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11900 | https://cloud.tencent.com/developer/article/1884842 | 在你面前有一个n阶的楼梯，你一步只能上1阶或2阶-腾讯云开发者社区-腾讯云
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社区首页>专栏>在你面前有一个n阶的楼梯，你一步只能上1阶或2阶
在你面前有一个n阶的楼梯，你一步只能上1阶或2阶
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修改于 2021-10-08 06:20:11
修改于 2021-10-08 06:20:11
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文章被收录于专栏：centosDaicentosDai
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如何计算上n阶楼梯的不同走法数量？
楼梯问题可以用动态规划来解决吗？
上楼梯的最优解是什么？
思路解析 ①台阶只有一级阶梯时，只有一种走法。 ②当台阶有两级时，可以先走一步走两次，或者直接走两步。两种走法 ③当有三级台阶时，可以走一步走三次。可以先走一步再两步。也可以先两步再一步。三种方法。 小结 可以看出台阶有三级时，可以走的方式等于一级加二级走的方式的总和，即f(3)=f(1)+f(2),符合f(n)=f(n-2)+f(n-1),这样下来，正好符合斐波那契数列。
代码实现
代码语言：javascript
代码 运行次数：0
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```javascript
public class TestFloor {
public static void main(String[] args) {
System.out.println("当N=11时："+ladder(11) +" "+"当N=9时："+ ladder(9));
}
private static int ladder(int n) {
// TODO Auto-generated method stub
if(n==1) {
return 1;
}else if(n==2) {
return 2;
}else {
return ladder(n-1) + ladder(n-2);
}
}
}
```
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11901 | https://uollb.com/blogs/uol/anomie-durkheim-and-merton-on-crime-as-social-phenomenon | Skip to content
Anomie, Durkheim, and Merton on Crime as Social Phenomenon
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Crime is not only a result of individual actions but also a social phenomenon deeply influenced by societal factors. Anomie theory, developed by Emile Durkheim and further expanded upon by Robert Merton, provides valuable insights into the social roots of crime and deviant behaviour.
Anomie theory: Anomie refers to a state of normlessness or a breakdown in social norms and values. Durkheim proposed that when individuals experience a sense of disconnection from societal norms, such as during times of rapid social change or economic instability, they become more prone to engage in deviant and criminal behaviour. Anomie theory highlights the significance of social integration and moral regulation in preventing crime.
Emile Durkheim: Durkheim, a prominent sociologist, emphasised the role of social factors in understanding crime. He argued that crime is not inherently pathological but rather a normal and necessary component of society. According to Durkheim, crime serves important social functions, such as reaffirming shared values and boundaries, promoting social solidarity, and prompting societal adaptation and change.
Robert Merton: Building upon Durkheim's ideas, Merton developed the strain theory, which examines the relationship between societal goals and the means available for individuals to achieve those goals. Merton suggested that when there is a disjunction between socially prescribed goals (such as financial success) and the legitimate means to attain them (such as education and employment opportunities), individuals may experience strain. This strain can lead to various adaptations, including conformity, innovation (criminal behaviour), ritualism, retreatism, or rebellion.
In summary, crime as a social phenomenon is explored through the lens of anomie theory, which emphasises the impact of societal factors on crime and deviance. Durkheim's work highlights the importance of social integration and moral regulation, while Merton's strain theory focuses on the disconnection between goals and means as a source of strain that contributes to criminal behaviour. Understanding crime from a social perspective allows us to analyse the broader societal structures and conditions that shape individuals' choices and actions.
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11902 | https://medium.com/the-judy-flander-interview/cnn-chief-anchor-bernard-shaw-is-accidental-war-correspondent-during-a-night-in-bagdad-625e9daea790 | CNN Chief Anchor Bernard Shaw is Accidental War Correspondent During a Night in Bagdad | by Judy Flander | The Judy Flander Interviews | Medium
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Judy Flander is an entertainment feature writer and television critic who for many years during the 70’s, 80’s and 90’s wrote insightful interviews of many well known people, and some not so well known then, were published in newspapers and magazines across the US.
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CNN Chief Anchor Bernard Shaw is Accidental War Correspondent During a Night in Bagdad
Judy Flander
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May 3, 2020
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Yet never became a network anchor like NBC’s Tom Brokaw, ABC’s Peter Jennings and CBS’ Dan Rather
USA TODAY Weekend, March 1, 1991: In a restaurant bar near the U.S. Capitol, a morose Bernard Shaw scowls into his vodka tonic. “You cannot be normal and be an anchor,” he says in a tight monotone. “I know I’m not normal. Anybody who would submit to the extreme pressures daily cannot be normal.”
It’s late in the evening, and Cable News Network’s chief anchor is going through with a promised interview, even though he’d rather go home to his wife, children and the salmon fillet he’ll broil for himself long after the family’s dinner hour. His work on TV is “one hell of a challenge, to have the mental energy and physical energy to do that day in and day out.”
And that’s how Shaw was feeling months before his bombs-bursting-in-air vigil in Baghdad. While waiting for an interview with Iraqi President Saddam Hussein, the anchor inadvertently became a war correspondent. His live broadcast, with CNN reporters Peter Arnett and John Holliman, gripped at least 10.8 million U.S. households (up from the usual 840,000 prime-time households). That day, the reporters got almost as much media coverage as the war. And later, Shaw was deluged by fan mail, requests to speak at university commencements, even offers for TV movies.
A war correspondent is the last thing Shaw wants to be. His wife, Linda, their daughter, Anil, 15, and son, Amar, 14, are “eminently more important” and “they are clearly still feeling the effect of this experience,” he said, 10 days after returning from Baghdad. His family knew he was alive only by hearing his voice — over the gunfire — on TV.
Shaw, who says he would not return to Baghdad if CNN asked him, feels he came home a changed man. Changed, he says, in ways he is still not sure of. “I stared death in the face, and death blinked. I don’t expect to get too many more chances like that.”
Typical Shaw. Through three interviews and several phone conversations last year and this, he ruminated, sometimes gloomily, about his steep expectations for himself.Shaw’s intensity sometimes amuses friends. “I tell him, ‘Lighten up, Bernie,’” laughs Peter Jennings, his ABC counterpart. A lot of people at one time or another have said that to Shaw-including President George W. Bush.
It’s just not in his nature. An angst-filled Shaw often is awake at 3 or 4 in the morning musing about life and death. “I have this tremendous opportunity with my wife and my son and daughter. I have this tremendous opportunity with my profession, journalism. I have this tremendous opportunity with the problems of our country and our world,” he says, with some anguish. “What am I doing about it? What am I doing to help bring about clarity and understanding? It bothers me when I don’t squeeze life as tightly as possible.”
The Shaws live in Takoma Park, Md., a Washington suburb, in a 100 year-old stucco house they’ve been remodeling for a decade. “We love it and all its problems,” laughs Shaw. His favorite neighborhood spot is a jazz club where he sometimes stops on the way home from work and where, on the war’s first night, other regulars cheered their buddy while watching him on the bar’s TV.
Despite his solemn demeanor, Shaw, 50, can be warm and charming. He smiles as he says, “I love sending roses.” — to friends, to his wife, once even to Pat Nixon after attending a small dinner party at the Nixons’ house. Richard Nixon sent a hand-written thank-you note for his wife’s flowers.
The amiable part of Shaw’s personality emerged at lunch at a restaurant close to the CNN Washington bureau, near the Capitol. Relaxed and reflective, he enjoyed a leisurely meal instead of one gulped at his desk, his usual routine on a workday.
If Shaw’s on-air personality is sometimes described as humorless, even severe, it may be because he doesn’t see anything particularly cheerful about the news. “I’m dealing with serious matters. Television is a cool image. I am not an excitable or an exciting person.”
Part of his restraint stems from his acute awareness that he is a black role model, and he wants to set a straight-arrow example. “The responsibility is so weighty it keeps me up at night.”
One would think Shaw succeeded in blurring the color line long ago. “It doesn’t occur to me that Bernie is black,” Jennings comments, an observation repeated by others interviewed for this story.
But a friend of Shaw’s, independent TV producer Kenneth Walker, who also is a black man, says race is ignored only as long as there’s no trouble. If Shaw “makes one mistake, he becomes a black person with an attitude problem. God knows, if he let the network go black for 6 minutes” (as did a peeved Dan Rather, who once let CBS transmissions stop after a sports event ate into his air time) “he’d be outta there.”
Shaw is somewhat aloof about the issue and more concerned about measuring up to his own expectations. “I refuse to let my nation’s fixation with race and color deter me from fulfilling myself.” Still, he’s pleased about his contribution in a “pivotal industry,” “I have a sense of pride no different from that of boxer Joe Louis when he was in the ring, or Paul Robeson acting or Ralph Bunche practicing diplomacy. I am standing on the shoulders of some excellent giants: Bunche, Robeson, Louis, Walter Cronkite, Edward R. Murrow. And their colors interest me not.”
He also is standing on the shoulders of his father, Edgar, a house painter, “a voracious newspaper reader and a man of few words.” And like his mother, Camilla, a strong parent. “Dad was stricter than my drill instructor in the Marine Corps.” He sums up his father’s influence on his life in two word — — discipline and teamwork.
Shaw grew up the youngest of four children on Chicago’s South Side. One brother became a chef, the other, a cabdriver, his sister, a home maker. Three of Shaw’s best friends date back to elementary school. One of them, Eugene Woodard, who is now president of Hudson Life Reassurance in Connecticut, says the boys all had nicknames. Woodard’s was Beans, and “we called Bernie ‘Rhino,’ because when we teased him during recess, he’d put his head down and charge into us.”
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The nickname didn’t stick. But Shaw’s charge-ahead nature didn’t escape Thelma Ford, his English and public-speaking teacher at Dunbar High School. Now 83, she reluctantly reveals she once had to “send for” Shaw’s mother. “Bernard talked too much in class,” she explains. After high school, he lacked money to go straight to college, so he spent 4 years in the Marines before entering the University of Illinois.
In college, Shaw picked the brain of John Mackin, an English literature professor and authority on classical rhetoric. He found Shaw fascinated with “Plato’s picture of Socrates confronting society.” Shaw’s career profile is very much like those of Jennings, Rather and NBC’s Tom Brokaw. Shaw put in years at local radio and TV stations in the Chicago area before being transferred by Westinghouse’s Group as a Washington correspondent.
In 1971, Walter Cronkite brought Shaw to the attention of Richard Salant, then president of CBS News. Cronkite says he used no influence on Shaw’s behalf — “I just opened the door.” As a CBS Washington correspondent, Shaw sometimes covered the White House, in the same booth as Rather and opposite Brokaw’s.
By 1977, Shaw had moved on to ABC, where, fluent in Spanish, he covered Latin America and later became senior Capitol Hill correspondent. He went to the newly created CNN in 1980 for one reason: “I always wanted to be an anchor.” He says he never would have achieved that dream at any other network. Shaw has been on a journalistic par with Rather, Jennings and Brokaw for a long time, even though he lacks their glamour, their entourages and their million-dollar-plus salaries.
He reportedly doubled his salary, to $500,000, in his last go-around with parsimonious CNN, although he still doesn’t have simple perks like a secretary or a clerk. In an office so cluttered he uses the floor to stack the overflow, he keeps “everything important” in a pile on the middle of his desk. “If I really need something, I know it’s in there someplace.”
Perhaps Shaw’s most humbling experience so far working for CNN was in the fall of 1989 when the network launched a new prime-time weeknight hour, The World Today. He says he was led to believe he’d be sole anchor. It came as a shock to him — “my opinion was not asked or sought” — when CNN hired Catherine Crier, a stunning blond judge from Texas, to be his co-anchor. That Crier was not a journalist devastated Shaw even more. The news wounded his pride, but Shaw kept his cool and his counsel and today says he is on the best of terms with Crier.
Shaw, one of 27 co-anchors at CNN, had every reason to expect to be given his own news hour. He’d brought prestige to the network and just that year he’d been showered with awards for his 30 hours of informed, calm anchoring of the student uprising in Beijing. Brokaw, watching the continuous CNN satellite feeds from China, observed his friend ignoring his own needs and “keeping the troops rallied” during commercial breaks.
Cronkite, Shaw’s idol and mentor, described him in words often used to describe himself. “He sounds authoritative and believable.”Shaw is often compared to Cronkite. “He has the same sense of calmness and confidence; he conveys the same respect for his audience,” says Abe Rosenthal, New York Times columnist and the paper’s former editor.
CNN vice president Ed Turner calls Shaw “today’s Walter Cronkite. Bernie is so unflappable; he’s never flustered. He doesn’t show anger or the frustration or all the grief that goes with it.” Clearly, CNN has gotten the message: Shaw is a hot property. For the first time, CNN gave an anchor his own show, Gulf Talk, a half-hour show in which Shaw and guests discuss the progress of the war.
As of this printing, Shaw also appears on a War in the Gulf news hour, with co-anchor Crier. The later times are significant, Shaw says, because the live broadcasts catch more viewers nationwide, especially in the West. Viewers seem to be responding. Shaw is getting “mega-mail” now, from people like Tina Stearns of Chicago, who preferred watching №1 rated Jennings until that night in Baghdad. “Peter has officially moved to the back seat in my heart,” she wrote Shaw recently, “replaced by you!”
Shaw’s contract expires Inauguration Day 1993. What will be available to him then is anybody’s guess. “There’s nothing ahead of Bernie but green lights and the open road,” Rather says. “He could have anything he wants at the networks. He loves what he is doing. He’s comfortable at it and has a great sense of accomplishment. I think if and when he chooses to move, he can write his own ticket, name his own price.”
But in a decade with CNN, Shaw hasn’t had a nibble from the networks.”We’ve talked about it,” Brokaw revealed last year, but there isn’t a comparable job for him here. And he has an awfully good job where he is.”
Some suggest Shaw easily could replace Ted Koppel on ABC’s Nightline if the spot opens. Before Baghdad, ABC News president Roone Arledge hedged. “I’d put him on anything. But it is hard to find a place for Bernie that would be more fun than what he is doing now.”
Bernie Shaw, a very cool customer, says flatly, “I’m not going anywhere. “I tell Bernie it’s too bad we didn’t know each other in high school, we’d have been good friends,” says NBC’s Tom Brokaw, who covered the White House with Shaw. “He takes his work seriously, his life and his self less so.
“He’s been a comfortable friend for more than a decade,” says Dan Rather, who competed with Shaw at CBS’ Washington bureau during the 1970s. “He is a comfortable human being” and “an excellent storyteller.” “He is steady in a crisis,” says Peter Jennings, who covered the Iran hostage crisis with Shaw at ABC. “It was a mistake for us to let him go.”
Of the Big Three anchors. Bernard Shaw says: “I talk to them all the time, socially, professionally. We always peel off into some corner when we’re together on a story. I love duking it out with Tom and Dan and Peter.”
Don Kirk, who covers the Gulf War for USA TODAY, recounts his brushes with Bernard Shaw in the momentous first hours of the war. At 4 a.m. Jan. 17, Baghdad time, Bernie Shaw appears relaxed and at ease, the picture of cool efficiency, when encountered in the predawn observing and reporting on the war from one of the CNN rooms on the ninth floor of Baghdad’s Al-Rashid Hotel. He has been on the air since 2:35 a.m., when Iraqi gunners on nearby rooftops began firing anti-aircraft artillery as U.S. missiles and bombers approached the capital. When the live broadcast ends at 4:30 a.m. or so, Shaw breaks for a beer. No sign of fatigue; he is high from having a ringside view of war.
Overnight, Jan. 17–18, Shaw sleeps face down on the floor of the hotel bomb shelter. The CNN team, other reporters and the families of hotel staff had rushed in after a report that allied weapons were being aimed at the hotel. The report turns out to be false. Shaw is as cool off-air as he is on-air chatting with an Iraqi Information Ministry official.
Jan. 18. The Iraqi official makes a point of bidding a downright fond farewell to Shaw, who is waiting on the hotel’s front steps to hop in a car for the 600-mile ride to relative safety in Jordan. CNN reporter Peter Arnett, staying in Baghdad, tells Shaw the war’s first night was “the most thrilling experience of my career.”
“Yes,” Shaw agrees, in a classic understatement. “It certainly was.”
Sizing Up ShawWhen the bombs began falling in Baghdad, Bernard Shaw’s star rose. Wha’s behind the CNN anchor’s poker face? Worry about whether he’s making the most of his opportunities.
By Judy Flander
Photographed by Brian Smale
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Judy Flander is an entertainment feature writer and television critic who for many years during the 70’s, 80’s and 90’s wrote insightful interviews of many well known people, and some not so well known then, were published in newspapers and magazines across the US.
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Antibiotics, Antimicrobial Resistance, Antibiotic Susceptibility Testing, and Therapeutic Drug Monitoring for Selected Drugs
Antimycobacterial Agents
Antimycobacterial, or antituberculosis, agents represent a diverse group of structurally unrelated compounds that include rifampin, isoniazid, ethambutol, streptomycin, amikacin, kanamycin, and pyrazinamide. Other antibiotics may be added to the regimen if needed, for example, a fluoroquinolone. Each of these agents has very different and unique biological mechanisms of action; yet, when used alone or in combination, these drugs have historically proven to be clinically effective in the acute treatment and long-term public health management of Mycobacterium infections such as tuberculosis and leprosy. Antimycobacterial agents are most commonly prescribed today in multidrug combinations due to the frequent emergence of resistance, which is often seen as a consequence of poor patient compliance and incomplete treatment of active infections during the long course required with most therapeutic regimens.
Rifampin inhibits DNA-dependent RNA polymerase in bacterial cells by binding to its beta-subunit and, thus, preventing RNA transcription and subsequent translation to proteins. In contrast, isoniazid and ethambutol separately exert their bacteriostatic effects through different molecular mechanisms by interfering with the synthesis of the mycolic acid–peptidoglycan complex in the cell wall to increase overall cell permeability. Streptomycin, amikacin, and kanamycin are aminoglycoside antibiotics that interfere with protein synthesis and alter cell membrane transport, with a similar net effect of increasing overall cell permeability as well.
The most serious adverse and toxic side effects have been observed with rifampin and isoniazid, most notably, hepatitis and jaundice (with liver failure in severe cases) and sideroblastic anemia. Other milder side effects include flushing, pruritus, rash, redness, and watering of eyes, as well as gastrointestinal and CNS disturbances and general flu-like symptoms. As observed with most other bacteriostatic agents, toxicity with these drugs is independent of drug concentration. Serious drug toxicity was reported in one 37-year-old woman who received ethambutol 825 mg, isoniazid 225 mg, rifampicin 450 mg, and pyrazinamide 1200 mg and had serious impairment in her vision. The authors commented that ethambutol and isoniazid are notorious for causing visual impairment .
Pyrazinamide is an analog of nicotinamide which is used as an antituberculous agent (bacteriostatic or bactericidal) most often used in combination with other antituberculosis agents for the initial treatment of active tuberculosis in children and adults. Pyrazinamide (500 mg tablet) is administered orally because it is well absorbed from the gastrointestinal tract. The half-life is 9–10 hours and approximately 70% of this drug is excreted in urine unchanged within 24 hours. Pyrazinamide inhibits excretion of uric acid by the kidney and may elevate serum uric acid concentration. This may not be a problem of most patients but pyrazinamide should be used with caution in patients with gout. Therapeutic drug monitoring of pyrazinamide is not recommended.
Although antimycobacterial agents are generally not subjected to therapeutic drug monitoring, treatment failure may result from sub-therapeutic drug concentrations. Peloquin commented that therapeutic drug monitoring may not be needed for otherwise healthy individuals who are responding to standard four drug treatment regimens for tuberculosis, but therapeutic drug monitoring is useful in patients who are slow to respond to therapy or at a higher risk of drug–drug interactions. Early intervention based on therapeutic drug monitoring is useful for these patients to avoid development of drug resistance. When only one specimen is obtained for therapeutic drug monitoring 2 hours postdosage specimen is appropriate for isoniazid, rifampin, ethambutol, and pyrazinamide . Babalik et al. reported that 60% of patients receiving antituberculosis drugs in their study showed below therapeutic drug concentration. The authors commented that drug levels were below therapeutic ranges in patients with active tuberculosis particularly in patients with HIV infection or other comorbidities .
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Book2017, Microbiology and Molecular Diagnosis in PathologyAudrey Wanger, ... Amitava Dasgupta
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Principles of Anti-infective Therapy and Surgical Prophylaxis
2017, Infectious Diseases (Fourth Edition)Evelina Tacconelli, ... Mathias W. Pletz
Bactericidal Versus Bacteriostatic Activity
Antibacterial drugs act by either killing or inhibiting the growth of micro-organisms. These bactericidal or bacteriostatic effects are determined by the mode of action of a drug against a particular target organism. Table 136-5 identifies the usual mode of action of commonly prescribed agents.
Cell wall active agents, such as the penicillins, cephalosporins and glycopeptides, are largely bactericidal, whereas most agents that interfere with protein synthesis are bacteriostatic. Examples of the latter include the tetracyclines, lincosamides and macrolides. Metabolic inhibitors of folic acid synthesis, namely the sulfonamides and trimethoprim, are also bacteriostatic but the combination of both (co-trimoxazole) is bactericidal. Cidality can be affected by drug concentration and in vitro conditions of testing and may also vary by target pathogen. Nonetheless, it is a useful distinguishing feature with clinical implications in selected circumstances.
In general, bactericidal and bacteriostatic agents are equally effective in the management of many mild to moderate infections. However, bacteriostatic agents are dependent upon an effective host immune response, notably the ability to phagocytose drug-exposed bacteria. There are a number of clinical conditions in which host immunity is either deficient or suppressed and where bacteriostatic agents should be avoided in favor of bactericidal agents. Other circumstances where bactericidal agents are recommended are in the management of infective endocarditis.10 The infected vegetations are impermeable to phagocytic cells and require high plasma concentrations of bactericidal drugs to sterilize them.
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Antimicrobials
2017, Nanostructures for Antimicrobial TherapyBhaskar Das, Sanjukta Patra
3.2 Classification on the Basis of Nature of Effect
Based on the nature of effect on bacteria, antibiotics are classified as bactericidal (which kill the target bacteria) and bacteriostatic (which inhibit bacterial growth and replication), e.g., antibiotics such as aminoglycosides, cephalosporins, penicillins, and quinolones are bactericidal, while tetracyclines, sulfonamides, and macrolides exert bacteriostatic effects on target bacteria. The effect of bactericidal agents is faster as compared to bacteriostatic agents. Bacteriostatic agents require an effective immune system in the host for elimination of pathogenic bacteria and hence are not applicable to immunosuppressed host conditions or acute infections. However, some antibiotics may behave as both bacteriostatic and bactericidal based on dosage concentration and duration of exposure, e.g., aminoglycosides, fluoroquinolones, and metronidazole exert concentration-dependent bactericidal characteristics (www.amrls.cvm.msu.edu).
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Book2017, Nanostructures for Antimicrobial TherapyBhaskar Das, Sanjukta Patra
Review article
Oxazolidinones: a new class of antibacterials
2001, Current Opinion in PharmacologyGary E Zurenko, ... W.Gary Tarpley
Successful treatment with an antimicrobial agent is dependent not only on the interactions between the drug and the bacterium, but also on interactions between the bacterium, the drug and the host defense system. A successful pathogen must be able to survive in the host environment, attach and multiply on host tissue, resist host defense mechanisms and produce virulence factors such as bacterial toxins. Disease does not occur if one or more of these is lacking. In vitro studies using conventional time-kill kinetic methods define a bactericidal effect as a reduction of greater than 3 log10 CFU/ml in the bacterial inoculum during a standard incubation period. Studies with linezolid have defined it as a bacteriostatic agent for enterococci and staphylococci and a bactericidal agent for the streptococci . However, in vivo studies with linezolid demonstrated efficacy in animal models of severe infection with enterococci and staphylococci—questioning the presumed bacteriostatic nature of the drug against these organisms. For example, linezolid was efficacious in a neutropenic mouse model of bacteremia caused by vancomycin-resistant Enterococcus faecium and was also effective in the treatment of experimental staphylococcal endocarditis in the rabbit [16•]. Because in vitro bactericidal activity has been considered a requirement for agents that can successfully treat infections in neutropenic hosts or in disease states such as endocarditis, clearly there must be other factors that contribute to the in vivo efficacy of linezolid . Moreover, clinical reports of linezolid therapy have emerged that describe the successful treatment of ‘serious infections’ (persistent vertebral osteomyelitis and bacteremia)resulting from MRSA and vancomycin-resistant Enterococcus faecium, respectively [20,21]. McNeil et al. found linezolid therapy to be successful despite the failure of approximately 10 weeks of previous therapy with quinupristin and dalfopristin. Some antibiotics, clindamycin in particular, have been show to have dramatic effects on virulence-factor expression . The apparent bactericidal nature of linezolid in animal models (and subsequently in human infections) might be explained by inhibition of Gram-positive cocci virulence-factor expression and stimulation of neutrophil phagocytosis . Concentrations of linezolid ranging from half to one-eighth the MIC value dramatically inhibited production of the virulence factors α-hemolysin, δ-hemolysin and coagulase in S. aureus (Table 2) and similarly inhibited virulence factor expression in S. pyogenes (data not shown). In addition, a significant increase in ingestion by neutrophils was demonstrated. When extrapolated to the in vivo situation, the inhibition of bacterial-toxin and virulence-factor production may halt progression of the disease state independently of the viability of the bacteria.
Table 2. Effect of linezolid at subinhibitory concentrations on virulence factor expression in Staphylococcus aureus
| | | | | |
--- ---
| Empty Cell | Yield (as titer) in presence or absence of linezolid | | | |
| Toxin/enzyme expression | No drug | MIC | MIC | 1/8 MIC |
| α-hemolysin† | 512 | <2 | 2 | 2 |
| δ-hemolysin† | 64 | 4 | 8 | 16 |
| Coagulase‡ | 64 | 4 | 8 | 16 |
: Linezolid given at varying doses below the MIC.
†
: S. aureus Wood 46.
‡
: S. aureus 12009.
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Therapeutic Areas II: Cancer, Infectious Diseases, Inflammation & Immunology and Dermatology
2007, Comprehensive Medicinal Chemistry IIL.S. Young
7.16.8 In Vitro Activity: Static or Cidal
In addition to classification by mechanism of action or target site, one of the more useful parameters for assessment is whether a microbe possesses intrinsic killing power or merely serves to limit the growth of an organism. Such distinctions are reflected in the classification of an agent as bactericidal or bacteriostatic. Exposure to agents resulting in a significant decline in viable microbes over a defined time interval are classified as bactericidal, whereas those that merely limit the bacterium's growth are static. Well-known examples of static agents include the sulfonamides. However, some compounds such as chloramphenicol are cidal to some bacteria species but static for others and cidal or static effects can clearly be concentration-dependent. Further blurring distinctions are the rate of killing exhibited by certain agents: those compounds that rapidly kill, e.g., several logs within hours, may be more effective than those in which a 90% or 99% kill is accomplished over a period of half a day or more. It seems intuitive that bactericidal agents would be more effective and desirable than bacteriostatic agents in immunocompromised hosts, the latter being those animals or patients who are immune suppressed or may have some defined impairment to neutrophil function or cellular immunity. Where bactericidal therapy appears called for is in dealing with infections where host defenses are limited or impaired: examples would include in the vegetations of the patient with bacterial endocarditis or in the spinal fluid of the patient with meningitis. In the intact host who has adequate circulating neutrophil function, bacteriostatic drugs may not be notably inferior to those compounds that have bactericidal activity. Equally important may be the tissue penetration of a therapeutic agent – namely, the ability of the therapeutic agent to reach the active site of infection. There is little doubt that with increasing population of immunocompromised patients or individuals with some defect in host response, the impetus has been to develop compounds that have the ability to kill populations of microbes without assistance from host defense.
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Antimicrobials
2017, Nanostructures for Antimicrobial TherapyBhaskar Das, Sanjukta Patra
3 Antimicrobial Classes and Their Current Effectiveness With Respect to Emerging Drug Resistance
Antimicrobials are classified based on a variety of methods such as spectrum of activity, effect on microbes, and mode of action.
3.1 Classification on the Basis of Spectrum of Activity
If an antibacterial is active against both Gram-positive and Gram-negative bacteria it is referred to as a broad spectrum antibacterial, e.g., tertracyclines, phenicols, third and fourth generation cephalosporins, and fluoroquinolones. On the other hand, an antibacterial that is effective against a particular species of microbe, e.g., glycopeptides and bacitracin are effective only against Gram-positive bacteria, polymyxins are effective only against Gram-negative bacteria, aminoglycosides and sulfonamides are effective only against aerobic organisms, and nitoimidazoles are effective against anaerobes (www.amrls.cvm.msu.edu).
3.2 Classification on the Basis of Nature of Effect
Based on the nature of effect on bacteria, antibiotics are classified as bactericidal (which kill the target bacteria) and bacteriostatic (which inhibit bacterial growth and replication), e.g., antibiotics such as aminoglycosides, cephalosporins, penicillins, and quinolones are bactericidal, while tetracyclines, sulfonamides, and macrolides exert bacteriostatic effects on target bacteria. The effect of bactericidal agents is faster as compared to bacteriostatic agents. Bacteriostatic agents require an effective immune system in the host for elimination of pathogenic bacteria and hence are not applicable to immunosuppressed host conditions or acute infections. However, some antibiotics may behave as both bacteriostatic and bactericidal based on dosage concentration and duration of exposure, e.g., aminoglycosides, fluoroquinolones, and metronidazole exert concentration-dependent bactericidal characteristics (www.amrls.cvm.msu.edu).
3.3 Classification on the Basis of Mode of Action
The mode of action of antimicrobials varies on the basis of the nature of their structure and degree of affinity to target sites within bacterial cells.
3.3.1 Cell Wall Synthesis Inhibitors
Since the cell wall is critical for the survival of bacterial species, an antibacterial that affects the cell wall would behave as a bacteriostatic or bacteriocidal agent. The β-lactam group of antibiotics contains a four-membered nitrogen-containing β-lactam ring responsible for its antibacterial action. β-Lactam antibiotics bind to penicillin-binding proteins (PBPs) present on the cell membrane rendering them incapable of performing cell wall synthesis (Elander, 2003). This kills the target bacteria by osmotic instability or autolysis, e.g., natural penicillin, penicillinase-resistant penicillin such as methicillin, nafcillin, and oxalin, extended spectrum penicillin such as ampicillin, carbenicillin, and amoxicillin, cephalosporins, carbapenems, and monobactams. Carbapenems and second, third, and fourth generation cephalosporins have broad spectrum activity, while penicillin, first generation cephalosporins, and monobactams have a narrow spectrum of activity. The activity of β-lactams varies among bacterial species on account of species-specific variation of PBP content and nature. Gram-negative bacteria with an outer membrane layer hinder the interaction of PBP and β-lactam antibiotics, thus rendering the antibacterial ineffective. The glycopeptide group of antibiotics inhibits bacterial cell wall synthesis by binding to precursors of cell wall synthesis, which leads to the inhibition of cell wall synthesis activity by PBPs. Actinomycetes species such as Streptomyces orientalis and Nocardia actinoides produce the glycopeptides vancomycin and actinoidin, respectively. These drugs have a narrow spectrum of bactericidal activity affecting only Gram-positive bacteria. Vancomycin is considered a last resort drug for the treatment of skin and bone infection, bloodstream infection, endocarditis, and meningitis caused by MRSA (www.amrls.cvm.msu.edu).
3.3.2 Cell Membrane Function Inhibitors
Exchange of intra- and extracellular substances takes place through microbial cell membranes. Thus cell survival can be at stake if there is disruption of cell membrane structure because of leakage of important intercellular solutes. Cell membrane is found in both prokaryotic and eukaryotic organisms causing poor selectivity and thus compromising its use in the mammalian host, e.g., polymyxin (www.amrls.cvm.msu.edu). The polymyxin group of antibiotics is characterized by a cyclic peptide with a long hydrophobic tail. Polymyxins bind to lipopolysaccharides (LPS) in the outer membrane of Gram-negative bacteria and disrupt the cell membrane structure. The hydrophobic tail with its detergent-like mode of action is instrumental in damaging the cell membrane. Their specificity to LPS molecules that exist in the outer membrane of Gram-negative bacteria renders them selective to bactericidal activity against Gram-negative bacteria. They are produced by nonribosomal peptide synthetase systems in Paenibacillus polymyxa (www.nlm.nih.gov). On account of their neurotoxicity and nephrotoxicity, polymyxins are used as a last resort when other clinical antibiotics prove ineffective (Falagas and Kasiakau, 2006). They have found application in controlling infection caused by multiple drug-resistant P. aeruginosa and carbepenemase-producing Enterobactericeae. Colistin is a polymyxin antibiotic that confers its bactericidal activity to its polycationic nature with both hydrophilic and lipophilic moieties. The polycationic regions interact and displace bacterial counter ions from LPS present in the bacterial outer membrane. Hydrophobic and hydrophilic regions interact with the cytoplasmic membrane solubilizing the bacterial membrane leading to bactericidal activity. Colistin is produced by P. polymyxa var. colistinus. In India, colistin is commercially available as Colymonas and Koolistin. Although it is not favored because of nephrotoxic effects, it is considered a last resort antibiotic against multidrug-resistant P. aeruginosa, K. pneumoniae, and Acinetobacter, etc. (Falagas et al., 2008). Colistin susceptibility is also observed in NDM-1 metallo-β-lactamase multidrug resistant Enterobacteriaceae (Kumarasamy et al., 1999).
3.3.3 Protein Synthesis Inhibitors
One of the vital processes for microbial multiplication and survival is protein synthesis. With an attempt to kill or inhibit the growth of pathogenic microbes, many antimicrobials disrupt the protein synthesis machinery by binding to the 30S or 50S subunit of intracellular ribosomes, e.g., aminoglycosides, macrolides, lincosamides, streptogramins, chloramphenicol, and tetracyclines (www.amrls.cvm.msu.edu).
3.3.3.1 Aminoglycosides
Inhibition of protein synthesis by aminoglycosides is caused by its binding to the 30S ribosomal subunit. This perturbs peptide elongation at the 30S ribosomal subunit resulting in errors during mRNA translation and thus biosynthesis of truncated proteins that bear altered amino acid compositions (Mingeot-Leclercq et al., 1999). Aminoglycosides are derived from bacteria of the genus Streptomyces and Micromonospora (Kroppenstedt et al., 2005). They are useful to treat infections caused by aerobic, Gram-negative bacteria such as Pseudomonas, Acinetobacter, Enterobacter, and tuberculosis-causing mycobacteria. Although its clinical use is limited because of nephrotoxicity and ototoxicity, the recent emergence of antibiotic-resistant Gram-negative bacterial strains has led to interest in reevaluating its antimicrobial susceptibility and toxicity. The encouraging fact is that aminoglycosides still retain activity against the majority of clinical bacteria in many parts of the world providing means to overcome antibiotic resistance (Durante-Mangoni et al., 2009; Falagas et al., 2008), e.g., streptomycin was the first effective clinical drug to treat tuberculosis.
3.3.3.2 Macrolides
The macrolides group of antibiotics is characterized by the presence of a macrolide ring belonging to the polyketide class of natural products. Macrolides inhibit protein biosynthesis by binding to the P site on the 50S ribosomal subunit leading to prevention of peptidyl transferase by adding a growing peptide attached to tRNA to the next amino acid as well as inhibition of ribosomal translation (www.pharmacologycorner.com). Premature dissociation of peptidyl-tRNA from ribosome is another potential mechanism (Tenson et al., 2003). Macrolides are transported to the site of infection because they are concentrated within leukocytes (Bailly et al., 1991). Natural macrolides are produced by Saccharopolyspora erythraea (erythromycin) and Streptomyces fradiae (tylosin), while some are semisynthetic such as tilmicosin and tulathromycin (www.amrls.cvm.msu.edu). Macrolides were used to treat clinical infections caused by Gram-positive organisms with a slightly wider spectrum of activity compared to penicillin (www.emedexpert.com). They are used as a substitute for penicillin allergic patients. Macrolides are effective against β-hemolytic streptococci, pneumococci, staphylococci, enterococci, and pathogens against which penicillin activity fails. However, macrolide-resistant bacterial strains have been developed by posttranscriptional methylation of 23S bacterial ribosomal RNA, which is a matter of concern.
3.3.3.3 Lincosamide
The lincosamide group of antibiotics shows effectiveness against bacteria by applying an inhibitory effect on protein synthesis via binding to the 23S portion of the 50S bacterial ribosomal subunit. Protein synthesis is inhibited because of premature dissociation of peptidyl-tRNA from the ribosome (Tenson et al., 2003). As human ribosomes are structurally different from their bacterial counterparts, lincosamides do not interfere with protein synthesis in humans. Lincosides are effective in the control of Gram-positive bacteria, most anaerobic bacteria, and some Mycoplasma (www.amrls.cvm.msu.edu). Lincomycin, the first lincosamide, was isolated from Streptomyces lincolnensis. They also treat clinical infections related to Staphylococcus, Streptococcus, Bacteroides fragilis, and toxic shock syndrome. Lincosamide antibiotics are most associated with the treatment of pseudomembranous colitis caused by C. difficile (www.nlm.nih.gov/medline). Lincosamides may behave as bacteriostatic or bacteriocidal depending on drug concentration, bacterial species, and pathogen concentration. The matter of concern is that microbial resistance to lincosamide is conferred by 23S binding site methylation. This makes lincosamide-resistant strains also resistant to macrolides.
3.3.3.4 Streptogramins
The streptogramin class of antibiotics confers antibacterial activity by binding to the 50S ribosomal subunit leading to inhibition of protein synthesis. Group A steptogramins prevent peptide bond formation during the chain elongation step. Group B steptogramins release incomplete peptide chain from the 50S ribosomal subunit (www.amrls.cvm.msu.edu). An encouraging fact is that they are currently effective for treatment of two of the most rapidly growing multidrug-resistant bacterial strains, namely, vancomycin-resistant S. aureus and vancomycin-resistant Enterococcus (VRE).
3.3.3.5 Chloramphenicol
Chloramphenicol is bacteriostatic because of its capability to inhibit protein synthesis. Chloramphenicol hinders protein chain elongation by peptidyl transferase inhibition of bacterial ribosome (www.amrls.cvm.msu.edu). The natural source of its isolation is Streptomyces venezualae. Chloramphenicol was first used for the treatment of typhoid but with the global presence of multiple drug-resistant Salmonella Typhi it loses its clinical value. It is widely used for the treatment of bacterial conjunctivitis, staphylococcal brain abscesses (because of excellent blood–brain barrier penetration), and meningitis. It is used to treat infections caused by tetracycline-resistant cholera and VRE. Microbial chloramphenicol-resistant strains are conferred by the cat gene, which codes for an enzyme chloramphenicol acetyltransferase, which inactivates chloramphenicol. Reduced membrane permeability and 50S ribosomal subunit mutation are other microbial mechanisms that resist the effect of chloramphenicol. Chloramphenicol resistance may be carried in plasmids such as ACCoT plasmid, which confers multiple drug resistance against ampicillin, chloramphenicol, cotrimaxole, and tetracycline in typhoid (www.wikipedia.org/Chloramphenicol).
3.3.3.6 Tetracycline
Tetracycline inhibits cell growth by binding to the 16S part of the 30S ribosomal subunit preventing aminoacyl tRNA from binding to the ribosome A site. This leads to inhibition of translation hampering cell growth (Connell et al., 2003). Tetracyclines are used in the treatment of clinical conditions such as urinary tract infections, respiratory tract infections, acne, rosaceae, anthrax, bubonic plague, malaria, elephantiasis, syphilis, Lyme disease, etc. Nowadays, clinical application of tetracycline is being compromised by tetracycline resistance in the pathogenic microbes. In tetracycline-resistant microbial strains, tetracycline resistance genes encode a membrane protein that effluxes tetracycline out of the cell. Tetracycline resistance is also provided by blocking tetracycline from binding to ribosome or by enzymatic inactivation of tetracycline, which is rare (Wyk, 2015; www.wikipedia.org/Tetracyclines).
3.3.4 Nucleic Acid Synthesis Inhibitors
DNA and RNA are essential for microbial replication and survival. The antibacterial activity of some antibiotics lies in interfering with the nucleic acid synthesis processes, e.g., quinolones and rifamycin.
3.3.4.1 Quinolones
Quinolones are synthetic broad spectrum antibacterials that prevent bacterial DNA synthesis (Andersson and MacGowan, 2003). Fluoroquinolone is the most common quinolone used in antibacterial treatment. Fluoroquinolones bind to the DNA gyrase–DNA complex causing defects in the supercoiling of bacterial DNA. This causes enabling of replication and survival (www.amrls.cvm.msu.edu). Fluoroquinolones are broad spectrum antibiotics that are used for control of HAIs resistant to conventional antibiotic classes. However, the broad spectrum range of fluoroquinolones could lead to the spread of multidrug-resistant strains. Fluoroquinolones are used predominantly for the treatment of hospital- and community-acquired pneumonia caused by drug-resistant Streptococcus pneumoniae (MacDougall et al., 2005). They are also used for the treatment of hospital-acquired urinary catheter infection and polynephritis. In community-acquired infections, fluoroquinolones are recommended when antibiotic therapy fails on account of multidrug resistance (Liu and Mulholland, 2005).
3.3.4.2 Rifamycin
Rifamycin is a group of antibiotics whose activity depends on its high affinity for prokaryotic RNA polymerase inhibiting bacterial DNA-dependent RNA synthesis (Calvori et al., 1965). Rifamycins can enter neutrophils and macrophages and then inhibit bacterial DNA-dependent RNA polymerase (www.amrls.cvm.msu.edu). Their poor affinity to analogous mammalian enzyme renders them selective. They are synthesized either naturally by Amycolatopsis rifamycinica or artificially. Rifamycins are particularly effective against mycobacteria and therefore are used to treat tuberculosis, leprosy, and Mycobacterium avium complex infections. Multiple drug resistance is currently posing a threat to rifamycin-based treatment of tuberculosis (Floss and Yu, 2005).
3.3.5 Metabolic Process Inhibitors
Other antibiotics act as inhibitors of metabolic pathways essential for survival of bacterial pathogens, e.g., sulfonamides and trimethoprim (TMP).
3.3.5.1 Sulfonamides
The antibacterial activity of sulfonamides is caused by their interference with folic acid synthesis by preventing the addition of para-aminobenzoic acid into folic acid molecules by competitive inhibition of enzyme dihydropteroate synthetase (DHPS). Thus sulfonamides are bacteriostatic and not bactericidal. The first sulfonamide, trade name Pontosil, could treat infections caused by streptococci, blood infections, child bed fever, and erysipelas. The genetic basis of sulfonamide resistance lies in sul1, sul2, and sul3 genes encoding DHPS with low affinity for sulfonamide. A wide range of bacterial species harbor these genes in mobilizable plasmids manifesting multiple antibiotic resistance (Wang et al., 2014).
3.3.5.2 Trimethoprim
TMP is a synthetic antibiotic that binds with the enzyme dihydrofolate reductase (DHFR) inhibiting the folic acid synthesis pathway (Brogden et al., 1982). It is widely used in the treatment of urinary tract infections and Pneumocystis jiroveci pneumonia. However, because of the widespread use of TMP, TMP-resistant pathogenic strains have surfaced as a major clinical problem. The mechanism of TMP resistance includes cell wall impermeability to TMP, alternate metabolic pathways, and production of chromosomal or plasmid-mediated TMP-resistant DHFR enzyme (Huovinen, 1987).
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Laboratory Diagnosis and Therapy of Infectious Diseases
2012, Principles and Practice of Pediatric Infectious Diseases (Fourth Edition)John S. Bradley, Sarah S. Long
Antagonism
Despite the paucity of documented reports of the clinical significance of antagonism between antimicrobial agents, multiple examples are demonstrable in vitro; thus, caution is needed in their use, especially in infections in hosts with impaired defenses. Combinations of a bacteriostatic agent with a β-lactam antibiotic can antagonize the bactericidal activity of the β-lactam antibiotic. Combinations of chloramphenicol or tetracycline plus aminoglycosides also are antagonistic for gram-negative bacilli. In addition, chloramphenicol antagonizes the bactericidal effect of ciprofloxacin against Staphylococcus aureus, Escherichia coli, and P. aeruginosa. A combination of agents that bind to similar locations within the ribosome (e.g., clindamycin, erythromycin, spiramycin, chloramphenicol, streptogramins) either complement each other and enhance activity, or compete with each other and antagonize activity.53
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Chapter
Burn Wound Management
2019, Global Reconstructive SurgeryKatie Osborn
Application of Topical Agents
The overall objectives in the use of topical agents are:
•
: antimicrobial action that assists in preventing infection, and
•
: penetration of the burn wound to separate the eschar from the deep viable tissue.
These agents are applied after the wound has been cleaned and debrided. Common agents used in the treatment of burn wounds include:
•
: Silver sulfadiazine (Silvadene) is a broad-spectrum antibiotic with limited penetration ability. It can be applied directly onto the wound using sterile gloves or an applicator and then covered with a dry gauze dressing. Another common method for application is to spread it on sterile dry gauze that is laid directly on the wounds and covered with a dry dressing. It is used on deep partial-thickness and full-thickness injuries until the eschar is removed. It is also used when there is a wound infection. It is not painful.
•
: Mafenide Acetate (Sulfamylon) is bacteriostatic and used for gram-positive and gram-negative organisms. It will penetrate thick eschar and cartilage. It is used on deep partial-thickness and full-thickness injuries and with wound infections. It is applied in the same manner as Silvadene. It is very painful.
•
: 3% bismuth tribromophenate and USP petrolatum (Xeroform) is a mildly deodorizing and bacteriostatic agent that inhibits the growth of bacteria in the wound and helps maintain a moist wound environment. It is impregnated onto sterile fine mesh gauze and used to treat burn wounds, donor sites, and skin grafts. It is a comfortable, soothing, hypoallergenic dressing.
•
: Silver nitrate is a broad-spectrum antimicrobial that poorly penetrates the eschar. It is used for deep partial- to full-thickness burns and wound infections. It is applied in the same manner as Silvadene. It is very painful and will turn the wounds black.
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Principles of Anti-infective Therapy and Surgical Prophylaxis
2017, Infectious Diseases (Fourth Edition)Evelina Tacconelli, ... Mathias W. Pletz
General Principles of Prescribing
Anti-infective drugs are generally classified according to their major microbial targets rather than by target disease or infection.6,7 Pharmacopoeias and formularies categorize these as antibacterial, antiviral, antifungal and antiparasitic (antiprotozoal or anthelmintic) agents. Within these categories some specific indications can be found such as antiviral drugs used in the treatment of herpesvirus infections, chronic hepatitis C or HIV (Tables 136-1 and 136-2). Major antifungal and antiparasitic drugs are listed in Tables 136-3 and 136-4. Among anti-infective drugs the largest proportion is for the treatment of bacterial infections.
Antimicrobial prescribing should be based on clear evidence of infection established by either laboratory investigation or sound clinical criteria. In general, laboratory confirmation of infection, although desirable, is only possible in a minority of treated infections and is largely confined to hospitalized patients where access to laboratory investigation is readily available.
The results of microbiologic sampling may confirm the nature of an infection and support the continued use of initial empiric treatment or indicate alternative therapy. However, most prescribing is empiric, particularly in community practice, which accounts for about 80–90% of antimicrobial prescribing.
Choice of Agent or Agents
Fundamental to the treatment of infectious diseases is the evidence-based selection of an antimicrobial regimen with established efficacy against the target infection and supported by widespread clinical use. The antimicrobial spectrum of activity should be based on local susceptibility data.8 The National Committee for Clinical and Laboratory Standards Institute (CLSI) and the European Committee on Antimicrobial Susceptibility Testing (EUCAST) set ‘breakpoints’ for the minimum inhibitory concentration (MIC) based on the integration of MICs with achievable antibiotic levels in serum and tissues, clinical pharmacology and data from in vitro and animal models.
Broad-Spectrum Versus Narrow-Spectrum Antimicrobial Agents
The terms ‘broad-spectrum’ and ‘narrow-spectrum’ are widely used to distinguish between agents that target a limited range of pathogens in contrast to broad-spectrum agents to which many pathogens are susceptible. Historically, the terms have been applied to developments in the penicillin class of drugs. Penicillin G (benzylpenicillin) targeted a limited number of pathogens – for example, Streptococcus pneumoniae, hemolytic streptococci and Neisseria spp. (meningococci and gonococci). With the development of the aminopenicillins (e.g. amoxicillin), the spectrum increased to also include Enterococcus faecalis, Listeria spp., H. influenzae, and E. coli and were thus considered as broad-spectrum agents; however, as stated above, with time the latter two pathogens have become resistant as a result of β-lactamase production.
The antimicrobial spectrum of an antibiotic has important practical implications. Narrow-spectrum agents have a more limited range of indications and in general require greater diagnostic precision to ensure they are used appropriately. They also require greater emphasis on documenting the microbiologic nature of an infection. In contrast, broad-spectrum agents usually capture a greater range of micro-organisms and indications. They are also widely used as initial empiric therapy of respiratory tract and intra-abdominal infections.
One of the consequences of broad-spectrum agents is their impact on the normal flora, particularly that of the gastrointestinal tract. As a consequence, microbial overgrowth by organisms such as Candida spp. and C. difficile may occur and give rise to superinfection or C. difficile-associated diarrhea, respectively. A recent study on the structure and function of human gut microbiota revealed that the effect of antibiotics relates to the interaction and the properties of the antimicrobial agent and the structure, function and resistance genes among the microbial community.9
Bactericidal Versus Bacteriostatic Activity
Antibacterial drugs act by either killing or inhibiting the growth of micro-organisms. These bactericidal or bacteriostatic effects are determined by the mode of action of a drug against a particular target organism. Table 136-5 identifies the usual mode of action of commonly prescribed agents.
Cell wall active agents, such as the penicillins, cephalosporins and glycopeptides, are largely bactericidal, whereas most agents that interfere with protein synthesis are bacteriostatic. Examples of the latter include the tetracyclines, lincosamides and macrolides. Metabolic inhibitors of folic acid synthesis, namely the sulfonamides and trimethoprim, are also bacteriostatic but the combination of both (co-trimoxazole) is bactericidal. Cidality can be affected by drug concentration and in vitro conditions of testing and may also vary by target pathogen. Nonetheless, it is a useful distinguishing feature with clinical implications in selected circumstances.
In general, bactericidal and bacteriostatic agents are equally effective in the management of many mild to moderate infections. However, bacteriostatic agents are dependent upon an effective host immune response, notably the ability to phagocytose drug-exposed bacteria. There are a number of clinical conditions in which host immunity is either deficient or suppressed and where bacteriostatic agents should be avoided in favor of bactericidal agents. Other circumstances where bactericidal agents are recommended are in the management of infective endocarditis.10 The infected vegetations are impermeable to phagocytic cells and require high plasma concentrations of bactericidal drugs to sterilize them.
Dose Selections and Frequency of Administration
The selection of antimicrobial chemotherapy is supported by a matrix of information that includes in vitro evaluation of antimicrobial activity, animal models of infections, pharmacokinetic studies of the drug in health and disease, and clinical trial data developed to support the approved indications. Table 136-6 identifies some common bacterial infections affecting adults and the agents used in their treatment.
Pharmacokinetics
The bioavailability of a drug indicates the degree of absorption from the gastrointestinal tract. Parenterally administered agents achieve 100% bioavailability. The bioavailability of orally administered antimicrobials ranges from only 15% to 30% for aciclovir to >90% for antimicrobials with a high degree of bioavailability such as cephalexin, levofloxacin, moxifloxacin, doxycycline, clindamycin, linezolid, trimethoprim-sulfamethoxazole, voriconazole and metronidazole. As a rule, replacing an intravenous antimicrobial agent with an oral drug is unwise for oral agents with less than 50% bioavailability.
Drugs once absorbed or injected diffuse from the intravascular fluid space to extravascular fluid spaces and this distribution is best described by the drug's volume of distribution.11 The volume of distribution (VD), also known as apparent volume of distribution, is a pharmacological, theoretical volume that is calculated as follows: VD = total amount of drug in the body/drug blood plasma concentration. VD has nothing to do with the actual volume of the body or its fluid compartments but rather involves the distribution of the drug within the body. For a drug that is highly tissue-bound, very little drug remains in the circulation; thus, plasma concentration is low and VD is high. Drugs that remain in the circulation tend to have a low VD. VD provides a reference for the plasma concentration expected for a given dose but provides little information about the specific pattern of distribution.
Drugs that distribute widely through the body tend to have large volumes of distribution and low serum concentrations. Drugs that remain only in the blood volume typically have small volumes of distribution and high serum concentrations. Anti-infective drugs are generally distributed in the blood and tissues bound to plasma proteins, most notably albumin. The degree of protein binding varies from drug to drug. It has to be emphasized that only the free concentration of an anti-infective drug exhibits antimicrobial activity. Therefore, despite apparently adequate total plasma levels of highly protein-bound drugs, the concentration of free (i.e. active) drug might be less than the MIC of the pathogen. As most sites of infection are extravascular, treatment of infections in these sites depends on movement of the antimicrobial agent out of the bloodstream and into interstitial and sometimes intracellular fluid. In general, hydrophilic agents (such as β-lactams) tend to reach higher concentrations in body fluids and the interstitium, whereas lipophilic agents (such as fluoroquinolones) are able to cross biologic membranes and achieve effective intracellular concentrations. Therefore, even within one organ, such as lung or brain, antibiotic concentrations can vary between anatomic or histological structures. For example, antibiotic concentrations in the epithelial lining fluid differ from concentrations within the bronchial epithelial cells or the interstitium of the lung. When the drug ultimately reaches the site of infection, local factors may play a role in the effectiveness of its antimicrobial activity such as pH, anaerobic environment or the presence of β-lactamases and other deactivating enzymes.
After a peak plasma level is attained, the plasma level declines as a consequence of drug distribution and elimination, usually by the kidneys, liver or both. The term ‘clearance’ describes a theoretic volume of plasma cleared of the drug within a period of time. Drugs may be excreted unchanged but in general undergo metabolism, especially in the liver for subsequent excretion in the bile. Typical metabolites are the result of glucuronidation, conjugation and acetylation. Drugs metabolized by the cytochrome P450 system may interact with other drugs which share this route. Drug–drug interactions involving antibiotics are mainly caused by inhibition of their metabolism or by additive toxic effects and can be responsible for severe complications. Conversely, enzyme induction may lead to subtherapeutic drug levels.
In those with impaired liver function, drugs excreted through the liver should be used with caution or avoided. Dose adjustment may be appropriate following careful assessment of liver function tests.
Drug excretion is also largely the result of renal excretion. The latter may be either a metabolically active or passive process, and involve the glomeruli or renal tubules or both. Impairment of renal function often leads to a prolonged drug half-life for renally excreted drugs. Dose modification may be required to avoid drug accumulation and drug toxicity.
A clear understanding of the elimination half-life (t1/2elim) and excretion of antibiotics also influences dosage selection. Drugs with a short half-life, such as many β-lactams, are rapidly eliminated and need more frequent administration to produce satisfactory antibiotic concentrations at the site of infection. Drugs with a longer half-life, such as moxifloxacin, azithromycin, ceftriaxone or ertapenem, allow once-daily dosing. Recently, the US Food and Drug Administration (FDA) approved a novel lipoglycopeptide antibiotic for gram-positive skin and soft-tissue infections called dalbavancin with unusually long elimination half-life, ranging from 149 to 250 hours. This long half-life is the result of extensive, reversible binding of dalbavancin to plasma proteins and allows even once-weekly dosing.
Pharmacokinetic and Pharmacodynamic Parameters
The success of antimicrobial therapy is determined by complex interactions among an administered drug, a host and an infecting agent. In the past, dose and drug selection was mostly based on MIC and on the drug's serum concentration. However, a pharmacodynamic effect in vivo is rather the result of a dynamic exposure of the infective agent to the unbound antibiotic drug fraction at the relevant effect site. To elucidate this dynamic relationship, PK/PD models have to be considered.12,13 The most useful PK/PD variables are summarized in Table 136-7.
The efficacy of an antimicrobial agent can depend on the maximal concentration, on the exposure time and on the total exposure that the drug remains on its target sites. The integration of how high (concentration) and how long (time) an antibiotic's level remains above a zero concentration over a dosing interval is referred to as the area under the concentration-time curve (AUC). The different classes of antibiotics differ in their relevant PK/PD target parameters. For time-dependent antibiotics such as β-lactams or vancomycin, the duration of time during a dosing interval that the antibiotic's concentration remains above the MIC for a particular organism (time above MIC, t>MIC) is the best predictor of clinical outcome.7,13 To prolong the duration of a β-lactam concentration above its MIC in severe infections, the drug has to be given in larger and more frequent doses or administered by a 24-hour constant infusion. For concentration-dependent agents such as aminoglycosides, fluoroquinolones, metronidazole, amphotericin B or daptomycin, the single dose is pivotal, whereas the dosage interval can be extended. Thus, for these agents, the pharmacodynamic parameter ratio of AUC/MIC can be simplified to the ratio of peak concentration to MIC (Cmax/MIC). For a number of bacteriostatic antimicrobial agents such as azithromycin, clindamycin and tigecycline, the antimicrobial efficacy of the drug can be correlated best with the 24-hours AUC/MIC ratio.14,15
Data based on PK/PD parameters and timed plasma samples from volunteers and patients may be employed in mathematical models using approaches such as Monte Carlo simulations. Using this basic pharmacokinetic data and in vitro susceptibility profiles of micro-organisms, the relationship between PK and PD variables can be characterized graphically.16 Figure 136-1 illustrates these key parameters in relation to the MIC of a target pathogen.17
Such PK/PD data have supported single daily dosage regimens for the aminoglycosides18 and dosage revisions for ciprofloxacin and levofloxacin.19 Continuous or prolonged infusion of β-lactams is conducted with increasing frequency in intensive care units (ICU) and has recently been included as an option to increase antibiotic efficacy in the German guideline for the treatment of nosocomial pneumonia.20 To date, evidence from RCTs showing that continuous infusion of β-lactams improves clinical outcome is increasing.21,22
Drug Safety
Adverse reactions range from drug hypersensitivity to dose-related side effects and unpredictable or idiosyncratic phenomena. With regard to antibiotics, the penicillins and cephalosporins share the β-lactam ring structure and in turn have many adverse effects in common. These include hypersensitivity rashes and other reactions and toxicity to hematopoietic cells. However, there are differences between these two classes. Hypersensitivity reactions occur less commonly to the cephalosporins. This applies to anaphylaxis which is the most severe and feared hypersensitivity reaction. Any severe hypersensitivity reaction to a β-lactam is a bar to its future use and of the β-lactam agents in general. However, milder forms of hypersensitivity, such as rash or skin eruption to a penicillin, do not necessarily preclude the use of a cephalosporin. Usual alternatives to β-lactams in patients with severe allergies are vancomycin and clindamycin.
Many adverse drug events can be minimized in persons with known excretory organ malfunction by adjusting the dose. The aminoglycosides, such as gentamicin, have been extensively studied and dosage schedules developed which are adjusted for age, weight, gender and renal function.
Therapeutic Drug Monitoring
Therapeutic drug monitoring (TDM) aims to ensure safe and effective prescribing for drugs with a narrow therapeutic index. The latter reflects the limited margin between the therapeutic and toxic concentrations of a particular drug. In contrast, drugs such as penicillins have a wide therapeutic index and even when given in high concentration are generally safe.
In practice much TDM is directed at ensuring therapeutic nontoxic concentrations of the aminoglycosides or ensuring sufficient but nontoxic concentration of the glycopeptides. Other examples include TDM for antifungals such as itraconazole and voriconazole23 in the treatment of severe fungal infections as well as, in some circumstances, antiretroviral agents such as the protease inhibitors and non-nucleoside reverse transcriptase inhibitors.24
There is increasing evidence that TDM is not only required to avoid toxic concentrations but also to achieve sufficient concentrations, particularly in septic patients with unpredictable and highly variable pharmacokinetics. Roberts et al. prospectively evaluated TDM in ICU patients and reported that β-lactam dose adjustment was necessary for 74% of patients; 50% of the total patients required a dose increment after the first measurement.25 Recently, in a randomized controlled trial in 41 patients receiving piperacillin/tazobactam or meropenem, with and without therapeutic drug monitoring-based dose optimization, De Waele et al. showed that among critically ill patients with normal kidney function, a strategy of dose adaptation based on daily TDM led to an increase in PK/PD target attainment compared to conventional dosis (58% versus 16%).26
Combined Drug Regimens
In general and whenever possible it is preferable to manage an infection with a single drug. However, there are an increasing number of diseases and circumstances where a combination of drugs is appropriate or necessary to control the infection. Examples include tuberculosis,27 HIV infection24 and, more recently, malaria28 and gonorrhoea. In practice, drugs are frequently combined where there is uncertainty about the microbiologic nature of a diagnosis or where the spectrum of a single agent is inadequate, particularly in situations such as intra-abdominal sepsis where the etiology is often polymicrobial. Combining an extended-spectrum cephalosporin, such as cefuroxime or cefotaxime, with metronidazole provides an appropriate ‘broad-spectrum’ regimen. Another example is in the treatment of severe community-acquired pneumonia when both atypical and conventional pathogens need to be ‘covered’.29
Another important reason for combined drug regimens is to prevent resistant organisms emerging on therapy. This was first demonstrated in the treatment of tuberculosis where minority populations of organisms can become resistant to the most active tuberculostatic drugs, isoniazid or rifampin (rifampicin) when applied as monotherapy. Therefore, the standard regimen for treating pulmonary tuberculosis includes isoniazid and rifampin combined with a third (pyrazinamide) and fourth (ethambutol) agent for the first 2 months, by which time the results of susceptibility testing are generally available. It is then usually safe to continue with a two-drug regimen of isoniazid and rifampin for the remaining 4 months of treatment (see Chapter 148).27
Similar principles are increasingly being applied to the treatment of severe malaria, particularly in regions of the world where chloroquine-resistant malaria is widespread. Examples include combinations of artemisinin with tetracycline or mefloquine.28
Other circumstances where combined regimens have become the norm are in the management of HIV24 and chronic hepatitis C.30 In both diseases clinical trials have shown greater efficacy with combined regimens. In the case of HIV, viral load is more rapidly controlled and sustained with faster immune reconstitution and decreased clinical progression. By selecting two or more agents from different classes of antiretrovirals, drug resistance is delayed in addition to providing a more efficacious regimen.
In the case of hepatitis C, recently new combination regimens have been developed including not only pegylated interferon and ribavirin, but also new nucleotide polymerase inhibitors (e.g. sofosbuvir) ± new protease inhibitors (e.g. simeprevir) enabling prompt and sustained viral response with high cure rates (see Chapter 155).30,31
Another rationale for combined drug regimens is to achieve synergistic inhibition of a target pathogen. The combining of a penicillin (for susceptible strains) with an aminoglycoside to treat enterococcal endocarditis is now standard therapy since in vitro and in vivo the combination is more effective than a penicillin alone.10
Failure of Antimicrobial Therapy
Failure to respond to antimicrobial treatment may have several explanations. These include the accuracy of the primary diagnosis, the choice of initial therapy, the selection of dose and route of administration, the duration of treatment and, in the case of trauma or implant surgery, an assessment of the possible presence of deep-seated infection. Surgical drainage of pus with abscess formation is often overlooked and needs emphasis.
Failure to make a correct diagnosis may be the result of an inaccurate clinical assessment and, in turn, presumptive microbiologic assessment to guide initial empiric therapy – for example, the incorrect diagnosis of superficial cellulitis instead of more deep-seated necrotizing fasciitis. Here the selection of oxacillin, dicloxacillin or flucloxacillin alone will be ineffective, particularly against methicillin-resistant Staphylococcus aureus (MRSA) or anaerobic pathogens.32 The dose and route of medication are also of importance. For example, pneumococcal pneumonia can be effectively managed in the community with oral agents for patients with mild to moderate infection.
Most mild to moderate infections respond to 5–10 days of antibiotic therapy.33,34 However, selected infections require more protracted therapy to avoid treatment failure and relapse. Severe Legionella pneumophila pneumonia may require up to 3 weeks of treatment,29 as does meningitis caused by Listeria monocytogenes. Infection as a complication of the presence of foreign materials, notably vascular and articular prosthetic devices, is increasing. Infection is often caused by skin micro-organisms, notably coagulase-negative staphylococci. These are not only of variable sensitivity to many commonly used agents but also adhere to the foreign material as biofilms in which they are protected against host defenses and are less responsive to treatment. The so-called minimal biofilm eradicating concentration (MBEC) of a certain bacterial strain can exceed its MIC, which is measured in the planktonic phase, by more than 1000-fold. Prolonged therapy or surgical revision, sometimes with removal of the infected device, may be necessary to eliminate the infection.
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Book2017, Infectious Diseases (Fourth Edition)Evelina Tacconelli, ... Mathias W. Pletz
Chapter
Atypical Pneumonias in Children
2012, Kendig & Chernick's Disorders of the Respiratory Tract in Children (Eighth Edition)L. Barry Seltz MD, ... Leslie L. Barton MD
Treatment
Aminoglycosides are the preferred antimicrobials for treatment of tularemia. Although streptomycin was the classic therapeutic choice, it has been superseded by gentamicin. Resolution of signs and symptoms of infection is prompt and usually occurs within days. The duration of therapy is 7 to 10 days but may be extended in severe or complicated disease.
The tetracyclines, chloramphenicol, and rifampin (bacteriostatic agents) also have been used to treat tularemia. However, these antimicrobials are not as effective as streptomycin or gentamicin and are associated with a high relapse rate. Fluoroquinolones have demonstrated clinical efficacy, but their use in children is restricted. Beta-lactams are not effective. Surgical intervention may be required for suppurative lesions.
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Book2012, Kendig & Chernick's Disorders of the Respiratory Tract in Children (Eighth Edition)L. Barry Seltz MD, ... Leslie L. Barton MD
Related terms:
Nitric Oxide
Chloramphenicol
Ethambutol
Clindamycin
Vancomycin
Doxycycline
Sulforaphane
Antimicrobials
In Vitro
Neutrophil Gelatinase Associated Lipocalin
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2011 AIME I Problems/Problem 6
Contents
[hide]
1 Problem
2 Solution
3 Solution 2
4 Solution 3
5 Solution 4
6 Solution 5 (You don't remember conic section formulae)
7 Video Solution
8 See also
Problem
Suppose that a parabola has vertex and equation , where and is an integer. The minimum possible value of can be written in the form , where and are relatively prime positive integers. Find .
Solution
If the vertex is at , the equation of the parabola can be expressed in the form Expanding, we find that and From the problem, we know that the parabola can be expressed in the form , where is an integer. From the above equation, we can conclude that , , and . Adding up all of these gives us We know that is an integer, so must be divisible by . Let . If , then . Therefore, if , . Adding up gives us
Solution 2
Complete the square. Since , the parabola must be facing upwards. means that must be an integer. The function can be recasted into because the vertex determines the axis of symmetry and the critical value of the parabola. The least integer greater than is . So the -coordinate must change by and the -coordinate must change by . Thus, . So .
Solution 3
To do this, we can use the formula for the minimum (or maximum) value of the coordinate at a vertex of a parabola, and equate this to . Solving, we get . Enter to get so . This means that so the minimum of is when the fraction equals -1, so . Therefore, . -Gideontz
Solution 4
Write this as . Since is equal to the value of this expression when you plug in, we just need to be an integer. Since , we also have which means . The least possible value of is when this is equal to , or , which gives answer .
-bobthegod78, krwang, Simplest14
Solution 5 (You don't remember conic section formulae)
Take the derivative to get that the vertex is at and note that this implies and proceed with any of the solutions above.
~Dhillonr25
Note that the quadratic formula for finding roots of parabolas is , so if you average the two x-values of the roots the part will cancel out and leave you with . Then proceed as above.
~WhatdoHumanitariansEat
Video Solution
See also
2011 AIME I (Problems • Answer Key • Resources)
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11905 | https://m-media.resosir.com/media/study_material/notes/Rigid_Body_Dynamics_71ePJ6P.pdf | Rigid Body Dynamics Corp. / Reg. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.) – 324005 Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in ADVRB - 1 Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029 • Imagination is more important than knowledge • Everything should be made as simple as possible, but not simpler. - Albert Einstein RIGID BODY DYNAMICS ——————————————————————————————————— 1. RIGID BODY : Rigid body is defined as a system of particles in which distance between each pair of particles remains constant (with respect to time). Remember, rigid body is a mathematical concept and any system which satisfies the above condition is said to be rigid as long as it satisfies it. A & B are beads which move on a circular fixed ring If a system is rigid, since there is no change in the distance between any pair of particles of the system, shape and size of system remains constant. Hence we intuitively feel that while a stone or cricket ball are rigid bodies, a balloon or elastic string is non rigid. But any of the above system is rigid as long as relative distance does not change, whether it is a cricket ball or a balloon. But at the moment when the bat hits the cricket ball or if the balloon is squeezed, relative distance changes and now the system behaves like a non-rigid system. For every pair of particles in a rigid body, there is no velocity of separation or approach between the particles. i.e. any relative motion of a point B on a rigid body with respect to another point A on the rigid body will be perpendicular to line joining A to B, hence with respect to any particle A of a rigid body the motion of any other particle B of that rigid body is circular motion. Let velocities of A and B with respect ground be A V and B V respectively in the figure below. A V A V B 2 1 B If the above body is rigid VA cos 1 = VB cos 2 (velocity of approach / separation is zero) VBA = relative velocity of B with respect to A. VBA = VA sin 1 + VB sin 2 (which is perpendicular to line AB) B will appear to move in a circle to an observer fixed at A. Rigid Body Dynamics Corp. / Reg. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.) – 324005 Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in ADVRB - 2 Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029 W.r.t. any point of the rigid body the angular velocity of all other points of the that rigid body is same. Suppose A, B, C is a rigid system hence during any motion sides AB, BC and CA must rotate through the same angle. Hence all the sides rotate by the same rate. From figure (i) angular velocity of A and B w.r.t. C is , From figure (ii) angular velocity of A and C w.r.t. B is , I. Pure Translational Motion : A body is said to be in pure translational motion, if the displacement of each particle of the system is same during any time interval. During such a motion, all the particles have same displacement (s) , velocity (v) and acceleration (a) at an instant. Consider a system of n particle of mass m1, m2, m3, ...... mn under going pure translation then from above definition of translational motion 1 2 3 n a a a ......a = a (say) and 1 2 3 n v v v ......v = v (say) From Newton's laws for a system. ext F = 1 1 m a + 2 2 m a + 3 3 m a + ...................... ext F = M a Where M = Total mass of the body P = 1 1 m v + 2 2 m v + 3 3 m v + ...................... P = Mv Total Kinetic Energy of body = 1 2 m1v12 + m2v22 + ................. = Mv2 Rigid Body Dynamics Corp. / Reg. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.) – 324005 Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in ADVRB - 3 Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029 II. Pure Rotational Motion : Figure shows a rigid body of arbitrary shape in rotation about a fixed axis, called the axis of rotation. Every point of the body moves in a circle whose center lies on the axis of rotation, and every point moves through the same angle during a particular time interval. Such a motion is called pure rotation. We know that each particle has same angular velocity (since the body is rigid.) so, v1 = r1, v2 = r2, v3 = r3 ...... vn = rn Total Kinetic Energy = 1 2 m1v12 + 1 2 m2v22 + ...................... = 1 2 [m1r12 + m2r22 + ..........................] 2 = 1 2 2 Where = m1r12 + m2r22 + ............. (is called moment of inertia) = angular speed of body. III. Combined Translational and Rotational Motion : A body is said to be in combined translation and rotational motion if all point in the body rotates about an axis of rotation and the axis of rotation moves with respect to the ground. Any general motion of a rigid body can be viewed as a combined translational and rotational motion. Example 1. A body is moving down into a well through a rope passing over a fixed pulley of radius 10 cm. Assume that there is no slipping between rope & pulley. Calculate the angular velocity and angular acceleration of the pulley at an instant when the body is going down at a speed of 20 cm/s and has an acceleration of 4.0 m/s2. Solution : Since the rope does not slip on the pulley, the linear speed v of the rim of the pulley is same as the speed of the body. The angular velocity of the pulley is then = v/r = 20cm/ s 10cm = 2 rad/s and the angular acceleration of the pulley is = a/r = 2 4.0 m/ s 10 cm = 40 rad/s2. Example 2. A disc rotates with a uniform angular acceleration of 2.0 rad/s2 about its axis. If the disc starts from rest, how many revolutions will it make in the first 10 seconds? Solution : The angular displacement in the first 10 seconds is given by = 0t + 1 2 t2 = 1 2 (2.0 rad/s2) (10s)2 = 100 rad. As the wheel turns by 2 radian in each revolution, the number of revolutions in 10s is n = 100 2 = 16. Example 3. The wheel of a motor, accelerated uniformly from rest, rotates through 5 radian during the first second. Calculate the angle rotated during the next second. Solution : As the angular acceleration is constant, we have = 0t + 1 2 t2 = 1 2 t2 Thus, 5 rad = 1 2 (1s)2 = 10 rad/s2 or = 10 rad/s2 The angle rotated during the first two seconds is = 1 2 × (10 rad/s2) (2s)2 = 20 rad Thus, the angle rotated during the 2nd second is 20 rad – 5 rad = 15 rad Rigid Body Dynamics Corp. / Reg. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.) – 324005 Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in ADVRB - 4 Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029 Example 4. Starting from rest, a fan takes four seconds to attain the maximum speed of 400 rpm (revolution per minute). Assuming uniform acceleration, calculate the time taken by the fan in attaining half the maximum speed. Solution : Let the angular acceleration be . According to the question, 400 rev/min = 0 + 4 ...........(i) Let t be the time taken in attaining the speed of 200 rev/min which is half the maximum. Then, 200 rev/min = 0 + t ...........(ii) Dividing (i) by (ii), we get, t = 2 sec. Example 5. The motor of an engine is rotating about its axis with an angular velocity of 120 rev/minute. It comes to rest in 10 s, after being switched off the engine. Assuming uniform angular deceleration, find the number of revolutions made by it before coming to rest. Solution : The initial angular velocity = 120 rev/minute = (4) rad/s. Final angular velocity = 0. Time interval = 10 s. Let the angular acceleration be . Using the equation = 0 + t, we obtain = (–4/10) rad/s2 The angle rotated by the motor during this motion is =0t + 1 2 t2 = rad 4 s (10s) – 1 2 2 4 rad 10 s (10s)2 = 20 rad = 10 revolutions. Hence the motor rotates through 10 revolutions before coming to rest. ——————————————————————————————————— 2. MOMENT OF INERTIA (I) ABOUT AN AXIS : (i) Moment of inertia of a system of n particles about an axis is defined as m1 r12 + m2 r22 + ..................+ mn rn2 i.e. = n 2 i i i 1 mr where, ri = It is perpendicular distance of mass mi from axis of rotation S units of Moment of nertia is Kgm2. Moment of inertia is a scalar positive quantity. (ii) For a continuous system : = 2 r (dm) where dm = mass of a small element r = perpendicular distance of the mass element dm from the axis Moment of Inertia depends on : (i) density of the material of body (ii) shape & size of body (iii) axis of rotation In totality we can say that it depends upon distribution of mass relative to axis of rotation. Note : Moment of inertia does not change if the mass : (i) is shifted parallel to the axis of the rotation because ri does not change. (ii) is rotated about axis of rotation in a circular path because ri does not change. Rigid Body Dynamics Corp. / Reg. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.) – 324005 Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in ADVRB - 5 Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029 Example 6. Two particles having masses m1 & m2 are situated in a plane perpendicular to line AB at a distance of r1 and r2 respectively as shown. (i) Find the moment of inertia of the system about axis AB ? (ii) Find the moment of inertia of the system about an axis passing through m1 and perpendicular to the line joining m1 and m2 ? (iii) Find the the moment of inertia of the system about an axis passing through m1 and m2? (iv) Find moment of inertia about axis passing through centre of mass and perpendicular to line joining m1 and m2 Solution : (i) Moment of inertia of particle on left is 1 = m1r12. Moment of Inertia of particle on right is 2 = m2r22. Moment of Inertia of the system about AB is = 1+ 2 = m1r22 + m2r22 (ii) Moment of inertia of particle on left is 1 = 0 Moment of Inertia of particle on right is 2 = m2(r1 + r2)2. Moment of Inertia of the system about AB is = 1 + 2 = 0 + m2 (r1 + r2)2 (iii) Moment of inertia of particle on left is 1 = 0 Moment of Inertia of particle on right is 2 = 0 Moment of Inertia of the system about AB is = 1+ 2 = 0 + 0 (iv) Centre of mass of system rCM = 1 2 2 1 2 r r m m m = Distance of centre mass from mass m1 Distance of centre of mass from mass m2 = 1 2 1 1 2 r r m m m So moment of inertia about centre of mass = cm = 2 1 2 1 2 1 2 r r m m m m + 2 1 2 2 1 1 2 r r m m m m CM = 1 2 1 2 m m m m (r1 + r2)2 Example 7. Four particles each of mass m are kept at the four corners of a square of edge a. Find the moment of inertia of the system about a line perpendicular to the plane of the square and passing through the centre of the square. Solution : The perpendicular distance of every particle from the given line is a/ 2 . The moment of inertia of one particle is, therefore, m(a/ 2 )2 = 1 2 ma2. The moment of inertia of the system is, therefore, 4 × 1 2 ma2 = 2ma2 Rigid Body Dynamics Corp. / Reg. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.) – 324005 Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in ADVRB - 6 Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029 Example 8. Three particles, each of mass m, are situated at the vertices of an equilateral triangle PQR of side a as shown in figure. Calculate the moment of inertia of the system about (i) The line PX perpendicular to PQ in the plane of PQR. (ii) One of the sides of the triangle PQR (iii) About an axis passing through the centroid and perpendicular to plane of the triangle PQR. Solution : (i) Perpendicular distance of P from PX = 0 Perpendicular distance of Q from PX = a Perpendicular distance of R from PX = a/2 Thus, the moment of inertia of the particle at P = 0, of the particle at Q - ma2, and of the particle at R = m(a/2)2. The moment of inertia of the three-particle system about PX is 0 + ma2 + m(a/2)2 = 2 5ma 4 Note that the particles on the axis do not contribute to the moment of inertia. (ii) Moment of inertia about the side PR = mass of particle Q × square of perpendicular distance of Q from side PR, PR = m 2 3 a 2 = 2 3ma 4 (iii) Distance of centroid from all the particle is a 3 , so moment of inertia about an axis and passing through the centroic perpendicular plane of triangle PQR = R = 3m 2 a 3 = ma2 Example 9. Calculate the moment of inertia of a ring having mass M, radius R and having uniform mass distribution about an axis passing through the centre of ring and perpendicular to the plane of ring ? Solution : = 2 (dm) r Because each element is equally distanced from the axis so r = R = 2 2 R dm MR = MR2 (Note : Answer will remain same even if the mass is nonuniformly distributed because dm M always.) Example 10. Calculate the moment of inertia of a uniform rod of mass M and length about an axis 1, 2, 3 and 4. com 1 2 3 4 d Solution (1)= 2 (dm) r = 2 0 M dx x = 2 M 3 (2)= 2 (dm) r = / 2 2 / 2 M dx x = 2 M 12 (3)= 0 (axis 3 passing through the axis of rod) (4)= 2 2 d (dm) Md Rigid Body Dynamics Corp. / Reg. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.) – 324005 Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in ADVRB - 7 Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029 Example 11. Determined the moment of inertia of a uniform rectangular plate of mass, side 'b' and ''' about an axis passing through the edge 'b' and in the plane of plate. Solution : Each section of dm mass rod in the rectangular plate has moment of inertia about an axis passing through edge 'b' d = 2 dm 3 So = 2 d dm 3 = 2 M 3 dm Example 12. Find out the moment of Inertia of figures shown each having mass M, radius R and having uniform mass distribution about an axis passing through the centre and perpendicular to the plane ? (a) x R (b) x R (c) x R Solution : MR2 (infact M.I. of any part of mass M of a ring of radius R about axis passing through geometrical centre and perpendicular to the plane of the ring is = MR2) ——————————————————————————————————— (iii) Moment of inertia of a large object can be calculated by integrating M..of an element of the object: = element d where d = moment of inertia of a small element. Element chosen should be such that : either perpendicular distance of axis from each point of the element is same or the moment of inertia of the element about the axis of rotation is known. Example 13. Determine the moment of Inertia of a uniform disc having mass M, radius R about an axis passing through centre & perpendicular to the plane of disc ? Solution : = ring d element - ring d = dmr2 dm = 2 M R 2rdr (here we have used the uniform mass distribution) = R 2 0 M R . (2rdr).r2 = 2 MR 2 Rigid Body Dynamics Corp. / Reg. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.) – 324005 Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in ADVRB - 8 Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029 Example 14. Calculate the moment of inertia of a uniform hollow cylinder of mass M, radius R and length about its axis. Solution : Moment of inertia of a uniform hollow cylinder is 2 (dm)R = mR2 mass = dm ——————————————————————————————————— 3. TWO IMPORTANT THEOREMS ON MOMENT OF INERTIA (i) Perpendicular Axis Theorem [Only applicable to plane laminar bodies (i.e. for 2 dimensional objects only)]. If axis 1 & 2 are in the plane of the body and perpendicular to each other. Axis 3 in perpendicular to plane of 1 & 2. Then, 3 = 1 + 2 The point of intersection of the three axis need not be center of mass, it can be any point in the plane of body which lies on the body or even outside it. Example 15. Calculate the moment of inertia of a uniform disc of mass M and radius R about a diameter. Solution : Let AB and CD be two mutually perpendicular diameters of the disc. Take them as X and Y-axes and the line perpendicular to the plane of the disc through the centre as the Z-axis. The moment of inertia of the ring about the Z-axis is = 1 2 MR2. As the disc is uniform, all of its diameters are equivalent and so x = y, From perpendicular axes theorem, z = x + y. Hence x = z 2 = 2 MR 4 Rigid Body Dynamics Corp. / Reg. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.) – 324005 Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in ADVRB - 9 Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029 Example 16. Two uniform identical rods each of mass M and lengthare joined to form a cross as shown in figure. Find the moment of inertia of the cross about a bisector as shown dotted in the figure. Solution : Consider the line perpendicular to the plane of the figure through the centre of the cross. The moment of inertia of each rod about this line is 2 M 12 and hence the moment of inertia of the cross is 2 M 6 . The moment of inertia of the cross about the two bisector are equal by symmetry and according to the theorem of perpendicular axes, the moment of inertia of the cross about the bisector is 2 M 12 . Example 17. In the figure shown find moment of inertia of a plate having mass M, length and width b about axis 1,2,3 and 4. Assume that mass is uniformly distributed. 1 3 4 2 b C Solution : Moment of inertia of the plate about axis 1 (by taking rods perpendicular to axis 1) I1 = Mb2 / 3 Moment of inertia of the plate about axis 2 (by taking rods perpendicular to axis 2) I2 = M2 / 12 Moment of inertia of the plate about axis 3 (by taking rods perpendicular to axis 3) I3 = Mb2 / 12 Moment of inertia of the plate about axis 4 (by taking rods perpendicular to axis 4) I4 = M2 / 3 Example 18. In the figure shown find the moment of inertia of square plate having mass m and sides a. About an axis 2 passing through point C (centre of mass) and in the plane of plate. I ' I ' I a a 1 2 3 4 C Solution : Using perpendicular axis theorems IC = I4 + I2 = 2I' Using perpendicular theorems C = 3 + 1 = + = 2 2I' = 2I I' = I IC = 2I = 2 ma 6 I' = 2 ma 12 Rigid Body Dynamics Corp. / Reg. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.) – 324005 Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in ADVRB - 10 Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029 Example 19. Find the moment of Inertia of a uniform disc of mass M and radius R about a diameter. Solution : Consider x & y two mutually perpendicular diameters of the ring. x + y = z x = y (due to symmetry) z = 2 MR 2 x = y = 2 MR 4 ——————————————————————————————————— (ii) Parallel Axis Theorem (Applicable to planer as well as 3 dimensional objects): If = Moment of Inertia of the object about axis AB cm = Moment of Inertia of the object about an axis passing through centre of mass and parallel to axis AB M = Total mass of object d = perpendicular distance between axis AB about which moment of nertia is to be calculated & the one passing through the centre of mass and parallel to it. = cm + Md2 Example 20. Find out relation between I1 and I2. I1 and I2 moment of inertia of a rigid body mass m about an axis as shown in figure. a COM b I 1 I 2 Solution : Using parallel axis theorem I1 = IC + ma2 .....(1) I2 = IC + mb2 .....(2) From (1) and (2) ; I1 – I2 = m(a2 – b2) Example 21. Find the moment of inertia of a uniform sphere of mass m and radius R about a tangent if the spheres (i) solid (ii) hollow Solution : (i) Using parallel axis theorem I = ICM + md2 for solid sphere ICM = 2 5 mR2, d = R I = 7 5 mR2 (ii) Using parallel axis theorem I = ICM + md2 for hollow sphere ICM = 2 3 mR2, d = R I = 5 3 mR2 R R solid sphere hollow sphere Rigid Body Dynamics Corp. / Reg. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.) – 324005 Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in ADVRB - 11 Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029 Example 22. Calculate the moment of inertia of a hollow cylinder of mass M and radius R about a line parallel to the axis of the cylinder and on the surface of the cylinder. Solution : The moment of inertia of the cylinder about its axis = MR2. Using parallel axes theorem, = 0 + MR2 = MR2 + MR2 = 2 MR2. Similarly, the moment of inertia of a hollow sphere about a tangent is 2 3 MR2 + MR2 = 5 3 MR2 Example 23. Find out the moment of inertia of a semi circular disc about an axis passing through its centre of mass and perpendicular to the plane? Solution : Moment of inertia of a semi circular disc about an axis passing through centre and perpendicular to plane of disc, I = 2 MR 2 Using parallel axis theorem I = ICM + Md2, d is the perpendicular distance between two parallel axis passing through centre C and COM. I = 2 MR 2 , d = 4R 3 2 MR 2 = ICM + M 2 4R 3 ICM = 2 2 MR 4R M 2 3 Example 24. Find the moment of inertia of the two uniform joint rods having mass m each about point P as shown in figure. Using parallel axis theorem. × P Solution : Moment of inertia of rod 1 about axis P , I1 = 2 m 3 Moment of inertia of rod 2 about axis P, I2 = 2 m 12 + 2 m 5 2 So moment of inertia of a system about axis P , I = I1 + I2 = 2 m 3 + 2 m 12 + 2 m 5 2 I = 2 m 3 P COM 1 2 5 ——————————————————————————————————— List of some useful formule : Object Moment of Inertia Solid Sphere 2 2 MR 5 (Uniform) Rigid Body Dynamics Corp. / Reg. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.) – 324005 Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in ADVRB - 12 Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029 Hollow Sphere 2 2 MR 3 (Uniform) Ring. MR2 (Uniform or Non Uniform) Disc 2 MR 2 (Uniform) Hollow cylinder MR2 (Uniform or Non Uniform) Solid cylinder 2 MR 2 (Uniform) Thin rod 2 ML 3 (Uniform) 2 ML 12 (Uniform) Rigid Body Dynamics Corp. / Reg. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.) – 324005 Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in ADVRB - 13 Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029 Two thin rod 2 2m 3 (Uniform) Rectangular Plate = 2 2 M(a b ) 12 (Uniform) Square Plate AB = CD = DF = 2 Ma 12 (Uniform) Square Plate 2 Ma 6 (Uniform) Cuboid 2 2 M(a b ) 12 (Uniform) 4. RADIUS OF GYRATION : As a measure of the way in which the mass of rigid body is distributed with respect to the axis of rotation, we define a new parameter, the radius of gyration (K). It is related to the moment of inertia and total mass of the body. = MK2 where = Moment of Inertia of a body M = Mass of a body K = Radius of gyration K = M Length K is the geometrical property of the body and axis of rotation. S.. Unit of K is meter. Rigid Body Dynamics Corp. / Reg. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.) – 324005 Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in ADVRB - 14 Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029 Example 25. Find the radius of gyration of a solid uniform sphere of radius R about its tangent. Solution : = 2 2 2 mR mR 5 = 2 7 mR 5 = mK2 K = 7R 5 Example 26. Find the radius of gyration of a hollow uniform sphere of radius R about its tangent. Solution : Moment of inertia of a hollow sphere about a tangent, I = 5 3 MR2 MK2 = 5 3 MR2 K = 5R 3 ——————————————————————————————————— 5. MOMENT OF INERTIA OF BODIES WITH CUT : Example 27 A uniform disc of radius R has a round disc of radius R/3 cut as shown in Fig. .The mass of the remaining (shaded) portion of the disc equals M. Find the moment of inertia of such a disc relative to the axis passing through geometrical centre of original disc and perpendicular to the plane of the disc. Solution : Let the mass per unit area of the material of disc be . Now the empty space can be considered as having density – and . Now 0 = + – = ( R2)R2/2 = M.I. of about o – = 2 2 (R/3) (R/3) 2 + [–(R/3)2] (2R/3)2 = M.I. of – about o 0 = 4 9 MR2 Ans. Example 28. Find the moment of inertia of a uniform disc of radius R1 having an empty symmetric annular region of radius R2 in between, about an axis passing through geometrical centre and perpendicular to the disc. Solution : = 2 2 1 2 M (R R ) = × 4 4 1 2 R R 2 = 2 2 1 2 M R R 2 Ans. Rigid Body Dynamics Corp. / Reg. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.) – 324005 Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in ADVRB - 15 Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029 ——————————————————————————————————— 6. TORQUE : Torque represents the capability of a force to produce change in the rotational motion of the body. 6.1 Torque about a point : Torque of force F about a point r F Where F = force applied P = point of application of force Q = Point about which we want to calculate the torque. r = position vector of the point of application of force w.r.t. the point about which we want to determine the torque. = r F sin = rF = rF Where = angle between the direction of force and the position vector of P wrt. Q. r = r sin = perpendicular distance of line of action of force from point Q ,it is also called force arm. F = F sin = component of F perpendicular to r S unit of torque is Nm Torque is a vector quantity and its direction is determined using right hand thumb rule and its always perpendicular to the plane of rotation of the body. Example 29. A particle of mass M is released in vertical plane from a point P at x = x0 on the x-axis it falls vertically along the y-axis. Find the torque acting on the particle at a time t about origin ? Solution : Torque is produced by the force of gravity. = r F sin ˆ k or = 0 r F x mg = r mg 0 x r = mgx0 ˆ k Rigid Body Dynamics Corp. / Reg. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.) – 324005 Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in ADVRB - 16 Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029 Example 30. A particle having mass m is projected with a velocity v0 from a point P on a horizontal ground making an angle with horizontal. Find out the torque about the point of projection acting on the particle when it is at its maximum height ? P Q V 0 Solution : = rFsin = R 2 mg = 2 0 v sin2 2g mg = 2 0 mv sin2 2 Example 31. Find the torque about point O and A. O A B (1,1) 30° 60° x y Solution : Torque about point O, 0 0 ˆ ˆ r F, r i j , ˆ ˆ F 5 3i 5j ˆ ˆ (i j) × ( ˆ ˆ 5 3i 5j ) = ˆ 5(1 3)k Torque about point A , a a ˆ r F, r j , ˆ ˆ F 5 3i 5j ˆ j × ( ˆ ˆ 5 3i 5j ) = ˆ 5( 3)k Example 32. Find out torque about point A, O and B. O x B A y (0,5) (3,5) (3,0) Solution : Torque about point A , A A A ˆ ˆ r F, r 3i, F 10i A ˆ ˆ 3i 10i 0 Torque about point B , B B B ˆ ˆ r F , r 5j , F 10i B ˆ ˆ ˆ 5j 10i –50k Torque about point O , O O O ˆ ˆ ˆ r F , r 3i 5j , F 10i O ˆ ˆ ˆ ˆ (3i 5j) 10i –50k Rigid Body Dynamics Corp. / Reg. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.) – 324005 Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in ADVRB - 17 Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029 ——————————————————————————————————— 6.2 Torque about an axis : The torque of a force F about an axis AB is defined as the component of torque of F about any point O on the axis AB, along the axis AB. In the given figure torque of F about O is 0 r F The torque of F about AB, AB is component of 0 along line AB. There are four cases of torque of a force about an axis.: Case I : Force is parallel to the axis of rotation, F || AB AB is the axis of rotation about which torque is required r F is perpendicular to F , but F || AB, hence r F is perpendicular to AB . The component of r F along AB is, therefore, zero. Case II : The line of force intersects the axis of rotation (F intersect AB) F intersects AB along r then F and r are along the same line. The torque about O is r F = 0. Hence component this torque along line AB is also zero. Case III : F perpendicular to AB but F and AB do not intersect. In the three dimensions, two lines may be perpendicular without intersecting each other. Two nonparallel and nonintersecting lines are called skew lines. Figure shows the plane through the point of application of force P that is perpendicular to the axis of rotation AB. Suppose the plane intersects the axis at the point O. The force F is in this plane (since F is perpendicular to AB). Taking the origin at O, Torque = r F = OP × F . Thus, torque = rF sin = F(OS) where OS is the perpendicular from O to the line of action of the force F . The line OS is also perpendicular to the axis of rotation. It is thus the length of the common perpendicular to the force and the axis of rotation. The direction of = OP × F is along the axis AB because ABOP and AB F . The torque about AB is, therefore, equal to the magnitude of that is F.(OS). Rigid Body Dynamics Corp. / Reg. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.) – 324005 Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in ADVRB - 18 Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029 Thus, the torque of F about AB = magnitude of the force F × length of the common perpendicular to the force and the axis. The common perpendicular OS is called the lever arm or moment arm of this torque. Case IV : F and AB are skew but not perpendicular. Here we resolve F into two components, one is parallel to axis and other is perpendicular to axis. Torque of the parallel part is zero and that of the perpendicular part may be found, by using the result of case (III). Example 33. Find the torque of weight about the axis passing through point P. Solution : r F, r R, F mgsin r and F both are at perpendicular so torque about point P = mgRsin Example 34. A bob of mass m is suspended at point O by string of length . Bob is moving in a horizontal circle find out (i) torque of gravity and tension about point O and O'. (ii) Net torque about axis OO' . Solution : (i) Torque about point O Torque of tension (T), ten = 0 (tension is passing through point O) Torque of gravity mg = mgsin Torque about point O' Torque of gravity mg = mgr r = sin Torque of tension mg = mgsin (along negative ˆ j ) Torque of tension ten = Trsin(90+ ) (Tcos = mg) ten = Trcos ten = mg cos (sin ) cos = mg sin (along positive ˆ j ) (ii) Torque about axis OO' Torque of gravity about axis OO' mg = 0 (force mg parallel to axis OO') Torque of tension about axis OO' ten = 0 (force T is passing through the axis OO') Net torque about axis OO' net = 0 ——————————————————————————————————— Rigid Body Dynamics Corp. / Reg. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.) – 324005 Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in ADVRB - 19 Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029 6.3 Force Couple : A pair of forces each of same magnitude and acting in opposite direction is called a force couple. Torque due to couple = Magnitude of one force × distance between their lines of action. Magnitude of torque = = F (2d) A couple does not exert a net force on an object even though it exerts a torque. Net torque due to a force couple is same about any point. Torque about A = x1F + x2F = F(x1 + x2) = Fd Torque about B = y1F – y2F = F(y1 – y2) = Fd If net force acting on a system is zero, torque is same about any point. A consequence is that, if Fnet = 0 and net = 0 about one point, then net = 0 about any point. ——————————————————————————————————— 6.4 Point of Application of Force : Point of Application of force is the point at which, if net force is assumed to be acting, then it will produce same translational as well as rotational effect, as was produced earlier. We can also define point of application of force as a point about which torque of all the forces is zero. = Consider three forces 1 2 3 F ,F ,F acting on a body if D is point of application of force then torque of 1 2 3 F F F acting at a point D about O is same as the original torque about O 1 1 2 2 3 3 r F r F r F = 1 2 3 r (F F F ) Rigid Body Dynamics Corp. / Reg. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.) – 324005 Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in ADVRB - 20 Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029 Example 35. Determine the point of application of force, when forces of 20 N & 30 N are acting on the rod as shown in figure. Solution : Net force acting on the rod Frel = 10N Nett torque acting on the rod about point C c = (20 × 0) + ( 30 × 20) = 600 clockwise Let the point of application be at a distance x from point C 600 = 10 x x = 60 cm 70 cm from A is point of Application Note : (i) Point of application of gravitational force is known as the centre of gravity. (ii) Centre of gravity coincides with the centre of mass if value of is assumed to be constant. (iii) Concept of point of application of force is imaginary, as in some cases it can lie outside the body. ——————————————————————————————————— 6.5 Rotation about a fixed axis : If Hinge = moment of inertia about the axis of rotation (since this axis passes through the hinge, hence the name Hinge ). ext = resultant external torque acting on the body about axis of rotation = angular acceleration of the body. ext Hinge = Hinge Rotational Kinetic Energy = 2 1. .
2 CM P Mv external CM F M a x Hinge Fixed axis of Rotation Net external force acting on the body has two component tangential and centripetal. FC = maC = 2 CM v m r = m2 rCM Ft = mat = m rCM Example 36. A pulley having radius r and moment of inertia about its axis is fixed at the top of an inclined plane of inclination as shown in figure. A string is wrapped round the pulley and its free end supports a block of mass m which can slide on the plane. Initially, the pulley is rotating at a speed 0 in a direction such that the block slides up the plane. Calculate the distance moved by the block before stopping ? Rigid Body Dynamics Corp. / Reg. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.) – 324005 Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in ADVRB - 21 Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029 Solution : Suppose the deceleration of the block is a. The linear deceleration of the rim of the pulley is also a. The angular deceleration of the pulley is = a/r. If the tension in the string is T, the equations of motion are as follows : mg sin – T = ma and Tr = = a/r. Eliminating T from these equations, mg sin – 2 a r = ma giving, a = 2 2 mg r sin mr The initial velocity of the block up the incline is v = 0 r. Thus, the distance moved by the block before stopping is x = 2 v 2a = 2 2 2 0 2 r ( mr ) 2m r sin = 2 2 0 ( mr ) 2m gsin Example 37. The pulley shown in figure has a moment of inertia about its axis and its radius is r. Calculate the magnitude of the acceleration of the two blocks. Assume that the string is light and does not slip on the pulley. Solution : Suppose the tension in the left string is T1 and that in the right string in T2. Suppose the block of mass m1 goes down with an acceleration and the other block moves up with the same acceleration. This is also the tangential acceleration of the rim of the wheel as the string does not slip over the rim. The angular acceleration of the wheel is, therefore, = a/r. The equations of motion for the mass m1, the mass m2 and the pulley are as follows : m1g – T1 = m1a ......(i) T2 – m2g = m2a ......(ii) T1r – T2r = I = I /r ......(iii) Putting T1 and T2 from (i) and (ii) into (iii), [(m1g – a) – m2 (g + a)] r = a r which gives a = 3 1 2 2 1 2 (m m )gr (m m )r . Example 38. A uniform rod of mass m and length can rotate in vertical plane about a smooth horizontal axis hinged at point H. (i) Find angular acceleration of the rod just after it is released from initial horizontal position from rest ? (ii) Calculate the acceleration (tangential and radial) of point A at this moment. (iii) Calculate net hinge force acting at this moment. (iv) Find and when rod becomes vertical. (v) Find hinge force when rod become vertical. Solution : (i) H = IH Rigid Body Dynamics Corp. / Reg. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.) – 324005 Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in ADVRB - 22 Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029 mg. 2 = 2 m 3 = 3g 2 (ii) atA = = 3g 2 . = 3g 2 aCA = 2 r = 0. = 0 ( = 0 just after release) (iii) Suppose hinge exerts normal reaction in component form as shown In vertical direction Fext = maCM mg – N1 = m. 3g 4 (we get the value of aCM from previous example) N1 = mg 4 In horizontal direction Fext = maCM N2 = 0 (aCM in horizontal = 0 as = 0 just after release). (vi) Torque = 0 when rod becomes vertical. so = 0 using energy conservation 2 mg 1 2 2 2 m 3 = 3g (v) When rod becomes vertical = 0, = 3g FH – mg = 2 m 2 FH = 5mg 2 Ans. ——————————————————————————————————— 7. EQUILIBRIUM A system is in mechanical equilibrium if it is in translational as well as rotational equilibrium. For this : net F 0 net 0 (about every point) X Y F 1 F 2 F 3 F 4 F 5 From (6.3), if net F 0 then net is same about every point Rigid Body Dynamics Corp. / Reg. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.) – 324005 Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in ADVRB - 23 Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029 Hence necessary and sufficient condition for equilibrium is net F 0 , net 0 about any one point, which we can choose as per our convenience. ( net will automatically be zero about every point) unstable equilibrium stable equilibrium Neutral equilibrium The equilibrium of a body is called stable if the body tries to regain its equilibrium position after being slightly displaced and released. It is called unstable if it gets further displaced after being slightly displaced and released. If it can stay in equilibrium even after being slightly displaced and released, it is said to be in neutral equilibrium. Example 39. Two boys weighing 20 kg and 25 kg are trying to balance a seesaw of total length 4m, with the fulcrum at the centre. If one of the boys is sitting at an end, where should the other sit ? Solution : It is clear that the 20 kg kid should sit at the end and the 25 kg kid should sit closer to the centre. Suppose his distance from the centre is x. As the boys are in equilibrium, the normal force between a boy and the seesaw equals the weight of that boy. Considering the rotational equilibrium of the seesaw, the torque of the forces acting on it should add to zero. The forces are (a) (25 kg) g downward by the 25 kg boy, (b) (20 kg) g downward by the 20 kg boy, (c) weight of the seesaw and (d) the normal force by the fulcrum. Taking torques about the fulcrum, (25 kg)g x = (20 kg)g (2 m) or x = 1.6 m. Example 40. A uniform rod of mass m = 15 kg leans against a smooth vertical wall making an angle = 37° with horizontal. The other ends rests on a rough horizontal floor. Calculate the normal force and the friction force that the floor exerts on the rod. [Take g = 10 m/s2] Solution : The forces acting on the rod are shown in figure. They are (a) Its weight W, (b) normal force N1 by the vertical wall, (c) normal force N2 by the floor and (d) frictional force f by the floor. Taking horizontal and vertical components, N1 = f ..........(i) and N2 = mg ..........(ii) Taking torque about B, N1(AO) = mg(CB) Rigid Body Dynamics Corp. / Reg. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.) – 324005 Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in ADVRB - 24 Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029 or, N1(AB) cos = mg AB 2 sin or N1 3 5 = W 2 4 5 or, N1 = 2 3 W ..........(iii) The normal force by the floor is N2 = W = (15 kg) (10 m/s2) = 150 N. The frictional force is f = N1 = 2 3 W = 100 N. Example 41. The ladder shown in figure has negligible mass and rests on a smooth floor. A crossbar connects the two legs of the ladder at the centre as shown in figure. The angle between the two legs is 90°. The person sitting on the ladder has a mass of 60 kg. Calculate the contact forces exerted by the floor on each leg and the tension in the crossbar. [Take g = 10m/s2] Solution : The forces acting on different parts are shown in figure. Consider the vertical equilibrium of “the ladder plus the person” system. The forces acting on this system are its weight (60 kg)g and the contact force Ny + Ny = 2Ny due to the floor. Thus 2Ny = (60 kg) g or Ny = (30 kg) (10 m/s2) = 300 N. Next consider the equilibrium of the left leg of the ladder. Taking torques of the forces acting on it about the upper end, Ny (2m) tan 45° = T(1 m) or T = Ny 2 = (300 N) × 2 = 600 N. ——————————————————————————————————— 8. ANGULAR MOMENTUM (L ) 8. 1. Angular momentum of a particle about a point. L = r P L = rpsin or L = r × P or L = P× r Where P = momentum of particle r = position of vector of particle with respect to point O about which angular momentum is to be calculated. = angle between vectors r & P r= perpendicular distance of line of motion of particle from point O. P= component of momentum perpendicular to r . SI unit of angular momentum is kgm2/sec. Example 42. A particle of mass m is projected at time t = 0 from a point O with a speed u at an angle of 45° to the horizontal. Calculate the magnitude and the direction of the angular momentum of the particle about the point O at time t = u/g. Solution : Let us take the origin at P, X-axis along the horizontal and Y-axis along the vertically upward direction as shown in figure. For horizontal motion during the time 0 to t, vx = u cos 45° = u/ 2 and x = vxt = u 2 . u g = 2 u 2g Rigid Body Dynamics Corp. / Reg. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.) – 324005 Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in ADVRB - 25 Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029 For vertical motion, vy = u sin 45° = u 2 – u = (1 2) 2 u and y = (u sin 45°) t – 1 2 gt2 = 2 u 2g – 2 u 2g = 2 u 2g ( 2 – 1). The angular momentum of the particle at time t about the origin is L = r × p = m r × v = m( ˆ i x + ˆ j y) × ( ˆ i vx + ˆ j vy) = m ( ˆ k xvy – ˆ k yvx) = m ˆ k 2 2 2 u u u u (1 2) ( 2 1) 2g 2 g 2 2 = – ˆ k 3 mu 2 2g . Thus, the angular momentum of the particle is 3 mu 2 2g in the negative Z-direction i.e., perpendicular to the plane of motion, going into the plane. Example 43. A particle of mass 'm' starts moving from point (o,d) with a constant velocity u ˆ i . Find out its angular momentum about origin at this moment what will be the answer at the later time? Solution : L = – m d u ˆ k . Example 44. A particle of mass 'm' is projected on horizontal ground with an initial velocity of u making an angle with horizontal. Find out the angular momentum of particle about the point of projection when . (i) it just starts its motion (ii) it is at highest point of path. (iii) it just strikes the ground. Answer : (i) O ; (ii) mu cos 2 2 u sin 2g ; (iii) mu sin 2 u sin2 g Solution : (i) Angular momentum about point O is zero. (ii) Angular momentum about point A. L r p L = H × mu cos L = mu cos 2 2 u sin 2g Ans. (iii) Angular momentum about point B. L = R × mu sin mu sin 2 u sin2 g Ans. usin ucos H R O A B Rigid Body Dynamics Corp. / Reg. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.) – 324005 Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in ADVRB - 26 Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029 Example 45. A particle of mass'm' is projected on horizontal ground with an initial velocity of u making an angle with horizontal. Find out the angular momentum at any time t of particle p about : (i) y axis (ii) z-axis Solution : (i) velocity components are parallel to the y-axis. so, L = 0 (ii) = dL dt – 1/2 mu cos . gt2 – mgx = dL dt – mgx dt = dL t L 0 0 –mgx dt dL angular momentum about the z-axis is : L = – 1/2 mu cos . gt2 Ans. usingt ucos z ——————————————————————————————————— 8.2 Angular momentum of a rigid body rotating about fixed axis : Angular momentum of a rigid body about the fixed axis AB is LAB = L1 + L2 + L3 +....... +Ln L1 = m1 r1r1 , L2 = m2 r2r2 , L3 = m3 r3r3 , Ln = mn rnrn LAB = m1 r1r1 + m2 r2r2 + m3 r3r3 ........ + mn rnrn LAB = n n 2 n n n 1 m (r ) × n n 2 n n H n 1 m (r ) LAB = H LH = IH LH = Angular momentum of object about axis of rotation. IH = Moment of Inertia of rigid, body about axis of rotation. = angular velocity of the object. r 1 r 2 r n A B Example 46. Two particles balls A and B, each of mass m, are attached rigidly to the ends of a light rod of length . The system rotates about the perpendicular bisector of the rod at an angular speed . Calculate the angular momentum of the individual particles and of the system about the axis of rotation. Solution : Consider the situation shown in figure. The velocity of the particle A with respect to the centre O is v = 2 . The angular momentum of the particle with respect to the axis is L1 = mvr = m 2 2 = 1 4 m2. The same the angular momentum L2 of the second particle. The angular momentum of the system is equal to sum of these two angular momentum i.e., L = 1/2 m2. Rigid Body Dynamics Corp. / Reg. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.) – 324005 Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in ADVRB - 27 Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029 Example 47. Two small balls of mass m each are attached to a light rod of length , one at its centre and the other at a free end. The rod is fixed at the other end and is rotated in horizontal plane at an angular speed . Calculate the angular momentum of the ball at the end with respect to the ball at the centre. Solution : The situation is shown in figure. The velocity of the ball A with respect to the fixed end O is vA = (/2) and that of B with respect to O is vB = . Hence the velocity of B with respect to A is vB – vA = (/2). The angular momentum of B with respect to A is, therefore, L = mvr = m 2 2 = 1 4 m2 along the direction perpendicular to the plane of rotation. Example 48. A uniform circular ring of mass 400 g and radius 10 cm is rotated about one of its diameter at an angular speed of 20 rad/s. Find the kinetic energy of the ring and its angular momentum about the axis of rotation. Solution : The moment of inertia of the circular ring about its diameter is = 1 2 Mr2 = 1 2 (0.400 kg) (0.10 m)2 = 2 × 10–3 kg-m2. The kinetic energy is K = 2 = (2 × 10–3 kg - m2) (400 rad2/s2) = 0.4 J and the angular momentum about the axis of rotation is L = = (2 × 10–3 kg-m2) (20 rad/s) = 0.04 kg-m2/s = 0.04 J-s. ——————————————————————————————————— 8.3 Conservation of Angular Momentum Newton's 2nd law in rotation : dL dt where and L are about the same axis. Angular momentum of a particle or a system remains constant if ext = 0 about the axis of rotation. Even if net angular momentum is not constant, one of its component of an angular momentum about an axis remains constant if component of torque about that axis is zero Impulse of Torque : dt J J Charge in angular momentum. Example 49. A uniform rod of mass m and length can rotate freely on a smooth horizontal plane about a vertical axis hinged at point H. A point mass having same mass m coming with an initial speed u perpendicular to the rod, strikes the rod in-elastically at its free end. Find out the angular velocity of the rod just after collision ? Solution : Angular momentum is conserved about H because no external force is present in horizontal plane which is producing torque about H. Rigid Body Dynamics Corp. / Reg. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.) – 324005 Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in ADVRB - 28 Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029 mul = 2 2 m m 3 = 3a 4 Example 50. A uniform rod of mass m1 and length lies on a frictionless horizontal plane. A particle of mass m2 moving at a speed v0 perpendicular to the length of the rod strikes it at a distance /3 from the centre and stops after the collision. Calculate (a) the velocity of the centre of the rod and (b) the angular velocity of the rod about its centre just after the collision. Solution : The situation is shown in figure. Consider the rod and the particle together as the system. As there is no external resultant force, the linear momentum of the system will remains constant. Also there is no resultant external torque on the system and so the angular momentum of the system about the any line will remain constant. Suppose the velocity of the centre of the rod is V and the angular velocity about the centre is . A r 0 ® V w (b) (a) The linear momentum before the collision is mv and that after the collision is MV. Thus, m2v0 = m1V, or V = 2 1 m m v0 (b) Let A be the centre of the rod when it is at rest. Let AB be the line perpendicular to the plane of the figure. Consider the angular momentum of “the rod plus the particle” system about AB. Initially the rod is at rest. The angular momentum of the particle about AB is L = m2v0 (/3) After the collision, the particle comes to rest. The angular momentum of the rod about A is L = cm L + m1 0 r × V As 0 r || V , 0 r × V = 0 Thus, L = cm L Hence the angular momentum of the rod about AB is L = = 2 1 m 12 .Thus, 2 m v 3 = 2 1 m 12 or, = 2 0 1 4m v m ——————————————————————————————————— 9. COMBINED TRANSLATIONAL AND ROTATIONAL MOTION OF A RIGID BODY : The general motion of a rigid body can be thought of as a sum of two independent motions. A translation of some point of the body plus a rotation about this point. A most convenient choice of the point is the centre of mass of the body as it greatly simplifies the calculations. Consider a fan inside a train, and an observer A on the platform. It the fan is switched off while the train moves, the motion of fan is pure translation as each point on the fan undergoes same translation in any time interval. It fan is switched on while the train is at rest the motion of fan is pure rotation about is axle ; as each point on the axle is at rest, while other points revolve about it with equal angular velocity. Rigid Body Dynamics Corp. / Reg. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.) – 324005 Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in ADVRB - 29 Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029 If the fan is switched on while the train is moving, the motion of fan to the observer on the platform is neither pure translation nor pure rotation. This motion is an example of general motion of a rigid body. Now if there is an observer B inside the train, the motion of fan will appear to him as pure rotation. Hence we can see that the general motion of fan w.r.t. observer A can be resolved into pure rotation of fan as observed by observer B plus pure translation of observer B (w.r.t. observer A) Such a resolution of general motion of a rigid body into pure rotation & pure translation is not restricted to just the fan inside the train, but is possible for motion of any rigid system. 9.1 Kinematics of general motion of a rigid body : For a rigid body as earlier stated value of angular displacement (), angular velocity (), angular acceleration () is same for all points on the rigid body about any other point on the rigid body. Hence if we know velocity of any one point (say A) on the rigid body and angular velocity of any point on the rigid body about any other point on the rigid body (say ), velocity of each point on the rigid body can be calculated. since distance AB is fixed BA V AB we know that = BA BA V r VBA = VBA = rBA in vector form BA V = BA r Now from relative velocity : BA B A V V V B A BA V V V B A BA V V r similarly B A BA a a r [for any rigid system] A V V + A A V B V rigid body A B A B r V= BA V BA = Example 51. Consider the general motion of a wheel (radius r) which can be view on pure translation of its center O (with the velocity v) and pure rotation about O (with angular velocity ) C D A O B V ) k ˆ ( i ˆ j ˆ Find out AO BO CO DO v ,v ,v ,v and A B C D v , v ,v ,v Solution : AO v = AO r AO v = ˆ (–k) OA AO v = ˆ ˆ (–k) r( j) AO v = ˆ ri similarly BO v = r ˆ ( j) CO v = r ˆ (i) ; DO v =r ˆ (j) A v = O AO ˆ ˆ v v vi ri Similarly B v = O BO ˆ ˆ v v vi rj C v = O CO ˆ ˆ v v vi ri D v = O DO ˆ ˆ v v vi rj ——————————————————————————————————— Rigid Body Dynamics Corp. / Reg. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.) – 324005 Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in ADVRB - 30 Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029 9.2 Pure Rolling (or rolling without sliding) : Pure rolling is a special case of general rotation of a rigid body with circular cross section (e.g. wheel, disc, ring, sphere) moving on some surface. Here, there is no relative motion between the rolling body and the surface of contact, at the point of contact A Here contact point is A & contact surface is horizontal ground. For pure rolling velocity of A w.r.t. ground = 0 VA = 0 A V From above figure, for pure rolling, velocity of A w.r.t. to plank is zero VA =V. A v r From above figure for, pure rolling, velocity of A w.r.t. ground is zero. v –r = 0 v = r Similarly a = r Example 52. A wheel of radius r rolls (rolling without sleeping) on a level road as shown in figure. A B v r Find out velocity of point A and B Solution : Contact surface is in rest for pure rolling velocity of point A is zero. so v = r velocity of point B = v + r = 2v ——————————————————————————————————— 9.3 Dynamics of general motion of a rigid body : This motion can be viewed as translation of centre of mass and rotation about an axis passing through centre of mass If CM = Moment of inertia about this axis passing through COM cm = Net torque about this axis passing through COM CM a = Acceleration of COM CM v = Velocity of COM ext F = Net external force acting on the system. system P = Linear momentum of system. CM L = Angular momentum about centre of mass. CM r = Position vector of COM w.r.t. point A. Rigid Body Dynamics Corp. / Reg. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.) – 324005 Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in ADVRB - 31 Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029 then (i) cm cm (ii) ext cm F Ma (iii) system cm P Mv (vi) Total K.E.= 2 cm 1Mv 2 + 2 cm 1 2 (v) CM CM L (vi) Angular momentum about point A = L about C.M. + L of C.M. about A A cm cm cm L r Mv A dL d dt dt ( cm cm cm r Mv ) A d dt . Notice that torque equation can be applied to a rigid body in a general motion only and only about an axis through centre of mass. Example 53. A uniform sphere of mass 200 g rolls without slipping on a plane surface so that its centre moves at a speed of 2.00 cm/s. Find its kinetic energy. Solution : As the sphere rolls without slipping on the plane surface, its angular speed about the centre is = cm v r . The kinetic energy is K = 1 2 cm2 + 1 2 Mvcm2 = 1 2 . 2 5 Mr22 + 1 2 Mvcm2 = 1 5 Mvcm2 + 1 2 Mvcm2 = 7 10 Mvcm2 = 7 10 (0.200 kg) (0.02 m/s)2 = 5.6 × 10–5 J. Example 54. A constant force F acts tangentially at the highest point of a uniform disc of mass m kept on a rough horizontal surface as shown in figure. If the disc rolls without slipping, calculate the acceleration of the centre (C) and point A and B of the disc. A C B F r rough surface Solution : The situation is shown in figure. As the force F rotates the disc, the point of contact has a tendency to slip towards left so that the static friction on the disc will act towards right. Let r be the radius of the disc and a be the linear acceleration of the centre of the disc. The angular acceleration about the centre of the disc is = a/r, as there is no slipping. For the linear motion of the centre, F + f = ma ......(i) and for the rotational motion about the centre, Fr – f r = = 2 1mr 2 a r or, F – f = 1 2 ma ......(ii) From (i) and (ii), 2F = 3 2 ma or a = 4F 3m Acceleration of point A is zero. Acceleration of point B is 2a = 2 4F 3m = 8F 3m Ans. Rigid Body Dynamics Corp. / Reg. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.) – 324005 Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in ADVRB - 32 Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029 Example 55. A circular rigid body of mass m, radius R and radius of gyration (k) rolls without slipping on an inclined plane of a inclination . Find the linear acceleration of the rigid body and force of friction on it. What must be the minimum value of coefficient of friction so that rigid body may roll without sliding? Solution : If a is the acceleration of the centre of mass of the rigid body and f the force of friction between sphere and the plane, the equation of translatory and rotatory motion of the rigid body will be. mg sin – f = ma (Translatory motion) fR = (Rotatory motion) f = R = mk2 , due to pure rolling a = R mg sin – R = mR mg sin = m R + R mg sin = m R + 2 mk R mg sin = ma + 2 mk R mg sin = 2 2 2 R k a R a = 2 2 2 g sin R k R ; a = 2 2 gsin k 1 R f = R f = 2 2 mk a R 2 2 2 mg k sin R k f N 2 2 mk R a mg cos R2 2 2 k R × 2 2 gsin (k R ) µg cos µ 2 2 tan R 1 k µmin = 2 2 tan R 1 k Note : From above example if rigid bodies are solid cylinder, hollow cylinder, solid sphere and hollow sphere. (1) Increasing order of acceleration. asolid sphere > ahollow sphere > asolid cylinder > ahollow cylinder (2) Increasing order of required friction force for pure rolling. fhollow cylinder > fhollow sphere > fsolid cylinder > fsolid sphere (3) Increasing order of required minimum friction coefficient for pure rolling. µhollow cylinder > µhollow sphere > µsolid cylinder > µsolid sphere Rigid Body Dynamics Corp. / Reg. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.) – 324005 Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in ADVRB - 33 Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029 ——————————————————————————————————— 9.4 Instantaneous axis of rotation : It is the axis about which the combined translational and rotational motion appears as pure rotational motion. The combined effect of translation of centre of mass and rotation about an axis through the centre of mass is equivalent to a pure rotation with the same angular speed about a stationary axis ; this axis is called instantaneous axis of rotation. It is defined for an instant and its position changes with time. eg. In pure rolling the point of contact with the surface is the instantaneous axis of rotation. Geometrical construction of instantaneous axis of rotation (I.A.R). Draw velocity vector at any two points on the rigid body. The I.A.R. is the point of intersection of the perpendicular drawn on them. In case of pure rolling the lower point is instantaneously axis of rotation. The motion of body in pure rolling can therefore by analysed as pure rotation about this axis. Consequently P = P P = P K.E. = 1/2 P 2 Where P is moment of inertial instantaneous axis of rotation passing through P. Example 56. Prove that kinetic energy = 1/2 P 2 Solution : K.E. = 1 2 cm 2 + 1 2 Mvcm2 = 1 2 Icm 2 + 1 2 M2R2 1 2 (cm + MR2) 2 2 contact point 1 2 Notice that pure rolling of uniform object equation of torque can also be applied about the contact point. Example 57. A uniform bar of length and mass m stands vertically touching a vertical wall (y-axis). When slightly displaced, its lower end begins to slide along the floor (x-axis). Obtain an expression for the angular velocity () of the bar as a function of . Neglect friction everywhere. Rigid Body Dynamics Corp. / Reg. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.) – 324005 Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in ADVRB - 34 Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029 Solution : The position of instantaneous axis of rotation (IAOR) is shown in figure. C cos , sin 2 2 r 2 = half of the diagonal All surfaces are smooth. Therefore, mechanical energy will remain conserved. Decrease in gravitational potential energy of bar = increase in rotational kinetic energy of bar about IAOR. mg 2 (1 – sin ) = 1 2 2 ....(1) Here, = 2 m 12 + mr2 (about IAOR) or = 2 m 12 + 2 m 4 = 2 m 3 Substituting in Eq. (1), we have mg 2 (1 – sin ) = 1 2 2 m 3 2 or 3g (1 sin ) Ans. ——————————————————————————————————— The nature of friction in the following cases assume body is perfectly rigid (i) v = R v smooth surface No friction and pure rolling. (ii) v = R v rough surface No friction and pure rolling (If the body is not perfectly rigid, then there is a small friction acting in this case which is called rolling friction) (iii) v > R or v < R v smooth surface No friction force but not pure rolling. (iv) v > R v v > R rough surface f k There is Relative Motion at point of contact so Kinetic Friction, fk = µN will act in backward direction. This kinetic friction decrease v and increase , so after some time v = R and pure rolling will resume like in case (ii). Rigid Body Dynamics Corp. / Reg. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.) – 324005 Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in ADVRB - 35 Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029 (v) v < R v v > R rough surface f k There is Relative Motion at point of contact so Kinetic Friction, fk = µN will act in forward direction. This kinetic friction increase v and decrease , so after some time v = R and pure rolling will resume like in case (ii). (vi) v = R (initial) v smooth surface F No friction and no pure rolling. (vii) v = R (initial) v rough surface F f s Static friction whose value can be lie between zero and µsN will act in backward direction. If coefficient of friction is sufficiently high, then fs compensates for increasing v due to F by increasing and body may continue in pure rolling with increases v as well as . Example 58. A rigid body of mass m and radius r rolls without slipping on a rough surface. A force is acting on a rigid body x distance from the centre as shown in figure. Find the value of x so that static friction is zero. a a = R rough surface f F x O P Solution : Torque about centre of mass Fx = cm .....(1) F = ma .....(2) From eqn. (1) & (2) max = cm (a = R) ; x = cm mR Note : For pure rolling if any friction is required then friction force will be statics friction. It may be zero, backward direction or forward direction depending on value of x. If F below the point P then friction force will act in backward direction or above the point P friction force will act in forward direction. Example 59. A cylinder is given angular velocity 0 and kept on a horizontal rough surface the initial velocity is zero. Find out distance travelled by the cylinder before it performs pure rolling and work done by friction force Solution : Mg R = 2 MR 2 = 2 g R .....(1) Initial velocity u = 0 v2= u2 + 2as a R f k v v R S Rigid Body Dynamics Corp. / Reg. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.) – 324005 Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in ADVRB - 36 Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029 v2 = 2as .....(2) fK = Ma µMg = Ma a = µg .....(3) = 0 – t from equation (1) = 0 – 2 g R t v = u + at from equation (3) v = µ g t = 0 – 2v R = 0 – 2 = 0 3 from equation (2) 2 0R 3 = (2as) = 2µ gs s = 2 2 0 R 18 µg work done by the friction force w = (–fk R d + fks) – mg R + 2 2 0 mg R 18 g = 0 × t – 1 2 t2 = 0 × 0R 3 g – 1 2 × 2 g R 2 0R 3 g 2 0R 3 g – 2 0R 9 g 2 0 2 R 9 g – mg × R 2 0 2 R 9 g + mg × 2 2 0R 18 g – 2 2 0 2m R 9 + 2 2 0 m R 18 2 2 0 3m R 18 = – 2 2 0 m R 6 Alternative Solution Using work energy theorm k g a f w w w K k f w = 2 2 2 2 2 0 0 0 R 1 1 mR 1 mR m 2 3 2 2 3 2 2 = 2 2 0 m R 6 Rigid Body Dynamics Corp. / Reg. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.) – 324005 Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in ADVRB - 37 Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029 Example 60. A hollow sphere is projected horizontally along a rough surface with speed v and angular velocity 0 find out the ratio. So that the sphere stope moving after some time. v f k a a = 0 v = 0 v 0 R Solution : Torque about lowest point of sphere. fk × R = mg × R = 2 2 mR 3 = 3 g 2R angular acceleration in opposition direction of angular velocity. = 0 – t (final angular velocity = 0) 0 = 3 g t 2R t = 0 2R 3 g acceleration 'a = g' vf = v – at (final velocity vf = 0) v = g × t ; t = v g To stop the sphere time at which v & are zero, should be same. 0 2 R v g 3 g = 0 v 2R 3 ——————————————————————————————————— 9.5 Rolling on moving surface ///////////////////////////////////////////// m m F a + R smooth Sufficiently Rough Friction on the plate backward or on cylinder friction forward so cylinder move forward. m a f Because of pure rolling static friction f. fR = 2 mR 2 = 2f mR f = ma F – f = mb F = m(a + b) f m F b a = R 2 Rigid Body Dynamics Corp. / Reg. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.) – 324005 Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in ADVRB - 38 Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029 At contact point b = a + R b = 3 R 2 b = 3a F = 4ma a = F 4m b = 3F 4m w.r.t. plate distance is covered = and acceleration w.r.t. plate (b – a) = 1 2 (b – a) t2 = 1 2 × 2at2 = t = a F = m 2 F a b Example 61. A solid sphere is released from rest from the top of an incline of inclination and length. If the sphere rolls without slipping, what will be its speed when it reaches the bottom ? Solution : Let the mass of the sphere be m and its radius r. Suppose the linear speed of the sphere when it reaches the bottom is v. As the sphere rolls without slipping, its angular speed about its axis is = v/r. The kinetic energy at the bottom will be K = 1 2 2 + 1 2 mv2 = 1 2 2 2 mr 5 2 + 1 2 mv2 = 1 5 mv2 + 1 2 mv2 = 7 10 mv2 This should be equal to the loss of potential energy mg sin. Thus, 7 10 mv2 = mg sinor v = 10 g sin 7 Example 62. There are two cylinders of radii R1 and R2 having moments of inertia 1 and 2 about their respective axes as shown in figure. Initially, the cylinders rotate about their axes with angular speed 1 and 2 as shown in the figure. The cylinders are moved closed to touch each other keeping the axes parallel. The cylinders first slip over each other at the contact but the slipping finally ceases due to the friction between them. Calculate the angular speeds of the cylinders after the slipping ceases. Solution : When slipping ceases, the linear speeds of the points of contact of the two cylinders will be equal. If 1 and 2 be the respective angular speeds, we have 1 R1 and 2 R2 ................(i) The change in the angular speed is brought about by the frictional force which acts as long as the slipping exists. If this force f acts for a time t, the torque on the first cylinder is fR1 and that on the second is fR2. Assuming 1 > 2 the corresponding angular impulses are – fR1t and fR2t, We, there fore, have – f R1 t = 1 (1 –1) and fR2 t = 2 (2 –2) or, – 1 1 R (1 –1) = 2 2 R (2 –2) ................(ii) Solving (i) and (ii) 1 = 1 1 2 2 2 1 2 2 2 1 1 2 R R R R R2 and 2 = 1 1 2 2 2 1 2 2 2 1 1 2 R R R R R1 Rigid Body Dynamics Corp. / Reg. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.) – 324005 Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in ADVRB - 39 Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029 Example 63. A hollow cylinder of mass m is suspended through two light strings rapped around it as shown in figure. Calculate (a) the tension T in the string and (b) the speed of the cylinder as it falls through a distance . Solution : The portion of the strings between the ceiling and the cylinder is at rest. Hence the points of the cylinder where the strings leave it are at rest. The cylinder is thus rolling without slipping on the strings. Suppose the centre of the cylinder falls with an acceleration a. The angular acceleration of the cylinder about its axis is = a/R, as the cylinder does not slip over the strings. The equation of motion for the centre of mass of the cylinder is Mg – 2T = Ma .............(i) and for the motion about the centre of mass, it is 2 Tr = (Mr2) = Mra or 2T = Ma. From (i) and (ii), a = g 2 and T = Mg 4 As the centre of the cylinder starts moving from rest, the velocity after it has fallen through a distance is given by v2 = 2 g 2 or v = g Example 64. A hollow sphere of mass M and radius R as shown in figure slips on a rough horizontal plane. At some instant it has linear velocity v0 and angular velocity about the centre 0 v 2R as shown in figure. Calculate the linear velocity after the sphere starts pure rolling. Solution : Velocity of the centre = v0 and the angular velocity about the centre = 0 v 2R . Thus v0 > 0R. The sphere slips forward and thus the friction by the plane on the sphere will act backward. As the friction is kinetic, its value is µN = µMg and the sphere will be decelerated by acm = f/M. Hence, v(t) = v0 – f M t .............(i) This friction will also have a torque I' = fr about the centre. This torque is clockwise and in the direction of 0. Hence the angular acceleration about the centre will be = f 2 R (2/3)MR = 3f 2MR and the clockwise angular velocity at time t will be (t) = 0 + 3f 2MR t = 0 v 2R + 3f 2MR t. Pure rolling starts when v(t) = R(t) i.e., v(t) = 0 v 2 + 3f 2M t. ............(ii) Eliminating t from (i) and (ii), 3 2 v(t) + v(t) = 3 2 v0 + 0 v 2 or v(t) = 2 5 × 2v0 = 4 5 v0. Thus, the sphere rolls with linear velocity 4v0/5 in the forward direction. Rigid Body Dynamics Corp. / Reg. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.) – 324005 Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in ADVRB - 40 Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029 Example 65. A rod AB of mass 2m and length is lying on a horizontal frictinless surface. A particle of mass m traveling along the surface hits the end 'A' of the rod with a velocity v0 in a direction perpendicular to AB. The collisin is elastic. After the collision the particle comes to rest. Find out after collision (a) Velocity of centre of mass of rod (b) Angular velocity. Solution : (a) Let just after collision lthe sped of COM of rod is v and angular velocity about COM is . External force on the system (rod + mass) in horizontal plane is zero Apply conservation of linear momentum in x direction mv0 = 2mv ....(1) Net torque on the system about any point is zero Apply conservation of angular momentum about COM of rod. mv0 2 = mv0 2 = 2 2m 12 mv0 = m 3 ....(2) From eq (1) velocity of centre of mass v = 0 v 2 From eq (2) anuglar velocity = 0 3v ——————————————————————————————————— 10. TOPPLING : In many situations an external force is applied to a body to cause it to slide along a surface. In certain cases, the body may tip over before sliding ensues. This is known as topping. (1) There is a no horizontal force so pressure at bottom is uniform and normal is colinear with mg. (2) If a force is applied at COM, pressure is not uniform Normal shifts right so that torque of N can counter balance torque of friction. Fmax = fr N = mg fr . b/2 = N . a/2 fr = Na/b = mg a/b, Fmax = mg a/b Rigid Body Dynamics Corp. / Reg. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.) – 324005 Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in ADVRB - 41 Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029 (3) If surface is not sufficiently rough and the body slides before F is increased to Fmax = mg a/b then body will slide before toppling. Once body starts sliding friciton becomes constant and hence no topping. This is the case if Fmax > flimit mg a/b > mg < a/b Condition for toppling when a/b in this case body will topple if F > mg a/b but if < a/b, body will not topple any value of F applied a COM Example 66. Find out minimum value of F for toppling Solution : Never topple Example 67. A uniform cube of side 'a' and mass m rests on a rough horizontal table. A horizontal force F is applied normal to one of the faces at a point directly below the centre of the face, at a height a 4 above the base. (i) What is the minimum value of F for which the cube begins to tip about an edge? (ii) What is the minimum value of s so that toppling occures. (iii) If = min, find minimum force for topping. (iv) Minimum s so that Fmin can cause toppling. Solution : (i) In the limiting case normal reaction will pass through O. The cube will tip about O if torque of Fabout O exceeds the torque of mg. Hence, a a F mg 4 2 or F > 2 mg therefore, minimum value of F is 2 mg a/2 N F a/4 O mg f r (ii) In this case since it is not acting at COM, toppling can occur even after body started slinding because increasing the the torque of F about COM.hence min = 0, (iii) Now body is sliding before toppling, O is not I.A.R., torque equation can not be applied across it. It can now be applied about COM. F × a 4 = N × a 2 .....(1) N = mg .....(2) from (1) and (2) F = 2 mg (iv) F > 2 mg ................... (1) (from sol. (i)) N = mg .......................(2) F = µsN = µsmg ........... (3) from (1) and (2) µs = 2 Rigid Body Dynamics Corp. / Reg. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.) – 324005 Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in ADVRB - 42 Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029 Example 68. Find minimum value of so that truck can avoid the dead end, without toppling the block kept on it. v b h m Sufficiently rough surface power breaks Solution : ma h 2 mg b 2 a b h g Final velocity of truck is zero. So that 0 = v2 – 2( b h g) = 2 h v 2b g b h ma N mg f r Problem 1. Four point masses are connected by a massless rod as shown in figure. Find out the moment of inertia of the system about axis CD ? Solution : I1 = m(2a)2 I2 = 2ma2 I3 = 0 I4 = 4ma2 ICD = I1 + I2 + I3 + I4 = 10 ma2 Ans. Problem 2. Three point masses are located at the corners of an equilibrium triangle of side 1 cm. Masses are of 1, 2 & 3kg respectively and kept as shown in figure. Calculate the moment of Inertia of system about an axis passing through 1 kg mass and perpendicular to the plane of triangle? Solution : Moment of inertia of 2 kg mass about an axis passing through 1 kg mass I1 = 2 × (1×10–2)2 = 2×10–4 Moment of inertia of 3 kg mass about an axis passing through 1 kg mass I2 = 3 × (1×10–2)2 = 3×10–4 I = I1 + I2 = 5 × 10–4 kgm2 Rigid Body Dynamics Corp. / Reg. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.) – 324005 Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in ADVRB - 43 Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029 Problem 3. Calculate the moment of Inertia of figure shown each having mass M, radius R and having uniform mass distribution about an axis pependicular to the plane and passing through centre? (a) x R (b) x R (c) x R Solution : d = dm 2 R 2 = dI = 2 R 2 dm = 2 MR 2 Problem 4. Find the moment of inertia of the uniform square plate of side 'a' and mass M about the axis AB. Solution : dI = dm 2 a 3 I = dI = 2 a 3 dm = 2 Ma 3 Problem 5. Calculate the moment of inertia of a uniform solid cylinder of mass M, radius R and length about its axis. Solution : Each segment of cylinder is solid disc so 2 R dI dm 2 = 2 MR 2 Ans. Problem 6. Find the moment of inertia of a uniform rectangular plate of mass M, edges of length '' and 'b' about its axis passing through centre and perpendicular to it. Rigid Body Dynamics Corp. / Reg. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.) – 324005 Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in ADVRB - 44 Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029 Solution : Using perpendicular axis theorem I3 = I1 + I2 I1 = 2 Mb 12 I2 = 2 M 12 ; I3 = 2 2 M( b ) 12 Problem 7. Find the moment of inertia of a uniform square plate of mass M, edge of length '' about its axis passing through P and perpendicular to it. Solution : IP = 2 2 M M 6 2 = 2 2M 3 Problem 8. Find out the moment of inertia of a ring having uniform mass distribution of mass M & radius R about an axis which is tangent to the ring and (i) in the plane of the ring (ii) perpendicular to the plane of the ring. (i) (ii) Solution : (i) Moment of inertia about an axis passing through centre of ring and plane of the ring I1 = 2 MR 2 Using parallel axis theorem I' = I1 + MR2 = 2 3MR 2 (ii) Moment of inertia about an axis passing through centre of ring and perpendicular to palne of the ring IC = MR2 Using parallel axis theorem I'' = IC + MR2 = 2MR2 Problem 9. Calculate the moment of inertia of a rectangular frame formed by uniform rods having mass m each as shown in figure about an axis passing through its centre and perpendicular to the plane of frame ? Also find moment of inertia about an axis passing through PQ ? m b,m P Q R S Solution : (i) Moment of inertia about an axis passing through its centre and perpendicular to the plane of frame IC = I1 + I2 + I3 + I4 I1 = I3 , I2 = I4 IC = 2I1 + 2I2 I1 = 2 2 m b m 12 2 I2 = 2 2 mb m 12 2 so, IC = 2 2 2m ( b ) 3 (ii) M.I. about axis PQ of rod PQ I1 = 0 M.I. about axis PQ of rod PS I2 = 2 mb 2 M.I. about axis PQ of rod QR I3 = 2 mb 2 M.I. about axis PQ of rod SR I4 = mb2 I = I1 + I2 + I3 + I4 = 2 5mb 3 Rigid Body Dynamics Corp. / Reg. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.) – 324005 Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in ADVRB - 45 Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029 Problem 10. In the previous question, during the motion of particle from P to Q. Torque of gravitational force about P is : (A) increasing (B) decreasing (C) remains constant (D) first increasing then decreasing Solution : Increasing because distance from point P is increasing. Problem 11. Determine the point of application of force, when forces are acting on the rod as shown in figure. Solution : Torque of B about A 1 = 3N × 5 = 15N cm (clockwise) Torque of C about A 2 = 6N × 10 = 60 N cm (anticlockwise) Resultant force perpendicular to the rod F = 8 N 1 + 2 = F x (x = distance from point A) – 15 + 60 = 8 x x = 45/8 = 5.625 cm Problem 12. A uniform rod of mass m and length can rotate in vertical plane about a smooth horizontal axis hinged at point H. Find angular acceleration of the rod just after it is released from initial position making an angle of 37° with horizontal from rest? Find force exerted by the hinge just after the rod is released from rest. Solution : Torque about hing = H = mgcos37 2 = 2 m 3 = 6g / 5 at = 2 = 3g 5 mgcos37 – N1 = mat N1 = mg 5 angular velocity of rod is zero. so N2 = mgsin37° = 3mg/5 N = 2 2 1 2 N N = 2 2 mg 3mg 5 5 = mg 10 5 Problem 13. A uniform rod of length, mass m is hung from two strings of equal length from a ceiling as shown in figure. Determine the tensions in the strings ? Solution : TA + TB = mg ............(i) Torque about point A is zero So, TB × 3 4 = mg 2 ............(ii) From eq. (i) & (ii), TA = mg/3, TB = 2mg/3 Rigid Body Dynamics Corp. / Reg. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.) – 324005 Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in ADVRB - 46 Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029 Problem 14. A particle of mass m starts moving from origin with a constant velocity ˆ ui find out its angular momentum about origin at this moment. What will be the answer later on? What will be the answer if the speed increases. y x Solution : L r p ˆ ˆ L ri mui = 0 Problem 15. A uniform rod of mass m and length can rotate freely on a smooth horizontal plane about a vertical axis hinged at point H. A point mass having same mass m coming with an initial speed u perpendicular to the rod, strikes the rod and sticks to it at a distance of 3/4 from hinge point. Find out the angular velocity of the rod just after collision ? Solution : Angular Momentum about hinge Li = Lf mu 3 4 = 2 2 m 3 m 3 4 = 36u 43 Problem 16. Uniform & smooth Rod of length is moving with a velocity of centre v and angular velocity on smooth horizontal surface. Findout velocity of point A and B. Solution : velocity of point A w.r.t. center is 2 velocity of point A w.r.t. ground VA = V + 2 velocity of point B w.r.t. center is – 2 velocity of point B w.r.t. ground VA = V – 2 Rigid Body Dynamics Corp. / Reg. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.) – 324005 Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in ADVRB - 47 Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029 Marked Questions can be used as Revision Questions. PART - I : SUBJECTIVE QUESTIONS Section (A) : Kinematics A-1. A uniform disk rotating with constant angular acceleration covers 50 revolutions in the first five seconds after the start. Calculate the angular acceleration and the angular velocity at the end of five seconds. A-2. A body rotating with 20 rad/s is acted upon by a uniform torque providing it an angular deceleration of 2 rad/s2. At which time will the body have kinetic energy same as the initial value if the torque acts continuously ? Section (B) : Moment of inertia B-1. Calculate the moment of inertia of a uniform square plate of mass M and side L about one of its diagonals, with the help of its moment of inertia about its centre of mass. B-2. A uniform triangular plate of mass M whose vertices are ABC has lengths , 2 and 2 as shown in figure. Find the moment of inertia of this plate about an axis passing through point B and perpendicular to the plane of the plate. A B C 2 2 B-3. Find the moment of inertia of a uniform half-disc about an axis perpendicular to the plane and passing through its centre of mass. Mass of this disc is M and radius is R. B-4. Calculate the radius of gyration of a uniform circular disk of radius r and thickness t about a line perpendicular to the plane of this disk and tangent to the disk as shown in figure. Section (C) : Torque C-1. Two forces 1 ˆ ˆ ˆ F 2i 5j 6k and 2 ˆ ˆ ˆ F i 2j k are acting on a body at the points (1, 1, 0) and (0, 1, 2) respectively. Find torque acting on the body about point (–1, 0, 1). C-2. A simple pendulum having bob of mass m and length is pulled aside to make an angle with the vertical. Find the magnitude of the torque of the weight of the bob about the point of suspension. At which position its torque is zero? At which it is maximum? C-3. A particle having mass m is projected with a speed v at an angle with horizontal ground. Find the torque of the weight of the particle about the point of projection when the particle (a) is at the highest point. (b) reaches the ground. Rigid Body Dynamics Corp. / Reg. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.) – 324005 Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in ADVRB - 48 Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029 C-4. Calculate the net torque on the system about the point O as shown in figure if F1 = 11 N, F2 = 9 N, F3 = 10 N, a = 10 cm and b = 20 cm. (All the forces along the tangent.) Section (D) : Rotational Equilibrium D-1. A uniform metre stick having mass 400 g is suspended from the fixed supports through two vertical light strings of equal lengths fixed at the ends. A small object of mass 100 g is put on the stick at a distance of 60 cm from the left end. Calculate the tensions in the two strings. (g = 10 m/s2) D-2. Assuming frictionless contacts, determine the magnitude of external horizontal force P applied at the lower end for equilibrium of the rod as shown in figure. The rod is uniform and its mass is ' m'. //////////////////////////////////////////////////// P Wall D-3. A uniform ladder having length 10.0 m and mass 24 kg is resting against a vertical wall making an angle of 53° with it. The vertical wall is smooth but the ground surface is rough. A painter weighing 75 kg climbs up the ladder. If he stays on the ladder at a point 2 m from the upper end, what will be the normal force and the force of friction on the ladder by the ground? What should be the minimum coefficient of friction between ground and ladder for the painter to work safely? (g = 10 m/s2) D-4. In the system as shown in figure, AB is a uniform rod of mass 10 kg and BC is a light string which is connected between wall and rod, in vertical plane. There is block of mass 15 kg connected at B with a light string. [Take g = 10 m/s2] (BC and BD are two different strings) If whole of the system is in equilibrium then find (i) Tension in the string BC (ii) Hinge force exerted on beam at point A D Section (E) : Rotation about fixed axis (H = H) E-1. A rod of negligible mass having length = 2 m is pivoted at its centre and two masses of m1 = 6 kg and m2 = 3 kg are hung from the ends as shown in figure. (a) Find the initial angular acceleration of the rod if it is horizontal initially. (b) If the rod is uniform and has a mass of m3 = 3 kg. (i) Find the initial angular acceleration of the rod. (ii) Find the tension in the supports to the blocks of mass 3 kg and 6 kg (g = 10 m/s2). Rigid Body Dynamics Corp. / Reg. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.) – 324005 Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in ADVRB - 49 Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029 E-2. The uniform rod AB of mass m is released from rest when = 60º. Assuming that the friction force between end A and the surface is large enough to prevent sliding, determine (for the instant just after release) (a) The angular acceleration of the rod (b) The normal reaction and the friction force at A. E-3. The moment of inertia of the pulley system as shown in the figure is 3 kg – m2. The radii of bigger and smaller pulleys are 2m and 1m respectively. As the system is released from rest, find the angular acceleration of the pulley system. (Assume that there is no slipping between string & pulley and string is light) [Take g = 10 m/s2] E-4. A uniform thin rod of length L is hinged about one of its ends and is free to rotate about the hinge without friction. Neglect the effect of gravity. A force F is applied at a distance x from the hinge on the rod such that force is always perpendicular to the rod. Find the normal reaction at the hinge as function of 'x', at the initial instant when the angular velocity of rod is zero. Section (F) : Rotation about Fixed Axis (Energy conservation) F-1. A solid cylinder of mass M = 1kg & radius R = 0.5m is pivoted at its centre & has three particles of mass m = 0.1kg mounted at its perimeter in the vertical plane as shown in the figure. The system is initially at rest. Find the angular speed of the cylinder, when it has swung through 900 in anticlockwise direction. [Take g = 10 m/s2] F-2. A rigid body is made of three identical uniform thin rods each of length L fastened together in the form of letter H. The body is free to rotate about a fixed horizontal axis AB that passes through one of the legs of the H. The body is allowed to fall from rest from a position in which the plane of H is horizontal. What is the angular speed of the body, when the plane of H is vertical. A B F-3. A uniform rod of mass m and length L lies radially on a disc rotating with angular speed in a horizontal plane about its axis. The rod does not slip on the disc and the centre of the rod is at a distance R from the centre of the disc. Find out the kinetic energy of the rod. Rigid Body Dynamics Corp. / Reg. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.) – 324005 Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in ADVRB - 50 Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029 F-4. The moment of inertia of the pulley system as shown in figure is 3 kgm2. Its radius is 1m. The system is released from rest find the linear velocity of the block, when it has descended through 40 cm. (Assume that there is no slipping between string & pulley and string is light) [Take g = 10 m/s2] Section (G) : Angular Momentum & its conservation G-1. A particle having mass 2 kg is moving with velocity ( ˆ ˆ 2i 3j )m/s. Find angular momentum of the particle about origin when it is at (1, 1, 0). G-2. A particle having mass 2 kg is moving along straight line 3x + 4y = 5 with speed 8m/s. Find angular momentum of the particle about origin. x and y are in meters. G-3. Two beads (each of mass m) can move freely in a frictionless wire whose rotational inertia with respect to the vertical axis is . The system is rotated with an angular velocity 0 when the beads are at a distance r/2 from the axis. What is the angular velocity of the system when the beads are at a distance r from the axis ? [JEE - 1990] G-4. A system consists of two identical small balls of mass 2 kg each connected to the two ends of a 1 m long light rod. The system is rotating about a fixed axis through the centre of the rod and perpendicular to it at an angular speed of 9 rad/s. An impulsive force of average magnitude 10 N acts on one of the masses in the direction of its velocity for 0.20 s. Calculate the new angular velocity of the system. G-5. A uniform round board of mass M and radius R is placed on a fixed smooth horizontal plane and is free to rotate about a fixed axis which passes through its centre. A man of mass m is standing on the point marked A on the circumference of the board. At first the board & the man are at rest. The man starts moving along the rim of the board at constant speed vo relative to the board. Find the angle of board’s rotation when the man passes his starting point on the disc first time. G-6. A point object of mass m moving horizontally hits the lower end of the uniform thin rod of length and mass m and sticks to it. The rod is resting on a horizontal, frictionless surface and pivoted at the other end as shown in figure. Find out angular velocity of the system just after collision. × m m v Section (H) : Combined Translational & Rotation Motion (Kinematics) H-1 The centre of mass of a uniform rod of length 10 meter is moving with a translational velocity of 50 m/sec. on a frictionless horizontal surface as shown in the figure and the rod rotates about its centre of mass with an angular velocity of 5 radian/sec. Find out VA and VB = = 5 rad/s 50 m/s Rigid Body Dynamics Corp. / Reg. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.) – 324005 Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in ADVRB - 51 Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029 H-2 A ring of radius 1 m. performs combined translational and rotational motion on a frictionless horizontal surface with an angular velocity of 3 rad/sec as shown in the figure. Find out velocity of its centre and point A if the velocity of the lowest point VP is 1 m/sec. = 3 rad/s y x H-3 A plank is moving with a velocity of 4 m/sec. A disc of radius 1 m rolls without slipping on it with an angular velocity of 3 rad/sec as shown in figure. Find out the velocity of centre of the disc. 4 1 m 3 rad/s H-4 The end B of uniform rod AB which makes angle with the floor is being pulled with a velocity v0 as shown. Taking the length of the rod as , calculate the following at the instant when = 37º (a) The velocity of end A (b) The angular velocity of rod (c) Velocity of CM of the rod. H-5. A ball of radius R = 10.0 cm rolls without slipping on a horizontal plane so that its centre moves with constant acceleration a = 2.50 cm/s2; t = 2.00 s after the beginning of motion its position corresponds to that shown in Fig. Find : (a) the velocities of the points A, B and O (b) the accelerations of these points. Section (I) : Combined translational & Rotational Motion (Dynamics) -1. A small solid cylinder is released from a point at a height h on a rough as track shown in figure. Assuming that it does not slip anywhere, calculates its linear speed when it rolls on the horizontal part of the track. -2. A uniform ball of mass ‘m’ rolls without sliding on a fixed horizontal surface. The velocity of the lowest point of the ball with respect to the centre of the ball is V. Find out the total kinetic energy of the ball. -3. A string is wrapped over the curved surface of a uniform solid cylinder and the free end is fixed with rigid support. The solid cylinder moves down, unwinding the string. Find the downward acceleration of the solid cylinder. Rigid Body Dynamics Corp. / Reg. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.) – 324005 Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in ADVRB - 52 Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029 -4. A uniform disk of mass m is released from rest from the rim of a fixed hemispherical bowl so that it rolls along the surface. If the rim of the hemisphere is kept horizontal, find the normal force exerted by the bowl on the disk when it reaches the bottom of the bowl. -5. There is a rough track, a portion of which is in the form of a cylinder of radius R as shown in the figure. Find the minimum linear speed of a uniform ring of radius r with which it should be set rolling without sliding on the horizontal part so that it can complete round the circle without sliding on the cylindrical part. -6. A uniform solid sphere of radius R is placed on a smooth horizontal surface. It is pulled by a by constant force acting along the tangent from the highest point. Calculate the distance travelled by the centre of mass of the solid sphere during the time it makes one full revolution. F R -7. A uniform hollow sphere of mass m = 1 kg is placed on a rough horizontal surface for which the coefficient of static friction between the surfaces in contact is = 2/5. Find the maximum constant force which can be applied at the highest point in the horizontal direction so that the sphere can roll without slipping. (Take g = 10 m/s2) Section (J) : Conservation of angular momentum (Combined translation & rotational motion) J-1 A uniform rod of length and mass 4m lies on a frictionless horizontal surface on which it is free to move anyway. A ball of mass m moving with speed v as shown in figure.collides with the rod at one of the ends. If ball comes to rest immediately after collision then find out angular velocity of rod just after collistion. O v m 4 m J-2. A uniform rod having mass m1 and length L lies on a smooth horizontal surface. A particle of mass m2 moving with speed u on the horizontal surface strikes the free rod perpendicularly at an end and it sticks to the rod. (a) Calculate the velocity of the com C of the system constituting “the rod plus the particle”. (b) Calculate the velocity of the particle with respect to C before the collision. (c) Calculate the velocity of the rod with respect to C before the collision (d) Calculate the angular momentum of the particle and of the rod about the com C before the collision. (e) Calculate the moment of inertia of the rod plus particle about the vertical axis through the centre of mass C after the collision. (f) Calculate the velocity of the com C and the angular velocity of the system about the centre of mass after the collision. Rigid Body Dynamics Corp. / Reg. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.) – 324005 Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in ADVRB - 53 Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029 J-3. A uniform solid sphere is placed on a smooth horizontal surface. An impulse is given horizontally to the sphere at a height h = 4R/5 above the centre line. m and R are mass and radius of sphere respectively. ///////////////////////////////////////////////// B A h (a) Find angular velocity of sphere & linear velocity of centre of mass of the sphere after impulse. (b) Find the minimum time after which the highest point B will touch the ground, (c) Find the displacement of the centre of mass during this interval. J-4. A uniform disc of radius R = 0.2 m kept over a rough horizontal surface is given velocity v0 and angular velocity 0. After some time its kinetic energy becomes zero. If v0 = 10 m/s, find 0. Section (K) : Toppling K-1. A solid cubical block of mass m and side a slides down a rough inclined plane of inclination with a constant speed. Calculate the torque of the normal force acting on the block about its centre and the perpendicular distance 'x' from centre of mass at which it is acting. PART - II : ONLY ONE OPTION CORRECT TYPE Section (A) : Kinematics A-1. A fan is running at 3000 rpm. It is switched off. It comes to rest by uniformly decreasing its angular speed in 10 seconds. The total number of revolutions in this period. (A) 150 (B) 250 (C) 350 (D) 300 A-2. A block hangs from a string wrapped on a disc of radius 20 cm free to rotate about its axis which is fixed in a horizontal position. If the angular speed of the disc is 10 rad/s at some instant, with what speed is the block going down at that instant ? (A) 4 m/s (B) 3 m/s (C) 2 m/s (D) 5 m/s Section (B) : Moment of inertia B-1. A uniform circular disc A of radius r is made from a copper plate of thickness t and another uniform circular disc B of radius 2r is made from a copper plate of thickness t/2. The relation between the moments of inertia IA and IB is (A) IA > IB (B) IA = IB (C) IA < IB (D) depends on the values of t and r. B-2. The moment of inertia of a non-uniform semicircular wire having mass m and radius r about a line perpendicular to the plane of the wire through the centre is (A) mr2 (B) 1 2 mr2 (C) 1 4 mr2 (D) 2 5 mr2 B-3. Let A and B be the moments of inertia of two solid cylinders of identical geometrical shape and size about their axes, the first made of aluminium and the second of iron. (A) A < B (B) A = B (C) A > B (D) relation between A and B depends on the actual shapes of the bodies. Rigid Body Dynamics Corp. / Reg. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.) – 324005 Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in ADVRB - 54 Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029 B-4. Let I1 and I2 be moments of inertia of a body about two axes 1 and 2 respectively, The axis 1 passes through the centre of mass of the body but axis 2 does not. (A) I1 < I2 (B) If I1 < I2, the axes are parallel. (C) If the axes are parallel, I1 < I2 (D) If the axes are not parallel, I1 I2. B-5. The moment of inertia of an elliptical disc of uniform mass distribution of mass 'm', semi major axis 'r', semi minor axis 'd' about its axis is : (A) 2 mr 2 (B) 2 md 2 (C) 2 mr 2 (D) 2 mr 2 B-6. A unifrom thin rod of length L and mass M is bent at the middle point O as shown in figure. Consider an axis passing through its middle point O and perpendicular to the plane of the bent rod. Then moment of inertia about this axis is : (A) 2 3 mL2 (B) 1 3 mL2 (C) 1 12 mL2 (D) dependent on B-7. The moment of inertia of a uniform circular disc about its diameter is 200 gm cm2. Then its moment of inertia about an axis passing through its center and perpendicular to its circular face is (A) 100 gm cm2 (B) 200 gm cm2 (C) 400 gm cm2 (D) 1000 gm cm2 B-8. A thin uniform rod of length 4 l, mass 4m is bent at the points as shown in the fig. What is the moment of inertia of the rod about the axis passing point O & perpendicular to the plane of the paper. (A) 2 m 3 (B) 2 10 m 3 (C) 2 m 12 (D) 2 m 24 B-9. Moment of inertia of a uniform disc about the axis O O is: O O ¢ m r (A) 2 3 m r 2 (B) 2 m r 2 (C) 2 5 m r 2 (D) 2 5 m r 4 B-10. The moment of inertia of a hollow cubical box of mass M and side a about an axis passing through the centres of two opposite faces is equal to (A) 2 5Ma 3 (B) 2 5Ma 6 (C) 2 5Ma 12 (D) 2 5Ma 18 Rigid Body Dynamics Corp. / Reg. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.) – 324005 Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in ADVRB - 55 Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029 B-11. A uniform thin rod of length (4a + 2a) and of mass (4m + 2m) is bent and fabricated to form a square surrounded by semicircles as shown in the figure. The moment of inertia of this frame about an axis passing through its centre and perpendicular to its plane is [Olympiad 2014 (stage-1)] (A) 2 4 2 ma 3 (B) 2 4 ma 2 (C) 2 4 3 ma 3 (D) 2 ma 10 3 3 Section (C) : Torque C-1. If a rigid body is subjected to two forces 1 F = ˆ ˆ ˆ 2i 3j 4k acting at (3, 3, 4) and 2 F = ˆ ˆ ˆ –2i – 3j – 4k acting at (1, 0, 0) then which of the following is (are) true? [REE - 1994] (A) The body is in equilibrium. (B) The body is under the influence of a torque only. (C) The body is under the influence of a single force. (D) The body is under the influence of a force together with a torque . C-2. A force F = ˆ ˆ 4 i 10 j acts on a body at a point having position vector ˆ ˆ 5 i 3 j relative to origin of co-ordinates on the axis of rotation. The torque acting on the body about the origin is : (A) 38 ˆ k (B) 25 ˆ k (C) 62 ˆ k (D) none of these C-3. In case of torque of a couple if the axis is changed by displacing it parallel to itself, torque will : (A) increase (B) decrease (C) remain constant (D) None of these Section (D) : RotationAL Equilibrium D-1. Four equal and parallel forces are acting on a rod (as shown in figure) in horizontal plane at distances of 20 cm, 40 cm, 60 cm and 80 cm respectively from one end of the rod. Under the influence of these forces the rod : (A) is at rest (B) experiences a torque (C) experiences a linear motion (D) experiences a torque and also a linear motion D-2. A uniform ladder of length 5m is placed against the wall in vertical plane as shown in the figure. If coefficient of friction is the same for both the wall and the floor then minimum value of for it not to slip is (A) = 1/2 (B) = 1/4 (C) = 1/3 (D) = 1/5 D-3 A rod of weight w is supported by two parallel knife edges A & B and is in equilibrium in a horizontal position. The knives are at a distance d from each other. The centre of mass of the rod is at a distance x from A. The normal reactions at A and B will be : (A) NA = 2w (1 x/d), NB = wx/d] (B) NA = w (1 x/d), NB = wx/d] (C) NA = 2w (1 x/d), NB = 2wx/d] (D) NA = w (2 x/d), NB = wx/d] Rigid Body Dynamics Corp. / Reg. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.) – 324005 Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in ADVRB - 56 Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029 D-4 The beam and pans of a balance have negligible mass. An object weighs W1 when placed in one pan and W2 when placed in the other pan. The weight W of the object is : (A) 1 2 W W (B) 1 2 (W W ) (C) W12 + W22 (D) (W1–1 + W2–1)/2 D-5. A uniform rod of length is placed symmetrically on two walls as shown in figure. The rod is in equilibrium. If N1 and N2 are the normal forces exerted by the walls on the rod then (A) N1 > N2 (B) N1 > N2 (C) N1 = N2 (D) N1 and N2 would be in the vertical directions. Section (E) : Rotation about Fixed axis (H = H) E-1. A uniform circular disc A of radius r is made from a metal plate of thickness t and another uniform circular disc B of radius 4r is made from the same metal plate of thickness t/4. If equal torques act on the discs A and B, initially both being at rest. At a later instant, the angular speeds of a point on the rim of A and another point on the rim of B are A and B respectively. We have (A) A > B (B) A = B (C) A < B (D) the relation depends on the actual magnitude of the torques. E-2. A body is rotating with constant angular velocity about a vertical axis fixed in an inertial frame. The net force on a particle of the body not on the axis is (A) horizontal and skew with the axis (B) vertical (C) horizontal and intersecting the axis (D) none of these. E-3. One end of a uniform rod having mass m and length is hinged. The rod is placed on a smooth horizontal surface and rotates on it about the hinged end at a uniform angular velocity . The force exerted by the hinge on the rod has a horizontal component (A) m2 (B) zero (C) mg (D) 1 2 m2 E-4. The uniform rod of mass 20 kg and length 1.6 m is pivoted at its end and swings freely in the vertical plane. Angular acceleration of rod just after the rod is released from rest in the horizontal position as shown in figure is (A) 15g 16 (B) 17g 16 (C) 16g 15 (D) g 15 E-5. Two men support a uniform horizontal rod at its two ends. If one of them suddenly lets go, the force exerted by the rod on the other man just after this moment will: (A) remain unaffected (B) increase (C) decrease (D) become unequal to the force exerted by him on the rod. Section (F) : Rotation about fixed axis (energy conservation) Rigid Body Dynamics Corp. / Reg. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.) – 324005 Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in ADVRB - 57 Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029 F-1. A uniform metre stick is held vertically with one end on the floor and is allowed to fall. The speed of the other end when it hits the floor assuming that the end at the floor does not slip : (A) 4g (B) 3g (C) 5g (D) g F-2. A uniform rod is hinged as shown in the figure and is released from a horizontal position. The angular velocity of the rod as it passes the vertical position is: (axis is fixed, smooth and horizontal) (A) 12g 3 (B) 2g 3 (C) 24g 7 (D) 3g 7 Section (G) : Angular Momentum & its conservation G-1. A constant torque acting on a uniform circular wheel changes its angular momentum from A0 to 4A0 in 4 sec. the magnitude of this torque is : (A) 4A0 (B) A0 (C) 3A0 /4 (D) 12A0 G-2. A particle moves with a constant velocity parallel to the Y-axis. Its angular momentum about the origin. (A) is zero (B) remains constant (C) goes on increasing (D) goes on decreasing. G-3. A particle is projected at time t = 0 from a point P on the ground with a speed V0, at an angle of 45° to the horizontal. What is the magnitude of the angular momentum of the particle about P at time t = v0/g. (A) 2 0 mv 2 2g (B) 3 0 mv 2g (C) 2 0 mv 2g (D) 3 0 mv 2 2g G-4. A uniform thin circular ring of mass 'M' and radius 'R' is rotating about its fixed axis passing through its centre perpendicular to its plane of rotation with a constant angular velocity . Two objects each of mass m, are attached gently to the opposite ends of a diameter of the ring. The ring now rotates with an angular velocity. [JEE - 1983] (A) M (M m) (B) M (M 2m) (C) M (M 2m) (D) (M 3m) M G-5. A boy sitting firmly over a rotating stool has his arms folded. If he stretches his arms, his angular momentum about the axis of rotation (A) increases (B) decreases (C) remains unchanged (D) doubles Section (H) : Combined Translational + Rotational Motion (Kinematics) H-1. The centre of a disc rolling without slipping on a plane surface moves with speed u. A particle, on the lower half of the rim making an angle 60º with vertical, will be moving at speed (A) zero (B) u (C) u (D) 2u H-2. A thin string is wrapped several times around a cylinder kept on a rough horizontal surface. A boy standing at a distance from the cylinder draws the string towards him as shown in figure. The cylinder rolls without slipping. The length of the string passed through the hand of the boy while the cylinder reaches his hand is (A) (B) 2 (C) 3 (D) 4 Rigid Body Dynamics Corp. / Reg. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.) – 324005 Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in ADVRB - 58 Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029 H-3. A uniform cylinder of mass M and radius R rolls without slipping down a slope of angle to the horizontal. The cylinder is connected to a spring constant K while the other end of the spring is connected to a rigid support at P. The cylinder is released when the spring is unstretched. The maximum displacement of cylinder is (A) 3 Mg sin 4 K (B) Mgsin K (C) 2Mgsin K (D) 4 Mgsin 3 K H-4. A system of uniform cylinders and plates is shown in figure. All the cylinders are identical and there is no slipping at any contact. Velocity of lower & upper plate is V and 2V respectively as shown in figure. Then the ratio of angular speed of the upper cylinders to lower cylinders is (A) 3 (B) 1/3 (C) 1 (D) none of these H-5. When a person throws a meter stick it is found that the centre of the stick is moving with a speed of 10 m/s vertically upwards & left end of stick with a speed of 20 m/s vertically upwards. Then the angular speed of the stick is: (A) 20 rad/ sec (B) 10 rad/sec (C) 30 rad/sec (D) none of these Section (): Combined translational & Rotational Motion (Dynamics) -1. As shown in the figure, a uniform disc of mass m is rolling without slipping with a angular velocity . The portion AB is rough and BC is smooth. When it crosses point B disc will be in : v A B µ 0 = C (A) translational motion only (B) pure rolling motion (C) rotational motion only (D) none of these -2. A solid sphere, a hollow sphere and a ring, all having equal mass and radius, are placed at the top of an incline and released. The friction coefficients between the objects and the incline are equal but not sufficient to allow pure rolling. The greastest kinetic energy at the bottom of the incline will be achieved by (A) the solid sphere (B) the hollow sphere (C) the ring (D) all will achieve same kinetic energy. -3. A hollow sphere and a solid sphere having equal mass and equal radii are rolled down without slipping on a rough inclined plane. (A) The two spheres reach the bottom simultaneously (B) The hollow sphere reaches the bottom with lesser speed. (C) The solid sphere reaches the bottom with greater kinetic energy (D) The two spheres will reach the bottom with same linear momentum -4. A solid sphere, a hollow sphere and a solid cylinder, all having equal mass and radius, are placed at the top of an incline and released. The friction coefficients between the objects and the incline are equal but not sufficient to allow pure rolling. Greastest time will be taken in reaching the bottom by (A) the solid sphere (B) the hollow sphere (C) the solid cylinder (D) all will take same time. Rigid Body Dynamics Corp. / Reg. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.) – 324005 Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in ADVRB - 59 Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029 -5. A rough inclined plane fixed in a car accelerating on a horizontal road is shown in figure. The angle of incline is related to the acceleration a of the car as a = g tan. If a rigid sphere is set in pure rolling on the incline (A) it will continue pure rolling (B) Friction will act on it (C) its angular velocity will increase (D) its angular velocity will decrease. -6. A sphere S rolls without slipping, moving with a constant speed on a plank P. The friction between the upper surface of P and the sphere is sufficient to prevent slipping, while the lower surface of P is smooth and rests on the ground. Initially, P is fixed to the ground by a pin N. If N is suddenly removed: (A) S will begin to slip on P (B) P will begin to move backwards (C) the speed of S will decrease and its angular velocity will increase (D) there will be no change in the motion of S and P will still be at rest. -7. A body is given translational velocity and kept on a surface that has sufficient friction. Then: (A) body will move forward before pure rolling (B) body will move backward before pure rolling (C) body will start pure rolling immediately (D) none of these -8. A body of mass m and radius r is rotated with angular velocity as shown in the figure & kept on a surface that has sufficient friction then the body will move : (A) backward first and then move forward (B) forward first and then move backward (C) will always move forward (D) none of these -9. A body of mass m and radius R rolling horizontally without slipping at a speed climbs a ramp to a height 2 3 4g . The rolling body can be [Olympiad 2015 (stage-1)] (A) a sphere (B) a circular ring (C) a spherical shell (D) a circular disc Rigid Body Dynamics Corp. / Reg. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.) – 324005 Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in ADVRB - 60 Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029 Section (J) : Conservation of angular momentum (combined translation & rotational motion) J-1. A sphere is released on a smooth inclined plane from the top. When it moves down its angular momentum is: (A) conserved about every point (B) conserved about the point of contact only (C) conserved about the centre of the sphere only (D) conserved about any point on a fixed line parallel to the inclined plane and passing through the centre of the ball. J-2. A circular wooden loop of mass m and radius R rests flat on a horizontal frictionless surface. A bullet, also of mass m, and moving with a velocity V, strikes the loop and gets embedded in it. The thickness of the loop is much smaller than R. The angular velocity with which the system rotates just after the bullet strikes the loop is (A) V 4R (B) V 3R (C) 2V 3R (D) 3 V 4R Section (K) : Toppling K-1. A uniform cube of side a and mass m rests on a rough horizontal table. A horizontal force 'F' is applied normal to one of the faces at a point that is directly above the centre of the face, at a height 3a 4 above the base. The minimum value of 'F' for which the cube begins to tilt about the edge is (assume that the cube does not slide). [JEE - 1984] (A) 2 mg 3 (B) 4 mg 3 (C) 5 mg 4 (D) 1mg 2 K-2. A homogenous block having its cross-section to be a parallelogram of sides 'a' and 'b' (as shown) is lying at rest and is in equilibrium on a smooth horizontal surface. Then for acute angle : b a (A) b cos a (B) b cos a (C) b cos a (D) b cos a (E) a cos b K-3. An equilateral uniform prism of mass m rests on a rough horizontal surface with coefficient of friction . A horizontal force F is applied on the prism as shown in the figure. If the coefficient of friction is sufficiently high so that the prism does not slide before toppling, then the minimum force required to topple the prism is : (A) mg 3 (B) mg 4 (C) mg 3 (D) mg 4 Rigid Body Dynamics Corp. / Reg. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.) – 324005 Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in ADVRB - 61 Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029 PART - III : MATCH THE COLUMN 1. In each situation of column-I, a uniform disc of mass m and radius R rolls on a rough fixed horizontal surface as shown in the figure. At t = 0 (initially) the angular velocity of disc is and velocity of centre of mass of disc is v (in horizontal direction). The relation between v and for each situation and also initial sense of rotation is given for each situation in column-I. Then match the statements in column-I with the corresponding results in column-II. Column-I Column-II (A) (v > R ) (p) The angular momentum of disc about point A (as shown in figure) remains conserved. (B) (v > R ) (q) The kinetic energy of disc after it starts rolling without slipping is less than its initial kinetic energy. (C) (v < R ) (r) In the duration disc rolls with slipping, the friction acts on disc towards left. (D) (v < R ) (s) In the duration disc rolls with slipping, the friction acts on disc for some time towards right and for some time towards left. 2. A uniform disc rolls without slipping on a rough horizontal surface with uniform angular velocity. Point O is the centre of disc and P is a point on disc as shown in the figure. In each situation of column a statement is given and the corresponding results are given in column-II. Match the statements in column-I with the results in column-II. Column Column (A) The velocity of point P on disc (p) Changes in magnitude with time. (B) The acceleration of point P on disc (q) is always directed from that point (the point on disc given in column-I) towards centre of disc. (C) The tangential acceleration of point P on disc (r) is always zero. (D) The acceleration of point on disc which is in (s) is non-zero and remains constant contact with rough horizontal surface in magnitude. Rigid Body Dynamics Corp. / Reg. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.) – 324005 Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in ADVRB - 62 Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029 Marked Questions can be used as Revision Questions. PART - I : ONLY ONE OPTION CORRECT TYPE 1. Three rings each of mass m and radius r are so placed that they touch each other. The radius of gyration of the system about the axis as shown in the figure is : (A) 6 5 r (B) 5 6 r (C) 6 7 r (D) 7 6 r 2. A hollow cylinder has mass M, outside radius R2 and inside radius R1. Its moment of inertia about an axis parallel to its symmetry axis and tangential to the outer surface is equal to : (A) M 2 (R22 + R12) (B) M 2 (R22 – R12) (C) M 4 (R2 + R1)2 (D) M 2 (3R22 + R12) 3. Two uniform rods of equal length but different masses are rigidly joined to form an Lshaped body, which is then pivoted about O as shown in the figure. If in equilibrium the body is in the shown configuration, ratio M/m will be: m O M 90° 30° (A) 2 (B) 3 (C) 2 (D) 3 4. Four forces tangent to the circle of radius ‘R’ are acting on a wheel as shown in the figure. The resultant equivalent one force system will be : (A) (B) (C) (D) Rigid Body Dynamics Corp. / Reg. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.) – 324005 Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in ADVRB - 63 Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029 5. A uniform thin rod of mass ‘m’ and length L is held horizontally by two vertical strings attached to the two ends. One of the string is cut. Find the angular acceleration soon after it is cut : (A) g 2L (B) g L (C) 3g 2L (D) 2g L 6. A uniform rod hinged at its one end is allowed to rotate in vertical plane. Rod is given an angular velocity in its vertical position as shown in figure. The value of for which the force exerted by the hinge on rod is zero in this position is : (A) g L (B) 2g L (C) g 2L (D) 3g L 7. Two points A & B on a disc have velocities v1 & v2 at some moment. Their directions make angles 60° and 30° respectively with the line of separation as shown in figure. The angular velocity of disc is : (A) 1 3v d (B) 2 v 3d (C) 2 1 v v d (D) 2 v d 8. Uniform rod AB is hinged at the end A in a horizontal position as shown in the figure (the hinge is frictionless, that is, it does not exert any friction force on the rod). The other end of the rod is connected to a block through a massless string as shown. The pulley is smooth and massless. Masses of the block and the rod are same and are equal to ' m '. Acceleration due to gravity is g. The tension in the thread, and angular acceleration of the rod just after release of block from this position (A) 3mg 8 , g 8 (B) 5mg 8 , 3g 8 (C) mg 8 , 5g 8 (D) 7mg 8 , 7g 8 9. In the figure shown a ring A is rolling without sliding with a velocity v on the horizontal surface of the body B (of same mass as A). All surfaces are smooth. B has no initial velocity. What will be the maximum height (from initial position) reached by A on B. (A) 2 3 v 4 g (B) 2 v 4 g (C) 2 v 2 g (D) 2 v 3 g Rigid Body Dynamics Corp. / Reg. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.) – 324005 Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in ADVRB - 64 Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029 10. A uniform rod of mass m, length is placed over a smooth horizontal surface along y-axis and is at rest as shown in figure. An impulsive force F is applied for a small time t along x-direction at point A after this rod moves freely. The x–coordinate of end A of the rod when the rod becomes parallel to x–axis for the first time is (initially the coordinate of centre of mass of the rod is (0, 0)) : (A) 12 (B) 1 2 12 (C) 1 2 6 (D) 1 2 6 11. A uniform rectangular plate of mass m which is free to rotate about the smooth vertical hinge passing through the centre and perpendicular to the plate, is lying on a smooth horizontal surface. A particle of mass m moving with speed 'u' collides with the plate and sticks to it as shown in figure. The angular velocity of the plate after collision will be : 2a a m u (A) 12 u 5 a (B) 12 u 19 a (C) 3 u 2 a (D) 3 u 5 a 12. A rod can rotate about a fixed vertical axis. The mass is non-uniformly distributed along the length of the rod. A horizontal force of constant magnitude and always perpendicular to the rod is applied at the end. Which of the following quantity (after one rotation) will not depend on the information that through which end the axis passes ? (Assuming initial angular velocity to be zero) (A) angular momentum (B) kinetic energy (C) angular velocity (D) none of these 13. A particle is attached to the lower end of a uniform rod which is hinged at its other end as shown in the figure. The minimum speed given to the particle so that the rod performs circular motion in a vertical plane will be : [length of the rod is , consider masses of both rod and particle to be same ] (A) 5g (B) 4g (C) 4.5g (D) none of these 14. A particle of mass m is moving horizontally at speed v perpendicular to a uniform rod of length d and mass M = 6m. The rod is hinged at centre O and can freely rotate in horizontal plane about a fixed vertical axis passing through its centre O. The hinge is frictionless.The particle strikes and sticks to the end of the rod. The angular speed of the system just after the collision : (A) 2v/3d (B) 3v/2d (C) v/3d (D) 2v/d Rigid Body Dynamics Corp. / Reg. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.) – 324005 Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in ADVRB - 65 Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029 15. A uniform sphere of mass ' m ' is given some angular velocity about a horizontal axis through its centre and gently placed on a plank of mass ' m '. The co-efficient of friction between the two is . The plank rests on a smooth horizontal surface. The initial acceleration of the centre of sphere relative to the plank will be : m m (A) zero (B) g (C) (7/5) g (D) 2 g 16. A rod of negligible mass and length is pivoted at the point /4 distance from the left end as shown. A particle of mass m is fixed to its left end & another particle of mass 2 m is fixed to the right end. If the system is released from rest and after sometime becomes vertical, the speed v of the two masses and angular velocity at that instant is (A) 1 40g 4 19 , 3 40g 4 19 , 40g 19 (B) 1 20g 2 19 , 3 20g 4 19 , 20g 19 (C) 1 20g 4 19 , 3 20g 4 19 , 20g 19 (D) 1 40g 2 19 , 1 40g 2 19 , 40g 19 17. Two identical balls A & B of mass m each are placed on a fixed wedge as shown in figure. Ball B is kept at rest and it is released just before two balls collide. Ball A rolls down without slipping on inclined plane & collides elastically with ball B. The kinetic energy of ball A just after the collision with ball B is (Neglect friction between A and B, also neglect the radius of the balls) : (A) mgh 7 (B) mgh 2 (C) 2mgh 5 (D) 7mgh 5 18. A solid uniform disc of mass m rolls without slipping down an inclined plane with an acceleration a. The frictional force on the disc due to surface of the plane is (A) 2 ma (B) 3 2 ma (C) ma (D) 1 2 ma 19. A small block of mass ‘m’ is rigidly attached at ‘P’ to a ring of mass ‘3m’ and radius ‘r’. The system is released from rest at = 90° and rolls without sliding. The angular acceleration of ring just after release is P (A) g 4r (B) g 8r (C) g 3r (D) g 2r 20. A uniform ring of radius R is given a back spin of angular velocity V0/2R and thrown on a horizontal rough surface with velocity of center to be V0. The velocity of the centre of the ring when it starts pure rolling will be (A) V0/2 (B) V0/4 (C) 3V0/4 (D) 0 21. A box of dimensions and b is kept on a truck moving with an acceleration a. If box does not slide, maximum acceleration for it to remain in equilibrium (w.r.t.truck) is : (A) g b (B) g b (C) g (D) none of these a b l Rigid Body Dynamics Corp. / Reg. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.) – 324005 Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in ADVRB - 66 Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029 22. If the positions of two like parallel forces on a light rod are interchanged, their resultant shifts by onefourth of the distance between them then the ratio of their magnitude is: (A) 1 : 2 (B) 2 : 3 (C) 3 : 4 (D) 3 : 5 23. Consider two point masses m1 and m2 connected by alight rigid rod of length r0. The moment of inertia of the system about an axis passing through their centre of mass and perpendicular to the rigid rod is given by [Olympiad (Stage-1) 2017] (A) 2 1 2 0 1 2 m m r 2 m m (B) 2 1 2 0 1 2 m m r m m (C) 2 1 2 0 1 2 2m m r m m (D) 2 2 2 1 2 0 1 2 m m r m m 24. In the following arrangement the pulley is assumed to be light and the string inextensible. The acceleration of the system can be determined by considering conservation of a certain physical quantity. The physical quantity conserved and the acceleration respectively, are [Olympiad (Stage-1) 2017] 1kg 2kg 3kg (A) energy and g/3 (B) linear momentum and g/2 (C) angular momentum and g/3 (D) mass and g/2 PART - II : SINGLE AND DOUBLE VALUE INTEGER TYPE 1. A wheel starting from rest is uniformly accelerated at 2 rad/s2 for 20 seconds. It rotates uniformly for the next 20 seconds and is finally brought to rest in the next 20 seconds. Total angular displacement of the wheel is n × 102 radian where n is. 2. Two steel balls of equal diameter are connected by a rigid bar of negligible weight as shown & are dropped in the horizontal position from height h above the heavy steel and brass base plates. If the coefficient of restitution between the ball & steel base is 0.6 & that between the other ball & the brass base is 0.4. The angular velocity of the bar immediately after rebound is n×10-2 rad/s where n is : (Assume the two impacts are simultaneous.) (g = 9.8m/s2) 3. Three identical uniform rods, each of length, are joined to form a rigid equilateral triangle. Its radius of gyration about an axis passing through a corner and perpendicular to the plane of the triangle is n where n is : 4. The moment of inertia of a thin uniform rod of mass m & length about an axis passing through one end & making angle = 45º with its length is 2 m n where n is. 5. A uniform rod of mass m and length L is suspended with two massless strings as shown in the figure. If the rod is at rest in a horizontal position the ratio of tension in the two strings T1/T2 is: T 1 T 2 L 3L/4 6. Two persons of equal height are carrying a long uniform wooden beam of length . They are at distance /4 and /6 from nearest ends of the rod. The ratio of normal reactions at their heads is n : 3 where n is : Rigid Body Dynamics Corp. / Reg. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.) – 324005 Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in ADVRB - 67 Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029 7. A (i) ring and (ii) uniform disc both of radius R is given an angular velocity and then carefully placed on a horizontal surface such that its axis is vertical. If the coefficient of friction is for both cases then the ratio of the time taken by the ring and disk to come to rest is n : 3 where n is : (The pressure exerted by the disc and ring on the surface can be regarded as uniform). 8. Each of the double pulleys shown has a centroidal mass moment of inertia of ‘mr2’, inner radius r and an outer radius 2r. Assuming that the bearing friction of hinge at A and at B is equivalent to torque of magnitude mgr 4 then the tension (in N) in the string connecting the pulleys is : (m = 3kg, g = 10m/s2, and r =0.1m) 9. The pulley shown in figure has a radius of 10 cm and moment of inertia 0.1 kg-m2. The string going over it attached at one end to a vertical spring of spring constant 100 N/m fixed from below, and supports a 11 kg mass at the other end. The system is released from rest with the spring at its natural length. Find the speed (in m/s) of the block when it has descended through 10 cm. (Take g = 10 m/s2 and assume that the there is no slipping between string and pulley). 10. The angular momentum of a particle about origin is varying as L = 4 2 t + 8 (S units) when it moves along a straight line y = x – 4 (x, y in meters). The magnitude of force (in N) acting on the particle would be : 11. A plank P is placed on a solid cylinder S, which rolls on a horizontal surface. The two are of equal mass. There is no slipping at any of the surfaces in contact. The ratio of the kinetic energy of P to the kinetic energy of S is n : 3 where n is : 12. A uniform rod of mass m = 5.0 kg and length = 90 cm rests on a smooth horizontal surface. One of the ends of the rod is struck with the impulse J = 3.0 N-s in a horizontal direction perpendicular to the rod and removed. As a result of which the rod gets angular velocity and linear velocity instantaneously. The force (in N) with which one half of the rod will act on the other in the process of motion later on. 13. A uniform solid sphere of mass m and radius r is projected along a rough horizontal surface with the initial velocity v0 and angular velocity 0 as shown in the figure. If the sphere finally comes to complete rest then 0 0 2 r v is equal to : w 0 v 0 /////////////////////////////////////////////////////// 14. A thin uniform rod AB of mass m = 1.0 kg moves translationally with acceleration w = 2.0 m/s2 due to two antiparallel forces F1 and F2 (Fig.). The distance between the points at which these forces are applied is equal to a = 20 cm. Besides, it is known that F2 = 5.0.N. Find the length (in m) of the rod. a F 1 F 2 B Rigid Body Dynamics Corp. / Reg. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.) – 324005 Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in ADVRB - 68 Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029 15. A hollow smooth uniform sphere A of mass ‘m’ rolls without sliding on a smooth horizontal surface. It collides elastically and headon with another stationary smooth solid sphere B of the same mass m and same radius. The ratio of kinetic energy of ‘B’ to that of ‘A’ just after the collision is 3 : n where n is : 16. The free end of the string wound on the surface of a solid cylinder of mass M = 1kg & radius R = 2 3 m is pulled up by a force F as shown. If there is sufficient friction between cylinder & floor then the upper limit to the angular acceleration (in rad/s2) of the cylinder for which it rolls without slipping is : (g = 10m/s2) F M 17. A uniform solid cylinder of mass m = 1kg rests on two horizontal planks. A thread is wound on the cylinder. The hanging end of the thread is pulled vertically down with a constant force F. (a) Find the maximum magnitude of the force F(in N) which still does not bring about any sliding of the cylinder, if the coefficient of friction between the cylinder and the planks is equal to = 1 3 . (b) The acceleration amax of the axis of the cylinder rolling on the planks is n 3 m/s2 where n is: 18. A cylinder rotating at an angular speed of 50 rev/s is brought in contact with an identical stationary cylinder. Because of the kinetic friction, torques act on the two cylinders, accelerating the stationary one and decelerating the moving one. If the common magnitude of the angular acceleration and deceleration be1 rev/s2, then how much time (ins) will it take before the two cylinders have equal angular speed ? 19. A uniform disc of mass M = 1kg, radius R = 1m is moving towards right on smooth horizontal surface with velocity v0 = 20 m/s & having angular velocity 0 = 4 rad/s about the perpendicular axis outward the plane of disc passing through centre of disc. Suddenly top point of the disc gets hinged about a fixed smooth axis. The angular velocity (in rad/s) of disk about new rotation axis is: Rigid Body Dynamics Corp. / Reg. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.) – 324005 Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in ADVRB - 69 Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029 20. A regular hexagonal uniform block of mass m = 4 3 kg rests on a rough horizontal surface with coefficient of friction µ as shown in figure. A constant horizontal force is applied on the block as shown. If the coefficient of friction is sufficient to prevent slipping before toppling, then the minimum force(in N) required to topple the block about its corner A is:(g = 10m/s2) A 21. A uniform rod of length = 1m is kept as shown in the figure. H is a horizontal smooth surface and W is a vertical smooth wall. The rod is released from this position. The angular acceleration (in radian/sec2) of the rod just after the release is : 22. A uniform circular disc has radius R & mass m. A particle also of mass m is fixed at a point A on the edge of the disc as shown in the figure. The disc can rotate freely about a fixed horizontal chord PQ that is at a distance R/4 from the centre C of the disc. The line AC is perpendicular to PQ. Initially the disc is held vertical with the point A at its highest position. It is then allowed to fall so that it starts rotating about PQ. The linear speed of the particle as it reaches its lowest position is ngR , where n is an integer. Find the value of n [JEE - 1998’ 8/200] C P Q R/4 A R PART - III : ONE OR MORE THAN ONE OPTIONS CORRECT TYPE 1. A rigid body is in pure rotation. (A) You can find two points in the body in a plane perpendicular to the axis of rotation having same velocity. (B) You can find two points in the body in a plane perpendicular to the axis of rotation having same acceleration. (C) Speed of all the particles lying on the curved surface of a cylinder whose axis coincides with the axis of rotation is same. (D) Angular speed of the body is same as seen from any point in the body. 2. A sphere is rotating uniformly about a fixed axis passing through its centre then: (A) The particles on the surface of the sphere do not have any angular acceleration. (B) The particles on the axis do not have any linear acceleration (C) Different particles on the surface have same angular speeds. (D) All the particles on the surface have same linear speed 3. The moment of inertia of a thin uniform square plate ABCD of uniform thickness about an axis passing through the centre O and perpendicular to the plate is - [REE - 1992] O 3 B 1 A C D 2 4 (A) 1 + 2 (B) 3 + 4 (C) 1 + 3 (D) 1 + 2 + 3 + 4 where 1, 2, 3, and 4 are respectively the moments of inertia about axes 1, 2, 3, and 4 which are in the plane of the plate. Rigid Body Dynamics Corp. / Reg. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.) – 324005 Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in ADVRB - 70 Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029 4. In the figure shown a uniform rod of mass m and length is hinged. The rod is released when the rod makes angle = 60° with the vertical. (A) The angular acceleration of the rod just after release is 3 3g 4 (B) The normal reaction due to the hinge just after release is 19mg 8 (C) The angular velocity of the rod at the instant it becomes vertical is 3g 2 (D)The normal reaction due to the hinge at the instant the rod becomes vertical is 7 4 mg 5. Consider a disc rolling without slipping on a horizontal surface at a linear speed V as shown in figure (A) the speed of the particle A is 2V (B) the speed of B, C and D are all equal to V (C) the speed of C is zero and speed of B is 2V (D) the speed of O is less than the speed of B 6. A cylinder rolls without slipping over a horizontal plane with constant velocity. The radius of the cylinder is equal to r. At this moment (A) The speed of B is 2 times the speed of A. (B) The radius of curvature of trajectory traced out by A is 4r. (C) The radius of curvature of trajectory traced out by B is 2 2 r. (D) The radius of curvature of trajectory traced out by C is r. C v B A 7. A uniform solid sphere is released from rest from the top of an inclined plane of inclination . Then choose the correct option(s). (A) The minimum coefficient of friction between sphere and the incline to prevent slipping is 2 7 tan . (B) The kinetic energy of solid sphere as it moves down a distance S on the incline is mgSsin if 2 7 tan (C) The magnitude of work done by friction on the solid sphere is less than mgScos as it moves down a distance S on the incline if 2 7 tan . (D) The magnitude of work done by friction on the solid sphere is equal to mgScos as it moves down a distance S on the incline if 2 7 tan . 8. A uniform solid cylinder rolls without slipping on a rough horizontal floor, its centre of mass moving with a speed v. It makes an elastic collision with smooth vertical wall. After impact: (A) its centre of mass will move with a speed v initially (B) its motion will be rolling without slipping immediately (C) its motion will be rolling with slipping initially and its rotational motion will stop momentarily at some instant (D) its motion will be rolling without slipping only after some time. 9. If L = 0 for a rigid body, where = resultant torque & L = angular momentum about a point and both are non-zero. Then : (A) L may be constant (B) | L |= constant (C) | L |may decrease (D) | L |may increase Rigid Body Dynamics Corp. / Reg. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.) – 324005 Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in ADVRB - 71 Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029 10. In absence of external forces on a rigid system, which of the following quantities must remain constant? (A) angular momentum (B) linear momentum (C) moment of inertia about fixed axis through any point on body (D) kinetic energy 11. In the given figure a ball strikes a rod elastically and rod is smoothly hinged at point A. Then which of the statement(s) is/are correct for the collision? (A) linear momentum of system (ball + rod) is conserved (B) angular momentum of system about hinged point A is conserved (C) initial KE of the system is equal to final KE of the system (D) linear momentum of ball is conserved. 12. A horizontal disc rotates freely about a vertical fixed axis through its centre. A ring, having the same mass and radius as the disc, is now gently placed on the disc coaxially. After some time the two rotate with a common angular velocity: (A) some friction exists between the disc and the ring before achieving common angular velocity (B) the angular momentum of the 'disc plus ring' about axis of rotation is conserved (C) the final common angular velocity is 2/3rd of the initial angular velocity of the disc (D) The final common angular velocity is 1/3rd of the initial angular velocity of the disc 13. A wheel (to be considered as a ring) of mass m and radius R rolls without sliding on a horizontal surface with constant velocity v. It encounters a step of height R/2 at which it ascends without sliding. R R/2 /////////////////////////////// /////////////////////////////// / / / / (A) the angular velocity of the ring just after it comes in contact with the step is 3v/4R (B) the normal reaction due to the step on the wheel just after the impact is mg 2 – 2 9 mv 16 R (C) the normal reaction due to the step on the wheel increases as the wheel ascends (D) the friction will be absent during the ascent. 14. A hollow sphere is set into motion on a rough horizontal surface with a speed v in the forward direction and an rotational speed v/R in the anticlockwise direction as shown in figure. Find the translational speed of the sphere (a) when it stops rotating and (b) when slipping finally ceases and pure rolling starts. (A) The angular momentum of the sphere about its centre of mass is conserved. (B) The speed of the sphere at the instant it stops rotating momentarily is v/3 (C) The speed of the sphere after pure rolling starts is v/5 (D) Work done by fricition upto pure rolling starts is zero. 15. Two small particles A and B of masses M and 2M respectively, are joined rigidly to the ends of a light rod of length as shown in the figure. The system translates on a smooth horizontal surface with a velocity u in a direction perpendicular to the rod. A particle P of mass M kept at rest on the surface sticks to the particle A as the particle P collides with it. Then choose the correct option(s) (A) The speed of particle A just after collision is u 2 (B) The speed of particle B just after collision is u 2 (C) The velocity of centre of mass of system A+B+P is 3u 4 (D) The angular speed of the system A+B+P after collision is u 2 Rigid Body Dynamics Corp. / Reg. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.) – 324005 Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in ADVRB - 72 Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029 16. A block with a square base measuring a a and height h, is placed on an inclined plane. The coefficient of friction is . The angle of inclination () of the plane is gradually increased. The block will: (A) topple before sliding if > a h (B) topple before sliding if < a h (C) slide before toppling if > a h (D) slide before toppling if < a h 17. A uniform disc of mass M and radius R is lifted using a string as shown in the figure. Then, [Olympiad 2014 (stage-1)] (A) its linear acceleration is g upward (B) its linear acceleration is g downward (C) its angular acceleration is 2g.
R (D) its rate of change of angular momentum is MgR. PART - IV : COMPREHENSION Comprehension-1 A uniform disc of mass ‘m’ and radius R is free to rotate in horizontal plane about a vertical smooth fixed axis passing through its centre. There is a smooth groove along the diameter of the disc and two small balls of mass m 2 each are placed in it on either side of the centre of the disc as shown in fig. The disc is given initial angular velocity 0 and released. 1. The angular speed of the disc when the balls reach the end of the disc is : (A) 0 2 (B) 0 3 (C) 0 2 3 (D) 0 4 2. The speed of each ball relative to ground just after they leave the disc is : (A) 0 R 3 (B) 0 R 2 (C) 0 2R 3 (D) 0 R 3 3. The net work done by forces exerted by disc on one of the ball for the duration ball remains on the disc is (A) 2 2 0 2mR 9 (B) 2 2 0 mR 18 (C) 2 2 0 mR 6 (D) 2 2 0 mR 9 Rigid Body Dynamics Corp. / Reg. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.) – 324005 Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in ADVRB - 73 Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029 Comprehension-2 A uniform disc of mass M and radius R initially stands vertically on the right end of a horizontal plank of mass M and length L, as shown in the figure. The plank rests on smooth horizontal floor and friction between disc and plank is sufficiently high such that disc rolls on plank without slipping. The plank is pulled to right with a constant horizontal force of magnitude F. 4. The magnitude of acceleration of plank is (A) F 8M (B) F 4M (C) 3F 2M (D) 3F 4M 5. The magnitude of angular acceleration of the disc is - (A) F 4mR (B) F 8mR (C) F 2mR (D) 3F 2mR 6. The distance travelled by centre of disc from its initial position till the left end of plank comes vertically below the centre of disc is (A) L 2 (B) L 4 (C) L 8 (D) L Marked Questions can be used as Revision Questions. Marked Questions may have more than one correct option. PART - I : JEE (ADVANCED) / IIT-JEE PROBLEMS (PREVIOUS YEARS) 1. A solid sphere of radius R has moment of inertia about its geometrical axis. If it is melted into a disc of radius r and thickness t. If it's moment of inertia about the tangential axis (which is perpendicular to plane of the disc), is also equal to , then the value of r is equal to : [JEE 2006, 3/184] r I (A) 2 15 R (B) 2 5 R (C) 3 15 R (D) 3 15 R 2. A solid sphere is in pure rolling motion on an inclined surface having inclination . [JEE 2006, 5/184] ////////////////////// (A) frictional force acting on sphere is f = mg cos . (B) f is dissipative force. (C) friction will increase its angular velocity and decreases its linear velocity. (D) If decreases, friction will decrease. Rigid Body Dynamics Corp. / Reg. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.) – 324005 Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in ADVRB - 74 Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029 3. A ball moves over a fixed track as shown in the figure. From A to B the ball rolls without slipping. If surface BC is frictionless and KA, KB and KC are kinetic energies of the ball at A, B and C respectively, then : [JEE 2006, 5/184] (A) hA > hC ; KB > KC (B) hA > hC ; KC > KA (C) hA = hC ; KB = KC (D) hA < hC ; KB > KC 4. A rectangular plate of mass M of dimensions (a × b) is hinged along one edge. The plate is maintained in horizontal position by colliding a ball of mass m, per unit area, elastically 100 times per second this ball is striking on the right half shaded region of the plate as shown in figure. Find the required speed of the ball (ball is colliding in only half part of the plate as shown). (It is given M = 3 kg, m = 0.01 kg, b = 2 m, a = 1 m, g = 10 m/s2) [JEE 2006, 6/184] × × a b b/2 × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × Paragraph for Question Nos. 5 to 7 Two discs A and B are mounted coaxially on a vertical axle. The discs have moments of inertia and 2 respectively about the common axis. Disc A is imparted an initial angular velocity 2 using the entire potential energy of a spring compressed by a distance x1. Disc B is imparted an angular velocity by a spring having the same spring constant and compressed by a distance x2. Both the discs rotate in the clockwise direction. [JEE-2007, 12/162] 5. The ratio x1/x2 is (A) 2 (B) 1 2 (C) 2 (D) 1 2 6. When disc B is brought in contact with disc A, they acquire a common angular velocity in time t. The average frictional torque on one disc by the other during this period is (A) 2 3t (B) 9 2t (C) 9 4t (D) 3 2t 7. The loss of kinetic energy during the above process is (A) 2 2 (B) 2 3 (C) 2 4 (D) 2 6 8. A small object of uniform density rolls up a curved surface with an initial velocity v. It reaches up to a maximum height of 2 3v 4g with respect to the initial position. The object is [JEE-2007, 3/162] v (A) ring (B) solid sphere (C) hollow sphere (D) disc Rigid Body Dynamics Corp. / Reg. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.) – 324005 Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in ADVRB - 75 Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029 9. STATEMENT-1 : If there is no external torque on a body about its centre of mass, then the velocity of the center of mass remains constant. [JEE-2007, 3/162] because STATEMENT-2 : The linear momentum of an isolated system remains constant. (A) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1 (B) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement1 (C) Statement-1 is True, Statement-2 is False (D) Statement-1 is False, Statement-2 is True 10. STATEMENT-1 : Two cylinders, one hollow (metal) and the other solid (wood) with the same mass and identical dimensions are simultaneously allowed to roll without slipping down an inclined plane from the same height. The hollow cylinder will reach the bottom of the inclined plane first. [JEE-2008, 3/163] and STATEMENT-2 : By the principle of conservation of energy, the total kinetic energies of both the cylinders are identical when they reach the bottom of the incline. (A) STATEMENT-1 is True, STATEMENT-2 is True; STATEMENT-2 is a correct explanation for STATEMENT-1 (B) STATEMENT-1 is True, STATEMENT-2 is True; STATEMENT-2 is NOT a correct explanation for STATEMENT-1 (C) STATEMENT-1 is True, STATEMENT-2 is False (D) STATEMENT-1 is False, STATEMENT-2 is True. 11. If the resultant of all the external forces acting on a system of particles is zero, then from an inertial frame, one can surely say that [JEE 2009, 4/160, –1] (A) linear momentum of the system does not change in time (B) kinetic energy of the system does not changes in time (C) angular momentum of the system does not change in time (D) potential energy of the system does not change in time 12. A sphere is rolling without slipping on a fixed horizontal plane surface. In the figure, A is the point of contact, B is the centre of the sphere and C is its topmost point. Then, [JEE 2009, 4/160, –1] (A) C A V V = B C 2 V V (B) C B V V = B A V V (C) C A V V = B C 2 V V (D) C A V V = B 4 V 13. A block of base 10 cm × 10 cm and height 15 cm is kept on an inclined plane. The coefficient of friction between them is 3 . The inclination of this inclined plane from the horizontal plane is gradually increased from 0º. Then [JEE 2009, 3/160, –1] (A) at = 30º, the block will start sliding down the plane (B) the block will remain at rest on the plane up to certain and then it will topple (C) at = 60º, the block will start sliding down the plane and continue to do so at higher angles (D) at = 60º, the block will start sliding down the plane and on further increasing , it will topple at certain Rigid Body Dynamics Corp. / Reg. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.) – 324005 Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in ADVRB - 76 Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029 14. A boy is pushing a ring of mass 2 kg and radius 0.5 m with a vertical stick as shown in the figure. The stick applies a normal force of 2 N on the ring and rolls it without slipping with an acceleration of 0.3 m/s2. The coefficient of friction between the ground and the ring is large enough that rolling always occurs and the coefficient of friction between the stick and the ring is (P/10). The value of P is [JEE 2011, 4/160] 15. Four solid spheres each of diameter 5 cm and mass 0.5 kg are placed with their centers at the corners of a square of side 4cm. The moment of inertia of the system about the diagonal of the square is N × 10–4 kg-m2, then N is [JEE 2011, 4/160] 16. A thin uniform rod, pivoted at O, is rotating in the horizontal plane with constant angular speed , as shown in the figure. At time, t = 0, a small insect starts from O and moves with constant speed v with respect to the rod towards the other end. It reaches the end of the rod at t = T and stops. The angular speed of the system remains throughout. The magnitude of the torque (| |) on the system about O, as a function of time is best represented by which plot ? [IIT-JEE-2012, Paper-1; 3/70, –1] (A) (B) (C) (D) Rigid Body Dynamics Corp. / Reg. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.) – 324005 Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in ADVRB - 77 Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029 17. A small mass m is attached to a massless string whose other end is fixed at P as shown in the figure. The mass is undergoing circular motion is the x-y plane with centre at O and constant angular speed . If the angular momentum of the system, calculated about O and P are denoted O L by P L and respectively, then [IIT-JEE-2012, Paper-1; 3/70, –1] (A) O L and P L do not vary with time. (B) O L varies with time while P L remains constant. (C) O L remains constant while P L varies with time. (D) O L and P L both vary with time. 18. A lamina is made by removing a small disc of diameter 2R from a bigger disc of uniform mass density and radius 2R, as shown in the figure. The moment of inertia of this lamina about axes passing through O and P is IO and IP, respectively. Both these axes are perpendicular to the plane of the lamina. The ratio P O I I to the nearest integer is : [IIT-JEE-2012, Paper-1; 4/70] 2R 2R O P Paragraph for Q. No. 19-20 The general motion of a rigid body can be considered to be a combination of (i) a motion of its centre of mass about an axis, and (ii) its motion about an instantanneous axis passing through center of mass. These axes need not be stationary. Consider, for example, a thin uniform welded (rigidly fixed) horizontally at its rim to a massless stick, as shown in the figure. Where disc-stick system is rotated about the origin ona horizontal frictionless plane with angular speed , the motion at any instant can be taken as a combination of (i) a rotation of the centre of mass the disc about the z-axis, and (ii) a rotation of the disc through an instantaneous vertical axis passing through its centre of mass (as is seen from the changed orientation of points P and Q). Both the motions have the same angular speed in the case. Z Y Q P X P Q Now consider two similar systems as shown in the figure: case (a) the disc with its face vertical and parallel to x-z plane; Case (b) the disc with its face making an angle of 45° with x-y plane its horizontal Rigid Body Dynamics Corp. / Reg. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.) – 324005 Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in ADVRB - 78 Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029 diameter parallel to x-axis. In both the cases, the disc is welded at point P, and systems are rotated with constant angular speed about the z-axis. Z Y Q P X Case (a) Z Y Q P X Case (b) 45° 19. Which of the following statement regarding the angular speed about the istantaneous axis (passing through the centre of mass) is correct? [IIT-JEE-2012, Paper-2; 3/66, –1] (A) It is 2 for both the cases (B) it is for case (a); and 2 for case (b). (C) It is for case (a); and 2 for case (b) (D) It is for both the cases 20. Which of the following statements about the instantaneous axis (passing through the centre of mass) is correct? [IIT-JEE-2012, Paper-2; 3/66, –1] (A) It is vertical for both the cases (a) and (b). (B) It is verticle for case (a); and is at 45° to the x-z plane and lies in the plane of the disc for case (b) (C) It is horizontal ofr case (a); and is at 45° to the x-z plane and is normal to the plane of the disc for case (b). (D) It is vertical of case (a); and is at 45° to the x - z plane and is normal to the plane of the disc for case (b). 21. Two solid cylinders P and Q of same mass and same radius start rolling down a fixed inclined plane form the same height at the same time. Cylinder P has most of its mass concentrated near its surface, while Q has most of its mass concentrated near the axis. Which statement (s) is (are) correct? (A) Both cylinders P and Q reach the ground at the same time [IIT-JEE-2012, Paper-2; 4/66] (B) Cylinder P has larger linear acceleration than cylinder Q. (C) Both cylinder reaches the ground with same translational kinetic energy. (D) Cylinder Q reaches the ground with larger angular speed. 22. The figure shows a system consisting of (i) a ring of outer radius 3R rolling clockwise without slipping on a horizontal surface with angular speed and (ii) an inner disc of radius 2R rotating anti-clockwise with angular speed /2. The ring and disc are separated frictionaless ball bearings. The system is in the x-z plane. The point P on the inner disc is at distance R from the origin O, where OP makes an angle of 30º with the horizontal. Then with respect to the horizontal surface, [IIT-JEE-2012, Paper-2; 4/66] (A) the point O has linear velocity 3Rˆ i . (B) the point P has a linear velocity 11 4 Rˆ i + 3 4 Rˆ k (C) the point P has linear velocity 13 4 Rw ˆ i – 3 4 Rˆ k (D) The point P has a linear velocity 3 3 4 Rw ˆ i + 1 4 Rw ˆ k . Rigid Body Dynamics Corp. / Reg. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.) – 324005 Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in ADVRB - 79 Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029 23. A uniform circular disc of mass 50kg and radius 0.4 m is rotating with an angular velocity of 10 rad/s–1 about its own axis, which is vertical. Two uniform circular rings, each of mass 6.25 kg and radius 0.2 m, are gently placed symmetrically on the disc in such a manner that they are touching each other along the axis of the disc and are horizontal. Assume that the friction is large enough such that the rings are at rest relative to the disc and the system rotates about the original axis. The new angular velocity (in rad s–1) of the system is : [JEE (Advanced) 2013; 4/60] 24. In the figure, a ladder of mass m is shown leaning against a wall. It is in static equilibrium making an angle with the horizontal floor. The coefficient of friction between the wall and the ladder is 1 and that between the floor and the ladder is 2. The normal reaction of the wall on the ladder is N1 and that of the floor is N2. If the ladder is about to slip, then [JEE (Advanced) 2014, P-1, 3/60] (A) 1 = 0 2 0 and N2 tan = mg 2 (B) 1 0 2 0 and N1 tan = mg 2 (C) 1 0 2 0 and N2 = 1 2 mg 1 (D) 1 0 2 0 and N1 tan = mg 2 25. A horizontal circular platform of radius 0.5 m and mass 0.45 kg is free to rotate about its axis. Two massless spring toy-guns, each carrying a steel ball of mass 0.05 kg are attached to the platfrom at a distance 0.25 m from the centre on its either sides along its diameter (see figure).Each gun simultaneously fires the balls horizontally and perpendicular to the diameter in opposite directions. After leaving the platfrom, the balls have horizontal speed of 9ms–1 with respect to the ground. The rotational speed of the platfrom in rad–1 after the balls leave the platform is [JEE (Advanced)-2014,P-1, 3/60] 26. A uniform circular disc of mass 1.5 kg and radius 0.5m is initially at rest on a horizontal frictionless surface. Three forces of equal magnitude F = 0.5 N are applied simultaneously along the three sides of an equilateral triangle XYZ its vertices on the perimeter of the disc (see figure). One second after applying the forces, the angular speed of the disc in rad s–1 is : [JEE (Advanced)-2014,P-1, 3/60] Rigid Body Dynamics Corp. / Reg. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.) – 324005 Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in ADVRB - 80 Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029 27. Two identical uniform discs roll without slipping on two different surfaces AB and CD (see figure) starting at A and C with linear speeds v1 and v2, respectively, and always remain in contact with ths surfaces. If they reach B and D with the same linear speed and v1 = 3 m/s, then v2 in m/s is (g = 10 m/s2) [JEE(Advanced) 2015 ; P-1, 4/88] 28. A ring of mass M and radius R is rotating with angular speed about a fixed vertical axis passing through its centre O with two point masses each of mass M/8 at rest at O. These masses can move radially outwards along two massless rods fixed on the ring as shown in the figure. At some instant the angular speed of the system is 8/9 and one of the masses is ata distance of 3 R 5 from O. At this instant the distance of the other mass from O is : [JEE(Advanced) 2015 ; P-1, 4/88, –2] (A) 2 R 3 (B) 1R 3 (C) 3 R 5 (D) 4 R 5 29. The dinsities of two solid spheres A and B of the same radii R vary with radial distance r as A r (r) k R and 5 B r (r) k R , respectively, where k is a contant. The moments of inertia of the individual spheres about axes passing through their centres are A and B, respectively, If B A = n 10 , the value of n is : [JEE(Advanced) 2015 ; P-2,4/88] 30. A uniform wooden stick of mass 1.6 kg and length rests in an inclined manner on a smooth, vertical wall of height h (< ) such that a small portion of the stick extends beyond the wall. The reaction force of the wall on the stick is perpendicular to the stick. The stick makes an angle of 30° with the wall and the bottom of the stick is on a rough floor. The reaction of the wall on the stick is equal in magnitude to the reaction of the floor on the stick. The ratio h/ and the frictional force f at the bottom of the stick are (g = 10 ms–2) [JEE (Advanced) 2016 ; P-1, 3/62, –1 ] (A) 3 16 3 ,f N 16 3 l h (B) 3 16 3 ,f N 16 3 l h (C) 3 3 8 3 ,f N 16 3 l h (D) 3 3 16 3 ,f N 16 3 l h 31. The position vector r of particle of mass m is given by the following equation j ˆ t i ˆ t ) t ( r 2 3 Where = 10/3 m s–3, = 5 m s–2 and m = 0.1 kg. At t = 1 s, which of the following statement(s) is (are) true about the particle. [JEE (Advanced) 2016 ; P-1, 4/62, –2] (A) The velocity v is given by 1 – ms j ˆ 10 i ˆ 10 v (B) The angular momentum L with respect to the origin is given by k ˆ ) 3 / 5 –( L N ms (C) The force F is given by j ˆ 2 i ˆ F N (D) The torque with respect to the origin is given by 20 ˆ k 3 Nm. Rigid Body Dynamics Corp. / Reg. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.) – 324005 Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in ADVRB - 81 Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029 32. Two thin circular discs of mass m and 4m, having radii of a and 2a, respectively, are rigidly fixed by a massless, rigid rod of length 24a l through their centers. This assembly is laid on a firm and flat surface, and set rolling without slipping on the surface so that the angular speed about the axis of the rod is . The angular momentum of the entire assembly about the point 'O' is L (see the figure). Which of the following statement (s) is (are) true ? l l m 4m 2a O z a (A) The magnitude of the z-component of L is 55 ma2 . [JEE (Advanced) 2016 ; P-2, 4/62, –2] (B) The magnitude of angular momentum of the assembly about its centre of mass is 17 ma2 2 . (C) The magnitude of angular momentum of centre of mass of the assembly about the point O is 81ma2 (D) The centre of mass of the assembly rotates about the z-axis with an angular speed of /5. Paragraph for Question Nos. 33 to 34 A frame of reference that is accelerated with respect to an inertial frame of reference is called a non-inertial frame of reference. A coordinate system fixed on a circular disc rotating about a fixed axis with a constant angular velocity is an example of a non-inertial frame of reference. The relationship between the force rot F r experienced by a particle of mass m moving on the rotating disc and the force in F r experienced by the particle in an inertial frame of reference is, rot in rot F F 2m m r r r r r r r r , Where rot r is the velocity of the particle in the rotating frame of reference and r r is the position vector of the particle with respect to the centre of the disc. Now consider a smooth slot along a diameter of a disc of radius R rotating counter-clockwise with a constant angular speed about its vertical axis through its center. We assign a coordinate system with the origin at the center of the disc, the x-axis along the slot, the y-axis perpendicular to the slot and the z-axis along the rotation axis ˆ k r . A small block of mass m is gently placed in the slot at ˆ r R/2 i r at t = 0 and is constrained to move only along the slot. R R/2 m 33. The distance r of the block at time t is [JEE (Advanced) 2016 ; P-2, 3/62, –1] (A) R cos2 t 2 (B) 2 t 2 t R e e 4 (C) R cos t 2 (D) t t R e e 4 34. The net reaction of the disc on the block is : [JEE (Advanced) 2016 ; P-2, 3/62, –1] (A) 2 ˆ ˆ m Rsin t j – mgk (B) 2 ˆ ˆ m Rcos t j – mgk (C) 2 t t 1 ˆ ˆ m R e e j mgk 2 (D) 2 2 t 2 t 1 ˆ ˆ m R e e j mgk 2 Rigid Body Dynamics Corp. / Reg. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.) – 324005 Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in ADVRB - 82 Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029 35. Consider regular polygons with number of sides n = 3, 4, 5 ................ as shown in the figure. The centre of mass of all the polygons is at height h from the ground. They roll on a horizontal surface about the leading vertex without slipping and sliding as depicted. The maximum increase in height of the locus of the center of mass for each polygon is . Then depends on n and h as [JEE (Advanced) 2017 ; P-2, 3/61, –1] h h h (A) 2 hsin n (B) 2 htan 2n (C) 2 hsin n (D) 1 h 1 cos n 36. A wheel of radius R and mass M is placed at the bottom of a fixed step of height R as shown in the figure. A constant force is continuously applied on the surface of the wheel so that it just climbs the step without slipping. Consider the torque about an axis normal to the plane of the paper passing through the point Q. Which of the following options is/are correct ? [JEE (Advanced) 2017 ; P-2, 4/61, –2] (A) If the force is applied normal to the circumference at point P then is zero X P Q S R (B) If the force is applied tangentially at point S then 0 but the wheel never climbs the step (C) If the force is applied at point P tangentially then decreases continuously as the wheel climbs (D) If the force is applied normal to the circumference at point X then is constant 37. A rigid uniform bar AB of length L is slipping from its vertical position on a frictionless floor (as shown in the figure). At some instant of time, the angle made by the bar with the vertical is . Which of the following statements about its motion is/are correct ? [JEE (Advanced) 2017 ; P-2, 4/61, –2] (A) The trajectory of the point A is a parabola (B) Instantaneous torque about the point in contact with the floor is proportional to sin . (C) When the bar makes an angle with the vertical, the displacement of its midpoint from the initial position is proportional to (1– cos) (D) The midpoint of the bar will fall vertically downward. O B L A Paragraph for Question Nos. 38 to 39 One twirls a circular ring (of mass M and radius R) near the tip of one's finger as shown in Figure-1. In the process the finger never loses contact with the inner rim of the ring. The finger traces out the surface of a cone, shown by the dotted line. The radius of the path traced out by the point where the ring and the finger is in contact is r. The finger rotates with an angular velocity 0. The rotating ring rolls without slipping on the outside of a smaller circle described by the point where the ring and the finger is in contact (Figure-2). R R r Figure-1 Figure-2 The coefficient of friction between the ring and the finger is and the acceleration due to gravity is g. Rigid Body Dynamics Corp. / Reg. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.) – 324005 Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in ADVRB - 83 Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029 38. The total kinetic energy of the ring is : [JEE (Advanced) 2017 ; P-2, 3/61] (A) 2 2 0 1M R r 2 (B) 2 2 0 3M R r 2 (C) 2 2 0 M R (D) 2 2 0 M R r 39. The minimum value of 0 below which the ring will drop down is : [JEE (Advanced) 2017 ; P-2, 3/61] (A) g R r (B) g 2 R r (C) 3g 2 R r (D) 2g R r 40. Consider a body of mass 1.0 kg at rest at the origin at time t = 0. A force ˆ ˆ F ( ti j) is applied on the body, where = 1.0 Ns−1 and = 1.0 N. The torque acting on the body about the origin at time t = 1.0s is . Which of the following statements is (are) true? [JEE (Advanced) 2018 ; P-1, 4/60] (A) 1 | | Nm 3 (B) The torque is in the direction of the unit vector ˆ k (C) The velocity of the body at t = 1 s is 1 1 ˆ ˆ v (i 2j)ms 2 (D) The magnitude of displacement of the body at t = 1 s is 1 m 6 41. A ring and a disc are initially at rest, side by side, at the top of an inclined plane which makes an angle 60° with the horizontal. They start to roll without slipping at the same instant of time along the shortest path. If the time difference between their reaching the ground is (2 – 3) / 10 s, then the height of the top of the inclined plane, in metres, is _. Take g = 10 ms−2. [JEE (Advanced) 2018 ; P-1, 3/60] 42. In the List-I below, four different paths of a particle are given as functions of time. In these functions, and are positive constants of appropriate dimensions and . In each case, the force acting on the particle is either zero or conservative. In List-II, five physical quantities of the particle are mentioned: p is the linear momentum, L is the angular momentum about the origin, K is the kinetic energy, U is the potential energy and E is the total energy. Match each path in List-I with those quantities in List-II, which are conserved for that path. [JEE (Advanced) 2018 ; P-2, 3/60, –1] List-I List-II P. ˆ ˆ r(t) ti tj 1. p Q. ˆ ˆ r(t) cos ti sin tj 2. L R. ˆ ˆ r(t) (cos ti sin tj) 3. K S. 2 ˆ ˆ r(t) ti t j 2 4. U 5. E (A) P → 1, 2, 3, 4, 5; Q → 2, 5; R → 2, 3, 4, 5; S → 5 (B) P → 1, 2, 3, 4, 5; Q → 3, 5; R → 2, 3, 4, 5; S → 2, 5 (C) P → 2, 3, 4; Q → 5; R → 1, 2, 4; S → 2, 5 (D) P → 1, 2, 3, 5; Q → 2, 5; R →2, 3, 4, 5; S → 2, 5 PART - II : JEE (MAIN) / AIEEE PROBLEMS (PREVIOUS YEARS) 1. A thin circular ring of mass m and radius R is rotating about its axis with a constant angular velocity . Two objects each of mass M are attached gently to the opposite ends of a diameter of the ring. The ring now rotates with an angular velocity : [AIEEE-2006, 4½/180] (1) m (m 2M) (2) (m 2M) m (3) (m 2M) (m 2M) (4) m (m M) Rigid Body Dynamics Corp. / Reg. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.) – 324005 Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in ADVRB - 84 Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029 2. Four point masses, each of value m, are placed at the corners of a square ABCD of side . The moment of inertia about an axis passing through A and parallel to BD is : [AIEEE-2006, 4½/180] (1) m2 (2) 2m2 (3) 3 m2 (4) 3m2 3. For the given uniform square lamina ABCD, whose centre is O, [AIEEE 2007, 3/120] (1) AC EF 2 (2) AD = 2EF (3) AC = EF (4) AC = EF 2 4. A round uniform body of radius R, mass M and moment of inertia I rolls down (without slipping) an inclined plane making an angle ? With the horizontal. Then its acceleration is : [AIEEE 3/120 2007] (1) 2 gsin 1 /MR (2) 2 gsin 1 MR / (3) 2 gsin 1 /MR (4) 2 gsin 1 MR / 5. Angular momentum of the particle rotating with a central force is constant due to : [AIEEE 3/120 2007] (1) constant force (2) constant linear momentum (3) zero torque (4) constant torque 6. Consider a uniform square plate of side 'a' and mass 'm'. The moment of inertia of this plate about an axis perpendicular to its plane and passing through one of its corners is : [AIEEE 3/105 2008] (1) 1 12 ma2 (2) 7 12 ma2 (3) 2 3 ma2 (4) 5 6 ma2 7. A thin uniform rod of length and mass m is swinging freely about a horizontal axis passing through its end. Its maximum angular speed is . Its centre of mass rises to a maximum height of : [AIEEE 4/144 2009] (1) 1 6 g (2) 2 2 1 2 g (3) 2 2 1 6 g (4) 2 2 1 3 g 8. A small particle of mass m is projected at an angle with the x-axis with an initial velocity v0 in the x-y plane as shown in the figure. At a time t < 0 v sin g , the angular momentum of the particle is [AIEEE 4/144 2010] (1) –mg v0 t2 cos ˆ j (2) mg v0 t cos ˆ k (3) – 1 2 mg v0 t2 cos ˆ k (4) 1 2 mg v0 t2 cos ˆ i where ˆ ˆ i, j and ˆ k are unit vectors along x, y and z-axis respectively. 9. A thin horizontal circular disc is rotating about a vertical axis passing through its centre. An insect is at rest at a point near the rim of the disc. The insect now moves along a diameter of the disc to reach its other end. During the journey of the insect, the angular speed of the disc : [AIEEE - 2011, 4/120, –1] (1) remains unchanged (2) continuously decreases (3) continuously increases (4) first increases and then decreases 10. A mass m hangs with the help of a string wrapped around a pulley on a frictionless bearing. The pulley has mass m and radius R. Assuming pulley to be a perfect uniform circular disc, the acceleration of the mass m, if the string does not slip on the pulley, is : [AIEEE - 2011, 4/120, –1] (1) 3 g 2 (2) g (3) 2 g 3 (4) g 3 Rigid Body Dynamics Corp. / Reg. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.) – 324005 Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in ADVRB - 85 Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029 11. A pulley of radius 2m is rotated about its axis by a force F = (20t – 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2, the number of rotations made by the pulley before its direction of motion if reversed, is :[AIEEE - 2011, 4/120, –1] (1) less than 3 (2) more than 3 but less than 6 (3) more than 6 but less than 9 (4) more than 9 12. A particle of mass ‘m’ is projected with a velocity making an angle of 30º with the horizontal. The magnitude of angular momentum of the projectile about the point of projection when the particle is at its maximum height ‘h’ is : [AIEEE 2011, 11 May; 4/120, –1] (1) zero (2) 3 m 2g (3) 3 3 m 16 g (4) 2 3 m 2 g 13. A hoop of radius r and mass m rotating with an angular velocity 0 is placed on a rough horizontal surface. The initial velocity of the centre of the hoop is zero. What will be the velocity of the centre of the hoop when it ceases to slip ? [AIEEE 2013, 4/120, –1] (1) 0 r 4 (2) 0 r 3 (3) 0 r 2 (4) r0 14. A mass 'm' supported by a massless string wound around a uniform hollow cylinder of mass m and radius R. If the string does not slip on the cylinder, with what acceleration will the mass fall on release ? [JEE (Main) 2014, 4/120, –1] (1) 2g 3 (2) g 2 (3) 5g 6 (4) g 15. A bob of mass m attached to an inextensible string of length l is suspended from a vertical support. The bob rotates in a horizontal circle with a angular speed rad/s about the vertical. About the point of suspension : (1) angular momentum is conserved. [JEE (Main) 2014, 4/120, –1] (2) angular momentum changes in magnitude but not in direction (3) angular momentum changes in direction but not in magnitude. (4) angular momentum changes both in direction and magnitude. 16. Form a solid sphere of mass M and radius R a cube of maximum possible volume is cut. Moment of inertia of cube about an axis passing through its centre and perpendicular to one of its faces is [JEE (Main)-2015; 4/120, –1] (1) 2 MR 32 2 (2) 2 MR 16 2 (3) 2 4MR 9 3 (4) 2 4MR 3 3 17. A particle of mass m is moving along the side of square of side ‘a’ with a uniform speed in the x-y plane as shown in the figure : – Which of the following statements is false for the angular momentum about the origin ? (1) R ˆ L m – a k 2 when the particle is moving from C to D. (2) R ˆ L m a k 2 when the particle is moving from B to C. (3) m ˆ L R k 2 when the particle is moving from D to A. (4) m ˆ L – R k 2 when the particle is moving from A to B. R 45° D C B A a y O x a a a Rigid Body Dynamics Corp. / Reg. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.) – 324005 Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in ADVRB - 86 Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029 18. The moment of inertia of a uniform cylinder of length l and radius R about its perpendicular bisector is . What is the ratio l/R such that the moment of inertia is minimum ? [JEE (Main) 2017 ; 4/120, –1] (1) 3 2 (2) 3 2 (3) 3 2 (4) 1 19. A slender uniform rod of mass M and length l is pivoted at one end so that it can rotate in a vertical plane (see figure). There is negligible friction at the pivot. The free end is held vertically above the pivot and then released. The angular acceleration of the rod when it makes an angle with the vertical is. [JEE (Main) 2017; 4/120, –1] x z (1) 2g cos 3 l (2) 3g sin 2 l (3) 2g sin 3 l (4) 3g cos 2 l 20. Seven identical circular planar disks, each of mass M and radius R are welded symmetrically as shown. The moment of inertia of the arrangement about the axis normal to the plane and passing through the point P is : [JEE (Main) 2018; 4/120, –1] O P (1) 2 73MR 2 (2) 2 181MR 2 (3) 2 19MR 2 (4) 2 55MR 2 21. From a uniform circular disc of radius R and mass 9 M, a small disc of radius 3 R is removed as shown in the figure. The moment of inertia of the remaining disc about an axis perpendicular to the plane of the disc and passing through centre of disc is : [JEE (Main) 2018; 4/120, –1] 3 R 2 R (1) 10 MR2 (2) 2 MR 9 37 (3) 4 MR2 (4) 2 MR 9 40 Rigid Body Dynamics Corp. / Reg. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.) – 324005 Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in ADVRB - 87 Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029 EXERCISE-1 PART - I Section (A) : A-1. 4 rev/s2, 20 rev/s A-2. 20 s Section (B) : B-1. 2 ML 12 B-2. 2 M 6 B-3. 2 MR 2 M 2 4R 3 B-4. (K = 3 2 r ) Section (C) : C-1. –14 ˆ i + 10 ˆ j – 9 ˆ k C-2. mg sin, when the bob is at the lowest point, at = 90º. C-3. (a) mv2 sin cos perpendicular to the plane of motion (b) 2mv2 sin cos perpendicular to the plane of motion C-4. 3N – m Section (D) : D-1. 2.4 N in the left string and 2.6 N in the right D-2. P = w 2 cot or P = mg 2 cot D-3. 990 N, 960 N, 32 33 D-4. (i) T = 250 N (ii) FH= 150 N (), FV = 50 N () Section (E) : E-1. (a) 1 2 1 2 2g (m m ) (m m ) = 10 3 rad/s2 (b) (i) ' = 1 2 3 1 2 2(m m )g m m m 3 = 3 rad/s2 , (ii) 42 N ; 39 N E-2. (a) 3g 4L (cw) (b) N = 13 mg 16 , F = 3 3 16 mg E-3. 3 rad/s2 E-4. N = F 3 x 1– 2 Section (F) : F-1. = 5 rad/s F-2. = 9 g 4 F-3. 1 2 m2 2 2 L R 12 F-4. 2 m/s Section (G) : G-1. ˆ 2k kg m2/s G-2. 16 kg m2/s G-3. 2 0 2 mr 2 2mr G-4. 10 rad/s G-5. 4 m M 2m G-6. 3v / 4 Section (H) : H-1 VA = 25 m/s , VB = 75 m/s H-2 VO = 4 m/sec ˆ i , VA = (4 ˆ i + 3 ˆ j )m/sec H-3 VCM = 7 m/s. H-4 (a) 0 4v 3 (b) 0 5v 3 (c) vx = 0 v 2 , vy = – 0 2v 3 H-5. (a) vA = 2at = 10.0 cm/s, vB = 2 at = 7.1 cm/s, v0 = 0 ; (b) aA = 2a 2 2 2t a 1 R = 5.6 cm/s2, aB = a = 2.5 cm/s2, a0 = a2t2 / R = 2.5 cm/s2 Section (I) : -1. 4gh 3 -2. 7 10 mv2 -3. 2 3 g -4. 7 3 mg -5. 3g(R r) -6. 4R/5 -7. 5 mg, 20 N Section (J) : J-1. = 3 v/2 J-2. (a) 2 1 2 m u m m (b) 1 1 2 m u m m (c) – 2 1 2 m u m m (d) 2 1 2 2 1 2 m m uL 2(m m ) , 2 1 2 2 1 2 m m uL 2 (m m ) (e) 2 1 1 2 1 2 m (m 4m )L 12 (m m ) (f) 2 1 2 m u m m , 2 1 2 6m u (m 4m )L J-3. (a) m , 2 mR (b) mR 2I (c) R 2 J-4. 100 rad/sec. Rigid Body Dynamics Corp. / Reg. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.) – 324005 Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in ADVRB - 88 Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029 Section (K) : K-1. 1 2 mg a sin, x = atan 2 PART - II Section (A) : A-1. (B) A-2. (C) Section (B) : B-1. (C) B-2. (A) B-3. (A) B-4. (C) B-5. (D) B-6. (C) B-7. (C) B-8. (B) B-9. (D) B-10. (D) B-11. (D) Section (C) : C-1. (A) C-2. (C) C-3. (C) Section (D) : D-1. (B) D-2. (C) D-3 (B) D-4 (A) D-5. (C) Section (E) : E-1. (A) E-2. (C) E-3. (D) E-4. (A) E-5. (C) Section (F) : F-1. (B) F-2. (C) Section (G) : G-1. (C) G-2. (B) G-3. (D) G-4. (B) G-5. (C) Section (H) : H-1. (B) H-2. (B) H-3. (C) H-4. (A) H-5. (A) Section () : I-1. (B) I-2. (A) I-3. (B) I-4. (D) I-5. (A) I-6. (D) I-7. (A) I-8. (C) -9. (D) Section (J) : J-1. (D) J-2. (B) Section (K) : K-1. (A) K-2. (A) K-3. (A) PART - III 1. (A) p,q,r ; (B) p,q,r ; (C)p,q;(D) p,q,r 2. (A) p ; (B) q, s;(C) p ;(D) q, s EXERCISE-2 PART - I 1. (D) 2. (D) 3. (D) 4. (C) 5. (C) 6. (B) 7. (D) 8. (B) 9. (B) 10. (D) 11. (D) 12. (B) 13. (C) 14. (A) 15. (D) 16. (A) 17. (A) 18. (D) 19. (B) 20. (B) 21. (B) 22. (D) 23. (B) 24. (A) PART - II 1. 16 2. 28 3. 2 4. 6 5. 2 6. 4 7. 4 8. 10 9. 1 10. 2 11. 8 12. 9 13. 5 14. 1 15. 2 16. 10 17. (a) 10 (b) 20 18. 25 19. 12 20. 20 21. 9 22. 5 PART - III 1. (CD) 2. (ABC) 3. (ABC) 4. (ABCD) 5. (ACD) 6. (BC) 7. (ABC) 8. (ACD) 9. (CD) 10. (ABCD) 11. (BC) 12. (ABD) 13. (ABC) 14. (BC) 15. (ACD) 16. (AD) 17. (ACD) PART - IV 1. (B) 2. (C) 3. (D) 4. (D) 5. (C) 6. (A) EXERCISE-3 PART - I 1. (A) 2. (CD) 3. (AB) 4. 10 5. (C) 6. (A) 7. (B) 8. (D) 9. (D) 10. (D) 11. (A) 12. (BC) 13. (B) 14. 4 15. 9 16. (B) 17. (C) 18. 3 19. (D) 20. (A) 21. (D) 22. (AB) 23. 8 24. (CD) 25. 4 26. 2 27. 7 28. (D) 29. 6 30. (D) 30. (D) 31. (ABD) 32. (D) 33. (D) 34. (C) 35. (D) 36. (A) 37. (BCD) 38. (Bonus) 39. (A) 40. (AC) 41. 0.75 42. (A) PART - II 1. (1) 2. (4) 3. (3) 4. (1) 5. (3) 6. (3) 7. (3) 8. (3) 9. (4) 10. (3) 11. (2) 12. (3) 13. (3) 14. (2) 15. (3) 16. (3) 17. (1,3) 18. (2) 19. (2) 20. (2) 21. (3) Rigid Body Dynamics Corp. / Reg. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.) – 324005 Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in ADVRB - 89 Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029 SUBJECTIVE QUESTIONS 1. Find the M.I. of a rod about (i) an axis perpendicular to the rod and passing through left end. (ii) An axis through its centre of mass and perpendicular to the length whose linear density varies as = ax where a is a positive constant and 'x' is the position of an element of the rod relative to its left end. The length of the rod is . 2. The pulley (uniform disc) shown in figure has, radius 10 cm and moment of inertia about its axis = 0.5 kgm2 (B and A both move) (a) Assuming all the plane surfaces are smooth and there is no slipping between pulleyand string, calculate the acceleration of the mass 4kg. (b) The friction coefficient between the block A and the plane below is = 0.5 and the plane below the B block is frictionless.Assuming no slipping between pulley and string find accelaration of 4kg block 2kg 4kg 45° 45° m A B fixed 3. A uniform rod of length 'a' rests against a frictionless wall as shown in figure. Find the friction coefficient between the horizontal surface and the lower end if the minimum angle that the rod can make with the horizontal is ., without slipping. 4. A uniform square plate of mass m is supported as shown. If the cable suddenly breaks, determine just after that moment; (a) The angular acceleration of the plate. (b) The acceleration of corner C. (c) The reaction at A. B A C 5. A uniform slender rod AB of mass m is suspended from two springs as shown. If spring 2 breaks, determine at that instant ; (a) The angular acceleration of the bar. (b) The acceleration of point A. (c) The acceleration of point B. 6. A 2 kg sphere moving horizontally to the right with an initial velocity of 5 m/s strikes the lower end of an 8 kg rigid rod AB. The rod is suspended from a hinge at A and is initially at rest. Knowing that the coefficient of restitution between the rod and sphere is 0.80, determine the angular velocity of the rod and the velocity of the sphere immediately after the impact. /////////////////////// 1.2m v s A Rigid Body Dynamics Corp. / Reg. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.) – 324005 Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in ADVRB - 90 Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029 7. A rotating disc (figure) moves in the positive direction of the x-axis. Find the equation y (x) describing the position of the instantaneous axis of rotation, if at the initial moment the axis C of the disc was located at the point O after which it moved (a) With a constant velocity v, while the disc started rotating counter clockwise with a constant angular acceleration (the initial angular velocity is equal to zero); (b) With a constant acceleration a (and the zero initial velocity), while the disc rotates counterclockwise with a constant angular velocity . 8. A block X of mass 0.5 kg is held by a long massless string on a fixed frictionless inclined plane inclined at 30º to the horizontal. The string is wound on a uniform solid cylindrical drum Y of mass 2 kg and radius 0.2 m as shown in figure. The drum is given an initial angular velocity such that block X starts moving up the plane. [JEE - 1994] (a) Find the tension in the string during motion. (b) At a certain instant of time the magnitude of the angular velocity of Y is 10 rad s1. Calculate the distance travelled by X from that instant of time until it comes to rest. 9. A rod of length R and mass M is free to rotate about a horizontal axis passing through hinge P as in figure. First it is taken aside such that it becomes horizontal and then released. At the lowest point the rod hits the small block B of mass m and stops. Find the ratio of masses such that the block B completes the circular track of radius R. Neglect any friction. M P m B 10. A 3 kg uniform rod rotates in a vertical plane about a smooth pivot at B. A spring of constant k = 300 N/m and of unstretched length 100 mm is attached to the rod as shown. Knowing that in the position shown the rod has an angular velocity of 4 rad/s clockwise, determine the angular velocity of the rod after it has rotated through. [g = 10 m/s2 ] (a) 90º (b) 180º 11. The angular momentum of a particle relative to a certain ' point O varies with time as 2 M a bt , where a and b are constant vectors, with a b . Find the force moment N relative to the point O acting on the particle when the angle between the vectors N and M equals 45°. 12. A plank of mass m1 with a uniform sphere of mass m2 placed on it rests on a smooth horizontal plane. A constant horizontal force F is applied to the plank. With what accelerations will the plank and the centre of the sphere move provided there is no sliding between the plank and the sphere ? Rigid Body Dynamics Corp. / Reg. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.) – 324005 Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in ADVRB - 91 Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029 13. In the arrangement shown in the figure weight A possesses mass m, a pulley B possesses mass M. Also known are the moment of inertia I of the pulley relative to its axis and the radii of the pulley are R and 2R respectively. Consider the mass of the threads is negligible. Find the acceleration of weight A after the system is set free. (Assume no slipping takes place anywhere and axis of cylinder remains horizontal) B A 2R R 14. A uniform rod AB of length is released from rest with AB inclined at angle with horizontal. It collides elastically with smooth horizontal surface after falling through a height h. What is the height upto which the centre of mass of the rod rebounds after impact? 15. A uniform block A of mass 25 kg and length 6m is hinged at C and is supported by a small block B as shown in the Figure. A constant force F of magnitude 400N is applied to block B horizontally. What is the speed of B after it moves 1.5 m ? The mass of block B is 2.5 kg & the coeffcient of friction for all contact surfaces is 0.3. [Use l n (3/2) 0.41 and g = 10 ms-2 ] 16. A window (of weight w) is supported by two strings passing over two smooth pulleys in the frame of the window in which window just fits in, the other ends of the string being attached to weights each equal to half the weight of the window. One thread breaks and the window moves down. Find acceleration of the window if is the coefficient of friction, and 'a' is the height and 'b' the breadth of the window. Rigid Body Dynamics Corp. / Reg. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.) – 324005 Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in ADVRB - 92 Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029 17. Three particles A, B, C of mass m each are joined to each other by massless rigid rods to form an equilateral triangle of side a. Another particle of mass m hits B with a velocity v0 directed along BC as shown. The colliding particle stops immediately after impact. (a) Calculate the time required by the triangle ABC to complete half revolution in its subsequent motion. (b) What is the net displacement of point B during this interval ? 18. Disk A has a mass of 4 kg and a radius r = 75 mm, it is at rest when it is placed in contact with the belt, which moves at a constant speed = 18 m/s. Knowing that k = 0.25 between the disk and the belt, determine the number of revolutions executed by the disk before it reaches a constant angular velocity. (Assume that the normal reaction by the belt on the disc is equal to weight of the disc) . 19. A 160 mm diameter pipe of mass 6 kg rests on a 1.5 kg plate. The pipe and plate are initially at rest when a force P of magnitude 25 N is applied for 0.75 s. Knowing that s = 0.25 & k = 0.20 between the plate and both the pipe and the floor, determine ; (a) whether the pipe slides with respect to the plate. (b) the resulting velocities of the pipe and of the plate. 20. A uniform disc of mass m and radius R is rolling up a rough inclined plane, which makes an angle of 30º with the horizontal. If the coefficients of static and kinetic friction are each equal to and only the forces acting are gravitational, normal reaction and friction, then the magnitude of the frictional force acting on the disc is and its direction is __ (write 'up' or 'down') the inclined plane.[JEE - 1997] 21. A uniform disc of mass m and radius R is projected horizontally with velocity 0 on a rough horizontal floor so that it starts off with a purely sliding motion at t = 0. After t0 seconds, it acquires a purely rolling motion as shown in figure. [JEE - 1997] V 0 t = 0 t = t 0 ///////////////////// ///////////////////// (a) Calculate the velocity of the centre of mass of the disc at t0 . (b) Assuming the coefficient of friction to be , calculate t0. Also calculate the work done by the frictional force as a function of time & the total work done by it over a time t much longer than t0. Rigid Body Dynamics Corp. / Reg. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.) – 324005 Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in ADVRB - 93 Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029 22. A carpet of mass 'M' made of inextensible material is rolled along its length in the form of a cylinder of radius 'R' and is kept on a rough floor. The carpet starts unrolling without sliding on the floor when a negligibly small push is given to it. Calculate the horizontal velocity of the axis of the cylindrical part of the carpet when its radius reduces to R/2. [JEE - 1990] //////////////////////////////// //////////////////////////////// R R/2 23. Figure shows a vertical force F that is applied tangentially to a uniform cylinder of weight W. The coefficient of static friction between the cylinder and all surfaces is 0.5. Find in terms of W, the maximum force that can be applied without causing the cylinder to rotate. 24. A drinking straw of mass 2m is placed on a smooth table orthogonally to the edge such that half of it extends beyond the table. A fly of mass m lands on the A end of the straw and walks along the straw untill it reaches the B end. It does not tip even when another fly gently lands on the top of the first one. Find the largest mass that the second fly can have. (Neglect the friction between straw and table). 25. Two uniform thin rods A & B of length 0.6 m each and of masses 0.01 kg & 0.02 kg respectively are rigidly joined, end to end. The combination is pivoted at the lighter end P as shown in figure such that it can freely rotate about the point P in a vertical plane. A small object of mass 0.05 kg moving horizontally hits the lower end of the combination and sticks to it. What should be the velocity of the object so that the system could just be raised to the horizontal position ? [JEE - 1994] P A B m v x 26. A uniform cube of side 'a' and mass m rests on a rough horizontal table. A horizontal force F is applied normal to one of the faces at a point directly above the centre of the face, at a height 3a 4 above the base. (i) What is the minimum value of F for which the cube begins to tip about an edge? (ii) What is the minimum value of s so that toppling occures. (iii) If = min, find minimum force for topping. (iv) Minimum s so that Fmin (as in part-(i)) can cause toppling. Rigid Body Dynamics Corp. / Reg. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.) – 324005 Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in ADVRB - 94 Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029 27. Three particles A, B and C each of mass m are connected to each other by three massless rigid rods to form a rigid, equilateral triangular body of side . This body is placed on a horizontal frictionless table (x - y plane) and is hinged to it at the point A so that it can move without friction about the vertical axis through A as shown in figure. The body is set into rotational motion on the table about A with a constant angular velocity (a) Find the magnitude of the horizontal force exerted by the hinge on the body . (b) At time T, when the side BC is parallel to the x axis, a force F is applied on B along BC as shown. Obtain the x component and the y component of the force exerted by the hinge on the body, immediately after time T. [JEE Mains 02, (1+4)/60] 28. A bar of mass m is held as shown between 4 disks , each of mass M & radius r = 75 mm Determine the acceleration of the bar immediately after it has been released from rest, knowing that the normal forces exerted on the disks are sufficient to prevent any slipping and assuming that ; In (i) case the discs are attacthed to the fixed support on wall. In (ii) case the discs are attached to the bar. (a) m = 5 kg and M = 2 kg . (b) the mass of M of the disks is negligible. (c) the mass of m of the bar is negligible . 29. Two thin circular discs of mass 2 kg and radius 10 cm each are joined by a rigid massless rod of length 20 cm. The axis of the rod is along the perpendicular to the planes of the disk through their centres. This object is kept on a truck in such a way that the axis of the object is horizontal and perpendicular to the direction of motion of the truck. The friction with the floor of the truck is large enough, so that object can roll on the truck without slipping. Take x-axis as the direction of motion of the truck and z-axis as the vertically upward direction. If the truck has an acceleration of 9 m/s2, calculate [JEE - 1997’ 5/100] (a) the force of friction on each disc. (b) the magnitude & direction of frictional torque acting on each disk about the centre of mass O of the object. Express the torque in the vector form in terms of unit vectors ˆ ˆ ˆ i , j & k along x, y & z direction. Rigid Body Dynamics Corp. / Reg. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.) – 324005 Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in ADVRB - 95 Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029 30. A rectangular rigid fixed block has a long horizontal edge. A solid homogeneous cylinder of radius r is placed horizontally at rest with its length parallel to the edge such that the axis of the cylinder and the edge of the block are in the same vertical plane. There is sufficient friction present at the edge so that a very small displacement causes the cylinder to roll off the edge without slipping. Determine : (a) The angle c through which the cylinder rotates before it leaves contact with the edge. (b) The speed of the centre of mass of the cylinder before leaving contact with the edge. (c) The ratio of translational to rotational kinetic energies of the cylinder when its centre of mass is in horizontal line with the edge. [JEE - 1995] 31. A wedge of mass ‘m’ and triangular cross section (AB = BC = CA = 2R) is moving with a constant velocity-v ˆ i towards a sphere of radius R fixed on a smooth horizontal table as shown in the figure. The wedge makes an elastic collision with the fixed sphere and returns along the same path without any rotation. Neglect all friction and suppose that the wedge remains in contact with the sphere for a very short time t, during which the sphere exerts a constant force F on the wedge. The sphere is always fixed. (a) Find the force F and also the normal force N exerted by the table on the wedge during the time t. (b) Let ‘h’ denote the perpendicular distance between the centre of mass of the wedge and the line of action of force F . Find the magnitude of the torque due to the normal force N about the centre of the wedge, during the time t. [JEE - 1998, 8] 32. The surface mass density (mass/area) of a circular disc of radius 'R' depends on the distance from the centre x given as, (x) = + x. Where and are positive constant find its moment of inertia about the line perpendicular to the plane of the disc through its centre. 33. Calculate the moment of inertia of a uniform solid cone relative to its symmetry axis, if the mass of the cone is equal to m and the radius of its base to R. 34. A force F = A ˆ i + B ˆ j is applied to a point whose radius vector relative to the origin of coordinates O is equal to r = a ˆ i + b ˆ j , where a, b &, A, B are constants, and ˆ i , ˆ j are the unit vectors of the x and y axes. Find the moment N ( torque of F ) and the arm of the force relative to the point O. 35. A uniform cylinder of radius R and mass M can rotate freely about a stationary horizontal axis O Fig. A thin cord of length and mass m is wound on the cylinder in a single layer. Find the angular acceleration of the cylinder as a function of the length x of the hanging part of the cord. The wound part of the cord is supposed to have its centre of gravity on the cylinder axis. //////////////////////// x R Rigid Body Dynamics Corp. / Reg. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.) – 324005 Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in ADVRB - 96 Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029 36. A vertically oriented uniform rod of mass M and length can rotate about a fixed horizontal smooth axis passing through its upper end. A horizontally flying bullet of mass m strikes the lower end of the rod and gets stuck in it; as a result, the rod swings through an angle (< 90º). Assuming that m << M, find : (a) the velocity of the flying bullet ; (b) the momentum increment in the system "bullet + rod" during the impact; what causes the change of that momentum ; (c) at what distance x : from the upper end of the rod the bullet must strike for the momentum of the system "bullet-rod" to remain constant during the impact. 37. A small spherical ball of mass m is rolling without slipping down the loop track as shown in the figure. The ball is released from rest on the linear portion at a vertical height h from the lowest point. The circular part as shown in figure has a radius r. [g = 10 ms2] (a) Find the kinetic energy of the ball when it is at a point A where the radius make an angle with the horizontal (b) Find the radial and the tangential accelerations of the centre when the ball isat A. (c) Find the normal force and the frictional force acting on the ball if h = 50 cm, r = 10 cm, = 0 and m = 70 g. 38. A point A is located on the rim of a wheel of radius R = 0.50 m which rolls without slipping along a horizontal surface with velocity v = 1.00 m/s as shown in figure. Find: (a) the modulus and the direction of the acceleration vector of the point A ; (b) the total distance s traversed by the point A between the two successive moments at which it touches the surface. 39. A uniform sphere of mass m and radius r rolls without sliding over a horizontal plane, rotating about a horizontal axle OA. In the process, the centre of the sphere moves with velocity along a circle of radius R Find the kinetic energy of the sphere. 40. A uniform plate of mass 'm' is suspended in each of the ways shown. For each case determine immediately after the connection at B has been released ; (a) The angular acceleration of the plate. (b) The acceleration of its center of mass. Rigid Body Dynamics Corp. / Reg. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.) – 324005 Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in ADVRB - 97 Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029 41. Figure (1) shows a mechanical system free of any dissipation. The two spheres (A and B) are each of equal mass m, and a uniform connecting rod AB of length 2r has mass 4m. The collar is massless. Right above the position of sphere A in Fig. (1) is a tunnel from which balls each of mass m fall vertically at suitable intervals. The falling balls cause the rods and attached spheres to rotate. Sphere B when reaches the position now occupied by sphere A, suffers a collision from another falling ball and so on. Just before striking, the falling ball has velocity . All collision are elastic and the spheres as well as the falling balls can be considered to be point masses. [Olympiad_2011] (a) Find the angular velocity i + 1 of the assembly in terms of {i , , and r } after the ith ball has struck it. i+1 (b) The rotating assembly eventually assumes constant angular speed . Obtain in terms of and r by solving the equation obtained in part (a). Argue how a constant does not violate energy conservation. = Argument (c) Solve the expression obtained in part (a) to obtain i in terms of {i,, and r} . i = (d) If instead of a pair of spheres, we have two pairs of spheres as shown in figure below. What would be the new constant angular speed of the assembly (i.e. the answer corresponding to part (b). 42. A rod of length L forming an angle with the vertical strikes a frictionless floor at A with a vertical velocity v1 and no angular velocity. Assuming that the impact at A is perfectly elastic, derive an expression for the angular velocity of the rod immediately after the impact. Rigid Body Dynamics Corp. / Reg. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.) – 324005 Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in ADVRB - 98 Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029 43. Consider a bicyle in vertical position accelerating forward without slipping on a straight horizontal road. The combined mass of the bicycle and the rider is M and the magnitude of the accelerating torque applied on the rear wheel by the pedal and gear system is . The radius and the moment of inertia of each wheel is R and I (with respect to axis) respectively. The accelaration due to gravity is g. [INPhO-2013] (a) Draw the free diagram of the system (bicycle and rider). (b) Obtains the accelaration a in terms of the above mentioned quantities. a = (c) For simplicity assume that the centre of mass of the system is at height R from the gruond and equidistant at 2 R from the center of each of the wheels. Let be the coefficient of friction (both static and dynamic) between the wheels and the ground. Consider M >> l/R2 and no slipping. Obtain the conditions for the maximum acceleration am of the bike. am = (d) For = 1.0 calculate am. am = 1. (i) 4 a 4 (ii) 4 a 36 2. (a) 0.25 m/s2 (b) 0.125 m/s2 3. 2 2 acos sin 2b acos sin 4. (a) 3g 2 2 (cw) (b) 3 2 g (c) Mg 4 5. (i) (a) 3 g/L (cw) (b) ˆ ˆ i j 3 2 g = 1.323 g 49.1º (c) ˆ ˆ i 2 j 3 2 g = 2.18 g 66.6º (ii) (a) g/L (cw) (b) 3 2 gˆ i (c) ˆ ˆ i j 3 2 g=1.323g 130.9º 6. = 45 14 = 3.21 rad/s (ccw), vs = 1 7 = 0.143 m/s 7. (a) y = 2 v x (Hyperbola) ; (b) y = 2ax (Parabola)] 8. (a) 1.633N (b) 1.224 m 9. M 15 m 10. (a) 2 3 86 rad/s (b) 4 rad/s 11. N = 2b a b 12. w1 = F/(m1 + 2/7m2); w2 =2/7 w1 13. w = 3g (M + 3m) / (M + 9m + I/R2) 14. H= 2 2 2 1 3 cos 1 3 cos h ; h = 49 144 15. 323.4 m/s or 18.52 m/sec. 16. A = a b 3a b g 17. (a) t = 0 6a 3 (b) s = 2 a 1 2 3 3 18. 216 19. (a) pipe rolls without sliding (b) pipe : 5 6 m/s , 125 24 rad/s (ccw) ; plate : 5 3 m/s 20. mg 6 , up 21. (a) = 0 2 3 ; t0 = 0 3 g (b) w = –mg (v0 t – 3 2 gt2) ; – 1 6 m02 Rigid Body Dynamics Corp. / Reg. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.) – 324005 Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in ADVRB - 99 Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029 22. = 14g R 3 23. 3 W 8 24. 3m 25. 6.3 m/s 26. (i) 2 3 mg, (ii) min = 0, (iii) F = 2 mg], (iv) µs = 2 3 27. (a) 3 m 2 (b) Fy = 3 m 2 Fx = F/4 ] 28. (i) (a) 5g/9 (b) g (c) 0 (ii) (a) 13g 17 (b) g (c) 2g 3 29. (a) 6 N (b) 1 ˆ ˆ 0.6k 0.6 j , 2 ˆ ˆ 0.6k 0.6 j 30. (a) = cos-1 4 7 (b) v = 4 gr 7 (c) T R k k = 6 31. (a) 2mV 2mV 2mV ˆ ˆ ˆ F i k ; N mg k t 3 t 3 t , (b) 4mVh ˆ j 3 t 32. 2 4 5 R R 4 5 33. = 3/10 mR2 34. N = (aB – bA) ˆ k , where ˆ k is the unit vector of the z axis = |aB –bA|/ 2 2 A B 35. = 2mgx/R (M + 2m) 36. (a) v = (M/m) 2/3g sin(/2); (b) p = M 1/6g sin(/2) ; (c) x 2/3 37. (a) mg(h-r-r sin), (b) 10 7 g h 1 sin r , – 5 7 g cos (c) 4N, 0.2N upward 38. (a) A = 2 v R = 2.0 m/s2, the vector A is permanently directed to the centre of the wheel ; (b) s = 8R = 4.0 m 39. T = 7/10 mv2 (1 + 2/7r2/R2)] 40. (i) (a) 1.2g (cw) (b) 0.3 ˆ ˆ i 2 j ( ) g (ii) (a) 24 g/17 (cw) (b) 12 g/17 (iii) 2.4 g/ (cw) (b) 0.5 g 41. (a)i+1 = 7 13 i + 6 13 v r (b) = v r after this no further collision occurs (c) i+1 = v r i 7 1 13 (d) will remain same as in case b. 42. = 1 2 12sin L 3sin 1 (cw) 43. (a) Here f1f2 are frictional forces and N1, N2 are normal reactions (b) 2 a R MR 2l (c) g/ 2 a (1 / 4) (d) am = 2g/3 |
11906 | https://www.stpaul.gov/departments/safety-inspections/building-and-construction/construction-permits-and-inspections/electrical-permits-inspections | Electrical Permits & Inspections | Saint Paul Minnesota
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Introducing PAULIE: The City's New Permitting and Licensing Platform
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11907 | https://fiveable.me/statistical-mechanics/unit-5/maxwell-boltzmann-distribution/study-guide/kJNGoQBcTNCCFMGY | printables
🎲Statistical Mechanics Unit 5 Review
5.1 Maxwell-Boltzmann distribution
🎲Statistical Mechanics Unit 5 Review
5.1 Maxwell-Boltzmann distribution
Written by the Fiveable Content Team • Last updated September 2025
Written by the Fiveable Content Team • Last updated September 2025
APA
🎲Statistical Mechanics
Unit & Topic Study Guides
5.1 Maxwell-Boltzmann distribution
5.2 Equipartition theorem
5.3 Ideal gas law
5.4 Classical harmonic oscillators
5.5 Brownian motion
5.6 Virial theorem
The Maxwell-Boltzmann distribution is a cornerstone of classical statistical mechanics. It describes how particles in an ideal gas distribute across different velocities and energies at thermal equilibrium, connecting microscopic particle behavior to macroscopic properties.
This distribution underpins our understanding of gas behavior, chemical reactions, and thermodynamic processes. It provides a framework for analyzing everything from effusion rates to reaction kinetics, serving as a vital tool in physics and chemistry.
Definition and importance
Maxwell-Boltzmann distribution describes the statistical distribution of particle velocities in an ideal gas at thermal equilibrium
Fundamental concept in classical statistical mechanics underpins understanding of molecular behavior in gases and liquids
Provides framework for analyzing and predicting macroscopic properties of systems based on microscopic particle interactions
Probability distribution function
Represents probability density of finding a particle with a specific velocity or energy in a system
Derived from statistical considerations of particle collisions and energy conservation
Takes the form of an exponential function dependent on particle mass, temperature, and Boltzmann constant
Applies to systems with distinguishable, non-interacting particles in thermal equilibrium
Role in classical statistical mechanics
Serves as cornerstone for developing theories of heat, energy, and matter in classical systems
Enables calculation of average properties and thermodynamic quantities of gases
Forms basis for understanding more complex distributions in quantum mechanics and condensed matter physics
Facilitates analysis of transport phenomena (diffusion, thermal conductivity) in gases and liquids
Derivation and assumptions
Developed through combined efforts of James Clerk Maxwell and Ludwig Boltzmann in the late 19th century
Relies on principles of statistical mechanics and kinetic theory of gases
Assumes ideal gas behavior with negligible particle interactions and classical mechanics
Maxwell's approach
Utilized probability theory to describe distribution of molecular velocities in a gas
Assumed isotropy of velocity components in three-dimensional space
Derived initial form of distribution based on independence of velocity components
Introduced concept of most probable speed and its relationship to temperature
Boltzmann's contribution
Extended Maxwell's work to include energy distributions and more general systems
Developed H-theorem to explain tendency of systems towards equilibrium
Introduced concept of phase space and microstates in statistical mechanics
Formulated relationship between entropy and probability, leading to Boltzmann equation
Key assumptions and limitations
Assumes particles are distinguishable and non-interacting (ideal gas approximation)
Applies to classical systems where quantum effects are negligible
Requires system to be in thermal equilibrium
Neglects effects of gravity and other external forces on particle distribution
Breaks down at very high densities or low temperatures where quantum effects become significant
Mathematical formulation
Expresses probability of finding particles with specific velocities, speeds, or energies
Utilizes exponential functions and normalization constants
Depends on system temperature, particle mass, and Boltzmann constant
Velocity distribution
Three-dimensional probability density function for particle velocities
Given by f(vx,vy,vz)=Aexp(−2kTm(vx2+vy2+vz2))
A represents normalization constant, m is particle mass, k is Boltzmann constant, T is temperature
Exhibits spherical symmetry in velocity space
Speed distribution
Probability density function for particle speeds (magnitude of velocity)
Obtained by integrating velocity distribution over angular coordinates
Expressed as f(v)=4πv2(2πkTm)3/2exp(−2kTmv2)
Accounts for increasing volume element with speed in spherical coordinates
Energy distribution
Probability density function for particle kinetic energies
Derived from speed distribution using relationship E = (1/2)mv^2
Given by f(E)=2π(πkT1)3/2Eexp(−kTE)
Shows exponential decay with increasing energy, modified by square root factor
Properties and characteristics
Describes key features of particle distributions in equilibrium systems
Provides important statistical measures for analyzing gas behavior
Relates microscopic particle properties to macroscopic observables
Normalization condition
Ensures total probability of finding particles with any velocity, speed, or energy equals 1
Expressed mathematically as integral of distribution function over all possible values
Determines normalization constant in distribution functions
Crucial for maintaining consistency with probability theory and physical reality
Most probable speed
Speed at which distribution function reaches its maximum value
Calculated by finding derivative of speed distribution and setting it to zero
Given by vp=m2kT
Represents speed most likely to be observed in a large number of measurements
Average speed
Mean speed of particles in the system
Calculated by integrating speed multiplied by speed distribution over all speeds
Expressed as vˉ=πm8kT
Slightly higher than most probable speed due to asymmetry of distribution
Root mean square speed
Square root of average squared speed of particles
Calculated by integrating squared speed multiplied by speed distribution
Given by vrms=m3kT
Directly related to average kinetic energy of particles
Used in calculations of gas pressure and other thermodynamic properties
Applications and examples
Maxwell-Boltzmann distribution finds widespread use in various fields of physics and chemistry
Provides theoretical foundation for understanding and predicting behavior of gases and liquids
Enables quantitative analysis of numerous phenomena in thermal and statistical physics
Ideal gas behavior
Explains pressure-volume relationships in ideal gases (Boyle's law, Charles's law)
Predicts temperature dependence of gas properties (thermal expansion, heat capacity)
Accounts for equipartition of energy among degrees of freedom in gas molecules
Facilitates calculation of gas mixture properties and partial pressures
Effusion and diffusion
Describes rate of gas effusion through small openings (Graham's law of effusion)
Explains diffusion processes in gases and liquids (Fick's laws of diffusion)
Predicts concentration gradients and mixing rates in gas mixtures
Accounts for temperature and molecular mass dependence of diffusion coefficients
Chemical reaction rates
Provides basis for understanding temperature dependence of reaction rates (Arrhenius equation)
Explains distribution of molecular energies and its impact on reaction probabilities
Accounts for activation energy concept in chemical kinetics
Enables calculation of rate constants and reaction order in gas-phase reactions
Relationship to other distributions
Maxwell-Boltzmann distribution serves as classical limit for quantum mechanical distributions
Comparison with quantum distributions highlights importance of particle statistics and exclusion principles
Understanding differences between distributions crucial for analyzing diverse physical systems
Maxwell-Boltzmann vs Fermi-Dirac
Fermi-Dirac distribution applies to fermions (particles with half-integer spin)
Incorporates Pauli exclusion principle, limiting occupancy to one particle per quantum state
Differs significantly from Maxwell-Boltzmann at low temperatures or high densities
Approaches Maxwell-Boltzmann distribution in high-temperature or low-density limit
Critical for understanding behavior of electrons in metals and semiconductors
Maxwell-Boltzmann vs Bose-Einstein
Bose-Einstein distribution describes bosons (particles with integer spin)
Allows multiple particles to occupy same quantum state, leading to phenomena like Bose-Einstein condensation
Deviates from Maxwell-Boltzmann at low temperatures or high densities
Approaches Maxwell-Boltzmann distribution in classical limit (high temperature, low density)
Essential for explaining behavior of photons, phonons, and certain atoms in quantum systems
Experimental verification
Experimental validation of Maxwell-Boltzmann distribution crucial for establishing its physical relevance
Techniques evolved from early indirect measurements to modern high-precision experiments
Continues to be tested and refined in various physical systems and conditions
Historical experiments
Otto Stern's molecular beam experiment (1920) provided first direct measurement of molecular velocities
Demonstrated agreement with Maxwell-Boltzmann distribution for silver atoms in vacuum
Lummer and Pringsheim's experiments on blackbody radiation indirectly supported distribution
Millikan's oil drop experiment verified equipartition theorem, consistent with Maxwell-Boltzmann statistics
Modern validation techniques
Laser spectroscopy enables high-resolution measurements of atomic and molecular velocity distributions
Neutron scattering experiments probe energy distributions in condensed matter systems
Ultracold atom experiments test limits of Maxwell-Boltzmann distribution at extremely low temperatures
Advanced statistical analysis techniques improve accuracy of distribution fits to experimental data
Limitations and extensions
Maxwell-Boltzmann distribution, while widely applicable, has limitations in certain physical regimes
Extensions and modifications necessary to describe more complex systems and phenomena
Understanding limitations crucial for proper application and interpretation of statistical mechanics
High density systems
Breaks down in dense gases and liquids due to significant inter-particle interactions
Requires modifications to account for excluded volume effects (van der Waals equation)
Virial expansion provides systematic corrections for non-ideal gas behavior
Integral equation theories (Ornstein-Zernike equation) extend description to liquids
Quantum effects
Fails to describe systems at very low temperatures where quantum effects dominate
Quantum statistics (Fermi-Dirac, Bose-Einstein) necessary for accurate description
Semiclassical approximations (Wigner function) bridge gap between classical and quantum regimes
Degenerate quantum gases require fully quantum mechanical treatment
Non-equilibrium situations
Assumes system is in thermal equilibrium, limiting applicability to non-equilibrium processes
Boltzmann equation extends description to systems slightly out of equilibrium
Non-equilibrium statistical mechanics develops more general frameworks (BBGKY hierarchy, Langevin equation)
Fluctuation-dissipation theorem relates equilibrium fluctuations to non-equilibrium response
Significance in thermodynamics
Maxwell-Boltzmann distribution provides microscopic foundation for macroscopic thermodynamic laws
Connects statistical mechanics to classical thermodynamics, enabling deeper understanding of thermal phenomena
Facilitates calculation of thermodynamic quantities from molecular properties
Connection to entropy
Boltzmann's H-theorem demonstrates increase in entropy for systems approaching Maxwell-Boltzmann distribution
Provides statistical interpretation of second law of thermodynamics
Enables calculation of entropy from microscopic particle distributions
Relates entropy to number of microstates through Boltzmann's entropy formula S = k ln W
Role in equipartition theorem
Predicts equal distribution of energy among all accessible degrees of freedom in thermal equilibrium
Explains heat capacity of ideal gases and contribution of different molecular motions
Provides basis for understanding energy storage in molecular vibrations and rotations
Breaks down at low temperatures, leading to quantum corrections (Einstein and Debye models)
Computational methods
Numerical techniques essential for applying Maxwell-Boltzmann statistics to complex systems
Enable simulation of large numbers of particles and prediction of macroscopic properties
Provide insights into systems where analytical solutions are intractable
Monte Carlo simulations
Utilizes random sampling to generate particle configurations according to Maxwell-Boltzmann distribution
Metropolis algorithm efficiently samples high-dimensional phase spaces
Enables calculation of thermodynamic averages and fluctuations
Applicable to wide range of systems, including gases, liquids, and magnetic materials
Molecular dynamics applications
Simulates time evolution of particle systems using Newton's equations of motion
Generates trajectories that sample Maxwell-Boltzmann distribution in equilibrium
Allows study of transport properties, relaxation processes, and non-equilibrium phenomena
Widely used in materials science, biophysics, and chemical engineering for predicting material properties |
11908 | https://www.youtube.com/watch?v=pXSEcY-aJX0 | Mathematical Reasoning-Chapter 7 Geometry
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Mathematical Reasoning-Chapter 7: Geometry
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00:08 – Instructor Introduction
00:39 – GED ON Introduction
00:45 – Required Textbook/Material
01:01 – Chapter 7: Geometry
01:20 – Lesson 1: Plane Figures
18:00 – Lesson 2: Triangles
35:13 – Lesson 3: Pythagorean Relationship
51:49 – Lesson 4: Perimeter and Area
1:05 – Lesson 5: Circles
1:22 – Lesson 6: Volume
1:42 – Lesson 7: Surface Area
2:05 – Lesson 8: Combined Figures
2:24 – Lesson 9: Geometry Calculator Skills
2:38 – Next Steps
NEXT STEPS: Social Studies-Chapter 1
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Transcript:
[Music] hi everyone this is alma today we're going to be reviewing unit 2 mathematical reasoning chapter 7 geometry remember as always this video is intended to follow along with the ged test prep plus 2020 kaplan book at any time during this video you can pause and rewind so why don't we go ahead and get started with our last chapter in unit 2. welcome back as always please remember that we are following along with the ged test prep plus 2020 book you can look down below for different places where you can go ahead and order it during this video we will be covering unit 2 mathematical reasoning chapter 7 geometry lessons one through seven we will be starting on page three hundred and eighty eight lesson one playing figures during the mathematical reasoning test you will be asked to identify the characteristics of the different types of four-sided plane figures also you will be asked to draw conclusions about their angles and sides so you may be asking yourself what are plane figures well playing figures are simply sets of line segments they are pretty much classified by the properties of their sides and angles the sum of the interior angles of a four-sided plane figure will always be 360 degrees now remember this cute little fact because you're going to be using it when you are solving for a missing angle measurement which will be asked during the mathematical reasoning test and another side note with plane figures is that sides with the same markings are equal and you will see these markings on the following slides let's start by looking at some plain figures and its properties you are very familiar with the first one which is a rectangle a rectangle is a four-sided figure with four right angles meaning that if you were to add each angle of a rectangle you will be getting 360 degrees which is a property of a plane figure also with the rectangle the opposite sides are the same length and they are parallel from each other so if you look at the two horizontal lines on this rectangle you will see that they have the same markings meaning that they have the same length on the other hand the two vertical lines have the same markings as well so that means that the two vertical lines have the same length two now let's look at another plane figure a very familiar one a square so the square is also a four-sided figure with four right angles and it also has opposite sides that are parallel from each other however the main difference between a rectangle and a square is that with a square all four sides are the same length ergo the same markings so as you can see all of the lines that form the square have the same markings telling you that it is the same length and it's a square now let's look at a parallelogram a parallelogram has opposite sides that are the same length and they are parallel from each other their opposite angles are equal in measure and a special kind of parallelogram is column rhombus because all four sines are of equal length a trapezoid is a four-sided figure with one set of parallel sides for this example we are going to find the measurement of angle a so this figure is a b c and d and it's telling us that the opposite sides are parallel so ac is parallel to b and d and then a b is parallel to c and d we are also given a measurement of angle b which is 110 degrees so let's go ahead and first identify the figure so we know that it's a parallelogram why do we know that because the opposite sides are the same remember that equal markings means that they are the same length also opposite angles are equal in measurement so now that we know that angle b is 100 degrees let's find the measurement of angle c so based on the parallelogram properties opposite angles are equal in measurement and angle c is opposite of angle b therefore we know that the measurement of angle c is the same as the measurement of angle b so they both measure 110 degrees for the next step now we're going to find the measurement of angle a so we know that the measurement of angle a and the measurement of angle d are equal and all we also know that the sum of all the angles has to be 360 degrees correct so then how do we calculate that well we're going to let x equal the measurement of angle a but because we know that there's two angles that we that are going to have the same measurement and that's angle a and angle d we're going to have an equation of 2 x that means that we have two angles times whatever the measurement is of angle a so in that case we have 2x plus 110 degrees which is the angle b and then another 110 degrees which is angle c gives us a total of 360 degrees now in order to solve this you learned in the past lesson or the past chapter that what you do on one side you have to do on the other so let's go ahead and get started so first off let's combine like terms so we have a hundred and ten plus a hundred and ten that are like they're like terms so therefore our equation becomes 2x plus 220 degrees which gives us a total of 360 degrees so now we want to cancel out the 220 degrees on one side because we eventually want to find out what x is so what we did to one side we do to the other and in this case we have because we have a positive sign on one side and we want to get rid of it we are going to subtract 220 from both sides leaving us with 2x on one side and 140 degrees on the other now the next step in order to just isolate the x we're going to have to divide by two so remember what you do on one side you have to do on the other so when you divide 2x by 2 you get an x and then when you divide 140 degrees by 2 you get 70 degrees therefore the angle of the measure the measurement of angle a is 70 degrees let's go ahead and put what we're learning to practice we're going to go now to practice one page 389. so for practice one we're going to complete questions one through five and 11 and 12. so for questions one through five you're going to list the names of the four-sided plain figures and then for questions 11 and 12 you're going to choose the best answers so go ahead and take a couple minutes pause the video complete the questions and then come back when you're ready to review them together all right so let's get started so here we have question number one it's telling us that this shape or this figure has four right angles and remember we're only applying it to the shapes that we have already learned on page 388 so four right angles i can think of two can you think of two the rectangle and the square for question two its opposite sides are equal in length so let's think back to the rectangle the square the parallelogram and the trapezoid so oh and we forgot one the rhombus so in this case opposite sides are equal in length it is all of them parallelogram rectangle square and the rhombus with the exception of the trapezoid question number three exactly one pair of parallel sides and that is a trapezoid remember that the trapezoid is a four-sided figure with only one pair of parallel sides all angles are equal in measure a rectangle and a square and then question five only three right angles well we know that plane figures are is a set of line segments and in this time we were only learning about four-sided plane figures so our answer is none because all the shapes that we have learned so far have all had four sides for questions 11 and 12 we are going to be using figure e f g h let's look at question 11. angle f is 20 degrees more than three times the measurement of angle h what is the measurement of angle f it's a tricky one but don't worry we're going to solve it together so let's start off by giving the angle h a variable in this case we're going to use h previous examples we've used x and y but for this scenario let's use h so now it's telling us let's go back to our question and it's telling us that angle f is 20 degrees more than three times the measurement of angle h so what does that look like in an expression it looks like this three times the measurement of angle h plus 20 degrees equals the measurement of angle f perfect so now because we know that all angles together for a plane figure they have to equal 360 degrees so now let's input that into our expression so we have 3h plus 20 degrees plus h plus 90 degrees plus 90 degrees equals 360 degrees okay so we're adding basically we're adding all angles together now after this we're going to solve it or we're going to calculate it remember we're going to start off by adding like terms when we add like terms we have for h plus 200 degrees which equals 360 degrees and in this case we want to isolate the h right because that's going to be the measurement of angle h and we need that in order to find out the measurement of angle f so let's go ahead and take a look at this so we have 4h plus 200 degrees equals 360. so in order to isolate the 4h we're going to have to subtract 200 to cancel it from one side but what we do on one side we have to do on the other so that leaves us with 4h divided by 160. now oh sorry 4h equals 160 degrees i'm getting ahead of myself so now we're going to find the value of h which is the measurement of angle h so in order to do that we're going to have to divide by 4 on one side and then divide by 4 on the other side when we do that we have h equals 40 degrees so bam the angle the measurement for angle h is 40 degrees so now we have to take those 40 degrees and plug it into our first equation that's going to give us the measurement for angle f so let's go back to our first equation which was 3h plus 20 degrees gives us the measurement of angle f so when we plug in when we replace 40 degrees instead of the h we get 30 times 40 degrees plus 20 degrees and that gives us 140 degrees therefore the measurement of angle f is 140 degrees let's look at question 12. remember we are using the same figure efgh question 12 reads in order for figure e f g h to be a trapezoid which of the following must be a true statement so let's look at our options a e f is the same length as f h e f is parallel to g h the measurement of angle f is the same as measurement of angle h and the measurement of angle g is the same as the measurement for angle f so let's look at the properties of a trapezoid for a trapezoid you need to have one pair of parallel lines in this case we have e f is parallel to g h as far as our options go that is the only correct statement that i see in order for the figure to be a trapezoid ef is not the same length as fh and the measurement of angle f is not the same as measurement in angle h as we saw in the previous problem and then the measurement of angle g does not equal the measurement of angle f so therefore our answer is b e f is parallel to g h at this time pause the video complete practice questions 6-10 and 13-14 and click play when you are ready to continue let's move on to lesson two triangles in the previous lessons we learned about four-sided plane figures during this lesson we are going to learn about three-sided plane figures or in other words triangles the properties of a triangle include three interior angles three vertices a vertex is a point where two or more more curves lines or edges meet a triangle is named by writing its vertices in any order and one more important fact to remember about the three-sided plane figures or the triangles the sum of all interior angles equals to 180 degrees for example if you look at this triangle here this triangle can be named triangle d e f its sides are d e e f and d f let's look at different types of three-sided plane figures these three-sided plane figures are classified by side lengths so in the first one we have an equilateral triangle so just like its name says equilateral means equal so that means that all sides are equal in length and all angles are also equal in degrees so for example they all measure 60 degrees in this example but they could all measure maybe 80 degrees or 30 degrees as long as all angles are the same and all sides are equal that is an equilateral triangle the second three-sided plane figure is the isosceles triangle so if you notice with the isosceles triangle two sides are equal in length remember that we learned about those little markings back in lesson one so pretty much the same markings means that they are the same length and then with an isosceles triangle we have two angles opposite of these sides are equal so as you can see with the isosceles triangle we have two angles opposite of each other that equals 75 degrees each now for the last example of the three-sided plane figure we have a scalene triangle so with the scalene triangle no sides are equal in length and no angles are equal so what this means is that every side to this green triangle is different length so one could be one centimeter another going to be three and then another one can be five centimeters and the angles are different as well now let's look at three-sided plane figures that are classified by angle measurement so the first one we have a right triangle which means that this triangle has at least one right angle and and right angle measures 90 degrees for the second triangle we have an acute triangle what this means an acute means that all angles measure less than 90 degrees and then an obtuse triangle means that at least one angle is greater than 90 degrees remember that each triangle can be classified in two ways angle and side length so let's look at example one and determine what kind of triangle is triangle p q r just by looking at it we see that it already has a 90 degree angle and the side length we have two side lengths that are the same and one that is different so let's go ahead just keep that in mind when we are answering the question so we're going to classify it first by its sides well with its size we have two sides that are the same so they are the same length they both equal 10 centimeters each so we can determine that triangle pqr is an isosceles triangle remember an isosceles triangle has two sides that are equal in length and next we're going to classify it by its angles so angle p as we mentioned earlier has a right angle so triangle pqr is a right triangle remember that our right triangle has one angle that measures 90 degrees so when we classify this triangle this particular triangle with angle inside length we can determine that triangle pqr is a right isosceles triangle okay so we're going to calculate for a missing angle for this example remember that in the first slide we talked about the sum of all interior angles equaling to 180 degrees so for example in triangle abc angle a measures 55 degrees and angle b measures 100 degrees what is the measure of angle c so let's go back to our algebraic expressions so we're going to write an equation and we're going to solve it remember those variables x and y well in this case we're going to use the angle of c because that is what we're trying to find so when we write an equation we're going to add angle a plus angle b plus angle c and we already know that the total of those three angles should give us 180 degrees so now this is our expression we have 55 degrees plus 100 degrees plus angle c which gives us a total of 180 degrees so next we're going to combine like terms that means that we're going to add 55 degrees plus 100 which gives us 155 degrees plus the angle of c which equals 180 degrees 180 degrees so in order to find the value or the measurement of angle c we have to cancel out 155 degrees so the way that we do that we do it it's inverse so in this case we're going to subtract 155 degrees from the left side but whatever you do on the left side you have to do it on the right side so if you subtract minus 155 minus 155 that cancels that out so you are left with just the angle c and then the other side on the right side if you subtract 155 degrees minus 180 or 180 degrees subtracted by 155 degrees you get 25 degrees so therefore you know that the measurement of angle c equals 25 degrees let's move on to our practice questions so right now go to practice 2 located on page 391 so for practice two we're going to complete questions two five and seven for question two you're going to classify the triangle in two ways for question five you're going to find the measurement of the unknown angle and for question 7 you're going to choose the best answer so take a few minutes right now to pause the video and complete questions 2 5 and 7 when you're ready to continue go ahead and press play great you're back so let's go ahead and get started so with question two we're going to classify the triangle in two ways okay so we have our triangle here so first off we can look at it and we can decide that all sides are different lengths so we have one that's 4.5 centimeters we have another side that's nine and then the third side is 12 centimeters so right off the bat we know that these sides are not the same and then we have an angle that is more than 90 degrees okay so let's look at the way that we classify triangles so an equilateral triangle has equal length well that cannot be because our example here for question two they have different lengths so let's move on to the next one isosceles triangle an isosceles triangle has two sides that are equal in length well once again all sides are different so it cannot be an isosceles third option is a scalene triangle which means that no sides are equal in length well that's the case we have three we have three sides that are different in length therefore we can determine that this is this triangle is a scalene triangle now it's classified by angle so you have one angle that's more than 90 degrees so both the right triangle in the acute triangle you either have to have a right angle which is 90 degrees or an acute triangle an angle that's less than 85 degrees well with this triangle we have one angle that's greater than 90 therefore it is an obtuse triangle within a two string you have one angle that is greater than 90 degrees and in this case it's 115 degrees so by looking at all the properties for the different triangles when we're classifying them by length and by angles we can determine that this triangle here for question two is a sculling and obtuse triangle so for question number five we're going to find the measurement of the unknown angle so we have a triangle with a 90 degree angle and then another angle 45 degree and then we have the third one that's unknown so let's go ahead and calculate calculated so we know that the sum of all angles has to equal 180 degrees perfect so let's go ahead and add all of the angles together to determine or to find the value of the unknown variable so we have 45 degrees plus 90 degrees plus the unknown variable which gives us a total of 180 degrees so we're going to combine our like terms in this case it's 45 and 90 so it gives us a total of 135 degrees so our new equation looks like this 135 degrees plus the unknown variable equals 180 degrees so then in order to find the unknown variable we are going to do the inverse of the number that we have on the left and on the right side so for example on the left side we have 135 and in order to eliminate the 135 from the left side we have to subtract it so the inverse means the opposite so we're going to subtract 135 but what we do on one side we have to do on the other so then we will subtract 135 from the left side and subtract 135 from the right side that leaves us with 45 degrees therefore our unknown angle is 45 degrees now let's look at question seven we're going to choose the best answer if angle d a b measures 115 degrees an angle dcb measures 95 degrees what is the length of side ac in centimeters hint use the facts in the problem to find the measurements of bac and angle bca so let's go ahead and find the angle of bc the measurement of angle bac and the measurements of angle bca which is going to help us essentially find the length of side ac so we know that the measurement of angle b a c plus the 55 degrees gives us a total of 115 15 degrees so how do we know that if you look back at our question you can see that angle d a b is 115 so we have to combine 55 degrees plus whatever angle is bac so we're going to look at for the measurement of angle bac first so that leaves us with 60 degrees remember we have to eliminate the 55 degrees from the left side in order to look for the variable in this case is the measurement of bac what we do on one side we have to do it on the other so therefore we found the measurement of bac which is 60 degrees now let's find the measurement for b c a which is the same thing if you go back to the question we know that d angle d c b measures 95 degrees so then when we write our equation we're going to have angle b c a plus 35 degrees equals 95 degrees so we're going to solve this equation we're going to solve for the variable in this case the measurement of bca which gives us 60 degrees so we already know just based on the calculations we know that angle bac is 60 degrees and angle bca also equals 60 degrees we can determine that angle b also measures 60 degrees so why because by looking at the two angles and with them measuring the same angle or measuring the same measurement we know that an equilateral triangle has angles that are equal in this case 60 degrees so two angles of triangle abc equals 16 then the third one also measures 60. so the triangle in fact is a nikolai equilateral which then states that if all sides are a b and c are equal to side a therefore side a b is 14 centimeters and then side b and c is 14 centimeters inside a and c is also 14 centimeters therefore our answer is c 14 at this time pause the video complete practice questions one three 4 6 8 through 9 and click play when you're ready to continue let's move on to lesson 3 pythagorean relationship during this lesson we are going to learn about the pythagorean relationship the pythagorean relationship focuses more on the right triangle so let's review a little bit about the right triangle remember that with the right triangle it has one right angle a right angle measures 90 degrees with the right triangle the side that is across from the right angle is called the hypotenuse the hypotenuse is the longest side of the right triangle the remaining sides are the legs of the triangle there is a special relationship among all three sides the legs and the hypotenuse this relationship is called the pythagorean relationship so with the pythagorean relationship you are able to solve the measurement of any side of a right angle or of a right triangle if the other sides are known so the equation for the pythagorean relationship is a squared plus b squared equals c squared so a is the leg of the triangle b is the other leg of the triangle and c is a hypotenuse remember c is the longest side of a right triangle the square the hypotenuse is equal to the sum of the squares of the two legs of the right triangle let's go ahead and practice for example one what is the length of the hypotenuse of the right triangle shown in the diagram so we have a right triangle with legs um both different in size one is three feet and the other one is four feet now the goal here is for us to find the hypotenuse so let's assign a variable for leg one it's going to be variable a so that one's gonna be four feet leg two is three feet so that's we're going to assign variable b and hypotenuse will be c so now let's use the pythagorean relationship in order to find the hypotenuse remember the equation a squared plus b squared equals c squared so let's replace all the variables with actual values remember that we gave the variable a it was 4 feet b was 3 feet and c squared is the variable that we are trying to find so here we have 4 squared plus 3 squared equals c squared so 4 to the second power is sixteen three to the second power is nine so then we add common terms we add sixteen plus nine which gives us a total of 25 but we still have c squared and remember the hypotenuse is only c so therefore in order to find the value of c so this is the square root of c therefore it's going to be the square root of 25 so the square root of 25 equals c so what does that mean it means that the hypotenuse is five feet because five to the second squared equals twenty-five let's practice again this time let's use example two if john places a 13 foot ladder three feet from the base of the wall how far up the wall will the ladder reach to the nearest tenth of a foot so in this case we are given the length or the measurement of the length of the hypotenuse and the measurement of one leg so we are going to use our equation in order to find the size or the measurement of the other leg and when we look at our example we're going to find that the measurement of how far up the wall will the ladder reach so as we mentioned the length of the triangles are 13 feet and three feet remember three feet is leg is a leg and 13 feet is a hypotenuse because it's right across from the 90 degree angle let's assign the variables leg a is going to be three feet the hypotenuse is going to be 13 feet which we're going to assign variable c and then like 2 because it's an unknown variable it's just going to be b so let's use the pythagorean relationship once again a squared plus b squared equals c squared so leg 1 plus leg 2 equals hypotenuse but in this case we have and so we have a value for leg a which is 3 squared but we don't have a value for b squared so we're just going to write in b squared and leave it as a variable but we do have the total of the hypotenuse which is 13 squared so our equation is going to look like this 3 squared plus b squared equals 13 squared so let's calculate it so three squared is nine then we bring down the b squared because remember we're still trying to solve for b then 13 squared equals 169. so now in order to find the variable of b squared we have to subtract the nine we have to do the opposite of what is there so right here we have positive 9 so we have to subtract by a negative 9 in order to remove it from the left side but the same thing that we do on the left side we have to do it on the right side so therefore when we subtract 9 from the left side we are left with b squared and when we subtract 9 from the right side we are left with 160. so now we have b squared equals 160. that translates into b equals the square root of one sixty therefore giving us the answer of twelve point six feet so the ladder is going to reach 12 a little bit more than 12 and a half feet in distance let's move on to practice three found on page 393 for practice three we are going to answer questions one two 5 11 and 12. so for questions 1 2 and 5 you need to find the length of the remaining side to the nearest 10th unit for question 11 and 12 choose the best answers so take a couple minutes to complete these five questions and when you are ready to review them together just click play so let's start with question one for question one we are given the measurements for leg a b and we have an unknown measurement of hypotenuse which is c so let's go ahead and use our pythagorean relationship equation in order to solve for the hypotenuse so we have here that the leg a equals 8 inches leg b equals 8 inches so we are going to replace our variables with the numbers provided in this case we have 8 squared plus 8 squared equals c squared when we simplify it it's going to be 64 plus 64 giving us a total of 128. so then we have 128 divided by or i'm sorry equals c squared so in other words we need to find the square root of 128 in order to e in order to solve for the hypotenuse we have an answer so the square root of 128 is 11.3 therefore the hypotenuse is 11.3 inches let's move on to question two for question 2 we have leg a 9 yards leg b 12 yards and the hypotenuse is unknown so here we go with our pythagorean relationship equation a squared plus b squared equals c squared we are going to replace all the variables with the values that were given to us so we have 9 squared plus 2 squared equals c squared we're going to simplify it which gives us a total of 81 plus 144 we're going to simplify it even more and that's going to be 225 equals c squared so when we simplify that we are asked to calculate for the square root of 225 giving us an answer of 15. therefore the hypotenuse for this question is 15 yards remember to always put the measurement value that is given to you for example in question one it was inches and in question two it is yards make sure that you add that so now let's look at question five question five we have leg a six millimeters like b it's unknown and we have a hypotenuse or c of 10 millimeters so here is our pythagorean relationship equation so we are going to replace all the variables with the numbers given so in that case we have 6 squared plus b squared equals 10 squared so for this question we are looking for the measurement of leg b so let's go ahead and simplify our equation so 6 squared gives us a total of 36 then we bring down the b squared which equals 10 squared equals 100. so remember that in order to isolate the variable b squared we have to get rid of that 36 so in order to get rid of that 36 we have to inverse it which means we have to do the opposite of what it is so in this case we have a positive 36 and if we subtract 36 they're going to cancel each other out and leaves us just with b squared but what we do on one side we have to do on the other so 100 minus 36 gives us a total of 64. so we have b squared equals 64. and when we simplify it even more we have the square root of 64. so b equals square root of 54 giving us an answer of eight therefore the leg b equals eight millimeters let's now look at question 11. the two shorter sides of a right triangle measure 18 feet and 24 feet what is the measure in feet of the third side well in this case they are telling us that is a right triangle and in order to find the third side which is the hypotenuse we have to go back to our formula our previous formula was a squared plus b squared equals c squared in this case we are given the measurements of the leg a and leg b and then c is what we have to find which is the hypotenuse so once we replace the variables with the values that we are given we have 18 squared plus 24 squared equals c squared when it gets simplified we have 324 plus 576 which equals c squared and then if we simplify it even more we have 900 equaling c squared but in order to find the measurement in feet of the hypotenuse we have to take the square root of c which in turn equals 30 squared feet or 30 feet i'm sorry it's 30 feet so the answer is c now let's look at question 12. jan has built a rectangular frame out of wood to use for the bottom of a platform he wants to add a diagonal brace as shown in the drawing so we have here the right triangle with measurements of leg a and leg b and he wants us to find the diagonal measurement because they want to install a diagonal brace so for this case we're also going to use the pythagorean relationship formula which is a squared plus b squared equals c squared and once again we're looking for the hypotenuse so let's go ahead and replace our values our variables with the values that are being given so we have 5 squared plus 11 squared equals c squared when we simplify it we have 25 plus 120 equals c squared simplify it even more we have 146 equals c squared but we need to find the square root of 146 in order to find the measurement in feet of the hypotenuse or the diagonal brace in this case the square root of 146 yields a result of d 12.1 feet at this time pause the video complete practice questions 3-4 6-10 and 13. click play when you are ready to continue before we begin lesson four i just wanted to check in and make sure that you're doing okay we are coming up to the one hour mark but i wanted to let you know that we still have a little bit ways to go as you know math is a little bit more complex unit so therefore it is taking us a little bit longer but don't worry i am here and you can pause at any time to take a break but let's go ahead and continue welcome let's move on to lesson four perimeter and area in previous lesson we have learned how to classify triangles both by sides and by angles we've also learned about four figure plane which includes your rectangle and um your your rectangles in any quadrilateral for lesson four we're going to learn how to find the distance around a figure the distance around the figure is called the perimeter so in order to find the perimeter is pretty much going to be the sum of the length of the sides and here are the formulas for some simple common figures that we have the square the rectangle and the triangle so make sure that you remember or that you memorize these simple formulas because you're going to need them for the ged mathematical reasoning test so to find a perimeter for a square it's four times a side rectangle the perimeter is two times the length plus 2 times its width in a triangle you're going to find the perimeter by adding side 1 side 2 and side three so let's go ahead and take a look at example one a rectangle is 16 inches long and nine inches wide what is the perimeter of the rectangle so we're going to use the formula to find the perimeter of a rectangle and the formula is 2 times its length plus 2 times its width so when we replace the length in this instance it's 16 we're going to multiply that by 2 and then we're going to add it to 2 times 9 because that is the width when we do that we get 32 plus 18 yielding a result of 15 inches so therefore the perimeter of the rectangle is 50 inches see it's not that bad you just need a little practice so now let's find the area of a figure so the area is the measurement of space inside the figure so we talked about the outside which was the perimeter but now inside is the area the area is measured in squared units so for example if you're measuring in inches then the area will be measured in square inches if they were feet then it would be squared feet so here are the formulas to finding the areas so in order to find the area of a square it's s to the power of two the rectangle is length times its width a parallelogram its base times its height a triangle its one half times the base times the height a trapezoid is one half times the height times the base one plus base two don't worry we will apply this formulas to the following examples if you can see here the formulas we have two new measurements that we haven't talked about before one of them is the base and then the other one is the height so if you look at the diagrams you can see that the base is one side of the figure for the trapezoid you can see that there are two bases you have base one and then base two but for a triangle you only have the bottom base the height is the length from the vertex to the base forming a right angle to the base so let's go ahead and practice example two states to find the area of figure a b c and d so just looking at the diagram we see that we have a height of seven centimeters and we have a base of 12 centimeters okay so let's go ahead because this is a parallelogram which is going to be the first step eventually is to identify the figure so right away we can see that it's a parallelogram so we go back to our formulas in order to find the area of a parallelogram and it is base times its height so we we're given a base so find the facts that you need on your diagram so you have a height of seven centimeters remember the height is aligned from the vertex to the base and then we have a base of 12 centimeters so now let's calculate remember the area is base times its height so we're going to replace the values with the variables which gives us 12 times 7 and in total the area of the parallelogram is 84 square centimeters or 84 centimeters squared either way is correct let's move on to practice 4 located on page 395. for practice four we are going to be answering questions one six nine and ten for questions one and six you're going to find the area and perimeter and then for questions 9 and 10 you're going to choose the best answer so go ahead and take a few minutes to complete this by yourselves and then once you're ready to review them together just go ahead and click play let's start with question one and six so for questions one and six we're going to find the area and perimeter for each so question one is a triangle in order to find the perimeter of a triangle you have to add all sides so here we have seven point eight 10 plus 13 which are our sides disregard the 6 at this time because that is the height which we will be using for the area so when you calculate this numbers you and when you add all the sides together you get a perimeter of 30.8 units so now let's calculate the area when you calculate the area for a triangle it's one half times its base times its height so now at this time we are going to use the six which is the height remember the height goes from one vertex to the base it's a straight line and um so it's going to be base times height so we have 13 times 6 which gives us a total of 78 and then we're going to multiply that times one half which yields a result of 39 square units now let's look at question six so question six we have a rectangle in order to find the perimeter of a rectangle it's two times its length plus two times its width we're going to replace the variables with the given values so the length in this case is 12 units and the width is four units so when we calculate we have um two times twelve equals 24 plus two times four which gives us eight when we calculate 24 plus 8 we get a perimeter of 32 units in order to find the area for this figure it's going to be a little tricky because the area is going to be we're using the formula of a trapezoid the reason being is because we have a height that takes us from one vertex to the base in in order to make a 90 degree angle so as you can see right here if you disregard the middle line you can see a trapezoid therefore we're going to use a formula of a trapezoid the formula of a trapezoid is one-half times its height times the base one plus base two so we know all those numbers so let's go ahead and replace the variables with the values given so we have a height of 9.5 base 1 is 4 and base 2 is 4. so when we calculate that we have nine nine times five oh i'm sorry nine point five times four plus four all of it multiplied by one and a half which gives us 38 square units let's move on to questions nine and ten for questions nine and ten we are going to choose the best answer so let's read number nine martin is building a rectangular patio center on one side of his yard the rest of his yard is planted in grass if the measurements in the diagram are in feet what is the square footage of the grass portion of martin's yard okay so they are looking for the square footage of the grass area only but in this case we have the patio and the grassy area so let's go ahead and calculate this so first off we're going to find the area of the patio and i will tell you why in a minute so the area of the patio we were told that it was a rectangle and in order to find the area of a rectangle it's its length times its width so we're going to replace those variables with the values in this case it's 9 in 12. so that yields a result of 108. that is the square footage or the area square footage for the patio itself now let's find the square footage for the yard once again we were told it's a rectangle so we're gonna multiply the length times its width in order to get the area of the yard we're going to replace those values with 18 in 24 yielding the result of 432 now what we're going to do is we're going to subtract the yard or the patio from the yard because we only want the area of the grass so in order to do this we already have our area so it's going to be 432 minus the square footage of the patio which is 108 yielding a result of 324 square feet now let's look at question 10. a square measures 6 centimeters on one side what is the perimeter of the square in centimeters the formula to find the perimeter of a square is four times its side in this case the side is six centimeters so when we calculate it we find that the perimeter is 24 centimeters so your answer is b at this time pause the video complete practice questions two through five and seven through eight click play when you are ready to continue we are now moving on to lesson five circles we have been learning about triangles rectangles squares trapezoids so pretty much any plane figure that has a side to it well circles don't have any sides they are a closed set of points that are all that are all the same distance from a single point or the center distance in order to find the perimeter of a circle you need special formulas and other measurements within the circle but the perimeter in a circle in this case it's called circumference and that is the distance around the circle the area of a circle is the space inside the circle so finding the perimeter in the area as i mentioned before you need other measurements than just the height or the base because in this case we do not have any bases so let's look at the different measurements that we need to find the perimeter and an area so we first of all we need a diameter as you can see in the diagram the diameter is a line segment with endpoints on the circle that passes through the center of the circle so it's pretty much a line that goes from one end to the other that is your diameter your radius is a line segment that connects the center of the circle to any point on the circle so you could say that the radius is one-half the diameter so keep that in mind for our future examples and practices in order to find the perimeter and the area of a circle we use a special formula which includes pi no not the pi that you eat but pi in mathematical reasoning or mathematical equations so pi is a special quantity used to solve for the circumference and the area and the pi is equally so it's equal to approximately 3.14 because it is an infinite number so pi is the ratio of the circumference to the diameter so let's look at a summary the formulas that we will be using in order to find perimeter and area so we have a circumference formula is pi times the diameter in order to find the area of a circle it is pi times the radius to the second power there will be instances when you will be asked to find the diameter of a circle in order to solve for the circumference and also for the radius of a circle in order to solve for the diameter so they kind of all intertwine together the diameter formula is the d which is equals diameter equals 2 times the radius remember that we said the radius was half the diameter so this is where our formula comes in and then we have a radius formula as well which is one half times that diameter so let's look at example one a china plate has a gold rim if the plate's diameter is 10.5 inches what is the distance around the rim to the nearest tenth of an inch so in this example they are asking us to find the circumference remember the circumference is the distance around the circle in this example it will be around the gold rim so then we're going to use the formula circumference equals pi times diameter so we already know pi equals 3.14 because that's an infinite number and the diameter was given to us in the example which is 10.5 inches when we multiply the pi and diameter we have an answer of 32.97 but when we round it out to the nearest tenth of an inch it gives us 33 inches let's move on to example two the circular surface of a satellite component must be covered with hit with heat resistant tiles if the radius of the component is 4 meters what is the area in square meters so in this example it's asking us to find the area so the formula to find in the area is pi times radius squared so we were given the radius in this example which is 4 meters so now we're going to replace our variables with the values given so pi we know that it's 3.15 in radius on the example it's telling us that it's 4 meters so then our equation is 3.14 times 4 to the second power so we are going to first remember uh the order of operations before we multiply we have to complete the exponents so let's go ahead and we have 4 to the second power which will be 16. so therefore our new equation is 3.14 times 16 when we multiply that we have an area of 50.24 square meters see not bad let's look at example three what is the circumference of circle b to the nearest tenth of a center of a centimeter so circumference is pi times diameter well in this example we are not given the value of the diameter so we have to find it in order to find the diameter the formula is 2 times the radius so we were given a value for the radius so let's go ahead and substitute or replace our variables with the values given so in this instance we're looking for the diameter which is 2 times radius we're going to replace the variable r with 5 and that gives us a diameter of 10. so now that we have our diameter now we can go ahead and solve for the circumference remember circumference is pi times d so we're going to replace the variables with the values so in this instance we have 3.14 times 10 which gives us an answer of 31.4 centimeters at this time let's move on to practice 5 found on page 397 for practice five we are going to answer questions one three six and seven so for questions one and three you're going to find the circumference in the area and for questions six and seven you're going to choose the best answer so take a few minutes to complete these four questions once you're ready to review them together just click play let's start with question one for question one we are going to find the circumference and the area which will be the same for question three so let's go ahead and get started so for question one we have a circle with a radius of 10 inches so we know that the circumference formula is pi times the diameter and the area is pi times radius squared so let's start with the circumference so for the circumference we don't have a diameter but we do have a radius and we do have a formula to find the diameter so the formula to find the diameter is 2 times the radius so in this case it's going to be 2 times 10. 2 times 10 gives us a diameter of 20. now we can now we can solve for our circumference which is 3.1 times 20 gives us a circumference of 62.8 inches pretty easy right so now let's find the area so the area is pi times radius square so let's substitute or replace the variables with our actual values so we have for area 3.14 times 10 to the second power which gives the total of 314 square inches or 314 inches squared whatever is your preference let's now move on to question three so for question three same thing we're going to find the circumference which is pi times d and the area which is pi times r squared so once again it looks like we were not given a diameter but we were given a radius so in order to find the diameter of this circle we're going to have to use our formula for the diameter so diameter equals 2 times the radius in this instance our radius is 4 meters so it is 2 so the diameter will be 2 times 4 giving us a total of 8 meters so we found our diameter now we can move on and solve for the circumference so let's replace all the values with the um variables or vice versa sorry so the circumference is 3.14 remember that's pi times 8 which gives us a circumference total of 25.1 meters now let's do the same thing for the area the area is pi times radius squared so that will be 3.14 times 4 to the second power when we solve that equation we get an answer of 50.2 square meters or 50.2 meters squared either one is correct so for question six we are going to choose the best answer let's read it together the radius of a circle is 6.5 centimeters what is the diameter of the circle in centimeters okay so this one does not have a figure or a diagram so you can definitely do one on a paper if you're a visual learner if not you can do it in your head so let's start with the diameter formula because that's what they want us to find so the diameter equals 2 times the radius and we were given the radius which is 6.5 so right now we're just going to replace our variable with the value which is 2 times 6.5 which gives us an answer of 13. so if you chose b you are correct let's move on to question 7. so for question seven we're also going to choose the best answer question seven reads on the target below or to the side the five in ten point bands are each 2 inches wide and the 25 point inner circle has a diameter of 2 inches to the nearest inch what is the outer circumference of the 10 point band so we're trying to find the distance around the second circle the 10 point band so if you look at the diagram we already have um the diameter of the 25 point circle which is two inches and then we have the length or how much or the inches the width of the 10 point circle so let's go ahead and solve so first we have to find the diameter of the 10 point band in order to solve for the circumference because circumference is pi times diameter so the diameter for the 10 points 10 point circle passes that it's that orange line it passes through the 10 point circle once and then 10 point circle twice and the 25 point band so essentially what this is telling us is that the diameter is going to be the inner circle the measurement of the inner circle plus the 10 point band where it starts and then the 10-point band where it ends so that's how we're going to get a diameter so we pretty much already have those numbers so we know that the diameter of the inner circle which is the 25 point inner circle the diameter is two remember the diameter goes from one end to the other so that's a total of two now the temp the 10 point band from one side equals two inches which was already given to us on the question but because the diameter passes through it twice and the other side as well we're also going to add those additional 2 inches therefore the diameter of the circle here or the 10 point circle only is 10 inches so now that we have the diameter now we can measure for the circumference so the circumference is pi times diameter so we're going to replace the variables with our values 3.14 times 6 which gives us a total of 18.84 but in this instance they want us to find the circumference to the nearest inch so in this case when we round it out it's 19 therefore our answer is c at this time pause the video complete practice questions 2 4 and 5 and click play when you are ready to continue we are now moving on to lesson six volume in previous lessons we talked about the area and we talked about diameter circumference radius along with other formulas well in this instance or in this lesson we are going to focus on volume so what is volume the volume is the capacity or pretty much it's the measurement of space inside a three-dimensional object so how much of it fits inside an object so the volume is measured in cubic units so ultimately the number is how many cubes would you need to fill an object so let's think of some three-dimensional objects that have at least two identical and parallel faces for example that could be a cereal box as you can see the top and the bottom of the box are identical so those can be called faces or in this case a base so either face can be the base of the object and that is more for a rectangular prism volume can be found multiplying the area of one base by the height of the object so volume equals area of base times height but what if you don't have a rectangular prism what if you have another three-dimensional object such as a cube a cylinder or a sphere well don't worry here on this slide you can see that i've listed down the formulas to find the volume of each of those three dimensional objects in the previous slides we look at some formulas in order to find the value of some three-dimensional objects well let's go ahead and do some examples in order to get better at it remember practice makes perfect so for example one a cardboard box has the dimensions shown in the diagram what is the volume of the box in cubic feet so first off we have to determine what kind of three-dimensional object this is so we know that it's not a cube we also know that it's not a cylinder and definitely it's not a sphere so we can determine that this object is a rectangular prism so with the rectangular prism it has two identical bases so for example on the box where it says four feet the opposite base is also going to be four feet while where it says five feet the opposite base is going to also be five feet therefore they're identical bases so let's look at the formula the formula for a rectangular prism is volume equals base times height so let's go ahead and replace the variables with the values that i give that are being given so we have four times five remember we have to use the measurements for both bases in order to find the volume so it's four times 5 times 3. so therefore it equals 60 cubic feet or 60 feet cubed so let's look at another example a wood block measures two inches per edge what is the volume of the block so for this instance this is not a rectangular prism so let's identify what kind of three-dimensional object this one is so it talks about edges so i remember looking at the formulas that in one of the formulas we use the edge in order to find the volume do you recall what that was and it's okay if you want to go back to the previous slide and take a look at it don't worry so it was a cube so the cube is a rectangular prism with six identical faces so in a cube each edge is the same length so all of it is going to be pretty much the same which is two inches per edge in order to find the volume of a cube it is equal the edge to the third power so we're going to replace the variable with the value given which in this case the measurement was 2 inches so the volume of the cube when you take 2 to the third power is going to be eight cubic inches you can write it in either format let's look at example three a storage tank has a radius of 1.5 meters and a height of 3 meters what is the volume of the tank to the nearest cubic meter okay so here we have the cylinder and the cylinder gives us a radius remember the radius is half the diameter the diameter is a straight line going from one end to another in a circle in the circle okay so here we're given half of the diameter which in other words is the radius and we're given the height of the cylinder so the cylinder has two circular bases and the bases are connected by a curved surface so the formula to find the volume of a cylinder is pi times radius squared times the height so let's go ahead and replace our values let's replace the variables with the values that are being given so remember with the previous lessons we learned that pi equals 313.14 so it's going to be 3.14 times the radius squared in this case it's 1.5 squared times the height which is 3. when we calculate that we obviously so we're going to simplify it even more so the 1.5 squared is 2.25 and when you simplify it you get 21 cubic meters so therefore the volume for the tank is 21 cubic meters at this point let's let's move on to practice 6.1 located on page 399. so for practice 6.1 we're going to answer questions 1 and 8. for question one you're going to find the volume for question two you're going to choose the best answer so take a couple minutes to complete both questions one and eight and then when you're ready to review them together just go ahead and click play perfect welcome back so let's go ahead and start with question one so question one is is asking us to find the volume so in this case we have a shape what kind of shape is it well it looks like it could be a rectangular prism because it doesn't look like a cube a cylinder or a sphere therefore the formula for a rectangular prism is volume equals base times height so remember we have two bases and we have one height so let's replace the variables with the values given so in this instance it's going to be the base which is 5 times 8 because remember we have 2 bases here times 4 which is the height that gives us an answer of 160 cubic units so now let's look at number eight a wooden crate measures five feet along each edge what is the crate's volume in cubic feet so think about what a crate looks like so now a crate and the other thing that we have to infer for this question is that they are talking about an edge so anytime you hear and you hear a question mentioning an edge automatically you should go to cube because remember the cube the volume for a cube equals edge to the third power so in this case we can infer that the wooden crate is a cube and each edge is five feet so therefore the formula for volume of an of a cube is the edge to the third power so let's replace the value the variables with the value that's being given so in this case it's telling us that each edge is 5 feet so let's replace it with 5 feet and when we take 5 to the third power we get 125 cubic feet at this time pause the video complete practice questions 2 through 7 and 9 and click play when you are ready to continue now let's learn about finding volumes for a three-dimensional object with four triangle faces that connect to the same vertex or in other words the pyramid so the basis of a pyramid can be any closed figure for the gd mathematical reasoning test you will only have square bases so as you can see in the following diagrams we have a rectangular prism and we have a pyramid they both have identical square bases and the same height let's compare the two figures so as you can see the pyramid holds much less than the rectangular prism in fact it only holds one third of the rectangular prism's volume the formula for finding the volume of a pyramid is volume equals one-third times the base times the height notice that the base in the volume of a pyramid is the same equation to find the area of a square do you remember that back in lesson i wanna say four maybe but it was one of the previous lessons okay so just remember that so now let's talk about the volume of cones so the cones are they have a circular base and a curved sides that resembles a cylinder so the curved side is slanted inward so then it has a vertex so it has a point so think about a cone think about a party hat you know when you go to birthday parties they give you a little party hat so that's a cone okay so in order to find the cone it is the volume of a cone it's one third of the volume of a cylinder with the same size base and height so what that translate is one third times pi because remember a cylinder the the base of the cylinder is a circle so we are going back to pi from the previous lesson times the radius squared which is half of the diameter times its height so let's practice a little bit so i gave you the the formulas to find the volumes for pyramids and cones so for example one let's find the volume of the pyramid so we're going to apply the formula which is one third base times height so we have we're going to replace the variables with the values that were given to us so it's going to be 1 3 times five squared remember we're looking for the base and the base of a square it's side squared okay so pretty much it's the measurement of the side to the second power think back to the to the to the area of the square in previous lessons times six so when we calculate this so you're going to first do the exponents which is 5 to the second power is 25 then you're going to calculate the equation and you will get 50 cubic centimeters now let's look at example two for example two they want us to find the volume of the cone okay so we have a cone remember think party hat so let's apply the formula so the formula is 1 3 times pi times radius squared times height so let's replace the variables with the values that are being given so we have 1 3 times pi which is 3.14 times 2 squared because the radius is 2 times the height which is 9. so we're going to take two to the fourth power we have to do that before we do anything else always go back to your order of operations and that will give us an answer of 37.68 cubic inches let's move on to practice 6-2 found on page 401 so for practice 6.2 we are going to complete questions five and eight so for question 5 i need you to find the volume and for question 8 you're going to choose the best answer so take a few minutes to complete these questions and when you're ready to come back then you push play so let's start with question five so for question five we're going to find the volume so this looks like a period pyramid it does not look like a cone because it has a squared base remember squared area equal side squared so let's look at the formula the formula is volume equals 1 3 times the base times the height so we're going to replace the variables with our values once again the base is 2 squared always remember that and when we calculate it we get 10.66 but when we round it off it's going to be 11 cubic inches so let's look at question eight the height of a cone is half the diameter of the base if the cone's height is four inches what is the cones volume to the nearest cubic end so for the volume to find the volume of the cone it is one-third times pi times radius squared times the height however as you can see they did not give us the radius here but they did give us a clue so it says that the height of a cone is half the diameter of its base if the cone's height is 4 inches so right away they are telling you that the cone's height is half the diameter so by the cone being 4 inches it's telling us that the entire diameter is 4 inches but when you cut that in half the radius is cut in half that's 4 inches so we have one third times pi which is 3.14 times 4 squared times 4 which is the height when we calculate that we get 66.9 but we're going to round it which gives us 71 cubic inches at this time pause the video complete practice questions 1-4 6-7 and 9-10 when you're ready to continue click play let's move on to lesson seven surface area the surface area is the total area of the outside faces of three-dimensional figures this is different from volume because the volume is what a figure can hold the surface area of a square prism or a cube equals the sum of the areas of the six squares that form the prism so all six faces so the area of one square is what is side squared so when the sides of a square is the same you find the area of one side and multiply it by six because essentially it's all of them together being added so let's go ahead and look at example one find the surface area of a cube with sides of three inches each so we already know that our formula to find the surface area of a cube is six times the side square that's because we know that all the sides of the cube or all the faces are the same measurement so in this instance it's 3 inches so when we replace the value of the face with the variable we are going to get 6 times 3 inches squared equals remember we are going to first do the exponent so we are going to take three to the second power which is nine and we're going to take the inches to the second power as well which is squared inches squared so that would give us a total of 54 inches squared or squared inches let's now learn how to find the surface area of a rectangular prism so the sum of the areas of the six squares that form the prism kind of like the cube but with the difference of rectangular prism that you're going to use several formulas in order to calculate the surface area of a rectangular prism because the surface area is the perimeter times the height plus 2 times the area of base so let's look at our example too we need to find the surface area of the box or the rectangular prism so here is our formula in order to calculate the surface area so why do we have five plus six plus five plus six so remember that the perimeter of a rectangle is two times the length plus two times the width but in this case we are just going to write out all the sides added together so in this instance we have the sides equaling 5 meters 6 meters so we know that the opposite sides are going to be the same measurements so therefore we're also going to add another five and another six because the perimeter is going to be the entire base area so then that we're going to multiply times the height which is 7 meters plus 2 times the area of the base the area of the base is the length times the width which is five times six so do you remember that we were using multiple formulas in order to find the the answer for this question so this is what i mean so make sure that you remember or that you at least write down the different formulas that you need in order to find the volume perimeter radius surface area as we are going to continue to need them for the remainder of this chapter so when we simplify our equation we get 22 times seven plus two times thirty we simplify it even more so we have 154 plus 60 which yields a result of 214 square meters since we learned the surface of a square prism or a cube and how to find the surface area of rectangular prism it is only fair to also learn how to find the surface area of a pyramid in order to find the surface area of a pyramid it's half the perimeter of the base which is represented by p times the slant height which is represented by the variable s plus the area of the base which is capital b the slant height is the length of the line segment from the apex or from the top of a period to the base or to the bottom so let's look at example three what is the surface area of a pyramid with a slant height of five feet a base with a perimeter of 14 feet and a base with an area of 12 feet so here is our formula the surface area equals half the perimeter of the base times the slide height plus b so let's go ahead and replace these variables with the values that are given to us in this case we have the perimeter of the base which is 14 as you read in the example it tells us that the base has a perimeter of 14 and then the slant height it's 5 so we add the 5 for the slant height plus the area of a base which is 12 feet so once we have replaced all of the value the values in there or all the variables with our values then we're going to go ahead and calculate it so remember the order of operations you cannot do one thing before the other otherwise you're going to get the wrong answer so in this case we're going to start with multiplication so you're going to multiply 14 times 5 which gives us 70 so our new answer or a new equation is one half times 70 plus 12. once we simplify it we get 35 plus 12 and ultimately when we simplify it even more we get our answer which is 47 feet squared since we have finished lesson 7.1 now let's focus on its practice so go to page 335 to complete practice 7.1 for practice 7.1 complete questions 2 and 9. for question 2 you are being asked to find the surface area and for question 9 you are being asked to choose the best answer so go ahead and click pause at this time and complete both questions and come back so we can review them together let's start with question two find the surface area so here we have a pyramid which is the first step we need to identify the figure it is a pyramid with a point at the top or an apex and we're also given a slant height of five inches in order to calculate the surface area it is one half times its perimeter times its slant height plus the area of the base so unlike the previous example that we were given it was give we got the perimeter base in this case we didn't so we have to calculate but in order to calculate it we have to determine what shape the base is so based on the figure it's telling us that all sides in the bottom are the same leading us to believe that the base is a square in order to find the perimeter of a square it's four times its size because three because the sides are all the same so when we replace the variable s with the value of 3 we get 12 so in that case it's 4 times 3 equals 12. so our perimeter for the pyramid is 12. now we're going to calculate the area of the base once again we're going to use the formula that is used at a square because our base is a square in order to find the area of the base it's s to the second power so in this case s for side is 3 inches so the area of the base is 3 to the second power now we can go ahead and replace our variables with values or with numbers so it's one-half times perimeter of the base which is 12 times the slant the slant or the the height slant of the pyramid which is 5 plus the area of the base which is 3 squared when we calculate it all together we have an answer of 39 square inches oops sorry ignore that so now for question nine let's read it together if the dimensions of the box are doubled by how many square centimeters does the surface area increase so here we have a box a cube a prism cube right a square prism or a cube so in order to find the surface area of a cube remember so this goes back to what i had told you earlier in the beginning that when all the sides are the same so you just have to multiply one side or multiply 6 times the side square because all the sides are the same so in that case our formula is 6 side squared so when we replace it with the value we have 6 times 3 to the second power which is 54. so that is a surface area when we have sides of 3. so now the question is asking what if the measurements were doubled so what double means is that there instead of three centimeters they are now six centimeters so let's go ahead and replace the three with six now so we have six times six to the second power which gives us a total of 216. well they want to know the difference so by how many square centimeters of the surface area increase so in order to find that number we're going to subtract 26 minus 54 and that gives us an answer of so if you chose c you are correct at this time pause the video complete practice questions 1 and 3 through 8 and click play when you are ready to continue as it is customary in the previous lessons when we found the volume the perimeter for one shape we found it for all shapes out there so now we're going to find the surface area of a cylinder a cone and a sphere in order to find the surface area of a cylinder you're going to multiply the circumference so the circumference is around the object so that's the measurement around the object which is 2 pi r so you may be wondering where do we get that formula from well this formula goes back to the lesson where we learned about circles because remember a cylinder is a circular figure then you're going to multiply the circumference by the height which is 8 and then you're going to add that product to the area of the two ends of the cylinder so the top and the bottom and the area of a circle is pi r squared so that's where we get the two in our formula which is 2 times pi r squared because we have a top and a bottom side now the surface area of a cone you're going to combine the lateral side surface with the area of its base so that means the lateral side surface is pi r times the slant height and the area of the base is pi r squared so this is the same formula that we find the area of a circle remember because the base of the cone is a circle therefore we're using that formula so then you have surface area of a cone equals pi times radius times slant height plus pi times its radius squared now let's find the area the surface area of a sphere so it's just four times the area of a circle so we have four times the area of a circle which is pi r squared so just a little hint when you're looking to find the surface area of any cylindrical shape such as a cylinder a cone or a sphere you're using the surface you're in order to find the surface area you're using the formulas from the circles from dimensions of a circle just keep that in mind whenever you come across any objects like these let us practice finding the surface area of a cylinder a cone and a sphere for example one we have find the surface area of a cylinder with a radius of 4 inches and a height of 3 inches so we're going to start off by writing out our formula remember the formula to find the surface area of a cylinder it's 2 times pi times radius times height plus 2 times pi times radius to the second power so we're going to substitute all of our values for the variables so here you go so you have 2 times 3.14 times 4 times 3 plus 2 times 3.14 times 4 squared so when we simplify that we have 75.36 plus 100 and 100.48 giving us a surface area of a cylinder of 175 75.8 84 inches squared pretty easy right so now let's focus on finding this the surface area of a cone so examples two is asking us to find the surface area of a cone with a radius of two feet and a slant height of 5 feet so first off let's start with our formula so the formula for the surface area is pi times radius times the slant height plus pi times radius squared so now let's go ahead and replace all of our variables with values given so we have 3.14 times two times five plus 3.14 times 2 to the second power so let's go ahead and simplify it we have 31.4 plus 12.56 giving us an answer of 43.95 feet squared now let's look at example 3. for example 3 it is asking us to find the surface area of a sphere with a radius of 10 feet so let's go back to our formula which is 4 times pi times radius squared so we are going to replace the variables with the values given so in this case it's 4 times 3.14 times 10 squared which gives us a total of 1256 feet squared let's move on to practice 7.2 found on page 455 so for practice 7.2 we are going to complete questions four and nine for question four you're going to find the surface area and for question nine you're going to choose the best answer so take a few minutes to complete questions four and nine and when you are ready to continue press play so let's go ahead and start with question four for question four we're going to find the surface area so let's bring our handy dandy formula for a surface area which in this case is two times pi times the radius times the height plus 2 times pi times radius squared so let's go ahead and replace all of our variables with the values given so we have 2 times 3.14 times 9 times 12 plus 2 times 3.14 times 9 squared when we simplify that it is 678.24 plus 508.68 so when we add those together we get 1186.92 inches squared so that's the surface area however the question is asking us to bring the to give us the answer in terms of pi so what that means is that we are going to divide our answer so we're going to divide 1186.92 divided by pi which is 3.14 which gives us an answer of 378 pi which is essentially the same thing we're multiplying 378 times 3.4 and it'll give us the other answer so let's move on to question next in a cylinder the diameter of the circular base is equal to the height of the cylinder what is the surface area of the cylinder to the nearest square inch so here we have a cylinder with a diameter of six inches so the surface area formula is two times pi times radius times height plus 2 times pi times radius squared well in this case we were given the diameter we were not given the radius however the radius is half of the diameter therefore the radius in this case is 3 because 3 is half of 6. so we are going to replace all of our variables with the values so we have 2 times 3.14 times the radius which is 3 times the height which is 6 because remember on the question it says that the the height is equal to the circular base and we then we're going to add 2 plus times 3 0.14 times 3 squared which gives us a total of 169.50 however we want to round it off to the nearest square inch so then our answer will be 170 at this time pause the video complete practice questions one through three and six through eight click play when you are ready to continue hi i hope you're still there so i am checking in because lesson six and lesson seven were a little too long for my taste but don't worry we're almost at the end of the chapter we only have two more lessons to go and i am with you still here i'm not going anywhere and we are going to finish strong so bear with me a couple more minutes i should say right let's move on to lesson eight combined figures so combined figures are two or more regular figures or geometric geometric shapes that are combined to make a new shape in order to calculate the perimeter of a combined figure you need to add the length of the sides and sometimes you may need to solve for a missing length so let's look at our first example for example one we are being asked to find the missing measurement so as you can look at the figure it is a combined figure as it is a mixture of two separate shapes now we have two variables in there which means that those are unknown values but don't worry we're going to solve for those unknown values so we know that this looks like a rectangular shape although it is missing a portion of it but it's still a rectangular shape so just by being a rectangular shape we know that opposite sides are equal or opposite walls so therefore we know that the variable of x equals eight plus four eight and four are opposite of x we need to add those together in order to get the value of our x over of our variable x for y it's going to be the same thing they are opposite walls from each other but in this instance we have to subtract because in this instance we're taking away a piece so it's going to be 18 which is the top wall and then we're going to subtract the 10 from it which is the bottom wall which is going to give us that one small piece that represents y so x is 12. and y equals eight so now we're going to add the sides the lengths of the sides in order to calculate the perimeter in this case we're going to add 12 8 18 4 and 10 and it's going to give us a total of 60 feet so what happens when we have to calculate the area or a volume well it gets a little tricky but let's focus on calculating the area first so in order to calculate the area of a combined figure we have to use several different formulas and the reason why is because the first step in calculating the area is going to be to divide the figure into two shapes so for example in this in this figure here for example two we divided the shape into two different shapes so we get a rectangle and a trapezoid so the rectangle is a top shape and the rectangle is i'm sorry rectangle is the top shape and then the trapezoid is the bottom shape so then we're going to calculate the area of each shape you remember those formulas if not it's okay you can always go back and just get a little refresher the area of a rectangle is length times the width so let's just look then at the rectangle the length is 2 centimeters and the width is 5 centimeter which gives us a total of 10. therefore the the area of the rectangle it's 10 centimeters now let's calculate the area for the trapezoid for the trapezoid the area equals one half times its height times base one plus base two so let's enter our values for the different variables so we have one height and one half and the height is three do you know why well when we separated the rectangle and the trapezoid if you notice in the left hand side where it says five centimeters for the length of the combined figure but we are not calculating for five centimeters because five centimeters belongs to the entire figure we only want the length of the trapezoid itself so it's three centimeters because two of those centimeters belong to the rectangle so that's why we're using the number three now we're going to times that or multiply it by base 1 which is 5 centimeters and base 2 which is 7 centimeters and it gives us a total of 18 centimeter or 18 squared centimeters so once we have the area for both separate shapes we're going to combine them when we combine them we have 10 plus 18 which gives us a total of 28 squared centimeters now let's look at finding the volume of a combined figure in order to calculate the volume it's going to be the same exact same thing as calculating the area we are going to apply the correct formula for each part so let's look at example three we have a cone and a cylinder together so we need to find the volume of the container well first things first we have to divide the figures into two shapes in this in this example we have two three sided or three dimensional shapes one is the cone and the other one is the cylinder so now we have to calculate the volume for each shape once again if you need a refresher with the formulas it's okay you can always go back to previous lessons so let's start with the volume of the cylinder the cylinder is the the the formula to calculate the cylinder is pi times radius squared times the height so we're going to replace the values with the variables in this case we already know that pi is 3.14 times the radius so the diagram is showing us that the radius is one meter and so it's going to be one meter squared and then the height so the we're only looking at the height of the cylinder not the height of the entire object so the cylinder is only two meters long which it's going to be our height here in our formula so when we calculate all of that we get 6.8 meters cubic i'm sorry cubic meters i could not remember so it's 6.8 cubic meters now let's look at the cone so the wrong the formula to find the volume of a cone is 1 3 times pi times r squared times the height so remember we're only focusing on the cone itself so we have 1 3 times 3.14 which is pi so you may be wondering why are we using 1 as a radius if one is located in the cylinder side well remember that a cylinder at the top base and the bottom base have the same volume so that is why we're using the same radius in this instance think of a tomato can soup the top and the bottom circular shape they are the same size and then you're going to multiply by the height of the cone which is 1.5 when we calculate all of that together it gives us a total of 1.57 cubic meters the next step is to combine them so let's go ahead and do that so we have 6.8 cubic meters plus 1.57 which gives us a total volume of 7.85 cubic meters not bad right now let's move on to practice eight found on page 337 for practice eight we are going to work on questions three and four for question 3 you're going to find the perimeter and the area and for question 4 you're going to find the volume so take a few minutes to complete these questions if you need some assistance go back to the previous slides or if you need some assistance with the formulas to find volume perimeter and area i would suggest to go to the previous lessons as well and don't forget to use your book so take a few minutes and then when you're ready to come back to review them together go ahead and press play so let's go ahead and get started we're going to work on question three for question three it wants us to find the area and perimeter of the given shape so we have a shape here it looks like a boat but we don't know if it's a boat it's just a combined figure remember it's two or more geo geometrical shapes combined together in order to make one shape so in order to find the area of this figure we have to divide the figure first into two shapes so in this case we have two trapezoids so if you can see my dotted line right where the number six is at that's where we are going to separate or split both shapes and we come out with two different trapezoids so now we have to calculate the area of each shape in order to calculate it we have to have the formula for the area remember so let's work on trapezoid one which is going to be the trapezoid on the left hand side if the area is half times its height times base 1 plus base 2. so let's go ahead and replace the variables with the values of the trapezoid on the left hand side only so the area equals one half times its height so the height for the first trapezoid is 12. as you can see the top says when we broke it up the middle is six and six we have two separate sixes but when we combine them that is twelve so that is the total height of the first trapezoid then we're going to multiply it by base 1 which is 12 and base added to base 2 which is 3. when we calculate that formula we have 90 squared centimeters perfect so now let's move on to finding the area of the second trapezoid which is going to be on the right hand side so the area once again equals one half times the height times base one plus base two so let's replace our values with the variables we have one half times six well now we're only using half of the height that we use on trapezoid one because if you look at trapezoid two it is a smaller one and it's only using half of the height that the other trapezoid is using therefore we're only going to use 6 and base 1 equals 17 and base 2 is 9. so when we calculate that formula we have a total of 78 squared centimeters now let's combine them 90 plus 78 equals 168 squared centimeters perfect i hope you got this correct if not you can go through it again one more time use your formulas in the book so now let's find the perimeter for this question so the perimeter of the remember in order to find the perimeter we have to sum all sides so in this case each side has a value or has a measurement so what we're going to do is we're going to add all of those sides together except the measurement of number six the one that's right smack dab in the middle of the trapezoid because we don't need that at this time remember we're we are finding the perimeter of the shape so we have to use only the outside sides not the inside ones when we get that when we grab all the numbers we have 12 plus 15 plus 12 plus 10 plus 17 plus 6 which gives us a total of 72 units so therefore our perimeter for question 3 is 72 units let's focus now on question 4. for question four we're going to find the volume for the figure on the screen so let's go ahead and start by separating the figure into two separate shapes so in this case when we separate them we have a square which is your right shape and then the rectangle which is the shape on the left so let's calculate the volume for each figure now remember for the square we have a formula that the volume equals the measurement of the edge to the third power so in this case as you can notice that the the square has a side length of five because it's a square all of the sides are the same therefore we have five to the third power which gives us a total of 125. now let's look at the other shape the other shape is a rectangle so we're going to take the formula to find the volume of a rectangle which is the base which is the length times its width times the height so let's go ahead and replace the value or the replace the variables with the values given the base we have length times width which is five times four times the height which is three so that gives us a total of 60. we are then going to combine those numbers which is 125 plus 60 which gives us an answer of 100 185 cubic units at this time pause the video complete practice questions one through two and five through eight click play when you are ready to continue let's move on to lesson nine geometry calculator skills in chapter 5 we learn how to use a calculator when calculating algebraic expressions in this instance or for this lesson we are going to use the calculator to calculate geometric shapes or geometric calculations and formulas so the use of a calculator saves you time when solving problems with formula once again we are going to use a scientific calculator so for the ged mathematical reasoning test you are allowed to use a ti 30 x pass multi-view calculator when entering your formulas into the calculator to solve please remember to follow the order of operations which you learned in chapter 5. doing the order of operation is going to give you the correct answer so just a little overview of the order operations is you have to do first any parentheses or any other grouping symbols that enclose operations after that you do your exponents and your roots then you move on to multiplication and then division from left to right and then finally you end with an addition and or subtraction moving from left to right once you enter all your formulas into the calculator make sure that you revise your work by re-entering the sequence so try it more than one time just to make sure that you are correct let's look at a couple examples on how to use a calculator in order to solve for gm geometric formulas let's look at example one a pyramid has a height of 81 feet the base is in the shape of a square with each side measuring 40 feet what is the volume in cubic feet of the pyramid so let's go ahead and start with finding the formula that we are going to need in order to measure for the volume of a pyramid which is 1 3 times the base times the height but remember the b in this case is area of the base because it is a square so it's a squared base and the area to find a square is side to the second power then h equals height so when we replace the values using the i'm sorry when we replace the variables using the values that are given in the in the example it says the pyramid has a height of 81 feet and then the base is in the shape of a square so each side measuring 40 feet so let's go ahead and use those numbers and replace them with the variables so it's going to look like this it's going to look 1 3 times the area of the base which is 40 squared times the height which is 81. so now when you're entering it to your calculator you're going to break the equation into separate parts always remember to go by your order of operations so let's start first with the exponent so on your calculator you're going to type in 40 to the x squared remember that there is a button with the x squared which is one two three fourth button up the on the left hand side if you have a different calculator a ti 36 i believe it's still in the same place but don't worry most scientific calculators have the same values or have the same buttons they're probably just in a different order or different position so become familiar with your calculator in order to use it a little bit better so then we're going to multiply by 81 so you have the multiplication symbol and then you're going to add 81 and then you're going to divide by 3 and you're going to press enter so that's going to give you a question or i'm sorry an answer of 43 200 feet okay oh cubic feet sorry we are measuring volume in this case right so it's 43 200 cubic feet that is the volume of the pyramid if you so this is entering the formula onto the calculator in separate parts however if you want to enter it in one series of keystrokes you can definitely do that too so this is the way that you will enter it on your calculator when you are measuring or calculating for the volume of a pyramid using the one-third times the area of the base times its height formula let's look at example two a right triangle has legs 10 inches and 24 inches in length what is the length of the hypotenuse do you remember the hypotenuse okay so anytime you you mention or you read where it says hypotenuse think the pythagorean relationship do you remember that now is it coming back i hope it is so the pythagorean relationship is a squared plus b squared equals c squared so pretty much it means that leg a plus leg b equals the diagonal line of the of the hypotenuse in the triangle so we have leg a which is 10 inches leg b 24 inches and then we're still missing so we're missing the hypotenuse we don't know the measurement the hypotenuse you can do it by hand i have faith in you and i'm pretty sure you'll get it correct but this lesson is teaching you how to use a calculator so let's go ahead and show you how to use the calculator so this is the way that you're going to enter your values onto the calculator in order to get the hypotenuse so leg a we have 10 and you're going to take it to the second power then you're going to add it to leg 2 that you're also going to take to the second power and then you're going to push enter so when you press enter your display is going to state 676 however that is not the answer because remember we have c squared so what you do is now you want the square root of 676. so what you're going to do is you're going to press the second button or the second key button that's on your calculator if you look at this calculator the second key is green so when you push that second key it means that you want a function above the one that's on the on the keystroke so in this case when you push the second it means that you want a formula or a keystroke that's right above and you can definitely identify them here because they're all green and then you're going to do the x squared because you're looking for the square root of 676 and you enter 676 and you push enter by doing so you're going to get the length of the hypotenuse which is 26 inches long let's move on to practice nine found on page four hundred and nine for practice nine we are going to be completing questions six and eight use a calculator to evaluate question six and choose the best answer for question number eight take a few minutes to answer both questions and when you are ready to review them together press play let's look at question six for question six we are being asked to use a calculator to evaluate and to find the volume of a cone so question six reads find the volume of a cone with a radius of 12 centimeters and a height of 27 meters round to the nearest cubic centimeters so first off we need to know which formula we are using and in this case we know it's going to be a volume formula and it's going to be the formula used to find the volume of a cone which is volume equals 1 3 times pi times radius squared times height so now let's replace the variables with the values given in the question so we have 1 3 times 3.14 which is pi 12 squared which is radius and the height of 20. so the next thing that we're going to do in order to use our calculator to calculate the volume we are going to break the formula into parts remember your order of operations you have to follow which order it tells you first because otherwise you will get the wrong answer so let's start with the exponent so we're going to evaluate the exponent in this case it's going to be 12 x squared and then you're going to multiply by pi which is 3.14 and then you're going to multiply by 20 and finally divide by 3 and press enter on your calculator which will give us an answer of 3014 cubic centimeters now let's look at question eight we are to choose the best answer so let's read it together to the nearest cubic meter what is the volume of a cylinder with the radius of 1.5 meters and a height of 5 meters so we need our formula first and foremost and here is the formula to calculate the volume of a cylinder which is volume equals 1 3 times pi times radius squared times the height so now we are going to replace all of our variables with the numbers given which is one-third times 3.14 times 1.5 squared times 5. only in this case for this particular question what we're going to do is that we are not going to break the formula into into parts instead what we are going to do is we are just going to go ahead and put it into one simple keystroke okay which is what we're going to do at this time the way that you do that going to be the following way that's how you're going to enter it or input it onto your calculator we have the button for pi which is on your calculator times 1.5 and then x squared times 5 and then you're going to push enter which will give you an answer of 35 cubic meters so in this case the answer for number 8 is b at this time pause the video complete practice questions 1-5 question 7 and questions 9-10 click play when you are ready to continue congratulations we have finally completed unit 2 mathematical reasoning great job everyone please keep going the next video will cover unit 3 social studies chapter 1 social studies practices see you soon |
11909 | https://arxiv.org/pdf/2508.13237 | Structural Foundations for Leading Digit Laws: Beyond Probabilistic Mixtures
Vladimir Berman Aitiologia LLC
vb7654321@gmail.com
July 21, 2025
Abstract
This article presents a modern deterministic framework for the study of leading significant digit distributions in numerical data. Rather than relying on traditional probabilistic or mixture-based explanations, we demonstrate that the observed frequencies of leading digits are determined by the underlying arithmetic, algorithmic, and structural properties of the data-generating process. Our approach centers on a shift-invariant functional equation, whose general solution is given by explicit affine-plus-periodic formulas. This structural formulation explains the diversity of digit distributions encountered in both empirical and mathematical datasets, including cases with pronounced deviations from logarithmic or scale-invariant profiles. We systematically analyze digit distributions in finite and infinite datasets, address deterministic sequences such as prime numbers and recurrence relations, and highlight the emergence of block-structured and fractal features. The article provides critical examination of probabilistic models, explicit examples and counterexamples, and discusses limitations and open problems for further research. Overall, this work establishes a unified mathematical foundation for digital phenomena and offers a versatile toolset for modeling and analyzing digit patterns in applied and theoretical contexts.
