mark Ott).
Because the definitions of both the mole and the atomic mass unit are based on the same reference substance, 12C, the molar mass of any substance is numerically equivalent to its atomic or formula weight in amu. Per the amu definition, a single 12C atom weighs 12 amu (its atomic mass is 12 amu). The former definition of the mole was that a mole was 12 g of 12C contains 1 mole of 12C atoms (its molar mass is 12 g/mol). This relationship holds for all elements, since their atomic masses are measured relative to that of the amu-reference substance, 12C. Extending this principle, the molar mass of a compound in grams is likewise numerically equivalent to its formula mass in amu (Figure ). On May 20, 2019 the definition was permanently changed to Avogadro's number: a mole is of any object, from atoms to apples.1
Figure : Each sample contains molecules or formula units—1.00 mol of the compound or element. Clock-wise from the upper left: 130.2 g of C8H17OH (1-octanol, formula mass 130.2 amu), 454.4 g of HgI2 (mercury(II) iodide, formula mass 454.4 amu), 32.0 g of CH3OH (methanol, formula mass 32.0 amu) and 256.5 g of S8 (sulfur, formula mass 256.5 amu). (credit: Sahar Atwa).
Table : Mass of one mole of elements
| Element | Average Atomic Mass (amu) | Molar Mass (g/mol) | Atoms/Mole |
| C | 12.01 | 12.01 | |
| H | 1.008 | 1.008 | |
| O | 16.00 | 16.00 | |
| Na | 22.99 | 22.99 | |
| Cl | 33.45 | 35.45 | |
While atomic mass and molar mass are numerically equivalent, keep in mind that they are vastly different in terms of scale, as represented by the vast difference in the magnitudes of their respective units (amu versus g). To appreciate the enormity of the mole, consider a small drop of water after a rainfall. Although this represents just a tiny fraction of 1 mole of water (~18 g), it contains more water molecules than can be clearly imagined. If the molecules were distributed equally among the roughly seven billion people on earth, each person would receive more than 100 billion molecules.
Video : The mole is used in chemistry to represent of something, but it can be difficult to conceptualize such a large number. Watch this video and then complete the “Think” questions that follow. Explore more about the mole by reviewing the information under “Dig Deeper.”
The relationships between formula mass, the mole, and Avogadro’s number can be applied to compute various quantities that describe the composition of substances and compounds. For example, if we know the mass and chemical composition of a substance, we can determine the number of moles and calculate number of atoms or molecules in the sample. Likewise, if we know the number of moles of a substance, we can derive the number of atoms or molecules and calculate the substance’s mass.
Example : Deriving Moles from Grams for an Element
According to nutritional guidelines from the US Department of Agriculture, the estimated average requirement for dietary potassium is 4.7 g. What is the estimated average requirement of potassium in moles?
Solution
The mass of K is provided, and the corresponding amount of K in moles is requested. Referring to the periodic table, the atomic mass of K is 39.10 amu, and so its molar mass is 39.10 g/mol. The given mass of K (4.7 g) is a bit more than one-tenth the molar mass (39.10 g), so a reasonable “ballpark” estimate of the number of moles would be slightly greater than 0.1 mol.
The molar amount of a substance may be calculated by dividing its mass (g) by its molar mass (g/mol):
The factor-label method supports this mathematical approach since the unit “g” cancels and the answer has units of “mol:”
The calculated magnitude (0.12 mol K) is consistent with our ballpark expectation, since it is a bit greater than 0.1 mol.
Exercise : Beryllium
Beryllium is a light metal used to fabricate transparent X-ray windows for medical imaging instruments. How many moles of Be are in a thin-foil window weighing 3.24 g?
Answer
: 0.360 mol
Example : Deriving Grams from Moles for an Element
A liter of air contains mol argon. What is the mass of Ar in a liter of air?
Solution
The molar amount of Ar is provided and must be used to derive the corresponding mass in grams. Since the amount of Ar is less than 1 mole, the mass will be less than the mass of 1 mole of Ar, approximately 40 g. The molar amount in question is approximately one-one thousandth (~10−3) of a mole, and so the corresponding mass should be roughly one-one thousandth of the molar mass (~0.04 g):
In this case, logic dictates (and the factor-label method supports) multiplying the provided amount (mol) by the molar mass (g/mol):
The result is in agreement with our expectations, around 0.04 g Ar.
Exercise
What is the mass of 2.561 mol of gold?
Answer
: 504.4 g
Example : Deriving Number of Atoms from Mass for an Element
Copper is commonly used to fabricate electrical wire (Figure ). How many copper atoms are in 5.00 g of copper wire?
Figure : Copper wire is composed of many, many atoms of Cu. (credit: Emilian Robert Vicol)
Solution
The number of Cu atoms in the wire may be conveniently derived from its mass by a two-step computation: first calculating the molar amount of Cu, and then using Avogadro’s number (NA) to convert this molar amount to number of Cu atoms:
Considering that the provided sample mass (5.00 g) is a little less than one-tenth the mass of 1 mole of Cu (~64 g), a reasonable estimate for the number of atoms in the sample would be on the order of one-tenth NA, or approximately 1022 Cu atoms. Carrying out the two-step computation yields:
The factor-label method yields the desired cancellation of units, and the computed result is on the order of 1022 as expected.
Exercise
A prospector panning for gold in a river collects 15.00 g of pure gold. How many Au atoms are in this quantity of gold?
Answer
: atoms
Example : Deriving Moles from Grams for a Compound
Our bodies synthesize protein from amino acids. One of these amino acids is glycine, which has the molecular formula C2H5O2N. How many moles of glycine molecules are contained in 28.35 g of glycine?
Solution
We can derive the number of moles of a compound from its mass following the same procedure we used for an element in Example :
The molar mass of glycine is required for this calculation, and it is computed in the same fashion as its molecular mass. One mole of glycine, C2H5O2N, contains 2 moles of carbon, 5 moles of hydrogen, 2 moles of oxygen, and 1 mole of nitrogen:
The provided mass of glycine (~28 g) is a bit more than one-third the molar mass (~75 g/mol), so we would expect the computed result to be a bit greater than one-third of a mole (~0.33 mol). Dividing the compound’s mass by its molar mass yields:
This result is consistent with our rough estimate.
Exercise
How many moles of sucrose, , are in a 25-g sample of sucrose?
Answer
: 0.073 mol
Example : Deriving Grams from Moles for a Compound
Vitamin C is a covalent compound with the molecular formula C6H8O6. The recommended daily dietary allowance of vitamin C for children aged 4–8 years is 1.42 × 10−4 mol. What is the mass of this allowance in grams?
Solution
As for elements, the mass of a compound can be derived from its molar amount as shown:
The molar mass for this compound is computed to be 176.124 g/mol. The given number of moles is a very small fraction of a mole (~10−4 or one-ten thousandth); therefore, we would expect the corresponding mass to be about one-ten thousandth of the molar mass (~0.02 g). Performing the calculation, we get:
This is consistent with the anticipated result.
Exercise
What is the mass of 0.443 mol of hydrazine, ?
Answer
: 14.2 g
Example : Deriving the Number of Molecules from the Compound Mass
A packet of an artificial sweetener contains 40.0 mg of saccharin (C7H5NO3S), which has the structural formula:
Given that saccharin has a molar mass of 183.18 g/mol, how many saccharin molecules are in a 40.0-mg (0.0400-g) sample of saccharin? How many carbon atoms are in the same sample?
Solution
The number of molecules in a given mass of compound is computed by first deriving the number of moles, as demonstrated in Example , and then multiplying by Avogadro’s number:
Using the provided mass and molar mass for saccharin yields:
The compound’s formula shows that each molecule contains seven carbon atoms, and so the number of C atoms in the provided sample is:
Exercise
How many molecules are contained in 9.213 g of this compound? How many hydrogen atoms?
Answer
Video : A preview of some of the uses we will have for moles in upcoming units
Summary
Video : Watch this video for a review of relative atomic mass and isotopes.
An atom consists of a small, positively charged nucleus surrounded by electrons. The nucleus contains protons and neutrons; its diameter is about 100,000 times smaller than that of the atom. The mass of one atom is usually expressed in atomic mass units (amu), which is referred to as the atomic mass. An amu is defined as exactly of the mass of a carbon-12 atom and is equal to 1.6605 10−24 g.
Protons are relatively heavy particles with a charge of 1+ and a mass of 1.0073 amu. Neutrons are relatively heavy particles with no charge and a mass of 1.0087 amu. Electrons are light particles with a charge of 1− and a mass of 0.00055 amu. The number of protons in the nucleus is called the atomic number (Z) and is the property that defines an atom’s elemental identity. The sum of the numbers of protons and neutrons in the nucleus is called the mass number and, expressed in amu, is approximately equal to the mass of the atom. An atom is neutral when it contains equal numbers of electrons and protons.
Isotopes of an element are atoms with the same atomic number but different mass numbers; isotopes of an element, therefore, differ from each other only in the number of neutrons within the nucleus. When a naturally occurring element is composed of several isotopes, the atomic mass of the element represents the average of the masses of the isotopes involved. A chemical symbol identifies the atoms in a substance using symbols, which are one-, two-, or three-letter abbreviations for the atoms.
Looking Beyond
Video : Remember our exploration into the size of an atom last week? This video goes deeper into investigating the size of the subatomic particles we just discussed.
Footnotes
Read more about the redefinition of SI units including the kilogram here (Laura Howe, CE&N, Nov. 16, 2018).
Key Equations
Glossary
anion
: negatively charged atom or molecule (contains more electrons than protons)
atomic mass
: average mass of atoms of an element, expressed in amu
atomic mass unit (amu)
: (also, unified atomic mass unit, u, or Dalton, Da) unit of mass equal to of the mass of a 12C atom
atomic number (Z)
: number of protons in the nucleus of an atom
cation
: positively charged atom or molecule (contains fewer electrons than protons)
chemical symbol
: one-, two-, or three-letter abbreviation used to represent an element or its atoms
Dalton (Da)
: alternative unit equivalent to the atomic mass unit
fundamental unit of charge
: (also called the elementary charge) equals the magnitude of the charge of an electron (e) with e = 1.602 10−19 C
ion
: electrically charged atom or molecule (contains unequal numbers of protons and electrons)
mass number (A)
: sum of the numbers of neutrons and protons in the nucleus of an atom
mole
: amount of substance containing the same number of atoms, molecules, ions, or other entities as the number of atoms in exactly 12 grams of 12C
molar mass
: mass in grams of 1 mole of a substance
unified atomic mass unit (u)
: alternative unit equivalent to the atomic mass unit
Contributors and Attributions
Paul Flowers (University of North Carolina - Pembroke), Klaus Theopold (University of Delaware) and Richard Langley (Stephen F. Austin State University) with contributing authors. Textbook content produced by OpenStax College is licensed under a Creative Commons Attribution License 4.0 license. Download for free at
Adelaide Clark, Oregon Institute of Technology
Fuse School, Open Educational Resource free of charge, under a Creative Commons License: Attribution-NonCommercial CC BY-NC (View License Deed:
Crash Course Chemistry, Crash Course is a division of Complexly and videos are free to stream for educational purposes.
TED-Ed’s commitment to creating lessons worth sharing is an extension of TED’s mission of spreading great ideas. Within TED-Ed’s growing library of TED-Ed animations, you will find carefully curated educational videos, many of which represent collaborations between talented educators and animators nominated through the TED-Ed website.
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2.1.3: The Structure of the Atom and How We Represent It (Problems)
2.1.5: Calculating Atomic Masses (Problems) |
6959 | https://zief0002.github.io/epsy-5261/07-03-confidence-intervals-extras.html | Table of contents
21.1 Assumptions for the Confidence Interval
21.2 Unbiased Sampling and CIs
21.2.1 Deeper Understanding of Unbiasedness and Accuracy
21.3 Confidence Level and Accuracy
21.3.1 Changing the Confidence Level
21.3.2 Using R to Obtain a 99% CI
21.4 Improving Precision of the CI
Effect Size
21 Confidence Intervals: Assumptions, Accuracy, and Precision
21 Confidence Intervals: Assumptions, Accuracy, and Precision
In this chapter you will learn about the assumptions that we need to make in order for the results of a confidence interval to be statistically valid. You will also learn about confidence levels and factors that affect the precision of a confidence interval.
21.1 Assumptions for the Confidence Interval
The assumptions underlying the use of a confidence interval are exactly the same as those underlying hypothesis tests. So, for using a CI to estimate a single mean, the assumptions are:
The distribution of values in the population is normally distributed.
The values in the population are independent from each other.
Whereas the assumptions underlying the use of a CI to estimate a single proportion are:
The values in the population follow a binomial distribution. This is true so long as there are only two values the attribute can take on (e.g., “Yes” or “No”).
The values in the population are independent from each other.
The quantities and are both greater than 10, where n is the sample size and is the sample proportion value.
The assumptions for a CI to estimate the differences in means between two groups are:
The distribution of values in both populations is normally distributed.
The values in both population are independent from each other.
Both populations have the same variance.
And, those for a CI to estimate the differences in proportions between two groups are:
The values in the two populations both follow a binomial distribution. This is true so long as there are only two values the attribute can take on (e.g., “Yes” or “No”) for each group being compared.
The values in the population are independent from each other.
For both groups, the quantities and are both greater than 10, where n is the sample size used in the group and is the sample proportion value.
21.2 Unbiased Sampling and CIs
The key to generalization and accuracy in the CI is that the sample was chosen using an unbiased sampling method—namely random sampling. Without this, the resulting interval is biased and the interval given is far less likely to be accurate in estimating the population parameter. In the metaphor introduced in Chapter 17, a biased method would mean that Arthur’s search locations for their keys was systematically in the wrong place. If we extend this idea to confidence intervals, the search location (sample statistic) would produce a confidence interval (search area) that is also systematically in the wrong place. Figure 21.1 shows this idea visually.
In this picture the search area, which represents the estimate of where Arthur thought they lost their keys, is inaccurate—the actual keys (population parameter) are not inside the bounds of the search area (inside the CI). This is because the search location (sample statistic) was initially produced by a biased method.
For example, in Chapter 20 we computed a CI for the difference in the average amount of change in IQ scores between users who started using marijuana as adults and those who started using marijuana as teens. The CI, , indicated the difference in the average amount of change in IQ scores between people who became persistent marijuana users as adults than those that became persistent marijuana users as teens is between 1.11 and 11.26. Unfortunately the sample of users was not chosen randomly—they were all volunteers. Because our sample was not chosen using an unbiased sampling method, it is highly likely that the CI produced is inaccurate. That is, the true population difference in the average amount of change in IQ scores between people who became persistent marijuana users as adults than those that became persistent marijuana users as teens is PROBABLY NOT between 1.11 and 11.26.
21.2.1 Deeper Understanding of Unbiasedness and Accuracy
To more deeply understand how an unbiased method affects accuracy of the CI, let’s consider another thought experiment. Imagine a population that is normally distributed with a mean of 10 and a standard deviation of 2. Now imagine that we are randomly sampling 25 observations from that population and using those 25 sampled observations to produce a CI that would be used to estimate the value of the population mean. We will repeat this process, say 100 times. A visual of this thought experiment is shown in Figure 21.2.
A plot of the 100 CIs is shown in Figure 21.3. The population mean is shown as the vertical line at 10. Some of the CIs included the value of 10. For example, the first sample produced a CI of . It included 10 as a plausible value. This CI is the lowest CI on the plot. It is colored grey since it included the value of 10 as a plausible value. Similarly the second sample also included 10 as a plausible value, it’s CI was , so it is also colored grey (it is the second lowest CI in the plot). Some of the confidence intervals in Figure 21.3 are red. Those CIs do not include 10 as a plausible value.
Now let’s carry out a thought experiment in which a biased sampling method is used. In this sampling method, the sample mean is, on average, higher than the population mean of 10. We will use this method to again draw 100 different samples of size 25. And again, we will produce a CI for each of the 100 samples.
In Figure 21.4 the CIs that include the population value of 10 are again colored grey and those that do not are colored red. Using an biased sampling method, we see that more of the CIs are red—they do not include 10 as a plausible value for the population mean, despite 10 being the actual population mean!
In both Figure 21.3 and Figure 21.4 we can see that some of the CIs include 10 and some don’t. But, on average, the unbiased sampling method produces many more CIs that include 10 than the biased sampling method does. Because of this, we would say that the unbiased sampling method produces more accurate estimates of the sample mean. Accuracy refers to whether the CI includes the population parameter for most CIs that could be produced. It does not mean that any one CI will be accurate or not. It is just more likely that a CI will include the population parameter if the sample that produced the CI was selected using an unbiased method.
In practice we do not draw many samples—we have one sample. This sample is used to create a single CI that we use to estimate the mean. Unlike in the thought experiment, in practice, we do not know what the value of the population parameter actually is. So we don’t know whether our CI includes this value (is a gray CI) or doesn’t (is a red CI). If we used an unbiased sampling method to select our sample, our CI is more likely to be gray than red.
21.3 Confidence Level and Accuracy
In the previous section you saw that even if you use an unbiased sampling method (e.g., random sampling) you still occasionally draw a sample that produces a red CI—it does not include the population parameter as a plausible value. In fact, looking back at Figure 21.3, we see that this will happen about 5% of the time (five of the 100 samples produced a red CI). This is because we produced a confidence interval using a method that will include the population parameter for 95% of the intervals it creates, so long as the sampling method used to select those samples is unbiased. Another way to say this is the method we used to create the CIs was accurate 95% of the time.
This accuracy rate is referred to as the confidence level. In practice we say that we created a 95% Confidence Interval. But be careful, using this language. Here are some caveats as you start to use this terminology:
The 95% doesn’t refer to the single CI, but rather to the accuracy rate across producing many, many CIs.
Across many different samples, we expect 95% of the CIs that are created to be grey and 5% to be red. In practice we never draw more than a single sample so there is ever only a single interval. This interval is either grey (it included the parameter in the interval) or red (it did not). Because of this, for a single sample we CANNOT SAY:
There is a 95% chance that the population parameter is in the interval. (WRONG! 😢)
It either is in the interval (100%) or it’s not (0%). In practice, we hope that our interval is grey and includes the true population parameter inside the interval, but there is not way to know if it does. If we used an unbiased sampling method to select the sample, then we know that 95% of all possible samples we could draw would produce a grey interval, and we are hoping that our interval is from a sample that produces a grey interval. (Again, this is far more likely if the sample is chosen using an unbiased sampling method.)
21.3.1 Changing the Confidence Level
We can change the method we use to produce the CI to raise or lower the accuracy level. For example, rather than creating a 95% CI, we could use a method that is 99% accurate (99% of the samples will produce a grey CI). In order to do this, we change the multiplication factor in the formula for the CI. For example, the following was the formula we introduced to compute a CI, which produces a 95% CI:
To create a 99% CI we use:
And to produce a 68% CI we use:
The higher the confidence level, the larger the multiplication factor in the margin of error. Think about how this impacts the CI. For example, in our marijuana case study the 95% CI was . A 99% CI for this difference is . The margin of error is bigger which makes the interval width wider. That is, we have more uncertainty (i.e., less precision) in our estimate.
To see how the accuracy changes when using a 99% CI (rather than a 95% CI) we can carry out a thought experiement. In this thought experiment (using an unbiased sampling method), we will draw 100 different samples of size 25 and produce a 99% CI for each of the 100 samples. The result is shown in Figure 21.5.
When we use a 99% CI only one of the 100 samples produced a CI that did not include the population mean of 10. Across the 100 sampes 99 of them (99%) included the population mean of 10. That implies that using a 99% CI is more accurate than using a 95% CI. Again note, that each individual interval either included the mean of 10 (was gray) or did not (was red)—the accuracy rate refers to how well the method does across many random samples.
The drawback to changing the accuracy level from 95% to 99% is that the intervals are wider. This, again, means that we have more uncertainty in the value of the population parameter because there are more plausible values in the interval. There is always a tradeoff between accuracy and precision; increasing one decreases the other. In practice, social science researchers tend to use and report 95% CIs.
21.3.2 Using R to Obtain a 99% CI
To change the confidence level for our CI, we need to include the argument conf.level= in the t_test() function. For example, to obtain a 99% CI for the difference in the average amount of change in IQ scores between users who started using marijuana as adults and those who started using marijuana as teens, we use the following syntax:
```
Load libraries
library(ggformula)
library(mosaic)
library(tidyverse)
Import data
cannabis <- read_csv("
View data
cannabis
```
| |
| ABCDEFGHIJ0123456789 |
| iq_change | cannabis_dep | | | |
--- ---
| 4 | Adult | | | |
| -10 | Adult | | | |
| -3 | Adult | | | |
| -1 | Adult | | | |
| -3 | Adult | | | |
| -7 | Adult | | | |
| 5 | Adult | | | |
| -7 | Adult | | | |
| 9 | Adult | | | |
| -7 | Adult | | | |
Next
1234
Previous
1-10 of 37 rows
```
One-sample t-test
Include conf.level=
my_t <- t.test(
iq_change ~ cannabis_dep,
data = cannabis,
var.equal = TRUE,
conf.level = .99
)
Compute CI
confint(my_t)
```
| |
| ABCDEFGHIJ0123456789 |
| mean in group Adult | mean in group Teen | lower | upper | level |
--- ---
| -2.071429 | -8.26087 | -0.6230182 | 13.0019 | 0.99 |
1 row
You can set the conf.level= argument to whichever confidence level you would like. The default is conf.level=.95, so not including the conf.level= argument will produce a 95% CI.
21.4 Improving Precision of the CI
Ideally, the confidence interval will be precise enough to be useful to the researcher. For example, say a researcher was interested in estimating the mean starting salary for a teacher in a particular state. If the CI they produced suggested the average starting salary was between $20,000 and $80,000, that would not be very useful. While it gives a set of plausible values, the precision of the interval is too low (too much uncertainty) to be useful in determining the average starting salary. On the other hand, a CI that estimated the average starting salary to be between $49,800 and $50,200 is quite useful. The high amount of precision in the estimate reduces the amount of uncertainty around the actual mean value.
The amount of precision in the CI is all governed by the margin of error. In our keys metaphor, the margin of error is represented by the size of the area that Arthur is searching.1 In Figure 21.6 and Figure 21.7, the two scenarios show vastly different search areas for Arthur’s keys. The left-hand figure has a smaller search area reflecting higher precision and less uncertainty about where the keys are. The right-hand figure has a larger search area which reflects lower precision and more uncertainty about where the keys are.
Numerically, the margin of error is half of the range of the confidence interval and is expressed as the part of the CI that gets added and subtracted from the sample estimate. For example the two CIs presented in the previous paragraph have a margin of error of $30,000 and $200, respectively. Their formulas to produce the CIs would be:
Recall that the way we obtain the margin of error (when we are estimating the 95% CI for a single mean) is:
where the value 2 was a multiplication factor related to the confidence level, SD is the sample standard deviation, and n is the sample size. To improve precision, we need to make the margin of error smaller (less uncertainty). How can we do this?
Use a smaller multiplication factor than 2 (i.e., make the confidence level smaller)
Have a smaller SD
Have a larger sample size
In the social sciences, since we always use a 95% CI, the multiplication factor will be 2, so we can’t really manipulate this without sacrificing accuracy. The SD is a function of the data, so there is really no way to manipulate this either (at least not easily). The only factor we as researchers can change is the sample size, and sometimes we don’t have control over that either.
Thinking about this relationship between sample size and precision of the estimate prior to collecting any data can help you plan a more meaningful study. For example, say you wanted to be able to estimate the starting salary of teachers so that the margin of error was no more than $1,000 (a range of $2,000 in the CI). How big a sample size might you need? You could substitute the desired margin of error into our formula and solve for n.
To solve for n, we also need an estimate for the SD we might expect in the data we collect. Often good guesses for this can be found in previous studies or the substantive literature. In our example, based on previous data, we expect that a reasonable guess for the SD might be about $10,000. Including this in our formula and solving for sample size:
To obtain a 95% CI with a margin of error of $1,000, we need a sample of salaries from 400 starting teachers (assuming that the SD in the data is $10,000).
Technically it would be the radius of the search area.↩︎
20 Confidence Intervals Using R
22 Standardized Effects for Means |
6960 | https://study.com/learn/lesson/video/tangent-ratio-formula-examples-triangle.html | Tangent Ratio of a Triangle | Formula & Examples - Video | Study.com
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What is the Tangent Ratio?
The tangent ratio is one of the most important concepts in trigonometry that relates the angles and sides of a right triangle. It is the ratio of the length of the opposite side to the length of the adjacent side for a given angle.
In simple terms, if you know the opposite and adjacent sides, you can calculate the tangent of an angle. This ratio is often written as (t a n θ), where θ is the angle between sides.
How to Find the Tangent of an Angle?
To find the tangent of an angle, use the formula:
t a n θ=o p p o s i t e a d j a c e n t
For example, if the opposite side is 3 units and the adjacent side is 4 units, then the tangent ratio is:
t a n θ=3 4=0.75
This indicates that the opposite side is 0.75 times the length of the adjacent side.
Example 1: Finding the Tangent Ratio
Let's consider a triangle where the opposite side is 3 and the adjacent side is 4. The tangent ratio for this triangle is:
t a n θ=3 4=0.75
No matter how large the triangle is, as long as the angle θ is the same, the tangent ratio will always remain 0.75.
Example 2: Finding an Unknown Side Using the Tangent Ratio
Suppose you have a triangle where the angle θ is 51 degrees, the adjacent side is 12 units, and you need to find the opposite side (x). The equation is like:
t a n(51)=x 12
Solving for x, we get
x=12×t a n(51)=12×1.2349=14.82.
Example 3: Finding the Angle Using the Tangent Ratio
Suppose the opposite side is of 11 inches and the adjacent side is 20 inches, but you need to find the angle θ. So, use the inverse tangent function:
t a n θ=11 20=0.55
Using your calculator to find the inverse tangent:
θ=t a n−1(0.55)=28.81∘.
Read Tangent Ratio of a Triangle | Formula & Examples Lesson
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6961 | https://en.wikipedia.org/wiki/Fick%27s_laws_of_diffusion | Jump to content
Fick's laws of diffusion
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From Wikipedia, the free encyclopedia
Mathematical descriptions of molecular diffusion
For the technique of measuring cardiac output, see Fick principle.
Fick's laws of diffusion describe diffusion and were first posited by Adolf Fick in 1855 on the basis of largely experimental results. They can be used to solve for the diffusion coefficient, D. Fick's first law can be used to derive his second law which in turn is identical to the diffusion equation.
Fick's first law: Movement of particles from high to low concentration (diffusive flux) is directly proportional to the particle's concentration gradient.
Fick's second law: Prediction of change in concentration gradient with time due to diffusion.
A diffusion process that obeys Fick's laws is called normal or Fickian diffusion; otherwise, it is called anomalous diffusion or non-Fickian diffusion.
History
[edit]
In 1855, physiologist Adolf Fick first reported his now well-known laws governing the transport of mass through diffusive means. Fick's work was inspired by the earlier experiments of Thomas Graham, which fell short of proposing the fundamental laws for which Fick would become famous. Fick's law is analogous to the relationships discovered at the same epoch by other eminent scientists: Darcy's law (hydraulic flow), Ohm's law (charge transport), and Fourier's law (heat transport).
Fick's experiments (modeled on Graham's) dealt with measuring the concentrations and fluxes of salt, diffusing between two reservoirs through tubes of water. It is notable that Fick's work primarily concerned diffusion in fluids, because at the time, diffusion in solids was not considered generally possible. Today, Fick's laws form the core of our understanding of diffusion in solids, liquids, and gases (in the absence of bulk fluid motion in the latter two cases). When a diffusion process does not follow Fick's laws (which happens in cases of diffusion through porous media and diffusion of swelling penetrants, among others), it is referred to as non-Fickian.
Fick's first law
[edit]
Fick's first law relates the diffusive flux to the gradient of the concentration. It postulates that the flux goes from regions of high concentration to regions of low concentration, with a magnitude that is proportional to the concentration gradient (spatial derivative), or in simplistic terms the concept that a solute will move from a region of high concentration to a region of low concentration across a concentration gradient. In one (spatial) dimension, the law can be written in various forms, where the most common form (see) is in a molar basis:
where
J is the diffusion flux, of which the dimension is the amount of substance per unit area per unit time. J measures the amount of substance that will flow through a unit area during a unit time interval,
D is the diffusion coefficient or diffusivity. Its dimension is area per unit time,
is the concentration gradient,
φ (for ideal mixtures) is the concentration, with a dimension of amount of substance per unit volume,
x is position, the dimension of which is length.
D is proportional to the squared velocity of the diffusing particles, which depends on the temperature, viscosity of the fluid and the size of the particles according to the Stokes–Einstein relation. The modeling and prediction of Fick's diffusion coefficients is difficult. They can be estimated using the empirical Vignes correlation model or the physically-motivated entropy scaling. In dilute aqueous solutions the diffusion coefficients of most ions are similar and have values that at room temperature are in the range of (0.6–2)×10−9 m2/s. For biological molecules the diffusion coefficients normally range from 10−10 to 10−11 m2/s.
In two or more dimensions we must use ∇, the del or gradient operator, which generalises the first derivative, obtaining
where J denotes the diffusion flux.
The driving force for the one-dimensional diffusion is the quantity −∂φ/∂x, which for ideal mixtures is the concentration gradient.
Variations of the first law
[edit]
Another form for the first law is to write it with the primary variable as mass fraction (yi, given for example in kg/kg), then the equation changes to
where
the index i denotes the ith species,
Ji is the diffusion flux of the ith species (for example in mol/m2/s),
Mi is the molar mass of the ith species,
ρ is the mixture density (for example in kg/m3).
The is outside the gradient operator. This is because
where ρsi is the partial density of the ith species.
Beyond this, in chemical systems other than ideal solutions or mixtures, the driving force for the diffusion of each species is the gradient of chemical potential of this species. Then Fick's first law (one-dimensional case) can be written
where
the index i denotes the ith species,
c is the concentration (mol/m3),
R is the universal gas constant (J/K/mol),
T is the absolute temperature (K),
μ is the chemical potential (J/mol).
The driving force of Fick's law can be expressed as a fugacity difference:
where is the fugacity in Pa. is a partial pressure of component i in a vapor or liquid phase. At vapor liquid equilibrium the evaporation flux is zero because .
Derivation of Fick's first law for gases
[edit]
Four versions of Fick's law for binary gas mixtures are given below. These assume: thermal diffusion is negligible; the body force per unit mass is the same on both species; and either pressure is constant or both species have the same molar mass. Under these conditions, Ref. shows in detail how the diffusion equation from the kinetic theory of gases reduces to this version of Fick's law:
where Vi is the diffusion velocity of species i. In terms of species flux this is
If, additionally, , this reduces to the most common form of Fick's law,
If (instead of or in addition to ) both species have the same molar mass, Fick's law becomes
where is the mole fraction of species i.
Fick's second law
[edit]
Fick's second law predicts how diffusion causes the concentration to change with respect to time. It is a partial differential equation which in one dimension reads
where
φ is the concentration in dimensions of , example mol/m3; φ = φ(x,t) is a function that depends on location x and time t,
t is time, example s,
D is the diffusion coefficient in dimensions of , example m2/s,
x is the position, example m.
In two or more dimensions we must use the Laplacian Δ = ∇2, which generalises the second derivative, obtaining the equation
Fick's second law has the same mathematical form as the Heat equation and its fundamental solution is the same as the Heat kernel, except switching thermal conductivity with diffusion coefficient :
Derivation of Fick's second law
[edit]
Fick's second law can be derived from Fick's first law and the mass conservation in absence of any chemical reactions:
Assuming the diffusion coefficient D to be a constant, one can exchange the orders of the differentiation and multiply by the constant:
and, thus, receive the form of the Fick's equations as was stated above.
For the case of diffusion in two or more dimensions Fick's second law becomes
which is analogous to the heat equation.
If the diffusion coefficient is not a constant, but depends upon the coordinate or concentration, Fick's second law yields
An important example is the case where φ is at a steady state, i.e. the concentration does not change by time, so that the left part of the above equation is identically zero. In one dimension with constant D, the solution for the concentration will be a linear change of concentrations along x. In two or more dimensions we obtain
which is Laplace's equation, the solutions to which are referred to by mathematicians as harmonic functions.
Example solutions and generalization
[edit]
Fick's second law is a special case of the convection–diffusion equation in which there is no advective flux and no net volumetric source. It can be derived from the continuity equation:
where j is the total flux and R is a net volumetric source for φ. The only source of flux in this situation is assumed to be diffusive flux:
Plugging the definition of diffusive flux to the continuity equation and assuming there is no source (R = 0), we arrive at Fick's second law:
If flux were the result of both diffusive flux and advective flux, the convection–diffusion equation is the result.
Example solution 1: constant concentration source and diffusion length
[edit]
A simple case of diffusion with time t in one dimension (taken as the x-axis) from a boundary located at position x = 0, where the concentration is maintained at a value n0 is
where erfc is the complementary error function. This is the case when corrosive gases diffuse through the oxidative layer towards the metal surface (if we assume that concentration of gases in the environment is constant and the diffusion space – that is, the corrosion product layer – is semi-infinite, starting at 0 at the surface and spreading infinitely deep in the material). If, in its turn, the diffusion space is infinite (lasting both through the layer with n(x, 0) = 0, x > 0 and that with n(x, 0) = n0, x ≤ 0), then the solution is amended only with coefficient 1/2 in front of n0 (as the diffusion now occurs in both directions). This case is valid when some solution with concentration n0 is put in contact with a layer of pure solvent. (Bokstein, 2005) The length is called the diffusion length and provides a measure of how far the concentration has propagated in the x-direction by diffusion in time t (Bird, 1976).
As a quick approximation of the error function, the first two terms of the Taylor series can be used:
If D is time-dependent, the diffusion length becomes
This idea is useful for estimating a diffusion length over a heating and cooling cycle, where D varies with temperature.
Example solution 2: Brownian particle and mean squared displacement
[edit]
Another simple case of diffusion is the Brownian motion of one particle. The particle's Mean squared displacement from its original position is:
where is the dimension of the particle's Brownian motion. For example, the diffusion of a molecule across a cell membrane 8 nm thick is 1-D diffusion because of the spherical symmetry; However, the diffusion of a molecule from the membrane to the center of a eukaryotic cell is a 3-D diffusion. For a cylindrical cactus, the diffusion from photosynthetic cells on its surface to its center (the axis of its cylindrical symmetry) is a 2-D diffusion.
The square root of MSD, , is often used as a characterization of how far the particle has moved after time has elapsed. The MSD is symmetrically distributed over the 1D, 2D, and 3D space. Thus, the probability distribution of the magnitude of MSD in 1D is Gaussian and 3D is a Maxwell-Boltzmann distribution.
Generalizations
[edit]
In non-homogeneous media, the diffusion coefficient varies in space, D = D(x). This dependence does not affect Fick's first law but the second law changes:
In anisotropic media, the diffusion coefficient depends on the direction. It is a symmetric tensor Dji = Dij. Fick's first law changes to it is the product of a tensor and a vector: For the diffusion equation this formula gives The symmetric matrix of diffusion coefficients Dij should be positive definite. It is needed to make the right-hand side operator elliptic.
For inhomogeneous anisotropic media these two forms of the diffusion equation should be combined in
The approach based on Einstein's mobility and Teorell formula gives the following generalization of Fick's equation for the multicomponent diffusion of the perfect components: where φi are concentrations of the components and Dij is the matrix of coefficients. Here, indices i and j are related to the various components and not to the space coordinates.
The Chapman–Enskog formulae for diffusion in gases include exactly the same terms. These physical models of diffusion are different from the test models ∂tφi = Σj Dij Δφj which are valid for very small deviations from the uniform equilibrium. Earlier, such terms were introduced in the Maxwell–Stefan diffusion equation.
For anisotropic multicomponent diffusion coefficients one needs a rank-four tensor, for example Dij,αβ, where i, j refer to the components and α, β = 1, 2, 3 correspond to the space coordinates.
Applications
[edit]
Equations based on Fick's law have been commonly used to model transport processes in foods, neurons, biopolymers, pharmaceuticals, porous soils, population dynamics, nuclear materials, plasma physics, and semiconductor doping processes. The theory of voltammetric methods is based on solutions of Fick's equation. On the other hand, in some cases a "Fickian (another common approximation of the transport equation is that of the diffusion theory)" description is inadequate. For example, in polymer science and food science a more general approach is required to describe transport of components in materials undergoing a glass transition. One more general framework is the Maxwell–Stefan diffusion equations
of multi-component mass transfer, from which Fick's law can be obtained as a limiting case, when the mixture is extremely dilute and every chemical species is interacting only with the bulk mixture and not with other species. To account for the presence of multiple species in a non-dilute mixture, several variations of the Maxwell–Stefan equations are used. See also non-diagonal coupled transport processes (Onsager relationship).
Fick's flow in liquids
[edit]
When two miscible liquids are brought into contact, and diffusion takes place, the macroscopic (or average) concentration evolves following Fick's law. On a mesoscopic scale, that is, between the macroscopic scale described by Fick's law and molecular scale, where molecular random walks take place, fluctuations cannot be neglected. Such situations can be successfully modeled with Landau-Lifshitz fluctuating hydrodynamics. In this theoretical framework, diffusion is due to fluctuations whose dimensions range from the molecular scale to the macroscopic scale.
In particular, fluctuating hydrodynamic equations include a Fick's flow term, with a given diffusion coefficient, along with hydrodynamics equations and stochastic terms describing fluctuations. When calculating the fluctuations with a perturbative approach, the zero order approximation is Fick's law. The first order gives the fluctuations, and it comes out that fluctuations contribute to diffusion. This represents somehow a tautology, since the phenomena described by a lower order approximation is the result of a higher approximation: this problem is solved only by renormalizing the fluctuating hydrodynamics equations.
Sorption rate and collision frequency of diluted solute
[edit]
Adsorption, absorption, and collision of molecules, particles, and surfaces are important problems in many fields. These fundamental processes regulate chemical, biological, and environmental reactions. Their rate can be calculated using the diffusion constant and Fick's laws of diffusion especially when these interactions happen in diluted solutions.
Typically, the diffusion constant of molecules and particles defined by Fick's equation can be calculated using the Stokes–Einstein equation. In the ultrashort time limit, in the order of the diffusion time a2/D, where a is the particle radius, the diffusion is described by the Langevin equation. At a longer time, the Langevin equation merges into the Stokes–Einstein equation. The latter is appropriate for the condition of the diluted solution, where long-range diffusion is considered. According to the fluctuation-dissipation theorem based on the Langevin equation in the long-time limit and when the particle is significantly denser than the surrounding fluid, the time-dependent diffusion constant is:
where (all in SI units)
kB is the Boltzmann constant,
T is the absolute temperature,
μ is the mobility of the particle in the fluid or gas, which can be calculated using the Einstein relation (kinetic theory),
m is the mass of the particle,
t is time.
For a single molecule such as organic molecules or biomolecules (e.g. proteins) in water, the exponential term is negligible due to the small product of mμ in the ultrafast picosecond region, thus irrelevant to the relatively slower adsorption of diluted solute.
The adsorption or absorption rate of a dilute solute to a surface or interface in a (gas or liquid) solution can be calculated using Fick's laws of diffusion. The accumulated number of molecules adsorbed on the surface is expressed by the Langmuir-Schaefer equation by integrating the diffusion flux equation over time as shown in the simulated molecular diffusion in the first section of this page:
A is the surface area (m2).
is the number concentration of the adsorber molecules (solute) in the bulk solution (#/m3).
D is diffusion coefficient of the adsorber (m2/s).
t is elapsed time (s).
is the accumulated number of molecules in unit # molecules adsorbed during the time .
The equation is named after American chemists Irving Langmuir and Vincent Schaefer.
Briefly as explained in,
the concentration gradient profile near a newly created (from ) absorptive surface (placed at ) in a once uniform bulk solution is solved in the above sections from Fick's equation,
where C is the number concentration of adsorber molecules at (#/m3).
The concentration gradient at the subsurface at is simplified to the pre-exponential factor of the distribution
And the rate of diffusion (flux) across area of the plane is
Integrating over time,
The Langmuir–Schaefer equation can be extended to the Ward–Tordai Equation to account for the "back-diffusion" of rejected molecules from the surface:
where is the bulk concentration, is the sub-surface concentration (which is a function of time depending on the reaction model of the adsorption), and is a dummy variable.
Monte Carlo simulations show that these two equations work to predict the adsorption rate of systems that form predictable concentration gradients near the surface but have troubles for systems without or with unpredictable concentration gradients, such as typical biosensing systems or when flow and convection are significant.
A brief history of diffusive adsorption is shown in the right figure. A noticeable challenge of understanding the diffusive adsorption at the single-molecule level is the fractal nature of diffusion. Most computer simulations pick a time step for diffusion which ignores the fact that there are self-similar finer diffusion events (fractal) within each step. Simulating the fractal diffusion shows that a factor of two corrections should be introduced for the result of a fixed time-step adsorption simulation, bringing it to be consistent with the above two equations.
A more problematic result of the above equations is they predict the lower limit of adsorption under ideal situations but is very difficult to predict the actual adsorption rates. The equations are derived at the long-time-limit condition when a stable concentration gradient has been formed near the surface. But real adsorption is often done much faster than this infinite time limit i.e. the concentration gradient, decay of concentration at the sub-surface, is only partially formed before the surface has been saturated or flow is on to maintain a certain gradient, thus the adsorption rate measured is almost always faster than the equations have predicted for low or none energy barrier adsorption (unless there is a significant adsorption energy barrier that slows down the absorption significantly), for example, thousands to millions time faster in the self-assembly of monolayers at the water-air or water-substrate interfaces. As such, it is necessary to calculate the evolution of the concentration gradient near the surface and find out a proper time to stop the imagined infinite evolution for practical applications. While it is hard to predict when to stop but it is reasonably easy to calculate the shortest time that matters, the critical time when the first nearest neighbor from the substrate surface feels the building-up of the concentration gradient. This yields the upper limit of the adsorption rate under an ideal situation when there are no other factors than diffusion that affect the absorber dynamics:
where:
is the adsorption rate assuming under adsorption energy barrier-free situation, in unit #/s,
is the area of the surface of interest on an "infinite and flat" substrate (m2),
is the concentration of the absorber molecule in the bulk solution (#/m3),
is the diffusion constant of the absorber (solute) in the solution (m2/s) defined with Fick's law.
This equation can be used to predict the initial adsorption rate of any system; It can be used to predict the steady-state adsorption rate of a typical biosensing system when the binding site is just a very small fraction of the substrate surface and a near-surface concentration gradient is never formed; It can also be used to predict the adsorption rate of molecules on the surface when there is a significant flow to push the concentration gradient very shallowly in the sub-surface.
This critical time is significantly different from the first passenger arriving time or the mean free-path time. Using the average first-passenger time and Fick's law of diffusion to estimate the average binding rate will significantly over-estimate the concentration gradient because the first passenger usually comes from many layers of neighbors away from the target, thus its arriving time is significantly longer than the nearest neighbor diffusion time. Using the mean free path time plus the Langmuir equation will cause an artificial concentration gradient between the initial location of the first passenger and the target surface because the other neighbor layers have no change yet, thus significantly lower estimate the actual binding time, i.e., the actual first passenger arriving time itself, the inverse of the above rate, is difficult to calculate. If the system can be simplified to 1D diffusion, then the average first passenger time can be calculated using the same nearest neighbor critical diffusion time for the first neighbor distance to be the MSD,
where:
(unit m) is the average nearest neighbor distance approximated as cubic packing, where is the solute concentration in the bulk solution (unit # molecule / m3),
is the diffusion coefficient defined by Fick's equation (unit m2/s),
is the critical time (unit s).
In this critical time, it is unlikely the first passenger has arrived and adsorbed. But it sets the speed of the layers of neighbors to arrive. At this speed with a concentration gradient that stops around the first neighbor layer, the gradient does not project virtually in the longer time when the actual first passenger arrives. Thus, the average first passenger coming rate (unit # molecule/s) for this 3D diffusion simplified in 1D problem,
where is a factor of converting the 3D diffusive adsorption problem into a 1D diffusion problem whose value depends on the system, e.g., a fraction of adsorption area over solute nearest neighbor sphere surface area assuming cubic packing each unit has 8 neighbors shared with other units. This example fraction converges the result to the 3D diffusive adsorption solution shown above with a slight difference in pre-factor due to different packing assumptions and ignoring other neighbors.
When the area of interest is the size of a molecule (specifically, a long cylindrical molecule such as DNA), the adsorption rate equation represents the collision frequency of two molecules in a diluted solution, with one molecule a specific side and the other no steric dependence, i.e., a molecule (random orientation) hit one side of the other. The diffusion constant need to be updated to the relative diffusion constant between two diffusing molecules. This estimation is especially useful in studying the interaction between a small molecule and a larger molecule such as a protein. The effective diffusion constant is dominated by the smaller one whose diffusion constant can be used instead.
The above hitting rate equation is also useful to predict the kinetics of molecular self-assembly on a surface. Molecules are randomly oriented in the bulk solution. Assuming 1/6 of the molecules has the right orientation to the surface binding sites, i.e. 1/2 of the z-direction in x, y, z three dimensions, thus the concentration of interest is just 1/6 of the bulk concentration. Put this value into the equation one should be able to calculate the theoretical adsorption kinetic curve using the Langmuir adsorption model. In a more rigid picture, 1/6 can be replaced by the steric factor of the binding geometry.
The bimolecular collision frequency related to many reactions including protein coagulation/aggregation is initially described by Smoluchowski coagulation equation proposed by Marian Smoluchowski in a seminal 1916 publication, derived from Brownian motion and Fick's laws of diffusion. Under an idealized reaction condition for A + B → product in a diluted solution, Smoluchovski suggested that the molecular flux at the infinite time limit can be calculated from Fick's laws of diffusion yielding a fixed/stable concentration gradient from the target molecule, e.g. B is the target molecule holding fixed relatively, and A is the moving molecule that creates a concentration gradient near the target molecule B due to the coagulation reaction between A and B. Smoluchowski calculated the collision frequency between A and B in the solution with unit #/s/m3:
where:
is the radius of the collision,
is the relative diffusion constant between A and B (m2/s),
and are number concentrations of A and B respectively (#/m3).
The reaction order of this bimolecular reaction is 2 which is the analogy to the result from collision theory by replacing the moving speed of the molecule with diffusive flux. In the collision theory, the traveling time between A and B is proportional to the distance which is a similar relationship for the diffusion case if the flux is fixed.
However, under a practical condition, the concentration gradient near the target molecule is evolving over time with the molecular flux evolving as well, and on average the flux is much bigger than the infinite time limit flux Smoluchowski has proposed. Before the first passenger arrival time, Fick's equation predicts a concentration gradient over time which does not build up yet in reality. Thus, this Smoluchowski frequency represents the lower limit of the real collision frequency.
In 2022, Chen calculates the upper limit of the collision frequency between A and B in a solution assuming the bulk concentration of the moving molecule is fixed after the first nearest neighbor of the target molecule. Thus the concentration gradient evolution stops at the first nearest neighbor layer given a stop-time to calculate the actual flux. He named this the critical time and derived the diffusive collision frequency in unit #/s/m3:
where:
is the area of the cross-section of the collision (m2),
is the relative diffusion constant between A and B (m2/s),
and are number concentrations of A and B respectively (#/m3),
represents 1/, where d is the average distance between two molecules.
This equation assumes the upper limit of a diffusive collision frequency between A and B is when the first neighbor layer starts to feel the evolution of the concentration gradient, whose reaction order is 2+1/3 instead of 2. Both the Smoluchowski equation and the JChen equation satisfy dimensional checks with SI units. But the former is dependent on the radius and the latter is on the area of the collision sphere. From dimensional analysis, there will be an equation dependent on the volume of the collision sphere but eventually, all equations should converge to the same numerical rate of the collision that can be measured experimentally. The actual reaction order for a bimolecular unit reaction could be between 2 and 2+1/3, which makes sense because the diffusive collision time is squarely dependent on the distance between the two molecules.
These new equations also avoid the singularity on the adsorption rate at time zero for the Langmuir-Schaefer equation. The infinity rate is justifiable under ideal conditions because when you introduce target molecules magically in a solution of probe molecule or vice versa, there always be a probability of them overlapping at time zero, thus the rate of that two molecules association is infinity. It does not matter that other millions of molecules have to wait for their first mate to diffuse and arrive. The average rate is thus infinity. But statistically this argument is meaningless. The maximum rate of a molecule in a period of time larger than zero is 1, either meet or not, thus the infinite rate at time zero for that molecule pair really should just be one, making the average rate 1/millions or more and statistically negligible. This does not even count in reality no two molecules can magically meet at time zero.
Biological perspective
[edit]
The first law gives rise to the following formula:
where
P is the permeability, an experimentally determined membrane "conductance" for a given gas at a given temperature,
c2 − c1 is the difference in concentration of the gas across the membrane for the direction of flow (from c1 to c2).
Fick's first law is also important in radiation transfer equations. However, in this context, it becomes inaccurate when the diffusion constant is low and the radiation becomes limited by the speed of light rather than by the resistance of the material the radiation is flowing through. In this situation, one can use a flux limiter.
The exchange rate of a gas across a fluid membrane can be determined by using this law together with Graham's law.
Under the condition of a diluted solution when diffusion takes control, the membrane permeability mentioned in the above section can be theoretically calculated for the solute using the equation mentioned in the last section (use with particular care because the equation is derived for dense solutes, while biological molecules are not denser than water. Also, this equation assumes ideal concentration gradient forms near the membrane and evolves):
where:
is the total area of the pores on the membrane (unit m2),
transmembrane efficiency (unitless), which can be calculated from the stochastic theory of chromatography,
D is the diffusion constant of the solute unit m2⋅s−1,
t is time unit s,
c2, c1 concentration should use unit mol m−3, so flux unit becomes mol s−1.
The flux is decay over the square root of time because a concentration gradient builds up near the membrane over time under ideal conditions. When there is flow and convection, the flux can be significantly different than the equation predicts and show an effective time t with a fixed value, which makes the flux stable instead of decay over time. A critical time has been estimated under idealized flow conditions when there is no gradient formed. This strategy is adopted in biology such as blood circulation.
Semiconductor fabrication applications
[edit]
The semiconductor is a collective term for a series of devices. It mainly includes three categories：two-terminal devices, three-terminal devices, and four-terminal devices. The combination of the semiconductors is called an integrated circuit.
The relationship between Fick's law and semiconductors: the principle of the semiconductor is transferring chemicals or dopants from a layer to a layer. Fick's law can be used to control and predict the diffusion by knowing how much the concentration of the dopants or chemicals move per meter and second through mathematics.
Therefore, different types and levels of semiconductors can be fabricated.
Integrated circuit fabrication technologies, model processes like CVD, thermal oxidation, wet oxidation, doping, etc. use diffusion equations obtained from Fick's law.
CVD method of fabricate semiconductor
[edit]
The wafer is a kind of semiconductor whose silicon substrate is coated with a layer of CVD-created polymer chain and films. This film contains n-type and p-type dopants and takes responsibility for dopant conductions. The principle of CVD relies on the gas phase and gas-solid chemical reaction to create thin films.
The viscous flow regime of CVD is driven by a pressure gradient. CVD also includes a diffusion component distinct from the surface diffusion of adatoms. In CVD, reactants and products must also diffuse through a boundary layer of stagnant gas that exists next to the substrate. The total number of steps required for CVD film growth are gas phase diffusion of reactants through the boundary layer, adsorption and surface diffusion of adatoms, reactions on the substrate, and gas phase diffusion of products away through the boundary layer.
The velocity profile for gas flow is:
where:
is the thickness,
is the Reynolds number,
x is the length of the substrate,
v = 0 at any surface,
is viscosity,
is density.
Integrated the x from 0 to L, it gives the average thickness:
To keep the reaction balanced, reactants must diffuse through the stagnant boundary layer to reach the substrate. So a thin boundary layer is desirable. According to the equations, increasing vo would result in more wasted reactants. The reactants will not reach the substrate uniformly if the flow becomes turbulent. Another option is to switch to a new carrier gas with lower viscosity or density.
The Fick's first law describes diffusion through the boundary layer. As a function of pressure (p) and temperature (T) in a gas, diffusion is determined.
where:
is the standard pressure,
is the standard temperature,
is the standard diffusitivity.
The equation tells that increasing the temperature or decreasing the pressure can increase the diffusivity.
Fick's first law predicts the flux of the reactants to the substrate and product away from the substrate:
where:
is the thickness ,
is the first reactant's concentration.
In ideal gas law , the concentration of the gas is expressed by partial pressure.
where
is the gas constant,
is the partial pressure gradient.
As a result, Fick's first law tells us we can use a partial pressure gradient to control the diffusivity and control the growth of thin films of semiconductors.
In many realistic situations, the simple Fick's law is not an adequate formulation for the semiconductor problem. It only applies to certain conditions, for example, given the semiconductor boundary conditions: constant source concentration diffusion, limited source concentration, or moving boundary diffusion (where junction depth keeps moving into the substrate).
Invalidity of Fickian diffusion
[edit]
Even though Fickian diffusion has been used to model diffusion processes in semiconductor manufacturing (including CVD reactors) in early days, it often fails to validate the diffusion in advanced semiconductor nodes (< 90 nm). This mostly stems from the inability of Fickian diffusion to model diffusion processes accurately at molecular level and smaller. In advanced semiconductor manufacturing, it is important to understand the movement at atomic scales, which is failed by continuum diffusion. Today, most semiconductor manufacturers use random walk to study and model diffusion processes. This allows us to study the effects of diffusion in a discrete manner to understand the movement of individual atoms, molecules, plasma etc.
In such a process, the movements of diffusing species (atoms, molecules, plasma etc.) are treated as a discrete entity, following a random walk through the CVD reactor, boundary layer, material structures etc. Sometimes, the movements might follow a biased-random walk depending on the processing conditions. Statistical analysis is done to understand variation/stochasticity arising from the random walk of the species, which in-turn affects the overall process and electrical variations.
Food production and cooking
[edit]
The formulation of Fick's first law can explain a variety of complex phenomena in the context of food and cooking: Diffusion of molecules such as ethylene promotes plant growth and ripening, salt and sugar molecules promotes meat brining and marinating, and water molecules promote dehydration. Fick's first law can also be used to predict the changing moisture profiles across a spaghetti noodle as it hydrates during cooking. These phenomena are all about the spontaneous movement of particles of solutes driven by the concentration gradient. In different situations, there is different diffusivity which is a constant.
By controlling the concentration gradient, the cooking time, shape of the food, and salting can be controlled.
See also
[edit]
Advection
Churchill–Bernstein equation
Diffusion
False diffusion
Gas exchange
Mass flux
Maxwell–Stefan diffusion
Nernst–Planck equation
Osmosis
Citations
[edit]
^ Vallero, Daniel A. (2024). "Physical transport of air pollutants". Air Pollution Calculations. pp. 163–190. doi:10.1016/B978-0-443-13987-1.00017-X. ISBN 978-0-443-13987-1.
^ Fick A (1855). "Ueber Diffusion". Annalen der Physik (in German). 94 (1): 59–86. Bibcode:1855AnP...170...59F. doi:10.1002/andp.18551700105.
Fick A (1855). "On liquid diffusion". The London, Edinburgh, and Dublin Philosophical Magazine and Journal of Science. 10 (63): 30–39. doi:10.1080/14786445508641925.
^ Philibert J (2005). "One and a Half Centuries of Diffusion: Fick, Einstein, before and beyond" (PDF). Diffusion Fundamentals. 2: 1.1–1.10. doi:10.62721/diffusion-fundamentals.2.187. Archived from the original (PDF) on 5 February 2009.
^ Vázquez JL (2006). "The Porous Medium Equation". Mathematical Theory. Oxford Univ. Press.
^ Gorban AN, Sargsyan HP, Wahab HA (2011). "Quasichemical Models of Multicomponent Nonlinear Diffusion". Mathematical Modelling of Natural Phenomena. 6 (5): 184–262. arXiv:1012.2908. doi:10.1051/mmnp/20116509. S2CID 18961678.
^ Atkins P, de Paula J (2006). Physical Chemistry for the Life Science.
^ Conlisk, A. Terrence (2013). Essentials of Micro- and Nanofluidics. p. 43. doi:10.1017/CBO9781139025614. ISBN 978-0-521-88168-5.
^ Vignes, Alain (May 1966). "Diffusion in Binary Solutions. Variation of Diffusion Coefficient with Composition". Industrial & Engineering Chemistry Fundamentals. 5 (2): 189–199. doi:10.1021/i160018a007. ISSN 0196-4313.
^ Schmitt, Sebastian; Hasse, Hans; Stephan, Simon (17 March 2025). "Entropy scaling for diffusion coefficients in fluid mixtures". Nature Communications. 16 (1): 2611. arXiv:2409.17615. Bibcode:2025NatCo..16.2611S. doi:10.1038/s41467-025-57780-z. ISSN 2041-1723. PMC 11914492. PMID 40097384.
^ Williams FA (1985). "Appendix E". Combustion Theory. Benjamin/Cummings.
^ Taylor R, Krishna R (1993). Multicomponent mass transfer. Wiley Series in Chemical Engineering. Vol. 2. John Wiley & Sons. ISBN 978-0-471-57417-0.[page needed]
^ Brogioli D, Vailati A (January 2001). "Diffusive mass transfer by nonequilibrium fluctuations: Fick's law revisited". Physical Review E. 63 (1 Pt 1) 012105. arXiv:cond-mat/0006163. Bibcode:2000PhRvE..63a2105B. doi:10.1103/PhysRevE.63.012105. PMID 11304296. S2CID 1302913.
^ Bian X, Kim C, Karniadakis GE (August 2016). "111 years of Brownian motion". Soft Matter. 12 (30): 6331–6346. Bibcode:2016SMat...12.6331B. doi:10.1039/c6sm01153e. PMC 5476231. PMID 27396746.
^ a b Pyle JR, Chen J (2 November 2017). "Photobleaching of YOYO-1 in super-resolution single DNA fluorescence imaging". Beilstein Journal of Nanotechnology. 8: 2296–2306. doi:10.3762/bjnano.8.229. PMC 5687005. PMID 29181286.
^ a b Langmuir I, Schaefer VJ (1937). "The Effect of Dissolved Salts on Insoluble Monolayers". Journal of the American Chemical Society. 29 (11): 2400–2414. Bibcode:1937JAChS..59.2400L. doi:10.1021/ja01290a091.
^ a b Ward AF, Tordai L (1946). "Time-dependence of Boundary Tensions of Solutions I. The Role of Diffusion in Time-effects". Journal of Chemical Physics. 14 (7): 453–461. Bibcode:1946JChPh..14..453W. doi:10.1063/1.1724167.
^ a b c d e f g h
Chen J (January 2022). "Simulating stochastic adsorption of diluted solute molecules at interfaces". AIP Advances. 12 (1): 015318. Bibcode:2022AIPA...12a5318C. doi:10.1063/5.0064140. PMC 8758205. PMID 35070490.
^ Pandey S, Gautam D, Chen J (16 July 2024). "Measuring the Adsorption Cross Section of YOYO-1 to Immobilized DNA Molecules". Journal of Physical Chemistry B. 128 (29): 7254–7262. doi:10.1021/acs.jpcb.4c03359. PMC 11286311. PMID 39014882.
^ a b c d Chen J (December 2022). "Why Should the Reaction Order of a Bimolecular Reaction be 2.33 Instead of 2?". The Journal of Physical Chemistry A. 126 (51): 9719–9725. Bibcode:2022JPCA..126.9719C. doi:10.1021/acs.jpca.2c07500. PMC 9805503. PMID 36520427.
^ Smoluchowski M (1916). "Drei Vorträge über Diffusion, Brownsche Molekularbewegung und Koagulation von Kolloidteilchen". Zeitschrift für Physik (in German). 17: 557–571, 585–599. Bibcode:1916ZPhy...17..557S.
^ Nosek TM. "Section 3/3ch9/s3ch9_2". Essentials of Human Physiology. Archived from the original on 24 March 2016.
^ Zhou L, Nyberg K, Rowat AC (September 2015). "Understanding diffusion theory and Fick's law through food and cooking". Advances in Physiology Education. 39 (3): 192–197. doi:10.1152/advan.00133.2014. PMID 26330037. S2CID 3921833.
Further reading
[edit]
Berg HC (1977). Random Walks in Biology. Princeton.
Bird RB, Stewart WE, Lightfoot EN (1976). Transport Phenomena. John Wiley & Sons.
Bokshtein BS, Mendelev MI, Srolovitz DJ, eds. (2005). Thermodynamics and Kinetics in Materials Science: A Short Course. Oxford: Oxford University Press. pp. 167–171.
Crank J (1980). The Mathematics of Diffusion. Oxford University Press.
Fick A (1855). "On liquid diffusion". Annalen der Physik und Chemie. 94: 59. – reprinted in Fick, Adolph (1995). "On liquid diffusion". Journal of Membrane Science. 100: 33–38. doi:10.1016/0376-7388(94)00230-v.
Smith WF (2004). Foundations of Materials Science and Engineering (3rd ed.). McGraw-Hill.
External links
[edit]
Fick's equations, Boltzmann's transformation, etc. (with figures and animations)
Fick's Second Law on OpenStax
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The Spark Of Life: Discharge Physics As A Key Aspect Of The Miller–Urey Experiment
By Keith Cowing
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Frontiers in Physics
April 10, 2024
Filed under biochemistry, Frontiers in Physics, Miller-Urey experiment, origin of life, prebiotic chemistry
The Miller–Urey experiment demonstrated the possibility of producing biomolecules from the chemical components of the primordial atmosphere, using an electric discharge.
It profoundly influenced the development of prebiotic chemistry and astrobiology. The essential aspect of the experiment is the action of the electric field on a gaseous mixture, which produces chemically active species. These last react to ultimately form biomolecules.
Sketch of the Miller-Urey experiment, showing vapor circulation and discharge electrodes. Picture courtesy of Dr. Domenico Aceto, CNR.
In this work the hypotheses and methods used to describe the chemical activation of a gas by an electric field, used in computational chemistry and in the physics of electrical discharges, are contrasted, showing that the second provides a much more realistic description of the primary events.
A future model should combine the insights of the two communities to bring forth a faithful and insightful description of the experiment, from the primary events to the formation of biomolecules.
The spark of life: discharge physics as a key aspect of the Miller–Urey experiment, Frontiers in Physics (open access)
Frontiers in Physics ( IF 3.1 ) Pub Date: 2024-04-05 , DOI:10.3389/fphy.2024.1392578Savino Longo
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6963 | https://artofmaths.wordpress.com/wp-content/uploads/2014/06/inequalities-a-mathematical-olympiad-approach.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, .
.
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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.
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Andreescu, T., Feng, Z., Lee, G., Problems and Solutions from Around the World. Mathematical Olympiads 2000-2001. MAA, 2003.
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Bulajich, R., G´ omez Ortega, J.A., Geometr´ ıa. Ejercicios y Problemas.
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208 Bibliography Kazarinoff, N., Geometric Inequalities. New Math. Library, MAA. 1961.
Larson, L., Problem-Solving Through Problems.
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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 |
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| Introduction to additive combinatorics (TCC)In a nutshell | | | --- | | Starting: | Tuesday, 15th October 2013 | | Dates: | 15th, 22nd October, 5th, 12th, 19th, 26th November | | Time slot: | Tuesdays, 9am-11am | | Level: | graduate course (in the Taught Course Centre) | | Prerequisites: | some basic Fourier analysis, linear algebra and probability | | Office hours: | Friday, 2pm (on google hangout - please email me if you wish to participate!) | | | Synopsis | Syllabus | Organisation | Assessment | BibliographySynopsis This course serves as a first introduction to additive combinatorics, a subject that has a substantial history but has gained much attention in recent years as a result of numerous high-profile breakthroughs such as the Green-Tao theorem on arithmetic progressions in the primes. Historically, additive combinatorics addresses problems regarding the additive structure of the integers. For example, what can we say about a finite set A of integers if we know that its sumset A+A={a+a': a, a' in A} is small? It turns out that such sets are highly structured, and we shall see various quantitative results to this effect in the 'toy setting' of F_p^n. We shall also study other additive structures such as 3-term arithmetic progressions in dense subsets of this vector space. The aforementioned results about arithmetic structure are intimately related to structural results about large graphs, and in particular Szemerédi's celebrated regularity lemma. After proving it and deriving some graph-theoretic as well as arithmetical consequences, we shall give some examples of applications to theoretical computer science. The methods employed throughout are a mix of analytic and algebraic techniques, together with some elementary probabilistic arguments and some hands-on combinatorics. For a precise list of topics please see the detailed syllabus below. The prerequisites for this course are minimal. We shall need the fundamental notions of (discrete) Fourier analysis as well as basic concepts from linear algebra and discrete probability. Feel free to contact me at julia.wolf at bristol.ac.uk with any questions prior to the start of the course. Return to top Syllabus | | | | --- | Date | Content | Recommended exercises | | 15/10/2013 | Introduction to discrete Fourier analysis: Meshulam's theorem and an arithmetic removal lemma | 1.9, 1.11, 1.14, 1.24 | | 22/10/2013 | The structure of sets with small sumset: Plünnecke's inequality and the Freiman-Ruzsa theorem | 2.3, 2.14 | | 05/11/2013 | Sumsets contain large subspaces: Bogolyubov's lemma and Chang's theorem | 3.3, 3.9, 3.11, 3.14 | | 12/11/2013 | Additive energy and small sumset: the Balog-Szemerédi-Gowers theorem | 4.2, 4.5, 4.10 | | 19/11/2013 | Szemerédi's regularity lemma for graphs: triangle removal and other applications | 5.9, 5.19 | | 26/11/2013 | Selected applications of additive combinatorics to theoretical computer science | 6.2, 6.3, 6.10 | Return to top Organisation This course will not take place on Tuesday, 29th October 2013. Return to top Assessment At the end of the course, participants will choose from a list of original research articles and write up an exposition of the chosen result. This exposition should place the result in the context of what has been discussed in the course, and should be sufficiently detailed for other course participants to be able to follow the main steps of the argument. The completion of the short weekly problem sets is optional but strongly encouraged. Return to top Bibliography The following reading material may be helpful to course participants. You may also wish to consult my personal roadmap for interesting surveys and original research articles in this general area. | | | | --- | [BTW] | B. Barak, L. Trevisan, A. Wigderson (2007): Lecture notes on Additive Combinatorics and Computer Science. | website | | [G] | B.J. Green (2005): Finite fields models in additive combinatorics. | arXiv | | [N] | M.B. Nathanson (1996): Additive Number Theory: Inverse problems and the geometry of sumsets, Springer Graduate Texts in Mathematics. | google books | | [TV] | T. Tao and V. Vu (2006): Additive Combinatorics, Cambridge Univ. Press. | google books | Return to top |
| This page was last updated 23rd October 2013. | © 2003-2025 Julia Wolf | |
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What is an Algebraic Expression?
ElevatEd
Math
May 20, 2025
Have you ever caught your child gazing blankly at those enigmatic jumbles of numbers, letters, and symbols on their math homework? Those enigmatic combinations resemble a code, but they're algebraic expressions - the foundation of advanced mathematics! What may appear as mathematical hieroglyphics to parents is a rich language that enables us to describe real-world situations in elegant, solvable forms.
And the best part? Once you decipher the code, these expressions unlock problem-solving techniques that reach far beyond the math classroom!
Learning algebraic expressions isn't all about getting through the next test - it's about learning critical thinking skills that serve students well in all areas of study and in life.
In this all-encompassing guide, we'll discover what an algebraic expression is, dissect its parts, and observe how these mathematical tools tackle real-world issues in surprisingly pragmatic ways!
Join our Math Course & Master Math Concepts Like a Pro!
What is an Algebraic Expression? A Simple Definition
Essentially, an algebraic expression is a math sentence that has variables, constants, and operation symbols (such as addition, subtraction, multiplication, or division). Unlike equations, algebraic expressions never have equal signs - they're more mathematical sentences than sentences.
In order to see what an algebraic expression is, let's define it in simple terms:
Components of Algebraic Expressions
Variables: Letters (e.g., x, y, z) used to denote unknown values
Constants: Fixed numerical values (like 7, 42, or 1.5)
Co-efficients: Values that you multiply variables with (in 5x, 5 is a coefficient)
Operators: Mathematical operators (+, -, ×, ÷, ^) applied in operations
Termini: Dividers with + or - signs for the term modifiers
Also Read - Practical Examples of Algebraic Expression Addition.
Types of Algebraic Expressions
To know an algebraic expression is to be familiar with the different types depending on the number of terms that they possess:
| | | |
---
| Type | Definition | Example |
| Monomial | Expression with one term | 7x |
| Binomial | Expression with two terms | 3y + 5 |
| Trinomial | Expression with three terms | x² + 4x + 4 |
| Polynomial | Expression with one or more terms | 2x³ - 5x² + x - 7 |
The Real-World Power of Algebraic Expressions
Algebraic expressions can seem abstruse, yet they help us model and solve problems from the real world in countless ways:
Practical Applications:
Shopping: Determining the cost of a collection of items (3c + 5d is buying 3 cookies for price c and 5 donuts for price d)
Sports: Total points computation (2p + 3q is 2-point and 3-point shot in basketball)
Business: Calculating profit (R - C is revenue minus costs)
Science: Identifying relationships among variables (v = d/t for velocity, distance, and time)
| |
| Fun Fact! - Did you know that the ancient Babylonians were performing algebra to solve problems as far back as 2000 BCE? But they didn't use symbols like we do today - everything was in words! Can you imagine how long your math homework would be without algebraic expressions! |
Also Read - Applications of Algebraic Expressions in Real Life.
How to Work with Algebraic Expressions
Algebraic expressions mastery requires several key skills:
Simplifying Expressions
Add or subtract like terms (terms that have the same variable raised to the same power) to simplify expressions.
Example: 3x + 5 + 2x - 7 simplifies to 5x - 2
Evaluating Expressions
Replace values with variables to determine the value of the expression.
If x = 4, then 2x + 3 = 2(4) + 3 = 11
Translating words to expressions
Convert word descriptions to mathematical equations.
Example: "2 times a number + 7" is 2n + 7
Also Read - 5 Tips to Simplify Any Algebraic Expression.
Typical Issues with Algebraic Equations
Algebraic expressions are challenging for most students since they:
Confuse expressions with equations
Randomise the order of operations.
Add unlike terms wrongly
Misread negative signs
At 98thPercentile, our mathematics curriculum confronts these challenges head-on with practice in visualisation strategies, real-life applications, and step-by-step guided practice. Our instructors are masters at taking the abstract and making it concrete so that students truly understand and feel confident working with it.
Building Solid Algebraic Foundations
It is crucial to comprehend algebraic expressions because they are the foundation of advanced mathematical concepts. In 98thPercentile, our creative math classes build this comprehension in a structured way, taking students from basic algebraic concepts to advanced problem-solving techniques through engaging, challenging lessons.
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FAQs
Q1: What is a basic definition of an algebraic expression?
Ans: An algebraic expression is a numerical statement that is made up of numbers, letters (variables), and operations (such as adding or multiplying) that are equal to something. Without equal signs, however, unlike with equations, expressions don't include equal signs. For instance, 3x + 5 is an expression that means "three times some number plus five."
Q2: How are algebraic expressions different from equations?
Ans: The primary distinction is that an algebraic expression is a value, whereas an equation is where two expressions are equal to one another. The expression 2x + 3 alone, but the equation 2x + 3 = 11 is a statement that can be solved. We explicitly teach students in 98thPercentile this distinction.
Q3: How can I help my child manage algebraic expressions more effectively?
Ans: Consistent practice with diverse examples is of great help. Our 98thPercentile classes offer guided practice with real-time feedback to assist students in gaining confidence in algebraic expressions through interactive lessons and real-life applications at their level.
Q4: Why is it necessary for my child to study algebraic expressions?
Ans: Algebraic expressions develop higher-order thinking, logical reasoning, and problem-solving skills that extend beyond mathematics. Algebraic expressions are the foundation for advanced-level math courses and prepare students for STEM-related careers. My curriculum at 98thPercentile emphasizes these connections to allow students to see the relevance of what they learn.
Q5: At what age should one introduce children to algebraic expressions?
Ans: While standard algebra typically starts around the time of 7th grade, we begin algebraic thinking earlier at 98thPercentile by using pattern identification and small variable ideas. Developing a concept of algebraic expression gradually before being formally taught is more sensible.
Related Articles
Mastering Algebraic Expressions and Equations
An Algebraic Expression and Its Types
Algebraic Equations Guide | 98thPercentile
Introduction to Algebraic Expressions Addition
Step-by-Step Guide to Adding Algebraic Expressions
Addition and Subtraction of Algebraic Functions: A Foundation of Math
Strategies for Subtraction of Algebraic Expressions
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6966 | https://open.maricopa.edu/fundamentalsoforganicchemistry/chapter/9-2-halogenation-reaction-of-alkanes/ | Skip to content
22 Halogenation Reaction of Alkanes
Chapter 22 Learning Objectives
Predict the products of halogenation reaction of alkanes.
Be able to draw the mechanism of radical halogenation of alkanes.
Explain the thermodynamics of halogenation reactions.
22.1 Reaction of Alkanes with Halogens
When alkanes react with halogen (Cl2 or Br2), with heat or light, hydrogen atom of the alkane is replaced by halogen atom and alkyl halide is produced as product. This can be generally shown as:
A specific example is:
Such type of reaction can be called as substitution because hydrogen is substituted by halogen; can also be called halogenation because halogen is introduced into the product. For this book, both terms are used in this chapter, interchangeably.
The net reaction for halogenation seems straightforward, the mechanism is more complicated though, it go through multiple steps that include initiation, propagation and termination.
We will take the example of mono-chlorination of methane, for the discussion of reaction mechanism.
CH4+ Cl2 → CH3Cl + HCl
22.2 Mechanism for Mono-chlorination of Methane
Initiation: Production of radical
With the energy provided from heat or light, chlorine molecule dissociates homolytically, each chlorine atom takes one of the bonding electrons, and two highly reactive chlorine radicals, Cl•, are produced.
Propagation: Formation of product and regeneration of radical
The propagation step involve two sub-steps. In the 1st step, the Cl• takes a hydrogen atom from the methane molecule (this is also called as hydrogen abstractionby Cl•), and C-H single bond breaks homolytically. A new σ bond is formed by Cl and H each donate one electron and HCl is produced as the side product. The CH3 radical, CH3•, the critical intermediate for the formation of product in next step, is formed as well.
In the 2nd step, the CH3• abstracts a chlorine atom to give final CH3Cl product, together with another Cl•. The regenerated Cl• can attack another methane molecule and cause the repetition of step 1, then step 2 is repeated, and so forth. Therefore the regeneration of the Cl• is particularly significant, it makes the propagation step self-repeat hundreds or thousands of time. The propagation step is therefore called the self-sustaining step, only small amount of Cl• is required at the beginning to promote the process.
Initiation and propagation are productive steps for the formation of product. This type of sequential, step-wise mechanism in which the earlier step generate the intermediate that cause the next step of the reaction to occur, is call the chain reaction.
The chain reaction will not continue forever though, because of the termination steps.
Termination: Consumption of radicals
When two radicals in the reaction mixture meet with each other, they combine to form a stable molecule. The combination of radicals lead to the decreasing of the number of radicals available to propagating the reaction, and the reaction slows and stops eventually, so the combination process is called termination step. A few examples of termination are given above, other combinations are possible as well.
Examples: Identifying the steps in a radical mechanism
Label the following steps as initiation, propagation or termination.
Answers: 1. Propagation 2. Initiation 3. Termination 4. Termination 5. Propagation
The propagation steps are the core steps in halogenation. The energy level diagram helps to provide further understanding of the propagation process.
The 1st step in propagation is endothermic, while the energy absorbed can be offset by the 2nd exothermic step. Therefore the overall propagation is exothermic process and the products are in lower energy level the than reactants.
The reaction heat (enthalpy) for each of the propagation step can also be calculated by referring to the homolytic bond dissociation energies (Table X.1). For such calculation, energy absorbed for bond-breaking step, so the bond energy was given “+” sign, and energy released for bond-forming step, and the “-” sign applied.
| | | | | | |
--- --- --- |
| Bond | kJ/mol | Bond | kJ/mol | Bond | kJ/mol |
| A — B → A • + B • | | | | | |
| F — F | 159 | H —Br | 366 | CH3 — I | 240 |
| Cl — Cl | 243 | H — I | 298 | CH3CH2 —H | 421 |
| Br — Br | 193 | CH3 — H | 440 | CH3CH2 —F | 444 |
| I — I | 151 | CH3 — F | 461 | CH3CH2 —Cl | 353 |
| H — F | 570 | CH3 — Cl | 352 | CH3CH2 — Br | 295 |
| H — Cl | 432 | CH3 — Br | 293 | CH3CH2 — I | 233 |
Table Homolytic Bond Dissociation Energies for Some Single Bonds
Examples
Calculation reaction energy for the propagation step of mono-chlorination of methane (referring to the corresponding bond energies in Table 9.1.)
Solution:
Step 1: H — CH3 + •Cl → CH3• + H — Cl
The H — CH3 bond broken, absorb energy, so +440 kJ
The H — Cl bond formed, release energy, so – 432 kJ
ΔH1 = +440 + (-432) = +8 kJ
Step 2: Cl — Cl + CH3• → CH3 — Cl + •Cl
The Cl — Cl bond broken, absorb energy, so +243 kJ
The CH3 — Cl formed, release energy, so -352kJ
ΔH2 = +243 + (-352) = – 109kJ
ΔHpropagation = ΔH1 + ΔH2 = +8 + ( – 109 ) = – 101kJ
The calculated data does match with the data from the energy diagram.
22.3 Reactivity Comparison of Halogenation
The energy changes for halogenation (substitution) with the other halogens can be calculated in the similar way, the results are summarized in Table 9.2.
| | | | | |
--- ---
| Reaction | F2 | Cl2 | Br2 | I2 |
| Step 1: H — CH3 + •X → CH3• + H — X | -130 | +8 | +74 | -142 |
| Step 2: X — X + CH3• → CH3 — X + • X | -322 | -109 | -100 | -89 |
| Overall propagation: H — CH3 + X — X → CH3 — X + HX | -452 | -101 | -26 | +53 |
Table Enthalpy of the Propagation Steps in Mono-halogenation of Methane (kJ/mol)
The data above indicate that the halogen radicals have different reactivity, fluorine is most reactive and iodine is least reactive. The iodine radical is very unreactive with overall “+” enthalpy, so iodine does not react with alkane at all. On the other side, the extreme high reactivity of fluorine is not a benefit either, the reaction for fluorine radical is so vigorous or even dangerous with lots heat released, and it is not practical to apply this reaction for any application because it is hard to control it. So Cl2 and Br2, with reactivity in the medium range, are used for halogen substitutions of alkanes. Apparently Cl2 is more reactive than Br2, and this leads to the different selectivity and application between the two halogens. |
6967 | https://www.quora.com/How-do-we-compare-sets | Something went wrong. Wait a moment and try again.
Types of Set
Elementary Set Theory
Set Operations
Applications of Sets
Sets (mathematics)
General Set Theory
Examples of Sets
Set Theory
5
How do we compare sets?
Jon Obermark
M.S. in Mathematics & Logic, University of Illinois at Urbana-Champaign (Graduated 1993)
·
Author has 6.1K answers and 1.5M answer views
·
1y
The point of set theory is to be a general model of all semantics. So there is meant to be some way of comparing sets that captures or models every way we ever compare things.
The default comparison operation between sets is obviously inclusion, either the element operation or the subset operation, and we have the model of ordinals where these coincide. Counting is the next most common way of comparing things in general, and set theory supports it with the structure of cardinal numbers, usually thought of as minimal ordinals that cannot be covered by smaller ordinals.
Next we often move up t
The point of set theory is to be a general model of all semantics. So there is meant to be some way of comparing sets that captures or models every way we ever compare things.
The default comparison operation between sets is obviously inclusion, either the element operation or the subset operation, and we have the model of ordinals where these coincide. Counting is the next most common way of comparing things in general, and set theory supports it with the structure of cardinal numbers, usually thought of as minimal ordinals that cannot be covered by smaller ordinals.
Next we often move up to sets with operators, and we identify structure-preserving maps. This gives us category theory. These often identify substructures that are ‘generated’ by the subsets via the operators creating lattices of containing and contained structures that can break down complex structures by pulling out smaller structures that are embedded in a way that respects the generation process. This gives us all the ideas of abstract algebra.
By elaborating the relations and their algebraic properties we create ways of describing structures that embed our intended meanings into sets and relations (which are also sets), producing structures like logics or networks or spaces, which model the important aspects of real semantics. Then we can discuss the actual semantics by discussing how the model embeds the inclusion operations between the sets.
So the answer kind of has to be any way you please, because you can build up bizarre combinations of sets that encode whatever scenario you want to model, and reduce your problem to identifying the right inclusion (or element, but that can be reduced to inclusion) relationships between the sets.
Related questions
What is the difference between disjoint sets and non-equivalent sets?
How can you compare two sets without counting every element in both sets?
What is the definition of "equal sets"? What is the relationship between equal sets and their elements? How do equal sets compare to other sets?
Is an empty set an element of every set?
What are universal set, union of sets, intersection of sets, subset, power set, empty set, equivalent set, joint set, and disjoint set?
David Harden
Bachelor of Science in Mathematics, Massachusetts Institute of Technology (Graduated 2004)
·
Author has 1.4K answers and 453.8K answer views
·
1y
This question is way too broad to be answered in a specific way. Most mathematical objects are defined as sets, so most ways of comparing two entities of the same type would work as answers.
If you mean a purely set-theoretic way of comparing them, then we can talk about elementhood, subsethood, and comparing cardinality. If you want to refine comparing cardinality, then we can compare elementhood posets to more than the first level of depth.
Senia Sheydvasser
PhD in Mathematics
·
Upvoted by
Terry Moore
, M.Sc. Mathematics, University of Southampton (1968) and
Hunter Johnson
, PhD Mathematics, University of Maryland, College Park (2008) · Author has 2.5K answers and 39.9M answer views
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2y
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Are there infinite sets compared to which other infinite sets are infinitesimal?
I don’t know of any formal meaning of comparing the sizes of (general) sets such that one set can be said to be “infinitesimal” compared to another set. I suppose that I could guess at what a reasonable definition ought to be—a suggestion might be that a set X is “infinitesimal” compared to a set Y if the cardinality of Y is greater than the cardinality of NX, where N are the natural numbers.
But, honestly, I don’t think it matters. For any reasonable definition of such a term, you are going to get that every set is infinitesimal compared to some other set. It’s not just that
I don’t know of any formal meaning of comparing the sizes of (general) sets such that one set can be said to be “infinitesimal” compared to another set. I suppose that I could guess at what a reasonable definition ought to be—a suggestion might be that a set X is “infinitesimal” compared to a set Y if the cardinality of Y is greater than the cardinality of NX, where N are the natural numbers.
But, honestly, I don’t think it matters. For any reasonable definition of such a term, you are going to get that every set is infinitesimal compared to some other set. It’s not just that the cardinalities of sets grow arbitrarily large—that wouldn’t be enough to draw such a conclusion. It is that the collection of cardinalities of sets is too damn big to be a set—for any set of sets (no matter how infinite!), there exists a set with cardinality larger than any set in that set.
David Joyce
Ph.D. in Mathematics, University of Pennsylvania (Graduated 1979)
·
Upvoted by
Tom McFarlane
, M.S. Mathematics, University of Washington (1994) and
José Ilhano Silva
, M. S. Mathematics & Differential Geometry, Federal University of Ceará (2017) · Author has 9.9K answers and 68.4M answer views
·
5y
Related
Can a set be a member of another set?
In Zermelo-Fraenkel set theory, the only thing that can be a member of a set is another set. That means something like {“Mary”, “John”, “Sue”} is not a set in Zermelo-Fraenkel set theory.
Naturally, once you realize that, you wonder if you can’t have things that aren’t sets in a set, then what can be a set in Zermelo-Fraenkel set theory? Well, there’s the empty set. That’s the set that has no elements. It’s denoted ∅. Once you have the empty set, you can have a set with it in it, namely {∅}. Now you’ve got two sets, ∅ and {∅}, and you can get some more, namel
In Zermelo-Fraenkel set theory, the only thing that can be a member of a set is another set. That means something like {“Mary”, “John”, “Sue”} is not a set in Zermelo-Fraenkel set theory.