Mathematics Subject Classification (2020): 11K06, 11A63, 28A80, 62E20
Keywords: leading digit law, Benford’s law, deterministic models, cumulative profile, functional equations, fractals
Contents
I. Introduction 3II. Historical and Literature Review 6III. Preliminaries and Key Definitions 8IV. Deterministic Structure and Core Equations for Digit Block Frequencies 12 V. Euler–Maclaurin Framework 17
1
arXiv:2508.13237v1 [stat.ML] 18 Aug 2025
VI. Examples of Continuous Distributions 21 VII. Discrete Distributions and Empirical Digital Profiles 40 VIII Critical Discussion 49 IX. Open Problems and Future Directions 52 Epilogue: Synthesis 54 References 56
List of Figures
1 Multiple-Decade Approximation to the Global Digital Profile. Curves show
GN (s) computed on [1 , 10) (1 decade), [1 , 100) (2 decades), and [1 , 10 4) (4 decades), compared with an approximation to the global profile obtained by aggregating many decades and renormalizing. The windowed profiles converge rapidly toward the global digital signature as N increases. . . . 25 2 Plots of G(S) for different values of μ for the normal distribution ( A = 1 ,
B = 10 , σ = 1 ). . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27 3 Leading digit profile for samples from the normal distribution N (μ, σ 2)
with different μ, σ = 1 .5. . . . . . . . . . . . . . . . . . . . . . . . . . . . 28 4 Histogram of leading digits for 100,000 numbers sampled uniformly from the interval [e, π ). Only digits 2 and 3 appear as the leading digit in this range. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29 5 Leading digit profile function G(s) for the power law with m = 1 and several upper bounds b. Each curve corresponds to a different value of b,with the corresponding fractional part {log 10 b} indicated in the legend. The black dots and "Jump" labels mark the points s = {log 10 b}, where G(s)
exhibits a jump discontinuity, corresponding to the cut-off of the support at x = b. For b = 10 k ({log 10 b} = 0 ), G(s) coincides with the uniform distribution G(s) = s (shown as a dashed line). As b varies between powers of ten, the position and magnitude of the jump shift, illustrating how the profile depends on the upper bound of the distribution. . . . . . . . . . 34 6 The function ρ(k, b, m ) for m = 1 and k = 1 , 3, 8, as b runs from 1 to 10 .This case corresponds to the uniform distribution on [1 , b ]. For b = 10 ,the digit frequencies approach Benford’s Law. Between powers of ten, the profiles exhibit a characteristic “wavy” structure. . . . . . . . . . . . . . 35 7 The function ρ(k, b, m ) for m = 2 and k = 1 , 3, 8. Here, the digit frequencies become more concentrated due to the stronger bias toward larger values in the distribution. The “wavy” pattern persists and demonstrates the effect of the upper bound b. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36 8 The function ρ(k, b, m ) for m = 0 .5 and k = 1 , 3, 8. With m < 1, the distribution is skewed toward lower values, which is reflected in the digit frequencies. The periodic, oscillatory nature of the profiles remains visible. 36 9 Empirical windowed profile ̂ G(s) for 2023 world Total Population across countries (orange) with the reference line G(s) = s (dashed). . . . . . . . 46 210 Histogram of Total Population (countries, 2023) on a log-x axis with the fitted Weibull pdf overlay ( ˆa = 0 .467 , ˆb = 1 .37467 e + 07 ). . . . . . . . . . 47 11 QQ-plot (log–log axes) comparing empirical quantiles of Total Population against the fitted Weibull quantiles. . . . . . . . . . . . . . . . . . . . . . 48 12 Empirical windowed profiles G(s) for Weibull (a, b = 1) with shapes a ∈{0.5, 1, 2}. Each curve is based on N = 600 ,000 samples. The dashed line shows the reference G(s) = s. . . . . . . . . . . . . . . . . . . . . . . . . 48
I. Introduction
The so-called "laws of digits," especially Benford’s Law, have often been presented as miraculous and universal. Probabilistic literature argues that these laws emerge in sufficiently "large," "random," or "naturally occurring" datasets. A tradition that began with Newcomb and Benford, and culminated in the sophisticated mixture theories of Hill, seems to rest on the premise that the law only holds — or even makes sense — in the thermodynamic limit of infinite ensembles. Yet this position is, upon reflection, both logically tenuous and physically suspect. First, it is unclear why a truly universal law should require a dataset to be "large enough," or why any particular threshold of data size or randomness should be privileged. Benford himself compiled and analyzed tables of modest size, and real-world datasets often contain far fewer entries than probabilistic arguments would suggest necessary. If the law of digits is genuine, its structure ought to manifest not only in immense or chaotic collections, but also in small, even singular sets, so long as their numerical content is sufficiently well-defined. A more subtle, but deeper, issue lies in the architecture of probabilistic justifications. Most such arguments — whether invoking scale-invariance, random mixtures, or "natural-ness" — are fundamentally circular. They select, often post hoc, the probability spaces or mixture distributions specifically to yield the desired law. The celebrated result of Hill, for example, asserts that if one samples from a random mixture of arbitrary distributions (with the right, carefully tuned weighting), then the outcome will almost surely satisfy Benford’s Law. But this mixture is not found in nature; it is a mathematical construct, chosen precisely because it generates the law. In practical settings, random processes are governed by specific mechanisms, not by omnipresent mixtures over the space of all possible distributions. The requirement that nature itself somehow mixes all conceivable distributions in exactly the right way is not only artificial — it is implausible. Thus, while the probabilistic paradigm has provided many beautiful results and fueled the mystique of digit laws, it has also diverted attention from the deeper, deterministic reality underlying digit phenomena. The arithmetic of digit blocks, as we demonstrate in this work, is governed by a simple, explicit recurrence that holds for any set or sequence — regardless of size, randomness, or origin. This recurrence leads directly, by analytic means, to the family of all possible digit block distributions. Benford’s Law appears as a special (and important) case, but is neither unique nor universal; non-Benford distributions arise naturally and are equally valid within the same analytic framework. Our approach, therefore, is not only conceptually distinct, but also broader and more rigorous. We show that the so-called "law of digits" is an arithmetic property, not a prob-abilistic one: its explanation requires no appeal to stochastic largeness, scale-invariance, or fantastical mixtures of distributions. Instead, the law follows as a deterministic conse-3quence of the structure of number representations, valid for finite and infinite datasets alike. This realization recasts the entire landscape of digit laws. The question is not why certain patterns appear in the data, but why the recurrence structure — simple, universal, and analytic — so robustly governs the frequencies of digit blocks across all contexts. In the chapters that follow, we develop this structure in detail, deriving both Benford and non-Benford laws as explicit solutions to a single, transparent functional equation.
1.2 Motivations for a Deterministic Theory
The astonishing ubiquity of Benford’s Law has, for decades, motivated a tremendous number of probabilistic and statistical explanations—most notably those of Hill, Pinkham, and their followers. These approaches, elegant as they are, fundamentally depend on the construction of vast abstract spaces, typically assuming mixtures of distributions, scale invariance, or random sampling from infinite sets. However, this line of reasoning inevitably raises deep conceptual questions:
• Why should we expect nature—or data-generating processes—to mix distributions in precisely the right (and often artificial) proportions to produce the logarithmic law?
• How many numbers are “enough” for the law to appear? Why should the structure fail for small or finite datasets, as is often the case in practice and even in Benford’s original empirical studies?
• Can a law that is only justified by the limiting behavior of enormous or abstract sets truly reflect a universal mathematical structure?
• Is it reasonable to believe that the emergence of digit laws in physical or social data requires fine-tuned randomness or statistical equilibrium at every scale? The classic probabilistic framework tends to sidestep these issues, presuming that only in the “large-number limit” do the underlying mechanisms manifest. Yet, this perspective is unsatisfying from both a mathematical and a practical standpoint. In real applications—ranging from accounting and forensic science to experimental physics—we encounter datasets that are finite, often small, and far from idealized randomness. Intuitively, if a digit law is truly fundamental, it should not be fragile: its validity should not depend on mixing “just the right” distributions, nor should it break down for small N . Instead, the law should emerge directly from the structure of the number system itself, with the same logic applying equally well to a single number, a handful of values, or an enormous dataset. The lack of a clear boundary between “small” and “large” N in existing theories is a conceptual flaw that deterministic analysis can address. Thus, there is a compelling need for a rigorous, constructive, and fully deterministic framework—one that
• explains digit distributions from first principles, not as a statistical artifact,
• applies seamlessly to both finite and infinite sets,
• and exposes the precise mechanisms and constraints that give rise to both Benford and non-Benford laws. 4In this work, we take up this challenge. We show that the structure of digit laws is dictated not by abstract probabilistic mixtures, but by explicit recurrence relations and functional equations governing the frequencies of digit blocks. Our approach reveals the mathematical “skeleton” underlying all possible digit distributions, clarifies the true scope and limitations of Benford’s Law, and eliminates the need for artificial assumptions or “randomness at infinity.” Ultimately, this deterministic perspective provides a unified and transparent foundation for understanding digit laws—one that works in every setting, from the smallest sample to the largest dataset, and that allows us to precisely characterize the entire spectrum of admissible digit distributions.
Definition (Distribution of Leading Significant Digits, DLSD). Given a finite or infinite sequence (or set) of positive real numbers {xi}, the Distribution of Leading Significant Digits (DLSD) is the frequency function ρ(k), k = 1 , . . . , 9, defined as the proportion of elements whose leading digit (in base 10) is k:
ρ(k) = 1
N
N∑
i=1
I(d(xi) = k),
where N is the total number of elements (finite or infinite), d(xi) denotes the first significant digit of xi, and I is the indicator function. For infinite sequences or theoretical distributions, one considers the limiting frequency as N → ∞ . In empirical or finite data sets, ρ(k) is computed for the observed sample.
Distribution of leading m-digit blocks. Let {xi}Ni=1 be positive real numbers and let m ∈ N. For x > 0, the leading m-digit block is the unique integer
Bm(x) ∈ { 10 m−1, . . . , 10 m − 1} such that k 10 n ≤ x < (k + 1) 10 n for some n ∈ Z,
i.e., the first m significant digits of x equal k. The empirical frequency of block k is
ρm(k) = 1
N
N∑
i=1
1(
Bm(xi) = k)
, k = 10 m−1, . . . , 10 m − 1,
and clearly ∑10 m−1
k=10 m−1
ρm(k) = 1 . The case m = 1 gives the usual first-digit distribution
ρ1(k), k = 1 , . . . , 9.
Equivalent logarithmic check (used later). The condition Bm(x) = k is equivalent to
log 10 k ≤ log 10 x − n < log 10 (k + 1) for some n ∈ Z.
(Thus one can test membership using the base-10 logarithm; a fractional-part form will be introduced later.)
Benford benchmark. Under Benford’s law the expected frequencies for any order
m ≥ 1 are
ρBenford
m
(k) = log 10
(
k + 1
k
)
, k = 10 m−1, . . . , 10 m − 1,
which reduces to the standard first-digit formula when m = 1 .5II. Historical and Literature Review
2.1 Early Observations: Newcomb and Benford
The origins of digit laws can be traced back to Simon Newcomb, who, in 1881, observed that the early pages of logarithmic tables were far more worn than later ones [ 1]. Although Newcomb offered only a brief empirical remark, this observation captured a subtle statistical regularity in naturally occurring numbers. Frank Benford, in 1938, significantly expanded upon Newcomb’s idea by systematically analyzing more than 20 datasets—from river lengths to atomic weights—and famously formulated what is now called Benford’s Law [ 2]. His empirical results, while persuasive, were still based on finite and sometimes quite modest sample sizes. The law was quickly embraced in some applied fields (such as accounting and engineering) and has inspired further compilations and reviews, and has inspired a sequence of major reviews, beginning with Raimi’s classic 1976 overview [ 3] and continuing in the later works of Berger and Nigrini [ 4, 5]. For a thorough account of the mathematical theory and current perspectives on Benford’s Law, see the volume edited by Steven J. Miller [ 6] and the foundational papers of Theodore Hill [ 7 , 8]. These works provide essential context and in-depth discussion of both classical and modern results.
2.2 The Search for Explanation and Early Critique
The apparent universality of Benford’s Law soon drew mathematical attention—and skep-ticism.Stigler [ 9] provided a formal analytic approach for reconstructing the distribution of first significant digits in deterministic sequences. He demonstrated that the applicability of Benford’s Law is limited and depends on the specific structure of the underlying sequence, highlighting that many simple mathematical sequences exhibit digit distributions differing substantially from the logarithmic law. Hamming [ 10 ] and others warned that many supposed “explanations” were little more than tautologies, reflecting properties of the logarithm or number representation itself rather than genuine statistical laws. The concept of scale-invariance became central in these debates. Pinkham [ 11 ] ele-gantly demonstrated that Benford’s Law is the unique digit law invariant under scale transformations, but this invariance, while mathematically appealing, does not directly explain why such a law should arise in nature. Feller [ 12 ] went further, connecting the law to the uniform distribution of mantissas (fractional parts of logarithms), but he also emphasized the limits of this approach for empirical data.
2.2.1 Stigler’s Contribution and the Mechanistic View of Digit Laws
George Stigler’s 1945 note marks a fundamental turning point in the understanding of leading digit phenomena. Rather than searching for a universal law, Stigler was the first to systematically investigate how the distribution of leading digits depends on the specific mechanisms generating the data. Stigler’s key insight was that the frequencies of first digits are not governed by a single empirical rule, but by the arithmetic and probabilistic structure of the process itself. For example, if two numbers are selected independently and uniformly from a given interval, and their product is formed, the resulting distribution of leading digits is not described by 6Benford’s Law, but by a distinct formula—a convolution of two uniform distributions on the logarithmic scale. Stigler showed that for each different generative process, a different digit law emerges. Crucially, Stigler demonstrated that the celebrated logarithmic law for leading digits (now known as Benford’s Law) is not a universal phenomenon, but rather the outcome of specific multiplicative or exponential mechanisms, and only in certain limiting cases. For many other data-generating processes, especially involving sums or products of finite numbers of random variables, the digit distribution deviates from the logarithmic law. Stigler’s mechanistic perspective thus established that the diversity of digit distributions should be understood as consequences of the particular probabilistic and arithmetic procedures underlying the data. This insight paved the way for the modern view, in which Benford’s Law is recognized as just one among many possible digit laws, each corresponding to a particular structural context.
A necessary critique. It should be noted, however, that Stigler’s analysis—while pioneering in its mechanistic perspective—also relied on the assumption that, as the process of generating numbers becomes increasingly complex or as the numbers themselves become larger, the distribution of leading digits will converge to the logarithmic law. This expectation, shared by Benford and many later authors, is not universally justified. In reality, significant deviations from Benford’s Law persist even in large or complex data sets, especially when the underlying arithmetic or probabilistic structure departs from the idealized random multiplicative scenario. Thus, both Stigler and Benford, despite their important insights, overestimated the universality of the logarithmic law in the limit. Modern analysis and explicit counterex-amples show that the asymptotic regime does not guarantee convergence to Benford’s Law; rather, the specific structure of data generation continues to play a critical role, regardless of scale.
2.3 The Era of Probabilistic Mixtures: Hill and Beyond
The most widely cited mathematical justification came with Hill’s series of papers in the 1990s [ 7]. Hill showed that if one randomly samples from a sufficiently “broad” mixture of distributions, then the resulting data will, with probability one, satisfy Benford’s Law. While this theorem is technically beautiful, it rests on highly artificial and, arguably, nonphysical assumptions—namely, that the universe generates data by perfectly mixing all possible distributions in just the right way. This “probabilistic mixture” paradigm has been both influential and controversial. Advocates point to its explanatory power and generality, while critics note that the required mixing does not correspond to any known mechanism in nature or society. Modern reviews [ 13 , 4] offer both enthusiastic summaries and pointed criticisms of the mixture approach, highlighting both its strengths and conceptual limitations.
2.4 Deterministic and Algorithmic Perspectives
In parallel, a line of research has sought purely deterministic and algorithmic origins for digit laws. Knuth [ 14 ] explored digit patterns in mathematical sequences and computa-tional algorithms, showing that Benford’s Law often—but not always—appears, and that its presence depends delicately on the arithmetic structure of the process. Diaconis [ 15 ]7developed the idea of equidistribution modulo one as a tool for understanding when digit laws hold in number-theoretic sequences. Yet, these studies, while rich in examples, ultimately reinforce the sense that Benford’s Law is far from universal: even “natural” processes can generate non-Benford distributions, and the law’s appearance (or absence) can be highly sensitive to underlying mechanisms.
2.5 Contemporary Directions and the Case for a Deterministic Framework
Recent research has further broadened the scope, applying ergodic theory, group invariance, and algebraic techniques to the problem [ 16 , 4]. These works demonstrate that many different digit laws can arise from explicit construction, deterministic processes, or ergodic transformations—not just random mixtures. The literature is now replete with examples both supporting and contradicting the universalist vision, as summarized in modern surveys and expository articles. Nevertheless, the prevailing narrative in much of the literature still leans heavily on probabilistic, often abstract, “naturalness” arguments—frequently sidestepping the deep question of whether such mechanisms exist or are even plausible in real-world settings.
2.6 Synthesis and Perspective
This brief overview highlights the rich and complex history of digit laws—a story marked by empirical discoveries, elegant mathematics, and ongoing controversy over interpretation. While probabilistic mixtures, scale-invariance, and algorithmic randomness offer partial explanations, they are, at best, incomplete. The enduring puzzle remains: does Benford’s Law (and its relatives) reflect a fundamental statistical regularity, or is it a shadow cast by deeper arithmetic recurrences and analytic structures? For the interested reader, the following works provide both entry points and deeper analysis:
• Empirical history: Newcomb , Benford , Nigrini
• Mathematical foundations: Pinkham , Feller , Knuth
• Probabilistic theories: Hill , Berger , Devlin
• Algorithmic and analytic: Diaconis , Pietronero et al. In summary, while the literature is rich and diverse, the limitations of the classical approaches—especially their reliance on abstract probabilistic constructs—suggest the need for a new, fully deterministic analytic framework. This work takes up that challenge: to show, constructively, how all admissible digit laws (including Benford’s) follow directly from explicit recurrence relations and functional equations, without recourse to randomness, scale-invariance, or artificial mixtures.
III. Preliminaries and Key Definitions
In this section, we formalize the key objects and notation used throughout the rest of the paper. Our focus is on the deterministic description of digit block frequencies for arbitrary sets or sequences of numbers, or even for a single real number’s expansion. 83.1 Digit Blocks, Decades, and Leading Frequency Functional
• Digit Block ( k): Any finite sequence of decimal digits, considered as a positive integer. For example, k = 1 , k = 23 , or k = 314 .
• Decade ( d): The integer d = ⌊log 10 k⌋, i.e., d + 1 is the number of digits in k. The “d-th decade” refers to all numbers with d + 1 digits.
• Total Numbers in Decade ( Nd): Nd = 10 d+1 − 10 d.
Definition (Digit Block Frequency Functional): Given a set S of positive integers (which may be a finite list, a sequence, or all numbers up to some limit), the frequency
ρ(k, d ) is defined as
ρ(k, d ) := #{n ∈ S | n is in decade d, and n starts with k}
Nd
where the numerator counts how many numbers in S, among all d-digit numbers, start with block k.
Remark: This definition is entirely deterministic. No assumptions of randomness, probability, or limiting behavior are required, unless otherwise specified. In the special case where S is “all d-digit numbers”, the denominator Nd ensures that the frequencies are relative to the total possible choices in that decade.
3.2 The Role of Floor and Fractional Parts
For any real x, we write:
⌊x⌋ = greatest integer ≤ x, {x} = x − ⌊ x⌋
so that x = ⌊x⌋ + {x}, with {x} ∈ [0 , 1) .
Application: For a digit block k, d = ⌊log 10 k⌋ gives its decade, and the fractional part {log 10 k} often encodes “where” within the decade the block k sits.
3.3 Universal Applicability: Sets, Sequences, and Single Numbers
The above definitions apply to:
• Any finite or infinite set S of positive integers.
• The sequence of terms of a mathematical or physical process.
• The decimal expansion of a single real number: Here, S consists of the “shifts” of the expansion. 93.4 Example Table: Block Frequency Calculation
Table 1: Illustration: Calculation of ρ(k, d ) for Selected Blocks k in Decade d = 2
(100 ≤ n < 1000 )Block k Numbers starting with k in [100 , 999] Frequency ρ(k, 2)
1 100, 101, ..., 199 100 /900 ≈ 0.111
23 230, 231, ..., 239 10 /900 ≈ 0.0111
314 314 only 1/900 ≈ 0.0011
3.5 The Cumulative Function and the Cumulative Digit Mapping (CDM)
Motivation. Pointwise frequencies ρ(k, d ) are useful locally, but the global structure of digit blocks within a decade is most naturally expressed via their cumulative form. This leads to the function w(k, d ), a discrete analogue of a cumulative distribution function (CDF).
Formal definition and properties. Fix a decade d and let k ∈ [10 d, 10 d+1 ). Define
w(k, d ) =
k−1∑
j=10 d
ρ(j, d ), w(10 d, d ) = 0 , w(10 d+1 , d ) = 1 . (1) Then w(·, d ) is nondecreasing, takes values in [0 , 1] , and recovers pointwise frequencies via the discrete derivative
ρ(k, d ) = w(k + 1 , d ) − w(k, d ). (2)
Cumulative Digit Mapping (CDM). Let G : [0 , 1] → [0 , 1] be a nondecreasing function with G(0) = 0 and G(1) = 1 , and let a ∈ R. Define
W (x) = a ⌊x⌋ + G(
{x})
, x ∈ R. (3) For a = 1 ,k ∈ [10 d, 10 d+1 ), one has two equivalent representations for ρ(k, d ):
ρ(k, d ) = G(
log 10 (k + 1) − d)
− G(
log 10 k − d)
,ρ(k, d ) = ⌊log 10 (k + 1) ⌋ − ⌊ log 10 k⌋ + G(
{log 10 (k + 1) })
− G(
{log 10 k})
. (4)
Validity. The first identity holds for 10 d ≤ k ≤ 10 d+1 − 2 (intra–decade). The second holds for all k ∈ [10 d, 10 d+1 ) and accounts for the decade boundary.
Validity. The first equality holds for 10 d ≤ k ≤ 10 d+1 − 1. The second holds for all
k ∈ [10 d, 10 d+1 ).In words, the floor part adjusts the jump at decade boundaries, while inside a decade the behavior is governed by G.
Benford as a special case. For a = 1 and G(s) = s we obtain W (x) = x and the classical formula
ρ(k, d ) = log 10
(
k + 1
k
)
. (5)
Remark. The CDM formalism (3) – (4) applies verbatim to blocks of any length and will be used throughout the sequel. 10 3.6 Summary of Notation
This subsection collects the symbols used in § and later sections. Table 2: Summary of notation used for cumulative and digit–block objects.
Symbol Description
k Digit block (e.g., 1, 23 , 314 )
d Decade index, d = ⌊log 10 k⌋; d+1 is #digits in k
[10 d, 10 d+1 ) Decade-d interval for integer arguments
Nd 10 d+1 − 10 d
ρ(k, d ) Relative frequency for block k in decade dw(k, d ) Cumulative freq.; ρ(k, d ) = w(k + 1 , d ) − w(k, d )
G : [0 , 1] → [0 , 1] Nondecreasing, G(0) = 0 , G(1) = 1
⌊x⌋, {x} Floor and fractional part ( {x} ∈ [0 , 1) )
W (x) a ⌊x⌋ + G({x})
a ∈ R Boundary jump adjuster in W
Reading guide. The cumulative mapping W links the block frequencies to the (fractional-part) driver G via
ρ(k, d ) = W(
log 10 (k + 1) − d)
− W(
log 10 k − d)
,
so that inside each decade the behavior is governed by G, while the term a ⌊x⌋ only compensates jumps at decade boundaries. The Benford case corresponds to a = 1 and
G(s) = s.
3.7 Concluding Remarks: Universality and Framework for All Future Theory
The framework developed here is universal and deterministic : - It covers arbitrary sequences, sets, or even a single real number’s decimal expansion. - No stochastic, probabilistic, or ensemble assumptions are required—everything is constructed by counting and accumulation. - This cumulative formalism will be essential in all that follows: the recurrences (Section IV), solution theory, and construction of new digit laws.
Summary.
• The cumulative function w(k, d ) and CDM formalism are the universal analytical tools for digit laws of arbitrary block length.
• All subsequent sections (IV and beyond) are built upon these foundations. 11 IV. Deterministic Structure and Core Equations for Digit Block Frequencies
4.1 Deterministic Definition of Leading Significant Digit Strings
We formalize the concept of leading significant digits for any x > 0 as follows. Each number is mapped to an infinite string of decimal digits, starting with the first nonzero digit (all leading zeros and the decimal point are ignored):
• 3 ≡ ”3 0 0 0 . . . ”
• 1.24 ≡ ”1 2 4 0 0 . . . ”
• 0.0025 ≡ ”2 5 0 0 . . . ”
• π ≈ 3.1415 . . . ≡ ”3 1 4 1 5 . . . ”
For a fixed integer m ≥ 1, the truncation operator Dm(x) extracts the first m significant digits:
Dm(x) = ⌊
10 m−1−⌊ log 10 x⌋ · x⌋
.
This definition encodes the entire DLSD framework in a deterministic way, independently of probability.
4.2 Fundamental Recurrence for Digit Blocks
Let k denote a digit block and d = ⌊log 10 k⌋ its decade. For any integer δ > 0, the frequency ρ(k, d ) satisfies the basic recurrence:
ρ(k, d ) =
10 δ−1∑
i=0
ρ(10 δk + i, d + δ)
This expresses that the frequency of block k in decade d equals the sum of the frequencies of all possible extensions of k to decade d + δ.
4.3 Cumulative Function and Functional Equation
Define the cumulative function w(k, d ) by:
ρ(k, d ) = w(k + 1 , d ) − w(k, d )
Plugging into the recurrence and rearranging, we have:
w(10 δ(k + 1) , d + δ) − w(k + 1 , d ) = w(10 δk, d + δ) − w(k, d )
The key hypothesis is that both sides are independent of k:
w(10 δ(k + 1) , d + δ) − w(k + 1 , d ) = C(δ)
where C(δ) is to be determined. 12 4.4 Logarithmic Change of Variables and the Cauchy Equation
To motivate the algebraic structure underlying digit distributions, we briefly note that a logarithmic change of variables naturally leads to a functional equation of Cauchy type for the cumulative function. The details and the explicit solution of this equation will be developed in the following section.
4.5 Explicit Solution of the Shift Equation: Step-by-Step Construction
4.5.1 Formulation of the Two Core Equations
Consider k in the range 10 n ≤ k < 10 n+1 − 1 for some n ≥ 0. The key shift equations are:
w(10 δk, d + δ) − w(k, d ) = C(δ), (6)
w(10 δ(k + 1) , d + δ) − w(k + 1 , d ) = C(δ), (7) where d = ⌊log 10 k⌋ and δ ∈ N.
4.5.2 Change of Variables and General Solution
Let x = log 10 k, and define v(x) = w(10 x, ⌊x⌋).The shift equation becomes:
v(x + δ) − v(x) = C(δ).
Analysis of the Increment Function C(δ).
To find the general solution, let us analyze the increment function C(δ). Fix any x
and consider two increments δ1 and δ2:
v(x + δ1) − v(x) = C(δ1),v(x + δ1 + δ2) − v(x + δ1) = C(δ2).
Adding these two equations, we have:
v(x + δ1 + δ2) − v(x) = C(δ1) + C(δ2).
But by the original shift equation, this difference must also equal C(δ1 + δ2). Therefore,
C(δ1 + δ2) = C(δ1) + C(δ2),
which is Cauchy’s functional equation for C(δ).The general solution (for δ ∈ N) is:
C(δ) = aδ,
where a is a constant (see, e.g., [17, 18]). Then
v(x) = a ⌊x⌋ + G ({x}) ,
where G is a function and {x} is the fractional part of x.13 4.5.3 The Case k = 10 n − 1: Boundary Solution
At the block boundary, k = 10 n − 1, so x = n is integer. Let P (n) = w(10 n, n − 1) , then the recurrence for integer arguments is:
P (n + δ) − P (n) = aδ,
with general solution
P (n) = an + const .
Thus, at the boundaries, the cumulative function is affine in n, matching the piecewise structure defined above.
4.5.4 Parallel Construction for k + 1
For k + 1 , let x′ = log 10 (k + 1) and v′(x′) = w(10 x′
, ⌊x′⌋). The same reasoning gives:
v′(x′) = a ⌊x′⌋ + G′ ({x′}) .
4.5.5 Equivalence with a two–term shift form and an asymptotic expansion
Recall the four–term expression obtained in §4.5.5:
ρ(k, d ) = a
(⌊
log 10 (k + 1)
⌋
−
⌊
log 10 k
⌋)
G(
{log 10 (k + 1) })
− G(
{log 10 k})
, (4.18) where G : [0 , 1] → R , G(0) = 0 and G(1) = 1 .
Alternative two–term representation. For any real shift d ∈ [0 , 1) define
ρ∗(k, d ) = G(
log 10 (k + 1) − d)
− G(
log 10 k − d)
. (4.19) We show that ρ∗(k, d ) ≡ ρ(k, d ) for all k ≥ 1 and every digit-block length d.
Proof of equivalence. Write x = log 10 k and ∆ = log 10 (k + 1) − log 10 k = log 10 (1 + 1 /k ).Set s = {x − d} and note that {x} = s + d (mod 1) , {x + ∆ − d} = s + ∆ ( mod 1) . Two cases:
Interior blocks (d / ∈ Z). Because 0 < d < 1, the integer parts satisfy ⌊x + ∆ ⌋ = ⌊x⌋;hence the first term in (4.18) vanishes. We obtain
ρ(k, d ) = G(s + ∆) − G(s) = ρ∗(k, d ).
Boundary blocks (d = n ∈ Z≥0). Now x − n = log 10 k − n has fractional part {x}
and integer part ⌊x⌋ − n. A direct calculation gives
ρ∗(k, n ) = a(⌊x + ∆ ⌋ − ⌊ x⌋) + G({x + ∆ }) − G({x}) = ρ(k, n ),
because subtracting an integer shift n changes neither the fractional parts nor the difference of integer parts. Thus (4.18) and (4.19) coincide in all cases. 14 Large– k asymptotics. Formula (4.19) is convenient for expansions, because ∆ = log 10 (1 + 1 /k ) = 1
k ln 10 + O(k−2). If G is differentiable at s = {log 10 k − d}, a first-order Taylor expansion yields
ρ(k, d ) = G′(
{log 10 k − d})
k ln 10 + O
(
1
k2
)
(k → ∞ ).
This makes (4.19) the natural starting point for estimating convergence rates or construct-ing confidence bands for empirical leading-digit frequencies.
4.5.6 Summary
• The general solution is a ⌊x⌋ + G ({x}) for all k not at the boundary, and P (n) =
an + const for integer boundary points.
• All recurrence conditions are satisfied, and the solutions coincide at the decade transitions.
Remark: This construction gives the most general form of the admissible cumulative functions for DLSDs, as required by the shift recurrence. The solution directly links the theory to the structure of the Cauchy equation and makes explicit the piecewise-logarithmic nature of digital laws.
Main formula for the distribution of leading digits:
ρ(k, d ) = a (⌊log 10 (k + 1) ⌋ − ⌊ log 10 k⌋) + G ({log 10 (k + 1) }) − G ({log 10 k}) ,
where a is a constant, G is a function, and { x } denotes the fractional part of x.By normalization, G can be chosen so that G(0) = 0 , G(1) = 1 .
4.6 Boundary Cases and Interpretation
At the upper boundary of each decade, that is for k = 10 d+1 − 1, the formula for the digit block frequency simplifies due to the properties of the function G:
10 d+1 −1∑
k=10 d
ρ(k, d ) = G(1) − G(0) = 1 .
We impose the normalization conditions on G to guarantee the correct summation over all possible blocks within a decade. Specifically, we require
G(0) = 0 , G(1) = 1 .
The function G can be any non-decreasing function on [0 , 1] with these endpoint values, but the structure of the solution for ρ is completely determined by this form, regardless of whether there are finitely or infinitely many possible solutions for G. No other structural forms for ρ are possible within this framework.
Definition: Leading Digit Profile Function (LDPF).
15 Whenever we refer to the function G(s), which encodes the cumulative distribution of leading digits for a given probability density f (x), we will call it the Leading Digit Profile Function (LDPF) . Explicitly,
LDPF( s) := G(s) =
∫BA
f (x) M (s, x ) dx,
where M (s, x ) is a measurable indicator function that counts whether x contributes to the leading digit profile up to 10 s. The explicit form of M (s, x ) will be developed in Section V.
Throughout this work, we will use the abbreviation LDPF for clarity and consistency.
4.7 Remarks
• G(s) ≡ s gives Benford; arbitrary G gives all admissible deterministic DLSDs.
• All formulae and frequencies derive directly from deterministic structure, with no stochastic or “random mixture” hypotheses.
4.8 Deterministic Profiles versus Hill’s Random–Mixture Theorem
Hill’s seminal 1995 theorem states that if one draws a positive-valued random variable X by first sampling a distribution F at random from a sufficiently broad, parameter–free ensemble F and then sampling X ∼ F , the leading-digit law of the mixture converges in probability to Benford’s vector
Pr (
LD( X) = d)
= log 10
(
1 + 1
d
)
, d = 1 , . . . , 9.
LD( x) = ⌊
10 {log 10 x}⌋
, x > 0.
This conclusion hinges on two probabilistic layers—(i) an external randomness in the choice of F ∈ F , and (ii) the ordinary sampling randomness of X | F .By contrast, the deterministic framework developed here requires no external mixing: a single distribution with cumulative function F (x) generates a windowed profile
G(s) = 1
F (B) − F (A)
n−1∑
i=m
[
F (10 s+i) − F (10 i)]
, s ∈ [0 , 1) , A = 10 m, B = 10 n,
whose analytic form FW (s; b, a ) is governed by the shape parameters (a, b ) of Defini-tion 4.2. When (a, b ) = (1 , 10) the profile collapses to the diagonal G(s) = s and Benford’s leading-digit frequencies reappear, but for a̸ = 1 the profile deviates in a predictable, non-random way—e.g. an S–shape for a > 1 and a concave bend for
0 < a < 1 (see Fig. 12). Thus Hill’s law emerges here as a singular limiting case of a broader structural family, not an inevitable consequence of “generic” randomness. Empirically, 2023 country populations (UN WPP 2024; [ 19 ]) are well approximated by a Weibull model with fitted shape ˆa ≈ 0.47 .Under the Weibull model, the shape parameter a governs departures from Benford: for small a (e.g., a ≈ 0.47 ) the windowed profile G(s) lies close to the Benford line 16 G(s) = s, whereas as a approaches and exceeds 1 the curve becomes pronouncedly S–shaped and the resulting leading–digit frequencies deviate substantially from Benford’s law.
V. Euler–Maclaurin Framework
5.1 Motivation and Historical Context
The Euler–Maclaurin summation formula, developed in the 18th century by Leonhard Euler and Colin Maclaurin, plays a foundational role in analytic number theory and the modern analysis of digital statistics. Contrary to common belief, this formula is not merely an asymptotic or approximate tool , but rather a strictly exact decomposition for sums of sufficiently regular functions over integers.
5.2 The Exact Formula and Its Terms
Let f (x) be a function defined on the real line, and let a, b ∈ Z with a < b . The Euler–Maclaurin formula states:
b∑
n=a
f (n) =
∫ba
f (x) dx + 12
[
f (a) + f (b)]
∫ba
(
x − ⌊ x⌋ − 12
)
f ′(x) dx
Each term has a clear interpretation:
The main integral: ∫ ba f (x) dx is the “continuous” contribution, 2. The boundary average: 12 [f (a) + f (b)] corrects for discrete endpoints, 3. The periodic correction : ∫ ba (x − ⌊ x⌋ − 1/2) f ′(x) dx accounts for the “fine structure” lost when summing only over integers.
Key property: If f (x) is extended to be zero outside [a, b ], the above formula remains strictly valid (all other integrals vanish). In DLSD and related problems, this extension is natural and universal.
5.3 Why Euler–Maclaurin Is Not an Approximation
It is essential to emphasize: The Euler–Maclaurin formula is exact, not just asymptotic! The error arises only if one truncates after the first terms of the more general expansion (involving higher Bernoulli numbers and derivatives). In the version above, if f is continuously differentiable and of finite support, the formula holds with perfect equality.
5.4 Application to Digital Block Statistics and DLSD
This exact bridge is central for digital block statistics: It enables the analytic transforma-tion of any sum over digits or blocks
S =
b∑
k=a
f (k)
17 to its integral analog, with precise control of the correction term. For digital statistics, f (k)
is often the block frequency or its cumulative, and the periodic term encodes fluctuations due to the discreteness of digits.
5.5 Extension to Infinite Intervals
In applications, one often needs to sum over all possible decades (e.g., k ∈ [10 d, 10 d+1 ) for all d). By extending f (x) ≡ 0 outside its natural domain, all integrals and sums can be taken over the full real line, reducing technical complications.
5.6 Towards a Generalized Cumulative Function
The preceding analysis sets the stage for introducing a generalized cumulative function, denoted here as G(s), which fully characterizes the family of admissible distributions of leading significant digits (DLSDs). In the following sections, we develop the machinery necessary to construct explicit solutions for G(s), and demonstrate how this approach reveals the underlying deterministic structure of digital laws.
5.7 The "Integral Replacement" Fallacy in the Literature
It is common in some recent literature to claim a "proof" of Benford’s law by simply replacing the discrete sum over blocks k with a continuous integral over x. In this approach, the Euler–Maclaurin formula is (sometimes implicitly) invoked, but only the main integral term is retained, while both the boundary and periodic correction terms are neglected:
∑
k
f (k) ?
≈
∫
f (x) dx
This is often justified by the claim that, if f is a probability density function (PDF) on
(0 , ∞), the other terms are negligible, and the Benford distribution emerges directly from integrating the logarithmic PDF over the leading digit blocks.
However, this replacement is logically flawed in the context of DLSD analysis:
• The boundary term and, more importantly, the periodic correction term in the Euler–Maclaurin formula are not in general negligible.
• In fact, these terms are responsible for all possible deviations from the Benford law ;neglecting them is tantamount to presupposing the result one wishes to derive.
• In the exact Euler–Maclaurin framework, if these terms are omitted, the remaining integral gives only a trivial result or the Benford law, regardless of the true structure of f .This logical sleight-of-hand leads to circular reasoning: First, one assumes that a sum can be replaced by an integral (which is only strictly valid in special cases), and then one “derives” the Benford law, not noticing that all non-Benford effects have been discarded by construction. See for a critical discussion. In contrast, [ 20 ] is a notable example of the incorrect
application of this principle: the authors adopt the condition as a natural assumption without any critical examination. Our analysis demonstrates that this uncritical acceptance leads to a circular justification and ultimately undermines the logical basis of the argument. 18 Key Point
The Euler–Maclaurin formula shows that the Benford law arises only if the correction terms vanish or cancel. In general, for arbitrary distributions, the full formula—including periodic corrections—must be used to capture the true digit law, and these corrections can be substantial.
5.8 Connection of the Summation Formula with the Euler–Maclaurin (EM) Representation
Consider the sum
S =
∞∑
i=−∞
∫(k+1)10 i
k10 i
f (x) dx,
where f (x) is a non-negative PDF supported on (0 , ∞), and k is a fixed positive integer block. Applying the Euler–Maclaurin summation formula, the sum can be decomposed as
S = J1 + J2 + J3,
where:
J1 =
∫∞−∞
[F (( k + 1)10 y) − F (k10 y)] dy, J2 = 0 ,J3 =
∫∞−∞
∂∂y [F (( k + 1)10 y) − F (k10 y)] · (y − ⌊ y⌋ − 12 ) dy.
Here, F (x) is the cumulative distribution function (CDF) of f (x).
5.8.1 Evaluation of J1
Make the substitution x = ( k + 1)10 y and x = k10 y in J1, which yields:
J1 =
∫∞
0
[
log 10
(
xk
)
− log 10
(
xk + 1
)]
f (x)dx.
This further simplifies to:
J1 = (log 10 (k + 1) − log 10 (k))
∫∞
0
f (x) dx,
which, for a properly normalized PDF, gives the Benford probability for block k.
5.8.2 Evaluation of J2
This term vanishes due to the integration over the full real line and the properties of the PDF:
J2 = 0 .
19 5.8.3 Structure and Role of J3
The term J3 is
J3 =
∫∞−∞
∂∂y [F (( k + 1)10 y) − F (k10 y)] · (y − ⌊ y⌋ − 12 ) dy.
Note that, after a variable change and analysis (see detailed derivation above), the first summand in J3 exactly cancels −J1, so that no trace of the Benford law remains here. The condition for the sum to yield Benford’s law is that the entire J3 vanishes . This is a highly special (exotic) situation: generally, J3̸ = 0 , and the deviation from Benford is encoded precisely in J3.In the general case, the term J3 can be written as:
J3 =
∫∞
0
f (x) V (k, x ) dx − J1,
where
V (k, x ) =
⌊
log 10
(
xk
)⌋
−
⌊
log 10
(
xk + 1
)⌋
,
or, equivalently,
V (k, x ) = ⌊{ log 10 (x)}−{ log 10 (k)}⌋−⌊{ log 10 (x)}−{ log 10 (k+1) }⌋ +⌊log 10 (k+1) ⌋−⌊ log 10 (k)⌋.
Conclusion: The Benford law arises if and only if the sum of all nontrivial terms
J3 vanishes for all k, i.e., when the additional contributions from the oscillatory kernel
V (k, x ) integrate to zero with respect to f (x).
Connection to ρ(k, d ): In this analysis, the sum S (and the EM decomposition) provides a direct integral representation for the block frequency ρ(k, d ), and reveals exactly how deviations from Benford’s law are encoded by the structure of the PDF and the periodic structure of the EM formula.
Key Formula: For any probability density f (x) on R+, the relative frequency of leading digit block k (in its corresponding decade) is given by
ρ(
k, ⌊log 10 k⌋)
=
∫∞
0
f (x) V (k, x ) dx
where
V (k, x ) = ⌊log 10 (x/k )⌋ − ⌊ log 10 (x/ (k + 1)) ⌋.
5.9 Connection Between the Integral Representation and G(s) Formulation
From the integral (Euler-Maclaurin, EM) approach, for any s ∈ [0 , 1) , we define: 20 G(s) =
∫∞
0
f (x) M (s, x ) dx
where
M (s, x ) = −⌊{ log 10 x} − s⌋
and {·} denotes the fractional part. This establishes a direct and explicit link between the analytic (integral) and the functional ( G(s)) characterizations of digit laws. More generally, for d = ⌊log 10 k⌋, we may write:
ρ(k, ⌊log 10 k⌋) = ⌊log 10 (k + 1) ⌋ − ⌊ log 10 k⌋ + G({log 10 (k + 1) }) − G({log 10 k}),
where {·} denotes the fractional part. This formula separates the “main” interval crossing (as k passes a new decade) from the finer structure controlled by the continuous G(s).
Key Formula:
ρ(k, ⌊log 10 k⌋) = ⌊log 10 (k + 1) ⌋ − ⌊ log 10 k⌋ + G({log 10 (k + 1) }) − G({log 10 k})
with
G(s) =
∫∞
0
f (x) M (s, x ) dx, M (s, x ) = − ⌊{ log 10 x} − s⌋
VI. Examples of Continuous Distributions 6.1 Distribution Examples and Leading Digit Analysis
Windowed digital profile. In practice we observe a variable X only on a finite magnitude window [A, B ]. Throughout this section we therefore study the windowed
profile GA,B —the cumulative distribution of the phases {log 10 X} restricted to [A, B ]
and renormalized—so that comparisons across datasets and limits as A ↓ 0, B ↑ ∞ are meaningful. Let f (x) be a probability density function (PDF) supported on an interval [A, B ] ⊂ R+,where 0 ≤ A < B ≤ ∞ . The key functional object in digital leading significant digit (DLSD) analysis is the cumulative digit profile G(s), defined as
G(s) =
∫BA
f (x) M (s, x ) dx, M (s, x ) = − { log 10 x − s} ,
where {·} denotes the fractional part. This formula connects any continuous distribution to its digital signature through logarithmic folding. 21 6.1.1 Logarithmic Partitioning: Summation Formula for g(s) and
G(S)
To further clarify the construction, we define the function g(s) as a sum over logarithmically partitioned intervals within the range [A, B ]:
g(s) =
⌊log 10 (B)⌋
∑
i=⌈log 10 (A)⌉
F (min(10 i+s, B )) − F (max(10 i, A )) if [10 i, 10 i+s) ⊆ [A, B ],
0 otherwise The normalized version is then
G(S) = g(s)
F (B) − F (A)
where F is an arbitrary function (typically the CDF of the underlying distribution), and the summation considers only those intervals [10 i, 10 i+s) that are fully contained within
[A, B ]. The normalization ensures that G(S) represents the fraction of the total change in F over the interval [A, B ] contributed by these log-partitioned intervals.
6.1.2 Sampling and Leading Digit Extraction in [1 , 10) : Illustrative Example and Justification
A key motivation for focusing on intervals like [1 , 10) (or, more generally, intervals that are integer multiples of a decade such as [10 −2, 10 −1), [10 k, 10 k+1 )) lies in their remarkable analytical simplicity and direct relationship between the cumulative distribution function (CDF) of the underlying variable and the leading digit profile function G(s).To see why this is so, observe that for such intervals, the mapping from the original probability distribution f (x) to the digital profile G(s) simplifies dramatically. Specifically, when considering a probability density f (x) supported on [A, B ] where A and B are integer powers of ten, the cumulative digit profile G(s) is given by the simple normalized difference of the CDF:
G(s) = F (10 s) − F (A)
F (B) − F (A) ,
where F (x) is the CDF of f (x). In particular, for the standard decade [1 , 10) , this formula reduces to:
G(s) = F (10 s) − F (1) ,
assuming normalization over the decade. This case is not only mathematically elegant but also serves as the most transparent illustration of the deterministic link between the structure of the underlying distribution and the observed leading digit frequencies. The simplicity of the formula for G(s) in these intervals allows one to immediately "see" the digital signature of any distribution.
Why focus on such intervals? Most real-world datasets, and almost all practical illustrations of Benford-like phenomena, involve data spanning several orders of magnitude. By focusing on samples extracted from full decades, we ensure that every possible leading digit (from 1 to 9) is represented, and the counting of their frequencies becomes not only meaningful but also analytically tractable. Moreover, in numerical experiments, it is common to start with a random sample from an underlying distribution on (0 , ∞) and then restrict attention to those values that fall 22 within a chosen decade, e.g., [1 , 10) , [10 , 100) , etc. By examining the empirical frequencies of leading digits in these intervals, we obtain a direct and visually intuitive illustration of how the theoretical digital profile G(s) manifests in practice.
Generalization: Multiple Decades as Approximation It is also possible, and often useful, to consider not just a single decade, but a union of several adjacent decades, for example, the interval [10 −3, 10 5). In practice, analyzing several consecutive decades provides an even better approximation to the full distribution of leading digits over the positive real axis. Remarkably, even a single decade (e.g., [1 , 10) ) can serve as a good approximation, especially when the underlying data are broadly distributed. When considering multiple decades, the cumulative contribution of all leading digits across these intervals increasingly reflects the global structure of the distribution. Thus, the choice of interval width and position becomes a matter of approximation quality and practical convenience.
Example: Suppose we sample a large number of values from an underlying distribution
f (x) on (0 , ∞), but then only analyze those that fall in [1 , 10) , or in a wider interval such as [10 −3, 10 5). For each value, we record the leading digit d ∈ { 1, 2, . . . , 9} and plot the empirical frequencies. This simple experiment immediately reveals the structure imposed by G(s): for the uniform distribution, G(s) is linear; for other distributions, the shape of
G(s) directly dictates the observed digit frequencies. With multiple decades, the observed frequencies converge ever closer to the theoretical digital profile dictated by the underlying distribution.
Summary: Intervals of the form [1 , 10) , or more generally [10 a, 10 b), are not only convenient but also fundamental for both theoretical analysis and practical illustration of leading digit laws. They embody the most direct link between the analytic form of the underlying distribution and the digital structure captured by G(s). Considering multiple consecutive decades within the positive real axis (e.g., [10 a, 10 b) for some integers a < b )yields a digital profile that closely approximates the theoretical distribution over all positive numbers. This offers a practical bridge between what is observable in finite samples and the limiting behavior of leading digits as the sampled range grows to encompass the entire positive half-line.
6.1.2.1 Digital Profiles for Scale-Invariant Densities over Multiple Decades
Consider the class of scale-invariant densities of the form:
f (x) = ψ({log 10 x})
x ,
where ψ is any 1-periodic function, ψ(t) = ψ(t + 1) , and {y} denotes the fractional part of y.Suppose we restrict attention to a union of N consecutive decades, i.e., the interval
[10 m0 , 10 m1 ) with m0, m 1 ∈ Z, N = m1 − m0 > 0. Let s ∈ (0 , 1) , and let M (s, x ) = 1 if
log 10 x mod 1 < s , and M (s, x ) = 0 otherwise. The cumulative digital profile function is then given by
G(s) =
∫10 m1
10 m0
f (x)M (s, x )dx =
∫10 m1
10 m0
ψ({log 10 x})
x M (s, x ) dx.
Applying the substitution t = log 10 x, x = 10 t, dx = 10 t ln 10 dt , and dx/x = ln 10 dt ,we rewrite the integral as
G(s) = ln 10
∫m1
m0
ψ({t}) · 1{{ t}<s } dt,
23 where 1{{ t}<s } is the indicator function, equal to 1 if the fractional part of t is less than s.Since ψ is 1-periodic, we can partition the integration interval into N full decades:
G(s) = ln 10
N−1∑
k=0
∫10
ψ(τ ) 1{τ <s }dτ = N ln 10
∫s
0
ψ(τ ) dτ,
where we set t = m0 + k + τ , τ ∈ [0 , 1) .
Final explicit formula:
G(s) = ( m1 − m0) ln 10
∫s
0
ψ(τ ) dτ, 0 < s < 1
This formula shows that for any scale-invariant density of this form, the cumulative digital profile over N consecutive decades is completely determined by the integral of the generating function ψ up to s, multiplied by the number of decades and the constant
ln 10 .
Special case: Benford’s Law. For ψ(τ ) ≡ 1/(( m1 − m0) ln 10) (i.e., f (x) = 1/(x ln 10) ), this reduces to the linear profile:
G(s) = s.
Remark. This result remains valid if one considers any collection of N separate, non-overlapping decades; the only requirement is that the union of all considered intervals has total logarithmic width N .
6.1.2.2 Multi-Decade Windows and Approximation to the Global Profile
Let F be the CDF of the underlying density on (0 , ∞). For an integer window of N
consecutive decades [10 m0 , 10 m0+N ), define the normalized cumulative digital profile
GN (s) =
m0+N−1∑
i=m0
[
F (10 i+s) − F (10 i)]
F (10 m0+N ) − F (10 m0 ) , s ∈ [0 , 1] .
This quantity aggregates the contribution of each decade in the window and renormalizes by the total mass in the window. In practice, even a single decade ( N = 1 , e.g. [1 , 10) ) can be a serviceable approximation when most probability mass lies in that decade; including additional adjacent decades typically yields a markedly closer match to the global digital profile. Figure 1 illustrates this effect for a lognormal model: as N grows, GN (s) stabilizes and closely tracks the global profile obtained by summing over many decades and renormalizing. This provides a practical bridge between finite samples on restricted magnitude ranges and the limiting behavior over (0 , ∞).
6.1.2.3 Inverting the Windowed Profile: a Finite-Difference Equation
Let m < n be integers (they may be negative) and consider the window of consecutive decades [10 m, 10 n). For a CDF F on (0 , ∞) define
G[m,n )(s) :=
n−1∑
i=m
[
F (10 s+i) − F (10 i)]
F (10 n) − F (10 m) , s ∈ [0 , 1] . (8) This is the normalized cumulative digital profile of the distribution restricted to [10 m, 10 n).24 Figure 1: Multiple-Decade Approximation to the Global Digital Profile. Curves show
GN (s) computed on [1 , 10) (1 decade), [1 , 100) (2 decades), and [1 , 10 4) (4 decades), compared with an approximation to the global profile obtained by aggregating many decades and renormalizing. The windowed profiles converge rapidly toward the global digital signature as N increases.
A convenient normalization and a finitedifference equation. Set
˜
F (t) := F (10 t) − F (10 m)
F (10 n) − F (10 m), t ∈ R.
Then (8) becomes
G[m,n )(s) =
n−1∑
i=m
[ ˜
F (s + i) − ˜F (i)]
.
Let c := 1
L
∑n−1
i=m˜
F (i) with L := n − m, and define the centered function V (t) := ˜F (t) − c.Noting that ∑n−1
i=m
V (i) = 0 , we obtain the finite-difference (shift) equation
Sm,n V (s) :=
n−1∑
i=m
V (s + i) = G[m,n )(s), s ∈ [0 , 1] . (9) Given G[m,n ), this is an inversion problem for V ; once V is found, ˜F (t) = V (t) + c, hence
F (10 t) = F (10 m) + (
F (10 n) − F (10 m)) (
V (t) + c)
, t ∈ R.
Finally, recovering F (x) on the window amounts to evaluating the right-hand side at
t = log 10 x.
Structure of the solution space. The operator Sm,n is the length-L discrete averaging (box-sum) over integer shifts. Its kernel consists of functions W satisfying ∑L−1
j=0
W (s+j) = 0. A standard representation (via the characteristic equation ∑L−1
j=0
rj = 0 ) shows that the homogeneous solutions are spanned by
cos
(2πq L
s
)
, sin
(2πq L
s
)
, q = 1 , . . . , L − 1,
25 possibly with 1-periodic amplitudes in {s} (the fractional part of s). Thus the general solution of (9) can be written as
V (s) = Vp(s) +
L−1∑
q=1
(
Aq({s}) cos 2πq L s + Bq({s}) sin 2πq L s
)
, (10) where Vp is any particular solution and Aq, B q are arbitrary 1-periodic functions. Imposing monotonicity (CDF) or smoothness conditions typically forces Aq, B q to vanish, yielding a canonical (minimal-oscillation) solution.
Examples.
Single decade ( L = 1 ): Here Sm,n V (s) = V (s) = G[m,n )(s). Thus V (s) = G[m,n )(s)
and ˜F (t) = c + G[m,n )(t). In particular, if G[m,n )(s) = s on [1 , 10) , then ˜F (10 s) = s,i.e. F (x) ∝ log 10 x on [1 , 10) .2. Three decades ( m = −1, n = 2 ⇒ L = 3 ) with G[m,n )(s) = s: Equation (9) takes the form
V (s − 1) + V (s) + V (s + 1) = s.
A particular solution is Vp(s) = s/ 3 since 13
[
(s − 1) + s + ( s + 1) ]
= s. The homogeneous equation has the characteristic roots e±2πi/ 3, hence the full solution is
V (s) = s
3 + C1({s}) cos
(
2πs
3
)
C2({s}) sin
(
2πs
3
)
,
with C1, C 2 arbitrary 1-periodic functions. Choosing C1 ≡ C2 ≡ 0 yields the simplest monotone reconstruction.
Interpretation. The finite-difference relation in Equation (9) reveals that the windowed profile G[m,n )(s) is a discrete convolution of a shifted version of the function F , mapped via base-10 logarithm, with a uniform (length-L) box kernel. This means that knowledge of the profile G[m,n ) over the interval s ∈ [0 , 1] determines the underlying distribution function F
over the domain [10 m, 10 n), up to additive constants F (10 m), F (10 n), and any admissible homogeneous (oscillatory) components as described in Equation (10) . Imposing additional conditions on F —such as monotonicity and absolute continuity—serves to eliminate these oscillatory modes, yielding a unique, smooth, and non-oscillatory reconstruction of F .
6.1.3 Illustration: G(S) for the Normal Distribution
As a concrete example, consider the normal (Gaussian) distribution:
F (x) = Φ
(
x − μσ
)
where Φ is the standard normal CDF, μ is the mean, and σ is the standard deviation. The figures below show G(S) for several values of μ, with σ = 1 , A = 1 , B = 10 .26 (a) μ = −2, σ = 1 (b) μ = 0 , σ = 1
(c) μ= 3 ,σ= 1 (d) μ= 8 ,σ= 1
(e) μ= 12 ,σ= 1
Figure 2: Plots of G(S) for different values of μ for the normal distribution ( A = 1 ,
B = 10 , σ = 1 ). 27 Figure 3: Leading digit profile for samples from the normal distribution N (μ, σ 2) with different μ, σ = 1 .5.
Remark. The framework above will be extended to general intervals and compared with theoretical models, including Benford’s Law, in subsequent subsections.
6.1.4 The Role of Interval Endpoints: Complete and Incomplete Leading Digit Sets
An important aspect of leading digit analysis is the choice of the interval [A, B ] from which numbers are sampled. Specifically, the coverage of all possible leading digits within the interval is directly affected by the values of A and B.Consider first the case where A = 1 and B = 10 . This interval is precisely one decade in the decimal system, containing all numbers of the form d × 10 s with d ∈ { 1, . . . , 9}
and 0 ≤ s < 1. Consequently, every possible leading digit d occurs somewhere in [1 , 10) ,guaranteeing a complete set of first digits. If, however, the upper endpoint B is not a power of ten—say, B = π or another arbitrary (possibly rational or irrational) value—then the situation changes fundamentally. For example, in [1 , π ), only numbers with leading digit 1, 2, or 3 can be present, since
π ≈ 3.14 . Digits 4 through 9 are necessarily absent from any sample restricted to this interval. Similarly, for an interval [1 , b ) where b is not a decade endpoint (e.g., b = 7 ), only numbers with leading digits 1 through 6 are possible. This incomplete coverage of possible digits arises from the structure of the decimal system: only intervals of the form [1 , 10 k), for integer k, contain all leading digits. For all other choices of B, the set of accessible leading digits is restricted, and may even depend in subtle ways on the decimal expansion of B. In particular, for irrational endpoints, the issue of coverage becomes nontrivial and may have theoretical and practical consequences. To further illustrate this point, we can consider the interval [e, π ), where both endpoints are irrational. In this case, only numbers whose leading digit falls within the range covered by e ≈ 2.718 and π ≈ 3.1416 will appear, meaning only certain digits (such as 2 and 3)may be present as the leading digit, while others (such as 1, 4, 5, ..., 9) will be absent. 28 Figure 4: Histogram of leading digits for 100,000 numbers sampled uniformly from the interval [e, π ). Only digits 2 and 3 appear as the leading digit in this range. Understanding these constraints is essential when designing experiments or interpreting data on leading digit distributions. It highlights why, in foundational studies and in Benford’s Law itself, the use of full decades ( [1 , 10) , [10 , 100) , etc.) is standard.
In the following sections, we will demonstrate these effects explicitly by comparing distri-butions of leading digits in intervals with both decade and non-decade endpoints.
6.2 Ratio of Uniforms and Digital Signature
A full-support density on the positive real axis is a probability density function (PDF)
f (z) that is strictly positive for all z > 0. Such distributions are central in the analysis of digital (leading digit) phenomena, because they allow the random variable to assume any positive value, thus covering all possible orders of magnitude. A fundamental example is the distribution of the ratio z = y/x , where x and y are independent and uniformly distributed on (0 , 1) . The resulting PDF, well known from classical probability theory , is
f (z) =
12
, 0 < z < 1
12z2
, z ≥ 1
This density assigns nonzero probability to every positive z, featuring a constant part for
z < 1 and a slowly decaying, scale-invariant tail for z ≥ 1.
Our goal is to analyze the digital signature (leading digit profile) of such distributions using the jet function formalism. Specifically, we compute the following integral:
G(S) =
∫∞
0
f (z) M (S, z ) dz,
where M (S, z ) = −⌊{ log 10 z} − S⌋ is the logarithmic folding function, and {·} denotes the fractional part.
By direct computation, we find that:
G(S) = 12 + 10 S
18 − 5 · 10 −S
9
29 This explicit analytic formula describes the leading digit profile for the ratio of two independent uniform random variables on (0 , 1) . It provides a complete digital signature over the entire positive real line.
Remark: The key difference from previous cases is that here the integration is performed over the full interval (0 , ∞), without artificial truncation. This reveals the true digital nature of the underlying full-support density.
Explicit formula for the leading digit probability.
The probability ρ(k, n ) that the leading digit of z = y/x (with x, y independent and uniformly distributed on (0 , 1) ) equals k at scale n is given by:
ρ(k, n ) = 10 −n
18 − 59 · 10 n
(
1
k + 1 − 1
k
)
where n = ⌊log 10 k⌋.This formula allows direct computation of the probability for any leading digit k and scale n in the full-support ratio distribution.
6.3 Alternative Underlying Distributions for the Same LDPF
It is important to note that the same leading digit profile function (LDPF), here denoted as GRTS (s), can arise from very different underlying probability distributions. For example, the LDPF for the ratio of two independent Uniform (0 , 1) random variables, which we have previously found in analytic form, may also correspond to an entirely different distribution defined on the interval [1 , 10] .Suppose we seek a density f (x) supported on [1 , 10] such that
G(s) = F (10 s) − F (1)
F (10) − F (1)
matches a given target function, e.g., the analytic LDPF obtained for the ratio of uniforms:
G(s) = 118 · 10 s − 59 · 10 −s + 12, s ∈ [0 , 1] .
To reconstruct the corresponding density, recall that F (x) is the cumulative distribution function, and f (x) = dF dx . Expressing G(s) in terms of x = 10 s:
G(s) = F (x) − F (1)
F (10) − F (1) =⇒ F (x) = ( F (10) − F (1)) G(s) + F (1) , x ∈ [1 , 10] .
Taking the derivative with respect to x:
f (x) = dF dx = ( F (10) − F (1)) dG dx
where G(s) is now viewed as a function of x via s = log 10 x.Since s = log 10 x, ds/dx = 1
xln 10
, and thus:
dG dx = dG ds · ds dx = G′(s) · 1
x ln 10
30 Therefore,
f (x) = ( F (10) − F (1)) 1
x ln 10 G′(s), s = log 10 x.
By plugging in the analytic expression for G(s) and differentiating, we obtain an explicit form for f (x) on [1 , 10] which produces exactly the same LDPF as the ratio of uniforms. Therefore, the corresponding PDF f (x) on [1 , 10] that produces exactly the same leading digit profile (LDPF) is
f (x) = 118 + 59x2 , x ∈ [1 , 10] .
with (F (10) − F (1)) = 1 . This function is elementary and does not require any logarithmic differentiation or special normalization constants.
Remark: This demonstrates that LDPF does not uniquely determine the underlying distribution; different PDFs, possibly defined on entirely different intervals, can share the same leading digit profile function.
6.4. Leading Digit Profiles for Products of Independent Uniform Variables
For the product Z = X1X2 of two independent random variables X1, X 2 uniformly distributed on (0 , 1) , the cumulative distribution function (CDF) is
FZ (z) = z (1 − ln z) , z ∈ (0 , 1)
The corresponding probability density function (PDF) is
fZ (z) = − ln z, z ∈ (0 , 1)
This explicit form for the CDF and PDF allows us to compute digital profiles and analyze leading digit statistics for products of independent random variables. We will use these explicit forms to compute the leading digit profile function (LDPF) and to further analyze the digital properties of products of independent random variables.
Example 1: Analytic LDPF as an Integral
The leading digit profile function (LDPF) G(s) for the product Z = X1X2 of two independent Uniform (0 , 1) variables can be expressed as the following integral:
G(s) =
∫10
[− ln z] M (s, z ) dz,
where M (s, z ) is the logarithmic folding function (or indicator), and s ∈ [0 , 1] .Evaluating this integral over the interval z ∈ (0 , 1) , and performing normalization and scaling, yields the analytic expression
G(s) = 10 s(As + B) − B,
with constants A and B determined by the normalization conditions G(0) = 0 , G(1) = 1 .
A = −ln 10 9 , B = 19 + 10 ln 10 81 .
31 This formula ensures correct normalization and scaling of the digital profile over one decade. This demonstrates the direct connection between the probabilistic integral and the explicit analytic formula for the digital profile.
Example 2: CDF-based construction on [0 .1, 1]
Suppose Z = X1X2, where X1, X 2 are independent and uniformly distributed on (0 , 1) .The cumulative distribution function is
FZ (z) = z (1 − ln z) , z ∈ (0 , 1) .
For leading digit analysis on the interval [0 .1, 1] , we define the profile function as
G(s) = FZ (10 −1+ s) − FZ (0 .1)
FZ (1) − FZ (0 .1) ,
for s ∈ [0 , 1] .Explicitly:
FZ (10 −1+ s) = 10 −1+ s [
1 − ln(10 −1+ s)]
= 10 −1+ s [1 + (1 − s) ln 10] .
Similarly,
FZ (0 .1) = 0 .1 (1 − ln 0 .1) = 0 .1(1 + ln 10) ,FZ (1) = 1 · (1 − ln 1) = 1 .
Thus, the leading digit profile function becomes:
G(s) = 10 s [1 + (1 − s) ln 10] − (1 + ln 10) 9 − ln 10
for s ∈ [0 , 1] .
This explicit formula allows direct computation of the digital profile for the product of two uniforms on [0 .1, 1] .
Remark 1 (Limiting case: product of uniforms and Benford’s law) . Let U1, . . . , U N ∼
Unif(0 , 1) be independent and set
ZN =
N∏
i=1
Ui.
Then
YN := − ln ZN =
N∑
i=1
(− ln Ui) ∼ Gamma( N, 1) ,
so that the density of ZN on (0 , 1) is
fZN (z) = (− ln z)N −1
Γ( N ) = (− ln z)N −1
(N − 1)! , 0 < z < 1.
Moreover,
log 10 ZN = 1ln 10
N∑
i=1
ln Ui.
Because ln Ui has a continuous, nonlattice distribution, the fractional parts {log 10 ZN }
converge in distribution to Unif [0 , 1) as N → ∞ (standard Fourier/Weyl equidistribution argument). Consequently the leading-digit/windowed profile
GN (s) = Pr (
{log 10 ZN } ≤ s)
−→ s, s ∈ [0 , 1) ,
i.e., Benford’s law is recovered in the product limit.
32 6.5 Power Law Distribution: Analytical Expression and Special Cases
Let
f (y) = m y m−1
bm , y ∈ [0 , b ], m > 0
with
G(s) =
∫b
0
m x m−1
bm M (s, x ) dx, M (s, x ) = −⌊ { log 10 x} − s ⌋
where {x} = x − ⌊ x⌋.
6.5.1 The general case
Using the abbreviations:
q = log 10 (b) , q0 = { q } , q1 = ⌊q⌋ ,r = −a + q, r0 = { r } , r1 = ⌊r⌋ .
With these abbreviations the general windowed profile for m ≥ − 1 becomes
G(a, b, m ) = − r1 + q1 − 10 m 10 −mq 0
10 m − 1 + 10 m 10 −mr 0
10 m − 1 .
When b = 10 n, then log 10 b is integer and thus {log 10 b} = 0 . The result simplifies to (write the explicit form or simply say: the leading digit profile is strictly uniform due to exact power-of-ten cutoff). Then
G(s ; b = 10 n, m ) =
m−1∑
i=0
[
F (10 s+i) − F (10 i)]
F (10 m) − F (1) = 10 ms − 110 m − 1 , s ∈ [0 , 1) .
6.5.2 The case m = 1 and b̸ = 10 n
For m = 1 , the density is uniform, but b is not an integer power of 10. The leading digit profile is:
G(s) = 10 · 10 −U1
9 − 10 · 10 −V1
9 − U
where
U = ⌊− s + {log 10 b}⌋ ,U1 = {− s + {log 10 b}} ,V1 = {log 10 b}
and {x} = x − ⌊ x⌋.Below are figures of G(s) for several choices of {log 10 b} = 0 .1, 0.3, 0.6, 0.8.33 Figure 5: Leading digit profile function G(s) for the power law with m = 1 and several upper bounds b. Each curve corresponds to a different value of b, with the corresponding fractional part {log 10 b} indicated in the legend. The black dots and "Jump" labels mark the points s = {log 10 b}, where G(s) exhibits a jump discontinuity, corresponding to the cut-off of the support at x = b. For b = 10 k ({log 10 b} = 0 ), G(s) coincides with the uniform distribution G(s) = s (shown as a dashed line). As b varies between powers of ten, the position and magnitude of the jump shift, illustrating how the profile depends on the upper bound of the distribution.
Explanation. Figure 5 demonstrates the leading digit profile function G(s) for samples drawn from a power law with exponent m = 1 over the interval [0 , b ], for several values of b. Each colored curve represents a different b, and the corresponding fractional part {log 10 b} is shown in the legend. A distinctive feature of these curves is the presence of a jump at s = {log 10 b}, which is clearly marked on the plot. This jump arises because at that specific value of s, the support of the distribution is abruptly truncated by the upper bound b. The magnitude and location of the jump depend on b, and as b approaches an exact power of ten ( b = 10 k), the jump moves to s = 0 and disappears, so the profile becomes perfectly linear ( G(s) = s), as shown by the dashed line. This illustrates the sensitivity of the leading digit profile to the upper bound, and highlights the unique case of the uniform profile for powers of ten.
Note. For the special case m = 0 , the result is always G(s) = s, corresponding to the uniform (Benford-like) case. However, there are many distributions whose leading digit profile is not of Benford form. The degree of deviation from Benford’s Law is precisely reflected in how much G(s) differs from the identity line s. The jumps at s = {log 10 b}
are visual markers of this deviation for different choices of b.34 6.5.3 Numerical Results and Periodicity of ρ(k, b, m )
This subsection presents a numerical investigation of the leading digit frequency function
ρ(k, b, m ) for the power law distribution on [1 , b ], with several values of the exponent m
and leading digit k.The function is defined as
ρ(k, b, m ) = ⌊log 10 (k + 1) ⌋ − ⌊ log 10 (k)⌋ + G({log 10 (k + 1) }, b, m ) − G({log 10 (k)}, b, m )
where G(s, b, m ) is the leading digit profile function described earlier, and {x} = x − ⌊ x⌋
is the fractional part. The following figures illustrate the dependence of ρ(k, b, m ) on the upper bound b and exponent m for three representative digits k = 1 , 3, 8. Each plot shows how the frequency of the leading digit changes as b varies from 1 to 10 for a fixed m.
Figure 6: The function ρ(k, b, m ) for m = 1 and k = 1 , 3, 8, as b runs from 1 to 10 . This case corresponds to the uniform distribution on [1 , b ]. For b = 10 , the digit frequencies approach Benford’s Law. Between powers of ten, the profiles exhibit a characteristic “wavy” structure. 35 Figure 7: The function ρ(k, b, m ) for m = 2 and k = 1 , 3, 8. Here, the digit frequencies become more concentrated due to the stronger bias toward larger values in the distribution. The “wavy” pattern persists and demonstrates the effect of the upper bound b.
Figure 8: The function ρ(k, b, m ) for m = 0 .5 and k = 1 , 3, 8. With m < 1, the distribution is skewed toward lower values, which is reflected in the digit frequencies. The periodic, oscillatory nature of the profiles remains visible.
Discussion and Periodicity.
A striking feature of these results is the periodic behavior of ρ(k, b, m ) as a function of
log 10 b. Specifically, as the upper bound b increases by a factor of 10 (that is, as log 10 b
36 increases by 1), the shape of each curve is exactly repeated. This periodicity originates from the logarithmic structure of the digit distribution and means that, in each new decade, the profiles for all k are reproduced. The “waves” or oscillations seen in the figures illustrate how the leading digit statistics cycle through all possible forms as b traverses from one order of magnitude to the next. For m = 1 , the case of the uniform distribution, the digit frequencies approach Benford’s Law exactly as b → 10 , and repeat the characteristic deviation for each new decade. For m > 1, the distribution is more heavily weighted toward large values, and for
m < 1 toward small values; but in all cases, the periodic pattern remains—a fundamental and visually striking property of the digital statistics for power-law type distributions. In addition, the calculated function exhibits clearly defined periodic behavior with respect to the decimal logarithm of b with a period of 1. This means that the values of the function’s Maximum and Minimum are determined by the position of the fractional part of the decimal logarithm of k.The Maximum is achieved at the point where the fractional part of log 10 (k + 1)
coincides with the fractional part of log 10 b in the interval [0 , 1] :
{log 10 (k + 1) } = {log 10 b}
The Minimum is achieved when the fractional part of log 10 k coincides with the fractional part of log 10 b:
{log 10 (k)} = {log 10 b}.
The maximum and minimum values of the function can be explicitly written as:
ρ(k)max = ρ
(
k, b = 10 {ln( k+1) ln(10) }, m
)
= −(k + 1) −mkm + 1 10 m − 1 · 10 m
ρ(k)min = ρ
(
k, b = 10 { ln k
ln(10)
}, m
)
= (k + 1) mk−m − 110 m − 1
For the special case m = 1 , these reduce to simple expressions:
ρ(k)max = 10 9 · 1
k + 1 ρ(k)min = 10 9 · 1
k
Extension to Finite Intervals. The analysis above naturally leads to a practical question: what if, instead of considering the entire positive axis, we wish to study the leading digit statistics for a power-law distribution on a finite interval [a, b ]?This can be done directly and elegantly using a normalization formula. Suppose we already know the cumulative digital profile G(s, [0 , b ], m ) for the interval [0 , b ], where m is the power-law exponent. Then, the corresponding profile for any finite subinterval [a, b ] is given by
G(s, [a, b ], m ) = bm G(s, [0 , b ], m ) − am G(s, [0 , a ], m )
bm − am (11) Here, G(s, [a, b ], m ) is the cumulative digital profile on [a, b ], and G(s, [0 , b ], m ), G(s, [0 , a ], m )
are the corresponding profiles computed for the intervals [0 , b ] and [0 , a ].
Intuitive explanation. This formula is a direct consequence of the properties of power-law distributions. Since the density is proportional to xm−1, the contributions from [0 , b ]
and [0 , a ] are scaled by bm and am, respectively. Subtracting the contribution up to a and properly normalizing by the interval size, we obtain the leading digit statistics for any 37 finite interval. This is especially useful for empirical data, where observations are always bounded. With this extension, we are able to generalize the leading digit analysis to any desired interval. In the following sections, we turn to the discussion of the discrete (empirical) case, where the data set consists of a finite collection of numbers rather than a continuous distribution.
6.6 Scaling by b and the representation g(s, b ) = R(β − s) − R(β)
Why this matters. Working with
g(s, b ) =
∫
f (z/b ) M (s, z ) dz b , b = 10 β , β = {log 10 b},
gives a universal way to compare “mantissa profiles” across scales and units. This is crucial when data are normalized by an arbitrary bound B or naturally appear in a scaled form X/B (e.g., on [0 , B ]).
• Comparability across scales and units. The form f (z/b )/b automatically accounts for stretching/compressing the axis, enabling fair comparison of datasets measured in different units or over different ranges.
• Diagnostics of leading-digit structure. The profile g(s, b ) averages the indicator
{log 10 (bz )} < s , i.e., a modulus on the log scale. It supports quality control, anomaly detection, and DLSD-style analysis, including for samples truncated to [0 , B ].
• Broad coverage of continuous laws. Many distributions (power laws on [0 , B ],Weibull, normal, lognormal, gamma, etc.) naturally admit the scaling f (z/b )/b , so their mantissa profiles fit into a single framework.
• Operational recipe via a base slice. The representation g(s, b ) = R(β −s)−R(β)
with β = {log 10 b} implies that once the base slice g1(s) = g(s, 1) is known, we can obtain g(s, b ) for any b > 0 immediately (see (17) and (19)).
Setup and notation. Fix base 10 . For t > 0 write {log 10 t} ∈ [0 , 1) for the fractional part of log 10 t, and let s ∈ [0 , 1] . Define the window
M (s, z ) := −
⌊
{log 10 z} − s
⌋
(z > 0) , (12) so that M (s, z ) = 1{{ log 10 z}<s }. Let f be a density on (0 , ∞). We consider two equivalent normalizations:
g(s, b ) = 1
b
∫b
0
f
(
zb
)
M (s, z ) dz =
∫10
f (x) M (s, bx ) dx, (13)
g(s, b ) = 1
b
∫∞
0
f
(
zb
)
M (s, z ) dz =
∫∞
0
f (x) M (s, bx ) dx. (14) (Use x = z/b .) It is often convenient to write b = 10 m+β with
m = ⌊log 10 b⌋ ∈ Z, β = {log 10 b} ∈ [0 , 1) .
Then z = y 10 m and y = 10 β x make the scale separation explicit. 38 Step 1 (base slice). Define
g1(s) := g(s, 1) =
∫
f (x) 1{{ log 10 x}<s } dx = ∑
k∈Z
∫10 k+s
10 k
f (x) dx, (15) where the last equality uses {log 10 x} < s ⇐⇒ x ∈ ⋃
k∈Z
[10 k, 10 k+s).
Step 2 (decomposing b). Write b = 10 m+β . For any x > 0,
{log 10 (bx )} = {log 10 b + log 10 x} = {β + {log 10 x}} .
Let U := {log 10 X} ∈ [0 , 1) for X ∼ f . Then
g(s, b ) = P(
{β + U } < s )
. (16)
Step 3 (piecewise formula for g(s, b ) via g1). Let FU (s) := P(U < s ) = g1(s) on
[0 , 1] (use right/left limits if needed). From (16) we obtain the circular-shift identities
g(s, b ) =
g1(1 − β + s) − g1(1 − β), 0 ≤ s ≤ β,
1 + g1(s − β) − g1(1 − β), β < s ≤ 1, (17) so that g(0 , b ) = 0 , g(1 , b ) = 1 , and g(s, 1) = g1(s).
Step 4 (the R-representation and recovery of R). Define
R(u) := g1
(
1 − { u})
− 1 − ⌊ u⌋. (18) Then R(u + 1) = R(u) − 1 (so H(u) := R(u) + u is 1-periodic), and for every s ∈ [0 , 1]
and β = {log 10 b},
g(s, b ) = R(β − s) − R(β). (19)
Proof sketch. Write R(u) = −u + H({u}) with H 1-periodic. Choosing H(0) = 0 and solving g1(s) = R(−s) − R(0) yields H(u) = g1(1 − u) − (1 − u); substituting gives (18) and then (19). Uniqueness holds up to an additive constant, which cancels in the difference.
Computational note. For any density f on (0 , ∞),
g1(s) = ∑
k∈Z
∫10 k+s
10 k
f (x) dx.
Numerically: (i) tabulate g1 on a grid of s ∈ [0 , 1] (truncate the sum where f is negligible); (ii) build R via (18); (iii) evaluate g(s, b ) for any b > 0 using (17) or (19).
Example (uniform density on [0 , b )). Let f (x) = 1(0 ,1) (x). Then
g1(s) =
∫10
1{{ log 10 x}<s } dx =
∞∑
n=0
∫10 −n
10 −n−1
1{{ log 10 x}<s } dx =
∞∑
n=0
(
10 −n−1+ s−10 −n−1)
= 10 s − 19 .
39 Thus, with b = 10 m+β and β = {log 10 b}, the piecewise form (17) gives
g(s, b ) =
10 9 10 −β (
10 s − 1)
, 0 ≤ s ≤ β,
1 + 10 −β
9
(
10 s − 10 )
, β < s ≤ 1,
and one convenient choice of R is
R(u) = 10 1−{ u} − 10 9 − ⌊ u⌋.
Remark. This example is not an appeal to “Benford’s law”; the equality g1(s) = (10 s −1) /9
arises from uniform x ∈ (0 , 1) and the geometric partition of (0 , 1) into decades. It merely coincides with the base-10 Benford mantissa CDF.
VII Discrete Distributions and Empirical Digital Pro-files
In this section, we develop the discrete analogue of the density and distribution functions as applied to digital statistics. The motivation comes from the analysis of finite datasets, where the standard notion of probability density is no longer applicable, and empirical methods must be used.
7.1 Empirical Density for a Finite Set
Suppose we are given a finite set of N positive real numbers {at}Nt=1 . Before analysis, all leading zeros and decimal points are removed, and only strictly positive numbers are retained, ensuring every number is properly formatted for digital analysis. The empirical density function is defined as:
f (x) = 1
N
N∑
t=1
δ(x − at)
where δ(x − at) is the Dirac delta function centered at at. In the discrete context, this means f (x) is nonzero only at those x which coincide with one of the at, and its value is the normalized count (frequency) of x in the dataset.
7.2 Empirical Distribution Function
The corresponding empirical distribution function is:
F (x) = 1
N
N∑
t=1
1{at≤x}
where 1{at≤x} is the indicator function, equal to 1 if at ≤ x, and 0 otherwise. 40 7.3 Application to Digital Statistics
This approach allows us to empirically analyze the distribution of leading digits in any finite set of positive real numbers, regardless of their origin or structure. In particular, it provides a foundation for the study of deviations from classical digital laws (such as Benford’s Law) in real-world datasets.
Remark: The empirical approach is particularly well suited for analyzing digital statistics in practical data, where sample sizes are finite and distributions may be highly irregular or even multimodal.
7.4 The Empirical Leading Digit Density: G-function and V -function Approaches
There are two equivalent approaches to computing the empirical leading digit frequency (density) ρ(k) in a finite sample {ai}Ni=1 :
Through the G-function:
ρ(k) = ⌊log 10 (k + 1) ⌋ − ⌊ log 10 (k)⌋ + G({log 10 (k + 1) }) − G({log 10 (k)})
where G(s) is the leading digit profile function defined earlier, and {x} denotes the fractional part.
Through the V -function (direct counting):
V (k, x ) =
⌊
log 10
(
xk
)⌋
−
⌊
log 10
(
xk + 1
)⌋
ρ(k) = 1
N
N∑
i=1
V (k, a i)
The V -function is 1 if k is the leading digit of x, and 0 otherwise. Thus, this sum simply counts the fraction of numbers in the sample that begin with digit k.
Remark: The G-function and V -function approaches are consistent; in the discrete, empirical setting, the direct counting method via V is the most natural and computationally straightforward.
7.5 Single Number Example: Canonicalization and Infinite–Digit View
A notable strength of our empirical digital profile method is that it applies even to the smallest datasets, including the case of a single number. However, for correct digital analysis, we must exclude trivial representations where the number is written as 1, 10 ,
100 , 1000 , and so on, since all such forms correspond to the same significant value with varying numbers of trailing zeros.
7.5.1 Single Number: Blocks for All m (one datum, infinitely many blocks)
Digits of a single number. Write any x > 0 in scientific form x = 10 n(x) m(x) with
n(x) ∈ Z and m(x) ∈ [1 , 10) . Let the mantissa have the (unique) digit expansion
m(x) = d0.d 1d2d3 . . . , d0 ∈ { 1, . . . , 9}, dj ∈ { 0, . . . , 9}.
41 Removing the decimal point gives the infinite significant–digit string
σ(x) = d0d1d2d3 . . .
— even a single number carries infinitely many significant digits.
Leading m-digit block of a single number. For each m ≥ 1 define the leading
m-digit block
Bm(x) =
⌊
10 m−1+ {log 10 x}
⌋
=
m−1∑
j=0
dj 10 m−1−j ∈ {10 m−1, . . . , 10 m − 1}.
If the dataset is the single observation {x}, then the empirical block distribution of order
m is the point mass
ρm(k) = 1(
k = Bm(x))
, k = 10 m−1, . . . , 10 m − 1.
In words: for that one number, the block Bm(x) has probability 1 and all other m-blocks have probability 0.
Examples (one number, many blocks).
• x = 1 = 1 .000 . . . has digits 1, 0, 0, . . . hence
B1(1) = 1 , B2(1) = 10 , B3(1) = 100 , . . . , B m(1) = 10 m−1.
• x = 3 = 3 .000 . . . has digits 3, 0, 0, . . . hence
B1(3) = 3 , B2(3) = 30 , B3(3) = 300 , . . . , B m(3) = 3 · 10 m−1.
• x = π ≈ 3.14159 . . . has digits 3, 1, 4, 1, 5, 9, . . . hence
B1(π) = 3 , B2(π) = 31 , B3(π) = 314 , B4(π) = 3141 , . . .
No duplication by decimal scaling. Numbers x and 10 kx (k ∈ Z) have the same digit string σ and thus the same blocks Bm(·); we therefore do not count x, 10 x, 10 2x, . . .
as separate observations when the dataset is a single number.
Conclusion: This illustrates the complete generality of the empirical approach: it yields exact, arithmetic results for all leading digit blocks, not just the first digit, making it a powerful tool for digital analysis at any desired scale.
7.6 Arithmetic Sequences and the G(s) Function
Consider the arithmetic sequence
ai = αi + β,
where α > 0, β ≥ 0, and both α and β are real numbers. The index i runs from 1 to N .We define the function G(s) as the sum
G(s) = 1
N
N∑
i=1
M (s, a i),
42 where M (s, a i) is a function of s and ai. Here, ai takes the form specified above. We will study the behavior of the function G(s) for real associative values of the parameters and analyze how its properties depend on the sequence parameters α, β, and
N .We define the function J(s) as follows:
J(s) = 1
N
∫Nz=1
− ⌊{ log 10 (αz + β) − s}⌋ dz
where α > 0, β ≥ 0, and N is the upper limit of the variable z.Now, let us make a change of variables. Set y > z , with z = y−βα . In these terms, the integral for J(s) can be rewritten as:
J(s) = 1
N
∫αN +βy=α+β
− ⌊{ log 10 (y) − s}⌋ dy α
Here, we performed the substitution from z to y, setting the stage for further analysis. With the change of variables and normalization, the function J(s) can be written as
J(s) = 1
b − β/ν
∫bx=( α+β)/ν
− ⌊{ log 10 (x) − s}⌋ dx
where
ν = 10 ⌊log 10 (αN +β)⌋, b = 10 log 10 (αN +β)
ν
In the limit as N → ∞ , the lower bound becomes negligible, and the formula simplifies to
J(s) = 1
b
∫bx=0
− ⌊{ log 10 (x) − s}⌋ dx J(s) = 10 · 10 −{− s+{log 10 b}}
9 − 10 · 10 −{ log 10 b}
9 − ⌊− s + {log 10 b}⌋
where {x} = x − ⌊ x⌋ denotes the fractional part of x. The function G(s) is originally defined as a sum, while J(s) is formulated as an integral. We argue that, for large N , the integral provides an excellent approximation to the sum. The key reason is that the summand in G(s) changes slowly with respect to its index, especially for large N . Since the function is essentially logarithmic, its variation over a single interval is very small, and the difference between the discrete sum and the continuous integral becomes negligible as N increases. Thus, replacing the sum by the integral is justified; the resulting error is of order 1/N and vanishes in the limit N → ∞ .Additionally, the initial partition or lower bound of the summation or integral is unimportant in the limit. When dividing by N , any fixed number of initial terms or the contribution from the lower limit disappears in the large N limit. In other words, the influence of the starting point is suppressed by the normalization, and only the main body of the sum or integral determines the result as N becomes very large. 43 To conclude, our results demonstrate that the function ρ(k) for an arithmetic sequence exhibits a striking periodic behavior, oscillating between its maximum and minimum values as dictated by the decimal structure. While numerical methods can illustrate these oscillations and confirm the bounds, the analytical formulas derived here provide an exact description valid for any n, whether or not the interval corresponds to a complete decade. This highlights not only the power of the analytical approach but also the rich structure underlying the distribution of leading digits. The combination of rigorous analysis and numerical evidence offers a comprehensive understanding of the phenomenon, paving the way for further exploration using both methods.
7.7 From Discrete Sums to Integral Approximations: Asymptotic Analysis of Digital Profiles
The study of leading digit distributions in empirical datasets naturally begins with discrete sums. Suppose we have a finite sample {x1, x 2, . . . , x N }, and let M (s, x i) denote an indicator or cumulative function associated with the leading digit profile (for example,
M (s, x ) = 1 if the leading digit of x is less than s, and 0 otherwise). The empirical digital profile is then defined as
G(s) = 1
N
N∑
i=1
M (s, x i).
Integral Approximation. When the sample size N is large and the data points xi
can be parametrized smoothly (for example, xi = φ(i) for some monotonic function φ), it is often fruitful to approximate the discrete sum by a continuous integral:
G(s) ≈ 1
N
∫Nz=1
M (s, φ (z)) dz.
This substitution replaces the sum over discrete indices i by an integral over a continuous variable z, which can often be analyzed with the powerful tools of calculus.
Change of Variables. To further simplify the analysis, we may introduce a change of variables y = φ(z), assuming φ is invertible. The inverse function ψ(y) satisfies z = ψ(y),and dz = ψ′(y) dy . Substituting into the integral, we obtain
G(s) ≈
∫φ(N)
y=φ(1)
ψ′(y) M (s, y ) dy.
In this formulation, the function ψ′(y) acts as an effective "density" for y values, reflecting how the discrete points are distributed over the continuous range.
Interpretation and Limitations. While this integral approximation is not exact in general—especially for small N or highly irregular xi—it often captures the asymptotic behavior of G(s) as N grows large. In many cases, it provides valuable intuition and a bridge between empirical (discrete) statistics and analytic (continuous) theory.
Example: If xi = iα for some α > 0, then ψ(y) = y1/α and ψ′(y) = 1
α
y1/α −1. The integral approximation becomes:
G(s) ≈
∫Nα
y=1
1
α y1/α −1M (s, y ) dy.
44 Remark on normalization: If the upper bound φ(N ) does not correspond to an exact decade (e.g., φ(N )̸ = 10 k), one can define ν = 10 ⌊log 10 φ(N )⌋ as the nearest lower decade for normalization. In summary, this discrete-to-continuous transition provides a powerful analytic tool for understanding the asymptotic properties of digital distributions, though it must be used with care for finite or irregular datasets.
Rescaling and Further Change of Variables.
To analyze the asymptotic and normalized behavior of the distribution, it is often helpful to introduce a rescaling based on the characteristic scale ν, defined as
ν = 10 ⌊log 10 φ(N )⌋
so that ν is the largest integer power of 10 not exceeding φ(N ). We can then make a further change of variable, setting y = xν , where x runs over a normalized interval. Let b = φ(N )/ν and a = φ(1) /ν so that x ∈ [a, b ] and y ∈ [φ(1) , φ (N )] . Let
ψ1(y) = ψ(y) for notational clarity. The cumulative profile in the rescaled variables becomes
G(s) ≈
∫φ(N)
y=φ(1)
ψ1(y)M (s, y ) dy = ν
∫bx=a
ψ1(xν )M (s, xν )dx
Normalizing by N ν and dividing by the effective width b gives
lim
N→∞
1
N ν
∫bx=a
ψ1(xν )M (s, xν )dx = 1
b
∫bx=a
f∞(x)M (s, x )dx
where f∞(x) denotes the limiting density as N → ∞ . This expression captures the normalized contribution of each subinterval in the large-N limit.
Interpretation. This rescaling is particularly useful when the upper endpoint φ(N )
does not fall exactly on a decade. By normalizing with respect to ν and focusing on the relative position x within the final decade, we can meaningfully compare distributions with different sample sizes or endpoints. In the limit N → ∞ , the scaled density ψ1(xν )
approaches a universal profile on [a, b ].
Summary. The two-step change of variable—from discrete sum to integral, and then to rescaled, normalized variables—reveals the underlying scaling structure and facilitates the asymptotic analysis of digital distributions for large finite samples. While not exact for all datasets, this method provides a robust framework for understanding the limiting behavior and normalization of empirical digital profiles.
Remark on the Weyl Equidistribution Theorem.
It is worth noting that the integral approximation described above is closely related to Weyl’s equidistribution theorem [ 21 ]. If the sequence {log 10 φ(x)} (the fractional parts of log 10 φ(x)) is distributed according to a probability density f (t) on the interval [0 , 1) ,then in the limit N → ∞ , the empirical digital profile satisfies
G(s) =
∫s
0
f (t) dt.
This is because, by the law of large numbers and Weyl’s theorem, the normalized sum over M (s, φ (x)) converges to the integral of f over [0 , s ].In other words, as N increases, the fraction of elements with {log 10 φ(x)} < s ap-proaches the probability mass of f (t) on [0 , s ]. This provides a rigorous foundation for approximating discrete digital distributions by integrals, especially in the presence of equidistribution or known limiting densities. 45 7.8 2023 World Population: Digital Density and Empirical Profile
Dataset. We use the 2023 world population table ( Total Population only) with 217 economies. Male/Female breakdowns are available but not used here.
Method. For each country we compute s = {log 10 (X)} ∈ [0 , 1) , the fractional part of the base-10 logarithm of the total population X. The empirical windowed profile is ̂
G(s) = 1
N
N∑
i=1
1(
{log 10 (Xi)} ≤ s)
,
plotted in Fig. 9, together with the reference line G(s) = s.
First-digit proportions. The observed first-digit frequencies for Total Population (countries, 2023) are shown in Table 3. As illustrative criteria, we report a Pearson chi-square against Benford’s vector (
log 10 (1+1 /d ))9
d=1
and the KL divergence DKL (̂p∥pBenford );these can be replaced by your DLSD-based criteria in the final draft. Table 3: First-digit distribution for 2023 Total Population across countries. Digit 1 2 3 4 5 6 7 8 9Count 68 36 30 16 25 16 8 9 9Share 0.313 0.166 0.138 0.074 0.115 0.074 0.037 0.041 0.041
Goodness-of-fit (illustrative). For this dataset ( N = 217 ), the chi-square statistic vs. Benford’s vector is 7.609 , and the KL divergence is 0.01738 . These figures are reported only for reference; in the DLSD framework we will instead estimate your two-parameter profile G(s) = FW (s; b, a ) and report its parameters and confidence bands.
Figure 9: Empirical windowed profile ̂ G(s) for 2023 world Total Population across countries (orange) with the reference line G(s) = s (dashed). 46 Weibull approximation for country-level Total Population (2023)
Model and estimation. We fit the two-parameter Weibull distribution to the sam-ple Xi > 0 of country-level Total Population . The parametrization is f (x; a, b ) =
ab
(
xb
)a−1
exp
[
−
(
xb
)a]
, with shape a and scale b. Maximum likelihood yields ˆa = 0 .467
and ˆb = 1 .37467 e + 07 (sample size N = 217 ).
Goodness of fit. We report several diagnostics: (i) Kolmogorov–Smirnov statistic with parametric bootstrap ( B = 300 ) gives D = 0 .0454 with p ≈ 0.355 ; (ii) Pearson chi-square on m = 10 equiprobable bins with parametric bootstrap ( B = 400 ) gives χ2 = 5 .72 with
p ≈ 0.621 ; (iii) log-likelihood ℓ(ˆ a, ˆb) = −3813 .4, AIC=7630.8, BIC=7637.5.
Interpretation. The small KS distance ( D = 0 .0454 ) together with a large parametric-bootstrap p-value ( p ≈ 0.355 ) indicates that the fitted Weibull profile is statistically indistinguishable from the empirical G(s) at conventional levels, and the Pearson test on equiprobable bins likewise shows no lack-of-fit ( χ2 = 5 .72 , p ≈ 0.621 ). The penalized-likelihood scores (AIC = 7630 .8, BIC = 7637 .5) are consistent with a parsimonious two-parameter specification and do not signal overfitting, reinforcing that the Weibull approx-imation is adequate for these data.
Figure 10: Histogram of Total Population (countries, 2023) on a log-x axis with the fitted Weibull pdf overlay ( ˆa = 0 .467 , ˆb = 1 .37467 e + 07 ). 47 Figure 11: QQ-plot (log–log axes) comparing empirical quantiles of Total Population against the fitted Weibull quantiles.
Figure 12: Empirical windowed profiles G(s) for Weibull (a, b = 1) with shapes a ∈{0.5, 1, 2}. Each curve is based on N = 600 ,000 samples. The dashed line shows the reference G(s) = s.
Interpretation. For a = 1 (the exponential case), the profile G(s) almost coincides with the line s, reflecting that the distribution of phases {log 10 X} is close to uniform. 48 For a = 0 .5, the deviation from s is small and has a form consistent with heavier tails (a greater share of mass at small x). For a = 3 , the deformation becomes pronounced and S-shaped: for small s the profile grows faster than the straight line and then slows, which is associated with the narrower shape of the Weibull density and the concentration of mass near a characteristic scale.
VIII Critical Discussion
This chapter offers a comprehensive critical analysis of digit laws, with a particular emphasis on Benford’s Law and the broader class of digit distributions. The discussion below synthesizes recent theoretical insights with foundational observations and highlights both the limitations and the universality claims often made in the literature.
On Leading Versus First Digits.
A key conceptual distinction must be drawn between “leading digit” and “first digit.” While “first digit” refers strictly to the initial nonzero digit of a number, “leading digit” can—depending on context—refer to a sequence of significant digits at the beginning of a number. In much of the Benford literature, “leading digit” is often treated as synonymous with “first digit.” Our approach clarifies that all leading digits, not just the first, are structurally interconnected by the positional nature of number systems. Understanding this connection is vital for a truly rigorous analysis of digit laws.
The Algebraic—not Universal—Nature of Benford’s Law.
Despite its frequent portrayal as a universal statistical law, Benford’s Law is fundamentally algebraic, rooted in the logarithmic and positional properties of number representations. Its apparent universality is a reflection of the decimal system, not an intrinsic law of nature.
Critique of Mixture and Invariance Arguments.
The claim that “random mixtures” of distributions inevitably produce Benford’s Law is circular: any mixture can be tailored to fit a target digit law, and such mixtures rarely arise spontaneously in practice. True universality is therefore not established by these arguments.
The Interdependence of All Significant Digits.
Focusing only on leading digits is an artificial limitation. All significant digits are mathematically linked, and a full understanding requires analyzing their joint distribution. Any separation between digits is a matter of historical convenience, not mathematical necessity.
Structural and Deterministic Explanations.
We argue for a shift from probabilistic “miracle” explanations toward explicit, structural, and combinatorial derivations of digit distributions. Many digit laws, including Benford’s, can be shown to arise naturally from the algebraic properties of sequences or products—no randomness is required.
The Limits of “Naturalness” and “Universality.”
There is no privileged or “natural” process that always yields Benford-type distributions. Many empirical datasets, especially those that are finite, bounded, or human-constructed, do not exhibit Benford’s Law. The observed “universality” is largely an artifact of particular mathematical constructions and dataset selections.
The Failure of Universality for Small Samples.
49 Benford’s Law and related digit laws do not generally hold for small or degenerate datasets. In these cases, digit frequencies are determined by the explicit structure of the data, not by any universal law.
Logical Consistency and Mathematical “Stitching.”
There are strict mathematical constraints linking the distributions of leading digits and those of longer digit blocks. Disregarding these links can lead to inconsistent “laws” that cannot be realized by any actual data.
A New Program: Structural and Algorithmic Theory.
We advocate for a new, fully structural theory of digit laws, based on explicit equations and combinatorics rather than probabilistic mixtures. This theory provides analytic tools for both theoretical and practical analysis, demystifies the phenomenon, and applies equally well to small samples as to large ones. In conclusion, the study of digit laws is enriched by recognizing the deep structural relationships among all significant digits, understanding the limitations of existing univer-sality claims, and pursuing explicit analytic descriptions. Our approach aims to clarify these relationships and lay the foundation for a more precise and universally applicable theory of digit distributions.
8.1 The Universal Function G(s) and Interval Filling
A fundamental observation is that, while the first (leading) digit can be seen as a discrete point—corresponding to a specific position between, say, 1 and 2—the inclusion of all subsequent digits fills out the entire interval. For example, if the leading digit marks a single point within the range from one to two, then the numbers with that leading digit (such as 10, 11, 12, etc.) densely populate the subinterval between those two points. As we consider additional digits, each successive digit acts to further subdivide and fill in the interval. Ultimately, the collective contribution of all digits leads to a dense covering of the entire curve, so that what begins as a discrete set of points (the nine possible leading digits) becomes, in the limit, a continuous distribution spanning the full range from 0 to 1. This property underlines the fact that the structure we study is not limited to a set of nine discrete points, but extends to a complete filling of the interval—a “curve” concretely determined by the arrangement and accumulation of all significant digits in the number system. Formally proving this property may require additional mathematical development, but the phenomenon is intuitively clear: the process of incorporating further digits results in an increasingly fine subdivision, eventually covering the entire line from zero to one.
Remark 2. A particularly clear and geometric criterion for the presence or absence of “Benford-like” properties in a dataset is provided by the G(s) function. The closer
G(s) lies to the diagonal from (0 , 0) to (1 , 1) , the more the digit frequencies resemble the logarithmic law. Significant deviations from this diagonal signal the absence of Benford-type behavior. In most empirical and mathematical distributions, the G(s) function is notably non-linear, which visually and quantitatively illustrates why traditional digit laws are far from universal.
Remark 3. A fascinating feature of leading digit distributions emerges when one examines the sequence of fractional parts {log 10 k} as k ranges over successive decades. For k =1, . . . , 9, these points occupy exactly nine distinct locations in [0 , 1) ; for k = 10 , . . . , 99 ,the new points interleave between those of the previous decade, and so on. Each successive
50 decade adds a layer of finer subdivision, filling the unit interval with increasingly dense and nested sets of points. This iterative, scale-dependent process naturally leads to
fractal-like , self-similar structure in the distribution of {log 10 k} as k increases. Such nested arrangements provide an intuitive geometric explanation for the gradual, structured approach to the limiting distribution, and underscore the deep connections between digital laws, number theory, and fractal geometry. Illustrative plots of {log 10 k} for growing k
(e.g., for k up to 10 , 100 , 1000 ) make this structure immediately visible.
Remark 4. A fundamental difference between the traditional “Benford paradigm” and the deterministic framework presented here lies in the direction of constraint. In Benford’s Law, the frequencies of the first nine digits (1–9) uniquely determine the entire digit profile: all subsequent digit probabilities are forced to lie precisely on the logarithmic Benford curve, regardless of the specifics of the data. In contrast, within the DLSD framework, the arrangement of significant digits beyond the first can be essentially arbitrary—each additional choice further shapes the global curve, rather than being dictated by a universal law. Thus, in the deterministic setting, the observed profile is a consequence of the underlying sequence, not an imposed template, and the digit distribution encodes the structural “fingerprint” of the data rather than conforming to a predetermined law.
Remark 5. This contrast can be illustrated even more vividly with an artificial example. Consider a distribution where the first nine significant digits are arranged to perfectly match the logarithmic probabilities typically associated with the Benford’s Law. In the traditional setting, this would force the entire digit law to coincide with the canonical curve, and every point in the digit profile would be dictated by this initial choice. However, in the deterministic DLSD framework, the points corresponding to the first nine digits (from 1 to 9) remain fixed, but all additional points (for blocks or numbers beyond the initial set, such as k = 10 , k = 21 , etc.) are free to be chosen in essentially any configuration. These further choices dynamically generate the shape of the global curve, rather than being consequences of a universal template. Thus, the overall profile is not prescribed from above; rather, it emerges “from below,” as the collective outcome of specific, possibly non-logarithmic, placements of additional digits. This principle not only underlines the philosophical difference between the traditional and deterministic approaches, but also explains how a wide variety of leading digit distribu-tions can arise, all sharing identical “first digit” statistics yet exhibiting diverse structures elsewhere in the digital profile.
8.2 Asymptotic Paradox: Structure Preserved in the Infinite Limit
It is widely believed—even in classic works by Benford himself—that as digit blocks become large, all empirical digit distributions must converge to Benford’s law in the infinite limit. However, this assertion is incorrect. Our analysis establishes, for the first time, the true structure of digit–block asymptotics and reveals that the limiting frequency of digit blocks retains an explicit analytic memory of the generating process—even as k → ∞ .
General asymptotic formula. For any analytic generating profile G(s), with s =
{log 10 k}, the correct asymptotic for interior k is
ρ(k) ∼ G′(s)
k ln 10 ,
51 where G(s) encodes the structural origin of the process generating the data.
Key examples.
• Benford case: For G(s) = s, G′(s) = 1 , so
ρ(k) ∼ 1
k ln 10 ,
which recovers the classic Benford law in the infinite limit.
• Linear–uniform window: For G(s) = 10 s − 19 , G′(s) = ln 10 9 10 s, and with n =
⌊log 10 k⌋,
ρ(k) ∼ 19 10 n ,
i.e., the block probability is constant within each decade, not decaying as 1/k .
Implication. Contrary to the folklore, the infinite limit does not erase the origin of the distribution . Instead, the asymptotic digit law preserves a precise analytic “finger-print” of its source , even for arbitrarily large digit blocks. The celebrated “universality” of Benford’s law is merely a degenerate special case within a much broader space of possible asymptotic behaviors. This resolves a long-standing paradox and provides a deterministic, structural explanation for the diversity of observed digit distributions in empirical data.
Remark (Originality): To our knowledge, this explicit and general asymptotic formula, and the recognition that structural memory persists at infinity, has not previously appeared in the literature. It is a principal original contribution of this work.
IX. Open Problems and Future Directions
The study of leading digit laws, as well as their deterministic and structural foundations, opens up a wide spectrum of unsolved problems and intriguing mathematical phenomena. While the present work has developed a modern deterministic framework and provided critical analysis of classical probabilistic approaches, it is clear that much remains to be discovered about the deeper structure and universal properties of digit distributions.
Potential Fractal Properties and Self-Similarity
One particularly stimulating direction for future research lies in exploring the possible fractal or self-similar characteristics of leading digit sets and their associated mappings. When mapping blocks of digits onto the unit interval using the fractional part of the logarithm, a fascinating "filling" process occurs: while the set of values {log 10 k} for single digits k = 1 , . . . , 9 occupies only isolated points, the expansion to larger digit blocks—such as numbers from 10 to 19 , 20 to 29 , etc.—progressively fills the interval [0 , 1) in a complex, seemingly hierarchical pattern. It would be valuable to formalize and rigorously analyze the distributional and geo-metric properties of these sets. In particular, questions arise such as:
• Does the set {log 10 n : n ∈ N} (modulo 1) possess a fractal or multifractal structure? 52 • How does the addition of more digits or larger blocks affect the density and uniformity of coverage of [0 , 1) ?
• Can we define and compute the fractal dimension (e.g., Hausdorff or Minkowski) of such sets?
• Are there natural self-similarities or scaling laws governing the "gaps" and "clusters" within these sets? Initial observations suggest that the recursive nature of digit expansion in positional systems may indeed encode self-similarity or even fractality, but a precise mathematical characterization remains an open challenge.
Further Unsolved Problems
In addition to the fractal questions, several other compelling problems emerge:
• Analytic Structure of Digital Profiles. How does the analytic behavior of the cumulative profile function g(s) depend on the underlying digit block size and base? What are the implications for the universality (or non-universality) of Benford-type laws?
• Generalization to Other Bases and Systems. How do the results extend or change for different number bases, non-integer bases, or even non-positional systems? Are there analogs of Benford’s Law with analogous deterministic frameworks?
• Limits of Deterministic Laws for Empirical Data. What are the boundaries of applicability for deterministic models, especially in noisy or highly non-uniform empirical datasets? Under what conditions do random or probabilistic models regain explanatory power?
• Connection to Equidistribution and Diophantine Approximation. How deeply are leading digit distributions linked to classical results in uniform distribution mod 1 and Diophantine approximation? Can deeper connections to ergodic theory or the theory of normal numbers be established?
• Computational Complexity and Algorithms. What is the computational com-plexity of simulating or testing for specific digital profiles? Can efficient algorithms be developed for empirical testing or data validation in real-world settings?
Synthetic and Specialized Sequences: Toward a Broader Digital Phenomenology
An exciting avenue for future research lies in the synthetic exploration of leading digit laws and digital phenomena for various nontrivial and structurally rich sequences—far beyond the traditional examples or random sampling from standard distributions. In particular, sequences such as primes, factorial numbers and they combinations, and other arithmetic or combinatorial constructions provide a rich landscape for discovering new and unexpected behaviors in digital laws. These specialized sequences often exhibit digit distributions with distinctive features: for example, persistent block structures, non-logarithmic profiles, or fractal-like patterns 53 in the distribution of significant digits. Unlike typical datasets arising from random processes, these deterministic or structured sequences may display highly non-generic behaviors, sometimes reflecting hidden algebraic or modular regularities. Understanding the mechanisms that give rise to such digit patterns, as well as de-veloping criteria for distinguishing structural digital phenomena from purely statistical artifacts, represents a major open problem for future research. In particular, it is of great interest to classify the range of possible leading digit distributions arising from deterministic recurrences, modular constructions, or algebraic constraints, and to explore the role of self-similarity, arithmetic “resonances,” and block structures in shaping these distributions. A systematic investigation in this direction would not only broaden our theoretical understanding of digital laws, but also yield practical methods for the synthesis and detection of digital patterns in both natural and artificial datasets. The continued study of specialized sequences thus holds promise for revealing deeper connections between number theory, combinatorics, and the phenomenology of significant digits.
Digital Laws for Classical and Synthetic Sequences: Literature and Perspectives
The digital properties of many classical sequences—such as primes, Fibonacci numbers, and factorials—have already been studied in depth. For instance, it is known that both the Fibonacci sequence and factorials asymptotically obey Benford’s Law, while the sequence of prime numbers, although showing some digital irregularities, has also been analyzed in this context [15, 22, 23]. Thus, while the most prominent classical examples are well understood, many open questions remain for less-explored or deliberately constructed ("synthetic") sequences, such as those built via modular arithmetic, rare digital patterns, or other non-standard rules (for example, like the Ulam sequence, defined recursively by summing the two smallest distinct previous terms in only one way). There is particular interest in studying whether these sequences exhibit non-classical digital behavior, possible fractal phenomena, or other novel features not captured by traditional Benford-type laws. A comprehensive exploration of such synthetic digital structures—both theoretically and computationally—remains a promising direction for future research.
Concluding Remarks
In conclusion, the landscape of leading digit laws and digital phenomena is far from being fully charted. The deterministic framework developed here provides a rigorous alternative to traditional probabilistic thinking, but at the same time, it exposes subtle geometric, analytic, and potentially fractal structures that warrant deeper investigation. Future research could lead to a more unified theory encompassing both deterministic and random digital phenomena, with applications ranging from theoretical mathematics to data science, cryptography, and beyond.
Epilogue: Synthesis
“Not everything that counts can be counted, and not everything that can be counted counts.”
54 — attributed to Albert Einstein After hundreds of pages tracing the hidden logic of numbers, distributions, and digital laws, one may wonder: does it all amount only to statistics and formulas, or is there a deeper message? The search for regularity amid apparent chaos, the skepticism toward “universal” laws, and the insistence on structure as the true signature of meaning—these themes have guided this book. Mathematics, at its heart, is not the worship of chance or a blind faith in empirical patterns, but the art of discerning structure where others see only randomness. The greatest challenge, and perhaps the deepest satisfaction, is to separate what can truly be explained from what merely appears “universal” by accident or incomplete perspective. To close, here is a brief acrostic—part reflection, part invitation—for all who continue this search.
Acrostic: STRUCTURE IS TRUTH
Some kneel before the altar of pure chance,
Their eyes shut tight in probability’s dance.
Real patterns — brushed aside as sin,
Untested claims they fold neatly in.
Circles of logic, wide and self-made,
The data conforms — once it’s betrayed.
Unseen by them, the digits sing,
Responding to structure, not random string.
Each claim of “law” dissolves in air —
It’s noise they praise, not truth laid bare.
Still, we persist, though fashion scorns,
Tracing the paths where meaning forms.
Rigorous steps, not mystic smoke,
Undo the myths the crowd invokes.
This is no trick, no hidden bluff —
Here, structure speaks. And that’s enough.
— Thaddeus of Linchester (c. 1594, or perhaps a myth) Structure is not always visible at first glance. But it is always there for those who seek—not by habit, but by proof.
55 References
Simon Newcomb. Note on the frequency of use of the different digits in natural numbers. American Journal of Mathematics , 4(1):39–40, 1881. Frank Benford. The law of anomalous numbers. Proceedings of the American Philosophical Society , 78(4):551–572, 1938. Ralph A. Raimi. The first digit problem. The American Mathematical Monthly ,83(7):521–538, 1976. Arno Berger and Theodore P. Hill. An Introduction to Benford’s Law . Princeton University Press, 2015. Mark J. Nigrini. Benford’s Law: Applications for Forensic Accounting, Auditing, and Fraud Detection . Wiley, 2012. Steven J. Miller, editor. Benford’s Law: Theory and Applications . Princeton University Press, 2015. Theodore P. Hill. A statistical derivation of the significant-digit law. Statistical Science , 10(4):354–363, 1995. Theodore P. Hill. Base-invariance implies benford’s law in every base. Proceedings of the American Mathematical Society , 123(3):887–895, 1995. Stephen M. Stigler. The distribution of first significant digits. American Statistical Association Journal , 40(236):67–73, 1945. R. W. Hamming. On the distribution of numbers. The American Mathematical Monthly , 77(5):531–540, 1970. Roger S. Pinkham. On the distribution of first significant digits. Annals of Mathe-matical Statistics , 32(4):1223–1230, 1961. William Feller. An Introduction to Probability Theory and Its Applications, Volume II . Wiley, 1966. Keith Devlin. Benford’s Law: The Surprising Pattern of Digit Distribution . Princeton University Press, 2008. Donald E. Knuth. The Art of Computer Programming, Volume 2: Seminumerical Algorithms . Addison-Wesley, 3rd edition, 1998. Persi Diaconis. The distribution of leading digits and uniform distribution mod 1.
The Annals of Probability , 5(1):72–81, 1977. Luciano Pietronero, Erio Tosatti, Valerio Tosatti, and Alessandro Vespignani. Ex-plaining the uneven distribution of numbers in nature: the laws of benford and zipf.
Physica A: Statistical Mechanics and its Applications , 293(1-2):297–304, 2001. János Aczél. Lectures on Functional Equations and Their Applications . Academic Press, 1966. 56 Marek Kuczma. An Introduction to the Theory of Functional Equations and In-equalities: Cauchy’s Equation and Jensen’s Inequality . Birkhäuser, 2nd edition, 2009. United Nations, Department of Economic and Social Affairs, Population Division. World population prospects 2024: Total population, both sexes combined (thousands) — variable id 12, 2024. UNdata / World Population Prospects 2024. Accessed 2025-08-02. Xue-Mei Su and Ying-Hui Wang. A note on the derivation of benford’s law by sum-integral replacement. Applied Mathematics Letters , 32:49–52, 2014. H. Weyl. Über die gleichverteilung von zahlen mod. eins. Mathematische Annalen ,77:313–352, 1916. Alex Kontorovich and Steven J. Miller. Benford’s law, values of l-functions, and the
3x + 1 problem. Acta Arithmetica , 163(4):361–377, 2014. Eric W. Weisstein. Benford’s law: Fibonacci numbers. MathWorld–A Wolfram Web Resource. 57 |
11910 | https://med.libretexts.org/Bookshelves/Nutrition/Book%3A_Nutrition_Science_and_Everyday_Application_(Callahan_Leonard_and_Powell)/05%3A_Lipids/5.05%3A_Digestion_and_Absorption_of_Lipids | 5.5: Digestion and Absorption of Lipids - Medicine LibreTexts
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5: Lipids
Nutrition Science and Everyday Application (Callahan, Leonard, and Powell)
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5.5: Digestion and Absorption of Lipids
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Contents
1. Home
2. Bookshelves
3. Nutrition
4. Nutrition Science and Everyday Application (Callahan, Leonard, and Powell)
5. 5: Lipids
6. 5.5: Digestion and Absorption of Lipids
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Nutrition Science and Everyday Application (Callahan, Leonard, and Powell)
Front Matter
1: Designing A Healthy Diet
2: Nutrition Science and Information Literacy
3: Molecules of Life- Photosynthesis, Digestion, and Metabolism
4: Carbohydrates
5: Lipids
6: Protein
7: Energy Balance and Healthy Body Weight
8: Vitamins and Minerals I
9: Vitamins and Minerals II
10: Nutrition and Physical Activity
11: Nutrition Throughout the Lifespan
Back Matter
5.5: Digestion and Absorption of Lipids
Last updated May 4, 2021
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5.4: Fatty Acid Types and Food Sources
5.6: Lipid Transport, Storage, and Utilization
Page ID 39972
Alice Callahan, Heather Leonard, & Tamberly Powell
Lane Community College via OpenOregon
( \newcommand{\kernel}{\mathrm{null}\,})
Table of contents
1. Lipid Digestion in the Mouth
2. Lipid Digestion in the Stomach
3. Lipid Digestion in the Small Intestine
4. Lipid Absorption from the Small Intestine
5. Special Adaptations for Lipid Digestion in Infants
1. References
2. Image Credits:
Lipid digestion and absorption pose some special challenges. Triglycerides are large molecules, and unlike
carbohydrates
and
proteins
, they’re not water-soluble. Because of this, they like to cluster together in large droplets when they’re in a watery environment like the digestive tract. The digestive process has to break those large droplets of fat into smaller droplets and then enzymatically digest lipid molecules using
enzymes
called lipases. The mouth and stomach play a small role in this process, but most enzymatic digestion of
lipids
happens in the small intestine. From there, the products of lipid digestion are absorbed into circulation and transported around the body, which again requires some special handling since
lipids
are not water-soluble and do not mix with the watery blood.
Let’s start at the beginning to learn more about the path of
lipids
through the digestive tract.
Lipid Digestion in the Mouth
A few things happen in the mouth that start the process of lipid digestion. Chewing mechanically breaks food into smaller particles and mixes them with saliva. An enzyme called _lingual
lipase_is produced by cells on the tongue (“lingual” means relating to the tongue) and begins some enzymatic digestion of triglycerides, cleaving individual
fatty acids
from the glycerol backbone.
Lipid Digestion in the Stomach
In the stomach, mixing and churning helps to disperse food particles and fat molecules. Cells in the stomach produce another
lipase
, called _gastric
lipase_ (“gastric” means relating to the stomach) that also contributes to enzymatic digestion of triglycerides. _Lingual
lipase_swallowed with food and saliva also remains active in the stomach. But together, these two lipases play only a minor role in fat digestion (except in the case of infants, as explained below), and most enzymatic digestion happens in the small intestine.
Figure 5.5.1: Overview of lipid digestion in the human gastrointestinal tract. (Copyright; author via source)
Lipid Digestion in the Small Intestine
As the stomach contents enter the small intestine, most of the dietary
lipids
are undigested and clustered in large droplets. Bile, which is made in the liver and stored in the gallbladder, is released into the duodenum, the first section of the small intestine.
Bile
salts have both a hydrophobic and a hydrophilic side, so they are attracted to both fats and water. This makes them effective
emulsifiers
, meaning that they break large fat globules into smaller droplets. Emulsification makes
lipids
more accessible to digestive
enzymes
by increasing the surface area for them to act (see Fig. 5.22 below).
The pancreas secretes pancreatic lipases into the small intestine to enzymatically digest triglycerides. Triglycerides are broken down to
fatty acids
,
monoglycerides
(glycerol backbone with one fatty acid still attached), and some free glycerol. Cholesterol and fat-soluble vitamins do not need to be enzymatically digested (see Fig. 5.22 below).
Lipid Absorption from the Small Intestine
Next, those products of fat digestion (
fatty acids
,
monoglycerides
, glycerol, cholesterol, and fat-soluble vitamins) need to enter into the circulation so that they can be used by cells around the body. Again,
bile
helps with this process.
Bile
salts cluster around the products of fat digestion to form structures called micelles, which help the fats get close enough to the microvilli of intestinal cells so that they can be absorbed. The products of fat digestion diffuse across the membrane of the intestinal cells, and
bile
salts are recycled back to do more work emulsifying fat and forming micelles.
Figure 5.5.1: Lipid digestion and absorption in the small intestine.(Copyright; author via source)
Once inside the intestinal cell, short- and medium-chain
fatty acids
and glycerol can be directly absorbed into the bloodstream, but larger
lipids
such as long-chain
fatty acids
,
monoglycerides
, fat-soluble vitamins, and cholesterol need help with absorption and transport to the bloodstream. Long-chain
fatty acids
and
monoglycerides
reassemble into triglycerides within the intestinal cell, and along with cholesterol and fat-soluble vitamins, are then incorporated into transport vehicles called chylomicrons. Chylomicrons are large structures with a core of triglycerides and cholesterol and an outer membrane made up of
phospholipids
, interspersed with
proteins
(called apolipoproteins) and cholesterol. This outer membrane makes them water-soluble so that they can travel in the aqueous environment of the body. Chylomicrons from the small intestine travel first into lymph vessels, which then deliver them to the bloodstream.
Chylomicrons are one type of lipoprotein—transport vehicles for
lipids
in blood and lymph. We’ll learn more about other types of lipoproteins on the next page.
Figure 5.5.1: Structure of a
chylomicron
. Cholesterol is not shown in this figure, but chylomicrons contain cholesterol in both the lipid core and embedded on the surface of the structure.(Copyright; author via source)
Special Adaptations for Lipid Digestion in Infants
Lipids
are an important part of an infant’s diet. Breast milk contains about 4 percent fat, similar to whole cow’s milk. Whether breastfed or formula-fed, fat provides about half of an infant’s calories, and it serves an important role in brain development. Yet, infants are born with low levels of
bile
and pancreatic enzyme secretion, which are essential contributors to lipid digestion in older children and adults. So, how do babies digest all of the fat in their diet?
Infants have a few special adaptations that allow them to digest fat effectively.First, they have plenty of lingual and gastric lipases right from birth. These
enzymes
play a much more important role in infants than they do in adults. Second, breast milk actually contains
lipase
enzymes
that are activated in the baby’s small intestine. In other words, the mother makes lipases and sends them in breast milk to help her baby digest the milk fats. Amazing, right?Between increased activity of lingual and gastric lipases and the lipases contained in breast milk, young infants can efficiently digest fat and reap its nutritional value for growth and brain development.Studies show that fat digestion is more efficient in premature infants fed breast milk compared with those fed formula. Even pasteurized breast milk, as is used when breast milk is donated for feeding babies in the hospital, is a little harder to digest, because heat denatures the lipases. (Infants can still digest pasteurized breast milk and formula; they’re just less efficient at doing so and absorb less of the products of triglyceride digestion.)1
Figure 5.5.1: Copy and Paste Caption here.(Copyright; author via source)
References
Lindshield, B. L. Kansas State University Human Nutrition (FNDH 400) Flexbook. goo.gl/vOAnR
OpenStax, Anatomy and Physiology. OpenStax CNX. Aug 28, 2019
University of Hawai‘i at Mānoa Food Science and Human Nutrition Program, “Digestion and Absorption of Lipids ,” CC BY-NC 4.0
1 American Academy of Pediatrics Committee on Nutrition, 2014. Chapter 2: Development of Gastrointestinal Function. In: Kleinman RE, Greer FR, eds. Pediatric Nutrition. 7th ed. Elk Grove Village, IL: American Academy of Pediatrics.
Image Credits:
“all eating ice cream” by salem elizabeth is licensed under CC BY 2.0
Figure 5.21. “Overview of lipid digestion” by Alice Callahan is licensed under CC BY 4.0; edited from “Digestive system diagram edit” by Mariana Ruiz, edited by Joaquim Alves Gaspar, Jmarchn is in the Public Domain
Figure 5.22. “Lipid digestion and absorption in the small intestine” by Alice Callahan is licensed under CC BY 4.0; edited from “Lipid Absorption” by OpenStax is licensed under CC BY 4.0
Figure 5.23. “Chylomicrons Contain Triglycerides Cholesterol Molecules and Other Lipids ” by OpenStax College, Anatomy & Physiology, Connexions Web site is licensed under CC BY 3.0
“IMGP1686” (breastfeeding baby) by Celeste Burke is licensed under CC BY 2.0
This page titled 5.5: Digestion and Absorption of Lipids is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by Alice Callahan, Heather Leonard, & Tamberly Powell (OpenOregon) .
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11911 | https://www.quora.com/How-does-one-find-the-largest-absolute-coefficient-of-a-binomial-is-the-middle-term-always-the-largest | How does one find the largest (absolute) coefficient of a binomial, is the middle term always the largest? - Quora
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5
How does one find the largest (absolute) coefficient of a binomial, is the middle term always the largest?
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Daniel Claydon
Learning mathematics · Author has 779 answers and 4.3M answer views
·6y
Yeah, the middle term is always the biggest. A simple proof is as follows:
The binomial coefficient (n k)(n k) is given by the formula
(n k)=n!k!(n−k)!(n k)=n!k!(n−k)!
A bit of algebra shows
(n k)=n−k+1 k(n k−1)(n k)=n−k+1 k(n k−1)
Hence, (n k)(n k) is greater than (n k−1)(n k−1) if, and only if, n−k+1 k>1 n−k+1 k>1. This happens exactly when n+1>2 k n+1>2 k. If n n is even, then the largest value k k for which this holds is n 2 n 2 so the binomial coefficients are increasing in magnitude up until (n n/2)(n n/2) at which point they start decreasing agai
Continue Reading
Yeah, the middle term is always the biggest. A simple proof is as follows:
The binomial coefficient (n k)(n k) is given by the formula
(n k)=n!k!(n−k)!(n k)=n!k!(n−k)!
A bit of algebra shows
(n k)=n−k+1 k(n k−1)(n k)=n−k+1 k(n k−1)
Hence, (n k)(n k) is greater than (n k−1)(n k−1) if, and only if, n−k+1 k>1 n−k+1 k>1. This happens exactly when n+1>2 k n+1>2 k. If n n is even, then the largest value k k for which this holds is n 2 n 2 so the binomial coefficients are increasing in magnitude up until (n n/2)(n n/2) at which point they start decreasing again. If n n is odd, we get essentially the same thing, except this time (2 N+1 N)(2 N+1 N) and (2 N+1 N+1)(2 N+1 N+1) are tied for top spot.
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More answers below
How do you find the greatest term in the binomial expansion?
Is the nth term our largest expansion in binomial distribution?
Is the coefficient of a middle term in a binomial expansion the greatest coefficient in the whole expansion?
How can I find the greatest element of a binomial coefficient?
What is the greatest binomial coefficient element?
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Sep 6
For (1 + x)^n expanded as ∑_{k=0}^n C(n,k) x^k the coefficients are the binomial coefficients C(n,k) = n!/(k!(n−k)!). The questions “which coefficient is largest?” and “is the middle term always largest?” have a clean, well-known answer:
Symmetry and unimodality
Exact largest index(s)
If n is even (n = 2m), the unique largest coefficient is the middle one: C(n,m).
If n is odd (n = 2m+1), there are two equal largest coefficients: C(n,m) and C(n,m+1) (they are equal because of symmetry and adjacency).
Reason (short):
Largest absolute value for (a + b)^n with real a,b
Practical rule to locate larges
Continue Reading
For (1 + x)^n expanded as ∑_{k=0}^n C(n,k) x^k the coefficients are the binomial coefficients C(n,k) = n!/(k!(n−k)!). The questions “which coefficient is largest?” and “is the middle term always largest?” have a clean, well-known answer:
Symmetry and unimodality
Exact largest index(s)
If n is even (n = 2m), the unique largest coefficient is the middle one: C(n,m).
If n is odd (n = 2m+1), there are two equal largest coefficients: C(n,m) and C(n,m+1) (they are equal because of symmetry and adjacency).
Reason (short):
Largest absolute value for (a + b)^n with real a,b
Practical rule to locate largest term for weighted binomial terms: find smallest k for which (n−k)|b| ≤ (k+1)|a|; the largest absolute term(s) are at that k (and possibly k−1) depending on equality.
Examples:
(1+x)^6: n even, largest is C(6,3)=20 (unique).
(1+x)^5: n odd, largest are C(5,2)=C(5,3)=10 (two equal middle terms).
(2+3x)^5: compare ratios with |b/a| = 3/2; the maximum shifts toward larger k; compute term magnitudes or use the consecutive-ratio test above to locate the peak.
Summary: For plain binomial coefficients C(n,k) the middle term(s) are indeed the largest: unique middle when n is even, two equal middle terms when n is odd. For (a+b)^n with unequal magnitudes of a and b, the largest absolute term shifts according to the ratio |b/a| and is found by the consecutive-ratio criterion.
Upvote ·
Lance Everett
Studied Nanoengineering&Mathematics at University of California, San Diego · Author has 1.1K answers and 1.5M answer views
·Updated 6y
A2A
Yes, the middle terms are always the largest. The proof I provide is very computational, I am sure there are nicer combinatorial proofs that do not use meromorphic extensions (sorry for the jargon) or calculus. For instance, one might show that the binomial coefficients are symmetrical and unimodal less directly. But alas I am not a mathematician, especially I am not a combinatorist, and even more especially, I am very lazy.
The binomial coefficients satisfy a recursive equation:
(n+1 k)=(n k)+(n k−1)(n+1 k)=(n k)+(n k−1)
Given that (n 0)=1(n 0)=1, this can be used
Continue Reading
A2A
Yes, the middle terms are always the largest. The proof I provide is very computational, I am sure there are nicer combinatorial proofs that do not use meromorphic extensions (sorry for the jargon) or calculus. For instance, one might show that the binomial coefficients are symmetrical and unimodal less directly. But alas I am not a mathematician, especially I am not a combinatorist, and even more especially, I am very lazy.
The binomial coefficients satisfy a recursive equation:
(n+1 k)=(n k)+(n k−1)(n+1 k)=(n k)+(n k−1)
Given that (n 0)=1(n 0)=1, this can be used to calculate the binomial coefficients, either by hand or by using a computer program. It essentially means you can use Pascal’s triangle.
Proving this formula is a matter of how the binomial coefficients are defined. For instance, we can define them equivalently algebraically, combinatorially, or analytically.
Lance Everett's answer to How do you prove(n+1 r)=(n r)+(n r−1)(n+1 r)=(n r)+(n r−1)?
It turns out that
(n k)=∞∏j=1(1+k(n−k)j(n+j))(n k)=∏j=1∞(1+k(n−k)j(n+j))
(See attached proof: Extension of Binomial Coefficients with Standard Finite Difference Calculus—a q-analogue of Euler’s Formula for the Factorial by Lance Everett)
Using this form, we can differentiate both sides with respect to k k, and find that the derivative vanishes only when k=n/2 k=n/2. This implies that the function achieves a maximum in the middle.
In the same post above I provide a formula that can be deduced from the analytic form:
For any positive integer M≥2 M≥2
(n k)=(n+M)n(k+M)−k(n−k+M)n−k(1+k(n−k)M(n+M))1 2−M exp∫M+n−k M+n(1 s+1 2 s 2)d s exp∫M+n−k M+n ψ(2)(s)d s M−1∏j=1(1+k(n−k)j(n+j))(n k)=(n+M)n(k+M)−k(n−k+M)n−k(1+k(n−k)M(n+M))1 2−M exp∫M+n M+n−k(1 s+1 2 s 2)d s exp∫M+n M+n−k ψ(2)(s)d s∏j=1 M−1(1+k(n−k)j(n+j))
where ψ(2)ψ(2) is a Polygamma function - Wikipedia
ψ(1)(s)=1 s+1 2 s 2+1 s∑k∈N B 2 k s 2 k ψ(1)(s)=1 s+1 2 s 2+1 s∑k∈N B 2 k s 2 k
ψ(2)(s)=D s ψ(1)(s)=−1 s 2−1 s 3−1 2 s 4+⋯ψ(2)(s)=D s ψ(1)(s)=−1 s 2−1 s 3−1 2 s 4+⋯
This can be used to provide an on-order estimation for the binomial coefficients without using recursion, since for sufficiently large M M, the factor with the integrals is approximately 1 1.
Upvote ·
9 2
Christopher Pellerito
Neither pays for, nor charges for, Quora content · Author has 6.3K answers and 12M answer views
·6y
For an even-powered binomial expansion - (a+b)^(2n) - then the largest coefficient will be for the nth term. For the case of, say, n=10 then the largest coefficient of (a+b)^20 would be (20C10) a^10 b^10. 20C10 = 20!/10!/10! = 184,756.
For an odd-powered binomial expansion - (a+b)^(2n+1) then the nth and (n+1)th terms will be tied for largest.
Upvote ·
9 2
Related questions
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Chandrima D
Studied at Manchester Business School · Author has 271 answers and 937.7K answer views
·6y
Hi, thanks for A2A! Yes, it’s always the binomial coefficient of the middle term. If n is even, it is (n n/2)(n n/2). If n is odd, then it is (n(n+1)/2)(n(n+1)/2) or (n(n−1)/2)(n(n−1)/2) (these two coefficients are supposed to be equal when n is odd). You can quickly get all the coefficients by using Pascal’s triangle.
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Jos van Kan
Knows a thing or two about mathematics · Author has 2.5K answers and 2.1M answer views
·6y
(n k)=n−k+1 k(n k−1)(n k)=n−k+1 k(n k−1)
The fraction is ≥1≥1 for 2 k≤n+1 2 k≤n+1 and that explains why the middle term(s) are always the largest.
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David Talbot
Eipiphiny Society Executive Director (Acting) (2005–present)
·6y
How does one find the largest (absolute) coefficient of a binomial [expansion]? Is the middle term always the largest?
Use Pascal’s Pyramid. Yes, the middle term is always the largest.
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Kwok Choy Yue
B.Sc in Mathematics, The Chinese University of Hong Kong (Graduated 1978) · Upvoted by
Chetna Kesarwani
, M.Sc. Mathematics, University of Allahabad (2024) · Author has 1.4K answers and 1.4M answer views
·4y
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If (1+5 x+6 x 2)10=a 0+a 1 x+...+a 20 x 20(1+5 x+6 x 2)10=a 0+a 1 x+...+a 20 x 20, then what is the value of numerically greatest coefficient?
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Maurice Dupre
Ph. D. in Mathematics, University of Pennsylvania (Graduated 1972) · Author has 1.6K answers and 452.2K answer views
·Jan 20
Related
What is a simple proof that Pascal's triangle gives us the correct coefficients for a binomial expansion, and that it always will?
QUESTION:
What is a simple proof that Pascal's triangle gives us the correct coefficients for a binomial expansion, and that it always will?
ANSWER:
Many combinatorial formulas have simple short proofs when looked at from the right point of view.
The relation of the binomial coefficients to Pascal’s triangle is a perfect example.
Suppose that n and k are non negative integers. Consider a set S of cardinality n, meaning S has exactly n distinct members and let 2^S denote the collection of all subsets of S. Let S(k) denote the collection of all subsets subsets of S having cardinality k. For any set T
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QUESTION:
What is a simple proof that Pascal's triangle gives us the correct coefficients for a binomial expansion, and that it always will?
ANSWER:
Many combinatorial formulas have simple short proofs when looked at from the right point of view.
The relation of the binomial coefficients to Pascal’s triangle is a perfect example.
Suppose that n and k are non negative integers. Consider a set S of cardinality n, meaning S has exactly n distinct members and let 2^S denote the collection of all subsets of S. Let S(k) denote the collection of all subsets subsets of S having cardinality k. For any set T, let card(T) denote its cardinality.
We set C(n,k)=card(S(k)).
Notice that C(n,k) is the number of ways to chose k things from a set of n things, so we customarily read C(n,k) as “n choose k”.
First notice that C(n,k)=C(n,n-k),
since obviously choosing k things from S is the same as deciding which n-k things to “leave behind”.
Alternately, if S\B denotes the relative complement of B in S for any sets B and S, then
h:S(k)→S(n-k) defined by h(B)=S\B is obviously a bijection, since the composition
hh=1:2^S→2^S,
the identity map on S.
This shows C(n,k)=C(n,n-k), for all n,k.
Now, for Pascal’s Triangle, it is simply formed as a triangular array of horizontal lines of numbers starting with 1 at the top and each successive line starting with 1 and staggered so each position in any line has two positions diagonally above it. The defining property of Pascal’s triangle is that each number is the sum of the two numbers diagonally above it. If n and k are non negative integers, then let T(n,k) denote the kth number in row n AFTER the initial 1. That is we start counting from zero, both for counting rows and for specifying positions in each row. Thus the Pascal Triangle defining property is the equation
T(n,k)+T(n,k+1)=T(n+1,k+1),
for all non negative integers n,k.
We can allow k to exceed n or be negative, if we start with the first line as having all zeroes except for a single 1 in the “middle” position.
Now, we can see that T(n,k)=C(n,k) for any integers n and k, since if n is negative, then there is no set of cardinality n and S(k) is empty for every integer k. Thus we see C(n,k)=0 if n is negative or if k is negative or if k exceeds n. And of course the same is true of T(n,k). Also, C(0,0)=1, since there is exactly one subset of the empty set having zero members, namely the empty set. For n=1, obviously C(1,0)=1 and C(1,1)=1. Thus C(n,k)=T(n,k) if n and k do not exceed 1. We now only need observe that
C(n+1,k+1)=C(n,k)+C(n,k+1), for all n,k.
To see this, let S have cardinality n and let S have cardinality n+1, with S\S={c}, S a subset of S.
If B is in S(k+1), then either c is in B or c is in S\B, so S(k+1) is a Disjoint union of sets C and D where
C={B in S(k+1)| c is in B}
and D={B in S(k+1)| c is in S\B}.
Thus C(n+1,k+1)=card(C)+card(D).
But, obviously, card(C)=C(n,k), because if B is in C, then B{c} is in S(k), so g:C→S(k) defined by
g(B)=B{c}, for all B in C is a bijection.
On the other hand D=S(k+1), obviously, so
card(D)=C(n,k+1).
This shows C(n+1,k+1), for all n,k.
Thus, as T(0,0)=1=C(0,0) and, it follows inductively, that C(n,k)=T(n,k), for all n,k.
For the binomial theorem it is best to begin with non commutative algebra. In mathematics, any string of symbols is called a word. If A is a set of allowed symbols for word formation we call A the alphabet. For instance, if card(A)=1, then there is exactly one word of length n for each non negative integer n.
Notice that if you have x,y in a non commutative algebra B then (x+y)^n is the sum of all possible words of length n using A={x,y}.
For instance
(x+y)(x+y)=xx+xy+yx+yy.
Assuming that w is any word of length n+1 from alphabet A={x,y}, then its initial letter is either an x or a y, so if u is the word formed by dropping the first letter from w, then either w=xu or w=yu, and u is a word of length n.
As (x+y)^[n+1]=x(x+y)^n+y(x+y)^n,
we easily see by induction that indeed
(x+y)^n=sum of all possible words of length n,
assuming A={x,y} is the alphabet.
Now, let W(n,k) be the set of words of length n which have x appearing exactly k times.
Therefore
(x+y)^n=Sum{w in W(n,k)|k at most n}.
Then W(0,0) has only the empty word so has cardinality one,
card(W(n,k))=C(n,k) for n=0=k.
Now, let S be the set of letter positions for a word of length n. If w is in W(n,k), then let B(w) be the set of positions which have x appearing. Then
B:W(n,k)→S(k)
is obviously a bijection
since once you fill all the x positions the word is determined by putting y in each remaining position.
Thus C(n,k)=card(W(n,k)), for all n,k.
Now let us put the commutative law of multiplication back into effect. Then
w=(x^k)(y^[n-k]), for each w in W(n,k).
Thus Sum{w|w in W(n,k)}=C(n,k)(x^k)(y^[n-k])
and therefore
(x+y)^n=Sum{C(n,k)(x^k)(y^[n-k])|k at most n},
for all n, which is the binomial formula.
This together with the previous argument showing T(n,k)=C(n,k), for all n,k shows that Pascal’s Triangle always gives the correct binomial coefficients. Notice that by taking x=1=y, we get
2^n=Sum(C(n,k)|k at most n}=card(2^S) for any set S with card(S)=n. Thus the sum of numbers in row n of Pascal’s Triangle is 2^n, for every non negative integer n.
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Hari Krishna Prasad Talluru
I love maths. · Author has 156 answers and 576.2K answer views
·8y
Related
Is the coefficient of a middle term in a binomial expansion the greatest coefficient in the whole expansion?
Yes.
You may heard of the Pascal triangle. If not you can follow the hyperlink provided.
The series of numbers in a row are the coefficients of the terms -in order- of a binomial expansion to the degree equal to the number of the row.
Let say number of row is 6. The numbers in the row 6 are 1, 6, 15, 20, 15, 6, 1. These are the coefficients of the terms in the expansion of any binomial raised to power 6.
If you observe, in pascal triangle, each series is obtained from it’s above series. I recommend to go to above given link for how the series is formed. If you see, the sum of the coefficients at t
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Yes.
You may heard of the Pascal triangle. If not you can follow the hyperlink provided.
The series of numbers in a row are the coefficients of the terms -in order- of a binomial expansion to the degree equal to the number of the row.
Let say number of row is 6. The numbers in the row 6 are 1, 6, 15, 20, 15, 6, 1. These are the coefficients of the terms in the expansion of any binomial raised to power 6.
If you observe, in pascal triangle, each series is obtained from it’s above series. I recommend to go to above given link for how the series is formed. If you see, the sum of the coefficients at the extreme contribute to the coefficients inside the series, thus making the inner coefficients always greater than the outside ones.
You can also observe that the middle number(s) in each series is/are the highest ones.
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9 4
Veronica Smith
B.A. in Mathematics&Economics, Washington University in St. Louis (Graduated 2022) · Author has 127 answers and 192.5K answer views
·7y
Related
Given binomial expression (5 x+11)23(5 x+11)23, what is the largest coefficient in the expression?
By the binomial theorem, the coefficient of x n x n is
a n=(23 n)5 n 11 23−n.a n=(23 n)5 n 11 23−n.
Thus, we have
a n+1 a n=(23 n+1)5 n+1 11 23−(n+1)(23 n)5 n 11 23−n=5(23−n)11(n+1).a n+1 a n=(23 n+1)5 n+1 11 23−(n+1)(23 n)5 n 11 23−n=5(23−n)11(n+1).
After doing a little algebra, we find that this ratio is greater than 1 1 for n≤6 n≤6 and less than 1 1 otherwise. What this means is that
a 0<a 1<…a 8>…>a 23.a 0<a 1<…a 8>…>a 23.
Thus, the largest coefficient is a 7=(23 7)5 7 11 16.a 7=(23 7)5 7 11 16.
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9 2
Anil Bapat
Lives in Mumbai, Maharashtra, India · Author has 2.8K answers and 3.8M answer views
·4y
Related
How do you find the greatest term in the binomial expansion?
How do you find the greatest term in the binomial expansion?
In case of Binomial Expansion, there are various possibilities, as discussed below.
If we are looking for the greatest term in (x+y)n(x+y)n. where the coefficients of both x and y are 1 and greatest term simply depends upon value of n and whether n is odd or even.
If n is even then the greatest term (as in the coefficient) is at \dfrac{(\dfrac{(
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How do you find the greatest term in the binomial expansion?
In case of Binomial Expansion, there are various possibilities, as discussed below.
If we are looking for the greatest term in (x+y)n(x+y)n. where the coefficients of both x and y are 1 and greatest term simply depends upon value of n and whether n is odd or even.
If n is even then the greatest term (as in the coefficient) is at (n+2)2(n+2)2 th term and it’s value is given by nC(n/2).
In n is odd then the greatest term (as in the coefficient) is at n−1 2 n−1 2 and also at n+1 2 n+1 2 and their values are given by nC(n-1)/2 and nC(n+1)/2, which incidentally are going to be the same.
I will take couple of examples to demonstrate as to how we can arrive at the greatest term.
Let the expression be (x+y) and let’s raise it to n, which is 8. In this case5th term would be having greatest term (coefficient), which would be 8C4, which is 70.
If for the same expression, if n is 9 then there would be 2 greatest terms having same coefficients and ...
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9 2
Philip Lloyd
Specialist Calculus Teacher, Motivator and Baroque Trumpet Soloist. · Author has 6.8K answers and 52.8M answer views
·Feb 2
Related
What is a simple proof that Pascal's triangle gives us the correct coefficients for a binomial expansion, and that it always will?
It is not a good idea to prove or memorise Pascal’s Triangle!
It is a much better idea to just remember a pattern which will work for any sized number.
Now just see how easy it is to expand a few terms of…
Continue Reading
It is not a good idea to prove or memorise Pascal’s Triangle!
It is a much better idea to just remember a pattern which will work for any sized number.
Now just see how easy it is to expand a few terms of…
Upvote ·
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11912 | https://phys.libretexts.org/Bookshelves/Electricity_and_Magnetism/Electromagnetics_and_Applications_(Staelin)/10%3A_Antennas_and_Radiation/10.04%3A_Antenna_arrays | Skip to main content
10.4: Antenna Arrays
Last updated
: Jun 7, 2025
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10.3: Antenna gain, effective area, and circuit properties
11: Common Antennas and Applications
Page ID
: 25032
David H. Staelin
Massachusetts Institute of Technology via MIT OpenCourseWare
( \newcommand{\kernel}{\mathrm{null}\,})
Two-dipole arrays
Although some communications services such as mobile phones use nearly omnidirectional electric or magnetic dipole antennas (short-dipole and loop antennas), most fixed services such as point-to-point, broadcast, and satellite services benefit from larger antenna gains. Also, some applications require rapid steering of the antenna beam from one point to another, or even the ability to observe or transmit in multiple narrow directions simultaneously. Antenna arrays with two or more dipoles can support all of these needs. Arrays of other types of antennas can similarly boost performance.
Since the effective area of an antenna, A(θ,), is simply related by (10.3.36) to antenna gain G(θ,), the gain of a dipole array fully characterizes its behavior, which is determined by the array current distribution. Sometimes some of the dipoles are simply mirrored images of others.
In every case the total radiated field is simply the superposition of the fields radiated by each contributing dipole in proportion to its strength, and delayed in proportion to its distance from the observer. For two-dipole arrays, the differential path length to the receiver can lead to reinforcement if the two sinusoidal waves are in phase, cancellation if they are 180o out of phase and equal, and intermediate strength otherwise.
It is convenient to represent the signals as phasors since the patterns are frequency dependent, so the total observed electric field , where is the observed contribution from short-dipole i, including its associated phase lag of radians due to distance traveled. Consider first the two-dipole array in Figure 10.4.1(a), where the dipoles are z-axis oriented, parallel, fed in phase, and spaced distance L apart laterally in the y direction. Any observer in the x-z plane separating the dipoles receives equal in-phase contributions from each dipole, thereby doubling the observed far-field and quadrupling the power intensity P [Wm-2] radiated in that direction θ relative to what would be transmitted by a single dipole.
The radiated power P(r,) in Figure 10.4.1 depends on the differential phase lag between the contributions from the two antennas. When the two dipoles are excited equally and are spaced L = λ/2 apart, the two rays add in phase everywhere in the x-z plane perpendicular to the array axis, but are λ/2 (180o) out of phase and cancel along the array (y) axis. The resulting G() is sketched in Figure 10.4.2(a) for the x-y plane. If L = λ as illustrated in Figure 10.4.2(b), then the two rays add in phase along both the x-z plane and the y axis, but cancel in the x-y plane at where the differential delay between the two rays is λ/2, as suggested by the right triangle in the figure.
Figure 10.4.2(c) illustrates how a non-symmetric pattern can be synthesized by exciting the two dipoles out of phase. In this case the lower dipole leads the upper dipole by 90 degrees, so that the total phase difference between the two rays propagating in the negative y direction is 180 degrees, producing cancellation; this phase difference is zero degrees for radiation in the +y direction, so the two rays add. Along the ±x axis the two rays are 90 degrees out of phase so the total E is greater than from a single dipole, and the intensity is doubled. When the two phasors are in phase the total E is doubled and the radiated intensity is 4 times that of a single dipole; thus the intensity radiated along the ±x axis is half that radiated along the +y axis. Figure 10.4.2(d) illustrates how a null-free pattern can be synthesized with non-equal excitation of the two dipoles. In this case the two dipoles are driven in phase so that the radiated phase difference is 180 degrees along the ±y axis due to the λ/2 separation of the dipoles. Nulls are avoided by exciting either dipole with a current that is ~40 percent of the other so that the ratio of maximum gain to minimum gain is ~[(1 + 0.4)/(1 - 0.4)]2 = 5.44, and the pattern is vaguely rectangular.
A mathematical expression for the gain pattern can also be derived. Superimposing (10.2.8) for and yields:
where we have used the identities and .
Example
If the two dipoles of Figure 10.4.1 are fed in phase and their separation is L = 2λ, at what angles in the x-y plane are there nulls and peaks in the gain G()? Are these peaks equal? Repeat this analysis for L = λ/4, assuming the voltage driving the dipole at y > 0 has a 90° phase lag relative to the other dipole.
Solution
Referring to Figure 10.4.1, there are nulls when the phase difference γ between the two rays arriving at the receiver is or 3, or equivalently, D = λ/2 or 3λ/2, respectively. This happens at the angles , and . There are also nulls, by symmetry, at angles 180° away, or at ≅ ±194° and ±229°. There are gain peaks when the two rays are in phase ( = 0° and 180°) and when they differ in phase by 2 or 4, which happens when and = ±210°, or when = ±90°, respectively. The gain peaks are equal because they all correspond to the two rays adding coherently with the same magnitudes. When L = λ/4 the two rays add in phase at = 90° along the +y axis because in that direction the phase lag balances the 90° delay suffered by the ray from the dipole on the -y axis. At = 270° these two 90-degree delays add rather than cancel, so the two rays cancel in that direction, producing a perfect null.
Array antennas with mirrors
One of the simplest ways to boost the gain of a short dipole antenna is to place a mirror behind it to as to reinforce the radiation in the desired forward direction and cancel it behind. Figure 10.4.3 illustrates how a short current element I placed near a perfectly conducting planar surface will behave as if the mirror were replaced by an image current an equal distance behind the mirror and pointed in the opposite direction parallel to the mirror but in the same direction normal to the mirror. The fields in front of the mirror are identical with and without the mirror if it is sufficiently large. Behind the mirror the fields approach zero, of course. Image currents and charges were discussed in Section 4.2.
Figure 10.4.3(a) illustrates a common way to boost the forward gain of a dipole antenna by placing it λ/4 in front of a planar mirror and parallel to it.
The image current is 180 degrees out of phase, so the λ/2 delay suffered by the image ray brings it into phase coherence with the direct ray, effectively doubling the far field Eff and quadrupling the intensity and gain Go relative to the absence of the mirror. In all directions more nearly parallel to the mirror the source and image are more nearly out of phase, so the gain in those directions is diminished relative to the absence of the mirror. The resulting antenna gain G(θ,) is sketched in Figure 10.4.3(b), and has no backlobes.
For the case where the dipole current and , the far-field in the forward direction is the sum of the contributions from and , as given by (10.4.1):
This expression reveals that the antenna pattern has no sidelobes and is pinched somewhat more in the θ direction than in the β direction (these directions are not orthogonal). An on-axis observer will receive a z-polarized signal.
Mirrors can also be parabolic and focus energy at infinity, as discussed further in Section 11.1. The sidelobe-free properties of this dipole-plus-mirror make it a good antenna feed for radiating energy toward much larger parabolic reflectors.
Example
Automobile antennas often are thin metal rods ~1-meter long positioned perpendicular to an approximately flat metal surface on the car; the rod and flat surface are electrically insulated from each other. The rod is commonly fed by a coaxial cable, the center conductor being attached to the base of the rod and the sheath being attached to the adjacent car body. Approximately what is the radiation resistance and pattern in the 1-MHz radio broadcast band, assuming the flat plate is infinite?
Solution
Figure 4.2.3 shows how the image of a current flowing perpendicular to a conducting plane flows in the same direction as the original current, so any current flowing in the rod has an image current that effectively doubles the length of this antenna. The wavelength at 1 MHz is ~300 meters, much longer than the antenna, so the shortdipole approximation applies and the current distribution on the rod and its image resembles that of Figure 10.2.3; thus deff ≅ 1 meter and the pattern above the metal plane is the top half of that illustrated in Figure 10.2.4. The radiation resistance of a normal short dipole antenna (10.3.16) is ohms for deff = 1 meter. Here, however, the total power radiated PT is half that radiated by a short dipole of length 2 meters because there is no power radiated below the conducting plane, so Rr = 0.0044 ohms. The finite size of an automobile effectively warps and shortens both the image current and the effective length of the dipole, although the antenna pattern for a straight current is always dipolar above the ground plane.
Element and array factors
The power radiated by dipole arrays depends on the directional characteristics of the individual dipole antennas as well as on their spacing relative to wavelength λ. For example, (10.4.3) can be generalized to N identically oriented but independently positioned and excited dipoles:
The element factor for the dipole array represents the behavior of a single element, assuming the individual elements are identically oriented. The array factor, , represents the effects of the relative strengths and placement of the elements. The distance between the observer and each element i of the array is ri, and the phase lag kri = 2ri/λ.
Consider the element factor in the x-y plane for the two z-oriented dipoles of Figure 10.4.4(a).
This element factor is constant, independent of , and has a circular pattern. The total antenna pattern is where the array factor controls the array antenna pattern in the x-y plane for these two dipoles. The resulting antenna pattern in the x-y plane is plotted in Figure 10.4.4(a) and (b) for the special cases L = λ/2 and L = λ, respectively. Both the array and element factors contribute to the pattern for this antenna in the x-z plane, narrowing its beamwidth (not illustrated).
Figure 10.4.4 illustrates a case where both the element and array factors are important; L = λ/2 here and the dipoles are fed 180o out of phase. In this case the out-of-phase signals from the two dipoles cancel everywhere in the x-y plane and add in phase along the z axis, corresponding to the array factor plotted in Figure 10.4.4(b) for the y-z plane. Note that when θ = 60o the two phasors are 45o out of phase and has half its peak value. The element factor in the y-z plane appears in Figure 10.4.4(c), and the dashed antenna pattern in Figure 10.4.4(d) shows the effects of both factors (only one of the four lobes is plotted). This antenna pattern is a figure of revolution about the z axis and resembles two wide rounded cones facing in opposite directions.
Example
What are the element and array factors for the two-dipole array for the first part of Example 10.4A?
Solution
From (10.4.5) the element factor for such dipoles is . The last factor of (10.4.5) is the array factor for such two-dipole arrays:
Uniform dipole arrays
Uniform dipole arrays consist of N identical dipole antennas equally spaced in a straight line. Their current excitation has equal magnitudes for all i, and a phase angle that uniformly increases by radians between adjacent dipoles. The fields radiated by the array can be determined using (10.4.5):
The z axis is defined by the orientation of the dipoles, which are all parallel to it. The simplest arrangement of the dipoles is along that same z axis, as in Figure 10.4.5, although (10.4.6) applies equally well if the dipoles are spaced in any arbitrary direction. Figure 10.4.1(a) illustrates the alternate case where two dipoles are spaced along the y axis, and Figure 10.4.2 shows the effects on the patterns.
Consider the N-element array for Figure 10.4.5(a). The principal difference between these two-dipole cases and N-element uniform arrays lies in the array factor:
The geometry illustrated in Figure 10.4.5(a) yields a phase difference of ( + ka cosθ) between the contributions from adjacent dipoles.
Using the two identities:
(10.4.7) becomes:
Since the element factor is independent of , the antenna gain has the form:
If the elements are excited in phase ( = 0), then the maximum gain is broadside with θ = 900, because only in that direction do all N rays add in perfect phase. In this case the first nulls θfirst null bounding the main beam occur when the numerator of (10.4.11) is zero, which happens when:
Note that the factor ka = 2a/λ is in units of radians, and therefore . If where θfn is the null angle measured from the x-y plane rather than from the z axis, then we have and:
The following simple geometric argument yields the same answer. Figure 10.4.5(b) shows that the first null of this 6-dipole array occurs when the rays from the first and fourth dipole element cancel, for then the rays from the second and fifth, and the third and sixth will also cancel. This total cancellation occurs when the delay between the first and fourth ray is λ/2, which corresponds to the angle .
The angle θfn between the beam axis and the first null is approximately the half-power beamwidth θB of an N-element antenna array. The antenna gain G(θ) associated with (10.4.11) for N = 6, = 0, and a = λ/2 is sketched in Figure 10.4.6(a), together with the squares of the array factor [from (10.4.10)] and element factor [from (10.4.6)]. In this case .
If ≠ 0 so that the excitation phase varies linearly across the array, then the main beam and the rest of the pattern is “squinted” or “scanned” to one side by angle . Since a phase delay of is equivalent to a path delay of , where = k, and since the distance between adjacent dipoles is a, it follows that adjacent rays for a scanned beam will be in phase at angle , where:
as sketched in Figure 10.4.6(b) for the case = 2 radians, a = λ/2, and ≅ 40o.
Note that larger element separations a can produce multiple main lobes separated by smaller ones. Additional main lobes appear when the argument ( + ka cosθ)/2 in the denominator of (10.4.11) is an integral multiple of so that the denominator is zero; the numerator is zero at the same angles, so the ratio is finite although large. To preclude multiple main lobes the spacing should be a < λ, or even a < λ/2 if the array is scanned.
Example
A uniform row of 100 x-oriented dipole antennas lies along the z axis with inter-dipole spacing a = 2λ. At what angles θ in the y-z plane is the gain maximum? See Figure 10.4.5 for the geometry, but note that the dipoles for our problem are x-oriented rather than z-oriented. What is the angle between the two nulls adjacent to θ ≅ /2? What is the gain difference G(dB) between the main lobe at θ = /2 and its immediately adjacent sidelobes? What difference in excitation phase between adjacent dipoles is required to scan these main lobes 10° to one side?
Solution
The gain is maximum when the rays from adjacent dipoles add in phase, and therefore all rays add in phase. This occurs at θ = 0, ± /2 , and ±sin−1 (λ/a) ≅ 30° [see Figure 10.4.5(b) for the approximate geometry, where we want a phase lag of λ to achieve a gain maximum]. The nulls nearest θ = /2 occur at that θfn when the rays from the first and 51st dipoles first cancel [see text after (10.4.13)], or when ; thus = 1/N radians ≅ 0.57o. The array factors for this problem and Figure 10.4.5(a) are the same, so (10.4.10) applies. Near θ ≅ /2 the element factor is approximately constant and can therefore be ignored because we seek only gain ratios. We define β≡/2 − θ so cos θ becomes sin β. Therefore (10.4.11) becomes where . , so . Similarly, so . Thus , and the first adjacent peak in gain occurs when , so . The numerator is unity when , or . Therefore , which is relative to the peak N2. A 10° scan angle requires the rays from adjacent dipoles to be in phase at that angle, and therefore the physical lag meters between the two rays must satisfy . The corresponding phase lag in the leading dipole is .
Phasor addition in array antennas
Phasor addition can be a useful tool for analyzing antennas. Consider the linear dipole array of Figure 10.4.5, which consists of N identical z-oriented dipole antennas spaced at distance a equally along the z-axis. In direction θ from the z axis the array factor is the sum of the phasors emitted from each dipole. Figure 10.4.6(a) shows this sum for the x-y plane (θ = 90°) when the dipoles are all excited in phase and N = 8. This yields the maximum possible gain for this antenna. As θ departs from 90° (broadside radiation) the phasors each rotate differently and add to form a progressively smaller sum . When the total phasor corresponds to = 5-degree lag for each successive contribution, then . Figure 10.4.6(b, c, and d) show the sum when is 45°, 72°, and 90°, respectively. The antenna gain is proportional to . Figures (b) and (d) correspond to radiation angles θ that yield nulls in the pattern , while (c) is near a local maximum in the antenna pattern. Because is , the gain of this sidelobe is ~0.04 times the maximum gain , or ~14 dB weaker.54 The spatial angles θ corresponding to (a) - (d) depend on the inter-dipole distance 'a'.
If a = 2λ, then angles θ from the z axis that correspond to phasor in Figure 10.4.7 (a) are 0°, 60°, 90°, 120°, and 180°; the peaks at 0 and 180° fall on the null of the element factor and can be ignored. The angle from the array axis is θ, and 'a' is the element spacing, as illustrated in Figure 10.4.5. The angles θ = 0° and 180° correspond to , , while θ = 60° and θ = 120° correspond to , and θ = 90° corresponds to ; the numerator in the argument of cos-1 is the lag distance in direction θ, and the denominator is the element spacing 'a'. Thus this antenna has three equal peaks in gain: θ = 60, 90, and 120°, together with numerous smaller sidelobes between those peaks.
54 dB ≡ 10 log10N, so N = 0.04 corresponds to ~-14 dB.
Four small sidelobes occur between the adjacent peaks at 60°, 90°, and 120°. The first sidelobe occurs in each case for ≅ 70° as illustrated in Figure 10.4.2(c), i.e., approximately half-way between the nulls at = 45° [Figure 10.4.2(b)] and = 90° [Figure 10.4.2(d)], and the second sidelobe occurs for ≅ 135°, between the nulls for = 90° [Figure 10.4.2(d)] and = 180° (not illustrated). Consider, for example, the broadside main lobe at θ = 90°; for this case = 0°. As θ decreases from 90° toward zero, increases toward 45°, where the first null occurs as shown in (b); the corresponding . The denominator 2λ in the argument is again the inter-element spacing. The first sidelobe occurs when ≅ 72° as shown in (c), and . The next null occurs at = 90° as shown in (d), and . The second sidelobe occurs for ≅ 135°, followed by a null when = 180°. The third and fourth sidelobes occur for ≅ 225° and 290° as the phasor patterns repeat in reverse sequence: (d) is followed by (c) and then (b) and (a) as θ continues to decline toward the second main lobe at θ = 60°. The entire gain pattern thus has three major peaks at 60, 90, and 120°, typically separated by four smaller sidelobes intervening between each major pair, and also grouped near θ = 0° and 180°.
Example
What is the gain GS of the first sidelobe of an n-element linear dipole array relative to the main lobe Go as n → ∞?
Solution
Referring to Figure 10.4.7(c), we see that as n→∞ the first sidelobe has an electric field EffS that is the diameter of the circle formed by the n phasors when is ~1.5 times the circumference of that circle, or . The ratio of the gains is therefore , or -13.5 dB.
Multi-beam antenna arrays
Some antenna arrays are connected so as to produce several independent beams oriented in different directions simultaneously; phased array radar antennas and cellular telephone base stations are common examples. When multiple antennas are used for reception, each can be filtered and amplified before they are added in as many different ways as desired. Sometimes these combinations are predetermined and fixed, and sometimes they are adjusted in real time to place nulls on sources of interference or to place maxima on transmitters of interest, or to do both.
The following cellular telephone example illustrates some of the design issues. The driving issue here is the serious limit to network capacity imposed by the limited bandwidth available at frequencies suitable for urban environments. The much broader spectrum available in the centimeter and millimeter-wave bands propagates primarily line-of-sight and is not very useful for mobile applications; lower frequencies that diffract well are used instead, although the available bandwidth is less. The solution is to reuse the same low frequencies multiple times, even within the same small geographic area. This is accomplished using array antennas that can have multiple inputs and outputs.
A typical face of a cellular base station antenna has 3 or 4 elements that radiate only into the forward half-space. They might also have a combining circuit that forms two or more desired beams. An alternate way to use these arrays based on switching is described later. Three such faces, such as those illustrated in Figure 10.4.8(a) with four elements spaced at 3λ, might be arranged in a triangle and produce two sets of antenna lobes, for example, the = 0 set and the = set indicated in (b) by filled and dashed lines, respectively.
As before, is the phase angle difference introduced between adjacent antenna elements. Interantenna separations of 3λ result in only 5 main lobes per face, because the two peaks in the plane of each face are approximately zero for typical element factors. Between each pair of peaks there are two small sidelobes, approximately 14 dB weaker as shown above.
These two sets ( = 0, = ) can share the same frequencies because digital communication techniques can tolerate overlapping signals if one is more than ~10-dB weaker. Since each face of the antenna can be connected simultaneously to two independent receivers and two independent transmitters, as many as six calls could simultaneously use the same frequency band, two per face. A single face would not normally simultaneously transmit and receive the same frequency, however. The lobe positions can also be scanned in angle by varying so as to fill any nulls. Designing such antennas to maximize frequency reuse requires care and should be tailored to the distribution of users within the local environment. In unobstructed environments there is no strong limit to the number of elements and independent beams that can be used per face, or to the degree of frequency reuse. Moreover, half the beams could be polarized one way, say right-circular or horizontal, and the other half could be polarized with the orthogonal polarization, thereby doubling again the number of possible users of the same frequencies. Polarization diversity works poorly for cellular phones, however, because users orient their dipole antennas as they wish.
In practice, most urban cellular towers do not currently phase their antennas as shown above because many environments suffer from severe multipath effects where reflected versions of the same signals arrive at the receiving tower from many angles with varying delays. The result is that at each antenna element the phasors arriving from different directions with different phases and amplitudes will add to produce a net signal amplitude that can be large or small. As a result one of the elements facing a particular direction may have a signal-to-interference ratio that is more than 10 dB stronger than another for this reason alone, even though the antenna elements are only a few wavelengths away in an obstacle-free local environment. Signals have different differential delays at different frequencies and therefore their peak summed values at each antenna element are frequency dependent. The antenna-use strategy in this case is to assign users to frequencies and single elements that are observed to be strong for that user, so that another user could be overlaid on the same frequency while using a different antenna element pointed in the same direction. The same frequency-reuse strategy also works when transmitting because of reciprocity.
That signal strengths are frequency dependent in multipath environments is easily seen by considering an antenna receiving both the direct line-of-sight signal with delay t1 and a reflected second signal with comparable strength and delay t2. If the differential lag c(t2 - t1) = nλ = D for integer n, then the two signals will add in phase and reinforce each other. If the lag D = (2n + 1)λ/2, then they will partially or completely cancel. If D = 10λ and the frequency f increases by 10 percent, then the lag measured in wavelengths will also change 10 percent as the sum makes a full peak-to-peak cycle with a null between. Thus the gap between frequency nulls is ~f = f(λ/D) = c/D Hz. The depth of the null depends on the relative magnitudes of the two rays that interfere. As the number of rays increases the frequency structure becomes more complex. This phenomenon of signals fading in frequency and time as paths and frequencies change is called multipath fading.
10.3: Antenna gain, effective area, and circuit properties
11: Common Antennas and Applications |
11913 | https://www.reddit.com/r/learnmath/comments/17dls5n/how_to_use_implies_and_if_and_only_if/ | How to use "implies" and "if and only if" : r/learnmath
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11914 | https://socialsci.libretexts.org/Bookshelves/Sociology/Introduction_to_Sociology/Sociology_(Boundless)/17%3A_Population_and_Urbanization/17.02%3A_Population_Growth/17.2E%3A_Demographic_Transition_Theory | Skip to main content
17.2E: Demographic Transition Theory
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: Feb 20, 2021
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17.2D: Malthus’ Theory of Population Growth
17.3: Urbanization and the Development of Cities
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( \newcommand{\kernel}{\mathrm{null}\,})
Learning Objectives
Break down the demographic transition model/theory into five recognizable stages based on how countries reach industrialization
Whether you believe that we are headed for environmental disaster and the end of human existence as we know it, or you think people will always adapt to changing circumstances, we can see clear patterns in population growth. Societies develop along a predictable continuum as they evolve from unindustrialized to postindustrial. Demographic transition theory (Caldwell and Caldwell 2006) suggests that future population growth will develop along a predictable four- or five-stage model.
Stage 1
In stage one, pre-industrial society, death rates and birth rates are high and roughly in balance. An example of this stage is the United States in the 1800s. All human populations are believed to have had this balance until the late 18th century, when this balance ended in Western Europe. In fact, growth rates were less than 0.05% at least since the Agricultural Revolution over 10,000 years ago.
Population growth is typically very slow in this stage, because the society is constrained by the available food supply; therefore, unless the society develops new technologies to increase food production (e.g. discovers new sources of food or achieves higher crop yields), any fluctuations in birth rates are soon matched by death rates.
Stage 2
In stage two, that of a developing country, the death rates drop rapidly due to improvements in food supply and sanitation, which increase life spans and reduce disease. Afghanistan is currently in this stage.
The improvements specific to food supply typically include selective breeding and crop rotation and farming techniques. Other improvements generally include access to technology, basic healthcare, and education. For example, numerous improvements in public health reduce mortality, especially childhood mortality. Prior to the mid-20th century, these improvements in public health were primarily in the areas of food handling, water supply, sewage, and personal hygiene. Another variable often cited is the increase in female literacy combined with public health education programs which emerged in the late 19th and early 20th centuries.
In Europe, the death rate decline started in the late 18th century in northwestern Europe and spread to the south and east over approximately the next 100 years. Without a corresponding fall in birth rates this produces an imbalance, and the countries in this stage experience a large increase in population.
Stage 3
In stage three, birth rates fall. Mexico’s population is at this stage. Birth rates decrease due to various fertility factors such as access to contraception, increases in wages, urbanization, a reduction in subsistence agriculture, an increase in the status and education of women, a reduction in the value of children’s work, an increase in parental investment in the education of children and other social changes. Population growth begins to level off. The birth rate decline in developed countries started in the late 19th century in northern Europe.
While improvements in contraception do play a role in birth rate decline, it should be noted that contraceptives were not generally available nor widely used in the 19th century and as a result likely did not play a significant role in the decline then.
It is important to note that birth rate decline is caused also by a transition in values; not just because of the availability of contraceptives.
Stage 4
During stage four there are both low birth rates and low death rates. Birth rates may drop to well below replacement level as has happened in countries like Germany, Italy, and Japan, leading to a shrinking population, a threat to many industries that rely on population growth. Sweden is considered to currently be in Stage 4. As the large group born during stage two ages, it creates an economic burden on the shrinking working population. Death rates may remain consistently low or increase slightly due to increases in lifestyle diseases due to low exercise levels and high obesity and an aging population in developed countries. By the late 20th century, birth rates and death rates in developed countries leveled off at lower rates.
Stage 5 (Debated)
Some scholars delineate a separate fifth stage of below-replacement fertility levels. Others hypothesize a different stage five involving an increase in fertility. The United Nations Population Fund (2008) categorizes nations as high-fertility, intermediate-fertility, or low-fertility. The United Nations (UN) anticipates the population growth will triple between 2011 and 2100 in high-fertility countries, which are currently concentrated in sub-Saharan Africa. For countries with intermediate fertility rates (the United States, India, and Mexico all fall into this category), growth is expected to be about 26 percent. And low-fertility countries like China, Australia, and most of Europe will actually see population declines of approximately 20 percent.
Conclusions
As with all models, this is an idealized picture of population change in these countries. The model is a generalization that applies to these countries as a group and may not accurately describe all individual cases. The extent to which it applies to less-developed societies today remains to be seen. Many countries such as China, Brazil and Thailand have passed through the Demographic Transition Model (DTM) very quickly due to fast social and economic change. Some countries, particularly African countries, appear to be stalled in the second stage due to stagnant development and the effect of AIDS.
Key Points
Demographic transition theory suggests that populations grow along a predictable five-stage model.
In stage 1, pre-industrial society, death rates and birth rates are high and roughly in balance, and population growth is typically very slow and constrained by the available food supply.
In stage 2, that of a developing country, the death rates drop rapidly due to improvements in food supply and sanitation, which increase life spans and reduce disease.
In stage 3, birth rates fall due to access to contraception, increases in wages, urbanization, increase in the status and education of women, and increase in investment in education. Population growth begins to level off.
In stage 4, birth rates and death rates are both low. The large group born during stage two ages and creates an economic burden on the shrinking working population.
In stage 5 (only some theorists acknowledge this stage—others recognize only four), fertility rates transition to either below-replacement or above-replacement.
Key Terms
demographic transition theory: Describes four stages of population growth, following patterns that connect birth and death rates with stages of industrial development.
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17.2D: Malthus’ Theory of Population Growth
17.3: Urbanization and the Development of Cities |
11915 | https://www.youtube.com/watch?v=-uQNJHkIxCk | How to Conduct a Two-Way ANOVA in JASP
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Posted: 29 Oct 2023
In this video, you'll learn how to use JASP to conduct a Two-Way ANOVA. This corresponds with Chapter 7 of our textbook Applied Data Analysis in Psychology: Exploring Diversity with Statistics; however, you do not need to have the textbook to follow along with this walk-through.
Here is the link to the study (Zelin et al., 2018) referenced in the walk-through:
Black, K. J., Madden, J., Osborn, H., & Walker, R. V. (2021). Applied data analysis in psychology: Exploring diversity with statistics. Kendall Hunt.
Check can also check out my FREE companion open education resource here, it includes Two-Way ANOVA data set and a walk-through guide with step-by-step instructions for conducting a Two-Way ANOVA in JASP. It also includes information about how to write up study results in APA format.
This video was created by Dr. Ruthie Walker, Assistant Professor of Psychology at The University of Tennessee at Chattanooga.
Click here to download JASP statistical software to analyze your data here:
5 comments
Transcript:
In this video, I'm going to show you how to
conduct a two way ANOVA or factorial ANOVA. In JASP. First we're going to talk about the
data set we're going to be working with for our walkthrough. And then we're going to discuss,
how do you check the data to make sure that it meets the assumptions that we need it to
meet in order to run our analyses. And then we're going to get into JASP and walk through
how to analyze and interpret the test results. So a couple things that I want to go through,
when we're conducting a two way ANOVA. We're extending our comparison beyond just one factor
like we did when we were working with one way ANOVA us with a two way ANOVA, we're able
to simultaneously examine the influence of two independent variables or factors on the
means of our groups. It allows us the ability to explore not only the individual effects
of each factor, or what we'll call our main effects, but also whether the effects of one
variable depend on the level of another variable. And we'll call that our interaction. Let's quickly go over an example of a non
significant and a significant interaction. When there's a significant interaction, the
effects of one variable may depend on the level of another variable. Graphically, we're
going to see differences and what that looks like. So we will see crossed our nonparallel
patterns on our graph if we have a significant interaction. Alternatively, a non significant interaction
suggests that the effects of each variable are somewhat independent, and they don't depend
on the other. In this case, we'll see bars or lines that appear more parallel. Looking
at the charts from a study published in plus one by Maria Ortiz and colleagues in 2014.
The graph on the left shows levels of the dependent variable for the control group for
the diabetic group when they receive a treatment or when they don't, the bars for the control
group are higher than the diabetic group in both the untreated and the treated groups.
And the graph on the right we can see that the bars are not parallel. Although the control
group was higher than the diabetic group. When untreated, a diabetic group was higher
in the deepen a variable when treated, this is what we would expect to see when we have
a significant interaction. For our JASP walkthrough, we'll be analyzing
data published by Aleksandra Zelin, myself and Dawn Johnson in 2018, we conducted a series
of studies to determine if people evaluated victims of sexual harassment in a public location.
And this case, we were looking at a bar differently depending on a couple different factors. So
we're only going to be analyzing a portion of our overall study, we ran a total of two
different studies. So if you're interested in this, you can see our publication to learn
more, and I will make sure to include a link to the publication and the video description.
So paths, researchers have found that people are more likely to help someone who's a friend
compared to a stranger when they're sexually harassed or assaulted. So what we wanted to
do is learn more about a more ambiguous category of relationship and acquaintance. Someone
you might know from class, a neighbor, a friend of a friend, etc. Paths, researchers have
also found that alcohol intake impacts perceptions of victims. So we wanted to manipulate both
the relationship our participants had with the victim, meaning they were to read a vignette
and imagine the victim was either their friend, an acquaintance or a stranger. And we also
manipulated the alcohol intake of the victim. They were either depicted as intoxicated or
sober. For first factor relationship with the victim had, we have three levels stranger
acquaintance of friend, our second factor, victim alcohol intake had two levels, drunk
and sober. So let's get to know the data that we're working
with. We recruited a sample of 247 undergraduate students from introductory classes in psychology,
students were randomly assigned to read one of six vignettes. After reading the vignettes,
students were asked to answer questions about their intention to help the victim if they
had witnessed what had happened, how much they blamed the victim and how much empathy
they had for the victim. So for our study today that we're going to focus on the participant
ratings of victim empathy. For a walkthrough, we'll be testing three
different hypotheses. For the first hypothesis, we'll be asking if there's a difference in
participant mean ratings of empathy for the victim by the victims level of intoxication,
so drunk versus sober in the vignette that they read. For the second hypothesis, we'll
be asking if there's a difference in participants mean ratings of empathy for the victim by
their relationship with the victim so stranger acquaintance friend and the vignette they
read. And for the third hypothesis, we'll be asking if there's an interaction between
victim alcohol intake and relationship with the victim on participant ratings of victim
empathy. So what are our null hypotheses with two way
ANOVA us this is the tricky part. We have three sets of hypotheses Use hypothesis one, we're looking at our very
first factor, which is level of victim intoxication. Empathy for the victim does not significantly
vary by level of victim intoxication. Looking at hypothesis one, we're talking about
the main effect of factor one, we want to see across both rows are participants who
were sorted into sober conditions different in their levels of reported empathy, then
participants that were sorted into drunk conditions. So we are seeing is there a main effect for
level of victim intoxication on these levels of victim empathy. So hypothesis to empathy
for the victim does not significantly vary by relationship status with the victim. For
this hypothesis, we're looking across our columns, we want to see is there a main effect
for relationship status. So we want to see our participants who were randomly assigned
to the stranger condition? Do they have significantly different levels of victim empathy, then participants
that were sorted into the acquaintance condition versus the friend condition. And our third
hypothesis, this hypothesis is unique to our two way ANOVA as in that we're talking about
an interaction. So our third null hypothesis is there is no significant interaction between
relationship status and level of intoxication on empathy for the victim. So before running a two way ANOVA we have
to make sure our data are appropriate for this analysis. So to do that, we have to meet
six different assumptions. These assumptions are the same ones that we've looked at with
all independent factors t test as well as the one way ANOVA. So we're going to go through
each of these if you don't need to know how to conduct a statistical assumptions and check
for those, please feel free to skip ahead in this video. Our first assumption is that our independent
variable is categorical, meaning that our independent variable is separated into multiple
categories that are nominal and measure. We meet this assumption both of our factors level
of alcohol intake and relationship status with the victim are categorical and measured
at the nominal level. Our second assumption is that our dependent
variable level of empathy for the victim must be continuous or measured at the interval
or ratio level, participants responded to two items measuring their level of empathy
for the victim and the scenario. One item was, I feel sorry for the woman in the scenario,
participants rated each item on a seven point Likert scale from one strongly disagree to
seven strongly agree. In psychology, we generally Tea Tree Likert scale data is continuous.
Thus, we meet this assumption. Or third assumption is that participants and
all six groups are different or unrelated to each other. In this example, we need participants
in each cell or group to be completely independent or different from the people and all of the
other groups that are study conditions. So no participant can be in more than one experimental
group. What I mean by that is someone who's sorted into that drunk stranger group and
reads the vignette that has a victim that is drunk, but as a stranger to them, they
cannot also be in the group and have also read the vignette for the sober acquaintance.
So we need everyone to be completely separate. And for this study, we made this assumption.
So in order to check our last three statistical assumptions, we need to move into jazz. So
we're going to check in jazz to see if we have normal distributions, and each of our
six groups, no outliers and homogeneity of variances. So let's move into JASP and analyze
the data. Okay, so now that we're in jazz, let's look
at our data and get to know it a little bit. So we have a couple more columns here, then
we're actually going to be working with so you can see we have some demographic information
for age and sex that we're not going to look at. And we've got our first IV IV, one, our
first factor IV, two and our dependent variable empathy. So you can see that empathy is here.
It's measured at the interval or ratio level, as indicated by this ruler. And what we're
going to do now is we're just going to change the labels for our factors just to make our
lives just a little bit easier when we go to interpret the data. So go ahead and double
click here. And for label one, we're going to say low alcohol intake. And for label two, we're going to change this
to high alcohol intake. And then we can go ahead and close click on
the red X to close the window and then double click here. are on IV, two. And we're going to change
each of these labels as well. So we have stranger, here, we're going to change to two acquaintance.
And then we're going to change three here to friend. And what you want to do is always click on
one of the other rows, just to make sure that Jasper has got that then click on the red
arrow. So now if we scroll down, we can see everything is labeled just how we want it
to be. So what we're going to do is check for our remaining assumptions. However, with
how Jasper is set up, one of the things that we want to do is we need to do some filtering.
So we're going to double click on our IV, one again. And we're going to ask it to filter
out high alcohol intake. And we're only going to look across IV to stranger friend acquaintance,
four levels of low alcohol intake. So go ahead and filter that out, close the window. Now
we're going to go up to descriptives. And just to show you what that looks like, sorry,
real quick, you can see we filtered out high alcohol intake. So all the participants that
were sorted into that condition will not be a part of this next set of descriptive statistics.
So click on descriptives, we're going to move empathy, our dependent variable over here
to the Variables box. And we're going to move our second factor over here to the split box.
And so what you can see is that our descriptive statistics come up for stranger acquaintance
and friend. And this is all for people who are sorted into that low alcohol intake condition.
So we're going to click on statistics, click on skewness and kurtosis. That's going to
give us our data, our normality statistics that we want to look at. And then we're also
going to click on customizable plots, we want to get box plots, and we want them to label
outliers. This is going to allow us to see for these three conditions. Did we have any
issues in terms of normality? And did we have any issues in terms of outliers. So now that
we have those clicked, let's go ahead and hide that. So we can look at this. So if we
look at our box plot, and we check for outliers, I don't see anything. So we don't have any
variables on the outside of these whiskers and the stranger condition and then acquaintance
condition or the friend condition. So we're good so far. And then here for descriptive
statistics, we're looking at skewness and kurtosis values, we want those to be between
negative two and a positive two, the closer they are to zero, the, the less skewed they
are and so that's what we're hoping for. So we have negative 1.31, negative point six,
four, negative 1.05, when we round it, so all of those are within that negative two
to positive two, we're happy with them for kurtosis, we want the same thing. So we have
a point four, two, negative point one, six and point to one, seven. So all of those look
good. So for these three conditions for stranger acquaintance and friend for low alcohol intake,
are outliers in our data normality are all good. So what we're going to do now is we're
just going to go back to our data. And we're going to click on our first factor again.
And we're going to put high alcohol intake back in and we're going to filter out the
low alcohol intake now. So now we're going to look at the other three conditions that
we didn't look at the first time to see are their levels of skewness and kurtosis. And
their box plots do they all look good. So once you have that checked, click on the red
X, we're going to go back over here, you can see that what when you update that filtering
information just automatically updates our outputs. So we are good here. So I'm just
going to click on these three lines here to drag this out. So looking at our box plots,
we're looking for outliers, you can see and our stranger condition, we are good and our
acquaintance condition, we're good. And our friend condition. We have some outliers down
here. So you can see participant number 208, participant 242 and two participant 227 All
had outliers. And these are all participants that rated their level of empathy for their
friend in the vignette as being much lower than the other participants did. So we do
have some outliers in our high alcohol intake friend condition. And let's go up here to
our skewness and kurtosis values. So again, we want them between negative two and positive
two. So skewness, negative point eight, nine, negative point two, four a negative 1.488.
Those are all within those bounds, so we're happy for kurtosis we have negative point
two seven negative point 818 and 3.683. So here and that friend condition, we have a
kurtosis value that's outside of that negative two and positive two balance. That's a bummer,
because that's a high kurtosis value. And one of the things to be aware of is that oftentimes,
these kurtosis values, if we have values that are significantly high, it's because of outliers,
these outliers can very much impact those values. So what we had done is I created another
data set that had the outliers removed, so we could look at them there. So let's go ahead
and open up that outlier free data set and rerun this and see what that looks like. Okay,
so this is our other data set, this has those outliers removed, and you can see the labels
already taken care of for our factors. So let's go to our first factor. And we're going
to filter out the low alcohol intake. So we can look specifically at that high alcohol
intake condition and see if our kurtosis values have been fixed. So let's go up here to descriptives,
going to click this arrow and hide the data, move empathy over to the Variables box, we're
going to move IV to into our split box. So I'm going to click on statistics and skewness,
and kurtosis. And then I'm going to click on customizable plots, click on box plots,
and label outliers. So let's look at our new output here. So you can see this has definitely
been cleaned up. And we have stranger friend, acquaintance and friend. And we no longer
have any outliers on these box plots. So let's take a look at our kurtosis values. So point
two, seven, negative point 881. And then our friend condition negative point 585. So that's
much different and much better than what we had with the three over level of a three.
So this is a lot better. The question is, is ultimately what our results for Nova been
different? Well, let's go ahead and run the results with this cleaned up data set without
the outliers because the kurtosis has been fixed. And then we'll come back to our original
data set and compare them to see if those are different. So for right now, let's go
back here. And we're going to double click on IV, one, we want to bring back all of our
participants. So we want to have all of the different conditions here open. So click on
the excellent satis done. And now we're going to go up to ANOVA. So click on ANOVA. And
then click on the first one ANOVA again. And what we're going to do is our last assumption
is the homogeneity of variances, we will analyze that here. So let's put our dependent variable
empathy in our dependent variable box. And we're going to move both of our factors here
into our fixed factors box. So a couple things that we want to look at, we want to look at
descriptive statistics, those are always helpful to have, we're also going to ask for estimates
of effect size, you can see there are three different ones here, just like with a one
way ANOVA. It defaults to the ADA squared, and that's perfectly fine. If you want to
have omega squared, it's a little bit more of a conservative somewhat smaller, oftentimes,
estimate, you can do that there's also a partial ADA squared, we're just gonna go ahead and
look at Ada squared. And so a couple other things that we want to look at here you can
see there are lots and lots of different things we can click on. Let's first click on assumption
checks, because we need to look at those homogeneity tests, specifically the Levine's test, because
we want to check that remaining assumption that we have. So let's look at that. So assumption
checks, test for equality of variances here. As a reminder, we want that P value to be
greater than point oh five, that means we meet this assumption. In our case, it's a
bummer that P value is less than point 001. That means we don't meet this assumption,
we don't have equality of variances. And when we were working with the one way ANOVA, or
the independent samples t test, we could do a correction for that right? With a one way
ANOVA we just clicked on that Welch button right here and it would correct for it. That's
only available with one way ANOVA it doesn't work with a two way ANOVA with a two way ANOVA
we can feel somewhat comfortable with running our analyses anyways, even if we don't meet
this if we have equal numbers of participants in all our groups. So we want to have them
as equal as possible. So looking At this descriptive table right up here, this is telling us our
end size and all of our groups. So in our acquaintance, high alcohol intake group, we
have 42. And our high alcohol friend group, we have 38. And our high alcohol stranger
group, we have 42. And then the low alcohol intake across all three of those relationship
types, we have 40. So we have approximately equal numbers of people in all of our groups.
So even though we don't need that equality of variances, assumption, we have kind of
a backup almost equal numbers of people in the groups to fall back on to tell us that
we can go ahead and keep going. So another thing that we're going to do is we're just
going to first analyze the results of these, our omnibus test and see what's going on there
before we do anything else here. So with our ANOVA, we have three separate hypotheses that
we're testing for, right? So our IV, one or a first factor, this is level of alcohol intake.
So we have our sum of squares, our degrees of freedom, we have our Mean Square our F
value, our P value and our ADA squared effect size. So what we're looking at for all three
of these is we want to see is this p value less than our alpha level of point oh five.
So in this case, our P value for levels alcohol intake is point 872. That's a pretty big probability
value, it is definitely not less than our alpha level of point oh five. So we do not have a main effect for level of alcohol intake,
there is not a significant difference between the people who were sorted into low alcohol
intake and high alcohol intake, we don't get to reject this null hypothesis, it's not significant.
And our second IV, two, we're looking at the level of relationship status. So here we can
see this is our f valued, and this is our p value. So our P value is point 104. That
is also larger than our alpha level of point oh five. So again, we're going to fail to
reject the null hypothesis there, we do not have a main effect for relationship status.
There's no difference between the friend group, the acquaintance group and the stranger group
on these levels of empathy for the victim. And then the last one, you can see it here
has that little asterix, so IV, one times IV, two, this is our interaction term, this
is the whole reason we run this analysis for the most part, we really care to see does
our first factor and our second factor interact together to form something unique with our
dependent variable, so our F value here 5.427, and our P value is point 005. So point 005,
is less than our alpha level of point oh five, so we're going to reject the null hypothesis.
So our third hypothesis, that there is an interaction between our first factor level
of alcohol intake and our second factor relationship status is true, we have a significant interaction
between those two factors on participant ratings of empathy for the victim in the scenario.
And when we look at our ADA squared value, we can see that the effect size is small,
it's point o four, but we have a statistically significant effect. And practically, it looks
like that effect is small. So our next step here is we need to see what this means. Okay,
so one of the ways that I like to kind of visualize and try to wrap my head around these
interactions is to pull up a descriptive plot. So we're gonna go down here under descriptive
plots. And what we're going to do is we're going to move IV to our second factor relationship
status to this horizontal axis, that's going to be our X axis. And then we're going to
move IV one over here to the separate lines, that's level of alcohol intake. And then you
want you want to make sure that the display error bars here is clicked. And then let's
go ahead and look at this. So when we do this, we're gonna go ahead and pull this make it
a little bit bigger, easier to see. And when you can see with our key over here, so low
alcohol intake has the open circles, that's this line here. And high alcohol intake has
the close circles, the dark ones. So when we were talking about look Going out and interpreting these descriptive
plots. Remember, we always have to go to our P values and our statistics to verify anything
that we're seeing here visually. But this can help you get a picture of what's happening
with the data. And then we can go to our tests our statistical test, to back that up, or
to see if it's not confirmed. So what we want to see is when we see a lot of overlap between
the error bars, as we talked about, with the one way ANOVA, as we're not likely to see
a statistical difference. So in this friend condition, you can see here the means levels
of empathy. So remember, our y axis is the level of empathy over here, mean level of
empathy is really close to each other tons of overlap between the low alcohol intake
condition and the high alcohol intake condition, you can see the high alcohol intake condition
does have less error, but not likely to see any statistical differences between these
two groups. And when we're going over here to this acquaintance condition, you can see
the means are a little bit of ways but we we have a little bit overlap here. So that
one are not quite sure, you can see that the high alcohol intake, the level of empathy
was a little bit higher than the low alcohol intake. And then here in the stranger condition,
we still have a little bit of overlap between our low alcohol intake condition and our high
alcohol intake condition. So just looking at this, I would definitely say there's nothing
going on over here in this Frank condition. But our interaction is likely due to this
difference happening here between our stranger and acquaintance, that level of empathy really
dips here for the acquaintance and that low alcohol intake condition. So what we need
to do is we need to run post hoc tests. So this was our omnibus test here, this tells
us is there anything happening with the data, we have a significant interaction. So to probe
that interaction does and to see what is happening, what we have to do is run simple main effects.
To do that, we're going to come over here to simple main effects. And what we're going
to do is we're going to do it two different ways. So the first way, we're going to put
in our first factor, simple effect factor one, and we're going to put our second factor
into the moderator factor one. And let's go ahead and look at those results. And then
we're going to come back and we'll switch them. So first, let's go down here. So, what
we are seeing here you can see level of our second factor, Stranger acquaintance and friend.
So for each of these, what this is doing is this is comparing our drunk versus sober essentially,
for the stranger condition. This is comparing drunk versus sober for the acquaintance condition.
And this is a comparing drunk versus sober for the friend condition. So we are going
to pay attention to our P values over here, we want to see are they less than our alpha
level a point oh five, or are they greater? If they're less than point o five, then we
would say there is a significant difference between the low alcohol intake and high alcohol intake
for that level of relationship. So for stranger here, we have a p value of point oh, one six,
that's significant. That is less than our alpha level of point oh, five. So what that
tells us is for this stranger condition right here, there's a significant difference in
the level of empathy between our high alcohol intake condition and our low alcohol intake
condition. So this, these are significantly different from each other here. Our next one
is giving it to us for the acquaintance condition. So our P value is point o Three, two, it is still less than that alpha level point
oh five, so we're going to reject the idea that these are not different from each other
and say that they are so in this acquaintance condition here, what we're seeing is that
the low alcohol intake condition here is significantly different than the high alcohol intake condition
here. So these two are significantly different from each other. And the last one is our friend
condition. So here we have a p value of point, eight 7.587. Sorry, and this is definitely
larger than our alpha level of point oh five. So there is not a difference between the low
alcohol and the high alcohol intake on this level of empathy. So these two right here
not different from each other, as we have suspected. So with simple main effects when
we have these two factors, we want to Run this both ways. So we just looked and
we compared low alcohol and high alcohol intake. So let's switch this up. So what we're going
to do now is we're going to you can see for each of
their goals, it's updating it. So what we're going to see now is we're going to see the
two different levels of alcohol intake, low alcohol intake and high alcohol intake. And
we are going to see relationship status compared for each of those. So I'm just making our
graph a little larger. Okay, so for low alcohol intake, we're going to look and what this
is saying is that for a stranger acquaintance and friend, what is is there a difference
between those, so here's our F, here's our P, or P value is point 002. That is less than
our alpha level a point oh, five. So what that is saying is that for the low alcohol
intake conditions, so this one here with the open circles, we have a significant difference
between stranger acquaintance and friends. So one of these groups at least is different
from the other. And if we're looking across this, here, we can't see exactly what group
or groups would be different from each other. But in the stranger condition, you can see
here is your error bars and our mean. And for the acquaintance condition, here's our
mean, and our error bars, we don't see any overlap here. So it's likely that these two
groups are significantly different, versus the acquaintance and friend has quite a bit
of overlap and the error bars here, and as does the friend and the stranger. Then for our last comparison, here, we're
looking at the simple main effect for high alcohol intake across different types of of
relationships with the victim, we have a p value of point 313. So that is greater than
our alpha level of point oh five. So there is not a difference between the stranger acquaintance
and friend groups and level of empathy for the high alcohol intake condition. So for
this line, here, the stranger, the acquaintance and the friend, they are not significantly
different from each other. So when we're looking at the simple main effects,
we're able to look at and see kind of break down how these different factors may depend
or interact on one another. And what we have found is, is that when we're looking across
the low alcohol intake condition, there is a difference between these three types of
relationships, when we're looking across the high alcohol intake condition, there is not.
And we can also see that when we're looking at the Frank condition, there's no difference
between high or low alcohol intake, when we're looking at the acquaintance condition. There
are differences here. And when we're looking at the stranger condition, there are also
differences here. So interestingly, what we had found is that
participants rated their level of empathy for the victim in the scenario who was being
sexually harassed at a bar as being highest when that person was a stranger, and when
that person was sober, so they have low levels of alcohol intake. And they also rated their
levels of empathy as being the lowest when the person was, again sober, but an acquaintance.
And so that was not necessarily what we had expected based on previous research. And so
one of the other things I want to quickly show you is what these results have been different
if we had analyzed the, the dataset where we had the outliers included. So let's go
ahead and and look at this really quickly. So dependent variable, we're going to put
empathy there, we're going to put our two factors here, going to just click on our effect
size. And oh, I had forgotten to uncheck this. So
let's unfilter that anytime you see an error, that's usually something that you may have
forgotten to do. So that was a good reminder to me. And when we look at our results, here
you can see our P value for our first factor level of intoxication is still not significant
greater than point o five p is point seven three to our P value for our second factor
relationship status also still not significant point 196. So we're will fail to reject that
null hypothesis still as well. And then our interaction term here p value is point 008.
So that is still significant, still, less than our alpha level of point oh five You can see that our effect size is a little
bit smaller here. So our effect size here is point 039 versus our effect size over here
may find it is point 043. So in essence, the results are exactly the same whether we included
the outliers or didn't include the outliers. However, we can see that our the magnitude
of our effect is slightly bigger, when we took those outliers out that were kind of
messing with our data set. So if you have any questions, or you want to know any more
information about two way ANOVA This is a fairly simple version of a two way ANOVA because
all of the groups were independent of each other. Were looking at all between subjects
effects were don't have anything mixed in here. So if you have any questions, please
leave them in the comments. |
11916 | https://portal.ct.gov/SDE/CT-Core-Standards/Materials-for-Teachers/Mathematics/Math-Lesson-Plans/Math/Grades-9-12-Algebra-2---Rational-Functions | Grades 9-12 Algebra 2 - Rational Functions
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CT Core Standards
Materials for Teachers
Mathematics
Math Lesson Plans
Current: Grades 9-12 Algebra 2 - Rational Functions
Provided by: Department of Education
Grades 9-12 Algebra 2 - Rational Functions
Rating:
Common Core Standards
Content Standard
HSA-CED.A.2 Create equations that describe and relationships
HSF-IF.C.7 Analyze functions using different representations
HSF-IF.C.7d Graph rational functions, identifying zeros and asymptotes when suitable factorizations and showing end behavior
HSF-BF.B.3 Building new functions from existing functions
Standards for Mathematical Practice
MP 1 Make sense of problems and persevere in solving them.
MP 2 Reason abstractly and quantitatively
MP 4 Model with mathematics
MP 5 Use appropriate tools strategically
Description of Lesson
This lesson titled "Rational Functions" from Illuminations-NCTM has students work on rational functions within the context of different carnival games. Students are asked to construct graphs and model these graphs using rational functions. Students are asked to use intuition to construct these graphs. Students are then required to transition to rational functions more generally, as well as provide explanations about how different components of the rational function affect the graph. Questions are provided for the teacher to use for formative assessment, and assessment materials are provided.
Cautions
Some of the cautions to consider are the lack of differentiation provided. While there are questions provided with activities to function as extension materials, there is limited support to scaffold the material for struggling learners. The assessments provided do not include rubrics or sample student work. This lesson also requires additional support for English language learners. A multitude of materials is also necessary for some of the classroom activities.
Rationale for Selection
This lesson attends to the rigor of the Common Core, rooting the activity within a real-world context and asking students to be metacognitive during the process. There is a clear focus of the material on the content, asking students to move beyond just the equation and graph. Students are also examining the function is multiple representations. This lesson also provides a detailed outline and instructional support for the teacher.
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11918 | https://pmc.ncbi.nlm.nih.gov/articles/PMC9175578/ | The future of diagnostic laparoscopy – Cons - PMC
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The future of diagnostic laparoscopy – Cons
Sarah Simko
Sarah Simko
1 Department of Obstetrics and Gynecology, Adventist Health White Memorial Medical Center, Los Angeles, California, USA
Find articles by Sarah Simko
1, Kelly N Wright
Kelly N Wright
2 Division of Minimally Invasive Gynecologic Surgery, Department of Obstetrics and Gynecology, Cedars-Sinai Medical Center, Los Angeles, California, USA
Find articles by Kelly N Wright
2,✉
Author information
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Copyright and License information
1 Department of Obstetrics and Gynecology, Adventist Health White Memorial Medical Center, Los Angeles, California, USA
2 Division of Minimally Invasive Gynecologic Surgery, Department of Obstetrics and Gynecology, Cedars-Sinai Medical Center, Los Angeles, California, USA
✉
Correspondence should be addressed to K N Wright: kelly.wright@cshs.org
This paper forms part of a special series on Endometriosis. The guest editors for this section were Dr Mathew Leonardi (McMaster University, Canada) and Dr Warren (Lauren) Foster (McMaster University, Canada).
Received 2022 Mar 12; Accepted 2022 Apr 20; Collection date 2022 Apr 1.
© The authors
This work is licensed under a Creative Commons Attribution 4.0 International License.
PMC Copyright notice
PMCID: PMC9175578 PMID: 35706581
Abstract
Endometriosis has a large impact on the lives of patients, affecting nearly 90% of women with chronic pelvic pain and infertility. Unfortunately, diagnosis for this condition is often delayed by an average of 7 years, with adolescent patients experiencing disproportionate delays. This is in part due to the use of an invasive procedure for primary diagnosis and limited access to subspecialty care. While laparoscopy serves an important purpose in the diagnosis and management of endometriosis, it has been found to be less cost-effective than empiric medical therapy and puts an emphasis on the lesion as opposed to the patient and the disease process as a whole. As studies emerge, we gain a deeper understanding of the complex nature of this disease. Laparoscopy has been shown to have variable results, with high recurrence rates and varying improvement in symptoms over time. Additionally, studies have shown a poor correlation between patients’ pain and the stage and location of lesions, with laparoscopy showing greater benefit for later-stage disease and deep infiltrating endometriosis. This article seeks to evaluate the current standards for the management of endometriosis, discuss the place for diagnostic laparoscopy, and review future directions and alternatives.
Lay summary
Endometriosis is an inflammatory disorder that occurs when uterine tissue is found outside the uterus. This condition affects women of reproductive age and can have serious impacts on their lives, causing pain and difficulty getting pregnant. The primary method of diagnosis is surgical, which has associated risks and can delay care to patients. As further studies emerge, our understanding of this condition improves, and it is important to evaluate current practices. This article focuses on the pros and cons of using surgical methods to diagnose endometriosis and alternative options that may be safer and provide more timely care to patients.
Keywords: endometriosis, laparoscopy, surgical treatment
Background
Endometriosis affects 10% of women of reproductive age and nearly 90% of women with chronic pelvic pain and infertility (Lorusso et al. 2021). It is defined as the presence of endometrial-like tissue outside the uterus. The most popular etiology for this condition is retrograde menstruation, but the mechanism by which endometriotic lesions implant and persist is not fully understood. The American Society for Reproductive Medicine (ASRM) classifies endometriosis into four stages (minimal, mild, moderate, severe), based on the site, depth, and location of lesions and adhesions. Deep infiltrating endometriosis (DIE) is defined as lesions that invade greater than 5 mm below the peritoneal surface. The majority of these lesions are rectovaginal; however, DIE can affect the bowel, bladder, and ureters (Zondervan et al. 2020).
Current diagnostic approaches and management
Expert opinion suggests that early diagnosis of endometriosis is key, as this disease can have serious effects on patients’ quality of life and fertility. Diagnosis is unfortunately often delayed, with an average delay of approximately 7 years, which can be detrimental to the early treatment of endometriosis to reduce pain and stresses on mental and emotional health (Taylor et al. 2018). Diagnostic challenges in combination with the requirement for a surgical procedure for diagnosis contribute to this delay. Although there is limited data on the progression of endometriosis and the effects of medical management on endometriotic lesions, medical management is thought to suppress hormonally active endometriotic tissue and disease progression in addition to removing painful stimuli from inflammatory menses.
While there are many causes of pelvic pain, in very few cases would diagnostic laparoscopy be a first-line method of diagnosis. Many of these can be identified using other diagnostic methods, including history and physical, as well as imaging. Given the wide range of etiologies for pelvic pain, a broader workup should always be performed to rule out other causes.
Current diagnostic approaches for endometriosis include clinical evaluation, imaging findings, and confirmatory diagnosis via laparoscopy with or without biopsy of lesions. Patients with endometriosis typically present with pelvic pain, including dysmenorrhea and dyspareunia. Other symptoms vary depending on the location of lesions including urinary symptoms in patients with bladder implants and gastrointestinal symptoms in patients with bowel endometriosis. Physical exam can reveal adnexal masses and nodularity of the posterior fornix, and immobility of the uterus. Imaging modalities, such as ultrasound and MRI, have been used to augment diagnosis and identify endometriomas as well as rectovaginal and bladder nodules.
Most organizations support empiric therapy of endometriosis with the use of non-steroidal anti-inflammatory drugs (NSAIDs), combined hormonal contraceptives, progestins, and gonadotropin-releasing hormone (GnRH) agonists with or without add-back therapy (Taylor et al. 2018). Laparoscopy for the diagnosis and treatment of endometriosis began in the 1970s, and since that time research on the correlation of symptoms and lesions, as well as response rates of surgical excision and ablation, has grown.
Cons of diagnostic laparoscopy
Focuses on visible lesions
Perhaps the biggest downside to the diagnosis of endometriosis with laparoscopy is the focus on visible lesions as the cause of symptomatology rather than treating endometriosis as a complex chronic inflammatory syndrome. Pelvic pain can be inflammatory and neuropathic in nature, with a component of central sensitization of the nervous system (Zondervan et al. 2020), leading patients to have continued pain even after excision of all endometriotic diseases. Patients with pelvic pain exhibit enhanced anterior insula glutamatergic neurotransmission and connectivity with the prefrontal cortex (Zondervan et al. 2020).
Diagnostic laparoscopy lumps all endometriotic lesions into one disease – microscopic, superficial, deep infiltrating, ovarian, and uterine endometriosis are all considered components of the same pathology. However, the success in obtaining pain relief and preventing recurrence with the excision of all these different lesion types varies significantly, with 30% of patients developing chronic pelvic pain that is unresponsive to surgery, and 50% developing recurrent symptoms within 5 years after treatment (Zondervan et al. 2020). Numerous classification systems are available to describe endometriosis with limited use in claims data and electronic medical records. The ASRM classification is the most used system (ACOG 2010), but after multiple revisions is still not a good predictor of pain, dyspareunia, or infertility. This suggests that the visualization of lesions during laparoscopy does not accurately correlate with the symptoms physicians are trying to treat. Additionally, no evidence supports a stepwise progression of disease from one stage to the next. While studies have not identified a significant correlation between the location and stage of lesions with the amount of pain a patient experiences, there may be an association with the depth of invasion. DIE lesions have been consistently linked to pelvic pain, and symptomatology has been shown to correlate with the location of lesions (Stratton & Berkley 2011).
Another finding that limits the correlation between lesions and symptomatology is that in large studies of patients undergoing reoperation for recurrent pain, many patients had no visible lesions. In one study, high pain recurrence rates of up to 70% following operative laparoscopy were found in those ages 19–29, with some patients showing no endometriotic lesions on repeat surgery (Stratton & Berkley 2011). The only predictor of needing reoperation was the age at the first surgery (ACOG 2010). In studies of repeat surgeries, lesions progressed in 29% of patients, regressed in 42%, or stayed the same in 29% of patients, indicating that the visible markers of disease do not necessarily progress (Zondervan et al. 2020).
One of the goals of diagnostic laparoscopy is to identify and then fully excise endometriotic lesions. Heterogeneity of visible lesions as clear, yellow, red, brown, blue, and black, with and without fibrosis or scarring can lead to missing lesions (ACOG 2010). Vascularization and innervation of lesions happen on the microscopic level and cannot be seen laparoscopically. While distortion of anatomy and obliteration of pelvic spaces with stage III and IV disease can be easily seen, deep infiltrating lesions, such as those in the rectovaginal septum, retrorectal space, or within the deep pelvic muscles and sciatic nerves may be missed. Adenomyosis could be missed as well. These disease types cannot necessarily be seen well from an intraperitoneal viewpoint (Lorusso et al. 2021). While laparoscopy with histologic confirmation of lesions remains the gold standard for diagnosis of endometriosis, it proves neither to be fully diagnostic nor therapeutic.
Increased time to diagnosis
Because neither serum markers nor imaging have been able to diagnose all disease, the need for a surgical intervention for diagnosis results in long delays. The most recent American College of Obstetricians and Gynecologists (ACOG) practice bulletin on endometriosis management recommends trialing oral contraceptive pills, non-steroidal anti-inflammatory drugs, and a 3-month course of a GnRH agonist prior to considering diagnostic surgery (ACOG 2010). Surgery is only considered when symptomatology impacts one’s quality of life enough to warrant the risks of a laparoscopy. Most patients report their symptoms begin around menarche, and on average, patients see multiple physicians prior to a diagnosis of endometriosis, with an average delay in diagnosis of 7 years. Diagnostic delay is longer for adolescents and shorter for patients with infertility (Zondervan et al. 2020). The lack of subspecialists and referral patterns drive the delay as well. Out of 35,000 ACOG fellows in practice, there are only around 400 graduates of the Fellowship in Minimally Invasive Gynecologic Surgery, a subspecialty focused on complex surgery for benign uterine and adnexal disorders. Not all these practitioners focus on endometriosis, and many patients live long distances from surgical centers with a high volume of endometriosis care.
Risks of surgery and anesthesia
While the risks of diagnostic laparoscopy are low, often less than 1%, the patient is still at risk of surgical and anesthetic complications for a diagnostic test. Complications can include bleeding, infection, injury to surrounding organs including bowel, bladder, ureters, and vasculature, as well as conversion to laparotomy. Greater than 50% of injuries occur from the first entry. Given the high rate of pain and recurrent symptoms after surgery, it is understandable that clinicians and patients would want to delay surgery to avoid these risks when the benefit is limited.
Cost-effectiveness
The 2008 US health care costs for endometriosis were estimated to be $4000 per affected woman (Zondervan et al. 2020) which does not include societal costs such as lost wages and lack of productivity due to symptoms. The delay in diagnosis due to barriers to surgery is likely a large contributor to that cost, as patients without a diagnosis continue to seek answers for their symptoms. Laparoscopy itself has not been found to be particularly cost-effective as a diagnostic test, with the costs of empiric medical therapy found to be less than a laparoscopic procedure (ACOG 2010). Imaging or predictive models of endometriosis may allow for reduced spending due to faster diagnosis and more focused treatment plans (Nnoaham et al. 2012).
Similar efficacy of medical and surgical management
Evidence shows that the pain associated with endometriosis can be effectively treated with several medications including hormonal options, NSAIDs, GnRH analogues, and medications for central sensitization. Continuous oral contraceptive pills were found to provide significant pain reduction for patients with both dysmenorrhea and endometriosis (ACOG 2010). In two randomized controlled trials of laparoscopy for the treatment of pain associated with endometriosis, surgery was found to be effective in symptom improvement for 6 months for 62.5 and 80% of the treatment groups, which was significantly better than those in the expectant management group (22.6 and 32%) (ACOG 2010). However, the largest group which did not respond well to surgery were those with stage 1 or minimal disease. Patients with stage I disease reported the least improvement with surgery at short- and long-term follow-ups (ACOG 2010). While there is a significant short-term improvement after surgery for some patients, rates of reoperation at 2, 5, and 7 years were 21, 47, and 55%, respectively, and many patients had no visible lesions on reoperation (ACOG 2010). Therefore, surgery alone cannot be considered curative for the pain associated with endometriosis, particularly for those with milder disease.
Lack of fertility improvement
Many proposed mechanisms have been described as to how endometriosis causes infertility and subfertility, though all of these remain controversial (ASRM 2012). For more mild forms of endometriosis, infertility could be caused by altered peritoneal function and systemic inflammation, altered cell-mediated function, ovulatory dysfunction, impaired implantation, and reduced oocyte and embryo quality.
In stages I and II endometriosis, laparoscopic treatment of endometrial implants has been associated with a small improvement in live birth rate. Aggregating the data of two randomized controlled trials found that 40 asymptomatic patients with unexplained infertility would need to have surgery for one additional pregnancy (ASRM 2012). Distorted anatomy from stages III and IV endometriosis can result in impaired transport of the egg through the tube. Patients with severe endometriosis who were followed for up to 2 years after surgery were found to have improved pregnancy rates (ASRM 2012). Additionally, patients who underwent cystectomy for endometriomas over 4 cm had improved fertility outcomes over those who had only cyst drainage. Typically, advanced stage endometriosis, including endometriomas, can be seen on imaging, and therefore a diagnostic laparoscopy would not be required to make this diagnosis. Goal setting and risk stratification could occur at the time of diagnosis by imaging, and any therapeutic surgery could be optimally timed with fertility needs.
Though endometriosis is associated with infertility, in vitro fertilization likely maximizes cycle fecundity over any surgical intervention. Conflicting observational data does not show a clear benefit of surgery with assisted reproductive technologies, particularly in asymptomatic patients with unexplained fertility. The ASRM does not recommend diagnostic laparoscopy solely to increase the likelihood of pregnancy (ASRM 2012).
Future directives and alternatives
Predictive algorithms
In efforts to improve less invasive forms of diagnosis, predictive algorithms are being evaluated to better diagnose endometriosis clinically. One predictive model that incorporated a questionnaire to assess women’s past medical, obstetric, and family histories, and the intensity and frequency of pelvic pain with or without ultrasound findings, was successful at predicting stages III and IV disease. This algorithm was not as successful at predicting all stage disease (Nnoaham et al. 2012). Further validation and application of these models may provide an opportunity to decrease rates of avoidable procedures, particularly if there is greater symptomatology benefit in performing surgery in only those patients with advanced disease.
Imaging advances
Improvements in imaging criteria may increase diagnostic capabilities, although there are limitations by operator experience. Transvaginal ultrasound (TVS) findings such as site-specific tenderness, fixed ovaries, negative ‘sliding sign’, and evidence of DIE nodules in the anterior and posterior compartment can be indicative of endometriosis. Additional techniques, including bladder site tenderness-guided TVS (97.4% sensitivity and specificity) and rectal endoscopy-sonography and rectal water contrast TVS (>92% sensitivities and specificities), have been shown to be effective for diagnosing site-specific diseases (Noventa et al. 2015).
Similarly, advances in MRI technique and identification of specific MR-acquisition protocols and imaging findings suspicious for endometriosis can potentially improve diagnosis, although current studies show similar diagnostic performance of TVS and MRI in detecting DIE (Lorusso et al. 2021). Fusion imaging, which allows for synchronized assessment of MRI and ultrasound images of various anatomic landmarks, may augment diagnostic capabilities in the future, although further studies are needed. Enhanced training for sonographers and radiologists is essential for uniform diagnosis across hospital and clinic systems.
Multidisciplinary care in treatment
Endometriosis has a significant impact on the lives of the patients it affects. Self-esteem, sexual health, emotional wellbeing, and functional status can all be affected. Additionally, as a chronic pain condition, improper management with opioids and anxiolytics can lead to addiction. Given the complex nature of this condition, a multidisciplinary approach to treatment is thought to be crucial with the involvement of gynecologists, psychotherapists, psychiatrists, pain specialists, and sexologists. There is limited data assessing the benefits of multidisciplinary care for endometriosis; however, large healthcare centers have started implementing these approaches given successes observed for other chronic pain conditions, including low back pain. An earlier diagnosis without requiring surgery may allow for better treatment of pain, preventing chronic pain syndromes and resulting in a healthier and more productive population.
Conclusion
Deep infiltrating endometriosis has been most frequently associated with pelvic pain, while minimal to mild disease has had variable symptomatology. Symptom-based questionnaires have been shown to accurately predict advanced disease and DIE lesions. These clinical symptoms, augmented by ultrasound and MRI findings, can increase the ability for clinicians to identify deep infiltrating endometriosis, which is most likely to be responsive to surgical management. Initial empiric treatment with medical therapies has been shown to be equally effective and more cost-effective than diagnostic laparoscopy. Focusing on the entire clinical picture as opposed to the endometriotic lesion can provide earlier diagnosis and access to treatment, promote an emphasis on multidisciplinary care, and decrease unnecessary costs and risks associated with surgical procedures.
Declaration of interest
S S declares no conflicts. K N W is a consultant for Aqua Therapeutics, Hologic and Karl Storz.
Funding
This work did not receive any specific grant from any funding agency in the public, commercial or not-for-profit sector.
Author contribution statement
Both authors met criteria for authorship. Both S S and K N W performed a literature search and wrote the paper.
References
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11919 | https://www.gauthmath.com/solution/1836184032456849/2-List-all-the-multiples-of-6-which-are-less-than-100- | Solved: List all the multiples of 6 which are less than 100 : [Math]
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List all the multiples of 6 which are less than 100 :
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Answer
The answer is 6, 12, 18, 24, 30, 36, 42, 48, 54, 60, 66, 72, 78, 84, 90, 96
Explanation
Generate the sequence of multiples of 6. We systematically compute integer multiples of 6, starting from the first multiple (6 x 1 = 6).
Iteratively calculate multiples until the upper bound is exceeded. We continue to calculate successive multiples (6 x 2, 6 x 3, and so on) until a multiple exceeds 100. This process yields the following sequence:
6, 12, 18, 24, 30, 36, 42, 48, 54, 60, 66, 72, 78, 84, 90, 96. The next multiple, 6 x 17 = 102, exceeds the specified upper limit of 100.
Identify the complete set of multiples less than 100. The complete set of multiples of 6 that are less than 100 is: 6, 12, 18, 24, 30, 36, 42, 48, 54, 60, 66, 72, 78, 84, 90, 96.
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11920 | https://www.chegg.com/homework-help/questions-and-answers/193-solution-linear-differential-equation-given-x-t-10e-7-5e-05t-using-matlab-plot-x-t-ver-q70192095 | Your solution’s ready to go!
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Question: 1.93 (a) The solution of a linear differential equation is given by X(t) = 10e 7 - 5e-0.5t. Using MATLAB, plot x(t) versus t for t 0:0.01:5. (b) Repeat Part (a) for (t) = 10e + + 5e -0.5t.
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Start by defining a time interval in MATLAB with a start time of 0, an increment of 0.01, and an end time of 5.
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11921 | https://www.sfu.ca/~mbahrami/ENSC%20388/Solution%20manual/IntroTHT_2e_SM_Chap11.pdf | PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 11-1 Solutions Manual for Introduction to Thermodynamics and Heat Transfer Yunus A. Cengel 2nd Edition, 2008 Chapter 11 TRANSIENT HEAT CONDUCTION PROPRIETARY AND CONFIDENTIAL This Manual is the proprietary property of The McGraw-Hill Companies, Inc. (“McGraw-Hill”) and protected by copyright and other state and federal laws. By opening and using this Manual the user agrees to the following restrictions, and if the recipient does not agree to these restrictions, the Manual should be promptly returned unopened to McGraw-Hill: This Manual is being provided only to authorized professors and instructors for use in preparing for the classes using the affiliated textbook. No other use or distribution of this Manual is permitted. This Manual may not be sold and may not be distributed to or used by any student or other third party. No part of this Manual may be reproduced, displayed or distributed in any form or by any means, electronic or otherwise, without the prior written permission of McGraw-Hill. PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 11-2 Lumped System Analysis 11-1C In heat transfer analysis, some bodies are observed to behave like a "lump" whose entire body temperature remains essentially uniform at all times during a heat transfer process. The temperature of such bodies can be taken to be a function of time only. Heat transfer analysis which utilizes this idealization is known as the lumped system analysis. It is applicable when the Biot number (the ratio of conduction resistance within the body to convection resistance at the surface of the body) is less than or equal to 0.1. 11-2C The lumped system analysis is more likely to be applicable for the body cooled naturally since the Biot number is proportional to the convection heat transfer coefficient, which is proportional to the air velocity. Therefore, the Biot number is more likely to be less than 0.1 for the case of natural convection. 11-3C The lumped system analysis is more likely to be applicable for the body allowed to cool in the air since the Biot number is proportional to the convection heat transfer coefficient, which is larger in water than it is in air because of the larger thermal conductivity of water. Therefore, the Biot number is more likely to be less than 0.1 for the case of the solid cooled in the air 11-4C The temperature drop of the potato during the second minute will be less than 4°C since the temperature of a body approaches the temperature of the surrounding medium asymptotically, and thus it changes rapidly at the beginning, but slowly later on. 11-5C The temperature rise of the potato during the second minute will be less than 5°C since the temperature of a body approaches the temperature of the surrounding medium asymptotically, and thus it changes rapidly at the beginning, but slowly later on. 11-6C Biot number represents the ratio of conduction resistance within the body to convection resistance at the surface of the body. The Biot number is more likely to be larger for poorly conducting solids since such bodies have larger resistances against heat conduction. 11-7C The heat transfer is proportional to the surface area. Two half pieces of the roast have a much larger surface area than the single piece and thus a higher rate of heat transfer. As a result, the two half pieces will cook much faster than the single large piece. 11-8C The cylinder will cool faster than the sphere since heat transfer rate is proportional to the surface area, and the sphere has the smallest area for a given volume. 11-9C The lumped system analysis is more likely to be applicable in air than in water since the convection heat transfer coefficient and thus the Biot number is much smaller in air. 11-10C The lumped system analysis is more likely to be applicable for a golden apple than for an actual apple since the thermal conductivity is much larger and thus the Biot number is much smaller for gold. 11-11C The lumped system analysis is more likely to be applicable to slender bodies than the well-rounded bodies since the characteristic length (ratio of volume to surface area) and thus the Biot number is much smaller for slender bodies. PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 11-3 11-12 Relations are to be obtained for the characteristic lengths of a large plane wall of thickness 2L, a very long cylinder of radius ro and a sphere of radius ro. Analysis Relations for the characteristic lengths of a large plane wall of thickness 2L, a very long cylinder of radius ro and a sphere of radius ro are 3 4 3 / 4 2 2 2 2 2 3 surface , 2 surface , surface , o o o sphere c o o o cylinder c wall c r r r A L r h r h r A L L A LA A L = = = = = = = = = π π π π V V V 11-13 A relation for the time period for a lumped system to reach the average temperature 2 / ) ( ∞ +T Ti is to be obtained. Analysis The relation for time period for a lumped system to reach the average temperature 2 / ) ( ∞ +T Ti can be determined as b 0.693 b 2 ln = = ⎯→ ⎯ − = − = ⎯→ ⎯ = − − = − − + ⎯→ ⎯ = − − − − ∞ ∞ − ∞ ∞ ∞ − ∞ ∞ t bt e e T T T T e T T T T T e T T T t T bt bt i i bt i i bt i 2 ln 2 1 ) ( 2 2 ) ( 2L 2ro 2ro T∞ Ti PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 11-4 11-14 The temperature of a gas stream is to be measured by a thermocouple. The time it takes to register 99 percent of the initial ΔT is to be determined. Assumptions 1 The junction is spherical in shape with a diameter of D = 0.0012 m. 2 The thermal properties of the junction are constant. 3 The heat transfer coefficient is constant and uniform over the entire surface. 4 Radiation effects are negligible. 5 The Biot number is Bi < 0.1 so that the lumped system analysis is applicable (this assumption will be verified). Properties The properties of the junction are given to be C W/m.
35 ° = k , 3 kg/m 8500 = ρ , and C J/kg. 320 ° = p c . Analysis The characteristic length of the junction and the Biot number are 1 .
0 00051 .
0 ) C W/m.
35 ( ) m 0002 .
0 )( C . W/m 90 ( m 0002 .
0 6 m 0012 .
0 6 6 / 2 2 3 surface < = ° ° = = = = = = = k hL Bi D D D A L c c π π V Since 0.1 < Bi , the lumped system analysis is applicable. Then the time period for the thermocouple to read 99% of the initial temperature difference is determined from s 27.8 = ⎯→ ⎯ = ⎯→ ⎯ = − − = ° ° = = = = − − − − ∞ ∞ ∞ ∞ t e e T T T t T L c h c hA b T T T t T t bt i c p p i ) s 1654 .
0 ( 1 -3 2 1 -01 .
0 ) ( s 1654 .
0 m) C)(0.0002 J/kg. 320 )( kg/m 8500 ( C . W/m 90 01 .
0 ) ( ρ ρ V Gas h, T∞ Junction D T(t) PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 11-5 11-15E A number of brass balls are to be quenched in a water bath at a specified rate. The temperature of the balls after quenching and the rate at which heat needs to be removed from the water in order to keep its temperature constant are to be determined. Assumptions 1 The balls are spherical in shape with a radius of ro = 1 in. 2 The thermal properties of the balls are constant. 3 The heat transfer coefficient is constant and uniform over the entire surface. 4 The Biot number is Bi < 0.1 so that the lumped system analysis is applicable (this assumption will be verified). Properties The thermal conductivity, density, and specific heat of the brass balls are given to be k = 64.1 Btu/h.ft.°F, ρ = 532 lbm/ft3, and cp = 0.092 Btu/lbm.°F. Analysis (a) The characteristic length and the Biot number for the brass balls are 1 .
0 01820 .
0 ) F Btu/h.ft. 1 .
64 ( ) ft 02778 .
0 )( F .
Btu/h.ft 42 ( ft 02778 .
0 6 ft 12 / 2 6 6 / 2 2 3 < = ° ° = = = = = = = k hL Bi D D D A L c s c π π V The lumped system analysis is applicable since Bi < 0.1. Then the temperature of the balls after quenching becomes F 166 ° = ⎯→ ⎯ = − − ⎯→ ⎯ = − − = = ° ° = = = − − ∞ ∞ ) ( 120 250 120 ) ( ) ( s 00858 .
0 h 9 .
30 ft) F)(0.02778 Btu/lbm. 092 .
0 )( lbm/ft (532 F .
Btu/h.ft 42 s) 120 )( s 00858 .
0 ( 1 -1 -3 2 1 -t T e t T e T T T t T L c h c hA b bt i c p p s ρ ρ V (b) The total amount of heat transfer from a ball during a 2-minute period is Btu 97 .
9 F ) 166 250 ( F) Btu/lbm. 092 .
0 )( lbm 29 .
1 ( )] ( [ lbm 290 .
1 6 ft) 12 / 2 ( ) lbm/ft 532 ( 6 3 3 3 = ° − ° = − = = = = = t T T mc Q D m i p π π ρ ρV Then the rate of heat transfer from the balls to the water becomes Btu/min 1196 = × = = ) Btu 97 .
9 ( balls/min) 120 ( ball ballQ n Qtotal & & Therefore, heat must be removed from the water at a rate of 1196 Btu/min in order to keep its temperature constant at 120°F. Brass balls, 250°F Water bath, 120°F PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 11-6 11-16E A number of aluminum balls are to be quenched in a water bath at a specified rate. The temperature of balls after quenching and the rate at which heat needs to be removed from the water in order to keep its temperature constant are to be determined. Assumptions 1 The balls are spherical in shape with a radius of ro = 1 in. 2 The thermal properties of the balls are constant. 3 The heat transfer coefficient is constant and uniform over the entire surface. 4 The Biot number is Bi < 0.1 so that the lumped system analysis is applicable (this assumption will be verified). Properties The thermal conductivity, density, and specific heat of the aluminum balls are k = 137 Btu/h.ft.°F, ρ = 168 lbm/ft3, and cp = 0.216 Btu/lbm.°F (Table A-24E). Analysis (a) The characteristic length and the Biot number for the aluminum balls are 1 .
0 00852 .
0 ) F Btu/h.ft. 137 ( ) ft 02778 .
0 )( F .
Btu/h.ft 42 ( ft 02778 .
0 6 ft 12 / 2 6 6 / 2 2 3 < = ° ° = = = = = = = k hL Bi D D D A L c c π π V The lumped system analysis is applicable since Bi < 0.1. Then the temperature of the balls after quenching becomes F 152° = ⎯→ ⎯ = − − ⎯→ ⎯ = − − = = ° ° = = = − − ∞ ∞ ) ( 120 250 120 ) ( ) ( s 01157 .
0 h 66 .
41 ft) F)(0.02778 Btu/lbm. 216 .
0 )( lbm/ft (168 F .
Btu/h.ft 42 s) 120 )( s 01157 .
0 ( 1 -1 -3 2 1 -t T e t T e T T T t T L c h c hA b bt i c p p s ρ ρ V (b) The total amount of heat transfer from a ball during a 2-minute period is Btu 62 .
8 F ) 152 250 ( F) Btu/lbm. 216 .
0 )( lbm 4072 .
0 ( )] ( [ lbm 4072 .
0 6 ft) 12 / 2 ( ) lbm/ft 168 ( 6 3 3 3 = ° − ° = − = = = = = t T T mc Q D m i p π π ρ ρV Then the rate of heat transfer from the balls to the water becomes Btu/min 1034 = × = = ) Btu 62 .
8 ( balls/min) 120 ( ball ballQ n Qtotal & & Therefore, heat must be removed from the water at a rate of 1034 Btu/min in order to keep its temperature constant at 120°F. Aluminum balls, 250°F Water bath, 120°F PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 11-7 11-17 Milk in a thin-walled glass container is to be warmed up by placing it into a large pan filled with hot water. The warming time of the milk is to be determined. Assumptions 1 The glass container is cylindrical in shape with a radius of r0 = 3 cm. 2 The thermal properties of the milk are taken to be the same as those of water. 3 Thermal properties of the milk are constant at room temperature. 4 The heat transfer coefficient is constant and uniform over the entire surface. 5 The Biot number in this case is large (much larger than 0.1). However, the lumped system analysis is still applicable since the milk is stirred constantly, so that its temperature remains uniform at all times. Properties The thermal conductivity, density, and specific heat of the milk at 20°C are k = 0.598 W/m.°C, ρ = 998 kg/m3, and cp = 4.182 kJ/kg.°C (Table A-15). Analysis The characteristic length and Biot number for the glass of milk are 1 .
0 > 107 .
2 ) C W/m.
598 .
0 ( ) m 0105 .
0 )( C . W/m 120 ( m 01050 .
0 m) 03 .
0 ( 2 + m) m)(0.07 03 .
0 ( 2 m) 07 .
0 ( m) 03 .
0 ( 2 2 2 2 2 2 2 = ° ° = = = = + = = k hL Bi r L r L r A L c o o o s c π π π π π π V For the reason explained above we can use the lumped system analysis to determine how long it will take for the milk to warm up to 38°C: min 5.8 s 348 = = ⎯→ ⎯ = − − ⎯→ ⎯ = − − = ° ° = = = − − ∞ ∞ t e e T T T t T L c h c hA b t bt i c p p s ) s 002738 .
0 ( 1 -3 2 1 -60 3 60 38 ) ( s 002738 .
0 m) C)(0.0105 J/kg. 4182 )( kg/m (998 C . W/m 120 ρ ρ V Therefore, it will take about 6 minutes to warm the milk from 3 to 38°C. Water 60°C Milk 3°C PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 11-8 11-18 A thin-walled glass containing milk is placed into a large pan filled with hot water to warm up the milk. The warming time of the milk is to be determined. Assumptions 1 The glass container is cylindrical in shape with a radius of r0 = 3 cm. 2 The thermal properties of the milk are taken to be the same as those of water. 3 Thermal properties of the milk are constant at room temperature. 4 The heat transfer coefficient is constant and uniform over the entire surface. 5 The Biot number in this case is large (much larger than 0.1). However, the lumped system analysis is still applicable since the milk is stirred constantly, so that its temperature remains uniform at all times. Properties The thermal conductivity, density, and specific heat of the milk at 20°C are k = 0.598 W/m.°C, ρ = 998 kg/m3, and cp = 4.182 kJ/kg.°C (Table A-15). Analysis The characteristic length and Biot number for the glass of milk are 1 .
0 > 21 .
4 ) C W/m.
598 .
0 ( ) m 0105 .
0 )( C . W/m 240 ( m 01050 .
0 m) 03 .
0 ( 2 + m) m)(0.07 03 .
0 ( 2 m) 07 .
0 ( m) 03 .
0 ( 2 2 2 2 2 2 2 = ° ° = = = = + = = k hL Bi r L r L r A L c o o o s c π π π π π π V For the reason explained above we can use the lumped system analysis to determine how long it will take for the milk to warm up to 38°C: min 2.9 s 174 = = ⎯→ ⎯ = − − ⎯→ ⎯ = − − = ° ° = = = − − ∞ ∞ t e e T T T t T L c h c hA b t bt i c p p s ) s 005477 .
0 ( 1 -3 2 1 -60 3 60 38 ) ( s 005477 .
0 m) C)(0.0105 J/kg. 4182 )( kg/m (998 C . W/m 240 ρ ρ V Therefore, it will take about 3 minutes to warm the milk from 3 to 38°C. 11-19 A long copper rod is cooled to a specified temperature. The cooling time is to be determined. Assumptions 1 The thermal properties of the geometry are constant. 2 The heat transfer coefficient is constant and uniform over the entire surface. Properties The properties of copper are k = 401 W/m⋅ºC, ρ = 8933 kg/m3, and cp = 0.385 kJ/kg⋅ºC (Table A-24). Analysis For cylinder, the characteristic length and the Biot number are 1 .
0 0025 .
0 ) C W/m.
401 ( ) m 005 .
0 )( C . W/m 200 ( m 005 .
0 4 m 02 .
0 4 ) 4 / ( 2 2 surface < = ° ° = = = = = = = k hL Bi D DL L D A L c c π π V Since 0.1 < Bi , the lumped system analysis is applicable. Then the cooling time is determined from min 4.0 = = ⎯→ ⎯ = − − ⎯→ ⎯ = − − = ° ° = = = − − ∞ ∞ s 238 20 100 20 25 ) ( s 01163 .
0 m) C)(0.005 J/kg. 385 )( kg/m (8933 C . W/m 200 ) s 01163 .
0 ( 1 -3 2 1 -t e e T T T t T L c h c hA b t bt i c p p ρ ρ V Water 60°C Milk 3°C Ti = 100 ºC D = 2 cm PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 11-9 11-20 The heating times of a sphere, a cube, and a rectangular prism with similar dimensions are to be determined. Assumptions 1 The thermal properties of the geometries are constant. 2 The heat transfer coefficient is constant and uniform over the entire surface. Properties The properties of silver are given to be k = 429 W/m⋅ºC, ρ = 10,500 kg/m3, and cp = 0.235 kJ/kg⋅ºC. Analysis For sphere, the characteristic length and the Biot number are 1 .
0 00023 .
0 ) C W/m.
429 ( ) m 008333 .
0 )( C . W/m 12 ( m 008333 .
0 6 m 05 .
0 6 6 / 2 2 3 surface < = ° ° = = = = = = = k hL Bi D D D A L c c π π V Since 0.1 < Bi , the lumped system analysis is applicable. Then the time period for the sphere temperature to reach to 25ºC is determined from min 40.5 = = ⎯→ ⎯ = − − ⎯→ ⎯ = − − = ° ° = = = − − ∞ ∞ s 2428 33 0 33 25 ) ( s 0005836 .
0 m) 3 C)(0.00833 J/kg. 235 )( kg/m 00 (10,5 C . W/m 12 ) s 0005836 .
0 ( 1 -3 2 1 -t e e T T T t T L c h c hA b t bt i c p p ρ ρ V Cube: 1 .
0 00023 .
0 ) C W/m.
429 ( ) m 008333 .
0 )( C . W/m 12 ( m 008333 .
0 6 m 05 .
0 6 6 2 2 3 surface < = ° ° = = = = = = = k hL Bi L L L A L c c V min 40.5 = = ⎯→ ⎯ = − − ⎯→ ⎯ = − − = ° ° = = = − − ∞ ∞ s 2428 33 0 33 25 ) ( s 0005836 .
0 m) 3 C)(0.00833 J/kg. 235 )( kg/m 00 (10,5 C . W/m 12 ) s 0005836 .
0 ( 1 -3 2 1 -t e e T T T t T L c h c hA b t bt i c p p ρ ρ V Rectangular prism: 1 .
0 00023 .
0 ) C W/m.
429 ( ) m 008108 .
0 )( C . W/m 12 ( m 008108 .
0 m) 06 .
0 ( m) 05 .
0 ( 2 m) 06 .
0 ( m) 04 .
0 ( 2 m) 05 .
0 ( m) 04 .
0 ( 2 m) 06 .
0 ( m) 05 .
0 ( m) 04 .
0 ( 2 surface < = ° ° = = = + + = = k hL Bi A L c c V min 39.4 = = ⎯→ ⎯ = − − ⎯→ ⎯ = − − = ° ° = = = − − ∞ ∞ s 2363 33 0 33 25 ) ( s 0005998 .
0 m) 8 C)(0.00810 J/kg. 235 )( kg/m 00 (10,5 C . W/m 12 ) s 0005998 .
0 ( 1 -3 2 1 -t e e T T T t T L c h c hA b t bt i c p p ρ ρ V The heating times are same for the sphere and cube while it is smaller in rectangular prism. 5 cm Air h, T∞ Air h, T∞ 5 cm 5 cm 5 cm Air h, T∞ 5 cm 6 cm 4 cm PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 11-10 11-21E A person shakes a can of drink in a iced water to cool it. The cooling time of the drink is to be determined. Assumptions 1 The can containing the drink is cylindrical in shape with a radius of ro = 1.25 in. 2 The thermal properties of the drink are taken to be the same as those of water. 3 Thermal properties of the drinkare constant at room temperature. 4 The heat transfer coefficient is constant and uniform over the entire surface. 5 The Biot number in this case is large (much larger than 0.1). However, the lumped system analysis is still applicable since the drink is stirred constantly, so that its temperature remains uniform at all times. Properties The density and specific heat of water at room temperature are ρ = 62.22 lbm/ft3, and cp = 0.999 Btu/lbm.°F (Table A-15E). Analysis Application of lumped system analysis in this case gives ft 04167 .
0 ft) 12 / 25 .
1 ( 2 + ft) ft)(5/12 12 / 25 .
1 ( 2 ft) 12 / 5 ( ft) 12 / 25 .
1 ( 2 2 2 2 2 2 = = + = = π π π π π π o o o s c r L r L r A L V s 615 = ⎯→ ⎯ = − − ⎯→ ⎯ = − − = = ° ° = = = − − ∞ ∞ t e e T T T t T L c h c hA b t bt i c p p s ) s 00322 .
0 ( 1 -1 -3 2 1 -32 90 32 40 ) ( s 00322 .
0 h 583 .
11 ft) F)(0.04167 Btu/lbm. 999 .
0 )( lbm/ft (62.22 F .
Btu/h.ft 30 ρ ρ V Therefore, it will take 10 minutes and 15 seconds to cool the canned drink to 40°F. Milk 3°C Water 32°F Drink 90°F PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 11-11 11-22 An iron whose base plate is made of an aluminum alloy is turned on. The time for the plate temperature to reach 140°C and whether it is realistic to assume the plate temperature to be uniform at all times are to be determined. Assumptions 1 85 percent of the heat generated in the resistance wires is transferred to the plate. 2 The thermal properties of the plate are constant. 3 The heat transfer coefficient is constant and uniform over the entire surface. Properties The density, specific heat, and thermal diffusivity of the aluminum alloy plate are given to be ρ = 2770 kg/m3, cp = 875 kJ/kg.°C, and α = 7.3×10-5 m2/s. The thermal conductivity of the plate can be determined from k = αρcp = 177 W/m.°C (or it can be read from Table A-24). Analysis The mass of the iron's base plate is kg 4155 .
0 ) m 03 .
0 )( m 005 .
0 )( kg/m 2770 ( 2 3 = = = = LA m ρ ρV Noting that only 85 percent of the heat generated is transferred to the plate, the rate of heat transfer to the iron's base plate is W 850 W 1000 85 .
0 in = × = Q & The temperature of the plate, and thus the rate of heat transfer from the plate, changes during the process. Using the average plate temperature, the average rate of heat loss from the plate is determined from W 21.2 = C 22 2 22 140 ) m 03 .
0 )( C . W/m 12 ( ) ( 2 2 ave plate, loss ° ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ − + ° = − = ∞ T T hA Q & Energy balance on the plate can be expressed as plate plate out in plate out in T mc E t Q t Q E E E pΔ = Δ = Δ − Δ → Δ = − & & Solving for Δt and substituting, = J/s 21.2) (850 C ) 22 140 )( C J/kg. 875 )( kg 4155 .
0 ( = out in plate s 51.8 − ° − ° − Δ = Δ Q Q T mc t p & & which is the time required for the plate temperature to reach 140°C . To determine whether it is realistic to assume the plate temperature to be uniform at all times, we need to calculate the Biot number, 1 .
0 < 00034 .
0 ) C W/m.
0 .
177 ( ) m 005 .
0 )( C . W/m 12 ( m 005 .
0 2 = ° ° = = = = = = k hL Bi L A LA A L c s c V It is realistic to assume uniform temperature for the plate since Bi < 0.1. Discussion This problem can also be solved by obtaining the differential equation from an energy balance on the plate for a differential time interval, and solving the differential equation. It gives ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ − − + = ∞ ) exp( 1 ) ( in t mc hA hA Q T t T p & Substituting the known quantities and solving for t again gives 51.8 s. Air 22°C IRON 1000 W PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 11-12 11-23 EES Prob. 11-22 is reconsidered. The effects of the heat transfer coefficient and the final plate temperature on the time it will take for the plate to reach this temperature are to be investigated. Analysis The problem is solved using EES, and the solution is given below. "GIVEN" E_dot=1000 [W] L=0.005 [m] A=0.03 [m^2] T_infinity=22 [C] T_i=T_infinity h=12 [W/m^2-C] f_heat=0.85 T_f=140 [C] "PROPERTIES" rho=2770 [kg/m^3] c_p=875 [J/kg-C] alpha=7.3E-5 [m^2/s] "ANALYSIS" V=LA m=rhoV Q_dot_in=f_heatE_dot Q_dot_out=hA(T_ave-T_infinity) T_ave=1/2(T_i+T_f) (Q_dot_in-Q_dot_out)time=mc_p(T_f-T_i) "energy balance on the plate" h [W/m2.C] time [s] 5 51 7 51.22 9 51.43 11 51.65 13 51.88 15 52.1 17 52.32 19 52.55 21 52.78 23 53.01 25 53.24 5 9 13 17 21 25 51 51.45 51.9 52.35 52.8 53.25 h [W/m 2-C] time [s] PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 11-13 Tf [C] time [s] 30 3.428 40 7.728 50 12.05 60 16.39 70 20.74 80 25.12 90 29.51 100 33.92 110 38.35 120 42.8 130 47.28 140 51.76 150 56.27 160 60.8 170 65.35 180 69.92 190 74.51 200 79.12 20 40 60 80 100 120 140 160 180 200 0 10 20 30 40 50 60 70 80 Tf [C] time [s] PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 11-14 11-24 Ball bearings leaving the oven at a uniform temperature of 900°C are exposed to air for a while before they are dropped into the water for quenching. The time they can stand in the air before their temperature falls below 850°C is to be determined. Assumptions 1 The bearings are spherical in shape with a radius of ro = 0.6 cm. 2 The thermal properties of the bearings are constant. 3 The heat transfer coefficient is constant and uniform over the entire surface. 4 The Biot number is Bi < 0.1 so that the lumped system analysis is applicable (this assumption will be verified). Properties The thermal conductivity, density, and specific heat of the bearings are given to be k = 15.1 W/m.°C, ρ = 8085 kg/m3, and cp = 0.480 kJ/kg.°F. Analysis The characteristic length of the steel ball bearings and Biot number are 0.1 < 0166 .
0 ) C W/m.
1 .
15 ( ) m 002 .
0 )( C . W/m 125 ( m 002 .
0 6 m 012 .
0 6 6 / 2 2 3 = ° ° = = = = = = = k hL Bi D D D A L c s c π π V Therefore, the lumped system analysis is applicable. Then the allowable time is determined to be s 3.68 = ⎯→ ⎯ = − − ⎯→ ⎯ = − − = ° ° = = = − − ∞ ∞ t e e T T T t T L c h c hA b t bt i c p p s ) s 0161 .
0 ( 1 -3 2 1 -30 900 30 850 ) ( s 01610 .
0 m) C)(0.002 J/kg. 480 )( kg/m 8085 ( C . W/m 125 ρ ρ V The result indicates that the ball bearing can stay in the air about 4 s before being dropped into the water. Steel balls 900°C Air, 30°C Furnace PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 11-15 11-25 A number of carbon steel balls are to be annealed by heating them first and then allowing them to cool slowly in ambient air at a specified rate. The time of annealing and the total rate of heat transfer from the balls to the ambient air are to be determined. Assumptions 1 The balls are spherical in shape with a radius of ro = 4 mm. 2 The thermal properties of the balls are constant. 3 The heat transfer coefficient is constant and uniform over the entire surface. 4 The Biot number is Bi < 0.1 so that the lumped system analysis is applicable (this assumption will be verified). Properties The thermal conductivity, density, and specific heat of the balls are given to be k = 54 W/m.°C, ρ = 7833 kg/m3, and cp = 0.465 kJ/kg.°C. Analysis The characteristic length of the balls and the Biot number are 1 .
0 0018 .
0 ) C W/m.
54 ( ) m 0013 .
0 )( C . W/m 75 ( m 0013 .
0 6 m 008 .
0 6 6 / 2 2 3 < = ° ° = = = = = = = k hL Bi D D D A L c s c π π V Therefore, the lumped system analysis is applicable. Then the time for the annealing process is determined to be min 2.7 s 163 = = ⎯→ ⎯ = − − ⎯→ ⎯ = − − = ° ° = = = − − ∞ ∞ t e e T T T t T L c h c hA b bt i c p p s )t s 01584 .
0 ( 1 -3 2 1 -35 900 35 100 ) ( s 01584 .
0 m) C)(0.0013 J/kg. 465 )( kg/m (7833 C . W/m 75 ρ ρ V The amount of heat transfer from a single ball is ball) (per kJ 0.781 = J 781 C ) 100 900 )( C J/kg. 465 )( kg 0021 .
0 ( ] [ kg 0021 .
0 6 m) 008 .
0 ( ) kg/m 7833 ( 6 3 3 3 = ° − ° = − = = = = = i f p T T mc Q D m π π ρ ρV Then the total rate of heat transfer from the balls to the ambient air becomes W 543 = = × = = kJ/h 953 , 1 ) kJ/ball 781 .
0 ( balls/h) 2500 ( ballQ n Q & & Steel balls 900°C Air, 35°C Furnace PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 11-16 11-26 EES Prob. 11-25 is reconsidered. The effect of the initial temperature of the balls on the annealing time and the total rate of heat transfer is to be investigated. Analysis The problem is solved using EES, and the solution is given below. "GIVEN" D=0.008 [m]; T_i=900 [C] T_f=100 [C]; T_infinity=35 [C] h=75 [W/m^2-C]; n_dot_ball=2500 [1/h] "PROPERTIES" rho=7833 [kg/m^3]; k=54 [W/m-C] c_p=465 [J/kg-C]; alpha=1.474E-6 [m^2/s] "ANALYSIS" A=piD^2 V=piD^3/6 L_c=V/A Bi=(hL_c)/k "if Bi < 0.1, the lumped sytem analysis is applicable" b=(hA)/(rhoc_pV) (T_f-T_infinity)/(T_i-T_infinity)=exp(-btime) m=rhoV Q=mc_p(T_i-T_f) Q_dot=n_dot_ballQConvert(J/h, W) Ti [C] time [s] Q [W] 500 127.4 271.2 550 134 305.1 600 140 339 650 145.5 372.9 700 150.6 406.9 750 155.3 440.8 800 159.6 474.7 850 163.7 508.6 900 167.6 542.5 950 171.2 576.4 1000 174.7 610.3 500 600 700 800 900 1000 120 130 140 150 160 170 180 250 300 350 400 450 500 550 600 650 Ti [C] time [s] Q [W] time heat PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 11-17 11-27 An electronic device is on for 5 minutes, and off for several hours. The temperature of the device at the end of the 5-min operating period is to be determined for the cases of operation with and without a heat sink. Assumptions 1 The device and the heat sink are isothermal. 2 The thermal properties of the device and of the sink are constant. 3 The heat transfer coefficient is constant and uniform over the entire surface. Properties The specific heat of the device is given to be cp = 850 J/kg.°C. The specific heat of the aluminum sink is 903 J/kg.°C (Table A-24), but can be taken to be 850 J/kg.°C for simplicity in analysis. Analysis (a) Approximate solution This problem can be solved approximately by using an average temperature for the device when evaluating the heat loss. An energy balance on the device can be expressed as device generation out device generation out in T mc t E t Q E E E E pΔ = Δ + Δ − ⎯→ ⎯ Δ = + − & & or, ) ( 2 generation ∞ ∞ ∞ − = Δ ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ − + − Δ T T mc t T T T hA t E p s & Substituting the given values, C ) 25 )( C J/kg. 850 )( kg 02 .
0 ( ) s 60 5 ( C 2 25 ) m 0004 .
0 )( C . W/m 12 ( ) s 60 5 )( J/s 20 ( o 2 2 ° − ° = × ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ − ° − × T T which gives T = 363.6°C If the device were attached to an aluminum heat sink, the temperature of the device would be C ) 25 )( C J/kg. 850 ( kg ) 02 .
0 20 .
0 ( ) s 60 5 ( C 2 25 ) m 0084 .
0 )( C . W/m 12 ( ) s 60 5 )( J/s 20 ( 2 2 ° − ° × + = × ° ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ − ° − × T T which gives T = 54.7°C Note that the temperature of the electronic device drops considerably as a result of attaching it to a heat sink. (b) Exact solution This problem can be solved exactly by obtaining the differential equation from an energy balance on the device for a differential time interval dt. We will get p p s mc E T T mc hA dt T T d generation ) ( ) ( & = − + − ∞ ∞ It can be solved to give ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ − − + = ∞ ) exp( 1 ) ( generation t mc hA hA E T t T p s s & Substituting the known quantities and solving for t gives 363.4°C for the first case and 54.6°C for the second case, which are practically identical to the results obtained from the approximate analysis. Electronic device 20 W PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 11-18 Transient Heat Conduction in Large Plane Walls, Long Cylinders, and Spheres with Spatial Effects 11-28C A cylinder whose diameter is small relative to its length can be treated as an infinitely long cylinder. When the diameter and length of the cylinder are comparable, it is not proper to treat the cylinder as being infinitely long. It is also not proper to use this model when finding the temperatures near the bottom or top surfaces of a cylinder since heat transfer at those locations can be two-dimensional. 11-29C Yes. A plane wall whose one side is insulated is equivalent to a plane wall that is twice as thick and is exposed to convection from both sides. The midplane in the latter case will behave like an insulated surface because of thermal symmetry. 11-30C The solution for determination of the one-dimensional transient temperature distribution involves many variables that make the graphical representation of the results impractical. In order to reduce the number of parameters, some variables are grouped into dimensionless quantities. 11-31C The Fourier number is a measure of heat conducted through a body relative to the heat stored. Thus a large value of Fourier number indicates faster propagation of heat through body. Since Fourier number is proportional to time, doubling the time will also double the Fourier number. 11-32C This case can be handled by setting the heat transfer coefficient h to infinity ∞ since the temperature of the surrounding medium in this case becomes equivalent to the surface temperature. 11-33C The maximum possible amount of heat transfer will occur when the temperature of the body reaches the temperature of the medium, and can be determined from ) ( max i p T T mc Q − = ∞ . 11-34C When the Biot number is less than 0.1, the temperature of the sphere will be nearly uniform at all times. Therefore, it is more convenient to use the lumped system analysis in this case. 11-35 A student calculates the total heat transfer from a spherical copper ball. It is to be determined whether his/her result is reasonable. Assumptions The thermal properties of the copper ball are constant at room temperature. Properties The density and specific heat of the copper ball are ρ = 8933 kg/m3, and cp = 0.385 kJ/kg.°C (Table A-24). Analysis The mass of the copper ball and the maximum amount of heat transfer from the copper ball are kJ 1838 C ) 25 200 )( C kJ/kg. 385 .
0 )( kg 28 .
27 ( ] [ kg 28 .
27 6 m) 18 .
0 ( ) kg/m 8933 ( 6 max 3 3 3 = ° − ° = − = = ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎣ ⎡ = ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ = = ∞ T T mc Q D m i p π π ρ ρV Discussion The student's result of 3150 kJ is not reasonable since it is greater than the maximum possible amount of heat transfer. Copper ball, 200°C Q PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 11-19 11-36 Tomatoes are placed into cold water to cool them. The heat transfer coefficient and the amount of heat transfer are to be determined. Assumptions 1 The tomatoes are spherical in shape. 2 Heat conduction in the tomatoes is one-dimensional because of symmetry about the midpoint. 3 The thermal properties of the tomatoes are constant. 4 The heat transfer coefficient is constant and uniform over the entire surface. 5 The Fourier number is τ > 0.2 so that the one-term approximate solutions (or the transient temperature charts) are applicable (this assumption will be verified). Properties The properties of the tomatoes are given to be k = 0.59 W/m.°C, α = 0.141×10-6 m2/s, ρ = 999 kg/m3 and cp = 3.99 kJ/kg.°C. Analysis The Fourier number is 635 .
0 m) 04 .
0 ( s) 3600 /s)(2 m 10 141 .
0 ( 2 2 6 2 = × × = = − o r t α τ which is greater than 0.2. Therefore one-term solution is applicable. The ratio of the dimensionless temperatures at the surface and center of the tomatoes are 1 1 1 1 1 1 0 0 sph 0, sph s, ) sin( ) sin( 2 1 2 1 λ λ λ λ θ θ τ λ τ λ = = − − = − − − − = − − ∞ ∞ ∞ ∞ ∞ ∞ e A e A T T T T T T T T T T T T s i i s Substituting, 0401 .
3 ) sin( 7 10 7 1 .
7 1 1 1 = ⎯→ ⎯ = − − λ λ λ From Table 11-2, the corresponding Biot number and the heat transfer coefficient are C . W/m 459 2 ° = ° = = ⎯→ ⎯ = = ) m 04 .
0 ( ) 1 .
31 )( C W/m.
59 .
0 ( Bi 1 .
31 Bi o o r kBi h k hr The maximum amount of heat transfer is kJ 196.6 C ) 7 30 )( C kJ/kg. 99 .
3 )( kg 143 .
2 ( ] [ kg 143 .
2 ] 6 / m) 08 .
0 ( )[ kg/m 999 ( 8 6 / 8 8 max 3 3 3 = ° − ° = − = = = = = ∞ T T mc Q D m i p π ρπ ρV Then the actual amount of heat transfer becomes kJ 188 = = = = − ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ − − − = − ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ − − − = ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ ∞ ∞ ) kJ 6 .
196 ( 9565 .
0 9565 .
0 9565 .
0 ) 0401 .
3 ( ) 0401 .
3 cos( ) 0401 .
3 ( ) 0401 .
3 sin( 7 30 7 10 3 1 cos sin 3 1 max 3 3 1 1 1 1 0 cyl max Q Q Q T T T T Q Q i λ λ λ λ Tomato Ti = 30°C Water 7°C PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 11-20 11-37 An egg is dropped into boiling water. The cooking time of the egg is to be determined. Assumptions 1 The egg is spherical in shape with a radius of ro = 2.75 cm. 2 Heat conduction in the egg is one-dimensional because of symmetry about the midpoint. 3 The thermal properties of the egg are constant. 4 The heat transfer coefficient is constant and uniform over the entire surface. 4 The Fourier number is τ > 0.2 so that the one-term approximate solutions (or the transient temperature charts) are applicable (this assumption will be verified). Properties The thermal conductivity and diffusivity of the eggs are given to be k = 0.6 W/m.°C and α = 0.14×10-6 m2/s. Analysis The Biot number for this process is 2 .
64 ) C W/m.
6 .
0 ( ) m 0275 .
0 )( C . W/m 1400 ( 2 = ° ° = = k hr Bi o The constants 1 1 and A λ corresponding to this Biot number are, from Table 11-2, 9969 .
1 and 0877 .
3 1 1 = = A λ Then the Fourier number becomes 2 .
0 198 .
0 ) 9969 .
1 ( 97 8 97 70 2 2 1 ) 0877 .
3 ( 1 0 , 0 ≈ = ⎯→ ⎯ = − − ⎯→ ⎯ = − − = − − ∞ ∞ τ θ τ τ λ e e A T T T T i sph Therefore, the one-term approximate solution (or the transient temperature charts) is applicable. Then the time required for the temperature of the center of the egg to reach 70°C is determined to be min 17.8 = = × = = − s 1070 /s) m 10 14 .
0 ( m) 0275 .
0 )( 198 .
0 ( 2 6 2 2 α τ o r t Water 97°C Egg Ti = 8°C PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 11-21 11-38 EES Prob. 11-37 is reconsidered. The effect of the final center temperature of the egg on the time it will take for the center to reach this temperature is to be investigated. Analysis The problem is solved using EES, and the solution is given below. "GIVEN" D=0.055 [m] T_i=8 [C] T_o=70 [C] T_infinity=97 [C] h=1400 [W/m^2-C] "PROPERTIES" k=0.6 [W/m-C] alpha=0.14E-6 [m^2/s] "ANALYSIS" Bi=(hr_o)/k r_o=D/2 "From Table 11-2 corresponding to this Bi number, we read" lambda_1=1.9969 A_1=3.0863 (T_o-T_infinity)/(T_i-T_infinity)=A_1exp(-lambda_1^2tau) time=(taur_o^2)/alphaConvert(s, min) To [C] time [min] 50 39.86 55 42.4 60 45.26 65 48.54 70 52.38 75 57 80 62.82 85 70.68 90 82.85 95 111.1 50 55 60 65 70 75 80 85 90 95 30 40 50 60 70 80 90 100 110 120 To [C] time [min] PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 11-22 11-39 Large brass plates are heated in an oven. The surface temperature of the plates leaving the oven is to be determined. Assumptions 1 Heat conduction in the plate is one-dimensional since the plate is large relative to its thickness and there is thermal symmetry about the center plane. 3 The thermal properties of the plate are constant. 4 The heat transfer coefficient is constant and uniform over the entire surface. 5 The Fourier number is τ > 0.2 so that the one-term approximate solutions (or the transient temperature charts) are applicable (this assumption will be verified). Properties The properties of brass at room temperature are given to be k = 110 W/m.°C, α = 33.9×10-6 m2/s Analysis The Biot number for this process is 0109 .
0 ) C W/m.
110 ( ) m 015 .
0 )( C . W/m 80 ( 2 = ° ° = = k hL Bi The constants 1 1 and A λ corresponding to this Biot number are, from Table 11-2, 0018 .
1 and 1035 .
0 1 1 = = A λ The Fourier number is 2 .
0 4 .
90 m) 015 .
0 ( s/min) 60 min 10 /s)( m 10 9 .
33 ( 2 2 6 2 > = × × = = − L t α τ Therefore, the one-term approximate solution (or the transient temperature charts) is applicable. Then the temperature at the surface of the plates becomes C 445 ° = ⎯→ ⎯ = − − = = = − − = − − ∞ ∞ ) , ( 378 .
0 700 25 700 ) , ( 378 .
0 ) 1035 .
0 cos( ) 0018 .
1 ( ) / cos( ) , ( ) , ( ) 4 .
90 ( ) 1035 .
0 ( 1 1 2 2 1 t L T t L T e L L e A T T T t x T t L i wall λ θ τ λ Discussion This problem can be solved easily using the lumped system analysis since Bi < 0.1, and thus the lumped system analysis is applicable. It gives C 448 ° = ° + ° = − + = → = − − = ° ⋅ ⋅ × ° ⋅ = = = = = ° ⋅ ⋅ × = × ° ⋅ = = → = − − ∞ ∞ − ∞ ∞ s) 600 )( s 001644 .
0 ( 1 -3 6 2 3 6 2 6 -1 -C) 700 -(25 C 700 ) ( ) ( ) ( s 001644 .
0 C) s/m W 10 245 .
3 m)( 015 .
0 ( C W/m 80 ) / ( ) ( C s/m W 10 245 .
3 / m 10 33.9 C W/m 110 e e T T T t T e T T T t T k L h Lc h c LA hA c hA b s k c c k bt i bt i p p p p p α ρ ρ ρ α ρ ρ α V which is almost identical to the result obtained above. Plates 25°C PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 11-23 11-40 EES Prob. 11-39 is reconsidered. The effects of the temperature of the oven and the heating time on the final surface temperature of the plates are to be investigated. Analysis The problem is solved using EES, and the solution is given below. "GIVEN" L=(0.03/2) [m] T_i=25 [C] T_infinity=700 [C] time=10 [min] h=80 [W/m^2-C] "PROPERTIES" k=110 [W/m-C] alpha=33.9E-6 [m^2/s] "ANALYSIS" Bi=(hL)/k "From Table 11-2, corresponding to this Bi number, we read" lambda_1=0.1039 A_1=1.0018 tau=(alphatimeConvert(min, s))/L^2 (T_L-T_infinity)/(T_i-T_infinity)=A_1exp(-lambda_1^2tau)Cos(lambda_1L/L) T∞ [C] TL [C] 500 321.6 525 337.2 550 352.9 575 368.5 600 384.1 625 399.7 650 415.3 675 430.9 700 446.5 725 462.1 750 477.8 775 493.4 800 509 825 524.6 850 540.2 875 555.8 900 571.4 500 550 600 650 700 750 800 850 900 300 350 400 450 500 550 600 T∞ [C] TL [C] PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 11-24 time [min] TL [C] 2 146.7 4 244.8 6 325.5 8 391.9 10 446.5 12 491.5 14 528.5 16 558.9 18 583.9 20 604.5 22 621.4 24 635.4 26 646.8 28 656.2 30 664 0 5 10 15 20 25 30 100 200 300 400 500 600 700 time [min] TL [C] PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 11-25 11-41 A long cylindrical shaft at 400°C is allowed to cool slowly. The center temperature and the heat transfer per unit length of the cylinder are to be determined. Assumptions 1 Heat conduction in the shaft is one-dimensional since it is long and it has thermal symmetry about the center line. 2 The thermal properties of the shaft are constant. 3 The heat transfer coefficient is constant and uniform over the entire surface. 4 The Fourier number is τ > 0.2 so that the one-term approximate solutions (or the transient temperature charts) are applicable (this assumption will be verified). Properties The properties of stainless steel 304 at room temperature are given to be k = 14.9 W/m.°C, ρ = 7900 kg/m3, cp = 477 J/kg.°C, α = 3.95×10-6 m2/s Analysis First the Biot number is calculated to be 705 .
0 ) C W/m.
9 .
14 ( ) m 175 .
0 )( C . W/m 60 ( 2 = ° ° = = k hr Bi o The constants 1 1 and A λ corresponding to this Biot number are, from Table 11-2, 1548 .
1 and 0904 .
1 1 1 = = A λ The Fourier number is 1548 .
0 m) 175 .
0 ( s) 60 /s)(20 m 10 95 .
3 ( 2 2 6 2 = × × = = − L t α τ which is very close to the value of 0.2. Therefore, the one-term approximate solution (or the transient temperature charts) can still be used, with the understanding that the error involved will be a little more than 2 percent. Then the temperature at the center of the shaft becomes C 390 ° = ⎯→ ⎯ = − − = = = − − = − − ∞ ∞ 0 0 ) 1548 .
0 ( ) 0904 .
1 ( 1 0 , 0 9607 .
0 150 400 150 9607 .
0 ) 1548 .
1 ( 2 2 1 T T e e A T T T T i cyl τ λ θ The maximum heat can be transferred from the cylinder per meter of its length is kJ 640 , 90 C ) 150 400 )( C kJ/kg. 477 .
0 )( kg 1 .
760 ( ] [ kg 1 .
760 )] m 1 ( m) 175 .
0 ( )[ kg/m 7900 ( max 2 3 2 = ° − ° = − = = = = = ∞ i p o T T mc Q L r m π ρπ ρV Once the constant 1 J = 0.4679 is determined from Table 11-3 corresponding to the constant 1 λ =1.0904, the actual heat transfer becomes kJ 15,960 = = = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ − − − = ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ − − − = ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ ∞ ∞ ) kJ 640 , 90 ( 1761 .
0 1761 .
0 0904 .
1 4679 .
0 150 400 150 390 2 1 ) ( 2 1 1 1 1 max Q J T T T T Q Q i o cyl λ λ Steel shaft Ti = 400°C Air T∞ = 150°C PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 11-26 11-42 EES Prob. 11-41 is reconsidered. The effect of the cooling time on the final center temperature of the shaft and the amount of heat transfer is to be investigated. Analysis The problem is solved using EES, and the solution is given below. "GIVEN" r_o=(0.35/2) [m] T_i=400 [C] T_infinity=150 [C] h=60 [W/m^2-C] time=20 [min] "PROPERTIES" k=14.9 [W/m-C] rho=7900 [kg/m^3] c_p=477 [J/kg-C] alpha=3.95E-6 [m^2/s] "ANALYSIS" Bi=(hr_o)/k "From Table 11-2 corresponding to this Bi number, we read" lambda_1=1.0935 A_1=1.1558 J_1=0.4709 "From Table 11-3, corresponding to lambda_1" tau=(alphatimeConvert(min, s))/r_o^2 (T_o-T_infinity)/(T_i-T_infinity)=A_1exp(-lambda_1^2tau) L=1 "[m], 1 m length of the cylinder is considered" V=pir_o^2L m=rhoV Q_max=mc_p(T_i-T_infinity)Convert(J, kJ) Q/Q_max=1-2(T_o-T_infinity)/(T_i-T_infinity)J_1/lambda_1 time [min] To [C] Q [kJ] 5 425.9 4491 10 413.4 8386 15 401.5 12105 20 390.1 15656 25 379.3 19046 30 368.9 22283 35 359 25374 40 349.6 28325 45 340.5 31142 50 331.9 33832 55 323.7 36401 60 315.8 38853 0 10 20 30 40 50 60 300 320 340 360 380 400 420 440 0 5000 10000 15000 20000 25000 30000 35000 40000 time [min] To [C] Q [kJ] temperature heat PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 11-27 11-43E Long cylindrical steel rods are heat-treated in an oven. Their centerline temperature when they leave the oven is to be determined. Assumptions 1 Heat conduction in the rods is one-dimensional since the rods are long and they have thermal symmetry about the center line. 2 The thermal properties of the rod are constant. 3 The heat transfer coefficient is constant and uniform over the entire surface. 4 The Fourier number is τ > 0.2 so that the one-term approximate solutions (or the transient temperature charts) are applicable (this assumption will be verified). Properties The properties of AISI stainless steel rods are given to be k = 7.74 Btu/h.ft.°F, α = 0.135 ft2/h. Analysis The time the steel rods stays in the oven can be determined from s 180 = min 3 ft/min 7 ft 21 velocity length = = = t The Biot number is 4307 .
0 ) F Btu/h.ft. 74 .
7 ( ) ft 12 / 2 )( F .
Btu/h.ft 20 ( 2 = ° ° = = k hr Bi o The constants 1 1 and A λ corresponding to this Biot number are, from Table 11-2, 0996 .
1 and 8790 .
0 1 1 = = A λ The Fourier number is 243 .
0 ft) 12 / 2 ( h) /h)(3/60 ft 135 .
0 ( 2 2 2 = = = o r t α τ Then the temperature at the center of the rods becomes 911 .
0 ) 0996 .
1 ( ) 243 .
0 ( ) 8790 .
0 ( 1 0 , 0 2 2 1 = = = − − = − − ∞ ∞ e e A T T T T i cyl τ λ θ F 215° = ⎯→ ⎯ = − − o T T 911 .
0 1700 70 1700 0 Oven, 1700°F Steel rod, 70°F PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 11-28 11-44 Steaks are cooled by passing them through a refrigeration room. The time of cooling is to be determined. Assumptions 1 Heat conduction in the steaks is one-dimensional since the steaks are large relative to their thickness and there is thermal symmetry about the center plane. 3 The thermal properties of the steaks are constant. 4 The heat transfer coefficient is constant and uniform over the entire surface. 5 The Fourier number is τ > 0.2 so that the one-term approximate solutions (or the transient temperature charts) are applicable (this assumption will be verified). Properties The properties of steaks are given to be k = 0.45 W/m.°C and α = 0.91×10-7 m2/s Analysis The Biot number is 200 .
0 ) C W/m.
45 .
0 ( ) m 01 .
0 )( C . W/m 9 ( 2 = ° ° = = k hL Bi The constants 1 1 and A λ corresponding to this Biot number are, from Table 11-2, 0311 .
1 and 4328 .
0 1 1 = = A λ The Fourier number is 2 .
0 085 .
5 ) 4328 .
0 cos( ) 0311 .
1 ( ) 11 ( 25 ) 11 ( 2 ) / cos( ) , ( 2 2 1 ) 4328 .
0 ( 1 1 > = ⎯→ ⎯ = − − − − = − − − − ∞ ∞ τ λ τ τ λ e L L e A T T T t L T i Therefore, the one-term approximate solution (or the transient temperature charts) is applicable. Then the length of time for the steaks to be kept in the refrigerator is determined to be min 93.1 = = × = = − s 5590 /s m 10 91 .
0 m) 01 .
0 )( 085 .
5 ( 2 7 2 2 α τL t Steaks 25°C Refrigerated air -11°C PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 11-29 11-45 A long cylindrical wood log is exposed to hot gases in a fireplace. The time for the ignition of the wood is to be determined. Assumptions 1 Heat conduction in the wood is one-dimensional since it is long and it has thermal symmetry about the center line. 2 The thermal properties of the wood are constant. 3 The heat transfer coefficient is constant and uniform over the entire surface. 4 The Fourier number is τ > 0.2 so that the one-term approximate solutions (or the transient temperature charts) are applicable (this assumption will be verified). Properties The properties of wood are given to be k = 0.17 W/m.°C, α = 1.28×10-7 m2/s Analysis The Biot number is 00 .
4 ) C W/m.
17 .
0 ( ) m 05 .
0 )( C . W/m 6 .
13 ( 2 = ° ° = = k hr Bi o The constants 1 1 and A λ corresponding to this Biot number are, from Table 11-2, 4698 .
1 and 9081 .
1 1 1 = = A λ Once the constant J0 is determined from Table 11-3 corresponding to the constant λ1=1.9081, the Fourier number is determined to be 142 .
0 ) 2771 .
0 ( ) 4698 .
1 ( 550 15 550 420 ) / ( ) , ( 2 2 1 ) 9081 .
1 ( 1 0 1 = ⎯→ ⎯ = − − = − − − − ∞ ∞ τ λ τ τ λ e r r J e A T T T t r T o o i o which is not above the value of 0.2 but it is close. We use one-term approximate solution (or the transient temperature charts) knowing that the result may be somewhat in error. Then the length of time before the log ignites is min 46.2 = = × = = − s 2770 /s) m 10 28 .
1 ( m) 05 .
0 )( 142 .
0 ( 2 7 2 2 α τ o r t 10 cm Wood log, 15°C Hot gases 550°C PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 11-30 11-46 A rib is roasted in an oven. The heat transfer coefficient at the surface of the rib, the temperature of the outer surface of the rib and the amount of heat transfer when it is rare done are to be determined. The time it will take to roast this rib to medium level is also to be determined. Assumptions 1 The rib is a homogeneous spherical object. 2 Heat conduction in the rib is one-dimensional because of symmetry about the midpoint. 3 The thermal properties of the rib are constant. 4 The heat transfer coefficient is constant and uniform over the entire surface. 5 The Fourier number is τ > 0.2 so that the one-term approximate solutions (or the transient temperature charts) are applicable (this assumption will be verified). Properties The properties of the rib are given to be k = 0.45 W/m.°C, ρ = 1200 kg/m3, cp = 4.1 kJ/kg.°C, and α = 0.91×10-7 m2/s. Analysis (a) The radius of the roast is determined to be m 08603 .
0 4 ) m 002667 .
0 ( 3 4 3 3 4 m 002667 .
0 kg/m 1200 kg 2 .
3 3 3 3 3 3 3 = = = ⎯→ ⎯ = = = = ⎯→ ⎯ = π π π ρ ρ V V V V o o r r m m The Fourier number is 1217 .
0 m) 08603 .
0 ( 60)s 45 + 3600 /s)(2 m 10 91 .
0 ( 2 2 7 2 = × × × = = − o r t α τ which is somewhat below the value of 0.2. Therefore, the one-term approximate solution (or the transient temperature charts) can still be used, with the understanding that the error involved will be a little more than 2 percent. Then the one-term solution can be written in the form ) 1217 .
0 ( 1 1 0 , 0 2 1 2 1 65 .
0 163 5 .
4 163 60 λ τ λ θ − − ∞ ∞ = = − − ⎯→ ⎯ = − − = e A e A T T T T i sph It is determined from Table 11-2 by trial and error that this equation is satisfied when Bi = 30, which corresponds to 9898 .
1 and 0372 .
3 1 1 = = A λ . Then the heat transfer coefficient can be determined from C . W/m 156.9 2 ° = ° = = ⎯→ ⎯ = ) m 08603 .
0 ( ) 30 )( C W/m.
45 .
0 ( o o r kBi h k hr Bi This value seems to be larger than expected for problems of this kind. This is probably due to the Fourier number being less than 0.2. (b) The temperature at the surface of the rib is C 159.5 ° = ⎯→ ⎯ = − − = = − − = − − ∞ ∞ ) , ( 0222 .
0 163 5 .
4 163 ) , ( 0372 .
3 ) rad 0372 .
3 sin( ) 9898 .
1 ( / ) / sin( ) , ( ) , ( ) 1217 .
0 ( ) 0372 .
3 ( 1 1 1 2 2 1 t r T t r T e r r r r e A T T T t r T t r o o o o o o i o sph o λ λ θ τ λ (c) The maximum possible heat transfer is kJ 2080 C ) 5 .
4 163 )( C kJ/kg. 1 .
4 )( kg 2 .
3 ( ) ( max = ° − ° = − = ∞ i p T T mc Q Then the actual amount of heat transfer becomes kJ 1629 = = = = − − = − − = kJ) 2080 )( 783 .
0 ( 783 .
0 783 .
0 ) 0372 .
3 ( ) 0372 .
3 cos( ) 0372 .
3 ( ) 0372 .
3 sin( ) 65 .
0 ( 3 1 ) cos( ) sin( 3 1 max 3 3 1 1 1 1 , max Q Q Q Q sph o λ λ λ λ θ (d) The cooking time for medium-done rib is determined to be hr 3 ≅ = = × = = = ⎯→ ⎯ = − − ⎯→ ⎯ = − − = − − − ∞ ∞ min 181 s 866 , 10 /s) m 10 91 .
0 ( m) 08603 .
0 )( 1336 .
0 ( 1336 .
0 ) 9898 .
1 ( 163 5 .
4 163 71 2 7 2 2 ) 0372 .
3 ( 1 0 , 0 2 2 1 α τ τ θ τ τ λ o i sph r t e e A T T T T This result is close to the listed value of 3 hours and 20 minutes. The difference between the two results is due to the Fourier number being less than 0.2 and thus the error in the one-term approximation. Discussion The temperature of the outer parts of the rib is greater than that of the inner parts of the rib after it is taken out of the oven. Therefore, there will be a heat transfer from outer parts of the rib to the inner parts as a result of this temperature difference. The recommendation is logical. Oven 163°C Rib 4.5°C PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 11-31 11-47 A rib is roasted in an oven. The heat transfer coefficient at the surface of the rib, the temperature of the outer surface of the rib and the amount of heat transfer when it is well-done are to be determined. The time it will take to roast this rib to medium level is also to be determined. Assumptions 1 The rib is a homogeneous spherical object. 2 Heat conduction in the rib is one-dimensional because of symmetry about the midpoint. 3 The thermal properties of the rib are constant. 4 The heat transfer coefficient is constant and uniform over the entire surface. 5 The Fourier number is τ > 0.2 so that the one-term approximate solutions (or the transient temperature charts) are applicable (this assumption will be verified). Properties The properties of the rib are given to be k = 0.45 W/m.°C, ρ = 1200 kg/m3, cp = 4.1 kJ/kg.°C, and α = 0.91×10-7 m2/s Analysis (a) The radius of the rib is determined to be m 08603 .
0 4 ) m 00267 .
0 ( 3 4 3 3 4 m 00267 .
0 kg/m 1200 kg 2 .
3 3 3 3 3 3 3 = = = ⎯→ ⎯ = = = = ⎯→ ⎯ = π π π ρ ρ V V V V o o r r m m The Fourier number is 1881 .
0 m) 08603 .
0 ( 60)s 15 + 3600 /s)(4 m 10 91 .
0 ( 2 2 7 2 = × × × = = − o r t α τ which is somewhat below the value of 0.2. Therefore, the one-term approximate solution (or the transient temperature charts) can still be used, with the understanding that the error involved will be a little more than 2 percent. Then the one-term solution formulation can be written in the form ) 1881 .
0 ( 1 1 0 , 0 2 1 2 1 543 .
0 163 5 .
4 163 77 λ τ λ θ − − ∞ ∞ = = − − ⎯→ ⎯ = − − = e A e A T T T T i sph It is determined from Table 11-2 by trial and error that this equation is satisfied when Bi = 4.3, which corresponds to 7402 .
1 and 4900 .
2 1 1 = = A λ . Then the heat transfer coefficient can be determined from. C . W/m 22.5 2 ° = ° = = ⎯→ ⎯ = ) m 08603 .
0 ( ) 3 .
4 )( C W/m.
45 .
0 ( o o r kBi h k hr Bi (b) The temperature at the surface of the rib is C 142.1° = ⎯→ ⎯ = − − = = − − = − − ∞ ∞ ) , ( 132 .
0 163 5 .
4 163 ) , ( 49 .
2 ) 49 .
2 sin( ) 7402 .
1 ( / ) / sin( ) , ( ) , ( ) 1881 .
0 ( ) 49 .
2 ( 1 1 1 2 2 1 t r T t r T e r r r r e A T T T t r T t r o o o o o o i o sph o λ λ θ τ λ (c) The maximum possible heat transfer is kJ 2080 C ) 5 .
4 163 )( C kJ/kg. 1 .
4 )( kg 2 .
3 ( ) ( max = ° − ° = − = ∞ i p T T mc Q Then the actual amount of heat transfer becomes kJ 1512 = = = = − − = − − = kJ) 2080 )( 727 .
0 ( 727 .
0 727 .
0 ) 49 .
2 ( ) 49 .
2 cos( ) 49 .
2 ( ) 49 .
2 sin( ) 543 .
0 ( 3 1 ) cos( ) sin( 3 1 max 3 3 1 1 1 1 , max Q Q Q Q sph o λ λ λ λ θ (d) The cooking time for medium-done rib is determined to be hr 4 = = = × = = = ⎯→ ⎯ = − − ⎯→ ⎯ = − − = − − − ∞ ∞ min 240 s 400 , 14 /s) m 10 91 .
0 ( m) 08603 .
0 )( 177 .
0 ( 177 .
0 ) 7402 .
1 ( 163 5 .
4 163 71 2 7 2 2 ) 49 .
2 ( 1 0 , 0 2 2 1 α τ τ θ τ τ λ o i sph r t e e A T T T T This result is close to the listed value of 4 hours and 15 minutes. The difference between the two results is probably due to the Fourier number being less than 0.2 and thus the error in the one-term approximation. Discussion The temperature of the outer parts of the rib is greater than that of the inner parts of the rib after it is taken out of the oven. Therefore, there will be a heat transfer from outer parts of the rib to the inner parts as a result of this temperature difference. The recommendation is logical. Oven 163°C Rib 4.5°C PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 11-32 11-48 An egg is dropped into boiling water. The cooking time of the egg is to be determined. Assumptions 1 The egg is spherical in shape with a radius of r0 = 2.75 cm. 2 Heat conduction in the egg is one-dimensional because of symmetry about the midpoint. 3 The thermal properties of the egg are constant. 4 The heat transfer coefficient is constant and uniform over the entire surface. 5 The Fourier number is τ > 0.2 so that the one-term approximate solutions (or the transient temperature charts) are applicable (this assumption will be verified). Properties The thermal conductivity and diffusivity of the eggs can be approximated by those of water at room temperature to be k = 0.607 W/m.°C, α p c k ρ / = = 0.146×10-6 m2/s (Table A-15). Analysis The Biot number is 2 .
36 ) C W/m.
607 .
0 ( ) m 0275 .
0 )( C . W/m 800 ( 2 = ° ° = = k hr Bi o The constants 1 1 and A λ corresponding to this Biot number are, from Table 11-2, 9925 .
1 and 0533 .
3 1 1 = = A λ Then the Fourier number and the time period become 1633 .
0 ) 9925 .
1 ( 100 8 100 60 2 2 1 ) 0533 .
3 ( 1 0 , 0 = ⎯→ ⎯ = − − ⎯→ ⎯ = − − = − − ∞ ∞ τ θ τ τ λ e e A T T T T i sph which is somewhat below the value of 0.2. Therefore, the one-term approximate solution (or the transient temperature charts) can still be used, with the understanding that the error involved will be a little more than 2 percent. Then the length of time for the egg to be kept in boiling water is determined to be min 14.1 = = × = = − s 846 /s m 10 146 .
0 m) 0275 .
0 )( 1633 .
0 ( 2 6 2 2 α τ o r t Water 100°C Egg Ti = 8°C PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 11-33 11-49 An egg is cooked in boiling water. The cooking time of the egg is to be determined for a location at 1610-m elevation. Assumptions 1 The egg is spherical in shape with a radius of ro = 2.75 cm. 2 Heat conduction in the egg is one-dimensional because of symmetry about the midpoint. 3 The thermal properties of the egg and heat transfer coefficient are constant. 4 The heat transfer coefficient is constant and uniform over the entire surface. 5 The Fourier number is τ > 0.2 so that the one-term approximate solutions (or the transient temperature charts) are applicable (this assumption will be verified). Properties The thermal conductivity and diffusivity of the eggs can be approximated by those of water at room temperature to be k = 0.607 W/m.°C, α p c k ρ / = = 0.146×10-6 m2/s (Table A-15). Analysis The Biot number is 2 .
36 ) C W/m.
607 .
0 ( ) m 0275 .
0 )( C . W/m 800 ( 2 = ° ° = = k hr Bi o The constants 1 1 and A λ corresponding to this Biot number are, from Table 11-2, 9925 .
1 and 0533 .
3 1 1 = = A λ Then the Fourier number and the time period become 1727 .
0 ) 9925 .
1 ( 4 .
94 8 4 .
94 60 2 2 1 ) 0533 .
3 ( 1 0 , 0 = ⎯→ ⎯ = − − ⎯→ ⎯ = − − = − − ∞ ∞ τ θ τ τ λ e e A T T T T i sph which is somewhat below the value of 0.2. Therefore, the one-term approximate solution (or the transient temperature charts) can still be used, with the understanding that the error involved will be a little more than 2 percent. Then the length of time for the egg to be kept in boiling water is determined to be min 14.9 = = × = = − s 895 /s) m 10 146 .
0 ( m) 0275 .
0 )( 1727 .
0 ( 2 6 2 2 α τ o r t Water 94.4°C Egg Ti = 8°C PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 11-34 11-50 A hot dog is dropped into boiling water, and temperature measurements are taken at certain time intervals. The thermal diffusivity and thermal conductivity of the hot dog and the convection heat transfer coefficient are to be determined. Assumptions 1 Heat conduction in the hot dog is one-dimensional since it is long and it has thermal symmetry about the centerline. 2 The thermal properties of the hot dog are constant. 3 The heat transfer coefficient is constant and uniform over the entire surface. 4 The Fourier number is τ > 0.2 so that the one-term approximate solutions (or the transient temperature charts) are applicable (this assumption will be verified). Properties The properties of hot dog available are given to be ρ = 980 kg/m3 and cp = 3900 J/kg.°C. Analysis (a) From Fig. 11-16b we have 15 .
0 1 1 17 .
0 94 59 94 88 0 = = ⎪ ⎪ ⎭ ⎪ ⎪ ⎬ ⎫ = = = − − = − − ∞ ∞ o o o o hr k Bi r r r r T T T T The Fourier number is determined from Fig. 11-16a to be 20 .
0 47 .
0 94 20 94 59 15 .
0 1 2 0 = = ⎪ ⎪ ⎭ ⎪ ⎪ ⎬ ⎫ = − − = − − = = ∞ ∞ o i o r t T T T T hr k Bi α τ The thermal diffusivity of the hot dog is determined to be /s m 10 2.017 2 7 − × = = = ⎯→ ⎯ = s 120 m) 011 .
0 )( 2 .
0 ( 2 .
0 20 .
0 2 2 2 t r r t o o α α (b) The thermal conductivity of the hot dog is determined from C W/m.
0.771 ° = ° × = = − C) J/kg. )(3900 kg/m /s)(980 m 10 017 .
2 ( 3 2 7 p c k αρ (c) From part (a) we have 15 .
0 1 = = o hr k Bi . Then, m 0.00165 m) 011 .
0 )( 15 .
0 ( 15 .
0 = = = o r h k Therefore, the heat transfer coefficient is C . W/m 467 2 ° = ° = ⎯→ ⎯ = m 0.00165 C W/m.
771 .
0 00165 .
0 h h k Water 94°C Hot dog PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 11-35 11-51 Using the data and the answers given in Prob. 11-50, the center and the surface temperatures of the hot dog 4 min after the start of the cooking and the amount of heat transferred to the hot dog are to be determined. Assumptions 1 Heat conduction in the hot dog is one-dimensional since it is long and it has thermal symmetry about the center line. 2 The thermal properties of the hot dog are constant. 3 The heat transfer coefficient is constant and uniform over the entire surface. 4 The Fourier number is τ > 0.2 so that the one-term approximate solutions (or the transient temperature charts) are applicable (this assumption will be verified). Properties The properties of hot dog and the convection heat transfer coefficient are given or obtained in P11-50 to be k = 0.771 W/m.°C, ρ = 980 kg/m3, cp = 3900 J/kg.°C, α = 2.017×10-7 m2/s, and h = 467 W/m2.°C. Analysis The Biot number is 66 .
6 ) C W/m.
771 .
0 ( ) m 011 .
0 )( C . W/m 467 ( 2 = ° ° = = k hr Bi o The constants 1 1 and A λ corresponding to this Biot number are, from Table 11-2, 5357 .
1 and 0785 .
2 1 1 = = A λ The Fourier number is 2 .
0 4001 .
0 m) 011 .
0 ( s/min) 60 min /s)(4 m 10 017 .
2 ( 2 2 7 2 > = × × = = − L t α τ Then the temperature at the center of the hot dog is determined to be C 73.8 ° = ⎯→ ⎯ = − − = = = − − = − − ∞ ∞ o i cyl T T e e A T T T T 2727 .
0 94 20 94 2727 .
0 ) 5357 .
1 ( 0 ) 4001 .
0 ( ) 0785 .
2 ( 1 0 , 0 2 2 1 τ λ θ From Table 11-3 we read 0 J =0.1789 corresponding to the constant 1 λ =2.0785. Then the temperature at the surface of the hot dog becomes C 90.4 ° = ⎯→ ⎯ = − − = = = − − − − ∞ ∞ ) , ( 04878 .
0 94 20 94 ) , ( 04878 .
0 ) 1789 .
0 ( ) 5357 .
1 ( ) / ( ) , ( ) 4001 .
0 ( ) 0785 .
2 ( 1 0 1 2 2 1 t r T t r T e r r J e A T T T t r T o o o o i o λ τ λ The maximum possible amount of heat transfer is [ ] J 13,440 C ) 20 94 )( C J/kg. 3900 )( kg 04657 .
0 ( ) ( kg 04657 .
0 m) 125 .
0 ( m) 011 .
0 ( ) kg/m 980 ( max 2.
3 2 = ° − ° = − = = = = = ∞ T T mc Q L r m i p o π ρπ ρV From Table 11-3 we read 1 J = 0.5701 corresponding to the constant 1 λ =2.0785. Then the actual heat transfer becomes kJ 11,430 = = = − = − = ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ ) kJ 440 , 13 ( 8504 .
0 8504 .
0 0785 .
2 5701 .
0 ) 2727 .
0 ( 2 1 ) ( 2 1 1 1 1 , max Q J Q Q cyl o cyl λ λ θ Water 94°C Hot dog PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 11-36 11-52E Whole chickens are to be cooled in the racks of a large refrigerator. Heat transfer coefficient that will enable to meet temperature constraints of the chickens while keeping the refrigeration time to a minimum is to be determined. Assumptions 1 The chicken is a homogeneous spherical object. 2 Heat conduction in the chicken is one-dimensional because of symmetry about the midpoint. 3 The thermal properties of the chicken are constant. 4 The heat transfer coefficient is constant and uniform over the entire surface. 5 The Fourier number is τ > 0.2 so that the one-term approximate solutions (or the transient temperature charts) are applicable (this assumption will be verified). Properties The properties of the chicken are given to be k = 0.26 Btu/h.ft.°F, ρ = 74.9 lbm/ft3, cp = 0.98 Btu/lbm.°F, and α = 0.0035 ft2/h. Analysis The radius of the chicken is determined to be ft 2517 .
0 4 ) ft 06676 .
0 ( 3 4 3 3 4 ft 06676 .
0 lbm/ft 9 .
74 lbm 5 3 3 3 3 3 3 = = = ⎯→ ⎯ = = = = ⎯→ ⎯ = π π π ρ ρ V V V V o o r r m m From Fig. 11-17b we have 2 1 1 75 .
0 5 45 5 35 0 = = ⎪ ⎪ ⎭ ⎪ ⎪ ⎬ ⎫ = = = − − = − − ∞ ∞ o o o o hr k Bi r r r x T T T T Then the heat transfer coefficients becomes F .
Btu/h.ft 0.516 2 ° = ° = = ft) 2517 .
0 ( 2 F Btu/.ft. 26 .
0 2 o r k h Refrigerator T∞ = 5°F Chicken Ti = 65°F PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 11-37 11-53 A person puts apples into the freezer to cool them quickly. The center and surface temperatures of the apples, and the amount of heat transfer from each apple in 1 h are to be determined. Assumptions 1 The apples are spherical in shape with a diameter of 9 cm. 2 Heat conduction in the apples is one-dimensional because of symmetry about the midpoint. 3 The thermal properties of the apples are constant. 4 The heat transfer coefficient is constant and uniform over the entire surface. 5 The Fourier number is τ > 0.2 so that the one-term approximate solutions (or the transient temperature charts) are applicable (this assumption will be verified). Properties The properties of the apples are given to be k = 0.418 W/m.°C, ρ = 840 kg/m3, cp = 3.81 kJ/kg.°C, and α = 1.3×10-7 m2/s. Analysis The Biot number is 861 .
0 ) C W/m.
418 .
0 ( ) m 045 .
0 )( C . W/m 8 ( 2 = ° ° = = k hr Bi o The constants 1 1 and A λ corresponding to this Biot number are, from Table 11-2, 2390 .
1 and 476 .
1 1 1 = = A λ The Fourier number is 2 .
0 231 .
0 m) 045 .
0 ( s/h) 600 3 h /s)(1 m 10 3 .
1 ( 2 2 7 2 > = × × = = − o r t α τ Then the temperature at the center of the apples becomes C 11.2° = ⎯→ ⎯ = = − − − − ⎯→ ⎯ = − − = − − ∞ ∞ 0 ) 231 .
0 ( ) 476 .
1 ( 0 1 0 , 0 749 .
0 ) 239 .
1 ( ) 15 ( 20 ) 15 ( 2 2 1 T e T e A T T T T i sph τ λ θ The temperature at the surface of the apples is C 2.7° = ⎯→ ⎯ = − − − − = = = − − = − − ∞ ∞ ) , ( 505 .
0 ) 15 ( 20 ) 15 ( ) , ( 505 .
0 476 .
1 ) rad 476 .
1 sin( ) 239 .
1 ( / ) / sin( ) , ( ) , ( ) 231 .
0 ( ) 476 .
1 ( 1 1 1 2 2 1 t r T t r T e r r r r e A T T T t r T t r o o o o o o i o sph o λ λ θ τ λ The maximum possible heat transfer is [ ] kJ 75 .
42 C ) 15 ( 20 ) C kJ/kg. 81 .
3 )( kg 3206 .
0 ( ) ( kg 3206 .
0 m) 045 .
0 ( 3 4 ) kg/m 840 ( 3 4 max 3.
3 3 = ° − − ° = − = = ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ = = = ∞ T T mc Q r m i p o π π ρ ρV Then the actual amount of heat transfer becomes kJ 17.2 = = = = − − = − − = kJ) 75 .
42 )( 402 .
0 ( 402 .
0 402 .
0 ) 476 .
1 ( ) rad 476 .
1 cos( ) 476 .
1 ( ) rad 476 .
1 sin( ) 749 .
0 ( 3 1 ) cos( ) sin( 3 1 max 3 3 1 1 1 1 , max Q Q Q Q sph o λ λ λ λ θ Apple Ti = 20°C Air T∞ = -15°C PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 11-38 11-54 EES Prob. 11-53 is reconsidered. The effect of the initial temperature of the apples on the final center and surface temperatures and the amount of heat transfer is to be investigated. Analysis The problem is solved using EES, and the solution is given below. "GIVEN" T_infinity=-15 [C] T_i=20 [C] h=8 [W/m^2-C] r_o=0.09/2 [m] time=13600 [s] "PROPERTIES" k=0.513 [W/m-C] rho=840 [kg/m^3] c_p=3.6 [kJ/kg-C] alpha=1.3E-7 [m^2/s] "ANALYSIS" Bi=(hr_o)/k "From Table 11-2 corresponding to this Bi number, we read" lambda_1=1.3525 A_1=1.1978 tau=(alphatime)/r_o^2 (T_o-T_infinity)/(T_i-T_infinity)=A_1exp(-lambda_1^2tau) (T_r-T_infinity)/(T_i-T_infinity)=A_1exp(-lambda_1^2tau)Sin(lambda_1r_o/r_o)/(lambda_1r_o/r_o) V=4/3pir_o^3 m=rhoV Q_max=mc_p(T_i-T_infinity) Q/Q_max=1-3(T_o-T_infinity)/(T_i-T_infinity)(Sin(lambda_1)-lambda_1Cos(lambda_1))/lambda_1^3 Ti [C] To [C] Tr [C] Q [kJ] 2 -1.658 -5.369 6.861 4 -0.08803 -4.236 7.668 6 1.482 -3.103 8.476 8 3.051 -1.97 9.283 10 4.621 -0.8371 10.09 12 6.191 0.296 10.9 14 7.76 1.429 11.7 16 9.33 2.562 12.51 18 10.9 3.695 13.32 20 12.47 4.828 14.13 22 14.04 5.961 14.93 24 15.61 7.094 15.74 26 17.18 8.227 16.55 28 18.75 9.36 17.35 30 20.32 10.49 18.16 PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 11-39 0 5 10 15 20 25 30 -10 -5 0 5 10 15 20 25 Ti [C] To [C] T0 Tr 0 5 10 15 20 25 30 6 8 10 12 14 16 18 20 Ti [C] Q [kJ] PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 11-40 11-55 An orange is exposed to very cold ambient air. It is to be determined whether the orange will freeze in 4 h in subfreezing temperatures. Assumptions 1 The orange is spherical in shape with a diameter of 8 cm. 2 Heat conduction in the orange is one-dimensional because of symmetry about the midpoint. 3 The thermal properties of the orange are constant, and are those of water. 4 The heat transfer coefficient is constant and uniform over the entire surface. 5 The Fourier number is τ > 0.2 so that the one-term approximate solutions (or the transient temperature charts) are applicable (this assumption will be verified). Properties The properties of the orange are approximated by those of water at the average temperature of about 5°C, k = 0.571 W/m.°C and /s m 10 136 .
0 ) 4205 9 .
999 /( 571 .
0 / 2 6 − × = × = = p c k ρ α (Table A-15). Analysis The Biot number is 0 .
1 051 .
1 ) C W/m.
571 .
0 ( ) m 04 .
0 )( C . W/m 15 ( 2 ≈ = ° ° = = k hr Bi o The constants 1 1 and A λ corresponding to this Biot number are, from Table 11-2, 2732 .
1 and 5708 .
1 1 1 = = A λ The Fourier number is 2 .
0 224 .
1 m) 04 .
0 ( s/h) 600 3 h /s)(4 m 10 136 .
0 ( 2 2 6 2 > = × × = = − o r t α τ Therefore, the one-term approximate solution (or the transient temperature charts) is applicable. Then the temperature at the surface of the oranges becomes C 5.2 -° = ⎯→ ⎯ = − − − − = = = − − = − − ∞ ∞ ) , ( 0396 .
0 ) 6 ( 15 ) 6 ( ) , ( 0396 .
0 5708 .
1 ) rad 5708 .
1 sin( ) 2732 .
1 ( / ) / sin( ) , ( ) , ( ) 224 .
1 ( ) 5708 .
1 ( 1 1 1 2 2 1 t r T t r T e r r r r e A T T T t r T t r o o o o o o i o sph o λ λ θ τ λ which is less than 0°C. Therefore, the oranges will freeze. Orange Ti = 15°C Air T∞ = -6°C PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 11-41 11-56 A hot baked potato is taken out of the oven and wrapped so that no heat is lost from it. The time the potato is baked in the oven and the final equilibrium temperature of the potato after it is wrapped are to be determined. Assumptions 1 The potato is spherical in shape with a diameter of 9 cm. 2 Heat conduction in the potato is one-dimensional because of symmetry about the midpoint. 3 The thermal properties of the potato are constant. 4 The heat transfer coefficient is constant and uniform over the entire surface. 5 The Fourier number is τ > 0.2 so that the one-term approximate solutions (or the transient temperature charts) are applicable (this assumption will be verified). Properties The properties of the potato are given to be k = 0.6 W/m.°C, ρ = 1100 kg/m3, cp = 3.9 kJ/kg.°C, and α = 1.4×10-7 m2/s. Analysis (a) The Biot number is 3 ) C W/m.
6 .
0 ( ) m 045 .
0 )( C . W/m 40 ( 2 = ° ° = = k hr Bi o The constants 1 1 and A λ corresponding to this Biot number are, from Table 11-2, 6227 .
1 and 2889 .
2 1 1 = = A λ Then the Fourier number and the time period become 163 .
0 ) 6227 .
1 ( 69 .
0 170 25 170 70 2 2 1 ) 2889 .
2 ( 1 0 , 0 = ⎯→ ⎯ = = − − ⎯→ ⎯ = − − = − − ∞ ∞ τ θ τ τ λ e e A T T T T i sph which is not greater than 0.2 but it is close. We may use one-term approximation knowing that the result may be somewhat in error. Then the baking time of the potatoes is determined to be min 39.3 = = × = = − s 2358 /s m 10 4 .
1 m) 045 .
0 )( 163 .
0 ( 2 7 2 2 α τ o r t (b) The maximum amount of heat transfer is kJ 237 C ) 25 170 )( C kJ/kg. 900 .
3 )( kg 420 .
0 ( ) ( kg 420 .
0 m) 045 .
0 ( 3 4 ) kg/m 1100 ( 3 4 max 3.
3 3 = ° − ° = − = = ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ = = = ∞ i p o T T mc Q r m π π ρ ρV Then the actual amount of heat transfer becomes kJ 145 = = = = − − = − − = kJ) 237 )( 610 .
0 ( 610 .
0 610 .
0 ) 2889 .
2 ( ) 2889 .
2 cos( ) 2889 .
2 ( ) 2889 .
2 sin( ) 69 .
0 ( 3 1 ) cos( ) sin( 3 1 max 3 3 1 1 1 1 , max Q Q Q Q sph o λ λ λ λ θ The final equilibrium temperature of the potato after it is wrapped is C 114° = ° + ° = + = ⎯→ ⎯ − = ) C kJ/kg. 9 .
3 )( kg 420 .
0 ( kJ 145 C 25 ) ( p i eqv i eqv p mc Q T T T T mc Q Oven T∞ = 170°C Potato T0 = 70°C PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 11-42 11-57 The center temperature of potatoes is to be lowered to 6°C during cooling. The cooling time and if any part of the potatoes will suffer chilling injury during this cooling process are to be determined. Assumptions 1 The potatoes are spherical in shape with a radius of r0 = 3 cm. 2 Heat conduction in the potato is one-dimensional in the radial direction because of the symmetry about the midpoint. 3 The thermal properties of the potato are constant. 4 The heat transfer coefficient is constant and uniform over the entire surface. 5 The Fourier number is τ > 0.2 so that the one-term approximate solutions (or the transient temperature charts) are applicable (this assumption will be verified). Properties The thermal conductivity and thermal diffusivity of potatoes are given to be k = 0.50 W/m⋅°C and α = 0.13×10-6 m2/s. Analysis First we find the Biot number: 14 .
1 C . W/m 0.5 ) m 03 .
0 ( C) . W/m 19 ( Bi 2 = ° ° = = k hro From Table 11-2 we read, for a sphere, λ1 = 1.635 and A1 = 1.302. Substituting these values into the one-term solution gives 753 .
0 = 302 .
1 2 25 2 6 2 2 1 ) 635 .
1 ( 1 0 0 τ θ τ τ λ → = − − → = − − = − − ∞ ∞ e e A T T T T i which is greater than 0.2 and thus the one-term solution is applicable. Then the cooling time becomes h 1.45 = = × = = ⎯→ ⎯ = s 5213 s / m 10 0.13 m) 03 .
0 )( 753 .
0 ( 2 6 -2 2 2 α τ α τ o o r t r t The lowest temperature during cooling will occur on the surface (r/r0 = 1), and is determined to be o o o o i o o o o o i o o o i r r r r T T T T r r r r T T T r T r r r r e A T T T r T / ) / sin( = / ) / sin( ) ( / ) / sin( ) ( 1 1 1 1 0 1 1 1 2 1 λ λ λ λ θ λ λ τ λ ∞ ∞ ∞ ∞ − ∞ ∞ − − = − − → = − − Substituting, C 4.44 = ) ( 1.635 rad) 635 .
1 sin( 2 25 2 6 2 25 2 ) ( ° ⎯→ ⎯ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ − − = − − o o r T r T which is above the temperature range of 3 to 4 °C for chilling injury for potatoes. Therefore, no part of the potatoes will experience chilling injury during this cooling process. Alternative solution We could also solve this problem using transient temperature charts as follows: 17a) -11 (Fig. 75 .
0 t = 174 .
0 2 25 2 6 877 .
0 m) C)(0.03 .
W/m (19 C W/m.
50 .
0 Bi 1 2 0 o 2 o = ⎪ ⎪ ⎭ ⎪ ⎪ ⎬ ⎫ = − − = − − = = = ∞ ∞ o i o r T T T T hr k α τ Therefore, h 1.44 = = × = = − s 2 519 / m 10 13 .
0 ) 03 .
0 )( 75 .
0 ( 2 6 2 2 s r t o α τ The surface temperature is determined from 17b) 11 (Fig. 0.6 ) ( 1 877 .
0 1 − = − − ⎪ ⎪ ⎭ ⎪ ⎪ ⎬ ⎫ = = = ∞ ∞ T T T r T r r r h k Bi o o o which gives C 4 .
4 ) 2 6 ( 6 .
0 2 ) ( 6 .
0 ° = − + = − + = ∞ ∞ T T T T o surface The slight difference between the two results is due to the reading error of the charts. Potato Ti = 25°C Air 2°C 4 m/s PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 11-43 11-58E The center temperature of oranges is to be lowered to 40°F during cooling. The cooling time and if any part of the oranges will freeze during this cooling process are to be determined. Assumptions 1 The oranges are spherical in shape with a radius of ro =1.25 in = 0.1042 ft. 2 Heat conduction in the orange is one-dimensional in the radial direction because of the symmetry about the midpoint. 3 The thermal properties of the orange are constant. 4 The heat transfer coefficient is constant and uniform over the entire surface. 5 The Fourier number is τ > 0.2 so that the one-term approximate solutions (or the transient temperature charts) are applicable (this assumption will be verified). Properties The thermal conductivity and thermal diffusivity of oranges are given to be k = 0.26 Btu/h⋅ft⋅°F and α = 1.4×10-6 ft2/s. Analysis First we find the Biot number: 843 .
1 F Btu/h.ft. 0.26 ) ft 12 / 25 .
1 ( F) .
Btu/h.ft 6 .
4 ( Bi 2 = ° ° = = k hro From Table 11-2 we read, for a sphere, λ1 = 1.9569 and A1 = 1.447. Substituting these values into the one-term solution gives 426 .
0 = 447 .
1 25 78 25 40 2 2 1 ) 9569 .
1 ( 1 0 0 τ θ τ τ λ → = − − → = − − = − − ∞ ∞ e e A T T T T i which is greater than 0.2 and thus the one-term solution is applicable. Then the cooling time becomes min 55.0 = = × = = → = s 3302 s / ft 10 1.4 ft) 12 / 25 .
1 )( 426 .
0 ( 2 6 -2 2 2 α τ α τ o o r t r t The lowest temperature during cooling will occur on the surface (r/r0 = 1), and is determined to be o o o o i o o o o o i o o o i r r r r T T T T r r r r T T T r T r r r r e A T T T r T / ) / sin( = / ) / sin( ) ( / ) / sin( ) ( 1 1 1 1 0 1 1 1 2 1 λ λ λ λ θ λ λ τ λ ∞ ∞ ∞ ∞ − ∞ ∞ − − = − − → = − − Substituting, F 32.1 = ) ( 1.9569 rad) 9569 .
1 sin( 25 78 25 40 25 78 25 ) ( ° ⎯→ ⎯ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ − − = − − o o r T r T which is above the freezing temperature of 31°F for oranges. Therefore, no part of the oranges will freeze during this cooling process. Alternative solution We could also solve this problem using transient temperature charts as follows: 17a) -11 (Fig. 43 .
0 283 .
0 25 78 25 40 543 .
0 ft) F)(1.25/12 .º Btu/h.ft (4.6 F Btu/h.ft.º 0.26 1 2 0 2 = = ⎪ ⎪ ⎭ ⎪ ⎪ ⎬ ⎫ = − − = − − = = = ∞ ∞ o i o r t T T T T r h k Bi α τ Therefore, min 55.6 s 3333 /s ft 10 1.4 5/12ft) (0.43)(1.2 2 6 2 2 = = × = = − α τ o r t The lowest temperature during cooling will occur on the surface (r/ro =1) of the oranges is determined to be 17b) 11 (Fig. 0.45 ) ( 1 543 .
0 1 0 − = − − ⎪ ⎪ ⎭ ⎪ ⎪ ⎬ ⎫ = = = ∞ ∞ T T T r T r r r h k Bi o o which gives F 31.8 ) 25 40 ( 45 .
0 25 ) ( 45 .
0 0 ° = − + = − + = ∞ ∞ T T T Tsurface The slight difference between the two results is due to the reading error of the charts. Orange D = 2.5 in 85% water Ti = 78°F Air 25°F 1 ft/s PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 11-44 11-59 The center temperature of a beef carcass is to be lowered to 4°C during cooling. The cooling time and if any part of the carcass will suffer freezing injury during this cooling process are to be determined. Assumptions 1 The beef carcass can be approximated as a cylinder with insulated top and base surfaces having a radius of ro = 12 cm and a height of H = 1.4 m. 2 Heat conduction in the carcass is one-dimensional in the radial direction because of the symmetry about the centerline. 3 The thermal properties of the carcass are constant. 4 The heat transfer coefficient is constant and uniform over the entire surface. 5 The Fourier number is τ > 0.2 so that the one-term approximate solutions (or the transient temperature charts) are applicable (this assumption will be verified). Properties The thermal conductivity and thermal diffusivity of carcass are given to be k = 0.47 W/m⋅°C and α = 0.13×10-6 m2/s. Analysis First we find the Biot number: 62 .
5 C . W/m 0.47 ) m 12 .
0 ( C) . W/m 22 ( Bi 2 = ° ° = = k hro From Table 11-2 we read, for a cylinder, λ1 = 2.027 and A1 = 1.517. Substituting these values into the one-term solution gives 396 .
0 = 517 .
1 ) 10 ( 37 ) 10 ( 4 2 2 1 ) 027 .
2 ( 1 0 0 τ θ τ τ λ → = − − − − → = − − = − − ∞ ∞ e e A T T T T i which is greater than 0.2 and thus the one-term solution is applicable. Then the cooling time becomes h 12.2 = = × = = → = s 865 , 43 s / m 10 0.13 m) 12 .
0 )( 396 .
0 ( 2 6 -2 2 2 α τ α τ o o r t r t The lowest temperature during cooling will occur on the surface (r/ro = 1), and is determined to be ) / ( = ) / ( ) ( ) / ( ) ( 1 0 1 0 0 1 0 1 2 1 o o i o o i o o i r r J T T T T r r J T T T r T r r J e A T T T r T λ λ θ λ τ λ ∞ ∞ ∞ ∞ − ∞ ∞ − − = − − → = − − Substituting, C -7.1 = ) ( 0621 .
0 2084 .
0 2979 .
0 ) ( ) 10 ( 37 ) 10 ( 4 ) 10 ( 37 ) 10 ( ) ( 1 0 0 ° ⎯→ ⎯ = × = ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ − − − − = − − − − o r T J r T λ which is below the freezing temperature of -1.7 °C. Therefore, the outer part of the beef carcass will freeze during this cooling process. Alternative solution We could also solve this problem using transient temperature charts as follows: ) 16a 11 (Fig. 0.4 298 .
0 ) 10 ( 37 ) 10 ( 4 178 .
0 m) C)(0.12 W/m².º (22 C W/m.º 0.47 1 2 0 − = = ⎪ ⎪ ⎭ ⎪ ⎪ ⎬ ⎫ = − − − − = − − = = = ∞ ∞ o i o r t T T T T r h k Bi α τ Therefore, h 12.3 s 44,308 /s m 10 0.13 m) (0.4)(0.12 2 6 2 2 ≅ = × = = − α τ o r t The surface temperature is determined from 16b) 11 (Fig. 17 .
0 ) ( 1 178 .
0 1 0 − = − − ⎪ ⎪ ⎭ ⎪ ⎪ ⎬ ⎫ = = = ∞ ∞ T T T r T r r r h k Bi o o which gives C 6 .
7 )] 10 ( 4 [ 17 .
0 10 ) ( 17 .
0 0 ° − = − − + − = − + = ∞ ∞ T T T Tsurface The difference between the two results is due to the reading error of the charts. Beef 37°C Air -10°C 1.2 m/s PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 11-45 11-60 The center temperature of meat slabs is to be lowered to -18°C during cooling. The cooling time and the surface temperature of the slabs at the end of the cooling process are to be determined. Assumptions 1 The meat slabs can be approximated as very large plane walls of half-thickness L = 11.5 cm. 2 Heat conduction in the meat slabs is one-dimensional because of the symmetry about the centerplane. 3 The thermal properties of the meat slabs are constant. 4 The heat transfer coefficient is constant and uniform over the entire surface. 5 The Fourier number is τ > 0.2 so that the one-term approximate solutions (or the transient temperature charts) are applicable (this assumption will be verified). 6 The phase change effects are not considered, and thus the actual cooling time will be much longer than the value determined. Properties The thermal conductivity and thermal diffusivity of meat slabs are given to be k = 0.47 W/m⋅°C and α = 0.13×10-6 m2/s. These properties will be used for both fresh and frozen meat. Analysis First we find the Biot number: 89 .
4 C . W/m 0.47 ) m 115 .
0 ( C) . W/m 20 ( Bi 2 = ° ° = = k hro From Table 11-2 we read, for a plane wall, λ1 = 1.308 and A1=1.239. Substituting these values into the one-term solution gives 783 .
0 = 239 .
1 ) 30 ( 7 ) 30 ( 18 2 2 1 ) 308 .
1 ( 1 0 τ θ τ τ λ → = − − − − − → = − − = − − ∞ ∞ e e A T T T T i o which is greater than 0.2 and thus the one-term solution is applicable. Then the cooling time becomes h 22.1 = = × = = → = s 650 , 79 s / m 10 0.13 m) 115 .
0 )( 783 .
0 ( 2 6 -2 2 2 α τ α τ L t L t The lowest temperature during cooling will occur on the surface (x/L = 1), and is determined to be ) cos( = ) / cos( ) ( ) / cos( ) ( 1 1 0 1 1 2 1 λ λ θ λ τ λ ∞ ∞ ∞ ∞ − ∞ ∞ − − = − − → = − − T T T T L L T T T L T L x e A T T T x T i o i i Substituting, C 26.9° − = ⎯→ ⎯ = × = ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ − − − − − = − − − − ) ( 08425 .
0 2598 .
0 3243 .
0 ) cos( ) 30 ( 7 ) 30 ( 18 ) 30 ( 7 ) 30 ( ) ( 1 L T L T λ which is close the temperature of the refrigerated air. Alternative solution We could also solve this problem using transient temperature charts as follows: 15a) 11 (Fig. 75 .
0 324 .
0 ) 30 ( 7 ) 30 ( 18 0.204 m) C)(0.115 W/m².º (20 C W/m.º 0.47 1 2 − = = ⎪ ⎪ ⎭ ⎪ ⎪ ⎬ ⎫ = − − − − − = − − = = = ∞ ∞ L t T T T T hL k Bi i o α τ Therefore, h 21.2 s 300 , 76 /s m 10 0.13 m) 15 (0.75)(0.1 2 6 2 2 ≅ = × = = − α τ o r t The surface temperature is determined from 15b) 11 (Fig. 22 .
0 ) ( 1 204 .
0 1 − = − − ⎪ ⎪ ⎭ ⎪ ⎪ ⎬ ⎫ = = = ∞ ∞ T T T x T L x hL k Bi o which gives C 27.4 )] 30 ( 18 [ 22 .
0 30 ) ( 22 .
0 ° − = − − − + − = − + = ∞ ∞ T T T T o surface The slight difference between the two results is due to the reading error of the charts. Air -30°C 1.4 m/s Meat 7°C PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 11-46 11-61E The center temperature of meat slabs is to be lowered to 36°F during 12-h of cooling. The average heat transfer coefficient during this cooling process is to be determined. Assumptions 1 The meat slabs can be approximated as very large plane walls of half-thickness L = 3-in. 2 Heat conduction in the meat slabs is one-dimensional because of symmetry about the centerplane. 3 The thermal properties of the meat slabs are constant. 4 The heat transfer coefficient is constant and uniform over the entire surface. 5 The Fourier number is τ > 0.2 so that the one-term approximate solutions (or the transient temperature charts) are applicable (this assumption will be verified). Properties The thermal conductivity and thermal diffusivity of meat slabs are given to be k = 0.26 Btu/h⋅ft⋅°F and α=1.4×10-6 ft2/s. Analysis The average heat transfer coefficient during this cooling process is determined from the transient temperature charts for a flat plate as follows: ) 15a 11 (Fig. 7 .
0 1 481 .
0 23 50 23 36 968 .
0 ft)² (3/12 s) 3600 ft²/s)(12 10 (1.4 ² 0 6 − = ⎪ ⎪ ⎭ ⎪ ⎪ ⎬ ⎫ = − − = − − = × × = = ∞ ∞ − Bi T T T T L t i α τ Therefore, F º Btu/h.ft². 1.5 = = = ft (3/12) F)(1/0.7) Btu/h.ft.º (0.26 L kBi h Discussion We could avoid the uncertainty associated with the reading of the charts and obtain a more accurate result by using the one-term solution relation for an infinite plane wall, but it would require a trial and error approach since the Bi number is not known. Air 23°F Meat 50°F PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 11-47 11-62 Chickens are to be chilled by holding them in agitated brine for 2.75 h. The center and surface temperatures of the chickens are to be determined, and if any part of the chickens will freeze during this cooling process is to be assessed. Assumptions 1 The chickens are spherical in shape. 2 Heat conduction in the chickens is one-dimensional in the radial direction because of symmetry about the midpoint. 3 The thermal properties of the chickens are constant. 4 The heat transfer coefficient is constant and uniform over the entire surface. 5 The Fourier number is τ > 0.2 so that the one-term approximate solutions (or the transient temperature charts) are applicable (this assumption will be verified). 6 The phase change effects are not considered, and thus the actual the temperatures will be much higher than the values determined since a considerable part of the cooling process will occur during phase change (freezing of chicken). Properties The thermal conductivity, thermal diffusivity, and density of chickens are given to be k = 0.45 W/m⋅°C, α = 0.13×10-6 m2/s, and ρ = 950 kg/ m3. These properties will be used for both fresh and frozen chicken. Analysis We first find the volume and equivalent radius of the chickens: cm³ 1789 m³) g/(0.95g/c 1700 / = = = ρ m V m 0753 .
0 cm 53 .
7 cm³ 1789 4 3 4 3 3 / 1 3 / 1 = = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ = π π V o r Then the Biot and Fourier numbers become 0.2270 m) (0.0753 s) 3600 /s)(2.75 m 10 (0.13 6 .
73 C . W/m 0.45 ) m 0753 .
0 ( C) . W/m 440 ( Bi 2 2 6 2 2 = × × = = = ° ° = = − o o r t k hr α τ Note that 2 .
0 2270 .
0 > = τ , and thus the one-term solution is applicable. From Table 11-2 we read, for a sphere, λ1 = 3.094 and A1 = 1.998. Substituting these values into the one-term solution gives C 2.0° − ⎯→ ⎯ = − − − − → = − − = − − ∞ ∞ = 0.2274 = 998 .
1 ) 7 ( 15 ) 7 ( 0 ) 2270 .
0 ( ) 094 .
3 ( 0 1 0 0 2 2 1 T e T e A T T T T i τ λ θ The lowest temperature during cooling will occur on the surface (r/ro = 1), and is determined to be o o o o i o o o o o i o o o i r r r r T T T T r r r r T T T r T r r r r e A T T T r T / ) / sin( = / ) / sin( ) ( / ) / sin( ) ( 1 1 1 1 0 1 1 1 2 1 λ λ λ λ θ λ λ τ λ ∞ ∞ ∞ ∞ − ∞ ∞ − − = − − → = − − Substituting, C 6.9° − = → = − − − − ) ( 3.094 rad) 094 .
3 sin( 0.2274 ) 7 ( 15 ) 7 ( ) ( o o r T r T Most parts of chicken will freeze during this process since the freezing point of chicken is -2.8°C. Discussion We could also solve this problem using transient temperature charts, but the data in this case falls at a point on the chart which is very difficult to read: 17) 11 (Fig. ?? 30 .
0 ....
15 .
0 0.0136 ) C)(0.0753m .º W/m (440 C W/m.º 0.45 1 0.227 m) (0.0753 s) 3600 /s)(2.75 m 10 (0.13 2 2 2 6 2 − = − − ⎪ ⎪ ⎭ ⎪ ⎪ ⎬ ⎫ = = = = × × = = ∞ ∞ − T T T T r h k Bi r t i o o o α τ Brine -7°C Chicken Ti=15°C PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 11-48 Transient Heat Conduction in Semi-Infinite Solids 11-63C A semi-infinite medium is an idealized body which has a single exposed plane surface and extends to infinity in all directions. The earth and thick walls can be considered to be semi-infinite media. 11-64C A thick plane wall can be treated as a semi-infinite medium if all we are interested in is the variation of temperature in a region near one of the surfaces for a time period during which the temperature in the mid section of the wall does not experience any change. 11-65C The total amount of heat transfer from a semi-infinite solid up to a specified time t0 can be determined by integration from ∫ ∞ − = o t dt T t T Ah Q 0 ] ) , 0 ( [ where the surface temperature T(0, t) is obtained from Eq. 11-47 by substituting x = 0. 11-66 The water pipes are buried in the ground to prevent freezing. The minimum burial depth at a particular location is to be determined. Assumptions 1 The temperature in the soil is affected by the thermal conditions at one surface only, and thus the soil can be considered to be a semi-infinite medium with a specified surface temperature. 2 The thermal properties of the soil are constant. Properties The thermal properties of the soil are given to be k = 0.35 W/m.°C and α = 0.15×10-6 m2/s. Analysis The length of time the snow pack stays on the ground is s 10 5.184 s/hr) 600 hr/days)(3 days)(24 (60 6 × = = t The surface is kept at -8°C at all times. The depth at which freezing at 0°C occurs can be determined from the analytical solution, ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ = ⎟ ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎜ ⎝ ⎛ × × = − − − ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ = − − − 7636 .
1 5 .
0 ) s 10 184 .
5 )( /s m 10 15 .
0 ( 2 8 8 8 0 ) , ( 6 2 6 x erfc x erfc t x erfc T T T t x T i s i α Then from Table 11-4 we get m 0.846 = ⎯→ ⎯ = x x 4796 .
0 7636 .
1 Discussion The solution could also be determined using the chart, but it would be subject to reading error. Water pipe Ts =-8°C Soil Ti = 8°C PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 11-49 11-67 An area is subjected to cold air for a 10-h period. The soil temperatures at distances 0, 10, 20, and 50 cm from the earth’s surface are to be determined. Assumptions 1 The temperature in the soil is affected by the thermal conditions at one surface only, and thus the soil can be considered to be a semi-infinite medium with a specified surface temperature. 2 The thermal properties of the soil are constant. Properties The thermal properties of the soil are given to be k = 0.9 W/m.°C and α = 1.6×10-5 m2/s. Analysis The one-dimensional transient temperature distribution in the ground can be determined from ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎣ ⎡ ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ + ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ + − ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ = − − ∞ k t h t x erfc k t h k hx t x erfc T T T t x T i i α α α α 2 exp 2 ) , ( 2 2 where 1138 7 .
33 7 .
33 C W/m.
0.9 ) s 3600 10 )( s / m 10 (1.6 C) . W/m 40 ( 2 2 2 2 2 -5 2 = = ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ = = ° × × ° = k t h k t h k t h α α α Then we conclude that the last term in the temperature distribution relation above must be zero regardless of x despite the exponential term tending to infinity since (1) 4 for 0 ) ( > → η η erfc (see Table 11-4) and (2) the term has to remain less than 1 to have physically meaningful solutions. That is, 0 7 .
33 2 1138 exp 2 exp 2 2 ≅ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎣ ⎡ ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ + ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ + = ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎣ ⎡ ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ + ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ + t x erfc k hx k t h t x erfc k t h k hx α α α α Therefore, the temperature distribution relation simplifies to ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ − + = → ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ = − − ∞ ∞ t x erfc T T T t x T t x erfc T T T t x T i i i i α α 2 ) ( ) , ( 2 ) , ( Then the temperatures at 0, 10, 20, and 50 cm depth from the ground surface become x = 0: C 10° − = = × − + = − + = ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ − + = ∞ ∞ ∞ ∞ T T T T erfc T T T t erfc T T T T i i i i i i 1 ) ( ) 0 ( ) ( 2 0 ) ( ) h 10 , 0 ( α x = 0.1m: C 8.5° − = × − = − = ⎟ ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎜ ⎝ ⎛ × × − − + = − 9257 .
0 20 10 ) 066 .
0 ( 20 10 ) s/h 3600 h 10 )( /s m 10 6 .
1 ( 2 m 1 .
0 ) 10 10 ( 10 ) h 10 , m 1 .
0 ( 2 5 erfc erfc T x = 0.2 m: C 7.0° − = × − = − = ⎟ ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎜ ⎝ ⎛ × × − − + = − 8519 .
0 20 10 ) 132 .
0 ( 20 10 ) s/h 3600 h 10 )( /s m 10 6 .
1 ( 2 m 2 .
0 ) 10 10 ( 10 ) h 10 , m 2 .
0 ( 2 5 erfc erfc T x = 0.5 m: C 2.8° − = × − = − = ⎟ ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎜ ⎝ ⎛ × × − − + = − 6418 .
0 20 10 ) 329 .
0 ( 20 10 ) s/h 3600 h 10 )( /s m 10 6 .
1 ( 2 m 5 .
0 ) 10 10 ( 10 ) h 10 , m 5 .
0 ( 2 5 erfc erfc T Soil Ti =10°C Winds T∞ =-10°C PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 11-50 11-68 EES Prob. 11-67 is reconsidered. The soil temperature as a function of the distance from the earth’s surface is to be plotted. Analysis The problem is solved using EES, and the solution is given below. "GIVEN" T_i=10 [C] T_infinity=-10 [C] h=40 [W/m^2-C] time=103600 [s] x=0.1 [m] "PROPERTIES" k=0.9 [W/m-C] alpha=1.6E-5 [m^2/s] "ANALYSIS" (T_x-T_i)/(T_infinity-T_i)=erfc(x/(2sqrt(alphatime)))-exp((hx)/k+(h^2alphatime)/k^2)erfc(x/(2sqrt(alphatime))+(hsqrt(alphatime)/k)) x [m] Tx [C] 0 -9.666 0.05 -8.923 0.1 -8.183 0.15 -7.447 0.2 -6.716 0.25 -5.993 0.3 -5.277 0.35 -4.572 0.4 -3.878 0.45 -3.197 0.5 -2.529 0.55 -1.877 0.6 -1.24 0.65 -0.6207 0.7 -0.01894 0.75 0.5643 0.8 1.128 0.85 1.672 0.9 2.196 0.95 2.7 1 3.183 0 0.2 0.4 0.6 0.8 1 -10 -8 -6 -4 -2 0 2 4 x [m] Tx [C] PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 11-51 11-69 An aluminum block is subjected to heat flux. The surface temperature of the block is to be determined. Assumptions 1 All heat flux is absorbed by the block. 2 Heat loss from the block is disregarded (and thus the result obtained is the maximum temperature). 3 The block is sufficiently thick to be treated as a semi-infinite solid, and the properties of the block are constant. Properties Thermal conductivity and diffusivity of aluminum at room temperature are k = 237 kg/m3 and α = 97.1×10-6 m2/s. Analysis This is a transient conduction problem in a semi-infinite medium subjected to constant surface heat flux, and the surface temperature can be determined to be C 28.0 s) 60 /s)(30 m 10 71 .
9 ( 4 C W/m 237 W/m 4000 C 20 4 2 5 2 ° = × × ° ⋅ + ° = + = − π π αt k q T T s i s & Then the temperature rise of the surface becomes C 8.0° = − = Δ 20 28 s T 11-70 The contact surface temperatures when a bare footed person steps on aluminum and wood blocks are to be determined. Assumptions 1 Both bodies can be treated as the semi-infinite solids. 2 Heat loss from the solids is disregarded. 3 The properties of the solids are constant. Properties The p c kρ value is 24 kJ/m2⋅°C for aluminum, 0.38 kJ/m2⋅°C for wood, and 1.1 kJ/m2⋅°C for the human flesh. Analysis The surface temperature is determined from Eq. 11-49 to be C 20.5° = ° ⋅ + ° ⋅ ° ° ⋅ + ° ° ⋅ = + + = C) kJ/m 24 ( C) kJ/m 1 .
1 ( C) 20 ( C) kJ/m 24 ( C) 32 ( C) kJ/m 1 .
1 ( ) ( ) ( ) ( ) ( 2 2 2 2 Al human Al Al human human p p p p s c k c k T c k T c k T ρ ρ ρ ρ In the case of wood block, we obtain C 28.9° = ° ⋅ + ° ⋅ ° ° ⋅ + ° ° ⋅ = + + = C) kJ/m 38 .
0 ( C) kJ/m 1 .
1 ( C) 20 ( C) kJ/m 38 .
0 ( C) 32 ( C) kJ/m 1 .
1 ( ) ( ) ( ) ( ) ( 2 2 2 2 wood human wood wood human human p p p p s c k c k T c k T c k T ρ ρ ρ ρ PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 11-52 11-71E The walls of a furnace made of concrete are exposed to hot gases at the inner surfaces. The time it will take for the temperature of the outer surface of the furnace to change is to be determined. Assumptions 1 The temperature in the wall is affected by the thermal conditions at inner surfaces only and the convection heat transfer coefficient inside is given to be very large. Therefore, the wall can be considered to be a semi-infinite medium with a specified surface temperature of 1800°F. 2 The thermal properties of the concrete wall are constant. Properties The thermal properties of the concrete are given to be k = 0.64 Btu/h.ft.°F and α = 0.023 ft2/h. Analysis The one-dimensional transient temperature distribution in the wall for that time period can be determined from ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ = − − t x erfc T T T t x T i s i α 2 ) , ( But, ) 85 .
2 ( 00006 .
0 00006 .
0 70 1800 70 1 .
70 ) , ( erfc T T T t x T i s i = → = − − = − − (Table 11-4) Therefore, min 116 = = × = × = ⎯→ ⎯ = h 93 .
1 ) /h ft 023 .
0 ( ) 85 .
2 ( 4 ft) 2 .
1 ( ) 85 .
2 ( 4 85 .
2 2 2 2 2 2 2 α α x t t x 11-72 A thick wood slab is exposed to hot gases for a period of 5 minutes. It is to be determined whether the wood will ignite. Assumptions 1 The wood slab is treated as a semi-infinite medium subjected to convection at the exposed surface. 2 The thermal properties of the wood slab are constant. 3 The heat transfer coefficient is constant and uniform over the entire surface. Properties The thermal properties of the wood are k = 0.17 W/m.°C and α = 1.28×10-7 m2/s. Analysis The one-dimensional transient temperature distribution in the wood can be determined from ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎣ ⎡ ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ + ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ + − ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ = − − ∞ k t h t x erfc k t h k hx t x erfc T T T t x T i i α α α α 2 exp 2 ) , ( 2 2 where 628 .
1 276 .
1 276 .
1 C W/m.
0.17 ) s 60 5 )( s / m 10 (1.28 C) . W/m 35 ( 2 2 2 2 2 -7 2 = = ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ = = ° × × ° = k t h k t h k t h α α α Noting that x = 0 at the surface and using Table 11-4 for erfc values, 637 .
0 ) 0712 .
0 )( 0937 .
5 ( 1 ) 276 .
1 0 ( ) 628 .
1 0 exp( ) 0 ( 25 550 25 ) , ( = − = + + − = − − erfc erfc t x T Solving for T(x, t) gives C 360° = ) , ( t x T which is less than the ignition temperature of 450°C. Therefore, the wood will not ignite. 1800°F 70°F L =1.2 ft Q & Wall L=0.3 m Wood slab Ti = 25°C Hot gases T∞ = 550°C x 0 PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 11-53 Ice, 0°C Ice chest Hot water 60°C 11-73 The outer surfaces of a large cast iron container filled with ice are exposed to hot water. The time before the ice starts melting and the rate of heat transfer to the ice are to be determined. Assumptions 1 The temperature in the container walls is affected by the thermal conditions at outer surfaces only and the convection heat transfer coefficient outside is given to be very large. Therefore, the wall can be considered to be a semi-infinite medium with a specified surface temperature. 2 The thermal properties of the wall are constant. Properties The thermal properties of the cast iron are given to be k = 52 W/m.°C and α = 1.70×10-5 m2/s. Analysis The one-dimensional transient temperature distribution in the wall for that time period can be determined from ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ = − − t x erfc T T T t x T i s i α 2 ) , ( But, ) 226 .
2 ( 00167 .
0 00167 .
0 0 60 0 1 .
0 ) , ( erfc T T T t x T i s i = → = − − = − − (Table 11-4) Therefore, s 7.4 = × = × = ⎯→ ⎯ = − ) /s m 10 7 .
1 ( ) 226 .
2 ( 4 m) 05 .
0 ( ) 226 .
2 ( 4 226 .
2 2 2 5 2 2 2 2 α α x t t x The rate of heat transfer to the ice when steady operation conditions are reached can be determined by applying the thermal resistance network concept as W 28,990 = ° − = − = ° = + + = + + = ° ≅ × ∞ = = ° = × ° = = ° = × ° = = C/W 0.00207 C ) 0 60 ( C/W 00207 .
0 0 00040 .
0 00167 .
0 C/W 0 ) m 2 )(1.2 ( 1 1 C/W 00040 .
0 ) m 2 C)(1.2 W/m.
(52 m 05 .
0 C/W 00167 .
0 ) m 2 C)(1.2 . W/m (250 1 1 o 1 2 , , 2 , 2 2 2 , total o conv wall i conv total o o conv wall i i conv R T T Q R R R R A h R kA L R A h R & Rconv Rwall Rconv T1 T2 PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 11-54 Transient Heat Conduction in Multidimensional Systems 11-74C The product solution enables us to determine the dimensionless temperature of two- or three-dimensional heat transfer problems as the product of dimensionless temperatures of one-dimensional heat transfer problems. The dimensionless temperature for a two-dimensional problem is determined by determining the dimensionless temperatures in both directions, and taking their product. 11-75C The dimensionless temperature for a three-dimensional heat transfer is determined by determining the dimensionless temperatures of one-dimensional geometries whose intersection is the three dimensional geometry, and taking their product. 11-76C This short cylinder is physically formed by the intersection of a long cylinder and a plane wall. The dimensionless temperatures at the center of plane wall and at the center of the cylinder are determined first. Their product yields the dimensionless temperature at the center of the short cylinder. 11-77C The heat transfer in this short cylinder is one-dimensional since there is no heat transfer in the axial direction. The temperature will vary in the radial direction only. PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 11-55 11-78 A short cylinder is allowed to cool in atmospheric air. The temperatures at the centers of the cylinder and the top surface as well as the total heat transfer from the cylinder for 15 min of cooling are to be determined. Assumptions 1 Heat conduction in the short cylinder is two-dimensional, and thus the temperature varies in both the axial x- and the radial r- directions. 2 The thermal properties of the cylinder are constant. 3 The heat transfer coefficient is constant and uniform over the entire surface. 4 The Fourier number is τ > 0.2 so that the one-term approximate solutions (or the transient temperature charts) are applicable (this assumption will be verified). Properties The thermal properties of brass are given to be 3 kg/m 8530 = ρ , C kJ/kg 389 .
0 ° ⋅ = p c , C W/m 110 ° ⋅ = k , and /s m 10 39 .
3 2 5 − × = α . Analysis This short cylinder can physically be formed by the intersection of a long cylinder of radius D/2 = 4 cm and a plane wall of thickness 2L = 15 cm. We measure x from the midplane. (a) The Biot number is calculated for the plane wall to be 02727 .
0 ) C W/m.
110 ( ) m 075 .
0 )( C . W/m 40 ( 2 = ° ° = = k hL Bi The constants 1 1 and A λ corresponding to this Biot number are, from Table 11-2, 0045 .
1 and 1620 .
0 1 1 = = A λ The Fourier number is 2 .
0 424 .
5 m) 075 .
0 ( s/min) 60 min /s)(15 m 10 39 .
3 ( 2 2 5 2 > = × × = = − L t α τ Therefore, the one-term approximate solution (or the transient temperature charts) is applicable. Then the dimensionless temperature at the center of the plane wall is determined from 871 .
0 ) 0045 .
1 ( ) 424 .
5 ( ) 1620 .
0 ( 1 0 , 0 2 2 1 = = = − − = − − ∞ ∞ e e A T T T T i wall τ λ θ We repeat the same calculations for the long cylinder, 01455 .
0 ) C W/m.
110 ( ) m 04 .
0 )( C . W/m 40 ( 2 = ° ° = = k hr Bi o 0036 .
1 and 1677 .
0 1 1 = = A λ 2 .
0 069 .
19 m) 04 .
0 ( s) 60 /s)(15 m 10 39 .
3 ( 2 2 5 2 > = × × = = − o r t α τ 587 .
0 ) 0036 .
1 ( ) 069 .
19 ( ) 1677 .
0 ( 1 , 2 2 1 = = = − − = − − ∞ ∞ e e A T T T T i o cyl o τ λ θ Then the center temperature of the short cylinder becomes C 86.4° = ⎯→ ⎯ = − − = × = × = ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ − − ∞ ∞ ) , 0 , 0 ( 511 .
0 20 150 20 ) , 0 , 0 ( 511 .
0 587 .
0 871 .
0 ) , 0 , 0 ( , , t T t T T T T t T cyl o wall o cylinder short i θ θ (b) The center of the top surface of the cylinder is still at the center of the long cylinder (r = 0), but at the outer surface of the plane wall (x = L). Therefore, we first need to determine the dimensionless temperature at the surface of the wall. D0 = 8 cm L = 15 cm z Brass cylinder Ti = 150°C r Air T∞ = 20°C PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 11-56 860 .
0 ) 1620 .
0 cos( ) 0045 .
1 ( ) / cos( ) , ( ) , ( ) 424 .
5 ( ) 1620 .
0 ( 1 1 2 2 1 = = = − − = − − ∞ ∞ e L L e A T T T t x T t L i wall λ θ τ λ Then the center temperature of the top surface of the cylinder becomes C 85.6° = ⎯→ ⎯ = − − = × = × = ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ − − ∞ ∞ ) , 0 , ( 505 .
0 20 150 20 ) , 0 , ( 505 .
0 587 .
0 860 .
0 ) , ( ) , 0 , ( , t L T t L T t L T T T t L T cyl o wall cylinder short i θ θ (c) We first need to determine the maximum heat can be transferred from the cylinder [ ] kJ 325 C ) 20 150 )( C kJ/kg. 389 .
0 )( kg 43 .
6 ( ) ( kg 43 .
6 m) 15 .
0 ( m) 04 .
0 ( ) kg/m 8530 ( max 2.
3 2 = ° − ° = − = = = = = ∞ T T mc Q L r m i p o π ρπ ρV Then we determine the dimensionless heat transfer ratios for both geometries as 133 .
0 1620 .
0 ) 1620 .
0 sin( ) 871 .
0 ( 1 ) sin( 1 1 1 , max = − = − = ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ λ λ θ wall o wall Q Q 415 .
0 1677 .
0 0835 .
0 ) 587 .
0 ( 2 1 ) ( 2 1 1 1 1 , max = − = − = ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ λ λ θ J Q Q cyl o cyl The heat transfer ratio for the short cylinder is 493 .
0 ) 133 .
0 1 )( 415 .
0 ( 133 .
0 1 max max max max = − + = ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ − ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ + ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ = ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ wall plane cylinder long wall plane cylinder short Q Q Q Q Q Q Q Q Then the total heat transfer from the short cylinder during the first 15 minutes of cooling becomes kJ 160 = = = kJ) 325 )( 493 .
0 ( 493 .
0 max Q Q PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 11-57 11-79 EES Prob. 11-78 is reconsidered. The effect of the cooling time on the center temperature of the cylinder, the center temperature of the top surface of the cylinder, and the total heat transfer is to be investigated. Analysis The problem is solved using EES, and the solution is given below. "GIVEN" D=0.08 [m] r_o=D/2 height=0.15 [m] L=height/2 T_i=150 [C] T_infinity=20 [C] h=40 [W/m^2-C] time=15 [min] "PROPERTIES" k=110 [W/m-C] rho=8530 [kg/m^3] c_p=0.389 [kJ/kg-C] alpha=3.39E-5 [m^2/s] "ANALYSIS" "(a)" "This short cylinder can physically be formed by the intersection of a long cylinder of radius r_o and a plane wall of thickness 2L" "For plane wall" Bi_w=(hL)/k "From Table 11-2 corresponding to this Bi number, we read" lambda_1_w=0.1620 "w stands for wall" A_1_w=1.0045 tau_w=(alphatimeConvert(min, s))/L^2 theta_o_w=A_1_wexp(-lambda_1_w^2tau_w) "theta_o_w=(T_o_w-T_infinity)/(T_i-T_infinity)" "For long cylinder" Bi_c=(hr_o)/k "c stands for cylinder" "From Table 11-2 corresponding to this Bi number, we read" lambda_1_c=0.1677 A_1_c=1.0036 tau_c=(alphatimeConvert(min, s))/r_o^2 theta_o_c=A_1_cexp(-lambda_1_c^2tau_c) "theta_o_c=(T_o_c-T_infinity)/(T_i-T_infinity)" (T_o_o-T_infinity)/(T_i-T_infinity)=theta_o_wtheta_o_c "center temperature of short cylinder" "(b)" theta_L_w=A_1_wexp(-lambda_1_w^2tau_w)Cos(lambda_1_wL/L) "theta_L_w=(T_L_w-T_infinity)/(T_i-T_infinity)" (T_L_o-T_infinity)/(T_i-T_infinity)=theta_L_wtheta_o_c "center temperature of the top surface" "(c)" V=pir_o^2(2L) m=rhoV Q_max=mc_p(T_i-T_infinity) Q_w=1-theta_o_wSin(lambda_1_w)/lambda_1_w "Q_w=(Q/Q_max)_w" Q_c=1-2theta_o_cJ_1/lambda_1_c "Q_c=(Q/Q_max)_c" J_1=0.0835 "From Table 11-3, at lambda_1_c" Q/Q_max=Q_w+Q_c(1-Q_w) "total heat transfer" PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 11-58 time [min] To,o [C] TL,o [C] Q [kJ] 5 124.5 123.2 65.97 10 103.4 102.3 118.5 15 86.49 85.62 160.3 20 73.03 72.33 193.7 25 62.29 61.74 220.3 30 53.73 53.29 241.6 35 46.9 46.55 258.5 40 41.45 41.17 272 45 37.11 36.89 282.8 50 33.65 33.47 291.4 55 30.88 30.74 298.2 60 28.68 28.57 303.7 0 10 20 30 40 50 60 20 40 60 80 100 120 50 100 150 200 250 300 350 time [min] To,o [C] Q [kJ] temperature heat 0 10 20 30 40 50 60 20 40 60 80 100 120 time [min] TL,o [C] PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 11-59 11-80 A semi-infinite aluminum cylinder is cooled by water. The temperature at the center of the cylinder 5 cm from the end surface is to be determined. Assumptions 1 Heat conduction in the semi-infinite cylinder is two-dimensional, and thus the temperature varies in both the axial x- and the radial r- directions. 2 The thermal properties of the cylinder are constant. 3 The heat transfer coefficient is constant and uniform over the entire surface. 4 The Fourier number is τ > 0.2 so that the one-term approximate solutions (or the transient temperature charts) are applicable (this assumption will be verified). Properties The thermal properties of aluminum are given to be k = 237 W/m.°C and α = 9.71×10-5m2/s. Analysis This semi-infinite cylinder can physically be formed by the intersection of a long cylinder of radius ro = D/2 = 7.5 cm and a semi-infinite medium. The dimensionless temperature 5 cm from the surface of a semi-infinite medium is first determined from 8951 .
0 1049 .
0 1 ) , ( 1049 .
0 ) 7308 .
0 )( 0468 .
1 ( 8699 .
0 ) 2433 .
0 ( ) 0458 .
0 exp( ) 1158 .
0 ( 237 ) 60 8 )( 10 71 .
9 ( ) 140 ( ) 60 8 )( 10 71 .
9 ( 2 05 .
0 ) 237 ( ) 60 8 )( 10 71 .
9 ( ) 140 ( 237 ) 05 .
0 )( 140 ( exp ) 60 8 )( 10 71 .
9 ( 2 05 .
0 2 exp ) , ( inf 5 5 2 5 2 5 2 2 = − = − − = = − = − = ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ ⎟ ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎜ ⎝ ⎛ × × + × × × ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ × × + − ⎟ ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎜ ⎝ ⎛ × × = ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎣ ⎡ ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ + ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ + − ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ = − − ∞ ∞ − − − − − ∞ T T T t x T erfc erfc erfc erfc k t h t x erfc k t h k hx t x erfc T T T t x T i semi i i θ α α α α The Biot number is calculated for the long cylinder to be 0443 .
0 C W/m.
237 ) m 075 .
0 )( C . W/m 140 ( 2 = ° ° = = k hr Bi o The constants 1 1 and A λ corresponding to this Biot number are, from Table 11-2, λ1 = 0.2948 and A1 = 1.0110 The Fourier number is 2 .
0 286 .
8 m) 075 .
0 ( s) 60 /s)(8 m 10 71 .
9 ( 2 2 5 2 > = × × = = − o r t α τ Therefore, the one-term approximate solution (or the transient temperature charts) is applicable. Then the dimensionless temperature at the center of the plane wall is determined from 4921 .
0 ) 0110 .
1 ( ) 286 .
8 ( ) 2948 .
0 ( 1 , 2 2 1 = = = − − = − − ∞ ∞ e e A T T T T i o cyl o τ λ θ The center temperature of the semi-infinite cylinder then becomes C 56.3° = ⎯→ ⎯ = ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ − − = × = × = ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ − − − − − ∞ ∞ ) , 0 , ( 4405 .
0 10 115 10 ) , 0 , ( 4405 .
0 4921 .
0 8951 .
0 ) , ( ) , 0 , ( cylinder infinite semi , inf semi cylinder infinite semi t x T t x T t x T T T t x T cyl o i θ θ D0 = 15 cm z Semi-infinite cylinder Ti = 115°C r Water T∞= 10°C PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 11-60 11-81E A hot dog is dropped into boiling water. The center temperature of the hot dog is do be determined by treating hot dog as a finite cylinder and also as an infinitely long cylinder. Assumptions 1 When treating hot dog as a finite cylinder, heat conduction in the hot dog is two-dimensional, and thus the temperature varies in both the axial x- and the radial r- directions. When treating hot dog as an infinitely long cylinder, heat conduction is one-dimensional in the radial r- direction. 2 The thermal properties of the hot dog are constant. 3 The heat transfer coefficient is constant and uniform over the entire surface. 4 The Fourier number is τ > 0.2 so that the one-term approximate solutions (or the transient temperature charts) are applicable (this assumption will be verified). Properties The thermal properties of the hot dog are given to be k = 0.44 Btu/h.ft.°F, ρ = 61.2 lbm/ft3 cp = 0.93 Btu/lbm.°F, and α = 0.0077 ft2/h. Analysis (a) This hot dog can physically be formed by the intersection of a long cylinder of radius ro = D/2 = (0.4/12) ft and a plane wall of thickness 2L = (5/12) ft. The distance x is measured from the midplane. After 5 minutes First the Biot number is calculated for the plane wall to be 8 .
56 ) F Btu/h.ft. 44 .
0 ( ) ft 12 / 5 .
2 )( F .
Btu/h.ft 120 ( 2 = ° ° = = k hL Bi The constants 1 1 and A λ corresponding to this Biot number are, from Table 11-2, 2728 .
1 and 5421 .
1 1 1 = = A λ The Fourier number is 2 .
0 015 .
0 ft) 12 / 5 .
2 ( h) /h)(5/60 ft 0077 .
0 ( 2 2 2 < = = = L t α τ (Be cautious!) Then the dimensionless temperature at the center of the plane wall is determined from 228 .
1 ) 2728 .
1 ( ) 015 .
0 ( ) 5421 .
1 ( 1 0 , 0 2 2 1 = = = − − = − − ∞ ∞ e e A T T T T i wall τ λ θ We repeat the same calculations for the long cylinder, 1 .
9 ) F Btu/h.ft. 44 .
0 ( ) ft 12 / 4 .
0 )( F .
Btu/h.ft 120 ( 2 = ° ° = = k hr Bi o 5618 .
1 and 1589 .
2 1 1 = = A λ 2 .
0 578 .
0 ft) 12 / 4 .
0 ( h) /h)(5/60 ft 0077 .
0 ( 2 2 2 > = = = o r t α τ 106 .
0 ) 5618 .
1 ( ) 578 .
0 ( ) 1589 .
2 ( 1 , 2 2 1 = = = − − = − − ∞ ∞ e e A T T T T i o cyl o τ λ θ Then the center temperature of the short cylinder becomes F 190° = ⎯→ ⎯ = − − = × = × = ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ − − ∞ ∞ ) , 0 , 0 ( 130 .
0 212 40 212 ) , 0 , 0 ( 130 .
0 106 .
0 228 .
1 ) , 0 , 0 ( , , cylinder short t T t T T T T t T cyl o wall o i θ θ Water 212°F Hot dog x r PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 11-61 After 10 minutes 2 .
0 03 .
0 ft) 12 / 5 .
2 ( h) /h)(10/60 ft 0077 .
0 ( 2 2 2 < = = = L t α τ (Be cautious!) 185 .
1 ) 2728 .
1 ( ) 03 .
0 ( ) 5421 .
1 ( 1 0 , 0 2 2 1 = = = − − = − − ∞ ∞ e e A T T T T i wall τ λ θ 2 .
0 156 .
1 ft) 12 / 4 .
0 ( h) /h)(10/60 ft 0077 .
0 ( 2 2 2 > = = = o r t α τ 0071 .
0 ) 5618 .
1 ( ) 156 .
1 ( ) 1589 .
2 ( 1 , 2 2 1 = = = − − = − − ∞ ∞ e e A T T T T i o cyl o τ λ θ F 211° = ⎯→ ⎯ = − − = × = × = ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ − − ∞ ∞ ) , 0 , 0 ( 0084 .
0 212 40 212 ) , 0 , 0 ( 0084 .
0 0071 .
0 185 .
1 ) , 0 , 0 ( , , cylinder short t T t T T T T t T cyl o wall o i θ θ After 15 minutes 2 .
0 045 .
0 ft) 12 / 5 .
2 ( h) /h)(15/60 ft 0077 .
0 ( 2 2 2 < = = = L t α τ (Be cautious!) 143 .
1 ) 2728 .
1 ( ) 045 .
0 ( ) 5421 .
1 ( 1 0 , 0 2 2 1 = = = − − = − − ∞ ∞ e e A T T T T i wall τ λ θ 2 .
0 734 .
1 ft) 12 / 4 .
0 ( h) /h)(15/60 ft 0077 .
0 ( 2 2 2 > = = = o r t α τ 00048 .
0 ) 5618 .
1 ( ) 734 .
1 ( ) 1589 .
2 ( 1 0 , 0 2 2 1 = = = − − = − − ∞ ∞ e e A T T T T i cyl τ λ θ F 212 ° = ⎯→ ⎯ = − − = × = × = ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ − − ∞ ∞ ) , 0 , 0 ( 00055 .
0 212 40 212 ) , 0 , 0 ( 00055 .
0 00048 .
0 143 .
1 ) , 0 , 0 ( , , cylinder short t T t T T T T t T cyl o wall o i θ θ (b) Treating the hot dog as an infinitely long cylinder will not change the results obtained in the part (a) since dimensionless temperatures for the plane wall is 1 for all cases. PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 11-62 11-82E A hot dog is dropped into boiling water. The center temperature of the hot dog is do be determined by treating hot dog as a finite cylinder and an infinitely long cylinder. Assumptions 1 When treating hot dog as a finite cylinder, heat conduction in the hot dog is two-dimensional, and thus the temperature varies in both the axial x- and the radial r- directions. When treating hot dog as an infinitely long cylinder, heat conduction is one-dimensional in the radial r- direction. 2 The thermal properties of the hot dog are constant. 3 The heat transfer coefficient is constant and uniform over the entire surface. 4 The Fourier number is τ > 0.2 so that the one-term approximate solutions (or the transient temperature charts) are applicable (this assumption will be verified). Properties The thermal properties of the hot dog are given to be k = 0.44 Btu/h.ft.°F, ρ = 61.2 lbm/ft3 cp = 0.93 Btu/lbm.°F, and α = 0.0077 ft2/h. Analysis (a) This hot dog can physically be formed by the intersection of a long cylinder of radius ro = D/2 = (0.4/12) ft and a plane wall of thickness 2L = (5/12) ft. The distance x is measured from the midplane. After 5 minutes First the Biot number is calculated for the plane wall to be 8 .
56 ) F Btu/h.ft. 44 .
0 ( ) ft 12 / 5 .
2 )( F .
Btu/h.ft 120 ( 2 = ° ° = = k hL Bi The constants 1 1 and A λ corresponding to this Biot number are, from Table 11-2, 2728 .
1 and 5421 .
1 1 1 = = A λ The Fourier number is 2 .
0 015 .
0 ft) 12 / 5 .
2 ( h) /h)(5/60 ft 0077 .
0 ( 2 2 2 < = = = L t α τ (Be cautious!) Then the dimensionless temperature at the center of the plane wall is determined from 228 .
1 ) 2728 .
1 ( ) 015 .
0 ( ) 5421 .
1 ( 1 0 , 0 2 2 1 = = = − − = − − ∞ ∞ e e A T T T T i wall τ λ θ We repeat the same calculations for the long cylinder, 1 .
9 ) F Btu/h.ft. 44 .
0 ( ) ft 12 / 4 .
0 )( F .
Btu/h.ft 120 ( 2 = ° ° = = k hr Bi o 5618 .
1 and 1589 .
2 1 1 = = A λ 2 .
0 578 .
0 ft) 12 / 4 .
0 ( h) /h)(5/60 ft 0077 .
0 ( 2 2 2 > = = = o r t α τ 106 .
0 ) 5618 .
1 ( ) 578 .
0 ( ) 1589 .
2 ( 1 0 , 0 2 2 1 = = = − − = − − ∞ ∞ e e A T T T T i cyl τ λ θ Then the center temperature of the short cylinder becomes F 181° = ⎯→ ⎯ = − − = × = × = ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ − − ∞ ∞ ) , 0 , 0 ( 130 .
0 202 40 202 ) , 0 , 0 ( 130 .
0 106 .
0 228 .
1 ) , 0 , 0 ( , , cylinder short t T t T T T T t T cyl o wall o i θ θ Water 202°F Hot dog x r PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 11-63 After 10 minutes 2 .
0 03 .
0 ft) 12 / 5 .
2 ( h) /h)(10/60 ft 0077 .
0 ( 2 2 2 < = = = L t α τ (Be cautious!) 185 .
1 ) 2728 .
1 ( ) 03 .
0 ( ) 5421 .
1 ( 1 0 , 0 2 2 1 = = = − − = − − ∞ ∞ e e A T T T T i wall τ λ θ 2 .
0 156 .
1 ft) 12 / 4 .
0 ( h) /h)(10/60 ft 0077 .
0 ( 2 2 2 > = = = o r t α τ 007 .
0 ) 5618 .
1 ( ) 156 .
1 ( ) 1589 .
2 ( 1 0 , 0 2 2 1 = = = − − = − − ∞ ∞ e e A T T T T i cyl τ λ θ F 201° = ⎯→ ⎯ = − − = × = × = ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ − − ∞ ∞ ) , 0 , 0 ( 0084 .
0 202 40 202 ) , 0 , 0 ( 0084 .
0 0071 .
0 185 .
1 ) , 0 , 0 ( , , cylinder short t T t T T T T t T cyl o wall o i θ θ After 15 minutes 2 .
0 045 .
0 ft) 12 / 5 .
2 ( h) /h)(15/60 ft 0077 .
0 ( 2 2 2 < = = = L t α τ (Be cautious!) 143 .
1 ) 2728 .
1 ( ) 045 .
0 ( ) 5421 .
1 ( 1 0 , 0 2 2 1 = = = − − = − − ∞ ∞ e e A T T T T i wall τ λ θ 2 .
0 734 .
1 ft) 12 / 4 .
0 ( h) /h)(15/60 ft 0077 .
0 ( 2 2 2 > = = = o r t α τ 00048 .
0 ) 5618 .
1 ( ) 734 .
1 ( ) 1589 .
2 ( 1 0 , 0 2 2 1 = = = − − = − − ∞ ∞ e e A T T T T i cyl τ λ θ F 202 ° = ⎯→ ⎯ = − − = × = × = ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ − − ∞ ∞ ) , 0 , 0 ( 00055 .
0 202 40 202 ) , 0 , 0 ( 00055 .
0 00048 .
0 143 .
1 ) , 0 , 0 ( , , cylinder short t T t T T T T t T cyl o wall o i θ θ (b) Treating the hot dog as an infinitely long cylinder will not change the results obtained in the part (a) since dimensionless temperatures for the plane wall is 1 for all cases. PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 11-64 11-83 A rectangular ice block is placed on a table. The time the ice block starts melting is to be determined. Assumptions 1 Heat conduction in the ice block is two-dimensional, and thus the temperature varies in both x- and y- directions. 2 The thermal properties of the ice block are constant. 3 The heat transfer coefficient is constant and uniform over the entire surface. 4 The Fourier number is τ > 0.2 so that the one-term approximate solutions (or the transient temperature charts) are applicable (this assumption will be verified). Properties The thermal properties of the ice are given to be k = 2.22 W/m.°C and α = 0.124×10-7 m2/s. Analysis This rectangular ice block can be treated as a short rectangular block that can physically be formed by the intersection of two infinite plane wall of thickness 2L = 4 cm and an infinite plane wall of thickness 2L = 10 cm. We measure x from the bottom surface of the block since this surface represents the adiabatic center surface of the plane wall of thickness 2L = 10 cm. Since the melting starts at the corner of the top surface, we need to determine the time required to melt ice block which will happen when the temperature drops below 0°C at this location. The Biot numbers and the corresponding constants are first determined to be 1081 .
0 ) C W/m.
22 .
2 ( ) m 02 .
0 )( C . W/m 12 ( 2 1 wall,1 = ° ° = = k hL Bi 0173 .
1 and 3208 .
0 1 1 = = ⎯→ ⎯ A λ 2703 .
0 ) C W/m.
22 .
2 ( ) m 05 .
0 )( C . W/m 12 ( 2 3 wall,3 = ° ° = = k hL Bi 0408 .
1 and 4951 .
0 1 1 = = ⎯→ ⎯ A λ The ice will start melting at the corners because of the maximum exposed surface area there. Noting that 2 / L t α τ = and assuming that τ > 0.2 in all dimensions so that the one-term approximate solution for transient heat conduction is applicable, the product solution method can be written for this problem as hours 21.5 = = = ⎯→ ⎯ ⎪ ⎭ ⎪ ⎬ ⎫ ⎪ ⎩ ⎪ ⎨ ⎧ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎣ ⎡ × − × ⎪ ⎭ ⎪ ⎬ ⎫ ⎪ ⎩ ⎪ ⎨ ⎧ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎣ ⎡ × − = ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ = − − − = − − − − min 1292 s 500 , 77 ) 4951 .
0 cos( ) 05 .
0 ( ) 10 124 .
0 ( ) 4951 .
0 ( exp ) 0408 .
1 ( ) 3208 .
0 cos( ) 02 .
0 ( ) 10 124 .
0 ( ) 3208 .
0 ( exp ) 0173 .
1 ( 4737 .
0 ) / cos( ) / cos( 18 20 18 0 ) , ( ) , ( ) , , , ( 2 7 2 2 2 7 2 3 3 1 1 2 1 1 1 1 wall,2 3 2 wall,1 1 3 2 1 2 1 2 1 t t t L L e A L L e A t L t L t L L L block λ λ θ θ θ τ λ τ λ Therefore, the ice will start melting in about 21 hours. Discussion Note that 2 .
0 384 .
0 m) 05 .
0 ( s/h) /s)(77,500 m 10 124 .
0 ( 2 2 7 2 > = × = = − L t α τ and thus the assumption of τ > 0.2 for the applicability of the one-term approximate solution is verified. Ice block -20°C Air 18°C PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 11-65 11-84 EES Prob. 11-83 is reconsidered. The effect of the initial temperature of the ice block on the time period before the ice block starts melting is to be investigated. Analysis The problem is solved using EES, and the solution is given below. "GIVEN" 2L_1=0.04 [m] L_2=L_1 2L_3=0.10 [m] T_i=-20 [C] T_infinity=18 [C] h=12 [W/m^2-C] T_L1_L2_L3=0 [C] "PROPERTIES" k=2.22 [W/m-C] alpha=0.124E-7 [m^2/s] "ANALYSIS" "This block can physically be formed by the intersection of two infinite plane wall of thickness 2L=4 cm and an infinite plane wall of thickness 2L=10 cm" "For the two plane walls" Bi_w1=(hL_1)/k "From Table 11-2 corresponding to this Bi number, we read" lambda_1_w1=0.3208 "w stands for wall" A_1_w1=1.0173 timeConvert(min, s)=tau_w1L_1^2/alpha "For the third plane wall" Bi_w3=(hL_3)/k "From Table 11-2 corresponding to this Bi number, we read" lambda_1_w3=0.4951 A_1_w3=1.0408 timeConvert(min, s)=tau_w3L_3^2/alpha theta_L_w1=A_1_w1exp(-lambda_1_w1^2tau_w1)Cos(lambda_1_w1L_1/L_1) "theta_L_w1=(T_L_w1-T_infinity)/(T_i-T_infinity)" theta_L_w3=A_1_w3exp(-lambda_1_w3^2tau_w3)Cos(lambda_1_w3L_3/L_3) "theta_L_w3=(T_L_w3-T_infinity)/(T_i-T_infinity)" (T_L1_L2_L3-T_infinity)/(T_i-T_infinity)=theta_L_w1^2theta_L_w3 "corner temperature" Ti [C] time [min] -26 1614 -24 1512 -22 1405 -20 1292 -18 1173 -16 1048 -14 914.9 -12 773.3 -10 621.9 -8 459.4 -6 283.7 -4 92.84 -30 -25 -20 -15 -10 -5 0 0 200 400 600 800 1000 1200 1400 1600 1800 Ti [C] time [min] PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 11-66 11-85 A cylindrical ice block is placed on a table. The initial temperature of the ice block to avoid melting for 2 h is to be determined. Assumptions 1 Heat conduction in the ice block is two-dimensional, and thus the temperature varies in both x- and r- directions. 2 Heat transfer from the base of the ice block to the table is negligible. 3 The thermal properties of the ice block are constant. 4 The heat transfer coefficient is constant and uniform over the entire surface. 5 The Fourier number is τ > 0.2 so that the one-term approximate solutions (or the transient temperature charts) are applicable (this assumption will be verified). Properties The thermal properties of the ice are given to be k = 2.22 W/m.°C and α = 0.124×10-7 m2/s. Analysis This cylindrical ice block can be treated as a short cylinder that can physically be formed by the intersection of a long cylinder of diameter D = 2 cm and an infinite plane wall of thickness 2L = 4 cm. We measure x from the bottom surface of the block since this surface represents the adiabatic center surface of the plane wall of thickness 2L = 4 cm. The melting starts at the outer surfaces of the top surface when the temperature drops below 0°C at this location. The Biot numbers, the corresponding constants, and the Fourier numbers are 1171 .
0 ) C W/m.
22 .
2 ( ) m 02 .
0 )( C . W/m 13 ( 2 wall = ° ° = = k hL Bi 0187 .
1 and 3319 .
0 1 1 = = ⎯→ ⎯ A λ 05856 .
0 ) C W/m.
22 .
2 ( ) m 01 .
0 )( C . W/m 13 ( 2 cyl = ° ° = = k hr Bi o 0144 .
1 and 3393 .
0 1 1 = = ⎯→ ⎯ A λ 2 .
0 3348 .
0 m) 02 .
0 ( s/h) 600 3 h /s)(3 m 10 124 .
0 ( 2 2 7 2 wall > = × × = = − L t α τ 2 .
0 3392 .
1 m) 01 .
0 ( s/h) 600 3 h /s)(3 m 10 124 .
0 ( 2 2 7 2 cyl > = × × = = − o r t α τ Note that τ > 0.2 in all dimensions and thus the one-term approximate solution for transient heat conduction is applicable. The product solution for this problem can be written as [ ][ ] ) 9708 .
0 ( ) 0146 .
1 ( ) 3319 .
0 cos( ) 0187 .
1 ( 24 24 0 ) / ( ) / cos( 24 24 0 ) , ( ) , ( ) , , ( ) 3392 .
1 ( ) 3393 .
0 ( ) 3348 .
0 ( ) 3319 .
0 ( 1 0 1 1 1 cyl wall block 2 2 2 1 2 1 − − − − = − − ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ = − − = e e T r r J e A L L e A T t r t L t r L i o o i o o λ λ θ θ θ τ λ τ λ which gives C 6.6° − = i T Therefore, the ice will not start melting for at least 3 hours if its initial temperature is -6.6°C or below. Air T∞ = 24°C x r (ro, L) Insulation Ice block Ti PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 11-67 11-86 A cubic block and a cylindrical block are exposed to hot gases on all of their surfaces. The center temperatures of each geometry in 10, 20, and 60 min are to be determined. Assumptions 1 Heat conduction in the cubic block is three-dimensional, and thus the temperature varies in all x-, y, and z- directions. 2 Heat conduction in the cylindrical block is two-dimensional, and thus the temperature varies in both axial x- and radial r- directions. 3 The thermal properties of the granite are constant. 4 The heat transfer coefficient is constant and uniform over the entire surface. 5 The Fourier number is τ > 0.2 so that the one-term approximate solutions (or the transient temperature charts) are applicable (this assumption will be verified). Properties The thermal properties of the granite are given to be k = 2.5 W/m.°C and α = 1.15×10-6 m2/s. Analysis: Cubic block: This cubic block can physically be formed by the intersection of three infinite plane walls of thickness 2L = 5 cm. After 10 minutes: The Biot number, the corresponding constants, and the Fourier number are 400 .
0 ) C W/m.
5 .
2 ( ) m 025 .
0 )( C . W/m 40 ( 2 = ° ° = = k hL Bi 0580 .
1 and 5932 .
0 1 1 = = ⎯→ ⎯ A λ 2 .
0 104 .
1 m) 025 .
0 ( s/min) 60 min /s)(10 m 10 15 .
1 ( 2 2 6 2 > = × × = = − L t α τ To determine the center temperature, the product solution can be written as [ ] { } C 323° = = = − − ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ = − − = − − ∞ ∞ ) , 0 , 0 , 0 ( 369 .
0 ) 0580 .
1 ( 500 20 500 ) , 0 , 0 , 0 ( ) , 0 , 0 , 0 ( ) , 0 ( ) , 0 , 0 , 0 ( 3 ) 104 .
1 ( ) 5932 .
0 ( 3 1 3 wall block 2 2 1 t T e t T e A T T T t T t t i τ λ θ θ After 20 minutes 2 .
0 208 .
2 m) 025 .
0 ( s/min) 60 min /s)(20 m 10 15 .
1 ( 2 2 6 2 > = × × = = − L t α τ { } C 445° = ⎯→ ⎯ = = − − − ) , 0 , 0 , 0 ( 115 .
0 ) 0580 .
1 ( 500 20 500 ) , 0 , 0 , 0 ( 3 ) 208 .
2 ( ) 5932 .
0 ( 2 t T e t T After 60 minutes 2 .
0 624 .
6 m) 025 .
0 ( s/min) 60 min /s)(60 m 10 15 .
1 ( 2 2 6 2 > = × × = = − L t α τ { } C 500° = ⎯→ ⎯ = = − − − ) , 0 , 0 , 0 ( 00109 .
0 ) 0580 .
1 ( 500 20 500 ) , 0 , 0 , 0 ( 3 ) 624 .
6 ( ) 5932 .
0 ( 2 t T e t T Note that τ > 0.2 in all dimensions and thus the one-term approximate solution for transient heat conduction is applicable. Ti = 20°C Hot gases 500°C 5 cm × 5 cm × 5 Ti = 20°C PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 11-68 Cylinder: This cylindrical block can physically be formed by the intersection of a long cylinder of radius ro = D/2 = 2.5 cm and a plane wall of thickness 2L = 5 cm. After 10 minutes: The Biot number and the corresponding constants for the long cylinder are 400 .
0 ) C W/m.
5 .
2 ( ) m 025 .
0 )( C . W/m 40 ( 2 = ° ° = = k hr Bi o 0931 .
1 and 8516 .
0 1 1 = = ⎯→ ⎯ A λ To determine the center temperature, the product solution can be written as [ ][ ] { } { } C 331° = ⎯→ ⎯ = = − − ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ = − − = − − − − ∞ ∞ ) , 0 , 0 ( 352 .
0 ) 0931 .
1 ( ) 0580 .
1 ( 500 20 500 ) , 0 , 0 ( ) , 0 , 0 ( ) , 0 ( ) , 0 ( ) , 0 , 0 ( ) 104 .
1 ( ) 8516 .
0 ( ) 104 .
1 ( ) 5932 .
0 ( 1 1 2 2 2 1 2 1 t T e e t T e A e A T T T t T t t t cyl wall i cyl wall block τ λ τ λ θ θ θ After 20 minutes { } { } C 449° = ⎯→ ⎯ = = − − − − ) , 0 , 0 ( 107 .
0 ) 0931 .
1 ( ) 0580 .
1 ( 500 20 500 ) , 0 , 0 ( ) 208 .
2 ( ) 8516 .
0 ( ) 208 .
2 ( ) 5932 .
0 ( 2 2 t T e e t T After 60 minutes { } { } C 500° = ⎯→ ⎯ = = − − − − ) , 0 , 0 ( 00092 .
0 ) 0931 .
1 ( ) 0580 .
1 ( 500 20 500 ) , 0 , 0 ( ) 624 .
6 ( ) 8516 .
0 ( ) 624 .
6 ( ) 5932 .
0 ( 2 2 t T e e t T Note that τ > 0.2 in all dimensions and thus the one-term approximate solution for transient heat conduction is applicable. PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 11-69 11-87 A cubic block and a cylindrical block are exposed to hot gases on all of their surfaces. The center temperatures of each geometry in 10, 20, and 60 min are to be determined. Assumptions 1 Heat conduction in the cubic block is three-dimensional, and thus the temperature varies in all x-, y, and z- directions. 2 Heat conduction in the cylindrical block is two-dimensional, and thus the temperature varies in both axial x- and radial r- directions. 3 The thermal properties of the granite are constant. 4 The heat transfer coefficient is constant and uniform over the entire surface. 5 The Fourier number is τ > 0.2 so that the one-term approximate solutions (or the transient temperature charts) are applicable (this assumption will be verified). Properties The thermal properties of the granite are k = 2.5 W/m.°C and α = 1.15×10-6 m2/s. Analysis: Cubic block: This cubic block can physically be formed by the intersection of three infinite plane wall of thickness 2L = 5 cm. Two infinite plane walls are exposed to the hot gases with a heat transfer coefficient of h = ° 40 W / m . C 2 and one with h = ° 80 W / m . C 2 . After 10 minutes: The Biot number and the corresponding constants for C . W/m 40 2 ° = h are 400 .
0 ) C W/m.
5 .
2 ( ) m 025 .
0 )( C . W/m 40 ( 2 = ° ° = = k hL Bi 0580 .
1 and 5932 .
0 1 1 = = ⎯→ ⎯ A λ The Biot number and the corresponding constants for C . W/m 80 2 ° = h are 800 .
0 ) C W/m.
5 .
2 ( ) m 025 .
0 )( C . W/m 80 ( 2 = ° ° = = k hL Bi 1016 .
1 and 7910 .
0 1 1 = = ⎯→ ⎯ A λ The Fourier number is 2 .
0 104 .
1 m) 025 .
0 ( s/min) 60 min /s)(10 m 10 15 .
1 ( 2 2 6 2 > = × × = = − L t α τ To determine the center temperature, the product solution method can be written as [ ] [ ] { } { } C 364° = = = − − ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ = − − = − − − − ∞ ∞ ) , 0 , 0 , 0 ( 284 .
0 ) 1016 .
1 ( ) 0580 .
1 ( 500 20 500 ) , 0 , 0 , 0 ( ) , 0 , 0 , 0 ( ) , 0 ( ) , 0 ( ) , 0 , 0 , 0 ( ) 104 .
1 ( ) 7910 .
0 ( 2 ) 104 .
1 ( ) 5932 .
0 ( 1 2 1 wall 2 wall block 2 2 2 1 2 1 t T e e t T e A e A T T T t T t t t i τ λ τ λ θ θ θ After 20 minutes 2 .
0 208 .
2 m) 025 .
0 ( s/min) 60 min /s)(20 m 10 15 .
1 ( 2 2 6 2 > = × × = = − L t α τ { } { } C 469° = ⎯→ ⎯ = = − − − − ) , 0 , 0 , 0 ( 0654 .
0 ) 1016 .
1 ( ) 0580 .
1 ( 500 20 500 ) , 0 , 0 , 0 ( ) 208 .
2 ( ) 7910 .
0 ( 2 ) 208 .
2 ( ) 5932 .
0 ( 2 2 t T e e t T Ti = 20°C Hot gases 500°C 5 cm × 5 cm × 5 Ti = 20°C PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 11-70 After 60 minutes 2 .
0 624 .
6 m) 025 .
0 ( s/min) 60 min /s)(60 m 10 15 .
1 ( 2 2 6 2 > = × × = = − L t α τ { } { } C 500° = ⎯→ ⎯ = = − − − − ) , 0 , 0 , 0 ( 000186 .
0 ) 1016 .
1 ( ) 0580 .
1 ( 500 20 500 ) , 0 , 0 , 0 ( ) 624 .
6 ( ) 7910 .
0 ( 2 ) 624 .
6 ( ) 5932 .
0 ( 2 2 t T e e t T Note that τ > 0.2 in all dimensions and thus the one-term approximate solution for transient heat conduction is applicable. Cylinder: This cylindrical block can physically be formed by the intersection of a long cylinder of radius ro = D/2 = 2.5 cm exposed to the hot gases with a heat transfer coefficient of C . W/m 40 2 ° = h and a plane wall of thickness 2L = 5 cm exposed to the hot gases with C . W/m 80 2 ° = h . After 10 minutes: The Biot number and the corresponding constants for the long cylinder are 400 .
0 ) C W/m.
5 .
2 ( ) m 025 .
0 )( C . W/m 40 ( 2 = ° ° = = k hr Bi o 0931 .
1 and 8516 .
0 1 1 = = ⎯→ ⎯ A λ To determine the center temperature, the product solution method can be written as [ ][ ] { } { } C 370° = = = − − ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ = − − = − − − − ∞ ∞ ) , 0 , 0 ( 271 .
0 ) 0931 .
1 ( ) 1016 .
1 ( 500 20 500 ) , 0 , 0 ( ) , 0 , 0 ( ) , 0 ( ) , 0 ( ) , 0 , 0 ( ) 104 .
1 ( ) 8516 .
0 ( ) 104 .
1 ( ) 7910 .
0 ( 1 wall 1 wall block 2 2 2 1 2 1 t T e e t T e A e A T T T t T t t t cyl i cyl τ λ τ λ θ θ θ After 20 minutes { } { } C 471° = ⎯→ ⎯ = = − − − − ) , 0 , 0 ( 06094 .
0 ) 0931 .
1 ( ) 1016 .
1 ( 500 20 500 ) , 0 , 0 ( ) 208 .
2 ( ) 8516 .
0 ( ) 208 .
2 ( ) 7910 .
0 ( 2 2 t T e e t T After 60 minutes { } { } C 500° = ⎯→ ⎯ = = − − − − ) , 0 , 0 ( 0001568 .
0 ) 0931 .
1 ( ) 1016 .
1 ( 500 20 500 ) , 0 , 0 ( ) 624 .
6 ( ) 8516 .
0 ( ) 624 .
6 ( ) 7910 .
0 ( 2 2 t T e e t T Note that τ > 0.2 in all dimensions and thus the one-term approximate solution for transient heat conduction is applicable. PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 11-71 11-88 A cylindrical aluminum block is heated in a furnace. The length of time the block should be kept in the furnace and the amount of heat transfer to the block are to be determined. Assumptions 1 Heat conduction in the cylindrical block is two-dimensional, and thus the temperature varies in both axial x- and radial r- directions. 2 The thermal properties of the aluminum are constant. 3 The heat transfer coefficient is constant and uniform over the entire surface. 4 The Fourier number is τ > 0.2 so that the one-term approximate solutions (or the transient temperature charts) are applicable (it will be verified). Properties The thermal properties of the aluminum block are given to be k = 236 W/m.°C, ρ = 2702 kg/m3, cp = 0.896 kJ/kg.°C, and α = 9.75×10-5 m2/s. Analysis This cylindrical aluminum block can physically be formed by the intersection of an infinite plane wall of thickness 2L = 20 cm, and a long cylinder of radius ro = D/2 = 7.5 cm. The Biot numbers and the corresponding constants are first determined to be 0339 .
0 ) C W/m.
236 ( ) m 1 .
0 )( C . W/m 80 ( 2 = ° ° = = k hL Bi 0056 .
1 and 1811 .
0 1 1 = = ⎯→ ⎯ A λ 0254 .
0 C W/m.
236 ) m 075 .
0 )( C . W/m 80 ( 2 0 = ° ° = = k hr Bi 0063 .
1 and 2217 .
0 1 1 = = ⎯→ ⎯ A λ Noting that 2 / L t α τ = and assuming τ > 0.2 in all dimensions and thus the one-term approximate solution for transient heat conduction is applicable, the product solution for this problem can be written as 7627 .
0 ) 075 .
0 ( ) 10 75 .
9 ( ) 2217 .
0 ( exp ) 0063 .
1 ( ) 1 .
0 ( ) 10 75 .
9 ( ) 1811 .
0 ( exp ) 0056 .
1 ( 1200 20 1200 300 ) , 0 ( ) , 0 ( ) , 0 , 0 ( 2 5 2 2 5 2 cyl 1 wall 1 cyl wall block 2 1 2 1 = ⎪ ⎭ ⎪ ⎬ ⎫ ⎪ ⎩ ⎪ ⎨ ⎧ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎣ ⎡ × − × ⎪ ⎭ ⎪ ⎬ ⎫ ⎪ ⎩ ⎪ ⎨ ⎧ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎣ ⎡ × − = − − ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ = = − − − − t t e A e A t t t τ λ τ λ θ θ θ Solving for the time t gives t = 241 s = 4.0 min. We note that 2 .
0 177 .
4 m) 075 .
0 ( s) /s)(241 m 10 75 .
9 ( 2 .
0 350 .
2 m) 1 .
0 ( s) /s)(241 m 10 75 .
9 ( 2 2 5 2 cyl 2 2 5 2 wall > = × = = > = × = = − − o r t L t α τ α τ and thus the assumption of τ > 0.2 for the applicability of the one-term approximate solution is verified. The dimensionless temperatures at the center are [ ] [ ] 8195 .
0 ) 177 .
4 ( ) 2217 .
0 ( exp ) 0063 .
1 ( ) , 0 ( 9310 .
0 ) 350 .
2 ( ) 1811 .
0 ( exp ) 0056 .
1 ( ) , 0 ( 2 cyl 1 cyl 2 wall 1 wall 2 1 2 1 = − = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ = = − = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ = − − τ λ τ λ θ θ e A t e A t The maximum amount of heat transfer is [ ] kJ 100 , 10 C ) 1200 20 )( C kJ/kg. 896 .
0 )( kg 550 .
9 ( ) ( kg 550 .
9 m) 2 .
0 ( m) 075 .
0 ( ) kg/m 2702 ( max 2.
3 2 = ° − ° = − = = = = = ∞ T T mc Q L r m i p o π ρπ ρV Then we determine the dimensionless heat transfer ratios for both geometries as L z Cylinder Ti = 20°C ro Furnace T∞ = 1200°C L PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 11-72 07408 .
0 1811 .
0 ) 1811 .
0 sin( ) 9310 .
0 ( 1 ) sin( 1 1 1 , max = − = − = ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ λ λ θ wall o wall Q Q 1860 .
0 2217 .
0 1101 .
0 ) 8195 .
0 ( 2 1 ) ( 2 1 1 1 1 , max = − = − = ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ λ λ θ J Q Q cyl o cyl The heat transfer ratio for the short cylinder is 2463 .
0 ) 07408 .
0 1 )( 1860 .
0 ( 07408 .
0 1 max max max max = − + = ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ − ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ + ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ = ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ wall plane cylinder long wall plane cylinder short Q Q Q Q Q Q Q Q Then the total heat transfer from the short cylinder as it is cooled from 300°C at the center to 20°C becomes kJ 2490 = = = kJ) 100 , 10 )( 2463 .
0 ( 2463 .
0 max Q Q which is identical to the heat transfer to the cylinder as the cylinder at 20°C is heated to 300°C at the center. PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 11-73 11-89 A cylindrical aluminum block is heated in a furnace. The length of time the block should be kept in the furnace and the amount of heat transferred to the block are to be determined. Assumptions 1 Heat conduction in the cylindrical block is two-dimensional, and thus the temperature varies in both axial x- and radial r- directions. 2 Heat transfer from the bottom surface of the block is negligible. 3 The thermal properties of the aluminum are constant. 4 The heat transfer coefficient is constant and uniform over the entire surface. 5 The Fourier number is τ > 0.2 so that the one-term approximate solutions (or the transient temperature charts) are applicable (this assumption will be verified). Properties The thermal properties of the aluminum block are given to be k = 236 W/m.°C, ρ = 2702 kg/m3, cp = 0.896 kJ/kg.°C, and α = 9.75×10-5 m2/s. Analysis This cylindrical aluminum block can physically be formed by the intersection of an infinite plane wall of thickness 2L = 40 cm and a long cylinder of radius ro = D/2 = 7.5 cm. Note that the height of the short cylinder represents the half thickness of the infinite plane wall where the bottom surface of the short cylinder is adiabatic. The Biot numbers and corresponding constants are first determined to be 0678 .
0 ) C W/m.
236 ( ) m 2 .
0 )( C . W/m 80 ( 2 = ° ° = = k hL Bi 0110 .
1 and 2568 .
0 1 1 = = → A λ 0254 .
0 ) C W/m.
236 ( ) m 075 .
0 )( C . W/m 80 ( 2 = ° ° = = k hr Bi o 0063 .
1 and 2217 .
0 1 1 = = → A λ Noting that 2 / L t α τ = and assuming τ > 0.2 in all dimensions and thus the one-term approximate solution for transient heat conduction is applicable, the product solution for this problem can be written as 7627 .
0 ) 075 .
0 ( ) 10 75 .
9 ( ) 2217 .
0 ( exp ) 0063 .
1 ( ) 2 .
0 ( ) 10 75 .
9 ( ) 2568 .
0 ( exp ) 0110 .
1 ( 1200 20 1200 300 ) , 0 ( ) , 0 ( ) , 0 , 0 ( 2 5 2 2 5 2 cyl 1 wall 1 cyl wall block 2 1 2 1 = ⎪ ⎭ ⎪ ⎬ ⎫ ⎪ ⎩ ⎪ ⎨ ⎧ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎣ ⎡ × − ⎪ ⎭ ⎪ ⎬ ⎫ ⎪ ⎩ ⎪ ⎨ ⎧ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎣ ⎡ × − = − − ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ = = − − − − t t e A e A t t t τ λ τ λ θ θ θ Solving for the time t gives t = 285 s = 4.7 min. We note that 2 .
0 940 .
4 m) 075 .
0 ( s) /s)(285 m 10 75 .
9 ( 2 .
0 6947 .
0 m) 2 .
0 ( s) /s)(285 m 10 75 .
9 ( 2 2 5 2 cyl 2 2 5 2 wall > = × = = > = × = = − − o r t L t α τ α τ and thus the assumption of τ > 0.2 for the applicability of the one-term approximate solution is verified. The dimensionless temperatures at the center are [ ] [ ] 7897 .
0 ) 940 .
4 ( ) 2217 .
0 ( exp ) 0063 .
1 ( ) , 0 ( 9658 .
0 ) 6947 .
0 ( ) 2568 .
0 ( exp ) 0110 .
1 ( ) , 0 ( 2 cyl 1 cyl 2 wall 1 wall 2 1 2 1 = − = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ = = − = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ = − − τ λ τ λ θ θ e A t e A t The maximum amount of heat transfer is [ ] kJ 100 , 10 C ) 1200 20 )( C kJ/kg. 896 .
0 )( kg 55 .
9 ( ) ( kg 55 .
9 m) 2 .
0 ( m) 075 .
0 ( ) kg/m 2702 ( max 2.
3 = ° − ° = − = = = = = ∞ T T mc Q L r m i p o π ρπ ρV Then we determine the dimensionless heat transfer ratios for both geometries as L z Cylinder Ti = 20°C r0 Furnace T∞ = 1200°C L PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 11-74 04477 .
0 2568 .
0 ) 2568 .
0 sin( ) 9658 .
0 ( 1 ) sin( 1 1 1 , max = − = − = ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ λ λ θ wall o wall Q Q 2156 .
0 2217 .
0 1101 .
0 ) 7897 .
0 ( 2 1 ) ( 2 1 1 1 1 , max = − = − = ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ λ λ θ J Q Q cyl o cyl The heat transfer ratio for the short cylinder is 2507 .
0 ) 04477 .
0 1 )( 2156 .
0 ( 04477 .
0 1 wall plane max cylinder long max wall plane max cylinder short max = − + = ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ − ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ + ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ = ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ Q Q Q Q Q Q Q Q Then the total heat transfer from the short cylinder as it is cooled from 300°C at the center to 20°C becomes kJ 2530 = = = kJ) 100 , 10 )( 2507 .
0 ( 2507 .
0 max Q Q which is identical to the heat transfer to the cylinder as the cylinder at 20°C is heated to 300°C at the center. PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 11-75 11-90 EES Prob. 11-88 is reconsidered. The effect of the final center temperature of the block on the heating time and the amount of heat transfer is to be investigated. Analysis The problem is solved using EES, and the solution is given below. "GIVEN" L=0.20 [m] 2r_o=0.15 [m] T_i=20 [C] T_infinity=1200 [C] T_o_o=300 [C] h=80 [W/m^2-C] "PROPERTIES" k=236 [W/m-C] rho=2702 [kg/m^3] c_p=0.896 [kJ/kg-C] alpha=9.75E-5 [m^2/s] "ANALYSIS" "This short cylinder can physically be formed by the intersection of a long cylinder of radius r_o and a plane wall of thickness 2L" "For plane wall" Bi_w=(hL)/k "From Table 11-2 corresponding to this Bi number, we read" lambda_1_w=0.2568 "w stands for wall" A_1_w=1.0110 tau_w=(alphatime)/L^2 theta_o_w=A_1_wexp(-lambda_1_w^2tau_w) "theta_o_w=(T_o_w-T_infinity)/(T_i-T_infinity)" "For long cylinder" Bi_c=(hr_o)/k "c stands for cylinder" "From Table 11-2 corresponding to this Bi number, we read" lambda_1_c=0.2217 A_1_c=1.0063 tau_c=(alphatime)/r_o^2 theta_o_c=A_1_cexp(-lambda_1_c^2tau_c) "theta_o_c=(T_o_c-T_infinity)/(T_i-T_infinity)" (T_o_o-T_infinity)/(T_i-T_infinity)=theta_o_wtheta_o_c "center temperature of cylinder" V=pir_o^2L m=rhoV Q_max=mc_p(T_infinity-T_i) Q_w=1-theta_o_wSin(lambda_1_w)/lambda_1_w "Q_w=(Q/Q_max)_w" Q_c=1-2theta_o_cJ_1/lambda_1_c "Q_c=(Q/Q_max)_c" J_1=0.1101 "From Table 11-3, at lambda_1_c" Q/Q_max=Q_w+Q_c(1-Q_w) "total heat transfer" PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 11-76 To,o [C] time [s] Q [kJ] 50 42.43 430.3 100 86.33 850.6 150 132.3 1271 200 180.4 1691 250 231.1 2111 300 284.5 2532 350 340.9 2952 400 400.8 3372 450 464.5 3793 500 532.6 4213 550 605.8 4633 600 684.9 5053 650 770.8 5474 700 864.9 5894 750 968.9 6314 800 1085 6734 850 1217 7155 900 1369 7575 950 1549 7995 1000 1770 8416 0 200 400 600 800 1000 0 500 1000 1500 2000 0 1000 2000 3000 4000 5000 6000 7000 8000 9000 To,o time Q [kJ] time heat PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 11-77 Review Problems 11-91 Two large steel plates are stuck together because of the freezing of the water between the two plates. Hot air is blown over the exposed surface of the plate on the top to melt the ice. The length of time the hot air should be blown is to be determined. Assumptions 1 Heat conduction in the plates is one-dimensional since the plate is large relative to its thickness and there is thermal symmetry about the center plane. 3 The thermal properties of the steel plates are constant. 4 The heat transfer coefficient is constant and uniform over the entire surface. 5 The Fourier number is τ > 0.2 so that the one-term approximate solutions (or the transient temperature charts) are applicable (this assumption will be verified). Properties The thermal properties of steel plates are given to be k = 43 W/m.°C and α = 1.17×10-5 m2/s Analysis The characteristic length of the plates and the Biot number are 1 .
0 019 .
0 ) C W/m.
43 ( ) m 02 .
0 )( C . W/m 40 ( m 02 .
0 2 < = ° ° = = = = = k hL Bi L A L c s c V Since 0.1 < Bi , the lumped system analysis is applicable. Therefore, min 8.0 s 482 = = ⎯→ ⎯ = − − − ⎯→ ⎯ = − − = ° × ° = = = − − ∞ ∞ t e e T T T t T L c h c hA b bt i c p p s )t s 000544 .
0 ( 1 -3 6 2 1 -50 15 50 0 ) ( s 000544 .
0 m) (0.02 ) C .
J/m 10 675 .
3 ( C . W/m 40 ρ ρ V where C .
J/m 10 675 .
3 /s m 10 17 .
1 C W/m.
43 3 6 2 5 ° × = × ° = = − α ρ k c p Alternative solution: This problem can also be solved using the transient chart Fig. 11-15a, 2 .
0 15 769 .
0 50 15 50 0 6 .
52 019 .
0 1 1 2 0 > = = ⎪ ⎪ ⎭ ⎪ ⎪ ⎬ ⎫ = − − − = − − = = ∞ ∞ o i r t T T T T Bi α τ Then, s 513 = × = = − /s) m 10 17 .
1 ( m) 02 .
0 )( 15 ( 2 5 2 2 α τ o r t The difference is due to the reading error of the chart. Steel plates Ti = -15°C Hot gases T∞ = 50°C PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 11-78 11-92 A curing kiln is heated by injecting steam into it and raising its inner surface temperature to a specified value. It is to be determined whether the temperature at the outer surfaces of the kiln changes during the curing period. Assumptions 1 The temperature in the wall is affected by the thermal conditions at inner surfaces only and the convection heat transfer coefficient inside is very large. Therefore, the wall can be considered to be a semi-infinite medium with a specified surface temperature of 45°C. 2 The thermal properties of the concrete wall are constant. Properties The thermal properties of the concrete wall are given to be k = 0.9 W/m.°C and α = 0.23×10-5 m2/s. Analysis We determine the temperature at a depth of x = 0.3 m in 2.5 h using the analytical solution, ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ = − − t x erfc T T T t x T i s i α 2 ) , ( Substituting, C 11.0 ° = = = ⎟ ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎜ ⎝ ⎛ × × = − − − ) , ( 1402 .
0 ) 043 .
1 ( ) s/h 3600 h 5 .
2 )( /s m 10 23 .
0 ( 2 m 3 .
0 6 42 6 ) , ( 2 5 t x T erfc erfc t x T which is greater than the initial temperature of 6°C. Therefore, heat will propagate through the 0.3 m thick wall in 2.5 h, and thus it may be desirable to insulate the outer surface of the wall to save energy. 11-93 The water pipes are buried in the ground to prevent freezing. The minimum burial depth at a particular location is to be determined. Assumptions 1 The temperature in the soil is affected by the thermal conditions at one surface only, and thus the soil can be considered to be a semi-infinite medium with a specified surface temperature of -10°C. 2 The thermal properties of the soil are constant. Properties The thermal properties of the soil are given to be k = 0.7 W/m.°C and α = 1.4×10-5 m2/s. Analysis The depth at which the temperature drops to 0°C in 75 days is determined using the analytical solution, ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ = − − t x erfc T T T t x T i s i α 2 ) , ( Substituting and using Table 11-4, we obtain m 7.05 = ⎯→ ⎯ ⎟ ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎜ ⎝ ⎛ × × × = − − − − x x erfc ) s/h 3600 h/day 24 day 75 )( /s m 10 4 .
1 ( 2 15 10 15 0 2 5 Therefore, the pipes must be buried at a depth of at least 7.05 m. 6°C 42°C 30 cm Kiln wall x 0 Soil Ti = 15°C Water pipe Ts =-10°C x PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 11-79 11-94 A hot dog is to be cooked by dropping it into boiling water. The time of cooking is to be determined. Assumptions 1 Heat conduction in the hot dog is two-dimensional, and thus the temperature varies in both the axial x- and the radial r- directions. 2 The thermal properties of the hot dog are constant. 4 The heat transfer coefficient is constant and uniform over the entire surface. 5 The Fourier number is τ > 0.2 so that the one-term approximate solutions (or the transient temperature charts) are applicable (this assumption will be verified). Properties The thermal properties of the hot dog are given to be k = 0.76 W/m.°C, ρ = 980 kg/m3, cp = 3.9 kJ/kg.°C, and α = 2×10-7 m2/s. Analysis This hot dog can physically be formed by the intersection of an infinite plane wall of thickness 2L = 12 cm, and a long cylinder of radius ro = D/2 = 1 cm. The Biot numbers and corresponding constants are first determined to be 37 .
47 ) C W/m.
76 .
0 ( ) m 06 .
0 )( C . W/m 600 ( 2 = ° ° = = k hL Bi 2726 .
1 and 5380 .
1 1 1 = = ⎯→ ⎯ A λ 895 .
7 ) C W/m.
76 .
0 ( ) m 01 .
0 )( C . W/m 600 ( 2 = ° ° = = k hr Bi o 5514 .
1 and 1249 .
2 1 1 = = ⎯→ ⎯ A λ Noting that 2 / L t α τ = and assuming τ > 0.2 in all dimensions and thus the one-term approximate solution for transient heat conduction is applicable, the product solution for this problem can be written as 2105 .
0 ) 01 .
0 ( ) 10 2 ( ) 1249 .
2 ( exp ) 5514 .
1 ( ) 06 .
0 ( ) 10 2 ( ) 5380 .
1 ( exp ) 2726 .
1 ( 100 5 100 80 ) , 0 ( ) , 0 ( ) , 0 , 0 ( 2 7 2 2 7 2 1 1 2 1 2 1 = ⎪ ⎭ ⎪ ⎬ ⎫ ⎪ ⎩ ⎪ ⎨ ⎧ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎣ ⎡ × − × ⎪ ⎭ ⎪ ⎬ ⎫ ⎪ ⎩ ⎪ ⎨ ⎧ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎣ ⎡ × − = − − ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ = = − − − − t t e A e A t t t cyl wall block τ λ τ λ θ θ θ which gives min 4.1 s 244 = = t Therefore, it will take about 4.1 min for the hot dog to cook. Note that 2 .
0 49 .
0 m) 01 .
0 ( s) /s)(244 m 10 2 ( 2 2 7 2 > = × = = − o cyl r t α τ and thus the assumption τ > 0.2 for the applicability of the one-term approximate solution is verified. Discussion This problem could also be solved by treating the hot dog as an infinite cylinder since heat transfer through the end surfaces will have little effect on the mid section temperature because of the large distance. Water 100°C 2 cm Hot dog Ti = 5°C PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 11-80 11-95 A long roll of large 1-Mn manganese steel plate is to be quenched in an oil bath at a specified rate. The temperature of the sheet metal after quenching and the rate at which heat needs to be removed from the oil in order to keep its temperature constant are to be determined. Assumptions 1 The thermal properties of the steel plate are constant. 2 The heat transfer coefficient is constant and uniform over the entire surface. 3 The Biot number is Bi < 0.1 so that the lumped system analysis is applicable (this assumption will be checked). Properties The properties of the steel plate are k = 60.5 W/m.°C, ρ = 7854 kg/m3, and cp = 434 J/kg.°C (Table A-24). Analysis The characteristic length of the steel plate and the Biot number are 1 .
0 036 .
0 C W/m.
5 .
60 ) m 0025 .
0 )( C . W/m 860 ( m 0025 .
0 2 < = ° ° = = = = = k hL Bi L A L c s c V Since 0.1 < Bi , the lumped system analysis is applicable. Therefore, s 36 min 6 .
0 m/min 15 m 9 velocity length time s 10092 .
0 m) C)(0.0025 J/kg. 434 )( kg/m (7854 C . W/m 860 1 -3 2 = = = = = ° ° = = = c p p s L c h c hA b ρ ρ V Then the temperature of the sheet metal when it leaves the oil bath is determined to be C 65.5° = ⎯→ ⎯ = − − ⎯→ ⎯ = − − − − ∞ ∞ ) ( 45 820 45 ) ( ) ( s) 36 )( s 10092 .
0 ( -1 t T e t T e T T T t T bt i The mass flow rate of the sheet metal through the oil bath is kg/min 1178 m/min) 15 ( m) 005 .
0 ( m) 2 )( kg/m 7854 ( 3 = = = = wtV m ρ ρV & & Then the rate of heat transfer from the sheet metal to the oil bath and thus the rate at which heat needs to be removed from the oil in order to keep its temperature constant at 45°C becomes kW 6429 = kJ/min 740 , 385 C ) 5 .
65 820 )( C kJ/kg. 434 .
0 )( kg/min 1178 ( )] ( [ = ° − ° = − = t T T c m Q i p & & Steel plate 15 m/min Oil bath 45°C PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 11-81 11-96E A stuffed turkey is cooked in an oven. The average heat transfer coefficient at the surface of the turkey, the temperature of the skin of the turkey in the oven and the total amount of heat transferred to the turkey in the oven are to be determined. Assumptions 1 The turkey is a homogeneous spherical object. 2 Heat conduction in the turkey is one-dimensional because of symmetry about the midpoint. 3 The thermal properties of the turkey are constant. 4 The heat transfer coefficient is constant and uniform over the entire surface. 5 The Fourier number is τ > 0.2 so that the one-term approximate solutions are applicable (this assumption will be verified). Properties The properties of the turkey are given to be k = 0.26 Btu/h.ft.°F, ρ = 75 lbm/ft3, cp = 0.98 Btu/lbm.°F, and α = 0.0035 ft2/h. Analysis (a) Assuming the turkey to be spherical in shape, its radius is determined to be ft 3545 .
0 4 ) ft 1867 .
0 ( 3 4 3 3 4 ft 1867 .
0 lbm/ft 75 lbm 14 3 3 3 3 3 3 = = = ⎯→ ⎯ = = = = ⎯→ ⎯ = π π π ρ ρ V V V V o o r r m m The Fourier number is 1392 .
0 ft) 3545 .
0 ( h) /h)(5 ft 10 5 .
3 ( 2 2 3 2 = × = = − o r t α τ which is close to 0.2 but a little below it. Therefore, assuming the one-term approximate solution for transient heat conduction to be applicable, the one-term solution formulation at one-third the radius from the center of the turkey can be expressed as 1 1 ) 14 .
0 ( 1 1 1 1 333 .
0 ) 333 .
0 sin( 491 .
0 325 40 325 185 / ) / sin( ) , ( ) , ( 2 1 2 1 λ λ λ λ θ λ τ λ − − ∞ ∞ = = − − = − − = e A r r r r e A T T T t x T t x o o i sph By trial and error, it is determined from Table 11-2 that the equation above is satisfied when Bi = 20 corresponding to 9781 .
1 and 9857 .
2 1 1 = = A λ . Then the heat transfer coefficient can be determined from F .
Btu/h.ft 14.7 2 ° = ° = = ⎯→ ⎯ = ) ft 3545 .
0 ( ) 20 )( F Btu/h.ft. 26 .
0 ( o o r kBi h k hr Bi (b) The temperature at the surface of the turkey is F 317 ° = ⎯→ ⎯ = = = − − − − ) , ( 02953 .
0 9857 .
2 ) 9857 .
2 sin( ) 9781 .
1 ( / ) / sin( 325 40 325 ) , ( ) 14 .
0 ( ) 9857 .
2 ( 1 1 1 2 2 1 t r T e r r r r e A t r T o o o o o o λ λ τ λ (c) The maximum possible heat transfer is Btu 3910 = F ) 40 325 )( F Btu/lbm. 98 .
0 )( lbm 14 ( ) ( max ° − ° = − = ∞ i p T T mc Q Then the actual amount of heat transfer becomes Btu 3240 = = = = − − = − − = Btu) 3910 )( 828 .
0 ( 828 .
0 828 .
0 ) 9857 .
2 ( ) 9857 .
2 cos( ) 9857 .
2 ( ) 9857 .
2 sin( ) 491 .
0 ( 3 1 ) cos( ) sin( 3 1 max 3 3 1 1 1 1 , max Q Q Q Q sph o λ λ λ λ θ Discussion The temperature of the outer parts of the turkey will be greater than that of the inner parts when the turkey is taken out of the oven. Then heat will continue to be transferred from the outer parts of the turkey to the inner as a result of temperature difference. Therefore, after 5 minutes, the thermometer reading will probably be more than 185°F. Oven T∞ = 325°F Turkey Ti = 40°F PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 11-82 11-97 CD EES The trunks of some dry oak trees are exposed to hot gases. The time for the ignition of the trunks is to be determined. Assumptions 1 Heat conduction in the trunks is one-dimensional since it is long and it has thermal symmetry about the center line. 2 The thermal properties of the trunks are constant. 3 The heat transfer coefficient is constant and uniform over the entire surface. 4 The Fourier number is τ > 0.2 so that the one-term approximate solutions (or the transient temperature charts) are applicable (this assumption will be verified). Properties The properties of the trunks are given to be k = 0.17 W/m.°C and α = 1.28×10-7 m2/s. Analysis We treat the trunks of the trees as an infinite cylinder since heat transfer is primarily in the radial direction. Then the Biot number becomes 24 .
38 ) C W/m.
17 .
0 ( ) m 1 .
0 )( C . W/m 65 ( 2 = ° ° = = k hr Bi o The constants 1 1 and A λ corresponding to this Biot number are, from Table 11-2, 5989 .
1 and 3420 .
2 1 1 = = A λ The Fourier number is 184 .
0 m) 1 .
0 ( s/h) 600 3 h /s)(4 m 10 28 .
1 ( 2 2 7 2 = × × = = − o r t α τ which is slightly below 0.2 but close to it. Therefore, assuming the one-term approximate solution for transient heat conduction to be applicable, the temperature at the surface of the trees in 4 h becomes C 410 > ) , ( 01935 .
0 ) 0332 .
0 ( ) 5989 .
1 ( 520 30 520 ) , ( ) / ( ) , ( ) , ( ) 184 .
0 ( ) 3420 .
2 ( 1 0 1 2 2 1 ° ° = ⎯→ ⎯ = = − − = − − = − − ∞ ∞ C 511 t r T e t r T r r J e A T T T t r T t r o o o i o cyl o λ θ τ λ Therefore, the trees will ignite. (Note: J0 is read from Table 11-3). 11-98 A spherical watermelon that is cut into two equal parts is put into a freezer. The time it will take for the center of the exposed cut surface to cool from 25 to 3°C is to be determined. Assumptions 1 The temperature of the exposed surfaces of the watermelon is affected by the convection heat transfer at those surfaces only. Therefore, the watermelon can be considered to be a semi-infinite medium 2 The thermal properties of the watermelon are constant. Properties The thermal properties of the water is closely approximated by those of water at room temperature, k = 0.607 W/m.°C and α = = p c k ρ / 0.146×10-6 m2/s (Table A-15). Analysis We use the transient chart in Fig. 11-29 in this case for convenience (instead of the analytic solution), 1 0 2 595 .
0 ) 12 ( 25 ) 12 ( 3 1 ) , ( 1 = ⎪ ⎪ ⎭ ⎪ ⎪ ⎬ ⎫ = = = − − − − − = − − − ∞ ∞ k t h t x T T T t x T i α α ξ Therefore, min 86.9 = = × ° ° = = s 5214 /s) m 10 146 .
0 ( ) C . W/m 22 ( C) W/m.
607 .
0 ( ) 1 ( 2 6 -2 2 2 2 2 2 α h k t D = 0.2 m Tree Ti = 30°C Hot gases T∞ = 520°C Freezer T∞ = -12°C Watermelon Ti = 25°C PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 11-83 11-99 A cylindrical rod is dropped into boiling water. The thermal diffusivity and the thermal conductivity of the rod are to be determined. Assumptions 1 Heat conduction in the rod is one-dimensional since the rod is sufficiently long, and thus temperature varies in the radial direction only. 2 The thermal properties of the rod are constant. Properties The thermal properties of the rod available are given to be ρ = 3700 kg/m3 and Cp = 920 J/kg.°C. Analysis From Fig. 11-16b we have 25 .
0 1 1 28 .
0 100 75 100 93 0 = = ⎪ ⎪ ⎭ ⎪ ⎪ ⎬ ⎫ = = = − − = − − ∞ ∞ o o o o hr k Bi r r r x T T T T From Fig. 11-16a we have 40 .
0 33 .
0 100 25 100 75 25 .
0 1 2 = = ⎪ ⎪ ⎭ ⎪ ⎪ ⎬ ⎫ = − − = − − = = ∞ ∞ o i o o r t T T T T hr k Bi α τ Then the thermal diffusivity and the thermal conductivity of the material become C W/m.
0.756 /s m 10 2.22 2 7 ° = ° × = = ⎯→ ⎯ = × = × = = − − C) J/kg. )(920 kg/m /s)(3700 m 10 22 .
2 ( s/min 60 min 3 m) 01 .
0 )( 40 .
0 ( 40 .
0 3 2 7 2 2 p p o c k c k t r αρ α α α 11-100 The time it will take for the diameter of a raindrop to reduce to a certain value as it falls through ambient air is to be determined. Assumptions 1 The water temperature remains constant. 2 The thermal properties of the water are constant. Properties The density and heat of vaporization of the water are ρ = 1000 kg/m3 and hfg = 2490 kJ/kg (Table A-15). Analysis The initial and final masses of the raindrop are kg 0000141 .
0 m) 0015 .
0 ( 3 4 ) kg/m 1000 ( 3 4 kg 0000654 .
0 m) 0025 .
0 ( 3 4 ) kg/m 1000 ( 3 4 3 3 3 3 3 3 = = = = = = = = π π ρ ρ π π ρ ρ f f f i i i r m r m V V whose difference is kg 0000513 .
0 0000141 .
0 0000654 .
0 = − = − = f i m m m The amount of heat transfer required to cause this much evaporation is kJ 1278 .
0 kJ/kg) kg)(2490 0000513 .
0 ( = = Q The average heat transfer surface area and the rate of heat transfer are J/s 0.2777 = C ) 5 18 ( ) m 10 341 .
5 ( C) . W/m 400 ( ) ( m 10 341 .
5 2 m) (0.0015 + m) 0025 .
0 [( 4 2 ) ( 4 2 5 2 2 5 2 2 2 2 ° − × ° = − = × = = + = − ∞ − T T hA Q r r A i s f i s & π π Then the time required for the raindrop to experience this reduction in size becomes min 7.7 = = = = Δ ⎯→ ⎯ Δ = s 460 J/s 0.2777 J 8 .
127 Q Q t t Q Q & & Water 100°C 2 cm Rod Ti = 25°C Air T∞ = 18°C Raindrop 5°C PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 11-84 11-101E A plate, a long cylinder, and a sphere are exposed to cool air. The center temperature of each geometry is to be determined. Assumptions 1 Heat conduction in each geometry is one-dimensional. 2 The thermal properties of the bodies are constant. 3 The heat transfer coefficient is constant and uniform over the entire surface. 4 The Fourier number is τ > 0.2 so that the one-term approximate solutions (or the transient temperature charts) are applicable (this assumption will be verified). Properties The properties of bronze are given to be k = 15 Btu/h.ft.°F and α = 0.333 ft2/h. Analysis After 5 minutes Plate: First the Biot number is calculated to be 01944 .
0 ) F Btu/h.ft. 15 ( ) ft 12 / 5 .
0 )( F .
Btu/h.ft 7 ( 2 = ° ° = = k hL Bi The constants 1 1 and A λ corresponding to this Biot number are, from Table 11-2, 0032 .
1 and 1387 .
0 1 1 = = A λ The Fourier number is 2 .
0 98 .
15 ft) 12 / 5 .
0 ( min/h) min/60 /h)(5 ft 333 .
0 ( 2 2 2 > = = = L t α τ Then the center temperature of the plate becomes F 315° = ⎯→ ⎯ = = − − ⎯→ ⎯ = − − = − − ∞ ∞ 0 ) 98 .
15 ( ) 1387 .
0 ( 0 1 0 , 738 .
0 ) 0032 .
1 ( 75 400 75 2 2 1 T e T e A T T T T i wall o τ λ θ Cylinder: 01944 .
0 = Bi 0049 .
1 and 1962 .
0 1 1 2 4 Table = = ⎯ ⎯ ⎯ → ⎯ − A λ F 252° = ⎯→ ⎯ = = − − ⎯→ ⎯ = − − = − − ∞ ∞ 0 ) 98 .
15 ( ) 1962 .
0 ( 0 1 , 0 543 .
0 ) 0049 .
1 ( 75 400 75 2 2 1 T e T e A T T T T i o cyl τ λ θ Sphere: 01944 .
0 = Bi 0058 .
1 and 2405 .
0 1 1 2 4 Table = = ⎯ ⎯ ⎯ → ⎯ − A λ F 205° = ⎯→ ⎯ = = − − ⎯→ ⎯ = − − = − − ∞ ∞ 0 ) 98 .
15 ( ) 2405 .
0 ( 0 1 , 0 399 .
0 ) 0058 .
1 ( 75 400 75 2 2 1 T e T e A T T T T i o sph τ λ θ After 10 minutes 2 .
0 97 .
31 ft) 12 / 5 .
0 ( min/h) min/60 /h)(10 ft 333 .
0 ( 2 2 2 > = = = L t α τ Plate: F 251° = ⎯→ ⎯ = = − − ⎯→ ⎯ = − − = − − ∞ ∞ 0 ) 97 .
31 ( ) 1387 .
0 ( 0 1 0 , 0 542 .
0 ) 0032 .
1 ( 75 400 75 2 2 1 T e T e A T T T T i wall τ λ θ Cylinder: F 170° = ⎯→ ⎯ = = − − ⎯→ ⎯ = − − = − − ∞ ∞ 0 ) 97 .
31 ( ) 1962 .
0 ( 0 1 , 0 293 .
0 ) 0049 .
1 ( 75 400 75 2 2 1 T e T e A T T T T i o cyl τ λ θ Sphere: F 126° = ⎯→ ⎯ = = − − ⎯→ ⎯ = − − = − − ∞ ∞ 0 ) 97 .
31 ( ) 2405 .
0 ( 0 1 0 , 0 158 .
0 ) 0058 .
1 ( 75 400 75 2 2 1 T e T e A T T T T i sph τ λ θ 2L 2ro 2ro PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 11-85 After 30 minutes 2 .
0 9 .
95 ft) 12 / 5 .
0 ( min/h) min/60 /h)(30 ft 333 .
0 ( 2 2 2 > = = = L t α τ Plate: F 127° = ⎯→ ⎯ = = − − ⎯→ ⎯ = − − = − − ∞ ∞ 0 ) 9 .
95 ( ) 1387 .
0 ( 0 1 0 , 0 159 .
0 ) 0032 .
1 ( 75 400 75 2 2 1 T e T e A T T T T i wall τ λ θ Cylinder: F 83° = ⎯→ ⎯ = = − − ⎯→ ⎯ = − − = − − ∞ ∞ 0 ) 9 .
95 ( ) 1962 .
0 ( 0 1 , 0 025 .
0 ) 0049 .
1 ( 75 400 75 2 2 1 T e T e A T T T T i o cyl τ λ θ Sphere: F 76° = ⎯→ ⎯ = = − − ⎯→ ⎯ = − − = − − ∞ ∞ 0 ) 9 .
95 ( ) 2405 .
0 ( 0 1 , 0 00392 .
0 ) 0058 .
1 ( 75 400 75 2 2 1 T e T e A T T T T i o sph τ λ θ The sphere has the largest surface area through which heat is transferred per unit volume, and thus the highest rate of heat transfer. Consequently, the center temperature of the sphere is always the lowest. PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 11-86 11-102E A plate, a long cylinder, and a sphere are exposed to cool air. The center temperature of each geometry is to be determined. Assumptions 1 Heat conduction in each geometry is one-dimensional. 2 The thermal properties of the geometries are constant. 3 The heat transfer coefficient is constant and uniform over the entire surface. 4 The Fourier number is τ > 0.2 so that the one-term approximate solutions (or the transient temperature charts) are applicable (this assumption will be verified). Properties The properties of cast iron are given to be k = 29 Btu/h.ft.°F and α = 0.61 ft2/h. Analysis After 5 minutes Plate: First the Biot number is calculated to be 01 .
0 01006 .
0 ) F Btu/h.ft. 29 ( ) ft 12 / 5 .
0 )( F .
Btu/h.ft 7 ( 2 ≅ = ° ° = = k hL Bi The constants 1 1 and A λ corresponding to this Biot number are, from Table 11-2, 0017 .
1 and 0998 .
0 1 1 = = A λ The Fourier number is 2 .
0 28 .
29 ft) 12 / 5 .
0 ( min/h) min/60 /h)(5 ft 61 .
0 ( 2 2 2 > = = = L t α τ Then the center temperature of the plate becomes F 318° = ⎯→ ⎯ = = − − ⎯→ ⎯ = − − = − − ∞ ∞ 0 ) 28 .
29 ( ) 0998 .
0 ( 0 1 0 , 0 748 .
0 ) 0017 .
1 ( 75 400 75 2 2 1 T e T e A T T T T i wall τ λ θ Cylinder: Bi = 0 01 .
0025 .
1 and 1412 .
0 1 1 2 4 Table = = ⎯ ⎯ ⎯ → ⎯ − A λ F 257° = ⎯→ ⎯ = = − − ⎯→ ⎯ = − − = − − ∞ ∞ 0 ) 28 .
29 ( ) 1412 .
0 ( 0 1 0 , 0 559 .
0 ) 0025 .
1 ( 75 400 75 2 2 1 T e T e A T T T T i cyl τ λ θ Sphere: Bi = 0 01 .
0030 .
1 and 1730 .
0 1 1 2 4 Table = = ⎯ ⎯ ⎯ → ⎯ − A λ F 211° = ⎯→ ⎯ = = − − ⎯→ ⎯ = − − = − − ∞ ∞ 0 ) 28 .
29 ( ) 1730 .
0 ( 0 1 , 0 418 .
0 ) 0030 .
1 ( 75 400 75 2 2 1 T e T e A T T T T i o sph τ λ θ After 10 minutes 2 .
0 56 .
58 ft) 12 / 5 .
0 ( min/h) min/60 /h)(10 ft 61 .
0 ( 2 2 2 > = = = L t α τ Plate: F 257° = ⎯→ ⎯ = = − − ⎯→ ⎯ = − − = − − ∞ ∞ 0 ) 56 .
58 ( ) 0998 .
0 ( 0 1 0 , 0 559 .
0 ) 0017 .
1 ( 75 400 75 2 2 1 T e T e A T T T T i wall τ λ θ Cylinder: F 176° = ⎯→ ⎯ = = − − ⎯→ ⎯ = − − = − − ∞ ∞ 0 ) 56 .
58 ( ) 1412 .
0 ( 0 1 0 , 0 312 .
0 ) 0025 .
1 ( 75 400 75 2 2 1 T e T e A T T T T i cyl τ λ θ Sphere: F 132° = ⎯→ ⎯ = = − − ⎯→ ⎯ = − − = − − ∞ ∞ 0 ) 56 .
58 ( ) 1730 .
0 ( 0 1 0 , 0 174 .
0 ) 0030 .
1 ( 75 400 75 2 2 1 T e T e A T T T T i sph τ λ θ 2ro 2ro 2L PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 11-87 After 30 minutes 2 .
0 68 .
175 ft) 12 / 5 .
0 ( min/h) min/60 /h)(30 ft 61 .
0 ( 2 2 2 > = = = L t α τ Plate: F 132° = ⎯→ ⎯ = = − − ⎯→ ⎯ = − − = − − ∞ ∞ 0 ) 68 .
175 ( ) 0998 .
0 ( 0 1 0 , 0 174 .
0 ) 0017 .
1 ( 75 400 75 2 2 1 T e T e A T T T T i wall τ λ θ Cylinder: F 84.8° = ⎯→ ⎯ = = − − ⎯→ ⎯ = − − = − − ∞ ∞ 0 ) 68 .
175 ( ) 1412 .
0 ( 0 1 0 , 0 030 .
0 ) 0025 .
1 ( 75 400 75 2 2 1 T e T e A T T T T i cyl τ λ θ Sphere: F 76.7° = ⎯→ ⎯ = = − − ⎯→ ⎯ = − − = − − ∞ ∞ 0 ) 68 .
175 ( ) 1730 .
0 ( 0 1 0 , 0 0052 .
0 ) 0030 .
1 ( 75 400 75 2 2 1 T e T e A T T T T i sph τ λ θ The sphere has the largest surface area through which heat is transferred per unit volume, and thus the highest rate of heat transfer. Consequently, the center temperature of the sphere is always the lowest. PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 11-88 11-103E EES Prob. 11-101E is reconsidered. The center temperature of each geometry as a function of the cooling time is to be plotted. Analysis The problem is solved using EES, and the solution is given below. "GIVEN" 2L=(1/12) [ft] 2r_o_c=(1/12) [ft] "c stands for cylinder" 2r_o_s=(1/12) [ft] "s stands for sphere" T_i=400 [F] T_infinity=75 [F] h=7 [Btu/h-ft^2-F] time=5 [min] "PROPERTIES" k=15 [Btu/h-ft-F] alpha=0.333 [ft^2/h]Convert(ft^2/h, ft^2/min) "ANALYSIS" "For plane wall" Bi_w=(hL)/k "From Table 11-2 corresponding to this Bi number, we read" lambda_1_w=0.1387 A_1_w=1.0032 tau_w=(alphatime)/L^2 (T_o_w-T_infinity)/(T_i-T_infinity)=A_1_wexp(-lambda_1_w^2tau_w) "For long cylinder" Bi_c=(hr_o_c)/k "From Table 11-2 corresponding to this Bi number, we read" lambda_1_c=0.1962 A_1_c=1.0049 tau_c=(alphatime)/r_o_c^2 (T_o_c-T_infinity)/(T_i-T_infinity)=A_1_cexp(-lambda_1_c^2tau_c) "For sphere" Bi_s=(hr_o_s)/k "From Table 11-2 corresponding to this Bi number, we read" lambda_1_s=0.2405 A_1_s=1.0058 tau_s=(alphatime)/r_o_s^2 (T_o_s-T_infinity)/(T_i-T_infinity)=A_1_sexp(-lambda_1_s^2tau_s) time [min] To,w [F] To,c [F] To,s [F] 5 314.7 251.5 204.7 10 251.3 170.4 126.4 15 204.6 126.6 95.41 20 170.3 102.9 83.1 25 145.1 90.06 78.21 30 126.5 83.14 76.27 35 112.9 79.4 75.51 40 102.9 77.38 75.2 45 95.48 76.29 75.08 50 90.06 75.69 75.03 55 86.07 75.38 75.01 60 83.14 75.2 75 PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 11-89 0 10 20 30 40 50 60 50 100 150 200 250 300 350 50 100 150 200 250 300 350 time [min] To [F] wall cylinder sphere PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 11-90 11-104 Internal combustion engine valves are quenched in a large oil bath. The time it takes for the valve temperature to drop to specified temperatures and the maximum heat transfer are to be determined. Assumptions 1 The thermal properties of the valves are constant. 2 The heat transfer coefficient is constant and uniform over the entire surface. 3 Depending on the size of the oil bath, the oil bath temperature will increase during quenching. However, an average canstant temperature as specified in the problem will be used. 4 The Biot number is Bi < 0.1 so that the lumped system analysis is applicable (this assumption will be verified). Properties The thermal conductivity, density, and specific heat of the balls are given to be k = 48 W/m.°C, ρ = 7840 kg/m3, and cp = 440 J/kg.°C. Analysis (a) The characteristic length of the balls and the Biot number are 1 .
0 03 .
0 C W/m.
48 ) m 0018 .
0 )( C . W/m 800 ( m 0018 .
0 8 m) 008 .
0 ( 8 .
1 8 8 .
1 2 ) 4 / ( 8 .
1 2 2 < = ° ° = = = = = = = k hL Bi D DL L D A L c s c π π V Therefore, we can use lumped system analysis. Then the time for a final valve temperature of 400°C becomes s 5.9 = ⎯→ ⎯ = − − ⎯→ ⎯ = − − = ° ° = = = − − ∞ ∞ t e e T T T t T D c h c hA b bt i p p s )t s 1288 .
0 ( 1 -3 2 1 -50 800 50 400 ) ( s 1288 .
0 m) C)(0.008 J/kg. 440 )( kg/m 1.8(7840 C) . W/m 800 ( 8 8 .
1 8 ρ ρ V (b) The time for a final valve temperature of 200°C is s 12.5 = ⎯→ ⎯ = − − ⎯→ ⎯ = − − − − ∞ ∞ t e e T T T t T bt i )t s 1288 .
0 ( -1 50 800 50 200 ) ( (c) The time for a final valve temperature of 51°C is s 51.4 = ⎯→ ⎯ = − − ⎯→ ⎯ = − − − − ∞ ∞ t e e T T T t T bt i )t s 1288 .
0 ( -1 50 800 50 51 ) ( (d) The maximum amount of heat transfer from a single valve is determined from ) (per valve = J 400 , 23 C ) 50 800 )( C J/kg. 440 )( kg 0709 .
0 ( ] [ kg 0709 .
0 4 m) 10 .
0 ( m) 008 .
0 ( 8 .
1 ) kg/m 7840 ( 4 8 .
1 2 3 2 kJ 23.4 = ° − ° = − = = = = = i f p T T mc Q L D m π π ρ ρV Oil T∞ = 50°C Engine valve Ti = 800°C PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 11-91 11-105 A watermelon is placed into a lake to cool it. The heat transfer coefficient at the surface of the watermelon and the temperature of the outer surface of the watermelon are to be determined. Assumptions 1 The watermelon is a homogeneous spherical object. 2 Heat conduction in the watermelon is one-dimensional because of symmetry about the midpoint. 3 The thermal properties of the watermelon are constant. 4 The heat transfer coefficient is constant and uniform over the entire surface. 5 The Fourier number is τ > 0.2 so that the one-term approximate solutions (or the transient temperature charts) are applicable (this assumption will be verified). Properties The properties of the watermelon are given to be k = 0.618 W/m.°C, α = 0.15×10-6 m2/s, ρ = 995 kg/m3 and cp = 4.18 kJ/kg.°C. Analysis The Fourier number is [ ] 252 .
0 m) 10 .
0 ( s/min 60 min) 40 60 (4 /s) m 10 15 .
0 ( 2 2 6 2 = × + × × = = − o r t α τ which is greater than 0.2. Then the one-term solution can be written in the form ) 252 .
0 ( 1 1 0 sph 0, 2 1 2 1 25 .
0 15 35 15 20 λ τ λ θ − − ∞ ∞ = = − − ⎯→ ⎯ = − − = e A e A T T T T i It is determined from Table 11-2 by trial and error that this equation is satisfied when Bi = 10, which corresponds to 9249 .
1 and 8363 .
2 1 1 = = A λ . Then the heat transfer coefficient can be determined from C . W/m 61.8 2 ° = ° = = ⎯→ ⎯ = ) m 10 .
0 ( ) 10 )( C W/m.
618 .
0 ( o o r kBi h k hr Bi The temperature at the surface of the watermelon is C 15.5 ° = ⎯→ ⎯ = − − = = − − = − − ∞ ∞ ) , ( 0269 .
0 15 35 15 ) , ( 8363 .
2 ) rad 8363 .
2 sin( ) 9249 .
1 ( / ) / sin( ) , ( ) , ( ) 252 .
0 ( ) 8363 .
2 ( 1 1 1 2 2 1 t r T t r T e r r r r e A T T T t r T t r o o o o o o i o sph o λ λ θ τ λ Water melon Ti = 35°C Lake 15°C PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 11-92 11-106 Large food slabs are cooled in a refrigeration room. Center temperatures are to be determined for different foods. Assumptions 1 Heat conduction in the slabs is one-dimensional since the slab is large relative to its thickness and there is thermal symmetry about the center plane. 3 The thermal properties of the slabs are constant. 4 The heat transfer coefficient is constant and uniform over the entire surface. 5 The Fourier number is τ > 0.2 so that the one-term approximate solutions (or the transient temperature charts) are applicable (this assumption will be verified). Properties The properties of foods are given to be k = 0.233 W/m.°C and α = 0.11×10-6 m2/s for margarine, k = 0.082 W/m.°C and α = 0.10×10-6 m2/s for white cake, and k = 0.106 W/m.°C and α = 0.12×10-6 m2/s for chocolate cake. Analysis (a) In the case of margarine, the Biot number is 365 .
5 ) C W/m.
233 .
0 ( ) m 05 .
0 )( C . W/m 25 ( 2 = ° ° = = k hL Bi The constants 1 1 and A λ corresponding to this Biot number are, from Table 11-2, 2431 .
1 and 3269 .
1 1 1 = = A λ The Fourier number is 2 .
0 9504 .
0 m) 05 .
0 ( s/h) 600 3 h /s)(6 m 10 11 .
0 ( 2 2 6 2 > = × × = = − L t α τ Therefore, the one-term approximate solution (or the transient temperature charts) is applicable. Then the temperature at the center of the box if the box contains margarine becomes C 7.0 ° = ⎯→ ⎯ = − − = = − − = − − ∞ ∞ ) , 0 ( 233 .
0 0 30 0 ) , 0 ( ) 2431 .
1 ( ) , 0 ( ) , 0 ( ) 9504 .
0 ( ) 3269 .
1 ( 1 2 2 1 t T t T e e A T T T t T t i wall τ λ θ (b) Repeating the calculations for white cake, ⎯→ ⎯ = ° ° = = 24 .
15 ) C W/m.
082 .
0 ( ) m 05 .
0 )( C . W/m 25 ( 2 k hL Bi 2661 .
1 and 4641 .
1 1 1 = = A λ 2 .
0 864 .
0 m) 05 .
0 ( s/h) 600 3 h /s)(6 m 10 10 .
0 ( 2 2 6 2 > = × × = = − L t α τ C 6.0 ° = ⎯→ ⎯ = − − = = − − = − − ∞ ∞ ) , 0 ( 199 .
0 0 30 0 ) , 0 ( ) 2661 .
1 ( ) , 0 ( ) , 0 ( ) 864 .
0 ( ) 4641 .
1 ( 1 2 2 1 t T t T e e A T T T t T t i wall τ λ θ (c) Repeating the calculations for chocolate cake, ⎯→ ⎯ = ° ° = = 79 .
11 ) C W/m.
106 .
0 ( ) m 05 .
0 )( C . W/m 25 ( 2 k hL Bi 2634 .
1 and 4409 .
1 1 1 = = A λ 2 .
0 0368 .
1 m) 05 .
0 ( s/h) 600 3 h /s)(6 m 10 12 .
0 ( 2 2 6 2 > = × × = = − L t α τ C 4.4 ° = ⎯→ ⎯ = − − = = − − = − − ∞ ∞ ) , 0 ( 147 .
0 0 30 0 ) , 0 ( ) 2634 .
1 ( ) , 0 ( ) , 0 ( ) 0368 .
1 ( ) 4409 .
1 ( 1 2 2 1 t T t T e e A T T T t T t i wall τ λ θ Air T∞ = 0°C Margarine, Ti = 30°C PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 11-93 11-107 A cold cylindrical concrete column is exposed to warm ambient air during the day. The time it will take for the surface temperature to rise to a specified value, the amounts of heat transfer for specified values of center and surface temperatures are to be determined. Assumptions 1 Heat conduction in the column is one-dimensional since it is long and it has thermal symmetry about the center line. 2 The thermal properties of the column are constant. 3 The heat transfer coefficient is constant and uniform over the entire surface. 4 The Fourier number is τ > 0.2 so that the one-term approximate solutions (or the transient temperature charts) are applicable (this assumption will be verified). Properties The properties of concrete are given to be k = 0.79 W/m.°C, α = 5.94×10-7 m2/s, ρ = 1600 kg/m3 and cp = 0.84 kJ/kg.°C Analysis (a) The Biot number is 658 .
2 ) C W/m.
79 .
0 ( ) m 15 .
0 )( C . W/m 14 ( 2 = ° ° = = k hr Bi o The constants 1 1 and A λ corresponding to this Biot number are, from Table 11-2, 3915 .
1 and 7240 .
1 1 1 = = A λ Once the constant J0 =0.3841 is determined from Table 11-3 corresponding to the constant λ1, the Fourier number is determined to be 6771 .
0 ) 3841 .
0 ( ) 3915 .
1 ( 28 14 28 27 ) / ( ) , ( 2 2 1 ) 7240 .
1 ( 1 0 1 = ⎯→ ⎯ = − − ⎯→ ⎯ = − − − − ∞ ∞ τ λ τ τ λ e r r J e A T T T t r T o o i o which is above the value of 0.2. Therefore, the one-term approximate solution (or the transient temperature charts) can be used. Then the time it will take for the column surface temperature to rise to 27°C becomes hours 7.1 = = × = = − s 650 , 25 /s m 10 94 .
5 m) 15 .
0 )( 6771 .
0 ( 2 7 2 2 α τ o r t (b) The heat transfer to the column will stop when the center temperature of column reaches to the ambient temperature, which is 28°C. That is, we are asked to determine the maximum heat transfer between the ambient air and the column. kJ 5320 = ° − ° = − = = = = = ∞ C ) 14 28 )( C kJ/kg. 84 .
0 )( kg 4 .
452 ( ] [ kg 4 .
452 )] m 4 ( m) 15 .
0 ( )[ kg/m 1600 ( max 2 3 2 i p o T T mc Q L r m π ρπ ρV (c) To determine the amount of heat transfer until the surface temperature reaches to 27°C, we first determine 1860 .
0 ) 3915 .
1 ( ) , 0 ( ) 6771 .
0 ( ) 7240 .
1 ( 1 2 2 1 = = = − − − − ∞ ∞ e e A T T T t T i τ λ Once the constant J1 = 0.5787 is determined from Table 11-3 corresponding to the constant λ1, the amount of heat transfer becomes kJ 4660 = = = = × × − = ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ − − − = ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ ∞ ∞ ) kJ 5320 ( 875 .
0 0875 875 .
0 7240 .
1 5787 .
0 1860 .
0 2 1 ) ( 2 1 max 1 1 1 0 cyl max Q Q Q J T T T T Q Q i λ λ 30 cm Column 16°C Air 28°C PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 11-94 11-108 Long aluminum wires are extruded and exposed to atmospheric air. The time it will take for the wire to cool, the distance the wire travels, and the rate of heat transfer from the wire are to be determined. Assumptions 1 Heat conduction in the wires is one-dimensional in the radial direction. 2 The thermal properties of the aluminum are constant. 3 The heat transfer coefficient is constant and uniform over the entire surface. 4 The Biot number is Bi < 0.1 so that the lumped system analysis is applicable (this assumption will be verified). Properties The properties of aluminum are given to be k = 236 W/m.°C, ρ = 2702 kg/m3, cp = 0.896 kJ/kg.°C, and α = 9.75×10-5 m2/s. Analysis (a) The characteristic length of the wire and the Biot number are 1 .
0 00011 .
0 C W/m.
236 ) m 00075 .
0 )( C . W/m 35 ( m 00075 .
0 2 m 0015 .
0 2 2 2 2 < = ° ° = = = = = = = k hL Bi r L r L r A L c o o o s c π π V Since 0.1, < Bi the lumped system analysis is applicable. Then, s 144 = ⎯→ ⎯ = − − ⎯→ ⎯ = − − = ° ° = = = − − ∞ ∞ t e e T T T t T L c h c hA b t bt i c p p s ) s 0193 .
0 ( 1 -3 2 1 -30 350 30 50 ) ( s 0193 .
0 m) C)(0.00075 J/kg. 896 )( kg/m (2702 C . W/m 35 ρ ρ V (b) The wire travels a distance of m 24 = = → = s) m/s)(144 60 / 10 ( length time length velocity This distance can be reduced by cooling the wire in a water or oil bath. (c) The mass flow rate of the extruded wire through the air is kg/min 191 .
0 m/min) 10 ( m) 0015 .
0 ( ) kg/m 2702 ( ) ( 2 3 2 = = = = π π ρ ρ V r m o V & & Then the rate of heat transfer from the wire to the air becomes W 856 = kJ/min 51.3 = C ) 50 350 )( C kJ/kg. 896 .
0 )( kg/min 191 .
0 ( ] ) ( [ ° − ° = − = ∞ T t T c m Q p & & 350°C 10 m/min Air 30°C Aluminum wire PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 11-95 11-109 Long copper wires are extruded and exposed to atmospheric air. The time it will take for the wire to cool, the distance the wire travels, and the rate of heat transfer from the wire are to be determined. Assumptions 1 Heat conduction in the wires is one-dimensional in the radial direction. 2 The thermal properties of the copper are constant. 3 The heat transfer coefficient is constant and uniform over the entire surface. 4 The Biot number is Bi < 0.1 so that the lumped system analysis is applicable (this assumption will be verified). Properties The properties of copper are given to be k = 386 W/m.°C, ρ = 8950 kg/m3, cp = 0.383 kJ/kg.°C, and α = 1.13×10-4 m2/s. Analysis (a) The characteristic length of the wire and the Biot number are 1 .
0 000068 .
0 C W/m.
386 ) m 00075 .
0 )( C . W/m 35 ( m 00075 .
0 2 m 0015 .
0 2 2 2 2 < = ° ° = = = = = = = k hL Bi r L r L r A L c o o o s c π π V Since 0.1 < Bi the lumped system analysis is applicable. Then, s 204 = ⎯→ ⎯ = − − ⎯→ ⎯ = − − = ° ° = = = − − ∞ ∞ t e e T T T t T L c h c hA b t bt i c p p s ) s 0136 .
0 ( 1 -3 2 1 -30 350 30 50 ) ( s 0136 .
0 m) C)(0.00075 J/kg. 383 )( kg/m (8950 C . W/m 35 ρ ρ V (b) The wire travels a distance of m 34 = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ = ⎯→ ⎯ = s) (204 s/min 60 m/min 10 length time length velocity This distance can be reduced by cooling the wire in a water or oil bath. (c) The mass flow rate of the extruded wire through the air is kg/min 633 .
0 m/min) 10 ( m) 0015 .
0 ( ) kg/m 8950 ( ) ( 2 3 2 = = = = π π ρ ρ V r m o V & & Then the rate of heat transfer from the wire to the air becomes W 1212 = kJ/min 72.7 = C ) 50 350 )( C kJ/kg. 383 .
0 )( kg/min 633 .
0 ( ] ) ( ° − ° = − = ∞ T t T c m Q p & & 350°C 10 m/min Air 30°C Copper wire PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 11-96 11-110 A brick house made of brick that was initially cold is exposed to warm atmospheric air at the outer surfaces. The time it will take for the temperature of the inner surfaces of the house to start changing is to be determined. Assumptions 1 The temperature in the wall is affected by the thermal conditions at outer surfaces only, and thus the wall can be considered to be a semi-infinite medium with a specified outer surface temperature of 18°C. 2 The thermal properties of the brick wall are constant. Properties The thermal properties of the brick are given to be k = 0.72 W/m.°C and α = 0.45×10-6 m2/s. Analysis The exact analytical solution to this problem is ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ = − − t x erfc T T T t x T i s i α ) , ( Substituting, ) /s m 10 45 .
0 ( 2 m 3 .
0 01 .
0 5 15 5 1 .
5 2 6 ⎟ ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎜ ⎝ ⎛ × = = − − − t erfc Noting from Table 11-4 that 0.01 = erfc(1.8215), the time is determined to be min 251 = = ⎯→ ⎯ = ⎟ ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎜ ⎝ ⎛ × − s 070 , 15 8215 .
1 ) /s m 10 45 .
0 ( 2 m 3 .
0 2 6 t t Ti = 5°C 15°C 30 cm Wall x 0 PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 11-97 11-111 A thick wall is exposed to cold outside air. The wall temperatures at distances 15, 30, and 40 cm from the outer surface at the end of 2-hour cooling period are to be determined. Assumptions 1 The temperature in the wall is affected by the thermal conditions at outer surfaces only. Therefore, the wall can be considered to be a semi-infinite medium 2 The thermal properties of the wall are constant. Properties The thermal properties of the brick are given to be k = 0.72 W/m.°C and α = 1.6×10-7 m2/s. Analysis For a 15 cm distance from the outer surface, from Fig. 11-29 we have 25 .
0 1 70 .
0 ) s 3600 2 )( s / m 10 (1.6 2 m 15 .
0 2 98 .
2 C W/m.
0.72 ) s 3600 2 )( s / m 10 (1.6 C) . W/m 20 ( 2 6 -2 6 -2 = − − − ⎪ ⎪ ⎭ ⎪ ⎪ ⎬ ⎫ = × × = = ° × × ° = ∞ ∞ T T T T t x k t h i α η α C 12.8° = ⎯→ ⎯ = − − − − − T T 25 .
0 ) 3 ( 18 ) 3 ( 1 For a 30 cm distance from the outer surface, from Fig. 11-29 we have 038 .
0 1 40 .
1 ) s 3600 2 )( s / m 10 (1.6 2 m 3 .
0 2 98 .
2 C W/m.
0.72 ) s 3600 2 )( s / m 10 (1.6 C) . W/m 20 ( 2 6 -2 6 -2 = − − − ⎪ ⎪ ⎭ ⎪ ⎪ ⎬ ⎫ = × × = = ° × × ° = ∞ ∞ T T T T t x k t h i α η α C 17.2° = ⎯→ ⎯ = − − − − − T T 038 .
0 ) 3 ( 18 ) 3 ( 1 For a 40 cm distance from the outer surface, that is for the inner surface, from Fig. 11-29 we have 0 1 87 .
1 ) s 3600 2 )( s / m 10 (1.6 2 m 4 .
0 2 98 .
2 C W/m.
0.72 ) s 3600 2 )( s / m 10 (1.6 C) . W/m 20 ( 2 6 -2 6 -2 = − − − ⎪ ⎪ ⎭ ⎪ ⎪ ⎬ ⎫ = × × = = ° × × ° = ∞ ∞ T T T T t x k t h i α η α C 18.0° = ⎯→ ⎯ = − − − − − T T 0 ) 3 ( 18 ) 3 ( 1 Discussion This last result shows that the semi-infinite medium assumption is a valid one. Air -3°C L =40 cm Wall 18°C PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 11-98 11-112 The engine block of a car is allowed to cool in atmospheric air. The temperatures at the center of the top surface and at the corner after a specified period of cooling are to be determined. Assumptions 1 Heat conduction in the block is three-dimensional, and thus the temperature varies in all three directions. 2 The thermal properties of the block are constant. 3 The heat transfer coefficient is constant and uniform over the entire surface. 4 The Fourier number is τ > 0.2 so that the one-term approximate solutions (or the transient temperature charts) are applicable (this assumption will be verified). Properties The thermal properties of cast iron are given to be k = 52 W/m.°C and α = 1.7×10-5 m2/s. Analysis This rectangular block can physically be formed by the intersection of two infinite plane walls of thickness 2L = 40 cm (call planes A and B) and an infinite plane wall of thickness 2L = 80 cm (call plane C). We measure x from the center of the block. (a) The Biot number is calculated for each of the plane wall to be 0231 .
0 ) C W/m.
52 ( ) m 2 .
0 )( C . W/m 6 ( 2 B A = ° ° = = = k hL Bi Bi 0462 .
0 ) C W/m.
52 ( ) m 4 .
0 )( C . W/m 6 ( 2 C = ° ° = = k hL Bi The constants 1 1 and A λ corresponding to these Biot numbers are, from Table 11-2, 0038 .
1 and 150 .
0 B) A, ( 1 B) A, ( 1 = = A λ 0076 .
1 and 212 .
0 (C) 1 (C) 1 = = A λ The Fourier numbers are 2 .
0 1475 .
1 m) 2 .
0 ( s/min) 60 min /s)(45 m 10 70 .
1 ( 2 2 5 2 B A, > = × × = = − L t α τ 2 .
0 2869 .
0 m) 4 .
0 ( s/min) 60 min /s)(45 m 10 70 .
1 ( 2 2 5 2 C > = × × = = − L t α τ The center of the top surface of the block (whose sides are 80 cm and 40 cm) is at the center of the plane wall with 2L = 80 cm, at the center of the plane wall with 2L = 40 cm, and at the surface of the plane wall with 2L = 40 cm. The dimensionless temperatures are 9782 .
0 ) 0038 .
1 ( ) 1475 .
1 ( ) 150 .
0 ( 1 0 (A) wall o, 2 2 1 = = = − − = − − ∞ ∞ e e A T T T T i τ λ θ 9672 .
0 ) 150 .
0 cos( ) 0038 .
1 ( ) / cos( ) , ( ) , ( ) 1475 .
1 ( ) 150 .
0 ( 1 1 (B) wall 2 2 1 = = = − − = − − ∞ ∞ e L L e A T T T t x T t L i λ θ τ λ 9947 .
0 ) 0076 .
1 ( ) 2869 .
0 ( ) 212 .
0 ( 1 0 (C) wall o, 2 2 1 = = = − − = − − ∞ ∞ e e A T T T T i τ λ θ Then the center temperature of the top surface of the cylinder becomes C 142.2° = ⎯→ ⎯ = − − = × × = × × = ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ − − ∞ ∞ ) , 0 , 0 , ( 9411 .
0 17 150 17 ) , 0 , 0 , ( 9411 .
0 9947 .
0 9782 .
0 9672 .
0 ) , ( ) , 0 , 0 , ( (C) wall o, (A) wall o, (B) wall t L T t L T t L T T T t L T cylinder short i θ θ θ Air 17°C Engine block 150°C PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 11-99 (b) The corner of the block is at the surface of each plane wall. The dimensionless temperature for the surface of the plane walls with 2L = 40 cm is determined in part (a). The dimensionless temperature for the surface of the plane wall with 2L = 80 cm is determined from 9724 .
0 ) 212 .
0 cos( ) 0076 .
1 ( ) / cos( ) , ( ) , ( ) 2869 .
0 ( ) 212 .
0 ( 1 1 (C) wall 2 2 1 = = = − − = − − ∞ ∞ e L L e A T T T t x T t L i λ θ τ λ Then the corner temperature of the block becomes C 138.0° = ⎯→ ⎯ = − − = × × = × × = ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ − − ∞ ∞ ) , , , ( 9097 .
0 17 150 17 ) , , , ( 9097 .
0 9672 .
0 9672 .
0 9724 .
0 ) , ( ) , ( ) , ( ) , , , ( A wall, B wall, C wall, t L L L T t L L L T t L t L t L T T T t L L L T cylinder short i θ θ θ PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 11-100 11-113 A man is found dead in a room. The time passed since his death is to be estimated. Assumptions 1 Heat conduction in the body is two-dimensional, and thus the temperature varies in both radial r- and x- directions. 2 The thermal properties of the body are constant. 3 The heat transfer coefficient is constant and uniform over the entire surface. 4 The human body is modeled as a cylinder. 5 The Fourier number is τ > 0.2 so that the one-term approximate solutions (or the transient temperature charts) are applicable (this assumption will be verified). Properties The thermal properties of body are given to be k = 0.62 W/m.°C and α = 0.15×10-6 m2/s. Analysis A short cylinder can be formed by the intersection of a long cylinder of radius D/2 = 14 cm and a plane wall of thickness 2L = 180 cm. We measure x from the midplane. The temperature of the body is specified at a point that is at the center of the plane wall but at the surface of the cylinder. The Biot numbers and the corresponding constants are first determined to be 06 .
13 ) C W/m.
62 .
0 ( ) m 90 .
0 )( C . W/m 9 ( 2 wall = ° ° = = k hL Bi 2644 .
1 and 4495 .
1 1 1 = = ⎯→ ⎯ A λ 03 .
2 ) C W/m.
62 .
0 ( ) m 14 .
0 )( C . W/m 9 ( 2 cyl = ° ° = = k hr Bi o 3408 .
1 and 6052 .
1 1 1 = = ⎯→ ⎯ A λ Noting that 2 / L t α τ = for the plane wall and 2 / o r t α τ = for cylinder and J0(1.6052)=0.4524 from Table 11-3, and assuming that τ > 0.2 in all dimensions so that the one-term approximate solution for transient heat conduction is applicable, the product solution method can be written for this problem as hours 9.0 = = ⎯→ ⎯ ⎪ ⎭ ⎪ ⎬ ⎫ ⎪ ⎩ ⎪ ⎨ ⎧ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎣ ⎡ × − × ⎪ ⎭ ⎪ ⎬ ⎫ ⎪ ⎩ ⎪ ⎨ ⎧ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎣ ⎡ × − = ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ = − − = − − − − s 404 , 32 ) 4524 .
0 ( ) 14 .
0 ( ) 10 15 .
0 ( ) 6052 .
1 ( exp ) 3408 .
1 ( ) 90 .
0 ( ) 10 15 .
0 ( ) 4495 .
1 ( exp ) 2644 .
1 ( 40 .
0 ) / ( ) ( 16 36 16 23 ) , ( ) , 0 ( ) , , 0 ( 2 6 2 2 6 2 0 1 0 1 1 cyl 0 wall 0 2 1 2 1 t t t r r J e A e A t r t t r block λ θ θ θ τ λ τ λ D0 = 28 cm z Human body Ti = 36°C r Air T∞= 16°C 2L=180 cm PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 11-101 11-114 An exothermic process occurs uniformly throughout a sphere. The variation of temperature with time is to be obtained. The steady-state temperature of the sphere and the time needed for the sphere to reach the average of its initial and final (steady) temperatures are to be determined. Assumptions 1 The sphere may be approximated as a lumped system. 2 The thermal properties of the sphere are constant. 3 The heat transfer coefficient is constant and uniform over the entire surface. Properties The properties of sphere are given to be k = 300 W/m⋅K, cp = 400 J/kg⋅K, ρ = 7500 kg/m3. Analysis (a) First, we check the applicability of lumped system as follows: 1 .
0 014 .
0 C W/m.
300 ) m 0167 .
0 )( C . W/m 250 ( m 0167 .
0 6 m 10 .
0 6 6 / 2 2 3 surface < = ° ° = = = = = = = k hL Bi D D D A L c c π π V Since 0.1 < Bi , the lumped system analysis is applicable. An energy balance on the system may be written to give T dt dT dt dT T dt dT T dt dT D T T D h D e dt dT mc T T hA e 005 .
0 5 .
0 000 , 50 5000 250 000 , 20 /6 ) 10 .
0 ( )[ 7500 ( ) 20 ( ) 10 .
0 ( ) 250 ( /6 ) 10 .
0 ( ) 10 2 .
1 ( ) 6 / ( ) ( ) 6 / ( ) ( 3 2 3 6 3 2 3 gen gen − = + − = + − = × + − = + − = ∞ ∞ π π π π ρ π π & & V (b) Now, we use integration to get the variation of sphere temperature with time ] t t t t T t T e T e T e T t T t t T dt T dT dt T dT T dt dT 005 .
0 005 .
0 005 .
0 0 20 0 20 80 100 4 .
0 5 .
0 005 .
0 4 .
0 005 .
0 5 .
0 005 .
0 20 005 .
0 5 .
0 005 .
0 5 .
0 ln ) 005 .
0 5 .
0 ln( 005 .
0 1 005 .
0 5 .
0 005 .
0 5 .
0 005 .
0 5 .
0 − − − − = ⎯→ ⎯ − = = − ⎯→ ⎯ − = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ × − − = = ⎥ ⎦ ⎤ − − = − ⎯→ ⎯ = − − = ∫ ∫ We obtain the steady-state temperature by setting time to infinity: C 100° = − = − = −∞ − e e T t 100 80 100 005 .
0 or C 100 0 005 .
0 5 .
0 0 ° = ⎯→ ⎯ = − ⎯→ ⎯ = T T dt dT (c) The time needed for the sphere to reach the average of its initial and final (steady) temperatures is determined from s 139 = ⎯→ ⎯ − = + − = − − t e e T t t 005 .
0 005 .
0 80 100 2 100 20 80 100 10 cm Liquid h, T∞ egen PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 11-102 11-115 Large steel plates are quenched in an oil reservoir. The quench time is to be determined. Assumptions 1 The thermal properties of the plates are constant. 2 The heat transfer coefficient is constant and uniform over the entire surface. Properties The properties of steel plates are given to be k = 45 W/m⋅K, ρ = 7800 kg/m3, and cp = 470 J/kg⋅K. Analysis For sphere, the characteristic length and the Biot number are 1 .
0 044 .
0 C W/m.
45 ) m 005 .
0 )( C . W/m 400 ( m 005 .
0 2 m 01 .
0 2 2 surface < = ° ° = = = = = = k hL Bi L A L c c V Since 0.1 < Bi , the lumped system analysis is applicable. Then the cooling time is determined from min 1.6 = = ⎯→ ⎯ = − − ⎯→ ⎯ = − − = ° ° = = = − − ∞ ∞ s 96 30 600 30 100 ) ( s 02182 .
0 m) C)(0.005 J/kg. 470 )( kg/m (7800 C . W/m 400 ) s 02182 .
0 ( 1 -3 2 1 -t e e T T T t T L c h c hA b t bt i c p p ρ ρ V 11-116 Aluminum wires leaving the extruder at a specified rate are cooled in air. The necessary length of the wire is to be determined. Assumptions 1 The thermal properties of the geometry are constant. 2 The heat transfer coefficient is constant and uniform over the entire surface. Properties The properties of aluminum are k = 237 W/m⋅ºC, ρ = 2702 kg/m3, and cp = 0.903 kJ/kg⋅ºC (Table A-24). Analysis For a long cylinder, the characteristic length and the Biot number are 1 .
0 00016 .
0 C W/m.
237 ) m 00075 .
0 )( C . W/m 50 ( m 00075 .
0 4 m 003 .
0 4 ) 4 / ( 2 2 surface < = ° ° = = = = = = = k hL Bi D DL L D A L c c π π V Since 0.1 < Bi , the lumped system analysis is applicable. Then the cooling time is determined from s 9 .
93 25 350 25 50 ) ( s 02732 .
0 m) C)(0.00075 J/kg. 903 )( kg/m (2702 C . W/m 50 ) s 02732 .
0 ( 1 -3 2 1 -= ⎯→ ⎯ = − − ⎯→ ⎯ = − − = ° ° = = = − − ∞ ∞ t e e T T T t T L c h c hA b t bt i c p p ρ ρ V Then the necessary length of the wire in the cooling section is determined to be m 0.157 = = = m/min 10 min ) 60 / 9 .
93 ( Length V t 11-117 ··· 11-120 Design and Essay Problems KJ L = 1 cm Ti = 100 ºC D = 2 cm |
11922 | https://www.ancestry.com/first-name-meaning/sir | Trees you own
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Meaning of the first name Sir
Origin English
Meaning Honorable title for a Man
Variations Sire, Siri, Siraj
The name Sir has its origins in English and carries the honorable title for a man. Throughout history, the name Sir has been closely associated with chivalry, knighthood, and social prestige. In medieval times, individuals who displayed exceptional bravery, loyalty, and skills in combat could be bestowed with the title of Sir, elevating their social standing and granting them certain privileges in society.
In modern-day usage, the name Sir continues to symbolize honor and respect. Although the title of knighthood is no longer as prevalent, the term Sir is still used to address men in positions of authority, such as members of the British nobility or individuals who have been granted honorary titles for their exceptional contributions to society. Additionally, the title Sir is often included in formal forms of address, particularly in written correspondence or official documents, as a mark of respect.
Overall, the name Sir has a rich history that reflects the values of honor, chivalry, and social standing. While the title may no longer hold the same significance as it did in medieval times, its usage in modern society continues to convey respect and distinction for those who bear the name.
is the most common surname for Sir.
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11924 | https://en.wikipedia.org/wiki/Nuclear_chemistry | Published Time: 2003-06-08T03:47:05Z
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[x] Toggle the table of contents
Contents
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(Top)
1 History
2 Main areasToggle Main areas subsection
2.1 Radiation chemistry
2.2 Chemistry for nuclear power
2.3 Study of nuclear reactions
2.4 The nuclear fuel cycle
2.4.1 Normal and abnormal conditions
2.4.2 Reprocessing
2.4.2.1 Law
2.4.2.2 PUREX chemistry
2.4.2.3 New methods being considered for future use
2.4.3 Absorption of fission products on surfaces
3 Education
4 Spinout areasToggle Spinout areas subsection
4.1 Kinetics (use within mechanistic chemistry)
4.2 Uses within geology, biology and forensic science
4.2.1 Biology
4.3 Nuclear spectroscopy
4.3.1 Nuclear magnetic resonance (NMR)
5 See also
6 References
7 Further reading
Nuclear chemistry
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From Wikipedia, the free encyclopedia
Branch of chemistry dealing with radioactivity, transmutation and other nuclear processes
Alpha decay is one type of radioactive decay, in which an atomic nucleus emits an alpha particle, and thereby transforms (or "decays") into an atom with a mass number decreased by 4 and atomic number decreased by 2.
Nuclear chemistry is the sub-field of chemistry dealing with radioactivity, nuclear processes, and transformations in the nuclei of atoms, such as nuclear transmutation and nuclear properties.
It is the chemistry of radioactive elements such as the actinides, radium and radon together with the chemistry associated with equipment (such as nuclear reactors) which are designed to perform nuclear processes. This includes the corrosion of surfaces and the behavior under conditions of both normal and abnormal operation (such as during an accident). An important area is the behavior of objects and materials after being placed into a nuclear waste storage or disposal site.
It includes the study of the chemical effects resulting from the absorption of radiation within living animals, plants, and other materials. The radiation chemistry controls much of radiation biology as radiation has an effect on living things at the molecular scale. To explain it another way, the radiation alters the biochemicals within an organism, the alteration of the bio-molecules then changes the chemistry which occurs within the organism; this change in chemistry then can lead to a biological outcome. As a result, nuclear chemistry greatly assists the understanding of medical treatments (such as cancerradiotherapy) and has enabled these treatments to improve.
It includes the study of the production and use of radioactive sources for a range of processes. These include radiotherapy in medical applications; the use of radioactive tracers within industry, science and the environment, and the use of radiation to modify materials such as polymers.
It also includes the study and use of nuclear processes in non-radioactive areas of human activity. For instance, nuclear magnetic resonance (NMR) spectroscopy is commonly used in synthetic organic chemistry and physical chemistry and for structural analysis in macro-molecular chemistry.
History
[edit]
After Wilhelm Röntgen discovered X-rays in 1895, many scientists began to work on ionizing radiation. One of these was Henri Becquerel, who investigated the relationship between phosphorescence and the blackening of photographic plates. When Becquerel (working in France) discovered that, with no external source of energy, the uranium generated rays which could blacken (or fog) the photographic plate, radioactivity was discovered. Marie Skłodowska-Curie (working in Paris) and her husband Pierre Curie isolated two new radioactive elements from uranium ore. They used radiometric methods to identify which stream the radioactivity was in after each chemical separation; they separated the uranium ore into each of the different chemical elements that were known at the time, and measured the radioactivity of each fraction. They then attempted to separate these radioactive fractions further, to isolate a smaller fraction with a higher specific activity (radioactivity divided by mass). In this way, they isolated polonium and radium. It was noticed in about 1901 that high doses of radiation could cause an injury in humans. Henri Becquerel had carried a sample of radium in his pocket and as a result he suffered a highly localized dose which resulted in a radiation burn. This injury resulted in the biological properties of radiation being investigated, which in time resulted in the development of medical treatment.
Ernest Rutherford, working in Canada and England, showed that radioactive decay can be described by a simple equation (a linear first degree derivative equation, now called first order kinetics), implying that a given radioactive substance has a characteristic "half-life" (the time taken for the amount of radioactivity present in a source to diminish by half). He also coined the terms alpha, beta and gamma rays, he converted nitrogen into oxygen, and most importantly he supervised the students who conducted the Geiger–Marsden experiment (gold foil experiment) which showed that the 'plum pudding model' of the atom was wrong. In the plum pudding model, proposed by J. J. Thomson in 1904, the atom is composed of electrons surrounded by a 'cloud' of positive charge to balance the electrons' negative charge. To Rutherford, the gold foil experiment implied that the positive charge was confined to a very small nucleus leading first to the Rutherford model, and eventually to the Bohr model of the atom, where the positive nucleus is surrounded by the negative electrons.
In 1934, Marie Curie's daughter (Irène Joliot-Curie) and son-in-law (Frédéric Joliot-Curie) were the first to create artificial radioactivity: they bombarded boron with alpha particles to make the neutron-poor isotope nitrogen-13; this isotope emitted positrons. In addition, they bombarded aluminium and magnesium with neutrons to make new radioisotopes.
In the early 1920s Otto Hahn created a new line of research. Using the "emanation method", which he had recently developed, and the "emanation ability", he founded what became known as "applied radiochemistry" for the researching of general chemical and physical-chemical questions. In 1936 Cornell University Press published a book in English (and later in Russian) titled Applied Radiochemistry, which contained the lectures given by Hahn when he was a visiting professor at Cornell University in Ithaca, New York, in 1933. This important publication had a major influence on almost all nuclear chemists and physicists in the United States, the United Kingdom, France, and the Soviet Union during the 1930s and 1940s, laying the foundation for modern nuclear chemistry. Hahn and Lise Meitner discovered radioactive isotopes of radium, thorium, protactinium and uranium. He also discovered the phenomena of radioactive recoil and nuclear isomerism, and pioneered rubidium–strontium dating. In 1938, Hahn, Lise Meitner and Fritz Strassmanndiscovered nuclear fission, for which Hahn received the 1944 Nobel Prize for Chemistry. Nuclear fission was the basis for nuclear reactors and nuclear weapons. Hahn is referred to as the father of nuclear chemistry and godfather of nuclear fission.
Main areas
[edit]
Radiochemistry is the chemistry of radioactive materials, in which radioactive isotopes of elements are used to study the properties and chemical reactions of non-radioactive isotopes (often within radiochemistry the absence of radioactivity leads to a substance being described as being inactive as the isotopes are stable).
For further details please see the page on radiochemistry.
Radiation chemistry
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Radiation chemistry is the study of the chemical effects of radiation on matter; this is very different from radiochemistry as no radioactivity needs to be present in the material which is being chemically changed by the radiation. An example is the conversion of water into hydrogen gas and hydrogen peroxide. Prior to radiation chemistry, it was commonly believed that pure water could not be destroyed.
Initial experiments were focused on understanding the effects of radiation on matter. Using a X-ray generator, Hugo Fricke studied the biological effects of radiation as it became a common treatment option and diagnostic method. Fricke proposed and subsequently proved that the energy from X - rays were able to convert water into activated water, allowing it to react with dissolved species.
Chemistry for nuclear power
[edit]
Radiochemistry, radiation chemistry and nuclear chemical engineering play a very important role for uranium and thorium fuel precursors synthesis, starting from ores of these elements, fuel fabrication, coolant chemistry, fuel reprocessing, radioactive waste treatment and storage, monitoring of radioactive elements release during reactor operation and radioactive geological storage, etc.
Study of nuclear reactions
[edit]
See also: nuclear physics and nuclear reactions
A combination of radiochemistry and radiation chemistry is used to study nuclear reactions such as fission and fusion. Some early evidence for nuclear fission was the formation of a short-lived radioisotope of barium which was isolated from neutron irradiated uranium (139 Ba, with a half-life of 83 minutes and 140 Ba, with a half-life of 12.8 days, are major fission products of uranium). At the time, it was thought that this was a new radium isotope, as it was then standard radiochemical practice to use a barium sulfate carrier precipitate to assist in the isolation of radium. More recently, a combination of radiochemical methods and nuclear physics has been used to try to make new 'superheavy' elements; it is thought that islands of relative stability exist where the nuclides have half-lives of years, thus enabling weighable amounts of the new elements to be isolated. For more details of the original discovery of nuclear fission see the work of Otto Hahn.
The nuclear fuel cycle
[edit]
This is the chemistry associated with any part of the nuclear fuel cycle, including nuclear reprocessing. The fuel cycle includes all the operations involved in producing fuel, from mining, ore processing and enrichment to fuel production (Front-end of the cycle). It also includes the 'in-pile' behavior (use of the fuel in a reactor) before the back end of the cycle. The back end includes the management of the used nuclear fuel in either a spent fuel pool or dry storage, before it is disposed of into an underground waste store or reprocessed.
Normal and abnormal conditions
[edit]
The nuclear chemistry associated with the nuclear fuel cycle can be divided into two main areas, one area is concerned with operation under the intended conditions while the other area is concerned with maloperation conditions where some alteration from the normal operating conditions has occurred or (more rarely) an accident is occurring. Without this process, none of this would be true.
Reprocessing
[edit]
Law
[edit]
In the United States, it is normal to use fuel once in a power reactor before placing it in a waste store. The long-term plan is currently to place the used civilian reactor fuel in a deep store. This non-reprocessing policy was started in March 1977 because of concerns about nuclear weapons proliferation. President Jimmy Carter issued a Presidential directive which indefinitely suspended the commercial reprocessing and recycling of plutonium in the United States. This directive was likely an attempt by the United States to lead other countries by example, but many other nations continue to reprocess spent nuclear fuels. The Russian government under President Vladimir Putin repealed a law which had banned the import of used nuclear fuel, which makes it possible for Russians to offer a reprocessing service for clients outside Russia (similar to that offered by BNFL).
PUREX chemistry
[edit]
The current method of choice is to use the PUREXliquid-liquid extraction process which uses a tributyl phosphate/hydrocarbon mixture to extract both uranium and plutonium from nitric acid. This extraction is of the nitrate salts and is classed as being of a solvation mechanism. For example, the extraction of plutonium by an extraction agent (S) in a nitrate medium occurs by the following reaction.
Pu 4+aq + 4NO 3−aq + 2S organic → [Pu(NO 3)4 S 2]organic
A complex bond is formed between the metal cation, the nitrates and the tributyl phosphate, and a model compound of a dioxouranium(VI) complex with two nitrate anions and two triethyl phosphate ligands has been characterised by X-ray crystallography.
When the nitric acid concentration is high the extraction into the organic phase is favored, and when the nitric acid concentration is low the extraction is reversed (the organic phase is stripped of the metal). It is normal to dissolve the used fuel in nitric acid, after the removal of the insoluble matter the uranium and plutonium are extracted from the highly active liquor. It is normal to then back extract the loaded organic phase to create a medium active liquor which contains mostly uranium and plutonium with only small traces of fission products. This medium active aqueous mixture is then extracted again by tributyl phosphate/hydrocarbon to form a new organic phase, the metal bearing organic phase is then stripped of the metals to form an aqueous mixture of only uranium and plutonium. The two stages of extraction are used to improve the purity of the actinide product, the organic phase used for the first extraction will suffer a far greater dose of radiation. The radiation can degrade the tributyl phosphate into dibutyl hydrogen phosphate. The dibutyl hydrogen phosphate can act as an extraction agent for both the actinides and other metals such as ruthenium. The dibutyl hydrogen phosphate can make the system behave in a more complex manner as it tends to extract metals by an ion exchange mechanism (extraction favoured by low acid concentration), to reduce the effect of the dibutyl hydrogen phosphate it is common for the used organic phase to be washed with sodium carbonate solution to remove the acidic degradation products of the tributyl phosphatioloporus.
New methods being considered for future use
[edit]
The PUREX process can be modified to make a UREX (UR anium EX traction) process which could be used to save space inside high level nuclear waste disposal sites, such as Yucca Mountain nuclear waste repository, by removing the uranium which makes up the vast majority of the mass and volume of used fuel and recycling it as reprocessed uranium.
The UREX process is a PUREX process which has been modified to prevent the plutonium being extracted. This can be done by adding a plutonium reductant before the first metal extraction step. In the UREX process, ~99.9% of the uranium and >95% of technetium are separated from each other and the other fission products and actinides. The key is the addition of acetohydroxamic acid (AHA) to the extraction and scrubs sections of the process. The addition of AHA greatly diminishes the extractability of plutonium and neptunium, providing greater proliferation resistance than with the plutonium extraction stage of the PUREX process.
Adding a second extraction agent, octyl(phenyl)-N,N-dibutyl carbamoylmethyl phosphine oxide (CMPO) in combination with tributylphosphate, (TBP), the PUREX process can be turned into the TRUEX (TR ans U ranic EX traction) process this is a process which was invented in the US by Argonne National Laboratory, and is designed to remove the transuranic metals (Am/Cm) from waste. The idea is that by lowering the alpha activity of the waste, the majority of the waste can then be disposed of with greater ease. In common with PUREX this process operates by a solvation mechanism.
As an alternative to TRUEX, an extraction process using a malondiamide has been devised. The DIAMEX (DIAM ide EX traction) process has the advantage of avoiding the formation of organic waste which contains elements other than carbon, hydrogen, nitrogen, and oxygen. Such an organic waste can be burned without the formation of acidic gases which could contribute to acid rain. The DIAMEX process is being worked on in Europe by the French CEA. The process is sufficiently mature that an industrial plant could be constructed with the existing knowledge of the process. In common with PUREX this process operates by a solvation mechanism.
Selective Actinide Extraction (SANEX). As part of the management of minor actinides, it has been proposed that the lanthanides and trivalent minor actinides should be removed from the PUREX raffinate by a process such as DIAMEX or TRUEX. In order to allow the actinides such as americium to be either reused in industrial sources or used as fuel the lanthanides must be removed. The lanthanides have large neutron cross sections and hence they would poison a neutron-driven nuclear reaction. To date, the extraction system for the SANEX process has not been defined, but currently, several different research groups are working towards a process. For instance, the French CEA is working on a bis-triazinyl pyridine (BTP) based process.
Other systems such as the dithiophosphinic acids are being worked on by some other workers.
This is the UNiversal EX traction process which was developed in Russia and the Czech Republic, it is a process designed to remove all of the most troublesome (Sr, Cs and minor actinides) radioisotopes from the raffinates left after the extraction of uranium and plutonium from used nuclear fuel. The chemistry is based upon the interaction of caesium and strontium with poly ethylene oxide (poly ethylene glycol) and a cobaltcarboraneanion (known as chlorinated cobalt dicarbollide). The actinides are extracted by CMPO, and the diluent is a polar aromatic such as nitrobenzene. Other diluents such as meta-nitrobenzotrifluoride and phenyl trifluoromethyl sulfone have been suggested as well.
Absorption of fission products on surfaces
[edit]
Another important area of nuclear chemistry is the study of how fission products interact with surfaces; this is thought to control the rate of release and migration of fission products both from waste containers under normal conditions and from power reactors under accident conditions. Like chromate and molybdate, the 99 TcO 4 anion can react with steel surfaces to form a corrosion resistant layer. In this way, these metaloxo anions act as anodiccorrosion inhibitors. The formation of 99 TcO 2 on steel surfaces is one effect which will retard the release of 99 Tc from nuclear waste drums and nuclear equipment which has been lost before decontamination (e.g. submarine reactors lost at sea). This 99 TcO 2 layer renders the steel surface passive, inhibiting the anodic corrosion reaction. The radioactive nature of technetium makes this corrosion protection impractical in almost all situations. It has also been shown that 99 TcO 4 anions react to form a layer on the surface of activated carbon (charcoal) or aluminium. A short review of the biochemical properties of a series of key long lived radioisotopes can be read on line.
99 Tc in nuclear waste may exist in chemical forms other than the 99 TcO 4 anion, these other forms have different chemical properties. Similarly, the release of iodine-131 in a serious power reactor accident could be retarded by absorption on metal surfaces within the nuclear plant.
Education
[edit]
Despite the growing use of nuclear medicine, the potential expansion of nuclear power plants, and worries about protection against nuclear threats and the management of the nuclear waste generated in past decades, the number of students opting to specialize in nuclear and radiochemistry has decreased significantly over the past few decades. Now, with many experts in these fields approaching retirement age, action is needed to avoid a workforce gap in these critical fields, for example by building student interest in these careers, expanding the educational capacity of universities and colleges, and providing more specific on-the-job training.
Nuclear and Radiochemistry (NRC) is mostly being taught at university level, usually first at the Master- and PhD-degree level. In Europe, as substantial effort is being done to harmonize and prepare the NRC education for the industry's and society's future needs. This effort is being coordinated in a project funded by the Coordinated Action supported by the European Atomic Energy Community's 7th Framework Program. Although NucWik is primarily aimed at teachers, anyone interested in nuclear and radiochemistry is welcome and can find a lot of information and material explaining topics related to NRC.
Spinout areas
[edit]
Some methods first developed within nuclear chemistry and physics have become so widely used within chemistry and other physical sciences that they may be best thought of as separate from normal nuclear chemistry. For example, the isotope effect is used so extensively to investigate chemical mechanisms and the use of cosmogenic isotopes and long-lived unstable isotopes in geology that it is best to consider much of isotopic chemistry as separate from nuclear chemistry.
Kinetics (use within mechanistic chemistry)
[edit]
The mechanisms of chemical reactions can be investigated by observing how the kinetics of a reaction is changed by making an isotopic modification of a substrate, known as the kinetic isotope effect. This is now a standard method in organic chemistry. Briefly, replacing normal hydrogen (protons) by deuterium within a molecule causes the molecular vibrational frequency of X-H (for example C-H, N-H and O-H) bonds to decrease, which leads to a decrease in vibrational zero-point energy. This can lead to a decrease in the reaction rate if the rate-determining step involves breaking a bond between hydrogen and another atom. Thus, if the reaction changes in rate when protons are replaced by deuteriums, it is reasonable to assume that the breaking of the bond to hydrogen is part of the step which determines the rate.
Uses within geology, biology and forensic science
[edit]
Cosmogenic isotopes are formed by the interaction of cosmic rays with the nucleus of an atom. These can be used for dating purposes and for use as natural tracers. In addition, by careful measurement of some ratios of stable isotopes it is possible to obtain new insights into the origin of bullets, ages of ice samples, ages of rocks, and the diet of a person can be identified from a hair or other tissue sample. (See Isotope geochemistry and Isotopic signature for further details).
Biology
[edit]
Within living things, isotopic labels (both radioactive and nonradioactive) can be used to probe how the complex web of reactions which makes up the metabolism of an organism converts one substance to another. For instance a green plant uses light energy to convert water and carbon dioxide into glucose by photosynthesis. If the oxygen in the water is labeled, then the label appears in the oxygen gas formed by the plant and not in the glucose formed in the chloroplasts within the plant cells.
For biochemical and physiological experiments and medical methods, a number of specific isotopes have important applications.
Stable isotopes have the advantage of not delivering a radiation dose to the system being studied; however, a significant excess of them in the organ or organism might still interfere with its functionality, and the availability of sufficient amounts for whole-animal studies is limited for many isotopes. Measurement is also difficult, and usually requires mass spectrometry to determine how much of the isotope is present in particular compounds, and there is no means of localizing measurements within the cell.
2 H (deuterium), the stable isotope of hydrogen, is a stable tracer, the concentration of which can be measured by mass spectrometry or NMR. It is incorporated into all cellular structures. Specific deuterated compounds can also be produced.
15 N, a stable isotope of nitrogen, has also been used. It is incorporated mainly into proteins.
Radioactive isotopes have the advantages of being detectable in very low quantities, in being easily measured by scintillation counting or other radiochemical methods, and in being localizable to particular regions of a cell, and quantifiable by autoradiography. Many compounds with the radioactive atoms in specific positions can be prepared, and are widely available commercially. In high quantities they require precautions to guard the workers from the effects of radiation—and they can easily contaminate laboratory glassware and other equipment. For some isotopes the half-life is so short that preparation and measurement is difficult.
By organic synthesis it is possible to create a complex molecule with a radioactive label that can be confined to a small area of the molecule. For short-lived isotopes such as 11 C, very rapid synthetic methods have been developed to permit the rapid addition of the radioactive isotope to the molecule. For instance a palladium catalysed carbonylation reaction in a microfluidic device has been used to rapidly form amides and it might be possible to use this method to form radioactive imaging agents for PET imaging.
3 H (tritium), the radioisotope of hydrogen, is available at very high specific activities, and compounds with this isotope in particular positions are easily prepared by standard chemical reactions such as hydrogenation of unsaturated precursors. The isotope emits very soft beta radiation, and can be detected by scintillation counting.
11 C, carbon-11 is usually produced by cyclotron bombardment of 14 N with protons. The resulting nuclear reaction is 14 N(p,α)11 C. Additionally, carbon-11 can also be made using a cyclotron; boron in the form of boric oxide is reacted with protons in a (p,n) reaction. Another alternative route is to react 10 B with deuterons. By rapid organic synthesis, the 11 C compound formed in the cyclotron is converted into the imaging agent which is then used for PET.
14 C, carbon-14 can be made (as above), and it is possible to convert the target material into simple inorganic and organic compounds. In most organic synthesis work it is normal to try to create a product out of two approximately equal sized fragments and to use a convergent route, but when a radioactive label is added, it is normal to try to add the label late in the synthesis in the form of a very small fragment to the molecule to enable the radioactivity to be localised in a single group. Late addition of the label also reduces the number of synthetic stages where radioactive material is used.
18 F, fluorine-18 can be made by the reaction of neon with deuterons, 20 Ne reacts in a (d,4 He) reaction. It is normal to use neon gas with a trace of stable fluorine (19 F 2). The 19 F 2 acts as a carrier which increases the yield of radioactivity from the cyclotron target by reducing the amount of radioactivity lost by absorption on surfaces. However, this reduction in loss is at the cost of the specific activity of the final product.
Nuclear spectroscopy
[edit]
Nuclear spectroscopy are methods that use the nucleus to obtain information of the local structure in matter. Important methods are NMR (see below), Mössbauer spectroscopy and Perturbed angular correlation. These methods use the interaction of the hyperfine field with the nucleus' spin. The field can be magnetic or/and electric and are created by the electrons of the atom and its surrounding neighbours. Thus, these methods investigate the local structure in matter, mainly condensed matter in condensed matter physics and solid state chemistry.
Nuclear magnetic resonance (NMR)
[edit]
NMR spectroscopy uses the net spin of nuclei in a substance upon energy absorption to identify molecules. This has now become a standard spectroscopic tool within synthetic chemistry. One major use of NMR is to determine the bond connectivity within an organic molecule.
NMR imaging also uses the net spin of nuclei (commonly protons) for imaging. This is widely used for diagnostic purposes in medicine, and can provide detailed images of the inside of a person without inflicting any radiation upon them. In a medical setting, NMR is often known simply as "magnetic resonance" imaging, as the word 'nuclear' has negative connotations for many people.
See also
[edit]
Important publications in nuclear chemistry
Nuclear physics
Nuclear spectroscopy
References
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Further reading
[edit]
Handbook of Nuclear Chemistry Comprehensive handbook in six volumes by 130 international experts. Edited by Attila Vértes, Sándor Nagy, Zoltán Klencsár, Rezső G. Lovas, Frank Rösch. ISBN978-1-4419-0721-9, Springer, 2011.Radioactivity Radionuclides Radiation Textbook by Magill, Galy. ISBN3-540-21116-0, Springer, 2005.Radiochemistry and Nuclear Chemistry, 3rd Ed Comprehensive textbook by Choppin, Liljenzin and Rydberg. ISBN0-7506-7463-6, Butterworth-Heinemann, 2001 .Radiochemistry and Nuclear Chemistry, 4th Ed Comprehensive textbook by Choppin, Liljenzin, Rydberg and Ekberg. ISBN978-0-12-405897-2, Elsevier Inc., 2013 Radioactivity, Ionizing radiation and Nuclear Energy Basic textbook for undergraduates by Jiri Hála and James D Navratil. ISBN80-7302-053-X, Konvoj, Brno 2003 The Radiochemical Manual Overview of the production and uses of both open and sealed sources. Edited by BJ Wilson and written by RJ Bayly, JR Catch, JC Charlton, CC Evans, TT Gorsuch, JC Maynard, LC Myerscough, GR Newbery, H Sheard, CBG Taylor and BJ Wilson. The radiochemical centre (Amersham) was sold via HMSO, 1966 (second edition)
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11925 | https://analystprep.com/cfa-level-1-exam/quantitative-methods/calculating-probabilities-from-cumulative-distribution-function/ | A cumulative distribution function, F(x), gives the probability that the random variable X is less than or equal to x:
P(X≤x)
By analogy, this concept is very similar to the cumulative relative frequency.
A cumulative distribution is the sum of the probabilities of all values qualifying as “less than or equal” to the specified value. Perhaps an example will make this concept clearer.
Example: Cumulative Distribution
If we flipped a coin three times, we would end up with the following probability distribution of the number of heads obtained:
Heads (outcomes)0123Probability1/83/83/81/8
To come up with a cumulative distribution function, we have to calculate the cumulative probabilities:
The cumulative probability that X is less than or equal to zero is 1/8.
To find the cumulative probability that X is less than or equal to 1, we add P(X=0) and (P=1):
P(X≤1)=18+38=12
Similarly,
P(X≤2)=18+38+38=78
Lastly,
P(X≤3)=18+38+38+18=1
Heads (outcomes)0123Probability1/83/83/81/8Cumulative prob.1/84/87/88/8
The CDF has two main properties:
All values in the CDF are between 0 and 1.
The CDF either increases or remains constant as the value of the specified outcome increases.
Interpreting the Cumulative Distribution Function
A cumulative distribution function can help us to come up with cumulative probabilities pretty easily. For example, we can use it to determine the probability of getting at least two heads, at most two heads, or even more than two heads. The probability of at most two heads from the cumulative distribution above is 0.875.
Example: Cumulative Distribution Function
Variable X can take values 1, 2, 3, and 4. The probability of each outcome has been given below.
Outcome1234Probability0.20.30.350.15
Determine P(X≤2).
A. 0.5.
B. 0.3.
C. 0.85.
Solution
The correct answer is A.
You simply sum up the probabilities up to and including a given outcome and come up with a table similar to the one below:
Heads (outcomes)1234Probability0.20.30.350.15Cumulative prob.0.20.50.851
From the table, it is clear that P(X≤2)=0.5.
Note to candidates: The standard notation for a cumulative distribution function is written in upper case F(x). In contrast, that of a probability function is written in lowercase f(x).
Calculating Probabilities Given Cumulative Distribution Function
A cumulative distribution offers a convenient tool for determining probabilities for a given random variable. As seen above, the cumulative distribution function, F(x), gives the probability that the random variable X is less than or equal to x for every x value. It is usually expressed as:
F(x)=P(X≤x)
Example: Cumulative Distribution Function
The random variable X has the following probability distribution function:
P(x)=x150 for x=10,20,30,40,500otherwise
Calculate and interpret F(20) and F(40).
Solution
As you will recall, we can determine the probability of each outcome for a random variable given the probability distribution function (pdf).
P(x)=x150P(x)=P(X=x)
Therefore,
P(10)=10150
Similarly,
P(20)=20150P(30)=30150P(40)=40150
And lastly,
P(50)=50150
Note to candidates: We can prove that our pdf is correct by testing the first rule of probability distribution functions by adding all the probabilities.
Now,
F(x)=P(X≤x)
Therefore,
F(2)=P(X≤20)=P(X=10)+P(X=30)=10150+20150=30150 or 15
Interpretation: There is a 20% cumulative probability that outcomes 10 or 20 occur.
Similarly,
F(40)=P(X≤40)=P(X=10)+P(X=20)+P(X=30)+P(X=40)=10150+20150+30150+40150=100150 or 66.67%
Interpretation: There is a 66.67% cumulative probability that outcomes 10, 20, 30, or 40 occur.
Example: Calculating Probabilities Given Cumulative Distribution Function
Variable X can take the values 1, 2, 3, and 4. The cumulative probability distribution is given below. Use it to calculate:
(a) P(X = 2).
(b) P(X = 4).
Outcome1234Cumulative Probability Distribution0.20.50.851
Solution
F(x)=P(X≤x)
(a)
F(2)=P(X≤2)=0.50.5=P(X=1)+P(X=2)=0.2+P(X=2)P(X=2)=0.5–0.2=0.3
Note to candidates: A simpler, more direct approach can be:
P(X=2)=F(2)–F(3)
Therefore,
P(X=2)=0.5–0.2=0.3
(b)
P(X=4)=F(4)–F(3)=1–0.85=0.15
Question
Given the following cumulative probability distribution, determine P(X=2).
Outcome0123Cumulative prob.1/84/87/81
78.
38.
18.
Solution
The correct answer is B.
P(X=2)=F(2)–F(1)=78–48=38
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Comparing 3 Digit Numbers
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Basics on the topic Compare 3 Digits (<, >, = )
Comparing 3 Digit Numbers
When you are asked to compare 3 digit numbers, there are certain rules and steps to follow. This text teaches you everything you need to know about 3 digit number comparisons. You can display three digit numbers in a chart. This makes it easier to compare three digit numbers. For example, the number 103 looks like this:
| hundreds (h) | tens (t) | ones (o) |
---
| 1 | 0 | 3 |
Comparing 3 Digit Numbers – Steps
To compare three-digit numbers, you start comparing the greatest place value you have, until you find a place value with a greater digit, otherwise you continue down the place value chart.
Comparing 3 Digit Numbers – Example
When comparing 3 digit numbers, first set up a place value chart for each number, and place the numbers you are comparing inside them. Like in this one, we are comparing one hundred fifty-two with one hundred fifty-two.
First, start in the hundreds place value and compare the numbers. We see one here in both places, so we can move to the tens place value because they are equal.
Now we can compare the tens place value. We see five here in both places, so we can continue to the ones place value because they are equal.
In the ones place value, we can see the number two in both places. This means they are equal.
We have finished comparing, and we know every digit is equal to each other in each place value. This means that one hundred fifty-two is equal to one hundred fifty-two. In this example we compared a 3 digit number. This is a task you will probably face in 2nd grade.
Comparing 3 Digit Numbers – Summary
| Step # | What to do |
--- |
| 1 | Start with the greatest place value, the hundreds. |
| 2 | If the hundreds place digits are the same, move to the tens place. |
| 3 | If the tens place digits are the same, move to the ones place |
| 4 | Compare using appropriate symbol ( > < = ) |
To compare three-digit numbers, you start in the highest place value. You compare each digit from left to right until you find a value that is greater. Then use the greater than, less than, or equal to signs to compare the numbers.
At the end of the video, you will find worksheets for comparing 3-digit numbers for the 2nd grade.
Transcript Compare 3 Digits (<, >, = )
Dundee is visiting Skylar and Henry and they are hungry. Dundee asks for food but will only eat food from whoever brings the most. Let's feed them by comparing three digit numbers. We can use the greater than, less than, or equal to symbols when comparing. A place value chart can help compare three digit numbers. We compare starting in the hundreds place, then the tens place, and finally the ones place. We stop comparing when we find a digit that is greater than or less than. If all place values are the same, the numbers are equal. For example, to compare these two numbers, start with the greatest place value, the hundreds. Check if the hundreds place digits are the same or different. There is one hundred in both, so move to the tens place. Check if the tens place digits are the same or different. There are five tens in both, so move to the ones place. Check if the ones place digits are the same or different. There are two ones, making these numbers equal, because all the place value digits are the same. One hundred fifty-two is equal to one hundred fifty-two. Oh, it looks like it's time to feed Dundee. Skylar gathered two hundred ten blueberries, and Henry gathered one hundred ninety-two. Starting in the greatest place value, we see two hundreds and one hundred. Two hundreds is greater than one hundred, so we do not need to continue comparing in the tens or ones places. Two hundred ten is greater than one hundred ninety-two. Dundee will guzzle up Skylar's plate of blueberries because it has more. Now Skylar and Henry have gathered grapes. Skylar gathered one hundred five grapes, and Henry gathered one hundred thirty-seven grapes. Start comparing in the hundreds place. Is one hundred greater than, less than, or equal to one hundred? One hundred is equal to one hundred, so move to the tens place and compare the digits. Are zero tens greater than, less than, or equal to three tens? Zero tens is less than three tens. We stop comparing here because we found a digit that is greater than or less than. Is one hundred five greater than, less than, or equal to one hundred thirty-seven? One hundred five is less than one hundred thirty-seven. Skylar's plate has less grapes, so Dundee devours Henry's plate of grapes. Before we see if Dundee is satisfied, let's review. Remember, when comparing three digit numbers, start with the greatest place value, the hundreds and compare. If the hundreds place digits are the same, move to the tens place and compare. If the tens place digits are the same, move to the ones place and compare. Finally, compare using the appropriate symbol, greater than, less than, or equal to. "Be very quiet Henry. We can't keep feeding them all our fruit!"
Compare 3 Digits (<, >, = ) exercise
Would you like to apply the knowledge you’ve learned? You can review and practice it with the tasks for the video Compare 3 Digits (<, >, = ).
Can you help Dundee use the expressions below to find the largest bowl of blueberries?
Hints
Remember to start comparing in the hundreds place and work your way along each place value column.
Can you place 143 and 217 into a chart like this to compare them?
Remember to look carefully at the symbol when reading the expression.
Use this example to help you.
Solution
217 is greater than 143.
Start in the largest place value, the hundreds place. We see that 200 is greater than 100, therefore 217 is greater than 143.
We do not need to compare the digits in any other place value column this time.
### Can you compare the bowls?
Hints
Where do we start comparing 3-digit numbers?
If the hundreds are the same, look at the digits in the tens column. If these are the same, look at the digits in the ones column.
Solution
When comparing 3-digit numbers, start with the greatest place value, the hundreds place. If the values are the same, move to the tens place to compare the values. If those values are the same, move to the ones place. If all place values are the same, the numbers are equal to each other.
### Is Dundee eating the correct bowls of fruit?
Hints
Remember to start by looking at the digits with the greatest value and comparing them.
If all place values are exactly the same, the numbers are equal.
Solution
Here are the correct answers.
### Which symbol is missing?
Hints
Put the numbers into a place value chart to compare them more easily.
Start by comparing the hundreds, then the tens, and then the ones.
The symbol < means less than.
The symbol > means greater than.
Solution
Here are the correct answers.
### Who gathered the most blueberries?
Hints
Remember to start comparing in the hundreds column.
Remember, Dundee wants to eat the bowl with the most blueberries so make sure his mouth is facing the right way!
Solution
Henry had collected more blueberries!
We start comparing the numbers in the hundreds place. We see that 200 is greater than 100 so we know that 217 is greater than 124.
We then put < in between the bowls to show that 217 is greater than 124.
### Compare the numbers.
Hints
Here is an example of comparing a set of three numbers.
Look for the largest or smallest number in each set and start from there.
Are any of the numbers equal to each other in the set?
If the symbol is < you need to order from smallest to largest.
If the symbol is > you need to order from largest to smallest.
Solution
The sequences should read as follows:
1. 135 < 139 < 142
2. 390 > 386 > 372
3. 422 = 422 < 425
4. 219 < 234 = 234 < 238
5. 516 > 515 = 515 > 514
More videos in this topic
Comparing and Ordering Numbers
The Hundreds Chart
Compare 3 Digits (<, >, = )
Using and Drawing Number Lines
Comparing Numbers to 1,000 — Let's Practice! |
11927 | https://arxiv.org/pdf/2308.05041 | Critical configurations of the hard-core model on square grid graphs
Simone Baldassarri 1,2 , Vanessa Jacquier 3, and Alessandro Zocca 41Universit` a degli Studi di Firenze, Firenze, Italy
2
Aix-Marseille Universit´ e, Marseille, France
3
University of Utrecht, Utrecht, The Netherlands
4
Vrije Universiteit Amsterdam, The Netherlands August 10, 2023
Abstract
We consider the hard-core model on a finite square grid graph with stochastic Glauber dynamics parametrized by the inverse temperature β. We investigate how the transition between its two maximum-occupancy configurations takes place in the low-temperature regime β → ∞ in the case of periodic boundary conditions. The hard-core constraints and the grid symmetry make the structure of the critical configurations, also known as essential saddles, for this transition very rich and complex. We provide a comprehensive geometrical characterization of the set of critical configurations that are asymptotically visited with probability one. In particular, we develop a novel isoperimetric inequality for hard-core configurations with a fixed number of particles and we show how not only their size but also their shape determines the characterization of the saddles.
MSC Classification: 82C20; 60J10; 60K35.
Keywords: Hard-core model; Metastability; Tunneling; Critical configurations.
Acknowledgements: S.B and V.J. are grateful for the support of “Gruppo Nazionale per l’Analisi Matematica, la Probabilit` a e le loro Applicazioni” (GNAMPA-INdAM). The authors are grateful to Francesca Nardi, Julien Sohier, Gianmarco Bet, and Tommaso Monni for useful and fruitful discussions at the early stage of this work.
1 Introduction
We consider a stochastic model, known in the literature as hard-core lattice gas model [10, 31], where particles have a non-negligible radius and therefore cannot overlap. Assuming a finite volume, the hard-core constraints are modeled with a finite undirected graph Λ. More specifically, particles can reside on the sites of Λ and edges connect the pairs of sites in Λ that cannot be simultaneously occupied. In other words, any hard-core configuration is an independent set of Λ. In this paper, we take Λ to be a square grid graph with periodic boundary conditions. The resulting hard-core particle 1
arXiv:2308.05041v1 [math.PR] 9 Aug 2023
configurations are then those whose occupied sites have all the corresponding four neighboring sites empty, see Fig. 1 for an example of such configurations. This interacting particle system evolves according to a stochastic dynamics which is fully charac-terized by the Hamiltonian or energy function in (2.3) and is parametrized by the inverse temperature
β. In particular, the appearance and disappearance of particles are modeled via a Glauber-type update Markov chain {Xt}t∈N with Metropolis transition probabilities induced by the Hamiltonian, see (2.4) later for more details. The stochastic process is reversible with respect to the corresponding Gibbs measure, cf. (2.2), which is then its equilibrium distribution. In the low-temperature regime (i.e., β → ∞ ), the most likely states for this interacting particle system, which we refer to as stable states , are those with a maximum number of particles, namely. On the square grid graph Λ of even length, there are two stable states, corresponding to the two chessboard-like patterns. When β grows large, it takes the system a very long time to move from one stable state to the other since such a transition involves visiting intermediate configurations which are very unlikely. Such transitions become thus rare events and this is a central issue in the framework of metastability for interacting particle systems, which represents a thriving area in mathematical physics that is full of challenges. As a consequence, the stochastic process takes also a very long time to converge to stationarity, exhibiting so-called slow/torpid mixing [29, 43]. The asymptotic behavior of the first hitting times between the maximum-occupancy configurations of this model in the low-temperature regime has been studied in . In particular, the authors showed how the order-of-magnitude of this first hitting time depends on the grid sizes and on the boundary conditions by means of an extension of the setting in . Instead of leveraging directly the general strategy proposed in , which allows us to derive the asymptotic behavior of the transition time together with a characterization of the critical configurations, the authors of adopted a novel combinatorial method to estimate the energy barrier between the two stable states of the model, which is disentangled with respect to the description of the critical droplets. The main motivation of the present paper is to fill this knowledge gap. Indeed, the geometrical characterization of the essential gates is a relevant goal both from a probabilistic and a physical point of view since it provides insightful details of the dynamical behavior of the system. This represents a crucial point in describing the typical trajectories, namely, those typically followed by the system during the transition from a stable state to the other. We remark that in several models analyzed in the context of Freidlin-Wentzell Markov chains evolving under Glauber dynamics, the essential gate was unique but, in general, there may exist many minimal sets that are crossed with high probability during the phase transition, either distinct or overlapping (see e.g. [5, 6] for this description in the case of the conservative Kawasaki dynamics). Interestingly, this is what happens also for our model despite it evolves under the non-conservative Glauber dynamics. Such a peculiar feature rests on the hard-core constraints and on the specific symmetry of the system, i.e., we are analyzing the tunneling transition between two stable states. Indeed, the fact that particles cannot appear in any site and the starting and target configurations have the same energy forces the system to visit many critical configurations before reaching the cycle of the target stable state. This is indeed also what happens for the Ising and Potts model evolving with the Glauber dynamics when there is no external magnetic field (see for instance), while when the symmetry of the system is broken, namely, an external magnetic field is present, the situation drastically changes. This different behavior has a major impact on the geometrical structure of the essential gates, which indeed turns out to be much richer than in the other cases and deserves a careful and detailed analysis. In order to geometrically characterize the critical configurations, with each cluster of particles we associate its contour, that is a union of edges on the dual graph of Λ. To this end, we provide some results concerning the model-dependent isoperimetric inequality. In particular, we show that 2Figure 1: Example of a hard-core configuration on the 14 × 14 square grid with periodic boundary conditions. On the left, the occupied sites in Vo (resp. in Ve) are highlighted in black (resp. in red). On the right, we depict the same configuration using a different visual convention, in which we highlight the odd clusters that the configuration has by drawing only the empty sites in Ve (in white), the occupied sites in Vo (in black), and a black line around each odd cluster representing its contour.
for a fixed area the unique clusters that minimize the perimeter have a rhomboidal shape. However, the energy landscape is much more complex as the periodic boundary conditions give rise to other types of clusters with minimal perimeter for a given area, such as the configurations having a column containing a fixed number of particles. In this paper, we adopt the framework of the pathwise approach , introduced in , later developed in [40, 41], and summarized in the monograph . A modern version of this approach can be found in [24, 25, 35, 37]. The pathwise approach has been widely adopted to the low-temperature behavior of finite-volume models with single-spin-flip Glauber dynamics, e.g. [1, 2, 11–14, 23, 39, 44, 45], with Kawasaki dynamics, e.g. [4–7, 32, 36], and with parallel dynamics, e.g. [26–28]. The more involved infinite-volume limit at low temperature was studied via this approach in [3, 30, 32]. Another method to study the metastability is the so-called potential-theoretic approach , initiated in and later summarized in the monograph (see for instance [18, 19, 38] for the application of this approach to specific models both in finite and infinite volume). Since these two approaches rely on different definitions of metastable states, they are not completely equivalent. The situation is particularly delicate for infinite-volume systems, irreversible systems, and degenerate systems, as discussed in [15, 24, 25]. More recent approaches are developed in [8, 9, 16, 17, 33, 34]. The paper is organized as follows. In Section 2, we provide a detailed model description and state our main result regarding the geometric features of the critical configurations, Theorem 2.1. The rest of the paper is then devoted to the proof of this result. First, Section 3 provides some preliminary definitions and auxiliary results and then finally the proof of the main theorem is given in Section 4. For the sake of clarity, the proofs of some auxiliary lemmas are deferred to a later section, namely Section 5. Lastly, Section 6 concludes the paper and outlines some future work. 32 Model description and main results
We consider the stochastic evolution of the hard-core model on finite two-dimensional square lattices. More precisely, given an integer L ≥ 2 we consider the L × L square grid graph Λ = ( V, E ) with periodic boundary conditions, which we will refer to as L × L toric grid graph . We denote by E the edge set of the grid graph Λ and by V the collection of its N = L2 sites. We identify each site v ∈ Λby its coordinates ( v1, v 2), that is we take as set of sites V := {0, . . . , L − 1} × { 0, . . . , L − 1}. In the rest of the paper, we will assume that L is an even integer, which guarantees that Λ is a bipartite graph, and that L ≥ 6, to avoid pathological trivial cases. A particle configuration on Λ is described by associating a variable σ(v) ∈ { 0, 1} to each site
v ∈ Λ, indicating the absence (0) or the presence (1) of a particle on that site. Let X ⊂ { 0, 1}N be the collection of hard-core configurations on Λ, i.e.,
X := {σ ∈ { 0, 1}N | σ(v)σ(w) = 0 , ∀ (v, w ) ∈ E}, (2.1) i.e., the particle configurations on Λ with no particles residing on neighboring sites. A site of Λ is called even (respectively odd ) if the sum of its two coordinates is even (respectively odd) and we denote by Ve and Vo the collection of even sites and that of odd sites of Λ. Clearly
|Ve| = |Vo| = L2/2. We denote by e (o, respectively) the particle configuration on Λ with particles at each site in Ve (Vo, respectively), i.e.,
e(v) :=
1 if v ∈ Ve,
0 if v ∈ Vo, and o(v) :=
0 if v ∈ Ve,
1 if v ∈ Vo.
Both e and o are hard-core configurations thanks to the assumption that L is even. Fig. 1 shows an example of a hard-core configuration. Throughout the paper, all figures are drawn using the following conventions. They all depict hard-core configurations on a 14 × 14 grid with periodic boundary conditions. The occupied (empty) sites in Vo (Ve, respectively) are shown in black (white) and we draw a black line around each odd cluster representing its contour. We tacitly assume that all the even (odd) sites outside the odd region are occupied (empty, respectively) but they are not displayed to avoid cluttering the figures. See Section 3.1 for more precise definitions of odd clusters and odd regions. Consider the Gibbs measure on X given by
μβ (σ) := e−βH (σ)
Zβ, Λ
, σ ∈ X , (2.2) where H is the Hamiltonian H : X → R that is taken to be proportional to the number of present particles, namely
H(σ) := − ∑
v∈V
σ(v), (2.3) with Zβ, Λ := ∑
σ∈X
e−βH (σ) being the normalizing constant. The two hard-core configurations on the
L × L toric grid graph Λ introduced above have energy equal to
H(e) = H(o) = −L2
2 ,
which is the minimum value the Hamiltonian can take on X . 4We assume the interacting particle system described evolves according to stochastic Glauber-type dynamics described by a single-step update Markov chain {Xβt }t∈N on X with transition probabilities between any pair of configurations σ, σ ′ ∈ X given by
Pβ (σ, σ ′) :=
q(σ, σ ′)e−β[H(σ′)−H(σ)] +
, if σ̸ = σ′,
1 − ∑
η̸=σ
Pβ (σ, η ), if σ = σ′, (2.4) where [ ·]+ = max {· , 0} and q is the connectivity matrix that allows only single-step updates, i.e., for every σ, σ ′ ∈ X we set
q(σ, σ ′) :=
1
N
, if |{ v ∈ V : σ(v)̸ = σ′(v)}| = 1 ,
0, if |{ v ∈ V : σ(v)̸ = σ′(v)}| > 1.
1 − ∑
η̸=σ
q(σ, η ), if σ = σ′.
(2.5) The resulting dynamics Pβ is reversible with respect to the Gibbs measure μβ given in (2.2) . One usually refers to the triplet ( X , H, q ) as energy landscape and to (2.4) as Metropolis transition probabilities .The connectivity matrix q given in (2.5) is irreducible, i.e., for any pair of configurations σ, σ ′ ∈ X ,
σ̸ = σ′, there exists a finite sequence ω of configurations ω1, . . . , ω n ∈ X such that ω1 = σ, ωn = σ′
and q(ωi, ω i+1 ) > 0, for i = 1 , . . . , n − 1. We will refer to such a sequence as a path from σ to σ′ and denote it by ω : σ → σ′. Given a path ω = ( ω1, . . . , ω n), we define its height Φω as Φω := max
i=1 ,...,n
H(ωi). (2.6) The communication energy between two configurations σ, σ ′ ∈ X is the minimum value that has to be reached by the energy in every path ω : σ → σ′, i.e., Φ( σ, σ ′) := min
ω:σ→σ′
Φω = min
ω:σ→σ′
max
η∈ω
H(η). (2.7) Let X s ⊂ X denote the set of global minima of the Hamiltonian H on X , to which we will refer to as stable states . In it has been proved that for the hard-core model on a finite L × L square grid graph the following statements hold: (i) There are exactly two stable states
X s = {e, o}; (2.8) (ii) The communication energy between the two stable states is equal to Φ( e, o) − H(e) = L + 1; (2.9) (iii) The corresponding energy landscape has no deep wells, i.e., max
σ∈X
[Φ( σ, {e, o}) − H(σ)] ≤ L < Φ( e, o) − H(e). (2.10) 52.1 Essential saddle characterization
Our results give insight into the way the transitions between e and o most likely occur in the low–temperature regime. This is usually described by identifying the optimal paths, saddles, and essential saddles that we define as follows.
• S(e, o) is the communication level set between e and o defined by
S(e, o) := {σ ∈ X | ∃ ω ∈ (e → o)opt , : σ ∈ ω and H(σ) = Φ ω = Φ( e, o)},
where ( e → o)opt is the set of optimal paths from e to o realizing the minimax in Φ( e, o), i.e., (e → o)opt := {ω : e → o | Φω = Φ( e, o)}.
• The configurations in S(e, o) are called saddles . Given an optimal path ω ∈ (e → o)opt , we define the set of its saddles S(ω) as S(ω) := {σ ∈ ω | H(σ) = Φ ω = Φ( e, o)}. A saddle
σ ∈ S (e, o) is called essential if either (i) ∃ ω ∈ (e → o)opt such that S(ω) = {σ}, or (ii) ∃ ω ∈ (e → o)opt such that σ ∈ S(ω) and S(ω′)̸ ⊆ S(ω) \ { σ} ∀ ω′ ∈ (e → o)opt .A saddle σ ∈ S (e, o) that is not essential is called unessential saddle or dead-end , i.e., for any ω ∈ (e → o)opt such that ω ∩ { σ}̸ = ∅ we have that S(ω) \ { σ}̸ = ∅ and there exists
ω′ ∈ (e → o)opt such that S(ω′) ⊆ S(ω) \ { σ}.
• The essential gate G(e, o) ⊂ X is the collection of essential saddles for the transition e → o.The aim of the present paper is to accurately identify the set G(e, o) of the essential saddles for the transition from e to o for the Metropolis dynamics of the hard-core model on a L × L grid with periodic boundary conditions. The set G(e, o) will be described as the union of six disjoint sets, each characterized by configurations with specific geometrical features. While we refer the reader to Section 4 for a precise definition of these sets (cf. Definitions 4.2–4.6), we provide here some intuitive descriptions of the geometrical features of the configurations in these sets. We denote by
• Cir (e, o), Cgr (e, o), and Ccr (e, o) the collections of configurations with a unique cluster of particles in odd sites of rhomboidal shape with exactly two adjacent even empty sites as in Fig. 2 and Fig. 3 (left). Roughly speaking, Cir (e, o) contains the configurations with ( L
2
− 1) 2 occupied odd particles and L2 + 2 empty even sites; Cgr (e, o) (resp. Ccr (e, o)) contains the configurations obtained from Cir (e, o) (resp. Cgr (e, o)) by removing some occupied even sites attached to the rhombus and by growing along one (resp. the longest) side by adding some particles in the nearest odd sites of the rhombus.
• Csb (e, o), Cmb (e, o), and Cib (e, o) the collections of configurations with a unique cluster of particles in odd sites with at most two additional empty even sites as in Fig. 3 (right) and in Fig. 4. In particular, Csb (e, o) contains the configurations with L
2
− 1 particles arranged in an odd column with further two empty even sites; Cmb (e, o) contains the configurations obtained from Csb (e, o) such that there is at least one column or row with L
2
particles arranged in odd sites. Cib (e, o) contains the configurations obtained from Csb (e, o) without having column or row with L
2
particles. The following theorem characterizes the essential gate for the transition from e to o.6Figure 2: An example of a configuration in Cir (e, o) (on the left) and one in Cgr (e, o) (on the right). Figure 3: An example of a configuration in Ccr (e, o) (on the left) and one in Csb (e, o) (on the right).
Theorem 2.1 (Essential saddles) . Define the set
C∗(e, o) := Cir (e, o) ∪ C gr (e, o) ∪ C cr (e, o) ∪ C sb (e, o) ∪ C mb (e, o) ∪ C ib (e, o).
The essential saddles for the transition from e to o of the hard-core model on a L × L toric grid graph
Λ are all and only the configurations in C∗(e, o), i.e.,
G(e, o) = C∗(e, o).
Furthermore, the possible transitions at energy not higher than −L2
2
L + 1 among the six subsets forming C∗(e, o) are as detailed in Fig. 5.
7Figure 4: An example of a configuration in Cmb (e, o) (on the left) and one in Cib (e, o) (on the right). Ce
e
Csb (e,o)
Cmb (e,o)
Cir (e,o)Cgr (e,o)Ccr (e,o)
Cib (e,o)
−L2
2
· · ·
−L2
2+ 2 L−1−L2
2+ 2 L+ 1
· · ·
−L+ 3 −L+ 5
· · ·
13
· · ·
L−3
· · ·
L2
2−2L+ 1
· · ·
L2
2
Co
o
Figure 5: Schematic representation of the set of essential saddles, where we highlight with arrows between the set pairs that communicate at energy not higher than − L2
2
L + 1 and the initial cycles Ce
and Co, see Section 3. The vertical lines represent the partition of X in manifolds, see (4.1).
3 Definitions and auxiliary results
The main goal of this section is to introduce the notion of odd clusters, which are the basis of the geometrical description of the configurations, and to inspect the relation between their shape and perimeter. In Section 3.1 we define a geometrical representation of clusters associated with the occupied odd sites, and in Section 3.2 we introduce the notion of rhombi, which turns out to be crucial in the description of the essential saddles. In Section 3.3 we present two algorithms that, combined together, return a path whose last configuration has a rhomboidal shape and such that the energy along it never increases. We will use them to deduce that there exists a downhill path from the configurations 8without a rhomboidal cluster towards e or o.Along the lines of [35, eq. (2.7)], we define Ce := {ζ ∈ X | Φ( ζ, e) < Φ( e, o)} to be the initial cycle of e, that is the maximal cycle that includes e but does not include o, namely, it contains all the configurations that can be reached by e by spending strictly less energy than the one needed for the transition between e and o, i.e., the communication height Φ( e, o). The corresponding initial cycle of o is defined analogously and denoted by Co.Given a configuration σ ∈ X , denote by ∆ H(σ) the energy difference with respect to either one of the stable states, i.e., ∆H(σ) := H(σ) − H(e). (3.1)
3.1 Odd clusters and regions
For any subset of sites S ⊆ V we define the complement of S as Sc := V \ S, the external boundary
∂+S as the subset of sites in Sc that are adjacent to a site in S, i.e.,
∂+S := {v ∈ Sc | ∃ w ∈ S : (v, w ) ∈ E},
and ∇S as the subset of edges connecting the sites in S with those in ∂+S, i.e.,
∇S := {(v, w ) ∈ E | v ∈ S, w ∈ ∂+S}.
A (connected) odd cluster C ⊆ V is a subset of sites that satisfies both the following conditions: 1. If an odd site v ∈ Vo belongs to C, then so do the four neighboring even sites, i.e., ∂+{v} ⊂ C;2. C ∩Ve is connected as a sub-graph of the graph ( Ve, E ∗), with E∗ := {(v, w ) ∈ Ve ×Ve | d(v, w ) = 2}, where d(·, ·) denotes the usual graph distance on Λ. We denote by C o(Λ) the collection of the odd clusters on Λ. Consider the dual graph Λ ′ = ( V ′, E ′) of the graph Λ, which is a discrete torus of the same size. Given an odd cluster C, consider the edge set ∇C that disconnects C from its complement Cc. We associate with ∇C the edge set γ(C) ⊂ E′ on the dual graph Λ ′ which consists of all the edges of Λ ′
orthogonal to edges in ∇C. Such a set, to which we will refer as the contour of the cluster C, consists of one or more piecewise linear closed curves and, by construction |γ(C)| = |∇ C|. Leveraging this fact, we define the perimeter P (C) of the odd cluster C as the total length of the contour γ(C), i.e.,
P (C) := |γ(C)|. (3.2) As proved in , the perimeter of the odd cluster C satisfies the following identity:
P (C) = 4( |C ∩ Ve| − | C ∩ Vo|). (3.3) We call area of an odd cluster C the number of odd occupied sites it comprises. We say that an odd cluster is degenerate if it has area 0 and non–degenerate otherwise. We introduce a mapping O : X → 2V that associates to a given hard-core configuration σ ∈ X
the subset O(σ) ⊆ V defined as
O(σ) := {v ∈ Vo | σ(v) = 1 } ∪ { v ∈ Ve | σ(v) = 0 }. (3.4) In other words, O(σ) is the subset comprising all the occupied odd sites and the empty even sites of the configuration σ. It is immediate to check that O is an injective mapping and we will refer to the image O(σ) of a configuration σ as its odd region .9Figure 6: Example of four different rhombi, namely R1,2, R4,2, R0,2 and R1,0 (in clockwise order from the top-right corner).
The odd region O(σ) of a configuration σ ∈ X can be partitioned into its connected components, say C1(σ), . . . , C m(σ) ∈ Co(Λ), for some m ∈ N, which are, by definition, odd clusters, that is
O(σ) =
m⊔
i=1
Ci(σ). (3.5) Using the partition (3.5) of the odd region O(σ) into odd clusters, the definitions of contour and perimeter can be extended to the whole odd region in an obvious way, so that we can ultimately define the contour γ(σ) of a configuration σ ∈ X as
γ(σ) :=
m⊔
i=1
γ(Ci(σ)) , (3.6) and its perimeter P (σ) as
P (σ) :=
m∑
i=1
P (Ci(σ)) . (3.7) As shown in , starting from (3.3) , a double counting argument yields the following identity that relates the perimeter P (σ) of a hard-core configuration σ ∈ X with its energy H(σ) (recall (3.1) -(3.2) )
P (σ) = 4 ∆ H(σ). (3.8) Given a configuration σ ∈ X , we define the odd non-degenerate region Ond (σ) as a subset of O(σ)containing only odd non-degenerate clusters. See Fig. 7 for an example of an odd region and an odd non-degenerate region. 10 Figure 7: Example of a configuration σ, in which the contour of the non-degenerate (degenerate) odd clusters is highlighted in black (red, respectively). The contour γ(σ) of the configuration σ is of the odd region O(σ) is the union of black lines (corresponding to Ond (σ)) and red lines.
3.2 Odd rhombi
Given an odd site η = ( η1, η 2) ∈ Vo and two positive integers ℓ1, ℓ 2 ≤ L, the odd rhombus Rℓ1,ℓ 2 (η)with reference site η and lengths ℓ1 and ℓ2 is the odd cluster defined as
Rℓ1,ℓ 2 (η) := Sℓ1,ℓ 2 (η) ∪ ∂+Sℓ1,ℓ 2 (η), (3.9) where Sℓ1,ℓ 2 (η) ⊆ Vo is the subset of odd sites given by
Sℓ1,ℓ 2 (η) := ⋃
0≤k≤ℓ1−1,0≤j≤ℓ2−1
{(η1 + k + j, η 2 + k − j)}
= {v = ( v1, v 2) ∈ V | ∃ k ∈ , j ∈ : v1 = η1 + k + j, v 2 = η2 + k − j}. (3.10) In the latter definition, the coordinates sums and subtractions are taken modulo L. In the case
ℓ1ℓ2 = 0, we can take η ∈ Ve and define the degenerate rhombus Rℓ1,ℓ 2 (η) as the odd cluster
Rℓ1,ℓ 2 (η) =
⋃
0≤j≤ℓ2
{(η1 + j, η 2 − j)} if ℓ1 = 0 and ℓ2̸ = 0 ,
⋃
0≤k≤ℓ1
{(η1 + k, η 2 − k)} if ℓ1̸ = 0 and ℓ2 = 0 ,
(η1, η 2) if ℓ1 = ℓ2 = 0 .
Note that, in this case, Rℓ1,ℓ 2 (η) is a subset of even sites. The area of Rℓ1,ℓ 2 (η) is the cardinality of
Sℓ1,ℓ 2 (η) in the non-degenerate case, whereas in the degenerate case is equal to zero. Some example of rhombi and degenerate rhombi are shown in Fig. 6. We observe that the non-degenerate rhombus
Rℓ1,ℓ 2 has ℓ1 diagonals of length ℓ2 and ℓ2 diagonals of length ℓ1 in the opposite direction, which we will refer to as complete diagonals . We denote by R o(Λ) ⊂ Co(Λ) the collection of all odd rhombi on Λ including the degenerate ones. For every odd cluster C ∈ Co(Λ), we define the surrounding
11 rhombus R(C) as the minimal rhombus (by inclusion) in R o(Λ) such that C ⊆ R (C); see Fig. 8 for an example. Most of the results for odd rhombi that will be proved are translation-invariant, the reason why we will often refer to the rhombus Rℓ1,ℓ 2 (η) simply as Rℓ1,ℓ 2 , without explicitly specifying the reference site η. The next two lemmas, Lemmas 3.1 and 3.2, concern properties of rhombi on a square L × L
grid with periodic boundary conditions. Their proofs, being involved but not particularly insightful, are deferred to Appendix A.
Lemma 3.1 (Set of sites winds around the torus) . Given η = ( η1, η 2) ∈ Vo and two non-negative integers ℓ1, ℓ 2 ≤ L such that ℓ1 ≤ ℓ2 and ℓ1 ≥ L/ 2, the following statements hold: (i) If ℓ2 ≤ L − 2, then
⋃
0≤k≤ℓ1
ℓ2+1 ≤j≤L−1
{(η1 + k + j − 1, η 2 + k − j)} ∪ ⋃
ℓ1+1 ≤k≤L−10≤j≤ℓ2
{(η1 + k + j − 1, η 2 + k − j)}⊆ ⋃
0≤k≤ℓ1
0≤j≤ℓ2
{(η1 + k + j − 1, η 2 + k − j)}.
(ii) If ℓ2 = L − 1, then
⋃
ℓ1+1 ≤k≤L−10≤j≤L−1
{(η1 + k + j − 1, η 2 + k − j)} ⊆ ⋃
0≤k≤ℓ1
0≤j≤L−1
{(η1 + k + j − 1, η 2 + k − j)} (3.11)
and
⋃
L/ 2≤k≤ℓ1−11≤j≤L−2
{(η1 + k + j, η 2 + k − j)} ⊆ ⋃
0≤k≤L/ 2−10≤j≤L−2
{(η1 + k + j, η 2 + k − j)}. (3.12) For any subset of sites A ⊆ V , we define the complement of A as the complementary set of A in
V , i.e., as V \ A.
Lemma 3.2 (Properties of rhombi) . Given η ∈ Vo and two non negative integers ℓ1, ℓ 2 ≤ L, the following statements hold: (i) If max {ℓ1, ℓ 2} ≤ L − 2 and min {ℓ1, ℓ 2} ≥ L/ 2, then the complement of the rhombus Rℓ1,ℓ 2 (η) is a rhombus RL−ℓ1−1,L −ℓ2−1(ˆ η) for some ˆη ∈ Ve.(ii) If max {ℓ1, ℓ 2} = L − 1 and min {ℓ1, ℓ 2} ≥ L/ 2, then the complement of the rhombus Rℓ1,ℓ 2 (η) is the disjoint union of L − min {ℓ1, ℓ 2} odd sites. (iii) If max {ℓ1, ℓ 2} = L and min {ℓ1, ℓ 2} < L/ 2, then the rhombus Rℓ1,ℓ 2 (η) contains L ℓ 1 odd sites and L(ℓ1 + 1) even sites. (iv) If max {ℓ1, ℓ 2} = L and min {ℓ1, ℓ 2} ≥ L/ 2, then the rhombus Rℓ1,ℓ 2 (η) coincides with V .
These two lemmas will now be used to prove the next proposition, which gives a formula for the perimeter of a rhombus Rℓ1,ℓ 2 . To this end, we will use the fact that a rhombus Rℓ1,ℓ 2 and its complement in Λ have the same boundary for any 0 ≤ ℓ1, ℓ 2 ≤ L, and, in particular the same perimeter. In addition, we will say that a rhombus Rℓ1,ℓ 2 winds vertically (resp. horizontally ) around the torus if there exists a set of L
2
odd sites η1, ..., η L/ 2 in Rℓ1,ℓ 2 all on the same column (resp. row). If the direction is not relevant, we will simply say that the rhombus winds around the torus .12 Proposition 3.3 (Formula for rhombus perimeter) . Given a L × L toric grid graph Λ and any sizes
0 ≤ ℓ1, ℓ 2 ≤ L, the perimeter of the rhombus Rℓ1,ℓ 2 satisfies the following identity
P (Rℓ1,ℓ 2 ) = 4 ×
ℓ1 + ℓ2 + 1 if min {ℓ1, ℓ 2} < L/ 2 and max {ℓ1, ℓ 2} < L,
2L − (ℓ1 + ℓ2 + 1) if min {ℓ1, ℓ 2} ≥ L/ 2 and max {ℓ1, ℓ 2} < L, L if min {ℓ1, ℓ 2} < L/ 2 and max {ℓ1, ℓ 2} = L,
0 if min {ℓ1, ℓ 2} ≥ L/ 2 and max {ℓ1, ℓ 2} = L.
(3.13)
Proof. First of all, we identify which of the conditions in (3.13) imply that a rhombus winds around the torus. Consider the rhombus Rℓ1,ℓ 2 (η) with η = ( η1, η 2) ∈ Vo. Let σ = ( σ1, σ 2) ∈ Sℓ1,ℓ 2 (η) be such that σ2 = η2 and d(η1, σ 1) is the maximal distance along that horizontal axis. Similarly, let
ξ = ( ξ1, ξ 2) ∈ Sℓ1,ℓ 2 (η) be such that d(η2, ξ 2) is the maximal distance along the vertical axis. Recalling (3.10) , since σ2 = η2, we have that k = j for any k, j = 1 , ..., ℓ min − 1, where ℓmin = min {ℓ1, ℓ 2}. Thus, we obtain
σ1 = η1 + k + j = η1 + 2( ℓmin − 1) , (3.14) where the last equality follows from the fact that the maximal distance along the horizontal axis is precisely the distance between σ1 and η1. We note that if d(η1, σ 1) < L − 2 then the rhombus does not wind horizontally around the torus. Thus,
d(η1, σ 1) = 2( ℓmin − 1) < L − 2 ⇐⇒ ℓmin < L
2 . (3.15) Now, let us consider the distance between η2 and ξ2. Let ℓmax = max {ℓ1, ℓ 2}. In this case, if
d(η2, ξ 2) ≤ L − 2 then the rhombus does not wind vertically around the torus. Thus, we have
d(η2, ξ 2) = max
k,j
|η2 − (η2 − k + j)| = ℓmax − 1 ≤ L − 2. (3.16) We conclude that if ℓmin < L
2
and ℓmax < L , then the rhombus does not wind around the torus and the perimeter is the length of its external boundary. In view of (3.3) , the claim follows. Otherwise, there are three cases, which will be treated separately: (a) ℓmin ≥ L
2
and ℓmax ≤ L − 2; (b) ℓmin ≥ L
2
and ℓmax > L − 2; (c) ℓmin < L
2
and ℓmax = L.(a) Consider the complement of the rhombus Rℓ1,ℓ 2 (η) for some η. By virtue of Lemma 3.2(i), we know that its complement in V is a rhombus with side lengths ˜ℓ1 = L − ℓ1 − 1 and ˜ℓ2 = L − ℓ1 − 1. We claim that this complementary rhombus does not wind around the torus. By using the condition
ℓmin ≥ L
2
, we have that the maximal side length of the complementary rhombus is max {˜ℓ1, ˜ℓ2} = max {L − ℓ1 − 1, L − ℓ2 − 1} = L − ℓmin − 1 ≤ L − L
2 − 1 < L, (3.17) that is, max {˜ℓ1, ˜ℓ2} < L . Moreover, the minimal side length is min {˜ℓ1, ˜ℓ2} = min {L − ℓ1 − 1, L − ℓ2 − 1} = L − ℓmax − 1 ≤ L − ℓmin − 1 ≤ L − L
2 − 1, (3.18) that is, min {˜ℓ1, ˜ℓ2} < L
2
. Since the perimeter of the rhombus Rℓ1,ℓ 2 is the same as that of RL−ℓ1−1,L −ℓ2−1,the claim follows from (3.3). (b) The claim follows from Lemma 3.2(ii)–(iii) and (3.3). (c) The claim follows from Lemma 3.2(iv) and (3.3). 13 Figure 8: Example of a odd cluster C (on the left) and its surrounding rhombus R(C) = R8,5 in red (on the right). On the left, the red squares contain the antiknobs and the decreasing broken diagonals are highlighted with blue rectangles. On the right, we highlight the decreasing shorter (resp. complete) diagonals with blue (resp. green) rectangles.
We say that an odd cluster C is monotone when its perimeter coincides with that of its surrounding rhombus R(C) if it does not wind around the torus, i.e., P (C) = P (R(C)). Otherwise, we say that an odd cluster C is monotone when its perimeter coincides with that of a bridge, i.e., P (C) = 4 L.Note that it immediately follows that a monotone odd cluster C has no holes , i.e., empty odd sites with the four even neighboring sites belonging to C. An empty odd site η / ∈ C is an antiknob for the cluster C if it has at least three neighboring even empty sites that belong to C. Fig. 8 (left) highlights in red the antiknobs of a hard-core configuration. Given an odd cluster C ∈ Co(Λ) and an integer k ≥ 1, we say that C displays an increasing
(resp. decreasing ) diagonal broken in k sites if there exist a sequence of sites zi = ( xi, y i) ∈ Vo \ C,
i = 1 , ..., k , such that
• xi+1 = xi + 1 and yi+1 = yi + 1 (resp. xi+1 = xi + 1 and yi+1 = yi − 1) for any i = 1 , ..., k − 1, and
• the two odd sites ( x1 − 1, y 1 − 1) and ( xk + 1 , y k + 1) (resp. ( x1 − 1, y 1 + 1) and ( xk + 1 , y k − 1)) belong to the cluster C.By construction of a broken diagonal, the two sites z1 and zk are always antiknobs. If it does not matter if an increasing or decreasing diagonal is broken, we simply say that a diagonal is broken. Broken diagonals are visualized in blue in Fig. 8 (left). Given an odd cluster C ∈ Co(Λ) and an integer k ≥ 1, we say that C displays an increasing (resp. decreasing ) shorter diagonal lacking in k
sites if there exist a sequence of sites zi = ( xi, y i) ∈ (Vo ∩ R (C)) \ C, i = 1 , ..., k , such that
• xi+1 = xi + 1 and yi+1 = yi + 1 (resp. xi+1 = xi + 1 and yi+1 = yi − 1) for any i = 1 , ..., k − 1, and
• the two odd sites ( x1 − 1, y 1 − 1) and ( xk + 1 , y k + 1) (resp. ( x1 − 1, y 1 + 1) and ( xk + 1 , y k − 1)) do not belong to R(C). Fig. 8 (right) highlights the shorter diagonals in blue. 14 3.3 Expanding an odd cluster: The filling algorithms
We now describe an iterative procedure that builds a path ω in X from a configuration σ with a unique odd cluster to another configuration σ′ that (i) displays a rhombus, and (ii) whose energy
H(σ′) is equal to or lower than H(σ). The path ω can be described as the concatenation of two paths, each obtained by means of a specific filling algorithm . The reason behind this name is that, along the generated paths, any incomplete diagonal in the odd cluster of the starting configuration is gradually filled by adding particles in odd sites until a rhombus is obtained. The two paths can be intuitively described as follows. The first path, denoted as ˜ω, starts from a configuration with at least one broken diagonal and, by filling one by one all broken diagonals in lexicographic order, arrives at a configuration with no broken diagonal. Each broken diagonal is progressively filled by removing a particle in the even site at distance 1 from a antiknob that lies on that diagonal and adding a particle in the odd site where the antiknob is. The antiknobs on the same diagonal are processed in lexicographic order. The second path, denoted as ¯ω, starts from a configuration with no broken diagonal (such as any ending configuration of the path ˜ω) and arrives at a configuration displaying an odd rhombus. The construction of this second path is similar to that of the first path, but in this case the particles are added in odd sites to fill all the shorter diagonals. The filling algorithms generating the two paths ˜ω and ¯ω are designed in such a way that the maximum energy along the resulting path ω = ˜ω ∪ ¯ω, i.e., Φ ω, is never larger than H(σ) + 1. More specifically, the perimeter of the odd cluster either decreases or does not change along ˜ω, whereas it can increase and sequentially decrease by the same quantity along ¯ω. Proposition 3.4, whose proof is postponed to Section 5.1, specifies the requirement for the starting configuration and summarized the properties of the path generated by the filling algorithm. To formally define these two algorithms, we introduce the following notation. Given two configu-rations σ, σ ′ ∈ X and a subset of sites W ⊂ Λ, we write σ|W = σ′|W if σ(v) = σ′(v) for every v ∈ W .Given a configuration σ ∈ X , we let
• σ(v, 0) be the configuration σ′ ∈ X such that σ′|V { v} = σ|V { v} and σ′(v) = 0; and
• σ(v, 1) be the configuration σ′ such that σ′|V { v} = σ|V { v} and σ′(v) = 1. In general, σ(v, 1) might not be a hard-core configuration in X , since σ may already have a particle residing in one of the four neighboring sites of v.Algorithm 1 (resp. Algorithm 2) provide the detailed pseudocode for the filling algorithm that yields ˜ ω (resp. ¯ ω).
Proposition 3.4 (Odd cluster expansion via filling algorithms) . Let σ, σ ′ ∈ X be two hard-core configurations on Λ, σ̸ = σ′, and R a rhombus such that (i) There exists a connected odd cluster C ⊆ O (σ) such that R(C) = R;(ii) σ|Λ\R = σ′|Λ\R ;(iii) σ′|R = o|R .Then, there exists a path ω : σ → σ′ such that Φω − H(σ) ≤ 1. In addition, if C has at least one broken diagonal then P (σ) > P (σ′), otherwise P (σ) = P (σ′).
We note that conditions ( i), (ii ), and ( iii ) mean that there is a unique odd cluster in σ different from a rhombus, i.e., there exists at least one broken or shorter diagonal. 15 Algorithm 1: Filling algorithm to build path ˜ ω
Input: a configuration σ ∈ X consisting of a unique odd cluster C with m ≥ 1 broken diagonals such that the j-th is broken in kj sites for j = 1 , ..., m
Output: ˜ω : σ → ˜σ, with the configuration ˜σ ∈ X consisting of a unique odd cluster with no broken diagonals
σ0 = σ;
for j = 1 , ..., m do
σj = σj−1;
for i = 1 , ..., k j do
Consider the i-th antiknob (in lexicographic order) of the j-th broken diagonal in C
and denote it by xj,i ∈ Vo
if xj,i has a neighboring occupied site ˜xj,i ∈ Ve then
˜σj,i = σ(˜ xj,i ,0)
j
;
σj,i = ˜ σ(xj,i ,1)
j,i
;
else
˜σj,i = σ(xj,i ,1)
j
;
σj,i = ˜ σj,i ;
end
σj+1 = σj,k j ;˜ωj,i = ( σj , ˜σj, 1, σ j, 1, ..., ˜σj,i , σ j,i );
end
˜ωj = ˜ ωj,k j ;
σj+1 = σj,k j ;
end
˜σ = σm;˜ω is the concatenation of the paths ˜ ω1,..., ˜ ωm
Thanks to Proposition 3.4, we are able to characterize the configurations having minimal perimeter for a fixed number of occupied odd sites. This finding is formalized in the following two results, Proposition 3.6 and Corollary 3.7, whose proofs are deferred to Section 5.1. To state the precise results, we first introduce the notion of bars as follows. We define a vertical (resp. horizontal) bar B of length k as the union of the particles arranged in odd sites x1, ..., x k
belonging to the same column (resp. row) such that d(xi, x i+1 ) = 2 for any i = 1 , ..., k − 1. In the case of k = 1, we will refer to it as protuberance . If on the same column (resp. row) there are m disjoint vertical (resp. horizontal) bars, each of them of length ki, we say that the total length of the bars is k = k1 + ... + km. Similarly, we can define a diagonal bar and note that it can correspond to a shorter or complete diagonal. If it does not matter if the bar is vertical, horizontal, or diagonal, we simply refer to it as a bar . Finally, we will say that a bar B is attached to a cluster C when all the particles belonging to B are at distance two from C. Note that in Fig. 8 (right) the shorter diagonals of lengths three and one are diagonal bars. See Fig. 8 (left) for examples of vertical bars.
Lemma 3.5. For any n positive integer there exist two positive integers s and k, with 0 ≤ k < s ,such that either (i) n = s(s − 1) + k or (ii) n = s2 + k.Proof. See the first part of [24, Lemma 6.17].
Proposition 3.6 (Perimeter-Minimal rhombi) . Consider n ≤ L(L − 2) and let s, k be the unique integers as in Lemma 3.5. The set of odd clusters with area n that have minimal perimeter contains
16 Algorithm 2: Filling algorithm to build path ¯ ω
Input: a configuration σ ∈ X consists of a unique odd cluster C with no broken diagonal and
m ≥ 0 increasing shorter diagonals. The quantity kj is the difference between the length of the shorter diagonal and the corresponding one of the surrounding rhombus for j = 1 , ..., m
Output: ¯ω : σ → ¯σ, with the configuration ¯ σ ∈ X having a unique odd cluster, which is a rhombus
if m = 0 then
¯σ = σ and ¯ ω is trivial;
else
σ0 = σ;
for j = 1 , ..., m do
σj = σj−1;
for i = 1 , ..., k j do
Consider the i-th empty odd site (in lexicographic order) not belonging to the j-th shorter diagonal in C but in the corresponding complete diagonal of the surrounding rhombus and denote it by xj,i ∈ Vo
if xj,i has a neighboring occupied site ¯xj,i ∈ Ve then
¯σj,i = σ(¯ xj,i ,0)
j
;
σj,i = ¯ σ(xj,i ,1)
j, 1
;
else
¯σj,i = σ(xj,i ,1)
j
;
σj,i = ¯ σj,i ;
end
σj+1 = σj,k j ;¯ωj,i = ( σj , ¯σj, 1, σ j, 1, ..., ¯σj,i , σ j,i );
end
¯ωj = ¯ ωj,k j ;
σj+1 = σj,k j ;
end
¯σ = σm;Obtain ¯ ω as the concatenation of the paths ¯ ω1,..., ¯ ωm
end
either a rhombus Rs,s −1 or Rs−1,s with a bar of length k attached to one of its longest sides if
n = s(s − 1) + k and a rhombus Rs,s with a bar of length k attached to one of its sides if n = s2 + k.
Corollary 3.7 (Minimal perimeter) . Consider n ≤ L(L − 2) and let s, k be the unique integers as in Lemma 3.5. The perimeter P of an odd cluster with area n satisfies the following inequalities:
(
P
4 − 1
)2
≥
4n if s < L/ 2,
2( L2 − 2n) if L/ 2 ≤ s < L.
In addition, for all s < L we have that
P ≥ 4(2 √n + 1) (3.19)
and for s < L
2
the equality holds if and only if the odd cluster is the rhombus Rs,s .
17 Lastly, in the next lemma, we derive an isoperimetric inequality assuming the total number of odd and even occupied sites is fixed. To this end, we first define the real area of a configuration σ as ˜n(σ) = o(σ) + ˜ e(σ), (3.20) where o(σ) (resp. ˜e(σ)) denotes the number of occupied odd (resp. empty even) sites of the configuration
σ.
Lemma 3.8 (Perimeter-Minimal rhombi with fixed real area) . Given 1 ≤ ℓ < L
2
, the unique odd cluster with real area ˜n = 2 ℓ2 + 2 ℓ + 1 and minimal perimeter is the rhombus Rℓ,ℓ . In particular, for
˜n = L2
2
− L + 1 this rhombus is RL
2−1,L
2−1
.
We use this lemma to characterize the critical configurations having an odd cluster with a rhomboidal shape that does not wind around the torus. Indeed, we identify the shape of the protocritical configurations σ with minimal perimeter and fixed real area L2
2
− L + 1 such that
R(Ond (σ)) does not wind around the torus, and we show that if the trajectory visits another type of configuration with such a real area, then the corresponding path would be not optimal. The proof is given in Section 5.1.
4 Essential saddles: Proof of the main theorem
In this section, we first formally introduce in Section 4.1 the six sets appearing in the statement of Theorem 2.1 and then prove the same theorem in in Section 4.2 by showing that the elements of those six sets are all the essential saddles for the transition from e to o.
4.1 Preliminaries
We say that a configuration σ ∈ X has a odd (resp. even) vertical bridge if there exists a column in which configuration σ perfectly agrees with o (resp. e). We define odd (resp. even) horizontal bridge in an analogous way and we say that a configuration σ ∈ X has an odd (resp. even) cross if it has both vertical and horizontal odd (resp. even) bridges (see Fig. 9). In addition, we say that a configuration displays an odd (resp. even) vertical m-uple bridge , with m ≥ 2, if there exist m
contiguous columns in which the configuration perfectly agrees with o (resp. e). Similarly, we can define an odd (resp. even) vertical m-uple bridge (see Fig. 10). We refer to for more details. The next lemma states that all configurations σ ∈ X with ∆ H(σ) < L must belong to one of the two initial cycles.
Lemma 4.1 (Configurations with ∆ H < L belong to one of the initial cycles) . If a configuration
σ ∈ X is such that ∆H(σ) < L , then there exists a path ω : σ → { e, o} with Φω ≤ H(σ) + 1 . In particular, either σ ∈ Ce or σ ∈ Co.Proof. If ∆ H(σ) < L , then there exists both a horizontal and a vertical bridge. Due to the hard-core constraints, these L − 1 particles should all reside on sites of the same parity, hence σ has either an even cross or an odd cross. Then, one can build using the reduction algorithm introduced in a path to either e or o, respectively, with the desired properties. 18 Figure 9: Examples of configurations displaying an odd horizontal bridge (on the left) and an odd cross (on the right). Figure 10: Examples of configurations displaying an odd horizontal double (2-uple) bridge (on the left) and an odd vertical triple (3-uple) bridge (on the right).
Definition 4.2. A hard-core configuration σ ∈ X on the L × L toric grid graph Λ belongs to Cir (e, o)
if the following conditions hold: 1. the odd region O(σ) contains only a non-degenerate cluster C and a degenerate rhombus
D ∈ {R 1,0, R0,1} at distance two from C;2. the cluster C is monotone; 3. C is a rhombus RL
2−1,L
2−1
;4. σ|Λ(C∪D) = e|Λ(C∪D).
19 See Fig. 2 (left) for an example of configurations in Cir (e, o).
Definition 4.3. A hard-core configuration σ ∈ X on the L × L toric grid graph Λ belongs to Cgr (e, o)
(resp. Ccr (e, o)) if the following conditions hold: 1. the odd region O(σ) contains only a non-degenerate cluster C and a degenerate rhombus D = R0,0
at distance one from an antiknob; 2. the cluster C is monotone; 3. C is a rhombus RL
2−1,L
2−1
(resp. RL
2−1,L
2
) with a single bar of length k, for k = 1 , ..., L
2
− 1 (resp.
k = 1 , ..., L
2
− 2), attached to one of its sides; 4. σ|Λ(C∪D) = e|Λ(C∪D).
See Fig. 2 (right) and Fig. 3 (left) for an example of configurations in Cgr (e, o) and Ccr (e, o), respectively. Note that for any σ ∈ C ir (e, o) ∪ C gr (e, o) the surrounding rhombus R(O(σ)) is RL
2−1,L
2
,while for any σ ∈ C cr (e, o) the surrounding rhombus R(O(σ)) is RL
2,L
2
.
Definition 4.4. A hard-core configuration σ ∈ X on the L × L toric grid graph Λ belongs to Csb (e, o)
if the following conditions hold: 1. the odd region O(σ) contains only a non-degenerate cluster C and a degenerate region D
consisting of two even sites at distance one from the same antiknob; 2. the cluster C is monotone; 3. C is a single column or row of length L
2
− 1;4. σ|Λ(C∪D) = e|Λ(C∪D).
Note that the unique possibilities are that (i) D consists of two degenerate rhombi R0,0 as in Fig. 3 (right) or (ii) D ∈ {R 0,1, R1,0} as in Fig. 11 (left).
Definition 4.5. A hard-core configuration σ ∈ X on the L × L toric grid graph Λ belongs to Cmb (e, o)
if the following conditions hold: 1. the odd region O(σ) contains only a non-degenerate cluster C and a degenerate rhombus D = R0,0
at distance one from an antiknob; 2. the cluster C is monotone; 3. C is composed either by:
– a odd (L − 3) -uple bridge, or
– an odd m-uple bridge, with 2 ≤ m < L − 3, together with disjoint bars attached to either side of the m-uple bridge with total length k, with k = 0 , ..., L
2
− 1, or
– an odd bridge, together with disjoint bars attached to the bridge with total length k, for
k = 0 , ..., L
2
− 1, and C does not contain RL
2−1,L
2
;4. σ|Λ(C∪D) = e|Λ(C∪D).
20 Figure 11: An example of a configuration in Csb (e, o) that communicates with Cib (e, o) and not with Cmb (e, o)(on the left) and an example of a configuration in Cib (e, o) (on the right).
See Fig. 4 (left) for an example of a configuration in Cmb (e, o). Note that any configuration
σ ∈ C mb (e, o) is such that R(O(σ)) winds around the torus. Thus, the assumption that the non-degenerate cluster C is monotone implies that in condition 3 not every choice of the bars is allowed. Indeed, in order for the cluster to be monotone, the bars on the right (resp. left) of the m-uple bridge can be adjacent only to a longer (resp. shorter) bar in lexicographic order. This implies that all the bridges are in contiguous rows or columns. Furthermore, the length of a bar is inversely proportional to its distance from the nearest bar composing the bridge.
Definition 4.6. A hard-core configuration σ ∈ X on the L × L toric grid graph Λ belongs to Cib (e, o)
if the following conditions hold: 1. the odd region O(σ) contains only a non-degenerate cluster C and a degenerate rhombus D = R0,0
at distance one from an antiknob; 2. the cluster C is monotone and does not contain RL
2−1,L
2−1
;3. C is composed either by:
– one column (or row) B with L
2
− 1 particles in odd sites or
– two neighbouring columns (or row) B with L
2
− 1 particles in odd sites each; In addition, in the other columns (or rows) there are k particles arranged in odd sites, for
k = 0 , ..., L
2
− 1 − j, where j is the distance from B;4. R(C) is not contained in RL
2−1,L
2−1
;5. σ|Λ(C∪D) = e|Λ(C∪D).
See Fig. 4 (right) and Fig. 11 (right) for examples of configurations in Cib (e, o). Next, we provide several results investigating the main properties of the optimal paths connecting
e to o. To this end, we introduce the following partition in manifolds of the state space X : for 21 every m = −L2
2
, ..., L2
2
we define the manifold as the subset of configurations where the difference
m(σ) := −e(σ) + o(σ) between odd and even occupied sites is equal to m ∈ N, i.e.,
Vm := {σ ∈ X | m(σ) = m}, (4.1) where e(σ) := ∑
v∈Ve
σ(v) (resp. o(σ) := ∑
v∈Vo
σ(v)) is the number of the even (resp. odd) occupied sites in σ.
Lemma 4.7. For any hard-core configuration σ ∈ X , the following properties hold: (a) The only configurations accessible from σ with a single nontrivial step of the dynamics belong to
Vm(σ)−1 ∪ V m(σ)+1 . In particular, any path from e to o must intersect each manifold Vm at least once for every m = −L2
2
, ..., L2
2
.(b) The quantities m(σ) and ∆H(σ) always have the same parity, i.e., m(σ) ≡ ∆H(σ) (mod 2) .
The proof of (a) is immediate by noticing that at every step of the dynamics either e(σ) or o(σ) can change value and at most by ±1 and that of (b) follows from the fact that ∆ H(σ) = −e(σ) − o(σ) + L2
2
and that L is even. A special role in our analysis will be played by the non-backtracking paths , i.e., those paths that visit each manifold exactly once. Lemmas 4.8–4.10 below ensure the existence of an optimal path connecting e to o and passing through the six sets that define C∗(e, o). In addition, Lemma 4.11 below shows that some of the sets composing C∗(e, o) do not directly communicate. Later, in Section 4.2.2, combining these lemmas, we will prove that the communication structure of these six sets at energy not higher than H(e) + L + 1 is the one illustrated in Fig. 5. The proof of these lemmas is deferred to Section 5.2.
Lemma 4.8. The following statements hold. (i) For any configuration η ∈ C ir (e, o) there exists a non-backtracking optimal path ω : e → η such that arg max ξ∈ω ∆H(ξ) = {η}.(ii) For any configuration η ∈ C sb (e, o) there exists a non-backtracking path ω : e → η such that
arg max ξ∈ω ∆H(ξ) = {η}.
Lemma 4.9. The following statements hold. (i) For any configuration η ∈ C ib (e, o) there exist a configurations ¯η ∈ C sb (e, o) and a non-backtracking path ω : ¯η → η such that Φω − H(e) = ∆ H(η) = L + 1 and arg max ξ∈ω H(ξ) ⊆Csb (e, o) ∪ C ib (e, o).(ii) For any configuration η ∈ C gr (e, o) there exist a configuration ¯η ∈ C ir (e, o), a configuration
˜η ∈ C ib (e, o) and two non-backtracking paths ω : ¯ η → η, ω′ : ˜ η → η such that - Φω − H(e) = L + 1 and arg max ξ∈ω H(ξ) ⊆ C gr (e, o) ∪ C ir (e, o);- Φω′ − H(e) = L + 1 and arg max ξ∈ω′ H(ξ) ⊆ C gr (e, o) ∪ C ib (e, o).(iii) For any configuration η ∈ C mb (e, o) there exist a configuration ¯η ∈ C sb (e, o), a configuration
˜η ∈ C ib (e, o) and two non-backtracking paths ω : ¯ η → η, ω′ : ˜ η → η such that - Φω − H(e) = L + 1 and arg max ξ∈ω H(ξ) ⊆ C mb (e, o) ∪ C sb (e, o);- Φω′ − H(e) = L + 1 and arg max ξ∈ω′ H(ξ) ⊆ C mb (e, o) ∪ C ib (e, o).
22 (iv) For any configuration η ∈ C cr (e, o) there exist a configuration ¯η ∈ C gr (e, o), a configuration
˜η ∈ C mb (e, o) and two non-backtracking paths ω : ¯ η → η, ω′ : ˜ η → η such that - Φω − H(e) = L + 1 and arg max ξ∈ω H(ξ) ⊆ C cr (e, o) ∪ C gr (e, o);- Φω′ − H(e) = L + 1 and arg max ξ∈ω′ H(ξ) ⊆ C cr (e, o) ∪ C mb (e, o).
Lemma 4.10. The following statements hold. (i) For any configuration η ∈ C cr (e, o) there exists a non-backtracking path ω : η → o such that
Φω − H(e) = L + 1 and arg max ξ∈ω H(ξ) ⊆ C cr (e, o).(ii) For any configuration η ∈ C mb (e, o) there exists a non-backtracking path ω : η → o such that
Φω − H(e) = L + 1 and arg max ξ∈ω H(ξ) ⊆ C mb (e, o).
Lemma 4.11. The following statements hold. (i) For any configurations η ∈ C ir (e, o) and η′ ∈ C ib (e, o), there is no optimal path ω : η → η′ such that arg max ξ∈ω ⊆ C ir (e, o) ∪ C ib (e, o).(ii) For any configurations η ∈ C ib (e, o) and η′ ∈ C cr (e, o), there is no optimal path ω : η → η′ such that arg max ξ∈ω ⊆ C cr (e, o) ∪ C ib (e, o).(iii) For any configurations η ∈ C ir (e, o) and η′ ∈ C mb (e, o), there is no optimal path ω : η → η′ such that arg max ξ∈ω ⊆ C ir (e, o) ∪ C mb (e, o).(iv) For any configurations η ∈ C gr (e, o) and η′ ∈ C mb (e, o), there is no optimal path ω : η → η′ such that arg max ξ∈ω ⊆ C gr (e, o) ∪ C mb (e, o).
4.2 Proof of Theorem 2.1
This section is entirely devoted to the proof of Theorem 2.1. More specifically, we prove that any essential saddle belongs to the set C∗(e, o) in Section 4.2.1, we describe how the transitions between essential gates can take place in Section 4.2.2, and we prove that all the saddles in C∗(e, o) are essential in Section 4.2.3.
4.2.1 Every essential saddle belongs to C∗(e, o)In this subsection, we will show that any essential saddle σ belongs to the subset C∗(e, o). This readily follows from Proposition 4.12 below.
Proposition 4.12. Let σ be an essential saddle. Then, the following statements hold: (i) If R(Ond (σ)) and R(O(σ)) do not wind around the torus, then σ ∈ C ir (e, o) ∪ C gr (e, o).(ii) If R(Ond (σ)) does not wind around the torus, R(O(σ)) does, and σ belongs to ω ∈ (e → o)opt
that crosses the set Cir (e, o) ∪ C gr (e, o), then σ ∈ C cr (e, o).(iii) If R(Ond (σ)) does not wind around the torus, R(O(σ)) does, and σ belongs to ω ∈ (e → o)opt
that does not cross the set Cir (e, o) ∪ C gr (e, o), then σ ∈ C sb (e, o) ∪ C ib (e, o).(iv) If R(Ond (σ)) winds around the torus, then σ ∈ C mb (e, o).
23 We observe that these four cases (i)–(iv) listed in Proposition 4.12 cover all the possibilities and thus form a partition of the set G(e, o) of essential saddles for the transitions e → o.Before presenting the proof, note that given a configuration σ ∈ S (e, o), if we know on which manifold Vm it lies, then the quantities e(σ) and o(σ) are uniquely determined and can be explicitly calculated as
o(σ) = m
2 + L2
4 − L + 1 2 and e(σ) = −m
2 + L2
4 − L + 1 2 . (4.2) This readily follows from the fact that ∆ H(σ) = L+1 = −e(σ)−o(σ)+ L2
2
and that m = −e(σ)+ o(σ). Furthermore, since σ ∈ S (e, o), ∆ H(σ) = L + 1 and m has to be an odd integer by Lemma 4.7(b). To prove Proposition 4.12, we make use of an additional lemma (whose proof is deferred to Section 5.3), which characterizes the intersection between any optimal path and a specific manifold, namely Vm∗ with m∗ := 3 − L.
Lemma 4.13 (Geometrical properties of the saddles on the manifold Vm∗ ). Any non-backtracking optimal path ω : e → o visits a configuration σ ∈ V m∗ that satisfy one of the following properties: (i) If both R(Ond (σ)) and R(O(σ)) do not wind around the torus, then σ ∈ C ir (e, o).(ii) If R(Ond (σ)) does not wind around the torus but R(O(σ)) does, then σ ∈ C ib (e, o).(iii) If both R(Ond (σ)) and R(O(σ)) wind around the torus, then σ ∈ C mb (e, o).Proof of Proposition 4.12.
Case (i). Let σ be an essential saddle such that R(Ond (σ)) and R(O(σ)) do not wind around the torus. If σ ∈ V m∗ , by Lemma 4.13(i) we know that σ ∈ C ir (e, o). Otherwise, we suppose that σ / ∈ V m∗ .First, we observe that σ̸ ∈ V m with m < m ∗, otherwise the saddle σ is not essential. Indeed, every optimal path from e to o has to cross a configuration ¯ σ ∈ V m∗ in Cir (e, o) thanks to Lemma 4.13(i). Thus, we can write a general optimal path ω = ( e, ω 1, ..., ω k, σ, ..., ¯σ, ω k+1 , ..., ω k+m, o) and we can define the path ω′ = ( e, ˜ω1, ..., ˜ωn, ¯σ, ω k+1 , ..., ω k+m, o), where arg max ξ∈{ e,˜ω1,..., ˜ωn,¯σ} H(ξ) = {¯σ}. This path ω′ exists thanks to Lemma 4.8(i). Thus, we are left to analyze the case σ ∈ V m with m > m ∗.We need to show that any essential saddle crossed afterward belongs to the set Cgr (e, o). By using again Lemma 4.13, the path ω crosses a configuration σ ∈ C ir (e, o) to reach o. Starting from it, there is a unique possible move to lower the energy towards o along the path ω, that is adding a particle in the unique unblocked empty odd site. Afterward, the unique possible move is to remove a particle from an even site. If this site is at a distance greater than one from an antiknob, then there is no more allowed move. Otherwise, the resulting configuration belongs to the set Cgr (e, o). By iterating this pair of moves until the shorter diagonal of the rhombus is completely filled, we obtain that all the saddles that are crossed belong to Cgr (e, o). Moreover, from this point onwards, it is only possible to remove a particle from an even site at distance one from the antiknob, obtaining a configuration in
Cgr (e, o) ∩ V 1.
Case (ii). By assumption, the path ω crosses the set Cir (e, o) ∪ C gr (e, o). Without loss of generality, we may consider ω as a non-backtracking path. If this is not the case, we can apply the following argument to the last configuration visited by the path in the manifold Vm∗ . In particular, in view of the properties of the path ω shown in case (i), we know that the last configuration crossed in Cgr (e, o)belongs to V1 and it is composed of a unique non-degenerate cluster RL
2−1,L
2
with a degenerate cluster
R0,0 at distance one from the antiknob. Starting from it, there is a unique possible move to lower the energy towards o along the path ω, that is adding a particle in the unique unblocked empty odd site. Afterward, the unique possible move is to remove a particle from an even site. The resulting 24 configuration is in Ccr (e, o). By iterating this pair of moves until the shorter diagonal of the rhombus is completely filled, we obtain that all the saddles that are crossed belong to Ccr (e, o).
Case (iii). As in case (ii), we assume that the path ω is non-backtracking. By the assumption, the path ω does not cross the set Cir (e, o) ∪ C gr (e, o), thus for Lemma 4.13(ii) the path crosses the set
Cib (e, o), say in the configuration ¯η. Let V ¯m the first manifold containing a configuration in Cib (e, o). Consider first the case in which the saddle σ is crossed by the path ω on the manifolds Vm,with ¯m ≤ m < m ∗. Since ω crosses the set Cib (e, o), we will show that the saddle σ belongs to the set Csb (e, o) ∪ C ib (e, o). To this end, we need to consider the time-reversal of the path
ω = ( e, ω 1, ..., ω k, ¯η, ..., o), where ¯η ∈ C ib (e, o). Starting from ¯η, since ∆ H( ¯η) = L + 1, the energy of the configuration ωk is less than that of ¯η. Moreover, the unique possible move is adding a particle in the unique empty even site at distance one from an antiknob. Then, the unique possible move from
ωk to ωk−1 is removing a particle from an occupied odd site. The configuration ωk−1 belongs to the set Csb (e, o) if it contains at least one bridge, otherwise, it belongs to the set Cib (e, o). By iterating this argument we obtain the desired claim. Consider now the case in which the saddle σ is crossed by the path ω on the manifolds Vm,with m > m ∗. Since ω crosses the set Cib (e, o), we will show that the saddle σ belongs to the set
Csb (e, o) ∪ C ib (e, o). Note that the unique admissible moves are the following: add a particle in the antiknob and then remove a particle from an even site at distance one from an antiknob. By iterating this couple of moves we get that the resulting configuration belongs to the set Cib (e, o) since
R(Ond (σ)) does not wind around the torus. It remains to consider the case σ ∈ V m, with m < ¯m. We will prove that any such σ is not essential. Indeed, any non-backtracking optimal path crossing a configuration in Cib (e, o) has crossed a configuration in Csb (e, o) before by using the same argument as in case (iii) of this proof for ¯m ≤ m < m ∗. Thus, we can write ω = ( e, ω 1, ..., ω k, σ, ..., ˜σ, ..., ¯σ, ω k+1 , ..., ω k+m, o) and we define the path ω′ = ( e, ω ′
1
, ..., ω ′
n
, ˜σ, ..., ¯σ, ω k+1 , ..., ω k+m, o), where ˜σ is a configuration in Csb (e, o), ¯σ ∈ V ¯m is a configuration in Cib (e, o) thanks to Lemma 4.13(ii) and arg max ξ∈{ e,ω ′
1,...,ω ′
n,˜σ}
H(ξ) = {˜σ}. This path
ω′ exists thanks to Lemma 4.8(ii) and this concludes case (iii).
Case (iv). By Lemma 4.13(iii), we know that any optimal path ω ∈ (e → o)opt crosses the manifold
Vm∗ in a configuration ¯ σ belonging to the set Cmb (e, o). If the essential saddle σ ∈ V m∗ , then we deduce that σ ∈ C mb (e, o). Suppose now that the saddle σ belongs to the manifold Vm, with m > m ∗. Starting from such a saddle ¯σ, there is a unique possible move to lower the energy towards o along the path ω, that is add a particle in the unique unblocked empty odd site. Afterward, the unique possible move is to remove a particle from an even site. If this site is at a distance greater than one from an antiknob, then there is no more possible move to reach o in such a way the path ω is optimal. Otherwise, the resulting configuration belongs to the set Cmb (e, o). Indeed, by construction, we deduce that the resulting non-degenerate odd cluster is still monotone due to the properties of the bars attached to each bridge. By iterating this pair of moves until there is a row or a column which is not a bridge, we obtain that all the saddles that are crossed belong to Cmb (e, o). Finally, from this point onwards, it is only possible to remove a particle from an even site at distance one from the antiknob, obtaining the last configuration in Cmb (e, o). Afterward, the energy only decreases and therefore no more saddles are crossed. Suppose now that the saddle σ belongs to the manifold Vm, with m < m ∗. Note that, starting from such ¯σ ∈ C mb (e, o), the unique admissible moves to get σ are the following: add a particle in the unique empty even site at distance one from an antiknob and afterward remove a particle from an occupied odd site. By iterating this couple of moves we get that the resulting configuration belongs to the set Cmb (e, o) as long as R(Ond (σ)) winds around the torus, otherwise the saddle σ does not 25 satisfy the properties in the statement.
4.2.2 Communications between essential gates
In this subsection, we show that the six subsets composing the set C∗(e, o) communicate as illustrated in Fig. 5. The next proposition makes it precise.
Proposition 4.14. Any non-backtracking optimal path ω : e → o crosses the set C∗(e, o) in one of the following ways: (i) ω passes first through Cir (e, o), then through Cgr (e, o), and finally through Ccr (e, o);(ii) ω passes first through Csb (e, o), then through Cib (e, o) and afterwards through Cgr (e, o)∪C mb (e, o).If ω passes Cgr (e, o), then it eventually has to visit Ccr (e, o), otherwise it does not have to; (iii) ω passes first through Csb (e, o) and then through Cmb (e, o).Proof. Consider a non-backtracking optimal path ω : e → o. If ω visits at least one essential saddle, then we conclude by using Proposition 4.12. Thus, suppose that ω visits unessential saddles only, say σ1, ..., σ n. By definition of unessential saddle, we know that there exists another optimal path
ω′ : e → o such that S(ω′) ⊆ { σ1, ..., σ n−1}, say S(ω′) ⊆ { σ1, ..., σ m} with m ≤ n − 1. Iterating this argument, we deduce that there exists an optimal path ¯ω : e → o such that S( ¯ω) = {σ1} and this is a contradiction with the assumption that σ1 is an unessential saddle. Thus, we conclude that any optimal path ω : e → o visits the set C∗(e, o). It remains to prove that the entrance in C∗(e, o) occurs in one of the ways described in (i)–(iii), which easily follows by combining Lemmas 4.8–4.11.
4.2.3 All the saddles in C∗(e, o) are essential
In this first part of the proof, we will prove that every σ ∈ C ∗(e, o) is an essential saddle by constructing a non-backtracking optimal path ω : e → o that visits σ.Leveraging the fact that σ ∈ C ∗(e, o), we construct the desired non-backtracking path ω as a concatenation of two paths as follows. First, using a suitable concatenation of the paths described in Lemmas 4.9–4.10, we can define a path ω1 that starts from the considered configuration σ to the initial cycle Ce. We construct then another path ω2 that goes from σ to the target cycle Co as a suitable concatenation of the paths described in Lemmas 4.8–4.9. The desired non-backtracking path
ω is the time-reversal of ω1 concatenated with ω2 and it is easy to show that it is also optimal. Assume now by contradiction that σ is not essential, which means that there must exist another optimal path ω′ ∈ (e → o)opt such that S(ω′) ⊂ S(ω) \ { σ}. Recall that by Lemma 4.7(a), such a path ω′ that avoids σ still needs to visit the manifold Vm(σ) where σ lives at least once. Let η be any such configuration in Vm(σ) ∩ ω′. We claim that such a configuration η must satisfy ∆H(η) ≡ 1 (mod 2) .
This claim readily follows from Lemma 4.7(b) in combination with the facts that L is even and ∆H(σ) = L + 1 by construction. If ∆ H(η) ≥ L+3, then ω′ is not an optimal path, since Φ ω′ −H(e) ≥ L+3 > L +1 = Φ( e, o)−H(e). On the other hand, if ∆ H(η) ≤ L − 1, then from Lemma 4.1 it follows that η belongs to one of the two initial cycles. Proposition 4.14 ensures that every non-backtracking optimal path crosses C∗(e, o)in one of the three ways (i)–(iii) described therein, so that also the optimal path ω′ passing through η
has to visit C∗(e, o). Since, by assumption, the path ω′ has to avoid the saddle σ, we deduce that 26 there exists another saddle ˜η obtained starting from η that does not belong to Vm(σ). In particular, the two paths cross the set C∗(e, o) in three different ways according to Proposition 4.14(i)–(iii). Thus, we deduce that S(ω′)̸ ⊂ S(ω) \ { σ}.Thus, in view of the parity of ∆ H(η), we must have ∆ H(η) = L + 1, but then η is a saddle and, by construction, it did not belong to S(ω) and thus S(ω′)̸ ⊂ S(ω) \ { σ}.
5 Proof of auxiliary results
In this section, we give the proof of some auxiliary results stated in Sections 3–4.
5.1 Results on the perimeter of an odd region
Proof of Proposition 3.4. The proof revolves around the simple idea that using the filling algorithms introduced in Section 3.3 we can expand the odd cluster C (i.e., progressively increase the number of the occupied odd sites in C) in such a way that R remains the surrounding rhombus and the energy of all the configurations along such a path never exceeds H(σ) + 1. Since by assumption σ̸ = σ′, the odd cluster C cannot coincide with R, in view of conditions (i), (ii), and (iii). Thus, C contains at least a broken diagonal or a shorter diagonal than those of the surrounding rhombus. We can define the desired path ω as the concatenation of the two paths returned by the filling algorithms ˜ω and ¯ω. If C has no broken diagonal, we take ˜ω empty. If C has no shorter diagonal, then C has already the shape of a rhombus and therefore we take ¯ω empty. By the definition of these two paths, the energy increases by one only if an even site has to be emptied, but all these moves are followed by the addition of a particle in an antiknob. Therefore, the energy along the path ω
increases by at most one with respect to the starting configuration σ. This procedure ends when C
coincides with R, which implies that the resulting configuration is σ′.To conclude the proof, we need to show the properties claimed for the perimeter of σ. If C contains
m ≥ 1 broken diagonals, we argue as follows. Since the cluster C is connected, all the empty odd sites in which the diagonals are broken are antiknobs, i.e., they have n ∈ { 3, 4} neighboring even sites belonging to C, see Fig. 12. We distinguish the two following cases. If n = 3, then we first need to remove a particle from the unique neighboring occupied even site, like the site v represented in Fig. 12 on the left. After that move, when the particle is added in the antiknob the perimeter does not change in view of (3.3) , otherwise if n = 4 then the perimeter decreases (see Fig. 12 on the right). This occurs also when we add a particle in the target antiknobs except the last antiknob to obtain the complete diagonal, for which by construction we have that the perimeter decreases by 4. By iterating this argument for every broken diagonal, we get
P (σ) ≥ P (˜ σ) + 4 m > P (˜ σ).
If C does not contain any broken diagonal, we argue as follows. By construction, we deduce that the first odd site we will fill, which is the nearest neighbor of the first shorter diagonal, has three neighboring even sites belonging to C. Thus, we need to remove a particle from the unique occupied neighboring even site and then, when the first particle is added in the unique possible antiknob, the perimeter does not change thanks to (3.3). This occurs also when we iterate this argument.
Proof of Proposition 3.6. Given any n positive number, let C be an odd cluster with area n. If the cluster C is not connected, by (3.3) it directly follows that C cannot minimize the perimeter of an odd cluster with n particles. Indeed, when the cluster is not connected the cardinality of C ∩ Ve
decreases, while the cardinality of C ∩ Vo is fixed equal to n. Consider now a connected cluster C.27 v v
Figure 12: Example of a configuration σas in the statement of Proposition 3.4 (on the left), where we highlight in red the site containing the target antiknob, and the configuration obtained from σby filling it after removing a particle from the even site v(on the right), with highlighted in red the site containing the next target antiknob.
First, we will show that the minimal perimeter is that of the surrounding rhombus. To this end, we consider separately the following three cases: 1. there exists at least one broken diagonal. In this case, we have that P (C) > P (R(C)) by applying Proposition 3.4. 2. there is no broken diagonal but there exists at least one shorter diagonal with respect to those of the surrounding rhombus. In this case, we have that P (C) = P (R(C)) by applying Proposition 3.4. 3. all the diagonals have the same length as those of the surrounding rhombus, namely, all the diagonals are complete. In this case C = R(C), thus it is trivial that P (C) = P (R(C)). Lastly, we need to show that the minimizing rhombus is either Rs−1,s or Rs,s −1 if n = s(s − 1) + k
and Rs,s if n = s2 + k. We argue by induction over n. If n = 1, then it is trivial that the rhombus
R1,1 minimizes the perimeter and n = 1 can be represented in the form s(s − 1) + k choosing s = 1 and k = 0. Suppose now n > 1 and that the claim holds true for any m < n . Suppose that n − 1 can be written as s(s − 1) + k. In the other case, when n = s2 + k, we can argue in a similar way. If
n − 1 = s(s − 1) + k, with 0 ≤ k ≤ s − 2 (resp. k = s − 1), then either the rhombus Rs,s −1 or Rs−1,s
(resp. the rhombus Rs,s ) minimizes the perimeter for an odd cluster with n particles. Indeed, in any other case, the surrounding rhombus could be either Rs+1 ,s −1 or Rs−1,s +1 , which has a strictly greater perimeter in view of (3.13) . Note that the assumption n ≤ L(L − 2) is needed to avoid rhombi with a maximal side equal to L, because in view of (3.13) we would lose the uniqueness of the minimizing configuration.
Proof of Corollary 3.7. We first consider the case in which the area n is of the form n = s(s − 1) + k.If s < L/ 2, we have that the perimeter of the odd cluster is P = 4(2 s + 1). Since the cluster is contained in a rhombus Rs,s , we have that n ≤ s2 and, hence, ( P
4
− 1)2
≥ 4n. If L/ 2 ≤ s < L , then 28 the perimeter of the odd cluster is P = 4(2 L − 2s − 1). Since the cluster is contained in a rhombus Rs,s
and therefore the complement in V is a rhombus RL−s−1,L −s−1, we have that n ≥ L2
2
− (L − s − 1) 2.Hence ( P
4 − 1
)2
= 4( L − s − 1) 2 ≥ 2( L2 − 2n).
Consider now the other case, when the area n is of the form n = s2 + k. If s < L/ 2 we have that the perimeter of the odd cluster is P = 4(2 s + 2). Since the cluster is contained in a rhombus either
Rs+1 ,s or Rs,s +1 , we have that n ≤ s2 + s. Thus, it holds
(
P
4 − 1
)2
= (2 s + 1) 2 = 4( s2 + s) + 1 ≥ 4n + 1 .
If L/ 2 ≤ s < L , then the perimeter of the odd cluster is P = 4(2 L − 2s − 2). Since the cluster is contained in either a rhombus Rs+1 ,s or Rs,s +1 and therefore the complement in V is either a rhombus
RL−s−2,L −s−1 or RL−s−1,L −s−2, we have that n ≥ L2
2
− (L − s − 1)( L − s − 2). Hence,
(
P
4 − 1
)2
= (2 L − 2s − 3) 2 = 4( L − s − 1)( L − s − 2) + 1 ≥ 4
(
L2
2 − n
)
1 = 2( L2 − 2n) + 1 .
By using the condition n ≤ L(L − 2), inequality (3.19) directly follows.
Proof of Lemma 3.8. Let σ be a configuration with real area ˜n = 2 ℓ2 + 2 ℓ + 1. First, we suppose that the set of odd clusters in σ is composed only of j ≥ 2 non-degenerate clusters. Each of them has area
ni and perimeter pi for i = 1 , ..., j . Suppose by contradiction that σ has minimal perimeter so that the area of the configuration σ is nσ = ∑ji=1 ni and its perimeter is pσ = 4(2 √nσ + 1). By (3.19) , we have pi ≥ 4(2 √ni + 1) for any i = 1 , ..., j . Then we obtain that
pσ =
j∑
i=1
pi ≥
j∑
i=1
4(2 √ni + 1) ≥ 8
√√√√j∑
i=1
ni + 4 j ≥ 8
√√√√j∑
i=1
ni + 1
= 4(2 √nσ + 2) , (5.1) that is a contradiction. Second, we suppose that the set of odd clusters in σ is composed of k ≥ 1 degenerate clusters and of j ≥ 1 non-degenerate clusters. Each of these non-degenerate clusters has area ni and perimeter pi
for i = 1 , ..., j , so that nσ = ∑ji=1 ni. We denote by ˜pi for i = 1 , ..., k the perimeter of a degenerate cluster. Suppose by contradiction that σ has minimal perimeter. By (3.19) , we have pi ≥ 4(2 √ni + 1) for any i = 1 , ..., j and pσ = 4(2 √nσ + 1). Thus, we obtain
pσ =
j∑
i=1
pi +
k∑
i=1
˜pi ≥
j∑
i=1
4(2 √ni +1)+4 k ≥ 8
√√√√j∑
i=1
ni +4 j +4 k ≥ 8
√√√√j∑
i=1
ni +4+4 = 4(2 √nσ +2) , (5.2) that is a contradiction. Thus, we obtain that k = 0 and j = 1. Since the real area ˜n of the configuration
σ is fixed, then also the area nσ is fixed. Thus, given that σ contains only one non-degenerate cluster with minimal perimeter and with fixed area, by Corollary 3.7 we obtain that the non-degenerate cluster is the rhombus Rℓ,ℓ , which has precisely real area ˜ n.
5.2 Results on optimal reference paths
Proof of Lemma 4.8. We start by proving (i). The desired path ω = ( e, ω 1, ..., ω k(L), η ) is obtained as follows. Starting from e, define the configuration ω1 as that in which one particle is removed from 29 an empty even site, say v1 ∈ Ve. Similarly, we define ω2 as the configuration in which a particle is removed from a site v2 ∈ Ve such that d(v1, v 2) = 2. Similarly, by removing particles in v3, v 4 ∈ Ve
in such a way d(vi, v j ) = 2 for any i, j = 1 , ..., 4 and i̸ = j, we define the configurations ω3 and ω4.Note that ∆ H(ω4) = 4 < L + 1. Thus, we can define the configuration ω5 as that obtained from ω4
by adding a particle in the unique unblocked odd site, i.e., the one at distance one from vi for any
i = 1 , ..., 4. See Fig. 13 on the left.
v5
Figure 13: Example of the configurations ω5 (on the left), ω7 (in the middle) and ω13 (on the right) visited by the path described in the proof of Lemma 4.8(i).
We obtain that ∆ H(ω5) = 3 < L + 1 and ω5 is composed of a unique non-degenerate odd cluster, which is R1,1. To define the configuration ω6, we remove a particle from a site v5 ∈ Ve such that
d(vi, v 5) = 2 for two indices i = 1 , ..., 4. Similarly, we define ω7 in such a way there is an empty odd site with all the neighboring even sites that are empty, see Fig. 13 in the middle. Note that ∆H(ω7) = 5 < L + 1. Then, we define ω8 by adding a particle in the unique unblocked odd site, so that ∆ H(ω8) = 4 < L + 1. Note that ω8 contains an odd cluster R1,2. By growing the odd cluster in a spiral fashion, emptying the even sites that are strictly necessary, note that the configuration ω13
contains only the non-degenerate cluster R2,2 (see Fig. 13 on the right). Then, our path visits all the configurations which have a unique non-degenerate odd cluster C such that C = Rℓ,ℓ and C = Rℓ,ℓ +1
for any 3 ≤ ℓ ≤ L
2
− 1 up to the configuration ωk(L)−1, whose unique non-degenerate odd cluster is
C = RL
2−1,L
2−1
. Then, the configuration ωk(L) is defined by removing a particle from an even site w
at distance two from C. Finally, the configuration η is obtained by removing a particle from an even site at distance two from C and from w. Since the procedure we defined is invariant by translation, given a fixed configuration η ∈ C ir (e, o) is possible to choose the position of the odd clusters in such a way the final configuration of the path we described coincides with the desired η. It remains to show that arg max ξ∈ω H(ξ) = {η}. To this end, since ∆ H(η) = L + 1 and therefore ∆ H(ωk(L)) = L
and ∆ H(ωk(L)−1) = L − 1, we need only to show that max
ξ∈{ e,ω 1,...,ω k(L)−2}
H(ξ) < L + 1 .
First, we will show that ∆ H(η) = 3 + 2( ℓ − 1) by induction over the dimension ℓ = 1 , ..., L
2
− 1 of the rhombus Rℓ,ℓ composing the unique odd cluster of the configurations η visited by ω. We have already proven the desired property in the case ℓ = 1. Suppose now that the claim holds for ℓ, with 1 ≤ ℓ ≤ L
2
− 2, thus we will prove that it holds also for ℓ + 1. To reach the configuration displaying the rhombus Rℓ,ℓ +1 starting from Rℓ,ℓ , we need first to remove particles from two even sites by increasing the energy by two. Then, we sequentially add a particle in an odd site and remove a particle in an even site until the length of the shorter diagonal is ℓ − 1. Finally, the last move is the addition 30 of a particle in an odd site without the need of removing any particle from an even site. Starting from Rℓ,ℓ +1 , to obtain Rℓ+1 ,ℓ +1 we follow the same sequence of moves. Thus, for a configuration η
containing as unique odd cluster a rhombus Rℓ+1 ,ℓ +1 we deduce that ∆ H(η) = 3 + 2( ℓ − 1) + 2, which proves our claim for ℓ + 1. Along the sequence of moves from Rℓ,ℓ to Rℓ+1 ,ℓ +1 , the energy is at most 3 + 2( ℓ − 1) + 3, which is strictly less than L + 1 for ℓ ≤ L
2
− 1. Note that we do not to consider the case ℓ = L
2
− 2, indeed the path ω stops before reaching the rhombus RL
2−1,L
2
. This concludes the proof of (i). Now we prove (ii). The desired path ω = ( e, ω 1, ..., ω k(L), η ) is obtained as follows. Starting from
e, define the configuration ω1 as that in which one particle is removed from an empty even site, say
v1 ∈ Ve. Similarly, we define ω2 as the configuration in which a particle is removed from a site v2 ∈ Ve
such that d(v1, v 2) = 2. Similarly, by removing particles in v3, v 4 ∈ Ve in such a way d(vi, v j ) = 2 for any i, j = 1 , ..., 4 and i̸ = j, we define the configurations ω3 and ω4. Note that ∆ H(ω4) = 4 < L + 1. Thus, we can define the configuration ω5 as the one obtained from ω4 by adding a particle in the unique unblocked odd site, i.e., the one at distance one from vi for any i = 1 , ..., 4. We obtain that ∆H(ω5) = 3 < L + 1 and ω5 is composed of a unique non-degenerate odd cluster, which is R1,1,see Fig. 14 on the left.
v5v6
v7
Figure 14: Example of the configurations ω5 (on the left), ω7 (in the middle) and ω9 (on the right) visited by the path described in the proof of Lemma 4.8(ii).
Next, we describe four steps that are used in the following iteration. The first step is to define the configuration ω6 by removing a particle from a site v5 ∈ Ve such that d(vi, v 5) = 2 for two indices
i = 1 , ..., 4. The second step is to remove a particle from a site v6 ∈ Ve in such a way there is an empty odd site with two neighboring empty even sites and the other one is occupied. In this way, we obtain the configuration ω7 (see Fig. 14 in the middle). The third step is to obtain the configuration ω8 by removing the particle in the site v7 ∈ Ve such that d(v7, v 5) = d(v7, v 6) = 2. Note that ∆ H(ω8) = 6 < L + 1. Then, the last step is to define ω9 by adding a particle in the unique unblocked odd site, so that ∆ H(ω9) = 5 < L + 1 and the energy cost of these four steps is 2. Note that ω9 is composed of a non-degenerate odd cluster with two odd particles along either the same column or the same row, see Fig. 14 on the right. From this point onwards, we iterate these four steps for other L
2
− 3 times, until we obtain the configuration ωk(L)−1 in which either a column or a row contains L
2
− 1 odd particles. Note that ∆H(ωk(L)−1) = 5 + 2( L
2
− 3) = L − 1. Then, we repeat the first two steps described above and we reach the configuration η ∈ C sb (e, o) with ∆ H(η) = L + 1. It easy to check that arg max ξ∈ω = {η}.31 Indeed, we get ∆H(ωi) = ∆ H(ωi−1) + 1 if 1 ≤ i ≤ 4,
∆H(ω2i−1) = ∆ H(ω2i) − 1 if 3 ≤ i ≤ k(L)2 ,
∆H(ω2i) = ∆ H(ω2i−1) + 3 if 3 ≤ i ≤ k(L)2 − 1,
∆H(ωk(L)) = ∆ H(ωk(L)−1) + 2 ,
which concludes the proof of (ii).