Naturally, once you realize that, you wonder if you can’t have things that aren’t sets in a set, then what can be a set in Zermelo-Fraenkel set theory? Well, there’s the empty set. That’s the set that has no elements. It’s denoted ∅. Once you have the empty set, you can have a set with it in it, namely {∅}. Now you’ve got two sets, ∅ and {∅}, and you can get some more, namely, {{∅}} and {∅,{∅}}. You might think that this is all pointless, but that’s Zermelo-Fraenkel set theory for you, and, surprisingly, it’s not pointless. You can base mathematics on it.
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Related questions
Can you explain the difference between an ordered set and a well-ordered set?
If A belongs to set A and set B, does it imply that A is a subset of B?
What is the difference between a set and an ordered set?
What are the different kinds of sets?
What are some examples of sets that have a difference?
David Kra
Achieved client, employer, & my program goals for 50+ years.
·
Author has 18.8K answers and 22.9M answer views
·
5y
Related
Are ordered sets automatically comparable?
No, you might need to supply an ordering function or operator for the sets and their elements.
If I understand most of wikipedia’s article on partially ordered sets, the answer is, “NO” for several reasons:
Not all ordered sets have an arithmetic value that can compare them.
Not all elements of ordered sets have an arithmetic value that can compare them using system supplied operators.
Even some ordered sets of numbers, the ordering might not be <, =, >. It might be something like “divides evenly into.”
File:Infinite lattice of divisors.svg - Wikipedia
No, you might need to supply an ordering function or operator for the sets and their elements.
If I understand most of wikipedia’s article on partially ordered sets, the answer is, “NO” for several reasons:
Not all ordered sets have an arithmetic value that can compare them.
Not all elements of ordered sets have an arithmetic value that can compare them using system supplied operators.
Even some ordered sets of numbers, the ordering might not be <, =, >. It might be something like “divides evenly into.”
File:Infinite lattice of divisors.svg - Wikipedia
Henk Brozius
studied math and one of the subjects was settheory.
·
Author has 3.6K answers and 2.2M answer views
·
1y
Related
What are the kinds of sets?
A set can be:
Empty
Transitive
Supertransitive
Inductive
A natural number
An ordinal
Regular
……
These are not “
the
kinds of sets” but only “
some
kinds of sets”. There are a lot more and nothing stops you or others from inventing a new kind of set.
A set
A
is empty if it has no elements, i.e. if the statement
x
∈
A
is false for every set
x
. The axiom of extensionality allows us to prove that there is at most one empty set. This allows us to give it a special notation which is
∅
. There is another axiom of ZF stating that an empty set exists.
2. A set
A
is transitive if each of the elements of
A
is
A set can be:
Empty
Transitive
Supertransitive
Inductive
A natural number
An ordinal
Regular
……
These are not “
the
kinds of sets” but only “
some
kinds of sets”. There are a lot more and nothing stops you or others from inventing a new kind of set.
A set
A
is empty if it has no elements, i.e. if the statement
x
∈
A
is false for every set
x
. The axiom of extensionality allows us to prove that there is at most one empty set. This allows us to give it a special notation which is
∅
. There is another axiom of ZF stating that an empty set exists.
2. A set
A
is transitive if each of the elements of
A
is also a subset of
A
. So if
a
∈
A
∧
b
∈
a
→
b
∈
A
.
3. A set
A
is supertransitive if all subsets of all elements of
A
are elements of
A
. So if
a
∈
A
∧
b
⊆
a
→
b
∈
A
.
4. A set
A
is inductive if
∅
∈
A
and for every
a
∈
A
we also have
a
∪
{
a
}
∈
A
.
5. A set
n
is a natural number if it is an element of every inductive set.
6. A set
a
is regular if every set
B
that contains
a
as an element also contains an element
b
such that
B
and
b
have no element in common.
7. A set
α
is an ordinal within the universe of regular sets if it is transitive and also its elements are transitive.
8. ……
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Zeeshan Ahmed
BS Mathematics from University of Education Lahore
·
2y
Related
How can you show that two sets are equal?
Two sets are said to be equal when they have same & equal number of elements.
Note: The order doesn't matter as long as the elements are same
i.e A={1,2,3} , B={3,2,1}
Samuel Gomes da Silva
Ph.D. in Mathematics & Set Theory, University of São Paulo (USP) (Graduated 2004)
·
Upvoted by
Nathan Hannon
, Ph. D. Mathematics, University of California, Davis (2021) · Author has 706 answers and 280.1K answer views
·
2y
Related
Are there infinite sets compared to which other infinite sets are infinitesimal?
There is no such a thing of a set being infinitesimal with respect to other in Set Theory, but maybe you would like to know about large cardinals.
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Arpit Agrawal
miss Mathematics out of my life
·
Author has 381 answers and 1.6M answer views
·
6y
Related
Can a set belong to another set?
Yes, it can. A set can be a collection of any type of entities, which can be sets themselves. For example, A is a set of numbers while B is a set of sets here:
A = {1, 2, 3, 4, 5}
B = { {1, 2}, {3, 4} }
As per Wikipedia:
In mathematics, a set is a collection of distinct objects, considered as an object in its own right.
Finally, we know there’s a special set associated to all sets, and this set is called a Power Set. A power set is the set of all subsets of the set. For example if A is a set = {1, 2, 3}. All subsets of A will be {1} , {2}, {3}, {1, 2}, {1, 3}, {2, 3} and {1, 2, 3}, and finally empt
Yes, it can. A set can be a collection of any type of entities, which can be sets themselves. For example, A is a set of numbers while B is a set of sets here:
A = {1, 2, 3, 4, 5}
B = { {1, 2}, {3, 4} }
As per Wikipedia:
In mathematics, a set is a collection of distinct objects, considered as an object in its own right.
Finally, we know there’s a special set associated to all sets, and this set is called a Power Set. A power set is the set of all subsets of the set. For example if A is a set = {1, 2, 3}. All subsets of A will be {1} , {2}, {3}, {1, 2}, {1, 3}, {2, 3} and {1, 2, 3}, and finally empty set called ∅ . Power set for a set is denoted by P(A).
In this case, P(A) = {∅, {1} , {2}, {3}, {1, 2}, {1, 3}, {2, 3}, {1, 2, 3}}
Ankur Kalra
Rating Analyst at CRISIL
·
10y
Related
If set A = {1,2,3,4} & set B = {4,5,6}, then what is set A-set B?
Let's consider a fruit say an Apple to understand this concept better.
Now consider the complete apple as set A. It is composed of lot many elements.
Suppose you eat a portion of the apple or in other words you eat some elements of the set A.
So the elements you ate are now in your stomach. Let's consider the elements you ate as part of set B. I'm saying that the elements are part of set B.
Reason: There is a possibility that you might have eaten or tasted some other fruits or vegetables, etc. as well thus adding their elements to the set B. Hence set B can be a much bigger set.
Now let's relate
Let's consider a fruit say an Apple to understand this concept better.
Now consider the complete apple as set A. It is composed of lot many elements.
Suppose you eat a portion of the apple or in other words you eat some elements of the set A.
So the elements you ate are now in your stomach. Let's consider the elements you ate as part of set B. I'm saying that the elements are part of set B.
Reason: There is a possibility that you might have eaten or tasted some other fruits or vegetables, etc. as well thus adding their elements to the set B. Hence set B can be a much bigger set.
Now let's relate the concept with the question above.
Set A comprises of four elements i.e. 1, 2, 3, 4.
Now you ate a part of set A and got 4 in the set B. Thus you subtracted 4 from set A. But set B has 2 other elements 5 and 6 which would have come from other fruits or vegetables.
Hence Set A - Set B = {1,2,3}
In short, Set A - Set B is a Mathematical operation which comprises of all those elements that are present in set A except for those which are part of both sets A and B.
Stewart Valin
Athlete. Writer Community volunteer.
·
Upvoted by
Klaus Ole Kristiansen
, M.Sc. Mathematics, University of Copenhagen (1992) · Author has 4.7K answers and 3.1M answer views
·
2y
Related
What are two examples of a set?
Bonus set
Bonus set
Bonus set
Bonus set
Joe Eaton
Distinguished Systems Engineer for Graph and Data at NVIDIA (2022–present)
·
Author has 563 answers and 410.9K answer views
·
2y
Related
What is the definition of "equal sets"? What is the relationship between equal sets and their elements? How do equal sets compare to other sets?
The set theory version of equal sets is that every element in set A is contained in set B, and every element in set B is contained in set A. This means that every element in either of the sets is in both.
This has other corollaries or consequences, such as the sets have the same number of elements, and each element is uniquely identifiable.
Equal sets compare to other sets in exactly the same way, for all intents of set theory they are the same.
Adamya Prakhar Goyal
Studied at Birla Institute of Technology and Science, Pilani - Pilani Campus
·
Author has 141 answers and 581.2K answer views
·
10y
Related
If set A = {1,2,3,4} & set B = {4,5,6}, then what is set A-set B?
Others have quite adequately answered the question, but I'd like to add to it.
Let A and B be two sets.
Steps to perform operation A−B.
Look at each element of A one-by-one. See if it is present in B or not. If it is present in B, mark it (in set A).
Now, remove all the elements of set A that you have marked in above step. The remaining set is your answer.
So, for the example asked proceed as follows.
Start with the first element of set A i.e. "1". Is it in set B? No.
Go to the next element of set A ("2"). Is it in set B? No.
Go to the next element of set A ("3"). Is it in set B? No.
Go to
Others have quite adequately answered the question, but I'd like to add to it.
Let A and B be two sets.
Steps to perform operation A−B.
Look at each element of A one-by-one. See if it is present in B or not. If it is present in B, mark it (in set A).
Now, remove all the elements of set A that you have marked in above step. The remaining set is your answer.
So, for the example asked proceed as follows.
Start with the first element of set A i.e. "1". Is it in set B? No.
Go to the next element of set A ("2"). Is it in set B? No.
Go to the next element of set A ("3"). Is it in set B? No.
Go to the next element of set A ("4"). Is it in set B? Yes. Mark it in set A.
Now, remove all marked elements of set A. There was only one, i.e. "4".
The remaining set is {1, 2, 3}. This is the answer for A-B.
NOTE: A−B≠B−A.
To perform B-A, do the "opposite". Look at each element of set B and see if it occurs in set A or not. If it occurs in set A, mark it (in set B). Now, simply remove the elements that you have marked in set B. The remaining set if your answer.
So, for the example asked proceed as follows.
Start with the first element of set B i.e. "4". Is it in set A? Yes. Mark it.
Go to the next element of set B ("5"). Is it in set A? No.
Go to the next element of set B ("6"). Is it in set A? No.
Now, remove all marked elements of set B. There was only one, i.e. "4".
The remaining set is {5, 6}. This is the answer for B-A.
A more general method to go from A-B to B-A or vice-versa.
In the first step of marking the common elements, mark them in both the sets. It doesn't take extra effort as you are comparing once anyways.
Now, to get A-B, simply remove the marked elements of set A, while to get B-A simply remove the marked elements of set B.
In mathematical terms,
A−B=A−(A∩B)
B−A=B−(B∩A)
Note that, (A∩B)=(B∩A) that is, you need not check the common terms of A and B twice. Simply check them starting with either A or B and the other one too will be taken care of (it's pretty obvious if you think about it).
The symbol ∩ denotes "intersection", i.e. the set of all common elements of A and B.
Related questions
What is the difference between disjoint sets and non-equivalent sets?
How can you compare two sets without counting every element in both sets?
What is the definition of "equal sets"? What is the relationship between equal sets and their elements? How do equal sets compare to other sets?
Is an empty set an element of every set?
What are universal set, union of sets, intersection of sets, subset, power set, empty set, equivalent set, joint set, and disjoint set?
Can you explain the difference between an ordered set and a well-ordered set?
If A belongs to set A and set B, does it imply that A is a subset of B?
What is the difference between a set and an ordered set?
What are the different kinds of sets?
What are some examples of sets that have a difference?
What is the difference between cardinality and size when it comes to sets?
In which case will the union of two sets be equal to an empty set?
How do we use sets?
What is the difference between a disjoint set and an empty set?
Can two sets have the same number of elements but be different sets? Is there any relationship between them?
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Auxins and their roles in Growth and Development
April 19, 2024
Home
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Auxins and their roles in Growth and Development
Auxins are a class of plant hormones that play a fundamental role in various aspects of plant growth and development. They are involved in processes such as cell elongation, apical dominance, root initiation, vascular tissue differentiation, and tropisms (growth responses to external stimuli). Here’s an overview of the roles of auxins in plant growth and development:
1. Cell Elongation and Expansion:
Stimulating Cell Growth: Auxins promote cell elongation, primarily in stems and young leaves. They facilitate the stretching of cells, leading to increased length of plant organs.
Cell Wall Loosening: Auxins activate enzymes that break down the rigid cell wall components, allowing the cell to expand and elongate.
2. Apical Dominance and Lateral Bud Inhibition:
Apical Dominance: Auxins produced by the apical bud inhibit the growth of lateral buds (buds found in the leaf axils), ensuring that the main stem (apical bud) grows more vigorously than the lateral branches.
3. Root Initiation and Growth:
Root Formation: Auxins promote the initiation and development of roots, both lateral and adventitious (formed from non-root tissues), aiding in the establishment of the root system.
Root Elongation: Auxins influence root elongation by stimulating cell division and expansion in the root tips.
4. Vascular Tissue Differentiation:
Xylem and Phloem Development: Auxins are critical in promoting the differentiation of vascular tissues, particularly xylem (responsible for water transport) and phloem (responsible for nutrient transport).
5. Tropisms:
Phototropism: Auxins are responsible for the growth response of plants to light, causing the bending of plant parts towards a light source.
Gravitropism: Auxins regulate the growth response of plants to gravity, causing roots to grow towards gravity and stems to grow against it.
6. Leaf Abscission and Senescence:
Leaf Senescence:Auxins influence the timing of leaf senescence, the aging and shedding of leaves. They can delay senescence, allowing the plant to retain functional leaves for a longer period.
7. Fruit Development:
Fruit Growth and Development: Auxins play a role in fruit development, influencing fruit size, shape, and overall growth.
8. Floral Initiation and Development:
Flowering: Auxins are involved in the initiation and development of flowers, influencing the formation of floral organs and overall flower structure.
9. Wound Healing and Regeneration:
Wound Healing:Auxinshelp in wound healing and regeneration of damaged plant tissues by stimulating cell division and tissue growth.
10. Interaction with Other Hormones:
Synergy with Cytokinins: Auxins often work in synergy with cytokinins, another class of plant hormones, to regulate various growth processes, striking a balance between cell division and cell elongation.
Understanding the roles of auxins and their interactions with other hormones is crucial for manipulating plant growth and development in agriculture. By optimizing auxin levels and their application, farmers can effectively modulate plant growth, increase yields, and improve crop quality.
Talk to your OMEX representativetoday to learn more about our auxin-containing products and how you can use them to promote growth and development and enhance productivity of your crop.
Dr. Abdel El Hadrami
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Calcium effect on Kochia and Thistle
June 3, 2024
Calcium, an essential plant nutrient, can influence the growth and health of various plants, including weeds like Kochia and thistle. However, calcium does not act as a selective herbicide to control these weeds directly. Instead, its effects are more related to plant health and vigor, which indirectly impacts the growth and competitiveness of weeds. Here’s […]
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6970 | https://www.vedantu.com/question-answer/a-roundrobin-tournament-is-played-amongst-class-10-maths-cbse-60cdaef6b9d60b2eae1a034d | A round-robin tournament is played amongst \[n\]-players. Each pair of players plays one game that culminates in a win for one player. A win earns \[1\] point; a loss \[0\] points. At the culmination of the tournament player \[i\] has \[{a_i}\] points \[\left( {0 \leqslant {a_i} \leqslant n - 1} \right)\]. Suppose that \[b = \left( {{b_1},{b_2},{b_3}, \ldots \ldots ,{b_n}} \right)\] is a given \[n\]-tuple of non-negative integers with \[{b_1} \leqslant {b_2} \leqslant \ldots \leqslant {b_n}\].. Prove that the \[n\]-tuple \[b\] could have arisen from a round-robin tournament as the set of final scores iffI.\[\sum\limits_{i = 1}^n {{b_i}} = \left( {\begin{array}{{20}{c}}n \\ 2 \end{array}} \right)\]II.\[\sum\limits_{i = 1}^k {{b_i} \geqslant \left( {\begin{array}{{20}{c}}k \\ 2 \end{array}} \right)} \] for \[1 \leqslant k \leqslant n\]
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A round-robin tournament is played amongst n-players. Each pair of players plays one game that culminates in a win for one player. A win earns 1 point; a loss 0 points. At the culmination of the tournament player i has a i points (0⩽a i⩽n−1). Suppose that b=(b 1,b 2,b 3,……,b n) is a given n-tuple of non-negative integers with b 1⩽b 2⩽…⩽b n.. Prove that the n-tuple b could have arisen from a round-robin tournament as the set of final scores iff
I.∑i=1 n b i=(n 2)
II.∑i=1 k b i⩾(k 2) for 1⩽k⩽n
Answer
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Hint:Here, in the question, we have been given that a round-robin tournament is played. And we have given some conditions for the same such as points-structure. And at last we have been given equations and are asked to prove which of the given equations holds true as a set of final scores of the round-robin tournament. We will first understand how the round robin tournament is played and how it’s structured.
Complete step-by-step answer:
A round-robin tournament is a tournament where each player/team plays against every other player/team exactly once.
And since there are n players, every player plays (n−1) matches. That is why a player can have a maximum of (n−1) points as given in the question.
Now we will understand the calculation of total points of a tournament.
Player 1 plays (n−1) matches, player 2 similarly plays (n−1) matches, but the total matches played by player 1&2 will be [(n−1)+(n−1)−1] because there is one match common between both the players. Similarly, player 3 will play (n−1) matches in total but the sum of matches played by all three players will be [(n−1)+[(n−1)−1]+[(n−1)−2]]. And moving so on, the additional matches played by the n t h player will be 0 as all the matches has already been counted till (n−1)t h player.
So, if we count the total number of matches in a tournament, it will be a series of sum of first (n−1) natural numbers.
Now we know that,
Sum of first n natural numbers = n(n+1)2
Therefore, Sum of first (n−1) natural numbers = n(n−1)2
Hence, the total number of matches played in tournament = n(n−1)2 and assuming that there is no tie, there will be a winner and a loser in each match, therefore,
Total number of wins = Total number of points scored in tournament = n(n−1)2.
Taking first option, ∑i=1 n b i=(n 2)
L.H.S. = ∑i=1 n b i
Since b i represents the points scored by each player, the sum of points scored by each player will be equal to the total points scored in a tournament.
Therefore, L.H.S. = n(n−1)2
Now, R.H.S. = (n 2)
=n C 2=n!2!(n−2)!=n(n−1)(n−2)!2(n−2)!=n(n−1)2
= L.H.S.
Taking second option, ∑i=1 k b i⩾(k 2)
We have L.H.S. = ∑i=1 k b i for 1⩽k⩽n
Clearly, it represents the sum of points scored by k players where 1⩽k⩽n. And also, we are given b 1⩽b 2⩽…⩽b n which further implies that the sum will be greater than or equal to the sum of first (k−1) natural numbers.
Now, Sum of first (k−1) natural numbers = k(k−1)2
Therefore, ∑i=1 k b i⩾(k 2)
Note:Whenever we face such types of questions, it is important that we understand the meaning and structure of round-robin tournaments. Whatever we have discussed here as a theory part is actually what we use for combination. A match is played between two players and we select the number of ways n players can play a match irrespective of any order.
If still getting confused, one can easily understand it numerically by assuming any natural value of n.
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6971 | https://cemc.uwaterloo.ca/sites/default/files/documents/2024/2024Gauss7Contest.pdf | The CENTRE for EDUCATION in MATHEMATICS and COMPUTING cemc.uwaterloo.ca Gauss Contest Grade 7 (The Grade 8 Contest is on the reverse side) Wednesday, May 15, 2024 (in North America and South America) Thursday, May 16, 2024 (outside of North America and South America) Time: 1 hour ©2024 University of Waterloo Calculating devices are allowed, provided that they do not have any of the following features: (i) internet access, (ii) the ability to communicate with other devices, (iii) information previously stored by students (such as formulas, programs, notes, etc.), (iv) a computer algebra system, (v) dynamic geometry software.
Instructions 1. Do not open the contest booklet until you are told to do so.
2. You may use rulers, compasses and paper for rough work.
3. Be sure that you understand the coding system for your answer sheet. If you are not sure, ask your teacher to explain it.
4. This is a multiple-choice test. Each question is followed by five possible answers marked A, B, C, D, and E. Only one of these is correct. When you have made your choice, enter the appropriate letter for that question on your answer sheet.
5. Scoring: Each correct answer is worth 5 in Part A, 6 in Part B, and 8 in Part C.
There is no penalty for an incorrect answer.
Each unanswered question is worth 2, to a maximum of 10 unanswered questions.
6. Diagrams are not drawn to scale. They are intended as aids only.
7. When your supervisor instructs you to start, you will have sixty minutes of working time.
The name, school and location of some top-scoring students will be published on the website, cemc.uwaterloo.ca. On this website, you will also be able to find copies of past Contests and excellent resources for enrichment, problem solving and contest preparation.
Grade 7 Scoring: There is no penalty for an incorrect answer.
Each unanswered question is worth 2, to a maximum of 10 unanswered questions.
Part A: Each correct answer is worth 5.
1.
When the four digits of 2024 are added, the result is (A) 10 (B) 8 (C) 224 (D) 44 (E) 16 2.
If n = 5, the value of n + 2 is (A) 5 (B) 6 (C) 7 (D) 8 (E) 9 3.
Which of the following shapes has a vertical line of symmetry?
(A) (B) (C) (D) (E) 4.
Students at Gauss Middle School were asked to choose their favourite school day.
The results are shown in the circle graph.
Which day was chosen by exactly one-quarter of the students?
(A) Monday (B) Tuesday (C) Wednesday (D) Thursday (E) Friday 25% Wednesday 10% Monday 15% Tuesday 20% Thursday 30% Friday 5.
A square with side length 5 has an area of (A) 5 (B) 10 (C) 15 (D) 16 (E) 25 6.
In the diagram, ∠PQR is a straight angle. The value of x is (A) 54 (B) 14 (C) 24 (D) 64 (E) 34 146° x° P Q R 7.
Katie completed two laps of a track without stopping. The first lap took 3 minutes and 45 seconds, and the second lap took 4 minutes and 35 seconds. What was her total time?
(A) 8 minutes 30 seconds (B) 7 minutes 50 seconds (C) 8 minutes 50 seconds (D) 8 minutes 20 seconds (E) 7 minutes 40 seconds 8.
The sequence of the five symbols ⃝◀⊠△⋆repeats to form the pattern: ⃝◀⊠△⋆⃝◀⊠△⋆· · · If the pattern is continued, the 23rd symbol in the pattern is (A) ⃝ (B) ◀ (C) ⊠ (D) △ (E) ⋆ 9.
Olivia cuts a 42 cm length of string into 2 cm pieces. Jeffcuts a 42 cm length of string into 3 cm pieces. How many more pieces of string does Olivia have than Jeff?
(A) 7 (B) 8 (C) 4 (D) 6 (E) 5 Grade 7 10.
A number is randomly chosen from the list 1, 2, 3, 4, 5, 6, 7, 8, 9. The probability that the chosen number is divisible by 2, or by 3, or by both 2 and 3, is (A) 4 9 (B) 5 9 (C) 6 9 (D) 7 9 (E) 8 9 Part B: Each correct answer is worth 6.
11.
In the subtraction of the two-digit numbers shown, the letters P and Q each represent a single digit.
8 P −Q 6 4 9 The value of P + Q is (A) 7 (B) 9 (C) 11 (D) 8 (E) 10 12.
The length of a rectangle is twice its width. The perimeter of the rectangle is 120 cm.
The width of the rectangle is (A) 20 cm (B) 60 cm (C) 30 cm (D) 50 cm (E) 10 cm 13.
Eloise purchased a number of water hand pumps to give to a charity. The mean (average) price was $85 per water pump. If Eloise spent a total of $765, how many water pumps did she purchase?
(A) 7 (B) 8 (C) 9 (D) 10 (E) 6 14.
The number 385 has three prime factors. The sum of these prime factors is (A) 21 (B) 26 (C) 25 (D) 23 (E) 22 15.
A circle has radius 2. If the radius of the circle is tripled, the area of the original circle divided by the area of the new circle is (A) 1 3 (B) 1 6 (C) 1 9 (D) 1 2 (E) 1 8 16.
Brett and Juanita each have a glass containing 300 mL of water. Brett pours half of his water out and then Juanita pours 20% of her water into Brett’s glass. What volume of water is now in Brett’s glass?
(A) 210 mL (B) 360 mL (C) 180 mL (D) 330 mL (E) 240 mL 17.
A circular spinner is divided into 12 identical unshaded sections and 3 identical shaded sections, as shown. Each unshaded section is 3 times the size of each shaded section. An arrow is attached to the centre of the spinner.
The arrow is spun once. What is the probability that the arrow stops in a shaded section?
(A) 1 15 (B) 1 5 (C) 1 12 (D) 1 13 (E) 1 4 18.
The Gaussbot factory assembles robots. Each robot comes in one of three colours: red, blue, or green. Each robot also has a number stamped on its head: 1, 2, 3, or 4.
The nth robot assembled is the first robot to have the same colour and the same number as a previously assembled robot. What is the greatest possible value of n?
(A) 11 (B) 12 (C) 13 (D) 7 (E) 8 Grade 7 19.
Five different integers in a list have a median of 10 and a range of 7. What is the smallest possible integer in the list?
(A) 4 (B) 5 (C) 6 (D) 7 (E) 8 20.
A standing desk has 31 height settings, numbered from the lowest height, 1, to the highest height, 31. Since the desk is not working properly, when the up button is pressed, the desk goes up 6 settings at a time if possible, otherwise it does not move.
When the down button is pressed, the desk goes down 4 settings at a time if possible, otherwise it does not move. If the desk starts at setting number 1, how many of the 31 settings will the desk be able to stop at?
(A) 14 (B) 16 (C) 9 (D) 15 (E) 10 Part C: Each correct answer is worth 8.
21.
Five different integers are selected from 1 to 6 and one integer is placed into each of the five squares shown. The integers are placed so that the sum of the three integers in the vertical column is 7, and the sum of the three integers in the horizontal row is 11. Which integer does not appear in any square?
(A) 3 (B) 4 (C) 2 (D) 6 (E) 5 22.
In the diagram, 17 toothpicks are used to make a 2 by 3 grid of squares. Of the toothpicks used, 10 are outer toothpicks and 7 are inner toothpicks. Suppose that toothpicks are used to make a 20 by 24 grid of squares.
To the nearest percent, what percentage of toothpicks used are inner toothpicks?
(A) 88% (B) 95% (C) 93% (D) 70% (E) 91% 23.
A rectangular prism has integer edge lengths and has a volume of V . The six faces of the prism are painted and then the prism is cut into 1 by 1 by 1 cubes. Of these cubes, 50 cubes have no paint on them. What is the mean (average) of all possible values of V ?
(A) 224 (B) 310 (C) 396 (D) 288 (E) 348 24.
A three-digit integer is an integer from 100 to 999, inclusive. A three-digit integer is called Tiny if no rearrangement of its digits gives a three-digit integer that is smaller.
For example, 138, 207 and 566 are Tiny, but 452, 360 and 727 are not. How many three-digit integers are Tiny?
(A) 255 (B) 201 (C) 212 (D) 234 (E) 219 25.
Suppose that w, x, y, z, (x + y), (x + z), (234 + z), and (234 −z) are 8 different prime numbers. If w +x+y = 234, and each of y and z is less than 50, the value of w −y is (A) 226 (B) 150 (C) 210 (D) 174 (E) 222 |
6972 | https://polypad.amplify.com/lesson/clock-angles | All Lesson Plans
Clock Angles
Overview and Objective
In this lesson, students will use the geared clock on Polypad to explore the angles made by the hour and minute hands of an analog clock.
Warm-Up
Start by discussing with students the angle of a full turn. Share this canvas with students and invite them to determine the angle measurement between two consecutive minutes and the angle measurement between two consecutive hours. There is a 6-degree angle between the minutes and there is a 30-degree angle between the marks of two consecutive hours.
You may also use the fraction slice of 1/12 to create each 30-degree angle.
Ask students to create a straight angle by arranging the hour and minute hands of the red clock. Let them share their answers.
Clarify with the students that It doesn't matter whether it is 6 am or 6 pm; the angle is the same, 180o, for both. (The easiest answer is 6:00, but since the hands turn smoothly, it will happen 11 more times in 12 hours.)
Discuss if 12:30 creates a straight angle or not. You can use the protractor tool to make a demonstration or let them measure on their own. You may use the same method to disprove 3:45 or 9:15.
On the same canvas, ask students to create the angles of 90o, 30o, and 60ousing the hands of the hour and minute of given geared clocks.
You can use the toggle seconds option at the bottom center toolbar after selecting a clock on the canvas to hide and show the second hand. The given clocks only show the hour and minute hands for students to measure the angles without confusion.
There are many possible answers to these questions. At this point, we will concentrate on the ones that are easy to find, like 1:00 or 11:00 for 30 degrees or 3:00 and 9:00 for 90 degrees.
Main Activity
Use the demonstration canvas to ask students the (smaller) angle between the hour and minute hands of the clock at 2:30. Start with asking students where each hand points out at 2:30.
Discuss with students that at 2:30, the minute hand points to 6 and the hour hand points halfway between the 2 and the 3.
You can use the fraction slice of 1/12 to mark each full 30o angle between the hands. We have 3 lots of 30o and a half of 30o. So the total angle is 3 x 30+15 = 105o.
You may then ask about 12:30, which shows 165o on the clock.
After students have an initial understanding of the angles created by the hour hand in relation to the minute hand, share this canvas with students and allow students time to work on the angle at 10:15.
Take some time to talk about possible steps, perhaps using the following prompts for discussion.
How many full 30 degrees lots are there between the hands?
Where the hour hand points exactly considering the minute hand covered 15 minutes of its 60-minute cycle?
Clarify with the students that the hour hand is 1/4 of the way from "10" to "11". Therefore the angle between 11 and where the hour hand points is 3/4 of 30o.
30 x 3/4=22.5 so the total angle is 4 x 30+22.5=142.5 degrees.
You might like to encourage students to estimate the angles before they calculate so that they can check if their final answers make sense.
The next question on the canvas is 6:45. Discuss where each hand points out at 6:45. Students need to be able to comment that hour hand is at 3/4 of the way between 6 and 7. So, we need to calculate 1/4 of 30 degrees to find the angle between the hands.
30 x 1/4=7.5. So, the total angle is 2 x 30+7.5=67.5 degrees. They can use the protractor to measure the angle.
According to the needs of students, you may increase the number of examples with the quarter and half hours. Then, if the students are ready for more complicated questions, you can move on to the next clocks on the canvas.
Closure
Share some student work with the class. Invite students to share which approaches they found most useful when answering these questions. To close the lesson, you may group students as pairs and share this canvas to ask the angle between the clock hands at 01:36.
Solutions
Support and Extension
For students ready for additional extension in this lesson, consider asking questions like finding the angle at 1:33 or 11:57. You may also ask them to find a time that has 180 degrees between the hour and minute hands other than 6:00 o'clock.
For students needing additional support with these ideas, you can increase the number of examples for each benchmark question.
Polypads for This Lesson
To assign these to your classes in Mathigon, save a copy to your Mathigon account.
Click here to learn how to share Polypads with students and how to view their work. |
6973 | https://www.tellmeinspanish.com/vocab/body-parts-in-spanish/ | Body Parts in Spanish: Beginner & Advanced Words - Tell Me In Spanish
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Body Parts in Spanish: Beginner & Advanced Words
Written by Daniela Sanchez in VocabularyLast Updated January 28, 2025
Whether you are an advanced or beginner learner, knowing the different body parts in Spanish can help you describe yourself and explain physical discomfort, among other useful applications.
So, in this guide, you’ll find:
List of Parts of the Body in Spanish
Basic terms
Advanced terms
Spanish Grammar: Body Parts & Definite Articles
Bonus: Idioms with Parts of the Body
Next Steps
Download PDF
In addition to real-life examples, I’ll also provide you with audio so you can check the pronunciation of these words.
List of Body Parts in Spanish
Below is a list of terms for parts of the body in Spanish. These words have been classified into beginner and advanced levels.
As a quick reminder, here are some of the most common Spanish verbs used to talk about body parts:
Doler: To hurt
Estirar: To stretch
Lastimar: To hurt
Rascar: To scratch
Quebrar / Romper: To break
Sonarse: Blow your nose
Tocar: To touch
Tener: To have
In the sections below, you’ll see more examples of these verbs.
Take Note:Although they share the same translation, doler and lastimar have slightly different functions in Spanish. The verb doler(works like ‘gustar’) describes aches and pains, whereas lastimar refers to injuries.
Basic body parts
Here is a list of common body parts you must know:
Axila: Armpit
Barbilla: Chin
Boca: Mouth
Brazo: Arm
Cabeza: Head
Cabello: Hair
Cejas: Eyebrows
Cintura: Waist
Codo: Elbow
Cuello: Neck
Dedo: Finger / Toe
Diente: Tooth
Espalda: Back
Estómago: Stomach
Frente: Forehead
Garganta: Throat
Hombro: Shoulder
Labio: Lip
Lengua: Tongue
Mano: Hand
Mejilla: Cheek
Muñeca: Wrist
Nariz: Nose
Ojo: Eye
Ombligo: Belly button
Oreja: Ear
Pantorrilla: Calf
Pecho: Chest
Pesteñas: Eyelids
Pie: Foot
Pierna: Leg
Rodilla: Knee
Talón: Heel
Tobillo: Ankle
Uña: Nail
Tip:‘Frente’, ‘mano’, and ‘nariz’ are feminine words in Spanish.
Check these sentences:
Mi papá se lastimó la espalda.
My dad hurt his back.
Sus ojos son azules.
Her eyesare blue.
Emma tiene unas pestañas muy largas.
Emma has such long eyelashes.
A nosotros nos duele mucho la cabeza.
Our head hurts a lot.
Charlie tiene una cicatriz en el pecho.
Charlie has a scar on his chest.
Take Note: In Spanish, we use more specific terms when talking about animal body parts. Some examples of these terms include cola(tail), pata(leg) and hocico(snout).
Advanced Spanish body parts
If you’re ready to level up your vocabulary, the following list contains more advanced terms for external and internal Spanish body parts:
Antebrazo: Forearm
Bazo: Spleen
Cadera: Hip
Cerebro: Brain
Columna: Spine / Backbone
Corazón: Heart
Costilla: Rib
Cráneo: Skull
Clavícula: Collarbone
Encía: Gum
Espinilla: Shin
Hígado: Liver
Hueso: Bone
Ingle: Groin
Intestino: Bowels
Lóbulo: Earlobe
Mándibula: Jaw
Músculo: Muscle
Muslo: Thigh
Nudillo: Knuckle
Nuca: Back of the neck
Paladar: Palate
Párpado: Eyelid
Planta del pie: Sole
Pómulo: Cheekbone
Pulmón: Lung
Púpila: Pupil
Riñón: Kidney
Sien: Temple
Vejiga: Bladder
Vena: Vein
Here are some sentences:
Betty se lastimó las encías.
Betty hurt her gums.
Nunca me he quebrado un hueso.
I have never broken a bone.
Miriam está enferma del corazón.
Miriam has heart disease.
Ellas quieren tonificar sus músculos.
They want to tone their muscles.
Body Parts & Spanish Definite Articles
To sound more natural, there is a small grammar rule that you must keep in mind when using these terms.
Unlike English, in Spanish, we rarely use possessive adjectives when talking about body parts. With this set of words, we use definite articles (el, los, la, las) to express possession. Check these sentences:
Incorrect
Me doblé mi tobillo.
I twisted my ankle.
Cynthia se cortó su cabello.
Cynthia cut her hair.
Correct
Me doblé el tobillo.
I twisted my ankle.
Cynthia se cortó el cabello.
Cynthia cut her hair.
In Spanish, using possessives with body parts can be overly-repetitive. The verb conjugation, pronouns, and context are enough to indicate who that body part belongs to.
However, you must use a possessive adjective when:
Using the verb serto describe body parts
It’s unclear who you’re talking about
Micara es redonda.
Myface is round.
¿De qué color sontus ojos?
What color areyoureyes?
Brian cayó rendido a suspies.
Brian fell head over heels for her.
Bonus: Spanish Idioms with Body Parts
The following idiomatic expressions with body parts not only are great to practice these terms, but will also help you sound more natural:
Costar un ojo de la cara: To cost an arm and a leg
Dar la cara: Face the consequences
Dar la espalda: Turn your back on
Echar una mano: Give a hand
Estar hasta las narices / la coronilla: To be fed up
Hacer oídos sordos: Turn a blind eye
Meter la pata: Put your foot in it
No chuparse el dedo: Not be born yesterday
No tener pelos en la lengua: Not mince your words
No tener ni pies ni cabeza: To make no sense
Poner el dedo en la llaga:To hit a raw nerve
Quedarse de brazos cruzados: To sit back and do nothing
Ser uña y mugre: Be joined at the hip
Tener el corazón en un puño: Have your heart in your mouth
Tener entre ceja y ceja: Have it in it for
Tomar el pelo: Play for a fool
Traer algo entre manos: To be up to something
Spanish Body Parts Additional Resources
If you’re serious about your Spanish, here are some additional resources to help you master this topic and improve your command of the language.
First, get familiar with the rules to form plural words in Spanish so you can use these terms correctly. Definite articles not only work with body parts, but also with different types of nouns. So, make sure you understand the rules to use these words.
To increase your vocabulary, check this list of basic nouns you must know.
Download the Body Parts in Spanish PDF
Memorizing all of the Spanish words for body parts can take a while. Feel free to download a copy of the PDF for this guide with a complete list of body parts in Spanish along with a visual graphic breakdown you can use to study and add more words to your vocabulary.
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6974 | https://www.vcalc.com/wiki/quadrilateral-area-with-diagonal | Typesetting math: 100%
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Quadrilateral Area from Sides and Diagonal
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A=f(a,b,c,d,D)
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| (a)Length of side a | | | --- | | | | | (m) meter (mm) millimeter (cm) centimeter (km) kilometer (in) inch (ft) foot (mi) mile (mil) mil (foot_survey_us) survey foot (U.S.) | | |
| (b)Length of side b | | | --- | | | | | (m) meter (mm) millimeter (cm) centimeter (km) kilometer (in) inch (ft) foot (yd) yard (mi) mile (mil) mil (foot_survey_us) survey foot (U.S.) | | |
| (c)Length of side c | | | --- | | | | | (m) meter (mm) millimeter (cm) centimeter (km) kilometer (in) inch (ft) foot (yd) yard (mi) mile (mil) mil (foot_survey_us) survey foot (U.S.) | | |
| (d)Length of side d | | | --- | | | | | (m) meter (mm) millimeter (cm) centimeter (km) kilometer (in) inch (ft) foot (yd) yard (mi) mile (mil) mil (foot_survey_us) survey foot (U.S.) | | |
| (D)Length of the diagonal | | | --- | | | | | (m) meter (mm) millimeter (cm) centimeter (km) kilometer (in) inch (ft) foot (yd) yard (mi) mile (mil) mil (foot_survey_us) survey foot (U.S.) | | |
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| Tags | Plane Geometry (2D) Area Polygon quadrilateral Verified |
| UUID | cc93386d-2f38-11e6-9770-bc764e2038f2 |
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The Area of Quadrilateral calculator computes the area of a simple quadrilateral given the four sides and the length of a diagonal between opposing corners.
INSTRUCTIONS: Choose units and enter the following:
(a) Length of side a
(b) Length of side b
(c) Length of side c
(d) Length of side d
(D) Length of diagonal (See Diagram)
Area of a Quadrilateral (A): The area (A) is returned in square meters. However, this can be automatically converted to many other area units (e.g. square inches) via the pull-down menu.
A quadrilateral is a polygon that has four sides, four vertices (corners), and four angles. The sum of the interior angles of any quadrilateral is always 360 degrees. Quadrilaterals can vary widely in shape and properties, including:
rectangles: squared corners and pairs of equal sides
squares: squared corners and equal sides
parallelograms: two sets of parallel sides
trapezoids: one set of parallel sides
rhombuses: four equal sides
kites: two pairs of equal and adjacent sides
Quadrilateral Calculators
Quadrilateral Area from Sides and Diagonal
Quadrilateral Area from Sides and Angles
Quadrilateral Perimeter
Quadrilateral Semi-Perimeter
Quadrilateral Volume
Quadrilateral Weight
Rectangle Area (A = ω ⋅ h)
Rectangle Diagonal
Trapezoid Area:From on top, bottom lengths and separation
Trapezoid Area from Length of Sides
Height of a Trapezoid
Angles of a Trapezoid
Area of a Kite (or Rhombus)
Regular Polygon Information
A regular polygon is a geometric shape with three or more straight sides where every side is the same length and every angle between connecting sides are the same angle. Because of the symmetry of the regular polygon, all the vertices of the polygon can be constructed to touch a circle in which the regular polygon is inscribed and all the chords that are polygon sides will then obviously be of equal length . Likewise, because of the regular polygon's symmetry, a circle constructed to be inscribed in a regular polygon and touching the polygon will touch the regular polygon at the midpoint of the polygon side. As shown in the pictures, Figure 1 and Figure 2, lines from the regular polygon's vertices to the circle's center form n isosceles triangles of equal area.
Regular Polygon Area Calculators
Area of Triangle - Three Sides
Area of a Square - Four Sides
Area of Pentagon - Five Sides
Area of Hexagon - Six Sides
Area of Heptagon - Seven Sides
Area of Octagon - Eight Sides
Area of Nonagon - Nine Sides
Area of Decagon - Ten Sides
Area of Hendecagon - Eleven Sides
Area of Dodecagon - Twelve Sides
Area of Polygon - Any number of sides (n).
This equation, Quadrilateral Area from Sides and Diagonal, references 1 page
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6975 | https://www.ohchr.org/Documents/Publications/RuleoflawMonitoringen.pdf | UNITED NATIONS Office of the United Nations High Commissioner for Human Rights RULE-OF-LAW TOOLS FOR POST-CONFLICT STATES Monitoring legal systems OFFICE OF THE UNITED NATIONS HIGH COMMISSIONER FOR HUMAN RIGHTS RULE-OF-LAW TOOLS FOR POST-CONFLICT STATES Monitoring legal systems United Nations New York and Geneva, 2006 ii NOTE The designations employed and the presentation of the material in this publication do not imply the expression of any opinion whatsoever on the part of the Secretariat of the United Nations concerning the legal status of any country, territory, city or area, or of its authorities, or concern-ing the delimitation of its frontiers or boundaries.
✵ Material contained in this publication may be freely quoted or reprinted, provided credit is given and a copy of the publication containing the reprinted material is sent to the Office of the United Nations High Commissioner for Human Rights, Palais des Nations, 8-14 avenue de la Paix, CH‑1211 Geneva 10, Switzerland.
✵ This publication has been produced with the financial assistance of the European Union. The views expressed herein can in no way be taken to reflect the official opinion of the European Union.
HR/PUB/06/3 iii CONTENTS Page Foreword . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . v Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 I. MONITORING METHODOLOGY . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5 A. The objective of monitoring legal systems in post-conflict contexts . . . . . . . . . . . . . . . . . . . . . . . . 5 B. Using a human rights analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7 C. International and regional standards . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8 D. Identifying and monitoring the domestic law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10 E. Monitoring to identify trends and systemic problems in practice . . . . . . . . . . . . . . . . . . . . . . . . . . 11 II. MONITORING CAPACITY . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13 A. Mandate . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13 B. Monitoring qualifications . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14 C. The role of monitoring . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16 D. Emergencies, interventions or action in specific cases . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18 E. Manuals and internal training . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18 F. Monitoring sustainability and exit strategies . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19 G. Knowledge and information resources and development . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20 III. THE MONITORING PROGRAMME . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23 A. Initial engagement in monitoring . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23 B. Selecting and updating priorities for monitoring . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24 C. Developing and maintaining relationships . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27 D. Case tracking and formulating indicators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28 E. Systems for monitoring . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30 F. Protection of information and information sharing . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31 iv Page IV. CASE MONITORING . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33 A. Case monitoring strategy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33 B. Interviewing . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34 C. Detention visits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35 D. File investigations and collecting documentation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37 E. Observing trial and investigatory proceedings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38 V. ANALYSIS, REPORTING AND RECOMMENDATIONS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41 A. The value of reporting . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41 B. External report publication and dissemination . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41 C. Internal reporting . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42 D. Individual case or issue reports . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43 E. Thematic reports . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 44 F. Periodic reviews . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45 G. Formulating recommendations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46 H. Follow-up and support to other rule-of-law aspects of peacekeeping . . . . . . . . . . . . . . . . . . . . 47 Annex Suggested guidelines for monitoring the justice system in a peacekeeping operation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49 Foreword The Office of the United Nations High Commissioner for Human Rights (OHCHR) has increasingly recognized the need to enhance its assistance in United Nations-wide efforts to work quickly and effectively to re-establish the rule of law and the administration of justice in post-conflict missions. Countries emerging from conflict and crisis are vulnerable to weak or non-existent rule of law, in-adequate law enforcement and justice administration capacity, and increased instances of human rights violations. This situation is often exacerbated by a lack of public confidence in State authorities and a shortage of resources. In 2003, OHCHR, as the United Nations focal point for coordinating system-wide attention for human rights, democracy and the rule of law, began to develop rule-of-law tools so as to ensure sustainable, long-term institutional capacity within United Nations missions and transitional administrations to respond to these demands. These rule-of-law tools will provide practical guidance to field missions and transitional administrations in critical transitional justice and rule of law-related areas. Each tool can stand on its own, but also fits into a coherent operational perspective. The tools are intended to outline the basic principles involved in: Mapping the Justice Sector, Prosecution Initiatives, Truth Commissions, Vetting and Monitoring Legal Systems.
This publication specifically addresses human rights monitoring of the justice system through the creation of a methodology. This tool is intended to reflect a comprehensive overview of the princi-ples, techniques and approaches involved in legal systems monitoring, principles which have been primarily garnered from previous experience and lessons learned from United Nations, Organiza-tion for Security and Co-operation in Europe and NGO legal systems monitoring programmes. The objective of this tool is to provide a framework for developing a monitoring programme to analyse institutions and the justice system as a whole from which good practices can be reinforced and bad practices or deficiencies addressed.
Clearly, this document cannot dictate strategic and programmatic decision-making, which needs to be made in the field in the light of the particular circumstances within each post-conflict environ-ment. However, the tool is meant to provide field missions and transitional administrations with the fundamental information required to develop a legal systems monitoring methodology, in line with international human rights standards and best practices.
The creation of these tools is only the beginning of the substantive engagement of OHCHR in tran-sitional justice policy development. I wish to express my appreciation and gratitude to all those who have contributed to the preparation of this important initiative.
Louise Arbour United Nations High Commissioner for Human Rights vi Acknowledgements OHCHR wishes to thank the individuals and organizations that provided comments, suggestions and support for the preparation of this tool. In particular, it would like to gratefully acknowl-edge the consultant who had primary responsibility for developing the tool, Shelley Inglis.
Special thanks are due to the European Commission, whose financial contribution made it pos-sible to carry out this project and publish the rule-of-law tools.
1 Introduction Post-conflict justice systems are characterized by severe dysfunction, low levels of human and material resources, destroyed infrastructure and lack of public trust. The past failure of the legal system to protect individual rights, prosecute violators and balance executive power is often either a direct cause of, or a substantial contributing factor to, the conflict. A history of cor-ruption, discrimination and abuse of power within the institutions of justice can destroy public confidence and perpetuate lawlessness and chaos.
Developing a justice system that protects human rights and promotes the rule of law is a criti-cal aspect of securing peace and preventing future conflict. As such, it can form a vital part of any peacekeeping operation and its mandate. The effective transformation of a dysfunctional justice system into one that meets fundamental international standards is an extremely long and difficult process. Effective reform requires a detailed and ongoing assessment of the func-tioning of the system and a clear picture of its problems. Consistent monitoring of the legal system for compliance with the domestic law and international standards of fairness provides decision makers with the accurate and specific information they need to strategically target resources for reform based on real rather than perceived challenges.
Transforming dysfunctional justice systems requires comprehensive institution-building and re-form. Monitoring legal systems implies a holistic approach that looks at the overall functioning of the system and its institutions. The police, the prisons, lawyers and the courts, among others, each have a critical and autonomous role to play to ensure a fair and effective process. Every justice institution must function efficiently and provide just results while working interdepen- dently with the others. If one institution or actor fails, the system is damaged and balancing mechanisms are needed to ensure recourse and redress.
1 The term “justice” refers to “an ideal of accountability and fairness in the protection and vindication of rights and the preven-tion and punishment of wrongs. Justice implies regard for the rights of the accused, for the interests of victims and for the well-being of society at large. It is a concept rooted in all national cultures and traditions and, while its administration usually implies formal judicial mechanisms, traditional dispute resolution mechanisms are equally relevant. The international com-munity has worked to articulate collectively the substantive and procedural requirements for the administration of justice….” Report of the Secretary-General on the rule of law and transitional justice in conflict and post-conflict societies (S/2004/616, para. 7).
2 The “rule of law” is defined as “a principle of governance in which all persons, institutions and entities, public and private, including the State itself, are accountable to laws that are publicly promulgated, equally enforced and independently adjudi-cated, and which are consistent with international human rights norms and standards. It requires, as well, measures to ensure adherence to the principles of supremacy of law, equality before the law, accountability to the law, fairness in the application of the law, separation of powers, participation in decision-making, legal certainty, avoidance of arbitrariness and procedural and legal transparency.” (S/2004/616, para. 6). 2 Monitoring of the legal system is aimed at identifying widespread patterns or trends in viola-tions of international standards and the systemic issues that hinder compliance in order to support effective reform initiatives. To assess compliance with the international human rights and relevant standards it is necessary to analyse the domestic law and daily practice in cases. The right to a fair legal process, access to justice and fair treatment, the independence of the judiciary, and the proper administration of justice lie at the heart of a fair and effective justice system.
High levels of insecurity in post-conflict or transitioning environments can hamper a whole reconstruction and peacebuilding effort. Establishing a base for the rule of law, beginning with public safety and accountability, is a foremost concern. Consequently, the monitoring of the justice system may focus primarily on the criminal justice process, particularly in the initial stages of a peacekeeping operation. Areas of civil-law protection are also often integrally linked to just and equitable development. Property law and family relations, including child protection, can have particular relevance for insecure post-conflict contexts and the use of informal or alterna-tive (usually unregulated) traditional methods and practices for resolving otherwise criminal and civil disputes can be prevalent.
Publicizing well-documented human rights concerns and systemic problems within the justice system, and informal justice mechanisms, can be essential not only for targeting reform and resources but also for generating the political will to do so. Monitoring should provide a com-prehensive picture of the strengths and weaknesses of the functioning of the justice system and assess the impact, positive and negative, of the reform efforts made. This will assist the authorities, civil society and the peacekeeping operation to evaluate progress towards the es-tablishment of the rule of law and the protection of human rights.
The objective of monitoring legal systems is to provide an analysis of the institutions and the system as a whole so that good practices can be reinforced and bad practices or deficiencies addressed. Monitoring is a tool not an end. Its value is reflected in the integrity and coherence of the information, the accuracy and depth of the analysis, and the relevancy and practicality of the recommendations. The recommendations should detail ways in which problems facing the justice system may be effectively addressed, for instance through legal and policy reform, setting up and developing institutions including accountability mechanisms, the provision of remedies, training and capacity-building, administrative and logistical support, and resource allocation.
This publication covers the following elements of monitoring legal systems, particularly in the context of peacekeeping: • Monitoring methodology and objective, applying a human rights analysis, using the in-ternational and domestic legal standards, and identifying trends and systemic issues in practice 3 • Developing a monitoring capacity, including the mandate and role of monitors, qualifica-tions and training, and knowledge and information resources • Establishing a monitoring programme, including selecting priorities and approaches, main-taining relationships, case tracking and systems that support effective monitoring • Case monitoring, including formulating a strategy, collecting and verifying information through interviews, case file reviews and attending court or other proceedings • Analysis, reporting and recommendations, including the development of internal and ex-ternal reporting, formulating recommendations and ensuring appropriate follow-up and support to other rule-of-law aspects of the peacekeeping operation It is important to recognize that this publication reflects a comprehensive overview of the prin-ciples, techniques and approaches involved in monitoring legal systems. While it is focused on the monitoring of a legal system as a whole, the principles, techniques and approaches can be applied to the monitoring of a discrete judicial mechanism, such as the Special Court for Sierra Leone, the Iraqi Special Tribunal or the Extraordinary Chamber in the Cambodian court system. It may not be possible to use or meet all aspects of this guidance in every situation or in every monitoring programme, although significant efforts should be made to strive to meet these standards. Ultimately, those responsible for legal system monitoring programmes in post-con-flict settings should set their priorities and goals based on the situation and implement these with professionalism, integrity and respect for the national context.
4 5 I. MONITORING METHODOLOGY A. The objective of monitoring legal systems in post-conflict contexts The objective of monitoring is to improve the legal system’s compliance with the law, includ-ing applicable international and regional standards, and to support the rule of law. Interna-tional and regional treaty, non-treaty and other standards provide the legitimate benchmarks by which to evaluate progress in the (re‑)establishment or strengthening of a justice system in a post-conflict environment. For the United Nations, the universally applicable standards adopted under its auspices must serve as the normative basis for activities in support of justice and the rule of law.3 The fundamental principles include non-discrimination and equal treatment, such as access to justice and fair treatment of victims, access to judicial remedy and/or redress established by law and treatment according to the domestic law, substantive and procedural fairness of proceedings, no impunity for crimes under international law, and the independent and impartial administration of justice.
Post-conflict environments suffer from devastated and often completely dysfunctional or dis-criminatory legal frameworks and institutions of justice. In many cases, the justice institutions, police, judiciary, prosecutorial and defence services, may need to be restructured or developed from scratch. In any post-conflict situation, it is unrealistic to expect immediate compliance by the institutions of justice or informal mechanisms with all international standards. However, there may be genuine efforts on the part of the Government and the society to move towards compliance and to integrate these standards into legal and justice reform plans. The monitor-ing of legal systems is a tool both to highlight for the authorities, society and other interested groups where specific issues with compliance lie and to evaluate the genuineness and effec-3 The Secretary-General has stated in his report “The normative foundation for our work in advancing the rule of law is the Charter of the United Nations itself, together with the four pillars of the modern international legal system: international human rights law; international humanitarian law; international criminal law; and international refugee law. This includes the wealth of United Nations human rights and criminal justice standards developed in the last half-century. These represent universally applicable standards adopted under the auspices of the United Nations and must therefore serve as the normative basis for all United Nations activities in support of justice and the rule of law.” (S/2004/161, para. 9).
6 tiveness of the efforts made to address these. The conclusions drawn from reliable monitor-ing are intended to assist Governments and the relevant authorities to produce targeted and meaningful changes that will make the justice system overall fairer and more effective. Without an accurate, in-depth and coherent picture of the actual functioning of the system, it is dif-ficult to formulate reform strategies that are targeted and effective while maximizing limited resources.
A part of this work is identifying the obstacles that post-conflict justice systems face in their ef-forts to achieve compliance with national law and international standards. Systemic institutional or administrative problems, such as insufficient or misused financial or material resources, can effectively hinder compliance. Although improved infrastructure, telecommunications, person-nel and material resources do not guarantee that the system will be fair and effective, they are necessary. For instance, effective and fair administration requires court clerks and others with the skills to manage files, keep adequate records, process timely summons and so on in order to ensure that a case goes to trial within a reasonable time, evidence is provided to the defence in a timely manner, and witnesses appear at trial to provide testimony and to be examined by the accused. Another example is that disproportionately low salaries for judges can contribute to corruption, affecting the impartiality of judicial processes.
Legal system monitoring should therefore cover the whole array of issues involved in establish-ing justice institutions that can function in accordance with the domestic law and in compliance with fundamental international standards of justice. This includes monitoring budgetary and financial allocations, administrative oversight and accountability or disciplinary mechanisms, ju-dicial appointment processes, human resources policies and staff allocation and training, law dissemination to justice officials and general publication, methods for ensuring appropriate interaction between the institutions and actors involved in justice, including the police, pros-ecutors, defence counsel and judges, as well as with related institutions and groups, such as referral networks for victims and witnesses, hospitals and protection programmes. All of these affect the ability of the formal justice system to provide the public with a fair, efficient and just resolution of cases.
In many post-conflict contexts, traditional practices are commonly used to resolve disputes. The monitors should be in a position to examine the extent to which these function, how and for which purposes or cases, and using which practices and norms. They should report on these practices and make recommendations on the ways in which the authorities and com-munities can address issues of concern, including whether these practices can be legalized as alternative dispute mechanisms, if not regulated, or be brought into conformity with domestic law protections and guarantees and compliance with basic international principles of justice. Although these are complex issues which may be raised through a monitoring programme, it should be recognized that changes to deeply embedded traditional practices may come about only as a result of broad-scale public and community consultation (including with traditionally 7 disempowered groups such as women) and firm commitment on the part of the authorities, community and religious leaders, and civil society.
B. Using a human rights analysis The human rights approach to legal system monitoring involves analysing both domestic law and domestic practice for compliance with international standards. Where the domestic law meets international standards, the focus is on whether the actual practice conforms to the domestic law. 1. Analysis of the law (de jure) for compliance with standards. This involves determining whether: - The guarantees provided in the law meet international or regional standards. For ex-ample, the domestic criminal procedure law states that an accused person may be detained for three months before being brought before a judge or judicial authority, but this length of time does not meet the requirements under international law of prompt review by a judicial authority. Another example may be that the law requires judges to be appointed by the executive branch for limited durations, which can violate international standards concerning judicial independence.
- The law lacks the guarantees or remedial responses that are required by international human rights standards. A common example is if the law does not provide for a ha-beas corpus mechanism to ensure the right to challenge one’s detention.
2. Analysis of the practice (de facto) of the law or its application to a particular case involves examining whether: - A failure to apply the domestic law violates international human rights standards. Such would be the case, for example, if the domestic criminal procedure code provides that witnesses must be present at trial for their testimony to be used as evidence and the accused is entitled to examine them, but in practice no witnesses are presented at trial, only their out-of-court statements are considered by the court and used against the accused. This practice fails to apply the domestic law and may constitute a violation of the right of the accused to challenge the witnesses against them under international human rights standards.
- The misapplication of the domestic law constitutes a violation of international human rights standards. For example, if the law states that the prosecutor and the accused shall be present for the review of detention orders, but in practice only the prosecu-tor attends, possibly resulting in the violation of the equality-of-arms principle under international law.
8 C. International and regional standards Four sources of international and regional standards may be used to monitor the legal system: international treaty standards; regional treaty and non-treaty standards; international custom-ary law; and international non-treaty standards.
International treaty standards. It is necessary to identify the treaties to which the host country is a party—which means that the Government or persons acting in a governmental capacity are bound to uphold the obligations established in them.4 In addition, it is important to identify the role of international law within the national legal system, i.e., whether treaties once ratified form part of national law.
International humanitarian, criminal and human rights law are all relevant for legal system moni-toring. Monitoring teams may be operating in areas where conflict persists and international hu-manitarian law will apply. It is critical to note that international human rights law always applies, including during conflict, although many rights can be derogated from in these circumstances.
Many of the human rights instruments and their articles reinforce each other, and the monitor-ing team should use all relevant sources to strengthen the analysis of whether a human rights violation exists. The key international human rights treaties establish committees to monitor their implementation by the States that are a party to them. The general comments of these committees provide authoritative guidance on the interpretation of the standards detailed in the conventions. As such, knowledge of these comments and of the findings of the committees can be useful in understanding the standards and in developing an in-depth analysis of whether they have been violated or abused. In addition to the primary human rights conventions used for legal monitoring, other conventions, such as the United Nations Convention against Trans- national Organized Crime and its Protocols, should be understood and used where relevant.
Regional treaty and non-treaty standards. Regional instruments cover Africa, the Americas and Europe, but notably not Asia.5 Courts and commissions have been created to interpret the re-gional instruments and provide judgements or opinions on their application. Where applicable, 4 International treaty standards include: - International Covenant on Civil and Political Rights - Convention on the Rights of the Child - Convention on the Elimination of All Forms of Discrimination against Women - Convention against Torture and Other Cruel, Inhuman or Degrading Treatment or Punishment - International Convention on the Elimination of All Forms of Racial Discrimination 5 The primary regional instruments are: - The African Charter on Human and Peoples’ Rights - The American Convention on Human Rights - The European Convention for the Protection of Human Rights and Fundamental Freedoms 9 the jurisprudence of these bodies can be extremely useful in detailing the meaning of these regional standards and how they apply to specific sets of facts. This jurisprudence can contrib-ute to the monitors’ substantive analysis of the legal system’s compliance with these and similar standards. There may be regional non-treaty standards and principles that can be used, such as in Asia the Beijing Statement of Principles of the Independence of the Judiciary adopted in August 1995.
Customary international law. Monitors may also look to another source of obligations if the Gov-ernment is not otherwise bound by treaties. Customary international law comprises norms con-sidered to be so widely recognized by the community of nations that they are deemed binding on all States. Some of the rights set forth in the Universal Declaration of Human Rights—which in articles 7 to 11 sets out the rights of equality before the law, an effective remedy for violations and a fair trial—are considered by some international legal scholars to be customary norms. Other customary norms are the prohibitions on genocide, slavery and war crimes. International non-treaty standards. These are not ratified by States, but are considered to pro-vide some essential guarantees for the administration of justice and fair processes. As a result of their adoption by the United Nations General Assembly, they carry significant political weight. Non-treaty standards should be used by the monitoring team to further develop its analysis of any possible violation or abuse. Since these standards often provide more specific guidelines than those contained in human rights conventions, they can be useful for articulating the pre-cise actions that authorities can take to improve compliance. For example, non-treaty standards specify what guarantees the independence of the judiciary, the procedures, services and actions required to meet the needs of victims of crime, the health and safety measures to be taken for detainees, and the duties that lawyers have to their clients. In this regard, these standards can also serve as helpful guidance for generating specific recommendations to the authorities to improve compliance with human rights norms.6 The Basic Principles on the Role of Lawyers, 6 The significant non-treaty standards related to monitoring the justice system are: - The Universal Declaration of Human Rights - The Declaration on the Elimination of Violence Against Women - Declaration of Basic Principles of Justice for Victims of Crime and Abuse of Power - Body of Principles for the Protection of All Persons under Any Form of Detention or Imprisonment - Standard Minimum Rules for the Treatment of Prisoners - Basic Principles on the Role of Lawyers - Guidelines on the Role of Prosecutors - Basic Principles on the Independence of the Judiciary - Safeguards guaranteeing protection of the rights of those facing the death penalty - United Nations Guidelines for the Prevention of Juvenile Delinquency - United Nations Rules for the Protection of Juveniles Deprived of their Liberty - United Nations Standard Minimum Rules for Non-custodial Measures - United Nations Standard Minimum Rules for the Administration of Juvenile Justice - Basic Principles on the Use of Force and Firearms by Law Enforcement Officials - Code of Conduct for Law Enforcement Officials 10 for example, can be used to recommend the development of a code of ethics and to detail the principles it should contain.
D. Identifying and monitoring the domestic law The applicable domestic law can be difficult to identify after a conflict. Peace agreements, Secu-rity Council resolutions and other transitional framework documents usually provide guidance on what the applicable law should be. However, they often do not address the hard questions, such as which specific versions or amendments of the law and regulations should be considered in effect. The monitor team, and the peacekeeping operation, should have a clear position on what is the applicable law.
Despite the potential difficulties of determining the applicable national law, it must serve as the backbone of the monitoring analysis. Treating individuals according to the law is a fundamental principle embedded in the international standards. In many post-conflict circumstances, the national laws may be relatively compliant with international standards, at least on paper. The substantial problems often arise in the practice, misapplication or failure to apply the law.
Many post-conflict countries suffer from a lack of formal, uniform and integrated legal systems. There are three sources of domestic law that may be relevant to monitoring their justice sys-tems. Some systems may actually be a hybrid of two or three sources of law. These are: • Constitutions, statutory or civil law. A constitution may provide the overall framework for the justice system and the legislative process as well as spell out the status of international law in the national justice system. For the criminal justice system, the primary laws of concern are the criminal law (or code), criminal procedure law and the law on penal sanc-tions. In addition, there are usually a number of special laws addressing key aspects of the criminal process, including juveniles codes, laws on policing or investigations and special security laws. For the civil-law process, the civil procedure law and administrative laws are critical along with laws on property, landownership and management, family laws, con-tract and employment and labour laws. To monitor the justice process, it is necessary to know the structure and powers of the courts and the actors. There are usually laws that detail the jurisdiction and administration of the court system and the roles and entitle-ments of parties to cases, police, prosecutors, lawyers and judges.
• Religious law. Forms of religious law may be integrated into legal frameworks or be used outside of the legal context. The use of religious law, both on a procedural and on a substantive level, should be monitored and documented in cases where it is applied incon-sistently with the statutory codes and where its application violates international human rights standards.
• Traditional laws and practices. The use of customary practices or traditions is common in many countries. The form these practices take depends on ethnicity, tribal affiliation and location of the community. Although general practices vary widely, these traditional 11 norms usually attempt to maintain community cohesion and structure and disproportion-ately disadvantage less powerful groups, in particular women. Traditional mechanisms of dispute resolution, often some form of a council of elders or tribal chiefs, can be used to resolve disputes and conflicts that are otherwise characterized in law as criminal offences or civil matters. They may, but often do not, enjoy a formal legal status. Their use can be pervasive and, rather than the formal justice system, can be more commonly relied on within and by individual communities. These practices should be monitored like all other dispute resolution processes for compliance with international standards. In particular, if these practices are unregulated, the extent to which Government structures and official actors participate in or sanction their use should be consistently monitored. It is critical that, despite practical difficulties, the legal system monitors integrate the monitoring of traditional laws and practices into their programme in order to provide a holistic picture of the extent to which basic principles of justice are being complied with in the resolution of disputes or conflicts within communities.
E. Monitoring to identify trends and systemic problems in practice Legal systems monitoring is a key aspect of the human rights analysis. It is not possible to monitor all aspects of practice, particularly with regard to cases. The approach is to identify emblematic practices, procedures, policies or cases in the legal system and seek to address these. In cases, this requires a methodology that extends beyond monitoring individual cases as single events for the purpose of identifying violations of human rights in the particular case and supporting individual remedies. Violations in individual cases can be the result of an isolated incident or of the behaviour of particular actors, for example, a corrupt judge or an insensi-tive police officer. To credibly identify systemic problems that require attention or reform, it is necessary to monitor groups of cases and identify trends and patterns within them in order to draw overall conclusions on the functioning of the system and formulate recommendations for widespread reforms.
Identifying trends or practices of violations requires monitoring individual cases of a similar type in a comprehensive manner and then comparing them to identify common abuses or problems. In this process, each case needs to be investigated individually through the collection of sufficient and credible information to draw a conclusion as to compliance with international standards and enable comparison. Such monitoring can have an impact on the conduct of indi-vidual cases as the presence and attention of a monitor to an individual case can affect officials’ responses to it. By monitoring for systemic problems the monitoring team can have a far wider impact on the human rights and rule-of-law situation as a whole, as well as having a positive effect on individual cases. In this way, similar violations may be prevented in the future.
To conduct such an analysis, as a general rule, every case must be monitored comprehensively. In cases involving a fair trial analysis, “trial” is defined as the entire legal process from the mo-12 ment of inception of the case through to its final resolution. It is not only the public proceed-ing where evidence is presented. Merely attending an actual courtroom trial without studying the case as a whole rarely results in identifying human rights violations, except when the trial practice is clearly and significantly egregious. This narrow approach also fails to address pretrial rights and may raise questions about the depth of the monitoring enterprise. To obtain effective and credible monitoring results, it is therefore important to monitor a case from inception, for instance, in criminal cases from arrests.
It is critical to understand that this type of legal monitoring involves investigating cases from a substantive as well as a procedural perspective in order to assess their overall fairness. This includes assessing the independence and impartiality of the tribunal, the effectiveness of the defence and the prosecution in the light of the evidence in criminal cases, and the indepen- dence and impartiality of the tribunal and the remedy in civil cases. It means that the monitor-ing team needs to evaluate, to a limited degree, the extent of the evidence, its sufficiency and the investigative or legal options in the case. Such an approach stems from the recognition that substantive monitoring is necessary to investigate effectively cases involving possible failures to prosecute or inadequate prosecutions, law enforcement, prosecutorial or judicial bias, miscon-duct and other subjective factors that can affect the fairness of the proceeding. Such cases are more difficult to monitor and document but they can often be at the root of systemic dysfunc-tion and unfairness in the administration of justice in post-conflict environments. This approach requires a sophisticated legal system monitoring programme with well-trained and highly expe-rienced monitors as their conclusions are, by their nature, more vulnerable to criticism.
13 II. MONITORING CAPACITY A. Mandate Various forms of legal system monitoring, including particular aspects such as fair trial observa-tion, have been undertaken in peacekeeping operations. A monitoring capacity has been lo-cated in different institutions in peacekeeping, for example, in the Organization for Security and Co-operation in Europe (OSCE) as a pillar of the United Nations Interim Administration Mission in Kosovo and in the Judicial System Assessment Programme in the United Nations Mission in Bosnia and Herzegovina, and in a hybrid national and international non-governmental organi-zation, the Judicial System Monitoring Programme, in Timor-Leste. In prior missions, the human rights components of United Nations operations would undertake some legal monitoring as part of their broader human rights monitoring mandate. In recent missions, such as in Haiti and Liberia, however, the legal monitoring team has been specifically designated and located within the peacekeeping operation itself. This recognizes that monitoring and reporting on the legal system in post-conflict environments are essential to support the development of the justice system to protect human rights and promote the rule of law.
The mandate is the starting point for any legal system monitoring programme. Ideally, it should be laid out in the Security Council resolution establishing the peacekeeping operation and be reaffirmed and made more specific in the national legal or regulatory system, through either the promulgation of a new law or through the signing of memorandums of understanding (MOUs) between the monitoring team and the relevant authorities (usually the judiciary, ministry of justice, ministry of the interior, etc.). Unlike the use of letters from ministers or other ad hoc means, this provides monitors with a consistent and reliable basis for asserting their authoriza-tion to monitor.
The monitoring mandate should address a number of critical issues, such as the parameters of the programme in relation to other monitoring teams, its relative independence, its pri-mary objectives and the obligation to issue public reports, the extent of its access and so on. The mandate will naturally shape the development and establishment of the monitoring programme.
14 The most important aspect of the mandate is the nature and the extent of the monitors’ access. Sound monitoring—the ability to collect accurate and extensive information—can require ac-cess to locations (e.g., prisons), people (prisoners), documents (court/police files) and proceed-ings (closed trials involving accused juveniles) which are often not or not completely open to the public. Access should be as unhindered and unrestricted as possible.
Access to investigatory proceedings and related evidence in cases which are otherwise not public can raise particular problems and reasonable exceptions to unlimited access may be expected. For example, agreements on access may stipulate that judges are entitled to deny access to monitors in hearings where one of the parties raises a reasonable objection to their presence or in cases where, during the proceeding, information may be revealed that could jeopardize the security of the investigations, military operations or the life and safety of one of the parties or persons involved in the case. Exceptions to unhindered access should be narrowly tailored and foresee other ways for monitors to gain the relevant information. Decisions on exceptions should provide appropriate notice and reasoning to the monitors. The monitoring team and individual monitors should have corresponding obligations to ensure the security of confidential information.
Unhindered or unrestricted access is particularly important with regard to prisons or any places of detention (official or unofficial), as detainees are at heightened risk of torture, inhuman and degrading treatment. The mandate should stipulate that monitors have access to all parts of places of detention and to all detainees, and that they may speak with prisoners individually and confidentially. The more clearly monitors’ access is laid out, the more authority they will have to reinforce these entitlements when facing opposition. Even where broad access is detailed in binding documents, monitors may still face denials of access or undue restrictions. Such in-stances should be reported to those responsible for the monitoring team, who should address them immediately.
Severely limited access can hinder the monitoring programme’s overall effectiveness and may indicate the need for the peacekeeping operation to take a different programmatic approach, for example to target more resources towards a mentoring programme instead of a monitoring programme, or to develop a more limited legal system observation capacity than a fully fledged monitoring programme.
B. Monitoring qualifications The trustworthiness of the monitoring conclusions is a reflection of the quality of the moni-tors and the programme. Monitors need to have a certain level of legitimacy and credibility in the eyes of those they are monitoring and those responsible for the administration of justice. It is important that local governments (or the responsible authorities) should not be able to disregard the conclusions and findings or put off initiatives to target reform that would ad-15 dress specific problems because of questions concerning the monitoring team. At the same time, it is for the monitoring programme to decide on its staff and to support their professional development.
Sound and effective monitoring of the legal system requires an integrated knowledge, experi-ence and skill set: • Knowledge in a range of technical areas relating to the law, in particular the domestic legal framework, the administration of the justice system, particular procedural and substantive laws, the international legal framework and standards, particularly human rights stan- dards, and the ways in which international standards work in practice; • Practical experience with justice systems, in particular of working within a system at some level and with a dysfunctional justice system, experience working in and sensitivity to post-conflict environments and with traumatized individuals and communities, and with the particular relevant communities, cultures and contexts; • Specific skills for gathering information through effective interviewing, legal analysis, coherent writing skills and effective diplomacy and communication skills to appropri-ately and successfully deal with victims, accused persons and Government or justice officials.
The monitoring programme may have international monitors, national monitors or a combi-nation of both. It is critical to have national lawyers or legal experts within the programme even if there are international monitors too. In some cases where there is a national capacity, international monitors may not be needed at all, such as in the OSCE monitoring programme in Bosnia and Herzegovina. The extent of national involvement in monitoring will depend on issues such as the capacity of the monitoring team, perceptions of objectivity and impar- tiality, safety and guarantees of confidentiality. The approach will depend on the particular circumstances of the host country or territory where the peacekeeping operation is taking place.
If the international monitors do not speak the language, language assistants or interpreters are often a necessary and highly critical part of the monitoring team. Legal language is considered technical in most places and, in post-conflict contexts, usually the province of an elite. Individu-als with particular knowledge of technical legal terminology, such as law students, are often necessary for effective legal monitoring interpretation. For the best monitoring results, other considerations should also be taken into account in the selection of interpreters and can be relevant for international and national monitors as well. These include prior history and associa-tion with armed groups or perpetrators during conflict, levels of prejudice and sensitivity to rural and other lifestyles, and discriminatory attitudes concerning gender. In order to develop trust with a range of victims and accused persons, it is important to ensure a balance of gender and ethnicity (religious or other social group) within the monitoring staff proportionate to that of the local communities.
16 C. The role of monitoring The underlying principle of monitoring the justice system is to reinforce the proper role and re-sponse of the system by encouraging those in it and those in authority to fulfil their obligations. It is not, however, to substitute for it where it is inadequate. The reasoning behind this principle should be apparent. It may be less apparent, however, that this has specific implications for the monitoring team and, in particular, monitors’ conduct. In post-conflict settings, monitors can feel a natural desire to address injustices directly in the face of a severely dysfunctional and unfair system. As a result, monitors may cobble together ad hoc and temporary solutions to individual problems, which serve to reinforce systemic failure. They can seek to make “right” a court case that they believe was decided unjustly, inappropriately interfering with the indepen- dence and impartiality of the tribunal. They can develop a close relationship with a prosecutor and provide legal advice on cases, thus bringing into question the impartiality and objectivity of the monitoring programme. Although monitors may have good intentions, their inappropriate actions can have severe consequences for the monitoring programme, the justice system and the expectations of those seeking redress.
Those responsible for the monitoring team must proactively address these issues through the development of monitoring protocols or codes of conduct during the initial establishment of a monitoring programme. Clear protocols or guidelines will serve not only to protect the integrity of the monitoring process, but in many instances can also help to ensure the safety of monitors and those with whom they have contact.
Circumstances in each individual post-conflict or peacekeeping setting will be different and situ-ations that require a regulatory response from the monitoring programme will evolve. Protocols and codes of conduct should be reviewed and updated to reflect changes on the ground.
Based on prior monitoring experiences, particular issues to address in protocols or codes of conduct include: • Regulating relations with the media, this may involve restricting monitors’ communica-tions with the press to background information and to information that would otherwise be public, such as the status of a case, verdict or evidence or information stemming from an open court proceeding. Leaking confidential or inappropriate information to the media can severely undercut the credibility of the monitoring team and its entitlement to unhin-dered or extensive access.
• Prohibiting the provision of legal advice or guidance to any party to a case, including to officials. Providing advice will jeopardize not only the credibility of the monitoring en-deavour but, more importantly, a party’s chances of success or the integrity of the judicial process. It is particularly important that monitors do not advise the accused. It is common for the police, prosecutors or judges to ask monitors for their opinions or suggestions on particular cases. Monitors should never participate in a case or proceeding by offer-17 ing information, nor should a monitor provide an opinion during the proceeding or on a case or attend a deliberation, even when asked. Any such action brings into question the monitor’s objectivity and neutrality and can be construed as inappropriate interference. Monitors must reinforce that it is the official’s role and responsibility to make decisions based on their professional judgement. When dealing with accused persons, victims or other parties, monitors should always make appropriate referrals and assist them in ac-cessing the services they need.
• Restricting interview formats and guidelines with the accused to facts pertaining to the process, not to the actual circumstances of the case nor with victims/witnesses who may provide evidence in court. This protects monitors from potentially being called as wit-nesses, particularly in systems where hearsay testimony is admitted. Monitors should not ask about the circumstances of, for example, the murder, the identity of the perpetrator or the details of what happened, but about when the arrest took place, what the police stated to the person on arrest, how the suspect was treated upon arrest, where he or she was taken, and similar matters. In some cases it may be necessary for the monitor to dis-cuss the circumstances of the alleged crime, not the specifics of the crime itself, in order to assess the accused person’s justification in defence or whether the crime stems from prior victimization. For example, in cases of women accused of domestic violence, the crime may stem from their own previous victimization from domestic violence or sexual assault. If cases are proceeding through the formal justice system to trial, monitors should gener-ally not speak to official case witnesses and should limit their interaction with victims.
• Developing limits on confidentiality for interviews with the accused and with officials. Generally, confidentiality should be assured for alleged victims of human rights violations. However, there should be clear limitations on this. For example, monitors should inform the accused that they will not keep confidential information about ongoing or planned criminal conduct. Monitors should have an obligation to report knowledge of current or planned criminal conduct. For officials, the general rule is that interviews do not involve any assurances of confidentiality. Exceptions to this rule may be if an official alleges that he is a victim of some misconduct or interference related to his position. For example, a prosecutor may allege that she has received death threats for her work on a case after refusing a bribe. Such allegations should be investigated separately and confidentiality related to those specific allegations may be assured.
• Setting limitations on relationships with justice officials (judges, prosecutors, lawyers, etc.). Monitors will spend a considerable amount of time with certain officials and will need to maintain positive working relationships with them. At the same time, monitors need to reinforce their independence by maintaining a professional distance and ensuring that they do not compromise their position. If a close relationship is developed with an official or group of officials, a monitor’s independence and objectivity are open to question. It is critical for monitors to understand that they are not mentors or advisers.
• Regulating relations, communications and information sharing with disciplinary boards, internal affairs mechanisms or other human rights institutions. Although legal systems 18 monitoring serves as a mechanism to ensure accountability, it is not an accountability or disciplinary mechanism per se. To protect the legitimacy of the monitoring process, par-ticularly with officials, the monitoring team should have a level of independence and in-form officials and others of when and under what circumstances information not available in reports or by other means will be shared with these other accountability mechanisms. Monitors should refer persons with complaints about official misconduct directly to these mechanisms. The response of the authorities to complaints to internal affairs and inspec-tion mechanisms, particularly within the police, should be monitored (as cases) to verify that the investigations and actions taken are adequate.
D. Emergencies, interventions or action in specific cases The monitoring programme will need to have a policy on whether and under what circum-stances monitors will take steps to support a change in the course of a specific case because they are aware of an egregious violation of an individual’s rights. Interventions in judicial cases should always be made with caution, professionalism and due regard for the principles of the independence and integrity of the justice system.
The monitoring programme should develop guidelines on monitors’ conduct in such cases for inclusion in the code of conduct. The guidelines should spell out whether a monitor can take immediate action by complaining to the appropriate authorities at the moment of observing or uncovering an egregious violation. At a minimum, the monitoring team should be in a posi-tion to rapidly document such a violation and ensure that the appropriate authorities are made aware of it and to follow up with them. This can be done through a special action report, writ-ten once all possible and easily accessible evidence of the violation has been verified and sent to the other persons concerned. Such specific interventions should be targeted at the appropriate supervising authorities, for example, the chief judge of the relevant court, a supervising judge at the supreme court or chief prosecutor, not at the trial judges or prosecutors. There should be immediate follow-up with the relevant authorities, the mission’s leadership and others to ensure that the relevant authorities have received and understood the information. Any action that the authorities take should be monitored and documented.
E. Manuals and internal training The realities facing peacekeepers, including the lack of local personnel after a prolonged or devastating conflict, mean that most monitors or staff will not have all the knowledge, experi-ence or skills required for monitoring. Monitoring programmes may develop specific manuals which can cover, among other things, the code of ethics for monitors, the domestic legal institutions and framework, the domestic law and practice, and the monitoring methodology and priorities. The OSCE monitoring programme in Bosnia and Herzegovina has developed such a manual.
19 Monitoring programmes can address such deficiencies and improve capacities within the moni-toring team by developing an internal training or capacity-building plan for all monitors (nation-al and international). The most beneficial training plan should result from a specific assessment of the strengths and weaknesses of the staff and the development of techniques and strategies to address these.
The plan should preferably capitalize on the strengths of the individual monitors and seek to support horizontal and equal knowledge and experience sharing within the group. For ex-ample, national staff may have considerable expertise in the local legal framework, legal and traditional practices, the functioning of the system before or during the conflict, cultural and linguistic norms and suggestions for conduct which will support more effective information gathering. For international staff, training should focus on the national legal system and the culture, sensitivities and priorities of the society. International staff may have more experience and understanding of international laws and standards, with peacekeeping contexts and inter-national organizations and approaches, and the ways that justice systems can function to meet standards of fairness and efficacy.
For certain skills and knowledge it will be necessary to seek external expertise. For example, university professors or supreme court judges may be needed to explain complicated aspects of the law, forensic or investigation experts to explain the process for medical evidence, women’s groups and other service providers to explain the particular ways that trauma manifests itself in certain groups, and international experts to train on particular monitoring techniques, such as interviewing skills or trial observation. The most effective training will promote interactive and participatory learning through the use of audio-visual and other training tools (for example, comparative videos of “fair” and “unfair” trials).
F. Monitoring sustainability and exit strategies The monitoring team should form, along with the appropriate decision makers and groups, a picture of the long-term sustainability of legal monitoring within the national context after the departure or downsizing of any peacekeeping operation. Such a picture is necessary for the monitoring programme to develop an external capacity-building plan and an exit strategy. Legal system monitoring can be undertaken by a range of groups or institutions, non-governmen-tal organizations, human rights institutions or through other quasi-governmental institutional arrangements. One example is the development of a hybrid national and international non-governmental organization for monitoring in Timor-Leste. The most appropriate solution will depend on the particular circumstances, including existing institutions and funding in the host country or territory.
Once a successor has been identified, the peacekeeping operation’s monitoring team should make the development of the professional capacity of that institution, group or programme 20 one of its principal objectives. A gradual handover from international to national monitors within the peacekeeping operation itself is one important technique for developing monitoring capacity and moving towards an exit strategy for international staff. A decision to hand over to national staff should depend on the particular context, including whether national monitors can be perceived to be impartial and credible. This process in and of itself may not ensure the long-term sustainability of legal system monitoring, however.
G. Knowledge and information resources and development A legal monitoring team needs information and a number of knowledge tools to support its ef-fective monitoring, analysis and reporting. All information and documentation can be placed in a monitoring library for reference, while some material will need to be available in the relevant languages for each monitor’s individual use.