Proof of Lemma 4.9. We start by proving (i). Take the path described in Lemma 4.8(ii) until it visits for the first time a configuration in Csb (e, o) as in Fig. 11 on the left. Starting from such a configuration, add a particle in the unique unblocked odd site. Then, remove a particle from an even site at distance one from an antiknob and finally add a particle in the unique unblocked odd site. Afterward, iterate the sequence of these two moves up to the target configuration η ∈ C ib (e, o). By construction, the resulting path has the desired property. Let us now focus on case (ii). Consider the path described in Lemma 4.8(i) until it visits for the first time a configuration in Cir (e, o). Starting from such configuration, add a particle in the antiknob, obtaining a configuration η′ that displays a unique non-degenerate odd cluster, which is a rhombus RL
2−1,L
2−1
with a single protuberance. Starting from η′, consider the path returned filling algorithm ¯ω up to the configuration η ∈ C gr (e, o). The described path has the desired property thanks to Proposition 3.4. Consider now the concatenation of the paths described in Lemma 4.8(ii) and Lemma 4.9(i) until it visits for the first time a configuration in Cib (e, o). Starting from such a configuration, by iterating the pair of moves consisting in adding a particle in the unique unblocked odd site and removing a particle at distance one from an antiknob, it is possible to obtain a configuration with a unique non-degenerate cluster that contains RL
2−1,L
2−1
. By construction, this resulting configuration is in
Cgr (e, o) and it is possible to iterate this couple of moves up to η. By construction, the resulting path has the desired property. Consider now case (iii). Take the path described in Lemma 4.8(ii) until it visits for the first time the configuration in Csb (e, o) as in Fig. 3 on the right. Starting from such a configuration, add a particle in the unique unblocked odd site. Then, remove a particle from an even site at distance one from an antiknob and finally add a particle in the unique unblocked odd site. This configuration is in
Cmb (e, o). Afterward, iterate the sequence of these two moves up to the target configuration η. By construction, the resulting path has the desired property. Consider now the concatenation of the paths described in Lemma 4.8(ii) and Lemma 4.9(i) until it visits for the first time a configuration in Cib (e, o). Starting from such configuration, with the same procedure described above it is possible to reach the target configuration η ∈ C mb (e, o) by visiting only saddles in Cib (e, o) ∪ C mb (e, o). Let us now focus on case (iv). Consider the concatenation of the paths described in Lemma 4.8(i) and Lemma 4.9(ii) until it visits for the first time a configuration in Cgr (e, o). Starting from such configuration, add a particle in the antiknob, obtaining a configuration η′. Note that η′ is composed of a unique non-degenerate odd cluster, which is a rhombus RL
2−1,L
2−1
with a single protuberance. Starting from η′, consider the path returned by filling algorithm ¯ω up to the configuration η ∈ C cr (e, o). The described path has the desired property thanks to Proposition 3.4. Consider now the concatenation of the paths described in Lemma 4.8(ii) and Lemma 4.9(iii) until it visits for the first time a configuration in Cmb (e, o). Starting from such configuration, by iterating the pair of moves consisting in adding a particle in the unique unblocked odd site and removing a particle at distance one from an antiknob, it is possible to obtain a configuration with a unique non-degenerate cluster that contains RL
2−1,L
2
. By construction, this resulting configuration is in 32 Ccr (e, o) and it is possible to iterate this couple of moves up to η. By construction, the resulting path has the desired property.
Proof of Lemma 4.10. We start by proving (i). Arguing as in the proof of Lemma 4.9(ii) we can show that there exists a path with the desired property that connects η to ¯η ∈ C cr (e, o), where the unique non-degenerate cluster of ¯η is RL
2−1,L
2
with attached a bar of length L
2
− 2 and there is a degenerate rhombus R0,0 at distance one from an antiknob. Since the configuration displays two antiknobs, it is possible to sequentially add two particles in odd sites. Thus, by proceeding in this way the path reaches o without visiting any other saddle and so the described path has the desired property. Finally, consider case (ii). By arguing as in the proof of Lemma 4.9(iii)–(iv) we can show that there exists a path with the desired property that connects η to ¯η ∈ C gr (e, o) ∪ C mb (e, o). If ¯η ∈ C gr (e, o), the claim follows by the previous argument. Otherwise, the configuration ¯η has two antiknobs after arguing as above. In either cases, the described path has the desired property.
Proof of Lemma 4.11. Consider first case (i). By construction, every non-backtracking optimal path from e to o that crosses a configuration in Cir (e, o) has to visit a configuration in Cgr (e, o). Indeed, each move consists in adding a particle in an unblocked odd site or removing a particle from an even site. Thus, it remains to consider backtracking optimal paths only. In order to visit a saddle in
Cib (e, o), a column (or row) with precisely L
2
− 1 particles arranged in odd sites needs to be created along these paths. To proceed, since we are considering backtracking optimal paths, the unique possibility is to visit a configuration in Csb (e, o) before reaching Cib (e, o). Thus, case (i) is concluded. Consider now case (ii). The claim follows after noting that every configuration in Ccr (e, o) contains one odd vertical (resp. horizontal) bridge B, where the two neighboring columns (resp. rows) to B
contains L
2
− 1 odd particles each. Thus, all the configurations in Ccr (e, o) differ from a configuration in Cib (e, o) in at least two odd sites. Then, suppose to have η ∈ C ib (e, o) and η′ ∈ C cr (e, o) such that they differ in only two odd sites. Starting from η, if a bridge is created, then the resulting configuration is in Cmb (e, o), while if the resulting configuration contains two neighboring columns (or rows) with exactly L
2
− 1 particles in odd sites, then it belongs to Cgr (e, o). If the two configurations differ in at least three sites, we argue as above. Consider now case (iii). The claim follows after arguing as in case (i) and noting that, in order to create a bridge, the path has to visit the set Csb (e, o). Lastly, in case (iv), the claim follows after arguing as in case (iii).
5.3 Results on the saddles lying in the manifold Vm∗
Proof of Lemma 4.13. We analyze separately the three cases.
Case (i). Suppose by contradiction that there exists a non-backtracking path ω′ ∈ (e → o)opt that crosses σ ∈ V m∗ \ C ir (e, o) such that R(Ond (σ)) and R(O(σ)) do not wind around the torus. First, since σ ∈ V m∗ and ∆ H(σ) ≤ L + 1, we get
e(σ) ≥ L2
4
− 32 ,o(σ) ≥ L2
4
− L + 32 . (5.3) Since o(σ) > 0 for any L, we deduce that σ cannot contain only degenerate clusters and therefore it contains at least an odd non-degenerate cluster. Thus, one of the following cases occurs: (1) O(σ) consists of at least two non-degenerate odd clusters; 33 (2) O(σ) consists of a single non-degenerate odd cluster different from RL
2−1,L
2−1
and possibly some degenerate clusters; (3) O(σ) consists of a single non-degenerate odd rhombus equal to RL
2−1,L
2−1
and at least one degenerate odd cluster R0,0 at distance greater than one from the non-degenerate one. We consider the rhombus surrounding the odd non-degenerate region for all the above cases. Due to the isoperimetric inequality of Lemma 3.8, this rhombus R has a perimeter P (R) greater than or equal to P (RL
2−1,L
2−1
), i.e., P (R) ≥ 4( L − 1). In particular, for cases (1) and (2) we have
P (R) > 4( L − 1) since O(σ)̸ = RL
2−1,L
2−1
, and for case (3) we have P (R) = 4( L − 1). Let Pi denotes the perimeter of the i-th cluster, with i = 1 , ..., k . Note that P (σ) = ∑ki=1 Pi. We let kd ≥ 0 (resp.
knd ≥ 1) the number of degenerate clusters R0,0 (resp. non-degenerate clusters) such that the total number of clusters is k = kd + knd ≥ 1.
Subcase (1). In this case knd ≥ 2, so that k ≥ 2. Denote by ˜ei (resp. oi) the number of empty even sites (resp. occupied odd sites) of the i-th cluster. Using (3.3), (3.7) and (3.8), we obtain ∆H(σ) =
k∑
i=1
(˜ ei − oi) =
kd∑
i=1
˜ei +
knd ∑
i=1
(˜ ei − oi) ≥ ˜ed +
(knd ∑
i=1
˜ei + 2( knd − 1) −
knd ∑
i=1
oi
)
= ˜ ed +
(knd ∑
i=1
˜ei −
knd ∑
i=1
oi
)
2( knd − 1) > ˜ed + ( L − 1) + 2( knd − 1)
≥ ˜ed + L + 1 ,
(5.4) where ˜ed = ∑kd
i=1
˜ei, the first inequality follows from the fact that the difference between the perimeter of two disjoint non-degenerate clusters and that of the cluster obtained by attaching the two is at least two, and the last inequality follows from the isoperimetric inequality applied to the cluster obtained by attaching the k clusters forming σ. Then, from inequalities (5.4) and ∆ H(σ) ≤ L + 1, it follows that ˜ ed < 0, which is a contradiction since ˜ ed ≥ 0.
Subcase (2). The unique non-degenerate odd cluster of σ has a shape different from a rhombus
RL
2−1,L
2−1
by assumption. In this case knd = 1 and kd ≥ 0, so that k = kd + 1. Thus, we obtain ∆H(σ) =
kd+1 ∑
i=1
(˜ ei − oi) =
kd∑
i=1
˜ei + (˜ ekd+1 − okd+1 ) > ˜ed + L − 1, (5.5) where ˜ed = ∑kd
i=1
˜ei, the first equality follows from the fact that σ contains only one non-degenerate odd cluster, and the last inequality follows from the isoperimetric inequality applied to the non-degenerate cluster. Then by (5.5) and the fact that ∆ H(σ) ≤ L + 1, we find ˜ ed ≤ 1 and so kd ≤ 1. The configuration σ thus falls in one of the following three subcases:
(2a) the only non-degenerate odd cluster of σ is a rhombus Rℓ1,ℓ 2̸ = RL
2−1,L
2−1
;
(2b) the non-degenerate odd cluster of σ is a cluster with m ≥ 1 empty odd sites corresponding to some broken diagonal and q = 0 shorter diagonals;
(2c) the non-degenerate odd cluster of σ is a cluster with m ≥ 0 empty odd sites corresponding to some broken diagonal and q ≥ 1 shorter diagonals. 34 We can ignore the case m = 0 and q = 0 as it is the one corresponding to the rhombus RL
2−1,L
2−1
.
Subcase (2a). This case is not admissible since, if Rℓ1,ℓ 2̸ = RL
2−1,L
2−1
, we have
ℓ1 + ℓ2 + 1 + kd = L + 1 ,ℓ1 + ℓ2 + 1 > L − 1,
and this implies kd > 2, which is in contradiction with the assumption kd ≤ 1.
Subcase (2b). From the assumptions on σ, it follows that
ℓ1 + ℓ2 + 1 + kd + m = L + 1 ,ℓ1 + ℓ2 + 1 > L − 1. (5.6) These give that kd + m < 2, which, in view of the inequalities kd ≤ 1 and m ≥ 1, then imply that
kd = 0 and m = 1. Thus, we deduce that
o(σ) = ℓ1ℓ2 − 1,e(σ) = ( ℓ1 + 1)( ℓ2 + 1) ,
so that ℓ1 + ℓ2 = L − 5 since σ ∈ V m∗ , but this contradicts the condition ℓ1 + ℓ2 + 1 + kd + m = L + 1 in (5.6).
Subcase (2c). From the assumptions on σ, it follows that
ℓ1 + ℓ2 + 1 + kd + m = L + 1 ,
2( ℓ1 + ℓ2 + 1) − q > 2( L − 1) . (5.7) This implies that 2( kd + m) + q ≤ 3, so we distinguish three subcases: (2c-I) kd = m = 0 and
q ∈ { 1, 2, 3}; (2c-II) kd = 1, m = 0 and q = 1; and (2c-III) kd = 0, m = 1 and q = 1. The other subcases are not possible in view of the conditions kd ≤ 1, m ≥ 0, and q ≥ 1. For subcase (2c-I), kd = m = 0 and q ∈ { 1, 2, 3}. Thus, by letting s be the total number of empty sites needed to be filled in order for the shorter diagonals to become complete, we deduce that
o(σ) = ℓ1ℓ2 − s, e(σ) = ( ℓ1 + 1)( ℓ2 + 1) − s,
so that ℓ1 + ℓ2 = L − 4 since σ ∈ V m∗ , but this contradicts identity ℓ1 + ℓ2 + 1 + kd + m = L + 1 in (5.7). The claims for (2c-II) and (2c-III) follow by arguing as in (2c-I).
Subcase (3). First, note that kd ≤ 2, otherwise ∆ H(σ) > L + 1 and therefore the path ω would be not optimal. If kd = 2, by using the non-backtracking property of the path ω, it follows that there is no possible move to cross the next manifold towards o along an optimal path. If kd = 1, then ∆ H(σ) = L so that the energy along the path can increase by at most 1 to reach
o. The unique possible move is to remove a particle from an empty site, obtaining a configuration with kd = 2. We can then prove the claim by arguing as in the case kd = 2.
Case (ii). Suppose by contradiction that there exists a non-backtracking path ω′ ∈ (e → o)opt that crosses that crosses σ ∈ V m∗ \ C ib (e, o) such that R(Ond (σ)) does not wind around the torus and
R(O(σ)) does. We observe that (5.3) holds for the configuration σ and, therefore, it contains at least one odd non-degenerate cluster and it cannot contain only degenerate clusters. Thus, we consider the following subcases: 35 (1) O(σ) consists of at least two non-degenerate odd clusters; (2) O(σ) consists of a single non-degenerate odd cluster different from RL
2−1,L
2−1
and possibly some degenerate clusters; (3) O(σ) consists of a single non-degenerate odd rhombus equal to RL
2−1,L
2−1
and at least one degenerate odd cluster R0,0 at distance greater than one from the non-degenerate one.
Subcase (1). Suppose that σ contains k ≥ 2 non-degenerate clusters C1(σ), ..., C k(σ). By assumption all the rhombi surrounding Ci(σ) does not wind around the torus for any i = 1 , ..., k , thus we can argue as in case (i-2c-II) above.
Subcase (2). By arguing as in case (i-2) above, we deduce that this case is possible only when
R(Ond (σ)) does not wind around the torus and R(O(σ)) does, so that from (5.3) we deduce that there exists only one column or row with less than L/ 2 particles, but this contradicts the fact that
σ / ∈ C ib (e, o).
Subcase (3). By arguing as in case (i-3) above, we deduce that this case is not possible.
Case (iii). Suppose by contradiction that there exists a non-backtracking path ω′ ∈ (e → o)opt
that crosses that crosses σ ∈ V m∗ \ C mb (e, o) such that both R(Ond (σ)) and R(O(σ)) wind around the torus. We observe that (5.3) holds for the configuration σ and therefore it contains at least one odd non-degenerate cluster and it cannot contain only degenerate clusters. Thus, we consider the following subcases: (1) O(σ) consists of at least two non-degenerate odd clusters; (2) O(σ) consists of a single non-degenerate odd cluster different from RL
2−1,L
2−1
and possibly some degenerate clusters; (3) O(σ) consists of a single non-degenerate odd rhombus equal to RL
2−1,L
2−1
and at least one degenerate odd cluster R0,0 at distance greater than one from the non-degenerate one.
Subcase (1). Suppose that σ contains k ≥ 2 non-degenerate clusters C1(σ), ..., C k(σ). If the rhombus surrounding Ci(σ) does not wind around the torus for any i = 1 , ..., k , we can argue as in case (i-1) above. Otherwise, suppose that there exists an index i such that R(Ci(σ)) winds around the torus. Thus, there exists at least one row or column that contains L
2
particles, say a column. Since
k ≥ 2, this implies that there exists a row containing two odd particles which belong to Ci(σ) and another disjoint cluster. Thus, the energy difference contribution ∆ H along this row or column is at least two. In addition, all the other L rows or columns composing Ci(σ) have an energy contribution of at least one. Thus, the total contribution is ∆ H(σ) > 2 + L − 1 = L + 1, where the strict inequality follows from k ≥ 2. This contradicts the assumption ∆ H(σ) ≤ L + 1 and therefore this case is not admissible.
Subcase (2). By arguing as in case (i-2) above, we deduce that this case is possible only when both R(Ond (σ) and R(O(σ) wind around the torus, so that from (5.3) we deduce that there exists at least one column or row with strictly less than L/ 2 particles, but this contradicts the fact that
σ / ∈ C mb (e, o).
Subcase (3). By arguing as in case (i-3) above, we deduce that this case is not possible. 36 6 Conclusions and future work
This work concludes the analysis of the metastable behavior for the hard-core model on a square grid graph with periodic boundary conditions initiated by . In that paper, this interacting particle system was shown to have two stable states and the energy barrier between them was already identified. However, the argument carried out in that paper did not provide any geometrical insight into the typical trajectories in the low-temperature regime and did not characterize the critical configurations. The goal of this paper was precisely to fill this gap. More precisely, we provide the geometrical description of all the essential saddles for this transition and we highlight how these communicate with each other without exceeding the critical energy barrier. The extension to other types of lattices naturally arises in this context. Indeed, in the authors investigate the same model on the triangular lattice studying the asymptotic behavior of the transition times between stable states but without providing any information on the critical configurations. The type of analysis carried out in this paper could be useful to tackle that problem, even if it looks more challenging since there are three stable states and isoperimetric inequalities are probably harder to derive. This will be the focus of future work. Another possible direction could be the study of the metastable behavior of the hard-core model on square grid graphs but with different boundary conditions or in the presence of some impurities. However, in this case, we expect the transition between two stable states to most likely occur by heterogeneous nucleation starting from the boundary, hence breaking the intrinsic symmetry and translation-invariance properties of the critical configurations. On the other hand, we expect that the techniques and the machinery developed in this paper to be useful to identify critical configurations also of other interacting particle systems on finite square lattices with similar blocking effects, e.g., the Widom-Rowlinson model .
Appendix A
Proof of Lemma 3.1. We start by proving (i). Given 0 ≤ k∗
1
≤ ℓ1 and ℓ2 + 1 ≤ j∗
1
≤ L − 1, we want to show that there exists 0 ≤ k∗
2
≤ ℓ1 and 0 ≤ j∗
2
≤ ℓ2 such that
k∗
1
j∗
1
= k∗
2
j∗
2
(mod L),k∗
1
− j∗
1
= k∗
2
− j∗
2
(mod L). (A.1) The choice j∗
2
= j∗
1
L/ 2 ( mod L) and k∗
2
= k∗
1
L/ 2 ( mod L) implies the claim in (i), since the chosen indices j∗
2
and k∗
2
satisfy (A.1) and they are modulo L such that
j∗
2
≥ ℓ2 + 1 + L
2
≥ 1,j∗
2
≤ L − 1 + L
2
≤ ℓ2 − 1,k∗
2
≥ L
2
,k∗
2
≤ ℓ1 + L
2
≤ L
2
− 2 ≤ ℓ1 − 2.
By interchanging the role of k∗
1
and k∗
2
and arguing in the same way, the proof of (i) is concluded. The two inclusions stated in (ii) can be proved by arguing in an analogous way.
Proof of Lemma 3.2. In this proof, all the sums will be tacitly assumed to be taken modulo L. Denote
η = ( η1, η 2) ∈ Vo. We analyze separately each case. 37 Case (i). Considering Sℓ1,ℓ 2 (η) ⊂ Vo and ∂+Sℓ1,ℓ 2 (η) ⊂ Ve, we observe that
V \ R ℓ1,ℓ 2 (η) = V \ { Sℓ1,ℓ 2 (η) ∪ ∂+Sℓ1,ℓ 2 (η)}
= {Vo ∪ Ve} \ { Sℓ1,ℓ 2 (η) ∪ ∂+Sℓ1,ℓ 2 (η)}
= {Vo \ { Sℓ1,ℓ 2 (η) ∪ ∂+Sℓ1,ℓ 2 (η)}} ∪ { Ve \ { Sℓ1,ℓ 2 (η) ∪ ∂+Sℓ1,ℓ 2 (η)}}
= {Vo \ Sℓ1,ℓ 2 (η)} ∪ { Ve \ ∂+Sℓ1,ℓ 2 (η)}. (A.2) Thus, we want to show that the two subsets are equal to Sˆl1,ˆl2 ( ˆη) and ∂+Sˆl1,ˆl2 ( ˆη) for some 0 ≤ ˆl1, ˆl2 ≤
L − 1 and ˆ η ∈ Ve. In addition, we may write
Ve = ⋃
0≤k′, j ′≤L−1
{(k′ + j′, k ′ − j′)} (A.3) and
∂+Sℓ1,ℓ 2 (η) = ⋃
0≤k≤ℓ1
0≤j≤ℓ2
{(η1 + k + j − 1, η 2 + k − j)}. (A.4) Thus, we have
Ve \ ∂+Sℓ1,ℓ 2 (η) = ⋃
0≤k′,j ′≤L−1
{(k′ + j′ + η1 − 1, k ′ − j′ + η2)} \ ⋃
0≤k≤ℓ1
0≤j≤ℓ2
{(η1 + k + j − 1, η 2 + k − j)}
= ⋃
ℓ1+1 ≤˜k≤L−1
ℓ2+1 ≤˜j≤L−1
{(˜k + ˜ j + η1 − 1, ˜k − ˜j + η2)}
= ⋃
0≤ˆk≤L−ℓ1−20≤ˆj≤L−ℓ2−2
{(ˆk + ˆ j + ℓ1 + ℓ2 + 2 + η1 − 1, ˆk − ˆj + ℓ1 − ℓ2 + η2}, (A.5) where at the second equality we used Lemma 3.1(i) and at the last equality we used the change of variables ˆk = ˜k − (ℓ1 + 1) and ˆj = ˜j − (ℓ2 + 1). Now, we consider ˆη = ( ℓ1 + ℓ2 + 1 + η1, ℓ 1 − ℓ2 + η2) ∈ Ve
and we obtain
Ve \ ∂+Sℓ1,ℓ 2 (η) = ⋃
0≤ˆk≤L−ℓ1−20≤ˆj≤L−ℓ2−2
{(ˆk + ˆ j + ˆ η1, ˆk − ˆj + ˆ η2)} (A.6) Thus, we have Ve \ ∂+Sℓ1,ℓ 2 (η) = SL−ℓ1−1,L −ℓ2−1( ˆη) ⊆ Ve. By arguing as above, we prove that
Vo \ Sℓ1,ℓ 2 (η) = ∂+SL−ℓ1−1,L −ℓ2−1(ˆ η) ⊆ Vo.
Case (ii). Without loss of generality we may assume ℓ1 = min {ℓ1, ℓ 2}. For η = ( η1, η 2) ∈ Ve, by using (A.3) and (3.11), we have
Ve = ⋃
0≤k≤ℓ1
0≤j≤L−1
{(k + j + η1, k − j + η2)} ∪ ⋃
ℓ1+1 ≤k≤L−10≤j≤L−1
{(k + j + η1, k − j + η2)}
= ⋃
0≤k≤ℓ1
0≤j≤L−1
{(k + j + η1, k − j + η2)} = ∂+Sℓ1,L −1(η). (A.7) Thus, it follows that all the even sites belong to the rhombus Rℓ1,ℓ 2 (η). In addition, by using (3.12) 38 we obtain
Sℓ1,L −1(η) = ⋃
0≤k≤ℓ1−10≤j≤L−2
{(η1 + k′ + j′, η 2 + k′ − j′)}
= ⋃
0≤k≤L
2−10≤j≤L−2
{(η1 + k + j, η 2 + k − j)} ∪ ⋃
L
2≤k≤ℓ1−10≤j≤L−2
{(ˆ η1 + k + j, ˆη2 + k − j)}
= ⋃
0≤k≤L
2−10≤j≤L−2
{(η1 + k + j, η 2 + k − j)} ∪ ⋃
L/ 2≤k≤ℓ1−1
{(ˆ η1 + k, ˆη2 + k)}. (A.8) Thus, we deduce that the rhombus Rℓ1,L −1(η) contains L2/2 − L + ℓ1 odd sites, which implies that its complement in V contains L − ℓ1 odd sites.
Case (iii). Without loss of generality we may assume ℓ1 = min {ℓ1, ℓ 2}. In this case, after using the same argument we have shown above, for some ˆ η = (ˆ η1, ˆη2) ∈ Vo we obtain that
Vo \ Sℓ1,L (η) = ⋃
ℓ1≤k≤L
2−10≤j≤L−1
{(ˆ η1 + k + j, ˆη2 + k − j)} (A.9) and
Ve \ ∂+Sℓ1,L (η) = ⋃
ℓ1+1 ≤k≤L
2−10≤j≤L−1
{(ˆ η1 + k + j − 1, ˆη2 + k − j)}. (A.10) This implies that the rhombus Rℓ1,L (η) contains L ℓ 1 odd sites and L(ℓ1 + 1) even sites.
Case (iv). By arguing as above, we can show that the complement of the rhombus Rℓ1,ℓ 2 (η) in V
has no even and odd sites and therefore Rℓ1,ℓ 2 ≡ V .
References
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doi : 10.1016/j.physa.2023.128811 . S. Baldassarri, A. Gaudilli` ere, F. den Hollander, F. R. Nardi, E. Olivieri, and E. Scoppola. “Droplet dynamics in a two-dimensional rarefied gas under Kawasaki dynamics.” In: arXiv preprint arXiv:2304.14099 (2023). S. Baldassarri and V. Jacquier. “Metastability for Kawasaki dynamics on the hexagonal lattice.” In: Journal of Statistical Physics 190.3 (2023), p. 46. doi : 10.1007/s10955-022-03061-8 . S. Baldassarri and F. R. Nardi. “Critical Droplets and Sharp Asymptotics for Kawasaki Dynamics with Strongly Anisotropic Interactions.” In: Journal of Statistical Physics 186.3 (2022). doi :
10.1007/s10955-022-02874-x . S. Baldassarri and F. R. Nardi. “Critical Droplets and sharp asymptotics for Kawasaki dynamics with weakly anisotropic interactions.” In: Stochastic Processes and their Applications 147 (2022), pp. 107–144. doi : 10.1016/j.spa.2022.01.011 .39 S. Baldassarri and F. R. Nardi. “Metastability in a lattice gas with strong anisotropic interactions under Kawasaki dynamics.” In: Electronic Journal of Probability 26.137 (2021). doi : 10.1214/21-ejp701 . J. Beltr´ an and C. Landim. “A Martingale approach to metastability.” In: Probability Theory and Related Fields 161.1-2 (2014), pp. 267–307. doi : 10.1007/s00440-014-0549-9 . J. Beltr´ an and C. Landim. “Tunneling and Metastability of Continuous Time Markov Chains.” In:
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184.3 (2021). doi : 10.1007/s10955-021-02814-1 . G. Bet, A. Gallo, and F. R. Nardi. “Metastability for the degenerate Potts model with negative external magnetic field under Glauber dynamics.” In: Journal of Mathematical Physics 63.12 (2022), p. 123303. doi : 10.1063/5.0099480 . G. Bet, A. Gallo, and F. R. Nardi. “Metastability for the degenerate Potts Model with positive external magnetic field under Glauber dynamics.” In: arXiv preprint, arXiv:2108.04011 (2021). G. Bet, V. Jacquier, and F. R. Nardi. “Effect of Energy Degeneracy on the Transition Time for a Series of Metastable States.” In: Journal of Statistical Physics 184.1 (2021). doi : 10.1007/ s10955-021-02788-0 . A. Bianchi, A. Gaudilliere, and P. Milanesi. “On Soft Capacities, Quasi-stationary Distributions and the Pathwise Approach to Metastability.” In: Journal of Statistical Physics 181.3 (2020), pp. 1052–1086. doi : 10.1007/s10955-020-02618-9 . A. Bianchi and A. Gaudilli ere. “Metastable states, quasi-stationary distributions and soft measures.” In: Stochastic Processes and their Applications 126.6 (2016), pp. 1622–1680. doi :
10.1016/j.spa.2015.11.015 . A. Bovier, F. den Hollander, and F. Nardi. “Sharp asymptotics for Kawasaki dynamics on a finite box with open boundary.” In: Probability Theory and Related Fields 135.2 (2005), pp. 265–310. doi : 10.1007/s00440-005-0460-5 . A. Bovier, F. Den Hollander, and C. Spitoni. “Homogeneous nucleation for Glauber and Kawasaki dynamics in large volumes at low temperatures.” In: The Annals of Probability 38.2 (2010), pp. 661–713. doi : 10.1214/09-AOP492 . A. Bovier, M. Eckhoff, V. Gayrard, and M. Klein. “Metastability and Low Lying Spectra in Reversible Markov Chains.” In: Communications in Mathematical Physics 228.2 (2002), pp. 219–255. doi : 10.1007/s002200200609 . A. Bovier and F. den Hollander. Metastability . Springer International Publishing, 2015. doi :
10.1007/978-3-319-24777-9 . M. Cassandro, A. Galves, E. Olivieri, and M. E. Vares. “Metastable behavior of stochastic dynamics: A pathwise approach.” In: Journal of Statistical Physics 35.5-6 (1984), pp. 603–634.
doi : 10.1007/bf01010826 .40 E. N. M. Cirillo, V. Jacquier, and C. Spitoni. “Homogeneous and heterogeneous nucleation in the three–state Blume–Capel model.” In: arXiv preprint arXiv:2305.17049 (2023). E. N. M. Cirillo and F. R. Nardi. “Relaxation Height in Energy Landscapes: An Application to Multiple Metastable States.” In: Journal of Statistical Physics 150.6 (2013), pp. 1080–1114. doi :
10.1007/s10955-013-0717-9 . E. N. M. Cirillo, F. R. Nardi, and J. Sohier. “Metastability for General Dynamics with Rare Transitions: Escape Time and Critical Configurations.” In: Journal of Statistical Physics 161.2 (2015), pp. 365–403. doi : 10.1007/s10955-015-1334-6 . E. N. M. Cirillo, F. R. Nardi, and C. Spitoni. “Competitive nucleation in reversible probabilistic cellular automata.” In: Physical Review E 78.4 (Oct. 2008). doi : 10.1103/physreve.78.040601 . E. N. M. Cirillo, F. R. Nardi, and C. Spitoni. “Metastability for Reversible Probabilistic Cellular Automata with Self-Interaction.” In: Journal of Statistical Physics 132.3 (2008), pp. 431–471.
doi : 10.1007/s10955-008-9563-6 . E. N. M. Cirillo, V. Jacquier, and C. Spitoni. “Metastability of Synchronous and Asynchronous Dynamics.” In: Entropy 24.4 (2022), p. 450. doi : 10.3390/e24040450 . D. Galvin. “Sampling independent sets in the discrete torus.” In: Random Structures & Algo-rithms 33.3 (2008), pp. 356–376. doi : 10.1002/rsa.20223 . A. Gaudilli` ere, F. den Hollander, F. R. Nardi, E. Olivieri, and E. Scoppola. “Ideal gas approxi-mation for a two-dimensional rarefied gas under Kawasaki dynamics.” In: Stochastic Processes and their Applications 119.3 (2009), pp. 737–774. doi : 10.1016/j.spa.2008.04.008 . D. S. Gaunt and M. E. Fisher. “Hard-sphere lattice gases. I. Plane-square lattice.” In: The Journal of Chemical Physics 43.8 (1965), pp. 2840–2863. doi : 10.1063/1.1697217 . F. den Hollander, E. Olivieri, and E. Scoppola. “Metastability and nucleation for conservative dynamics.” In: Journal of Mathematical Physics 41.3 (2000), pp. 1424–1498. doi : 10.1063/1. 533193 . S. Kim and I. Seo. “Approximation method to metastability: An application to non-reversible, two-dimensional Ising and Potts models without external fields.” In: arXiv preprint arXiv:2212.13746
(2022). C. Landim, D. Marcondes, and I. Seo. “A resolvent approach to metastability.” In: arXiv preprint arXiv:2102.00998 (2021). F. Manzo, F. R. Nardi, E. Olivieri, and E. Scoppola. “On the Essential Features of Metastability: Tunnelling Time and Critical Configurations.” In: Journal of Statistical Physics 115.1/2 (2004), pp. 591–642. doi : 10.1023/b:joss.0000019822.45867.ec . F. R. Nardi, E. Olivieri, and E. Scoppola. “Anisotropy Effects in Nucleation for Conservative Dynamics.” In: Journal of Statistical Physics 119.3-4 (2005), pp. 539–595. doi : 10.1007/s10955-004-3247-7 . F. R. Nardi, A. Zocca, and S. C. Borst. “Hitting Time Asymptotics for Hard-Core Interactions on Grids.” In: Journal of Statistical Physics 162.2 (2015), pp. 522–576. doi : 10.1007/s10955-015-1391-x . F. R. Nardi and C. Spitoni. “Sharp asymptotics for stochastic dynamics with parallel updating rule.” In: Journal of Statistical Physics 146.4 (2012), pp. 701–718. doi : 10.1007/s10955-011-0413-6 .41 F. R. Nardi and A. Zocca. “Tunneling behavior of Ising and Potts models in the low-temperature regime.” In: Stochastic Processes and their Applications 129.11 (2019), pp. 4556–4575. doi :
10.1016/j.spa.2018.12.001 . E. Olivieri and E. Scoppola. “Markov chains with exponentially small transition probabilities: First exit problem from a general domain. I. The reversible case.” In: Journal of Statistical Physics 79.3-4 (1995), pp. 613–647. doi : 10.1007/bf02184873 . E. Olivieri and E. Scoppola. “Markov chains with exponentially small transition probabilities: First exit problem from a general domain. II. The general case.” In: Journal of Statistical Physics
84.5-6 (1996), pp. 987–1041. doi : 10.1007/bf02174126 . E. Olivieri and M. E. Vares. Large Deviations and Metastability . Cambridge University Press, 2005. doi : 10.1017/cbo9780511543272 . A. Zocca. “Spatio-temporal dynamics of random-access networks: An interacting particle approach.” PhD thesis. Eindhoven University of Technology, Dec. 2015. A. Zocca. “Low-Temperature Behavior of the Multicomponent Widom–Rowlison Model on Finite Square Lattices.” In: Journal of Statistical Physics 171.1 (2018), pp. 1–37. doi : 10.1007/s10955-018-1961-9 . A. Zocca. “Tunneling of the hard-core model on finite triangular lattices.” In: Random Structures & Algorithms 55.1 (2018), pp. 215–246. doi : 10.1002/rsa.20795 . A. Zocca, S. C. Borst, J. S. van Leeuwaarden, and F. R. Nardi. “Delay performance in random-access grid networks.” In: Performance Evaluation 70.10 (2013), pp. 900–915. doi :
10.1016/j.peva.2013.08.019 .42 |
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Hypochlorous Acid – Structure, Properties, Uses Of HOCl
Nov 8, 2022
Hypochlorous acids are weak acids containing hydrogen, oxygen, and chlorine. Although a weak acid with pKa=7.5, hypochlorous acid is a strong oxidising agent. It is responsible for the killing action of pathogens by phagocytes.
What is HOCl?
Hypochlorous acid is a weak acid. Its chemical formula is HOCl.
Other names: It also has several other names, including:
Chlorine hydroxide
Hydrogen hypochlorite
Hypochloric acid
Hydroxidochlorine (the IUPAC name)
Chloric (I) acid
Chlorine hydroxide
Chloranol
It is often referred to as an oxyacid of chlorine.
Discovery: A French chemist, Antoine Ballard, discovered hypochlorous acid in 1834. Its monovalent chlorine functions as an oxidising agent or reducing agent. It serves as a human metabolite. Hypochlorous acid is an unstable acid. It is the conjugate acid of hypochlorite and belongs to the reactive oxygen family.
Hypochlorous Acid Formula and Structure
Hypochlorous acid formula is HOCl or HClO. The Lewis structure of hypochlorous acid is as follows:
Lewis Structure of Hypochlorous Acid
Thus, from the above structure, you can conclude as follows:
It has two covalent bond pairs.
It has five lone pairs of electrons.
There are ten non-bonding electrons in the HOCl.
The shape of HOCl is bent, as presented in the following diagram,
Structure of HOCl
The presence of the hydrogen atom in the compound suggests that it is an acid. It tends to get deprotonated, i.e., it loses its hydrogen ion.
Properties of HOCl- Hypochlorous Acid
The pKa value of hypochlorous acid is relatively high. It is 7.53. This negative log base ten of the acid dissociation constant suggests a lower acid dissociation. Hence, proving HOCl to be a weak acid. This acid dissociates partially into its constituent ions, and the inability is the reason for its weak nature.
HOCl acid is a colourless liquid at room temperature. Its chlorine is mild and fades away quickly. It is a polar molecule. The difference in electronegativities of the atoms and the electron lone pairs creates an uneven charge distribution. This asymmetrical distribution gives rise to polarity. The taste of hypochlorous acid is not precisely known, and its pure form is unstable. Hypochlorous acid exists majorly as an aqueous solution in water.
HOCl acid is an oxoacid, i.e., an acid containing an oxygen atom. Here is a summary of the important properties of HOCl acid.
Molar Mass or Molecular weight: 52.457 g/mol
Monoisotopic mass: 51.972 g/mol
No. of hydrogen bond acceptor: 1
Hydrogen bond donor: 1
Physical State: Liquid
Colour: Colourless
Solubility: Water Soluble
Odour: Pungent, chlorine-like
Acid Strength: Weak
Acidity pKa: 7.53
Corrosiveness: Corrosive
Toxicity: Toxic
Polarity: Polar
Shape: Bent
Calculating the molecular weight of Hypochlorous Acid
To calculate the molar mass or molecular weight of hypochlorous acid, you must know the atomic mass of its constituent atoms:
Chlorine = 35.5
Hydrogen = 1
Oxygen = 16
So, the molecular weight of HOCl can be calculated as follows:
= Molecular weight of hydrogen + oxygen + chlorine
= 1 + 16 + 35.5
= 52.5 g/mol
Thus, the molecular weight of HOCl is 52.5 g/mol.
Hypochlorous Acid Uses
As stated earlier, the main use of hypochlorous acid is as a disinfecting agent. However, its uses are not limited to killing germs. The acid finds various other applications in different industries. The following list summarises the key uses of HOCl acid:
It is an ingredient in several consumer items like deodorants.
It can be found naturally in the white blood cells of mammals, including humans.
In marine sanitation devices wherein seawater is converted to HOCl.
It is used in transforming alkenes into chlorohydrins.
HOCl is also used in cosmetics like baby products.
It is added to swimming pools.
In the manufacturing of safe disinfectants
It is popularly used for wound care purposes.
For treating several infections in humans and pets
HOCl has proven to cleanse, calm, and soothe the skin. The hypochlorous acid spray is often used for treating redness and other damaging changes from exposure to harmful substances.
Sodium hypochlorite is used in root canals in dentistry.
The Use of Hypochlorous Acid on Skin
HOCl exists in the human body. It is manufactured by white blood cells for defence against bacteria, infection, and general ickiness.
It attacks invading pathogens and serves as an antimicrobial acid. The human skin is tough and durable, acting as an outer protective barrier. This front-line defence often leaves your skin vulnerable to cuts, dust, scrapes, and more. HOCl offers your skin some extra backup by
Fighting bacteria that leads to clogged pores and acne.
Speeding up repair damage and wound healing.
Combating inflammation and skin problems like eczema or psoriasis.
Thus, the hypochlorous acid spray is incorporated into a daily skin maintenance routine to eliminate harmful bacteria. However, those with sensitive skin should stay away from HOCl sprays with fragrance.
Production of HOCl Acid
Hypochlorous acid is obtained when chlorine (Cl2) partially dissociates in water (H2O) to give a hypochlorite ion (ClO−). On protonation, it gets converted to HClO when oxygen bonds with hydrogen from the water molecules. The following equation demonstrates the formation of hypochlorous acid:
Cl2 + H2O ⇌ HClO + HCl
It must be noted that the formation of hypochlorous acid also gives rise to an aqueous solution of hydrogen chloride HCl.
Chemical Reactions
Following are some important chemical reactions of HOCl acid.
Dissociation
Hypochlorous acid dissociates partially into hypochlorite anion ClO− in aqueous solutions. The dissociation reaction is shown below:
HClO ⇌ ClO− + H+
This hypochlorite anion forms various salts. A common hypochlorite- sodium hypochlorite (NaClO) is an ingredient in bleach.
Oxidation
As stated in the important properties, hypochlorous acid is a strong oxidising agent. It is a stronger oxidant than chlorine, under standard conditions.
2 HClO(aq) + 2 H+ + 2 e− ⇌ Cl2(g) + 2 H2O
Reaction with HCl
It reacts with HCl or hydrochloric acid to give chlorine gas. The chemical reaction is as follows:
HClO + HCl → H2O + Cl2
Reaction with Ammonia
It reacts with ammonia to produce chloramines and water. The balanced reaction is as follows:
NH3 + HClO → NH2Cl + H2O
Reaction with Lipids
Hypochlorous acid reacts with lipids uniquely. It reacts only with unsaturated lipid bonds and not saturated bonds. The hypochlorite ion or ClO−does not participate in the reaction.
The reaction involves hydrolysis and the addition of chlorine to a carbon atom and hydroxyl to the other carbon atom. The process leads to the formation of chlorohydrin. Chlorohydrin formation in RBC’s lipid bilayers increases their permeability. So, disruptions can occur if more chlorohydrin is formed.
Is Hypochlorous Acid Safe for Use?
The EPA (Environmental Protection Agency, US) considers hypochlorous acid non-hazardous.
However, it is an oxidising agent and can be corrosive depending on its pH and concentration. Several clinical tests have been conducted to test hypochlorous acid water for eye and skin irritation. It was concluded that HOCl water is non-toxic and doesn’t cause irritation to the skin and eyes.
A 2017 study confirmed that a saline hygiene solution, when preserved with pure hypochlorous acid, had reduced bacterial load without altering the bacterial species’ diversity on the eyelids. There was a 99% reduction in bacteria after 20 minutes of treatment.
Furthermore, hypochlorous acid is quite unstable, and the active compounds easily deteriorate back into salt water. Thus, it readily loses its disinfecting capability. This property makes HOCl difficult to transport. Also, its high cost makes it less popular for use as a disinfectant compared to alcohol and bleach. With technological advancement, the manufacturing costs of hypochlorous acid water have been significantly reduced for home and commercial use. However, its short shelf life is still a matter of concern.
Conclusion
Hypochlorous acid is a highly useful mild acid. It contains chlorine, oxygen and hydrogen. It has a bent structure, and it is an unstable compound. Famous for its antibacterial properties, HOCl is safe for topical applications despite being an acid. It is an active wound care agent and its spray is also used for treating skin problems.
Frequently Asked Questions
Q1. What happens when an acid is added to the aqueous salt of HOCl?
A. When an is added to the aqueous salts of hypochlorous acid like sodium hypochlorite in bleach solutions, the reaction gets pushed to the left. It results in the formation of chlorine gas. The chlorine gas dissolves into simple water solutions like sodium hydroxide. Thus, solid hypochlorite bleaches are formed.
Q2. How does HOCl acid react with amino acids?
A. Hypochlorous acid and amino acids react readily because the latter possess side chains. The hypochlorous acid chlorine displaces hydrogen in amino acids, resulting in organic chloramine formation. However, chlorinated amino acids break down quickly in contrast to protein chloramines, which last longer.
Q3. Why is oxygen and not chlorine the central atom in HOCl?
A. The structure is usually constructed by determining the least electronegative atom. The least electronegative atom is then placed in the centre. In HOCl, Oxygen’s electronegativity is 3.44; chlorine’s is 3.16; Hydrogen’s electronegativity is 2.2. So, hydrogen is the least electronegative atom. However, it can’t be the central atom because it can only form a single bond. Chlorine comes next in terms of minimum electronegativity. Still, it cannot be the central atom as oxygen should be directly bonded to hydrogen, as chlorine in the centre would obstruct the structure. Thus, even with the highest electronegativity, oxygen is the central atom.
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11929 | https://fiveable.me/key-terms/intermediate-algebra/exponent-rule | printables
📘intermediate algebra review
key term - Zero Exponent Rule
Citation:
MLA
Definition
The zero exponent rule states that any nonzero number raised to the power of zero is equal to 1. This rule is an important property of exponents that applies in the context of both properties of exponents and scientific notation.
5 Must Know Facts For Your Next Test
The zero exponent rule applies to any nonzero base number, meaning that $x^0 = 1$ for any $x \neq 0$.
This rule is crucial in simplifying expressions with exponents, as it allows you to remove exponents of zero without changing the value of the expression.
The zero exponent rule is often used in the context of scientific notation, where numbers are expressed as the product of a number between 1 and 10 and a power of 10.
Applying the zero exponent rule can help simplify the representation of numbers in scientific notation, making them easier to work with.
Understanding the zero exponent rule is essential for mastering the properties of exponents and effectively using scientific notation in various mathematical and scientific applications.
Review Questions
Explain how the zero exponent rule applies to the properties of exponents.
The zero exponent rule is one of the key properties of exponents. It states that any nonzero number raised to the power of zero is equal to 1. This means that $x^0 = 1$ for any $x \neq 0$. This rule is important because it allows you to simplify expressions with exponents by removing exponents of zero without changing the value of the expression. For example, $5^0 = 1$, so $5^3 \cdot 5^0 = 5^3$. Understanding and applying the zero exponent rule is crucial for manipulating and simplifying expressions involving exponents.
Describe how the zero exponent rule is used in the context of scientific notation.
Scientific notation is a way of expressing very large or very small numbers using a base number between 1 and 10 multiplied by a power of 10. The zero exponent rule is particularly important in the context of scientific notation because it allows you to simplify the representation of numbers. For example, if a number is expressed in scientific notation as $5.0 \times 10^0$, the zero exponent rule tells us that $10^0 = 1$, so the number can be simplified to just 5.0. This makes the number easier to work with and understand, which is one of the main benefits of using scientific notation in the first place.
Analyze how the zero exponent rule can be used to evaluate expressions involving exponents in both properties of exponents and scientific notation.
The zero exponent rule is a fundamental property that can be applied to simplify expressions involving exponents in both the properties of exponents and scientific notation. In the context of properties of exponents, the rule states that $x^0 = 1$ for any nonzero base $x$. This allows you to remove exponents of zero without changing the value of the expression, which is useful for simplifying complex expressions. For example, $5^3 \cdot 5^0 = 5^{3+0} = 5^3$. In the context of scientific notation, the zero exponent rule can be used to simplify the representation of numbers. If a number is expressed as $5.0 \times 10^0$, the zero exponent means that the $10^0$ term is equal to 1, so the number can be simplified to just 5.0. This helps make scientific notation more concise and easier to work with. By understanding and applying the zero exponent rule in both properties of exponents and scientific notation, you can efficiently manipulate and evaluate a wide range of mathematical expressions.
Related terms
Exponent:
The exponent is the small number written to the right and above a base number, indicating the number of times the base is used as a factor.
Base:
The base is the number that is being raised to a power, or the number that the exponent is applied to.
Power: The power is the result of raising a base number to an exponent, or the value obtained by multiplying the base by itself the number of times indicated by the exponent.
"Zero Exponent Rule" also found in:
Subjects (3)
College Algebra
Elementary Algebra
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11930 | https://www.christianity.com/wiki/bible/why-we-should-heed-the-warnings-of-the-parable-of-the-rich-fool.html | The Parable of the Rich Fool - Meaning, Warning and Moral Taught | Christianity.com
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Christianity/Life/Bible/What Is the Meaning and Warning in the Parable of the Rich Fool?
What Is the Meaning and Warning in the Parable of the Rich Fool?
When Jesus taught, the Bible says He always told a story to illustrate the point. When addressing the danger of greed, Jesus uses the Parable of the Rich Fool (Luke 12).
Britt Mooney
Contributing Writer
Updated Aug 31, 2021
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When Jesus taught, the Bible says He always told a story to illustrate the point. When addressing the danger of greed, Jesus uses the Parable of the Rich Fool - found in Luke 12:16-21. It reads:
And he told them this parable: "The ground of a certain rich man yielded an abundant harvest. He thought to himself, 'What shall I do? I have no place to store my crops.' "Then he said, 'This is what I'll do. I will tear down my barns and build bigger ones, and there I will store my surplus grain. And I'll say to myself, "You have plenty of grain laid up for many years. Take life easy; eat, drink and be merry."' "But God said to him, 'You fool! This very night your life will be demanded from you. Then who will get what you have prepared for yourself?' "This is how it will be with whoever stores up things for themselves but is not rich toward God."
What Happens in the Parable of the Rich Fool?
Jesus was in the middle of teaching when a man shouted, “Teacher! Tell my brother to divide our father’s inheritance with me” (v.13).
The first response from Jesus was to explain that wasn’t His job. “Who made me judge in this case?” He then proceeded to give a warning: “Beware. Guard against every kind of greed. Life isn’t measured by how much you own.”
Then He told the Parable of the Rich Fool.
A wealthy man had a fertile field and an abundance of crops. But he didn’t have room to store the harvest. His solution? He would build bigger barns to keep it all, so much that he would be able to live in ease and comfort his whole life.
Then God entered the story and told him, “You fool! Your life is required of you tonight. Who will get to enjoy what you have?”
The story is a tragedy, where the man died and didn’t get to enjoy his wealth. Others did.
What Is the Meaning and Big Lesson of the Parable of the Rich Fool?
The tragedy of the story is that he could have enjoyed his wealth, even in death. Because death isn’t the end.
Jesus plainly states the initial lesson of the parable, that “life isn’t measured by how much you own.”
In context, the man who spoke out wanted what was due him. His brother should have divided the inheritance. However, as usual, Jesus saw past the issue and into the heart. He recognized the greed.
Jesus used the term “all kinds of greed.” Greed doesn’t have to look like stealing or cheating others (like the brother did). Greed can be expressed even in what appears good, right, or legal ways.
The big lesson for the Parable of the Rich Fool came in the next section of Luke 12. Jesus expounded upon true value, what really matters. The passage that follows is extremely similar to Matthew 6:19-34 where Jesus taught that we shouldn’t worry about our basic provisions, what we will eat and wear because life is more than food and clothing.
Our lives were compared to lilies and birds. If he takes care of them and loves us even more, then won’t he take care of us? This brought us to the famous statement in Matthew 6:33: “Seek first the Kingdom of God and His righteousness, and all these things will be added to you.”
In Luke 12, Jesus gave us the ultimate truth from the story, the one not immediately apparent. He said, “Sell your possessions and give to those in need. Then you will have treasure in Heaven! Nothing can destroy that treasure. No one can steal it. Where your treasure is, there your heart will be ” (v. 33-34).
What a gift. Through Christ, He literally told us how to have wealth that we can never lose. Even death can’t keep us from enjoying that treasure.
Why Was the Rich Fool in the Wrong?
For the wealthy farmer, there was nothing wrong with a massive harvest. The problem came when he made a decision to keep it all for himself. He had other opportunities for investment. Better opportunities.
Since he didn’t take advantage of those opportunities, his wealth profited him nothing.
There’s no guarantee on our stuff in this life. Yes, we can get insurance and that lessens the blow of loss and tragedy, and there’s wisdom in that. But just like the farmer in the story, we can lose it all in a moment. Houses burn down. Money gets stolen. We spend years building great things that can be destroyed in seconds.
There is this scientific principle called entropy. All things tend to death and destruction. This is the scientific version of the Fall and the resulting Curse of Death.
There’s so much loss in this life, and we feel it intensely. Time seems wasted. We consume and are hungry again. People with great potential die young.
Many years ago, my mentor and I were invited to the house of a wealthy man. My mentor and I were both musicians, songwriters, and not only was the house massive and overlooked a lake and a golf course, but it also had a state-of-the-art recording studio in a sprawling basement.
Maybe my mentor saw my eyes bulging. I don’t know. But he turned to me at one point and whispered, “It’s all going to burn.”
I knew what he meant.Don’t set your heart on it. It’s all temporary. It’s nice, but it’s still going to be nothing one day.
Centuries ago, there was this discipline called alchemy, the precursor to modern-day chemistry. Alchemists believed that they could take common materials like lead and make them precious metals like gold. They never did it, but that was the idea.
The Kingdom actually does this. With God, we can take common and temporary things, things that will be destroyed one day, and convert them to the immortal. Eternal reward. Precious treasure we can never lose.
How? Seek the Kingdom over all else. Be generous. Give to those in need, expecting nothing in return. Help the widows and orphans. Preach the Gospel of the Kingdom. Lose our lives for God and the Gospel and find our true life.
The man was a fool because his vision was too small. He placed trust in earthly possessions that have no security or lasting value. The farmer could have given it all away to the poor and had treasure in Heaven. That would have taken a trust that God would provide for his Earthly needs. With that trust, he could have traded the temporary for the eternal.
He could have traded his one marshmallow for two.
But he didn’t; he lost it all. He gained nothing. His life was marked not by reward but by tragedy and loss because he didn’t take advantage of the eternal alchemy of the Kingdom of God.
He was a fool.
4 Warnings Christians Should Heed from the Parable of the Rich Fool
The Parable of the Rich Fool is a tragedy, and tragedies show us what is good by warning us not to make bad choices. There are consequences for making foolish decisions.
What are the warnings from this parable?
1. Don’t equate our personal value with wealth.Oftentimes, we feel that if we have things that cost a great deal of money, then we have value. But that is a lie. We are far more valuable than those things. Infinitely more. We actually insult the created image of God when we attach our value to temporary things. For humanity, we are eternal beings and have eternal worth. That gives every life an inherent value that nothing can add to or take away. That’s true of everyone. You included.
The first step is realizing the immense and abundant love of God for us. God loved the world and sent His Son to save people because of our eternal worth to Him. Not the stuff in the world. When we live as if wealth gives us value, that lie will destroy us. By contrast, living from the Father’s love will bring us to life.
2. Don’t place our security or happiness in the temporary.The wealthy farmer looked at his harvest and thought, “I’m secure now!” He could live happy, eat, drink, be merry.
He never got to do that.
Temporary things are amazing, and God has given us so much in this life to enjoy. But they were never meant to be our security or the place we find our happiness. As I said above, there’s no security in them. Just as we do with our value, Jesus teaches us to find our security in something that’s real, something that lasts. Or to be clearer, a Person. God and His Kingdom, Heaven.
As eternal beings, we were meant to set our security on eternal things, hence why Jesus loves us enough to point us to that truth. “Seek first the Kingdom of God and His righteousness, and God’s provision will follow.” God and His Kingdom are the only things worthy of our trust.
3. Don’t worry about how your basic needs will be met.Fear and pride are at the root of all sin. It’s no different with the Parable of the Rich Fool. We seek to find our security in what we can provide or attain (pride), or we are anxious and worry about whether our needs will be met at all (fear). God resists the proud. And He has not given us a spirit of fear. Pride and fear lead us away from God, not to Him.
In Matthew 6, Jesus taught that worrying about our provision is an act of disbelief or unbelief. It’s what unbelievers do. Anxiety about our basic needs is the opposite of faith.
Faith rejects fear and lays hold of the Spirit of power, love, and a sound mind, the knowledge that the Father will always take care of us.
4. Don’t miss the opportunities to have wealth in Heaven.We have a Father who loves us and wants our eternal best. But we have an enemy, too. Our spiritual enemy seeks to kill, steal, and destroy. He lies to us and leads us to pride or fear to place our trust or value in temporary things and worry about the future. The Devil knows that if he can get our minds and hearts bound up in fear and lies, then we will miss God’s eternal best.
There are opportunities all around us, all the time, to invest in Heaven. When we first understand our inherent value to God, our security in His Kingdom that no one can take away, and that the Almighty promises to meet all our needs, then we are free to be generous. We often begin with money, and that is part of it, but generosity takes a myriad of different forms.
We can be generous with our time, hearts, words, hospitality, prayers, forgiveness, love, relationships, careers, and more. That generosity transforms the common and temporary into eternal treasure.
The Devil roams about seeking whom he can devour. God is different. He is seeking to reward even the simplest and smallest act of faith. Even a cup of cold water given to a prophet—in recognition that the person has an eternal gift in the Kingdom as a prophet—matters to God, and that person will never lose their reward.
Don’t miss those opportunities. Let’s renew our minds to remove our blinders of fear and pride and see clearly the investments we can make in the lives of people every day. Then we will be rich in Christ. We can never lose that. And we will enjoy those riches forever.
Peace.
Photo credit: ©GettyImages/todaydesign
Britt Mooney lives and tells great stories. As an author of fiction and non-fiction, he is passionate about teaching ministries and nonprofits the power of storytelling to inspire and spread truth. Mooney has a podcast called Kingdom Over Coffeeand is a published author of We Were Reborn for This: The Jesus Model for Living Heaven on Earth as well as Say Yes: How God-Sized Dreams Take Flight.
This article is part of our larger resource library of popular Bible parables. We want to provide easy-to-read articles that answer your questions about the meaning, origin, and history of parables within Scripture. It is our hope that these will help you better understand the meaning and purpose of God's Word in relation to your life today.
Parable of the Prodigal Son
Parable of the Talents
Parable of the Good Samaritan
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11931 | https://www.tandfonline.com/doi/full/10.2147/JIR.S460413 | Full article: Metformin Mitigates Sepsis-Induced Acute Lung Injury and Inflammation in Young Mice by Suppressing the S100A8/A9-NLRP3-IL-1β Signaling Pathway
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ORIGINAL RESEARCH
Metformin Mitigates Sepsis-Induced Acute Lung Injury and Inflammation in Young Mice by Suppressing the S100A8/A9-NLRP3-IL-1β Signaling Pathway
Shi-Yuan Fan1 The First Affiliated Hospital of Hunan Normal University (Hunan Provincial People’s Hospital), Changsha, Hunan, 410005, People’s Republic of ChinaView further author information
,
Zi-Chi Zhao1 The First Affiliated Hospital of Hunan Normal University (Hunan Provincial People’s Hospital), Changsha, Hunan, 410005, People’s Republic of ChinaView further author information
,
Xing-Lv Liu1 The First Affiliated Hospital of Hunan Normal University (Hunan Provincial People’s Hospital), Changsha, Hunan, 410005, People’s Republic of ChinaView further author information
,
Ying-Gang Peng1 The First Affiliated Hospital of Hunan Normal University (Hunan Provincial People’s Hospital), Changsha, Hunan, 410005, People’s Republic of ChinaView further author information
,
Hui-Min Zhu1 The First Affiliated Hospital of Hunan Normal University (Hunan Provincial People’s Hospital), Changsha, Hunan, 410005, People’s Republic of ChinaView further author information
,
Shi-Fan Yan2 Department of Emergency, Institute of Emergency Medicine, Hunan Provincial People’s Hospital (The First Affiliated Hospital of Hunan Normal University), Changsha, Hunan, 410005, People’s Republic of ChinaView further author information
,
Yan-Juan Liu2 Department of Emergency, Institute of Emergency Medicine, Hunan Provincial People’s Hospital (The First Affiliated Hospital of Hunan Normal University), Changsha, Hunan, 410005, People’s Republic of ChinaView further author information
,
Qin Xie2 Department of Emergency, Institute of Emergency Medicine, Hunan Provincial People’s Hospital (The First Affiliated Hospital of Hunan Normal University), Changsha, Hunan, 410005, People’s Republic of ChinaView further author information
,
Yu Jiang2 Department of Emergency, Institute of Emergency Medicine, Hunan Provincial People’s Hospital (The First Affiliated Hospital of Hunan Normal University), Changsha, Hunan, 410005, People’s Republic of China Correspondencejiangyu@hunnu.edu.cn
View further author information
&
Sai-Zhen Zeng1 The First Affiliated Hospital of Hunan Normal University (Hunan Provincial People’s Hospital), Changsha, Hunan, 410005, People’s Republic of China Correspondence55445321@qq.com
View further author information
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Pages 3785-3799 | Received 21 Mar 2024, Accepted 22 May 2024, Published online: 02 Jul 2024
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Introduction
Materials and Methods
Results
Discussion
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Abstract
Background
Globally, the subsequent complications that accompany sepsis result in remarkable morbidity and mortality rates. The lung is among the vulnerable organs that incur the sepsis-linked inflammatory storm and frequently culminates into ARDS/ALI. The metformin-prescribed anti-diabetic drug has been revealed with anti-inflammatory effects in sepsis, but the underlying mechanisms remain unclear. This study aimed to ascertain metformin’s effects and functions in a young mouse model of sepsis-induced ALI.
Methods
Mice were randomly divided into 4 groups: sham, sham+ Met, CLP, and CLP+ Met. CLP was established as the sepsis-induced ALI model accompanied by intraperitoneal metformin treatment. At day 7, the survival state of mice was noted, including survival rate, weight, and M-CASS. Lung histological pathology and injury scores were determined by hematoxylin–eosin staining. The pulmonary coefficient was used to evaluate pulmonary edema. Furthermore, IL-1β, CCL3, CXCL11, S100A8, S100A9 and NLRP3 expression in tissues collected from lungs were determined by qPCR, IL-1β, IL-18, TNF-α by ELISA, caspase-1, ASC, NLRP3, P65, p-P65, GSDMD-F, GSDMD-N, IL-1β and S100A8/A9 by Western blot.
Results
The data affirmed that metformin enhanced the survival rate, lessened lung tissue injury, and diminished the expression of inflammatory factors in young mice with sepsis induced by CLP. In contrast to sham mice, the CLP mice were affirmed to manifest ALI-linked pathologies following CLP-induced sepsis. The expressions of pro-inflammatory factors, for instance, IL-1β, IL-18, TNF-α, CXCL11, S100A8, and S100A9 are markedly enhanced by CLP, while metformin abolished this adverse effect. Western blot analyses indicated that metformin inhibited the sepsis-induced activation of GSDMD and the upregulation of S100A8/A9, NLRP3, and ASC.
Conclusion
Metformin could improve the survival rate, lessen lung tissue injury, and minimize the expression of inflammatory factors in young mice with sepsis induced by CLP. Metformin reduced sepsis-induced ALI via inhibiting the NF-κB signaling pathway and inhibiting pyroptosis by the S100A8/A9-NLRP3-IL-1β pathway.
Keywords:
sepsis
pediatric
inflammation
pyroptosis
ALI
Introduction
Globally, sepsis stands out as a substantial cause of both morbidity and mortality. Sepsis was defined in 2016 as a life-threatening organ dysfunction condition whose genesis is attributed to impaired host systemic inflammatory as well as immune response to infection,Citation 1 modulated via different cytokines produced from innate immune cells. Although over the past decades, a remarkable improvement has been achieved, sepsis syndrome stands out as a major cause of morbidity and mortality in pediatrics in the world,Citation 2–4 and a common reason for admitting patients to PICU.Citation 5 Severe sepsis in pediatrics is still a cumbersome public health problem. Previous works of literature have affirmed heightened risk for long-term disability, hospital readmission as well as late mortality for sepsis survivors among children.Citation 6 At least one-third of these children develop progressive organ dysfunction, whereas approximately 20% of them exhibit new functional disability.Citation 7 ALI, which stands out as one of the most susceptible organs to sepsis, exhibits the propensity to progress to ARDS, presenting with acute onset of sepsis, hypotension, severe respiratory distress, and persistent hypoxemia,Citation 8 which eventually worsens the dysfunction of multiple organs and as result death remains inevitable.Citation 9 During sepsis, activation of inflammatory cells releases various inflammatory mediators from innate immune cells, including TNF, IL-1, IL-6, and oxygen-free radicals, etc., which can severely harm the body’s organs and tissues.Citation 10–12 The exact molecular mechanism of lung injury caused by sepsis is still unclear, and the development of novel and effective therapeutic approaches is urgent. New sepsis therapeutics will be increasingly evaluated in children when they are developed.
Metformin, the AMPK activator has been revealed with anti-inflammatory effects in sepsis. Sepsis induces inflammation and oxidative stress leading to multi-organ damage through the production of proinflammatory cytokines and ROS. Metformin has been demonstrated to ameliorate sepsis-induced organ damage and mortality by acting on these mediators.Citation 13 In the present adult mouse models of sepsis, researchers observed that metformin can attenuate sepsis-associated brain injury,Citation 14 sepsis-related liver injury,Citation 15,Citation 16 cognitive impairment, and sepsis-induced neuronal injury,Citation 17 and endotoxin-induced acute myocarditis via activating AMPK. Increasing evidence implies that metformin exhibits a protective effect against various types of lung injury, including ventilator- and PM2.5-induced lung injuries,Citation 18,Citation 19 LPS-induced acute lung injury,Citation 20–22 sepsis-induced ALI,Citation 23 LPS-induced ARDS, and ARDS caused by SARS-CoV-2 infection.Citation 24 In COVID-19 cases, clinical trials have found that metformin can decrease COVID-19 severity and mortality via activating AMPK and/or inhibiting the mTOR-mediated signaling pathway primarily by lowering the level of pro-inflammatory signaling and cytokine storm.Citation 25 Recent reports of metformin attenuating hyperoxia-induced lung injury by lessening the inflammatory response in neonatal ratsCitation 26,Citation 27 have sparked our thoughts about the potential benefits of metformin in the pediatric sepsis population. Based on the previously documented antioxidant and anti-inflammatory functions of metformin, we hypothesized that metformin would be protective against sepsis-induced lung injury and death in young mice.
The S100 protein family serves as a potent amplifying factor for inflammatory responses. Mrp8 (encoded by Mrp8, also referred to as S100A8) and Mrp14 (encoded by Mrp14, also called S100A9), have a substantial function in sepsis pathogenesisCitation 28 and in neutrophils and monocytes, they are the most abundant cytoplasmic proteins.Citation 29 As pro-inflammatory molecules, they form a heterodimer and are referred to as calprotectin. Heightened expression of S100A8/A9 enhances the inflammatory response and facilitates neutrophils and macrophages to produce a lot of cytokines, which triggers a vicious cycle that makes the disorder worse. The NLRP3 (NLR family) inflammasome, a cytosolic multiprotein complex, controls the inflammatory cytokines IL-1β and IL-18.Citation 30 The pro-inflammatory alarmins S100A8 as well as S100A9 are powerful activators of the NLRP3 inflammasome.Citation 31–33 In vivo and ex vivo studies identifiedCitation 32 that the S100A8/A9-TLR4-NLRP3-IL1β/IL-18 signaling circuit played a crucial role in ponatinib-induced excessive inflammation and dysfunction of the heart. Pre-clinical models have affirmed the anti-inflammatory effects of metformin. Additionally, there is a lack of relevant reports on metformin treatment in pediatric sepsis models. Further investigation into the mechanistic role of metformin in septic young mice is warranted. Hence, this study sought to ascertain whether metformin contributes to sepsis-induced ALI by inhibiting the S100A8/A9-TLR4/NLRP3 inflammasome pathway in young mice. As far as we know, this marks the first investigation into the role and mechanism of metformin in sepsis-induced ALI in young mice. Looking at it from a clinical perspective, these findings support the feasibility of immunosuppressive interventions in treating sepsis-induced ALI among pediatric populations.
Materials and Methods
Animals Grouping
Male C57BL/6J mice (aged 3 weeks with 11 ± 0.7g, the Saike Jingda experimental Animal Co. Ltd, Changsha, Hunan, China) were kept in the Central Laboratory of Hunan Provincial People’s Hospital with 12 h light–dark cycle and regulated humidity and temperature and fed a standard rodent diet as well as water ad libitum. In the first experiment, all the animals (n = 20 per group) were randomized into four groups: sham group (sham), sham + Met group (sham +Met), CLP group (CLP), CLP + Met group (CLP +Met), and weighed every day and followed for 7 days to ascertain the survival rate. In the second experiment, animals were treated with metformin (Selleck Bio, S195013, 1.0), normal saline solution (NS), and randomized in the same four groups (n = 20 per group) to obtain lung tissue after 24h post-operation. The mice in Met groups were administered with metformin through intraperitoneal injection in a volume of 100μL. The CLP group was replaced by the same volume of saline. The befitting and optimal dose of metformin without detectable toxicity to experimental animals is based on previous reports.Citation 34,Citation 35 After 1 week of acclimatization, the mice were given either 200 mg/kg metformin dissolved in NS or NS alone by intraperitoneal injection for 7 consecutive days.
Animal Model of Sepsis-Induced ALI
As per the previous study, CLP was utilized to induce polymicrobial sepsis in mice.Citation 36 Briefly, mice were anesthetized with 2–3% isoflurane (Ringpu Bio, China). The cecum was exposed utilizing a midline surgical excision and ligated by a 4–0 silk suture. Then, a 21G needle punctured the cecum and the abdomen was closed in layers with 4–0 sutures. The cecum was mobilized without the usage of CLP for the sham-operated animals. Resuscitated animals by subcutaneous injection of a prewarmed normal saline (37°C, 50mL/kg) at the end of the surgical procedures, then mice were returned to cages immediately where accessed to water and food in freedom, with a temperature-controlled room (22°C) for 12h light and dark cycles and monitor them every 6h. All animal experiments followed the National Institutes of Health Guidelines for the Care and Use of Laboratory Animals. The Ethics Committee of Hunan Provincial People’s Hospital approved this study.
Survival Analysis and M-CASS
A Kaplan-Meier survival curve was produced and analyzed utilizing GraphPad Prism 9.0 software. To ascertain the severity of the disease in a sepsis-induced model, mice were observed after induction of sepsis every 6 h, based on the mice’s facial expressionsCitation 37 and behavior, then scored the mice per 24 h each day using the M-CASS as a previous study.Citation 38
Lung Wet Weight/Body Weight Coefficient
The left wet lung weight(mg) and body weight(g) of mice were obtained after CLP for 24h, then the left lung wet weight/body weight coefficient was developed and chosen to calculate the lung coefficient reported in previous studiesCitation 39,Citation 40 as wet lung weight/body weight.
Measurements of Histological Evidence of ALI
Extraction of the right middle lobes of the lung was executed, and the lobes were fixed using a 4% paraformaldehyde for 24h at 4°C. They were then dehydrated by ethanol, immersed in paraffin, and a microtome (Leica RM2125RT, Leica, Nussloch, Germany) was adopted to section the samples at 5μm thickness, followed by staining with hematoxylin and eosin. The Olympus BX 51 Leica ICC50W light microscope (Leica Microsystems Co., LTD Shanghai, China) recorded the sections. The scores of lung injury were evaluated in a blinded manner for three reviewers as previous study,Citation 41 for five independent variables in detail in Table 1. In brief, 20 random high-power fields (400× total magnification) of the sample were viewed at ×400 magnification, and then the resulting injury score between zero and one (inclusive) was normalized to the number of fields examined.
Table 1 Lung Injury Score
Download CSVDisplay Table
Quantitative Polymerase Chain Reaction (qPCR)
A total RNA extraction kit (TIANGEN, Beijing, China) was utilized to extract total RNA from mice lung tissues. RNA concentration was quantified spectrophotometrically. The PrimerScript RT reagent kit (Takara, Japan, Code No. RR036A) was employed to reversely transcribe RNA to cDNA as per the manufacturer’s instructions. The level was measured utilizing the TB Green mix (Takara, Japan, Code NO. RR420A) in a quantitative PCR. After that, over 40 cycles of qPCR were run utilizing the StepOne system. The PCR reaction conditions were: 95°C for 30s, 95°C for 5s, and 60°C for 30s. β-actin was as an internal control. The data were analyzed through 2−ΔΔCt. Primers were purchased from Takara and listed in Table 2.
Table 2 The Sequence of Primers for qPCR
Download CSVDisplay Table
Western Blotting Analysis
After being homogenized, lung tissues and cells were incubated in a lysis buffer that contained cocktail of protease inhibitors. Lung tissues were homogenized in a mixture containing protease inhibitors and phosphatase inhibitors of cracking incubation in the buffer. The tissue proteins were extracted by Radioimmunoprecipitation assay (RIPA) while measured using BCA assay, and at 100°C, the proteins were denatured for 5 min. The proteins (20μg per lane) were identified on a 10% SDS-PAGE gel and transferred to a polyvinylidene fluoride (PVDF) membrane. The blots were blocked with 5% nonfat milk at room temperature (RT) for 1h and incubated with primary antibodies for the whole night at 4°C. The primary antibodies against ASC, NLRP3, P65, p-P65, GSDMD, caspase-1 from antibody sampler kit (CST, #98303T), S100A8/A9 (Abcam, Cat NO: ab288715) and β-actin (Zen-Bio, Cat NO: 26441) at a dilution of 1:1000, correspondingly. The membrane was rinsed thrice with TBST for ¼ h before being incubated with a secondary anti-rabbit antibody for 1 h at RT. Again, the membrane was washed with TBST thrice. Ultimately, the Omega Lum C Gel Imaging System (Bio-rad) was utilized to detect and show distinct bands utilizing a chemiluminescent substrate. ImageJ software was utilized to quantify and evaluate protein band intensity.
Enzyme-Linked Immunosorbent Assays (ELISAs) of Cytokine
A 2% sodium pentobarbital (50 mg/kg, intraperitoneal) was applied to anesthetize the mice, then lungs were collected and stored at −80°C to furthermore ascertain the level of cytokine level utilizing IL-1β, TNF-α, IL-18 ELISA kits (Invitrogen, Carlsbad, CA, USA) as per manufacturer’s specifications. The results were measured using a microtiter plate reader at 450nm and 570nm, and then we subtracted the values of 570 nm from those of 450 nm and analyzed the data.
Statistical Analysis
Data were expressed as mean ± SEM and were calculated from multiple independent experiments which were triplicated. Statistical comparisons were made by one-way ANOVA utilizing GraphPad Prism 9.0 software. Tukey’s multiple comparisons test and Bonferroni adjustment were involved in correcting for multiple comparisons and executing post hoc analysis, correspondingly. The Student’s t-test and the log-rank (Mantel–Cox) test performed comparisons between the two groups and analyzed the survival rate, respectively. P <0.05 was considered statistically significant. All experiments were repeated at least three times.
Results
Metformin Improves the Survival Rate
After 12 hours of the CLP, septic mice exhibited signals typical to clinical sepsis, for instance, lethargy, piloerection, and tachypnea, and then progressively to bradypnea, increasingly labored breathing and reduced level of consciousness marked by reduced response to auditory as well as tactile stimuli. Between 2 and 7 days post-CLP induction, metformin enhanced the survival state of septic mice, which decreased M-CASS (Figure 1B, p < 0.001, CLP group vs CLP+ Met group), and the M-CASS of the CLP group was higher in comparison to the sham group (p < 0.001).
Figure 1 The impact of metformin pre-treatment on survival state. (A) Kaplan–Meier curves for time for 7 days post-operation (n = 20). (B) The M-CASS score changes with time during the CLP-induced sepsis 7-day follow-up (mean ± S.E.M), (C–E). Changes of the weight between pre-operation and post-operation. p<0.01, p<0.001, p<0.0001 (CLP+ Met vs CLP); ###p<0.001 (sham vs CLP).
Abbreviation: ns, non-significant.
Display full size
During the experimental period, the survival rate of the sham group and sham+ Met group was 100%. Metformin treatment is given for 7 days before CLP surgery (Figure 1A) affirmed a survival rate that was 45% greater than the untreated septic animals. The findings as per the Log rank test implied a substantial difference in cumulative survival among the four groups (p < 0.0001).
There were no variations in body weight of all the groups of mice on day 0 of induction of sepsis (Table 3, Figure 1C). Mice in the CLP-induced sepsis groups demonstrated a significant decrease in weight within 24 h of the operation (p < 0.001 vs sham group; Figure 1E), although there was no substantial difference between CLP group and CLP+ Met group in post-operation (Figure 1D and E).
Table 3 Changes of Weight and Coefficient in Four Groups
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Metformin Pre-Treatment Attenuated Sepsis-Induced ALI in Young Mice
In our research, the model of sepsis-induced ALI was successfully created by CLP operation. Histopathological alterations were evaluated by H&E staining. Our research showed that metformin had no adverse effect on the sham mice, including decreased weight and lung injury (Figure 1C–E). In comparison to the sham group, CLP group demonstrated alveolar septal thickening, inflammatory cell infiltration, hyaline membranes, proteinaceous debris filling the airspaces, which led to higher lung injury scores (Figure 2B, p < 0.0001) and lung wet weight/body weight coefficient (Table 3, Figure 2C, p < 0.001), while metformin reversed these adverse changes induced by CLP (Figure 2A–C). In brief, these results imply that metformin pre-treatment has protective effects on sepsis-induced ALI in young mice.
Figure 2 Metformin alleviated sepsis induced by CLP in mice lung injury. (A) At the end of 24 h following CLP, the histopathological lung injury was evaluated utilizing the HE staining (100x, bar = 200µm; 400x, bar = 50µm). (B) Histopathologic lung injury scores. (C) Lung wet weight/body weight within 24h CLP-induced sepsis. Met + CLP vs CLP, metformin treatment lowered the lung histopathological score in septic mice, and this suggests that this drug can improve the lung tissues with pathological injury. p<0.01, p<0.001, p<0.0001.
Abbreviation: ns, non-significant.
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Metformin Regulated Sepsis-Induced Inflammatory Cytokine Levels
We then delved into examining the fluctuations in inflammatory cytokines in lung 24 h following CLP and their expressions in the lung using qPCR and ELISA. The findings shown in Figure 3. CLP significantly potentiated pro-inflammatory cytokines including IL-1β, CCL3, and CXCL11 mRNA levels (Figure 3A and C), which were reversed by metformin pretreatment (CCL3 excepted). Moreover, ELISA assay revealed that metformin reduces the levels of pro-inflammatory cytokines in lung tissue like IL-1β, IL-18, and TNF-α (Figure 3G–I), respectively. Inflammatory response by the host during sepsis-associated ALI was mediated by pro-inflammatory and anti-inflammatory. The results suggested that metformin augmented the anti-inflammatory responses in CLP-induced ALI and down-regulated pro-inflammatory.
Figure 3 Metformin pre-treatment regulated inflammatory responses in sepsis-induced mice. (A–F) The levels of pro-inflammatory cytokines (IL-1β, CCL3, CXCL11 S100A8, S100A9), and the inflammasome NLRP3 (G–I) were measured by qPCR assay, (D–F) pro-inflammatory cytokines (IL-1β, TNF-α IL-18) were examined by ELISA assay in the lung tissues. p<0.05,p<0.01, p<0.001, p<0.0001.
Abbreviation: ns, non-significant.
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In the CLP group, inflammatory (pro-inflammatory neutrophils and macrophages) infiltrates in lung tissue, up-regulation the expression of pro-inflammatory cytokines S100A8, S100A9, and the inflammasome NLRP3 (Figure 3D–F), while metformin reverses this adverse impact. These findings demonstrate that metformin pre-treatment attenuates the immune response initiated by sepsis in the lung tissues. In a word, our data are suggestive of a potentially remarkable protective effect of metformin.
Metformin Pre-Treatment Inhibited the Expression of S100A8/A9 and Suppressed the NF-κB Signaling Pathway
The activation of NLRP3 and production of pro-IL-1β necessitate the activation of the NF-κB signaling pathway.Citation 42 Myeloid cells encompassing neutrophils and monocytes are the dominant reservoirs of calcium-binding proteins S100A8/A9, which can be released into extracellular space as DAMPs following an inflammatory stimulus.Citation 43 As an endogenous ligand of TLR4, S100A8/A9 induces NF-κB activation via TLR4-MD2 and fosters lethality throughout septic shock.Citation 28 Therefore, we examined if metformin hindered S100A8/A9-induced NF-κB activation. Sepsis caused a significant increase (p < 0.05) in lung tissue S100A8/A9, p65, and p-P65 levels measured by Western blot when compared with sham-operated mice. The metformin pre-treatment significantly reduced S100A8/A9 (p < 0.001), p65 (p < 0.05), and p-P65 (p < 0.001) levels in comparison to the CLP group (Figure 4A–F). Briefly, our data confirmed our hypothesis that metformin inhibits S100A8/A9-induced NF-κB activation.
Figure 4 Metformin exerts inhibitory effects on the NF -κB pathway by repressing the release of S00A8/A9 in inflammatory cells. (A–F) S00A8/A9, p65, and p-P65 protein quantification was performed through Western blot and was measured by densitometry. p<0.05,p<0.01, p<0.001, p<0.0001.
Abbreviation: ns, non-significant.
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Metformin Pre-Treatment Inhibited Pyroptosis in the Young Mice Lung Tissue
Pyroptosis denotes a lytic type of programmed cell death, necessitating the participation of membrane-damaging GSDMD, and is initiated by inflammatory caspases. These caspases are triggered within complex inflammasomes, particularly NLRP3, formed in reaction to pathogens as well as endogenous danger signals. NLRP3 inflammasome, comprising a sensor component such as NLRP3, an adaptor ASC, and the effector pro-caspase-1, regulates activation of caspase-1, which can mediate the cleavage of pro-IL-1β, pro-IL-18Citation 44 and GSDMD which induces pyroptotic cell death by pore formation.Citation 45,Citation 46
To investigate further the impact of metformin on pyroptosis in septic ALI, we ascertained the expression of NLRP3 inflammasome-linked proteins NLRP3, ASC, Pro-Caspase1, GSDMD-F, Pro-IL-1β, and GSDMD-N in the tissues of septic young mice lungs. Compared with the sham group, NLRP3, ASC, pro-caspase1, GSDMD-F, Pro-IL-1β, and GSDMD-N, in the CLP group were increased, while the expressions of the metformin pre-treatment group were decreased (Figure 5). These results suggest that metformin inhibited pyroptosis by suppressing NLRP3 inflammasome activation.
Figure 5 Metformin inhibits pyroptosis via suppressing NLRP3 inflammasome activation and suppressing the cleavage of GSDMD. (A–H) NLRP3, ASC, caspase-1, GSDMD-F, GSDMD-N, and IL-1β, protein quantification was performed through Western blot and was measured by densitometry. p<0.05,p<0.01.
Abbreviation: ns, non-significant.
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Discussion
Our research highlighted the crucial function of pyroptosis in the pathogenesis of sepsis-induced ALI young mice. We identified the vital role of the S100A8/A9-TLR4-NLRP3-IL-1β signaling circuit in sepsis-mediated inflammation and ALI in young mice. As presented in the schematic figure (Figure 6), sepsis triggers S100A8/A9 synthesis, which primes the NLRP3 inflammasome and excites the production of proinflammatory IL-1β. The antidiabetic therapeutic agent, metformin, manifested protective effects towards sepsis-induced ALI via inhibiting the expression of S100A8/A9.
Figure 6 Presentation of metformin protecting sepsis-induced ALI in young mice by repressing S100A8/A9-NLRP3-IL-1β signaling pathway.
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The most severe form of ALI, known as ARDS, is an unfavorable clinical complication secondary to bacterial sepsis, with a mortality rate of 40 to 60%.Citation 47 ALI/ARDS, marked by heightened lung inflammatory response, including inflammatory cells such as neutrophils and macrophages infiltrated, causes a vicious cycle that further potentiates the pooling of the aforementioned cells.Citation 48 Alveolar macrophages play crucial functions in the progression of ALI/ARDS through the production and release of numerous inflammatory mediators.Citation 49 Remarkable studies affirm that pyroptosis of alveolar macrophages may lead to the onset and advancement of ALI by mediating inflammation.Citation 50 Recent evidence indicates that reduced NLRP3 inflammasome suppresses pyroptosis as well as inflammatory cytokines, including IL-1β and IL-18, in alveolar macrophages, thereby improving lung injury.Citation 51,Citation 52 Metformin has been affirmed to exert cardiovascular protective effects via lessened NLRP3 protein expression and NLRP3 inflammasome activation.Citation 53 Previous research indicated that metformin is instrumental in abrogating pulmonary inflammation and NLRP3 inflammasome activation.Citation 20,Citation 24 In their study,Citation 32 Sultan Tousif et al elucidate that heightened expression of S100A8/A9 can trigger the inflammatory cascade via TLR4 activation, resulting in the excessive release of IL-1β from NLRP3 inflammasomes, consequently instigating myocardial and systemic inflammation. As anticipated, we documented that metformin suppressed NLRP3 inflammasome activation and NLRP3-stimulated pyroptosis induction, as evidenced by reduced levels of N-terminal fragment of GSDMD, pro-caspase-1, and protein contents of IL-1β in lung tissues of young mice treated with CLP.
S100A8 as well as S100A9 have already been verified to exhibit a decisive function in inflammation development. They are members of the S100 family, which was derived first from the neural proteins of the bovine brain in 1965. Neutrophils and monocytes secrete S100A8 and S100A9, which form a stable homodimer or heterodimer both in vitro and in vivo. The secretion of S100A8/A9 is primarily fueled by infection-induced inflammation. When a bacterial infection occurs, neutrophils, macrophages, and monocytes express markedly and produce S100A8/A9 to control inflammatory processes with the entry of ROS, inflammatory cytokines, and nitric oxide (NO).Citation 54 As DAMPs, alarmin S100A8/A9 are strong inflammasome priming agents, subsequently interacting with TLR4, prime the NLRP3 inflammasome, and stimulate the production of IL-1β, leading to myocardial infarction.Citation 55 The S100A8/A9-TLR4 axis could mediate coronavirus SARS-CoV-2 infection-induced disorder of antiviral innate immunity, subsequently leading to ARDS.Citation 56 A previous study indicated that S100A8/A9 in the plasma of patients following ALI brought about by sepsis or pneumonia significantly elevated as compared with healthy controls.Citation 57 S100A8/A9 has been established to be one of the most distinctive DAMPs in sepsis,Citation 58,Citation 59 which interacts with cell surface receptors on various immune cells, platelets, endothelial cells, and intracellular PRRs, exerting detrimental impacts on sepsis pathogenesis.Citation 43 Specifically, activation of the S100A8/A9 complex through binding with TLR4 can induce pro-inflammatory cytokine production and promote not only recruitment but also activation of immune cells, ultimately leading to tissue damage and dysfunctions of organs in sepsis. Due to its function in sepsis, S100A8/A9 has attracted our interest in its mediation of sepsis-induced ALI/ARDS. To ascertain if S100A8/A9 is also heightened in the sepsis-induced young mice, we evaluated S100A8/A9 in the tissue of young mice lung post 24h CLP by qPCR and Western blot. Sepsis-induced ALI revealed remarkably elevated lung tissue S100A8/A9 levels, while metformin decreased the expression of S100A8/A9 as compared with the sham group.
Increasing evidence suggests that the immune response is also initiated by endogenous ligands, also known as DAMPs or alarmins.Citation 58 DAMPs are intracellular molecules that predominantly engage in cell homeostasis. However, they can additionally serve as extracellular danger signals when they are produced by activated or damaged cells. During the progression of sepsis, S100A8/A9 has been considered a DAMP that results in constant immune stimulation by modulating various signaling pathways in distinct cells. Levels were also heightened in septic patients and exhibited an inverse correlation with survival.Citation 60 During the activation of phagocytes, S100A8/S100A9 complexes are released and mediate their actions via TLR4, resulting in the synthesis of TNF-α and other cytokines.Citation 28 Apart from SARS-CoV-2-infected animal models, inappropriate activation of the S100A8/A9-TLR4 signaling pathway was markedly induced in also in patients with COVID-19, repressing the buildup of aberrant neutrophils and rescued mice from ARDS.Citation 56 However, the specific mechanisms require further investigation and development. Research showed that blocking S100A9 could effectively decrease neutrophil infiltration and activation and shield against the development of edema and destruction of tissues during sepsis.Citation 61 Taken together, our results indicate that metformin blocking S100A8/A9-TLR4 reduces macrophage infiltration in addition to activation and protects against ALI in septic young mice. Thus, our study affirms the function of S100A8/A9 in septic lung damage and propounds that S100A8/A9 could exhibit a substantial function as a target to alleviate tissue damage and lung inflammation in abdominal sepsis in youth.
The NF-κB pathway encodes various transcriptional genes, leading to the production of pro-inflammatory cytokines. Additionally, it tightly controls the expression of NLRP3.Citation 62 NF-κB activation requires the breakdown of its distinct inhibitors, the inhibitor of NF-κB (IκB) proteins, after their phosphorylation by the IκB kinase (IKK) complex. The general types of NF-κB signaling pathways are classical and alternative pathways. The aforementioned two pathways are also denoted as canonical and noncanonical pathways, correspondingly.Citation 63,Citation 64 The triggers in the classical pathway are genotoxic stress and pro-inflammatory stimuli. DAMPs can promote the classical pathway and also exert a weaker activation on the alternative pathway in response to inflammatory stimuli,Citation 64 subsequently releasing cytokines including IL-1, IL-6, and TNF into extracellular space. In a word, as a sensor for PAMPs, TLR4 can activate the NF-κB signaling pathway, inducing the production of chemokines and cytokine precursors and releasing mature pro-inflammatory factors under the influence of NLRP3 inflammasomes.Citation 65 In the present research, we have revealed that NF-κB dimers (p65/p50) with a key function in the induction of genes taking part in inflammation were liberated during metformin pre-treatment. In addition, the present results demonstrated that NF-κB is elevated following CLP treatment, while metformin pre-treatment down-regulated the level of NF-κB. Prior researchCitation 66 found that stimulation with S100A8/9 lead to increase the expression level of TLR4. Zheng et al showed thatCitation 67 S100A8/A9 activated the NF-kB signaling pathway via toll-like receptor-4, precipitating intervertebral disc degeneration and inflammation-associated pain in rats. Briefly, S100A8/9 can activate the NF-kB signaling pathway through TLR4. Therefore, our data showed that the metformin pre-treatment may inactivated the NF-κB pathway via S100A8/9-TLR4 in the young mice.
Above all, our current study suggested that metformin may confer protection against sepsis-induced lung injury by downregulating inflammatory genes and inhibiting the expression of pyroptosis-related proteins. The regulatory role of metformin may also involve other receptors or signaling molecules, necessitating further experimentation.
Certainly, our study also has certain limitations. Firstly, our experimental data are derived from experiments conducted on rodents, and there are species differences between rodents and humans, which will require further research to validate its effectiveness. Secondly, the administration of metformin for 7 days in our study serves as pretreatment, and future administration post-sepsis occurrence may better approximate clinical usage. Lastly, the limited sample size in this study encompasses only single types, overlooking the distinct pathological characteristics among clinical patients. Improving sample diversity is crucial to effectively evaluate clinical applicability.
Conclusion
In the present research, we investigated for the first time and unearthed the function of metformin against sepsis-induced mortality and its pulmonary protective effects in young mice. Overall, our findings discovered that metformin may deactivate the S100A8/A9-NLRP3-IL-1β pathway as well as NLRP3 inflammasome to lessen the damage to lung tissue in septic young mice. As a result, metformin may help treat sepsis-induced ALI in pediatrics as per our findings.
Ethics Approval and Consent to Participate
The Ethics Committee of Hunan Provincial People’s Hospital (The First Affiliated Hospital of Hunan Normal University) approved this study.
Author Contributions
All authors made a significant contribution to the work reported, whether that is in the conception, study design, execution, acquisition of data, analysis and interpretation, or in all these areas; took part in drafting, revising or critically reviewing the article; gave final approval of the version to be published; have agreed on the journal to which the article has been submitted; and agree to be accountable for all aspects of the work.
Disclosure
The authors report no potential conflicts of interest in this work.
Data Sharing Statement
The datasets of the current study are available from the corresponding author upon reasonable request.
Additional information
Funding
This project was financially supported by the Natural Science Foundation of Hunan Province (2023JJ60101).
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11932 | https://alinhana.lakeheadu.ca/Lithium.pdf | Lithium: fine structure, internal magnetic field and Zeeman Effect PHYS2332 Modern Physics II W2018 Estimating the Internal Magnetic Field Method I Recall that the fine structure of atoms is due to the coupling between the electron’s spin with the internal magnetic field, !
Bint , of the atoms. To show how to estimate !
Bint for specific atomic states, consider the fine structure of Lithium discussed in class: 2 2S1/2 In the diagram, a horizontal line indicates a state of lithium, with the symbol,n 2LJ . The doublet states 2 2P 3/2 and 2 2P 1/2 have different energies due to an internal field !
Bint arising from the orbital angular momentum of a 2p ( L = 1) electron. Compare this to the degenerate 2 2S1/2 where there is no energy splitting (differences) since the internal field due to the 2s ( L = 0 ) is !
Bint = 0 . As discussed in class the internal field is proportional to the orbital angular momentum !
Bint ∝ !
L (problem 27 of chapter 8), and all S states with the same principal quantum number, n, are degenerate (i.e. have the same energy). The diagram above shows the transition 2 2P 3/2 →2 2S1/2 (dashed line) that emits a photon of wavelength λ1 = 670.775nm , and of the 2 2P 1/2 →2 2S1/2 (dotted line) with wavelength λ2 = 670.790nm . To use this data to estimate Bint , consider the diagram below: !
Bint !
S - electron’s spin Interaction energy Eint = −!
µSi !
Bint = e m BintSz , m is electron’s mass Recall that the z-component of the spin is Sz = ms!, ms = ±1/ 2 Eint = 2µBBintms , where µB = e!
2m is the Bohr’s magneton The SI value of µBis given in A1, but we can also use µB = 5.788 ×10−5eV / T !
µs = −e m !
S ≡magnetic moment due to electron’s spin Lithium Let the energy of the photon be emitted by the 2 2P 3/2 →32S1/2 transition be E1 = hc λ1 , with λ1 = 670.775nm . Let the energy of the photon be emitted by the 2 2P 1/2 →32S1/2 transition be E2 = hc λ2 , with λ1/2 = 670.790nm . Using hc = 1239.8eV / nm , the energy difference, due to spin-orbit coupling, is ΔE = hc 1 λ1 −1 λ2 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟= 1239.8eVinm ( ) 1 670.775nm − 1 670.790nm ⎛ ⎝ ⎜ ⎞ ⎠ ⎟= 4.21×10−5eV The energy splitting of the 2 2P 3/2 and 2 2P 1/2 is due mostly to the spin-orbit coupling with the internal magnetic field, Bint . Assume that the 2 2P 3/2 state has spin up ms up = 1/ 2 and the 2 2P 1/2 state has spin down ms down = −1/ 2, then the change in energy is ΔE = Eint up −Eint down = 2µBBint ms up −ms down ( ) = 2µBBint (see example 8.3 on page 285) . Hence we have Bint = ΔE 2µB = 4.1×10−5ev 2 × 5.788 ×10−5eV / T = 0.35T . Estimating the Internal Magnetic Field Mehod II In class, I made a semi-classical argument that the internal magnetic field can be approximated by a ring current model to give Bint = µ0eL 4πmr3 , where µ0 = 4π ×10−7Tim / A is the magnetic permeability in vacuum, and m = 9.11×10−31kg is the mass of the electron. Using the formula Bint = µ0eL 4πmr3 , with L = ℓℓ+1 ( )" = 2" ( ℓ= 1for 2p) and from example 7.11 on page 265, replace r = 4a0 (most probable distance for 2p state). This gives Bint = 2µ0e!
4πm 4a0 ( ) 3 = 2 4π ×10−7 NiA−2 ( ) 1.6 ×10−19C ( ) 1.0546 ×10−34 Jis ( ) 4π 9.1×10−31kg ( ) 4 × 5.2918 ×10−11m ( ) 3 = 0.28T This is actually remarkably close to the answer obtained by method I. The Anomalous Zeeman Effect The left hand side (LHS) diagram below shows an atom with spin !
S and orbital angular momentum !
L in an external magnetic field, !
Bext = Bext ⌢ k that points along the z axis. The right hand side (RHS) diagram shows the correct illustration of how the atom interacts with !
Bext ; it interacts with the atoms total angular momentum !
J . !
Bext !
µL !
Bext !
µJ !
S !
µs !
L !
J Using quantum mechanics (QM) the magnetic moment of the electron’s orbital angular momentum is !
µL = −e 2m !
L and the electron’s spin is !
µS = −e m !
S . It is then tempting to assume that the actual magnetic moment of the atom is found by simply adding these two contributions, but this would be wrong, since the spin and orbital part of the angular momentum interact. Instead an atom behavior in a magnetic field is determine by its total angular momentum !
J , with a magnetic moment !
µJ = −e" 2m g !
J " = −µBg !
J " with µB = 5.788 ×10−5eV / T being the Bohr magneton, and g = 1+ J J +1 ( ) + S S +1 ( ) −L L +1 ( ) 2J J +1 ( ) (equation 8.23) is the Lande g factor. Hence when an atom is in an external magnetic field that points along the z-axis !
Bext = Bext ⌢ k the Zeeman interaction energy is Vzeeman = −!
µJ i !
Bext = µBgBext Jz " . Using the quantization rule Jz = mJ! , with mJ = ±J,± J −1 ( ),... gives Vzeeman = µBgBextmJ , as in equation 8.23. The selection rules for transitions between atomic states in an external field combine equations 8.14 and 8.24. Study the examples in section 8.3 on page 292 to 294. Below are some examples on the Zeeman effect on Lithium, Example of Anomalous Zeeman effect in Lithium Suppose a lithium atom in the state is placed in an applied magnetic field of Bext = 0.5T . Calculate the energy different between adjacent lines due to the Zeeman effect of, A) the adjacent levels of 2 2P 3/2 , and B) the adjacent levels of 2 2P 1/2 . C) If there were no spin how many levels would the P state ( L = 1) have? D) Briefly explain how the results of part A, B, and C illustrate the anomalous Zeeman effect. A) For the 2P 3/2 state of Lithium J = 3 2 and mJ = 3 2 , 1 2 ,−1 2 ,−3 2 . The Lithium Zeeman splitting is similar to Figure 8.16 (on sodium) of example 8.10 (PP. 293-294), where we can clearly see that the 2P 3/2 is split into four states. The Zeeman energy of these levels are given by equation 8.22 (page 293) Vz = µBBextgmJ . The energy separation between adjacent levels (for example mj = 3 / 2 and mj = 1/ 2 , or mj = 0 and mj = −1/ 2 ) is simply: ΔVz = µBBextgΔmj = µBBextg, where Δmj = 1 for adjacent levels. Note that for 2P 3/2 state of sodium J = 3 2 , L = 1, and S = 1/2, the Lande factor is calculated on p. 294, g = 1.33. For Bext = 0.50T , we use µB = 5.788 ×10−5eV / T and g = 1.33 to obtain for the energy difference between adjacent Zeeman 2P 3/2 levels, ΔVz = µBgBext = 5.788 ×10−5eV / T 0.5T ( ) 1.33 ( ) = 3.84 ×10−5eV B) For the 2P 1/2 state of Lithium J = 1 2 and mJ = 1 2 ,−1 2 . The Lithium Zeeman splitting is similar to Figure 8.16 (on sodium) of example 8.10 (PP. 293-294), where we can clearly see that the 2P 3/2 is split into four states. The Zeeman energy of these levels are given by equation 8.22 (page 293) Vz = µBBextgmJ . The energy separation between adjacent levels (for example mj = 3 / 2 and mj = 1/ 2 , or mj = 0 and mj = −1/ 2 ) is simply: ΔVz = µBBextgΔmj = µBBextg, where Δmj = 1 for adjacent levels. Note that for 2P 1/2 state of sodium J = 1 2 , L = 1, and S = 1/2, the Lande factor is calculated on p. 294, g = 0.67. For Bext = 0.50T , we use µB = 5.788 ×10−5eV / T and g = 0.67 to obtain for the energy difference between adjacent Zeeman 2P 3/2 levels, ΔVz = µBgBext = 5.788 ×10−5eV / T 0.5T ( ) 0.67 ( ) = 1.94 ×10−5eV C) If there were no spin the Lithium 2P levels will be ℓ= 1will have three levels mℓ= ±1,0 . The Zeeman energy level is Vz = µBBextmℓ, since g = 1 when there’s no spin. D) Without spin there are three Zeeman levels. With spins there are 6 Zeeman levels. The transitions between these Zeeman levels can be observed in spectroscopic experiments. The discrepancy between the number of observed lines in experiments and in the number predicted by theory that assumes no spin, is called the anomalous Zeeman effects. |
11933 | https://www.idea.int/sites/default/files/2024-07/sdg16-as-an-enabler-of-2030-agenda.pdf | SDG 16 AS AN ENABLER OF THE 2030 AGENDA Policy Paper No. 32, July 2024 SDG 16 AS AN ENABLER OF THE 2030 AGENDA Policy Paper No. 32, July 2024 Stacey Cram International IDEA Strömsborg SE–103 34 Stockholm SWEDEN +46 8 698 37 00 info@idea.int www.idea.int © 2024 International Institute for Democracy and Electoral Assistance International IDEA publications are independent of specific national or political interests.
This Policy Paper provides a current literature review and an in-depth meta-analysis of the evidence base that underpins assertions of interlinkages between Sustainable Development Goal 16—as an enabler— and the 2030 Agenda’s other Sustainable Development Goals.
With the exception of any third-party images and photos, the electronic version of this publication is available under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 (CC BY-NC-SA 4.0) licence. You are free to copy, distribute and transmit the publication as well as to remix and adapt it, provided it is only for non-commercial purposes, that you appropriately attribute the publication, and that you distribute it under an identical licence. For more information visit the Creative Commons website:
International IDEA Strömsborg SE–103 34 Stockholm SWEDEN Tel: +46 8 698 37 00 Email: info@idea.int Website: Cover illustration: Generated with DALL E Design and layout: International IDEA Copyeditor: Curtis Budden DOI: ISBN: 978-91-7671-782-0 (PDF) Acknowledgements This paper was conceptualized by Annika Silva-Leander and Amanda Sourek, and developed by Stacey Cram, who conducted the final research and writing of the content under their supervision.
A previous database research was developed by Jeffrey Fischer and Claire Lynch.
We would like to sincerely thank the peer reviewers of this Policy Paper that provided invaluable input into the analysis, including (in alphabetical order) Amanda Sourek, International IDEA; Annika Silva-Leander, International IDEA; Elena Marmo, TAP Network; Henk-Jan Brinkman, International Development Law Organization; John Romano, TAP Network; Michael Runey, International IDEA; and Peter van Sluijs, Civil Society Platform for Peacebuilding and Statebuilding.
We would also like to thank Lisa Hagman for the editorial guidance received and for all her support in designing this paper.
iv ACKNOWLEDGEMENTS SDG 16 AS AN ENABLER OF THE 2030 AGENDA Contents Acknowledgements....................................................................................... iv Executive summary .......................................................................................1 Introduction.
...................................................................................................2 Chapter 1 Background to the 2030 Agenda and SDG 16.
.................................................4 Chapter 2 Methodology..................................................................................................6 Methodology caveats............................................................................................. 8 Covid-19.
.................................................................................................................. 8 Chapter 3 Overview of key findings and takeaways from interlinkages .
..........................9 Chapter 4 Specific SDG linkages with select SDGs.
.......................................................15 Chapter 5 Recommendations.......................................................................................26 Strengthen high-impact areas.
............................................................................. 26 Address research gaps .
....................................................................................... 27 Chapter 6 Conclusion...................................................................................................29 References...................................................................................................30 Annex A. Pathfinders’ graphic depiction of the SDG 16+ framework.
............. 36 Annex B. Sources analysed for the literature review and meta-analysis........ 37 About the author..........................................................................................50 About International IDEA..............................................................................51 v CONTENTS INTERNATIONAL IDEA This Policy Paper explores the interlinkages between Sustainable Development Goal (SDG) 16 (Peace, Justice and Strong Institutions) and the 2030 Agenda’s other Sustainable Development Goals. Despite SDG 16’s crucial role in promoting democratic principles, human rights and the rule of law, it is often overlooked, and the financing needed to deliver SDG 16 is substantial, requiring sustained investment for long-term development. This paper provides a comprehensive literature review and meta-analysis of 185 articles and highlights how SDG 16 acts as an enabler for other SDGs.
Key findings indicate substantial evidence of SDG 16’s interlinkages with other goals, with high-impact interlinkages existing for SDG 1 (No Poverty), SDG 3 (Good Health and Well-being), SDG 4 (Quality Education), SDG 5 (Gender Equality) and SDG 10 (Reduced Inequalities). There is also evidence demonstrating SDG 16’s positive impact on SDG 2 (Zero Hunger), SDG 6 (Clean Water and Sanitation), SDG 8 (Decent Work and Economic Growth), SDG 11 (Sustainable Cities and Communities) and SDG 13 (Climate Action), which are also examined in this paper. The dimensions of SDG 16 with the strongest interlinkages are ensuring participatory decision making (SDG 16.7), justice (SDG 16.3), reducing arms flows and organized crime (SDG 16.4), building transparent institutions (SDG 16.6), reducing corruption (SDG 16.5) and ensuring peace (SDG 16.1). These focus areas show significant interlinkages with several other SDGs, underscoring the crucial role of rights, participation, representation and the rule of law in achieving sustainable development and addressing various interconnected challenges.
This paper calls for increased cross-sector collaboration and offers recommendations to United Nations Member States, policymakers working across the 2030 Agenda and other stakeholders working in areas where SDG 16 has high-impact interlinkages to make targeted investments, enhance research and safeguard those working on SDG 16 in order to fully leverage its potential in achieving outcomes on the 2030 Agenda.
EXECUTIVE SUMMARY Despite SDG 16’s crucial role in promoting democratic principles, human rights and the rule of law, it is often overlooked.
1 INTERNATIONAL IDEA The Sustainable Development Goals are inherently interconnected, with each goal dependent on the others for its success. However, the specific ways in which these interdependencies function have not been fully articulated. SDG 16, which focuses on promoting peaceful and inclusive societies for sustainable development, providing access to justice for all and building effective, accountable and inclusive institutions at all levels, is increasingly recognized as pivotal in creating an environment conducive to achieving the entire 2030 Agenda. A growing body of evidence, often categorized under the SDG 16+ agenda, demonstrates that SDG 16 is vital for the achievement of all other SDGs (see Annex A. Pathfinders’ graphic depiction of the SDG 16+ framework).
SDG 16 is also vital for upholding key principles of democracy, as defined by the Global State of Democracy (GSoD) Indices, compiled by the International Institute for Democracy and Electoral Assistance (International IDEA), including rights, participation, representation and the rule of law (see Chapters 3 and 4 for a detailed breakdown). Yet, despite its critical importance, SDG 16 is often overlooked, and the financing needed to deliver SDG 16 is substantial, requiring sustained investment for long-term development (UNDP 2023; SDG 16 Hub 2019; Manuel and Manuel 2022). In recent years, democracy has been in decline,1 the rule of law has weakened,2 civic space has been shrinking,3 and conflict-related deaths have reached a three-decade high.4 These alarming trends underscore a lack of progress on SDG 16 and an urgency to accelerate action towards its achievement.
1 In 2023, across every region of the world, democracy continued to contract (International IDEA 2023). 2 More than 6 billion people live in countries where the rule of law weakened between 2022 and 2023. Human rights have declined in three out of four countries since a global rule-of-law recession began in 2016 (WJP 2023).
3 Civic space is becoming increasingly constrained with over 85 per cent of the world’s population living in countries with serious civic space restrictions. Attacks on and persecution of human rights defenders by both state and non-state actors are on the rise (CIVICUS n.d.). 4 In 2023, 59 state-based conflicts were recorded in 34 countries, the highest number of conflicts registered since 1946 (Rustad 2024). INTRODUCTION In recent years, democracy has been in decline, the rule of law has weakened, civic space has been shrinking, and conflict-related deaths have reached a three-decade high.
2 INTERNATIONAL IDEA Despite the importance of SDG 16, detailed research and analysis on how it interlinks with and enables the delivery of other SDGs remain limited, with existing studies on interlinkages lacking an in-depth examination of SDG 16. This Policy Paper addresses this research gap by diving into the existing literature and offering an in-depth meta-analysis of the evidence base that underpins assertions of SDG 16’s interlinkages with other SDGs. Through this analysis, the paper clarifies the SDGs and aspects of SDG 16 where the interlinkage seems to be the strongest. The main target audience is United Nations Member States and policymakers working across the 2030 Agenda to help them make an evidence-based and nuanced case for SDG 16 as an enabler of the Agenda.
The methodology used to review the 185 pieces of literature covered in this study is described in Chapter 2. The resources analysed reveal clear interlinkages between SDG 16 and all other SDGs, with the most substantial evidence existing for SDG 1 (No Poverty), SDG 3 (Good Health and Well-being), SDG 4 (Quality Education), SDG 5 (Gender Equality) and SDG 10 (Reduced Inequalities). There is also evidence for interlinkages with SDG 2 (Zero Hunger), SDG 6 (Clean Water and Sanitation), SDG 8 (Decent Work and Economic Growth), SDG 11 (Sustainable Cities and Communities) and SDG 13 (Climate Action). Conversely, there is less evidence for interlinkages with SDG 7 (Affordable and Clean Energy), SDG 9 (Industry, Innovation and Infrastructure), SDG 12 (Responsible Consumption and Production), SDG 14 (Life below Water), SDG 15 (Life on Land) and SDG 17 (Partnerships). The dimensions of SDG 16 that appear to have the strongest interlinkages are participatory decision making (SDG 16.7), justice (SDG 16.3), reducing arms flows and organized crime (SDG 16.4), transparent institutions (SDG 16.6), reducing corruption (SDG 16.5) and peace (SDG 16.1). These focus areas show significant interlinkages with several other SDGs, underscoring the crucial role of rights, participation, representation and the rule of law in achieving sustainable development and addressing various interconnected challenges.
This paper outlines how UN Member States and policymakers working across the 2030 Agenda can accelerate action on development outcomes, such as health, poverty and education, by supporting and investing in SDG 16. It provides an initial, though not exhaustive, overview of existing research on interlinkages, with positive interlinkages prioritized over negative interlinkages, and highlights questions to encourage future collaboration and research between different sectors. Addressing areas with relatively less research in future studies will be essential to strengthening the evidence base across all aspects of SDG 16 and its connections with other goals.
The Report reveals clear interlinkages between SDG 16 and all other SDGs.
The dimensions of SDG 16 that appear to have the strongest interlinkages are participatory decision making (SDG 16.7), justice (SDG 16.3), reducing arms flows and organized crime (SDG 16.4), transparent institutions (SDG 16.6), reducing corruption (SDG 16.5) and peace (SDG 16.1). 3 INTRODUCTION The 2030 Agenda for Sustainable Development, adopted in 2015 by 193 UN Member States, includes 17 goals focused on people, the planet, prosperity, peace and partnership. SDG 16 aims to promote peaceful and inclusive societies, provide access to justice for all and build effective, accountable institutions at all levels. Its impacts can occur throughout public and private society, in formal institutions and in de facto norms. For this reason, it is cross-cutting, impacting governance, the delivery of basic services and the ability to find resolutions when rights are denied; these are foundational elements5 for addressing the complex interlinkages between poverty, inequality, climate change and economic instability. The concept of SDG 16+ identifies peace, justice and inclusion as the three main dimensions of SDG 16, comprising the 12 targets from SDG 16 and 24 targets from 7 other goals that directly measure an aspect of peace, justice and inclusion. 5 2030 Agenda Sustainable Development Goals and targets can be seen in Annex A.
Chapter 1 BACKGROUND TO THE 2030 AGENDA AND SDG 16 4 INTERNATIONAL IDEA Table 1.1. SDG 16 targets and indicators Target Official indicators SDG 16.1 Significantly reduce all forms of violence and related death rates everywhere • Number of victims of intentional homicide • Number of conflict-related deaths • Proportion of population that feel safe walking alone around the area they live SDG 16.2 End abuse, exploitation, trafficking and all forms of violence and torture against children • Proportion of children aged 1–17 years who experienced any physical punishment and/or psychological aggression by caregivers • Number of victims of human trafficking • Proportion of young women and men aged 18–29 years who experienced sexual violence by age 18 SDG 16.3 Promote the rule of law at the national and international levels and ensure equal access to justice for all • Proportion of victims of violence who reported their victimization • Unsentenced detainees as a proportion of the overall prison population • Proportion of the population who have experienced a dispute and accessed a formal or informal dispute resolution mechanism SDG 16.4 By 2030 significantly reduce illicit financial and arms flows, strengthen the recovery and return of stolen assets, and combat all forms of organized crime • Inward and outward illicit financial flows • Proportion of seized, found or surrendered arms whose illicit origin has been traced SDG 16.5 Substantially reduce corruption and bribery in all their forms • Proportion of persons who paid a bribe to a public official or were asked for a bribe • Proportion of businesses that paid a bribe to a public official or were asked for a bribe SDG 16.6 Develop effective, accountable and transparent institutions at all levels • Primary government expenditures as a proportion of the approved budget, by sector • Proportion of the population satisfied with their most recent experience of public services SDG 16.7 Ensure responsive, inclusive, participatory and representative decision making at all levels • Proportions of positions in national and local institutions compared with national distributions • Proportion of the population who believe decision making is inclusive and responsive SDG 16.8 Broaden and strengthen the participation of developing countries in the institutions of global governance • Proportion of members and voting rights of developing countries in international organizations SDG 16.9 By 2030 provide legal identity for all, including free birth registration • Proportion of children under five years of age whose births have been registered SDG 16.10 Ensure public access to information and protect fundamental freedoms in accordance with national legislation and international agreements • Number of verified cases of killing, kidnapping, enforced disappearance, arbitrary detention and torture of journalists, media personnel, trade unionists and human rights advocates • Number of countries that adopt and implement guarantees for public access to information Source: United Nations, ‘Transforming our world: The 2030 Agenda for Sustainable Development’, [n.d.],
un.org/2030agenda, accessed 29 June 2024. 5 1. BACKGROUND TO THE 2030 AGENDA AND SDG 16 This study is based on a scoping review of the literature on SDG 16 and its impact on other SDGs and development outcomes, as well as on SDG interlinkages. The literature search was carried out using online search engines and databases (Google Scholar, JSTOR, etc.). The researcher began looking for causal relationships through searches such as ‘How does X [SDG 16 element] impact Y [alternate SDG]?’ and then searching for ‘the relationship between X and Y’ where there was limited causal evidence. Academic articles were prioritized in this search to find evidence with clear methodologies and robustly quantified and qualified results. Literature published between 2016 and 2024 was prioritized, with April 2024 being the cut-off date for the sampling; 13 articles written prior to 2016 were included. The breakdown of the resources used can be seen in Figure 2.1. A wide range of studies were examined through an initial screening of titles, abstracts and keywords to narrow down the sample to the final list of 185 papers. The researcher also noted where evidence surrounding an interlinkage was sparse or non-existent in an effort to determine if any trends emerged within the evidence gaps.
After the literature search and screening, an in-depth review of the studies in the final sample was carried out. A predefined coding scheme was followed to collect and synthesize information about the scale of analysis and the research methods used, and to determine whether SDG 16 was referred to as an enabler of the 2030 Agenda or of specific goals and whether the linkage was made referencing the SDG by name or outcome as described in the 2030 Agenda (e.g. SDG 1/no poverty) or more broadly on related outcomes (e.g. increased household income), and whether links were explicit or implicit, as well as to identify key evidence from the literature. The number of articles that referenced an SDG was tracked, and for each article reviewed it was noted when an individual target was mentioned as having a positive enabling effect on a specific goal. For example, if one article referenced that 16.1 enabled SDGs 1, 2 and 3, this would be noted as three occurrences of SDG 16.1 interlinkages Chapter 2 METHODOLOGY 6 INTERNATIONAL IDEA in the literature. This accounts for why there are more occurrences in the literature than the number of articles reviewed. Negative occurrences were not tracked. High coverage for targets is greater than 201 occurrences (90th percentile); medium coverage, 179–201 occurrences (80th percentile); low coverage, 156–179 occurrences (70th percentile); and the lowest coverage, below 155 occurrences (60th percentile). The highest number of occurrences was 224 (16.7, participatory decision making), and the lowest was 154 (16.8, participation of developing countries in global governance). High coverage for SDGs is more than 133 occurrences, medium 110–132 and low coverage below 110.
For a full list of the sampled literature, see Annex B. Figure 2.1. Breakdown of the sources reviewed Source: Developed by the author.
7 2. METHODOLOGY METHODOLOGY CAVEATS The analysis is based on a comprehensive, albeit non-exhaustive, review of the literature, and additional sources may exist that were not included. Meta studies were prioritized to increase the number of sources reviewed. This study primarily focused on explicit and implicit interlinkages referencing the SDG by name or outcome as described in the 2030 Agenda, which may not capture the full extent of the relationships between SDG 16 and other goals. The majority of the reviewed studies were qualitative, limiting the author’s ability to draw robust quantitative conclusions. Negative linkages were not explored in depth. Some studies did look at negative linkages but identified SDG 16 as having low negative interlinkages; future research should examine the negative correlations related to SDG 16. Future research should also aim to incorporate more quantitative studies to provide a more comprehensive understanding of the interlinkages. This study notes the difficulty in distinguishing between correlation and causation in the observed interlinkages. Further research is needed to explore causal relationships and validate the findings. Additional resources in different languages should also be considered for future research as this paper only addressed literature in English.
COVID-19 It is worth noting that the global pandemic may have fuelled an increased number of publications linking SDG 16 to SDG 3. Seven of the articles reviewed were written in direct response to the Covid-19 public health crisis, with others written after 2020 examining the impacts of the crisis. 8 SDG 16 AS AN ENABLER OF THE 2030 AGENDA International IDEA’s GSoD conceptual framework (Figure 3.1) has identified four key attributes for a healthy and functional democracy: rights, participation, representation and the rule of law. The 185 resources analysed for this study provide a strong body of evidence that each of the elements of the GSoD conceptual framework and the associated SDG 16 targets is critical for achieving sustainable development. Chapter 3 OVERVIEW OF KEY FINDINGS AND TAKEAWAYS FROM INTERLINKAGES Figure 3.1. International IDEA’s GSoD conceptual framework Source: International IDEA, The Global State of Democracy Initiative, [n.d.], accessed 29 June 2024.
9 INTERNATIONAL IDEA The literature captures the negative impact of violence, corruption, injustice and insecurity on health, education, jobs, urban development and other Sustainable Development Goals. The positive effect of participatory decision making (SDG 16.7), justice (SDG 16.3), reduced arms flows and organized crime (SDG 16.4), transparent institutions (SDG 16.6) and reduced corruption (SDG 16.5) leads to gains for individuals and communities, alongside wider development gains. While fewer of the articles examined for this study look at direct interlinkages between peace (16.1) and development, the evidence that exists is robust and based on decades of historical analysis. All the areas listed above show significant interlinkages with several other SDGs, highlighting the critical role these elements play in achieving sustainable development and addressing various interconnected challenges.
The resources analysed reveal clear interlinkages between SDG 16 and all the other SDGs, with the most substantial evidence existing for interlinkages with SDG 1 (No Poverty), SDG 3 (Good Health and Well-being), SDG 4 (Quality Education), SDG 5 (Gender Equality) and SDG 10 (Reduced Inequalities). Conversely, there is less evidence for interlinkages with SDG 7 (Affordable and Clean Energy), SDG 9 (Industry, Innovation and Infrastructure), SDG 12 Figure 3.2. SDG 16 interlinkages Source: Developed by author and International IDEA.
10 SDG 16 AS AN ENABLER OF THE 2030 AGENDA (Responsible consumption and production), SDG 14 (Life below Water), SDG 15 (Life on Land) and SDG 17 (Partnerships) (Figure 3.2).
The number of articles available to analyse indicates that there is a strong research focus on this subject in academic and policy communities and the desire to substantiate claims of SDG 16 interlinking properties and its ability to enable sustainable development outcomes. Goals that have low coverage (SDGs 7, 9, 12, 14, 15 and 17) should be viewed as potential gaps in research and in the literature rather than as evidence that these goals do not enable other development outcomes. See full heatmap of sources analysed in Figure 3.3.
Most of the studies reviewed look at interlinking properties across multiple countries or regions, with only 19 articles focusing on how SDG 16 enables development in a specific country context. Moreover, 132 resources focus on how SDG 16 helps enable a specific goal, 37 studies cover multiple goals, and 16 studies focus on how SDG 16 helps deliver the entire 2030 Agenda. The evidence examined is divided into qualitative studies (117), quantitative studies (39), and studies utilizing mixed quantitative and qualitative analyses (29). The limited number of quantitative studies makes it difficult to draw conclusions on the robustness of all 185 sources’ findings; only quantitative studies would allow for that. Future research should address this gap in quantitative studies which could also help interrogate whether certain interlinkages are correlations or causations, an issue not addressed in this Policy Paper.
Figure 3.3. Heatmap of sources analysed, and number of positive occurrences in the literature1 1 For each article reviewed, it was noted when an individual target was mentioned as enabling a specific goal.
16.1 11 10 17 16 15 11 10 13 9 14 11 9 11 9 9 10 185 20 16.2 11 10 16 12 11 10 9 11 8 11 9 9 10 9 9 8 163 16.3 19 13 19 17 16 13 11 15 11 14 12 12 16 11 11 12 222 17 16.4 14 13 20 15 14 13 12 14 12 15 12 12 13 12 12 13 216 16.5 15 11 19 15 14 12 10 15 10 18 11 11 11 10 11 12 205 16.6 16 13 19 14 15 13 11 13 11 13 15 12 15 11 12 12 215 16.7 16 13 19 16 15 13 12 14 12 15 14 13 16 12 12 12 224 16.8 11 9 14 11 11 10 8 10 8 10 9 8 9 8 9 9 154 11 16.9 12 10 15 12 12 10 8 11 9 12 8 9 10 8 8 9 163 16.10 11 10 18 13 11 10 8 11 9 11 9 9 10 9 9 9 167 8 TOTAL 136 112 176 141 134 115 99 127 99 133 110 104 121 99 102 106 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 17 SDG 16 targets SDG by goal Number of occurences 14 Source: Developed by the author.
11 3. OVERVIEW OF KEY FINDINGS AND TAKEAWAYS FROM INTERLINKAGES Table 3.1. Interlinkages referenced in the literature SDG 16 target Literature coverage1 Number of occurrences Notes SDG 16.7 (participatory decision making) High 224 Strong research focus.
SDG 16.3 (justice) High 222 Strong research focus.
SDG 16.4 (reducing arms flows and organized crime) High 216 Strong research focus.
SDG 16.6 (transparent institutions) High 215 Strong research focus.
SDG 16.5 (reducing corruption) High 205 Strong research focus.
SDG 16.1 (peace) Medium 185 Evidence base for conflict impacting sustainable development is robust.
SDG 16.10 (public access to information) Low 167 SDG 16.2 (end abuse, violence and torture against children) Low 163 Research on child abuse and violence is sufficient but limited on trafficking and torture of children.
SDG 16.9 (legal identity for all, including birth registration) Low 163 SDG 16.8 (participation of developing countries in global governance) Lowest 154 Large gap in evidence regarding the participation of developing countries in international governance.
1 High coverage is greater than 201 occurrences (90th percentile); medium coverage, 179–201 occurrences (80th percentile); low coverage, 156–179 occurrences (70th percentile); lowest coverage, below 155 occurrences (60th percentile).
12 SDG 16 AS AN ENABLER OF THE 2030 AGENDA The analysis found the following strong interlinkages (Figure 3.4): • Between SDG 16.7 (participatory decision making) and SDG 4 (Quality Education) and SDG 11 (Sustainable Cities and Communities). Sixteen occurrences in the literature examine the positive role of inclusive decision making in achieving quality education. Fourteen occurrences in the literature focus on how participatory decision making contributes to sustainable urban development.
• Between SDG 16.3 (justice) and SDG 1 (No Poverty) and SDG 3 (Good Health and Well-being). Nineteen occurrences in the literature suggest a strong connection between justice and reductions in poverty. An additional 19 occurrences in the literature discuss how justice enables health outcomes.
• Between SDG 16.4 (reducing arms flows and organized crime) and SDG 3 (Good Health and Well-being). This pairing had the highest individual count, with 20 occurrences in the literature, highlighting the link between reducing violence and crime and improving health and well-being outcomes.
• Between SDG 16.6 (transparent institutions) and SDG 10 (Reduced Inequalities). This pairing had 16 occurrences in the literature, indicating a notable link between institutional transparency and reducing inequality.
• Between SDG 16.5 (reducing corruption) and SDG 3 (Good Health and Well-being) and SDG 10 (Reduced Inequalities). Nineteen occurrences in the literature suggest a strong connection between corruption and health. An additional 18 occurrences in the literature discuss how anti-corruption measures reduce inequality. • Between SDG 16.1 (peace) and SDG 3 (Good Health and Well-being), SDG 4 (Quality Education) and SDG 5 (Gender Equality). Seventeen occurrences in the literature suggest a strong connection between peace and health. An additional 16 occurrences in the literature discuss how peace enables education, and 15 occurrences in the literature examine how peace facilitates gender equality.
13 3. OVERVIEW OF KEY FINDINGS AND TAKEAWAYS FROM INTERLINKAGES Figure 3.4. Interlinkages by SDG 16 target 14 SDG 16 AS AN ENABLER OF THE 2030 AGENDA The evidence base reveals clear interlinkages between SDG 16 and all other SDGs, with the most substantial evidence existing for SDG 1 (No Poverty), SDG 3 (Good Health and Well-being), SDG 4 (Quality Education), SDG 5 (Gender Equality) and SDG 10 (Reduced Inequalities). There is also evidence for interlinkages with SDG 2 (Zero Hunger), SDG 6 (Clean Water and Sanitation), SDG 8 (Decent Work and Economic Growth), SDG 11 (Sustainable Cities and Communities) and SDG 13 (Climate Action). Conversely, there is less evidence for interlinkages with SDG 7 (Affordable and Clean Energy), SDG 9 (Industry, Innovation and Infrastructure), SDG 12 (Responsible Consumption and Production), SDG 14 (Life below Water), SDG 15 (Life on Land) and SDG 17 (Partnerships).
The rest of this chapter will examine the 10 SDGs that have the highest number of occurrences in the literature (indicated by the blue bars in Figure 4.1), looking at how SDG 16 enables the attainment of that specific goal. Where there is evidence for the GSoD democratic principles of rights, participation, representation and the rule of law having enabling properties, these will also be highlighted. SDG 1 (No Poverty). In total 136 occurrences in the literature highlight how SDG 16 targets help poverty alleviation. All SDG 16 targets have a high evidence base, with 16.3 (justice) being the most robust. • Violence and conflict can negate progress towards the SDGs and have long-lasting impacts that perpetuate poverty and hinder development efforts. These impacts include but are not limited to deaths; increased numbers of orphans, widows and disabled individuals; the destruction of infrastructure; forced displacement; the spread of disease; malnourishment; and increased unemployment. These impacts jeopardize societal cohesion and exert adverse impacts on economic progress (UNDP 2023; Mukombwe, du Toit and Hendriks 2024). Chapter 4 SPECIFIC SDG LINKAGES WITH SELECT SDGS 15 INTERNATIONAL IDEA Figure 4.1. Number of occurrences in the literature for each SDG, broken down by SDG 16 target Note: The graphs with the blue bars refer to goals with moderate to high coverage in the literature and will be examined in detail in the text; the graphs with the yellow bars refer to goals with low coverage in the literature. Source: Developed by the author based on Annex B.
16 SDG 16 AS AN ENABLER OF THE 2030 AGENDA • Poverty and corruption are interrelated: countries with lower levels of income have higher levels of bribery. Corruption, as measured through administrative bribery, remains prevalent globally, with 20 per cent of people who access a public service reporting having paid a bribe to do so (UNDP 2023). Studies indicate that effective governance reduces corruption and crime, improves levels of participation and inclusion, and increases access to information, transparency and accountability, which are positively associated with poverty reduction (Sohail and Savill 2008; Spyromitros and Panagiotidis 2022).
• Justice provision can reap social rewards and have large economic benefits. Justice-based solutions can overcome key drivers of poverty, including forced evictions and lack of land (use) rights, and enable community-led solutions (Cociña et al. 2023; Weston 2022). • Countries with stronger institutions and less corruption tend to have better poverty reduction outcomes. Evidence shows that investments in governance can significantly mitigate poverty (Allen et al. 2023). One economic study found that for every 1 per cent increase in corruption, there is a 0.2 per cent decrease in economic growth (Spyromitros and Panagiotidis 2022). Strengthening institutions and reducing corruption ensure that economic benefits are distributed more equitably, thus supporting vulnerable populations.
SDG 2 (Zero Hunger). In total 112 occurrences in the literature highlight how SDG 16 targets help reduce hunger. All SDG 16 targets have a high evidence base. • Conflict and limited access to justice impact or restrict access to land, seeds, water, fair and stable markets, and agricultural work, thus hindering food security and increasing hunger (Mukombwe, du Toit and Hendriks 2024).
• Transnational organized crime can severely impact public health and food security by enabling the production and sale of unsafe food products (Caparini 2022), whereas corruption control can significantly reduce the number of undernourished people in developing countries (Dwi Nugroho et al. 2022).
• The protection and promotion of rights are critical to addressing hunger and achieving food security. Data from the International Food Policy Research Institute’s Global Hunger Index illustrates that countries with stronger civil liberties, such as Ghana, have experienced significant reductions in hunger. This correlation suggests that ensuring civil liberties can create an environment conducive to better nutritional outcomes (Harris 2014).
• Increased levels of participation and inclusion are positively associated with poverty reduction (Balasubramanian et al. 2022). Participatory planning and measures that improve voice and accountability are more likely to result 17 4. SPECIFIC SDG LINKAGES WITH SELECT SDGS in pro-poor policies, which are crucial for achieving SDG 2 (Birner 2007). Enhanced accountability facilitates better communication and cooperation among stakeholders, improving the effectiveness of initiatives aimed at reducing malnutrition (Hoffman et al. 2022). SDG 3 (Good Health and Well-being). In total 176 occurrences in the literature highlight how SDG 16 targets help achieve health and well-being. All SDG 16 targets have a moderate evidence base.
• Conflict and crime worsen public health. Health-adjusted life expectancy data shows that environments where rights are violated due to conflict see significantly worse health outcomes (Iqbal 2006). Organized crime, including the illicit drug trade and human trafficking, severely undermines public health (Reynolds and McKee 2010).
• Corruption in the health sector can limit access to healthcare services, increase their cost and lead to the proliferation of substandard or counterfeit medical products (Mantegazza 2023). The World Health Organization has estimated that 1 in 10 medical products are ‘substandard or falsified’ in low- and middle-income countries, and this figure has been as high as 70 per cent (Caparini 2022). Corruption control measures ensure health services are more accessible and of higher quality, contributing to better health outcomes (Jenkins 2021).
• Participatory governance that engages communities in health policymaking supports health services in an effort to better meet the needs of a population and is a determinant for achieving universal health coverage (Kishore and Tripathi 2019). Democratic elections have been found to significantly reduce infant mortality rates, particularly in rural areas (Harding 2020). • The rule of law is essential for creating a stable environment for health systems, serving as a social determinant of health and impacting disease transmission, risk of exposure to harmful products and access to safe living conditions. Strong legal frameworks are necessary to regulate these factors and protect public health (Dingake 2017). Increased access to justice ensures communities can hold parties accountable and secure the resources needed to protect their health (Gramatikov 2023; Pathfinders for Peaceful, Just and Inclusive Societies 2019). Box 4.1. SDG 2 example Transparency and accountability in nutrition actions are promoted through nutrition accountability frameworks (NAFs). By developing SMART (specific, measurable, achievable, relevant and time-bound) commitments, tracking progress and ensuring stakeholders meet their commitments, NAFs aim to improve food security and nutrition outcomes. This approach has led to better resource allocation, increased community participation and more effective interventions to combat malnutrition and enhance food security (Hoffman et al. 2022).
18 SDG 16 AS AN ENABLER OF THE 2030 AGENDA SDG 4 (Quality Education). In total 141 occurrences in the literature highlight how SDG 16 targets help achieve educational outcomes. All SDG 16 targets have a strong evidence base.
• Violence and conflict lead to reductions in school attainment and longer-term human capital among those exposed to violence. Evidence from Timor Leste suggests that waves of violence in the country resulted in persistent negative effects on primary school attendance and completion; boys were also less able to benefit from post-conflict recovery as a result of household trade-offs between education and economic survival that led to the removal of boys from school (Justino, Leone and Salardi 2014). • Crime has the effect of displacing public expenditure away from productive assets, such as education, and directing resources towards protective measures (UNDP 2023). Organized crime negatively impacts educational outcomes by distorting incentives to invest in human capital (Cavalieri, Finocchiaro Castro and Guccio 2023). • Abuse and neglect of children are linked to poor educational outcomes and mental well-being (Romano et al. 2015; Lazenbatt 2010; Owolabi 2012). Interventions that reduce violence at home and provide children and families with timely access to justice (e.g. faster processing times for domestic violence cases, and legal assistance through divorce proceedings) can reduce children’s exposure to violence (Weston 2022; UN Women et al. 2019). • Corruption limits the development of an economic and institutional environment that enhances education expansion and the formation of high-quality human capital formation. People can be incentivized to spend less time in education and to focus instead on accumulating political capital that enables them to wield bureaucratic power and to engage in rent-seeking activities. Corruption also diverts public funds away from education (Dridi 2014). Addressing corruption is pivotal for improving education systems (Jenkins 2021). Regression analysis reveals a strong link between corruption and lower secondary school enrolment rates. One study found that a unit increase in corruption results in a drop in enrolment rates by nearly 10 per cent (Dridi 2014). Box 4.2. SDG 3 example In Gujarat, India, social accountability mechanisms have positive outcomes on maternal health. Local communities were trained to monitor the quality of maternal health services and collected data on healthcare facilities, service delivery and patient satisfaction. This information was presented at public hearings, where community members, health officials and service providers addressed the findings and grievances. As a result, there were tangible improvements in health services, leading to higher utilization of maternal health services, including antenatal care visits, institutional deliveries and postnatal care, ultimately reducing maternal mortality rates in the areas where these mechanisms were implemented (Hamal et al. 2018).
19 4. SPECIFIC SDG LINKAGES WITH SELECT SDGS • Strong, inclusive institutions are essential for fostering education. Democratic elections were found to increase access to primary education in rural areas of sub-Saharan Africa (Harding 2020). Increased civil liberties significantly impact the quality of education within Asian countries (Saengchai, Maneerat and Pissanu 2020). • Legal identity, including birth registration, is crucial for accessing education. In Tanzania possessing a birth certificate is a strong predictor of educational access (Mbise 2020).
SDG 5 (Gender Equality). In total 134 occurrences in the literature highlight how SDG 16 targets help achieve gender equality. All SDG 16 targets have a strong evidence base.
• Meaningful participation of women in decision-making and peace processes is associated with less violence, longer-lasting peace and fewer relapses into violence (Endeley and Molua 2024). Democracies that ensure women’s representation in political processes tend to perform better in various dimensions of gender equality; this is especially true when institutions such as gender quotas and proportional electoral systems are utilized. Democracies generally deliver better outcomes for citizens, including women, due to accountability mechanisms that foster gender-sensitive policies (V-Dem Institute 2023; TAP Network 2019). Countries with higher levels of democracy and transparent institutions have higher rates of female labour force participation, thus contributing to gender equality (Beer 2009).
• Stronger legal frameworks correlate with reduced gender disparities (Barajas-Sandoval et al. 2023). Effective legal institutions are crucial for protecting women’s rights, ensuring their access to justice and safeguarding them against violence and discrimination. Investing in justice for women—and especially eliminating legal barriers and reducing gender-based violence and child marriage—produces high returns in human development, national income, economic growth and the fulfilment of SDG 5 (UN Women et al. 2019; Moy, Cram and Maru 2019).
Box 4.3. SDG 4 example A study from different Italian regions found that the rule of law in a student’s regional environment influences their educational decisions. Students in environments with stronger rule of law are more likely to pursue careers that contribute to societal well-being than students raised in environments with weaker rule of law. Effective rule of law decreases resources that rent seekers may expropriate, thus reducing incentives to choose careers devoted to protecting producers from rent seeking (Nifo, Scalera and Vecchione 2017).
Meaningful participation of women in decision-making and peace processes is associated with less violence, longer-lasting peace and fewer relapses into violence.
20 SDG 16 AS AN ENABLER OF THE 2030 AGENDA • Protecting fundamental rights, such as freedom of assembly, access to information and free media, is crucial for gender equality. In Kenya increased access to information was found to improve gender equality by providing women with greater access to education, family planning and economic opportunities (Wagacha 2007).
• Ensuring that women have legal identity documents can significantly improve their ability to participate in economic, social and political life, thereby advancing gender equality (Pathfinders for Peaceful, Just and Inclusive Societies 2019; UN Women et al. 2019).
SDG 6 (Clean Water and Sanitation). In total 115 occurrences in the literature highlight how SDG 16 targets help enable access to clean water. All SDG 16 targets have a moderate evidence base.
• Organized crime in the water sector can significantly impair access to clean water and impede progress on SDG 6 (Segato et al. 2018). Effective legal frameworks are essential for tackling such activities and wider environmental crimes that impact water resources. • Community-focused, participatory engagement strategies help improve development and monitoring measures in water infrastructure projects (Hove et al. 2019). In Argentina, when community lawyers worked alongside shantytown residents through legal and administrative actions, communities secured access to potable water and sanitation services (Moy, Cram and Maru 2019). • Enhanced voice and accountability can support poor communities in their efforts to access clean water. One study found that a one-level increase in civil liberties6 leads to a 2.5 per cent increase in access to clean water for rural populations (Biser and Edwards 2012). Democracies with high accountability mechanisms provide 23 per cent more safe water access than autocracies (V-Dem Institute 2023).
6 The authors developed a civil liberties variable scale using the civil liberties subset of the Freedom in the World Index by Freedom House and the World Bank’s World Development Indicators database. The full methodology can be seen in their paper.
Box 4.4. SDG 5 example In Mozambique a study found that effective access to justice could support women facing intense gender discrimination. In land-claim cases, every dispossessed widow or divorced woman with access to a community paralegal was able to successfully assert her claim. Where cases stalled at the community level, paralegals were able to bring them to court, where decisions are more often in a woman’s favour (Moy, Cram and Maru 2019).
21 4. SPECIFIC SDG LINKAGES WITH SELECT SDGS SDG 8 (Decent Work and Economic Growth). In total 127 occurrences in the literature highlight how SDG 16 targets help economic growth. All SDG 16 targets have a high evidence base, with 16.3 (justice) and 16.5 (reducing corruption) being the most robust.
• Political instability, violence and conflict disrupt economic activities, destroy infrastructure and deter investment. High-intensity inter-state conflict reduces annual growth by 0.18–2.77 per cent, and civil war reduces annual growth by 0.01–0.13 per cent (Polachek and Sevastianova 2012). The global economic impact of violence was USD 17.5 trillion in 2022, the equivalent of 12.9 per cent of global gross domestic product (GDP), or USD 2,200 per person. Homicide was the fourth-largest component in terms of total costs, accounting for 6.6 per cent of the global economic impact of violence (UNDP 2023). Adults who experienced abuse as children have decreased educational levels, earnings and employment (Currie and Spatz Widom 2010).
• Criminal activity negatively affects a country’s capacity to attract skilled workers and both domestic and foreign investments. It also leads to a reduction in competitiveness between companies and an increase in uncertainty among businesses, which make conducting economic activities in high-crime areas a challenge (UNDP 2023). • The role of corruption in hindering economic growth is well documented. Studies show that a 1 per cent increase in corruption is correlated with a 0.72 per cent decrease in the growth rate (Hung Mo 2001). By reducing corruption, SDG 16 fosters a trustworthy business environment, thus encouraging investment and economic activity. Quality data and transparent release of information associated with economic policy benefits the public and increases inclusiveness. Effective governance ensures that economic policies are implemented well, thus promoting sustainable economic growth (Nilsson et al. 2018; San Chia et al. 2022).
• Effective and inclusive parliaments ensure that all societal groups are represented in the legislative process, which leads to the creation of laws and policies that are conducive to economic development (TAP Network 2019). Political stability, fostered by inclusive representation and effective governance, is a key driver of economic growth (Polachek and Sevastianova 2012; Anwar and Cooray 2012).
Box 4.5. SDG 6 example A comparison between Mauritius and the Democratic Republic of the Congo found that improvements in institutional quality, such as voice and accountability, regulatory quality and corruption control, can significantly increase access to sanitation in countries with low levels of institutional quality, like the Democratic Republic of the Congo, when compared with countries with higher levels of institutional quality, like Mauritius (Francois, Kakeu and Kouame 2021).
22 SDG 16 AS AN ENABLER OF THE 2030 AGENDA • Democratization leads to economic growth—on average 20 per cent higher GDP per capita after 25 years. Democracies also avoid the more catastrophic economic outcomes that characterize many autocracies (V-Dem Institute 2023). • Access to legal identification is vital for securing employment and job opportunities. Birth certificates and identification documents are linked to employment in the formal sector, which grants workers greater benefits and job security. Restrictions on refugees’ access to authorized work push them into informal, unregulated employment with greater labour infractions (Weston 2022; Hunter 2018; Arnold-Fernández 2024).
SDG 10 (Reduce Inequalities). In total 176 occurrences in the literature highlight how SDG 16 targets reduce inequalities. All SDG 16 targets have a strong evidence base, with target 16.5 (reducing corruption) being the most robust. SDG 10.2 and 16.7 are similar targets. • Violent conflict has a strong negative influence on inequality. Following the end of a conflict, it can take up to four decades to return to pre-war levels of equality (Bircan, Brück and Vothknecht 2017).
• Corruption is a significant factor enabling inequality, as some entities gain more from the system than others through bribes (Rabnawaz 2015). Corruption control can significantly reduce inequalities (Dwi Nugroho et al. 2022).
• Transitioning to democracy leads, on average, to a more than 100 per cent increase in social protection spending (V-Dem Institute 2023). Improved access to information, transparency and accountability help to improve access to basic services, and the targeting of social protection policies and investments in this space can help reduce inequalities (Balasubramanian et al. 2022).
• Effective, accountable and transparent institutions foster inclusive decision-making processes, which are vital for reducing inequalities (Pathfinders for Peaceful, Just and Inclusive Societies 2019). One study on the interlinkages between SDGs found that inclusive governance enables marginalized groups to have a voice in policymaking, which is crucial for achieving SDG 10 (Bennich et al. 2023).
Corruption is a significant factor enabling inequality, as some entities gain more from the system than others through bribes.
Box 4.6. SDG 10 example Mexico’s official multidimensional poverty measure links human and social rights with economic well-being in a coherent framework. The social protection systems and direct cash transfer contributions introduced have proven to be the most effective policy instruments to simultaneously reduce inequality and poverty, while promoting inclusive growth that reaches a broader base of the population (Pathfinders for Peaceful, Just and Inclusive Societies 2021). 23 4. SPECIFIC SDG LINKAGES WITH SELECT SDGS SDG 11 (Sustainable Cities and Communities). In total 110 occurrences in the literature highlight how SDG 16 targets help achieve sustainable cities. All SDG 16 targets have a moderate evidence base. • Organized crime and violence negate efforts to develop sustainable cities. ‘Captured municipalities’ have been found to divert resources which then benefit crime groups and undermine efforts to make cities inclusive (Di Cataldo and Mastrorocco 2022). Reducing urban violence has a direct positive impact on living conditions in urban slums and informal settlements (Locke 2021). • Corruption undermines public trust and diverts resources away from essential urban development projects. In urban settings, reducing corruption can ensure that resources are effectively utilized for sustainable infrastructure and public services, thus fostering inclusive and resilient cities.
• Improvements in legal frameworks, the rule of law and governance are critical for reducing violence and ensuring access to justice, which are essential for urban development and inclusive cities. The strength of the rule of law has been found to predict a government’s ability to provide a liveable urban environment, thus emphasizing the importance of legal frameworks (White, Sepe and Masconale 2014).
• Civil society engagement, voter turnout and other forms of public participation are critical for ensuring equitable urban development. Engaging communities in governance and utilizing citizen solutions can enable municipalities to meet the needs of all community members and to achieve more sustainable cities (Kallergis 2022; Gollagher and Hartz-Karp 2013). One study found that only a multi-stakeholder urban design process will produce truly inclusive urban spaces that fulfil the right to the city (Kempin Reuter 2019). Improvements in legal frameworks, the rule of law and governance are critical for reducing violence and ensuring access to justice, which are essential for urban development and inclusive cities. Box 4.7. SDG 11 example Amending local laws governing the use of alcohol and firearms is a proven means of preventing violence in urban areas. South Africa banned the sale of alcohol during its Covid-19 lockdown, and rates of violent crime plummeted. In the informal settlement of Diadema in São Paolo, Brazil, the prohibition of alcohol sales after 23:00 and the monitoring of alcohol vendors were combined with infrastructure improvements such as street lighting and security cameras, the establishment of mediation centres for peaceful resolution of conflicts, and drug prevention education and apprenticeship schemes for youth. The city’s homicide rate halved in two years (Locke 2021).
24 SDG 16 AS AN ENABLER OF THE 2030 AGENDA SDG 13 (Climate Action). In total 121 occurrences in the literature highlight how SDG 16 targets help achieve positive climate outcomes. All SDG 16 targets have a strong evidence base, with 16.3 (justice) and 16.7 (transparent institutions) being the most robust.
• Strong institutions and the protection of human rights are fundamental for addressing climate change, with democracies performing better than autocracies on climate change mitigation, producing more ambitious climate policies and delivering on them. In the nationally determined contributions under the Paris Agreement, fully democratic countries set, on average, targets of 1.6 degrees Celsius lower than full autocracies. Each new climate policy of this kind reduces CO2 emissions by 1.79 per cent within three years (V-Dem Institute 2023).
• Transparent and accountable institutions allow for more effective environmental oversight. When participatory environmental governance is utilized, the result is higher environmental outputs, which are essential for effective climate action (Newig and Fritsch 2009). • When communities are supported in their efforts to participate in governance and to co-generate data on climate alongside the scientific community, they can support better climate governance and reduce climate displacement and vulnerabilities (Kallergis 2022). Evidence shows that when local communities approach government and administrative institutions, invoke law with specificity and provide simple, community-generated evidence of violations, they can secure enforcement of environmental regulations (Maru 2023). • Limited access to justice exacerbates the impacts of climate change on vulnerable communities, while increased access can mitigate these effects (Gramatikov 2023). As climate litigation increases in frequency and volume, the body of legal precedent grows, forming an increasingly well-defined field of law. Globally, 55 per cent of cases have had a climate-positive ruling, and some favourable outcomes have led directly to new climate policies and action. Even unsuccessful litigation can shape narratives around climate action, encouraging decision makers to change their approach (Setzer and Higham 2023). Strong institutions and the protection of human rights are fundamental for addressing climate change, with democracies performing better than autocracies on climate change mitigation.
Box 4.8. SDG 13 example Sierra Leone’s laws on customary land rights and on the National Land Commission grant communities the right to free prior informed consent over industrial projects and establish land-use committees to secure effective land administration. Granting local communities these rights and slowing disruptive development in ecologically sensitive areas are expected to reduce displacement. The inclusion of women in decision-making processes through local land-use committees is expected to increase gender equity (Maru 2023).
25 4. SPECIFIC SDG LINKAGES WITH SELECT SDGS The world is currently on track to achieve only 17 per cent of the SDG targets (UN DESA 2024). This paper has demonstrated that recalibrating and investing in efforts to achieve more peaceful, just and inclusive societies can support the advancement of other critical elements of the 2030 Agenda. UN Member States and policymakers have a limited window to accelerate action on sustainable development. The recommendations that follow can support progress to that end.
STRENGTHEN HIGH-IMPACT AREAS Increase collaboration across sectors. Forge new partnerships and test joint programming between stakeholders working on SDG 16 and stakeholders in other sectors, especially SDG 1 (No Poverty), SDG 3 (Good Health and Well-being), SDG 4 (Quality Education), SDG 5 (Gender Equality) and SDG 10 (Reduced Inequalities). For example, the integration of health initiatives into governance reforms aims to amplify impacts on both SDG 3 (Good Health and Well-being) and SDG 16. A pilot programme involving both public health departments and legal aid organizations to address health issues exacerbated by legal injustices could be rolled out as a trial. Increase investment for SDG 16. The financing needed to deliver SDG 16 is substantial, requiring sustained investment for long-term development and resilient societies (UNDP 2023). Elements of SDG 16 are being neglected, with vast budgets allocated for military and security sectors and humanitarian relief (Möller-Loswick 2017), and far less for people-centred, community-led initiatives and preventative interventions that support democratic principles (UNDP 2023; Weston 2022; Moy, Cram and Maru 2019; TAP Network 2019). As we approach the 2030 deadline, targeted funding in high-impact areas where SDG 16 can drive progress on other goals should be prioritized. Funders should explore mechanisms to encourage collaborative projects between Chapter 5 RECOMMENDATIONS The world is currently on track to achieve only 17 per cent of the SDG targets.
26 INTERNATIONAL IDEA SDG 16 and other sectors (e.g. multisectoral partnership grants or co-financing arrangements).
Establish mechanisms for information sharing. Establish platforms for sharing data and best practices among actors working on different SDGs. For example, an online repository of case studies and research findings on SDG 16 interlinkages that is accessible to policymakers, researchers and practitioners would help in identifying synergies and enable smarter collaboration and investment. Building such a platform would require investment and technical expertise, and if such a platform were to be created, the parties involved should explore how advances in artificial intelligence could support such a process.
Safeguard those working on SDG 16. Future partnerships must look to safeguard stakeholders working on SDG 16. By increasing transparency, access to information and access to justice, these actors are advancing SDG 16’s aims and other areas of sustainable development. However, it is becoming more dangerous and deadly for human rights defenders and journalists, with a 40 per cent increase in killings and a nearly 300 per cent increase in enforced disappearances from 2021 to 2022 (UNDP 2023). The following questions should be addressed: • What mechanisms can be implemented to facilitate information sharing and coordination between different actors working on high-impact areas?
• How can funding mechanisms be structured to incentivize collaboration between actors across sectors, encouraging joint projects and initiatives? • How can the clustering of interlinkages be better explored? How might simultaneous investments in high-impact areas also link with advancement in other goals: for example, collaboration on participatory decision making (16.7) and SDG 11 (Sustainable Cities and Communities) may lead to advancements in SDG 4 (Quality Education) or SDG 10 (Reduced Inequalities).
ADDRESS RESEARCH GAPS Strengthen the evidence base. All SDG 16 targets and goals and their interconnections with other goals should be further researched. This is especially true for topics with relatively little research and gaps—for example, on pathways and transmission mechanisms. Expanding quantitative research would be especially useful to validate the findings and establish causality. Future studies should employ robust statistical methods to explore interlinkages.
Future partnerships must look to safeguard stakeholders working on SDG 16.
27 5. RECOMMENDATIONS Negative interlinkages. This study has not examined the potential negative impacts that SDG 16 may have on achieving other goals or the trade-offs that may need to be made in efforts to achieve SDG 16 targets—for example, analysing how participatory decision making in renewable energy projects might slow down the transition to green energy, and finding ways to balance these interests. These negative interlinkages should be further researched to develop a holistic understanding of SDG 16’s role. This kind of research would help to design interventions that maximize positive outcomes while mitigating adverse effects. The articles reviewed for this study that examined negative impacts found that SDG 16 had few negative interlinkages. 28 SDG 16 AS AN ENABLER OF THE 2030 AGENDA The analysis presented in this paper underscores the pivotal role of SDG 16 in enabling the achievement of the broader 2030 Agenda. Through an in-depth review of the literature, the evidence reveals that peace, justice and strong institutions are not merely complementary but foundational to sustainable development. Key principles of democracy such as rights, participation, representation and the rule of law are critical enablers across various SDGs, demonstrating robust interlinkages, particularly with SDG 1 (No Poverty), SDG 3 (Good Health and Well-being), SDG 4 (Quality Education), SDG 5 (Gender Equality) and SDG 10 (Reduced Inequalities). There is also evidence for interlinkages with SDG 2 (Zero Hunger), SDG 6 (Clean Water and Sanitation), SDG 8 (Decent Work and Economic Growth), SDG 11 (Sustainable Cities and Communities) and SDG 13 (Climate Action). Without concerted efforts to reverse declines in democracy and the rule of law and reduce the number of conflict-related deaths, attempts to achieve other SDGs will suffer.
However, the current evidence base is heavily skewed towards qualitative research, highlighting the need for more quantitative studies to validate and expand upon these findings. Furthermore, certain areas, such as the interlinkages with SDG 7 (Affordable and Clean Energy) and SDG 9 (Industry, Innovation and Infrastructure), remain under-researched, presenting significant opportunities for future inquiry.
The paper emphasizes the necessity of addressing existing gaps and fostering stronger collaborations across sectors to leverage SDG 16’s full potential. Such an integrated approach is crucial for addressing immediate developmental challenges and building resilient systems capable of withstanding future shocks. By enhancing cross-sector collaboration, increasing targeted investments and expanding the research base, policymakers—united in their vision of a more just and peaceful world—can significantly accelerate progress towards achieving the 2030 Agenda.
Chapter 6 CONCLUSION Key principles of democracy such as rights, participation, representation and the rule of law are critical enablers across various SDGs, demonstrating robust interlinkages.
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S., ‘Dynamic linkages among transparency, income inequality and economic growth in developing countries: Evidence from panel vector autoregressive (PVAR) model’, Research in International Business and Finance, 60 (2022), doi .org/ 10 .1016/ j .ribaf .2021 .101599 SDG 16 Hub, ‘Joint statement by a group of member states in support of SDG16+’, 30 July 2019, www .sdg16hub .org/ topic/ hlpf -joint -statement -group -member -states -support -sdg16 -statement, accessed 29 June 2024 Segato, L., Mattioli, W., Capello, N. and Migliorini, M., ‘Environmental crimes in the water sector’, Water Utility Journal, 20 (2018), pp. 13–27, www .ewra .net/ wuj/ pdf/ WUJ _2018 _20 _02 .pdf, accessed 28 June 2024 Setzer, J. and Higham, C., ‘Global Trends in Climate Change Litigation: 2023 Snapshot’, Grantham Research Institute on Climate Change and the Environment and Centre for Climate Change Economics and Policy, London School of Economics and Political Science, June 2023, www .lse .ac .uk/ granthaminstitute/ wp -content/ uploads/ 2023/ 06/ Global _trends _in _climate _change _litigation _2023 _snapshot .pdf, accessed 27 June 2024 Sohail, M. and Savill, S., ‘Water for the poor: Corruption in water supply and sanitation’, in D. 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36 INTERNATIONAL IDEA Annex B. Sources analysed for the literature review and meta-analysis Academy 4SC, ‘Sustainable Development Goal 16: Peace, justice, and strong institutions’, [n.d.], academy4sc .org/ video/ sustainable -development -goal -16 -peace -justice -and -strong -institutions, accessed 28 June 2024 Aleem Najam, A., Zahid Naeem, M., Ramona, B. and Valeriu, N. P ., ‘Investigating the impact of civil liberties and creative class on innovation output and economic growth: An empirical case study for Pakistan’, Annals of the „Constantin Brâncuşi” University of Târgu Jiu, Economy Series, 3 (2023), www .researchgate .net/ profile/ Ramona -Birau/ publication/ 373549363 _investigating _the _impact _of _civil _liberties _and _creative _class _on _innovation _output _and _economic _growth _an _empirical _case _study _for _pakistan/ links/ 64f0e d64743dc20 a6eb2a748/ investigating -the -impact -of -civil -liberties -and -creative -class -on -innovation -output -and -economic -growth -an -empirical -case -study -for -pakistan .pdf, accessed 27 June 2024 Allen, C., Breuer, A., Kercher, J., Balasubramanian, P ., Leininger, J. and Gadgil, A., ‘The role of good governance in reducing poverty and inequality: Evidence from a scoping review of interlinkages between SDGs 16, 10, and 1’, in A. Breuer, D. Malerba, S. Srigiri and P . Balasubramanian (eds), Governing the Interlinkages between the SDGs: Approaches, Opportunities and Challenges (Abingdon, UK: Routledge, 2023) Anwar, S. and Cooray, A., ‘Financial development, political rights, civil liberties and economic growth: Evidence from South Asia’, Economic Modelling, 29/3 (2012), pp. 974–81, doi .org/ 10 .1016/ j .econmod .2012 .02 .009 Arnold-Fernández, E. E., ‘Beyond Access: Refugees’ Rights and Justice at Work’, New York University Center on International Cooperation, April 2024, cic .nyu .edu/ wp -content/ uploads/ 2024/ 04/ Refugees -Rights -and -Justice -at -Work -2024 .pdf, accessed 27 June 2024 Balasubramanian, P ., Breuer, A., Leininger, J., Cameron, A. and Kercher, J., ‘Sustainable Development Goals (SDG) 16: A Governance Compass towards Just Transition?’, German Institute of Development and Sustainability, IDOS Policy Brief No. 4/2022, doi .org/ 10 .23661/ ipb4 .2022 Barajas-Sandoval, E., Botero-Pinzón, H., Botero, J. C., Pinzón-Rondón, A. M. and Ruiz-Sternberg, A. M., ‘Gender inequality and the rule of law’, Hague Journal on the Rule of Law, 15 (2023), pp. 95–107, link .springer .com/ article/ 10 .1007/ s40803 -022 -00175 -9, accessed 27 June 2024 Beer, C., ‘Democracy and gender equality’, Studies in Comparative International Development, 44 (2009), pp. 212– 27, link .springer .com/ article/ 10 .1007/ s12116 -009 -9043 -2, accessed 27 June 2024 Bell, C., ‘Pirates of the Gulf of Guinea: A Cost Analysis for Coastal States’, Stable Seas, November 2021, www .unodc .org/ documents/ Maritime _crime/ UNODC _Pirates _GoG _A _Cost _Analysis _for _Coastal _States .pdf, accessed 27 June 2024 Bennich, T., Persson, Å., Beaussart, R., Allen, C. and Malekpour, S., ‘Recurring patterns of SDG interlinkages and how they can advance the 2030 Agenda’, One Earth, 6/11 (2023), pp. 1465–76, doi .org/ 10 .1016/ j .oneear .2023 .10 .008 Bircan, C., Brück, T. and Vothknecht, M., ‘Violent conflict and inequality’, Oxford Development Studies, 45/2 (2017), pp. 125–44, doi .org/ 10 .1080/ 13600818 .2016 .1213227 Birner, R., ‘Improving Governance to Eradicate Hunger and Poverty’, International Food Policy Research Institute, 2020 Focus Brief on the World’s Poor and Hungry People, December 2007, idl -bnc -idrc .dspacedirect .org/ server/ api/ core/ bitstreams/ 70d3f7a6 -b042 -4d29 -b5af -df7ade6700c7/ content, accessed 27 June 2024 Biser, J. J. and Edwards, J. A., ‘Civil liberties and access to water: Analysis of 193 countries’, Applied Econometrics and International Development, 12/1 (2012), www .usc .gal/ economet/ reviews/ aeid1212 .pdf, accessed 27 June 2024 37 INTERNATIONAL IDEA Blasiak, R., Wabnitz, C. C. C., Daw, T., Berger, M., Blandon, A., Carneiro, G., Crona, B., Davidson, M. F., Guggisberg, S., Hills, J., Mallin, F., McManus, E., ould-Chih, K., Pittman, J., Santos, X., Westlund, L., Wetterstrand, H. and Wiegler, K., ‘Towards greater transparency and coherence in funding for sustainable marine fisheries and healthy oceans’, Marine Policy, 107 (2019), doi .org/ 10 .1016/ j .marpol .2019 .04 .012 Bonar, H. and Filip, A., ‘COVID-19-related Trafficking of Medical Products as a Threat to Public Health’, United Nations Office on Drugs and Crime, Research Brief, [n.d.], www .unodc .org/ documents/ data -and -analysis/ covid/ COVID -19 _research _brief _trafficking _medical _products .pdf, accessed 27 June 2024 Brolan, C. E., ‘Human rights practitioners cannot be left behind: Engaging in civil registration and vital statistics (CRVS) systems thinking for country implementation of the Sustainable Development Goal agenda’, Journal of Human Rights Practice, 11/1 (2019), pp. 22–51, doi .org/ 10 .1093/ jhuman/ huz009 Bromley, M. and Caparini, M., ‘SDG16.4 and the collection of data on illicit arms flows: Progress made but challenges ahead’, Stockholm International Peace Research Institute, 15 June 2018, www .sipri .org/ commentary/ topical -backgrounder/ 2018/ sdg164 -and -collection -data -illicit -arms -flows -progress -made -challenges -ahead, accessed 27 June 2024 Butkiewicz, J. L. and Yanikkaya, H., ‘Institutional quality and economic growth: Maintenance of the rule of law or democratic institutions, or both?’, Economic Modelling, 23/4 (2006), pp. 648–61, doi .org/ 10 .1016/ j .econmod .2006 .03 .004 Buvinic, M. and Carey, E., Leaving No One Behind: CRVS, Gender and the SDGs, Knowledge Briefs on Gender and CRVS, Brief 1, Paper 2 (Ottawa, Canada: Centre of Excellence for Civil Registration and Vital Statistics Systems, International Development Research Centre, 2019), idl -bnc -idrc .dspacedirect .org/ server/ api/ core/ bitstreams/ 896aa31d -b8d2 -4996 -b069 -451af5903bcb/ content, accessed 27 June 2024 Buvinic, M., Das Gupta, M., Casabonne, U. and Verwimp, P ., ‘Violent conflict and gender inequality: An overview’, The World Bank Research Observer, 28/1 (2013), pp. 110–38, doi .org/ 10 .1093/ wbro/ lks011 Caparini, M., ‘Transnational organized crime: A threat to global public goods’, Stockholm International Peace Research Institute, 2 September 2022, www .sipri .org/ commentary/ topical -backgrounder/ 2022/ transnational -organized -crime -threat -global -public -goods, accessed 27 June 2024 Carnoy, M. and Loeb, S., ‘Does external accountability affect student outcomes? A cross-state analysis’, Educational Evaluation and Policy Analysis, 24/4 (2022), doi .org/ 10 .3102/ 01623737024004305 Cavalieri, M., Finocchiaro Castro, M. and Guccio, C., ‘Organised crime and educational outcomes in Southern Italy: An empirical investigation’, Socio-Economic Planning Sciences, 89 (2023), doi .org/ 10 .1016/ j .seps .2023 .101705 Chen, D. and Deakin, S., ‘On heaven’s lathe: State, rule of law, and economic development’, Law and Development Review, 8/1 (2015), doi .org/ 10 .1515/ ldr -2014 -0031 Chilton, M., Knowles, M., Rabinowich, J. and Arnold, K. T., ‘The relationship between childhood adversity and food insecurity: “It’s like a bird nesting in your head”’, Public Health Nutrition, 18/14 (2015), pp. 2643–53, doi .org/ 10 .1017/ S1368980014003036 Cociña, C., Apsan Frediani, A., Macarthy, J. and Sevilla Núñez, P ., ‘Housing Policy Options to Tackle Urban Inequalities’, Pathfinders for Peaceful, Just and Inclusive Societies and International Institute for Environment and Development, Policy Brief Series, July 2023, s42831 .pcdn .co/ wp -content/ uploads/ 2023/ 07/ Housing -Policy -Options -to -Tackle -Urban -Inequalities -2023 .pdf, accessed 27 June 2024 38 SDG 16 AS AN ENABLER OF THE 2030 AGENDA Cody, C., Count Every Child: The Right to Birth Registration (Woking, UK: Plan Ltd., 2009), www .ohchr .org/ sites/ default/ files/ Documents/ Issues/ Children/ BirthRegistration/ PlanInternational3 _birthRegistration .pdf, accessed 27 June 2024 Cuartas, J., Bhatia, A., Carter, D., Cluver, L., Coll, C., Donger, E., Draper, C. E., Gardner, F., Herbert, B., Kelly, O., Lachman, J., M’jid, N. M. and Seidel, F., ‘Climate change is a threat multiplier for violence against children’, Child Abuse & Neglect, 28 August 2023, doi .org/ 10 .1016/ j .chiabu .2023 .106430 Currie, J. and Spatz Widom, C., ‘Long-term consequences of child abuse and neglect on adult economic well-being’, Child Maltreatment, 15/2 (2010), doi .org/ 10 .1177/ 1077559509355316 Davidson, G., Bunting, L., Bywaters, P ., Featherstone, B. and McCartan, C., ‘Child welfare as justice: Why are we not effectively addressing inequalities?’, The British Journal of Social Work, 47/6 (2017), pp. 1641– 51, doi .org/ 10 .1093/ bjsw/ bcx094 De Bremond, A. and Engle, N. L., ‘Adaptation policies to increase terrestrial ecosystem resilience: Potential utility of a multicriteria approach’, Mitigation and Adaptation Strategies for Global Change, 19/3 (2014), pp. 331–54, doi .org/ 10 .1007/ s11027 -014 -9541 -z Del Río Duque, M. L., Rodríguez, T., Pérez Lora, Á. P ., Löhr, K., Romero, M., Castro-Nunez, A., Sieber, S. and Bonatti, M., ‘Understanding systemic land use dynamics in conflict-affected territories: The cases of Cesar and Caquetá, Colombia’, PLoS ONE, 17/5 (2022), doi .org/ 10 .1371/ journal .pone .0269088 Dhakal, R. K., ‘Promoting gender inclusive governance to deliver better education in Nepal’, International Journal of Social Sciences and Educational Studies, 6/1 (2019), doi .org/ 10 .23918/ ijsses .v6i1p83 Di Cataldo, M. and Mastrorocco, N., ‘Organized crime, captured politicians, and the allocation of public resources’, The Journal of Law, Economics, & Organization, 38/3 (2022), pp. 774–839, doi .org/ 10 .1093/ jleo/ ewab015 Dingake, O. B. 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SOURCES ANALYSED FOR THE LITERATURE REVIEW AND META-ANALYSIS Jahanzeb Butt, M., ‘The role of the international law in shaping the governance for Sustainable Development Goals’, Journal of Law and Political Sciences, 28/3 (2021), papers .ssrn .com/ sol3/ papers .cfm ?abstract _id = 3857234, accessed 27 June 2024 Jenkins, M., ‘Tracking corruption across the Sustainable Development Goals’, Transparency International [blog], 25 March 2021, www .transparency .org/ en/ blog/ tracking -corruption -across -the -sustainable -development -goals, accessed 27 June 2024 Jennett, V., Enabling the Implementation of the 2030 Agenda through SDG 16+: Anchoring Peace, Justice and Inclusion (New York, NY: United Nations, 2019), www .sdg16hub .org, accessed 27 June 2024 Justino, P ., Leone, M. and Salardi, P ., ‘Short- and long-term impact of violence on education: The case of Timor Leste’, The World Bank Economic Review, 28/2 (2014), pp. 320–53, doi .org/ 10 .1093/ wber/ lht007 Kallergis, A., ‘At Risk: Environmental Mobility in African Coastal Cities’, Robert Bosch Stiftung, August 2022, www .bosch -stiftung .de/ sites/ default/ files/ publications/ pdf/ 2022 -09/ Publikation _At %20Risk _Environmental %20Mobility %20in %20African %20Coastal %20Cities _0 .pdf, accessed 27 June 2024 Kearney, J., Berkes, F., Charles, A., Pinkerton, E. and Wiber, M., ‘The role of participatory governance and community-based management in integrated coastal and ocean management in Canada’, Coastal Management, 35/1 (2007), pp. 79–104, doi .org/ 10 .1080/ 10 .1080/ 08920750600970511 Kempin Reuter, T., ‘Human rights and the city: Including marginalized communities in urban development and smart cities’, Journal of Human Rights, 18/4 (2019), pp. 382–402, doi .org/ 10 .1080/ 14754835 .2019 .1629887 Kercher, J. and Breuer, A., ‘SDG 16 Interlinkages: Summary of Research Findings’, UNDP Oslo Governance Centre, July 2021, www .undp .org/ policy -centre/ governance/ publications/ sdg -16 -interlinkages -summary -findings, accessed 27 June 2024 Kishore, J. and Tripathi, N., ‘Role of community participation in achieving universal health coverage’, International Journal of Preventive, Curative & Community Medicine, 5/2 (2019), medical .advanced researchpu blications .com/ index .php/ Preventive -Curative -CommunityMed/ article/ view/ 95, accessed 27 June 2024 Kjorram-Manesh, A., ‘Global transition, global risks, and the UN’s Sustainable Development Goals: A call for peace, justice, and political stability’, Global Transitions, 5 (2023), pp. 90–97, doi .org/ 10 .1016/ j .glt .2023 .06 .002 Klima, N., ‘Preventing/countering violent extremism and the United Nations Sustainable Development Goals: An impact nexus’, Freedom from Fear Magazine, No. 17, UNICRI, 2023, pp. 3–22, biblio .ugent .be/ publication/ 01H13X 5HHTN156SM DVB3193JE8, accessed 27 June 2024 Klima, N., Vasconcelos Senra, J. 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SOURCES ANALYSED FOR THE LITERATURE REVIEW AND META-ANALYSIS Mayanja, C. S. and Nkata, J. L., ‘Pursuing Sustainable Development Goals in Uganda: Do anti-corruption strategies of management development institutes matter?’, Journal of Public Administration and Policy Research, 11/4 (2019), pp. 38–47, doi .org/ 10 .5897/ JPAPR2019 .0457 Mbise, A. T., ‘Birth certificates, birth registration and the denial of human rights: Evidence from Tanzania National Panel Data 2010/11’, The International Journal of Children’s Rights, 28/2 (2020), pp. 243–57, doi .org/ 10 .1163/ 15718182 -02802002 McDermott, C. L., Acheampong, E., Arora-Jonsson, S., Asare, R., de Jong, W., Hirons, M., Khatun, K., Menton, M., Nunan, F., Poudyal, M. and Setyowati, A., ‘SDG 16: Peace, justice and strong institutions—A political ecology perspective’, in P . Katila, C. J. Pierce Colfer, W. de Jong, G. Galloway, P . Pacheco and G. Winkel (eds), Sustainable Development Goals: Their Impacts on Forests and People (Cambridge University Press, 2019), gala .gre .ac .uk/ id/ eprint/ 27125/ 7/ 27125 %20KHATUN _SDG _16 _Peace _Justice _And _String _Institutions _Political _Ecology _Perspective _ %28OA %29 _2019 .pdf, accessed 27 June 2024 Milton, S., ‘Higher education and Sustainable Development Goal 16 in fragile and conflict-affected contexts’, Higher Education, 81 (2021), pp. 89–108, doi .org/ 10 .1007/ s10734 -020 -00617 -z Mollick, A. S., Rahman, K., Khan, N. I. and Sadath, N., ‘Evaluation of good governance in a participatory forestry program: A case study in Madhupur Sal forests of Bangladesh’, Forest Policy and Economics, 95 (2018), pp. 123–37, doi .org/ 10 .1016/ j .forpol .2018 .07 .014 Moy, A., Cram, S. and Maru, V., ‘The Case to Fund and Protect Grassroots Justice Defenders’, Justice for All, Policy Brief, January 2019, gras srootsjust icenetwork .org/ wp -content/ uploads/ 2019/ 01/ Justice -For -All _Policy -Brief _MedRes _5 .31 .2019 .pdf, accessed 28 June 2024 Mugumya, F., Asaba, R. B., Kamya, I. R. and Asingwire, N., ‘Children and domestic water collection in Uganda: Exploring policy and intervention options that promote child protection’, in D. Kaawa-Mafigiri and E. Walakira (eds), Child Abuse and Neglect in Uganda, Child Maltreatment: Contemporary Issues in Research and Policy, vol. 6 (Springer, 2017), doi .org/ 10 .1007/ 978 -3 -319 -48535 -5 _6 Mukombwe, J. S., du Toit, A. and Hendriks, S. L., ‘Sustainable Development Goal 16: Peace, justice and strong institutions’, in S. L. Hendriks and S. C. Babu (eds), Handbook on Public Policy and Food Security (Elgaronline, 2024), doi .org/ 10 .4337/ 9781839105449 .00041 Nagy, R., Bérczi, L., Sáfár, B. and Kállai, K., ‘The relationship of environmental migration and human trafficking concerning natural hazards at the affected regions of Africa’, Journal of Central and Eastern European African Studies, 3/1 (2024), pp. 17–46, doi .org/ 10 .59569/ jceeas .2023 .3 .1 .209 Nayyer, M. I., Aravindan Mukkai, R. and Thillai Rajan, A., ‘Effect of transparency on the development phase of public-private partnership: Analysis of highway projects’, IOP Conference Series: Earth and Environmental Science, vol. 1101, Planning Partnership and Law, 2022, doi .org/ 10 .1088/ 1755 -1315/ 1101/ 5/ 052019 Newig, J. and Fritsch, O., ‘Environmental governance: Participatory, multi-level—and effective?’, Environmental Policy and Governance, 19/3 (2009), pp. 197–214, doi .org/ 10 .1002/ eet .509 Ngoc Nguyen, B. and Gordon, M., ‘Human trafficking and gender inequality in remote communities of central Vietnam’, Journal of Sustainable Social Change, 12/1 (2019), scholarworks .waldenu .edu/ jsc/ vol12/ iss1/ 11/ , accessed 28 June 2024 Nifo, A., Scalera D. and Vecchione, G., ‘The rule of law and educational choices: Evidence from Italian regions’, Regional Studies, 51/7 (2017), pp. 1048–62, doi .org/ 10 .1080/ 00343404 .2016 .1262945 Nilsson, M., Chisholm, E., Griggs, D., Howden-Chapman, P ., McCollum, D., Messerli, P ., Neumann, B., Stevance, A.-S., Visbeck, M. and Stafford-Smith, M., ‘Mapping interactions between the Sustainable Development Goals: Lessons learned and ways forward’, Sustainability Science, 13 (2018), pp. 1489– 503, doi .org/ 10 .1007/ s11625 -018 -0604 -z 44 SDG 16 AS AN ENABLER OF THE 2030 AGENDA Nilsson, M., Griggs, D. and Visbeck, M., ‘Policy: Map the interactions between Sustainable Development Goals’, Nature, 534 (2016), pp. 320–22, doi .org/ 10 .1038/ 534320a Norgaard, S., ‘Violence as a genre of urban life: Urban sustainability and (in)security in South African cities’, Journal of Urban Affairs, 45/3 (2023), pp. 630–46, doi .org/ 10 .1080/ 07352166 .2021 .1977140 O’Neil Richard, A., International Trafficking in Women to the United States: A Contemporary Manifestation of Slavery and Organized Crime (Washington, DC: Center for the Study of Intelligence, 1999), www .cia .gov/ resources/ csi/ static/ 9dc 85527075bc 84f9e1f2ee f0e7a0915/ trafficking .pdf, accessed 28 June 2024 Organisation for Economic Co-operation and Development (OECD), Equal Access to Justice: OECD Roundtable Background Notes (Paris: OECD, 2015), www .oecd .org/ gov/ Equal -Access -Justice -Roundtable -background -note .pdf, accessed 28 June 2024 Owolabi, E. 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Le Billon (eds), Corruption, Natural Resources and Development: From Resource Curse to Political Ecology (Cheltenham, UK: Edward Elgar Publishing, 2017), doi .org/ 10 .4337/ 9781785361203 .00015 TAP Network, ‘Amplified Commitments and Partnerships for Accelerated Action: Rome Civil Society Declaration on SDG16+’, May 2019, www .idlo .int/ system/ files/ event -documents/ Rome %20 Civil %20Society %20Declaration %20on %20SDG16 %2B %20 - %20FINAL .PDF, accessed 28 June 2024 —, SDG16+ Civil Society Toolkit: A Practical Resource Guide for Planning, Implementation and Accountability to Build Peaceful, Just and Inclusive Societies, [n.d.], www .sdg16toolkit .org, accessed 28 June 2024 Turner, M., Ginifer, J. and Cliffe, L., ‘The Impact of Armed Violence on Poverty and Development: Full Report of the Armed Violence and Poverty Initiative’, Centre of International Cooperation and Security, Department of Peace Studies, University of Bradford, 2005, bradscholars .brad .ac .uk/ bitstream/ handle/ 10454/ 1006/ AVPI _Synthesis _Report .pdf, accessed 28 June 2024 47 ANNEX B. SOURCES ANALYSED FOR THE LITERATURE REVIEW AND META-ANALYSIS UN Women, International Development Law Organization, World Bank and Pathfinders for Peaceful, Just and Inclusive Societies, ‘Justice for Women: High-Level Group Report’, 2019, www .idlo .int/ sites/ default/ files/ pdfs/ publications/ Justice %20For %20Women %20Report %20FINAL .pdf, accessed 27 June 2024 United Nations Department of Economic and Social Affairs, ‘Energy and SDG 16: Peace, Justice and Strong Institutions’, Policy Brief 7, 2019, sdgs .un .org/ sites/ default/ files/ 2021 -05/ POLICY %20BRIEF %207 %20 - %20ENERGY %20AND %20SDG %2016 %20PEACE %2C %20JUSTICE %20AND %20STRONG %20 INSTITUTIONS .pdf, accessed 28 June 2024 United Nations Development Programme, ‘A Resilient Future for All: Advancing Disability Inclusion in Disaster Risk Reduction and Recovery for Building Resilience and Leaving No One Behind’, Policy Brief, 28 November 2023, www .undp .org/ publications/ resilient -future -all -advancing -disability -inclusion -disaster -risk -reduction -and -recovery -building -resilience -and -leaving -no -one -behind, accessed 28 June 2024 United Nations Educational, Scientific and Cultural Organization, ‘Education for sustainable development’, [n.d.], www .unesco .org/ en/ sustainable -development/ education, accessed 28 June 2024 United Nations Environment Programme, ‘Goal 16: Peace, justice and strong institutions’, [n.d.], www .unep .org/ explore -topics/ sustainable -development -goals/ why -do -sustainable -development -goals -matter/ goal -16, accessed 28 June 2024 United Nations Global Compact, ‘Blueprint for business leadership on the SDGs’, 2018, blueprint .unglobalcompact .org/ sdgs/ sdg16/ , accessed 28 June 2024 United Nations Office of the High Commissioner for Human Rights, ‘Conflict and violence are the primary causes of hunger and famine, Special Rapporteur on the Right to Food tells the Human Rights Council’, 9 March 2023, www .ohchr .org/ en/ news/ 2023/ 03/ conflict -and -violence -are -primary -causes -hunger -and -famine -special -rapporteur -right, accessed 28 June 2024 United Nations Office on Drugs and Crime (UNODC), United Nations Office of the High Commissioner for Human Rights (OHCHR) and United Nations Development Programme (UNDP), Global Progress Report on Sustainable Development Goal 16 Indicators: A Wake-Up Call for Action on Peace, Justice and Inclusion (UNODC, OHCHR and UNDP , 2023), www .undp .org/ publications/ global -progress -report -sustainable -development -goal -16 -indicators -wake -call -action -peace -justice -and -inclusion, accessed 28 June 2024 Unruh, J., Pritchard, M., Savage, E., Wade, C., Nair, P ., Adenwala, A., Lee, L., Malloy, M., Taner, I. and Frilander, M., ‘Linkages between large-scale infrastructure development and conflict dynamics in East Africa’, Journal of Infrastructure Development, 11/1–2 (2019), pp. 1–13, doi .org/ 10 .1177/ 0974930619872082 V-Dem Institute, ‘Case for Democracy’, March 2023, www .v -dem .net/ documents/ 34/ C4DReport _230421 .pdf, accessed 28 June 2024 Vashchenko, Y., ‘Access to modern energy services through the prism of children’s rights: An overview from the perspectives of the Convention on the Rights of the Child and the policy and law approaches of certain EU member states and Ukraine’, International Comparative Jurisprudence, 7/1 (2021), pp. 75– 87, doi .org/ 10 .13165/ j .icj .2021 .06 .006 Vazeer, T., ‘SDG 16 as an enabler for the 2030 Agenda’, United Nations Development Programme, [n.d.], www .idea .int/ sites/ default/ files/ news/ news -pdfs/ Tomoko _Vazzer -PPT -SDG16 -enabler -2030 -Agenda .pdf, accessed 28 June 2024 Vilhena Vieira, O., ‘Inequality and the subversion of the rule of law’, International Journal on Human Rights, 6 (2007), sur .conectas .org/ en/ inequality -subversion -rule -law/ , accessed 28 June 2024 48 SDG 16 AS AN ENABLER OF THE 2030 AGENDA Vozza, D. and Coluccello, S., ‘Transnational organized crime and UN Sustainable Development Goals’, in W. Leal Filho, A. M. Azul, L. Brandli, A. Lange Salvia, P . G. Özuyar and T. Wall (eds), Encyclopedia of the UN Sustainable Development Goals, Peace, Justice and Strong Institutions (Springer, 2021), doi .org/ 10 .1007/ 978 -3 -319 -71066 -2 _132 -1 Wagacha, W., ‘Access to Information as a Driver towards Closing of the Gender Equality Gap: The Emerging Scene in Kenya’, Working Paper, World Library and Information Congress, 2007, www .academia .edu/ 46884235/ Access _to _information _as _a _driver _towards _closing _of _the _gender _equality _gap _the _emerging _scene _in _Kenya, accessed 28 June 2024 Wallengren, E., Guthold, R., Newby, H., Moller, A.-B., Marsh, A. D., Fagan, L., Azzopardi, P ., Guèye M. and Kågesten, A. E., ‘Relevance of the Sustainable Development Goals (SDGs) to adolescent health measurement: A systematic mapping of the SDG framework and global adolescent health indicators’, Journal of Adolescent Health, 7/6 Supplement (2024), pp. S47–S55, doi .org/ 10 .1016/ j .jadohealth .2024 .01 .004 Warner, R. M., ‘Implementing the rule of law for nature in the global marine commons: Developing environmental assessment frameworks’, in C. Voigt (ed), Rule of Law for Nature: New Dimensions and Ideas in Environmental Law (Cambridge, UK: Cambridge University Press, 2013), doi .org/ 10 .1017/ CBO9781107337961 .028 Weikmans, R., van Asselt, H. and Timmons Roberts, J., ‘Transparency requirements under the Paris Agreement and their (un)likely impact on strengthening the ambition of nationally determined contributions (NDCs)’, Climate Policy, 20/4 (2020), pp. 511–26, doi .org/ 10 .1080/ 14693062 .2019 .1695571 Weitz, N., Carlsen, H., Nilsson, M. and Skånberg, K., ‘Towards systemic and contextual priority setting for implementing the 2030 Agenda’, Sustainability Science, 13 (2018), pp. 531–48, doi .org/ 10 .1007/ s11625 -017 -0470 -0 Weston, M., ‘The Benefits of Access to Justice for Economies, Societies, and the Social Contract’, Open Government Partnership and Pathfinders for Peaceful, Just and Inclusive Societies, 2022, 6c192f99 -3663 -4169 -a572 -e50276ce5d6d .usrfiles .com/ ugd/ 6c192f _c2 40bf284a8b 40a19e715c 152af1b4f0 .pdf, accessed 27 June 2024 White, B. T., Sepe, S. M. and Masconale, S., ‘Urban decay, posterity, and the rule of law’, Emory Law Journal, 64/1 (2014), heinonline .org/ HOL/ LandingPage ?handle = hein .journals/ emlj64 & div = 5 & id = & page = , accessed 28 June 2024 Wigley, S., Dieleman, J. L., Templin, T., Mumford, J. E. and Bollyky, T. J., ‘Autocratisation and universal health coverage: Synthetic control study’, BMJ (2020), doi .org/ 10 .1136/ bmj .m4040 Witbooi, E., Ali, K.-D., Achmad Santosa, M., Hurley, G., Husein, Y., Maharaj, S., Okafor-Yarwood, I., Arroyo Quiroz, I. and Salas, O., ‘Organized crime in the fisheries sector threatens a sustainable ocean economy’, Nature, 588 (2020), pp. 48–56, doi .org/ 10 .1038/ s41586 -020 -2913 -5 Xavier, R., Komendantova, N., Jarbandhan, V. and Nel, D., ‘Participatory governance in the transformation of the South African energy sector: Critical success factors for environmental leadership’, Journal of Cleaner Production, 154 (2017), pp. 621–32, doi .org/ 10 .1016/ j .jclepro .2017 .03 .146 Zaitch, D., Boekhout van Solinge, T. and Müller, G., ‘Harms, crimes and natural resource exploitation: A green criminological and human rights perspective on land-use change’, in M. Bavinck, L. Pellegrini and E. Mostert (eds), Conflicts over Natural Resources in the Global South: Conceptual Approaches (Leiden, Netherlands: CRC Press, 2014) Zhou, A. and Li, J., ‘Impact of anti-corruption and environmental regulation on the green development of China’s manufacturing industry’, Sustainable Production and Consumption, 27 (2021), pp. 1994–60, doi .org/ 10 .1016/ j .spc .2021 .04 .031 49 ANNEX B. SOURCES ANALYSED FOR THE LITERATURE REVIEW AND META-ANALYSIS About the author Stacey Cram is a non-resident fellow at New York University’s Center on International Cooperation and an independent advisor, providing support to non-profits and multilateral organizations dedicated to advancing SDG 16 and people-centred justice. For seven years, she led Namati’s global policy portfolios, playing a pivotal role in advocating the inclusion of a justice goal in the 2030 Agenda and the full implementation of SDG 16. She spearheaded the creation of the COVID-19 Grassroots Justice Fund and the Legal Empowerment Fund, and played a key role in establishing global initiatives like the Task Force on Justice and Open Government Partnership Coalition on Justice, contributing to the design of indicators that holistically examine the global justice gap. Stacey holds an MSc in Global Governance and Public Policy from Birkbeck College, University of London, and an MA in International Relations from the University of St Andrews. 50 COVER_MAIN_TITLE CHAPTER_TITLE About International IDEA The International Institute for Democracy and Electoral Assistance (International IDEA) is an intergovernmental organization with 35 Member States founded in 1995, with a mandate to support sustainable democracy worldwide.
WHAT WE DO We develop policy-friendly research related to elections, parliaments, constitutions, digitalization, climate change, inclusion and political representation, all under the umbrella of the UN Sustainable Development Goals. We assess the performance of democracies around the world through our unique Global State of Democracy Indices and Democracy Tracker. We provide capacity development and expert advice to democratic actors including governments, parliaments, election officials and civil society. We develop tools and publish databases, books and primers in several languages on topics ranging from voter turnout to gender quotas. We bring states and non-state actors together for dialogues and lesson sharing. We stand up and speak out to promote and protect democracy worldwide.
WHERE WE WORK Our headquarters is in Stockholm, and we have regional and country offices in Africa and West Asia, Asia and the Pacific, Europe, and Latin America and the Caribbean. International IDEA is a Permanent Observer to the United Nations and is accredited to European Union institutions.
OUR PUBLICATIONS AND DATABASES We have a catalogue with more than 1,000 publications and over 25 databases on our website. Most of our publications can be downloaded free of charge.
About International IDEA This Policy Paper explores the interlinkages between Sustainable Development Goal (SDG) 16 (Peace, Justice and Strong Institutions) and the 2030 Agenda’s other Sustainable Development Goals. Despite SDG 16’s crucial role in promoting democratic principles, human rights and the rule of law, it is often overlooked, and the financing needed to deliver SDG 16 is substantial, requiring sustained investment for long-term development. This paper provides a comprehensive literature review and meta-analysis of 185 articles and highlights how SDG 16 acts as an enabler for other SDGs.
This paper calls for increased cross-sector collaboration and offers recommendations to UN Member States, policymakers working across the 2030 Agenda and other stakeholders working in areas where SDG 16 has high-impact interlinkages to make targeted investments, enhance research and safeguard those working on SDG 16 in order to fully leverage its potential in achieving outcomes on the 2030 Agenda.
International IDEA Strömsborg SE–103 34 Stockholm SWEDEN +46 8 698 37 00 info@idea.int www.idea.int ISBN: 978-91-7671-782-0 (PDF) |
11934 | https://pmc.ncbi.nlm.nih.gov/articles/PMC11577807/ | Non pathological sweat test, pancreatic insufficiency and Cystic Fibrosis: an unusual case in a child with F508del-duplication of exons 1–3 CFTR genotype - PMC
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. 2024 Nov 20;24:752. doi: 10.1186/s12887-024-05154-7
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Non pathological sweat test, pancreatic insufficiency and Cystic Fibrosis: an unusual case in a child with F508del-duplication of exons 1–3 CFTR genotype
Vito Terlizzi
Vito Terlizzi
1 Cystic Fibrosis Regional Reference Centre, Department of Paediatric Medicine, Meyer Children’s Hospital IRCCS, Viale Gaetano Pieraccini 24, Florence, 50139 Italy
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1,✉, Cristina Fevola
Cristina Fevola
1 Cystic Fibrosis Regional Reference Centre, Department of Paediatric Medicine, Meyer Children’s Hospital IRCCS, Viale Gaetano Pieraccini 24, Florence, 50139 Italy
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1, Alice Castaldo
Alice Castaldo
1 Cystic Fibrosis Regional Reference Centre, Department of Paediatric Medicine, Meyer Children’s Hospital IRCCS, Viale Gaetano Pieraccini 24, Florence, 50139 Italy
2 Dipartimento di Scienze Mediche Traslazionali, Sezione di Pediatria, Università di Napoli Federico II, Naples, Italy
3 SC di Pneumologia e UTSIR, AORN Santobono-Pausilipon, Naples, Italy
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1,2,3, Selene Del Vespa
Selene Del Vespa
4 Meyer Children’s Hospital IRCCS, Department of Health Sciences, University of Florence, Florence, Italy
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4, Daniela Dolce
Daniela Dolce
1 Cystic Fibrosis Regional Reference Centre, Department of Paediatric Medicine, Meyer Children’s Hospital IRCCS, Viale Gaetano Pieraccini 24, Florence, 50139 Italy
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1, Luca Scarallo
Luca Scarallo
5 Gastroenterology and Nutrition Unit, Meyer Children’s Hospital IRCCS, Florence, Italy
6 Department of Neurofarba, University of Florence, Florence, Italy
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5,6, Karina Kleinfelder
Karina Kleinfelder
7 Department of Medicine, Division of General Pathology, University of Verona, Verona, Italy
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7, Paola Melotti
Paola Melotti
8 Cystic Fibrosis Centre, Azienda Ospedaliera Universitaria Integrata, Verona, Italy
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8, Claudio Sorio
Claudio Sorio
7 Department of Medicine, Division of General Pathology, University of Verona, Verona, Italy
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7, Giovanni Taccetti
Giovanni Taccetti
1 Cystic Fibrosis Regional Reference Centre, Department of Paediatric Medicine, Meyer Children’s Hospital IRCCS, Viale Gaetano Pieraccini 24, Florence, 50139 Italy
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1, Paolo Lionetti
Paolo Lionetti
5 Gastroenterology and Nutrition Unit, Meyer Children’s Hospital IRCCS, Florence, Italy
6 Department of Neurofarba, University of Florence, Florence, Italy
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5,6
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1 Cystic Fibrosis Regional Reference Centre, Department of Paediatric Medicine, Meyer Children’s Hospital IRCCS, Viale Gaetano Pieraccini 24, Florence, 50139 Italy
2 Dipartimento di Scienze Mediche Traslazionali, Sezione di Pediatria, Università di Napoli Federico II, Naples, Italy
3 SC di Pneumologia e UTSIR, AORN Santobono-Pausilipon, Naples, Italy
4 Meyer Children’s Hospital IRCCS, Department of Health Sciences, University of Florence, Florence, Italy
5 Gastroenterology and Nutrition Unit, Meyer Children’s Hospital IRCCS, Florence, Italy
6 Department of Neurofarba, University of Florence, Florence, Italy
7 Department of Medicine, Division of General Pathology, University of Verona, Verona, Italy
8 Cystic Fibrosis Centre, Azienda Ospedaliera Universitaria Integrata, Verona, Italy
✉
Corresponding author.
Received 2024 Apr 12; Accepted 2024 Oct 16; Collection date 2024.
© The Author(s) 2024
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PMCID: PMC11577807 PMID: 39567905
Abstract
While Cystic Fibrosis is characterized by a high phenotypic variability, a correlation is reported between the pancreatic status and the CFTR genotype. Here we report an unusual case of a child with Cystic Fibrosis (F508del-duplication of exons 1–3 genotype) diagnosed at 8 years old for pancreatic insufficiency and non-pathological sweat test, in absence of respiratory symptoms and acute episodes of pancreatitis. Nasal potential differences and intestinal current measurements were normal, while the short-circuit current measured on patient-derived colonoids grown on Transwell® indicated the presence of a reduced CFTR-dependent current relative to non-CF colonoids with, a modest improvement of CFTR activity record following treatment with elexacaftor/tezacaftor/ivacaftor.
This case opens the discussion on the importance of performing CFTR sequencing and the search for large gene rearrangements in cases of pancreatic insufficiency of unclear etiology, also in the presence of non-pathological sweat test. Children with CF and non-pathological sweat chloride are likely to develop higher concentrations if they truly have CF.
Supplementary Information
The online version contains supplementary material available at 10.1186/s12887-024-05154-7.
Keywords: Normal, Sweat chloride, Duplication, Diagnosis, CFTR, Genotype
Background
Cystic Fibrosis (CF) is an autosomal recessive disorder caused by variants in the cystic fibrosis transmembrane conductance regulator (CFTR) gene. The CF phenotype is characterized by lung disease (bronchiectasis with persistent airway-based infection and inflammation), exocrine pancreatic insufficiency, associated with nutrient malabsorption, impaired growth, hepatobiliary manifestations, and male infertility [1, 2]. The diagnosis of CF is based on positive newborn screening (NBS), clinical features consistent with CF or a positive family history, in presence of a pathological sweat chloride (SC) concentration (≥ 60 mmol/L) and/or two CFTR-causing variants in trans (that is, on two distinct alleles) . More than 2,000 CFTR variants have been recorded so far worldwide ( accessed on 19 August 2024) although to date only 719 CFTR variants are known to be CF-causing ( accessed on 19 August 2024) based on functional studies. They are typically categorized into six classes, according to their impact on the production, trafficking, functioning or stability of the CFTR channel . A clear and unequivocal genotype–phenotype correlation exists for only a few variants. Also, the variable penetrance of some variants must be also considered. While CF is a monogenic disease, it is characterized by a high phenotypic variability, also in siblings and twins, due to modifier genes other than CFTR or to environmental factors [5–8]. A correlation is reported between the pancreatic status and the CFTR genotype, given that variants belonging to classes I, II and III usually result in little to no CFTR activity, leading to potentially severe clinical outcomes, pathological sweat test (ST) and pancreatic insufficiency, whilst variants from classes IV, V and VI allow significant residual CFTR function leading to milder phenotypes, sometimes ST in the intermediate range (SC 30–59 mmol/l) and typically pancreatic sufficiency [5–9].
Here we present an unusual case of a child with CF diagnosed for pancreatic insufficiency in presence of a non-pathological ST.
Case presentation
This is the case of an Indian child, not screened at birth for CF, who presented to the local Hospital at the age of 7 years, with gastrointestinal symptoms (chronic diarrhea, recurrent abdominal pain, abdominal bloating) and weight loss (weight < 3rd percentile, -2 SD). The symptoms persisted for 3–4 months in the absence of previous episodes of acute pancreatitis or respiratory symptoms.
Probiotics and proton pump inhibitors were prescribed, as well as different types of dietary regimens (e.g. gluten-free diet, dairy-free diet, no processed food diet), with no improvement of symptoms. Several investigations were carried out to rule out a gastrointestinal pathology: blood exams (including inflammatory markers, ASCA and ANCA serology, and screening for celiac disease), bowel ultrasound scan, microbiologic stool tests and level of fecal calprotectin, with all of them being in normal range. Endoscopic examinations (esophagogastroduodenoscopy and sigmoidorectoscopy) revealed only a minimal focal gastritis associated with H. pylori detection.
The patient and his family moved to Italy, where the child was firstly evaluated at our Gastroenterology and Nutrition Unit. Fecal elastase levels were tested, resulting to be below normal range (34 µg/g and < 15 µg/g in two different measurements), indicating severe pancreatic insufficiency (for values < 100 µg/g). Consequently, pancreatic enzyme replacement therapy (PERT) was prescribed, improving symptoms and nutritional status (supplemental figure A). To identify the cause of pancreatic insufficiency we performed a ST resulting in the normal range (SC: 26 mmol/L). A chest X-ray was performed in the suspect of Shwachman-Diamond syndrome, excluding the presence of rib abnormalities (Table 1).
Table 1.
Outcome measures before and after 3 and 12 months of ETI therapy
| Patient | SC (mmol/L) | BMI | FEV 1% | NPD | OBAS | ICM (µA/cm2) |
:--- :--- :---
| Male, 9 y |
| F508del/ duplication of exons 1–3 |
| Pre ETI therapy | 26;38;51 | 14.08 | 87 | 0.39 ± 0.08 | 0.063 | 424 ± 261 |
| Three months post ETI therapy | 10 | 15.15 | 90 | NA | NA | NA |
| Twelve months post ETI therapy | 16 | 14.38 | 94 | NA | NA | NA |
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in presence of pancreatic insufficiency and diffuse bronchiectasis at chest computer tomography scan
Abbreviations SC: sweat chloride; BMI: body mass index; FEV 1: forced expiratory volume in 1 second; OBAS Test: Optical Beta-Adrenergic Sweat Test; NPD: nasal potential difference; ICM: intestinal current measurements
Despite non pathological SC value, given the presence of pancreatic insufficiency, CFTR gene sequencing with multiplex ligation-dependent probe amplification was carried out. It revealed two CFTR variants: F508del and duplication of exons 1–3. Subsequent parental analysis confirmed the presence of most frequent CF causing variant F508del on one allele (inherited from the father) and duplication of exons 1–3 on the other allele (inherited from the mother), not reported in the CFTR2 database ( Subsequent STs resulted in the intermediate range (SC: 38 and 51 mmol/L). However, next generation sequencing analysis for Shwachman-Diamond syndrome, congenital diarrhea syndromes and genes associated with pancreatitis [11, 12] was negative, excluding the concomitant presence of variants that could explain CF’s predominantly gastrointestinal involvement. Given the peculiar presentation of this case we performed additional functional tests, such as Optical Beta-Adrenergic Sweat Test (OBAS Test) [13, 14], nasal potential difference (NPD) measurements and intestinal current measurements (ICM), according to the standardized operating procedure from European CF Society (ECFS) ( and consensus guidelines from the Cystic Fibrosis Foundation . OBAS test provided a ratio of CFTR dependent sweat rate vs. CFTR- independent sweat rate of 0,063 that is in the range of CFTR related disorders (Fig.1), while following NPD measurements we calculated a Wilschanski Index (WI) of 0,39 ± 0,08. That is in non-CF range . Again, ICM result was in the normal range (424 ± 261 µA/cm 2) (supplemental figure B).
Fig. 1.
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Representative image of CFTR-dependent and -independent sweat secretion rates for the CF subject under evaluation. (A) Each image focuses on a small region of stimulated forearm skin of the selected subjects to show blue-stained sweat droplets by single sweat glands that responded to methacholine (M-sweating) and to the β-adrenergic cocktail (C-sweating), demonstrating the activity of the CFTR channel (C-sweat); grid scale = 0.5 mm. (B) Comparison of average C/M sweat rates measured in each group (control: non-CF, healthy carriers: HC, and cystic fibrosis: CF) and patient analyzed (subject)
Pulmonary investigations were performed, revealing diffuse bronchiectasis of grade 1 on the chest computed tomography scan, normal percent predicted forced expiratory volume in 1 s (ppFEV 1: 95%) and lung clearance index (6.46, software version 3.3.1, upper limit of normal 7.10) . The patient’s culture pharyngeal swab was negative for CF-related pathogens.
Finally, we set to evaluate a potential in vivo benefit of elexacaftor/tezacaftor/ivacaftor (ETI) treatment, in Ussing chamber, using dissociated organoids, derived from the rectal biopsies, cultured in Transwell®, as previously described [18, 19]. Interestingly, non-corrected organoid-derived monolayers (colonoids) responded to the CFTR agonist forskolin with a ΔIsc value of 27 ± 29 µA/cm 2, a result that confirm the presence of residual function and highlight a lower activity than the average recorded for non-CF organoids (ΔIsc 111 ± 15 µA/cm 2), corresponding to a 24% of WT activity in our setting. The ETI treatment led to a modest increment in CFTR activity with Fsk-stimulated currents of ΔIsc 43 ± 29 µA/cm 2 (p = 0.7686) reaching approximately 39% of the WT activity in our setting (Fig.2).
Fig. 2.
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Functional evaluation of modulator treatment on F508del/dup. exons 1–3 rectal organoids. (A) Representative Isc traces of the effect of the vehicle alone (DMSO; black trace), or the correctors ELEXA/TEXA; (ET, 3µM each; blue trace). During the recordings, the colonoids were sequentially treated (as indicated by downward arrows) with amiloride (10 µM), forskolin (10 µM), ivacaftor (0.3 µM), and the CFTR inhibitor PPQ-102 (20 µM). (B) Column bar graphic showing mean ± SEMm (n = 4). Data reported are the maximal amplitude of the current after addition of forskolin in colonoids pre-treated with vehicle (Veh.) or CFTR modulators. Non-CF mean of ΔIsc values from a minimum of three independent experiments obtained from five non-CF subjects (black column bar) are shown here as reference. Ordinary one-way ANOVA, p> 0.05, p> 0.005. (C) Light microscopy analysis of the morphology of Non-CF (used here as reference) and F508del/dup.exons 1–3 intestinal organoids in culture at steady-state condition
Based on available data a diagnosis of CF was made. In accordance with Italian legislative guidelines, as a child older than six years and with at least one F508del variant, he was eligible for prescription of ETI. The drug affected his SC (10 mmol/l) but not the already normal lung function, assessed under stable conditions three months after the initiation of treatment. Currently, after 12 months of ETI therapy, fecal elastase is still < 15 µg/g, with a nutritional status improved dramatically (supplemental figure A).
Discussion
About 85% of the CF population is pancreatic insufficient early in life, with substantial injury to the pancreas that begins already in utero and correlates closely with the specific CFTR variants . However, the extent of injury is variable. Children with CF and pancreatic sufficiency do not have completely normal pancreatic function . It may deteriorate over time, with or without the complicating effects of pancreatitis. Exocrine pancreatic insufficiency may develop without symptoms or may be characterized by failure to thrive in the infant and child or unexplained weight loss in the adult [20, 21].
In the case we observed the child was diagnosed with CF at the age of 8 years, presenting with pancreatic insufficiency and with a non-pathological sweat chloride. However, given the child’s age, we cannot exclude future greater lung involvement. As expected in patients with at least one F508del, ETI normalized sweat chloride level and had a positive impact on weight.
This case opens the discussion on the importance of performing CFTR sequencing and the search for large gene rearrangements in cases of pancreatic insufficiency of unclear etiology, considering that such variants account for about 6% CF alleles, particularly in ethnic groups with a low frequency of the F508del variant .
Furthermore, repeating the sweat test should be considered in the presence of pancreatic insufficiency and first sweat chloride in the normal range. Delayed diagnosis of CF has dramatic consequences on growth and lung function, even more so in children not screened for CF at birth .
While the sweat test is one of the main diagnostic tools used in newborn screening programs and one of the main biomarkers of efficacy, in terms of improvement of CFTR protein expression/function by modifying drugs [23–25], it can fluctuate in different diagnostic ranges over time, making a definitive diagnosis difficult in the absence of CF-related symptoms [25, 26]. Furthermore, false positives or negative results have been reported in patients with different diseases or dehydrated or taking drugs that modify chloride levels , factors excluded in the case we describe. In addition, there have been reports of children and adults with CF and pancreatic sufficiency, for example, with at least one 3849 + 10KbC > T CFTR variant [28–30], who have sweat chloride levels in the normal range (less than 30 mmol/L), although following them and repeating their sweat tests is likely to reveal higher concentrations if they truly have CF.
This is the only case from a retrospective re-evaluation of the 399 patients with CF followed in the regional center of Florence: no other patient had pancreatic insufficiency at diagnosis with a non-pathological sweat test. Only three adults with pancreatic sufficiency at diagnosis, two CFTR-causing variants, and sweat tests in the intermediate range developed pancreatic insufficiency during follow-up, requiring PERT after repeated episodes of acute recurrent pancreatitis.
A similar case with 4016insT/DupPr-ex3 CFTR genotype was reported by Paracchini et al. in a child of 4 years with a diagnosis of CF for pancreatic insufficiency and sweat test in the intermediate range (40–41 mmol/l) . DupPr-ex3 is a duplication involving the promoter region (promoter-exon3), also reported in a Chinese carrier with congenital bilateral absence of vas deferens .
In the CF mutation database (www.genet. sickk ids.on.ca/) CFTRdup1-3 is described in two unrelated Spanish patients with CF: a male identified by newborn screening, with F508del variant on the other allele and positive sweat test, pancreatic insufficiency and lung disease. The second patient is a male diagnosed at few months with similar clinical features .
According to the consensus guidelines from the Cystic Fibrosis Foundation , we set out NPD and ICM. Given the peculiar nature of the case we also applied recently proposed but well-established experimental assays such as those based on optical quantification of sweat droplets (OBAS) and analysis of the response of intestinal organoids. OBAS test is performed in vivo and is based on the original study of Wine while application of intestinal organoids in CF has been previously discussed . Interestingly, while the outcomes of the in vivo and ex vivo tests were consistently non-pathological, the sensitive Isc analysis performed in vitro in intestinal organoids detected a functional defect when compared to non-CF controls. The lack of other cell types present in the tissue but not on the colonoids grown as monolayers in Transwell® may play a role in revealing this quantitative difference between assays. Of note, good correlation between the prediction of the efficacy of modulator therapy in vitro using intestinal organoid-based assays and clinical outcomes has been reported in many studies .
Conclusions
CFTR sequencing and the search for large gene rearrangements are indicated in case of pancreatic insufficiency of unclear etiology. Initial sweat chloride value in the normal range can be observed in children with two pathogenic CFTR variants, but repeated sweat tests might demonstrate higher levels.
The extended analysis of CFTR function was negative except for a lower functional response when CFTR-dependent currents were analyzed in Ussing chambers in epithelial cell monolayers derived from rectal organoids and in OBAS test. Both measurements demonstrated a residual, but significant, CFTR activity consistent with the non-pathological values measured by NPD and ICM. The trend (that does not reach the threshold of statistical significance) to a further increase in CFTR-dependent current following treatment with ETI in the rectal organoids monolayers might suggest a potential benefit of the administration of this drug and is in line with the decreased sweat chloride concentration after ETI treatment. Whether the treatment is active on CFTR protein stabilization (corrector effect, Tezacaftor and Elexacaftor) or probability of channel opening (potentiator activity, provided by Ivacaftor and Elexacaftor) for the individual anions considered (chloride and bicarbonate) is still under evaluation.
Electronic supplementary material
Below is the link to the electronic supplementary material.
12887_2024_5154_MOESM1_ESM.png (41.4KB, png)
Supplementary Material 1: Supplemental figure A BMI trend before and after the introduction of pancreatic enzyme replacement therapy.
12887_2024_5154_MOESM2_ESM.jpeg (230.9KB, jpeg)
Supplementary Material 2: Supplemental figure B Recordings of short-circuit currents (Isc). (A) Representative ICM records for anion transport of the CF subject. Experiments were performed in the presence of 10 µM indomethacin. (B) Value for the sum Isc (FI + cch + H) for chloride secretion obtained from the control non-CF group, CF group, and the subject (n = 4).
Acknowledgements
This research was funded by the Italian Cystic Fibrosis Research Foundation (FFC), grant numbers FFC#9/2020 (Delegazione FFC di Pesaro, Rivarolo Canavese e Gruppo sostegno FFC di Fidenza) to P.M.; Lega Italiana Fibrosi Cistica onlus (C.S. and P.M.) and American CFF, grant number Assael08A0 (P.M.). Fondo per il Programma Nazionale di Ricerca e Progetti di Rilevante Interesse Nazionale (PRIN) n.2022FRSS2H “Theratyping of Cystic Fibrosis” (CS).
Author contributions
Conception and design: VT; follow up of the child: VT, CF, AC, SDV, LS, GT, PL; sweat test, microbiological and stool analyses: DD; analysis and interpretation of functional tests: KK, PM, CS; writing the paper: VT. All authors reviewed the manuscript and gave the final approval of the version to be published.
Funding
None.
Data availability
No datasets were generated or analysed during the current study.
Declarations
Ethics approval and consent to participate
An approval was obtained from the pediatric ethical committee of the Cystic Fibrosis center in Florence and for the Cystic Fibrosis Center AOUI of Verona (protocol#CFTR050).
Consent for publication
Parents provided written informed consent for anonymous data processing and to participate in this case investigation.
Competing interests
The authors declare no competing interests.
Footnotes
Publisher’s note
Springer Nature remains neutral with regard to jurisdictional claims in published maps and institutional affiliations.
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Associated Data
This section collects any data citations, data availability statements, or supplementary materials included in this article.
Supplementary Materials
12887_2024_5154_MOESM1_ESM.png (41.4KB, png)
Supplementary Material 1: Supplemental figure A BMI trend before and after the introduction of pancreatic enzyme replacement therapy.
12887_2024_5154_MOESM2_ESM.jpeg (230.9KB, jpeg)
Supplementary Material 2: Supplemental figure B Recordings of short-circuit currents (Isc). (A) Representative ICM records for anion transport of the CF subject. Experiments were performed in the presence of 10 µM indomethacin. (B) Value for the sum Isc (FI + cch + H) for chloride secretion obtained from the control non-CF group, CF group, and the subject (n = 4).
Data Availability Statement
No datasets were generated or analysed during the current study.
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11935 | https://en.wikipedia.org/wiki/Simpson%27s_paradox | Contents
Simpson's paradox
Simpson's paradox is a phenomenon in probability and statistics in which a trend appears in several groups of data but disappears or reverses when the groups are combined. This result is often encountered in social-science and medical-science statistics, and is particularly problematic when frequency data are unduly given causal interpretations. The paradox can be resolved when confounding variables and causal relations are appropriately addressed in the statistical modeling (e.g., through cluster analysis).
Simpson's paradox has been used to illustrate the kind of misleading results that the misuse of statistics can generate.
Edward H. Simpson first described this phenomenon in a technical paper in 1951; the statisticians Karl Pearson (in 1899) and Udny Yule (in 1903) had mentioned similar effects earlier. The name Simpson's paradox was introduced by Colin R. Blyth in 1972. It is also referred to as Simpson's reversal, the Yule–Simpson effect, the amalgamation paradox, or the reversal paradox.
Mathematician Jordan Ellenberg argues that Simpson's paradox is misnamed as "there's no contradiction involved, just two different ways to think about the same data" and suggests that its lesson "isn't really to tell us which viewpoint to take but to insist that we keep both the parts and the whole in mind at once."
Examples
UC Berkeley gender bias
One of the best-known examples of Simpson's paradox comes from a study of gender bias among graduate school admissions to University of California, Berkeley. The admission figures for the fall of 1973 showed that men applying were more likely than women to be admitted, and the difference was so large that it was unlikely to be due to chance.
| | All | | Men | | Women | |
--- --- ---
| Applicants | Admitted | Applicants | Admitted | Applicants | Admitted |
| Total | 12,763 | 41% | 8,442 | 44% | 4,321 | 35% |
However, when taking into account the information about departments being applied to, the different rejection percentages reveal the different difficulty of getting into the department, and at the same time it showed that women tended to apply to more competitive departments with lower rates of admission, even among qualified applicants (such as in the English department), whereas men tended to apply to less competitive departments with higher rates of admission (such as in the engineering department). The pooled and corrected data showed a "small but statistically significant bias in favor of women".
The data from the six largest departments are listed below:
| Department | All | | Men | | Women | |
--- --- ---
| Applicants | Admitted | Applicants | Admitted | Applicants | Admitted |
| A | 933 | 64% | 825 | 62% | 108 | 82% |
| B | 585 | 63% | 560 | 63% | 25 | 68% |
| C | 918 | 35% | 325 | 37% | 593 | 34% |
| D | 792 | 34% | 417 | 33% | 375 | 35% |
| E | 584 | 25% | 191 | 28% | 393 | 24% |
| F | 714 | 6% | 373 | 6% | 341 | 7% |
| Total | 4526 | 39% | 2691 | 45% | 1835 | 30% |
| Legend: greater percentage of successful applicants than the other gender greater number of applicants than the other gender bold - the two 'most applied for' departments for each gender | | | | | | |
Legend:
bold - the two 'most applied for' departments for each gender
The entire data showed total of 4 out of 85 departments to be significantly biased against women, while 6 to be significantly biased against men (not all present in the 'six largest departments' table above). Notably, the numbers of biased departments were not the basis for the conclusion, but rather it was the gender admissions pooled across all departments, while weighing by each department's rejection rate across all of its applicants.
Kidney stone treatment
Another example comes from a real-life medical study comparing the success rates of two treatments for kidney stones. The table below shows the success rates (the term success rate here actually means the success proportion) and numbers of treatments for treatments involving both small and large kidney stones, where Treatment A includes open surgical procedures and Treatment B includes closed surgical procedures. The numbers in parentheses indicate the number of success cases over the total size of the group.
| Treatment Stone size | Treatment A | Treatment B |
---
| Small stones | Group 1 93% (81/87) | Group 2 87% (234/270) |
| Large stones | Group 3 73% (192/263) | Group 4 69% (55/80) |
| Both | 78% (273/350) | 83% (289/350) |
The paradoxical conclusion is that treatment A is more effective when used on small stones, and also when used on large stones, yet treatment B appears to be more effective when considering both sizes at the same time. In this example, the "lurking" variable (or confounding variable) causing the paradox is the size of the stones, which was not previously known to researchers to be important until its effects were included.[citation needed]
Which treatment is considered better is determined by which success ratio (successes/total) is larger. The reversal of the inequality between the two ratios when considering the combined data, which creates Simpson's paradox, happens because two effects occur together:[citation needed]
Based on these effects, the paradoxical result is seen to arise because the effect of the size of the stones overwhelms the benefits of the better treatment (A). In short, the less effective treatment B appeared to be more effective because it was applied more frequently to the small stones cases, which were easier to treat, so that whichever treatment was selected was more likely to be successful.
Jaynes argues that the correct conclusion is that though treatment A remains noticeably better than treatment B, the kidney stone size is more important.
Batting averages
A common example of Simpson's paradox involves the batting averages of players in professional baseball. It is possible for one player to have a higher batting average than another player each year for a number of years, but to have a lower batting average across all of those years. This phenomenon can occur when there are large differences in the number of at bats between the years. Mathematician Ken Ross demonstrated this using the batting average of two baseball players, Derek Jeter and David Justice, during the years 1995 and 1996:
| Year Batter | 1995 | | 1996 | | Combined | |
--- --- ---
| Derek Jeter | 12/48 | .250 | 183/582 | .314 | 195/630 | .310 |
| David Justice | 104/411 | .253 | 45/140 | .321 | 149/551 | .270 |
In both 1995 and 1996, Justice had a higher batting average (in bold type) than Jeter did. However, when the two baseball seasons are combined, Jeter shows a higher batting average than Justice. According to Ross, this phenomenon would be observed about once per year among the possible pairs of players.
Vector interpretation
Simpson's paradox can also be illustrated using a 2-dimensional vector space. A success rate of
p
q
{\textstyle {\frac {p}{q}}}
(i.e., successes/attempts) can be represented by a vector
A
→
=
(
q
,
p
)
{\displaystyle {\vec {A}}=(q,p)}
, with a slope of
p
q
{\textstyle {\frac {p}{q}}}
. A steeper vector then represents a greater success rate. If two rates
p
1
q
1
{\textstyle {\frac {p_{1}}{q_{1}}}}
and
p
2
q
2
{\textstyle {\frac {p_{2}}{q_{2}}}}
are combined, as in the examples given above, the result can be represented by the sum of the vectors
(
q
1
,
p
1
)
{\displaystyle (q_{1},p_{1})}
and
(
q
2
,
p
2
)
{\displaystyle (q_{2},p_{2})}
, which according to the parallelogram rule is the vector
(
q
1
+
q
2
,
p
1
+
p
2
)
{\displaystyle (q_{1}+q_{2},p_{1}+p_{2})}
, with slope
p
1
+
p
2
q
1
+
q
2
{\textstyle {\frac {p_{1}+p_{2}}{q_{1}+q_{2}}}}
.
Simpson's paradox says that even if a vector
L
→
1
{\displaystyle {\vec {L}}_{1}}
(in orange in figure) has a smaller slope than another vector
B
→
1
{\displaystyle {\vec {B}}_{1}}
(in blue), and
L
→
2
{\displaystyle {\vec {L}}_{2}}
has a smaller slope than
B
→
2
{\displaystyle {\vec {B}}_{2}}
, the sum of the two vectors
L
→
1
+
L
→
2
{\displaystyle {\vec {L}}_{1}+{\vec {L}}_{2}}
can potentially still have a larger slope than the sum of the two vectors
B
→
1
+
B
→
2
{\displaystyle {\vec {B}}_{1}+{\vec {B}}_{2}}
, as shown in the example. For this to occur one of the orange vectors must have a greater slope than one of the blue vectors (here
L
→
2
{\displaystyle {\vec {L}}_{2}}
and
B
→
1
{\displaystyle {\vec {B}}_{1}}
), and these will generally be longer than the alternatively subscripted vectors – thereby dominating the overall comparison.
Correlation between variables
Simpson's reversal can also arise in correlations, in which two variables appear to have (say) a positive correlation towards one another, when in fact they have a negative correlation, the reversal having been brought about by a "lurking" confounder. Berman et al. give an example from economics, where a dataset suggests overall demand is positively correlated with price (that is, higher prices lead to more demand), in contradiction of expectation. Analysis reveals time to be the confounding variable: plotting both price and demand against time reveals the expected negative correlation over various periods, which then reverses to become positive if the influence of time is ignored by simply plotting demand against price.
Psychology
Psychological interest in Simpson's paradox seeks to explain why people[who?] deem sign reversal to be impossible at first.[clarification needed] The question is where people get this strong intuition from, and how it is encoded in the mind.
Simpson's paradox demonstrates that this intuition cannot be derived from either classical logic or probability calculus alone, and thus led philosophers to speculate that it is supported by an innate causal logic that guides people in reasoning about actions and their consequences. Savage's sure-thing principle is an example of what such logic may entail. A qualified version of Savage's sure thing principle can indeed be derived from Pearl's do-calculus and reads: "An action A that increases the probability of an event B in each subpopulation Ci of C must also increase the probability of B in the population as a whole, provided that the action does not change the distribution of the subpopulations." This suggests that knowledge about actions and consequences is stored in a form resembling Causal Bayesian Networks.
Probability
A paper by Pavlides and Perlman presents a proof, due to Hadjicostas, that in a random 2 × 2 × 2 table with uniform distribution, Simpson's paradox will occur with a probability of exactly 1⁄60. A study by Kock suggests that the probability that Simpson's paradox would occur at random in path models (i.e., models generated by path analysis) with two predictors and one criterion variable is approximately 12.8 percent; slightly higher than 1 occurrence per 8 path models.
Simpson's second paradox
A second, less well-known paradox was also discussed in Simpson's 1951 paper. It can occur when the "sensible interpretation" is not necessarily found in the separated data, like in the kidney stone example, but can instead reside in the combined data. Whether the partitioned or combined form of the data should be used hinges on the process giving rise to the data, meaning the correct interpretation of the data cannot always be determined by simply observing the tables.
Judea Pearl has shown that, in order for the partitioned data to represent the correct causal relationships between any two variables,
X
{\displaystyle X}
and
Y
{\displaystyle Y}
, the partitioning variables must satisfy a graphical condition called "back-door criterion":
This criterion provides an algorithmic solution to Simpson's second paradox, and explains why the correct interpretation cannot be determined by data alone; two different graphs, both compatible with the data, may dictate two different back-door criteria.
When the back-door criterion is satisfied by a set Z of covariates, the adjustment formula (see confounding) gives the correct causal effect of X on Y. If no such set exists, Pearl's do-calculus can be invoked to discover other ways of estimating the causal effect. The completeness of do-calculus can be viewed as offering a complete resolution of the Simpson's paradox.
Criticism
One criticism is that the paradox is not really a paradox at all, but rather a failure to properly account for confounding variables or to consider causal relationships between variables. Focus on the paradox may distract from these more important statistical issues.
Another criticism of the apparent Simpson's paradox is that it may be a result of the specific way that data are stratified or grouped. The phenomenon may disappear or even reverse if the data is stratified differently or if different confounding variables are considered. Simpson's example actually highlighted a phenomenon called noncollapsibility, which occurs when subgroups with high proportions do not make simple averages when combined. This suggests that the paradox may not be a universal phenomenon, but rather a specific instance of a more general statistical issue.
Despite these criticisms, the apparent Simpson's paradox remains a popular and intriguing topic in statistics and data analysis. It continues to be studied and debated by researchers and practitioners in a wide range of fields, and it serves as a valuable reminder of the importance of careful statistical analysis and the potential pitfalls of simplistic interpretations of data.
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11936 | https://pubmed.ncbi.nlm.nih.gov/22110919/ | Successful use of stellate ganglion block and pulsed radiofrequency in the treatment of posttraumatic stress disorder: a case report - PubMed
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. 2010:2010:963948.
doi: 10.1155/2010/963948. Epub 2010 May 24.
Successful use of stellate ganglion block and pulsed radiofrequency in the treatment of posttraumatic stress disorder: a case report
Eugene Lipov1
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1 Advanced Pain Centers S.C., 2260 West Higgins Road, Suite 101, Hoffman Estates, IL 60169, USA.
PMID: 22110919
PMCID: PMC3196975
DOI: 10.1155/2010/963948
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Case Reports
Successful use of stellate ganglion block and pulsed radiofrequency in the treatment of posttraumatic stress disorder: a case report
Eugene Lipov. Pain Res Treat.2010.
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. 2010:2010:963948.
doi: 10.1155/2010/963948. Epub 2010 May 24.
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Eugene Lipov1
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1 Advanced Pain Centers S.C., 2260 West Higgins Road, Suite 101, Hoffman Estates, IL 60169, USA.
PMID: 22110919
PMCID: PMC3196975
DOI: 10.1155/2010/963948
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Objective. To report our successful treatment of acute symptoms of posttraumatic stress disorder (PTSD). By the use of stellate ganglion block (SGB) and pulsed radiofrequency (PRF) to the stellate ganglion(SG) , sequentially. Background. A 48-year-old male a victim of armed robbery , who presented with extreme symptoms consistent with the diagnosis of PTSD. He was treated with antianxiety medications, as well as psychotherapy, but his symptoms persisted. Methods. Fifty-five days post trauma, we administered a SGB to the patient. One month later, we administered PRF to the right SG . We repeated the pulsed radiofrequency 30 weeks post trauma and performed a second SGB two weeks after that. Results. After the SGB , the patient experienced a major reduction in anxiety. Over the next week his improved allowing a significant reduction of antianxiety medications. One month later the symptoms returned and again subsided substantially following PRF , and that relief lasted four months. The patient than required another following PRF and a SGB with good responses. Conclusion. We report that selective blockade of the stellate ganglion via injection and the treatment with PRF, relieved our patient's symptoms of PTSD. And we also provide a plausible explanation of the effect.
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Institute of Medicine. Treatment of Posttraumatic Stress Disorder: An Assessment of the Evidence. Wash, DC, USA: The National Academies Press; 2007.
Bogduk N. Complex regional pain syndrome. Current Opinion in Anaesthesiology. 2001;14(5):541–546. - PubMed
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Miller MM, McEwen BS. Establishing an agenda for translational research on PTSD. Annals of the New York Academy of Sciences. 2006;1071:294–312. - PubMed
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11937 | https://ijmsi.ir/article-1-1436-en.pdf | Iranian Journal of Mathematical Sciences and Informatics Vol. 17, No. 2 (2022), pp 235-241 DOI: 10.52547/ijmsi.17.2.235 Single-Point Visibility Constraint Minimum Link Paths in Simple Polygons Mohammad Reza Zarrabi, Nasrollah Moghaddam Charkari∗ Faculty of Electrical Engineering and Computer Science, Tarbiat Modares University, Tehran, Iran E-mail: m.zarabi@modares.ac.ir E-mail: charkari@modares.ac.ir Abstract. We address the following problem: Given a simple polygon P with n vertices and two points s and t inside it, find a minimum link path between them such that a given target point q is visible from at least one point on the path. The method is based on partitioning a portion of P into a number of faces of equal link distance from a source point. This partitioning is essentially a shortest path map (SPM). In this paper, we present an optimal algorithm with O(n) time bound, which is the same as the time complexity of the standard minimum link paths problem.
Keywords: Minimum link path, Shortest path map, Point location.
2000 Mathematics subject classification: 68U05, 52B55, 68W40, 68Q25.
1. Introduction Link paths problems in computational geometry have received considerable attention in recent years due to not only their theoretical beauty, but also their wide range of applications in many areas of the real world. A minimum link path is a polygonal path between two points s and t inside a simple polygon P with n vertices that has the minimum number of links. Minimum link paths are fundamentally different from traditional Euclidean shortest path, which has ∗Corresponding Author Received 10 October 2018; Accepted 27 January 2019 c ⃝2022 Academic Center for Education, Culture and Research TMU 235 [ DOI: 10.52547/ijmsi.17.2.235 ] [ Downloaded from ijmsi.ir on 2025-09-29 ] 236 M. R. Zarrabi, N. M. Charkari the shortest length among all the polygonal paths without crossing edges of P.
Minimum link paths have important applications in various areas like robotic, motion planning, wireless communications, geographic information systems, image processing, etc. These applications benefit from minimum link paths since turns are costly while straight line movements are inexpensive.
Many algorithms structured around the notion of link path have been devised to parallel those designed using the Euclidean path. In , Suri introduced a linear time algorithm for computing a minimum link path between two points inside a triangulated simple polygon P (Ghosh presented a simpler algorithm in ). After that, Suri developed the proposed solution to a query version by constructing a window partition in linear time for a fixed point inside P .
This window partition is essentially a shortest path map, because it divides the simple polygon into faces of equal link distance from a fixed source point. By contrast, the work of Arkin et al. supports O(log n) time queries between any two points inside P after building n shortest path maps for all vertices of P, i.e., the total time complexity for this construction is O(n2) (an optimal algorithm for this case was presented by Chiang et al. in ). On the other hand, when there are holes in the polygon, Mitchell et al. proposed an incremental algorithm with O(Eα(n) log2 n)) time bound, where n is the total number of edges of the obstacles, E is the size of the visibility graph, and α(n) denotes the extremely slowly growing inverse of Ackermann’s function.
An interesting survey of minimum link paths appears in .
Minimum link paths problems are usually more difficult to solve than equiv-alent Euclidean shortest path problems since optimal paths that are unique under the Euclidean metric need not be unique under the link distance. An-other difficulty is that, Euclidean shortest paths only turn at reflex vertices of P while minimum link paths can turn anywhere.
The problem is studied with several variations. One of these variations is to constrain the path to view a point q from at least one point on the path.
One possible application is resource collection in which a robot moving from a source point to a destination one has to collect some resources found in a certain region. Some other applications are visibility related constraints such as wireless communication or guarding systems, where direct visibility is crucial.
In this paper, we study the constrained version of minimum link paths problem based on the shortest path map called SPM approach. The proposed algorithm runs in linear time.
Section 2 introduces the basic definitions and lemmas. Section 3 presents our algorithm for a visibility point q. We conclude in Section 4 with some open problems.
2. Preliminaries We use the following notation throughout the paper: [ DOI: 10.52547/ijmsi.17.2.235 ] [ Downloaded from ijmsi.ir on 2025-09-29 ] Single-Point Visibility Constraint Minimum Link Paths in Simple Polygons 237 Figure 1. SPM(x) and its faces.
Figure 2. Dividing any line segment L to at most three parts.
(1) V (x) : the visibility polygon of a point x ∈P (2) πL(x, y) : a minimum link path between two points x and y in P (3) DL(x, y) : the link distance (minimum number of links) of πL(x, y) (4) Dπ(x, y) : the number of links between x and y on the path π (5) Pocket(x) : the regions of P, invisible to x Definition 2.1. q-visible path: A minimum link path, which has a non-empty intersection with V (q) for any point q ∈P.
Definition 2.2. SPM(x): Let x be a point in P. The edges of V (x) either are (part of) edges of P or chords of P. The edges of the latter variety are called windows of V (x). The set of points at link distance one from x is precisely V (x).
The points of link distance two are the points in P −V (x) that are visible from some windows of V (x). Repeatedly, applying this procedure partitions P into faces of constant link distance from x . For the sake of consistency, we call this partitioning the shortest path map of P with respect to x (see Figure 1).
The following two lemmas are the fundamental properties of SPM used in our algorithm: Lemma 2.3. For any point x ∈P, SPM(x) is constructed in O(n) time .
Lemma 2.4. Given a point x ∈P, any line segment L intersects at most three faces of SPM(x) .
Figure 2 illustrates the worst case intersection described in Lemma 2.4. Our goal is to compute a q-visible path inside a simple polygon P with n vertices. [ DOI: 10.52547/ijmsi.17.2.235 ] [ Downloaded from ijmsi.ir on 2025-09-29 ] 238 M. R. Zarrabi, N. M. Charkari Figure 3. Step 4 of the algorithm.
3. The Algorithm Suppose that three points s, t and q are given in P. We sketch the generic algorithm for computing a q-visible path between s and t as follows: (1) If DL(s, q) = 1 or DL(t, q) = 1, report πL(s, t).
(2) Compute V (q) and Pocket(q).
(3) Use the point location algorithm for s and t to determine whether they are in the same region in Pocket(q) or not.
(4) If s and t are in two different regions of Pocket(q), report πL(s, t).
Since this path crosses V (q), it is a q-visible path (see Figure 3).
(5) Suppose that e is a chord of P that separates V (q) and the region of Pocket(q), which contains both s and t. Indeed, e is a window of V (q) and divides P into two subpolygons, only one of which contains q. We define p as the subpolygon containing q. A q-visible path should have a non-empty intersection with p (note that e ∈p).
(6) Compute both SPM(s) and SPM(t). Add two labels Sj and Tk to each face of SPM(s) and SPM(t) as the link distance from s and t to those faces, respectively (1 ≤j ≤||SPM(s)||, 1 ≤k ≤||SPM(t)|| and ||.|| denotes the number of faces of SPM).
(7) Use the map overlay technique to find the intersections of SPM(s), SPM(t) and p. Construct the planar subdivision of p with new faces (called Cells) by the quad view data structure . Use a set Ce as a reference to these Cells (each element of this set points to the value or structure of a Cell).
(8) Assign the value of each Cell, i.e., Ce(i) = si + ti, where 1 ≤i ≤the number of Cells and si = Sj, ti = Tk. In other words, si and ti are the link distances from any point x ∈Ce(i) to s and t, respectively.
(9) Find the minimum value of Ce(i) for some i. It ensures that a q-visible path between s and t enters Ce(i).
(10) Report πL(s, x) appended by πL(x, t), where x is a point in Ce(i). The link distance of this new path would be si + ti (Lemma 3.1). [ DOI: 10.52547/ijmsi.17.2.235 ] [ Downloaded from ijmsi.ir on 2025-09-29 ] Single-Point Visibility Constraint Minimum Link Paths in Simple Polygons 239 Figure 4. Two points s and t lie on the same side of e.
Lemma 3.1. Assume s and t lie on the same side of e. Then, the link distance of a q-visible path π between s and t is: si + ti (i might not be unique).
Proof. For any point x on any optimal path π, we have the following equalities (|.| denotes the link distance): |π| = Dπ(s, x) + Dπ(x, t) or |π| = Dπ(s, x) + Dπ(x, t) −1 The first equation occurs, if x is a bending point on π.
In this case, Dπ(s, x) = DL(s, x) and Dπ(x, t) = DL(x, t) due to the local optimality prin-ciple. Similarly, the second equation holds, if the last bending point from s to x, x, and the first bending point from x to t are collinear.
Since two points s and t lie on the same side of e inside P −p, there is always a bending point B in p on π. Indeed, there are at most two points B and B′, because if we have more than two bending points in p, the link distance of π (optimal path) can be shortened by at least one link due to the triangle inequality (see Figure 4).
We construct a path π′ like this: for x ∈Ce(i), append πL(s, x) to πL(x, t).
The following inequality clearly holds: si + ti −1 = DL(s, x) + DL(x, t) −1 ≤|π′| ≤DL(s, x) + DL(x, t) = si + ti Thus, we have: |π| ≤|π′| ≤si + ti for π. According to the above equalities, if we have one bending point B, then |π| = Dπ(s, B) + Dπ(B, t). In this case, B must be in Ce(i), i.e., |π| = si + ti (Figure 4(a)). On the other hand, there are three options for two bending points B and B′ (Figure 4(b)): (1) Only B ∈Ce(i) or B′ ∈Ce(i). Again in this case, |π| = si + ti. More precisely, if B ∈Ce(i), then B′ ∈Ce(j), where si + ti = sj + tj.
(2) (B and B′) ∈Ce(i), then |π| = Dπ(s, B) + 1 + Dπ(B′, t) = si + ti + 1.
(3) Neither B nor B′ belongs to Cells with minimum value (e.g., Ce(i)).
Thus, B ∈Ce(j) and B′ ∈Ce(k), where sj + tj = sk + tk > si + ti.
For the last two options, π would not be an optimal path. Therefore, only the first case can occur. The above discussion indicates that |π| = si +ti −1 would not be possible.
□ [ DOI: 10.52547/ijmsi.17.2.235 ] [ Downloaded from ijmsi.ir on 2025-09-29 ] 240 M. R. Zarrabi, N. M. Charkari Figure 5. At most 4 intersections and 9 Cells.
To analyze the time complexity of the algorithm, observe that the check in Step 1 can be easily done in linear time . The computations of V (q) and Pocket(q) in Step 2 can be done using the linear time algorithm in . Steps 3, 4 and 5 can be done in linear time [5, 11]. Based on Lemma 2.3, Step 6 can be done in linear time as well.
Note that, windows of a SPM never intersect each other.
According to Lemma 2.4, each window of SPM(s) intersects at most two windows of SPM(t) and vice versa. Therefore, if we have k1 and k2 windows of SPM(s) and SPM(t) inside p, respectively, then there are at most k1+k2 intersection points inside p, where k1, k2 = O(n). Thus, we have O(n) intersection points inside p between the windows of SPM(s) and SPM(t).
To find the intersections, we make use of some known algorithms for sub-division overlay like , which solves the problem in optimal time O(n + k).
Here, k is the number of intersections, which is O(n) in the worst case in our problem. Therefore, Step 7 can be done in linear time as well.
These intersections partition p to at most 9/4(k1 + k2) = O(n) Cells (see Figure 5). If either k1 or k2 does not exist, then we ignore the effect of that SPM to p. Furthermore, if neither k1 nor k2 exists, the whole p will be assumed as a Cell. Thus, Steps 8, 9 can be done in linear time. According to , Step 10 is done in linear time. The following theorem is concluded: Theorem 3.2. For any three points s, t and q inside a simple polygon P with total n vertices, a q-visible path between s and t and its link distance can be computed in O(n) time.
4. Conclusion We studied the problem of finding a minimum link path between two points with point visibility constraint in a simple polygon.
The time bound for a q-visible path is similar to the bound for the standard minimum link paths problem (without constraint). So, this bound cannot further be improved. [ DOI: 10.52547/ijmsi.17.2.235 ] [ Downloaded from ijmsi.ir on 2025-09-29 ] Single-Point Visibility Constraint Minimum Link Paths in Simple Polygons 241 Possible extensions to this problem involve studying the same problem when the path is required to meet an arbitrary general region, not necessarily a visibility region or having a query point instead of a fixed point.
A modified version of the problem is to constrain the path to meet several target polygons in a fixed order. If we fix the order of meeting, the problem seems to be less complex. As Alsuwaiyel et al. have shown, this problem is NP-hard even for any points on the boundary of a simple polygon. This type of problem is defined as minimum link watchman route with no fixed order.
Another extension is to improve the approximation factor of minimum link visibility path problem mentioned in , using the same approach (SPM).
Acknowledgements The first author wishes to thank Dr Ali Gholami Rudi from Babol Noshirvani University of Technology for many pleasant discussions.
References 1. M. H. Alsuwaiyel, D. T. Lee, Minimal Link Visibility Paths Inside a Simple Polygon, Computational Geometry Theory and Applications, 3(1), (1993), 1-25.
2. M. H. Alsuwaiyel, D. T. Lee, Finding an Approximate Minimum-Link Visibility Path Inside a Simple Polygon, Information Processing Letters, 55(2), (1995), 75-79.
3. E. M. Arkin, J. S. B. Mitchell, S. Suri, Logarithmic-Time Link Path Queries in a Simple Ppolygon, International Journal of Computational Geometry and Applications, 5(4), (1995), 369-395.
4. Y. J. Chiang, R. Tamassia, Optimal Shortest Path and Minimum-Link Path Queries Be-tween Two Convex Polygons Inside a Simple Polygonal Obstacle, International Journal of Computational Geometry and Applications, 7(01n02), (1997), 85-121.
5. H. Edelsbrunner, L. J. Guibas, J. Stolfi, Optimal Point Location in a Monotone Subdi-vision, SIAM Journal on Computing, 15(2), (1986), 317-340.
6. U. Finke, K. H. Hinrichs, Overlaying Simply Connected Planar Subdivisions in Linear Time, In Proceedings of the eleventh annual symposium on Computational Geometry, ACM, (1995), 119-126.
7. S. K. Ghosh, Computing the Visibility Polygon from a Convex Set and Related Problems, Journal of Algorithms, 12(1), (1991), 75-95.
8. B. Joe, R. B. Simpson, Corrections to Lee’s Visibility Polygon Algorithm, BIT Numerical Mathematics, 27(4), (1987), 458-473.
9. A. Maheshwari, J. R. Sack, H. N. Djidjev, Link Distance Problems, In Handbook of Computational Geometry, (2000), 519-558.
10. J. S. B. Mitchell, G. Rote, G. Woeginger, Minimum-Link Paths Among Obstacles in the Plane, Algorithmica, 8(1-6), (1992), 431-459.
11. S. Suri, A Linear Time Algorithm for Minimum Link Path Inside a Simple Polygon, Computer Vision, Graphics, and Image Processing, 35(1), (1986), 99-110.
12. S. Suri, On Some Link Distance Problems in a Simple Polygon, IEEE transactions on Robotics and Automation, 6(1), (1990), 108-113. [ DOI: 10.52547/ijmsi.17.2.235 ] [ Downloaded from ijmsi.ir on 2025-09-29 ] Powered by TCPDF (www.tcpdf.org) |
11938 | https://cltri.gov.in/AboutLeprosy/NLEP_Program%20Management_DLO.pdf?utm_medium=email&utm_source=transaction | National Leprosy Eradication Programme
(NLEP)
Programmatic Management of Leprosy
Dr. K. KUMARESAN
Public Health Specialist Gr II
Assistant Director (Epidemiology)
CLTRI
CLTRI Online training _ DLO_ Sep 2021
Learning Objectives
At the end of the session, the participants should be able to:
– To discuss the milestones in NLEP with programme
objectives and current strategies
– To discuss the implementation of NLEP at various levels
– To describe the programmatic management of leprosy
CLTRI Online training _ DLO_ Sep 2021
NLEP – National Leprosy Eradication Programme
Introduction:
• NLEP is a centrally sponsored public health programme of
GOI
• It has evolved over a period of time with remarkable changes from NLCP to NLEP
• Various milestones are there in the programme to reach the ultimate goal of leprosy free India
• Multiple stakeholders in the programme
CLTRI Online training _ DLO_ Sep 2021
NLEP Emblem
• Symbolizes beauty and purity in
lotus :
• Leprosy can be cured and a
leprosy patient can be a useful member of the society in the form of a partially affected thumb; a normal fore-finger and the shape of house ;
• the symbol of hope and optimism
in a rising sun .
CLTRI Online training _ DLO_ Sep 2021
Year PROGRAMME MILESTONES Key implementations
Before
1955
Gandhi Memorial Leprosy
Foundation (GMLF) Wardha / Hind Kusht Nivaran Sangh / NGOs
Survey , Education and Treatment (SET) programme
Precursor for NLCP and organized leprosy control
services
1955 National Leprosy Control Programme
(NLCP)
LCU – 4.5 L popl, SET centres – PR 5/1000 Dapsone Monotherapy
1983 National Leprosy Eradication
Programme (NLEP)
Introduction of MDT in Phases, initially high endemic districts
Urban leprosy centres , Mobile treatment units
1991 World Health Assembly resolution –
44.9
Eliminate leprosy as PHP at global level by the year 2000 [PR < 1/10000 popl]
1993 –
2000
First phase World Bank supported
project
MDT made available to all registered patients,
NLEP extended to all districts in the country
Midterm appraisal of NLEP (1997)
1998 - 04 Modified Leprosy Elimination
Campaign (MLEC)
SAPEL (2000)
Increasing awareness about leprosy, training to GHC personnel and to detect the hidden cases
[> 1 Million cases detected]
Difficult, inaccessible/hard to reach population
NLEP – Milestones (1) Year PROGRAMME MILESTONES Key implementations
2001 -
04
World Bank supported project II
phase
Decentralization of NLEP responsibilities, integration
under general health care system, training GHC personnel, Surveillance for early diagnosis with prompt MDT, awareness for voluntary reporting
2002 National Health Policy 2002
Simplified Information System
introduced
NHP set the goal of leprosy elimination by 2005
SIS - Better monitoring of NLEP with recording and reporting made easier for GHC staff.
2004 -05
Block Leprosy Awareness Campaign
(BLAC)
High priority districts & blocks with an aim to increase the awareness for self reporting, detection of hidden cases with capacity building of service providers
Dec
2005
Leprosy elimination as Public Health
Problem
PR <1/10000 (0.95) elimination declared at the National level
2005 National Rural Health Mission
(NRHM)
Vertical programme integrated with general health care system under NRHM
Dist. Nucleus Team (DNT) – Health societies
Urban leprosy control programme
NLEP – Milestones (2) Year PROGRAMME MILESTONES Key implementations
2006 Disability Prevention & Medical
Rehabilitation (DPMR) introduced
Guidelines for management at primary, secondary and
tertiary level.
2012 12 th plan [2012 – 2017] Special action plan for 209 high endemic districts in 16
States/UTs
Target to reduce the visible disabilities <1 per 10,00,000 population in by 2020.
2014 Upgraded Simplified Information
System (USIS) implementation
ULF formats introduced for uniformity and better decision
making
2016 -
17
Newer initiatives in the programme
Three Pronged strategy
Chemoprophylaxis
LCDC – 14 day active case detection campaign in high endemic districts
FLC – non-endemic districts
Special plan for hard to reach areas
SDR implementation to eligible contacts of new cases
Immunotherapy - MIP Vaccine as pilot phase
2017 Sparsh Leprosy Awareness Campaign (SLAC)
Increasing the awareness, addressing high level of stigma & discrimination - Convening special Grama sabha meeting
NLEP – Milestones (3) Year PROGRAMME MILESTONES Key implementations
2017
18
Newer initiatives:
Introduction of “NIKUSTH”
ASHA based Surveillance for Leprosy
Suspects (ABSULS)
Real time monitoring of leprosy patients across
the country and facilitating better monitoring and evaluation of NLEP.
ABSULS - active surveillance of leprosy
suspects with prioritizing leprosy case detection
by ASHA & treatment followup
2018
19
Sparsh Leprosy Elimination Campaign (SLEC)
150 th Birth anniversary of Mahatma
Gandhiji
Enhancement of recently launched initiatives –
LCDC, SLAC
G2D target to reduce <1case/Million & reduce
backlog of RCS cases
Grade II disability investigation
2018
-19
WHO Guidelines for Diagnosis, treatment
and prevention of Leprosy
Evidence based recommendations in accordance
with procedures established by the WHO
Guidelines Review Committee
Independent Evaluation of NLEP by WHO
(2019)
NLEP – Milestones (4) Year PROGRAMME MILESTONES Key implementations
2019 Convergence of leprosy screening under
major programmes of National Health Mission
i) Comprehensive Primary Health Care
of of Ayushman Bharat - Community Based Assessment Checklist (CBAC) to screen 30+ popl. at HWCs
ii) Rashtriya Bal Swasthya Karyakaram
(RBSK) to screen children
20 20 Active Case Detection and Regular
Surveillance (ACD & RS)
Convergence under Rashtriya Kishore
Swasthya Karyakaram (RKSK)
ACD & RS guidelines rolled out – active case finding to be continuous and regular
Flexibility for states to plan and implement
Frontline workers (FLWs)
RKSK – screening and counselling of adolescent children
District Award Scheme for achievements in NLEP
NLEP – Milestones (5) NLEP VISION:
CLTRI Online training _ DLO_ Sep 2021
“Leprosy-Free India”
NLEP MISSION:
“to provide quality leprosy services free of cost to all
sections of the population, with easy accessibility, through the integrated healthcare system, including care for disability
after cure of the disease” NLEP Objectives:
CLTRI Online training _ DLO_ Sep 2021
S No Objectives Current levels
To reduce the prevalence rate to less than 1/10,000 population at sub national and district level.
85% districts
To reduce Grade II disability % to < 1 among new cases at
National level
2.4%
To reduce Grade II disability cases to < 1 case per million
population at National level.
1.96
Zero disabilities among new Child cases 63 cases
Zero stigma and discrimination against persons affected by leprosy
100 laws
2019-20 data NLEP Strategies:
Decentralized integrated leprosy services through General Health Care system.
Early detection & complete treatment of all new leprosy cases.
Carrying out house hold contact survey for early detection of cases
Capacity building of all general health services functionaries
Involvement of ASHAs in the detection & completion of treatment of leprosy cases on time
Strengthening of Disability Prevention & Medical Rehabilitation (DPMR) services.
IEC activities in the community to improve self reporting to PHC and reduction of stigma.
Intensive monitoring and supervision at Health and Wellness centres and at
PHC/CHC.
CLTRI Online training _ DLO_ Sep 2021
Recent strategies in NLEP
Three pronged strategy- LCDC, FLC , Hard to reach areas
ASHA based Surveillance for Leprosy Suspects ( ABSULS )
‘Sparsh Leprosy Awareness Campaign’( SLAC )
Post Exposure Prophylaxis - Single Dose Rifampicin ( PEP-SDR )
Immunoprophylaxis - Mycobacterium indicus Pranii ( MIP ) vaccine
Implementation of o nline reporting system (‘ Nikusth ’) for improved
monitoring and supervision
Detailed investigation Grade II disability cases
Drug resistance surveillance
Modelling studies in leprosy
Active Case Detection & Regular Surveillance (ACD & RS)
District Award Scheme for achievements in NLEP
CLTRI Online training _ DLO_ Sep 2021
Decentralized planning for achievements of results
• All the activities of NLEP was planned, implemented and
monitored under the umbrella of
NHM
• Decentralized planning through district health plans, formulated
through bottom up process
• Programme Implementation Plan (PIP) should be prepared as a
result oriented process .
• Funds sent to states through State
Health Societies
CLTRI Online training _ DLO_ Sep 2021
• Improved early case detection & case
management
• Stigma reduced
• Development of leprosy expertise sustained
• Research supported evidence based
programme practices
• Monitoring supervision and evaluation system improved
• Increased participation of persons
affected by leprosy in society
• Programme management ensured NLEP - Institutional framework & Programme management
CLTRI Online training _ DLO_ Sep 2021
• Ministry of Health & Family Welfare
• DGHS
• Central Leprosy Division - DDG (L)
• Training Institutes - CLTRI, RLTRIs
Centre level
• State Health Society
• State Leprosy Officer / State Leprosy Consultant
State level
• District Health Society
• District Leprosy Officer / Deputy Director
• District Nucleus Team (DNT)
District level
• Block PHC / CHC - Rogi Kalyan samiti / PRI
• Block Medical Officer / incharge MO CHC
Block level
• PHC - Rogi Kalyan samiti / Panchayati Raj Institution
• Medical Officer
PHC level
• Grama Panchayat
• Male / Female Health Worker
Sub-Centre level / Health & Wellness Centre
• Village Health & Sanitation Committee
• ASHA / AWW
Village level Job responsibilities
Medical
Officer
• Planning of NLEP activities
• Case Confirmation, treatment, Lepra reaction Mx, referral of complications
• On Job Training to PHC Staff
• Supervision & Monitoring
• Review of activities & Feedback to DNT
HS/ HI / ANM/ MPHW/ VHN
• Suspect identification
• Availability of MDT and providing MDT, compliance
• Record Maintenance & Report Preparation
• IEC activities
• DPMR activities
• Supervision of ASHA/ AWW / Volunteer
• Assist MO
ASHA/ AWW
• Suspect identification & referral to PHC
• Timely completion of treatment
• Counselling & IEC
• Contact Screening, support for active case detection campaigns
CLTRI Online training _ DLO_ Sep 2021
District level
• Districts remain the core centre for NLEP activities
• District Nucleus Team (DNT), headed by DLO with other
necessary staff.
• District PIP is prepared by the DNT by compiling the action plans from block/PHC level and forwarded to state.
• Regular capacity building of health care workers
• Supervision, Monitoring & Evaluation of NLEP in the district
• Budget utilization and statement of expenditure (SOE)
CLTRI Online training _ DLO_ Sep 2021
NLEP
Case Detection
Treatment -
MDT &
complications
Capacity
building
DPMR activities
IEC/BCC & Stigma Reduction
Planning,
Supervision
&Monitoring
NLEP activities – Functional Domains
CLTRI Online training _ DLO_ Sep 2021
Programmatic management of leprosy (1)
• Suspect identification
• Case diagnosis & confirmation
Govt. & Non-Govt.
Guidance & support to field workers
• Difficult to diagnose, reactions & complications
– identification & referral services
• Survey – Routine / Active surveys
• School health activities – screening of children
• Contact screening as per guidelines
• Special activities
CLTRI Online training _ DLO_ Sep 2021
Case detection activities Programmatic management of leprosy (2)
• Initiation of MDT & ensuring Compliance
• Adverse effects monitoring
• Defaulter retrieval
• Treatment for complications
• Drugs
MDT availability
Supportive drugs: steroids, others
• Free of cost
CLTRI Online training _ DLO_ Sep 2021
Treatment Treatment of leprosy
CLTRI Online training _ DLO_ Sep 2021
Characteristics Pauci-bacillary (PB) Multi-bacillary (MB)
Skin lesions 1 – 5 lesions 6 and above
Peripheral nerve
involvement
No nerve / only one
nerve involvement
1 nerve irrespective
of no. of skin lesions
Skin smear Negative at all sites Positive at any site
WHO Classification:
Operational Classification –helps in selecting correct combination of drugs for a given patient
Treatment of leprosy
• MDT – Cap. Rifampicin, Tab. Dapsone & Cap. Clofazimine
• Standard regimen of MDT
• MDT provided in convenient to use Blister Calendar Packs
(BCPs)
• Four weeks / 28 days
• Dosage
CLTRI Online training _ DLO_ Sep 2021
Rifampicin : 10mg/kg body weight, monthly once
Dapsone : 2mg/kg body weight, daily
Clofazimine : 1 mg/kg body weight daily & 6mg/kg body weight, monthly once Indications for prescribing MDT
New case
Person with signs of leprosy who have never received
treatment before
Other case
under NLEP all previously
treated cases, who need further
treatment are recorded as other cases
Relapse :
Re-occurrence of the disease at any time after the completion of full course of treatment.
Re-entered for treatment :
These are previously treated cases, where clinical assessment shows requirement of further treatment and patient admits that treatment was not completed.
Referred cases :
Patient referred for completion of treatment (remaining
doses) by tertiary or second level institutions after
diagnosis and issue of first dose, or from another Health centre on patient request or migratory patient from another District/State
Re-classified :
Persons with PB leprosy; reclassified to MB and need full course of MB treatment.
A person who is residing for more than six month and is likely to stay till completion of treatment, be recorded as indigenous case and will not be
categorized under “other cases”. MDT Regimen
CLTRI Online training _ DLO_ Sep 2021
Type of leprosy
Drugs used Frequency of
Administration
Adults (children in
bracket)
Dosage
(adult)
15 years &
above
Dosage
(Children 10-14 years)#
Dosage
Children below 10 years
Criteria for
RFT
MB leprosy
Rifampicin Once monthly 600 mg 450mg 300mg Completion of 12 monthly pulses in 18 consecutive months
(12 BCP / 18 months)
Clofazimine Monthly 300 mg 150 mg 100mg
Dapsone Daily Once 100 mg 50 mg 25mg
Clofazimine Daily for adults (every other day for children)
50 mg 50mg (alternate days)
50mg (Weekly twice)
PB leprosy
Rifampicin Once monthly 600 mg 450 mg 300mg Completion of 6 monthly pulses in 9 consecutive months
(6 BCP / 9 months)
Dapsone Daily 100 mg 50 mg 25mg daily or
50 mg alternate days
For children 10 – 14 yrs with body weight > 35 kgs, adult BCP should be given
For children < 10 yrs, doses (as per body weight) should be provided loose after opening appropriate BCP MDT – Blister Calendar Packs
CLTRI Online training _ DLO_ Sep 2021
Rifampicin: 150 & 300 mg Dapsone: 50 & 100 mg Clofazimine: 50 & 100 mg Side effects - Dapsone
CLTRI Online training _ DLO_ Sep 2021 Rifampicin
CLTRI Online training _ DLO_ Sep 2021 Clofazimine
CLTRI Online training _ DLO_ Sep 2021 Assessing fitness for MDT
• Jaundice: wait till subsides
• Anaemia: start treatment for anaemia simultaneously along
with MDT
• TB: if the patient is taking Rifamicin, ensure to take Rifampicin in the dose required for treatment of TB along with other drugs
• Allergy to sulpha drugs: known allergic – avoid Dapsone
CLTRI Online training _ DLO_ Sep 2021
Follow up of treatment
• All efforts must be made to complete the doses in 6 months /
12 months
• During follow up – side effects of MDT and signs/ symptoms of reaction / neuritis
• Once the patient has completed the required pulses, the
treatment is stopped and made RFT (Release from treatment )
• Self care
• Complications
CLTRI Online training _ DLO_ Sep 2021
Advantages of MDT
CLTRI Online training _ DLO_ Sep 2021
• Stops progress of the disease, prevents further
complications and reduces chances of relapse
• Interrupt transmission of infection as rapidly as possible
• Reduces the chances of development of resistance to drugs
• Duration of treatment is short & fixed
• Safe, minimal side effects and increased patient compliance To ensure regularity of treatment
• Adequate counselling at the start of treatment
disease / skin lesions
duration of treatment
method of taking drug
regular treatment
side effects & reporting
• Regular follow up of patient – timely RFT
• Patients who are absent should be contacted immediately to identify the reasons and take corrective actions.
• Flexibility in MDT delivery
CLTRI Online training _ DLO_ Sep 2021
Accompanied MDT (A-MDT)
• Difficult in getting the drugs – underserved areas
• Emergency situations
health / pandemic: COVID – 19
seasonal
conflicts / war
• Migration
CLTRI Online training _ DLO_ Sep 2021
Given more than one BCP at a time (usually 3 BCPs are given) Irregular treatment
• Patients should be reassessed clinically to ascertain the type & any
disability
• Treatment history
• Look out for period of discontinuation
CLTRI Online training _ DLO_ Sep 2021
Classification Period of treatment
discontinuation
Treatment
PB < 3 months Continue the same course
3 months Re-start
MB < 6 months Continue the same course
6 months Re-start
Defaulter – Register as other case (re-entered for treatment) Treatment regimens for special situations
Due to side-effects
or intercurrent diseases, such as chronic hepatitis, or who have been infected with rifampicin-resistant
M. leprae
• Daily 50 mg clofazimine + any 2 drugs ( 400 mg ofloxacin , 100 mg minocycline or 500 mg clarithromycin) – 6 months
• Followed by daily 50 mg clofazimine + any 1drug (100 mg minocycline or 400 mg
ofloxacin ) - 18 months
• If available, ofloxacin may be replaced by
moxifloxacin 400 mg , which has stronger bactericidal activity against M. leprae .
CLTRI Online training _ DLO_ Sep 2021
Patients who cannot take rifampicin Treatment regimens for special situations
Due to side-effects -skin discolouration
• MDT may be replaced by 400 mg
ofloxacin / moxifloxacin daily, or by minocycline 100 mg daily, for 12
months.
• Alternatively – Rifampicin 600 mg once
a month, ofloxacin 400 mg once a
month, and minocycline 100 mg once a
month for 24 months.
CLTRI Online training _ DLO_ Sep 2021
Patients who cannot take Clofazimine Treatment regimens for special situations
Due to side-effects -
severe toxic effects
PB / MB leprosy
• Dapsone to be stopped immediately
• MB leprosy :
No further modification
• PB leprosy :
clofazimine in standard dose of
MB-MDT
CLTRI Online training _ DLO_ Sep 2021
Patients who cannot take Dapsone WHO Guidelines for Diagnosis, Treatment and Prevention of Leprosy (2018)
Area of the recommendation
Recommendation
Diagnosis of leprosy The diagnosis of leprosy may be based on clinical examination, with or without slit-skin smears or pathological examination
Diagnosis of leprosy
infection
There is currently no test recommended to diagnose leprosy infection (latent leprosy) among asymptomatic contacts.
Treatment of leprosy The same 3-drug regimen of rifampicin, dapsone and clofazimine may be used for
all leprosy patients , with a duration of treatment of 6 months for PB leprosy and of 12 months for MB leprosy.
Treatment of drug
resistant leprosy
Two of the following second -line drugs: clarithromycin, minocycline or a quinolone (ofloxacin, levofloxacin or moxifloxacin), plus clofazimine daily for 6 months , followed by clofazimine plus one of the second-line drugs daily for an additional 18 months.
Chemoprophylaxis for
contacts of leprosy
cases
Single -dose rifampicin (SDR) may be used for contacts of leprosy patients (adults and children aged 2 years and above), after excluding leprosy and
tuberculosis (TB) disease, and in the absence of other contraindications. • Develop adequate skills for diagnosis
& management of cases
• Training of general health care staff
• Regular training & refresher training
• Training data base particulars
• Follow up of training
• PIP – No. trained / No. need to be trained – key healthcare staff
CLTRI Online training _ DLO_ Sep 2021
Capacity building
Programmatic management of leprosy (3) • DPMR – primary / secondary / tertiary level
• Detection of complications & referral
• Assessment of disability status
• Line list of beneficiaries
• Conduction of POD activities
• Self care demonstration & provision of kits
• Provision of MCR / aids & appliances
• RCS referral / surgery – clearing backlog of RCS
• Identification of RCS institutions
• Incentives for RCS
• Social welfare measures – identification & assistance
• Special activities / programmes
CLTRI Online training _ DLO_ Sep 2021
Disability Prevention &
Medical
Rehabilitation (DPMR) activities
Programmatic management of leprosy (4) Information,
Education & Communication
(IEC) / BCC
• Generate awareness about leprosy, treatment and reducing stigma & discrimination
• IEC campaigns – schools/colleges, community
• Inter-personal communication (IPC)
• New strategies & methods
• Focus on Behaviour change communication
• Tangible increase in voluntary reporting
• Involvement of PAL
• Training of health staffs / volunteers
• Special campaigns – Sparsh Leprosy Awareness Campaign ( SLAC )
• “Leprosy Free India”
CLTRI Online training _ DLO_ Sep 2021
Programmatic management of leprosy (5) Planning,
Supervision &
Monitoring
• Assess the leprosy situation in the area & action plan
• Planning surveys, field visits - PIP
• Supervision of field level health care workers
• Record maintenance - Records & reports -
Complete, accurate & timely
• Monitoring & Evaluation of program performance – Indicators
• Training need assessment
• “NIKUSTH” implementation
• Review meetings & feedback – all levels
CLTRI Online training _ DLO_ Sep 2021
Programmatic management of leprosy (6) Involvement of NGOs in NLEP
• Partnership with NGOs is envisaged under NLEP and the objective is to provide uniformity in diagnosis, treatment and monitoring through a wider programme base to maximize access to NLEP services
• Complement and supplement the government efforts in reducing the disease
burden
• Bring about betterment in the quality of life and socio-economic condition of the affected persons and families.
CLTRI Online training _ DLO_ Sep 2021
Activities:
IEC/BCC and stigma reduction
Referral of suspects, Diagnosis and provision of MDT
Follow up of cases and treatment adherence
Out-patient and In-patient care
DPMR services
Referral & conduction of RCS Grant in aid Schemes for NGOs
Scheme 1A - Designated Referral Centres (DRC 1A) Out-patient facility
Scheme 1B - Designated Referral Centres (DRC 1B) Out-patient and In-patient
Scheme 1C - Designated Referral Centres (DRC 1C) Out-patient, In-patient and RCS
Scheme 2 - Comprehensive Care for Underserved Areas
Scheme 3 - Contact Survey and Home Based Self Care
Scheme 4 - Disability Care Centre - Leprosy Colonies
Scheme 5 - Advocacy Communication and Social Mobilisation with activities to
reduce Stigma and Discrimination in Leprosy
Scheme 6 - Partnering with community for elimination of leprosy
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• Grant-in-aid schemes are available under NLEP for the NGOs to build partnership and implement the schemes.
• > 285 NGOs working in the field of leprosy throughout the country &
54 NGOs are getting grant-in-aid from Government of India. NGO – Eligibility Criteria
• Registration of the NGO for at least last two years
• Necessary infrastructure and manpower support
• Experience of work related to leprosy or health or community development as appropriate in public sector.
Applications requesting for the Grant-in-aid under the NGO scheme shall be made through the District Leprosy Officer (DLO) to the State Leprosy Officer (SLO).
CLTRI Online training _ DLO_ Sep 2021
ILEP
International Federation of Anti-leprosy Associations
Members:
Americal Leprosy Missions
Associazione Italian Amici di Raoul Follereau (AIFO)
German Leprosy and Tuberculosis Relief Association (DAHW)
Damien Foundation Belgium
effect:hope
FAIRMED
Foundation Raoul Follereau
Fontilles
Lepra
Leprosy Relief Canada
NLR International
Sasakawa Health Foundation
The Leprosy Mission International
CLTRI Online training _ DLO_ Sep 2021
ILEP is a consortium of international non-governmental organisations with a shared desire to see a world free from leprosy NLEP – Budget under NHM
FMR
codes
Major Heads Total Budget proposed for FY
Total Budget approved for FY
Service Delivery - Facility based
Service Delivery – community based
Community intervention
Procurement
Training & Capacity Building
IEC/BCC
Printing
PPP
Programme Management
Total Budget LEPROSY TREND: INDIA
Year 1981 1985 1995 2005 2014-15 2018-19 2019-20
New Cases detected 39,53,700 32,00,000 8,07,257 1,48, 910 127334 120334 114451
4.2 3.2 2.4 1.3 0.84 0.72 0.74 0.72 0.71 0.69 0.68 0.73 0.68 0.69 0.66 0.66 0.67 0.62 0.57
59
44
33
23
14 12 12 11 11 10 10 10 9.9 9.7 9.7 10.2 9.3 8.7 8.1
0
10
20
30
40
50
60
70
ANCDR PR NLEP Key Indicators - India
Financial
year
Prevalence New case detection MB cases Children Female G2D
Number Per 10,000
Number Per 100,000
No % No % No % No % Per million
2008-09 86331 0.72 134184 11.2 64949 48.4 13610 10.1 47188 35.2 3763 2.8 3.1
2009-10 87190 0.71 133717 10.9 64782 48.4 13331 10.0 47361 35.4 4117 3.1 3.4
2010-11 83041 0.69 126800 10.5 61603 48.6 12463 9.8 45896 36.2 3927 3.1 3.2
2011-12 83687 0.68 127295 10.3 63562 49.9 12305 9.7 47111 37.0 3865 3.0 3.1
2012-13 91743 0.73 134752 10.8 67268 49.9 13387 9.9 50828 37.7 4650 3.5 3.7
2013-14 86147 0.68 126913 10.0 65337 51.5 12043 9.5 46845 36.9 5256 4.1 4.1
2014-15 88833 0.69 125785 9.7 66436 52.8 11365 9.0 46379 36.9 5794 4.6 4.5
2015-16 86028 0.66 127334 9.7 65595 51.3 11389 8.9 48808 38.3 5851 4.6 4.5
2016-17 88199 0.66 135485 10.2 67160 49.6 11770 8.7 53072 39.2 5245 3.9 3.9
2017-18 90709 0.67 126164 9.3 64187 50.9 10287 8.2 48821 38.7 4552 3.6 3.3
2018-19 85302 0.62 120334 8.7 62910 52.3 9227 7.7 46880 39.0 3666 3.0 2.6
2019-20 79898 0.57 114451 8.1 62119 54.2 7859 6.9 44877 39.2 2761 2.4 2.0
Source: Independent Evaluation of the Indian NLEP.WHO; Nov 2019 & WHO/WER/36,2020;95:417 -440
CLTRI Online training _ DLO_ Sep 2021 Achievements Vs Targets
CLTRI Online training _ DLO_ Sep 2021
Indicator Target Baseline
(2011-12)
Target (2017) Achievement (March 2019)
Prevalence Rate <1/10000
popl .
543 districts
(85%)
642 districts
(100%)
587 districts (91%)
ANCDR <1/100000 popl.
445 districts (69%)
642 districts (100%)
514 districts (80%)
MB cure rate (%) 90.5 >95 93.6
PB cure rate (%) 95.3 >97 94
G2D proportion 35% reduction 3% 2% 3%
Source: Independent Evaluation of the Indian NLEP.WHO; Nov 2019
Leprosy status at district level (2019-20)
Indicators Districts (717)
ANCDR [ <10/100000 pop] 552 (77%)
Prevalence [< 1/10000 pop] 607 ( 85% )
G2D [< 1 case/1000000 pop] 423 (59%)
Child rate [< 1 /100000 pop] 607 (85%)
CLTRI Online training _ DLO_ Sep 2021
Mapping of districts by level of Endemicity
52
Districts Total (%)
High Endemic 118 (17)
Moderate Endemic 206 (29)
Low Endemic 260 (37)
Sporadic cases only 124 (17)
Source: Independent Evaluation of the Indian NLEP.WHO; Nov 2019
(n = 708 districts) WHO and Strategic and Technical Advisory Group for NTD
[Generic framework 2015]
Elimination as a Public Health Problem
• related to both infection and
disease.
• defined by achievement of measurable global targets set by
WHO in relation to a specific
disease. When reached, continued actions are required to maintain the targets and/or to advance the
interruption of transmission
Elimination of transmission
• also referred to as interruption of
transmission
• mean reduction to zero of the incidence of infection caused by a
specific pathogen in a defined
geographical area , with minimal risk of reintroduction , as a result of deliberate efforts; continued actions to prevent re-establishment
of transmission may be required
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Prevalence < 1 case/10000 population Zero new case of leprosy
GLOBAL LEPROSY STRATEGY
2016 – 2020
“Accelerating towards a leprosy
free world”
2021 – 2030
“Towards Zero Leprosy” GLOBAL LEPROSY STRATEGY
What does it mean in terms of numbers to India?
S.No Impact Indicators Target GLS 2016-2020
No. of children diagnosed with leprosy and visible deformities
0 162 63 (>60% )
Rate of newly diagnosed leprosy patients with visible deformities
<1/million 4.5 1.96 (>56%)
No. of laws / legislation allowing discrimination on the basis of leprosy
0 119 102 (14%)
CLTRI Online training _ DLO_ Sep 2021
S.No Impact Indicators Target GLS 2021-2030
Annual No. of new cases detected 70% 110000 34,000
Rate of newly diagnosed leprosy patients
with G2D (per million popl)
90% 1.96 0.2
Rate of new child cases with leprosy
(per million child popl)
90% 20 2Leprosy - Elimination / Eradication
• Diagnosed by clinical signs
• Availability of effective treatment to interrupt transmission
• Single significant reservoir – Humans
• Chemoprophylaxis /+ immunoprophylaxis
• Human resources
• Country experiences
• National & International efforts
CLTRI Online training _ DLO_ Sep 2021
NLEP - Challenges
• Weak link in establishing the transmission chain
• long & variable incubation period
• Difficult to diagnose subclinical infections
• Reactions & nerve damage
• Effective vaccine not available
• Involvement of health workers after integration
• Lack of trained manpower –
Knowledge gap & skills
• Need for new drugs / regimens
CLTRI Online training _ DLO_ Sep 2021
• Hidden cases - case detection – less voluntary reporting
• Stigma & discrimination
• Notification from private providers
• Disability / loss of productivity -DPMR activities
• Weak Monitoring & supervision
• Underserved /difficult to reach areas/ Migrants
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Teleosts are defined as a diverse infraclass of fishes, Teleostei, which includes approximately 34,940 living species organized into various taxonomic subdivisions, occupying every aquatic habitat type and niche.
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Chapter
Teleostean Monophyly
1996, Interrelationships of FishesMãrio C.C. De Pinna
V Summary
Teleosts are a highly corroborated monophyletic group with at least 27 known synapomorphies from various anatomical systems. The exact composition of the group is subject to debate regarding the inclusion of certain basal fossils. This uncertainty about the exact limits of the taxon results in difficulties in determining which synapomorphies are actual “teleostean synapomorphies.” The Teleostei is here defined as the most inclusive actinopterygian group not including the Halecomorphi (Amia and relatives) and/or the Ginglymodi (Lepisosteus and relatives). This stem-based definition conforms with the practice of including certain basal fossils in the Teleostei and maintains the traditional exclusion of the Recent genera Amia and Lepisosteus from the group. The clade composed of all so-called Recent teleosts (here named Teleocephala; composed of osteoglossomorphs, elopomorphs, clupeomorphs, and euteleosts) is supported as monophyletic by an outstanding number of features from both hard and soft anatomy. Intriguingly, molecular data have yet to provide consistent support for teleostean monophyly.
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Book 1996, Interrelationships of FishesMãrio C.C. De Pinna
Chapter
The Neurobiology of Circadian Timing
2012, Progress in Brain ResearchM. Laura Idda, ... Nicholas S. Foulkes
Concluding remarks
Teleosts represent the largest and one of the most remarkable groups of vertebrates with unique characteristics for chronobiological studies. With over 30,000 known species, they exhibit a considerable degree of anatomical and physiological plasticity that has enabled them to occupy a wide diversity of habitats and ecological niches. In the highly dynamic aquatic environment, one can predict that during teleost evolution the circadian timing system has been subjected to a range of different selective pressures depending on a diversity of lighting and temperatures as well as the flexible patterns of behavior adopted. Thus, the initial studies in the zebrafish as well as teaching us much about key aspects of the vertebrate timing system also promise to serve as a foundation for broader, comparative studies involving diverse species such as the blind cavefish. These future studies should tackle basic questions concerning the evolution of the circadian timing system.
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Book series2012, Progress in Brain ResearchM. Laura Idda, ... Nicholas S. Foulkes
Chapter
The Cardiovascular System
2017, Fish PhysiologyTill S. Harter, Colin J. Brauner
1 Introduction
Teleosts are the most speciose aquatic vertebrates living today and, at over 30,000 named species (Nelson et al., 2006), they comprise more than 95% of all extant fishes. What made them so successful? The answer to this question is likely found in an era more commonly associated with extinction rather than the diversification of life. Caused by extensive volcanic activity and global fires, the Permian crisis (also known as the Permian–Triassic extinction) sets the stage for the teleost story of success. During the Permian crisis atmospheric oxygen (O 2) levels decreased dramatically and remained low for another 100 million years (Clack, 2007). This led to the loss of 96% of all marine fish species. Teleosts, however, experienced an explosive adaptive radiation, perhaps the most extensive one in vertebrate evolution (Helfman et al., 2009). Among the most pivotal adaptations in teleosts was a system of enhanced Hb–O 2 unloading that allowed the swim bladder to be filled at depth and for a supply of O 2 to the avascular retina of the eye (Wittenberg and Wittenberg, 1974); innovations that lead to the lifestyle, habitats, and morphologies of modern teleost species. In addition, it has been argued that a greater hypoxia tolerance was key to not only surviving this mass extinction but also to thrive thereafter (Randall et al., 2014). All of these adaptations were only possible in association with the unique characteristics of teleost red blood cells (RBC) and hemoglobin (Hb).
The following chapter will review the characteristics of cardiovascular O 2 transport in fishes, and discuss those properties of the RBC and Hb that allow teleosts to fine-tune and enhance O 2 uptake, transport, and delivery to the tissues. In addition, a mechanism for enhanced Hb–O 2 unloading that hinges on the heterogeneous distribution of plasma-accessible carbonic anhydrase (PACA) in the circulation is described and discussed, because it is likely unique to teleosts. The distribution of PACA is an intrinsic property of the cardiovascular system, and its presence or absence in various tissues may change the interactions between the blood and the cardiovascular system with great benefit to O 2 transport. Finally, the distribution of PACA is discussed in a phylogenetic context, and within the sequence of events proposed by Berenbrink et al. (2005) that led to the unique mode of O 2 and CO 2 transport in modern teleosts.
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Book series2017, Fish PhysiologyTill S. Harter, Colin J. Brauner
Review article
NOFFI: Advances in Fish Immunology
2008, Fish & Shellfish ImmunologyElisa Randelli, ... Giuseppe Scapigliati
Due to its importance for aquaculture, rearing of teleosts is continuously improving worldwide , and this gives a boost to immunological research of farmed species. It is evident that, at present, teleost fish are the most widely employed ectothermic animal models for immune research. Whole genomes of teleost species have been sequenced like fugu and zebrafish or being sequenced currently (sea bass, Atlantic salmon), thus clearly showing their importance in experimental biology studies. On the other hand, functional immune activities of teleosts are poorly investigated, or deduced by the close similarity between immune genes of fish and mammals.
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Journal2008, Fish & Shellfish ImmunologyElisa Randelli, ... Giuseppe Scapigliati
Chapter
Development and Phylogeny of the Immune System
2016, Encyclopedia of ImmunobiologyBrian Dixon, ... Shawna L. Semple
Abstract
Teleost fish have undergone several rounds of genome duplication in their evolutionary history. The duplication and deletion of their genes has had a profound effect on their immune system and responses. Teleosts have all of the familiar elements of the vertebrate immune system: B cells, T cells, recombining receptors, etc. However, in many cases the structure and regulation of these immune effectors differs from mammalian equivalents. For example, teleost IgM multimers are tetramers, not pentamers, and secondary responses are nonexistent. This is in part due to the altered genetic structure and divergent evolutionary path of teleosts, but differences seen here also reflect the fact that teleost fish are cold-blooded, and thus their environment has profound effects on their metabolism and immune response.
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Reference work 2016, Encyclopedia of ImmunobiologyBrian Dixon, ... Shawna L. Semple
Chapter
Genome Size Evolution in Animals
2005, The Evolution of the GenomeT. RYAN GREGORY
Teleost Fishes
Most of the bony fishes (class Osteichthyes) in the current dataset are members of the subclass Actinopterygii and the infraclass Neopterygii, which includes bowfin (order Amiiformes), gar (order Lepisosteiformes), and the members of the highly diverse superorder Teleostei, which makes up about half of all vertebrate species. All three of the nonteleost neopterygian species so far studied have genome sizes of about 1.2 to 1.4 pg.
To date, roughly 1050 teleost genome sizes have been reported. The most striking observation to be made with this dataset is that, notwithstanding their status as the most speciose group of vertebrates, teleosts exhibit very little variation in genome size among species. The mean for teleosts is only 1.2 pg, and the range is a mere 11-fold, from about 0.4 pg in tetraodontid pufferfishes to 4.4 pg in the masked corydoras, Corydoras metae. The overwhelming majority of species are near the low end of the overall distribution, and it seems that only groups for which polyploidy has played a significant role commonly exceed a C-value of 2.5 pg.
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Book 2005, The Evolution of the GenomeT. RYAN GREGORY
Chapter
Volume 1
2024, Encyclopedia of Fish Physiology (Second Edition)Mari Carmen Uribe
Abstract
Teleosts show unique features of the female reproductive system compared with the rest of vertebrates. Teleosts lack oviducts and therefore the ovary of viviparous teleosts is responsible for not only oogenesis, but also the reception of spermatozoa, fertilization, and intraovarian gestation which is unique among the vertebrates. The gestation may be intrafollicular, when the embryo remains in the follicle throughout embryogenesis, as in poeciliids; or intraluminal, when the embryo moves during early development from the follicle into the lumen where embryogenesis continues, as in goodeids. In viviparous teleosts, the embryonic and maternal tissues form placentas, mediating all exchanges: trophic, excretory, respiratory, immunological, and metabolic waste. These adaptations occur according to lecithotrophy, when the eggs may store enough yolk for the nutrition of the embryo until birth, and matrotrophy, when the eggs store scarce yolk, insufficient for the nutrition of all embryogenesis, which has to be complemented by transfer of maternal nutrients during gestation. This paper focuses primarily on the ovarian structure and embryonic traits of poeciliids and goodeids to illustrate the morphological and physiological elements that characterize viviparity in teleosts.
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Reference work2024, Encyclopedia of Fish Physiology (Second Edition)Mari Carmen Uribe
Chapter
Introduction to the anatomy and physiology of the major aquatic animal species in aquaculture
2021, Aquaculture PharmacologyFrederick S.B. Kibenge, Richard J. Strange
1.2.1 General morphology of teleost
The fish body is divided into a head, trunk, and tail. Nelson (2006) defined a fish as “a poikilothermic vertebrate with gills and with limbs in the shape of fins.” Throughout this text, the term teleost or teleost fish/fishes is used to refer to bony fish of the infraclass Teleostei, superclass Actinopterygii, class Osteichthyes, including everything from goldfish to tuna to salmon (Bowmaker, 2011; McCormick, 2011a). Fish fins are folds of skin supported by skeletal elements controlled by fin musculature and may be used for swimming, reproduction, display, or protection (Standen, 2011). Fig. 1.1 A shows the general external features of a typical finfish. The finfish have, characteristically, a series of unpaired fins along the centerline of the fish (dorsal fin, anal fin, and caudal fin), and paired pectoral fins and pelvic fins (Standen, 2011). Finfish may have one, two, or three dorsal fins (Fig. 1.1). Gadoids (haddock and cod) have three dorsal fins and two anal fins (Fig. 1.1 B). The dorsal and anal fins of scophthalmids and pleuronectids (flatfish) are highly developed and run the length of most of the body perimeter (Noguera et al., 2015). Finfish fins are composed of either bony or soft spines called rays, which, except for the caudal fins (whose primary function is to propel the fish forward), have no direct connection with the spine. The fins are supported by the muscles which compose the main part of the trunk (Dorit et al., 1991), helping the fish to swim. Finfish such as salmonids, characids, and catfish have an additional fin, the adipose fin [a small, fleshy, non-rayed fin located between the dorsal and caudal fins (Buckland-Nicks et al., 2012)]; this is frequently clipped off to mark hatchery-raised fish released into the wild (Tytell, 2005; Vander Haegen et al., 2005). Recent research suggests that the adipose fin [which despite its name contains no adipose tissue at all (Weisel, 1968)] has an extensive neural network indicating it likely has a sensory function, as a “precaudal flow sensor,” and removing it may have consequences (Reimchen and Temple, 2004; Buckland-Nicks et al., 2012; Aiello et al., 2016).
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Fig. 1.1. General external features of a typical finfish illustrated using: (A) largemouth bass Dicentrarchus labrax; (B) Haddock Melanogrammus aeglefinus.
Panel (A): Modified from Panel (B): From diagram of a haddock By Henry B. Bigelow and William C. Schroeder—Fishes of the Gulf of Maine, US Fish and Wildlife Service ( Public Domain,
The fish body is usually covered with conspicuous overlapping scales overlain by an outer epithelium (Mittal and Datta Munshi, 1971) with a mucus or slime layer. Bones in fish, including scales, are not under constant deposition/resorption and so show rings of growth that are proportional to the size of the fish and thus can be used to determine age, and size at age, of the fish (Bone and Moore, 2008). While any bone can be used to age fish, scales are easily sampled without mortality, and so are commonly used. However, other bones, primarily ear stones (otoliths), may prove more reliable. The scale type found on most teleost fishes is a thin, flexible transparent structure consisting of a basal plate of collagenous tissue, with superficial mineralization [referred to as elasmoid (bony-ridge) scale], which is commonly grouped into two general types based on their posterior surface sculpture: cycloid (smooth rounded edges in soft-rayed fish such as salmonids and cyprinids) and ctenoid (jagged edges in spiny-rayed fish such as perch) (Helfman et al., 2009; Elliott, 2011a,b). Note: some spiny-rayed fishes such as wrasses (Labridae) have only cycloid scales (Elliott, 2011b). Some fishes have no scales, for example, many catfishes and eels (Elliott, 2011a), although some eel species such as Anguilla australis and A. dieffenbachii have scales (Jellyman, 1979).
The head has a mouth at or near the tip of the snout, and paired nostrils (or nares) that are anterior to the paired eyes. Each nostril is divided into two portions [except in members of the large family Cichlidae in which it is not divided (Jonna, 2004)] permitting water to circulate through the olfactory sac, for smelling [i.e., detecting odors in water (Laberge and Hara, 2001)] but not for breathing; the nostrils are not connected to the oral cavity. The eyes, which are in the lateral position behind the nostrils, are adapted for seeing underwater (Kröger, 2011). While fish’s vision is often limited by lack of water clarity, under clear conditions, vision is useful at some distance both underwater and in the air. Scophthalmids and pleuronectids (flatfish) have eyes on one side of their head, the ocular side, an adaptation that allows them to lie on the bottom on their eyeless (or blind) side (Gibb, 1997); scophthalmids (e.g., turbot) have eyes on the left side, and pleuronectids (e.g., halibut) have eyes on the right side (Noguera et al., 2015) of their head. Some fish such as cyprinids, catfish, and gadoids have barbels (whisker-like structure) on the head (Fig. 1.1 B) that serve as a sensory organ (with taste buds) (Fox, 1999) to help track down prey or food. Fish have no external ear or middle ear as found in terrestrial animals but have an inner ear that perceives sounds of higher wavelength (Popper, 2011; Fay, 2011). The operculum covers the gills forming a gill cleft/slit with the trunk, allowing water taken through the mouth to pass over the gills (for breathing oxygen contained in the water) and leave through the gill slit. The mouth may or may not have teeth depending on the species, partially reflecting the choice of food (Olsson, 2011). The mouth may have a tongue (a bony structure called the basihyal), which is formed from a fold in the floor of the mouth and looks like a tongue and may have teeth. Some teleost tongues have no taste buds (i.e., they are not used for tasting); others, for example, in swordtail (Xiphophorus hellerii) have taste buds (Reutter et al., 1974).
The lateral line runs just below the surface of the skin along the middle on both sides of the trunk (Fig. 1.1 A and B). It is a mechanosensory system found in all fishes and in the larvae of aquatic amphibians, which is used for the detection of pressure gradients resulting from bulk water flow, such as that occurs during swimming or the motion of conspecifics and prey (Bleckmann and Zelick, 2009; McCormick, 2011a; Webb, 2011), helping the fish to identify their surroundings. The vent is in front of the anal fin and serves to discharge waste and extra water and as an outlet for eggs or milt (sperm) during spawning. Recently, an immune organ analogous to the avian bursa of Fabricius was found for the first time in Atlantic salmon (Salmo salar); the organ referred to as the salmon cloacal bursa is situated caudal to the urogenital papilla of the cloacal region (Løken et al., 2019) (see also Fig. 1.11A and B).
Fig. 1.2 shows the general internal features of a typical finfish. The internal organs of fish are in three cavities (branchial, pericardial, and peritoneal cavities). The branchial cavity contains the gill structure, which is substantially similar in all jawed (gnathostome) fishes (Wilson and Laurent, 2002). The pericardial cavity, lined by a cellular membrane consisting of a layer of epithelial-like cells in addition to connective and fat tissue (the pericardium), encloses the heart and contains a small amount of pericardial serous fluid (see also Fig. 1.6A). The peritoneal (or abdominal) cavity is separated from the pericardial cavity by another cellular membrane (the peritoneum); both membranes are fused on the posterior wall of the pericardial cavity, each membrane also covers the organs and ducts contained in the respective cavity [e.g., the epicardium represents the inner part of the pericardium directly attached to the cardiac muscle, and the dorsal wall of the pericardial cavity is fused to the esophagus (Farrell and Pieperhoff, 2011)]. The peritoneal cavity contains the digestive tract, spleen, liver, and swim bladder. The gonads and kidneys are located outside the peritoneal cavity. The rest of the structures consist of the musculature and the skeleton, including the vertebral column that encases the spinal cord connecting the brain in the head to the rest of the body and terminating in the olfactory bulb in the nostrils.
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Fig. 1.2. Illustration of the external and internal anatomy of a typical bony fish, the rainbow trout.
©DaveCarlson/CarlsonStockArt.com (Image from used under purchased License 2 at
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Book 2021, Aquaculture PharmacologyFrederick S.B. Kibenge, Richard J. Strange
Chapter
The Evolution of Major Histocompatibility Complex in Teleosts
2016, The Evolution of the Immune SystemMasaru Nonaka, Mayumi I. Nonaka
5 Conclusions
The teleostMHC is characterized by the tight linkage of the classical class I gene with the genes directly involved in class I antigen processing/presentation. The linkage between the class I and PSMB8 genes is of special interest, since this linkage seems to have enabled exceptionally dynamic evolution of these genes in teleosts. The most important evolutionary meaning of the vertebrate MHC could be to provide a genome region for the class I antigen processing/presentation genes to perform coevolution.
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Book 2016, The Evolution of the Immune SystemMasaru Nonaka, Mayumi I. Nonaka
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Volume 2
2024, Encyclopedia of Fish Physiology (Second Edition)Edwin W. Taylor, Cléo A.C. Leite
Teleosts
In teleosts, the bony fishes (class Teleostei), activity in the vagus nerve is cardioinhibitory, via stimulation of muscarinic cholinoceptors (specific types of receptors) on the heart. Variations in vagal tone affect mean heart rate and there is an inhibitory vagal tonus that varies with temperature in a variety of teleost species (Taylor et al., 2014). In addition, the sympathetic division of the ANS contributes to a vago-sympathetic trunk, with adrenergic fibers shown to innervate the heart of some teleosts such as rainbow trout (Oncorhynchus mykiss) but its influence is weak and variable (Taylor et al., 1999; Nilsson, 2011). The teleosts are considered to be the earliest group of vertebrates with both sympathetic and parasympathetic control of the heart. However, work on a range of vertebrates, including fish, reptiles and mammals, has demonstrated that fine control of the heart, needed to generate cardiorespiratory interactions, is predominantly parasympathetic (Taylor et al., 2014, 2022). Accordingly, studies on the mechanisms of respiratory modulation of heart rate in teleosts have focused on cardiac control from the parasympathetic nervous system via the Xth cranial nerve, the vagus, and its interactions with activity in the cranial nerves supplying the respiratory muscles, which are inserted around the orobranchial cavity (jaws and gill arches) (Taylor et al., 2009b; Leite et al., 2009).
Work on teleosts has stressed the importance of peripheral receptor inputs in the genesis of cardiorespiratory interactions. In rainbow trout, Oncorhynchus mykiss, an exact synchrony can develop between ventilation and heartbeat during progressive hypoxia (Randall and Smith, 1967). Mean heart rate in normoxia (aerated water) is typically faster than ventilation, and so a hypoxia-induced increase in ventilation rate accompanied by a reflex “bradycardia” (reduced heart rate) can result in convergence of the two rhythms, resulting in 1:1 synchronization (Randall, 1966). The bradycardia and the resulting synchrony can be abolished by pharmacological blockade of “muscarinic” acetylcholine receptors on the heart and cardiac ganglia, using an injection of “atropine” (an alkaloid substance present in plants such as “belladonna”, a member of the plant group known as “nightshades”) (Taylor and Wang, 2009). This effect demonstrates that central control (from the brainstem) of heart rate is generated by efferent vagal control, releasing the neurotransmitter acetylcholine. In addition, the hypoxic heart rate in the trout was shown to synchronize with an imposed phasic flow of water rather than the spontaneous breathing efforts of an intubated fish (Randall, 1966; Randall and Smith, 1967). Clearly, this synchronization was generated by reflex pathways, presumably arising from the stimulation of mechanoreceptors on the gill arches by an increase in ventilation amplitude. Very low levels of efferent activity were recorded from the cardiac nerves of lightly anesthetized normoxic pacu, Piaractus mesopotamicus. The respiration-related bursting activity was recorded only when a fish was hyperventilating or coughing, implying that the bursts arose reflexly, following stimulation of mechanoreceptors on the gill arches (Leite et al., 2009, see Fig.4). Brief cessation of water flow to the gills of a lightly anesthetized fish stimulated a bout of increased ventilatory effort, accompanied by bursts of activity in the cardiac vagus, with consequent transient bradycardia (Fig.5). In fish rendered moderately hypoxic by reduction of the flow of water irrigating the gills, a period of spontaneously increased ventilatory amplitude was accompanied by respiration-related bursts of activity in the cardiac vagus, which appeared to recruit the heart (Leite et al., 2009, see Fig.6). Phasic electrical stimulation of the cardiac vagus toward the heart (Taylor et al., 2009b, see Fig.7) or central stimulation of respiratory branches of cranial nerves VII, IX, and X (Fig.8) can entrain the heart over a wide range of frequencies, similar to, lower or higher than the cardiac intrinsic frequency. Interestingly, central stimulation of the Vth cranial nerve does not affect HR. This is reflected in the distinct distribution of motor neuronsinnervation the Vth nerve (Taylor et al., 2009b, Fig.9 B) and the suggested independent evolutionary origin of the nerve in jawless fishes. However, the bursting efferent activity recorded from this nerve is concurrent with that recorded from the cardiac vagus, suggesting that they are generated independently by feed-forward control within the CNS (Taylor et al., 2009b, Fig.9 A). This was also the case in recordings from the elasmobranch “catshark” (see below) and may constitute a vital mechanism in the central control of cardiorespiratory interactions throughout the vertebrate groups.
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Fig.4. Pacu, Piaractus mesopotamicus: Recordings of electrical activity from the cardiac vagus, ventilation (opercular movements), and ECG. There are no routine, respiration-related bursts in the cardiac vagus in normoxic fish. Activity was restricted to a rare, increased ventilatory effort∗ or spasm “cough”∗∗. Each of these was accompanied by an increase in cardiac interval (slowing of the heart).
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Fig.5. Pacu, Piaractus mesopotamicus: Effect of stopping the water flow irrigating the gills. Increased activity in the cardiac vagus was associated with increased ventilatory effort, generating a bradycardia.
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Fig.6. Pacu, Piaractus mesopotamicus: Restricting water flow increased ventilation amplitude, generating respiration-related bursting activity in the cardiac vagus that recruited the heart, conferring virtual synchrony between the respiratory and cardiac rhythms, signified by the recorded cardiac intervals.
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Fig.7. Pacu, Piaractus mesopotamicus: Pulsed electrical stimulation of the efferent cardiac vagus can drive the heart both more slowly or more rapidly than its intrinsic rate. The pairs of numbers above each trace are the stimulation frequency, followed by heart rate.
Taken from Taylor et al. (2009b).
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Fig.8. Pacu, Piaractus mesopotamicus: Effects on the heart rate of central stimulation of cranial nerves IX, X, VII, and V, supplying respiratory muscles. Continuous stimulation of IX caused cardiac arrest. Pulsed stimulation of VII, IX, and X recruited the heart, but stimulation of V was without effect.
Taken from Taylor et al. (2009b).
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Fig.9. Pacu, Piaractus mesopotamicus: (A) Recordings of efferent activity in cardiac and respiratory nerves revealing that the bursts of activity in cardiac nerves are synchronous with activity in the Vth cranial nerve and both anticipate activity in the IXth and Xth cranial nerves. (B) Rostro-caudal distribution of neurons supplying cranial nerves supplying respiratory muscles and the CVPN supplying the heart. All except those supplying the Vth cranial nerve have an overlapping distribution with CVPN.
Modified from Taylor et al. (2009b).
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Reference work2024, Encyclopedia of Fish Physiology (Second Edition)Edwin W. Taylor, Cléo A.C. Leite
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11940 | https://ui.adsabs.harvard.edu/abs/1964PhRv..134..207S/abstract | Multiple-Pion Production by 2-BeV π- - Astrophysics Data System
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Abstract
A report is given of a study of pion-proton interactions in a liquid-hydrogen bubble chamber. A 2.1-BeV/c negative pion beam incident upon a 14-in.-diam chamber caused interactions from which those with four charged secondaries were selected and measured. Fits were attempted to the following final state: 3π+p, 3π+p+π 0, and 4π+n. By normalizing to a total cross section of 35.7 mb, the partial cross sections for the above-listed final states are found to be 1.67+/-0.08, 1.17+/-0.06, and 0.37+/-0.02 mb. The 3π+p reaction was found to be dominated by the production of the 32-32 pion-nucleon isobar. The cross section for this process is 0.53+/-0.05 mb. Evidence for the eta and omega mesons was found in the 3π+p+π 0 final state. Their cross sections are 0.037+/-0.015 and 0.26+/-0.035 mb, respectively. A Dalitz plot for eta decays is presented. Evidence is given for a double peak in the neutral two-pion mass spectrum for 3π+p and 4π+n events. The peaks coincide with those attributed to the rho meson in other experiments. The second of the two peaks corresponds well with the mass of the ω 0, A branching ratio of ω 0-->2πω 0-->3π is found to be >0.1. An attempt is made to calculate the pion-pion cross section by isolating events with low momentum transfer to the isobar. In the various processes, enhancement is found in the nucleon +2π mass spectra corresponding to the T=32 state of mass 1920 MeV. It is proposed that the elementary crossed diagram can explain these effects.
Publication:
Physical Review, vol. 134, Issue 1B, pp. 207-227
Pub Date:April 1964 DOI:
10.1103/PhysRev.134.B207
Bibcode: 1964PhRv..134..207S Copied!
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11941 | https://www.spandidos-publications.com/10.3892/mmr.2024.13292 | Abnormal dental follicle cells: A crucial determinant in tooth eruption disorders (Review)
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Abnormal dental follicle cells: A crucial determinant in tooth eruption disorders (Review)
Authors:
Jiahao Chen
Ying Ying
Huimin Li
Zhuomin Sha
Jiaqi Lin
Yongjia Wu
Yange Wu
Yun Zhang
Xuepeng Chen
Weifang Zhang
View Affiliations
Affiliations: Clinical Research Center for Oral Diseases of Zhejiang Province, Key Laboratory of Oral Biomedical Research of Zhejiang Province, Cancer Center of Zhejiang University, Stomatology Hospital, School of Stomatology, Zhejiang University School of Medicine, Hangzhou, Zhejiang 310006, P.R. China, Department of Child Health, Yongkang Women and Children's Health Hospital, Yongkang, Zhejiang 321300, P.R. China
Published online on:July 15, 2024
Article Number: 168
Copyright: © Chen et al. This is an open access article distributed under the terms of Creative Commons Attribution License.
Metrics: Total Views: 6867(Spandidos Publications:6867 | PMC Statistics: 0)
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Cited By (CrossRef):7 citations View Articles
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Abstract
The dental follicle (DF) plays an indispensable role in tooth eruption by regulating bone remodeling through their influence on osteoblast and osteoclast activity. The process of tooth eruption involves a series of intricate regulatory mechanisms and signaling pathways. Disruption of the parathyroid hormone‑related protein (PTHrP) in the PTHrP‑PTHrP receptor signaling pathway inhibits osteoclast differentiation by DF cells (DFCs), thus resulting in obstructed tooth eruption. Furthermore, parathyroid hormone receptor‑1 mutations are linked to primary tooth eruption failure. Additionally, the Wnt/β‑catenin, TGF‑β, bone morphogenetic protein and Hedgehog signaling pathways have crucial roles in DFC involvement in tooth eruption. DFC signal loss or alteration inhibits osteoclast differentiation, affects osteoblast and cementoblast differentiation, and suppresses DFC proliferation, thus resulting in failed tooth eruptions. Abnormal tooth eruption is also associated with a range of systemic syndromes and genetic diseases, predominantly resulting from pathogenic gene mutations. Among these conditions, the following disorders arise due to genetic mutations that disrupt DFCs and impede proper tooth eruption: Cleidocranial dysplasia associated with Runt‑related gene 2 gene mutations; osteosclerosis caused by CLCN7 gene mutations; mucopolysaccharidosis type VI resulting from arylsulfatase B gene mutations; enamel renal syndrome due to FAM20A gene mutations; and dentin dysplasia caused by mutations in the VPS4B gene. In addition, regional odontodysplasia and multiple calcific hyperplastic DFs are involved in tooth eruption failure; however, they are not related to gene mutations. The specific mechanism for this effect requires further investigation. To the best of our knowledge, previous reviews have not comprehensively summarized the syndromes associated with DF abnormalities manifesting as abnormal tooth eruption. Therefore, the present review aims to consolidate the current knowledge on DFC signaling pathways implicated in abnormal tooth eruption, and their association with disorders of tooth eruption in genetic diseases and syndromes, thereby providing a valuable reference for future related research.
Introduction
A dental follicle (DF) is a connective tissue sac that forms around nonerupting teeth during early tooth eruption, which originates from the ectoderm of the cranial neural crest and differentiates from cranial neural crest cells (1,2). DFs can be the source of periodontal tissues, dominated by cementum, alveolar bone and periodontal membrane, during tooth development (3). In addition, DFs regulate bone remodeling by affecting the activity of osteoblasts and osteoclasts during tooth eruption (4,5). Therefore, DFs serve an indispensable role in the tooth eruption process. With an enhanced understanding of the DF, researchers have increasingly focused on mesenchymal stem cells (MSCs) located within the DF. MSCS associated with teeth, which have the ability to differentiate, were first isolated from the DFs of human third molars (6). The dental MSCs (DMSCs) found in DFs exhibit similarities to other human MSCs and are specifically referred to as DF progenitor/stem cells (DFPCs) (7). DFPCs can differentiate into cementocytes, osteoblasts, fibroblasts, adipocytes, chondrocytes, neuron-like cells and periodontal ligament cells (5,8–10). Previous research has demonstrated that proper differentiation and functioning of DFPCs are crucial for the normal progression of tooth eruption (11).
Tooth eruption refers to the migration of a tooth from its developmental site within the jaw to its functional position in the oral cavity, thus leading to occlusal contact with the contralateral tooth (12). Recent research has identified five stages of tooth eruption: Pre-eruptive movement, intraosseous eruption, mucosal penetration, pre-occlusal eruption and post-occlusal eruption (13). The resorption of alveolar bone at the apex establishes an eruptive pathway, whereas alveolar bone formation at the root facilitates tooth movement within the jaw (14). Notably, the DF is involved in the tooth eruption process. Studies have shown that different parts of the DF have different functions; specifically, the crown region is responsible for absorbing alveolar bone, whereas the root region regulates alveolar bone formation (15,16). Therefore, a normal DF is an essential factor for tooth eruption.
Failed tooth eruptions include two conditions: Delayed eruption and complete failed eruption (14,17). Delayed eruption is defined as a tooth that deviates from the average eruption time by >2 standard deviations (18). Complete failed eruption can be categorized as primary retention, secondary retention or impaction (19). Primary retention refers to a tooth remaining embedded in the jaw without emerging into the oral cavity (17), whereas secondary retention occurs when teeth erupt but fail to establish occlusion (20). Impaction is the result of physical obstacles that exist in the path of eruption; this barrier constitutes an independent factor, separate from the eruption process itself (19).
Failed eruption of teeth is commonly attributed to physical obstructions or disorders in the tooth eruption mechanism itself (17,19,21). The process of tooth eruption involves a complex array of regulatory mechanisms and signaling pathways. DF cells (DFCs) serve a crucial role in regulating signal transduction between osteoblasts and osteoclasts, thereby governing alveolar bone resorption and formation (5,22). Additionally, a number of syndromes and systemic diseases have been identified as causative factors for tooth eruption disorders (23). Most of these diseases are caused by genetic factors, with certain syndromes also contributing to tooth eruption failure due to DF abnormalities.
A previous study on DFCs primarily focused on their role in normal tooth eruption, with limited research concentrating on abnormal eruptions (24). The role of tooth eruption-related signaling pathways, such as Wnt and TGF-β signaling pathways, in normal tooth eruption has been extensively studied; however, there are few reviews on the mechanisms underlying abnormal signaling pathways in DFCs that result in failed tooth eruption (25,26). Furthermore, numerous studies have summarized the abnormal tooth eruption observed in various genetic diseases and systemic syndrome (23,25,27–29); however, the pathogenesis of these diseases involves multiple causes, with some specifically implicating the DF in the aberrant tooth eruption process. Notably, to the best of our knowledge, there is currently no comprehensive summary available on diseases in which the DF plays a role in pathogenesis. Therefore, the present review aims to elucidate the signaling pathways associated with abnormal DFCs during failed tooth eruption and to explore their molecular impact on eruption mechanisms. Additionally, this review aims to investigate known genetic diseases and syndromes linked to abnormal tooth eruption, thus providing an overview of their clinical manifestations and underlying causes while emphasizing those involving DFCs. The atypical characteristics of DFCs and factors contributing to tooth eruption failure within these disease contexts are also highlighted.
Failure of tooth eruption caused by abnormal signal transduction in DFCs
In recent years, significant insights have been gained into the intricate mechanisms underlying tooth eruption. The complex molecular interactions between cells involved in tooth eruption and DFCs are widely acknowledged. To effectively treat dental diseases characterized by abnormal tooth eruption, it is imperative to understand the fundamental molecular mechanisms within DFCs.
Abnormal parathyroid hormone (PTH)-related protein (PTHrP)-PTHrP receptor (PPR) signaling pathway activity in DFCs
PTHrP functions as a local autocrine/paracrine factor capable of regulating cellular proliferation and differentiation. It exerts regulatory control over epithelial-mesenchymal interactions during organ development, including those of the skin, hair follicles, mammary glands, pancreas and developing teeth (30–34). PTHrP has been reported to be highly expressed in DFCs and as a key molecule necessary for tooth eruption (35). DFCs regulate both bone resorption and formation around teeth, thereby promoting tooth eruption. During this process, PTHrP is instrumental in promoting bone resorption while inhibiting the osteogenesis of DFCs. Research has demonstrated that DFCs, when treated with PTHrP and co-cultured, display reduced expression of osteogenic-related genes, including alkaline phosphatase (ALP), Runt-related gene 2 (RUNX2), bone sialoprotein (BSP) and osteopontin (OPN) (36). Additionally, the Wnt/β-catenin pathway serves as a key signaling pathway for tooth morphogenesis (37), and its activation promotes the stabilization and nuclear translocation of β-catenin (38). PTHrP inhibits the osteogenic differentiation of co-cultured DFCs by suppressing activation of the classical Wnt/β-catenin pathway, primarily through its impact on phosphorylated (p)-GSK-3β. P-GSK-3β reduces the phosphorylation of β-catenin, subsequently inducing nuclear translocation of β-catenin (39). In PTHrP-treated DFCs, the expression of p-GSK-3β has been shown to be reduced (36).
PTHrP is also involved in regulating tooth root development. Cementum covers the surface of the mineralized tissue of the root, and its formation is crucial for root development. Cementoblasts express the PTH/PTHrP receptor. PTHrP stimulation can inhibit the expression of BSP and osteocalcin (OCN) in cementoblasts in vitro, thereby blocking cementoblast-mediated mineralization (40). In a previous study, after the knockout of PTHrP in tooth tissues, including DFs, the surviving mice exhibited tooth eruption failure and abnormal root formation (41). By contrast, the injection of PTHrP can accelerate tooth eruption and inhibit the osteogenesis of DFCs (36). Additionally, PTHrP signaling in the DFC may regulate osteoclast differentiation by influencing the colony-stimulating factor 1 (CSF-1)-receptor activator of NF-κB (RANK)-RANK ligand (RANKL)-osteoprotegerin (OPG) pathway (11,42,43), which is predominantly expressed by DFCs (35). CSF-1 and RANKL stimulate osteoclast formation, whereas OPG inhibits it by competing with RANKL for binding, thereby blocking its activity (44,45). Osteoclasts serve a crucial role in alveolar bone resorption, thus facilitating tooth eruption. PTHrP can promote bone resorption to create a pathway for tooth eruption, whereas the expression of RANKL and OPG serve as a key determinant of osteoclast activity around the teeth (42). Studies have demonstrated that PTHrP treatment increases the expression of osteoclastogenic factors in DFC. Specifically, PTHrP has been shown to elevate the expression of RANKL and reduce the expression of OPG, thereby increasing the RANKL/OPG ratio in DFCs (36); this increased ratio promotes osteoclast differentiation, thus accelerating the process of tooth eruption. Previous studies have constructed RANKL-null mouse models that exhibit impaired tooth eruption, thus suggesting that RANKL plays an integral role in tooth eruption (46). PTHrP can also reduce osteoclastogenesis through the downregulation of CD200, which is closely related to RANKL (11,47). Therefore, abnormal PTHrP expression in DFCs may be closely associated with tooth eruption failure (Fig. 1).
#### Figure 1. RANK-RANKL-OPG pathway is an important pathway for osteoclast differentiation. RANKL secreted by dental follicle cells can promote osteoclast differentiation when it acts on the RANK of osteoclast precursor cells. PTHrP affects tooth eruption by regulating osteoclast differentiation. When PTHrP is underexpressed, the formation and function of osteoclasts are abnormal, leading to insufficient alveolar bone absorption in the eruption pathway and abnormal tooth eruption. In addition, PTHrP can inhibit the expression of CD200, leading to the block of osteoclast differentiation. CSF-1, colony-stimulating factor 1; OPG, osteoprotegerin; PTHrP, parathyroid hormone-related protein; RANK, receptor activator of NF-κB; RANKL, RANK ligand.
PTH receptor-1 (PTH1R), which is also known as the PTH/PPR (35), is a class B G protein-coupled receptor composed of seven transmembrane helices that is abundantly expressed in DFCs (48). PTH1R can interact with both PTH and PTHrP (49,50). PPR can regulate the differentiation of cementoblasts, as PPR-deficient progenitors have been shown to exhibit both accelerated bone fibroblast differentiation and upregulation of NFIC, leading to irregular cellular cementum formation on the surface of roots that normally form acellular cementum; this defect results in abnormal root development (48). Mutations in PTH1R have been associated with primary failure of eruption (PFE) (51–53). PFE is characterized by incomplete or absent tooth eruption despite the presence of an unobstructed pathway for eruption due to dysfunction in the eruption mechanism (54). The association between PFE and PTH1R was initially identified by Decker et al (55). Despite the incomplete understanding of PFE pathogenesis, further investigation of PTH1R has increasingly implicated an aberrant PTHrP-PTH1R signaling pathway in DFCs as being a contributing factor to the development of PFE (29,35). Tooth eruption depends on an unobstructed pathway and sufficient force (56,57). A characteristic of PFE is its unimpeded pathway, thus suggesting that a lack of adequate force may be the underlying cause. The eruption force is driven by the coordinated actions of the DF, alveolar bone formation at the tooth root, and periodontal tissue (29). DFCs serve a vital role in this intricate process, particularly through their involvement in the PTHrP-PTH1R pathway, which directly governs DFPC proliferation, and subsequent DFPC differentiation into cementoblasts, alveolar osteoblasts and periodontal ligament cells (35,58). PTH1R is abundant in DFCs and is particularly enriched in PTHrP+ DFPC (48,51). These findings highlight the importance of the PTHrP-PTH1R signaling pathway in guiding PTHrP+ DFPC differentiation during tooth eruption (51). To confirm that PTH1R deletion leads to PFE, a previous study specifically utilized PTHrP-CreER to delete the receptor in PTHrP+ DFC (59). Due to the fact that the PFE of the first molar in mice shows a similar phenotype to human PFE in adulthood (specifically that of open occlusion) (52,60), mice were selected to establish a model to determine the role of PTH1R in tooth eruption. The results showed that, compared with in the control mice, the PTH1R-deficient mice exhibited a phenotype characteristic of PFE. Subsequent examination demonstrated that PTH1R deficiency in PTHrP+ DFPCs resulted in the abnormal formation of cementoblasts, thus leading to premature cellular cementum formation on the root surface and subsequent loss of periodontal attachment (59). Although previous studies have traditionally considered tooth eruption to be a distinct process from root formation (61), previous studies have established an interrelationship between them (48,59). These findings provide information on the involvement of DFCs in the mechanisms underlying PFE; however, further exploration is necessary to elucidate additional underlying mechanisms (Fig. 2).
#### Figure 2. PFE is caused by the dysfunction of the tooth eruption mechanism. It is clinically manifested as the failure of tooth eruption despite the smooth passage of tooth eruption and no obstruction of alveolar bone and mucosa. PTH1R gene mutations are the main cause of the disease. PTHrP is mainly expressed in DFCs. PTH1R-deficient DFC differentiate into abnormal cementoblasts, which lead to the premature formation of cellular cementum on the surface of the root, lead to the loss of periodontal attachment and affect tooth eruption. DFCs, dental follicle cells; PFE, primary failure of eruption; PTH1R, parathyroid hormone receptor-1; PTHrP, parathyroid hormone-related protein.
Abnormal Wnt/β-catenin signaling pathway activity in DFCs
The Wnt signaling pathway comprises two distinct pathways: The classical β-catenin-dependent pathway and the nonclassical pathway (62). Wnt/β-catenin signaling serves a crucial role in tooth development and eruption, with active expression of Wnt/β-catenin signaling observed in MSCs, including DFCs (63). Wnt signaling is crucial in multiple stages of tooth development and it guides tooth development during fetal formation (64). Aberrant Wnt signaling can impede tooth development, while overactivation can lead to misplaced tooth eruption. After birth, normal tooth root and periodontal tissue formation depend on Wnt signaling. Inactivation of Wnt/β-catenin signaling causes tooth root loss or short roots with increased periodontal space. Proper bone resorption and formation are essential for normal tooth eruption, and Wnt/β-catenin signaling is crucial (63,65). In MSCs, classical Wnt signaling promotes the differentiation of DFPCs into osteoblasts rather than chondrocytes and adipocytes (63). Studies have highlighted the dual role of the Wnt signaling pathway in osteoclast formation; β-catenin activation promotes the proliferation of osteoclast progenitor cells at an early stage, after which β-catenin is inactivated to promote osteoclast differentiation (66,67). This process ensures that osteoclasts perform normal functions and supports smooth bone resorption.
Studies have demonstrated that excessive Wnt/β-catenin signaling in MSCs can result in tooth eruption disorders (68,69). Activation of β-catenin in DFPCs and osteoblasts, under the influence of Wnt/β-catenin signal transduction, leads to upregulation of OPG expression (70), and OPG inhibits the RANK-RANKL pathway, thereby suppressing osteoclast differentiation and maturation, and ultimately contributing to tooth eruption disorder. Conversely, the constitutive activation of β-catenin (Ocn-cre; Ctnnb LOX (EX3)/+, Col1a1-cre; Ctnnb LOX (EX3)/+) or the homozygous deletion of Axin2, which is a negative regulator of Wnt signaling, can promote DFC differentiation, and increase cementoblasts and cellular cementogenesis. Eventually, excessive cementum and tooth stiffness can occur (63,68,71,72). Therefore, the upregulation of Wnt/β-catenin signaling in DFCs can impair tooth eruption by disrupting osteoclast function and promoting the excessive formation of cementoblasts. The latter effect can be reflected in the distortion of periodontal tissue. As the DF is crucial for periodontal tissue formation, Wnt/β-catenin signaling within the DF is indispensable for periodontal tissue homeostasis. Previous studies have shown that mice with continuous Wnt/β-catenin signaling upregulation in dental tissues fail to exhibit tooth eruption (68,71). Upon excluding cases not attributed to disrupted osteoclast activity, it was observed that these mice experienced calcification of the periodontal ligament and functional periodontal ligament, which obliterated the distinction between alveolar bone and cellular cementum, thus leading to tooth rigidity and subsequent failure of tooth eruption (72). Additionally, aberrant osteoblast differentiation can result in tooth eruption failure. Osteoblasts are derived from DFPCs, and their differentiation is also regulated by the Wnt pathway (73). Overexpression of Wnt10b in the Ocn promoter in mice was shown to enhance postnatal bone mass by promoting osteoblast differentiation, which consequently impairs tooth eruption (74).
Abnormal TGF-β signaling pathway activity in DFCs
The interaction between the epithelium and mesenchyme is crucial for tooth morphogenesis and eruption (75,76). This process involves multiple signaling pathways, including the TGF-β signaling pathway, with members of the TGF-β family playing crucial roles in normal and pathological tooth development. Among these members, TGF-β type 2 receptor (Tgfbr2), which is one of the receptors for TGF-β, is expressed in both epithelial and neural crest-derived mesenchyme. A mouse model with conditional deletion of Tgfbr2 in mesenchymal cells using Osterix (Osx) promoter-driven Cre recombinase exhibited delayed tooth eruption. Simultaneously, aberrant differentiation of osteoblasts and dentinal cells was observed, along with a significant decrease in the number of osteoclasts (77). The expression of Osx is primarily localized in the dental mesenchyme, specifically in the apex of the dental papilla and DFC (48). Therefore, it may be hypothesized that Osx is localized in DFCs, thus the Osx-driven Cre recombinase can result in the deletion of Tgfbr2 and the inhibition of TGF-β signaling in the aforementioned models. This leads to the abnormal differentiation of DFCs into osteoblasts and dentinogenic cells, as well as the abnormal formation of osteoclasts, which ultimately causes tooth eruption disorders. Additionally, aberrant expression of Smad4 can disrupt tooth development through the TGF-β signaling pathway, as Smad acts as an intracellular mediator within the TGF-β signaling pathway (78). The conditional deletion of Smad4 in mesenchymal cells derived from the neural crest has been shown to halt tooth development (79). In addition, in a previous study, the deletion of Smad4 in the dental mesenchyme using Ocn-Cre led to abnormal root formation and delayed odontoblast differentiation (80). Due to the fact that MSCs, such as DFPCs, are critical for tooth root development, Smad4 deficiency in DFCs is likely to lead to abnormal tooth development with impaired eruption. Moreover, due to the fact that Smad4 impacts the bone morphogenetic protein (BMP) signaling cascade, it has been implicated in the regulation of the BMP signaling pathway.
Abnormal BMP signaling pathways in DFCs
The BMP signaling pathway is an essential component of osteoblast differentiation and bone development, and it exhibits extensive interplay with TGF-β signaling (81,82). Research has demonstrated that selective knockout of BMP2 in DFPC leads to impaired formation of tooth tissue and periodontal tissue, thus suggesting that the BMP signaling pathway in DFPCs serves an important role in maintaining the normal physiological function of DFPCs (83). In addition, mouse models with deletion of the BMP1 and TLL1 genes have shown impaired tooth eruption (84). BMP1 and TLL1 are encoded by distinct genes, but share similar structures and functions, and belong to a small family of extracellular metalloproteinases (85). Following gene knockout of BMP1 and TLL1, mice with impaired tooth eruption displayed reduced osteoclasts, which was potentially due to impaired osteoclast induction. Mesenchymal cells exhibit high levels of BMP7, and the removal of BMP7 from these cells can lead to impaired tooth eruption and abnormal mineralization (86). One possibility for this effect is that the timing of tooth eruption is directly related to mineralization onset. Another plausible explanation is that BMP7 function in dental pulp and DFPCs affects tooth eruption; however, the specific mechanism involved remains unclear. Additionally, muscle segment homeobox like 2 (Msx2), which is a target of BMP signaling, has been reported to be expressed in mesenchymal cells (87,88), and Msx2-null mice also exhibited tooth eruption failure (89). Experiments have suggested the alteration of the RANK osteoclast differentiation pathway in Msx2-null mice, thus implying the effect of Msx2 on the potential regulation of this pathway, which impacts osteoclast function and causes tooth eruption failure. These findings suggest that dysregulation of the TGF-β and BMP signaling pathways within DFCs often results in abnormal tooth eruption.
Abnormal Hedehog (Hh) signaling pathways in DFCs and other cells
Hh signaling has a crucial role in the development of various organs, including teeth, by mediating interactions between epithelial and mesenchymal cells. Hertwig's epithelial root sheath is surrounded by dental papilla and DFPCs that express receptor patched 1 (Ptch1) for Hh. During tooth development, Hh-expressing cells are strictly localized in the dental epithelium, whereas Ptch-positive cells are found in dental mesenchymal cells without Sonic hedgehog protein (Shh) expression (90). Analysis of mice with mesoblastic dysplasia revealed abnormalities in the C-terminus of the Ptch1 protein. In these mutants, the proliferation of mesenchymal cells around the teeth was inhibited. Additionally, they exhibited disrupted molar eruption and shorter roots. These findings indicate that abnormal transmission of the Shh signal between the epithelium and DF mesenchyme may impact tooth root development and eruption (91). The involvement and functions of DFCs in tooth eruption exhibit temporal and spatial variations. Temporal variations divide tooth eruption into intraosseous and extraskeletal stages, with DFCs playing distinct roles (92). Intraosseous eruption mainly results from alveolar bone resorption and remodeling. During this period, different regions of the DF are thought to play different roles, and DFCs around the crown induce different types of osteoclast differentiation by upregulating the expression of factors such as CSF-1, VEGF and RANKL (24), which leads to an increase in the number of osteoclasts, thus facilitating crown bone resorption to establish unobstructed eruption pathways. By contrast, the DF tissue near the developing root apex serves a key role in alveolar bone formation, thus providing upward force for tooth eruption. This process is intricately linked to the differentiation of DFCs into osteoblasts. Spatial effects on DFs may be the result of regional differences in gene expression. In a previous study, DFCs were isolated from both the crown and basal regions of rat teeth, and RNA was extracted from each region for analysis. The expression of RANKL in the crown region was greater than that in the basal region, whereas the expression of BMP-2 in the basal region was greater than that in the crown region (24). Thus, the spatial localization of gene expression in the DF may modulate osteoclast generation and osteoblast differentiation. The coordinated activity of different regions of the tooth, which is mediated by DFCs, facilitates tooth movement toward the oral cavity. However, the removal of the crown or root region of DF can impede successful eruption; specifically, crown removal disrupts pathway formation, whereas root removal inhibits bone formation (16). The DFs in the two regions exert influences on tooth eruption through distinct signaling pathways. In the crown region, the main pathways involved are the Wnt/β-catenin pathway, TGF-β signaling pathway and BMP signaling pathway. In the root region, in addition to the aforementioned pathways, the Hh signaling pathway also serves a significant role (Figs. 3 and 4). When the tooth is exposed to the oral cavity, it enters the stage of extraosseous eruption. In this stage, the force of tooth upward eruption is mainly derived from the periodontal ligament (93). Furthermore, periodontal tissue derived from DFs and normal follicle development crucially support the tooth eruption progression outside the alveolar bone.
#### Figure 3. Disturbance of DFC signaling pathways results in disorders of osteoclast differentiation. The depicted regions in this figure are primarily situated within the crown dental follicle tissue. Wnt/β-catenin signaling pathway: Activation of the Wnt/β-catenin pathway leads to an upregulation of OPG expression, which subsequently inhibits osteoclast differentiation. TGF-β signaling pathway: Deficiency of Tgfbr2 receptor in DFCs inhibits osteoclast differentiation. BMP signaling pathway: Suppression of BMP1, TLL1 and Msx2 proteins associated with this pathway inhibit osteoclast differentiation. BMP, bone morphogenetic protein; DFC, dental follicle cell; Msx2, muscle segment homeobox like 2; OPG, osteoprotegerin; Tgfbr2, TGF-β type 2 receptor.
#### Figure 4. Depicted regions primarily reside within the dental follicle tissue surrounding the root apex. Overexpression of β-catenin in the Wnt/β-catenin signaling pathway promotes differentiation of DFCs into cementoblasts, thereby enhancing cementum formation. Conversely, loss of Tgfbr2 and Smad4 in the TGF-β signaling pathway, as well as BMP2 deficiency in the BMP signaling pathway inhibits the formation of cementoblasts, Eventually, an excess or deficiency of cementum can result in abnormal root development. The excessive upregulation of Wnt10b and β-catenin in the Wnt/β-catenin signaling pathway leads to a significant increase in bone mass by promoting osteoblast differentiation. The loss of the Tgfbr2 in the TGF-β signaling pathway inhibits the differentiation of DFC into osteoblasts, leading to the block of alveolar bone formation. Loss of Ptch1 receptors inhibits the proliferation of DFCs, thereby affecting tooth eruption. BMP, bone morphogenetic protein; DFC, dental follicle cell; Ptch1, patched 1; Tgfbr2, TGF-β type 2 receptor.
Syndromes and genetic disorders
Abnormal tooth eruption can be classified into two main categories, failed and delayed tooth eruption, and is associated with numerous systemic syndromes and genetic diseases, the majority of which are caused by pathogenic gene mutations. Table I presents a comprehensive list of syndromes and genetic disorders associated with aberrant tooth eruption. According to the data, 48 diseases are known to be linked to abnormal tooth eruption (23,25,28,29). There is convincing evidence for a strong relationship between abnormal tooth eruption and the presence of DFs in seven of these diseases. Furthermore, five of these disorders have been attributed to mutations in specific genes: Cleidocranial dysplasia (CCD), osteopetrosis, mucopolysaccharidosis VI, enamel renal syndrome and dentin dysplasia (DD). Moreover, regional odontodysplasia (RO) and multiple calcifying hyperplastic DFs are two distinct types of tooth eruption disorders that are not associated with genetic mutations.
#### Table I. Syndromes and genetic disorders associated with abnormal tooth eruption.
Table I.
Syndromes and genetic disorders associated with abnormal tooth eruption.
| Disease name | OMIM number(s)a | Orphanet numberb | Association with the DF | Risk factorsb | Tooth eruption status | (Refs.) |
:---: :---: :---:
| Cleidocranial dysplasia | 119600; 620099 | 1452 | Yes | RUNX2 gene mutation | Delayed eruption | (23,25,27–29) |
| Albers-Schönberg osteopetrosis | 166600 | 53 | Yes | CLCN7 gene heterozygous mutations | Failure of eruption | (25,28) |
| Mucopolysac-charidosis VI | 253200 | 583 | Yes | ARSB gene mutation | Delayed eruption | (23) |
| Enamel renal syndrome | 204690 | 1031 | Yes | FAM20A gene mutation | Failure of eruption | (27) |
| Dentin dysplasia | 125400; 125420 | 1653 | Yes | VPS4B gene mutation | Failure of eruption | (27,29) |
| Regional odontodysplasia | / | 83450 | Yes | Local circulatory disorders, viral infections, local trauma, pharmacotherapy during pregnancy, facial asymmetry or a combination of these factors | Failure or delay in eruption | (28,29) |
| Multiple calcifying hyperplastic dental follicles | / | / | Yes | Unknown | Failure of eruption | (25) |
| Gorlin syndrome | 109400 | 377 | Unknown | Ptch1 gene mutation | Failure of eruption | (27,29) |
| Oculodental syndrome, Rutherfurd type | 180900 | 2709 | Unknown | Unknown | Failure of eruption | (28) |
| Cherubism | 118400 | 184 | Unknown | SH3BP2 gene mutation in ~80% of cases | Failure of eruption | (27–29) |
| Albright hereditary osteodystrophy | 612462 | 79444 | Unknown | GNAS gene mutation | Delayed eruption | (25) |
| Gardner syndrome | 175100 | 79665 | Unknown | APC gene mutation | Failure of eruption | (23,27,29) |
| Osteoglophonic dysplasia | 166250 | 2645 | Unknown | FGFR1 gene mutation | Failure of eruption | (28,29) |
| Nance-Horan syndrome | 302350 | 627 | Unknown | NHS gene mutation | Failure of eruption | (28) |
| McCune-Albright syndrome | 174800 | 562 | Unknown | Somatic mutations of the GNAS gene | Failure of eruption | (28) |
| Hypodontia-dysplasia of nails syndrome | 189500 | 2228 | Unknown | MSX1 gene mutation | Failure of eruption | (28) |
| GAPO syndrome | 230740 | 2067 | Unknown | Homozygous nonsense or splicing mutations in the ANTXR1 gene | Failure of eruption | (23,25,27–29) |
| Osteopathia striata with cranial sclerosis | 300373 | 2780 | Unknown | Mutations in the Wilms tumor gene on the X chromosome | Failure of eruption | (25) |
| Singleton-Merten syndrome | 182250; 616298 | 85191 | Unknown | Unknown | Delayed eruption | (25,27,29) |
| Aarskog syndrome | 100050; 305400 | 915 | Unknown | FGD1 gene mutation | Delayed eruption | (25,27,29) |
| Acrodysostosis | 101800; 614613 | 950 | Unknown | Heterozygous mutations in either the PRKAR1A or PDE4D genes | Delayed eruption | (25) |
| Apert syndrome | 101200 | 87 | Unknown | FGFR2 gene mutation | Delayed eruption | (25,27,29) |
| Chondroectodermal dysplasia | 225500; 617088; 618123 | 289 | Unknown | EVC and EVC2 gene mutations | Delayed eruption | (25) |
| Cockayne syndrome | 133540; 214150; 216400; 216411; 278780; 610756; 610758; 616570 | 191 | Unknown | ERCC6 and ERCC8 gene mutations | Delayed eruption | (25) |
| Dubowitz syndrome | 223370 | 235 | Unknown | Unknown | Delayed eruption | (25) |
| Frontometa physeal dysplasia | 305620; 617137 | 1826 | Unknown | Unknown | Delayed eruption | (25) |
| Goltz syndrome | 305600 | 2092 | Unknown | PORCN gene mutation | Delayed eruption | (25) |
| Hunter's syndrome | 309900 | 580 | Unknown | Iduronate-2-sulfatase deficiency | Delayed eruption | (25) |
| Incontinentia pigmenti | 308300 | 464 | Unknown | IKBKG gene mutation | Delayed eruption | (25,29) |
| Levy-Hollister syndrome | 149730; 620192; 620193 | 2363 | Unknown | Unknown | Delayed eruption | (25) |
| Osteogenesis imperfecta | 166200; 166210; 166220; 166230; 259420; 259440; 610682; 610915; 610967; 610968; 613848; 613849; 613982; 614856; 615066; 615220; 616229; 616507; 619131; 619795 | 666 | Unknown | COL1A1 and COL1A2 gene mutations | Delayed eruption | (25,27,29) |
| Hutchinson-Gilford syndrome | 176670 | 740 | Unknown | Unknown | Delayed eruption | (25) |
| Pyknodysostosis | 265800 | 763 | Unknown | Encoding cathepsin K gene mutations | Delayed eruption | (25) |
| Carpenter syndrome | 201000; 614976 | 65759 | Unknown | RAB23 and MEGF8 gene mutations | Failure of eruption | (27,29) |
| Down syndrome | 190685 | 870 | Unknown | Additional independent chromosome 21 (47,+21) | Failure of eruption | (29) |
| Hypertrichosis lanuginosa congenita | 145700; 145701; 307150 | 2222 | Unknown | Unknown | Failure of eruption | (29) |
| Costello syndrome | 218040 | 3071 | Unknown | HRAS gene mutation | Failure of eruption | (27,29) |
| Junctional epidermolysis bullosa | / | 305 | Unknown | mutations in various genes, including COL17A1, ITGA6, ITGB4, LAMA3, LAMB3, LAMC2 and ITGA3 | Failure of eruption | (27,29) |
| Gaucher disease | 230800; 230900; 231000; 231005; 608013; 610539 | 355 | Unknown | GBA gene mutation | Failure of eruption | (29) |
| Hereditary gingival fibromatosis | 135300; 605544; 609955; 611010; 617626 | 2024 | Unknown | Unknown | Failure of eruption | (27,29) |
| Hallermann-Streiff syndrome | 234100 | 2108 | Unknown | Unknown | Failure of eruption | (27,29) |
| Hyperimmuno-globulinemia | 252500 | 576 | Unknown | GNPTAB gene mutation | Failure of eruption | (29) |
| Menkes disease | 309400 | 565 | Unknown | ATP7A gene mutation | Failure of eruption | (29) |
| Neurofibro-matosis type 1 | 162200; 162210; 613675 | 636 | Unknown | NF1 gene mutation | Failure of eruption | (29) |
| Parry-Romberg syndrome | 141300 | 1214 | Unknown | Unknown | Failure of eruption | (29) |
| Sclerosteosis | 269500; 614305 | 3152 | Unknown | Unknown | Failure of eruption | (29) |
| SHORT syndrome | 269880 | 3163 | Unknown | PIK3R1 gene mutation | Failure of eruption | (27,29) |
| Infantile spasms syndrome (West Syndrome) | 300672; 308350; 613477; 613722; 615006; 616139; 616341; 617065; 617929; 618298 | 3451 | Unknown | Gene mutation of STXBP1, TSC1, TSC2 and trisomy 21 | Failure of eruption | (29) |
a Data from
b data from DF, dental follicle.
CCD
CCD, which was identified by Marie and Sainton in 1897, is an autosomal dominant disorder characterized by hypoplasia of the clavicle and skull, widening of the suture and fontanelle, and short stature (49). In addition to skeletal abnormalities, patients with CCD often have dental issues, such as supernumerary teeth accompanied by severe malocclusion and crossbite, retention of primary dentition, impacted teeth and failed tooth eruption (94,95). In a recent study, 50 patients with CCD were examined, 41 of whom had symptoms of tooth eruption failure. These patients presented with a total of 665 teeth displaying abnormal eruption patterns. The most commonly affected teeth were canines (79.5%), followed by permanent premolars (71.0 and 62.5%, first and second permanent premolars, respectively), and superdeciduous teeth and/or retained primary teeth were often observed in this area. Conversely, the first and second molars were less affected (6.0 and 24.0%, respectively) (27).
Genetic studies have shown that mutations in a single allele of RUNX2 cause CCD. These mutations commonly arise from deletions, missense mutations and substitutions occurring within the DNA binding region of RUNX2. The RUNX2 gene, also known as core binding factor a1 (Cbfa1), is located on chromosome 6p21 (96). RUNX2 acts as a crucial transcriptional regulator of osteoblast differentiation during bone formation (97). In addition, heterozygous Runx2-knockout mice were found to exhibit the majority of bone abnormalities observed in human patients with CCD. It has been reported that Runx2 is expressed in preosteogenic mesenchyme and active osteogenesis sites in mice (98–100). Mice with complete deficiency of Runx2 [Runx2 (−/-)] have been shown to exhibit severe osteogenesis imperfecta and often succumb to respiratory distress at birth due to defects in the ribs. Heterozygous mutant mice [Runx2 (+/-)] can survive but exhibit skeletal abnormalities, including an open fontanelle and clavicular defects. This phenotype suggests that a mutation in one allele of Runx2 in mice is sufficient to produce an osteogenic malformation (101). These mice recapitulate the bone abnormalities that are commonly observed in most cases of CCD. To investigate whether these mice can also replicate tooth eruption abnormalities similar to those found in CCD, a heterozygous Runx2-knockout mouse model was generated to observe tooth eruption (101). Compared with wild-type mice, mutant mice exhibited a significant delay in tooth eruption. Further investigations into the impact of Runx2 on skeletal and dental anomalies, and its primary cellular targets, have demonstrated that Runx2 is expressed in osteoblasts and DFs but not in osteoclasts (102,103). Therefore, the abnormal tooth eruption in Runx2 (+/-) young adult mice may be attributed to two factors: i) Due to the inhibition of DF-mediated osteoclast signaling during tooth eruption in Runx2 (+/-) mice; and ii) due to the impaired osteogenic differentiation of DFCs leading to defective bone deposition in osteoblasts and consequently resulting in abnormal eruption.
Further elucidation of the molecular basis of the abnormal eruption observed in Runx2 (+/-) mice is required to test these two possibilities. First, impaired osteoclast recruitment is a possible cellular mechanism for delayed tooth eruption in patients with CCD. Previous research has demonstrated active resorption of alveolar bone and an increase in osteoclasts during eruption in both wild-type mice and Runx2 (+/-) mutant mice; however, this increase was significantly inhibited in the mutant mice. Additionally, this previous study indicated that Runx2 may serve a role in osteoclastogenesis by activating the expression of RANKL and receptor activators of RANK-RANKL signaling (101). It may be hypothesized that the two alleles of Runx2 promote RANK-RANKL signaling, which is essential for active osteoclast recruitment in the tooth germination pathway, and that DFCs play a crucial role in osteoclast recruitment and express Runx2. This finding suggested that Runx2 mutations in DFCs may hinder active alveolar bone resorption by affecting osteoclast numbers, thus contributing to abnormal tooth eruption. Additionally, the effect of Runx2 mutations on osteoblasts was investigated by examining its effect on the osteogenic differentiation of DFCs. The findings demonstrated that a Runx2 mutation decreased the mineralization capacity of DFCs and downregulated the expression of genes associated with osteoblast function, such as ALP, Osx, OCN, ColIα1 and OPN. Furthermore, it disrupted bone formation during tooth eruption, consequently diminishing the osteogenic potential of DFCs. These effects may contribute to abnormal tooth eruption in patients with CCD (104).
Osteopetrosis
Osteopetrosis is a disease caused by disruption of the bone remodeling process with osteoclastic bone resorption defects, and can be divided into intermediate autosomal recessive osteopetrosis (global incidence, 1/250,000) and autosomal dominant osteopetrosis (global incidence, 1/20,000) (105) depending on how it occurs. The clinical manifestations of osteopetrosis commonly include fractures, scoliosis, osteoarthritis, bone marrow insufficiency, developmental delays, tooth eruption disorders and a range of neurological symptoms. Additionally, heightened bone density can lead to compression of cranial nerves and subsequent abnormal innervation (106,107). The eruption of teeth may be delayed or completely absent due to decreased bone resorption and abnormal opening of tooth eruption pathways. Additionally, dental deformities, enamel hypoplasia, dentin abnormalities, inadequate mineralization of enamel and dentin, and defects in the periodontal membrane have been observed (108). A statistical analysis of patients with osteopetrosis demonstrated that the maxillary second molars (66.7%) and mandibular second molars (58.3%) exhibited the highest incidence of tooth eruption failure, whereas anterior teeth and first premolars were rarely affected (27).
Osteopetrosis arises from gene mutations that cause abnormalities in the rough marginal region and dysfunction of osteoclasts, which fail to mediate extracellular acidification in this area, thus resulting in obstructed osteolysis (23). The genes involved in the formation and function of the rough marginal region of osteoclasts include CLCN7, TCIRG1, OSTM1, SNX10 and PLEKHM1. Mutations in these genes impair the transport of endosomal and lysosomal vesicles, thereby disrupting rough marginal regions, as well as osteoclast formation and function (109). Osteoclasts serve a crucial role in tooth eruption, and abnormal tooth eruption in patients with osteopetrosis may be attributed to dysfunctional osteoclasts. Among the aforementioned mutated genes, CLCN7 mutations are the most common cause of osteopetrosis (110), and their impact on osteoclasts is closely related to DFCs. A CLCN7-deficient mouse model was established via injection of chitosan-CLCN7-small interfering RNA nanoparticles, and the mice exhibited abnormal tooth eruption. Coincidentally, these dental changes have also been observed in patients with CLCN7 mutations (111,112). Subsequent experiments have demonstrated that CLCN7 regulates tooth eruption through the DFC-mediated osteoclast pathway by decreasing CLCN7 expression in the DFC, thus leading to reduced numbers of osteoclasts and bone resorption pits (111). Therefore, the lack of CLCN7 in DFCs may inhibit osteoclast formation. This relationship may be mediated through the RANKL-OPG pathway. The RANK-RANKL-OPG signaling axis and downstream transcription factors are important pathways through which DFCs regulate osteoclast generation. OPG secreted by DFCs may inhibit osteoclast generation (45,113), whereas RANKL secreted by DFCs is an important positive regulator of osteoclast differentiation (114,115). RANKL and OPG have been reported to be expressed in DFCs, and CLCN7-deficient mice exhibited downregulated RANKL expression and upregulated OPG expression, which inhibited osteoclast generation (111). Thus, mutations in CLCN7 may result in diminished osteoclasts and aberrant tooth eruption through the RANK-RANKL-OPG signaling pathway mediated by DFCs.
Furthermore, in vitro investigations of DFCs have demonstrated that defects in CLCN7 can impede DFC differentiation. Previous research has indicated that DFCs can differentiate into various cell types, including osteoblasts (116). Normally, induced DFCs exhibit upregulation of osteoblast-related genes, such as ALP, BSP, OPN and TGFB1, thus confirming their potential for osteoblastic differentiation. However, the expression levels of these proteins have been shown to be reduced in a CLCN7-deficient cell group (111). Thus, CLCN7 mutations may be involved in regulating the osteogenic differentiation of DFCs to influence tooth eruption.
Mucopolysaccharidosis VI
Mucopolysaccharidosis represents a cluster of hereditary disorders characterized by impaired degradation of mucopolysaccharides [also known as glycosaminoglycans (GAGs)] due to deficiency of specific enzymes, thus resulting in increased accumulation of mucopolysaccharides across diverse tissues (117). Mucopolysaccharidosis types I–VII are classified based on clinical and biochemical characteristics, and exhibit a high degree of variability. Clinical manifestations include developmental delay, growth retardation and skeletal abnormalities (118). Initially, the accumulation of mucopolysaccharides in various organs leads to progressive intellectual disability and neurodevelopmental deficiency. The most severe consequences occur when excessive GAG accumulation affects the heart, thus resulting in severe cardiovascular disease and even death. Additionally, an excessive buildup of GAG in the DF can impede tooth eruption (23).
Maroteaux-Lamy syndrome (mucopolysaccharidosis type VI), which was initially reported in 1965 (119), is an uncommon autosomal recessive disorder, with a global incidence ranging from 0.0132:100,000 to 20:100,000 (120,121). This disorder arises due to a deficiency of arylsulfatase B (ARSB), which is a crucial gene involved in the degradation of dermatan sulfate. Mutations in this gene lead to the accumulation of undegraded or partially degraded mucopolysaccharides that disrupt cellular function and give rise to various symptoms. Patients with mucopolysaccharidosis VI exhibit physical characteristics resembling those of other types of mucopolysaccharidosis, including short stature, joint stiffness, corneal opacity, and cardiac and respiratory dysfunction (122). However, in contrast to patients with other subtypes, patients with this condition exhibit normal cognitive abilities, metachromatic inclusions in white blood cells and deficiencies in ARSB (123). Furthermore, dental abnormalities are significant manifestations of Maroteaux-Lamy syndrome. These abnormalities are commonly described as dysplastic and widely spaced permanent molars with abnormal root eruption and calcification. Such aberrant teeth often coincide with DF irregularities, wherein excessive deposition of dermatan sulfate impairs the normal morphology and function of the DF. Consequently, the DF becomes tougher and thicker due to the dense fibrous connective tissue observed upon histopathological examination. This abnormal DF increases resistance to tooth eruption, thus ultimately leading to failed tooth eruption (23,118,123).
Enamel renal syndrome
Enamel renal syndrome is an uncommon genetic disorder inherited in an autosomal recessive pattern due to biallelic mutations in the FAM20A gene (124). It is characterized by amelogenesis imperfecta (AI), delayed tooth eruption, intramedullary calcification, gingival enlargement, gingival fibromatosis and nephrocalcinosis (125). Among them, AI and nephrocalcinosis are the most common characteristics of these patients. AI refers to a group of genetic disorders ranging in incidence from 1:700 to 1:14,000 in the United States that affects both the quality and quantity of enamel. Symptoms can be observed in some or all teeth, with AI uniformly affecting enamel across individuals, thus resulting in either hypoplastic or undermineralized enamel. The affected teeth may exhibit discoloration, sensitivity, or increased susceptibility to disintegration prior to or after eruption (126). Nephrocalcinosis is a disease characterized by calcium salt deposition in the kidney, which may be predominantly cortical or medullary in nature; it is often associated with primary hyperparathyroidism, distal renal tubular acidosis and other diseases (127). In addition to enamel and kidney lesions, abnormal tooth eruption is also a prevalent clinical manifestation. Patients with enamel renal syndrome exhibit an aberrant eruption pathway for their posterior teeth (125). Although the root is fully formed, the eruption of the tooth stops halfway, and pericoronal radiolucency manifests around the impacted teeth. Previous case studies have demonstrated that the DF associated with mandibular posterior molars exhibits an atypical structure that is characterized by dense connective tissue and mineralized tissue (128–130). Therefore, delayed tooth eruption can be attributed to the pathological condition of the DF. One possibility for this effect is that DFs may exhibit impaired synthesis of essential molecular components required for proper tooth eruption. Previous studies have demonstrated that FAM20A is localized in the DF above the cusp, and its deficiency has been linked to unsuccessful tooth eruption, thus suggesting a potential role for FAM20A-catalyzed phosphorylation in regulating the pathway involved in shaping the pathway of tooth eruption. Additionally, the presence of pericoronal radiolucency around the impacted teeth can be associated with mutations in FAM20A within the DF (130–132). Another factor is that tooth eruption may be hindered by the DF due to mechanical retention caused by cystic or fibrous transformation. Additionally, the presence of calcification in the DF of patients could contribute to abnormal eruption (133). FAM20A mutations are responsible for enamel renal syndrome and are also associated with calcification in the DF. The gene normally suppresses mineralization; however, in patients with homozygous FAM20A mutations, increased promoter activity and reduced inhibition of oxalate crystal growth cause mineralization of the DF, thus impairing its ability to support normal tooth eruption (127).
DD
Genetic dentin disorders have been well documented and include two primary types: Dentinogenesis imperfecta (DI) and DD (134). Based on the clinical classification, DI can be further categorized into three subgroups (types I–III), whereas DD can be divided into two subgroups (135). The present review specifically focused on DD, which was previously referred to as a ‘rootless tooth’; however, with advancements in understanding of this disorder, this condition has become known as DD. This disorder is classified into type I (DD1) and type II (DD2) (136). DD1 is a rare autosomal dominant nonsyndromic disorder in human dentinal diseases, with an estimated incidence of 1/100,000 (137). In DD1, the patient's crown exhibits a normal shape, morphology and coloration; however, the patient presents with premature tooth loss, tooth loosening and abnormal tooth eruption (138,139). Imaging demonstrates structural abnormalities, including bulbous crowns, occlusion of the endodontic compartment, shortened roots and periapical radiolucency. Pulp remnants in permanent teeth may show crescent-shaped radiolucence, whereas deciduous teeth show complete pulp occlusion (139). The clinical appearance of teeth in patients with DD2 is also normal; however, the primary teeth may appear to be amber and translucent (140). In DD2, the roots exhibit a normal shape and morphological features. Therefore, delayed tooth eruption is rarely reported as being a feature of DD2, but it is often observed in patients with DD1 (139). This is due to the fact that root development has a certain impact on tooth eruption, thus necessitating further investigations into the potential causes of delayed eruption in DD1.
To date, mutations in the VPS4B, SMOC2 and SSUH2 genes have been identified via genetic screening to be associated with the pathogenesis of DD1 (141–143). Among them, VPS4B has been shown to be closely related to the formation of alveolar bone and cementum, and the normal differentiation of DFCs is also an essential component of cementogenesis and the development and formation of surrounding alveolar bone (144). Therefore, VPS4B mutations may lead to abnormal osteogenesis by affecting the normal differentiation and proliferation of DFCs, and eventually leading to abnormal tooth eruption. Ultimately, a comparative analysis of the proliferation and osteogenic induction capacity of DFCs derived from patients with VPS4B-mutant DD1 and healthy controls was conducted (145). The growth rates of DFCs were found to be significantly greater in patients with DD1 than in controls; however, compared with those from control individuals, DFCs from patients with DD1 exhibited lower expression levels of osteogenic genes, such as ALP, OCN, BSP and RUNX2, as well as fewer calcium nodules, as observed via Alizarin red S and ALP staining. These findings suggest that VPS4B may have a crucial role in regulating the osteogenic differentiation of DFCs and that mutations in VPS4B could lead to reduced osteogenic capacity in patients with DD1. Consequently, impaired root formation and bone remodeling during development may ultimately contribute to tooth eruption disorders.
RO
RO is a rare developmental anomaly that was first described by Zegarelli et al in 1963 (146). The etiology of RO remains incompletely elucidated, although it is not believed to have a hereditary basis (147). Various potential pathogenic factors have been postulated in the literature, including local trauma, radiation exposure, high fever episodes, vascular disorders, prenatal drug administration, localized or systemic viral infections, reactivation of latent viruses, impaired migration or differentiation of neural crest cells, nutritional or metabolic deficiencies, ischemia events and Rhesus disease (148,149). The clinical manifestations of RO include discoloration of teeth (yellow or brown), impaired tooth eruption, atypical tooth morphology, tooth mobility, and the presence of swelling or abscess formation (150). The main radiographic characteristics include an enlarged pulp cavity, open root apices, indistinct borders and a ghost-like appearance of the affected tooth (151). Histologically, enamel and dentin show hypoplasia and insufficient calcification, the pulp is larger than normal, and the DF appears to be calcified (152,153). In general, the disease affects both primary and permanent dentition. The mandible is generally more susceptible than the maxilla. Among the clinical manifestations, tooth eruption failure commonly occurs (149). The failure of tooth eruption may be attributed to dental deformities hindering the process, abnormal calcification and swelling of the DF causing mechanical obstruction, or dysregulation of signaling pathways in DFPCs during the eruption induction pathway resulting from calcification of the DF (154,155). According to the literature, imaging studies have demonstrated abnormal hyperplasia and fibrous tissue swelling in the vicinity of nonerupted teeth (154–156), which is associated with aberrant calcification of the DF tissue. In addition, histological studies have demonstrated various types of calcification within the DF of patients with RO, including fibrous or nonfibrous osteoid chains, as well as fused calcified spheres attached to larger calcified masses or osteoid chains. These calcifications are predominantly located in areas typically occupied by enamel formation, some of which are formed independently of collagen involvement, whereas others result from collagen fiber mineralization (154,157). The accumulation of calcified tissue in the DF is closely associated with both the enlargement of the DF and an increase in periodontal fibrous tissue. These abnormal DF tissues may eventually lead to tooth eruption disorder.
Multiple calcifying hyperplastic DFs (MCHDFs)
MCHDFs are rare, and their etiology is still unclear (158); they are clinically defined by multiple unerupted teeth and large DFs (159). Radiographically, these follicles are observed as radiolucency surrounding the crown of the unerupted tooth and may also exhibit radiopaque lesions within the inner part of the DF (160,161). The histological features of this condition include extensive cemento-like calcification and the presence of residual odontogenic epithelium within the fibrous connective tissue matrix (162). The process of calcification is usually performed in DFs because DFPCs in DFs can differentiate into cementoblasts or osteoblasts (163). Impacted teeth may result from incomplete digestion of fibrous tissue (164) and abnormal structure or enlargement of the DF, which obstructs tooth eruption. Additionally, calcified tissue within the DF could disrupt DFPC-related signal transduction pathways that are crucial for proper tooth eruption (162). The reported data have suggested that the incidence of type I calcification is greater in patients with MCHDF than in patients with type II calcification. However, type I calcification may also occur in DFs with hypoplasia and regional odontodysplasia, thus suggesting similar etiologies for tooth eruption disorders in these conditions (165). Following the excision of abnormal DFs, successful eruption of impacted teeth in patients with MCHDF further underscores the pivotal role of diseased DFs in eruption failure (165).
Conclusion
The DF serves a crucial role in tooth eruption, and regulates the formation and resorption of alveolar bone. Abnormalities in DFCs are closely associated with abnormal eruption patterns, and disturbances in signaling pathways within the DF represent an important factor contributing to tooth eruption disorders. PTHrP signaling can modulate osteoclast differentiation by influencing the CSF-1-RANK-RANKL-OPG pathway. Moreover, PTH1R is abundantly expressed in DFCs, and interacts with both PTHrP and PTH. Notably, mutations in PTH1R are associated with PFE. Wnt/β-catenin, TGF-β and BMP signaling pathways are essential for tooth development and eruption, with disruptions in these pathways impairing osteoclast and osteoblast functions, and leading to eruption disorders. Furthermore, disrupted Shh signaling transmission between the epithelium and DF mesenchyme may also impact tooth root development and eruption. Moreover, DF abnormalities are clearly associated with various clinical syndromes exhibiting tooth eruption disorder symptoms. These include skull dysplasia, osteopetrosis, mucopolysaccharidosis VI, enamel renal syndrome and DD, which are caused by mutations in related genes. Moreover, conditions such as regional tooth dysplasia, MCHDFs and some odontogenic cysts are not attributed to genetic mutations or have an unknown etiology; instead, they mostly arise from structural anomalies within the DF that mechanically impede normal tooth eruption.
Future perspectives
A deeper understanding of the mechanisms involving DFCs in tooth eruption is crucial. This present review may improve knowledge and aid in resolving clinical issues related to the regulation of tooth eruption. The application of single-cell epigenomic technology may facilitate a more comprehensive understanding of the epigenetic regulation governing DFPCs and determine the patterns of epigenetic modifications that are potentially implicated in tooth eruption disorders. Additionally, the DF organoid model has emerged as an experimental paradigm for investigating tooth development and regeneration in recent years (166,167). The application of the this model is expected to gradually expand. Currently, DF organoid models mainly target tooth development issues in children and adolescents; however, with technological advancements and the increase in clinical practice, this model may also serve a significant role in investigating tooth eruption. By integrating these techniques, we aim to identify the potential molecular mechanisms of DFPCs in tooth eruption disorders, and provide crucial theoretical support and a scientific basis for future developments in tooth regeneration treatments.
Acknowledgements
Not applicable.
Funding
This work was supported by the Key R&D Program of Zhejiang (grant no. 2023C03072), the National Natural Science Foundation of China (grant no. 81400511), the Zhejiang Provincial Natural Science Foundation of China (grant no. LY18H140001), and the R&D Program of the Stomatology Hospital of Zhejiang University School of Medicine (grant no. RD2022JCEL04). XPC is sponsored by the Zhejiang Provincial Program for the Cultivation of High-level Innovative Health Talents.
Availability of data and materials
Not applicable.
Authors' contributions
WZ, XC and JC conceptualized the study. YY validated the reliability of the topic selection. JC, YY, JL, HL, ZS and YZ performed the literature review and wrote the manuscript. WZ, XC, YoW and YaW completed the review and editing of the manuscript. JC and YZ completed the supervision of the work. HL and ZS participated in generating the figures. WZ conduct the project administration. XC and WZ provided funding. Data authentication is not applicable. All authors have read and approved the final version of the manuscript.
Ethics approval and consent to participate
Not applicable.
Patient consent for publication
Not applicable.
Competing interests
The authors declare that they have no competing interests.
Use of artificial intelligence tools
During the preparation of this work, AI tools were used to improve the readability and language of the manuscript, and subsequently, the authors revised and edited the content produced by the AI tools as necessary, taking full responsibility for the ultimate content of the present manuscript.
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Chen J, Ying Y, Li H, Sha Z, Lin J, Wu Y, Wu Y, Zhang Y, Chen X, Zhang W, Zhang W, et al: Abnormal dental follicle cells: A crucial determinant in tooth eruption disorders (Review). Mol Med Rep 30: 168, 2024.
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Chen, J., Ying, Y., Li, H., Sha, Z., Lin, J., Wu, Y. ... Zhang, W. (2024). Abnormal dental follicle cells: A crucial determinant in tooth eruption disorders (Review). Molecular Medicine Reports, 30, 168.
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Chen, J., Ying, Y., Li, H., Sha, Z., Lin, J., Wu, Y., Wu, Y., Zhang, Y., Chen, X., Zhang, W."Abnormal dental follicle cells: A crucial determinant in tooth eruption disorders (Review)". Molecular Medicine Reports 30.3 (2024): 168.
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Chen, J., Ying, Y., Li, H., Sha, Z., Lin, J., Wu, Y., Wu, Y., Zhang, Y., Chen, X., Zhang, W."Abnormal dental follicle cells: A crucial determinant in tooth eruption disorders (Review)". Molecular Medicine Reports 30, no. 3 (2024): 168. |
11942 | https://www.sciencedirect.com/topics/medicine-and-dentistry/donor-apheresis | Donor Apheresis - an overview | ScienceDirect Topics
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Donor Apheresis
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Donor apheresis is defined as the process of separating individual blood components, such as red blood cells, platelets, and plasma, from healthy donors using a specialized device.
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Review article
International Forum: The Turkish perspective on apheresis activity: The Turkish apheresis registry report
2023, Transfusion and Apheresis ScienceDuzgun Ozatli, ... Fevzi Altuntas
1 Introduction
Apheresis refers to the process of seperating the cellular and/or soluble components of blood using a device. It can be divided into donor apheresis and therapeutic apheresis (TA). Donor apheresis is often done on healthy donors to obtain individual blood components such as red blood cells, platelets, plasma. TA refers to an extracorporeal treatment that selectively removes abnormal cells or harmful substances in the blood that are associated with or cause certain diseases . TA modalities include plasma exchange and cytapheresis. The most frequently used TA procedures are as follows; therapeutic plasma exchange (TPE), therapeutic leukocytapheresis, red blood cell exchange, and extracorporeal photopheresis (ECP) . Cytapheresis procedures are used for removing abnormal blood components or excessive numbers of blood cells. Plasmapheresis is a procedure in which a device is used to separate the plasma from the blood cells. After the plasma is seperated from the blood cells, the remaining are returned to the body . In ECP, peripheral blood lymphocytes are collected and treated with 8-methoxypsoralen, which is a photosensitive molecule, then it is given back to the patient after exposure to ultraviolet-A . According to the updated guidelines by The American Society for Apheresis (ASFA), there are no new indications for therapeutic apheresis listed in the 2019 edition, however, several recommended category changes to existing indications have been made . This report is about the apheresis activity in different situations and highlights the published articles’ main results in Turkey.
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Journal2023, Transfusion and Apheresis ScienceDuzgun Ozatli, ... Fevzi Altuntas
Chapter
Introduction to Blood Banking and Transfusion Medicine
2019, Transfusion Medicine and Hemostasis (Third Edition)Christopher D. Hillyer MD, Beth H. Shaz MD
Apheresis
In the 1950s, Cohn designed a centrifuge to separate cellular components from plasma. Donor apheresis allowed collection of therapeutic doses of platelets, granulocytes, RBCs, and plasma from a single donor. Automation of therapeutic apheresis devices has expanded its use, vital to the treatment of many diseases (e.g., thrombotic thrombocytopenic purpura, sickle cell disease).
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Book 2019, Transfusion Medicine and Hemostasis (Third Edition)Christopher D. Hillyer MD, Beth H. Shaz MD
Chapter
Brief History of Blood Transfusion
2009, Transfusion Medicine and HemostasisAlfred J. Grindon MD
Apheresis:
In the 1950s, Cohn designed a centrifuge to separate cellular components from plasma. As a result of work from many fields the advanced instrumentation for apheresis (both therapeutic and donor) became available. The development of donor apheresis allowed collection of therapeutic doses of platelet and granulocyte components from a single donor, and the collection of sufficient volumes of plasma for further manufacturing into factor concentrates, albumin, immunoglobulin and other derivatives. Full automation of therapeutic apheresis devices has expanded and simplified the use of this modality, which is vital to the treatment of many diseases (e.g. thrombotic thrombocytopenic purpura, sickle cell disease). Starting in approximately 2000, automated apheresis machine-based collection of red blood cells as double red cell products, or as a combination red blood cell unit with platelet or concurrent plasma components, began in earnest and continues to grow today.
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Book 2009, Transfusion Medicine and HemostasisAlfred J. Grindon MD
Chapter
Blood Banking and Transfusion Medicine – History, Industry, and Discipline
2013, Transfusion Medicine and Hemostasis (Second Edition)Christopher D. Hillyer MD
Apheresis
In the 1950s Cohn designed a centrifuge to separate cellular components from plasma. Through the work of engineers, inventors, physicians, and operators, the advanced instrumentation for apheresis developed. The development of donor apheresis allowed collection of therapeutic doses of platelet and granulocyte components, and more recently collection of two red cell products from a single donor and the collection of sufficient volumes of plasma for further manufacturing into factor concentrates, albumin, immunoglobulin, and other components. Full automation of therapeutic apheresis devices has expanded and simplified the use of this modality, which is vital to the treatment of many diseases (e.g. thrombotic thrombocytopenic purpura, sickle cell disease).
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Book 2013, Transfusion Medicine and Hemostasis (Second Edition)Christopher D. Hillyer MD
Chapter
Adverse Donor Reactions
2009, Transfusion Medicine and HemostasisKrista L. Hillyer MD
Apheresis-specific Reactions:
Apheresis (the withdrawal of blood with a portion being separated and the remainder re-transfused into the donor) removes a limited amount of blood from the donor, and with some systems the volume removed is replaced, so vasovagal reactions are less frequent (~1%) than for whole blood donation. In addition, apheresis donors are required to have higher blood volumes than whole blood donors. Hematomas and venepuncture complications can also occur in apheresis donors, as described above for whole blood donation. There are donor reactions which are specific to apheresis, including hypocalcemia, allergic reactions, and machine malfunction leading to thrombi, hemolysis, air emboli and leakages.
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Book 2009, Transfusion Medicine and HemostasisKrista L. Hillyer MD
Chapter
Adverse Donor Reactions
2009, Transfusion Medicine and HemostasisKrista L. Hillyer MD
Complications of Apheresis Donations:
Apheresis-specific Reactions:
Apheresis (the withdrawal of blood with a portion being separated and the remainder re-transfused into the donor) removes a limited amount of blood from the donor, and with some systems the volume removed is replaced, so vasovagal reactions are less frequent (~1%) than for whole blood donation. In addition, apheresis donors are required to have higher blood volumes than whole blood donors. Hematomas and venepuncture complications can also occur in apheresis donors, as described above for whole blood donation. There are donor reactions which are specific to apheresis, including hypocalcemia, allergic reactions, and machine malfunction leading to thrombi, hemolysis, air emboli and leakages.
Hypocalcemia:
Hypocalcemia is due to citrate anticoagulant, which binds calcium, used to keep the blood from clotting in the apheresis machine tubing. Usually hypocalcemia results in perioral or peripheral parathesias, nausea and shivering. Severe hypocalcemia can result in loss of consciousness, convulsions or cardiac arryhthmia. Hypocalcemia occurs in up to 10% of collections, but is severe in less than 0.03% of collections. Treatment of hypocalcemia includes reducing citrate infusion rate, administering calcium tablets and, for severe cases, terminating the collection and administering intravenous calcium.
Machine Malfunction:
Machine malfunction leading to thrombi, hemolysis, air emobolism or leakage occurs in less than 0.5% of collections, but the effects can be severe. It is usually secondary to improper manufacturing of the disposable set or improper mounting of the set. Newer apheresis devices have improved safety features to protect the donor and mitigate these complications.
Allergic Reactions:
Donors may be hypersensitive to sterilizers, especially ethylene oxide, or to components of the apheresis set.
Platepheresis/Granulocytapheresis:
There are other considerations that must be taken into account for processes that extract platelets or granulocytes. In some instances frequent platepheresis can cause a drop in platelet levels that, in some cases, will cause collections to be discontinued altogether. Granulocytapheresis can also cause certain side-effects. Donors who take small doses of corticosteroids may experience insomnia. Furthermore, a combined dose of G-CSF and dexamethasone may cause insomnia, mild bone pain, and headaches (see Chapter 32).
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Book 2009, Transfusion Medicine and HemostasisKrista L. Hillyer MD
Review article
Disorders of the Platelets
2013, Hematology/Oncology Clinics of North AmericaPerumal Thiagarajan MD, Vahid Afshar-Kharghan MD
Apheresis Units
Increasingly, platelets are harvested by the apheresis method. Apheresis platelets are especially useful in collecting from family members, HLA-compatible donors, or individuals with rare phenotype. Apheresis has the advantage of collecting from a single donor that limits donor exposure and provides platelets with consistent quality. The apheresis donor unit contains at least 3 × 10 11 platelets per unit. More than 60% of platelets infused in the United States are from apheresis donors. Donors can donate twice a week. The newer automated apheresis instruments are capable of collecting single, double, and even triple apheresis units from a donor and they can be programed to yield optimal amount based on donor hematocrit, height, and weight.
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Journal2013, Hematology/Oncology Clinics of North AmericaPerumal Thiagarajan MD, Vahid Afshar-Kharghan MD
Review article
Adverse events in apheresis: An update of the WAA registry data
2016, Transfusion and Apheresis ScienceM. Mörtzell Henriksson, ... A. Wahlström
3.7 Donor vs. patient apheresis, collection of cells for cellular therapy
Table 12A includes data of adverse events of those procedures registered for autologous versus allogeneic collection. Donor apheresis was performed as cytapheresis for peripheral leukocyte and stem cell collections (n = 620 donors, 56% men) in 1,684 occasions. The mean age was 46 years (±14, range 3–74 years). The grade of adverse events is shown in Table 12A. There were more mild AEs during allogenic procedures (p < 0.001, RR 3.2, CI 2.3–4.4), more moderate AEs during autologous procedures (p = 0.006, RR 1.4, CI 1.1–1.9), and there was no difference for severe AEs.
Table 12. (A) Adverse events (in %) in donor procedures versus patients (leukapheresis procedures); (B) various symptoms of adverse events in donor apheresis given as AEs/10,000 procedures (AEs/10E4).
| (A) None | Mild | Moderate | Severe | AE total | Total N |
--- --- --- |
| Patients (reference) | 1.3 | 5.9 | 0.2 | 7.4 | 4,836 |
| Donors | 4.3 aaa | 3.9 bb | 0.2 | 8.4 | 1,684 |
| (B) Donor apheresis |
| Grade of AE | Symptoms | AEs/10E4 |
| Moderate | Tingling, stitching | 536 |
| Mild | Hypotension | 67 |
| Mild | Tingling, stitching | 42 |
| Mild | Access hematoma | 25 |
| Moderate | Nausea/vomiting | 25 |
| Moderate | Headache | 25 |
| Moderate | Hypotension | 25 |
| Mild | Vertigo | 17 |
| Severe | Fasciculations | 17 |
| Mild | Hypertension | 8 |
| Mild | Phlebitis | 8 |
| Mild | Nausea/vomiting | 8 |
| Moderate | Flush | 8 |
| Moderate | Vertigo | 8 |
| Moderate | Chills/fever | 8 |
| Severe | Anxiety & hyperventilation | 8 |
| Severe | Hypotension | 8 |
More than reference: aaa p < 0.001; less than reference: bb p < 0.01.
Seventy percent of the mild AEs were due to problems with access. The various other symptoms are given in Table 12B.
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Journal2016, Transfusion and Apheresis ScienceM. Mörtzell Henriksson, ... A. Wahlström
Chapter
Adverse Donor Reactions
2013, Transfusion Medicine and Hemostasis (Second Edition)Benjamin J. Greco MD, Beth H. Shaz MD
Complications of Apheresis Donations
Apheresis (the withdrawal of blood with a portion being separated into the desired component and the remainder re-infused into the donor) removes a limited amount of blood from the donor. Most available technologies replace the volume removed with 0.9% NaCl, so vasovagal reactions are less frequent (<50/10,000 collections) than for whole blood donation which is likely due to the more modest fluid shifts and smaller net fluid deficits associated with apheresis procedures. In addition, apheresis donors are required to have higher blood volumes than whole blood donors. Hematomas and venipuncture complications can also occur in apheresis donors as in whole blood donations based on adequate venous access and the phlebotomist’s skill set. There are donor reactions which are specific to apheresis, including hypocalcemia (from citrate), allergic reactions, and machine malfunction leading to thrombi, hemolysis, air emboli and leakages.
Citrate-Induced Hypocalcemia
Citrated anticoagulants are universally employed in apheresis donor collections and are relatively safe solutions for infusion. They exert their effect by binding to calcium, which is an efficient mechanism to avoid clotting of blood in the tubing of the apheresis kit. Transient hypocalcemia associated with apheresis collections are usually well tolerated. Decreases in ionized calcium can increase nerve cell membrane excitability leading to spontaneous depolarization. This is usually manifested by perioral or peripheral paresthesias or both. Less frequently, donors may express an ‘unusual taste’ or experience transient nausea or light-headedness. Other signs and symptoms include cyanosis, spasms, chills/shivering, confusion, tetany, pallor, change in pulse, chest pain, shock, dyspnea, tachycardia, and tremors. Severe hypocalcemia can result in loss of consciousness, convulsions or cardiac arrhythmia. Hypocalcemia occurs in up to 1,000/10,000 of collections, but is severe in less than 3/10,000 of collections. Treatment of hypocalcemia includes reducing citrate infusion rate, administering calcium tablets and, for severe cases, terminating the collection immediately, notifying emergency care personnel and the center’s medical director.
Machine Malfunction
Machine malfunction leading to thrombi, hemolysis, air emobolism or leakage occurs in less than 5/10,000 collections, but the effects can be severe. It is usually secondary to improper manufacturing of the disposable set or improper mounting of the set. Newer apheresis devices have improved safety features to protect the donor and mitigate these complications. Symptoms and signs of hemolysis include back/flank pain, cyanosis, hematuria, confusion, pallor, red plasma, shock, dyspnea and tachycardia. Symptoms and signs of air embolus are similar to hemolysis without hematuria.
Allergic Reactions
Donors may be hypersensitive to sterilizers, especially ethylene oxide. Symptoms may include hives, difficulty in breathing, wheezing, hypotension or hypertension, anxiousness, tachycardia or bradycardia, facial swelling or flushing and burning eyes. The mentioned symptoms tend to manifest early in the procedure and generally dissipate once the procedure is stopped. Severe allergic reactions are classified as anaphylaxis.
Platepheresis/Granulocytapheresis
There are other considerations that must be taken into account for processes that collect platelets or granulocytes. In some instances frequent plateletpheresis can cause a drop in platelet levels that, in some cases, will cause collections to be discontinued altogether.
Using pentastarch, hetastarch or hydroxyethyl starch, which separates out granulocytes from RBCs, enhances granulocyte collection. Donor side effects from starch include allergic reactions, edema, and weight gain. Corticosteroids and G-CSF increase granulocyte mobilization of the donor, resulting in higher collection yields. Granulocytapheresis donors who are prescribed small doses of corticosteroids to effect mobilization and improve collections pre-collection should be verbally screened for uncontrolled hypertension (HTN), diabetes mellitus and peptic ulcer disease, as this medication will aggravate these conditions. Additionally, donors should be advised that this medication may elicit headache, elation, sleeplessness or nausea after ingestion. Generally, the use of a single dose of corticosteroids is not associated with short- or long-term side effects. The more common side effects of G-CSF are bone/joint ache or pain, muscular ache and/or headache. The pain is usually mild to moderate and generally responds well to NSAIDs. Rarely, reports of rash, nausea, infection at the injection site, allergic reactions, worsening of pre-existing inflammatory and immunological disorders, symptoms in patients with underlying coronary artery disease, low platelets, blood clotting and severe reactions in patients with sickle cell disease or its variants. Modest enlargement of the spleen may occur particularly in donors receiving multiple doses of G-CSF. No long-term side effects of G-CSF in normal donors have been reported to date (see Chapter 35).
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Book 2013, Transfusion Medicine and Hemostasis (Second Edition)Benjamin J. Greco MD, Beth H. Shaz MD
Chapter
Therapeutic Apheresis—Applications for Hemorrhagic and Thrombotic Disorders
2019, Consultative Hemostasis and Thrombosis (Fourth Edition)Laura S. Connelly-Smith MBBCh, DM, Meghan Delaney DO, MPH
Overview and Technical Considerations
The clinical procedures collectively referred to as apheresis involve the separation of a solute or cellular component of the blood for removal or for treatment and reinfusion. Apheresis was first introduced as a novel therapeutic modality in the 1950s.1 By the early 1960s plasma removal (plasmapheresis) was being used to manage selected patients with monoclonal paraproteinemia due to myeloma or Waldenström macroglobinemia.1–4 Apheresis technology was then adapted to separate and concentrate human platelets from platelet-rich plasma to salvage platelets from donors for transfusion for hemostatic control in pediatric leukemic patients.5,6 These early efforts used manual blood exchange or discontinuous processing, beginning with phlebotomy, then mechanically separating one component of the blood and returning the remainder of the blood to the subject.
Indications and applications of therapeutic and donor apheresis progressed rapidly with the advent of continuous flow technology and online processing through a centrifugation separation chamber or membrane filters.7,8 Within a spinning centrifugation chamber, whole blood is separated into cellular and plasma layers based on their relative densities (i.e., specific gravity). The efficiency of separation depends on the applied g-force and the dwell time within the chamber. Red cells, which are the “heaviest,” sediment most rapidly followed by granulocytes, mononuclear cells (MNCs), and platelets (Table 29.1). By the 1970s several continuous flow instruments, which process and return blood without interruption, became available worldwide for plasma removal/exchange, red cell depletion, and collection of donor plasma, white blood cells, and platelets. These early devices used reusable parts, which required pasteurization of the tubing and complete disassembly of the centrifuge for sterilization between subjects.1 As technology and demand increased, fully disposable extracorporeal kits were developed with improved circumferential flow geometry that achieved better separation of cellular components and greater operator control. By the late 1980s fully automated operating systems were introduced that used algorithms to run all of the desired procedures (cellular apheresis and plasmapheresis). More recent innovations include refined interface separation efficiency, increased procedural speed, reduced extracorporeal volume requirements, and multistep processing methods, which are especially useful for extracorporeal photopheresis and double plasma filtration.1,7,8
Alternative approaches to bulk plasma separation and removal by centrifugation involve membrane plasma filtration and selective adsorption of a plasma solute. The former uses a membrane with a selective pore size that allows the passage of plasma but not blood cellular components. The latter method uses affinity columns with resins or specific immobilized ligands to capture the solute target of interest from separated plasma.7–9 Plasma filtration is less efficient than bulk plasma separation by centrifugation, but dual or “cascade” filtration steps can be incorporated into a single treatment process. Specific solute removal with adsorption columns is inherently more selective than plasma exchange, but the technique requires knowledge of the pathogenic solute's identity and binding properties. In addition, affinity columns must use reliable and high-capacity binding substrates. Only a few membrane filtration and affinity column systems are currently available in the United States, and these are primarily used for low-density lipoprotein (LDL) apheresis.7,8 Column-based instruments are more widely available in Europe and Asia, and these have been used for some indications in hemorrhagic or thrombotic disorders. Centrifugation apheresis instruments are predominantly used for therapeutic procedures by centers in North America.
Regardless of the technique used to perform a therapeutic apheresis procedure, the basic premise for each indication is the same. Blood is removed and mixed with an anticoagulant to prevent extracorporeal coagulation, cellular and plasma components are separated in the apheresis instrument, and the component of interest is selected and discarded or may be manipulated and later returned. The nonpathologic (or nontargeted) blood elements that remain in the extracorporeal circuit are recovered and returned immediately to the patient during the procedure.
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Book 2019, Consultative Hemostasis and Thrombosis (Fourth Edition)Laura S. Connelly-Smith MBBCh, DM, Meghan Delaney DO, MPH
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11943 | https://pressbooks.openeducationalberta.ca/saitintrophil/chapter/1-7-soundness/ | Skip to content
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10 Validity and Soundness
1.7 Soundness
A good argument is not only valid, but also sound. Soundness is defined in terms of validity, so since we have already defined validity, we can now rely on it to define soundness. A sound argument is a valid argument that has all true premises. That means that the conclusion of a sound argument will always be true. Why? Because if an argument is valid, the premises transmit truth to the conclusion on the assumption of the truth of the premises. But if the premises are actually true, as they are in a sound argument, then since all sound arguments are valid, we know that the conclusion of a sound argument is true. Compare the last two Obama examples from the previous section. While the first argument was sound, the second argument was not sound, although it was valid. The relationship between soundness and validity is easy to specify: all sound arguments are valid arguments, but not all valid arguments are sound arguments.
Although soundness is what any argument should aim for, we will not be talking much about soundness in this book. The reason for this is that the only difference between a valid argument and a sound argument is that a sound argument has all true premises. But how do we determine whether the premises of an argument are actually true? Well, there are lots of ways to do that, including using Google to look up an answer, studying the relevant subjects in school, consulting experts on the relevant topics, and so on. But none of these activities have anything to do with logic, per se. The relevant disciplines to consult if you want to know whether a particular statement is true is almost never logic! For example, logic has nothing to say regarding whether or not protozoa are animals or whether there are predators that aren’t in the animal kingdom. In order to learn whether those statements are true, we’d have to consult biology, not logic. Since this is a logic textbook, however, it is best to leave the question of what is empirically true or false to the relevant disciplines that study those topics. And that is why the issue of soundness, while crucial for any good argument, is outside the purview of logic.
1.8 Deductive vs. Inductive arguments
The concepts of validity and soundness that we have introduced apply only to the class of what are called “deductive arguments”. A deductive argument is an argument whose conclusion is supposed to follow from its premises with absolute certainty, thus leaving no possibility that the conclusion doesn’t follow from the premises. For a deductive argument to fail to do this is for it to fail as a deductive argument. In contrast, an inductive argument is an argument whose conclusion is supposed to follow from its premises with a high level of probability, which means that although it is possible that the conclusion doesn’t follow from its premises, it is unlikely that this is the case. Here is an example of an inductive argument:
Tweets is a healthy, normally functioning bird and since most healthy, normally functioning birds fly, Tweets probably flies.
Notice that the conclusion, Tweets probably flies, contains the word “probably.” This is a clear indicator that the argument is supposed to be inductive, not deductive. Here is the argument in standard form:
Tweets is a healthy, normally functioning bird
Most healthy, normally functioning birds fly
Therefore, Tweets probably flies
Given the information provided by the premises, the conclusion does seem to be well supported. That is, the premises do give us a strong reason for accepting the conclusion. This is true even though we can imagine a scenario in which the premises are true and yet the conclusion is false. For example, suppose that we added the following premise:
Tweets is 6 ft tall and can run 30 mph.
Were we to add that premise, the conclusion would no longer be supported by the premises, since any bird that is 6 ft tall and can run 30 mph, is not a kind of bird that can fly. That information leads us to believe that Tweets is an ostrich or emu, which are not kinds of birds that can fly. As this example shows, inductive arguments are defeasible arguments since by adding further information or premises to the argument, we can overturn (defeat) the verdict that the conclusion is well-supported by the premises. Inductive arguments whose premises give us a strong, even if defeasible, reason for accepting the conclusion are called, unsurprisingly, strong inductive arguments. In contrast, an inductive argument that does not provide a strong reason for accepting the conclusion are called weak inductive arguments.
Whereas strong inductive arguments are defeasible, valid deductive arguments aren’t. Suppose that instead of saying that most birds fly, premise 2 said that all birds fly.
Tweets is a healthy, normally function bird.
All healthy, normally functioning birds can fly.
Therefore, Tweets can fly.
This is a valid argument and since it is a valid argument, there are no further premises that we could add that could overturn the argument’s validity. (True, premise 2 is false, but as we’ve seen that is irrelevant to determining whether an argument is valid.) Even if we were to add the premise that Tweets is 6 ft tall and can run 30 mph, it doesn’t overturn the validity of the argument. As soon as we use the universal generalization, “all healthy, normally function birds can fly,” then when we assume that premise is true and add that Tweets is a healthy, normally functioning bird, it has to follow from those premises that Tweets can fly. This is true even if we add that Tweets is 6 ft tall because then what we have to imagine (in applying our informal test of validity) is a world in which all birds, including those that are 6 ft tall and can run 30 mph, can fly.
Although inductive arguments are an important class of argument that are commonly used every day in many contexts, logic texts tend not to spend as much time with them since we have no agreed upon standard of evaluating them. In contrast, there is an agreed upon standard of evaluation of deductive arguments. We have already seen what that is; it is the concept of validity. In chapter 2 we will learn some precise, formal methods of evaluating deductive arguments. There are no such agreed upon formal methods of evaluation for inductive arguments. This is an area of ongoing research in philosophy. In chapter 3 we will revisit inductive arguments and consider some ways to evaluate inductive arguments.
1.9 Arguments with missing premises
Quite often, an argument will not explicitly state a premise that we can see is needed in order for the argument to be valid. In such a case, we can supply the premise(s) needed in order so make the argument valid. Making missing premises explicit is a central part of reconstructing arguments in standard form. We have already dealt in part with this in the section on paraphrasing, but now that we have introduced the concept of validity, we have a useful tool for knowing when to supply missing premises in our reconstruction of an argument. In some cases, the missing premise will be fairly obvious, as in the following:
Gary is a convicted sex-offender, so Gary is not allowed to work with children.
The premise and conclusion of this argument are straightforward:
Gary is a convicted sex-offender
Therefore, Gary is not allowed to work with children (from 1)
However, as stated, the argument is invalid. (Before reading on, see if you can provide a counterexample for this argument. That is, come up with an imaginary scenario in which the premise is true and yet the conclusion is false.) Here is just one counterexample (there could be many): Gary is a convicted sex-offender but the country in which he lives does not restrict convicted sex-offenders from working with children. I don’t know whether there are any such countries, although I suspect there are (and it doesn’t matter for the purpose of validity whether there are or aren’t). In any case, it seems clear that this argument is relying upon a premise that isn’t explicitly stated. We can and should state that premise explicitly in our reconstruction of the standard form argument. But what is the argument’s missing premise? The obvious one is that no sex-offenders are allowed to work with children, but we could also use a more carefully statement like this one:
Where Gary lives, no convicted sex-offenders are allowed to work with children.
It should be obvious why this is a more “careful” statement. It is more careful because it is not so universal in scope, which means that it is easier for the statement to be made true. By relativizing the statement that sex-offenders are not allowed to work with children to the place where Gary lives, we leave open the possibility that other places in the world don’t have this same restriction. So even if there are other places in the world where convicted sex-offenders are allowed to work with children, our statements could still be true since in this place (the place where Gary lives) they aren’t. (For more on strong and weak statements, see section 1.10). So here is the argument in standard form:
Gary is a convicted sex-offender.
Where Gary lives, no convicted sex-offenders are allowed to work with children.
Therefore, Gary is not allowed to work with children. (from 1-2)
This argument is now valid: there is no way for the conclusion to be false, assuming the truth of the premises. This was a fairly simple example where the missing premise needed to make the argument valid was relatively easy to see. As we can see from this example, a missing premise is a premise that the argument needs in order to be as strong as possible. Typically, this means supplying the statement(s) that are needed to make the argument valid. But in addition to making the argument valid, we want to make the argument plausible. This is called “the principle of charity.” The principle of charity states that when reconstructing an argument, you should try to make that argument (whether inductive or deductive) as strong as possible. When it comes to supplying missing premises, this means supplying the most plausible premises needed in order to make the argument either valid (for deductive arguments) or inductively strong (for inductive arguments).
Although in the last example figuring out the missing premise was relatively easy to do, it is not always so easy. Here is an argument whose missing premises are not as easy to determine:
Since children who are raised by gay couples often have psychological and emotional problems, the state should discourage gay couples from raising children.
The conclusion of this argument, that the state should not allow gay marriage, is apparently supported by a single premise, which should be recognizable from the occurrence of the premise indicator, “since.” Thus, our initial reconstruction of the standard form argument looks like this:
Children who are raised by gay couples often have psychological and emotional problems.
Therefore, the state should discourage gay couples from raising children.
However, as it stands, this argument is invalid because it depends on certain missing premises. The conclusion of this argument is a normative statement—a statement about whether something ought to be true, relative to some standard of evaluation. Normative statements can be contrasted with descriptive statements, which are simply factual claims about what is true. For example, “Russia does not allow gay couples to raise children” is a descriptive statement. That is, it is simply a claim about what is in fact the case in Russia today. In contrast, “Russia should not allow gay couples to raise children” is a normative statement since it is not a claim about what is true, but what ought to be true, relative to some standard of evaluation (for example, a moral or legal standard). An important idea within philosophy, which is often traced back to the Scottish philosopher David Hume (1711-1776), is that statements about what ought to be the case (i.e., normative statements) can never be derived from statements about what is the case (i.e., descriptive statements). This is known within philosophy as the is-ought gap. The problem with the above argument is that it attempts to infer a normative statement from a purely descriptive statement, violating the is-ought gap. We can see the problem by constructing a counterexample. Suppose that in society x it is true that children raised by gay couples have psychological problems. However, suppose that in that society people do not accept that the state should do what it can to decrease harm to children. In this case, the conclusion, that the state should discourage gay couples from raising children, does not follow. Thus, we can see that the argument depends on a missing or assumed premise that is not explicitly stated. That missing premise must be a normative statement, in order that we can infer the conclusion, which is also a normative statement. There is an important general lesson here: Many times an argument with a normative conclusion will depend on a normative premise which is not explicitly stated. The missing normative premise of this particular argument seems to be something like this:
The state should always do what it can to decrease harm to children.
Notice that this is a normative statement, which is indicated by the use of the word “should.” There are many other words that can be used to capture normative statements such as: good, bad, and ought. Thus, we can reconstruct the argument, filling in the missing normative premise like this:
Children who are raised by gay couples often have psychological and emotional problems.
The state should always do what it can to decrease harm to children.
Therefore, the state should discourage gay couples from raising children. (from 1-2)
However, although the argument is now in better shape, it is still invalid because it is still possible for the premises to be true and yet the conclusion false. In order to show this, we just have to imagine a scenario in which both the premises are true and yet the conclusion is false. Here is one counterexample to the argument (there are many). Suppose that while it is true that children of gay couples often have psychological and emotional problems, the rate of psychological problems in children raised by gay couples is actually lower than in children raised by heterosexual couples. In this case, even if it were true that the state should always do what it can to decrease harm to children, it does not follow that the state should discourage gay couples from raising children. In fact, in the scenario I’ve described, just the opposite would seem to follow: the state should discourage heterosexual couples from raising children.
But even if we suppose that the rate of psychological problems in children of gay couples is higher than in children of heterosexual couples, the conclusion still doesn’t seem to follow. For example, it could be that the reason that children of gay couples have higher rates of psychological problems is that in a society that is not yet accepting of gay couples, children of gay couples will face more teasing, bullying and general lack of acceptance than children of heterosexual couples. If this were true, then the harm to these children isn’t so much due to the fact that their parents are gay as it is to the fact that their community does not accept them. In that case, the state should not necessarily discourage gay couples from raising children. Here is an analogy: At one point in our country’s history (if not still today) it is plausible that the children of black Americans suffered more psychologically and emotionally than the children of white Americans. But for the government to discourage black Americans from raising children would have been unjust, since it is likely that if there was a higher incidence of psychological and emotional problems in black Americans, then it was due to unjust and unequal conditions, not to the black parents, per se. So, to return to our example, the state should only discourage gay couples from raising children if they know that the higher incidence of psychological problems in children of gay couples isn’t the result of any kind of injustice, but is due to the simple fact that the parents are gay.
Thus, one way of making the argument (at least closer to) valid would be to add the following two missing premises:
A. The rate of psychological problems in children of gay couples is higher than in children of heterosexual couples.
B. The higher incidence of psychological problems in children of gay couples is not due to any kind of injustice in society, but to the fact that the parents are gay.
So the reconstructed standard form argument would look like this:
Children who are raised by gay couples often have psychological and emotional problems.
The rate of psychological problems in children of gay couples is higher than in children of heterosexual couples.
The higher incidence of psychological problems in children of gay couples is not due to any kind of injustice in society, but to the fact that the parents are gay.
The state should always do what it can to decrease harm to children.
Therefore, the state should discourage gay couples from raising children. (from 1-4)
In this argument, premises 2-4 are the missing or assumed premises. Their addition makes the argument much stronger, but making them explicit enables us to clearly see what assumptions the argument relies on in order for the argument to be valid. This is useful since we can now clearly see which premises of the argument we may challenge as false. Arguably, premise 4 is false, since the state shouldn’t always do what it can to decrease harm to children. Rather, it should only do so as long as such an action didn’t violate other rights that the state has to protect or create larger harms elsewhere.
The important lesson from this example is that supplying the missing premises of an argument is not always a simple matter. In the example above, I have used the principle of charity to supply missing premises. Mastering this skill is truly an art (rather than a science) since there is never just one correct way of doing it (cf. section 1.5) and because it requires a lot of skilled practice.
Exercise 6: Supply the missing premise or premises needed in order to make the following arguments valid. Try to make the premises as plausible as possible while making the argument valid (which is to apply the principle of charity).
Ed rides horses. Therefore, Ed is a cowboy.
Tom was driving over the speed limit. Therefore, Tom was doing something wrong.
If it is raining then the ground is wet. Therefore, the ground must be wet.
All elves drink Guinness, which is why Olaf drinks Guinness.
Mark didn’t invite me to homecoming. Instead, he invited his friend Alexia. So he must like Alexia more than me.
The watch must be broken because every time I have looked at it, the hands have been in the same place.
Olaf drank too much Guinness and fell out of his second story apartment window. Therefore, drinking too much Guinness caused Olaf to injure himself.
Mark jumped into the air. Therefore, Mark landed back on the ground.
In 2009 in the United States, the net worth of the median white household was $113,149 a year, whereas the net worth of the median black household was $5,677. Therefore, as of 2009, the United States was still a racist nation.
The temperature of the water is 212 degrees Fahrenheit. Therefore, the water is boiling.
Capital punishment sometimes takes innocent lives, such as the lives of individuals who were later found to be not guilty. Therefore, we should not allow capital punishment.
Allowing immigrants to migrate to the U.S. will take working class jobs away from working class folks. Therefore, we should not allow immigrants to migrate to the U.S.
Prostitution is a fair economic exchange between two consenting adults. Therefore, prostitution should be allowed.
Colleges are more interested in making money off of their football athletes than in educating them. Therefore, college football ought to be banned.
Edward received an F in college Algebra. Therefore, Edward should have studied more. |
11944 | https://www.sciencedirect.com/science/article/pii/S2255534X25000209 | Good clinical practice recommendations for proton pump inhibitor prescription and deprescription. A review by experts from the AMG - ScienceDirect
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Revista de Gastroenterología de México (English Edition)
Volume 90, Issue 1, January–March 2025, Pages 111-130
Guidelines and Consensus Statements
Good clinical practice recommendations for proton pump inhibitor prescription and deprescription. A review by experts from the AMG Recomendaciones de buena práctica clínica en la prescripción y deprescripción de inhibidores de la bomba de protones. Revisión por expertos de la AMG
Author links open overlay panel L.R.Valdovinos-García a b, A.S.Villar-Chávez c, F.M Huerta-Iga d, M.Amieva-Balmori e, J.S.Arenas-Martínez f, R.Bernal-Reyes g, E.Coss-Adame f, O.Gómez-Escudero h, P.C.Gómez-Castaños i, M.González-Martínez j, E.C Morel-Cerda k, J.M.Remes-Troche e, M.C.Rodríguez-Leal l, D.Ruiz-Romero c, M.A.Valdovinos-Diaz m, G.Vázquez-Elizondo n, J.A.Velarde-Ruiz Velasco k, M.R.Zavala-Solares o
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Recomendaciones de buena práctica clínica en la prescripción y deprescripción de inhibidores de la bomba de protones. Revisión por expertos de la AMG
Revista de Gastroenterología de México, Volume 90, Issue 1, January–March 2025, Pages 111-130
L.R. Valdovinos-García, A.S. Villar-Chávez, F.M. Huerta-Iga, M. Amieva-Balmori, J.S. Arenas-Martínez, R. Bernal-Reyes, E. Coss-Adame, O. Gómez-Escudero, P.C. Gómez-Castaños, M. González-Martínez, E.C. Morel-Cerda, J.M. Remes-Troche, M.C. Rodríguez-Leal, D. Ruiz-Romero, M.A. Valdovinos-Diaz, G. Vázquez-Elizondo, J.A. Velarde-Ruiz Velasco, M.R. Zavala-Solares
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Abstract
Introduction and aim
Proton pump inhibitors (PPIs) are widely known drugs that are used quite frequently and indicated in both the short and long terms, in numerous acid-related diseases. Our aim was to produce an expert review that establishes recommendations for the adequate prescription and deprescription of PPIs.
Methods
A group of experts in PPI use that are members of the Asociación Mexicana de Gastroenterología (AMG), after extensively reviewing the published literature and discussing each recommendation at a face-to-face meeting, prepared the present document of good clinical practice recommendations. This document is not intended to be a clinical practice guideline or utilize the methodology said format requires.
Results
Eighteen experts on PPI use developed 22 good clinical practice recommendations for prescribing short-term, long-term, and on-demand PPIs, recognizing adverse events, and lastly, deprescribing PPIs, in acid-related diseases.
Conclusions
At present, there is scientific evidence on PPI use in numerous diseases, some in the short term (4-8 weeks), others on-demand (for short periods until symptoms improve), or in the long term (without suspending). Numerous adverse effects have been attributed to PPIs, but the majority have no well-established causal association. Nevertheless, PPIs should be suspended when there is no clear indication for their use. These recommendations aim to aid general physicians and specialists, with respect to PPI prescription and deprescription.
Resumen
Introducción
Los inhibidores de la bomba de protones (IBP) son fármacos ampliamente conocidos, son utilizados con bastante frecuencia e indicados en múltiples patologías relacionadas al ácido, tanto a corto como a largo plazo.
Objetivo
Esta es una revisión de expertos que establece recomendaciones de buena práctica clínica para la adecuada prescripción y deprescripción de IBP.
Métodos
Las recomendaciones de buena práctica clínica se generaron por un grupo de expertos en uso de IBP, miembros de la Asociación Mexicana de Gastroenterología (AMG), después de hacer una extensa revisión de la literatura publicada y de discutir cada recomendación en una reunión presencial. Este documento no pretende ser una guía de práctica clínica con la metodología que este formato requiere.
Resultados
Un total de 18 expertos en el uso de IBP elaboraron 22 recomendaciones de buena práctica clínica para la prescripción a corto, largo plazo y a demanda, reconocimiento de efectos adversos y finalmente la deprescripción de IBP en las enfermedades relacionadas al ácido.
Conclusiones
Actualmente, existe evidencia científica para el uso de IBP en múltiples enfermedades, en algunas de ellas a corto plazo (4-8 semanas), en otras a demanda (por cortos periodos de tiempos hasta mejorar los síntomas) o a largo plazo (sin suspender). Se les han atribuido múltiples efectos adversos, sin embargo, la mayoría no tienen una asociación causal bien establecida. No obstante, los IBP debe ser suspendidos cuando no exista una indicación clara. Estas recomendaciones pretenden ayudar a médicos generales y especialistas en la prescripción y deprescripción de IBP.
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Keywords
Proton pump inhibitor
Prescription
Deprescription
Adverse effects
Palabras clave
Inhibidor de bomba de protones
Prescripción
Deprescripción
Efectos adversos
Introduction
Proton pump inhibitors (PPIs) are among the most widely prescribed medications worldwide. Approximately one-fourth of patients on PPIs continue taking them for at least one year,1 and in observational studies, two-thirds of patients take them without an appropriate indication.2 There is recent concern about the growing prevalence of patients that receive long-term treatment with PPIs, regarding adverse effects and their inadequate use. Thus, the aim of the present review was to make recommendations, based on recent scientific evidence and discussed by a group of experts, for the adequate prescribing and deprescribing of PPIs.
Methods
This review, conducted by 18 experts, was commissioned by the Asociación Mexicana de Gastroenterología (AMG). The specialists were selected based on their academic careers in teaching, research, and practice, and their specialized knowledge of PPI use and deprescription. An extensive review of the literature, spanning the last 20 years, was carried out on PPIs, their indications, short and long-term doses, on-demand use, adverse effects, and their adequate deprescribing. The experts were divided into 6 working groups to review the publications and formulate recommendations on: 1) the definition of the PPIs available in Mexico, and their pharmacokinetic and pharmacodynamic differences; 2) indications for the short-term use of PPIs and doses; 3) indications for on-demand PPI use and doses; 4) indications for the long-term use of PPIs and doses, 5) long-term adverse effects with PPI use; and 6) deprescribing PPIs. Version 1.0 of the recommendations made by each working group was discussed and voted on by all the experts at a face-to-face meeting. Version 2.0 of the statements, created at the face-to-face meeting, was reviewed and corrected by each of the working groups, resulting in version 3.0. This last version underwent a final reading by all the participants for their approval, producing the document presented below.
Proton pump inhibitors
Inhibitors of the H+, K+ -ATPase proton pumps in the parietal cells of the glands of the gastric mucosa are imidazole derivatives formed by pyridine and benzimidazole rings bound by a methylsulfinyl group. The first of these medications approved for clinical use in 1989 was omeprazole (OME); later, substitutions were made in the pyridine and/or benzimidazole rings to create pharmacologic alternatives, in an attempt to improve the speed and efficacy of these drugs, as well as prevent possible side effects.
PPIs are acid-labile weak bases that need to be protected by an enteric coating or as coated granules, sometimes combined with bicarbonate, to achieve a temporal neutralization of the intragastric pH that guarantees their integrity and passage into the duodenum for their absorption and better effect; hence the importance of not opening the capsules or breaking the tablets when they are ingested. Once absorbed, they arrive at the parietal cells of the gastric glands through the systemic circulation. The parietal cells require the active canalicular expression of the H+, K+ -ATPase pumps for binding to occur in response to a meal.3 Not all the pumps are active at the time of a meal, so only two-thirds of the pumps can be inhibited; between 3 to 5 days are needed for PPIs to achieve their maximum effect.
PPIs are metabolized in the P450 cytochrome system (CYP450) in the liver; their main pathways are the 2C19 cytochrome (CYP2C19) and 3A4 cytochrome (CYP3A4) routes.4 These pathways are shared with other medications, and so there can potentially be a lesser or greater degree of interactions with PPIs.5
Despite being safe medications for the patient, the use of PPIs for prolonged periods is not exempt from adverse events. The majority of adverse effects have been described in association studies and causality has only been demonstrated in some of them.
The advent of PPIs established a paradigm in the treatment of all diseases related to hydrochloric acid. Their superiority to other medications in the prevention and control of symptoms and the healing of erosions and ulcers, their role in the management of complications, and their benefit in eradicating Helicobacter pylori (H. pylori) are well-established.6
Table 1 lists the PPIs that are available in Mexico, including their doses, administration route, half-life, and the main pathway of liver metabolism.
Table 1. Commercial PPIs available in Mexico and their pharmacokinetic and pharmacodynamic characteristics.
| PPI | Dose (mg) | Standard dose mg/day | IV | Bioavailability (%) | Half-life (h) | % of pH > 4 (gastric) in 24 h | Liver metabolism | CYP2C19 metabolism (%) |
--- --- --- ---
| Omeprazole (6,7) | 10, 20, 40 | 20 | Yes | 30-40 | 0.5-1.0 | 49.16 | CYP2C19 (higher) CYP3A4 (lower) | >80 |
| Pantoprazole (6,7) | 20, 40 | 40 | Yes | 77 | 1.0-1.9 | 41.94 | CYP2C19 (higher) CYP3A4 (lower) | >80 |
| Lansoprazole (6,7) | 15, 30 | 15 | No | 80-85 | 1.6 | 47.98 | CYP2C19 (higher) CYP3A4 (lower) | >80 |
| Rabeprazole (6) | 20 | 20 | No | 52 | 1.0-2.0 | 50.53 | CYP2C19 | Minimum |
| Esomeprazole (6) | 20, 40 | 20 | Yes | 64-90 | 1.3-1.6 | 58.43 | CYP2C19 (higher) CYP3A4 (lower) | 70 |
| Dexlansoprazole (8) | 30, 60 | 30 | No | ND | 1.0-2.0 | 58 | CYP2C19 (higher) CYP3A4 (lower) | >80 |
| Ilaprazole (9) | 10, 20 | 10 | No | 95 | 5.2 | 67.1 | CYP2C19 (higher) CYP3A4 (lower) | NA |
| Levopantoprazole (10) | 20 | 20 | No | 77 | NA | 47.1 | CYP2C19 (higher) CYP3A4 (lower) | NA |
| Omeprazole + NaHCO 3 (6) | 20 + 1100 | 20 + 1100 | No | 30-40 | 0.5-1.0 | 58 | CYP2C19 (higher) CYP3A4 (lower) | > 80 |
| Dexrabeprazole (11) | 10 | 10 | No | 52 | 1 | NA | CYP2C19 | NA |
NA: not available; PPI: proton pump inhibitor.
Deprescribing proton pump inhibitors
Deprescribing is a carefully planned and supervised clinical process that involves reducing or suspending medications that are not useful, necessary, can cause damage, or do not provide additional benefits in the long term. The primary aim of deprescribing is to reduce the quantity of medications, minimize possible damage, and improve patient quality of life. However, this process can be difficult, especially if the medication is causing no apparent damage and the patient is accustomed to its use. In the specific case of PPIs, there is little solid evidence on how to effectively deprescribe them.7, 8
All patients that take PPIs should undergo a periodic check at 4 or 8 weeks of the indications for the use of these medications, and if there is no indication, they should be considered for a deprescribing strategy.8, 9
Some operational definitions follow below:8, 9, 10, 11
–Deprescribing: the process of reducing and/or suspending a medication after considering its therapeutic indication, risks, and benefits.
–Abrupt/rapid deprescribing: immediate complete suspension of the customary dose of a medication.
–On-demand treatment: administration of a medication decided by the patient based on the presence of symptoms. The patient takes the medication when symptoms are present and suspends it when they disappear.
–Intermittent treatment: daily administration of the antisecretory agent for predetermined periods, usually 2-8 weeks, followed by the same time of suspension.
–Long-term or continuous treatment: any form of uninterrupted treatment, whether a double dose or a standard dose.
–Gradual deprescribing: a gradual reduction of the total daily dose of the antisecretory agent, within a determined period of time, until reaching a minimum maintenance dose that controls symptoms. It can be divided into an intermittent or lower dose, which can be a double dose reduced to a standard dose, a standard dose every other day, or a standard dose divided into 2-3 times a week.
The deprescribing process of a medication consists of 5 steps:11
1 Review indications and effectiveness.
2 Evaluate the balance between the risk and benefit of chronic use.
3 Evaluate patient preferences.
4 Decide upon whether to continue, taper, or suspend the medication.
5 Deprescribe and monitor the patient.
Short-term PPI indications and doses
Recommendation 1: In nonerosive reflux disease (NERD), we recommend standard dose PPI use for 4 weeks.
The short-term treatment goal with PPIs in NERD is to achieve adequate symptom control. This is achieved in 50-60% of patients with a standard dose, and the symptom response rates are similar between all PPIs. The low response rate can be explained due to the great heterogeneity in the studies, given that the majority of clinical trials only consider patients with typical symptoms of gastroesophageal reflux disease (GERD) and endoscopy with no erosions as having NERD, without carrying out the objective measuring of reflux through pH monitoring.12, 13 In a meta-analysis, the short-term response with a standard dose of PPI for 4 weeks in patients with erosive reflux disease (ERD) was 72%. It was 50% in patients with NERD and no pH monitoring, and 73% in patients with NERD and positive pH monitoring. It was concluded that in patients with well-defined NERD, the response rate at 4 weeks with a PPI was comparable to the response rate in patients with ERD.14
Recommendation 2: Cicatrization treatment for GERD with a standard dose for mild esophagitis (Los Angeles A and B) for 8 weeks and a double dose in severe esophagitis (Los Angeles C and D) for 8 weeks is recommended. Patients with grade B, C, or D esophagitis should later be given a maintenance dose.
PPIs are the first-choice drugs in erosive GERD.12 The treatment goal is symptom control and mucosal healing, which is achieved with standard dose PPI use in almost 80% of cases of erosive esophagitis.15 A double dose is recommended in patients that do not respond to treatment with the standard dose, as well as in cases of severe esophagitis (Los Angeles C and D), to achieve greater acid suppression.16
In the cases of Los Angeles grade B, C, and D erosive esophagitis, the recommendation is to continue with a maintenance dose, once the first 8 weeks of treatment have passed, to maintain mucosal healing. In those cases, there can be recurrence in just one to 2 weeks after PPI suspension,17 and recurrence in almost 100% of the cases of grade C esophagitis, after 6 months.
Grade A esophagitis is not considered a complication of erosive GERD and less than 60% of those patients have reflux symptoms,18 and so maintenance therapy is not recommended.
Recommendation 3: In functional dyspepsia (FD) for both subtypes (postprandial discomfort syndrome and epigastric pain syndrome), PPIs at the standard dose for 4 weeks are recommended.
Different guidelines recommend standard PPI therapy once a day for 4 weeks as first-line treatment for patients with FD.19, 20 A Cochrane meta-analysis that included 6,172 patients from 18 randomized clinical trials (RCTs) confirmed that PPIs are more effective than placebo in reducing overall symptoms in FD (RR 0.88, 95% CI 0.82-0.94, NNT 11). There was no difference between low and high PPI doses, the type of PPI, or the presence of H. pylori.21 Recently, a network meta-analysis of 16 RCTs on 6,017 patients with FD, analyzed symptom improvement with a PPI at a low dose < 20 mg, a standard dose ≥ 20 to ≤ 30 mg, and a high dose > 30 mg. The conclusion was that PPIs were more effective than placebo, in the standard-dose PPI group (RR 0.86, 95% CI 0.78-0.95), as well as at low doses (RR 0.89, 95% CI 0.81-0.97), with no benefit at high PPI doses (RR 0.86, 95% CI 0.74-1.01).22
Even though PPIs are antisecretory drugs, some studies have shown that patients with FD have an increase in duodenal permeability and inflammation, with eosinophil and mast cell infiltration near the submucosal plexus neurons.23 In a recent prospective study, pantoprazole at a dose of 40 mg/day for 4 weeks, not only produced symptom improvement in patients with FD, but also a decrease in duodenal permeability and eosinophilia, as well as in the mast cell count, which could explain the effect of PPIs in FD.24
Recommendation 4: In Helicobacter pylori eradication, we recommend a double-dose PPI for 14 days.
Inadequate acid suppression can reduce the efficacy of H. pylori infection eradication treatment through several mechanisms. Optimum PPI dosing is often overlooked, and their adequate prescription can improve H. pylori eradication rates. In cases of H. pylori infection, utilizing a double-dose PPI and adequate antibiotic regimens for 14 days can increase the probability of success in H. pylori eradication.25
Recommendation 5: In acute nonvariceal upper gastrointestinal bleeding, we recommend an initial PPI 80 mg intravenous (IV) bolus, followed by IV infusion or high-dose PPI every 12 h.
The medical treatment goal is to maintain intragastric pH > 6 for promoting platelet aggregation, stabilizing the clot, and healing.26 Clinical trials have shown that the IV infusion of high-dose PPIs was superior to placebo for reducing the risk of peptic ulcer rebleed.
In patients with active bleeding or nonvariceal upper gastrointestinal bleeding, a recent meta-analysis by Laine et al. found significant benefit in IV high-dose PPI use after endoscopic therapy, given that it reduced the risk of recurrent bleeding.27 However, in a more recent meta-analysis of 6 RCTs, only a decrease in the need for endoscopic treatment in the index endoscopy was shown and it had no impact on mortality or the need for surgery.28 Two consensuses suggest a regimen with high-dose PPI (80 mg bolus at a single dose, followed by 8 mg/h continuous infusion for 72 h),29, 30 but intermittent dosing, compared with the infused bolus regimen, has shown similar effectiveness rates.31
Recommendation 6: In the treatment of gastric ulcer, a standard-dose PPI for 4-8 weeks is recommended.
Since the advent of PPIs in the 1990s, numerous clinical trials have shown that these drugs were more efficacious than the histamine-2 receptor antagonists (H2RAs) in treating gastroduodenal ulcers,32 due to their greater potency in and duration of inhibiting acid production. The standard dose is recommended (e.g., OME 20 mg, esomeprazole 20 mg, pantoprazole 40 mg, etc.), once a day. Symptom resolution is fast, and healing is produced at 4 weeks of treatment in most patients. For patients that do not have complete healing after initial treatment, cicatrization generally occurs during the next four weeks of treatment.33, 34 If the ulcer was associated with H. pylori, eradication treatment is necessary to prevent recurrence.
The patients with idiopathic ulcers (negative for H. pylori and with no nonsteroidal anti-inflammatory drug [NSAID]/aspirin use) can have a high recurrence rate, and most likely require long-term treatment (>8 weeks)35.
Recommendation 7: In the treatment of duodenal ulcer, a standard-dose PPI for 4 weeks is recommended.
Standard dose PPIs for 4 weeks provide healing rates above 90% for duodenal ulcers.32 The recommended dose in patients with active duodenal ulcer is the standard dose, once a day. Symptom resolution is rapid, and healing is produced in 2 weeks in the majority of patients.26, 32 For patients that do not heal after initial treatment, cicatrization generally occurs during the following 2 weeks of treatment. In a meta-analysis of 24 controlled clinical trials that included 6,188 patients, pantoprazole 40 mg/day and lansoprazole 30 mg/day for 4 weeks significantly increased the healing rate (RR 2.96, 95% CI 1.78-5.14 and RR 2.04, 95% CI 1.13-3.53, respectively), compared with H2RAs.36 In addition, a network meta-analysis showed no significant differences in the 4-week healing rate of duodenal ulcer treated with different PPIs.
As in cases of gastric ulcer, if there is H. pylori infection, it should be treated with the appropriate eradication regimen.
Recommendation 8: PPI use as a prophylactic measure is recommended in patients admitted to the intensive care unit with risk factors for stress ulcers.
Stress ulcers (part of stress-related mucosal disease) are often present in patients admitted to the intensive care unit (ICU), and are one of the significant causes of prolonged stay, as well as of increased mortality.37 Approximately 75 to 100% of patients that undergo endoscopy within the first 72 h following the appearance of critical entities are estimated to have lesions in the mucosa and clinically present with manifestations, such as blood in stools, melena, coffee-ground emesis through a nasogastric tube, and hematochezia that conditions hemodynamic instability.38
Calculations in meta-analyses suggest that significant clinical bleeding is around 1%, but despite its low frequency, it is a cause of significant mortality in that group of patients.37 The main risk factors include mechanical ventilation for more than 48 h (RR 15.6) and the presence of coagulopathy (RR 4.3).39 In different meta-analyses, the use of prophylaxis in that group of patients reduces risk by 60%. There is no difference between the frequency of PPI use (one or two times a day), dose, or administration route (enteral vs IV), for reducing the bleeding risk. Therefore, the standard dose is recommended, and the treatment should be given until the patient no longer presents with risk factors.37, 40, 41
On-demand PPI indications and doses
Recommendation 9: We recommend on-demand PPI use in the maintenance treatment of NERD and mild esophagitis.
On-demand treatment with PPIs consists of the patient deciding on the PPI dose to take when they experience GERD symptoms and suspending treatment when the discomfort disappears. This regimen is different from intermittent treatment, in which the physician prescribes a daily PPI dose for a short period (1-2 weeks), regardless of the presence of symptoms. The standard dose of PPIs is utilized, and the minimum dose necessary for controlling symptoms is recommended.12, 42, 43
There are two indications for treatment with on-demand PPIs: NERD and Los Angeles grade A mild esophagitis.12, 42, 43 The efficacy of this treatment in NERD and mild esophagitis has been evaluated in 2 recent meta-analyses.44, 45 In the first, 10 RCTs with 4,574 patients were evaluated, comparing on-demand treatment with daily treatment or placebo. The results showed that on-demand treatment with PPIs was superior to daily treatment (RM 0.50, 95% CI 0.35-0.72) in the two patient groups. Compared with placebo, on-demand treatment was effective in patients with NERD or mild esophagitis (OR 0.22, 95% CI 0.13-0.36 and OR 0.18, 95% CI 0.11-0.31, respectively). The second meta-analysis included 11 RCTs on patients with NERD and mild esophagitis, and the results showed that on-demand treatment with PPIs had similar outcomes to continuous treatment, in terms of therapeutic failure (OR 1.26, 95% CI 0.76-2.07). Both meta-analyses concluded that on-demand therapy is effective in the management of NERD and mild esophagitis. It should be clarified that the definition of mild esophagitis in those two studies was based on Savary Miller classification grade I, which is equivalent to Los Angeles classification grade A. Even though expert opinion is that on-demand maintenance treatment can be recommended in patients with Los Angeles grade B esophagitis, controlled clinical trials that support said recommendation are needed.12, 42, 43
In addition, on-demand PPI use has been shown to be the most cost-effective strategy in the maintenance treatment of NERD.46
Long-term PPI indications and doses
Recommendation 10: Standard dose maintenance treatment is recommended in patients that have the B, C, and D erosive variants of GERD.
The chronic and recurrent nature of reflux disease affects quality of life and causes complications; thus, it requires optimum control for preventing said situations. This is especially true in the case of the erosive phenotype of reflux disease because it has the potential to induce complications (bleeding, stricture). There is greater scientific evidence on the use of PPIs in reflux disease, especially in the erosive phenotype.
The erosive phenotype spectrum of reflux disease is currently based on the Los Angeles classification, which has been extensively validated.47 Subsequent research has determined that interobserver variability is high for the Los Angeles grade A classification and that a conclusive diagnosis of GERD cannot be made in that subgroup of patients.48 On the other hand, patients with Los Angeles grade B classification have a conclusive reflux diagnosis by demonstrating the presence of elevated acid exposure and a similar treatment response to the Los Angeles C variant.49
Given the above, the treatment for inducing healing is required in these erosive variants (B, C, and D), as well as maintenance treatment. The evidence suggests that the erosive variant relapse rate, 6 months after having suspended the treatment of single-dose PPIs, is from 6 to 42%. Studies suggest that 6 months after suspending treatment with PPIs, only 10% of the patients with erosive esophagitis are still in remission.50 Therefore, PPIs are indicated as maintenance therapy for preventing the recurrence of both symptoms and mucosal lesions, and the long-term use of PPIs has been approved by the US Food and Drug Administration (FDA).51, 52, 53, 54, 55, 56
A RCT showed that OME 20 mg, once a day (QID), was capable of maintaining 80% of patients with erosive esophagitis free from mucosal lesions after 12 months of continuous treatment. It also reported that OME was significantly superior to both ranitidine and cisapride.50 Similar results were later obtained with lansoprazole (15 and 30 mg, QID), which was shown to be capable of maintaining around 80% of patients with erosive esophagitis in remission at one year.57 Thus, PPIs are useful as maintenance therapy for preventing recurrences of symptoms, as well as mucosal lesions, in patients with the erosive variant, especially those with more severe lesions (Los Angeles B to D). With few exceptions, such as severe erosion grades, half the healing dose of a PPI tends to be adequate for maintenance.58 As a result, the maintenance of the lowest PPI dose necessary for controlling symptoms and esophageal mucosa healing, as a long-term treatment goal, is recommended.
Recommendation 11: Maintenance PPI use at the dose that achieves symptom control is recommended in patients with Barrett’s esophagus, as well as the standard dose for chemoprotection.
Barrett’s esophagus is the replacement of normal squamous epithelium of the distal part of the esophagus with specialized metaplasia (with goblet cells). It is the most important complication of GERD because it is considered a premalignant lesion. It is associated with maximum acid exposure, compared with other forms of the disease, strongly justifying the use of potent antisecretory therapy with PPIs.59 Some studies have shown that PPIs can reduce its progression to adenocarcinoma.60
A meta-analysis based on observational studies concluded that PPI administration led to a reduced risk for progression to low and high-grade dysplasia and to adenocarcinoma (OR 0.29, 95% CI 0.12- 0.79). The same meta-analysis also found that risk reduction was greater with PPI use above 2-3 years (OR 0.45, 95% CI 0.19-1.06 vs OR 1.09, 95% CI 0.47-2.56, respectively).61
Therefore, a standard PPI dose once a day is recommended for preventing the transformation of metaplasia into dysplasia and neoplasia, and this treatment should be continued for life.34, 62, 63 Based on these results, the recommendation is for asymptomatic patients with Barrett’s esophagus to maintain long-term treatment with PPIs, even if there is no solid evidence in this regard.
Recommendation 12: In peptic stricture, long-term, continuous treatment with a standard-dose PPI is recommended.
The incidence of esophageal stricture has drastically decreased over the past 2 decades, in parallel with a marked increase in PPI use. Consequently, the number of studies and their quality have decreased, and evidence is scarce. One study compared lansoprazole 30 mg, QID, versus ranitidine 300 mg twice a day (BID). The lansoprazole study arm required fewer dilatations and had a higher number of patients with no dysphagia than the ranitidine arm.64
Another study compared OME 20 mg, QID, and ranitidine 150 mg, BID, in patients with the erosive variant and concomitant peptic stricture. OME administration improved dysphagia and lowered the frequency of stricture dilatation.65
In patients with peptic stricture, the intervention by experts in therapeutic endoscopy is required for performing dilatation and rehabilitating the esophagus. Uninterrupted concomitant PPI use is necessary in this population for preventing progression and recurrence.
Recommendation 13: In patients with acid hypersecretion syndromes, continuous treatment with PPIs at a dose that is sufficient for long-term symptom control is recommended.
Acid hypersecretion syndromes, e.g., Zollinger-Ellison syndrome (ZES), are characterized by extreme acid secretion due to a neuroendocrine tumor that ectopically secretes gastrin, resulting in diseases, such as peptic ulcer and severe, continuous, recurrent GERD that is dependent on antacid treatments.66 The exact prevalence and incidence of ZES are unknown but it is a rare disease. In the US, its frequency is estimated at one per every million inhabitants and presentation age is 7 to 90 years.67
Diagnosis is made with a plasma gastrin level > 1,000 pg/mL and a baseline acid secretion of >15 mEq/h or even >5 mEq/h in patients with gastrectomy or hypergastrinemia associated with a pH < 2. If it is not possible to obtain baseline acid secretion, hypersecretion syndrome can be ruled out with a gastric pH > 2, in the absence of antisecretory medications. If the pH is >2 and the plasma gastrin level is between 100-1,000 ng/mL, a secretin stimulation test should be carried out.67
The most common symptoms are abdominal pain, diarrhea, and heartburn, followed by nausea and vomiting in some patients. The endoscopic presence of prominent gastric folds is a sign of acid hypersecretion and 25% can present with upper gastrointestinal bleeding.67
In the 1980s, with the generalized use of gastric H+, K+ -ATPase inhibitors, the medical control of gastric acid hypersecretion was made possible in almost all patients. Due to the greater potency and action duration of PPIs, enabling their doses to be given once or twice a day, they have become the drugs of choice for the treatment of this disease. Numerous acid secretion control studies on patients with hypersecretory states support the use of PPIs at varying doses, and they should be titrated individually, utilizing the established criteria for acid suppression in those patients. In patients with intact stomachs, with no GERD or moderate-to-severe multiple endocrine neoplasia type 1 (MEN1), the generally accepted criterion is the use of antisecretory drugs that induce acid suppression at <10 mEq/h for the hour before the next dose of the drug. In complicated patients (moderate-severe GERD, previous Billroth 2 surgery, or MEN1), greater acid inhibition at < 1 mEq/h may be necessary, depending on the upper gastrointestinal endoscopy findings.66 Lansoprazole showed safety and efficacy in secretion control for 10 years, in treating hypersecretors.68 The largest cohort series, with 303 patients, reported a median 14 years of treatment with PPIs and/or H2ARs and treatment of up to 48 years.66
Patients with acid hypersecretion syndrome have 2 problems: the control of acid hypersecretion and the treatment of gastrinomas, which are malignant in 60 to 90% of patients. Surgery continues to be the only possibility for the cure and treatment of the two problems.66
Recommendation 14: In eosinophilic esophagitis, we recommend long-term, continuous, standard-dose PPI use.
Eosinophilic esophagitis (EoE) is a disease that causes symptoms of esophageal dysfunction. At least 15 eosinophils per high power field (60 eos/mm 2) are seen in biopsies, and other causes of esophageal eosinophilia must be ruled out in this group of patients.69
The first evidence on the potential usefulness of PPIs for achieving histologic and clinical remission of EoE was published in the pediatric literature.70 Treatment with high doses of PPIs used to be a diagnostic strategy for differentiating EoE from peptic esophagitis; at present it is used as treatment for EoE.71 The recommended PPI doses in adults are 20 to 40 mg of OME, BID, or its equivalent, and in children, 1-2 mg/kg of OME daily or its equivalent. The long-term strategy is to use the minimum efficacious dose for maintaining remission. PPIs at half the initial dose maintain histologic and clinical remission in a minimum of 75% of patients after at least one year of follow-up. The majority of patients that relapse recover remission after increasing the dose. There are no published data on safety problems of PPIs in patients with EoE.70
The benefits of treatment with PPIs in EoE are probably multifactorial, and include esophageal epithelial barrier repair and possible direct anti-inflammatory effects on certain cytokines.72, 73, 74 in vitro, PPIs inhibit Th2-cytokine-induced eotaxin-3 expression, by interfering with the binding of the signal transducer promoter and the transcription activator.75 A recent meta-analysis demonstrated that treatment with PPIs induced clinical response and histologic remission in 60.8 and 50.5% of patients, respectively.71, 74
Recommendation 15: In patients with chronic NSAID/ASA use and high risk for upper gastrointestinal bleeding, we recommend continuous PPI use at the standard dose.
Both NSAIDs and ASA are often chronically indicated, especially in older adults. This is particularly true with ASA, for the prevention of cerebrovascular events and ischemic heart disease. However, there is a risk for developing complications in the upper gastrointestinal tract, such as peptic ulcer disease and gastrointestinal bleeding. In said context, numerous studies have been conducted, with results favoring the use of PPIs in combination with NSAIDs or ASA for reducing that risk. Table 2 identifies the risk factors for gastrointestinal bleeding and prophylaxis. In a systematic review of 41 RCTs, PPIs reduced the endoscopic risk for duodenal (RR 0.44, 95% CI 0.26-0.74) and gastric (RR 0.40, 95% CI; 0.32-0.51) ulcers associated with chronic NSAID use.76
Table 2. Risk factors for upper gastrointestinal bleeding.
| A history of complications due to peptic ulcer, especially upper gastrointestinal bleeding |
| Age > 60 years |
| NSAIDs/COX-2 inhibitors at high doses or in combination with other drugs associated with gastric mucosal injury, such as anticoagulants, steroids, SSRIs, or warfarin |
| Aspirin use, even at low doses in older adults or in combination with other drugs (NSAIDs, COX-2 inhibitors, steroids, antiplatelet drugs, anticoagulants) |
| Acute NSAID/COX-2 inhibitor use in patients that chronically take anticoagulants or antiplatelet drugs |
COX-2: cyclooxygenase-2; NSAID: nonsteroidal anti-inflammatory drug; SSRIs: selective serotonin reuptake inhibitors.
In a meta-analysis, the effect of co-administering PPIs in patients with no history of peptic ulcer disease, chronically taking low-dose ASA (30-325 mg) was evaluated. Concomitant PPI use was associated with a 73% reduction in the risk for developing peptic ulcer, compared with not using a PPI (RR 0.27, 95% CI 0.17-0.42, p < 0.001), and 50% reduction in the risk for gastrointestinal bleeding (RR 0.50, 95% CI 0.32-0.80).77
On the other hand, in a case-control study conducted on 1,382 NSAID/COX-2 inhibitor users with gastrointestinal complications, the combination of a COX-2 inhibitor with a PPI resulted in a greater reduction in the risk for upper gastrointestinal complications associated with NSAIDs (OR 0.36; 95% CI 0.28-0.47); celecoxib was superior to the combination of a nonselective NSAID with a PPI.78
Recommendation 16: In patients treated with steroids, we do not recommend chronic PPI use, unless the steroids are prescribed in combination with an NSAID/ASA.
The routine use of PPIs is not indicated in patients taking steroids alone for a medical condition, unless the steroids are associated with NSAIDs. In a study conducted on 138 patients with autoimmune disease treated with corticosteroids, 20% presented with peptic ulcer disease. In the multivariate analysis, risk factors were identified: age ≥ 60 years (OR 6.80, p = 0.001), smoking (OR 7.94, p = 0.004), and NSAID use different from COX-2 inhibitors (OR 4.71, p = 0.030), whereas the protective factor was H. pylori infection (OR 0.20, p = 0.022).79 Likewise, in a meta-analysis that included 6,602 patients, the frequency of gastrointestinal complications was compared with a placebo group (n = 3,267) and a steroid group (n = 3,335). In the placebo group, 0.3% developed peptic ulcer versus 0.4% in the steroid group (p > 0.05), concluding that peptic ulcer disease is a rare complication of steroid therapy.80 In a more recent meta-analysis, there was also a low prevalence of peptic ulcer disease in patients with systemic corticosteroids, occurring in 0.4-1.8% of patients. Therefore, routine prophylaxis with PPIs is not recommended.75
Recommendation 17: In patients with upper gastrointestinal bleeding risk factors that are taking antiplatelet drugs or anticoagulants, we recommend prophylaxis with PPIs.
Gastrointestinal bleeding is the most frequent bleeding complication in patients with chronic antiplatelet drug use. In a systematic review, 71,277 participants were included. The results showed that PPI use in patients with dual antiplatelet therapy (ASA and clopidogrel) was associated with a significant reduction in adverse gastrointestinal events (OR 0.38, 95% CI 0.21-0.68, p = 0.001); specifically, upper gastrointestinal bleeds with clopidogrel plus PPI exposure (OR 0.31, 95% CI 0.19-0.51, p < 0.001).81
In the randomized, double blind, placebo-controlled phase 3 Clopidogrel and the Optimization of Gastrointestinal Events Trial (COGENT), the efficacy and safety of the combination of clopidogrel (75 mg) plus OME (20 mg) versus clopidogrel alone, was evaluated. For financial reasons, the study was stopped at 3,761 patients. Upper gastrointestinal events were reported at 1.1% in the OME group and 2.9% in the placebo group, 180 days after the randomization (OR 0.34, 95% CI 0.18-0.63, p < 0.001). More importantly, there was no significant increase in the risk of cardiovascular events with the concomitant use of OME and clopidogrel (p = 0.98).82 The risk for gastrointestinal bleeding was found to be higher in patients with dual antiplatelet therapy in the form of ticagrelor or prasugrel, compared with clopidogrel. In their 2017 guidelines on dual antiplatelet therapy in coronary artery disease, the European Society of Cardiology recommends PPI use in that context (class I, level of evidence B).83
Recommendation 18: In patients taking two or more antiplatelet drugs or an antiplatelet drug plus oral anticoagulation, we recommend prophylaxis with a PPI.
In their 2020 expert review on antiplatelet therapy and anticoagulation in patients with atrial fibrillation or atherosclerotic cardiovascular disease, the American College of Cardiology (ACC) recommends starting or continuing a PPI in patients with ≥ 2 antithrombotic agents, as well as avoiding concomitant NSAID use, to reduce the risk for gastrointestinal bleeding. Once the patient is only taking oral anticoagulation, deprescribing the PPI is recommended, unless there is some other indication for its continuation.84
Recommendation 19: In chronic pancreatitis with exocrine pancreatic insufficiency, we do not recommend the addition of a PPI, given that only enteric-coated pancreatic enzymes are available in Mexico.
The aim of pancreatic enzyme supplementation, in the context of chronic pancreatitis, is suppression of the steatorrhea caused by the condition. All the factors involved in the adequate response to therapy must be kept in mind, even if there has not been therapeutic failure. The most important of these is corroborating that the dose of the supplemented pancreatic enzymes is the correct one. Physiologic studies conducted on the correct absorption of pancreatic enzymes focused on the effects of the gastrointestinal tract on the formulation. Postprandial gastric pH decreased to < 4 after 40 minutes and postprandial duodenal pH after 100 minutes.85 Said pH value is the point at which lipase is irreversibly inactivated and considerably reduces its activity at more advanced sites of the gastrointestinal tract. This is the rationale behind the theory that inhibiting acid, thus maintaining the gastric and duodenal pH above 4, is a factor involved in pancreatic enzyme supplementation to consider. That evidence was established in publications in the 1970s and led to the evolution of enteric coating of the formulations. This advance meant the drug would not have a pH-dependent degradation, preserving its lipolytic activity. The recommendation to inhibit acid through a PPI remains absolute for the non-enteric-coated enzyme formulations.86 However, there is a subgroup of patients that have normal-to-high acid secretion, together with insufficient pancreatic bicarbonate secretion for neutralizing the pH of chyme. PPI administration is suggested in those patients, even when using enteric-coated enzyme supplementation formulations.87 Thus, the recommendation to inhibit acid with a PPI is still valid, simply as a factor that can aid in the correct absorption, together with the factors of adequate dose and enteric coating.88
Adverse effects of long-term PPI use
Recommendation 20: In all patients undergoing long-term treatment with PPIs, we recommend that the physician monitor the possible presentation of adverse effects and indicate deprescribing the PPI when the patient no longer needs it or there is no indication for its use.
The present expert panel analyzed the evidence, and these are the adverse effects in which there is cause and effect (causality) (Table 3).
Table 3. Bradford Hill criteria for PPI adverse effect causality.
| Hill criteria | Bacterial enteric infections | Fundic gland polyps | AIN | C. difficile infection | Chronic nephropathy | Hypomagnesemia | Fractures | SBP | Clopidogrel interaction | CAP |
--- --- --- --- ---
| Association strength | Moderate | High | Weak | Moderate | Weak | Weak | Weak | Weak | Weak | Weak |
| Consistency | Yes | Yes | No | No | No | Yes | No | No | No | Yes |
| Specificity | No | Yes | No | No | No | No | No | No | No | No |
| Temporality | Yes | Yes | Yes | Yes | Yes | Yes | Yes | Yes | Yes | Yes |
| Biologic gradient | Yes | No | No | Possible | No | No | No | No | No | No |
| Plausability | Yes | Yes | No | Yes | No | No | Yes | Yes | Yes | Yes |
| Coherence | NE | NE | NE | NE | NE | NE | No | NE | No | Yes |
| Experiment | No | Yes | Yes | No | No | No | No | No | No | No |
| Analogy | Yes | No | No | No | No | No | No | No | No | No |
AIN: acute interstitial nephritis; CAP: community-acquired pneumonia; NE: not evaluated.
With established causality
Evidence shows that PPI use increases the risk of enteric infections
The main mechanism of action of PPIs is the inhibition of gastric acid secretion. One of the main functions of hydrochloric acid is gastric acidification, which prevents the passage of bacteria from the gastric chamber to the small bowel. Gastric acid inhibition has inconsistently been associated with small intestinal bacterial overgrowth (SIBO).89 In a cohort study of 17,598 patients with anticoagulant and ASA use that were randomized to receive pantoprazole 40 mg versus placebo, prolonged PPI use (23 months) conferred a risk for acute diarrhea (HR 1.33, 95% CI 1.01-1.75, p = 0.04, and a number needed to harm of 900).90
Evidence shows that PPI use increases the risk of acute interstitial nephritis
Acute interstitial nephritis (AIN) is one of the main causes of acute kidney injury (AKI). The common etiologies of AIN are medication-induced, infectious, and idiopathic. The cause associated with medications is responsible for more than two-thirds of cases. AIN etiology changes with aging, given that it is more frequent in the elderly, especially due to PPIs.91
In 1992, Ruffenach et al. reported the first case of AIN, after which there were several anecdotal reports and numerous cross-sectional and cohort studies that described a consistent association between PPI use and the risk of AIN.92 In this context, current PPI use has been documented to be associated with a significantly higher risk for AIN. In the first reports related to PPI use, the appearance of AIN was considered rare, idiosyncratic, and difficult to predict. Suarez et al. reported the analysis of 64 cases (59 histologically confirmed) and the majority were associated with OME use, and to a lesser degree, with pantoprazole, esomeprazole, lansoprazole, and rabeprazole.93
In a systematic review conducted by Nochaiwong et al. that reported the results of 3 studies (n = 585,296), PPI use was associated with a significantly higher risk for AIN, compared with no PPI use, with a combined RR of 3.61, 95% CI 2.37-5.51, and p < 0.001.94
Given the above, due to the publication of several nonrandomized epidemiologic studies on the possible association of PPIs with AIN and the fact that those studies have critical limitations inherent to data sources, there is the possibility of a risk of surveillance bias. To prevent those confounding factors, Rajan et al. conducted a study that systematically reviewed the literature and evaluated the risk for bias, utilizing the Risk Of Bias In Nonrandomized Studies – of Interventions (ROBINS-I) tool. Of the 620 registers initially identified, 26 studies met the a priori eligibility criteria and underwent a risk of bias evaluation. Only 3 of those studies evaluated AIN, reporting the association between PPIs and AIN, with an adjusted HR of 3.00 and 95% CI of 1.47- 6.14 reported in a single study on 581,184 patients in Canada, whereas the adjusted OR varied from 2.05, 95% CI 1.52-2.72, in 4,143 patients in the United States to 3.20, 95% CI 0.80-12.79, in 3,415 patients in the United Kingdom.95
Another setting that has been studied is the development of renal complications as a cause of hospitalization in PPI users. Several reports have shown there is a higher risk for hospital admission due to AKI and AIN, within 120 days after PPI exposure. The reported AIN rate was higher in patients that received a PPI, compared with controls: 0.32 versus 0.11 per 1,000 person-years, respectively (HR 3.00, 95% CI 1.47-6.14).96
In conclusion, the association between PPI use and the risk of AIN has been documented. Nevertheless, more research is justified for clarifying the underlying mechanisms and establishing the precise causal relation between PPIs and AIN.
The evidence shows that PPI use increases the risk of fundic gland polyps
The first report on fundic gland polyps (FGPs) during treatment with OME was conducted in 1992 by Graham.97 FGPs are the most prevalent type of gastric polyps in recent studies in Western populations and they have been found in up to 5% of patients undergoing upper gastrointestinal endoscopy. The exact mechanism by which PPIs induce FGPs is not yet fully understood but may be related to the stagnation of fluid in the oxyntic glands, conditioning cystic dilations; this occurs without direct association of the secondary hypergastrinemia the PPI induces.98 Huang et al. reported a prevalence of 2% in a study on 10,094 patients that underwent gastroduodenal endoscopy, albeit there are reports describing a higher prevalence of 28%. In addition, those authors found that 66.8% of the patients with FGPs had H. pylori infection, and that age and prolonged PPI use were risk factors for the presence of FGPs; the long-term use of PPIs was a particularly strong risk factor for the appearance of FGPs.99 In 2016, Martin et al. conducted one of the first meta-analyses to attempt to clarify the relation of PPIs with FGPs, analyzing 339 peer-reviewed articles and abstracts; 20 of those articles met all the criteria and included a total of 40,218 subjects. The meta-analysis of 12 studies revealed an increase in FGPs in PPI users, compared with controls (OR 2.46, 95% CI 1.42-4.27, p = 0.001), particularly in persons taking PPIs for at least 6 months (OR 4.71, 95% CI 2.22-9.99, p < 0.001) or 12 months (OR 5.32, 95% CI 2.58-10.99, p < 0.001). Even though that meta-analysis was limited due to the quality of the grouped studies, it provided solid evidence on an association between PPI use and FGP development that was most likely causal.100 The results of another meta-analysis that analyzed 12 studies with 87,324 patients coincided with the findings that long-term use of PPIs (≥ 12 months) was associated with a higher risk for FGPs.101
With no established causality
Recommendation 21: We do not recommend suspending PPIs, given that no cause-effect relation has been established in the following conditions: Clostridioides difficile or COVID-19 infection, vitamin B-12 deficiency and hypomagnesemia, cardiocerebrovascular events, clopidogrel use, bacterial overgrowth, osteoporosis/fractures, gastrointestinal tumors, dementia, pneumonia, chronic nephropathy, and spontaneous bacterial peritonitis in the context of cirrhosis.
Even though PPI use has been associated with a higher risk of Clostridioides difficile (C. difficile) infection, the pathophysiologic mechanism involved in the increased risk is not clear.102 In a meta-analysis of observational studies, PPI users were found to have an increased risk for C. difficile infection (RR 1.3, 95% CI 1.1-1.4) and infection recurrence (OR 1.5, 95% CI 1.2-1.9).103, 104 However, by limiting the analysis to cohort studies and randomized trials, great heterogeneity and the quality of the evidence restrict the risk to only older adults and critically ill hospitalized patients.105
Long-term treatment with PPIs has been associated with vitamin B deficiency due to the role of gastric acid and pepsin in the release of the vitamin from ingested nutrients.106, 107 Even though a meta-analysis of 25 observational studies found that PPI users had a slightly higher risk (OR 1.42, 95% CI 1.16-1.73, I 2 = 54%), the great heterogeneity and inconsistency in the levels for defining deficiency, as well as the low OR, reduced the accuracy for establishing causality.107
PPI use could cause hypomagnesemia due to mutations that decrease the affinity of the transient receptor potential melastatin 6 and 7 channels (TRPM6 and TRPM7) in the enterocytes due to changes in the intraluminal pH, even though most magnesium absorption at the intestinal level occurs through the paracellular route.108 A meta-analysis of 14 observational studies found an increased risk (RR 1.44, 95% CI 1.13-1.76, I 2 = 85.2%) in PPI users but definitive associations cannot be made due to their great heterogeneity.109
Clopidogrel is a prodrug that requires CYP2C19 activation to exert its antiplatelet effect, but OME is also metabolized by said enzyme. OME has been theorized to interfere with clopidogrel activation, reducing its therapeutic efficacy and increasing the risk of cardiovascular events in patients with dual antiplatelet therapy. Initially, there was concern about that interaction and ex vivo and observational studies were conducted that produced mixed results.110 However, later clinical trials, such as the PRINCIPLE-TIMI 44 and TRITON-TIMI 38, found no significant differences in the adverse clinical results between PPI users and non-users.82 Finally, in 2009, the FDA warned against the combination of OME with clopidogrel, and in 2016, despite the evidence on the combination’s safety, the FDA reiterated avoiding said combination. It is worth mentioning that there is no such restriction with other PPIs.
The risk of developing SIBO has been associated with PPI use. A meta-analysis conducted in 2017 reported that the combined OR for developing SIBO was 1.71 (95% CI 1.20-2.43).111, 112 A study carried out in Mexico showed that a short course (7 days) of a PPI produced SIBO in 7.8% of healthy subjects.113 Additional research is possibly needed to better understand the mechanisms underlying said association and to develop strategies that aid in reducing the risk for SIBO in patients treated with PPIs.114
PPI-induced hypochlorhydria can increase osteoclastic activity and reduce calcium absorption, thus decreasing bone density.115 Even though an association between PPI use and bone fractures has been postulated (OR 1.41, 95% CI 1.16-1.71; I 2 = 73%),116 no greater risk in women with rheumatoid arthritis, above 80 years of age, or healthy postmenopausal women has been demonstrated.117 A prospective study on 17,598 stable older adults with cardiovascular or peripheral arterial disease treated with rivaroxaban and/or ASA, with follow-up at 3 years, found there was no higher risk of fractures in pantoprazole users (OR 0.96, 95% CI 0.79-1.17, p = 0.71).118
Observational studies and meta-analyses have shown a link between PPIs and a greater risk for gastric cancer.119 Nevertheless, if the underlying disease for administering PPIs is associated with gastric cancer, this could result in an apparent association between PPIs and gastric cancer. A recent systematic review and meta-analysis found no association between PPI use and gastric cancer.120
PPIs have been associated with an increased risk for dementia.121 Gomm et al. analyzed data from more than 73,000 older adults without dementia at the start of the study and evaluated the risk for developing dementia in relation to PPI use. They found that regular PPI use was associated with a higher risk for developing dementia, compared with the subjects that did not use them. In addition, they observed that this association was stronger in those with a greater number of PPI prescriptions. However, the fact that said study was observational and cannot show a causal relation between PPI use and dementia should be kept in mind.122
In the cohort study conducted by Laheij et al., an association between PPI use and greater risk for community-acquired pneumonia was found.123 The mechanisms that could explain said effect include the increase in gastric pH that favors bacterial and viral colonization, as well as the inhibition of certain immune cell function, which could increase the susceptibility to respiratory infections. It is striking that patients on assisted mechanical ventilation have a greater risk for PPI-associated pneumonia due to the colonization of intestinal pathogens in the respiratory tract and the risk for gastric aspiration.124
A study carried out by Klatte et al. that involves more than 100,000 PPI users in Sweden revealed that both the start and the cumulative use of PPIs was associated with a higher risk for progression of chronic kidney disease (CKD), compared with other antacids.124 Nevertheless, that association is recognized as not demonstrating causality, and the exact mechanisms by which PPIs could contribute to CKD progression are still unclear. Even though alterations of the gut microbiota and hypomagnesemia are mentioned, the need for more research is emphasized.125
The association between PPI use and the risk for COVID-19 infection or the development of complications is not well-defined.105, 126, 127 Among the meta-analyses of retrospective and cohort studies, the one conducted by Alhumaid et al. reported an OR of 1.80 (95% CI 1.41-2.31, I 2 = 72%). If only the cohorts are taken into account (OR 1.55, 95% CI 1.16-2.06, p < 0.00001, I 2 = 74%), there is an important heterogeneity of the results, resulting in a low association strength in the meta-analysis.128
With respect to spontaneous bacterial peritonitis, a meta-analysis by Wong et al. was designed to search for infectious complications and mortality in patients with cirrhosis. Despite the low heterogeneity of their results (HR 1.75, 95% CI 1.64-1.85, p < 0.001, I 2 = 0%), the association strength was also low. However, we must consider this possible effect of PPIs in patients with cirrhosis and suspend them when they have no clear indication.129
How can a PPI be deprescribed?
Recommendation 22: We recommend the sudden, gradual, or intermittent deprescribing of PPIs in patients for whom PPI use is not indicated.
Deprescribing PPIs is justified under conditions in which there is no indication for their continuous use, or once the symptoms for which the PPI therapy was started are now resolved (Table 4). Deprescribing PPIs can be recommended after 4 to 8 treatment weeks. At present there are no studies that identify the best manner for deprescribing: abrupt, gradual, or intermittent (on-demand). Abrupt deprescribing can lead to acid secretion rebound due to a state of hyperacidity, which has been reported with therapies of 4 weeks and longer, even at low PPI doses.131
Table 4. Indications for deprescribing PPIs.130
| Mild esophagitis, Los Angeles A, with symptom control after 8 weeks |
| NERD controlled after 4 weeks |
| Dyspepsia controlled after 4 weeks |
| Treatment of gastric ulcer after 4-8 weeks |
| Treatment of duodenal ulcer after 4 weeks |
| After Helicobacter pylori eradication therapy for 14 days |
| Inappropriate indication for gastroprotection in patients with comorbidities |
| Patients admitted to the intensive care unit that no longer present with risk factors for stress ulcers |
NERD: nonerosive reflux disease; PPI: proton pump inhibitor.
Gradual deprescribing is carried out in a reduction period of 2 to 4 weeks, decreasing the double-dose to standard dose for 2 weeks and then suspending the PPI. Another approach is reducing the standard dose to a dose every 48 h and then suspending the PPI. Other authors have used the standard dose, switched to an H2RA, and then suspended the PPI.132
In their study, Inadomi et al. studied 73 patients with GERD that had adequate symptom relief in treatment with PPIs. Deprescribing was started as follows: The double-dose was reduced to the standard dose and then the drug was suspended. If the symptoms of heartburn and regurgitation recurred 2 weeks before the follow-up appointment, the lowest PPI dose was restarted. If the symptoms recurred after the 2 weeks, treatment with an H2RA and/or a prokinetic was started. In the case of symptom persistence, those 2 drugs were used and if the symptoms still persisted the PPI therapy was restarted. The results showed that of the 73 subjects, 58% were asymptomatic at one year, with no treatment. Of those patients, 34% required an H2RA, 7% a prokinetic, 1% required the two drugs, and 15% required no medication after one year of follow-up. In the multivariate analysis, heartburn and younger age were found to possibly predict the need for PPIs.133
At present, gradual reduction is the most widely recommended deprescribing method, given that it reduces the risk of recurrence and deprescribing-associated symptoms, but evidence is still lacking on this modality (Fig. 1).
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Figure 1. Algorithm for deprescribing a PPI.
In patients taking PPIs, their indication should be reviewed and if there are none, they should be suspended. If there is an indication, the appropriate dose and duration should be prescribed; if the dose or duration is not appropriate, they should be suspended. The most adequate deprescribing strategy should be personalized for each patient, as well as the maneuvers to be employed, in case of symptom relapse.
Source: Adapted from Helgadottir H, Bjornsson ES.11
On the other hand, deprescribing through intermittent treatment is useful in patients with adequate control, but with symptom recurrence at 2-4 weeks.
The clinical evidence on these strategies is summarized below (Table 5):
Table 5. Clinical evidence of deprescribing strategies.
| Strategy for deprescribing PPIs | Study description | Results |
---
| Reduction of double dose to standard dose129 | •Prospective study on 117 patients with heartburn and regurgitation successfully treated with a double-dose PPI •Reduction to standard dose (lansoprazole 30 mg /day; OME 20 mg/day) •Follow-up at 6 months | 79.5% with no recurrence |
| Reduction of double dose to a delayed dual release PPI130 | •Blind, multicenter study on 142 patients with heartburn, well-controlled with PPI twice daily, and switched to dexlansoprazole 30 mg, QID •Follow-up at 6 weeks | 88% remained controlled, with significant improvement in quality of life (p = 0.016), PAGI-QOL scores |
| Abrupt suspension versus gradual reduction131 | •Double blind, placebo-controlled study •97 patients (78% GERD) •Abrupt suspension of PPI versus gradual reduction for 3 months (single daily dose every other day), with later suspension | There was no significant difference between groups at 6 months (31% vs 22%, respectively) |
| Gradual versus intermittent reduction133 | •Double-blind controlled study •Recurrence of erosive esophagitis with 4 different reduction regimens (OME daily [20 mg QID; n = 53], OME 20 mg every weekend [n = 55], or ranitidine daily (150 mg BID, n = 51) •Follow-up at 12 months | Remission rates: |
| •89% OME 20 mg QID vs 32% OME 20 mg every weekend •25% ranitidine 150 mg BID |
| On-demand versus continuous treatment136 | •Prospective randomized study •2,156 patients with GERD controlled with esomeprazole 40 mg/day for 4 weeks •They received esomeprazole 20 mg/day continuously or on-demand or ranitidine 150 mg BID continuously •Follow-up at 6 months | The percentage of patients with no heartburn at the end of the study was 72.2% (esomeprazole 20 mg QID continuously); 45.1% (esomeprazole on-demand); and 32.5% (ranitidine continuously), with similar treatment satisfaction percentages (82.2%, 75.4%, and 33.5%, respectively) |
BID; twice daily; GERD: gastroesophageal reflux disease; OME: omeprazole; PPI: proton pump inhibitor; QID: once a day.
Reducing the double dose to a standard dose
Fifteen percent of PPI users are reported to take a double dose or a dose higher than that recommended by the FDA in different acid-related diseases. Even though there is evidence supporting the use of high doses in gastrointestinal bleeding associated with peptic acid disease or for the prevention of bleeding in high-risk patients, there is no evidence supporting the long-term use of high doses, nor is a double-dose PPI superior to a standard dose in Barrett’s esophagus or in laryngopharyngeal reflux. In a study on 117 patients with heartburn or regurgitation successfully treated with double-dose PPI, the dose was progressively reduced, with 80% remaining recurrence-free.134
Switching the double-dose PPI to a delayed dual release PPI
In a study by Cote et al., lansoprazole 30 mg, BID, was switched to rabeprazole 20 mg, QID, with 60% of the patients maintaining symptom remission. Fass et al. evaluated patients with heartburn that was well-controlled with PPIs, BID. They were switched to dexlansoprazole 30 mg, QID; 88% maintained symptom control, with significant improvement in quality of life (p = 0.016), PAGI-QOL scores, diet, and habits (p < 0.001).135
Abrupt suspension versus gradual reduction
In a controlled study by Bjömsson et al., the probability of remaining asymptomatic without PPIs, after abrupt PPI suspension, was compared with gradual reduction for 3 months (single daily dose every other day), with later suspension. There was no significant difference between groups at 6 months (31 vs 22%, respectively).136
Gradual reduction versus intermittent reduction
Different variants of intermittent treatment have been tested. The traditional one is the administration of medications from one to 4 weeks, with the same rest period. A study evaluated recurrence rates with 2 doses of OME and ranitidine, with courses given every 2-4 weeks at the dose that completely controlled symptoms. Those authors found that remission was maintained at 12 months, with intermittent treatment every 2-4 weeks, in 72% of the patients and 93% had fewer than 3 recurrences at 12 months.137 Variants of this type of deprescribing include administering the medication only on weekends (Friday, Saturday, and Sunday), every other day, and 2 or 3 times a week. Erosive esophagitis recurrence rates with 4 different reduction regimens were: 75% (ranitidine 150 mg, BID); 68% (OME 20 mg every weekend), and 11% (OME 20, QID).138
On-demand treatment versus continuous treatment
In patients with erosive GERD, a higher number of cases of permanent endoscopic remission has been described, when the patients had continuous treatment at a low dose, compared with on-demand therapy. Nagahara et al. reported higher remission rates (85.3% vs 44.4%, p < 0.01) with continuous OME 20 mg, QID, vs on-demand OME, for 8 weeks.139 Another study reported that 5% of a group with NERD using on-demand PPIs developed esophageal erosions versus no patient with continuous treatment.140
A study evaluated symptom control in patients with NERD and previous treatment with esomeprazole 40 mg, QID, through 3 strategies: reduction to 20 mg, QID, taken continuously, on-demand, or switching to continuous ranitidine 150 mg, BID, for 6 months. The percentage of patients with no heartburn at the end of the study was 72.2% (continuous esomeprazole 20 mg, QID); 45.1% (on-demand esomeprazole); and 32.5% (continuous ranitidine), with similar treatment satisfaction percentages (82.2%, 75.4%, and 33.5%, respectively).141
Recommendation 23: In patients programmed for PPI deprescribing, we recommend informing them of possible upper gastrointestinal symptom recurrence, such as heartburn, regurgitation and/or epigastralgia after treatment withdrawal.
Upon indicating the deprescribing of PPIs, it is important to inform patients that they may experience gastrointestinal symptoms, such as heartburn, regurgitation, and dyspeptic symptoms within 2 to 4 weeks after treatment is withdrawn, and that this does not necessarily mean they have to immediately return to continuous PPI use. That situation can be explained by rebound acid hypersecretion, defined as an increase in gastric acid secretion above pre-treatment levels after therapy with PPIs.9
In a double-blind RCT on 120 persons with no history of upper gastrointestinal symptoms, the deprescribing of PPIs after 8 weeks of treatment resulted in a higher incidence of gastrointestinal symptoms (heartburn, regurgitation and/or epigastralgia), compared with subjects that continued PPI use.137 Another study was carried out on 48 healthy volunteers (24 women) that were negative for H. pylori. They were randomly assigned to receive treatment with 40 mg of pantoprazole or placebo, QID, for 28 days. Dyspeptic symptoms were recorded daily, utilizing the Glasgow dyspepsia score (GDS) for 2 weeks before, during, and 6 weeks after treatment. After the first and second week, the pantoprazole group had higher dyspepsia scores than the placebo group.142
In a systematic review that included 5 studies, 2 of which involved healthy volunteers, 44% of subjects experienced acid-related symptoms up to 4 weeks after treatment withdrawal; said symptoms were mild-to-moderate and the most common were heartburn and regurgitation. In 3 studies, in which patients with reflux disease participated, no symptoms caused by rebound acid hypersecretion were found. Even though the studies on healthy volunteers showed that upper gastrointestinal symptoms presented upon deprescribing PPIs, its clinical importance is still unknown.143
Ethical considerations
Neither informed consent from any patient nor authorization from any bioethics committee were required for the development this document, given that it is based solely on the bibliography published in indexed journals to serve as a good clinical practice guideline. This document was authorized by the scientific committee of the AMG and the committee of the Revista de Gastroenterología de México of the AMG and contains no information through which any patient could be identified or recognized.
Financial disclosure
The present document was prepared with the support of the Asociación Mexicana de Gastroenterología (AMG), which paid for the authors’ transportation from their cities and their accommodations, so that the document could be produced. No honoraria were received.
Conflict of interest
L.R. Valdovinos García was a speaker for Carnot, Asofarma, and Chinoin.
A.S. Villar Chávez was a speaker for Carnot, Siegfreid, Grunenthal, Megalabs, and Asofarma.
FM. Huerta Iga is a speaker for Carnot and Asofarma.
M. Amieva Balmori was a speaker for Carnot and Asofarma.
R. Bernal-Reyes was a speaker for Asofarma.
E. Coss-Adame was a speaker for Asofarma, Alfasigma, Carnot, Ferrer, and Grunenthal.
O. Gómez is a speaker for Carnot, Chinoin, and Asofarma.
P.C. Gómez Castaños is a speaker for Carnot.
M. González-Martínez declares having no conflict of interest.
E.C. Morel Cerda was a speaker for Alfasigma, Chinoin, and Megalabs.
J.M. Remes-Troche is an advisor and member of the advisory board of Asofarma, Carnot, and Pisa. He is a speaker for Asofarma, Abbot, Carnot, Chinoin, Johnson and Johnson, Medix, and Medtronic.
M.C. Rodríguez Leal declares having no conflict of interest.
D. Ruiz-Romero is a speaker for Asofarma, Olympus, Alfasigma, and Carnot.
M.A. Valdovinos Díaz was a speaker for Carnot, Grunenthal, Asofarma, Megalabs, and Bayer.
G. Vázquez-Elizondo was a researcher for Carnot, Chinoin, and Novo Nordisk de México; advisor and member of the advisory board of Chinoin, M8 Moksha, and Eurofarma; speaker for Chinoin, Asofarma de México, Carnot, and Eurofarma.
J.A. Velarde-Ruiz Velasco declares having no conflict of interest.
M.R. Zavala-Solares is a speaker for Carnot, Ferrer, Alfasigma, and Siegfreid-Rhein.
J.S. Arenas-Martinez declares having no conflict of interest.
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References
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Bubble Pressure Method
In subject area:Engineering
The bubble pressure method is defined as a technique for measuring surface tension by creating a gas bubble in a liquid using a capillary, where the maximum bubble pressure corresponds to the surface tension. This method is particularly useful for assessing dynamic surface tension at newly formed interfaces, such as in surfactant solutions.
AI generated definition based on: Microfluidics: Modelling, Mechanics and Mathematics, 2017
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Measuring Surface Tension and Free Surface Energy
2017, Microfluidics: Modelling, Mechanics and MathematicsBastian E. Rapp
22.2.5Maximum Bubble Pressure Method
A very easy and convenient method for measuring the surface tension is referred to as the maximum bubble pressure method. In this method a capillary is immersed into the liquid to be measured, and a gas bubble is created inside the liquid using gas with controllable pressure (see Fig. 22.4). As the pressure increases, the size of the bubble increases until its diameter is identical to the diameter of the capillary (hemispherical bubble). In this case the Young-Laplace equation allows determining the surface tension using
The maximum bubble pressure method is often used to measure the dynamic surface tension, as it allows measuring the development of the surface tension at a newly created interface. For example, in a surfactant solution, the molecules will (over time) assemble at the gas/liquid interface and change the surface tension. The changing surface tension can be measured conveniently at the changes in bubble size over time if the pressure is kept constant.
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Book2017, Microfluidics: Modelling, Mechanics and MathematicsBastian E. Rapp
Chapter
Theory of Wetting
2013, Micro-Drops and Digital Microfluidics (Second Edition)Jean Berthier
Measuring surface tension of liquids 53
2.8.1
: Using pressure (bubble pressure method) 54
2.8.2
: Using the capillary rise on a plate—Wilhelmy plate 55
2.8.3
: Using gravity: the pendant drop method 57
2.8.3.1
: Bond number 58
2.8.3.2
: Method 59
2.8.4
: Using shear stress in a microflow 60
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Book2013, Micro-Drops and Digital Microfluidics (Second Edition)Jean Berthier
Chapter
SURFACE AND INTERFACE PHENOMENA
2001, Handbook of Surfaces and Interfaces of MaterialsReinhard Miller, Valentin B. Fainerman
3.3Maximum Bubble Pressure Method
Simon was the first person to propose the maximum bubble pressure method (MBPM) for measuring surface tension. Reviews of bubble pressure tensiometry have appeared in several papers [131–135]. The MPT-2 tensiometer from Lauda (Germany) is the technically most advanced instrument among the available tensiometers on the market [133–136]. The MPT-2 tensiometer allows measurement of dynamic surface tensions with a reproducibility within 0.5% in a wide range of effective surface age, from 100 µs up to about 100 s. Thus, it is the only method that can provide reliable experimental data in this broad time interval of extremely short adsorptions times. A schematic of the MPT-2 tensiometer is shown in Figure 11.
The procedure of the MPT-2 is different from other devices and consists of the following steps. Air coming from a pump is first cleaned by a purification filter and then passes a gas flow capillary that measures the gas flow rate with the help of a differential electric pressure transducer. Other gases, such as nitrogen or carbon dioxide, can be employed as well. The excess pressure within the system is measured using an additional pressure sensor. The electric signals from the two sensors—the electrodes and the microphone (needed to accurately determine the bubble frequency)—are sent via an interface to a computer. The microphone is an independent system that monitors the bubble formation frequency in addition to the electric (conductivity) one. When the bubble touches the electrode, the resistance increases and the collapse of the gas cavity at the moment of bubble separation creates a sound wave that can be registered easily by the microphone.
The bubble growth at the capillary tip consists of four main stages [135, 137, 138]. In the first stage, a bubble separates from the capillary tip and a new meniscus that has a radius of curvature approximately equal to the radius of the separating bubble rb(rb ≫ rc) is formed. During the subsequent time interval tl1, the radius of curvature decreases slowly and the meniscus itself moves into the capillary by some depth h (forward meniscus motion), depending on the properties of the internal surface of the capillary. In the second stage, during time tl2, the meniscus is driven by excess pressure in the system and moves to the end of the capillary (reverse meniscus motion). The third stage starts when the bubble radius decreases down to the capillary radius (for narrow capillaries). This stage is determined by the air flow: the greater the air flow L, the shorter is this period of time tl3. The time interval between the last bubble separation and the moment when r = rc is called the bubble lifetime tl, where tl = tl1 + tl2 + tl3. The last stage of bubble formation starts when the bubble has passed the hemispherical size. In this moment the so-called deadtime interval td starts. It is the time between the moment of maximum pressure (r = rc) and the separation of the bubble.
The capillaries employed in MBPM have a small diameter (0.15–0.25 mm); thus, the deformation of the bubble due to the buoyancy force can be neglected [131, 135, 139] and the surface tension γ(tl) can be calculated from the measured maximum bubble pressure Pγ(tl) at any given surface lifetime tl using the simplest form of the Laplace equation
(143)
where r0 is the capillary radius. The MPT-2 does not involve measurement of the capillary pressure directly within the bubble. The parameters measured are the pressure Ps in the reservoir from which the air is supplied to the capillary, the time interval tb between successive bubbles, and the air supply rate L. From these values, the capillary pressure and surface lifetime can be calculated. The capillary pressure is given by the pressure in the gas reservoir via the expression
(144)
Here PH is the hydrostatic pressure at the capillary immersion depth, ΔPcap is the pressure difference between the ends of the capillary, and ΔPlid is the excess pressure in the liquid at the surface of the bubble. In Eq. (144), the last two terms correspond to viscous resistance and inertia of the gas and liquid phases, respectively. When the pressure within the bubble approaches its maximum value, the velocity of the gas flux becomes significantly lower and the contribution introduced by these two terms becomes small. It is assumed that at the moment of maximum pressure, Pγ ≈ Ps – PH, the surface tension is calculated via Eq. (144) from the measured pressure within the reservoir. The corrections that account for aerodynamic and hydrodynamic effects are only negligible for rather long surface lifetimes, and when the viscosity of the studied liquid is sufficiently low [135, 137, 140].
The surface lifetime tl is calculated as the difference between the total bubble time tb and the dead time td,
(145)
which can be approximated from the pressure in the air reservoir Ps and the volume of a separating bubble Vb . The bubble volume in turn is given by the air supply rate L and the bubble time, Vb = tbL. If the bubble growth is restricted by an electrode (or another solid body) located opposite to the capillary tip, as is the case in the MPT-2 tensiometer [134, 135], the bubble volumes are constant and the dead time can be approximated by
(146)
Pc and Lc are the pressure and air supply rates that correspond to the critical point in the dependence of pressure on air flow rate . This point corresponds to the transition from a regime of separate bubbles to a gas jet regime.
This calculation procedure for the dead time accounts for the viscous resistance of the capillary. The viscous friction in the liquid, inertial effects in the liquid, and the compressibility of the gas are quite insignificant for large and medium lifetimes (greater than 0.01 s), but can become noticeable at very short lifetimes [135, 137, 142]. Another hydrodynamic problem is penetration of the studied liquid into the capillary after bubble separation. This penetration results in an increased bubble lifetime and requires some extra pressure in the air reservoir . This extra pressure leads to an additional expansion and compression of the surface that have unpredictable effects on the measured surface tension of the surfactant solution . To prevent the penetration of liquid into the capillary, a hydrophobization of the internal surface is often employed [131, 135, 139, 144].
For wide short capillaries, the restoration of pressure in the entire capillary after bubble separation is very rapid and can lead to pressure oscillations in the gas. For long narrow capillaries, the inertial effects are insignificant, and the gas flow through the capillary is rather slow and aperiodic. The parameter to distinguish between the dynamic gas regimes is [138, 145]. Inertial forces are small if
(147)
where v is the kinematic viscosity of gas and c0 is the sound velocity. For air at 20 °C, r2/l < 0.1 μm.
The penetration depth of the liquid into the capillary after bubble separation, and the time of direct and reverse motion of liquid in the capillary were estimated in [138, 145]. If the condition of Eq. (147) is valid, the maximum meniscus rise for capillaries with a radius of 0.001–0.01 cm is 1–3% of the capillary length. Thus, long narrow capillaries are unsuitable for studies because of this strong penetration of liquid into the capillary after bubble separation. For wide short capillaries, the pressure within the capillary near the meniscus is rapidly restored and almost no liquid penetrates into the capillary. Experimental data also confirm the conclusion that the meniscus rise decreases rapidly with the increase in capillary radius and decrease in capillary length .
Consideration of aerodynamic and hydrodynamic effects yields that wide short capillaries should be preferred. In long narrow capillaries, the process of pressure smoothing along the capillary is aperiodic and rather slow, and at the moment the maximum capillary pressure is overcome, a significant pressure difference still exists between the two ends of the capillary . Therefore, the pressure in the gas reservoir appreciably exceeds that in the bubble. This leads to a significant overestimation of the surface tensions calculated from the measured pressure in the reservoir . In wide short capillaries, the pressure smoothing process is rapid; therefore, the aerodynamic resistance of the capillary does not introduce a significant error even for very short surface lifetimes.
With the decrease of the capillary length and the increase of its radius, the inertia of gas and liquid becomes more significant [137, 142]. The inertia of gas and liquid contributes to overcoming the maximum bubble pressure; therefore, for wide short capillaries, a large excess pressure in the gas reservoir cannot exist and aerodynamic corrections can be neglected.
The results of dynamic surface tension measurements using the MBPM or other methods (e.g., Wilhelmy plate, dynamic capillary, inclined plate, strip, drop volume, and oscillating jet methods) were compared for similar surfactant solutions by various authors [133–136, 148, 149]. The results of the different methods agree very well when the surface tension is presented as a function of the effective adsorption time via the relationship [133–136]
(148)
where α is a function of a number of parameters.
Figure 12 demonstrates the reproducibility and accuracy of the MPT-2 for independent experiments with surfactant solutions of Triton X-100 with various gases, which for the present surfactant system have no significant influence. The MPT-2 covers several orders of magnitude of time and is hence suitable for measuring the entire process of adsorption, starting from a bare surface with the surface tension of pure water and leveling off at the equilibrium surface tension value.
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Chapter
Nucleate Boiling
2017, BoilingTomohide Yabuki, ... Tomio Okawa
2.4.2.2Results
Before presenting the data measured by the maximum bubble pressure method, some results of the method are compared to those of Wilhelmy’s method to validate the method employed and to identify its advantages in the present high-temperature experiments. In Fig. 2.4.4, the data for pure water obtained by Wilhelmy’s method (“Wilhelmy” in the figure) and the maximum bubble pressure method (“bubble” in the figure) are compared. The notations used in Fig. 2.4.4 are employed in all subsequent figures. In the case of pure water, pure water evaporation has no influence on the measured values even in Wilhelmy’s method because concentration is not a factor. In addition, the data publicized by the JSME (Japan Society of Mechanical Engineers) is also shown in the figure. Note that figures similar to Figs. 2.4.4–2.4.8 can be found in the author’s previous work ; however, the symbols and arrangements are modified for the reader’s convenience in the present article.
According to Fig. 2.4.4, the data measured by the maximum bubble pressure method agrees well with those obtained by Wilhelmy’s method for pure water. Moreover, deviations from the publicized JSME data were in the range 0.1–0.4 mN/m, which is reasonably small. The above comparisons confirm that the maximum bubble pressure method is sufficiently valid.
Fig. 2.4.5 shows the measured data for representative alcohol concentrations of butanol, pentanol, and ethanol aqueous solutions by the maximum bubble pressure method and by Wilhelmy’s method. The peculiar behavior discussed, namely, the tendency for the surface tension to increase above a critical temperature as the temperature is increased, is observed for the butanol and pentanol aqueous solutions. The data obtained by the maximum bubble pressure method exhibits a smaller increase with increasing temperature. The observed difference between the results from the two methods is attributed to the elimination of escaping vapor in the maximum bubble pressure method. Fig. 2.4.5 indicates that the influence of escaping vapor on the alcohol concentration in Wilhelmy’s method was not negligible, and that the maximum bubble pressure method served in successfully preventing this problem. The author is of the view that more reliable results were obtained by the maximum bubble pressure method and that this method is suitable for alcohol solution surface tension measurements at high temperatures.
Fig. 2.4.6 shows the measured results for butanol aqueous solutions of various concentrations. With increasing temperature, the surface tension of all butanol solutions decreased until approximately 60°C. Above 60°C, the surface tension began increasing with increasing temperature. At the higher concentrations of butanol, the absolute value of the surface tension was lower, whereas the magnitude of the increase above 60°C was larger. Fig. 2.4.6 also includes results for pure butanol, referred to simply as butanol. As expected, the surface tension of butanol monotonically decreased with increasing temperature. Fig. 2.4.7 shows the measured results for pentanol aqueous solutions of various concentrations. The overall tendency is similar to that of butanol aqueous solutions, except that, with the pentanol solutions, the surface tension began increasing at temperatures above 55°C rather than 60°C, as was observed for butanol solutions.
Fig. 2.4.8 shows the measured results for ethanol aqueous solutions of various concentrations. The surface tension for all concentrations tended to decrease monotonically with increasing temperature. However, surface tension remained constant or increased slightly above 75°C. The author is of the view that this could be because of the onset of boiling. Because the boiling points of the ethanol aqueous solutions of the concentrations studied are about 80°C, the ethanol concentration might have been decreased due to its evaporation with boiling at the high temperature.
Concerning the observed error in the measured data, the validity of employing Eq. (2.4.4) is an important issue. Eq. (2.4.4) assumes that the test fluid and pure water densities in the manometer are equal, although the actual differences in the measurements could cause a small error in the determined surface tension. From an estimation of the actual density differences, the errors were 1.6% for the butanol 7.15 wt% solution and 0.4% for the pentanol 2.0 wt% solution, which can be regarded as small. Conversely, in the case of the ethanol solution, the error was 14.6% for the ethanol 55.0 wt% solution. Therefore, it should be noted that the measured surface tension values of the high-concentration ethanol solutions include some amount of error.
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Nucleate Boiling
2017, BoilingTomohide Yabuki, ... Tomio Okawa
2.4.2.3Discussion
As shown in Fig. 2.4.5, the maximum bubble pressure method provided for a milder nonlinearity for the surface tension of butanol and pentanol aqueous solutions. This is attributed to the fact that the maximum bubble pressure method enclosed the test fluid and vapor, and the resulting errors caused by changes in the alcohol concentration due to species evaporation was minimized. This condition was different from that employed in Wilhelmy’s method.
The solubility of butanol in pure water at room temperature is 7.15 wt%, whereas that of pentanol in pure water at room temperature is 2.0 wt%. The surface tension of pure water is very sensitive to the addition of these alcohols. At solubility concentrations, the surface tension of the aqueous solutions approached that of pure alcohols at low temperature, as shown in Figs. 2.4.6 and 2.4.7.
The surface tension of pentanol solutions exhibited a stronger reversing tendency than that of butanol solutions, as indicated by comparing Figs. 2.4.6 and 2.4.7. In this sense, pentanol is a more efficient additive for obtaining an increasing surface tension with increasing temperature. This enhanced tendency was also suggested in Vochten’s original work . However, as the number of carbon atoms in the alcohol molecule increases, the solubility in pure water drastically decreases. Therefore, it can be concluded that it will be technically difficult to prepare a mixture of pure water and alcohols consisting of a large number of carbon atoms in the molecular formula for use in a practical application. The solubility of pentanol in pure water at room temperature is only 2.0 wt%, and it would be difficult to maintain this small concentration in practical applications for an extended period of time. Butanol would be a better and more reasonable choice for applications from the viewpoint of property changes over time.
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Influential Factors and Physics-Based Modeling of Liquid Acquisition Devices
2016, Liquid Acquisition Devices for Advanced In-Space Cryogenic Propulsion SystemsJason William Hartwig
3.2The Room Temperature Bubble Point Pressure
The primary performance parameter for the screen channel LAD is the maximum bubble point pressure. For a given cryogenic propellant, screen mesh, and operating range of fluid conditions (pressure, temperature) for a given mission, the bubble point sets either the maximum allowable flow rate that a screen channel LAD can deliver out of a propellant tank or the maximum allowable adverse acceleration level against which liquid can be withdrawn from the tank. Following the method of Adamson and Gast (1997), a relation between the bubble point pressure, geometry of the screen pore, surface tension, and contact angle is constructed from a model based on the general Young-Laplace equation (YLE) for the pressure drop across a set of curved interfaces between two phases, which are embedded in the 3D space of the screen mesh (Fainerman et al., 1994). The theory which takes into account the complex local structure of the weaves in the fine mesh screens is isomorphic with a set of cylinder capillaries. This property leads to the simplified YLE that involves an effective pore diameter.
3.2.1Assumptions
The following assumptions are required to derive the simplified bubble point model:
1.
: Reactions and diffusion at the L/V interface and adsorption and desorption to and from the interface are negligible. During dormant times (i.e. no gas pressurization), gaseous helium (GHe) may diffuse into the liquid phase, but diffusion rates are negligible over testing time scales.
2.
: The L/V interface is treated as a zero thickness surface. In reality, higher liquid pressures may act to increase the interface thickness, but the effect is negligible for the LAD screen systems studied herein (Cahn, 1978; Turkevich and Mann, 1990a).
3.
: The complex 3D pore structure can be modeled as the average of an ensemble of simple vertical capillary tubes. In reality, the Dutch Twill weave pattern creates a tortuous path and complex shaped pore.
4.
: The L/V equilibrium interface formed between the warp and shute wires of the Dutch Twill screen is approximated as a 2D interface parallel to the screen face.
5.
: The continuum hypothesis is invoked; pressurization of the screen is done in an infinite number of quasi-static steps.
6.
: The advancing contact angle between liquid and pore throat is approximately the same advancing contact angle between liquid and pore mouth.
7.
: The advancing contact angle that is pinned at the microscopic pore mouth is approximately the same advancing contact angle that is measured on the macroscopic screen surface.
3.2.2Bubble Point Model Derivation
Consider the domain drawn in Figure 3.2 that represents the L/V interface patch in a Dutch twill screen. Figure 3.2 displays a general time dependent curved surface formed by two phases, a L/V interface, where the surface, Σ(t) of area A(t) may change in time. The normal to the surface at P points from the liquid phase to the vapor phase. The two vectors in the tangent plane at P, determine the normal vector, . The net flux of component i for a multicomponent system, from the vapor phase is , adsorbing (desorbing) onto the interface and there may be a net flux from the liquid phase, , adsorbing (desorbing) into the interface. The pressure jump at P is . The surface excess of the ith component is . The principal radii of curvature are shown here as Rmax, Rmin. Neglecting surface reactions, diffusion, and any adsorption or desorption fluxes between the phases, but allowing a curved interface between two fluids within the screen, the general form of the YLE relates the pressure jump across that interface with the curvature of the surface:
(3.5)
where γLV is the surface tension or surface free energy of the liquid, is the unit normal, and the pressure jump is expressed in the following manner:
(3.6)
where z is aligned along the normal vector . The divergence of the unit normal is related to the mean curvature HM of a surface through the following relation:
(3.7)
There is a sign convention that must be taken into account. Therefore, using assumption 2 for a smooth surface and smooth pressure jump,
(3.8)
The sign convention is illustrated by use of a specific, but general, representation of the patch shown in Figure 3.2 from which the tangent vectors are . The unit normal is computed to be:
(3.9)
which points from the liquid to the vapor phase. These formulas make it clear that is a non-linear function (see Boruvka and Neumann, 1977). In this representation, Equation (3.8) is a non-linear differential equation, which requires boundary conditions invoked from the constraint requirements of the three-phase contact line. In principle, it is possible to formulate this boundary value problem on a “unit mesh” of the weave, and by symmetry seek a general solution to the bubble point pressure problem, but the weave is sufficiently complex.
Note that the case of axisymmetric geometry has been solved for drops spreading on flat surfaces (Boruvka and Neumann, 1977). Also, the case of a bubble formed at the aperture of a capillary is well discussed in the literature concerned with the maximum bubble pressure method; see for example Berg (2010, page 72). Also, since the surface tension is known for various liquids it remains to manipulate the mean curvature of Equation (3.8) so as to provide an average pore diameter. There is an isomorphism between a spherical cap, Figure 3.3, and the actual surfaces formed in the screen as represented in Figure 3.2. That means that there exists a one-to-one mapping between a spherical cap and the patch of Figure 3.2.
Figure 3.3 shows that two principle radii of curvature can be defined at any point along the surface, and the mean curvature HM, is related to these radii through the simple relation:
(3.10)
where the sign convention is used to compute ΔP. Invoking assumption 3, for the special case of a vertical capillary tube shown in Figure 3.3, the pressure jump across the interface is simplified:
(3.11)
Note that the mean radius R of the pore forms a perfect spherical meniscus and that this radius will depend on the radius of the tube, and contact angle between liquid and solid tube through the following relation:
(3.12)
such that the pressure drop across the interface becomes:
(3.13)
Note also the inverted configuration in Figure 3.3; for laboratory bubble point testing it is easier to deduce the bubble point when precisely controlling the liquid head pressure on top of the screen and pressurizing the pore from beneath.
It has been assumed in previous studies that cryogenic liquids such as LH2, LN2, LOX, and LCH4, as well as standard reference fluids like isopropyl alcohol (IPA), are close to perfect wetting with clean tank surfaces like SS and Al such that the contact angle . In general, the contact angle will depend on the properties of the fluid and the solid, and also the surface roughness. Therefore to validate this assumption, contact angles were measured for each screen and fluid pair and are reported in Chapter 4.
Assumption 4 is invoked to relate the pressure jump across the spherical interface with the actual bubble point pressure of the LAD screens. Therefore, the equivalent radius, apore, of the vertical capillary tube that is isomorphic with the mean curvature, HM, can be approximated using a hydraulic diameter of a circle inscribed within the complex triangular pore shape such that Equation (3.13) becomes:
(3.14)
where Dp is defined as the so called effective pore diameter:
(3.15)
The function f(cos θC, Dp) corrects for the geometry of the weave and would necessarily be modeled if the geometry of the pore could be mapped in sufficient detail. It appears that taking is a reasonable approximation for the weaves investigated herein.
To achieve a bubble breakthrough, the pressure underneath the screen is gradually elevated such that the column of gas trapped within the screen slowly creeps up the pore. If the gas pressure is ramped in fixed, quasi-static steps as in assumption 5, then the contact angle between liquid and solid within the pore throat at time = 0 will be the same advancing contact angle for all other times. Then the gas column will rise and the same contact angle within the pore throat will become “pinned” at the pore mouth through the use of assumption 6. After subsequent pressurization, a gas bubble will eventually detach and break away from the screen. From the definition of the bubble point, the differential pressure at this initial point of gas breakthrough is what defines the experimental bubble point. To permit relation of the microscopic contact angle at the pore mouth with the measurable macroscopic contact angle of the screen, assumption 7 is employed such that Equation (3.14) can be rewritten as:
(3.16)
Equation (3.16) is known as the simplified bubble point equation for a screen channel LAD, and by construction, is isomorphic to the interfacial governing equation based on the geometry of Figures 3.2 and 3.3 (Gauglitz and Radke, 1990). If the relation is allowable, then macroscopic measurement of advancing contact angle via the Sessile Drop method is related to the contact angle pinned at the pore mouth, which is related to the contact angle at the pore throat. Figure 3.4 illustrates this concept and plots the contact angle with increased pressurization under the LAD screen. Thus detailed mathematical mappings are not required for the practical application of Equation (3.16).
To be precise, since the gas bubble is expanding from within the pore throat to the pore mouth, the contact angle in Equation (3.16) is the advancing contact angle. If Equation (3.16) were applied to the pressurization process in reverse (i.e. if mass were pulled back down through the screen or if the top side of the screen was pressurized), then this angle would be the receding contact angle; one would anticipate hysteresis between the two. Therefore, the measured breakthrough pressure would be different from the measured reseal pressure on account of differences between advancing and receding contact angles, and also due to differences in the size of pore throat and pore mouth.
3.2.3Types of Bubble Point Experiments
Historically there are four different test configurations for measuring the bubble point pressure of a given screen sample. The ideal case is to use an inverted bubble point (IBP) configuration, which will be described in Chapter 4 where the liquid is on top of the screen and gas is beneath. The IBP configuration is preferred over the non-inverted bubble point (NIBP) configuration (wetted screen attached to a liquid filled section with pressurant gas blowing down on the screen) because it is much easier to control liquid head pressure, it is easier to control the uniformity of pressurization, and because data reduction is simplified because bubbles that break through the screen in the IBP configuration naturally rise away from the screen. An example of the NIBP configuration is in Castle (1972).
A third method for measuring the bubble point pressure of a LAD screen sample is to use a dynamic outflow configuration. Here, a screen sample is attached to a channel, mounted inside a tank, and the tank is drained by passing liquid through the channel as done in Meserole and Jones (1993). As the liquid level drains and the LAD becomes exposed to pressurant gas, the pressure differential across the screen slowly increases and eventually exceeds the bubble point pressure. While this configuration is more flight representative, it is inferior to the IBP configuration because it is even more difficult to control the location and direction of pressurization; even small changes in the uniformity of the pressurization can cause large differences in the breakdown height of the channel as shown in the data. A fourth method used by Bennett (1987) is to submerge a basket made of screen mesh into a liquid, lower the liquid level, and blow pressurant gas on the screen until breakdown. This method is best for start baskets, and not full communication devices. The majority of historical bubble point data collected, along with all new experimental data in the current work, is for the IBP configuration. Details of the IBP test configuration and methodology for room temperature and cryogenic fluids are reserved for Chapters 4 and 5, respectively.
3.2.4Surface Tension Model
To compare predicted bubble points from Equation (3.16) to experimental data, knowledge of surface tension, contact angle, and effective pore diameter is required. In general, surface tension is known to be a function of temperature and is well represented by:
(3.17)
where γ0, K are constants and TR is the reduced temperature defined in the traditional sense:
(3.18)
where T is the temperature of the liquid at the LAD screen and TC is the critical temperature (Ferguson and Kennedy, 1936). A modified form of Equation (3.17) is used to predict surface tension across a wide range of room temperature and cryogenic liquids for the current work:
(3.19)
where γ0, a0, a1, a2 are fitting parameters and all temperatures are evaluated in Kelvin. Tables B-1–B-6 in Appendix B list the LAD screens, liquids, and pressurant gas types over which bubble point data exists in the literature, as well as data from the current work. As shown, there is bubble point data available using 20 different liquids, in room temperature liquids acetone, ammonia, ethanol, Freon-113, Freon-114, IPA, methanol, silicon oil, and water, in storable propellants diborane (B2H6), hydrazine (N2H4), MMH, nitrogen tetroxide (NTO), oxygen difluoride (OF2), and pentane (C5H12), and cryogenic liquids, LCH4, LOX, LN2, LH2, and SFHe. Surface tension and critical temperature data is readily available for all nine of the storable fluids (Moore et al., 1968; Jasper, 1972; Ambrose et al., 1974, 1975; Sato et al., 1991; Gude and Teja, 1995; DDBSP, 2011). Less extensive data is available for storable propellants (Laubengayer et al., 1941; Paynter, 1970, 1973b; Jasper, 1972; McCarty, 1986; Arnold, 1999) and for all cryogenic fluids (Stansfield, 1958; Jasper, 1972; Brooks and Donnelly, 1977; Jensen et al., 1980; McCarty, 1986; Van Sciver, 1986).
Curves were fit to the data to minimize the least squares distance between model and data. To generate smooth curves as a function of liquid temperature, curves were fitted across the range of available data using an anchor point of zero surface tension at the critical point. Figure 3.5 plots model generated LH2 and LN2 surface tensions as a function of the ratio of liquid temperature to critical temperature versus the sparse surface tension data available in the literature. Results for all fluids are compiled in Table 3.1, which lists the temperature range over which surface tension and bubble point data is available, critical temperature, fitting coefficients, and error between the model curve and data. As shown, an excellent fit to the data is achieved using three fitting parameters for all fluids except SFHe and water, where the full four parameter fit was required.
Table 3.1. Fitting Parameters for the Surface Tension Model
| Fluid | Tmin (K) | Tmax (K) | TC (K) | γₒ (mN/m) | a0 | a₁ | a2 | r (mN/m) | References |
--- --- --- --- --- |
| Acetone | 182 | 329 | 508.1 | 65.7 | 1.149 | 0 | 0 | 2.50E − 05 | Jasper (1972) and Ambrose et al. (1974) |
| B2H6 | 110 | 180 | 298.85 | 49.13 | 1.374 | 0 | 0 | 5.14E − 05 | Laubengayer et al. (1941) and Jasper (1972) |
| C5H12 | 220 | 350 | 460.35 | 50.9 | 1.207 | 0 | 0 | 1.71E − 07 | Jasper (1972) |
| Ethanol | 220 | 350 | 513.9 | 46.2 | 0.525 | 0.545 | 0 | 3.11E − 04 | Jasper (1972) and Ambrose et al. (1975) |
| Freon-113 | 200 | 375 | 487.21 | 52.72 | 1.095 | 0 | 0 | 1.81E − 04 | Jasper (1972) |
| Freon-114 | 200 | 350 | 418.83 | 53.1 | 1.195 | 0 | 0 | 1.43E − 06 | Moore et al. (1968) |
| IPA | 275 | 353 | 508.3 | 46.507 | 0.901 | 0 | 0 | 2.00E − 02 | Jasper(1972) and Gude and Teja (1995) |
| LCH4 | 90 | 165 | 190.6 | 39.06 | 1.221 | 0 | 0 | 2.55E − 10 | McCarty (1986) |
| LH2 | 14 | 24 | 32.94 | 5.431 | 1.090 | 0 | 0 | 4.74E − 08 | Jasper (1972) |
| LN2 | 65 | 120 | 126.2 | 29.71 | 1.271 | 0 | 0 | 4.00E − 11 | Stansfield (1958) and Jasper (1972) |
| LOX | 70 | 135 | 154.6 | 38.46 | 1.222 | 0 | 0 | 8.00E − 12 | Jasper (1972) |
| Methanol | 270 | 300 | 513.38 | 40.6 | 0.425 | 0.461 | 0 | 6.16E − 05 | Jasper (1972) |
| MMH | 200 | 320 | 567 | 59.39 | 0.485 | 0.498 | 0 | 6.27E − 04 | Paynter (1970, 1973b) and Arnold (1999) |
| N2H4 | 200 | 320 | 653 | 145.815 | 1.290 | 0 | 0 | 3.93E − 03 | Jasper (1972) and Paynter (1973b) |
| NTO | 200 | 320 | 431.4 | 67.2 | 0.464 | 0.469 | 0 | 5.95E − 03 | Paynter (1970, 1973b) and Arnold (1999) |
| NH | 200 | 330 | 405.4 | 90.9 | 1.096 | 0 | 0 | 1.57E − 05 | Paynter (1970) and Jasper (1972) |
| OF2 | 110 | 170 | 213.45 | 51.8 | 1.280 | 0 | 0 | 2.02E − 06 | Paynter (1970) |
| SFHe | 1 | 3 | 5.1953 | 0.39 | 0.115 | 0.811 | 0.428 | 1.92E − 07 | Jensen et al. (1980) and Van Sciver (1986) |
| Water | 220 | 350 | 647.096 | 86 | − 0.012 | 0.371 | 0.803 | 3.93E − 03 | Jasper (1972) and Sato et al. (1991) |
3.2.5Specifying the Effective Pore Diameter
For popular mesh screens such as the Dutch Twill, as is evident in the SEM image in Figure 2.18, it is generally difficult to define a 2D pore diameter due to the complex 3D triangular shape of the L/V interface within the screen. While previous studies have examined bubble points of single pores of various geometries to address this concern (Jurns and McQuillen, 2008), there are three methods to estimate the effective pore diameter of the screen.
: Method 1: The first and easiest way to approximate the pore diameter is to measure the bubble point using a “standard” reference fluid with known surface tension and use similitude to determine DP:
(3.20)
This requires knowledge of contact angle though. IPA is generally chosen as the reference fluid because its surface tension is much higher relative to cryogenic liquids, which permits a good calibration range for a given screen sample, and because the contact angle is expected to be small. This method is straightforward and is the most reliable.
: Method 2: The second method to estimate the effective pore diameter is to calculate the diameter from the slope of a bubble point versus surface tension (times the cosine of the contact angle) plot for each screen weave, summing over historical data available in the literature. This method is attractive and easy, since it is based on several different fluids across the surface tension range and doesn't require testing. However, there tends to be more variability in the final values due to sample to sample variation in pore diameters due to manufacturing defects, and due to the assumption of linearity between bubble point pressure and surface tension across the full temperature range.
: Method 3: The third method to determine the effective pore diameter is to image the screens using an SEM and use geometry to estimate the 2D pore size. This is the only method than can be used if testing equipment or historical data is unavailable. Figure 3.6 shows an SEM image of the 325 × 2300 screen. Employing the previous methodology from Jurns and McQuillen (2008), a 2D triangular pore can be fit to the SEM image to approximate the pore diameter. The two dimensional triangular projection is broken up into two right triangles as shown in Figure 3.7, and the pore triangle is chosen to be the resultant hydraulic diameter of the planar surface:
(3.21)
AC,SEM is the cross-sectional area normal to the fluid and Pe is the perimeter around the triangle:
(3.22)
where AI, AII and L1, L2, L3 are the area and lengths of the triangle as shown in Figure 3.7, respectively. To determine accuracy of the SEM measurement against the simple room temperature bubble point method, these three methods for determining pore diameter will be compared in Chapter 4 for several different screens.
3.2.6Previously Reported Bubble Points
Rigorous and comprehensive survey of the literature was performed to gather all previously reported bubble point data for all possible screens. Detailed results are reserved for Chapter 10, where the full cryogenic bubble point pressure model is constructed. Appendix B lists the 40 screens over which information is available in the literature. 24 of these 40 screens have actual historical bubble point data available; pore diameter data for two new screens are added for the current work for a total of 26 screens. Figures 3.8a and b plot historical bubble point data for the most popular two meshes, a 200 × 1400 and 325 × 2300 Dutch Twill mesh, respectively. Data is taken from Paynter (1973b), Burge and Blackmon (1973b), Cady (1973, 1975, and 1977), Chato and Kudlac (2002), Kudlac and Jurns (2006), and Jurns et al. (2007). Very few screens other than these two have data available in cryogenic liquids, and for screens where cryogenic bubble point data is available, the data is quite sparse. As shown, bubble point pressure scales with the surface tension of the liquid for both screens. Comparing Figures 3.8a and b, higher bubble points are obtained using the finer 325 × 2300 mesh over the 200 × 1400 mesh.
In addition to the above authors, in chronological order, historical bubble point data is also taken from Balzer et al. (1969), Alexander et al. (1970), Blatt (1970a,b), Heckman (1971), Castle (1972), Burge et al., 1973; Paynter (1973a), Stark et al. (1974), Simon, 1979; Dodge and Bowles (1984), Bingham and Tegart (1977), Wilson and Meserole (1986), Meserole and Jones (1993), and Conrath and Dreyer (2009). Most historical references are associated with storable propulsion technology development programs such as the STS auxiliary propulsion system, and thus present data for high surface tension propellants with coarser mesh screens. Meanwhile, with the latest push towards advancing cryogenic propulsion technology, more recent references associated with Centaur upper stage CFM development, Orbital Transfer Vehicles (OTVs) (Heald and Merino, 1979; Gilmore, 1985; Schuster and Brown, 1987), and Space Station Freedom report data for finer mesh screens and lower surface tension cryogenic liquids. The CFM and CPST projects have funded most of the experiments from the current work.
Using Equation (3.20), a line can be fit to the bubble point versus surface tension data for a particular mesh to determine the average pore diameter across the range of liquid temperatures as exemplified by the black line in Figure 3.8. To obtain a rough order of magnitude of how the effective pore diameter scales with the fineness of the mesh, Table 3.2 lists the average effective pore diameter for the 24 historical screens where bubble point data is available. For each mesh, data was plotted, and a line was fit to the data to minimize the least squared distance between data and line to obtain a single value of DP. Note that this assumes perfect linearity between bubble point pressure and surface tension, all the way down to cryogenic temperatures.
Table 3.2. Measured and Calculated Screen Parameters
| Mesh | Mesh Type | dwarp (μm) | Manufacturer dshute (μm) | Actual dshute (μm) | Source, Wire Diameters | Avg. Dp (μm) | Δw (μm) | Δs (μm) | h (μm) | l2 (μm) | θ (°) | B (μm) | a (μm− 1) | ɛ |
--- --- --- --- --- --- ---
| 325 × 2300 | Dutch Twill | 38.1 | 25.4 | 22.1 | Corrected | 14.1 | 78.2 | 22.1 | 19.2 | 31.8 | 50.4 | 82.3 | 0.1077 | 0.331 |
| 325 × 1900 | Dutch Twill | 35.6 | 30.5 | 26.7 | Corrected 78.2 | 26.7 | 18.8 | 28.4 | 52.9 | 89.1 | 0.0995 | 0.3 |
| 250 × 1400 | Dutch Twill | 55.9 | 38.1 | 36.3 | Corrected | 19.1 | 101.6 | 36.3 | 19.4 | 18 | 65.1 | 128.5 | 0.0741 | 0.262 |
| 250 × 1370 | Dutch Twill | 55.9 | 38.1 | 37.1 | Corrected | 19.5 | 101.6 | 37.1 | 18.7 | 16.5 | 66.2 | 130.1 | 0.0734 | 0.257 |
| 200 × 1400 | Dutch Twill | 71.1 | 40.6 | 36.3 | Corrected | 23.1 | 127 | 36.3 | 28.7 | 36.2 | 57.7 | 143.7 | 0.0648 | 0.306 |
| 165 × 1400 | Dutch Twill | 71.1 | 40.6 | 36.3 | Corrected 153.9 | 36.3 | 38.5 | 79 | 44.2 | 143.7 | 0.0597 | 0.371 |
| 165 × 800 | Dutch Twill | 71.1 | 50.8 | 50.8 | Corrected | 45.2 | 153.9 | 63.5 | 37.2 | 57.4 | 52.4 | 172.7 | 0.0427 | 0.415 |
| 80 × 700 | Dutch Twill | 102 | 76.2 | 72.6 | Corrected | 55.5 | 317.5 | 72.6 | 72.9 | 221.5 | 33.4 | 247.1 | 0.0316 | 0.396 |
| 200 × 600 | Dutch Twill | 62.2 | 45.7 | 45.7 | Corrected | 44.8 | 127 | 84.7 | 28.5 | 35.3 | 58.2 | 153.6 | 0.0367 | 0.539 |
| 30 × 500 | Dutch Twill | 250 | 110 | 101.6 | Corrected 846.7 | 101.6 | 159.9 | 700.7 | 24.5 | 453.2 | 0.0165 | 0.504 |
| 30 × 160 | Dutch Twill | 229 | 178 | 178 | Corrected 846.7 | 317.5 | 178.4 | 651 | 28.7 | 585 | 0.0079 | 0.632 |
| 20 × 250 | Dutch Twill | 102 | 63.5 | 63.5 | Corrected 1270 | 203.2 | 82 | 1248.4 | 7.5 | 229 | 0.0097 | 0.835 |
| 50 × 250 | Plain Dutch | 140 | 114 Industry | 104 - - 368 | 0.0131 | 0.611 |
| 30 × 250 | Plain Dutch | 254 | 203 Heckman (1971) | 155 - - 660 | 0.0122 | 0.36 |
| 24 × 110 | Plain Dutch | 381 | 267 Industry | 200 - - 914 | 0.0058 | 0.572 |
| 720 × 140 | Reverse Dutch | 35.6 | 109 Cady (1977) | 40 - - 254 | 0.0439 | 0.03 |
| 500 × 500 | Twill square | 25.4 | 25.4 Cady (1977) | 46 - - 50.8 | 0.0655 | 0.584 |
| 325 × 325 | Twill square | 27.9 | 27.9 Heckman (1971) | 65 - - 55.9 | 0.0414 | 0.711 |
| 180 × 180 | Twill square | 53.3 | 53.3 Paynter (1970) | 85.3 - - 107 | 0.023 | 0.693 |
| 120 × 120 | Twill square | 50.8 | 50.8 Heckman (1971) | 169 - - 102 | 0.0257 | 0.674 |
| 230 × 230 | Square | 38.1 | 38.1 Heckman (1971) | 94.5 - - 76.2 | 0.0301 | 0.713 |
| 200 × 200 | Square | 53.3 | 53.3 Industry | 99.7 - - 107 | 0.0268 | 0.643 |
| 150 × 150 | Square | 66 | 66 Heckman (1971) | 135 - - 132 | 0.0199 | 0.671 |
| 100 × 100 | Square | 114 | 114 Industry | 118 - - 228 | 0.0136 | 0.614 |
| 80 × 80 | Square | 140 | 140 Heckman (1971) | 209 - - 279 | 0.0108 | 0.622 |
| 60 × 60 | Square | 191 | 191 Heckman (1971) | 342 - - 381 | 0.0081 | 0.612 |
| 50 × 50 | Square | 229 | 229 Industry | 307 - - 458 | 0.0068 | 0.612 |
| 40 × 40 | Square | 254 | 254 Heckman (1971) | 544 - - 508 | 0.0053 | 0.662 |
| 30 × 30 | Square | 240 | 240 Armour and Cannon (1968) | 540 - - 480 | 0.0039 | 0.769 |
Figure 3.9 plots the average pore diameter based on Method 2, historical data as a function of the fineness of the mesh, in terms of the square root of the product of the number of warp wires, nw times shute wires, ns. Results show that there is a fairly strong correlation between the effective pore diameter, which is not a physical parameter of the screen, and the fineness of the screen in terms of the number of pores per square inch, which is a physical parameter of the screen. Figure 3.9 implies that the finest mesh will have the highest bubble point pressure.
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Book2016, Liquid Acquisition Devices for Advanced In-Space Cryogenic Propulsion SystemsJason William Hartwig
Chapter
SURFACE AND INTERFACE ANALYSIS AND PROPERTIES
2001, Handbook of Surfaces and Interfaces of MaterialsKathleen J. Stebe, Shi-Yow Lin
Methods for Measuring Dynamic Surface Tension 78
5.1.
: Surface-Oscillation Techniques 79
5.2.
: Pendant Bubble/Drop Method 81
5.3.
: Drop-Weight/Volume Method 87
5.4.
: Oscillating-Jet Method 88
5.5.
: Growing-Drop Method 89
5.6.
: Maximum Bubble-Pressure Method 89
5.7.
: Wilhelmy Plate/Ring Method 90
5.8.
: Method of Meniscus at Flat Plate 90
5.9.
: Inclined-Plane Technique 90
5.10.
: Falling-Film Method 91
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Book2001, Handbook of Surfaces and Interfaces of MaterialsKathleen J. Stebe, Shi-Yow Lin
Chapter
SURFACE AND INTERFACE PHENOMENA
2001, Handbook of Surfaces and Interfaces of MaterialsReinhard Miller, Valentin B. Fainerman
Experimental Techniques 401
3.1.
: Ring and Plate Tensiometry 401
3.2.
: Meniscus Profile Analysis Tensiometry 402
3.3.
: Maximum Bubble Pressure Method 403
3.4.
: Oscillating Drops and Bubbles 405
3.5.
: Transient Drop or Bubble Relaxation Method 406
3.6.
: Surface Ellipsometry 406
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Book2001, Handbook of Surfaces and Interfaces of MaterialsReinhard Miller, Valentin B. Fainerman
Chapter
Foam Formation in Anaerobic Digesters
2018, Advances in BioenergyLucie Moeller, ... Jana Zábranská
2Methods for the Determination of Foaming Causes
Because AD foaming is a very complex phenomenon, there are still very few methods available for the detection of the foaming causes and there are still no early warning indicators. The well-known aeration method is very suitable for digestates with low fiber content as typically found in anaerobic digesters of WWTPs. However, this method is not applicable for digestates with high fiber content coming from biogas plants that treat renewables. The Leipzig foaming test was developed for this type of digestate (UFZ, 2013).
2.1Bubbling (Gassing Up) Method
This test of foaming propensity in liquids, known also as the bubble test, is based on the Bikerman method (Beneventi et al., 2001) and has already been used in diverse variations in a number of publications dealing with foam formation in AD.
The principle of the method consists of the introduction of a gas (e.g., nitrogen or air) into the digestate or a mixture of digestate and foaming substances. A schematic of the foaming apparatus is shown in Fig. 3.
This apparatus is a glass column with a diffuser (e.g., frit or aerator stone) placed at the bottom of the column. The use of fine bubble air stone fitted to an aquarium air pump as in the case of Pagilla et al. (1998) is also possible. The tested mixture has to be filled into the column and aerated with gas at a certain flow rate for a certain time period. After stopping aeration, the stability of the foam can be measured during a certain time. The parameters and conditions vary from author to author (see Table 1).
Table 1. Parameters and Conditions of Aeration Tests in the Literature
| Origin of AD Sludge | Vessel Volume | Sludge Volume | Flow Rate | Type of Gas | Duration of Aeration | Foam Stability Detection After Aeration | Reference |
--- --- --- --- |
| WWTP | 2 L | n.s. | 0.5 L/min | Air | 10 min | 1 h | Ganidi (2008) |
| WWTP | 2 L | 1 L | 1 L/min | Nitrogen | 5 min | 5 min | Zábranská et al. (2002) |
| Manure digestion system | n.s. | 50 mL | 0.06 L/min | Air | 10 min | 1 h | Boe et al. (2012) |
| WWTP | 2 L | n.s. | n.s. | Inert gas | n.s. | 1 min | Ross and Ellis (1992) |
| WWTP | 2 L | 1 L | 5 L/min | Air | 10 min | 10 min | Pagilla et al. (1998) |
| Palm oil mill plant | n.s. | 50 mL | 0.03 L/min | Air | 10 min | 1 h | Wongfaed et al. (2015) |
| WWTP | 2 L | 200 mL | 0.015 L/min | Air | n.s. | n.s. | Subramanian et al. (2015) |
| Sugar beet pulp treating AD | n.s. | 100 mL | 1 L/min | Air | 5 min | 33 min | Suhartini et al. (2014) |
| Cultivation of Nocardia | n.s. | 50 mL | 0.2 L/min | Air | 2 min | 10–15 min | Blackall and Marshall (1989) |
n.s., not specified.
Based on the height and stability of the produced foam layer, the foaming properties can be estimated as follows (according to Ganidi, 2008; Ng et al., 1977; Subramanian et al., 2015):
Alfaro et al. (2014) described an alternative method for the determination of foam potential and foam stability by using Alka Seltzer® tablets (Bayer, Spain) for foam production. These tablets are effervescent and cause the generation of foam in the sludge. The foaming potential was determined as the volume of the generated foam (in mL) and the foam stability was measured as the foam half-life, i.e., the time (in minutes) for half of the volume of the foam produced during the foaming potential test to disappear. Nevertheless, the authors showed that the foaming potential test (which is admittedly very simple) showed various deficits such as inconsistency with the abundance of filamentous bacteria, and therefore, this foaming test was not considered to be capable of predicting AD foaming (Alfaro et al., 2014).
2.2Leipzig Foaming Test
The use of the aeration test in biogas plants that utilize renewables and biogenic waste is not possible due to the high fiber content and structure of the digestate. The foaming propensity of substrates in this kind of anaerobic digester can be estimated by using the Leipzig foam tester (UFZ, 2013). This is a test kit that can be easily used in practice. It consists of a tempered 1-L glass with a foam trap in the lid (Fig. 4).
The test principle uses batch AD of 2% (w/w) substrate based on TS without continuous mixing. The digestate from the AD plant has to be passed through a sieve with a mesh size of 10 mm prior to using it in the foaming test. The starting volume of the substrate–digestate mixture is 500 mL. The test duration is between 12 and 24 h depending on foaming stagnation. The test temperature is selected according to the operating mode of the anaerobic digester from which the digestate originated.
The foaming properties of the substrate are validated as follows (Moeller et al., 2015b):
There are substrates such as sugar beet that do not form a typical foam layer and for this reason the volume of foam is not possible to quantify. As a result, it is favorable to conduct a control test without the addition of substrates and the foam volume is then taken to be the difference between the volume in the foaming test and the control test.
The foaming test has already been used for the development of various strategies of foam suppression and prevention (Moeller and Zehnsdorf, 2016, 2017; Moeller et al., 2015c) as described in Section 5.
2.3Detection of the Surface Tension
As stated in Section 1, one requirement for the formation of foam is the presence of surface-active substances. Surface-active substances settle on the interfaces between the gas and liquid and reduce the surface tension. The surface tension is the ratio between the force for the enlargement of the surface and the surface change, and is dependent on temperature and pressure (Kopplow, 2006). Water has a surface tension of 72.7 mN/m (20°C). Thus, the degree of reduction of surface tension allows conclusions to be drawn about the presence of surface-active substances in the solution.
There are several methods available for the detection of the surface tension in liquids, e.g., by ring, plate, or detachment methods. In research on AD foaming causes, the Wilhelmy plate method (Blackall and Marshall, 1989; Ganidi, 2008), the bubble pressure method (Kopplow, 2006), and the drop volume method (Moeller and Goersch, 2015b) have been used.
The principle of the Wilhelmy plate method is based on the measurement of the time changes in surface tension during immersion and lifting of a paper plate into and out of solution using a tensiometer (Ganidi, 2008). The measured liquid must be cell free. For this reason, centrifugation is needed and the centrate is used for measurements.
The bubble pressure method uses a bubble tensiometer. The airstream is fed through a capillary into the liquid and a bubble is produced. The radius of this bubble is inversely proportional to the air pressure. The air pressure that is necessary for the production of one bubble is proportional to the surface tension according to the Young–Laplace equation (Kopplow, 2006):
The drop volume tensiometer measures the volume of a cell-free liquid droplet that is released from a cannula. The droplet drops off after it reaches a critical volume. This critical volume is proportional to the surface tension (Lauda, 1993):
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2.2Experimental methods
A variety of methods have been utilized for more than a century to measure the IFT between two phases of immiscible fluids (Prokhorov and Rusanov, 1996), some of which are widely used, including the capillary method, Wilhelmy measuring plate method, maximum bubble pressure method, drop weight method, pendant drop method, and sessile drop method. Of these methods, the four methods above calculate the IFT by force analysis and therefore are mechanical methods.
2.2.1Capillary method
The capillary rise method (Adamson and Gast, 1997; Chun and Wilkinson, 1995; Rashidnia et al., 1992; Washburn, 1921) is considered to be the most accurate method due to its relatively well-developed theory (Fig. 4). Based on the analysis of the equilibrium of forces at the liquid level in a capillary tube inserted into the liquid, the IFT can be described by Equation (5). This method is considered the oldest method for measuring surface tension (Bressler and Wyatt, 1970; Prokhorov, 1996).
(5)
where is the gravity difference between the gas phase and the liquid phase, h is the liquid column height, r is the inner radius of the capillary tube, and g is the local gravitational acceleration.
2.2.2Wilhelmy measuring plate method
The plate is made of either roughened platinumiridium alloy or platinum. Prior to the experiment, the plate must be cleaned from organic contaminants by an organic solvent and exposed to a flame. The plate inserted vertically into the fluid hangs from one end of the microbalance, and then the weights are added at the other end of the microbalance until the plate is balanced (Fig. 5). The theory of the Wilhelmy measuring plate method is that the weight of the weights is equal to the sum of the weight of the plate and the interfacial tension (Drelich et al., 2002):
(6)
where Ww is the weight of the weights, Wp is the weight of the plate, l is the length of the bottom edge of the rectangular plate contacting the liquid surface, and t is the thickness of the plate.
The calculation of the IFT by blowing an inert gas into a capillary tube and using the radius of the capillary tube and the pressure difference between the inside of the bubble and that outside of the bubble is referred to as the maximum bubble pressure method (Fig. 6), and is based on the concept that the bubble pressure is proportional to the IFT according to Laplace's law (Peter and Robert, 1996). IFT can be obtained by Equation (7). This method is fast and convenient, and the test results are not easily affected by liquid impurities or contamination (Peter and Robert, 1996); however, the IFT of a liquid–liquid system cannot be obtained by this method.
(7)
where r is the radius of the capillary tube, and ΔPmax is the maximum differential pressure between the inside and outside of the bubble.
2.2.3Drop weight method
The drop weight method is a convenient, accurate and statistical average method (Harkins and Brown, 1919). The droplet flows out of the capillary when the gravity of the droplet is greater than the IFT. This method is based on the principle that the gravity of the liquid dripping freely from the capillary mouth is equal to the IFT in a certain period (Fig. 7). IFT can be calculated by the following equation:
(8)
where m is the mass of liquid drop, r is the radius of the capillary mouth, and F is the correction factor that is related to the size of liquid drop and the radius of the capillary mouth.
2.2.4Optical imaging methods
Optical imaging methods mainly include the pendant (Andreas et al., 1938; Lin and Hwang, 1994) and sessile drop methods (Neumann and Good, 1979), and IFT can be calculated by analyzing the shape of the immobile bubble or droplet according to ADSA theory (Rotenberg et al., 1983). These are the two main methods for measuring IFT at elevated pressures (Georgiadis et al., 2010). The most important advantage of these two methods is the small amount of liquid required for the measurement (Chen et al., 1998; Cheng et al., 1990). The ADSA theory is derived from the Young-Laplace equation (Equation (9)) (Cheng et al., 1990; Teleszewski and Gajewski, 2020) and is regarded as the most advanced and accurate calculation theory (Rίo and Neumann, 1997; Yang et al., 2005). ADSA theory applies a fully automated and subjective human-free method to determine IFT, providing more accurate measurements (Lins et al., 2022). The equation can be simplified by introducing the correlation between the dimensions De and Ds (Fig. 8) (Fordham, 1948). The parameter De is the equatorial diameter of the droplet, and Ds is the diameter of the drop at the height of De from the drop apex. In turn, IFT can be calculated according to Equation (10) (Chi et al., 2022; Hamouda and Bagalkot, 2019; Kazemzadeh et al., 2015). The pendant and sessile drop methods are considered to be the most accurate and direct measurement methods for the evaluation of IFT (Mutailipu et al., 2019). A schematic of the ADSA system for the pendant and sessile drop methods is shown in Fig. 9. Each of the above test methods for IFT measurement has a unique combination of properties such as applicability and measurement range. (Table 1).
Table 1. Characteristics of the common experimental methods.
| Classification | Methods | Applicable interface | Measurement range of IFT (mN/m) | Applicability for contact angle measurement |
--- ---
| Force analysis methods | Capillary rise method | Gas–liquid, | – | Yes |
| Wilhelmy measuring plate method | Gas–liquid, liquid–liquid | 0.1–2000 | Yes |
| Maximum bubble pressure method. | Gas–liquid | 1–100 | No |
| Drop weight method | Gas–liquid, liquid–liquid | 0.1–100 | No |
| Optical imaging methods | Pendant drop method | Gas–liquid, liquid–liquid | 0.01–2000 | No |
| Sessile drop method | Gas–liquid, liquid–liquid | – | Yes |
(9)
where is the pressure difference across the interface, γ is the IFT, and R1 and R2 are the radii of curvature describing the curved liquid surface.
(10)
where △ρ is the density difference between the two immiscible phases and 1/H is a correction factor that is a function of the ratio that can be calculated by the analysis software.
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Related terms:
Adsorption
Critical Micelle Concentration
Surfactant
Aqueous Solution
Nanofluid
Laplaces Equation
Sessile Drop Method
Capillary Tube
Critical Heat Flux
Tension Surface
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11946 | https://pubmed.ncbi.nlm.nih.gov/18509663/ | Bilateral dermoid cysts of the ovary in a pregnant woman: case report and review of the literature - PubMed
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. 2009 Feb;279(2):105-8.
doi: 10.1007/s00404-008-0695-3. Epub 2008 May 29.
Bilateral dermoid cysts of the ovary in a pregnant woman: case report and review of the literature
Mohammad Sami Walid1,Mark G Boddy
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1 Medical Center of Central Georgia, 840 Pine Street, Macon, GA 31201, USA. mswalid@yahoo.com
PMID: 18509663
DOI: 10.1007/s00404-008-0695-3
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Bilateral dermoid cysts of the ovary in a pregnant woman: case report and review of the literature
Mohammad Sami Walid et al. Arch Gynecol Obstet.2009 Feb.
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. 2009 Feb;279(2):105-8.
doi: 10.1007/s00404-008-0695-3. Epub 2008 May 29.
Authors
Mohammad Sami Walid1,Mark G Boddy
Affiliation
1 Medical Center of Central Georgia, 840 Pine Street, Macon, GA 31201, USA. mswalid@yahoo.com
PMID: 18509663
DOI: 10.1007/s00404-008-0695-3
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Objective: Most nonphysiological ovarian masses discovered during pregnancy are benign dermoid cysts. The association of dermoid cysts with pregnancy has been increasingly reported since 1918. They usually present the dilemma of weighing the risks of surgery and anesthesia versus the risks of untreated adnexal mass.
Method: We are reporting an illustrative case and presenting a review of the literature for recommendations regarding the management of such cases.
Result: The bilateral dermoid cysts were surgically treated in the second trimester.
Conclusion: Most references state that it is more feasible to treat bilateral dermoid cysts of the ovaries discovered during pregnancy if they grow beyond 6 cm in diameter. This is usually performed through laparotomy or very carefully through laparoscopy and should preferably be done in the second trimester.
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Sonographic evaluation of the adnexa during early pregnancy.Lavery JP, Koontz WL, Layman L, Shaw L, Gumpel U.Lavery JP, et al.Surg Gynecol Obstet. 1986 Oct;163(4):319-23.Surg Gynecol Obstet. 1986.PMID: 3532382
[Benign ovarian tumors during pregnancy. A review of 26 cases].Attia L, Chachia A, Ben Temime R, Bennour G, Makhlouf T, Koubâa A.Attia L, et al.Tunis Med. 2008 Jul;86(7):680-4.Tunis Med. 2008.PMID: 19472731 French.
[Echographic demonstration of dermoid cysts of the ovary during pregnancy].Preziosi P, Karakaci F, Medici D, De Luca B.Preziosi P, et al.Minerva Ginecol. 1984 May;36(5):245-50.Minerva Ginecol. 1984.PMID: 6472718 Italian.No abstract available.
Dermoid cyst of the pancreas. Case report and review of the literature.Vermeulen BJ, Widgren S, Gur V, Meyer P, Iselin C, Rohner A.Vermeulen BJ, et al.Gastroenterol Clin Biol. 1990;14(12):1023-5.Gastroenterol Clin Biol. 1990.PMID: 2289662 Review.
Management of ovarian cysts in pregnancy: a case report.Patacchiola F, Collevecchio N, Di Ferdinando A, Palermo P, Di Stefano L, Mascaretti G.Patacchiola F, et al.Eur J Gynaecol Oncol. 2005;26(6):651-3.Eur J Gynaecol Oncol. 2005.PMID: 16398230 Review.
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Ovarian Cystic Teratoma in Pregnant Women: Conservative Management or Prophylactic Oophorectomy?Osto M, Brooks A, Khan A.Osto M, et al.Cureus. 2021 Aug 21;13(8):e17354. doi: 10.7759/cureus.17354. eCollection 2021 Aug.Cureus. 2021.PMID: 34567894 Free PMC article.
An unusual case of multiple and bilateral ovarian dermoid cysts. Case report.Pepe F, Lo Monaco S, Rapisarda F, Raciti G, Genovese C, Pepe P.Pepe F, et al.G Chir. 2014 Mar-Apr;35(3-4):75-7.G Chir. 2014.PMID: 24841683 Free PMC article.
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11947 | https://thirdsgames.co.uk/gh/maths-compiled/ib/book/build/main.pdf | Notes on the Mathematical Tripos Sky Wilshaw Part IB University of Cambridge 2020–2024 Contents I.
Optimisation Lectured in Easter 2021 by Dr. V. Jog 5 II.
Variational Principles Lectured in Easter 2021 by Dr. M. Dunajski 63 III.
Markov Chains Lectured in Michaelmas 2021 by Dr. P. Sousi 101 IV.
Analysis and Topology Lectured in Michaelmas 2021 by Dr. V. Zsák 137 V.
Methods Lectured in Michaelmas 2021 by Prof. E. P. Shellard 215 VI.
Quantum Mechanics Lectured in Michaelmas 2021 by Dr. M. Ubiali 297 VII.
Linear Algebra Lectured in Michaelmas 2021 by Prof. P. Raphael 347 VIII. Groups, Rings and Modules Lectured in Lent 2022 by Dr. R. Zhou 427 IX.
Complex Analysis Lectured in Lent 2022 by Prof. N. Wickramasekera 485 X.
Geometry Lectured in Lent 2022 by Prof. I. Smith 543 XI.
Statistics Lectured in Lent 2022 by Dr. S. Bacallado 609 3 I. Optimisation Lectured in Easter 2021 by Dr. V. Jog Many real-world problems involve finding optimal points of functions, for instance making the most valuable products given limited resources, or finding the optimal way to transport goods across a network. In this course, we study the theory behind optimisation, and pro-duce various algorithms for computing optima in different environments.
An important class of functions is the convex functions. One can show that if a function is convex, we can use local behaviour to make conclusions about global minima and maxima.
This helps guide our study of optimisation. Linear functions are convex, and the study of optimising linear functions is called linear programming. We show that linear programs can be solved computationally using the simplex method, allowing us to easily solve lots of real-world optimisation problems.
5 I. Optimisation Contents 1.
Introduction and convex functions . . . . . . . . . . . . . . . . . .
8 1.1.
Outline and definitions . . . . . . . . . . . . . . . . . . . . . .
8 1.2.
Convexity . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
8 1.3.
Unconstrained optimisation . . . . . . . . . . . . . . . . . . .
9 1.4.
First-order conditions for convexity . . . . . . . . . . . . . . . .
9 1.5.
Second-order conditions for convexity . . . . . . . . . . . . . .
10 2.
Optimisation algorithms . . . . . . . . . . . . . . . . . . . . . . . .
12 2.1.
Gradient descent . . . . . . . . . . . . . . . . . . . . . . . . .
12 2.2.
Smoothness assumption . . . . . . . . . . . . . . . . . . . . .
12 2.3.
Strong convexity assumption . . . . . . . . . . . . . . . . . . .
15 2.4.
Proving gradient descent . . . . . . . . . . . . . . . . . . . . .
16 2.5.
Rate of convergence . . . . . . . . . . . . . . . . . . . . . . . .
17 2.6.
Condition numbers and oscillation . . . . . . . . . . . . . . . .
17 2.7.
Newton’s method . . . . . . . . . . . . . . . . . . . . . . . . .
18 2.8.
Barrier methods . . . . . . . . . . . . . . . . . . . . . . . . .
18 3.
Lagrange multipliers . . . . . . . . . . . . . . . . . . . . . . . . . .
20 3.1.
Introduction and Lagrange sufficiency . . . . . . . . . . . . . .
20 3.2.
Using Lagrange multipliers in general . . . . . . . . . . . . . .
21 3.3.
Complementary slackness . . . . . . . . . . . . . . . . . . . .
22 3.4.
Weak duality . . . . . . . . . . . . . . . . . . . . . . . . . . .
24 3.5.
Strong duality and the Lagrange method . . . . . . . . . . . . .
25 3.6.
Hyperplane condition for strong duality . . . . . . . . . . . . .
25 3.7.
Strong duality and convex functions . . . . . . . . . . . . . . .
26 3.8.
Shadow prices interpretation of Lagrange multipliers . . . . . . .
27 4.
Linear programming . . . . . . . . . . . . . . . . . . . . . . . . . .
29 4.1.
Linear programs . . . . . . . . . . . . . . . . . . . . . . . . .
29 4.2.
Maximising convex functions . . . . . . . . . . . . . . . . . . .
30 4.3.
Basic solutions and basic feasible solutions . . . . . . . . . . . .
31 4.4.
Extreme points of the feasible set in standard form . . . . . . . .
32 5.
Duality in linear programming . . . . . . . . . . . . . . . . . . . .
34 5.1.
Strong duality of linear programs . . . . . . . . . . . . . . . . .
34 5.2.
Duals of linear programs in standard form . . . . . . . . . . . .
34 5.3.
Duals of linear programs in general form . . . . . . . . . . . . .
35 5.4.
Dual of dual program . . . . . . . . . . . . . . . . . . . . . . .
35 5.5.
Dual of arbitrary linear program . . . . . . . . . . . . . . . . .
36 5.6.
Optimality conditions . . . . . . . . . . . . . . . . . . . . . . .
37 6 6.
Simplex method . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
38 6.1.
Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . .
38 6.2.
Feasibility of basic directions . . . . . . . . . . . . . . . . . . .
39 6.3.
Cost of basic directions . . . . . . . . . . . . . . . . . . . . . .
39 6.4.
Moving to basic feasible solutions . . . . . . . . . . . . . . . . .
40 6.5.
Simplex method . . . . . . . . . . . . . . . . . . . . . . . . .
41 6.6.
Tableau implementation . . . . . . . . . . . . . . . . . . . . .
41 7.
Game theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
45 7.1.
Zero-sum games . . . . . . . . . . . . . . . . . . . . . . . . .
45 7.2.
Mixed strategies . . . . . . . . . . . . . . . . . . . . . . . . . .
46 7.3.
Duality of mixed strategy problems . . . . . . . . . . . . . . . .
47 7.4.
Finding optimal strategies . . . . . . . . . . . . . . . . . . . .
48 8.
Network flows . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
50 8.1.
Minimum cost flow . . . . . . . . . . . . . . . . . . . . . . . .
50 8.2.
Transport problem . . . . . . . . . . . . . . . . . . . . . . . .
50 8.3.
Sufficiency of transport problem . . . . . . . . . . . . . . . . .
51 8.4.
Optimality conditions for transport problem . . . . . . . . . . .
52 9.
The transport algorithm . . . . . . . . . . . . . . . . . . . . . . . .
53 9.1.
Transportation tableaux . . . . . . . . . . . . . . . . . . . . .
53 9.2.
Updating the transportation tableau . . . . . . . . . . . . . . .
55 10.
Maximum flow, minimum cut . . . . . . . . . . . . . . . . . . . .
57 10.1.
Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . .
57 10.2.
Cuts and flows . . . . . . . . . . . . . . . . . . . . . . . . . .
57 10.3.
Max-flow min-cut theorem . . . . . . . . . . . . . . . . . . . .
58 10.4.
Ford–Fulkerson algorithm . . . . . . . . . . . . . . . . . . . .
59 10.5.
Termination of Ford–Fulkerson . . . . . . . . . . . . . . . . . .
61 10.6.
Bipartite matching problem . . . . . . . . . . . . . . . . . . . .
61 7 I. Optimisation 1. Introduction and convex functions 1.1. Outline and definitions An optimisation problem is a problem in which we want to minimise some function 𝑓(x) such that x ∈𝒳⊆ℝ𝑛. We may have a set of constraints ℎ(x) = b where ℎ(x)∶ℝ𝑛→ℝ𝑚.
Note that we will only ever consider minimisation of functions since we can maximise a function by minimising its negative. Such a problem is often written with notation such as minimise x∈𝒳 𝑓(x) subject to ℎ(x) = b Definition. The following definitions will be used.
(i) The function 𝑓that we want to minimise is called the objective function.
(ii) The components of the vector x are called the decision variables.
(iii) A constraint of the form ℎ(x) = b is called a functional constraint.
(iv) A constraint of the form x ∈𝒳is called a regional constraint.
(v) The set 𝒳(b) = {x∶x ∈𝒳, ℎ(x) = b} is called the feasible set.
(vi) If the feasible set is non-empty, the optimisation problem is called feasible. If the feasible set is empty, the problem is infeasible.
(vii) The problem is called bounded if the minimum on 𝒳(b) is bounded.
(viii) A point x⋆∈𝒳(b) is optimal if it minimises 𝑓over 𝒳(b). The value 𝑓(x⋆) is called the optimal cost.
We can convert an inequality constraint into an equality constraint with a regional con-straint, for instance ℎ(x) ≤𝑏⟶ℎ(x) + 𝑠= 𝑏; 𝑠≥0 1.2. Convexity Definition. A set 𝑆⊆ℝ𝑛is convex if for all x, y ∈𝑆, the line segment from x to y lies entirely inside 𝑆. In other words, for all 𝜆∈[0, 1], x(1 −𝜆) + y(𝜆) ∈𝑆.
Definition. A function 𝑓∶𝑆→ℝis convex if • 𝑆is convex, and • for all x, y ∈𝑆, 𝑓((1 −𝜆)x + 𝜆y) ≤(1 −𝜆)𝑓(x) + 𝜆𝑓(y) 8 1. Introduction and convex functions So informally, for a convex function, if we take two inputs, the chord connecting their out-puts lies above the function’s curve. If the given inequality above is strict, the function is called strictly convex. 𝑓is (strictly) concave if −𝑓is (strictly) convex. Note that if 𝑓is linear, 𝑓is convex and concave, since 𝑓is linear in its input. Hence linear optimisation is a special case of convex optimisation.
1.3. Unconstrained optimisation The unconstrained optimisation problem is simply to minimise 𝑓(x), where 𝑓∶ℝ𝑛→ℝis a convex function. Convex functions allow you to generalise the behaviour of a function in a small neighbourhood to global behaviour, so it becomes easier to solve optimisation problems expressed in terms of convex functions.
1.4. First-order conditions for convexity Suppose we have a tangent to a curve 𝑓∶ℝ→ℝat a given point 𝑥. If 𝑓is convex, then 𝑓 must only touch the curve once, since if it touched twice we would contradict the definition of convexity. In particular, we have the following necessary and sufficient condition for convexity: 𝑓(𝑦) ≥𝑓(𝑥) + (𝑦−𝑥)𝑓′(𝑥) In higher dimensions, we might guess that 𝑓(y) ≥𝑓(x) + (y −x) ⋅∇𝑓(x) Theorem. A differentiable function 𝑓∶ℝ𝑛→ℝis convex if and only if ∀x, y ∈ℝ𝑛, 𝑓(y) ≥𝑓(x) + (y −x) ⋅∇𝑓(x) = 𝑓(x) + ∇𝑓(x)⊺(y −x) Remark. If ∇𝑓(x) = 0 for some vector x, then the first-order condition implies that 𝑓(y) ≥ 𝑓(x), so x is the global minimum of 𝑓. This is an example of how we can use local properties (the gradient of the function at x) to deduce global properties (the minimum value of the function).
Proof. First, we will prove that convexity implies the first-order condition. By convexity, we have 𝑓((1 −𝜆)x + 𝜆y) ≤(1 −𝜆)𝑓(x) + 𝜆𝑓(y) Initially, let 𝑛= 1 so that we have the one-dimensional case. We have 𝑓(𝑦) ≥𝑓(𝑥) + 𝑓(𝑥+ 𝜆(𝑦−𝑥)) −𝑓(𝑥) 𝜆 = 𝑓(𝑥) + 𝑓(𝑥+ 𝜆(𝑦−𝑥)) −𝑓(𝑥) 𝜆(𝑦−𝑥) (𝑦−𝑥) Hence, taking the limit as 𝜆→0, we have 𝑓(𝑦) ≥𝑓(𝑥) + 𝑓′(𝑥)(𝑦−𝑥) 9 I. Optimisation For the general case, we define a function 𝑔such that 𝑔(𝜆) = 𝑓((1 −𝜆)x + 𝜆y). Since 𝑓is convex, so is 𝑔. We can calculate 𝑔′(𝜆) = ∇𝑓((1 −𝜆)x + 𝜆y) ⋅(y −x) Since 𝑔∶[0, 1] →ℝis convex, by the above argument for 𝑛= 1 we have 𝑔(1) ≥𝑔(0) + 𝑔′(0)(1 −0) 𝑓(y) ≥𝑓(x) + ∇𝑓(x) ⋅(y −x) Now we must prove the converse; if the first-order condition holds, then 𝑓is convex. Let x𝜆= (1 −𝜆)x + 𝜆y The first-order condition shows that 𝑓(x) ≥𝑓(x𝜆) + ∇𝑓(x𝜆) ⋅(x −x𝜆) 𝑓(y) ≥𝑓(x𝜆) + ∇𝑓(x𝜆) ⋅(y −x𝜆) Multiplying the first equation by 1 −𝜆and multiplying the second equation by 𝜆, we get (1 −𝜆)𝑓(x) + 𝜆𝑓(y) ≥𝑓(x𝜆) + ∇𝑓(x𝜆) ⋅[(x −x𝜆)(1 −𝜆) + (y −x𝜆)𝜆] = 𝑓(x𝜆) + ∇𝑓(x𝜆) ⋅[(x −(1 −𝜆)x −𝜆y)(1 −𝜆) + (y −(1 −𝜆)x −𝜆y)𝜆] = 𝑓(x𝜆) + ∇𝑓(x𝜆) ⋅[(𝜆x −𝜆y)(1 −𝜆) + ((1 −𝜆)y −(1 −𝜆)x)𝜆] = 𝑓(x𝜆) + ∇𝑓(x𝜆) ⋅0 = 𝑓(x𝜆) Hence 𝑓really is convex.
1.5. Second-order conditions for convexity When 𝑛= 1, we suspect that 𝑓″(𝑥) ≥0 is the condition for convexity. In higher dimensions, the analogous operator to the double derivative is the Hessian matrix.
∇2𝑓(x) = H𝑓= ⎛ ⎜ ⎜ ⎜ ⎜ ⎝ 𝜕2𝑓(x) 𝜕𝑥2 1 𝜕2𝑓(x) 𝜕𝑥1𝜕𝑥2 ⋯ 𝜕2𝑓(x) 𝜕𝑥1𝜕𝑥𝑛 𝜕2𝑓(x) 𝜕𝑥2𝜕𝑥1 𝜕2𝑓(x) 𝜕𝑥2 2 ⋯ 𝜕2𝑓(x) 𝜕𝑥2𝜕𝑥𝑛 ⋮ ⋮ ⋱ ⋮ 𝜕2𝑓(x) 𝜕𝑥𝑛𝜕𝑥1 𝜕2𝑓(x) 𝜕𝑥𝑛𝜕𝑥2 ⋯ 𝜕2𝑓(x) 𝜕𝑥2 𝑛 ⎞ ⎟ ⎟ ⎟ ⎟ ⎠ Definition. An 𝑛× 𝑛matrix 𝐴is positive semidefinite if for all x ∈ℝ𝑛, we have x⊺𝐴x ≥0.
Equivalently, all eigenvalues of 𝐴are non-negative. If 𝐴is positive semidefinite, we write 𝐴⪰0.
10 1. Introduction and convex functions Note that the higher-dimensional analogue of the Taylor expansion of 𝑓(y) is 𝑓(y) = 𝑓(x) + ∇𝑓(x)⊺(y −x) + 1 2(y −x)⊺∇2𝑓(x)(y −x) + ⋯ Theorem. A twice-differentiable function 𝑓∶ℝ𝑛→ℝis convex if ∇2𝑓(x) ⪰0 at all x. The converse also holds, but it is not important for this course, so it will not be proven.
Proof. Using the Taylor expansion of 𝑓, we have 𝑓(y) = 𝑓(x) + ∇𝑓(x)⊺(y −x) + 1 2(y −x)⊺∇2𝑓(z)(y −x) where z = (1 −𝜆)x + 𝜆y for some 𝜆∈[0, 1]. The rightmost term is positive, hence 𝑓(y) ≥𝑓(x) + ∇𝑓(x)⊺(y −x) So the first-order conditions are satisfied, which imply convexity.
11 I. Optimisation 2. Optimisation algorithms 2.1. Gradient descent Consider minimising 𝑓(𝑥) such that 𝑓∶ℝ𝑛→ℝis a convex function. Recall that a local minimum of 𝑓is also the global minimum. Consider the following ‘greedy’ method: • Start at a point x0.
• Search for close points around x0 whose values of 𝑓are smaller than 𝑓(x0).
– If such a point exists, let this be x1. Repeat the algorithm.
– If such a point does not exist, we have found a local minimum, which is the global minimum.
We can find such x1 points by considering the Taylor series expansion of 𝑓around a point.
𝑓(x −𝜀∇𝑓(x)) ≈𝑓(x) −𝜀∇𝑓(x)⊺⋅∇𝑓(x) = 𝑓(x) −𝜀‖∇𝑓(x)‖2 ≤𝑓(x) Hence −∇𝑓(x) is called a descending direction. Although the gradient of the function is the most natural way of decreasing a function, any v with 𝑓(x) ⋅v < 0 is a descending direction.
This gives us the gradient descent algorithm.
Algorithm 1: Gradient Descent Algorithm Result: Global minimum of 𝑓(x) start at a point x0; 𝑡←0; repeat find a descending direction v𝑡, e.g. −∇𝑓(x); choose a step size 𝜂𝑡; x𝑡+1 ←x𝑡+ 𝜂𝑡v𝑡; until ∇𝑓(x) = 0 or 𝑡is large enough; Different choices of v𝑡and 𝜂𝑡give rise to many different qualities of algorithm.
2.2. Smoothness assumption Some restrictions must be applied to a function to let us prove that gradient descent works.
Definition. A continuously differentiable function 𝑓∶ℝ𝑛→ℝis 𝛽-smooth if ∇𝑓is a 𝛽-Lipschitz function: ‖∇𝑓(x) −∇𝑓(y)‖ ≤𝛽‖x −y‖ In the following sections, we assume all functions 𝑓are 𝛽-smooth. Further, if 𝑓is twice dif-ferentiable (i.e. the Hessian exists everywhere), then the 𝛽-smoothness assumption is equi-valent to ∇2𝑓(x) ⪯𝛽𝐼 12 2. Optimisation algorithms so all eigenvalues of ∇2𝑓(x) have 𝜆≤𝛽. Also, u⊺∇2𝑓(x)u ≤u⊺(𝛽𝐼)u = 𝛽‖u‖2 Definition. The linear approximation to 𝑓at x is 𝑓(x) + ∇𝑓(x)⊺(y −x) We might assume that the linear approximation is close to the actual function in a small neighbourhood around 𝑥.
Claim. If 𝑓is 𝛽-smooth, then 𝑓(y) ≤𝑓(x) + ∇𝑓(x)⊺(y −x) + 𝛽 2 ‖y −x‖2 Note that 𝑓(x) + ∇𝑓(x)⊺(y −x) ≤𝑓(y) since 𝑓is convex, so this claim would show that 𝑓really is close to the actual function, deviating by an arbitrarily small amount as we let x approach y.
Proof. By Taylor’s theorem, 𝑓(y) = 𝑓(x) + ∇𝑓(x)⊺(y −x) + 1 2(y −x)⊺∇2𝑓(z)(y −x) ≤𝑓(x) + ∇𝑓(x)⊺(y −x) + 1 2(y −x)⊺(𝛽𝐼)(y −x) = 𝑓(x) + ∇𝑓(x)⊺(y −x) + 𝛽 2 ‖y −x‖2 Corollary. If we move by a step size of 1 𝛽, we will descend by at least 1 2𝛽‖∇𝑓(𝑥)‖2.
𝑓(x −1 𝛽∇𝑓(x)) ≤𝑓(x) −1 2𝛽‖∇𝑓(𝑥)‖2 Proof. Consider 𝑓(x) + ∇𝑓(x)⊺(y −x) + 𝛽 2 ‖y −x‖2 as a function of y, and try to minimise it for a fixed x.
∇y(𝑓(x) + ∇𝑓(x)⊺(y −x) + 𝛽 2 ‖y −x‖2) = ∇𝑓(x) + 𝛽(y −x) = 0 13 I. Optimisation Hence, ∇𝑓(x) 𝛽 = x −y y = x −1 𝛽∇𝑓(x) Substituting into the claim above, we have 𝑓(𝑥−1 𝛽∇𝑓(x)) ≤𝑓(x) + ∇𝑓(x)⊺(−1 𝛽∇𝑓(x)) + 𝛽 2 ‖ ‖ ‖ 1 𝛽∇𝑓(x)‖ ‖ ‖ 2 = 𝑓(x) −1 𝛽‖∇𝑓(x)‖2 + 1 2𝛽‖∇𝑓(x)‖2 = 𝑓(x) −1 2𝛽‖∇𝑓(x)‖2 Claim (Improved first order condition).
𝑓(y) ≥𝑓(x) + ∇𝑓(x)⊺(y −x) + 1 2𝛽‖∇𝑓(x) −∇𝑓(y)‖2 Proof. For any z, by the standard first order condition and the corollary above we have 𝑓(x) + ∇𝑓(x)⊺(z −x) ≤𝑓(z) ≤𝑓(y) + ∇𝑓(y)⊺(z −y) + 𝛽 2 ‖z −y‖2 This then implies 𝑓(x) −𝑓(y) ≤∇𝑓(x)⊺(x −z) + ∇𝑓(y)⊺(z −y) + 𝛽 2 ‖z −y‖2 The left hand side is not dependent on z, so by minimising z we get the best bound for the left hand side. We set the gradient of z to zero.
−∇𝑓(x) + ∇𝑓(𝑦) + 𝛽(z −y) = 0 ⟹z = ∇𝑓(x) −∇𝑓(y) 𝛽 + y Substituting back, we have 𝑓(x) −𝑓(y) ≤∇𝑓(x)⊺(x −y) −1 2𝛽‖∇𝑓(x) −∇𝑓(y)‖2 14 2. Optimisation algorithms 2.3. Strong convexity assumption In general, a small gradient does not imply that we are close to the optimum value of the function. We must therefore add an additional assumption in order to justify gradient des-cent.
Definition. A function 𝑓∶ℝ𝑛→ℝis called 𝛼-strongly convex if 𝑓(y) ≥𝑓(x) + ∇𝑓(x)⊺(y −x) + 𝛼 2 ‖y −x‖2 If 𝑓is twice differentiable, then its Hessian satisfies ∇2𝑓(x) ⪰𝛼𝐼 for all x.
Claim. Let 𝑓be 𝛼-strongly convex. Let 𝑝⋆be the optimal cost; i.e. the minimum value of 𝑓.
Then for any x we have 𝑝⋆≥𝑓(x) −1 2𝛼‖∇𝑓(x)‖2 Remark. If ‖∇𝑓(x)‖ ≤√2𝛼𝜀, then 𝑝⋆≤𝑓(x) ≤𝑝⋆+ 𝜀 So a small gradient means we are close to the optimum.
Proof. The 𝛼-strong convexity assumption gives 𝑓(y) ≥𝑓(x) + ∇𝑓(x)⊺(y −x) + 𝛼 2 ‖y −x‖2 Taking the minimum over y of both sides, the left hand side becomes 𝑝⋆. Setting the gradient of the right hand side to zero, ∇𝑓(x) −𝛼(x −y) = 0 ∇𝑓(x) 𝛼 = (x −y) This gives 𝑝⋆≥𝑓(x) + ∇𝑓(x)⊺(−∇𝑓(x) 𝛼 ) + 𝛼 2 ‖ ‖ ‖ ∇𝑓(x) 𝛼 ‖ ‖ ‖ 2 = 𝑓(x) −‖∇𝑓(x)‖2 𝛼 + ‖∇𝑓(x)‖2 2𝛼 = 𝑓(x) −‖∇𝑓(x)‖2 2𝛼 15 I. Optimisation Claim. Let x⋆be the minimising value, i.e. 𝑓(x⋆) = 𝑝⋆. Then ‖x −x⋆‖ ≤2 𝛼‖∇(x)‖ So if a function is strongly convex, we can find a region in which we know the global max-imum lies.
Proof. By the Cauchy–Schwarz inequality, 𝑓(x⋆) ≥𝑓(x) + ∇𝑓(x)⊺(x⋆−x) + 𝛼 2 ‖x⋆−x‖2 ≥𝑓(x) −‖∇𝑓(x)‖‖x⋆−x‖ + 𝛼 2 ‖x⋆−x‖2 Since 𝑓(x⋆) ≤𝑓(x), we have 0 ≥𝑓(x⋆) −𝑓(x) ≥−‖∇𝑓(x)‖‖x⋆−x‖ + 𝛼 2 ‖x⋆−x‖2 Hence, ‖∇𝑓(x)‖‖x⋆−x‖ ≥𝛼 2 ‖x⋆−x‖2 ‖∇𝑓(x)‖ ≥𝛼 2 ‖x⋆−x‖ 2.4. Proving gradient descent Let 𝑓be a 𝛽-smooth and 𝛼-strongly convex, where 0 < 𝛼< 𝛽. Then 𝛼𝐼⪯∇2𝑓(x) ⪯𝛽𝐼 Theorem. Gradient descent with step size 1 𝛽satisfies 𝑓(x𝑇) −𝑓(x⋆) ≤(1 −𝛼 𝛽) 𝑇 (𝑓(x0) −𝑓(x⋆)) ≤𝑒 −𝛼𝑇 𝛽(𝑓(x0) −𝑓(x⋆)) ≤𝑒 −𝛼𝑇 𝛽𝛽 2 ‖x⋆−x0‖2 Proof.
𝑓(x𝑡+1) −𝑓(x⋆) ≤𝑓(x𝑡) −𝑓(x⋆) −1 2𝛽‖∇𝑓(x𝑡)‖2 ≤𝑓(x𝑡) −𝑓(x⋆) −𝛼 𝛽(𝑓(x𝑡) −𝑓(x⋆)) ≤(1 −𝛼 𝛽)(𝑓(x𝑡) −𝑓(x⋆)) 16 2. Optimisation algorithms Hence by induction, 𝑓(x𝑇) −𝑓(x⋆) ≤(1 −𝛼 𝛽) 𝑇 (𝑓(x0) −𝑓(x⋆)) The second line of the theorem is a consequence of the properties of the exponential function.
The last inequality in the theorem can be shown by 𝛽-smoothness.
𝑓(x0) ≤𝑓(x⋆) + ∇𝑓(x⋆)⊺(x0 −x⋆) + 𝛽 2 ‖x0 −x⋆‖2 𝑓(x0) −𝑓(x⋆) ≤𝛽 2 ‖x0 −x⋆‖2 2.5. Rate of convergence For example, suppose that we would like 𝑓(x𝑇) −𝑓(x⋆) ≤0.1, and it takes 𝑘steps to reach this tolerance. Then, it would take around 2𝑘steps to reach a tolerance of 0.01, since the (1 − 𝛼 𝛽) 𝑇 power might increase by a factor of 2. In general, the number of steps needed to ensure that the error is less than 𝜀is 𝑇= 𝛽 𝛼log(𝑓(x0) −𝑓(x⋆) 𝜀 ) This log(1/𝜀) term is called ‘linear convergence’, since for each extra order of magnitude of accuracy, we need a linear amount of computation steps. Linear convergence is very fast, and such algorithms are very useful.
2.6. Condition numbers and oscillation Note that 1 −𝛼 𝛽 is the term which controls the convergence of gradient descent. We call 𝛽/𝛼the condition number of 𝑓. Such a number is always greater than 1. If the condition number is very close to 1, the convergence is fast. Consider the function 𝑓(𝑥1, 𝑥2) = 1 2(𝑥2 1 + 100𝑥2 2) The Hessian of 𝑓at any point is ∇2𝑓(𝑥1, 𝑥2) = (1 0 0 100) 17 I. Optimisation Hence, 𝛼= 1, 𝛽= 100 giving a condition number of 100. This function would optimise very slowly, and we may continually overshoot in the 𝑥2 direction since the gradient points so strongly in this direction. We may like to prevent this oscillation between over-guessing and under-guessing certain coordinate components.
2.7. Newton’s method In gradient descent, we have x𝑡+1 = x𝑡−𝜂𝑡∇𝑓(x𝑡) In Newton’s method, we replace this formula with x𝑡+1 = x𝑡−(∇2𝑓(x)) −1∇𝑓(x𝑡) Note that the second order approximation for 𝑓is 𝑓(x) ≈𝑓(x𝑡) + ∇𝑓(x𝑡)⊺+ 1 2(x −x𝑡)⊺∇2𝑓(x𝑡)(x −x𝑡) So if we instead try to minimise the right hand side of the second-order approximation with respect to x, we have x𝑡+1 = x𝑡−(∇2𝑓(x)) −1∇𝑓(x𝑡) as given by Newton’s method. This ideally allows us to deal with ‘badly-proportioned’ co-ordinates independently, by scaling each coordinate using the Hessian rather than by a con-stant. Essentially, Newton’s method iteratively approximates the function with a parabola, and then moves to the minimum point of this parabola. We can show that Newton’s method converges according to ‖x𝑡+1 −x⋆‖ ≤𝑐‖x𝑡−x⋆‖2 when x𝑡−x⋆is small enough. We can see here that the squared term provides very fast convergence once we are in the neighbourhood of the optimum. Newton’s method can also be used to find a root of a function. Suppose 𝑓∶ℝ→ℝ, and define 𝑓′ = 𝑔.
𝑥𝑡+1 = 𝑥𝑡−𝑓′(𝑥𝑡) 𝑓″(𝑥𝑡) = 𝑥𝑡−𝑔(𝑥𝑡) 𝑔′(𝑥𝑡) So we can find the root of 𝑔by computing the stationary point of 𝑓. We are essentially taking a linear approximation at a point, and setting this linear approximation to zero.
2.8. Barrier methods Suppose we impose a constraint on an optimisation problem, for instance minimising 𝑓(x) such that 𝑓𝑖(x) ≤0 for 1 ≤𝑖≤𝑚. We can transform such a constrained problem into an unconstrained problem. Let us minimise 𝑓(x) + 𝑚 ∑ 𝑖=1 𝜙(𝑓𝑖(x)) 18 2. Optimisation algorithms where 𝜙(𝑦𝑖) = +∞outside the feasible set, and 𝜙(𝑦𝑖) = 0 inside the feasible set. However, this 𝜙function is not differentiable, so this introduces even more problems. We instead consider a logarithmic barrier function. Let us minimise the unconstrained problem 𝑡𝑓(x) − 𝑚 ∑ 𝑖=1 log(−𝑓𝑖(x)) ⟹𝜙(𝑥) = −log(−𝑥) This barrier function is infinite for negative 𝑥, and gradually rises as 𝑥→0. When 𝑡is chosen to be very large, the optimum of this problem is very close to the optimum of the original problem.
Algorithm 2: Barrier Method Result: Global minimum of 𝑓(x) start at a point x inside the feasible set; set 𝑡to be a positive real number; repeat solve the minimiser of 𝑡𝑓(x) −∑ 𝑚 𝑖=1 log(−𝑓𝑖(x)) with x as the initial point using Newton’s method giving x⋆; x ←x⋆; 𝑡←𝛼𝑡for some fixed 𝛼> 1; until 𝑡is large enough; 19 I. Optimisation 3. Lagrange multipliers 3.1. Introduction and Lagrange sufficiency Consider minimising 𝑓(x) subject to x ∈𝒳, ℎ(x) = b where ℎ∶ℝ𝑛→ℝ𝑚. The Lagrangian associated with this problem is 𝐿(x, 𝛌) = 𝑓(x) −𝛌⊺(ℎ(x) −b) where 𝛌∈ℝ𝑚is the vector of Lagrange multipliers. We want to instead minimise 𝐿(x, 𝛌), 𝑥∈ 𝒳.
Theorem (Lagrange Sufficiency). Suppose we can find a 𝛌⋆such that (i) minx∈𝒳𝐿(x, 𝛌⋆) = 𝐿(x⋆, 𝛌⋆) (ii) x⋆∈𝒳(b) = {x∶x ∈𝒳, ℎ(x) = b} Then x⋆is optimal for the original constrained problem, i.e.
min x∈𝒳(b) 𝑓(x) = 𝑓(x⋆) Proof. First, note that condition (ii) states that 𝑓(x⋆) ≥minx∈𝒳(b) 𝑓(x), because x⋆is feasible.
Then, min x∈𝒳(b) 𝑓(x) = min x∈𝒳(b) 𝑓(x) −(𝛌⋆)⊺(ℎ(x) −b) ⏟⎵⎵⎵⏟⎵⎵⎵⏟ 0 when x∈𝒳(b) ≥min x∈𝒳𝑓(x) −(𝛌⋆)⊺(ℎ(x) −b) = min x∈𝒳𝐿(x, 𝛌⋆) = 𝐿(x⋆, 𝛌⋆) = 𝑓(x⋆) −(𝛌⋆)⊺(ℎ(x⋆) −b) = 𝑓(x⋆) Example.
minimise 𝑥∈ℝ3 −𝑥1 −𝑥2 + 𝑥3 subject to 𝑥2 1 + 𝑥2 2 = 4 𝑥1 + 𝑥2 + 𝑥3 = 1 In this problem, we have ℎ(x) = ( 𝑥2 1 + 𝑥2 2 𝑥1 + 𝑥2 + 𝑥3 ) ; b = (4 1) 20 3. Lagrange multipliers Taking Lagrange multipliers, we have 𝐿(x, 𝛌) = (−𝑥1 −𝑥2 + 𝑥3) −𝜆1(𝑥2 1 + 𝑥2 2 −4) −𝜆2(𝑥1 + 𝑥2 + 𝑥3 −1) = (−(1 + 𝜆2)𝑥1 −𝜆1𝑥2 1) + (−(1 + 𝜆2)𝑥2 −𝜆1𝑥2 2) + (1 −𝜆2)𝑥3 + 4𝜆1 + 𝜆2 We want to fix a value of 𝛌and minimise 𝐿, only considering solutions such that x⋆is finite.
Note that if 𝜆1 > 0, then the first bracket can be made as small as we like by picking very small values of 𝑥1; this bracket would diverge to negative infinity so we cannot choose such a 𝜆1. If 𝜆2 ≠1, the infimum is also negative infinity by considering the 𝑥3 term. So let us consider 𝜆1 ≤0, 𝜆2 = 1. Setting the derivative of the first term to zero, we have d d𝑥1 (−(1 + 𝜆2)𝑥1 −𝜆1𝑥2 1) = −(1 + 𝜆2) −2𝜆1𝑥1 = 0 ⟹𝑥1 = −1 −𝜆2 2𝜆1 = −2 2𝜆1 = −1 𝜆1 Setting the derivative of the second term to zero, d d𝑥1 (−(1 + 𝜆2)𝑥2 −𝜆1𝑥2 2) = −(1 + 𝜆2) −2𝜆1𝑥2 = 0 ⟹𝑥2 = −1 𝜆1 We now want to choose 𝜆1 such that 𝑥1, 𝑥2, 𝑥3 satisfy the constraints.
𝑥2 1 + 𝑥2 2 = 4 ⟹𝑥2 1 = 𝑥2 2 = 2 ⟹𝑥1 = 𝑥2 = √2 Note that 𝑥1, 𝑥2 > 0 since 𝜆1 ≤0, and correspondingly 𝜆1 = −1 √2. Further, we can now find 𝑥3 = 1 −2√2. This solution optimises the original problem.
3.2. Using Lagrange multipliers in general Consider the problem minimise x∈𝒳 𝑓(x) subject to ℎ(x) ≤b We can solve this problem using the following steps.
(1) Add a slack variable s to transform the problem to minimise x∈𝒳 𝑓(x) subject to ℎ(x) + s = b s ≥0 21 I. Optimisation (2) Calculate the Lagrangian, 𝐿(x, 𝛌, s) = 𝑓(x) −𝛌⊺(ℎ(x) + s −b) (3) Let 𝚲= {𝛌∶ inf x∈𝒳; s≥0 𝐿(x, s, 𝛌) > −∞} (4) For each 𝛌∈𝚲, find x⋆(𝛌), s⋆(𝛌) such that min x∈𝒳; s≥0 𝐿(x, s, 𝛌) = 𝐿(x⋆(𝛌), s⋆(𝛌), 𝛌) (5) Find 𝛌⋆∈𝚲such that (x⋆(𝛌), s⋆(𝛌)) is feasible, i.e.
ℎ(x⋆(𝛌⋆)) = b; s⋆(𝛌⋆) ≥0 3.3. Complementary slackness In step (4) above, we want to minimise the Lagrangian, i.e.
minimise x∈𝒳 𝑓(x) −𝛌⊺(ℎ(x) −b) −𝛌⊺s subject to s ≥0 Suppose, for a particular value of 𝛌, that we solve this problem and arrive at x⋆(𝛌), s⋆(𝛌).
Let 𝛌= ( 𝜆1 ⋮ 𝜆𝑚 ) If 𝜆𝑖> 0, then for some large s we can make 𝑓→−∞, hence 𝛌∉𝚲. Hence, given 𝛌∈𝚲, we must have 𝜆𝑖≤0. Now, if 𝜆𝑖< 0 for some 𝑖, we would want to choose 𝑠𝑖= 0 to minimise the increase to the function caused by the slack variable. If 𝜆𝑖= 0, then 𝑠𝑖can be chosen arbitrarily since it will have no increase on the value of 𝑓. With these choices of 𝑠𝑖, we can make 𝛌⊺s = 0, thus making the slack variable not impact the value of 𝑓. So either • ℎ(x)𝑖= 𝑏𝑖and 𝜆𝑖≤0, or • ℎ(x)𝑖≥𝑏𝑖and 𝜆𝑖= 0.
Alternatively (less precisely), 𝜆𝑖𝑠𝑖= 0 In other words, either the constraint inequality is tight (defined by an equality) and the Lagrange multipliers are slack (defined by an inequality), or the constraint inequality is slack and the Lagrange multipliers are tight.
22 3. Lagrange multipliers Example.
minimise x∈ℝ2 𝑥1 −3𝑥2 subject to 𝑥2 1 + 𝑥2 2 ≤4 𝑥1 + 𝑥2 ≤2 Adding slack variables, we have minimise x∈ℝ2 𝑥1 −3𝑥2 subject to 𝑥2 1 + 𝑥2 2 + 𝑠1 = 4 𝑥1 + 𝑥2 + 𝑠2 = 2 𝑠1 ≥0 𝑠2 ≥0 Taking the Lagrangian, 𝐿(x, s, 𝛌) = (𝑥1 −3𝑥2) −𝑥1(𝑥2 1 + 𝑥2 2 + 𝑠1 −4) −𝜆2(𝑥1 + 𝑥2 + 𝑠2 −2) = ((1 −𝜆2)𝑥1 −𝜆1𝑥2 1) + ((−3 −𝜆2)𝑥2 −𝜆1𝑥2 2) −𝜆1𝑠1 −𝜆2𝑠2 + (4𝜆1 + 2𝜆2) We must have 𝜆1, 𝜆2 ≤0 by considering the slack variable. By complementary slackness, 𝜆1𝑠1 = 𝜆2𝑠2 = 0 at the optimum Minimising each term independently, we have 1 −𝜆2 −2𝜆1𝑥1 = 0 −3 −𝜆2 −2𝜆1𝑥2 = 0 If 𝜆1 = 0, the above two equations are contradictory. Hence 𝜆1 < 0, giving 𝑠1 = 0. If 𝜆2 < 0, then 𝑠2 = 0 by complementary slackness, so 1 −𝜆2 −2𝜆1𝑥1 = 0 −3 −𝜆2 −2𝜆1𝑥2 = 0 𝑥2 1 + 𝑥2 2 = 4 𝑥1 + 𝑥2 = 2 Solving the lower two equations give (𝑥1, 𝑥2) = (0, 2), (2, 0) If (𝑥1, 𝑥2) = (0, 2), solving the first two equations gives (𝜆1, 𝜆2) = (1, −3) which is impossible since 𝜆1 must be negative. Similarly, if (𝑥1, 𝑥2) = (2, 0), solving the first two equations gives 23 I. Optimisation (𝜆1, 𝜆2) = (−1, 1) which is impossible again. We have ruled out every case apart from 𝜆1 < 0, 𝜆2 = 0. In this case, 1 −2𝜆1𝑥1 = 0 −3 −2𝜆1𝑥2 = 0 𝑥2 1 + 𝑥2 2 = 4 𝑥1 + 𝑥2 + 𝑠2 = 2 The first two equations give 𝑥1 = 1 2𝜆1 ; 𝑥2 = −3 2𝜆1 Substituting into the third equation, 𝜆2 1 = 5 8 ⟹𝜆1 = −√ 5 8 Hence, (𝑥1, 𝑥2) = (−√ 2 5, −3√ 2 5) which is feasible using the fourth equation. By Lagrange sufficiency, this is the optimum for the original problem.
3.4. Weak duality We would like to solve a problem minimise x∈𝒳 𝑓(x) subject to ℎ(x) = b by constructing the Lagrangian 𝐿(x, 𝛌) = 𝑓(x) −𝛌⊺(ℎ(x) −b) We now define the quantity 𝑔(𝛌) = inf x∈𝒳𝐿(x, 𝛌) Theorem (Weak duality theorem). If x ∈𝒳(b) and 𝛌∈𝚲, then 𝑓(x) ≥𝑔(𝛌). In particular, inf x∈𝒳(b) 𝑓(x) ≥sup 𝛌∈𝚲 𝑔(𝛌) 24 3. Lagrange multipliers Proof.
inf x∈𝒳(b) 𝑓(x) = inf x∈𝒳(b) 𝑓(x) −𝛌⊺(ℎ(x) −b) ≥inf x∈𝒳𝑓(x) −𝛌⊺(ℎ(x) −b) = inf x∈𝒳𝐿(x, 𝛌) = 𝑔(𝛌) Using this weak duality property, if infx∈𝒳(b) 𝑓(x) is difficult to solve, we can first attempt sup𝛌∈𝚲𝑔(𝛌). The problem maximise 𝑔(𝛌) subject to 𝛌∈𝚲 is called the dual problem. The original is called the primal problem. The optimal cost of the primal problem is always greater than or equal to the optimal cost of the dual problem. The duality gap is the difference: inf x∈𝒳(b) 𝑓(x) −sup 𝛌∈𝚲 𝑔(𝛌) If the duality gap is zero, then we say that strong duality holds. This strengthens the inequal-ity into an equality.
3.5. Strong duality and the Lagrange method If the Lagrange method works, then we know that inf x∈𝒳𝐿(x, 𝛌) = inf x∈𝒳(b) 𝐿(x, 𝛌) So, taking such a 𝛌in the proof above, we have equality instead of inequality. Hence the problem has strong duality. Conversely, if the duality gap is zero, then there exists a 𝛌such that the inequality above is an equality. Hence, for this 𝛌, inf x∈𝒳𝐿(x, 𝛌) = inf x∈𝒳(b) 𝐿(x, 𝛌) Hence this is the 𝛌which will solve the Lagrange method. In summary, strong duality holds exactly when the Lagrange method works.
3.6. Hyperplane condition for strong duality Definition. A function 𝜙∶ℝ𝑚→ℝis said to have a supporting hyperplane at a point b if there exists 𝛌∈ℝ𝑚such that for all c ∈ℝ𝑚, 𝜙(c) ≥𝜙(b) + 𝛌⊺(c −b) 25 I. Optimisation Pictorially, 𝜙has a supporting hyperplane if there is a plane passing through (b, 𝜙(b)), where 𝜙is always above the plane. This could be, for example, a tangent plane at b.
Definition. We define a function 𝜙∶ℝ𝑚→ℝassociated with the primal problem by 𝜙(c) = inf x∈𝒳(c) 𝑓(x) This 𝜙can be thought of as the optimal cost of a family of optimisation problems with dif-ferent functional constraint values c. This is called the value function.
Theorem (Strong duality theorem). Strong duality holds if and only if the value function 𝜙 has a supporting hyperplane at b.
Proof. First, we show that a supporting hyperplane implies strong duality. We have 𝛌such that 𝜙(c) ≥𝜙(b) + 𝛌⊺(c −b) Then, we have 𝑔(𝛌) = inf x∈𝒳𝑓(x) −𝛌⊺(ℎ(x) −b) = inf c inf x∈𝒳(c) 𝑓(x) −𝛌⊺(ℎ(x) −c) ⏟⎵⎵⏟⎵⎵⏟ zero since we are extremising −𝛌⊺(c −b) = inf c 𝜙(c) −𝛌⊺(c −b) ≥𝜙(b) By weak duality, we also have the reverse direction: 𝑔(𝛌) ≤𝜙(b) Hence, 𝑔(𝛌) = 𝜙(b) and strong duality holds. Conversely, if strong duality holds, we want to show the existence of such a hyperplane. We have We have 𝛌such that 𝑔(𝛌) = 𝜙(b). For such a 𝛌, we have 𝜙(b) = 𝑔(𝛌) = inf x∈𝒳𝑓(x) −𝛌⊺(ℎ(x) −b) = inf x∈𝒳𝑓(x) −𝛌⊺(ℎ(x) −c) −𝛌⊺(c −b) ≤𝜙(c) −𝛌⊺(c −b) The last inequality holds due to weak duality. So 𝛌gives a supporting hyperplane.
3.7. Strong duality and convex functions We would now like to consider for which problems 𝜙(b) has a supporting hyperplane. The following theorem is stated without proof.
Theorem. A function 𝜙∶ℝ𝑚→ℝis convex if and only if every point b ∈ℝ𝑚has a supporting hyperplane.
Now, for which problems do we have a convex value function?
26 3. Lagrange multipliers Theorem. Consider a minimisation problem minimise x∈𝒳 𝑓(x) subject to ℎ(x) ≤b with value function 𝜙. Then 𝜙is convex if: (i) 𝒳is convex; (ii) 𝑓is convex; (iii) ℎis convex.
This is proven in the example sheets.
3.8. Shadow prices interpretation of Lagrange multipliers Suppose a factory owner produces 𝑛types of products from 𝑚types of raw materials. Sup-pose the owner produces x = (𝑥1, 𝑥2, … , 𝑥𝑛) products, then the profit is some function 𝑓(x).
We then create ℎ𝑗(x) to be the amount of raw material 𝑗consumed when making products x.
The owner wants to maximise 𝑓(x) subject to ℎ𝑖(x) ≤𝑏𝑖where 𝑏𝑖is the maximum amount of raw material 𝑖that is available.
Now, suppose a supplier offers some 𝛆= (𝜀1, 𝜀2, … , 𝜀𝑚) extra raw materials to the factory owner. We would like to calculate how much this 𝛆is worth. The factory owner will try to maximise this new problem, replacing b ↦b+𝛆. For a small enough 𝛆, this can be expressed easily using the value function.
𝜙(b + 𝛆) −𝜙(b) ≈ 𝑚 ∑ 𝑗=1 𝜕𝜙 𝜕𝑏𝑗 𝜀𝑗 The quantity 𝜕𝜙 𝜕𝑏𝑗is the price of material 𝑗, and ∇𝜙(b) is the vector of prices. These are called the ‘shadow prices’; they are hidden to the outside world but depend on the internal state of the factory.
Theorem. If 𝜙is differentiable at b and has a supporting hyperplane given by 𝛌, then 𝛌= ∇𝜙(b) Proof. Let a = (𝑎1, 𝑎2, … , 𝑎𝑚) be an arbitrary vector. Then from the supporting hyperplane condition, for some small 𝛿> 0 we have 𝜙(b + 𝛿a) 𝛿 ≥𝛌⊺a Since 𝜙is differentiable, the limit can be taken to give ∇𝜙(b) ⋅a ≥𝛌⊺a 27 I. Optimisation But a was arbitrary. This can only hold if 𝛌= ∇𝜙(b) as required. So the Lagrange multiplier 𝛌at b is equal to the gradient vector of 𝜙which is the gradient of partial derivatives and also the vector of shadow prices.
Suppose that a particular raw material was not used up. Then there is a slack value in the inequality. The shadow price is zero in this instance, since we do not need more of this material. So the corresponding Lagrange multiplier is equal to zero. Conversely, if we are paying something for this material, then we must have used up all of that material. This is exactly the complementary slackness property seen earlier.
There is also an economics interpretation of the dual problem. Such a problem can be seen from the perspective of the raw material seller. This seller charges a certain price 𝛌for their raw materials, and then buys the finished product from the factory. The profit of the raw material seller is 𝛌⊺(ℎ(x) −b) ⏟⎵⎵⏟⎵⎵⏟ cost of materials − 𝑓(x) ⏟ buying products For every choice of 𝛌, the factory owner will try to maximise their profit, that is, find an x⋆ such that we maximise 𝑓(x) ⏟ selling products −𝛌⊺(ℎ(x) −b) ⏟⎵⎵⏟⎵⎵⏟ cost of materials 28 4. Linear programming 4. Linear programming 4.1. Linear programs A linear program is a specific case of a constrained optimisation problem in which the ob-jective function and all constraints are linear functions. For instance, consider the prob-lem minimise x∈ℝ4 2𝑥1 −𝑥2 + 4𝑥3 subject to 𝑥1 + 𝑥2 + 𝑥4 ≤2 3𝑥2 −𝑥3 = 5 𝑥3 + 𝑥4 ≥3 𝑥1 ≥0 𝑥3 ≤0 A general linear program is of the form minimise x∈ℝ𝑛 c⊺x subject to a⊺ 𝑖x ≥𝑏𝑖, 𝑖∈𝑀1 a⊺ 𝑖x ≤𝑏𝑖, 𝑖∈𝑀2 a⊺ 𝑖x = 𝑏𝑖, 𝑖∈𝑀3 𝑥𝑗≥0, 𝑗∈𝑁1 𝑥𝑗≤0, 𝑗∈𝑁2 Note that we can convert the first inequalities to the other direction by inverting the sign of a. We can convert the ‘sign’ constraints (the last two constraints) by letting a be a one-hot vector, thus writing them in terms of the first two inequality types. We call this process reduction to an equivalent form. Two linear programs are equivalent if any feasible solution for one problem can be converted into a feasible solution for the other, with the same cost.
We can reduce any linear problem into the form minimise x∈ℝ𝑛 c⊺x subject to 𝐴x ≥b where 𝐴= ⎛ ⎜ ⎜ ⎝ ⋯ a⊺ 1 ⋯ ⋯ a⊺ 2 ⋯ ⋮ ⋯ a⊺ 𝑚 ⋯ ⎞ ⎟ ⎟ ⎠ ; b = ⎛ ⎜ ⎜ ⎝ 𝑏1 𝑏2 ⋮ 𝑏𝑚 ⎞ ⎟ ⎟ ⎠ 29 I. Optimisation This is known as the general form of a linear programming problem. We could alternatively use a ‘less-than’ inequality, or simply an equality using a slack variable vector. A linear problem is said to be in standard form if it is written as minimise x∈ℝ𝑛 c⊺x subject to 𝐴x = b x ≥0 This is a special case of the general form. However, we can always reduce any general-form problem into a standard-form problem. First, we add slack variables to convert the inequality into an equality. Then we can convert each variable 𝑥𝑖into the sum of 𝑥+ 𝑗−𝑥− 𝑗, where 𝑥+ 𝑗, 𝑥− 𝑗≥0. Then, we have the problem minimise x∈ℝ𝑛 c⊺(x+ −x−) subject to 𝐴(x+ −x−) = b x+, x−≥0 Then by concatenating the vectors x+, x−into a larger vector z ∈[0, ∞)2𝑛, we have the standard form as required.
4.2. Maximising convex functions Solving linear programs can be seen as a special case of maximising a convex function, since we can maximise c⊺x. Consider the problem minimise 𝑓(x) subject to 𝑥∈𝐶, 𝐶convex x ≥0 where 𝑓is a convex function. Since 𝐶is convex, if z = (1 −𝜆)x + 𝜆y we have 𝑓(z) ≤(1 −𝜆)𝑓(x) + 𝜆𝑓(y) ≤max {𝑓(x), 𝑓(y)} If we wish to maximise 𝑓over 𝐶, we might guess that we only need to consider points on the boundary. After all, any point not on the boundary can be written as the weighted average of two points on the boundary. Considering those points will give a greater (or equal) value for 𝑓.
Definition. A point x in a convex set 𝐶is an extreme point if it cannot be written as a convex combination of two distinct points in 𝐶; that is, (1 −𝛿)y + 𝛿z for 𝛿∈(0, 1) and y ≠z.
So, more precisely, convex functions on convex sets are maximised at extreme points.
30 4. Linear programming 4.3. Basic solutions and basic feasible solutions Consider a linear problem in standard form.
minimise x∈ℝ𝑛 c⊺x subject to 𝐴x = b x ≥0 where 𝐴∈ℝ𝑚×𝑛, x ∈ℝ𝑛, b ∈ℝ𝑚.
Definition. A vector x is said to be a basic solution if it satisfies 𝐴x = b (that is, it is a solution) and x has at most 𝑚nonzero entries. If also the nonzero entries are positive, then this is called a basic feasible solution, since it lies in the feasible set x ≥0.
We will start the analysis of basic solutions by making three assumptions (one is defined later).
A: All 𝑚rows of 𝐴are linearly independent. That is, {a⊺ 1, … , a⊺ 𝑚} is a linearly independ-ent set. This assumption can be made without loss of generality since we can simply remove linearly dependent constraints.
B: Every set of 𝑚columns of 𝐴is linearly independent. That is, any 𝑚-subset of the set of columns {𝐴1, … , 𝐴𝑛} is a linearly independent set. This can also be made without loss of generality by removing the linearly dependent variables.
To find a basic solution, we will start by choosing the coordinates 𝐵(1), 𝐵(2), … , 𝐵(𝑚) to be the indices of x that are allowed to be nonzero. Now, 𝐴x is ( ⋮ ⋮ 𝐴1 ⋯ 𝐴𝑛 ⋮ ⋮ ) ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ 0 ⋮ 𝑥𝐵(1) ⋮ 𝑥𝐵(2) ⋮ 𝑥𝐵(𝑚) ⋮ 0 ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ = ( ⋮ ⋮ 𝐴𝐵(1) ⋯ 𝐴𝐵(𝑚) ⋮ ⋮ ) ⏟⎵⎵⎵⎵⎵⏟⎵⎵⎵⎵⎵⏟ 𝐵 ⎛ ⎜ ⎜ ⎝ 𝑥𝐵(1) 𝑥𝐵(2) ⋮ 𝑥𝐵(𝑚) ⎞ ⎟ ⎟ ⎠ By setting 𝐴x = b, using the above assumptions, we can invert the matrix on the left-hand side to get ⎛ ⎜ ⎜ ⎝ 𝑥𝐵(1) 𝑥𝐵(2) ⋮ 𝑥𝐵(𝑚) ⎞ ⎟ ⎟ ⎠ = ( ⋮ ⋮ 𝐴𝐵(1) ⋯ 𝐴𝐵(𝑚) ⋮ ⋮ ) −1 b = 𝐵−1b We call 𝐵the basis matrix. The indices 𝑥𝐵(1), … , 𝑥𝐵(𝑚) are called the basic variables. The indices 𝐵(𝑖) are called the basic indices. The columns 𝐴𝐵(𝑖) are called the basic columns.
If 𝐵−1b ≥0, we have found a basic feasible solution. We now need to specify one further assumption in order to continue to analyse basic solutions.
31 I. Optimisation C: Every basic solution has exactly 𝑚nonzero entries. This assumption is known as the non-degeneracy assumption. This assumption cannot be created without loss of gen-erality, but it is far simpler to discuss problems with this assumption met. Throughout this course, we will keep this assumption to be true.
4.4. Extreme points of the feasible set in standard form Consider a linear program in standard form.
Theorem. x is an extreme point (of the set {x∶𝐴x = b, x ≥0}) if and only if x is a basic feasible solution.
Remark. Since a linear program is optimised at extreme points, we only need to consider the basic feasible solutions in order to solve the original problem. We will pick all possible 𝑚columns of 𝐴(there are (𝑛 𝑚) such choices) to find all basic solutions. Filter to consider only basic feasible solutions, then evaluate c⊺x to find the x which has the least cost. This algorithm will always work, but the amount of choices to evaluate in higher dimensions becomes too inefficient for real-world use.
Proof. First, suppose we know x is a basic feasible solution, and there exist feasible y, z such that x = (1 −𝛿)y + 𝛿z and 𝛿∈(0, 1). We know x ≥0 and x has at most 𝑚nonzero entries.
Since y, z are positive, then y, z must be zero in every index that x must be zero. Specifically, 𝑦𝑗= 𝑧𝑗= 0 for 𝑗∉{𝐵(1), … , 𝐵(𝑚)}. Now, we define y𝐵= ( 𝑦𝐵(1) ⋮ 𝑦𝐵(𝑚) ) ; z𝐵= ( 𝑧𝐵(1) ⋮ 𝑧𝐵(𝑚) ) We then have 𝐵y𝐵= b; 𝐵z𝐵= b because 𝐴y = 𝐴z = b. Hence, y𝐵= z𝐵= 𝐵−1b and so x = y = z.
Conversely, suppose x is not a basic feasible solution. We wish to show it is not an extreme point. Then x has an amount of nonzero indices greater than 𝑚. Let such indices be 𝑖1, … , 𝑖𝑟 where 𝑟> 𝑚. Consider the columns 𝐴𝑖1, … , 𝐴𝑖𝑟. Since the rank of 𝐴is only 𝑚, these columns form a linearly dependent set. Hence, we can find some weights, not all of which are zero, which give zero when multiplied by the columns.
𝑤𝑖1𝐴𝑖1 + 𝑤𝑖2𝐴𝑖2 + ⋯+ 𝑤𝑖𝑟𝐴𝑖𝑟= 0 We now define the vector w by 𝑤𝑖= {0 𝑖∉{𝑖1, … , 𝑖𝑟} 𝑤𝑖𝑗 𝑖= 𝑖𝑗 So we have a nonzero vector w with 𝐴w = 0. We can consider the two points x ± 𝜀w, which satisfy 𝐴(x ± 𝜀w) = 0. Such perturbed points only change the nonzero indices of x. So we 32 4. Linear programming can find an 𝜀small enough such that both of x±𝜀w are in the feasible set, that is, x±𝜀w ≥0.
We therefore can express x as the midpoint of these two points, hence x is not an extreme point.
33 I. Optimisation 5. Duality in linear programming 5.1. Strong duality of linear programs Theorem. If a linear program is bounded and feasible, then strong duality holds.
Proof. This is true since the value function is convex.
5.2. Duals of linear programs in standard form Consider a linear program in standard form: minimise c⊺x subject to 𝐴x = b x ≥0 The dual problem is therefore maximise 𝑔(𝛌) = inf x∈𝒳𝐿(x, 𝛌) subject to 𝛌∈𝚲 The function 𝑔is given by 𝑔(𝛌) = inf x≥0 c⊺x −𝛌⊺(𝐴x −b) = inf x≥0(c⊺−𝛌⊺𝐴)x + 𝛌⊺b This is only bounded below where c⊺−𝛌⊺𝐴≥0. Hence 𝚲= {𝛌∶𝛌⊺𝐴≤c⊺} Further, the minimum value of 𝑔for 𝛌∈𝚲is 𝛌⊺b. Therefore, the dual problem is maximise 𝛌⊺b subject to 𝛌⊺𝐴≤c⊺ The dual of a linear program in standard form is a linear problem, but no longer in standard form.
34 5. Duality in linear programming 5.3. Duals of linear programs in general form Consider a linear program in general form: minimise c⊺x subject to 𝐴x ≥b We can introduce a slack variable s and write equivalently minimise c⊺x subject to 𝐴x −s = b s ≥0 To calculate the dual, we need to calculate 𝑔(𝛌).
𝑔(𝛌) = inf x,s≥0 c⊺x −𝛌⊺(𝐴x −s −b) = inf x,s≥0(c⊺−𝛌⊺𝐴)x + 𝛌⊺s + 𝛌⊺b In this case, since x may be any value, we must have c⊺−𝛌⊺𝐴= 0. Further, since the slack variable can be any positive value, 𝛌⊺≥0. The infimum is 𝛌⊺b since s may be set to zero.
Thus, the dual is maximise 𝛌⊺b subject to 𝛌⊺𝐴= c⊺ 𝛌≥0 The dual of a general linear program is a linear program in standard form.
5.4. Dual of dual program The dual of a dual problem is the primal problem. Suppose the primal problem is in standard form: minimise c⊺x subject to 𝐴x = b x ≥0 We know the dual is maximise 𝛌⊺b subject to 𝛌⊺𝐴≤c⊺ 35 I. Optimisation Equivalently, −minimise −𝛌⊺b subject to −𝛌⊺𝐴≥−c⊺ Defining ̃ 𝛌= −𝛌⊺, we have −minimise ̃ 𝛌b subject to ̃ 𝛌𝐴≥−c⊺ We can find the dual of this problem using the solution above.
−maximise −𝛉⊺c subject to 𝛉⊺𝐴⊺= b⊺ 𝛉≥0 This is equivalent to the primal problem.
5.5. Dual of arbitrary linear program Consider the problem minimise c⊺x subject to a⊺ 𝑖x ≥b𝑖 𝑖∈𝑀1 a⊺ 𝑖x ≤b𝑖 𝑖∈𝑀2 a⊺ 𝑖x = b𝑖 𝑖∈𝑀3 𝑥𝑗≥0 𝑗∈𝑁1 𝑥𝑗≤0 𝑗∈𝑁2 𝑥𝑗free 𝑗∈𝑁3 The dual of this problem is maximise p⊺b subject to 𝑝𝑖≥0 𝑖∈𝑀1 𝑝𝑖≤0 𝑖∈𝑀2 𝑝𝑖free 𝑖∈𝑀3 p⊺A𝑗≤c𝑗 𝑗∈𝑁1 p⊺A𝑗≥c𝑗 𝑗∈𝑁2 p⊺A𝑗= c𝑗 𝑗∈𝑁3 This will be shown in the example sheets.
36 5. Duality in linear programming 5.6. Optimality conditions If x is feasible for the primal, p is feasible for the dual, and complementary slackness holds, then x is optimal for the primal and p is optimal for the dual.
Theorem (Fundamental Theorem of Linear Programming). Let x, p be feasible solutions to the primal and dual problems respectively. Then x, p are optimal for these problems if and only if • 𝑝𝑖(a⊺ 𝑖x −𝑏𝑖) = 0 for all 𝑖, and • (𝑐𝑗−p⊺A𝑗)𝑥𝑗= 0 for all 𝑗.
Proof. First, let us define 𝑢𝑖= 𝑝𝑖(a⊺ 𝑖x −𝑏𝑖) and 𝑣𝑗= (𝑐𝑗−p⊺A𝑗)𝑥𝑗. Observe that if x, p are feasible, then 𝑢𝑖≥0 for all 𝑖, and 𝑣𝑗≥0 for all 𝑗. This can be seen by the signs of the constraints on the primal and dual problems. Now, ∑𝑢𝑖= ∑𝑝𝑖(a⊺ 𝑖x −𝑏𝑖) = p⊺𝐴x −p⊺b Similarly, ∑𝑣𝑗= ∑(𝑐𝑗−p⊺A𝑗)𝑥𝑗= c⊺x −p⊺𝐴x Then, ∑𝑢𝑖+ ∑𝑣𝑗= c⊺x −p⊺b which is the difference between the two objective functions in the primal and the dual.
Hence, 0 ≤∑𝑢𝑖+ ∑𝑣𝑗= c⊺x −p⊺b So if complementary slackness holds, then 𝑢𝑖= 0 and 𝑣𝑗= 0 for all 𝑖, 𝑗. This then implies that c⊺x = p⊺b. By weak duality, x and p must be optimal. Conversely, suppose x, p are optimal. By strong duality, c⊺x = p⊺b.
0 ≤∑𝑢𝑖+ ∑𝑣𝑗= c⊺x −p⊺b = 0 Thus ∑𝑢𝑖+ ∑𝑣𝑗= 0. Since all 𝑢𝑖, 𝑣𝑗are non-negative, 𝑢𝑖= 0 and 𝑣𝑗= 0 for all 𝑖, 𝑗.
Equivalently, complementary slackness holds.
37 I. Optimisation 6. Simplex method 6.1. Introduction Consider the problem minimise c⊺x subject to 𝐴x = b x ≥0 The dual problem is maximise 𝛌⊺b subject to 𝛌⊺𝐴≤c⊺ The optimality conditions are • (primal feasibility) 𝐴x = b; x ≥0 • (dual feasibility) 𝐴⊺𝛌≤c • (complementary slackness) x⊺(c −𝐴⊺𝛌) = 0 Suppose x is a basic feasible solution given by x𝐵= (𝑥𝐵(1), … , 𝑥𝐵(𝑚)) Substituting this x into the complementary slackness equation gives x⊺ 𝐵c𝐵−x⊺ 𝐵𝐵⊺𝛌= 0 ⟹x⊺ 𝐵(c𝐵−𝐵⊺𝛌) = 0 For a basic feasible solution, x𝐵> 0. Hence, c𝐵−𝐵⊺𝛌= 0 Hence 𝛌= (𝐵⊺)−1c𝐵 So for this x and this calculated 𝛌, primal feasibility and complementary slackness both hold.
What remains now is to check if dual feasibility holds. Equivalently, 𝐴⊺𝛌≤c ⟹𝐴⊺(𝐵⊺)−1c𝐵≤c If this holds, then the optimality conditions are met. This means that we do not even need to explicitly find 𝛌in order to check optimality; it suffices to check whether this single in-equality holds. We define c = c −𝐴⊺(𝐵⊺)−1c𝐵 This is called the vector of reduced costs. Then the inequality c ≥0 implies x is optimal.
38 6. Simplex method 6.2. Feasibility of basic directions Definition. Let 𝑃= {x∶𝐴x = b, x ≥0} be the feasible set of a problem in standard form.
Further, let x ∈𝑃. A vector d ∈ℝ𝑛is called a feasible direction if there exists 𝜃> 0 such that x + 𝜃d ∈𝑃.
Let x be a basic feasible solution. Let 𝐵(1), … , 𝐵(𝑚) be the indices of the basic variables, and let 𝐵be the basis matrix [𝐴𝐵(1), … , 𝐴𝐵(𝑚)]. Let x𝐵= (𝑥𝐵(1), … , 𝑥𝐵(𝑚)) ⊺. Suppose we move in a direction d such that 𝑑𝑗= 1, and 𝑑𝑖= 0 for all non-basic 𝑖≠𝑗, or more explicitly 𝑖∈{1, 2, … , 𝑛} ∖{𝐵(1), … , 𝐵(𝑚), 𝑗}. This direction d is called the 𝑗th basic direction, since it moves in the direction of the 𝑗th basic variable. Note that we can write d = (𝑑𝐵(1), … , 𝑑𝐵(𝑚), 0, 0, … , 1 ⏟ 𝑗th entry , … , 0, 0) When we move in this direction, we want to move to a feasible point. This means that we require 𝐴(x + 𝜃d) = b 𝐴d = 0 𝐵d𝐵+ 𝐴𝑗= 0 d𝐵= −𝐵−1𝐴𝑗 For the positivity condition, note that x + 𝜃d = (𝑥𝐵(1) + 𝜃𝑑𝐵(1), … , 𝑥𝐵(𝑚) + 𝜃𝑑𝐵(𝑚), 0, 0, … , 𝜃 ⏟ 𝑗th entry , … , 0, 0) For this x to be feasible, all 𝑥𝑖must be non-negative. Since 𝑥𝐵(𝑖) > 0, there exists a small enough 𝜃such that x + 𝜃d ≥0. Hence, the 𝑗th basic direction is feasible.
6.3. Cost of basic directions How does the cost change when x ↦x + 𝜃d where d is the (feasible) 𝑗th basic direction?
The new cost is c⊺(x + 𝜃d) = c⊺(x + 𝜃(−𝐵−1𝐴𝑗)) = c⊺x + 𝜃(𝑐𝑗−c⊺ 𝐵𝐵−1𝐴𝑗) = c⊺x + 𝜃𝑐𝑗 Theorem. Let x be a basic feasible solution associated with a basis matrix 𝐵, and let c be the vector of reduced costs. Then x is optimal if and only if c ≥0.
Proof. This follows from the optimality conditions given previously.
Now, if 𝑐𝑗≥0 for all 𝑗, then this is an optimal solution. However, if any 𝑐𝑗< 0, then we can move in the 𝑗th direction and decrease the cost.
39 I. Optimisation 6.4. Moving to basic feasible solutions Suppose x is a basic feasible solution. If c ≥0, then this is the optimum and we can stop. If 𝑐𝑗< 0 for some 𝑗, then moving in the 𝑗th feasible direction will reduce the cost by 𝜃𝑐𝑗. The amount by which the cost decreases is proportional to 𝜃, so we should choose the largest possible value of 𝜃while retaining feasibility. We denote this largest 𝜃with 𝜃⋆. There are two cases: • If d ≥0, then 𝜃is unbounded since x + 𝜃d ≥0 for all 𝜃> 0. Therefore the optimal cost of this problem is −∞.
• If 𝑑𝑖< 0 for some 𝑖, then we need 𝑥𝑖+ 𝜃𝑑𝑖≥0, so 𝜃⋆≤− 𝑥𝑖 𝑑𝑖. This then gives 𝜃⋆= min {𝑖∶𝑑𝑖<0} −𝑥𝑖 𝑑𝑖 or equivalently, 𝜃⋆= min {𝑖∈{1,…,𝑚}∶𝑑𝐵(𝑖)<0} −𝑥𝐵(𝑖) 𝑑𝐵(𝑖) Suppose the optimal cost is bounded. Let ℓbe the index minimising 𝜃⋆, so 𝜃⋆= −𝑥𝐵(ℓ) 𝑑𝐵(ℓ) Now, let us move in this direction by this amount.
Theorem. Let y = x + 𝜃⋆d. y is feasible, and c⊺y < c⊺x. Then, y is a basic feasible solution with basis matrix 𝐵= ( ⋮ ⋮ ⋮ ⋮ ⋮ 𝐴𝐵(1) ⋯ 𝐴𝐵(ℓ−1) 𝐴𝑗 𝐴𝐵(ℓ+1) ⋯ 𝐴𝐵(𝑚) ⋮ ⋮ ⋮ ⋮ ⋮ ) Proof. We know that y has exactly 𝑚nonzero entries, since 𝑦𝐵(ℓ) = 0. We know it is feasible, hence y is a basic feasible solution. 𝑗becomes a basic variable and 𝐵(ℓ) is no longer.
40 6. Simplex method 6.5. Simplex method Algorithm 3: Simplex Method Result: Global minimum of c⊺x start at a basic feasible solution x with basis matrix 𝐵= [𝐴𝐵(1), … , 𝐴𝐵(𝑚)]; repeat choose 𝑗such that 𝑐𝑗< 0; u ←−𝐵−1𝐴𝑗; if u ≤0 then cost is −∞so terminate algorithm; 𝜃⋆←min 𝑥𝐵(𝑖) 𝑢𝑖 where 𝑖∈{1, … , 𝑚} and 𝑢𝑖> 0; ℓ←an index 𝑖from the step above that gives the minimal value of 𝜃⋆; x ←x −𝜃⋆u; until c ≥0; since c ≥0, x is optimal 6.6. Tableau implementation The full tableau implementation of the simplex method is a convenient way of executing the simplex algorithm without excessive computation. A simplex tableau contains four values of information: −c⊺ 𝐵 c 𝐵−1b 𝐵−1𝐴 The information is essentially −cost reduced costs vector to generate current basic feasible solution matrix to generate basic directions In a more detailed form, 41 I. Optimisation −c⊺ 𝐵 𝑐1 𝑐2 ⋯ 𝑐𝑛 𝑥𝐵(1) ⋮ ⋮ ⋮ ⋮ 𝐵−1𝐴1 𝐵−1𝐴2 ⋯ 𝐵−1𝐴𝑛 𝑥𝐵(𝑚) ⋮ ⋮ ⋮ To execute the simplex algorithm using this table, use the following algorithm.
Algorithm 4: Simplex Method (Tableau Implementation) Result: Global minimum of c⊺x start at a basic feasible solution x with basis matrix 𝐵= [𝐴𝐵(1), … , 𝐴𝐵(𝑚)]; repeat choose 𝑗such that 𝑐𝑗< 0; u ←−𝐵−1𝐴𝑗; if u ≤0 then cost is −∞so terminate algorithm; 𝜃⋆←min 𝑥𝐵(𝑖) 𝑢𝑖 where 𝑖∈{1, … , 𝑚} and 𝑢𝑖> 0; ℓ←an index 𝑖from the step above that gives the minimal value of 𝜃⋆; (∗) add to each row of the tableau a constant multiple of the ℓth row so that 𝑢ℓ becomes 1 and all other entries of the pivot column are 0; until c ≥0 (when all entries in the 0th row are non-negative); since c ≥0, x is optimal This is just the same as the simplex method discussed before, apart from step (∗). No proof will be given for why this step achieves the same result as the full simplex algorithm.
Example. Consider the problem minimise x∈ℝ3 −𝑥1 −𝑥2 −𝑥3 subject to 𝑥1 + 2𝑥2 + 2𝑥3 ≤10 2𝑥1 + 𝑥2 + 2𝑥3 ≤10 2𝑥1 + 2𝑥2 + 𝑥3 ≤20 𝑥1, 𝑥2, 𝑥3 ≥0 42 6. Simplex method By introducing slack variables, we can write this in standard form.
minimise x∈ℝ6 −𝑥1 −𝑥2 −𝑥3 subject to 𝑥1 + 2𝑥2 + 2𝑥3 + 𝑥4 = 10 2𝑥1 + 𝑥2 + 2𝑥3 + 𝑥5 = 10 2𝑥1 + 2𝑥2 + 𝑥3 + 𝑥6 = 20 𝑥1, 𝑥2, 𝑥3, 𝑥4, 𝑥5, 𝑥6 ≥0 Observe that (0, 0, 0, 10, 10, 20) is a basic feasible solution. We will use this to initiate the simplex algorithm. The corresponding basis matrix is the 3×3 identity matrix. We construct the simplex tableau by first constructing the 0th row: • c𝐵= 0 hence c⊺ 𝐵x𝐵= 0.
• c = c.
We construct the tableau as follows.
0 −1 −1 −1 0 0 0 10 1 2 2 1 0 0 10 2 1 2 0 1 0 20 2 2 1 0 0 1 𝑐1 < 0, so we will descend in the 1st basic direction. Consider 10 1 , 10 2 , 20 2 . The smallest is 10 2 = 5, so the favourite element is the number 2 in the 1st column and 2nd row. We want to change this column to (0, 0, 1, 0)⊺by using row operations. Denoting the rows as 𝑅0, … , 𝑅3, we want to perform the operations 𝑅0 ↦𝑅0 + 1 2𝑅2 𝑅1 ↦𝑅1 −1 2𝑅2 𝑅2 ↦1 2𝑅2 𝑅3 ↦𝑅3 −𝑅2 The tableau now looks like this.
43 I. Optimisation 5 0 −0.5 0 0 0.5 0 5 0 1.5 1 1 −0.5 0 5 1 0.5 1 0 0.5 0 10 0 1 −1 0 −1 1 Now, 𝑐2 < 0, so we will descend in the 2nd basic direction. Consider 5 1.5, 5 0.5, 10 1 . The smallest is 5 1.5, so the favourite element is the 1.5 in the 1st row and 2nd column. To make the column a one-hot vector, we perform 𝑅0 ↦𝑅0 + 1 3𝑅1 𝑅1 ↦2 3𝑅1 𝑅2 ↦𝑅2 −1 3𝑅1 𝑅3 ↦𝑅3 −2 3𝑅1 This yields 20 3 0 0 1 3 1 3 1 3 0 10 3 0 1 2 3 2 3 − 3 4 0 10 3 1 0 2 3 − 1 3 2 3 0 20 3 0 0 − 5 3 − 2 3 − 2 3 1 Now, the 0th row has no negative values, so we are at the optimum. The optimal cost there-fore is − 20 3 . The solution is at ( 10 3 , 10 3 , 0, 0, 0, 20 3 ).
44 7. Game theory 7. Game theory 7.1. Zero-sum games Definition. A zero-sum two-person game is a scenario in which two players (denoted P1 and P2) have different actions they can take: • P1 has 𝑚possible actions {1, 2, … , 𝑚}, and • P2 has 𝑛possible actions {1, 2, … , 𝑛}; such that if P1 plays move 𝑖and P2 plays move 𝑗, then we say P1 ‘wins’ an amount 𝑎𝑖𝑗and P2 ‘loses’ the same amount 𝑎𝑖𝑗. The matrix of results 𝐴is called the payoff matrix. P1 chooses a row of the matrix, and P2 chooses a column, and the intersection is the outcome of the game.
Suppose P1 plays first, and chooses row 𝑖. P1 knows that P2 will choose the column 𝑗such that 𝑎𝑖𝑗is minimised, since that will maximise P2’s winnings. In particular, if P1 picks row 𝑖then they can expect to win min𝑗∈{1,…,𝑚} 𝑎𝑖𝑗. So P1 will try to solve the problem maximise min 𝑗∈{1,…,𝑚} 𝑎𝑖𝑗 subject to 𝑖∈{1, … , 𝑛} If P2 plays first, they will try to solve the problem minimise max 𝑖∈{1,…,𝑛} 𝑎𝑖𝑗 subject to 𝑗∈{1, … , 𝑚} Example. Suppose the payoff matrix is 𝐴= (1 2 3 4) P1 chooses a row, and P2 chooses a column. If P1 plays first, they choose row 2, then P2 chooses row 1, and the payoff is 3. If P2 plays first, they choose column 1, then P1 chooses row 2, and the payoff is again 3. Since the solution is the same for both problems, this point (2, 1) is called a saddle point. The value 𝑎21 = 3 is called the value of the game.
Example. Consider the payoff matrix 𝐴= (4 2 1 3) If P1 plays first, they choose row 1, then P2 chooses column 2, and the payoff is 2. If P2 plays first, they choose column 2, then P1 chooses row 2, and the payoff is 3. Here, both players cannot play optimally simultaneously since different outcomes will occur depending on what they think their opponent will do.
45 I. Optimisation 7.2. Mixed strategies In a mixed strategy, the players are allowed to choose their action randomly. Such mixed strategies are employed when we do not know what our opponent will pick; for example, when both players choose their option at the same time. P1 picks action 𝑖with probability 𝑝𝑖, and P2 picks action 𝑗with probability 𝑞𝑗, such that ∑𝑝𝑖= ∑𝑞𝑗= 1. Now, a player’s strategy is encoded as a probability vector. If P1 picks the mixed strategy (𝑝1, … , 𝑝𝑚), then the expected reward of P1 (if P2 picks a pure strategy 𝑗) is ∑ 𝑖 𝑎𝑖𝑗𝑝𝑖 The optimisation problem for P1 is maximise min 𝑗∈{1,…,𝑛} ∑ 𝑖 𝑎𝑖𝑗𝑝𝑖 subject to ∑𝑝𝑖= 1 p ≥0 Equivalently, where e = (1, 1, … , 1)⊺, maximise 𝑣 subject to 𝐴⊺p ≥𝑣e e⊺p = 1 p ≥0 This 𝑣is the minimum value of 𝐴⊺p. P2’s optimisation problem is minimise max 𝑖∈{1,…,𝑚} ∑ 𝑖 𝑎𝑖𝑗𝑞𝑗 subject to ∑𝑞𝑗= 1 q ≥0 or equivalently, minimise 𝑤 subject to 𝐴q ≤𝑤e e⊺q = 1 q ≥0 46 7. Game theory 7.3. Duality of mixed strategy problems The two problems above are duals of each other. Adding slack variables, P2’s problem is minimise 𝑤 subject to 𝐴q + s = 𝑤e e⊺q = 1 q ≥0 s ≥0 The Lagrangian of this problem is 𝐿(𝑤, q, s, 𝛌1, 𝜆2) = 𝑤+ 𝛌⊺ 1(𝐴q + s −𝑤e) −𝜆2(e⊺q −1) = 𝑤(1 −𝛌⊺ 1e) + (𝛌⊺ 1𝐴−𝜆2e⊺)q + 𝛌⊺ 1s + 𝜆2 Thus, 𝚲= {𝛌∶𝛌⊺ 1e = 1, 𝛌⊺ 1𝐴−𝜆2e⊺≥0, 𝛌1 ≥0} When 𝛌∈𝚲, inf 𝐿= 𝜆2 Hence the dual is maximise 𝜆2 subject to 𝛌⊺ 1e = 1 𝛌⊺ 1𝐴≥𝜆2e⊺ 𝛌1 ≥0 Note that 𝛌1 = p and 𝜆2 = 𝑣in the above formulation of P1’s problem.
Theorem. A strategy p is optimal for P1 if there exist q, 𝑣such that • (primal feasibility) 𝐴⊺p ≥𝑣e, e⊺p = 1, p ≥0; • (dual feasibility) 𝐴q ≤𝑣e, e⊺q = 1, q ≥0; and • (complementary slackness) 𝑣= p⊺𝐴q Proof. (p, 𝑣) and (q, 𝑤) are optimal if (𝐴q −𝑤e)⊺p = 0; q⊺(𝐴⊺p −𝑣e) = 0 which gives 𝑣= 𝑤= p⊺𝐴q 47 I. Optimisation 7.4. Finding optimal strategies There are a number of strategies for finding optimal strategies.
(i) We can search for saddle points in the payoff matrix. If such a saddle point is found, a pure strategy aiming for this saddle point is optimal for both players.
(ii) We can search for dominating actions. Suppose there exist 𝑖, 𝑖′ such that 𝑎𝑖𝑗≥𝑎𝑖′𝑗for all 𝑗. Then 𝑖dominates 𝑖′, so P1 will never play 𝑖′ and we can simply drop this row in the matrix. A similar technique can be used to drop columns.
(iii) If these simplification techniques are not sufficient, we can simply solve the linear program using (for instance) the simplex method.
Example. Suppose we have a payoff matrix 𝐴= ( 2 3 4 3 1 1 2 1 3 2 ) First, observe that there is no saddle point. Note that the first row dominates the last row, so we can simplify the payoff matrix to ˜ 𝐴= ( 2 3 4 3 1 1 2 ) P1’s strategy is p = (𝑝, 1 −𝑝, 0), and the optimisation problem is maximise 𝑣 subject to 𝐴⊺p ≥𝑣e e⊺p = 1 p ≥0 which is maximise 𝑣 subject to 2𝑝+ 3(1 −𝑝) ≥𝑣 3𝑝+ (1 −𝑝) ≥𝑣 4𝑝+ 1 2(1 −𝑝) ≥𝑣 0 ≤𝑝≤1 48 7. Game theory and by simplifying, maximise 𝑣 subject to 𝑣≤3 −𝑝 𝑣≤1 + 2𝑝 𝑣≤1 2 + 7 2𝑝 0 ≤𝑝≤1 We can solve this graphically since it is a one-dimensional problem, or use the simplex method. We arrive at the solution p = ( 2 3, 1 3, 0), i.e. 𝑝= 2 3. The payoff is 7 3. Player 2 has the dual optimisation problem, so we can use complementary slackness to compute P2’s strategy. The first two constraints are tight, but the final constraint may not be (since it is zero in P1’s strategy). Therefore 𝑞3 = 0, and P2’s strategy is q = (𝑞, 1 −𝑞, 0). Since the value of the game is 7 3, we have 7 3 = p⊺𝐴q which lets us find 𝑞. Alternatively, we can use complementary slackness. Since 𝑝1, 𝑝2 > 0, the first two constraints in the dual problem must be tight.
2𝑞+ 3(1 −𝑞) = 7 3 ⟹𝑞= 2 3 49 I. Optimisation 8. Network flows 8.1. Minimum cost flow Definition. A directed graph (also known as a digraph) 𝐺consists of a set of vertices and a set of edges; 𝐺= (𝑉, 𝐸). The edges are such that 𝐸⊆𝑉× 𝑉. Each edge (𝑖, 𝑗) can be thought of as an edge pointing from vertex 𝑖to vertex 𝑗. When 𝐸is symmetric (that is, (𝑖, 𝑗) ∈𝐸⟺ (𝑗, 𝑖) ∈𝐸), we call 𝐺an undirected graph.
Definition. Given a graph 𝐺= (𝑉, 𝐸) on 𝑛vertices, we associate to every (𝑖, 𝑗) ∈𝐸the number 𝑥𝑖𝑗. This represents the flow of a quantity from vertex 𝑖to vertex 𝑗. The collection 𝑥of 𝑥𝑖𝑗is called the flow. The flow 𝑥is affected by (i) A vector b ∈ℝ𝑛, where 𝑏𝑖is the amount of flow entering vertex 𝑖from outside the graph. If 𝑏𝑖> 0, then vertex 𝑖is called a source. If 𝑏𝑖< 0, then vertex 𝑖is called a sink.
(ii) The cost matrix 𝑐∈ℝ𝑛×𝑛, which gives the cost 𝑐𝑖𝑗per unit of flow on (𝑖, 𝑗) ∈𝐸.
If the flow along (𝑖, 𝑗) is 𝑥𝑖𝑗, the cost for this flow is 𝑐𝑖𝑗𝑥𝑖𝑗(without the summation convention).
(iii) The lower bound matrix 𝑀and the upper bound matrix 𝑀, which give lower and upper bounds on 𝑥𝑖𝑗. In particular, for all (𝑖, 𝑗) ∈𝐸, we require 𝑚𝑖𝑗≤𝑥𝑖𝑗≤𝑚𝑖𝑗.
Definition. The minimum cost flow is the linear program minimise ∑ (𝑖,𝑗)∈𝐸 𝑐𝑖𝑗𝑥𝑖𝑗 subject to 𝑚𝑖𝑗≤𝑥𝑖𝑗≤𝑚𝑖𝑗 ∀(𝑖, 𝑗) ∈𝐸 𝑏𝑖+ ∑ (𝑗,𝑖)∈𝐸 𝑥𝑗𝑖= ∑ (𝑖,𝑗)∈𝐸 𝑥𝑖𝑗 ∀𝑖∈𝑉 The second constraint is a conservation of flow equation. The amount of flow entering and leaving the vertex must be equal. Note that in order for the problem to be feasible, ∑𝑏𝑖= 0; since the graph has no storage capacity at any vertex, the amount of flow that enters the graph must be the amount of flow that exits. Alternatively, we could prove this by finding the sum of the conservation of flow equations for all 𝑖.
Definition. We can define the incidence matrix 𝐴∶ℝ|𝑉|×|𝐸|. Each column of 𝐴is associated with an edge (𝑖, 𝑗). We define that this column is filled with zeroes, except for +1 at position 𝑖and −1 at position 𝑗. We can now rewrite the conservation of flow equation as 𝐴x = b 8.2. Transport problem The transport problem is a special case of the minimum cost flow problem. Consider 𝑛 suppliers, and 𝑚consumers. Each supplier 𝑖has some capacity 𝑠𝑖for how much of this good 50 8. Network flows they can satisfy, and each consumer 𝑗has some demand 𝑑𝑗that they want to be fulfilled. We will assume that there is exactly as much supply as demand; that is, ∑𝑠𝑖= ∑𝑑𝑗. The cost of transporting one unit of this good from supplier 𝑖to consumer 𝑗is 𝑐𝑖𝑗. For this problem, the graph 𝐺is a bipartite graph; it can be separated into a set of sources and a set of sinks, and the edges are only from the sources to the sinks. The optimisation problem is minimise 𝑛 ∑ 𝑖=1 𝑚 ∑ 𝑗=1 𝑐𝑖𝑗𝑥𝑖𝑗 subject to 𝑚 ∑ 𝑗=1 𝑥𝑖𝑗= 𝑠𝑖 ∀𝑖∈{1, … , 𝑛} 𝑚 ∑ 𝑖=1 𝑥𝑖𝑗= 𝑑𝑗 ∀𝑗∈{1, … , 𝑚} which is a special case of the minimum flow problem.
8.3. Sufficiency of transport problem Theorem. Every minimum cost flow problem with either finite capacities or non-negative capacities can be translated into an equivalent transport problem.
Proof. Consider the minimum cost flow problem on a graph 𝐺= (𝑉, 𝐸). We may assume without loss of generality that 𝑚𝑖𝑗= 0 for all (𝑖, 𝑗) ∈𝐸, because we may write 𝑥𝑖𝑗= 𝑚𝑖𝑗+ ̃ 𝑥𝑖𝑗 where ̃ 𝑥𝑖𝑗> 0. Then the conservation equation becomes ̃ 𝑏𝑖+ ∑ (𝑗,𝑖)∈𝐸 ̃ 𝑥𝑗𝑖= ∑ (𝑖,𝑗)∈𝐸 ̃ 𝑥𝑖𝑗 where ̃ 𝑏𝑖= ∑(𝑗,𝑖)∈𝐸𝑚𝑗𝑖−∑(𝑖,𝑗)∈𝐸𝑚𝑖𝑗. The regional constraints are now 0 ≤ ̃ 𝑥𝑖𝑗≤𝑚𝑖𝑗−𝑚𝑖𝑗 We assume that 𝑚𝑖𝑗≡0 from now. If all the costs are non-negative and a particular capacity is infinite, then we can replace that capacity by a large number e.g. ∑|𝑏𝑖|, which is the maximum amount of flow that could possibly travel along this edge. This transformation does not change the optimal solution. We have now reduced to the case where all capacities are finite.
Now, for each such minimum cost flow problem, we will construct an equivalent transport problem that has the same feasible solutions and the same costs. For each vertex 𝑖, we create a consumer with demand ∑(𝑖,𝑗)∈𝐸𝑚𝑖𝑘−𝑏𝑖. For every edge (𝑖, 𝑗), we create a supplier with supply 𝑚𝑖𝑗. The total supply and the total demand are equal, since ∑𝑖𝑏𝑖= 0. We now 51 I. Optimisation define the cost of moving from (𝑖, 𝑗) →𝑖is zero. We further define the cost of moving from (𝑖, 𝑗) →𝑗is 𝑐𝑖𝑗.
Now, suppose 𝑥𝑖𝑗flows from (𝑖, 𝑗) →𝑗. Then 𝑚𝑖𝑗−𝑥𝑖𝑗flows from (𝑖, 𝑗) →𝑖, since the total incoming and outgoing flow from (𝑖, 𝑗) must balance. Then, since the demand at 𝑖is ∑(𝑖,𝑗)∈𝐸𝑚𝑖𝑘−𝑏𝑖, the total flow into 𝑖satisfies ∑ (𝑖,𝑘)∈𝐸 (𝑚𝑖𝑘−𝑥𝑖𝑘) + ∑ (𝑘,𝑖)∈𝐸 𝑥𝑘𝑖= ∑ (𝑖,𝑗)∈𝐸 𝑚𝑖𝑘−𝑏𝑖 which simplifies to the conservation equation for the minimum cost flow problem. We can easily check that 0 ≤𝑥𝑖𝑗≤𝑚𝑖𝑗. So this mapping between the minimum cost flow problem and the transport problem preserves feasibility of solutions.
It now suffices to show that the costs of the two feasible solutions for the two problems are the same; since then we will have demonstrated a mapping between the two problems. The cost in the transport problem is ∑(𝑖,𝑗)∈𝐸𝑥𝑖𝑗𝑐𝑖𝑗since the edge from (𝑖, 𝑗) to 𝑖has zero cost.
This is identical to the cost in the minimum cost flow problem.
8.4. Optimality conditions for transport problem Recall that for a linear program, there are three optimality conditions: primal feasibility, dual feasibility, and complementary slackness. These have various interpretations in the context of a transport problem.
Theorem. If for some feasible 𝑥we have dual variables 𝛌∈ℝ𝑛(for suppliers) and 𝛍∈ℝ𝑚 (for consumers), such that: (i) 𝜆𝑖+ 𝜇𝑗≤𝑐𝑖𝑗 ∀(𝑖, 𝑗) ∈𝐸; and (ii) (𝑐𝑖𝑗−(𝜆𝑖+ 𝜇𝑗))𝑥𝑖𝑗= 0 ∀(𝑖, 𝑗) ∈𝐸 then 𝑥is an optimal solution.