Examples of reference material and knowledge tools that should be collected are: • Founding documents of the mission, such as Security Council resolutions, mandates and MOUs • Background information on the administration of the justice system before and during the conflict, which can be found in human rights, media, legal and other reports or docu-ments • International laws in the relevant languages, including non-treaty standards • Compilations of relevant jurisprudence of international and regional human rights courts or committees (note that the United Nations Human Rights Committee is discussing a new general comment on fair trial rights) • International Commission of Jurists materials, Amnesty International fair trials manual and other monitoring assistance tools, interviewing guidelines and best practices7 • The domestic laws, including regulations and administrative laws • Compilation of local court jurisprudence, legal reports, cases or other analyses of the court system in the national context 7 Some human rights monitoring and other documents that the monitoring team should have access to: - Lawyers Committee for Human Rights, What is a fair trial? A basic guide to legal standards and practice (New York, 2000) - International Commission of Jurists, International Commission of Jurists Trial Observation Manual (Châtelaine, 2002) - University of Oslo Norwegian Institute of Human Rights, Manual on human rights monitoring: an introduction for human rights field officers, 2nd ed. (Oslo, 2001) - Office of the United Nations High Commissioner for Human Rights, Training manual on human rights monitoring, Professional training series No. 7 (Geneva, 2001) - Kathryn English and Adam Stapleton, The human rights handbook, a practical guide to monitoring human rights (Colchester, University of Essex, Human Rights Centre, 1995) - Helsinki Foundation for Human Rights, Human Rights Monitoring (Warsaw, 2001) - United Nations Office for Drug Control and Crime Prevention, Guide for policy makers on the implementation of the United Nations Declaration of Basic Principles of Justice for Victims of Crime and Abuse of Power and Handbook on justice for victims on the use and application of the Declaration of Basic Principles of Justice for Victims of Crime and Abuse of Power (Vienna, 1999) 21 • Reports, research or studies on the situation in the country and in particular thematic areas, such as gender-based violence, health, etc.
• Policies, codes of conduct and regulations of the relevant institutions of the justice sector • Media/press reports and articles on the justice system A monitoring capacity also requires the development of knowledge and information resources for monitors on the ground. If these resources do not already exist or are not developed by other components of the peacekeeping operation, they may need to be collected by the moni-toring team itself. These include: • Maps of the country and identification of locations such as detention centres, courts, po-lice stations, universities and other institutions • Information on the available services, shelters, counselling and health and medical facilities • Lists of the human rights organizations, women’s groups and other relevant civil society organizations, their contact information and locations • Organization charts of the police, courts, prosecutors and bar associations and lists of personnel (levels and ranks, competencies, salaries, etc.) • Descriptions of the various institutional authorities of the Government • Contacts of media outlets In post-conflict contexts these resources will require consistent updating and improvement, as new laws and policies are promulgated, new officials and judges are appointed, and new ser- vices and programmes are set up. Although monitors may be expected to know much about the area in which they monitor, they should not be expected to know about all the changes and new developments that take place within the justice and related sectors. Those responsible for the monitoring team need to ensure that, to the extent possible, monitors have the information that they need to monitor the justice system effectively.
22 23 III. THE MONITORING PROGRAMME A. Initial engagement in monitoring The challenges of initiating a legal systems monitoring programme can be overwhelming, particu-larly in post-conflict environments experiencing a total lack of resources, destroyed infrastructure, low capacity, no unified justice system and numerous problems with the applicable law. There is often the simple question of where to begin. It is usually impossible to monitor each case and every court in the justice system, including, where relevant, those being decided through informal justice processes. In peacekeeping operations, the personnel and financial resources needed for a comprehensive monitoring programme are rarely available. In these conditions, it is important not to attempt to monitor a large volume of cases or issues simultaneously because this will result in a superficial and less credible analysis of the problems and human rights issues.
Monitoring programmes should be established, where circumstances allow, during the initial phases of the mission for a number of reasons. Firstly, the monitoring programme is critical to observing and revealing the actual functioning of the courts so that sound and thoughtful deci-sions on resource allocation and reform can be made. Secondly, it is important for officials and others to accept that monitoring takes place and integrate it into their work.
If the justice system is barely functioning when the peacekeeping operation starts, the moni-toring team may want to establish itself by conducting an overall assessment or mapping of the system, identifying the actors engaged, the state of facilities and resources, administrative oversight, the extent of dissemination of the applicable laws to justice officials, and the level of overall functioning. This can accomplish a number of goals simultaneously. It will assist the monitors in developing the relationships and knowledge resources they need to do their work effectively. It can support a positive and cooperative perception about monitoring among the officials and actors in the system. Moreover, the results can provide donors and the operation’s leadership with needed information for identifying initial targeted or quick-impact projects. Before undertaking such an assessment, the monitoring team should ensure that there have not been similar assessments before. Prior assessments should be used and built upon; duplication of effort should be avoided.
24 It is important that, once an initial assessment of administrative and structural issues has been undertaken, these issues continue to be monitored within the context of cases and not sepa-rately, if the justice system is processing cases. By focusing on the conduct of cases, the moni-toring team reinforces the notion that improved resources, technology and infrastructure help the system to be more efficient, effective and fair. This link needs to be reflected in the monitor-ing approach by assessing, through examining the conduct of cases, whether the system has the resources and materials it needs to do the job properly.
Another useful tool in the initial phases is to undertake public opinion surveys to assess what the common perceptions and beliefs are concerning the institutions of justice and the func-tioning of the legal system. Such surveys can also track to what extent the population or certain groups use formal mechanisms or, alternatively, traditional practices and for which disputes and problems. The results of such initial surveys can then provide the baseline for comparison and evaluation as the programme develops and seeks to monitor the effectiveness of reform efforts and changes in public perception, reliance and use of the legal system and the justice institutions. (The use of statistics and the collection of data are discussed in section III.D below.) All justice systems in post-conflict contexts suffer from a number of ills. These are usually poor administration of justice, corruption, executive interference or lack of independence, and/or bias in the judiciary. These types of endemic problems severely jeopardize the legitimacy of the rule of law and can be at the root of institutional dysfunction. If inappropriate legislation or administrative and financial mismanagement are contributing factors, they can be fairly easily tracked, analysed and diagnosed through the monitoring of budgetary allocations, judicial and prosecutorial oversight and appointment mechanisms, and administrative laws and regulations. However, if these problems are endemic on the level of individual practice in particular cases, they can be extremely difficult to monitor systematically.
Monitors may receive individual complaints about such problems in cases which should be investigated and verified. However, it is rare that enough cases of this type will be collected through individual complaints to illustrate that there is a pattern or practice. These issues can be uncovered more systematically through the regular monitoring of priority cases, as discussed below, particularly “regular” crimes. Once identified, incidents of corruption, executive or other improper interference should be investigated in their own right and compared to draw conclu-sions on systemic practice.
B. Selecting and updating priorities for monitoring The monitoring team need not monitor every case or every law in order to draw useful and valid conclusions on systemic human rights and rule-of-law problems. The key is to strategically identify and prioritize the types of cases and legal issues to monitor and then to undertake 25 comprehensive monitoring to uncover exactly what is happening with these cases and with the implementation of the law.
Identifying and prioritizing cases can enhance the monitoring programme in several ways. The process of selecting priorities assists in grouping cases in order to compare, identify and report on systemic issues and patterns or trends in human rights violations. Prioritizing cases can focus monitors who might otherwise be overwhelmed by the number of cases and the particular human rights issues they need to address. Setting clear priorities for the allocation of their time and resources enables monitors to develop more comprehensive and credible informa-tion, enhancing the integrity and value of their conclusions. In general, selecting priorities can ensure more effective monitoring and clearly defined aims in terms of reporting and making recommendations for change.
The selection of priority cases and issues will depend on a number of factors particular to the peacekeeping context, including the caseload and the functioning of the justice system and the resources of the monitoring programme. The establishment of particular transitional justice8 mech-anisms, such as hybrid courts, truth commissions or reparations programmes, may have an impact on monitoring priorities. Non-judicial mechanisms, such as truth commissions, can be monitored for compliance with their statutes and international standards (e.g., due process, witness protec-tion, naming of perpetrators) and reported on. In addition, the prevalence of the use of informal or traditional or customary mechanisms for justice will depend on the peacekeeping context. Where these practices are pervasive, such as the traditional dispute resolution mechanisms, gacacas, in Rwanda, monitoring of them should be factored into the establishment of priorities.
There are a number of factors that should be considered and synthesized to create a list of monitoring priorities. Some serve as indicators of cases where human rights violations are more probable. All are relevant to post-conflict contexts: • Membership of a vulnerable group as victim or perpetrator or party to a case. Vulner-able groups are populations that are particularly susceptible to human rights abuses in all societies, particularly those emerging from conflict. Typically they are: children, ethnic or religious minorities, refugees and internally displaced persons, and persons with mental disabilities. Other vulnerable groups include immigrants, the elderly and persons with a non-traditional sexual orientation. Due to gender inequality women as a group may be considered vulnerable to particular violations based on their sex, such as sexual exploita-8 See the report of the Secretary-General on the rule of law and transitional justice in conflict and post-conflict societies (S/2004/616, para. 8). “Transitional justice” is defined as comprising “the full range of processes and mechanisms associated with a society’s attempts to come to terms with a legacy of large-scale past abuses, in order to ensure accountability, serve justice and achieve reconciliation. These may include both judicial and non-judicial mechanisms, with differing levels of inter-national involvement (or none at all) and individual prosecutions, reparations, truth-seeking, institutional reform, vetting and dismissals, or a combination thereof.” 26 tion or gender-based violence. In addition, women who are members of vulnerable groups experience a heightened (or twofold) risk of human rights violations. Due to their vulner-ability or status, these groups tend to be discriminated against and require enhanced or distinct responses by the authorities under human rights principles. This discrimination often manifests itself in criminal, property and family law cases. For many of these groups, the Government may have specific international human rights obligations.
• Cases that stem from the human rights history of the country. The identification of cases in this category will depend on the particular circumstances of the country’s history of violations of humanitarian law and gross violations of human rights. The clearest cases are those concerning accountability for past violations, such as cases for war crimes, genocide and crimes against humanity, including violations particular to women. Other such cases may be harder to identify and may include current human rights violations by persons believed to be former perpetrators, politically motivated proceedings, property disputes related to the return of refugees and ethnic violence against returning groups. These is-sues can be found in both criminal- and civil-law cases. Naturally, there can be an overlap between these cases and others based on membership of a vulnerable group, such as with ethnic minorities associated with the previous regime.
• Types of crimes/civil cases. To obtain a clear picture of systemic violations in the justice system, it is usually necessary to monitor “regular” cases, that is, those that may not have particular indicators of human rights issues. With this approach, cases involving certain types of crimes with particular significance—such as murder, grave assault or injury, rape, trafficking or other severe criminal conduct—may be designated for monitoring. In the civil-law realm, certain types of cases such as property and landownership cases, divorce and child custody may be designated as priorities. Some of these crime designations will clearly overlap with those connected to vulnerable groups and, in some instances, reflect the human rights history of the country. For example, rape is usually perpetrated against women and often as a tool of war, or trafficking in persons is perpetrated against women and children during and after the conflict.
• Classification by human rights standard. It can be helpful to prioritize cases for monitoring on the basis of a particular human rights standard. This may be a preferred approach dur-ing the initial establishment of the programme and/or the initial functioning of the new justice system. This is especially true in the area of fair pretrial rights, which are often the most pressing issues in the immediate aftermath of conflict and where there are specific guarantees compliance with which is easy to monitor. For example, a number of detain-ees’ cases can be monitored for their compliance with pretrial rights, such as the prohibi-tion of unlawful arrest and detention, the right to be brought promptly before a judge, the right to a defence lawyer and to prepare a defence, and the right to be free from torture, inhuman and degrading treatment.
Priorities for monitoring should be consistently reviewed and updated to ensure that the monitor-ing programme is taking into account changes in the system, developments and actions taken by 27 Government and others, in particular training programmes, and the human rights environment. Monitoring of priority cases does not end with an intervention, the publication of a report or the advocacy of particular recommendations. The effectiveness of the authorities’ actions taken in response to concerns raised by the monitoring team creates new monitoring priorities. Using this approach will ensure that the monitoring team is consistently following up and assessing improvements or changes in the system’s fairness and efficiency based on reform initiatives.
C. Developing and maintaining relationships The success of legal system monitoring depends largely on maintaining regular contacts and relationships with the key actors engaged in and responsible for the system. These relation-ships should be mutually beneficial; for the monitoring team to be able to accomplish its work and for the actors to work towards improving the system’s functioning. Such relationships should be maintained by the monitoring team at all levels. At a central level, these actors are leaders and decision makers (minister of justice, interior, defence), chief justices and adminis-trators of courts, police commissioners and chief prosecutors, large human rights and victim support groups, bar associations and chief medical and related professionals and religious or tribal leaders, where relevant. For monitors in the field, they are police investigators, military or other law enforcement groups or officials, prosecutors, judges, prison governors and, where relevant, community elders or other political actors active within the geographic area covered by the monitoring. They also include defence lawyers, advocacy and victims’ NGOs, community groups and medical professionals. Positive working relationships with these non-governmental actors can be essential for gathering information that officials may be unwilling to provide as well as to gauge civil society’s perception of the justice system.
Official actors, such as the police, prosecutors and judges, are sources of information critical to the monitoring process and are also responsible for compliance with the law and international standards. Similarly, their supervisors or Government authorities are not only responsible for the overall system’s human rights record but also key leaders and allies in efforts to effect reform and change to improve compliance with international standards, including through the promo-tion of fundamental principles such as transparency and accountability. These dual functions can create a certain tension in their relationships with monitoring teams. Monitoring is particu-larly sensitive for officials directly involved in a case who, like witnesses, can provide relevant information, but also usually play a role in the specific violations. Consequently, maintaining positive working relationships with officials at all levels involves a substantial degree of subtlety, diplomacy and good judgement on the part of programme supervisors and monitors.
To maintain these good relationships with all institutions and actors within the system while effec-tively monitoring official conduct, it is critical for the monitoring team to have a clear understand-ing of exactly how the institutions and the system should ideally function and the parameters of each official’s role. The monitoring team needs to address appropriately the common tendency of 28 officials in a dysfunctional post-conflict justice system to exaggerate the limits of their own role and to blame other components of the system. This tendency is usually system-wide; between police and prosecutors, prosecutors and judges, judges and administrators, and so on.
Monitoring should serve to reinforce the integrated functioning of the system as a whole rather than assess specific and targeted culpability. For example, a judge may rightly claim that an inadequate investigation is the fault of the prosecutor requiring monitoring and analysis. At the same time, the monitoring team needs to acknowledge and address the fact that the judge may also be responsible where he or she has the authority to remedy the inadequacy by, for example, returning the case for further investigation or releasing the accused from detention if an investigation is not completed within a certain time limit.
The monitors must also recognize good practices and areas of progress and regularly communicate these to the authorities, officials and other actors. This can serve a number of purposes. It illus-trates that the monitoring programme is impartial and objective and that it exists for the purpose of assistance, not obstruction. Praising good practices reinforces the credibility of the monitoring programme and can make it easier to raise concerns regarding bad practices or regressive actions.
The most important aspect of maintaining positive working relationships with key actors is for the monitors to be consistent and clear with officials and non-governmental groups about their activities, assessments and reports. In particular, officials who are either directly involved in a specific case or who supervise it should be informed, through meetings and similar means, of the monitoring process, any interventions, the timing and purpose of reports, and any other significant action. Supervisors of the monitoring programme should ensure regular meetings with the justice system authorities, chief judges, ministers of justice, police commissioners, chief prosecutors and others to discuss issues raised through the monitoring process and to follow up on recommendations and actions taken in response.
D. Case tracking and formulating indicators Tracking cases, trends or other information statistically can be a key part of a successful moni-toring programme. Such tracking should be used to support credible and sound monitoring and evaluation, not to collect data and information for the system itself. The ministry of justice, the police, detention authorities and others should be responsible for collecting and maintaining data and producing statistics on cases, detainees, crime reporting, etc. Any needs, issues or obstacles that the authorities face in meeting these obligations (for example, capacity-build-ing needs in data and information management) can be identified and changes recommended through effective monitoring.
For the monitoring programme, collecting data and tracking cases can serve a number of im-portant purposes: 29 • Representing the number and kind of cases monitored. This assists in illustrating the number of cases being monitored in the light of the total caseload and the extent to which the concerns identified may represent systemic or structural problems because of their prevalence.
• Supporting assertions of trends and systemic human rights concerns. Collecting and ana-lysing statistics on cases can provide a quantitative picture of the functioning of the justice system. Hard numbers can be used to refute allegations that each case is different or as-sumptions that certain issues are not critical. Statistics can be particularly important for Government institutions and donors that do not like to base allocation of resources on anecdotal evidence and, as such, help to target resources towards the concerns that have been clearly and firmly established through monitoring.
• Evaluating the impact of monitoring and any improvements in the system’s compli-ance over time. From its inception, the monitoring programme should be seeking to evaluate its impact on the system’s compliance with international standards. This can be done through the development and use of indicators. Clusters of indicators, including quantitative ones, such as case tracking data or salaries and resource allocation, and qualitative or subjective indicators should be used to monitor particular sets of results. Accessing a variety of information is often difficult in environments where there are few data, such as in post-conflict situations. If the monitoring team has additional resources and the necessary capacity, it may plan to undertake surveys, facilitate focus groups and use other techniques to assess impact and the functioning of the justice system. These techniques should be used across different target groups, the general public, legal ac-tors and those working in the system, and others, such as parties, defendants, victims and witnesses.
As with case monitoring, it is very difficult to track every case. Therefore, it may be pref-erable to track only cases that lie within the priorities set for monitoring. A case tracking system can take a number of forms. However, at its simplest it can be a database with basic information on the cases of persons detained and being monitored. This information can be collected and updated through the regular monitoring of detention facilities and the files of detainees and accused persons. For example, the information tracked can include the identity of the accused person or party (given either by name or code), date of birth, gen-der, residence, defence counsel, date of alleged offence or claim, type of offence or claim, date of arrest or detention, place of detention, dates of extension of detention, detaining authority, date of indictment or charge, hearings or trial dates, verdict, sentence and appeal outcome.
The data can then be compiled and statistically analysed to reflect a number of human rights issues and indicators. For instance with the above information, reports can address the length of pretrial detention, disproportionately high conviction rates, appropriateness and proportion-ality of sentences, and percentages and trends in groups detained or prosecuted (e.g., numbers 30 of juveniles facing serious charges or women in detention for sex-related offences) or vulner-able groups filing property or other claims, and in offences or claims, such as high numbers in sexual violence or kidnapping prosecutions or high numbers of property cases. Other types of data may be collected through a range of means, including surveys, focus groups, anecdotal research and so on, if the monitoring team has access to the particular expertise and resources needed.
E. Systems for monitoring The initial stages of post-conflict or peacekeeping operations tend to be characterized by cha-otic institutions, debilitated infrastructure and telecommunications, and insecurity. In such an environment, it is important for the monitoring team to establish systems to ensure that it has the basic information needed to do the job. For example, the monitoring team needs to be aware of any court proceedings, of their nature, their dates and times, but often there is no official court scheduling process or none is functioning. To address this, a system needs to be developed which, to the extent possible, is consistently and clearly defined so that the monitor-ing team can learn about proceedings. The type of system developed, e.g., whether a particular official agrees to contact the team or whether the team regularly checks with a certain official, court clerk or other, will depend on the circumstances.
The monitoring team needs to address systematically the identification of cases, visits to prisons or police detention facilities, the process for collecting case tracking information, the scheduling of court proceedings, whether pretrial or trial, and the incidences of informal dispute resolu-tions, where relevant. If the use of informal dispute resolution based on customary or traditional law or practices is prevalent, the monitoring team may need to develop a specific strategy to ensure access to the processes and information, and to be aware of any proceedings or meet-ings in advance. It may be necessary to use only national monitors. If access to these traditional decision-making practices is extremely difficult for any staff of the monitoring programme, the team may need to establish a system for monitoring and reporting with national non-govern-mental organizations.
Visits to prisons or police detention facilities can be critical for investigating compliance with the pretrial rights of detainees. They can also serve as an important source of information on and identification of new cases, particularly those which have not yet proceeded to the court. Much case tracking information can be gathered through regular and systematic visits to detention facilities, particularly in the first phases of establishing a programme. It is important for visits to be regular (weekly, biweekly, etc. depending on the circumstances), but ideally not scheduled in advance with prison authorities. Detention visits with individuals can raise expectations and cause significant confusion and disruption for detainees. Legal system monitoring teams may forgo individual detainee interviews if they can collect the necessary information from other sources, such as human rights monitors, defence lawyers or NGOs.
31 F. Protection of information and information sharing A legal system monitoring team needs to be able to secure its confidential and sensitive infor-mation. This is particularly true if it has broad access to documents and information that are not otherwise public. This requires establishing a system that tracks and secures case files and other information.
The information collected through legal system monitoring, including its reports, are products of the peacekeeping operation. They can be extremely useful to the mission’s leadership, includ-ing the Special Representative of the Secretary-General, and can assist in raising the political profile of efforts in the justice sector and in the promotion of the rule of law. Ensuring that the Special Representative has credible information can be especially important with regard to is-sues and cases that may have particular significance, politically and socially, for the host country or any of its communities’ efforts to support ongoing peace. In addition, the monitoring team’s assessments of the development of the legal system and the promotion of the rule of law may be critical information for the Secretary-General and the Security Council, for instance in the review of the mission’s mandate. The monitoring team should seek to ensure an appropri-ate mechanism to feed its information and analysis to the Special Representative and into the Secretary-General’s reports on the mission to the Security Council. In determining the type of information that is used for these purposes, the supervisors of the monitoring programme need to take into account the impact of any publication on the monitoring team’s access entitlements and on the politicization of the monitoring process.
Furthermore, as part of a United Nations peacekeeping operation, the monitoring programme’s information may be subject to information-sharing arrangements with other international insti-tutions, in particular the International Criminal Court. Based on the principle of complementarity, the Court may need to make specific assessments of the ability or willingness of a national justice system in a country hosting a peacekeeping operation to prosecute persons accused of crimes under international law in a manner consistent with international due process standards. In such circumstances, the information and analysis of the monitoring programme may be particularly useful to the Office of the Court’s Prosecutor. The supervisors of the legal system monitoring programme should know what arrangements the United Nations has made with the Interna-tional Criminal Court and, where indicated by the particular circumstance of the peacekeeping operation, establish contact with an appropriate representative of the Court.
32 33 IV. CASE MONITORING A. Case monitoring strategy There are no rigid rules for monitoring cases. It is not possible to foresee the nuances and vari-ations that characterize each individual case. Cases will vary widely in all respects: in terms of the facts, what stage the case is at, the relevant law, the extent and adequacy of the system’s response and so on. All these factors will affect what sources of information exist and the se-lection of techniques to collect the information. As a result, monitors need to decide in each individual case what the monitoring strategy will be. A case monitoring strategy should include: identifying which sources of information are available, in which order these sources should be tapped, which techniques for information collection to use, what facts to elicit, which proceed-ings should be observed and so on. The formulation of a particular case monitoring strategy will be up to the individual monitors and depend on any constraints on their work. It should be updated and modified throughout the monitoring process, for example, after key interviews, review of the case file and attendance at critical court or other proceedings.
The actual monitoring of cases is highly challenging and involves collecting substantial informa-tion from multiple sources, assessing the credibility of statements and evidence, and integrating and analysing the information obtained. The integrity of a monitoring team is based primarily on the accuracy and veracity of its information and its conclusions. Conclusions about violations of international standards by the authorities, or persons acting on their behalf, that are based on insufficient or partially reliable information can severely undermine the credibility of the monitoring programme itself. The more credible the monitoring conclusions of violations of the law, the more likely it is that they can bring about positive changes.
Collecting information from multiple sources makes for more comprehensive monitoring and serves to verify individual sources. The reliability of the information collected depends on the extent to which it is consistent with material collected from independent or unrelated sources. There are three principal ways to gather the information needed to monitor a legal systems case effectively: interviewing; reviewing case files and collecting other documentation; and attend-ing trials or other proceedings.
34 In some cases, not all these approaches will be available, and only certain techniques can be used to collect information, such as interviewing victims and witnesses and assessing the cred-ibility of the victim’s story through witness testimony. However, if a number of sources are avail-able, it is critical for monitors to gather information from every source. For example, detainees’ claims that they are unlawfully detained should be verified by a review of their court case files. Likewise, claims by officials, such as prosecutors, police and judges, need to be checked and corroborated. In many cases the simplest way to do this is to review the court or police file for documentation of the fact asserted by the official.
To the extent possible, monitors should make every effort to collect all available information from different sources, using a range of techniques for fact-finding, before drawing any conclu-sions as to compliance in a case. Monitors should always be able to identify the source of their information and the process used for verification.
B. Interviewing Collecting information through interviewing is a core monitoring technique. Interviewing vic-tims, including the accused, and witnesses is the most common way to gather first-hand in-formation about a case involving violations of international standards. Interviewing officials directly involved in a case, such as prosecutors, judges, defence lawyers and police officers, is equally important. This can include other actors, such as community elders, in cases that pro-ceed through a traditional process. Interviews of other persons with second-hand information concerning the case, such as victim counsellors or NGOs, can be helpful to assess the credibility of the other information gathered and to develop leads for other interviews.
Monitoring the legal system demands the highest standards of accuracy and credibility of the information relied on. When choosing whom to interview and how to assess the information, legal systems monitors need to recognize that indirect information is less reliable than direct testimony, and be particularly careful of regarding hearsay or indirect information as fact. Wide-spread allegations of violations or causes of system dysfunction based on rumours, hearsay and assumptions can be common in post-conflict contexts. If such information is repeated by a number of unrelated sources, this may lend it more credence. However, indirect testimony should always be confirmed through other fact-finding techniques, such as reviewing case documentation or police investigations, unless this is impossible.
Gathering accurate and credible information depends on the use of sound interviewing tech-niques. Such techniques assist in eliciting the necessary information, internally verifying the interviewee’s statements, corroborating independent information and creating leads to other sources of information. The techniques used for interviewing will depend on the interviewee and the circumstances of the interview. Some common guidelines exist for conducting any general type of interview, but there are special techniques for interviewing victims, witnesses, 35 accused persons, Government officials (including judges, prosecutors and police) and other ex-perts or actors with knowledge of the case (such as medical experts, doctors, defence lawyers or other advocates and NGOs).9 Special considerations apply to interviews with victims, wit-nesses and accused persons who are members of vulnerable groups, such as women victims of sexual violence or exploitation, children or the mentally ill.
The monitoring team may consider developing standardized formats for certain types of inter-views that are undertaken in all cases or for certain issues. For example, formats may be useful for interviews with detainees or for interviews seeking information solely on the lawfulness of arrest and detention. Although formats cannot be strictly followed due to unforeseeable vari-ations in every interview, they can assist in ensuring consistency of the information gathered in similar cases. Such consistency is important for integrating and analysing information to develop conclusions on trends or patterns of practice.
With or without formats, monitors should be trained in good practices for interviewing par-ticularly vulnerable groups in order to ensure the reliability of the information elicited and the integrity of the monitoring process. There are a number of human rights monitoring guides which detail good practices for interviewing.10 As legal system monitoring focuses on the way that the system and the institutions within it work, it requires rigorous fact checking and cross-checking against any files, documentation, evidence or other proof of the system’s or individual actors’ conduct or inaction.
C. Detention visits Visits to prisons or detention facilities can be a crucial aspect of legal system monitoring. Their purpose is to assess whether detainees are unlawfully arrested and detained and whether they have been given access to justice in the ways provided for by domestic and international hu-man rights law. The pretrial rights of accused persons include the right to humane conditions of detention and to be free from torture, inhuman and degrading treatment. Investigating the use of torture or inhuman treatment may be directly related to the detainee’s case, for example, if a confession used in the legal case was a result of such treatment.
However, monitoring conditions of detention and treatment generally is a specialized part of human rights monitoring that is not specifically covered here. Often the International Commit-tee of the Red Cross (ICRC) will undertake visits to assess detention conditions and treatment, as will other human rights monitoring components of the peacekeeping operation. Overlaps in mandate should be addressed by the legal system monitoring team in a cooperative manner 90 See Office of the United Nations High Commissioner for Human Rights, Training manual on human rights monitoring… 10 Ibid.
36 to maximize the use of scarce mission resources and minimize confusion and inconvenience for detainees and detention authorities.
However, in many instances, visiting detention centres to collect information will be necessary. The first step is for monitors to request to review the register of the prison. Registers should include important information such as the date of detention, the detaining authority, the dates of review or extension of detention and any trial or other court hearing dates. In addition, registers should have information regarding medical examinations of the detainees. Monitors will probably want to review the register again after interviewing specific detainees to note the information for their cases.
Monitors should interview a prison authority about the functioning of the prison and the prison’s interaction with the other components of the justice system, such as the police, other detaining authorities, prosecutors, etc. Monitors should include questions on whether judges or other re-sponsible authorities visit the facility and, if so, how often and what the visits consist of. Particular questions should be asked about the frequency of family or other visits, the detainee’s access to the outside world and the provision of medical treatment. Contact information for any doctor or other professional who visits the prison should be noted. Monitors should also ask about access to defence counsel, facilities for defence communication and conduct of counsel. The particular assertions of prison officials should always be checked in interviews with detainees. A general assessment of the number, attitude and conduct of prison guards should be undertaken.
Interviews with detainees should take place only in confidential settings (out of the hearing of, but not necessarily the sight of, prison guards). Monitors should be particularly careful not to place detainees at risk of reprisal from the prison officials for speaking with monitors. Regular monitoring should ensure continual observation of this important issue.
When allegations of torture or other inhuman or degrading treatment while in detention arise, monitors should make sure to ask specific questions about any medical or other treatment provided and about the medical professional involved. Allegations concerning such treatment need immedi-ate verification and, where appropriate, action according to the monitoring programme’s protocols. If allegations concern other victims, immediate access to these individuals should be sought.
Accused persons should be interviewed about their criminal cases to assess the lawfulness of their detention and whether their rights to a fair trial have been respected. Most accused per-sons want to convince monitors that they are innocent and therefore deserve assistance. Moni-tors should explain that all detainees and accused persons are entitled to these rights regardless of the question of their culpability.
Monitors need to ask specific questions about the details of the arrest and police questioning, the detention of the persons and the criminal case against them, as well as the information they 37 have received and the opportunities they have to meet their defence counsel in a confidential setting and/or to develop an adequate defence. These inquiries should include asking detainees to show the monitor any written documentation they have been given. Their description of the events should always be checked with their defence counsel and against the information in the prison register and the court or police file. Questions should also focus on whether the detainee, his or her family or others have been subject to pressure, requests for bribes or “fa-vours” or other inappropriate conduct from any justice officials (including prison guards) or any defence lawyer.
Communicating with the authorities to support the release of an unlawfully detained individual can be an important function. Monitors should therefore take immediate action to corroborate credible and serious allegations of unlawful arrest and detention by reviewing documentation on the case and, if necessary and mandated by the monitoring programme’s protocols, take appropriate steps to raise official attention to the case with a view to supporting the individual’s release.
D. File investigations and collecting documentation There are multiple reasons for reviewing case files of the police, the court and the defence in cases proceeding through the formal justice systems. They include: • Confirming facts and assertions made by interviewees, such as accused persons and jus-tice officials.
• Verifying the system’s compliance with domestic law and international standards, includ-ing the pretrial rights of accused persons in criminal cases (as detailed in article 9 of the International Covenant on Civil and Political Rights and, for juveniles, in article 37 of the Convention on the Rights of the Child).
• Assessing whether the authorities are investigating crimes or processing cases that have been reported or any indications of unequal treatment. (There can be many reasons for a failure to investigate, including inadequate work ethics, corruption, discrimination or inappropriate interference into the process, and all possibilities need to be investigated and analysed.) • Obtaining information from other persons with knowledge of the case who may be in-terviewed.
• Preparing for the observation of hearings or trials.
Files will usually have a wide range of documents, including orders of the court or other of-ficials, official decisions and evidentiary documentation, such as statements or medical reports. Monitors should review files in criminal cases to assess compliance with procedural guarantees for the accused under domestic law, compliance with substantive justice issues such as adequa-cy and legitimacy of investigation by the authorities and timeliness of actions, and compliance with equality of arms and right to defence.
38 Police files or records of others with the authority to arrest should be reviewed for indications of any basis for an arrest, the circumstances of the arrest and any safeguards provided to the arrested individual, or any other investigative actions undertaken at the police stage. In cases where there has been no formal prosecution, police files can be critical. The police file should reveal what information or evidence the police had that could have led to a prosecution and where the case ended. Such information can be essential for following up cases where the authorities failed to adequately investigate and prosecute alleged perpetrators. For example, there is often a trend in failures to investigate and prosecute incidents of domestic violence in post-conflict contexts. Such trends may be monitored by reviewing the files to investigate the complaints made to the police and the police’s investigative actions as well as by interviewing victims, NGOs, witnesses and others, if appropriate.
Depending on the programme’s mandate and access entitlements, monitors may be able to review court or police case files but not make copies. So the monitor will need to take copious notes of their contents, particularly names, contact information and important dates, orders and decisions. These notes or exact copies of case files should then be organized, and any gaps in the information should be filled through interviews with officials, medical experts or other relevant actors.
The collection of other documentation in particular cases is possible, but unusual. Under certain circumstances, monitors may have the opportunity to gather critical documentation or assess the physical scene of an alleged violation, such as a private home or a prison cell where torture is alleged to have taken place, and observe the extent to which the physical surroundings cor-roborate the testimony of the victim or witness. In particular, this may occur if the system is failing to respond to a case. However, as the purpose of legal systems monitoring is to observe the adequacy of the response of the system itself (or the lack of it), the monitoring team needs to be careful not to collect and keep confidential documentation that would otherwise be evidence in a case or to disturb any evidence at a scene that may be the subject of a criminal investigation.
E. Observing trial and investigatory proceedings Where cases are being processed through formal justice institutions, directly observing legal proceedings is essential to drawing conclusions on the legal system’s compliance with national law and international standards.
All proceedings should be monitored to ensure compliance with both the procedural and sub-stantive aspects of justice based on national law and international standards. Criminal proceed-ings can be monitored to assess a number of human rights issues, including most commonly compliance with the national law and fair trial standards as articulated in article 14 of the International Covenant on Civil and Political Rights and, for juveniles, article 40 of the Conven-39 tion on the Rights of the Child. Other concerns may be observed such as a failure to adequately prosecute the accused, indicating a bias and concerns of impunity. These concerns can include inappropriate or unequal treatment of victims and witnesses by the court, particularly in sexual offence cases. In criminal and civil proceedings, one of the fundamental principles is equality before the law and equal treatment. Particularly in cases involving women, proceedings should be assessed for compliance with this principle. In civil cases, the independence and impartiality of the tribunal are equally imperative.
All proceedings are important. Monitors should plan to be present at all of them, including any pretrial hearings and continuing trials. However, in complex cases, there may be several pretrial hearings and it may not be possible to attend them all. Identifying the hearings that may be important, such as those of critical witnesses or the victim, and attending these are judgements that monitors need to make based on the information they have—and should form part of their monitoring strategy. Before attending, monitors should already have obtained significant information about the case from interviews of the parties to the case, the accused and the prosecutor, reviews of the case file and so on.
When attending a proceeding, monitors should sit in a position that reflects their independence and impartiality, that is, a position that clearly demonstrates that they are not a party to the proceeding. If appropriate, monitors may want to have their role explained to all those present, particularly in pretrial hearings which are not open to the public. They should ensure that they are able to hear and see everything.
Monitors should take notes of everything that is said and presented during the proceedings. This practice demonstrates their professionalism and indicates that an independent record of the proceedings is being made. As there may be no formal record of the details of such pro-ceedings in post-conflict justice systems, monitors must ensure that their notes are comprehen-sive and accurate.
Conduct outside of the courtroom, before and after the trial or hearing, can be important in terms of assessing the case as a whole. This conduct includes interactions between prosecutors and witnesses, defence lawyers, police officers, victims and witnesses, accused persons, family members, between judges themselves and between judges and prosecutors. Monitors should be constantly observing such interactions and noting anything that raises concerns.
40 41 V. ANALYSIS, REPORTING AND RECOMMENDATIONS A. The value of reporting Information has limited usefulness unless it is disseminated. In legal system monitoring, facts of case proceedings should be collected, actions of the authorities analysed against the domestic law and then the case as a whole viewed against the relevant international standards. These different aspects of monitoring should all be contained in a report that lays out clearly the basis for the conclusions about any violations of international standards by the justice system or its institutions. Specific recommendations should be made to remedy them and prevent future oc-currences. The value of these reports naturally depends on the integrity and coherence of the information, the accuracy and depth of the analysis, and the relevance and practicality of the recommendations.
Reports should provide a comprehensive picture of the strengths and weaknesses of the func-tioning of the justice system. This means that they should clearly detail the positive steps that the authorities are taking or particular areas where compliance with recognized standards is relatively good. Acknowledging such achievements or positive practices reflects well on the integrity and sincerity of the monitoring team and reinforces the maintenance of the system’s good practices. An example of this is the inclusion, in a report documenting widespread viola-tions of pretrial rights, of the conclusion that detainees have consistent and mostly unfettered access to the outside world and family in line with international standards.
B. External report publication and dissemination Publishing external monitoring reports can be a sensitive process. Special considerations ap-ply in determining whether a report should be made public or remain internal (confidential) in so far as it is distributed only to the Government and other relevant justice actors or authori-ties. The mandate of the monitoring programme should indicate whether it is responsible for publishing public reports. Such an option can provide the team with certain leverage with the authorities and, more importantly, it can be a useful tool for raising public awareness about the justice process and human rights. To appropriately fulfil these functions, public reports need 42 to be formulated carefully and differently than those purely for Government consumption. In particular, they should be drafted with an awareness-raising perspective, providing technical information and analysis in plain language and visual formats which are attractive to laypersons. Any public reports should be sent to the relevant Government institutions for response and comment before publication, and changes should be made if appropriate.
It is important, where appropriate, for external reports (whether public or internal) to be provid-ed to all relevant actors and institutions within the justice system and related to it, for example, the justice ministry, the supreme court, the interior ministry, the defence ministry, the attorney general, the president’s or prime minister’s office—even if the reports do not directly address violations by some of these particular components of the system. It is also important to provide these reports to the donor community. The widespread dissemination of reports can, in itself, reinforce both the interconnected nature of a properly functioning system and the responsibility of all actors within the system to ensure compliance with international standards. The timing of publication or dissemination of any reports needs to be handled carefully and involve a consid-eration of the potential impact on the individual cases mentioned in them should they still be proceeding through the justice system.
A simple procedural mechanism for the dissemination of reports to, and follow-up meetings with, these institutions should be established. Ensuring follow-up on reports, and, in particular, on the implementation of the recommendations with all relevant actors and institutions in the peacekeeping context is critical and can form the basis for the monitoring programme’s overall success.
C. Internal reporting Internal reporting methods within the monitoring programme and peacekeeping operation should generally be developed to ensure that monitors are collecting the same types of infor-mation, verifying that information and using a consistent legal analysis. Standardized require-ments contribute to a commonality of approach, consistency and accuracy of information. Criti-cally, they facilitate the comparison of similar cases to identify trends. Monitors’ internal reports are vital as they usually form the basis of the programme’s external reports.
The monitoring programme should establish a regular internal reporting schedule. It is impor-tant, however, that monitors are required to write only reports that can be used, and not to create a substantial amount of unread and unnecessary documents. Legal monitoring reports should not be summaries of cases, as is often the case in human rights monitoring programmes. Reports on legal monitoring need to be specific in their presentation of the compliance of the justice system with law and standards based on a comprehensive examination of the facts, evidence, proceedings and so on. It is best for monitors to complete their internal reports soon after collecting and verifying the information.
43 For case monitoring, it is recommended that monitors keep complete files containing all the information collected, interviews, case file reviews, observations of proceedings and other documentation or information, and an internal report (or series of reports) compiled on the individual case.
One example of an internal report is an individual case monitoring report which reflects the monitoring of the entire case, entailing an analysis of the treatment of the case as a whole, based on the monitor’s complete investigation. A case report should contain basic elements such as background, facts of the case, case analysis, conclusions, any interventions made and comments or recommendations. The case analysis should indicate the domestic law (standards used or standards missing from domestic law) relevant to the case, apply the facts of the case to domes-tic standards (Did the facts support the use of the law? Was the law ignored or violated? Etc.) leading to conclusions as to whether these constitute a violation of international standards or otherwise a failure to meet standards. Of course, the content of the report will vary considerably and depend on the particulars of the case and also on its nature as a whole (namely, whether it has proceeded through the formal justice system or the informal mechanisms, or whether the justice sector has failed to respond to it). A case monitoring report can be written before the case comes to a final conclusion, for example, to cover violations of pretrial rights up until indict-ment. Once cleared by monitoring supervisors, these reports may be disseminated outside the programme as a part of interventions in individual cases with the authorities, if appropriate.
D. Individual case or issue reports An external individual case or issue report should be compiled in significant cases or on sub-stantial issues and published soon after the conclusion or at some other specific stage (e.g., once the verdict is pronounced, during a parliamentarian review of a new piece of legislation). These reports analyse the authorities’ compliance with international human rights standards in an individual case.
They are especially useful in cases that have a particular significance for society and reflect the challenges of ensuring justice in the transitional period or after the conflict—commonly known as landmark cases. Obvious examples are prosecutions for genocide, war crimes or crimes against humanity or first prosecutions for significant cases of trafficking, incest or prop-erty disputes.
Individual issue reports can highlight de jure violations concerning existing or draft laws. For example, the report may analyse a law going through the parliament for its compliance with international standards and highlight the ways in which it violates human rights norms.
Because of the nature of individual reports, naming names and giving specifics of the case is what lends them weight. Recommendations can be targeted at remedying violations in the 44 particular case or, if appropriate, in similar cases. In issuing reports on draft legislation, the recommendations should lay out the specific ways in which the draft law can be brought into compliance with international standards. Decisions on the timing of these reports are important and should take into account their potential impact on the process in question, particularly on court cases if there may be appellate proceedings.
These reports are similar to an individual case report compiled by the monitor, except that they provide more synthesized information, and more detailed analysis and recommendations. The format of an external individual case or issue report should vary from the internal report by detailing first the international standards and then the domestic law at issue. In case reports, these sections should be followed by the facts and analysis of the particular case and then recommendations.
E. Thematic reports A common type of legal systems monitoring report is a thematic report which centres on a comparison of the conduct of similar cases and draws conclusions on a set of systemic prob-lems or violations (for example, rights in detention, juveniles in serious crime cases, women defendants, the conduct of sexually related cases, the administration of justice, the treatment of property cases concerning returnees). These reports can be formulated at any time when a suitable number of similar cases have been monitored and the violations documented. The-matic reports involve a more sophisticated analysis than individual reports, and they are more difficult to write. Formulating a thematic report entails going through a substantial number of individual cases, each with its own analysis of violations in law and practice, and identifying trends in these cases.
These types of reports do not focus on naming names, but on the general conduct of groups of cases and identified patterns. Accordingly, they can be more appropriate for groups of cases that are highly sensitive or involve considerable security concerns. Although the facts of the cases remain important, pseudonyms should be used for the victims and general titles for the actors, such as police officer, prosecutor, judge, community elder or governor. In most instances, other identifying factors such as the location where certain cases arose should be included to indicate regional or other similarities or distinctions.
These reports can be particularly useful in highlighting both de jure and de facto violations. Some examples of violations that can be covered by thematic reports are: trends in cases where the system fails to respond (e.g., property crimes and civil cases are reported by return-ees but the cases are not investigated, prosecuted or taken through the system); identifiable widespread systemic problems in ensuring certain rights (e.g., lack of defence counsel and no provision for detainees to develop a defence); absence of a law to prosecute for a commonly occurring offence (e.g., the law does not provide for the prosecution of domestic violence); 45 a law exists but is inadequate to address a commonly occurring crime (e.g., the law covers smuggling but not trafficking); a widespread practice of not enforcing the law (e.g., failure of courts to review the basis for pretrial detention according to law); the use of other practices that supplant the use of law in groups of cases (e.g., domestic violence is resolved by informal customary methods rather than through the justice system); widespread unlawful arrests and detention and so on.
Case tracking information can be an important aspect of thematic reports and serves to sup-port conclusions drawn. For example, if the theme is pretrial rights, the statistics on lengths of pretrial detention should support allegations that prolonged pretrial detentions are taking place.
The conclusions of thematic reports are based on the identification of widespread trends, pat-terns and practices. Accordingly, the recommendations in thematic reports should be broader than those in individual reports and include ways to systemically address violations, issues and concerns.
F. Periodic reviews Periodic monitoring reviews of the justice system can be published to cover a particular time period, such as every 6 or 12 months, to provide an overall picture of the functioning of the justice and related systems and track the development of the justice system over time. This type of report can provide a critical tool to highlight progress made or areas of declining compli-ance, the impact of policy, legal or other developments over the time period monitored, and the extent to which recommendations made by the monitoring team in previous reports have been implemented. The use of indicators and statistical information, including quantitative and qualitative data and the results of surveys, in these reports is very important.
These reviews usually cover the work of the monitoring team within that period and include its conclusions on the general compliance of the system with international standards. Their prepa-ration is similar to that of thematic reports, but they take longer due to their comprehensive nature.
Periodic reviews should include: • General developments in the justice system, such as new legislation or policies, newly established courts, newly appointed judges • Case tracking information during the time period, any trends supported by statistics • Cases monitored and violations or issues observed and documented within the period (es-sentially a collection of the substance of thematic reports) • Reviews of any reform initiatives or actions undertaken by authorities or others in response to concerns raised by monitors 46 • Assessments of the impact of those actions, both on individual cases and on systemic problems • Assessment of the progress made by the justice system in complying with international standards using indicators and other benchmarks • Recommendations for further action G. Formulating recommendations Recommendations which lay out ways in which identified problems may be realistically ad-dressed can be the most critical aspect of the legal systems monitoring report. In short, they should be constructive and practical. Ideally, specific recommendations should be formulated for each of the concerns or issues raised in the report. In general, the scope of the recommenda-tions as a whole should be broad enough to address all aspects of ensuring justice in line with international standards. For example, recommendations may be targeted at: • Remedies for egregious violations or unethical conduct in particular cases or groups of cases • Legal reform initiatives that bring domestic law into compliance with international law or create legal remedies to provide a human rights-based response (the formulation of new laws or reform of old laws or internal regulatory and administrative frameworks) • Policy reform initiatives, governmental plans and strategies, the creation of Government structures and positions, task forces and working groups to address areas in need of re-form in the justice sector • Institutional or programmatic development, new courts (e.g., juvenile, family) and alterna-tive dispute mechanisms and institutions (e.g., property and housing directorates), minis-tries and departments, health and medical forensic institutions, social and economic insti-tutions, specialized detention facilities, including juvenile centres, and witness protection facilities or programmes • Mechanisms of accountability, discipline, professional ethics (e.g., internal affairs com-ponents, judicial inspection units, anti-corruption commissions), internal accounting and monitoring mechanisms, bar associations, codes of ethics, standards of practice • Training and awareness-raising, academic curricula development, capacity-building initia-tives, human rights training programmes, public information campaigns, advocacy and media institutions • Services and treatment development (psychosocial, legal, social and medical), funding and support of non-governmental or governmental social and child-protection services and networks, shelters and alternatives to detention • Administrative support systems, structural mechanisms or resource development and al-location to overcome obstacles to human rights compliance (transcription services, renova-tions of prisons, courts, police stations, hiring and training of clerks, security services, etc.) • Legal education and police training, capacity-building and training programmes and in-stitutions for judicial, police and legal actors to enhance knowledge of domestic law and international law 47 • Special or hybrid courts, institutions and mechanisms (tribunals, commissions) that sup-port the justice system, or extend it, and thereby improve compliance Recommendations may be addressed to all sectors of society and all actors, including non-gov-ernmental actors, intergovernmental actors, donor Governments and communities. It is impor-tant that they cover essential independent components of the system that assist in its function-ing, such as the stimulation of an active defence bar, the creation and funding of a wide range of services to victims, and an increase in medical and forensic expertise.
Recommendations should be clear and comprehensible. They should address each authority or actor by name, explain the actions that should be taken and provide a clear time frame for those actions. Due to the holistic nature of justice systems, there should be corresponding recommen-dations for all components so that the overall functioning of the system is improved.
In order to be meaningful, recommendations should reflect the reality of the situation and take into account the practical needs of officials in terms of achieving compliance with recognized standards. Although recommendations on such matters as resource allocation and practical necessities (building reconstruction, computers, case files, etc.) will not directly improve compli-ance with international standards, they are essential for making compliance more likely.
The impact and consequences of recommendations must be thought through clearly and com-pletely, with consideration given to all possible requirements for implementation and the re-sults. This includes realistically assessing the financial and budgetary implications of the recom-mendations, individually and taken as a whole.
Moreover, international standards concerning human rights and the administration of justice are complex, and what may seem to provide a remedy for one situation may well give rise to other concerns. For example, in recommending the creation of a special court, related mat-ters—such as equality before the law, treatment and processes for appeals—must be taken into account. Similarly, in the development of policy recommendations, potential consequences must be addressed. For example, the imposition of stricter policies on borders to control traf-ficking in human beings has shown to potentially result in infringements of women’s rights to travel and freedom of movement. In sum, all possible human rights-related consequences should be anticipated in the development of recommendations.
H. Follow-up and support to other rule-of-law aspects of peacekeeping The release of reports and recommendations requires follow-up consultations with the authori-ties and with other relevant actors, including civil society, the media, educational institutions and donors. This follow-up serves to explain the monitoring process and its findings to relevant actors, including the public if relevant, and to support the implementation of recommendations 48 for reform initiatives. As previously mentioned, the extent to which the report is recognized and the recommendations are implemented should be integrated into the monitoring programme’s priorities moving forward.
The primary responsibility of a legal systems monitoring programme is to monitor, report and make recommendations. The programme is, however, uniquely positioned to support other as-pects of the peacekeeping mission in its joint efforts with the Government to undertake reforms in the area of justice. This can take place through the provision of up-to-date information on the justice system to other components of the operation and its partners, participation in working groups, task forces and other mechanisms of coordination, strategic and action planning and, if appropriate, policy development, support to training efforts, particularly through the provi-sion of concrete case examples and information from which effective and meaningful training materials and approaches can be formulated, and through comments on draft legislation and regulations to ensure conformity with international standards and cohesion with domestic law and practice. Supervisors should plan for the monitors to participate in these activities, where necessary and relevant, and should ensure appropriate time and resource allocation which does not substantially impact on the success of the monitoring programme.
49 Annex SUGGESTED GUIDELINES FOR MONITORING THE JUSTICE SYSTEM IN A PEACEKEEPING OPERATION The following are guidelines for peacekeeping operations to follow in attempting to monitor and report on the local legal system. They are not rigid rules, but rather seek to orient and guide a legal system monitoring team and its interaction with representatives of the legal system and with international donors. Their underlying principle is that the operation’s primary objective regarding legal reform is to reinforce the justice system in its functioning, not to attempt to replace it with the operation’s own actions.
1. A monitoring programme should have representatives at the headquarters and each re-gional or field office. The headquarters should have a programme chief and at least one reporting and analysis officer. Each regional or field office should have designated monitors who are responsible for evaluating the performance of the judicial system in the area cov-ered by the office. The monitors should be lawyers, ideally with experience as practitioners in a judicial system and in working in a post-conflict environment.
2. The chief of the programme and each of the field monitors should meet every local judicial official and legal actor in their area of responsibility and introduce themselves and explain that they will be observing the workings of the justice system with the goal of identifying ways to improve and reinforce the administration of justice. The mandate of the monitoring programme should be clearly explained to all local stakeholders and, based on this, the extent of access to documentation and information should be determined. The tone should not be confrontational or judgemental, but rather one of concern, interest and willingness to help. The monitors should also regularly ascertain the views of local human rights or other organizations about how the justice system functions: its weaknesses and problems or its strengths and successes.
3. The chief of the programme should ensure that all monitors have the basic legal texts and an understanding of the laws and the history of the legal system. The basis for the analysis of the legal system monitoring team should be compliance with the domestic law and its compliance with international standards for the administration of justice. Systems should be established at the headquarters and field levels to support a regular pattern of identifying and tracking cases entering and proceeding through the system, attending court sessions, and talking to judicial officials and local legal actors and organizations.
4. Determining the legality of arrests, the judicial status of prisoners and their access to justice must also be included in the work programme. If the operation contains an inter-50 national police component (United Nations Civilian Police (CIVPOL)) and/or prison officers, then lines of communication, including regular sharing of information, must be established and maintained with them.
5. The monitoring team, after gaining an overview of the system, should select priority types of cases, issues or implementation of legislation for all monitors to follow systemically. The priorities should be based on an analysis of key challenges within the system, particularly with regard to ensuring law and order in the post-conflict environment and the political situation and ramifications of cases for the peace or transitional process. Such cases could involve prosecutions for serious human rights abuses or violations of the laws of war. These priorities should be regularly reviewed and modified to reflect changes on the ground and to move the programme’s analysis to other issues. The monitoring team should not attempt to cover all courts and all cases within the system at one time.
6. Field monitors should be responsible for regularly sending standardized information and analysis to headquarters. Reporting templates can facilitate standardized information collection. The headquarters should be responsible for synthesizing the reports, high-lighting key issues, patterns and trends, and incorporating this into the operation’s regular reporting (daily, weekly or monthly). Once sufficient information is gathered to draw systemic conclusions on certain cases, concerns or themes, the monitoring team should be responsible for publishing thematic or periodic review reports either for judicial and Government and international donor consumption or, if appropriate, for the public at large.
7. The chief of the monitoring programme should have regular meetings with the chief of the judiciary, the minister of justice and other officials in the Government to discuss issues raised by the legal system monitors. The team should seek to provide immediate feedback to the minister and the chief of the judiciary based on information from the field. In the field, monitors should, at appropriate moments depending on the judgement of the chief and the head of the field office, set up meetings with local judicial officials and legal actors to convey the operation’s concerns, questions or suggestions. This should be done in a way that seeks to reinforce the proper application of the law.
8. The monitoring team needs to be in a position to assess for the peacekeeping operation and the international donor community whether ongoing problems in the administration of justice arise principally from a lack of will or from outside interference/intimidation, or from a lack of knowledge or resources necessary for judges and legal actors to fulfil their obligations. This is valuable as various training efforts aim first of all to improve the capacity and performance of judicial officials and legal actors. The monitors will be uniquely placed to evaluate whether the training of judges and prosecutors is taking root or whether adjust-ments need to be made.
51 9. It is important to note any progress in the administration of justice. If the courts are working more efficiently, if judges are truly free to decide cases on the merits, and if the prosecutors and police are following the law on arrest and detention, then the peacekeep-ing operation should make this known. If the system has improved in some aspect, then recognizing this will increase the operation’s credibility and relationships because it cannot be accused of only highlighting problems or criticizing the host Government. More impor-tantly, the operation should seek to reinforce those in the Government, in the judiciary or legal community desiring change and encourage them to take even further measures to improve the administration of justice.
10. Periodic reports on the justice system should reflect the operation’s assessment of the needs of the legal system, personnel and the impact that assistance efforts are hav-ing, positive or negative. Reports should focus on substantive issues and seek to identify the problems that prevent the justice system in each region from functioning properly. Reports should contain specific examples with as much detail as possible to describe the real state of the justice system. Within the framework of the priorities of the monitoring programme, some potential issues to be spotlighted in the periodic or thematic reports are: (i) Lack of trained judicial officials; assess the impact of any training given to judicial officials: performance improved, same, worse. How? Why? Try to identify reasons for any change or failure to change.
(ii) Lack of infrastructure, materials or equipment necessary to perform duties. Are the materials appropriate to the environment, e.g., have computers been sent to a court that does not have electricity? Are there vehicles or are they in need of repair?
(iii) Lack of will to fulfil obligations, conduct investigations, create dossiers, show up for work on time.
(iv) Interference by the military, police, local officials or other Government agents in judicial matters.
(v) Corruption/extortion, bribery or other inducements intended to affect a judicial decision.
(vi) Evidence of bias (ethnic, religious, racial, national or social class) in any judicial deci-sions. Are people from different groups treated differently for the same acts/of-fences?
(vii) Threats against judicial officials, from whom and for what reasons.
(viii) Access by defence lawyers to their clients: immediate, in conditions respecting con-fidentiality.
(ix) Are arrest warrants legally issued and executed? Is it possible to challenge pretrial detention decisions in a court and to have this decision reviewed by a higher court? Are pretrial or pre-charge detention periods longer than the law allows and, if so, why?
(x) Timeliness and conduct of trials and other judicial proceedings.
52 (xi) Number of judges and prosecutors serving, vacancies, transfers and salaries and working conditions and number of available lawyers to defend the indigent (struc-tures for legal aid?).
(xii) Appropriate use of Government resources for the court system and comparative analysis between budgetary allocations to the ministry of justice and the judiciary as opposed to other Government functions.
(xiii) Involvement of and oversight by the authorities, such as the ministry of justice or the chief of the judiciary. Does anyone from the ministry of justice ever come to inspect the courts, prosecutor’s office or prisons? Is there any sign of the inspector general from the justice minister exerting some control or oversight over judicial officers’ performance? Have any judges, prosecutors or lawyers been punished for failing to perform their professional duties?
11. The monitoring team should gather and use statistics whenever possible to illustrate problems, trends or improvements in its reports. Ideally, it should identify specific indica-tors to measure any progress made in the administration of justice. Public surveys can be very useful to evaluate and track the perception of the justice system by the population at large.
12. Reports should combine a diagnosis based on a comparison of a number of similar cases or issues with a proactive, problem-solving approach, and include practicable and appropriate recommendations to address the problems identified. Identifying the root causes of the deficiencies should usually take priority over focusing on incidental mani-festations of malfeasance. The recommendations made in reports require follow-up by the monitoring team. The development of programmes by the peacekeeping opera-tion, the donor community and technical organizations that are feasible and achievable, based on available resources, should also be an outgrowth of solid reporting.
53 54 |
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StatPearls [Internet]. Treasure Island (FL): StatPearls Publishing; 2025 Jan-.
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Treasure Island (FL): StatPearls Publishing; 2025 Jan-.
Oral Contraceptive Pills
Danielle B. Cooper; Preeti Patel.
Author Information and Affiliations
Authors
Danielle B. Cooper1; Preeti Patel.
Affiliations
1 LSU Health Sciences Center
Last Update: February 29, 2024.
Continuing Education Activity
Currently, 3 types of oral contraceptive pills are broadly prescribed: combined estrogen-progesterone, progesterone-only, and continuous or extended-use pills. The birth control pill is the most commonly prescribed form of contraception in the US. Approximately 25% of women aged 15 to 44 who currently use contraception reported using the pill as their method of choice. The most commonly prescribed pill is the combined hormonal pill with estrogen and progesterone. Progesterone is the hormone that prevents pregnancy, and the estrogen component controls menstrual bleeding. Birth control pills are widely used to avoid pregnancy; however, given their side effects in populations at risk, several other methods of contraception are in use as well. Each of these methods is considered based on prior history, medications in use, and concurrent medical diseases such as clotting disorders.
This activity reviews the indications and contraindications, pharmacology, and various formulations of oral contraceptives and highlights the role of the interprofessional team in educating patients about birth control. Given the process of matching patients with the preferred method of contraception, the team consisting of primary care or ob/gyn is likely the most effective in recommending OCPs.
Objectives:
Differentiate between types and formulations of oral contraceptives, considering factors such as hormonal composition, dosage, and administration schedule.
Screen patients for potential contraindications, including medical conditions, medications, and lifestyle factors affecting oral contraceptive safety and efficacy.
Apply strategies to optimize oral contraceptive effectiveness, such as counseling on correct pill-taking techniques and addressing common misconceptions or concerns.
Collaborate with an interprofessional team, such as gynecologists, pharmacists, and nurses, to ensure comprehensive and coordinated patient care using oral contraceptives.
Access free multiple choice questions on this topic.
Indications
The birth control pill is the most commonly prescribed form of contraception in the US. Approximately 25% of women aged 15 to 44 who currently use contraception reported using the pill as their method of choice. Oral contraceptive pills (OCPs) are either combined estrogen-progesterone (also called combined oral contraceptive pill-COC) or progesterone-only pill (POP).
The most commonly prescribed pill is the COC pill. Progesterone is the hormone that prevents pregnancy, and the estrogen component controls menstrual bleeding. Birth control pills are primarily used to avoid pregnancy. The type of medicine used estimates the effectiveness of these oral contraceptive medicines.
Perfect use, in which the method is used consistently and correctly every time, means less than one woman out of 100 will become pregnant in the first year.
Typical use, meaning the method may not always be used consistently or correctly, results in a failure rate of 9 women out of 100 that will become pregnant during the first year of using this method. Unfortunately, due to human error, the quoted failure rate for COC pills is typically 9%.
Most women take OCPs to prevent pregnancy, but 14% use them for non-contraceptive reasons. OCPs can be used to address other health conditions, particularly menstrual-related disorders such as menstrual pain, irregular menstruation, fibroids, endometriosis-related pain, and menstrual-related migraines. The FDA has formally approved combined pills for acne for specific brands.
Strong epidemiologic evidence supports a 50% reduction in the risk of endometrial cancer among women who have used COCs compared with those who have never used COCs. This effect lasts for up to 20 years. COC use decreases the risk of ovarian cancer by 27%; the longer the duration of use, the greater the risk reduction. OCs have also been reported to reduce the risk of colon cancer by 18%. Some formulations have indications for the treatment of acne and hirsutism.
Mechanism of Action
Progesterone is primarily responsible for preventing pregnancy. The primary mechanism of action is the prevention of ovulation; they inhibit follicular development and prevent ovulation. Progestogen-negative feedback works at the hypothalamus to decrease the pulse frequency of the gonadotropin-releasing hormone. This, in turn, will reduce follicle-stimulating hormone (FSH) secretion and decrease luteinizing hormone (LH). If the follicle isn't developing, an increase in the estradiol levels (the follicle makes estradiol) does not occur. The progestogen-negative feedback and lack of estrogen-positive feedback on LH secretion stop the mid-cycle LH surge. Ovulation is prevented when no follicle develops, and no LH surge occurs to release the follicle.
Estrogen has some effect on inhibiting follicular development because of the negative feedback on the anterior pituitary with slowed FSH secretion; it is just not as prominent as progesterone's effect. Another primary mechanism of action is progesterone's ability to inhibit sperm from penetrating through the cervix and upper genital tract by making the cervical mucous unfriendly.
Combined Oral Contraceptive (COC)
The usual estrogen component is combined with a different generation of progestin components with varying degrees of androgenic and progestogenic potential. The combination is prescribed based on desirable effects and risk of adverse events with the progestin component and dose of estrogen and progestin levels.
Estrogen component: Estradiol, ethinylestradiol, estetrol, mestranol
First-generation progestin: Norethindrone acetate, ethynodiol diacetate, lynestrenol, norethynodrel
Second-generation progestin: Levonorgestrel, dl-norgestrel
Third-generation progestin: Norgestimate, gestodene, desogestrel
Unclassified progestin: Drospirenone, cyproterone acetate
Progesterone-Only Pill
While multiple types of progestin pills are available in the US, most frequently, formulations have drospirenone or norethindrone. Drospirenone suppresses ovulation and also has anti-mineralocorticoid activity. While norethindrone primarily acts by thickening cervical mucus to inhibit sperm penetration, suppressing ovulation, and decreasing the mid-cycle LH and FSH peaks, it slows the ovum's movement through fallopian tubes and alters endometrium thickness. Some progestin compounds have more potent antiandrogenic properties and, therefore, are more effective in treating polycystic ovary syndrome, hirsutism, and acne.
Administration
Combined Oral Contraceptive
Choice of COC: The ethinyl estradiol dose is usually less than 50 mcg in this combination of pills. The pills can be either monophasic (same dose of both components in the active pills) or multiphasic (varying doses weekly of both or either component in the active pills). Depending on the withdrawal bleeding desired by the patient and clinically recommended, it can be prescribed as a cyclic (monthly bleeding), extended cyclic (every 3 months bleeding), or continuous dosing(no bleeding).
Cyclic formulations: The cyclic formulations have active hormone pills for 21 to 24 days, followed by 7 to 4 days of hormone-free pills.
Extended cycle formulations: The extended cycle formulations have active hormone pills daily for 3 months, followed by a placebo week.
Continuous use formulation: This can be manipulated by using only active pills from monthly formulations for 1 year, which will functionally stop all menstrual bleeding. The most common complication from the extended cycle is breakthrough bleeding. Any combined oral contraceptive pill formulation can be used in this manner, but typically, monophasic pills are the easiest to manipulate.
Initiation: Combined OCPs are recommended at approximately the same time each day. Avoid taking them greater than 24 hours apart as this could affect efficacy. Two methods of initiating COC for women are recognized per their priority as follows:
First-day start: Pills are started on the first day of menses, and this is considered the best strategy as contraceptive efficacy is faster than other methods.
Quick start: Pills are started on any day medicine is given to women. When patients initiate this method, they are not protected from pregnancy in the first 7 days, and an additional form of birth control is recommended.
Sunday start: Pills are started on the first Sunday after the period begins. When patients initiate the contraceptive pills on the Sunday start method, they are unprotected from pregnancy in the first 7 days. An additional form of birth control is recommended during this period.
Abortion/lost pregnancy:After abortion or losing a pregnancy in the first 2 trimesters, start OC pills within the first 7 days if contraception is desired.
Postpartum: Avoid COC for the first 21 days following postpartum, as this group has a higher risk of venous thromboembolism (VTE). Postpartum women with a higher risk of VTE should not be started on COC for the first 42 days following delivery, irrespective of breastfeeding. Women age 35 years and above, BMI 30 kg/m2, smoking, preeclampsia, immobility, thrombophilia, transfusion at delivery, peripartum cardiomyopathy, history of VTE, postpartum hemorrhage, or postcesarean delivery are at risk of VTE.
Breastfeeding considerations: Avoid COC for the first 42 days following postpartum if a woman is breastfeeding, as hormones impact lactation. Introducing estrogen before 3 weeks of postpartum might increase the risk of thromboembolism in these women. In breastfed infants, cases of breast enlargement are reported mostly with estrogen doses higher than are currently used, which were reversible. WHO recommends performing a risk-benefit assessment for using COC between 6 weeks to 6 months postpartum in breastfeeding women.
Missed doses: If a patient misses a tablet, take the missed tablet as soon as they remember and the next tablet at the usual time (taking two pills in 1 day). If the patient misses 2 tablets in a row in the first or second week, take 2 tablets the day the patient remembers and 2 pills the next day, then resume 1 per day. Use additional forms of contraception until the patient begins a new cycle. Check the package insert for accurate information on managing missed doses.
Emergency contraception: Emergency contraception is recommended in the first week of the cycle if unprotected intercourse occurs and two or more COC pills are missed (except ulipristal acetate).
Progesterone Only Pill
Choice of progesterone pill: The CDC has provided guidelines for users of progesterone-only pills (POPs) who have other medical conditions. Most women can use POP, which appears to be chosen for women who have contraindications to COC or want to avoid the estrogen component in the contraceptive pill.
Norethindrone is commercially available as 0.35 mg tablets, and the dose is lower than that of COC pills. The drug is available in a packet of 28 hormone pills, which should be taken continuously, and no hormone-free pills are in the pack.
Drospirenone is commercially available as 4 mg drospirenone, and the package contains 24 hormone tablets with 4 hormone-free pills.
Initiation: The pill must be taken at the same time each day to maximize contraceptive efficacy. Use backup contraception if the patient starts POPs more than 5 days from the onset of menses. POP can be started on any day of the menstrual cycle, but recommendations exist to begin on the first day of menses. Use a backup contraceptive (eg, condoms) method for the first 48 hours following initiation if POP is started within the first 5 days of menses.
Missed dose: Women who miss taking a norethindrone POP dose by more than 3 hours or had vomiting or severe diarrhea within 3 hours of taking a POP are advised to take the missed pill as soon as they remember and the next tablet at the scheduled time. Additional contraception (eg, condoms) for 48 hours after the late dose is also recommended.
Emergency contraception: Emergency contraception can be offered (except ulipristal acetate) to women who have unprotected intercourse within 48 hours of initiating POP or missed pills where backup contraception or abstinence was advised.
Adverse Effects
Most side effects of OCPs are mild and disappear with continued use or switching to another pill formulation. The most common adverse effect of combined OCPs is breakthrough bleeding. Women will also complain of nausea, headaches, abdominal cramping, breast tenderness, and increased vaginal discharge or decreased libido. Nausea can be avoided by taking the medication at night before sleep. Most other consequences will resolve with time or switching OCP to a different preparation.
Women who have a pre-existing cardiovascular condition or smoke should not use OCs. OC progestogens can impair glucose metabolism in healthy adult women for the first 6 months of therapy. Women with diabetes might need to increase insulin intake to regulate blood glucose levels within the desired range. OCPs can cause hypertension in 4% to 5% of healthy women and exacerbate hypertension in about 9% to 16% of women with pre-existing hypertension.
Four studies on teenage women found a small negative effect of combined OCPs on the acquisition of bone mineral density. In addition, COC use increases the risk of VTE, especially during the first year of initiation. VTE risk increases with high ethinyl estradiol dose and 3rd and 4th generation progestin.
In a meta-analysis, which included 28 publications, reported COC users were at higher risk of ischemic stroke (relative risk 1.7, 95% CI 1.5 to 1.9) and myocardial infarction (relative risk 1.6, 95% CI 1.2 to 2.1) when compared with non-users. The risks did not depend on the type or generation of progestagen. Data analysis showed the risk of ischemic stroke or myocardial infarction increases with higher doses of estrogen. This risk was highest when pills had more than 50 μg of estrogen. Most preparations now contain less than 50 μg of estrogen, making COC substantially safer. COC pill containing 30 μg of estrogen and levonorgestrel is the safest oral form of combined oral contraceptive.
These pills do not protect against any sexually transmitted infections (STIs). Thus, using a condom is highly recommended, especially if a new partner is involved. Regular patient monitoring is necessary to ensure serious side effects are mitigated. In one study, women who used COCs for one year of "continuous use" were monitored for return of fertility, and 97% of women had spontaneous menses within 90 days after discontinuing COCs. Exogenous estrogen components may induce or exacerbate symptoms of angioedema in women with a history of hereditary angioedema. Chloasma may occasionally occur, especially in subjects with a known history of chloasma gravidarum. If women tend to develop chloasma, avoid exposure to ultraviolet radiation and the sun while taking COCs.
POP users have reported acne flare and follicular ovarian cysts. Changes in menses and unscheduled, irregular bleeding are the most commonly reported adverse effects of POPs. Progestogen-only OCs have fewer systemic side effects than combined OCs but often cause menstrual changes. Their long-term effects are not yet known. Both COC and POP have significant drug interactions, so patient medication history should be taken thoroughly before prescribing OC, including supplements. For example, co-administering OC with antiseizure medicines (phenytoin, carbamazepine, oxcarbazepine, lamotrigine, barbiturates) can result in a lower level of OC in blood and reduce the effectiveness of OC.
Contraindications
CDC and WHO have set criteria for women who want to initiate COC or POP. Patients need to be screened for contraindications before starting COC or POP, summarized below, and CDC criteria are listed in The 2016 U.S. Medical Eligibility Criteria for Contraceptive Use.
Combined Oral Contraceptive
A few absolute and relative contraindications of oral contraceptive use are recognized. For example, women with uncontrolled hypertension should not initiate oral contraceptive use until their hypertension is managed; people with diabetes may experience some hyperglycemia when starting OCs, but these are issues that can be addressed.
However, OCPs are contraindicated in smokers (more than 15 cigarettes per day) over age 35 due to a significant risk for cardiovascular events, specifically deep vein thromboembolism. The risk of VTE increases among OC users 3 to 9/10,000 woman-years compared with nonusers who are not pregnant and not taking hormones (1-5/10,000 woman-years). The risk is more significant in those aged over 35 and smoking.
Women with hypertension (systolic BP greater than or equal to 140 mm Hg or diastolic BP ≥90 mmHg as per the CDC), breast cancer, known ischemic heart disease, migraines with auras, endometrial cancer, cirrhosis, hepatocellular adenoma, or malignant hepatoma are contraindicated to use combined hormone contraceptive pill.
Women with a history of VTE, stroke, breast or endometrial cancer, and valvular heart disease (history of subacute bacterial endocarditis, the risk for atrial fibrillation, or pulmonary hypertension) should not use OCPs as these conditions represent unacceptable health risks to the users.
Women who have 2 or more risk factors for cardiovascular diseases, such as older age, diabetes, hypertension, or smoking, should also not use COC as the risk outweighs the benefits.
Women with thrombogenic mutations like prothrombin mutation, factor V Leiden, protein C, protein S, and antithrombin deficiencies should not use COC. Too many alternatives are available for the patient to prevent pregnancy and not increase the patient’s health risk.
Progesterone Only Pill
Women with suspected or known pregnancy
Patient with undiagnosed abnormal uterine bleeding
Patients with breast cancer or suspected
Acute liver disease, benign or malignant liver tumors, or severe cirrhosis
History of bariatric surgeries
Women taking certain antiseizure medications
Monitoring
COC
The patients should be counseled thoroughly on potential adverse reactions before initiating hormone contraceptive pills and informed to report signs and symptoms of serious adverse reactions to achieve better adherence and treatment outcomes. A healthy woman taking COCs should visit her primary care clinician annually for a blood pressure check and routine medical care. Monitor blood pressure in women with well-controlled hypertension being managed medically. Monitor prediabetic and diabetic women periodically, as hormone contraceptives may impair glucose intolerance and are usually dose-dependent.
POP
Prescribers should monitor baseline weight and BMI before initiating the POP.
Drospirenone
Ithas an anti-mineralocorticoid activity that might cause hyperkalemia in high-risk patients (3 mg drospirenone has comparable anti-mineralocorticoid activity as 25 mg dose of spironolactone), so women with diseases that predispose to hyperkalemia or who are taking medications that increase serum potassium concentration should be monitored for serum potassium level in the first treatment cycle.
Medications that may increase serum potassium levels include potassium supplementation, potassium-sparing diuretics, angiotensin-converting enzyme inhibitors, angiotensin-II receptor antagonists, aldosterone antagonists, heparin, and non-steroidal anti-inflammatory drugs. In addition, long-term users of CYP3A4 inhibitors concomitantly administered with OC should be monitored for serum potassium levels. Potent CYP3A4 inhibitors include protease inhibitors (eg, indinavir, boceprevir), azole antifungals (eg, ketoconazole, voriconazole, itraconazole), and clarithromycin.
Toxicity
If a patient takes too many OCPs, the most likely complications are severe headaches, nausea, or vomiting. No antidote is available to treat this condition; the symptoms are treated with antiemetics and analgesics. If the patient has other risk factors significant for increased VTE, one may temporarily consider a prophylactic anticoagulant medication. High doses of estrogen and progesterone are treatment options for menorrhagia that have led to severe or symptomatic anemia.
No reports of severe effects are apparent from an overdose, including ingestion by children. However, overdosage may cause withdrawal bleeding in females and nausea. Drospirenone is a spironolactone analog with anti-mineralocorticoid properties, so monitor serum concentration of potassium and sodium and evidence of metabolic acidosis in overdose cases. Contact the local poison control center for the protocol if an overdose is suspected or confirmed.
Enhancing Healthcare Team Outcomes
Oral contraceptive pills provide patients with the option to prevent pregnancy. If the patient has medical conditions that increase their risk of taking combined OC or progestin OC, the prescriber should inform patients of many alternatives to prevent pregnancy. OCs are a choice made by the patient and her clinician after adequate counseling of risks and benefits. Significant non-contraceptive uses of hormonal contraceptives should be considered when counseling the patient about her options. Many OC formulations can provide menstrual regularity, treating both menorrhagia and dysmenorrhea. They can be utilized to induce amenorrhea for lifestyle considerations. The birth control pill is prescribed by many healthcare professionals, including physicians, advanced practice practitioners, and specialists like obstetricians, internists, and gynecologists.
Patients' health benefits can influence the choice of contraceptive pills, so pharmacists should help prescribers choose affordable options for individual patients. Nurses should educate the patient regarding the potential adverse reactions, monitor the patient's vitals at each visit, and collaborate with the prescribing clinician regarding any concerns. Pharmacists should perform medication reconciliation, counsel patients thoroughly about proper oral contraceptive dosing and use, various start methods, how to address missed doses, and emphasize using a backup contraceptive method for recommended medicine if contraception is desired. The interprofessional team can achieve maximum therapeutic benefit for women who want to use oral contraceptive drugs, leading to optimal outcomes with the fewest adverse events.
Review Questions
Access free multiple choice questions on this topic.
Comment on this article.
References
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: Disclosure: Danielle Cooper declares no relevant financial relationships with ineligible companies.
: Disclosure: Preeti Patel declares no relevant financial relationships with ineligible companies.
Copyright © 2025, StatPearls Publishing LLC.
This book is distributed under the terms of the Creative Commons Attribution-NonCommercial-NoDerivatives 4.0 International (CC BY-NC-ND 4.0) ( ), which permits others to distribute the work, provided that the article is not altered or used commercially. You are not required to obtain permission to distribute this article, provided that you credit the author and journal.
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Cooper DB, Patel P. Oral Contraceptive Pills. [Updated 2024 Feb 29]. In: StatPearls [Internet]. Treasure Island (FL): StatPearls Publishing; 2025 Jan-.
In this Page
Continuing Education Activity
Indications
Mechanism of Action
Administration
Adverse Effects
Contraindications
Monitoring
Toxicity
Enhancing Healthcare Team Outcomes
Review Questions
References
Related information
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PubMed Central citations
PubMed
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Water symmetry operations
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| Figure 2.1. Rotation by 180° is a symmetry operation of the water molecule. | Figure 2.1. Rotation by 180° is a symmetry operation of the water molecule. |
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| Figure 2.3. C2 is a symmetry operation of both water and hydrogen peroxide, but hydrogen peroxide has no plane of mirror symmetry. The orientation of the symmetry axis in H2O2 is perpendicular to the 0-0 bond and bisects the angle between the H-0-0 and O-O-H planes. | Figure 2.3. C2 is a symmetry operation of both water and hydrogen peroxide, but hydrogen peroxide has no plane of mirror symmetry. The orientation of the symmetry axis in H2O2 is perpendicular to the 0-0 bond and bisects the angle between the H-0-0 and O-O-H planes. |
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| Fig. 3.21 Normal modes of vibration of [he water molecule (a) symmetrical stretching mode. A,i (b> bending mode. <4, (c) andsyiranetncal stretching mode. Bt, and their transformations under C symmetry operations. | [Fig. 3.21 Normal modes of vibration of he water molecule (a) symmetrical stretching mode. A,i (b> bending mode. <4, (c) andsyiranetncal stretching mode. Bt, and their transformations under C symmetry operations. |
![Fig. 3.21 Normal modes of vibration of he water molecule (a) symmetrical stretching mode. A,i (b> bending mode. <4, (c) andsyiranetncal stretching mode. Bt, and their transformations under C symmetry operations.
| | |
--- |
| Figure 6-17. The C2v symmetry operations applied to the two hydrogen I s orbitals of water as basis functions. | Figure 6-17. The C2v symmetry operations applied to the two hydrogen I s orbitals of water as basis functions. |
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Lesson Explainer: Direction Angles and Direction Cosines Mathematics • Third Year of Secondary School
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In this explainer, we will learn how to find direction angles and direction cosines for a given vector in space.
We know, in three-dimensional coordinate space, we have the -, -, and -axes. These are perpendicular to one another as seen in the diagram below. The unit vectors , , and act in the -, -, and -directions respectively.
Definition: The Direction Angles
Given a vector , the angles that this vector makes with the -, -, and -axes, respectively, are , , and . These are known as the direction angles and are written .
These direction angles lead us to a definition for the direction cosines. We know, in right-angled trigonometry, the cosine of any angle is equal to the length of the side adjacent to the angle divided by the length of the hypotenuse:
Definition: The Direction Cosines
The direction cosines are the cosines of the three direction angles, , , and :
, , and are the -, -, and -components of vector , and is the magnitude or norm of this vector, where
We can rearrange the three equations such that
In our first example, we will demonstrate how to find a vector given its direction angles and magnitude.
Example 1: Finding a Vector given Its Norm and Direction Angles
Find vector whose norm is 41 and whose direction angles are .
Answer
The norm of a vector is its magnitude or length, and the direction angles , , and are the angles between the vector and the -, -, and -axes respectively.
Using the formulae for the direction cosines, we know that, for a vector ,
Multiplying each of these equations by , the components of are as follows:
Therefore, .
Before looking at our next example, we will consider a formula that links the direction cosines.
Consider the direction cosines as follows:
Squaring both sides of these three equations gives us
Adding these three equations, we get
We also know that
So,
This means that the right-hand side of the equation is equal to 1.
Therefore,
Formula: Property of Direction Cosines for a Three-Dimensional Vector
If , , and are the three direction angles and , , and are their corresponding direction cosines, then
Example 2: Finding the Third Direction Angle of a Vector
Suppose that , , and are the direction angles of a vector. Which of the following, to the nearest hundredth, is ?
Answer
In order to answer this question, we will use the fact that if the three direction angles of a vector are , , and , then
If we let , , and , we have,
Taking the square root of both sides of our equation,
Taking the inverse cosine of both sides,
So, the value of , to the nearest hundredth, is .
In our next example, we will demonstrate how to calculate the direction cosines of a vector.
Example 3: Finding the Direction Cosines of a Vector
Find the direction cosines of the vector .
Answer
Given any vector with components , , and , the direction cosines are where
Substituting , , and , we have
Therefore,
The direction cosines of the vector are .
Example 4: Finding the Direction Angles of a Vector
Find the direction angles of the vector .
Answer
Given any vector with components , , and , the direction cosines are where
Substituting , , and , we have
Therefore, which simplify to
Taking the inverse cosine of both sides of these three equations,
The direction angles of the vector are .
Example 5: Finding the Direction Angles of a Vector
Find the measure of the direction angles of the vector , represented by the given figure, corrected to one decimal place.
Answer
We begin by writing the vector in terms of its three components, where we will define 1 unit to be 1 cm.
In the -direction, we travel 8 cm, so the -component is 8. In the -direction, we travel 19 cm, so the -component is 19. In the -direction, we travel 9 cm, so the -component is 9:
The magnitude of the vector is where , , and are the -, -, and -components of vector :
Given any vector with components , , and and direction angles , , and ,
So,
Therefore, the direction angles of the vector are , , .
We will finish this explainer by recapping some of the key points.
Key Points
The direction angles , , and are the angles between a vector and the -, -, and -axes respectively.
The direction cosines are the cosines of the three direction angles, , , and , such that
This means that , , and are as follows:
The following formula links the three direction cosines:
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6980 | https://www.youtube.com/watch?v=oaK7DkBFDxo | Work Problems in Algebra (Man-Hours)
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This video explains how to easily solve work problems in algebra using man-time analysis.
Problem 1. A job could be done by eleven workers in 15 days. Five workers started the job. They were reinforced with four more workers at the beginning of the 6th day. Find the total number of days it took them to finish the job.
Problem 2. Twelve men can finish the job in 16 days. Five men were working at the start and after 8 days 3 men were added. How many days will it take to finish the job?