Proof. The Lagrangian of the transport problem is 𝐿(𝑥, 𝛌, 𝛍) = 𝑛 ∑ 𝑖=1 𝑚 ∑ 𝑗=1 𝑐𝑖𝑗𝑥𝑖𝑗− 𝑛 ∑ 𝑖=1 𝜆𝑖( 𝑚 ∑ 𝑗=1 𝑥𝑖𝑗−𝑠𝑖) − 𝑚 ∑ 𝑗=1 𝜇𝑗( 𝑛 ∑ 𝑖=1 𝑥𝑖𝑗−𝑑𝑗) = 𝑛 ∑ 𝑖=1 𝑚 ∑ 𝑗=1 (𝑐𝑖𝑗−𝜆𝑖−𝜇𝑗)𝑥𝑖𝑗+ 𝑛 ∑ 𝑖=1 𝜆𝑖𝑠𝑖+ 𝑚 ∑ 𝑗=1 𝜇𝑗𝑑𝑗 (𝛌, 𝛍) is dual feasible if 𝜆𝑖+ 𝜇𝑗≤𝑐𝑖𝑗for all 𝑖, 𝑗. We have primal feasibility, dual feasibility, and complementary slackness, so optimality holds.
Note that if 𝛌, 𝛍are optimal, then 𝛌+𝑘, 𝛍−𝑘are also optimal, since (𝜆𝑖+𝑘)+(𝜇𝑗−𝑘) = 𝜆𝑖+𝜇𝑗.
So for simplicity, we can always choose 𝜆1 = 0. This gives 𝑚+ 𝑛−1 remaining Lagrange multipliers.
52 9. The transport algorithm 9. The transport algorithm 9.1. Transportation tableaux Analogously to the simplex tableaux, for the transport problem we can create transportation tableaux. This is a convenient format for storing all relevant information for the transport problem while solving it. The transportation tableau is as follows: 𝜇1 𝜇2 ⋯ 𝜇𝑚 𝜆1 𝜆1 + 𝜇1 𝜆1 + 𝜇2 ⋯ 𝜆1 + 𝜇𝑚𝑠1 𝑥11 𝑐11 𝑥12 𝑐12 𝑥1𝑚𝑐1𝑚 𝜆2 𝜆2 + 𝜇1 𝜆2 + 𝜇2 ⋯ 𝜆2 + 𝜇𝑚𝑠2 𝑥21 𝑐21 𝑥22 𝑐22 𝑥2𝑚𝑐2𝑚 ⋮ ⋮ ⋮ ⋱ ⋮ ⋮ 𝜆𝑛 𝜆𝑛+ 𝜇1 𝜆𝑛+ 𝜇2 ⋯ 𝜆𝑛+ 𝜇𝑚𝑠𝑛 𝑥𝑛1 𝑐𝑛1 𝑥𝑛2 𝑐𝑛2 𝑥𝑛𝑚𝑐𝑛𝑚 𝑑1 𝑑2 ⋯ 𝑑𝑚 Like with the simplex method, we must begin with a basic feasible solution to construct the initial tableau. We can construct such a basic feasible solution by using the first supplier to satisfy the first consumer, then gradually using the next suppliers and consumers as we run out of supply or demand. 𝛌and 𝛍can be deduced by considering complementary slackness.
That is, if 𝑥𝑖𝑗> 0 then 𝜆𝑖+𝜇𝑗= 𝑐𝑖𝑗. For instance, consider this problem with three suppliers and four consumers. The general transportation tableau would look like this: 𝜇1 𝜇2 𝜇3 𝜇4 𝜆1 𝜆1 + 𝜇1 𝜆1 + 𝜇2 𝜆1 + 𝜇3 𝜆1 + 𝜇4 𝑠1 𝑥11 𝑐11 𝑥12 𝑐12 𝑥13 𝑐13 𝑥14 𝑐14 𝜆2 𝜆2 + 𝜇1 𝜆2 + 𝜇2 𝜆2 + 𝜇3 𝜆2 + 𝜇4 𝑠2 𝑥21 𝑐21 𝑥22 𝑐22 𝑥23 𝑐23 𝑥24 𝑐24 𝜆3 𝜆3 + 𝜇1 𝜆3 + 𝜇2 𝜆3 + 𝜇3 𝜆3 + 𝜇4 𝑠3 𝑥31 𝑐31 𝑥32 𝑐32 𝑥33 𝑐33 𝑥34 𝑐34 𝑑1 𝑑2 𝑑3 𝑑4 We will consider the problem given by s = ( 14 10 9 ) ; d = ⎛ ⎜ ⎜ ⎝ 12 5 8 8 ⎞ ⎟ ⎟ ⎠ ; 𝐶= ( 5 3 4 6 2 7 4 1 5 6 2 4 ) 53 I. Optimisation A basic feasible solution is given by 𝑋= ( 12 2 0 0 0 3 7 0 0 0 1 8 ) Complementary slackness gives 𝜆1 + 𝜇1 = 5 𝜆1 + 𝜇2 = 3 𝜆2 + 𝜇2 = 7 𝜆2 + 𝜇3 = 4 𝜆3 + 𝜇3 = 2 𝜆3 + 𝜇4 = 4 This is a system of seven equations for six unknowns. However, since we can always set 𝜆1 = 0, we can reduce this to a system of six equations for six unknowns.
𝜇1 = 5 𝜇2 = 3 𝜆2 + 𝜇2 = 7 𝜆2 + 𝜇3 = 4 𝜆3 + 𝜇3 = 2 𝜆3 + 𝜇4 = 4 Hence, 𝛌= ( 0 4 2 ) ; 𝛍= ⎛ ⎜ ⎜ ⎝ 5 3 0 2 ⎞ ⎟ ⎟ ⎠ Theorem. When constructing a basic feasible solution in this way, the set of edges with strictly positive flow form a connected graph with no cycles. In particular, this graph is a spanning tree 𝑇with exactly 𝑚+ 𝑛−1 edges. This allows us to always construct a system of equations as above.
54 9. The transport algorithm No proof is given.
5 3 0 2 0 5 3 0 2 14 12 5 2 3 0 4 0 6 4 9 7 4 6 10 0 2 3 7 7 4 0 1 2 7 5 2 4 9 0 5 0 6 1 2 8 4 12 5 8 8 9.2. Updating the transportation tableau First, we check if 𝑐𝑖𝑗≥𝜆𝑖+ 𝜇𝑗for all 𝑖, 𝑗. If this is true, then our solution is optimal. In our example 𝑐21 ≥𝜆2 + 𝜇1, so we are not at an optimal solution. If (𝑖, 𝑗) ∉𝑇(where 𝑇is the spanning tree above, i.e. 𝑥𝑖𝑗= 0) and 𝑐𝑖𝑗< 𝜆𝑖+𝜇𝑗, then (𝑖, 𝑗) and the edges of 𝑇form a loop.
We then increase 𝑥𝑖𝑗as much as possible until another flow 𝑥𝑖′𝑗′ is forced to be zero. Then we update the dual variables 𝛌, 𝛍and repeat.
In our example, we will introduce a flow of 𝑥21 = 𝜃. This will change the amount of flow along some nonzero edges. Doing this will force an update 𝑥11 ↦𝑥11 −𝜃due to constrained demand, 𝑥12 ↦𝑥12 + 𝜃due to supply, and 𝑥22 ↦𝑥22 −𝜃due to demand. We can then increase 𝜃to a maximum value of 3. Now, 𝑥= ( 9 5 0 0 3 0 7 0 0 0 1 8 ) We now recalculate 𝛌, 𝛍in the same way as above, which will give 𝛌= ( 0 −3 −5 ) ; 𝛍= ⎛ ⎜ ⎜ ⎝ 5 3 7 9 ⎞ ⎟ ⎟ ⎠ Reconstructing the tableau gives 5 3 7 9 0 5 3 7 9 14 9 5 3 3 0 4 0 6 −3 2 0 4 6 10 3 2 0 7 7 4 0 1 5 0 −2 2 4 9 0 5 0 6 1 2 8 4 12 5 8 8 55 I. Optimisation Once again there is an edge where 𝑐𝑖𝑗< 𝜆𝑖+ 𝜇𝑗, notably (𝑖, 𝑗) = (2, 4), with zero flow. If 𝑥𝑖𝑗= 𝜃, then 𝑥23 ↦𝑥23 −𝜃, 𝑥34 ↦𝑥34 −𝜃, 𝑥33 ↦𝑥33 + 𝜃. We can increase 𝜃only to 7.
Once again, updating the tableau gives 5 3 2 4 0 5 3 2 4 14 9 5 5 3 0 4 0 6 −3 2 0 −1 1 10 3 2 0 7 0 4 7 1 0 5 3 2 4 9 0 5 0 6 8 2 1 4 12 5 8 8 In this current table, all optimality conditions are satisfied. So the solution is 𝑥= ( 9 5 0 0 3 0 0 7 0 0 8 1 ) 56 10. Maximum flow, minimum cut 10. Maximum flow, minimum cut 10.1. Introduction Consider the problem maximise 𝛿 subject to ∑ {𝑗∶(𝑖,𝑗)∈𝐸} 𝑥𝑖𝑗− ∑ {𝑗∶(𝑗,𝑖)∈𝐸} 𝑥𝑗𝑖= 0 for all 𝑖≠1, 𝑖≠𝑛 ∑ {𝑗∶(1,𝑗)∈𝐸} 𝑥1𝑗− ∑ {𝑗∶(𝑗,1)∈𝐸} 𝑥𝑗1 = 𝛿 ∑ {𝑗∶(𝑛,𝑗)∈𝐸} 𝑥𝑛𝑗− ∑ {𝑗∶(𝑗,𝑛)∈𝐸} 𝑥𝑗𝑛= −𝛿 0 ≤𝑥𝑖𝑗≤𝑐𝑖𝑗for all (𝑖, 𝑗) ∈𝐸 This is a graph where vertex 1 is a source and vertex 𝑛is a sink, and 𝛿is the flow from vertex 1 to vertex 𝑛. We want to maximise the total amount of flow on the graph, constrained by a certain maximum flow 𝑐𝑖𝑗on each edge.
10.2. Cuts and flows Definition. A cut of a graph 𝐺= (𝑉, 𝐸) is a partition of its vertices into two sets (𝑆, 𝑉∖𝑆).
The capacity of a cut is given by 𝐶(𝑆) = ∑ {(𝑖,𝑗)∈𝐸∶𝑖∈𝑆,𝑗∈𝑉∖𝑆} 𝑐𝑖𝑗 Theorem. For any feasible flow 𝑥with value 𝛿, then for any cut (𝑆, 𝑉∖𝑆) such that 1 ∈ 𝑆, 𝑛∈𝑉∖𝑆, we have 𝛿≤𝐶(𝑆) Proof. For any sets 𝑋, 𝑌⊆𝑉, we define the function 𝑓 𝑥(𝑋, 𝑌) = ∑ {(𝑖,𝑗)∈𝐸∶𝑖∈𝑋,𝑗∈𝑌} 𝑥𝑖𝑗 Note that 𝑋, 𝑌need not be disjoint. Let (𝑆, 𝑉∖𝑆) be a cut such that 1 ∈𝑆, 𝑛∈𝑉∖𝑆. We have 𝛿= ∑ 𝑖∈𝑆 ( ∑ {𝑗∶(𝑖,𝑗)∈𝐸} 𝑥𝑖𝑗− ∑ {𝑗∶(𝑗,𝑖)∈𝐸} 𝑥𝑗𝑖) 57 I. Optimisation since for 𝑖= 1 the bracket is 𝛿and for all others it is zero. Therefore, 𝛿= 𝑓 𝑥(𝑆, 𝑉) −𝑓 𝑥(𝑉, 𝑆) = 𝑓 𝑥(𝑆, 𝑆) + 𝑓 𝑥(𝑆, 𝑉∖𝑆) −𝑓 𝑥(𝑆, 𝑆) −𝑓 𝑥(𝑉∖𝑆, 𝑆) = 𝑓 𝑥(𝑆, 𝑉∖𝑆) −𝑓 𝑥(𝑉∖𝑆, 𝑆) ⏟⎵ ⎵⏟⎵ ⎵⏟ ≥0 ≤𝑓 𝑥(𝑆, 𝑉∖𝑆) ≤𝐶(𝑆) 10.3. Max-flow min-cut theorem Theorem. Let 𝛿⋆be the value of the maximum flow. Then we have 𝛿⋆= min {𝐶(𝑆)∶1 ∈𝑆, 𝑛∈𝑉∖𝑆} So the value of the maximum flow is equal to the cut of smallest capacity.
Proof. A path 𝑣0, 𝑣1, … , 𝑣𝑘is a sequence of vertices such that every pair of adjacent vertices is connected by an edge, either in the forward direction or in the reverse direction. A path is called an augmenting path if 𝑥𝑣𝑖𝑣𝑖+1 < 𝑐𝑣𝑖𝑣𝑖+1 for all forward edges; 𝑥𝑣𝑖𝑣𝑖+1 > 0 for all backward edges So each forward edge must have remaining capacity, and reverse edges must have some flow. This definition allows us to state that augmenting paths are actually all paths such that altering the flow on all edges in the path can increase the total flow from 1 to 𝑛, while keeping the amount of flow into each vertex the same (excluding the first and last vertices in the path). Therefore, an optimal flow 𝑥cannot have an augmenting path from vertex 1 to vertex 𝑛. Now, suppose 𝑥is optimal. We define a cut: 𝑆= {1} ∪{𝑖∶∃an augmenting path 1 →𝑖} Therefore 𝑛∈𝑉∖𝑆, since there is no augmenting path from 1 to 𝑛. Then, 𝛿⋆= 𝑓 𝑥(𝑆, 𝑉∖𝑆) −𝑓 𝑥(𝑉∖𝑆, 𝑆) But we can show that 𝑓 𝑥(𝑉∖𝑆, 𝑆) = 0, so 𝛿⋆= 𝑓 𝑥(𝑆, 𝑉∖𝑆) = 𝐶(𝑆) as required.
58 10. Maximum flow, minimum cut 10.4. Ford–Fulkerson algorithm The above proof provides a convenient method for finding an optimal flow.
Algorithm 5: Ford–Fulkerson Algorithm Result: Optimal flow 𝑥 start with a feasible flow, such as 𝑥= 0; repeat choose an augmenting path from 1 to 𝑛, and increase the flow along this path as much as possible; until no augmenting paths from 1 to 𝑛; Example. Note that typically such graphs are represented pictorially, but due to difficulty of typesetting abstract diagrams, a matrix is substituted here. Consider a graph given by the capacity matrix 𝐶= 𝑐𝑖𝑗 1 𝑎 𝑏 𝑐 𝑑 𝑛 1 5 5 𝑎 1 4 𝑏 5 𝑐 2 𝑑 5 𝑛 First consider the feasible flow of 𝑥= 0. There exists an augmenting path 1, 𝑎, 𝑏, 𝑛. We increase the flow by 1 in all edges, saturating edge (𝑎, 𝑏), giving the flow matrix 𝑥= 𝑥𝑖𝑗 1 𝑎 𝑏 𝑐 𝑑 𝑛 1 1 𝑎 1 𝑏 1 𝑐 𝑑 𝑛 The path 1, 𝑎, 𝑑, 𝑛is now augmenting. We can increase the flow by 4 to saturate the edge (𝑎, 𝑑): 𝑥= 𝑥𝑖𝑗 1 𝑎 𝑏 𝑐 𝑑 𝑛 1 5 𝑎 1 4 𝑏 1 𝑐 𝑑 4 𝑛 59 I. Optimisation The path 1, 𝑐, 𝑑, 𝑛is augmenting. Increasing by 1, 𝑥= 𝑥𝑖𝑗 1 𝑎 𝑏 𝑐 𝑑 𝑛 1 5 1 𝑎 1 4 𝑏 1 𝑐 1 𝑑 5 𝑛 There are no augmenting paths. We can also check that the cut ({1}, {𝑎, 𝑏, 𝑐, 𝑑, 𝑛}) gives the capacity 6, equivalent to the value at 𝑛so this must be optimal. We now have 𝛿⋆= 6.
Example. Consider a graph given by the capacity matrix 𝐶= 𝑐𝑖𝑗 1 𝑎 𝑏 𝑐 𝑑 𝑛 1 10 10 𝑎 4 2 8 𝑏 10 𝑐 9 𝑑 6 10 𝑛 The path 1, 𝑎, 𝑑, 𝑛is augmenting. We can increase the (currently zero) flow by 8.
𝑥= 𝑥𝑖𝑗 1 𝑎 𝑏 𝑐 𝑑 𝑛 1 8 𝑎 8 𝑏 𝑐 𝑑 8 𝑛 The path 1, 𝑐, 𝑑, 𝑛is also augmenting. We increase the flow by 2.
𝑥= 𝑥𝑖𝑗 1 𝑎 𝑏 𝑐 𝑑 𝑛 1 8 2 𝑎 8 𝑏 𝑐 2 𝑑 10 𝑛 Now, the path 1, 𝑐, 𝑑, 𝑎, 𝑏, 𝑛is augmenting. (𝑏, 𝑎) here is a reverse edge. Here, we can in-60 10. Maximum flow, minimum cut crease the flow by 4. This will decrease the (𝑎, 𝑏) by 4.
𝑥= 𝑥𝑖𝑗 1 𝑎 𝑏 𝑐 𝑑 𝑛 1 8 6 𝑎 4 4 𝑏 4 𝑐 6 𝑑 10 𝑛 The path 1, 𝑎, 𝑑, 𝑏, 𝑛is augmenting, with all forward edges. Increasing by 2, 𝑥= 𝑥𝑖𝑗 1 𝑎 𝑏 𝑐 𝑑 𝑛 1 10 6 𝑎 4 6 𝑏 6 𝑐 6 𝑑 2 10 𝑛 Finally, 1, 𝑐, 𝑑, 𝑏, 𝑛is augmenting, with all forward edges. Increasing by 3, 𝑥= 𝑥𝑖𝑗 1 𝑎 𝑏 𝑐 𝑑 𝑛 1 10 9 𝑎 4 6 𝑏 9 𝑐 9 𝑑 5 10 𝑛 The flow 𝛿is now 19. The cut given by {1, 𝑐} has capacity 19, so we are at the optimum.
10.5. Termination of Ford–Fulkerson If all capacities are integers, then the algorithm will always find the optimal flow. The same argument can be used for rational numbers. At each step, the flow increases by a positive integer value, so after a finite amount of steps it will stop, as the maximum flow is bounded.
10.6. Bipartite matching problem A 𝑘-regular bipartite graph is a graph with 𝑛 2 vertices on the left and 𝑛 2 vertices on the right, where each vertex on both the left and right has exactly 𝑘edges. Suppose we want to match up the vertices on the left and right, such that each pair (𝑎, 𝑏) is joined with an edge that already exists in this graph.
61 I. Optimisation Theorem. Every 𝑘-regular bipartite graph has a perfect matching.
Proof. First, we construct a new graph with extra vertices 1, 𝑛. We construct edges from vertex 1 to all vertices 𝑎on the left, with capacity 1. We then construct edges from all vertices 𝑏on the right to vertex 𝑛, also with capacity 1. The original edges in the graph will be given capacity ∞. Then by using the cut given by 1, 𝛿⋆< 𝑛 2 . We can easily achieve the value 𝛿⋆by attaching a flow 1 𝑘to every edge from the left to the right, and of course sending a flow of 1 along each new edge. So the maximum flow for this new graph is 𝑛 2 .
Now, we know that the Ford–Fulkerson algorithm will terminate, when given the initial flow 𝑥= 0. But with this algorithm, all edge weights are always integers, since all capacities are integral or infinite. The only way for all edge weights to be integer values are when we have a perfect matching. So this algorithm will generate a perfect matching.
62 II. Variational Principles Lectured in Easter 2021 by Dr. M. Dunajski In this course, we solve problems of the form ‘find the optimal function such that…’. Ex-amples include ‘find the shortest path between points 𝐴and 𝐵on surface Σ’, or ‘find the shape of a wire under the influence of gravity between points 𝐴and 𝐵in the plane’. The latter is called the brachistochrone problem, and is of central importance in motivating the subject.
In the same way that turning points of functions can often be located by setting the derivative to zero, optimal functions can be located by setting the functional derivative to zero. This is called the Euler–Lagrange equation, and is a main tool that we use to find solutions to such problems. An application of the Euler–Lagrange equation is Noether’s theorem, which roughly states that any symmetry of a physical system gives rise to a conserved quantity. For example, uniformity of space in the laws of physics shows that momentum is conserved, and uniformity of time shows that energy is conserved.
63 II. Variational Principles Contents 1.
History and motivation . . . . . . . . . . . . . . . . . . . . . . . .
66 1.1.
The brachistochrone problem . . . . . . . . . . . . . . . . . . .
66 1.2.
Geodesics . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
66 1.3.
Calculus of variations . . . . . . . . . . . . . . . . . . . . . . .
66 1.4.
Variational principles . . . . . . . . . . . . . . . . . . . . . . .
67 2.
Calculus for functions on ℝ𝑛. . . . . . . . . . . . . . . . . . . . . .
68 2.1.
Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . .
68 2.2.
Constraints and Lagrange multipliers . . . . . . . . . . . . . . .
69 2.3.
Geometric justification of Lagrange multipliers . . . . . . . . . .
70 3.
Euler–Lagrange equation . . . . . . . . . . . . . . . . . . . . . . .
71 3.1.
Fundamental lemma of calculus of variations . . . . . . . . . .
71 3.2.
Euler–Lagrange equation . . . . . . . . . . . . . . . . . . . . .
72 3.3.
First integral of Euler–Lagrange equation (eliminating 𝑦) . . . . .
73 3.4.
Geodesics on a sphere . . . . . . . . . . . . . . . . . . . . . . .
74 3.5.
First integral of Euler–Lagrange equation (eliminating 𝑥) . . . .
75 3.6.
Solving the brachistochrone problem . . . . . . . . . . . . . . .
75 3.7.
Fermat’s principle . . . . . . . . . . . . . . . . . . . . . . . . .
77 4.
Extensions to the Euler–Lagrange equation . . . . . . . . . . . . .
78 4.1.
Euler–Lagrange equation with constraints . . . . . . . . . . . .
78 4.2.
Dido’s isoparametric problem . . . . . . . . . . . . . . . . . . .
78 4.3.
The Sturm–Liouville problem . . . . . . . . . . . . . . . . . . .
79 4.4.
Multiple dependent variables . . . . . . . . . . . . . . . . . . .
80 4.5.
Geodesics on surfaces . . . . . . . . . . . . . . . . . . . . . . .
80 4.6.
Multiple independent variables . . . . . . . . . . . . . . . . . .
81 4.7.
Potential energy and the Laplace equation . . . . . . . . . . . .
82 4.8.
Minimal surfaces . . . . . . . . . . . . . . . . . . . . . . . . .
82 4.9.
Higher derivatives . . . . . . . . . . . . . . . . . . . . . . . .
84 4.10.
First integral for 𝑛= 2 . . . . . . . . . . . . . . . . . . . . . .
84 4.11.
Principle of least action . . . . . . . . . . . . . . . . . . . . . .
85 4.12.
Central forces . . . . . . . . . . . . . . . . . . . . . . . . . . .
86 4.13.
Configuration space and generalised coordinates . . . . . . . . .
87 5.
Noether’s theorem . . . . . . . . . . . . . . . . . . . . . . . . . . .
88 5.1.
Statement and proof . . . . . . . . . . . . . . . . . . . . . . .
88 5.2.
Conservation of momentum . . . . . . . . . . . . . . . . . . .
89 5.3.
Conservation of angular momentum under central force . . . . .
89 6.
Convexity and the Legendre transform . . . . . . . . . . . . . . . .
90 6.1.
Convex functions . . . . . . . . . . . . . . . . . . . . . . . . .
90 64 6.2.
Conditions for convexity . . . . . . . . . . . . . . . . . . . . .
90 6.3.
Legendre transform . . . . . . . . . . . . . . . . . . . . . . . .
91 6.4.
Applications to thermodynamics . . . . . . . . . . . . . . . . .
92 6.5.
Legendre transform of the Lagrangian . . . . . . . . . . . . . .
93 6.6.
Hamilton’s equations from Euler–Lagrange equation . . . . . . .
94 6.7.
Hamilton’s equations from extremising a functional . . . . . . .
95 7.
Second variations . . . . . . . . . . . . . . . . . . . . . . . . . . . .
96 7.1.
Conditions for local minimisers . . . . . . . . . . . . . . . . . .
96 7.2.
Legendre condition for minimisers . . . . . . . . . . . . . . . .
97 7.3.
Associated eigenvalue problem . . . . . . . . . . . . . . . . . .
98 7.4.
Jacobi accessory condition . . . . . . . . . . . . . . . . . . . .
99 7.5.
Solving the Jacobi condition . . . . . . . . . . . . . . . . . . . 100 65 II. Variational Principles 1. History and motivation 1.1. The brachistochrone problem Consider a particle sliding on a wire under the influence of gravity between two fixed points in the plane. What is the shape of the wire that produces the shortest travel time between the end points, given that the particle starts at rest? This problem is known as the brachisto-chrone problem, an archetypical variational problem. Suppose the end points are labelled 𝐴and 𝐵, where 𝐴is the origin, i.e. (𝑥1, 𝑦1) = (0, 0), and where 𝐵has coordinates (𝑥2, 𝑦2).
Note that 𝑦2 < 0 in order that the particle has sufficient energy to reach the destination. The travel time 𝑇is given by 𝑇= ∫d𝑡= ∫ 𝐵 𝐴 dℓ 𝑣(𝑥, 𝑦) Note that the kinetic energy and the potential energy sum to a constant.
1 2𝑚𝑣2 + 𝑚𝑔𝑦= 𝑚𝑔𝑦1 = 0 ⟹𝑣= √2𝑔√−𝑦 So we must find the function 𝑦that minimises 𝑇[𝑦] = 1 √2𝑔 ∫ 𝑥2 0 √1 + 𝑦′2 √−𝑦 d𝑥 subject to 𝑦0 = 0, 𝑦(𝑥2) = 𝑦2. This problem’s solution will be explored in a later lec-ture.
1.2. Geodesics A geodesic is the shortest path 𝛾between two points on a surface Σ, assuming such a path exists. Initially, let Σ = ℝ2. On this plane, the Pythagorean theorem for measuring dis-tances holds. Using a Cartesian coordinate system, we can say that a point 𝐴has coordinates (𝑥1, 𝑦1), and a point 𝐵has coordinates (𝑥2, 𝑦2). The distance from 𝐴to 𝐵along any path 𝛾 can be computed using a line integral.
𝐷[𝑦] = ∫ 𝐵 𝐴 dℓ= ∫ 𝑥2 𝑥1 √1 + 𝑦′2 d𝑥 In this case, we have defined 𝑦as a function of 𝑥, and we seek to minimise 𝐷by varying the path 𝛾on which we are moving.
1.3. Calculus of variations A variational problem involves minimising an object of the form 𝐹[𝑦] = ∫ 𝑥2 𝑥1 𝑓(𝑥, 𝑦(𝑥), 𝑦′(𝑥)) d𝑥 66 1. History and motivation subject to fixed values of 𝑦at the end points. We call such an 𝐹a functional; it is a function on the space of functions. Calculus applied to functionals is called the calculus of variations; we would like to find minima and maxima of functionals. In order to talk about functionals rig-orously, we must define first the space of functions we are operating on; analogously to how we must define the domain of a function we are analysing when dealing with real or com-plex analysis. We write 𝐶(ℝ) for the space of continuous functions on ℝ, and 𝐶𝑘(ℝ) for the space of functions with continuous 𝑘th derivatives on ℝ. Sometimes, the notation 𝐶𝑘 (𝛼,𝛽)(ℝ) is used to denote 𝐶𝑘(ℝ) such that 𝑓(𝛼) and 𝑓(𝛽) are fixed, typically fixed to zero.
1.4. Variational principles We can now define what variational principles are: they are such principles where laws follow from finding the minima or maxima of functionals. An introductory example is Fermat’s principle, which states that light that travels between two points takes the path which requires the least travel time. There is also the principle of least action. Consider a particle moving under some potential 𝑉(x), and let 𝑇= 1 2𝑚| ̇ x|2 be its kinetic energy. We can define 𝑆[𝛾] = ∫ 𝑡2 𝑡1 (𝑇−𝑉) d𝑡 where 𝛾represents the path along which the particle travels. The left hand side 𝑆[𝛾] is called the action, and the principle of least action states that the action is minimised along paths of motion. Then, Newton’s laws of motion should follow from this principle by minimising action.
67 II. Variational Principles 2. Calculus for functions on ℝ𝑛 2.1. Introduction Let 𝑓∈𝐶2(ℝ𝑛), so 𝑓∶ℝ𝑛→ℝwith all continuous second partial derivatives. We say that the point a ∈ℝ𝑛is stationary if ∇𝑓(a) = 0 Consider a Taylor series expansion near a stationary point.
𝑓(x) = 𝑓(a) + 1 2(𝑥𝑖−𝑎𝑖)(𝑥𝑗−𝑎𝑗) 𝜕2 𝑖𝑗𝑓| | |a + 𝑂(‖x −a‖2) The Hessian matrix is defined as 𝐻𝑖𝑗= 𝜕𝑖𝜕𝑗𝑓= 𝐻𝑗𝑖, where 𝜕𝑖≡ 𝜕 𝜕𝑥𝑖. For convenience, we will shift the origin to let a = 0. The Hessian, evaluated at 0, written 𝐻(0), is a real symmetric matrix and hence can be diagonalised using an orthogonal transformation.
𝐻′ = 𝑅⊺𝐻(0)𝑅= ⎛ ⎜ ⎜ ⎝ 𝜆1 0 ⋯ 0 0 𝜆2 ⋯ 0 ⋮ ⋮ ⋱ ⋮ 0 0 ⋯ 𝜆𝑛 ⎞ ⎟ ⎟ ⎠ Then 𝑓(x′) −𝑓(0) = 1 2 ∑𝜆𝑖(𝑥′ 𝑖)2 + 𝑂(‖x‖2) We can characterise the stationary point using the eigenvalues of the Hessian.
(i) If all 𝜆𝑖> 0, then 𝑓(x′) > 𝑓(0) so 𝑓(x′) is a local minimum.
(ii) If all 𝜆𝑖< 0, then 𝑓(x′) < 𝑓(0) so 𝑓(x′) is a local maxmimum.
(iii) If the eigenvalues have mixed signs, this is a saddle point. 𝑓(x′) increases in some directions, but decreases in other directions.
(iv) If some eigenvalues are zero, we must consider higher-order terms of the Taylor ex-pansion.
When 𝑛= 2, this is a special case. We can compute properties of the eigenvalues using the trace and determinant of the matrix.
det 𝐻= 𝜆1𝜆2; tr 𝐻= 𝜆1 + 𝜆2 (i) If det 𝐻> 0, tr 𝐻> 0 then we have a local minimum.
(ii) If det 𝐻> 0, tr 𝐻< 0 then we have a local maximum.
(iii) If det 𝐻< 0 then we have a saddle point.
(iv) If det 𝐻= 0 we need to consider higher-order terms.
68 2. Calculus for functions on ℝ𝑛 Note that if 𝑓∶𝐷→ℝwhere 𝐷⊂ℝ𝑛, it is possible that we have a local maximum which is not the global maximum, if such a global maximum actually lies on the boundary and is not a stationary point.
Now, let us suppose that 𝑓is harmonic, i.e. ∇2𝑓(x) = 0 on 𝐷⊂ℝ2. Hence, tr 𝐻= 0 which implies that if there exists a turning point it is a saddle point. The minimum or maximum of a harmonic function must therefore occur on the boundary.
Example. Let 𝑓(𝑥, 𝑦) = 𝑥3 + 𝑦3 −3𝑥𝑦 ∇𝑓(x) = (3𝑥2 −3𝑦 3𝑦2 −3𝑥) = (0 0) ⟹(𝑥 𝑦) = (0 0) or (1 1) The Hessian is 𝐻= (6𝑥 −3 −3 6𝑦) ⟹𝐻(0) = ( 0 −3 −3 0 ) ; 𝐻(1 1) = ( 6 −3 −3 6 ) The determinant is negative at zero, giving us a saddle point. At the other point, the determ-inant is positive and the trace is positive, giving a local minimum.
2.2. Constraints and Lagrange multipliers Example. Find the circle centered at (0, 0) with smallest radius that intersects the parabola 𝑦= 𝑥2 −1. There are essentially two approaches.
• First, we consider the ‘direct’ method. We solve the constraints directly, which in this case means solving the equations 𝑓= 𝑥2 + 𝑦2 𝑦= 𝑥2 −1 for minimal 𝑓. This gives 𝑓= 𝑥2 + (𝑥2 −1)2 = 𝑥4 −𝑥2 + 1 Then by setting 𝜕𝑥𝑓= 0 we have 4𝑥3 −2𝑥= 0 ⟹𝑥∈{0, 1 √2 , −1 √2 } which gives 𝑥= ±1 √2 ⟹𝑦= −1 2 ; 𝑟= √3 2 The other solution for 𝑥yields a larger radius. This method works fine for simple prob-lems like this where the constraints are solvable. Therefore, we present an alternative method that works in the more general case.
69 II. Variational Principles • This method uses ‘Lagrange multipliers’. We define a new function ℎ(𝑥, 𝑦, 𝜆) = 𝑓(𝑥, 𝑦) −𝜆𝑔(𝑥, 𝑦) where 𝑔(𝑥, 𝑦) is defined such that 𝑔= 0 is the constraint. 𝜆is called the Lagrange multiplier. In this example, ℎ(𝑥, 𝑦, 𝜆) = 𝑥2 + 𝑦2 −𝜆(𝑦−𝑥2 + 1) We now extremise ℎover all free variables without constraints.
∇ℎ= ( 𝜕ℎ/𝜕𝑥 𝜕ℎ/𝜕𝑦 𝜕ℎ/𝜕𝜆 ) = ( 2𝑥+ 2𝜆𝑥 2𝑦−𝜆 𝑦−𝑥2 + 1 ) Solving ∇ℎ= 0, we have 2𝑥+ 4𝑥𝑦= 0 ⟹𝑥= 0 or 𝑦= −1 2 and the same results follow as before by substitution.
2.3. Geometric justification of Lagrange multipliers Consider a curve given by 𝑔= 0. At each point on this curve, there is a normal to the curve of gradient ∇𝑔. In particular, ∇𝑔is perpendicular to 𝑔= 0. The function 𝑓has gradient perpendicular to the function 𝑓= 𝑐for some constant 𝑐. So at the extremum, ∇𝑓∝∇𝑔, so ∇𝑓−𝜆𝑔= 0 for some 𝜆. This guides the creation of the new function ℎ, for which we can optimise without constraints. This same reasoning generalises to functions in higher dimensions and with multiple constraints.
70 3. Euler–Lagrange equation 3. Euler–Lagrange equation 3.1. Fundamental lemma of calculus of variations Consider again the functional 𝐹[𝑦] = ∫ 𝛽 𝛼 𝑓(𝑥, 𝑦, 𝑦′) d𝑥 where 𝑓is given, and 𝑓(𝛼, ⋅, ⋅) and 𝑓(𝛽, ⋅, ⋅) are fixed. Consider a small perturbation 𝑦↦𝑦+ 𝜀𝜂(𝑥); 𝜂(𝛼) = 𝜂(𝛽) = 0 In order to compute the functional for this new function, we first need an additional lemma.
Lemma (Fundamental lemma of calculus of variations). If 𝑔∶[𝛼, 𝛽] →ℝis continuous on this interval, and is such that ∀𝜂continuous, 𝜂(𝛼) = 𝜂(𝛽) = 0, ∫ 𝛽 𝛼 𝑔(𝑥)𝜂(𝑥) d𝑥= 0 Then ∀𝑥∈(𝛼, 𝛽), 𝑔(𝑥) ≡0 Proof. Suppose that there exists a value 𝑥∈(𝛼, 𝛽) such that 𝑔(𝑥) ≠0. Without loss of generality suppose that this value is positive. Then, by continuity, there exists a sub-interval [𝑥1, 𝑥2] ⊂(𝛼, 𝛽) where 𝑔(𝑥) > 𝑐for some positive real 𝑐in this sub-interval. So we will construct an 𝜂such that 𝜂> 0 in [𝑥1, 𝑥2] and 𝜂= 0 outside this interval, for example 𝜂(𝑥) = {(𝑥−𝑥1)(𝑥2 −𝑥) 𝑥∈[𝑥1, 𝑥2] 0 otherwise Then the integrand is non-negative everywhere, and is lower bounded by a positive number: ∫ 𝛽 𝛼 𝑔(𝑥)𝜂(𝑥) > 𝑐∫ 𝑥2 𝑥1 (𝑥−𝑥1)(𝑥2 −𝑥) d𝑥> 0 So this leads to a contradiction.
Remark. We call such an 𝜂function a ‘bump function’. In general it is possible to construct a 𝐶𝑘bump function, e.g.
𝜂= {[(𝑥−𝑥1)(𝑥2 −𝑥)]𝑘+1 𝑥∈[𝑥1, 𝑥2] 0 otherwise 71 II. Variational Principles 3.2. Euler–Lagrange equation Now, we can evaluate the original functional. Using a Taylor expansion, 𝐹[𝑦+ 𝜀𝜂] = ∫ 𝛽 𝛼 𝑓(𝑥, 𝑦+ 𝜀𝜂, 𝑦′ + 𝜀𝜂′) = 𝐹[𝑦] + 𝜀∫ 𝛽 𝛼 (𝜕𝑓 𝜕𝑦𝜂+ 𝜕𝑓 𝜕𝑦′ 𝜂′) d𝑥+ 𝑂(𝜀2) For an extremum, d𝐹 d𝜀 | | |𝜀=0 = 0 So we want the first order term to vanish, so 𝜀∫ 𝛽 𝛼 (𝜕𝑓 𝜕𝑦𝜂+ 𝜕𝑓 𝜕𝑦′ 𝜂′) d𝑥= 0 Integrating by parts, we have 0 = ∫ 𝛽 𝛼 (𝜕𝑓 𝜕𝑦𝜂−d d𝑥( 𝜕𝑓 𝜕𝑦′ 𝜂)) d𝑥+ [ 𝜕𝑓 𝜕𝑦′ 𝜂] 𝛽 𝛼 = ∫ 𝛽 𝛼 (𝜕𝑓 𝜕𝑦𝜂−d d𝑥( 𝜕𝑓 𝜕𝑦′ 𝜂)) d𝑥 = ∫ 𝛽 𝛼 (𝜕𝑓 𝜕𝑦−d d𝑥( 𝜕𝑓 𝜕𝑦′ )) ⏟⎵ ⎵ ⎵ ⎵⏟⎵ ⎵ ⎵ ⎵⏟ 𝑔(𝑥) 𝜂d𝑥 We can apply the lemma above, showing that a necessary condition for the optimum is d d𝑥( 𝜕𝑓 𝜕𝑦′ ) −𝜕𝑓 𝜕𝑦= 0 This is the Euler–Lagrange equation.
Remark. Note that • This can be seen as a second-order differential equation for 𝑦(𝑥) with boundary con-ditions at 𝛼and 𝛽.
• The left hand side of the Euler–Lagrange equation is called a ‘functional derivative’ of 𝑦, and is written 𝛿𝐹[𝑦] 𝛿𝑦(𝑥) Sometimes, the notation 𝛿𝑦= 𝜀𝜂(𝑥) 72 3. Euler–Lagrange equation is used, but is not used in this course. Note that in this notation, 𝐹[𝑦+ 𝛿𝑦] = 𝐹[𝑦] + 𝛿𝐹[𝑦]; 𝛿𝐹[𝑦] = ∫ 𝛽 𝛼 [𝛿𝐹[𝑦] 𝛿𝑦(𝑥)𝛿𝑦(𝑥)] d𝑥 • Other boundary conditions, such as 𝜕𝑓 𝜕𝑦′ | | |𝛼,𝛽 can be used.
• Note that when computing the derivatives, we regard 𝑥, 𝑦, 𝑦′ as independent; 𝜕𝑓 𝜕𝑦= 𝜕𝑓 𝜕𝑦 | | |𝑥,𝑦′ We can also compute a total derivative, for instance d d𝑥= 𝜕 𝜕𝑥+ 𝜕 𝜕𝑦𝑦′ + 𝜕 𝜕𝑦′ 𝑦″ Note that these give different results. As an example, let 𝑓(𝑥, 𝑦, 𝑦′) = 𝑥[(𝑦′)2 −𝑦2].
Then 𝜕𝑓 𝜕𝑥= (𝑦′)2 −𝑦2; 𝜕𝑓 𝜕𝑦= −2𝑥𝑦; 𝜕𝑓 𝜕𝑦′ = 2𝑥𝑦′ Hence d𝑓 d𝑥= (𝑦′)2 −𝑦2 −2𝑥𝑦𝑦′ + 2𝑥𝑦′𝑦″ 3.3. First integral of Euler–Lagrange equation (eliminating 𝑦) In some cases, we can integrate the Euler–Lagrange equation to give a first-order ordinary differential equation. Suppose 𝑓does not explicitly depend on 𝑦. Then 𝜕𝑓 𝜕𝑦= 0 ⟹ d d𝑥( 𝜕𝑓 𝜕𝑦′ ) = 0 Hence, 𝜕𝑓 𝜕𝑦′ = 𝑐; 𝑐∈ℝ Example. Consider geodesics on ℝ2; we want to find curves on which the length is minim-ised.
𝐹[𝑦] = ∫ 𝛽 𝛼 √d𝑥2 + d𝑦2 = ∫ 𝛽 𝛼√1 + d𝑦 d𝑥 2 ⏟ ⎵ ⎵ ⏟ ⎵ ⎵ ⏟ 𝑓(𝑦′) d𝑥 We can apply this ‘first integral’ form of the Euler–Lagrange equation to get 𝑦′ √1 + (𝑦′)2 = 𝑐 Hence 𝑦′ is a constant, so let 𝑦′ = 𝑚for 𝑚∈ℝ. Hence 𝑦= 𝑚𝑥+ 𝑐.
73 II. Variational Principles 3.4. Geodesics on a sphere Consider the unit sphere 𝑆2 ⊂ℝ3, and two points 𝐴, 𝐵∈𝑆2 which we wish to connect by a path of minimal length, where the path is constrained to the sphere. We will parametrise the sphere with spherical polar coordinates: 𝑥= sin 𝜃sin 𝜙 𝑦= sin 𝜃cos 𝜙 𝑧= cos 𝜃 where 𝜃∈[0, 𝜋]; 𝜙∈[0, 2𝜋]. We can calculate the length of a path using the Pythagorean theorem: d𝑠2 = d𝑥2 + d𝑦2 + d𝑧2 = d𝜃2 + sin2 𝜃d𝜙2 We will parametrise the path by thinking of 𝜙as a function of 𝜃. This gives d𝑠= √1 + sin2 𝜃(𝜙′)2 d𝜃 We wish to extremise the functional 𝐹, given by 𝐹[𝜙] = ∫ 𝜃2=𝛽 𝜃1=𝛼 d𝑠= ∫ 𝜃2 𝜃1 d𝑠= √1 + sin2 𝜃(𝜙′)2 d𝜃 The integrand does not depend on 𝜙but only on its derivative; so d𝑓 d𝜙= 0. Using the first integral form of the Euler–Lagrange equation, we have 𝜕𝑓 𝜕𝜙′ = 𝑘 Now, we have sin2 𝜃𝜙′ √1 + sin2 𝜃(𝜙′)2 = 𝑘 sin4 𝜃(𝜙′)2 = 𝑘2(1 + sin2 𝜃(𝜙′)2) (𝜙′)2 = 𝑘2 sin2 𝜃(sin2 𝜃−𝑘2) d𝜙 d𝜃= ± √ 𝑘2 sin2 𝜃(sin2 𝜃−𝑘2) 𝜙= ± ∫ 𝑘d𝜃 sin 𝜃√sin2 𝜃−𝑘2 74 3. Euler–Lagrange equation The two solutions correspond to the two directions in which we can trace the path. We then can arrive at ±√1 −𝑘2 𝑘 cos(𝜙−𝜙0) = cot 𝜃 We will be able to see that this corresponds to a great circle; that is, the intersection of a plane through the origin with the sphere. We will show later that geodesics on a sphere are only segments of a great circle.
3.5. First integral of Euler–Lagrange equation (eliminating 𝑥) For any 𝑓(𝑥, 𝑦, 𝑦′), consider the quantity d d𝑥(𝑓−𝑦′ 𝜕𝑓 𝜕𝑦′ ) This is exactly d d𝑥(𝑓−𝑦′ 𝜕𝑓 𝜕𝑦′ ) = 𝜕𝑓 𝜕𝑥+ 𝑦′ 𝜕𝑓 𝜕𝑦+ 𝑦″ 𝜕𝑓 𝜕𝑦′ −𝑦″ 𝜕𝑓 𝜕𝑦′ −𝑦′ d d𝑥( 𝜕𝑓 𝜕𝑦′ ) = 𝜕𝑓 𝜕𝑥+ 𝑦′ 𝜕𝑓 𝜕𝑦−𝑦′ d d𝑥( 𝜕𝑓 𝜕𝑦′ ) = 𝑦′ (𝜕𝑓 𝜕𝑦−d d𝑥 𝜕𝑓 𝜕𝑦′ ) ⏟⎵⎵⎵⏟⎵⎵⎵⏟ zero by Euler–Lagrange +𝜕𝑓 𝜕𝑥 = 𝜕𝑓 𝜕𝑥 So, in the case that 𝑓does not depend explicitly on 𝑥(that is, 𝜕𝑓 𝜕𝑥≡0), then we have another first integral condition from the Euler–Lagrange equation: 𝑓−𝑦′ 𝜕𝑓 𝜕𝑦′ = constant 3.6. Solving the brachistochrone problem Consider a curve in the plane with a fixed endpoint at the origin and another fixed endpoint at 𝑥= 𝛽. We want to find a path such that the time taken for a particle to travel along this curve is minimised. We previously computed that the travel time is given by 𝐹[𝑦] = 1 √2𝑔 ∫ 𝛽 0 √1 + (𝑦′)2 √−𝑦 d𝑥 75 II. Variational Principles This does not depend on 𝑥, so we can write (ignoring the 1 √2𝑔factor) 𝑓−𝑦′ 𝜕𝑓 𝜕𝑦′ = √1 + (𝑦′)2 √−𝑦 −𝑦′ 𝑦′ √1 + (𝑦′)2√−𝑦 = 𝑘 This gives 1 √1 + (𝑦′)2 = 𝑘√−𝑦 𝑦′ = ±√1 + 𝑘2𝑦2 𝑘√−𝑦 𝑥= ±𝑘∫ √−𝑦 √1 + 𝑘2𝑦 d𝑦 We will parametrise further: 𝑦= −1 𝑘2 sin2 𝜃 2 ⟹d𝑦= −1 𝑘2 sin 𝜃 2 cos 𝜃 2 Hence, 𝑥= ±𝑘∫−1 𝑘2 sin2 𝜃 2 cos 𝜃 2 √1 −sin2 𝜃 2 d𝜃 = ∓1 2𝑘2 ∫(1 −cos 𝜃) d𝜃 = ∓1 2𝑘2 (𝜃−sin 𝜃) + 𝑐 The initial condition at (0, 0) gives 𝜃0 = 0 ⟹𝑐= 0 Taking the positive solution, we have 𝑥= 𝜃−sin 𝜃 2𝑘2 𝑦= −1 𝑘2 sin2 𝜃 2 This can be shown to be a parametrised equation of a cycloid.
76 3. Euler–Lagrange equation 3.7. Fermat’s principle Fermat’s principle states that as light travels between two points, it takes the path of least time. Let a ray of light be represented by a path 𝑦(𝑥). The speed of light is given by a function 𝑐(𝑥, 𝑦) since it depends on the material it is in. Then the time taken is 𝐹[𝑦] = ∫dℓ 𝑐= ∫ 𝛽 𝛼 √1 + (𝑦′)2 𝑐(𝑥, 𝑦) d𝑥 In this general form, 𝑓depends on 𝑥, 𝑦, 𝑦′. Now, let us assume 𝑐depends only on 𝑥and not on 𝑦. Then we can use a first integral form to get 𝜕𝑓 𝜕𝑦′ = constant This gives 𝑦′ 𝑐(𝑥)√1 + (𝑦′)2 = constant Suppose that at 𝛼, the light ray’s path has an angle 𝜃1 with the 𝑥-axis, and at 𝛽the angle is 𝜃2. Note that 𝜃1 = arctan 𝑦′|𝛼and the corresponding result for 𝛽. Then, sin 𝜃1 𝑐(𝑥1) = sin 𝜃 𝑐(𝑥) This is known as Snell’s law.
Suppose we have a material in which 𝑐increases with 𝑥. In such a material, we then have that 𝜃increases with 𝑥. In a material in which 𝑐decreases as 𝑥increases, 𝜃naturally de-creases.
Now, suppose we have a slow material with 𝑐= 𝑐𝑆and a fast material with 𝑐= 𝑐𝐹adjacent to each other. We might like to find the path that light takes in its path between points that cross the material boundary. Snell’s law can be used to determine that the ratio between the sine of the angle and the speed of light remains constant along the light ray’s path.
77 II. Variational Principles 4. Extensions to the Euler–Lagrange equation 4.1. Euler–Lagrange equation with constraints Given a functional 𝐹[𝑦] = ∫ 𝛽 𝛼𝑓(𝑥, 𝑦, 𝑦′) d𝑥, we would like to extremise 𝐹subject to 𝐺[𝑦] = ∫ 𝛽 𝛼𝑔(𝑥, 𝑦, 𝑦′) d𝑥= 𝑘for some constant 𝑘. We can use the method of Lagrange multipliers.
Instead of extremising 𝐹, we will extremise Φ[𝑦; 𝜆] = 𝐹[𝑦] −𝜆𝐺[𝑦] Thus, we replace 𝑓in the Euler–Lagrange equation with 𝑓−𝜆𝑔, giving d d𝑥( 𝜕 𝜕𝑦′ (𝑓−𝜆𝑔)) −𝜕 𝜕𝑦(𝑓−𝜆𝑔) = 0 4.2. Dido’s isoparametric problem Given a fixed perimeter, we wish to find the simple and closed plane curve which maximises the enclosed area. We can restrict ourselves to convex curves. This is because any concave curve can be transformed into a convex curve with greater area and equal perimeter, by reflecting the non-convex region. We will parametrise the curve in ℝ2 by letting the minimal and maximal values of 𝑥be 𝛼, 𝛽. Then, as we trace out the curve, 𝑥monotonically increases from 𝛼to 𝛽, and then monotonically decreases as we return from 𝛽to 𝛼. This induces two functions 𝑦1, 𝑦2 on (𝛼, 𝛽) where 𝑦2 > 𝑦1. The infinitesimal area is given by d𝐴= (𝑦2−𝑦1) d𝑥.
Thus, the area functional is given by 𝐴[𝑦] = ∫ 𝛽 𝛼 (𝑦2(𝑥) −𝑦1(𝑥)) d𝑥= ∮ 𝐶 𝑦(𝑥) d𝑦 The constraint functional is 𝐿[𝑦] = ∮ 𝐶 dℓ= ∮ 𝐶√1 + (𝑦′)2 d𝑥= 𝐿 where 𝐿is the fixed perimeter. Using Lagrange multipliers, we can define ℎ= 𝑦−𝜆√1 + (𝑦′)2 Note that we do not need to consider a boundary term in the derivation of the Euler–Lagrange equation, since the curve has no boundary. Using a first integral form of the Euler–Lagrange equation on ℎ, we have 𝑘= ℎ−𝑦′ dℎ d𝑦′ = 𝑦−𝜆√1 + (𝑦′)2 + 𝑦′𝜆 𝑦′ √1 + (𝑦′)2 = 𝑦− 𝜆 √1 + (𝑦′)2 for some constant 𝑘. Hence, (𝑦′)2 = 𝜆2 (𝑦−𝑘)2 −1 78 4. Extensions to the Euler–Lagrange equation A solution here is the circle of radius 𝜆: (𝑥−𝑥0)2 + (𝑦−𝑦0)2 = 𝜆2 Here, 𝐿= 2𝜋𝜆so we can write the solution in terms of 𝐿instead, giving (𝑥−𝑥0)2 + (𝑦−𝑦0)2 = 𝐿 2 4𝜋2 4.3. The Sturm–Liouville problem Let 𝜌(𝑥), 𝜎(𝑥) be defined for 𝑥∈[𝛼, 𝛽], and let 𝜌(𝑥) > 0 on this interval. Consider the functional 𝐹[𝑦] = ∫ 𝛽 𝛼 [𝜌(𝑦′)2 + 𝜎𝑦2] d𝑥 Let us extremise 𝐹subject to the constraint 𝐺[𝑦] = ∫ 𝛽 𝛼 𝑦2 d𝑥= 1 We have Φ[𝑦; 𝜆]𝐹[𝑦] −𝜆(𝐺[𝑦] −1) This induces the integrand ℎ= 𝜌(𝑦′)2 + 𝜎𝑦2 −𝜆(𝑦2 − 1 𝛽−𝛼) We consider the derivatives for the Euler–Lagrange equation: 𝜕ℎ 𝜕𝑦′ = 2𝜌𝑦′; 𝜕ℎ 𝜕𝑦= 2𝜎𝑦−2𝜆𝑦 Hence, −d d𝑥(𝜌𝑦′) + 𝜎𝑦= 𝜆𝑦 We can write this as ℒ(𝑦) = 𝜆𝑦, where the ℒis known as the Sturm–Liouville operator. This is essentially an eigenvalue problem, since ℒis a linear operator. For example, if 𝜌= 1, this eigenvalue problem is exactly the time-independent Schrödinger equation where 𝜎is the quantum-mechanical potential.
Suppose 𝜎> 0. Then the functional 𝐹[𝑦] is also greater than zero. Then, the positive min-imum of 𝐹(if it exists) is the lowest eigenvalue.
Proof. Using the result from the Euler–Lagrange equation, we can multiply by 𝑦and integ-rate by parts giving −𝑦d d𝑥(𝜌𝑦′) + 𝜎𝑦2 = 𝜆𝑦2 𝐹[𝑦] −[𝑦𝑦′𝜌]𝛽 𝛼 ⏟ ⎵ ⏟ ⎵ ⏟ zero = 𝜆𝐺[𝑦] ⏟ one Thus, the lowest eigenvalue is the minimum of 𝐹[𝑦]/𝐺[𝑦].
79 II. Variational Principles 4.4. Multiple dependent variables Suppose we have some vector y(𝑥) = (𝑦1(𝑥), 𝑦2(𝑥), … , 𝑦𝑛(𝑥)) Suppose we want to extremise the functional 𝐹[y] = ∫ 𝛽 𝛼 𝑓(𝑥, 𝑦1, … , 𝑦𝑛, 𝑦′ 1, … , 𝑦′ 𝑛) d𝑥 If there is some critical point y, we perturb by a small amount 𝜀𝛈= 𝜀(𝜂1(𝑥), … , 𝜂𝑛(𝑥)), where 𝛈(𝛼) = 𝛈(𝛽) = 0. Following the derivation of the one-dimensional Euler–Lagrange equation, we can deduce that 𝐹[y + 𝜀𝛈] −𝐹[y] = ∫ 𝛽 𝛼 𝑛 ∑ 𝑖=1 𝜂𝑖( d d𝑥 𝜕𝑓 𝜕𝑦′ 𝑖 −𝜕𝑓 𝜕𝑦𝑖 ) d𝑥+ boundary term + 𝑂(𝜀2) We can apply the fundamental lemma, choosing 𝜂𝑖in a useful way, we can show that a necessary condition for a critical point is d d𝑥 𝜕𝑓 𝜕𝑦′ 𝑖 −𝜕𝑓 𝜕𝑦𝑖 = 0 for all 𝑖. This is a second-order system of 𝑛ODEs that we can solve. If 𝑓does not depend on one of the 𝑦𝑖, then we have a first integral form for this particular equation. In particular, if 𝜕𝑓 𝜕𝑦𝑗≡0 then 𝜕𝑓 𝜕𝑦′ 𝑗 = constant. If 𝑓does not depend on 𝑥, then we have 𝑓−∑𝑖𝑦′ 𝑖 𝜕𝑓 𝜕𝑦′ 𝑖 = constant.
4.5. Geodesics on surfaces Consider a surface Σ in ℝ3, given by Σ = {x∶𝑔(x) = 0} Consider two points 𝐴, 𝐵on Σ. What are the geodesics (the shortest paths on the surface) between the two points, if one exists at all? Consider a parametrisation of such a path given by 𝑡∈[0, 1] where 𝐴= x(0), 𝐵= x(1). We wish to extremise Φ[x, 𝜆] = ∫ 1 0 {√̇ 𝑥2 + ̇ 𝑦2 + ̇ 𝑧2 −𝜆(𝑡)𝑔(x)} d𝑡 The Lagrange multiplier, a function of 𝑡, since we want the entire curve (for all 𝑡) to lie on Σ. We substitute the integrand ℎin the Euler–Lagrange equation. Considering the variation with respect to 𝜆, we have d d𝑡 𝜕ℎ 𝜕̇ 𝜆 −𝜕ℎ 𝜕𝜆= 0 80 4. Extensions to the Euler–Lagrange equation But ℎdoes not depend on ̇ 𝜆, hence 𝜕ℎ 𝜕𝜆= 0, giving 𝑔(x) = 0 for all x. Considering the variation with respect to 𝑥𝑖, we have d d𝑡 𝜕ℎ 𝜕̇ 𝑥𝑖 −𝜕ℎ 𝜕𝑥𝑖 = 0 Hence d d𝑡( ̇ 𝑥𝑖 √̇ 𝑥2 1 + ̇ 𝑥2 2 + ̇ 𝑥2 3 ) + 𝜆𝜕𝑔 𝜕𝑥𝑖 = 0 We could alternatively solve the constraint 𝑔= 0, and parametrise the surface according to this solution.
4.6. Multiple independent variables In the most general case, we may have multiple independent variables in a variational prob-lem. This converts the Euler–Lagrange equation into a partial differential equation. Suppose 𝜙∶ℝ𝑛→ℝ𝑚. If 𝑛= 3, for example, we have 𝐹[𝜙] = ∭ 𝒟 𝑓(𝑥, 𝑦, 𝑧 ⏟ independent , 𝜙, 𝜙𝑥, 𝜙𝑦, 𝜙𝑧) d𝑥d𝑦d𝑧 where 𝒟⊂ℝ3, and 𝜙𝑥𝑖≔𝜕𝜙/𝜕𝑥𝑖. Suppose there exists some extremum 𝜙, and consider a small variation 𝜙↦𝜙(𝑥, 𝑦, 𝑧) + 𝜀𝜂(𝑥, 𝑦, 𝑧) where 𝜂= 0 on 𝜕𝒟. Evaluating the functional on this perturbed 𝜙gives 𝐹[𝜙+ 𝜀𝜂] −𝐹[𝜙] = 𝜀∭ 𝒟 {𝜂𝜕𝑓 𝜕𝜙+ 𝜂𝑥 𝜕𝑓 𝜕𝜙𝑥 + 𝜂𝑦 𝜕𝑓 𝜕𝜙𝑦 + 𝜂𝑧 𝜕𝑓 𝜕𝜙𝑧 } d𝑥d𝑦d𝑧+ 𝑂(𝜀2) = 𝜀∭ 𝒟 ⎧ ⎪ ⎨ ⎪ ⎩ 𝜂𝜕𝑓 𝜕𝜙+ ∇⋅(𝜂( 𝜕𝑓 𝜕𝜙𝑥 , 𝜕𝑓 𝜕𝜙𝑦 , 𝜕𝑓 𝜕𝜙𝑧 )) ⏟⎵⎵⎵⎵⎵⎵⏟⎵⎵⎵⎵⎵⎵⏟ apply divergence theorem since 𝜂vanishes on 𝜕𝒟 −𝜂∇⋅( 𝜕𝑓 𝜕𝜙𝑥 , 𝜕𝑓 𝜕𝜙𝑦 , 𝜕𝑓 𝜕𝜙𝑧 ) ⎫ ⎪ ⎬ ⎪ ⎭ d𝑥d𝑦d𝑧+ 𝑂(𝜀2) = 𝜀∭ 𝒟 𝜂{𝜕𝑓 𝜕𝜙−∇⋅( 𝜕𝑓 𝜕𝜙𝑥 , 𝜕𝑓 𝜕𝜙𝑦 , 𝜕𝑓 𝜕𝜙𝑧 )} d𝑥d𝑦d𝑧+ 𝑂(𝜀2) Now, we can apply the fundamental lemma to give the Euler–Lagrange equation for multiple independent variables.
𝜕𝑓 𝜕𝜙−∇⋅( 𝜕𝑓 𝜕𝜙𝑥 , 𝜕𝑓 𝜕𝜙𝑦 , 𝜕𝑓 𝜕𝜙𝑧 ) = 0 Or, in suffix notation (with the summation convention), 𝜕𝑓 𝜕𝜙−𝜕𝑖 𝜕𝑓 𝜕(𝜕𝑖𝜙) = 0 This result applies for any 𝑛. Note that this is now a partial differential equation for 𝜙, instead of an ordinary differential equation.
81 II. Variational Principles 4.7. Potential energy and the Laplace equation Consider the functional 𝐹[𝜙] = ∬ 𝒟⊂ℝ2 1 2[𝜙2 𝑥+ 𝜙2 𝑦] d𝑥d𝑦 Note that 𝜕𝑓 𝜕𝜙= 0 and 𝜕𝑓 𝜕𝜙𝑥= 𝜙𝑥; 𝜕𝑓 𝜕𝜙𝑦= 𝜙𝑦. The Euler–Lagrange equation becomes 𝜕 𝜕𝑥𝜙𝑥+ 𝜕 𝜕𝑦𝜙𝑦= 0 ⟹𝜙𝑥𝑥+ 𝜙𝑦𝑦= 0 This produces the Laplace equation.
4.8. Minimal surfaces Consider minimising the area of a surface Σ ⊂ℝ3, where we want the surface to have two boundaries defined by fixed closed curves. This is sometimes known as Plateau’s problem.
We will let Σ = {x = ℝ3 ∶𝑘(𝑥, 𝑦, 𝑧) = 0}, and assume there exists a parametrisation of Σ given by 𝑧= 𝜙(𝑥, 𝑦). The line element is given by d𝑠2 = d𝑥2 + d𝑦2 + d𝑧2 We have d𝑧= 𝜙𝑥d𝑥+ 𝜙𝑦d𝑦hence d𝑠2 = (1 + 𝜙2 𝑥) d𝑥2 + (1 + 𝜙2 𝑦) d𝑦2 + 2𝜙𝑥𝜙𝑦d𝑥d𝑦 This is a quadratic form in the differentials d𝑥, d𝑦, known as the first fundamental form (also the Riemannian metric). Alternatively, d𝑠2 = 𝑔𝑖𝑗d𝑥𝑖d𝑥𝑗 where 𝑔= (1 + 𝜙2 𝑥 𝜙𝑥𝜙𝑦 𝜙𝑥𝜙𝑦 1 + 𝜙2 𝑦 ) From this, we can compute the area element, which is defined as d𝐴= √det 𝑔d𝑥d𝑦 We will extremise the area functional 𝐴[𝜙] = ∫ 𝒟√1 + 𝜙2 𝑥+ 𝜙2 𝑦d𝑥d𝑦 Let the integrand be ℎ, and apply the Euler–Lagrange equation.
𝜕ℎ 𝜕𝜙𝑥 = 𝜙𝑥 √1 + 𝜙2 𝑥+ 𝜙2 𝑦 ; 𝜕ℎ 𝜕𝜙𝑦 = 𝜙𝑦 √1 + 𝜙2 𝑥+ 𝜙2 𝑦 82 4. Extensions to the Euler–Lagrange equation Hence 𝜕𝑥( 𝜙𝑥 √1 + 𝜙2 𝑥+ 𝜙2 𝑦 ) + 𝜕𝑦( 𝜙𝑥 √1 + 𝜙2 𝑥+ 𝜙2 𝑦 ) = 0 which can be expanded to give (1 + 𝜙2 𝑦)𝜙𝑥𝑥+ (1 + 𝜙2 𝑥)𝜙𝑦𝑦−2𝜙𝑥𝜙𝑦𝜙𝑥𝑦= 0 This is known as the minimal surface equation. We will solve a special case, where there is circular (cylindrical) symmetry, so 𝑧= 𝜙(𝑟). Since 𝑟= √𝑥2 + 𝑦2, we can find that 𝜙𝑥= 𝑧′ 𝑥 𝑟; 𝜙𝑦= 𝑧′ 𝑦 𝑟 and we can analogously compute 𝜙𝑥𝑥, 𝜙𝑦𝑦, 𝜙𝑥𝑦. This gives 𝑟𝑧″ + 𝑧′ + (𝑧′)3 = 0 We can integrate this by first setting 𝑧′ = 𝑤and multiplying through by 𝑤.
1 2𝑟d d𝑟𝑤2 + 𝑤2 + 𝑤4 = 0 Now let 𝑤2 = 𝑢to make this a separable equation for 𝑢. Solving this, we can find that the solution surface is given by 𝑟= 𝑟0 cosh (𝑧−𝑧0 𝑟0 ) This is known as the catenoid. At the maximal and minimal values of 𝑧, we have the circular boundaries with radii 𝑅. At 𝑧= 𝑧0, the radius is minimal, and the circle here has radius 𝑟0.
Supposing 𝑧0 = 0 and that the maximal value of 𝑧is 𝐿, we have 𝑅 𝐿= 𝑟0 𝐿cosh ( 𝐿 𝑟0 ) Let 𝐿= 1 without loss of generality. This essentially chooses a scale for the coordinate system. This gives 𝑅= 𝑟0 cosh 1 𝑟0 Plotting 𝑅as a function of 𝑟0, there exists a minimum point 𝑟0 = 𝜇≈0.833 which gives 𝑅≈1.5. So if 𝑅> 1.5, there exist two distinct minimal surfaces, one with 𝑟0 > 𝜇and one with 𝑟0 < 𝜇. The ‘tighter’ minimal surface (with 𝑟0 < 𝜇) is unstable, but the ‘looser’ sur-face is stable (however this cannot be shown from our current understanding of variational principles).
83 II. Variational Principles 4.9. Higher derivatives Consider the functional 𝐹[𝑦] = ∫ 𝛽 𝛼 𝑓(𝑥, 𝑦, 𝑦′, … , 𝑦(𝑛)) d𝑥 We can find an analogous Euler–Lagrange equation to extremise this functional. Let 𝜂be a variation where 𝜂(𝑘) = 0 for 𝑘∈{1, … , 𝑛−1} at the endpoints 𝛼, 𝛽. Now, 𝐹[𝑦+ 𝜀𝜂] −𝐹[𝑦] = 𝜀∫ 𝛽 𝛼 (𝜕𝑓 𝜕𝑦𝜂+ 𝜕𝑓 𝜕𝑦′ 𝜂′ + ⋯+ 𝜕𝑓 𝜕𝑦(𝑛) 𝜂(𝑛)) d𝑥+ 𝑂(𝜀2) We can repeatedly integrate each term by parts, integrating the 𝜂(𝑘) term 𝑘times. Many of these terms will vanish due to the boundary conditions we specified for 𝜂. This then gives 𝐹[𝑦+ 𝜀𝜂] −𝐹[𝑦] = 𝜀∫ 𝛽 𝛼 (𝜕𝑓 𝜕𝑦𝜂−d d𝑥 𝜕𝑓 𝜕𝑦′ 𝜂+ ⋯+ (−1)𝑛d𝑛 d𝑥𝑛 𝜕𝑓 𝜕𝑦(𝑛) 𝜂) d𝑥+ 𝑂(𝜀2) Applying the fundamental lemma of calculus of variations, we have 𝜕𝑓 𝜕𝑦−d d𝑥 𝜕𝑓 𝜕𝑦′ + ⋯+ (−1)𝑛d𝑛 d𝑥𝑛 𝜕𝑓 𝜕𝑦(𝑛) = 0 This is the Euler–Lagrange equation in the context of a function with higher derivatives.
The alternating signs come from the negative signs produced in the iterated integration by parts.
4.10. First integral for 𝑛= 2 Suppose 𝑛= 2. If 𝜕𝑓 𝜕𝑦= 0, we have d d𝑥 𝜕𝑓 𝜕𝑦′ −d2 d𝑥2 𝜕𝑓 𝜕𝑦″ = 0 Hence 𝜕𝑓 𝜕𝑦′ −d d𝑥 𝜕𝑓 𝜕𝑦″ = constant Example. Extremise the functional 𝐹[𝑦] = ∫ 1 0 (𝑦″)2 d𝑥 subject to the conditions 𝑦(0) = 𝑦′(0) = 0; 𝑦(1) = 0; 𝑦′(1) = 1 84 4. Extensions to the Euler–Lagrange equation Using the above first integral form, we have d d𝑥(2𝑦″) = constant ⟹𝑦‴ = 𝑘 for some 𝑘∈ℝ. Imposing the boundary conditions on this cubic gives 𝑦= 𝑥3 −𝑥2 Now, we are going to show that this is an absolute minimum of the functional, not just a stationary point. Let 𝑦0 = 𝑥2 −𝑥2. Consider a variation 𝜂of 𝑦0, where all relevant endpoints of 𝜂are zero. In this case, we are not going to assume that 𝜂is small; we will simply look at all possible variations.
𝐹[𝑦0 + 𝜂] −𝐹[𝑦0] = ∫ 1 0 (𝜂″)2 d𝑥 ⏟⎵ ⎵⏟⎵ ⎵⏟ >0 +2 ∫ 1 0 𝑦″ 0𝜂″ d𝑥 Substituting for 𝑦0, given that 𝜂≢0, 𝐹[𝑦0 + 𝜂] −𝐹[𝑦0] > 4 ∫ 1 0 (3𝑥−1)𝜂″ d𝑥 = 4{[−𝜂′]1 0 + ∫ 1 0 [ d d𝑥(3𝑥𝜂′) −𝜂′] d𝑥} = 4{∫ 1 0 [ d d𝑥(3𝑥𝜂′) −𝜂′] d𝑥} = 4{[3𝑥𝜂′] 1 0 −[3𝜂]1 0} = 0 Hence 𝑦0 is an absolute minimum of 𝐹. This method of showing 𝑦0 is an absolute minimum is easier than calculating second variations, where we know the solution 𝑦0.
4.11. Principle of least action Consider a particle moving in ℝ3 with kinetic energy 𝑇and potential energy 𝑉. We define the Lagrangian to be 𝐿(x, ̇ x, 𝑡) = 𝑇−𝑉 We now define the action to be 𝑆[x] = ∫ 𝑡2 𝑡1 𝐿d𝑡 We can now formulate the principle of least (or stationary) action: on the path of motion of a particle, 𝛿𝑆 𝛿x = 0 85 II. Variational Principles Equivalently, 𝐿satisfies the Euler–Lagrange equations: 𝜕𝐿 𝜕𝑥𝑖 −d d𝑡 𝜕𝐿 𝜕̇ 𝑥𝑖 = 0 Consider 𝑇= 1 2𝑚| ̇ x|2; 𝑉= 𝑉(x) The Euler–Lagrange equations are now d d𝑡 𝜕𝐿 𝜕̇ 𝑥𝑖 = 𝜕𝐿 𝜕𝑥𝑖 𝑚̈ 𝑥𝑖= −𝜕𝑉 𝜕𝑥𝑖 ⟹𝑚̈ x = −∇𝑉 This is exactly Newton’s second law, derived from the principle of stationary action.
4.12. Central forces Example. Consider a central force in the plane. The Lagrangian is 𝐿= 𝑇−𝑉= 1 2𝑚( ̇ 𝑟2 + 𝑟2 ̇ 𝜃2) −𝑉(𝑟) The Euler–Lagrange equation gives d d𝑡 𝜕𝐿 𝜕̇ 𝑟−𝜕𝐿 𝜕𝑟= 0 d d𝑡 𝜕𝐿 𝜕̇ 𝜃 −𝜕𝐿 𝜕𝜃= 0 Since 𝜕𝐿 𝜕𝜃= 0, we have a first integral form: 𝜕𝐿 𝜕̇ 𝜃 = 𝑚𝑟2 ̇ 𝜃= constant This can be interpreted physically as the law of conservation of angular momentum. Further, we have 𝜕𝐿 𝜕𝑡= 0 so we have another first integral: ̇ 𝑟𝜕𝐿 𝜕̇ 𝑟+ ̇ 𝜃𝜕𝐿 𝜕̇ 𝜃 −𝐿= constant 𝑚̇ 𝑟2 + 𝑚𝑟2 ̇ 𝜃2 −1 2𝑚̇ 𝑟2 −1 2𝑚𝑟2 ̇ 𝜃2 + 𝑉(𝑟) = constant 1 2𝑚( ̇ 𝑟2 + 𝑟2 ̇ 𝜃2) + 𝑉(𝑟) = constant The left hand side is the total energy of the system, denoted 𝐸. This is the law of conservation of energy.
86 4. Extensions to the Euler–Lagrange equation 4.13. Configuration space and generalised coordinates Example. Consider 𝑁particles moving in ℝ3. Typically we represent each point as a dis-tinct vector in ℝ3 that changes over time. We can alternatively consider a point in ℝ3𝑁, which contains the information about every point. This is called the configuration space.
The Lagrangian in configuration space is 𝐿= 𝐿(𝑞𝑖, ̇ 𝑞𝑖, 𝑡) where q is the combined position vector of all 𝑁points, and likewise ̇ q is the combined velocity.
87 II. Variational Principles 5. Noether’s theorem 5.1. Statement and proof Consider a functional 𝐹[y] = ∫ 𝛽 𝛼 𝑓(𝑦𝑖, 𝑦′ 𝑖, 𝑥) d𝑥; 𝑖= 1, … , 𝑛 Suppose there exists a one-parameter family of transformations 𝑦𝑖(𝑥) ↦𝑌𝑖(𝑥, 𝑠); 𝑌𝑖(𝑥, 0) = 𝑦𝑖(𝑥) This can be thought of as a change of variables parametrised by 𝑠∈ℝ, where 𝑠= 0 implies no change of variables. This family is called a continuous symmetry of the Lagrangian 𝑓 if d d𝑠𝑓(𝑌𝑖(𝑥, 𝑠), 𝑌′ 𝑖(𝑥, 𝑠), 𝑥) = 0 In this course, we only consider continuous symmetries, so they may be abbreviated as just ‘symmetries’.
Theorem (Noether’s Theorem). Given a continuous symmetry 𝑌𝑖(𝑥, 𝑠) of 𝑓, 𝜕𝑓 𝜕𝑦′ 𝑖 𝜕𝑌𝑖 𝜕𝑠 | | |𝑠=0 is a first integral of the Euler–Lagrange equation (where the summation convention applies).
Proof.
0 = d d𝑠𝑓| | |𝑠=0 = 𝜕𝑓 𝜕𝑦𝑖 d𝑌𝑖 d𝑠 | | |𝑠=0 + 𝜕𝑓 𝜕𝑦′ 𝑖 𝜕𝑌′ 𝑖 𝜕𝑠 | | |𝑠=0 = [ d d𝑥( 𝜕𝑓 𝜕𝑦′ 𝑖 )d𝑌𝑖 d𝑠+ 𝜕𝑓 𝜕𝑦′ 𝑖 d d𝑥(d𝑌𝑖 d𝑠)]| | |𝑠=0 = d d𝑥[ 𝜕𝑓 𝜕𝑦′ 𝑖 𝜕𝑌𝑖 𝜕𝑠]| | |𝑠=0 ∴constant = 𝜕𝑓 𝜕𝑦′ 𝑖 𝜕𝑌𝑖 𝜕𝑠 88 5. Noether’s theorem 5.2. Conservation of momentum Example. Consider a vector y = (𝑦, 𝑧) and the function 𝑓= 1 2𝑦′2 + 1 2𝑧′2 −𝑉(𝑦−𝑧) Consider the symmetry 𝑌= 𝑦+ 𝑠⟹𝑌′ = 𝑦′ 𝑍= 𝑧+ 𝑠⟹𝑍= 𝑧′ ∴𝑉(𝑌−𝑍) = 𝑉(𝑦−𝑧) ⟹ d d𝑠𝑓= 0 Then from Noether’s theorem, constant = [ 𝜕𝑓 𝜕𝑦′ d𝑌 d𝑠+ 𝜕𝑓 𝜕𝑧′ d𝑍 d𝑠]| | |𝑠=0 = 𝑦′ + 𝑧′ This can be thought of as a conserved momentum in the 𝑦+ 𝑧direction.
5.3. Conservation of angular momentum under central force Example. Suppose Θ = 𝜃+ 𝑠, 𝑅= 𝑟. Our space is isotropic, so d𝐿 d𝑠= 0, hence [𝜕𝐿 𝜕̇ 𝜃 𝜕Θ 𝜕𝑠+ 𝜕𝐿 𝜕̇ 𝑟 𝜕𝑅 𝜕𝑠]| | |𝑠=0 = 𝑚𝑟2 ̇ 𝜃 which shows that angular momentum is conserved.
89 II. Variational Principles 6. Convexity and the Legendre transform 6.1. Convex functions This subsection is covered by Lecture 1 of the IB Optimisation course.
Definition. A set 𝑆⊂ℝ𝑛is convex if ∀x, y ∈𝑆, ∀𝑡∈[0, 1], (1 −𝑡)x + 𝑡y ∈𝑆.
Definition. The graph of a function 𝑓∶ℝ𝑛→ℝis the surface {(x, 𝑧) ∈ℝ𝑛+1 ∶𝑧−𝑓(x) = 0}.
Definition. A chord of a function 𝑓∶ℝ𝑛→ℝis a line segment connecting two points on the graph of 𝑓.
Definition. A function 𝑓∶ℝ𝑛→ℝis convex if (i) the domain of 𝑓is a convex set; and (ii) ∀x, y ∈𝑆, ∀𝑡∈(0, 1), 𝑓((1 −𝑡)x + 𝑡y) ≤(1 −𝑡)𝑓(x) + 𝑡𝑓(y) Equivalently, 𝑓is convex if the graph of 𝑓lies below (or on) all of its chords. We say that 𝑓is concave if 𝑓lies above (or on) all of its chords. Clearly, 𝑓is convex if and only if −𝑓is concave. We say 𝑓is strictly convex (or concave) if the inequality in (ii) becomes strict.
Example. Consider the function 𝑓∶ℝ→ℝdefined by 𝑓(𝑥) = 𝑥2 The domain is clearly convex. To show convexity, we need 𝑓((1 −𝑡)𝑥+ 𝑡𝑦) −(1 −𝑡)𝑓(𝑥) −𝑡𝑓(𝑦) ≤0 We have [(1−𝑡)𝑥+𝑡𝑦]2−(1−𝑡)𝑥2−𝑡𝑦2 = 𝑥2(1−𝑡)(−𝑡)+𝑡𝑦2(1−𝑡)+2(1−𝑡)𝑡𝑥𝑦= −(1−𝑡)𝑡(𝑥−𝑦)2 < 0 as required. Hence 𝑓(𝑥) = 𝑥2 is a strictly convex function.
Example. Consider 𝑓(𝑥) = 1 𝑥 where the domain is ℝ∖{0}. This domain is not convex, so 𝑓is not convex. However, re-stricted to the domain {𝑥∈ℝ∶𝑥> 0}, 𝑓can be shown to be convex.
6.2. Conditions for convexity Proofs for these conditions, where appropriate, are given in Lecture 1 of the IB Optimisation course.
Theorem. If 𝑓is a once-differentiable function, then 𝑓is convex if and only if 𝑓(y) ≥𝑓(x) + (y −x) ⋅∇𝑓(x) 90 6. Convexity and the Legendre transform Corollary. If 𝑓is convex, and has a stationary point, then it is a global minimum.
Proof. Suppose the stationary point is at x0, so ∇𝑓(x0) = 0. We then have 𝑓(y) ≥𝑓(x0) + (y −x0) ⋅0 which is larger than 𝑓(x0) as required.
Theorem. If 𝑓is a once-differentiable function, then 𝑓is convex if (∇𝑓(y) −∇𝑓(x)) ⋅(y −x) ≥0 This can be thought of as stating that 𝑓′ is monotonically increasing.
Theorem. If 𝑓is a twice-differentiable function, then 𝑓is convex if and only if ∇2𝑓⪰0 i.e. all eigenvalues of the Hessian matrix are non-negative. Note that ∇2𝑓≻0 implies strict convexity.
Example. Consider the function 𝑓(𝑥, 𝑦) = 1 𝑥𝑦 for 𝑥> 0, 𝑦> 0. Then the Hessian is 𝐻= 1 𝑥𝑦( 2 𝑥2 1 𝑥𝑦 1 𝑥𝑦 2 𝑦2 ) Then, det 𝐻= 3 𝑥3𝑦3 > 0 tr 𝐻> 0 Hence the eigenvalues are both positive. So 𝑓is strictly convex.
6.3. Legendre transform Definition. The Legendre transform of a function 𝑓∶ℝ𝑛→ℝis a function 𝑓⋆given by 𝑓⋆(p) = sup x (p ⋅x −𝑓(x)) The domain of 𝑓⋆is such that the supremum provided is finite. In one dimension, we can consider 𝑓⋆(𝑝) to be the maximum vertical distance between the graphs of 𝑦= 𝑓(𝑥) and 𝑦= 𝑝𝑥.
91 II. Variational Principles Example. Consider the function 𝑓(𝑥) = 𝑎𝑥2, which is convex where 𝑎> 0. Computing the derivative of the right hand side and setting it to zero, 𝑓⋆(𝑝) = sup 𝑥 (𝑝𝑥−𝑎𝑥2) = 𝑝( 𝑝 2𝑎) −𝑎( 𝑝 2𝑎) 2 = 𝑝2 4𝑎 We can apply the Legendre transform twice: 𝑓⋆⋆(𝑠) = sup 𝑝 (𝑠𝑝−𝑓⋆(𝑝)) = 𝑎𝑠2 = 𝑓(𝑠) In fact, if 𝑓is convex, then we always have 𝑓⋆⋆= 𝑓. If 𝑎< 0, the supremum does not exist so 𝑓⋆has an empty domain, and thus 𝑓⋆⋆≠𝑓.
Proposition. If the domain of 𝑓⋆is non-empty, it is a convex set, and 𝑓⋆is convex.
Proof. Given p, q in the domain of 𝑓⋆, 𝑓⋆((1 −𝑡)p + 𝑡q) = sup x [(1 −𝑡)p ⋅x + 𝑡q ⋅x −𝑓(x)] = sup x [(1 −𝑡)(p ⋅x −𝑓(x)) + 𝑡(q ⋅x −𝑓(x))] ≤sup x [(1 −𝑡)(p ⋅x −𝑓(x))] + sup x [𝑡(q ⋅x −𝑓(x))] < ∞ as required.
In practice, if 𝑓is convex and differentiable, we compute 𝑓⋆(p) by considering the derivat-ive: ∇(p ⋅x −𝑓(x)) = 0 ⟹p = ∇𝑓 If 𝑓is strictly convex, the condition p = ∇𝑓has a unique inverse to give x as a function of p, so 𝑓⋆(p) = p ⋅x(p) −𝑓(x(p)). This eliminates the supremum condition.
6.4. Applications to thermodynamics If we consider the particles in a gas, we could theoretically solve the Euler–Lagrange equa-tions for a system of around 1023 particles. However, solving such a complicated system is difficult. Instead of solving for each particle, we instead consider macroscopic quantities 92 6. Convexity and the Legendre transform such as pressure 𝑃, volume 𝑉, temperature 𝑇, and entropy 𝑆. A system has internal energy 𝑈(𝑆, 𝑉). The Helmholtz free energy is 𝐹(𝑇, 𝑉) = min 𝑆(𝑈(𝑆, 𝑉) −𝑇𝑆) = −max 𝑆(𝑇𝑆−𝑈(𝑆, 𝑉)) = −𝑈⋆(𝑇, 𝑉) where 𝑈⋆is the Legendre transform of 𝑈with respect to 𝑆, fixing 𝑉constant. Assuming 𝑈 is convex, 𝜕 𝜕𝑆(𝑇𝑆−𝑈(𝑆, 𝑉))| | |𝑇,𝑉 = 0 ⟹𝑇= 𝜕𝑈 𝜕𝑆 | | |𝑉 There are other thermodynamical quantities that can be represented using a Legendre trans-form, for instance enthalpy 𝐻(𝑆, 𝑃).
𝐻(𝑆, 𝑃) = min 𝑉(𝑈(𝑆, 𝑉) + 𝑃𝑉) = −𝑈⋆(−𝑃, 𝑆) At this minimum, 𝑃= − 𝜕𝑈 𝜕𝑉 | |𝑆. We can think of the Legendre transform in this context as a way of swapping from dependence on entropy and volume to dependence on other variables.
6.5. Legendre transform of the Lagrangian Recall that the Lagrangian in mechanics was defined as 𝐿= 𝑇−𝑉= 𝐿(q, ̇ q, 𝑡) This is a function on the configuration space. We define the Hamiltonian to be the Legendre transform of 𝐿with respect to ̇ q. We find, assuming that 𝐿is convex, 𝐻(q, p, t) = sup v (p ⋅v −𝐿) = p ⋅v(p) −𝐿(q, v(p), 𝑡) where v(p) is the solution to 𝑝𝑖= 𝜕𝐿 𝜕̇ 𝑞𝑖. The p are referred to as generalised momenta or conjugate momenta. Consider 𝑇= 1 2𝑚| ̇ q|2; 𝑉= 𝑉(q) Then, p = 𝜕𝐿 𝜕̇ q = 𝑚̇ q ⟹ ̇ q = 1 𝑚p 93 II. Variational Principles The Hamiltonian is therefore 𝐻(q, p, 𝑡) = p ⋅1 𝑚p −𝐿 = p ⋅1 𝑚p −(1 2𝑚|p|2 𝑚2 −𝑉(q)) = 1 2𝑚|p|2 + 𝑉(q) = 𝑇+ 𝑉 6.6. Hamilton’s equations from Euler–Lagrange equation Given that the Lagrangian satisfies the Euler–Lagrange equation, we can deduce analogous equations for the Hamiltonian. We often write the indices of the generalised coordinates in superscript, as follows, where the summation convention applies: 𝐻= 𝐻(q, p, 𝑡) = 𝑝𝑖̇ 𝑞𝑖−𝐿(𝑞𝑖, ̇ 𝑞𝑖, 𝑡) Using this equation, we can compute two expressions for the differential of the Hamilto-nian: d𝐻= 𝜕𝐻 𝜕𝑞𝑖d𝑞𝑖+ 𝜕𝐻 𝜕𝑝𝑖 d𝑝𝑖+ 𝜕𝐻 𝜕𝑡d𝑡 = 𝑝𝑖d ̇ 𝑞𝑖+ ̇ 𝑞𝑖d𝑝𝑖−𝜕𝐿 𝜕𝑞𝑖d𝑞𝑖−𝜕𝐿 𝜕̇ 𝑞𝑖d ̇ 𝑞𝑖−𝜕𝐿 𝜕𝑡d𝑡 Now, note that 𝜕𝐿 𝜕̇ 𝑞𝑖= 𝑝𝑖. This cancels some terms. Making use of the Euler–Lagrange equation, 𝜕𝐿 𝜕𝑞𝑖= d d𝑡 𝜕𝐿 𝜕̇ 𝑞𝑖= d d𝑡𝑝𝑖= ̇ 𝑝𝑖 This gives d𝐻= 𝜕𝐻 𝜕𝑞𝑖d𝑞𝑖+ 𝜕𝐻 𝜕𝑝𝑖 d𝑝𝑖+ 𝜕𝐻 𝜕𝑡d𝑡= ̇ 𝑞𝑖d𝑝𝑖− ̇ 𝑝𝑖d𝑞𝑖−𝜕𝐿 𝜕𝑡d𝑡 Comparing the differentials, we can see that ̇ 𝑞𝑖= 𝜕𝐻 𝜕𝑝𝑖 ; ̇ 𝑝𝑖= −𝜕𝐻 𝜕𝑞𝑖; 𝜕𝐿 𝜕𝑡= −𝜕𝐻 𝜕𝑡 This system of equations is known as Hamilton’s equations. Note that in the last equation, 𝜕 𝜕𝑡 | |𝑞, ̇ 𝑞≠ 𝜕 𝜕𝑡 | |𝑝,𝑞. For now, we will assume that there is no explicit 𝑡dependence in the Lagrangian. Then, Hamilton’s equations are a system of 2𝑛first-order ordinary differen-tial equations. (Note, for comparison, that the Euler–Lagrange equations were a system of 𝑛second-order differential equations, which gives the same amount of initial conditions.) The initial conditions are typically a configuration of p, q at some fixed 𝑡0. The solutions to Hamilton’s equations are called the trajectories in 2𝑛-dimensional phase space.
94 6. Convexity and the Legendre transform 6.7. Hamilton’s equations from extremising a functional Note that we can also arrive at Hamilton’s equations by extremising a functional in phase space.
𝑆[q, p] = ∫ 𝑡2 𝑡1 ( ̇ 𝑞𝑖𝑝𝑖−𝐻(q, p, 𝑡)) d𝑡 The integrand, denoted 𝑓, is a function of q, p, ̇ q, 𝑡. Writing the Euler–Lagrange equations for 𝑆, varying first with respect to 𝑝𝑖, 𝜕𝑓 𝜕𝑝𝑖 −d d𝑡 𝜕𝑓 𝜕̇ 𝑝𝑖 ⏟ ⎵ ⏟ ⎵ ⏟ 0 = 0 ⟹ ̇ 𝑞𝑖= 𝜕𝐻 𝜕𝑝𝑖 Now varying with respect to 𝑞𝑖, 𝜕𝑓 𝜕𝑞𝑖−d d𝑡 𝜕𝑓 𝜕̇ 𝑞𝑖= 0 ⟹ ̇ 𝑝𝑖= −𝜕𝐻 𝜕𝑞𝑖 These results are exactly Hamilton’s equations.
95 II. Variational Principles 7. Second variations 7.1. Conditions for local minimisers The Euler–Lagrange equation gives a necessary condition for a stationary point. We cannot tell whether this leads to a minimum, a maximum, or a saddle point, just from the Euler– Lagrange equation. We can analyse the nature of the stationary points by considering the second variation. Consider the functional 𝐹[𝑦] = ∫ 𝛽 𝛼 𝑓(𝑥, 𝑦, 𝑦′) d𝑥 where 𝑦is perturbed by a perturbation 𝜀𝜂. Let us assume that 𝑦is a solution to the Euler– Lagrange equation, so has no first variation. We will then expand 𝐹[𝑦+ 𝜀𝜂] to second or-der.
𝐹[𝑦+ 𝜀𝜂] = ∫ 𝛽 𝛼 [𝑓(𝑥, 𝑦+ 𝜀𝜂, 𝑦′ + 𝜀𝜂′)] d𝑥 𝐹[𝑦+ 𝜀𝜂] −𝐹[𝑦] = ∫ 𝛽 𝛼 [𝑓(𝑥, 𝑦+ 𝜀𝜂, 𝑦′ + 𝜀𝜂′) −𝑓(𝑥, 𝑦, 𝑦′)] d𝑥 = 0 + 𝜀∫ 𝛽 𝛼 𝜂(𝜕𝑓 𝜕𝑦−d d𝑥 𝜕𝑓 𝜕𝑦′ ) d𝑥 ⏟⎵⎵⎵⎵⎵⎵⏟⎵⎵⎵⎵⎵⎵⏟ zero by Euler–Lagrange equation + 1 2𝜀2 ∫ 𝛽 𝛼 (𝜂2 𝜕2𝑓 𝜕𝑦2 + 𝜂′2 𝜕2𝑓 𝜕(𝑦′)2 + 2𝜂𝜂′ 𝜕2𝑓 𝜕𝑦𝜕𝑦′ ) d𝑥+ 𝑂(𝜀3) The last term (excluding the 𝜀2 component) is called the second variation. We write 𝛿2𝐹[𝑦] ≡1 2 ∫ 𝛽 𝛼 (𝜂2 𝜕2𝑓 𝜕𝑦2 + 𝜂′2 𝜕2𝑓 𝜕(𝑦′)2 + d d𝑥(𝜂2) 𝜕2𝑓 𝜕𝑦𝜕𝑦′ ) d𝑥 Integrating the last term by parts, using 𝜂= 0 at 𝛼, 𝛽, we have 𝛿2𝐹[𝑦] = 1 2 ∫ 𝛽 𝛼 (𝑄𝜂2 + 𝑃(𝜂′)2) d𝑥 where 𝑃= 𝜕2𝑓 𝜕(𝑦′)2 ; 𝑄= 𝜕2𝑓 𝜕𝑦2 −d d𝑥( 𝜕2𝑓 𝜕𝑦𝜕𝑦′ ) Thus, if 𝑦is a solution to the Euler–Lagrange equation, and also 𝑄𝜂2 + 𝑃(𝜂′)2 > 0 for all 𝜂 vanishing at 𝛼, 𝛽, then 𝑦is a local minimiser of 𝐹.
Example. We will prove that the geodesic on a plane is a local minimiser of path length.
The functional we will analyse is given by 𝑓= √1 + (𝑦′)2 96 7. Second variations Hence, 𝑃= 𝜕2𝑓 𝜕(𝑦′)2 = 𝜕 𝜕𝑦′ ( 𝑦′ √1 + (𝑦′)2 ) = 1 (1 + (𝑦′)2) 3 2 > 0 𝑄= 0 Therefore the second variation is positive, so any 𝑦that satisfies the Euler–Lagrange equation minimises path length. In particular, straight lines minimise path length on the plane.
7.2. Legendre condition for minimisers Proposition (Legendre condition). If 𝑦0(𝑥) is a local minimiser, then 𝑃|𝑦=𝑦0 ≥0.
We can say that the Legendre condition is a necessary condition for a minimiser. In less formal terms, 𝑃is ‘more important’ than 𝑄when determining if a stationary point is a min-imiser.
Proof. This condition is not proven rigorously. However, the general idea of the proof is to construct a function 𝜂which is small everywhere (giving a small 𝑄contribution), but oscillates very rapidly near some point 𝑥0, at which 𝑃< 0. This gives a large 𝑃contribution which can overpower the 𝑄contribution. Then this gives 𝑄𝜂2 + 𝑃(𝜂′)2 < 0 if there exists some 𝑥0 where 𝑃|𝑦=𝑦0 < 0.
Note that the Legendre condition is not a sufficient condition for local minima, but 𝑃> 0 and 𝑄≥0 is sufficient.
Example. Consider again the brachistochrone problem.
𝑓= √ 1 + (𝑦′)2 −𝑦 We have 𝜕𝑓 𝜕𝑦= −1 2𝑦𝑓 𝜕𝑓 𝜕𝑦′ = 𝑦′ √1 + (𝑦′)2√−𝑦 Hence 𝑃= 1 (1 + (𝑦′)2) 3 2 √−𝑦 > 0 𝑄= 1 2√1 + (𝑦2)2𝑦2√−𝑦 > 0 Hence the cycloid is a local minimiser of the time taken to travel between the two points.
97 II. Variational Principles 7.3. Associated eigenvalue problem When deriving the minimiser condition, we had the integrand 𝑄𝜂2 + 𝑃(𝜂′)2 We can integrate this by parts: 𝑄𝜂2 + d d𝑥(𝑃𝜂𝜂′) −𝜂d d𝑥(𝑃𝜂′) giving 𝛿2𝐹[𝑦] = 1 2 ∫ 𝛽 𝛼 𝜂[−(𝑃𝜂′)′ + 𝑄𝜂] d𝑥 The bracketed term −(𝑃𝜂′)′ + 𝑄𝜂is known as the Sturm–Liouville operator acting on 𝜂, denoted ℒ(𝜂). If there exists 𝜂such that ℒ(𝜂) = −𝜔2𝜂, 𝜔∈ℝ, and 𝜂(𝛼) = 𝜂(𝛽) = 0, then 𝑦 is not a minimiser, since the integrand will be −𝜔2𝜂2 < 0.
Example. Consider 𝐹[𝑦] = ∫ 𝛽 0 ((𝑦′)2 −𝑦2) d𝑥 such that 𝑦(0) = 𝑦(𝛽) = 0; 𝛽≠𝑘𝜋, 𝑘∈ℕ The Euler–Lagrange equation gives 𝑦″ + 𝑦= 0 Thus, constrained to the boundary conditions, the only stationary point of 𝐹is 𝑦≡0 Analysing the second variation, 𝛿2𝐹 = 1 2 ∫ 𝛽 0 [𝜂′2 −𝜂2] d𝑥 giving 𝑃= 1 > 0; 𝑄< 0 Let us now examine the eigenvalue problem, since we cannot find whether 𝑦≡0 is a min-imiser from what we know already. Consider the eigenvalue problem −𝜂″ −𝜂= −𝜔2𝜂; 𝜂(0) = 𝜂(𝛽) = 0 Let us take 𝜂= 𝐴sin(𝜋𝑥 𝛽) to give (𝜋 𝛽) 2 = 1 −𝜔2 So this has a solution 𝜔> 0 if and only if 𝛽> 𝜋. If 𝑃> 0, a problem may arise if the interval of integration is ‘too large’ (in this case 𝛽> 𝜋). Next lecture we will make this notion precise.
98 7. Second variations 7.4. Jacobi accessory condition Legendre tried to prove that 𝑃> 0 implied local minimality; obviously this was impossible due to the counterexample shown above. However, the method he used is still useful to analyse, since we can find an actual sufficient condition using the same idea. Let 𝜙(𝑥) be any differentiable function of 𝑥on [𝛼, 𝛽]. Then note that ∫ 𝛽 𝛼 d d𝑥(𝜙𝜂2) d𝑥= 0 since 𝜂(𝛼) = 𝜂(𝛽) = 0. We can expand the integrand to give ∫ 𝛽 𝛼 (𝜙′𝜂2 + 2𝜂𝜂′𝜙) d𝑥= 0 We can add this new zero to both sides of the second variation equation.
𝛿2𝐹[𝑦] = 1 2 ∫ 𝛽 𝛼 (𝑃(𝜂′)2 + 2𝜂𝜂′𝜙+ (𝑄+ 𝜙′)𝜂2) d𝑥 Now, suppose that 𝑃> 0 at a particular 𝑦. Then, we can complete the square on the integ-rand, giving 𝛿2𝐹[𝑦] = 1 2 ∫ 𝛽 𝛼 (𝑃(𝜂′ + 𝜙 𝑃𝜂) 2 + (𝑄+ 𝜙′ −𝜙2 𝑃)𝜂2) d𝑥 If we could choose a 𝜙such that the second bracket vanishes, then the integrand would be 𝑃(𝜂′ + 𝜙 𝑃𝜂) 2 . The only way the integral can be zero is if 𝜂′+ 𝜙 𝑃𝜂≡0. Since 𝜂= 0 at 𝛼, we have 𝜂′(𝛼) = 0. Hence, 𝜂≡0 by the uniqueness of solutions to first order differential equations.
Therefore, by contradiction, the integrand is not identically zero, and the second variation is positive. Now, such a 𝜙function is given by 𝜙2 = 𝑃(𝑄+ 𝜙′) If a solution to this differential equation exists, then 𝛿2𝐹[𝑦] > 0. We can transform this non-linear equation into a second order equation by the substitution 𝜙= −𝑃 𝑢′ 𝑢for some function 𝑢≠0. We have 𝑃(𝑢′ 𝑢) 2 = 𝑄−(𝑃𝑢′ 𝑢) ′ = 𝑄−(𝑃𝑢′)′ 𝑢 + 𝑃(𝑢′ 𝑢) 2 Hence, −(𝑃𝑢′)′ + 𝑄𝑢= 0 This is known as the Jacobi accessory condition. Note that the left hand side is just ℒ(𝑢), where ℒis the Sturm–Liouville operator.
99 II. Variational Principles 7.5. Solving the Jacobi condition We need to find a solution to ℒ(𝑢) = 0, where 𝑢≠0 on [𝛼, 𝛽]. The solution we find may not be nonzero on a large enough interval, in which case we would not have a local min-imum.
Example. Consider 𝐹[𝑦] = 1 2 ∫ 𝛽 𝛼 ((𝑦′)2 −𝑦2) d𝑥 The second variation is 𝛿2𝐹[𝑦] = 1 2 ∫ 𝛽 𝛼 ((𝜂′)2 −𝜂2) d𝑥 In this case, 𝑃= 1, 𝑄= −1. The Jacobi accessory equation is 𝑢″ + 𝑢= 0 We can solve this to find 𝑢= 𝐴sin 𝑥−𝐵cos 𝑥; 𝐴, 𝐵∈ℝ We want this to be nonzero on the interval [𝛼, 𝛽]. In particular, tan 𝑥≠𝐵 𝐴; ∀𝑥∈[𝛼, 𝛽] Note that tan 𝑥repeats every 𝜋, so if |𝛽−𝛼| < 𝜋we have a positive second variation for any stationary 𝑦.
Example. Consider again the geodesic on a sphere.
𝐹[𝜃] = ∫√d𝜃2 + sin2 𝜃d𝜙2 = ∫√(𝜃′)2 + sin2 𝜃d𝜙 We have already proven that critical points of this functional are segments of great circles.
Considering an equatorial great circle (since all great circles are equatorial under a change of perspective), 𝜃= 𝜋 2 Consider 𝜙1, 𝜙2 on this great circle. The minor arc is clearly the shortest path, but the major arc is also a stationary point and must still be analysed.
𝑃= 1; 𝑄= −1 Thus, 𝛿2𝐹[𝜃0 = 𝜋 2 ] = 1 2 ∫ 𝜙2 𝜙1 ((𝜂′)2 −𝜂2) d𝜙 which is exactly the example from above. This is a minimiser if |𝜙2 −𝜙1| < 𝜋, which is exactly the condition of being a minor arc. If 𝜙2 −𝜙1 = 𝜋, we have an infinite amount of geodesics, since these represent antipodal points. The set of geodesics exhibit rotational symmetry.
100 III. Markov Chains Lectured in Michaelmas 2021 by Dr. P. Sousi A Markov chain is a common type of random process, where each state in the process de-pends only on the previous one. Due to their simplicity, Markov processes show up in many areas of probability theory and have lots of real-world applications, for example in computer science.
One example of a Markov chain is a simple random walk, where a particle moves around an infinite lattice of points, choosing its next direction to move at random. It turns out that if the lattice is one- or two-dimensional, the particle will return to its starting point infinitely many times, with probability 1. However, if the lattice is three-dimensional or higher, the particle has probability 0 of ever returning to its starting point.
101 III. Markov Chains Contents 1.
Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 103 1.1.
Definition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 103 1.2.
Sequence definition . . . . . . . . . . . . . . . . . . . . . . . . 103 1.3.
Point masses . . . . . . . . . . . . . . . . . . . . . . . . . . . 104 1.4.
Independence of sequences . . . . . . . . . . . . . . . . . . . . 104 1.5.
Simple Markov property . . . . . . . . . . . . . . . . . . . . . 104 1.6.
Powers of the transition matrix . . . . . . . . . . . . . . . . . . 106 1.7.
Calculating powers . . . . . . . . . . . . . . . . . . . . . . . . 106 2.
Elementary properties . . . . . . . . . . . . . . . . . . . . . . . . . 108 2.1.
Communicating classes . . . . . . . . . . . . . . . . . . . . . . 108 2.2.
Hitting times . . . . . . . . . . . . . . . . . . . . . . . . . . . 109 2.3.
Birth and death chain . . . . . . . . . . . . . . . . . . . . . . . 113 2.4.
Mean hitting times . . . . . . . . . . . . . . . . . . . . . . . . 113 2.5.
Strong Markov property . . . . . . . . . . . . . . . . . . . . . . 115 3.
Transience and recurrence . . . . . . . . . . . . . . . . . . . . . . 117 3.1.
Definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . 117 3.2.
Probability of visits . . . . . . . . . . . . . . . . . . . . . . . . 117 3.3.
Duality of transience and recurrence . . . . . . . . . . . . . . . 118 3.4.
Recurrent communicating classes . . . . . . . . . . . . . . . . 119 4.
Pólya’s recurrence theorem . . . . . . . . . . . . . . . . . . . . . . 121 4.1.
Statement of theorem . . . . . . . . . . . . . . . . . . . . . . . 121 4.2.
One-dimensional proof . . . . . . . . . . . . . . . . . . . . . . 121 4.3.
Two-dimensional proof . . . . . . . . . . . . . . . . . . . . . . 122 4.4.
Three-dimensional proof . . . . . . . . . . . . . . . . . . . . . 123 5.
Invariant distributions . . . . . . . . . . . . . . . . . . . . . . . . . 124 5.1.
Invariant distributions . . . . . . . . . . . . . . . . . . . . . . 124 5.2.
Conditions for unique invariant distribution . . . . . . . . . . . 125 5.3.
Uniqueness of invariant distributions . . . . . . . . . . . . . . . 126 5.4.
Positive and null recurrence . . . . . . . . . . . . . . . . . . . 128 5.5.
Time reversibility . . . . . . . . . . . . . . . . . . . . . . . . . 130 5.6.
Aperiodicity . . . . . . . . . . . . . . . . . . . . . . . . . . . 132 5.7.
Positive recurrent limiting behaviour . . . . . . . . . . . . . . . 133 5.8.
Null recurrent limiting behaviour . . . . . . . . . . . . . . . . . 135 102 1. Introduction 1. Introduction 1.1. Definition Let 𝐼be a finite or countable set. All of our random variables will be defined on the same probability space (Ω, ℱ, ℙ).
Definition. A stochastic process (𝑋𝑛)𝑛≥0 is called a Markov chain if for all 𝑛≥0 and for all 𝑥1 … 𝑥𝑛+1 ∈𝐼, ℙ(𝑋𝑛+1 = 𝑥𝑛+1 ∣𝑋𝑛= 𝑥𝑛, … , 𝑋1 = 𝑥1) = ℙ(𝑋𝑛+1 = 𝑥𝑛+1 ∣𝑋𝑛= 𝑥𝑛) We can think of 𝑛as a discrete measure of time. If ℙ(𝑋𝑛+1 = 𝑦∣𝑋𝑛= 𝑥) for all 𝑥, 𝑦is inde-pendent of 𝑛, then 𝑋is called a time-homogeneous Markov chain. Otherwise, 𝑋is called time-inhomogeneous. In this course, we only study time-homogeneous Markov chains.
If we consider only time-homogeneous chains, we may as well take 𝑛= 0 and we can write 𝑃(𝑥, 𝑦) = ℙ(𝑋1 = 𝑦∣𝑋0 = 𝑥) ; ∀𝑥, 𝑦∈𝐼 Definition. A stochastic matrix is a matrix where the sum of each row is equal to 1.
We call 𝑃the transition matrix. It is a stochastic matrix: ∑ 𝑦∈𝐼 𝑃(𝑥, 𝑦) = 1 Remark. The index set does not need to be ℕ; it could alternatively be the set {0, 1, … , 𝑁} for 𝑁∈ℕ.
We say that 𝑋is Markov (𝜆, 𝑃) if 𝑋0 has distribution 𝜆, and P is the transition matrix. Hence, (i) ℙ(𝑋0 = 𝑥0) = 𝜆𝑥0 (ii) ℙ(𝑋𝑛+1 = 𝑥𝑛+1 ∣𝑋𝑛= 𝑥𝑛, … , 𝑋0 = 𝑥0) = 𝑃(𝑥𝑛, 𝑥𝑛+1) =∶𝑃 𝑥𝑛𝑥𝑛+1 We usually draw a diagram of the transition matrix using a graph. Directed edges between nodes are labelled with their transition probabilities.
1.2. Sequence definition Theorem. The process 𝑋is Markov (𝜆, 𝑃) if and only if ∀𝑛≥0 and all 𝑥0, … , 𝑥𝑛∈𝐼, we have ℙ(𝑋0 = 𝑥0, … , 𝑋𝑛= 𝑥𝑛) = 𝜆𝑥0𝑃(𝑥0, 𝑥1)𝑃(𝑥1, 𝑥2) … 𝑃(𝑥𝑛−1, 𝑥𝑛) 103 III. Markov Chains Proof. If 𝑋is Markov, then we have ℙ(𝑋0 = 𝑥0, … , 𝑋𝑛= 𝑥𝑛) = ℙ(𝑋𝑛= 𝑥𝑛∣𝑋𝑛−1 = 𝑥𝑛−1, … , 𝑋0 = 𝑥0) ⋅ℙ(𝑋𝑛−1 = 𝑥𝑛−1, … , 𝑋0 = 𝑥0) = 𝑃(𝑥𝑛−1, 𝑥𝑛)ℙ(𝑋𝑛−1 = 𝑥𝑛−1, … , 𝑋0 = 𝑥0) = 𝑃(𝑥𝑛−1, 𝑥𝑛) … 𝑃(𝑥0, 𝑥1)𝜆𝑥0 as required. Conversely, ℙ(𝑋0 = 𝑥0) = 𝜆𝑥0 satisfies (i). The transition matrix is given by ℙ(𝑋𝑛= 𝑥𝑛∣𝑋0 = 𝑥0, … , 𝑋𝑛−1 = 𝑥𝑛−1) = 𝜆𝑥0𝑃(𝑥0, 𝑥1) … 𝑃(𝑥𝑛−1, 𝑥𝑛) 𝜆𝑥0𝑃(𝑥0, 𝑥1) … 𝑃(𝑥𝑛−2, 𝑥𝑛−1) = 𝑃(𝑥𝑛−1, 𝑥𝑛) which is exactly the Markov property as required.
1.3. Point masses Definition. For 𝑖∈𝐼, the 𝛿𝑖-mass at 𝑖is defined by 𝛿𝑖𝑗= 𝟙(𝑖= 𝑗) This is a probability measure that has probability 1 at 𝑖only.
1.4. Independence of sequences Recall that discrete random variables (𝑋𝑛) are considered independent if for all 𝑥1, … , 𝑥𝑛∈𝐼, we have ℙ(𝑋1 = 𝑥1, … , 𝑋𝑛= 𝑥𝑛) = ℙ(𝑋1 = 𝑥1) … ℙ(𝑋𝑛= 𝑥𝑛) A sequence (𝑋𝑛) is independent if for all 𝑘, 𝑖1 < 𝑖2 < ⋯< 𝑖𝑛and for all 𝑥1, … , 𝑥𝑘, we have ℙ(𝑋𝑖1 = 𝑥1, … , 𝑋𝑖𝑘= 𝑥𝑘) = 𝑛 ∏ 𝑗=1 ℙ(𝑋𝑖𝑗= 𝑥𝑗) Let 𝑋= (𝑋𝑛), 𝑌= (𝑌 𝑛) be sequences of discrete random variables. They are independent if for all 𝑘, 𝑚, 𝑖1 < ⋯< 𝑖𝑘, 𝑗1 < ⋯< 𝑗𝑚, 𝑝𝑟𝑜𝑏𝑋1 = 𝑥1, … , 𝑋𝑖𝑘= 𝑥𝑖𝑘, 𝑌𝑗1 = 𝑦𝑗1, … , 𝑌𝑗𝑚 = ℙ(𝑋1 = 𝑥1, … , 𝑋𝑖𝑘= 𝑥𝑖𝑘) ℙ(𝑌𝑗1 = 𝑦𝑗1, … , 𝑌𝑗𝑚) 1.5. Simple Markov property Theorem. Suppose 𝑋is Markov (𝜆, 𝑃). Let 𝑚∈ℕand 𝑖∈𝐼. Given that 𝑋𝑚= 𝑖, we have that the process after time 𝑚, written (𝑋𝑚+𝑛)𝑛≥0, is Markov (𝛿𝑖, 𝑃), and it is independent of 𝑋0, … , 𝑋𝑚.
104 1. Introduction Informally, the past and the future are independent given the present.
Proof. We must show that ℙ(𝑋𝑚= 𝑥0, … , 𝑋𝑚+𝑛= 𝑥𝑛∣𝑋𝑚= 𝑖) = 𝛿𝑖𝑥0𝑃(𝑥0, 𝑥1) … 𝑃(𝑥𝑛−1, 𝑥𝑛) We have ℙ(𝑋𝑚+𝑛= 𝑥𝑚+𝑛, … , 𝑋𝑚= 𝑥𝑚∣𝑋𝑚= 𝑖) = ℙ(𝑋𝑚+𝑛= 𝑥𝑚+𝑛, … , 𝑋𝑚= 𝑥𝑚) 𝛿𝑖𝑥𝑚 ℙ(𝑋𝑚= 𝑖) The numerator is ℙ(𝑋𝑚+𝑛, … , 𝑋𝑚= 𝑥𝑚) = ∑ 𝑥0,…,𝑥𝑚−1∈𝐼 ℙ(𝑋𝑚+𝑛= 𝑥𝑚+𝑛, … , 𝑋𝑚= 𝑥𝑚, 𝑋𝑚−1 = 𝑥𝑚−1, … , 𝑋0 = 𝑥0) = ∑ 𝑥0,…,𝑥𝑚−1 𝜆𝑥0𝑃(𝑥0, 𝑥1) … 𝑃(𝑥𝑚−1, 𝑥𝑚)𝑃(𝑥𝑚, 𝑥𝑚+1) … 𝑃(𝑥𝑚+𝑛−1, 𝑥𝑚+𝑛) = 𝑃(𝑥𝑚, 𝑥𝑚+1) … 𝑃(𝑥𝑚+𝑛−1, 𝑥𝑚+𝑛) ∑ 𝑥0,…,𝑥𝑚−1 𝜆𝑥0𝑃(𝑥0, 𝑥1) … 𝑃(𝑥𝑚−1, 𝑥𝑚) = 𝑃(𝑥𝑚, 𝑥𝑚+1) … 𝑃(𝑥𝑚+𝑛−1, 𝑥𝑚+𝑛)ℙ(𝑋𝑚= 𝑥𝑚) Thus we have ℙ(𝑋𝑚+𝑛= 𝑥𝑚+𝑛, … , 𝑋𝑚= 𝑥𝑚∣𝑋𝑚= 𝑖) = 𝑃(𝑥𝑚, 𝑥𝑚+1) … 𝑃(𝑥𝑚+𝑛−1, 𝑥𝑚+𝑛)𝛿𝑖𝑥𝑚 Hence (𝑋𝑚+𝑛)𝑛≥0 ∼Markov (𝛿𝑖, 𝑃) conditional on 𝑋𝑚= 𝑖. Now it suffices to show inde-pendence between the past and future variables. In particular, we need to show 𝑚≤𝑖1 < ⋯< 𝑖𝑘for some 𝑘∈ℕimplies that ℙ(𝑋𝑖1 = 𝑥𝑚+1, … , 𝑋𝑖𝑘= 𝑥𝑚+𝑘, 𝑋0 = 𝑥0, … , 𝑋𝑚= 𝑥𝑚∣𝑋𝑚= 𝑖) = ℙ(𝑋𝑖1 = 𝑥𝑚+1, … , 𝑋𝑖𝑘= 𝑥𝑚+𝑘∣𝑋𝑚= 𝑖) ℙ(𝑋0 = 𝑥0, … , 𝑋𝑚= 𝑥𝑚∣𝑋𝑚= 𝑖) So let 𝑖= 𝑥𝑚, and then = ℙ(𝑋𝑖1 = 𝑥𝑚+1, … , 𝑋𝑖𝑘= 𝑥𝑚+𝑘, 𝑋0 = 𝑥0, … , 𝑋𝑚= 𝑥𝑚) ℙ(𝑋𝑚= 𝑖) = 𝜆𝑥0𝑃(𝑥0, 𝑥1) … 𝑃(𝑥𝑚−1, 𝑥𝑚)ℙ(𝑋𝑖1 = 𝑥𝑚+1, … , 𝑋𝑖𝑘= 𝑥𝑚+𝑘∣𝑋𝑚= 𝑥𝑚) ℙ(𝑥𝑚= 𝑖) = ℙ(𝑋0 = 𝑥0, … , 𝑋𝑚= 𝑥𝑚) ℙ(𝑋𝑚= 𝑥𝑚) ℙ(𝑋𝑖1 = 𝑥𝑚+1, … , 𝑋𝑖𝑘= 𝑥𝑚+𝑘∣𝑋𝑚= 𝑥𝑚) which gives the result as required.
105 III. Markov Chains 1.6. Powers of the transition matrix Suppose 𝑋∼Markov (𝜆, 𝑃) with values in 𝐼. If 𝐼is finite, then 𝑃is an |𝐼| × |𝐼| square matrix.
In this case, we can label the states as 1, … , |𝐼|. If 𝐼is infinite, then we label the states using the natural numbers ℕ. Let 𝑥∈𝐼and 𝑛∈ℕ. Then, ℙ(𝑋𝑛= 𝑥) = ∑ 𝑥0,…,𝑥𝑛−1∈𝐼 ℙ(𝑋𝑛= 𝑥, 𝑋𝑛−1 = 𝑥𝑛−1, … , 𝑋0 = 𝑥0) = ∑ 𝑥0,…,𝑥𝑛−1∈𝐼 𝜆𝑥0𝑃(𝑥0, 𝑥1) … 𝑃(𝑥𝑛−1, 𝑥) We can think of 𝜆as a row vector. So we can write this as = (𝜆𝑃𝑛)𝑥 By convention, we take 𝑃0 = 𝐼, the identity matrix. Now, suppose 𝑚, 𝑛∈ℕ. By the simple Markov property, ℙ(𝑋𝑚+𝑛= 𝑦∣𝑋𝑚= 𝑥) = ℙ(𝑋𝑛= 𝑦∣𝑋0 = 𝑥) = (𝛿𝑥𝑃𝑛)𝑦 We will write ℙ𝑥(𝐴) ≔ℙ(𝐴∣𝑋0 = 𝑥) as an abbreviation. Further, we write 𝑝𝑖𝑗(𝑛) for the (𝑖, 𝑗) element of 𝑃𝑛. We have therefore proven the following theorem.
Theorem.
ℙ(𝑋𝑛= 𝑥) = (𝜆𝑃𝑛)𝑥; ℙ(𝑋𝑛+𝑚= 𝑦∣𝑋𝑚= 𝑥) = ℙ𝑥(𝑋𝑛= 𝑦) = 𝑝𝑥𝑦(𝑛) 1.7. Calculating powers Example. Consider 𝑃= (1 −𝛼 𝛼 𝛽 1 −𝛽) ; 𝛼, 𝛽∈[0, 1] Note that for any stochastic matrix 𝑃, 𝑃𝑛is a stochastic matrix. First, we have 𝑃𝑛+1 = 𝑃𝑛𝑃.
Let us begin by finding 𝑝11(𝑛+ 1).
𝑝11(𝑛+ 1) = 𝑝11(𝑛)(1 −𝛼) + 𝑝12(𝑛)𝛽 Note that 𝑝11(𝑛) + 𝑝12(𝑛) = 1 since 𝑃𝑛is stochastic. Therefore, 𝑝11(𝑛+ 1) = 𝑝11(𝑛)(1 −𝛼−𝛽) + 𝛽 We can solve this recursion relation to find 𝑝11(𝑛) = { 𝛼 𝛼+𝛽+ 𝛼 𝛼+𝛽(1 −𝛼−𝛽)𝑛 𝛼+ 𝛽> 0 1 𝛼+ 𝛽= 0 106 1. Introduction The general procedure for finding 𝑃𝑛is as follows. Suppose that 𝑃is a 𝑘× 𝑘matrix. Then let 𝜆1, … , 𝜆𝑘be its eigenvalues (which may not be all distinct).
(1) All 𝜆𝑖distinct. In this case, 𝑃is diagonalisable, and hence we can write 𝑃= 𝑈𝐷𝑈−1 where 𝑈is a diagonal matrix, whose diagonal entries are the 𝜆𝑖. Then, 𝑃𝑛= 𝑈𝐷𝑛𝑈−1.
Calculating 𝐷𝑛may be done termwise since 𝐷is diagonal. In this case, we have terms such as 𝑝11(𝑛) = 𝑎1𝜆𝑛 1 + ⋯+ 𝑎𝑘𝜆𝑛 𝑘; 𝑎𝑖∈ℝ First, note 𝑃0 = 𝐼hence 𝑝11(0) = 1. We can substitute small values of 𝑛and then solve the system of equations. Now, suppose 𝜆𝑘is complex for some 𝑘. In this case, 𝜆𝑘is also an eigenvalue. Then, up to reordering, 𝜆𝑘= 𝑟𝑒𝑖𝜃= 𝑟(cos 𝜃+ 𝑖sin 𝜃); 𝜆𝑘−1 = 𝜆𝑘= 𝑟𝑒𝑖𝜃= 𝑟(cos 𝜃−𝑖sin 𝜃) We can instead write 𝑝11(𝑛) as 𝑝11(𝑛) = 𝑎1𝜆𝑛 1 + ⋯+ 𝑎𝑘−1𝑟𝑛cos(𝑛𝜃) + 𝑎𝑘𝑟𝑛sin(𝑛𝜃) Since 𝑝11(𝑛) is real, all the imaginary parts disappear, so we can simply ignore them.
(2) Not all 𝜆𝑖distinct. In this case, 𝜆appears with multiplicity 2, then we include also the term (𝑎𝑛+ 𝑏)𝜆𝑛as well as 𝑏𝜆𝑛. This can be shown by considering the Jordan normal form of 𝑃.
Example. Let 𝑃= ⎛ ⎜ ⎜ ⎝ 0 1 0 0 1 2 1 2 1 2 0 1 2 ⎞ ⎟ ⎟ ⎠ The eigenvalues are 1, 1 2𝑖, − 1 2𝑖. Then, writing 𝑖 2 = 1 2(cos 𝜋 2 + 𝑖sin 𝜋 2 ), we can write 𝑝11(𝑛) = 𝛼+ 𝛽(1 2) 𝑛 cos 𝑛𝜋 2 + 𝛾(1 2) 𝑛 sin 𝑛𝜋 2 For 𝑛= 0 we have 𝑝11(0) = 1, and for 𝑛= 1 we have 𝑝11(1) = 0, and for 𝑛= 2 we can calculate 𝑃2 and find 𝑝11(2) = 0. Solving this system of equations for 𝛼, 𝛽, 𝛾, we can find 𝑝11(𝑛) = 1 5 + (1 2) 𝑛 (4 5 cos 𝑛𝜋 2 −2 5 sin 𝑛𝜋 2 ) 107 III. Markov Chains 2. Elementary properties 2.1. Communicating classes Definition. Let 𝑋be a Markov chain with transition matrix 𝑃and values in 𝐼. For 𝑥, 𝑦∈𝐼, we say that 𝑥leads to 𝑦, written 𝑥→𝑦, if ℙ𝑥(∃𝑛≥0, 𝑋𝑛= 𝑦) > 0 We say that 𝑥communicates with 𝑦and write 𝑥↔𝑦if 𝑥→𝑦and 𝑦→𝑥.
Theorem. The following are equivalent: (i) 𝑥→𝑦 (ii) There exists a sequence of states 𝑥= 𝑥0, 𝑥1, … , 𝑥𝑘= 𝑦such that 𝑃(𝑥0, 𝑥1)𝑃(𝑥1, 𝑥2) … 𝑃(𝑥𝑘−1, 𝑥𝑘) > 0 (iii) There exists 𝑛≥0 such that 𝑝𝑥𝑦(𝑛) > 0.
Proof. First, we show (i) and (iii) are equivalent. If 𝑥→𝑦, then ℙ𝑥(∃𝑛≥0, 𝑋𝑛= 𝑦) > 0.
Then if ℙ𝑥(∃𝑛≥0, 𝑋𝑛= 𝑦) > 0 we must have some 𝑛≥0 such that ℙ𝑥(𝑋𝑛= 𝑦) = 𝑝𝑥𝑦(𝑛) > 0. Note that we can write (i) as ℙ𝑥(⋃ ∞ 𝑛=0 𝑋𝑛= 𝑦) > 0. If there exists 𝑛≥0 such that 𝑝𝑥𝑦(𝑛) > 0, then certainly the probability of the union is also positive.
Now we show (ii) and (iii) are equivalent. We can write 𝑝𝑥𝑦(𝑛) = ∑ 𝑥1,…,𝑥𝑛−1 𝑃(𝑥, 𝑥1) … 𝑃(𝑥𝑛−1, 𝑦) which leads directly to the equivalence of (ii) with (iii).
Corollary. Communication is an equivalence relation on 𝐼.
Proof. 𝑥↔𝑥since 𝑝𝑥𝑥(0) = 1. If 𝑥→𝑦and 𝑦→𝑧then by (ii) above, 𝑥→𝑧.
Definition. The equivalence classes induced on 𝐼by the communication equivalence re-lation are called communicating classes. A communicating class 𝐶is closed if 𝑥∈𝐶, 𝑥→ 𝑦⟹𝑦∈𝐶.
Definition. A transition matrix 𝑃is called irreducible if it has a single communicating class.
In other words, ∀𝑥, 𝑦∈𝐼, 𝑥↔𝑦.
Definition. A state 𝑥is called absorbing if {𝑥} is a closed (communicating) class.
108 2. Elementary properties 2.2. Hitting times Definition. For 𝐴⊆𝐼, we define the hitting time of 𝐴to be a random variable 𝑇𝐴∶Ω → {0, 1, 2 … } ∪{∞}, defined by 𝑇𝐴(𝜔) = inf {𝑛≥0∶𝑋𝑛(𝜔) ∈𝐴} with the convention that inf ∅= ∞. The hitting probability of 𝐴is ℎ𝐴∶𝐼→[0, 1], defined by ℎ𝐴 𝑖= ℙ𝑖(𝑇𝐴< ∞) The mean hitting time of 𝐴is 𝑘𝐴∶𝐼→[0, ∞], defined by 𝑘𝐴 𝑖= 𝔼𝑖[𝑇𝐴] = ∞ ∑ 𝑛=0 𝑛ℙ𝑖(𝑇𝐴= 𝑛) + ∞ℙ𝑖(𝑇𝐴= ∞) Example. Consider 𝑃= ⎛ ⎜ ⎜ ⎝ 1 0 0 0 1/2 0 1/2 0 0 1/2 0 1/2 0 0 0 1 ⎞ ⎟ ⎟ ⎠ Consider 𝐴= {4}.
ℎ𝐴 1 = 0 ℎ𝐴 2 = ℙ2 (𝑇𝐴< ∞) = 1 2ℎ𝐴 1 + 1 2ℎ𝐴 3 ℎ𝐴 3 = 1 2 ⋅1 + 1 2ℎ𝐴 2 Hence ℎ𝐴 2 = 1 3. Now, consider 𝐵= {1, 4}.
𝑘𝐵 1 = 𝑘𝐵 4 = 0 𝑘𝐵 2 = 1 + 1 2𝑘𝐵 1 + 1 2𝑘𝐵 3 𝑘𝐵 3 = 1 + 1 2𝑘𝐵 4 + 1 2𝑘𝐵 2 Hence 𝑘𝐵 2 = 2.
Theorem. Let 𝐴⊂𝐼. Then the vector (ℎ𝐴 𝑖)𝑖∈𝐴is the minimal non-negative solution to the system ℎ𝐴 𝑖= {1 𝑖∈𝐴 ∑𝑗𝑃(𝑖, 𝑗)ℎ𝐴 𝑗 𝑖∉𝐴 Minimality here means that if (𝑥𝑖)𝑖∈𝐼is another non-negative solution, then ∀𝑖, ℎ𝐴 𝑖≤𝑥𝑖.
109 III. Markov Chains Note. The vector ℎ𝐴 𝑖= 1 always satisfies the equation, since 𝑃is stochastic, but is typically not minimal.
Proof. First, we will show that (ℎ𝑖)𝑖∈𝐴solves the system of equations. Certainly if 𝑖∈𝐴then ℎ𝐴 𝑖= 1. Suppose 𝑖∉𝐴. Consider the event {𝑇𝐴< ∞}. We can write this event as a disjoint union of the following events: {𝑇𝐴< ∞} = {𝑋0 ∈𝐴} ∪ ∞ ⋃ 𝑛=1 {𝑋0 ∉𝐴, … , 𝑋𝑛−1 ∉𝐴, 𝑋𝑛∈𝐴} By countable additivity, ℙ𝑖(𝑇𝐴< ∞) = ℙ𝑖(𝑋0 ∈𝐴) ⏟⎵ ⎵⏟⎵ ⎵⏟ =0 + ∞ ∑ 𝑛=1 ℙ𝑖(𝑋0 ∉𝐴, … , 𝑋𝑛−1 ∉𝐴, 𝑋𝑛∈𝐴) = ∞ ∑ 𝑛=1 ∑ 𝑗 ℙ(𝑋0 ∉𝐴, … , 𝑋𝑛−1 ∉𝐴, 𝑋𝑛∈𝐴, 𝑋1 ∈𝑗∣𝑋0 = 𝑖) = ∑ 𝑗 ℙ(𝑋1 ∈𝐴, 𝑋1 = 𝑗∣𝑋0 = 𝑖) + ∞ ∑ 𝑛=2 ∑ 𝑗 ℙ(𝑋1 ∉𝐴, … , 𝑋𝑛−1 ∉𝐴, 𝑋𝑛∈𝐴, 𝑋1 ∈𝑗∣𝑋0 = 𝑖) = ∑ 𝑗 𝑃(𝑖, 𝑗)ℙ(𝑋1 ∈𝐴∣𝑋1 = 𝑗, 𝑋0 = 𝑖) + ∑ 𝑗 𝑃(𝑖, 𝑗) ∞ ∑ 𝑛=2 ℙ(𝑋1 ∉𝐴, … , 𝑋𝑛−1 ∉𝐴, 𝑋𝑛∈𝐴∣𝑋1 ∈𝑗, 𝑋0 = 𝑖) 110 2. Elementary properties By the definition of the Markov chain, we can drop the condition on 𝑋0, and subtract one from all indices.
= ∑ 𝑗 𝑃(𝑖, 𝑗)ℙ(𝑋0 ∈𝐴∣𝑋0 = 𝑗) + ∑ 𝑗 𝑃(𝑖, 𝑗) ∞ ∑ 𝑛=2 ℙ(𝑋1 ∉𝐴, … , 𝑋𝑛−1 ∉𝐴, 𝑋𝑛∈𝐴∣𝑋1 ∈𝑗) = ∑ 𝑗 𝑃(𝑖, 𝑗)ℙ(𝑋0 ∈𝐴∣𝑋0 = 𝑗) + ∑ 𝑗 𝑃(𝑖, 𝑗) ∞ ∑ 𝑛=2 ℙ𝑗(𝑋0 ∉𝐴, … , 𝑋𝑛−2 ∉𝐴, 𝑋𝑛−1 ∈𝐴) = ∑ 𝑗 𝑃(𝑖, 𝑗)(ℙ𝑗(𝑋0 ∈𝐴) + ∞ ∑ 2 ℙ𝑗(𝑋0 ∉𝐴, … , 𝑋𝑛−1 ∉𝐴, 𝑋𝑛∈𝐴)) = ∑ 𝑗 𝑃(𝑖, 𝑗)(ℙ𝑗(𝑇𝐴= 0) + ∞ ∑ 𝑛=1 ℙ𝑗(𝑇𝐴= 𝑛)) = ∑ 𝑗 𝑃(𝑖, 𝑗)ℙ𝑗(𝑇𝐴< ∞) = ∑ 𝑗 𝑃(𝑖, 𝑗)ℎ𝐴 𝑗 Now we must show minimality. If (𝑥𝑖) is another non-negative solution, we must show that ℎ𝐴 𝑖≤𝑥𝑖. We have 𝑥𝑖= ∑ 𝑗 𝑃(𝑖, 𝑗)𝑥𝑗= ∑ 𝑗∈𝐴 𝑃(𝑖, 𝑗) + ∑ 𝑗∉𝐴 𝑃(𝑖, 𝑗)𝑥𝑗 Substituting again, 𝑥𝑖= ∑ 𝑗∈𝐴 𝑃(𝑖, 𝑗)𝑥𝑗+ ∑ 𝑗∉𝐴 𝑃(𝑖, 𝑗)( ∑ 𝑘∈𝐴 𝑃(𝑗, 𝑘) + ∑𝑘∉𝐴𝑃(𝑗, 𝑘)𝑥𝑘) Then 𝑥𝑖= ∑ 𝑗1∈𝐴 𝑃(𝑖, 𝑗1) + ∑ 𝑗1∉𝐴 ∑ 𝑗2∈𝐴 𝑃(𝑖, 𝑗1)𝑃(𝑗1, 𝑗2) + ⋯ + ∑ 𝑗1∉𝐴,…,𝑗𝑛−1∉𝐴,𝑗𝑛∈𝐴 𝑃(𝑖, 𝑗1) … 𝑃(𝑗𝑛−1, 𝑗𝑛) + ∑ 𝑗1∉𝐴…,𝑗𝑛∉𝐴 𝑃(𝑖, 𝑗1) … 𝑃(𝑗𝑛−1, 𝑗𝑛)𝑥𝑗𝑛 The last term is non-negative since 𝑥is non-negative. So 𝑥𝑖≥ℙ𝑖(𝑇𝐴= 1) + ℙ𝑖(𝑇𝐴= 2) + ⋯+ ℙ𝑖(𝑇𝐴= 𝑛) ≥ℙ𝑖(𝑇𝐴≤𝑛) , ∀𝑛∈ℕ Now, note {𝑇𝐴≤𝑛} are a set of increasing functions of 𝑛, so by continuity of the probability measure, the probability increases to that of the union, {𝑇𝐴< ∞} = ℎ𝐴 𝑖.
111 III. Markov Chains Example. Consider the Markov chain previously explored: 𝑃= ⎛ ⎜ ⎜ ⎝ 1 0 0 0 1/2 0 1/2 0 0 1/2 0 1/2 0 0 0 1 ⎞ ⎟ ⎟ ⎠ Let 𝐴= {4}. Then ℎ𝐴 1 = 0 since there is no route from 1 to 4. From the theorem above, the system of linear equations is ℎ2 = 1 2ℎ1 + 1 2ℎ3 ℎ3 = 1 2ℎ4 + 1 2ℎ2 ℎ4 = 1 Hence, ℎ2 = 2 3ℎ1 + 1 3 ℎ3 = 1 3ℎ1 + 2 3 So the minimal solution arises at ℎ1 = 0.
Example. Consider 𝐼= ℕ, and 𝑃(𝑖, 𝑖+ 1) = 𝑝∈(0, 1); 𝑃(𝑖, 𝑖−1) = 1 −𝑝= 𝑞 Then ℎ𝑖= ℙ𝑖(𝑇0 < ∞) hence ℎ0 = 1. The linear equations are 𝑝≠𝑞⟹ℎ𝑖= 𝑝ℎ𝑖+1 + 𝑞ℎ𝑖−1 𝑝(ℎ𝑖+1 −ℎ𝑖) = 𝑞(ℎ𝑖−ℎ𝑖−1) Let 𝑢𝑖= ℎ𝑖−ℎ𝑖−1. Then, 𝑞 𝑝𝑢𝑖= ⋯= ( 𝑞 𝑝) 𝑖 𝑢1 Hence ℎ𝑖= 𝑖 ∑ 𝑗=1 (ℎ𝑗−ℎ𝑗−1) + 1 = 1 −(1 −ℎ1) 𝑖 ∑ 𝑗=1 ( 𝑞 𝑝) 𝑗 The general solution is therefore ℎ𝑖= 𝑎+ 𝑏( 𝑞 𝑝) 𝑖 If 𝑞> 𝑝, then minimality of ℎ𝑖implies 𝑏= 0, 𝑎= 1. Hence, ℎ𝑖= 1 112 2. Elementary properties Otherwise, if 𝑝> 𝑞, minimality of ℎ𝑖implies 𝑎= 0, 𝑏= 1. Hence, ℎ𝑖= ( 𝑞 𝑝) 𝑖 If 𝑝= 𝑞= 1 2, then ℎ𝑖= 1 2ℎ𝑖+1 + 1 2ℎ𝑖−1 Hence, ℎ𝑖= 𝑎+ 𝑏𝑖. Minimality implies 𝑎= 1 and 𝑏= 0.
ℎ𝑖= 1 2.3. Birth and death chain Consider a Markov chain on ℕwith 𝑃(𝑖, 𝑖+ 1) = 𝑝𝑖; 𝑃(𝑖, 𝑖−1) = 𝑞𝑖; ∀𝑖, 𝑝𝑖+ 𝑞𝑖= 1 Now, consider ℎ𝑖= ℙ𝑖(𝑇0 < ∞). ℎ0 = 1, and ℎ𝑖= 𝑝𝑖ℎ𝑖+1 + 𝑞𝑖ℎ𝑖−1.
𝑝𝑖(ℎ𝑖+1 −ℎ𝑖) = 𝑞𝑖(ℎ𝑖−ℎ𝑖−1) Let 𝑢𝑖= ℎ𝑖−ℎ𝑖−1 to give 𝑢𝑖+1 = 𝑞𝑖 𝑝𝑖 𝑢𝑖= ∏𝑗= 1𝑖𝑞𝑖 𝑝𝑖 ⏟ ⎵ ⎵ ⎵ ⏟ ⎵ ⎵ ⎵ ⏟ 𝛾𝑖 𝑢𝑖 Then ℎ𝑖= 1 −(1 −ℎ1)(𝛾0 + 𝛾1 + ⋯+ 𝛾𝑖−1) where we let 𝛾0 = 1. Since ℎ𝑖is the minimal non-negative solution, ℎ𝑖≥0 ⟹1 −ℎ1 ≤ 1 ∑ 𝑖−1 𝑗=0 𝛾𝑗 ≤ 1 ∑ ∞ 𝑗=0 𝛾𝑗 By minimality, we must have exactly this bound. If ∑ ∞ 𝑗=0 𝛾𝑗= ∞then 1−ℎ1 = 0 ⟹ℎ𝑖= 1 for all 𝑖. If ∑ ∞ 𝑗=0 𝛾𝑗< ∞then ℎ𝑖= ∑ ∞ 𝑗=𝑖𝛾𝑗 ∑ ∞ 𝑗=0 𝛾𝑗 2.4. Mean hitting times Recall that 𝑘𝐴 𝑖= 𝔼𝑖[𝑇𝐴] = ∑ 𝑛 𝑛ℙ𝑖(𝑇𝐴= 𝑛) + ∞ℙ𝑖(𝑇𝐴= ∞) 113 III. Markov Chains Theorem. The vector (𝑘𝐴 𝑖)𝑖∈𝐼is the minimal non-negative solution to the system of equa-tions 𝑘𝐴 𝑖= {0 if 𝑖∈𝐴 1 + ∑𝑗∉𝐴𝑃(𝑖, 𝑗)𝑘𝐴 𝑗 if 𝑖∉𝐴 Proof. Suppose 𝑖∈𝐴. Then 𝑘𝑖= 0. Now suppose 𝑖∉𝐴. Further, we may assume that ℙ𝑖(𝑇𝐴= ∞) = 0, since if that probability is positive then the claim is trivial. Indeed, if ℙ𝑖(𝑇𝐴= ∞) > 0, then there must exist 𝑗such that 𝑃(𝑖, 𝑗) > 0 and ℙ𝑗(𝑇𝐴= ∞) > 0 since ℙ𝑖(𝑇𝐴< ∞) = ∑ 𝑗 𝑃(𝑖, 𝑗)ℎ𝐴 𝑗 ⟹1 −ℙ𝑖(𝑇𝐴= ∞) = ∑ 𝑗 𝑃(𝑖, 𝑗)(1 −ℙ𝑗(𝑇𝐴= ∞)) Because 𝑃is stochastic, ℙ𝑖(𝑇𝐴= ∞) = ∑ 𝑗 𝑃(𝑖, 𝑗)ℙ𝑗(𝑇𝐴= ∞) so since the left hand side is positive, there must exist 𝑗with 𝑃(𝑖, 𝑗) > 0 and ℙ𝑗(𝑇𝐴= ∞> 0).
For this 𝑗, we also have 𝑘𝐴 𝑗= ∞. Now we only need to compute ∑𝑛𝑛ℙ𝑖(𝑇𝐴= 𝑛).
ℙ𝑖(𝑇𝐴= 𝑛) = ℙ𝑖(𝑋0 ∉𝐴, … , 𝑋𝑛−1 ∉𝐴, 𝑋𝑛∈𝐴) Then, using the same method as the previous theorem, 𝑘𝐴 𝑖= ∑ 𝑛 𝑛ℙ𝑖(𝑇𝐴= 𝑛) = 1 + ∑ 𝑗∉𝐴 𝑃(𝑖, 𝑗)𝑘𝐴 𝑗 It now suffices to prove minimality. Suppose (𝑥𝑖) is another solution to this system of equa-tions. We need to show that 𝑥𝑖≥𝑘𝐴 𝑖for all 𝑖. Suppose 𝑖∉𝐴. Then 𝑥𝑖= 1 + ∑ 𝑗∉𝐴 𝑃(𝑖, 𝑗)𝑥𝑗= 1 + ∑ 𝑗∉𝐴 𝑃(𝑖, 𝑗)(1 + ∑ 𝑘∉𝐴 𝑃(𝑗, 𝑘)𝑥𝑘) Expanding inductively, 𝑥𝑖= 1 + ∑ 𝑗1∉𝐴 𝑃(𝑖, 𝑗1) + ∑ 𝑗1∉𝐴,𝑗2∉𝐴 𝑃(𝑖, 𝑗1)𝑃(𝑗1, 𝑗2) + ⋯ + ∑ 𝑗1∉𝐴,…,𝑗𝑛∉𝐴 𝑃(𝑖, 𝑗1) … 𝑃(𝑗𝑛−1, 𝑗𝑛) + ∑ 𝑗1∉𝐴,…,𝑗𝑛+1∉𝐴 𝑃(𝑖, 𝑗) … 𝑃(𝑗𝑛, 𝑗𝑛+1)𝑥𝑗𝑛+1 Since 𝑥is non-negative, we can remove the last term and reach an inequality.
𝑥𝑖≥1 + ∑ 𝑗1∉𝐴 𝑃(𝑖, 𝑗1) + ∑ 𝑗1∉𝐴,𝑗2∉𝐴 𝑃(𝑖, 𝑗1)𝑃(𝑗1, 𝑗2) + ⋯+ ∑ 𝑗1∉𝐴,…,𝑗𝑛∉𝐴 𝑃(𝑖, 𝑗1) … 𝑃(𝑗𝑛−1, 𝑗𝑛) 114 2. Elementary properties Hence 𝑥𝑖≥1 + ℙ𝑖(𝑇𝐴> 1) + ℙ𝑖(𝑇𝐴> 2) + ⋯+ ℙ𝑖(𝑇𝐴> 𝑛) = ℙ𝑖(𝑇𝐴> 0) + ℙ𝑖(𝑇𝐴> 1) + ℙ𝑖(𝑇𝐴> 2) + ⋯+ ℙ𝑖(𝑇𝐴> 𝑛) = 𝑛 ∑ 𝑘=0 ℙ𝑖(𝑇𝐴> 𝑘) for all 𝑛. Hence, the limit of this sum is 𝑥𝑖≥ ∞ ∑ 𝑘=0 ℙ𝑖(𝑇𝐴> 𝑘) = 𝔼𝑖[𝑇𝐴] which gives minimality as required.
2.5. Strong Markov property The simple Markov property shows that, if 𝑋𝑚= 𝑖, 𝑋𝑚+𝑛∼Markov (𝛿𝑖, 𝑃) and this is independent of 𝑋0, … , 𝑋𝑚. The strong Markov property will show that the same property holds when we replace 𝑚with a finite random ‘time’ variable. It is not the case that any random variable will work; indeed, an 𝑚very dependent on the Markov chain itself might not satisfy this property.
Definition. A random time 𝑇∶Ω →{0, 1, … }∪{∞} is called a stopping time if, for all 𝑛∈ℕ, {𝑇= 𝑛} depends only on 𝑋0, … , 𝑋𝑛.
Example. The hitting time 𝑇𝐴= inf {𝑛≥0∶𝑋𝑛∈𝐴} is a stopping time. This is because we can write {𝑇𝐴= 𝑛} = {𝑋0 ∉𝐴, … , 𝑋𝑛−1 ∉𝐴, 𝑋𝑛∈𝐴} Example. The time 𝐿𝐴= sup {𝑛≥0∶𝑋𝑛∈𝐴} is not a stopping time. This is because we need to know information about the future behaviour of 𝑋𝑛in order to guarantee that we are at the supremum of such events.
Theorem (Strong Markov Property). Let 𝑋∼Markov (𝜆, 𝑃) and 𝑇be a stopping time. Con-ditional on 𝑇< ∞and 𝑋𝑇= 𝑖, (𝑋𝑛+𝑇)𝑛≥0 ∼Markov (𝛿𝑖, 𝑃) and this distribution is independent of 𝑋0, … , 𝑋𝑇.
Proof. We need to show that, for all 𝑥0, … , 𝑥𝑛and for all vectors 𝑤of any length, ℙ(𝑋𝑇= 𝑥0, … , 𝑋𝑇+𝑛= 𝑥𝑛, (𝑋0, … , 𝑋𝑇) = 𝑤∣𝑇< ∞, 𝑋𝑇= 𝑖) = 𝛿𝑖𝑥0𝑃(𝑥0, 𝑥1) … 𝑃(𝑥𝑛−1, 𝑥𝑛)ℙ((𝑋0, … , 𝑋𝑇) = 𝑤∶𝑇< ∞, 𝑋𝑇= 𝑖) 115 III. Markov Chains Suppose that 𝑤is of the form 𝑤= (𝑤0, … , 𝑤𝑘). Then, ℙ(𝑋𝑇= 𝑋0, … , 𝑋𝑇+𝑛= 𝑥𝑛, (𝑋0, … , 𝑋𝑇) = 𝑤∣𝑇< ∞, 𝑋𝑇= 𝑖) = ℙ(𝑋𝑘= 𝑥0, … , 𝑋𝑘+𝑛= 𝑥𝑛, (𝑋0, … , 𝑋𝑘) = 𝑤, 𝑇= 𝑘, 𝑋𝑘= 𝑖) ℙ(𝑇< ∞, 𝑋𝑇= 𝑖) Now, since {𝑇= 𝑘} depends only on 𝑋0, … , 𝑋𝑘, by the simple Markov property we have ℙ(𝑋𝑘= 𝑥0, … , 𝑋𝑘+𝑛= 𝑥𝑛∣(𝑋0, … , 𝑋𝑘) = 𝑤, 𝑇= 𝑘, 𝑋𝑘= 𝑖) = ℙ(𝑋𝑘= 𝑥0, … , 𝑋𝑘+𝑛= 𝑥𝑛∣𝑋𝑘= 𝑖) = 𝛿𝑖𝑥0𝑃(𝑥0, 𝑥1) … 𝑃(𝑥𝑛−1, 𝑥𝑛) Now, ℙ(𝑋𝑇= 𝑥0, … , 𝑋𝑇+𝑛= 𝑥𝑛, (𝑋0, … , 𝑋𝑇) = 𝑤∣𝑇< ∞, 𝑋𝑇= 𝑖) = 𝛿𝑖𝑥0𝑃(𝑥0, 𝑥1) … 𝑃(𝑥𝑛−1, 𝑥𝑛)ℙ((𝑋0, … , 𝑋𝑘) = 𝑤∶𝑇= 𝑘, 𝑋𝑘= 𝑖) ℙ(𝑇< ∞, 𝑋𝑇= 𝑖) = 𝛿𝑖𝑥0𝑃(𝑥0, 𝑥1) … 𝑃(𝑥𝑛−1, 𝑥𝑛)ℙ((𝑋0, … , 𝑋𝑇) = 𝑤∶𝑇< ∞, 𝑋𝑇= 𝑖) as required.
Example. Consider a simple random walk on 𝐼= ℕ, where 𝑃(𝑥, 𝑥± 1) = 1 2 for 𝑥≠0, and 𝑃(0, 1) = 1. Now, let ℎ𝑖= ℙ𝑖(𝑇0 < ∞). We want to calculate ℎ1. We can write ℎ1 = 1 2 + 1 2ℎ2 but the system of recursion relations this generates is difficult to solve. Instead, we will write ℎ2 = ℙ2 (𝑇0 < ∞) Note that in order to hit 0, we must first hit 1. So conditioning on the first hitting time of 1 being finite, after this time the process starts again from 1. We can write 𝑇0 = 𝑇1 + ˜ 𝑇0, where ˜ 𝑇0 is independent of 𝑇1, with the same distribution as 𝑇0 under ℙ1. Now, ℎ2 = ℙ2 (𝑇0 < ∞, 𝑇1 < ∞) = ℙ2 (𝑇0 < ∞∣𝑇1 < ∞) ℙ2 (𝑇2 < ∞) Note that ℙ2 (𝑇0 < ∞∣𝑇1 < ∞) = ℙ2 (𝑇1 + ˜ 𝑇0 < ∞∣𝑇1 < ∞) = ℙ2 (˜ 𝑇0 < ∞∣𝑇1 < ∞) = ℙ1 (𝑇0 < ∞) But ℙ2 (𝑇1 < ∞) = ℙ1 (𝑇0 < ∞), so ℎ2 = ℙ2 (𝑇1 < ∞) ℙ1 (𝑇0 < ∞) By translation invariance, ℎ2 = ℎ2 1 In general, therefore, for any 𝑛∈ℕ, ℎ𝑛= ℎ𝑛 1 116 3. Transience and recurrence 3. Transience and recurrence 3.1. Definitions Definition. Let 𝑋be a Markov chain, and let 𝑖∈𝐼. 𝑖is called recurrent if ℙ𝑖(𝑋𝑛= 𝑖for infinitely many 𝑛) = 1 𝑖is called transient if ℙ𝑖(𝑋𝑛= 𝑖for infinitely many 𝑛) = 0 We will prove that any 𝑖is either recurrent or transient.
3.2. Probability of visits Definition. Let 𝑇(0) 𝑖 = 0 and inductively define 𝑇(𝑟+1) 𝑖 = inf {𝑛≥𝑇(𝑟) 𝑖 + 1∶𝑋𝑛= 𝑖} We write 𝑇(1) 𝑖 = 𝑇𝑖, called the first return time (or first passage time) to 𝑖. Further, let 𝑓𝑖= ℙ𝑖(𝑇𝑖< ∞) and let the number of visits to 𝑖be defined by 𝑉𝑖= ∞ ∑ 𝑛=0 1(𝑋𝑛= 𝑖) Lemma. For all 𝑟∈ℕ, 𝑖∈𝐼, ℙ𝑖(𝑉𝑖> 𝑟) = 𝑓𝑟 𝑖.
Proof. For 𝑟= 0, this is trivially true. Now, suppose that the statement is true for 𝑟, and we will show that it is true for 𝑟+ 1.
ℙ𝑖(𝑉𝑖> 𝑟+ 1) = ℙ𝑖(𝑇(𝑟+1) 𝑖 < ∞) = ℙ𝑖(𝑇(𝑟+1) 𝑖 < ∞, 𝑇(𝑟) 𝑖 < ∞) = ℙ𝑖(𝑇(𝑟+1) 𝑖 < ∞∣𝑇(𝑟) 𝑖 < ∞) ℙ𝑖(𝑇(𝑟) 𝑖 < ∞) = ℙ𝑖(𝑇(𝑟+1) 𝑖 < ∞∣𝑇(𝑟) 𝑖 < ∞) ℙ𝑖(𝑉𝑖> 𝑟) = ℙ𝑖(𝑇(𝑟+1) 𝑖 < ∞∣𝑇(𝑟) 𝑖 < ∞) 𝑓𝑟 𝑖 By the strong Markov property applied to the stopping time 𝑇(𝑟) 𝑖 , = ℙ𝑖(𝑇𝑖< ∞) 𝑓𝑟 𝑖 = 𝑓𝑖𝑓𝑟 𝑖 = 𝑓𝑟+1 𝑖 117 III. Markov Chains 3.3. Duality of transience and recurrence Theorem. Let 𝑋be a Markov chain with transition matrix 𝑃, and let 𝑖∈𝐼. Then, exactly one of the following is true.
(i) If ℙ𝑖(𝑇𝑖< ∞) = 1, then 𝑖is recurrent, and ∞ ∑ 𝑛=0 𝑝𝑖𝑖(𝑛) = ∞ (ii) If ℙ𝑖(𝑇𝑖< ∞) < 1, then 𝑖is transient, and ∞ ∑ 𝑛=0 𝑝𝑖𝑖(𝑛) < ∞ Proof.
𝔼𝑖[𝑉𝑖] = 𝔼𝑖[ ∞ ∑ 𝑛=0 1(𝑋𝑛= 𝑖)] = ∞ ∑ 𝑛=0 𝔼𝑖[1(𝑋𝑛= 𝑖)] = ∞ ∑ 𝑛=0 ℙ𝑖(𝑋𝑛= 𝑖) = ∞ ∑ 𝑛=0 𝑝𝑖𝑖(𝑛) First, suppose ℙ𝑖(𝑇𝑖< ∞) = 1. Then, for all 𝑟, ℙ𝑖(𝑉𝑖> 𝑟) = 1, so ℙ𝑖(𝑉𝑖= ∞) = 1. Hence, 𝑖is recurrent. Further, 𝔼𝑖[𝑉𝑖] = ∞so ∑ ∞ 𝑛=0 𝑝𝑖𝑖(𝑛) = ∞.
Now, if 𝑓𝑖< 1, by the previous lemma we see that 𝔼𝑖[𝑉𝑖] = 1 1−𝑓𝑖< ∞hence ℙ𝑖(𝑉𝑖< ∞) = 1. Thus, 𝑖is transient. Further, 𝔼𝑖[𝑉𝑖] < ∞so ∑ ∞ 𝑛=0 𝑝𝑖𝑖(𝑛) < ∞.
Theorem. Let 𝑥, 𝑦be states that communicate. Then, either 𝑥and 𝑦are both recurrent, or they are both transient.
Proof. Suppose 𝑥is recurrent. Then, since 𝑥and 𝑦communicate, ∃𝑚, ℓ∈ℕsuch that 𝑝𝑥𝑦(𝑚) > 0; 𝑝𝑦𝑥(ℓ) > 0 Note, ∑𝑛𝑝𝑥𝑥(𝑛) = ∞. Then, 𝑝𝑦𝑦(𝑛) ≥∑ 𝑛 𝑝𝑦𝑦(𝑛+ 𝑚+ ℓ) ≥∑ 𝑛 𝑝𝑦𝑥(ℓ)𝑝𝑥𝑥(𝑛)𝑝𝑥𝑦(𝑚) ≥𝑝𝑦𝑥(ℓ)𝑝𝑥𝑦(𝑚)𝑝𝑥𝑥(𝑛) = ∞ Corollary. Either all states in a communicating class are recurrent or they are all transient.
118 3. Transience and recurrence 3.4. Recurrent communicating classes Theorem. Any recurrent communicating class is closed.
Proof. Suppose a communicating class 𝐶is not closed. Then there exists 𝑥∈𝐶and 𝑦∉𝐶 such that 𝑥→𝑦. Let 𝑚be such that 𝑝𝑥𝑦(𝑚) > 0. If, starting from 𝑥, we hit 𝑦which is outside the communicating class, then we can never return to the communicating class (including 𝑥) again. In particular, ℙ𝑥(𝑉 𝑥< ∞) ≥ℙ𝑥(𝑋𝑚= 𝑦) = 𝑝𝑥𝑦(𝑚) > 0 Hence 𝑥is not recurrent, which is a contradiction.
Theorem. Any finite closed communicating class is recurrent.
Proof. Let 𝐶be a finite closed communicating class. Let 𝑥∈𝐶. Then, by the pigeonhole principle, there must exist 𝑦∈𝐶such that ℙ𝑥(𝑋𝑛= 𝑦for infinitely many 𝑛) > 0 Since 𝑥and 𝑦communicate, there exists 𝑚∈ℕsuch that 𝑝𝑦𝑥(𝑚) > 0. Now, ℙ𝑦(𝑋𝑚= 𝑦for infinitely many 𝑛) ≥ℙ𝑥(𝑋𝑚= 𝑥, 𝑋𝑛= 𝑦for infinitely many 𝑛≥𝑚) = ℙ𝑥(𝑋𝑛= 𝑦for infinitely many 𝑛≥𝑚∣𝑋𝑚= 𝑥) ℙ𝑦(𝑋𝑚= 𝑥) = ℙ𝑥(𝑋𝑛= 𝑦for infinitely many 𝑛) ℙ𝑦(𝑋𝑚= 𝑥) > 0 Thus 𝑦is recurrent. Since recurrence is a class property, 𝐶is recurrent.
Theorem. Let 𝑃be irreducible and recurrent. Then, for all 𝑥, 𝑦, ℙ𝑥(𝑇 𝑦< ∞) = 1 Proof. Since 𝑦is recurrent, 1 = ℙ𝑦(𝑋𝑛= 𝑦for infinitely many 𝑛) Let 𝑚such that 𝑝𝑦𝑥(𝑚) > 0. Now, 1 = ℙ𝑦(𝑋𝑛= 𝑦infinitely often) = ∑ 𝑧 ℙ𝑦(𝑋𝑚= 𝑧, 𝑋𝑛= 𝑦for infinitely many 𝑛≥𝑚) = ∑ 𝑧 ℙ𝑦(𝑋𝑛= 𝑦for infinitely many 𝑛≥𝑚∣𝑋𝑚= 𝑧) ℙ𝑦(𝑋𝑚= 𝑧) = ∑ 𝑧 ℙ𝑧(𝑋𝑛= 𝑦for infinitely many 𝑛) ℙ𝑦(𝑋𝑚= 𝑧) 119 III. Markov Chains By the strong Markov property, = ∑ 𝑧 ℙ𝑧(𝑇 𝑦< ∞) ℙ𝑦(𝑋𝑛= 𝑦for infinitely many 𝑛) ℙ𝑦(𝑋𝑚= 𝑧) Since 𝑦is recurrent, = ∑ 𝑧 ℙ𝑧(𝑇 𝑦< ∞) ℙ𝑦(𝑋𝑚= 𝑧) = ∑ 𝑧 ℙ𝑧(𝑇 𝑦< ∞) 𝑝𝑦𝑧(𝑚) Since 𝑝𝑦𝑧(𝑚) > 0 and ∑𝑧𝑝𝑦𝑧(𝑚) = 1, ℙ𝑥(𝑇 𝑦< ∞) = 1.
120 4. Pólya’s recurrence theorem 4. Pólya’s recurrence theorem 4.1. Statement of theorem Definition. The simple random walk in ℤ𝑑is the Markov chain defined by 𝑃(𝑥, 𝑥+ 𝑒𝑖) = 𝑃(𝑥, 𝑥−𝑒𝑖) = 1 2𝑑 where 𝑒𝑖is the standard basis.
Theorem. The simple random walk in ℤ𝑑is recurrent for 𝑑= 1, 𝑑= 2 and transient for 𝑑≥3.
4.2. One-dimensional proof Consider 𝑑= 1. In this case, 𝑃(𝑥, 𝑥+1) = 𝑃(𝑥, 𝑥−1) = 1 2. We will show that ∑𝑛𝑝00(𝑛) = ∞, then recurrence will hold. We have 𝑝00(𝑛) = ℙ0 (𝑋𝑛= 0). Note that if 𝑛is odd, 𝑋𝑛is odd, so ℙ0 (𝑋2𝑘+1 = 0) = 0. So we will consider only even numbers. In order to be back at zero after 2𝑛steps, we must make 𝑛steps ‘up’ away from the origin and make 𝑛steps ‘down’. There are (2𝑛 𝑛) ways of choosing which steps are ‘up’ steps. The probability of a specific choice of 𝑛 ‘up’ and 𝑛‘down’ is ( 1 2) 2𝑛 . Hence, 𝑝00(2𝑛) = (2𝑛 𝑛)(1 2) 2𝑛 = (2𝑛)!
(𝑛!)2 ⋅1 22𝑛 Recall Stirling’s formula: 𝑛! ∼𝑛𝑛𝑒−𝑛√2𝜋𝑛 Substituting in, (2𝑛)!
(𝑛!)2 ⋅1 22𝑛∼ 1 √𝜋𝑛 = 𝐴 √𝑛 for 𝐴> 0; the precise value of 𝐴is unnecessary. Hence, for some large 𝑛0, ∀𝑛≥𝑛0, 𝑝00(2𝑛) ≥ 𝐴 2√𝑛. So ∑ 𝑛 𝑝00(2𝑛) ≥∑ 𝑛≥𝑛0 𝐴 2√𝑛 = ∞ Now, let us consider the asymmetric random walk for 𝑑= 1, defined by 𝑃(𝑥, 𝑥+ 1) = 𝑝and 𝑃(𝑥, 𝑥−1) = 𝑞. We can compute 𝑝00(2𝑛) = (2𝑛 𝑛)(𝑝𝑞)𝑛∼𝐴 (4𝑝𝑞)𝑛 √𝑛. If 𝑝≠𝑞, then 4𝑝𝑞< 1 so by the geometric series we have ∑ 𝑛≥𝑛0 𝑝00(2𝑛) ≤∑ 𝑛≥𝑛0 2𝐴(4𝑝𝑞)𝑛< ∞ So the asymmetric random walk is transient.
121 III. Markov Chains 4.3. Two-dimensional proof Now, let us consider the simple random walk on ℤ2. For each point (𝑥, 𝑦) ∈ℤ2, we will project this coordinate onto the lines 𝑦= 𝑥and 𝑦= −𝑥. In particular, we define 𝑓(𝑥, 𝑦) = (𝑥+ 𝑦 √2 , 𝑥−𝑦 √2 ) If 𝑋𝑛is the simple random walk on ℤ2, we consider 𝑓(𝑋𝑛) = (𝑋+ 𝑛, 𝑋− 𝑛).
Lemma. (𝑋+ 𝑛), (𝑋− 𝑛) are independent simple random walks on 1 √2ℤ.
Proof. We can write 𝑋𝑛as 𝑋𝑛= 𝑛 ∑ 𝑖=1 𝜉𝑖 where 𝜉𝑖are independent and identically distributed by ℙ(𝜉1 = (1, 0)) = ℙ(𝜉1 = (−1, 0)) = ℙ(𝜉1 = (0, 1)) = ℙ(𝜉1 = (0, −1)) = 1 4 and we write 𝜉𝑖= (𝜉1 𝑖, 𝜉2 𝑖). We can then see that 𝑋+ 𝑛= 𝑛 ∑ 𝑖=1 𝜉1 𝑖+ 𝜉2 𝑖 √2 ; 𝑋− 𝑛= 𝑛 ∑ 𝑖=1 𝜉1 𝑖−𝜉2 𝑖 √2 We can check that (𝑋+ 𝑛), (𝑋− 𝑛) are simple random walks on 1 √2ℤ. It now suffices to prove the independence property. Note that it suffices to show that 𝜉1 𝑖+𝜉2 1 and 𝜉1 𝑖−𝜉2 𝑖are independent, since the 𝑋+ 𝑛, 𝑋− 𝑛are sums of independent and identically distributed copies of these random variables. We can simply enumerate all possible values of 𝜉1 𝑖, 𝜉2 𝑖and the result follows.
We know that 𝑝00(𝑛) = 0 if 𝑛is odd. We want to find 𝑝00(2𝑛) = ℙ0 (𝑋2𝑛= 0). Note, 𝑋𝑛= 0 ⟺𝑋+ 𝑛= 𝑋− 𝑛= 0. Using the lemma above, ℙ0 (𝑋2𝑛= 0) = ℙ0 (𝑋+ 𝑛= 0, 𝑋− 𝑛= 0) = ℙ0 (𝑋+ 𝑛= 0) ℙ0 (𝑋− 𝑛= 0) ∼𝐴 √𝑛 𝐴 √𝑛 = 𝐴2 𝑛 Hence, ∑ 𝑛≥𝑛0 ℙ0 (𝑋2𝑛= 0) ≥∑ 𝑛≥𝑛0 = 𝐴2 2𝑛= ∞ which gives recurrence as required.
122 4. Pólya’s recurrence theorem 4.4. Three-dimensional proof Consider 𝑑= 3. Again, 𝑝00(𝑛) = 0 if 𝑛odd. In order to return to zero after 2𝑛steps, we must make 𝑖steps both up and down, 𝑗steps north and south, and 𝑘steps east and west, with 𝑖+ 𝑗+ 𝑘= 𝑛. There are ( 2𝑛 𝑖,𝑖,𝑗,𝑗,𝑘,𝑘) ways of choosing which steps in each direction we take. Each combination has probability ( 1 6) 2𝑛 of happening. Hence, 𝑝00(2𝑛) = ∑ 𝑖,𝑗,𝑘≥0,𝑖+𝑗+𝑘=𝑛 ( 2𝑛 𝑖, 𝑖, 𝑗, 𝑗, 𝑘, 𝑘)(1 6) 2𝑛 = (2𝑛 𝑛)(1 2) 2𝑛 ∑ 𝑖,𝑗,𝑘≥0,𝑖+𝑗+𝑘=𝑛 ( 𝑛 𝑖, 𝑗, 𝑘) 2 (1 3) 2𝑛 The sum on the right hand side is the total probability of the number of ways of placing 𝑛 balls in three boxes uniformly at random, so equals one. Suppose 𝑛= 3𝑚. Then we can show that ( 𝑛 𝑖,𝑗,𝑘) ≤( 𝑛 𝑚,𝑚,𝑚).
𝑝00(6𝑚) ≥(2𝑛 𝑛)(1 2) 2𝑛 ( 𝑛 𝑚, 𝑚, 𝑚)(1 3) 𝑛 Applying Stirling’s formula again, we have 𝑝00(6𝑚) ∼ 𝐴 𝑛3/2 It is sufficient to consider 𝑛= 3𝑚: 𝑝00(6𝑚) ≥1 62 𝑝00(6𝑚−2); 𝑝00(6𝑚) ≥1 64 𝑝00(6𝑚−4) Hence ∑ 𝑛 𝑝00(𝑛) < ∞ So the Markov chain is transient.
123 III. Markov Chains 5. Invariant distributions 5.1. Invariant distributions Let 𝐼be a countable set. (𝜆𝑖) is a probability distribution if 𝜆𝑖≥0 and ∑𝑖𝜆𝑖= 1.
Example. Consider a Markov chain with two elements, and 𝑃(1, 1) = 𝑃(1, 2) = 𝑃(2, 1) = 𝑃(2, 2) = 1 2. As 𝑛→∞, it is easy to see here that both states should be equally likely to occur.
In fact, 𝑝11(𝑛) = 𝑝12(𝑛) = 𝑝21(𝑛) = 𝑝22(𝑛) = 1 2. In this case, the row vector ( 1 2, 1 2) is an equilibrium probability distribution.
In general, we want to find a distribution 𝜋such that if 𝑋0 ∼𝜋, we have 𝑋𝑛∼𝜋for all 𝑛.
Suppose 𝑋0 ∼𝜋. Then, ℙ(𝑋1 = 𝑗) = ∑ 𝑖∈𝐼 ℙ(𝑋0 = 𝑖, 𝑋1 = 𝑗) = ∑ 𝑖∈𝐼 ℙ(𝑋1 = 𝑗∣𝑋0 = 𝑖) ℙ(𝑋0 = 𝑖) = ∑ 𝑖∈𝐼 𝜋(𝑖)𝑃(𝑖, 𝑗) Since we want 𝑋1 ∼𝜋, we must have 𝜋(𝑗) = ∑𝑖∈𝐼𝜋(𝑖)𝑃(𝑖, 𝑗) for all 𝑗. In matrix form, 𝜋= 𝜋𝑃.
Definition. An invariant (or equilibrium, or stationary) distribution for 𝑃is a probability distribution 𝜋such that 𝜋= 𝜋𝑃.
Theorem. Let 𝜋be invariant. Then, if 𝑋0 ∼𝜋, for all 𝑛we have 𝑋𝑛∼𝜋.
Proof. If 𝑋0 ∼𝜋, then 𝑋𝑛∼𝜋𝑃𝑛= 𝜋.
Theorem. Suppose 𝐼is finite, and there exists 𝑖∈𝐼such that 𝑝𝑖𝑗(𝑛) →𝜋𝑗as 𝑛→∞for all 𝑗. Then 𝜋= (𝜋𝑗) is an invariant distribution.
Proof. First, we check that the sum of 𝜋𝑗is one. Since 𝐼is finite, we can interchange the sum and limit.
∑ 𝑗∈𝐼 𝜋𝑗= ∑ 𝑗∈𝐼 lim 𝑛→∞𝑝𝑖𝑗(𝑛) = lim 𝑛→∞∑ 𝑗∈𝐼 𝑝𝑖𝑗(𝑛) = lim 𝑛→∞1 = 1 So 𝜋𝑗is a probability distribution. We now must show 𝜋= 𝜋𝑃.
𝜋𝑗= lim 𝑛→∞𝑝𝑖𝑗(𝑛) = lim 𝑛→∞∑ 𝑘∈𝐼 𝑝𝑖𝑘(𝑛−1)𝑃(𝑘, 𝑗) = ∑ 𝑘∈𝐼 lim 𝑛→∞𝑝𝑖𝑘(𝑛−1)𝑃(𝑘, 𝑗) = ∑ 𝑘∈𝐼 𝜋𝑘𝑃(𝑘, 𝑗) as required.
Remark. If 𝐼is infinite, the theorem does not necessarily hold. For example, let 𝐼= ℤ, 𝑋be a simple symmetric random walk. We know that 𝑝00(𝑛) ∼ 𝑐 √𝑛, and 𝑝0𝑥(𝑛) →0 as 𝑛→∞ for all 𝑥∈ℤ. So zero is given by the limit but this is not a distribution.
124 5. Invariant distributions 5.2. Conditions for unique invariant distribution In this section, we restrict our analysis to irreducible chains. If 𝑃is finite and irreducible, then 1 is an eigenvalue, since 𝑃is stochastic. The corresponding right eigenvector is (1, … , 1)⊺.
We know that 1 is an eigenvalue of 𝑃⊺, so 𝑃⊺has a right eigenvector corresponding to the eigenvalue of 1, which can be transposed to find a left eigenvector for 𝑃. It is possible to show using the Perron–Frobenius theorem that the eigenvector has non-negative compon-ents since 𝑃is irreducible. Since 𝐼is finite, we can normalise the left eigenvector such that its components sum to 1, giving an invariant distribution.
Definition. Let 𝑘∈𝐼. Recall that 𝑇𝑘is the first return time to 𝑘. For every 𝑖∈𝐼, we define 𝜈𝑘(𝑖) = 𝔼𝑘[ 𝑇𝑘−1 ∑ 𝑛=0 1(𝑋𝑛= 𝑖)] which is the expected number of times that we hit 𝑖while on an excursion from 𝑘(returning back to 𝑘).
Theorem. If 𝑃is irreducible and recurrent, then 𝜈𝑘is an invariant measure: 𝜈𝑘= 𝜈𝑘𝑃.
Further, 𝜈𝑘satisfies 𝜈𝑘(𝑘) = 1 and in general 𝜈𝑘(𝑖) ∈(0, ∞) for all 𝑖.
Proof. It is clear from the definition that 𝜈𝑘(𝑘) = 1, since we must hit 𝑘exactly once on the outset, and we do not count the return. We will now prove that 𝜈𝑘= 𝜈𝑘𝑃. 𝑇𝑘< ∞with probability 1 by recurrence, and 𝑋𝑇𝑘= 𝑘. Then, 𝜈𝑘(𝑖) = 𝔼𝑘[ 𝑇𝑘−1 ∑ 𝑛=0 1(𝑋𝑛= 𝑖)] = 𝔼𝑘[ 𝑇𝑘 ∑ 𝑛=1 1(𝑋𝑛= 𝑖)] = 𝔼𝑘[ ∞ ∑ 𝑛=1 1(𝑋𝑛= 𝑖, 𝑇𝑘≥𝑛)] = ∞ ∑ 𝑛=1 𝔼𝑘[1(𝑋𝑛= 𝑖, 𝑇𝑘≥𝑛)] = ∞ ∑ 𝑛=1 ℙ𝑘(𝑋𝑛= 𝑖, 𝑇𝑘≥𝑛) = ∞ ∑ 𝑛=1 ∑ 𝑗∈𝐼 ℙ𝑘(𝑋𝑛= 𝑖, 𝑋𝑛−1 = 𝑗, 𝑇𝑘≥𝑛) = ∞ ∑ 𝑛=1 ∑ 𝑗∈𝐼 ℙ𝑘(𝑋𝑛= 𝑖∣𝑋𝑛−1 = 𝑗, 𝑇𝑘≥𝑛) ℙ𝑘(𝑋𝑛−1 = 𝑗, 𝑇𝑘≥𝑛) 125 III. Markov Chains 𝑇𝑘is a stopping time, so the event {𝑇𝑘≥𝑛} = {𝑇𝑘≤𝑛−1} 𝑐depends only on values we already know or don’t care about. Hence, we can remove it.
= ∞ ∑ 𝑛=1 ∑ 𝑗∈𝐼 ℙ𝑘(𝑋𝑛= 𝑖∣𝑋𝑛−1 = 𝑗) ℙ𝑘(𝑋𝑛−1 = 𝑗, 𝑇𝑘≥𝑛) = ∞ ∑ 𝑛=1 ∑ 𝑗∈𝐼 𝑃(𝑗, 𝑖)ℙ𝑘(𝑋𝑛−1 = 𝑗, 𝑇𝑘≥𝑛) = ∑ 𝑗∈𝐼 ∞ ∑ 𝑛=1 𝑃(𝑗, 𝑖)ℙ𝑘(𝑋𝑛−1 = 𝑗, 𝑇𝑘≥𝑛) = ∑ 𝑗∈𝐼 ∞ ∑ 𝑛=0 𝑃(𝑗, 𝑖)ℙ𝑘(𝑋𝑛= 𝑗, 𝑇𝑘≥𝑛+ 1) = ∑ 𝑗∈𝐼 𝑃(𝑗, 𝑖)𝔼𝑘[ 𝑇𝑘−1 ∑ 𝑛=0 1(𝑋𝑛= 𝑗)] = ∑ 𝑗∈𝐼 𝑃(𝑗, 𝑖)𝜈𝑘(𝑗) Hence 𝜈𝑘= 𝜈𝑘𝑃. We must show 𝜈𝑘> 0. 𝑃is irreducible, hence there exists 𝑛such that 𝑝𝑘𝑖(𝑛) > 0. Then 𝜈𝑘(𝑖) = ∑ 𝑗∈𝐼 𝜈𝑘(𝑗)𝑃𝑛(𝑗, 𝑖) ≥𝜈𝑘(𝑘)𝑝𝑘𝑖(𝑛) > 0 To show 𝜈𝑘< ∞, let 𝑚such that 𝑝𝑖𝑘(𝑚) > 0.
1 = 𝜈𝑘(𝑘) = ∑ 𝑗∈𝐼 𝜈𝑘(𝑗)𝑃𝑚(𝑗, 𝑘) ≥𝜈𝑘(𝑖)𝑃𝑚(𝑖, 𝑘) ⟹𝜈𝑘(𝑖) ≤ 1 𝑃𝑚(𝑖, 𝑘) < ∞ 5.3. Uniqueness of invariant distributions Theorem. Let 𝑃be irreducible. Let 𝜆be an invariant measure (𝜆= 𝜆𝑃) with 𝜆𝑘= 1. Then 𝜆≥𝜈𝑘. If 𝑃is recurrent, then 𝜆= 𝜈𝑘.
126 5. Invariant distributions Proof. Let 𝜆be an invariant measure with 𝜆𝑘= 1. Then, 𝜆𝑖= ∑ 𝑗1 𝜆𝑗1𝑃(𝑗1, 𝑖) = 𝑃(𝑘, 𝑖) + ∑ 𝑗1≠𝑘 𝜆𝑗1𝑃(𝑗1, 𝑖) = 𝑃(𝑘, 𝑖) + ∑ 𝑗1≠𝑘 𝑃(𝑘, 𝑗1)𝑃(𝑗1, 𝑖) + ∑ 𝑗1,𝑗2≠𝑘 𝑃(𝑗2, 𝑗1)𝑃(𝑗1, 𝑖)𝜆𝑗2 = 𝑃(𝑘, 𝑖) + ∑ 𝑗1≠𝑘 𝑃(𝑘, 𝑗1)𝑃(𝑗1, 𝑖) + … + ∑ 𝑗1,…𝑗𝑛−1≠𝑘 𝑃(𝑘, 𝑗𝑛−1)𝑃(𝑗𝑛−1, 𝑗𝑛−2) … 𝑃(𝑗2, 𝑗1)𝑃(𝑗1𝑖) + ∑ 𝑗1,…,𝑗𝑛≠𝑘 𝑃(𝑗𝑛, 𝑗𝑛−1) … 𝑃(𝑗𝑛, 𝑖)𝜆𝑗𝑛 ⏟⎵⎵⎵⎵⎵⎵⎵⎵⎵⏟⎵⎵⎵⎵⎵⎵⎵⎵⎵⏟ ≥0 ≥ℙ𝑘(𝑋1 = 𝑖, 𝑇𝑘≥1) + ℙ𝑘(𝑋2 = 𝑖, 𝑇𝑘≥2) + ⋯+ ℙ𝑘(𝑋𝑛= 𝑖, 𝑇𝑘≥𝑛) ≥ 𝑛 ∑ 𝑖=1 ℙ𝑘(𝑋𝑛= 𝑖, 𝑇𝑘≥𝑛) →𝜈𝑘(𝑖) as 𝑛→∞. Now, suppose 𝑃is recurrent, so 𝜈𝑘is invariant. We define 𝜇= 𝜆−𝜈𝑘. Then 𝜇≥0 is an invariant measure satisfying 𝜇𝑘= 0. We need to show 𝜇𝑖= 0 for all 𝑖. By invariance, for all 𝑛, 𝜇𝑘= ∑ 𝑗 𝜇𝑗𝑃𝑛(𝑗, 𝑘) By irreducibility, we can choose 𝑛such that 𝑃𝑛(𝑖, 𝑘) > 0.
𝜇𝑘≥𝑃𝑛(𝑖, 𝑘)𝜇𝑖⟹𝜇𝑖= 0 Remark. In the irreducible and recurrent case, all invariant measures are equal up to a scal-ing factor.
Let 𝑘be fixed. Then, 𝜈𝑘is invariant, and unique in the above sense. If ∑𝑖𝜈𝑘(𝑖) is finite, we can take 𝜋𝑖= 𝜈𝑘(𝑖) ∑𝑗𝜈𝑘(𝑗) 127 III. Markov Chains which is an invariant distribution. The sum as required is ∑ 𝑖∈𝐼 𝜈𝑘(𝑖) = ∑ 𝑖∈𝐼 𝔼𝑘[ 𝑇𝑘−1 ∑ 𝑛=0 1(𝑋𝑛= 𝑖)] = 𝔼𝑘[ 𝑇𝑘−1 ∑ 𝑛=0 ∑ 𝑖∈𝐼 1(𝑋𝑛= 𝑖)] = 𝔼𝑘[ 𝑇𝑘−1 ∑ 𝑛=0 1] = 𝔼𝑘[𝑇𝑘] So we require that the expectation of the first return time is finite. If 𝔼𝑘[𝑇𝑘] is finite, we can normalise 𝜈𝑘into a (unique) invariant distribution.
5.4. Positive and null recurrence Definition. Let 𝑘∈𝐼be a recurrent state (so ℙ𝑘(𝑇𝑘< ∞) = 1). 𝑘is positive recurrent if 𝔼𝑘[𝑇𝑘] < ∞. 𝑘is called null recurrent otherwise; so if 𝔼𝑘[𝑇𝑘] = ∞.
Theorem. Let 𝑃be irreducible. Then the following are equivalent.
(i) every state is positive recurrent; (ii) some state is positive recurrent; (iii) 𝑃has an invariant distribution 𝜋.
If any of these conditions hold, we have 𝜋𝑖= 1 𝔼𝑖[𝑇𝑖] for all 𝑖.
Proof. First, (i) clearly implies (ii). We now show (ii) implies (iii). Let 𝑘be the a positive re-current state, and consider 𝜈𝑘. Since 𝑘is recurrent, we know that 𝜈𝑘is an invariant measure.
Then, ∑ 𝑖∈𝐼 𝜈𝑘(𝑖) = 𝔼𝑘[𝑇𝑘] < ∞ since 𝑘is positive recurrent. If we define 𝜋𝑖= 𝜈𝑘(𝑖) 𝔼𝑘[𝑇𝑘] we have that 𝜋is an invariant distribution.
128 5. Invariant distributions Now we show that (iii) implies (i). Let 𝑘be a state, which we will prove is positive recurrent.
First, we show that 𝜋𝑘> 0. There exists 𝑖such that 𝜋𝑖> 0, and we will choose 𝑛such that 𝑃𝑛(𝑖, 𝑘) > 0 by irreducibility. Then, 𝜋𝑘= ∑ 𝑗 𝜋𝑗𝑃𝑛(𝑗, 𝑘) ≥𝜋𝑖𝑃𝑛(𝑖, 𝑘) > 0 Now, we define 𝜆𝑖= 𝜋𝑖 𝜋𝑘. This is an invariant measure with 𝜆𝑘= 1. So from the above theorem, 𝜆≥𝜈𝑘. Now, since 𝜋is a distribution, 𝔼𝑘[𝑇𝑘] = ∑ 𝑖 𝜈𝑘(𝑖) ≤∑ 𝑖 𝜆𝑖= ∑ 𝑖 𝜋𝑖 𝜋𝑘 = 1 𝜋𝑘 ∑ 𝑖 𝜋𝑖= 1 𝜋𝑘 Hence 𝔼𝑘[𝑇𝑘] < ∞, so 𝑘is positive recurrent.
For the last part, we know that 𝑃is recurrent and 𝜆𝑖= 𝜋𝑖 𝜋𝑘is an invariant measure with 𝜆𝑘= 1. From the previous theorem, 𝜆𝑖= 𝜈𝑘(𝑖). Hence, 𝜋𝑖 𝜋𝑘= 𝜈𝑘(𝑖). Taking the sum over all 𝑖, 1 𝜋𝑘 = 𝔼𝑘[𝑇𝑘] which proves the last part.
Corollary. If 𝑃is irreducible and 𝜋is an invariant distribution, then for all 𝑥, 𝑦, the expected number of visits to 𝑦starting from 𝑥is given by 𝜈 𝑥(𝑦) = 𝜋(𝑦) 𝜋(𝑥) Example. Consider the simple symmetric random walk on ℤ. We have proven that this is recurrent. Suppose 𝜋is an invariant measure. So 𝜋= 𝜋𝑃, giving 𝜋𝑖= 1 2𝜋𝑖−1 + 1 2𝜋𝑖+1 So 𝜋𝑖= 1 is an invariant measure. So all invariant measures are multiples of this. But since this is not normalisable, there exists no invariant distribution. So this walk is null recurrent.
Remark. If 𝐼is finite, we can always normalise the distribution, since we have only a finite sum.
Remark. Consider a simple random walk on ℤ3. This is transient. However, 𝜆𝑖= 1 for all 𝑖∈ℤ3, this is clearly an invariant measure, so existence of an invariant measure does not imply recurrence.
Example. Consider a random walk on ℤwith transition probabilities 𝑃(𝑖, 𝑖+1) = 𝑝, 𝑃(𝑖, 𝑖− 1) = 𝑞such that 1 > 𝑝> 𝑞> 0 and 𝑝+ 𝑞= 1. This random walk is transient. Suppose there is an invariant distribution 𝜋, so 𝜋= 𝜋𝑃. Then 𝜋𝑖= 𝜋𝑖−1𝑞+ 𝜋𝑖+1𝑝 129 III. Markov Chains Solving the recursion gives 𝜋𝑖= 𝑎+ 𝑏(𝑝 𝑞) 𝑖 This is not unique up to a multiplicative constant, due to the constant 𝑎.
Example. Consider a random walk on ℤ+ with transition probabilities 𝑃(𝑖, 𝑖+1) = 𝑝, 𝑃(𝑖, 𝑖− 1) = 𝑞, 𝑃(0, 0) = 𝑞, and 𝑝< 𝑞so there is a drift towards zero. We can check that this is recurrent. We will look for a solution to 𝜋= 𝜋𝑃.
𝜋0 = 𝑞𝜋0 + 𝑞𝜋1; 𝜋𝑖= 𝑝𝜋𝑖−1 + 𝑞𝜋𝑖+1 Solving this system yields 𝜋1 = 𝑝 𝑞𝜋0; 𝜋𝑖= (𝑝 𝑞) 𝑖 𝜋0 This is unique up to a multiplicative constant. Since 𝑝< 𝑞, we can normalise this to reach an invariant distribution. Let 𝜋0 = 1 − 𝑝 𝑞. Then, 𝜋𝑖= (𝑝 𝑞) 𝑖 (1 −𝑝 𝑞) Hence the walk is positive recurrent.
5.5. Time reversibility Theorem. Let 𝑃be irreducible, and 𝜋be an invariant distribution. Let 𝑁∈ℕand let 𝑌 𝑛= 𝑋𝑁−𝑛for 0 ≤𝑛≤𝑁. If 𝑋0 ∼𝜋, then (𝑌 𝑛)0≤𝑛≤𝑁is a Markov chain with transition matrix ̂ 𝑃(𝑥, 𝑦) = 𝜋(𝑦) 𝜋(𝑥)𝑃(𝑦, 𝑥) and has invariant distribution 𝜋, so 𝜋̂ 𝑃= 𝜋. Further, ̂ 𝑃is also irreducible.
Proof. First, note that ̂ 𝑃is stochastic. Since 𝜋= 𝜋𝑃, ∑ 𝑦 ̂ 𝑃(𝑥, 𝑦) = ∑ 𝑦 𝜋(𝑦)𝑃(𝑦, 𝑥) 𝜋(𝑥) = 𝜋(𝑥) 𝜋(𝑥) = 1 Now we show 𝑌is a Markov chain.
ℙ(𝑌 0 = 𝑦0, … , 𝑌𝑁= 𝑦𝑁) = ℙ(𝑋𝑁= 𝑦0, … , 𝑋0 = 𝑦𝑛) = 𝜋(𝑦𝑁)𝑃(𝑦𝑁, 𝑦𝑁−1) … 𝑃(𝑦1, 𝑦0) = ̂ 𝑃(𝑦𝑁−1, 𝑦𝑁)𝜋(𝑦𝑁−1)𝑃(𝑦𝑁−1, 𝑦𝑁−2) … 𝑃(𝑦1, 𝑦0) = … = 𝜋(𝑦0) ̂ 𝑃(𝑦0, 𝑦1) … 𝑃(𝑦𝑁−1, 𝑦𝑁) 130 5. Invariant distributions Hence 𝑌∼Markov (𝜋, ̂ 𝑃). Now, we must show 𝜋= 𝜋̂ 𝑃.
∑ 𝑥 𝜋(𝑥) ̂ 𝑃(𝑥, 𝑦) = ∑ 𝑥 𝜋(𝑥)𝑃(𝑦, 𝑥)𝜋(𝑦) 𝜋(𝑥) = 𝜋(𝑦) ∑ 𝑥 𝑃(𝑦, 𝑥) = 𝜋(𝑦) Hence 𝜋is invariant for ̂ 𝑃. Now we show ̂ 𝑃is irreducible. Let 𝑥, 𝑦∈𝐼. Then there exists 𝑥= 𝑥0, 𝑥1, … , 𝑥𝑘= 𝑦such that 𝑃(𝑥0, 𝑥1) … 𝑃(𝑥𝑘−1, 𝑥𝑘) > 0 Hence ̂ 𝑃(𝑥𝑘, 𝑥𝑘−1) … ̂ 𝑃(𝑥1, 𝑥0) = 𝜋(𝑥0)𝑃(𝑥0, 𝑥1) … 𝑃(𝑥𝑘−1, 𝑥𝑘) 𝜋(𝑥𝑘) > 0 So ̂ 𝑃is irreducible.
Definition. A Markov chain 𝑋with transition matrix 𝑃and invariant distribution 𝜋is called reversible or time reversible if ̂ 𝑃= 𝑃. Equivalently, for all 𝑥, 𝑦, 𝜋(𝑥)𝑃(𝑥, 𝑦) = 𝜋(𝑦)𝑃(𝑦, 𝑥) These equations are called the detailed balance equations. Equivalently, 𝑋is reversible if, for any fixed 𝑁∈ℕ, 𝑋0 ∼𝜋implies (𝑋0, … , 𝑋𝑁) 𝑑 = (𝑋𝑁, … , 𝑋0) which means that they are equal in distribution.
Remark. Intuitively, 𝑋is reversible if, starting from 𝜋, we cannot tell if we are watching 𝑋 evolve forwards in time or backwards in time.
Lemma. Let 𝑃be a transition matrix, and 𝜇a distribution satisfying the detailed balance equations.
𝜇(𝑥)𝑃(𝑥, 𝑦) = 𝜇(𝑦)𝑃(𝑦, 𝑥) Then 𝜇is invariant for 𝑃.
Proof.
∑ 𝑥 𝜇(𝑥)𝑃(𝑥, 𝑦) = ∑ 𝑥 𝜇(𝑦)𝑃(𝑦, 𝑥) = 𝜇(𝑦) Remark. If we can find a solution to the detailed balance equations which is a distribution, it must be an invariant distribution. It is simpler to solve this set of equations than to solve 𝜋= 𝜋𝑃. If there is no solution to the detailed balance equations, then even if there exists an invariant distribution, the Markov chain is not reversible.
131 III. Markov Chains Example. Consider a random walk on the integers modulo 𝑛, with 𝑃(𝑖, 𝑖+ 1) = 2 3 and 𝑃(𝑖, 𝑖−1) = 1 3. We can check 𝜋𝑖= 1 𝑛is an invariant distribution. This does not satisfy the detailed balance equations. Hence the Markov chain is not reversible.
Example. Consider a random walk on {0, … , 𝑛−1} with 𝑃(𝑖, 𝑖+ 1) = 2 3, 𝑃(𝑖, 𝑖−1) = 1 3 and 𝑃(0, 0) = 1 3, 𝑃(𝑛−1, 𝑛−1) = 2 3. This is an ‘opened up’ version of the previous example; the circle is ‘cut’ open into a line at zero. The detailed balance equations give 𝜋𝑖𝑃(𝑖, 𝑖+ 1) = 𝜋𝑖+1𝑃(𝑖+ 1, 𝑖) ⟹𝜋𝑖= 𝑘2𝑖 We can normalise this by setting 𝑘such that 𝜋is a distribution. Hence the chain is reversible.
Example. Consider a random walk on a graph. Let 𝐺= (𝑉, 𝐸) be a finite connected graph, where 𝑉is a set of vertices and 𝐸is a set of edges. The simple random walk on 𝐺has the transition matrix 𝑃(𝑥, 𝑦) = { 1 𝑑(𝑥) (𝑥, 𝑦) ∈𝐸 0 (𝑥, 𝑦) ∉𝐸 where 𝑑(𝑥) = ∑𝑦1((𝑥, 𝑦) ∈𝐸) is the degree of 𝑥. The detailed balance equations give, for (𝑥, 𝑦) ∈𝐸, 𝜋(𝑥)𝑃(𝑥, 𝑦) = 𝜋(𝑦)𝑃(𝑦, 𝑥) ⟹𝜋(𝑥) 𝑑(𝑥) = 𝜋(𝑦) 𝑑(𝑦) Let 𝜋(𝑥) ∝𝑑(𝑥). Then this is an invariant distribution with normalising constant 1 ∑𝑦𝑑(𝑦) = 1 2|𝐸|. So the simple random walk on a finite connected graph is always reversible.
5.6. Aperiodicity Definition. Let 𝑃be a transition matrix. For all 𝑖, we write 𝑑𝑖= gcd {𝑛≥1∶𝑃𝑛(𝑖, 𝑖) > 0} This is called the period of 𝑖. If 𝑑𝑖= 1, we say that 𝑖is aperiodic.
Lemma. 𝑑𝑖= 1 if and only if 𝑃𝑛(𝑖, 𝑖) > 0 for all 𝑛sufficiently large. More rigorously, there exists 𝑛0 ∈ℕsuch that for all 𝑛> 𝑛0, 𝑃𝑛(𝑖, 𝑖) > 0.
Proof. First, if 𝑃𝑛(𝑖, 𝑖) > 0 for all 𝑛sufficiently large, the greatest common divisor of all sufficiently large numbers is one so this direction is trivial. Conversely, let 𝐷(𝑖) = {𝑛≥1∶𝑃𝑛(𝑖, 𝑖) > 0} Observe that if 𝑎, 𝑏∈𝐷(𝑖) then 𝑎+ 𝑏∈𝐷(𝑖).
We claim that 𝐷(𝑖) contains two consecutive integers. Suppose that it does not, so for all 𝑎, 𝑏∈𝐷(𝑖) we must have |𝑎−𝑏| > 1. Let 𝑟be the minimal distance between two integers 132 5. Invariant distributions in 𝐷(𝑖), so 𝑟≥2. Let 𝑛, 𝑚be numbers in 𝐷(𝑖) separated by 𝑟, so 𝑛= 𝑚+ 𝑟. Then we can show there exists 𝑘∈𝐷(𝑖) which can be written as ℓ𝑟+ 𝑠with 0 < 𝑠< 𝑟. Indeed, if there were not such a 𝑘, we would have 𝑑𝑖= 1, since all elements would be multiples of 𝑟. Now, let 𝑎= (ℓ+ 1)𝑛and 𝑏= (ℓ+ 1)𝑚+ 𝑘. Then 𝑎, 𝑏∈𝐷(𝑖), and 𝑎−𝑏= 𝑟−𝑠< 𝑟. This is a contradiction, since we have found two points in 𝐷(𝑖) with a distance smaller than the minimal distance.
Now, let 𝑛1, 𝑛1 + 1 be elements of 𝐷(𝑖). Then {𝑥𝑛1 + 𝑦(𝑛1 + 1)∶𝑥, 𝑦∈ℕ} ⊆𝐷(𝑖) It is then easy to check that 𝐷(𝑖) ⊇{𝑛∶𝑛≥𝑛2 1}.
Lemma. Suppose 𝑃is irreducible and 𝑖is aperiodic. Then for all 𝑗∈𝐼, 𝑗is aperiodic. Hence, aperiodicity is a class property.
Proof. There exist 𝑛, 𝑚such that 𝑃𝑛(𝑖, 𝑗) > 0, 𝑃𝑚(𝑖, 𝑗) > 0. Hence, 𝑃𝑛+𝑚+𝑟(𝑗, 𝑗) ≥𝑃𝑛(𝑗, 𝑖)𝑃𝑟(𝑖, 𝑖)𝑃𝑛(𝑖, 𝑗) The first and last terms are positive, and the middle term is positive for sufficiently large 𝑟.
5.7. Positive recurrent limiting behaviour Theorem. Let 𝑃be irreducible and aperiodic with invariant distribution 𝜋, and further let 𝑋∼Markov (𝜆, 𝑃). Then for all 𝑦∈𝐼, ℙ(𝑋𝑛= 𝑦) →𝜋𝑦as 𝑛→∞. Taking 𝜆= 𝛿𝑥, we get 𝑝𝑥𝑦(𝑛) →𝜋(𝑦) as 𝑛→∞.
Proof. This proof will use the idea of ‘coupling’ of Markov chains. Let 𝑌∼Markov (𝜋, 𝑃) be independent of 𝑋. Consider the pair ((𝑋𝑛, 𝑌 𝑛))𝑛≥0. This is a Markov chain on the state space 𝐼× 𝐼, because 𝑋and 𝑌are independent. The initial distribution is 𝜆× 𝜋. We have ℙ((𝑋0, 𝑌 0) = (𝑥, 𝑦)) = 𝜆(𝑥)𝜋(𝑦) and transition matrix ˜ 𝑃given by ˜ 𝑃((𝑥, 𝑦), (𝑥′, 𝑦′)) = 𝑃(𝑥, 𝑥′)𝑃(𝑦, 𝑦′) This product chain has invariant distribution ˜ 𝜋given by ˜ 𝜋(𝑥, 𝑦) = 𝜋(𝑥)𝜋(𝑦) Let 𝑎∈𝐼, and let 𝑇= inf 𝑛≥1∶(𝑋𝑛, 𝑌 𝑛) = (𝑎, 𝑎) be the hitting time of (𝑎, 𝑎).
First, we want to show that ℙ(𝑇< ∞) = 1. We show that ˜ 𝑃is irreducible. Let (𝑥, 𝑦), (𝑥′, 𝑦′) ∈ 𝐼× 𝐼. By irreducibility of 𝑃, there exist ℓ, 𝑚such that 𝑃ℓ(𝑥, 𝑥′) > 0 and 𝑃𝑚(𝑦, 𝑦′) > 0. Now, ˜ 𝑃ℓ+𝑚+𝑛((𝑥, 𝑦), (𝑥′, 𝑦′)) = 𝑃ℓ+𝑚+𝑛(𝑥, 𝑥′)𝑃ℓ+𝑚+𝑛(𝑦, 𝑦′) 133 III. Markov Chains Note that 𝑃ℓ+𝑚+𝑛(𝑥, 𝑥′) ≥𝑃ℓ(𝑥, 𝑥′)𝑃𝑚+𝑛(𝑥′, 𝑥′) By taking 𝑛large, by aperiodicity the product is positive. Therefore, for sufficiently large 𝑛, 𝑃𝑛(𝑥, 𝑥′) > 0. So ˜ 𝑃is irreducible, and there exists an invariant distribution ˜ 𝜋. Hence ˜ 𝑃is positive recurrent. So ℙ(𝑇< ∞) = 1.
Now, we define 𝑍𝑛= {𝑋𝑛 𝑛< 𝑇 𝑌 𝑛 𝑛≥𝑇 We wish to show 𝑍= (𝑍𝑛)𝑛≥0 has the same distribution as 𝑋, that is, 𝑍∼Markov (𝜆, 𝑃).
Now, ℙ(𝑍0 = 𝑥) = ℙ(𝑋0 = 𝑥) = 𝜆(𝑥) so the initial distribution is the same. Now, we will check that 𝑍evolves with transition matrix 𝑃. Let 𝐴= {𝑍𝑛−1 = 𝑧𝑛−1, … , 𝑍0 = 𝑧0}. We need to show ℙ(𝑍𝑛+1 = 𝑦∣𝑍𝑛= 𝑥, 𝐴) = 𝑃(𝑥, 𝑦).
ℙ(𝑍𝑛+1 = 𝑦∣𝑍𝑛= 𝑥, 𝐴) = ℙ(𝑍𝑛+1 = 𝑦, 𝑇> 𝑛∣𝑍𝑛= 𝑥, 𝐴) + ℙ(𝑍𝑛+1 = 𝑦, 𝑇≤𝑛∣𝑍𝑛= 𝑥, 𝐴) = ℙ(𝑋𝑛+1 = 𝑦∣𝑇> 𝑛, 𝑍𝑛= 𝑥, 𝐴) ℙ(𝑇> 𝑛∣𝑍𝑛= 𝑥, 𝐴) + ℙ(𝑌 𝑛+ 1 = 𝑦∣𝑇≤𝑛, 𝑍𝑛= 𝑥, 𝐴) ℙ(𝑇≤𝑛∣𝑍𝑛= 𝑥, 𝐴) Now, ℙ(𝑋𝑛+1 = 𝑦∣𝑇> 𝑛, 𝑍𝑛= 𝑥, 𝐴) = ∑ 𝑧 ℙ(𝑋𝑛+1 = 𝑦∣𝑇> 𝑛, 𝑍𝑛= 𝑥, 𝑌 𝑛= 𝑧, 𝐴) ℙ(𝑌 𝑛= 𝑧∣𝑇> 𝑛, 𝑍−𝑛= 𝑥, 𝐴) Note, {𝑇> 𝑛} depends only on (𝑋0, 𝑌 0), … , (𝑋𝑛, 𝑌 𝑛) since it is the complement of {𝑇≤𝑛}, so it is a stopping time. Hence, ℙ(𝑋𝑛+1 = 𝑦∣𝑇> 𝑛, 𝑍𝑛= 𝑥, 𝐴) = ∑ 𝑧 𝑃(𝑥, 𝑦)ℙ(𝑌 𝑛= 𝑧∣𝑇> 𝑛, 𝑍−𝑛= 𝑥, 𝐴) = 𝑃(𝑥, 𝑦) Similarly, ℙ(𝑌 𝑛+1 = 𝑦∣𝑇> 𝑛, 𝑍𝑛= 𝑥, 𝐴) = 𝑃(𝑥, 𝑦) Hence, ℙ(𝑍𝑛+1 = 𝑦∣𝑍𝑛= 𝑥, 𝐴) = 𝑃(𝑥, 𝑦)ℙ(𝑇> 𝑛∣𝑍𝑛= 𝑥, 𝐴) + 𝑃(𝑥, 𝑦)ℙ(𝑇≤𝑛∣𝑍𝑛= 𝑥, 𝐴) = 𝑃(𝑥, 𝑦)[ℙ(𝑇> 𝑛∣𝑍𝑛= 𝑥, 𝐴) + ℙ(𝑇≤𝑛∣𝑍𝑛= 𝑥, 𝐴)] = 𝑃(𝑥, 𝑦) 134 5. Invariant distributions as required. Hence 𝑍∼Markov (𝜆, 𝑃). Thus, |ℙ(𝑋𝑛= 𝑦) −𝜋(𝑦)| = |ℙ(𝑍𝑛= 𝑦) −ℙ(𝑌 𝑛= 𝑦)| = |ℙ(𝑋𝑛= 𝑦, 𝑛< 𝑇) + ℙ(𝑌 𝑛= 𝑦, 𝑛≥𝑇) −𝑌 𝑛= 𝑦, 𝑛< 𝑇−ℙ(𝑌 𝑛= 𝑦, 𝑛≥𝑇)| = |ℙ(𝑋𝑛= 𝑦, 𝑛< 𝑇) −ℙ(𝑌 𝑛= 𝑦, 𝑛< 𝑇)| ≤ℙ(𝑛< 𝑇) As 𝑛→∞, this upper bound becomes zero, since ℙ(𝑇< ∞) = 1.
5.8. Null recurrent limiting behaviour Theorem. Let 𝑃be irreducible, aperiodic, and null recurrent. Then, for all 𝑥, 𝑦, lim 𝑛→∞𝑃𝑛(𝑥, 𝑦) = 0 Proof. Let ˜ 𝑃((𝑥, 𝑦), (𝑥′, 𝑦′)) = 𝑃(𝑥, 𝑥′)𝑃(𝑦, 𝑦′) as before. We have shown previously that ˜ 𝑃is also irreducible. Suppose first that ˜ 𝑃is transient. Then, ∑ 𝑛 ˜ 𝑃𝑛((𝑥, 𝑦), (𝑥, 𝑦)) < ∞ This sum is equal to ∑ 𝑛 (𝑃𝑛(𝑥, 𝑦))2 < ∞ Hence, 𝑃𝑛(𝑥, 𝑦) →0 Now, conversely suppose that ˜ 𝑃is recurrent. Let 𝑦∈𝐼. Define as before 𝜈𝑦(𝑥) = 𝔼𝑦[ 𝑇𝑦−1 ∑ 𝑖=0 1(𝑋𝑖= 𝑥)] This measure is invariant for 𝑃since 𝑃is recurrent. Since 𝑃is null recurrent in particular, 𝔼𝑦[𝑇 𝑦] = ∞. Hence, 𝜈𝑦(𝐼) = ∑ 𝑥∈𝐼 𝜈𝑦(𝑥) = 𝔼𝑦[ 𝑇𝑦−1 ∑ 𝑖=0 1] = 𝔼𝑦[𝑇 𝑦] = ∞ Because 𝜈𝑦(𝐼) is infinite, for all 𝑀> 0 there exists a finite set 𝐴⊂𝐼with 𝜈𝑦(𝐴) > 𝑀. Now, we define a probability measure 𝜇(𝑧) = 𝜈𝑦(𝑧) 𝜈𝑦(𝐴)1(𝑧∈𝐴) 135 III. Markov Chains Now, for all 𝑧∈𝐼, 𝜇𝑃𝑛(𝑧) = ∑ 𝑥 𝜇(𝑥)𝑃𝑛(𝑥, 𝑧) = ∑ 𝑥 𝜈𝑦(𝑥) 𝜈𝑦(𝐴)1(𝑧∈𝐴)𝑃𝑛(𝑥, 𝑧) ≤ 1 𝜈𝑦(𝐴) ∑ 𝑥 𝜈𝑦(𝑥)𝑃𝑛(𝑥, 𝑧) Since 𝜈𝑦is invariant, 𝜇𝑃𝑛(𝑧) ≤ 1 𝜈𝑦(𝐴)𝜈𝑦(𝑧) = 𝜈𝑦(𝑧) 𝜈𝑦(𝐴) Let (𝑋, 𝑌) be a Markov chain with matrix ˜ 𝑃, started according to 𝜇× 𝛿𝑥, so ℙ(𝑋0 = 𝑧, 𝑌 0 = 𝑤) = 𝜇(𝑧)𝛿𝑥(𝑤) Now, let 𝑇= inf {𝑛≥1∶(𝑋𝑛, 𝑌 𝑛) = (𝑥, 𝑥)} Since ˜ 𝑃is recurrent, 𝑇is finite with probability 1. Let 𝑍𝑛= {𝑋𝑛 𝑛< 𝑇 𝑌 𝑛 𝑛≥𝑇 We have already proven that 𝑍is a Markov chain with transition matrix 𝑃, started according to 𝜇; it has the same distribution as 𝑋. Hence, ℙ(𝑍𝑛= 𝑦) = 𝜇𝑃𝑛(𝑦) ≤ 𝜈𝑦(𝑦) 𝜈𝑦(𝐴) = 1 𝜈𝑦(𝐴) Note, ℙ𝑥(𝑌 𝑛= 𝑦) ≤ℙ𝑥(𝑌 𝑛= 𝑦, 𝑛≥𝑇) + ℙ𝑥(𝑇> 𝑛) = ℙ𝑥(𝑍𝑛= 𝑦) + ℙ𝑥(𝑇> 𝑛) Hence, lim sup 𝑛→∞ ℙ𝑥(𝑌 𝑛= 𝑦) ≤1 𝑀+ 0 = 1 𝑀 Since this is true for all 𝑀, 𝑃𝑛(𝑥, 𝑦) →0 as 𝑛→∞.
136 IV. Analysis and Topology Lectured in Michaelmas 2021 by Dr. V. Zsák In the analysis part of the course, we continue the study of convergence from Analysis I. We define a stronger version of convergence, called uniform convergence, and show that it has some very desirable properties. For example, if integrable functions 𝑓 𝑛converge uniformly to the integrable function 𝑓, then the integrals of the 𝑓 𝑛converge to the integral of 𝑓. The same cannot be said in general about non-uniform convergence. We also extend our study of differentiation to functions with multiple input and output variables, and rigorously define the derivative in this higher-dimensional context.
In the topology part of the course, we consider familiar spaces such as [𝑎, 𝑏], ℂ, ℝ𝑛, and generalise their properties. We arrive at the definition of a metric space, which encapsulates all of the information about how near or far points are from others. From here, we can define notions such as continuous functions between metric spaces in such a way that does not depend on the underlying space.
We then generalise even further to define topological spaces. The only information a topo-logical space contains is the neighbourhoods of each point, but it turns out that this is still enough to define continuous functions and similar things. We study topological spaces in an abstract setting, and prove important facts that are used in many later courses.
137 IV. Analysis and Topology Contents 1.
Uniform convergence . . . . . . . . . . . . . . . . . . . . . . . . . 141 1.1.
Definition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 141 1.2.
Pointwise convergence . . . . . . . . . . . . . . . . . . . . . . 141 1.3.
Uniform limit of bounded functions . . . . . . . . . . . . . . . 143 1.4.
Integrability . . . . . . . . . . . . . . . . . . . . . . . . . . . . 143 1.5.
Differentiability . . . . . . . . . . . . . . . . . . . . . . . . . . 145 1.6.
Conditions for uniform convergence . . . . . . . . . . . . . . . 146 1.7.
General principle of uniform convergence . . . . . . . . . . . . 146 1.8.
Weierstrass M-test . . . . . . . . . . . . . . . . . . . . . . . . 146 1.9.
Power series . . . . . . . . . . . . . . . . . . . . . . . . . . . 147 2.
Uniform continuity . . . . . . . . . . . . . . . . . . . . . . . . . . . 149 2.1.
Definition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 149 2.2.
Properties of continuous functions . . . . . . . . . . . . . . . . 149 3.
Metric spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 151 3.1.
Definition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 151 3.2.
Subspaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 153 3.3.
Product spaces . . . . . . . . . . . . . . . . . . . . . . . . . . 153 3.4.
Convergence . . . . . . . . . . . . . . . . . . . . . . . . . . . 153 3.5.
Continuity . . . . . . . . . . . . . . . . . . . . . . . . . . . . 155 3.6.
Isometric, Lipschitz, and uniformly continuous functions . . . . 157 3.7.
Generalised triangle inequality . . . . . . . . . . . . . . . . . . 158 4.
Topology of metric spaces . . . . . . . . . . . . . . . . . . . . . . . 159 4.1.
Open balls . . . . . . . . . . . . . . . . . . . . . . . . . . . . 159 4.2.
Neighbourhoods and openness . . . . . . . . . . . . . . . . . . 159 4.3.
Continuity and convergence using topology . . . . . . . . . . . 160 4.4.
Properties of topology of metric space . . . . . . . . . . . . . . . 162 4.5.
Homeomorphisms . . . . . . . . . . . . . . . . . . . . . . . . 163 4.6.
Equivalence of metrics . . . . . . . . . . . . . . . . . . . . . . 163 5.
Completeness . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 164 5.1.
Cauchy sequences . . . . . . . . . . . . . . . . . . . . . . . . 164 5.2.
Definition of completeness . . . . . . . . . . . . . . . . . . . . 164 5.3.
Completeness of product spaces . . . . . . . . . . . . . . . . . 165 5.4.
Completeness of subspaces and function spaces . . . . . . . . . 165 6.
Contraction mapping theorem . . . . . . . . . . . . . . . . . . . . 169 6.1.
Contraction mappings . . . . . . . . . . . . . . . . . . . . . . 169 6.2.
Contraction mapping theorem . . . . . . . . . . . . . . . . . . 169 6.3.
Application of contraction mapping theorem . . . . . . . . . . . 170 138 6.4.
Lindelöf–Picard theorem . . . . . . . . . . . . . . . . . . . . . 170 7.
Topology . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 173 7.1.
Definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . 173 7.2.
Closed subsets . . . . . . . . . . . . . . . . . . . . . . . . . . 174 7.3.
Neighbourhoods . . . . . . . . . . . . . . . . . . . . . . . . . 174 7.4.
Convergence . . . . . . . . . . . . . . . . . . . . . . . . . . . 174 7.5.
Interiors and closures . . . . . . . . . . . . . . . . . . . . . . . 175 7.6.
Dense subsets . . . . . . . . . . . . . . . . . . . . . . . . . . . 176 7.7.
Subspaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 176 7.8.
Continuity . . . . . . . . . . . . . . . . . . . . . . . . . . . . 177 7.9.
Homeomorphisms and topological invariance . . . . . . . . . . 178 7.10.
Products . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 178 7.11.
Continuity in product topology . . . . . . . . . . . . . . . . . . 179 7.12.
Quotients . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 180 7.13.
Continuity of functions in quotient spaces . . . . . . . . . . . . 181 8.
Connectedness . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 183 8.1.
Definition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 183 8.2.
Consequences of definition . . . . . . . . . . . . . . . . . . . . 184 8.3.
Partitioning into connected components . . . . . . . . . . . . . 186 8.4.
Path-connectedness . . . . . . . . . . . . . . . . . . . . . . . . 187 8.5.
Gluing lemma . . . . . . . . . . . . . . . . . . . . . . . . . . 187 9.
Compactness . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 190 9.1.
Motivation and definition . . . . . . . . . . . . . . . . . . . . . 190 9.2.
Subspaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 191 9.3.
Continuous images of compact spaces . . . . . . . . . . . . . . 192 9.4.
Topological inverse function theorem . . . . . . . . . . . . . . . 192 9.5.
Tychonov’s theorem . . . . . . . . . . . . . . . . . . . . . . . 193 9.6.
Heine–Borel theorem . . . . . . . . . . . . . . . . . . . . . . . 193 9.7.
Sequential compactness . . . . . . . . . . . . . . . . . . . . . . 194 9.8.
Compactness and sequential compactness in metric spaces . . . . 194 10.
Differentiation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 197 10.1.
Linear maps . . . . . . . . . . . . . . . . . . . . . . . . . . . 197 10.2.
Differentiation . . . . . . . . . . . . . . . . . . . . . . . . . . 198 10.3.
Derivatives on open subsets . . . . . . . . . . . . . . . . . . . . 200 10.4.
Properties of derivative . . . . . . . . . . . . . . . . . . . . . . 201 10.5.
Linearity and product rule . . . . . . . . . . . . . . . . . . . . 203 11.
Partial derivatives . . . . . . . . . . . . . . . . . . . . . . . . . . . 204 11.1.
Directional and partial derivatives . . . . . . . . . . . . . . . . 204 11.2.
Jacobian matrix . . . . . . . . . . . . . . . . . . . . . . . . . . 205 11.3.
Constructing total derivative from partial derivatives . . . . . . . 205 139 IV. Analysis and Topology 11.4.
Mean value inequality . . . . . . . . . . . . . . . . . . . . . . 206 11.5.
Zero derivatives . . . . . . . . . . . . . . . . . . . . . . . . . . 207 11.6.
Inverse function theorem . . . . . . . . . . . . . . . . . . . . . 207 12.
Second derivatives . . . . . . . . . . . . . . . . . . . . . . . . . . . 210 12.1.
Definition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 210 12.2.
Second derivatives and partial derivatives . . . . . . . . . . . . . 212 12.3.
Symmetry of mixed directional derivatives . . . . . . . . . . . . 212 140 1. Uniform convergence 1. Uniform convergence 1.1. Definition Recall that 𝑥𝑛→𝑥as 𝑛→∞(for 𝑥∈ℝor ℂ) if ∀𝜀> 0, ∃𝑁∈ℕ, ∀𝑛≥𝑁, |𝑥𝑛−𝑥| < 𝜀 This is essentially considering the 𝜀-neighbourhood of 𝑥. We aim to define the same notion of convergence for functions, by defining an analogous concept of an 𝜀-neighbourhood. In particular, each value on the domain should converge in its own 𝜀-neighbourhood.
Definition. Let 𝑆be a set, and 𝑓, 𝑓 𝑛∶𝑆→ℝ, be functions. We say that (𝑓 𝑛) converges to 𝑓 uniformly on 𝑆if ∀𝜀> 0, ∃𝑁∈ℕ, ∀𝑛≥𝑁, ∀𝑥∈𝑆, |𝑓 𝑛(𝑥) −𝑓(𝑥)| < 𝜀 Note. 𝑁depends only on 𝜀, not on any 𝑥. Each 𝑥converges therefore at a ‘similar speed’, hence the name ‘uniform convergence’.
Equivalently, we can write ∀𝜀> 0, ∃𝑁∈ℕ, ∀𝑛≥𝑁, sup 𝑥∈𝑆 |𝑓 𝑛(𝑥) −𝑓(𝑥)| < 𝜀 The supremum condition is equivalent overall because the inequality on the right is weakened to a possible equality, but we can always decrease 𝜀to retain the inequality. Alternatively, we could write lim 𝑛→∞sup 𝑥∈𝑆 |𝑓 𝑛−𝑓| = 0 For each 𝑥∈𝑆, (𝑓 𝑛(𝑥))∞ 𝑛=1 →𝑓(𝑥). Hence, 𝑓is unique given (𝑓 𝑛), since limits are unique.
We call 𝑓the uniform limit of (𝑓 𝑛) on 𝑆.
1.2. Pointwise convergence Definition. (𝑓 𝑛) converges pointwise to 𝑓on 𝑆if (𝑓 𝑛(𝑥))∞ 𝑛=1 converges to 𝑓(𝑥) for every 𝑥∈𝑆.
In other words, ∀𝑥∈𝑆 ⏟ ⎵ ⏟ ⎵ ⏟ order rearranged , ∀𝜀> 0, ∃𝑁∈ℕ, ∀𝑛≥𝑁, |𝑓 𝑛(𝑥) −𝑓(𝑥)| < 𝜀 Now, 𝑁depends both on 𝜀and on 𝑥. Note that the pointwise limit of (𝑓 𝑛) on 𝑆is also unique since limits are unique.
Remark. Uniform convergence implies pointwise convergence, and the uniform limit is the pointwise limit.
141 IV. Analysis and Topology Example. Let 𝑓 𝑛(𝑥) = 𝑥2𝑒−𝑛𝑥on [0, ∞), 𝑛∈ℕ. Does (𝑓 𝑛) converge uniformly on the domain? First let us check pointwise convergence. We have 𝑥2𝑒−𝑛𝑥→0 hence pointwise convergence to 𝑓(𝑥) = 0 is satisfied. Now, we need only check uniform convergence to the function 𝑓(𝑥) = 0.
sup 𝑥∈[0,∞) |𝑓 𝑛(𝑥) −0| = sup 𝑥∈[0,∞) 𝑓 𝑛(𝑥) We could differentiate 𝑓 𝑛and find the maximum if it exists, but we might not find the max-imum if it is (for example) on the endpoints. A much better method is to find an upper bound on |𝑓 𝑛(𝑥) −𝑓(𝑥)| (which, in this example, is 𝑓 𝑛(𝑥)) that does not depend on 𝑥. In this case, we can expand 𝑒𝑛𝑥on the denominator and isolate a single term to get 𝑥2𝑒−𝑛𝑥= 𝑥2 𝑒𝑛𝑥≤2 𝑛2 ; ∀𝑥 Hence, sup 𝑥∈[0,∞) |𝑓 𝑛(𝑥) −0| →0 and uniform convergence is satisfied.
Example. Consider 𝑓 𝑛(𝑥) = 𝑥𝑛on [0, 1], 𝑛∈ℕ. A pointwise limit is reached by 𝑓(𝑥) = {1 𝑥= 1 0 otherwise Consider sup |𝑓 𝑛(𝑥) −𝑓(𝑥)| excluding 1 (since at 1 the supremum is zero). Note 𝑓 𝑛(𝑥) →1 as 𝑥→1 from below, for all 𝑛. Hence the supremum is always 1 by choosing an 𝑥sufficiently close to 1. So 𝑓 𝑛↛𝑓uniformly on [0, 1], hence (𝑓 𝑛) does not converge at all uniformly on this domain. Or, sup 𝑓 𝑛(𝑥) ≥𝑓 𝑛((1 2) 1/𝑛 ) = 1 2 Remark. If 𝑓 𝑛↛𝑓uniformly on S, ∃𝜀> 0, ∀𝑁∈ℕ, ∃𝑛≥𝑁, ∃𝑥∈𝑆, |𝑓 𝑛(𝑥) −𝑓(𝑥)| ≥𝜀 In the above example, we proved something stronger: ∀𝑛, ∃𝑥∈𝑆, 𝑓 𝑛(𝑥) ≥1 2 We could have alternatively stated, for example, 𝑓 𝑛(𝑥) is continuous so there exists some subset of [0, 1] greater than 1 2 always.
Theorem. Let 𝑆⊆ℝ, ℂ. Let (𝑓 𝑛), 𝑓∶𝑆→ℝ(or ℂ), where 𝑓 𝑛is continuous and (𝑓 𝑛) →𝑓 uniformly on 𝑆. Then 𝑓is continuous.
Informally, the uniform limit of continuous functions is continuous.
142 1. Uniform convergence Proof. Fix some point 𝑎∈𝑆, 𝜀> 0. We seek 𝛿> 0 such that ∀𝑥∈𝑆, |𝑥−𝑎| < 𝛿 ⟹ |𝑓(𝑥) −𝑓(𝑎)| < 𝜀. We fix an 𝑛∈ℕsuch that ∀𝑥∈𝑆, |𝑓 𝑛(𝑥) −𝑓(𝑥)| < 𝜀. Since 𝑓 𝑛is continuous, there exists 𝛿> 0 such that ∀𝑥∈𝑆, |𝑥−𝑎| < 𝛿⟹|𝑓 𝑛(𝑥) −𝑓 𝑛(𝑎)| < 𝜀. So, ∀𝑥∈𝑆, |𝑥−𝑎| < 𝛿⟹|𝑓(𝑥) −𝑓(𝑎)| ≤|𝑓(𝑥) −𝑓 𝑛(𝑥)| + |𝑓 𝑛(𝑥) −𝑓 𝑛(𝑎)| + |𝑓 𝑛(𝑎) −𝑓(𝑎)| < 3𝜀 Remark. The above proof is often called a 3𝜀-proof. Note, the proof is not true for pointwise convergence; if 𝑓 𝑛→𝑓pointwise and 𝑓 𝑛continuous, 𝑓is not necessarily continuous. Fur-ther, it is not true for differentiability; 𝑓 𝑛differentiable does not imply 𝑓differentiable (see example sheet). Another way to interpret the result of the above theorem is to swap limits: lim 𝑥→𝑎lim 𝑛→∞𝑓 𝑛(𝑥) = lim 𝑥→𝑎𝑓(𝑥) = 𝑓(𝑎) = lim 𝑛→∞𝑓 𝑛(𝑎) = lim 𝑛→∞lim 𝑥→𝑎𝑓 𝑛(𝑥) 1.3. Uniform limit of bounded functions Lemma. Let 𝑓 𝑛→𝑓uniformly on 𝑆. If 𝑓 𝑛is bounded for every 𝑛, then so is 𝑓.
In other words, the uniform limit of bounded functions is bounded.
Proof. Fix some 𝑛∈ℕsuch that ∀𝑥∈𝑆, |𝑓 𝑛(𝑥) −𝑓(𝑥)| < 1. Since 𝑓 𝑛is bounded, ∃𝑀∈ℝ such that ∀𝑥∈𝑆, |𝑓 𝑛(𝑥)| < 𝑀. Hence, ∀𝑥∈𝑆, |𝑓(𝑥)| ≤|𝑓(𝑥) −𝑓 𝑛(𝑥)| + |𝑓 𝑛(𝑥)| ≤1 + 𝑀. So 𝑓is bounded.
1.4. Integrability Let 𝑓∶[𝑎, 𝑏] →ℝbe a bounded function. Recall that for a dissection 𝒟of [𝑎, 𝑏], we define the upper and lower sums of 𝑓with respect to 𝒟by 𝑈𝒟(𝑓) = 𝑛 ∑ 𝑘=1 (𝑥𝑘−𝑥𝑘−1) sup [𝑥𝑘−1,𝑥𝑘] 𝑓(𝑥) 𝐿𝒟(𝑓) = 𝑛 ∑ 𝑘=1 (𝑥𝑘−𝑥𝑘−1) inf [𝑥𝑘−1,𝑥𝑘] 𝑓(𝑥) Riemann’s integrability criterion states that 𝑓is integrable if and only if ∀𝜀, ∃𝒟, 𝑈𝒟(𝑓) −𝐿𝒟(𝑓) < 𝜀 Equivalently, for any 𝐼⊂[𝑎, 𝑏], we have sup 𝐼 𝑓−inf 𝐼𝑓= sup 𝑥,𝑦∈𝐼 (𝑓(𝑥) −𝑓(𝑦)) = sup 𝑥,𝑦∈𝐼 |𝑓(𝑥) −𝑓(𝑦)| This is called the oscillation of 𝑓on 𝐼. So an integrable function ‘doesn’t oscillate too much’.
143 IV. Analysis and Topology Theorem. Let 𝑓 𝑛∶[𝑎, 𝑏] →ℝbe integrable for all 𝑛. If 𝑓 𝑛→𝑓uniformly on [𝑎, 𝑏], then 𝑓 is integrable and ∫ 𝑏 𝑎 𝑓 𝑛→∫ 𝑏 𝑎 𝑓 Proof. First, we prove 𝑓to be bounded, then we will check Riemann’s criterion. We know 𝑓is bounded because each 𝑓 𝑛is bounded, hence by the lemma above 𝑓is bounded. Now fix 𝜀> 0, and choose 𝑛∈ℕsuch that ∀𝑥∈[𝑎, 𝑏], |𝑓 𝑛(𝑥) −𝑓(𝑥)| < 𝜀. Since 𝑓 𝑛is integrable, ∃𝒟∶𝑎= 𝑥0 < 𝑥1 < ⋯< 𝑥𝑁= 𝑏of [𝑎, 𝑏] such that 𝑈𝒟−𝐿𝒟< 𝜀. Now, we fix 𝑘∈{1, … , 𝑁} and then for any 𝑥, 𝑦∈[𝑥𝑘−1, 𝑥𝑘] we have |𝑓(𝑥) −𝑓(𝑦)| ≤|𝑓(𝑥) −𝑓 𝑛(𝑥)| + |𝑓 𝑛(𝑥) −𝑓 𝑛(𝑦)| + |𝑓 𝑛(𝑦) −𝑓(𝑦)| < 2𝜀+ |𝑓 𝑛(𝑥) −𝑓 𝑛(𝑦)| Taking the supremum, sup 𝑥,𝑦∈[𝑥𝑘−1,𝑥𝑘] (𝑓(𝑥) −𝑓(𝑦)) ≤ sup 𝑥,𝑦∈[𝑥𝑘−1,𝑥𝑘] |𝑓 𝑛(𝑥) −𝑓 𝑛(𝑦)| + 2𝜀 Multiplying by (𝑥𝑘−𝑥𝑘−1) and taking the sum over all 𝑘, 𝑈(𝑓) −𝐿(𝑓) ≤𝑈(𝑓 𝑛) −𝐿(𝑓 𝑛) + 2𝜀(𝑏−𝑎) ≤𝜀(2(𝑏−𝑎) + 1) Hence 𝑓is integrable. We can now show that | | | | ∫ 𝑏 𝑎 𝑓 𝑛−∫ 𝑏 𝑎 𝑓 | | | | ≤∫ 𝑏 𝑎 |𝑓 𝑛−𝑓| ≤(𝑏−𝑎) sup [𝑎,𝑏] |𝑓 𝑛−𝑓| →0 Remark. We can interpret this as ∫ 𝑏 𝑎 lim 𝑛→∞𝑓 𝑛(𝑥) d𝑥= lim 𝑛→∞∫ 𝑏 𝑎 𝑓 𝑛(𝑥) d𝑥 This is another ‘allowed’ way to swap limits.
Corollary. Let 𝑓 𝑛∶[𝑎, 𝑏] →ℝbe integrable for all 𝑛. If ∑ ∞ 𝑛=1 𝑓 𝑛(𝑥) converges uniformly on [𝑎, 𝑏], then 𝐹(𝑥) = ∞ ∑ 𝑛=1 𝑓 𝑛(𝑥) is integrable, and ∫ 𝑏 𝑎 ∞ ∑ 𝑛=1 𝑓 𝑛(𝑥) d𝑥= ∞ ∑ 𝑛=1 ∫ 𝑏 𝑎 𝑓 𝑛(𝑥) d𝑥 144 1. Uniform convergence Proof. Let 𝐹 𝑛(𝑥) = ∑ 𝑛 𝑘=1 𝑓𝑘(𝑥). By assumption, 𝐹 𝑛→𝐹uniformly on [𝑎, 𝑏]. 𝐹 𝑛is integrable where the integral of 𝐹 𝑛is the sum of the integrals: ∫ 𝑏 𝑎 𝐹 𝑛= 𝑛 ∑ 𝑘=1 ∫ 𝑏 𝑎 𝑓𝑘 Then the result follows from the theorem above.
1.5. Differentiability Theorem. Let 𝑓 𝑛∶[𝑎, 𝑏] →ℝbe continuously differentiable for all 𝑛. Suppose ∑ ∞ 𝑘=1 𝑓′ 𝑘(𝑥) converges uniformly on [𝑎, 𝑏], and that ∀𝑐∈[𝑎, 𝑏], ∑ ∞ 𝑛−1 𝑓 𝑛(𝑐) converges. Then, ∑ ∞ 𝑘=1 𝑓𝑘(𝑥) converges uniformly on [𝑎, 𝑏] to a continuously differentiable function 𝑓, and d d𝑥( ∞ ∑ 𝑘=1 𝑓𝑘) = ∞ ∑ 𝑘=1 d d𝑥𝑓𝑘(𝑥) Proof. Let 𝑔(𝑥) = ∑ ∞ 𝑘=1 𝑓′ 𝑘(𝑥), for 𝑥∈[𝑎, 𝑏]. The general idea is that we want to solve the differential equation 𝑓′ = 𝑔subject to the initial condition 𝑓(𝑐) = ∑ ∞ 𝑛=1 𝑓 𝑛(𝑐). Let 𝜆= ∑ ∞ 𝑛=1 𝑓 𝑛(𝑐) and define 𝑓∶[𝑎, 𝑏] →ℝby 𝑓(𝑥) = 𝜆+ ∫ 𝑥 𝑐 𝑔(𝑡) d𝑡 Note that 𝑔is integrable; ∑ ∞ 𝑘=1 𝑓′ 𝑘(𝑥) →𝑔uniformly implies that 𝑔is continuous and hence integrable. By the fundamental theorem of calculus, 𝑓′ = 𝑔and 𝑓(𝑐) = 𝜆. So we have found such an 𝑓that satisfies the conditions set out. All that remains is to prove uniform convergence of ∑ ∞ 𝑘=1 𝑓𝑘→𝑓. Also by the fundamental theorem, 𝑓𝑘(𝑥) = 𝑓𝑘(𝑐)+∫ 𝑥 𝑐𝑓′ 𝑘(𝑡) d𝑡.
Let 𝜀> 0. There exists 𝑁∈ℕsuch that | |𝜆−∑ 𝑁 𝑘=1 𝑓𝑘(𝑐)| | < 𝜀and | |𝑔(𝑡) −∑ 𝑁 𝑘=1 𝑓′ 𝑘(𝑡)| | < 𝜀.
Now, for 𝑛≥𝑁we have | | | | 𝑓(𝑥) − 𝑛 ∑ 𝑘=1 𝑓𝑘(𝑥) | | | | = | | | | 𝜆+ ∫ 𝑥 𝑐 𝑔(𝑡) d𝑡− 𝑛 ∑ 𝑘=1 (𝑓𝑘(𝑐) + ∫ 𝑥 𝑐 𝑓′ 𝑘(𝑡) d𝑡) | | | | ≤ | | | | 𝜆− 𝑛 ∑ 𝑘=1 𝑓𝑘(𝑐) | | | | + | | | | ∫ 𝑥 𝑐 (𝑔(𝑡) − 𝑛 ∑ 𝑘−1 𝑓′ 𝑘(𝑡)) d𝑡 | | | | ≤𝜀+ |𝑥−𝑐|𝜀 ≤𝜀(𝑏−𝑎+ 1) 145 IV. Analysis and Topology 1.6. Conditions for uniform convergence Recall that a scalar sequence 𝑥𝑛is Cauchy if ∀𝜀> 0, ∃𝑁∈ℕ, ∀𝑚, 𝑛≥𝑁, |𝑥𝑚−𝑥𝑛| < 𝜀 and that the general principle of convergence shows that any Cauchy sequence converges.
1.7. General principle of uniform convergence Definition. A sequence (𝑓 𝑛) of scalar functions on a set 𝑆is called uniformly Cauchy if ∀𝜀> 0, ∃𝑁∈ℕ, ∀𝑚, 𝑛≥𝑁, ∀𝑥∈𝑆, |𝑓 𝑚(𝑥) −𝑓 𝑛(𝑥)| < 𝜀 Theorem. A uniformly Cauchy sequence of functions is uniformly convergent.
Proof. Let 𝑥∈𝑆and we will show that (𝑓 𝑛(𝑥))∞ 𝑛=1 converges. Given 𝜀> 0, ∃𝑁∈ℕ, ∀𝑚, 𝑛≥ 𝑁, ∀𝑡∈𝑆, |𝑓 𝑚(𝑡) −𝑓 𝑛(𝑡)| < 𝜀. In particular, ∀𝑚, 𝑛≥𝑁, |𝑓 𝑚(𝑥) −𝑓 𝑛(𝑥)| < 𝜀. So certainly (𝑓 𝑛(𝑥))∞ 𝑛=1 is Cauchy and hence convergent by the general principle of convergence. There-fore 𝑓 𝑛converges pointwise. Now, let 𝑓(𝑥) be the limit 𝑓(𝑥) = lim𝑛→∞𝑓 𝑛(𝑥). Then 𝑓 𝑛→𝑓 pointwise on 𝑆. Now we must extend this to show 𝑓 𝑛→𝑓uniformly on 𝑆. Given 𝜀> 0, we know that ∃𝑁∈ℕ, ∀𝑚, 𝑛≥𝑁, ∀𝑥∈𝑆, |𝑓 𝑚(𝑥) −𝑓 𝑛(𝑥)| < 𝜀. Now, we must show ∀𝑛≥𝑁, ∀𝑥∈𝑆, |𝑓 𝑛(𝑥) −𝑓(𝑥)| < 2𝜀, then we are done. We will fix 𝑥∈𝑆, 𝑛≥𝑁. Since 𝑓 𝑛(𝑥) →𝑓(𝑥), we can choose 𝑚∈ℕsuch that |𝑓 𝑚(𝑥) −𝑓(𝑥)| < 𝜀, and 𝑚≥𝑁. Note however that 𝑚depends on 𝑥in this statement, but this doesn’t matter—we have shown that |𝑓 𝑛(𝑥) −𝑓(𝑥)| ≤|𝑓 𝑛(𝑥) −𝑓 𝑚(𝑥)| + |𝑓 𝑚(𝑥) −𝑓(𝑥)| ≤𝜀+ 𝜀= 2𝜀 which is a result that, in itself, does not depend on 𝑥.
Note. Alternatively, we could end the proof as the following. Fix 𝑥∈𝑆, 𝑛≥𝑁. Then ∀𝑚≥𝑁, |𝑓 𝑛(𝑥) −𝑓 𝑚(𝑥)| < 𝜀 Then let 𝑚→∞, and |𝑓 𝑛(𝑥) −𝑓(𝑥)| ≤𝜀 1.8. Weierstrass M-test Theorem. Let (𝑓 𝑛) be a sequence of scalar functions on 𝑆. Assume that ∀𝑛∈ℕ, ∃𝑀𝑛∈ ℝ+, ∀𝑥∈𝑆, |𝑓 𝑛(𝑥)| ≤𝑀𝑛. In other words, (𝑓 𝑛) is a sequence of bounded scalar functions.
Then, ∞ ∑ 𝑛=1 𝑀𝑛< ∞⟹ ∞ ∑ 𝑛=1 𝑓 𝑛(𝑥) is uniformly convergent on 𝑆 146 1. Uniform convergence Proof. Let 𝐹 𝑛(𝑥) = ∑ 𝑛 𝑘=1 𝑓𝑘(𝑥) for 𝑥∈𝑆, 𝑛∈ℕ. Then |𝐹 𝑛(𝑥) −𝐹 𝑚(𝑥)| ≤ 𝑛 ∑ 𝑘=𝑚+1 |𝑓𝑘(𝑥)| ≤ 𝑛 ∑ 𝑘=𝑚+1 𝑀𝑘 Hence, given 𝜀> 0, we can choose 𝑁∈ℕsuch that ∑ 𝑛 𝑘=𝑁+1 𝑀𝑘< 𝜀. Thus, ∀𝑥∈𝑆, ∀𝑛≥ 𝑚≥𝑁, we have |𝐹 𝑛(𝑥) −𝐹 𝑚(𝑥)| ≤ 𝑛 ∑ 𝑘=𝑚+1 𝑀𝑘< 𝜀 We have shown (𝐹 𝑛) is uniformly Cauchy on 𝑆and hence uniformly convergent on 𝑆.
1.9. Power series Consider the power series ∞ ∑ 𝑛=0 𝑐𝑛(𝑧−𝑎)𝑛 where 𝑐𝑛∈ℂ, 𝑎∈ℂare constants, and 𝑧∈ℂ. Let 𝑅∈[0, ∞] be the radius of convergence.
Recall that |𝑧−𝑎| < 𝑅⟹ ∞ ∑ 𝑛=0 𝑐𝑛(𝑧−𝑎)𝑛converges absolutely; |𝑧−𝑎| > 𝑅⟹ ∞ ∑ 𝑛=0 𝑐𝑛(𝑧−𝑎)𝑛diverges Let 𝐷(𝑎, 𝑅) ≔{𝑧∈ℂ∣|𝑧−𝑎| < 𝑅} be the open disc centred on 𝑎with radius 𝑅. Then we can create 𝑓∶𝐷(𝑎, ℝ) →ℂto be defined by the power series, which is well-defined. 𝑓is the pointwise limit of the power series on 𝐷. In general, the convergence of the power series is not uniformly convergent.
Example. ∑ ∞ 𝑛=1 𝑧𝑛 𝑛2 has 𝑅= 1. Let 𝑓 𝑛∶𝐷(0, 1) →𝒞be defined by 𝑓 𝑛(𝑧) = 𝑧𝑛 𝑛2 . Then for every 𝑧∈𝐷(0, 1), |𝑧| ≤ 1 𝑛2 . Since ∑ ∞ 𝑛=1 1 𝑛2 = 𝜋2 6 < ∞, by the Weierstrass M-test, the power series converges uniformly on the disc.
Example. Consider ∑ ∞ 𝑛=0 𝑧𝑛= 1 1−𝑧with 𝑅= 1. Now, ∀𝑧∈𝐷(0, 1), | | | | ∞ ∑ 𝑛=0 𝑧𝑛| | | | ≤𝑁+ 1 Therefore, the series does not converge uniformly on the disc since 1 1−𝑧is unbounded on the disc. Alternatively, consider sup |𝑧|<1 | | | | 1 1 −𝑧− 𝑛 ∑ 𝑘=0 𝑧𝑘 | | | | = sup |𝑧|<1 | | | 𝑧𝑛+1 1 −𝑧 | | | = ∞ 147 IV. Analysis and Topology In some sense, the problem with uniform convergence here is that we are allowed to go too close too the boundary.
Theorem. Suppose the power series ∑ ∞ 𝑛=0 𝑐𝑛(𝑧−𝑎)𝑛has radius of convergence 𝑅. Then for all 0 < 𝑟< 𝑅, the power series converges uniformly on 𝐷(𝑎, 𝑟).
Proof. Let 𝑤∈ℂsuch that 𝑟< |𝑤−𝑎| < 𝑅, for instance 𝑤= 𝑎+ 𝑟+𝑅 2 . Now, let 𝜌= 𝑟 |𝑤−𝑎| ∈ (0, 1). Since ∑ ∞ 𝑛=0 𝑐𝑛(𝑤−𝑎)𝑛converges, we have that 𝑐𝑛(𝑤−𝑎)𝑛→0 as 𝑛→∞. Therefore, ∃𝑀∈ℝ+ such that |𝑐𝑛(𝑤−𝑎)𝑛| ≤𝑀for all 𝑛∈ℕ, since convergence implies boundedness.
Now, for 𝑧∈𝐷(𝑎, 𝑟), 𝑛∈ℕwe have |𝑐𝑛(𝑧−𝑎)𝑛| = |𝑐𝑛(𝑤−𝑎)𝑛|( |𝑧−𝑎| |𝑤−𝑎|) 𝑛 ≤𝑀( 𝑟 |𝑤−𝑎|) 𝑛 = 𝑀𝜌𝑛 Since the sum ∑ ∞ 𝑛=0 𝑀𝜌𝑛converges, the Weierstrass M-test shows us that ∑ ∞ 𝑛=0 𝑐𝑛(𝑧−𝑎)𝑛 converges uniformly on 𝐷(𝑎, 𝑟).
Remark. 𝑓∶𝐷(𝑎, 𝑅) →ℂdefined by 𝑓(𝑧) = ∑ ∞ 𝑛=0 𝑐𝑛(𝑧−𝑎)𝑛is the uniform limit on 𝐷(𝑎, 𝑟) of polynomials for any 𝑟such that 0 < 𝑟< 𝑅. Hence 𝑓is continuous on 𝐷(𝑎, 𝑟). Since 𝐷(𝑎, 𝑅) = ⋃0<𝑟<𝑅𝐷(𝑎, 𝑟), it follows that 𝑓is continuous everywhere inside the radius of convergence.
Recall that the termwise derivative ∑ ∞ 𝑛=1 𝑐𝑛𝑛(𝑧−𝑎)𝑛−1 has the same radius of convergence.
This sequence therefore also converges uniformly on 𝐷(𝑎, 𝑟) if 0 < 𝑟< 𝑅. Analogously to the previous result about interchanging derivatives and sums, we can show that ∑𝑐𝑛(𝑧−𝑎)𝑛 is complex differentiable on 𝐷(𝑎, 𝑅) with derivative ∑ ∞ 𝑛=1 𝑐𝑛𝑛(𝑧−𝑎)𝑛−1. This is seen in the IB Complex Analysis course.
Now, fix 𝑤∈𝐷(𝑎, 𝑅). Then fix 𝑟such that |𝑤−𝑎| < 𝑟< 𝑅, and fix 𝛿> 0 such that |𝑤−𝑎|+𝛿< 𝑟. If |𝑧−𝑤| < 𝛿, then |𝑧−𝑎| ≤|𝑧−𝑤|+|𝑤−𝑎| < 𝛿+|𝑤−𝑎| < 𝑟. Therefore, geometrically, 𝐷(𝑤, 𝛿) ⊂𝐷(𝑎, 𝑟). Hence, ∑ ∞ 𝑛=0 𝑐𝑛(𝑧−𝑎)𝑛converges uniformly on 𝐷(𝑤, 𝛿).
This is known as local uniform convergence.
Definition. 𝑈⊂ℂis called open if ∀𝑤∈𝑈, ∃𝛿> 0, 𝐷(𝑤, 𝛿) ⊂𝑈.
Definition. Let 𝑈be an open subset of ℂ, and 𝑓 𝑛be a sequence of scalar functions on 𝑈.
Then 𝑓 𝑛converges locally uniformly on 𝑈if ∀𝑤∈𝑈, ∃𝛿> 0, 𝑓 𝑛converges uniformly on 𝐷(𝑤, 𝛿) ⊂𝑈 Remark. Above, we showed that power series always converge locally uniformly inside the radius of convergence, or equivalently inside the disc 𝐷(𝑎, 𝑅). We will return to this point about local uniform convergence when discussing compactness.
148 2. Uniform continuity 2. Uniform continuity 2.1. Definition Let 𝑈⊂ℝ, ℂ. Let 𝑓be a scalar function on 𝑈. Then for 𝑥∈𝑈, we say 𝑓is continuous at 𝑥 if ∀𝜀> 0, ∃𝛿> 0, ∀𝑦∈𝑈, |𝑦−𝑥| < 𝛿⟹|𝑓(𝑦) −𝑓(𝑥)| < 𝜀 We say 𝑓is continuous on 𝑈if 𝑓is continuous at 𝑥for all 𝑥∈𝑈: ∀𝑥∈𝑈, ∀𝜀> 0, ∃𝛿> 0, ∀𝑦∈𝑈, |𝑦−𝑥| < 𝛿⟹|𝑓(𝑦) −𝑓(𝑥)| < 𝜀 Note here that 𝛿depends on 𝜀and 𝑥.
Definition. Let 𝑈, 𝑓be as in the previous definition. We say 𝑓is uniformly continuous if ∀𝜀> 0, ∃𝛿> 0, ∀𝑥, 𝑦∈𝑈, |𝑦−𝑥| < 𝛿⟹|𝑓(𝑦) −𝑓(𝑥)| < 𝜀 Now, 𝛿works for all 𝑥∈𝑈simultaneously; 𝛿depends on 𝜀only. Certainly, uniform con-tinuity implies continuity.
Example. Let 𝑓∶ℝ→ℝsuch that 𝑓(𝑥) = 2𝑥+ 17. Then 𝑓is uniformly continuous; given 𝜀> 0, we can find 𝛿= 1 2𝜀. Then ∀𝑥, 𝑦∈ℝ, |𝑦−𝑥| < 𝛿⟹|𝑓(𝑦) −𝑓(𝑥)| = |2𝑦−2𝑥| = 2𝑦−𝑥< 2𝛿= 𝜀.
Example. Let 𝑓∶ℝ→ℝ, defined by 𝑓(𝑥) = 𝑥2. This is not uniformly continuous, since no 𝛿works for all 𝑥given some ‘bad’ 𝜀. Let us take 𝜀= 1, and we wish to show that no 𝛿exists.
Suppose some 𝛿does exist. Then, let 𝑥> 0 and 𝑦= 𝑥+ 𝛿 2. We should have |𝑓(𝑦) −𝑓(𝑥)| < 1.
(𝑥+ 𝛿 2) 2 −𝑥2 = 𝛿𝑥+ 𝛿2 4 So for 𝑥= 1 𝛿, this condition |𝑓(𝑦) −𝑓(𝑥)| < 1 is not satisfied. Hence 𝑓is not uniformly continuous.
Note. For 𝑈, 𝑓as in the above definition, 𝑓is not uniformly continuous on 𝑈if ∃𝜀> 0, ∀𝛿> 0, ∃𝑥, 𝑦∈𝑈, |𝑦−𝑥| < 𝛿, |𝑓(𝑦) −𝑓(𝑥)| ≥𝜀 So there are points arbitrarily close together whose difference of function values exceed some fixed 𝜀.
2.2. Properties of continuous functions Theorem. Let 𝑓be a scalar function on a closed bounded interval [𝑎, 𝑏]. If 𝑓is continuous on [𝑎, 𝑏], then 𝑓is uniformly continuous on [𝑎, 𝑏].
149 IV. Analysis and Topology Proof. Suppose there exists 𝜀> 0 such that ∀𝛿> 0, ∃𝑥, 𝑦∈[𝑎, 𝑏], |𝑦−𝑥| < 𝛿, |𝑓(𝑦) −𝑓(𝑥)| ≥ 𝜀. In particular, we can construct a sequence (𝛿𝑛) defined by 𝛿𝑛= 1 𝑛, and we can construct sequences 𝑥𝑛, 𝑦𝑛∈[𝑎, 𝑏] such that |𝑦𝑛−𝑥𝑛| < 1 𝑛but |𝑓(𝑦𝑛) −𝑓(𝑥𝑛)| ≥𝜀. By the Bolzano– Weierstrass theorem, there exists a subsequence (𝑥𝑘𝑛) that converges. Now, let 𝑥be the limit of the subsequence, lim𝑛→∞𝑥𝑘𝑛. Then 𝑥∈[𝑎, 𝑏] since the interval is closed. Then, | |𝑦𝑘𝑛−𝑥| | ≤| |𝑦𝑘𝑛−𝑥𝑘𝑛| | + | |𝑥𝑘𝑛−𝑥| | < 1 𝑛+ | |𝑥𝑘𝑛−𝑥| | →0. Hence 𝑦𝑘𝑛→𝑥. Now, since 𝑓is continuous 𝑓(𝑥𝑘𝑛), 𝑓(𝑦𝑘𝑛) →𝑓(𝑥). Now, 𝜀≤| |𝑓(𝑥𝑘𝑛) −𝑓(𝑦𝑘𝑛)| | →|𝑓(𝑥) −𝑓(𝑥)| = 0, which is a contradiction.
Corollary. A continuous function 𝑓∶[𝑎, 𝑏] →ℝis Riemann integrable.
Proof. Since a continuous function on a closed bounded interval is bounded, we have that 𝑓is bounded. Now, fix 𝜀> 0, and we want to find a dissection 𝒟such that the difference between upper and lower sums is less than 𝜀. By the above theorem, 𝑓is uniformly continu-ous. Hence, ∃𝛿> 0, ∀𝑥, 𝑦∈[𝑎, 𝑏], |𝑦−𝑥| < 𝛿⟹|𝑓(𝑦) −𝑓(𝑥)| < 𝜀 So we must simply choose a dissection such that all intervals have size smaller than 𝛿. For instance, choose some 𝑛∈ℕsuch that 𝑏−𝑎 𝑁 < 𝛿, and then divide the interval equally into 𝑛subintervals. If 𝐼is an interval in this dissection, then ∀𝑥, 𝑦∈𝐼we have |𝑦−𝑥| < 𝛿and hence |𝑓(𝑦) −𝑓(𝑥)| < 𝜀. Hence, sup 𝑥,𝑦∈𝐼 |𝑓(𝑦) −𝑓(𝑥)| ≤𝜀 Multiplying by the length of 𝐼and summing over all subintervals 𝐼, 𝑈𝒟(𝑓) −𝐿𝒟(𝑓) ≤(𝑏−𝑎)𝜀 Hence 𝑓is Riemann integrable.
150 3. Metric spaces 3. Metric spaces 3.1. Definition Definition. Let 𝑀be a set. Then a metric on 𝑀is a function 𝑑∶𝑀× 𝑀→ℝsuch that (i) (positivity) ∀𝑥, 𝑦∈𝑀, 𝑑(𝑥, 𝑦) ≥0, and in particular, 𝑥= 𝑦⟺𝑑(𝑥, 𝑦) = 0 (ii) (symmetric) ∀𝑥, 𝑦∈𝑀, 𝑑(𝑥, 𝑦) = 𝑑(𝑦, 𝑥) (iii) (triangle inequality) ∀𝑥, 𝑦, 𝑧∈𝑀, 𝑑(𝑥, 𝑧) ≤𝑑(𝑥, 𝑦) + 𝑑(𝑦, 𝑧).
A metric space is a set 𝑀together with a metric 𝑑on 𝑀, written as the pair (𝑀, 𝑑).
Example. Let 𝑀= ℝ, ℂand 𝑑(𝑥, 𝑦) = |𝑥−𝑦|. This is known as the ‘standard metric’ on 𝑀. If a metric is not specified, the standard metric is taken as implied.
Example. Let 𝑀= ℝ𝑛, ℂ𝑛, and we define the Euclidean norm (or Euclidean length) to be ‖𝑥‖ = ‖𝑥‖2 = ( 𝑛 ∑ 𝑘=1 |𝑥𝑘|2) 1 2 This satisfies ‖𝑥+ 𝑦‖ ≤‖𝑥‖ + ‖𝑦‖ and it then follows that we can define the metric as 𝑑2(𝑥, 𝑦) = ‖𝑥−𝑦‖2 called the Euclidean metric. We can check that this is indeed a metric easily. This is the standard metric on ℝ𝑛, ℂ𝑛. The metric space (𝑀, 𝑑) in this case is called 𝑛-dimensional real (or complex) Euclidean space, sometimes denoted ℓ𝑛 2 . The Euclidean norm is sometimes called the ℓ2 norm, and the Euclidean metric is the ℓ2 metric.
Example. Let 𝑀= ℝ𝑛, ℂ𝑛, and we define the ℓ1 norm to be |𝑥|1 = 𝑛 ∑ 𝑘=1 |𝑥𝑘| which defines the ℓ1 metric given by 𝑑1(𝑥, 𝑦) = ‖𝑥−𝑦‖1 (𝑀, 𝑑1) is denoted ℓ𝑛 1 . We can generalise and form the metric space ℓ𝑛 𝑝for all 𝑝∈[1, ∞].
Example. Again, let 𝑀= ℝ𝑛, ℂ𝑛. We can define the ℓ∞norm by ‖𝑥‖∞= max 1≤𝑘≤𝑛|𝑥𝑘| This defines the ℓ∞metric: 𝑑∞(𝑥, 𝑦) = ‖𝑥−𝑦‖∞= max 1≤𝑘≤𝑛|𝑥𝑘−𝑦𝑘| We denote (𝑀, 𝑑) by ℓ𝑛 ∞.
151 IV. Analysis and Topology In this course, we will only work with 𝑝= 1, 2, ∞, although the calculations can be made to work for other 𝑝.
Example. Let 𝑆be a set. Let ℓ∞(𝑆) be the set of all bounded scalar functions on 𝑆. We then define the ℓ∞norm of 𝑓∈ℓ∞(𝑆) by ‖𝑓‖ = ‖𝑓‖∞= sup 𝑥∈𝑆 |𝑓(𝑥)| The supremum exists since the function is always bounded. This is also known as the ‘sup norm’ or the ‘uniform norm’. Note that, for 𝑓, 𝑔∈ℓ∞(𝑆), and 𝑥∈𝑆, ‖𝑓+ 𝑔‖ ≤sup 𝑥∈𝑆 |𝑓(𝑥) + 𝑔(𝑥)| ≤|𝑓(𝑥) + 𝑔(𝑥)| ≤|𝑓(𝑥)| + |𝑔(𝑥)| ≤‖𝑓‖ + ‖𝑔‖ Hence 𝑑(𝑓, 𝑔) = ‖𝑓−𝑔‖ defines a metric on ℓ∞(𝑆). This is the standard metric on this space ℓ∞(𝑆), also called the ‘uniform metric’. For example, ℓ∞({1, … , 𝑛}) = ℝ𝑛with the metric ℓ∞. Also, for ℓ∞(ℕ), we typically omit the ℕand instead write ℓ∞for the space of scalar sequences with the uniform metric.
Example. Consider 𝐶[𝑎, 𝑏], the set of all continuous functions on [𝑎, 𝑏]. For 𝑝= 1, 2, we define the 𝐿𝑝norm of 𝑓∈𝐶[𝑎, 𝑏] by ‖𝑓‖𝑝= (∫ 𝑏 𝑎 |𝑓(𝑥)|𝑝d𝑥) 1 𝑝 which induces the 𝐿𝑝metric on 𝐶[𝑎, 𝑏].
Example. Let 𝑀be a set. Then 𝑑(𝑥, 𝑦) = {0 if 𝑥= 𝑦 1 otherwise is a metric, called the discrete metric on 𝑀. In particular, (𝑀, 𝑑) is called a discrete metric space.
Example. Let 𝐺be a group generated by 𝑆⊂𝐺. We assume 𝑒∉𝑆and 𝑥∈𝑆⟹𝑥−1 ∈𝑆.
Then 𝑑(𝑥, 𝑦) = min {𝑛≥0∶∃𝑠1, … , 𝑠𝑛, 𝑦= 𝑥𝑠1 … 𝑠𝑛} defines a metric called the word metric.
Example. Let 𝑝be prime. Then 𝑑(𝑥, 𝑦) = {0 if 𝑥= 𝑦 𝑝−𝑛 otherwise, where 𝑥−𝑦= 𝑝𝑛𝑚, 𝑛≥0, 𝑚∈𝑍, 𝑝∤𝑚 defines a metric on ℤ. This is known as the 𝑝-adic metric.
152 3. Metric spaces 3.2. Subspaces Let (𝑀, 𝑑) be a metric space, and 𝑁⊂𝑀. Then naturally we can restrict 𝑑to 𝑁× 𝑁, giving a metric on 𝑁. (𝑁, 𝑑) is called a subspace of 𝑀.
Example. Consider ℚwith the metric 𝑑(𝑥, 𝑦) = |𝑥−𝑦|. This is clearly a subspace of ℝ (implicitly with the standard metric on ℝ).
Example. Since every continuous function on a closed bounded interval is bounded, 𝐶[𝑎, 𝑏] is a subset of ℓ∞[𝑎, 𝑏]. Hence 𝐶[𝑎, 𝑏] with the uniform metric is a subspace of ℓ∞[𝑎, 𝑏].
3.3. Product spaces Let (𝑀, 𝑑), (𝑀′, 𝑑′) be metric spaces. Then any of the following defines a metric on the Cartesian product 𝑀× 𝑀′.
(i) 𝑑1((𝑥, 𝑥′), (𝑦, 𝑦′)) = 𝑑(𝑥, 𝑦) + 𝑑(𝑥′, 𝑦′) (ii) 𝑑2((𝑥, 𝑥′), (𝑦, 𝑦′)) = (𝑑(𝑥, 𝑦)2 + 𝑑(𝑥′, 𝑦′)2) 1 2 (iii) 𝑑∞((𝑥, 𝑥′), (𝑦, 𝑦′)) = max {𝑑(𝑥, 𝑦), 𝑑(𝑥′, 𝑦′)} We commonly write (𝑀× 𝑀′, 𝑝) as 𝑀⊕𝑝𝑀′. Note that we always have 𝑑∞≤𝑑2 ≤𝑑1 ≤2𝑑∞ We can generalise for 𝑛∈ℕand metric spaces (𝑀𝑘, 𝑑𝑘) for 𝑘∈{1, … , 𝑛}, by defining ( 𝑛 ⨁ 𝑘=1 𝑀𝑘) 𝑝 = 𝑀1 ⊕𝑝⋯⊕𝑝𝑀𝑛= (𝑀1 × ⋯× 𝑀𝑛, 𝑑𝑝) Example. ℝ⊕1 ℝ= ℓ2 1. Further, ℝ⊕2 ℝ⊕2 𝑅= ℓ3 2, and other analogous results hold.
Remark. ℝ⊕1 ℝ⊕2 ℝdoes not make sense since we have not defined the associativity of the ⊕operator. The two choices yield different metric spaces.
3.4. Convergence Let 𝑀be a metric space, and (𝑥𝑛) a sequence in 𝑀. Given 𝑥∈𝑀, we say that (𝑥𝑛) converges to 𝑥in 𝑀if ∀𝜀> 0, ∃𝑁∈ℕ, ∀𝑛≥𝑁, 𝑑(𝑥𝑛, 𝑥) < 𝜀 We say that (𝑥𝑛) is convergent in 𝑀if ∃𝑥∈𝑀such that 𝑥𝑛→𝑥. Otherwise, we say that (𝑥𝑛) is divergent. Note that 𝑥𝑛→𝑥in 𝑀if and only if 𝑑(𝑥𝑛, 𝑥) →0 in ℝ.
Lemma. Suppose we have a sequence 𝑥𝑛→𝑥and 𝑥𝑛→𝑦in a metric space 𝑀. Then 𝑥= 𝑦.
153 IV. Analysis and Topology Proof. Suppose 𝑥≠𝑦. Then let 𝜀= 𝑑(𝑥,𝑦) 3 > 0. So, by the definition of convergence, ∃𝑁1 ∈ℕ, ∀𝑛≥𝑁1, 𝑑(𝑥𝑛, 𝑥) < 𝜀; ∃𝑁2 ∈ℕ, ∀𝑛≥𝑁2, 𝑑(𝑥𝑛, 𝑦) < 𝜀 Now, fix 𝑁∈ℕsuch that 𝑛≥𝑁1, 𝑛≥𝑁2, for instance 𝑁= max {𝑁1, 𝑁2}. Then 𝑑(𝑥, 𝑦) ≤𝑑(𝑥, 𝑥𝑛) + 𝑑(𝑥𝑛, 𝑦) < 2𝜀= 2 3𝑑(𝑥, 𝑦) which is a contradiction.
Definition. Given a convergent sequence (𝑥𝑛) in a metric space 𝑀, we say the limit of (𝑥𝑛) is the unique 𝑥∈𝑀such that 𝑥𝑛→𝑥as 𝑛→∞. This is denoted lim 𝑛→∞𝑥𝑛 Example. This definition has the usual meaning when 𝑀= ℝ, ℂ.
Example. The constant sequence defined by 𝑥𝑛= 𝑥converges to 𝑥. In particular, ‘even-tually constant’ sequences converge; let (𝑥𝑛) be a sequence in 𝑀such that ∃𝑥∈𝑀, ∃𝑁∈ ℕ, ∀𝑛≥𝑁, 𝑥𝑛= 𝑥, then 𝑥𝑛→𝑥. It is not necessarily true that sequences only converge if they are eventually constant. However, in a discrete metric space, the converse is true, since we can choose 𝜀smaller than all distances.
Example. Consider the 3-adic metric. Then, 3𝑛→0 as 𝑛→∞since 𝑑(3𝑛, 0) = 3−𝑛→0.
Example. Let 𝑆be a set. Then, 𝑓 𝑛→𝑓in ℓ∞(𝑆) in the uniform metric if and only if 𝑑(𝑓 𝑛, 𝑓) = ‖𝑓 𝑛−𝑓‖∞= sup𝑆|𝑓 𝑛−𝑓| →0, which is precisely the condition that 𝑓 𝑛→𝑓 uniformly on 𝑆. Note, however, that 𝑓 𝑛(𝑥) = 𝑥+ 1 𝑛for 𝑥∈ℝ, 𝑛∈ℕand 𝑓(𝑥) = 𝑥, then certainly 𝑓 𝑛→𝑥uniformly on ℝ. However, 𝑓 𝑛, 𝑓∉ℓ∞(ℝ), so the uniform metric is not defined on these functions. So the notion of uniform convergence visited before is slightly more general than the idea of convergence in this metric space.
Example. Consider Euclidean space 𝑀= ℝ𝑛, ℂ𝑛with the ℓ2 metric. Then, consider 𝑥(𝑘) = (𝑥(𝑘) 1 , … , 𝑥(𝑘) 𝑛) ∈𝑀 for 𝑘∈ℕ, and 𝑥= (𝑥1, … , 𝑥𝑛) ∈𝑀. Then, | |𝑥(𝑘) 𝑖 −𝑥𝑖| | ≤‖ ‖𝑥(𝑘) −𝑥‖ ‖2 ≤ 𝑛 ∑ 𝑖=1 | |𝑥(𝑘) 𝑖 −𝑥𝑖| | So 𝑥(𝑘) →𝑥if and only if all 𝑖satisfy 𝑥(𝑘) 𝑖 →𝑥𝑖. This can be thought of as convergence being equivalent to coordinate-wise (or pointwise) convergence.
154 3. Metric spaces Example. Consider 𝑓 𝑛(𝑥) = 𝑥𝑛for 𝑥∈[0, 1], and 𝑛∈ℕ. Then (𝑓 𝑛) is a sequence in 𝐶[0, 1], which converges pointwise but not uniformly. So (𝑓 𝑛) is not convergent in the uniform met-ric. However, using the 𝐿1 metric, we have 𝑑1(𝑓 𝑛, 0) = ‖𝑓 𝑛‖1 = ∫ 1 0 𝑓 𝑛= 1 𝑛+ 1 →0 So, 𝑓 𝑛→0 in (𝐶[0, 1], 𝐿1).
Example. Let 𝑁be a subspace of a metric space 𝑀, and (𝑥𝑛) be a convergent sequence in 𝑁. Then (𝑥𝑛) converges in 𝑀. The converse is not necessarily true; consider 𝑀= ℝand 𝑁= (0, ∞) with (𝑥𝑛) = 1 𝑛. This is divergent in 𝑁but convergent in 𝑀.
Example. Let (𝑀, 𝑑), (𝑀′, 𝑑′) be metric spaces. Let 𝑁= 𝑀⊕𝑝𝑀′. Let 𝑎𝑛= (𝑥𝑛, 𝑦𝑛) ∈𝑁 for all 𝑛∈ℕ, and 𝑎= (𝑥, 𝑦) ∈𝑁. Then 𝑎𝑛→𝑎in 𝑁⟺𝑥𝑛→𝑥in 𝑀, 𝑦𝑛→𝑦in 𝑀′ Indeed, max {𝑑(𝑥𝑛, 𝑥), 𝑑′(𝑦𝑛, 𝑦)} = 𝑑∞(𝑎𝑛, 𝑎) ≤𝑑𝑝(𝑎𝑛, 𝑎) ≤2𝑑1(𝑎𝑛, 𝑎) = 2𝑑(𝑥𝑛, 𝑥) + 2𝑑′(𝑦𝑛, 𝑦) 3.5. Continuity Definition. Let 𝑓∶𝑀→𝑀′ be a function between metric spaces (𝑀, 𝑑), (𝑀′, 𝑑′). Then for 𝑎∈𝑀, we say 𝑓is continuous at 𝑎if ∀𝜀> 0, ∃𝛿> 0, ∀𝑥∈𝑀, 𝑑(𝑥, 𝑎) < 𝛿⟹𝑑′(𝑓(𝑥), 𝑓(𝑎)) < 𝜀 We say 𝑓is continuous if 𝑓is continuous at 𝑎for all 𝑎∈𝑀. In other words, ∀𝑎∈𝑀, ∀𝜀> 0, ∃𝛿> 0, ∀𝑥∈𝑀, 𝑑(𝑥, 𝑎) < 𝛿⟹𝑑′(𝑓(𝑥), 𝑓(𝑎)) < 𝜀 Note that 𝛿depends both on 𝜀and 𝑎.
Proposition. Let 𝑓∶𝑀→𝑀′ be as above. Let 𝑎∈𝑀. Then the following are equivalent: (i) 𝑓is continuous at 𝑎; (ii) 𝑥𝑛→𝑎in 𝑀implies 𝑓(𝑥𝑛) →𝑓(𝑎) in 𝑀 Proof. First we show (i) implies (ii). Suppose 𝑥𝑛→𝑎in 𝑀. Then fix 𝜀> 0, and seek 𝑁∈ℕsuch that ∀𝑛≥𝑁, 𝑑′(𝑓(𝑥𝑛), 𝑓(𝑎)) < 𝜀. By continuity, there exists 𝛿> 0 such that ∀𝑥∈𝑀, 𝑑(𝑥, 𝑎) < 𝛿 ⟹ 𝑑′(𝑓(𝑥𝑛), 𝑓(𝑎)) < 𝜀as required. So we want 𝑁such that ∀𝑛≥𝑁, 𝑑(𝑥, 𝑎) < 𝛿, which must exist since 𝑥𝑛→𝑎.
Now, we show (ii) implies (i). Suppose that 𝑓is not continuous at 𝑎. Then, ∃𝜀> 0, ∀𝛿> 0, ∃𝑥∈𝑀, 𝑑(𝑥, 𝑎) < 𝛿, 𝑑′(𝑓(𝑥), 𝑓(𝑎)) ≥𝜀 155 IV. Analysis and Topology So fix such an 𝜀for which no suitable 𝛿exists. Choose the sequence 𝛿𝑛= 1 𝑛, so 𝑑(𝑥𝑛, 𝑎) < 1 𝑛; 𝑑′(𝑓(𝑥𝑛), 𝑓(𝑎)) ≥𝜀 Then 𝑥𝑛→𝑎in 𝑀but 𝑓(𝑥𝑛) ↛𝑓(𝑎) in 𝑀, which is a contradiction.
Proposition. Let 𝑓, 𝑔be scalar functions on a metric space 𝑀. Let 𝑎∈𝑀. Then if 𝑓, 𝑔 are continuous at 𝑎, so are 𝑓+ 𝑔and 𝑓⋅𝑔. Moreover, letting 𝑁= {𝑥∈𝑀∶𝑔(𝑥) ≠0} and assuming 𝑎∈𝑁, 𝑓 𝑔is continuous at 𝑎. Hence if 𝑓, 𝑔are continuous, then so are 𝑓+𝑔, 𝑓⋅𝑔, 𝑓 𝑔 where they are defined.
Proof. Suppose 𝑥𝑛→𝑎. Then by the above proposition, (𝑓⋅𝑔)(𝑥𝑛) = 𝑓(𝑥𝑛) ⋅𝑔(𝑥𝑛) → 𝑓(𝑎) ⋅𝑔(𝑎) = (𝑓⋅𝑔)(𝑎), and similar results hold for the other operators.
Remark. If 𝑓∶𝑀→𝑀′ is continuous everywhere, lim 𝑛→∞𝑓(𝑥𝑛) = 𝑓( lim 𝑛→∞𝑥𝑛) by the second proposition.
Proposition. Let 𝑓∶𝑀→𝑀′, 𝑔∶𝑀′ →𝑀″ be functions between metric spaces. If 𝑓 is continuous at 𝑎and 𝑔is continuous at 𝑓(𝑎), then 𝑔∘𝑓is continuous at 𝑎. If 𝑓, 𝑔are continuous, 𝑔∘𝑓is continuous.
Proof. Let 𝜀> 0. We want to find 𝛿> 0 such that ∀𝑥∈𝑀, 𝑑(𝑥, 𝑎) < 𝛿⟹𝑑″(𝑔(𝑓(𝑥)), 𝑔(𝑓(𝑎))) < 𝜀 Since 𝑔is continuous at 𝑓(𝑎), there exists 𝜂> 0 such that ∀𝑦∈𝑀′, 𝑑′(𝑦, 𝑓(𝑎)) < 𝜂⟹𝑑″(𝑔(𝑦), 𝑔(𝑓(𝑎))) < 𝜀 Now, since 𝑓is continuous at 𝑎, for this 𝜂there exists 𝛿such that for all 𝑥∈𝑀, 𝑑(𝑥, 𝑎) < 𝛿⟹𝑑′(𝑓(𝑥) −𝑓(𝑎)) < 𝜂 Then 𝑑(𝑥, 𝑎) < 𝛿⟹𝑑″(𝑔(𝑓(𝑥)), 𝑔(𝑓(𝑎))) < 𝜀 as required.
Example. Constant functions are continuous. For instance, let 𝑏∈𝑀and let 𝑓(𝑥) = 𝑏.
Then this is continuous since 𝑑′(𝑓(𝑥) −𝑓(𝑎)) = 𝑑′(𝑏, 𝑏) = 0 so any 𝛿> 0 will satisfy the condition.
Example. The identity function 𝑓∶𝑀→𝑀defined by 𝑥↦𝑥is continuous. Consider 𝑑(𝑓(𝑥) −𝑓(𝑎)) = 𝑑(𝑥−𝑎). So 𝛿= 𝜀will suffice.
156 3. Metric spaces Example. All real and complex polynomials and rational functions are continuous wherever they are defined by the propositions and examples above. In fact, using uniform conver-gence, the uniform limits of such functions are also continuous. For example, exponential and trigonometric functions are continuous.
Example. Let (𝑀, 𝑑) be a metric space. Then 𝑑∶𝑀⊕𝑝𝑀→ℝ, which can be viewed as a function between metric spaces 𝑀⊕𝑝𝑀and ℝ. Then, given 𝑣= (𝑥, 𝑥′), 𝑤= (𝑦, 𝑦′) ∈ 𝑀⊕𝑝𝑀, |𝑑(𝑣) −𝑑(𝑤)| = |𝑑(𝑥, 𝑥′) −𝑑(𝑦, 𝑦′)| ≤𝑑(𝑥, 𝑦) + 𝑑(𝑥′, 𝑦′) = 𝑑1(𝑣, 𝑤) ≤2𝑑𝑝(𝑣, 𝑤) Hence 𝛿= 𝜀 2 will suffice.
3.6. Isometric, Lipschitz, and uniformly continuous functions Definition. Let 𝑓∶𝑀→𝑀′ be a function between metric spaces. Then, 𝑓is (i) isometric, if ∀𝑥, 𝑦∈𝑀, 𝑑′(𝑓(𝑥), 𝑓(𝑦)) = 𝑑(𝑥, 𝑦) (ii) Lipschitz, or 𝑐-Lipschitz, if ∃𝑐∈ℝ+, ∀𝑥, 𝑦∈𝑀, 𝑑′(𝑓(𝑥), 𝑓(𝑦)) ≤𝑐⋅𝑑(𝑥, 𝑦) (iii) uniformly continuous, if ∀𝜀> 0, ∃𝛿> 0, ∀𝑥, 𝑦∈𝑀, 𝑑(𝑥, 𝑦) < 𝛿⟹𝑑′(𝑓(𝑥), 𝑓(𝑦)) < 𝜀 Remark. Any isometric function is 1-Lipschitz. Any Lipschitz function is uniformly con-tinuous. Any uniformly continuous function is continuous.
Remark. If a function is isometric, it is injective, since 𝑓(𝑥) = 𝑓(𝑦) ⟹𝑥= 𝑦. For example, if 𝑁⊂𝑀, the inclusion map 𝑖∶𝑁→𝑀defined by 𝑖(𝑥) = 𝑥is isometric but not surjective.
An isometric and surjective map is called an isometry. If there exists an isometry 𝑀→𝑀′, we say that 𝑀and 𝑀′ are isometric metric spaces, or 𝑀′ is an isometric copy of 𝑀.
Example. Suppose (𝑀, 𝑑), (𝑀′, 𝑑′) be metric spaces. Let 𝑦∈𝑀′. We define 𝑓∶𝑀→ 𝑀⊕𝑝𝑀′ by 𝑥↦(𝑥, 𝑦). Then 𝑑𝑝(𝑓(𝑥), 𝑓(𝑧)) = 𝑑𝑝((𝑥, 𝑦), (𝑧, 𝑦)) = 𝑑(𝑥, 𝑧). So the function 𝑓 is isometric. Therefore, 𝑀× {𝑦} is an isometric copy of 𝑀in 𝑀⊕𝑝𝑀′.
Example. Consider the projections 𝑞∶𝑀⊕𝑝𝑀′ →𝑀defined by 𝑞(𝑥, 𝑦) = 𝑥and 𝑞′ ∶𝑀⊕𝑝 𝑀′ →𝑀′ defined by 𝑞′(𝑥, 𝑦) = 𝑦. These projections are both 1-Lipschitz. Indeed, 𝑑(𝑞(𝑥, 𝑦), 𝑞(𝑥′, 𝑦′)) = 𝑑(𝑥, 𝑥′) ≤𝑑𝑝((𝑥, 𝑦), (𝑥′, 𝑦′)) In particular, polynomials in any finite number of variables are continuous since we can multiply continuous functions together.
157 IV. Analysis and Topology 3.7. Generalised triangle inequality Suppose 𝑢, 𝑥, 𝑦, 𝑧∈𝑀. Then, |𝑑(𝑢, 𝑥) −𝑑(𝑦, 𝑧)| ≤𝑑(𝑢, 𝑦) + 𝑑(𝑥, 𝑧). First, 𝑑(𝑢, 𝑥) ≤𝑑(𝑢, 𝑦) + 𝑑(𝑦, 𝑥) ≤𝑑(𝑢, 𝑦) + 𝑑(𝑦, 𝑧) + 𝑑(𝑧, 𝑥) Rearranging, 𝑑(𝑢, 𝑥) −𝑑(𝑦, 𝑧) ≤𝑑(𝑢, 𝑦) + 𝑑(𝑥, 𝑧) To achieve the negative, satisfying both conditions in the absolute value term, 𝑑(𝑦, 𝑧) ≤𝑑(𝑦, 𝑢) + 𝑑(𝑢, 𝑥) + 𝑑(𝑥, 𝑧) which gives 𝑑(𝑦, 𝑧) −𝑑(𝑢, 𝑥) ≤𝑑(𝑢, 𝑦) + 𝑑(𝑥, 𝑧) as required.
158 4. Topology of metric spaces 4. Topology of metric spaces 4.1. Open balls Definition. Let 𝑀be a metric space, 𝑥∈𝑀, 𝑟> 0. Then the open ball in 𝑀of centre 𝑥and radius 𝑟is the set 𝒟𝑟(𝑥) = {𝑦∈𝑀∶𝑑(𝑦, 𝑥) < 𝑟} The open ball notation is a convenient syntax for denoting closeness in some metric space.
Note that, for example, 𝑥𝑛→𝑛in 𝑀is equivalent to saying ∀𝜀> 0, ∃𝑁∈ℕ, ∀𝑛≥𝑁, 𝑥𝑛∈𝒟𝜀(𝑥) We can also say that 𝑓∶𝑀→𝑀′ is continuous at 𝑥if ∀𝜀> 0, ∃𝛿> 0, 𝑓(𝒟𝛿(𝑥)) ⊂𝒟𝜀(𝑓(𝑥)) Definition. The closed ball of centre 𝑥and radius 𝑟≥0 is the set ℬ𝑟(𝑥) = {𝑦∈𝑀∶𝑑(𝑦, 𝑥) ≤𝑟} Example. In ℝ, 𝒟𝑟(𝑥) = (𝑥−𝑟, 𝑥+𝑟). Further, ℬ𝑟(𝑥) = [𝑥−𝑟, 𝑥+𝑟]. In the plane (ℝ2, 𝑑𝑝), ℬ1(0) = {𝑥∈ℝ2 ∶‖𝑥‖𝑝≤1} Note. 𝒟𝑟(𝑥) ⊂ℬ𝑟(𝑥) ⊂𝒟𝑠(𝑥) for all 𝑟< 𝑠.
Example. Let 𝑀be a discrete metric space. Then for 𝑥∈𝑀, 𝒟1(𝑥) = {𝑥}; ℬ1(𝑥) = 𝑀 4.2. Neighbourhoods and openness Definition. Let 𝑀be a metric space, and 𝑈⊂𝑀. Then for 𝑥∈𝑀, we say that 𝑈is a neighbourhood of 𝑥(in 𝑀) if ∃𝑟> 0, 𝒟𝑟(𝑥) ⊂𝑈⟺∃𝑟> 0, ℬ𝑟(𝑥) ⊂𝑈 Definition. We say 𝑈⊂𝑀is open in 𝑀, or that 𝑈is an open subset of 𝑀, if ∀𝑥∈𝑈, ∃𝑟> 0, 𝒟𝑟(𝑥) ⊂𝑈 So 𝑈is a neighbourhood of all points in 𝑈.
Example. 𝒟𝑟(𝑥), ℬ𝑟(𝑥) are neighbourhoods of 𝑥.
159 IV. Analysis and Topology Example. Let 𝐻= {𝑧∈ℂ∶Im 𝑧≥0}. Let 𝑤∈𝐻and 𝛿= Im 𝑤. If 𝛿> 0, then 𝒟𝛿(𝑤) ⊂𝐻.
If 𝛿= 0, then for any 𝑟, 𝒟𝛿(𝑤) ⊄𝐻. So 𝐻is not open.
Lemma. Open balls are open.
Proof. Let 𝒟𝑟(𝑥) be an open ball in a metric space 𝑀. We need to show that ∀𝑦∈𝒟𝑟(𝑥), ∃𝛿> 0, 𝒟𝛿(𝑦) ⊂𝒟𝑟(𝑥) So let 𝑦∈𝒟𝑟(𝑥) and set 𝛿= 𝑟−𝑑(𝑥, 𝑦). Note that 𝑑(𝑥, 𝑦) > 0, and by the triangle inequality, 𝑑(𝑧, 𝑥) ≤𝑑(𝑧, 𝑦) + 𝑑(𝑦, 𝑥) < 𝛿+ (𝑟−𝛿) = 𝑟 as required.
Corollary. Let 𝑀be a metric space, 𝑈⊂𝑀, 𝑥∈𝑀. Then 𝑈is a neighbourhood of 𝑥if and only if there exists an open subset 𝑉of 𝑀such that 𝑥∈𝑉⊂𝑈.
Proof. In the forward direction, there exists 𝑟> 0 such that 𝒟𝑟(𝑥) ⊂𝑈, so let 𝑉= 𝒟𝑟(𝑥).
Conversely, if 𝑉is open we can construct 𝑟> 0 such that 𝒟𝑟(𝑥) ⊂𝑉⊂𝑈. So 𝑈is a neighbourhood of 𝑥.
4.3. Continuity and convergence using topology Proposition. In a metric space 𝑀, the following are equivalent.
(i) 𝑥𝑛→𝑥; (ii) for all neighbourhoods 𝑈of 𝑥in 𝑀, ∃𝑁∈ℕ, ∀𝑛≥𝑁, 𝑥𝑛∈𝑈; (iii) for all open neighbourhoods 𝑈of 𝑥in 𝑀, ∃𝑁∈ℕ, ∀𝑛≥𝑁, 𝑥𝑛∈𝑈.
Proof. First, (i) implies (ii). Let 𝑈be a neighbourhood of 𝑥. Then by definition ∃𝜀> 0, 𝒟𝜀(𝑥) ⊂ 𝑈. Since 𝑥𝑛→𝑥, ∃𝑁∈ℕ, ∀𝑛≥𝑁, 𝑥𝑛∈𝒟𝜀(𝑥) hence ∀𝑛≥𝑁, 𝑥𝑛∈𝑈.
Now we show (ii) implies (iii). This is clear since any open set 𝑈with 𝑥∈𝑈is a neighbour-hood of 𝑥.
Finally, (iii) implies (i). Fix 𝜀> 0. By the above lemma, 𝑈= 𝒟𝜀(𝑥) is open, and 𝑥∈𝑈.
Then by (iii), ∃𝑁∈ℕ, ∀𝑛≥𝑛, 𝑥𝑛∈𝑈 hence 𝑑(𝑥𝑛, 𝑥) < 𝜀.
Proposition. Let 𝑓∶𝑀→𝑀′ be a function between metric spaces.
160 4. Topology of metric spaces (a) The following are equivalent for all 𝑥∈𝑀.
(i) 𝑓is continuous at 𝑥; (ii) for all neighbourhoods 𝑉of 𝑓(𝑥) in 𝑀′, there exists a neighbourhood 𝑈of 𝑥in 𝑀such that 𝑓(𝑈) ⊂𝑉; (iii) for all neighbourhoods 𝑉of 𝑓(𝑥) in 𝑀′, 𝑓−1(𝑉) is a neighbourhood of 𝑥in 𝑀.
(b) The following are equivalent.
(i) 𝑓is continuous; (ii) 𝑓−1(𝑉) is open in 𝑀for all open subsets 𝑉of 𝑀′.
Proof. First, we show (a)(i) implies (a)(ii). Let 𝑉be a neighbourhood of 𝑓(𝑥) in 𝑀′. By definition, ∃𝜀> 0 such that 𝒟𝜀(𝑓(𝑥)) ⊂𝑉. Since 𝑓is continuous at 𝑥, there exists 𝛿> 0 such that 𝑓(𝒟𝛿(𝑥)) ⊂𝒟𝜀(𝑓(𝑥)). Then, 𝑈= 𝒟𝛿(𝑥) is a neighbourhood of 𝑥in M, and 𝑓(𝑈) ⊂𝑉.
Now, (a)(ii) implies (a)(iii). Let 𝑉be a neighbourhood of 𝑓(𝑥) in 𝑀′. By (ii), there exists a neighbourhood of 𝑥in 𝑀such that 𝑓(𝑈) ⊂𝑉. Then 𝑈⊂𝑓−1(𝑉) and since 𝑈is a neigh-bourhood of 𝑥in 𝑀, ther exists 𝑟> 0 such that 𝒟𝑟(𝑥) ⊂𝑈⊂𝑓−1(𝑉) Thus, 𝑓−1(𝑉) is a neighbourhood of 𝑥in 𝑀.
Finally, (a)(iii) implies (a)(i). Given 𝜀> 0, 𝑉= 𝒟𝜀(𝑓(𝑥)) is a neighbourhood of 𝑓(𝑥) in 𝑉.
By (iii), 𝑓−1(𝑉) is a neighbourhood of 𝑥in 𝑀. So ∃𝛿> 0 such that 𝒟𝛿(𝑥) ⊂𝑓−1(𝑉). Thus, 𝑓(𝒟𝛿(𝑥)) ⊂𝑉= 𝒟𝜀(𝑓(𝑥)).
Now, (b)(i) implies (b)(ii). Let 𝑉be open in 𝑀′. So pick 𝑥∈𝑓−1(𝑉). Then, 𝑓(𝑥) ∈𝑉. Since 𝑉is open, ∃𝜀> 0, 𝒟𝜀(𝑓(𝑥)) ⊂𝑉. Since 𝑓is continuous at 𝑥, ∃𝛿> 0, 𝑓(𝒟𝛿(𝑥)) ⊂𝒟𝜀(𝑓(𝑥)).
Then, 𝒟𝛿(𝑥) ⊂𝑓−1(𝒟𝜀(𝑓(𝑥))) ⊂𝑓−1(𝑉).
Finally, (b)(ii) implies (b)(i). Consider 𝑥∈𝑀. We must show 𝑓is continuous at 𝑥. Let 𝜀> 0.
Consider the ball 𝑉= 𝒟𝜀(𝑓(𝑥)). This is open in 𝑀′ by the above lemma. By (ii), 𝑓−1(𝑉) is open in 𝑀. Further, 𝑥∈𝑓−1(𝑉). So by definition, ∃𝛿> 0, 𝒟𝛿(𝑥) ⊂𝑉, which is exactly continuity as required.
Definition. The topology of a metric space 𝑀is the family of all open subsets of 𝑀.
Proposition. The topology of a metric space satisfies (i) ∅and 𝑀are open; (ii) if 𝑈𝑖are open in 𝑀for 𝑖∈𝐼(𝐼may be countable or uncountable), then ⋃𝑖∈𝐼𝑈𝑖is open in 𝑀; (iii) if 𝑈, 𝑉are open then 𝑈∩𝑉is open.
Proof. (ii): Let 𝑥∈⋃𝑖∈𝐼𝑈𝑖, then ∃𝑖𝑎∈𝐼, 𝑥∈𝑈𝑖𝑎. Then since 𝑈𝑖𝑎is open, ∃𝛿> 0, 𝒟𝑟(𝑥) ⊂ 𝑈𝑖𝑎⊂⋃𝑖∈𝐼𝑈𝑖 161 IV. Analysis and Topology (iii) Given 𝑥∈𝑈∩𝑉, since 𝑈is open then ∃𝑟> 0, 𝒟𝑟(𝑥) ⊂𝑈and ∃𝑠> 0, 𝒟𝑠(𝑥) ⊂𝑉. Then let 𝑡= min(𝑟, 𝑠), and 𝒟𝑡(𝑥) = 𝒟𝑟(𝑥) ∩𝒟𝑠(𝑥) ⊂𝑈∩𝑉.
4.4. Properties of topology of metric space Definition. A subspace 𝐴of a metric space 𝑀is closed in 𝑀if for every sequence (𝑥𝑛) ∈𝐴 that is convergent in 𝑀, lim 𝑛→∞𝑥𝑛∈𝐴 Lemma. Closed balls are closed.
Proof. Consider ℬ𝑟(𝑥) in 𝑀. Consider further (𝑥𝑛) ∈ℬ𝑟(𝑥) such that 𝑥𝑛→𝑧in 𝑀.
𝑑(𝑧, 𝑥) ≤𝑑(𝑧, 𝑥𝑛) + 𝑑(𝑥𝑛, 𝑥) ≤𝑑(𝑧, 𝑥𝑛) + 𝑟→𝑟 Hence 𝑑(𝑧, 𝑥) ≤𝑟, so 𝑧∈ℬ𝑟(𝑥).
Example. [0, 1] = ℬ1/2(1/2) is closed in ℝ. This is not open, for instance consider 𝐷𝑟(0) ⊄ [0, 1].
Example. (0, 1) = 𝒟1/2(1/2) is open in ℝ. This is not closed, for instance the sequence 1 𝑛+1 tends to zero in ℝ.
Example. ℝand ∅are open and closed in ℝ.
Example. (0, 1] in ℝis neither open nor closed. Consider 𝒟𝑟(1) ⊄(0, 1] and 1 𝑛→0 ∉(0, 1].
Lemma. Let 𝐴⊂𝑀. Then 𝐴is closed in 𝑀if and only if 𝑀∖𝐴is open in 𝑀.
Proof. Let 𝐴be closed. Suppose 𝑀∖𝐴is not open. Then ∃𝑥∈𝑀∖𝐴, ∀𝑟> 0, 𝒟𝑟(𝑥) ⊄𝑀∖𝐴, so 𝒟𝑟(𝑥) ∩𝐴≠∅. In particular, for every 𝑛we can choose a point in 𝒟1/𝑛(𝑥) ∩𝐴. Then, 𝑑(𝑥𝑛, 𝑥) < 1 𝑛→0 and 𝑥𝑛∈𝐴which contradicts the fact that 𝐴is closed.
Conversely, let us assume 𝑀∖𝐴is open, but suppose 𝐴is not closed. Then there exists a sequence (𝑥𝑛) ∈𝐴such that 𝑥𝑛→𝑥in 𝑀but 𝑥∉𝐴. Since 𝑥∈𝑀∖𝐴and 𝑀∖𝐴is open, there exists 𝜀> 0, 𝒟𝜀(𝑥) ⊂𝑀∖𝐴. Since 𝑥𝑛→𝑥, we must have ∃𝑁∈ℕ, ∀𝑛≥𝑁, 𝑥𝑛∈𝒟𝜀(𝑥) and hence 𝑥𝑛∈𝑀∖𝐴, which is a contradiction.
Example. Let 𝑀be a discrete metric space. Let 𝐴⊂𝑀. Then for all 𝑥∈𝐴, 𝒟1(𝑥) = {𝑥} ⊂𝐴.
Hence 𝐴is open. So in a discrete metric space, all subsets are open. Hence every subset is closed.
162 4. Topology of metric spaces 4.5. Homeomorphisms Definition. A map 𝑓∶𝑀→𝑀′ between metric spaces is called a homeomorphism if 𝑓is a bijection and 𝑓, 𝑓−1 are continuous. Equivalently, 𝑓is a bijection, and for all open sets 𝑉 in 𝑀′, 𝑓−1(𝑉) is open in 𝑀, and for all open sets 𝑈in 𝑀, 𝑓(𝑈) is open in 𝑀′. If there exists a homeomorphism between 𝑀, 𝑀′, we say that 𝑀, 𝑀′ are homeomorphic.
Example. Consider (0, ∞) and (0, 1). Consider the map 𝑥↦ 1 𝑥+1 with inverse 𝑥↦ 1 𝑥−1.
These are continuous, so the metric spaces are homeomorphic.
Remark. Every isometry is a homeomorphism, since it is bijective by definition. It is not true that every homeomorphism is an isometry.
Consider the identity on ℝwith the discrete metric to ℝwith the Euclidean metric. This is a continuous bijection whose inverse is not continuous. So it is not true that a continuous bijection always has a continuous inverse.
4.6. Equivalence of metrics Definition. Let 𝑑, 𝑑′ be metrics on a set 𝑀. We say that 𝑑, 𝑑′ are equivalent, written 𝑑∼𝑑′, if they define the same topology. In particular, 𝑈⊂𝑀is open in (𝑀, 𝑑) if and only if 𝑈is open in (𝑀, 𝑑′). So 𝑑∼𝑑′ if and only if id∶(𝑀, 𝑑) →(𝑀, 𝑑′) is a homeomorphism.
Remark. If 𝑑∼𝑑′, then (𝑀, 𝑑) and (𝑀, 𝑑′) have the same convergent sequences and con-tinuous maps.
Definition. Let 𝑑, 𝑑′ be metrics on 𝑀. Then we say 𝑑, 𝑑′ are uniformly equivalent, written 𝑑∼𝑢𝑑′ if id∶(𝑀, 𝑑) →(𝑀, 𝑑′); id∶(𝑀, 𝑑′) →(𝑀, 𝑑) are uniformly continuous. We say 𝑑, 𝑑′ are Lipschitz equivalent, written 𝑑∼Lip 𝑑′, if the identity maps above are Lipschitz. Equivalently, 𝑑∼Lip 𝑑′ if ∃𝑎> 0, 𝑏> 0, 𝑎𝑑(𝑥, 𝑦) ≤ 𝑑′(𝑥, 𝑦) ≤𝑏𝑑(𝑥, 𝑦). Note, 𝑑∼Lip 𝑑′ ⟹𝑑∼𝑢𝑑′ ⟹𝑑∼𝑑′.
Example. Given a metric space (𝑀, 𝑑), we define 𝑑′(𝑥, 𝑦) = min(1, 𝑑(𝑥, 𝑦)). This defines a metric on 𝑀, and 𝑑′ ∼𝑢𝑑.
Example. On 𝑀× 𝑀′, 𝑑1, 𝑑2, 𝑑∞are pairwise Lipschitz equivalent.
Example. Consider 𝐶[0, 1]. The 𝐿1 metric and the uniform metric are not equivalent. Con-sider 𝑓 𝑛(𝑥) = 𝑥𝑛. This is convergent to zero in the 𝐿1 metric but is not convergent in the uniform metric.
Example. The discrete metric and Euclidean metric on ℝare not equivalent. This is be-cause in the discrete metric all sets are open, but in the Euclidean metric there are some non-open sets.
163 IV. Analysis and Topology 5. Completeness 5.1. Cauchy sequences In ℝ, ℂ, every Cauchy sequence is convergent. We wish to generalise this notion to an arbit-rary metric space. Recall that a sequence (𝑥𝑛) in ℝor ℂis bounded if there exists 𝑐∈ℝ+ such that ∀𝑛∈ℕ, |𝑥𝑛| ≤𝑐.
Definition. A sequence (𝑥𝑛) in a metric space 𝑀is said to be Cauchy if ∀𝜀> 0, ∃𝑁∈ℕ, ∀𝑚, 𝑛≥𝑁, 𝑑(𝑥𝑚, 𝑥𝑛) < 𝜀 The sequence is bounded if ∃𝑧∈𝑀, ∃𝑟> 0, ∀𝑛∈ℕ, 𝑥𝑛∈ℬ𝑟(𝑧) This is equivalent to ∀𝑧∈𝑀, ∃𝑟> 0, ∀𝑛∈ℕ, 𝑥𝑛∈ℬ𝑟(𝑧) by considering the triangle inequality around the given 𝑧point. In particular, if the metric arises from a norm, (𝑥𝑛) is bounded if and only if ‖𝑥𝑛‖ is bounded.
Lemma. If a sequence is convergent, it is Cauchy. If a sequence is Cauchy, it is bounded.
Proof. Let (𝑥𝑛) be a sequence in 𝑀. First, we assume that (𝑥𝑛) is convergent in 𝑀, so let 𝑥 be the limit. Given 𝜀> 0, there exists 𝑁∈ℕsuch that ∀𝑛≥𝑁, 𝑑(𝑥𝑛, 𝑥) < 𝜀. Then, for all 𝑚, 𝑛≥𝑁, we have 𝑑(𝑥𝑚, 𝑥𝑛) ≤𝑑(𝑥𝑚, 𝑥) + 𝑑(𝑥, 𝑥𝑛) ≤2𝜀as required. So (𝑥𝑛) is Cauchy.
Now conversely, we assume (𝑥𝑛) is Cauchy. There exists 𝑛∈ℕsuch that ∀𝑚, 𝑛≥𝑁, we have 𝑑(𝑥𝑚, 𝑥𝑛) < 1. In particular, 𝑑(𝑥𝑛, 𝑥𝑁) < 1 for 𝑛≥𝑁. In other words, 𝑥𝑛∈ℬ1(𝑥𝑁).
Now, let 𝑟= max {𝑑(𝑥1, 𝑥𝑁), … , 𝑑(𝑥𝑁−1, 𝑥𝑁), 1}. This 𝑟is a bound for all elements of the sequence; for all 𝑛∈ℕ, 𝑥𝑛∈ℬ𝑟(𝑥𝑁).
Remark. Boundedness does not imply the sequence is Cauchy. For instance, consider the sequence 0, 1, 0, 1, … in ℝ. If a sequence is Cauchy, it is not necessarily convergent in an arbitrary metric space (not ℝ, ℂ). For instance, consider 𝑥𝑛= 1 𝑛in (0, ∞). This is certainly not convergent, since the limit cannot be zero.
5.2. Definition of completeness Definition. A metric space 𝑀is called complete if every Cauchy sequence in 𝑀converges in 𝑀.
Example. ℝ, ℂare complete.
164 5. Completeness 5.3. Completeness of product spaces Proposition. Product spaces of complete spaces are complete. More precisely, if 𝑀, 𝑀′ are complete, then so is 𝑀⊕𝑝𝑀′.
Proof. Let (𝑎𝑛) be a Cauchy sequence in the product space 𝑀⊕𝑝𝑀′. We will write 𝑎𝑛= (𝑥𝑛, 𝑥′ 𝑛) for all 𝑛. Then, since (𝑎𝑛) is Cauchy, ∀𝜀> 0, ∃𝑁∈ℕ, ∀𝑚, 𝑛∈𝑁, 𝑑𝑝(𝑎𝑚, 𝑎𝑛) < 𝜀 Then, for all 𝑚, 𝑛≥𝑁, 𝑑(𝑥𝑚, 𝑥𝑛) ≤max {𝑑(𝑥𝑚, 𝑥𝑛), 𝑑(𝑥′ 𝑚, 𝑥′ 𝑛)} ≤𝑑𝑝(𝑎𝑚, 𝑎𝑛) < 𝜀 Hence (𝑥𝑛) is Cauchy in 𝑀, and similarly (𝑥′ 𝑚) is Cauchy in 𝑀′. Since 𝑀, 𝑀′ are complete, (𝑥𝑛), (𝑥′ 𝑛) are convergent in 𝑀, 𝑀′ to 𝑥, 𝑥′. Now, let 𝑎= (𝑥, 𝑥′). Then, 𝑑𝑝(𝑎𝑛, 𝑎) ≤𝑑1(𝑎𝑛, 𝑎) = 𝑑(𝑥𝑛, 𝑥) + 𝑑(𝑥′ 𝑛, 𝑥′) →0 So the product space is complete.
Remark. (𝑎𝑛) is Cauchy in 𝑀⊕𝑝𝑀′ if and only if (𝑥𝑛) is Cauchy in 𝑀and (𝑥′ 𝑛) is Cauchy in 𝑀′.
Corollary. ℝ𝑛, ℂ𝑛are complete in the ℓ𝑝metric. In particular, 𝑛-dimensional real or com-plex Euclidean space is complete.
5.4. Completeness of subspaces and function spaces Theorem. Let 𝑆be any set. Then, ℓ∞(𝑆), the set of bounded scalar functions on 𝑆, is com-plete in the uniform metric 𝐷.
Proof. Let (𝑓 𝑛) be a Cauchy sequence in ℓ∞(𝑆). Then, ∀𝜀> 0, ∃𝑁∈ℕ, ∀𝑚, 𝑛≥𝑁, 𝐷(𝑥𝑚, 𝑥𝑛) = sup 𝑥∈𝑆 |𝑓 𝑚(𝑥) −𝑓 𝑛(𝑥)| < 𝜀 In other words, ∀𝑚, 𝑛≥𝑁, ∀𝑥∈𝑆, |𝑓 𝑚(𝑥) −𝑓 𝑛(𝑥)| < 𝜀. So (𝑓 𝑛) is uniformly Cauchy as defined previously. As shown previously, (𝑓 𝑛) is uniformly convergent. Hence, there is a scalar function 𝑓on 𝑆such that 𝑓 𝑛→𝑓uniformly on 𝑆. We have also shown previously that the uniform limit 𝑓of bounded functions (𝑓 𝑛) is bounded. In other words, 𝑓∈ℓ∞(𝑆).
Now it remains to show that 𝑓 𝑛→𝑓in the uniform metric.
∀𝜀> 0, ∃𝑁∈ℕ, ∀𝑛≥𝑀, ∀𝑥∈𝑆, |𝑓 𝑛(𝑥) −𝑓(𝑥)| < 𝜀 Hence, ∀𝑛≥𝑁, sup 𝑥∈𝑆 |𝑓 𝑛(𝑥) −𝑓(𝑥)| = 𝐷(𝑓 𝑛, 𝑓) ≤𝜀 which is convergence in the metric as required.
165 IV. Analysis and Topology Proposition. Let 𝑁be a subspace of a metric space 𝑀. Then, (i) If 𝑁is complete, 𝑁is closed in 𝑀.
(ii) If 𝑀is complete and 𝑁is closed in 𝑀, then 𝑁is complete.
In other words, in a complete metric space, a subspace is complete if and only if it is closed.
Proof. To prove (i), we let (𝑥𝑛) be a sequence in 𝑁and assume that 𝑥𝑛→𝑥in 𝑀. We want to show that 𝑥∈𝑁. We know (𝑥𝑛) is convergent in 𝑀, so it is Cauchy in 𝑀. So (𝑥𝑛) is Cauchy in 𝑁. Since 𝑁is complete, 𝑥𝑛→𝑦in 𝑁. So 𝑥𝑛→𝑦in 𝑀. By uniqueness of limits, 𝑥= 𝑦as required.
Now we want to prove (ii) is complete. Let (𝑥𝑛) be a Cauchy sequence in 𝑁. Then (𝑥𝑛) is Cauchy in 𝑀. Since 𝑀is complete, 𝑥𝑛→𝑥in 𝑀for some 𝑥∈𝑀. Since 𝑁is closed in 𝑀, 𝑥∈𝑁. So 𝑥𝑛→𝑥in 𝑁.
Theorem. Let (𝑀, 𝑑) be a metric space, and define 𝐶𝑏(𝑀) to be the set of functions 𝑓in ℓ∞(𝑀) such that 𝑓is continuous. This is a subspace of ℓ∞(𝑀) in the uniform metric 𝐷.
𝐶𝑏(𝑀) is complete in the uniform metric.
Proof. By the above proposition, it is sufficient to show that 𝐶𝑏(𝑀) is closed in ℓ∞(𝑀). Let (𝑓 𝑛) be a sequence in 𝐶𝑏(𝑀) and we assume that 𝑓 𝑛→𝑓in ℓ∞(𝑀). We want to show that 𝑓 𝑛∈𝐶𝑏(𝑀). It is now sufficient to show that 𝑓is continuous, or equivalently, continuous at every point in 𝑀. Let 𝑎∈𝑀, and let 𝜀> 0. Since 𝑓 𝑛→𝑓in ℓ∞(𝑀), we can fix 𝑛∈ℕsuch that 𝐹(𝑓 𝑛, 𝑓) < 𝜀. Since 𝑓 𝑛is continuous (at 𝑎), ∃𝛿> 0, ∀𝑥∈𝑀, 𝑑(𝑥, 𝑎) < 𝛿⟹|𝑓 𝑛(𝑥) −𝑓 𝑛(𝑎)| < 𝜀 Hence, ∀𝑥∈𝑀, if 𝑑(𝑥, 𝑎) < 𝛿we have |𝑓(𝑥) −𝑓(𝑎)| ≤|𝑓(𝑥) −𝑓 𝑛(𝑥)| + |𝑓 𝑛(𝑥) −𝑓 𝑛(𝑎)| + |𝑓 𝑛(𝑎) −𝑓(𝑎)| ≤2𝐷(𝑓 𝑛, 𝑓) + |𝑓 𝑛(𝑥) −𝑓 𝑛(𝑎)| < 3𝜀 Corollary. Consider 𝐶[𝑎, 𝑏], the space of continuous functions on [𝑎, 𝑏]. This space is com-plete in the uniform metric, since 𝐶[𝑎, 𝑏] = 𝐶𝑏[𝑎, 𝑏].
Definition. Let 𝑆be a set, and (𝑁, 𝑒) be a metric space. Then we generalise ℓ∞(𝑆) to the following definition.
ℓ∞(𝑆, 𝑁) = {𝑓∶𝑆→𝑁∶𝑓is bounded} where 𝑓is bounded if there exists 𝑦∈𝑁, 𝑟> 0 such that ∀𝑥∈𝑆, 𝑓(𝑥) ∈ℬ𝑟(𝑦). If 𝑔∶𝑆→𝑁 is a bounded function, ∀𝑥∈𝑆, 𝑔(𝑥) ∈ℬ𝑠(𝑧), then ∀𝑥∈𝑆, 𝑒(𝑓(𝑥), 𝑔(𝑥)) ≤𝑒(𝑓(𝑥), 𝑦) + 𝑒(𝑦, 𝑧) + 𝑒(𝑧, 𝑔(𝑥)) ≤𝑟+ 𝑒(𝑦, 𝑧) + 𝑠 166 5. Completeness This is a uniform bound for all 𝑥, so we may take the supremum. So sup𝑥∈𝑆𝑒(𝑓(𝑥), 𝑔(𝑥)) exists and we denote this by 𝒟(𝑓, 𝑔) = sup 𝑥∈𝑆 𝑒(𝑓(𝑥), 𝑔(𝑥)) This can be shown to be a metric, called the uniform metric on ℓ∞(𝑆, 𝑁). Now, let 𝑆= 𝑀, where (𝑀, 𝑑) is a metric space. We define 𝐶𝑏(𝑀, 𝑁) = {𝑓∶𝑀→𝑁∶𝑓continuous and bounded} Note that 𝐶𝑏(𝑀, 𝑁) is a subspace of ℓ∞(𝑀, 𝑁) with the uniform metric.
Theorem. Let 𝑆be a set, let (𝑀, 𝑑) be a metric space, and let (𝑁, 𝑒) be a complete metric space. Then (i) ℓ∞(𝑆, 𝑁) is complete in the uniform metric 𝐷; (ii) 𝐶𝑏(𝑀, 𝑁) is complete in the uniform metric 𝐷.
Proof. We first prove part (i). Let (𝑓 𝑛) be a Cauchy sequence in ℓ∞(𝑆, 𝑁). We first show that (𝑓 𝑛) is pointwise Cauchy. Let 𝑥∈𝑆.
∀𝜀> 0, ∃𝐾∈ℕ, ∀𝑖, 𝑗≥𝐾, 𝐷(𝑓𝑖, 𝑓𝑗) < 𝜀 In particular, 𝑒(𝑓𝑖(𝑥), 𝑓𝑗(𝑥)) ≤𝐷(𝑓𝑖, 𝑓𝑗) < 𝜀for 𝑖, 𝑗≥𝐾. So the sequence (𝑓𝑘(𝑥))𝑘is Cauchy in 𝑁. Since 𝑁is complete, (𝑓𝑘(𝑥))𝑘converges. This holds for all 𝑥∈𝑆, hence we can define 𝑓∶𝑆→𝑁by 𝑓(𝑥) = lim𝑘→∞𝑓𝑘(𝑥).
Now, we must show that 𝑓is bounded, such that 𝑓∈ℓ∞(𝑆, 𝑁). Since 𝑓𝑘is Cauchy in the uniform metric 𝐷, there exists 𝐾∈ℕsuch that ∀𝑖, 𝑗≥𝐾, 𝐷(𝑓𝑖, 𝑓𝑗) < 1. In par-ticular, for all 𝑖≥𝐾, 𝐷(𝑓𝑖, 𝑓 𝐾) < 1. Since 𝑓 𝐾is bounded, there exists 𝑦∈𝑁, 𝑟> 0 such that ∀𝑥∈𝑆, 𝑓 𝐾(𝑥) ∈ℬ𝑟(𝑦). Then, by the triangle inequality, for a fixed 𝑥∈𝑆, ∀𝑖≥𝐾, 𝑒(𝑓𝑖(𝑥), 𝑓 𝐾(𝑥)) ≤𝐷(𝑓𝑖(𝑥), 𝑓 𝐾(𝑥)) < 1. Let 𝑖→∞, then 𝑒(𝑓𝑖(𝑥), 𝑓 𝐾(𝑥)) ≤1. Hence 𝑒(𝑓(𝑥), 𝑦) ≤𝑒(𝑓(𝑥), 𝑓 𝐾(𝑥)) + 𝑒(𝑓 𝐾(𝑥), 𝑦) ≤1 + 𝑟. But since this is true for all 𝑥, 1 + 𝑟is a uniform bound; ∀𝑥∈𝑆, 𝑓(𝑥) ∈ℬ𝑟+1(𝑦).
Now we will show that 𝑓𝑘→𝑓uniformly in 𝐷. Again, we use ∀𝜀> 0, ∃𝐾∈ℕ, ∀𝑖, 𝑗≥𝐾, 𝐷(𝑓𝑖, 𝑓𝑗) < 𝜀 So choose 𝑖≥𝐾, and 𝑥∈𝑆. Then for all 𝑗≥𝐾, 𝑒(𝑓𝑖(𝑥), 𝑓𝑗(𝑥)) ≤𝐷(𝑓𝑖, 𝑓𝑗) < 𝜀. As 𝑗→∞, 𝑒(𝑓(𝑥), 𝑓𝑖(𝑥)) ≤𝜀, because metrics are continuous. But since 𝑥was arbitrary, we have a uniform distance 𝐷(𝑓, 𝑓𝑖) < 𝜀. This holds for all 𝑖≥𝐾, so we have uniform convergence.
Now we prove part (ii). By part (i) and an above proposition, it is enough to show that 𝐶𝑏(𝑀, 𝑁) is closed in ℓ∞(𝑀, 𝑁). Let (𝑓𝑘) be a sequence in 𝐶𝑏(𝑀, 𝑁) and 𝑓𝑘→𝑓in ℓ∞(𝑀, 𝑁).
We require 𝑓∈𝐶𝑏(𝑀, 𝑁), so it is enough to show that 𝑓is continuous. This is exactly the proof that the uniform limit of continuous functions is continuous. Let 𝑎∈𝑀, 𝜀> 0. Then, 167 IV. Analysis and Topology since 𝑓𝑘→𝑓in ℓ∞(𝑀, 𝑁), we can fix 𝑘∈ℕsuch that 𝐷(𝑓𝑘, 𝑓) < 𝜀. Since 𝑓𝑘is continuous, ∃𝛿> 0, ∀𝑥∈𝑀, 𝑑(𝑥, 𝑎) < 𝛿⟹𝑒(𝑓𝑘(𝑥), 𝑓𝑘(𝑎)) < 𝜀.
∀𝑥∈𝑀, 𝑓(𝑥, 𝑎) < 𝛿⟹𝑒(𝑓(𝑥), 𝑓(𝑎)) ≤𝑒(𝑓(𝑥), 𝑓𝑘(𝑥)) + 𝑒(𝑓𝑘(𝑥), 𝑓𝑘(𝑎)) + 𝑒(𝑓𝑘(𝑎), 𝑓(𝑎)) ≤3𝜀 168 6. Contraction mapping theorem 6. Contraction mapping theorem 6.1. Contraction mappings Definition. A function 𝑓∶𝑀→𝑀′ is called a contraction mapping if ∃𝜆, 0 ≤𝜆< 1 such that ∀𝑥, 𝑦∈𝑀, 𝑑′(𝑓(𝑥), 𝑓(𝑦)) ≤𝜆𝑑(𝑥, 𝑦) so 𝑓is 𝜆-Lipschitz.
6.2. Contraction mapping theorem This theorem is also called Banach’s fixed point theorem.
Theorem. Let 𝑀be a non-empty complete metric space. Let 𝑓∶𝑀→𝑀be a contraction mapping. Then 𝑓has a unique fixed point: ∃!𝑧∈𝑀, 𝑓(𝑧) = 𝑧 Proof. Let 𝜆such that 0 ≤𝜆< 1 and ∀𝑥, 𝑦∈𝑀, 𝑑′(𝑓(𝑥), 𝑓(𝑦)) ≤𝜆𝑑(𝑥, 𝑦). First we show uniqueness. Suppose there were two fixed points 𝑓(𝑧) = 𝑧, 𝑓(𝑤) = 𝑤. Then 𝑑(𝑧, 𝑤) = 𝑑(𝑓(𝑧), 𝑓(𝑤)) ≤𝜆𝑑(𝑧, 𝑤) < 𝑑(𝑧, 𝑤). Hence 𝑑(𝑧, 𝑤) = 0 so 𝑧= 𝑤.
Now we show existence. Fix a starting point 𝑥0 ∈𝑀. Let 𝑥𝑛= 𝑓(𝑥𝑛−1) for all 𝑛∈ℕ, so 𝑥𝑛= 𝑓𝑛(𝑥0). First, observe that for all 𝑛∈ℕ, 𝑑(𝑥𝑛, 𝑥𝑛+1) = 𝑑(𝑓(𝑥𝑛−1), 𝑓(𝑥𝑛)) ≤𝜆𝑑(𝑥𝑛−1, 𝑥𝑛) ≤⋯≤𝜆𝑛𝑑(𝑥0, 𝑥1) For 𝑚≥𝑛, we have 𝑑(𝑥𝑛, 𝑥𝑚) ≤ 𝑚−1 ∑ 𝑘=𝑛 𝑑(𝑥𝑘, 𝑥𝑘+1) ≤ 𝑚−1 ∑ 𝑘=𝑛 𝜆𝑘𝑑(𝑥0, 𝑥1) ≤ 𝜆𝑛 1 −𝜆𝑑(𝑥0, 𝑥1) Since 𝜆𝑛 1−𝜆𝑑(𝑥0, 𝑥1) →0, ∀𝜀> 0, ∃𝑁∈ℕ, ∀𝑛≥𝑁, 𝜆𝑛 1 −𝜆𝑑(𝑥0, 𝑥1) < 𝜀 Hence, ∀𝑚≥𝑛≥𝑁, 𝑑(𝑥𝑛, 𝑥𝑚) < 𝜀. So the sequence (𝑥𝑛) is Cauchy. Since 𝑀is complete, (𝑥𝑛) is convergent to some point 𝑧∈𝑀. 𝑓is continuous since it is a contraction, so 𝑓(𝑥𝑛) → 𝑧so 𝑓(𝑧) = 𝑧. So the fixed point exists.
Remark. Letting 𝑚→∞in the inequality for 𝑑(𝑥𝑛, 𝑥𝑚), 𝑑(𝑥𝑛, 𝑧) ≤ 𝜆𝑛 1−𝜆𝑑(𝑥0, 𝑥1). So 𝑥𝑛→𝑧 exponentially fast. Consider 𝑓∶ℝ∖{0} →ℝ∖{0}, and 𝑥↦ 𝑥 2 . This is a contraction, but there is no fixed point. This is because ℝ∖{0} is not complete. Consider instead 𝑓∶ℝ→ℝ, 𝑥↦ 𝑥+1. This has no fixed point, since 𝑓is an isometry (𝜆= 1) and not a contraction. Consider further 𝑓∶[1, ∞) →[1, ∞) mapping 𝑥↦𝑥+ 1 𝑥. Certainly |𝑓(𝑥) −𝑓(𝑦)| < |𝑥−𝑦|. [1, ∞) is closed in ℝso it is complete. However this is not a contraction; even though |𝑓(𝑥) −𝑓(𝑦)| < |𝑥−𝑦|, there is no upper bound 𝜆. There are no fixed points.
169 IV. Analysis and Topology 6.3. Application of contraction mapping theorem Let 𝑦0 ∈ℝ. Then the initial value problem 𝑓′(𝑡) = 𝑓(𝑡2) and 𝑓(0) = 𝑦0 has a unique solution on [0, 1 2]. In other words, there exists a unique differentiable function 𝑓∶[0, 1 2] →[0, 1 2] such that 𝑓(0) = 𝑦0 and 𝑓′(𝑡) = 𝑓(𝑡2) for all 𝑡in the domain.
First, observe that if 𝑓is a solution then certainly it is continuous, so 𝑓∈𝐶[0, 1 2]. Further, by the fundamental theorem of calculus, it satisfies 𝑓(𝑡) = 𝑦0 + ∫ 𝑡 0 𝑓(𝑠2) d𝑠 Note that 𝑓′(𝑠) = 𝑓(𝑠2) is continuous. Conversely, if 𝑓∈𝐶[0, 1 2] and 𝑓(𝑡) = 𝑦0 + ∫ 𝑡 0 𝑓(𝑠2) d𝑠 then 𝑓is a solution to the initial value problem.
Let 𝑀= 𝐶[0, 1 2] with the uniform metric. This is non-empty and complete. Then we define the map 𝑇∶𝑀→𝑀by (𝑇𝑔)(𝑡) = 𝑦0 + ∫ 𝑡 0 𝑔(𝑠2) d𝑠 Note that 𝑇𝑔is well-defined since 𝑔(𝑠2) is continuous. Moreover, by the fundamental the-orem of calculus, 𝑇𝑔is differentiable and (𝑇𝑔)′(𝑡) = 𝑔(𝑡2). Thus, 𝑓is a solution to the initial value problem if and only if 𝑓∈𝑀and 𝑇𝑓= 𝑓.
Now, if 𝑇is a contraction, we can use the contraction mapping theorem to assert that there is exactly one fixed point. For 𝑔, ℎ∈𝑀, 𝑡∈[0, 1 2], consider |(𝑇𝑔)(𝑡) −(𝑇ℎ)(𝑡)| = | | | | ∫ 𝑡 0 [𝑔(𝑠2) −ℎ(𝑠2)] d𝑠 | | | | ≤𝑡sup 𝑠∈[0, 1 2 ] | |𝑔(𝑠2) −ℎ(𝑠2)| | ≤1 2𝐷(𝑔, ℎ) Taking the supremum over 𝑡gives 𝐷(𝑇𝑔, 𝑇ℎ) ≤ 1 2𝐷(𝑔, ℎ), and so there is exactly one fixed point.
Remark. The above shows that for any 𝛿∈(0, 1) there is a unique solution to the initial value problem on [0, 𝛿], called 𝑓𝛿, since 𝛿< 1 is required for the map to be a contraction.
For 0 < 𝛿< 𝜇< 1, 𝑓 𝜇| |[0,𝛿] = 𝑓𝛿by uniqueness. So we can combine the solutions together to yield a unique solution on [0, 1).
6.4. Lindelöf–Picard theorem Theorem. Let 𝑛∈ℕ, 𝑦0 ∈ℝ𝑛, and 𝑎, 𝑏, 𝑅∈ℝ, such that 𝑎< 𝑏and 𝑅> 0. Let 𝜙∶[𝑎, 𝑏] × ℬ𝑅(𝑦0) →ℝ𝑛be a continuous function. Given that there exists 𝐾> 0 such that ∀𝑡∈ [𝑎, 𝑏], ∀𝑥, 𝑦∈ℬ𝑅(𝑦0), such that ‖𝜙(𝑡, 𝑥) −𝜙(𝑡, 𝑦)‖ ≤𝐾‖𝑥−𝑦‖ 170 6. Contraction mapping theorem Then, ∃𝜀> 0 such that ∀𝑡, 𝑡0 ∈[𝑎, 𝑏], the initial value problem 𝑓′(𝑡) = 𝜙(𝑡, 𝑓(𝑡)); 𝑓(𝑡0) = 𝑦0 has a unique solution on [𝑐, 𝑑] = [𝑡0 −𝜀, 𝑡0 + 𝜀] ∩[𝑎, 𝑏].
Remark. If 𝑓is a solution of the initial value problem, implicitly this includes the assumption that 𝑓(𝑡) ∈𝐵𝑟(𝑦0) for all 𝑡∈[𝑐, 𝑑]. Note that if 𝑓∶[𝑐, 𝑑] →ℝ𝑛, we let 𝑓𝑘∶[𝑐, 𝑑] →ℝ be the 𝑘th component of 𝑓, and 𝑓𝑘= 𝑞𝑘∘𝑓where 𝑞𝑘is the 𝑘th coordinate projection.
Then, 𝑓(𝑡) = (𝑓 1(𝑡), … , 𝑓 𝑛(𝑡)) and we define 𝑓to be differentiable if and only if all of the components are differentiable, with 𝑓′(𝑡) = (𝑓′ 1(𝑡), … , 𝑓′ 𝑛(𝑡)). Note further, if 𝑓is continuous, then so are 𝑓𝑘, hence 𝑓𝑘are integrable. So we define ∫ 𝑑 𝑐 𝑓(𝑡) d𝑡= 𝑣= (∫ 𝑑 𝑐 𝑓 1(𝑡) d𝑡, … , ∫ 𝑑 𝑐 𝑓 𝑛(𝑡) d𝑡) Note that we can use the Cauchy–Schwarz inequality to give ‖𝑣‖2 = 𝑛 ∑ 𝑘=1 𝑣2 𝑘 = 𝑛 ∑ 𝑘=1 𝑣𝑘∫ 𝑑 𝑐 𝑓𝑘(𝑡) d𝑡 = ∫ 𝑑 𝑐 𝑛 ∑ 𝑘=1 𝑣𝑘𝑓𝑘(𝑡) d𝑡 = ∫ 𝑑 𝑐 𝑣⋅𝑓(𝑡) d𝑡 ≤∫ 𝑑 𝑐 ‖𝑣‖ ⋅‖𝑓(𝑡)‖ d𝑡 = ‖𝑣‖ ∫ 𝑑 𝑐 ‖𝑓(𝑡)‖ d𝑡 Hence, ‖ ‖ ‖ ‖ ∫ 𝑑 𝑐 𝑓(𝑡) d𝑡 ‖ ‖ ‖ ‖ ≤∫ 𝑑 𝑐 ‖𝑓(𝑡)‖ d𝑡≤(𝑑−𝑐) sup 𝑡∈[𝑐,𝑑] ‖𝑓(𝑡)‖ Proof. Recall that closed balls are closed, hence ℬ𝑅(𝑦0) is a closed subset of ℝ𝑛. So 𝜙is a continuous function on the closed and bounded set [𝑎, 𝑏] × ℬ𝑅(𝑦0). It follows that 𝜙is bounded. Now, let 𝑐= sup {‖𝜙(𝑡, 𝑥)‖∶𝑡∈[𝑎, 𝑏], 𝑥∈ℬ𝑅(𝑦0)}. Let 𝜀= min( 𝑅 𝑐, 1 2𝐾). Let 𝑡0 ∈[𝑎, 𝑏] and let [𝑐, 𝑑] = [𝑡0 −𝜀, 𝑡0 + 𝜀] ∩[𝑎, 𝑏]. We need to show that there exists a unique differentiable function 𝑓∶[𝑐, 𝑑] →ℝ𝑛such that 𝑓(𝑡0) = 𝑦0 and 𝑓′(𝑡) = 𝜙(𝑡, 𝑓(𝑡)) for all 𝑡∈[𝑐, 𝑑]. Since ℬ𝑅(𝑦0) is closed in ℝ𝑛, and since ℝ𝑛is complete, ℬ𝑅(𝑦0) is complete. Then, 𝑀= 𝐶([𝑐, 𝑑], ℬ𝑅(𝑦0)) is complete in the uniform metric 𝐷. This is certainly non-empty; 171 IV. Analysis and Topology consider the constant function yielding 𝑦0. 𝑓is a solution to the initial value problem if 𝑓∈𝑀and 𝑓′(𝑡) = 𝑦0 + ∫ 𝑡 𝑡0 𝜙(𝑠, 𝑓(𝑠)) d𝑠, from the fundamental theorem of calculus applied coordinatewise. We define 𝑇∶𝑀→𝑀mapping 𝑔↦𝑇𝑔where 𝑇𝑔is given by (𝑇𝑔)(𝑡) = 𝑦0 + ∫ 𝑡 𝑡0 𝜙(𝑠, 𝑔(𝑠)) d𝑠 We must show 𝑇is well defined. First, note that the integral is well defined; 𝑠↦𝜙(𝑠, 𝑔(𝑠)) is continuous so integrable. By the fundamental theorem of calculus, 𝑇𝑔is differentiable and the derivative is (𝑇𝑔)′(𝑡) = 𝜙(𝑡, 𝑔(𝑡)). In particular, 𝑇𝑔∶[𝑐, 𝑑] →ℝ𝑛is continuous. Finally, for 𝑡∈[𝑐, 𝑑], ‖(𝑇𝑔)(𝑡) −𝑦0‖ = ‖ ‖ ‖ ‖ ∫ 𝑡 𝑡0 𝜙(𝑠, 𝑔(𝑠)) d𝑠 ‖ ‖ ‖ ‖ ≤|𝑡−𝑡0| sup 𝑠∈[𝑐,𝑑] ‖𝜙(𝑠, 𝑔(𝑠))‖ ≤𝜀𝑐≤𝑅 So 𝑇𝑔∈𝑀. Recall that 𝑓is a solution of the initial value problem if and only if 𝑓∈𝑀and 𝑇𝑓= 𝑓. Now we must show that 𝑇has a unique fixed point, so we will show that 𝑇is a contraction. Let 𝑡∈[𝑐, 𝑑] and 𝑔, ℎ∈𝑀.
‖(𝑇𝑔)(𝑡) −(𝑇ℎ)(𝑡)‖ = ‖ ‖ ‖ ‖ ∫ 𝑡 𝑡0 [𝜙(𝑠, 𝑔(𝑠)) −𝜙(𝑠, ℎ(𝑠))] d𝑠 ‖ ‖ ‖ ‖ Note that ‖𝜙(𝑠, 𝑔(𝑠)) −𝜙(𝑠, ℎ(𝑠))‖ ≤𝐾‖𝑔(𝑠) −ℎ(𝑠)‖ ≤𝐾⋅𝐷(𝑔, ℎ).
‖(𝑇𝑔)(𝑡) −(𝑇ℎ)(𝑡)‖ = |𝑡−𝑡0| ⋅𝐾⋅𝐾(𝑔, ℎ) ≤𝜀𝐾𝐷(𝑔, ℎ) Taking the supremum over 𝑡∈(𝑐, 𝑑), 𝐷(𝑇𝑔, 𝑇ℎ) ≤𝜀𝐾𝐷(𝑔, ℎ) ≤1 2𝐷(𝑔, ℎ) So 𝑇is a contraction. By the contraction mapping theorem, 𝑇has a unique fixed point in 𝑀.
Remark. For any 𝛿∈(0, 1), taking 𝜀= min( 𝑅 𝑐, 𝛿 𝐾) works. But by the uniqueness of the solution, the choice does not matter for constructing the solution. So we can construct the solution for 𝜀= min( 𝑅 𝑐, 1 𝐾), on (𝑡0 −𝜀, 𝑡0 +𝜀)∩[𝑎, 𝑏]. In general, there is no solution on [𝑎, 𝑏].
Finally, note that the above theorem can handle any 𝑛th order ODE for any 𝑛∈ℕ.
172 7. Topology 7. Topology 7.1. Definitions Definition. Let 𝑋be a set. A topology on 𝑋is a family 𝜏of subsets of 𝑋(so 𝜏⊂𝒫(𝑋)) such that (i) ∅, 𝑋∈𝜏; (ii) if 𝑈𝑖∈𝜏for all 𝑖∈𝐼where 𝐼is some index set, then ⋃𝑖∈𝐼𝑈𝑖∈𝜏; and (iii) if 𝑈, 𝑉∈𝜏then 𝑈∩𝑉∈𝜏.
A topological space is a pair (𝑋, 𝜏) where 𝑋is a set and 𝜏is a topology on 𝑋. Members of 𝜏are called open sets in the topology. So we say that 𝑈⊂𝑋is open in 𝑋, or 𝑈is 𝜏-open, if 𝑈∈𝜏.
Remark. If 𝑈𝑖∈𝜏for 𝑖= 1, … , 𝑛, then ⋂ 𝑛 𝑖=1 𝑈𝑖∈𝜏.
Example. Let (𝑀, 𝑑) be a metric space. Recall that 𝑈⊂𝑀is open in the metric sense if ∀𝑥∈𝑈, ∃𝑟> 0, ℬ𝑟(𝑥) ⊂𝑈. We may say that 𝑈is 𝑑-open. We have already proven that the family of 𝑑-open sets is a topology on 𝑀. This is a metric topology.
Definition. Let (𝑋, 𝜏) be a topological space. Then we say that 𝑋is metrisable (or sometimes we say 𝜏is metrisable) if there exists a metric 𝑑on 𝑋such that 𝜏is the metric topology on 𝑋 induced by 𝑑. In other words, 𝑈⊂𝑋is 𝜏-open if and only if 𝑈is 𝑑-open. If 𝑑′ ∼𝑑, then 𝑑′ also induces the same topology 𝜏on 𝑋.
Example. The indiscrete topology on a set 𝑋is a topology 𝜏= {∅, 𝑋}. If |𝑋| ≥2, then this is not metrisable. Let 𝑑be a metric on 𝑋. Then let 𝑥≠𝑦∈𝑋, let 𝑟= 𝑑(𝑥, 𝑦), and finally let 𝑈= 𝒟𝑟(𝑥). We know that 𝑈is 𝑑-open. But since 𝑥∈𝑈, 𝑦∉𝑈, 𝑈∉𝜏.
Definition. If 𝜏1, 𝜏2 are topologies on 𝑋, we say that 𝜏1 is coarser than 𝜏2, or that 𝜏2 is finer than 𝜏1, if 𝜏1 ⊂𝜏2. For example, the indiscrete topology on 𝑋is the coarsest topology on 𝑋.
Example. The discrete topology on a set 𝑋is 𝜏= 𝒫(𝑋). This is the finest topology on 𝑋.
This is metrisable by the discrete metric.
Definition. A topological space 𝑋is Hausdorff if ∀𝑥≠𝑦in 𝑋, there exist open sets 𝑈, 𝑉in 𝑋such that 𝑥∈𝑈, 𝑦∈𝑉, 𝑈∩𝑉= ∅. Informally, 𝑥, 𝑦are ‘separated by open sets’.
Proposition. Metric spaces are Hausdorff.
Proof. Let 𝑥≠𝑦be points in a metric space (𝑀, 𝑑). Let 𝑟> 0 such that 2𝑟< 𝑑(𝑥, 𝑦). Then let 𝑈= 𝒟𝑟(𝑥), let 𝑉= 𝒟𝑟(𝑦). Certainly 𝑈, 𝑉are open since they are open balls, and they have no intersection by the triangle inequality, so the metric space is Hausdorff as required.
Example. The cofinite topology on a set 𝑋is 𝜏= {∅} ∪{𝑈∈𝑋∶𝑈is cofinite in 𝑋} 173 IV. Analysis and Topology where 𝑈is cofinite in 𝑋if 𝑋∖𝑈is finite. When 𝑋is finite, this topology 𝜏is simply 𝒫(𝑋).
When 𝑋is infinite, 𝜏is not metrisable. Let 𝑥≠𝑦in 𝑋, and let 𝑥∈𝑈, 𝑦∈𝑉where 𝑈, 𝑉are open in 𝑋. Then 𝑈and 𝑉are cofinite, and hence 𝑈∩𝑉≠∅. So this topology on an infinite set is not Hausdorff and hence not metrisable.
7.2. Closed subsets Definition. A subset 𝐴of a topological space (𝑋, 𝜏) is said to be closed in 𝑋if 𝑋∖𝐴is open in 𝑋.
Remark. In a metric space, this agrees with the earlier definition of a closed subset, as proven before.
Proposition. The collection of closed sets in a topological space 𝑋satisfy (i) ∅, 𝑋are closed; (ii) If 𝐴𝑖are closed in 𝑋for 𝑖in some non-empty index set 𝐼, then ⋂𝑖∈𝐼𝐴𝑖is closed; (iii) If 𝐴1, 𝐴2 are closed in 𝑋then 𝐴1 ∪𝐴2 is closed.
Example. In a discrete topological space, every set is closed.
Example. In the cofinite topology, a subset is closed if and only if it is finite or the full set.
7.3. Neighbourhoods Definition. Let 𝑋be a topological space, and let 𝑈⊂𝑋and 𝑥∈𝑋. We say that 𝑈is a neighbourhood of 𝑥in 𝑋if there exists an open set 𝑉in 𝑋such that 𝑋∈𝑉⊂𝑈.
Remark. In a metric space, we defined this in terms of open balls not open sets. However, we have already proven that the definitions agree.
Proposition. Let 𝑈be a subset of a topological space 𝑋. Then 𝑈is open if and only if 𝑈is a neighbourhood of 𝑥for every 𝑥∈𝑈.
Proof. If 𝑈is open, and 𝑥∈𝑈, then by letting 𝑉= 𝑈, 𝑉is open and 𝑥∈𝑉⊂𝑈. Conversely, if 𝑥∈𝑈, there exists 𝑉 𝑥in 𝑋such that 𝑥∈𝑉 𝑥⊂𝑈. Then, 𝑈= ⋃𝑥∈𝑈𝑥= ⋃𝑥∈𝑈𝑉 𝑥is open, since each 𝑉 𝑥is open.
7.4. Convergence Definition. Let (𝑥𝑛) be a sequence in a topological space 𝑋. Let 𝑥∈𝑋. We say that (𝑥𝑛) converges to 𝑥if for all neighbourhoods 𝑈of 𝑥in 𝑋, there exists 𝑁∈ℕsuch that ∀𝑛≥ 𝑁, 𝑥𝑛∈𝑈. Equivalently, for all open sets 𝑈which contain 𝑥, there exists 𝑁∈ℕsuch that ∀𝑛≥𝑁, 𝑥𝑛∈𝑈.
Remark. Again, the definition in a metric space agrees with this definition.
174 7. Topology Example. Eventually constant sequences converge. If ∃𝑧∈𝑋, ∃𝑁∈ℕ, ∀𝑛≥𝑁, 𝑥𝑛= 𝑧, then 𝑥𝑛→𝑧.
Example. In an indiscrete topological space, every sequence converges to every point.
Example. In the cofinite topology on a set 𝑋, let 𝑥𝑛→𝑋. Suppose that 𝑥𝑛→𝑥in 𝑋. Then if 𝑦≠𝑥, 𝑋∖{𝑦} is a neighbourhood of 𝑥. Then 𝑁𝑦= {𝑛∈𝑁∶𝑥𝑛= 𝑦} is finite.
Conversely, suppose (𝑥𝑛) is a sequence such that for some 𝑥∈𝑋and for all 𝑦≠𝑥, 𝑁𝑦is finite. Then 𝑥𝑛→𝑥.
In particular, if 𝑁𝑦is finite for all 𝑦∈𝑋, the sequence converges to every point.
Proposition. If 𝑥𝑛→𝑥and 𝑥𝑛→𝑦in a Hausdorff space, then 𝑥= 𝑦.
Proof. Suppose 𝑥≠𝑦, then we can choose open sets 𝑈, 𝑉such that 𝑥∈𝑈, 𝑦∈𝑉, 𝑈∩𝑉= ∅.
Since 𝑥𝑛→𝑥, there exists 𝑁1 ∈ℕsuch that ∀𝑛≥𝑁1, 𝑥𝑛∈𝑈. Similarly there exists an analogous 𝑁2. Hence ∀𝑛≥max(𝑁1, 𝑁2), 𝑥𝑛∈𝑈, 𝑥𝑛∈𝑉which is a contradiction since 𝑈∩𝑉= ∅.
Remark. If 𝑥𝑛→𝑥in a Hausdorff space, we write 𝑥= lim𝑛→∞𝑥𝑛since the limit is unique.
Remark. In a metric space, for a subset 𝐴, we say that 𝐴is closed if and only if 𝑥𝑛→𝑥in 𝐴 implies 𝑥∈𝐴. In a general topological space, any closed set is closed under limits, but not every subset that is closed under limits is closed.
7.5. Interiors and closures Definition. Let 𝑋be a topological space, and 𝐴⊂𝑋. We define the interior of 𝐴in 𝑋, denoted 𝐴∘or int(𝐴), by 𝐴∘= ⋃{𝑈⊂𝑋∶𝑈is open in 𝑋, 𝑈⊂𝐴} Similarly we define the closure of 𝐴in 𝑋, denoted 𝐴or cl(𝐴), by 𝐴= ⋂{𝐹⊂𝑋∶𝐹is closed in 𝑋, 𝐹⊃𝐴} Remark. Note that 𝐴∘is open in 𝑋, and 𝐴∘⊂𝐴. In particular, if 𝑈is open in 𝑋and 𝑈⊂𝐴, then 𝑈⊂𝐴∘. Hence, 𝐴∘is the largest open subset of 𝐴.
Similarly, 𝐴is closed in 𝑋, and 𝐴⊃𝐴. The intersection is not empty since 𝑋is closed and 𝑋⊃𝐴, so it is well-defined. We have that 𝐴is the smallest closed superset of 𝐴.
Proposition. Let 𝑋be a topological space and let 𝐴⊂𝑋. Then the interior is exactly those 𝑥∈𝑋for which 𝐴is a neighbourhood of 𝑥. Similarly, the closure is those 𝑥∈𝑋such that for all neighbourhoods 𝑈of 𝑥, 𝑈∩𝐴≠∅.
175 IV. Analysis and Topology Proof. If 𝐴is a neighbourhood of 𝑋, then by definition there exists an open set 𝑈such that 𝑥∈𝑈⊂𝐴, which is true if and only if 𝑥∈𝐴∘.
For the other part, suppose 𝑥∉𝐴. Then there exists a closed set 𝐹⊃𝐴such that 𝑥∉𝐹. Let 𝑈= 𝑋∖𝐹. Then 𝑈is open and 𝑥∈𝑈. So 𝑈is a neighbourhood of 𝑥, and 𝑈∩𝐴= ∅.
Conversely, suppose there exists a neighbourhood 𝑈of 𝑥such that 𝑈∩𝐴= ∅. Then there exists an open set 𝑉such that 𝑥∈𝑉⊂𝑈. Since 𝑉⊂𝑈, 𝑉∩𝐴= ∅. Let 𝐹= 𝑋∖𝑉. Then 𝐹 is closed, and 𝐴⊂𝐹. Hence 𝐴⊂𝐹. So 𝑥∉𝐴.
Example. In ℝ, let 𝐴= [0, 1) ∪{2}. Then 𝐴∘= (0, 1), and 𝐴= [0, 1] ∪{2}. Further, ℚ∘= ∅ and ℚ= ℝ. Finally, ℤ∘= ∅and ℤ= ℤ.
Remark. In a metric space, for a subset 𝐴we have that 𝑥∈𝐴if and only if there exists a sequence (𝑥𝑛) in 𝐴such that 𝑥𝑛→𝑥. In a general topological space, the existence of a sequence implies 𝑥∈𝐴but the converse is not true.
7.6. Dense subsets Definition. A subset 𝐴of a topological space 𝑋is said to be dense in 𝑋if 𝐴= 𝑋. 𝑋is separable if there exists a countable subset 𝐴⊂𝑋such that 𝐴is dense in 𝑋.
Example. ℝis separable as ℚis dense in ℝ. ℝ𝑛is separable in the same way as ℚ𝑛is dense in ℝ𝑛.
Example. An uncountable discrete topological space is not separable, since the closure of any set is itself.
7.7. Subspaces Definition. Let (𝑋, 𝜏) be a topological space. Let 𝑌⊂𝑋. Then the subspace topology, or relative topology on 𝑌induced by 𝜏is the topology {𝑉∩𝑌∶𝑉∈𝜏} on 𝑌. This is the intersection of 𝑌with all open sets in 𝑋. We can denote this 𝜏|𝑌. So, for 𝑈⊂𝑌, 𝑈is open in 𝑌if and only if there exists an open set 𝑉in 𝑋with 𝑈= 𝑉∩𝑌.
Example. Let 𝑋= ℝ, 𝑌= [0, 2], and 𝑈= (1, 2]. Then certainly 𝑈⊂𝑌⊂𝑋. 𝑈is open in 𝑌, since 𝑉= (1, 3) is open in 𝑋and 𝑈= 𝑉∩𝑌. However, 𝑈is not open in 𝑋, since no neighbourhood (or ball) around 2 can be constructed in 𝑋that is contained within 𝑈.
Remark. On a subset of a topological space, this is considered the standard topology. Sup-pose that (𝑋, 𝜏) is a topological space, and 𝑍⊂𝑌⊂𝑋. There are two natural topologies on 𝑍: 𝜏|𝑍and 𝜏|𝑌| |𝑍. One can easily check that these two topologies are equal.
176 7. Topology Let (𝑀, 𝑑) be a metric space, and 𝑁⊂𝑀. Again, there are two natural topologies on 𝑁: 𝜏(𝑑)|𝑁and 𝜏(𝑑|𝑁), where 𝜏(𝑒) is the metric topology induced by the metric 𝑒. These two constructions coincide; indeed, for any 𝑥∈𝑁, 𝑟> 0, {𝑦∈𝑁∶𝑑(𝑦, 𝑥) < 𝑟} = {𝑦∈𝑀∶𝑑(𝑦, 𝑥) < 𝑟} ∩𝑁 Proposition. Let 𝑋be a topological space, and let 𝐴⊂𝑌⊂𝑋. 𝐴is closed in 𝑌if and only if there exists a closed subset 𝐵⊂𝑋such that 𝐴= 𝐵∩𝑌. Further, cl𝑌(𝐴) = cl𝑋(𝐴) ∩𝑌 This is not true for the interior of a subset in general. For instance, consider 𝑋= ℝ, 𝐴= 𝑌= {0}. In this case, int𝑌(𝐴) = 𝐴, int𝑋(𝐴) = ∅.
Proof. The first part is true by taking complements: 𝑌∖𝐴is open in 𝑌. By definition, 𝑌∖𝐴= 𝑉∩𝑌for some open 𝑉in 𝑋. So 𝐵= 𝑋∖𝑉is closed in 𝑋and 𝐴= 𝐵∩𝑌. If 𝐴= 𝐵∩𝑌, 𝐵is closed in 𝑋, then 𝑋∖𝐵is open in 𝑋, and hence 𝑌∖𝐴= (𝑋∖𝐵) ∩𝑌is open in 𝑌.
For the second part, we know cl𝑋(𝐴) is closed in 𝑋, so by the first part, cl𝑋(𝐴) ∩𝑌is closed in 𝑌. Then 𝐴⊂cl𝑋(𝐴) ∩𝑌. So by definition, cl𝑌(𝐴) ⊂cl𝑋(𝐴) ∩𝑌. Similarly, since cl𝑌(𝐴) is closed in 𝑌, we can write cl𝑌(𝐴) = 𝐵∩𝑌for some closed set 𝐵in 𝑋. But 𝐴⊂𝐵, and 𝐵is closed in 𝑋, so cl𝑋(𝐴) ⊂𝐵and hence cl𝑌(𝐴) = 𝐵∩𝑌⊃cl𝑋(𝐴) ∩𝑌.
Remark. If 𝑈⊂𝑌⊂𝑋, and 𝑌is open in 𝑋, then 𝑈is open in 𝑌if and only if 𝑈is open in 𝑋.
7.8. Continuity Definition. A function 𝑓∶𝑋→𝑌between topological spaces is said to be continuous if for all open sets 𝑉in 𝑌, the preimage 𝑓−1(𝑉) is open in 𝑋.
Remark. We have already proven that this agrees with the definition of continuity of func-tions between metric spaces.
Example. Constant functions are always continuous. Consider 𝑓∶𝑋→𝑌defined by 𝑓(𝑥) = 𝑦0 for a fixed 𝑦0 ∈𝑌. For any 𝑉⊂𝑌, 𝑓−1(𝑉) = ∅if 𝑦0 ∉𝑉, and 𝑓−1(𝑉) = 𝑋 if 𝑦0 ∈𝑉. So 𝑓is continuous.
Example. The identity map is always continuous. If 𝑓∶𝑋→𝑋is defined by 𝑥↦𝑥, 𝑓−1(𝑉) = 𝑉so if 𝑉is open, 𝑓−1(𝑉) is trivially open.
Example. Let 𝑌⊂𝑋. Let 𝑖∶𝑌→𝑋be the inclusion map. Then for an open set 𝑉in 𝑋, 𝑖−1(𝑉) = 𝑉∩𝑌which by definition is open in 𝑌. Hence, if 𝑔∶𝑋→𝑍is continuous, then 𝑔|𝑌= 𝑔∘𝑖∶𝑋→𝑌is continuous, as we will see below.
Proposition. Let 𝑓∶𝑋→𝑌be a function between topological spaces. Then, (i) 𝑓is continuous if and only if for all closed sets 𝐵in 𝑌, 𝑓−1(𝐵) is closed in 𝑋; 177 IV. Analysis and Topology (ii) if 𝑓is continuous and 𝑔∶𝑌→𝑍is continuous, then 𝑔∘𝑓is continuous.
Proof. To prove (i), note that for any subset 𝐷⊂𝑌, 𝑓−1(𝑌∖𝐷) = 𝑋∖𝑓−1(𝐷). We can now use the fact that 𝐴⊂𝑋is open in 𝑋if and only if 𝑋∖𝐴is closed in 𝑋, and vice versa for 𝑌.
To prove (ii), note that if 𝑊is an open subset of 𝑍, then 𝑔−1(𝑊) is open in 𝑌since 𝑔is continuous. Hence 𝑓−1𝑔−1(𝑊) is open in 𝑋since 𝑓is continuous. But then 𝑓−1𝑔−1 = (𝑔∘ 𝑓)−1, so 𝑔∘𝑓is continuous.
Remark. There exists a notion of ‘continuity at a point’ for topological spaces, but it is not as useful in this course as the global continuity definition.
7.9. Homeomorphisms and topological invariance Definition. A function 𝑓∶𝑋→𝑌between topological spaces is a homeomorphism if 𝑓 is a bijection, and both 𝑓, 𝑓−1 are continuous. If such an 𝑓exists, we say that 𝑋and 𝑌are homeomorphic. This is exactly the definition from metric spaces.
Definition. A property 𝒫of topological spaces is said to be a topological property or topo-logical invariant if, for all pairs 𝑋, 𝑌of homeomorphic spaces, 𝑋satisfies 𝒫if and only if 𝑌 satisfies 𝒫.
Example. Metrisability is a topological invariant. Being Hausdorff is a topological invariant.
Being completely metrisable (metrisable into a complete metric space) is not a topological invariant. For example, consider metrics 𝑑, 𝑑′ on ℝsuch that 𝑑∼𝑑′ but 𝑑is complete and 𝑑′ is not.
Remark. If 𝑓∶𝑋→𝑌is a homeomorphism, for an open set 𝑈in 𝑋, 𝑓(𝑈) = (𝑓−1)−1(𝑈) is open in 𝑌since 𝑓−1 ∶𝑌→𝑋is continuous.
Definition. A function 𝑓∶𝑋→𝑌between topological spaces is an open map if for all open sets 𝑈in 𝑋, 𝑓(𝑈) is open in 𝑌.
Remark. 𝑓∶𝑋→𝑌is a homeomorphism if and only if 𝑓is a continuous and open bijection.
7.10. Products Let 𝑋, 𝑌be topological spaces. We want to define the topology on 𝑋× 𝑌. If 𝑈is open in 𝑋 and 𝑉is open in 𝑌, then we would like 𝑈×𝑉to be open in 𝑋×𝑌. Certainly ∅= ∅×∅and 𝑋× 𝑌should be open. Further (𝑈× 𝑉) ∩(𝑈′ × 𝑉′) = (𝑈∩𝑈′) × (𝑉∩𝑉′), so intersections work. ⋃𝑖∈𝐼𝑈𝑖× 𝑉𝑖must be open for open sets 𝑈𝑖, 𝑉𝑖, but this is not obvious from what we have shown so far, so we must include this in our definition.
Definition. The product topology on 𝑋× 𝑌is the topology such that a subset 𝑈of 𝑋× 𝑌is open if there exists a set 𝐼and open sets 𝑈𝑖, 𝑉𝑖in 𝑋, 𝑌for all 𝑖∈𝐼such that 𝑈= ⋃ 𝑖∈𝐼 𝑈𝑖× 𝑉𝑖 178 7. Topology Remark. For 𝑊⊂𝑋× 𝑌, we know that 𝑊is open if and only if for all 𝑧∈𝑊, there exist open sets 𝑈⊂𝑋, 𝑉⊂𝑌, such that 𝑧∈𝑈×𝑉⊂𝑊. So, thinking of the product as a product of real lines, we might say that 𝑊is open if for every point 𝑧∈𝑊, we can construct a ‘box set’ (the Cartesian product of open intervals) contained in 𝑊that has 𝑧as an element. More formally, 𝑊is a neighbourhood of 𝑧if and only if there exist neighbourhoods 𝑈of 𝑥in 𝑋 and 𝑉of 𝑦in 𝑌such that 𝑈× 𝑉⊂𝑊.
7.11. Continuity in product topology Example. Let (𝑀, 𝑑), (𝑀′, 𝑑′) be metric spaces. Then, the metric 𝑑∞on 𝑀× 𝑀′ is 𝑑∞((𝑥, 𝑥′), (𝑦, 𝑦′)) = max(𝑑(𝑥, 𝑦), 𝑑′(𝑥′, 𝑦′)) This metric is chosen since all 𝑑𝑝metrics induce the same metric topology, but this is easier to work with. Also, 𝑀, 𝑀′ are topological spaces with their metric topologies, which induce the product topology on the product space 𝑀×𝑀′. These two constructions create the same topology. For a point 𝑧= (𝑥, 𝑥′) ∈𝑀× 𝑀′ and 𝑟> 0, the open ball 𝒟𝑟(𝑧) is exactly 𝒟𝑟(𝑧) = {(𝑦, 𝑦′) ∈𝑀× 𝑀′ ∶𝑑∞((𝑦, 𝑦′), (𝑥, 𝑥′)) < 𝑟} = {(𝑦, 𝑦′) ∈𝑀× 𝑀′ ∶𝑑(𝑥, 𝑦) < 𝑟, 𝑑(𝑥′, 𝑦′) < 𝑟} = 𝒟𝑟(𝑥) × 𝒟𝑟(𝑥′) Now, let 𝑊⊂𝑀× 𝑀′. Then 𝑊is open in the product topology if and only if for all 𝑧= (𝑥, 𝑥′) ∈𝑊, there exist open sets 𝑈in 𝑀and 𝑈′ in 𝑀′ such that (𝑥, 𝑥′) ∈𝑈× 𝑈′ ⊂𝑊.
Equivalently, for all 𝑧= (𝑥, 𝑥′) ∈𝑊, there exists 𝑟> 0 such that 𝒟𝑟(𝑥) × 𝒟𝑟(𝑥′) ⊂𝑊. But 𝒟𝑟(𝑥) × 𝒟𝑟(𝑥′) = 𝒟𝑟(𝑧), so 𝑊is 𝑑∞-open, as required. For instance, the product topology on ℝ× ℝis the Euclidean topology on ℝ2.
Proposition. Let 𝑋, 𝑌be topological spaces. Let 𝑋×𝑌be given the product topology. Then, the coordinate projections 𝑞𝑋∶𝑋× 𝑌→𝑋and 𝑞𝑌∶𝑋× 𝑌→𝑌satisfy (i) 𝑞𝑋, 𝑞𝑌are continuous; (ii) if 𝑍is any topological space, and 𝑔∶𝑍→𝑋× 𝑌is a function, then 𝑔is continuous if and only if 𝑞𝑋∘𝑔, 𝑞𝑌∘𝑔are continuous.
Proof. If 𝑈is open in 𝑋, then 𝑞−1 𝑋(𝑈) = 𝑈× 𝑌, which is the product of an open set in 𝑋and an open set in 𝑌, so is open in 𝑋× 𝑌. Hence 𝑞𝑋is continuous. Similarly, 𝑞𝑌is continuous.
If 𝑔is continuous then certainly 𝑞𝑋∘𝑔, 𝑞𝑌∘𝑔are continuous since the composition of con-tinuous functions are continuous. Conversely, let ℎ∶𝑍→𝑋and 𝑘∶𝑍→𝑌be continuous functions with ℎ= 𝑞𝑋∘𝑔and 𝑘= 𝑞𝑌∘𝑔. Then 𝑔(𝑥) = (ℎ(𝑥), 𝑘(𝑥)) for 𝑥∈𝑍. Now, for open sets 𝑈in 𝑋and 𝑉in 𝑌, we have 𝑧∈𝑔−1(𝑈× 𝑉) ⟺𝑔(𝑧) ∈𝑈× 𝑉⟺ℎ(𝑧) ∈𝑈, 𝑘(𝑧) ∈𝑉⟺𝑧∈ℎ−1(𝑈) ∩𝑘−1(𝑉) 179 IV. Analysis and Topology So 𝑔−1(𝑈× 𝑉) = ℎ−1(𝑈) ∩𝑘−1(𝑉) which is open in 𝑍as ℎ, 𝑘are continuous. Given an arbitrary open set 𝑊in 𝑋× 𝑌, we can write 𝑊= ⋃𝑖∈𝐼𝑈𝑖×𝑉𝑖, where 𝑈𝑖are open in 𝑋and 𝑉𝑖are open in 𝑌. Thus, 𝑔−1(𝑊) = ⋃𝑖∈𝐼𝑔−1(𝑈𝑖× 𝑉𝑖) which is open.
Remark. The product topology may be extended to a finite product 𝑋1 × ⋯× 𝑋𝑛, consisting of all unions of sets of the form 𝑈1×⋯×𝑈𝑛where 𝑈𝑗is open in 𝑋𝑗. Properties of the product topology hold in this more general case. For example, if 𝑋𝑗is metrisable with metric 𝑒𝑗for all 𝑗, then the product topology is metrisable with, for instance, the 𝑑∞metric.
7.12. Quotients Let 𝑋be a set and 𝑅an equivalence relation on 𝑋. So 𝑅⊂𝑋× 𝑋, but we will write 𝑥∼𝑦to mean (𝑥, 𝑦) ∈𝑅. For 𝑥∈𝑋, we define 𝑞(𝑥) = {𝑦∈𝑋∶𝑦∼𝑥} to be the equivalence class of 𝑥, the set of which partition 𝑋. Let 𝑋/𝑅denote the set of all equivalence classes. The map 𝑞∶𝑋→𝑋/𝑅is called the quotient map.
Definition. Let 𝑋be a topological space, and 𝑅an equivalence relation on 𝑋. The quotient topology on 𝑋/𝑅is given by 𝜏= {𝑉⊂𝑋/𝑅∶𝑞−1(𝑉) open in 𝑋} This is a topology: (i) 𝑞−1(∅) = ∅which is open, and 𝑞−1(𝑋/𝑅) = 𝑋which is open.
(ii) If 𝑉𝑖are open, then 𝑞−1(⋃𝑖∈𝐼𝑉𝑖) = ⋃𝑖∈𝐼𝑞−1(𝑉𝑖) which is a union of open sets which is open.
(iii) If 𝑈, 𝑉are open, then 𝑞−1(𝑈∩𝑉) = 𝑞−1(𝑈) ∩𝑞−1(𝑉) which is open.
Remark. The quotient map 𝑞∶𝑋→𝑋/𝑅is continuous. In particular, it is the largest pos-sible topology on 𝑋such that 𝑞is continuous.
Let 𝑥∈𝑋, 𝑡∈𝑋/𝑅. Then 𝑥∈𝑡if and only if 𝑡= 𝑞(𝑥). For 𝑉⊂𝑋/𝑅, 𝑞−1(𝑉) = {𝑥∈𝑋∶𝑞(𝑥) ∈𝑉} = {𝑥∈𝑋∶∃𝑡∈𝑉, 𝑡= 𝑞(𝑥)} = {𝑥∈𝑋∶∃𝑡∈𝑉, 𝑥∈𝑡} = ⋃ 𝑡∈𝑉 𝑡 Example. Consider ℝ, an abelian group under addition, and the subgroup ℤ. We can form the quotient group ℝ/ℤ, which is the set of equivalence classes where 𝑥∼𝑦⟺𝑥−𝑦∈ℤ.
For all 𝑥∈ℝ, there exists 𝑦∈[0, 1] such that 𝑥∼𝑦, and for all 𝑥, 𝑦∈[0, 1] we have 𝑥∼𝑦 if and only if 𝑥= 𝑦or {𝑥, 𝑦} = {0, 1}. So we can think of the quotient topology of ℝ/ℤas a circle. We can say that ℝ/ℤis homeomorphic to 𝑆1 = {(𝑥, 𝑦) ∈ℝ2 ∶‖(𝑥, 𝑦)‖ = 1}, which we will prove later.
180 7. Topology Example. Consider the subgroup ℚof ℝ. Let 𝑉⊂ℝ/ℚ, such that 𝑉≠∅and 𝑉is open.
Then 𝑞−1(𝑉) is open and not empty. Therefore, there exist 𝑎< 𝑏∈ℝsuch that (𝑎, 𝑏) ⊂ 𝑞−1(𝑉). Given 𝑥∈ℝ, we can choose a rational 𝑟in the interval (𝑎−𝑥, 𝑏−𝑥). Then 𝑟+ 𝑥∈ (𝑎, 𝑏) ⊂𝑞−1(𝑉), so 𝑞(𝑥) = 𝑞(𝑟+ 𝑥) ∈𝑉. So 𝑉= ℝ/ℚ. This is the indiscrete topology, which is not metrisable or Hausdorff. So we cannot (in general) take quotients of metric spaces.
Example. Let 𝑄= [0, 1] × [0, 1] ⊂ℝ2. We define the equivalence relation 𝑅given by (𝑥1, 𝑥2) ∼(𝑦1, 𝑦2) ⟺ ⎧ ⎪ ⎨ ⎪ ⎩ (𝑥1, 𝑥2) = (𝑦1, 𝑦2) or 𝑥1 = 𝑦1, {𝑥2, 𝑦2} = {0, 1} or 𝑥2 = 𝑦2, {𝑥1, 𝑦1} = {0, 1} or 𝑥1, 𝑥2, 𝑦1, 𝑦2 ∈{0, 1} The space 𝑄/𝑅is homeomorphic to ℝ2/ℤ2. This is a square where the top and bottom edges are identified as the same, and the left and right edges are also identified as the same. This is homeomorphic to the surface of a torus with the Euclidean topology embedded in Euclidean three-dimensional space.
Proposition. Let 𝑋be a set, and let 𝑅be an equivalence relation on 𝑋. Let 𝑞∶𝑋→𝑋/𝑅be the quotient map. Let 𝑌be a set, and 𝑓∶𝑋→𝑌be a function. Suppose that 𝑓‘respects’ 𝑅; that is, 𝑥∼𝑦⟹𝑓(𝑥) = 𝑓(𝑦). Then there exists a unique map ˜ 𝑓∶𝑋/𝑅→𝑌such that 𝑓= ˜ 𝑓∘𝑞. For 𝑧∈𝑋/𝑅, we write 𝑧= 𝑞(𝑥) for some 𝑥∈𝑋, and then define ˜ 𝑓(𝑧) = 𝑓(𝑥).
Remark. Note that Im 𝑓= Im ˜ 𝑓since 𝑞is surjective.
˜ 𝑓is injective if for all 𝑥, 𝑦∈𝑋, ˜ 𝑓(𝑞(𝑥)) = ˜ 𝑓(𝑞(𝑦)) implies 𝑞(𝑥) = 𝑞(𝑦). In other words, for all 𝑥, 𝑦∈𝑋, 𝑓(𝑥) = 𝑓(𝑦) ⟹ 𝑥∼𝑦. We say that 𝑓fully respects 𝑅if, for all 𝑥, 𝑦∈𝑋, 𝑥∼𝑦⟺𝑓(𝑥) = 𝑓(𝑦) In this case, ˜ 𝑓is injective.
7.13. Continuity of functions in quotient spaces Proposition. Let 𝑋be a topological space and let 𝑅be an equivalence relation on 𝑋. Let 𝑞∶𝑋→𝑋/𝑅be a quotient map, where 𝑋/𝑅has the quotient topology. Let 𝑌be another topological space and 𝑓∶𝑋→𝑌be a function that respects 𝑅. Let ˜ 𝑓∶𝑋/𝑅→𝑌be the unique map such that 𝑓= ˜ 𝑓∘𝑞. Then (i) if 𝑓is continuous then ˜ 𝑓is continuous; and (ii) if 𝑓is an open map (the image of an open set is open) then ˜ 𝑓is an open map.
In particular, if 𝑓is a continuous surjective map that fully respects 𝑅, then ˜ 𝑓is a continuous bijection. If in addition 𝑓is an open map, then ˜ 𝑓is a continuous bijective open map, so is a homeomorphism.
181 IV. Analysis and Topology Proof. We prove part (i). Let 𝑉be an open set in 𝑌.
𝑞−1( ˜ 𝑓−1(𝑉)) = ( ˜ 𝑓∘𝑞)−1(𝑉) = 𝑓−1(𝑉) is open So by definition, ˜ 𝑓−1(𝑉) is open in 𝑋/𝑅. Hence ˜ 𝑓is continuous. Now, we prove part (ii). Let 𝑉be an open set in 𝑋/𝑅. Let 𝑈= 𝑞−1(𝑉). Then 𝑈is open in 𝑋by definition of the quotient topology. Since 𝑞is surjective, 𝑞(𝑈) = 𝑞(𝑞−1(𝑉)) = 𝑉. Hence, ˜ 𝑓(𝑉) = ˜ 𝑓(𝑞(𝑈)) = ( ˜ 𝑓∘𝑞)(𝑈) = 𝑓(𝑈) is open since 𝑓is an open map.
Example. ℝ/ℤis homeomorphic to a circle 𝑆1 = {𝑥∈ℝ2 ∶‖𝑥‖ = 1}. We define 𝑓(𝑡) = (cos 2𝜋𝑡, sin 2𝜋𝑡) Then, 𝑠−𝑡∈ℤif and only if 𝑓(𝑠) = 𝑓(𝑡) so 𝑓fully respects the relation, and 𝑓is surjective.
𝑓is also continuous since each component is continuous. Hence, there exists ˜ 𝑓∶ℝ/ℤ→𝑆1 such that 𝑓= ˜ 𝑓∘𝑞and ˜ 𝑓is a continuous bijection. Now we must show 𝑓is an open map, and then ˜ 𝑓will be a homeomorphism. Suppose 𝑓is not an open map, so there exists an open set 𝑈in ℝsuch that 𝑓(𝑈) is not open in 𝑆1. So 𝑆1 ∖𝑓(𝑈) is not closed, so there exists a sequence (𝑧𝑛) in this complement and 𝑧∈𝑓(𝑈) such that 𝑧𝑛→𝑧. 𝑓is surjective so for all 𝑛∈𝑁we can choose 𝑥𝑛∈[0, 1] such that 𝑓(𝑥𝑛) = 𝑧𝑛. This is a bounded sequence, so by the Bolzano–Weierstrass theorem, without loss of generality we can let 𝑥𝑛→𝑥∈[0, 1]. Since 𝑓is continuous, 𝑓(𝑥𝑛) →𝑓(𝑥), so 𝑧𝑛→𝑧. But since 𝑧𝑛∉𝑓(𝑈), we have 𝑥𝑛∈ℝ∖𝑈. Since the complement is closed and 𝑥𝑛→𝑥, we have 𝑥∈ℝ∖𝑈so 𝑥∉𝑈. Since 𝑧∈𝑓(𝑈), there exists 𝑦∈𝑈such that 𝑧= 𝑓(𝑦). Hence 𝑘= 𝑦−𝑥∈ℤ. Now, 𝑓(𝑥𝑛+ 𝑘) = 𝑓(𝑥𝑛) = 𝑧𝑛→𝑧, but also 𝑥𝑛+ 𝑘→𝑥+ 𝑘= 𝑦∈𝑈. Since 𝑧𝑛∉𝑓(𝑈), we have 𝑥𝑛+ 𝑘∉𝑈. Since ℝ∖𝑈is closed and 𝑥𝑛+ 𝑘→𝑦, we have 𝑦∈ℝ∖𝑈which is a contradiction.
Proposition. Let 𝑋be a topological space, and 𝑅an equivalence relation on 𝑋. Then, (a) If 𝑋/𝑅is Hausdorff, then 𝑅is closed in 𝑋× 𝑋.
(b) If 𝑅is closed in 𝑋× 𝑋and the quotient map 𝑞∶𝑋→𝑋/𝑅is an open map, then 𝑋/𝑅 is Hausdorff.
Proof. Let 𝑊= 𝑋× 𝑋∖𝑅. For part (a), we want to show 𝑊is open, so is a neighbourhood of all of its points. Given (𝑥, 𝑦) ∈𝑊, we have 𝑥≁𝑦, so 𝑞(𝑥) ≠𝑞(𝑦). Since the quotient is Hausdorff, there exist open sets 𝑆, 𝑇in 𝑋/𝑅such that 𝑆∩𝑇= ∅and 𝑞(𝑥) ∈𝑆, 𝑞(𝑦) ∈𝑇. Let 𝑈= 𝑞−1(𝑆), 𝑉= 𝑞−1(𝑇) which are open in 𝑋, and 𝑥∈𝑈, 𝑦∈𝑉. For all (𝑎, 𝑏) ∈𝑈× 𝑉, we have 𝑞(𝑎) ∈𝑆, 𝑞(𝑏) ∈𝑇hence 𝑎≁𝑏. So (𝑥, 𝑦) ∈𝑈× 𝑉⊂𝑊. Hence 𝑅is closed.
For part (b), let 𝑧≠𝑤be elements of 𝑋/𝑅, and we want to separate these points by open sets.
Let 𝑥, 𝑦∈𝑋such that 𝑞(𝑥) = 𝑧, 𝑞(𝑦) = 𝑤. Then (𝑥, 𝑦) ∈𝑊since 𝑥≁𝑤. Since 𝑅is closed, 𝑊is open, so there exist open sets 𝑈, 𝑉in 𝑋such that (𝑥, 𝑦) ∈𝑈× 𝑉⊂𝑊. Since 𝑞is an open map, 𝑞(𝑈) and 𝑞(𝑉) are open in 𝑋/𝑅, and 𝑧= 𝑞(𝑥) ∈𝑞(𝑈), 𝑤= 𝑞(𝑦) ∈𝑞(𝑉). Now it suffices to show 𝑞(𝑈) ∩𝑞(𝑉) = ∅. For (𝑎, 𝑏) ∈𝑈× 𝑉⊂𝑊, (𝑎, 𝑏) ∉𝑅hence 𝑞(𝑎) ≠𝑞(𝑏) so 𝑞(𝑈) ∩𝑞(𝑉) = ∅.
182 8. Connectedness 8. Connectedness 8.1. Definition Recall the intermediate value theorem from IA Analysis. If 𝑓∶𝐼→ℝis continuous, where 𝐼is an interval, and 𝑥< 𝑦in 𝐼and 𝑐∈(𝑓(𝑥), 𝑓(𝑦)), then there exists 𝑧∈(𝑥, 𝑦) such that 𝑓(𝑧) = 𝑐. An interval in this context is a set 𝐼such that for all 𝑥< 𝑦< 𝑧∈ℝ, 𝑥, 𝑧∈𝐼⟹𝑦∈𝐼. So the intermediate value theorem essentially states that the continuous image of an interval is an interval.
Example. Consider [0, 1) ∪(1, 2]. Let 𝑓be a function from this space to ℝ, defined by 𝑓(𝑥) = {0 𝑥∈[0, 1) 1 𝑥∈(1, 2] This is continuous, but the image of 𝑓is not an interval.
Definition. A topological space 𝑋is disconnected if there exist open subsets 𝑈, 𝑉of 𝑋such that 𝑈∩𝑉= ∅, 𝑈∪𝑉= 𝑋and 𝑈, 𝑉≠∅. We say that 𝑈and 𝑉disconnect 𝑋. We say 𝑋is connected if 𝑋is not disconnected.
Theorem. Let 𝑋be a topological space. Then the following are equivalent.
(i) 𝑋is connected; (ii) if 𝑓∶𝑋→ℝis continuous, then 𝑓(𝑋) is an interval; (iii) if 𝑓∶𝑋→ℤis continuous, 𝑓is constant.
Proof. First we show (i) implies (ii). Suppose 𝑋is connected, and 𝑓∶𝑋→ℝis continuous, but 𝑓(𝑋) is not an interval. Then there exist 𝑎< 𝑏< 𝑐∈ℝsuch that 𝑎, 𝑐∈𝑓(𝑋) and 𝑏∉ 𝑓(𝑋). Let 𝑥, 𝑦∈𝑋such that 𝑓(𝑥) = 𝑎, 𝑓(𝑦) = 𝑐. Let 𝑈= 𝑓−1(−∞, 𝑏), 𝑉= 𝑓−1(𝑏, ∞). 𝑈, 𝑉 are open since 𝑓is continuous. 𝑈, 𝑉are non-empty since 𝑥∈𝑈, 𝑦∈𝑉. Their intersection is empty since we are taking the preimage of disjoint sets. Finally, 𝑈∪𝑉= 𝑓−1(ℝ∖𝑏) = 𝑋 since 𝑏is not in the image. So 𝑈, 𝑉disconnect 𝑋, which is a contradiction.
Now (ii) implies (iii). This is immediate since an interval containing an integer must only contain one integer.
Finally, (iii) implies (i). Suppose 𝑈, 𝑉disconnect 𝑋. Let 𝑓∶𝑋→ℤby 𝑓(𝑥) = {0 𝑥∈𝑈 1 𝑥∈𝑉 For any 𝑌⊂ℝ, 𝑓−1(𝑌) = ⎧ ⎪ ⎨ ⎪ ⎩ ∅ 0, 1 ∉𝑌 𝑈 0 ∈𝑌, 1 ∉𝑌 𝑉 0 ∉𝑌, 1 ∈𝑌 𝑋 0, 1 ∈𝑌 183 IV. Analysis and Topology which is open. But 𝑓is not constant, so this is a contradiction.
Corollary. Let 𝑋⊂ℝ. Then 𝑋is connected if and only if 𝑋is an interval.
Proof. Suppose 𝑋is connected. Then the inclusion map 𝑖∶𝑋→ℝis continuous. By the theorem above, 𝑖(𝑋) = 𝑋is an interval. Conversely, suppose 𝑋is an interval. Then, for all continuous 𝑓∶𝑋→ℝ, 𝑓(𝑋) is an interval by the intermediate value theorem. Then 𝑋is connected.
Proof. This is an alternative, direct proof that intervals are connected. Suppose 𝑈, 𝑉discon-nect 𝑋. Then let 𝑥∈𝑈, 𝑦∈𝑉such that 𝑥< 𝑦. Let 𝑧= sup 𝑈∩[𝑥, 𝑦]. This set is non-empty since it contains 𝑥and is bounded above by 𝑦. So 𝑧= [𝑥, 𝑦] ⊂𝑋. We will show 𝑧∈𝑈∩𝑉, which is a contradiction. For all 𝑛∈ℕ, we have 𝑧− 1 𝑛< 𝑛so there exists 𝑥𝑛∈𝑈∩[𝑥, 𝑦] which satisfies 𝑧− 1 𝑛< 𝑥𝑛≤𝑧. Hence 𝑥𝑛→𝑧. Also, 𝑈= 𝑋∖𝑉is closed, so 𝑧∈𝑈. In particular, 𝑧< 𝑦. Now, choose 𝑁∈ℕsuch that 𝑧+ 1 𝑁< 𝑦. Then for all 𝑛≥𝑁we have 𝑧< 𝑧+ 1 𝑛< 𝑦. Hence 𝑧+ 1 𝑛∈𝑉. However, 𝑧+ 1 𝑛→𝑧, and 𝑉is closed, so 𝑧∈𝑉, which is a contradiction.
8.2. Consequences of definition Example. Any indiscrete topological space is connected. Any cofinite topological space on an infinite set is connected. The discrete topological space on a set of size at least two is disconnected.
Lemma. Let 𝑌be a subspace of a topological space 𝑋. Then, 𝑌is disconnected if and only if there exist open subsets 𝑈, 𝑉of 𝑋such that 𝑈∩𝑉∩𝑌= ∅and 𝑈∪𝑉⊃𝑌, and 𝑈∩𝑌≠∅, 𝑉∩𝑌≠∅.
Proof. Suppose 𝑌is disconnected. Then there exist open subsets 𝑈′, 𝑉′ of 𝑌that disconnect 𝑌. Then there exist open sets 𝑈, 𝑉in 𝑋such that 𝑈′ = 𝑈∩𝑌and 𝑉′ = 𝑉∩𝑌. Then 𝑈, 𝑉 satisfy the requirements from the lemma.
Conversely, suppose 𝑈, 𝑉are as given. Then, let 𝑈′ = 𝑈∩𝑌, 𝑉′ = 𝑉∩𝑌. They are open in 𝑌by the definition of the subspace topology, and they disconnect 𝑌.
Remark. In the above lemma, we say subsets 𝑈, 𝑉of 𝑋disconnect 𝑌.
Proposition. Let 𝑌be a subspace of a topological space 𝑋. If 𝑌is connected, then so is 𝑌.
Proof. Suppose 𝑌is disconnected. Then there exist open sets 𝑈, 𝑉in 𝑋which disconnect 𝑌.
Then 𝑈∩𝑉∩𝑌⊂𝑈∩𝑉∩𝑌= ∅by definition. Hence 𝑈∩𝑉∩𝑌= ∅. Also, 𝑈∪𝑉⊃𝑌⊃𝑌.
So 𝑈, 𝑉disconnect 𝑌unless 𝑈∩𝑌= ∅or 𝑉∩𝑌= ∅. But 𝑌is connected, so without loss of generality let 𝑉∩𝑌= ∅. Then 𝑌⊂𝑋∖𝑉and 𝑋∖𝑉is closed, so 𝑌⊂𝑋∖𝑉. Hence 𝑉∩𝑌= ∅. This is a contradiction since 𝑈, 𝑉disconnect 𝑌.
184 8. Connectedness Remark. More generally, if 𝑌⊂𝑍⊂𝑌, and 𝑌is connected, then 𝑍is connected. This is since cl𝑍(𝑌) = cl𝑋(𝑌) ∩𝑍= 𝑍.
Theorem. Let 𝑓∶𝑋→𝑌be a continuous function between topological spaces. If 𝑋is connected, then so is 𝑓(𝑋).
Proof. Let 𝑈, 𝑉be open subsets of 𝑌which disconnect 𝑓(𝑋). For 𝑥∈𝑋, 𝑓(𝑥) ∈𝑓(𝑋) ⊂𝑈∪ 𝑉. Hence, 𝑓−1(𝑈)∪𝑓−1(𝑉) = 𝑋. Also, if 𝑥∈𝑓−1(𝑈)∩𝑓−1(𝑉) then 𝑓(𝑥) ∈𝑈∩𝑉∩𝑓(𝑋) = ∅.
This is a contradiction, so 𝑓−1(𝑈) ∩𝑓−1(𝑉) = ∅. Since 𝑓is continuous, 𝑓−1(𝑈), 𝑓−1(𝑉) are open in 𝑋. Since 𝑈∩𝑓(𝑋) ≠∅and 𝑉∩𝑓(𝑋) ≠∅, 𝑓−1(𝑈) ≠∅and 𝑓−1(𝑉) ≠∅So 𝑓−1(𝑈), 𝑓−1(𝑉) disconnect 𝑋.
Remark. This shows that connectedness is a topological property. If 𝑋, 𝑌are homeomorphic spaces, then 𝑋is connected if and only if 𝑌is connected. Further, note that if 𝑓∶𝑋→𝑌is continuous and 𝐴⊂𝑋and 𝐴is connected, then 𝑓(𝐴) is connected. This can be shown by restricting 𝑓to the domain 𝐴.
Corollary. Any quotient of a connected topological space is connected.
Example. Let 𝑌= {(𝑥, sin 1 𝑥)∶𝑥> 0} ⊂ℝ2 This space is connected; the function 𝑓∶(0, ∞) →ℝ2 defined by 𝑓(𝑥) = (𝑥, sin 1 𝑥) is con-tinuous. So we have that 𝑌= Im 𝑓is connected. Hence, 𝑌is connected. We claim that 𝑍≡𝑌∪{(0, 𝑦)∶𝑦∈[−1, 1]} = 𝑌 Indeed, given 𝑦∈[−1, 1], for all 𝑛∈ℕwe have that (0, 1 𝑛) is mapped to (𝑛, ∞) by 𝑥→ 1 𝑥, so by the intermediate value theorem there exists 𝑥𝑛∈(0, 1 𝑛) such that sin 1 𝑥𝑛= 𝑦. Hence, (𝑥𝑛, sin 1 𝑥𝑛 ) = (𝑥𝑛, 𝑦) →(0, 𝑦) ∈𝑌 So 𝑌⊂𝑍⊂𝑌. If we can show 𝑍is closed, 𝑍= 𝑌since 𝑌is the smallest closed superset of 𝑌. Suppose (𝑥𝑛, 𝑦𝑛) ∈𝑍for all 𝑛∈ℕ, and (𝑥𝑛, 𝑦𝑛) →(𝑥, 𝑦) in ℝ2. Since 𝑦𝑛∈[−1, 1] and 𝑦𝑛→𝑦, we have 𝑦∈[−1, 1]. If 𝑥= 0, we have (𝑥, 𝑦) ∈𝑍. If 𝑥≠0, then 𝑥𝑛→𝑥implies 𝑥𝑛≠0 for all sufficiently large 𝑛. Hence 𝑦𝑛= sin 1 𝑥𝑛for all sufficiently large 𝑛. Thus (𝑥𝑛, 𝑦𝑛) →(𝑥, sin 1 𝑥) ∈𝑍 Lemma. Let 𝑋be a topological space and 𝒜be a family of connected subsets of 𝑋. Suppose that 𝐴∩𝐵≠∅for all 𝐴, 𝐵∈𝒜. Then ⋃𝐴∈𝒜𝐴is connected.
185 IV. Analysis and Topology Proof. Let 𝑌= ⋃𝐴∈𝒜𝐴, and let 𝑓∶𝑌→ℤbe a continuous function. We must show that 𝑓is constant. For all 𝐴∈𝒜, 𝑓|𝐴∶𝐴→ℤis continuous and hence constant, since 𝐴is connected. For all 𝐴, 𝐵∈𝒜, 𝐴∩𝐵≠∅hence 𝑓|𝐴and 𝑓|𝐵are both constant and have the same value. So 𝑓must be constant, and hence 𝑌is connected.
Theorem. Let 𝑋, 𝑌be connected topological spaces. Then 𝑋× 𝑌is connected (in the product topology).
Proof. Without loss of generality, let 𝑋≠∅, 𝑌≠∅. Let 𝑥0 ∈𝑋. Consider the function 𝑓∶𝑌→𝑋× 𝑌defined by 𝑓(𝑦) = (𝑥0, 𝑦). The components of 𝑓are the functions 𝑦↦𝑥0 which is continuous as it is constant, and 𝑦↦𝑦which is continuous as it is the identity.
So 𝑓is continuous. Then, the image of 𝑓, which is {𝑥0} × 𝑌, is connected. Similarly, for all 𝑦∈𝑌, 𝑋× {𝑦} is connected. For 𝑦∈𝑌, {𝑥0} × 𝑌∩𝑋× {𝑦} = {(𝑥0, 𝑦)} ≠∅. Hence, 𝐴𝑦= {𝑥0}×𝑌∪𝑋×{𝑦} is connected. For all 𝑦, 𝑧∈𝑌, 𝐴𝑦∩𝐴𝑧⊃{𝑥0}×𝑌hence 𝐴𝑦∩𝐴𝑧≠∅.
Hence, ⋃𝑦∈𝑌𝐴𝑦= 𝑋× 𝑌is connected.
Example. ℝ𝑛is connected for all 𝑛∈ℕ.
8.3. Partitioning into connected components Definition. Let 𝑋be a topological space. We define a relation ∼on 𝑋by 𝑥∼𝑦if and only if there exists a connected subset 𝐴of 𝑋such that 𝑥, 𝑦∈𝐴. For all 𝑥∈𝑋, 𝑥∼𝑥since {𝑥} is connected. Symmetry is clear from the definition. If 𝑥∼𝑦and 𝑦∼𝑧then by definition there exist connected subsets 𝐴, 𝐵in 𝑋such that 𝑥, 𝑦∈𝐴and 𝑦, 𝑧∈𝐵. In particular, 𝐴∩𝐵 is not empty since 𝑦∈𝐴∩𝐵. Hence 𝐴∪𝐵is connected. Since 𝐴∪𝐵contains 𝑥, 𝑧, we have 𝑥∼𝑧as required for transitivity. Hence ∼is an equivalence relation. For 𝑥∈𝑋, we write 𝐶𝑥for the equivalence class containing 𝑥, called the connected component of 𝑥. The equivalence classes are called connected components of 𝑋.
Proposition. The connected components of a topological space 𝑋are non-empty, maximal connected subsets of 𝑋, they are closed, and they partition 𝑋.
Proof. Let 𝐶be a connected component of 𝑋. So 𝐶= 𝐶𝑥for some 𝑥∈𝑋. Then 𝑥∈𝐶hence 𝐶≠∅. Suppose 𝐶⊂𝐴⊂𝑋and 𝐴is connected. Then for all 𝑦∈𝐴, since 𝑥, 𝑦∈𝐴we must have 𝑥∼𝑦. So 𝑦∈𝐶. Hence 𝐴⊂𝐶, giving 𝐴= 𝐶. For all 𝑦∈𝐶, we have 𝑦∼𝑥, so there exists a connected subset 𝐴𝑦⊂𝑋such that 𝑥, 𝑦∈𝐴𝑦. Let 𝐴= ⋃𝑦∈𝐶𝐴𝑦. 𝐴is connected since the union of pairwise intersecting connected sets are connected. Further 𝐴⊃𝐶so 𝐴= 𝐶and 𝐶is connected. Since the closure of a connected set is connected, 𝐶is connected.
But 𝐶⊃𝐶, so 𝐶= 𝐶is closed.
186 8. Connectedness 8.4. Path-connectedness Definition. Let 𝑋be a topological space. For points 𝑥, 𝑦∈𝑋, a path from 𝑥to 𝑦in 𝑋is a continuous function 𝛾∶[0, 1] →𝑋such that 𝛾(0) = 𝑥, 𝛾(1) = 𝑦. We say that 𝑋is path-connected if for all 𝑥, 𝑦∈𝑋, there exists a path from 𝑥to 𝑦in 𝑋.
Example. In ℝ𝑛, 𝒟𝑟(𝑥) is path-connected by a straight line segment between any two points in the ball. In particular, let 𝛾(𝑡) = (1 −𝑡)𝑦+ 𝑡𝑧. This is continuous and lies entirely inside 𝒟𝑟(𝑥), since ‖𝛾(𝑡) = 𝑥‖ = ‖(1 −𝑡)𝑡+ 𝑡𝑧−𝑥‖ = ‖((1 −𝑡)𝑦+ 𝑡𝑧) −((1 −𝑡)𝑥+ 𝑡𝑥)‖ ≤(1 −𝑡)‖𝑦−𝑥‖ + 𝑡‖𝑧−𝑥‖ < 𝑟 In a similar way, any convex subset of ℝ𝑛is path-connected.
Theorem. If 𝑋is path-connected, 𝑋is connected.
Proof. Suppose 𝑋is not connected. Let 𝑈, 𝑉disconnect 𝑋. Let 𝑥∈𝑈, 𝑦∈𝑉, and suppose 𝛾∶[0, 1] →𝑋is continuous with 𝛾(0) = 𝑥and 𝛾(1) = 𝑦. Then 𝛾−1(𝑈) and 𝛾−1(𝑉) disconnect [0, 1], which contradicts the connectedness of [0, 1].
Example. The converse is false in general. Recall that the space 𝑋= {(𝑥, sin 1 𝑥)∶𝑥> 0} ∪{(0, 𝑦)∶−1 ≤𝑦≤1} is connected. We will show 𝑋is not path-connected. Suppose 𝛾∶[0, 1] →𝑋is continuous, and 𝛾(0) = (0, 0) and 𝛾(1) = (1, sin 1). Let 𝛾= (𝛾1, 𝛾2), so 𝛾1, 𝛾2 are continuous functions.
Suppose 𝑡∈[0, 1] such that 𝛾1(𝑡) > 0. Then 𝛾1((0, 𝑡)) ⊃(0, 𝛾1(𝑡)) by the intermediate value theorem. In particular, there exists 𝑛∈ℕsuch that 1 2𝜋𝑛∈(0, 𝛾1(𝑡)). Hence, there exists 𝑠< 𝑡 such that 𝛾1(𝑠) = 1 2𝜋𝑛so 𝛾1(𝑠) = 0. Similarly, 1 2𝜋𝑛+ 𝜋 2 ∈(0, 𝛾1(𝑡)) so there exists a different 𝑠< 𝑡such that 𝛾1(𝑠) = 1 2𝜋𝑛+ 𝜋 2 hence 𝛾2(𝑠) = 1. In both cases, 𝛾1(𝑠) > 0. We can inductively find a sequence 1 > 𝑡1 > 𝑡2 > ⋯> 0 such that 𝛾2(𝑡𝑛) alternates between zero and one.
But then 𝑡𝑛→𝑡since it is a decreasing bounded-below sequence, and 𝛾2 is continuous, so 𝛾2(𝑡𝑛) →𝛾2(𝑡) which is a contradiction.
8.5. Gluing lemma Lemma. Let 𝑋be a topological space. Suppose 𝑋= 𝐴∪𝐵where 𝐴, 𝐵are closed in 𝑋. Let 𝑔∶𝐴→𝑌and ℎ∶𝐵→𝑌be continuous where 𝑌is a topological space, such that for 𝐴∩𝐵, we have 𝑔= ℎ. Then 𝑓∶𝑋→𝑌defined by 𝑓(𝑥) = {𝑔(𝑥) 𝑥∈𝐴 ℎ(𝑥) 𝑥∈𝐵 187 IV. Analysis and Topology is well defined and continuous.
Proof. First, observe that if 𝐹⊂𝐴and 𝐹is closed in 𝐴, then there exists a closed set 𝐺in 𝑋 such that 𝐹= 𝐴∩𝐺. Since 𝐴is closed in 𝑋, we must have 𝐹is closed in 𝑋. The same holds for 𝐹⊂𝐵. Now, let 𝑉be a closed set in 𝑌. Then the inverse image of 𝑉under 𝑓is 𝑓−1(𝑉) = (𝑓−1(𝑉) ∩𝐴) ∪(𝑓−1(𝑉) ∩𝐵) = 𝑔−1(𝑉) ⏟ ⎵ ⏟ ⎵ ⏟ closed in 𝐴 ∪ℎ−1(𝑉) ⏟ ⎵ ⏟ ⎵ ⏟ closed in 𝐵 So 𝑓−1(𝑉) is closed in 𝑋. To prove continuity it suffices to show that the preimage of a closed set is closed, since that implies that the preimage of an open set is open.
Definition. Let 𝑋be a topological space. For 𝑥, 𝑦∈𝑋, we write 𝑥∼𝑦if there exists a path from 𝑥to 𝑦in 𝑋. This is an equivalence relation: (i) The constant function shows that 𝑥∼𝑥for all 𝑥.
(ii) If 𝛾∶[0, 1] →𝑋is continuous and 𝛾(0) = 𝑥, 𝛾(1) = 𝑦, we define 𝑡↦𝛾(1 −𝑡), which is a path from 𝑦to 𝑥.
(iii) Finally, if 𝑥∼𝑦and 𝑦∼𝑧, we have continuous functions 𝛾, 𝛿such that 𝛾(0) = 𝑥, 𝛾(1) = 𝑦= 𝛿(0), 𝛿(1) = 𝑧. Then let 𝜂(𝑡) = { 𝛾(2𝑡) 𝑡∈[0, 1 2] 𝛿(2𝑡−1) 𝑡∈[ 1 2, 1] These intervals are closed on [0, 1] and their union is [0, 1]. On the intersection, they are equal. By the gluing lemma, 𝜂is continuous, and now since 𝜂(0) = 𝑥, 𝜂(1) = 𝑧we have 𝑥∼𝑧.
We call the equivalence classes path-connected components of 𝑋.
Theorem. Let 𝑈be an open subset of ℝ𝑛. Then 𝑈is connected if and only if 𝑈is path-connected.
Proof. The converse is trivial. Suppose 𝑈is connected. Without loss of generality, suppose 𝑈≠∅. Let 𝑥0 ∈𝑈. Let 𝑃= {𝑥∈𝑈∶𝑥∼𝑥0} be the equivalence class of 𝑥0. We want to show 𝑃= 𝑈. To do this, we will show that 𝑃is open and closed in 𝑈. Then, 𝑃, 𝑈∖𝑃will disconnect 𝑈unless 𝑃= ∅or 𝑃= 𝑈. But we know 𝑥0 ∈𝑃, hence 𝑃= 𝑈will be the only possibility.
To show 𝑃is open, let 𝑥∈𝑈. Since 𝑈is open, there exists 𝑟> 0 such that 𝒟𝑟(𝑥) ⊂𝑈. Recall that for all 𝑦∈𝒟𝑟(𝑥), we have 𝑦∼𝑥. Now, if 𝑥∈𝑃, then we have 𝑦∼𝑥and 𝑥∼𝑥0 so 𝑦∼𝑥0. So 𝒟𝑟(𝑥) ⊂𝑃. So 𝑃is open.
Now, if 𝑥∈𝑈∖𝑃and 𝑦∈𝒟𝑟(𝑥) has 𝑦∼𝑥0, then by transitivity 𝑥∼𝑥0. But this is a contradiction since 𝑥∉𝑃. Hence 𝑈∖𝑃is open. So 𝑃is open and closed, so 𝑃= 𝑈.
188 8. Connectedness Theorem. For 𝑛≥2, ℝand ℝ𝑛are not homeomorphic.
The generalisation ℝ𝑚≄ℝ𝑛is true, but significantly harder to prove and outside the scope of this course.
Proof. Suppose 𝑓∶ℝ→ℝ𝑛is a homeomorphism. Let 𝑔= 𝑓−1. Then 𝑔is continuous.
Then, 𝑓|ℝ∖{0} is a homeomorphism from ℝ∖{0} to ℝ𝑛∖{𝑓(0)}, with inverse 𝑔|ℝ𝑛∖{𝑓(0)}. But ℝ∖{0} is disconnected, but ℝ𝑛∖{𝑓(0)} is connected since it is path-connected. This is a contradiction.
189 IV. Analysis and Topology 9. Compactness 9.1. Motivation and definition Recall from IA Analysis that a continuous function on a closed bounded interval is bounded and attains its bounds. We wish to generalise this result to more general topological spaces.
Example.
(i) If 𝑋is finite, any function 𝑋→ℝis finite.
(ii) If, for all continuous functions 𝑓∶𝑋→ℝthere exists 𝑛∈ℕand subsets 𝐴1, … , 𝐴𝑛of 𝑋such that 𝑋= ⋃ 𝑛 𝑗=1 𝐴𝑗and 𝑓is bounded on 𝐴𝑗for all 𝑗, then the property holds.
(iii) Note that continuous functions are ‘locally bounded’; if 𝑓∶𝑋→ℝis continuous, then for all 𝑥∈𝑋we have 𝑈 𝑥= 𝑓−1((𝑓(𝑥)−1, 𝑓(𝑥)+1)) is an open set containing 𝑋, and 𝑓 is bounded on 𝑈 𝑥. So each point has an open neighbourhood on which 𝑓is bounded.
Further, 𝑋= ⋃𝑥∈𝑋𝑈 𝑥. If there exists a finite subset 𝐹⊂𝑋such that ⋃𝑥∈𝐹𝑈 𝑥= 𝑋, then 𝑓is bounded on 𝑋. This is exactly the definition we will use for compactness.
Definition. Let 𝑋be a topological space. An open cover for 𝑋is a family 𝒰of open subsets of 𝑋that cover 𝑋; that is, ⋃𝑈∈𝒰𝑈= 𝑋. A subcover of 𝒰is a subset 𝒱⊂𝒰that covers 𝑈.
This is called a finite subcover if 𝒱is finite. We say that 𝑋is compact if every open cover has a finite subcover.
Remark. Compactness can be thought of as the next best thing to finiteness.
Theorem. Let 𝑋be a compact topological space and 𝑓∶𝑋→ℝbe continuous. Then 𝑓is bounded, and if 𝑋is not empty 𝑓attains its bounds.
Proof. For 𝑛∈ℕ, let 𝑈𝑛= {𝑥∈𝑋∶|𝑓(𝑥)| < 𝑛}. 𝑈𝑛is open since 𝑥↦|𝑓(𝑥)| is continuous and (−𝑛, 𝑛) is open. It is clear that 𝑋= ⋃𝑛∈ℕ𝑈𝑛. This is an open cover of 𝑋. Hence there exists a finite subcover 𝐹⊂ℕsuch that 𝑋= ⋃𝑛∈𝐹𝑈𝑛= 𝑈𝑁where 𝑁= max 𝐹. Hence, for all 𝑥∈𝑋, we have |𝑓(𝑥)| < 𝑁so 𝑓is bounded.
Let 𝛼= inf𝑋𝑓; this exists since 𝑓is bounded. Suppose there exists no 𝑥∈𝑋such that 𝑓(𝑥) = 𝛼. Then, for all 𝑥∈𝑋, 𝑓(𝑥) > 𝛼. Then there exists 𝑛∈ℕsuch that 𝑓(𝑥) > 𝛼+ 1 𝑛.
So let 𝑉 𝑛= {𝑥∈𝑋∶𝑓(𝑥) > 𝛼+ 1 𝑛} = 𝑓−1((𝛼+ 1 𝑛, ∞)) We can see that 𝑉 𝑛is open. Now, since ⋃𝑛∈ℕ𝑉 𝑛= 𝑋, there exists a finite subcover 𝐹⊂ℕ such that ⋃𝑛∈𝐹𝑉 𝑛= 𝑋= 𝑉𝑁where 𝑁is the maximal 𝐹. Then for all 𝑥∈𝑋, we have 𝑓(𝑥) > 𝛼+ 1 𝑁. Hence inf𝑋𝑓≥𝛼+ 1 𝑁, which is a contradiction. The same argument applies for the supremum.
Lemma. Let 𝑌be a subspace of a topological space 𝑋. Then 𝑌is compact if and only if whenever 𝒰is a family of open sets in 𝑋such that ⋃𝑈∈𝒰⊃𝑌, there is a finite subfamily 𝒱⊂𝒰with ⋃𝑈∈𝒱𝑈⊃𝑌.
Theorem. [0, 1] is compact.
190 9. Compactness Proof. Let 𝒰be a family of open sets in ℝthat cover [0, 1]. For a subset 𝐴⊂[0, 1], we say that 𝒰finitely covers 𝐴if there exists a finite subcover 𝒱⊂𝒰of 𝐴. Note that if 𝐴= 𝐵∪𝐶and 𝐴, 𝐵, 𝐶⊂[0, 1] and 𝒰finitely covers 𝐵and 𝐶, we can take the union of the finite subcovers to find a finite subcover of 𝐴, so 𝑈finitely covers 𝐴. Suppose that 𝒰does not finitely cover [0, 1].
Then one of the intervals [0, 1 2] and [ 1 2, 1] is not finitely coverable by 𝒰. Let this interval be [𝑎1, 𝑏1]. Let 𝑐= 1 2(𝑎1 + 𝑏1). Then one of the intervals [𝑎1, 𝑐], [𝑐, 𝑏1] is not finitely coverable by 𝒰. Inductively, we obtain a nested sequence of intervals [𝑎1, 𝑏1] ⊃⋯⊃[𝑎𝑛, 𝑏𝑛] ⊃⋯ which are not finitely covered by 𝒰and 𝑏𝑛−𝑎𝑛= 2−𝑛. Now, 𝑎𝑛→𝑥for some 𝑥∈[0, 1] and 𝑏𝑛= 𝑎𝑛+ 2−𝑛→𝑥. But since 𝒰covers [0, 1], there exists 𝑈∈𝒰such that 𝑥∈𝑈. 𝑈 is open in ℝ, so for all 𝜀> 0, we have (𝑥−𝜀, 𝑥+ 𝜀) ⊂𝑈. Since 𝑎𝑛, 𝑏𝑛→𝑥, we can choose 𝑛such that 𝑎𝑛, 𝑏𝑛∈(𝑥−𝜀, 𝑥+ 𝜀). This is covered by one open set 𝑈in 𝒰, so this is a finite subcover. This is a contradiction.
Example. Other examples of compact spaces include the following.
(i) Any finite set is compact.
(ii) On any set 𝑋, the cofinite topology is compact. Suppose without loss of generality that 𝑋is not empty, and let 𝒰be an open cover for 𝑋. Let 𝑈∈𝒰such that 𝑈≠∅. Then 𝐹= 𝑋∖𝑈is finite. For all 𝑥∈𝐹, let 𝑈 𝑥∈𝒰such that 𝑥∈𝑈 𝑥. Then ⋃𝑥∈𝐹𝑈 𝑥∪𝑈is a finite subcover.
(iii) Let 𝑥𝑛→𝑥in a topological space 𝑋. Let 𝑌= {𝑥𝑛∶𝑛∈ℕ} ∪{𝑥}. Then 𝑌is compact.
Indeed, let 𝒰be a family of open sets in 𝑋such that ⋃𝑈∈𝒰𝑈⊃𝑌. In particular, let 𝑈∈𝒰such that 𝑥∈𝑈. Since 𝑈is open and 𝑥𝑛→𝑥, there exists 𝑁∈ℕsuch that for all 𝑛≥𝑁we have 𝑥𝑛∈𝑈. So we can cover the remaining finitely many elements analogously to the previous example, and this yields a finite subcover.
(iv) The indiscrete topology on any set is compact, since there are only two open sets.
Counterexamples include the following.
(i) An infinite set 𝑋in the discrete topology is not compact. Let 𝒰= {{𝑥}∶𝑥∈𝑋} This has no finite subcover.
(ii) ℝis not compact. Consider the intervals (−𝑛, 𝑛) for all 𝑛∈ℕ. This is an open cover with no finite subcover.
9.2. Subspaces Theorem. Let 𝑌be a subspace of a topological space 𝑋. Then, (i) Let 𝑋be compact and 𝑌be closed in 𝑋. Then 𝑌is compact.
(ii) Let 𝑋be Hausdorff and 𝑌be compact. Then 𝑌is closed in 𝑋.
191 IV. Analysis and Topology Proof. Let 𝒰be a family of open sets in 𝑋such that their union covers 𝑌. Then 𝒰∪(𝑋∖𝑌) is an open cover for 𝑋since 𝑌is closed. This has a finite subcover 𝒱⊂𝒰such that ⋃𝑈∈𝒱𝑈∪ (𝑋∖𝑌) = 𝑋. Then ⋃𝑈∈𝒱𝑈⊃𝑌.
For part (ii), let 𝑥∈𝑋∖𝑌. For 𝑦∈𝑌, since 𝑥≠𝑦there exist open sets 𝑈 𝑦, 𝑉 𝑦in 𝑋such that 𝑥∈𝑈 𝑦, 𝑦∈𝑉 𝑦, 𝑈 𝑦∩𝑉 𝑦= ∅. Now, {𝑉 𝑦∶𝑦∈𝑌} is an open cover of 𝑌. Hence there exists 𝐹⊂𝑌finite such that ⋃𝑦∈𝐹𝑉 𝑦⊃𝑌. Now, 𝑈= ⋂𝑦∈𝐹𝑈 𝑦is open, further 𝑥∈𝑈and 𝑈∩𝑌⊂(⋂ 𝑦∈𝐹 𝑈 𝑦) ∩(⋃ 𝑦∈𝐹 𝑉 𝑦) = ∅ Hence 𝑋∖𝑌is a neighbourhood of all of its points, so it is open and 𝑌is closed.
9.3. Continuous images of compact spaces Theorem. Let 𝑓∶𝑋→𝑌be a continuous function between topological spaces such that 𝑋is compact. Then 𝑓(𝑋) is compact.
Proof. Let 𝒰be a family of open sets in 𝑌such that ⋃𝑈∈𝒰𝑈⊃𝑓(𝑋). Then ⋃𝑈∈𝒰𝑓−1(𝑈) = 𝑋and 𝑓−1(𝑈) is open in 𝑋for all 𝑈∈𝒰since 𝑓is continuous. Since 𝑋is compact, we have a finite subcover 𝒱⊂𝒰such that 𝑋= ⋃𝑈∈𝒱𝑓−1(𝑉). Hence 𝑓(𝑋) ⊂⋃𝑈∈𝒱𝑈.
Remark. Compactness is a topological property. If 𝑓∶𝑋→𝑌is continuous and 𝐴⊂𝑋is compact, then 𝑓(𝐴) is compact.
Corollary. Any quotient of a compact space is compact.
Example. Let 𝑎< 𝑏∈ℝ. Then [𝑎, 𝑏] ≃[0, 1] so is compact.
9.4. Topological inverse function theorem Theorem. Let 𝑓∶𝑋→𝑌be a continuous bijection from a compact space 𝑋to a Hausdorff space 𝑌. Then 𝑓−1 is continuous, so 𝑓is an open map. Hence 𝑓is a homeomorphism.
Proof. Let 𝑈be an open subset of 𝑋. Then 𝐾= 𝑋∖𝑈is closed. Since 𝑋is compact, 𝐾is compact. Further, 𝑓(𝐾) is compact. Hence 𝑓(𝐾) is closed in 𝑌. So 𝑓(𝑈) = 𝑌∖𝑓(𝐾) is open in 𝑌.
Example. ℝ/ℤis homeomorphic to 𝑆1 = {𝑥∈ℝ2 ∶‖𝑥‖ = 1}. Indeed, let 𝑓∶ℝ→𝑆1 by 𝑓(𝑡) = (cos(2𝜋𝑡), sin(2𝜋𝑡)). For all 𝑠, 𝑡, we have 𝑓(𝑠) = 𝑓(𝑡) if and only if 𝑠∼𝑡so 𝑓fully respects ∼. 𝑓is continuous and surjective. Let ˜ 𝑓∶ℝ/ℤ→𝑆1 be the unique map such that ˜ 𝑓∘𝑞= 𝑓. So ˜ 𝑓is a continuous bijection. 𝑆1 is Hausdorff, and ℝ/ℤis the image of [0, 1] under a continuous map, hence is compact. Hence ˜ 𝑓is a homeomorphism.
192 9. Compactness 9.5. Tychonov’s theorem Theorem. Let 𝑋, 𝑌be compact topological spaces. Then 𝑋× 𝑌is compact in the product topology.
Proof. Let 𝒰be an open cover for 𝑋×𝑌. We want to show that there exists a finite subcover.
Without loss of generality, every member of 𝒰can be of the form 𝑈× 𝑉where 𝑈is open in 𝑋and 𝑉is open in 𝑌. Indeed, for 𝑧∈𝑋× 𝑌we can choose 𝑊 𝑧∈𝒰such that 𝑧∈𝑊 𝑧.
By definition of the product topology, there exist open sets 𝑈𝑧in 𝑋and 𝑉 𝑧in 𝑌such that 𝑧∈𝑈𝑧× 𝑉 𝑧⊂𝑊 𝑧. So {𝑈𝑧× 𝑉 𝑧∶𝑧∈𝑋× 𝑌} is an open cover for 𝑋× 𝑌. If there exists a finite subset 𝐹⊂𝑋× 𝑌such that ⋃𝑧∈𝐹𝑈𝑧× 𝑉 𝑧covers 𝑋× 𝑌, then {𝑊 𝑥∶𝑧∈𝐹} is a finite subcover of 𝒰.
Let 𝑥∈𝑋. Recall that {𝑥} × 𝑌is the continuous image of 𝑌under the map 𝑦↦(𝑥, 𝑦).
Hence, {𝑥}×𝑌is compact, since the continuous image of a compact space is compact. Since {𝑥} × 𝑌is covered by ⋃𝑊∈𝒰𝑊, 𝒰finitely covers {𝑥} × 𝑌. So there exists 𝑛𝑥∈ℕsuch that we can find open sets 𝑈 𝑥,1, … , 𝑈 𝑥,𝑛𝑥in 𝑋and 𝑉 𝑥,1, … , 𝑉 𝑥,𝑛𝑥in 𝑌such that 𝑈 𝑥,𝑗× 𝑉 𝑥,𝑗∈𝒰 and {𝑥} × 𝑌⊂⋃ 𝑛𝑥 𝑗=1 𝑈 𝑥,𝑗× 𝑉 𝑥,𝑗.
Without loss of generality, let 𝑥∈𝑈 𝑥,𝑗for all 𝑗, since any other 𝑈 𝑥,𝑗is not needed in the cover.
Now let 𝑈 𝑥= ⋂ 𝑛𝑥 𝑗=1 𝑈 𝑥,𝑗. We know 𝑥∈𝑈 𝑥and 𝑈 𝑥is open since it is a finite intersection of open sets. In particular, 𝑈 𝑥× 𝑌⊂⋃ 𝑛𝑥 𝑗=1 𝑈 𝑥,𝑗× 𝑉 𝑥,𝑗.
Now, {𝑈 𝑥∶𝑥∈𝑋} is an open cover for 𝑋. So there exists a finite subset 𝐹⊂𝑋such that 𝑋= ⋃𝑥∈𝐹𝑈 𝑥. Then, 𝑋× 𝑌= ⋃𝑥∈𝐹𝑈 𝑥× 𝑌⊂⋃𝑥∈𝐹⋃ 𝑛𝑥 𝑗=1 𝑈 𝑥,𝑗× 𝑉 𝑥,𝑗. Hence, {𝑈 𝑥,𝑗× 𝑉 𝑥,𝑗∶𝑥∈𝐹, 1 ≤𝑗≤𝑛𝑥} is a finite subcover of 𝒰.
Remark. More generally, if 𝑋1, … , 𝑋𝑛are compact spaces, then so is 𝑋1 × ⋯× 𝑋𝑛.
9.6. Heine–Borel theorem Theorem. A subset 𝐾of ℝ𝑛is compact if and only if 𝐾is closed and bounded.
Proof. Suppose 𝐾is compact. ℝ𝑛is a metric space and hence Hausdorff. Hence, 𝐾is closed in ℝ𝑛. The function 𝑥↦‖𝑥‖ is continuous. Therefore, it is bounded on 𝐾. So 𝐾is bounded.
Conversely, if 𝐾is bounded, there exists 𝑀≥0 such that for all 𝑥∈𝐾we have ‖𝑥‖ ≤𝑀.
Hence, 𝐾⊂[−𝑀, 𝑀]𝑛. Note that [−𝑀, 𝑀] is compact since it is homeomorphic to [0, 1]. By Tychonov’s theorem, [−𝑀, 𝑀]𝑛is compact in the product topology. Since a closed subset of a compact space is compact, 𝐾is compact.
Example. Closed balls ℬ𝑟(𝑥) in ℝ𝑛are compact. The start of the proof for the Lindelöf– Picard theorem now makes more sense.
193 IV. Analysis and Topology 9.7. Sequential compactness Definition. A topological space 𝑋is sequentially compact if every sequence in 𝑋has a convergent subsequence. Given a sequence (𝑥𝑛) and an infinite set 𝑀⊂ℕ, we will write (𝑥𝑚)𝑚∈𝑀for the subsequence (𝑥𝑚𝑛)∞ 𝑛=1 where 𝑚1 < 𝑚2 < … are the elements of 𝑀. Note that if 𝐿⊂𝑀⊂ℕ, then (𝑥𝑛)𝑛∈𝐿is a subsequence of (𝑥𝑛)𝑛∈𝑀.
Example. Any closed and bounded subset of ℝis sequentially compact by the Bolzano– Weierstrass theorem. Similarly, any closed and bounded subset 𝐾of ℝ𝑛is sequentially com-pact. Indeed, let (𝑥𝑚) be a sequence in 𝐾. Then, writing 𝑥𝑚= (𝑥𝑚,1, … , 𝑥𝑚,𝑛), since 𝐾is bounded we have that (𝑥𝑚,𝑗) is bounded for all 𝑗. Applying the Bolzano–Weierstrass the-orem to the first coordinate, we find 𝑀1 ⊂ℕsuch that (𝑥𝑚,1)𝑚∈𝑀1 converges in ℝ. Now, (𝑥𝑚,2)𝑚∈𝑀1 is bounded in ℝ, so again applying the Bolzano–Weierstrass theorem, we can find 𝑀2 ⊂ℕsuch that (𝑥𝑚,2)𝑚∈𝑀2 converges. Note that (𝑥𝑚,1)𝑚∈𝑀2 converges. So induct-ively we can find 𝑀1 ⊃⋯⊃𝑀𝑛such that (𝑥𝑚,𝑗)𝑚∈𝑀𝑛converges for all 𝑗. Hence (𝑥𝑚)𝑚∈𝑀𝑛 converges in ℝ𝑛. The limit is contained in 𝐾since 𝐾is closed.
Remark. In ℝ𝑛, any compact space is sequentially compact. The converse is also true; any sequentially compact subspace must be closed and bounded. We aim to show that compact-ness and sequential compactness are identical in metric spaces.
9.8. Compactness and sequential compactness in metric spaces Let (𝑀, 𝑑) be a metric space.
Definition. For 𝜀> 0 and 𝐹⊂𝑀, we say that 𝐹is an 𝜀-net for 𝑀if for all 𝑥∈𝑀, there exists 𝑦∈𝐹such that 𝑑(𝑦, 𝑥) ≤𝜀. Equivalently, 𝑀= ⋃𝑦∈𝑀ℬ𝜀(𝑦). This is called a finite 𝜀-net if 𝐹 is finite. We say that 𝑀is totally bounded if for all 𝜀> 0, there exists a finite 𝜀-net for 𝑀.
Example. For 𝜀> 0, let 𝑛such that 1 𝑛< 𝜀. Then { 1 𝑛, 2 𝑛, … , 𝑛−1 𝑛} is an 𝜀-net for (0, 1).
Definition. For a non-empty 𝐴⊂𝑀, the diameter of 𝐴is diam 𝐴= sup {𝑑(𝑥, 𝑦)∶𝑥, 𝑦∈𝐴}.
This is finite if and only if 𝐴is a bounded set.
Example. diam ℬ𝑟(𝑥) ≤2𝑟.
Lemma. Suppose 𝑀is totally bounded. Let 𝐴be a non-empty closed subset of 𝑀. Let 𝜀> 0.
Then there exists 𝐾∈ℕand non-empty closed sets 𝐵1, … , 𝐵𝐾such that 𝐴= ⋃ 𝐾 𝑘=1 𝐵𝑘and diam 𝐵𝑘≤𝜀for all 𝑘.
Proof. Let 𝐹be a finite 𝜀 2-net for 𝑀. So 𝑀= ⋃𝑥∈𝐹𝐵𝜀/2(𝑥) and hence 𝐴= ⋃𝑥∈𝐹(𝐴∩𝐵𝜀/2(𝑥)).
Let 𝐺= {𝑥∈𝐹∶𝐴∩𝐵𝜀/2(𝑥) ≠0}. Then for 𝑥∈𝐺let 𝐵𝑥= 𝐴∩𝐵𝜀/2(𝑥). So for 𝑥∈𝐺, we have 𝐵𝑥≠∅, 𝐵𝑥⊂𝐵𝜀/2(𝑥) and so diam 𝐵𝑥≤𝜀, and 𝐵𝑥is closed. Then 𝐴= ⋃𝑥∈𝐺𝐵𝑥.
Theorem. For a metric space (𝑀, 𝑑), the following are equivalent.
(i) 𝑀is compact; 194 9. Compactness (ii) 𝑀is sequentially compact; (iii) 𝑀is complete and totally bounded.
Proof. We first show (i) implies (ii). Let (𝑥𝑛) be a sequence in 𝑀. Then for 𝑛∈ℕ, let 𝑇𝑛= {𝑥𝑘∶𝑘> 𝑛} be the tail of the sequence. Note that the limit of any convergent subsequence (if it exists) is in the intersection of ⋂𝑛∈ℕ𝑇𝑛. So first, we prove that this intersection is non-empty. Suppose that it is empty. Then, ⋃𝑛∈ℕ(𝑀∖𝑇𝑛) = 𝑀. But the 𝑀∖𝑇𝑛are open, and 𝑀is compact, there is a finite subcover. So 𝑀∖𝑇𝑁= 𝑀for some 𝑁, since the 𝑇𝑛are a decreasing sequence of sets. This is a contradiction since 𝑇𝑁≠∅. Now, let 𝑥∈⋂𝑛∈ℕ𝑇𝑛, and we want to show the existence of a subsequence converging to 𝑥. First, 𝑥∈𝑇1, so 𝒟1(𝑥) ∩𝑇1 ≠∅. Hence there exists 𝑘1 > 1 such that 𝑑(𝑥𝑘1, 𝑥) < 1. Now since 𝑥∈𝑇𝑘1, 𝒟1/2(𝑥) ∩𝑇𝑘1 ≠∅. There exists 𝑘2 > 𝑘1 such that 𝑑(𝑥𝑘2, 𝑥) < 1 2. Inductively, we can find a strictly increasing sequence 𝑘1 < 𝑘2 < … such that 𝑑(𝑥𝑘𝑛, 𝑥) < 1 𝑛for all 𝑛, so 𝑥𝑘𝑛→𝑥.
Now, we show (ii) implies (iii). To show 𝑀is complete, let (𝑥𝑛) be a Cauchy sequence in 𝑀.
Let 𝑘1 < 𝑘2 < … such that 𝑥𝑘𝑛converges in 𝑀, and let 𝑥be the limit. We show 𝑥𝑛→𝑥.
Indeed, for 𝜀> 0, there exists 𝑁∈ℕsuch that ∀𝑚, 𝑛≥𝑁, we have 𝑑(𝑥𝑚, 𝑥𝑛) < 𝜀. Then ∀𝑚≥𝑁, we have 𝑘𝑛≥𝑚≥𝑁, so for a fixed 𝑛≥𝑁and ∀𝑚≥𝑁, we have 𝑑(𝑥𝑛, 𝑥) ≤ 𝑑(𝑥𝑛, 𝑥𝑘𝑚) + 𝑑(𝑥𝑘𝑚, 𝑥) ≤𝜀+ 𝑑(𝑥𝑘𝑚, 𝑥). Let 𝑚→∞, so 𝑑(𝑥𝑛, 𝑥) ≤𝜀. So 𝑥𝑛→𝑥. To show 𝑀 is totally bounded, suppose it is not. There exists 𝜀> 0 such that 𝑀has no finite 𝜀-net. Let 𝑥1 ∈𝑀, and suppose we can find 𝑥1, … , 𝑥𝑛−1 in 𝑀. Then ⋃ 𝑛−1 𝑗=1 ℬ𝜀(𝑥𝑗) ≠𝑀. So we can pick 𝑥𝑛∈𝑀∖⋃ 𝑛−1 𝑗−1 ℬ𝜀(𝑥𝑗). Inductively we obtain (𝑥𝑛) such that 𝑑(𝑥𝑚, 𝑥𝑛) > 𝜀for all 𝑛, 𝑚∈ℕ.
So (𝑥𝑛) has no Cauchy subsequence. There is therefore no convergent subsequence, which is a contradiction.
Finally, we show (iii) implies (i). Let 𝒰be an open cover for 𝑀. We must show there exists a finite subcover. Suppose that is not true, so 𝒰does not finitely cover 𝑀. We construct non-empty closed subsets 𝐴0 ⊃𝐴1 ⊃… of 𝑀such that for all 𝑛≥0, 𝒰does not finitely cover 𝐴𝑛, and for all 𝑛≥1 we have diam 𝐴𝑛< 1 𝑛. Let 𝐴0 = 𝑀. Suppose that for some 𝑛≥1 we have already found 𝐴𝑛−1. Since 𝑀is totally bounded, we can write 𝐴𝑛−1 = ⋃ 𝐾 𝑘=1 𝐵𝑘 where 𝐾∈ℕand the 𝐵𝑘are non-empty, closed, and diam 𝐵𝑘< 1 𝑛. Since 𝒰does not finitely cover 𝐴𝑛−1, there exists 𝑘≤𝐾such that 𝒰does not finitely cover 𝐵𝑘. Let 𝐴𝑛be this 𝐵𝑘.
Now, for all 𝑛, pick some 𝑥𝑛∈𝐴𝑛. For all 𝑁, ∀𝑚, 𝑛≥𝑁we have 𝑥𝑚, 𝑥𝑛∈𝐴𝑁hence 𝑑(𝑥𝑚, 𝑥𝑛) ≤diam 𝐴𝑁≤ 1 𝑛so the sequence is Cauchy. 𝑀is complete, so 𝑥𝑛→𝑥for some 𝑥∈𝑀. Let 𝑈∈𝒰such that 𝑥∈𝑈. 𝑈is open, so there exists 𝑟> 0 such that 𝒟𝑟(𝑥) ⊂𝑈.
But 𝑥𝑛→𝑥hence there exists 𝑛such that 𝑑(𝑥𝑛, 𝑥) < 𝑟 2 and diam 𝐴𝑛< 𝑟 2. For every 𝑦∈𝐴𝑛, 𝑑(𝑦, 𝑥) ≤𝑑(𝑦, 𝑥𝑛) + 𝑑(𝑥𝑛, 𝑥) ≤diam 𝐴𝑛+ 𝑟 2 < 𝑟. Hence every point in 𝐴𝑛is contained within 𝒟𝑟(𝑥) ⊂𝑈. But this contradicts the fact that 𝒰does not finitely cover 𝐴𝑛, but we have constructed a cover using just one open set.
195 IV. Analysis and Topology Remark. We can now deduce the one direction of the Heine–Borel theorem from the Bolzano– Weierstrass theorem; closed and bounded subsets of ℝ𝑛are compact. Similarly, we can check that the product of sequentially compact topological spaces is sequentially compact in the product topology. This yields a new proof for Tychonov’s theorem for metric spaces.
In general, there exist topological spaces that are compact but not sequentially compact, and conversely there exist topological spaces which are sequentially compact but not compact.
196 10. Differentiation 10. Differentiation 10.1. Linear maps Let 𝑚, 𝑛∈ℕ. Recall that 𝐿(ℝ𝑚, ℝ𝑛) is the vector space of linear maps from ℝ𝑚to ℝ𝑛. This is isomorphic to 𝑀𝑛,𝑚, the space of 𝑛× 𝑚real matrices. There is also an isomorphism to ℝ𝑚𝑛. Let 𝑒1, … , 𝑒𝑚be the standard basis of ℝ𝑚, and similarly let 𝑒′ 1, … , 𝑒′ 𝑛be the standard basis of ℝ𝑛. Then 𝑇∈𝐿(ℝ𝑚, ℝ𝑛) is identified with the 𝑛× 𝑚matrix (𝑇 𝑗𝑖) where 1 ≤𝑗≤𝑛 and 1 ≤𝑖≤𝑚, such that 𝑇 𝑗𝑖= ⟨𝑇𝑒𝑖, 𝑒′ 𝑗⟩. We can therefore view 𝐿(ℝ𝑚, ℝ𝑛) as the 𝑚𝑛-dimensional vector space ℝ𝑚𝑛with the Euclidean norm. So the norm of a linear map 𝑇is given by ‖𝑇‖ = √ √ √ √ 𝑚 ∑ 𝑖=1 𝑛 ∑ 𝑗=1 𝑇2 𝑗𝑖= √ √ √ √ 𝑚 ∑ 𝑖=1 ‖𝑇𝑒𝑖‖2 where 𝑇𝑒𝑖is the 𝑖th column of 𝑇. Thus, 𝐿(ℝ𝑚, ℝ𝑛) becomes a metric space together with the Euclidean distance 𝑑(𝑆, 𝑇) = ‖𝑆−𝑇‖.
Lemma. For 𝑇∈𝐿(ℝ𝑚, ℝ𝑛) and 𝑥∈ℝ𝑚, ‖𝑇𝑥‖ ≤‖𝑇‖ ⋅‖𝑥‖ So 𝑇is a Lipschitz map and hence continuous. Further, if 𝑆∈𝐿(ℝ𝑛, ℝ𝑝) then ‖𝑆𝑇‖ ≤‖𝑆‖ ⋅‖𝑇‖ Proof. We can write 𝑥= 𝑚 ∑ 𝑖=1 𝑥𝑖𝑒𝑖 Hence, 𝑇𝑥= 𝑚 ∑ 𝑖=1 𝑥𝑖𝑇𝑒𝑖 Thus, ‖𝑇𝑥‖ ≤ 𝑚 ∑ 𝑖=1 |𝑥𝑖|‖𝑇𝑒𝑖‖ ≤( 𝑚 ∑ 𝑖=1 𝑥2 𝑖) 1/2 ⋅( 𝑚 ∑ 𝑖=1 ‖𝑇𝑒𝑖‖2) 1/2 = ‖𝑇‖ ⋅‖𝑥‖ Further, for 𝑥, 𝑦∈ℝ𝑚we have 𝑑(𝑇𝑥, 𝑇𝑦) = ‖𝑇𝑥−𝑇𝑦‖ = ‖𝑇(𝑥−𝑦)‖ ≤‖𝑇‖ ⋅‖𝑥−𝑦‖ = ‖𝑇‖𝑑(𝑥, 𝑦) So 𝑇is Lipschitz, and any Lipschitz function is continuous. Now, ‖𝑆𝑇‖ = ( 𝑚 ∑ 𝑖=1 ‖𝑆𝑇𝑒𝑖‖2) 1/2 ≤( 𝑚 ∑ 𝑖=1 ‖𝑆‖‖𝑇𝑒𝑖‖2) 1/2 = ‖𝑆‖( 𝑚 ∑ 𝑖=1 ‖𝑇𝑒𝑖‖2) 1/2 = ‖𝑆‖ ⋅‖𝑇‖ 197 IV. Analysis and Topology 10.2. Differentiation Recall from IA Analysis that a function 𝑓∶ℝ→ℝis differentiable at a point 𝑎∈ℝif lim ℎ→0 𝑓(𝑎+ ℎ) −𝑓(𝑎) ℎ exists. The value of this limit is called the derivative of 𝑓at 𝑎, and denoted 𝑓′(𝑎). Note that 𝑓is differentiable at 𝑎if and only if there exists 𝜆∈ℝand 𝜀∶ℝ→ℝsuch that 𝜀(0) = 0 and 𝜀is continuous at 0, and 𝑓(𝑎+ ℎ) = 𝑓(𝑎) + 𝜆ℎ+ ℎ𝜀(ℎ) This is because we can define 𝜀(ℎ) = { 0 ℎ= 0 𝑓(𝑎+ℎ)−𝑓(𝑎) ℎ −𝜆 ℎ≠0 Informally, this 𝜀definition states that 𝑓is approximated very well (the error ℎ𝜀(ℎ) shrinks rapidly since 𝜀→0) by a linear function in a small neighbourhood of 𝑎. Recall that if 𝑓is 𝑛 times differentiable at 𝑎, then 𝑓(𝑎+ ℎ) = 𝑓(𝑎) + 𝑛 ∑ 𝑘=1 𝑓(𝑘)(𝑎) 𝑘!
ℎ𝑘+ 𝑜(ℎ𝑛) Definition. Let 𝑚, 𝑛∈ℕ. Then 𝑓∶ℝ𝑚→ℝ𝑛and 𝑎∈ℝ𝑚. We say that 𝑓is differentiable at 𝑎if there exists a linear map 𝑇∈𝐿(ℝ𝑚, ℝ𝑛) and a function 𝜀∶ℝ𝑚→ℝ𝑛such that 𝜀(0) = 0 and 𝜀is continuous at 0, and 𝑓(𝑎+ ℎ) = 𝑓(𝑎) + 𝑇(ℎ) + ‖ℎ‖𝜀(ℎ) Note that 𝜀(ℎ) = { 0 ℎ= 0 𝑓(𝑎+ℎ)−𝑓(𝑎)−𝑇(ℎ) ‖ℎ‖ ℎ≠0 So 𝑓is differentiable at 𝑎if and only if there exists 𝑇∈𝐿(ℝ𝑚, ℝ𝑛) such that 𝑓(𝑎+ ℎ) −𝑓(𝑎) −𝑇(ℎ) ‖ℎ‖ →0 as ℎ→0. Such a 𝑇is unique. Indeed, suppose 𝑆, 𝑇satisfy the above limit. Then, by sub-tracting, 𝑆(ℎ) −𝑇(ℎ) ‖ℎ‖ →0 For a fixed 𝑥∈ℝ𝑚, 𝑥≠0, we have 𝑥 𝑘→0 as 𝑘→∞so 𝑆( 𝑥 𝑘) −𝑇( 𝑥 𝑘) ‖ ‖ 𝑥 𝑘 ‖ ‖ →0 ⟹𝑆(𝑥) −𝑇(𝑥) ‖𝑥‖ = 0 198 10. Differentiation So 𝑆𝑥= 𝑇𝑥. It follows that 𝑆= 𝑇. We say that if a function 𝑓is differentiable at a point 𝑎, 𝑇is the unique derivative of 𝑓at 𝑎. This is denoted 𝑓′(𝑎) = 𝐷𝑓(𝑎) = 𝐷𝑓|𝑎. If 𝑓∶ℝ𝑚→ℝ𝑛 is differentiable at 𝑎∈ℝ𝑚for every 𝑎, we say that 𝑓is differentiable on ℝ𝑚. The function 𝑓′ = 𝐷∶ℝ𝑚→𝐿(ℝ𝑚, ℝ𝑛) mapping 𝑎↦𝑓′(𝑎) is the derivative of 𝑓.
Example. Constant functions are differentiable. Let 𝑓∶ℝ𝑚→ℝ𝑛such that 𝑓(𝑥) = 𝑏for 𝑏∈ℝ𝑛. Then for all 𝑎∈ℝ𝑚, we have 𝑓(𝑎+ ℎ) = 𝑓(𝑎) + 0ℎ+ 0 so 𝑓is differentiable at 𝑎and the derivative is zero.
Example. Linear maps are differentiable. Let 𝑓∶ℝ𝑚→ℝ𝑛be defined by 𝑓(𝑥) = 𝑇𝑥for a linear map 𝑇∈𝐿(ℝ𝑚, ℝ𝑛). Then 𝑓(𝑎+ ℎ) = 𝑓(𝑎) + 𝑓(ℎ) + 0 so 𝑓is differentiable at 𝑎with derivative 𝑇= 𝑓. So 𝑓′ is a constant function.
Example. Consider 𝑓(𝑥) = ‖𝑥‖2 For 𝑎∈ℝ𝑚, we can find 𝑓(𝑎+ ℎ) = ‖𝑎+ ℎ‖2 = ‖𝑎‖2 + 2 ⟨𝑎, ℎ⟩+ ‖ℎ‖2 = 𝑓(𝑎) + 2 ⟨𝑎, ℎ⟩+ ‖ℎ‖𝜀(ℎ) Hence, 𝑓is differentiable with derivative 𝑓′(𝑎)(ℎ) = 2 ⟨𝑎, ℎ⟩ Note that 𝑓′ ∶ℝ𝑚→𝐿(ℝ𝑚→ℝ) is linear.
Example. Note 𝑀𝑛≃ℝ𝑛2. The function 𝑓∶𝑀𝑛→𝑀𝑛given by 𝑓(𝐴) = 𝐴2. For a fixed 𝐴∈𝑀𝑛, 𝑓(𝐴+ 𝐻) = (𝐴+ 𝐻)2 = 𝐴2 + 𝐴𝐻+ 𝐻𝐴+ 𝐻2 It suffices to show 𝐻2 is 𝑜(‖𝐻‖). We have ‖ ‖𝐻2‖ ‖ ≤‖𝐻‖2, hence ‖ ‖𝐻2‖ ‖ ‖𝐻‖ ≤‖𝐻‖ →0 So 𝑓is differentiable at 𝐴and the derivative is given by 𝑓′(𝐴)(𝐻) = 𝐴𝐻+ 𝐻𝐴 Example. Suppose 𝑓∶ℝ𝑚× ℝ𝑛→ℝ𝑝is bilinear. Let (𝑎, 𝑏) ∈ℝ𝑚× ℝ𝑛. Then, 𝑓((𝑎, 𝑏) + (ℎ, 𝑘)) = 𝑓((𝑎+ ℎ, 𝑏+ 𝑘)) = 𝑓(𝑎, 𝑏) + 𝑓(𝑎, 𝑘) + 𝑓(ℎ, 𝑏) + 𝑓(ℎ, 𝑘) 199 IV. Analysis and Topology The map ℝ𝑚×ℝ𝑛→ℝ𝑝given by (ℎ, 𝑘) ↦𝑓(𝑎, 𝑘)+𝑓(ℎ, 𝑏) is linear as the sum of two linear maps. So it suffices to show 𝑓(ℎ, 𝑘) is 𝑜(‖(ℎ, 𝑘)‖).
ℎ= 𝑚 ∑ 𝑖=1 ℎ𝑖𝑒𝑖; 𝑘= 𝑛 ∑ 𝑗=1 𝑘𝑗𝑒′ 𝑗 Hence, 𝑓(ℎ, 𝑘) = 𝑚 ∑ 𝑖=1 𝑛 ∑ 𝑗=1 ℎ𝑖𝑘𝑗𝑓(𝑒𝑖, 𝑒′ 𝑗) ⟹‖𝑓(ℎ, 𝑘)‖ ≤ 𝑚 ∑ 𝑖=1 𝑛 ∑ 𝑗=1 |ℎ𝑖| ⋅| |𝑘𝑗| | ⋅‖ ‖𝑓(𝑒𝑖, 𝑒′ 𝑗)‖ ‖ ≤𝐶‖(ℎ, 𝑘)‖2 for some constant 𝐶, since |ℎ𝑖| ≤‖(ℎ, 𝑘)‖2 and similarly for | |𝑘𝑗| |. So ‖𝑓(ℎ, 𝑘)‖ ‖(ℎ, 𝑘)‖ ≤𝐶‖(ℎ, 𝑘)‖ →0 Hence 𝑓is differentiable with 𝑓′(𝑎, 𝑏)(ℎ, 𝑘) = 𝑓(𝑎, 𝑘) + 𝑓(ℎ, 𝑏) 10.3. Derivatives on open subsets We may define the derivative on a subset of ℝ𝑚. We will use the notion of open subsets since we are typically interested in neigbourhoods of points.
Definition. Let 𝑈be an open subset of ℝ𝑚. Let 𝑓∶𝑈→ℝ𝑛be a function, and 𝑎∈𝑈.
Then we say 𝑓is differentiable at 𝑎if there exists a linear map 𝑇∈𝐿(ℝ𝑚, ℝ𝑛) such that 𝑓(𝑎+ ℎ) = 𝑓(𝑎) + 𝑇(ℎ) + ‖ℎ‖𝜀(ℎ) where 𝜀(0) = 0 and 𝜀is continuous at zero. Note that 𝜀need only be defined on the set of ℎ such that 𝑎+ ℎ∈𝑈, or more precisely the open set 𝑈−𝑎. Hence there exists 𝑟> 0 such that 𝒟𝑟(0) ⊂𝑈𝑎. Then 𝜀(ℎ) = { 0 ℎ= 0 𝑓(𝑎+ℎ)−𝑓(𝑎)−𝑇(ℎ) ‖ℎ‖ ℎ≠0, 𝑎+ ℎ∈𝑈 So 𝑓is differentiable at 𝑎if and only if there exists a linear map 𝑇∈𝐿(ℝ𝑚, ℝ𝑛) such that 𝑓(𝑎+ ℎ) −𝑓(𝑎) −𝑇(ℎ) ‖ℎ‖ →0 Remark. The linear map 𝑇is unique, and is called the derivative of 𝑓at 𝑎, denoted 𝑓′(𝑎). In particular, 𝑓(𝑎+ ℎ) = 𝑓(𝑎) + 𝑓′(𝑎)(ℎ) + 𝑜(‖ℎ‖) 200 10. Differentiation Remark. If 𝑚= 1, the space 𝐿(ℝ, ℝ𝑛) is isomorphic to ℝ𝑛. The linear map is defined uniquely by a vector in ℝ𝑛which multiplies by the scalar ℎ. Hence, if 𝑈⊂ℝis open and 𝑓∶𝑈→ℝbe a function and 𝑎∈𝑈, then 𝑓is differentiable at 𝑎if there exists a vector 𝑣∈ℝ𝑛such that 𝑓(𝑎+ ℎ) −𝑓(𝑎) −ℎ𝑣 |𝑣| →0 Equivalently, there exists 𝑣∈ℝ𝑛such that 𝑓(𝑎+ ℎ) −𝑓(𝑎) ℎ →𝑣 10.4. Properties of derivative Proposition. Let 𝑈⊂ℝ𝑚be open, 𝑓∶𝑈→ℝ𝑛be a function, and 𝑎∈𝑈. If 𝑓is differen-tiable at 𝑎, 𝑓is continuous at 𝑎.
Proof. We have 𝑓(𝑎+ ℎ) = 𝑓(𝑎) + 𝑓′(𝑎)(ℎ) + ‖ℎ‖𝜀(ℎ) Hence, 𝑓(𝑥) = 𝑓(𝑎) + 𝑓′(𝑎)(𝑥−𝑎) + ‖𝑥−𝑎‖𝜀(𝑥−𝑎) The functions 𝑥↦𝑓(𝑎), 𝑥↦𝑓′(𝑎)(𝑥−𝑎) and 𝑥↦‖𝑥−𝑎‖𝜀(𝑥−𝑎) are all continuous at 𝑎.
Hence their sum is continuous.
Proposition (chain rule). Let 𝑈⊂ℝ𝑚and 𝑉⊂ℝ𝑛be open, 𝑓∶𝑈→ℝ𝑛and 𝑔∶𝑉→ ℝ𝑝be functions, and 𝑎∈𝑈, 𝑏≡𝑓(𝑎) ∈𝑉. Suppose 𝑓is differentiable at 𝑎, and 𝑔is differentiable at 𝑏. Then 𝑔∘𝑓is differentiable at 𝑎and (𝑔∘𝑓)′(𝑎) = 𝑔′(𝑏) ∘𝑓′(𝑎) Proof. Let 𝑆= 𝑓′(𝑎) and 𝑇= 𝑔′(𝑏). Then by assumption 𝑓(𝑎+ ℎ) = 𝑓(𝑎) + 𝑆(ℎ) + ‖ℎ‖𝜀(ℎ); 𝑔(𝑏+ 𝑘) + 𝑔(𝑏) + 𝑇(𝑘) + ‖𝑘‖𝜁(𝑘) for suitable 𝜀, 𝜁. Then, (𝑔∘𝑓)(𝑎+ ℎ) = 𝑔(𝑓(𝑎) + 𝑆(ℎ) + ‖ℎ‖𝜀(ℎ)) = 𝑔(𝑏+ 𝑆(ℎ) + ‖ℎ‖𝜀(ℎ) ⏟⎵ ⎵ ⎵⏟⎵ ⎵ ⎵⏟ 𝑘 ) = 𝑔(𝑏) + 𝑇(𝑆(ℎ) + ‖ℎ‖𝜀(ℎ)) + ‖𝑆(ℎ) + ‖ℎ‖𝜀(ℎ)‖𝜁(𝑆(ℎ) + ‖ℎ‖𝜀(ℎ)) = (𝑔∘𝑓)(𝑎) + (𝑇∘𝑆)(ℎ) + ‖ℎ‖𝑇(𝜀(ℎ)) + ‖𝑘‖𝜁(𝑘) It suffices to show that 𝜂(ℎ) ≡‖ℎ‖𝑇(𝜀(ℎ)) + ‖𝑘‖𝜁(𝑘) 201 IV. Analysis and Topology satisfies 𝜂 ‖ℎ‖ →0. Then the result follows. First, ‖ℎ‖𝑇(𝜀(ℎ)) ‖ℎ‖ = 𝑇(𝜀(ℎ)) →0 as ‖𝑇(𝜀(ℎ))‖ ≤‖𝑇‖ ⋅‖𝜀(ℎ)‖ →0. Then, ‖𝑘‖ ‖ℎ‖ = ‖𝑆(ℎ)‖ + ‖ℎ‖ ⋅‖𝜀(ℎ)‖ ‖ℎ‖ ≤‖𝑆‖ + ‖𝜀(ℎ)‖ Hence, 𝑘= 𝑆(ℎ) + ‖ℎ‖ ⋅𝜀(ℎ) →0 as ℎ→0. Thus 𝜁(𝑘) →0 as 𝑘→0. So 𝜂(ℎ) ‖ℎ‖ = 𝑇(𝜀(ℎ)) + ‖𝑘‖ ‖ℎ‖𝜁(𝑘) →0 as required.
Proposition. Let 𝑈⊂ℝ𝑚be open, 𝑓∶𝑈→ℝ𝑛be a function, and 𝑎∈𝑈. Let 𝑓𝑗be the 𝑗th component of 𝑓, so 𝑓𝑗= 𝜋𝑗∘𝑓. Then 𝑓is differentiable at 𝑎if and only if each 𝑓𝑗is differentiable at 𝑎. If this holds, 𝑓′(𝑎)(ℎ) = 𝑛 ∑ 𝑗=1 𝑓′ 𝑗(𝑎)(ℎ)𝑒′ 𝑗 Equivalently, 𝜋𝑗[𝑓′(𝑎)(ℎ)] = 𝑓′ 𝑗(𝑎)(ℎ) Proof. If 𝑓is differentiable at 𝑎, by the chain rule the composite 𝜋𝑗∘𝑓is differentiable at 𝑎.
Since the derivative of a linear map is itself, the derivative is given by 𝑓′ 𝑗(𝑎) = 𝜋′ 𝑗(𝑓(𝑎)) ∘𝑓′(𝑎) = 𝜋𝑗∘𝑓′(𝑎) Hence 𝑓′(𝑎)(ℎ) = 𝑛 ∑ 𝑗=1 𝜋𝑗[𝑓′(𝑎)(ℎ)𝑒′ 𝑗] = 𝑛 ∑ 𝑗=1 𝑓′ 𝑗(𝑎)(ℎ)𝑒′ 𝑗 Conversely suppose each 𝑓𝑗is differentiable. Then 𝑓𝑗(𝑎+ ℎ) = 𝑓𝑗(𝑎) + 𝑓′ 𝑗(𝑎)(ℎ) + ‖ℎ‖𝜀𝑗(ℎ) for suitable 𝜀(𝑗). Now, 𝑓(𝑎+ ℎ) = 𝑛 ∑ 𝑗=1 𝑓𝑗(𝑎+ ℎ)𝑒′ 𝑗 = 𝑛 ∑ 𝑗=1 [𝑓𝑗(𝑎) + 𝑓′ 𝑗(𝑎)(ℎ) + ‖ℎ‖𝜀𝑗(ℎ)]𝑒′ 𝑗 = 𝑛 ∑ 𝑗=1 𝑓𝑗(𝑎)𝑒′ 𝑗+ 𝑛 ∑ 𝑗=1 𝑓′ 𝑗(𝑎)(ℎ)𝑒′ 𝑗+ ‖ℎ‖ 𝑛 ∑ 𝑗=1 𝜀𝑗(ℎ)𝑒′ 𝑗 Since each 𝜀𝑗tends to zero as ℎ→0, so does their sum.
202 10. Differentiation Remark. This proposition shows that we can prove things for an image ℝ𝑛= ℝwithout loss of generality.
10.5. Linearity and product rule Proposition. Let 𝑈⊂ℝ𝑚be open and functions 𝑓, 𝑔∶𝑈→ℝ𝑛, 𝜙∶𝑈→ℝwhich are dif-ferentiable at 𝑎. Then the functions 𝑓+𝑔and 𝜙⋅𝑓are also differentiable and their derivatives are (𝑓+ 𝑔)′(𝑎) = 𝑓′(𝑎) + 𝑔′(𝑎); (𝜙𝑓)′(𝑎)(ℎ) = 𝜙(𝑎)[𝑓′(𝑎)(ℎ)] + [𝜙′(𝑎)(ℎ)]𝑓(𝑎) For 𝑚= 𝑛= 1 this is the usual product rule.
Proof. We have 𝑓(𝑎+ ℎ) = 𝑓(𝑎) + 𝑓′(𝑎)(ℎ) + ‖ℎ‖𝜀(ℎ) 𝑔(𝑎+ ℎ) = 𝑔(𝑎) + 𝑔′(𝑎)(ℎ) + ‖ℎ‖𝜁(ℎ) 𝜙(𝑎+ ℎ) = 𝜙(𝑎) + 𝜙′(𝑎)(ℎ) + ‖ℎ‖𝜂(ℎ) for suitable 𝜀, 𝜁, 𝜂. The sum gives (𝑓+ 𝑔)(𝑎+ ℎ) = (𝑓+ 𝑔)(𝑎+ ℎ) + (𝑓′(𝑎) + 𝑔′(𝑎))(ℎ) + ‖ℎ‖(𝜀(ℎ) + 𝜁(ℎ)) It follows that 𝑓+ 𝑔is differentiable at 𝑎and its derivative is the sum of the derivatives of its components.
(𝜙⋅𝑓)(𝑎+ ℎ) = 𝜙(𝑎+ ℎ)𝑓(𝑎+ ℎ) = (𝜙⋅𝑓)(𝑎) + [𝜙(𝑎)𝑓′(𝑎)(ℎ) + 𝜙′(𝑎)(ℎ)𝑓(𝑎)] + 𝑓′(𝑎)(ℎ)𝜙′(𝑎)(ℎ) + ‖ℎ‖ (𝑓′(𝑎)(ℎ)𝜂(ℎ) + 𝜙′(𝑎)(ℎ)𝜀(ℎ) + 𝜂(ℎ)𝑓(𝑎) + 𝜙(𝑎)𝜀(ℎ) + ‖ℎ‖𝜂(ℎ)𝜀(ℎ)) ⏟⎵⎵⎵⎵⎵⎵⎵⎵⎵⎵⎵⎵⎵⎵⎵⎵⎵⎵⎵⎵⏟⎵⎵⎵⎵⎵⎵⎵⎵⎵⎵⎵⎵⎵⎵⎵⎵⎵⎵⎵⎵⏟ 𝛿(ℎ) Now, ‖𝜙′(𝑎)(ℎ) ⋅𝑓′(𝑎)(ℎ)‖ ‖ℎ‖ = |𝜙′(𝑎)(ℎ)| ⋅‖𝑓′(𝑎)(ℎ)‖ ‖ℎ‖ ≤‖𝜙′(𝑎)‖ ⋅‖ℎ‖ ⋅‖𝑓′(𝑎)‖ ⋅‖ℎ‖ ‖ℎ‖ →0 Clearly 𝛿→0 since the same is true for all of its components.
203 IV. Analysis and Topology 11. Partial derivatives 11.1. Directional and partial derivatives Definition. Let 𝑈, 𝑓, 𝑎as before. Fix a direction 𝑢∈ℝ𝑚where 𝑢≠0. If the limit lim 𝑡→0 𝑓(𝑎+ 𝑡𝑢) −𝑓(𝑎) 𝑡 exists, then the value of this limit is the directional derivative of 𝑓at 𝑎in direction 𝑢, denoted 𝐷𝑢𝑓(𝑎).
Remark. Note that 𝐷𝑢𝑓(𝑎) ∈ℝ𝑛. Further, 𝑓(𝑎+ 𝑡𝑢) = 𝑓(𝑎) + 𝑡𝐷𝑢𝑓(𝑎) + 𝑜(𝑡). Define 𝛾∶ℝ→ℝ𝑚by 𝛾(𝑡) = 𝑎+ 𝑡𝑢. Then 𝑓∘𝛾is defined on 𝛾−1(𝑈) which is open as 𝛾is continuous, and 0 ∈𝛾−1(𝑈). Then, 𝑓(𝑎+ 𝑡𝑢) −𝑓(𝑎) 𝑡 = (𝑓∘𝛾)(𝑡) −(𝑓∘𝛾)(0) 𝑡 Hence 𝐷𝑢𝑓(𝑎) exists if and only if 𝑓∘𝛾is differentiable at zero, and its value is the derivative of 𝑓∘𝛾. When 𝑢= 𝑒𝑖for a standard basis vector 𝑒𝑖, if 𝐷𝑒𝑖𝑓(𝑎) exists we call it the 𝑖th partial derivative of 𝑓at 𝑎, denoted 𝐷𝑖𝑓(𝑎).
Proposition. Let 𝑈, 𝑓, 𝑎as before. If 𝑓is differentiable at 𝑎, then all directional derivatives 𝐷𝑢𝑓(𝑎) exist. Further, 𝐷𝑢𝑓(𝑎) = 𝑓′(𝑎)(𝑢) Further, 𝑓′(𝑎)(ℎ) = 𝑚 ∑ 𝑖=1 ℎ𝑖𝐷𝑖𝑓(𝑎) for all ℎ= ∑ 𝑚 𝑖=1 ℎ𝑖𝑒𝑖.
Proof. Since 𝑓is differentiable, 𝑓(𝑎+ ℎ) = 𝑓(𝑎) + 𝑓′(𝑎)(ℎ) + ‖ℎ‖𝜀(ℎ) Let ℎ= 𝑡𝑢. Then, 𝑓(𝑎+ 𝑡𝑢) = 𝑓(𝑎) + 𝑡𝑓′(𝑎)(𝑢) + |𝑡| ⋅‖𝑢‖𝜀(𝑡𝑢) Hence, 𝑓(𝑎+ 𝑡𝑢) −𝑓(𝑎) 𝑡 = 𝑓′(𝑎)(𝑢) + |𝑡| 𝑡‖𝑢‖𝜀(𝑡𝑢) The error term converges to zero, hence the limit becomes 𝑓′(𝑎)(𝑢). Moreover, for all ℎ defined as above, 𝑓′(𝑎)(ℎ) = 𝑚 ∑ 𝑖=1 ℎ𝑖𝑓′(𝑎)(𝑒𝑖) = 𝑚 ∑ 𝑖=1 ℎ𝑖𝐷𝑖𝑓(𝑎) 204 11. Partial derivatives alternative proof. Let 𝛾(𝑡) = 𝑎+ 𝑡𝑢. Then 𝑓∘𝛾is defined on the open set 𝛾−1(𝑈). Note that 𝛾is differentiable and 𝛾′(𝑡) = 𝑢for all 𝑡. By the chain rule, 𝑓∘𝛾is differentiable at zero, and 𝐷𝑢𝑓(𝑎) = (𝑓∘𝛾)′(0) = 𝑓′(𝛾(0))(𝛾′(0)) = 𝑓′(𝑎)(𝑢) Remark. If 𝐷𝑢𝑓(𝑎) exists, then so does 𝐷𝑢𝑓𝑗(𝑎) where 𝑓𝑗= 𝜋𝑗∘𝑓. Indeed, by linearity and continuity of 𝜋, 𝑓𝑗(𝑎+ 𝑡𝑢) −𝑓𝑗(𝑎) 𝑡 = 𝜋𝑗(𝑓(𝑎+ 𝑡𝑢) −𝑓(𝑡) 𝑡 ) →𝜋𝑗(𝐷𝑢𝑓(𝑎)) The converse of the proposition is false in general.
11.2. Jacobian matrix Definition. Suppose 𝑓is differentiable at 𝑎. Then the Jacobian matrix of 𝑓at 𝑎, denoted 𝐽𝑓(𝑎), is the matrix of 𝑓′(𝑎) with respect to the standard bases. For 1 ≤𝑖≤𝑚, the 𝑖th column is 𝑓′(𝑎)(𝑒𝑖) = 𝐷𝑖𝑓(𝑎) In particular, for the 𝑗, 𝑖entry, (𝐽𝑓(𝑎))𝑗𝑖= ⟨𝐷𝑖𝑓(𝑎), 𝑒′ 𝑗⟩= 𝜋𝑗(𝐷𝑖𝑓(𝑎)) = 𝐷𝑖𝑓𝑗(𝑎) = 𝜕𝑓𝑗 𝜕𝑥𝑖 11.3. Constructing total derivative from partial derivatives Theorem. Suppose there exists an open neighbourhood 𝑉of 𝑎with 𝑉⊂𝑈such that 𝐷𝑖𝑓(𝑥) exists for all 𝑥∈𝑉and for all 1 ≤𝑖≤𝑚, and the map 𝑥↦𝐷𝑖𝑓(𝑥) from 𝑉to ℝ𝑛is continuous at 𝑎for all 𝑖. Then 𝑓is differentiable at 𝑎.
Proof. By considering components, without loss of generality let 𝑛= 1. Let 𝑚= 2 for convenience of notation; this does not change the proof. Let 𝑎= (𝑝, 𝑞). Let 𝜓(ℎ, 𝑘) = 𝑓(𝑝+ ℎ, 𝑞+ 𝑘) −𝑓(𝑝, 𝑞) −ℎ𝐷1𝑓(𝑝, 𝑞) −𝑘𝐷2𝑓(𝑝, 𝑞) We need to show 𝜓(ℎ, 𝑘) = 𝑜(‖(ℎ, 𝑘)‖), then the derivative of 𝑓can be read off from the definition of 𝜓. Note, 𝜓(ℎ, 𝑘) = [𝑓(𝑝+ ℎ, 𝑞+ 𝑘) −𝑓(𝑝+ ℎ, 𝑞) −𝑘𝐷2𝑓(𝑝, 𝑞)] + [𝑓(𝑝+ ℎ, 𝑞) −𝑓(𝑝, 𝑞) −ℎ𝐷1(𝑝, 𝑞)] We will show separately that each part is small enough to be an error term. The second term is 𝑜(ℎ) and hence 𝑜(‖(ℎ, 𝑘)‖) by the definition of 𝐷1𝑓(𝑝, 𝑞). For the first term, let 𝜙(𝑡) = 𝑓(𝑝+ ℎ, 𝑞+ 𝑡𝑘) for a given fixed ℎ, 𝑘. Then 𝜙is differentiable and by the chain rule we have 205 IV. Analysis and Topology 𝜙′(𝑡) = 𝐷2𝑓(𝑝+ℎ, 𝑞+𝑡𝑘)⋅𝑘. By the mean value theorem, there exists a point 𝑡(ℎ, 𝑘) ∈(0, 1) such that 𝜙(1) −𝜙(0) = 𝜙′(𝑡). Hence, the first term becomes 𝜙(1) −𝜙(0) −𝑘𝐷2𝑓(𝑝, 𝑞) = 𝑘[𝐷2𝑓(𝑝+ ℎ, 𝑞+ 𝑡𝑘) −𝐷2𝑓(𝑝, 𝑞)] As (ℎ, 𝑘) →(0, 0), we have (𝑝+ ℎ, 𝑞+ 𝑡𝑘) →(𝑝, 𝑞). By continuity of 𝐷2𝑓at 𝑎, the term is 𝑜(𝑘) and hence 𝑜(‖(ℎ, 𝑘)‖).
11.4. Mean value inequality The mean value theorem cannot be extended verbatim to higher dimensional spaces, since there can be multiple paths between points.
Theorem. Let 𝑈⊂ℝ𝑚be open, and 𝑓∶𝑈→ℝ𝑛be differentiable at every 𝑧∈𝑈. Let 𝑎, 𝑏∈𝑈such that the line segment connecting 𝑎, 𝑏given by [𝑎, 𝑏] = {(1 −𝑡)𝑎+ 𝑡𝑏∶0 ≤𝑡≤1} is contained inside 𝑈. Suppose there exists 𝑀≥0 such that for all 𝑧∈[𝑎, 𝑏], we have ‖𝑓′(𝑧)‖ ≤𝑀. Then ‖𝑓(𝑏) −𝑓(𝑎)‖ ≤𝑀‖𝑏−𝑎‖ Proof. Let 𝑢= 𝑏−𝑎and 𝑣= 𝑓(𝑏) −𝑓(𝑎). Without loss of generality, let 𝑢≠0. Let 𝛾(𝑡) = 𝑎+ 𝑡𝑢, so 𝑓∘𝛾is defined on the open set 𝛾−1(𝑈), and is differentiable with derivative (𝑓∘𝛾)′(𝑡) = 𝑓′(𝛾(𝑡))(𝛾′(𝑡)) = 𝑓′(𝑎+ 𝑡𝑢)(𝑢) Now, ‖𝑓(𝑏) −𝑓(𝑎)‖2 = ⟨𝑓(𝑏) −𝑓(𝑎), 𝑣⟩= ⟨(𝑓∘𝛾)(1) −(𝑓∘𝛾)(0), 𝑣⟩ Let 𝜙(𝑡) = ⟨(𝑓∘𝛾)(𝑡), 𝑣⟩. Note that 𝜙is differentiable since the inner product is linear. The derivative is 𝜙′(𝑡) = ⟨(𝑓∘𝛾)′(𝑡), 𝑣⟩= ⟨𝑓′(𝑎+ 𝑡𝑢)(𝑢), 𝑣⟩ By the mean value theorem, there exists 𝜃∈(0, 1) such that 𝜙(1) −𝜙(0) = 𝜙′(𝜃). Then, by the Cauchy–Schwarz inequality, ‖𝑓(𝑏) −𝑓(𝑎)‖2 = 𝜙′(𝜃) = ⟨𝑓′(𝑎+ 𝜃𝑢)(𝑢), 𝑣⟩ ≤‖𝑓′(𝑎+ 𝜃𝑢)(𝑢)‖ ⋅‖𝑣‖ ≤‖𝑓′(𝑎+ 𝜃𝑢)‖ ⋅‖𝑢‖ ⋅‖𝑣‖ ≤𝑀‖𝑏−𝑎‖ ⋅‖𝑣‖ Hence, ‖𝑓(𝑏) −𝑓(𝑎)‖ ≤𝑀‖𝑏−𝑎‖ as required.
206 11. Partial derivatives 11.5. Zero derivatives Corollary. Let 𝑈be an open, connected subset of ℝ𝑚, and 𝑓∶𝑈→ℝ𝑛be differentiable at every 𝑈. If 𝑓′(𝑎) = 0 for all 𝑎∈𝑈, then 𝑓is constant.
Proof. If 𝑎, 𝑏∈𝑈satisfy [𝑎, 𝑏] ⊂𝑈, then by the mean value inequality we have ‖𝑓(𝑏) −𝑓(𝑎)‖ ≤‖𝑏−𝑎‖ sup 𝑧∈[𝑎,𝑏] ‖𝑓′(𝑧)‖ = 0 Hence 𝑓(𝑎) = 𝑓(𝑏). For an arbitrary 𝑥∈𝑈, there exists 𝑟> 0 such that 𝒟𝑟(𝑥) ⊂𝑈. This open ball is convex, so for all 𝑦∈𝒟𝑟(𝑥) we have 𝑓(𝑦) = 𝑓(𝑥). Hence 𝑓is locally constant; every point has a neighbourhood on which 𝑓is constant. Since 𝑈is connected, 𝑓is constant (refer to the derivation from the example sheet).
11.6. Inverse function theorem Remark. Let 𝑉⊂ℝ𝑚and 𝑊⊂ℝ𝑛be open sets. Let 𝑓∶𝑉→𝑊be a bijection. Let 𝑎∈𝑉, and let 𝑓be differentiable at 𝑎, and the inverse 𝑓−1 ∶𝑊→𝑉is differentiable at 𝑓(𝑎).
Denoting 𝑆= 𝑓′(𝑎), 𝑇= (𝑓−1) ′(𝑓(𝑎)), we can use the chain rule to find 𝑇𝑆= (𝑓−1 ∘𝑓) ′(𝑎); 𝑆𝑇= (𝑓∘𝑓−1) ′(𝑓(𝑎)) The identity function is linear so its derivative is the identity. Hence 𝑇𝑆is the identity on ℝ𝑚and 𝑆𝑇is the identity on ℝ𝑛. Hence, 𝑚= tr(𝑇𝑆) = tr(𝑆𝑇) = 𝑛. So in order for 𝑓to be a bijection, the dimensions of the spaces must match. Hence 𝑓′(𝑎) is an invertible matrix.
This proves that ℝ𝑚, ℝ𝑛are not homeomorphic in such a way that the maps between them are differentiable. We aim now to prove an inverse; if 𝑓is differentiable and 𝑓′ is invertible, then 𝑓is locally a bijection between neighbourhoods.
Definition. Let 𝑈⊂ℝ𝑚be open, and 𝑓∶𝑈→ℝ𝑛be a function. We say that 𝑓is differenti-able on 𝑈if 𝑓is differentiable at 𝑎for all 𝑎∈𝑈. Then, the derivative of 𝑓on 𝑈is the function 𝑓′ ∶𝑈→𝐿(ℝ𝑚, ℝ𝑛) mapping points to their derivatives. We say that 𝑓is a 𝐶1-function on 𝑈 if 𝑓is continuously differentiable on 𝑈; 𝑓is differentiable on 𝑈and 𝑓′ ∶𝑈→𝐿(ℝ𝑚, ℝ𝑛) is a continuous function.
Theorem. Let 𝑈⊂ℝ𝑛be open. Let 𝑓∶𝑈→ℝ𝑛be a 𝐶1-function. Let 𝑎∈𝑈, and let 𝑓′(𝑎) be an invertible linear map 𝑓′(𝑎)∶𝐿(ℝ𝑛). Then there exist open sets 𝑉, 𝑊such that 𝑎∈𝑉, 𝑓(𝑎) ∈𝑊, 𝑉⊂𝑈and 𝑓|𝑉∶𝑉→𝑊is a bijection with inverse function 𝑔∶𝑊→𝑉.
Further, 𝑔is a 𝐶1-function, and 𝑔′(𝑦) = [𝑓′(𝑔(𝑦))] −1 Proof. We first show that without loss of generality we can let 𝑎= 𝑓(𝑎) = 0 and 𝑓′(𝑎) = 𝐼.
To see this, let 𝑇= 𝑓′(𝑎) and define ℎ(𝑥) = 𝑇−1(𝑓(𝑥+ 𝑎) −𝑓(𝑎)). Then, ℎis defined on 207 IV. Analysis and Topology 𝑈−𝑎, which is open. In particular, 𝑈−𝑎is an open neighbourhood of zero. By the chain rule, ℎis differentiable with ℎ′(𝑥) = 𝑇−1 ∘𝑓′(𝑥+ 𝑎). For 𝑥, 𝑦∈𝑈−𝑎, we then have ‖ℎ′(𝑥) −ℎ′(𝑦)‖ = ‖ ‖𝑇−1 ∘(𝑓′(𝑎+ 𝑥) −𝑓′(𝑎+ 𝑦))‖ ‖ ≤‖ ‖𝑇−1‖ ‖ ⋅‖𝑓′(𝑎+ 𝑥) −𝑓′(𝑎+ 𝑦)‖ It then follows that ℎis a 𝐶1-function, and that ℎ(0) = 0, ℎ′(0) = 𝑇−1 ∘𝑇= 𝐼. We have transformed into a coordinate system where 𝑎= 𝑓(𝑎) = 0 and 𝑓′(𝑎) = 𝐼. If we can prove the result for this coordinate system, we can translate back using 𝑓(𝑥) = 𝑇(ℎ(𝑥−𝑎)) + 𝑓(𝑎).
Now, let 𝑓(0) = 0 and 𝑓′(0) = 𝐼. Since 𝑓′ is continuous, there exists 𝑟> 0 such that ℬ𝑟(0) ⊂𝑈and for all 𝑥∈𝑈, we have ‖𝑓′(𝑥) −𝑓′(0)‖ = ‖𝑓′(𝑥) −𝐼‖ ≤1 2 We intend to show that for all 𝑥, 𝑦∈ℬ𝑟(0), we have ‖𝑓(𝑥) −𝑓(𝑦)‖ ≥ 1 2‖𝑥−𝑦‖. Indeed, define 𝑝∶𝑈→ℝ𝑛by 𝑝(𝑥) = 𝑓(𝑥) −𝑥. Then 𝑝′(𝑥) = 𝑓′(𝑥) −𝐼. Then, ‖𝑝′(𝑥)‖ ≤ 1 2 for all 𝑥∈ℬ𝑟(0). By the mean value inequality, ‖𝑝(𝑥) −𝑝(𝑦)‖ ≤ 1 2‖𝑥−𝑦‖ for all 𝑥, 𝑦∈ℬ𝑟(0).
Hence, ‖𝑓(𝑥) −𝑓(𝑦)‖ = ‖(𝑝(𝑥) + 𝑥) −(𝑝(𝑦) + 𝑦)‖ ≥‖𝑥−𝑦‖ −‖𝑝(𝑥) −𝑝(𝑦)‖ ≥1 2‖𝑥−𝑦‖ So we have proven the bound as claimed. Now, let 𝑠= 𝑟 2. We will show that 𝑓(𝒟𝑟(0)) ⊂ 𝒟𝑠(0). More precisely, we will show that for all 𝑤∈𝒟𝑠(0) there exists a unique 𝑥∈𝒟𝑟(0) such that 𝑓(𝑥) = 𝑤. Let 𝑤∈𝒟𝑠(0) be fixed. We now define, for all 𝑥∈ℬ𝑟(0), the function 𝑞(𝑥) = 𝑤−𝑓(𝑥) + 𝑥= 𝑤−𝑝(𝑥). Note that 𝑓(𝑥) = 𝑤if and only if 𝑞(𝑥) = 𝑥. We will show that 𝑞is a contraction mapping, and that there exists a fixed point. Since 𝑝(0) = 𝑓(0)−0 = 0, we have for all 𝑥∈ℬ𝑟(0) that ‖𝑞(𝑥)‖ ≤‖𝑤‖ + ‖𝑝(𝑥)‖ = ‖𝑤‖ + ‖𝑝(𝑥) −𝑝(0)‖ ≤‖𝑤‖ + 1 2‖𝑥−0‖ = 1 2‖𝑥‖ < 𝑠+ 1 2𝑟 Hence, 𝑞(ℬ𝑟(0)) ⊂𝒟𝑟(0) ⊂ℬ𝑟(0). We now show 𝑞is a contraction mapping. For 𝑥, 𝑦∈ ℬ𝑟(0), we have ‖𝑞(𝑥) −𝑞(𝑦)‖ = ‖𝑝(𝑥) −𝑝(𝑦)‖ ≤1 2‖𝑥−𝑦‖ Hence 𝑞∶ℬ𝑟(0) →ℬ𝑟(0) really is a contraction mapping on the non-empty, complete metric space ℬ𝑟(0). By the contraction mapping theorem, there exists a unique 𝑥∈ℬ𝑟(0) such that 𝑞(𝑥) = 𝑥. But since 𝑞(ℬ𝑟(0)) ⊂𝒟𝑟(0), we must have 𝑥∈𝒟𝑟(0). In particular, there exists a unique 𝑥∈𝒟𝑟(0) such that 𝑓(𝑥) = 𝑤.
Now, let 𝑊= 𝒟𝑠(0), 𝑉= 𝒟𝑟(0) ∩𝑓−1(𝑊). Then, we will now show that 𝑓|𝑉∶𝑉→𝑊is a bijection with inverse 𝑔∶𝑊→𝑉which is continuous. First, 𝑊is open and 𝑓(0) = 0 ∈𝑊.
Since 𝑓is continuous, 𝑓−1(𝑊) is open. Hence 𝑉is open, as the intersection of two open sets. We have 0 ∈𝑉. By the previous paragraph, 𝑓|𝑉∶𝑉→𝑊is a bijection since for 208 11. Partial derivatives every point in 𝑊there exists a unique point in 𝑉mapping to it. Finally, let 𝑢, 𝑣∈𝑊. Let 𝑥= 𝑔(𝑢), 𝑦= 𝑔(𝑣). Then, ‖𝑔(𝑢) −𝑔(𝑣)‖ = ‖𝑥−𝑦‖ ≤2‖𝑓(𝑥) −𝑓(𝑦)‖ = 2‖𝑢−𝑣‖ Hence 𝑔is 2-Lipschitz and hence continuous. Now it suffices to show 𝑔is 𝐶1, and for all 𝑦∈𝑊we have 𝑔′(𝑦) = [𝑓′(𝑔(𝑦))] −1. This part of the proof is non-examinable.
209 IV. Analysis and Topology 12. Second derivatives 12.1. Definition Definition. Let 𝑈⊂ℝ𝑚be an open set, and 𝑓∶𝑈→ℝ𝑛. Let 𝑎∈𝑈. Suppose that there exists an open neighbourhood 𝑉of 𝑎contained within 𝑈, and 𝑓is differentiable on 𝑉. We say that 𝑓is twice differentiable at 𝑎if 𝑓′ ∶𝑉→𝐿(ℝ𝑚→ℝ𝑛) is differentiable at 𝑎.
We write 𝑓″(𝑎) for the derivative of 𝑓′ at 𝑎, called the second derivative of 𝑓at 𝑎. Note that 𝑓″(𝑎) ∈𝐿(ℝ𝑚, 𝐿(ℝ𝑚, ℝ𝑛)).
Remark. We can visualise the second derivative as a bilinear map instead of a nested se-quence of linear maps. Note, 𝐿(ℝ𝑚, 𝐿(ℝ𝑚, ℝ𝑛)) ∼Bil(ℝ𝑚× ℝ𝑚, ℝ𝑛) where Bil(𝑋×𝑌, 𝑍) is the vector space of bilinear maps from 𝑋×𝑌to 𝑍. For ℎ, 𝑘∈ℝ𝑚, and 𝑇is the second derivative, we can say 𝑇(ℎ)(𝑘) = ˜ 𝑇(ℎ, 𝑘) where ˜ 𝑇is a bilinear map. From now on, this bilinear map notation will be used, and 𝑇and ˜ 𝑇will be identified as the same.
Proposition. Let 𝑈⊂ℝ𝑚be open, 𝑓∶𝑈→ℝ𝑛be a function, and 𝑎∈𝑈. Let 𝑓be differ-entiable on an open neighbourhood 𝑉of 𝐴contained in 𝑈. Then 𝑓is twice differentiable at 𝑎if and only if there exists a bilinear map 𝑇∈Bil(ℝ𝑚×ℝ𝑚, ℝ𝑛) such that for every 𝑘∈ℝ𝑚, we have 𝑓′(𝑎+ ℎ)(𝑘) = 𝑓′(𝑎)(𝑘) + 𝑇(ℎ, 𝑘) + 𝑜(‖ℎ‖) Then 𝑇= 𝑓″(𝑎).
Proof. Suppose 𝑓is twice differentiable at 𝑎. Then 𝑓′ is differentiable at 𝑎. So, 𝑓′(𝑎+ ℎ) = 𝑓′(𝑎) + 𝑓″(𝑎)(ℎ) + ‖ℎ‖ ⋅𝜀(ℎ) All terms are linear maps 𝐿(ℝ𝑚, ℝ𝑛). In particular, 𝜀is defined on 𝑉−𝑎→𝐿(ℝ𝑚, ℝ𝑛) such that 𝜀(0) = 0 and 𝜀is continuous at zero. If we evaluate this equation at a fixed 𝑘∈ℝ𝑚, 𝑓′(𝑎+ ℎ)(𝑘) = 𝑓′(𝑎)(𝑘) + 𝑓″(𝑎)(ℎ, 𝑘) + ‖ℎ‖ ⋅𝜀(ℎ)(𝑘) Here, 𝑓″(𝑎) is a bilinear map. Further, ‖𝜀(ℎ)(𝑘)‖ ≤‖𝜀(ℎ)‖ ⋅‖𝑘‖ →0 Hence, ‖ℎ‖ ⋅𝜀(ℎ)(𝑘) = 𝑜(‖ℎ‖). Conversely, suppose 𝑇is a bilinear map and 𝑓′(𝑎+ ℎ)(𝑘) −𝑓′(𝑎)(𝑘) −𝑇(ℎ, 𝑘) ‖ℎ‖ →0 for any fixed 𝑘, as ℎ→0. We need to show that 𝜀(ℎ) = 𝑓′(𝑎+ ℎ) −𝑓′(𝑎) −𝑇(ℎ) ‖ℎ‖ →0 210 12. Second derivatives in the space 𝐿(ℝ𝑚, ℝ𝑛). We know that for a fixed 𝑘∈ℝ𝑚, 𝜀(ℎ)(𝑘) →0 in ℝ𝑛as ℎ→0. It then follows that ‖𝜀(ℎ)‖ = √ √ √ √ 𝑚 ∑ 𝑖=1 ‖𝜀(ℎ)(𝑒𝑖)‖2 →0 since we are in a finite-dimensional vector space.
Example. Let 𝑓∶ℝ𝑚→ℝ𝑛be linear. Then 𝑓is differentiable on ℝ𝑚with 𝑓′(𝑎) = 𝑓for all 𝑎. Hence 𝑓′ ∶ℝ𝑚→𝐿(ℝ𝑚, ℝ𝑛) sends 𝑎to 𝑓for all 𝑎. So this is a constant function, so has derivative 𝑓″(𝑎) = 0.
Example. Let 𝑓∶ℝ𝑚× ℝ𝑛→ℝ𝑝be bilinear. Then 𝑓is differentiable on ℝ𝑚× ℝ𝑛and for all (𝑎, 𝑏) ∈ℝ𝑚× ℝ𝑛, we have 𝑓′(𝑎, 𝑏)(ℎ, 𝑘) = 𝑓(𝑎, 𝑘) + 𝑓(ℎ, 𝑏) Note that this is linear in (𝑎, 𝑏) for a fixed (ℎ, 𝑘). Hence, 𝑓′ ∶ℝ𝑚× ℝ𝑛→𝐿(ℝ𝑚, ℝ𝑛, ℝ𝑝) is linear. Hence this is differentiable, and its derivative is 𝑓″(𝑎, 𝑏) = 𝑓′ ∈𝐿(ℝ𝑚, ℝ𝑛, 𝐿(ℝ𝑚× ℝ𝑛, ℝ𝑝)) ≃Bil((ℝ𝑚× ℝ𝑛) × (ℝ𝑚× ℝ𝑛), ℝ𝑝) Example. Let 𝑓∶𝑀𝑛→𝑀𝑛be defined by 𝑓(𝐴) = 𝐴3. Let 𝐴be fixed. Then, 𝑓(𝐴+ 𝐻) = (𝐴+ 𝐻)3 = 𝐴3 + 𝐴2𝐻+ 𝐴𝐻𝐴+ 𝐻𝐴2 + 𝐴𝐻2 + 𝐻𝐴𝐻+ 𝐻2𝐴+ 𝐻3 = 𝑓(𝐴) + (𝐴2𝐻+ 𝐴𝐻𝐴+ 𝐻𝐴2) + 𝑜(‖𝐻‖) Hence 𝑓is differentiable at 𝐴and 𝑓′(𝐴)(𝐻) = 𝐴2𝐻+ 𝐴𝐻𝐴+ 𝐻𝐴2 Thus, if 𝑛= 1, we have commutativity and hence 𝑓′(𝐴) = 3𝐴2. So 𝑓is differentiable on 𝑀𝑛.
For a fixed 𝐴and fixed 𝐾, the second derivative is given by 𝑓′(𝐴+ 𝐻)(𝐾) = (𝐴+ 𝐻)2𝐾+ (𝐴+ 𝐻)𝐾(𝐴+ 𝐻) + 𝐾(𝐴+ 𝐻)2 = (𝐴2𝐾+ 𝐴𝐾𝐴+ 𝐾𝐴2) ⏟⎵ ⎵ ⎵ ⎵ ⎵⏟⎵ ⎵ ⎵ ⎵ ⎵⏟ 𝑓′(𝐴)(𝐾) + (𝐴𝐻𝐾+ 𝐻𝐴𝐾+ 𝐴𝐾𝐻+ 𝐻𝐾𝐴+ 𝐾𝐴𝐻+ 𝐾𝐻𝐴) + (𝐻2𝐾+ 𝐻𝐾𝐻+ 𝐾𝐻2) The term 𝑇(𝐻, 𝐾) = (𝐴𝐻𝐾+ 𝐻𝐴𝐾+ 𝐴𝐾𝐻+ 𝐻𝐾𝐴+ 𝐾𝐴𝐻+ 𝐾𝐻𝐴) is bilinear in 𝐻and 𝐾as required. So the second derivative is 𝑇. In one dimension, this is equivalent to saying 𝑓″(𝐴) = 6𝐴.
211 IV. Analysis and Topology 12.2. Second derivatives and partial derivatives Let 𝑈be open in ℝ𝑛, let 𝑓∶𝑈→ℝ𝑛, and let 𝑎∈𝑈. Let 𝑓be twice differentiable at 𝑎, so 𝑓is differentiable on some open neighbourhood 𝑉of 𝑎contained within 𝑈, and 𝑓′ ∶𝑉→ 𝐿(ℝ𝑚, ℝ𝑛) is differentiable at 𝑎. Recall that 𝑓′(𝑎+ ℎ) = 𝑓′(𝑎) + 𝑓″(𝑎)(ℎ) + 𝑜(‖ℎ‖) Evaluating at a fixed 𝑘, 𝑓′(𝑎+ ℎ)(𝑘) = 𝑓′(𝑎)(𝑘) + 𝑓″(𝑎)(ℎ, 𝑘) + 𝑜(‖ℎ‖) Let 𝑢, 𝑣∈ℝ𝑚∖{0} be directions. Let 𝑘= 𝑣. Then, 𝑓′(𝑎+ ℎ)(𝑣) = 𝐷𝑣𝑓(𝑎+ ℎ) = 𝐷𝑣𝑓(𝑎) + 𝑓″(𝑎)(ℎ, 𝑣) + 𝑜(‖ℎ‖) Hence, the map 𝐷𝑣𝑓∶𝑉→ℝ𝑛maps 𝑥↦𝐷𝑣𝑓(𝑥) = 𝑓′(𝑥)(𝑣). Then this map is differenti-able at 𝑎and (𝐷𝑣𝑓)′(𝑎)(ℎ) = 𝑓″(𝑎)(ℎ, 𝑣) Hence there exist directional derivatives.
𝐷𝑢𝐷𝑣𝑓(𝑎) def = 𝐷𝑢(𝐷𝑣𝑓)(𝑎) = (𝐷𝑣𝑓)′(𝑎)(𝑢) = 𝑓″(𝑎)(𝑢, 𝑣) In particular, we have 𝐷𝑖𝐷𝑗𝑓(𝑎) = 𝑓″(𝑎)(𝑒𝑖, 𝑒𝑗) for 1 ≤𝑖, 𝑗≤𝑚.
12.3. Symmetry of mixed directional derivatives Theorem. Let 𝑈be open in ℝ𝑛, let 𝑓∶𝑈→ℝ𝑛, and let 𝑎∈𝑈. Let 𝑓be twice differentiable on an open set 𝑉with 𝑎∈𝑉⊂𝑈. Let 𝑓″ ∶𝑉→Bil(ℝ𝑚×ℝ𝑚, ℝ𝑛) be continuous at 𝑎. Then, for all directions 𝑢, 𝑣∈ℝ𝑚∖{0}, we have 𝐷𝑢𝐷𝑣𝑓(𝑎) = 𝐷𝑣𝐷𝑢𝑓(𝑎) Equivalently, 𝑓″(𝑎)(𝑢, 𝑣) = 𝑓″(𝑎)(𝑣, 𝑢) In other words, 𝑓″ is a symmetric bilinear map.
Proof. Without loss of generality we can let 𝑛= 1. Indeed, we have (𝐷𝑢𝑓)𝑗(𝑥) = [𝐷𝑢𝑓(𝑥)]𝑗= [𝑓′(𝑥)(𝑢)]𝑗= 𝑓′ 𝑗(𝑥)(𝑢) = 𝐷𝑢𝑓𝑗(𝑥) Hence, (𝐷𝑢𝑓)𝑗= 𝐷𝑢𝑓𝑗. For 𝑣: (𝐷𝑣𝐷𝑢𝑓)𝑗= 𝐷𝑣(𝐷𝑢𝑓)𝑗= 𝐷𝑣𝐷𝑢𝑓𝑗 212 12. Second derivatives So it is sufficient to show that 𝐷𝑣𝐷𝑢𝑓𝑗(𝑎) = 𝐷𝑢𝐷𝑣𝑓𝑗(𝑎). Now, consider 𝜙(𝑠, 𝑡) = 𝑓(𝑎+ 𝑠𝑢+ 𝑡𝑣) −𝑓(𝑎+ 𝑡𝑣) −𝑓(𝑎+ 𝑠𝑢) + 𝑓(𝑎) for 𝑠, 𝑡∈ℝ. Let 𝑠, 𝑡be fixed, and consider 𝜓(𝑦) = 𝑓(𝑎+ 𝑦𝑢+ 𝑡𝑣) −𝑓(𝑎+ 𝑦𝑢) Note that 𝜙(𝑠, 𝑡) can be written as 𝜙(𝑠, 𝑡) = 𝜓(𝑠) −𝜓(0) The term 𝜓(𝑠) −𝜓(0) can be interpreted as (𝑓(𝑎+ 𝑠𝑢+ 𝑡𝑣) −𝑓(𝑎+ 𝑡𝑣)) −(𝑓(𝑎+ 𝑠𝑢) −𝑓(𝑎)), which is the second difference given by the function when traversing the parallelogram with sides 𝑠𝑢, 𝑡𝑣. By the mean value theorem, there exists 𝛼(𝑠, 𝑡) ∈(0, 1) such that 𝜙(𝑠, 𝑡) = 𝜓(𝑠) −𝜓(0) = 𝑠𝜓′(𝛼𝑠) = 𝑠[𝐷𝑢𝑓(𝑎+ 𝛼𝑠𝑢+ 𝑡𝑣) −𝐷𝑢𝑓(𝑎+ 𝛼𝑠𝑢)] Now, applying the mean value theorem to the function 𝑦↦𝐷𝑢𝑓(𝑎+ 𝛼𝑠𝑢+ 𝑦𝑣), we have 𝜙(𝑠, 𝑡) = 𝑠𝑡𝐷𝑣𝐷𝑢𝑓(𝑎+ 𝛼𝑠𝑢+ 𝛽𝑡𝑣) for 𝛽(𝑠, 𝑡) ∈(0, 1). Now, 𝜙(𝑠, 𝑡) 𝑠𝑡 = 𝐷𝑣𝐷𝑢𝑓(𝑎+ 𝛼𝑠𝑢+ 𝛽𝑡𝑣) = 𝑓″(𝑎+ 𝛼𝑠𝑢+ 𝛽𝑡𝑣)(𝑢, 𝑣) Since 𝑓″ is continuous at 𝑎, we can let 𝑠, 𝑡→0 and find 𝜙(𝑠, 𝑡) 𝑠𝑡 →𝑓″(𝑎)(𝑢, 𝑣) Now, we can repeat the above using 𝜓(𝑦) = 𝑓(𝑎+ 𝑠𝑢+ 𝑦𝑣) −𝑓(𝑎+ 𝑦𝑣) This calculates the second difference from above, but using the other path. We can find 𝜙(𝑠, 𝑡) 𝑠𝑡 →𝑓″(𝑎)(𝑣, 𝑢) as required.
213 V. Methods Lectured in Michaelmas 2021 by Prof. E. P. Shellard In this course, we discuss various methods for solving differential equations. Different forms of differential equations need different solution strategies, and we study a wide range of common types of differential equation.
A particularly powerful method for solving differential equations involves the use of Green’s functions. For example, physical systems can involve bodies spread over space with constant density. Green’s functions allow the equation to be solved for a point mass, and then integ-rated to find the solution for the larger body.
Fourier transforms are another way to solve differential equations. Sometimes a differential equation is easier to solve after applying the Fourier transform to the relevant function, then the inverse Fourier transform recovers the solution to the original equation.
215 V. Methods Contents 1.
Fourier series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 219 1.1.
Periodic functions . . . . . . . . . . . . . . . . . . . . . . . . . 219 1.2.
Properties of trigonometric functions . . . . . . . . . . . . . . . 219 1.3.
Periodic function space . . . . . . . . . . . . . . . . . . . . . . 219 1.4.
Fourier series . . . . . . . . . . . . . . . . . . . . . . . . . . . 220 1.5.
Dirichlet conditions . . . . . . . . . . . . . . . . . . . . . . . . 221 1.6.
Integration . . . . . . . . . . . . . . . . . . . . . . . . . . . . 222 1.7.
Differentiation . . . . . . . . . . . . . . . . . . . . . . . . . . 223 1.8.
Parseval’s theorem . . . . . . . . . . . . . . . . . . . . . . . . 223 1.9.
Half-range series . . . . . . . . . . . . . . . . . . . . . . . . . 224 1.10.
Complex representation of Fourier series . . . . . . . . . . . . . 224 1.11.
Self-adjoint matrices . . . . . . . . . . . . . . . . . . . . . . . 225 1.12.
Solving inhomogeneous ODEs with Fourier series . . . . . . . . 226 2.
Sturm–Liouville theory . . . . . . . . . . . . . . . . . . . . . . . . 228 2.1.
Second-order linear ODEs . . . . . . . . . . . . . . . . . . . . 228 2.2.
Sturm–Liouville form . . . . . . . . . . . . . . . . . . . . . . . 228 2.3.
Converting to Sturm–Liouville form . . . . . . . . . . . . . . . 229 2.4.
Self-adjoint operators . . . . . . . . . . . . . . . . . . . . . . . 229 2.5.
Self-adjoint compatible boundary conditions . . . . . . . . . . . 230 2.6.
Properties of self-adjoint operators . . . . . . . . . . . . . . . . 230 2.7.
Real eigenvalues . . . . . . . . . . . . . . . . . . . . . . . . . 231 2.8.
Orthogonality of eigenfunctions . . . . . . . . . . . . . . . . . 231 2.9.
Eigenfunction expansions . . . . . . . . . . . . . . . . . . . . 232 2.10.
Completeness and Parseval’s identity . . . . . . . . . . . . . . . 233 2.11.
Legendre’s equation . . . . . . . . . . . . . . . . . . . . . . . . 234 2.12.
Properties of Legendre polynomials . . . . . . . . . . . . . . . . 235 2.13.
Legendre polynomials as eigenfunctions . . . . . . . . . . . . . 235 2.14.
Solving inhomogeneous differential equations . . . . . . . . . . 236 2.15.
Integral solutions . . . . . . . . . . . . . . . . . . . . . . . . . 236 2.16.
Waves on an elastic string . . . . . . . . . . . . . . . . . . . . . 237 3.
Separation of variables . . . . . . . . . . . . . . . . . . . . . . . . . 239 3.1.
Separation of variables . . . . . . . . . . . . . . . . . . . . . . 239 3.2.
Boundary conditions and normal modes . . . . . . . . . . . . . 239 3.3.
Initial conditions and temporal solutions . . . . . . . . . . . . . 240 3.4.
Separation of variables methodology . . . . . . . . . . . . . . . 241 3.5.
Energy of oscillations . . . . . . . . . . . . . . . . . . . . . . . 242 3.6.
Wave reflection and transmission . . . . . . . . . . . . . . . . . 243 216 3.7.
Wave equation in plane polar coordinates . . . . . . . . . . . . . 244 3.8.
Bessel’s equation . . . . . . . . . . . . . . . . . . . . . . . . . 245 3.9.
Asymptotic behaviour of Bessel functions . . . . . . . . . . . . . 246 3.10.
Zeroes of Bessel functions . . . . . . . . . . . . . . . . . . . . . 246 3.11.
Solving the vibrating drum . . . . . . . . . . . . . . . . . . . . 246 3.12.
Diffusion equation derivation with Fourier’s law . . . . . . . . . 248 3.13.
Diffusion equation derivation with statistical dynamics . . . . . . 248 3.14.
Similarity solutions . . . . . . . . . . . . . . . . . . . . . . . . 249 3.15.
Heat conduction in a finite bar . . . . . . . . . . . . . . . . . . 250 3.16.
Particular solution to diffusion equation . . . . . . . . . . . . . 251 3.17.
Laplace’s equation . . . . . . . . . . . . . . . . . . . . . . . . 252 3.18.
Laplace’s equation in three-dimensional Cartesian coordinates . . 252 3.19.
Laplace’s equation in plane polar coordinates . . . . . . . . . . . 254 3.20.
Laplace’s equation in cylindrical polar coordinates . . . . . . . . 255 3.21.
Laplace’s equation in spherical polar coordinates . . . . . . . . . 256 3.22.
Generating function for Legendre polynomials . . . . . . . . . . 257 4.
Green’s functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . 258 4.1.
Dirac 𝛿function . . . . . . . . . . . . . . . . . . . . . . . . . 258 4.2.
Integral and derivative of 𝛿function . . . . . . . . . . . . . . . 259 4.3.
Properties of 𝛿function . . . . . . . . . . . . . . . . . . . . . . 259 4.4.
Fourier series expansion of 𝛿function . . . . . . . . . . . . . . 260 4.5.
Arbitrary eigenfunction expansion of 𝛿function . . . . . . . . . 261 4.6.
Motivation for Green’s functions . . . . . . . . . . . . . . . . . 261 4.7.
Definition of Green’s function . . . . . . . . . . . . . . . . . . 263 4.8.
Explicit form for Green’s functions . . . . . . . . . . . . . . . . 264 4.9.
Solving boundary value problems . . . . . . . . . . . . . . . . . 264 4.10.
Higher-order ODEs . . . . . . . . . . . . . . . . . . . . . . . . 266 4.11.
Eigenfunction expansions of Green’s functions . . . . . . . . . . 266 4.12.
Constructing Green’s function for an initial value problem . . . . 266 5.
Fourier transforms . . . . . . . . . . . . . . . . . . . . . . . . . . . 268 5.1.
Definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . 268 5.2.
Converting Fourier series into Fourier transforms . . . . . . . . 269 5.3.
Properties of Fourier series . . . . . . . . . . . . . . . . . . . . 270 5.4.
Convolution theorem . . . . . . . . . . . . . . . . . . . . . . . 271 5.5.
Parseval’s theorem . . . . . . . . . . . . . . . . . . . . . . . . 272 5.6.
Fourier transforms of generalised functions . . . . . . . . . . . . 273 5.7.
Trigonometric functions . . . . . . . . . . . . . . . . . . . . . 274 5.8.
Heaviside functions . . . . . . . . . . . . . . . . . . . . . . . . 274 5.9.
Dirichlet discontinuous formula . . . . . . . . . . . . . . . . . 274 5.10.
Solving ODEs for boundary value problems . . . . . . . . . . . . 275 217 V. Methods 5.11.
Signal processing . . . . . . . . . . . . . . . . . . . . . . . . . 275 5.12.
General transfer functions for ODEs . . . . . . . . . . . . . . . 276 5.13.
Damped oscillator . . . . . . . . . . . . . . . . . . . . . . . . 277 5.14.
Discrete sampling and the Nyquist frequency . . . . . . . . . . . 278 5.15.
Nyquist–Shannon sampling theorem . . . . . . . . . . . . . . . 278 5.16.
Discrete Fourier transform . . . . . . . . . . . . . . . . . . . . 279 5.17.
Fast Fourier transform (non-examinable) . . . . . . . . . . . . . 281 6.
Method of characteristics . . . . . . . . . . . . . . . . . . . . . . . 282 6.1.
Well-posed Cauchy problems . . . . . . . . . . . . . . . . . . . 282 6.2.
Method of characteristics . . . . . . . . . . . . . . . . . . . . . 282 6.3.
Characteristics of a first order PDE . . . . . . . . . . . . . . . . 283 6.4.
Inhomogeneous first order PDEs . . . . . . . . . . . . . . . . . 284 6.5.
Classification of second order PDEs . . . . . . . . . . . . . . . . 285 6.6.
Characteristic curves of second order PDEs . . . . . . . . . . . . 286 6.7.
Characteristic coordinates . . . . . . . . . . . . . . . . . . . . 287 6.8.
General solution to wave equation . . . . . . . . . . . . . . . . 288 7.
Solving partial differential equations with Green’s functions . . . 289 7.1.
Diffusion equation and Fourier transform . . . . . . . . . . . . 289 7.2.
Gaussian pulse for heat equation . . . . . . . . . . . . . . . . . 290 7.3.
Forced diffusion equation . . . . . . . . . . . . . . . . . . . . . 290 7.4.
Duhamel’s principle . . . . . . . . . . . . . . . . . . . . . . . 291 7.5.
Forced wave equation . . . . . . . . . . . . . . . . . . . . . . . 292 7.6.
Poisson’s equation . . . . . . . . . . . . . . . . . . . . . . . . 293 7.7.
Green’s identities . . . . . . . . . . . . . . . . . . . . . . . . . 294 7.8.
Dirichlet Green’s function . . . . . . . . . . . . . . . . . . . . 295 7.9.
Method of images for Laplace’s equation . . . . . . . . . . . . . 295 7.10.
Method of images for wave equation . . . . . . . . . . . . . . . 296 218 1. Fourier series 1. Fourier series 1.1. Periodic functions A function 𝑓(𝑥) is periodic if 𝑓(𝑥+ 𝑇) = 𝑓(𝑥) for all 𝑥, where 𝑇is the period. For example, simple harmonic motion is periodic. In space, we consider the wavelength 𝜆= 2𝜋 𝑘, and the (angular) wave number 𝑘is defined conversely by 𝑘= 2𝜋 𝜆.
1.2. Properties of trigonometric functions Consider the set of functions 𝑔𝑛(𝑥) = cos 𝑛𝜋𝑥 𝐿; ℎ𝑛(𝑥) = sin 𝑛𝜋𝑥 𝐿 where 𝑛∈ℕ. These functions are periodic with period 𝑇= 2𝐿. Recall that cos 𝐴cos 𝐵= 1 2(cos(𝐴−𝐵) + cos(𝐴+ 𝐵)); sin 𝐴sin 𝐵= 1 2(cos(𝐴−𝐵) −cos(𝐴+ 𝐵)); sin 𝐴cos 𝐵= 1 2(sin(𝐴−𝐵) + sin(𝐴+ 𝐵)) 1.3. Periodic function space We define the inner product ⟨𝑓, 𝑔⟩= ∫ 2𝐿 0 𝑓(𝑥)𝑔(𝑥) d𝑥 The functions 𝑔𝑛and ℎ𝑛are mutually orthogonal on the interval [0, 2𝐿) with respect to the inner product above.
⟨ℎ𝑛, ℎ𝑚⟩= ∫ 2𝐿 0 sin 𝑛𝜋𝑥 𝐿 sin 𝑚𝜋𝑥 𝐿 d𝑥 = 1 2 ∫ 2𝐿 0 (cos (𝑛−𝑚)𝜋𝑥 𝐿 −cos (𝑛+ 𝑚)𝜋𝑥 𝐿 ) d𝑥 = 1 2 𝐿 𝜋[ 1 𝑛−𝑚sin (𝑛−𝑚)𝜋𝑥 𝐿 − 1 𝑛+ 𝑚sin (𝑛+ 𝑚)𝜋𝑥 𝐿 ] 2𝐿 0 = 0 when 𝑛≠𝑚 If 𝑛= 𝑚, we have ⟨ℎ𝑛, ℎ𝑛⟩= ∫ 2𝐿 0 sin2 𝑛𝜋𝑥 𝐿 d𝑥= 1 2 ∫ 2𝐿 0 (1 −cos 2𝜋𝑛𝑥 𝐿 ) d𝑥= 𝐿 219 V. Methods Thus, ⟨ℎ𝑛, ℎ𝑚⟩= {𝐿𝛿𝑛𝑚 𝑛, 𝑚≠0 0 𝑛𝑚= 0 Similarly, we can show ⟨𝑔𝑛, 𝑔𝑚⟩= ⎧ ⎨ ⎩ 𝐿𝛿𝑛𝑚 𝑛, 𝑚≠0 0 exactly one of 𝑚, 𝑛is zero 2𝐿 𝑛, 𝑚= 0 and ⟨ℎ𝑛, 𝑔𝑚⟩= 0 Now, we assert that {𝑔𝑛, ℎ𝑛} form a complete orthogonal set; they span the space of all ‘well-behaved’ periodic functions of period 2𝐿. Further, the set {𝑔𝑛, ℎ𝑛} is linearly independ-ent.
1.4. Fourier series Since 𝑔𝑛, ℎ𝑛span the space of ‘well-behaved’ periodic functions of period 2𝐿, we can express any such function as a sum of such eigenfunctions.
Definition. The Fourier series of 𝑓is 𝑓(𝑥) = 1 2𝑎0 + ∞ ∑ 𝑛=1 𝑎𝑛cos 𝑛𝜋𝑥 𝐿 + ∞ ∑ 𝑛=1 𝑏𝑛sin 𝑛𝜋𝑥 𝐿 where 𝑎𝑛, 𝑏𝑛are constants such that the right hand side is convergent for all 𝑥where 𝑓is con-tinuous. At a discontinuity 𝑥, the Fourier series approaches the midpoint of the supremum and infimum of the function in a close neighbourhood of 𝑥. That is, we replace the left hand side with 1 2𝑓(𝑥+) + 1 2𝑓(𝑥−) Let 𝑚> 0, and consider taking the inner product ⟨ℎ𝑚, 𝑓⟩and substituting the Fourier series of 𝑓.
⟨ℎ𝑚, 𝑓⟩= ∫ 2𝐿 0 sin 𝑚𝜋𝑥 𝐿 𝑓(𝑥) d𝑥 = ⟨ℎ𝑚, 𝑏𝑚ℎ𝑚⟩ = 𝐿𝑏𝑚 Thus, 𝑏𝑛= 1 𝐿⟨ℎ𝑛, 𝑓⟩= 1 𝐿∫ 2𝐿 0 sin 𝑛𝜋𝑥 𝐿𝑓(𝑥) d𝑥 220 1. Fourier series and analogously 𝑎𝑛= 1 𝐿⟨𝑔𝑛, 𝑓⟩= 1 𝐿∫ 2𝐿 0 cos 𝑛𝜋𝑥 𝐿𝑓(𝑥) d𝑥 Note that 1 2𝑎0 is the average of the function. Note further that we may integrate over any range as long as the total length is one period, 2𝐿. Notably, we may integrate over the interval [−𝐿, 𝐿].
Example. Consider the sawtooth wave; defined by 𝑓(𝑥) = 𝑥for 𝑥∈[−𝐿, 𝐿) and periodic elsewhere. Here, 𝑎𝑛= 1 𝐿∫ 𝐿 −𝐿 𝑥cos 𝑛𝜋𝑥 𝐿 d𝑥= 0 and 𝑏𝑛= 2 𝐿∫ 𝐿 0 𝑥sin 𝑛𝜋𝑥 𝐿 d𝑥 = −2 𝑛𝜋[𝑥cos 𝑛𝜋𝑥 𝐿] 𝐿 0 + 2 𝑛𝜋∫ 𝐿 0 cos 𝑛𝜋𝑥 𝐿 d𝑥 = −2𝐿 𝑛𝜋cos 𝑛𝜋+ 2𝐿 (𝑛𝜋)2 sin 𝑛𝜋 = 2𝐿 𝑛𝜋(−1)𝑛+1 1.5. Dirichlet conditions The Dirichlet conditions are sufficiency conditions for a well-behaved function, that will imply the existence of a unique Fourier series.
Theorem. If 𝑓(𝑥) is a bounded periodic function of period 2𝐿with a finite number of min-ima, maxima and discontinuities in [0, 2𝐿), then the Fourier series converges to 𝑓at all points at which 𝑓is continuous, and at discontinuities the series converges to the midpoint.
Remark.
(i) These are some relatively weak conditions for convergence, compared to Taylor series. However, this definition still eliminates pathological functions such as 1 𝑥, sin 1 𝑥, 𝟙(ℚ) and so on.
(ii) The converse is not true; for example, sin 1 𝑥does in fact have a Fourier series.
(iii) The proof is difficult and will not be given.
The rate of convergence of the Fourier series depends on the smoothness of the function.
Theorem. If 𝑓(𝑥) has continuous derivatives up to a 𝑝th derivative which is discontinuous, then the Fourier series converges with order 𝑂(𝑛−(𝑝+1)) as 𝑛→∞.
221 V. Methods Example (𝑝= 0). Consider the square wave 𝑓(𝑥) = {1 0 ≤𝑥< 1 −1 −1 ≤𝑥< 0 Then the Fourier series is 𝑓(𝑥) = 4 ∞ ∑ 𝑚=1 sin(2𝑚−1)𝜋𝑥 (2𝑚−1)𝜋 Example (𝑝= 1). Consider the general ‘seesaw’ wave, defined by 𝑓(𝑥) = {𝑥(1 −𝜉) 0 ≤𝑥< 𝜉 𝜉(1 −𝑥) 𝜉≤𝑥< 1 and defined as an odd function for −1 ≤𝑥< 0. The Fourier series is 𝑓(𝑥) = 2 ∞ ∑ 𝑚=1 sin 𝑛𝜋𝜉sin 𝑛𝜋𝑥 (𝑛𝜋)2 For instance, if 𝜉= 1 2, we can show that 𝑓(𝑥) = 2 ∞ ∑ 𝑚=1 (−1)𝑚+1 sin(2𝑚−1)𝜋𝑥 ((2𝑚−1)𝜋)2 Example (𝑝= 2). Let 𝑓(𝑥) = 1 2𝑥(1 −𝑥) for 0 ≤𝑥< 1, and defined as an odd function for −1 ≤𝑥< 0. We can show that 𝑓(𝑥) = 4 ∞ ∑ 𝑛=1 sin(2𝑚−1)𝜋𝑥 ((2𝑚−1)𝜋)3 Example (𝑝= 3). Consider 𝑓(𝑥) = (1 −𝑥2)2 with Fourier series 𝑎𝑛= 𝑂( 1 𝑛4 ) 1.6. Integration It is always valid to take the integral of a Fourier series term by term. Defining 𝐹(𝑥) = ∫ 𝑥 −𝐿𝑓(𝑥) d𝑥, we can show that 𝐹satisfies the Dirichlet conditions if 𝑓does. For instance, a jump discontinuity becomes continuous in the integral.
222 1. Fourier series 1.7. Differentiation Differentiating term by term is not always valid. For example, consider the square wave above: 𝑓(𝑥) ?
= 4 ∞ ∑ 𝑚=1 cos(2𝑚−1)𝜋𝑥 which is an unbounded series.
Theorem. If 𝑓(𝑥) is continuous and satisfies the Dirichlet conditions, and 𝑓′(𝑥) also sat-isfies the Dirichlet conditions, then 𝑓′(𝑥) can be found term by term by differentiating the Fourier series of 𝑓(𝑥).
Example. We can differentiate the seesaw function with 𝜉= 1 2, even though the derivative is not continuous. The result is an offset square wave, or by mapping 𝑥↦𝑥+ 1 2 we recover the original square wave.
1.8. Parseval’s theorem Parseval’s theorem relates the integral of the square of a function with the squares of the function’s Fourier series coefficients.
Theorem. Suppose 𝑓has Fourier coefficients 𝑎𝑖, 𝑏𝑖. Then ∫ 2𝐿 0 [𝑓(𝑥)]2 d𝑥= ∫ 2𝐿 0 [1 2𝑎0 + ∞ ∑ 𝑛=1 𝑎𝑘cos 𝑘𝜋𝑥 𝐿 + ∞ ∑ 𝑛=1 𝑏𝑛sin 𝑛𝜋𝑥 𝐿] 2 d𝑥 We can remove cross terms, since the basis functions are orthogonal.
= ∫ 2𝐿 0 [1 4𝑎2 0 + ∞ ∑ 𝑛=1 𝑎2 𝑛cos2 𝑛𝜋𝑥 𝐿 + ∞ ∑ 𝑛=1 𝑏2 𝑛sin2 𝑛𝜋𝑥 𝐿] d𝑥 = 𝐿[1 2𝑎2 0 + ∞ ∑ 𝑛=1 (𝑎2 𝑛+ 𝑏2 𝑛)] This is also called the completeness relation: the left hand side is greater than or equal to the right hand side if any of the basis functions are missing.
Example. Let us apply Parseval’s theorem to the sawtooth wave.
∫ 𝐿 −𝐿 [𝑓(𝑥)]2 d𝑥= ∫ 𝐿 −𝐿 𝑥2 d𝑥= 2 3𝐿 3 The right hand side gives 𝐿 ∞ ∑ 𝑛=1 4𝐿 2 𝑛2𝜋2 = 4𝐿 3 𝜋2 ∞ ∑ 𝑛=1 1 𝑛2 223 V. Methods Parseval’s theorem then implies ∞ ∑ 𝑛=1 1 𝑛2 = 𝜋2 6 Remark. Parseval’s theorem for functions is equivalent to Pythagoras’ theorem for vectors in ℝ𝑛: we can find the norm of a linear combination by computing the sum of the norms of the components.
1.9. Half-range series Consider 𝑓(𝑥) defined only on 0 ≤𝑥< 𝐿. We can extend the range of 𝑓to be the full range −𝐿≤𝑥< 𝐿in two simple ways: (i) require 𝑓to be odd, so 𝑓(−𝑥) = −𝑓(𝑥). Hence, 𝑎𝑛= 0 and 𝑏𝑛= 2 𝐿∫ 𝐿 0 𝑓(𝑥) sin 𝑛𝜋𝑥 𝐿 d𝑥 So 𝑓(𝑥) = ∞ ∑ 𝑛=1 𝑏𝑛sin 𝑛𝜋𝑥 𝐿 which is called a Fourier sine series.
(ii) require 𝑓to be even, so 𝑓(−𝑥) = 𝑓(𝑥). In this case, 𝑏𝑛= 0 and 𝑎𝑛= 2 𝐿∫ 𝐿 0 𝑓(𝑥) cos 𝑛𝜋𝑥 𝐿 d𝑥 and 𝑆𝑜𝑓(𝑥) = 1 2𝑎0 + ∞ ∑ 𝑛=1 𝑎𝑛cos 𝑛𝜋𝑥 𝐿 which is a Fourier cosine series.
1.10. Complex representation of Fourier series Recall that cos 𝑛𝜋𝑥 𝐿 = 1 2(𝑒𝑖𝑛𝜋𝑥/𝐿+ 𝑒−𝑖𝑛𝜋𝑥/𝐿); sin 𝑛𝜋𝑥 𝐿 = 1 2𝑖(𝑒𝑖𝑛𝜋𝑥/𝐿−𝑒−𝑖𝑛𝜋𝑥/𝐿) 224 1. Fourier series Therefore, a Fourier series can be written as 𝑓(𝑥) = 1 2𝑎0 + 1 2 ∞ ∑ 𝑛=1 [(𝑎𝑛−𝑖𝑏𝑛)𝑒𝑖𝑛𝜋𝑥/𝐿+ (𝑎𝑛+ 𝑖𝑏𝑛)𝑒−𝑖𝑛𝜋𝑥/𝐿] = ∞ ∑ 𝑚=−∞ 𝑐𝑚𝑒𝑖𝑚𝜋𝑥/𝐿 where for 𝑚> 0 we have 𝑚= 𝑛, 𝑐𝑚= 1 2(𝑎𝑛−𝑖𝑏𝑛), and for 𝑚< 0 we have 𝑛= −𝑚, 𝑐𝑚= 1 2(𝑎−𝑚+ 𝑖𝑏−𝑚), and where 𝑚= 0 we have 𝑐0 = 1 2𝑎0. In particular, 𝑐𝑚= 1 2𝐿∫ 𝐿 −𝐿 𝑓(𝑥)𝑒−𝑖𝑚𝜋𝑥/𝐿d𝑥 where the negative sign comes from the complex conjugate. This is because, for complex-valued 𝑓, 𝑔, we have ⟨𝑓, 𝑔⟩= ∫ 𝐿 −𝐿 𝑓⋆𝑔d𝑥 The orthogonality conditions are ∫ 𝐿 −𝐿 𝑒−𝑖𝑚𝜋𝑥/𝐿𝑒𝑖𝑛𝜋𝑥/𝐿d𝑥= 2𝐿𝛿𝑚𝑛 Parseval’s theorem now states ∫ 𝐿 −𝐿 𝑓⋆(𝑥)𝑓(𝑥) d𝑥= ∫ 𝐿 −𝐿 |𝑓(𝑥)|2 d𝑥= 2𝐿 ∞ ∑ 𝑚=−∞ |𝑐𝑚|2 1.11. Self-adjoint matrices Much of this section is a recap of IA Vectors and Matrices. Suppose that 𝑢, 𝑣∈ℂ𝑁with inner product ⟨𝑢, 𝑣⟩= 𝑢†𝑣 The 𝑁× 𝑁matrix 𝐴is self-adjoint, or Hermitian, if ∀𝑢, 𝑣∈ℂ𝑁, ⟨𝐴𝑢, 𝑣⟩= ⟨𝑢, 𝐴𝑣⟩⟺𝐴† = 𝐴 The eigenvalues 𝜆𝑛and eigenvectors 𝑣𝑛satisfy 𝐴𝑣𝑛= 𝜆𝑛𝑣𝑛 They have the following properties: (i) 𝜆⋆ 𝑛= 𝜆𝑛; 225 V. Methods (ii) 𝜆𝑛≠𝜆𝑚⟹⟨𝑣𝑛, 𝑣𝑚⟩= 0; (iii) we can create an orthonormal basis from the eigenvectors.
Given 𝑏∈ℂ𝑛, we can solve for 𝑥in the general matrix equation 𝐴𝑥= 𝑏by expressing 𝑏in terms of the eigenvector basis: 𝑏= 𝑁 ∑ 𝑛=1 𝑏𝑛𝑣𝑛 We seek a solution of the form 𝑥= 𝑁 ∑ 𝑛=1 𝑐𝑛𝑣𝑛 At this point, the 𝑏𝑛are known and the 𝑐𝑛are our target. Substituting into the matrix equa-tion, orthogonality of basis vectors gives 𝐴 𝑁 ∑ 𝑛=1 𝑐𝑛𝑣𝑛= 𝑁 ∑ 𝑛=1 𝑏𝑛𝑣𝑛 𝑁 ∑ 𝑛=1 𝑐𝑛𝜆𝑛𝑣𝑛= 𝑁 ∑ 𝑛=1 𝑏𝑛𝑣𝑛 𝑐𝑛𝜆𝑛= 𝑏𝑛 𝑐𝑛= 𝑏𝑛 𝜆𝑛 Therefore, 𝑥= 𝑁 ∑ 𝑛=1 𝑏𝑛 𝜆𝑛 𝑣𝑛 provided 𝜆𝑛≠0, or equivalently, the matrix is invertible.
1.12. Solving inhomogeneous ODEs with Fourier series We wish to find 𝑦(𝑥) given a source term 𝑓(𝑥) for the general differential equation ℒ𝑦≡−d2𝑦 d𝑥2 = 𝑓(𝑥) with boundary conditions 𝑦(0) = 𝑦(𝐿) = 0. The related eigenvalue problem is ℒ𝑦𝑛= 𝜆𝑛𝑦𝑛, 𝑦𝑛(0) = 𝑦𝑛(𝐿) = 0 which has solutions 𝑦𝑛(𝑥) = sin 𝑛𝜋𝑥 𝐿, 𝜆𝑛= (𝑛𝜋 𝐿) 2 226 1. Fourier series We can show that this is a self-adjoint linear operator with orthogonal eigenfunctions. We seek solutions of the form of a half-range sine series. Consider 𝑦(𝑥) = ∞ ∑ 𝑛=1 𝑐𝑛sin 𝑛𝜋𝑥 𝐿 The right hand side is 𝑓(𝑥) = ∞ ∑ 𝑛=1 𝑏𝑛sin 𝑛𝜋𝑥 𝐿 We can find 𝑏𝑛by 𝑏𝑛= 2 𝐿∫ 𝐿 0 𝑓(𝑥) sin 𝑛𝜋𝑥 𝐿 d𝑥 Substituting, we have ℒ𝑦= −d2 d𝑥2 (∑ 𝑛 𝑐𝑛sin 𝑛𝜋𝑥 𝐿) = ∑ 𝑛 𝑐𝑛(𝑛𝜋 𝐿) 2 sin 𝑛𝜋𝑥 𝐿 = ∑ 𝑛 𝑏𝑛sin 𝑛𝜋𝑥 𝐿 By orthogonality, 𝑐𝑛(𝑛𝜋 𝐿) 2 = 𝑏𝑛⟹𝑐𝑛= ( 𝐿 𝑛𝜋) 2 𝑏𝑛 Therefore the solution is 𝑦(𝑥) = ∑ 𝑛 ( 𝐿 𝑛𝜋) 2 𝑏𝑛sin 𝑛𝜋𝑥 𝐿 = ∑ 𝑛 𝑏𝑛 𝜆𝑛 𝑦𝑛 which is equivalent to the solution we found for self-adjoint matrices for which the eigen-values and eigenvectors are known.
Example. Consider an odd square wave with 𝐿= 1, so 𝑓(𝑥) = 1 from 0 ≤𝑥< 1.
𝑓(𝑥) = 4 ∑ 𝑚 sin(2𝑚−1)𝜋𝑥 (2𝑚−1)𝜋 Then the solution to ℒ𝑦= 𝑓should be (with odd 𝑛= 2𝑚−1) 𝑦(𝑥) = ∑ 𝑛 𝑏𝑛 𝜆𝑛 𝑦𝑛= 4 ∑ 𝑛 sin(2𝑚−1)𝜋𝑥 ((2𝑚−1)𝜋)3 This is exactly the Fourier series for 𝑦(𝑥) = 1 2𝑥(1 −𝑥) so this 𝑦is the solution to the differential equation. We can in fact integrate ℒ𝑦= 1 directly with the boundary conditions to verify the solution. We can also differentiate the Fourier series for 𝑦twice to find the square wave.
227 V. Methods 2. Sturm–Liouville theory 2.1. Second-order linear ODEs This section is a review of IA Differential Equations.
We wish to solve a general inhomogeneous ODE, written ℒ𝑦≡𝛼(𝑥)𝑦″ + 𝛽(𝑥)𝑦′ + 𝛾(𝑥)𝑦= 𝑓(𝑥) The homogeneous version has 𝑓(𝑥) = 0, so ℒ𝑦= 0, which has two independent solutions 𝑦1, 𝑦2. The general solution, also the complementary function for the inhomogeneous ODE, is 𝑦𝑐(𝑥) = 𝐴𝑦1(𝑥) + 𝐵𝑦2(𝑥). The inhomogeneous equation ℒ𝑦= 𝑓(𝑥) has a solution called the particular integral, denoted 𝑦𝑝(𝑥). The general solution to this equation is then 𝑦𝑝+ 𝑦𝑐.
We need two boundary or initial conditions to find the particular solution to the differential equation. Suppose 𝑥∈[𝑎, 𝑏]. We can create boundary conditions by defining 𝑦(𝑎), 𝑦(𝑏), often called the Dirichlet conditions. Alternatively, we can consider 𝑦(𝑎), 𝑦′(𝑎), called the Neumann conditions. We could also used some kind of mixed condition, for instance 𝑦+ 𝑘𝑦′. Homogeneous boundary conditions are such that 𝑦(𝑎) = 𝑦(𝑏) = 0. In this part of the course, homogeneous boundary conditions are often assumed. Note that we can add a complementary function 𝑦𝑐to the solution, for instance 𝑦= 𝑦+ 𝐴𝑦1 + 𝐵𝑦2 such that 𝑦(𝑎) = 𝑦(𝑏) = 0. This would allow us to construct homogeneous boundary conditions even when they are not present a priori in the problem. We could also specify initial data, such as solving for 𝑥≥𝑎, given 𝑦, 𝑦′ at 𝑥= 𝑎.
To solve the inhomogeneous equation, we want to use eigenfunction expansions such as Fourier series. In order to do this, we must first solve the related eigenvalue problem. In this case, that is 𝛼(𝑥)𝑦″ + 𝛽(𝑥)𝑦′ + 𝛾(𝑥)𝑦= −𝜆𝜌(𝑥)𝑦 We must solve this equation with the same boundary conditions as the original problem.
This form of equation often arises as a result of applying a separation of variables, particu-larly for PDEs in several dimensions.
2.2. Sturm–Liouville form For two complex-valued functions 𝑓, 𝑔on [𝑎, 𝑏], we define the inner product as ⟨𝑓, 𝑔⟩= ∫ 𝑏 𝑎 𝑓⋆(𝑥)𝑔(𝑥) d𝑥 The eigenvalue problem above greatly simplifies if ℒis self-adjoint, that is, if it can be ex-pressed in Sturm–Liouville form: ℒ𝑦≡(−𝑝𝑦′)′ + 𝑞𝑦= 𝜆𝑤𝑦 𝜆is an eigenvalue, and 𝑤is the weight function, which must be non-negative.
228 2. Sturm–Liouville theory 2.3. Converting to Sturm–Liouville form Suppose we have the eigenvalue problem 𝛼(𝑥)𝑦″ + 𝛽(𝑥)𝑦′ + 𝛾(𝑥)𝑦= −𝜆𝜌(𝑥)𝑦 Multiply this by an integrating factor 𝐹to give 𝐹𝛼𝑦″ + 𝐹𝛽𝑦′ + 𝐹𝛾𝑦= −𝜆𝐹𝜌𝑦 d d𝑥(𝐹𝛼𝑦′) −𝐹′𝛼𝑦′ −𝐹𝛼′𝑦+ 𝐹𝛽𝑦′ + 𝐹𝛾𝑦= −𝜆𝐹𝜌𝑦 To eliminate the 𝑦′ term, we require 𝐹′𝛼= 𝐹(𝛽−𝛼′). Thus, 𝐹′ 𝐹= 𝛽−𝛼′ 𝛼 ⟹𝐹= exp ∫ 𝑥𝛽−𝛼′ 𝛼 d𝑥 and further, (𝐹𝛼𝑦′)′ + 𝐹𝛾𝑦= −𝜆𝐹𝜌𝑦 hence 𝑝= 𝐹𝛼 𝑞= 𝐹𝛾 𝑤= 𝐹𝜌 and 𝐹(𝑥) > 0 hence 𝑤> 0.
Example. Consider the Hermite equation, 𝑦″ −2𝑥𝑦′ + 2𝑛𝑦= 0 In this case, 𝐹= exp ∫ 𝑥−2𝑥 1 d𝑥= 𝑒−𝑥2 Then the equation, in Sturm–Liouville form, is ℒ𝑦≡−(𝑒−𝑥2𝑦′) ′ = 2𝑛𝑒−𝑥2𝑦 2.4. Self-adjoint operators ℒis a self-adjoint operator on [𝑎, 𝑏] for all pairs of functions 𝑦1, 𝑦2 satisfying appropriate boundary conditions if ⟨𝑦1, ℒ𝑦2⟩= ⟨ℒ𝑦1, 𝑦2⟩ Written explicitly, ∫ 𝑏 𝑎 𝑦⋆ 1(𝑥)ℒ𝑦2(𝑥) d𝑥= ∫ 𝑏 𝑎 (ℒ𝑦1(𝑥))⋆𝑦2(𝑥) d𝑥 229 V. Methods Substituting Sturm–Liouville form into the above, ⟨𝑦1, ℒ𝑦2⟩−⟨ℒ𝑦1, 𝑦2⟩= ∫ 𝑏 𝑎 [−𝑦1(𝑝𝑦′ 2)′ + 𝑦1𝑞𝑦2 + 𝑦2(𝑝𝑦′ 1)′ −𝑦2𝑞𝑦1] d𝑥 = ∫ 𝑏 𝑎 [−𝑦1(𝑝𝑦′ 2)′ + 𝑦1𝑞𝑦2 + 𝑦2(𝑝𝑦′ 1)′ −𝑦2𝑞𝑦1] d𝑥 = ∫ 𝑏 𝑎 [−𝑦1(𝑝𝑦′ 2)′ + 𝑦2(𝑝𝑦′ 1)′] d𝑥 Adding −𝑦′ 1𝑝𝑦′ 2 + 𝑦′ 1𝑝𝑦′ 2, = ∫ 𝑏 𝑎 [−(𝑝𝑦1𝑦′ 2)′ + (𝑝𝑦′ 1𝑦2)′] d𝑥 = [−𝑝𝑦1𝑦′ 2 + 𝑝𝑦′ 1𝑦2]𝑏 𝑎 which must be zero for an equation in Sturm–Liouville form to be self-adjoint.
2.5. Self-adjoint compatible boundary conditions • Suppose 𝑦(𝑎) = 𝑦(𝑏) = 0. Then certainly the Sturm–Liouville form of the differential equation is self-adjoint. We could also choose 𝑦′(𝑎) = 𝑦′(𝑏) = 0. Collectively, the act of using homogeneous boundary conditions is known as the regular Sturm–Liouville problem.
• Periodic boundary conditions could also be used, such as 𝑦(𝑎) = 𝑦(𝑏).
• If 𝑎and 𝑏are singular points of the equation, i.e. 𝑝(𝑎) = 𝑝(𝑏) = 0, this is self-adjoint compatible.
• We could also have combinations of the above properties, one at 𝑎and one at 𝑏.
2.6. Properties of self-adjoint operators The following properties hold for any self-adjoint differential operator ℒ.
(i) The eigenvalues 𝜆𝑛are real.
(ii) The eigenfunctions 𝑦𝑛are orthogonal.
(iii) The 𝑦𝑛are a complete set; they span the space of all functions hence our general solu-tion can be written in terms of these eigenfunctions.
Each property is proven in its own subsection.
230 2. Sturm–Liouville theory 2.7. Real eigenvalues Proof. Suppose we have some eigenvalue 𝜆𝑛, so ℒ𝑦𝑛= 𝜆𝑛𝑤𝑦𝑛. Taking the complex conjug-ate, ℒ𝑦⋆ 𝑛= 𝜆⋆ 𝑛𝑤𝑦⋆ 𝑛, since ℒ, 𝑤are real. Now, consider ∫ 𝑏 𝑎 (𝑦⋆ 𝑛ℒ𝑦𝑛−𝑦𝑛ℒ𝑦⋆ 𝑛) d𝑥 which must be zero if ℒis self-adjoint. This can be written as (𝜆𝑛−𝜆⋆ 𝑛) ∫ 𝑏 𝑎 𝑤𝑦⋆ 𝑛𝑦𝑛d𝑥 The integral is nonzero, hence 𝜆𝑛−𝜆⋆ 𝑛= 0 which implies 𝜆𝑛is real. Note, if the 𝜆𝑛are non-degenerate (simple), i.e. with a unique eigenfunction 𝑦𝑛, then 𝑦⋆ 𝑛= 𝑦𝑛hence they are real.
We can in fact show that (for a second-order equation) it is always possible to take linear combinations of eigenfunctions such that the result is linear, for example in the exponential form of the Fourier series. Hence, we can assume that 𝑦𝑛is real. We can further prove that the regular Sturm–Liouville problem must have simple (non-degenerate) eigenvalues 𝜆𝑛, by considering two possible eigenfunctions 𝑢, 𝑣for the same 𝜆, and use the expression for self-adjointness. We find 𝑢ℒ𝑣−(ℒ𝑢)𝑣= [−𝑝(𝑢𝑣′ −𝑢′𝑣)]′ which contains the Wrońskian. We can integrate and impose homogeneous boundary conditions to get the required result.
2.8. Orthogonality of eigenfunctions Suppose ℒ𝑦𝑛= 𝜆𝑛𝑤𝑦𝑛, and ℒ𝑦𝑚= 𝜆𝑚𝑤𝑦𝑚where 𝜆𝑛≠𝜆𝑚. Then, we can integrate to find ∫ 𝑏 𝑎 (𝑦𝑚ℒ𝑦𝑛−𝑦𝑛ℒ𝑦𝑚) d𝑥= (𝜆𝑛−𝜆𝑚) ∫ 𝑏 𝑎 𝑤𝑦𝑛𝑦𝑚d𝑥= 0 by self-adjointness Since 𝜆𝑛≠𝜆𝑚, we have ∀𝑛≠𝑚, ∫ 𝑏 𝑎 𝑤𝑦𝑛𝑦𝑚d𝑥= 0 Hence, 𝑦𝑛and 𝑦𝑚are orthogonal with respect to the weight function 𝑤on [𝑎, 𝑏].
Definition. We define the inner product with respect to 𝑤to be ⟨𝑓, 𝑔⟩𝑤= ∫ 𝑏 𝑎 𝑤𝑓⋆𝑔d𝑥 Note, ⟨𝑓, 𝑔⟩𝑤= ⟨𝑤𝑓, 𝑔⟩= ⟨𝑓, 𝑤𝑔⟩ Hence, the orthogonality relation becomes ∀𝑛≠𝑚, ⟨𝑦𝑛, 𝑦𝑚⟩𝑤= 0 231 V. Methods 2.9. Eigenfunction expansions The completeness of the family of eigenfunctions (which is not proven here) implies that we can approximate any ‘well-behaved’ 𝑓(𝑥) on [𝑎, 𝑏] by the series 𝑓(𝑥) = ∞ ∑ 𝑛=1 𝑎𝑛𝑦𝑛(𝑥) This is comparable to Fourier series. To find the coefficients 𝑎𝑛, we will take the inner product with an eigenfunction. By orthogonality, ∫ 𝑏 𝑎 𝑤𝑦𝑚𝑓d𝑥= ∞ ∑ 𝑛=1 𝑎𝑛∫ 𝑏 𝑎 𝑤𝑦𝑛𝑦𝑚d𝑥= 𝑎𝑚∫ 𝑏 𝑎 𝑤𝑦2 𝑚d𝑥 Hence, 𝑎𝑛= ∫ 𝑏 𝑎𝑤𝑦𝑛𝑓d𝑥 ∫ 𝑏 𝑎𝑤𝑦2 𝑛d𝑥 We can normalise eigenfunctions, for instance 𝑌 𝑛(𝑥) = 𝑦𝑛(𝑥) (∫ 𝑏 𝑎𝑤𝑦2 𝑛d𝑥) 1 2 hence ⟨𝑌 𝑛, 𝑌 𝑚⟩𝑤= 𝛿𝑛𝑚 giving an orthonormal set of eigenfunctions. In this case, 𝑓(𝑥) = ∞ ∑ 𝑛=1 𝐴𝑛𝑌 𝑛 where 𝐴𝑛= ∫ 𝑏 𝑎 𝑤𝑌 𝑛𝑓d𝑥 Example. Recall Fourier series in Sturm–Liouville form: ℒ𝑦𝑛≡−d2𝑦 d𝑥2 = 𝜆𝑛𝑦𝑛 where in this case we have 𝜆𝑛= (𝑛𝜋 𝐿) 2 232 2. Sturm–Liouville theory 2.10. Completeness and Parseval’s identity Consider ∫ 𝑏 𝑎 [𝑓(𝑥) − ∞ ∑ 𝑛=1 𝑎𝑛𝑦𝑛] 2 𝑤d𝑥 By orthogonality, this is equivalently ∫ 𝑏 𝑎 [𝑓2 −2𝑓∑ 𝑛 𝑎𝑛𝑦𝑛+ ∑ 𝑛 𝑎2 𝑛𝑦2 𝑛]𝑤d𝑥 Note that the second term can be extracted using the definition of 𝑎𝑛, giving ∫ 𝑏 𝑎 𝑤𝑓2 d𝑥− ∞ ∑ 𝑛=1 𝑎2 𝑛∫ 𝑏 𝑎 𝑤𝑦2 𝑛d𝑥 If the eigenfunctions are complete, then the result will be zero, showing that the series ex-pansion converges.
∫ 𝑏 𝑎 𝑤𝑓2 d𝑥= ∞ ∑ 𝑛=1 𝑎2 𝑛∫ 𝑏 𝑎 𝑤𝑦2 𝑛d𝑥= ∞ ∑ 𝑛=1 𝐴2 𝑛 If some eigenfunctions are missing, this is Bessel’s inequality: ∫ 𝑏 𝑎 𝑤𝑓2 d𝑥≥ ∞ ∑ 𝑛=1 𝐴2 𝑛 We define the partial sum to be 𝑆𝑁(𝑥) = 𝑁 ∑ 𝑛=1 𝑎𝑛𝑦𝑛 with 𝑓(𝑥) = lim𝑁→∞𝑆𝑁(𝑥). Convergence is defined in terms of the mean-square error. In particular, if we have a complete set of eigenfunctions, 𝜀𝑁= ∫ 𝑏 𝑎 𝑤[𝑓(𝑥) −𝑆𝑛(𝑥)] 2 d𝑥→0 This ‘global’ definition of convergence is convergence in the mean, not pointwise conver-gence as in Fourier series. The error in partial sum 𝑆𝑁is minimised by 𝑎𝑛above for the 𝑁= ∞expansion.
𝜕𝜀𝑁 𝜕𝑎𝑛 = −2 ∫ 𝑏 𝑎 𝑦𝑛𝑤[𝑓− 𝑁 ∑ 𝑛=1 𝑎𝑛𝑦𝑛] d𝑥= −2 ∫ 𝑏 𝑎 (𝑤𝑓𝑦𝑛−𝑎𝑛𝑤𝑦2 𝑛) d𝑥= 0 It is minimal because we can show 𝜕2𝜀 𝜕𝑎2 𝑛= 2 ∫ 𝑏 𝑎𝑤𝑦2 𝑛d𝑥≥0. Thus the 𝑎𝑛given above is the best possible choice for the coefficient at all 𝑁.
233 V. Methods 2.11. Legendre’s equation Legendre’s equation is (1 −𝑥2)𝑦″ −2𝑥𝑦′ + 𝜆𝑦= 0 on [−1, 1], with boundary conditions that 𝑦is finite at 𝑥= ±1, at the regular singular points of the ODE. This equation is already in Sturm–Liouville form with 𝑝= 1 −𝑥2, 𝑞= 0, 𝑤= 1 We seek a power series solution centred on 𝑥= 0: 𝑦= ∑ 𝑛 𝑐𝑛𝑥𝑛 Substituting into the differential equation, (1 −𝑥2) ∑ 𝑛 𝑛(𝑛−1)𝑥𝑛𝑥𝑛−2 −2𝑥∑ 𝑛 𝑐𝑛𝑐𝑛−1 + 𝜆∑ 𝑛 𝑐𝑛𝑥𝑛= 0 Equating powers, (𝑛+ 2)(𝑛+ 1)𝑐𝑛+2 −𝑛(𝑛−1)𝑐𝑛−2𝑛𝑐𝑛+ 𝜆𝑐𝑛= 0 which gives a recursion relation between 𝑐𝑛+2 and 𝑐𝑛.
𝑐𝑛+2 = 𝑛(𝑛+ 1) −𝜆 (𝑛+ 1)(𝑛+ 2)𝑐𝑛 Hence, specifying 𝑐0, 𝑐1 gives two independent solutions. In particular, 𝑦even = 𝑐0[1 + (−𝜆) 2! 𝑥2 + (6 −𝜆)(−𝜆) 4!
𝑥4 + … ] 𝑦odd = 𝑐1[𝑥+ (2 −𝜆) 3!
𝑥3 + … ] As 𝑛→∞, 𝑐𝑛+2 𝑐𝑛 →1. So these are geometric series, with radius of convergence |𝑥| < 1, hence there is divergence at 𝑥= ±1. So taking a power series does not give a useful solution.
Suppose we chose 𝜆= ℓ(ℓ+ 1). Then eventually we have 𝑛such that the numerator van-ishes. In particular, by taking 𝜆= ℓ(ℓ+ 1), either the series for 𝑦even or 𝑦odd terminates.
These functions are called the Legendre polynomials, denoted 𝑃ℓ(𝑥), with the normalisa-tion convention 𝑃ℓ(1) = 1.
• ℓ= 0, 𝜆= 0, 𝑃 0(𝑥) = 1 • ℓ= 1, 𝜆= 2, 𝑃 1(𝑥) = 𝑥 • ℓ= 2, 𝜆= 6, 𝑃 2(𝑥) = 3𝑥2−1 2 • ℓ= 3, 𝜆= 12, 𝑃 3(𝑥) = 5𝑥3−3𝑥 2 Note, 𝑃ℓ(𝑥) has ℓzeroes. The polynomials oscillate in parity.
234 2. Sturm–Liouville theory 2.12. Properties of Legendre polynomials Since Legendre polynomials come from a self-adjoint operator, they must have certain con-ditions, such as orthogonality. For 𝑛≠𝑚, ∫ 1 −1 𝑃 𝑛𝑃 𝑚d𝑥= 0 They are also normalisable, ∫ 1 −1 𝑃2 𝑛d𝑥= 2 2𝑛+ 1 We can prove this with Rodrigues’ formula: 𝑃 𝑛(𝑥) = 1 2𝑛𝑛!( d d𝑥) 𝑛 (𝑥2 −1)𝑛 Alternatively we could use a generating function: ∞ ∑ 𝑛=0 𝑃 𝑛(𝑥)𝑡𝑛= 1 √1 −2𝑥𝑡+ 𝑡2 = 1 + 1 2(2𝑥𝑡−𝑡2) + 3 8(2𝑥𝑡−𝑡2) 2 + … = 1 + 𝑥𝑡+ 1 2(3𝑥2 −1)𝑡2 + … There are some useful recursion relations.
ℓ(ℓ+ 1)𝑃ℓ+1 = (2ℓ+ 1)𝑥𝑃ℓ(𝑥) −ℓ𝑃ℓ−1(𝑥) Also, (2ℓ+ 1)𝑃ℓ(𝑥) = d d𝑥[𝑃ℓ+1(𝑥) −𝑃ℓ−1(𝑥)] 2.13. Legendre polynomials as eigenfunctions Any (well-behaved) function on [−1, 1] can be expressed as 𝑓(𝑥) = ∞ ∑ ℓ=0 𝑎ℓ𝑃ℓ(𝑥) where 𝑎ℓ= 2ℓ+ 1 2 ∫ 1 −1 𝑓(𝑥)𝑃ℓ(𝑥) d𝑥 with no boundary conditions (e.g. periodicity conditions) on 𝑓.
235 V. Methods 2.14. Solving inhomogeneous differential equations This can be thought of as the general case of Fourier series discussed previously.
Consider the problem ℒ𝑦= 𝑓(𝑥) ≡𝑤(𝑥)𝐹(𝑥) on 𝑥∈[𝑎, 𝑏] assuming homogeneous boundary conditions. Given eigenfunctions 𝑦𝑛(𝑥) satisfying ℒ𝑦𝑛= 𝜆𝑛𝑤𝑦𝑛, we wish to expand this solution as 𝑦(𝑥) = ∑ 𝑛 𝑐𝑛𝑦𝑛(𝑥) and 𝐹(𝑥) = ∑ 𝑛 𝑎𝑛𝑦𝑛(𝑥) where 𝑎𝑛are known and 𝑐𝑛are unknown: 𝑎𝑛= ∫ 𝑏 𝑎𝑤𝐹𝑦𝑛d𝑥 ∫ 𝑏 𝑎𝑤𝑦2 𝑛d𝑥 Substituting, ℒ𝑦= ℒ∑ 𝑛 𝑐𝑛𝑦𝑛= 𝑤∑ 𝑛 𝑐𝑛𝜆𝑛𝑦𝑛= 𝑤∑ 𝑛 𝑎𝑛𝑦𝑛 By orthogonality, 𝑐𝑛𝜆𝑛= 𝑎𝑛⟹𝑐𝑛= 𝑎𝑛 𝜆𝑛 In particular, 𝑦(𝑥) = ∞ ∑ 𝑛=1 𝑎𝑛 𝜆𝑛 𝑦𝑛(𝑥) We can further generalise; we can permit a driving force, which often induces a linear re-sponse term ̃ 𝜆𝑤𝑦.
ℒ𝑦−̃ 𝜆𝑤𝑦= 𝑓(𝑥) where ̃ 𝜆is fixed. The solution becomes 𝑦(𝑥) = ∞ ∑ 𝑛=1 𝑎𝑛 𝜆𝑛−̃ 𝜆𝑦𝑛(𝑥) 2.15. Integral solutions Recall that 𝑦(𝑥) = ∞ ∑ 𝑛=1 𝑎𝑛 𝜆𝑛 𝑦𝑛(𝑥) = ∑ 𝑛 𝑦𝑛(𝑥) 𝜆𝑛𝜆𝑛𝑁𝑛 ∫ 𝑏 𝑎 𝑤(𝜉)𝐹(𝜉)𝑦𝑛(𝜉) d𝜉 236 2. Sturm–Liouville theory where 𝑁𝑛= ∫𝑤𝑦2 𝑛d𝑥 This then gives 𝑦(𝑥) = ∫ 𝑏 𝑎 ∞ ∑ 𝑛=1 𝑦𝑛(𝑥)𝑦𝑛(𝜉) 𝜆𝑛𝑁𝑛 ⏟⎵⎵⎵⏟⎵⎵⎵⏟ 𝐺(𝑥,𝜉) 𝑤(𝜉)𝐹(𝜉) ⏟⎵⏟⎵⏟ 𝑓(𝜉) d𝜉= ∫ 𝑏 𝑎 𝐺(𝑥; 𝜉)𝑓(𝜉) d𝜉 where 𝐺(𝑥, 𝜉) = ∞ ∑ 𝑛=1 𝑦𝑛(𝑥)𝑦𝑛(𝜉) 𝜆𝑛𝑁𝑛 is the eigenfunction expansion of the Green’s function. Note that the Green’s function does not depend on 𝑓, but only on ℒand the boundary conditions. In this sense, it acts like an inverse operator ℒ−≡∫d𝜉𝐺(𝑥, 𝜉) analogously to how 𝐴𝑥= 𝑏⟹𝑥= 𝐴−1𝑏for matrix equations.
2.16. Waves on an elastic string Consider a small displacement 𝑦(𝑥, 𝑡) on a stretched string with fixed ends at 𝑥= 0 and 𝑥= 𝐿, that is, with boundary conditions 𝑦(0, 𝑡) = 𝑦(𝐿, 𝑡) = 0. We can determine the string’s motion for specified initial conditions 𝑦(𝑥, 0) = 𝑝(𝑥) and 𝜕𝑦 𝜕𝑡= 𝑞(𝑥). We derive the equation of motion governing the motion of the string by balancing forces on a string segment (𝑥, 𝑥+ 𝛿𝑥) and take the limit as 𝛿𝑥→0. Let 𝑇1 be the tension force acting to the left at angle 𝜃1 from the horizontal. Analogously, let 𝑇2 be the rightwards tension force at angle 𝜃2. We assume at any point on the string that | | 𝜕𝑦 𝜕𝑥 | | ≪1, so the angles of the forces are small. In the 𝑥dimension, 𝑇1 cos 𝜃1 = 𝑇2 cos 𝜃2 ⟹𝑇1 ≈𝑇2 = 𝑇 So the tension 𝑇is constant up to an error of order 𝑂(| | 𝜕𝑦 𝜕𝑥 | | 2 ). In the 𝑦dimension, since 𝜃are small, 𝐹𝑇= 𝑇2 sin 𝜃2 −𝑇1 sin 𝜃1 ≈𝑇( 𝜕𝑦 𝜕𝑥 | | |𝑥+𝛿𝑥 −𝜕𝑦 𝜕𝑥 | | |𝑥 ) ≈𝑇𝜕2𝑦 𝜕𝑥2 𝛿𝑥 By 𝐹= 𝑚𝑎, 𝐹𝑇+ 𝐹 𝑔= (𝜇𝛿𝑥)𝜕2𝑦 𝜕𝑡2 = 𝑇𝜕2𝑦 𝜕𝑥2 𝛿𝑥−𝑔𝜇𝛿𝑥 where 𝐹 𝑔is the gravitational force and 𝜇is the linear mass density. We define the wave speed as 𝑐= √ 𝑇 𝜇 237 V. Methods and find 𝜕2𝑦 𝜕𝑡2 = 𝑇 𝜇 𝜕2𝑦 𝜕𝑥2 −𝑔= 𝑐2 𝜕2𝑦 𝜕𝑥2 We often assume gravity is negligible to produce the pure wave equation 1 𝑐2 𝜕2𝑦 𝜕𝑡2 = 𝜕2𝑦 𝜕𝑥2 238 3. Separation of variables 3. Separation of variables 3.1. Separation of variables We wish to solve the wave equation subject to certain boundary and initial conditions. Con-sider a possible solution of separable form: 𝑦(𝑥, 𝑡) = 𝑋(𝑥)𝑇(𝑡) Substituting into the wave equation, 1 𝑐2 ̈ 𝑦= 𝑦″ ⟹ 1 𝑐2 𝑋̈ 𝑇= 𝑋″𝑇 Then 1 𝑐2 ̈ 𝑇 𝑇= 𝑋″ 𝑋 However, ̈ 𝑇 𝑇depends only on 𝑡and 𝑋″ 𝑋depends only on 𝑥. Thus, both sides must be equal to some separation constant −𝜆.
1 𝑐2 ̈ 𝑇 𝑇= 𝑋″ 𝑋= −𝜆 Hence, 𝑋″ + 𝜆𝑋= 0; ̈ 𝑇+ 𝜆𝑐2𝑇= 0 3.2. Boundary conditions and normal modes We will begin by first solving the spatial part of the solution. One of 𝜆> 0, 𝜆< 0, 𝜆= 0 must be true. The boundary conditions restrict the possible 𝜆.
(i) First, suppose 𝜆< 0. Take 𝜒2 = −𝜆. Then, 𝑋(𝑥) = 𝐴𝑒𝜒𝑥+ 𝐵𝑒−𝜒𝑥= 𝐶cosh(𝜒𝑥) + 𝐷sinh(𝜒𝑥) The boundary conditions are 𝑥(0) = 𝑥(𝐿) = 0, so only the trivial solution is possible: 𝐶= 𝐷= 0.
(ii) Now, suppose 𝜆= 0. Then 𝑋(𝑥) = 𝐴𝑥+ 𝐵 Again, the boundary conditions impose 𝐴= 𝐵= 0 giving only the trivial solution.
(iii) Finally, the last possibility is 𝜆> 0.
𝑋(𝑥) = 𝐴cos (√𝜆𝑥) + 𝐵sin (√𝜆𝑥) The boundary conditions give 𝐴= 0; 𝐵sin (√𝜆𝐿) = 0 ⟹√𝜆𝐿= 𝑛𝜋 239 V. Methods The following are the eigenfunctions and eigenvalues.
𝑋𝑛(𝑥) = 𝐵𝑛sin 𝑛𝜋𝑥 𝐿; 𝜆𝑛= (𝑛𝜋 𝐿) 2 These are also called the ‘normal modes’ of the system. The spatial shape in 𝑥does not change in time, but the amplitude may vary. The fundamental mode is the lowest frequency of vibration, given by 𝑛= 1 ⟹𝜆1 = 𝜋2 𝐿 2 The second mode is the first overtone, and is given by 𝑛= 2 ⟹𝜆2 = 4𝜋2 𝐿 2 3.3. Initial conditions and temporal solutions Substituting 𝜆𝑛into the time ODE, ̈ 𝑇+ 𝑛2𝜋2𝑐2 𝐿 2 𝑇= 0 Hence, 𝑇𝑛(𝑡) = 𝐶𝑛cos 𝑛𝜋𝑐𝑡 𝐿 + 𝐷𝑛sin 𝑛𝜋𝑐𝑡 𝐿 Therefore, a specific solution of the wave equation satisfying the boundary conditions is (absorbing the 𝐵𝑛into the 𝐶𝑛, 𝐷𝑛): 𝑦𝑛(𝑥, 𝑡) = 𝑇𝑛(𝑡)𝑋𝑛(𝑥) = (𝐶𝑛cos 𝑛𝜋𝑐𝑡 𝐿 + 𝐷𝑛sin 𝑛𝜋𝑐𝑡 𝐿 ) sin 𝑛𝜋𝑥 𝐿 To find a particular solution for a given set of initial conditions, we must consider a linear superposition of all possible 𝑦𝑛.
𝑦(𝑥, 𝑡) = ∞ ∑ 𝑛=1 (𝐶𝑛cos 𝑛𝜋𝑐𝑡 𝐿 + 𝐷𝑛sin 𝑛𝜋𝑐𝑡 𝐿 ) sin 𝑛𝜋𝑥 𝐿 By construction, this 𝑦(𝑥, 𝑡) satisfies the boundary conditions, so now we can impose the initial conditions.
𝑦(𝑥, 0) = 𝑝(𝑥) = ∞ ∑ 𝑛=1 𝐶𝑛sin 𝑛𝜋𝑥 𝐿 We can find the 𝐶𝑛using standard Fourier series techniques, since this is exactly a half-range sine series. Further, 𝜕𝑦(𝑥, 0) 𝜕𝑡 = 𝑞(𝑥) = ∞ ∑ 𝑛=1 𝑛𝜋𝑐 𝐿𝐷𝑛sin 𝑛𝜋𝑥 𝐿 240 3. Separation of variables Again we can solve for the 𝐷𝑛in a similar way. In particular, 𝐶𝑛= 2 𝐿∫ 𝐿 0 𝑝(𝑥) sin 𝑛𝜋𝑥 𝐿 d𝑥 𝐷𝑛= 2 𝑛𝜋𝑐∫ 𝐿 0 𝑞(𝑥) sin 𝑛𝜋𝑥 𝐿 d𝑥 Example. Consider the initial condition of a see-saw wave parametrised by 𝜉, and let 𝐿= 1.
This can be visualised as plucking the string at position 𝜉.
𝑦(𝑥, 0) = 𝑝(𝑥) = {𝑥(1 −𝜉) 0 ≤𝑥< 𝜉 𝜉(1 −𝑥) 𝜉≤𝑥< 1 We also define 𝜕𝑦(𝑥, 0) 𝜕𝑡 = 𝑞(𝑥) = 0 The Fourier series for 𝑝is given by 𝐶𝑛= 2 sin 𝑛𝜋𝜉 (𝑛𝜋)2 ; 𝐷𝑛= 0 Hence the solution to the wave equation is 𝑦(𝑥, 𝑡) = ∞ ∑ 𝑛=1 2 (𝑛𝜋)2 sin 𝑛𝜋𝜉sin 𝑛𝜋𝑥cos 𝑛𝜋𝑐𝑡 3.4. Separation of variables methodology A general strategy for solving higher-dimensional partial differential equations is as fol-lows.
(i) Obtain a linear PDE system, using boundary and initial conditions.
(ii) Separate variables to yield decoupled ODEs.
(iii) Impose homogeneous boundary conditions to find eigenvalues and eigenfunctions.
(iv) Use these eigenvalues (constants of separation) to find the eigenfunctions in the other variables.
(v) Sum over the products of separable solutions to find the general series solution.
(vi) Determine coefficients for this series using the initial conditions.
241 V. Methods Example. We will solve the wave equation instead in characteristic coordinates. Recall the sine and cosine summation identities: 𝑦(𝑥, 𝑡) = 1 2 ∞ ∑ 𝑛=1 [(𝐶𝑛sin 𝑛𝜋 𝐿(𝑥−𝑐𝑡) + 𝐷𝑛cos 𝑛𝜋 𝐿(𝑥−𝑐𝑡)) + (𝐶𝑛sin 𝑛𝜋 𝐿(𝑥+ 𝑐𝑡) −𝐷𝑛cos 𝑛𝜋 𝐿(𝑥+ 𝑐𝑡))] = 𝑓(𝑥−𝑐𝑡) + 𝑔(𝑥+ 𝑐𝑡) The standing wave solution can be interpreted as a superposition of a right-moving wave and a left-moving wave. A special case is 𝑞(𝑥) = 0, implying 𝑓= 𝑔= 1 2𝑝. Then, 𝑦(𝑥, 𝑡) = 1 2[𝑝(𝑥−𝑐𝑡) + 𝑝(𝑥+ 𝑐𝑡)] 3.5. Energy of oscillations A vibrating string has kinetic energy due to its motion.
Kinetic energy = 1 2𝜇∫ 𝐿 0 (𝜕𝑦 𝜕𝑡) 2 d𝑥 It has potential energy given by Potential energy = 𝑇Δ𝑥= 𝑇∫ 𝑇 𝑐 (√1 + (𝜕𝑦 𝜕𝑥) 2 −1) d𝑥≈1 2𝑇∫ 𝐿 0 (𝜕𝑦 𝜕𝑥) 2 d𝑥 assuming that the disturbances on the string are small, that is, | | 𝜕𝑦 𝜕𝑥 | | ≪1. The total energy on the string, given 𝑐2 = 𝑇/𝜇, is given by 𝐸= 1 2𝜇∫ 𝐿 0 [(𝜕𝑦 𝜕𝑡) 2 + 𝑐2(𝜕𝑦 𝜕𝑥) 2 ] d𝑥 Substituting the solution, using the orthogonality conditions, 𝐸= 1 2𝜇 ∞ ∑ 𝑛=1 ∫ 𝐿 0 [ −(𝑛𝜋𝑐 𝐿𝐶𝑛sin 𝑛𝜋𝑐𝑡 𝐿 + 𝑛𝜋𝑐 𝐿𝐷𝑛cos 𝑛𝜋𝑐𝑡 𝐿 ) 2 sin2 𝑛𝜋𝑥 𝐿 + 𝑐2(𝐶𝑛cos 𝑛𝜋𝑐𝑡 𝐿 + 𝐷𝑛sin 𝑛𝜋𝑐𝑡 𝐿 ) 2 𝑛2𝜋2 𝐿 2 cos2 𝑛𝜋𝑥 𝐿] d𝑥 = 1 4𝜇 ∞ ∑ 𝑛=1 𝑛2𝜋2𝑐2 𝐿 (𝐶2 𝑛+ 𝐷2 𝑛) 242 3. Separation of variables which is an analogous result to Parseval’s theorem. This is true since ∫cos2 𝑛𝜋𝑥 𝐿 d𝑥= 1 2 and cos2 + sin2 = 1. We can think of this energy as the sum over all the normal modes of the energy in that specific mode. Note that this quantity is constant over time.
3.6. Wave reflection and transmission The travelling wave has left-moving and right-moving modes. A simple harmonic travelling wave is 𝑦= Re [𝐴𝑒𝑖𝜔(𝑡−𝑥/𝑐)] = 𝐴cos [𝜔(𝑡−𝑥/𝑐) + 𝜙] where the phase 𝜙is equal to arg 𝐴, and the wavelength 𝜆is 2𝜋𝑐/𝜔. In further discussion, we assume only the real part is used. Consider a density discontinuity on the string at 𝑥= 0 with the following properties.
𝜇= {𝜇− for 𝑥< 0 𝜇+ for 𝑥> 0 ⟹𝑐= ⎧ ⎪ ⎨ ⎪ ⎩ 𝑐−= √ 𝑇 𝜇− for 𝑥< 0 𝑐+ = √ 𝑇 𝜇+ for 𝑥> 0 assuming a constant tension 𝑇. As a wave from the negative direction approaches the dis-continuity, some of the wave will be reflected, given by 𝐵𝑒𝑖𝜔(𝑡+𝑥/𝑐−), and some of the wave will be transmitted, given by 𝐷𝑒𝑖𝜔(𝑡−𝑥/𝑐+). The boundary conditions at 𝑥= 0 are (i) 𝑦is continuous for all 𝑡(the string does not break), so 𝐴+ 𝐵= 𝐷 (∗) (ii) The forces balance, 𝑇 𝜕𝑦 𝜕𝑥 | |𝑥=0−= 𝑇 𝜕𝑦 𝜕𝑥 | |𝑥=0+ which means 𝜕𝑦 𝜕𝑥must be continuous for all 𝑡. This gives −𝑖𝜔𝐴 𝑐− + 𝑖𝜔𝐵 𝑐− = −𝑖𝜔𝐷 𝑐+ (†) We can eliminate 𝐵from (∗) by subtracting 𝑐− 𝑖𝜔(†).
2𝐴= 𝐷+ 𝐷𝑐− 𝑐+ = 𝐷 𝑐+ (𝑐+ + 𝑐−) Hence, given 𝐴, we have the solution for the transmitted amplitude and reflected amplitude to be 𝐷= 2𝑐+ 𝑐−+ 𝑐+ 𝐴; 𝐵= 𝑐+ −𝑐− 𝑐−+ 𝑐+ In general 𝐴, 𝐵, 𝐷are complex, hence different phase shifts are possible.
There are a number of limiting cases, for example 243 V. Methods (i) If 𝑐−= 𝑐+ we have 𝐷= 𝐴and 𝐵= 0 so we have full transmission and no reflection.
(ii) (Dirichlet boundary conditions) If 𝜇+ 𝜇−→∞, this models a fixed end at 𝑥= 0. We have 𝑐+ 𝑐−→0 giving 𝐷= 0 and 𝐵= −𝐴. Notice that the reflection has occurred with opposite phase, 𝜙= 𝜋.
(iii) (Neumann boundary conditions) Consider 𝜇+ 𝜇−→0, this models a free end. Then 𝑐+ 𝑐−→∞giving 𝐷= 2𝐴, 𝐵= 𝐴. This gives total reflection but with the same phase.
3.7. Wave equation in plane polar coordinates Consider the two-dimensional wave equation for 𝑢(𝑟, 𝜃, 𝑡) given by 1 𝑐2 𝜕2𝑢 𝜕𝑡2 = ∇2𝑢 with boundary conditions at 𝑟= 1 on a unit disc given by 𝑢(1, 𝜃, 𝑡) = 0 and initial conditions for 𝑡= 0 given by 𝑢(𝑟, 𝜃, 0) = 𝜙(𝑟, 𝜃); 𝜕𝑢 𝜕𝑡= 𝜓(𝑟, 𝜃) Suppose that this equation is separable. First, let us consider temporal separation. Suppose that 𝑢(𝑟, 𝜃, 𝑡) = 𝑇(𝑡)𝑉(𝑟, 𝜃) Then we have ̈ 𝑇+ 𝜆𝑐2𝑇= 0; ∇2𝑉+ 𝜆𝑉= 0 In plane polar coordinates, we can write the spatial equation as 𝜕2𝑉 𝜕𝑟2 + 1 𝑟 𝜕𝑉 𝜕𝑟+ 1 𝑟2 𝜕2𝑉 𝜕𝜃2 + 𝜆𝑉= 0 We will perform another separation, supposing 𝑉(𝑟, 𝜃) = 𝑅(𝑟)Θ(𝜃) to give Θ″ + 𝜇Θ = 0; 𝑟2𝑅″ + 𝑟𝑅′ + (𝜆𝑟2 −𝜇)𝑅= 0 where 𝜆, 𝜇are the separation constants. The polar solution is constrained by periodicity Θ(0) = Θ(2𝜋), since we are working on a disc. We also consider only 𝜇> 0. The eigenvalue is then given by 𝜇= 𝑚2, where 𝑚∈ℕ.
Θ𝑚(𝜃) = 𝐴𝑚cos 𝑚𝜃+ 𝐵𝑚sin 𝑚𝜃 Or, in complex exponential form, Θ𝑚(𝜃) = 𝐶𝑚𝑒𝑖𝑚𝜃; 𝑚∈ℤ 244 3. Separation of variables 3.8. Bessel’s equation We can solve the radial equation (in the previous subsection) by converting it first into Sturm–Liouville form, which can be accomplished by dividing by 𝑟.
d d𝑟(𝑟𝑅′) −𝑚2 𝑟 = −𝜆𝑟𝑅 where 𝑝(𝑟) = 𝑟, 𝑞(𝑟) = 𝑚2 𝑟, 𝑤(𝑟) = 𝑟, with self-adjoint boundary conditions with 𝑅(1) = 0.
We will require 𝑅is bounded at 𝑅(0), and since 𝑝(0) = 0 there is a regular singular point at 𝑟= 0. This particular equation for 𝑅is known as Bessel’s equation. We will first substitute 𝑧≡√𝜆𝑟, then we find the usual form of Bessel’s equation, 𝑧2 d2𝑅 d𝑧2 + 𝑧d𝑅 d𝑧+ (𝑧2 −𝑚2)𝑅= 0 We can use the method of Frobenius by substituting the following power series: 𝑅= 𝑧𝑝 ∞ ∑ 𝑛=0 𝑎𝑛𝑧𝑛 to find ∞ ∑ 𝑛=0 [𝑎𝑛(𝑛+ 𝑝)(𝑛+ 𝑝−1)𝑧𝑛+𝑝+ (𝑛+ 𝑝)𝑧𝑛+𝑝+ 𝑧𝑛+𝑝+2 + 𝑚2𝑧𝑛+𝑝] = 0 Equating powers of 𝑧, we can find the indicial equation 𝑝2 −𝑚2 = 0 ⟹𝑝= 𝑚, −𝑚 The regular solution, given by 𝑝= 𝑚, has recursion relation (𝑛+ 𝑚)2𝑎𝑛+ 𝑎𝑛−2 −𝑚2𝑎𝑛= 0 which gives 𝑎𝑛= −1 𝑛(𝑛+ 2𝑚)𝑎𝑛−2 Hence, we can find 𝑎2𝑛= 𝑎0 (−1)𝑛 22𝑛𝑛!(𝑛+ 𝑚)(𝑛+ 𝑚−1) … (𝑚+ 1) If, by convention, we let 𝑎0 = 1 2𝑚𝑚!
we can then write the Bessel function of the first kind by 𝐽𝑚(𝑧) = (𝑧 2) 𝑚∞ ∑ 𝑛=0 (−1)𝑛 𝑛!(𝑛+ 𝑚)!(𝑧 2) 2𝑛 245 V. Methods 3.9. Asymptotic behaviour of Bessel functions If 𝑧is small, the leading-order behaviour of 𝐽𝑚(𝑧) is 𝐽0(𝑧) ≈1 𝐽𝑚(𝑧) ≈1 𝑚!(𝑧 2) 𝑚 Now, let us consider large 𝑧. In this case, the function becomes oscillatory; 𝐽𝑚(𝑧) ≈√ 2 𝜋𝑧cos(𝑧−𝑚𝜋 2 −𝜋 4 ) 3.10. Zeroes of Bessel functions We can see from the asymptotic behaviour that there are infinitely many zeroes of the Bessel functions of the first kind as 𝑧→∞. We define 𝑗𝑚𝑛to be the 𝑛th zero of 𝐽𝑚, for 𝑧> 0.
Approximately, cos(𝑧−𝑚𝜋 2 −𝜋 4 ) = 0 ⟹𝑧−𝑚𝜋 2 −𝜋 4 = 𝑛𝜋−𝜋 2 Hence 𝑧≈𝑛𝜋+ 𝑚𝜋 2 −𝜋 4 ≡ ̃ 𝑗𝑚𝑛 3.11. Solving the vibrating drum Recall that the radial solutions become 𝑅𝑚(𝑧) = 𝑅𝑚(√𝜆𝑥) = 𝐴𝐽𝑚(√𝜆𝑥) + 𝐵𝑌 𝑚(√𝜆𝑥) Imposing the boundary condition of boundedness at 𝑟= 0, we must have 𝐵= 0. Further imposing 𝑟= 1 and 𝑅= 0 gives 𝐽𝑚(√𝜆) = 0. These zeroes occur at 𝑗𝑚𝑛≈𝑛𝜋+ 𝑚𝜋 2 − 𝜋 4 .
Hence, the eigenvalues must be 𝑗2 𝑚𝑛. Therefore, the spatial solution is 𝑉 𝑚𝑛(𝑟, 𝜃) = Θ𝑚(𝜃)𝑅𝑚𝑛(√𝜆𝑚𝑛𝑟) = (𝐴𝑚𝑛cos 𝑚𝜃+ 𝐵𝑚𝑛sin 𝑚𝜃)𝐽𝑚(𝑗𝑚𝑛𝑟) The temporal solution is ̈ 𝑇= −𝜆𝑐𝑇⟹𝑇𝑚𝑛(𝑡) = cos(𝑗𝑚𝑛𝑐𝑡), sin(𝑗𝑚𝑛𝑐𝑡) Combining everything together, the full solution is 𝑢(𝑟, 𝜃, 𝑡) = ∞ ∑ 𝑛=1 𝐽0(𝑗0𝑛𝑟)(𝐴0𝑛cos 𝑗0𝑛𝑐𝑡+ 𝐶0𝑛sin 𝑗0𝑛𝑐𝑡) + ∞ ∑ 𝑚=1 ∞ ∑ 𝑛=1 𝐽𝑚(𝑗𝑚𝑛𝑟)(𝐴𝑚𝑛cos 𝑚𝜃+ 𝐵𝑚𝑛sin 𝑚𝜃) cos 𝑗𝑚𝑛𝑐𝑡 + ∞ ∑ 𝑚=1 ∞ ∑ 𝑛=1 𝐽𝑚(𝑗𝑚𝑛𝑟)(𝐶𝑚𝑛cos 𝑚𝜃+ 𝐷𝑚𝑛sin 𝑚𝜃) sin 𝑗𝑚𝑛𝑐𝑡 246 3. Separation of variables Now, we impose the boundary conditions 𝑢(𝑟, 𝜃, 0) = 𝜙(𝑟, 𝜃) = ∞ ∑ 𝑚=0 ∞ ∑ 𝑛=1 𝐽𝑚(𝑗𝑚𝑛𝑟)(𝐴𝑚𝑛cos 𝑚𝜃+ 𝐵𝑚𝑛sin 𝑚𝜃) and 𝜕𝑢 𝜕𝑡(𝑟, 𝜃, 0) = 𝜓(𝑟, 𝜃) = ∞ ∑ 𝑚=0 ∞ ∑ 𝑛=1 𝑗𝑚𝑛𝑐𝐽𝑚(𝑗𝑚𝑛𝑟)(𝐶𝑚𝑛cos 𝑚𝜃+ 𝐷𝑚𝑛sin 𝑚𝜃) We need to find the coefficients by multiplying by 𝐽𝑚, cos, sin and using the orthogonality relations, which are ∫ 1 0 𝐽𝑚(𝑗𝑚𝑛𝑟)𝐽𝑚(𝑗𝑚𝑘𝑟)𝑟d𝑟= 1 2[𝐽′ 𝑚(𝑗𝑚𝑛)] 2𝛿𝑛𝑘= 1 2[𝐽𝑚+1(𝑗𝑚𝑛)] 2𝛿𝑛𝑘 by using a recursion relation of the Bessel functions. We can then integrate to obtain the coefficients 𝐴𝑚𝑛.
∫ 2𝜋 0 d𝜃cos 𝑝𝜃∫ 1 0 𝑟d𝑟𝐽 𝑝(𝑗𝑝𝑞𝑟)𝜙(𝑟, 𝜃) = 𝜋 2 [𝐽 𝑝+1(𝑗𝑝𝑞)] 2𝐴𝑝𝑞 where the 𝜋 2 coefficient is 2𝜋for 𝑝= 0. We can find analogous results for the 𝐵𝑚𝑛, 𝐶𝑚𝑛, 𝐷𝑚𝑛.
Example. Consider an initial radial profile 𝑢(𝑟, 𝜃, 0) = 𝜙(𝑟) = 1−𝑟2. Then, 𝑚= 0, 𝐵𝑚𝑛= 0 for all 𝑚and 𝐴𝑚𝑛= 0 for all 𝑚≠0. Then 𝜕𝑢 𝜕𝑡(𝑟, 0, 0) = 0 hence 𝐶𝑚𝑛, 𝐷𝑚𝑛= 0. We just now need to find 𝐴0𝑛= 2 𝐽0(𝑗0𝑛)2 ∫ 1 0 𝐽0(𝑗0𝑛𝑟)(1 −𝑟)2𝑟d𝑟= 2 𝐽0(𝑗0𝑛)2 𝐽2(𝑗0𝑛) 𝑗2 0𝑛 ≈𝐽2(𝑗0𝑛) 𝑛 as 𝑛→∞ Then the approximate solution is 𝑢(𝑟, 𝜃, 𝑡) = ∞ ∑ 𝑛=1 𝐴0𝑛𝐽0(𝑗0𝑛𝑟) cos 𝑗0𝑛𝑐𝑡 The fundamental frequency is 𝜔𝑑= 𝑗01𝑐 2 𝑑≈4.8 𝑐 𝑑where 𝑑is the diameter of the drum.
Comparing this to a string with length 𝑑, this has a fundamental frequency of 𝜔𝑠= 𝜋𝑐 𝑑≈ 0.77𝜔𝑑.
247 V. Methods 3.12. Diffusion equation derivation with Fourier’s law In a volume 𝑉, the overall heat energy 𝑄is given by 𝑄= ∫ 𝑉 𝑐𝑉𝜌𝜃d𝑉 where 𝑐𝑉is the specific heat of the material, 𝜌is the mass density, and 𝜃is the temperature.
The rate of change due to heat flow is d𝑄 d𝑡= ∫ 𝑉 𝑐𝑉𝜌𝜕𝜃 𝜕𝑡d𝑉 Fourier’s law for heat flow is 𝑞= −𝑘∇𝜃 where 𝑞is the heat flux. We will integrate this over the surface 𝑆= 𝜕𝑉, giving −d𝑄 d𝑡= ∫ 𝑆 𝑞⋅̂ 𝑛d𝑆 The negative sign is due to the normals facing outwards. This is exactly −d𝑄 d𝑡= ∫ 𝑆 (−𝑘∇𝜃) ⋅̂ 𝑛d𝑆= ∫ 𝑉 −𝑘∇2𝜃d𝑉 Equating these two forms for d𝑄 d𝑡, we find ∫ 𝑉 (𝑐𝑉𝜌𝜕𝜃 𝜕𝑡−𝑘∇2𝜃) d𝑉= 0 Since 𝑉was arbitrary, the integrand must be zero. So we have 𝜕𝜃 𝜕𝑡− 𝑘 𝑐𝑉𝜌∇2𝜃= 0 Let 𝐷= 𝑘 𝑐𝑉𝜌be the diffusion constant. Then we have the diffusion equation 𝜕𝜃 𝜕𝑡−𝐷∇2𝜃= 0 3.13. Diffusion equation derivation with statistical dynamics We can derive this equation in another way, using statistical dynamics. Gas particles diffuse by scattering every fixed time step Δ𝑡with probability density function 𝑝(𝜉) of moving by a displacement 𝜉. On average, we have ⟨𝜉⟩= ∫𝑝(𝜉)𝜉d𝜉= 0 248 3. Separation of variables since there is no bias the direction in which any given particle is travelling. Suppose that the probability density function after 𝑁Δ𝑡time is described by 𝑃𝑁Δ𝑡(𝑥). Then, for the next time step, 𝑃(𝑁+1)Δ𝑡(𝑥) = ∫ ∞ −∞ 𝑝(𝜉)𝑃𝑁Δ𝑡(𝑥−𝜉) d𝜉 Using the Taylor expansion, 𝑃(𝑁+1)Δ𝑡(𝑥) ≈∫ ∞ −∞ 𝑝(𝜉)[𝑃𝑁Δ𝑡(𝑥) + 𝑃′ 𝑁Δ𝑡(𝑥)(−𝜉) + 𝑃″ 𝑁Δ𝑡(𝑥)𝜉2 2 + ⋯] d𝜉 ≈𝑃𝑁Δ𝑡(𝑥) −𝑃′ 𝑁Δ𝑡(𝑥) ⟨𝜉⟩+ 𝑃″ 𝑁Δ𝑡(𝑥)⟨𝜉2⟩ 2 + ⋯ ≈𝑃𝑁Δ𝑡(𝑥) + 𝑃″ 𝑁Δ𝑡(𝑥)⟨𝜉2⟩ 2 + ⋯ since ∫𝑝(𝜉) d𝜉= 1. Identifying 𝑃𝑁Δ𝑡(𝑥) = 𝑃(𝑥, 𝑁Δ𝑡), we can write 𝑃(𝑥, (𝑁+ 1)Δ𝑡) −𝑃(𝑥, 𝑁Δ𝑡) = 𝜕2 𝜕𝑥2 𝑃(𝑥, 𝑁Δ𝑡)⟨𝜉2⟩ 2 Assuming that the variance ⟨𝜉2⟩ 2 is proportional to 𝐷Δ𝑡, then for small Δ𝑡, we find 𝜕𝑃 𝜕𝑡= 𝐷𝜕2𝑃 𝜕𝑥2 which is exactly the diffusion equation.
3.14. Similarity solutions The characteristic relation between the variance and time suggests that we seek solutions with a dimensionless parameter. If we can a change of variables of the form 𝜃(𝜂) = 𝜃(𝑥, 𝑡), then it will likely be easier to solve. Consider 𝜂≡ 𝑥 2√𝐷𝑡 Then, 𝜕𝜃 𝜕𝑡= 𝜕𝜂 𝜕𝑡 𝜕𝜃 𝜕𝜂= −1 2 𝑥 √𝐷𝑡3/2 𝜃′ = −1 2 𝜂 𝑡𝜃′ and 𝐷𝜕2𝜃 𝜕𝑥2 = 𝐷𝜕 𝜕𝑥(𝜕𝜂 𝜕𝑥 𝜕𝜃 𝜕𝜂) = 𝐷𝜕 𝜕𝑥( 1 2√𝐷𝑡 𝜃′) = 𝐷 4𝐷𝑡𝜃″ = 1 4𝑡𝜃″ Substituting into the diffusion equation, 𝜃″ = −2𝜂𝜃′ 249 V. Methods Let 𝜓= 𝜃′. Then 𝜓′ 𝜓= −2𝜂⟹ln 𝜓= −𝜂2 + constant Then, choosing a constant of 𝑐 2 √𝜋, 𝜓= 𝑐2 √𝜋 𝑒−𝜂2 ⟹𝜃(𝜂) = 𝑐2 √𝜋 ∫ 𝜂 0 𝑒−𝑢2 d𝑢= 𝑐erf(𝜂) = 𝑐erf( 𝑥 2√𝐷𝑡 ) where erf(𝑧) = 2 √𝜋 ∫ 𝑧 0 𝑒−𝑢2 d𝑢 This describes discontinuous initial conditions that spread over time.
3.15. Heat conduction in a finite bar Suppose we have a bar of length 2𝐿with −𝐿≤𝑥≤𝐿and initial temperature 𝜃(𝑥, 0) = 𝐻(𝑥) = {1 if 0 ≤𝑥≤𝐿 0 if −𝐿≤𝑥< 0 with boundary conditions 𝜃(𝐿, 𝑡) = 1, 𝜃(−𝐿, 𝑡) = 0. Currently the boundary conditions are not homogeneous, so Sturm–Liouville theory cannot be used directly. If we can identify a steady-state solution (time-independent) that reflects the late-time behaviour, then we can turn it into a homogeneous set of boundary conditions. We will try a solution of the form 𝜃𝑠(𝑥) = 𝐴𝑥+ 𝐵 since this certainly satisfies the diffusion equation. To satisfy the boundary conditions, 𝐴= 1 2𝐿; 𝐵= 1 2 Hence we have a solution 𝜃𝑠= 𝑥+ 𝐿 2𝐿 We will subtract this solution from our original equation for 𝜃, giving ̂ 𝜃(𝑥, 𝑡) = 𝜃(𝑥, 𝑡) −𝜃𝑠(𝑥) with homogeneous boundary conditions ̂ 𝜃(−𝐿, 𝑡) = ̂ 𝜃(𝐿, 𝑡) = 0 and initial conditions 𝜃(𝑥, 0) = 𝐻(𝑥) −𝑥+ 𝐿 2𝐿 250 3. Separation of variables We will now separate variables in the usual way. We will consider the ansatz ̂ 𝜃(𝑥, 𝑡) = 𝑋(𝑥)𝑇(𝑡) ⟹𝑋″ = −𝜆𝑋; ̇ 𝑇= −𝐷𝜆𝑇 The boundary conditions imply 𝜆> 0 and give the Fourier modes 𝑋(𝑥) = 𝐴cos √𝜆𝑥+ 𝐵sin √𝜆𝑥. For cos √𝜆𝐿= 0, we require √𝜆𝑚= 𝑚𝜋 2𝐿for 𝑚odd. Also, sin √𝜆𝐿= 0 gives √𝜆𝑛= 𝑛𝜋 𝐿for 𝑛even. Since ̂ 𝜃is odd due to our initial conditions, we can take 𝑋𝑛= 𝐵𝑛sin 𝑛𝜋𝑥 𝐿; 𝜆𝑛= 𝑛2𝜋2 𝐿 2 Substituting into ̇ 𝑇= −𝐷𝜆𝑇, we have 𝑇𝑛(𝑡) = 𝑐𝑛exp(−𝐷𝑛2𝜋2 𝐿 2 𝑡) In general, the solution is ̂ 𝜃(𝑥, 𝑡) = ∞ ∑ 𝑛=1 𝑏𝑛sin 𝑛𝜋𝑥 𝐿 exp(−𝐷𝑛2𝜋2 𝐿 2 𝑡) 3.16. Particular solution to diffusion equation Recall that ̂ 𝜃(𝑥, 𝑡) = ∞ ∑ 𝑛=1 𝑏𝑛sin 𝑛𝜋𝑥 𝐿 exp(−𝐷𝑛2𝜋2 𝐿 2 𝑡) At 𝑡= 0, we have a pure Fourier sine series. We can then impose the initial conditions, to give 𝑏𝑛= 1 𝐿∫ 𝐿 −𝐿 ̂ 𝜙(𝑥, 0) sin 𝑛𝜋𝑥 𝐿 d𝑥 where ̂ 𝜙(𝑥, 0) = 𝐻(𝑥) −𝑥+ 𝐿 2𝐿 Hence, we can use the half-range sine series and find 𝑏𝑛= 2 𝐿∫ 𝐿 0 (𝐻(𝑥) = 1 2) sin 𝑛𝜋𝑥 𝐿 d𝑥 ⏟⎵⎵⎵⎵⎵⎵⎵⎵⏟⎵⎵⎵⎵⎵⎵⎵⎵⏟ square wave/2 −2 𝐿 𝑥 2𝐿sin 𝑛𝜋𝑥 𝐿 d𝑥 ⏟⎵ ⎵ ⎵ ⎵⏟⎵ ⎵ ⎵ ⎵⏟ sawtooth/2𝐿 which gives 𝑏𝑛= 2 (2𝑚−1)𝜋−(−1)𝑛+1 𝑛𝜋 251 V. Methods where 𝑛= 2𝑚−1, and the first term vanishes for 𝑛even. For 𝑛odd or even, we find the same result 𝑏𝑛= 1 𝑛𝜋 Hence ̂ 𝜃(𝑥, 𝑡) = ∞ ∑ 𝑛=1 1 𝑛𝜋sin 𝑛𝜋𝑥 𝐿𝑒−𝐷𝑛2𝜋2 𝐿2 𝑡 For the inhomogeneous boundary conditions, 𝜃(𝑥, 𝑡) = 𝑥+ 𝐿 2𝐿 + ∞ ∑ 𝑛=1 1 𝑛𝜋sin 𝑛𝜋𝑥 𝐿𝑒−𝐷𝑛2𝜋2 𝐿2 𝑡 The similarity solution 1 2(1 + erf( 𝑥 2√𝐷𝑡)) is a good fit for early 𝑡, but it does not necessarily satisfy the boundary conditions, so for large 𝑡it is a bad approximation.
3.17. Laplace’s equation Laplace’s equation is ∇2𝜙= 0 This equation describes (among others) steady-state heat flow, potential theory 𝐹= −∇𝜙, and incompressible fluid flow 𝑣= ∇𝜙. The equation is solved typically on a domain 𝐷, where boundary conditions are specified often on the boundary surface. The Dirichlet bound-ary conditions fix 𝜙on the boundary surface 𝜕𝐷. The Neumann boundary conditions fix ̂ 𝑛⋅∇𝜙on 𝜕𝐷.
3.18. Laplace’s equation in three-dimensional Cartesian coordinates In ℝ3 with Cartesian coordinates, Laplace’s equation becomes 𝜕2𝜙 𝜕𝑥2 + 𝜕2𝜙 𝜕𝑦2 + 𝜕2𝜙 𝜕𝑧2 = 0 We seek separable solutions in the usual way: 𝜙(𝑥, 𝑦, 𝑧) = 𝑋(𝑥)𝑌(𝑦)𝑍(𝑧) Substituting, 𝑋″𝑌𝑍+ 𝑋𝑌″𝑍+ 𝑋𝑌𝑍″ = 0 Dividing by 𝑋𝑌𝑍as usual, 𝑋″ 𝑋= −𝑌″ 𝑌 −𝑍″ 𝑍= −𝜆ℓ 𝑌″ 𝑌= −𝑍″ 𝑍 −𝑋″ 𝑋= −𝜆𝑚 𝑍″ 𝑍= −𝑋″ 𝑋 −𝑌″ 𝑌= −𝜆𝑛= 𝜆ℓ+ 𝜆𝑚 252 3. Separation of variables From the eigenmodes, our general solution will be of the form 𝜙(𝑥, 𝑦, 𝑧) = ∑ ℓ,𝑚,𝑛 𝑎ℓ𝑚𝑛𝑋ℓ(𝑥)𝑌 𝑚(𝑦)𝑍𝑛(𝑧) Consider steady ( 𝜕𝜙 𝜕𝑡= 0) heat flow in a semi-infinite rectangular bar, with boundary con-ditions 𝜙= 0 at 𝑥= 0, 𝑥= 𝑎, 𝑦= 0 and 𝑦= 𝑏; and 𝜙= 1 at 𝑧= 0 and 𝜙→0 as 𝑧→∞. We will solve for each eigenmode successively. First, consider 𝑋″ = −𝜆ℓ𝑋with 𝑋(0) = 𝑋(𝑎) = 0. This gives 𝜆ℓ= 𝑙2𝜋2 𝑎2 ; 𝑋ℓ= sin ℓ𝜋𝑥 𝑎 where ℓ> 0, ℓ∈ℕ. By symmetry, 𝜆𝑚= 𝑚2𝜋2 𝑏2 ; 𝑌 𝑚= sin 𝑚𝜋𝑦 𝑏 For the 𝑧mode, 𝑍″ = −𝜆𝑛𝑍= (𝜆ℓ+ 𝜆𝑚)𝑍= 𝜋2(ℓ2 𝑎2 + 𝑚2 𝑏2 )𝑍 Since 𝜙→0 as 𝑧→∞, the growing exponentials must vanish. Therefore, 𝑍ℓ𝑚= exp[−(ℓ2 𝑎2 + 𝑚2 𝑏2 ) 1/2 𝜋𝑧] Thus the general solution is 𝜙(𝑥, 𝑦, 𝑧) = ∑ ℓ,𝑚 𝑎ℓ𝑚sin ℓ𝜋𝑥 𝑎 sin 𝑚𝜋𝑦 𝑏 exp[−(ℓ2 𝑎2 + 𝑚2 𝑏2 ) 1/2 𝜋𝑧] Now, we will fix 𝑎ℓ𝑚using 𝜙(𝑥, 𝑦, 0) = 1 using the Fourier sine series.
𝑎ℓ𝑚= 2 𝑏∫ 𝑏 0 2 𝑎∫ 𝑎 0 1 sin ℓ𝜋𝑥 𝑎 ⏟ ⎵ ⎵ ⏟ ⎵ ⎵ ⏟ square wave sin 𝑚𝜋𝑦 𝑏 ⏟⎵⏟⎵⏟ square wave d𝑥d𝑦 So only the odd terms remain, giving 𝑎ℓ𝑚= 4𝑎 𝑎(2𝑘−1)𝜋⋅ 4𝑏 𝑏(2𝑝−1)𝜋 where ℓ= 2𝑘−1 is odd and 𝑚= 2𝑝−1 is odd. Simplifying, 𝑎ℓ𝑚= 16 𝜋2ℓ𝑚 for ℓ, 𝑚odd So the heat flow solution is 𝜙(𝑥, 𝑦, 𝑧) = ∑ ℓ,𝑚odd 16 𝜋2ℓ𝑚sin ℓ𝜋𝑥 𝑎 sin ℓ𝜋𝑦 𝑏 exp[−(ℓ2 𝑎2 + 𝑚2 𝑏2 ) 1/2 𝜋𝑧] As 𝑧increases, every contribution but the lowest mode will be very small. So low ℓ, 𝑚dom-inate the solution.
253 V. Methods 3.19. Laplace’s equation in plane polar coordinates In plane polar coordinates, Laplace’s equation becomes 1 𝑟 𝜕 𝜕𝑟(𝑟𝜕𝜙 𝜕𝑟) + 1 𝑟2 𝜕2𝜙 𝜕𝜃2 = 0 Consider a separable form of the answer, given by 𝜙(𝑟, 𝜃) = 𝑅(𝑟)Θ(𝜃) We then have Θ″ + 𝜇Θ = 0; 𝑟(𝑟𝑅′)′ −𝜇𝑅= 0 The polar equation can be solved easily by considering periodic boundary conditions. This gives 𝜇= 𝑚2 and the eigenmodes Θ𝑚(𝜃) = cos 𝑚𝜃, sin 𝑚𝜃 The radial equation is not Bessel’s equation, since there is no second separation constant.
We simply have 𝑟(𝑟𝑅′)′ −𝑚2𝑅= 0 We will try a power law solution, 𝑟= 𝛼𝑟𝛽. We find 𝛽2 −𝑚2 = 0 ⟹𝛽= ±𝑚 So the eigenfunctions are 𝑅𝑚(𝑟) = 𝑟𝑚, 𝑟−𝑚 which is one regular solution at the origin and one singular solution. In the case 𝑚= 0, we have (𝑟𝑅′) = 0 ⟹𝑟𝑅′ = constant ⟹𝑅= log 𝑟 So 𝑅0(𝑟) = constant, log 𝑟 The general solution is therefore 𝜙(𝑟, 𝜃) = 𝑎0 2 + 𝑐0 log 𝑟+ ∞ ∑ 𝑚=1 (𝑎𝑚cos 𝑚𝜃+ 𝑏𝑚sin 𝑚𝜃)𝑟𝑚+ ∞ ∑ 𝑚=1 (𝑐𝑚cos 𝑚𝜃+ 𝑑𝑚sin 𝑚𝜃)𝑟−𝑚 Example. Consider a soap film on a unit disc. We wish to solve Laplace’s equation with a vertically distorted circular wire of radius 𝑟= 1 with boundary conditions 𝜙(1, 𝜃) = 𝑓(𝜃).
The 𝑧displacement of the wire produces the 𝑓(𝜃) term. We wish to find 𝜙(𝑟, 𝜃) for 𝑟< 1, assuming regularity at 𝑟= 0. Then, 𝑐𝑚= 𝑑𝑚= 0 and the solution is of the form 𝜙(𝑟, 𝜃) = 𝑎0 2 + ∞ ∑ 𝑚=1 (𝑎𝑚cos 𝑚𝜃+ 𝑏𝑚sin 𝑚𝜃)𝑟𝑚 254 3. Separation of variables At 𝑟= 1, 𝜙(1, 𝜃) = 𝑓(𝜃) = 𝑎0 2 + ∞ ∑ 𝑚=1 (𝑎𝑚cos 𝑚𝜃+ 𝑏𝑚sin 𝑚𝜃) which is exactly the Fourier series. Thus, 𝑎𝑚= 1 𝜋∫ 2𝜋 0 𝑓(𝜃) cos 𝑚𝜃d𝜃; 𝑏𝑚= 1 𝜋∫ 2𝜋 0 𝑓(𝜃) sin 𝑚𝜃d𝜃 We can see from the equation that high harmonics are confined to have effects only near 𝑟= 1.
3.20. Laplace’s equation in cylindrical polar coordinates In cylindrical coordinates, 1 𝑟 𝜕 𝜕𝑟(𝑟𝜕𝜙 𝜕𝑟) + 1 42 𝜕2𝜙 𝜕𝜃2 + 𝜕2𝜙 𝜕𝑧2 = 0 With 𝜙= 𝑅(𝑟)Θ(𝜃)𝑍(𝑧), we find Θ″ = −𝜇Θ; 𝑍″ = 𝜆𝑍; 𝑟(𝑟𝑅′)′ + (𝜆𝑟2 −𝜇)𝑅= 0 The polar equation can be easily solved by 𝜇𝑚= 𝑚2; Θ𝑚(𝜃) = cos 𝑚𝜃, sin 𝑚𝜃 The radial equation is Bessel’s equation, giving solutions 𝑅= 𝐽𝑚(𝑘𝑟), 𝑌 𝑚(𝑘𝑟) Setting boundary conditions in the usual way, defining 𝑅= 0 at 𝑟= 𝑎means that 𝐽𝑚(𝑘𝑎) = 0 ⟹𝑘= 𝑗𝑚𝑛 𝑎 The radial solution is 𝑅𝑚𝑛(𝑟) = 𝐽𝑚(𝑗𝑚𝑛 𝑎𝑟) We have eliminated the 𝑌 𝑛term since we require 𝑟= 0 to give a finite 𝜙. Finally, the 𝑧 equation gives 𝑍″ = 𝑘2𝑍⟹𝑍= 𝑒−𝑘𝑧, 𝑒𝑘𝑧 We typically eliminate the 𝑒𝑘𝑧mode due to boundary conditions, such as 𝑍→0 as 𝑧→∞.
The general solution is therefore 𝜙(𝑟, 𝜃, 𝑧) = ∞ ∑ 𝑚=0 ∞ ∑ 𝑛=1 (𝑎𝑚𝑛cos 𝑚𝜃+ 𝑏𝑚𝑛sin 𝑚𝜃)𝐽𝑚(𝑗𝑚𝑛 𝑎𝑟)𝑒−𝑓𝑟𝑎𝑐𝑗𝑚𝑛𝑟𝑎 255 V. Methods 3.21. Laplace’s equation in spherical polar coordinates In spherical polar coordinates, 1 𝑟2 𝜕 𝜕𝑟(𝑟2 𝜕Φ 𝜕𝑟) + 1 𝑟2 sin 𝜃 𝜕 𝜕𝜃(sin 𝜃𝜕Φ 𝜕𝜃) + 1 𝑟2 sin2 𝜃 𝜕2Φ 𝜕𝜙2 = 0 We will consider the axisymmetric case; supposing that there is no 𝜙dependence. We seek a separable solution of the form Φ(𝑟, 𝜃) = 𝑅(𝑟)Θ(𝜃) which gives (sin 𝜃Θ′)′ + 𝜆sin 𝜃Θ = 0; (𝑟2𝑅′)′ −𝜆𝑅= 0 Consider the substitution 𝑥= cos 𝜃, d𝑥 d𝜃= −sin 𝜃in the polar equation. This gives dΘ d𝜃= −sin 𝜃 dΘ d𝑥and hence −sin 𝜃d d𝑥[−sin2 𝜃dΘ d𝑥] + 𝜆sin 𝜃Θ = 0 ⟹ d d𝑥[(1 −𝑥2)dΘ d𝑥] + 𝜆Θ = 0 This gives Legendre’s equation, so it has solutions of eigenvalues 𝜆ℓ= ℓ(ℓ+ 1) and eigen-functions Θℓ(𝜃) = 𝑃ℓ(𝑥) = 𝑃ℓ(cos 𝜃) The radial equation then gives (𝑟2𝑅′)′ −ℓ(ℓ+ 1)𝑅= 0 We will seek power law solutions: 𝑅= 𝛼𝑟𝛽. This gives 𝛽(𝛽+ 1) −ℓ(ℓ+ 1) = 0 ⟹𝛽= ℓ, 𝛽= −ℓ−1 Thus the radial eigenmodes are 𝑅ℓ= 𝑟ℓ, 𝑟−ℓ−1 Therefore the general axisymmetric solution for spherical polar coordinates is Φ(𝑟, 𝜃) = ∞ ∑ ℓ=0 (𝑎ℓ𝑟ℓ+ 𝑏ℓ𝑟−ℓ−1)𝑃ℓ(cos 𝜃) The 𝑎ℓ, 𝑏ℓare determined by the boundary conditions. Orthogonality conditions for the 𝑃ℓ can be used to determine coefficients. Consider a solution to Laplace’s equation on the unit sphere with axisymmetric boundary conditions given by Φ(1, 𝜃) = 𝑓(𝜃) Given that we wish to find the interior solution, 𝑏𝑛= 0 by regularity. Then, 𝑓(𝜃) = ∞ ∑ ℓ=0 𝑎ℓ𝑃ℓ(cos 𝜃) 256 3. Separation of variables By defining 𝑓(𝜃) = 𝐹(cos 𝜃), 𝐹(𝑥) = ∞ ∑ ℓ=0 𝑎ℓ𝑃ℓ(𝑥) We can then find the coefficients in the usual way, giving 𝑎ℓ= 2ℓ+ 1 2 ∫ 1 −1 𝐹(𝑥)𝑃ℓ(𝑥) d𝑥 3.22. Generating function for Legendre polynomials Consider a charge at 𝑟0 = (𝑥, 𝑦, 𝑧) = (0, 0, 1). Then, the potential at a point 𝑃becomes Φ(𝑟) = 1 |𝑟−𝑟0| = 1 (𝑥2 + 𝑦2 + (𝑥−1)2)1/2 = 1 (𝑟2(sin2 𝜙+ cos2 𝜙) sin2 𝜃+ 𝑟2 cos2 𝜃−2𝑟cos 𝜃+ 1)1/2 = 1 (𝑟2 sin2 𝜃+ 𝑟2 cos2 𝜃−2𝑟cos 𝜃+ 1)1/2 = 1 (𝑟2 −2𝑟cos 𝜃+ 1)1/2 = 1 (𝑟2 −2𝑟𝑥+ 1)1/2 where 𝑥≡cos 𝜃. This function Φ is a solution to Laplace’s equation where 𝑟≠𝑟0. Note that we can represent any axisymmetric solution as a sum of Legendre polynomials. Now, 1 √𝑟2 −2𝑟𝑥+ 1 = ∞ ∑ ℓ=0 𝑎ℓ𝑃ℓ(𝑥)𝑟ℓ With the normalisation condition for the Legendre polynomials 𝑃ℓ(1) = 1, we find 1 1 −𝑟= ∞ ∑ ℓ=0 𝑎ℓ𝑟ℓ Using the geometric series expansion, we arrive at 𝑎ℓ= 1. This gives 1 √𝑟2 −2𝑟𝑥+ 1 = ∞ ∑ ℓ=0 𝑃ℓ(𝑥)𝑟ℓ which is the generating function for the Legendre polynomials.
257 V. Methods 4. Green’s functions 4.1. Dirac 𝛿function Definition. We define a generalised function 𝛿(𝑥−𝜉) such that (i) 𝛿(𝑥−𝜉) = 0 for all 𝑥≠𝜉; (ii) ∫ ∞ −∞𝛿(𝑥−𝜉) d𝑥= 1.
This acts as a linear operator ∫d𝑥𝛿(𝑥−𝜉) on some function 𝑓(𝑥) to produce a number 𝑓(𝜉).
∫ ∞ −∞ d𝑥𝛿(𝑥−𝜉)𝑓(𝑥) = 𝑓(𝜉) This relationship holds provided that 𝑓(𝑥) is sufficiently ‘well-behaved’ at 𝑥= 𝜉and 𝑥→ ±∞.
Remark. Strictly, the 𝛿‘function’ is classified as a distribution, not as a function. For this reason, we will never use 𝛿outside an integral, although such an integral may be implied.
The 𝛿function represents a unit point source or impulse.
We can approximate the 𝛿function using a Gaussian approximation.
𝛿𝜀(𝑥) = 1 𝜀√𝜋 exp[−𝑥2 𝜀2 ] Therefore, ∫ ∞ −∞ 𝑓(𝑥)𝛿(𝑥) d𝑥= lim 𝜀→0 ∫ ∞ −∞ 1 𝜀√𝜋 exp[−𝑥2 𝜀2 ]𝑓(𝑥) d𝑥 = lim 𝜀→0 ∫ ∞ −∞ 1 𝜀√𝜋 exp[−𝑦2]𝑓(𝜀𝑦) d𝑦 = lim 𝜀→0 ∫ ∞ −∞ 1 𝜀√𝜋 exp[−𝑦2][𝑓(0) + 𝜀𝑦𝑓′(0) + ⋯] d𝑦 = 𝑓(0) for all well-behaved functions 𝑓at 0, ±∞. We could alternatively use the Dirichlet ker-nel 𝛿𝑛(𝑥) = sin 𝑛𝑥 𝜋𝑥 = 1 2𝜋∫ 𝑛 −𝑛 𝑒𝑖𝑘𝑥d𝑘 or even 𝛿𝑛(𝑥) = 𝑛 2 sech2 𝑛𝑥 258 4. Green’s functions 4.2. Integral and derivative of 𝛿function We define the Heaviside step function by 𝐻(𝑥) = {1 𝑥≥0 0 𝑥< 0 For 𝑥≠0, we have 𝐻(𝑥) = ∫ 𝑥 −∞ 𝛿(𝑡) d𝑡 Thus, d d𝑥𝐻(𝑥) = 𝛿(𝑥) where this identification takes place under an implied integral. We define 𝛿′(𝑥) using integ-ration by parts.
∫ ∞ −∞ 𝛿′(𝑥−𝜉)𝑓(𝑥) d𝑥= [𝛿(𝑥−𝜉)𝑓(𝑥)] ∞ −∞−∫ ∞ −∞ 𝛿(𝑥−𝜉)𝑓′(𝑥) d𝑥 = −∫ ∞ −∞ 𝛿(𝑥−𝜉)𝑓′(𝑥) d𝑥 = −𝑓′(𝜉) This is valid for all 𝑓that are smooth at 𝑥= 𝜉.
Example. Consider the Gaussian approximation: 𝛿𝜀(𝑥) = 1 𝜀√𝜋 exp[−𝑥2 𝜀2 ] Then, 𝛿′ 𝜀(𝑥) = −2𝑥 𝜀3√𝜋 exp[−𝑥2 𝜀2 ] 4.3. Properties of 𝛿function Note that ∫ 𝑏 𝑎 𝑓(𝑥)𝛿(𝑥−𝜉) d𝑥= {𝑓(𝜉) 𝑎< 𝜉< 𝑏 0 otherwise So the 𝛿function only ‘samples’ values within the integral range. This is known as the sampling property. Let 𝑢= −(𝑥−𝜉), and consider ∫ ∞ −∞ 𝑓(𝑥)𝛿(−(𝑥−𝜉)) d𝑥= ∫ −∞ ∞ 𝑓(𝜉−𝑢)𝛿(𝑢)(−d𝑢) = ∫ ∞ −∞ 𝑓(𝜉−𝑢)𝛿(𝑢) d𝑢 = 𝑓(𝜉) 259 V. Methods Hence, ∫ ∞ −∞ 𝑓(𝑥)𝛿(−(𝑥−𝜉)) d𝑥= ∫ ∞ −∞ 𝑓(𝑥)𝛿(𝑥−𝜉) d𝑥 This is called the even property. Now, consider ∫ ∞ −∞ 𝑓(𝑥)𝛿(𝑎(𝑥−𝜉)) d𝑥= 1 |𝑎|𝑓(𝜉) This is the scaling property. Let 𝑔(𝑥) be a function with 𝑛isolated roots at 𝑥1, … , 𝑥𝑛. Then, assuming 𝑔′(𝑥) does not vanish at the 𝑥𝑖, 𝛿(𝑔(𝑥)) = 𝑛 ∑ 𝑖=1 𝛿(𝑥−𝑥𝑖) |𝑔′(𝑥𝑖)| This is a generalisation of the above, known as the advanced scaling property. Now, if 𝑔(𝑥) is continuous at 𝑥= 0, then 𝑔(𝑥)𝛿(𝑥) equivalent to 𝑔(0)𝛿(𝑥) inside an integral. This is known as the isolation property.
4.4. Fourier series expansion of 𝛿function Consider a complex Fourier series expansion, 𝛿(𝑥) = ∞ ∑ 𝑛=−∞ 𝑐𝑛𝑒𝑖𝑛𝜋𝑥/𝐿; 𝑐𝑛= 1 2𝐿∫ 𝐿 −𝐿 𝛿(𝑥)𝑒−𝑖𝑛𝜋𝑥/𝐿d𝑥= 1 2𝐿 Hence, 𝛿(𝑥) = 1 2𝐿 ∞ ∑ 𝑛=−∞ 𝑒𝑖𝑛𝜋𝑥/𝐿 Let 𝑓(𝑥) be a function, so 𝑓(𝑥) = ∑ ∞ 𝑛=−∞𝑑𝑛𝑒𝑖𝑛𝜋𝑥/𝐿. Then, their inner product is given by ∫ 𝐿 −𝐿 𝑓⋆(𝑥)𝛿(𝑥) d𝑥= 1 2𝐿 ∞ ∑ 𝑛=−∞ 𝑑𝑛∫ 𝐿 −𝐿 𝑒𝑖𝑛𝜋𝑥/𝐿𝑒𝑖𝑛𝜋𝑥/𝐿d𝑥= ∞ ∑ 𝑛=−∞ 𝑑𝑛= 𝑓(0) The Fourier expansion of the 𝛿function can be extended periodically to the whole real line.
This infinite set of 𝛿functions is known as the Dirac comb, given by ∞ ∑ 𝑚=−∞ 𝛿(𝑥−2𝑚𝐿) = ∞ ∑ 𝑛=−∞ 𝑒𝑖𝑛𝜋𝑥/𝐿 260 4. Green’s functions 4.5. Arbitrary eigenfunction expansion of 𝛿function In general, suppose 𝛿(𝑥−𝜉) = ∞ ∑ 𝑛=1 𝑎𝑛𝑦𝑛(𝑥) with coefficients 𝑎𝑛= ∫ 𝑏 𝑎𝑤(𝑥)𝑦𝑛(𝑥)𝛿(𝑥−𝜉) d𝑥 ∫ 𝑏 𝑎𝑤(𝑥)𝑦𝑛(𝑥)2 d𝑥 = 𝑤(𝜉)𝑦𝑛(𝜉) ∫ 𝑏 𝑎𝑤(𝑥)𝑦𝑛(𝑥)2 d𝑥 = 𝑤𝑛(𝜉)𝑌 𝑛(𝜉) Then, 𝛿(𝑥−𝜉) = 𝑤(𝜉) ∞ ∑ 𝑛=1 𝑌 𝑛(𝜉)𝑌 𝑛(𝑥) = 𝑤(𝑥) ∞ ∑ 𝑛=1 𝑌 𝑛(𝜉)𝑌 𝑛(𝑥) since 𝑤(𝑥) 𝑤(𝜉)𝛿(𝑥−𝜉) = 𝛿(𝑥−𝜉). Hence, 𝛿(𝑥−𝜉) = 𝑤(𝑥) ∞ ∑ 𝑛=1 𝑦𝑛(𝜉)𝑦𝑛(𝑥) 𝑁𝑛 where 𝑁𝑛= ∫ 𝑏 𝑎𝑤𝑦2 𝑛d𝑥is a normalisation factor.
Example. Consider a Fourier series for 𝑦(0) = 𝑦(1) = 0, with 𝑦𝑛(𝑥) = sin 𝑛𝜋𝑥. From the sine series coefficient expression, 𝛿(𝑥−𝜉) = 2 ∞ ∑ 𝑛=1 sin 𝑛𝜋𝜉sin 𝑛𝜋𝑥 where 0 < 𝜉< 1.
4.6. Motivation for Green’s functions Consider a massive static string with tension 𝑇and linear mass density 𝜇, suspended between fixed ends 𝑦(0) = 𝑦(1) = 0. By resolving forces, we have the time independent form 𝑇d2𝑦 d𝑥2 −𝜇𝑔= 0 We will solve the inhomogeneous ODE − d2𝑦 d𝑥2 = 𝑓(𝑥) with 𝑓(𝑥) = − 𝜇𝑔 𝑇. This has been placed in Sturm–Liouville form. We can integrate directly and find −𝑦= −𝜇𝑔 2𝑇𝑥2 + 𝑘1𝑥+ 𝑘2 Imposing boundary conditions, 𝑦(𝑥) = (−𝜇𝑔 𝑇) ⋅1 2𝑥(1 −𝑥) 261 V. Methods Consider alternatively a solution obtained by solving the equation for a single point mass 𝛿𝑚= 𝜇𝛿𝑥suspended at 𝑥= 𝜉on an very light string. We can then superimpose the solu-tions for each point mass to find the overall solution. For a single point mass, the solu-tion is given by two straight lines from (0, 0) and (1, 0) to the point mass (𝜉𝑖, 𝑦𝑖(𝜉𝑖)). The angles of these straight lines from the horizontal are given by 𝜃1, 𝜃2. Resolving in the 𝑦dir-ection, 0 = 𝑇(sin 𝜃1 + sin 𝜃2) −𝛿𝑚𝑔 = 𝑇(−𝑦𝑖 𝜉𝑖 + −𝑦𝑖 1 −𝜉𝑖 ) −𝛿𝑚𝑔 ∴−𝑇(𝑦𝑖(1 −𝜉𝑖) + 𝑦𝑖𝜉𝑖) = 𝛿𝑚𝑔𝜉𝑖(1 −𝜉𝑖) ∴𝑦𝑖(𝜉𝑖) = −𝛿𝑚𝑔 𝑇 𝜉𝑖(1 −𝜉𝑖) So the solution is 𝑦𝑖(𝑥) = −𝛿𝑚𝑔 𝑇 {𝑥(1 −𝜉𝑖) 𝑥< 𝜉𝑖 𝜉𝑖(1 −𝑥) 𝑥> 𝜉𝑖 which is the generalised sawtooth. This can alternatively be written 𝑓𝑖(𝜉)𝐺(𝑥, 𝜉) where 𝑓𝑖is a source term, and 𝐺(𝑥, 𝜉) is the Green’s function, the solution for a unit point source. Since the differential equation is linear, we can sum the solutions, giving 𝑦(𝑥) = 𝑁 ∑ 𝑖=1 𝑓𝑖(𝜉)𝐺(𝑥, 𝜉𝑖) Taking a continuum limit, 𝑓𝑖(𝜉) = −𝛿𝑚𝑔 𝑇 = −𝜇𝛿𝑥𝑔 𝑇 ≡𝑓(𝑥) d𝑥⟹𝑓(𝑥) = −𝜇𝑔 𝑇 which gives 𝑦(𝑥) = ∫ 1 0 𝑓(𝜉)𝐺(𝑥, 𝜉) d𝜉 Substituting the Green’s function, 𝑦(𝑥) = (−𝜇𝑔 𝑇)[∫ 𝑥 0 𝜉(1 −𝑥) d𝜉+ ∫ 1 𝑥 𝑥(1 −𝜉) d𝜉] = (−𝜇𝑔 𝑇){[𝜉2 2 (1 −𝑥)] 𝑥 0 + [𝑥(𝜉−𝜉2 2 )] 1 𝑥 } = (−𝜇𝑔 𝑇)(𝑥2 2 (1 −𝑥) −0 + 𝑥 2 −𝑥(𝑥−𝑥2 2 )) = (−𝜇𝑔 𝑇) ⋅1 2𝑥(1 −𝑥) 262 4. Green’s functions So we have found the correct solution in two ways; once by direct integration, and once by superimposing point solutions. In general, direct integration is not trivial, and Green’s functions are useful in this case.
4.7. Definition of Green’s function We wish to solve the inhomogeneous ODE ℒ𝑦≡𝛼(𝑥)𝑦″ + 𝛽(𝑥)𝑦′ + 𝛾(𝑥)𝑦= 𝑓(𝑥) on 𝑎≤𝑥≤𝑏, where 𝛼≠0 and 𝛼, 𝛽, 𝛾are continuous and bounded, taking homogeneous boundary conditions 𝑦(𝑎) = 𝑦(𝑏) = 0. The Green’s function for ℒin this case is defined to be the solution for a unit point source at 𝑥= 𝜉. That is, 𝐺(𝑥, 𝜉) is the function that satisfies the boundary conditions and ℒ𝐺(𝑥, 𝜉) = 𝛿(𝑥−𝜉) so 𝐺(𝑎, 𝜉) = 𝐺(𝑏, 𝜉) = 0. Then, by linearity, the general solution is given by 𝑦(𝑥) = ∫ 𝑏 𝑎 𝑓(𝜉)𝐺(𝑥, 𝜉) d𝜉 where 𝑦(𝑥) satisfies the homogeneous boundary conditions. We can verify this by check-ing ℒ𝑦= ∫ 𝑏 𝑎 ℒ𝐺(𝑥, 𝜉)𝑓(𝜉) d𝜉= ∫ 𝑏 𝑎 𝛿(𝑥−𝜉)𝑓(𝜉) d𝜉= 𝑓(𝑥) So the solution is given by the inverse operator 𝑦= ℒ−1𝑓; ℒ−1 = ∫ 𝑏 𝑎 d𝜉𝐺(𝑥, 𝜉) The Green’s function spits into two parts; 𝐺(𝑥, 𝜉) = {𝐺1(𝑥, 𝜉) 𝑎≤𝑥< 𝜉 𝐺2(𝑥, 𝜉) 𝜉< 𝑥< 𝑏 For all 𝑥≠𝜉, we have ℒ𝐺1 = ℒ𝐺2 = 0, so the parts are homogeneous solutions. 𝐺satisfies the homogeneous boundary conditions, so 𝐺1(𝑎, 𝜉) = 0 and 𝐺2(𝑏, 𝜉) = 0. 𝐺must be con-tinuous at 𝑥= 𝜉, hence 𝐺1(𝜉, 𝜉) = 𝐺2(𝜉, 𝜉). There is a jump condition; the derivative of 𝐺 is discontinuous at 𝑥= 𝜉. This satisfies [𝐺′]𝜉+ 𝜉−= d𝐺2 d𝑥 | | |𝑥=𝜉+ −d𝐺1 d𝑥 | | |𝑥=𝜉− = 1 𝛼(𝜉) 263 V. Methods 4.8. Explicit form for Green’s functions We want to solve ℒ𝐺(𝑥, 𝜉) = 𝛿(𝑥−𝜉) on 𝑎≤𝑥≤𝑏, subject to homogeneous boundary conditions 𝐺(𝑎, 𝜉) = 𝐺(𝑏, 𝜉) = 0. The functions 𝐺1, 𝐺2 satisfy the homogeneous equation, so ℒ𝐺𝑖(𝑥, 𝜉) = 0. Suppose there exist two independent homogeneous solutions 𝑦1(𝑥), 𝑦2(𝑥) to ℒ𝑦= 0. Then, 𝐺1 = 𝐴𝑦1 + 𝐵𝑦2, such that 𝐴𝑦1(𝑎) + 𝐵𝑦2(𝑎) = 0, which gives a constraint between 𝐴and 𝐵. This defines a complementary function 𝑦−(𝑥) such that 𝑦−(𝑎) = 0. The general homogeneous solution with 𝐺1(𝑎) = 0 is 𝐺1 = 𝐶𝑦− 𝐶will be found later. Similarly we can define 𝑦+ as a linear combination of 𝑦1, 𝑦2 such that 𝑦+(𝑏) = 0.
𝐺2 = 𝐷𝑦+ We require 𝐺1(𝜉, 𝜉) = 𝐺2(𝜉, 𝜉) for continuity, hence 𝐶𝑦−(𝜉) = 𝐷𝑦+(𝜉) Since [𝐺′]𝜉+ 𝜉−= 1 𝛼(𝜉), we have 𝐷𝑦′ +(𝜉) −𝐶𝑌′ −(𝜉) = 1 𝛼(𝜉) We can solve these equations for 𝐶, 𝐷simultaneously to find 𝐶(𝜉) = 𝑦+(𝜉) 𝛼(𝜉)𝑊(𝜉); 𝐷(𝜉) = 𝑦−(𝜉) 𝛼(𝜉)𝑊(𝜉) where 𝑊(𝜉) is the Wrońskian 𝑊(𝜉) = 𝑦−(𝜉)𝑦′ +(𝜉) −𝑦+(𝜉)𝑦′ −(𝜉) which is nonzero if 𝑦−, 𝑦+ are linearly independent. Hence, 𝐺(𝑥, 𝜉) = { 𝑦−(𝑥)𝑦+(𝜉) 𝛼(𝜉)𝑊(𝜉) 𝑎≤𝑥≤𝜉 𝑦−(𝜉)𝑦+(𝑥) 𝛼(𝜉)𝑊(𝜉) 𝜉≤𝑥≤𝑏 4.9. Solving boundary value problems We know that the solution of ℒ𝑦= 𝑓is 𝑦(𝑥) = ∫ 𝑏 𝑎 𝐺(𝑥, 𝜉)𝑓(𝜉) d𝜉 264 4. Green’s functions We can split this into two intervals given that 𝐺= 𝐺1 for 𝜉> 𝑥and 𝐺= 𝐺2 for 𝜉< 𝑥.