Watch the previous lesson - Work Problems (Inlet & Outlet Pipes):
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is work problem so this is the last video for work problems but this time specific to us or man-hour analysis so um from the word itself man days man hours so a number of manpower times the number of hours or the number of days that is your man days or man uh and the number of total working days so kunta akwang ibu tang naman power is i need 4k manpower for men tapos on total working days only proposal is say 20 days so 40 a 4 times 20 that is equal to 80 mondays for example billions sixth seventh eighth ninth tenth day so five days okay so 11th day [Music] okay so solving for x meaning to say class is x is equal to 11.25 days meaning to say class eleven point twenty five days extension extension of one point twenty five days parallel project so meaning the same project by is how many manpower do you need just to complete the 20 days the deadline just to comply the deadline so solving for x answer is x is equal to four point five percent someone on point to work problems nato so let's try to solve this problem uh we have a job can be done it could be done by 11 workers in 15 days five workers started a job they were reinforced with former workers at the beginning of the sixth day find the total number of days it took them to finish the job project that's 165 mondays so five workers started the job for pilaka days a long next statement they were reinforced with former workers at the beginning of the sixth day so what was about for day ones four days one two three four five limas they were reinforced with former workers mandal so five plus four nine and a cella times x will is 15.56 but this is not the final answer find the number of the total find the total number of days it took them to finish the job cunning 15.56 natto days five days to say on total number of days nato nanak has 5 plus 15.56 so the final answer is 20. 56 class oops so we have here 12 men can finish the job in 16 days five men were working at the start and eight days three men were added and after eight days three men were added how many days will it take to finish the job so nakai uh 12 katao initially 12 men times the emotions 16 days you have what um 192 command days for a certain project okay so anasha uh five men were working at the start so five spilaka days since the problem after eight days three men were dead so four days one two three four five six seven eight um five katao ang for days one to eight so eight nintendo last three men were admin dao so eight how many days will it take to finish the job 192 days to finish the 192 mondays requirement okay so solving for x you get a 19 days but this is not the final answer okay how many days will it take to finish the job so one is similar language previous problem so coming 19 days so the total number of days is 8 19 the answer is 27 days |
6981 | https://blog.csdn.net/qq_39499914/article/details/109599638 | 数学基础 - 第九章 不等式和不等式组_第九章不等式与不等式组单元教学计划-CSDN博客
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数学基础 - 第九章 不等式和不等式组
最新推荐文章于 2024-06-24 22:21:36 发布
原创 于 2020-11-10 15:19:23 发布·921 阅读
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本文介绍了不等式的基本概念及解集，并详细探讨了一元一次不等式的解法，包括其性质和如何求解一元一次不等式组。
9.1不等式
9.1.1不等式及其解集
由上可知,在前面问题中,汽车要在12:00之前驶过A地,车速必须大于 75 km/h.
一般地，一个含有未知数的不等式的所有的解，组成这个不等式的 解集(solution set)．求不等式的解集的过程叫做 解不等式.
9.1.2不等式的性质
一般地,不等式有以下性质：
不等式的性质1不等式两边加(或减)同一个数(或式子),不等号的方向不变.
不等式的性质2不等式的性质 2 不等式两边乘(或除以)同一个正数,不等号的方向不变.
不等式的性质3不等式两边乘(或除以)同一个负数,不等号的方向改变.
9.2一元一次不等式
可以发现，上述每个不等式都只含有一个未知数,并且未知数的次数是1.类似于一元一次方程,含有一个未知数,未知数的次数是1的不等式,叫做 一元一次不等式(linear inequality in one unknown).
从上节我们知道,不等式
x-7>26
的解集是
x>33
这个解集是通过“不等式两边都加7，不等号的方向不变”而得到的,事实上,这相当于由x-7>26得x>26+7．这就是说，解不等式时也可以“移项”，即把不等式一边的某项变号后移到另一边,而不改变不等号的方向.
一般地,利用不等式的性质,采取与解一元一次方程相类似的步骤,就可以求出一元一次不等式的解集.
9.2一元一次不等式组
类似于方程组，把这两个不等式合起来，组成 一个一元一次不等式组(system of linear inequalities in one unknown),
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一般地，形如 样式的函数，称为反比例函数，其中x是自变量，y是函数。 反比例函数图象 一次函数图像是一条直线，二次函数图像是一条抛物线，反比例函数图象举例如下： 可发现图象规律如下： 当k>0时，图象位于一、三象限；k<0时，图象位于二、四象限。 当k>0时，在一个象限内，y随x增大而减小。 当k<0时，在一个象限内，y随x增大而增大。 反比例函数的图象由两条曲线 组 成，它是双曲线。 另外一半情况下，反比例函数既关于y=x对称，又关于y=-...
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Optical Windows
Optical windows are transparent components made of materials like glass or crystal that are used to protect and control the flow of light in optical systems. IR windows and optics for use in FTIR Spectroscopy and CO2 lasers. This includes discs, rectangular plates and various other geometries.
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Fused Silica (UV) Brewster Windows
Fused Silica (UV) Brewster Windows
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Brewster windows are optical components designed to minimize reflection of light at a specific angle of incidence, known as Brewster's angle. These windows are crucial in laser systems and optical setups where controlling polarization and reducing unwanted reflections are essential. By allowing light with a specific polarization to pass through while minimizing reflections, Brewster windows play a pivotal role in enhancing optical performance and efficiency.
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| Wavelength range: 190-2,500nm | Coating: Uncoated |
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Polarization Mastery: Unraveling Brewster Windows and Their Optical Significance
In the realm of optics, where precision reigns supreme, Brewster windows emerge as a sophisticated innovation named after the visionary Sir David Brewster. This article embarks on an enlightening exploration of Brewster windows, delving into their crucial role in polarization control, reflection minimization, and their profound impact on optimizing optical systems.
Decoding Brewster Windows
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Polarization Precision
At the core of Brewster windows lies their remarkable ability to exert control over the polarization state of light. When light encounters a Brewster window at Brewster's angle, the reflected light becomes completely polarized, oriented perpendicular to the plane of incidence. This exceptional attribute finds wide-ranging applications in fields such as laser systems, optical communications, and spectroscopy, where meticulous polarization management is crucial for accurate measurements and optimal performance.
The Mathematics of Brewster's Angle
Brewster's angle, denoted as θB, is a fundamental optical angle that dictates the behavior of light when it strikes a dielectric surface. It is intricately tied to the refractive indices of the two media involved – the incident medium (usually air) and the refractive medium (such as glass or other dielectric materials). Mathematically, Brewster's angle is given by the arctangent of the ratio of the refractive indices: θB = arctan(n2 / n1), where n1 is the refractive index of the incident medium and n2 is the refractive index of the refractive medium.
The Brewster Effect: Polarization Control
At Brewster's angle, a remarkable phenomenon occurs: the reflected light becomes perfectly polarized. This means that the electric field vector of the reflected light is parallel to the plane of incidence. This property finds extensive use in controlling the polarization state of light. When light strikes a surface at Brewster's angle, the reflected light is polarized in a specific direction, enabling applications such as glare reduction and efficient optical communication.
Reflection Minimization: Brewster's Window
One of the most significant applications of Brewster's angle is in minimizing reflections. When light impinges on a dielectric surface at Brewster's angle, the intensity of the reflected light becomes zero for light polarized parallel to the plane of incidence. This property is exploited in the design of Brewster windows, which are placed in optical systems to minimize unwanted reflections. Brewster windows find extensive use in laser cavities, interferometers, and other optical setups where reflection reduction is critical for optimal performance.
Limitations and Considerations
While Brewster's angle offers remarkable benefits in polarization control and reflection reduction, it is important to note that it applies only to a specific polarization direction. For unpolarized light or light with different polarization orientations, the angle of incidence may need to be adjusted to achieve similar effects. Additionally, the refractive indices of materials can vary with wavelength, leading to the concept of the Brewster region, where Brewster's angle holds true for a range of wavelengths.
Conclusion
In the intricate realm of optics, Brewster's angle emerges as a mathematical gem with profound practical implications. Its ability to control polarization and minimize reflections has paved the way for innovations in optical technology, from laser systems to communications and beyond. As we navigate the intricate interplay of angles, refractive indices, and polarization states, Brewster's angle stands as a testament to the remarkable synergy between theory and application in the captivating world of optics.
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6983 | https://pmc.ncbi.nlm.nih.gov/articles/PMC9027645/ | Adnate Leaf-Base and the Origin of Ribs in Succulent Stems of Euphorbia L - PMC
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. 2022 Apr 15;11(8):1076. doi: 10.3390/plants11081076
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Adnate Leaf-Base and the Origin of Ribs in Succulent Stems of Euphorbia L.
Gustavo Arévalo-Rodrigues
Gustavo Arévalo-Rodrigues
1 Núcleo de Conservação da Biodiversidade, Instituto de Pesquisas Ambientais-PEFI, Sao Paulo 04301-902, SP, Brazil; arevalogustavo92@gmail.com (G.A.-R.); inescordeiro@sp.gov.br (I.C.)
Find articles by Gustavo Arévalo-Rodrigues
1, Fernanda Hurbath
Fernanda Hurbath
2 Unidade Passos, Universidade do Estado de Minas Gerais, Passos 37900-106, MG, Brazil; fhurbath@gmail.com
Find articles by Fernanda Hurbath
2, Erika Prado
Erika Prado
3 Departamento de Botânica, Instituto de Biociências, Universidade de São Paulo, Sao Paulo 05508-090, SP, Brazil; erikapradomaximo@usp.br (E.P.); isabellagalvao97@gmail.com (I.G.)
Find articles by Erika Prado
3, Isabella Galvão
Isabella Galvão
3 Departamento de Botânica, Instituto de Biociências, Universidade de São Paulo, Sao Paulo 05508-090, SP, Brazil; erikapradomaximo@usp.br (E.P.); isabellagalvao97@gmail.com (I.G.)
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3, Inês Cordeiro
Inês Cordeiro
1 Núcleo de Conservação da Biodiversidade, Instituto de Pesquisas Ambientais-PEFI, Sao Paulo 04301-902, SP, Brazil; arevalogustavo92@gmail.com (G.A.-R.); inescordeiro@sp.gov.br (I.C.)
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1, Diego Demarco
Diego Demarco
3 Departamento de Botânica, Instituto de Biociências, Universidade de São Paulo, Sao Paulo 05508-090, SP, Brazil; erikapradomaximo@usp.br (E.P.); isabellagalvao97@gmail.com (I.G.)
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3,
Editors: Milan S Stankovic, Paula Baptista, Petronia Carillo
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1 Núcleo de Conservação da Biodiversidade, Instituto de Pesquisas Ambientais-PEFI, Sao Paulo 04301-902, SP, Brazil; arevalogustavo92@gmail.com (G.A.-R.); inescordeiro@sp.gov.br (I.C.)
2 Unidade Passos, Universidade do Estado de Minas Gerais, Passos 37900-106, MG, Brazil; fhurbath@gmail.com
3 Departamento de Botânica, Instituto de Biociências, Universidade de São Paulo, Sao Paulo 05508-090, SP, Brazil; erikapradomaximo@usp.br (E.P.); isabellagalvao97@gmail.com (I.G.)
Correspondence: diegodemarco@usp.br
Roles
Milan S Stankovic: Academic Editor
Paula Baptista: Academic Editor
Petronia Carillo: Academic Editor
Received 2022 Mar 5; Accepted 2022 Apr 11; Collection date 2022 Apr.
© 2022 by the authors.
Licensee MDPI, Basel, Switzerland. This article is an open access article distributed under the terms and conditions of the Creative Commons Attribution (CC BY) license (
PMC Copyright notice
PMCID: PMC9027645 PMID: 35448804
Abstract
Stem succulence evolved independently in many plant lineages as an adaptation to arid environments. One of the most interesting cases is the convergence between Cactaceae and Euphorbia, which have anatomical adaptations mostly to increase photosynthetic capability and water storage. Our goal was to describe the shoot development in two succulent species of Euphorbia using light microscopy coupled with high-resolution X-ray-computed tomography. Collateral cortical bundles were observed associated with the stem ribs in both species. The analysis of vasculature demonstrated that these bundles are, in fact, leaf traces that run axially along a portion of the internode. That structural pattern is due to an ontogenetic alteration. During shoot development, the leaf-bases remain adnate to the stem near the SAM, forming an axial component. When the internode elongates, the leaf bundles stretch as cortical bundles. The meristematic activity associated with the bundles forms the stem ribs, as leaf veins near the node, and induce rib formation along the entire internode even in the portion where the leaf traces join the stele. In addition, heterochronic shifts are also involved in the evolution of the shoot system in these Euphorbia, being related to early deciduous reduced leaves and the transference of the main photosynthetic function to the stem. This study demonstrates for the first time the influence of leaf developmental shifts and stem rib formation in Euphorbia and sheds new light on the evolution of stem succulence.
Keywords: plant development, vasculature, leaf traces, structure, microtomography, Euphorbiaceae
1. Introduction
Among the numerous adaptive strategies found in arid environments, stem succulence plays a pivotal role in plant survival. This evolutionary innovation probably occurred in more than 30 families [1,2,3,4,5], as an important convergence which favored the occupation of environments lacking water by distinct angiosperm lineages . This strategy involves organ thickening by means of the expansion of parenchyma (chlorophyll and water or starch storage tissues) in the cortex or pith [6,7,8,9]. A most remarkable convergence case is observed comparing cactiform species of Euphorbia (Euphorbiaceae) and cacti (Cactaceae) [10,11,12,13,14], but it is not restricted to these two families. Stem succulence based on parenchyma expansion has also been reported for several unrelated genera, such as Caralluma, Duvalia, Echidnopsis, Hoodia, Huernia, Larryleachia, Pachypodium, Plumeria, Pseudolithos, and Stapelia in Apocynaceae, Othonna and Senecio in Asteraceae, Tylecodon in Crassulaceae, Pelargonium and Sarcocaulon in Geraniaceae, Dorstenia in Moraceae, and Cissus in Vitaceae [1,2,4,5,9,15,16,17].
Despite the fact that many valuable data on the origin and structure of the succulence are available for leaves of some families [18,19,20] and stems of Cactaceae [9,21,22,23,24], little is known about the succulent stems of Euphorbia (Euphorbiaceae). Even though some xeromorphic characters related to succulence are apparently of the same morphological nature, their development and evolutionary shifts are practically unexplored, particularly in groups with non-succulent ancestors , such as Euphorbia . The need for further studies to determine the origin of succulence in different lineages is reinforced by a study of Mauseth , who reported the lack of many features typically considered xeromorphic in succulent stems of 28 species from seven families other than Cactaceae.
Euphorbia L. is the largest genus of Euphorbiaceae and one of the largest within angiosperms , with about 2000 species occurring worldwide, especially in arid and semi-arid environments in the Tropics . The genus has many life forms, such as herbs, geophytes, trees and shrubs, which encompass the major diversity of succulent species of the genus and are morphologically characterized as xerophytic cactiform species and pencil-stem species . The cactiform species usually have organs modified into thorns and succulent and/or ribbed stems which are photosynthetic, at least in the young parts. The pencil-stem species have similar features, but the stems are thinner and permanently photosynthetic along the entire shoot system . Remarkably, species of Euphorbia with those morphologies usually have highly reduced leaves but some other species retained large foliage leaves, decreasing the dependence on the stem photosynthesis, a trait much rarer in Cactaceae [25,28,29].
The subgenus Euphorbia is recognized by the most intricate evolution of xeromorphic growth form with five independent origins in the genus and a wide diversity of species found in Africa and Madagascar. Fifteen species occur in the Neotropics , whose main monophyletic section is Euphorbia sect. Brasilienses V.W. Steinm. & Dorsey with five succulent pencil-stem species: E. attastoma var. attastoma Rizzini, E. attastoma var. xanthochlora Rizzini, E. holochlorina Rizzini, E. phosphorea Mart., E. sipolisii N.E.Br. and E. tetrangularis Hurbath & Cordeiro. These five species are shrubs, usually ramified, with succulent photosynthetic stems with 4–8 ribs, bearing reduced, early deciduous leaves. They are endemic to Brazil, occurring in environments with sandy soils such as “caatinga” (shrubland vegetation common to the arid climate of northeast Brazil) and rocky uplands .
Succulent stems usually have abundant water-storage tissue and chlorenchyma [6,7,8,32,33,34,35] and may have vascular modifications since the succulence is recurrently associated with anatomical changes in the vascular tissues [19,20,36,37,38,39,40]. Although the xeromorphic features of the cactiform Euphorbia are widely known, anatomical investigations of their stems and early deciduous leaves are lacking.
Among the few anatomical studies performed in the genus, the presence of cortical vascular bundles stands out as one of the striking unexpected features found in some species [9,30,32,41,42], also reported for some Cactaceae [9,43,44,45,46,47]. The main function assigned to these bundles is to improve the transport of photo-assimilates from the chlorenchyma to secondary vascular system and to provide mechanical support to the shoot [30,46,48]. Additionally, Solereder and Metcalfe & Chalk reported a high development of the cortex in succulent stems, mainly due to the activity of marginal meristem. However, its action mechanism remains unclear.
Our study aimed to investigate the development and vascular architecture of pencil stems and reduced leaves in Euphorbia in order to provide a model to understand the origin of some xeromorphic features in succulent stems of the genus.
2. Results
Euphorbia attastoma and E. tetrangularis are candelabriform shrubs with succulent branches of pencil-stem type. The leaves are reduced and restricted to the apex with early abscission (Figure 1A). Both species have ribbed stems, showing six ribs and spiral phyllotaxis in E. attastoma (Figure 1B) and four ribs and alternate, distichous leaves in E. tetrangularis (Figure 1C). The ribs are found only in the internodes, arranged in alternate groups of three projections, which coincide with the base of the leaves, which are simple and sessile.
Figure 1.
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Stem morphology and anatomy of Euphorbia attastoma (A,B,D,E) and E. tetrangularis (C,F–I). Cross sections (D,E,F–I). (A) Stem ribs in frontal view (arrows) with leaves restricted to the shoot apex. (B,C) Note the spiral leaves in (B) and alternate, distichous leaves with stipules in the leaf base (dashed circle) in (C). (D) Detail of a rib showing many layers of chlorenchyma and a papillate epidermis. (E,F) Stem with a prominent vascular bundle in the eustele in each radius of the ribs, numbering six in E. attastoma (E) and four in E. tetrangularis (F). Note the numerous branched laticifers (arrowhead in (F)). (G) Detail of a vascular bundle of the eustele. (H) Median portion of the leaf with chlorenchyma and vascular bundles. (I) Detail of the leaf axil with colleters (asterisk) and one of the glomeriform stipules (arrow).
2.1. Anatomy
The stem of both species has a uniseriate papillate epidermis with stomata and a cortex particularly thicker in the ribbed region due to the presence of dozens of layers of parenchyma (Figure 1D). The cortex is divided in two regions: an outer cortex formed by chlorenchyma with elongate cells near the epidermis and an inner cortex containing several layers of parenchyma with starch grains. The vascular system is arranged in an eustele of irregular shape (polygonal), whose angles vary according to the number of ribs (Figure 1E,F). Each angle has a large bundle, which is considerably larger than the other bundles (Figure 1G). In addition to the stele, vascular bundles are observed in the cortex, opposite to the ribs, whose number varies according to the number of ribs (Figure 2A–F). Branched laticifers occur throughout parenchyma and vascular bundles (Figure 1F).
Figure 2.
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Ontogeny and vasculature of the shoot in Euphorbia attastoma (A,C,D,F) and E. tetrangularis (B,E,G). (A) Origin of the leaf primordium (LP) in the flank of the shoot apical meristem (SAM). (B) Leaves from distinct nodes in different developmental stages. Note the parenchymatic expansion of the midrib of the leaf (dashed square). (C) Nodal region showing three leaf traces associated with their respective ribs and the leaf gap (arrowhead). (D) Reduced leaf with three bundles and the ribbed stem. (E) Detail of the anticlinal and periclinal division of the ground meristem cells during rib development. (F) Divergence of leaf traces from the eustele, evidencing their relationship with the stem ribs (arrows). Note the profuse multiplication of parenchyma in front of each trace, forming the ribs (dashed squares). (G) Leaf terminal venation with areoles composed of tracheids (dashed square). VB = vascular bundle.
The leaves are relatively small, varying from 6.7–11 × 2.2–3 mm and have a uniseriate epidermis with stomata and chlorenchyma (Figure 1H). They have three vascular bundles in the base, which subsequently branch in five to nine in the median portion (Figure 1H). Laterally to the base of the leaf blade, a pair of glanduliform stipules with papillate epidermis are formed together with a profusion of sessile colleters, which extend from one stipule to the other along the leaf axil (Figure 2B). These colleters are formed by a secretory palisade epidermis and a non-secretory parenchyma core (Figure 1I).
2.2. Shoot Ontogeny and Vasculature
The leaf primordium originates from the peripheral zone in the flank of shoot apical meristem (SAM). The initiation occurs in regular plastochrons, which vary according to the phyllotaxis (Figure 2A). During the expansion of the leaf primordium, its base remains united to the stem primordium, becoming part of the axis. The subsequent intercalary growth of the stem primordium elongates this adnate region, and the leaf-base is stretched along a portion of the internode, where the leaf traces are observed as stem collateral cortical bundles.
In this second stage of the shoot morphogenesis, the ground meristem in front of the collateral cortical bundles (leaf bundles) generates numerous layers of parenchyma (Figure 2E), which correspond to leaf veins. Ontogenetically, these veins are the ribs of the stem, formed by the cortical bundle and a large amount of parenchyma, which doubles the thickness of the cortex. In regions without ribs, the cortex of both species is 1.5–2 mm thick, but in the rib radius the cortex is about 4–4.5 mm. This relationship of the leaf base as an axial component can also be perceived through the analysis of vasculature (Figure 2 and Figure 3), but it is not restricted to the portion of leaf-base adnation. In fact, since the leaf-base becomes an axial component in the shoot apex, its incorporated vasculature induces the proliferation of the ground meristem basipetally, both in the region with cortical bundles and in the region without bundles, which, actually, corresponds to the most length of the internode.
Figure 3.
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Microtomography with 3D-reconstruction of the shoot of Euphorbia attastoma. (A) General view of the shoot. (B,C) Reconstructions of the area enclosed by the dashed lines in (A). (B) Detail of the leaf traces (LT in red) connected to the eustele (Eu in blue) surrounding the medullary parenchyma (P in gray). Note that the three leaf traces do not branch until reach the free portion of the leaf (L). (C) Nodal region showing the leaf traces (LT in red).
Leaf morphogenesis is rapid. The free portion of the leaf, found around 1.5–2 cm above the node, is comprised exclusively of the leaf blade. Early in leaf development, meristems differentiate into mature tissues, stopping the leaf expansion. As a consequence, the leaves remain small and are also early deciduous, being restricted to the apex portion of the shoot. Thus, the photosynthesis of the plant is supplied almost entirely by the stem with adnate leaf-bases.
Shoot vasculature varies according to evolutionary developmental shifts. In the nodal region, three leaf traces diverge from the stele and occupy a peripheral position in the cortex in the region of the stem rib, running parallel to the surface along the internode, profoundly altering the stem morphology. These traces remain unbranched until the base of the free portion of the leaf (Figure 3, Supplementary Material Video S1). In the leaf blade, the three bundles ramify, forming secondary and tertiary veins, which terminate with tracheids in the areoles (Figure 2G).
3. Discussion
The present study revealed the origin and nature of the ribs of the succulent stems in two Neotropical species of Euphorbia for the first time. Using developmental analysis and 3D-reconstruction techniques, we were able to demonstrate that the leaf-base is congenitally adnate to the stem, as an axial component, and the development of the three main veins of this base induces rib formation along the entire internode of the succulent pencil stems.
Succulent stems are common in Euphorbia, exhibiting ribs with reduced, early deciduous leaves. Rudimentary leaves occur in other New World lineages with ribbed stems of the subgenus Euphorbia, such as in E. pteroneura A. Berger (sect. Euphorbiastrum (Klotzsch & Garcke) Boiss.), a Mexican species (sensu Dorsey et al. ). Ribbed stems also occur in herbs, particularly in E. sect. Stachydium Boiss., the sister group of E. sect. Brasilienses, e.g., E. heterodoxa Müll. Arg., an endemic species from rocky outcrops in Northeastern of Brazil. Other succulent lineages within the genus also have rudimentary leaves with succulent ribbed and/or tuberculate stems. Those species belong to the Old World and are currently classified in the subgenus Euphorbia and E. subg. Athymalus Neck. ex Rchb. [27,49]. Given the diversity of species bearing ribbed stems in unrelated lineages, the main common trait is the presence of rudimentary leaves. This fact is certainly involved in the stem organogenesis in Euphorbia since the stem ribs are concomitantly produced with the leaf primordium, exhibiting an easily recognizable relationship between both organs.
Restriction of leaf blade development and transference of function to stem, which becomes the main photosynthetic organ of the plant, is usually related to succulent stems in arid environments [9,48,50,51,52,53]. The enlargement of the cortex of these stems appears to be related to the expansion of the photosynthetic tissue, in addition to the increment in water and starch storage [9,48,51,54,55,56,57,58].
The cortex may be extremely broad in Cactaceae due to cortical bundles which produce intrafascicular secondary phloem and xylem in many species [9,48,50]. On the other hand, we did not observe any expansion of the stem tissues in Euphorbia due to vascular proliferation of the cortical bundles. Actually, the non-separation of the leaf-base from the SAM flank during the leaf primordium development seems to be the reason for the succulence of the pencil stem in the genus. The maintenance of the leaf-base forming an axial component has already been reported for plants of other families .
Fusion of the leaf adaxial side to the stem is not an uncommon process for bud protection in xerophytes since it reduces the water loss to the environment. This process usually involves expansion of the leaf-base. In succulent species of Aizoaceae and Amaranthaceae, a large expansion of the leaf-base around the SAM and a reduction in the leaf blade have previously been reported [18,20,60]. This foliar characteristic appears to have evolved multiple times in these two families, coinciding with the aridification of southern Africa in the Late Miocene [60,61,62]. However, there is no fusion of the leaf-base to the stem, as observed in Euphorbia. Conversely, leaf fusion may be total in other lineages. In Podostemaceae, the leaf-base is completely united to the promeristem, originating shoots apparently devoid of SAM in the subfamily Podostemoideae .
In general, main xeromorphic features of the succulent stems are the increase in parenchyma tissue of the cortex or pith, reduction of leaf size and number, and establishment of mechanisms for water protection and its storage . Regardless of the origin of expanded cortical tissues of the succulent stems, the enlargement is associated with the presence of cortical bundles. The emergence of cortical bundles was a key innovation in the evolution of Cactoideae (Cactaceae), which allow some species to develop cortices up to 30 cm thick keeping all their tissues hydrated [48,64]. If the cortex is unvascularized, the slow transport of water by diffusion appears to limit the increase in cortex thickness [55,64].
Extrastelar vascular bundles, such as the collateral cortical bundles (leaf traces) observed in the Neotropical Euphorbia sect. Brasilienses, have also been reported for E. weberbaueri Mansf. (E. sect. Euphorbiastrum) (Klotzsch & Garcke) Boiss. , another Neotropical species with ribs very similar to those described herein, and also for African species as Euphorbia fortuita A.C. White, R.A. Dyer & B. Sloane, Euphorbia horrida Boiss., Euphorbia obesa Hook. F. and Euphorbia officinarum L. . Those additional bundles have been reported for 55 eudicot families, extending along the internodes, as cortical bundles or medullary bundles . In the case of cortical bundles, Howard observed that their relationship with the leaves varies depending on the group analyzed and can be found in the stem (1) unrelated to the leaf vasculature, (2) partially related to leaf vasculature or (3) completely related to leaf vasculature. This third condition is the one discovered here in Euphorbia attastoma and E. tetrangularis. This variation indicates that the mere occurrence of cortical bundles in the stem does not necessarily imply adnation of the leaf to the stem. The main example is the cacti, whose extensive set of cortical bundles, observed in addition to the leaf/bud traces, are truly cauline, derived from procambial cells originated in the cortical area near the shoot apical meristem [44,48,64].
Despite the fact that the origin of ribbed stems has not been investigated in other species of Euphorbia, further anatomical studies may reveal similarities in relation to our results, indicating the likely evolution of this character in other clades of the genus. Mauseth found that the leaf traces are related to the stem ribs in nine species of Euphorbia. If this relationship exists between this type of stem succulence and leaf traces in other species, we can assume that the number of ribs and their disposition depends on the phyllotaxis and plastochron. This hypothesis is supported by the difference observed between E. attastoma (six ribs) and E. tetrangularis (four ribs). This is related to the three leaf traces that diverge from the eustele in different times during development, indicating a case of heterochrony in closely phylogenetically related species. The ribs are surely associated with the stem thickness, allowing the candelabriform morphology observed in E. sect. Brasilienses and seems to be important to accommodate seasonal expansion and contraction of stem .
The evolutionarily shift of plants from one form into another involves modifications of the developmental patterns . In this study, we verified that during the initial development of leaf primordia, the leaf-base remains adnate to the developing stem in Euphorbia. This congenital fusion of tissues displaces the orientation of the leaf traces, which are oblique/horizontal and run directly to the leaf-base at the nodes in the vast majority of angiosperms , to run roughly parallel to the surface of the stem along a short portion of the internode (Figure 4).
Figure 4.
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Diagrams showing the leaf traces diverging from stele in Euphorbia attastoma (A) and E. tetrangularis (B). Note that the leaf traces (red) diverge in groups of three in both species.
Fusion of the leaf-base to the stem is not an uncommon event in angiosperms , but the developmental process observed in Euphorbia appears to be an evolutionary novelty and is described for the first time in this study. Despite the fact that the general leaf development is well known and widely analyzed in textbooks and articles, the origin of the stem is often neglected. After the formation of the leaf primordia in the shoot apex and the establishment of the nodal regions, there is an elongation of the internode by multiplication and elongation of the cells of the primary meristems . This second stage of the stem morphogenesis has been altered in the Euphorbia pencil stem. When the internode elongates, the leaf-base united to the stem also elongates (Figure 5), stretching the leaf traces along the internode. Consequently, the leaf is “displaced” to another region above its point of origin (Figure 4).
Figure 5.
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Schematic drawings of two stages of shoot development. (A) Regular shoot development. Note that the leaf-base participation in the axis is restricted to its point of origin at the node, even after the internode elongation. (B) Shoot of Euphorbia attastoma and E. tetrangularis. Note that adnate leaf-base forms an axial component, displacing the leaf to a region above the node. The fused portion elongates along with the internode. The origins of axial tissues were colored in one plastochron. Dashed line: node; blue: leaf-base component; yellow: cauline component; green: indiscriminant components.
The increment of foliar tissues into the stem considerably expanded the cortex, increasing its photosynthetic capacity (Figure 5). Additionally, there is an early maturation of the leaf meristems, interrupting the leaf blade development and producing very reduced leaves. This type of developmental shift of the tissue differentiation timing is known as heterochrony and has been shown to be one of the main processes in plant evolution . In addition, some heterochronic changes may lead to transfer of function (heterotopy) due to spatial relationships that change over time . This can be observed in the cactiform species of Euphorbia, whose stem becomes the main photosynthetic organ of the plant due to the early leaf abscission.
Despite our observation that cortical bundles (leaf traces) elongate for a short region of the stem above the node, these vascular bundles undoubtedly stimulate parenchyma proliferation, forming the ribs continuously along the entire internode. To date, it is not possible to infer which endogenous signal is responsible for this change in stem tissues. However, previous studies show that the hormonal stimulus responsible for the differentiation of procambium, and consequently the vascular bundles in shoots, is the auxin. Polar auxin gradients induce procambium differentiation along the path of its flow and may regulate vascular adaptation to the plant’s environment [70,71]. Thus, if auxin is related to the differentiation of cortical bundles, we can hypothesize that auxin is also related to rib differentiation regardless of the presence of cortical bundles from one node to the other, deviating from the expected transport pathway due to the adnation of the leaf base to the stem. Further immunocytochemical studies are needed to definitively verify this hypothesis and clarify the physiological regulation of succulence evolution in Euphorbia.
4. Materials and Methods
For this study two species of E. sect. Brasilienses were selected: E. attastoma (Hurbath 853, 854) and E. tetrangularis (Hurbath 844). The samples were collected from three individuals of each species cultivated in glasshouse at Instituto de Biociências at the Universidade de São Paulo. Vouchers of the species were deposited at SP Herbarium.
Shoots were fixed in FAA for 24 h (formalin, acetic acid, 50% ethanol 1:1:18 v:v) and then stored in 70% ethanol. Entire leaves were first cleared using 100% ethanol, then treated with 10% sodium hydroxide for 2 h, followed by 5% sodium hypochlorite. The leaves were stained with 1% safranin and mounted in Kaiser’s glycerin gelatin . For anatomical analyses, shoot apices were isolated, dehydrated in a butyl series , embedded in Paraplast® (Leica Microsystems, Wetzlar, Germany) and transversely or longitudinally sectioned using a Leica RM2145 rotary microtome. Sections 12 µm thick were stained with astra blue and safranin , and the slides were mounted in Permount® (Fisher Scientific, Pittsburgh, PA, USA). The photomicrographs were taken using a Leica DMLB light microscope coupled with a digital camera.
For high-resolution X-ray-computed tomography (HRXCT), we used the fixed shoots of E. attastoma, which were treated with phosphotungstic acid in 70% ethanol for one week. Subsequently, the samples were dehydrated in an ascending ethyl series containing 1% phosphotungstic acid (1:1; v:v) and wrapped with parafilm in a tube filled with 100% ethanol. Finally, the samples were scanned using a SkyScan 1176 microtomograph (Bruker, Billerica, MA, USA). The exposure time was approximately 4 h per sample and for the 3D reconstruction we used the software CTVox and CTVol, 3D Doctor (Able Software Corp., Lexington, KY, USA) and 3D Slicer. Image sequences and the video (Supplementary Material) were segmented automatically and manually.
The vasculature diagrams were made based on the microscopic observations and the high-resolution X-ray-computed tomography, using the software Cinema 4D (Friedrichsdorf, Hesse, Germany) and Adobe Photoshop (San Jose, CA, USA).
5. Conclusions
We demonstrate for the first time that the succulence of the pencil stem in Euphorbia is due to the incorporation of the leaf-base as an axial component during shoot development. This joining of tissues doubles the thickness of the cortex of stem and its amount of photosynthetic tissue. The elongation of the internode joined to the leaf-base transforms the leaf traces into cortical bundles that are associated with proliferation of the parenchyma, forming the stem ribs. A likely hormonal signal from the leaf is transported basipetally along the entire internode, inducing rib formation regardless of the presence of cortical bundles. Heterochronic shifts are also involved in the evolution of the very small leaves of Euphorbia, their early abscission and the transference of the main photosynthetic function to the stem. A major sampling is needed to evaluate the specific role of these features in the adaptive success of the Euphorbia and its diversification in arid environments.
Acknowledgments
This study is part of the PhD dissertation of the first author in the Programa de Pós-Graduação em Biodiversidade Vegetal e Meio Ambiente of the Instituto de Pesquisas Ambientais-PEFI, São Paulo, Brazil. The authors thank CNPq for financial support and Instituto de Biociências at Universidade de São Paulo and Instituto de Botânica, where this study took place.
Supplementary Materials
The following supporting information can be downloaded at: Video S1: Microtomographic reconstruction of the shoot of Euphorbia attastoma.
Click here for additional data file. (225.9KB, zip)
Author Contributions
Conceptualization: D.D.; data curation: G.A.-R., E.P., I.G. and D.D.; funding acquisition: G.A.-R. and I.C.; investigation: G.A.-R., F.H., E.P., I.G., I.C. and D.D.; methodology: G.A.-R., E.P., I.G. and D.D.; resources: G.A.-R., I.C. and D.D.; supervision: I.C. and D.D.; validation: G.A.-R., E.P., I.G., I.C. and D.D.; visualization: G.A.-R., E.P., I.G. and D.D.; writing—original draft: G.A.-R., I.C. and D.D.; writing—review and editing: G.A.-R., F.H., E.P., I.G., I.C. and D.D. All authors have read and agreed to the published version of the manuscript.
Funding
This research was funded by Conselho Nacional de Desenvolvimento Científico e Tecnológico (CNPq proc. #169737/2018-7 and #311275/2019-2) and Fundação de Apoio à Pesquisa do Estado de São Paulo (FAPESP proc. 11812-1/2015).
Institutional Review Board Statement
Not applicable.
Informed Consent Statement
Not applicable.
Data Availability Statement
All figures and tables of this manuscript have been unpublished and were made specifically for this article.
Conflicts of Interest
The authors declare no conflict of interest.
Footnotes
Publisher’s Note: MDPI stays neutral with regard to jurisdictional claims in published maps and institutional affiliations.
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6984 | https://www.sciencedirect.com/science/article/pii/S0012821X24005181 | Earth's precession rate evolution and rapid fall during the Late Proterozoic - ScienceDirect
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Abstract
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1. Introduction
2. Robust structures in k(t)
3. Tidal evolution of the Sun-Earth-Moon system
4. Mass redistribution
5. Solar system wide disruption
6. Atmospheric tides
7. Conclusions
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Appendix A. Earth-moon system Modelling equations
Appendix B. Hydrostatic equilibrium and the impact of density profile changes
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Earth and Planetary Science Letters
Volume 648, 15 December 2024, 119086
Earth's precession rate evolution and rapid fall during the Late Proterozoic
Author links open overlay panel David Waltham a, Mattias Green b
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•We develop techniques for estimating past Earth-Moon system parameters (from the observed Earth-axis precession history) for the general case of non-coplanar, non-circular orbits including the effects of both lunar and solar tides.
•We also develop techniques for evaluating alternatives to tidal drag as the cause of precession rate change (specifically mass-redistribution within the Earth and solar system wide disruption of orbital parameters).
•We show that published estimates of Earth's precession rate indicate a late Proterozoic, rapid drop of 8.4 ± 3.0 “/y.
•If this jump is due to enhanced tidal drag this implies a long-duration increase of tidal dissipation by a factor of at least two.
•The jump cannot be plausibly explained by mass redistribution within the Earth or by solar system wide disruption.
Abstract
We consider the use of cyclostratigraphic estimates of ancient Earth-axis precession rates, k, as a proxy for ancient length of day and ancient Earth-Moon distance. Analysis of published estimates for k indicate a statistically robust acceleration in the rate at which k fell during the Late Proterozoic. We investigate whether this accelerated fall-rate can be reasonably explained by an increase in tidal drag, at that time, or whether alternate explanations are needed. A tidal drag explanation requires an unusually large and long-lived resonance in Earth's oceans. However, alternate explanations are even less viable. A rearrangement of Earth's internal structure can be ruled out by the excessive geothermal heat production this would have caused, whilst mass redistribution due to Late Proterozoic glaciation can also be ruled out as the k-history did not return to its former trend after glaciation ended. Disruption of the Earth-Moon-Sun system by a nearby passing star is similarly unable to account for the observations since the required disruption is much too large to have happened without additional, clearly observable effects. We also consider a possible impact from thermally driven, atmospheric tides but reject this explanation as it would decelerate the fall in precession rather than accelerate it. These conclusions required development of novel techniques for inverse modelling the k-history to directly give (i) tidal-drag and (ii) the potential energy liberated by internal mass distribution.
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Keywords
Cyclostratigraphy
Precession
Earth-Moon distance
Length of day
Late proterozoic
Tidal Drag
1. Introduction
Estimating ancient day-length (and the related ancient Earth-Moon distance) has been the subject of research over many decades using a wide range of approaches involving interpreting corals (Wells, 1963), stromatolites (Pannella, 1972; Vanyo, 1985), bivalves (Pannella, 1976), tidalites (Sonett and Chan, 1998; Williams, 2000), banded iron formations (Walker and Zahnle, 1986) and Milanković cycles (see references in Table 2). The results are not only intrinsically interesting but are important for understanding ancient climates (Hays et al., 1976; Zachos et. al., 2001; Westerhold et. al., 2020; Dutkiewitcz et. al., 2024), ancient tides (Green et. al., 2017) and for accurate cyclostratigraphy (Hinnov 2000). Furthermore, the evolution of the Earth-Moon system may have played a role in enhancing Earth's habitability and is relevant to estimates of Earth's remaining habitable lifetime (Waltham, 2019).
Milankovitch-cycle-based estimates are emerging as the most promising route to robust and widely accepted estimates of ancient Earth-Moon-system properties (see Table 2 for references). In particular, the development of TimeOpt and related analyses (Meyers, 2015; Meyers and Malinverno, 2018; Malinverno and Meyers, 2024) as a robust technique for extracting Milankovitch cycles from ancient sediments has led to an explosion of estimates for the ancient precession rate (k) of Earth's rotational axis. Table 2 shows 22 recent estimates of ancient k and most of these values were obtained using TimeOpt methodology. The modern value (and associated uncertainty), also given in this table, is discussed in more detail later in the paper.
The observed evolution of k through time can be interpreted as the consequence of tidal-drag mediated transfer of angular momentum from the Earth's spin to the Moon's orbit and therefore yields estimates of ancient Earth-Moon separation and ancient length of day (LOD).
However, simplifying assumptions are often made when calculating LOD and Earth-Moon separation from k. For example, solar tides are often assumed to be sufficiently small that they can be ignored (Berger et. al., 1989). Even more often, it is assumed that the Earth and lunar orbits are circular and co-planar (Farhat et. al., 2022a; Waltham, 2015) or that eccentricity and inclination can be fixed at their present-day values (Berger et. al., 1989). It has also not been demonstrated that long-term (>10 6 years) changes in k are always dominated by tidal evolution of the Earth-Moon-system with no other processes affecting k significantly. This paper is concerned with relaxing these assumptions so that a wider range of possible k-influencing processes can be evaluated.
The fundamental controls on precession rate are well understood. Earth's axial precession rate is given by (Berger et al., 1989)(1)k=3 2 n e 2 Ω H((1−e e 2)−3/2+M m M s a e 3 a m 3(1−e m 2)−3/2(1−3 2 s i n 2 i))cos(o).
Furthermore, it is usually assumed that the Earth is close to hydrostatic equilibrium so that dynamic ellipticity, H, takes the form(2)H=K Ω 2 where K depends only upon the density distribution within the Earth (see Appendix A). Combining Eqs. (1) and (2) then yields(3)k=3 2 n e 2 K Ω((1−e e 2)−3/2+M m M s a e 3 a m 3(1−e m 2)−3/2(1−3 2 s i n 2 i))cos(o).
This paper is concerned with how, when and by how much the parameters in Eq. (3) changed over the last 2.5 Gy to give rise to the k-evolution shown in Table 2 (note that this paper uses ka, Ma and Ga to denote ages and ky, My and Gy to denote durations). Specifically, our paper: (i) demonstrates that the data in Table 2 robustly show a rapid fall in k in the late Proterozoic; (ii) develops a procedure for estimating tidal-evolution of the parameters in Eq. (3), from k, for non-coplanar, non-circular orbits including solar tides; (iii) quantifies non-tidal processes that might impact k; (iv) applies the resultant framework to interpretation of the late Proterozoic drop in k.
2. Robust structures in k(t)
2.1. Precession rate uncertainties
Obliquity varies on time-scales of tens of ky (e.g. see Laskar et al., 2004), which gives rise to relatively rapid fluctuations in precession rate around a much more slowly changing mean value, k 0. In particular, it follows from Eq. (3) that(4)σ k=σ o k 0 t a n(o)where σ k is the standard deviation in precession rate resulting from a standard deviation σ o in obliquity. Laskar et al. (2004, their Eq. (31)) give a mean (from a trend fitted to the last 1 My) modern obliquity of 23.270773⁰ (c.f. the exact present-day value of 23.2545° (Laskar et al. (2004, Eq. (33)) and mean modern precession rate of 50.467718 “/y (Laskar et al. (2004 Eq. (35)), c.f. present-day 50.29 “/y, Laskar et al. (2004, Table 1), whilst their 24-term quasi-periodic approximation to Earth-orbit inclination (values from Laskar et al. (2004, Table 5)) substituted into Laskar et al. (2004, Eq. (26)) implies a 1-standard deviation variation in obliquity of 0.503960⁰. Hence, Eq. (4) gives a present-day precession of k 0=50.47±0.19 “/y which is the value shown in Table 2.
Table 1. Symbols used in this paper.
| Symbol | Description |
--- |
| a | Orbit semi-major axis |
| C | Earth principal moment of inertia |
| δ | Sun-star minimum separation |
| D | Tidal dissipation rate |
| Δt | Tidal lag |
| e | Orbital eccentricity |
| ε | Ellipticity |
| ϕ | Latitude |
| G | Universal gravitation constant |
| H | Earth dynamical ellipticity |
| η | Radau parameter |
| i | Lunar orbit inclination to ecliptic |
| J | Jeffreys parameter |
| k | Earth axis precession rate |
| K | Constant relating spin-rate to dynamic ellipticity |
| k 0 | Present day average Earth-axis precession rate |
| L | Earth-spin angular momentum |
| M | Mass |
| M' | Reduced mass |
| n | Mean motion |
| N | Number of stellar close encounters |
| o | Earth axis obliquity to ecliptic |
| P | Period of time |
| R | Radius of the Earth or Moon (depending on subscript) |
| ρ | Density |
| r | Mean radius of ellipsoid |
| r' | Distance from Earth's centre |
| ρ o | Mean density within a given radius |
| S | Constant characterizing close encounter frequency |
| σ k | Uncertainty in precession rate |
| σ o | Uncertainty in obliquity |
| t | Time |
| Ω | Earth siderial spin rate |
| X | Earth-spin angular momentum perpendicular to ecliptic |
| subscript a | Values after k-jump |
| subscript b | Values before k-jump |
| subscript e | Earth values |
| subscript m | Lunar values |
| subscript s | Solar values |
Table 2. Precession rates, associated uncertainties and sources used in this paper and plotted in Fig. 1. The uncertainty in present day precession characterizes short-term fluctuations caused by obliquity variations (see text for details).
| Location | Age (Ma) | k ("/yr) | σ k | Reference |
--- ---
| Present Day | 0 | 50.47 | 0.19 | Laskar et al. (2004) |
| U1408-U1410 | 40 | 51.28 | 0.56 | De Vleeschouwer et al. (2023) |
| U1410 | 42.5 | 51.49 | 0.06 | Boulila and Hinnov (2022) |
| Site 1262 | 54.95 | 51.55 | 0.17 | Boulila and Hinnov (2022) |
| Walvis Ridge | 55 | 51.27 | 0.29 | Malinverno and Meyers (2024) |
| Guando | 245 | 56.70 | 2.26 | Huang et al. (2024) |
| Wujiaping Fm. | 259 | 55.86 | 1.3 | Zhou et al. (2022) |
| Lucaogou Fm. | 290 | 57.06 | 1.36 | Huang et al. (2024) |
| H-32 Iowa | 375 | 59.53 | 3.24 | Huang et al. (2024) |
| PozarCS | 410 | 59.72 | 1.89 | Huang et al. (2024) |
| Antocosti Island | 448 | 59.02 | 1.63 | Huang et al. (2024) |
| Pingliang Fm. | 455 | 59.71 | 1.29 | Zhou et al. (2022) |
| Liangjiashan | 470 | 59.21 | 1.29 | Huang et al. (2024) |
| Alum Shale | 493 | 62.76 | 2.81 | Huang et al. (2024) |
| Qiongzhusi Fm. | 524 | 59.94 | 0.25 | Zhang et al. (2022) |
| Duoshantuo | 570 | 63.49 | 2.92 | Huang et al. (2024) |
| Datangpo Fm. | 655 | 70.21 | 2.08 | Zhou et al. (2022) |
| Yemahe Fm. | 1215 | 84.15 | 1.74 | Zhou et al. (2024) |
| Wumishan Fm. | 1480 | 92.62 | 2.39 | Zhou et al. (2024) |
| Xiamaling Fm. | 1400 | 87.62 | 1.38 | Malinverno and Meyers (2024) |
| Chuanlinggou Fm. | 1634 | 96.40 | 1.94 | Zhou et al. (2024) |
| Chuanlinggou Fm | 1640 | 87.14 | 0.25 | Shi et al. (2023) |
| Joffre Falls | 2460 | 108.60 | 8.5 | Lantink et al. (2022) |
| Dales Gorge Mbr. | 2465 | 105.26 | 1.34 | Zhou et al. (2022) |
The uncertainties, in the remaining precession rates shown in Table 2, range from 0.06 to 8.5 “/y and were estimated as part of Milanković cycle analyses undertaken by the cited authors. These uncertainties are based upon the distribution of precession rates that give a reasonable match of observed peaks to modelled peaks in the spectra of sedimentary sequences. A range of uncertainties is expected because of differences in the signal to noise ratio of the cyclic signals in the sediments analyzed. However, the lowest uncertainties are below the lower-bound set by the obliquity variations discussed above. This seems unlikely and suggests that there are significant uncertainties in the uncertainties themselves.
2.2. Regression fits to the data
Next, we attempt to determine which structures in the k-history are robust features of Earth's history and which could be artefacts of the uncertainties in k.
Fig. 1 plots the data and 1 σ error bars from Table 2. Visually, the data show a trend for the Phanerozoic (i.e. 539–0 Ma) and the last Proterozoic data point (hereafter denoted the Phanerozoic trend) and a different trend for all the remaining Proterozoic data points (hereafter denoted the Proterozoic trend). Hence, there may be an anomalously rapid drop in k during the late Proterozoic. On the other hand, this apparent k-jump may be the consequence of noisy data. In that case, a single regression through all the data is appropriate (as shown by the dashed line in Fig. 1).
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Fig. 1. Precession rate data from Table 2. The plot also shows 1-segment (dashed) and 2-segment (undashed) best-fits to the data. The dotted line shows our estimate of the lowest-gradient jump between the two trends in the 2-segment model but note that a much shorter duration jump is also consistent with the data.
Statistical testing, of the 1-segment versus 2-segment models, can be undertaken using the Chow test (Chow, 1960) whose null-hypothesis is that the gradients and intercepts of the Proterozoic and Phanerozoic trends do not differ significantly. The result of this analysis is that the observed differences would only occur 0.0008 % of the time if the null hypothesis is true. Hence, the alternate hypothesis (i.e. the 2-segment model) is preferred.
We also undertook an F-test (Holický, 2013) to compare the 2-segment model residuals (i.e. model minus data) with the expected uncertainties (estimated from the rms value of the individual uncertainties, in Table 2, excluding the present day value). The F-test consists of calculating the variance ratio (residuals variance over expected variance) which should be close to one if the model gives rise to residuals of the expected typical size. An F-test therefore indicates if the model is overfitting the data or underfitting the data.
For the 2-segment model of Fig. 1, F = 0.59. The corresponding probability is found from the F-distribution along with 20 degrees of freedom for the residuals variance (23 data points minus two gradients and one intercept) and 22 degrees of freedom for the expected variance (23 data points minus one calculated rms value). The resulting cumulative probability is p = 0.88, i.e. well within the 5 % significance range of 0.025 <p< 0.975. Hence, the 2-segment model has residuals of the expected size.
An important consequence of the F-test above, is that there is no evidence that a more complex model is justified by the data. This does not mean that there are no other structures in Earth's precession history (e.g. a flattening around 1.5 Ga (Mitchell & Kirscher (2023)) or a staircase pattern (Huang et al., 2024)), just that these cannot be substantiated using only the data of Table 2.
Hence, the remainder of this paper assumes the 2-segment model and investigates possible explanations for a rapid-drop in k in the late Proterozoic. Given the limited data constraints, the k-drop may have occurred over an extended period (e.g. between 800 Ma and 400 Ma, as indicated by the dotted line in Fig. 1) or could have been much shorter (e.g. an almost instantaneous drop at any time between 570 Ma and 655 Ma).
Assuming a rapid jump at around 600 Ma, the Phanerozoic trend yields k a=63.3 ± 0.4 “/y whilst the Proterozoic trend yields k b=71.7 ± 2.5 “/y. Hence, for an instantaneous jump, the drop between the two trends would have been 8.4 ± 2.6 “/y. For an extended jump duration, the overall drop would have been bigger (i.e. from further up the Proterozoic trend to lower down the Phanerozoic trend) but, if we define the anomaly as the difference between the observed change in k and that expected from its “normal” evolution, then the change due to normal evolution should be subtracted from the jump. The consequence is that the anomaly size is almost unaffected by the assumed jump time or duration (the two effects would precisely cancel if the Proterozoic and Phanerozoic gradients were identical).
In summary, the key conclusion is that the data from Table 2 is well described by a linear regression through the Proterozoic, and a separate linear regression in the Phanerozoic, but with an anomalous fall of 8.4 ± 2.6 “/y between these trends. A late-Proterozoic rapid fall in precession is therefore statistically robust and the remainder of this paper is concerned with its possible causes as this gives us a concrete example of how to use the methods developed, below, for analyzing k-history.
3. Tidal evolution of the Sun-Earth-Moon system
We start our analysis of k-history by initially interpreting it as being entirely due to tidal-drag. We model that history using equations given in Néron de Surgy and Laskar (1997) (hereafter NdSL97). For convenience, these are set out in Appendix A.
The NdSL97 equations yield evolutionary histories for all the parameters in Eq. (3), rather than assuming that some may be neglected. Our main reason for using this complex modelling approach is that we look explicitly at the effects of changing parameters such as i, o and e m later in this paper. However, following NdSL97, we do assume constant values for a e and e e since tidal-drag mediated changes in these are insignificant.
A key parameter, required by the NdSL97 modelling equations, is the tidal lag (Δt) which characterizes the strength of tidal dissipation. This fluctuates through time due to plate-tectonically driven changes in ocean-basin configurations as well as the slowing rotation rate of the Earth (Hansen, 1982; Green et. al., 2017). However, the ancient tidal-lag history is not well constrained. Recent work (e.g. Farhat et al., 2022a) has tackled this issue by imposing a theoretical tidal-drag model (e.g. for a global or hemispherical ocean) with parameters that can be adjusted to obtain a fit to the observed k-evolution. In this paper, we take a different approach and introduce an inverse-modelling procedure for directly estimating the tidal lag required to match the observed rate of change in precession, i.e. Δt(t) is an output of our modelling rather than an input.
Rewriting Eq. (3) in terms of the variables used in NdSL97 (viz. L = CΩ, where C is Earth's moment of inertia, X=L cos(o) and cos(i)) gives(5)k=3 n e 2 K X 2 C[(1−e e 2)−3 2+M m 2 M s a e 3 a m 3(1−e m 2)−3 2(3 c o s 2 i−1)].
The variables that evolve due to tidal drag and, thereby, alter k are X, a m, e m and cos(i). Hence,(6)d k d t=f X d X d t+f a m d a m d t+f e m d e m d t+f c o s i d c o s(i)d t where f X, f am, f em, f cosi are moderately long expressions determined from Eq. (5) and set out in Appendix A. The other variables in Eq. (5) (i.e. n e, K, M m, M s, a e and e e) are not affected significantly by tidal drag although some will be looked at later under non-tidal-drag influences on k.
NdSL97 provides expressions for the derivatives on the right-hand side of Eq. (6) and they can be summarized as of the general form(7)d x d t=α x Δ t+β x where x is X, a m, e m or cos(i), and α x and β x are moderately long expressions (set out in Appendix A). Combining Eqs. (6) and (7) leads to(8)d k d t=γ Δ t+θ.
Here, γ and θ are expressions that are, again, set out in Appendix A. Eq. (8) gives the Δt history required to reproduce the observed dk/dt history provided γ (t) and θ(t) can be found (which are, ultimately, given by the NdSL97 equations). All other parameters (i.e. L, X, a m, e m and cos(i)) are updated simultaneously using the NdSL97 equations (see Appendix A for further details).
Modelling with the NdSL97 equations does not include planetary perturbations (since they only model the Sun-Earth-Moon system). Perturbations cause changes to Earth eccentricity and obliquity on time scales of tens of thousands to millions of years (Laskar et al., 2004) but these are also the time scales over which the sediments, used in Milankovitch cycle analyses, are deposited. Hence, the appropriate values of e e and o are their long-term averages. Laskar et al's (2004) 26-term quasi-periodic approximation for Earth's eccentricity implies an average eccentricity of 0.021566 (Table 4 of Laskar et. al. (2004) substituted into their Eq. (25)), c.f. its present-day value of 0.0167 (Murray and Dermott, 1999).
Importantly, the results from Eq. (8) are path dependent, i.e. the calculated parameters at any given age depend upon the entire k-history, since that time, rather than just upon k at that time. In contrast, a simple model which ignores solar tides and assumes circular, coplanar orbits does produce a unique LOD and a m for any given k but this is not the case for the more sophisticated model used here. Modelling the consequences of uncertainties in k therefore requires a Monte-Carlo approach—modelling a large ensemble of k-history realizations compatible with the data.
We generated these using the means and standard errors of the gradients and intercepts for the 2-segment model shown in Fig. 1. In addition, and to be maximally conservative about the increased tidal drag during the jump, we assumed that the jump began at 800 Ma and ended at 400 Ma, i.e. we assumed it took place over the maximum period consistent with the data.
The resulting range of tidal lag histories, after 1000 simulations, is plotted in Fig. 2 (upper left) which also shows the ±2σ range of results. The plot shows that tidal lag has smoothly increased, over most of the past 2.5 Gy, but that the rapid fall in k, during the late Proterozoic, implies a substantial increase in tidal-lag at the start of the k-jump (by a factor of 2).
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Fig. 2. Modelling results for the 2-segment fit to precession data, shown in Fig. 1. Dashed lines show the ±2σ ranges.
Fig. 2 also shows the resulting Earth-Moon system parameters (numerical values can be obtained from nm2.rhul.ac.uk/NdSL97). All parameters show smooth changes over time but with more rapid changes during the k-jump.
Associated with the evolution of Earth-Moon system properties is an evolution of the tidal dissipation. Dissipation, due to tides raised on Earth by the Moon, is (Murray and Dermott, 1999)(9)D m=0.5 M m′n m a m(Ω−n m)(d a m/d t)e t where M m’ is the reduced lunar mass (=M e M m/(M e+M m), (da m/dt)et is the rate of lunar recession due to tides on Earth (given by Eq. (A11)). The lunar mean motion, n m, is found from Kepler's 3rd Law(10)G(M e+M m)=n m 2 a m 3.
Similarly, tidal dissipation due to solar tides raised on Earth is(11)D s=0.5 M e′n e a e(Ω−n e)d a e d t where M e’ is the reduced Earth mass (=M e M s/(M e+M s)), da e/dt is the rate of Earth-Sun recession, due to tides on Earth (given by Eq. (A23)). Note that da e/dt was ignored earlier, as it has an insignificant effect on a e. However, the small value of da e/dt becomes significant, in Eq. (11), because it is multiplied by the reduced Earth mass and Earth-Sun distance which are much larger than the equivalent lunar values in Eq. (9).
The resulting total tidal dissipation (i.e. D m+D s) is shown in Fig. 3.
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Fig. 3. Evolution of tidal dissipation. Minimum and maximum total tidal dissipation correspond to the ±2σ ranges of the total (=solar + lunar) dissipation. The solar and lunar components (without uncertainties) are also shown.
Fig. 3 shows that the late Proterozoic precession-jump implies high tidal dissipation. The best estimate (∼5.5 TW) is higher than the modern dissipation rate (∼3.5 TW, Daher et. al., 2021; Green et. al., 2017) which is, itself, unusually high due to a temporary tidal resonance in the N. Atlantic (e.g. Green, 2010). However, the uncertainty in the size of the late Proterozoic k-jump does allow for a much smaller (but still substantial) increase in dissipation from 2 TW, before the jump, to around 3 TW during the jump.
The modelled late Proterozoic tidal dissipation also goes on for much longer (400 My) than the present day resonance (lifetime of ∼10My, Green et. al., 2022). However, as noted in Section 2, the time-duration of the jump could be shorter although that leads to a steeper k-gradient and correspondingly higher dissipation. Hence, the curves shown in Fig. 3 are best estimates (with uncertainty) of the lower-bound for dissipation; tidal dissipation may have been much higher.
One possible explanation for the increased tidal dissipation at this time is that the large reduction in ocean depth during the Late Proterozoic glaciations led to an amplification of the tides and associated enhanced dissipation rates (Wunsch, 2016). However, explicit numerical simulations of Cryogenian tides show the opposite—the tides were weak because of the supercontinent state of Cryogenian Earth (Green et al., 2020). It should be noted that the tidal model of Farhat et. al. (2022a) does reproduce a rapid fall in precession rate at around 600 Ma (see their Fig. 8). However, this is for a simplified “hemispherical ocean”, and it remains unclear whether a more realistic, dynamic ocean geometry can maintain strong tidal resonance for such extended periods of time.
The key conclusion of this section is therefore that explaining the accelerated fall of precession-rate in the late Proterozoic, by the usual mechanism of tidal-drag mediated lunar recession, requires unusually high tidal drag for an unusually long period of time. Given these possible issues with explaining the k-jump as due to enhanced tidal dissipation, it is sensible to investigate alternate mechanisms for rapid changes in precession.
4. Mass redistribution
It is widely recognized that Earth's precession rate is affected by mass redistribution due to glacial cycles (e.g. see Farhat et. al. (2022b) and references therein). However, the resulting changes in precession rate are temporary, i.e. they disappear once the ice sheets disappear. Hence, we cannot explain the permanent, late-Proterozoic k-drop by invoking mass redistribution due to large-scale glaciation. Instead, in this section, we investigate the possibility that the late Proterozoic fall in precession was caused by a permanent mass-redistribution, e.g. due to mass movements within and between the mantle and core.
This section therefore considers the possibility of a change in the density profile of the Earth leading to a change of K in Eq. (2). Note that, because a change in the internal structure will alter Earth's moment of inertia, this will also affect C and Ω.
To obtain a first pass estimate of the scale of mass redistribution required, and to investigate whether this is small enough to be plausible, we assume the jump in k is sufficiently rapid that a m, e m, i, and o can be approximated by their average values during the jump. Under these assumptions, Eq. (3) gives(12)k a K b Ω b=k b K a Ω a where subscripts b refer to values before the jump and subscripts a refer to values after the jump.
Parameters, at 600 Ma immediately after the jump, are given in Table 3. These are estimated assuming tidal-drag mediated evolution, since the jump, or (where appropriate) assuming modern values. The tidal-drag mediated values have been calculated using the methods of Section 3 but now assuming the Phanerozoic trend extends back to 600 Ma.
Table 3. Leftmost 5 columns are parameter values (and 1σ uncertainties) immediately after the k-jump, assuming this occurred at 600 Ma and that evolution was entirely due to tidal drag from 600 Ma until the present. Rightmost 3 columns show modern values that we assume have not changed since 600 Ma.
| k | 63.33 | ± | 0.41 | "/y | K | 6.157 | 10 5 s 2/rad 2 |
| Ω | 8.277 | ± | 0.028 | 10−5 rad s-1 | e e | 0.02 | |
| i | 5.2064 | ± | 0.0018 | ⁰ | n e | 1.99099 | 10−7 rad s-1 |
| a m | 367.37 | ± | 0.50 | Mm | a e | 1.496 | 10 11 m |
| o | 22.467 | ± | 0.022 | ⁰ | m s | 1.9889 | 10 30 kg |
| e m | 0.05153 | ± | 0.000176 | | m m | 7.3477 | 10 22 kg |
| Empty Cell | | | | | C | 8.0345 | 10 37 kg m |
Substituting k a, k b, K a and Ω a into Eq. (12) then gives K b Ω b = 57.7 ± 2.6 s rad-1.
Next, rearranging Eq. (B12) from Appendix B gives(13)4 G m e 2 K b Ω b 3 R C a Ω a=(5 2+η R−1)/(1+2 5 1+η R)2 where η R is η at Earth's surface and η is related to the density profile via(14)d d r{ρ 0 r 5 1+η}=5 ρ 0 r 4 where, ρ 0(r) is the mean density between the centre and r. The key point to note, here, is that the Radau parameter (η) is related to the density profile, i.e. η R can be calculated given any specific ρ(r) for the Earth.
Substituting the known values into the left-hand side of Eq. (13) gives 4.32±0.21 (dimensionless). Fig. 4 shows that the right-hand side of Eq. (13) is in this range when −0.45 <η R< 0.1 (c.f. modern value of 0.59 as discussed in Appendix A). Note that values of η R ∼ −1.0 are also possible. The implication is that η R (and hence, ρ(r)) needs to change substantially at the k-jump if mass redistribution is the cause. Unfortunately, ρ(r) is not unique given η R, i.e. we cannot be specific about the precise mass redistribution required.
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Fig. 4. The right hand side of Eq. (13). Dashed lines show the range of η R, required prior to the late Proterozoic k-jump, if the jump was caused by internal mass reorganization of Earth.
Instead, the plausibility of the required change in η R can be gauged by estimating the release of potential energy associated with mass redistribution sufficient to change η R from 0.1 (its maximum value before the jump) to 0.59 (the modern value assumed after the jump). Eq. (14) can be rewritten as(15)ρ¯R 5 1+η R=5∫0 R ρ o r 4 d r=15 4 π∫0 R M r d r where M(r) (=4 πr 3 ρ 0/3) is the mass contained within radius r. Hence(16)ρ¯R 5(1+η R b−1+η R a)=15 4 π∫0 R Δ M r d r where ΔM=M b−M a whilst subscripts a and b denote values before and after the k-jump, respectively. Mass redistribution can be approximated as the sum of N mass transfers from radius r U to radius r L, i.e. Eq. (16) can be rewritten as(17)ρ¯R 5(1+η R b−1+η R a)=15 4 π∑i=1 N Δ M i(∫r L r u r d r)i=15 4 π∑i=1 N Δ M i(r U 2−r L 2)i
We wish to estimate the potential energy released by this mass movement but, as we are only after an order of magnitude estimate, we make the approximation that potential energy change is of the form Mass x g 1/2 x height change where g 1/2 is the acceleration due to gravity at a point halfway between r L and r u. Furthermore, we approximate g 1/2 by the gravity in a uniform Earth, i.e. assume gravity increases linearly from the Earth's centre to its surface. Hence, the potential energy change is(18)Δ U=∑i=1 N Δ M i g i(r U−r L)i=∑i=1 N Δ M i g(r U+r L)i 2 R(r U−r L)i.where g i is the gravity acceleration at the corresponding halfway radius and g is the acceleration due to gravity at r = R. Eqs. (17) and (18) then give(19)Δ U=2 π ρ¯g R 4 15(1+η R b−1+η R a)
Hence, the release of potential energy is ∼10 31 J. This is the same order of magnitude as the energy released by core differentiation (Solomon, 1979). Note that it is unlikely that Earth's core formed over a few hundred million years at the end of the Proterozoic—Jacobsen (2005) and Rubie et. al. (2015) amongst others, show convincingly that core formation was almost certainly complete by 4Ga.
We have also been unable to conceive of any other mass redistribution processes that would have the required scale. An Ediacaran age for onset of inner-core growth, for example, as proposed by Bono et. al. (2019), would not have involved a sufficiently high mass transfer as inner-core growth largely involves a phase change (from liquid to solid) rather than a transfer of mass.
Furthermore, energy release of 10 31 J over a maximum jump duration of 400 My implies a minimum mean power output of ∼10 15 W. This compares to Earth's present day heat output of 4 .2 × 10 13 W. The mass transfer required to explain the late Proterozoic k-jump would therefore have released heat at a rate two orders of magnitude faster than Earth's present day geothermal output. It is not plausible that this would have had no major observable effects.
Hence, mass redistribution can be ruled out as the cause of the late Proterozoic k-jump.
5. Solar system wide disruption
The next explanation for the k-jump this paper considers, is major disruption of the solar system that altered eccentricities, inclinations or obliquity. Such changes might have been induced by close encounters with passing stars (Kaib and Raymond, 2024). Alternatively, they could be the result of resonant interactions in the solar system (e.g. see Zeebe and Lantink, 2024).
Recent results from the Gaia mission (Bailer-Jones, 2022) found 61 star trajectories giving close encounters (< 1 parsec) within ±6My of the present. If encounters are random both in time and space then the number of encounters should scale with time and with impact cross-section so that it is of the form(20)N=S δ 2 P where S is a constant, δ is encounter separation and P is period of time. The 61 encounters in 12 My for δ=1 parsec gives S = 61/12 parsecs−2 My−1. Rearranging for δ then gives that, over the last 2.5 Gy, the closest encounter (i.e. N = 1) will likely have been around 0.01 parsecs. For context, that is <100th of the current distance to the nearest star. Could there have been an encounter at that sort of distance in the late Proterozoic and could it have disrupted orbits sufficiently to produce the observed change in k? To investigate this, this section quantifies the changes in orbital parameters required to explain the k-jump. It then considers whether changes of the required magnitude are plausible.
Our starting point, as with the mass-redistribution investigation above, is to assume that there was a rapid change, at around 600 Ma, and that evolution has been purely tidal since that time. The relevant post-jump parameters were already given in Table 3.
Eq. (3) then allows the pre-jump parameters to be calculated assuming that k altered because one parameter changed. The results are summarized in Table 4. To be clear, the hypotheses being discussed are that the k-drop occurred when one of the parameters in Table 4 altered to the value in Table 3 (with all other parameters fixed at their Table 3 values). This only gives a feel for the scale of parameter changes needed (since, in reality, multiple parameters would have changed).
Table 4. Parameters before the jump if jump caused by wide-spread disruption of the Solar System.
| k | 71.7 | ± | 2.5 | "/y |
| e e | 0.46 | ± | 0.05 | |
| e m | 0.33 | ± | 0.04 | |
| o | no soln | | | ⁰ |
| i | no soln | | | ⁰ |
The conclusion from these analyses is that the higher precession rate, before the jump, could be the result of pre-jump high eccentricity in either the Earth or lunar orbits. However, even a zero inclination for the lunar orbit or Earth obliquity, prior to the jump, is insufficient to explain the high k-value.
The scale of eccentricity change required far exceeds that produced by resonance phenomena within the solar system (e.g. see Zeebe and Lantink, 2024) and, hence, close stellar encounters need to be considered instead.
Kaib and Raymond (2024) investigated the impact of close stellar encounters on the Solar System. From the point of view of this paper, the key conclusion was that substantial disruption results in increased eccentricities of planetary orbits. Close encounters are extremely unlikely to do the opposite and actually circularize orbits. But that is what is required to explain the k-jump (i.e. eccentricities are smaller in Table 3 (post jump) than in Table 4 (pre-jump)).
Furthermore, changes of the required scale in Earth's eccentricity can be ruled out because the pre-jump eccentricity would have had to have been of the order of 0.4 or more. Such a high eccentricity would have led to climatic impacts that would be obvious in the sedimentary record. It is also inherently unlikely that only Earth had an eccentric orbit which was then circularized; erratic behavior in one planet tends to be transferred to the orbits of other planets. Indeed, the easiest way to induce high eccentricity in Earth's orbit is to have high eccentricity in Jupiter's orbit (Horner et. al., 2019). High Jupiter eccentricity would also excite similar eccentricity in the orbits of the other inner planets. Hence, if Earth had eccentricity of order 0.4 then it's likely that the other inner planets would too, and this would have led to intersecting orbits between Venus, Earth and Mars leading to probable collisions.
Dramatic circularization of the lunar orbit is a little harder to eliminate. There is some evidence for a highly eccentric lunar orbit around 100–200 My after lunar formation (Garrick-Bethell et al., 2006) due to evection resonance (Touma and Wisdom, 1998) but it is not clear when and how any such high eccentricity was lost. More recent reanalysis has questioned these conclusions and suggested, instead, that the early moon probably had a low-eccentricity orbit (Keane and Matsuyma, 2014). Eccentricity can be decreased by tidal drag of the Moon due to tides raised on it by Earth, but this drag fell dramatically when the Moon's rotation became synchronous with its orbital period which is estimated to have occurred when the Earth-Moon distance reached around 35 Earth radii (i.e. within a few tens of millions of years after formation, e.g. see Ćuk et al., 2016). Hence, there is a lack of evidence for a Proterozoic Moon with high eccentricity and a lack of mechanisms for circularizing any such orbit.
Overall, a late Proterozoic k-drop, because of Solar System disruption, appears to be ruled out by a lack of mechanisms for producing changes of the required scale and a lack of evidence in the sedimentary record of the significant climate impacts which would have resulted from high eccentricities before the late Proterozoic.
6. Atmospheric tides
We end this paper by considering the possible impact of atmospheric tides (Thomson, 1882). Diurnally varying warming of Earth's atmosphere causes twin bulges in the atmosphere which are acted upon by lunar and solar tidal forces.
Critically, these thermally generated bulges are advanced by around 2 h relative to the sub-solar point. Hence, for solar atmospheric tides, there is a tidal-advance rather than the tidal-lag associated with ocean tides. This tide-advance leads to net transfer of angular momentum from Earth's orbit to Earth's rotation. Hence, solar atmospheric tides act to slow the rate at which LOD is increasing (and also slow the, tiny, rate at which Earth's orbit is expanding). For lunar atmospheric tides, on the other hand, the bulges vary over half a lunar orbit from having a large advance through to having a large lag. Hence, angular momentum transfer averages to zero over long times.
The overall impact of atmospheric tides on Earth-axis precession rate is therefore entirely due to its influence on Ω (there are impacts on a e and e e but these are insignificant). However, this impact is relatively unimportant at the present day (thermal tide torques are ∼5 % of gravitational tidal torques (Thomson, 1882; Lambeck, 2005)).
It has been suggested that, in the Proterozoic, thermal tides were enhanced by resonance (Zahnle and Walker, 1987; Bartlett and Stevenson, 2016; Wu et. al., 2023;Laskar et al., 2023; Farhat et. al., 2024). Under these conditions, the Moon would have continued to recede (due to gravitational tides) whilst changes in Ω would have slowed or even stopped (due to thermal tides offsetting gravitational tides). Inspection of Eq. (3) then shows that these effects would have slowed the rate at which k fell.
In contrast, explaining the late-Proterozoic accelerated k-drop by atmospheric tides would require not only resonance enhancement but also that the thermal bulge was retarded rather than advanced. We have not been able to devise a mechanism to achieve this and, hence, we reject the hypothesis that stengthened atmospheric tides explain the rapid k-drop in the late Proterozoic.
7. Conclusions
7.1. Novel procedures for analyzing ancient precession rates
This paper has developed:
a.An inverse-modelling procedure for estimating ancient tidal-drag (and the corresponding Earth-Moon-System parameters) for the case of non-coplanar, non-circular orbits with both lunar and solar tides. This procedure has been used to produce a 2.5 Gy evolutionary history of the Earth-Moon system that is derived from the observed precession history rather than being tied to a specific model for tidal-lag evolution.
b.A procedure for quantifying the mass redistribution, within the Earth, required to explain any observed change in precession rate.
c.A procedure for estimating the corresponding potential energy release (and hence geothermal impact) of the mass distribution quantified in (b).
7.2. Specific conclusions
This paper has used the techniques developed above to investigate the implications of published Earth-precession estimates. The key conclusions are:
a.Estimates of Earth's precession rate, using Milankovitch cycles in sediments, indicate a late Proterozoic rapid drop of 8.4 ± 3.0 “/y.
b.If the jump, in (a), is due to enhanced tidal drag then this implies an increase of tidal dissipation by a factor of at least two and possibly much more.
c.A tidal drag explanation, for the rapid k-drop, is plausible only if the drop is near the lower end of its possible range and took ∼400 My. Hence, a tidal-drag explanation requires support from further data and theoretical confirmation that tidal resonances can be maintained for longer periods than indicated by models of more recent tidal resonance.
d.The jump, in (a), cannot be plausibly explained by mass redistribution within the Earth as that would have required mass redistribution on the same scale as that for core-differentiation and it would have generated geothermal output 2 orders of magnitude greater than the present day.
e.The jump, in (a), cannot be plausibly explained by Solar System wide disruption (e.g. due to a nearby passing star). Disruption would not have generated changes of the right nature (i.e. decreased rather than increase eccentricities) and there is no evidence, in the sedimentary record, of the large scale climate impacts that earlier high eccentricity would have entailed.
f.The jump, in (a), cannot be caused by enhanced atmospheric tides as these would act to decelerate, rather than accelerate, changes in precession rate.
CRediT authorship contribution statement
David Waltham: Writing – review & editing, Writing – original draft, Software, Methodology, Formal analysis, Data curation, Conceptualization. Mattias Green: Writing – review & editing, Writing – original draft, Validation, Project administration, Funding acquisition, Conceptualization.
Declaration of competing interest
The authors declare that they have no known competing financial interests or personal relationships that could have appeared to influence the work reported in this paper.
Acknowledgements
Funding was provided by the Natural Environment Research Council (MATCH; NE/S009566/1). We'd like to thank Andrew Mitten for his substantial contributions to early discussions that led to this paper. Python code to perform the calculations can be found at We are very grateful to Linda Hinnov, Alberto Malinverno and an anonymous referee for thoughtful, detailed and constructive reviews which greatly improved the ms.
Appendix A. Earth-moon system Modelling equations
Eq. (5) was(5.1)k=3 n e 2 K X 2 C[(1−e e 2)−3/2+M m 2 M s a e 3 a m 3(1−e m 2)−3/2(3 c o s 2 i−1)].
Hence(6.1)d k d t=f X d X d t+f a m d a m d t+f e m d e m d t+f c o s i d c o s(i)d t where elementary calculus gives(A1)f X=BE+D a m 3(1−e m 2)−3/2(3 c o s 2 i−1)f a m=−3 B X D a m 4(1−e m 2)−3/2(3 c o s 2 i−1)(A3)f e m=3 B X D a m 3 e m(1−e m 2)−5/2(3 c o s 2 i−1)(A4)f c o s i=6 B X D a m 3(1−e m 2)−3/2 c o s i with(A5)B=3 n e 2 K/2 C(A6)D=M m a e 3/2 M s(A7)E=(1−e e 2)−3/2.
The derivatives required, in Eq. (6), can be obtained from the Earth-Moon-Sun modelling equations of NdSL97. There are two contributions to d X/d t. Firstly, solar tides give(A8)(d X/d t)s=−3 G M s 2 R 5 k 2 2 a e 6[2(1+15 2 e e 2)X C−2(1+27 2 e e 2)n e]Δ t.(where R is Earth's radius and k 2 is Earth's Love number). Lunar tides add a contribution of(A9)(d X/d t)m=−3 G M m 2 R 5 k 2 2 a m 6[(1+15 2 e m 2)(1+c o s 2 i)X C−2(1+27 2 e m 2)n m c o s i]Δ t
It will be convenient, below, to write the sum of Eq. (B8) and (B9) as(A10)d X/d t=(d X/d t)s+(d X/d t)m=(d X/d t)1 Δ t where (dX/dt)1 is dX/dt with Δt = 1.
For da m/dt there are again two contributions. Firstly, the contribution to lunar recession due to tides raised by the Moon on Earth is(A11)(d a m/d t)e t=6 G M m 2 R 5 k 2 M′m a m 7[(1+27 2 e m 2)X C n m c o s i−(1+23 e m 2)n e]Δ t
Secondly, there is a contribution due to tides raised on the Moon by Earth which has the form(A12)(d a m/d t)m t=57 G M e 2 R m 5 k 2 m e m 2 Δ t m M′m a m 7 where R m is the lunar radius, k 2 m is the Love number for the Moon and Δt m is the tidal lag for the moon (assumed constant in this paper). It will be convenient, later, to write the sum of these terms as(A13)d a m/d t=(d a m/d t)e t+(d a m/d t)m t=(d a m/d t)e t 1 Δ t+(d a m/d t)m t where (da m/dt)et1 is (da m/dt)et evaluated for Δt = 1.
The next derivative (de m/dt) similarly has contributions from lunar tides raised on Earth and from Earth tides raised on the moon. These are, respectively,(A14)(d e m/d t)e t=3 G M m 2 R 5 k 2 M′m a m 8[11 X 2 C n m c o s i−9]Δ t and(A15)(d e m/d t)m t=21 G M e 2 R m 5 k 2 m e m Δ t m 2 M′m a m 8.
Similarly to above, it is convenient to write the sum of these terms as(A16)d e m/d t=(d e m/d t)e t+(d e m/d t)m t=(d e m/d t)e t 1 Δ t+(d e m/d t)m t.
Moving onto the dcosi/dt term of Eq. (6), there is only one contribution (due to tides on Earth raised by the moon) and this takes the form(A17)d c o s i/d t=3 G M m 2 R 5 k 2 2 M′m a m 8[(1+8 e m 2)X C n m s i n 2 i]Δ t which, similarly to above, can be conveniently written in the form(A18)d c o s i/d t=(d c o s i/d t)1 Δ t.
Eqs. (6), (A10), (A13), (A16) and (A18) can now be combined to give(A19)d k/d t=[f x(d X/d t)1+f a m(d a m/d t)e t 1+f e m(d e m/d t)e t 1+f c o s i(d c o s i/d t)1]Δ t+[f a m(d a m/d t)m t+(d e m/d t)m t]which can be summarized as(8.1)d k d t=γ Δ t+θ where γ(t) and θ(t) are the terms in square brackets from Eq. (A19). Eq. (8) is used in the main text to determine Δt(t) from (dk/dt)(t).
In addition to the equations above, we also need to model evolution of Earth's obliquity and the tidally driven recession of Earth from the Sun. NdSL97 gives these as(A20)d L/d t=−3 G M s 2 R 5 k 2 Δ t 2 a e 6[2(1+15 2 e e 2)(1+X 2 L 2)L C−2(1+27 2 e e 2)X L n e].due to solar tides acting the Earth;(A21)d L/d t=−3 G M m 2 R 5 k 2 Δ t 2 a m 6[1 2(1+15 2 e m 2)(3−c o s 2 i+{3 c o s 2 i−1}X 2 L 2)L C−2(1+27 2 e m 2)X L n m c o s i]due to lunar tides acting on the Earth; and(A22)d L/d t=−3 G M m M s R 5 k 2 Δ t 4 a e 3 a m 3(1+3 2 e e 2)(1+3 2 e m 2)(3 c o s 2 i−1)(1−X 2 L 2)L C due to cross-tides (i.e. the effect of solar tides on the Moon and vice versa). Earth recession from the Sun is given by(A23)d a e/d t=6 G M e 2 R 5 k 2 M′e a e 7[(1+27 2 e e 2)X C n e−(1+23 e e 2)]Δ t.
This final expression is used when calculating Eq. (11) in the main text.
Appendix B. Hydrostatic equilibrium and the impact of density profile changes
In the hydrostatic theory of the figure of the Earth (Clairaut, 1743) the interior is assumed to consist of ellipsoidal shells of constant density whose ellipticity changes with distance, r, from the centre of the Earth (defined by the major semi-axis) according to(B1)ρ 0(d 2 ε d r 2−6 ε r 2)+6 ρ r(d ε d r+ε r)=0
(Jefferys, 1976, section 4.03 Eq. (15) for the case n = 2). Here, ρ 0 is the mean density between the centre and r (i.e.(B2)ρ 0=3 r 3∫0 r ρ a 2 d a where a is the major semi axis of successive ellipsoids), ρ is the density at r and ε is the ellipticity defined by(B3)r′=r(1+ε(1 3−s i n 2 ϕ))with r’ the distance from Earth's centre to the ellipsoid at latitude ϕ.
Jeffreys (1976) (following Radau, 1885) simplifies Eq. (B1) to (Jeffreys, 1976, section 4.03, Eq. 42)(B4)d d r{ρ 0 r 5 1+η}=5 ρ 0 r 4 where η is defined as(B5)η=r ε d ε d r.
The resultant dynamic ellipticity is given by (Jeffreys, 1976, section 4.03, Eq. 49)(B6)H=ε R−J 2 1−2 5 1+η R where R-subscripts indicate values at the Earth's surface and J is given by (Jeffreys, 1976, section 4.03, eq 29)(B7)J=Ω 2 R 3 G m e.
Here, R the semi-major axis of the ellipsoid at Earth's surface. The surface ellipticity can then be found from (Jeffreys, 1976, section 4.03, Eq. 30)(B8)R(d ε/d r)R+2 ε R=5 2 J.
Eqs. (B5) to (B8) combine to give(B9)H=(5 2+η R−1)/(1−2 5 1+η R).
Note that the term in square brackets is a function only of η R which, in turn, depends only upon the density distribution via Eq. (B4). Hence, H is proportional to Ω 2 for all 1-dimensional density models, with a constant of proportionality, K, that depends only on the density distribution.
In particular, assuming the PREM density profile for the modern Earth (Dziewonski and Anderson, 1981) yields η R = 0.59 (from (B4) and K = 6.11×10 5 s 2 (from Eq. (B9)) which may be compared to the value determined from modern k (using Eq. (3)) of K = 6.16×10 5 s 2. Note that, in this paper, we use the second of these estimates.
The foregoing mathematical development can then be used to evaluate the impact, on k, of a change in density profile. Note that such a change will alter Earth's ellipticity (via Eq. (B9) and its moment of inertia (and hence Ω).
The Earth's principal moment of inertia, C, is (Jefferys, 1976, section 4.03 Eq. 47)(B10)C=2 3(1−2 5 1+η R)m e R 2.
Conservation of angular momentum (if changes in Earth's density profile cause the k-jump) then requires that(B11)C b Ω b=C a Ω a
Where b-subscripts refer to values before the jump and a-subscripts refer to values after the jump. Equation (B9-B11) combine to yield(B12)A b Ω b=[3(5 2+η R−1)R C a Ω a]/[4(1−2 5 1+η R)2 G m e 2]which is used in the main text to determine η R.
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Code is available on GitHub and the details are given in the paper. No original data.
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How can I better understand manipulating "operators" in mathematical relations?
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Sometimes, (especially in physics), it's common to see mathematical relations manipulated and/or derived by separating "operators" from the things they "act on." I can usually keep up with and follow derivations when reading along in the book, but it bothers me that I don't really understand how it's "justified" to do that.
A classic example would be:
▽=⟨∂∂x,∂∂y,∂∂z⟩▽=⟨∂∂x,∂∂y,∂∂z⟩
We can "apply" the operator in different ways, and I can see intuitively how it works - it appears that "multiplying" the operator by a "thing" is what "applies". This is how we can write:
▽⋅F=∂U∂x+∂V∂y+∂W∂z▽⋅F=∂U∂x+∂V∂y+∂W∂z
if
F=U x^+V y^+W z^F=U x^+V y^+W z^
And similarly with ▽×▽× to define the curl.
But what exactly are we allowed to do with operators and what are we not allowed to do? Is there a name for this kind of treatment? What "are" operators and what rules do they obey? For example, it seems obvious that the "square root" of an operator wouldn't make any sense. Furthermore, it seems to be a given that "squaring" a derivative operator turns it into a "second derivative" operator.
How can I learn more about this?
multivariable-calculus
operator-theory
vector-analysis
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edited Oct 14, 2013 at 15:41
Thomas Andrews
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asked Oct 14, 2013 at 15:35
cemulatecemulate
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An operator is a function. That's all. It's a rule that takes inputs (usually numbers) and produces outputs (usually numbers).
We are allowed to do at least one thing with an function: apply it. If we have two functions f f and g g, we can always define a new function f∘g f∘g by applying g g and then f f, in other words, (f∘g)(x)=f(g(x))(f∘g)(x)=f(g(x)). This is called function composition.
Since we can compose two functions, we can compose a function with itself. In that case, we use the notation f 2 f 2 for f∘f f∘f, and f n f n for the obvious generalization. That's all that's going on when we "square" an operator.
When the outputs of two functions are things that can be added together, for instance if both functions have numbers as outputs, then we can define their sum f+g f+g as a function obtained by applying f f and g g and then summing the results: (f+g)(x)=f(x)+g(x)(f+g)(x)=f(x)+g(x). We can do the same with multiplication and division, and so on. If you think of ++ and ×× as functions themselves (accepting two inputs) then this is really just a special case of composition.
So there's nothing magical going on, all we ever do with operators (functions) is apply them to one another. Things like the Laplacian or Curl are nothing more than multiple layers of composed functions.
As to why differential operators, those are simply functions that map functions to functions. The first derivative, for instance, is something that takes a function and maps it to its derivative (another function). We can just compose that with itself to obtain the second derivative: composing it with itself means applying it twice.
As a final note, you mention square roots of functions: those actually do exist, you just need to look for a function which when applied twice in a row has the same effect as the function whose square root you want.
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answered Oct 14, 2013 at 16:27
Jack MJack M
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Thanks. Okay, I understand that you can do this algebra of functions, but if I just have some operator or map A:U→V A:U→V and I have an element u∈U u∈U, then when I write A u A u I mean the result of applying operator A A with input u u? Is writing the operator next to a member of its input type "implied application"?cemulate –cemulate 2013-10-14 18:12:54 +00:00 Commented Oct 14, 2013 at 18:12
@user7534 I can't speak for every possible case (especially since you're in physics, which I'm unfamiliar with), but yes, I would think so.Jack M –Jack M 2013-10-14 18:40:40 +00:00 Commented Oct 14, 2013 at 18:40
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You may consider an operator as a function of functions, i. e., an operator is a function which arguments are functions itself. For example, if f∈C 0([0,1)]f∈C 0([0,1)] you may declare a function T:C 0([0,1])→R T:C 0([0,1])→R which arguments are functions via
T(f):=∫1 0 f(x)d x.T(f):=∫0 1 f(x)d x.
So the term “operator” reflects that we're dealing with a function of functions as we call a function a “map” in a geometrical context.
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edited Oct 14, 2013 at 16:10
answered Oct 14, 2013 at 16:04
Michael HoppeMichael Hoppe
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From Wikipedia, the free encyclopedia
Tool to track locally defined data attached to the open sets of a topological space
This article is about sheaves on topological spaces. For sheaves on a site, see Grothendieck topology and Topos.
Look up sheaf in Wiktionary, the free dictionary.
In mathematics, a sheaf (pl.: sheaves) is a tool for systematically tracking data (such as sets, abelian groups, rings) attached to the open sets of a topological space and defined locally with regard to them. For example, for each open set, the data could be the ring of continuous functions defined on that open set. Such data are well-behaved in that they can be restricted to smaller open sets, and also the data assigned to an open set are equivalent to all collections of compatible data assigned to collections of smaller open sets covering the original open set (intuitively, every datum is the sum of its constituent data).
The field of mathematics that studies sheaves is called sheaf theory.
Sheaves are understood conceptually as general and abstract objects. Their precise definition is rather technical. They are specifically defined as sheaves of sets or as sheaves of rings, for example, depending on the type of data assigned to the open sets.
There are also maps (or morphisms) from one sheaf to another; sheaves (of a specific type, such as sheaves of abelian groups) with their morphisms on a fixed topological space form a category. On the other hand, to each continuous map there is associated both a direct image functor, taking sheaves and their morphisms on the domain to sheaves and morphisms on the codomain, and an inverse image functor operating in the opposite direction. These functors, and certain variants of them, are essential parts of sheaf theory.
Due to their general nature and versatility, sheaves have several applications in topology and especially in algebraic and differential geometry. First, geometric structures such as that of a differentiable manifold or a scheme can be expressed in terms of a sheaf of rings on the space. In such contexts, several geometric constructions such as vector bundles or divisors are naturally specified in terms of sheaves. Second, sheaves provide the framework for a very general cohomology theory, which encompasses also the "usual" topological cohomology theories such as singular cohomology. Especially in algebraic geometry and the theory of complex manifolds, sheaf cohomology provides a powerful link between topological and geometric properties of spaces. Sheaves also provide the basis for the theory of D-modules, which provide applications to the theory of differential equations. In addition, generalisations of sheaves to more general settings than topological spaces, such as the notion of a sheaf on a category with respect to some Grothendieck topology, have provided applications to mathematical logic and to number theory.
Definitions and examples
[edit]
In many mathematical branches, several structures defined on a topological space (e.g., a differentiable manifold) can be naturally localised or restricted to open subsets : typical examples include continuous real-valued or complex-valued functions, -times differentiable (real-valued or complex-valued) functions, bounded real-valued functions, vector fields, and sections of any vector bundle on the space. The ability to restrict data to smaller open subsets gives rise to the concept of presheaves. Roughly speaking, sheaves are then those presheaves, where local data can be glued to global data.
Presheaves
[edit]
See also: Presheaf (category theory)
Let be a topological space. A presheaf of sets on consists of the following data:
For each open set , there exists a set . This set is also denoted . The elements in this set are called the sections of over . The sections of over are called the global sections of .
For each inclusion of open sets , a function . In view of many of the examples below, the morphisms are called restriction morphisms. If , then its restriction is often denoted by analogy with restriction of functions.
The restriction morphisms are required to satisfy two additional (functorial) properties:
For every open set of , the restriction morphism is the identity morphism on .
If we have three open sets , then the composite .
Informally, the second axiom says it does not matter whether we restrict to in one step or restrict first to , then to . A concise functorial reformulation of this definition is given further below.
Many examples of presheaves come from different classes of functions: to any , one can assign the set of continuous real-valued functions on . The restriction maps are then just given by restricting a continuous function on to a smaller open subset , which again is a continuous function. The two presheaf axioms are immediately checked, thereby giving an example of a presheaf. This can be extended to a presheaf of holomorphic functions and a presheaf of smooth functions .
Another common class of examples is assigning to the set of constant real-valued functions on . This presheaf is called the constant presheaf associated to and is denoted .
Sheaves
[edit]
Given a presheaf, a natural question to ask is to what extent its sections over an open set are specified by their restrictions to open subsets of . A sheaf is a presheaf whose sections are, in a technical sense, uniquely determined by their restrictions.
Axiomatically, a sheaf is a presheaf that satisfies both of the following axioms:
(Locality) Suppose is an open set, is an open cover of with for all , and are sections. If for all , then .
(Gluing) Suppose is an open set, is an open cover of with for all , and is a family of sections. If all pairs of sections agree on the overlap of their domains, that is, if for all , then there exists a section such that for all .
In both of these axioms, the hypothesis on the open cover is equivalent to the assumption that .
The section whose existence is guaranteed by axiom 2 is called the gluing, concatenation, or collation of the sections . By axiom 1 it is unique. Sections and satisfying the agreement precondition of axiom 2 are often called compatible ; thus axioms 1 and 2 together state that any collection of pairwise compatible sections can be uniquely glued together. A separated presheaf, or monopresheaf, is a presheaf satisfying axiom 1.
The presheaf consisting of continuous functions mentioned above is a sheaf. This assertion reduces to checking that, given continuous functions which agree on the intersections , there is a unique continuous function whose restriction equals the . By contrast, the constant presheaf is usually not a sheaf as it fails to satisfy the locality axiom on the empty set (this is explained in more detail at constant sheaf).
Presheaves and sheaves are typically denoted by capital letters, being particularly common, presumably for the French word for sheaf, faisceau. Use of calligraphic letters such as is also common.
It can be shown that to specify a sheaf, it is enough to specify its restriction to the open sets of a basis for the topology of the underlying space. Moreover, it can also be shown that it is enough to verify the sheaf axioms above relative to the open sets of a covering. This observation is used to construct another example which is crucial in algebraic geometry, namely quasi-coherent sheaves. Here the topological space in question is the spectrum of a commutative ring , whose points are the prime ideals in . The open sets form a basis for the Zariski topology on this space. Given an -module , there is a sheaf, denoted by on the , that satisfies
: the localization of at .
There is another characterization of sheaves that is equivalent to the previously discussed.
A presheaf is a sheaf if and only if for any open and any open cover of , is the fibre product . This characterization is useful in construction of sheaves, for example, if are abelian sheaves, then the kernel of sheaves morphism is a sheaf, since projective limits commutes with projective limits. On the other hand, the cokernel is not always a sheaf because inductive limits do not necessarily commute with projective limits. One way to fix this is to consider Noetherian topological spaces; all open sets are compact so that the cokernel is a sheaf, since finite projective limits commutes with inductive limits.
Further examples
[edit]
Sheaf of sections of a continuous map
[edit]
Any continuous map of topological spaces determines a sheaf on by setting
Any such is commonly called a section of , and this example is the reason why the elements in are generally called sections. This construction is especially important when is the projection of a fiber bundle onto its base space. For example, the sheaves of smooth functions are the sheaves of sections of the trivial bundle.
Another example: the sheaf of sections of
is the sheaf which assigns to any the set of branches of the complex logarithm on .
Given a point and an abelian group , the skyscraper sheaf is defined as follows: if is an open set containing , then . If does not contain , then , the trivial group. The restriction maps are either the identity on , if both open sets contain , or the zero map otherwise.
Sheaves on manifolds
[edit]
On an -dimensional -manifold , there are a number of important sheaves, such as the sheaf of -times continuously differentiable functions (with ). Its sections on some open are the -functions . For , this sheaf is called the structure sheaf and is denoted . The nonzero functions also form a sheaf, denoted . Differential forms (of degree ) also form a sheaf . In all these examples, the restriction morphisms are given by restricting functions or forms.
The assignment sending to the compactly supported functions on is not a sheaf, since there is, in general, no way to preserve this property by passing to a smaller open subset. Instead, this forms a cosheaf, a dual concept where the restriction maps go in the opposite direction than with sheaves. However, taking the dual of these vector spaces does give a sheaf, the sheaf of distributions.
Presheaves that are not sheaves
[edit]
In addition to the constant presheaf mentioned above, which is usually not a sheaf, there are further examples of presheaves that are not sheaves:
Let be the two-point topological space with the discrete topology. Define a presheaf as follows: The restriction map is the projection of onto its first coordinate, and the restriction map is the projection of onto its second coordinate. is a presheaf that is not separated: a global section is determined by three numbers, but the values of that section over and determine only two of those numbers. So while we can glue any two sections over and , we cannot glue them uniquely.
Let be the real line, and let be the set of bounded continuous functions on . This is not a sheaf because it is not always possible to glue. For example, let be the set of all such that . The identity function is bounded on each . Consequently, we get a section on . However, these sections do not glue, because the function is not bounded on the real line. Consequently is a presheaf, but not a sheaf. In fact, is separated because it is a sub-presheaf of the sheaf of continuous functions.
Motivating sheaves from complex analytic spaces and algebraic geometry
[edit]
One of the historical motivations for sheaves have come from studying complex manifolds, complex analytic geometry, and scheme theory from algebraic geometry. This is because in all of the previous cases, we consider a topological space together with a structure sheaf giving it the structure of a complex manifold, complex analytic space, or scheme. This perspective of equipping a topological space with a sheaf is essential to the theory of locally ringed spaces (see below).
Technical challenges with complex manifolds
[edit]
One of the main historical motivations for introducing sheaves was constructing a device which keeps track of holomorphic functions on complex manifolds. For example, on a compact complex manifold (like complex projective space or the vanishing locus in projective space of a homogeneous polynomial), the only holomorphic functions
are the constant functions. This means there exist two compact complex manifolds which are not isomorphic, but nevertheless their rings of global holomorphic functions, denoted , are isomorphic. Contrast this with smooth manifolds where every manifold can be embedded inside some , hence its ring of smooth functions comes from restricting the smooth functions from , of which there exist plenty.
Another complexity when considering the ring of holomorphic functions on a complex manifold is given a small enough open set , the holomorphic functions will be isomorphic to . Sheaves are a direct tool for dealing with this complexity since they make it possible to keep track of the holomorphic structure on the underlying topological space of on arbitrary open subsets . This means as becomes more complex topologically, the ring can be expressed from gluing the . Note that sometimes this sheaf is denoted or just , or even when we want to emphasize the space the structure sheaf is associated to.
Tracking submanifolds with sheaves
[edit]
Another common example of sheaves can be constructed by considering a complex submanifold . There is an associated sheaf which takes an open subset and gives the ring of holomorphic functions on . This kind of formalism was found to be extremely powerful and motivates a lot of homological algebra such as sheaf cohomology since an intersection theory can be built using these kinds of sheaves from the Serre intersection formula.
Operations with sheaves
[edit]
Morphisms
[edit]
Morphisms of sheaves are, roughly speaking, analogous to functions between them. In contrast to a function between sets, which is simply an assignment of outputs to inputs, morphisms of sheaves are also required to be compatible with the local–global structures of the underlying sheaves. This idea is made precise in the following definition.
Let and be two sheaves of sets (respectively abelian groups, rings, etc.) on . A morphism consists of a morphism of sets (respectively abelian groups, rings, etc.) for each open set of , subject to the condition that this morphism is compatible with restrictions. In other words, for every open subset of an open set , the following diagram is commutative.
For example, taking the derivative gives a morphism of sheaves on ,
Indeed, given an (-times continuously differentiable) function (with in open), the restriction (to a smaller open subset ) of its derivative equals the derivative of .
With this notion of morphism, sheaves of sets (respectively abelian groups, rings, etc.) on a fixed topological space form a category. The general categorical notions of mono-, epi- and isomorphisms can therefore be applied to sheaves.
A morphism of sheaves on is an isomorphism (respectively monomorphism) if and only if there exists an open cover of such that are isomorphisms (respectively injective morphisms) of sets (respectively abelian groups, rings, etc.) for all . These statements give examples of how to work with sheaves using local information, but it's important to note that we cannot check if a morphism of sheaves is an epimorphism in the same manner. Indeed the statement that maps on the level of open sets are not always surjective for epimorphisms of sheaves is equivalent to non-exactness of the global sections functor—or equivalently, to non-triviality of sheaf cohomology.
Stalks of a sheaf
[edit]
Main article: Stalk (sheaf)
The stalk of a sheaf captures the properties of a sheaf "around" a point , generalizing the germs of functions.
Here, "around" means that, conceptually speaking, one looks at smaller and smaller neighborhoods of the point. Of course, no single neighborhood will be small enough, which requires considering a limit of some sort. More precisely, the stalk is defined by
the direct limit being over all open subsets of containing the given point . In other words, an element of the stalk is given by a section over some open neighborhood of , and two such sections are considered equivalent if their restrictions agree on a smaller neighborhood.
The natural morphism takes a section in to its germ at . This generalises the usual definition of a germ.
In many situations, knowing the stalks of a sheaf is enough to control the sheaf itself. For example, whether or not a morphism of sheaves is a monomorphism, epimorphism, or isomorphism can be tested on the stalks. In this sense, a sheaf is determined by its stalks, which are a local data. By contrast, the global information present in a sheaf, i.e., the global sections, i.e., the sections on the whole space , typically carry less information. For example, for a compact complex manifold , the global sections of the sheaf of holomorphic functions are just , since any holomorphic function
is constant by Liouville's theorem.
Turning a presheaf into a sheaf
[edit]
It is frequently useful to take the data contained in a presheaf and to express it as a sheaf. It turns out that there is a best possible way to do this. It takes a presheaf and produces a new sheaf called the sheafification or sheaf associated to the presheaf . For example, the sheafification of the constant presheaf (see above) is called the constant sheaf. Despite its name, its sections are locally constant functions.
The sheaf can be constructed using the étalé space of , namely as the sheaf of sections of the map
Another construction of the sheaf proceeds by means of a functor from presheaves to presheaves that gradually improves the properties of a presheaf: for any presheaf , is a separated presheaf, and for any separated presheaf , is a sheaf. The associated sheaf is given by .
The idea that the sheaf is the best possible approximation to by a sheaf is made precise using the following universal property: there is a natural morphism of presheaves so that for any sheaf and any morphism of presheaves , there is a unique morphism of sheaves such that . In fact, is the left adjoint functor to the inclusion functor (or forgetful functor) from the category of sheaves to the category of presheaves, and is the unit of the adjunction. In this way, the category of sheaves turns into a Giraud subcategory of presheaves. This categorical situation is the reason why the sheafification functor appears in constructing cokernels of sheaf morphisms or tensor products of sheaves, but not for kernels, say.
Subsheaves, quotient sheaves
[edit]
If is a subsheaf of a sheaf of abelian groups, then the quotient sheaf is the sheaf associated to the presheaf ; in other words, the quotient sheaf fits into an exact sequence of sheaves of abelian groups;
(this is also called a sheaf extension.)
Let be sheaves of abelian groups. The set of morphisms of sheaves from to forms an abelian group (by the abelian group structure of ). The sheaf hom of and , denoted by,
is the sheaf of abelian groups where is the sheaf on given by (note sheafification is not needed here). The direct sum of and is the sheaf given by , and the tensor product of and is the sheaf associated to the presheaf .
All of these operations extend to sheaves of modules over a sheaf of rings ; the above is the special case when is the constant sheaf .
Basic functoriality
[edit]
Main article: Image functors for sheaves
Since the data of a (pre-)sheaf depends on the open subsets of the base space, sheaves on different topological spaces are unrelated to each other in the sense that there are no morphisms between them. However, given a continuous map between two topological spaces, pushforward and pullback relate sheaves on to those on and vice versa.
Direct image
[edit]
The pushforward (also known as direct image) of a sheaf on is the sheaf defined by
Here is an open subset of , so that its preimage is open in by the continuity of .
This construction recovers the skyscraper sheaf mentioned above:
where is the inclusion, and is regarded as a sheaf on the singleton by .
For a map between locally compact spaces, the direct image with compact support is a subsheaf of the direct image. By definition, consists of those whose support is mapped properly. If is proper itself, then , but in general they disagree.
Inverse image
[edit]
The pullback or inverse image goes the other way: it produces a sheaf on , denoted out of a sheaf on . If is the inclusion of an open subset, then the inverse image is just a restriction, i.e., it is given by for an open in . A sheaf (on some space ) is called locally constant if by some open subsets such that the restriction of to all these open subsets is constant. On a wide range of topological spaces , such sheaves are equivalent to representations of the fundamental group .
For general maps , the definition of is more involved; it is detailed at inverse image functor. The stalk is an essential special case of the pullback in view of a natural identification, where is as above:
More generally, stalks satisfy .
Extension by zero
[edit]
"Extension by zero" redirects here. For uses in analysis, see Sobolev space § Extension by zero, and Extension of a function.
For the inclusion of an open subset, the extension by zero (pronounced "j lower shriek of F") of a sheaf of abelian groups on is the sheafification of the presheaf defined by
: if and otherwise.
For a sheaf on , this construction is in a sense complementary to , where is the inclusion of the complement of :
: for in , and the stalk is zero otherwise, while
: for in , and equals otherwise.
More generally, if is a locally closed subset, then there exists an open of containing such that is closed in . Let and be the natural inclusions. Then the extension by zero of a sheaf on is defined by .
Due to its nice behavior on stalks, the extension by zero functor is useful for reducing sheaf-theoretic questions on to ones on the strata of a stratification, i.e., a decomposition of into smaller, locally closed subsets.
Complements
[edit]
Sheaves in more general categories
[edit]
In addition to (pre-)sheaves as introduced above, where is merely a set, it is in many cases important to keep track of additional structure on these sections. For example, the sections of the sheaf of continuous functions naturally form a real vector space, and restriction is a linear map between these vector spaces.
Presheaves with values in an arbitrary category are defined by first considering the category of open sets on to be the posetal category whose objects are the open sets of and whose morphisms are inclusions. Then a -valued presheaf on is the same as a contravariant functor from to . Morphisms in this category of functors, also known as natural transformations, are the same as the morphisms defined above, as can be seen by unraveling the definitions.
If the target category admits all limits, a -valued presheaf is a sheaf if the following diagram is an equalizer for every open cover
of any open set :
Here the first map is the product of the restriction maps
and the pair of arrows the products of the two sets of restrictions
and
If is an abelian category, this condition can also be rephrased by requiring that there is an exact sequence
A particular case of this sheaf condition occurs for being the empty set, and the index set also being empty. In this case, the sheaf condition requires to be the terminal object in .
Ringed spaces and sheaves of modules
[edit]
Main articles: Ringed space and Sheaf of modules
In several geometrical disciplines, including algebraic geometry and differential geometry, the spaces come along with a natural sheaf of rings, often called the structure sheaf and denoted by . Such a pair is called a ringed space. Many types of spaces can be defined as certain types of ringed spaces. Commonly, all the stalks of the structure sheaf are local rings, in which case the pair is called a locally ringed space.
For example, an -dimensional manifold is a locally ringed space whose structure sheaf consists of -functions on the open subsets of . The property of being a locally ringed space translates into the fact that such a function, which is nonzero at a point , is also non-zero on a sufficiently small open neighborhood of . Some authors actually define real (or complex) manifolds to be locally ringed spaces that are locally isomorphic to the pair consisting of an open subset of (respectively ) together with the sheaf of (respectively holomorphic) functions. Similarly, schemes, the foundational notion of spaces in algebraic geometry, are locally ringed spaces that are locally isomorphic to the spectrum of a ring.
Given a ringed space, a sheaf of modules is a sheaf such that on every open set of , is an -module and for every inclusion of open sets , the restriction map is compatible with the restriction map : the restriction of is the restriction of times that of for any in and in .
Most important geometric objects are sheaves of modules. For example, there is a one-to-one correspondence between vector bundles and locally free sheaves of -modules. This paradigm applies to real vector bundles, complex vector bundles, or vector bundles in algebraic geometry (where consists of smooth functions, holomorphic functions, or regular functions, respectively). Sheaves of solutions to differential equations are -modules, that is, modules over the sheaf of differential operators. On any topological space, modules over the constant sheaf are the same as sheaves of abelian groups in the sense above.
There is a different inverse image functor for sheaves of modules over sheaves of rings. This functor is usually denoted and it is distinct from . See inverse image functor.
Finiteness conditions for sheaves of modules
[edit]
Finiteness conditions for module over commutative rings give rise to similar finiteness conditions for sheaves of modules: is called finitely generated (respectively finitely presented) if, for every point of , there exists an open neighborhood of , a natural number (possibly depending on ), and a surjective morphism of sheaves (respectively, in addition a natural number , and an exact sequence .) Paralleling the notion of a coherent module, is called a coherent sheaf if it is of finite type and if, for every open set and every morphism of sheaves (not necessarily surjective), the kernel of is of finite type. is coherent if it is coherent as a module over itself. Like for modules, coherence is in general a strictly stronger condition than finite presentation. The Oka coherence theorem states that the sheaf of holomorphic functions on a complex manifold is coherent.
The étalé space of a sheaf
[edit]
In the examples above it was noted that some sheaves occur naturally as sheaves of sections. In fact, all sheaves of sets can be represented as sheaves of sections of a topological space called the étalé space, from the French word étalé [etale], meaning roughly "spread out". If is a sheaf over , then the étalé space (sometimes called the étale space) of is a topological space together with a local homeomorphism such that the sheaf of sections of is . The space is usually very strange, and even if the sheaf arises from a natural topological situation, may not have any clear topological interpretation. For example, if is the sheaf of sections of a continuous function , then if and only if is a local homeomorphism.
The étalé space is constructed from the stalks of over . As a set, it is their disjoint union and is the obvious map that takes the value on the stalk of over . The topology of is defined as follows. For each element and each , we get a germ of at , denoted or . These germs determine points of . For any and , the union of these points (for all ) is declared to be open in . Notice that each stalk has the discrete topology as subspace topology. A morphism between two sheaves determine a continuous map of the corresponding étalé spaces that is compatible with the projection maps (in the sense that every germ is mapped to a germ over the same point). This makes the construction into a functor.
The construction above determines an equivalence of categories between the category of sheaves of sets on and the category of étalé spaces over . The construction of an étalé space can also be applied to a presheaf, in which case the sheaf of sections of the étalé space recovers the sheaf associated to the given presheaf.
This construction makes all sheaves into representable functors on certain categories of topological spaces. As above, let be a sheaf on , let be its étalé space, and let be the natural projection. Consider the overcategory of topological spaces over , that is, the category of topological spaces together with fixed continuous maps to . Every object of this category is a continuous map , and a morphism from to is a continuous map that commutes with the two maps to . There is a functor
sending an object to . For example, if is the inclusion of an open subset, then
and for the inclusion of a point , then
is the stalk of at . There is a natural isomorphism
,
which shows that (for the étalé space) represents the functor .
is constructed so that the projection map is a covering map. In algebraic geometry, the natural analog of a covering map is called an étale morphism. Despite its similarity to "étalé", the word étale [etal] has a different meaning in French. It is possible to turn into a scheme and into a morphism of schemes in such a way that retains the same universal property, but is not in general an étale morphism because it is not quasi-finite. It is, however, formally étale.
The definition of sheaves by étalé spaces is older than the definition given earlier in the article. It is still common in some areas of mathematics such as mathematical analysis.
Sheaf cohomology
[edit]
Main article: Sheaf cohomology
In contexts where the open set is fixed, and the sheaf is regarded as a variable, the set is also often denoted
As was noted above, this functor does not preserve epimorphisms. Instead, an epimorphism of sheaves is a map with the following property: for any section there is a covering where
of open subsets, such that the restriction are in the image of . However, itself need not be in the image of . A concrete example of this phenomenon is the exponential map
between the sheaf of holomorphic functions and non-zero holomorphic functions. This map is an epimorphism, which amounts to saying that any non-zero holomorphic function (on some open subset in , say), admits a complex logarithm locally, i.e., after restricting to appropriate open subsets. However, need not have a logarithm globally.
Sheaf cohomology captures this phenomenon. More precisely, for an exact sequence of sheaves of abelian groups
(i.e., an epimorphism whose kernel is ), there is a long exact sequenceBy means of this sequence, the first cohomology group is a measure for the non-surjectivity of the map between sections of and .
There are several different ways of constructing sheaf cohomology. Grothendieck (1957) introduced them by defining sheaf cohomology as the derived functor of . This method is theoretically satisfactory, but, being based on injective resolutions, of little use in concrete computations. Godement resolutions are another general, but practically inaccessible approach.
Computing sheaf cohomology
[edit]
Especially in the context of sheaves on manifolds, sheaf cohomology can often be computed using resolutions by soft sheaves, fine sheaves, and flabby sheaves (also known as flasque sheaves from the French flasque meaning flabby). For example, a partition of unity argument shows that the sheaf of smooth functions on a manifold is soft. The higher cohomology groups for vanish for soft sheaves, which gives a way of computing cohomology of other sheaves. For example, the de Rham complex is a resolution of the constant sheaf on any smooth manifold, so the sheaf cohomology of is equal to its de Rham cohomology.
A different approach is by Čech cohomology. Čech cohomology was the first cohomology theory developed for sheaves and it is well-suited to concrete calculations, such as computing the coherent sheaf cohomology of complex projective space . It relates sections on open subsets of the space to cohomology classes on the space. In most cases, Čech cohomology computes the same cohomology groups as the derived functor cohomology. However, for some pathological spaces, Čech cohomology will give the correct but incorrect higher cohomology groups. To get around this, Jean-Louis Verdier developed hypercoverings. Hypercoverings not only give the correct higher cohomology groups but also allow the open subsets mentioned above to be replaced by certain morphisms from another space. This flexibility is necessary in some applications, such as the construction of Pierre Deligne's mixed Hodge structures.
Many other coherent sheaf cohomology groups are found using an embedding of a space into a space with known cohomology, such as , or some weighted projective space. In this way, the known sheaf cohomology groups on these ambient spaces can be related to the sheaves , giving . For example, computing the coherent sheaf cohomology of projective plane curves is easily found. One big theorem in this space is the Hodge decomposition found using a spectral sequence associated to sheaf cohomology groups, proved by Deligne. Essentially, the -page with terms
the sheaf cohomology of a smooth projective variety , degenerates, meaning . This gives the canonical Hodge structure on the cohomology groups . It was later found these cohomology groups can be easily explicitly computed using Griffiths residues. See Jacobian ideal. These kinds of theorems lead to one of the deepest theorems about the cohomology of algebraic varieties, the decomposition theorem, paving the path for Mixed Hodge modules.
Another clean approach to the computation of some cohomology groups is the Borel–Bott–Weil theorem, which identifies the cohomology groups of some line bundles on flag manifolds with irreducible representations of Lie groups. This theorem can be used, for example, to easily compute the cohomology groups of all line bundles on projective space and grassmann manifolds.
In many cases there is a duality theory for sheaves that generalizes Poincaré duality. See Grothendieck duality and Verdier duality.
Derived categories of sheaves
[edit]
The derived category of the category of sheaves of, say, abelian groups on some space X, denoted here as , is the conceptual haven for sheaf cohomology, by virtue of the following relation:
The adjunction between , which is the left adjoint of (already on the level of sheaves of abelian groups) gives rise to an adjunction
: (for ),
where is the derived functor. This latter functor encompasses the notion of sheaf cohomology since for .
| Image functors for sheaves |
| direct image |
| inverse image |
| direct image with compact support |
| exceptional inverse image |
| |
| |
| Base change theorems |
| |
Like , the direct image with compact support can also be derived. By virtue of the following isomorphism parametrizes the cohomology with compact support of the fibers of :
:
This isomorphism is an example of a base change theorem. There is another adjunction
Unlike all the functors considered above, the twisted (or exceptional) inverse image functor is in general only defined on the level of derived categories, i.e., the functor is not obtained as the derived functor of some functor between abelian categories. If and X is a smooth orientable manifold of dimension n, then
:
This computation, and the compatibility of the functors with duality (see Verdier duality) can be used to obtain a high-brow explanation of Poincaré duality. In the context of quasi-coherent sheaves on schemes, there is a similar duality known as coherent duality.
Perverse sheaves are certain objects in , i.e., complexes of sheaves (but not in general sheaves proper). They are an important tool to study the geometry of singularities.
Derived categories of coherent sheaves and the Grothendieck group
[edit]
Another important application of derived categories of sheaves is with the derived category of coherent sheaves on a scheme denoted . This was used by Grothendieck in his development of intersection theory using derived categories and K-theory, that the intersection product of subschemes is represented in K-theory as
where are coherent sheaves defined by the -modules given by their structure sheaves.
Sites and topoi
[edit]
Main articles: Grothendieck topology and Topos
André Weil's Weil conjectures stated that there was a cohomology theory for algebraic varieties over finite fields that would give an analogue of the Riemann hypothesis. The cohomology of a complex manifold can be defined as the sheaf cohomology of the locally constant sheaf in the Euclidean topology, which suggests defining a Weil cohomology theory in positive characteristic as the sheaf cohomology of a constant sheaf. But the only classical topology on such a variety is the Zariski topology, and the Zariski topology has very few open sets, so few that the cohomology of any Zariski-constant sheaf on an irreducible variety vanishes (except in degree zero). Alexandre Grothendieck solved this problem by introducing Grothendieck topologies, which axiomatize the notion of covering. Grothendieck's insight was that the definition of a sheaf depends only on the open sets of a topological space, not on the individual points. Once he had axiomatized the notion of covering, open sets could be replaced by other objects. A presheaf takes each one of these objects to data, just as before, and a sheaf is a presheaf that satisfies the gluing axiom with respect to our new notion of covering. This allowed Grothendieck to define étale cohomology and ℓ-adic cohomology, which eventually were used to prove the Weil conjectures.
A category with a Grothendieck topology is called a site. A category of sheaves on a site is called a topos or a Grothendieck topos. The notion of a topos was later abstracted by William Lawvere and Miles Tierney to define an elementary topos, which has connections to mathematical logic.
History
[edit]
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The first origins of sheaf theory are hard to pin down – they may be co-extensive with the idea of analytic continuation[clarification needed]. It took about 15 years for a recognisable, free-standing theory of sheaves to emerge from the foundational work on cohomology.
1936 Eduard Čech introduces the nerve construction, for associating a simplicial complex to an open covering.
1938 Hassler Whitney gives a 'modern' definition of cohomology, summarizing the work since J. W. Alexander and Kolmogorov first defined cochains.
1943 Norman Steenrod publishes on homology with local coefficients.
1945 Jean Leray publishes work carried out as a prisoner of war, motivated by proving fixed-point theorems for application to PDE theory; it is the start of sheaf theory and spectral sequences.
1947 Henri Cartan reproves the de Rham theorem by sheaf methods, in correspondence with André Weil (see De Rham–Weil theorem). Leray gives a sheaf definition in his courses via closed sets (the later carapaces).
1948 The Cartan seminar writes up sheaf theory for the first time.
1950 The "second edition" sheaf theory from the Cartan seminar: the sheaf space (espace étalé) definition is used, with stalkwise structure. Supports are introduced, and cohomology with supports. Continuous mappings give rise to spectral sequences. At the same time Kiyoshi Oka introduces an idea (adjacent to that) of a sheaf of ideals, in several complex variables.
1951 The Cartan seminar proves theorems A and B, based on Oka's work.
1953 The finiteness theorem for coherent sheaves in the analytic theory is proved by Cartan and Jean-Pierre Serre, as is Serre duality.
1954 Serre's paper Faisceaux algébriques cohérents (published in 1955) introduces sheaves into algebraic geometry. These ideas are immediately exploited by Friedrich Hirzebruch, who writes a major 1956 book on topological methods.
1955 Alexander Grothendieck in lectures in Kansas defines abelian category and presheaf, and by using injective resolutions allows direct use of sheaf cohomology on all topological spaces, as derived functors.
1956 Oscar Zariski's report Algebraic sheaf theory
1957 Grothendieck's Tohoku paper rewrites homological algebra; he proves Grothendieck duality (i.e., Serre duality for possibly singular algebraic varieties).
1957 onwards: Grothendieck extends sheaf theory in line with the needs of algebraic geometry, introducing: schemes and general sheaves on them, local cohomology, derived categories (with Verdier), and Grothendieck topologies. There emerges also his influential schematic idea of 'six operations' in homological algebra.
1958 Roger Godement's book on sheaf theory is published. At around this time Mikio Sato proposes his hyperfunctions, which will turn out to have sheaf-theoretic nature.
At this point sheaves had become a mainstream part of mathematics, with use by no means restricted to algebraic topology. It was later discovered that the logic in categories of sheaves is intuitionistic logic (this observation is now often referred to as Kripke–Joyal semantics, but probably should be attributed to a number of authors).
See also
[edit]
Coherent sheaf
Gerbe
Stack (mathematics)
Sheaf of spectra
Perverse sheaf
Presheaf of spaces
Constructible sheaf
De Rham's theorem
Notes
[edit]
^ Eisenbud, David; Harris, Joe (6 April 2006), The Geometry of Schemes, GTM, New York, NY: Springer, pp. 11–18, ISBN 978-0-387-22639-2
^ Tennison, B. R. (1975), Sheaf theory, Cambridge University Press, MR 0404390
^ Bredon (1997, Chapter V, §1)
^ Demailly, Jean-Pierre. "Complex Analytic and Differential Geometry" (PDF). Archived (PDF) from the original on 28 August 2020.
^ Cartan, Henri. "Variétés analytiques complexes et cohomologie" (PDF). Archived (PDF) from the original on 8 October 2020.
^ a b "differential geometry - Holomorphic functions on a complex compact manifold are only constants". Mathematics Stack Exchange. Retrieved 2020-10-07.
^ Hawley, Newton S. (1950). "A Theorem on Compact Complex Manifolds". Annals of Mathematics. 52 (3): 637–641. doi:10.2307/1969438. JSTOR 1969438.
^ SGA 4 II 3.0.5
^ Iversen (1986, Chapter VII)
^ Ramanan (2005)
^ Hartshorne (1977), Theorem III.5.1.
^ Deligne, Pierre (1971). "Théorie de Hodge : II". Publications Mathématiques de l'IHÉS. 40: 5–57. doi:10.1007/BF02684692. S2CID 118967613.
^ Deligne, Pierre (1974). "Théorie de Hodge : III". Publications Mathématiques de l'IHÉS. 44: 5–77. doi:10.1007/BF02685881. S2CID 189777706.
^ Iversen (1986, Chapter VII, Theorem 1.4)
^ Kashiwara & Schapira (1994, Chapter III, §3.1)
^ de Cataldo & Migliorini (2010)
^ Grothendieck. "Formalisme des intersections sur les schema algebriques propres".
^ Steenrod, N. E. (1943). "Homology with Local Coefficients". Annals of Mathematics. 44 (4): 610–627. doi:10.2307/1969099. JSTOR 1969099.
^ Dieudonné, Jean (1989). A history of algebraic and differential topology 1900–1960. Birkhäuser. pp. 123–141. ISBN 978-0-8176-3388-2.
^ Cartan, Henri; Serre, Jean-Pierre (1953). "Un théorème de finitude concernant les variétés analytiques compactes". Comptes Rendus Hebdomadaires des Séances de l'Académie des Sciences de Paris. 237: 128–130. Zbl 0050.17701.
^ Serre, Jean-Pierre (1955), "Faisceaux algébriques cohérents" (PDF), Annals of Mathematics, Second Series, 61 (2): 197–278, doi:10.2307/1969915, ISSN 0003-486X, JSTOR 1969915, MR 0068874
^ Zariski, Oscar (1956), "Scientific report on the second summer institute, several complex variables. Part III. Algebraic sheaf theory", Bulletin of the American Mathematical Society, 62 (2): 117–141, doi:10.1090/S0002-9904-1956-10018-9, ISSN 0002-9904
^ Grothendieck, Alexander (1957), "Sur quelques points d'algèbre homologique", The Tohoku Mathematical Journal, Second Series, 9 (2): 119–221, doi:10.2748/tmj/1178244839, ISSN 0040-8735, MR 0102537
References
[edit]
Bredon, Glen E. (1997), Sheaf theory, Graduate Texts in Mathematics, vol. 170 (2nd ed.), Springer-Verlag, ISBN 978-0-387-94905-5, MR 1481706 (oriented towards conventional topological applications)
de Cataldo, Andrea Mark; Migliorini, Luca (2010). "What is a perverse sheaf?" (PDF). Notices of the American Mathematical Society. 57 (5): 632–4. arXiv:1004.2983. Bibcode:2010arXiv1004.2983D. MR 2664042.
Godement, Roger (2006) , Topologie algébrique et théorie des faisceaux, Paris: Hermann, ISBN 2705612521, MR 0345092
Grothendieck, Alexander (1957), "Sur quelques points d'algèbre homologique", The Tohoku Mathematical Journal, Second Series, 9 (2): 119–221, doi:10.2748/tmj/1178244839, ISSN 0040-8735, MR 0102537
Hirzebruch, Friedrich (1995), Topological methods in algebraic geometry, Classics in Mathematics, Springer-Verlag, ISBN 978-3-540-58663-0, MR 1335917 (updated edition of a classic using enough sheaf theory to show its power)
Iversen, Birger (1986), Cohomology of sheaves, Universitext, Springer, doi:10.1007/978-3-642-82783-9, ISBN 3-540-16389-1, MR 0842190
Kashiwara, Masaki; Schapira, Pierre (1994), Sheaves on manifolds, Grundlehren der Mathematischen Wissenschaften [Fundamental Principles of Mathematical Sciences], vol. 292, Springer-Verlag, ISBN 978-3-540-51861-7, MR 1299726 (advanced techniques such as the derived category and vanishing cycles on the most reasonable spaces)
Mac Lane, Saunders; Moerdijk, Ieke (1994), Sheaves in Geometry and Logic: A First Introduction to Topos Theory, Universitext, Springer-Verlag, ISBN 978-0-387-97710-2, MR 1300636 (category theory and toposes emphasised)
Martin, William T.; Chern, Shiing-Shen; Zariski, Oscar (1956), "Scientific report on the Second Summer Institute, several complex variables", Bulletin of the American Mathematical Society, 62 (2): 79–141, doi:10.1090/S0002-9904-1956-10013-X, ISSN 0002-9904, MR 0077995
Ramanan, S. (2005), Global calculus, Graduate Studies in Mathematics, vol. 65, American Mathematical Society, doi:10.1090/gsm/065, ISBN 0-8218-3702-8, MR 2104612
Seebach, J. Arthur; Seebach, Linda A.; Steen, Lynn A. (1970), "What is a Sheaf", American Mathematical Monthly, 77 (7): 681–703, doi:10.1080/00029890.1970.11992563, MR 0263073, S2CID 203043621
Serre, Jean-Pierre (1955), "Faisceaux algébriques cohérents" (PDF), Annals of Mathematics, Second Series, 61 (2): 197–278, doi:10.2307/1969915, ISSN 0003-486X, JSTOR 1969915, MR 0068874
Swan, Richard G. (1964), The Theory of Sheaves, Chicago lectures in mathematics (3 ed.), University of Chicago Press, ISBN 9780226783291 {{citation}}: ISBN / Date incompatibility (help) (concise lecture notes)
Tennison, Barry R. (1975), Sheaf theory, London Mathematical Society Lecture Note Series, vol. 20, Cambridge University Press, ISBN 978-0-521-20784-3, MR 0404390 (pedagogic treatment)
Rosiak, Daniel (2022). Sheaf theory through examples. Cambridge, Massachusetts. doi:10.7551/mitpress/12581.001.0001. ISBN 978-0-262-37042-4. OCLC 1333708310. S2CID 253133215.{{cite book}}: CS1 maint: location missing publisher (link) (introductory book with open access)
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6987 | https://www.youtube.com/watch?v=WH4MueuinYo | Locus of the image of the point (2,3) in the line (2 x-3 y+4)+k(x-2 y+3)=0, k ∈ R, is a
(1) circl...
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Locus of the image of the point (2,3) in the line (2 x-3 y+4)+k(x-2 y+3)=0, k ∈ R, is a
(1) circle of radius √(2)
(2) circle of radius √(3)
(3) straight line parallel to x-axis
(4) straight line parallel to y-axis
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हेलो बच्चों लेट्स रीड दिस क्वेश्चन द क्वेश्चन सेज लोकस ऑफ द इमेज ऑफ द पॉइंट 2 3 इन द लाइन 2x - 3y प् 4 प् के एकस - 2y प् 3 = 0 के बिलोस टू रियल नंबर इज अ फर्स्ट सर्कल ऑफ रेडियस रट सेकंड सर्कल ऑफ रेडियस रथ थर्ड स्ट्रेट लाइन पैरेलल टू एक्स एक्सिस फोथ स्ट्रेट लाइन पैरेलल टू वा एक्सिस तो हम देख लेते की कांसेप्ट की कांसेप्ट तो की कांसेप्ट है कि अगर आपके पास लाइन एलव हो एट ठीक है एव इ 0 और एट इ 0 दो लाइन की इक्वेशन लिखी हो तो एलव प्स के ए2 इ 0 एक और लाइन की इक्वेशन लिखी हो तो एलव और एलटू का एक इंटरसेक्शन पॉइंट मिलेगा यहां से अगर मैं इंटरसेक्शन इंटर सेक्शन पॉइंट ए मिले तो जो ए पॉइंट है वो इस एव प् के ए2 इ 0 को भी सेटिस्फाई करेगा a विल सेटिस्फाई विल सेटिस फाई दिस इक्वेशन आल्सो ठीक है सिंपल है अब देखो इसके अलावा हम कांसेप्ट देख लेते हैं कि डिस्टेंस फॉर्मूला क्या होता है डिस्टेंस डिस्टेंस फॉर्मूला तो डिस्टेंस फॉर्मूला होता है कि अगर आपके पास दो पॉइंट्स है x1 y1 x2 y2 तो उनके बीच का डिस्टेंस होता है x2 - x1 का होल स्क्वायर प् y2 - y1 का होल स्क्वायर ठीक है इसके अलावा हम कांसेप्ट देख लेते हैं कि अगर आपके पास कोई लाइन दिया हो और एक पॉइंट दिया हो ठीक है ये लाइन मान लिया ए है और ये पॉइंट a है तो a का इमेज इस लाइन के रिस्पेक्ट में लेट्स सपोज बी बना तो ए जो होगा वो ए के परपेंडिकुलर होगा और ये पॉइंट अगर मैं o मान लू तो हम यहां से बोल सकते हैं a = o ठीक है तो ये इन सारे कांसेप्ट का यूज करके हम क्वेश्चन को सॉल्व करेंगे अब हमारे पास इक्वेशन ऑफ लाइन दी हुई है 2x - 3y + 4+ k x - 2y + 3 = 0 तो यहां से हमें दो लाइंस और मिल रही है एक l1 मान ले तो वो मिल रहा है हमें 2x - 3y + 4 2x - 3y + 4 = 0 ठीक है दूसरा हमें मिल रहा है x - 2y + 3 l2 अगर हम इसको बोले तो x - 2y + 3 = जीरो ठीक है तो अब हमें दो इक्वेशन मिल गई तो हमने कांसेप्ट देखा कि l1 ए2 ऐसे दो इक्वेशन लेख उनका इंटरसेक्शन पॉइंट अगर ए मिल रहा है तो वो ए पॉइंट l1 प् के ए2 को भी सेटिस्फाई करेगा तो उसी कांसेप्ट के हिसाब से अगर मैं इसका इंटरसेक्शन पॉइंट निकालू तो कैसे निकलेगा यहां लिख ले इंटरसेक्शन इंटरसेक्शन पॉइंट ऑफ ए1 एंड ए2 इज ए ठीक है ए पॉइंट मा लिया अब यहां पर देखते हैं कैसे निकलेगा सिंपली इसको अगर मैं मान लू फर्स्ट इक्वेशन इसको मान लू सेकंड इक्वेशन तो मैं यहां क्या कर सकता हूं फर्स्ट इक्वेशन माइन सेकंड इक्वेशन 2 कर सकता हूं तो यहां से क्या बन रहा है 2x - 3y + 4 - 2x + 4y - 6 = 0 2x 2x कैंसिल आउट हो गया यहां आ गया y = 2 के बराबर y की वैल्यू टू आ गई अब इसको मैं फर्स्ट इक्वेशन में अगर पुट कर दूं यहां लिख दूं कि पुट वैल्यू ऑफ y इन फर्स्ट इक्वेशन तो क्या मिल जाएगा हमें 2x - 6 प् 4 = 0 यानी x की वैल्यू यहां से आ गई वन के बराबर ठीक है x1 आ गया y इ 2 आ गया यानी ए पॉइंट का कोऑर्डिनेट मैं लिख सकता हूं सिंपली वमा 2 अबर हम एक डायग्राम बनाते हैं छोटा सा तो मान लो ये ए लाइन है ठीक है ए ये l1 प् के ए2 वाली लाइन है ठीक है इ 0 तो यहां पर एक पॉइंट होगा a यहां पर एक पॉइंट होगा p और इसके रिस्पेक्ट में इमेज यहां पर बन रहा होगा q ठीक है p पॉइंट हमें दिया हुआ है 2 3 लोकस ऑफ इमेज ऑफ द पॉइंट 2 3 तो p पॉइंट को अगर हम मान ले 2 3 h q पॉइंट को h के तो बेसिकली h के का ही लोकस हमें बताना है ठीक है और a पॉइंट हमने निकाल लिया कितना 1 2 तो यहां लिख लेते हैं इसको 1 2 ए पॉइंट आ गया देखो इसको कनेक्ट कर दो अगर ठीक है यहां पर ये कनेक्ट कर दिया अब हमने एक कांसेप्ट देखा कि अगर प क प इमेज है बना रहा है एल लाइन के रिस्पेक्ट में q तो प क जो लाइन होगा वो परपेंडिकुलर होगा ए1 प् के2 = 0 लाइन के और यहां पर अगर पॉइंट है लेट्स सपोज o तो इ क होगा ठीक है तो इस कांसेप्ट के हिसाब से अगर हम देखे तो हम ले सकते हैं कि ट्रायंगल एओ ए एंड ट्रायंगल ए क और ओ ठीक है अब इसमें आप देखो तो एंगल प ए जो होगा वो एंगल ओ ए इ 90 डिग्री होगा एओ साइड जो होगा वो एओ साइड के बराबर होगा कॉमन साइड है और पीओ जो होगा वो क के बराबर होगा हमने कांसेप्ट में देख लिया तो यहां से आप देखो तो ये क्या हो रहा है ए ए क्रुए स क्रुंस क्राइटेरिया क्राइटेरिया का यूज कर सकते हैं और हम बोल सकते हैं कि एओ कांग्रा एओ के ठीक है अगर दोनों बराबर है तो हम यहां बोल सकते हैं कि बाई सी प सीटी यानी कांग्रुअस ऑफ कांग्रुअस कांग्रुअस बरा होगा ए a क के बराबर हो गया ठीक है अब हम क्या करेंगे कि a पॉइंट पता है p पॉइंट पता है a पॉइंट पता है q पॉइंट पता है तो डिस्टेंस फार्मूला का यूज कर लेंगे ए = a क तो ए हम लिख सकते हैं यहां पर रट अंडर x2 - x1 का स्क्वा यानी ए के लिए हो जाएगा 2 -1 का स्क्वायर यानी 1 का स्क्वायर प् 3 - 2 का स्क्वा यानी 1 का स्क्वायर इक्व टू रूट अंडर यहां बन गया h -1 काल स्क्वा प् k - 2 का होल स्क्वायर तो यहां से मैं डायरेक्ट लिख सकता हूं h - 1 स् + k - 2 स् = 2 तो ये हो गया हमारा अ यहां पे इक्वेशन बना तो अब इस चीज को आप ध्यान से देखो तो ये एक सर्कल के सर्कल का इक्वेशन लिखा हुआ है तो अगर मैं h और k को रिप्लेस कर दो x और y से तो ये बन जाएगा x - 1 का स् + y - 2 का स् = 2 और 2 को मैं लिख सकता हूं √2 का होल स्क्वायर तो ये बेसिकली हो गया सर्कल का इक्वेशन जहां पर रेडियस जो है रेडियस रेडियस इ इक्वल ू जाएगा रटू तो फाइनल ंस हम क्या बोलेंगे सो द फाइनल आंसर इ ऑप्शन वन सर्कल ऑफ रेडियस रटू होप यू अंडरस्टूड इट वेल बेस्ट ऑफ लक थैंक यू |
6988 | https://artofproblemsolving.com/wiki/index.php/Incenter?srsltid=AfmBOoqcB3j3OqZtjtbkoqfgYlv9tUn4U9xb7QnXpE7U9nn4v6n64obO | Art of Problem Solving
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Triangle ABC with incenterI, with angle bisectors (red), incircle (blue), and inradii (green)
The incenter of a triangle is the intersection of its (interior) angle bisectors. The incenter is the center of the incircle. Every nondegenerate triangle has a unique incenter.
Proof of Existence
Consider a triangle . Let be the intersection of the respective interior angle bisectors of the angles and . We observe that since lies on an angle bisector of , is equidistant from and ; likewise, it is equidistant from and ; hence it is equidistant from and and and therefore lies on an angle bisector of . Since it lies within the triangle , this is the interior angle bisector of . Since is equidistant from all three sides of the triangle, it is the incenter.
It should be noted that this proof parallels that for the existence of the circumcenter.
The proofs of existence for the excenters is the same, except that certain angle bisectors are exterior.
Properties of the Incenter
The incenter of any triangle lies within the orthocentroidal circle.
The unnormalised areal coordinates of the incenter are
Let be a point on the circumcircle of such that bisects . Then points , , and lie on a circle centered at .
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6989 | https://www.cs.upc.edu/~antoni/fix.pdf | Symmetry Breaking in Tournaments Antoni Lozano Logic and Programming Research Group, Technical University of Catalonia Abstract. We provide upper bounds for the determining number and the metric dimension of tournaments. A set of vertices S ⊆V (T ) is a determining set for a tournament T if every nontrivial automorphism of T moves at least one vertex of S, while S is a resolving set for T if every two distinct vertices in T have different distances to some vertex in S. We show that the minimum size of a determining set for an order n tournament (its determining number) is bounded by ⌊n/3⌋, while the minimum size of a resolving set for an order n strong tournament (its metric dimension) is bounded by ⌊n/2⌋. Both bounds are optimal.
1 Introduction The idea of somehow fixing the vertices of a graph in order to destroy all its non-trivial automorphisms has captured the attention of reserchers in recent years, leading to theoretical results for different graph families (trees , Kneser graphs , digraphs , wheels , or hypercubes ) and to applications in areas such as robotics or chemistry .
One can distinguish vertices in a graph by considering their distances to a given subset of fixed vertices: If every two distinct vertices of the graph have different distances to some vertex in the chosen subset, no nontrivial automorphism can be left. In a more nonconstructive approach, one can just fix a given set of vertices so that no nontrivial automorphism is possible in the graph without caring whether it is due to distances or not. The set fixed in the first approach is known as a resolving set [19, 11], while the set fixed in the second one has been called fixing set and determining set . In both cases, it is desirable to find a set of minimum size which “fixes” or “destroys” all nontrivial automorphisms in a graph, making the automorphism group of the resulting structure —say, a labeled graph— trivial.
This way, the sizes of the smallest determining and resolving sets for a graph can be seen as two parameters on its degree of symmetry.
Still another way that has been considered for fixing the vertices in a graph is by a coloring (called distinguishing coloring ) such that the only automor-phism which preserves colors is the identity. Again, the problem here is to find a distinguishing coloring where the number of colors is minimized. In order to make reference to this notion, Albertson and Collins introduced the term symmetry breaking, which will be used here to refer to the three preceeding approaches as a whole, while the associated parameters to be minimized will be called symmetry parameters.
Tournament graphs have been extensively studied (for classical references, see [12, 15]). Some of their well-known properties (for example, the fact that their automorphism groups have odd order and, hence, are solvable) may help to get stronger conclusions on symmetry breaking in tournaments than in general directed graphs. In this paper, we consider determining and resolving sets for tournaments and prove the existence of optimal upper bounds for their minimum sizes amount-ing to constant fractions of their order: 1 3 for determining sets, and 1 2 for resolving sets of strong tournaments.
1.1 Tournaments If D is a directed graph (digraph for short) V (D) and A(D) will denote the vertex set and the arc set of D. Given two distinct vertices u, v of D, we will write u ∈D instead of u ∈V (D) and uv ∈A(D) rather than (u, v) ∈A(D). The notation dD(x, y) (or just d(u, v) when D is understood) stands for the directed distance from u to v in D, that is, the length of the shortest directed path from u to v in D.
A tournament models the outcome of a competition where every player plays against each other in a 2-player match. Formally, a digraph T is a tournament if between every pair of distinct vertices u, v ∈T, we have uv ∈T or vu ∈T but not both. For any tournament T, define the following subsets of its vertices for any u, v ∈T: – Tu→= {w ∈T | uw ∈T} – T→u = {w ∈T | wu ∈T} – Tuv→= Tu→∩Tv→ – Tu→j = Tu→∩T→v – T→uv = T→u ∩T→v The indegree of a vertex u is |T→u| while its outdegree is |Tu→|. A tournament T is regular if the indegree (and, consequently, the outdegree) of all vertices is the same. In the case of a regular tournament T of order 2n + 1, it is easy to see that |Tu→| = |T→u| = n for every vertex u ∈T.
A tournament T is transitive if its vertices can be numbered from 1 up to n in such a way that every arc ij in T satisfies i < j. Therefore, there is a unique transitive tournament of order n up to isomophism, which is denoted by TTn. Since the existence of a directed 3-cycle —which we will denote by C3— contradicts the condition in the definition of TTn, a characterization of transitive tournaments, then, is precisely that they do not contain the subtournament C3. An almost transitive tournament of order n ≥3, denoted by TT ∗ n, is a tournament that differs from TTn in the fact that it contains the arc n1 instead of 1n, while the rest of the arcs remain unchanged.
A tournament is strong if all directed distances between vertices are defined. It is obvious that while TTn is not strong, TT ∗ n is a strong tournament.
2 1.2 Symmetry Parameters Given a digraph D, we denote by Γ(D) the automorphism group of D and by S1 the trivial group. We note the well-known fact that every automorphism in Γ(D) is an isometry, that is, for any u, v ∈D and φ ∈Γ(D), d(u, v) = d(φ(u), φ(v)).
We say that an automorphism φ fixes a vertex u if φ(u) = u and that it fixes a set S ⊆V (D) if it fixes every u ∈S. The set of automorphisms of D that fix the set S is a subgroup of Γ(D) called pointwise stabilizer of S in D, and we denote it by StabΓ (D)(S) (or just Stab(S) when D is understood). Given a vertex u of D, the set OD(u) = {φ(u) | φ ∈Γ(D)} is called the orbit of u in D. Orbits of D induce a partition in the set of vertices. Note that no automorphism can move a vertex from an orbit to another.
Definition 1. A set S ⊆V (D) for which StabΓ (D)(S) = S1 is called a determin-ing set of D, and the minimum cardinality of a determining set for D, denoted by Det(D), is called determining number of D.
A digraph with no nontrivial automorphisms is called rigid. Determining sets (also known as fixing sets) have been introduced independently by Harary and Boutin . As a simple example, a determining set for C3 is any set containing one vertex, while the empty set is a determining set for TTk, for any k, since any transitive tournament is rigid. Also note that any set with n −2 vertices of a tournament of order n is a determining set, since no automorphism can exchange the remaining two vertices (tournaments do not have involutions).
In a different approach introduced separately by Harary and Melter , and by Slater , the vertices in a graph can be distinguished according to their distances to a given subset of vertices.
Definition 2. We say that a vertex u of a digraph D resolves a pair of vertices v, w ∈D if d(v, u) ̸= d(w, u). We also say that S ⊆V (D) is a resolving set for D if every pair of vertices in D can be resolved by some vertex in S. The minimum size of a resolving set for D is referred to as the metric dimension of D, and is denoted by Dim(D).
While the metric dimension of a digraph does not need to be defined, the situation in the case of tournaments is different. We would like to stress the fact that Definition 2 does not require all directed distances to be defined (not even the directed distances to all vertices in the resolving set), but only the distances to the vertices which resolve each pair of distinct vertices.
Proposition 1. For any order n tournament T, Dim(T) ≤n −1.
Proof. It is well known that every tournament has a Hamiltonian path . Sup-pose, then, that u1 . . . un is a Hamiltonian path for a tournament T. Then, given two vertices ui, uj such that i < j, they can be resolved by uj since d(ui, uj) > d(uj, uj) = 0. Therefore, {u2, . . . , un} is a resolving set.
⊓ ⊔ 3 We observe that the upper bound of Proposition 1 is best possible for general tournaments.
Proposition 2. For every integer n, Dim(TTn) = n −1.
Proof. Let V (TTn) = {1, . . . , n} be the numbering of the vertices from the defini-tion of TTn given in Subsection 1.1 (that is, every arc ij satisfies i < j), and let S be a determining set for TTn. Then, given two vertices i, j such that i < j, we have that – vertex j resolves the pair since d(i, j) = 1 and d(j, j) = 0, – no vertex k < j can resolve it since distance d(j, k) is undefined, and – no vertex k > j can resolve it neither since d(i, k) = d(j, k) = 1.
We conclude that j is the only vertex which resolves a pair i, j such that i < j, and then, must belong to S. Therefore, the only vertex that can be left out of S is 1, and we have Dim(TTn) ≥n −1. On the other hand, Dim(TTn) ≤n −1 by Proposition 1, and the result is proved.
⊓ ⊔ To establish the relation between resolving and determining sets, we just follow the proof by Erwin and Harary (, Lemma 2), which can be applied without changes to digraphs and we reproduce here for completeness.
Proposition 3. Every resolving set for a digraph is also a determining set. There-fore, for any digraph D, Det(D) ≤Dim(D).
Proof. Let S be a resolving set for a digraph D. To get a contradiction, suppose that S is not a determining set for D and, then, Stab(S) ̸= S1. In this case, there must be a vertex v and an automorphism φ ∈Stab(S) such that v ̸= φ(v). Since S is a resolving set for D, there must be a vertex u ∈S for which d(v, u) ̸= d(φ(v), u).
However, d(v, u) = d(φ(v), φ(u)) = d(φ(v), u) where the first equality holds because φ is an isometry, and the second one because φ fixes u, and we have a contradiction.
⊓ ⊔ 1.3 Outline In Section 2, we prove that the determining number of every tournament of order n is tightly bounded by ⌊n/3⌋.
Section 3 is devoted to the metric dimension. We derive a tight upper bound of ⌊n/2⌋in the case of strong tournaments of order n (as we have just seen, there is no constant fraction upper bound in the general case). In order to prove this bound, we first answer a conjecture posed by Kannan, Naor, and Rudich on their concept of anchor (), and link it to the metric dimension. We also show in Section 3 that resolving sets and anchors are exactly the same concept in regular tournaments.
Finally, in the light of our results, we remind in Section 4 a conjecture posed by Albertson and Collins () on the distinguishing number of tournaments.
4 2 The Determining Number The determining number is 0, for exemple, for the transitive tournament of k vertices, TTk, (in fact, for any rigid tournament) but how large can it be for a general tournament? Define TTk(T) as TTk where each vertex has been substituted by a copy of T.
Proposition 4. For every n > 0, there is an order n tournament Tn such that Det(Tn) = ⌊n/3⌋.
Proof. If n = 3k for some integer k, we take Tn = TTk(C3). Consider any set S ⊆V (Tn) containing exactly one vertex from each copy of C3 in Tn. Since no vertex can be moved by an automorphism between different copies of C3, and every nontrivial automorphism of one of the copies must move all three vertices, we have that Stab(S) = S1, and S is a determining set for Tn with |S| = k. Observe that any set of cardinality smaller than k would fail to contain some vertex from one of the copies of C3, and that would make it possible to rotate C3 while fixing the rest of the vertices in Tn.
If n ≡d modulo 3 with d = 1 or d = 2, we take Tn as the tournament Tn−d defined above plus d new vertices pointing to all vertices in Tn−d. Since the new vertices must be fixed in any nontrivial automorphism of Tn, Det(Tn) = Det(Tn−d) = (n −d)/3 = ⌊n/3⌋.
⊓ ⊔ Now we will see that there is always a determining set with at most ⌊n/3⌋ vertices for an n order tournament. In the proof, we exploit the fact that an automorphism cannot move vertices between different orbits and, then, in order to construct a determining set for T, it is enough to construct determining sets for the subtournaments induced by the orbits, and then merge them. We will need the following proposition for digraphs, which is the direct translation of a similar one stated in for graphs and does not require a separate proof.
Proposition 5. Let O1, . . . , Ok be the vertex orbits of a digraph D. Let H1, . . . , Hk be the associated induced subtournaments. Let S1, . . . Sk be determining sets for H1, . . . , Hk. Then, S = S1 ∪· · · ∪Sk is a determining set for D.
Now, we can state the main theorem of this section.
Theorem 1. For every order n tournament T, Det(T) ≤⌊n/3⌋.
Proof. Given a tournament T of order n, we will show that the subtournament T[O] induced by an orbit O of T has a determining set SO of size at most ⌊|O|/3⌋. Therefore, the union of all determining sets for the different orbits S = S O is an orbit of T SO will satisfy that |S| ≤⌊n/3⌋while, by Proposition 5, S is a determining set for T. Thus, it is enough to prove the statement separetely for each orbit.
5 Suppose then that O is an orbit of T, and that |O| = m. Let H = T[O] be the subtournament of T induced by O. The fact that O is an orbit implies that H must be vertex transitive (there exists φ ∈Γ(H) such that φ(u) = v for any u, v ∈H) and, hence, regular.
We will proceed by induction on m. For m ≤2, the determining set can be empty since there is no nontrivial automorphism. For the inductive step, we will suppose that m ≥3, but note that in the case m = 3, the only tournaments are TT3 (which is already rigid) and C3 (which can be made rigid by fixing one vertex).
Now, for m ≥3, we consider three cases depending on the remainder of m when dividing by 3. All congruencies in the rest of the proof are taken modulo 3.
Case 1: m ≡0.
Let u be any vertex in H. Since H is regular, |H→u| = |Hu→|, which together with the fact that |H→u| + |Hu→| + 1 = m ≡0 leaves the only possibility that |H→u| = |Hu→| ≡1.
Therefore, there exists an integer r such that |H→u| = |Hu→| = 3r + 1. By induction hypothesis, both H→u and Hu→have determining sets of size r. Their union plus vertex u form a determining set for H of size 2r + 1. Since |H| = m = 2(3r + 1) + 1 = 6r + 3, the determining set has at most (in this case, exactly) ⌊m/3⌋vertices as expected.
Case 2: m ≡2.
Let u be any vertex in H. As in the previous case, |H→u| = |Hu→|, which now implies that |H→u| = |Hu→| ≡2. Therefore, there exists an integer r such that |H→u| = |Hu→| = 3r + 2. By induction hypothesis, both H→u and Hu→have determining sets of size r. Their union plus vertex u form a determining set for H of size 2r + 1. Since |H| = m = 2(3r + 2) + 1 = 6r + 5, the determining set has at most (in this case, less than) ⌊m/3⌋vertices.
Case 3: m ≡1.
In this case, for any u ∈H, |H→u| = |Hu→| ≡0. A determining set for H constructed inductively as in the previous cases would have more than ⌊m/3⌋ vertices. However, we can complete the proof using two subcases: – Subcase 3.1: For every arc uv ∈H, |Hu→v| ≡2.
Since we are assuming that m ≥3, let u, v, w be three vertices in H. Moreover, we can select them in such a way that uv, vw, and wu are in H, the reason being that H cannot be transitive in this subcase and, therefore, it must contain a 3-cycle. Now, H can be split into the following subsets: • Hu→v, Hv→w, and Hw→u, whose cardinalities are all congruent with 2 mod-ulo 3. Let |Hu→v| = 3r + 2, |Hv→w| = 3s + 2, and |Hw→u| = 3t + 2 for some r, s, t ≥0, • the set {u, v, w}, and 6 • the remaining vertices in H, say H′. Note that since m ≡1 and the number of vertices counted so far is multiple of 3, |H′| ≡1, say |H′| = 3z + 1 for some z.
u v w H v w H w u H u v Fig. 1. Subcase 3.1: |Hu→v| ≡|Hv→w| ≡|Hw→u| ≡2.
Summing up the previous numbers, we have: m = |Hu→v| + |Hv→w| + |Hw→u| + |{u, v, w}| + |H′| = = (3r + 2) + (3s + 2) + (3t + 2) + 3 + (3z + 1) = = 3(r + s + t + z) + 10.
On the other hand, we can assume the existence of determining sets for Hu→v, Hv→w, Hw→u, and H′ by induction hypothesis, of sizes r, s, t, and z, respec-tively. Their union plus the vertices u, v, and w is a determining set for H with less than ⌊m/3⌋vertices.
– Subcase 3.2: There is an arc uv ∈H for which |Hu→v| ̸≡2.
Let u, v be two such vertices. Now, we consider the partition of H into the sets: Hu→v, H→uv, Hv→u, Huv→, and {u, v}. Using the assumption (of Case 3) that for any w ∈H, |H→w| = |Hw→| ≡0, we can observe the following: i. |Hu→v ∪H→uv| ≡2, since the previous union plus v is exactly H→v.
ii. |H→uv ∪Hv→u| ≡0, since the previous union is exactly Hv→.
iii. |Hv→u ∪Huv→| ≡2, since the previous union plus v is exactly Hu→.
Now, we can see that there are only two possibilities: • |Hu→v| ≡1. Then, by i, |H→uv| ≡1; by ii, |Hv→u| ≡2; and by iii, |Huv→| ≡ 1. There must be some r, s, t, z in this case such that |Hu→v| = 3r + 1, |H→uv| = 3s + 1, |Hv→u| = 3t + 2, and |Huv→| = 3z + 1.
• |Hu→v| ≡0. Then, by i, |H→uv| ≡2; by ii, |Hv→u| ≡1; and by iii, |Huv→| ≡ 2. There must be som r, s, t, z in this case such that |Hu→v| = 3r, |H→uv| = 3s + 2, |Hv→u| = 3t + 1, and |Huv→| = 3z + 2.
7 Hu–›v Huv–› Hv–›u H–›uv u v 2 Fig. 2. Subcase 3.2: For some arc uv ∈H, |Hu→v| ̸≡2.
In any of the above two possibilities, H contains the four previous subsets plus u and v, giving |H| = 3(r + s + t + z) + 7. The union of the determining sets given by the induction hypothesis plus u and v gives a determining set for H of size r + s + t + z + 2, which is less than ⌊m/3⌋.
⊓ ⊔ Note that the upper bound ⌊n/3⌋given in Theorem 1 is tight by Proposition 4.
3 The Metric Dimension As we have seen in Section 1, if we do not consider any additional condition on a tournament of order n, its metric dimension can be as large as n−1. In this section, we show that just requiring a tournament to be strong, its metric dimension drops to at most ⌊n/2⌋.
We start with an example of a strong tournament having a large metric di-mension, and we will see later that it is indeed the largest possible for any strong tournament.
Proposition 6. For every n > 2, Dim(TT ∗ n) = ⌊n/2⌋.
Proof. Let V (TT ∗ n) = {1, . . . , n} be the numbering of the vertices from the defini-tion of TT ∗ n (that is, ij ∈T implies i < j except for the case of the arc n1, which belongs to TT ∗ n). Note that the even (odd) numbered vertices constitute a resolving set of size ⌊n/2⌋since every pair of odd (even) numbered vertices are distinguished by some even (odd) numbered vertex. This shows that Dim(TT ∗ n) ≤⌊n/2⌋.
To see why Dim(TT ∗ n) ≥⌊n/2⌋, we observe in the first place that it cannot happen that two vertices i, i + 1 in the range 1 < i < i + 1 < n do not belong to the resolving set, for no other vertex can resolve them. Moreover, if one of the vertices 2 or n −1 is not in the resolving set, then either 1 or n must be in. Both conditions imply that the resolving set must contain at least ⌊n/2⌋vertices.
⊓ ⊔ 8 Kannan, Naor, and Rudich introduced the notion of anchors for tourna-ments in order to get a simple algorithm for tournament isomorphism. Anchors can be seen as a sort of simplification of resolving sets having the advantage that they are always defined.
An anchor in a tournament T is a subset S ⊆V (T) such that for all vertices u, v ∈V (T) −S, u ̸= v, there exists a vertex w ∈S such that exactly one of uw and vw is an arc of T, in which case we say that w distinguishes the pair u, v.
Thus, if the vertices in an anchor are fixed, all vertices in the tournament would be fixed and no automorphism other than the identity would be possible; in this respect, the concept is similar to the other symmetry breaking concepts seen so far. Let Anchor(T) denote the size of the smallest anchor for T.
In , it is shown that for any order n tournament T, Anchor(T) ≤⌊2n/3⌋, and the authors conjecture that, in fact, Anchor(T) ≤⌊n/2⌋. We start proving here that their conjecture is correct.
Theorem 2. For every order n tournament T, Anchor(T) ≤⌊n/2⌋.
Proof. Let T be a tournament with n vertices. We will preceed by induction on n.
The statement is trivially true for n ≤2 by the definition of anchor. Now, suppose that n ≥3 and consider two cases depending on the parity of n: Case 1: n is even.
Select a vertex u from T. Now, V (T)−{u} can be split into the disjoint sets of vertices T→u and Tu→. Since |V (T) −{u}| is odd, one of |T→u| and |Tu→| is even and one is odd. Suppose without loss of generality that the first quantity is odd and equal to 2r + 1 and the second one is even and equal to 2s, for integers r, s.
Then, by induction hypothesis, T→u has an anchor A of size r, and Tu→has an anchor B of size s.
Note that the set of vertices S = A∪B ∪{u} has size r +s+1 = n/2. Further-more, S is an anchor since for any x, y / ∈S, we can consider three possibilities: – x, y ∈T→u. In this case, x and y are distinguished by some vertex in A.
– x, y ∈Tu→. Symmetrically, x and y are distinguished by some vertex in B.
– x ∈Tu→and y ∈T→u or viceversa. In this case, vertex u distinguishes x and y.
Case 2: n is odd.
Consider the parity of the sets Tu→, for u ∈V . If some of these sets has odd cardinality, then the corresponding set T→u must have odd cardinality too; suppose the respective cardinalities are 2r + 1 and 2s + 1 for some integers r, s.
By induction hypothesis, these two sets of vertices must have anchors of sizes r and s, respectively. Similarly to the previous case, the union of the anchors plus vertex u form an anchor for T of size r + s + 1. Note that, in this case, n = (2r + 1) + (2s + 1) + 1 = 2(r + s + 1) + 1, so the anchor has size ⌊n/2⌋, as required.
Consider now the case in which all sets Tu→, for u ∈V , have even cardinality.
Now, we consider two subcases: 9 – Subcase 2.1: There are two distinct vertices u, v in T s.t. |Tuv→| is odd.
We can assume, without loss of generality, that uv ∈T. We can now split the set of vertices Tv→into two sets: Tv→= Tv→u ∪Tuv→ Since the above union is disjoint, |Tuv→| is odd, and |Tv→| is even, then |Tv→u| must be odd. We can classify all the vertices in T different from u and v into the disjoint sets T→v −{u}, Tv→u, and Tuv→.
Since the last two sets have odd cardinality and the union of all three sets has cardinality n −2, which is odd, the first one must have odd cardinality too.
Suppose that, for some integers r, s, and t, |T→v −{u}| = 2r + 1, |Tv→u| = 2s + 1, |Tuv→| = 2t + 1.
u v Tuv Tv u T v-{u} Fig. 3. Subcase 2.1: |Tuv→|, |Tv→u|, and |T→v −{u}| are odd.
By induction hypothesis, T→v −{u}, Tv→u, and Tuv→must have anchors of respective sizes r, s, and t which we will call A, B, and C, resp. Note that the set S = A ∪B ∪C ∪{u, v} contains r + s + t + 2 vertices. Since n = 2(r + s + t) + 5 = 2(r + s + t + 2) + 1, |S| is exactly ⌊n/2⌋. Now, we will argue that S is an anchor for T. Given x, y / ∈S, we have the following possibilities for x and y: • In the case that both x and y belong to one of the sets T→v −{u}, Tv→u, or Tuv→, they can be distinguished by some vertex in the respective anchor: A, B, or C.
• If x or y is in T→v −{u} and the other one is in either Tv→u or Tuv→, they are distinguished by v.
• If x or y is in Tv→u and the other one in Tuv→, they are distinguished by u.
– Subcase 2.2: For all distinct vertices u, v in T, |Tuv→| is even.
Note that, in this case, the tournament T cannot be transitive since we can always select two “consecutive” vertices u, v in a transitive tournament such that |Tuv→| is odd. Then, T must contain a 3-cycle, similarly to Subcase 3.1 in the proof of Theorem 1.
10 Let u, v, and w be three vertices such that uv, vw, and wu are three arcs in T. Note that we can partition the vertices in Tu→, Tv→, and Tw→as follows: Tu→= Tu→v ∪Tuv→∪{v} Tv→= Tv→w ∪Tvw→∪{w} Tw→= Tw→u ∪Twu→∪{u} Our assumptions on cardinalities in this subcase applied to the above relations imply that the cardinalities of the sets Tu→v, Tv→w, and Tw→u must be odd, say |Tu→v| = 2r + 1, |Tv→w| = 2s + 1, and |Tw→u| = 2t + 1 for some integers r, s, and t. Note too that all three sets are disjoint. We now apply the induction hypothesis to the above sets and get an anchor A for Tu→v with size r, an anchor B for Tv→w with size s, and an anchor C for Tw→u with size t. Let now T ′ be the set containing the rest of the vertices, that is T ′ = V (T) −(Tu→v ∪Tv→w ∪Tw→u ∪{u, v, w}).
The set T ′ must have odd cardinality, say |T ′| = 2m + 1, and (by induction hypothesis) an anchor D of size m.
u v w Tv w Tw u Tu v Fig. 4. Subcase 2.2: |Tu→v|, |Tv→w|, and |Tw→u| are odd.
Consider the set S = A ∪B ∪C ∪D ∪{u, v, w}. Its size is r + s + t + m + 3, whereas n = 2(r+s+t+m)+7 = 2(r+s+t+m+3)+1; therefore, |S| = ⌊n/2⌋.
Now, we will argue that S constitutes an anchor for T. Given x, y / ∈S, we have the following possibilities for x and y: • In the case that both x and y belong to one of the sets Tu→v, Tv→w, Tw→u, or T ′, they can be distinguished by some vertex in the respective anchor: A, B, C, or D.
11 • In the case that x and y belong to different sets from Tu→v, Tv→w, Tw→u, then u, v, or w must distinguish x and y. For example, if x ∈Tu→v and y ∈Tv→w, they are distinguished by v, the rest of the cases being similar.
• In the case that one of x, y is in T ′ and the other one is in Tu→v, Tv→w, or Tw→u, then u, v, or w will distinguish x and y. For example, suppose that x ∈T ′ and y ∈Tu→v. Since vertices in T ′ are not in y ∈Tu→v, it means that either ux or xv cannot be an arc in T, which means that either xu or vx is an arch in T: in the first case, x and y are distinguished by u, in the second one, by v. The rest of the cases are similar.
⊓ ⊔ The previous upper bound for the size of anchors has some implications regard-ing the metric dimension. In the first place, we note the following relation between the two notions.
Proposition 7. Every anchor in a strong tournament is a resolving set.
Proof. Suppose that S is an anchor in a strong tournament T, and take two distinct vertices u, v ∈V (T) −S which are distinguished by a vertex w ∈S. According to the definition of anchor, T contains exactly one of the arcs uw and vw. If it contains uw (and then, wv), we have d(u, w) = 1 while d(v, w) exists (because T is strong) but must be greater than 1. Then, d(v, w) > d(u, w). The case when T contains the arc vw but not uw similarly implies that d(v, w) < d(u, w). In either case, d(u, w) ̸= d(v, w), and so w resolves u and v.
Since a resolving set for T must resolve any pair u, v of distinct vertices in T (not only in V (T) −S), suppose that at least one of them, say u, belongs to S. In this case, vertex u itself resolves the pair since d(u, u) = 0 < d(v, u).
⊓ ⊔ Now, combining Theorem 2 with Proposition 7, we get our main conclusion on the metric dimension of tournaments.
Corollary 1. For every order n strong tournament T, Dim(T) ≤⌊n/2⌋.
Note that the upper bound of ⌊n/2⌋for the metric dimension of strong tour-naments is tight by Proposition 6.
It is interesting to observe that resolving sets and anchors become the same notion when applied to regular tournaments.
Proposition 8. Let T be a regular tournament. Then: 1. Every anchor in T is a resolving set.
2. Every resolving set in T is an anchor.
Proof.
1. By the fact that regular tournaments are strong and Proposition 7.
12 2. Alspach proved in that each arc of a regular tournament of order n is contained in a cycle of each length k, k = 3, . . . , n. Taking k = 3, we can conclude any two distinct vertices u, v in T belong to a directed 3-cycle, and so we can assure that either d(u, v) = 1 or d(u, v) = 2. Then, a vertex w in a resolving set for T can only resolve two other vertices u, v by keeping distance 1 to one of them and 2 to the other; this means that T contains exactly one of the arcs uw, vw and, then, w distinguishes u and v. Thus, any resolving set is, in fact, an anchor.
⊓ ⊔ In the case of regular tournaments, we can also add a lower bound due to the fact that any anchor of an order n tournament must have at least ⌈log n⌉−1 vertices . Then, Dim(T) ≥⌈log n⌉−1 for every regular tournament T of order n.
4 A Conjecture Revisited The symmetry breaking notion of distinguishability introduced by Albertson and Collins in is very related to the notions studied here. A (not necessarily proper) d-coloring of a (di)graph G is said to be d-distinguishing if G has no nontrivial automorphism which preserves the colors. The distinguishing number D(G) of G is defined as the minimum d such that G has a d-distinguishing coloring. Adapting this notion to permutation groups, Seress showed that the distinguishing num-ber of solvable permutation groups is at most 5. Since the automorphism group of tournaments is solvable, we have: Observation. For every tournament T, D(T) ≤5.
However, Albertson and Collins state the following conjecture in .
Conjecture. If T is a tournament, then D(T) = 2.
By Theorem 3 in , we know that a tournament T contains a determining set whose induced subtournament is rigid if and only if D(T) = 2. We wonder if the determining sets constructed in Theorem 1 (more formally speaking, their induced subtournaments) can be made rigid by adding vertices in a way similar to the technique used in .
5 Acknowledgements The motivation for this research started while the author was visiting the Labora-toire de Rechereche en Informatique (Universit´ e Paris Sud) under an invitation of Miklos Santha. The author would like to thank Christoph D¨ urr for several helpful conversations on the topic of this paper, which has been partially supported by CICYT project TIN2007-68005-C04-03 and MICINN project TIN2008-06582-C03-01.
13 References 1. B. Alspach. Cycles of each length in regular tournaments. Canad. Math. Bull., 10: 283–286, 1967.
2. M. Albertson and D. Boutin. Using Determining Sets to Distinguish Kneser Graphs. Elec-tronic Journal of Combinatorics, 14: R20, 2007.
3. M. Albertson and K. Collins. Symmetry Breaking in Graphs. Electronic Journal of Combi-natorics, 3: R18, 1996.
4. M. Albertson and K. Collins. A Note on Breaking the Symmetries of Tournaments. Pro-ceedings of the Thirtieth Southeastern International Conference on Combinatorics, Graph Theory, and Computing. Congr. Numer. 136: 129-131, 1999.
5. D. Boutin. Identifying Graph Automorphisms Using Determining Sets. The Electronic Jour-nal of Combinatorics, 13, #R78, 2006.
6. G. Chartrand, L. Eroh, M. Johnson, O. Oellermann. Resolvability in graphs and the metric dimension of a graph. Discrete Applied Mathematics, 105: 99–113, 2000.
7. M. Chan. The distinguishing number of the augmented cube and hypercube powers. Discrete Mathematics, 308: 2330–2336, 2008.
8. G. Chartrand, M. Raines, P. Zhang. The Directed Distance Dimension of Oriented Graphs.
Mathematica Bohemica, 125(2): 155-168, 2000.
9. D. Erwin and F. Harary. Destroying Automorphisms by Fixing Nodes. Dicrete Mathematics, 306: 3244–3252, 2006.
10. F. Harary. Methods of Destroying the Symmetries of a Graph. Bull. Malaysian Math. Sci.
Soc., 24(2): 183–191, 2001.
11. F. Harary and R.A. Melter. On the metric dimension of a graph. Ars Combinatorica, 2: 191–195, 1976.
12. F. Harary and L. Moser. The Theory of Round Robin Tournamnets. The American Mathe-matical Monthly, 73(3): 231–246, 1966.
13. S. Kannan, M. Naor, and S. Rudich. Anchors in Tournaments. Technical Report TR 92-08, Dept. of Computer Science, University of Arizona, 1992.
14. S. Khuller, B. Raghavachari, A. Rosenfeld. Landmarks in Graphs. Discrete Appl. Math., 70: 217–229, 1996.
15. J. W. Moon. Topics on Tournaments. Holt, Rinehart and Winston, New York, 1968.
16. O. R. Oellermann, J. Peters-Fransen. The strong metric dimension of graphs and digraphs.
Discrete Applied Mathematics, 155: 356–364, 2007.
17. L. Redei. Ein kombinatoricher Satz. Acta Litt. Szeged, 7: 39–43, 1934.
18. ´ A. Seress. The minimal base size of primitive solvable permutation groups. London Math.
Soc., 53(2): 243–255, 1996.
19. P.J. Slater. Leaves of trees. Congr. Numer., 14: 549–559, 1975.
20. B. Shanmukha, B. Sooryanarayana, and K.S. Harinath. Metric Dimension of Wheels. Far East J. Appl. Math., 8(3): 217–229, 2002.
14 |
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Root-Mean Square-Arithmetic Mean-Geometric Mean-Harmonic mean Inequality
The Root-Mean Power-Arithmetic Mean-Geometric Mean-Harmonic Mean Inequality (RMP-AM-GM-HM) or Exponential Mean-Arithmetic Mean-Geometric Mean-Harmonic Mean Inequality (EM-AM-GM-HM) or Quadratic Mean-Arithmetic Mean-Geometric Mean-Harmonic Mean Inequality (QM-AM-GM-HM), is an inequality of the root-mean power, arithmetic mean, geometric mean, and harmonic mean of a set of positivereal numbers that says:
, where , and is the .
The geometric mean is the theoretical existence if the root mean power equals 0, which we couldn't calculate using radicals because the 0th root of any number is undefined when the number's absolute value is greater than or equal to 1. This creates the indeterminate form of . Then, we can say that the limit as x goes to 0 is the geometric mean of the numbers.
The quadratic mean's root mean power is 2 and the arithmetic mean's root mean power is 1, as and the harmonic mean's root mean power is -1 as . Similarly, there is a root mean cube (or cubic mean), whose root mean power equals 3.
When the root mean power approaches , the mean approaches the highest number. When the root mean power reaches , the mean approaches the lowest number.
with equality if and only if . This inequality can be expanded to the power mean inequality, and is also known as the Mean Inequality Chain.
As a consequence, we can have the following inequality: If are positive reals, then with equality if and only if ; which follows directly by cross multiplication from the AM-HM inequality. This is extremely useful in problem-solving.
The Root Mean Power of 2 is also known as the quadratic mean, and the inequality is therefore sometimes known as the QM-AM-GM-HM Inequality.
Proof
The inequality is a direct consequence of the Cauchy-Schwarz Inequality; Alternatively, the RMS-AM can be proved using Jensen's inequality: Suppose we let (We know that is convex because and therefore ). We have: Factoring out the yields: Taking the square root to both sides (remember that both are positive):
The inequality is called the AM-GM inequality, and proofs can be found here.
The inequality is a direct consequence of AM-GM; , so , so .
Therefore, the original inequality is true.
Geometric Proofs
The inequality is clearly shown in this diagram for
Desmos SlidersDesmos Equation NOTE: The Desmos equation will not show the line when the numbers are negative.
(Note how the RMS is "sandwiched" between the minimum and the maximum)
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The mitochondrial membrane permeability transition (MPT) is a Ca2+-dependent increase in the permeability of the mitochondrial membrane that leads to loss of Δψ, mitochondrial swelling, and rupture of the outer mitochondrial membrane.
From:Biochimica et Biophysica Acta (BBA) - Bioenergetics, 2006
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Review article
Mechanisms and pathways of mitochondrial outer membrane protein biogenesis
2021, Biochimica et Biophysica Acta (BBA) - BioenergeticsArushi Gupta, Thomas Becker
Abstract
Outer membrane proteins integrate mitochondria into the cellular environment. They warrant exchange of small molecules like metabolites and ions, transport proteins into mitochondria, form contact sites to other cellular organelles for lipid exchange, constitute a signaling platform for apoptosis and inflammation and mediate organelle fusion and fission. The outer membrane contains two types of integral membrane proteins. Proteins with a transmembrane β-barrel structure and proteins with a single or multiple α-helical membrane spans. All outer membrane proteins are produced on cytosolic ribosomes and imported into the target organelle. Precursors of β-barrel and α-helical proteins are transported into the outer membrane via distinct import routes. The translocase of the outer membrane (TOM complex) transports β-barrel precursors across the outer membrane and the sorting and assembly machinery (SAM complex) inserts them into the target membrane. The mitochondrial import (MIM) complex constitutes the major integration site for α-helical embedded proteins. The import of some MIM-substrates involves TOM receptors, while others are imported in a TOM-independent manner. Remarkably, TOM, SAM and MIM complexes dynamically interact to import a large set of different proteins and to coordinate their assembly into protein complexes. Thus, protein import into the mitochondrial outer membrane involves a dynamic platform of protein translocases.
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Journal2021, Biochimica et Biophysica Acta (BBA) - BioenergeticsArushi Gupta, Thomas Becker
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Strategy and Drug Research
2007, Comprehensive Medicinal Chemistry IIJ.A. Dykens
2.27.2.2 Membranes and Electron Transport Complexes
The outer, relatively permeable, mitochondrial membrane is reminiscent of endoplasmic reticulum in both lipid composition and transport physiology. Not so the inner membrane surrounding the central matrix space which contains the Kreb's cycle enzymes. The inner membrane contains approximately twice the amount of phosphatidylethanolamine, and half the amount of phosphatidylcholine, of the outer membrane, plus about 20% by weight cardiolipin (diphosphatidylglycerin), an atypical membrane lipid with four highly unsaturated acyl chains found in the cell almost exclusively in the mitochondrial inner membrane.49
Over 80% (by weight) of the inner membrane is composed of proteins, the highest density within the cell. In large measure, these proteins are the four respiratory complexes: C-I, NADH-CoQ oxidoreductase; C-II, succinate dehydrogenase-CoQ oxidoreductase; C-III, CoQ-cytochrome c oxidoreductase; and C-IV, cytochromec oxidase (COX), plus adenosine triphosphate (ATP) synthase, frequently referred to as Complex V, although unlike the other four complexes, it does not catalyze a redox reaction.
OxPhos includes two separate, but interdependent, processes: electron transfer system (ETS) and ATP synthase (Figures 3 and 4). The first entails the sequential redox transfer of electrons, removed from pyruvate in the Krebs cycle within the matrix, down a potential energy gradient via the four respiratory complexes, ultimately reducing molecular O 2 to H 2 O. In so doing, protons are translocated out of the matrix to yield a transmembrane electrochemical potential. In the second phase of OxPhos, the chemiosmotic potential generated by ETS drives ADP phosphorylation via ATP synthase.9
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Figure 3. Schematic of the four respiratory complexes within the inner mitochondrial membrane, with electron flow and some of the mid-point redox potentials indicated.
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Figure 4. Three respiratory complexes couple redox catalysis with proton translocation, thereby generating a transmembrane potential of ∼220 mV (matrix negative) due to both electrical and pH gradients. ATP synthase, also known as Complex V, couples the dissipation of the proton motive force to phosphorylation of ADP.
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Reference work2007, Comprehensive Medicinal Chemistry IIJ.A. Dykens
Review article
40 Cell Biology 2017
2017, Current Opinion in Plant BiologyMarkus Schwarzländer, Philippe Fuchs
The inner mitochondrial membrane—an organised tapestry to integrate electron transport with cell physiology
The IMM topologically separates the matrix from the cristal space (CS) and the peripheral intermembrane space (pIMS). Extensive folding of the CM gives rise to its substantial surface, allowing the IMM to host the respiratory machinery at high density. While IMM function as a capacitor for the proton motive force (PMF) has been thoroughly studied, analyses of dynamic structure–function relationships in living cells are a recent development pioneered in non-plant systems (e.g. [33•]), after classical TEM approaches had been largely correlative or descriptive.
Structure and architecture
The cristae of plant mitochondria typically do not align in any preferred orientation, and within the cellular context saccular cristae, that is, a condensed matrix configuration, appear to dominate (Figure 2). This seems in contrast with the often highly-ordered fine architecture of animal mitochondria (e.g. ). Condensed configuration has been linked to active electron transport in ATP-synthesizing or uncoupled isolated mammalian mitochondria [36,37]. For plants a clear connection between IMM structure and function is currently lacking, but will likely gain momentum from in vivo super-resolution techniques as well as novel electron microscopy approaches. There has been accumulating evidence already that the plant IMM features a remarkable degree of organisation and spatial specialization.
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Figure 2. Hypothetical concept of how spatial and functional separation of the inner mitochondrial membrane (IMM) may be organised in plants. Saccular cristae as frequently observed in plant mitochondria maintain a comparatively large intracristal space volume, which is connected to the peripheral intermembrane space (pIMS) by a cristal junction of a small relative diameter. Junction properties may be dynamically adjusted to modulate exchange between intracristal space and pIMS, potentially involving regulation by the ring-like oligomers of the mitochondrial contact site and cristae organising system (MICOS). Preferential localization of respiratory complexes and supercomplexes of the classical pathway [I–V, and cytochrome c (Cyt c)] to the cristal membrane and of the alternative pathway [internal and external alternative NAD(P)H dehydrogenases (ND) and alternative oxidase (AOX)] to the inner boundary section of the IMM. Uncoupling protein (UCP) and IMM transporters for the exchange of solutes with the cytosol are shown in the peripheral section of the inner mitochondrial membrane to allow for the efficient solute transport across both membrane systems. OMM: outer mitochondrial membrane; VDAC: voltage-dependent anion carrier; TOM: translocase of the outer mitochondrial membrane; TIM: translocase of the inner mitochondrial membrane.
Mitochondria in dry seed embryos had long been assumed to not contain an intact IMM, since TEM suggests that cristae were largely absent (e.g. ). They were hence referred to as promitochondria that require IMM biogenesis at seed imbibition before respiratory functionality can be established . Yet, a recent characterization found Arabidopsis embryo mitochondria to respire and to build up a membrane potential immediately at imbibition [40••]. Further, fully assembled respiratory complexes I, III, IV and V were detected in dry Arabidopsis seeds . Both findings are consistent with early observations of respiratory competent mitochondria isolated from dry sunflower seeds and demonstrate that respiratory competence and IMM integrity are preserved in the dry seed. The late embryogenesis abundant (LEA) protein M that is expressed in the matrix of pea seeds was found particularly effective in stabilizing integrity of cardiolipin-containing membranes [43,44]. That does not take away from the fact that extensive membrane expansion and biogenesis are certainly required for maturation to full functionality.
The IMM is strongly enriched with cardiolipin across species. The significance of cardiolipin in planta was recently confirmed genetically in a series of studies on Arabidopsis knockout lines lacking CARDIOLIPIN SYNTHASE 1 (CLS1). The cardiolipin free mutants are strongly impaired in development and a large proportion of their mitochondria display an abnormal architecture, that is, giant mitochondria of 2–6 μm long axis and poorly developed cristae [22•,45,46]. Mitochondria from heterotrophic CLS1 knockout cell cultures revealed a dramatic reduction in respiratory capacity, partially accounted for by a loss of cytochromec , illustrating the strong interdependence of membrane lipids and proteins in determining mitochondrial function.
Given the very high protein-to-lipid ratio of the IMM, particularly abundant proteins are good candidates for determining IMM architecture. In contrast to the chloroplast, mitochondrial ATP-synthase complexes can dimerize. Current evidence suggests that the dimeric form of ATP-synthase may even be the exclusive form to be present in vivo and that dimerisation is conserved in mitochondria across eukaryotes. The dimer structure and its ability to introduce curvature into the IMM was initially resolved in Polytomella and later confirmed to shape cristal rims across several species, with a particularly pointed angle found in potato [49,50•]. ATP-synthase was found exclusively as stable dimers in extracts of Polytomella and Chlamydomonas , while the monomeric form dominates in extracts of higher plants like Arabidopsis even at very mild solubilisation conditions [52••,53,54]. This implies lower dimer stability in higher plants also in vivo—a noteworthy correlation with the common occurrence of saccular cristae, in which pointed angles are less frequent. The cristal rim has been proposed to function as a proton trap, optimizing ATP-synthase function through precise spatial positioning and separation from the proton pumps . Highly preferential localization of the respiratory chain complexes in the CM was demonstrated in bovine heart tissue and HeLa cells by immune-gold labelling and their endogenous diffusion was restricted. At least a small proportion of ETC complexes appear to be present in the IBM considering the observation of a physical interaction with the IMM import protein Tim21 in yeast . Already over 30 years ago, an enrichment of ETC complexes in the cristae of Jerusalem artichoke mitochondria was observed by biochemical IMM sub-fractionation, leading to the early suggestion of lateral IMM heterogeneity in plants (Figure 2). High abundance of ETC complexes implies that the complexes are themselves involved in shaping cristae, probably without introducing any considerable angle, stabilizing more planar areas.
The mitochondrial contact site and cristae organising system (MICOS) was discovered as a protein complex that coordinates cristae and OMM–IMM tethering by bridging the pIMS in fungal and animal cells. Homologs of several MICOS proteins, including the two MICOS core proteins MIC60 and MIC10, were recently identified in plants, and found to cluster in a large mitochondrial transmembrane lipoprotein complex [60,61] (Figure 2). So far, only plant MIC60 was experimentally investigated and appears to fulfil bona fide MICOS functions, that is, cristae formation and organisation, but additionally plays a role in lipid trafficking. Substitution of mitochondrial phospholipids with plastidal glycolipids allows the cell to tap mitochondrial phosphate under phosphate deprivation [27,28••]. The topological bottleneck that is set by MICOS raises several fundamental questions about the organisation of the IMM. How can proteins mechanistically pass the junction as shaped by MICOS oligomers to change localization between CM and IBM? How can specific proteins be trafficked into a specific membrane section? To what extent can cristae junctions be limiting for solute exchange between CS and pIMS or even be conditionally regulated? The hypothetical option to open and close cristae junctions would give rise to a cristae subcompartment that can be topologically and functionally distinct from the IMS in protein inventory and physiology. Under such circumstances and considering enriched respiratory activity in the CM, the cristae may obtain bioenergetic properties similar to those of the chloroplast thylakoids, while saccular conformation can give rise to considerable volume of the CS. There is emerging evidence from animal cells that cristae-shaping proteins modulate mitochondrial bioenergetics and impact on respiratory supercomplex stability [62,63]. For instance, the cristae junction organiser and membrane fusion proteinOptic Atrophy 1 (OPA1) acts in cristae remodelling depending on energy substrate status . Yet, an obvious homologue of OPA1 is absent in plant genomes, suggesting that other proteins act in its place.
IMM structure has turned out to be highly ordered in mammals, and an analogous situation is becoming increasingly likely in plants. Despite their topology as an electrically continuous membrane, the IBM and the CM differ functionally and this may only be one facet of a much more sophisticated tapestry, involving membrane sections that are distinctly shaped and functionalized through higher-order-arrangements of proteins and lipids. Although direct evidence is still scarce for plant mitochondria, testable hypotheses can be derived from the assumption of structural and functional order, which we exemplify here for respiration and transport of plant mitochondria.
Respiratory chain and complexes
Any organisational subdivision of the IMM into IBM and CM will be mirrored by specific functionalization through different protein and lipid inventories. This may be exemplified with the respiratory chain proteins as well as different solute transporters. The respiratory chain complexes make up a substantial proportion of the IMM, which is even higher in the CM. While alternative electron transport proteins have been specifically studied in plant mitochondria, much of what is known about the function of the classical electron transport complexes in plants has been deduced from work in non-plant systems. Yet, this has been proving risky given the additional complexity that has been emerging for plant respiratory complexes. In general, plant mitochondrial complexes contain more subunits than their mammalian counterparts . For Complex II it was demonstrated that additional subunits have evolved to complement the truncation of conserved subunits to ensure membrane anchoring . The detection of a complete additional structural element comprised of γ-carbonic anhydrase (like) proteins in plant Complex I [66,67] that appears to act as a condensation point early in complex assembly [68–70] has given rise to the intriguing workbench concept of evolutionary integration of novel functions and activities into Complex I . This concept may be applicable also to the other abundant IMM proteins.
Firstly, a large proportion of the CM consists of respiratory complexes. For non-membrane proteins to associate with membranes, limitation of free lipid surface may dictate association with the most abundant membrane proteins. Secondly, through their activity, the complexes provide specific physiological microenvironments at the membrane surfaces with respect to pH, lipids, partner proteins and metabolites that may favour specific reactions. Although carbonic anhydrase activity, which would be required for a functional carbon concentration mechanism, has not been empirically shown for Complex I of higher plants, it may benefit from local alkaline pH as set by the proton pumping activity of Complex I. A stable lateral pH gradient of 0.3 pH units was demonstrated for the CS-side of the CM in HeLa cell mitochondria between Complex IV and the Complex V dimer as a result of proton fluxes [72••], and an analogous steady-state pH gradient can be expected at the matrix side. Thirdly, exploitation of respiratory chain complexes as platforms may allow generating higher order associations also for ‘soluble’matrix proteins that may not be detected as complex components due to weaker or only transient binding. This may allow for the membrane to contribute to matrix organisation via more or less stable protein–protein associations. Evidence comes from the finding that IMM associated Complex II can physically interact with several tricarboxylic acid (TCA) cycle proteins and fumarate (a product from Complex II) is metabolically channelled [73•]. Since no direct physical interaction between Complex II and fumarase was found, diffusion may be limited by the IMM surface itself to result in channelling between two non-interacting enzymes. Hence, the finding of different stochiometries between Complex I and Complex II in Arabidopsis depending on tissue type (high ratio in leaves and stem; low in root ) may not only reflect differences in the balance of electron entry into the chain, but also mirror wider differences in physical organisation of matrix enzymes.
Catalytic cooperation and substrate channelling by respiratory supercomplexes appears increasingly implausible . Alternative roles may be in modulating the capacity of complexes to act as workbenches for interactors and to structure the CM and its surface to support channelling of metabolic fluxes in the matrix (Figure 3). Removal of a stomatin-like protein (SLP1) that appear to interact with specific supercomplexes as oligomers in the inner membrane of Arabidopsis mitochondria, decreased supercomplex abundance, suggesting that the microenvironment provided by supercomplexes underlies organisation beyond molecular crowding . Two recent complexome profiling studies on Arabidopsis leaf mitochondria delivered a rich inventory of mitochondrial protein complexes, validating known assemblies, but also proposing several novel complexes [52••,77]. This included evidence that also alternative electron transport components, that is, alternative oxidase (AOX) proteins and alternative dehydrogenases (NDA and NDB), may form higher order structures as evidenced by complexes of 140, 150 and 700 kDa [52••], in line with an earlier report . Since the external NDs can oxidize NAD(P)H from the cytosol, and NDs and AOXs do not contribute to proton pumping, their exposure to the pIMS rather than the CS may be predicted. While this is hypothetical, lateral separation of electron transport pathways may offer an organisational means to the yet unresolved question of how electron partitioning and coupling may be regulated in plant mitochondria. (Controlled) lateral diffusion of ubiquinone between the IBM and the CM could nevertheless connect both pathways, similar to the diffusion of plastoquinone and plastocyanine to bridge the spatial separation of PSII and PSI in the chloroplast thylakoids. Lateral fine-localization of electron transport proteins in the IMM deserves testing in plants in a similar manner as in mammalian cells since mitochondrial electron transport can accept substrates from the matrix and the cytosol by a wide range of membrane associated-dehydrogenases (for a recent review see ).
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Figure 3. Hypothetical concept of how structured mitochondrial membrane surfaces may limit solute diffusion and support metabolic substrate channelling. Generation of constraint, and ordered, physical spaces through the protrusions of membrane proteins is illustrated for respiratory Complex I (CI) at the matrix side of the inner mitochondrial membrane (IMM). Note that there is no experimental evidence for physical supercomplex assemblies containing multiple copies of Complex I, but the principle of membrane surface structuring through proteins, protein complexes and supercomplexes can be generalized.
Transport and uncoupling
Mitochondrial chemiosmotic function was originally postulated based on an impermeable IMM, which allows ion transport exclusively in exchange for protons . Electrogenic uniport as an additional mode was initially not included since it would dissipate the membrane potential by ion cycling leading to uncoupling of oxidative phosphorylation. Yet, IMM uniport activity has since been demonstrated, which shifts the problem to the regulation required. Reports from several species indicate conditional K+ uniport in plant mitochondria (recently reviewed by ), which has been implicated in matrix volume control and protection from oxidative stress in mammals. The physical identities of the channels are subject to active current research in mammals and may give valuable clues also for plants. A long-standing debate about the existence of Ca 2+ uniport in plant mitochondria could recently be resolved by identification of homologs of the mammalian Ca 2+uniporter complex in plants that show channel activity. Electrogenic Ca 2+ uniport across the IMM in Arabidopsis was found to be regulated by MICU, a Ca 2+-binding EF-hand protein [82•]. That allows setting a strict threshold of external Ca 2+ concentration for activation of the uniporter. The pore itself is composed of MCU proteins , which were found as part of a complex in Arabidopsis mitochondria, suggesting operation as a tetra-mer or penta-mer [52••]. For a recent review on plant mitochondrial Ca 2+ transport and regulation see .
Several further transport activities could recently be linked with individual proteins, which has substantially increased the inventory of validated plant IMM transporters (for a review on plant mitochondrial transporters see ). Mitochondrial ATP/phosphate carriers (APCs) and coenzyme A (CoA) transporters were proposed to mediate matrix loading with ATP and CoA, based on in vitro transport and yeast complementation studies [86–88]. In vitro and in vivo data have raised the possibility that the mitochondrial ABC-transporter ATM3 exports glutathionepolysulfide from the matrix for cytosolic FeS-cluster biogenesis in Arabidopsis and rice [89,90].
The transport systems in the IMM may be expected to be spatially ordered, to allow for specific exchange of the matrix either with the cytosol/pIMS or the cristae. Transported compounds passing the OMM may preferentially pass the IBM. Alternatively, transport via the CM may provide additional options for integration en route within the pIMS and the CS. Similar to the cooperation of TIM complex components at the IBM with the TOM complex for matrix protein import, efficient solute exchange with the cytosol may be best implemented by close vicinity or physical interaction of the transporters in the OMM and IBM. This may be relevant to achieve high flux rates as required for respiratory metabolites or rapid signals, such as Ca 2+. Consistently, MCU was recently found to be preferentially localized at the IBM of cardiac muscle mitochondria . By contrast, co-factor transport as required for biogenesis may be less reliant on a diffusion-optimization. As such, the positioning and lateral shuttling of transporters within the IMM between IBM and CM may provide a means of fine-tuning transport activity.
Spatial separation maintains the possibility for IMM transport to tap into the membrane potential generated mainly in the CM due to the electrical continuity of the IMM. Linking organelle import/export and electron-transport-based proton pumping in the same membrane is a decisive difference to the chloroplast, where thylakoid membrane and inner envelope are energetically separated. This principle offers to link transport to the steep electrical potential, similar to the plasma membrane, yet with the additional challenge of maintaining the PMF for ATP synthesis even under fluctuating transport requirements. The flexibility to modulate coupling is particularly refined in plant mitochondria, as indicated by the alternative respiratory machinery, but different PMF-active transporters may add to this repertoire. The dual role of many IMM transporters, including the K+ and Ca 2+ uniporters, to consume PMF and to mediate solute flux makes unambiguously pinpointing of their physiological significance not always straightforward. Several transport processes have been suggested to allow for uncoupling and, in turn, protection against hyperpolarisation and avoidance of excessive ROS release by the respiratory chain.
A physiological role as uncoupler through passive proton translocation along the gradient has been described for plant uncoupling proteins . Overexpression of UNCOUPLING PROTEIN 1 (UCP1) in potato mitochondria reduced the PMF and increased the respiratory rate consistent with increased flux rates through the TCA cycle . Vice versa, an Arabidopsis UCP1 mutant showed signs of photorespiratory defects indicating limitations in mitochondrial substrate oxidation capacity . The role for plant UCPs as professional uncouplers has been controversially discussed, due to their ability to transport organic acids in vitro . Yet, mammalian thermogenic UCP1 has been found tightly regulated at the posttranslational level through sulfenylation , suggesting additional levels of control in planta that will need considering. Transients in membrane energization of Arabidopsis mitochondria have been proposed to represent an uncoupling mechanism at the level of individual organelles , but the identity of the operating channel is still elusive. Electrogenic non-proton ion uniport has been proposed as the underlying principle matching the characteristic increase of matrix pH resulting from the decrease in electrical potential . Absence of an Arabidopsis MscS-Like (MSL) protein belonging to a family of mechanosensitive ion channels was associated with hyperpolarisation of isolated mitochondria , but did not abolish the pulsing phenomena, suggesting several uncoupling mechanisms to operate in parallel. Limited diffusion in the CS and the pIMS as evidenced by lateral pH gradients [72••] render the exact localization of IMM transporters relevant for uncoupling efficiency. For instance passive proton translocation may be most effective across membrane domains where the gradient is the steepest, and uncoupling may be reduced by localizing the responsible protein to the IBM, as recently suggested for mammalian UCP4 that was found in the vicinity of VDAC rather than ATP-synthase dimers by superresolution microscopy . This adds an organisational concept of uncoupling to the already sophisticated network of respiratory modulation of plant mitochondria.
If the ATP-synthase dimer indeed represents the mitochondrial permeability transition pore (mtPTP), as recently proposed also for plants [101,102], their well-defined location in curved CM areas where the proton gradient is thought to be steepest [49,55] may make uncoupling particularly effective. Yet, the role of ATP-synthase in giving rise to the mtPTP remains controversial. The physiological meaning and the consequences of a proposed mode of transient pore opening in particular for the intricate structure-function tapestry of the IMM will require careful further appraisal.
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Journal2017, Current Opinion in Plant BiologyMarkus Schwarzländer, Philippe Fuchs
Review article
16th European Bioenergetics Conference 2010
2010, Biochimica et Biophysica Acta (BBA) - BioenergeticsKornelius Zeth
1 Introduction
The mitochondrial outer membrane originates from the uptake of an endosymbiotic precursor bacterium from the α-proteobacterial subgroup into a eukaryotic precursor cell . During the course of evolution the lipid composition of both membranes has changed significantly due to the different environmental requirements but the protein representatives carrying important functions remained partly the same. In yeast five integral outer membrane proteins of β-barrel topology were identified, three of which, scVDAC1, scVDAC2 and scTom40 belong to the 19-stranded β-barrel group. Sam50 displays a high sequence similarity to the bacterial precursor protein Omp85 and very probably belongs to a second 16-stranded β-barrel subgroup [2,3]. Another protein Mdm10 involved in the maturation of MOM protein complexes that is predicted to contain β-sheet structure is not present in higher eukaryotes and has also no sequence homologs among bacteria . Together, this ensemble of β-barrel proteins in complex with adjacent factors (TOM and SAM associated proteins) allows the communication and exchange between mitochondria and the surrounding cytosol and forms the platform for a variety of protein–protein interactions.
In the early 70's and 80's the mitochondrial porin has been discovered as a channel structurally and functionally resembling many bacterial porins[5,6]. At the same time, in 1976, the discovery of the switch between an anion- to a cation-selective state of lower conductance at low voltages in planar lipid membranes led to the term of the voltage dependent anion channel, VDAC . These investigations and selected following experiments represented the hallmark characteristics of VDAC family proteins and constrained the view onto the channel for about two decades. In 2008, the 3D structure of VDAC has been determined independently by three research groups using either high resolution NMR spectroscopy, X-ray crystallography or a hybrid approach combining information of both techniques [8–11]. All three structure models basically show the same architecture of a 19-stranded β-barrel with the N-terminal helix enclosed. This structure investigation was not only a breakthrough in the understanding of MOM structure biology but also forms the reliable basis for further biochemical and modeling experiments.
From sequence alignments it became indicative that Tom40 shares the same overall architecture as determined for VDAC proteins. The channel forms the integral part of the TOM complex which in S. cerevisiae is assembled by three preprotein receptor molecules Tom20, Tom22 and Tom70 and a number of small Tom proteins (Tom5, Tom6, Tom7) . Tom40 is the core component of the 400–500 kDa pore-like structure which allows preprotein passage into the intermembrane space where proteins are further targeted into the four compartments . Electron microscopy data provided the first structural insights into the entire (containing the Tom70 component) or core TOM complex with a dynamic equilibrium of two or three Tom40 subunits present in the protein ensemble [14,15]. The TOM reconstructions acquired so far represent only fungi–mitochondrial complexes (from S. cerevisiae and N. crassa) at medium resolution but the pore structures agree well with the Tom40 model we generated on the basis of VDAC structures, both in dimensions of vertical and horizontal size (our unpublished data). The major function of the Tom40 protein is to enable the passage and lateral release of partially folded or unfolded secondary structure elements [14,16]. From the dimensions of this pore model, a folded α-helix may easily pass through without previous unfolding but a β-hairpin structure would have to be unfolded before passage. This model of one pore forming protein per barrel wall is also supportive for the idea of e.g. Rapaport that mitochondrial outer membrane proteins comprising one C-terminal α-helix may be inserted into a rim of several Tom40 β-barrels and subsequently released laterally into the outer membrane .
Notably, the composition of proteins assembling the TOM and SAM complexes (see next paragraph) is not consistent between mitochondria of different phylogenetic lineages (for review see: [18,19]). For TOM a subset of three proteins, Tom40, Tom22 and Tom7 are ubiquitously found in all eukaryotes .
The third protein of barrel topology, Sam50, (which is also known as the Tob55 protein of the TOB complex: t opogenesis of mitochondrial o uter membrane b eta-barrel proteins) belongs to the sorting and assembly machinery, SAM, which consists in yeast of three proteins: the integral Sam50 pore and two Sam35 and Sam37 accessory proteins [21,22]. Sam50 and also the plant homolog Toc75 from chloroplasts both show a clear sequence similarity to the Omp85 evolutionary precursor protein from Gram-negative bacteria [19,23,24]. By contrast to the bacterial Omp85 homologs which N-terminally harbor five POTRA (Po lypeptide Tr anslocation A ssociated) domains, the Sam50 protein has evolved differently and contains only one of these scaffold domains . POTRA domains together with the integral membrane portion form the typical structural fingerprint of proteins belonging to the Omp85 or related membrane insertase families. Functionally, SAM complexes can recognize the β-signal stretch which consists of a four residue motif encoded with high similarity in the very last β-strand of any MOM protein . This arrangement is somewhat reminiscent to a sequence similarly observed in outer membrane proteins of bacteria where a handful of residues are strongly discriminative for functions including (I) insertion into the outer membrane and (II) binding to DegS and induction of the σE-dependent stress response [26,27]. This led to the conclusion of the C-termini of OMPs and β-barrel MOMs harboring specific motifs important for their stability.
Crystal structures of mammalian VDAC and the bacterial FhaC protein provide a solid basis to model related MOM proteins of β-barrel topology of sufficient similarity such as the human Tom40 (similar to VDAC) and Sam50 (on the basis of FhaC protein) [8–10,28]. In this article the relationship between these proteins in sequence and structure are discussed.
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Journal2010, Biochimica et Biophysica Acta (BBA) - BioenergeticsKornelius Zeth
Chapter
Mitochondria
2019, Plant Cell Biology (Second Edition)Randy Wayne
14.3 Composition of Mitochondria
The outer membrane of plant mitochondria comprises only 6.8% of the mitochondrial protein, whereas the inner mitochondrial membrane accounts for 30% of the total protein. The inner membrane is composed of 60% protein and 40% lipid on a dry weight basis, whereas the outer membrane is composed of 20% protein and 80% lipid on a dry weight basis. The outer mitochondrial membrane contains a channel protein called porin that is permeable to molecules that are 10,000 Da or less. It is similar to the pore protein in the outer membrane of the chloroplast envelope.
14.3.1 Proteins
Proteomic analysis shows that the mitochondria are composed of about 3000 proteins (Bykova and Moller, 2006; Taylor and Millar, 2015). The proteins used to identify the mitochondria in vitro are those associated with respiration. They include succinate dehydrogenase, cytochrome c oxidase, and fumarase (Hogeboom and Schneider, 1955), as well as several cytochromes, including a, a 3, c, and b 5 (Chance and Williams, 1955). In animal cells, which lack plastids, the enzymes involved in lipid metabolism are present on the outer mitochondrial membrane.
The marker enzymes for the outer membrane are the succinate:cytochrome c oxidoreductase and the NADH: cytochrome c oxidoreductase. The marker enzymes for the inner mitochondrial membrane are malate dehydrogenase and succinate: K 3 Fe(CN)6 oxidoreductase.
14.3.2 Lipids
In contrast to the galactolipid-rich chloroplasts, the mitochondria are rich in phospholipids (23%–27% by weight). Phospholipids account for as much as 90% of the total mitochondrial lipids. Mitochondria contain high concentrations of phosphatidylcholine (PC) and phosphatidylethanolamine (PE) as well as diphosphatidylglycerol (DPG=cardiolipin), a lipid that is endemic to mitochondria (Table 14.1). Cardiolipin is also found in the plasma membrane of bacteria, giving support for the hypothesis that mitochondria evolved from bacteria (see Chapter 15).
Table 14.1. Phospholipid Composition (% Phospholipid by Weight) of Sycamore Cells and Mitochondria
| Phospholipid | Cells | Mitochondria |
---
| Phosphatidylcholine | 47 | 43 |
| Phosphatidylethanolamine | 29 | 35 |
| Diphosphatidylglycerol a | 1.8 | 13 |
| Phosphatidylinositol | 16 | 6 |
| Phosphatidylglycerol | 5 | 3 |
a Also known as cardiolipin.
From Douce, R., 1985. Mitochondria in Higher Plants. Structure, Function, and Biogenesis. Academic Press, Orlando.
The lipid compositions of the inner and outer membranes of the mitochondria are distinct. In particular, the inner membrane contains cardiolipin, which is absent from the outer membrane. The inner membrane is also richer in PE and poorer in PC relative to the outer membrane (Table 14.2). The hydrocarbons derived from fatty acids in the outer membrane are slightly more saturated than the hydrocarbons derived from fatty acids in the inner membrane. The phospholipids of plant mitochondria are much less saturated than those from mammalian mitochondria, which exist at 37°C.
Table 14.2. Phospholipid Composition (% Phospholipid by Weight) of Inner and Outer Membranes of Sycamore and Mung Bean Mitochondria
| Empty Cell | Sycamore Cells | Mung Bean Cells |
--- |
| Lipid | IM | OM | IM | OM |
| Phosphatidylcholine | 41 | 54 | 29 | 68 |
| Phosphatidylethanolamine | 37 | 30 | 50 | 24 |
| Diphosphatidylglycerol | 14.5 | 0 | 17 | 0 |
| Phosphatidylinositol | 5 | 11 | 2 | 5 |
| Phosphatidylglycerol | 2.5 | 4.5 | 1 | 2 |
IM, inner membrane; OM, outer membrane.
From Douce, R., 1985. Mitochondria in Higher Plants. Structure, Function, and Biogenesis. Academic Press, Orlando.
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Book2019, Plant Cell Biology (Second Edition)Randy Wayne
Chapter
Mitochondrial Biology and Medicine
2019, Emery and Rimoin's Principles and Practice of Medical Genetics and Genomics (Seventh Edition)Douglas C. Wallace, ... Vincent Procaccio
10.2 Mitochondrial Biochemistry
Each cell harbors hundreds to thousands of double-membrane mitochondria. The outer mitochondrial membrane is perhaps the remnant of the phagocytic vesicle and the inner membrane that of the eubacterial plasma membrane. The matrix is the bacterial cytoplasm, and the intermembrane space separates the two organisms.
The highly invaginated inner membrane harbors the enzymes of OXPHOS on in-foldings called cristae. The cristae are closed at the junction with the intermembrane space to create “cristae lumens” . The mitochondria inner membrane has a unique composition that includes cardiolipin, and the mitochondrion has an independent bacteria-like lipidbiosynthesis system that produces the essential lipoic acid. The mitochondria are the metabolic hub of the cell. Central metabolic pathways include the tricarboxylic cycle (TCA, Krebs cycle), fatty acid β-oxidation, amino acid metabolism, and cholesterol and heme synthesis. Pyruvate from glycolysis enters the mitochondrion via the pyruvate carrier and is then cleaved by pyruvate dehydrogenase (PDH) to generate acetyl-CoA and reduced nicotinamide dinucleotide (NADH). Acetyl-CoA is also generated by β-oxidation, and the mitochondrial acetyl-CoA is condensed with oxaloacetate to generate citrate. Citrate can be exported into the cytosol where it is cleaved by ATP-citrate-lyase to acetyl-CoA and oxaloacetate. Cytosolic acetyl-CoA can be used for fatty acid synthesis or as substrates for protein and histone acetylation. Within the mitochondrion, citrate is metabolized to isocitrate, aconitate, α-ketoglutarate, succinyl-CoA, succinate, malate, and oxaloacetate via the TCA cycle. Isocitrate dehydrogenase (IDH) has three isoforms—two in the mitochondrion, an NADP+-linked IDH2 and an NAD+-linked IDH3, and one in the cytosol, an NADP+-linked form. IDH1, IDH1, and IDH2 are important in cancer genetics[21,22]. PDH, α-ketoglutarate dehydrogenase, and branched chain-keto acid dehydrogenase (BCKDH) are multiple polypeptide complexes that share common subunits and cofactors, including lipoic acid, and are modulated by Ca 2+. The integrity of the mitochondrial inner membrane requires special transport systems to move organic molecules in and out of the mitochondrion. This is accomplished in part by the 53 members of the solute carrier family 25 (SLC25), also called the mitochondrial carrier family, which transport carboxylates, amino acids, nucleotides, and cofactors across the inner mitochondrial membrane[24–26]. Mitochondria have also been proposed to contain an adenylyl cyclase and a nitric oxide (NO) synthase .
Each mitochondrion is a capacitor composed of a proton gradient across the mitochondria inner membrane generated by the electron transport chain (ETC). The ETC burns dietary calories (reducing equivalents) with oxygen in a stepwise process that transmits electrons from reduced to oxidized starting with NADH and succinate. NADH is oxidized by complex I (NADH:CoQ oxidoreductase or NADH dehydrogenase) and succinate is oxidized by complex II (succcinate:CoQ oxidoreductase, or succinate dehydrogenase). Complexes I and II then transfer the electrons to the lipid carrier, coenzyme Q 10 (CoQ), which transfers them to complex III. Complex III transfers the electrons to cytochrome c and cytochrome c transfers the electrons to complex IV (cytochrome c oxidase or COX). Complex IV combines four electrons with O 2 to generate 2H 2 O. As the electrons traverse complex I, III, and IV, the energy released is used to pump protons across the mitochondrial inner membrane, four through complex I and one each through complexes III and IV (Fig. 10.1A).
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Figure 10.1. Overview of the OXPHOS complexes I–V showing the molecular structures of the five OXPHOS complexes .
Each of the respiratory complexes is composed of multiple polypeptides. The molecular structures of complex I [29,30], complex II , complex III , and complex IV [33,34] have been established. Complex I is composed of 45 polypeptides, complex II of four, complex III of 11, and complex IV of 13 (Fig. 10.1). Each complex also contains various prosthetic groups that conduct electrons. Complex I harbors an FMN site to collect electrons from NADH, eight iron-sulfur (FeS) centers, and a CoQbinding site. Complex II harbors an FAD, an FeS, a cytochrome b, and a CoQ binding site. Complex III has the heme-based cytochromes b and c1, the Rieske FeS center, and two CoQ binding sites and the cytochrome c contains heme c. Complex IV encompasses cytochromes a + a 3, two Cu centers, and the oxygen reaction center . The respiratory chain is also assembled into super-complexes, the physiological function of which remains to be determined [36–39].
The proton gradient (ΔP=ΔΨ+Δμ H+) is used to energize multiple mitochondrial functions, the best known is the generation of ATP from ADP + Pi by complex V (proton-translocating ATP synthase or ATP synthase). Complex V is composed of 18 polypeptides. The ETC enzymes and the ATP synthase are arrayed within the cristae lumen membranes with the ETC charging the proton gradient within the cristae lumens and the ATP synthase utilizing the proton gradient to drive its spinning ring of “c” subunits within the inner membrane. Each c subunit has a negatively charged carboxyl group that picks up a proton from the cristae lumen via a half channel in the abutting membrane bound and static ATP6 protein. The c ring then rotates 360 degrees within the plane of the inner membrane until it comes back to the ATP6 subunit. The ATP6 protein has a second half proton channel open to the matrix through which the proton is released. The c-ring wheel has a γ subunit axial that protrudes inside the F1 (3α:3β) barrel. The barrel is fixed to the ATP6 inner membrane protein and is also static . The wheel spins at 300 Hz , and the spinning axial contacts the three β subunits sequentially, causing conformational changes to condense ADP+Pi to make ATP. The ATP synthase is an offset dimer , which is aggregated around the edges of the cristae lumens.
ATP generated in the matrix is exported to the intermembrane space by the adenine nucleotide translocators (ANTs) which belonging to the mitochondrial carrier protein family. There are four ANT isoforms in humans, ANT1 being expressed in the heart and muscle, with ANT1 mutations having been identified in human CPEO and mitochondrial myopathy and cardiomyopathy[43–45]. The intermembrane space communicates with the cytosol via the voltage-dependent anion channels (VDAC, porin).
Besides ATP synthesis, the proton gradient is used for multiple other processes. The best characterized of these is the uptake of cytosolic Ca 2+ though the mitochondria Ca 2+uniporter (MCU) complex [46,47]. The cristae lumens are closed at the intermembrane space by the MICOS + OpaI complex [48,49]. Opa1 can be cleaved by OMA1 resulting in the opening of the cristae and the release of cytochrome c and protons into the cytoplasm initiating apoptosis. This is associated with the activation of the inner membrane mitochondrial permeability transition pore (mtPTP) located within the inner membrane and resulting in depolarization of the proton gradient. This combination of events initiates the intrinsic pathway of apoptosis. The structure of the mtPTP, which interacts with the Ca 2+-activated cyclophilin D (cypD), is actively debated with current contenders being the ATP synthase dimers, ATP synthase c-ring [41,50–54], or the SPG7 hexamer .
The mitochondria are highly dynamic organelles undergoing fission via activated Drp1 and fusion mediated by Opa1, Mfn1, and Mnf2. The mitochondria replicate within the cytosol and the excess mitochondria are removed by mitophagy. Hence, the mitochondria are a colony of bacteria maintained in a metastable state via a balance between proliferation and degradation.
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Book2019, Emery and Rimoin's Principles and Practice of Medical Genetics and Genomics (Seventh Edition)Douglas C. Wallace, ... Vincent Procaccio
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Mitochondria and Bacterial Pathogens - Part B
2023, International Review of Cell and Molecular BiologyKrishnaveni Mohareer, Sharmistha Banerjee
2.2 Metabolites
The mitochondrial outer membrane possesses numerous transport proteins that regulate metabolite and coenzyme transport across mitochondria. Based on the nature of the solute, the different transporters include the SLC25A family, mitochondrial ABCB transporters, the mitochondrial pyruvate carrier (MPC), and the sideroflexin carriers. However, solutes less than 5 kDa are freely accessible to the mitochondrial outer membrane.
The end product of glycolysis, pyruvate, enters mitochondria through MPC and is irreversibly converted by pyruvate dehydrogenase to acetyl-CoA with the availability of coenzyme A, which is transported by SLC25A42. Malate shuttles across the mitochondrial membrane with the help of three carriers, namely, SLC25A10 (DIC or dicarboxylate carrier), SLC25A11 (OGC or oxoglutarate carrier), and SLC25A1 (CIC or tricarboxylate transport protein). Malate gets antiported in exchange for α-KG. Please refer to the review (Cunningham and Rutter, 2020) and references therein for more details.
The cholesterol levels in mitochondria are much lesser than either plasma membranes (∼40 fold) or ER membranes (∼2.5 fold) (Horvath and Daum, 2013) and tightly regulated, making them very sensitive to slight changes in cholesterol levels. Elevated cholesterol levels influence antioxidant levels, as reflected by decreased 2-oxoglutarate carrier activity (Coll et al., 2003) and decreased GSH. Besides, it also results in decreased oxidative phosphorylation, as reflected by ATP levels. The altered membrane fluidity also leads to mitochondrial membrane depolarization and could lead to cell death (Martin et al., 2014).
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Book series2023, International Review of Cell and Molecular BiologyKrishnaveni Mohareer, Sharmistha Banerjee
Chapter
Introduction
2016, The Membranes of Cells (Third Edition)Philip L. Yeagle
1.4.4 Mitochondrial Membranes
Mitochondria are bounded by two membranes, reminiscent of Gram-negative bacteria. Mitochondria have the inner mitochondrial membrane and the outer mitochondrial membrane, separated by the intermembrane space (analogous to the periplasmic space in Gram-negative bacteria—however, mitochondria have no equivalent to the cell wall). Inside the inner mitochondrial membrane is the matrix, which, among other constituents, contains a small (relative to nuclear) DNA that codes for a minority of the proteins found in the mitochondria (the majority of mitochondrial proteins, including membrane proteins, are transcribed from nuclear mRNA).
The outer membrane, like that of many bacteria, contains a class of membrane proteins distinct in structure from most other membrane proteins in other cell membranes. The mitochondrial outer membrane contains porins. The porins in both bacteria and in mitochondria are β-barrels. In some of these proteins, the center of the β-barrel forms a pore enabling passive transport across the membrane. The outer membrane also contains a protein apparatus for the import of proteins used in the mitochondria but synthesized from cytoplasmic mRNA. The β-barrel proteins will be explored in chapter “Membrane Proteins.”
The inner mitochondrial membrane has a much higher membrane protein content than most membranes. The inner mitochondrial membrane thus has the lowest lipid–protein ratio of eukaryotic cell membranes. The inner membrane also contains the lipid, cardiolipin, or diphosphatidylglycerol, which is found almost uniquely in that membrane (and not in the outer mitochondrial membrane, for example). The inner membrane is highly invaginated to form cristae that fold back and forth across the interior of the mitochondrion. This structure provides a large membrane surface area to support the major work of the mitochondrion, ATP synthesis, which occurs on the inner membrane.
Multiple mitochondria are found within a single eukaryotic cell, sometimes hundreds or more. Mitochondria are about the size of an E. coli bacterium. In fact it has been hypothesized that bacteria are the precursor to mitochondria. The endosymbiot hypothesis asserts that endocytosis of a bacterium by a eukaryote early in evolution led to the formation of the mitochondrion. This process would explain the two membranes of the mitochondria, the primary role of mitochondria in synthesizing ATP (by the same overall mechanism as in E. coli, for example), and the presence of a separate DNA in the mitochondria. Mitochondrial DNA is inherited separately from nuclear DNA (and plays unique roles in DNA identification of organisms, including humans). A similar process is hypothesized for the development of chloroplasts, after endocytosis of a prokaryote capable of photosynthesis.
ATP synthesis is perhaps the most important function of mitochondria (and chloroplasts). The amazing path of ATP synthesis flows through an intricate set of enzymes, many transmembrane proteins, located in the inner membrane. Some of these proteins will be examined in depth later in chapter “Membrane Transport.” This pathway is an excellent example of the motif that some cellular functions are crucially dependent on the structure of a membrane, in addition to utilizing membrane-bound enzyme complexes. The inner membrane supports the enzymes of the electron transport and oxidative phosphorylation pathways in specified protein complexes. Electron transport by the inner mitochondrial membrane crucially creates a transmembrane electrochemical gradient across the inner mitochondrial membrane. This gradient is then used to phosphorylate ADP and form ATP (and oxygen is reduced to water) through oxidative phosphorylation by a transmembrane protein complex in the inner mitochondrial membrane.
Not only do all these chemical reactions take place in and around the inner mitochondrial membrane, but also oxidative phosphorylation would not take place without a membrane and its structural features. The process leading to ATP synthesis creates, through a mechanism involving the movement of electrons through a series of coupled oxidation and reduction reactions, a proton gradient across the inner mitochondrial membrane. Embedded in that membrane is an enzyme that synthesizes ATP from ADP and P i, the ATP synthase. The ATP synthase obtains the energy for that energetically highly unfavorable reaction from the transmembrane proton gradient. The process consequently first drives protons across the inner mitochondrial membrane using the energy from the electron transport process. Then the protons are allowed to flow back across the membrane (and down a concentration gradient) through the ATP synthase in the process that produces ATP. The ATP synthase will be discussed in more detail in chapter “Membrane Transport.” The unequal proton concentrations (on either side of the inner mitochondrial membrane) and the pathway whereby that gradient is dissipated driven by thermodynamics are what provide the energy for ATP production. A membrane that is otherwise impermeable to protons is a required component of the major pathway of ATP synthesis in bacteria and in eukaryotes. Therefore ATP synthesis, upon which our life depends, is possible only because of a biological membrane and its unique properties that are the subject of this book.
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Book2016, The Membranes of Cells (Third Edition)Philip L. Yeagle
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The human islet amyloid polypeptide in protein misfolding disorders: Mechanisms of aggregation and interaction with biomembranes
2021, Chemistry and Physics of LipidsAdam El Saghir, ... Neville Vassallo
1.6 Mitochondrial membranes and IAPP
Being evolutionary derived from an endosymbiotic bacterium, mitochondria are enveloped by a double-membrane system, an outer mitochondrial membrane (OMM) and an inner mitochondrial membrane (IMM) (Roger et al., 2017). The mitochondrial double-membrane ensures that the cellular cytoplasm remains separate from mitochondrial contents, whilst creating two compartments, namely the internal mitochondrial matrix and the inter-membrane space. Cellular respiration occurs in the IMM, through the electron transport chain (ETC) via the process of oxidative phosphorylation (OXPHOS), which generates 80–90 % of ATP in most mammalian tissues. The ETC comprises five large membrane protein complexes (complexes I–V) and two principal electron carriers, ubiquinone and cytochromec. Synthesis of ATP is driven by creation of an electrochemical gradient across the IMM as a result of proton-pumping activity by complexes I, III and IV (known as the proton motive force), with the trans-membrane gradient being then harvested by the rotary action of F 0 F 1-ATPase (complex V). Thus, due to their central role in energy generation, mitochondria are often referred to as the cell’s “powerhouses” (Nicholls and Ferguson, 2013). Apart from being essential for cellular energy homeostasis, mitochondria play critical roles in other fundamental processes like the regulation of cell death by apoptosis, signaling pathways (including calcium signaling), maintenance of redox balance, and also house important biosynthetic pathways (Duchen and Szabadkai, 2010; Rossi et al., 2019).
A unique ancient phospholipid common to mitochondrial and bacterial membranes, and which is essential for mitochondrial bioenergetics functions, is cardiolipin (CL; 1,3-diphosphatidyl-sn-glycerol) (Baile et al., 2014; Horvath and Wittung-Stafshede, 2016). CL associates with several mitochondrial proteins, counting all respiratory complexes and most mitochondrial solute carriers (Claypool, 2009; Claypool et al., 2008; Schwall et al., 2012). It also acts as a “lipid stabilizer” of supramolecular respiratory supercomplexes, thus maximizing OXPHOSelectron transfer (Mileykovskaya and Dowhan, 2014; Vartak et al., 2013). In lieu of its bioenergetic functions, CL is mainly located in the IMM; however, transfer of CL between the mitochondrial membranes at membrane contact site complexes has been observed, especially in relation to apoptosis (Epand et al., 2007; Maguire et al., 2017; Schlattner et al., 2014).
Previous studies have suggested that decreases in insulin secretion and metabolic output in T2DM may be attributed to mitochondrial dysfunction (Fex et al., 2018). The fact that toxic hIAPP oligomers form intracellularly makes it plausible that they interact with organelle membranes – including those of mitochondria. Indeed, using cryo-immunogold labeling with an oligomer-specific antibody in hIAPP transgenic mice, in human insulinoma, and in pancreas from humans with and without T2DM, cytosolic hIAPP oligomers were strikingly observed adjacent to morphologically compromised mitochondria and disrupted mitochondrial membranes (Gurlo et al., 2010). Similarly, images of immunogold-labeled hIAPP in mouse pancreatic islet β-cells and human insulinoma, exhibited subcellular localization to mitochondria, and in particular to cristae of the IMM which feature high spontaneous negative curvature due to the presence of phosphatidylethanolamine (PE) and CL lipids in the IMM. Moreover, in the presence of membranes composed of physiologically relevant levels of negatively charged lipids, hIAPP induced membrane curvature and disrupted membrane integrity (Kegulian et al., 2015). Oligomerization of IAPP may facilitate access to the mitochondrial compartment as a result of increased concentration of peptide positive charges on the N-terminal α-helical domain, thereby acquiring a charge density comparable to that of cell-penetrating peptides (CPPs) (Magzoub and Miranker, 2012). Thus, membranes which exert a strong localizing effect such as mitochondrial membranes may support the toxic formation of hIAPP species. Observed mito-toxic effects of hIAPP on INS-1E rat insulinoma β-cells included mitochondrial fragmentation, loss of mitochondrial membrane potential and ATP depletion (Kegulian et al., 2015; Malishev et al., 2019; Ye et al., 2011). Hence, a loss of mitochondrial membrane integrity by proximate toxic hIAPP oligomers provides a novel mechanism for mitochondrial dysfunction in pancreatic β-cells in T2DM.
The CPP properties of hIAPP are also shared by the other amyloid disordered peptides. In fact, in neurodegenerative diseases such as AD and PD, a direct association has been found between oligomerization of Aβ, tau, and α-syn, and disrupted mitochondrial function (Briston and Hicks, 2018). Misfolded neurotoxic aggregates of Aβ 1-42, tau, α-syn and huntingtin have been shown to impair mitochondrial ETC complexes, increase generation of mitochondrial reactive oxygen species (ROS) and disrupt mitochondrial calcium handling. All this leads to a severe impairment of mitochondrial energy metabolism by limiting availability of ATP (Abramov et al., 2017). What is more, mitochondrial CL appears to promote ion-conducting nanopore formation in mitochondrial membranes by toxic tau and α-synuclein species, thereby leading to a loss of mitochondrial structural integrity (Camilleri et al., 2020; Ghio et al., 2019). Therefore, a better understanding of the effects of hIAPP on mitochondria should help uncover cross-talk between pathological pathways in AD, PD and T2DM.
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Journal2021, Chemistry and Physics of LipidsAdam El Saghir, ... Neville Vassallo
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15th European Bioenergetics Conference 2008
2008, Biochimica et Biophysica Acta (BBA) - BioenergeticsAlessandro Rimessi, ... Rosario Rizzuto
Mitochondria are delimited by two membranes. The outer mitochondrial membrane (OMM) is permeable to ions and small proteins (MW < 10 kDa) due to the abundance of a large conductance channel, known as mitochondrial porin or voltage-dependent anion channel (VDAC). It should be noted, however, that the channel appears to be gated in vivo, and permeability is controlled by ATP and other regulatory factors . For these reasons, Ca 2+ diffusion through the OMM was traditionally considered not to be a limiting factor in mitochondrial Ca 2+ uptake. Recent data showed that the availability and selective placement of VDAC channels at ER/mitochondria contact sites facilitate mitochondrial Ca 2+ accumulation, in keeping with the idea that the latter process requires the fast and efficient transfer of Ca 2+ microdomains from the mouth of the Ca 2+ channels located in neighbouring ER or plasma membranes to the transporters of the ion-impermeant inner membrane (IMM) [7,8]. The IMM is an ion-impermeable membrane, with a much larger extension of the OMM (and consequent formation of foldings into the internal space, known as cristae). The activity of respiratory chain complexes allows the translocation of H+ in the space between the two membranes, which consequently generates an electrochemical gradient (ΔμH) composed of a chemical (ΔpH) and electrical (ΔψH) component. In mitochondria, most of the ΔμH established by the respiratory chain is supposed to be in the form of ΔψH (∼180 mV), which provides a huge driving force for Ca 2+ entry into the organelle. Indeed, collapse of the ΔψH by protonophores, such as p-[trifluoromethoxyl]-phenyl-hydrazone (FCCP), abolishes mitochondrial Ca 2+ uptake.
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Journal2008, Biochimica et Biophysica Acta (BBA) - BioenergeticsAlessandro Rimessi, ... Rosario Rizzuto
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6992 | https://torstencurdt.com/tech/posts/modulo-of-negative-numbers/ | programming
Modulo of Negative Numbers
The modulo operator returns the remainder of a division. But things get a little more tricky when you throw negative numbers into the mix.
461
The modulo or often referred to as “mod” represents the remainder of a division. In 1801 Gauss published a book covering modular arithmetics. Later a widely accepted mathematical definition was given by Donald Knuth
```
mod(a, n) = a - n floor(a / n)
```
Doing an integer division and then multiplying it again means finding the biggest number smaller than a that is dividable by n without a remainder. Subtracting this from a yields the remainder of the division and by that the modulo.
But what does the modulu operation do? What can it be used for when coding?
Restricting Bounds
In programming, the modulo operator (% or sometimes mod) often is used to restrict an index to the bounds of an array or length limited data structure.
```
values = [ 3, 4, 5 ]
index = 5
value_at_index = values[ index % values.length ]
```
For the above example this means 5 mod 3 = 2 following the definition is 5 - floor(5/3)3 = 2. This means that no matter the value index has, the array bounds are met.
But is that really the case?
What happens if the dividend or the divisor is signed and holds a negative value?
Turns out the rules of modulo on negative numbers indeed depend on the language you are using.
How does modulus work in java?
How does modulus work in javascript?
How does modulus work in python?
While the code looks pretty much the same in most languages, printing the result shows languages in mostly two different camps.
Only two languages take a different stance: Dart and in particular Zig, which
distinguishes both cases as @rem(a,b) and @mod(a,b) and errors out on a negative divisor.
| Language | 13 mod 3 | -13 mod 3 | 13 mod -3 | -13 mod -3 |
--- ---
| C | 1 | -1 | 1 | -1 |
| C# | 1 | -1 | 1 | -1 |
| C++ | 1 | -1 | 1 | -1 |
| Elixir | 1 | -1 | 1 | -1 |
| Erlang | 1 | -1 | 1 | -1 |
| Go | 1 | -1 | 1 | -1 |
| Java | 1 | -1 | 1 | -1 |
| Javascript | 1 | -1 | 1 | -1 |
| Kotlin | 1 | -1 | 1 | -1 |
| Nim | 1 | -1 | 1 | -1 |
| PHP | 1 | -1 | 1 | -1 |
| Rust | 1 | -1 | 1 | -1 |
| Scala | 1 | -1 | 1 | -1 |
| Swift | 1 | -1 | 1 | -1 |
| Crystal | 1 | 2 | -2 | -1 |
| Haskell | 1 | 2 | -2 | -1 |
| Lua | 1 | 2 | -2 | -1 |
| Python | 1 | 2 | -2 | -1 |
| Ruby | 1 | 2 | -2 | -1 |
| Dart | 1 | 2 | 1 | 2 |
| Zig @rem | 1 | -1 | error | error |
| Zig @mod | 1 | 2 | error | error |
So if you use the modulo operator to ensure correct bounds for accessing a collection, beware that some languages need a little more diligence. A simple and efficient way is to check the sign.
```
int mod(a, b) {
c = a % b
return (c < 0) ? c + b : c
}
```
As another option, you could also apply the modulo twice.
```
int mod(a, b) {
(((a % b) + b) % b)
}
```
Even or Odd
Another pitfall to watch out for is when testing whether a number is odd or even using the modulo operator. Based on the above findings you should always compare against 0.
```
bool is_odd(int n) {
return n % 2 != 0; // could be 1 or -1
}
```
But anyone that has ever looked a layer below C will point out that using the modulo isn’t necessarily the best implementation for is_odd anyway. Multiplication and especially divisions are some of the most expensive instructions on a CPU. If you are dealing with 2-based numbers there is often a faster way.
```
x % 2n == x & (2n - 1) // for n>0
```
At least for a positive divisor, the modulo operation can be replaced with a simple bitwise and operation.
```
x % 2 == x & 1
x % 4 == x & 3
x % 8 == x & 7
...
```
Which allows for a much faster implementation of is_odd.
```
bool is_odd(int n) {
return n & 1 != 0;
}
```
In Summary
The modulo operator can be incredibly useful but developers need also to be aware of the above edge cases and when to use or not use it.
For a more detailed discussion see the wikipedia article.
If you enjoyed reading this article please give it a like.
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6993 | https://www.studocu.com/en-us/document/massachusetts-institute-of-technology/principles-of-applied-mathematics/symmetric-polynomials/103340106 | Symmetric Polynomials: Key Results and Applications in Geometry - Studocu
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Symmetric Polynomials: Key Results and Applications in Geometry
Lecture on the symmetricity of certain polynomials under the rational field.
Original title:
Symmetric-polynomials
Course
Principles Of Applied Mathematics (18. 310)
29 documents
University
Massachusetts Institute of Technology
Academic year:2015/2016
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Arkadii Slinko
1 Introduction
These notes begin with two basic results about symmetric polynomials: the Fundamental Theorem of Sym- metric Polynomials, which states that any symmetric polynomial can be expressed in terms of the appropriate elementary symmetric polynomials, and Muirhead’s inequality, which gives comparisions between the sizes of symmetric polynomials in a given set of variables.
Further on, we analyse relationships between properties of the roots of a cubic and properties of its coefficients (which are symmetric polynomials of the roots). This analysis is then used to derive a number of triangle- geometric inequalities.
We restrict ourselves throughout to polynomials in three variables. The results of Sections 2-3, however, hold (with appropriately generalized statements and proofs) for symmetric polynomials in arbitrarily many variables.
2 The Fundamental Theorem
A polynomial F (x, y, z) is said to be symmetric if
F (x, y, z) = F (x, z, y) = F (y, x, z) = F (y, z, x) = F (z, x, y) = F (z, y, x).
Example 1. F (x, y, z) = x 3 yz + xy 3 z + xyz 3.
The three polynomials
σ 1 (x, y, z) = x + y + z, σ 2 (x, y, z) = xy + yz + zx, σ 3 (x, y, z) = xyz.
are the best-known examples. For instance, they appear in Vieta’s theorem expressing the coefficients of a polynomial in terms of its roots. They are called elementary symmetric polynomials. Of course there are many other important symmetric polynomials, e.
∆(x, y, z) = (x − y) 2 (y − z) 2 (z − x) 2 ,
but the importance of the elementary symmetric polynomials stems from the main theorem of this section, which will assert that all symmetric polynomials can be expressed as polynomials in the elementary symmetric polynomials σ 1 , σ 2 , σ 3.
To start on the proof of this theorem, let us consider firstly the monomials xk 1 yk 2 zk 3 of a fixed degree n = k 1 + k 2 + k 3. We consider these monomials as words in the alphabet of the three letters x, y, z. Let us write down all the monomials in the lexicographic (vocabularly) order. For instance, for n = 3 we shall have:
x 3 > x 2 y > x 2 z > xy 2 > xyz > xz 2 > y 3 > y 2 z > yz 2 > z 3 ,
where the sign “>” means lexicographically earlier. In general, for two monomials of degree n we write
xk 1 yk 2 zk 3 > xm 1 ym 2 zm 3
if k 1 > m 1 or if k 1 = m 1 and k 2 > m 2 and say that the first monomial is earlier than the second one.
Let F (x, y, z) be a symmetric polynomial which monomials are all of the same degree n. Among them there is a monomial that are earlier of all other monomials of F (x, y, z). We shall call it the leading monomial of F (x, y, z).
Example 2. The leading monomial of
F (x, y, z) = x 3 yz + xy 3 z + xyz 3
is x 3 yz.
Lemma 1. A monomial xk 1 yk 2 zk 3 is a leading monomial of some symmetric polynomial if and only if k 1 ≥ k 2 ≥ k 3. If k 1 ≥ k 2 ≥ k 3 the monomial xk 1 yk 2 zk 3 is the leading monomial of σk 11 −k 2 σk 22 −k 3 σk 33.
Proof. If, for instance, k 1 < k 2 , then the monomial xk 1 yk 2 zk 3 cannot be the leading monomial of a symmetric polynomial F (x, y, z), because F (x, y, z) = F (y, x, z) contains also a monomial xk 2 yk 1 zk 3 which is earlier than xk 1 yk 2 zk 3. The case k 2 < k 3 is similar.
The leading monomial of σk 11 −k 2 σk 22 −k 3 σk 33 is equal to the product of the leading monomials of σk 1 1 −k 2 , σ 2 k 2 −k 3 and σk 33 , that is xk 1 −k 2 · (xy)k 2 −k 3 · (xyz)k 3 = xk 1 yk 2 zk 3.
Now we can prove the main theorem of the section.
Theorem 2 (Fundamental Theorem). For any symmetric polynomial F (x, y, z) there exists a (not necessarily symmetric) polynomial f (u, v, w) such that
F (x, y, z) = f (σ 1 , σ 2 , σ 3 ).
Proof. We sketch an algorithm for constructing f. Let xk 1 yk 2 zk 3 be the leading monomial of F (x, y, z). Then k 1 ≥ k 2 ≥ k 3 and the polynomial σ 1 k 1 −k 2 σ 2 k 2 −k 3 σ 3 k 3 has the same leading monomial. Consider
F 1 (x, y, z) = F (x, y, z) − σ 1 k 1 −k 2 σ 2 k 2 −k 3 σ 3 k 3.
This polynomial is also symmetric but its leading monomial is later than that of F (x, y, z). This process of reducing of leading monomials can be continued. Since we have only finite number of monomials of a fixed degree, finally we shall obtain the required presentation of F (x, y, z).
Example 3. In this example we shall not use the complete algorithm exposed in the proof of Theorem 2 since an easier way is possible. Consider
δ(x, y, z) = (x + y − z)(y + z − x)(z + x − y).
Then δ(x, y, z) = −σ 31 + 4σ 1 σ 2 − 8 σ 3.
The following computation shows that this is true:
δ(x, y, z) = (σ 1 − 2 z)(σ 1 − 2 x)(σ 1 − 2 y) = σ 31 − 2(x + y + z)σ 12 + 4(xy + yz + zx)σ 1 − 8 xyz = −σ 13 + 4σ 1 σ 2 − 8 σ 3.
Example 4. Here the complete algorithm cannot be avoided. We leave the routine computation to the reader.
∆(x, y, z) = − 4 σ 31 σ 3 + σ 21 σ 22 + 18σ 1 σ 2 σ 3 − 4 σ 32 − 27 σ 32.
Exercise 1. Express the following symmetric polynomials as polynomials of σ 1 , σ 2 , σ 3 :
(a) x 2 + y 2 + z 2 ;
Proof. We will prove this theorem for the case n = 3. The general case can be proved similarly. We assume that (α) 6 = (β) and not all the xi are equal. Let us consider the following three partial cases from which the general case will follow.
(a) Let (α) = (α 1 , α 2 , α 3 ), and ρ be a positive real number such that ρ < α 1 −α 2 and (α′) = (α 1 −ρ, α 2 +ρ, α 3 ). Then
3!
(
x(α) − x(α
′ ))
= (xα 1 1 xα 2 2 + xα 1 2 xα 2 1 − xα
′ 1 1 x
α′ 2 2 − x
α′ 2 1 x
α′ 1 2 )x
α 3 3
+(xα 1 1 xα 3 2 + xα 1 2 xα 3 1 − x α′ 1 1 x
α′ 2 3 − x
α′ 2 1 x
α′ 1 3 )x
α 3 2 +(xα 2 1 xα 3 2 + xα 2 2 xα 3 1 − x α′ 1 2 x
α′ 2 3 − x
α′ 2 2 x
α′ 1 3 )x
α 3 1.
We claim the right-hand side is positive. Indeed, certainly for any xi and xj ,
xα i 1 xα j 2 + xα i 2 xα j 1 − x α′ 1 i x
α′ 2 j − x
α′ 2 i x
α′ 1 j = (xixj )
α 2 [(xα 1 −α 2 −ρ i − x
α 1 −α 2 −ρ j )(x
ρ i − x
ρ j )
]
≥ 0 ,
since the two round-bracketed differences are either both non-negative or both non-positive. Moreover, since not all the xi are equal, at least one of x 1 6 = x 2 , x 1 6 = x 3 , x 2 6 = x 3 will hold, securing strict inequality. So x(α) > x(α ′ ) as required.
(b) Similarly, if ρ < α 2 − α 3 and (α′′) = (α 1 , α 2 − ρ, α 3 + ρ), then x(α) > x(α ′′) .
(c) The same argument also shows that if α 2 = α′ 2 , then, for any ρ ≤ (α 1 −α 2 ), and (α′′′) = (α 1 −ρ, α 2 , α 3 +ρ), then x(α) > x(α
′′′) .
Suppose now that α 2 < β 2. Then α 2 < β 1 , and ρ = α 1 − β 1 < α 1 − α 2. Then by (a), x(α) > x(α
′ ) , where (α′) = (α 1 − ρ, α 2 + ρ, α 3 ) = (β 1 , α 2 + ρ, α 3 ). As α 3 < β 3 , α 2 + ρ > β 2 , hence x(α
′) > x(β) by (b). Thus x(α) > x(β), as required. The case α 2 > β 2 is dealt with similarly using (b) and (a). If α 2 = β 2 , then the statement follows straight from (c).
Example 7. For all non-negative x 1 , x 2 , x 3
1 3 (
x 1 + x 2 + x 3 ) ≥
1
3 (
√
x 1 x 2 +
√
x 1 x 3 +
√
x 2 x 3 ) ≥ 3
√
x 1 x 2 x 3
or, after substitution a =
√
x 1 , b =
√
x 2 , c =
√
x 3 ,
1 3 (
a 2 + b 2 + c 2 ) ≥
1
3 (
ab + ac + bc) ≥
√ 3
a 2 b 2 c 2.
Exercise 4. Prove that for all non-negative a, b, c the following two inequalities hold:
a 3 + b 3 + c 3 ≥ a 2 (b + c) + b 2 (c + a) + c 2 (a + b), a 3 + b 3 + c 3 ≥ a 2 b + b 2 c + c 2 a.
Establish when these are equalities.
4 Cubic polynomials and their roots
Theorem 4. Let a, b, c be the roots (real or complex) of a cubic equation
x 3 − σ 1 x 2 + σ 2 x − σ 3 = 0. (1)
Then any symmetric function of the roots a, b, c is a polynomial function of the coefficients σ 1 , σ 2 , σ 3.
Proof. Follows from Vieta’s theorem and the Fundamental Theorem.
The value ∆(a, b, c) = (a − b) 2 (b − c) 2 (c − a) 2 ,
is known as the discriminant of the equation (1). We know it is equal to
∆(a, b, c) = − 4 σ 31 σ 3 + σ 21 σ 22 + 18σ 1 σ 2 σ 3 − 4 σ 32 − 27 σ 32.
Theorem 5. Let σ 1 , σ 2 , σ 3 be real. Then the equation (1) has exactly three (possibly repeated) real roots if and only if − 4 σ 31 σ 3 + σ 21 σ 22 + 18σ 1 σ 2 σ 3 − 4 σ 32 − 27 σ 32 ≥ 0. (2)
If σ 1 ≥ 0 , σ 2 ≥ 0 , σ 3 ≥ 0 all these roots are positive. These roots are the lengths of sides of a (nondegenerate) triangle if and only if in addition − σ 13 + 4σ 1 σ 2 − 8 σ 3 > 0. (3)
Proof. Let a, b, c be the roots of (1). If all of them are real, then obviously
∆(a, b, c) = − 4 σ 31 σ 3 + σ 21 σ 22 + 18σ 1 σ 2 σ 3 − 4 σ 32 − 27 σ 32 ≥ 0.
Suppose now that (2) holds, i ∆(a, b, c) ≥ 0. It is known that either all roots a, b, c are real or one of them is real and the other two are complex conjugates. We will show that the later case cannot occur. Indeed, if a is real, b = z, c = z, then
∆(a, b, c) = (a − z) 2 (z − z) 2 (z − a) 2 = [(a − z)(a − z)] 2 (2=(z)z) 2 · i 2 = (a 2 − 2 <(z) · a + |z| 2 ) 2 · 4(=(z)z) 2 · (−1) < 0 ,
where z = <(z) + i=(z). Moreover, (3) together with Example 3 shows that
δ(a, b, c) = (a + b − c)(b + c − a)(c + a − b) > 0.
Either all three factors are positive and we can construct a triangle or two factors are negative. The later case cannot happen since if, for instance, a + b − c < 0 and b + c − a < 0, adding these two inequalities we obtain 2 b < 0.
Corollary 6. A cubic equation x 3 + px + q = 0, (4)
where p and q are real, has three real roots if and only if ∆ = − 4 p 3 − 27 q 2 ≥ 0.
Note that any cubic with real coefficients can be reduced to a cubic of the form (4) by a simple substitution.
5 Triangle geometry
Recall that the lengths of the sides a, b, c of a triangle ABC are the roots of the cubic equation
x 3 − 2 px 2 + (p 2 + r 2 + 4rR)x − 4 prR = 0,
where p, r and R are the semiperimeter, inradius and circumradius respectively of ABC.
Theorem 7. Three positive numbers p, r, R are the semiperimeter, inradius and circumradius respectively of some triangle, if and only if (p 2 − 2 R 2 − 10 Rr + r 2 ) 2 ≤ 4 R(R − 2 r) 3. (5)
Proof. By the second part of Theorem 8,
r ≤
√
3
9
p ≤
1
2
R,
so we need only prove the inequalities involving S. Indeed, applying the second part of Theorem 8 again gives
S = pr ≥ 3
√
3 r 2 ,
that is, r ≤
1
3
√√
3 S =
√ 43
3
√
S. Also by Theorem 8’s first part,
S = pr ≤
3
√
3
2
Rr =
1
3
√
3
·
27
2
Rr ≤
1
3
√
3 (
Rr − 5 r 2 ) ≤
p 2 3
√
3
,
whence 1 3
√√
3 S ≤
√
3
9
p.
Example 8 (IMO, 1962). Let a, b, c be the sides of a triangle of an area S. Prove that
a 2 + b 2 + c 2 ≥ 4
√
3 S.
Solution. Using both parts of Theorem 8,
a 2 + b 2 + c 2 = σ 12 − 2 σ 2 = (2p) 2 − 2(p 2 + r 2 + 4rR) = 2 p 2 − 2 r 2 − 8 rR ≥ 32 rR − 12 r 2 − 8 rR = 24 rR − 12 r 2 ≥ 18 rR ≥ 18 r ·
2
3
√
3
p
= 4
√
3 S.
Many other problems can be solved by this technique.
Exercise 5. Prove that
(a) 36r 2 ≤ 20 rR − 4 r 2 ≤ ab + bc + ca ≤ 4(r + R) 2 ≤ 9 R 2 ;
(b) 24
√
3 r 3 ≤ 12
√
3 r 2 R ≤ abc ≤ 6
√
3 rR 2 ≤ 3
√
3 R 3 ;
(c)
√
3
R
≤
1
a
+ 1
b
+ 1
c
≤
√
3 R
4 r 2
;
(d)
1
R 2
≤
1
ab
+ 1
bc
+ 1
ca
≤
1
4 r 2
;
(e)
1
R 2
≤
1
a 2
+ 1
b 2
+ 1
c 2
≤
1
4 r 2
;
(f) 6 ≤ a p − a
+
b p − b
+
c p − c
= 4
R − 2 r r
;
(g) 9r 2 ≤ ha + hb + hc ≤
9
2
R;
(h)
2
R
≤
1
ha
+
1
h 6
+ 1
hc
= 1
r
;
(i) 9r ≤ ρa + ρb + ρc ≤
9
2
R;
(j)
2
R
≤
1
ρa
+ 1
ρb
+ 1
ρc
= 1
r
;
(k) 3
√
3
r R
≤ sin α + sin β + sin γ ≤
3
√
3
2
;
(l)
3 r R
≤ cos α + cos β + cos γ ≤
3
2.
Exercise 6. Prove that it is possible to construct a triangle from segments of lengths
(a) sin α, sin β, sin γ;
(b) cos 2 α 2 , cos 2 β 2 , cos 2 γ 2.
Exercise 7. Prove the following inequalities:
(a) (2a − p)(b − c) 2 + (2b − p)(c − a) 2 + (2c − p)(a − b) 2 ≥ 0;
(b) (a + b)(b + c)(c + a) ≤ 8 pR(R + 2r)
(c) ab(a + b) + bc(b + c) + ca(c + a) ≤ 8 pR(R + r)
(d) a 3 + b 3 + c 3 ≤ 8 p(R 2 − r 2 )
(e) (p − a)(p − b)(p − c) ≥ (2a − p)(2b − p)(2c − p)
(f) 20S 2 ≤ (a 2 + b 2 )(b + c − a) + (b 2 + c 2 )(c + a − b) + (c 2 + a 2 )(a + b − c) ≤ 8
[
(p − a) 4 + (p − b) 4 + (p − c) 4
]
+ 12S 2 ;
(g) if ABC is acute-angled, then
2 abc(abc + p(a 2 + b 2 + c 2 ) − (a 3 + b 3 + c 3 )) ≥ 5(a 2 + b 2 − c 2 )(b 2 + c 2 − a 2 )(c 2 + a 2 − b 2 );
(h) a 2 + b 2 + c 2 ≤
72 R 4
9 R 2 − 4 r 2
;
(i) (1 − cos α)(1 − cos β)(1 − cos γ) ≥ cos α · cos β · cos γ;
(j) (1 + cos 2α)(1 + cos 2β)(1 + cos 2γ) + cos 2α · cos 2β · cos 2γ ≥ 0.
6 Problems
(USA, 1977-79) Prove that
(a)
1
a
+ 1
b
+ 1
c
≥
9
2 p
;
(b) a 2 + b 2 + c 2 ≥
4 p 2 3 ; (c) p 2 ≥ 3
√
3 S;
(d) a 2 + b 2 + c 2 ≥ 4
√
3 S;
(IMO) Three roots of the equation
x 4 − px 3 + qx 2 − rx + s = 0
are tan A, tan B, tan C, where A, B, C are the angles of a triangle. Determine the fourth root as a function of p, q, r, s (only).
January 26, 2009 mathsolympiad.org
Symmetric Polynomials: Key Results and Applications in Geometry
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Symmetric Polynomials: Key Results and Applications in Geometry
Course: Principles Of Applied Mathematics (18. 310)
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University: Massachusetts Institute of Technology
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New Zealand Mathematical Olympiad Committee
Symmetric P olynomials
A rkadii Slinko
1 In tro duction
These notes b egin with t w o basic results ab out symmetric p olynomials:the F undamen tal Theorem of Sym-
metric P olynomials,which states that an y symmetric polynomial can b e expressed in terms of the appropriate
elementary symmetric p olynomials,and Muirhead’s inequalit y,whic h gives comparisions betw een the sizes of
symmetric p olynomials in a giv en set of v ariables.
F urther on,we analyse relationships betw een properties of the ro ots of a cubic and prop erties of its coefficients
(whic h are symmetric p olynomials of the ro ots).This analysis is then used to deriv e a num ber of triangle-
geometric inequalities.
W e restrict ourselv es throughout to p olynomials in three v ariables.The results of Sections 2-3,how ev er,hold
(with appropriately generalized statemen ts and pro ofs)for symmetric polynomials in arbitrarily many v ariables.
2 The F undamen tal Theorem
A p olynomial F(x,y,z)is said to be symmetric if
F(x,y,z)=F(x,z,y)=F(y,x,z)=F(y,z,x)=F(z,x,y)=F(z,y,x).
Example 1.F(x,y,z)=x 3 y z+xy 3 z+xy z 3.
The three p olynomials
σ 1(x,y,z)=x+y+z,
σ 2(x,y,z)=xy+y z+z x,
σ 3(x,y,z)=xy z.
are the b est-kno wn examples.F or instance,they app ear in Vieta’s theorem expressing the coefficients of a
p olynomial in terms of its roots.They are called elementary symmetric p olynomials.Of course there are man y
other imp ortan t symmetric p olynomials,e.g.
∆(x,y,z)=(x−y)2(y−z)2(z−x)2,
but the imp ortance of the elemen tary symmetric polynomials stems from the main theorem of this section,
whic h will assert that all symmetric p olynomials can b e expressed as polynomials in the elementary symmetric
p olynomials σ 1,σ 2,σ 3.
T o start on the pro of of this theorem,let us consider firstly the monomials x k 1 y k 2 z k 3 of a fixed degree n=
k 1+k 2+k 3.W e consider these monomials as w ords in the alphab et of the three letters x,y,z.Let us write
do wn all the monomials in the lexicographic(vocabularly)order.F or instance,for n=3 w e shall ha ve:
x 3>x 2 y>x 2 z>xy 2>xy z>xz 2>y 3>y 2 z>y z 2>z 3,
where the sign“>”means lexicographically earlier.In general,for t w o monomials of degree n we write
x k 1 y k 2 z k 3>x m 1 y m 2 z m 3
1
if k 1>m 1 or if k 1=m 1 and k 2>m 2 and sa y that the first monomial is earlier than the second one.
Let F(x,y,z)b e a symmetric polynomial which monomials are all of the same degree n.Among them there
is a monomial that are earlier of all other monomials of F(x,y,z).W e shall call it the le ading monomial of
F(x,y,z).
Example 2.The leading monomial of
F(x,y,z)=x 3 y z+xy 3 z+xy z 3
is x 3 y z.
Lemma 1.A monomial x k 1 y k 2 z k 3 is a le ading monomial of some symmetric p olynomial if and only if k 1≥
k 2≥k 3.If k 1≥k 2≥k 3 the monomial x k 1 y k 2 z k 3 is the le ading monomial of σ k 1−k 2
1 σ k 2−k 3
2 σ k 3
3.
Pr o of.If,for instance,k 1<k 2,then the monomial x k 1 y k 2 z k 3 cannot b e the leading monomial of a symmetric
p olynomial F(x,y,z),because F(x,y,z)=F(y,x,z)contains also a monomial x k 2 y k 1 z k 3 whic h is earlier than
x k 1 y k 2 z k 3.The case k 2<k 3 is similar.
The leading monomial of σ k 1−k 2
1 σ k 2−k 3
2 σ k 3
3 is equal to the pro duct of the leading monomials of σ k 1−k 2
1,σ k 2−k 3
2
and σ k 3
3,that is x k 1−k 2·(xy)k 2−k 3·(xy z)k 3=x k 1 y k 2 z k 3.
No w we can pro v e the main theorem of the section.
Theorem 2(F undamen tal Theorem).F or any symmetric p olynomial F(x,y,z)ther e exists a(not ne c essarily
symmetric)p olynomial f(u,v,w)such that
F(x,y,z)=f(σ 1,σ 2,σ 3).
Pr o of.W e sketc h an algorithm for constructing f.Let x k 1 y k 2 z k 3 be the leading monomial of F(x,y,z).Then
k 1≥k 2≥k 3 and the p olynomial σ k 1−k 2
1 σ k 2−k 3
2 σ k 3
3 has the same leading monomial.Consider
F 1(x,y,z)=F(x,y,z)−σ k 1−k 2
1 σ k 2−k 3
2 σ k 3
3.
This p olynomial is also symmetric but its leading monomial is later than that of F(x,y,z).This pro cess of
reducing of leading monomials can b e con tin ued.Since w e ha ve only finite n um b er of monomials of a fixed
degree,finally we shall obtain the required presentation of F(x,y,z).
Example 3.In this example we shall not use the complete algorithm exposed in the pro of of Theorem 2 since
an easier w ay is possible.Consider
δ(x,y,z)=(x+y−z)(y+z−x)(z+x−y).
Then
δ(x,y,z)=−σ 3
1+4 σ 1 σ 2−8 σ 3.
The follo wing computation shows that this is true:
δ(x,y,z)=(σ 1−2 z)(σ 1−2 x)(σ 1−2 y)
=σ 3
1−2(x+y+z)σ 2
1+4(xy+y z+z x)σ 1−8 xy z
=−σ 3
1+4 σ 1 σ 2−8 σ 3.
Example 4.Here the complete algorithm cannot b e a voided.W e lea ve the routine computation to the reader.
∆(x,y,z)=−4 σ 3
1 σ 3+σ 2
1 σ 2
2+18 σ 1 σ 2 σ 3−4 σ 3
2−27 σ 2
3.
Exercise 1.Express the follo wing symmetric p olynomials as p olynomials of σ 1,σ 2,σ 3:
(a)x 2+y 2+z 2;
2
(b)x 3+y 3+z 3;
(c)x 4+y 4+z 4;
(d)x 2 y+y 2 z+z 2 x+xy 2+y z 2+z x 2;
(e)x 2 y 2+y 2 z 2+z 2 x 2.
Exercise 2.Chec k Example 4.
Exercise 3.Express as a function of σ 1,σ 2,σ 3:
(a)1
x 2+1
y 2+1
z 2;
(b)1
xy+1
y z+1
z x.
3 Muirhead’s theorem
Let us consider the set S n of sequences(α)=(α 1,...,α n)with the follo wing tw o properties:
•α 1+α 2+...+α n=1,
•α 1≥α 2≥...≥α n≥0.
F or an y tw o sequences(α)and(β)from S w e say that(α)majorises(β)if
α 1+α 2+...+α r≥β 1+β 2+...+β r
for all 1≤r<n.W e denote this as(α)(β).If(α)(β)and(α)6=(β),we will write(α)(β).
Example 5.(1,0,0)(1
2,1
2,0)(1
2,1
3,1
6)(1
3,1
3,1
3).
No w we in troduce one more notation:for(α)from S n we denote
x(α)=1
n!(x α 1
1 x α 2
2...x α n
n+...),
where the dots denote all n!−1 terms obtained by permutations of α’s.
Example 6.Let n=3.Then
x(1,0,0)=1
6(x 1
1 x 0
2 x 0
3+x 1
1 x 0
3 x 0
2+x 0
1 x 1
2 x 0
3+x 0
1 x 1
2 x 0
3+x 0
1 x 0
2 x 1
3+x 0
1 x 0
2 x 1
3)
=1
3(x 1+x 2+x 3);
x(1
2,1
2,0)=1
6(x
1
2
1 x
1
2
2 x 0
3+x
1
2
1 x
1
2
2 x 0
3+x 0
1 x
1
2
2 x
1
2
3+x 0
1 x
1
2
2 x
1
2
3+x
1
2
1 x 0
2 x
1
2
3+x
1
2
1 x 0
2 x
1
2
3)
=1
3(√x 1 x 2+√x 1 x 3+√x 2 x 3);
x(1
3,1
3,1
3)=3
√x 1 x 2 x 3.
Theorem 3(Muirhead).If(α)(β),then the ine quality
x(α)≥x(β)
holds for al l non-ne gative x 1,...,x n.Ther e is e quality only when either(α)=(β)or al l the x i ar e e qual.
3
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Pr o of.W e will prov e this theorem for the case n=3.The general case can b e prov ed similarly.W e assume
that(α)6=(β)and not all the x i are equal.Let us consider the following three partial cases from whic h the
general case will follo w.
(a)Let(α)=(α 1,α 2,α 3),and ρ b e a positive real n umber such that ρ<α 1−α 2 and(α 0)=(α 1−ρ,α 2+ρ,α 3).
Then
3!x(α)−x(α 0)=(x α 1
1 x α 2
2+x α 2
1 x α 1
2−x α 0
1
1 x α 0
2
2−x α 0
2
1 x α 0
1
2)x α 3
3
+(x α 1
1 x α 2
3+x α 2
1 x α 1
3−x α 0
1
1 x α 0
2
3−x α 0
2
1 x α 0
1
3)x α 3
2
+(x α 1
2 x α 2
3+x α 2
2 x α 1
3−x α 0
1
2 x α 0
2
3−x α 0
2
2 x α 0
1
3)x α 3
1.
W e claim the righ t-hand side is p ositiv e.Indeed,certainly for an y x i and x j,
x α 1
i x α 2
j+x α 2
i x α 1
j−x α 0
1
i x α 0
2
j−x α 0
2
i x α 0
1
j=(x i x j)α 2(x α 1−α 2−ρ
i−x α 1−α 2−ρ
j)(x ρ
i−x ρ
j)≥0,
since the t wo round-brac k eted differences are either b oth non-negativ e or b oth non-positive.Moreov er,
since not all the x i are equal,at least one of x 1 6=x 2,x 1 6=x 3,x 2 6=x 3 will hold,securing strict inequality.
So x(α)>x(α 0)as required.
(b)Similarly,if ρ<α 2−α 3 and(α 00)=(α 1,α 2−ρ,α 3+ρ),then x(α)>x(α 00).
(c)The same argumen t also shows that if α 2=α 0
2,then,for any ρ≤(α 1−α 2),and(α 000)=(α 1−ρ,α 2,α 3+ρ),
then x(α)>x(α 000).
Supp ose no w that α 2<β 2.Then α 2<β 1,and ρ=α 1−β 1<α 1−α 2.Then by(a),x(α)>x(α 0),where
(α 0)=(α 1−ρ,α 2+ρ,α 3)=(β 1,α 2+ρ,α 3).As α 3<β 3,α 2+ρ>β 2,hence x(α 0)>x(β)by(b).Th us
x(α)>x(β),as required.The case α 2>β 2 is dealt with similarly using(b)and(a).If α 2=β 2,then the
statemen t follows straigh t from(c).
Example 7.F or all non-negativ e x 1,x 2,x 3
1
3(x 1+x 2+x 3)≥1
3(√x 1 x 2+√x 1 x 3+√x 2 x 3)≥3
√x 1 x 2 x 3
or,after substitution a=√x 1,b=√x 2,c=√x 3,
1
3(a 2+b 2+c 2)≥1
3(ab+ac+bc)≥3
√a 2 b 2 c 2.
Exercise 4.Pro v e that for all non-negativ e a,b,c the following t wo inequalities hold:
a 3+b 3+c 3≥a 2(b+c)+b 2(c+a)+c 2(a+b),
a 3+b 3+c 3≥a 2 b+b 2 c+c 2 a.
Establish when these are equalities.
4 Cubic p olynomials and their ro ots
Theorem 4.L et a,b,c b e the r o ots(r e al or c omplex)of a cubic e quation
x 3−σ 1 x 2+σ 2 x−σ 3=0.(1)
Then any symmetric function of the r o ots a,b,c is a p olynomial function of the c o efficients σ 1,σ 2,σ 3.
Pr o of.F ollows from Vieta’s theorem and the F undamen tal Theorem.
4
The v alue
∆(a,b,c)=(a−b)2(b−c)2(c−a)2,
is kno wn as the discriminant of the equation(1).W e kno w it is equal to
∆(a,b,c)=−4 σ 3
1 σ 3+σ 2
1 σ 2
2+18 σ 1 σ 2 σ 3−4 σ 3
2−27 σ 2
3.
Theorem 5.L et σ 1,σ 2,σ 3 b e r e al.Then the e quation(1)has exactly thr e e(p ossibly r ep e ate d)r e al r o ots if and
only if
−4 σ 3
1 σ 3+σ 2
1 σ 2
2+18 σ 1 σ 2 σ 3−4 σ 3
2−27 σ 2
3≥0.(2)
If σ 1≥0,σ 2≥0,σ 3≥0 al l these r o ots ar e p ositive.These r o ots ar e the lengths of sides of a(nonde gener ate)
triangle if and only if in addition
−σ 3
1+4 σ 1 σ 2−8 σ 3>0.(3)
Pr o of.Let a,b,c b e the roots of(1).If all of them are real,then ob viously
∆(a,b,c)=−4 σ 3
1 σ 3+σ 2
1 σ 2
2+18 σ 1 σ 2 σ 3−4 σ 3
2−27 σ 2
3≥0.
Supp ose no w that(2)holds,i.e∆(a,b,c)≥0.It is kno wn that either all ro ots a,b,c are real or one of them is
real and the other t wo are complex conjugates.W e will sho w that the later case cannot o ccur.Indeed,if a is
real,b=z,c=z,then
∆(a,b,c)=(a−z)2(z−z)2(z−a)2
=[(a−z)(a−z)]2(2=(z)z)2·i 2
=(a 2−2<(z)·a+|z|2)2·4(=(z)z)2·(−1)
<0,
where z=<(z)+i=(z).Moreov er,(3)together with Example 3 shows that
δ(a,b,c)=(a+b−c)(b+c−a)(c+a−b)>0.
Either all three factors are p ositiv e and w e can construct a triangle or tw o factors are negativ e.The later case
cannot happ en since if,for instance,a+b−c<0 and b+c−a<0,adding these tw o inequalities w e obtain
2 b<0.
Corollary 6.A cubic e quation
x 3+px+q=0,(4)
wher e p and q ar e r e al,has thr e e r e al r o ots if and only if∆=−4 p 3−27 q 2≥0.
Note that an y cubic with real co efficien ts can b e reduced to a cubic of the form(4)by a simple substitution.
5 T riangle geometry
Recall that the lengths of the sides a,b,c of a triangle AB C are the ro ots of the cubic equation
x 3−2 px 2+(p 2+r 2+4 r R)x−4 prR=0,
where p,r and R are the semip erimeter,inradius and circumradius resp ectiv ely of AB C.
Theorem 7.Thr e e p ositive numb ers p,r,R ar e the semip erimeter,inr adius and cir cumr adius r esp e ctively of
some triangle,if and only if
(p 2−2 R 2−10 Rr+r 2)2≤4 R(R−2 r)3.(5)
5
Pr o of.W e wan t to show that the roots of the cubic
x 3−2 px 2+(p 2+r 2+4 r R)x−4 prR=0
are the sides of a triangle.T o do this we use Theorem 5 about the relationship b etw een properties of a cubic’s
ro ots and properties of its co efficients.
First,we need to sho w that the ro ots of this cubic are real.Theorem 5 sa ys this happ ens if and only if the
cubic’s co efficien ts satisfy condition(2).Substituting and manipulating,we find this to be equiv alen t to the
condition(5)sp ecified.
Next w e need to show that all three roots of the cubic are p ositive;this is clear b y Theorem 5,since 2 p,
p 2+r 2+4 r R and 4 prR are all p ositive.
Finally w e will chec k that the roots of our cubic are the sides of a triangle,using Theorem 5’s further condition
(3).Indeed,for every positive p,r,R we ha v e
−(2 p)3+4(2 p)(p 2+r 2+4 r R)−8·4 prR=8 pr 2>0.
Corollary 8.F or any triangle AB C,
(a)27 r 2≤16 Rr−5 r 2≤p 2≤4 R 2+4 Rr+3 r 2≤27
7 R 2.
(b)3√3 r≤p≤3√3
2 R.
These ine qualities b e c ome e qualities if and only if the triangle AB C is e quilater al.
Pr o of.By Theorem 7,
2 R 2+10 Rr−r 2−2(R−2 r)p R(R−2 r)≤p 2≤2 R 2+10 Rr−r 2+2(R−2 r)p R(R−2 r).
It follo ws that
2 R 2+10 r R−r 2−2(R−2 R)p R(R−2 r)=16 rR−5 r 2+h(R−2 r)−p R(R−2 r)i 2
≥16 r R−5 r 2
and
2 R 2+10 r R−r 2+2(R−2 r)p R(R−2 r)=4 R 2+4 rR+3 r 2−h(R−2 r)−p R(R−2 r)i 2
≤4 R 2+4 r R+3 r 2.
In b oth cases equalit y o ccurs if R=2 r whic h means by Euler’s theorem that AB C is equilateral.Using R≥2 r
w e conclude further
16 r R−5 r 2≥27 r 2,
4 R 2+4 r R+3 r 2≤27
4 R 2.
This pro ves(a).Part(b)follo ws directly.
Theorem 9.If S is the ar e a of a triangle AB C,then
r≤
4
√3
3√S≤√3
9 p≤1
2 R.
If e quality o c curs in any of these,then AB C is e quilaterial.
6
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Principles Of Applied Mathematics18. 310Massachusetts Institute of Technology
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Discover more from: Principles Of Applied Mathematics18. 310Massachusetts Institute of Technology29 documents Go to course
4 653e2005 b8c74e3db10916 a8226c0f04 MIT18 310F13 Homework 10 Principles Of Applied Mathematics 100% (3)
2 Principles Of Applied Mathematics cheat-sheet Principles Of Applied Mathematics 100% (3)
3 Abf00d007c2ed257001 d191ed3c1f98d MIT18 310F13 Homework 1 Principles Of Applied Mathematics 100% (2)
14 2021 AMC 10 B Official Solutions and Problem Breakdown Principles Of Applied Mathematics 100% (1)
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Pr o of.By the second part of Theorem 8,
r≤√3
9 p≤1
2 R,
so w e need only prov e the inequalities in volving S.Indeed,applying the second part of Theorem 8 again gives
S=pr≥3√3 r 2,
that is,r≤1
3 q√3 S=
4
√3
3√S.Also by Theorem 8’s first part,
S=pr≤3√3
2 Rr=1
3√3·27
2 Rr≤1
3√3(16 Rr−5 r 2)≤p 2
3√3,
whence 1
3 q√3 S≤√3
9 p.
Example 8(IMO,1962).Let a,b,c b e the sides of a triangle of an area S.Pro v e that
a 2+b 2+c 2≥4√3 S.
Solution.Using b oth parts of Theorem 8,
a 2+b 2+c 2=σ 2
1−2 σ 2
=(2 p)2−2(p 2+r 2+4 r R)
=2 p 2−2 r 2−8 r R
≥32 r R−12 r 2−8 rR
=24 r R−12 r 2
≥18 r R
≥18 r·2
3√3 p
=4√3 S.
Man y other problems can b e solv ed by this tec hnique.
Exercise 5.Pro v e that
(a)36 r 2≤20 r R−4 r 2≤ab+bc+ca≤4(r+R)2≤9 R 2;
(b)24√3 r 3≤12√3 r 2 R≤abc≤6√3 r R 2≤3√3 R 3;
(c)√3
R≤1
a+1
b+1
c≤√3 R
4 r 2;
(d)1
R 2≤1
ab+1
bc+1
ca≤1
4 r 2;
(e)1
R 2≤1
a 2+1
b 2+1
c 2≤1
4 r 2;
(f)6≤a
p−a+b
p−b+c
p−c=4 R−2 r
r;
(g)9 r 2≤h a+h b+h c≤9
2 R;
7
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6994 | https://mathworld.wolfram.com/HyperbolicTangent.html | Hyperbolic Tangent
Download
Wolfram Notebook
By way of analogy with the usual tangent
| | |
--- |
| | (1) |
the hyperbolic tangent is defined as
| | | | |
--- --- |
| | | | (2) |
| | | | (3) |
| | | | (4) |
where
is the hyperbolic sine and is the hyperbolic cosine.
The notation
is sometimes also used (Gradshteyn and Ryzhik 2000, p. xxix).
is implemented in the Wolfram Language
as Tanh[z].
Special values include
| | | | |
--- --- |
| | | | (5) |
| | | | (6) |
where
is the golden ratio.
The derivative of is
| | |
--- |
| | (7) |
and higher-order derivatives are given by
| | |
--- |
| | (8) |
where
is an Eulerian number.
The indefinite integral is given by
| | |
--- |
| | (9) |
has Taylor series
| | | | |
--- --- |
| | | | (10) |
| | | | (11) |
(OEIS A002430 and A036279).
As Gauss showed in 1812, the hyperbolic tangent can be written using a continued
fraction as
| | |
--- |
| | (12) |
(Wall 1948, p. 349; Olds 1963, p. 138). This continued fraction is also known as Lambert's continued fraction
(Wall 1948, p. 349).
The hyperbolic tangent satisfies the second-order
ordinary differential equation
| | |
--- |
| | (13) |
together with the boundary conditions and .
See also
Bernoulli Number, Catenary, Correlation Coefficient--Bivariate
Normal Distribution, Fisher's z-'-Transformation,
Hyperbolic Cotangent, Hyperbolic
Functions, Inverse Hyperbolic Tangent,
Lorentz Group, Mercator
Projection, Oblate Spheroidal Coordinates,
Pseudosphere, Surface
of Revolution, Tangent, Tractrix
Explore with Wolfram|Alpha
More things to try:
inverse hyperbolic tangent of x
inverse hyperbolic tangent of .99
d/dx hyperbolic tangent(x)
References
Abramowitz, M. and Stegun, I. A. (Eds.). "Hyperbolic Functions." §4.5 in Handbook
of Mathematical Functions with Formulas, Graphs, and Mathematical Tables, 9th printing.
New York: Dover, pp. 83-86, 1972.Gradshteyn, I. S. and Ryzhik,
I. M. Tables
of Integrals, Series, and Products, 6th ed. San Diego, CA: Academic Press,
2000.Jeffrey, A. "Hyperbolic Identities." §2.5 in Handbook
of Mathematical Formulas and Integrals, 2nd ed. Orlando, FL: Academic Press,
pp. 117-122, 2000.Olds, C. D. Continued
Fractions. New York: Random House, 1963.Sloane, N. J. A.
Sequences A002430/M2100 and A036279
in "The On-Line Encyclopedia of Integer Sequences."Spanier,
J. and Oldham, K. B. "The Hyperbolic Tangent and Cotangent Functions." Ch. 30 in An
Atlas of Functions. Washington, DC: Hemisphere, pp. 279-284, 1987.Wall,
H. S. Analytic
Theory of Continued Fractions. New York: Chelsea, 1948.Zwillinger,
D. (Ed.). "Hyperbolic Functions." §6.7 in CRC
Standard Mathematical Tables and Formulae. Boca Raton, FL: CRC Press, pp. 476-481
1995.
Referenced on Wolfram|Alpha
Hyperbolic Tangent
Cite this as:
Weisstein, Eric W. "Hyperbolic Tangent."
From MathWorld--A Wolfram Resource.
Subject classifications
Find out if you already have access to Wolfram tech through your organization |
6995 | https://arxiv.org/html/2405.18789v2 | Published Time: Wed, 19 Jun 2024 00:16:14 GMT
An Overview of Some Single Machine Scheduling Problems: Polynomial Algorithms, Complexity and Approximability
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Why HTML?Report IssueBack to AbstractDownload PDF
Table of Contents
1 Introduction
2 Some Basics
3 Polynomially Solvable Cases, Complexity and Approximability Issues
3.1 Scheduling a Single Machine with Release Times and Due Dates to Minimize Maximum Job Lateness
3.2 Minimizing the Number of Late Jobs
3.2.1 Nonpreemptive Scheduling to Minimize the Number of Late Jobs: 1|r j;p j=p|∑U j conditional 1 subscript 𝑟 𝑗 subscript 𝑝 𝑗 conditional 𝑝 subscript 𝑈 𝑗 1|r_{j};p_{j}=p|\sum U_{j}1 | italic_r start_POSTSUBSCRIPT italic_j end_POSTSUBSCRIPT ; italic_p start_POSTSUBSCRIPT italic_j end_POSTSUBSCRIPT = italic_p | ∑ italic_U start_POSTSUBSCRIPT italic_j end_POSTSUBSCRIPT
3.2.2 Preemptive Scheduling to Maximize the Number of on Time Jobs: 1|r j;p m t n|∑U j 1|r_{j};pmtn|\sum U_{j}1 | italic_r start_POSTSUBSCRIPT italic_j end_POSTSUBSCRIPT ; italic_p italic_m italic_t italic_n | ∑ italic_U start_POSTSUBSCRIPT italic_j end_POSTSUBSCRIPT
3.3 Preemptive Scheduling to Minimize the Weighted Number of Late Jobs with Equal Length: 1|r j;p j=p;pmtn|∑w jU j conditional 1 subscript 𝑟 𝑗 subscript 𝑝 𝑗 𝑝 conditional 𝑝 𝑚 𝑡 𝑛 subscript 𝑤 𝑗 subscript 𝑈 𝑗 1|r_{j};p_{j}=p;pmtn|\sum w_{j}U_{j}1 | italic_r start_POSTSUBSCRIPT italic_j end_POSTSUBSCRIPT ; italic_p start_POSTSUBSCRIPT italic_j end_POSTSUBSCRIPT = italic_p ; italic_p italic_m italic_t italic_n | ∑ italic_w start_POSTSUBSCRIPT italic_j end_POSTSUBSCRIPT italic_U start_POSTSUBSCRIPT italic_j end_POSTSUBSCRIPT
3.4 Scheduling to Minimize the Weighted Number of Late jobs with Deadlines and Release/Due Date Intervals
3.5 Scheduling with Tardiness-based Objectives
3.6 Preemptive Scheduling in Overloaded Systems
3.7 Scheduling with Restarts
3.8 Scheduling with Financial Resource Constraints
4 Concluding Remarks
References
License: arXiv.org perpetual non-exclusive license
arXiv:2405.18789v2 [cs.DS] 17 Jun 2024
An Overview of Some Single Machine Scheduling Problems: Polynomial Algorithms, Complexity and Approximability
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Nodari Vakhania 1,, Frank Werner 2, Víctor Pacheco-Valencia 1
and Kevin Johedan Ramírez-Fuentes 1
1 Centro de Investigación en Ciencias
Universidad Autónoma del Estado de Morelos
Cuernavaca 62209, Mexico
2 Faculty of Mathematics, Otto-von-Guericke University,
39106 Magdeburg, Germany
Correspondence: nodari@uaem.mx
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Abstract:
Since the publication of the first scheduling paper in 1954, a huge number of works dealing with different types of single machine problems appeared. They addressed many heuristics and enumerative procedures, complexity results or structural properties of certain problems. Regarding surveys, often particular subjects like special objective functions are discussed, or more general scheduling problems were surveyed, where a substantial part is devoted to single machine problems. In this paper we present some results on polynomial algorithms, complexity and approximation issues, where the main focus is on results, which have been published during the last decades in papers, where at least one of the first two authors of this paper was involved. We hope that the reviewed results will stimulate further investigation in related research fields.
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Keywords: Single machine scheduling; Polynomial algorithms; Complexity; Approximation algorithms
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MSC classification: 90B35, 90C59
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1 Introduction
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Scheduling problems arise in many real-life applications including our everyday life. We commonly desire to optimize our time arranging different activities, where the time is limited and also the resources that we use are scarce. We wish to optimize some of our personal criteria but we also depend on the availability of the latter resources. As a simple but very common daily example, consider an apartment with a single bathroom, where a small family lives. The family members gets up at some fixed time, and all of them need to spend some time in the bathroom, which is a scarce resource (or machine). Typically such a (renewable) resource may serve a single object or job, a family member in our example. Every member of the family needs to leave the apartment at some fixed time to reach his/her job or school without delay. This time can be different for different members of the family. Here arises a small optimization problem that asks to schedule the time interval for each member of the family so as to meet the deadline of each member of the family.
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Generally, in a scheduling problem we are given jobs or tasks which are some requests that need to be performed by some resource, a machine or a processor. In our small example, bathroom requirements for each member of the family are jobs and the bathroom is a machine. We can give a vast amount of examples of jobs and machines in our practical life; e.g., a job in a factory or a software application in a computer or on a smart phone, or a lesson at school. Here machines, computers, smart phones, classrooms and teachers are resource examples of. Every jobs requires some time units on a resource, whereas a resource is able to process at most one job at a time. Indeed, one classroom cannot be used for two different classes simultaneously or a a teacher cannot give two or more classes at the same time. There is a restriction on the whole time window during which a factory or a teacher can work. It is a scheduling task to arrange the order for processing the jobs. Often there is some minimization or maximization criterion called an objective function. For example, a very common criterion is to minimize the maximum job or machine completion time or the makespan which is commonly denoted by C max subscript 𝐶 C_{\max}italic_C start_POSTSUBSCRIPT roman_max end_POSTSUBSCRIPT.
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We shall denote the given set of n 𝑛 n italic_n jobs by {1,2,…,n}1 2…𝑛{1,2,\dots,n}{ 1 , 2 , … , italic_n }. The processing time of job j 𝑗 j italic_j is denoted by p j subscript 𝑝 𝑗 p_{j}italic_p start_POSTSUBSCRIPT italic_j end_POSTSUBSCRIPT. Job j 𝑗 j italic_j may have some additional characteristics. For example, the time moment from when job j 𝑗 j italic_j is available to be processed is called its release or readiness time and is commonly denoted by r j subscript 𝑟 𝑗 r_{j}italic_r start_POSTSUBSCRIPT italic_j end_POSTSUBSCRIPT. In many applications, a job, once completed by the machine, needs to be delivered to the customer by an independent transportation unit, which may not be a scarce resource in a given model. Such a delivery time (or tail) of job j 𝑗 j italic_j is commonly denoted by q j subscript 𝑞 𝑗 q_{j}italic_q start_POSTSUBSCRIPT italic_j end_POSTSUBSCRIPT. The job delivery takes no machine time (a job is delivered by an independent agent to a customer), though q j subscript 𝑞 𝑗 q_{j}italic_q start_POSTSUBSCRIPT italic_j end_POSTSUBSCRIPT contributes to the completion time of job j 𝑗 j italic_j. Hence, the makespan crucially depends on the job delivery times. The due date d j subscript 𝑑 𝑗 d_{j}italic_d start_POSTSUBSCRIPT italic_j end_POSTSUBSCRIPT is the desirable completion time of job j 𝑗 j italic_j. Both above parameters specify the urgency of job j 𝑗 j italic_j: the more is q j subscript 𝑞 𝑗 q_{j}italic_q start_POSTSUBSCRIPT italic_j end_POSTSUBSCRIPT, the more urgent is job j 𝑗 j italic_j; likewise, the less is d j subscript 𝑑 𝑗 d_{j}italic_d start_POSTSUBSCRIPT italic_j end_POSTSUBSCRIPT, the more urgent is job j 𝑗 j italic_j.
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All above parameters are normally non-negative integral numbers. In some cases, job j 𝑗 j italic_j is not allowed to be completed after its due date. Then the term deadline instead of “due date” is commonly used. Thus, if deadlines are given, our objective is to find a schedule in which all jobs are completed before or at their deadlines. With due dates, our objective will be to minimize or maximize some due date oriented objective function. Let the lateness of a job in a schedule be the difference between its completion time and its due date. One of the common objective functions is the maximal job lateness L max subscript 𝐿 L_{\max}italic_L start_POSTSUBSCRIPT roman_max end_POSTSUBSCRIPT (which we have to minimize). Another common objective function is the number of jobs completed at or before their due dates ∑U j subscript 𝑈 𝑗\sum U_{j}∑ italic_U start_POSTSUBSCRIPT italic_j end_POSTSUBSCRIPT which we have to maximize (U j subscript 𝑈 𝑗 U_{j}italic_U start_POSTSUBSCRIPT italic_j end_POSTSUBSCRIPT is a 0-1 variable indicating whether job j 𝑗 j italic_j is completed on time or not in the schedule). We may allow to omit some jobs in the case when the total workload is more than the machine capacity. In many applications, job due dates are important since, if a job is not completed by its due date then its benefit or functionality becomes neglectable. Hence, we may wish to maximize the throughput, i.e.,the number of jobs completed on time by their due dates. We may also consider more general criteria to maximize some profits. The profit of job j 𝑗 j italic_j can be expressed by its weight which is commonly denoted by w j subscript 𝑤 𝑗 w_{j}italic_w start_POSTSUBSCRIPT italic_j end_POSTSUBSCRIPT. Job preemptions might be allowed or not. In a preemptive scheduling problem, a job can be interrupted on the machine and later resumed on the machine. In a non-preemptive scheduling problem, a job J j subscript 𝐽 𝑗 J_{j}italic_J start_POSTSUBSCRIPT italic_j end_POSTSUBSCRIPT, once scheduled on the machine, should be processed without any interruption until its full completion on the machine. The machine is called idle at some moment t 𝑡 t italic_t if no job is processed by the machine at that moment, otherwise it is said to be busy.
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Problem restrictions on the way in which the jobs are to be performed by machines define the set of feasible solutions. For example, a machine can process at most one job simultaneously; the precedence relations defined on the set of jobs, restrict the relative order in which the jobs can be scheduled. Scheduling problems are sometimes referred to as sequencing problems since the jobs need to be scheduled in a sequential manner (as no job overlapping is permitted). The precedence relations between the jobs can be represented by directed graphs which are often referred to as precedence (task) graphs. In such graphs, nodes represent jobs, and there is an arc (i,j)𝑖 𝑗(i,j)( italic_i , italic_j ) if and only if job i 𝑖 i italic_i immediately precedes job j 𝑗 j italic_j.
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A schedule prescribes to each job the starting time on the machine creating in this way a total order on the set of jobs by sequencing the jobs on the machine. Denote by s j S subscript superscript 𝑠 𝑆 𝑗 s^{S}{j}italic_s start_POSTSUPERSCRIPT italic_S end_POSTSUPERSCRIPT start_POSTSUBSCRIPT italic_j end_POSTSUBSCRIPT (c j S subscript superscript 𝑐 𝑆 𝑗 c^{S}{j}italic_c start_POSTSUPERSCRIPT italic_S end_POSTSUPERSCRIPT start_POSTSUBSCRIPT italic_j end_POSTSUBSCRIPT, respectively) the starting time (completion time, respectively) of job j 𝑗 j italic_j in a schedule S 𝑆 S italic_S. A feasible schedule is a schedule which satisfies all restrictions of the given problem. Since all problems imply resource restrictions, in any feasible schedule, if job j 𝑗 j italic_j is scheduled on the machine, then s j S≥s i S+p i superscript subscript 𝑠 𝑗 𝑆 superscript subscript 𝑠 𝑖 𝑆 subscript 𝑝 𝑖 s_{j}^{S}\geq s_{i}^{S}+p_{i}italic_s start_POSTSUBSCRIPT italic_j end_POSTSUBSCRIPT start_POSTSUPERSCRIPT italic_S end_POSTSUPERSCRIPT ≥ italic_s start_POSTSUBSCRIPT italic_i end_POSTSUBSCRIPT start_POSTSUPERSCRIPT italic_S end_POSTSUPERSCRIPT + italic_p start_POSTSUBSCRIPT italic_i end_POSTSUBSCRIPT, for any job i 𝑖 i italic_i scheduled earlier on the machine in S 𝑆 S italic_S (recal that the machine can handle only one job at any time). Besides, if j 𝑗 j italic_j has a release time then it cannot start before this time, i.e., s j S≥r j superscript subscript 𝑠 𝑗 𝑆 subscript 𝑟 𝑗 s_{j}^{S}\geq r_{j}italic_s start_POSTSUBSCRIPT italic_j end_POSTSUBSCRIPT start_POSTSUPERSCRIPT italic_S end_POSTSUPERSCRIPT ≥ italic_r start_POSTSUBSCRIPT italic_j end_POSTSUBSCRIPT; note that c j S=s j S+p j subscript superscript 𝑐 𝑆 𝑗 subscript superscript 𝑠 𝑆 𝑗 subscript 𝑝 𝑗 c^{S}{j}=s^{S}{j}+p_{j}italic_c start_POSTSUPERSCRIPT italic_S end_POSTSUPERSCRIPT start_POSTSUBSCRIPT italic_j end_POSTSUBSCRIPT = italic_s start_POSTSUPERSCRIPT italic_S end_POSTSUPERSCRIPT start_POSTSUBSCRIPT italic_j end_POSTSUBSCRIPT + italic_p start_POSTSUBSCRIPT italic_j end_POSTSUBSCRIPT whenever no job preemption is allowed. An optimal schedule is a one which minimizes the given objective (goal) function.
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Graham’s three-field notation is a common way to abbreviate a scheduling problem. The first, the second and the third fields define the machine environment, the job characteristics and the objective function, respectively. In a single-machine (single-processor) scheduling problem, a “1” is used in the first field in the three-field notation to specify a single machine environment. For example, 1|r j,q j|C max 1|r_{j},q_{j}|C_{\max}1 | italic_r start_POSTSUBSCRIPT italic_j end_POSTSUBSCRIPT , italic_q start_POSTSUBSCRIPT italic_j end_POSTSUBSCRIPT | italic_C start_POSTSUBSCRIPT roman_max end_POSTSUBSCRIPT represents the single machine scheduling problem with release times and tails to minimize the maximum full job completion time (the makespan).
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There exist a huge number of papers dealing with single machine problems. In particular, many heuristics and metaheuristic as well as enumerative algorithms have been suggested and tested for special single machine problems. Regarding surveys, the authors often restrict to special subjects and/or consider more general problem settings. As an example, there exist three very detailed surveys on scheduling problems with setup times or costs, see [4, 5, 6]. For problems with so-called p 𝑝 p italic_p-batching (i.e., several jobs can be processed in parallel in a batch), a survey has been given by Fowler and Mönch . The case of equal processing times has been surveyed by Kravchenko and Werner .
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Regarding single machine problems, often surveys are given for particular objective functions. As a few examples, we can mention e.g. a survey on the total weighted tardiness problem , on the weighted number of tardy jobs problem or again for more general problems, a survey on scheduling with late work criteria , or more recently on early and late work scheduling , where a substantial part is dedicated to single machine scheduling. While mostly a minimization problem is considered, Gafarov et al. reviewed algorithms and complexity issues for single machine maximization problems.
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Finally we note that single machine problems are also important for some practical multi-machine scheduling applications. Often there exists a bottleneck machine in a manufacturing environment. If one can solve the scheduling problem on the bottleneck machine optimally, this is at least a very good base for the overall schedule.
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In this paper, we review some results related to polynomial algorithms, the complexity and approximability issues. We explicitly mention that we do not survey metaheuristic and enumerative algorithms. In addition to some necessary basic results, the main focus is on the results which have been presented in the past in papers, where the authors of this work participated.
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The rest of this paper is organized as follows. In Section 2, we review some necessary basics for single machine scheduling problems. Then, in Section 3, we discuss polynomial algorithms, complexity and approximability issues for different classes of single machine scheduling problems. We also outline some subjects for future research, partially also for more general multi-machine scheduling problems. The paper ends with some concluding remarks in Section 4.
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2 Some Basics
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In this section, we give some basic notations, concepts and properties that have been used in the study of single machine scheduling problems. Single machine scheduling problems are trivial if every job has a single parameter, its processing time. For minimizing the makespan, it suffices to construct any feasible schedule without a machine idle time or a gap including jobs in any order:
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A gap on a machine is a longest time interval during which this machine is idle, i.e., it processes no job. It can be assumed that there occurs a 0-length gap (c j,t i)subscript 𝑐 𝑗 subscript 𝑡 𝑖(c_{j},t_{i})( italic_c start_POSTSUBSCRIPT italic_j end_POSTSUBSCRIPT , italic_t start_POSTSUBSCRIPT italic_i end_POSTSUBSCRIPT ) if job i 𝑖 i italic_i starts at its release time at the the completion time of job j 𝑗 j italic_j.
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A block is a consecutive part in a schedule which is preceded and succeeded by a gap.
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Single machine scheduling problems get less trivial if jobs are not simultaneously released, i.e., they have also different release times. Then, to avoid the creation of any redundant gap, we just need to keep the machine busy whenever this is possible. It is easy to see that this goal will be achieved if we just order the jobs according to non-decreasing release times. After this preprocessing step, the jobs can be included in this order, again, leaving no avoidable gap. This procedure will take O(nlogn)𝑂 𝑛 𝑛 O(n\log n)italic_O ( italic_n roman_log italic_n ) time due to the preprocessing step.
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If both, job release and due dates are given, then it makes sense to minimize the maximum job lateness (note that due dates are irrelevant for the makespan minimization). Suppose preemptions are allowed. Then we have a simple method to solve the problem optimally even on-line. We usually start at the earliest job release time and set initially the current time t 𝑡 t italic_t to that magnitude. Iteratively, among the available jobs, we schedule the next a most urgent job, one with the smallest due date. However, if during the execution of that job, another more urgent job gets released, the former job is interrupted at the release time of the latter job, which gets included at its release time. In this way, there will occur no forced delay for any urgent job by a non-urgent one, unlike the case without preemptions (in a given schedule S 𝑆 S italic_S, the delay of job j 𝑗 j italic_j is t j S−r j superscript subscript 𝑡 𝑗 𝑆 subscript 𝑟 𝑗 t_{j}^{S}-r_{j}italic_t start_POSTSUBSCRIPT italic_j end_POSTSUBSCRIPT start_POSTSUPERSCRIPT italic_S end_POSTSUPERSCRIPT - italic_r start_POSTSUBSCRIPT italic_j end_POSTSUBSCRIPT). The so constructed schedule is compact also in the sense that it minimizes the makespan. In particular, it has no avoidable machine idle time.
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Assume that job i 𝑖 i italic_i is processed at time r j subscript 𝑟 𝑗 r_{j}italic_r start_POSTSUBSCRIPT italic_j end_POSTSUBSCRIPT in the schedule S 𝑆 S italic_S, where d i>d j subscript 𝑑 𝑖 subscript 𝑑 𝑗 d_{i}>d_{j}italic_d start_POSTSUBSCRIPT italic_i end_POSTSUBSCRIPT > italic_d start_POSTSUBSCRIPT italic_j end_POSTSUBSCRIPT. Without allowing the interruption of i 𝑖 i italic_i, this job delays the starting of job j 𝑗 j italic_j; note that this delay will be less than p i subscript 𝑝 𝑖 p_{i}italic_p start_POSTSUBSCRIPT italic_i end_POSTSUBSCRIPT since upon the completion of job i 𝑖 i italic_i, job j 𝑗 j italic_j can immediately be scheduled. The delay of job j 𝑗 j italic_j may result in a non-optimal schedule. Indeed, this will be the case when job j 𝑗 j italic_j is too urgent with a “small” due-date. In an optimal schedule, it should be included with no or with a smaller delay.
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Now we describe a basic popular greedy non-preemptive algorithm to which we will refer as earliest due-date (ED) heuristic). It that has been widely used for solution of different single machine and other more complex scheduling problems. As the above described preemptive one, it is on-line. The heuristic was suggested by Jackson and later developed by Schrage . The first version is just the above described greedy algorithm with the pre-processing stage. The second version was adopted for jobs having both, release and due times. We describe this heuristic now. We let the current scheduling time t 𝑡 t italic_t be the minimal release time of an unscheduled job or the time when the machine completes the last job, whichever magnitude is larger. If no job is yet scheduled, we let t:=0 assign 𝑡 0 t:=0 italic_t := 0. Iteratively, among the jobs already released by time t 𝑡 t italic_t, the heuristic schedules one with the smallest due date. It breaks ties selecting a longest job.
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Let σ 𝜎\sigma italic_σ be the initial ED-schedule generated by the ED-heuristic for the originally given problem instance. Since no urgent job can be delayed by more than the processing time of another job, it is easy to see that
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L max(σ)<p max=max j∈Jp j.subscript 𝐿 𝜎 subscript 𝑝 subscript 𝑗 𝐽 subscript 𝑝 𝑗 L_{\max}(\sigma)<p_{\max}=\max_{j\in J}p_{j}.italic_L start_POSTSUBSCRIPT roman_max end_POSTSUBSCRIPT ( italic_σ ) < italic_p start_POSTSUBSCRIPT roman_max end_POSTSUBSCRIPT = roman_max start_POSTSUBSCRIPT italic_j ∈ italic_J end_POSTSUBSCRIPT italic_p start_POSTSUBSCRIPT italic_j end_POSTSUBSCRIPT .
In particular, σ 𝜎\sigma italic_σ is not necessarily optimal.
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Let us introduce some additional concepts that will be useful to understand better the problem. Let S 𝑆 S italic_S be an ED-schedule, one generated by the ED-heuristic.
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Among all jobs in the schedule S 𝑆 S italic_S, it is useful to distinguish the ones whose lateness attains L max(S)subscript 𝐿 𝑆 L_{\max}(S)italic_L start_POSTSUBSCRIPT roman_max end_POSTSUBSCRIPT ( italic_S ). In a given block B∈S 𝐵 𝑆 B\in S italic_B ∈ italic_S, let us call among them the last scheduled job from this block B 𝐵 B italic_B an overflow job in the schedule S 𝑆 S italic_S. Let o(S)𝑜 𝑆 o(S)italic_o ( italic_S ) be the earliest overflow job in the schedule S 𝑆 S italic_S, and let the critical block B(S)𝐵 𝑆 B(S)italic_B ( italic_S ) of S 𝑆 S italic_S be the block containing job o(S)𝑜 𝑆 o(S)italic_o ( italic_S ).
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Let now e 𝑒 e italic_e be a job from the block B(S)𝐵 𝑆 B(S)italic_B ( italic_S ) such that d e>d o(S)subscript 𝑑 𝑒 subscript 𝑑 𝑜 𝑆 d_{e}>d_{o(S)}italic_d start_POSTSUBSCRIPT italic_e end_POSTSUBSCRIPT > italic_d start_POSTSUBSCRIPT italic_o ( italic_S ) end_POSTSUBSCRIPT, i.e., it is less urgent than the overflow job o(S)𝑜 𝑆 o(S)italic_o ( italic_S ). Then job e 𝑒 e italic_e is referred to as an emerging job in the schedule S 𝑆 S italic_S. The latest scheduled emerging job in the block B(S)𝐵 𝑆 B(S)italic_B ( italic_S ) is called the delaying emerging job.
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The kernel K(S)𝐾 𝑆 K(S)italic_K ( italic_S ) in the schedule S 𝑆 S italic_S consists of the set of jobs, included in that schedule between the delaying emerging job l(S)𝑙 𝑆 l(S)italic_l ( italic_S ) and the overflow job o(S)𝑜 𝑆 o(S)italic_o ( italic_S ), not including job l(S)𝑙 𝑆 l(S)italic_l ( italic_S ) (but including job o(S)𝑜 𝑆 o(S)italic_o ( italic_S )). It immediately follows that the due date of any job of the kernel K(S)𝐾 𝑆 K(S)italic_K ( italic_S ) is no less than that of job o(S)𝑜 𝑆 o(S)italic_o ( italic_S )).
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The above definitions can straightforwardly be applied to any (not necessarily critical) block. Then there may exist more than one overflow job (at most one in in each block) in a schedule. Note that we will have the same number of kernels in that schedule.
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The earliest starting time of a kernel K 𝐾 K italic_K is given by
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𝒮(K)=min{r j|j∈K}.𝒮 𝐾 conditional subscript 𝑟 𝑗 𝑗 𝐾{\cal S}(K)=\min{r_{j}|j\in K}.caligraphic_S ( italic_K ) = roman_min { italic_r start_POSTSUBSCRIPT italic_j end_POSTSUBSCRIPT | italic_j ∈ italic_K } .
The starting time 𝒮(K,S)𝒮 𝐾 𝑆{\cal S}(K,S)caligraphic_S ( italic_K , italic_S ) of kernel K 𝐾 K italic_K in the schedule S 𝑆 S italic_S is the starting time of the first scheduled job of that kernel in the schedule S 𝑆 S italic_S.
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If the ED-schedule S 𝑆 S italic_S is not optimal, the overflow job o(S)𝑜 𝑆 o(S)italic_o ( italic_S ) is to be restarted earlier. By modifying artificially the initial problem parameters and applying the ED-heuristic to this modified instance, an alternative ED-schedule in which job o(S)𝑜 𝑆 o(S)italic_o ( italic_S ) is restarted earlier can be created, as we describe now. In the schedule S 𝑆 S italic_S, we look for the overflow job o(S)𝑜 𝑆 o(S)italic_o ( italic_S ), the kernel K(S)𝐾 𝑆 K(S)italic_K ( italic_S ) and the set of all emerging jobs. In an attempt to reduce L max(S)subscript 𝐿 𝑆 L_{\max}(S)italic_L start_POSTSUBSCRIPT roman_max end_POSTSUBSCRIPT ( italic_S ) (restart job o(S)𝑜 𝑆 o(S)italic_o ( italic_S ) earlier), we activate an emerging job l 𝑙 l italic_l for the kernel K(S)𝐾 𝑆 K(S)italic_K ( italic_S ), that is, we reschedule job l 𝑙 l italic_l after the kernel K(S)𝐾 𝑆 K(S)italic_K ( italic_S ). The activation is carried out in two steps. At step 1 the release time of job l 𝑙 l italic_l is increased artificially to it a magnitude, no-less than the largest job release time in the kernel K(S)𝐾 𝑆 K(S)italic_K ( italic_S ). At step, ED-heuristic is applied to the modified in this way problem instance. Since now the release time of job l 𝑙 l italic_l is no less than that of any job of the kernel K(S)𝐾 𝑆 K(S)italic_K ( italic_S ), and d l subscript 𝑑 𝑙 d_{l}italic_d start_POSTSUBSCRIPT italic_l end_POSTSUBSCRIPT is larger than the due date of any job from the kernel K(S)𝐾 𝑆 K(S)italic_K ( italic_S ), the ED-heuristic will give the priority to the jobs of the kernel K(S)𝐾 𝑆 K(S)italic_K ( italic_S ) and will reschedule job l 𝑙 l italic_l after all former jobs. The ED-schedule {S}l subscript 𝑆 𝑙{S}_{l}{ italic_S } start_POSTSUBSCRIPT italic_l end_POSTSUBSCRIPT, obtained in this way, is called a (weakly) complementary to S 𝑆 S italic_S schedule.
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Note that in the schedule S 𝑆 S italic_S, some emerging jobs might have been originally included after the kernel K(S)𝐾 𝑆 K(S)italic_K ( italic_S ). Then at least one such job will be included before the jobs of the kernel K(S)𝐾 𝑆 K(S)italic_K ( italic_S ) in the schedule {S}l subscript 𝑆 𝑙{S}{l}{ italic_S } start_POSTSUBSCRIPT italic_l end_POSTSUBSCRIPT by the ED-heuristic (since no job of the kernel K(S)𝐾 𝑆 K(S)italic_K ( italic_S ) is released at time s l(S)subscript 𝑠 𝑙 𝑆 s{l}(S)italic_s start_POSTSUBSCRIPT italic_l end_POSTSUBSCRIPT ( italic_S )). As a result, the kernel K(S)𝐾 𝑆 K(S)italic_K ( italic_S ) may not be restarted at its earliest starting time. To guarantee the restarting of the kernel K(S)𝐾 𝑆 K(S)italic_K ( italic_S ) at its earliest starting time, the above emerging jobs might be forced to remain after the kernel K(S)𝐾 𝑆 K(S)italic_K ( italic_S ). To reach this goal, we increase the release times of these jobs, similarly as we did it for job l 𝑙 l italic_l. The ED-schedule, obtained by the ED-heuristic for the modified in this way problem instance, is called a (strongly) complementary to S 𝑆 S italic_S schedule and is denoted by S l subscript 𝑆 𝑙 S_{l}italic_S start_POSTSUBSCRIPT italic_l end_POSTSUBSCRIPT. A complementary schedule can be abbreviated as a C-schedule.
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Note that there is exactly one job less included before the kernel K(S)𝐾 𝑆 K(S)italic_K ( italic_S ) in the schedule S l subscript 𝑆 𝑙 S_{l}italic_S start_POSTSUBSCRIPT italic_l end_POSTSUBSCRIPT compared to schedule S 𝑆 S italic_S. (The same condition does not held in {S}l subscript 𝑆 𝑙{S}{l}{ italic_S } start_POSTSUBSCRIPT italic_l end_POSTSUBSCRIPT; in fact, we may have more jobs scheduled before the kernel K(S)𝐾 𝑆 K(S)italic_K ( italic_S ) in the schedule {S}l subscript 𝑆 𝑙{S}{l}{ italic_S } start_POSTSUBSCRIPT italic_l end_POSTSUBSCRIPT than in the schedule S 𝑆 S italic_S.) Then, as it is easy to see, the overflow job o(S)𝑜 𝑆 o(S)italic_o ( italic_S ) will be completed earlier in the schedule S l subscript 𝑆 𝑙 S_{l}italic_S start_POSTSUBSCRIPT italic_l end_POSTSUBSCRIPT than it was completed in the schedule S 𝑆 S italic_S:
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Lemma 2.1
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Let S 𝑆 S italic_S be a strongly C-schedule and let l 𝑙 l italic_l be the emerging job scheduled in i 𝑖 i italic_i th position in in that schedule. Let, further j 𝑗 j italic_j be the job, immediately succeeding job l 𝑙 l italic_l in schedule S 𝑆 S italic_S. Then if d j<d l subscript 𝑑 𝑗 subscript 𝑑 𝑙 d_{j}<d_{l}italic_d start_POSTSUBSCRIPT italic_j end_POSTSUBSCRIPT < italic_d start_POSTSUBSCRIPT italic_l end_POSTSUBSCRIPT, there will occur a gap in between the i 𝑖 i italic_i th scheduled job and the kernel K(S)𝐾 𝑆 K(S)italic_K ( italic_S ) in the schedule S l subscript 𝑆 𝑙 S_{l}italic_S start_POSTSUBSCRIPT italic_l end_POSTSUBSCRIPT . Furthermore, there will arise a gap immediately before the earliest scheduled job of the kernel K(S)𝐾 𝑆 K(S)italic_K ( italic_S ) in the complementary schedule S l subscript 𝑆 𝑙 S_{l}italic_S start_POSTSUBSCRIPT italic_l end_POSTSUBSCRIPT, where l 𝑙 l italic_l is the live emerging job in the schedule S 𝑆 S italic_S. As a result, the kernel K(S)𝐾 𝑆 K(S)italic_K ( italic_S ) will start at the minimum release time of a job from that kernel.
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Proof. Assume first that job j 𝑗 j italic_j is scheduled in the i 𝑖 i italic_i th position in the schedule S l subscript 𝑆 𝑙 S_{l}italic_S start_POSTSUBSCRIPT italic_l end_POSTSUBSCRIPT and there arises no gap immediately before that job in that schedule. Then r j≤r l subscript 𝑟 𝑗 subscript 𝑟 𝑙 r_{j}\leq r_{l}italic_r start_POSTSUBSCRIPT italic_j end_POSTSUBSCRIPT ≤ italic_r start_POSTSUBSCRIPT italic_l end_POSTSUBSCRIPT holds, but since d j<d l subscript 𝑑 𝑗 subscript 𝑑 𝑙 d_{j}<d_{l}italic_d start_POSTSUBSCRIPT italic_j end_POSTSUBSCRIPT < italic_d start_POSTSUBSCRIPT italic_l end_POSTSUBSCRIPT, the ED-heuristic would schedule job j 𝑗 j italic_j instead of job l 𝑙 l italic_l at the moment t l(S)subscript 𝑡 𝑙 𝑆 t_{l}(S)italic_t start_POSTSUBSCRIPT italic_l end_POSTSUBSCRIPT ( italic_S ) in S 𝑆 S italic_S, which is a contradiction. We use a similar reasoning if instead of job j 𝑗 j italic_j another job k 𝑘 k italic_k with d k<d l subscript 𝑑 𝑘 subscript 𝑑 𝑙 d_{k}<d_{l}italic_d start_POSTSUBSCRIPT italic_k end_POSTSUBSCRIPT < italic_d start_POSTSUBSCRIPT italic_l end_POSTSUBSCRIPT is scheduled in the i 𝑖 i italic_i th position in schedule S l subscript 𝑆 𝑙 S_{l}italic_S start_POSTSUBSCRIPT italic_l end_POSTSUBSCRIPT. If d k≥d l subscript 𝑑 𝑘 subscript 𝑑 𝑙 d_{k}\geq d_{l}italic_d start_POSTSUBSCRIPT italic_k end_POSTSUBSCRIPT ≥ italic_d start_POSTSUBSCRIPT italic_l end_POSTSUBSCRIPT, since the schedule S l subscript 𝑆 𝑙 S_{l}italic_S start_POSTSUBSCRIPT italic_l end_POSTSUBSCRIPT is a strongly C-schedule, job k 𝑘 k italic_k must be an emerging job scheduled between job l 𝑙 l italic_l and kernel K(S)𝐾 𝑆 K(S)italic_K ( italic_S ) in S 𝑆 S italic_S. However, there is scheduled one job less before the kernel K(S)𝐾 𝑆 K(S)italic_K ( italic_S ) in the schedule S l subscript 𝑆 𝑙 S_{l}italic_S start_POSTSUBSCRIPT italic_l end_POSTSUBSCRIPT than in the schedule S 𝑆 S italic_S, and our claim easily follows.
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To distinguish the type of a gap in Lemma 2.1 from a “natural” gap that arises in the initial ED-schedule, the former type of a gap is sometimes referred to as an “artificial” one.
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3 Polynomially Solvable Cases, Complexity and Approximability Issues
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As earlier noted, the non-preemptive single machine scheduling problem becomes non-trivial, if, besides job the processing times, a job has at least one more parameter (as otherwise, from any permutation of the n 𝑛 n italic_n jobs, an optimal makespan schedule can be constructed in linear time by assigning the jobs from that permutation without leaving any idle time interval on the machine). In particular, all jobs are simultaneously released, any non-idle time schedule minimizes the makespan C max subscript 𝐶 C_{\max}italic_C start_POSTSUBSCRIPT roman_max end_POSTSUBSCRIPT.
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Minimizing maximum job lateness L max subscript 𝐿 L_{\max}italic_L start_POSTSUBSCRIPT roman_max end_POSTSUBSCRIPT is a bit more complex (recall that the lateness of job j 𝑗 j italic_j in a schedule S 𝑆 S italic_S is L j S=c j S−d j superscript subscript 𝐿 𝑗 𝑆 subscript superscript 𝑐 𝑆 𝑗 subscript 𝑑 𝑗 L_{j}^{S}=c^{S}{j}-d{j}italic_L start_POSTSUBSCRIPT italic_j end_POSTSUBSCRIPT start_POSTSUPERSCRIPT italic_S end_POSTSUPERSCRIPT = italic_c start_POSTSUPERSCRIPT italic_S end_POSTSUPERSCRIPT start_POSTSUBSCRIPT italic_j end_POSTSUBSCRIPT - italic_d start_POSTSUBSCRIPT italic_j end_POSTSUBSCRIPT). If only the job due dates are given (all the jobs being released at time 0), the Earliest Due date (ED) heuristic obtains an optimal schedule minimizing maximum job lateness (Jackson ). This heuristic schedules the jobs in non-decreasing order of their due dates without any idle time intervals (gaps). Vice versa, if only job release times are given, the Earliest Release time (ER) heuristic generates an optimal makespan schedule. In that heuristic, initially, the minimum job release time is assigned to the current scheduling time, and any job released at that time is assigned to the machine. Iteratively, the next scheduling time is defined as the maximum between the completion time of the latest so far assigned job and the minimum release time taken among the yet unscheduled jobs. Again, any (unscheduled) job released at the current scheduling time is assigned to the machine.
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Because of the job release times, a schedule created by the ER-heuristic may contain a gap. However, it may contain no gap that might be avoided. In particular, the total gap length in such a schedule is the minimal possible one. We can also use this heuristic if all the jobs have a common due date d 𝑑 d italic_d. In other words, all jobs are equally urgent. Then, as it is easy to see, an optimal schedule minimizing the maximum job lateness contains no gap that might be avoided, and it has the minimum possible total gap length. Therefore, the ER-heuristic minimizes also the maximum job lateness.
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It is easy to see that the ER-heuristic will no more minimize the maximum job lateness even if only two possible job due dates are allowed. Indeed, consider an instance with two jobs, job 1 released at time 0 and having the due date 11 and job 2 released at time 1 with the due date 6. The ER-heuristic will assign job 1 to the time interval [0,5)0 5[0,5)[ 0 , 5 ) and job 2 to the time interval [5,10)5 10[5,10)[ 5 , 10 ) resulting in the maximum job lateness 4 (that of job 2). Observe that this schedule contains no gap. At the same time, an optimal schedule does contain a gap [0,1)0 1[0,1)[ 0 , 1 ): it assigns job 2 to the time interval [1,6)1 6[1,6)[ 1 , 6 ) and then job 1 to the time interval [6,11)6 11[6,11)[ 6 , 11 ) completing both jobs on-time with the maximum job lateness equal to 0.
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3.1 Scheduling a Single Machine with Release Times and Due Dates to Minimize Maximum Job Lateness
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The general problem with release tames and due dates to minimize maximum lateness, abbreviated as 1|r j|L max 1 subscript 𝑟 𝑗 subscript 𝐿 1|r_{j}|L_{\max}1 | italic_r start_POSTSUBSCRIPT italic_j end_POSTSUBSCRIPT | italic_L start_POSTSUBSCRIPT roman_max end_POSTSUBSCRIPT, is strongly NP-hard . If we replace the due dates by tails with the objective to minimize maximum lateness, we get an equivalent problem 1|r j,q j|C max 1|r_{j},q_{j}|C_{\max}1 | italic_r start_POSTSUBSCRIPT italic_j end_POSTSUBSCRIPT , italic_q start_POSTSUBSCRIPT italic_j end_POSTSUBSCRIPT | italic_C start_POSTSUBSCRIPT roman_max end_POSTSUBSCRIPT. In this transformation, every due date d j subscript 𝑑 𝑗 d_{j}italic_d start_POSTSUBSCRIPT italic_j end_POSTSUBSCRIPT is replaced by the tail q j=A−d j subscript 𝑞 𝑗 𝐴 subscript 𝑑 𝑗 q_{j}=A-d_{j}italic_q start_POSTSUBSCRIPT italic_j end_POSTSUBSCRIPT = italic_A - italic_d start_POSTSUBSCRIPT italic_j end_POSTSUBSCRIPT, where A 𝐴 A italic_A is a sufficiently large constant (no less than the maximum job due date). So, the smaller is the due date, the larger is the tail. In other words, the urgency of every job is conserved with this transformation. It is straightforward to see that the two versions are equivalent. The full completion time of job j 𝑗 j italic_j in S 𝑆 S italic_S is given by 𝒞 j(S)=c j S+q j subscript 𝒞 𝑗 𝑆 superscript subscript 𝑐 𝑗 𝑆 subscript 𝑞 𝑗{\cal C}{j}(S)=c{j}^{S}+q_{j}caligraphic_C start_POSTSUBSCRIPT italic_j end_POSTSUBSCRIPT ( italic_S ) = italic_c start_POSTSUBSCRIPT italic_j end_POSTSUBSCRIPT start_POSTSUPERSCRIPT italic_S end_POSTSUPERSCRIPT + italic_q start_POSTSUBSCRIPT italic_j end_POSTSUBSCRIPT. The objective in that setting is to find an optimal makespan schedule, i.e., a feasible schedule S 𝑆 S italic_S with the minimal value of the maximal full job completion time.
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An efficient implicit enumerative algorithms for the problems 1|r j|L max 1 subscript 𝑟 𝑗 subscript 𝐿 1|r_{j}|L_{\max}1 | italic_r start_POSTSUBSCRIPT italic_j end_POSTSUBSCRIPT | italic_L start_POSTSUBSCRIPT roman_max end_POSTSUBSCRIPT and 1|r j,q j|C max 1|r_{j},q_{j}|C_{\max}1 | italic_r start_POSTSUBSCRIPT italic_j end_POSTSUBSCRIPT , italic_q start_POSTSUBSCRIPT italic_j end_POSTSUBSCRIPT | italic_C start_POSTSUBSCRIPT roman_max end_POSTSUBSCRIPT was suggested in McMahon and Florian and later, based on similar ideas, in Carlier . A number of other special cases are solvable in polynomial time. If all jobs are of unit length, the problem is again easy to solve, see Horn : indeed, while applying the ED-heuristic, no new (earlier unreleased) job can be released within the execution interval of a currently running job since job release times are integer and every job takes just a unit time of processing. As a result, no job may cause a forced delay of any more urgent job.
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As we have already seen, if all r j subscript 𝑟 𝑗 r_{j}italic_r start_POSTSUBSCRIPT italic_j end_POSTSUBSCRIPT s are equal, then scheduling the jobs in order of non-decreasing due dates gives an optimal schedule in O(nlogn)𝑂 𝑛 𝑛 O(n\log n)italic_O ( italic_n roman_log italic_n ) time . Similarly, if all d j subscript 𝑑 𝑗 d_{j}italic_d start_POSTSUBSCRIPT italic_j end_POSTSUBSCRIPT s are equal, then scheduling the jobs in order of non-decreasing release times is optimal. An O(nlogn)𝑂 𝑛 𝑛 O(n\log n)italic_O ( italic_n roman_log italic_n ) solution of the problem was proposed in for the special case when the release times, processing times and due-dates are restricted in a specific manner so that each r j subscript 𝑟 𝑗 r_{j}italic_r start_POSTSUBSCRIPT italic_j end_POSTSUBSCRIPT lies within a certain time interval defined by p j subscript 𝑝 𝑗 p_{j}italic_p start_POSTSUBSCRIPT italic_j end_POSTSUBSCRIPT and d j subscript 𝑑 𝑗 d_{j}italic_d start_POSTSUBSCRIPT italic_j end_POSTSUBSCRIPT.
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With arbitrary release times and due dates (or tails), if all processing times are equal (but not necessarily of unit length), it is still possible to find an optimal schedule in polynomial time, see Garey et al. . Even with two allowable job processing times p 𝑝 p italic_p and 2p 2 𝑝 2p 2 italic_p, for an arbitrary integer p 𝑝 p italic_p, the problem 1|p j∈{p,2p},r j,q j|C max conditional 1 subscript 𝑝 𝑗 𝑝 2 𝑝 subscript 𝑟 𝑗 conditional subscript 𝑞 𝑗 subscript 𝐶 1|p_{j}\in{p,2p},r_{j},q_{j}|C_{\max}1 | italic_p start_POSTSUBSCRIPT italic_j end_POSTSUBSCRIPT ∈ { italic_p , 2 italic_p } , italic_r start_POSTSUBSCRIPT italic_j end_POSTSUBSCRIPT , italic_q start_POSTSUBSCRIPT italic_j end_POSTSUBSCRIPT | italic_C start_POSTSUBSCRIPT roman_max end_POSTSUBSCRIPT remains polynomially solvable . In fact, in the algorithm from the latter reference the processing times of only some jobs (which become known during the execution of the algorithm) are restricted to p 𝑝 p italic_p or 2p 2 𝑝 2p 2 italic_p, whereas the rest of the jobs may have arbitrary processing times. The algorithm is obtained as a consequence of two enumerative algorithms for the problem 1|r j,q j|C max 1|r_{j},q_{j}|C_{\max}1 | italic_r start_POSTSUBSCRIPT italic_j end_POSTSUBSCRIPT , italic_q start_POSTSUBSCRIPT italic_j end_POSTSUBSCRIPT | italic_C start_POSTSUBSCRIPT roman_max end_POSTSUBSCRIPT. The first algorithm is exponential. The the second algorithm is a restriction of the first one and runs in time O(n 2logn)𝑂 superscript 𝑛 2 𝑛 O(n^{2}\log n)italic_O ( italic_n start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT roman_log italic_n ) under certain conditions which are to be held during its execution. Otherwise, another auxiliary procedure is to be applied to guarantee the optimality. An O(n 2lognlogP)𝑂 superscript 𝑛 2 𝑛 𝑃 O(n^{2}\log n\log P)italic_O ( italic_n start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT roman_log italic_n roman_log italic_P ) implementation of that procedure for the problem 1|p j∈{P,2P},r j,q j|C max conditional 1 subscript 𝑝 𝑗 𝑃 2 𝑃 subscript 𝑟 𝑗 conditional subscript 𝑞 𝑗 subscript 𝐶 1|p_{j}\in{P,2P},r_{j},q_{j}|C_{\max}1 | italic_p start_POSTSUBSCRIPT italic_j end_POSTSUBSCRIPT ∈ { italic_P , 2 italic_P } , italic_r start_POSTSUBSCRIPT italic_j end_POSTSUBSCRIPT , italic_q start_POSTSUBSCRIPT italic_j end_POSTSUBSCRIPT | italic_C start_POSTSUBSCRIPT roman_max end_POSTSUBSCRIPT yields an algorithm with the same time complexity for this setting.
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If the maximum job processing time p max subscript 𝑝 p_{\max}italic_p start_POSTSUBSCRIPT roman_max end_POSTSUBSCRIPT is bounded from above by a polynomial function in n 𝑛 n italic_n (i.e., P(n)=O(n k)𝑃 𝑛 𝑂 superscript 𝑛 𝑘 P(n)=O(n^{k})italic_P ( italic_n ) = italic_O ( italic_n start_POSTSUPERSCRIPT italic_k end_POSTSUPERSCRIPT )) and the maximum difference between the job release times is bounded by a constant R 𝑅 R italic_R, then the corresponding parameterized problem 1|p max≤P(n),|r j−r i|<R|L max conditional 1 subscript 𝑝 𝑚 𝑎 𝑥 𝑃 𝑛 subscript 𝑟 𝑗 subscript 𝑟 𝑖 bra 𝑅 subscript 𝐿 1|p_{max}\leq P(n),|r_{j}-r_{i}|<R\ |L_{\max}1 | italic_p start_POSTSUBSCRIPT italic_m italic_a italic_x end_POSTSUBSCRIPT ≤ italic_P ( italic_n ) , | italic_r start_POSTSUBSCRIPT italic_j end_POSTSUBSCRIPT - italic_r start_POSTSUBSCRIPT italic_i end_POSTSUBSCRIPT | < italic_R | italic_L start_POSTSUBSCRIPT roman_max end_POSTSUBSCRIPT can still be solved in polynomial time O(n k+1lognlogp max)𝑂 superscript 𝑛 𝑘 1 𝑛 subscript 𝑝 O(n^{k+1}\log n\log p_{\max})italic_O ( italic_n start_POSTSUPERSCRIPT italic_k + 1 end_POSTSUPERSCRIPT roman_log italic_n roman_log italic_p start_POSTSUBSCRIPT roman_max end_POSTSUBSCRIPT ), see Vakhania & Werner . Note that in this setting the general problem is parameterized by p max subscript 𝑝 p_{\max}italic_p start_POSTSUBSCRIPT roman_max end_POSTSUBSCRIPT and the maximum difference between two job release times.
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There are known other conditions that allow a polynomial time solution of the problem. For example, if the job parameters are embedded in a special manner, i.e., for any pair of jobs j,i 𝑗 𝑖 j,i italic_j , italic_i with r i>r j subscript 𝑟 𝑖 subscript 𝑟 𝑗 r_{i}>r_{j}italic_r start_POSTSUBSCRIPT italic_i end_POSTSUBSCRIPT > italic_r start_POSTSUBSCRIPT italic_j end_POSTSUBSCRIPT and d i<d j subscript 𝑑 𝑖 subscript 𝑑 𝑗 d_{i}<d_{j}italic_d start_POSTSUBSCRIPT italic_i end_POSTSUBSCRIPT < italic_d start_POSTSUBSCRIPT italic_j end_POSTSUBSCRIPT,
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d j−r j−p j≤d i−r i−p i subscript 𝑑 𝑗 subscript 𝑟 𝑗 subscript 𝑝 𝑗 subscript 𝑑 𝑖 subscript 𝑟 𝑖 subscript 𝑝 𝑖 d_{j}-r_{j}-p_{j}\leq d_{i}-r_{i}-p_{i}italic_d start_POSTSUBSCRIPT italic_j end_POSTSUBSCRIPT - italic_r start_POSTSUBSCRIPT italic_j end_POSTSUBSCRIPT - italic_p start_POSTSUBSCRIPT italic_j end_POSTSUBSCRIPT ≤ italic_d start_POSTSUBSCRIPT italic_i end_POSTSUBSCRIPT - italic_r start_POSTSUBSCRIPT italic_i end_POSTSUBSCRIPT - italic_p start_POSTSUBSCRIPT italic_i end_POSTSUBSCRIPT
holds, and with
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r i+p i≥r j+p j,subscript 𝑟 𝑖 subscript 𝑝 𝑖 subscript 𝑟 𝑗 subscript 𝑝 𝑗 r_{i}+p_{i}\geq r_{j}+p_{j},italic_r start_POSTSUBSCRIPT italic_i end_POSTSUBSCRIPT + italic_p start_POSTSUBSCRIPT italic_i end_POSTSUBSCRIPT ≥ italic_r start_POSTSUBSCRIPT italic_j end_POSTSUBSCRIPT + italic_p start_POSTSUBSCRIPT italic_j end_POSTSUBSCRIPT ,
d i≥d j subscript 𝑑 𝑖 subscript 𝑑 𝑗 d_{i}\geq d_{j}italic_d start_POSTSUBSCRIPT italic_i end_POSTSUBSCRIPT ≥ italic_d start_POSTSUBSCRIPT italic_j end_POSTSUBSCRIPT holds, then the resultant problem 1|r j,e m b e d d e d|L max 1|r_{j},embedded|L_{\max}1 | italic_r start_POSTSUBSCRIPT italic_j end_POSTSUBSCRIPT , italic_e italic_m italic_b italic_e italic_d italic_d italic_e italic_d | italic_L start_POSTSUBSCRIPT roman_max end_POSTSUBSCRIPT is solvable in O(nlogn)𝑂 𝑛 𝑛 O(n\log n)italic_O ( italic_n roman_log italic_n ) time, see Vakhania .
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In a maximal polynomially solvable special case of the problem 1|p j:divisible,r j|L max:conditional 1 subscript 𝑝 𝑗 𝑑 𝑖 𝑣 𝑖 𝑠 𝑖 𝑏 𝑙 𝑒 conditional subscript 𝑟 𝑗 subscript 𝐿 1|p_{j}:divisible,r_{j}|L_{\max}1 | italic_p start_POSTSUBSCRIPT italic_j end_POSTSUBSCRIPT : italic_d italic_i italic_v italic_i italic_s italic_i italic_b italic_l italic_e , italic_r start_POSTSUBSCRIPT italic_j end_POSTSUBSCRIPT | italic_L start_POSTSUBSCRIPT roman_max end_POSTSUBSCRIPT, the job processing times form mutually divisible numbers (say, powers of 2) .
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For the problem 1|r j,p j=p|L max conditional 1 subscript 𝑟 𝑗 subscript 𝑝 𝑗 conditional 𝑝 subscript 𝐿 𝑚 𝑎 𝑥 1|r_{j},p_{j}=p|L_{max}1 | italic_r start_POSTSUBSCRIPT italic_j end_POSTSUBSCRIPT , italic_p start_POSTSUBSCRIPT italic_j end_POSTSUBSCRIPT = italic_p | italic_L start_POSTSUBSCRIPT italic_m italic_a italic_x end_POSTSUBSCRIPT with constant processing times, Lazarev et al. presented two polynomial algorithms. The first procedure is based on trisection search and determines an optimal solution in O(Qnlogn)𝑂 𝑄 𝑛 𝑛 O(Qn\log n)italic_O ( italic_Q italic_n roman_log italic_n ) time, where 10−Q superscript 10 𝑄 10^{-Q}10 start_POSTSUPERSCRIPT - italic_Q end_POSTSUPERSCRIPT represents the accuracy of the parameters of the problem. The second approach uses an auxiliary problem, where the maximum completion time is minimized under the constraint that L max subscript 𝐿 𝑚 𝑎 𝑥 L_{max}italic_L start_POSTSUBSCRIPT italic_m italic_a italic_x end_POSTSUBSCRIPT is bounded by a particular value. Using this approach, the authors determine for the bi-criteria problem with the two criteria C max subscript 𝐶 𝑚 𝑎 𝑥 C_{max}italic_C start_POSTSUBSCRIPT italic_m italic_a italic_x end_POSTSUBSCRIPT and L max subscript 𝐿 𝑚 𝑎 𝑥 L_{max}italic_L start_POSTSUBSCRIPT italic_m italic_a italic_x end_POSTSUBSCRIPT and the Pareto set of optimal solutions in O(n 3logn)𝑂 superscript 𝑛 3 𝑛 O(n^{3}\log n)italic_O ( italic_n start_POSTSUPERSCRIPT 3 end_POSTSUPERSCRIPT roman_log italic_n ) time.
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With arbitrary job processing times, the problem remains (weakly) NP 𝑁 𝑃 NP italic_N italic_P-hard already with only two allowable job release times and due dates 1|r j∈{r 1,r 2},d j∈{d 1,d 2}|L max formulae-sequence conditional 1 subscript 𝑟 𝑗 subscript 𝑟 1 subscript 𝑟 2 subscript 𝑑 𝑗 conditional subscript 𝑑 1 subscript 𝑑 2 subscript 𝐿 1|r_{j}\in{r_{1},r_{2}},d_{j}\in{d_{1},d_{2}}|L_{\max}1 | italic_r start_POSTSUBSCRIPT italic_j end_POSTSUBSCRIPT ∈ { italic_r start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT , italic_r start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT } , italic_d start_POSTSUBSCRIPT italic_j end_POSTSUBSCRIPT ∈ { italic_d start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT , italic_d start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT } | italic_L start_POSTSUBSCRIPT roman_max end_POSTSUBSCRIPT, see Chinos and Vakhania . This is a minimal NP 𝑁 𝑃 NP italic_N italic_P-hard special case of the problem 1|r j,d j|L max 1|r_{j},d_{j}|L_{\max}1 | italic_r start_POSTSUBSCRIPT italic_j end_POSTSUBSCRIPT , italic_d start_POSTSUBSCRIPT italic_j end_POSTSUBSCRIPT | italic_L start_POSTSUBSCRIPT roman_max end_POSTSUBSCRIPT. The latter setting was later studied in Reynoso and Vakhania , where structural properties of this special case have led to a number of optimality conditions under which it can be solved optimally in O(nlogn)𝑂 𝑛 𝑛 O(n\log n)italic_O ( italic_n roman_log italic_n ) time were established. It was shown that, for bottleneck instances where none of these conditions apply, the problem can be reduced to the SUBSET SUM problem. The authors presented computational results accomplished for 50 million randomly generated problem instances, reporting that almost all these instances were solved optimally by one of the established optimality conditions. Moreover, the heuristic algorithms that incorporate the optimality conditions established in succeeded to solve the SUBSET SUM problem also optimally for almost all the 50 million instances. A very recent work considered a natural generalization of the latter problem with any fixed number of release times, see Escalona and Vakhania . For this setting, the optimality conditions were derived and again, it was shown how the problem can be reduced to the SUBSET SUM. It is a challenging question to what extent the above obtained results can be generalized for any fixed number of release and delivery times, a setting with much wider practical applications.
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In a more general objective function f max subscript 𝑓 𝑚 𝑎 𝑥 f_{max}italic_f start_POSTSUBSCRIPT italic_m italic_a italic_x end_POSTSUBSCRIPT defined as the maximum of non-decreasing cost functions f j subscript 𝑓 𝑗 f_{j}italic_f start_POSTSUBSCRIPT italic_j end_POSTSUBSCRIPT has been considered for the preemptive problem 1|r j,p r e c,p m t n|f max 1|r_{j},prec,pmtn|f_{max}1 | italic_r start_POSTSUBSCRIPT italic_j end_POSTSUBSCRIPT , italic_p italic_r italic_e italic_c , italic_p italic_m italic_t italic_n | italic_f start_POSTSUBSCRIPT italic_m italic_a italic_x end_POSTSUBSCRIPT. It is assumed that the cost functions fulfill the order relation f i(t)≤f j(t)subscript 𝑓 𝑖 𝑡 subscript 𝑓 𝑗 𝑡 f_{i}(t)\leq f_{j}(t)italic_f start_POSTSUBSCRIPT italic_i end_POSTSUBSCRIPT ( italic_t ) ≤ italic_f start_POSTSUBSCRIPT italic_j end_POSTSUBSCRIPT ( italic_t ) or f j(t)≤f i(t)subscript 𝑓 𝑗 𝑡 subscript 𝑓 𝑖 𝑡 f_{j}(t)\leq f_{i}(t)italic_f start_POSTSUBSCRIPT italic_j end_POSTSUBSCRIPT ( italic_t ) ≤ italic_f start_POSTSUBSCRIPT italic_i end_POSTSUBSCRIPT ( italic_t ) for any pair i,j 𝑖 𝑗 i,j italic_i , italic_j of jobs and for any time t 𝑡 t italic_t of the scheduling interval. In the case of tree-like precedence constraints, the derived algorithm has a time complexity of O(nlogn)𝑂 𝑛 𝑛 O(n\log n)italic_O ( italic_n roman_log italic_n ). In addition, a parallel algorithm is given with a time complexity of O(log 2n)𝑂 superscript 2 𝑛 O(\log^{2}n)italic_O ( roman_log start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT italic_n ) when using O(n 2logn)𝑂 superscript 𝑛 2 𝑛 O(n^{2}\log n)italic_O ( italic_n start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT roman_log italic_n ) processors.
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A special kind of parameterized analysis called the Variable Parameter (VP) analysis was recently suggested in . Based on this approach, an implicit enumeration algorithm and a polynomial time approximation scheme for the general setting 1|r j,q j|C max 1|r_{j},q_{j}|C_{\max}1 | italic_r start_POSTSUBSCRIPT italic_j end_POSTSUBSCRIPT , italic_q start_POSTSUBSCRIPT italic_j end_POSTSUBSCRIPT | italic_C start_POSTSUBSCRIPT roman_max end_POSTSUBSCRIPT were proposed. The variable parameter is defined as the number of emerging jobs. First, a partial solution in time O(nlogn)𝑂 𝑛 𝑛 O(n\log n)italic_O ( italic_n roman_log italic_n ) without these jobs is constructed. Then this partial solution is augmented to a complete optimal solution in time, which is exponential solely in the number of the variable parameter (the number of emerging jobs). The author also gives alternative time complexity expression for both, the implicit enumeration and the polynomial time approximation scheme, in which the exponential dependence is solely on the maximum processing time and the maximum delivery time of an emerging job. If one applies the fixed parameter analysis to these estimations, one gets a polynomial-time dependence. A very recent extensive experimental study showed that, in practice, the variable parameter is much less than n 𝑛 n italic_n. Moreover, the ratio of the variable parameter to n 𝑛 n italic_n asymptotically converges to 0 (see Ramirez and Vakhania ).
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The problem 1|r j|L max 1 subscript 𝑟 𝑗 subscript 𝐿 1|r_{j}|L_{\max}1 | italic_r start_POSTSUBSCRIPT italic_j end_POSTSUBSCRIPT | italic_L start_POSTSUBSCRIPT roman_max end_POSTSUBSCRIPT is important on its own and also because of its possible applications for the solution of other more complex scheduling problems and other optimization problems. For example, in the Multiple Traveling Salesman Problem (M-TSP), k 𝑘 k italic_k salesmen, each starting from a special point 0 0 called depot, has to visit a number of clients and return to the depot. The objective is to minimize the total cost of all k 𝑘 k italic_k tours (see, for example ). In the setting with time windows M-TSP-TW, each client has to be visited within a specified time interval. In the former basic setting, it is easy to find a feasible solution to the problem. However, in the latter setting with time windows, there may exist no feasible solution for a given k 𝑘 k italic_k. Hence, it is crucial to determine whether there exists a feasible solution for a given k 𝑘 k italic_k. More generally, one may be interested to calculate the minimum k 𝑘 k italic_k for which there exists a feasible solution. We can use our scheduling problem for this purpose.
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In the basic case, we wish to determine whether there exists a feasible solution with k=1 𝑘 1 k=1 italic_k = 1. For a given instance of M-TSP-TW with n+1 𝑛 1 n+1 italic_n + 1 clients, we create an instance of he problem 1|r j,p j=0|L max conditional 1 subscript 𝑟 𝑗 subscript 𝑝 𝑗 conditional 0 subscript 𝐿 1|r_{j},p_{j}=0|L_{\max}1 | italic_r start_POSTSUBSCRIPT italic_j end_POSTSUBSCRIPT , italic_p start_POSTSUBSCRIPT italic_j end_POSTSUBSCRIPT = 0 | italic_L start_POSTSUBSCRIPT roman_max end_POSTSUBSCRIPT with n 𝑛 n italic_n jobs, associating job j 𝑗 j italic_j with client j 𝑗 j italic_j, for each i=1,…,n 𝑖 1…𝑛 i=1,\dots,n italic_i = 1 , … , italic_n, and letting the processing time of each job to be 0. Note that this is a simplified version of the generic problem 1|r j|L max 1 subscript 𝑟 𝑗 subscript 𝐿 1|r_{j}|L_{\max}1 | italic_r start_POSTSUBSCRIPT italic_j end_POSTSUBSCRIPT | italic_L start_POSTSUBSCRIPT roman_max end_POSTSUBSCRIPT. In this version, the release and due times of each job are determined according to the time window of the corresponding client. We consider the feasibility version of the problem, where one looks for a schedule in which no job has a positive lateness. One can immediately see that, if there is no feasible solution to the scheduling problem, there is also no feasible solution for the corresponding instance of the problem M-TSP-TW with k=1 𝑘 1 k=1 italic_k = 1 (i.e., TSP-TW). In general, we let u 𝑢 u italic_u be an upper bound on the total number of salesmen, and carry out binary search within the interval [1,u]1 𝑢[1,u][ 1 , italic_u ] solving at each iteration an instance of the corresponding scheduling problem Pk|r j,p j=0|L max conditional 𝑃 𝑘 subscript 𝑟 𝑗 subscript 𝑝 𝑗 conditional 0 subscript 𝐿 Pk|r_{j},p_{j}=0|L_{\max}italic_P italic_k | italic_r start_POSTSUBSCRIPT italic_j end_POSTSUBSCRIPT , italic_p start_POSTSUBSCRIPT italic_j end_POSTSUBSCRIPT = 0 | italic_L start_POSTSUBSCRIPT roman_max end_POSTSUBSCRIPT with k 𝑘 k italic_k parallel identical machines and with trial k∈[1,u]𝑘 1 𝑢 k\in[1,u]italic_k ∈ [ 1 , italic_u ]. In this way, we can find the minimum k 𝑘 k italic_k for which there exists a feasible solution to the problem M-TSP-TW.
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3.2 Minimizing the Number of Late Jobs
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Another common objective function with given due dates is the number of late jobs: a job in a schedule is called late if it is completed after its due date in that schedule, otherwise it is called on time. In the problem of this subsection, we deal with n 𝑛 n italic_n jobs with release times and due dates which have to be scheduled on the machine so as to maximize the number of on time jobs, or equivalently, to minimize the number of late jobs. We consider both, preemptive and non-preemptive cases, and also cases when the jobs have an additional parameter called weight.
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3.2.1 Nonpreemptive Scheduling to Minimize the Number of Late Jobs: 1|r j;p j=p|∑U j conditional 1 subscript 𝑟 𝑗 subscript 𝑝 𝑗 conditional 𝑝 subscript 𝑈 𝑗 1|r_{j};p_{j}=p|\sum U_{j}1 | italic_r start_POSTSUBSCRIPT italic_j end_POSTSUBSCRIPT ; italic_p start_POSTSUBSCRIPT italic_j end_POSTSUBSCRIPT = italic_p | ∑ italic_U start_POSTSUBSCRIPT italic_j end_POSTSUBSCRIPT
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In this subsection, we deal with the non-preemptive version of the problem 1|r j|∑U j 1 subscript 𝑟 𝑗 subscript 𝑈 𝑗 1|r_{j}|\sum U_{j}1 | italic_r start_POSTSUBSCRIPT italic_j end_POSTSUBSCRIPT | ∑ italic_U start_POSTSUBSCRIPT italic_j end_POSTSUBSCRIPT. This problem was studied in Vakhania , where two polynomial time algorithms were proposed. The first one with the time complexity O(n 3logn)𝑂 superscript 𝑛 3 𝑛 O(n^{3}\log n)italic_O ( italic_n start_POSTSUPERSCRIPT 3 end_POSTSUPERSCRIPT roman_log italic_n ) solves optimally the problem if during its execution no job with some specially defined property occurs. The second algorithm is an adaptation of the first one for the special case of the problem when all jobs have the same length 1|r j;p j=p|∑U j conditional 1 subscript 𝑟 𝑗 subscript 𝑝 𝑗 conditional 𝑝 subscript 𝑈 𝑗 1|r_{j};p_{j}=p|\sum U_{j}1 | italic_r start_POSTSUBSCRIPT italic_j end_POSTSUBSCRIPT ; italic_p start_POSTSUBSCRIPT italic_j end_POSTSUBSCRIPT = italic_p | ∑ italic_U start_POSTSUBSCRIPT italic_j end_POSTSUBSCRIPT. The time complexity of this algorithm is O(n 3logn)𝑂 superscript 𝑛 3 𝑛 O(n^{3}\log n)italic_O ( italic_n start_POSTSUPERSCRIPT 3 end_POSTSUPERSCRIPT roman_log italic_n ) (it improved an earlier known fastest algorithm with the time complexity of O(n 5)𝑂 superscript 𝑛 5 O(n^{5})italic_O ( italic_n start_POSTSUPERSCRIPT 5 end_POSTSUPERSCRIPT ) by Chrobak et al. .
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3.2.2 Preemptive Scheduling to Maximize the Number of on Time Jobs: 1|r j;p m t n|∑U j 1|r_{j};pmtn|\sum U_{j}1 | italic_r start_POSTSUBSCRIPT italic_j end_POSTSUBSCRIPT ; italic_p italic_m italic_t italic_n | ∑ italic_U start_POSTSUBSCRIPT italic_j end_POSTSUBSCRIPT
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In the problem of this subsection, n 𝑛 n italic_n jobs with release times and due dates have to be scheduled preemptively on a single machine so as to minimize the number of late jobs, i.e., problem 1|r j;p m t n|∑U j 1|r_{j};pmtn|\sum U_{j}1 | italic_r start_POSTSUBSCRIPT italic_j end_POSTSUBSCRIPT ; italic_p italic_m italic_t italic_n | ∑ italic_U start_POSTSUBSCRIPT italic_j end_POSTSUBSCRIPT. We call a job late (on time, respectively) if it is completed after (at or before, respectively) its due date. We will say that job j 𝑗 j italic_j is feasibly scheduled if it starts no earlier than at time r j subscript 𝑟 𝑗 r_{j}italic_r start_POSTSUBSCRIPT italic_j end_POSTSUBSCRIPT and overlaps with no other job on the machine. A feasible schedule is one which feasibly schedules all jobs. The objective is to find a feasible schedule with a minimal number of late jobs, or equivalently, with the maximal number of on time jobs.
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This problem is known to be polynomially solvable. The first polynomial dynamic programming algorithm with a time complexity of O(n 5)𝑂 superscript 𝑛 5 O(n^{5})italic_O ( italic_n start_POSTSUPERSCRIPT 5 end_POSTSUPERSCRIPT ) and a space complexity of O(n 3)𝑂 superscript 𝑛 3 O(n^{3})italic_O ( italic_n start_POSTSUPERSCRIPT 3 end_POSTSUPERSCRIPT ) was developed by Lawler . The result was improved by Baptiste who suggested another dynamic programming algorithm with a time complexity of O(n 4)𝑂 superscript 𝑛 4 O(n^{4})italic_O ( italic_n start_POSTSUPERSCRIPT 4 end_POSTSUPERSCRIPT ) and a space complexity of O(n 2)𝑂 superscript 𝑛 2 O(n^{2})italic_O ( italic_n start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT ). The latter result was further improved by Vakhania , where an algorithm with the time and space complexities of O(n 3)𝑂 superscript 𝑛 3 O(n^{3})italic_O ( italic_n start_POSTSUPERSCRIPT 3 end_POSTSUPERSCRIPT ) and O(n)𝑂 𝑛 O(n)italic_O ( italic_n ), respectively, was proposed. Unlike the earlier algorithms, the latter algorithm is not based on dynamic programming. It uses the ED-heuristic for scheduling the jobs. The main task is to construct a schedule in which all jobs are scheduled on time, with the number of such jobs being maximal. As any late job can be scheduled arbitrarily late without affecting the objective function, jobs not included in the former schedule can be later added at the end of it in an arbitrary order. During the construction of the schedule, some jobs are finally disregarded, whereas some other jobs are omitted with the possibility to be included into a later created schedule.
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3.3 Preemptive Scheduling to Minimize the Weighted Number of Late Jobs with Equal Length: 1|r j;p j=p;pmtn|∑w jU j conditional 1 subscript 𝑟 𝑗 subscript 𝑝 𝑗 𝑝 conditional 𝑝 𝑚 𝑡 𝑛 subscript 𝑤 𝑗 subscript 𝑈 𝑗 1|r_{j};p_{j}=p;pmtn|\sum w_{j}U_{j}1 | italic_r start_POSTSUBSCRIPT italic_j end_POSTSUBSCRIPT ; italic_p start_POSTSUBSCRIPT italic_j end_POSTSUBSCRIPT = italic_p ; italic_p italic_m italic_t italic_n | ∑ italic_w start_POSTSUBSCRIPT italic_j end_POSTSUBSCRIPT italic_U start_POSTSUBSCRIPT italic_j end_POSTSUBSCRIPT
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In this section, we consider the weighted version of the problem of the previous section in which preemptions are allowed and all jobs have an equal length. Moreover, w j subscript 𝑤 𝑗 w_{j}italic_w start_POSTSUBSCRIPT italic_j end_POSTSUBSCRIPT is the non-negative integer weight of job j 𝑗 j italic_j. We minimize the weighted throughput which is the total weight of the completed jobs. Equivalently, we minimize the weighted number of late jobs. The problem ican be abbreviated as 1|r j;p j=p;pmtn|∑w jU j conditional 1 subscript 𝑟 𝑗 subscript 𝑝 𝑗 𝑝 conditional 𝑝 𝑚 𝑡 𝑛 subscript 𝑤 𝑗 subscript 𝑈 𝑗 1|r_{j};p_{j}=p;pmtn|\sum w_{j}U_{j}1 | italic_r start_POSTSUBSCRIPT italic_j end_POSTSUBSCRIPT ; italic_p start_POSTSUBSCRIPT italic_j end_POSTSUBSCRIPT = italic_p ; italic_p italic_m italic_t italic_n | ∑ italic_w start_POSTSUBSCRIPT italic_j end_POSTSUBSCRIPT italic_U start_POSTSUBSCRIPT italic_j end_POSTSUBSCRIPT.
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The first polynomial dynamic programming algorithm of the complexity O(n 10)𝑂 superscript 𝑛 10 O(n^{10})italic_O ( italic_n start_POSTSUPERSCRIPT 10 end_POSTSUPERSCRIPT ) for the problem was suggested by Baptiste. Baptiste et al. improved the above time complexity and proposed another dynamic programming algorithm which runs in O(n 4)𝑂 superscript 𝑛 4 O(n^{4})italic_O ( italic_n start_POSTSUPERSCRIPT 4 end_POSTSUPERSCRIPT ) time.
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Next, we briefly discuss some directions for further research. First we note that for the multi-processor case, the weighted version is NP 𝑁 𝑃 NP italic_N italic_P-complete . At the same time, the non-weighted version still remains open. In particular, it is not known whether the problem P|r j;p j=p;pmtn|∑U j conditional 𝑃 subscript 𝑟 𝑗 subscript 𝑝 𝑗 𝑝 conditional 𝑝 𝑚 𝑡 𝑛 subscript 𝑈 𝑗 P|r_{j};p_{j}=p;{pmtn}|\sum U_{j}italic_P | italic_r start_POSTSUBSCRIPT italic_j end_POSTSUBSCRIPT ; italic_p start_POSTSUBSCRIPT italic_j end_POSTSUBSCRIPT = italic_p ; italic_p italic_m italic_t italic_n | ∑ italic_U start_POSTSUBSCRIPT italic_j end_POSTSUBSCRIPT can be solved in polynomial time. (Even for 2 2 2 2 processors, one cannot the search to solely earliest deadline schedules. As an example, consider an instance with three jobs with feasible intervals (0,3)0 3(0,3)( 0 , 3 ), (0,4)0 4(0,4)( 0 , 4 ), and (0,5)0 5(0,5)( 0 , 5 ) and the processing time p=3 𝑝 3 p=3 italic_p = 3. There is a feasible schedule that completes all three jobs, whereas the earliest-deadline schedule completes only jobs 1 1 1 1 and 2 2 2 2.) For the multi-processor case, it is also interesting to study the preemptive version, where jobs are not allowed to migrate between processors.
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3.4 Scheduling to Minimize the Weighted Number of Late jobs with Deadlines and Release/Due Date Intervals
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In , Gordon et al. dealt with the problem, where in addition to release and due dates also deadlines are given. This means that certain jobs have to be completed on time. The authors considered the two cases of similarly ordered release and dues dates (i.e., r 1≤r 2≤…≤r n subscript 𝑟 1 subscript 𝑟 2…subscript 𝑟 𝑛 r_{1}\leq r_{2}\leq\ldots\leq r_{n}italic_r start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT ≤ italic_r start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT ≤ … ≤ italic_r start_POSTSUBSCRIPT italic_n end_POSTSUBSCRIPT and d 1≤d 2≤…≤d n subscript 𝑑 1 subscript 𝑑 2…subscript 𝑑 𝑛 d_{1}\leq d_{2}\leq\ldots\leq d_{n}italic_d start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT ≤ italic_d start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT ≤ … ≤ italic_d start_POSTSUBSCRIPT italic_n end_POSTSUBSCRIPT as well nested release and due date intervals (i.e., none of two such intervals do overlap, this they have at most joint point or one covers the other one). For these problems, a reduced problem is constructed, where the deadlines are eliminated, and it is shown that the optimal solution of the reduced problem can be used to find an optimal solution of the initial problem.
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In a later paper, Gordon et al. considered the preemptive version of the above problem with nested release and due date intervals. They derived necessary and sufficient conditions for the existence of a schedule in which all jobs are completed by their due dates. For the case of oppositely ordered processing times and job weights, a polynomial algorithm of the complexity O(nlogn)𝑂 𝑛 𝑛 O(n\log n)italic_O ( italic_n roman_log italic_n ) has been presented.
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3.5 Scheduling with Tardiness-based Objectives
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While the problem 1||∑T j 1||\sum T_{j}1 | | ∑ italic_T start_POSTSUBSCRIPT italic_j end_POSTSUBSCRIPT is known to be NP-hard, Gafarov et al. considered the special case of equal processing times, i.e., problem 1|r j,p j=p|∑w jT j conditional 1 subscript 𝑟 𝑗 subscript 𝑝 𝑗 conditional 𝑝 subscript 𝑤 𝑗 subscript 𝑇 𝑗 1|r_{j},p_{j}=p|\sum w_{j}T_{j}1 | italic_r start_POSTSUBSCRIPT italic_j end_POSTSUBSCRIPT , italic_p start_POSTSUBSCRIPT italic_j end_POSTSUBSCRIPT = italic_p | ∑ italic_w start_POSTSUBSCRIPT italic_j end_POSTSUBSCRIPT italic_T start_POSTSUBSCRIPT italic_j end_POSTSUBSCRIPT. After deriving some properties, they gave polynomial algorithms for two special cases of the problem. If the difference between the maximum due date and the minimum due date is not greater than the processing time p 𝑝 p italic_p, the problem can be solved by a modification of the algorithm by Baptiste in O(n 10)𝑂 superscript 𝑛 10 O(n^{10})italic_O ( italic_n start_POSTSUPERSCRIPT 10 end_POSTSUPERSCRIPT ) time. The special case with ordered weights and processing times according to w 1≤w 2≤…≤w n subscript 𝑤 1 subscript 𝑤 2…subscript 𝑤 𝑛 w_{1}\leq w_{2}\leq\ldots\leq w_{n}italic_w start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT ≤ italic_w start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT ≤ … ≤ italic_w start_POSTSUBSCRIPT italic_n end_POSTSUBSCRIPT and d 1≤d 2≤…≤d n subscript 𝑑 1 subscript 𝑑 2…subscript 𝑑 𝑛 d_{1}\leq d_{2}\leq\ldots\leq d_{n}italic_d start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT ≤ italic_d start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT ≤ … ≤ italic_d start_POSTSUBSCRIPT italic_n end_POSTSUBSCRIPT can be solved in O(n 9)𝑂 superscript 𝑛 9 O(n^{9})italic_O ( italic_n start_POSTSUPERSCRIPT 9 end_POSTSUPERSCRIPT ) time by a similar modification from .
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In , Lazarev and Werner presented a new general graphical algorithm which often improves the complexity or at least the running time of dynamic programming algorithms, and it can be used for efficient approximation algorithm. Compared to standard dynamic programming algorithms, the number of states can often substantially be reduced.
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In a standard dynamic programming algorithm for permutation problems, functions F l(t)subscript 𝐹 𝑙 𝑡 F_{l}(t)italic_F start_POSTSUBSCRIPT italic_l end_POSTSUBSCRIPT ( italic_t ) and the corresponding partial sequences have to be stored for any stage l=1,2,…,n 𝑙 1 2…𝑛 l=1,2,\ldots,n italic_l = 1 , 2 , … , italic_n and any time t=0,1,…,UB 𝑡 0 1…𝑈 𝐵 t=0,1,\ldots,UB italic_t = 0 , 1 , … , italic_U italic_B. Here UB 𝑈 𝐵 UB italic_U italic_B is an upper bound for the scheduling interval, e.g. the sum of all processing times (or the maximal due date). The values F 1(t)subscript 𝐹 1 𝑡 F_{1}(t)italic_F start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT ( italic_t ) typically represent the best function value when considering the first l 𝑙 l italic_l jobs and allowing at most t 𝑡 t italic_t time units for the processing of the scheduled (e.g. early) jobs. This results in a pseudo-polynomial algorithm of the complexity O(n∑p j)𝑂 𝑛 subscript 𝑝 𝑗 O(n\sum p_{j})italic_O ( italic_n ∑ italic_p start_POSTSUBSCRIPT italic_j end_POSTSUBSCRIPT ) .
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The graphical algorithm combines different states, where the best partial solution does not change and the corresponding objective function F l(t)subscript 𝐹 𝑙 𝑡 F_{l}(t)italic_F start_POSTSUBSCRIPT italic_l end_POSTSUBSCRIPT ( italic_t ) can be described by the same linear function. As a result, at each stage l 𝑙 l italic_l, there are certain intervals separated by the so-called breakpoints t 1 l<t 2 l<…t h l l subscript superscript 𝑡 𝑙 1 subscript superscript 𝑡 𝑙 2…subscript superscript 𝑡 𝑙 subscript ℎ 𝑙 t^{l}{1}<t^{l}{2}<\ldots t^{l}{h{l}}italic_t start_POSTSUPERSCRIPT italic_l end_POSTSUPERSCRIPT start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT < italic_t start_POSTSUPERSCRIPT italic_l end_POSTSUPERSCRIPT start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT < … italic_t start_POSTSUPERSCRIPT italic_l end_POSTSUPERSCRIPT start_POSTSUBSCRIPT italic_h start_POSTSUBSCRIPT italic_l end_POSTSUBSCRIPT end_POSTSUBSCRIPT, where the structure of the linear function belonging to the best partial solution changes. In addition, for the graphical algorithm, the times t 𝑡 t italic_t do not need to be integer.
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In , Gafarov et al. described the application of this graphical algorithm to the approximate solution of various single machine problems. In particular, the graphical algorithm is applied to 5 total tardiness problems, where a pseudo-polynomial algorithm exists, and the graphical algorithm yields fully polynomial-time approximation schemes yielding the best known running time.
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In , Gafarov et al. considered the maximization of total tardiness and by applying the graphical algorithm, they transformed a pseudo-polynomial algorithm into a polynomial one. More precisely, the complexity of O(n∑p j)𝑂 𝑛 subscript 𝑝 𝑗 O(n\sum p_{j})italic_O ( italic_n ∑ italic_p start_POSTSUBSCRIPT italic_j end_POSTSUBSCRIPT ) can be reduced to O(n 2)𝑂 superscript 𝑛 2 O(n^{2})italic_O ( italic_n start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT ) what settled the open complexity status of the problem max1|(nd)|∑T j 1 𝑛 𝑑 subscript 𝑇 𝑗\max 1|(nd)|\sum T_{j}roman_max 1 | ( italic_n italic_d ) | ∑ italic_T start_POSTSUBSCRIPT italic_j end_POSTSUBSCRIPT), where nd 𝑛 𝑑 nd italic_n italic_d means that no inserted idle times occur. The authors present a detailed numerical example and give also a practical application of such maximization problems, where wind turbines have to be mounted in several regions of a country.
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While for more general settings, such a graphical algorithm improves often the running time in comparison with existing algorithms, the complexity remains pseudo-polynomial. For instance, in , Lazarev et al. presented graphical algorithms with a complexity of O(n∑p j)𝑂 𝑛 subscript 𝑝 𝑗 O(n\sum p_{j})italic_O ( italic_n ∑ italic_p start_POSTSUBSCRIPT italic_j end_POSTSUBSCRIPT ) for a special case of a generalized tardiness problem corresponding to the minimization of late work as well as for the minimization of the weighted number of late jobs.
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In , Lazarev and Werner considered various special cases of the (unweighted) total tardiness problem 1||∑T j 1||\sum T_{j}1 | | ∑ italic_T start_POSTSUBSCRIPT italic_j end_POSTSUBSCRIPT. After proving that already the special case of oppositely ordered processing times and due dates, i.e. p 1≥p 2≥…≥p n subscript 𝑝 1 subscript 𝑝 2…subscript 𝑝 𝑛 p_{1}\geq p_{2}\geq\ldots\geq p_{n}italic_p start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT ≥ italic_p start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT ≥ … ≥ italic_p start_POSTSUBSCRIPT italic_n end_POSTSUBSCRIPT and d 1≤d 2≤…≤d n subscript 𝑑 1 subscript 𝑑 2…subscript 𝑑 𝑛 d_{1}\leq d_{2}\leq\ldots\leq d_{n}italic_d start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT ≤ italic_d start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT ≤ … ≤ italic_d start_POSTSUBSCRIPT italic_n end_POSTSUBSCRIPT, is NP 𝑁 𝑃 NP italic_N italic_P-hard in the ordinary sense, they gave several pseudo-polynomial and polynomial algorithms for special cases of this problem. In particular, O(n 2)𝑂 superscript 𝑛 2 O(n^{2})italic_O ( italic_n start_POSTSUPERSCRIPT 2 end_POSTSUPERSCRIPT ) algorithms have been given for the cases d j−d j−1>p j,j=2,3,…,n formulae-sequence subscript 𝑑 𝑗 subscript 𝑑 𝑗 1 subscript 𝑝 𝑗 𝑗 2 3…𝑛 d_{j}-d_{j-1}>p_{j},j=2,3,\ldots,n italic_d start_POSTSUBSCRIPT italic_j end_POSTSUBSCRIPT - italic_d start_POSTSUBSCRIPT italic_j - 1 end_POSTSUBSCRIPT > italic_p start_POSTSUBSCRIPT italic_j end_POSTSUBSCRIPT , italic_j = 2 , 3 , … , italic_n as well as for the case that the difference between the maximal and the minimal due date is not greater than 1. In both cases, the processing times do not need to be integer.
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In further research, the graphical algorithm can be applied to more general problems, where a pseudo-polynomial algorithm exists and Boolean variables are used for making yes/no decisions.
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3.6 Preemptive Scheduling in Overloaded Systems
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In the single machine scheduling problem in overloaded systems, each job has a release time, a deadline and a weight. The latter parameter reflects its profit. Any job can be preempted during the execution, and the objective is to maximize the overall profit. Two possibly models and the definition of the profit gained by the execution of task j 𝑗 j italic_j can be defined. In the standard model, a completed task j 𝑗 j italic_j gives the profit w jp j subscript 𝑤 𝑗 subscript 𝑝 𝑗 w_{j}p_{j}italic_w start_POSTSUBSCRIPT italic_j end_POSTSUBSCRIPT italic_p start_POSTSUBSCRIPT italic_j end_POSTSUBSCRIPT, whereas a non-completed task gives no profit. In the metered model, a task j 𝑗 j italic_j executed for t≤p j 𝑡 subscript 𝑝 𝑗 t\leq p_{j}italic_t ≤ italic_p start_POSTSUBSCRIPT italic_j end_POSTSUBSCRIPT time units gives the profit w jt subscript 𝑤 𝑗 𝑡 w_{jt}italic_w start_POSTSUBSCRIPT italic_j italic_t end_POSTSUBSCRIPT (this task may not be completed). For the off-line metered profit model, Chrobak et al. used a reduction to bipartite matchings and maximal flows to develop a polynomial-time algorithm. For the online metered case, they presented an algorithm with competitive ratio e/(e−1)≈1.5820 𝑒 𝑒 1 1.5820 e/(e-1)\approx 1.5820 italic_e / ( italic_e - 1 ) ≈ 1.5820 and proved a lower lower bound of 5−1≈1.236 5 1 1.236\sqrt{5}-1\approx 1.236 square-root start_ARG 5 end_ARG - 1 ≈ 1.236 on the competitive ratio of algorithms for this problem.
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3.7 Scheduling with Restarts
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In the on-line single machine scheduling problem with restarts, jobs arriving over time are characterized by processing, release and delivery times, which become known at the arrival (release) time of each job. The objective is to minimize the maximum full job completion time, the makespan. As earlier mentioned, the corresponding off-line problem is strongly NP-hard, but the preemptive version can be solved in polynomial time by the preemptive LDT-heuristic. For the non-preemptive case, the LDT-heuristic has a performance guarantee of 2, which is the best possible one unless we allow the machine to be idle when there is an unprocessed job available. Hoogeveen and Vestjens developed an on-line algorithm with the best possible worst-case ratio
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(5+1)/2≈1.61803 5 1 2 1.61803(\sqrt{5}+1)/2\approx 1.61803( square-root start_ARG 5 end_ARG + 1 ) / 2 ≈ 1.61803
in which a waiting strategy is incorporated. Van den Akker, Hoogeveen and Vakhania considered an extension of the standard model by allowing job restarts: if an urgent job comes in while the machine is busy, then the currently running job can be interrupted in favor of the former urgent job. In this model, unlike traditional preemptive models, the already processed part of the interrupted job is wasted, this job is to be started all over again. The algorithm proposed in has the best possible worst-case ratio of 1.5 1.5 1.5 1.5 (which is better than the above mentioned bound for the case without the restarts).
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Again, we give some possible directions for further research in connection with allowed restarts. As we just mentioned, with the restart model considered in , if a job is restarted, then the portion of this job that was already processed by the machine is lost. This may be avoided for some applications, where the already completed part of an abandoned job gives some benefit and this job can be restarted from the time moment when it was interrupted or at some earlier time. Hence, one may consider models with partial restarts when the already processed part of an interrupted job counts partly or completely. In particular, it is interesting to find out whether there exist better worst-case performance bounds for models with partial restarts.
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3.8 Scheduling with Financial Resource Constraints
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Gafarov et al. considered single machine problems with a non-renewable resource (abbreviated NR 𝑁 𝑅 NR italic_N italic_R) like e.g. money or fuel. This means that at any time t i,i=0,1,…,UB formulae-sequence subscript 𝑡 𝑖 𝑖 0 1…𝑈 𝐵 t_{i},i=0,1,\ldots,UB italic_t start_POSTSUBSCRIPT italic_i end_POSTSUBSCRIPT , italic_i = 0 , 1 , … , italic_U italic_B with t 0=0 subscript 𝑡 0 0 t_{0}=0 italic_t start_POSTSUBSCRIPT 0 end_POSTSUBSCRIPT = 0 and UB 𝑈 𝐵 UB italic_U italic_B as an upper bound on the scheduling interval, one gets an amount G(t i)≥0 𝐺 subscript 𝑡 𝑖 0 G(t_{i})\geq 0 italic_G ( italic_t start_POSTSUBSCRIPT italic_i end_POSTSUBSCRIPT ) ≥ 0 of the resource, while any job j 𝑗 j italic_j has a consumption of g j subscript 𝑔 𝑗 g_{j}italic_g start_POSTSUBSCRIPT italic_j end_POSTSUBSCRIPT at the starting time s j subscript 𝑠 𝑗 s_{j}italic_s start_POSTSUBSCRIPT italic_j end_POSTSUBSCRIPT of the job. Hence, the equality
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∑j=1 n g j=∑i=0 UB G(t i)superscript subscript 𝑗 1 𝑛 subscript 𝑔 𝑗 superscript subscript 𝑖 0 𝑈 𝐵 𝐺 subscript 𝑡 𝑖\sum_{j=1}^{n}g_{j}=\sum_{i=0}^{UB}G(t_{i})∑ start_POSTSUBSCRIPT italic_j = 1 end_POSTSUBSCRIPT start_POSTSUPERSCRIPT italic_n end_POSTSUPERSCRIPT italic_g start_POSTSUBSCRIPT italic_j end_POSTSUBSCRIPT = ∑ start_POSTSUBSCRIPT italic_i = 0 end_POSTSUBSCRIPT start_POSTSUPERSCRIPT italic_U italic_B end_POSTSUPERSCRIPT italic_G ( italic_t start_POSTSUBSCRIPT italic_i end_POSTSUBSCRIPT )
holds. One looks for an optimal permutation (j 1,j 2,…,j n)subscript 𝑗 1 subscript 𝑗 2…subscript 𝑗 𝑛(j_{1},j_{2},\ldots,j_{n})( italic_j start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT , italic_j start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT , … , italic_j start_POSTSUBSCRIPT italic_n end_POSTSUBSCRIPT ) for a particular objective function with the starting times s j 1,s j 2,…,s j n subscript 𝑠 subscript 𝑗 1 subscript 𝑠 subscript 𝑗 2…subscript 𝑠 subscript 𝑗 𝑛 s_{j_{1}},s_{j_{2}},\ldots,s_{j_{n}}italic_s start_POSTSUBSCRIPT italic_j start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT end_POSTSUBSCRIPT , italic_s start_POSTSUBSCRIPT italic_j start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT end_POSTSUBSCRIPT , … , italic_s start_POSTSUBSCRIPT italic_j start_POSTSUBSCRIPT italic_n end_POSTSUBSCRIPT end_POSTSUBSCRIPT of the jobs such that the resource constraints are satisfied, i.e., the inequality
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∑l=1 i g j l≤∑{G(t v)∣t v≤s j ifor allv≤s j i}superscript subscript 𝑙 1 𝑖 subscript 𝑔 subscript 𝑗 𝑙 conditional-set 𝐺 subscript 𝑡 𝑣 subscript 𝑡 𝑣 subscript 𝑠 subscript 𝑗 𝑖 for all 𝑣 subscript 𝑠 subscript 𝑗 𝑖\sum_{l=1}^{i}g_{j_{l}}\leq\sum{G(t_{v})\mid t_{v}\leq s_{j_{i}}\mbox{ for % all }v\leq s_{j_{i}}}∑ start_POSTSUBSCRIPT italic_l = 1 end_POSTSUBSCRIPT start_POSTSUPERSCRIPT italic_i end_POSTSUPERSCRIPT italic_g start_POSTSUBSCRIPT italic_j start_POSTSUBSCRIPT italic_l end_POSTSUBSCRIPT end_POSTSUBSCRIPT ≤ ∑ { italic_G ( italic_t start_POSTSUBSCRIPT italic_v end_POSTSUBSCRIPT ) ∣ italic_t start_POSTSUBSCRIPT italic_v end_POSTSUBSCRIPT ≤ italic_s start_POSTSUBSCRIPT italic_j start_POSTSUBSCRIPT italic_i end_POSTSUBSCRIPT end_POSTSUBSCRIPT for all italic_v ≤ italic_s start_POSTSUBSCRIPT italic_j start_POSTSUBSCRIPT italic_i end_POSTSUBSCRIPT end_POSTSUBSCRIPT }
holds for any i=1,2,…,n 𝑖 1 2…𝑛 i=1,2,\ldots,n italic_i = 1 , 2 , … , italic_n. Such problems play a role e.g. for government-financed organizations when a project can only be started after receiving the required funds.
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The authors focused on complexity results and proved that the problems 1|NR|f 1 𝑁 𝑅 𝑓 1|NR|f 1 | italic_N italic_R | italic_f with f∈{C max,∑C j,∑U j}𝑓 subscript 𝐶 𝑚 𝑎 𝑥 subscript 𝐶 𝑗 subscript 𝑈 𝑗 f\in{C_{max},\sum C_{j},\sum U_{j}}italic_f ∈ { italic_C start_POSTSUBSCRIPT italic_m italic_a italic_x end_POSTSUBSCRIPT , ∑ italic_C start_POSTSUBSCRIPT italic_j end_POSTSUBSCRIPT , ∑ italic_U start_POSTSUBSCRIPT italic_j end_POSTSUBSCRIPT } are strongly NP 𝑁 𝑃 NP italic_N italic_P-hard. When minimizing total tardiness ∑T j subscript 𝑇 𝑗\sum T_{j}∑ italic_T start_POSTSUBSCRIPT italic_j end_POSTSUBSCRIPT, the case of equal due dates (d j=d subscript 𝑑 𝑗 𝑑 d_{j}=d italic_d start_POSTSUBSCRIPT italic_j end_POSTSUBSCRIPT = italic_d) is already strongly NP 𝑁 𝑃 NP italic_N italic_P-hard. The following special cases with minimizing total tardiness were proven to be weakly NP 𝑁 𝑃 NP italic_N italic_P-hard:
1) equal processing times (p j=p subscript 𝑝 𝑗 𝑝 p_{j}=p italic_p start_POSTSUBSCRIPT italic_j end_POSTSUBSCRIPT = italic_p);
2) equal due dates (d j=d)subscript 𝑑 𝑗 𝑑(d_{j}=d)( italic_d start_POSTSUBSCRIPT italic_j end_POSTSUBSCRIPT = italic_d ) and equal consumptions (g j=g subscript 𝑔 𝑗 𝑔 g_{j}=g italic_g start_POSTSUBSCRIPT italic_j end_POSTSUBSCRIPT = italic_g).
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On the other hand, the special case of equal processing times (p j=p subscript 𝑝 𝑗 𝑝 p_{j}=p italic_p start_POSTSUBSCRIPT italic_j end_POSTSUBSCRIPT = italic_p) and ordered consumptions and due dates according to g 1≤g 2≤…≤g n subscript 𝑔 1 subscript 𝑔 2…subscript 𝑔 𝑛 g_{1}\leq g_{2}\leq\ldots\leq g_{n}italic_g start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT ≤ italic_g start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT ≤ … ≤ italic_g start_POSTSUBSCRIPT italic_n end_POSTSUBSCRIPT and d 1≤d 2≤…≤d n subscript 𝑑 1 subscript 𝑑 2…subscript 𝑑 𝑛 d_{1}\leq d_{2}\leq\ldots\leq d_{n}italic_d start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT ≤ italic_d start_POSTSUBSCRIPT 2 end_POSTSUBSCRIPT ≤ … ≤ italic_d start_POSTSUBSCRIPT italic_n end_POSTSUBSCRIPT is polynomially solvable.
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This paper considered also the budget scheduling problem with the objective to minimize the makespan. Instead of a resource consumption g j subscript 𝑔 𝑗 g_{j}italic_g start_POSTSUBSCRIPT italic_j end_POSTSUBSCRIPT for each job, two values g j−superscript subscript 𝑔 𝑗 g_{j}^{-}italic_g start_POSTSUBSCRIPT italic_j end_POSTSUBSCRIPT start_POSTSUPERSCRIPT - end_POSTSUPERSCRIPT and g j+superscript subscript 𝑔 𝑗 g_{j}^{+}italic_g start_POSTSUBSCRIPT italic_j end_POSTSUBSCRIPT start_POSTSUPERSCRIPT + end_POSTSUPERSCRIPT are given, where g j−superscript subscript 𝑔 𝑗 g_{j}^{-}italic_g start_POSTSUBSCRIPT italic_j end_POSTSUBSCRIPT start_POSTSUPERSCRIPT - end_POSTSUPERSCRIPT has the same meaning as g j subscript 𝑔 𝑗 g_{j}italic_g start_POSTSUBSCRIPT italic_j end_POSTSUBSCRIPT before. The value g j+superscript subscript 𝑔 𝑗 g_{j}^{+}italic_g start_POSTSUBSCRIPT italic_j end_POSTSUBSCRIPT start_POSTSUPERSCRIPT + end_POSTSUPERSCRIPT describes the additional earnings at the completion time of job j 𝑗 j italic_j. The authors showed that the problem 1|N R,g j−,g j+|C max 1|NR,g_{j}^{-},g_{j}^{+}|C_{max}1 | italic_N italic_R , italic_g start_POSTSUBSCRIPT italic_j end_POSTSUBSCRIPT start_POSTSUPERSCRIPT - end_POSTSUPERSCRIPT , italic_g start_POSTSUBSCRIPT italic_j end_POSTSUBSCRIPT start_POSTSUPERSCRIPT + end_POSTSUPERSCRIPT | italic_C start_POSTSUBSCRIPT italic_m italic_a italic_x end_POSTSUBSCRIPT belongs to the class APX 𝐴 𝑃 𝑋 APX italic_A italic_P italic_X of problems allowing a constant-factor approximation algorithm.
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For future work, due to the NP 𝑁 𝑃 NP italic_N italic_P-hardness of the majority of problems, it would be interesting to derive elimination rules and structural properties of optimal solutions which can be used in enumerative algorithms.
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4 Concluding Remarks
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In this paper, we reviewed some theoretical results on classical single machine scheduling problems. Since there exists a huge number of heuristic/metaheuristic and enumerative algorithms, our focus was on polynomial algorithms as well as complexity and approximability issues, discussed in the papers authored or co-authored by the authors of this work. The summarized polynomial special cases are also of interest when the conditions are not exactly satisfied and the resulting solution can be taken as a good heuristic one.
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For single machine scheduling problems, efficient solution methods can often be found easier than for the corresponding multiprocessor or shop scheduling problems. However, such methods may give a better insight into the latter more complex problems and may well serve as a basis for the development of solution methods for these problems. At the same time, optimal solutions to the single machine scheduling problems can directly be used to solve the corresponding multiprocessor or shop scheduling problems. For example, for such NP 𝑁 𝑃 NP italic_N italic_P-hard problems, strong lower bounds can be obtained based on the solutions of the corresponding single machine problems. Non-strict lower bounds, which are easier and faster to obtain, can be used in any search-tree based approximation algorithm, such as beam search, for example. Similarly, using strict lower bounds, we obtain exact branch-and-bound algorithms. Once implemented and tested, the above algorithms can be used as an algorithmic engine for the construction of decision support systems. Finally, we note that a number of algorithms for basic single machine problems can relatively easily be adopted using additional graph-completion mechanisms to solve the corresponding extended versions with precedence relations, transportation and setup times and other real-life job scheduling problems.
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References
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6996 | https://www.sciencedirect.com/science/article/abs/pii/S095183202300546X | Redundancy allocation problem with a mix of components for a multi-state system and continuous performance level components - ScienceDirect
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Cited by (16)
Reliability Engineering & System Safety
Volume 241, January 2024, 109632
Redundancy allocation problem with a mix of components for a multi-state system and continuous performance level components
Author links open overlay panel Mani Sharifi 2, Sharareh Taghipour 1 2
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Highlights
•We are presenting a modified UGF (MUGF) method which can calculate the availability of the MSS of the BS-CPL.
•The presented has a significantly low computational time.
•The BS-CPL for a RAPMC is considered in this paper for the first time.
•We adapted an effective GA to solve this complex problem in a significantly low computational time.
•We presented a two-stage parameter tuning that expedites the GA's convergence speed.
Abstract
This paper considered a new single-objective redundancy allocation problem with a mix of components (RAPMC) for a series-parallel system. The components are Binary-State Continuous Performance Level (BSCPL), which means each component has working or failed states, while the performance level of the components varies from its maximum performance level to zero. In RAPMC, the model can allocate a mix of components to each sub-system, so the model aims to maximize the system's availability during a fixed period of time by determining the optimal combination of the components in the sub-systems. The mathematical model of the system is presented in the first step. Then, a modified version of the universal generating function is proposed to calculate the sub-systems and the system's availability. Next, a novel genetic algorithm is adapted to solve the problem. Finally, the performance of the adapted GA is validated using the results of an exact method, Full enumeration. The adapted GA showed its applicability to solve the presented model in significantly low computational time.
Introduction
Many real-case engineering problems optimize the reliability of the system connected serially, like the electrical circuits or the hydro transmission systems. Two methods are generally considered to increase such system's reliability: reliability optimization and redundancy allocation problem (RAP) . Reliability optimization problem increases the system's reliability and improves the performance of the components and systems by considering some technical and organizational activities , which improves the components’ reliability. However, RAP increases the system's reliability by allocating redundant components to the system's sub-systems . Fyffe et al. presented the RAP for the first time in 1968. Their presented RAP aimed to maximize the system's reliability by determining the optimally allocated components to each sub-system under the system's cost and weight constraints.
After that, many different assumptions have been considered to draw the RAP nearer to the real-case condition. Fig.1 shows different features of the RAP and the position of the current research in the RAP literature.
In this paper, we presented a mathematical model that aims to optimize the system availability by determining the combination of allocated components to each sub-system. The system structure is series-parallel, meaning that the sub-systems are connected serially, while the components in each sub-system are connected in parallel. So, the failure of one sub-system causes the system's failure. In other words, the system works when all its sub-system works. Moreover, each sub-system works when at least one of its components works. To put it better, the sub-system fails when all its components fail. The considered RAP is the RAP with a mix of components (RAPMC), which means that a combination of the available components can be allocated to each sub-system . In terms of the component type, the components are non-identical; however, the model can allocate identical or non-identical components to each sub-system. The presented model is able to use the components with constant or time-dependent failure rates, which turned it into a general model. The components are the binary state, meaning they have working or failed states while they have a continuous performance level . So, in general, the components can be considered as a binary state or multi-state; however, in this paper, we prefer to say that the components are binary-state. Since the components have a continuous performance level, the sub-system and the system will have a discrete performance level, making the multi-state system (MSS). Finally, the considered redundancy strategy in this paper is active, which means that all available components start working together at the beginning of the system's mission horizon .
This paper considers the binary-state continuous performance level (BS-CPL) components. In order to solve a RAP with BS-CPL components, the general formulas are first presented to calculate the components, sub-systems, and system availability. Then the RAP is modeled using mixed-integer non-linear programming. Next, the presented model is solved using a novel genetic algorithm (GA). to have a fast convergence of the adapted GA, a novel two-stage response surface methodology (RSM) is used, and a full enumeration (FE) technique is used to validate the performance of the GA in terms of achieving to the optimal or near-optimal solutions.
The rest of the paper is organized as follows. Section 2 investigates the RAP literature, followed by the literature gap analysis and the paper's novelty. Section 3 in the problem definition, the mathematical model of the RAP, and calculating the system availability. In Section 4, the solving methodology is presented. Section 5 is the numerical example, and the last section is the conclusion and further studies.
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Section snippets
Literature review
This paper presents a RAPMC with BD-CPL for a system with a series-parallel configuration and active redundancy strategy. In series-parallel systems, the sub-systems are connected serially, which means the failure of one sub-system causes the system's failure. In other words, the system works when all its sub-system works. While in the sub-systems, the components are connected in parallel, which means the sub-system works when at least one of its components works. To put it better, the
Model assumptions
The assumption of the considered system and its component, as well as the model assumptions, are provided as follows:
•The system consists of s serially connected sub-systems,
•Different component types are available to be allocated to each sub-system,
•A combination of the available components can be allocated to each sub-system,
•In each sub-system, the components are connected in parallel,
•Each component has two discrete states as working and failed,
•Each component has a continuous working performance,
Solving methodology
Genetic algorithm is a robust meta-heuristic algorithm to solve the optimization problem (i.e., [52,53]), especially RAP (i.e., i.e., general GA , , , Parallel GA [57,58], NSGA-II and NRGA [23,, , , ]). So, in this paper we adapted a novel GA to solve the proposed RAP. Moreover, to validate the performance of the adapted GA, we used a full enumeration (FE) technique.
The main novel aspects of the proposed GA are considering three different types of crossover operators
Numerical example
In this section, firstly, the model parameters are presented. Then, the mathematical model, as well as the performance of the adapted GA in terms of achieving the optimal or near optimal solutions, is checked. Next, a large-scale problem is solved to show the structure of the solutions of the GA, and finally, the sensitivity analysis is conducted.
In Section 5.2, we solved 5 instances with 3 to 7 sub-systems for the validation of the model and algorithm. In Section 5.3, we solved an instance
Conclusion and further studies
This paper investigated the redundancy allocation problem of a series-parallel system with a mix of components. The component was the binary state, working or failed, and continuous performance level, which categorized the components and the system as multi-state. A mixed integer non-linear mathematical model was presented to optimize the system's availability by determining the combinations of the allocated model to each sub-system. Firstly, the exact formulas to calculate the component,
Funding Statement
The funding for this research was provided by the Canada Research Chair (CRC) program.
CRediT authorship contribution statement
Mani Sharifi: Conceptualization, Methodology, Software, Validation, Formal analysis, Investigation, Visualization, Writing – original draft. Sharareh Taghipour: Supervision, Project administration, Funding acquisition, Writing – review & editing.
Declaration of Competing Interest
The authors declare that they have no known competing financial interests or personal relationships that could have appeared to influence the work reported in this paper.
Dr. Mani Sharifi is a post-doctoral research fellow at the Reliability, Risk, and Maintenance Research Laboratory (RRMR Lab) at the Mechanical and Industrial Engineering Department of Toronto Metropolitan University. He holds a B.Sc. degree from Qazvin IAU, an M.Sc. degree from the south Tehran branch IAU, and a Ph.D. degree from Tehran Research and Science IAU in industrial engineering. He was the Managerial editor of the Journal of Optimization in Industrial Engineering. He joined Ryerson
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Developing k-out-of-n: G multilevel system with mixed redundancy strategy to protect DSP code using simplified swarm optimization
2025, Reliability Engineering and System Safety Show abstract Mitigating transient faults in aerospace software, particularly radiation-induced Single-Event Upsets (SEUs) affecting Digital Signal Processor (DSP) code, remains a critical challenge. Redundancy strategies are among the most effective approaches to address SEUs. This study introduces a novel program architecture based on a k-out-of-n: G system, incorporating mixed redundancy strategies to enhance the reliability of DSP code. As DSP code operates within a multilevel system, this is the first study to integrate a multilevel system with k-out-of-n: G modules using mixed redundancy strategies. This integration allows for diverse combinations of subsystem reliability calculations, making it strategically advantageous. To evaluate the reliability of redundant program modules in each subsystem, a modular continuous-time Markov chain (CTMC) was applied. To address the extensive combinations of k and n values alongside redundancy strategies, simplified swarm optimization (SSO) was employed for multilevel encoding and near-optimal solution discovery. Experiments on the Fast Fourier Transformation (FFT) program demonstrated the method's effectiveness compared to state-of-the-art approaches, further verifying its scalability and capability to establish a more stable and highly reliable DSP code architecture for large-scale problems.
### An exact algorithm for RAP with k-out-of-n subsystems and heterogeneous components under mixed and K-mixed redundancy strategies
2025, Advanced Engineering Informatics Show abstract Redundancy design is a widely used technique for enhancing system reliability across various industries, including aerospace and manufacturing. Consequently, the redundancy allocation problem (RAP) has attracted considerable attention in the field of reliability engineering. The RAP seeks to determine an optimal redundancy scheme for each subsystem under resource constraints to maximize system reliability. However, existing RAP models and exact algorithms are predominantly confined to simple 1-out-of-n subsystems or single optimization strategies, thereby limiting the optimization potential and failing to adequately address the engineering requirements. This paper introduces a model and an exact algorithm for RAP with k-out-of-n subsystems and heterogeneous components under mixed and K-mixed redundancy strategies. The model employs a continuous time Markov chain method to calculate subsystem reliability exactly. A dynamic programming (DP) algorithm based on super component and sparse node strategies is designed to obtain the exact solution for RAP. Numerical experiments confirm that all benchmark test problems reported in the literature are exactly solved by the proposed DP. The experiment results demonstrate that the proposed RAP model offers high flexibility and potential for reliability optimization. Additionally, owing to the generality of the problem considered, the proposed DP also exactly solves other RAP models with 1-out-of-n subsystems and simplified redundancy strategies, which provides a more generalized framework for redundancy optimization. Finally, the research’s applicability in reliability engineering is validated through an optimization case study of a natural gas compressor pipeline system.
### A genetic algorithm and particle swarm optimization for redundancy allocation problem in systems with limited number of non-cooperating repairmen
2024, Expert Systems with Applications Citation Excerpt :
Algorithm 5 presents the pseudocode of the proposed PSO algorithm, which has been implemented in MATLAB 2022b. Table 2 presents the basic parameters for the GA and PSO, which were established based on numerous publications on swarm (Huang, 2015; Huang et al., 2021; Mousavi et al., 2019; Yeh, 2019) and evolutionary algorithms (Keshavarz Ghorabaee et al., 2015; Sharifi & Taghipour, 2024; Sun et al., 2019). Both algorithms have a population and swarm size of 100, and run for 1000 generations/iterations. Show abstract This article considered the redundancy allocation problem (RAP) in repairable heterogeneous systems composed of k-out-of-n: G subsystems with a limited number of repairmen. The unit costs of the binary-state components within the subsystems remain constant over time. Exponential characteristics of independent failure and repair processes were assumed. To evaluate subsystem and system availability, a Continuous Time Markov Chain (CTMC) adapted to three standby modes (cold, warm, and hot) was developed. The characteristics of the redundant system and the assumption of a limited number of repairmen contribute to the existing scientific literature on RAP. Two nature-inspired metaheuristic approaches were developed to maximize the availability of the coherent system as an objective function to optimize redundancy with a cost constraint on component purchase. The proposed approaches were tested on a 15-unit large-scale system. The effectiveness and performance of the algorithms was evaluated as a function of the number of generations/iterations and population/swarm size. The genetic algorithm (GA) proved to be highly effective in finding the global optimum and outperformed particle swarm optimization (PSO) in terms of computational performance. Analysis of metaheuristic algorithms can offer valuable insights into future research on redundancy optimization. The sensitivity analysis of the large-scale system clearly demonstrates that the number of repairmen and cost constraints (up to three times the sum of the value of active components) have a significant impact on the system’s availability. Above this level, increasing spending on system building, regardless of the number of repairmen, is not justified by considerations of system availability.
### A novel importance measure considering multi-constraints for RAP optimization of 1-out-of-n subsystems with mixed redundancy strategy
2024, Reliability Engineering and System Safety Citation Excerpt :
Subsequently, the CTMC-based method is also introduced to RAP to accurately assess the reliability of subsystems under cold-standby and mixed strategies [11,12]. Along with developments in system characteristics (e.g., network structure , k-out-of-n subsystems , multi-state ) and redundancy methods (e.g., heterogeneous components , mixed strategy ), RAP has become increasingly complex. The challenging features of RAP, such as the nonlinear objective function and integer decision variables, incur exponentially increasing computational costs with the system scale. Show abstract Intelligent optimization algorithms are the mainstream approach to solving redundancy allocation problems (RAP) with challenging features. Since importance measures (IM) can identify critical components, the combination of IM-based local optimization and intelligent algorithms has wide applications in various optimization problems; however, it is less studied in RAP. Existing IMs also failed to address both the objective function and multiple constraints like cost and weight; this may result in an imprecise identification of critical subsystems for RAP optimization. This paper considers a RAP with a mixed strategy, i.e., active and standby strategies can be applied to a subsystem simultaneously. Two novel IMs are proposed based on a Lagrangian function: cost-centric RAP-based importance (CRI) and weight-centric RAP-based importance (WRI). CRI (WRI) reveals the comprehensive effect of cost (weight) consumption on the system reliability and other resources. A local optimization algorithm guided alternately by CRI and WRI is presented to adjust the redundancy level of subsystems; then, this algorithm is introduced into a genetic algorithm (GA) to determine the component types and redundancy level of all subsystems. Compared with other algorithms and previous studies, the superiority of the proposed hybrid GA is demonstrated via numerical experiments and a well-known benchmark example.
### A hierarchical k-out-of-n optimization model for enhancing reliability of fire alarm systems
2024, Process Safety and Environmental Protection Citation Excerpt :
For instance, if there are three possible types of sensors in a system, and the utilization of two sensors is permitted for each type, it is necessary to define 6 decision variables in the model to account for these possibilities. There has been numerous work which report the accuracy and efficiency of a GA in finding optimal solutions for systems with k-out-of-n configurations (Baladeh et al., 2017; Nath and Muhuri, 2022; Peiravi et al., 2022; Baladeh and Taghipour, 2023; Zhang et al., 2023; Sharifi and Taghipour, 2024). There are comparison studies to demonstrate the advantage of GA over other metaheuristic algorithms in solving RAPs (Peiravi et al., 2021; Sharifi et al., 2022). Show abstract Fire alarm systems are crucial for mitigating the risks and damages associated with fires in process industries. These systems heavily depend on the performance of fire detection sensors to provide early detection and warnings, facilitating prompt evacuation and intervention measures. In order to improve the reliability of sensor systems typically requires the use of redundancy allocation techniques, such as a k-out-of-n configuration. In k-out-of-n configuration, the signals from all sensors are directed to a voter, which compares them and transmits a final signal once at least k signals are matched. However, challenges arise due to diverse failure modes with distinct impacts on the system. Integrating sensors through a single-layer k-out-of-n configuration, when there is only one voter, fails to fully highlight the potential advantages of redundancy in sensor systems. This paper addresses the limitations of the conventional configurations by introducing a hierarchical k-out-of-n system in which sensors can be integrated across multiple layers. In the k-out-of-n hierarchical system, sensors are organized into distinct groups, each employing its own k-out-of-n sub-system. Subsequently, the outputs of these sensor groups can be integrated with other groups through their respective voters in higher layers of the hierarchy. Ultimately, at the top layer, all the voters from the preceding layers must be connected to a top voter to transmit a final signal. Furthermore, an optimization model is developed to minimize the system cost while maximizing reliability, considering the best selection of sensors and the system configuration. The proposed optimization model takes into account a multi-sensor system with two competing failure modes and incorporates the probability of sensors' failure modes in the system. The hierarchical k-out-of-n system provides an opportunity to validate sensors by comparing them within their designated groups, allowing for the identification of the location of a failure and the planning of appropriate actions accordingly. The proposed optimization model is applied to a case of a fire detection system in storage warehouse to demonstrate its advantages over the conventional model.
### A mathematical optimization model for determining safety integrity levels in process facilities
2024, Reliability Engineering and System Safety Show abstract In process industries, Safety Instrumented Functions (SIFs) are implemented to meet government-mandated risk tolerance criteria, which aim to ensure the safety of both individuals and society. The level of risk reduction achieved by a SIF is quantified by its Safety Integrity Level (SIL). Determining the target SILs is a challenging task for facility owners as they must balance the costs and benefits of risk reduction while complying with regulatory requirements. Furthermore, governments define risk tolerance criteria for an overall facility rather than for single scenarios. Therefore, determining a set of SILs for all SIFs becomes imperative to collectively reduce both individual and societal risks to tolerable levels intended for the overall facility. In this paper, we propose a mathematical optimization model to determine the most beneficial collection of SILs that guarantees meeting the risk tolerance criteria. Our model considers both individual and societal risk perspectives and effectively models an overall facility. The proposed model is applied to a process facility and the results are compared with the results from the existing methods for SIL determination. The results demonstrate the model's superior effectiveness in helping owners to select the most beneficial target SILs while ensuring compliance with government requirements.
View all citing articles on Scopus
Dr. Mani Sharifi is a post-doctoral research fellow at the Reliability, Risk, and Maintenance Research Laboratory (RRMR Lab) at the Mechanical and Industrial Engineering Department of Toronto Metropolitan University. He holds a B.Sc. degree from Qazvin IAU, an M.Sc. degree from the south Tehran branch IAU, and a Ph.D. degree from Tehran Research and Science IAU in industrial engineering. He was the Managerial editor of the Journal of Optimization in Industrial Engineering. He joined Ryerson University's Department of Mechanical & Industrial Engineering (MIE) in November 2018. His area of interest includes Reliability Engineering, Combinatorial Optimization, Statistical Optimization, as well as Production Scheduling.
Dr. Sharareh Taghipour is an Associate Professor and the RRMR Lab Director at the Department of Mechanical and Industrial Engineering at Toronto Metropolitan University. She is a Tier 2 Canada Research Chair in Physical Asset Management. Dr. Taghipour received her B.Sc. in Mathematics and Computer Science and her M.Sc. in Industrial Engineering from Sharif University of Technology, Iran. She received her Ph.D. in Industrial Engineering from the University of Toronto. Her research interests are mathematical modeling and optimization with the applications in reliability engineering, maintenance optimization, and production scheduling.
1
Canada Research Chair in Physical Asset Management, Ph.D., Associate Professor.
2
Department of Mechanical and Industrial Engineering, Toronto Metropolitan University (Formerly Ryerson University), 350 Victoria Street, Toronto, Ontario, Canada.
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MCAT Mnemonics: Diatomic Elements
Ken Tao
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Ken Tao is the MedSchoolCoach expert on MCAT, and discusses acronyms to help remember the diatomic elements: Have No Fear of Ice Cold Beer.
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Hey everyone. Welcome back to another MCAT Mnemonic Monday. My name is Ken and I’m an MCAT expert with MedSchoolCoach. Today, we have an MCAT general chemistry mnemonic and it is on the diatonic elements.
These are good to keep in mind because when you’re calculating the enthalpy change of a reaction using heat of formation you want to keep in mind that the diatomic elements exist in their diatomic forms and their standard state and therefore have a heat of formation of zero.
So our Mnemonic is: Have No Fear Of Ice Cold Beer.
So these are our seven diatomic elements: Hydrogen, Nitrogen, Flourine, Oxygen, Iodine, Chlorine, Iodine, and Bromine.
And what I particularly like about this mnemonic is you’ll notice that ice is a solid, and beer is a liquid. All of the diatomic elements are gases except iodine which is a solid, and bromine which is a liquid.
Related posts:
MCAT Mnemonics: Standing Waves
MCAT Mnemonics: Blots
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MCAT Mnemonics: Myopia and Hyperopia
Ken Tao
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Ken is nationally recognized as a premier MCAT mind. He has worked with thousands of undergraduate students as a graduate teaching assistant and MCAT instructor/tutor for the Princeton Review. At Princeton Review, Ken was the only tutor certified in all subjects, was one of the highest rated MCAT tutors ever and was a teacher trainer. Additionally, Ken worked to found Magoosh's MCAT division. He has written content for dozen's of MCAT books and guides. He is now the Director of MCAT at MedSchoolCoach
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6998 | https://math.stackexchange.com/questions/1280411/a-question-about-2-times-3-rectangles | Skip to main content
A question about (2×3)-rectangles.
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The following is a problem from TopCoder:
Problem. Given the width and the height of a rectangular grid, return the total number of non-square rectangles that can be found on the grid.
For example, for a grid with width 3 and height 3, i.e.,
there are
four (2×3)-rectangles,
six (1×3)-rectangles and
twelve (1×2)-rectangles.
There are thus a total number of 4+6+12=22 rectangles. Note that we don’t count (1×1)-, (2×2)- or (3×3)-rectangles because these are actually squares.
I have assumed that “2×3” means “2 rows and 3 columns”. Is my assumption correct?
Can someone help me to find how many rectangles there are in the rectangular grid below?
EDIT: The actual grid in the question is unknown, so a (5×7)-grid is being given as a working example.
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edited May 13, 2015 at 17:49
Berrick Caleb Fillmore
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asked May 13, 2015 at 15:03
James K JJames K J
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I would look for the total number of rectangles including squares, and then subtract the number of squares. A rectangle is given by its top-left-to-bottom-right diagonal, which is a pair of points.
– Arthur
Commented
May 13, 2015 at 15:05
Based on the counts, 2x3 must mean either 2 rows and 3 columns or 2 columns and 3 rows -- i.e., both orientations of the (non-square) rectangle. Likewise for 1x3 and 1x2.
– Barry Cipra
Commented
May 13, 2015 at 15:13
1
Arthur’s suggestion is good. To count the number of rectangles, count the number of ways to choose two points not in the same row or column. Each such pair of points determines a rectangle having a diagonal between those points. The number of such pairs of points is twice the number of rectangles (including squares), because each rectangle has two diagonals. (If you count ordered pairs of points, you will get four times the number of rectangles.)
– Steve Kass
Commented
May 13, 2015 at 15:14
Also, what do you mean by "the below rectangular grid"? Is there a picture missing?
– Barry Cipra
Commented
May 13, 2015 at 15:14
Tried to fix up the question but I don't think the requested image was available, so put a sample grid in place - @jamesjohnson92 can you confirm what the required grid dimensions are?
– Joffan
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May 13, 2015 at 16:11
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Notation
I have assumed that “2×3” means “2 rows and 3 columns”. Is my assumption correct?
It depends on the context. In general I'd agree, but since the counts you quote doesn't include a separate handling of the 3×2 case, in this situation I'd rather assume that 2×3 is meant to cover both orientations, both two rows and three columns and two columns and three rows.
Total count
Can someone help me to find how many rectangles there are in the rectangular grid below?
If all you are interested in is the total number of non-square rectangles, then it doesn't matter how you write down each individual count. In fact, I'd not iterate over all possible shapes, but instead use a different approach.
For a grid of m×n tiles, there are (m+1)×(n+1) tile corners. If you consider each of them as a possible corner of a rectangle, and define each rectangle by two opposite corners, then you have a total of
((m+1)(n+1))2
possible combinations of two points. Obviously, that number is way too large, because we counted some things we shouldn't, and counted others more than once. So let's address those issues.
You don't want either dimension of such a rectangle to be zero. So if you picked the first point, then there are m+n+1 possible positions where that second point would be in the same row and/or the same column as the first. Subtract that and you are at
(m+1)(n+1)((m+1)(n+1)−(m+n+1))=(m+1)(n+1)mn.
As Arthur pointed out in a comment, you can think of this as picking the second point from the m×n possible corner positions which you obtain by removing the “forbidden” row and column. At this point you have two points in distinct rows and columns, but nothing is sorted yet. So every possible rectangle is counted four times: any of its four corners might be the first point, with the opposite corner as the second point. So the number of rectangles is
14(m+1)(n+1)mn.
Now you still have to get rid of those squares. How many are there? Let's look for a pattern. There are m×n squares of size 1. There are (m−1)×(n−1) squares of size 2, and so on. The maximal size such a square can have is the smaller of the dimensions of your tile grid. So there are
∑i=1min(m,n)(m+1−i)(n+1−i)
squares in total. Assuming m≤n this simplifies to
∑i=1m(m+1−i)(n+1−i)=16m(m+1)(3n−m+1).
Now take the number of all rectangles, subtract all squares, and you are done. If you expand and then factor the result, you get a total count of
112m(m+1)(3n2+2m−3n−2)=112m(m+1)(3n(n−1)+2(m−1)).
For m=n=3 this indeed yields a count of 22. For m=5,n=7 (remember to ensure m≤n) this gives a count of 335 non-square rectangles. You can verify that count with a brute-force computation.
Bonus question
If you want to, it might be interesting to think about why that formula always results in an integer if m,n are integers.
Either m or m+1 is divisible by two, so their product is even. For the same reason, n(n−1) is even, and since 2(m−1) is even, too, (3n(n−1)+2(m−1)) must be even. So the whole thing is the product of two even numbers, and therefore divisible by four.
Exactly one of m,(m+1),(m−1) is divisible by three. In the first two cases, the whole product is obviously divisible by three. In the last case, (3n(n−1)+2(m−1)) is divisible by three since it's the sum of two integers which are divisible by three. As the whole thing is divisible by four and by three, it is divisible by twelve.
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edited May 13, 2015 at 18:26
answered May 13, 2015 at 17:14
MvGMvG
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Good answer! Well thought out, steps shown, and a clean formula at the end. Seriously, you've made a great answer.
– Asimov
Commented
May 13, 2015 at 17:26
1
Very good answer, but it could be slightly cleaner. (m+1)(n+1)−(m+n+1) simplifies to mn, with the interpretation that after picking one vertex, you delete a row and a column, so there are mn vertices left.
– Arthur
Commented
May 13, 2015 at 18:19
@Arthur: Good point. I deliberately avoided simplifications just for the sake of shorter formulas at that point, but with that interpretation this is indeed a win in both complexity and clarity. I incorporated that in my post.
– MvG
Commented
May 13, 2015 at 18:29
@MvG I have understood the equations. One more help, can you explain " how to get rid of squares" and why did you choose min(m,n) ?
– James K J
Commented
May 16, 2015 at 14:29
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Let the lattice be composed of m rows and n columns forming mn squares.
The lattice is actually composed by (m+1) horizontal and (n+1) vertical and lines. An object (it doesn’t matter where it is a rectangle or a square) is formed by any two horizontal lines and any two vertical lines. Thus, the number of such objects = (m+12)⋅(n+12) …. (1).
Let us count the total number of 1 x 1 squares. Notice that each these squares has a distinct right-bottom corner point. We found m x n those right-bottom corner points. This means there are are [m] . [n] square whose size is 1 x 1. See figure 1.
We use the similar fashion to count the number of 2 x 2 squares and there are [m – 1] . [n – 1] in total. See figure 2.
The other sizes of squares can be counted similarly and the counting stops when any one of the square bracket terms has reached a 1.
Thus, the total number of squares = [m] . [n] + [m – 1][n – 1] + [m – 2] . [n – 2] + … …… (2)
(2) has a regualar pattern and can therefore be re-written in some closed form.
The required total = (1) – (2)
For a 3 by 3 lattice, TOTAL = 36 – 14 = 22
For a 7 by 5 lattice, TOTAL = 6∗51∗2∗8∗71∗2–(.+.+….)=335.
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answered May 13, 2015 at 18:26
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6999 | https://www2.stat.duke.edu/courses/Spring16/sta532/lec/fish.pdf | Fisher Information & Efficiency Robert L. Wolpert Department of Statistical Science Duke University, Durham, NC, USA 1 Introduction Let f(x | θ) be the pdf of X for θ ∈Θ; at times we will also consider a sample x = {X1, · · · , Xn} of size n ∈N with pdf fn(x | θ) = Q f(xi | θ). In these notes we’ll consider how well we can estimate θ or, more generally, some function g(θ), by observing X or x.
Let λ(X | θ) = log f(X | θ) be the natural logarithm of the pdf, and let λ′(X | θ), λ′′(X | θ) be the first and second partial derivative with respect to θ (not X). In these notes we will only consider “regular” distributions, those with continuously differentiable (log) pdfs whose support doesn’t depend on θ and which attain a maximum in the interior of Θ (not at an edge) where λ′(X | θ) vanishes.
That will include most of the distributions we consider (normal, gamma, poisson, binomial, exponential, . . . ) but not the uniform.
The quantity λ′(X | θ) is called the “Score” (sometimes it’s called the “Score statistic” but really that’s a misnomer, since it usually depends on the parameter θ and statistics aren’t allowed to do that). For a random sample x of size n, since the logarithm of a product is the sum of the logarithms, the Score is the sum λ′ n(x | θ) = (∂/∂θ) log fn(x | θ) = P λ(Xn | θ).
Usually the MLE ˆ θ is found by solving the equation λ′ n(x | ˆ θ) = 0 for the Score to vanish, but today we’ll use it for other things. For fixed θ, and evaluated at the random variable X (or vector x), the quantity Z := λ′(X | θ) (or Zn := λ′ n(x | θ)) is a random variable; let’s find its mean and variance. First consider a single observation, or n = 1.
Let’s consider continuously-distributed random variable X (for discrete distributions, just replace the integrals below with sums). Since 1 = Z f(x | θ) dx for any pdf and for every θ ∈Θ, we can take a derivative to find 0 = ∂ ∂θ Z f(x | θ) dx = Z ∂ ∂θf(x | θ) f(x | θ) f(x | θ) dx = Z λ′(x | θ) f(x | θ) dx = Eθλ′(X | θ), (1) 1 so the Score always has mean zero. The same reasoning shows that, for random samples, Eθλ′ n(x | θ) = 0. The variance of the Score is denoted I(θ) = Eθ λ′(X | θ)2 (2) and is called the Fisher Information function. Differentiating (1) (using the product rule) gives us another way to compute it: 0 = ∂ ∂θ Z λ′(x | θ) f(x | θ) dx = Z λ′′(x | θ) f(x | θ) dx + Z λ′(x | θ) f ′(x | θ) dx = Z λ′′(x | θ) f(x | θ) dx + Z λ′(x | θ) f ′(x | θ) f(x | θ) f(x | θ) dx = Eθ λ′′(X | θ) + Eθ λ′(X | θ)2 = Eθ λ′′(X | θ) + I(θ) by (2), so I(θ) = Eθ −λ′′(X | θ) .
(3) Since λn(x | θ) = P λ(Xi | θ) is the sum of n iid RVs, the variance In(θ) = Eθ[λ′ n(x | θ)2] = n I(θ) for a random sample of size n is just n times the Fisher Information for a single observation.
1.1 Examples Normal: For the No(θ, σ2) distribution with fixed σ2 > 0, λ(x | θ) = −1 2 log(2πσ2) − 1 2σ2 (x −θ)2 λ′(x | θ) = 1 σ2 (x −θ) λ′′(x | θ) = −1 σ2 I(θ) = Eθ (X −θ)2 σ4 = 1 σ2 , the “precision” (inverse variance). The same holds for a random sample In(θ) = n/σ2 of size n. Thus the variance (σ2/n) of ˆ θn = ¯ xn is exactly 1/In(θ).
Poisson: For the Po(θ), λ(x | θ) = x log θ −log x! −θ λ′(x | θ) = x −θ θ λ′′(x | θ) = −x θ2 I(θ) = Eθ (X −θ)2 θ2 = 1 θ, so again ˆ θn = ¯ xn has variance 1/In(θ) = θ/n.
2 Bernoulli: For the Bernoulli distribution Bi(1, θ), λ(x | θ) = x log θ + (1 −x) log(1 −θ) λ′(x | θ) = x θ −1 −x 1 −θ λ′′(x | θ) = −x θ2 − 1 −x (1 −θ)2 I(θ) = Eθ X(1 −θ)2 + (1 −X)θ2 θ2(1 −θ)2 = 1 θ(1 −θ), so again ˆ θn = ¯ xn has variance 1/In(θ) = θ(1 −θ)/n.
Exponential: For X ∼Ex(θ), λ(x | θ) = log θ −xθ λ′(x | θ) = 1 θ −x λ′′(x | θ) = −1 θ2 I(θ) = Eθ (1 −Xθ)2 θ2 = 1 θ2, so In(θ) = n/θ2. This time the variance of ˆ θ = 1/¯ xn, θ2n2 (n−1)2(n−2), is bigger than 1/In(θ) = θ2/n (you can compute this by noticing that ¯ xn ∼Ga(n, nθ) has a Gamma distribution, and computing arbitrary moments E[Y p] = β−pΓ(α + p)/Γ(α) for any Y ∼Ga(α, β), p > −α).
It turns out that this lower bound always holds.
1.2 The Information Inequality Let T(X) be any statistic with finite variance, and denote its mean by m(θ) = EθT(X).
By the triangle inequality, the square of the covariance of any two random variables is never more than the product of their variances (this is just another way of saying the correlation is bounded by ±1). We’re going to apply that idea to the two random variables T(X) and λ′(X | θ) (whose variance is Vθ[λ′(X | θ)] = I(θ)): Vθ[T(X)] · Vθ[λ′(X | θ] ≥ Eθ [T(X) −m(θ)] · [λ′(X | θ) −0] 2 = Z [T(x)] · [λ′(x | θ)] f(x | θ) dx 2 = Z T(x) f ′(x | θ) dx 2 = m′(θ) 2 , 3 so Vθ[T(X)] ≥[m′(θ)]2 I(θ) or, for samples of size n, Vθ[Tn(x)] ≥[m′(θ)]2 n I(θ) .
For vector parameters θ ∈Θ ⊂Rd the Fisher Information is a matrix I(θ) = Eθ[∇λ(x | θ) ∇λ(x | θ)⊺] = Eθ[−∇2λ(x | θ)] where ∇f(θ) denotes the gradient of a real-valued function f : Θ →R, a vector whose components are the partial derivatives ∂f(θ)/∂θi; where x⊺denotes the 1×d transpose of the d×1 column vector x; and where ∇2f(θ) denotes the Hessian, or d × d matrix of mixed second partial derivatives.
If T(X) was intended as an estimator for some function g(θ), the MSE is Eθ [Tn(x) −g(θ)]2 = Vθ[Tn(x)] + [m(θ) −g(θ)]2 ≥[m′(θ)]2 n I(θ) + [m(θ) −g(θ)]2 = [β′(θ) + g′(θ)]2 n I(θ) + β(θ)2 where β(θ) ≡m(θ) −g(θ) denotes the bias. In particular, for estimating g(θ) = θ itself, Eθ [Tn(x) −θ]2 ≥[β′(θ) + 1]2 n I(θ) + β(θ)2 and, for unbiased estimators, the MSE is always at least: Eθ [Tn(x) −θ]2 ≥ 1 n I(θ).
These lower bounds were long thought to be discovered independently in the 1940s by statisticians Harold Cr´ amer and Calyampudi Rao, and so this is often called the “Cr´ amer-Rao Lower Inequality,” but ever since Erich Lehmann brought to everybody’s attention their earlier discovery by Maurice Fr´ echet in the 1870s they became called the “Information Inequality.” We saw in examples that the bound is exactly met by the MLEs for the mean in normal and Poisson examples, but the inequality is strict for the MLE of the rate parameter in an exponential (or gamma) distribution.
It turns out there is a simple criterion for when the bound will be “sharp,” i.e., for when an estimator will exactly attain this lower bound. The bound arose from the inequality ρ2 ≤1 for the covariance ρ of T(X) and λ′(X | θ); this inequality will be an equality precisely when ρ = ±1, i.e., when T(X) can be written as an affine function T(X) = u(θ)λ′(X | θ) + v(θ) of the Score. The 4 coefficients u(θ) and v(θ) can depend on θ, but not on X, but any θ-dependence has to cancel so that T(X) won’t depend on θ (because it was a statistic). In the Normal and Poisson examples the statistic T was X, which could indeed be written as an affine function of λ′(X | θ) = (X −θ)/σ2 for the Normal or λ′(X | θ) = (X −θ)/θ for the Poisson, while in the Exponential case 1/X cannot be written as an affine function of λ′(X | θ) = (1/θ −X).
1.2.1 Multivariate Case A multivariate version of the Information Inequality exists as well. If Θ ⊂Rk for some k ∈N, and if T : X →Rn is an n-dimensional statistic for some n ∈N for data X ∼f(x | θ) taking values in a space X of arbitrary dimension, define the mean function m : Rk →Rn by m(θ) := EθT(X) and its n × k Jacobian matrix by Jij := ∂mi(θ)/∂θj.
Then the multivariate Information Inequality asserts that Covθ[T(X)] ≥J I(θ)−1J⊺ where I(θ) := Covθ[∇θ log f(X | θ)] is the Fisher information matrix, where the notation “A ≥B” for n × n matrices A, B means that [A −B] is positive semi-definite, and where C⊺denotes the k × n transpose of an n × k matrix C. This gives lower bounds on the variance of z′T(X) for all vectors z ∈Rn and, in particular, lower bounds for the variance of components Ti(X).
Examples Normal Mean & Variance If both the mean µ and precision τ = 1/σ2 are unknown for normal variates Xi iid ∼No(µ, 1/τ), the Fisher Information for θ = (µ, τ) is I(θ) = −E " ∂2 ∂µ2 ℓ ∂2 ∂µ∂τ ℓ ∂2 ∂τ∂µℓ ∂2 ∂τ 2 ℓ # = −E −τ (X −µ) (X −µ) −τ −2/2 = τ 0 0 τ −2/2 where ℓ(θ) = 1 2[log(τ/2π) −τ(X −µ)2] is the log likelihood for a single observation.
Multivariate Normal Mean If the mean vector µ ∈Rk is unknown but the covariance matrix Σij = E(Xi −µi)(Xj −µj) is known for a multivariate normal RV X ∼No(µ, Σ), the (k × k)-Fisher Information matrix for µ is Iij(µ) = −E ∂2 ∂µi∂µj −1 2 log 2πΣ −1 2(X −µ)′Σ−1(X −µ) = Σ−1 ij, so (as in the one-dimensional case) the Fisher Information is just the precision (now a matrix).
Gamma If both the shape α and rate λ are unknown for gamma variates Xi iid ∼Ga(α, λ), the Fisher Infor-mation for θ = (α, λ) is I(θ) = −E " ∂2 ∂µ2 ℓ ∂2 ∂µ∂τ ℓ ∂2 ∂τ∂µℓ ∂2 ∂τ 2ℓ # = −E −ψ′(α) λ−1 λ−1 −αλ−2 = ψ′(α) −λ−1 −λ−1 αλ−2 5 where ℓ(θ) = α log λ + (α −1) log X −λX −log Γ(α) is the log likelihood for a single observation, and where ψ′(z) := [log Γ(z)]′′ is the “trigamma function” (Abramowitz and Stegun, 1964, 6.4), the derivative of the “digamma function” ψ(z) := [log Γ(z)]′ (Abramowitz and Stegun, 1964, 6.3).
1.3 Regularity The Information Inequality requires some regularity conditions for it to apply: • The Fisher Information exists— equivalently, the log likelihood ℓ(θ) := log f(x | θ) has partial derivatives with respect to θ for every x ∈X and θ ∈Θ, and they are in L2 X, f(x | θ) for every θ ∈Θ.
• Integration (wrt x) and differentiation (wrt θ) commute in the expression ∇θ Z T T(x) f(x | θ) dx = Z T T(x) ∇θf(x | θ) dx.
This latter condition will hold whenever the support {x : f(x | θ) > 0} doesn’t depend on θ, and log f(x | θ) has two continuous derivatives wrt θ everywhere.
These conditions both fail for the case of {Xi} iid ∼Un(0, θ), the uniform distribution on an interval [0, θ], and so does the conclusion of the Information Inequality.
In this case the MLE is ˆ θn = X∗ n(X) := max{Xi : 1 ≤i ≤n}, the sample maximum, whose mean squared error Eθ|ˆ θn −θ|2 = 2θ2 (n + 1)(n + 2) tends to zero at rate n−2 as n →∞, while the Information Inequality bounds that rate below by a multiple of n−1 for problems satisfying the Regularity Conditions.
2 Efficiency An estimator δ(X) of g(θ) is called efficient if it satisfies the Information Inequality exactly; oth-erwise its (absolute) efficiency is defined to be Eff(δ) = [β′(θ)+g′(θ)]2 I(θ) + β(θ)2 Eθ {[δ(X) −g(θ)]2} or, if the bias β(θ) ≡[Eθδ(X) −g(θ)] vanishes, = [g′(θ)]2 I(θ)Vθ[δ(X)].
It is asymptotically efficient if the efficiency for a sample of size n converges to one as n →∞. This happens for the MLE ˆ θn = 1/¯ xn for an Exponential distribution above, for example, whose bias is βn(θ) = E[ 1 ¯ xn −θ] = θ n−1 and whose absolute efficiency is Eff(ˆ θn) = [1/(n−1)+1]2 n/θ2 + θ2 (n−1)2 θ2n2 (n−1)2(n−2) + θ2 (n−1)2 = (n + 1)(n −2) (n + 2)(n −1).
This increases to one as n →∞, so ¯ θn is asymptotically efficient.
6 3 Asymptotic Relative Efficiency: ARE If one estimator δ1 of a quantity g(θ) has MSE Eθ|δ1(x) −g(θ)|2 ≈c1/n for large n, while another δ2 has MSE Eθ|δ2(x) −g(θ)|2 ≈c2/n with c1 < c2, then the first will need a smaller sample-size n1 = (c1/c2)n2 to achieve the same MSE as the second would achieve with a sample of size n2.
The ratio (c2/c1) is called the asymptotic relative efficiency (or ARE) of δ1 wrt δ2. For example, if c2 = 2c1, then δ1 needs c1/c2 = 0.5 times the sample size, and is (c2/c1) = 2 times more efficient than δ2.
We won’t go into it much in this course, but it’s interesting to know that the for estimating the mean θ of the normal distribution No(θ, σ2) the sample mean ˆ θ = ¯ xn achieves the Information Inequality lower bound of Eθ|¯ xn −θ|2 = σ2/n = 1/In(θ), while the sample median M(x) has approximate1 MSE Eθ|M(x−θ σ ) −θ|2 ≈πσ2/2n, so the ARE of the median to the mean is σ2/n πσ2/2(n + 2) →2 π ≈0.6366, so the sample median will require a sample-size about π/2 ≈1.57 times larger than the sample mean would for the same MSE. The median does offer a significant advantage over the sample mean, however— it is robust against model misspecification, e.g., it is relatively unaffected by a few percent (or up to half) of errors or contamination in the data. Ask me if you’d like to know more about that.
4 Change of Variables for Fisher Information The Fisher information function θ ⇝I(θ) depends on the particular way the model is param-eterized.
For example, the Bernoulli distribution can be parametrized in terms of the success probability p, the logistic η = log( p 1−p), or in angular form with θ = arcsin(√p).
The Fisher Information functions for these various choices will differ, but in a very specific way.
Consider a statistical model that can be parametrized in either of two ways, X ∼f(x | θ) = g(x | η), with θ = φ(η), η = ψ(θ) for one-dimensional parameter vectors θ and ψ, related by invertible differentiable 1:1 transforma-tions. Then the Fisher Information functions Iθ and Iη in these two parameterizations are related 1You can show this by considering Φ M( x−θ σ ) , which is the median of n iid uniform random variables and so has exactly a Be(α, β) distribution with α = β = n+1 2 distribution for odd n, hence variance 1/4(n + 2). Then use the “Delta method” to relate the variance of M( x−θ σ ) to that of Φ(M( x−θ σ )), V[Φ(M( x−θ σ ))] ≈[Φ′(0)]2V[M( x−θ σ )] = 1 2π V[M( x−θ σ )], so 1 4(n+2) ≈ 1 2π V[M( x−θ σ )] = 1 2πσ2 V[M(x)], or V[M(x)] ≈ πσ2 2(n+2).
7 by Iθ(θ) = E ( ∂ ∂θ log f(x | θ) 2) = E ( ∂ ∂θ log g(x | ψ(θ)) 2) = E ( ∂ ∂η log g(x | η)∂η ∂θ 2) = E ( ∂ ∂η log g(x | η) 2) ∂η ∂θ 2 = Iη(η) Jη θ 2, (4) the Fisher information Iη(η) in the η paramaterization times the square of the Jacobian Jη θ := ∂η/∂θ for changing variables.
For example, the Fisher Information for Bernoulli random variable in the usual success-probability parametrization was shown in Section (1.1) to be Ip(p) = 1/p(1−p). In the logistic parametrization it would be Iη(η) = eη (1 + eη)2 = Ip eη 1 + eη (Jp η)2, for Jacobian Jp η = ∂p/∂η = eη/(1 + eη)2 and inverse transformation p = eη/(1 + eη), while in the arcsin parameterization the Fisher Information Iθ(θ) = Ip sin2(θ) (Jp θ )2 = 4, a constant, for Jacobian Jp θ = ∂p/∂θ = 2 sin θ cos θ and inverse transformation p = sin2 θ.
4.1 Jeffreys’ Rule Prior From (4) it follows that the unnormalized prior distributions πθ J(θ) = Iθ(θ) 1/2 and πη J(η) = Iη(η) 1/2 are related by πθ J(θ) = πη J(η) |∂η/∂θ|, exactly the way they should be related for the change of variables η ⇝θ.
It was this invariance property that led Harold Jeffreys (1946) to propose πJ(θ) as a default “ob-jective” choice of prior distribution. Much later Jos´ e Bernardo (1979) showed that this is also the “Reference prior” that maximizes the entropic distance from the prior to the posterior, i.e., the information to be learned from an experiment; see (Berger et al., 2009, 2015) for a more recent and broader view. In estimation problems with one-dimensional parameters a Bayesian analysis using πJ(θ) ∝I(θ)1/2 is widely regarded as a suitable objective Bayesian approach.
8 4.2 Examples using Jeffreys’ Rule Prior Normal: For the No(θ, σ2) distribution with known σ2, the Fisher Information is I(θ) = 1/σ2, a constant (for known σ2), so πJ(θ) ∝1 is the improper uniform distribution on R and the posterior distribution for a sample of size n is πJ(θ | x) ∼No( ¯ Xn, σ2/n) with posterior mean ˆ θJ = ¯ Xn, the same as the MLE.
Poisson: For the Po(θ), the Fisher Information is I(θ) = 1/θ, so πJ(θ) ∝θ−1/2 is the improper conjugate Ga(1/2, 0) distribution and the posterior for a sample x of size n is πJ(θ | x) ∼Ga 1 2 + X Xi, n , with posterior mean ˆ θJ = ¯ Xn + 1/2n, asymptotically the same as the MLE.
Bernoulli: For the Bi(1, θ), the Fisher Information is I(θ) = 1/θ(1−θ), so πJ(θ) ∝θ−1/2(1−θ)−1/2 is the conjugate Be(1/2, 1/2) distribution and the posterior for a sample x of size n is πJ(θ | x) ∼Ga 1 2 + X Xi, 1 2 + n − X Xi , with posterior mean ˆ θJ = (P Xi + 1 2)/(n + 1). In the logistic parametrization the Jeffreys’ Rule prior is πJ(η) ∝1/(eη/2 + e−η/2) ∝sech(η/2), the hyperbolic secant, while the Jeffreys’ Rule prior is uniform πJ(θ) ∝1 in the arcsin parametrization, but each of these induces the Be(1 2, 1 2) for p under a change of variables.
Exponential: For the Ex(θ), the Fisher Information is I(θ) = 1/θ2, so the Jeffreys’ Rule prior is the scale-invariant improper πJ(θ) ∝1/θ on R+, with posterior density for a sample x of size n is πJ(θ | x) ∼Ga n, X Xi , with posterior mean ¯ θJ = 1/ ¯ Xn equal to the MLE.
4.3 Multidimensional Parameters When θ and η are d-dimensional a similar change-of-variables expression holds, but now Iθ, Iη and the Jacobian Jη θ are all d×d matrices: Iθ ij(θ) = E ∂ ∂θi log f(x | θ) ∂ ∂θj log f(x | θ) = E ∂ ∂θi log g(x | ψ(θ)) ∂ ∂θj log g(x | ψ(θ)) = E ( X k ∂ ∂ηk log g(x | η)∂ηk ∂θi ! X l ∂ ∂ηl log g(x | η) ∂ηl ∂θj !) Iθ(θ) = Jη θ ⊺Iη(η)Jη θ , 9 Jeffreys noted that now the prior distribution πJ(θ) ∝| det I(θ)|1/2 proportional to the square root of the determinant of the Jacobian matrix is invariant under changes of variables, but both he and later authors noted that πJ(θ) has some undesirable features in d ≥2 dimensions including the important example of No(µ, σ2) with unknown mean and variance. “Ref-erence priors” (Berger and Bernardo, 1992a,b) are currently considered the best choice for objective Bayesian analysis in multi-parameter problems, but they are challenging to compute and to work with. The tensor product of independent one-dimensional Jeffreys priors πJ(θ1)πJ(θ2) · · · πJ(θd) are frequently recommended as an acceptable alternative.
References Abramowitz, M. and Stegun, I. A., eds. (1964), Handbook of Mathematical Functions With For-mulas, Graphs, and Mathematical Tables, Applied Mathematics Series, volume 55, Washing-ton, D.C.: National Bureau of Standards, reprinted in paperback by Dover (1974); on-line at
Berger, J. O. and Bernardo, J. M. (1992a), “On the development of the Referrence Prior method,” in Bayesian Statistics 4, eds. J. M. Bernardo, J. O. Berger, A. P. Dawid, and A. F. M. Smith, Oxford, UK: Oxford University Press, pp. 35–49.
Berger, J. O. and Bernardo, J. M. (1992b), “Ordered group reference priors with application to a multinomial problem,” Biometrika, 79, 25–37.
Berger, J. O., Bernardo, J. M., and Sun, D. (2009), “The Formal Definition of Reference Priors,” Annals of Statistics, 37, 905–938.
Berger, J. O., Bernardo, J. M., and Sun, D. (2015), “Overall Objective Priors,” Bayesian Analysis, 10, 189–221, doi:10.1214/14-BA915.
Bernardo, J. M. (1979), “Reference posterior distributions for Bayesian inference (with discussion),” Journal of the Royal Statistical Society, Ser. B: Statistical Methodology, 41, 113–147.
Jeffreys, H. (1946), “An Invariant Form for the Prior Probability in Estimation Problems,” Pro-ceedings of the Royal Society of London. Series A, Mathematical and Physical Sciences, 186, 453–461, doi:10.1098/rspa.1946.0056.
Last edited: May 11, 2021 10 |
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