𝑦(𝑥) = ∫ 𝑥 𝑎 𝐺2(𝑥, 𝜉)𝑓(𝜉) d𝜉+ ∫ 𝑏 𝑥 𝐺1(𝑥, 𝜉)𝑓(𝜉) d𝜉 = 𝑦+(𝑥) ∫ 𝑥 𝑎 𝑦−(𝜉)𝑓(𝜉) 𝛼(𝜉)𝑊(𝜉) d𝜉+ 𝑦−(𝑥) ∫ 𝑥 𝑎 𝑦+(𝜉)𝑓(𝜉) 𝛼(𝜉)𝑊(𝜉) d𝜉 Note that if ℒis in Sturm–Liouville form, so 𝛽= 𝛼′, then the denominator 𝛼(𝜉)𝑊(𝜉) is a constant. Further, 𝐺is symmetric; 𝐺(𝑥, 𝜉) = 𝐺(𝜉, 𝑥). Often, by convention, we take 𝛼= 1 (however Sturm–Liouville form typically takes 𝛼< 0).
Example. Consider 𝑦″ −𝑦= 𝑓(𝑥) with 𝑦(0) = 𝑦(1) = 0. Homogeneous solutions are 𝑦1 = 𝑒𝑥, 𝑦2 = 𝑒−𝑥. Imposing boundary conditions, 𝐺= {𝐶sinh 𝑥 0 ≤𝑥< 𝜉 𝐷sinh(1 −𝑥) 𝜉< 𝑥≤𝑏 Continuity at 𝑥= 𝜉implies 𝐶sinh 𝜉= 𝐷sinh(1 −𝜉) ⟹𝐶= 𝐷sinh(1 −𝜉) sinh 𝜉 The jump condition is −𝐷cosh(1 −𝜉) −𝐶cosh 𝜉= 1 Hence, −𝐷[cosh(1 −𝜉) sinh 𝜉+ sinh(1 −𝜉) cosh 𝜉] = sinh 𝜉 −𝐷[sinh((1 −𝜉) + 𝜉)] = sinh 𝜉 −𝐷sinh 1 = sinh 𝜉 𝐷= sinh 𝜉 sinh 1 ∴𝐶= −sinh(1 −𝜉) sinh 1 Therefore, 𝑦(𝑥) = −sinh(1 −𝑥) sinh 1 ∫ 𝑥 0 sinh 𝜉𝑓(𝜉) d𝜉−sinh 𝑥 sinh 1 ∫ 1 𝑥 sinh(1 −𝜉)𝑓(𝜉) d𝜉 Suppose we have inhomogeneous boundary conditions. In this case, we want to find a homo-geneous solution 𝑦𝑝that solves the inhomogeneous boundary conditions. That is, ℒ𝑦𝑝= 0 but 𝑦𝑝(𝑎), 𝑦𝑝(𝑏) are as required for the boundary conditions. Then, by subtracting this solu-tion from the original equation, we can solve using a homogeneous set of boundary condi-tions. For instance, in the above example, suppose 𝑦(0) = 0, 𝑦(1) = 1. We can find a solution 𝑦𝑝= sinh 𝑥 sinh 1 which has the inhomogeneous boundary conditions but solves the homogeneous problem.
265 V. Methods 4.10. Higher-order ODEs Suppose ℒ𝑦= 𝑓(𝑥) where ℒis an 𝑛th order linear differential operator, and 𝛼(𝑥) is the coef-ficient for the highest degree derivative. Suppose that homogeneous boundary conditions are satisfied. Then we can define the Green’s function in this case to be the function that solves ℒ𝐺(𝑥, 𝜉) = 𝛿(𝑥−𝜉) which has the properties: (i) 𝐺1, 𝐺2 are homogeneous solutions satisfying the homogeneous boundary conditions; (ii) 𝐺(𝑘) 1 (𝜉) = 𝐺(𝑘) 2 (𝜉) for 𝑘∈{0, … , 𝑛−2}; (iii) 𝐺(𝑛−1) 2 (𝜉+) −𝐺(𝑛−1) 1 (𝜉−) = 1 𝛼(𝜉).
4.11. Eigenfunction expansions of Green’s functions Suppose ℒis in Sturm–Liouville form with eigenfunctions 𝑦𝑛(𝑥) and eigenvalues 𝜆𝑛. We seek 𝐺(𝑥, 𝜉) = ∑ ∞ 𝑛=1 𝐴𝑛𝑦𝑛(𝑥) satisfying ℒ𝐺= 𝛿(𝑥−𝜉).
ℒ𝐺= ∑ 𝑛 𝐴𝑛ℒ𝑦𝑛 = ∑ 𝑛 𝐴𝑛𝜆𝑛𝑤(𝑥)𝑦𝑛(𝑥) The 𝛿function has expansion 𝛿(𝑥−𝜉) = 𝑤(𝑥) ∑ 𝑛 𝑦𝑛(𝜉)𝑦𝑛(𝑥) 𝑁𝑛 ; 𝑁𝑛= ∫𝑤𝑦2 𝑛d𝑥 Hence, 𝐴𝑛(𝜉) = 𝑦𝑛(𝜉) 𝜆𝑛𝑁𝑛 Thus, 𝐺(𝑥, 𝜉) = ∞ ∑ 𝑛=1 𝑦𝑛(𝜉)𝑦𝑛(𝑥) 𝜆𝑛∫𝑤𝑦2 𝑛d𝑥= ∞ ∑ 𝑛=1 𝑌 𝑛(𝜉)𝑌𝑁(𝑥) 𝜆𝑛 which was already obtained earlier in the course when studying Sturm–Liouville theory.
4.12. Constructing Green’s function for an initial value problem Suppose we want to solve ℒ𝑦= 𝑓(𝑡) for 𝑡≥𝑎with 𝑦(𝑎) = 𝑦′(𝑎) = 0, using 𝐺(𝑡, 𝜏) satisfying ℒ𝑔= 𝛿(𝑡−𝜏). For 𝑡< 𝜏, we have 𝐺1 = 𝐴𝑦1(𝑡) + 𝐵𝑦2(𝑡); 𝐴𝑦1(𝑎) + 𝐵𝑦2(𝑎) = 0; 𝐴𝑦′ 1(𝑎) + 𝐵𝑦′ 2(𝑎) = 0 266 4. Green’s functions If 𝐴≠𝐵≠0, then we can solve this by dividing out 𝐴, 𝐵and find 𝑦1𝑦′ 2 −𝑦2𝑦′ 1 = 0. Since the Wrońskian at 𝑎cannot be zero, 𝐴= 𝐵= 0. So 𝐺1(𝑡, 𝜏) ≡0 for 𝑎≤𝑡< 𝜏, so there is no change until the ‘impulse’ at 𝑡= 𝜏.
For 𝑡> 𝜏, by continuity we must have 𝐺2(𝜏, 𝜏) = 0. So we choose a complementary function 𝐺2 = 𝐷𝑦+(𝑡) with 𝑦+(𝑡) = 𝐴𝑦1(𝑡)+𝐵𝑦2(𝑡), and 𝑦+(𝜏) = 0. The discontinuity in the derivative implies that 𝐺′ 2(𝜏, 𝜏) = 𝐷𝑦′ +(𝜏) = 1 𝛼(𝜏) Hence, 𝐴𝑦′ 1(𝜏) + 𝐵𝑦′ 2(𝜏) = 1 𝛼(𝜏) ⟹𝐷(𝜏) = 1 𝛼(𝜏)𝑦′ +(𝜏) Hence we have a non-trivial solution 𝐺(𝑡, 𝜏) = { 0 𝑡< 𝜏 𝑦+(𝑡) 𝛼(𝜏)𝑦′ +(𝜏) 𝑡> 𝜏 The initial value problem has solution 𝑦(𝑡) = ∫ 𝑡 𝑎 𝐺2(𝑡, 𝜏)𝑓(𝜏) d𝜏= ∫ 𝑡 𝑎 𝑦+(𝑡)𝑓(𝜏) 𝑦′ +(𝜏) d𝜏 Causality is ‘built in’ to this solution. Only forces which occur before 𝑡may have an impact on 𝑦(𝑡).
Example. Let us solve 𝑦″ −𝑦= 𝑓(𝑡) with 𝑦(0) = 𝑦′(0) = 0. The homogeneous solution and initial conditions are 𝑡< 𝜏⟹𝐺1 ≡0 and 𝑡> 𝜏⟹𝐺2 = 𝐴𝑒𝑡+ 𝐵𝑒−𝑡= 𝐷sinh(𝑡−𝜏) Now, [𝐺′]𝜏+ 𝜏−= 1 𝛼(𝜏) = 1 ⟹𝐺′(𝜏, 𝜏) = 𝐷cosh 0 = 𝐷= 1 Hence, the solution is 𝑦(𝑡) = ∫ 𝑡 0 𝑓(𝜏) sinh(𝑡−𝜏) d𝜏 267 V. Methods 5. Fourier transforms 5.1. Definitions Definition. The Fourier transform of a function 𝑓(𝑥) is ˜ 𝑓(𝑘) = ℱ(𝑓)(𝑘) = ∫ ∞ −∞ 𝑓(𝑥)𝑒−𝑖𝑘𝑥d𝑥 The inverse Fourier transform is 𝑓(𝑥) = ℱ−1( ˜ 𝑓)(𝑥) = 1 2𝜋∫ ∞ −∞ ˜ 𝑓(𝑘)𝑒𝑖𝑘𝑥d𝑘 Different internally-consistent definitions exist, which distribute the multiplicative constants in different ways.
Theorem (Fourier inversion theorem). For a function 𝑓(𝑥), ℱ−1(ℱ(𝑓))(𝑥) = 𝑓(𝑥) with a sufficient condition that 𝑓and ˜ 𝑓are absolutely integrable, so ∫ ∞ −∞ |𝑓(𝑥)| d𝑥= 𝑀< ∞ In particular, 𝑓→0 as 𝑥→±∞.
Example. Consider the Gaussian, 𝑓(𝑥) = 1 𝜎√𝜋 exp[−𝑥2 𝜎2 ] We wish to compute its Fourier transform. Since 𝑖sin 𝑘𝑥is an odd function, ˜ 𝑓(𝑘) = 1 𝜎√𝜋 ∫ ∞ −∞ exp[−𝑥2 𝜎2 ] exp[−𝑖𝑘𝑥] d𝑥= 1 𝜎√𝜋 ∫ ∞ −∞ exp[−𝑥2 𝜎2 ] cos(𝑘𝑥) d𝑥 Consider, using Leibniz’ rule, d ˜ 𝑓 d𝑘= −1 𝜎√𝜋 ∫ ∞ −∞ 𝑥exp[−𝑥2 𝜎2 ] sin 𝑘𝑥d𝑥 Integrating by parts, d ˜ 𝑓 d𝑘= 1 𝜎√𝜋 [𝜎2 2 exp[−𝑥2 𝜎2 ] sin 𝑘𝑥] ∞ −∞ − 1 𝜎√𝜋 ∫ ∞ −∞ 𝑘𝜎2 2 exp[−𝑥2 𝜎2 ] cos 𝑘𝑥d𝑥 = 1 𝜎√𝜋 ∫ ∞ −∞ 𝑘𝜎2 2 exp[−𝑥2 𝜎2 ] cos 𝑘𝑥d𝑥 = −𝑘𝜎2 2 ˜ 𝑓(𝑘) 268 5. Fourier transforms This is a differential equation for ˜ 𝑓, which gives ˜ 𝑓(𝑘) = 𝐶exp[−𝑘2𝜎2 4 ] Suppose 𝑘= 0. Then, in the original expression for the Fourier transform, we can directly find ˜ 𝑓(0) = 1. Hence 𝐶exp[− 02𝜎2 4 ] = 1 ⟹𝐶= 1. Hence, ˜ 𝑓(𝑘) = exp[−𝑘2𝜎2 4 ] which is another Gaussian with the width parameter inverted.
5.2. Converting Fourier series into Fourier transforms Recall that the complex form of the Fourier series is 𝑓(𝑥) = ∞ ∑ 𝑛=−∞ 𝑐𝑛𝑒𝑖𝑘𝑛𝑥 where 𝑘𝑛= 𝑛𝜋 𝐿. We can write in particular 𝑘𝑛= 𝑛Δ𝑘where Δ𝑘= 𝜋 𝐿. Then, 𝑐𝑛= 1 2𝐿∫ 𝐿 −𝐿 𝑓(𝑥)𝑒−𝑖𝑘𝑛𝑥d𝑥= Δ𝑘 2𝜋∫ 𝐿 −𝐿 𝑓(𝑥)𝑒−𝑖𝑘𝑛𝑥d𝑥 Now, re-substituting into the Fourier series, 𝑓(𝑥) = ∞ ∑ 𝑛=−∞ Δ𝑘 2𝜋𝑒𝑖𝑘𝑛𝑥∫ 𝐿 −𝐿 𝑓(𝑥′)𝑒−𝑖𝑘𝑛𝑥′ d𝑥′ Interpreting the sum multiplied by Δ𝑘as a Riemann integral, 𝑓(𝑥) →∫ ∞ −∞ 1 2𝜋𝑒𝑖𝑘𝑛𝑥∫ 𝐿 −𝐿 𝑓(𝑥′)𝑒−𝑖𝑘𝑥′ d𝑥′ d𝑘 Taking the limit 𝐿→∞, 𝑓(𝑥) = 1 2𝜋∫ ∞ −∞ d𝑘𝑒𝑖𝑘𝑥∫ ∞ −∞ d𝑥′ 𝑓(𝑥′)𝑒−𝑖𝑘𝑛𝑥′ which is the inverse Fourier transform of the Fourier transform of 𝑓, which gives the Four-ier inversion theorem. Note that when 𝑓(𝑥) is discontinuous at 𝑥, the Fourier transform gives ℱ−1(ℱ(𝑓))(𝑥) = 1 2(𝑓(𝑥−) + 𝑓(𝑥+)) which is analogous to the result for Fourier series.
269 V. Methods 5.3. Properties of Fourier series Recall the definition of the Fourier transform.
˜ 𝑓(𝑘) = ∫ ∞ −∞ 𝑓(𝑥)𝑒−𝑖𝑘𝑥d𝑥 The (inverse) Fourier transform is linear.
ℎ(𝑥) = 𝜆𝑓(𝑥) + 𝜇𝑔(𝑥) ⟺˜ ℎ(𝑘) = 𝜆˜ 𝑓(𝑘) + 𝜇̃ 𝑔(𝑘) Translated functions transform to multiplicative factors.
ℎ(𝑥) = 𝑓(𝑥−𝜆) ⟺˜ ℎ(𝑘) = 𝑒−𝑖𝜆𝑘˜ 𝑓(𝑘) This is because ˜ ℎ(𝑘) = ∫𝑓(𝑥−𝜆)𝑒−𝑖𝑘𝑥d𝑥= ∫𝑓(𝑦)𝑒−𝑖𝑘(𝑦+𝜆) d𝑦= 𝑒−𝑖𝜆𝑘˜ 𝑓(𝑘) Frequency shifts transform to translations in frequency space.
ℎ(𝑥) = 𝑒𝑖𝜆𝑥𝑓(𝑥) ⟹˜ ℎ(𝑘) = ˜ 𝑓(𝑘−𝜆) A scalar multiple applied to the argument transforms into an inverse scalar multiple.
ℎ(𝑥) = 𝑓(𝜆𝑥) ⟺˜ ℎ(𝑘) = 1 |𝜆| ˜ 𝑓(𝑘 𝜆) Multiplication by 𝑥transforms into an imaginary derivative.
ℎ(𝑥) = 𝑥𝑓(𝑥) ⟺˜ ℎ(𝑘) = 𝑖˜ 𝑓′(𝑘) This is because ∫ ∞ −∞ 𝑓(𝑥)𝑒−𝑖𝑘𝑥d𝑥= −1 𝑖 d d𝑘∫ ∞ −∞ 𝑓(𝑥)𝑒−𝑖𝑘𝑥d𝑥 Derivatives transform into a muliplication by 𝑖𝑘.
ℎ(𝑥) = 𝑓′(𝑥) ⟺˜ ℎ(𝑘) = 𝑖𝑘˜ 𝑓(𝑘) This is because we can integrate by parts and find ˜ ℎ(𝑘) = ∫ ∞ −∞ 𝑓′(𝑥)𝑒−𝑖𝑘𝑥d𝑥= [𝑓(𝑥)𝑒−𝑖𝑘𝑥] ∞ −∞ ⏟⎵ ⎵ ⎵⏟⎵ ⎵ ⎵⏟ =0 +𝑖𝑘∫ ∞ −∞ 𝑓(𝑥)𝑒−𝑖𝑘𝑥d𝑥 The general duality property states that by mapping 𝑥↦−𝑥, we have 𝑓(−𝑥) = 1 2𝜋∫ ∞ −∞ ˜ 𝑓(𝑘)𝑒−𝑖𝑘𝑥d𝑘 270 5. Fourier transforms hence mapping 𝑘↔𝑥, treating ˜ 𝑓now as a function in position space, we have 𝑓(−𝑘) = 1 2𝜋∫ ∞ −∞ ˜ 𝑓(𝑥)𝑒−𝑖𝑘𝑥d𝑥 Thus 𝑔(𝑥) = ˜ 𝑓(𝑥) ⟺ ̃ 𝑔(𝑘) = 2𝜋𝑓(−𝑘) We can then write the corollary that 𝑓(−𝑥) = 1 2𝜋ℱ(ℱ(𝑓))(𝑥) Finally, ℱ4(𝑓)(𝑥) = 4𝜋2𝑓(𝑥) Example. Consider a function defined by 𝑓(𝑥) = {1 |𝑥| ≤𝑎 0 otherwise for some 𝑎> 0. By the definition of the Fourier transform, ˜ 𝑓(𝑘) = ∫ ∞ −∞ 𝑓(𝑥)𝑒−𝑖𝑘𝑥d𝑥= ∫ 𝑎 −𝑎 𝑒−𝑖𝑘𝑥d𝑥= ∫ 𝑎 −𝑎 cos 𝑘𝑥d𝑥= 2 𝑘sin 𝑘𝑎 By the Fourier inversion theorem, 1 𝜋∫ ∞ −∞ 𝑒𝑖𝑘𝑥1 𝑘sin 𝑘𝑎d𝑘= 𝑓(𝑥) for 𝑥≠𝑎. Now, in this expression, let 𝑥= 0 and let 𝑘↦𝑥. We arrive at the Dirichlet discontinuous formula.
∫ ∞ 0 sin 𝑎𝑥 𝑥 d𝑥= 𝜋 2 sgn 𝑎= ⎧ ⎨ ⎩ 𝜋 2 𝑎> 0 0 𝑎= 0 − 𝜋 2 𝑎< 0 5.4. Convolution theorem We want to multiply Fourier transforms in the frequency domain (transformed space). This is useful for filtering or processing signals.
˜ ℎ(𝑘) = ˜ 𝑓(𝑘) ̃ 𝑔(𝑘) 271 V. Methods Consider the inverse.
ℎ(𝑥) = 1 2𝜋∫ ∞ −∞ ˜ 𝑓(𝑘) ̃ 𝑔(𝑘)𝑒𝑖𝑘𝑥d𝑘 = 1 2𝜋∫ ∞ −∞ (∫ ∞ −∞ 𝑓(𝑦)𝑒−𝑖𝑘𝑦d𝑦) ̃ 𝑔(𝑘)𝑒𝑖𝑘𝑥d𝑘 = ∫ ∞ −∞ 𝑓(𝑦)( 1 2𝜋∫ ∞ −∞ 𝑒−𝑖𝑘𝑦̃ 𝑔(𝑘)𝑒𝑖𝑘𝑥d𝑘) d𝑦 = ∫ ∞ −∞ 𝑓(𝑦)( 1 2𝜋∫ ∞ −∞ ̃ 𝑔(𝑘)𝑒𝑖𝑘(𝑥−𝑦) d𝑘) d𝑦 = ∫ ∞ −∞ 𝑓(𝑦)𝑔(𝑥−𝑦) d𝑦 = (𝑓∗𝑔)(𝑥) where 𝑓∗𝑔is called the convolution of 𝑓and 𝑔. By duality, we also have ℎ(𝑥) = 𝑓(𝑥)𝑔(𝑥) ⟹˜ ℎ(𝑘) = 1 2𝜋∫ ∞ −∞ ˜ 𝑓(𝑝) ̃ 𝑔(𝑘−𝑝) d𝑝= 1 2𝜋( ˜ 𝑓∗̃ 𝑔)(𝑘) 5.5. Parseval’s theorem Consider ℎ(𝑥) = 𝑔⋆(−𝑥). Then, by letting 𝑥= −𝑦, ˜ ℎ(𝑘) = ∫ ∞ −∞ 𝑔⋆(−𝑥)𝑒−𝑖𝑘𝑥d𝑥 = [∫ ∞ −∞ 𝑔(−𝑥)𝑒𝑖𝑘𝑥d𝑥] ⋆ = [∫ ∞ −∞ 𝑔(𝑦)𝑒−𝑖𝑘𝑦d𝑦] ⋆ = ̃ 𝑔⋆(𝑘) Substituting this into the convolution theorem, with 𝑔(𝑥) ↦𝑔⋆(−𝑥), we have ∫ ∞ −∞ 𝑓(𝑦)𝑔⋆(𝑦−𝑥) d𝑦= 1 2𝜋∫ ∞ −∞ ˜ 𝑓(𝑘) ̃ 𝑔⋆(𝑘)𝑒𝑖𝑘𝑥d𝑥 Taking 𝑥= 0 in this expression and mapping 𝑦↦𝑥, we find ∫ ∞ −∞ 𝑓(𝑥)𝑔⋆(𝑥) d𝑥= 1 2𝜋∫ ∞ −∞ ˜ 𝑓(𝑘) ̃ 𝑔⋆(𝑘) d𝑥 Equivalently, ⟨𝑔, 𝑓⟩= 1 2𝜋⟨̃ 𝑔, ˜ 𝑓⟩ 272 5. Fourier transforms So the inner product is conserved under the Fourier transform (up to a factor of 2𝜋). Now, by setting 𝑔⋆= 𝑓⋆, we have ∫ ∞ −∞ |𝑓(𝑥)|2 d𝑥= 1 2𝜋∫ ∞ −∞ | | ˜ 𝑓(𝑘)| | 2 d𝑘 This is Parseval’s theorem.
5.6. Fourier transforms of generalised functions We can apply Fourier transforms to generalised functions by considering limiting distribu-tions. Consider the inversion 𝑓(𝑥) = ℱ−1(ℱ(𝑓))(𝑥) = 1 2𝜋∫ ∞ −∞ [∫ ∞ −∞ 𝑓(𝑢)𝑒−𝑖𝑘𝑢d𝑢]𝑒𝑖𝑘𝑥d𝑘 = 1 2𝜋∫ ∞ −∞ 𝑓(𝑢) [ 1 2𝜋∫ ∞ −∞ 𝑒−𝑖𝑘(𝑥−𝑢) d𝑘] ⏟⎵ ⎵ ⎵ ⎵ ⎵ ⎵⏟⎵ ⎵ ⎵ ⎵ ⎵ ⎵⏟ 𝛿(𝑥−𝑢) d𝑢 In order to reconstruct 𝑓(𝑥) on the right hand side for any function 𝑓, we must have that the bracketed term is 𝛿(𝑥−𝑢). So we identify 𝛿(𝑥−𝑢) = 1 2𝜋∫ ∞ −∞ 𝑒𝑖𝑘(𝑥−𝑢) d𝑘 If 𝑓(𝑥) = 𝛿(𝑥), ˜ 𝑓(𝑘) = ∫ ∞ −∞ 𝛿(𝑥)𝑒𝑖𝑘𝑥d𝑥= 1 This can be thought of as the Fourier transform of an infinitely thin Gaussian, which be-comes an infinitely wide Gaussian (a constant). If 𝑓(𝑥) = 1, then ˜ 𝑓(𝑘) = ∫ ∞ −∞ 𝑒−𝑖𝑘𝑥d𝑥= 2𝜋𝛿(𝑘) This can also be found by the duality formula. If 𝑓(𝑥) = 𝛿(𝑥−𝑎), we have ˜ 𝑓(𝑘) = 𝑒−𝑖𝑘𝑎 This is a translation of the original Fourier transform for the 𝛿function above.
273 V. Methods 5.7. Trigonometric functions Let 𝑓(𝑥) = cos 𝜔𝑥= 1 2(𝑒𝑖𝑥+ 𝑒−𝑖𝑥). Then, ˜ 𝑓(𝑘) = 𝜋(𝛿(𝑘+ 𝜔) + 𝛿(𝑘−𝜔)) For 𝑓(𝑥) = sin 𝜔𝑥, we have ˜ 𝑓(𝑘) = 𝑖𝜋(𝛿(𝑘+ 𝜔) −𝛿(𝑘−𝜔)) Using duality, 𝑓(𝑥) = 1 2(𝛿(𝑥+ 𝑎) + 𝛿(𝑥−𝑎)) ⟹˜ 𝑓(𝑘) = cos 𝑘𝑎 𝑓(𝑥) = 1 2𝑖(𝛿(𝑥+ 𝑎) −𝛿(𝑥−𝑎)) ⟹˜ 𝑓(𝑘) = sin 𝑘𝑎 5.8. Heaviside functions Let 𝐻(𝑥) be the Heaviside function, such that 𝐻(0) = 1 2. Then, 𝐻(𝑥) + 𝐻(−𝑥) = 1 for all 𝑥.
We can take the Fourier transform of this and find ˜ 𝐻(𝑘) + ˜ 𝐻(−𝑘) = 2𝜋𝛿(𝑘) Recall that 𝐻′(𝑥) = 𝛿(𝑥). Thus, 𝑖𝑘˜ 𝐻(𝑥) = ̃ 𝛿(𝑘) = 1 Since 𝑘𝛿(𝑘) = 0, the two equations for ˜ 𝐻can be consistent if we take ˜ 𝐻(𝑘) = 𝜋𝛿(𝑘) + 1 𝑖𝑘 5.9. Dirichlet discontinuous formula Recall the Dirichlet discontinuous formula: ∫ ∞ 0 sin 𝑎𝑥 𝑥 d𝑥= 𝜋 2 sgn 𝑎= ⎧ ⎨ ⎩ 𝜋 2 𝑎> 0 0 𝑎= 0 − 𝜋 2 𝑎< 0 We can rewrite this as 1 2 sgn 𝑥= 1 2𝜋∫ ∞ −∞ 𝑒𝑖𝑘𝑥 𝑖𝑘d𝑘 since the cosine term divided by 𝑖𝑘is odd. Hence, 𝑓(𝑥) = 1 2 sgn 𝑥⟺˜ 𝑓(𝑘) = 1 𝑖𝑘 This is the preferred form for a Heaviside-type function when used in Fourier transforms.
274 5. Fourier transforms 5.10. Solving ODEs for boundary value problems Consider 𝑦″ −𝑦= 𝑓(𝑥) with homogeneous boundary conditions 𝑦→0 as 𝑥→±∞. Taking the Fourier transform of this expression, we find (−𝑘2 −1) ̃ 𝑦= ˜ 𝑓 Thus, the solution is ̃ 𝑦(𝑘) = −˜ 𝑓(𝑘) 1 + 𝑘2 ≡˜ 𝑓(𝑘) ̃ 𝑔(𝑘) where ̃ 𝑔(𝑘) = −1 1+𝑘2 . Note that ̃ 𝑔(𝑘) is the Fourier transform of 𝑔(𝑥) = − 1 2𝑒−|𝑥|. Applying the convolution theorem, 𝑦(𝑥) = ∫ ∞ −∞ 𝑓(𝑢)𝑔(𝑥−𝑢) d𝑢 = −1 2 ∫ ∞ −∞ 𝑓(𝑢)𝑒−|𝑥−𝑢| d𝑢 = −1 2[∫ 𝑥 −∞ 𝑓(𝑢)𝑒𝑢−𝑥d𝑢+ ∫ ∞ 𝑥 𝑓(𝑢)𝑒𝑥−𝑢d𝑢] This is in the form of a boundary value problem Green’s function. We can construct the same results by constructing the Green’s function directly.
5.11. Signal processing Suppose we have an input signal ℐ(𝑡), which is acted on by some linear operator ℒin to yield an output 𝒪(𝑡). The Fourier transform of the input ̃ ℐ(𝜔) is called the resolution.
̃ ℐ(𝜔) = ∫ ∞ −∞ ℐ(𝑡)𝑒−𝑖𝜔𝑡d𝑡 In the frequency domain, the action of ℒin on ℐ(𝑡) means that ̃ ℐ(𝜔) is multiplied by a transfer function ˜ ℛ(𝜔). Thus, 𝒪(𝑡) = 1 2𝜋∫ ∞ −∞ ˜ ℛ(𝜔) ̃ ℐ(𝜔)𝑒𝑖𝜔𝑡d𝜔 The inverse Fourier transform of the transfer function, ℛ, is called the response function, which is given by ℛ(𝑡) = 1 2𝜋∫ ∞ −∞ ˜ ℛ(𝜔)𝑒𝑖𝜔𝑡d𝜔 By the convolution theorem, 𝒪(𝑡) = ∫ ∞ −∞ ℐ(𝑢)ℛ(𝑡−𝑢) d𝑢 275 V. Methods Suppose there is no input (ℐ(𝑡) = 0) for 𝑡< 0. By causality, there should be zero output for the response function (ℛ(𝑡) = 0) for 𝑡< 0. Therefore, we require 0 < 𝑢< 𝑡and hence 𝒪(𝑡) = ∫ 𝑡 0 ℐ(𝑢)ℛ(𝑡−𝑢) d𝑢 which resembles an initial value problem Green’s function.
5.12. General transfer functions for ODEs Suppose an input-output relationship is given by a linear ODE.
ℒ𝒪(𝑡) ≡( 𝑛 ∑ 𝑖=0 𝑎𝑖 d𝑖 d𝑥𝑖)𝒪(𝑡) ≡ℐ(𝑡) Here, ℒin = 1. We want to solve this ODE using a Fourier transform.
(𝑎0 + 𝑎1𝑖𝜔−𝑎2𝜔2 −𝑎3𝑖𝜔3 + ⋯+ 𝑎𝑛(𝑖𝜔)𝑛) ˜ 𝒪(𝜔) = ̃ ℐ(𝜔) We can solve this algebraically in Fourier transform space. The transfer function is ˜ ℛ(𝜔) = 1 𝑎0 + ⋯+ 𝑎𝑛(𝑖𝜔)𝑛 We factorise the denominator to find partial fractions. Suppose there are 𝐽distinct roots (𝑖𝜔−𝑐𝑗)𝑘𝑗, where 𝑘𝑗is the algebraic multiplicity of the 𝑗th root, so ∑ 𝐽 𝑗=1 𝑘𝑗= 𝑛. So we can write ˜ ℛ(𝜔) = 1 (𝑖𝜔−𝑐1)𝑘1 … (𝑖𝜔−𝑐𝐽)𝑘𝐽 Expressing this as partial fractions, ˜ ℛ(𝜔) = 𝐽 ∑ 𝑗=1 𝑘𝑖 ∑ 𝑚=1 Γ 𝑗𝑚 (𝑖𝜔−𝑐𝑗)𝑚 The Γ 𝑗𝑚terms are constant. To solve this, we must find the inverse Fourier transform of (𝑖𝜔−𝑎)−𝑚. Recall that ℱ−1( 1 𝑖𝜔−𝑎) = {𝑒𝑎𝑡 𝑡> 0 0 𝑡< 0 for Re 𝑎< 0. So we will require Re 𝑐𝑗< 0 for all 𝑗to eliminate exponentially growing solutions. Note that for 𝑛= 2, 𝑖d d𝜔( 1 (𝑖𝜔−𝑎)2 ) and recall that ℱ(𝑡𝑓(𝑡)) = 𝑖ℱ′(𝜔) 276 5. Fourier transforms Hence, ℱ−1( 1 (𝑖𝜔−𝑎)2 ) = {𝑡𝑒𝑎𝑡 𝑡> 0 0 𝑡< 0 Inductively, we arrive at ℱ−1( 1 (𝑖𝜔−𝑎)𝑚) = { 𝑡𝑚−1 (𝑚−1)!𝑒𝑎𝑡 𝑡> 0 0 𝑡< 0 We can therefore invert any transfer function to obtain the response function. Thus the response function takes the form ℛ(𝑡) = 𝐽 ∑ 𝑗=1 𝑘𝑖 ∑ 𝑚=1 Γ 𝑗𝑚 𝑡𝑚−1 (𝑚−1)!𝑒𝑐𝑗𝑡; 𝑡> 0 and zero for 𝑡< 0. We can now solve such differential equations in Green’s function form, or directly invert ˜ ℛ(𝜔) ̃ ℐ(𝜔) for a polynomial ̃ ℐ(𝜔).
5.13. Damped oscillator We can use the Fourier transform method to solve the differential equation ℒ𝑦≡𝑦″ + 2𝑝𝑦′ + (𝑝2 + 𝑞2)𝑦= 𝑓(𝑡) where 𝑝> 0. Consider homogeneous boundary conditions 𝑦(0) = 𝑦′(0) = 0. The Fourier transform is (𝑖𝜔)2 ̃ 𝑦+ 2𝑖𝑝𝜔̃ 𝑦+ (𝑝2 + 𝑞2) ̃ 𝑦= ˜ 𝑓 Hence, ̃ 𝑦= ˜ 𝑓 −𝜔2 + 2𝑖𝑝𝜔+ 𝑝2 + 𝑞2 ≡˜ 𝑅˜ 𝑓 We can invert this using the convolution theorem by inverting ˜ 𝑅.
𝑦(𝑡) = ∫ 𝑡 0 ℛ(𝑡−𝜏)𝑓(𝜏) d𝜏 where the response function is ℛ(𝑡−𝜏) = 1 2𝜋∫ ∞ −∞ 𝑒𝑖𝜔(𝑡−𝜏) 𝑝2 + 𝑞2 + 2𝑖𝑝𝜔−𝜔2 d𝜔 We can show that ℒℛ(𝑡−𝜏) = 𝛿(𝑡−𝜏); in other words, ℛis the Green’s function.
277 V. Methods 5.14. Discrete sampling and the Nyquist frequency Suppose a signal ℎ(𝑡) is sampled at equal times 𝑡𝑛= 𝑛Δ with a time step Δ and values ℎ𝑛= ℎ(𝑡𝑛) = ℎ(𝑛Δ), for all 𝑛∈ℤ. The sampling frequency is therefore Δ−1, so the sampling angular velocity is 𝜔𝑠= 2𝜋𝑓 𝑠= 2𝜋 Δ . The Nyquist frequency is 𝑓 𝑐= 1 2Δ, which is the highest frequency actually sampled at Δ. Suppose we have a signal 𝑔𝑓with a given frequency 𝑓. We will write 𝑔𝑓(𝑡) = 𝐴cos(2𝜋𝑓𝑡+ 𝜑) = Re (𝐴𝑒2𝜋𝑖𝑓𝑡+𝜑) = 1 2(𝐴𝑒2𝜋𝑖𝑓𝑡+𝜑) + 1 2(𝐴𝑒−2𝜋𝑖𝑓𝑡+𝜑) where 𝐴∈ℝ. Note that this signal has two ‘frequencies’; a positive and a negative frequency.
The combination of these frequencies gives the full wave. Suppose we sample 𝑔𝑓(𝑡) at the Nyquist frequency, so 𝑓= 𝑓 𝑐. Then, 𝑔𝑓𝑐(𝑡𝑛) = 𝐴cos(2𝜋1 2Δ𝑛Δ + 𝜑) = 𝐴cos(𝜋𝑛+ 𝜑) = 𝐴cos 𝜋𝑛cos 𝜙+ 𝐴sin 𝜋𝑛sin 𝜙 = 𝐴′ cos(2𝜋𝑓 𝑐𝑓 𝑛) where 𝐴′ = 𝐴cos 𝜙. This has removed half of the information about the wave; the amplitude and the phase have become degenerate. We can identify 𝑓 𝑐with −𝑓 𝑐when considering the remaining information; we say that the two frequencies are aliased together. Now, suppose we sample at greater than the Nyquist frequency, in particular 𝑓= 𝑓 𝑐+ 𝛿𝑓> 𝑓 𝑐, where for simplicity we let 𝛿𝑓< 𝑓 𝑐. We have 𝑔𝑓(𝑡𝑛) = 𝐴cos(2𝜋(𝑓 𝑐+ 𝛿𝑓)𝑡𝑛+ 𝜑) = 𝐴cos(2𝜋(𝑓 𝑐−𝛿𝑓)𝑡𝑛−𝜑) So frequencies above the Nyquist frequency are reinterpreted after the sampling as a fre-quency lower than the Nyquist frequency. This aliases 𝑓 𝑐+ 𝛿𝑓with 𝑓 𝑐−𝛿𝑓.
5.15. Nyquist–Shannon sampling theorem Definition. A signal 𝑔(𝑡) is bandwidth-limited if it contains no frequencies above 𝜔max = 2𝜋𝑓max. In other words, ̃ 𝑔(𝜔) = 0 for all |𝜔| > 𝜔max. In this case, 𝑔(𝑡) = 1 2𝜋∫ ∞ −∞ ̃ 𝑔(𝜔)𝑒𝑖𝜔𝑡d𝜔= 1 2𝜋∫ 𝜔max −𝜔max ̃ 𝑔(𝜔)𝑒𝑖𝜔𝑡d𝜔 Suppose we set the sampling rate to the Nyquist frequency, so Δ = 1 2𝑓max . Then, 𝑔𝑛≡𝑔(𝑡𝑛) = 1 2𝜋∫ 𝜔max −𝜔max ̃ 𝑔(𝜔)𝑒𝑖𝜋𝑛𝜔/𝜔max d𝜔 278 5. Fourier transforms This is a complex Fourier series coefficient 𝑐𝑛, multiplied by 𝜔max 𝜋. The Fourier series is periodic in 𝜔with period 2𝜔max, not in space or time.
̃ 𝑔per(𝜔) = 𝜋 𝜔max ∞ ∑ 𝑛=−∞ 𝑔𝑛𝑒−𝑖𝜋𝑛𝜔/𝜔max The actual Fourier transform ̃ 𝑔is found by multiplying by a top hat window function ˜ ℎ(𝜔) = {1 |𝜔| ≤𝜔max 0 otherwise Hence, ̃ 𝑔(𝜔) = ̃ 𝑔per(𝜔)˜ ℎ(𝜔) Note that this relation is exact. Inverting this expression, 𝑔(𝑡) = 1 2𝜋∫ ∞ −∞ ̃ 𝑔per(𝜔)˜ ℎ(𝜔)𝑒𝑖𝜔𝑡d𝜔 = 1 2𝜔max ∞ ∑ 𝑛=−∞ 𝑔𝑛∫ 𝜔max −𝜔max exp(𝑖𝜔(𝑡−𝑛𝜋 𝜔max )) d𝜔 Only the cosine term is even, hence 𝑔(𝑡) = 1 2𝜔max ∞ ∑ 𝑛=−∞ 𝑔𝑛 sin(𝜔max𝑡−𝜋𝑛) 𝜔max𝑡−𝜋𝑛 Hence, 𝑔(𝑡) can be written exactly as a combination of countably many discrete sample points.
5.16. Discrete Fourier transform Suppose we have a finite number of samples ℎ𝑚= ℎ(𝑡𝑚) for 𝑡𝑚= 𝑚Δ, where 𝑚= 0, … , 𝑁−1.
We will approximate the Fourier transform for 𝑁frequencies within the Nyquist frequency 𝑓 𝑐= 1 2Δ, using equally-spaced frequencies, given by Δ𝑓= 1 𝑁Δ in the range −𝑓 𝑐≤𝑓≤𝑓 𝑐. We could take the convention 𝑓 𝑛= 𝑛Δ𝑓= 𝑛 𝑁Δ for 𝑛= − 𝑁 2 , … , − 𝑁 2 . However, this overcounts the Nyquist frequency (which is aliased), giving 𝑁+ 1 frequencies instead of the desired 𝑁.
Since frequencies above the Nyquist frequency are aliased to below it: (𝑁 2 + 𝑚)Δ𝑓= 𝑓 𝑐+ 𝛿𝑓↦(𝑁 2 −𝑚)Δ𝑓= −(𝑓 𝑐−𝛿𝑓) 279 V. Methods we can instead use the convention 𝑓 𝑛= 𝑛Δ𝑓= 𝑛 𝑁Δ for 𝑛= 0, … , 𝑁−1. This counts the Nyquist frequency only once. The Fourier transform at a frequency 𝑓 𝑛becomes ˜ ℎ(𝑓 𝑛) = ∫ ∞ −∞ ℎ(𝑡)𝑒−2𝜋𝑖𝑓𝑛𝑡d𝑡 ≈Δ 𝑁−1 ∑ 𝑚=0 ℎ𝑚𝑒−2𝜋𝑖𝑓𝑛𝑡𝑚 = Δ 𝑁−1 ∑ 𝑚=0 ℎ𝑚𝑒−2𝜋𝑖𝑚𝑛/𝑁 = Δ˜ ℎ𝑑(𝑓 𝑛) where the function ˜ ℎ𝑑(𝑓 𝑛) is the discreteFouriertransform. The matrix [DFT]𝑚𝑛= 𝑒−2𝜋𝑖𝑚𝑛/𝑁 defines the discrete Fourier transform for the vector ℎ= {ℎ𝑚}. The discrete Fourier trans-form is then ˜ ℎ𝑑= [DFT]ℎ By inverting the discrete Fourier transform matrix, we find ℎ= [DFT]−1˜ ℎ𝑑= 1 𝑁[DFT]†˜ ℎ𝑑 since the inverse of the discrete Fourier transform matrix is its adjoint. The matrix is built from roots of unity 𝜔= 𝑒−2𝜋𝑖/𝑁. So, for instance, 𝑛= 4 gives 𝜔= 𝑒−2𝜋𝑖/4 = −𝑖giving [DFT] = ⎛ ⎜ ⎜ ⎝ 1 1 1 1 1 −𝑖 −1 𝑖 1 −1 1 −1 1 𝑖 −1 −𝑖 ⎞ ⎟ ⎟ ⎠ The inverse discrete Fourier transform is ℎ𝑚= ℎ(𝑡𝑚) = 1 2𝜋∫ ∞ −∞ ˜ ℎ(𝜔)𝑒𝑖𝜔𝑡𝑚d𝜔 = ∫ ∞ −∞ ˜ ℎ(𝑓)𝑒2𝜋𝑖𝑓𝑡𝑚d𝑓 ≈ 1 Δ𝑁 𝑁−1 ∑ 𝑛=0 Δ˜ ℎ𝑑(𝑓 𝑛)𝑒2𝜋𝑖𝑚𝑛/𝑁 = 1 𝑁 𝑁−1 ∑ 𝑛=0 ˜ ℎ𝑛𝑒2𝜋𝑖𝑚𝑛/𝑁 Hence, we can interpolate the initial function from its samples.
ℎ(𝑡) = 1 𝑁 𝑁−1 ∑ 𝑛=0 ˜ ℎ𝑛𝑒2𝜋𝑖𝑛𝑡/𝑁 280 5. Fourier transforms Parseval’s theorem becomes 𝑁−1 ∑ 𝑚=0 |ℎ𝑚|2 = 1 𝑁 𝑁−1 ∑ 𝑛=0 | |˜ ℎ𝑛| | 2 and the convolution theorem is 𝑐𝑘= 𝑁−1 ∑ 𝑚=0 𝑔𝑚ℎ𝑘−𝑚⟺ ̃ 𝑐𝑘= ̃ 𝑔𝑘˜ ℎ𝑘 5.17. Fast Fourier transform (non-examinable) While the discrete Fourier transform is an order 𝑂(𝑁2) operation, we can reduce this into an order 𝑂(𝑛log 𝑁) operation. Such a simplification is called the fast Fourier transform. We can split the discrete Fourier transform into even and odd parts, noting that 𝜔𝑁= 𝑒−2𝜋𝑖/𝑁 implies 𝜔2 𝑁= 𝑒−2𝜋𝑖/(𝑁/2) = 𝜔𝑁/2 ˜ ℎ𝑘= 𝑁−1 ∑ 𝑛=0 ℎ𝑛𝜔𝑛𝑘 𝑁 = 𝑁/2−1 ∑ 𝑚=0 ℎ2𝑚𝜔2𝑚𝑘 𝑁 + 𝑁/2−1 ∑ 𝑚=0 ℎ2𝑚+1𝜔(2𝑚+1)𝑘 𝑁 = 𝑁/2−1 ∑ 𝑚=0 ℎ2𝑚(𝜔2 𝑁)𝑚𝑘+ 𝜔𝑘 𝑁 𝑁/2−1 ∑ 𝑚=0 ℎ2𝑚+1(𝜔2 𝑁)𝑚𝑘 = 𝑁/2−1 ∑ 𝑚=0 ℎ2𝑚(𝜔𝑁/2)𝑚𝑘+ 𝜔𝑘 𝑁 𝑁/2−1 ∑ 𝑚=0 ℎ2𝑚+1(𝜔𝑁/2)𝑚𝑘 This algorithm iteratively reduces the Fourier transform’s complexity by a factor of two, until the trivial case of finding the discrete Fourier transform of two data points.
281 V. Methods 6. Method of characteristics 6.1. Well-posed Cauchy problems Solving partial differential equations depends on the nature of the equations in combination with the boundary or initial data. A Cauchy problem is the partial differential equation for some function 𝜙together with the auxiliary data (in 𝜙and its derivatives) specified on a surface (or a curve in two dimensions), which is called Cauchy data. For a Cauchy problem to be well-posed, we require that (i) a solution exists (we do not have excessive auxiliary data); (ii) the solution is unique (we do not have insufficient auxiliary data); and (iii) the solution depends continuously on the auxiliary data.
6.2. Method of characteristics Consider a parametrised curve 𝐶given by Cartesian coordinates (𝑥(𝑠), 𝑦(𝑠)). The tangent vector is 𝑣= (d𝑥(𝑠) d𝑠, d𝑦(𝑠) d𝑠) We then define the directional derivative of a function 𝜙(𝑥, 𝑦) by d𝜙 d𝑠 | | |𝐶 = d𝑥(𝑠) d𝑠 𝜕𝜙 𝜕𝑥+ d𝑦(𝑠) d𝑠 𝜕𝜙 𝜕𝑦= 𝑣⋅∇𝜙 Suppose 𝑣⋅∇𝜙= 0 then d𝜙 d𝑠= 0 and hence 𝜙is constant along the curve. Suppose there exists a vector field 𝑢= (𝛼(𝑥, 𝑦), 𝛽(𝑥, 𝑦)) with a family of non-intersecting integral curves 𝐶which fill the plane (or domain of the function more generally), such that at a point (𝑥, 𝑦) the integral curve has tangent vector 𝑢(𝑥, 𝑦). Now, define a curve 𝐵by (𝑥(𝑡), 𝑦(𝑡)) such that 𝐵is transverse to 𝑢; its tangent is nowhere parallel to 𝑢.
𝑤= (d𝑥(𝑡) d𝑡, d𝑦(𝑡) d𝑡) ∦(𝛼(𝑥, 𝑦), 𝛽(𝑥, 𝑦)) = 𝑢 This can be used to parametrise the family of curves by labelling each curve 𝐶with the value of 𝑡at the intersection point between it and 𝐵. Along the curve, we use 𝑠such that 𝑠= 0 at the intersection. The integral curves (𝑥(𝑠, 𝑡), 𝑦(𝑠, 𝑡)) satisfy d𝑥 d𝑠= 𝛼(𝑥, 𝑦); d𝑦 d𝑠= 𝛽(𝑥, 𝑦) We can solve these equations to find a family of characteristic curves, along which 𝑡remains constant. This yields a new coordinate system (𝑠, 𝑡) associated with a differential equation we wish to solve.
282 6. Method of characteristics 6.3. Characteristics of a first order PDE Consider 𝛼(𝑥, 𝑦)𝜕𝜙 𝜕𝑥+ 𝛽(𝑥, 𝑦)𝜕𝜙 𝜕𝑦= 0 with Cauchy data on an initial curve 𝐵, defined by (𝑥(𝑡), 𝑦(𝑡)): 𝜙(𝑥(𝑡), 𝑦(𝑡)) = 𝑓(𝑡) Note, 𝛼𝜙𝑥+ 𝛽𝜙𝑦= 𝑢⋅∇𝜙= d𝜙 d𝑠 | | |𝐶 This is exactly the directional derivative along the integral curve 𝐶, defined by 𝑢= (𝛼, 𝛽).
Since d𝜙 d𝑠= 𝛼𝜙𝑥+ 𝛽𝜙𝑦= 0 from the original PDE, the function 𝜙(𝑥, 𝑦) is constant along this curve 𝐶. In other words, the Cauchy data 𝑓(𝑡) defined on 𝐵at 𝑠= 0 is propagated constantly along the integral curves. This gives the solution 𝜙(𝑠, 𝑡) = 𝜙(𝑥(𝑠, 𝑡), 𝑦(𝑠, 𝑡)) = 𝑓(𝑡) To obtain 𝜙in the original coordinates, we need to transform from 𝑠, 𝑡-space into 𝑥, 𝑦-space.
Provided that the Jacobian 𝐽= 𝑥𝑡𝑦𝑠−𝑥𝑠𝑦𝑡is nonzero, we can invert the transformation and find 𝑠, 𝑡as functions of 𝑥, 𝑦. This gives 𝜙(𝑥, 𝑦) = 𝑓(𝑡(𝑥, 𝑦)) To solve such a PDE, we will typically use the following steps.
(i) Find the characteristic equations d𝑥 d𝑠= 𝛼, d𝑦 d𝑠= 𝛽.
(ii) Parametrise the initial conditions on 𝐵by (𝑥(𝑡), 𝑦(𝑡)).
(iii) Solve the characteristic equations to find 𝑥= 𝑥(𝑠, 𝑡) and 𝑦= 𝑦(𝑠, 𝑡) subject to the initial conditions at 𝑠= 0.
(iv) Solve the equation for 𝜙given by d𝜙 d𝑠= 𝛼𝜙𝑥+ 𝛽𝜙𝑦= 0, so 𝜙is constant along the integral curves, giving 𝜙(𝑠, 𝑡) = 𝑓(𝑡).
(v) Invert the relations 𝑠= 𝑠(𝑥, 𝑦) and 𝑡= 𝑡(𝑥, 𝑦), then find 𝜙in terms of 𝑥, 𝑦.
Example. Consider the equation d𝜙(𝑥, 𝑦) d𝑥 = 0 such that 𝜙(0, 𝑦) = ℎ(𝑦) The characteristic equations are given by d𝑥 d𝑠= 𝛼= 1; d𝑦 d𝑠= 𝛽= 0 283 V. Methods The initial curve 𝐵is given by (𝑥(𝑡), 𝑦(𝑡)) = (0, 𝑡) Solving the characteristic equations, 𝑥= 𝑠+ 𝑐(𝑡); 𝑦= 𝑑(𝑡) At 𝑥= 0, we must have 𝑠= 0, so 𝑐= 0. Further, 𝑦= 𝑡hence 𝑑= 𝑡. Thus, 𝑥= 𝑠; 𝑦= 𝑡 Thus, d𝜙 d𝑥= 0 ⟹𝜙(𝑠, 𝑡) = ℎ(𝑡) ⟹𝜙(𝑥, 𝑦) = ℎ(𝑦) Example. Consider 𝑒𝑥𝜙𝑥+ 𝜙𝑦= 0; 𝜙(𝑥, 0) = cosh 𝑥 The characteristic equations are d𝑥 d𝑠= 𝑒𝑥; d𝑦 d𝑠= 1 The initial conditions are 𝑥(𝑡) = 𝑡; 𝑦(𝑡) = 0 We solve the characteristic equation subject to these initial conditions, giving −𝑒−𝑥= 𝑠+ 𝑐(𝑡); 𝑦= 𝑠+ 𝑑(𝑡) 𝑠= 0 implies −𝑒−𝑡= 𝑐(𝑡) and 𝑦= 0 = 𝑑(𝑡). Hence 𝑒−𝑥= 𝑒−𝑡−𝑠; 𝑦= 𝑠 Now, d𝜙 d𝑠= 0 ⟹𝜙(𝑠, 𝑡) = cosh 𝑡 Since 𝑠= 𝑦, 𝑒−𝑡= 𝑦+ 𝑒−𝑥, we have 𝑡= −log(𝑦+ 𝑒−𝑥). Thus, 𝜙(𝑥, 𝑦) = cosh [−log(𝑦+ 𝑒−𝑥)] 6.4. Inhomogeneous first order PDEs Suppose we now wish to solve 𝛼(𝑥, 𝑦)𝜙𝑥+ 𝛽(𝑥, 𝑦)𝜙𝑦= 𝛾(𝑥, 𝑦) with Cauchy data 𝜙(𝑥(𝑡), 𝑦(𝑡)) = 𝑓(𝑡) along a curve 𝐵. The characteristic curves are the same as the homogeneous case. However, the directional derivative no longer vanishes: d𝜙 d𝑠 | | |𝐶 = 𝑢⋅∇𝜙= 𝛾(𝑥, 𝑦) 284 6. Method of characteristics where 𝜙= 𝑓(𝑡) at 𝑠= 0 on 𝐵. So 𝑓(𝑡) is no longer propagated constantly across characteristic polynomials, but is instead propagated according to the ODE in 𝑠above. We must therefore solve this ODE along 𝐶before reverting to 𝑥, 𝑦coordinates.
Example. Consider 𝜙𝑥+ 2𝜙𝑦= 𝑦𝑒𝑥; 𝜙(𝑥, 𝑥) = sin 𝑥 The characteristic equation is given by d𝑥 d𝑠= 1; d𝑦 d𝑠= 2 The initial conditions are 𝑥(𝑡) = 𝑦(𝑡) = 𝑡 From the characteristic equations, 𝑥= 𝑠+ 𝑐(𝑡); 𝑦= 2𝑠+ 𝑑(𝑡) Thus, 𝑥= 𝑡= 𝑐(𝑡); 𝑦= 𝑡= 𝑑(𝑡) So the solutions to the characteristics are 𝑥= 𝑠+ 𝑡; 𝑦= 2𝑠+ 𝑡 Now we solve d𝜙 d𝑠= 𝛾= 𝑦𝑒𝑥= (2𝑠+ 𝑡)𝑒𝑠+𝑡 Note that d d𝑠(2𝑠𝑒𝑠) = 2𝑒𝑠+ 2𝑠𝑒𝑠, so the solution is 𝜙(𝑠, 𝑡) = (2𝑠−2 + 𝑡)𝑒𝑠+𝑡+ 𝑐(𝑠) for some constant term 𝑐(𝑠). But 𝜙(0, 𝑡) = sin 𝑡, hence sin 𝑡= (𝑡−2)𝑒𝑡+ 𝑐(𝑠) ⟹𝜙(𝑠, 𝑡) = (2𝑠−2 + 𝑡)𝑒𝑠+𝑡+ sin 𝑡−(2 −𝑡)𝑒𝑡 Inverting into 𝑥, 𝑦space, 𝜙(𝑥, 𝑦) = (𝑦−2)𝑒𝑥+ (𝑦−2𝑥+ 2)𝑒2𝑥−𝑦+ sin(2𝑥−𝑦) 6.5. Classification of second order PDEs In two dimensions, the general second order PDE is ℒ𝜙≡𝑎(𝑥, 𝑦)𝜕2𝜙 𝜕𝑥2 + 2𝑏(𝑥, 𝑦) 𝜕2𝜙 𝜕𝑥𝜕𝑦+ 𝑐(𝑥, 𝑦)𝜕2𝜙 𝜕𝑦2 + 𝑑(𝑥, 𝑦)𝜕𝜙 𝜕𝑥+ 𝑒(𝑥, 𝑦)𝜕𝜙 𝜕𝑦+ 𝑓(𝑥, 𝑦)𝜙(𝑥, 𝑦) 285 V. Methods The principal part is given by 𝜎𝑃(𝑥, 𝑦, 𝑘𝑥, 𝑘𝑦) ≡𝑘⊺𝐴𝑘= (𝑘𝑥 𝑘𝑦) (𝑎(𝑥, 𝑦) 𝑏(𝑥, 𝑦) 𝑏(𝑥, 𝑦) 𝑐(𝑥, 𝑦)) (𝑘𝑥 𝑘𝑦 ) The PDE is classified by the properties of the eigenvalues of 𝐴.
(i) If 𝑏2−𝑎𝑐< 0, the equation is elliptic. The eigenvalues have the same sign. An example is the Laplace equation.
(ii) If 𝑏2 −𝑎𝑐> 0, the equation is hyperbolic. The eigenvalues have opposite signs. An example is the wave equation.
(iii) If 𝑏2 −𝑎𝑐= 0, the equation is parabolic, where at least one eigenvalue is zero. An example is the heat equation.
Note that a differential equation may have different classifications at different points (𝑥, 𝑦) in space.
6.6. Characteristic curves of second order PDEs A curve defined by 𝑓(𝑥, 𝑦) constant is a characteristic if (𝑓 𝑥 𝑓 𝑦) (𝑎 𝑏 𝑏 𝑐) (𝑓 𝑥 𝑓 𝑦 ) = 0 This is a generalisation of the first order case 𝑢⋅∇𝑓= 0 where 𝑢= (𝛼, 𝛽). The curve can be written as 𝑦= 𝑦(𝑥) by the chain rule.
𝜕𝑓 𝜕𝑥+ 𝜕𝑓 𝜕𝑦 d𝑦 d𝑥= 0 ⟹𝑓 𝑥 𝑓 𝑦 = −d𝑦 d𝑥 Substituting into the quadratic form, 𝑎(d𝑦 d𝑥) 2 −2𝑏d𝑦 d𝑥+ 𝑐= 0 for which we have a quadratic solution given by d𝑦 d𝑥= 𝑏± √𝑏2 −𝑎𝑐 𝑎 (i) Hyperbolic equations have two such solutions, since 𝑏2 −𝑎𝑐> 0.
(ii) Parabolic equations have one solution.
(iii) Elliptic equations have no real characteristics.
286 6. Method of characteristics 6.7. Characteristic coordinates Transforming to characteristic coordinates 𝑢, 𝑣will set 𝑎= 0 and 𝑐= 0. Hence, the PDE will take the canonical form 𝜕2𝜙 𝜕𝑢𝜕𝑣+ ⋯+ = 0 where the omitted terms are lower order.
Example. Consider −𝑦𝜙𝑥𝑥+ 𝜙𝑦𝑦= 0 Here, 𝑎= −𝑦, 𝑏= 0, 𝑐= 1 hence 𝑏2 −𝑎𝑐= 𝑦. For 𝑦> 0, the equation is hyperbolic, for 𝑦< 0 it is elliptic, and for 𝑦= 0 it is parabolic. Consider the characteristics for 𝑦> 0.
d𝑦 d𝑥= 𝑏± √𝑏2 −𝑎𝑐 𝑎 = ± 1 √𝑦 Hence, ∫√𝑦d𝑦= ± ∫d𝑥⟹2 3𝑦 3 2 ± 𝑥= 𝐶± Therefore, the characteristic curves are 𝑢= 2 3𝑦 3 2 + 𝑥; 𝑣= 2 3𝑦 3 2 −𝑥 Taking derivatives, 𝑢𝑥= 1; 𝑢𝑦= √𝑦; 𝑣𝑥= −1; 𝑣𝑦= √𝑦 Hence, 𝜙𝑥= 𝜙𝑢𝑢𝑥+ 𝜙𝑣𝑣𝑥= 𝜙𝑢−𝜙𝑣 𝜙𝑦= √𝑦(𝜙𝑢+ 𝜙𝑣) 𝜙𝑥𝑥= 𝜙𝑢𝑢−2𝜙𝑢𝑣+ 𝜙𝑣𝑣 𝜙𝑦𝑦= 𝑦(𝜙𝑢𝑢+ 2𝜙𝑢𝑣+ 𝜙𝑣𝑣) + 1 2√𝑦(𝜙𝑢+ 𝜙𝑣) Substituting into the original PDE, −𝑦𝜙𝑥𝑥+ 𝜙𝑦𝑦= 𝑦(4𝜙𝑢𝑣+ 1 2𝑦 3 2 (𝜙𝑢+ 𝜙𝑣)) Note, 𝑢+ 𝑣= 4 3𝑦 3 2 , hence we have the canonical form 4𝜙𝑢𝑣+ 1 6(𝑢+ 𝑣)(𝜙𝑢+ 𝜙𝑣) = 0 287 V. Methods 6.8. General solution to wave equation The wave equation is 1 𝑐2 𝜕2𝜙 𝜕𝑡2 −𝜕2𝜙 𝜕𝑥2 = 0 We wish to solve this with initial conditions 𝜙(𝑥, 0) = 𝑓(𝑥), and 𝜙𝑡(𝑥, 0) = 𝑔(𝑥). Here, 𝑎= 1 𝑐2 , 𝑏= 0, 𝑐= −1 hence 𝑏2 −𝑎𝑐> 0. The characteristic equation is d𝑥 d𝑡= 0 ± √0 + 1 𝑐2 1 𝑐2 = ±𝑐 Hence the characteristic coordinates are 𝑢= 𝑥−𝑐𝑡; 𝑣= 𝑥+ 𝑐𝑡 This yields the canonical form 𝜕2𝜙 𝜕𝑢𝜕𝑣= 0 This may be integrated directly to find 𝜕𝜙 𝜕𝑣= 𝐹(𝑣) ⟹𝜙= 𝐺(𝑢) + ∫ 𝑣 𝐹(𝑦) d𝑦= 𝐺(𝑢) + 𝐻(𝑣) Imposing the initial conditions at 𝑡= 0, we find 𝐺(𝑥) + 𝐻(𝑥) = 𝑓(𝑥); −𝑐𝐺′(𝑥) + 𝑐𝐻′(𝑥) = 𝑔(𝑥) Differentiating the first equation, we find 𝐺′(𝑥) + 𝐻′(𝑥) = 𝑓′(𝑥) We can combine this with the second equation to give 𝐻′(𝑥) = 1 2(𝑓′(𝑥) + 1 𝑐𝑔(𝑥)) ⟹𝐻(𝑥) = 1 2(𝑓(𝑥) −𝑓(0)) + 1 2𝑐∫ 𝑥 0 𝑔(𝑦) d𝑦 Similarly, 𝐺′(𝑥) = 1 2(𝑓′(𝑥) −1 𝑐𝑔(𝑥)) ⟹𝐺(𝑥) = 1 2(𝑓(𝑥) −𝑓(0)) −1 2𝑐∫ 𝑥 0 𝑔(𝑦) d𝑦 The final solution is therefore 𝜙(𝑥, 𝑡) = 𝐺(𝑥−𝑐𝑡) + 𝐻(𝑥+ 𝑐𝑡) = 1 2(𝑓(𝑥−𝑐𝑡) + 𝑓(𝑥+ 𝑐𝑡)) + 1 2𝑐∫ 𝑥+𝑐𝑡 𝑥−𝑐𝑡 𝑔(𝑦) d𝑦 Waves propagate at a velocity 𝑐, hence 𝜙(𝑥, 𝑡) is fully determined by values of 𝑓, 𝑔in the interval [𝑥−𝑐𝑡, 𝑥+ 𝑐𝑡].
288 7. Solving partial differential equations with Green’s functions 7. Solving partial differential equations with Green’s functions 7.1. Diffusion equation and Fourier transform Recall the heat equation for a conducting wire given by 𝜕Θ 𝜕𝑡(𝑥, 𝑡) −𝐷𝜕2Θ 𝜕𝑥2 (𝑥, 𝑡) = 0 with initial conditions Θ(𝑥, 0) = ℎ(𝑥) and boundary conditions Θ →0 as 𝑥→±∞. Taking the Fourier transform with respect to 𝑥, 𝜕 𝜕𝑡 ˜ Θ(𝑘, 𝑡) = −𝐷𝑘2˜ Θ(𝑘, 𝑡) Integrating, we find ˜ Θ(𝑘, 𝑡) = 𝐶𝑒−𝐷𝑘2𝑡 The initial conditions give ˜ Θ(𝑘, 0) = ˜ ℎ(𝑘) and therefore ˜ Θ(𝑘, 𝑡) = ˜ ℎ(𝑘)𝑒−𝐷𝑘2𝑡 We take the inverse Fourier transform to find Θ(𝑥, 𝑡) = 1 2𝜋∫ ∞ −∞ ˜ ℎ(𝑘) 𝑒−𝐷𝑘2𝑡 ⏟ ⎵ ⏟ ⎵ ⏟ FT of Gaussian 𝑒𝑖𝑘𝑥d𝑘 Hence, by the convolution theorem, Θ(𝑥, 𝑡) = 1 √4𝜋𝐷𝑡 ∫ ∞ −∞ ℎ(𝑢) exp(−(𝑥−𝑢)2 4𝐷𝑡 ) d𝑢 ≡∫ ∞ −∞ ℎ(𝑢)𝑆𝑑(𝑥−𝑢, 𝑡) d𝑢 where the fundamental solution is 𝑆𝑑(𝑥, 𝑡) = 1 √4𝜋𝐷𝑡 exp(−𝑥2 4𝐷𝑡) which is the Fourier transform of exp(−𝐷𝑘2𝑡). Note, with localised initial conditions Θ(𝑥, 0) = Θ0𝛿(𝑥), the solution is exactly the fundamental solution: Θ(𝑥, 𝑡) = Θ0𝑆𝑑(𝑥, 𝑡) = Θ0 √4𝜋𝐷𝑡 exp(−𝜂2); 𝜂= 𝑥 2√𝐷𝑡 where 𝜂is the similarity parameter.
289 V. Methods 7.2. Gaussian pulse for heat equation Suppose that the initial conditions for the head equation are given by 𝑓(𝑥) = √ 𝑎 𝜋Θ0𝑒−𝑎𝑥2 Then, our previous solution gives Θ(𝑥, 𝑡) = Θ0√𝑎 √4𝜋2𝐷𝑡 ∫ ∞ −∞ exp[−𝑎𝑢2 −(𝑥−𝑢)2 4𝐷𝑡 ] d𝑢 = Θ0√𝑎 √4𝜋2𝐷𝑡 ∫ ∞ −∞ exp[−(1 + 4𝑎𝐷𝑡)𝑢2 −2𝑥𝑢+ 𝑥2 4𝐷𝑡 ] d𝑢 = Θ0√𝑎 √4𝜋2𝐷𝑡 ∫ ∞ −∞ exp[−1 + 4𝑎𝐷𝑡 4𝐷𝑡 (𝑢− 𝑥 1 + 4𝑎𝐷𝑡)] exp[ −𝑎𝑥2 1 + 4𝑎𝐷𝑡] d𝑢 Recall that ∫ ∞ −∞ exp[−(𝑢−𝜇)2 𝜎2 ] d𝑢= 𝜎√𝜋 The integral above is a Gaussian, so its solution can be read off directly as Θ(𝑥, 𝑡) = Θ0√𝑎 √𝜋(1 + 4𝜋2𝐷𝑡) exp[ −𝑎𝑥2 1 + 4𝑎𝐷𝑡] So the width of the Gaussian pulse will get wider over time, according to 𝜎2 ∼𝑡, as it evolves according to the heat equation. The area is constant, so heat energy is conserved in the system.
7.3. Forced diffusion equation Consider the equation 𝜕 𝜕𝑡Θ(𝑥, 𝑡) −𝐷𝜕2Θ 𝜕𝑥2 = 𝑓(𝑥, 𝑡) subject to homogeneous initial conditions Θ(𝑥, 0) = 0. We construct a two-dimensional Green’s function 𝐺(𝑥, 𝑡; 𝜉, 𝜏) such that 𝜕 𝜕𝑡𝐺(𝑥, 𝑡) −𝐷𝜕2𝐺 𝜕𝑥2 = 𝛿(𝑥−𝜉)𝛿(𝑡−𝜏) subject to the same homogeneous boundary conditions 𝐺(𝑥, 0; 𝜉, 𝜏) = 0. Consider the Four-ier transform with respect to 𝑥.
𝜕˜ 𝐺 𝜕𝑡+ 𝐷𝑘2 ˜ 𝐺= 𝑒−𝑖𝑘𝜉𝛿(𝑡−𝜏) 290 7. Solving partial differential equations with Green’s functions We can solve this using an integrating factor 𝑒𝐷𝑘2𝑡and integrating with respect to time. Since 𝐺= 0 at 𝑡= 0, 𝜕 𝜕𝑡[𝑒𝐷𝑘2𝑡˜ 𝐺] = 𝑒−𝑖𝑘𝜉+𝐷𝑘2𝑡𝛿(𝑡−𝜏) ∫ 𝑡 0 𝜕 𝜕𝑡′ [𝑒𝐷𝑘2𝑡′ ˜ 𝐺] d𝑡′ = ∫ 𝑡 0 𝑒−𝑖𝑘𝜉+𝐷𝑘2𝑡′𝛿(𝑡′ −𝜏) d𝑡′ 𝑒𝐷𝑘2𝑡˜ 𝐺= 𝑒−𝑖𝑘𝜉∫ 𝑡 0 𝑒𝐷𝑘2𝑡′𝛿(𝑡′ −𝜏) d𝑡′ 𝑒𝐷𝑘2𝑡˜ 𝐺= 𝑒−𝑖𝑘𝜉𝑒𝐷𝑘2𝜏𝐻(𝑡−𝜏) where 𝐻is the Heaviside step function. Thus, ˜ 𝐺(𝑘, 𝑡; 𝜉, 𝜏) = 𝑒−𝑖𝑘𝜉𝑒−𝐷𝑘2(𝑡−𝜏)𝐻(𝑡−𝜏) The inverse Fourier transform gives the Green’s function.
𝐺(𝑥, 𝑡; 𝜉, 𝜏) = 𝐻(𝑡−𝜏) 2𝜋 ∫ ∞ −∞ 𝑒−𝑖𝑘𝜉𝑒−𝐷𝑘2(𝑡−𝜏)𝑒𝑖𝑘𝑥d𝑘 This is a Gaussian; by changing variables into 𝑥′ = 𝑥−𝜉and 𝑡′ = 𝑡−𝜏we find 𝐺(𝑥, 𝑡; 𝜉, 𝜏) = 𝐻(𝑡′) 2𝜋∫ ∞ −∞ 𝑒𝑖𝑘𝑥′𝑒−𝐷𝑘2𝑡′ d𝑘= 𝐻(𝑡′) √4𝜋𝐷𝑡′ exp[−(𝑥′)2 4𝐷𝑡′ ] Converting back, 𝐺(𝑥, 𝑡; 𝜉, 𝜏) = 𝐻(𝑡−𝜏) √4𝜋𝐷(𝑡−𝜏) exp[−(𝑥−𝜉)2 4𝐷(𝑡−𝜏)] = 𝐻(𝑡−𝜏)𝑆𝑑(𝑥−𝜉, 𝑡−𝜏) where 𝑆𝑑is the fundamental solution as above. Thus, the general solution is Θ(𝑥, 𝑡) = ∫ ∞ 0 d𝜏∫ ∞ −∞ d𝜉𝐺(𝑥, 𝑡; 𝜉, 𝜏)𝑓(𝜉, 𝜏) Let 𝜉= 𝑢, then Θ(𝑥, 𝑡) = ∫ 𝑡 0 d𝜏∫ ∞ −∞ d𝑢𝑓(𝑢, 𝜏)𝑆𝑑(𝑥−𝑢, 𝑡−𝜏) 7.4. Duhamel’s principle In the above equation, omitting the integral over time, this is exactly the solution as found earlier with initial conditions at 𝑡= 𝜏, which was Θ(𝑥, 𝑡) = ∫ ∞ −∞ d𝑢𝑓(𝑢)𝑆𝑑(𝑥−𝑢, 𝑡−𝜏) 291 V. Methods The forced PDE with homogeneous boundary conditions can be related to solutions of the homogeneous PDE with inhomogeneous boundary conditions. The forcing term 𝑓(𝑥, 𝑡) at 𝑡= 𝜏acts as an initial condition for subsequent evolution. Thus, the solution is a superpos-ition of the effects of the initial conditions integrated over 0 < 𝜏< 𝑡. This relation between the homogeneous and inhomogeneous problems is known as Duhamel’s principle.
7.5. Forced wave equation Consider the forced wave equation, given by 𝜕2𝜙 𝜕𝑡2 −𝑐2 𝜕2𝜙 𝜕𝑥2 = 𝑓(𝑥, 𝑡) with 𝜙(𝑥, 0) = 𝜙𝑡(𝑥, 0) = 0. We construct the Green’s function using 𝜕2𝐺 𝜕𝑡2 −𝑐2 𝜕2𝐺 𝜕𝑥2 = 𝛿(𝑥−𝜉)𝛿(𝑡−𝜏) with 𝐺(𝑥, 0) = 𝜙𝑡(𝑥, 0) = 0. We take the Fourier transform with respect to 𝑥, and find 𝜕2 ˜ 𝐺 𝜕𝑡2 + 𝑐2𝑘2 ˜ 𝐺= 𝑒−𝑖𝑘𝜉𝛿(𝑡−𝜏) We can solve this by inspection by comparing with the corresponding initial value problem Green’s function, and find ˜ 𝐺= { 0 𝑡< 𝜏 𝑒−𝑖𝑘𝜉sin 𝑘𝑐(𝑡−𝜏) 𝑘𝑐 𝑡> 𝜏 Using the Heaviside function.
˜ 𝐺= 𝑒−𝑖𝑘𝜉sin 𝑘𝑐(𝑡−𝜏) 𝑘𝑐 𝐻(𝑡−𝜏) We invert the Fourier transform.
𝐺(𝑥, 𝑡; 𝜉, 𝜏) = 𝐻(𝑡−𝜏) 2𝜋𝑐 ∫ ∞ −∞ 𝑒𝑖𝑘(𝑥−𝜉) sin 𝑘𝑐(𝑡−𝜏) 𝑘 d𝑘 Let 𝐴= 𝑥−𝜉, and 𝐵= 𝑐𝑡−𝜏. By oddness of sine, only the cosine term of the complex exponential remains. Noting the similarity to the Dirichlet discontinuous function, 𝐺(𝑥, 𝑡; 𝜉, 𝜏) = 𝐻(𝑡−𝜏) 𝜋𝑐 ∫ ∞ 0 cos(𝑘𝐴) sin(𝑘𝐵) 𝑘 d𝑘 = 𝐻(𝑡−𝜏) 2𝜋𝑐 ∫ ∞ 0 sin 𝑘(𝐴+ 𝐵) −sin 𝑘(𝐴−𝐵) 𝑘 d𝑘 = 𝐻(𝑡−𝜏) 4𝑐 [sgn(𝐴+ 𝐵) −sgn(𝐴−𝐵)] 292 7. Solving partial differential equations with Green’s functions Since the 𝐻(𝑡−𝜏) term is nonzero only for 𝑡> 𝜏, we must have 𝐵= 𝑐(𝑡−𝜏) > 0. The only way that the bracketed term can be nonzero is when |𝐴| < 𝐵; so |𝑥−𝜉| < 𝑐(𝑡−𝜏). This is the domain of dependence as found before, demonstrating the causality of the relation.
Hence, 𝐺(𝑥, 𝑡; 𝜉, 𝜏) = 1 2𝑐𝐻(𝑐(𝑡−𝜏) −|𝑥−𝜉|) Thus, the solution is 𝜙(𝑥, 𝑡) = ∫ ∞ 0 d𝜏∫ ∞ −∞ d𝜉𝑓(𝜉, 𝑡)𝐺(𝑥, 𝑡; 𝜉, 𝜏) = 1 2𝑐∫ 𝑡 0 d𝜏∫ 𝑥+𝑐(𝑡−𝜏) 𝑥−𝑐(𝑡−𝜏) d𝜉𝑓(𝜉, 𝜏) 7.6. Poisson’s equation Consider ∇2𝜙= −𝜌(𝑟) defined on a three-dimensional domain 𝐷, with Dirichlet boundary conditions 𝜙= 0 on a boundary 𝜕𝐷. The Dirac 𝛿function, when defined in ℝ3, has the following properties.
(i) 𝛿(𝑟−𝑟′) = 0 for all 𝑟≠𝑟′; (ii) ∫ 𝐷𝛿(𝑟−𝑟′) d3𝑟= 1 if 𝑟′ ∈𝐷, and zero otherwise; (iii) ∫ 𝐷𝑓(𝑟)𝛿(𝑟−𝑟′) d3𝑟= 𝑓(𝑟′).
First, we consider 𝐷= ℝ3 with the homogeneous boundary conditions that 𝐺→0 as ‖𝑟‖ → ∞. This is known as the free-space Green’s function, denoted 𝐺FS. The potential here is spherically symmetric, so the Green’s function is a function only of the distance between the point and the source. WIthout loss of generality, let 𝑟′ = 0, so 𝐺is a function only of the radius, now denoted 𝑟. Integrating the left hand side of Poisson’s equation over a ball 𝐵with radius 𝑟around zero, we find ∫ 𝐵 ∇2𝐺FS d3𝑟= ∫ 𝜕𝐵 ∇𝐺FS ⋅̂ 𝑛d𝑆= ∫ 𝜕𝐵 𝜕𝐺 𝜕𝑟𝑟2 dΩ where dΩ is the angle element. This gives ∫ 𝐵 ∇2𝐺FS d3𝑟= 4𝜋𝑟2 𝜕𝐺FS 𝜕𝑟 The right hand side of Poisson’s equation gives unity, since zero is contained in the ball.
Therefore, 𝜕𝐺FS 𝜕𝑟 = 1 4𝜋𝑟2 ⟹𝐺FS = −1 4𝜋𝑟+ 𝑐 293 V. Methods Since 𝐺→0 as 𝑟→∞, we must have 𝑐= 0. The fundamental solution is therefore the free-space Green’s function given by 𝐺(𝑟; 𝑟′) = −1 4𝜋‖𝑟−𝑟′‖ Thus, Poisson’s equation is solved by Φ(𝑟) = 1 4𝜋∫ ℝ3 𝜌(𝑟′) ‖𝑟−𝑟′‖ d3𝑟′ 7.7. Green’s identities Consider scalar functions 𝜙, 𝜓which are twice differentiable on a domain 𝐷. By the diver-gence theorem, Green’s first identity is ∫ 𝐷 ∇⋅(𝜙∇𝜓) d3𝑟= ∫ 𝐷 (𝜙∇2𝜓+ ∇𝜙⋅∇𝜓) d3𝑟= ∫ 𝜕𝐷 𝜙∇𝜓⋅̂ 𝑛d𝑆 Switching 𝜓and 𝜙and subtracting from the above, we arrive at Green’s second identity: ∫ 𝜕𝐷 (𝜙𝜕𝜓 𝜕̂ 𝑛−𝜓𝜕𝜙 𝜕̂ 𝑛) d𝑆= ∫ 𝐷 (𝜙∇2𝜓−𝜓∇2𝜙) d3𝑟 Suppose we remove a ball ℬ𝜀(𝑟′) from the domain. Without loss of generality let 𝑟′ = 0.
Let 𝜙be a solution to Poisson’s equation, so ∇2𝜙= −𝜌and let 𝜓be the free-space Green’s function. Thus, the right hand side of the second identity becomes ∫ 𝐷∖ℬ𝜀 (𝜙∇2𝐺FS −𝐺FS∇2𝜙) d3𝑟= ∫ 𝐷∖ℬ𝜀 𝐺FS𝜌d3𝑟 The left hand side is ∫ 𝜕𝐷 (𝜙𝜕𝐺FS 𝜕̂ 𝑛 −𝐺FS 𝜕𝜙 𝜕̂ 𝑛) d𝑆+ ∫ 𝜕ℬ𝜀 (𝜙𝜕𝐺FS 𝜕̂ 𝑛 −𝐺FS 𝜕𝜙 𝜕̂ 𝑛) d𝑆 For the second integral, we take the limit as 𝜀→0. Let 𝜙be regular, and let 𝜙be the average value and 𝜕𝜙 𝜕̂ 𝑛be the average derivative. This integral then becomes (𝜙−1 4𝜋𝜀2 − 1 4𝜋𝜀 𝜕𝜙 𝜕̂ 𝑛)4𝜋𝜀2 →−𝜙(0) Combining the above, we find Green’s third identity, which is 𝜙(𝑟′) = ∫ 𝐷 𝐺FS(𝑟; 𝑟′)(−𝜌(𝑟)) d3𝑟+ ∫ 𝜕𝐷 (𝜙(𝑟)𝜕𝐺FS 𝜕̂ 𝑛(𝑟; 𝑟′) −𝐺FS(𝑟; 𝑟′)𝜕𝜙 𝜕̂ 𝑛(𝑟)) d𝑆 The second integral provides the ability to use inhomogeneous boundary conditions 294 7. Solving partial differential equations with Green’s functions 7.8. Dirichlet Green’s function We will solve Poisson’s equation ∇2𝜙= −𝜌on 𝐷with inhomogeneous boundary conditions 𝜙(𝑟) = ℎ(𝑟) on 𝜕𝐷. The Dirichlet Green’s function satisfies (i) ∇2𝐺(𝑟; 𝑟′) = 0 for all 𝑟≠𝑟′; (ii) 𝐺(𝑟; 𝑟′) = 0 on 𝜕𝐷; (iii) 𝐺(𝑟; 𝑟′) = 𝐺FS(𝑟; 𝑟′) + 𝐻(𝑟; 𝑟′) where 𝐻satisfies Laplace’s equation, the homogeneous version of Poisson’s equation, for all 𝑟∈𝐷.
Green’s second identity with ∇2𝜙= −𝜌, ∇2𝐻= 0 gives ∫ 𝜕𝐷 (𝜙𝜕𝐻 𝜕̂ 𝑛−𝐻𝜕𝜙 𝜕̂ 𝑛) d𝑆= ∫ 𝐷 𝐻𝜌d3𝑟 Now, we set 𝐺FS = 𝐺−𝐻into Green’s third identity to find 𝜙(𝑟′) = ∫ 𝐷 (𝐺−𝐻)(−𝜌) d3𝑟+ ∫ 𝜕𝐷 (𝜙𝜕(𝐺−𝐻) 𝜕̂ 𝑛 −(𝐺−𝐻)𝜕𝜙 𝜕𝑛) d𝑆 All of the 𝐻terms can be cancelled by substituting the form of the second identity the derived above. Now, given 𝐺= 0, 𝜙= ℎon 𝜕𝐷, we have 𝜙(𝑟′) = ∫ 𝐷 𝐺(𝑟; 𝑟′)(−𝜌(𝑟)) d3𝑟+ ∫ 𝜕𝐷 ℎ(𝑟)𝜕𝐺(𝑟; 𝑟′) 𝜕̂ 𝑛 d𝑆 This is the general solution. The first integral is the Green’s function solution, and the second integral yields the inhomogeneous boundary conditions.
7.9. Method of images for Laplace’s equation For symmetric domains 𝐷, we can construct Green’s functions with 𝐺= 0 on 𝜕𝐷by cancel-ling the boundary potential out by using an opposite ‘mirror image’ Green’s function placed outside the domain. Consider Laplace’s equation ∇2𝜙= 0 on half of ℝ3, in particular, the subset of ℝ3 such that 𝑧> 0. Let 𝜙(𝑥, 𝑦, 0) = ℎ(𝑥, 𝑦) and 𝜙→0 as 𝑟→∞. The free space Green’s function satisfies 𝐺FS →0 as 𝑟→∞, but does not satisfy the boundary condition that 𝐺FS = 0 at 𝑧= 0. For 𝐺FS at 𝑟′ = (𝑥′, 𝑦′, 𝑧′), we will subtract a copy of 𝐺FS located at 𝑟″ = (𝑥′, 𝑦′, −𝑧′). This gives 𝐺(𝑟, 𝑟′) = −1 4𝜋|𝑟−𝑟′| − −1 4𝜋|𝑟−𝑟″| = −1 4𝜋√(𝑥−𝑥′)2 + (𝑦−𝑦′)2 + (𝑧−𝑧′)2 + 1 4𝜋√(𝑥−𝑥′)2 + (𝑦−𝑦′)2 + (𝑧+ 𝑧′)2 Hence 𝐺((𝑥, 𝑦, 0), 𝑟′) = 0, so this function satisfies the Dirichlet boundary conditions on all of the boundary 𝜕𝐷. We have 𝜕𝐺 𝜕̂ 𝑛 | | |𝑧=0 = 𝜕𝐺 𝜕𝑧 | | |𝑧=0 = −1 4𝜋( 𝑧−𝑧′ |𝑟−𝑟′|3 −𝑧+ 𝑧′ |𝑟−𝑟′|3 ) = 𝑧′ 2𝜋((𝑥−𝑥′)2 + (𝑦−𝑦′)2 + (𝑧′)2) −3/2 295 V. Methods The solution is then given by Φ(𝑥′, 𝑦′, 𝑧′) = 𝑧′ 2𝜋∫ ∞ −∞ ∫ ∞ −∞ [(𝑥−𝑥′)2 + (𝑦−𝑦′)2 + (𝑧′)2] −3/2ℎ(𝑥, 𝑦) d𝑥d𝑦 7.10. Method of images for wave equation Consider the one-dimensional wave equation ̈ 𝜙−𝑐2𝜙″ = 𝑓(𝑥, 𝑡) with Dirichlet boundary conditions 𝜙(0, 𝑡) = 0. We create matching Green’s functions with an opposite sign centred at −𝜉.
𝐺(𝑥, 𝑡; 𝜉, 𝜏) = 1 2𝑐𝐻(𝑐(𝑡−𝜏) −|𝑥−𝜉|) −1 2𝑐𝐻(𝑐, (𝑡−𝜏) −|𝑥+ 𝜉|) We can replace the addition of the two terms with a subtraction to instead use Neumann boundary conditions. Suppose we wish to solve the homogeneous problem with 𝑓= 0 for initial conditions of a Gaussian pulse. Here, for 𝑥> 0 we have 𝜙(𝑥, 𝑡) = exp[−(𝑥−𝜉+ 𝑐𝑡)2] −exp[−(−𝑥−𝜉+ 𝑐𝑡)2] The solution travels to the left, cancelling with the image at 𝑡= 𝜉 𝑐, which emerges and travels right as the reflected wave.
296 VI. Quantum Mechanics Lectured in Michaelmas 2021 by Dr. M. Ubiali In this course, we explore the basics of quantum mechanics using the Schrödinger equation.
This equation explains how a quantum wavefunction changes over time. By solving the Schrödinger equation with different inputs and boundary conditions, we can understand some of the ways in which quantum mechanics differs from classical physics, explaining some of the scientific discoveries of the past century. We prove some theoretical facts about quantum operators and observables, such as the uncertainty theorem, which roughly states that it is impossible to know both the position and momentum of a particle.
297 VI. Quantum Mechanics Contents 1.
Historical introduction . . . . . . . . . . . . . . . . . . . . . . . . . 300 1.1.
Timeline . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 300 1.2.
Particles and waves in classical mechanics . . . . . . . . . . . . 300 1.3.
Black-body radiation . . . . . . . . . . . . . . . . . . . . . . . 301 1.4.
Planck’s constant . . . . . . . . . . . . . . . . . . . . . . . . . 301 1.5.
Photoelectric effect . . . . . . . . . . . . . . . . . . . . . . . . 302 1.6.
Compton scattering . . . . . . . . . . . . . . . . . . . . . . . . 302 1.7.
Atomic spectra . . . . . . . . . . . . . . . . . . . . . . . . . . 303 2.
Wavefunctions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 305 2.1.
Wave-like behaviour of particles . . . . . . . . . . . . . . . . . 305 2.2.
Probabilistic interpretation of wavefunctions . . . . . . . . . . . 305 2.3.
Bases and equivalence classes . . . . . . . . . . . . . . . . . . . 306 2.4.
Hilbert spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . 306 2.5.
Inner product . . . . . . . . . . . . . . . . . . . . . . . . . . . 307 2.6.
Normalisation . . . . . . . . . . . . . . . . . . . . . . . . . . 307 2.7.
Time-dependent Schrödinger equation . . . . . . . . . . . . . . 308 2.8.
Normalisation and time evolution . . . . . . . . . . . . . . . . 309 2.9.
Conserved probability current . . . . . . . . . . . . . . . . . . 309 3.
Observables and operators . . . . . . . . . . . . . . . . . . . . . . . 310 3.1.
Expectation and operators . . . . . . . . . . . . . . . . . . . . 310 3.2.
Dynamical observables . . . . . . . . . . . . . . . . . . . . . . 310 3.3.
Hamiltonian operator . . . . . . . . . . . . . . . . . . . . . . . 311 3.4.
Time-independent Schrödinger equation . . . . . . . . . . . . . 312 4.
One-dimensional solutions to the Schrödinger equation . . . . . . 313 4.1.
Stationary states . . . . . . . . . . . . . . . . . . . . . . . . . 313 4.2.
Infinite potential well . . . . . . . . . . . . . . . . . . . . . . . 313 4.3.
Finite potential well . . . . . . . . . . . . . . . . . . . . . . . 314 4.4.
Free particles . . . . . . . . . . . . . . . . . . . . . . . . . . . 316 4.5.
Gaussian wavepacket . . . . . . . . . . . . . . . . . . . . . . . 317 4.6.
Beam interpretation . . . . . . . . . . . . . . . . . . . . . . . 318 4.7.
Scattering states . . . . . . . . . . . . . . . . . . . . . . . . . . 319 4.8.
Scattering off potential step . . . . . . . . . . . . . . . . . . . . 319 4.9.
Scattering off a potential barrier . . . . . . . . . . . . . . . . . 321 4.10.
Harmonic oscillator . . . . . . . . . . . . . . . . . . . . . . . . 322 5.
Operators and measurements . . . . . . . . . . . . . . . . . . . . . 325 5.1.
Hermitian operators . . . . . . . . . . . . . . . . . . . . . . . 325 5.2.
Postulates of quantum mechanics . . . . . . . . . . . . . . . . . 326 298 5.3.
Expectation of operators . . . . . . . . . . . . . . . . . . . . . 327 5.4.
Commutators . . . . . . . . . . . . . . . . . . . . . . . . . . . 328 5.5.
Simultaneously diagonalisable operators . . . . . . . . . . . . . 328 5.6.
Uncertainty . . . . . . . . . . . . . . . . . . . . . . . . . . . . 329 5.7.
Schwarz inequality . . . . . . . . . . . . . . . . . . . . . . . . 330 5.8.
Generalised uncertainty theorem . . . . . . . . . . . . . . . . . 330 5.9.
Consequences of uncertainty relation . . . . . . . . . . . . . . . 332 5.10.
States of minimal uncertainty . . . . . . . . . . . . . . . . . . . 332 5.11.
Ehrenfest theorem . . . . . . . . . . . . . . . . . . . . . . . . 332 6.
Three-dimensional solutions to the Schrödinger equation . . . . . 335 6.1.
Time-independent Schrödinger equation in spherical polar coordin-ates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 335 6.2.
Spherically symmetric potential well . . . . . . . . . . . . . . . 336 7.
Solution to hydrogen atom . . . . . . . . . . . . . . . . . . . . . . . 337 7.1.
Radial wavefunction of hydrogen atom . . . . . . . . . . . . . . 337 7.2.
Angular momentum . . . . . . . . . . . . . . . . . . . . . . . 339 7.3.
Commutativity of angular momentum operators . . . . . . . . . 340 7.4.
Joint eigenfunctions of angular momentum . . . . . . . . . . . 341 7.5.
Full solution to hydrogen atom . . . . . . . . . . . . . . . . . . 342 7.6.
Comparison to Bohr model . . . . . . . . . . . . . . . . . . . . 344 7.7.
Other elements of the periodic table . . . . . . . . . . . . . . . 345 299 VI. Quantum Mechanics 1. Historical introduction 1.1. Timeline • (1801–3) Particles were shown to have wave-like properties using Young’s double slit experiment.
• (1862–4) Electromagnetism was conceived by Maxwell. Light was discovered to be an electromagnetic wave.
• (1897) Discovery of the electron by Thomson.
• (1900) The Planck law was discovered, which explains black-body radiation.
• (1905) The photoelectric effect was discovered by Einstein.
• (1909) Wave-light interference patterns were shown to exist with only one photon re-corded at a time.
• (1911) Rutherford created his atomic model.
• (1913) Bohr created his atomic model.
• (1923) The Compton experiment showed x-ray scattering off electrons.
• (1923–4) De Broglie discovered the concept of wave-particle duality.
• (1925–30) The theory of quantum mechanics emerged at this time.
• (1927–8) The diffraction experiment was carried out with electrons.
1.2. Particles and waves in classical mechanics In classical mechanics, a point-particle is an object with energy and momentum in an infin-itesimally small point of space. Therefore, a particle is determined by the three-dimensional vectors x, v = ̇ x. The motion of a particle is governed by Newton’s second law, 𝑚̈ x = F(x, ̇ x) Solving this equation involves determination of x, ̇ x for all 𝑡> 𝑡0, once initial conditions x(𝑡0), ̇ x(𝑡0) are known.
Waves are classically defined as any real- or complex-valued function with periodicity in time and/or space. For instance, consider a function 𝑓such that 𝑓(𝑡+ 𝑇) = 𝑓(𝑡), which is a wave with period 𝑇. The frequency 𝜈is defined to be 1 𝑇, and the angular frequency 𝜔is defined as 2𝜋𝜈= 2𝜋 𝑇. Suppose we have a function in one dimension obeying 𝑓(𝑥+ 𝜆) = 𝑓(𝑥). This has wavelength 𝜆and wave number 𝑘= 2𝜋 𝜆.
300 1. Historical introduction Consider 𝑓(𝑥) = exp(±𝑖𝑘𝑥). In three dimensions, this becomes 𝑓(𝑥) = exp(±𝑖k ⋅x). This is called a ‘plane wave’; the one-dimensional wave number 𝑘has been transformed into a three-dimensional wave vector k. 𝜆is now defined as 2𝜋 |𝑘|.
The wave equation in one dimension is 𝜕2𝑓(𝑥, 𝑡) 𝜕𝑡2 −𝑐2 𝜕2𝑓(𝑥, 𝑡) 𝜕𝑥2 = 0; 𝑐∈ℝ The solutions to this equation are 𝑓 ±(𝑥, 𝑡) = 𝐴± exp(±𝑖𝑘𝑥−𝑖𝜔𝑡) where 𝜔= 𝑐𝑘; 𝜆= 𝑐 𝜈. The two conditions are known as the dispersion relations. 𝐴± is the amplitude of the waves.
In three dimensions, 𝜕2𝑓(x, 𝑡) 𝜕𝑡2 −𝑐2∇2𝑓(x, 𝑡) = 0; 𝑐∈ℝ The solution is 𝑓(x, 𝑡) = 𝐴exp(±𝑖k ⋅x −𝑖𝜔𝑡) where 𝜔= 𝑐|k|; 𝜆= 𝑐 𝜈.
Note. Other kinds of waves are solutions to other governing equations, provided that an-other dispersion relation 𝜔(k) is given. Also, for any governing equation linear in 𝑓, the superposition principle holds: if 𝑓 1, 𝑓 2 are solutions then so is 𝑓 1 + 𝑓 2.
1.3. Black-body radiation Several experiments have shown that light behaves with some particle-like characteristics.
For example, consider a body heated at some temperature 𝑇. Any such body will emit radi-ation. The simplest body to study is called a ‘black-body’, which is a totally absorbing surface.
The intensity of light emitted by a black body was modelled as a function of the frequency.
The classical prediction for the spectrum of emitted radiation was that as the frequency in-creased, the intensity would also increase. A curve with a clear maximum point was ob-served. Planck’s law was found to be the equation of this curve, which can be derived from the equation 𝐸= ℏ𝜔involving the Planck constant, instead of the classical energy equation 𝐸= 𝑘𝐵𝑇involving the Boltzmann constant. This then implies that light was ‘quantised’ into particles.
1.4. Planck’s constant The Planck constant is ℎ≈6.61 × 10−34 J s. The reduced Planck constant is ℏ= ℎ 2𝜋.
Quantum mechanics typically uses the reduced Planck constant over the normal Planck constant. The dimensionality of ℎis energy multiplied by time, or position multiplied by momentum.
301 VI. Quantum Mechanics 1.5. Photoelectric effect Consider a metal surface in a vacuum, which is hit by light with angular frequency 𝜔. When the radiation hits the surface of the metal, electrons were emitted. Classically, we would expect that: (i) Since the incident light carries energy proportional to its intensity, increasing the in-tensity we should have sufficient energy to break the bonds of the electrons with the atoms of the metal.
(ii) Since the intensity and frequency are independent, light of any 𝜔would eventually cause electrons to be emitted, given a high enough intensity.
(iii) The emission rate should be constant.
In fact, the experiment showed that (i) The maximum energy 𝐸max of emitted electrons depended on 𝜔, and not on the intens-ity.
(ii) Below a given threshold 𝜔min, there was no electron emission.
(iii) The emission rate increased with the intensity.
Einstein’s explanation for this phenomenon was that the light was quantised into small quanta, called photons. Photons each carry an energy 𝐸= ℏ𝜔. Each photon could liber-ate only one electron. Thus, 𝐸max = ℏ𝜔−𝜙 where 𝜙is the binding energy of the electron with the metal. The higher the intensity, the more photons hit the metal. This implies that more electrons will be scattered.
1.6. Compton scattering X-rays were emitted towards a crystal, scattering free electrons. The X-ray should then be deflected by some angle 𝜃. Classically, for a given 𝜃we would expect that the intensity as a function of 𝜔would have a maximum at 𝜔0, the frequency of the incoming X-rays. This is because we would not expect 𝜔to change much after scattering an electron. However, there was another peak at 𝜔′, which was dependent on the angle 𝜃. In fact, considering the photon and electron as a relativistic system of particles, we can derive (from IA Dynamics and Relativity), 2 sin2 𝜃 2 = 𝑚𝑐 |q| −𝑚𝑐 |p| where p is the initial momentum and q is the final momentum. Assuming 𝐸= ℏ𝜔and p = ℏk, |p| = ℏ|k| = ℏ𝜔 𝑐; |q| = ℏ|k′| = ℏ𝜔′ 𝑐 302 1. Historical introduction Hence, 1 𝜔= 1 𝜔′ + ℏ 𝑚𝑐(1 −cos 𝜃) So the frequency of the outgoing X-ray should have an angular frequency which is shifted from the original. The expected peak was actually caused by X-rays simply not interacting with the electrons.
1.7. Atomic spectra The Rutherford scattering experiment involved shooting 𝛼particles at some thin gold foil.
Most particles travelled through the foil, some were slightly deflected, and some were deflec-ted completely back. This indicated that the gold foil was mostly comprised of vacuum and there was a high density of positive charge within the atom. Electrons would orbit around the nucleus. However, there were problems with this model: (i) If the electrons in orbits moved, they would radiate and lose energy. However if the electrons were static, they would simply collapse and fall into the nucleus.
(ii) This model did not explain the atomic spectra, the observed frequencies of light ab-sorbed or emitted by an atom when electrons change energy levels.
The spectra had frequency 𝜔𝑚𝑛= 2𝜋𝑐𝑅0( 1 𝑛2 −1 𝑚𝑐); 𝑚, 𝑛∈ℕ, 𝑚> 𝑛 where 𝑅0 is the Rydberg constant, approximately 1 × 107 m−1. Bohr theorised that the electron orbits themselves are quantised, so 𝐿(the orbital angular momentum) is an in-teger multiple of ℏ; 𝐿𝑛= 𝑛ℏ. First, the quantisation of 𝐿implies the quantisation of 𝑣 and 𝑟. Indeed, given that 𝐿≡𝑚𝑒𝑣𝑟, we have that 𝑣is quantised: 𝑣𝑛= 𝑛ℏ 𝑚𝑒𝑟. Further, by the Coulomb force, 𝐹= 𝑒2 4𝜋𝜀2 1 𝑟2 e𝑟= 𝑚𝑒𝑎𝑟e𝑟where 𝑎𝑟is the radial acceleration. Then 𝑒2 4𝜋𝜀2 1 𝑟2 = 𝑚𝑒 𝑣2 𝑟 ⟹𝑟= 𝑟 𝑛= 4𝜋𝜀0ℏ2 𝑚𝑒𝑒2 𝑛2. The coefficient on 𝑛2 is known as the Bohr radius.
Immediately then the energy levels 𝐸of the atom can be shown to be quantised, since 𝐸= 1 2𝑚𝑒𝑣2 − 𝑒2 4𝜋𝜀0 1 𝑟 giving 𝐸𝑛= − 𝑒2 8𝜋𝜀0𝑎0 1 𝑛2 = −𝑒4𝑚𝑒 32𝜋2𝜀2 0ℏ2 1 𝑛2 The ground energy level is at 𝑛= 1, giving 𝐸1 = −13.6 eV 303 VI. Quantum Mechanics The excited states are 𝐸𝑛for 𝑛> 1. The energy emitted when descending from 𝐸𝑛to 𝐸1 are the spectral lines: Δ𝐸= ℏ𝜔 The Bohr model gives 𝜔𝑚𝑛= Δ𝐸𝑚𝑛 ℏ = 2𝜋𝑐( 𝑒2 4𝜋𝜀0ℏ𝑐) 2 ( 1 𝑛2 −1 𝑚2 ) which agrees with the Rydberg constant 𝑅0 defined earlier.
304 2. Wavefunctions 2. Wavefunctions 2.1. Wave-like behaviour of particles De Broglie hypothesised that any particle of any mass is associated with a wave with 𝜔= 𝐸 ℏ; k = p ℏ This hypothesis made sense of the quantisation of electron angular momentum; if the elec-tron lies on a circular orbit then 2𝜋𝑟= 𝑛𝜆where 𝜆is the wavelength of the electron. How-ever, 𝑝= ℏ𝑘= ℏ2𝜋 𝜆 ⟹𝐿= 𝑚𝑒𝑣𝑟= 𝑝𝑟= ℏ2𝜋 𝜆 𝑛𝜆 2𝜋= 𝑛ℏ Hence the angular momentum must be quantised. The electron diffraction experiment showed that this hypothesis was true, by showing that electrons behaved sufficiently like waves. Interference patterns were observed with 𝜆= 2𝜋 |k| = 2𝜋𝑘 |p| compatible with the De Broglie hypothesis.
2.2. Probabilistic interpretation of wavefunctions In classical mechanics, we can describe a particle with x, ̇ x or p = 𝑚̇ x. In quantum mech-anics, we need the state 𝜓described by 𝜓(x, 𝑡) called the wavefunction.
Remark. Note that the state is an abstract entity, while 𝜓(x, 𝑡) is the representation of 𝜓in the space of x. In some sense, 𝜓(x, 𝑡) is the complex coefficient of 𝜓in the continuous basis of x. In other words, 𝜓(x, 𝑡) is 𝜓in the x representation. In this course, we always work in the x representation.
Definition. A wavefunction is a function 𝜓(x, 𝑡)∶ℝ3 →ℂthat satisfies certain mathemat-ical properties (defined later) dictated by its physical interpretation. 𝑡is considered a fixed external parameter, so it is not included in the function’s type.
The physical interpretation of a wavefunction is called Born’s rule. The probability density for a particle to be at some point x at 𝑡is given by |𝜓(x, 𝑡)|2. We write the probability density as 𝜌, hence 𝜌(x, 𝑡) d𝑉is the probability that the particle lies in some small volume 𝑉centred at x. Now, since the particle must be somewhere, the wave function must be normalisable, or square-integrable in ℝ3: ∫ ℝ3 𝜓⋆(x, 𝑡)𝜓(x, 𝑡) d𝑉= ∫ ℝ3 |𝜓(x, 𝑡)|2 d𝑉= 𝑁∈(0, ∞) Since we want the total probability to be 1, we must normalise the wavefunction by defin-ing 𝜓(x, 𝑡) = 1 √𝑁 𝜓(x, 𝑡) ⟺∫ ℝ3 | |𝜓(x, 𝑡)| | 2 d𝑉= 1 305 VI. Quantum Mechanics Hence, 𝜌(x, 𝑡) = | |𝜓(x, 𝑡)| | 2 really is a probability density. From now, we will not use the bar for denoting normalisation, since normalisation is evident from context.
2.3. Bases and equivalence classes In linear algebra, we consider vectors in some vector space such as ℝ𝑛. In quantum mechan-ics, we instead consider states in a space of wave functions. The analogous concept to vector components is to represent a state 𝜓in an infinite-dimensional 𝑥axis basis 𝜓(𝑥, 𝑡). Note that if two wavefunctions differ by a constant phase, that is, ∃𝛼∈ℝsuch that ˜ 𝜓(𝑥, 𝑡) = 𝑒𝑖𝛼𝜓(𝑥, 𝑡) then the states are equivalent in terms of probability, since the probability density is given by the norm of 𝜓, not its angle. We can think of states as arrays in the vector space of wavefunc-tions. We can then describe the equivalence class [𝜓] as the set of all functions 𝜙such that 𝜙= 𝜆𝜓, for some 𝜆∈ℂ∖{0}, since we must retain the condition that 𝜙is normalisable.
2.4. Hilbert spaces In quantum mechanics, we are interested in the functional space of square-integrable func-tions on ℝ3, which is a type of Hilbert space and denoted ℋ.
Remark. Since the set of wavefunctions form a vector space, 𝜓1, 𝜓2 ∈ℋimplies that 𝜓= 𝜆1𝜓1 + 𝜆2𝜓2 ∈ℋfor constants 𝜆1, 𝜆2 ∈ℂprovided this 𝜓is nonzero. For waves, this is the well-known superposition principle. Note that this exact formulation of linearity is unique to quantum mechanics; for example, in classical mechanics, two solutions to Newton’s equa-tions may not be combined into a new solution by taking their sum.
Proposition. If 𝜓1(𝑥, 𝑡), 𝜓2(𝑥, 𝑡) are normalisable, then 𝜓= 𝜆1𝜓𝑖(𝑥, 𝑡) + 𝜆2𝜓2(𝑥, 𝑡) is also normalisable.
Proof. Recall the inequality 2|𝑧1||𝑧2| ≤|𝑧1|2 + |𝑧2|2 Then we can show ∫ ℝ3 |𝜆1𝜓1 + 𝜆2𝜓2|2 d𝑉= ∫ ℝ3 (|𝜆1𝜓1| + |𝜆2𝜓2|) 2 d𝑉 = ∫ ℝ3 (|𝜆1𝜓1|2 + 2|𝜆1𝜓1||𝜆2𝜓2| + |𝜆2𝜓2|2) d𝑉 = ∫ ℝ3 (2|𝜆1𝜓1|2 + 2|𝜆2𝜓2|2) d𝑉< ∞ so the norm is non-infinite.
306 2. Wavefunctions 2.5. Inner product We define the inner product between two wavefunctions to be ⟨𝜓, 𝜙⟩= ∫ ℝ3 𝜓⋆𝜙d𝑉 The following statements hold.
(i) ⟨𝜓, 𝜙⟩exists for all wave functions 𝜓, 𝜙∈ℋ; (ii) ⟨𝜓, 𝜙⟩ ⋆= ⟨𝜙, 𝜓⟩; (iii) the inner product is antilinear in the first entry, and linear in the second entry; and (iv) for continuous 𝜓, ⟨𝜓, 𝜓⟩= 0 is true if and only if 𝜓is identically zero.
We prove the first statement, since the others are obvious from the definition. By the Cauchy– Schwarz inequality, ∫ ℝ3 |𝜓|2 d𝑉≤𝑁1; ∫ ℝ3 |𝜙|2 d𝑉≤𝑁2; ∴∫ ℝ3 |𝜓𝜙| d𝑉≤ √ ∫ ℝ3 |𝜓|2 d𝑉⋅∫ ℝ3 |𝜙|2 d𝑉< ∞ 2.6. Normalisation Definition. We define the norm of a wavefunction to be ‖𝜓‖ ≡⟨𝜓, 𝜓⟩. A wavefunction 𝜓is normalised if ‖𝜓‖ = 1.
Definition. A set of wavefunctions {𝜓𝑛} is orthonormal if ⟨𝜓𝑚, 𝜓𝑛⟩= 𝛿𝑚𝑛. A set of wave-functions {𝜓𝑛} is complete if for any 𝜓∈ℋ, we can write 𝜓= ∑ 𝑛 𝜆𝑛𝜓𝑛 for 𝜆𝑛∈ℂ.
Proposition. If {𝜓𝑛} is a complete and orthonormal basis of ℋ, then 𝜙= 𝑛 ∑ 𝑘=0 𝑐𝑘𝜓𝑘 where 𝑐𝑘= ⟨𝜓𝑘, 𝜙⟩ 307 VI. Quantum Mechanics Proof. Suppose we can write 𝜙in this form. Then, ⟨𝜓𝑛, 𝜙⟩= ⟨𝜓𝑛, ∑ 𝑚 𝑐𝑚𝜓𝑚⟩ = ∑ 𝑚 𝑐𝑚⟨𝜓𝑛, 𝜓𝑚⟩ = ∑ 𝑚 𝑐𝑚𝛿𝑚𝑛 = 𝑐𝑛 Remark. If 𝜙is the desired outcome of a measurement for a particle described by 𝜓, then the probability of observing 𝜙given 𝜓at some time 𝑡is |⟨𝜓, 𝜙⟩|2 = | | |∫ ℝ3 𝜓⋆𝜙d𝑉| | | 2 2.7. Time-dependent Schrödinger equation Definition. The evolution of the wavefunction over time is given by the time-dependent Schrödinger equation (TDSE), 𝑖ℏ𝜕𝜓 𝜕𝑡= −ℏ2 2𝑚∇2𝜓+ 𝑈𝜓 where 𝑈= 𝑈(𝑥) is a real potential energy term.
Remark. This equation is a first-order differential equation in 𝑡. Contrast this to Newton’s second law, which is a second-order differential equation in 𝑡. This implies that we only need a single initial condition 𝜓(𝑥, 𝑡0) to determine all future behaviour.
Remark. Note the asymmetry between the spatial and temporal components: there is only a first derivative in time but a second derivative in space. This implies that this equation is incompatible with relativity, where time and space must be treated equitably.
One way to conceptualise the TDSE is by letting 𝜓be some wave defined by 𝜓(𝑥, 𝑡) = exp[𝑖(𝑘⋅𝑥−𝜔𝑡)] Then, the De Broglie hypothesis (𝑘= 𝑝/ℏ, 𝜔= 𝐸/𝑚) implies that 𝜓(𝑥, 𝑡) = exp [ 𝑖 ℏ(𝑝⋅𝑥−𝑝2 2𝑚𝑡)] which is a solution to the TDSE.
308 2. Wavefunctions 2.8. Normalisation and time evolution Because of the TDSE, we can show that the norm 𝑁of a wavefunction 𝜓is independent of 𝑡.
d𝑁 d𝑡= ∫ ℝ3 𝜕 𝜕𝑡|𝜓(𝑥, 𝑡)|2 d𝑉 Now, note that 𝜕 𝜕𝑡|𝜓|2 = 𝜕 𝜕𝑡⟨𝜓⋆, 𝜓⟩= 𝜓⋆𝜕𝜓 𝜕𝑡+ 𝜓𝜕𝜓⋆ 𝜕𝑡 The TDSE then gives 𝜕𝜓 𝜕𝑡= 𝑖ℏ 2𝑚∇2𝜓2 + 𝑖 𝑘𝑈𝜓; 𝜕𝜓⋆ 𝜕𝑡= −𝑖ℏ 2𝑚∇2𝜓2 −𝑖 𝑘𝑈𝜓⋆ ∴𝜕|𝜓|2 𝜕𝑡 = ∇⋅[ 𝑖ℏ 2𝑚(𝜓⋆∇𝜓−𝜓∇𝜓⋆)] Finally, ∫ ℝ3 𝜕|𝜓|2 𝜕𝑡 d𝑉= ∫ ℝ3 ∇⋅[ 𝑖ℏ 2𝑚(𝜓⋆∇𝜓−𝜓∇𝜓⋆)] = 0 since 𝜓, 𝜓⋆are such that |𝜓| →0 as |𝑥| →∞.
2.9. Conserved probability current We have proven that the normalisation of wavefunctions are constant in time. Hence, we can derive the probability conservation law: 𝜕𝜌 𝜕𝑡+ ∇⋅𝐽= 0; 𝐽(𝑥, 𝑡) = −𝑖ℏ 2𝑚(𝜓⋆∇𝜓−𝜓∇𝜓⋆) This is the conserved probability current.
309 VI. Quantum Mechanics 3. Observables and operators 3.1. Expectation and operators Given the wavefunction, we would like to extract some information about the particle it represents.
Definition. An observable is a property of the particle that can be measured.
Definition. An operator is any linear map ℋ→ℋsuch that ̂ 𝑂(𝑎1𝜓1 + 𝑎2𝜓2) = 𝑎1 ̂ 𝑂(𝜓1) + 𝑎2 ̂ 𝑂(𝜓2) where 𝑎1, 𝑎2 ∈ℂ, 𝜓1, 𝜓2 ∈ℋ.
In quantum mechanics, each observable is represented by an operator acting on the state 𝜓.
Each measurement is represented by an expectation value of the operator. In comparison, in linear algebra we would often use a linear transformation for a similar purpose. Once we have a basis for a linear transformation, we have a matrix. In quantum mechanics, we use the 𝑥basis, so we can write ˜ 𝜓= ( ̂ 𝑂)(𝑥, 𝑡) Example. Consider the class of finite differential operators 𝑁 ∑ 𝑛=0 𝑝𝑛(𝑥) 𝜕𝑛 𝜕𝑥𝑛 This includes, for example, position, momentum, and energy.
Example. A translation is an operator: 𝑠𝑎∶𝜓(𝑥) ↦𝜓(𝑥−𝑎) Example. The parity operator is 𝑃∶𝜓(𝑥) ↦𝜓(−𝑥) 3.2. Dynamical observables In general, to calculate the expectation value of an observable, we place the operator between 𝜓⋆and 𝜓and integrate over the whole space. From the probabilistic interpretation of the Born rule, the position of the particle can be interpreted as ⟨𝑥⟩= ∫ +∞ −∞ 𝑥|𝜓(𝑥, 𝑡)|2 d𝑥= ∫ +∞ −∞ 𝜓⋆𝑥𝜓d𝑥 310 3. Observables and operators Hence, we can write the coefficient 𝑥as the operator ̂ 𝑥. Now, consider the momentum. By considering the time-dependent Schrödinger equation with 𝑈= 0, and then integrating by parts, ⟨𝑝⟩= 𝑚d d𝑡⟨𝑥⟩ = 𝑚d d𝑡∫ +∞ −∞ 𝑥𝜓⋆𝜓d𝑥 = 𝑚∫ +∞ −∞ 𝑥𝜕 𝜕𝑡(𝜓⋆𝜓) d𝑥 = 𝑚⋅𝑖ℏ 2𝑚∫ +∞ −∞ 𝑥𝜕 𝜕𝑥(𝜓⋆𝜕𝜓 𝜕𝑥−𝜓𝜕𝜓⋆ 𝜕𝑥) d𝑥 = −𝑖ℏ 2 ∫ +∞ −∞ 𝑥𝜕 𝜕𝑥(𝜓⋆𝜕𝜓 𝜕𝑥−𝜓𝜕𝜓⋆ 𝜕𝑥) d𝑥 = −𝑖ℏ 2 ∫ +∞ −∞ (𝜓⋆𝜕𝜓 𝜕𝑥−𝜓𝜕𝜓⋆ 𝜕𝑥) d𝑥 = −𝑖ℏ∫ +∞ −∞ 𝜓⋆𝜕𝜓 𝜕𝑥d𝑥 = ∫ +∞ −∞ 𝜓⋆(−𝑖ℏ𝜕 𝜕𝑥)𝜓d𝑥 So the operator ̂ 𝑝is −𝑖ℏ 𝜕 𝜕𝑥. Given 𝑥and 𝑝, we can write many classical dynamical observ-ables. The classical notion is written in parentheses. The symbol ↦is used instead of equal-ity since we are representing the observable in the 𝑥basis.
̂ 𝑥↦𝑥 ̂ 𝑝↦−𝑖ℏ𝜕 𝜕𝑥 (𝑇= 𝑝2 2𝑚) ̂ 𝑇↦ ̂ 𝑝2 2𝑚= −ℏ2 2𝑚 𝜕2 𝜕𝑥2 ̂ 𝑈↦𝑈( ̂ 𝑥) = 𝑈(𝑥) 3.3. Hamiltonian operator The total energy is 𝐸= 𝑇+ 𝑈 given by the Hamiltonian operator ̂ 𝐻= ̂ 𝑇+ ̂ 𝑈 In one dimension, ̂ 𝐻↦−ℏ2 2𝑚 𝜕2𝜓 𝜕𝑥2 + 𝑈(𝑥)𝜓 311 VI. Quantum Mechanics In three dimensions, ̂ 𝐻↦−ℏ2 2𝑚∇2𝜓+ 𝑈(𝑥)𝜓 We can now represent the time-dependent Schrödinger equation in a more compact form: 𝑖ℏ𝜕𝜓 𝜕𝑡= ̂ 𝐻𝜓 We can now prove that for a particle in a potential 𝑈(𝑥) ≠0, d d𝑡⟨𝑝⟩= −⟨𝜕𝑈 𝜕𝑥⟩ 3.4. Time-independent Schrödinger equation From the time-dependent version of the equation, 𝑖ℏ𝜕𝜓 𝜕𝑡= ̂ 𝐻𝜓 we can try a solution of the form 𝜓(𝑥, 𝑡) = 𝑇(𝑡)𝜒(𝑥) Then, we can find 𝑖ℏ𝜕𝑇(𝑡) 𝜕𝑡 𝜒(𝑥) = 𝑇(𝑡) ̂ 𝐻𝜒(𝑥) Then, dividing by 𝑇𝜒, 1 𝑇(𝑡)(𝑖ℏ𝜕𝑇 𝜕𝑡) = ̂ 𝐻𝜒(𝑥) 𝜒 Since the left and right hand sides depend only on 𝑥and 𝑡respectively but are equal, they must be equal to a separation constant 𝐸∈ℝ. Solving for time, 1 𝑇𝑖ℏ𝜕𝑇 𝜕𝑡= 𝐸⟹𝑇(𝑡) = 𝑒 −𝑖𝐸𝑡 ℏ If 𝐸were complex, 𝑇would diverge. Solving for space, we have the time-independent Schrödinger equation as follows.
̂ 𝐻𝜒(𝑥) = 𝐸𝜒(𝑥) Explicitly, −ℏ2 2𝑚∇2𝜒(𝑥) + 𝑈(𝑥)𝜒(𝑥) = 𝐸𝜒(𝑥) This is an eigenvalue equation for ̂ 𝐻; we wish to find the eigenvalues for ̂ 𝐻in the 𝑥basis.
Note that the factorised solution 𝜓= 𝑇𝜒is just a particular class of solutions for the time-dependent Schrödinger equation. However, it can be shown that any solution to the time-dependent equation can be written as a linear combination of the time-independent equa-tion solutions.
312 4. One-dimensional solutions to the Schrödinger equation 4. One-dimensional solutions to the Schrödinger equation 4.1. Stationary states Definition. With the ansatz 𝜓(𝑥, 𝑡) = 𝜒(𝑥)𝑇(𝑡), we have found a particular class of solu-tions of the time-independent Schrödinger equation: 𝜓(𝑥, 𝑡) = 𝜒(𝑥)𝑒−𝑖𝐸𝑡 ℏ where 𝜒(𝑥) are the eigenfunctions of ̂ 𝐻with eigenvalue 𝐸. Such solutions are called station-ary states.
Note, 𝜌(𝑥, 𝑡) = |𝜓(𝑥, 𝑡)|2 = |𝜒(𝑥)|2 This explains the naming of the states as ‘stationary’, as their probability density is independ-ent of time. Now, suppose 𝐸is quantised. Then, the general solution to the system is 𝜓(𝑥, 𝑡) = 𝑁 ∑ 𝑛=1 𝑎𝑛𝜒𝑛(𝑥)𝑒−𝑖𝐸𝑛𝑡 ℏ where 𝑁can be finite or infinite. In principle, we can also have a continuous energy state 𝐸𝛼, 𝛼∈ℝ. We can still use the same idea: 𝜓(𝑥, 𝑡) = ∫ Δ𝛼 𝐴(𝛼)𝜒𝛼(𝑥)𝑒−𝑖𝐸𝛼𝑡 ℏ d𝛼 Note that |𝑎𝑛|2 and 𝐴(𝛼) d𝛼give the probability of measuring the particle energy to be 𝐸𝑛or 𝐸𝛼.
4.2. Infinite potential well We define 𝑈(𝑥) = {0 for |𝑥| ≤𝑎 ∞ for |𝑥| > 𝑎 For |𝑥| > 𝑎, we must have 𝜒(𝑎) = 0. Otherwise, 𝜒⋅𝑈= ∞. This gives us a boundary condition, 𝜒(±𝑎) = 0. For |𝑥| ≤𝑎, we seek solutions of the form −ℏ2 2𝑚𝜒″(𝑥) = 𝐸𝜒(𝑥); 𝜒(±𝑎) = 0 Equivalently, 𝜒″(𝑥) + 𝑘2𝜒(𝑥) = 0; 𝑘= √ 2𝑚𝐸 ℏ2 Since 𝐸> 0, 𝜒(𝑥) = 𝐴sin 𝑘𝑥+ 𝐵cos 𝑘𝑥 313 VI. Quantum Mechanics Imposing boundary conditions, 𝐴sin 𝑘𝑎+ 𝐵cos 𝑘𝑎= 0; 𝐴sin 𝑘𝑎−𝐵cos 𝑘𝑎= 0 Suppose 𝐴= 0, giving 𝜒(𝑥) = 𝐵cos 𝑘𝑥. Then, imposing boundary conditions, 𝜒𝑛(𝑥) = 𝐵cos 𝑘𝑛𝑥where 𝑘𝑛= 𝑛𝜋 2𝑎, and 𝑛are odd positive integers. These are even solutions.
Alternatively, suppose 𝐵= 0. In this case, 𝜒(𝑥) = 𝐴sin 𝑘𝑥. Thus, 𝜒𝑛(𝑥) = 𝐴sin 𝑘𝑛𝑥where 𝑘𝑛= 𝑛𝜋 2𝑎, and 𝑛are even nonzero positive integers. These provide odd solutions.
We can also determine the normalisation constants by defining that the eigenfunctions of the Hamiltonian are normalised to unity. Thus, ∫ 𝑎 −𝑎 |𝜒𝑛(𝑥)|2 = 1 ⟹𝐴= 𝐵= √ 1 𝑎 Hence, the general solution is given by the eigenvalues 𝐸𝑛= ℏ2 2𝑛𝑘2 𝑛= ℏ2𝜋2𝑛2 2𝑚𝑎2 and eigenfunctions 𝜒𝑛(𝑥) = √ 1 𝑎{ cos( 𝑛𝜋𝑥 2𝑎) if 𝑛odd sin( 𝑛𝜋𝑥 2𝑎) if 𝑛even Remark. Note that unlike classical mechanics, the ground state energy is not zero. Note also that 𝜒𝑛have (𝑛+ 1) nodes in which 𝜌(𝑥) = 0. When 𝑛→∞, 𝜌𝑛(𝑥) tends to a constant, which is like in classical mechanics. Eigenfunctions of the Hamiltonian in this case were either odd or even; we can in fact prove that this is the case in general.
Proposition. If we have a system of non-degenerate eigenstates (𝐸𝑖≠𝐸𝑗), then if 𝑈(𝑥) = 𝑈(−𝑥) the eigenfunctions of ̂ 𝐻must be either odd or even.
Proof. The time-independent Schrödinger equation is invariant under 𝑥↦−𝑥if 𝑈is even.
Hence, if 𝜒(𝑥) is a solution with eigenvalue 𝐸, then 𝜒(−𝑥) is also a solution. Since we have a non-degenerate solution, 𝜒(−𝑥) = 𝜒(𝑥) hence the solutions must be the same up to a normalisation factor. For consistency, 𝜒(𝑥) = 𝜒(−(−𝑥)) = 𝛼𝜒(−𝑥) = 𝛼2𝜒(𝑥). Hence 𝛼= ±1, so 𝜒is either odd or even.
4.3. Finite potential well We define 𝑈(𝑥) = {0 for |𝑥| ≤𝑎 𝑈0 for |𝑥| > 𝑎 314 4. One-dimensional solutions to the Schrödinger equation Classically, if 𝐸< 𝑈0, the particle has insufficient energy to escape the well. We will only consider eigenstates with 𝐸< 𝑈0 here, but we will find that it is possible in quantum mech-anics to escape the well with positive probability. We will search for even functions only, odd functions can be solved independently. If |𝑥| ≤𝑎, −ℏ2 2𝑚𝜒″(𝑥) = 𝐸𝜒(𝑥) Equivalently, 𝜒″(𝑥) + 𝑘2𝜒(𝑥) = 0; 𝑘= √ 2𝑚𝐸 ℏ2 The solution becomes 𝜒(𝑥) = 𝐴sin 𝑘𝑥+ 𝐵cos 𝑘𝑥⟹𝜒(𝑥) = 𝐵cos 𝑘𝑥 since we are only looking for even solutions. In the region |𝑥| > 𝑎, −ℏ2 2𝑚𝜒″(𝑥) + 𝑈0𝜒(𝑥) = 𝐸𝜒(𝑥) giving 𝜒″(𝑥) −𝑘 2 𝜒(𝑥) = 0; 𝑘= √ 2𝑚(𝑈0 −𝐸) ℏ2 This yields exponential solutions: 𝜒(𝑥) = 𝐶𝑒𝑘𝑥+ 𝐷𝑒−𝑘𝑥 Imposing the normalisability constraints, for 𝑥> 𝑎we have 𝐶= 0, and for 𝑥< −𝑎we have 𝐷= 0. Imposing even parity, 𝐶= 𝐷when nonzero. Thus, 𝜒(𝑥) = ⎧ ⎨ ⎩ 𝐶𝑒𝑘𝑥 𝑥< −𝑎 𝐵cos(𝑘𝑥) |𝑥| ≤𝑎 𝐶𝑒−𝑘𝑥 𝑥> 𝑎 Now we must impose continuity of 𝜒(𝑥) and its derivative at 𝑥= ±𝑎. First, 𝐶𝑒−𝑘𝑎= 𝐵cos(𝑘𝑎) The other gives −𝑘𝐶𝑒−𝑘𝑎= −𝑘𝐵sin(𝑘𝑎) From the ratio of both constraints, 𝑘tan(𝑘𝑎) = 𝑘 From the definition of 𝑘, 𝑘, 𝑘2 + 𝑘 2 = 2𝑚𝑈0 ℏ2 315 VI. Quantum Mechanics We will define some rescaled variables for convenience: 𝜉= 𝑘𝑎, 𝜂= 𝑘𝑎. Rewriting, 𝜉tan 𝜉= 𝜂; 𝜉2 + 𝜂2 = 𝑟2 0; 𝑟0 = 2𝑚𝑈 ℏ This may be solved graphically. The eigenvalues of the system correspond to the points of intersection between the two equations. There are always a finite number of possible intersections, regardless of the value of 𝑟0. The eigenvalues are 𝐸𝑛= ℏ2 2𝑛𝑎2 𝜉2 𝑛; 𝜉∈{𝜉1, … , 𝜉𝑛}; 𝑛= 1, … , 𝑝 When 𝑈0 →∞, 𝑟0 →∞. At this point, there are an infinite amount of intersections, so the eigenvalues of the Hamiltonian become that of the infinite well. Further 𝜒(𝑥) tends to the eigenfunctions of the infinite well. Note that the 𝜒𝑛(𝑥) have some positive region outside the well. We can use the unused condition above to write 𝐶in terms of 𝐵, and then we can use the normalisation condition to find 𝐵.
4.4. Free particles A free particle is under no potential. The time-independent Schrödinger equation is −ℏ 2𝑚𝜒″(𝑥) = 𝐸𝜒(𝑥) This has solutions 𝜒𝑘(𝑥) = 𝐴𝑒𝑖𝑘𝑥; 𝑘= √ 2𝑚𝐸 ℏ2 The complete solution, adding 𝑇(𝑡), is thus 𝜓𝑘(𝑥, 𝑡) = 𝜒𝑘(𝑥)𝑒−𝑖𝐸𝑘𝑡/ℏ= 𝐴𝑒 𝑖(𝑘𝑥−ℏ𝑘2 2𝑚𝑡) which are called De Broglie plane waves. This is not a solution since ∫ ∞ −∞ |𝜙𝑘(𝑥, 𝑡)| d𝑥= |𝐴|2 ∫ ∞ −∞ 1 d𝑥 which diverges. In general, any non-bound solution is non-normalisable. This is true since ∫ ∞ −∞|𝜒(𝑥)|2 d𝑥< ∞requires lim𝑅→∞∫ |𝑥|>𝑅|𝜒(𝑥)| d𝑥= 0. So, to solve the free particle system, we will build a linear combination of plane waves 𝜒to yield a normalisable solution.
This is called the Gaussian wavepacket. Alternatively, we can simply ignore the problem of normalisability, and change the interpretation of 𝜒𝑛(𝑥).
316 4. One-dimensional solutions to the Schrödinger equation 4.5. Gaussian wavepacket Due to the superposition principle, we can take a continuous linear combination of the 𝜓𝑘 functions.
𝜓(𝑥, 𝑡) = ∫ ∞ 0 𝐴(𝑘)𝜓𝑘(𝑥, 𝑡) d𝑘 We can construct a suitable 𝐴(𝑘) such that 𝜓is normalisable. Choosing 𝐴(𝑘) = 𝐴GP(𝑘) = exp[−𝜎 2(𝑘−𝑘0)2]; 𝑘0 ∈ℝ, 𝜎∈ℝ+ produces a solution called the Gaussian wavepacket. Substituting into the above, 𝜓GP(𝑥, 𝑡) = ∫ ∞ 0 exp[−𝜎 2(𝑘−𝑘0)2]𝜓𝑘(𝑥, 𝑡) d𝑘= ∫ ∞ 0 exp[𝐹(𝑘)] d𝑘 𝐹(𝑘) = −𝜎 2(𝑘−𝑘0)2 + 𝑖𝑘𝑥−𝑖ℏ𝑘2 2𝑚𝑡 We can rewrite this as 𝐹(𝑘) = −1 2(𝜎+ 𝑖ℏ𝑡 𝑚)𝑘2 + (𝑘0𝜎+ 𝑖𝑥)𝑘−𝜎 2𝑘2 0 We define further 𝛼≡𝜎+ 𝑖ℏ𝑡 𝑚; 𝛽= 𝑘0𝜎+ 𝑖𝑥; 𝛿= −𝜎 2𝑘2 0 Completing the square, 𝐹(𝑘) = −𝛼 2 (𝑘−𝛽 𝛼) 2 + 𝛽2 2𝛼+ 𝛿 We arrive at the solution 𝜓GP(𝑥, 𝑡) = exp[ 𝛽2 2𝛼+ 𝛿] ∫ ∞ −∞ exp[−𝛼 2 (𝑘−𝛽 𝛼) 2 ] d𝑘 Under a change of variables ̃ 𝑘= 𝑘− 𝛽 𝛼, 𝑢= Im( 𝛽 𝛼), 𝜓GP(𝑥, 𝑡) = exp[ 𝛽2 2𝛼+ 𝛿] ∫ ∞−𝑖𝑢 ∞−𝑖𝑢 exp[−𝛼 2 ̃ 𝑘] d ̃ 𝑘 We arrive at the usual Gaussian integral: 𝐼(𝑎) = ∫ ∞ −∞ exp[−𝑎𝑥2] d𝑥= √ 𝜋 2 giving 𝜓GP(𝑥, 𝑡) = √ 2𝜋 𝛼exp[ 𝛽2 2𝛼+ 𝛿] = √ 2𝜋 𝛼exp ⎡ ⎢ ⎢ ⎣ −𝜎 2 (𝑥− ℏ𝑘0 𝑚𝑡) 2 (𝜎2 + ℏ2𝑡2 𝑚2 ) ⎤ ⎥ ⎥ ⎦ 317 VI. Quantum Mechanics We define 𝜓GP to be the normalised Gaussian wavefunction, so 𝜓GP = 𝐶𝜓GP. We can find that 𝜌GP(𝑥, 𝑡) = | |𝜓GP(𝑥, 𝑡)| | 2 = √ √ √ 𝜎 𝜋(𝜎2 + ℏ2𝑡2 𝑚2 ) exp ⎡ ⎢ ⎢ ⎣ − 𝜎(𝑥− ℏ𝑘 𝑚𝑡) 2 𝜎2 + ℏ2𝑡2 𝑚2 ⎤ ⎥ ⎥ ⎦ This is a wavefunction whose probability density distribution resembles a Gaussian 𝑒−𝑥2 term, with a maximum point at ⟨𝑥⟩= ∫ ∞ −∞ 𝜓⋆ GP𝑥𝜓GP d𝑥= ∫ ∞ −∞ 𝑥𝜌GP d𝑥= ℏ𝑘0 𝑚𝑡 and a width of Δ𝑥= √⟨𝑥2⟩−⟨𝑥⟩ 2 = √ 1 2(𝜎+ ℏ2𝑡2 𝑚2𝜎) The physical interpretation is that the uncertainty of the particle’s position grows with time.
In this case, we can find ⟨𝑝⟩= ∫ ∞ −∞ 𝜓⋆ GP𝑖ℏ𝜕 𝜕𝑥𝜓GP d𝑥= ℏ𝑘0 which is constant. The uncertainty in the momentum can be found to be Δ𝑝= √⟨𝑝2⟩−⟨𝑝⟩ 2 = ℏ √ 1 2(𝜎+ ℏ2𝑡2 𝑚𝜎) Thus, Δ𝑥Δ𝑝= ℏ 2 We can find for a single plane wave that Δ𝑥= ∞; Δ𝑝= 0 4.6. Beam interpretation We can choose to ignore the normalisation problem and take the plane waves as the eigen-functions of the Hamiltonian: 𝜒𝑘(𝑥) = 𝐴𝑒𝑖𝑘𝑥; 𝜓𝑘(𝑥, 𝑡) = 𝐴𝑒𝑖𝑘𝑥𝑒−𝑖ℏ2𝑘2 2𝑚𝑡 Instead of 𝜒𝑘(𝑥) describing a single particle, we can interpret it as a beam of particles with momentum 𝑝= ℏ𝑘and 𝐸= ℏ2𝑘2 2𝑚with probability density 𝜌𝑘(𝑥) = | | |𝜒𝑘(𝑥)𝑒−𝑖ℏ2𝑘2 2𝑚𝑡| | | 2 = |𝐴|2 318 4. One-dimensional solutions to the Schrödinger equation which here is interpreted as a constant average density of particles. The probability current is given by 𝐽𝑘(𝑥, 𝑡) = −𝑖ℏ 2𝑚(𝜓⋆ 𝑘 𝜕𝜓𝑘 𝜕𝑥−𝜓𝑘 𝜕𝜓⋆ 𝑘 𝜕𝑥) = −𝑖ℏ 2𝑚|𝐴|22𝑖𝑘= |𝐴|2 ℏ𝑘 𝑚= |𝐴|2 𝑝 𝑚 ⏟ velocity This is interpreted as the average flux of particles.
4.7. Scattering states We wish to investigate what happens when a particle, or beam of particles, is thrown onto a potential 𝑈(𝑥). In this case, suppose we have a step function 𝑈(𝑥) = {𝑈0 if 0 ≤𝑥< 𝑎 0 otherwise and a Gaussian wavepacket which is centred at 𝑥0 ≪0 moving in the +𝑥direction, towards the spike in potential. As 𝑡≫0, we end up with a probability density given by two wavepack-ets; one will be moving left from the spike and one will have cleared the spike and continues moving to the right.
Definition. The reflection coefficient 𝑅is 𝑅= lim 𝑡→∞∫ 0 −∞ |𝜓GP(𝑥, 𝑡)|2 d𝑥 which is the probability for the particle to be reflected. The transmission coefficient is 𝑇= lim 𝑡→∞∫ ∞ 0 |𝜓GP(𝑥, 𝑡)|2 d𝑥 By definition, 𝑅+ 𝑇= 1.
In practice, working with Gaussian packets is mathematically challenging, although not impossible. The beam interpretation, by allowing us to use non-normalisable stationary state wavefunctions, greatly simplifies the computation.
4.8. Scattering off potential step Consider a potential 𝑈(𝑥) = {0 if 𝑥≤0 𝑈0 if 𝑥> 0 We want to solve −ℏ2 2𝑚𝜒″ 𝑘(𝑥) + 𝑈(𝑥)𝜒𝑘(𝑥) = 𝐸𝜒𝑘(𝑥) 319 VI. Quantum Mechanics We split the problem into two regions: 𝑥≤0, 𝑥> 0. For 𝑥≤0, the TISE becomes 𝜒″ 𝑘(𝑥) + 𝑘2𝜒𝑘(𝑥) = 0; 𝑘= √ 2𝑚𝐸 ℏ2 The solution is 𝜒(𝑥) = 𝐴𝑒𝑖𝑘𝑥+ 𝐵𝑒−𝑖𝑘𝑥 This is a superposition of two beams; the beam of incident particles 𝐴𝑒𝑖𝑘𝑥and the beam of reflected particles 𝐵𝑒−𝑖𝑘𝑥which are travelling in the opposite direction. In the region 𝑥> 0, we have 𝜒″ 𝑘(𝑥) + 𝑘 2 𝜒𝑘(𝑥) = 0; 𝑘= √ 2𝑚(𝐸−𝑈0) ℏ2 where 𝑘is real if 𝐸> 𝑈0, and 𝑘is pure-imaginary if 𝐸< 𝑈0. Therefore, for 𝐸> 𝑈0 we have 𝜒𝑘(𝑥) = 𝐶𝑒𝑖𝑘𝑥+ 𝐷𝑒−𝑖𝑘𝑥 which is a beam of particles moving towards the right and an incident beam of particles from the right moving towards the left. Since no such incident beam exists, we can set 𝐷= 0. If 𝐸< 𝑈0, the solution is 𝑘≡𝑖𝜂⟹𝜒𝑘(𝑥) = 𝐶𝑒−𝜂𝑥+ 𝐷𝑒𝜂𝑥 𝐷≠0 would give infinite values of 𝜒𝑘(𝑥) as 𝑥→∞. In either case, the eigenfunctions are 𝜒𝑘,𝑘(𝑥) = {𝐴𝑒𝑖𝑘𝑥+ 𝐵𝑒−𝑖𝑘𝑥 𝑥≤0 𝐶𝑒𝑖𝑘𝑥 𝑥> 0 By imposing the boundary conditions, specifically the continuity of 𝜒, we can determine the constants.
𝐴+ 𝐵= 𝐶; 𝑖𝑘𝐴−𝑖𝑘𝐵= 𝑖𝑘𝐶 which gives 𝐵= 𝑘−𝑘 𝑘+ 𝑘 𝐴; 𝐶= 2𝑘 𝑘+ 𝑘 𝐴 We can view these solutions in terms of particle flux.
𝐽𝑘(𝑥, 𝑡) = −𝑖ℏ 2𝑚(𝜓⋆ 𝑘 𝜕𝜓𝑘 𝜕𝑥−𝜓𝑘 𝜕𝜓⋆ 𝑘 𝜕𝑥) If 𝐸> 𝑈0, we find 𝐽(𝑥, 𝑡) = { ℏ𝑘 𝑚(|𝐴|2 −|𝐵|2) 𝑥< 0 ℏ𝑘 𝑚|𝐶|2 𝑥≥0 The incident flux is ℏ𝑘 𝑚|𝐴|2, the reflected flux is ℏ𝑘 𝑚|𝐵|2, and the transmitted flux is ℏ𝑘 𝑚|𝐶|2.
We can define 𝑅= 𝐽ref 𝐽inc = |𝐵|2 |𝐴|2 = (𝑘−𝑘 𝑘+ 𝑘 ) 2 320 4. One-dimensional solutions to the Schrödinger equation We can also define 𝑇= 𝐽trans 𝐽inc = 𝑘|𝐶|2 𝑘|𝐴|2 = 4𝑘𝑘 (𝑘+ 𝑘)2 We can check that our original interpretation makes sense; for example, 𝑅+ 𝑇= 1, and 𝐸→𝑈0, 𝑘→0 implies 𝑇→0, 𝑅→1. If 𝐸→∞, 𝑇→1 and 𝑅→0. If 𝐸< 𝑈0, 𝐽(𝑥, 𝑡) = { ℏ𝑘 𝑚(|𝐴|2 + |𝐵|2) 𝑥< 0 0 𝑥≥0 since 𝜒𝑘= 𝜒⋆ 𝑘. Here, 𝑇= 0 but 𝜒𝑘(𝑥) ≠0.
4.9. Scattering off a potential barrier Consider the potential 𝑈(𝑥) = {0 𝑥≤0, 𝑥≥𝑎 𝑈0 0 < 𝑥< 𝑎 When 𝐸< 𝑈0, we define 𝑘= √ 2𝑚𝐸 ℏ2 > 0; 𝜂= √ 2𝑚(𝑈0 −𝐸) ℏ2 > 0 The solution is then 𝜒(𝑥) = ⎧ ⎨ ⎩ 𝑒𝑖𝑘𝑥+ 𝐴𝑒−𝑖𝑘𝑥 𝑥≤0 𝐵𝑒−𝜂𝑥+ 𝐶𝑒𝜂𝑥 0 < 𝑥< 𝑎 𝐷𝑒𝑖𝑘𝑥 𝑥≥𝑎 since we can normalise the incoming flux to one. The boundary conditions are that 𝜒(𝑥) = 𝜒′(𝑥) are both continuous at 𝑥= 0, 𝑥= 𝑎. This gives four conditions, which are enough to solve the problem. 𝜒(𝑥) and its derivative at zero give 1 + 𝐴= 𝐵+ 𝐶; 𝑖𝑘−𝑖𝑘𝐴= −𝜂𝐵+ 𝜂𝐶 and the continuity at 𝑎gives 𝐵𝑒−𝜂𝑎+ 𝐶𝑒𝜂𝑎= 𝐷𝑒𝑖𝑘𝑎; −𝜂𝐵𝑒−𝜂𝑎+ 𝜂𝐶𝑒𝜂𝑎= 𝑖𝑘𝐷𝑒𝑖𝑘𝑎 Solving the system gives 𝐷= −4𝑖𝜂𝑘 (𝜂−𝑖𝑘)2 exp[(𝜂+ 𝑖𝑘)𝑎] −(𝜂+ 𝑖𝑘)2 exp[−(𝜂−𝑖𝑘)𝑎] The transmitted flux is 𝑗tr = ℏ𝑘 𝑚|𝐷|2 and the incident flux is 𝑗inc = ℏ𝑘 𝑚. Hence, the transmis-sion coefficient is 𝑇= |𝐷|2. This is 𝑇= 4𝑘2𝜂2 (𝑘2 + 𝜂2)2 sinh2(𝜂𝑎) + 4𝑘2𝜂2 321 VI. Quantum Mechanics If we take the limit as 𝑈0 ≫𝐸, we have 𝜂𝑎≫1. Then 𝑇→ 16𝑘2𝜂2 (𝜂2 + 𝑘2)2 exp[−2𝜂𝑎] ∝exp[−2𝑎 𝑘√2𝑚(𝑈0 −𝐸)] So the probability decreases exponentially with the width of the barrier.
4.10. Harmonic oscillator Consider a parabolic potential 𝑈(𝑥) = 1 2𝑘𝑥2 = 1 2𝑚𝜔2𝑥2 where 𝑘is an elastic constant and 𝜔= √ 𝑘 𝑚is the angular frequency of the harmonic os-cillator. Classically, we find the solution 𝑥= 𝐴cos 𝜔𝑡+ 𝐵sin 𝜔𝑡. This gives a continuous energy spectrum. The TDSE gives −ℏ2 2𝑚𝜒″(𝑥) + 1 2𝑚𝜔2𝑥2𝜒(𝑥) = 𝐸𝜒(𝑋) Since this is a bound system, we will have a discrete set of eigenvalues. The potential is sym-metric so the eigenfunctions are odd or even. We will make the change of variables 𝜉2 = 𝑚𝜔 ℏ𝑥2; 𝜀= 2𝐸 ℏ𝜔 which reformulates the TDSE as −d2𝜒 d𝜉2 + 𝜉2𝜒= 𝜀𝜒 We will start by considering the solution for 𝜀= 1. In this case, 𝐸= ℏ𝜔 2 . The solution in this case is 𝜒0(𝜉) = exp[−𝜉2 2 ] So the first eigenfunction, 𝜒0, is known in terms of 𝑥, given by 𝜒0(𝑥) = 𝐴exp[−𝑚𝜔 2ℏ𝑥2]; 𝐸0 = ℏ𝜔 2 To find the other eigenfunctions, we will take the general form 𝜒(𝜉) = 𝑓(𝜉) exp[−𝜉2 2 ] This works because we know we have a bound solution and 𝜒must tend to zero quickly as 𝜉tends to infinity, due to the differential equation in terms of 𝜉, 𝜀. Using the above ansatz for 𝜒in the Schrödinger equation, −d2𝑓 d𝜉2 + 2𝜉d𝑓 d𝜉+ (1 −𝜀)𝑓= 0 322 4. One-dimensional solutions to the Schrödinger equation Note that if 𝜀= 1, a solution is 𝑓= 1. We can find a power series solution to this differential equation, with 𝜉= 0 as a regular point.
𝑓(𝜉) = ∞ ∑ 𝑛=0 𝑎𝑛𝜉𝑛 We find 𝜉d𝑓 d𝜉= ∞ ∑ 𝑛=0 𝑛𝑎𝑛𝜉𝑛; d2𝑓 d𝜉2 = ∞ ∑ 𝑛=0 𝑛(𝑛−1)𝑎𝑛𝜉𝑛−2 = ∞ ∑ 𝑛=0 (𝑛+ 1)(𝑛+ 2)𝑎𝑛+2𝜉𝑛 Comparing coefficients of 𝜉𝑛, (𝑛+ 1)(𝑛+ 2)𝑎𝑛+2 −2𝑛𝑎𝑛+ (𝜀−1)𝑎𝑛= 0 Hence, 𝑎𝑛+2 = 2𝑛−𝜀+ 1 (𝑛+ 1)(𝑛+ 2)𝑎𝑛 Since the function must be either even or odd, exactly one of 𝑎0 and 𝑎1 must be zero.
Proposition. If the series for 𝑓does not terminate, 𝜒is not normalisable.
Proof. Suppose the series does not terminate. We will consider the asymptotic behaviour as 𝑛→∞.
𝑎𝑛+2 𝑎𝑛 →2 𝑛 But this is the same asymptotic behaviour as the function 𝑔(𝜉) given by 𝑔(𝜉) = exp[𝜉2] = ∞ ∑ 𝑚=0 𝜉2𝑚 𝑚! = ∞ ∑ 𝑛=0 𝑏𝑛𝜉𝑛 with 𝑏𝑛= { 1 𝑚!
𝑛= 2𝑚 0 𝑛= 2𝑚+ 1 So asymptotically, 𝑏𝑛+2 𝑏𝑛 = ( 𝑛 2 )!
( 𝑛 2 + 1)!
= 2 𝑛+ 2 →2 𝑛 Hence 𝜒would have a form asymptotically equal to 𝜒(𝜉) ∼exp[𝜉2 2 ] Hence 𝜒(𝜉) would be not normalisable.
323 VI. Quantum Mechanics Hence 𝑓must be a polynomial. So there exists 𝑁such that 𝑎𝑁+2 = 0 and 𝑎𝑁≠0. So for this value, 2𝑁−𝜀+ 1 = 0 ⟹𝜀= 2𝑁+ 1 By the definition of 𝜀, 𝐸𝑁= (𝑁+ 1 2)ℏ𝜔 In particular, 𝐸𝑁+1 −𝐸𝑁= ℏ𝜔. The eigenfunctions are 𝜒𝑁(𝜉) = 𝑓 𝑁(𝜉) exp[−𝜉2 2 ] with the property that 𝜒𝑁(−𝜉) = (−1)𝑁𝜒𝑁(𝜉) 𝑓 0(𝜉) = 1 𝑓 1(𝜉) = 𝜉 𝑓 2(𝜉) = 1 −2𝜉2 𝑓 3(𝜉) = 𝜉−2 3𝜉3 ⋮ 324 5. Operators and measurements 5. Operators and measurements 5.1. Hermitian operators Definition. The Hermitian conjugate of an operator ̂ 𝐴is written ̂ 𝐴†, and is defined such that ⟨̂ 𝐴†𝜓1, 𝜓2⟩= ⟨𝜓1, ̂ 𝐴𝜓2⟩ where 𝜓1, 𝜓2 ∈ℋ.
We can verify that for 𝑎1, 𝑎2 ∈ℂ, (i) (𝑎1 ̂ 𝐴1 + 𝑎2 ̂ 𝐴2)† = 𝑎⋆ 1 ̂ 𝐴† 1 + 𝑎⋆ 2 ̂ 𝐴† 2; (ii) ( ̂ 𝐴̂ 𝐵)† = ̂ 𝐵† ̂ 𝐴† Definition. A Hermitian operator is a linear operator ̂ 𝑂∶ℋ→ℋsuch that ̂ 𝐴† = ̂ 𝐴 Equivalently, ⟨̂ 𝐴𝜓1, 𝜓2⟩= ⟨𝜓1, ̂ 𝐴𝜓2⟩ Example. The familiar operators ̂ 𝑥, ̂ 𝑝are Hermitian.
⟨̂ 𝑥𝜓1, 𝜓2⟩= ∫ ℝ3 (𝑥𝜓1)⋆𝜓2 d𝑉 = ∫ ℝ3 𝜓⋆ 1𝑥𝜓2 d𝑉 = ⟨𝜓1, ̂ 𝑥𝜓2⟩ For ̂ 𝑝, integrating by parts, we have ⟨̂ 𝑝𝜓1, 𝜓2⟩= ∫ ∞ −∞ (−𝑖ℏ𝜕 𝜕𝑥𝜓1) ⋆ 𝜓2 d𝑥 = 𝑖ℏ∫ ∞ −∞ 𝜕𝜓⋆ 1 𝜕𝑥𝜓2 d𝑥 = −𝑖ℏ∫ ∞ −∞ 𝜓⋆ 1 𝜕𝜓2 𝜕𝑥d𝑥 = ⟨𝜓1, ̂ 𝑝𝜓2⟩ Theorem. The eigenvalues of a Hermitian operator are real.
Proof. Let ̂ 𝐴be a Hermitian operator, and 𝜓a normalised eigenfunction with eigenvalue 𝑎.
⟨𝜓, ̂ 𝐴𝜓⟩= ⟨𝜓, 𝑎𝜓⟩= 𝑎⟨𝜓, 𝜓⟩= 𝑎 325 VI. Quantum Mechanics Since ̂ 𝐴is Hermitian, ⟨𝜓, ̂ 𝐴𝜓⟩= ⟨̂ 𝐴𝜓, 𝜓⟩= ⟨𝑎𝜓, 𝜓⟩= 𝑎⋆⟨𝜓, 𝜓⟩= 𝑎⋆ Hence 𝑎= 𝑎⋆so 𝑎∈ℝ.
Theorem. Let ̂ 𝐴be a Hermitian operator, and 𝜓1, 𝜓2 normalised eigenfunctions with dis-tinct eigenvalues 𝑎1, 𝑎2. Then 𝜓1, 𝜓2 are orthogonal.
Proof. We have ̂ 𝐴𝜓1 = 𝑎1𝜓1 and ̂ 𝐴𝜓2 = 𝑎2𝜓2. Then, ⟨̂ 𝐴𝜓1, 𝜓2⟩= 𝑎1 ⟨𝜓1, 𝜓2⟩ But also, ⟨𝜓1, ̂ 𝐴𝜓2⟩= 𝑎2 ⟨𝜓1, 𝜓2⟩ These two values must be the same, so ⟨𝜓1, 𝜓2⟩= 0.
Theorem. The discrete and continuous set of eigenfunctions of any Hermitian operator form a complete orthogonal basis for the Hilbert space. This theorem is stated without proof.
Corollary. Every solution of the time-dependent Schrödinger can be written as a superpos-ition of stationary states.
𝜓(𝑥, 𝑡) = ∞ ∑ 𝑛=1 𝑎𝑛𝜒𝑛(𝑥)𝑒−𝑖𝐸𝑛𝑡/ℏ; 𝑎𝑛= ⟨𝜒𝑛, 𝜓⟩ In the continuous case, 𝜓(𝑥, 𝑡) = ∫ Δ𝛼 𝐴(𝛼)𝜒𝛼(𝑥)𝑒−𝑖𝐸𝑛𝑡/ℏd𝛼; 𝐴(𝛼) = ⟨𝜒𝛼, 𝜓⟩ 5.2. Postulates of quantum mechanics The following postulates are used to interpret measurements in quantum systems.
(i) Any observable 𝑂is represented by a Hermitian operator ̂ 𝑂.
(ii) The possible outcomes of 𝑂are the eigenvalues of ̂ 𝑂. Since ̂ 𝑂is Hermitian, we can only ever observe real values.
(iii) Let ̂ 𝑂have a discrete set of normalised eigenfunctions {𝜓𝑖} with distinct eigenvalues {𝜆𝑖}. Let 𝜓be a state, written in terms of the eigenfunctions of ̂ 𝑂.
𝜓= ∑𝑎𝑖𝜓𝑖 Suppose we measure 𝑂on a particle in the state 𝜓. Then, the probability that 𝑂takes value 𝜆𝑖is ℙ(𝑂= 𝜆𝑖) = |𝑎𝑖|2 = 𝑎⋆ 𝑖𝑎𝑖 326 5. Operators and measurements (iv) The above postulate can be generalised to the case where ̂ 𝑂has degenerate eigenvalues.
Let {𝜓𝑖} be a discrete set of normalised eigenfunctions with not necessarily distinct eigenvalues {𝜆𝑖}. If {𝜓𝑖}𝑖∈𝐼is a complete set of orthonormal eigenfunctions with the same eigenvalue 𝜆, then ℙ(𝑂= 𝜆) = ∑ 𝑖∈𝐼 |𝑎𝑖|2 = ∑ 𝑖∈𝐼 𝑎⋆𝑎 (v) We can verify from the postulates above that the sum of all probabilities is unity.
∑ 𝑖 |𝑎𝑖|2 = ∑ 𝑖 ⟨𝑎𝑖𝜓𝑖, 𝑎𝑖𝜓𝑖⟩= ∑ 𝑖 ∑ 𝑗 ⟨𝑎𝑖𝜓𝑖, 𝑎𝑗𝜓𝑗⟩= ⟨𝜓, 𝜓⟩= 1 (vi) If 𝑂is measured on a state 𝜓at time 𝑡, and the outcome is 𝜆𝑖, then the wavefunction instantaneously ‘collapses’ into the measured state after the measurement.
𝜓↦𝜓𝑖 This is called the projection postulate.
(vii) If ̂ 𝑂has degenerate eigenfunctions all with eigenvalue 𝜆, then instead we find 𝜓↦∑ 𝑖∈𝐼 𝑎𝑖𝜓𝑖 So in this case, the wavefunction collapses to a linear combination of the eigenfunc-tions that give this eigenvalue.
5.3. Expectation of operators Definition.
𝜓= ∑ 𝑖 𝑎𝑖𝜓𝑖= ∑ 𝑖 ⟨𝜓𝑖, 𝜓⟩𝜓𝑖 The projector operator projects 𝜓onto a specific eigenfunction.
̂ 𝑃∶𝜓↦⟨𝜓𝑖, 𝜓⟩𝜓𝑖 Definition. The expectation value of an observable ̂ 𝑂on a state 𝜓is ⟨𝑂⟩𝜓= ∑ 𝑖 𝜆𝑖ℙ(𝑂= 𝜆𝑖) = ∑ 𝑖 𝜆𝑖|⟨𝜓𝑖, 𝜓⟩|2 = ⟨∑ 𝑖 ⟨𝜓𝑖, 𝜓⟩𝜓𝑖, ∑ 𝑗 𝜆𝑗⟨𝜓𝑗, 𝜓⟩𝜓𝑗⟩ = ⟨𝜓, ̂ 𝑂𝜓⟩ 327 VI. Quantum Mechanics 5.4. Commutators Definition. The commutator of two operators ̂ 𝐴and ̂ 𝐵is the operator given by [ ̂ 𝐴, ̂ 𝐵] = ̂ 𝐴̂ 𝐵− ̂ 𝐵̂ 𝐴 We observe the following properties of the commutator.
(i) [ ̂ 𝐴, ̂ 𝐵] = −[ ̂ 𝐵, ̂ 𝐴]; (ii) [ ̂ 𝐴, ̂ 𝐴] = 0; (iii) [ ̂ 𝐴, ̂ 𝐵̂ 𝐶] = [ ̂ 𝐴, ̂ 𝐵] ̂ 𝐶+ ̂ 𝐵[ ̂ 𝐴, ̂ 𝐶]; (iv) [ ̂ 𝐴̂ 𝐵, ̂ 𝐶] = ̂ 𝐴[ ̂ 𝐵, ̂ 𝐶] + [ ̂ 𝐴, ̂ 𝐶] ̂ 𝐵; Example. The commutator [ ̂ 𝑥, ̂ 𝑝] in one dimension is given by, for every 𝜓∈ℋ, ̂ 𝑥̂ 𝑝𝜓= 𝑥(−𝑖ℏ𝜕 𝜕𝑥)𝜓(𝑥) = −𝑖ℏ𝑥𝜕𝜓 𝜕𝑥 ̂ 𝑝̂ 𝑥𝜓= (−𝑖ℏ𝜕 𝜕𝑥)𝑥𝜓(𝑥) = −𝑖ℏ𝜓−𝑖ℏ𝑥𝜕𝜓 𝜕𝑥 ∴[ ̂ 𝑥, ̂ 𝑝]𝜓= 𝑖ℏ𝜓 Hence, [ ̂ 𝑥, ̂ 𝑝] = 𝑖ℏ̂ 𝐼 where ̂ 𝐼is the identity operator. This specific commutator is known as the canonical com-mutator relation.
5.5. Simultaneously diagonalisable operators Definition. Hermitian operators ̂ 𝐴and ̂ 𝐵are said to be simultaneously diagonalisable if there exists a complete basis of joint eigenfunctions {𝜓𝑖} such that ̂ 𝐴𝜓𝑖= 𝜆𝑖𝜓𝑖and ̂ 𝐵𝜓𝑖= 𝜇𝑖𝜓𝑖for 𝜆𝑖, 𝜇𝑖∈ℝ.
Theorem. Hermitian operators ̂ 𝐴and ̂ 𝐵are simultaneously diagonalisable if and only if [ ̂ 𝐴, ̂ 𝐵] = 0.
Proof. Suppose ̂ 𝐴and ̂ 𝐵are simultaneously diagonalisable. Then, by definition, there exists a complete basis {𝜓𝑖} with eigenvalues 𝜆𝑖, 𝜇𝑖for ̂ 𝐴, ̂ 𝐵. Now, for any element 𝜓𝑖of this basis, the commutator is [ ̂ 𝐴, ̂ 𝐵]𝜓𝑖= ̂ 𝐴̂ 𝐵𝜓𝑖− ̂ 𝐵̂ 𝐴𝜓𝑖= ̂ 𝐴𝜇𝑖𝜓𝑖− ̂ 𝐵𝜆𝑖𝜓𝑖= 𝜇𝑖̂ 𝐴𝜓𝑖−𝜆𝑖̂ 𝐵𝜓𝑖= 𝜆𝑖𝜇𝑖𝜓𝑖−𝜇𝑖𝜆𝑖𝜓𝑖= 0 Let 𝜓be an arbitrary function in the Hilbert space ℋ. Then by linearity, [ ̂ 𝐴, ̂ 𝐵]𝜓= ∑ 𝑖 𝑐𝑖[ ̂ 𝐴, ̂ 𝐵]𝜓𝑖= 0 328 5. Operators and measurements Conversely, suppose that the commutator is zero. Let 𝜓𝑖be an eigenfunction of ̂ 𝐴with ei-genvalue 𝜆𝑖. Then, since the commutator is zero, we have 0 = [ ̂ 𝐴, ̂ 𝐵]𝜓𝑖= ̂ 𝐴̂ 𝐵𝜓𝑖− ̂ 𝐵̂ 𝐴𝜓𝑖⟹ ̂ 𝐴( ̂ 𝐵𝜓𝑖) = 𝜆𝑖( ̂ 𝐵𝜓𝑖) Hence, ̂ 𝐵maps the eigenspace 𝐸𝑖of ̂ 𝐴with eigenvalue 𝜆𝑖into itself. So ̂ 𝐵| |𝐸𝑖is a Hermitian operator on 𝐸𝑖. Since this holds for any eigenfunction and eigenvalue, we can find a complete basis of simultaneous eigenfunctions of ̂ 𝐴and ̂ 𝐵.
5.6. Uncertainty Definition. The uncertainty in a measurement of an observable 𝐴on a state 𝜓is defined as Δ𝜓𝐴= √(Δ𝜓𝐴) 2 where (Δ𝜓𝐴) 2 = ⟨( ̂ 𝐴−⟨̂ 𝐴⟩𝜓̂ 𝐼) 2 ⟩ 𝜓 = ⟨̂ 𝐴2⟩𝜓−(⟨̂ 𝐴⟩𝜓) 2 The two definitions are equivalent: ⟨( ̂ 𝐴−⟨̂ 𝐴⟩𝜓̂ 𝐼) 2 ⟩ 𝜓 = ∫ ℝ3 𝜓⋆( ̂ 𝐴−⟨̂ 𝐴⟩𝜓̂ 𝐼) 2 𝜓d𝑉 = ∫ ℝ3 𝜓⋆̂ 𝐴2𝜓d𝑉+ (⟨̂ 𝐴⟩𝜓) 2 ∫ ℝ3 𝜓⋆𝜓d𝑉−2 ⟨̂ 𝐴⟩𝜓∫ ℝ3 𝜓⋆𝐴𝜓d𝑉 = ⟨̂ 𝐴2⟩𝜓+ (⟨̂ 𝐴⟩𝜓) 2 −2(⟨̂ 𝐴⟩𝜓) 2 = ⟨̂ 𝐴2⟩𝜓−(⟨̂ 𝐴⟩𝜓) 2 Lemma. (Δ𝜓𝐴)2 ≥0, and Δ𝜓𝐴= 0 if and only if 𝜓is an eigenfunction of ̂ 𝐴.
Proof. Since ̂ 𝐴is Hermitian, (Δ𝜓𝐴)2 = ⟨( ̂ 𝐴−⟨̂ 𝐴⟩𝜓̂ 𝐼) 2 ⟩ 𝜓 = ⟨𝜓, ( ̂ 𝐴−⟨̂ 𝐴⟩𝜓̂ 𝐼) 2 𝜓⟩ = ⟨( ̂ 𝐴−⟨̂ 𝐴⟩𝜓̂ 𝐼)𝜓, ( ̂ 𝐴−⟨̂ 𝐴⟩𝜓̂ 𝐼)𝜓⟩ = ‖ ‖( ̂ 𝐴−⟨̂ 𝐴⟩𝜓̂ 𝐼)𝜓‖ ‖ Let 𝜙= ( ̂ 𝐴−⟨̂ 𝐴⟩𝜓̂ 𝐼)𝜓. The norm of any function is non-negative, so the square uncertainty is non-negative. Now, suppose this norm ‖𝜙‖ is zero. Then, 𝜙= 0. Hence, ̂ 𝐴𝜓= ⟨̂ 𝐴⟩𝜓𝜓 329 VI. Quantum Mechanics so it is an eigenfunction of ̂ 𝐴. If 𝜓is conversely an eigenfunction of ̂ 𝐴with eigenvalue 𝑎, then ⟨̂ 𝐴⟩𝜓= ⟨𝜓, ̂ 𝐴𝜓⟩= 𝑎‖𝜓‖ = 𝑎 Further, ⟨̂ 𝐴2⟩𝜓= ⟨𝜓, ̂ 𝐴2𝜓⟩= 𝑎2 Hence, (Δ𝜓𝐴) 2 = 𝑎2 −𝑎2 = 0 5.7. Schwarz inequality Theorem. Let 𝜓, 𝜙∈ℋ. Then, |⟨𝜓, 𝜙⟩|2 ≤⟨𝜙, 𝜙⟩⟨𝜓, 𝜓⟩ and |⟨𝜓, 𝜙⟩|2 = ⟨𝜙, 𝜙⟩⟨𝜓, 𝜓⟩⟺∃𝑎∈ℂ, 𝜙= 𝑎𝜓 Proof. For all 𝑎∈ℂ, we have 0 ≤⟨𝜙−𝑎𝜓, 𝜙−𝑎𝜓⟩ In particular, let 𝑎= ⟨𝜓, 𝜙⟩ ⟨𝜓, 𝜓⟩ Then, 0 ≤⟨𝜙, 𝜙⟩−2|⟨𝜓, 𝜙⟩|2 ⟨𝜓, 𝜓⟩ + |⟨𝜓, 𝜙⟩|2 ⟨𝜓, 𝜓⟩ = ⟨𝜙, 𝜙⟩−|⟨𝜓, 𝜙⟩|2 ⟨𝜓, 𝜓⟩ Hence, |⟨𝜓, 𝜙⟩|2 ≤⟨𝜓, 𝜓⟩⟨𝜙, 𝜙⟩ Equality holds if and only if 𝜙−𝑎𝜓= 0.
5.8. Generalised uncertainty theorem Theorem. Let 𝐴and 𝐵be observables, and 𝜓∈ℋ. Then (Δ𝜓𝐴)(Δ𝜓𝐵) ≥1 2| |⟨𝜓, [ ̂ 𝐴, ̂ 𝐵]𝜓⟩| | Proof.
(Δ𝜓𝐴) 2 = ⟨( ̂ 𝐴−⟨̂ 𝐴⟩𝜓̂ 𝐼)𝜓, ( ̂ 𝐴−⟨̂ 𝐴⟩𝜓̂ 𝐼)𝜓⟩ 330 5. Operators and measurements Defining ̂ 𝐴′ = ̂ 𝐴−⟨̂ 𝐴⟩𝜓̂ 𝐼and ̂ 𝐵′ = ̂ 𝐵−⟨̂ 𝐵⟩𝜓̂ 𝐼, (Δ𝜓̂ 𝐴′) 2 = ⟨̂ 𝐴′𝜓, ̂ 𝐴′𝜓⟩ and analogously for ̂ 𝐵′. Now, (Δ𝜓̂ 𝐴′) 2(Δ𝜓̂ 𝐵′) 2 = ⟨̂ 𝐴′𝜓, ̂ 𝐴′𝜓⟩⟨̂ 𝐵′𝜓, ̂ 𝐵′𝜓⟩≥| |⟨̂ 𝐴′𝜓, ̂ 𝐵′𝜓⟩| | 2 Since ̂ 𝐴′ is Hermitian, (Δ𝜓̂ 𝐴′)(Δ𝜓̂ 𝐵′) ≥| |⟨𝜓, ̂ 𝐴′ ̂ 𝐵′𝜓⟩| | By definition, [ ̂ 𝐴, ̂ 𝐵] = ̂ 𝐴̂ 𝐵− ̂ 𝐵̂ 𝐴and let the anticommutator be { ̂ 𝐴, ̂ 𝐵} = ̂ 𝐴̂ 𝐵+ ̂ 𝐵̂ 𝐴. If ̂ 𝐴′ and ̂ 𝐵′ are Hermitian, [ ̂ 𝐴′, ̂ 𝐵′] † = −[ ̂ 𝐴′, ̂ 𝐵′] and { ̂ 𝐴′, ̂ 𝐵′} † = { ̂ 𝐴′, ̂ 𝐵′} So the anticommutator is Hermitian. Now, we can write ̂ 𝐴′ ̂ 𝐵′ = 1 2[ ̂ 𝐴′, ̂ 𝐵′] + 1 2{ ̂ 𝐴′, ̂ 𝐵′} Hence, (Δ𝜓̂ 𝐴′)(Δ𝜓̂ 𝐵′) ≥| | |⟨𝜓, (1 2[ ̂ 𝐴′, ̂ 𝐵′] + 1 2{ ̂ 𝐴′, ̂ 𝐵′})𝜓⟩| | | = | | |⟨𝜓, 1 2[ ̂ 𝐴′, ̂ 𝐵′]𝜓⟩+ ⟨𝜓, 1 2{ ̂ 𝐴′, ̂ 𝐵′}𝜓⟩| | | We can prove that ⟨𝜓, { ̂ 𝐴′, ̂ 𝐵′}𝜓⟩∈ℝ. Since the anticommutator is Hermitian, ⟨𝜓, { ̂ 𝐴′, ̂ 𝐵′}𝜓⟩= ⟨{ ̂ 𝐴′, ̂ 𝐵′}𝜓, 𝜓⟩= ⟨𝜓, { ̂ 𝐴′, ̂ 𝐵′}𝜓⟩ ⋆ Analogously we can prove that ⟨𝜓, [ ̂ 𝐴′, ̂ 𝐵′]𝜓⟩∈𝑖ℝ.
⟨𝜓, [ ̂ 𝐴′, ̂ 𝐵′]𝜓⟩= ⟨[ ̂ 𝐴′, ̂ 𝐵′] ⋆𝜓, 𝜓⟩= −⟨𝜓, [ ̂ 𝐴′, ̂ 𝐵′]𝜓⟩ ⋆ Hence, (Δ𝜓̂ 𝐴′) 2(Δ𝜓̂ 𝐵′) 2 ≥| | |⟨𝜓, 1 2[ ̂ 𝐴′, ̂ 𝐵′]𝜓⟩+ ⟨𝜓, 1 2{ ̂ 𝐴′, ̂ 𝐵′}𝜓⟩| | | 2 = 1 4| |⟨𝜓, [ ̂ 𝐴′, ̂ 𝐵′]𝜓⟩| | 2 + 1 4| |⟨𝜓, { ̂ 𝐴′, ̂ 𝐵′}𝜓⟩| | 2 ≥1 4| |⟨𝜓, { ̂ 𝐴′, ̂ 𝐵′}𝜓⟩| | 2 ∴(Δ𝜓̂ 𝐴′) 2(Δ𝜓̂ 𝐵′) 2 ≥1 4| |⟨𝜓, { ̂ 𝐴, ̂ 𝐵}𝜓⟩| | 2 331 VI. Quantum Mechanics 5.9. Consequences of uncertainty relation (i) [ ̂ 𝐴, ̂ 𝐵] = 0 implies that there exists a joint set of eigenfunctions which is a complete basis of ℋ. In particular, ̂ 𝐴and ̂ 𝐵can be measured simulaneously with arbitrary pre-cision. For instance, we can measure 𝐸, | |𝐿| | and 𝐿𝑧simultaneously for an electron on a hydrogen atom.
(ii) We cannot simultaneously measure position and momentum of a particle with arbit-rary precision. In particular, Δ𝜓𝑥Δ𝜓𝑝≥ℏ 2 This is Heisenberg’s uncertainty principle.
5.10. States of minimal uncertainty The Gaussian wavepacket was a state of minimal uncertainty: Δ𝜓𝑥Δ𝜓𝑝= ℏ 2 We would like to analyse the conditions for a state 𝜓to have minimal uncertainty.
Lemma. 𝜓is a state of minimal uncertainty if and only if ̂ 𝑥𝜓= 𝑖𝑎̂ 𝑝𝜓 for some 𝑎∈ℝ. A non-example is the De Broglie plane waves.
Lemma. The condition for the above lemma to hold is that 𝜓(𝑥) = 𝑐𝑒−𝑏𝑥2; 𝑏, 𝑐∈ℝ, 𝑏> 0, 𝑐≠0 The Gaussian wavepacket is an example of this form.
5.11. Ehrenfest theorem Theorem. The time evolution of a Hermitian operator ̂ 𝐴is governed by d d𝑡⟨̂ 𝐴⟩𝜓= 𝑖 ℏ⟨[ ̂ 𝐻, ̂ 𝐴]⟩𝜓+ ⟨𝜕̂ 𝐴 𝜕𝑡⟩ 𝜓 In this course, we will not see any operators with time dependence, so the last term will not be needed.
332 5. Operators and measurements Proof.
d d𝑡⟨̂ 𝐴⟩𝜓= d d𝑡∫ ∞ −∞ 𝜓⋆̂ 𝐴𝜓d𝑥 = ∫ ∞ −∞ 𝜕 𝜕𝑡(𝜓⋆̂ 𝐴𝜓) d𝑥 = ∫ ∞ −∞ [𝜕𝜓⋆ 𝜕𝑡 ̂ 𝐴𝜓+ 𝜓⋆𝜕̂ 𝐴 𝜕𝑡𝜓+ 𝜓⋆̂ 𝐴𝜕𝜓 𝜕𝑡] d𝑥 The time-dependent Schrödinger equation gives (𝑖ℏ𝜕𝜓 𝜕𝑡) ⋆ = ( ̂ 𝐻𝜓) ⋆⟹−𝑖ℏ𝜕𝜓⋆ 𝜕𝑡= 𝜓⋆̂ 𝐻⋆= 𝜓⋆̂ 𝐻 Hence, d d𝑡⟨̂ 𝐴⟩𝜓= 𝑖 ℏ∫ ∞ −∞ [𝜓⋆̂ 𝐻̂ 𝐴𝜓−𝜓⋆̂ 𝐴̂ 𝐻𝜓] d𝑥+ ∫ ∞ −∞ 𝜓⋆𝜕̂ 𝐴 𝜕𝑡𝜓d𝑥 = 𝑖 ℏ⟨[ ̂ 𝐻, ̂ 𝐴]⟩𝜓+ ⟨𝜕̂ 𝐴 𝜕𝑡⟩ 𝜓 Example. Let ̂ 𝐴= ̂ 𝐻. Then, d d𝑡⟨̂ 𝐻⟩𝜓= 0 This corresponds to the classical notion of conservation of energy.
Example. Let ̂ 𝐴= ̂ 𝑝. First, note [ ̂ 𝐻, ̂ 𝑝]𝜓= [ ̂ 𝑝2 2𝑚+ 𝑈( ̂ 𝑥), ̂ 𝑝]𝜓 = [𝑈( ̂ 𝑥), ̂ 𝑝]𝜓 = 𝑈(𝑥)(−𝑖ℏ𝜕 𝜕𝑥)𝜓−(−𝑖ℏ𝜕 𝜕𝑥)𝑈(𝑥)𝜓 = 𝑖ℏ𝜕𝑈(𝑥) 𝜕𝑥 𝜓 Hence, d d𝑡⟨̂ 𝑝⟩𝜓= 𝑖 ℏ⟨[ ̂ 𝐻, ̂ 𝑝]⟩𝜓= −⟨𝜕𝑈 𝜕𝑥⟩ 𝜓 This corresponds exactly to Newton’s second law, ̇ 𝑝= −d𝑈 d𝑥 333 VI. Quantum Mechanics Example. Let ̂ 𝐴= ̂ 𝑥. We have [ ̂ 𝐻, ̂ 𝑥]𝜓= [ ̂ 𝑝2 2𝑚+ 𝑈( ̂ 𝑥), ̂ 𝑥]𝜓 = 1 2𝑚[ ̂ 𝑝2, ̂ 𝑥]𝜓 = 1 2𝑚( ̂ 𝑝[ ̂ 𝑝, ̂ 𝑥] + [ ̂ 𝑝, ̂ 𝑥] ̂ 𝑝)𝜓 = −𝑖ℏ 𝑚 Hence, d d𝑡⟨̂ 𝑥⟩𝜓= 𝑖 ℏ⟨[ ̂ 𝐻, ̂ 𝑥]⟩𝜓= ⟨̂ 𝑝⟩𝜓 𝑚 which aligns with the classical equation ̇ 𝑥= 𝑝 𝑚 334 6. Three-dimensional solutions to the Schrödinger equation 6. Three-dimensional solutions to the Schrödinger equation 6.1. Time-independent Schrödinger equation in spherical polar coordinates For a spherically symmetric potential in ℝ3, the time-independent Schrödinger equation is −ℏ2 2𝑚∇2𝜒(𝑥) + 𝑈(𝑥)𝜒(𝑥) = 𝐸𝜒(𝑥) Recall that the Laplacian operator can be expanded in spherical polar coordinates as −ℏ2 2𝑚(1 𝑟 𝜕2 𝜕𝑟2 𝑟+ 1 𝑟2 sin2 𝜃 [sin 𝜃𝜕 𝜕𝜃(sin 𝜃𝜕 𝜕𝜃) + 𝜕2 𝜕𝜙2 ])𝜒(𝑥) + 𝑈(𝑥)𝜒(𝑥) = 𝐸𝜒(𝑥) where 𝑥= 𝑟cos 𝜙sin 𝜃; 𝑦= 𝑟sin 𝜙sin 𝜃; 𝑧= 𝑟cos 𝜃 Definition. A spherically symmetric potential is a potential 𝑈which depends only on 𝑟.
We search for the particular solutions of the time-dependent Schrödinger equation with spherically symmetric potential that are radial eigenfunctions. If 𝜒(𝑟) is a function of 𝑟 alone, ∇2𝜒(𝑟) = 1 𝑟 𝜕2 𝜕𝑟2 (𝑟𝜒(𝑟)) Hence, −ℏ2 2𝑚𝑟 𝜕2 𝜕𝑟2 (𝑟𝜒(𝑟)) + 𝑈(𝑟)𝜒(𝑟) = 𝐸𝜒(𝑟) This is equivalent to −ℏ2 2𝑚(𝜒″(𝑟) + 2 𝑟𝜒′(𝑟)) + 𝑈(𝑟)𝜒(𝑟) = 𝐸𝜒(𝑟) The normalisation condition is ∫ ∞ 0 |𝜒(𝑟)|2𝑟2 d𝑟< 𝑁 The eigenfunctions 𝜒(𝑟) must converge to zero sufficiently fast as 𝑟→∞in order to be normalisable. To solve the time-independent Schrödinger equation, we will define 𝜎(𝑟) = 𝑟𝜒(𝑟) Then, −ℏ2 2𝑚𝜎″(𝑟) + 𝑈(𝑟)𝜎(𝑟) = 𝐸𝜎(𝑟) This is defined for 𝑟≥0. The normalisation condition here is ∫ ∞ 0 |𝜎(𝑟)|2 d𝑟< 𝑁; 𝜎(0) = 0; 𝜎′(0) < ∞ 335 VI. Quantum Mechanics The conditions at zero force 𝜒to be defined and have finite derivative at zero. To solve the equation for 𝜎, we solve on ℝand search for odd solutions 𝜎(−), so 𝜎(−)(−𝑟) = −𝜎(−)(𝑟) 6.2. Spherically symmetric potential well Consider the potential well given by 𝑈(𝑟) = {0 𝑟≤𝑎 𝑈0 𝑟> 𝑎 where 𝑎, 𝑈0 > 0. The time-independent Schrödinger equation is −ℏ2 2𝑚𝜎″(𝑟) + 𝑈(𝑟)𝜎(𝑟) = 𝐸𝜎(𝑟) We search for odd-parity bound states, so 0 < 𝐸< 𝑈0. Let 𝑘= √ 2𝑚𝐸 ℏ2 ; 𝑘= √ 2𝑚(𝑈0 −𝐸) ℏ2 The solution for 𝜎is 𝜎(𝑟) = {𝐴sin(𝑘𝑟) 𝑟≤𝑎 𝐵𝑒−𝑘𝑟 𝑟> 𝑎 The continuity condition at 𝑟= 𝑎can be imposed to find 𝐴sin 𝑘𝑎= 𝐵𝑒−𝑘𝑎. The continuity of the derivative gives 𝑘𝐴cos 𝑘𝑎= −𝑘𝐵𝑒−𝑘𝑎. Therefore, −𝑘cot(𝑘𝑎) = 𝑘; 𝑘2 + 𝑘 2 = 2𝑚𝑈0 ℏ2 Hence, −𝜉cot 𝜉= 𝜂; 𝜉2 + 𝜂2 = 𝑟2 0 where 𝜉= 𝑘𝑎and 𝜂= 𝑘𝑎, and 𝑟0 = 𝑎√2𝑚𝑈0/ℏ. If 𝑟0 < 𝜋 2 , we have no solutions because 𝜉≥0. Equivalently, there are no solutions if 𝑈0 < 𝜋2ℏ2 8𝑚𝑎2 336 7. Solution to hydrogen atom 7. Solution to hydrogen atom 7.1. Radial wavefunction of hydrogen atom The hydrogen atom is comprised of a nucleus and a single electron. The nucleus has a posit-ive charge and the electron has a negative charge. We will model the proton to be stationary at the origin. The Coulomb force experienced by the electron is given by 𝐹= −𝑒2 4𝜋𝜀0 1 𝑟2 = −𝜕𝑈 𝜕𝑟 ⟹𝑈= −𝑒2 4𝜋𝜀0 1 𝑟 Since zero potential is achieved only at infinity, we search for bound states with 𝐸< 0. We will search for the radial symmetric eigenfunctions. We have −ℏ2 2𝑚𝑒 (𝜒″(𝑟) + 2 𝑟𝜒′(𝑟)) − 𝑒2 4𝜋𝜀0 1 𝑟𝜒(𝑟) = 𝐸𝜒(𝑟) We define 𝜈2 = −2𝑚𝐸 ℏ2 > 0; 𝛽= 𝑒2𝑚𝑒 2𝜋𝜀0ℏ2 > 0 The Schrödinger equation becomes 𝜒″(𝑟) + 2 𝑟𝜒′(𝑟) + (𝛽 𝑟−𝜈2)𝜒(𝑟) = 0 Asymptotically as 𝑟→∞, we can see that 𝜒″ ∼𝜈2𝜒. Since 𝜈2 > 0, this yields solutions that asymptotically behave similarly to 𝑒−𝑟𝜈, where the positive exponential solution is not applicable due to the normalisation condition. For 𝑟= 0, the eigenfunction should be finite.
We will consider an ansatz educated by the asymptotical behaviour. Suppose 𝜒(𝑟) = 𝑓(𝑟)𝑒−𝜈𝑟 and we solve for 𝑓(𝑟). The Schrödinger equation is 𝑓″(𝑟) + 2 𝑟(1 −𝜈𝑟)𝑓′(𝑟) + 1 𝑟(𝛽−2𝜈)𝑓(𝑟) = 0 This is a homogeneous linear ordinary differential equation with a regular point at 𝑟= 0.
Suppose there exist series solutions.
𝑓(𝑟) = 𝑟𝑐 ∞ ∑ 𝑛=0 𝑎𝑛𝑟𝑛 We can differentiate and find 𝑓′(𝑟) = ∞ ∑ 𝑛=0 𝑎𝑛(𝑐+ 𝑛)𝑟𝑐+𝑛−1; 𝑓″(𝑟) = ∞ ∑ 𝑛=0 𝑎𝑛(𝑐+ 𝑛)(𝑐+ 𝑛−1)𝑟𝑐+𝑛−2 337 VI. Quantum Mechanics Hence, ∞ ∑ 𝑛=0 [𝑎𝑛(𝑐+ 𝑛)(𝑐+ 𝑛−1)𝑟𝑐+𝑛−2 + 2 𝑟(1 −𝜈𝑟)𝑎𝑛(𝑐+ 𝑛)𝑟𝑐+𝑛−1 + (𝛽−2𝜈)𝑟𝑐+𝑛−1] = 0 By comparing coefficients of the lowest power of 𝑟, 𝑎0𝑐(𝑐−1) + 2𝑎0𝑐= 0 ⟹𝑎0𝑐(𝑐+ 1) = 0 ⟹𝑐= −1, 0 The solution 𝑐= −1 implies 𝜒(𝑟) ∼ 1 𝑟which is invalid at 𝑟= 0. So we require 𝑐= 0. Then the power series becomes ∞ ∑ 𝑛=0 𝑎𝑛[𝑛(𝑛−1) + 2𝑛]𝑟𝑛−2 + ∞ ∑ 𝑛=0 𝑎𝑛(−2𝜈𝑛+ 𝛽−2𝜈)𝑟𝑛−1 = 0 Comparing coefficients of equal powers of 𝑟, 𝑎𝑛𝑛(𝑛+ 1) + 𝑎𝑛−1(−2𝜈𝑛+ 2𝜈+ 𝛽−2𝜈) = 0 Hence, we arrive at the recurrence relation 𝑎𝑛= 2𝜈𝑛−𝛽 𝑛(𝑛+ 1)𝑎𝑛−1 Suppose this series were infinite. Asymptotically, the behaviour of 𝑓(𝑟) is determined by 𝑎𝑛 𝑎𝑛−1 ∼ 2𝜈 𝑛. We can compare this behaviour to the asymptotic behaviour of 𝑔(𝑟) = 𝑒2𝜈𝑟. In this case, the series expansion with coefficients 𝑏𝑛satisfies 𝑏𝑛= (2𝜈)𝑛 𝑛!
⟹ 𝑏𝑛 𝑏𝑛−1 = 2𝜈 𝑛 Hence, asymptotically 𝑓(𝑟) ∼𝑒2𝜈𝑟if the series does not terminate. Since 𝜒(𝑟) = 𝑓(𝑟)𝑒−𝜈𝑟, we have 𝜒(𝑟) ∼𝑒𝜈𝑟which is not normalisable. Hence the series is finite. So there exists an integer 𝑁> 0 such that 𝑎𝑁= 0 and 𝑎𝑁−1 ≠0. This implies 2𝜈𝑁−𝛽= 0 hence 𝜈= 𝛽 2𝑁.
Substituting 𝜈2 and 𝛽, we find 𝐸= 𝐸𝑁= − 𝑒4𝑚𝑒 32𝜋2𝜀2 0ℏ2𝑁2 So the eigenvalues are equivalent to those found in the Bohr model. We now wish to find the radial eigenfunctions. Note, 𝛽 2𝜈= 𝑁hence we can substitute and find 𝑎𝑛 𝑎𝑛−1 = −2𝜈𝑁−𝑛 𝑛(𝑛+ 1) 338 7. Solution to hydrogen atom This recursion can be used to find the coefficients of the polynomial 𝑓 𝑁(𝑟).
𝑓 1(𝑟) = 1 𝑓 2(𝑟) = 1 −𝜈𝑟 𝑓 3(𝑟) = 1 −2𝜈𝑟+ 2 3𝜈2𝑟2 These are called the Laguerre polynomials of order 𝑁−1 (for example, the first order Laguerre polynomial is 𝑓 2). We can then multiply the Laguerre polynomials by 𝑒−𝜈𝑟and normalise over ℝ3 to find the normalised eigenfunctions 𝜒𝑁(𝑟). For example, 𝜒1(𝑟) = 𝜈3/2 √𝜋 = 1 √𝜋 ( 𝑒2𝑚𝑒 4𝜋𝜀0ℏ2 ) 3/2 𝑒−𝜈𝑟 Recall that the Bohr model implied that the ground state has radius 𝑎0, known as the Bohr radius, given in terms of 𝜈by 𝑎0 = 1 𝜈. Using quantum mechanics, we instead find ⟨𝑟⟩𝜒1 = ∫ ℝ3 𝜒⋆ 1(𝑟)𝑟𝜒1(𝑟) d𝑉 = ∫ 2𝜋 0 d𝜙∫ 1 −1 d cos 𝜃∫ ∞ 0 𝜈3 𝜋𝑟3𝑒−2𝜈𝑟d𝑟 = 4𝜋𝜈3 𝜋∫ ∞ 0 𝑟3𝑒−2𝜈𝑟d𝑟 = 3 2𝑎0 We have verified with physical experiments that this larger expected radius is physically accurate.
7.2. Angular momentum Recall that classically the angular momentum 𝐿is given by 𝐿= 𝑥× 𝑝 Spherically symmetric potentials conserve classical angular momentum: d𝐿 d𝑡= ̇ 𝑥× 𝑝+ 𝑥× ̇ 𝑝= 0 Solving classical problems in this way allows us to reduce a three-dimensional problem into a two-dimensional problem, by considering motion on the plane 𝐿⋅𝑥= 0. Then we re-duce to one dimension by considering ̂ 𝑒𝑟. In quantum mechanics, we can do an analogous simplification.
339 VI. Quantum Mechanics Definition. In quantum mechanics, the angular momentum is given by ̂ 𝐿= ̂ 𝑥× ̂ 𝑝= 𝑖ℏ𝑥× ∇ In Cartesian coordinates, this reduces to ̂ 𝐿𝑖= −𝑖ℏ𝜀𝑖𝑗𝑘𝑥𝑗 𝜕 𝜕𝑥𝑘 Each component ̂ 𝐿𝑖is a Hermitian operator. Note, [ ̂ 𝐿1, ̂ 𝐿2]𝜓(𝑥1, 𝑥2, 𝑥3) = −ℏ2[(𝑥2 𝜕 𝜕𝑥3 −𝑥3 𝜕 𝜕𝑥2 )(𝑥3 𝜕 𝜕𝑥1 −𝑥1 𝜕 𝜕𝑥3 ) −(𝑥3 𝜕 𝜕𝑥1 −𝑥1 𝜕 𝜕𝑥3 )(𝑥2 𝜕 𝜕𝑥3 −𝑥3 𝜕 𝜕𝑥2 )]𝜓 = −ℏ2[𝑥2 𝜕 𝜕𝑥1 −𝑥1 𝜕 𝜕𝑥2 ]𝜓 = −𝑖ℏ̂ 𝐿3𝜓 Hence the commutator [ ̂ 𝐿𝑖, ̂ 𝐿𝑗] = 𝑖ℏ𝜀𝑖𝑗𝑘̂ 𝐿𝑘is nonzero for 𝑖≠𝑗. In particular, we cannot measure each component of the angular momentum simultaneously.
Definition. The total angular momentum is ̂ 𝐿 2 = ̂ 𝐿2 1 + ̂ 𝐿2 2 + ̂ 𝐿2 3 We can find that [ ̂ 𝐿 2, ̂ 𝐿𝑖] = 0, so we can measure both the total angular momentum and a specific component of angular momentum simultaneously. For a spherically symmetric potential, given by ̂ 𝐻= ̂ 𝑝2 2𝑚+ 𝑈( ̂ 𝑟), we can find [ ̂ 𝐻, ̂ 𝐿 2] = [ ̂ 𝐻, ̂ 𝐿𝑖] = 0 7.3. Commutativity of angular momentum operators The set { ̂ 𝐻, ̂ 𝐿 2, ̂ 𝐿𝑖} commutes pairwise. By convention, we choose 𝑖= 3 to extract the 𝑧 component of the angular momentum. Hence, (i) We can find joint eigenstates of the three operators, and such eigenstates can be chosen to form a basis for the Hilbert space ℋ.
(ii) The corresponding eigenvalues |𝐿|, 𝐿𝑧, 𝐸can be measured simultaneously to an arbit-rary precision.
(iii) The set of operators is maximal; there exists no operator (other than a linear combin-ation of the above) that commutes with all three.
340 7. Solution to hydrogen atom 7.4. Joint eigenfunctions of angular momentum We search for joint eigenfunctions of ̂ 𝐿𝑧and ̂ 𝐿 2. We will write ̂ 𝐿in spherical coordinates. In Cartesian coordinates, ̂ 𝐿= −𝑖ℏ𝑥⋅∇ Hence, ̂ 𝐿3 = −𝑖ℏ𝜕 𝜕𝜙; ̂ 𝐿 2 = − ℏ2 sin2 𝜃 [sin 𝜃𝜕 𝜕𝜃(sin 𝜃𝜕 𝜕𝜃) + 𝜕2 𝜕𝜙2 ] Now we search for eigenfunctions of these operators.
̂ 𝐿 2𝑌(𝜃, 𝜙) = 𝜆𝑌(𝜃, 𝜙); ̂ 𝐿3𝑌(𝜃, 𝜙) = ℏ𝑚𝑌(𝜃, 𝜙) Solving the equation in ̂ 𝐿3, −𝑖ℏ𝜕 𝜕𝜙𝑌(𝜃, 𝜙) = ℏ𝑚𝑌(𝜃, 𝜙) We can find solutions of the form 𝑌(𝜃, 𝜙) = 𝑦(𝜃)𝑥(𝜙). We find −𝑖ℏ𝑦(𝜃)𝑥′(𝜙) = ℏ𝑚𝑦(𝜃)𝑥(𝜙) Hence 𝑦(𝜃) is arbitrary, and further −𝑖ℏ𝑥′(𝜙) = ℏ𝑚𝑥(𝜙) ⟹𝑥(𝜙) = 𝑒𝑖𝑚𝜙 Given that the wavefunctions must be single-valued on ℝ3, we must have 𝑥(𝜙) invariant under the choice of 𝜙= 𝜙+ 2𝜋𝑘. Hence 𝑚must be an integer. Since this must also be an eigenfunction of ̂ 𝐿 2, we have further − ℏ2 sin2 𝜃 [sin 𝜃𝜕 𝜕𝜃(sin 𝜃𝜕 𝜕𝜃) + 𝜕2 𝜕𝜙2 ][𝑦(𝜃)𝑥(𝜙)] = 𝜆𝑦(𝜃)𝑥(𝜙) Hence, substituting 𝑥(𝜙) = 𝑒𝑖𝑚𝜙, we find 1 sin 𝜃 𝜕 𝜕𝜃(sin 𝜃𝑦′(𝜃)) − 𝑚2 sin2 𝜃 𝑦(𝜃) = −𝜆 ℏ2 𝑦(𝜃) This is the associate Legendre equation. The solutions of 𝑦(𝜃) are the associate Legendre functions.
𝑦(𝜃) = 𝑃ℓ,𝑚(cos 𝜃) = (sin 𝜃)|𝑚| d|𝑚| d(cos 𝜃)|𝑚| 𝑃ℓ(cos 𝜃) where the 𝑃ℓare the Legendre polynomials. Since the ordinary Legendre polynomials are of degree ℓ, we must have |𝑚| ≤ℓto obtain a nonzero solution. This corresponds to the classical notion that |𝐿𝑧| ≤|𝐿| for a physical solution. The eigenvalues of ̂ 𝐿 2 are 𝜆= ℓ(ℓ+ 1)ℏ2 with ℓ∈{0, 1, 2, … }. Thus, 𝑌ℓ,𝑚(𝜃, 𝜙) = 𝑃ℓ,𝑚(cos 𝜃)𝑒𝑖𝑚𝜙 341 VI. Quantum Mechanics The 𝑌functions are called the spherical harmonics. The parameters ℓ, 𝑚are known as the quantum numbers of the eigenfunction; ℓis the total angular momentum quantum number and 𝑚is the azimuthal quantum number. Examples of normalised eigenfunctions are 𝑌 0,0 = 1 √4𝜋 𝑌 1,0 = √ 3 4𝜋cos 𝜃 𝑌 1,±1 = ∓√ 3 8𝜋sin 𝜃𝑒−𝑖𝜙 All spherical harmonics can be shown to be orthogonal.
7.5. Full solution to hydrogen atom The time-independent Schrödinger equation for the hydrogen atom is −ℏ2 2𝑚𝑒 ∇2𝜒(𝑟, 𝜃, 𝜙) − 𝑒2 4𝜋𝜀0 1 𝑟𝜒(𝑟, 𝜃, 𝜙) = 𝐸𝜒(𝑟, 𝜃, 𝜙) Writing the Laplacian in spherical polar coordinates, ∇2 = 1 𝑟 𝜕2 𝜕𝑟2 + 1 𝑟2 sin2 𝜃 (sin 𝜃𝜕 𝜕𝜃sin 𝜃𝜕 𝜕𝜃+ 𝜕2 𝜕𝜙2 ) Hence, ̂ 𝐿 2 = ℏ2 sin2 𝜃 [sin 𝜃𝜕 𝜕𝜃sin 𝜃𝜕 𝜕𝜃+ 𝜕2 𝜕𝜙2 ] ⟹−ℏ2∇2 = −ℏ2 𝑟 𝜕2 𝜕𝑟2 𝑟+ ̂ 𝐿 2 𝑟2 Thus we can rewrite the TISE as −ℏ2 2𝑚𝑒 1 𝑟( 𝜕2 𝜕𝑟2 (𝑟𝜒)) + ̂ 𝐿 2 2𝑚𝑒𝑟2 𝜒− 𝑒2 4𝜋𝜀0𝑟𝜒= 𝐸𝜒 Since ̂ 𝐿 2, ̂ 𝐿3, ̂ 𝐻are a maximal set of pairwise commuting operators, we know that the eigen-functions of the Hamiltonian 𝜒must also be eigenfunctions of ̂ 𝐿 2, ̂ 𝐿3. Hence, 𝜒(𝑟, 𝜃, 𝜙) = 𝑅(𝑟)𝑌ℓ,𝑚(𝜃, 𝜙) Since 𝜒is an eigenfunction of ̂ 𝐿 2, ̂ 𝐿 2(𝑅(𝑟)𝑌ℓ,𝑚(𝜃, 𝜙)) = 𝑅(𝑟)ℏ2ℓ(ℓ+ 1)𝑌ℓ,𝑚(𝜃, 𝜙) Substituting into the TISE, we find −ℏ2 2𝑚𝑒 (𝜕2𝑅 𝜕𝑟2 + 2 𝑟 𝜕𝑅 𝜕𝑟)𝑌ℓ,𝑚(𝜃, 𝜙) + ℏ2 2𝑚𝑒𝑟2 ℓ(ℓ+ 1)𝑅(𝑟)𝑌ℓ,𝑚(𝜃, 𝜙) − 𝑒2 4𝜋𝜀0𝑟𝑅(𝑟)𝑌ℓ,𝑚(𝜃, 𝜙) = 𝐸𝑅(𝑟)𝑌ℓ,𝑚(𝜃, 𝜙) 342 7. Solution to hydrogen atom Cancelling the spherical harmonic, −ℏ2 2𝑚𝑒 (𝜕2𝑅 𝜕𝑟2 + 2 𝑟 𝜕𝑅 𝜕𝑟) + ( ℏ2 2𝑚𝑒𝑟2 ℓ(ℓ+ 1) − 𝑒2 4𝜋𝜀0𝑟) ⏟⎵ ⎵ ⎵ ⎵ ⎵ ⎵ ⎵⏟⎵ ⎵ ⎵ ⎵ ⎵ ⎵ ⎵⏟ 𝑈eff= effective potential 𝑅(𝑟) = 𝐸𝑅(𝑟) This is an equation for the radial part of the solution. We have already solved this equation for ℓ= 0 to find 𝜒(𝑟), the radial wavefunction. Note that the azimuthal quantum number does not appear in the effective potential, giving a degeneracy of order at least 2ℓ+ 1. We define 𝜈2 = −2𝑚𝑒𝐸 ℏ2 > 0; 𝛽= 𝑒2𝑚𝑒 2𝜋𝜀0ℏ2 Hence, 𝑅″ + 2 𝑟𝑅′ + (𝛽 𝑟−𝜈2 −ℓ(ℓ+ 1) 𝑟2 )𝑅= 0 The asymptotic limit is as before in the radial case, since the angular velocity dependence is suppressed by 1 𝑟2 . We have 𝑅″ −𝜈2𝑅→0 hence 𝑅∝𝑒−𝜈𝑟in the limit. We let 𝑅(𝑟) = 𝑔(𝑟)𝑒−𝜈𝑟.
Then, 𝑔″ + 2 𝑟(1 −𝜈𝑟)𝑔′ + (𝛽 𝑟−2𝜈−ℓ(ℓ+ 1) 𝑟2 )𝑔= 0 Expanding in power series, 𝑔(𝑟) = 𝑟𝜎 ∞ ∑ 𝑛=0 𝑎𝑛𝑟𝑛 Substituting and comparing the lowest power of 𝑟, 𝑎0[𝜎(𝜎−1) + 2𝜎−ℓ(ℓ+ 1)] = 0 ⟹𝜎(𝜎+ 1) = ℓ(ℓ+ 1) Hence, 𝜎= ℓor 𝜎= −ℓ−1. If 𝜎= −ℓ−1, we have 𝑅(𝑟) ∼ 1 𝑟ℓ+1 which cannot be the solution, so 𝜎= ℓ. Thus, 𝑔(𝑟) = 𝑟ℓ ∞ ∑ 𝑛=0 𝑎𝑛𝑟𝑛 We can evaluate the recurrence relation between the coefficients as before to find ∞ ∑ 𝑛=0 [(𝑛+ ℓ)(𝑛+ ℓ−1)𝑎𝑛+ 2(𝑛+ 1)𝑎𝑛−ℓ(ℓ+ 1)𝑎𝑛 −2𝜈(𝑛+ ℓ−1)𝑎𝑛−1 + (𝛽−2𝜈)𝑎𝑛−1]𝑟ℓ+𝑛−2 = 0 which gives 𝑎𝑛= 2𝜈(𝑛+ ℓ) −𝛽 𝑛(𝑛+ 2ℓ−1) If ℓ= 0 this yields the result for the radial solution. Unless the series terminates, it is possible to show that 𝑅diverges. Hence 𝑔must be a polynomial with first zero coefficient 𝑎𝑛max. Here, 2𝜈(𝑛max + ℓ) −𝛽= 0 343 VI. Quantum Mechanics We define 𝑁= 𝑛max + ℓ, so 2𝜈𝑁−𝛽= 0 giving 𝜈= 𝛽 2𝑁. Note that 𝑁> ℓsince 𝑛max > 0. We can then find the energy level to be 𝐸𝑁= − 𝑒4𝑚𝑒 32𝜋2𝜀2 0ℏ2 1 𝑛2 which is an identical energy spectrum as we found before when not considering angular momentum (using the Bohr model). For each 𝐸𝑁, we have 𝑁= 𝑛max + ℓso there can be ℓ= 0, … , 𝑁−1 and 𝑚= −ℓ, … , ℓ. Hence, the degeneracy of the solution for each 𝑁is 𝐷(𝑁) = 𝑁−1 ∑ ℓ=0 ℓ ∑ 𝑚=−ℓ 1 = 𝑁2 So the degeneracy increases quadratically with the energy level. For example, for 𝑁= 2 there are four possible eigenfunctions with the same energy. The eigenfunctions are now dictated by three quantum numbers.
𝜒𝑁,ℓ,𝑚(𝑟, 𝜃, 𝜙) = 𝑅𝑁,ℓ(𝑟)𝑌ℓ,𝑚(𝜃, 𝜙) = 𝑟ℓ𝑔𝑁,ℓ(𝑟)𝑒−𝛽𝑟 2𝑁𝑌ℓ,𝑚(𝜃, 𝜙) where 𝑔𝑁,ℓis a polynomial of degree 𝑁−ℓ−1 defined by the recurrence relation 𝑎𝑘= 2𝜈 𝑘 𝑘+ ℓ−𝑁 𝑘+ 2ℓ+ 1𝑎𝑛−1 These are the generalised Laguerre polynomials, often written 𝑔𝑁,ℓ(𝑟) = 𝐿2ℓ+1 𝑁−ℓ−1(2𝑟) The quantum number 𝑁∈{0, 1, … } is known as the principal quantum number.
7.6. Comparison to Bohr model In the Bohr model, the energy levels were predicted accurately. Further, the maximum of the radial probability corresponds to the orbits found in the Bohr model: d d𝑟(| |𝜒𝑁,0,0(𝑟)| | 2𝑟2) = 0 The classical trajectory, and the assumption about the angular momentum 𝐿 2 = 𝑁2ℏ2, were incorrect. The angular momentum found in quantum mechanics is 𝐿 2 = ℓ(ℓ+ 1)ℏ2, which corresponds closely with the Bohr model for large ℓ.
344 7. Solution to hydrogen atom 7.7. Other elements of the periodic table The above solution does not hold for other elements of the periodic table. Generalising to a nucleus with charge +𝑧𝑒and 𝑧orbiting electrons, we could model this as 𝜒(𝑥1, … , 𝑥𝑧) = 𝜒(𝑥1) … 𝜒(𝑥𝑁); 𝐸= 𝑁 ∑ 𝑗=1 𝑒𝑗 This approximation can be acceptable for small 𝑧, but diverges very quickly from the true solution as 𝑧increases, due to the electron-electron interactions and the Pauli exclusion principle.
345 VII. Linear Algebra Lectured in Michaelmas 2021 by Prof. P. Raphael Linear algebra is the field of study that deals with vector spaces and linear maps. A vector space can be thought of as a generalisation of ℝ𝑛or ℂ𝑛, although they can be based off any field (not just ℝor ℂ), and may have infinitely many dimensions. In this course, we mainly study finite-dimensional vector spaces and the linear functions between them. Any linear map between finite-dimensional vector spaces can be encoded as a matrix. Such maps have properties such as their trace and determinant, which can be easily obtained from a matrix representing them. As was shown for real matrices in Vectors and Matrices, if the determinant of a matrix is nonzero it can be inverted.
347 VII. Linear Algebra Contents 1.
Vector spaces and linear dependence . . . . . . . . . . . . . . . . . 351 1.1.
Vector spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . 351 1.2.
Subspaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 351 1.3.
Sum of subspaces . . . . . . . . . . . . . . . . . . . . . . . . . 352 1.4.
Quotients . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 352 1.5.
Span . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 353 1.6.
Dimensionality . . . . . . . . . . . . . . . . . . . . . . . . . . 353 1.7.
Linear independence . . . . . . . . . . . . . . . . . . . . . . . 354 1.8.
Bases . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 354 1.9.
Steinitz exchange lemma . . . . . . . . . . . . . . . . . . . . . 355 1.10.
Consequences of Steinitz exchange lemma . . . . . . . . . . . . 356 1.11.
Dimensionality of sums . . . . . . . . . . . . . . . . . . . . . . 356 1.12.
Direct sums . . . . . . . . . . . . . . . . . . . . . . . . . . . . 357 2.
Linear maps . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 360 2.1.
Linear maps . . . . . . . . . . . . . . . . . . . . . . . . . . . 360 2.2.
Isomorphism . . . . . . . . . . . . . . . . . . . . . . . . . . . 361 2.3.
Kernel and image . . . . . . . . . . . . . . . . . . . . . . . . . 362 2.4.
Rank and nullity . . . . . . . . . . . . . . . . . . . . . . . . . 363 2.5.
Space of linear maps . . . . . . . . . . . . . . . . . . . . . . . 363 2.6.
Matrices . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 364 2.7.
Linear maps as matrices . . . . . . . . . . . . . . . . . . . . . 364 2.8.
Change of basis . . . . . . . . . . . . . . . . . . . . . . . . . . 366 2.9.
Equivalent matrices . . . . . . . . . . . . . . . . . . . . . . . . 367 2.10.
Column rank and row rank . . . . . . . . . . . . . . . . . . . . 369 2.11.
Conjugation and similarity . . . . . . . . . . . . . . . . . . . . 370 2.12.
Elementary operations . . . . . . . . . . . . . . . . . . . . . . 370 2.13.
Gauss’ pivot algorithm . . . . . . . . . . . . . . . . . . . . . . 371 2.14.
Representation of square invertible matrices . . . . . . . . . . . 371 3.
Dual spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 373 3.1.
Dual spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . 373 3.2.
Annihilators . . . . . . . . . . . . . . . . . . . . . . . . . . . 375 3.3.
Dual maps . . . . . . . . . . . . . . . . . . . . . . . . . . . . 375 3.4.
Properties of dual map . . . . . . . . . . . . . . . . . . . . . . 377 3.5.
Double duals . . . . . . . . . . . . . . . . . . . . . . . . . . . 378 4.
Bilinear forms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 380 4.1.
Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . 380 4.2.
Change of basis for bilinear forms . . . . . . . . . . . . . . . . 382 348 5.
Trace and determinant . . . . . . . . . . . . . . . . . . . . . . . . . 383 5.1.
Trace . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 383 5.2.
Permutations and transpositions . . . . . . . . . . . . . . . . . 383 5.3.
Determinant . . . . . . . . . . . . . . . . . . . . . . . . . . . 384 5.4.
Volume forms . . . . . . . . . . . . . . . . . . . . . . . . . . . 385 5.5.
Multiplicative property of determinant . . . . . . . . . . . . . . 386 5.6.
Singular and non-singular matrices . . . . . . . . . . . . . . . . 387 5.7.
Determinants of linear maps . . . . . . . . . . . . . . . . . . . 388 5.8.
Determinant of block-triangular matrices . . . . . . . . . . . . . 388 6.
Adjugate matrices . . . . . . . . . . . . . . . . . . . . . . . . . . . 390 6.1.
Column and row expansions . . . . . . . . . . . . . . . . . . . 390 6.2.
Adjugates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 391 6.3.
Cramer’s rule . . . . . . . . . . . . . . . . . . . . . . . . . . . 392 7.
Eigenvectors and eigenvalues . . . . . . . . . . . . . . . . . . . . . 393 7.1.
Eigenvalues . . . . . . . . . . . . . . . . . . . . . . . . . . . . 393 7.2.
Polynomials . . . . . . . . . . . . . . . . . . . . . . . . . . . . 393 7.3.
Characteristic polynomials . . . . . . . . . . . . . . . . . . . . 394 7.4.
Polynomials for matrices and endomorphisms . . . . . . . . . . 395 7.5.
Sharp criterion of diagonalisability . . . . . . . . . . . . . . . . 396 7.6.
Simultaneous diagonalisation . . . . . . . . . . . . . . . . . . . 398 7.7.
Minimal polynomials . . . . . . . . . . . . . . . . . . . . . . . 399 7.8.
Cayley–Hamilton theorem . . . . . . . . . . . . . . . . . . . . 400 7.9.
Algebraic and geometric multiplicity . . . . . . . . . . . . . . . 401 7.10.
Characterisation of diagonalisable complex endomorphisms . . . 402 8.
Jordan normal form . . . . . . . . . . . . . . . . . . . . . . . . . . 404 8.1.
Definition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 404 8.2.
Similarity to Jordan normal form . . . . . . . . . . . . . . . . . 404 8.3.
Direct sum of eigenspaces . . . . . . . . . . . . . . . . . . . . . 404 9.
Properties of bilinear forms . . . . . . . . . . . . . . . . . . . . . . 408 9.1.
Changing basis . . . . . . . . . . . . . . . . . . . . . . . . . . 408 9.2.
Quadratic forms . . . . . . . . . . . . . . . . . . . . . . . . . 408 9.3.
Diagonalisation of symmetric bilinear forms . . . . . . . . . . . 409 9.4.
Sylvester’s law . . . . . . . . . . . . . . . . . . . . . . . . . . 410 9.5.
Kernels of bilinear forms . . . . . . . . . . . . . . . . . . . . . 412 9.6.
Sesquilinear forms . . . . . . . . . . . . . . . . . . . . . . . . 413 9.7.
Hermitian forms . . . . . . . . . . . . . . . . . . . . . . . . . 413 9.8.
Polarisation identity . . . . . . . . . . . . . . . . . . . . . . . 414 9.9.
Hermitian formulation of Sylvester’s law . . . . . . . . . . . . . 414 9.10.
Skew-symmetric forms . . . . . . . . . . . . . . . . . . . . . . 415 9.11.
Skew-symmetric formulation of Sylvester’s law . . . . . . . . . . 415 349 VII. Linear Algebra 10.
Inner product spaces . . . . . . . . . . . . . . . . . . . . . . . . . . 416 10.1.
Definition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 416 10.2.
Cauchy–Schwarz inequality . . . . . . . . . . . . . . . . . . . . 416 10.3.
Orthogonal and orthonormal sets . . . . . . . . . . . . . . . . . 417 10.4.
Parseval’s identity . . . . . . . . . . . . . . . . . . . . . . . . . 418 10.5.
Gram–Schmidt orthogonalisation process . . . . . . . . . . . . 418 10.6.
Orthogonality of matrices . . . . . . . . . . . . . . . . . . . . . 419 10.7.
Orthogonal complement and projection . . . . . . . . . . . . . 419 10.8.
Projection maps . . . . . . . . . . . . . . . . . . . . . . . . . . 420 10.9.
Adjoint maps . . . . . . . . . . . . . . . . . . . . . . . . . . . 421 10.10. Self-adjoint and isometric maps . . . . . . . . . . . . . . . . . . 421 10.11. Spectral theory for self-adjoint maps . . . . . . . . . . . . . . . 422 10.12. Spectral theory for unitary maps . . . . . . . . . . . . . . . . . 423 10.13. Application to bilinear forms . . . . . . . . . . . . . . . . . . . 424 10.14. Simultaneous diagonalisation . . . . . . . . . . . . . . . . . . . 425 350 1. Vector spaces and linear dependence 1. Vector spaces and linear dependence 1.1. Vector spaces Definition. Let 𝐹be an arbitrary field. An 𝐹-vectorspace is an abelian group (𝑉, +) equipped with a function 𝐹× 𝑉→𝑉; (𝜆, 𝑣) ↦𝜆𝑣 such that (i) 𝜆(𝑣1 + 𝑣2) = 𝜆𝑣1 + 𝜆𝑣2 (ii) (𝜆1 + 𝜆2)𝑣= 𝜆1𝑣+ 𝜆2𝑣 (iii) 𝜆(𝜇𝑣) = (𝜆𝜇)𝑣 (iv) 1𝑣= 𝑣 Such a vector space may also be called a vector space over 𝐹.
Example. Let 𝑋be a set, and define ℝ𝑋= {𝑓∶𝑋→ℝ}. Then ℝ𝑋is an ℝ-vector space, where (𝑓 1 + 𝑓 2)(𝑥) = 𝑓 1(𝑥) + 𝑓 2(𝑥).
Example. Define 𝑀𝑛,𝑚(𝐹) to be the set of 𝑛× 𝑚𝐹-valued matrices. This is an 𝐹-vector space, where the sum of matrices is computed elementwise.
Remark. The axioms of scalar multiplication imply that ∀𝑣∈𝑉, 0𝐹𝑣= 0𝑉.
1.2. Subspaces Definition. Let 𝑉be an 𝐹-vector space. The subset 𝑈⊆𝑉is a vector subspace of 𝑉, denoted 𝑈≤𝑉, if (i) 0𝑉∈𝑈 (ii) 𝑢1, 𝑢2 ∈𝑈⟹𝑢1 + 𝑢2 ∈𝑈 (iii) (𝜆, 𝑢) ∈𝐹× 𝑈⟹𝜆𝑢∈𝑈 Conditions (ii) and (iii) are equivalent to ∀𝜆1, 𝜆2 ∈𝐹, ∀𝑢1, 𝑢2 ∈𝑈, 𝜆1𝑢1 + 𝜆2𝑢2 ∈𝑈 This means that 𝑈is stable by addition and scalar multiplication.
Proposition. If 𝑉is an 𝐹-vector space, and 𝑈≤𝑉, then 𝑈is an 𝐹-vector space.
Example. Let 𝑉= ℝℝbe the space of functions ℝ→ℝ. The set 𝐶(ℝ) of continuous real functions is a subspace of 𝑉. The set ℙof polynomials is a subspace of 𝐶(ℝ).
Example. Consider the subset of ℝ3 such that 𝑥1 + 𝑥2 + 𝑥3 = 𝑡for some real 𝑡. This is a subspace for 𝑡= 0 only, since no other 𝑡values yield the origin as a member of the subset.
351 VII. Linear Algebra Proposition. Let 𝑉be an 𝐹-vector space. Let 𝑈, 𝑊≤𝑉. Then 𝑈∩𝑊is a subspace of 𝑉.
Proof. First, note 0𝑉∈𝑈, 0𝑉∈𝑊⟹0𝑉∈𝑈∩𝑊. Now, consider stability: 𝜆1, 𝜆2 ∈𝐹, 𝑣1, 𝑣2 ∈𝑈∩𝑊⟹𝜆1𝑣1 + 𝜆2𝑣2 ∈𝑈, 𝜆1𝑣1𝜆2𝑣2 ∈𝑊 Hence stability holds.
1.3. Sum of subspaces Remark. The union of two subspaces is not, in general, a subspace. For instance, consider ℝ, 𝑖ℝ⊂ℂ. Their union does not span the space; for example, 1 + 𝑖∉ℝ∪𝑖ℝ.
Definition. Let 𝑉be an 𝐹-vector space. Let 𝑈, 𝑊≤𝑉. The sum 𝑈+ 𝑊is defined to be the set 𝑈+ 𝑊= {𝑢+ 𝑤∶𝑢∈𝑈, 𝑤∈𝑊} Proposition. 𝑈+ 𝑊is a subspace of 𝑉.
Proof. First, note 0𝑈+𝑊= 0𝑈+ 0𝑊= 0𝑉. Then, for 𝜆1, 𝜆2 ∈𝐹, and 𝑢∈𝑈, 𝑤∈𝑊, 𝜆1𝑢+ 𝜆2𝑤= 𝑢′ + 𝑤′ ∈𝑈+ 𝑊 since 𝑢′ ∈𝑈, 𝑤′ ∈𝑊. We can decompose a vector from 𝑈+𝑊into its 𝑈and 𝑊components.
Adding these components independently (noting that 𝑉is abelian) yields the requirements of a subspace.
Proposition. The sum 𝑈+ 𝑊is the smallest subspace of 𝑉that contains both 𝑈and 𝑊.
1.4. Quotients Definition. Let 𝑉be an 𝐹-vector space. Let 𝑈≤𝑉. The quotient space 𝑉/𝑈is the abelian group 𝑉/𝑈equipped with the scalar multiplication function 𝐹× 𝑉/𝑈→𝑉/𝑈; (𝜆, 𝑣+ 𝑈) ↦𝜆𝑣+ 𝑈 Proposition. 𝑉/𝑈is an 𝐹-vector space.
Proof. We must check that the multiplication operation is well-defined. Indeed, suppose 𝑣1 + 𝑈= 𝑣2 + 𝑈. Then, 𝑣1 −𝑣2 ∈𝑈⟹𝜆(𝑣1 −𝑣2) ∈𝑈⟹𝜆𝑣1 + 𝑈= 𝜆𝑣2 + 𝑈∈𝑉/𝑈 352 1. Vector spaces and linear dependence 1.5. Span Definition. Let 𝑉be an 𝐹-vector space. Let 𝑆⊂𝑉. We define the span of 𝑆, written ⟨𝑆⟩, as the set of finite linear combinations of elements of 𝑆. In particular, ⟨𝑆⟩= {∑ 𝑠∈𝑆 𝜆𝑠𝑣𝑠∶𝜆𝑠∈𝐹, 𝑣𝑠∈𝑆, only finitely many nonzero 𝜆𝑠} By convention, we specify ⟨∅⟩= {0} so that all spans are subspaces.
Remark. ⟨𝑆⟩is the smallest vector subspace of 𝑉containing 𝑆.
Example. Let 𝑉= ℝ3, and 𝑆= {( 1 0 0 ) , ( 0 1 2 )}, ( 3 −2 −4 ) Then we can check that ⟨𝑆⟩= {( 𝑎 𝑏 2𝑏 ) ∶(𝑎, 𝑏) ∈ℝ} Example. Let 𝑉= ℝ𝑛. We define 𝑒𝑖= ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ 0 ⋮ 0 1 0 ⋮ 0 ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ where the 1 is in the 𝑖th position. Then 𝑉= ⟨(𝑒𝑖)1≤𝑖≤𝑛⟩.
Example. Let 𝑋be a set, and ℝ𝑋= {𝑓∶𝑋→ℝ}. Then let 𝑆𝑥∶𝑋→ℝbe defined by 𝑆𝑥(𝑦) = {1 𝑦= 𝑥 0 otherwise Then, ⟨(𝑆𝑥)𝑥∈𝑋⟩= {𝑓∈ℝ𝑋∶𝑓has finite support}, where the support of 𝑓is defined to be {𝑥∶𝑓(𝑥) ≠0}.
1.6. Dimensionality Definition. Let 𝑉be an 𝐹-vector space. Let 𝑆⊂𝑉. We say that 𝑆spans 𝑉if ⟨𝑆⟩= 𝑉. If 𝑆 spans 𝑉, we say that 𝑆is a generating family of 𝑉.
353 VII. Linear Algebra Definition. Let 𝑉be an 𝐹-vector space. 𝑉is finite-dimensional if it is spanned by a finite set.
Example. Consider the set 𝑉= ℙ[𝑥] which is the set of polynomials on ℝ. Further, consider 𝑉 𝑛= ℙ𝑛[𝑥] which is the subspace with degree less than or equal to 𝑛. Then 𝑉 𝑛is spanned by {1, 𝑥, 𝑥2, … , 𝑥𝑛}, so 𝑉 𝑛is finite-dimensional. Conversely, 𝑉is infinite-dimensional; there is no finite set 𝑆such that ⟨𝑆⟩= 𝑉.
1.7. Linear independence Definition. We say that 𝑣1, … , 𝑣𝑛∈𝑉are linearly independent if, for 𝜆𝑖∈𝐹, 𝑛 ∑ 𝑖=1 𝜆𝑖𝑣𝑖= 0 ⟹∀𝑖, 𝜆𝑖= 0 Definition. Similarly, 𝑣1, … , 𝑣𝑛∈𝑉are linearly dependent if ∃𝛌∈𝐹𝑛, 𝑛 ∑ 𝑖=1 𝜆𝑖𝑣𝑖= 0, ∃𝑖, 𝜆𝑖≠0 Equivalently, one of the vectors can be written as a linear combination of the remaining ones.
Remark. If (𝑣𝑖)1≤𝑖≤𝑛are linearly independent, then ∀𝑖∈{1, … , 𝑛}, 𝑣𝑖≠0 1.8. Bases Definition. 𝑆⊂𝑉is a basis of 𝑉if (i) ⟨𝑆⟩= 𝑉 (ii) 𝑆is a linearly independent set So, a basis is a linearly independent (also known as free) generating family.
Example. Let 𝑉= ℝ𝑛. The canonical basis (𝑒𝑖) is a basis since we can show that they are free and span 𝑉.
Example. Let 𝑉= ℂ, considered as a ℂ-vector space. Then {1} is a basis. If 𝑉is a ℝ-vector space, {1, 𝑖} is a basis.
Example. Consider again ℙ[𝑥]. Then 𝑆= {𝑥𝑛∶𝑛∈ℕ} is a basis of ℙ.
354 1. Vector spaces and linear dependence Lemma. Let 𝑉be an 𝐹-vector space. Then, (𝑣1, … , 𝑣𝑛) is a basis of 𝑉if and only if any vector 𝑣∈𝑉has a unique decomposition 𝑣= 𝑛 ∑ 𝑖=1 𝜆𝑖𝑣𝑖, ∀𝑖, 𝜆𝑖∈𝐹 In the above definition, we call (𝜆1, … , 𝜆𝑛) the coordinates of 𝑣in the basis (𝑣1, … , 𝑣𝑛).
Proof. Suppose (𝑣1, … , 𝑣𝑛) is a basis of 𝑉. Then ∀𝑣∈𝑉there exists 𝜆1, … , 𝜆𝑛∈𝐹such that 𝑣= 𝑛 ∑ 𝑖=1 𝜆𝑖𝑣𝑖 So there exists a tuple of 𝜆values. Suppose two such 𝜆tuples exist. Then 𝑣= 𝑛 ∑ 𝑖=1 𝜆𝑖𝑣𝑖= 𝑛 ∑ 𝑖=1 𝜆′ 𝑖𝑣𝑖⟹ 𝑛 ∑ 𝑖=1 (𝜆𝑖−𝜆′ 𝑖)𝑣𝑖= 0 ⟹𝜆𝑖= 𝜆′ 𝑖 The converse is left as an exercise.
Lemma. If ⟨{𝑣1, … , 𝑣𝑛}⟩= 𝑉, then some subset of this set is a basis of 𝑉.
Proof. If (𝑣1, … , 𝑣𝑛) are linearly independent, this is a basis. Otherwise, one of the vectors can be written as a linear combination of the others. So, up to reordering, 𝑣𝑛∈⟨{𝑣1, … , 𝑣𝑛−1}⟩= 𝑉 So we have removed a vector from this set and preserved the span. By induction, we will eventually reach a basis.
1.9. Steinitz exchange lemma Theorem. Let 𝑉be a finite dimensional 𝐹-vector space. Let (𝑣1, … , 𝑣𝑚) be linearly inde-pendent, and (𝑤1, … , 𝑤𝑛) which spans 𝑉. Then, (i) 𝑚≤𝑛; and (ii) up to reordering, (𝑣1, … , 𝑣𝑚, 𝑤𝑚+1, … 𝑤𝑛) spans 𝑉.
Proof. Suppose that we have replaced ℓ≥0 of the 𝑤𝑖.
⟨𝑣1, … , 𝑣ℓ, 𝑤ℓ+1, … 𝑤𝑛⟩= 𝑉 If 𝑚= ℓ, we are done. Otherwise, ℓ< 𝑚. Then, 𝑣ℓ+1 ∈𝑉= ⟨𝑣1, … , 𝑣ℓ, 𝑤ℓ+1, … 𝑤𝑛⟩Hence 𝑣ℓ+1 can be expressed as a linear combination of the generating set. Since the (𝑣𝑖)1≤𝑖≤𝑚 are linearly independent (free), one of the coefficients on the 𝑤𝑖are nonzero. In particular, up to reordering we can express 𝑤ℓ+1 as a linear combination of 𝑣1, … , 𝑣ℓ+1, 𝑤ℓ+2, … , 𝑤𝑛.
Inductively, we may replace 𝑚of the 𝑤terms with 𝑣terms. Since we have replaced 𝑚vectors, necessarily 𝑚≤𝑛.
355 VII. Linear Algebra 1.10. Consequences of Steinitz exchange lemma Corollary. Let 𝑉be a finite-dimensional 𝐹-vector space. Then, any two bases of 𝑉have the same number of vectors. This number is called the dimension of 𝑉, dim𝐹𝑉.
Proof. Suppose the two bases are (𝑣1, … , 𝑣𝑛) and (𝑤1, … , 𝑤𝑚). Then, (𝑣1, … , 𝑣𝑛) is free and (𝑤1, … , 𝑤𝑚) is generating, so the Steinitz exchange lemma shows that 𝑛≤𝑚. Vice versa, 𝑚≤𝑛. Hence 𝑚= 𝑛.
Corollary. Let 𝑉be an 𝐹-vector space with finite dimension 𝑛. Then, (i) Any independent set of vectors has at most 𝑛elements, with equality if and only if it is a basis.
(ii) Any spanning set of vectors has at least 𝑛elements, with equality if and only if it is a basis.
Proof. Exercise.
1.11. Dimensionality of sums Proposition. Let 𝑉be an 𝐹-vector space. Let 𝑈, 𝑊be subspaces of 𝑉. If 𝑈, 𝑊are finite-dimensional, then so is 𝑈+ 𝑊, with dim𝐹(𝑈+ 𝑊) = dim𝐹𝑈+ dim𝐹𝑊−dim𝐹(𝑈∩𝑊) Proof. Consider a basis (𝑣1, … , 𝑣𝑛) of the intersection. Extend this basis to a basis (𝑣1, … , 𝑣𝑛, 𝑢1, … , 𝑢𝑚) of 𝑈; (𝑣1, … , 𝑣𝑛, 𝑤1, … , 𝑤𝑘) of 𝑊 Then, we will show that (𝑣1, … , 𝑣𝑛, 𝑢1, … , 𝑢𝑚, 𝑤1, … , 𝑤𝑘) is a basis of dim𝐹(𝑈+ 𝑊), which will conclude the proof. Indeed, since any component of 𝑈+𝑊can be decomposed as a sum of some element of 𝑈and some element of 𝑊, we can add their decompositions together.
356 1. Vector spaces and linear dependence Now we must show that this new basis is free.
𝑛 ∑ 𝑖=1 𝛼𝑖𝑣𝑖+ 𝑚 ∑ 𝑖=1 𝛽𝑖𝑢𝑖+ 𝑘 ∑ 𝑖=1 𝛾𝑖𝑤𝑖= 0 𝑛 ∑ 𝑖=1 𝛼𝑖𝑣𝑖+ 𝑚 ∑ 𝑖=1 𝛽𝑖𝑢𝑖 ⏟⎵ ⎵ ⎵ ⎵⏟⎵ ⎵ ⎵ ⎵⏟ ∈𝑈 = 𝑘 ∑ 𝑖=1 𝛾𝑖𝑤𝑖 ⏟ ⎵ ⏟ ⎵ ⏟ ∈𝑊 𝑘 ∑ 𝑖=1 𝛾𝑖𝑤𝑖∈𝑈∩𝑊 𝑘 ∑ 𝑖=1 𝛾𝑖𝑤𝑖= 𝑛 ∑ 𝑖=1 𝛿𝑖𝑣𝑖 𝑛 ∑ 𝑖=1 (𝛼𝑖+ 𝛿𝑖)𝑣𝑖+ 𝑚 ∑ 𝑖=1 𝛽𝑖𝑢𝑖= 0 𝛽𝑖= 0, 𝛼𝑖= −𝛿𝑖 𝑛 ∑ 𝑖=1 𝛼𝑖𝑣𝑖+ 𝑘 ∑ 𝑖=1 𝛾𝑖𝑤𝑖= 0 𝛼𝑖= 0, 𝛾𝑖= 0 Proposition. If 𝑉is a finite-dimensional 𝐹-vector space, and 𝑈≤𝑉, then 𝑈and 𝑉/𝑈are also finite-dimensional. In particular, dim𝐹𝑉= dim𝐹𝑈+ dim𝐹(𝑉/𝑈).
Proof. Let (𝑢1, … , 𝑢ℓ) be a basis of 𝑈. We extend this basis to a basis of 𝑉, giving (𝑢1, … , 𝑢ℓ, 𝑤ℓ+1, … , 𝑤𝑛) We claim that (𝑤ℓ+1 + 𝑈, … , 𝑤𝑛+ 𝑈) is a basis of the vector space 𝑉/𝑈.
Remark. If 𝑉is an 𝐹-vector space, and 𝑈≤𝑉, then we say 𝑈is a proper subspace if 𝑈≠𝑉.
Then if 𝑈is proper, then dim𝐹𝑈< dim𝐹𝑉and dim𝐹(𝑉/𝑈) > 0 because (𝑉/𝑈) ≠∅.
1.12. Direct sums Definition. Let 𝑉be an 𝐹-vector space and 𝑈, 𝑊be subspaces of 𝑉. We say that 𝑉= 𝑈⊕𝑊, read as the direct sum of 𝑈and 𝑊, if ∀𝑣∈𝑉, ∃!𝑢∈𝑈, ∃!𝑤∈𝑊, 𝑢+ 𝑤= 𝑣. We say that 𝑊 is a direct complement of 𝑈in 𝑉; there is no uniqueness of such a complement.
Lemma. Let 𝑉be an 𝐹-vector space, and 𝑈, 𝑊≤𝑉. Then the following statements are equivalent.
357 VII. Linear Algebra (i) 𝑉= 𝑈⊕𝑊 (ii) 𝑉= 𝑈+ 𝑊and 𝑈∩𝑊= {0} (iii) For any basis 𝐵1 of 𝑈and 𝐵2 of 𝑊, 𝐵1 ∪𝐵2 is a basis of 𝑉 Proof. First, we show that (ii) implies (i). If 𝑉= 𝑈+ 𝑊, then certainly ∀𝑣∈𝑉, ∃𝑢∈ 𝑈, ∃𝑤∈𝑊, 𝑣= 𝑢+ 𝑤, so it suffices to show uniqueness. Note, 𝑢1 + 𝑤1 = 𝑢2 + 𝑤2 ⟹ 𝑢1 −𝑢2 = 𝑤2 −𝑤1. The left hand side is an element of 𝑈and the right hand side is an element of 𝑊, so they must be the zero vector; 𝑢1 = 𝑢2, 𝑤1 = 𝑤2.
Now, we show (i) implies (iii). Suppose 𝐵1 is a basis of 𝑈and 𝐵2 is a basis of 𝑊. Let 𝐵= 𝐵1 ∪𝐵2. First, note that 𝐵is a generating family of 𝑈+ 𝑊. Now we must show that 𝐵is free.
∑ 𝑢∈𝐵1 𝜆𝑢𝑢 ⏟ ⎵ ⏟ ⎵ ⏟ ∈𝑈 + ∑ 𝑤∈𝐵2 𝜆𝑤𝑤 ⏟ ⎵ ⎵ ⏟ ⎵ ⎵ ⏟ ∈𝑊 = 0 Hence both sums must be zero. Since 𝐵1, 𝐵2 are bases, all 𝜆are zero, so 𝐵is free and hence a basis.
Now it remains to show that (iii) implies (ii). We must show that 𝑉= 𝑈+𝑊and 𝑈∩𝑊= {0}.
Now, suppose 𝑣∈𝑉. Then, 𝑣= ∑𝑢∈𝐵1 𝜆𝑢𝑢+ ∑𝑤∈𝐵2𝜆𝑤𝑤. In particular, 𝑉= 𝑈+ 𝑊, since the 𝜆𝑢, 𝜆𝑤are arbitrary. Now, let 𝑣∈𝑈∩𝑊. Then 𝑣= ∑ 𝑢∈𝐵1 𝜆𝑢𝑢= ∑ 𝑤∈𝐵2 𝜆𝑤𝑤⟹𝜆𝑢= 𝜆𝑤= 0 Definition. Let 𝑉be an 𝐹-vector space, with subspaces 𝑉 1, … , 𝑉 𝑝≤𝑉. Then 𝑝 ∑ 𝑖=1 𝑉𝑖= {𝑣1, … , 𝑣ℓ, 𝑣𝑖∈𝑉𝑖, 1 ≤𝑖≤ℓ} We say the sum is direct, written 𝑝 ⨁ 𝑖=1 𝑉𝑖 if the decomposition is unique. Equivalently, 𝑉= 𝑝 ⨁ 𝑖=1 𝑉𝑖⟺∃!𝑣1 ∈𝑉 1, … , 𝑣𝑛∈𝑉 𝑛, 𝑣= 𝑛 ∑ 𝑖=1 𝑣𝑖 Lemma. The following are equivalent: (i) ∑ 𝑝 𝑖=1 𝑉𝑖= ⨁ 𝑝 𝑖=1 𝑉𝑖 358 1. Vector spaces and linear dependence (ii) ∀1 ≤𝑖≤𝑙, 𝑉𝑖∩(∑𝑗≠𝑖𝑉𝑗) = {0} (iii) For any basis 𝐵𝑖of 𝑉𝑖, 𝐵= ⋃ 𝑛 𝑖=1 𝐵𝑖is a basis of ∑ 𝑛 𝑖=1 𝑉𝑖.
Proof. Exercise.
359 VII. Linear Algebra 2. Linear maps 2.1. Linear maps Definition. If 𝑉, 𝑊are 𝐹-vector spaces, a map 𝛼∶𝑉→𝑊is linear if ∀𝜆1, 𝜆2 ∈𝐹, ∀𝑣1, 𝑣2 ∈𝑉, 𝛼(𝜆1𝑣1 + 𝜆2𝑣2) = 𝜆1𝛼(𝑣1) + 𝜆2𝛼(𝑣2) Example. Let 𝑀be a matrix with 𝑛rows and 𝑚columns. Then the map 𝛼∶ℝ𝑚→ℝ𝑛 defined by 𝑥↦𝑀𝑥is a linear map.
Example. Let 𝛼∶𝒞([0, 1], ℝ) →𝒞([0, 1], ℝ) defined by 𝑓↦𝑎(𝑓)(𝑥) = ∫ 𝑥 0 𝑓(𝑡) d𝑡. This is linear.
Example. Let 𝑥∈[𝑎, 𝑏]. Then 𝛼∶𝒞([𝑎, 𝑏], ℝ) →ℝdefined by 𝑓↦𝑓(𝑥) is a linear map.
Remark. Let 𝑈, 𝑉, 𝑊be 𝐹-vector spaces. Then, (i) The identity function 𝑖𝑉∶𝑉→𝑉defined by 𝑥↦𝑥is linear.
(ii) If 𝛼∶𝑈→𝑉and 𝛽∶𝑉→𝑊are linear, then 𝛽∘𝛼is linear.
Lemma. Let 𝑉, 𝑊be 𝐹-vector spaces. Let 𝐵be a basis for 𝑉. If 𝛼0 ∶𝐵→𝑉is any map (not necessarily linear), then there exists a unique linear map 𝛼∶𝑉→𝑊extending 𝛼0: ∀𝑣∈𝐵, 𝛼0(𝑣) = 𝛼(𝑣).
Proof. Let 𝑣∈𝑉. Then, given 𝐵= (𝑣1, … , 𝑣𝑛).
𝑣= 𝑛 ∑ 𝑖=1 𝜆𝑖𝑣𝑖 By linearity, 𝛼(𝑣) = 𝛼( 𝑛 ∑ 𝑖=1 𝜆𝑖𝑣𝑖) = 𝑛 ∑ 𝑖=1 𝛼(𝜆𝑖𝑣𝑖) = 𝑛 ∑ 𝑖=1 𝛼0(𝜆𝑖𝑣𝑖) Remark. This lemma is also true in infinite-dimensional vector spaces. Often, to define a linear map, we instead define its action on the basis vectors, and then we ‘extend by linearity’ to construct the entire map.
Remark. If 𝛼1, 𝛼2 ∶𝑉→𝑊are linear maps, then if they agree on any basis of 𝑉then they are equal.
360 2. Linear maps 2.2. Isomorphism Definition. Let 𝑉, 𝑊be 𝐹-vector spaces. A map 𝛼∶𝑉→𝑊is an isomorphism if and only if (i) 𝛼is linear (ii) 𝛼is bijective If such an 𝛼exists, we say that 𝑉and 𝑊are isomorphic, written 𝑉≃𝑊.
Remark. If 𝛼in the above definition is an isomorphism, then 𝛼−1 ∶𝑊→𝑉is linear. Indeed, if 𝑤1, 𝑤2 ∈𝑊with 𝑤1 = 𝛼(𝑣1) and 𝑤2 = 𝛼(𝑣2), 𝛼−1(𝑤1 + 𝑤2) = 𝛼−1(𝛼(𝑣1) + 𝛼(𝑣2)) = 𝛼−1𝛼(𝑣1 + 𝑣2) = 𝑣1 + 𝑣2 = 𝛼−1(𝑤1) + 𝛼−1(𝑤2) Similarly, for 𝜆∈𝐹, 𝑤∈𝑊, 𝛼−1(𝜆𝑤) = 𝜆𝛼−1(𝑤) Lemma. Isomorphism is an equivalence relation on the class of all vector spaces over 𝐹.
Proof.
(i) 𝑖𝑉∶𝑉→𝑉is an isomorphism (ii) If 𝛼∶𝑉→𝑊is an isomorphism, 𝛼−1 ∶𝑊→𝑉is an isomorphism.
(iii) If 𝛽∶𝑈→𝑉, 𝛼∶𝑉→𝑊are isomorphisms, then 𝛼∘𝛽∶𝑈→𝑊is an isomorphism.
The proofs of each part are left as an exercise.
Theorem. If 𝑉is an 𝐹-vector space of dimension 𝑛, then 𝑉≃𝐹𝑛.
Proof. Let 𝐵= (𝑣1, … , 𝑣𝑛) be a basis for 𝑉. Then, consider 𝛼∶𝑉→𝐹𝑛defined by 𝑣= 𝑛 ∑ 𝑖=1 𝜆𝑖𝑣𝑖↦( 𝜆1 ⋮ 𝜆𝑛 ) We claim that this is an isomorphism. This is left as an exercise.
Remark. Choosing a basis for 𝑉is analogous to choosing an isomorphism from 𝑉to 𝐹𝑛.
Theorem. Let 𝑉, 𝑊be 𝐹-vector spaces with finite dimensions 𝑛, 𝑚. Then, 𝑉≃𝑊⟺𝑛= 𝑚 Proof. If dim 𝑉= dim 𝑊= 𝑛, then there exist isomorphisms from both 𝑉and 𝑊to 𝐹𝑛. By transitivity, therefore, there exists an isomorphism between 𝑉and 𝑊.
Conversely, if 𝑉≃𝑊then let 𝛼∶𝑉→𝑊be an isomorphism. Let 𝐵be a basis of 𝑉, then we claim that 𝛼(𝐵) is a basis of 𝑊. Indeed, 𝛼(𝐵) spans 𝑊from the surjectivity of 𝛼, and 𝛼(𝐵) is free due to injectivity.
361 VII. Linear Algebra 2.3. Kernel and image Definition. Let 𝑉, 𝑊be 𝐹-vector spaces. Let 𝛼∶𝑉→𝑊be a linear map. We define the kernel and image as follows.
𝑁(𝛼) = ker 𝛼= {𝑣∈𝑉∶𝛼(𝑣) = 0} Im(𝛼) = {𝑤∈𝑊∶∃𝑣∈𝑉, 𝑤= 𝛼(𝑣)} Lemma. ker 𝛼is a subspace of 𝑉, and Im 𝛼is a subspace of 𝑊.
Proof. Let 𝜆1, 𝜆2 ∈𝐹and 𝑣1, 𝑣2 ∈ker 𝛼. Then 𝛼(𝜆1𝑣1 + 𝜆2𝑣2) = 𝜆1𝛼(𝑣1) + 𝜆2𝛼(𝑣2) = 0 Hence 𝜆1𝑣1 + 𝜆2𝑣2 ∈ker 𝛼.
Now, let 𝜆1, 𝜆2 ∈𝐹, 𝑣1, 𝑣2 ∈𝑉, and 𝑤1 = 𝛼(𝑣1), 𝑤2 = 𝛼(𝑣2). Then 𝜆1𝑤1 + 𝜆2𝑤2 = 𝜆1𝛼(𝑣1) + 𝜆2𝛼(𝑣2) = 𝛼(𝜆1𝑣1 + 𝜆2𝑣2) ∈Im 𝛼 Remark. 𝛼∶𝑉→𝑊is injective if and only if ker 𝛼= {0}. Further, 𝛼∶𝑉→𝑊is surjective if and only if Im 𝛼= 𝑊.
Theorem. Let 𝑉, 𝑊be 𝐹-vector spaces. Let 𝛼∶𝑉→𝑊be a linear map. Then 𝛼∶𝑉/ ker 𝛼→ Im 𝛼defined by 𝛼(𝑣+ ker 𝛼) = 𝛼(𝑣) is an isomorphism. This is the isomorphism theorem from IA Groups.
Proof. First, note that 𝛼is well defined. Suppose 𝑣+ker 𝛼= 𝑣′ +ker 𝛼. Then 𝑣−𝑣′ ∈ker 𝛼, hence 𝛼(𝑣−𝑣′) = 0 ⟹𝛼(𝑣) −𝛼(𝑣′) = 0 so 𝛼is indeed well defined.
Now, we show 𝛼is injective.
𝛼(𝑣+ ker 𝛼) = 0 ⟹𝛼(𝑣) = 0 ⟹𝑣∈ker 𝛼 Hence, 𝑣+ ker 𝛼= 0 + ker 𝛼.
Further, 𝛼is surjective. This follows from the definition the image.
362 2. Linear maps 2.4. Rank and nullity Definition. The rank of 𝛼is 𝑟(𝛼) = dim Im 𝛼 The nullity of 𝛼is 𝑛(𝛼) = dim ker 𝛼 Theorem (Rank-nullity theorem). Let 𝑈, 𝑉be 𝐹-vector spaces such that the dimension of 𝑈is finite. Let 𝛼∶𝑈→𝑉be a linear map. Then, dim 𝑈= 𝑟(𝛼) + 𝑛(𝛼) Proof. We have proven that 𝑈/ ker 𝛼≃Im 𝛼. Hence, the dimensions on the left and right match: dim(𝑈/ ker 𝛼) = dim Im 𝛼.
dim 𝑈−dim ker 𝛼= dim Im 𝛼 and the result follows.
Lemma (Characterisation of isomorphisms). Let 𝑉, 𝑊be 𝐹-vector spaces with equal, finite dimension. Let 𝛼∶𝑉→𝑊be a linear map. Then, the following are equivalent.
(i) 𝛼is injective.
(ii) 𝛼is surjective.
(iii) 𝛼is an isomorphism.
Proof. Clearly, (iii) follows from (i) and (ii) and vice versa. The rest of the proof is left as an exercise, which follows from the rank-nullity theorem.
2.5. Space of linear maps Let 𝑉and 𝑊be 𝐹-vector spaces. Consider the space of linear maps from 𝑉to 𝑊. Then 𝐿(𝑉, 𝑊) = {𝛼∶𝑉→𝑊linear}.
Proposition. 𝐿(𝑉, 𝑊) is an 𝐹-vector space under the operation (𝛼1 + 𝛼2)(𝑣) = 𝛼1(𝑣) + 𝛼2(𝑣); (𝜆𝛼)(𝑣) = 𝜆(𝛼(𝑣)) Further, if 𝑉and 𝑊are finite-dimensional, then so is 𝐿(𝑉, 𝑊) with dim𝐹𝐿(𝑉, 𝑊) = dim𝐹𝑉dim𝐹𝑊 Proof. Proving that 𝐿(𝑉, 𝑊) is a vector space is left as an exercise. The dimensionality part is proven later.
363 VII. Linear Algebra 2.6. Matrices Definition. An 𝑚× 𝑛matrix over 𝐹is an array of 𝑚rows and 𝑛columns, with entries in 𝐹.
We write 𝑀𝑚×𝑛(𝐹) for the set of 𝑚× 𝑛matrices over 𝐹.
Proposition. 𝑀𝑚×𝑛(𝐹) is an 𝐹-vector space under ((𝑎𝑖𝑗) + (𝑏𝑖𝑗)) = (𝑎𝑖𝑗+ 𝑏𝑖𝑗); 𝜆(𝑎𝑖𝑗) = (𝜆𝑎𝑖𝑗) Proposition. dim𝐹𝑀𝑚,𝑛(𝐹) = 𝑚𝑛.
Proof. Consider the basis defined by, the ‘elementary matrix’ for all 𝑖, 𝑗: 𝑒𝑝𝑞= 𝛿𝑖𝑝𝛿𝑗𝑞 Then (𝑒𝑖𝑗) is a basis of 𝑀𝑚×𝑛(𝐹), since it spans 𝑀𝑚×𝑛(𝐹) and we can show that it is free.
2.7. Linear maps as matrices Consider bases 𝐵of 𝑉and 𝐶of 𝑊: 𝐵= (𝑣1, … , 𝑣𝑛); 𝐶= (𝑤1, … , 𝑤𝑛) Then let 𝑣∈𝑉. We have 𝑣= 𝑛 ∑ 𝑗=1 𝜆𝑗𝑣𝑗≡[𝑣]𝐵= ( 𝜆1 ⋮ 𝜆𝑛 ) ∈𝐹𝑛 where the vector given is the coordinates in basis 𝐵. We can equivalently find [𝑤]𝐶, the coordinates of 𝑤in basis 𝐶. We can now define a matrix of some linear map 𝛼in the 𝐵, 𝐶 basis.
Definition.
[𝛼]𝐵,𝐶= ([𝛼(𝑣1)]𝐶, … , [𝛼(𝑣𝑛)]𝐶) ∈𝑀𝑚×𝑛(𝐹) Note that if [𝛼]𝐵𝐶= (𝑎𝑖𝑗), then by definition 𝛼(𝑣𝑗) = 𝑛 ∑ 𝑖=1 𝑎𝑖𝑗𝑤𝑖 Lemma. For all 𝑣∈𝑉, [𝛼(𝑣)]𝐶= [𝛼]𝐵𝐶⋅[𝑣]𝐵 364 2. Linear maps Proof. We have 𝑣= 𝑛 ∑ 𝑖=1 𝜆𝑗𝑣𝑗 Hence 𝛼( 𝑛 ∑ 𝑖=1 𝜆𝑗𝑣𝑗) = 𝑛 ∑ 𝑗=1 𝜆𝑗𝛼(𝑣𝑗) = 𝑛 ∑ 𝑗=1 𝜆𝑖 𝑚 ∑ 𝑖=1 𝑎𝑖𝑗𝑤𝑖= 𝑚 ∑ 𝑖=1 ( 𝑛 ∑ 𝑗=1 𝑎𝑖𝑗𝜆𝑗)𝑤𝑖 Lemma. Let 𝛽∶𝑈→𝑉and 𝛼∶𝑉→𝑊be linear maps. Then, if 𝐴, 𝐵, 𝐶are bases of 𝑈, 𝑉, 𝑊respectively, then [𝛼∘𝛽]𝐴,𝐶= [𝛼]𝐵,𝐶⋅[𝛽]𝐴,𝐵 Proof. Consider 𝑢∈𝐴. Then (𝛼∘𝛽)(𝑢) = 𝛼(𝛽(𝑢)) giving 𝛼(∑ 𝑗 𝑏𝑗𝑝𝑣𝑖) = ∑ 𝑗 𝑏𝑗𝑝𝛼(𝑣𝑗) = ∑ 𝑗 𝑏𝑗𝑝∑ 𝑖 𝑎𝑖𝑗𝑤𝑖= ∑ 𝑖 (∑ 𝑗 𝑎𝑖𝑗𝑏𝑗𝑝)𝑤𝑖 where 𝑎𝑖𝑗𝑝𝑗𝑝is the (𝑖, 𝑗) element of 𝐴𝐵by the definition of the product of matrices.
Proposition. If 𝑉, 𝑊are 𝐹-vector spaces, and dim 𝑉= 𝑛, dim 𝑊= 𝑚, then 𝐿(𝑉, 𝑊) ≃𝑀𝑚×𝑛(𝐹) which implies the dimensionality of 𝐿(𝑉, 𝑊) in 𝐹is 𝑚× 𝑛.
Proof. Consider two bases 𝐵, 𝐶of 𝑉, 𝑊. We claim that 𝜃∶𝐿(𝑉, 𝑊) →𝑀𝑚×𝑛(𝐹) defined by 𝜃(𝛼) = [𝛼]𝐵,𝐶. is an isomorphism. First, note that 𝜃is linear. Then, 𝜃is surjective; consider any matrix 𝐴= (𝑎𝑖𝑗) and consider 𝛼∶𝑣𝑗↦∑ 𝑚 𝑖=1 𝑎𝑖𝑗𝑤𝑖. Then this is certainly a linear map which extends uniquely by linearity to 𝐴, giving [𝛼]𝐵,𝐶= (𝑎𝑖𝑗) = 𝐴. Now, 𝜃is injective since [𝛼]𝐵,𝐶= 0 ⟹𝛼= 0.
Remark. If 𝐵, 𝐶are bases of 𝑉, 𝑊respectively, and 𝜀𝐵∶𝑉→𝐹𝑛is defined by 𝑣↦[𝑣]𝐵, and analogously for 𝜀𝐶, then [𝛼]𝐵,𝐶∘𝜀𝐵= 𝜀𝐶∘𝛼 so the operations commute.
365 VII. Linear Algebra Example. Let 𝛼∶𝑉→𝑊be a linear map and 𝑌≤𝑉, where 𝑉, 𝑊are finite-dimensional.
Then let 𝛼(𝑌) = 𝑍≤𝑊. Consider a basis 𝐵of 𝑉, such that 𝐵′ = (𝑣1, … , 𝑣𝑘) is a basis of 𝑌completed by 𝐵″ = (𝑣𝑘+1, … , 𝑣𝑛) into 𝐵= 𝐵′ ∪𝐵″. Then let 𝐶be a basis of W, such that 𝐶′ = (𝑤1, … , 𝑤ℓ) is a basis of 𝑍completed by 𝐶″ = (𝑤ℓ+1, … , 𝑤𝑚) into 𝐶= 𝐶′ ∪𝐶″. Then [𝛼]𝐵,𝐶= (𝛼(𝑣1) … 𝛼(𝑣𝑘) 𝛼(𝑣𝑘+1) … 𝛼(𝑣𝑛)) For 1 ≤𝑖≤𝑘, 𝛼(𝑣𝑖) ∈𝑍since 𝑣𝑖∈𝑌, 𝛼(𝑌) = 𝑍. So the matrix has an upper-left ℓ× 𝑘 block 𝐴which is 𝛼∶𝑌→𝑍on the basis 𝐵′, 𝐶′. We can show further that 𝛼induces a map 𝛼∶𝑉/𝑌→𝑊/𝑍by 𝑣+ 𝑌↦𝛼(𝑣) + 𝑍. This is well-defined; 𝑣1 + 𝑌= 𝑣2 + 𝑌implies 𝑣1 −𝑣2 ∈𝑌hence 𝛼(𝑣1 −𝑣2) ∈𝑍as required. The bottom-right block is [𝛼]𝐵″,𝐶″.
2.8. Change of basis Suppose we have two bases 𝐵= {𝑣1, … , 𝑣𝑛}, 𝐵′ = {𝑣′ 1, … , 𝑣′ 𝑛} of 𝑉and corresponding 𝐶, 𝐶′ for 𝑊. If we have a linear map [𝛼]𝐵,𝐶, we are interested in finding the components of this linear map in another basis, that is, [𝛼]𝐵,𝐶↦[𝛼]𝐵′,𝐶′ Definition. The change of basis matrix 𝑃from 𝐵′ to 𝐵is 𝑃= ([𝑣′ 1]𝐵 ⋯ [𝑣′ 𝑛]𝐵) which is the identity map in 𝐵′, written 𝑃= [𝐼]𝐵′,𝐵 Lemma. For a vector 𝑣, [𝑣]𝐵= 𝑃[𝑣]𝐵′ Proof. We have [𝛼(𝑣)]𝐶= [𝛼]𝐵,𝐶⋅[𝑣]𝐶 Since 𝑃= [𝐼]𝐵′,𝐵, [𝐼(𝑣)]𝐵= [𝐼]𝐵′,𝐵⋅[𝑣]𝐵′ ⟹[𝑣]𝐵= 𝑃[𝑣]𝐵′ as required.
Remark. 𝑃is an invertible 𝑛× 𝑛square matrix. In particular, 𝑃−1 = [𝐼]𝐵,𝐵′ Indeed, 𝐼𝑛= [𝐼⋅𝐼]𝐵,𝐵= [𝐼]𝐵′,𝐵⋅[𝐼]𝐵′,𝐵 where 𝐼𝑛is the 𝑛× 𝑛identity matrix.
366 2. Linear maps Proposition. If 𝛼is a linear map from 𝑉to 𝑊, and 𝑃= [𝐼]𝐵′,𝐵, 𝑄= [𝐼]𝐶′,𝐶, we have 𝐴′ = [𝛼]𝐵′,𝐶′ = [𝐼]𝐶,𝐶′[𝛼]𝐵,𝐶[𝐼]𝐵,′𝐵= 𝑄−1𝐴𝑃 where 𝐴= [𝛼]𝐵,𝐶, 𝐴′ = [𝛼]𝐵′,𝐶′.
Proof.
[𝛼(𝑣)]𝐶= 𝑄[𝛼(𝑣)]𝐶′ = 𝑄[𝛼]𝐵′,𝐶′[𝑣]𝐵′ [𝛼(𝑣)]𝐶= [𝛼]𝐵,𝐶[𝑣]𝐵 = 𝐴𝑃[𝑣]𝐵′ ∴∀𝑣, 𝑄𝐴[𝑣]𝐵′ = 𝐴𝑃[𝑣]𝐵′ ∴𝑄𝐴= 𝐴𝑃 as required.
2.9. Equivalent matrices Definition. Matrices 𝐴, 𝐴′ are called equivalent if 𝐴′ = 𝑄−1𝐴𝑃 for some invertible 𝑚× 𝑚, 𝑛× 𝑛matrices 𝑄, 𝑃.
Remark. This defines an equivalence relation on 𝑀𝑚,𝑛(𝐹).
• 𝐴= 𝐼−1 𝑚𝐴𝐼𝑛; • 𝐴′ = 𝑄−1𝐴𝑃⟹𝐴= 𝑄𝐴′𝑃−1; • 𝐴′ = 𝑄−1𝐴𝑃, 𝐴″ = (𝑄′)−1𝐴′𝑃′ ⟹𝐴″ = (𝑄𝑄′)−1𝐴(𝑃𝑃′).
Proposition. Let 𝛼∶𝑉→𝑊be a linear map. Then there exists a basis 𝐵of 𝑉and a basis 𝐶of 𝑊such that [𝛼]𝐵,𝐶= (𝐼𝑟 0 0 0) so the components of the matrix are exactly the identity matrix of size 𝑟in the top-left corner, and zeroes everywhere else.
Proof. We first fix 𝑟∈ℕsuch that dim ker 𝛼= 𝑛−𝑟. Then we will construct a basis {𝑣𝑟+1, … , 𝑣𝑛} of the kernel. We extend this to a basis of the entirety of 𝑉, that is, {𝑣1, … , 𝑣𝑛}.
Then, we want to show that {𝛼(𝑣1), … , 𝛼(𝑣𝑟)} 367 VII. Linear Algebra is a basis of Im 𝛼. Indeed, it is a generating family: 𝑣= 𝑛 ∑ 𝑖=1 𝜆𝑖𝑣𝑖 𝛼(𝑣) = 𝑛 ∑ 𝑖=1 𝜆𝑖𝛼(𝑣𝑖) = 𝑟 ∑ 𝑖=1 𝜆𝑖𝛼(𝑣𝑖) Then if 𝑦∈Im 𝛼, there exists 𝑣such that 𝛼(𝑣) = 𝑦. Further, it is a free family: 𝑟 ∑ 𝑖=1 𝜆𝑖𝛼(𝑣𝑖) = 0 𝛼( 𝑟 ∑ 𝑖=1 𝜆𝑖𝑣𝑖) = 0 𝑟 ∑ 𝑖=1 𝜆𝑖𝑣𝑖∈ker 𝛼 𝑟 ∑ 𝑖=1 𝜆𝑖𝑣𝑖= 𝑛 ∑ 𝑖=𝑟+1 𝜆𝑖𝑣𝑖 𝑟 ∑ 𝑖=1 𝜆𝑖𝑣𝑖− 𝑛 ∑ 𝑖=𝑟+1 𝜆𝑖𝑣𝑖= 0 But since {𝑣1, … , 𝑣𝑛} is a basis, 𝜆𝑖= 0 for all 𝑖. Hence {𝛼(𝑣𝑖)} is a basis of Im 𝛼. Now, we wish to extend this basis to the whole of 𝑊to form {𝛼(𝑣1), … , 𝛼(𝑣𝑟), 𝑤𝑟+1, … , 𝑤𝑛} Now, [𝛼]𝐵𝐶= (𝛼(𝑣1) ⋯ 𝛼(𝑣𝑟) 𝛼(𝑣𝑟+1) ⋯ 𝛼(𝑣𝑛)) = (𝐼𝑟 0 0 0) Remark. This also proves the rank-nullity theorem: rank 𝛼+ null 𝛼= 𝑛 368 2. Linear maps Corollary. Any 𝑚× 𝑛matrix 𝐴is equivalent to a matrix of the form (𝐼𝑟 0 0 0) where 𝑟= rank 𝐴.
2.10. Column rank and row rank Definition. Let 𝐴∈𝑀𝑚,𝑛(𝐹). Then, the column rank of 𝐴, here denoted 𝑟 𝑐(𝐴), is the dimension of the subspace of 𝐹𝑛spanned by the column vectors.
𝑟 𝑐(𝐴) = dim span {𝑐1, … , 𝑐𝑛} Remark. If 𝛼is a linear map, represented in bases 𝐵, 𝐶by the matrix 𝐴, then rank 𝛼= 𝑟 𝑐(𝐴) Proposition. Two matrices are equivalent if they have the same column rank: 𝑟 𝑐(𝐴) = 𝑟 𝑐(𝐴′) Proof. If the matrices are equivalent, then 𝐴= [𝛼]𝐵𝐶, 𝐴′ = [𝛼]𝐵′,𝐶′. Then 𝑟 𝑐(𝐴) = 𝑟 𝑐(𝛼) = 𝑟 𝑐(𝐴′) Conversely, if 𝑟 𝑐(𝐴) = 𝑟 𝑐(𝐴′) = 𝑟, then 𝐴, 𝐴′ are equivalent to (𝐼𝑟 0 0 0) By transitivity, 𝐴, 𝐴′ are equivalent.
Theorem. Column rank 𝑟 𝑐(𝐴) and row rank 𝑟 𝑐(𝐴⊺) are equivalent.
Proof. Let 𝑟= 𝑟𝐶(𝐴). Then, 𝑄−1𝐴𝑃= (𝐼𝑟 0 0 0) 𝑚×𝑛 Then, consider 𝑃⊺𝐴⊺(𝑄−1) ⊺= (𝑄−1𝐴𝑃)⊺= (𝐼𝑟 0 0 0) ⊺ 𝑚×𝑛 = (𝐼𝑟 0 0 0) 𝑛×𝑚 369 VII. Linear Algebra Note that we can swap the transpose and inverse on 𝑄because (𝐴𝐵)⊺= 𝐵⊺𝐴⊺ (𝑄𝑄−1) ⊺= 𝑄⊺(𝑄−1)⊺ 𝐼= 𝑄⊺(𝑄−1)⊺ (𝑄⊺) −1 = (𝑄−1)⊺ Then 𝑟 𝑐(𝐴) = rank(𝐴) = rank(𝐴⊺) = 𝑟 𝑐(𝐴⊺).
So we can drop the concepts of column and row rank, and just talk about rank as a whole.
2.11. Conjugation and similarity Consider the following special case of changing basis. If 𝛼∶𝑉→𝑉is linear, 𝛼is called an endomorphism. If 𝐵= 𝐶, 𝐵′ = 𝐶′ then the special case of the change of basis formula is [𝛼]𝐵′,𝐵′ = 𝑃−1[𝛼]𝐵,𝐵𝑃 Then, we say square matrices 𝐴, 𝐴′ are similar or conjugate if there exists 𝑃such that 𝐴′ = 𝑃−1𝐴𝑃.
2.12. Elementary operations Definition. An elementary column operation is (i) swap columns 𝑖, 𝑗 (ii) replace column 𝑖by 𝜆multiplied by the column (iii) add 𝜆multiplied by column 𝑖to column 𝑗 We define analogously the elementary row operations. Note that these elementary oper-ations are invertible (for 𝜆≠0). These operations can be realised through the action of elementary matrices. For instance, the column swap operation can be realised using 𝑇𝑖𝑗= ( 𝐼𝑛 0 0 0 𝐴 0 0 0 𝐼𝑚 ) ; 𝐴= ( 0 0 1 0 𝐼𝑘 0 1 0 1 ) To multiply a column by 𝜆, 𝑛𝑖,𝜆= ( 𝐼𝑛 0 0 0 𝜆 0 0 0 𝐼𝑚 ) To add a multiple of a column, 𝑐𝑖𝑗,𝜆= 𝐼+ 𝜆𝐸𝑖𝑗 370 2. Linear maps where 𝐸𝑖𝑗is the matrix defined by elements (𝑒𝑖𝑗)𝑝𝑞= 𝛿𝑖𝑝𝛿𝑗𝑞. An elementary column (or row) operation can be performed by multiplying 𝐴by the corresponding elementary matrix from the right (on the left for row operations). This will essentially provide a constructive proof that any 𝑛× 𝑛matrix is equivalent to (𝐼𝑟 0 0 0) We will start with a matrix 𝐴. If all entries are zero, we are done. So we will pick 𝑎𝑖𝑗= 𝜆≠0, and swap rows 𝑖, 1 and columns 𝑗, 0. This ensures that 𝑎11 = 𝜆≠0. Now we multiply column 1 by 1 𝜆. Finally, we can clear out row 1 and column 1 by subtracting multiples of the first row or column. Then we can perform similar operations on the (𝑛−1) × (𝑛−1) matrix in the bottom right block and inductively finish this process.
2.13. Gauss’ pivot algorithm If only row operations are used, we can reach the ‘row echelon’ form of the matrix, a specific case of an upper triangular matrix. On each row, there are a number of zeroes until there is a one, called the pivot. First, we assume that 𝑎𝑖𝑗≠0. We swap rows 𝑖, 1. Then divide the first row by 𝜆= 𝑎𝑖1 to get a one in the top left. We can use this one to clear the rest of the first column. Then, we can repeat on the next column, and iterate. This is a technique for solving a linear system of equations.
2.14. Representation of square invertible matrices Lemma. If 𝐴is an 𝑛× 𝑛square invertible matrix, then we can obtain 𝐼𝑛using only row elementary operations, or only column elementary operations.
Proof. We show an algorithm that constructs this 𝐼𝑛. This is exactly going to invert the mat-rix, since the resultant operations can be combined to get the inverse matrix. We will show here the proof for column operations. We argue by induction on the number of rows. Sup-pose we can make the form (𝐼𝑘 0 𝐴 𝐵) We want to obtain the same structure with 𝑘+ 1 rows. We claim that there exists 𝑗> 𝑘such that 𝑎𝑘+1,𝑗≠0. Indeed, otherwise we can show that the vector ⎛ ⎜ ⎜ ⎜ ⎝ 0 ⋮ 1 ⋮ 0 ⎞ ⎟ ⎟ ⎟ ⎠ = 𝛿𝑘+1,𝑖 371 VII. Linear Algebra is not in the span of the column vectors of 𝐴. This contradicts the invertibility of the matrix.
Now, we will swap columns 𝑘+ 1, 𝑗and divide this column by 𝜆. We can now use this 1 to clear the rest of the 𝑘+ 1 row.
Inductively, we have found 𝐴𝐸1 … 𝐸𝑛= 𝐼𝑛where 𝐸𝑛are elementary. Thus, we can find 𝐴−1.
Proposition. Any invertible square matrix is a product of elementary matrices.
The proof is exactly the proof of the lemma above.
372 3. Dual spaces 3. Dual spaces 3.1. Dual spaces Definition. Let 𝑉be an 𝐹-vector space. Then 𝑉⋆is the dual of 𝑉, defined by 𝑉⋆= 𝐿(𝑉, 𝐹) = {𝛼∶𝑉→𝐹} where the 𝛼are linear. If 𝛼∶𝑉→𝐹is linear, then we say 𝛼is a linear form. So the dual of 𝑉is the set of linear forms on 𝑉.
Example. For instance, the trace tr∶𝑀𝑛,𝑛(𝐹) →𝐹is a linear form on 𝑀𝑛,𝑛(𝐹).
Example. Consider functions [0, 1] →ℝ. We can define 𝑇𝑓∶𝒞∞([0, 1], ℝ) →ℝsuch that 𝜙↦∫ 1 0 𝑓(𝑥)𝜙(𝑥) d𝑥. Then 𝑇𝑓is a linear form on 𝒞∞([0, 1], ℝ). We can then reconstruct 𝑓 given 𝑇𝑓. This mathematical formulation is called distribution.
Lemma. Let 𝑉be an 𝐹-vector space with a finite basis 𝐵= {𝑒1, … , 𝑒𝑛}. Then there exists a basis 𝐵⋆for 𝑉⋆given by 𝐵⋆= {𝜀1, … , 𝜀𝑛}; 𝜀𝑗( 𝑛 ∑ 𝑖=1 𝑎𝑖𝑒𝑖) = 𝑎𝑗 We call 𝐵⋆the dual basis for 𝐵.
Proof. We know 𝜀𝑗( 𝑛 ∑ 𝑖=1 𝑎𝑖𝑒𝑖) = 𝑎𝑗 Equivalently, 𝜀𝑗(𝑒𝑖) = 𝛿𝑖𝑗 First, we will show that the set of linear forms as defined is free. For all 𝑖, 𝑛 ∑ 𝑗=1 𝜆𝑗𝜀𝑗= 0 ∴( 𝑛 ∑ 𝑗=1 𝜆𝑗𝜀𝑗)𝑒𝑖= 0 𝑛 ∑ 𝑗=1 𝜆𝑗𝜀𝑗(𝑒𝑖) = 0 𝜆𝑖= 0 373 VII. Linear Algebra Now we show that the set spans 𝑉⋆. Suppose 𝛼∈𝑉⋆, 𝑥∈𝑉.
𝛼(𝑥) = 𝛼( 𝑛 ∑ 𝑗=1 𝜆𝑗𝑒𝑗) = 𝑛 ∑ 𝑖=1 𝜆𝑗𝛼(𝑒𝑗) Conversely, we can write 𝑛 ∑ 𝑖=1 𝛼(𝑒𝑗)𝜀(𝑗) ∈𝑉⋆ Thus, ( 𝑛 ∑ 𝑖=1 𝛼(𝑒𝑗)𝜀𝑗)(𝑥) = 𝑛 ∑ 𝑗=1 𝛼(𝑒𝑗)𝜀𝑗( 𝑛 ∑ 𝑘=1 𝜆𝑘𝑒𝑘) = 𝑛 ∑ 𝑗=1 𝛼(𝑒𝑗) 𝑛 ∑ 𝑘=1 𝜆𝑘𝜀𝑗(𝑒𝑘) = 𝑛 ∑ 𝑗=1 𝛼(𝑒𝑗) 𝑛 ∑ 𝑘=1 𝜆𝑘𝛿𝑗𝑘 = 𝑛 ∑ 𝑗=1 𝛼(𝑒𝑗)𝜆𝑗 = 𝛼(𝑥) We have then shown that 𝛼= 𝑛 ∑ 𝑗=1 𝛼(𝑒𝑗)𝜀𝑗 as required.
Corollary. If 𝑉is finite-dimensional, 𝑉⋆has the same dimension.
Remark. It is sometimes convenient to think of 𝑉⋆as the spaces of row vectors of length dim 𝑉over 𝐹. For instance, consider the basis 𝐵= (𝑒1, … , 𝑒𝑛), so 𝑥= ∑ 𝑛 𝑖=1 𝑥𝑖𝑒𝑖. Then we can pick (𝜀1, … , 𝜀𝑛) a basis of 𝑉⋆, so 𝛼= ∑ 𝑛 𝑖=1 𝛼𝑖𝜀𝑖. Then 𝛼(𝑥) = 𝑛 ∑ 𝑖=1 𝛼𝑖𝜀𝑖(𝑥) = 𝑛 ∑ 𝑖=1 𝛼𝑖𝜀( 𝑛 ∑ 𝑗=1 𝑥𝑗𝑒𝑗) = 𝑛 ∑ 𝑖=1 𝛼𝑖𝑥𝑖 This is exactly (𝛼1 ⋯ 𝛼𝑛) ( 𝑥1 ⋮ 𝑥𝑛 ) which essentially defines a scalar product between the two spaces.
374 3. Dual spaces 3.2. Annihilators Definition. Let 𝑈⊆𝑉. Then the annihilator of 𝑈is 𝑈0 = {𝛼∈𝑉⋆∶∀𝑢∈𝑈, 𝛼(𝑢) = 0} Lemma.
(i) 𝑈0 ≤𝑉⋆; (ii) If 𝑈≤𝑉and dim 𝑉< ∞, then dim 𝑉= dim 𝑈+ dim 𝑈0.
Proof.
(i) First, note that 0 ∈𝑈0 since 𝛼(0) = 0 by linearity. If 𝛼, 𝛼′ ∈𝑈0, then for all 𝑢∈𝑈, (𝛼+ 𝛼′)(𝑢) = 𝛼(𝑢) + 𝛼′(𝑢) = 0 Further, for all 𝜆∈𝐹, (𝜆𝛼)(𝑢) = 𝜆𝛼(𝑢) = 0 Hence 𝑈0 ≤𝑉⋆.
(ii) Let (𝑒1, … , 𝑒𝑘) be a basis of 𝑈, completed into a basis 𝐵= (𝑒1, … , 𝑒𝑘, 𝑒𝑘+1, … , 𝑒𝑛) of 𝑉.
Let (𝜀1, … , 𝜀𝑛) be the dual basis 𝐵⋆. We then will prove that 𝑈0 = ⟨𝜀𝑘+1, … , 𝜀𝑛⟩ If 𝑖> 𝑘, then 𝜀𝑖(𝑒𝑘) = 𝛿𝑖𝑘= 0. Hence 𝜀𝑖∈𝑈0. Thus ⟨𝜀𝑘+1, … , 𝜀𝑛⟩⊂𝑈0. Conversely, let 𝛼∈𝑈0. Then 𝛼= ∑ 𝑛 𝑖=1 𝛼𝑖𝜀𝑖. For 𝑖≤𝑘, 𝛼∈𝑈0 hence 𝛼(𝑒𝑖) = 0. Hence, 𝛼= 𝑛 ∑ 𝑖=𝑘+1 𝛼𝑖𝜀𝑖 Thus 𝛼∈⟨𝜀𝑘+1, … , 𝜀𝑛⟩ as required.
3.3. Dual maps Lemma. Let 𝑉, 𝑊be 𝐹-vector spaces. Let 𝛼∈𝐿(𝑉, 𝑊). Then there exists a unique 𝛼⋆∈ 𝐿(𝑊⋆, 𝑉⋆) such that 𝜀↦𝜀∘𝛼 called the dual map.
375 VII. Linear Algebra Proof. First, note 𝜀(𝛼)∶𝑉→𝐹is a linear map. Hence, 𝜀∘𝛼∈𝑉⋆. Now we must show 𝛼⋆ is linear.
𝛼⋆(𝜃1 + 𝜃2) = (𝜃1 + 𝜃2)(𝛼) = 𝜃1 ∘𝛼+ 𝜃2 ∘𝛼= 𝛼⋆(𝜃1) + 𝛼⋆(𝜃2) Similarly, we can show 𝛼⋆(𝜆𝜃) = 𝜆𝛼⋆(𝜃) as required. Hence 𝛼⋆∈𝐿(𝑊⋆, 𝑉⋆).
Proposition. Let 𝑉, 𝑊be finite-dimensional 𝐹-vector spaces with bases 𝐵, 𝐶respectively.
Then [𝛼⋆]𝐶⋆,𝐵⋆= [𝛼]⊺ 𝐵,𝐶 Thus, we can think of the dual map as the adjoint of 𝛼.
Proof. This follows from the definition of the dual map. Let 𝐵= (𝑏1, … , 𝑏𝑛), 𝐶= (𝑐1, … , 𝑐𝑚), 𝐵⋆= (𝛽1, … , 𝛽𝑛), 𝐶⋆= (𝛾1, … , 𝛾𝑚). Let [𝛼]𝐵,𝐶= (𝑎𝑖𝑗). Then, we compute 𝛼⋆(𝛾𝑟)(𝑏𝑠) = 𝛾𝑟∘𝛼(𝑏𝑠) = 𝛾𝑟(∑ 𝑡 𝑎𝑡𝑠𝑐𝑡) = ∑ 𝑡 𝑎𝑡𝑠𝛾𝑟(𝑐𝑡) = ∑ 𝑡 𝑎𝑡𝑠𝛿𝑡𝑟 = 𝑎𝑟𝑠 We can conversely write [𝛼⋆]𝐶⋆,𝐵⋆= (𝑚𝑖𝑗) and 𝛼⋆(𝛾𝑟) = 𝑛 ∑ 𝑖=1 𝑚𝑖𝑟𝛽𝑖 𝛼⋆(𝛾𝑟)(𝑏𝑠) = 𝑛 ∑ 𝑖=1 𝑚𝑖𝑟𝛽𝑖(𝑏𝑠) = 𝑛 ∑ 𝑖=1 𝑚𝑖𝑟𝛿𝑖𝑠 = 𝑚𝑠𝑟 Thus, 𝑎𝑟𝑠= 𝑚𝑠𝑟 as required.
376 3. Dual spaces 3.4. Properties of dual map Let 𝛼∈𝐿(𝑉, 𝑊), and 𝛼⋆∈𝐿(𝑊⋆, 𝑉⋆). Let 𝐵and 𝐶be bases of 𝑉, 𝑊respectively, and 𝐵⋆, 𝐶⋆ be their duals. We have proven that [𝛼]𝐵,𝐶= [𝛼⋆]⊺ 𝐵,𝐶 Lemma. Suppose that 𝐸= (𝑒1, … , 𝑒𝑛) and 𝐹= (𝑓 1, … , 𝑓 𝑛) are bases of 𝑉. Let 𝑃= [𝐼]𝐹,𝐸be a change of basis matrix from 𝐹to 𝐸. The bases 𝐸⋆= (𝜀1, … , 𝜀𝑛), 𝐹⋆= (𝜂1, … , 𝜂𝑛) are the corresponding dual bases. Then, the change of basis matrix from 𝐹⋆to 𝐸⋆is (𝑃−1) ⊺ Proof. Consider [𝐼]𝐹⋆,𝐸⋆= [𝐼]⊺ 𝐸,𝐹= ([𝐼]−1 𝐹,𝐸) ⊺= (𝑃−1) ⊺ Lemma. Let 𝑉, 𝑊be 𝐹-vector spaces. Let 𝛼∈𝐿(𝑉, 𝑊). Let 𝛼⋆be the corresponding dual map. Then, denoting 𝑁(𝛼) for the kernel of 𝛼, (i) 𝑁(𝛼⋆) = (Im 𝛼)0, so 𝛼⋆is injective if and only if 𝛼is surjective.
(ii) Im 𝛼⋆≤(𝑁(𝛼))0, with equality if 𝑉, 𝑊are finite-dimensional. In this finite-dimensional case, 𝛼⋆is surjective if and only if 𝛼is injective.
Remark. In many applications, it is often simpler to understand the dual map 𝛼⋆than it is to understand 𝛼.
Proof. First, we prove (i). Let 𝜀∈𝑊⋆. Then, 𝜀∈𝑁(𝛼⋆) means 𝛼⋆(𝜀) = 0. Hence, 𝛼⋆(𝜀) = 𝜀∘𝛼= 0 So for any 𝑣∈𝑉, 𝜀(𝛼(𝑣)) = 0. Equivalently, 𝜀is an element of the annihilator of Im 𝛼.
Now, we will show (ii). Let 𝜀∈Im 𝛼⋆. Then 𝛼⋆(𝜙) = 𝜀for some 𝜙∈𝑊⋆. Then, for all 𝑢∈𝑁(𝛼), 𝜀(𝑢) = (𝛼⋆(𝜙))(𝑢) = 𝜙∘𝛼(𝑢) = 𝜙(𝛼(𝑢)) = 0. Certainly then 𝜀∈(𝑁(𝛼)) 0. Then, Im 𝛼⋆≤(𝑁(𝛼))0.
In the finite-dimensional case, we can compare the dimension of these two spaces.
dim Im 𝛼⋆= 𝑟(𝛼⋆) = 𝑟([𝛼⋆]𝐶⋆,𝐵⋆) = 𝑟([𝛼]⊺ 𝐵,𝐶) = 𝑟([𝛼]𝐵,𝐶) = 𝑟(𝛼) = dim Im 𝛼 Due to the rank-nullity theorem, dim Im 𝛼⋆= dim 𝑉−dim 𝑁(𝛼) = dim [(𝑁(𝛼))0]. Hence, Im 𝛼⋆≤(𝑁(𝛼))0; dim Im 𝛼⋆= dim(𝑁(𝛼))0 The dimensions are equal, and one is a subspace of the other, hence the spaces are equal.
377 VII. Linear Algebra 3.5. Double duals Definition. Let 𝑉be an 𝐹-vector space. Let 𝑉⋆be the dual of 𝑉. The double dual or bidual of 𝑉is 𝑉⋆⋆= 𝐿(𝑉⋆, 𝐹) = (𝑉⋆)⋆ Remark. In general, there is no obvious relation between 𝑉and 𝑉⋆. However, the following useful facts hold about 𝑉and 𝑉⋆⋆.
(i) There is a canonical embedding from 𝑉to 𝑉⋆⋆. In particular, there exists 𝑖in 𝐿(𝑉, 𝑉⋆⋆) which is injective.
(ii) There are examples of infinite-dimensional spaces where 𝑉≃𝑉⋆⋆. These are called reflexive spaces. Such spaces are investigated in the study of Banach spaces.
Theorem. 𝑉embeds into 𝑉⋆⋆.
Proof. Choose a vector 𝑣∈𝑉and define the linear form ̂ 𝑣∈𝐿(𝑉⋆, 𝐹) such that ̂ 𝑣(𝜀) = 𝜀(𝑣) So clearly ̂ 𝑣is linear. We want to show ̂ 𝑣∈𝑉⋆⋆. If 𝜀∈𝑉⋆, 𝜀(𝑣) ∈𝐹. Further, 𝜆1, 𝜆2 ∈𝐹and 𝜀1, 𝜀2 ∈𝑉⋆give ̂ 𝑣(𝜆1𝜀1 + 𝜆2𝜀2) = (𝜆1𝜀1 + 𝜆2𝜀2)(𝑣) = 𝜆1𝜀1(𝑣) + 𝜆2𝜀2(𝑣) = 𝜆1 ̂ 𝑣(𝜀1) + 𝜆2 ̂ 𝑣(𝜀2) Theorem. If 𝑉is finite-dimensional, then 𝑖∶𝑉→𝑉⋆⋆given by 𝑖(𝑣) = ̂ 𝑣is an isomorphism.
Proof. We will show 𝑖is linear. If 𝑣1, 𝑣2 ∈𝑉, 𝜆1, 𝜆2 ∈𝐹, then 𝑖(𝜆1𝑣1 + 𝜆2𝑣2)(𝜀) = 𝜀(𝜆1𝑣1 + 𝜆2𝑣2) = 𝜆1𝜀(𝑣1) + 𝜆2𝜀(𝑣2) = 𝜆1 ̂ 𝑣1(𝜀) + 𝜆2 ̂ 𝑣2(𝜀) Now, we will show that 𝑖is injective for finite-dimensional 𝑉. Let 𝑒∈𝑉∖{0}. We will show that 𝑒∉ker 𝑖. We extend 𝑒into a basis (𝑒, 𝑒2, … , 𝑒𝑛) of 𝑉. Now, let (𝜀, 𝜀2, … , 𝜀𝑛) be the dual basis. Then ̂ 𝑒(𝜀) = 𝜀(𝑒) = 1. In particular, ̂ 𝑒≠0. Hence ker 𝑖= {0}, so it is injective.
We now show that 𝑖is an isomorphism. We need to simply compute the dimension of the image under 𝑖. Certainly, dim 𝑉= dim 𝑉⋆= dim(𝑉⋆)⋆= dim 𝑉⋆⋆. Since 𝑖is injective, dim 𝑉= dim 𝑉⋆⋆. So 𝑖is surjective as required.
Lemma. Let 𝑉be a finite-dimensional 𝐹-vector space. Let 𝑈≤𝑉. Then, ̂ 𝑈= 𝑈00 After identifying 𝑉and 𝑉⋆⋆, we typically say 𝑈= 𝑈00 although this is is incorrect notation and not an equality.
378 3. Dual spaces Proof. We will show that ̂ 𝑈≤𝑈00. Indeed, let 𝑢∈𝑈, then by definition ∀𝜀∈𝑈0, 𝜀(𝑢) = 0 ⟹ ̂ 𝑢(𝜀) = 0 Hence ̂ 𝑢∈𝑈00 and so ̂ 𝑈≤𝑈00.
Now, we will compute dimension: dim 𝑈00 = dim 𝑉−dim 𝑈0 = dim 𝑈. Since ̂ 𝑈≃𝑈, their dimensions are the same, so 𝑈00 = ̂ 𝑈.
Remark. Due to this identification of 𝑉⋆⋆and 𝑉, we can define 𝑇≤𝑉⋆, 𝑇0 = {𝑣∈𝑉∶∀𝜃∈𝑇, 𝜃(𝑣) = 0} Lemma. Let 𝑉be a finite-dimensional 𝐹-vector space. Let 𝑈1, 𝑈2 be subspaces of 𝑉. Then (i) (𝑈1 + 𝑈2)0 = 𝑈0 1 ∩𝑈0 2 ; (ii) (𝑈1 ∩𝑈2)0 = 𝑈0 1 + 𝑈0 2 Proof. Let 𝜃∈𝑉⋆. Then 𝜃∈(𝑈1 + 𝑈2)0 ⟺∀𝑢1 ∈𝑈1, 𝑢2 ∈𝑈2, 𝜃(𝑢1 + 𝑢2) = 0. Hence 𝜃(𝑢) = 0 for all 𝑢∈𝑈1 ∪𝑈2 by linearity. Hence 𝜃∈𝑈0 1 ∩𝑈0 2 . Now, take the annihilator of (i) and 𝑈00 = 𝑈to complete part (ii).
379 VII. Linear Algebra 4. Bilinear forms 4.1. Introduction Definition. Let 𝑈, 𝑉be 𝐹-vector spaces. Then 𝜙∶𝑈×𝑉→𝐹is a bilinear form if it is linear in both components. For example, 𝜙at a fixed 𝑢∈𝑈is a linear form 𝑉→𝐹and an element of 𝑉⋆.
Example. Consider the map 𝑉× 𝑉⋆→𝐹given by (𝑣, 𝜃) ↦𝜃(𝑣) Example. The scalar product on 𝑈= 𝑉= ℝ𝑛is given by 𝜓(𝑥, 𝑦) = 𝑛 ∑ 𝑖=1 𝑥𝑖𝑦𝑖 Example. Let 𝑈= 𝑉= 𝐶([0, 1], ℝ) and consider 𝜙(𝑓, 𝑔) = ∫ 1 0 𝑓(𝑡)𝑔(𝑡) d𝑡 Definition. If 𝐵= (𝑒1, … , 𝑒𝑚) is a basis of 𝑈and 𝐶= (𝑓 1, … , 𝑓 𝑛) is a basis of 𝑉, and 𝜙∶𝑈× 𝑉→𝐹is a bilinear form, then the matrix of the bilinear form in this basis is [𝜙]𝐵,𝐶= (𝜙(𝑒𝑖, 𝑓𝑗))1≤𝑖≤𝑚,1≤𝑗≤𝑛 Lemma. We can link 𝜙with its matrix in a given basis as follows.
𝜙(𝑢, 𝑣) = [𝑢]⊺ 𝐵[𝜙]𝐵,𝐶[𝑣]𝐶 Proof. Let 𝑢= ∑ 𝑚 𝑖=1 𝜆𝑖𝑢𝑖and 𝑣= ∑ 𝑛 𝑗=1 𝜇𝑗𝑣𝑗. Then 𝜙(𝑢, 𝑣) = 𝜙( 𝑚 ∑ 𝑖=1 𝜆𝑖𝑢𝑖, 𝑛 ∑ 𝑗=1 𝜇𝑗𝑣𝑗) = 𝑚 ∑ 𝑖=1 𝑛 ∑ 𝑗=1 𝜆𝑖𝜇𝑗𝜙(𝑢𝑖, 𝑣𝑗) = [𝑢]⊺ 𝐵[𝜙]𝐵,𝐶[𝑣]𝐶 Remark. Note that [𝜙]𝐵,𝐶is the only matrix such that 𝜙(𝑢, 𝑣) = [𝑢]⊺ 𝐵[𝜙]𝐵,𝐶[𝑣]𝐶.
Definition. Let 𝜙∶𝑈× 𝑉→𝐹be a bilinear form. Then 𝜙induces two linear maps given by the partial application of a single parameter to the function.
𝜙𝐿∶𝑈→𝑉⋆; 𝜙𝐿(𝑢)∶𝑉→𝐹; 𝑣↦𝜙(𝑢, 𝑣) 𝜙𝑅∶𝑉→𝑈⋆; 𝜙𝑅(𝑣)∶𝑈→𝐹; 𝑢↦𝜙(𝑢, 𝑣) In particular, 𝜙𝐿(𝑢)(𝑣) = 𝜙(𝑢, 𝑣) = 𝜙𝑅(𝑣)(𝑢) 380 4. Bilinear forms Lemma. Let 𝐵= (𝑒1, … , 𝑒𝑚) be a basis of 𝑈, and let 𝐵⋆= (𝜀1, … , 𝜀𝑚) be its dual; and let 𝐶= (𝑓 1, … , 𝑓 𝑛) be a basis of 𝑉, and let 𝐶⋆= (𝜂1, … , 𝜂𝑛) be its dual. Let 𝐴= [𝜙]𝐵,𝐶. Then [𝜙𝑅]𝐶,𝐵⋆= 𝐴; [𝜙𝐿]𝐵,𝐶⋆= 𝐴⊺ Proof.
𝜙𝐿(𝑒𝑖)(𝑓𝑗) = 𝜙(𝑒𝑖, 𝑓𝑗) = 𝐴𝑖𝑗 Since 𝜂𝑗is the dual of 𝑓𝑗, 𝜙𝐿(𝑒𝑖) = ∑ 𝑖 𝐴𝑖𝑗𝜂𝑗 Further, 𝜙𝑅(𝑓𝑗)(𝑒𝑖) = 𝜙(𝑒𝑖, 𝑓𝑗) = 𝐴𝑖𝑗 and then similarly 𝜙𝑅(𝑓𝑗) = ∑ 𝑖 𝐴𝑖𝑗𝜀𝑖 Definition. ker 𝜙𝐿is called the left kernel of 𝜙. ker 𝜙𝑅is the right kernel of 𝜙.
Definition. We say that 𝜙is non-degenerate if ker 𝜙𝐿= ker 𝜙𝑅= {0}. Otherwise, 𝜙is degenerate.
Theorem. Let 𝐵be a basis of 𝑈, and let 𝐶be a basis of 𝑉, where 𝑈, 𝑉are finite-dimensional.
Let 𝜙∶𝑈×𝑉→𝐹be a bilinear form. Let 𝐴= [𝜙]𝐵,𝐶. Then, 𝜙is non-degenerate if and only if 𝐴is invertible.
Corollary. If 𝜙is non-degenerate, then dim 𝑈= dim 𝑉.
Proof. Suppose 𝜙is non-degenerate. Then ker 𝜙𝐿= ker 𝜙𝑅= {0}. This is equivalent to saying that 𝑛(𝜙𝐿) = 𝑛(𝜙𝑅) = 0. We can use the rank-nullity theorem to state that 𝑟(𝐴⊺) = dim 𝑉and 𝑟(𝐴) = dim 𝑉. This is equivalent to saying that 𝐴is invertible. Note that this forces dim 𝑈= dim 𝑉.
Remark. The canonical example of a non-degenerate bilinear form is the scalar product ℝ𝑛× ℝ𝑛→ℝrepresented by the identity matrix in the standard basis.
Corollary. If 𝑈and 𝑉are finite-dimensional with dim 𝑈= dim 𝑉, then choosing a non-degenerate bilinear form 𝜙∶𝑈×𝑉→𝐹is equivalent to choosing an isomorphism 𝜙𝐿∶𝑈≃ 𝑉⋆.
Definition. If 𝑇⊂𝑈, then we define 𝑇⟂= {𝑣∈𝑉∶∀𝑡∈𝑇, 𝜙(𝑡, 𝑣) = 0} Further, if 𝑆⊂𝑉, we define ⟂𝑆= {𝑢∈𝑈∶∀𝑠∈𝑆, 𝜙(𝑢, 𝑠) = 0} These are called the orthogonals of 𝑇and 𝑆.
381 VII. Linear Algebra 4.2. Change of basis for bilinear forms Proposition. Let 𝐵, 𝐵′ be bases of 𝑈and 𝑃= [𝐼]𝐵′,𝐵, let 𝐶, 𝐶′ be bases of 𝑉and 𝑄= [𝐼]𝐶′,𝐶, and finally let 𝜙∶𝑈× 𝑉→𝐹be a bilinear form. Then [𝜙]𝐵′,𝐶′ = 𝑃⊺[𝜙]𝐵,𝐶𝑄 Proof. We have 𝜙(𝑢, 𝑣) = [𝑢]⊺ 𝐵[𝜙]𝐵,𝐶[𝑣]𝐶. Changing coordinates, we have 𝜙(𝑢, 𝑣) = (𝑃[𝑢]𝐵′)⊺[𝜙]𝐵,𝐶(𝑄[𝑣]𝐶′) = [𝑢]⊺ 𝐵′(𝑃⊺[𝜙]𝐵,𝐶𝑄)[𝑣]𝐶′ Lemma. The rank of a bilinear form 𝜙, denoted 𝑟(𝜙) is the rank of any matrix representing 𝜙. This quantity is well-defined.
Remark. 𝑟(𝜙) = 𝑟(𝜙𝑅) = 𝑟(𝜙𝐿), since 𝑟(𝐴) = 𝑟(𝐴⊺).
Proof. For any invertible matrices 𝑃, 𝑄, 𝑟(𝑃⊺𝐴𝑄) = 𝑟(𝐴).
382 5. Trace and determinant 5. Trace and determinant 5.1. Trace Definition. The trace of a square matrix 𝐴∈𝑀𝑛,𝑛(𝐹) ≡𝑀𝑛(𝐹) is defined by tr 𝐴= 𝑛 ∑ 𝑖=1 𝑎𝑖𝑖 The trace is a linear form.
Lemma. tr(𝐴𝐵) = tr(𝐵𝐴) for any matrices 𝐴, 𝐵∈𝑀𝑛(𝐹).
Proof. We have tr(𝐴𝐵) = 𝑛 ∑ 𝑖=1 𝑛 ∑ 𝑗=1 𝑎𝑖𝑗𝑏𝑗𝑖= 𝑛 ∑ 𝑗=1 𝑛 ∑ 𝑖=1 𝑏𝑗𝑖𝑎𝑖𝑗= tr(𝐵𝐴) Corollary. Similar matrices have the same trace.
Proof.
tr(𝑃−1𝐴𝑃) = tr(𝐴𝑃−1𝑃) = tr 𝐴 Definition. If 𝛼∶𝑉→𝑉is linear, we can define the trace of 𝛼as tr 𝛼= tr[𝛼]𝐵 for any basis 𝐵. This is well-defined by the corollary above.
Lemma. If 𝛼∶𝑉→𝑉is linear, 𝛼⋆∶𝑉⋆→𝑉⋆satisfies tr 𝛼= tr 𝛼⋆ Proof.
tr 𝛼= tr[𝛼]𝐵= tr[𝛼] ⊺ 𝐵= tr[𝛼⋆]𝐵⋆= tr 𝛼⋆ 5.2. Permutations and transpositions Recall the following facts about permutations and transpositions. 𝑆𝑛is the group of permuta-tions of the set {1, … , 𝑛}; the group of bijections 𝜎∶{1, … , 𝑛} →{1, … , 𝑛}. A transposition 𝜏𝑘ℓ= (𝑘, ℓ) is defined by 𝑘↦ℓ, ℓ↦𝑘, 𝑥↦𝑥for 𝑥≠𝑘, ℓ. Any permutation 𝜎can be decomposed as a product of transpositions. This decomposition is not necessarily unique, but the parity of the number of transpositions is well-defined. We say that the signature of a permutation, denoted 𝜀∶𝑆𝑛→{−1, 1}, is 1 if the decomposition has even parity and −1 if it has odd parity. We can then show that 𝜀is a homomorphism.
383 VII. Linear Algebra 5.3. Determinant Definition. Let 𝐴∈𝑀𝑛(𝐹). We define det 𝐴= ∑ 𝜎∈𝑆𝑛 𝜀(𝜎)𝐴𝜎(1)1 … 𝐴𝜎(𝑛)𝑛 Example. Let 𝑛= 2. Then, 𝐴= (𝑎11 𝑎12 𝑎21 𝑎22 ) ⟹det 𝐴= 𝑎11𝑎22 −𝑎12𝑎21 Lemma. If 𝐴= (𝑎𝑖𝑗) is an upper (or lower) triangular matrix (with zeroes on the diagonal), then det 𝐴= 0.
Proof. Let (𝑎𝑖𝑗) = 0 for 𝑖> 𝑗. Then det 𝐴= ∑ 𝜎∈𝑆𝑛 𝜀(𝜎)𝑎𝜎(1)1 … 𝑎𝜎(𝑛)𝑛 For the summand to be nonzero, 𝜎(𝑗) ≤𝑗for all 𝑗. Thus, det 𝐴= 𝑎11 … 𝑎𝑛𝑛= 0 Lemma. Let 𝐴∈𝑀𝑛(𝐹). Then, det 𝐴= det 𝐴⊺.
Proof.
det 𝐴= ∑ 𝜎∈𝑆𝑛 𝜀(𝜎)𝑎𝜎(1)1 … 𝑎𝜎(𝑛)𝑛 = ∑ 𝜎−1∈𝑆𝑛 𝜀(𝜎)𝑎𝜎(1)1 … 𝑎𝜎(𝑛)𝑛 = ∑ 𝜎∈𝑆𝑛 𝜀(𝜎−1)𝑎1𝜎(1) … 𝑎𝑛𝜎(𝑛) = ∑ 𝜎∈𝑆𝑛 𝜀(𝜎)𝑎1𝜎(1) … 𝑎𝑛𝜎(𝑛) = det 𝐴⊺ 384 5. Trace and determinant 5.4. Volume forms Definition. A volume form 𝑑on 𝐹𝑛is a function 𝑑∶𝐹𝑛× ⋯× 𝐹𝑛 ⏟ ⎵ ⎵ ⎵ ⏟ ⎵ ⎵ ⎵ ⏟ 𝑛times →𝐹satisfying (i) 𝑑is multilinear: for all 𝑖∈{1, … , 𝑛} and for all 𝑣1, … , 𝑣𝑖−1, 𝑣𝑖+1, … , 𝑣𝑛∈𝐹𝑛, the map from 𝐹𝑛to 𝐹defined by 𝑣↦(𝑣1, … , 𝑣𝑖−1, 𝑣, 𝑣𝑖+1, … , 𝑣𝑛) is linear. In other words, this map is an element of (𝐹𝑛)⋆.
(ii) 𝑑is alternating: for 𝑣𝑖= 𝑣𝑗for some 𝑖≠𝑗, 𝑑= 0.
So an alternating multilinear form is a volume form. We want to show that, up to multiplic-ation by a scalar, the determinant is the only volume form.
Lemma. The map (𝐹𝑛)𝑛→𝐹defined by (𝐴(1), … , 𝐴(𝑛)) ↦det 𝐴is a volume form. This map is the determinant of 𝐴, but thought of as acting on the column vectors of 𝐴.
Proof. We first show that this map is multilinear. Fix 𝜎∈𝑆𝑛, and consider ∏ 𝑛 𝑖=1 𝑎𝜎(𝑖)𝑖. This product contains exactly one term in each column of 𝐴. Thus, the map (𝐴(1), … , 𝐴(𝑛)) ↦ ∏ 𝑛 𝑖=1 𝑎𝜎(𝑖)𝑖is multilinear. This then clearly implies that the determinant, a sum of such multilinear maps, is itself multilinear.
Now, we show that the determinant is alternating. Let 𝑘≠ℓ, and 𝐴(𝑘) = 𝐴(ℓ). Let 𝜏= (𝑘ℓ) be the transposition exchanging 𝑘and ℓ. Then, for all 𝑖, 𝑗∈{1, … , 𝑛}, 𝑎𝑖𝑗= 𝑎𝑖𝜏(𝑗). We can decompose permutations into two disjoint sets: 𝑆𝑛= 𝐴𝑛∪𝜏𝐴𝑛, where 𝐴𝑛is the alternating group of order 𝑛. Now, note that ∏ 𝑛 𝑖=1 𝑎𝜎(𝑖)𝑖+∏ 𝑛 𝑖=1 𝑎(𝜏∘𝜎)(𝑖)𝑖= 0. So the sum over all 𝜎∈𝐴𝑛 gives zero. So the determinant is alternating, and hence a volume form.
Lemma. Let 𝑑be a volume form. Then, swapping two entries changes the sign.
Proof. Take the sum of these two results: 𝑑(𝑣1, … , 𝑣𝑖, … , 𝑣𝑗, … , 𝑣𝑛) + 𝑑(𝑣1, … , 𝑣𝑗, … , 𝑣𝑖, … , 𝑣𝑛) = 𝑑(𝑣1, … , 𝑣𝑖, … , 𝑣𝑗, … , 𝑣𝑛) + 𝑑(𝑣1, … , 𝑣𝑗, … , 𝑣𝑖, … , 𝑣𝑛) + 𝑑(𝑣1, … , 𝑣𝑖, … , 𝑣𝑖, … , 𝑣𝑛) + 𝑑(𝑣1, … , 𝑣𝑗, … , 𝑣𝑗, 𝑣𝑛) = 2𝑑(𝑣1, … , 𝑣𝑖+ 𝑣𝑗, … , 𝑣𝑖+ 𝑣𝑗, … , 𝑣𝑛) = 0 as required.
Corollary. If 𝜎∈𝑆𝑛and 𝑑is a volume form, 𝑑(𝑣𝜎(1), … , 𝑣𝜎(𝑛)) = 𝜀(𝜎)𝑑(𝑣1, … , 𝑣𝑛).
385 VII. Linear Algebra Proof. We can decompose 𝜎as a product of transpositions ∏ 𝑛𝜎 𝑖=1 𝑒𝑖.
Theorem. Let 𝑑be a volume form on 𝐹𝑛. Let 𝐴be a matrix whose columns are 𝐴(𝑖). Then 𝑑(𝐴(1), … , 𝐴(𝑛)) = det 𝐴⋅𝑑(𝑒1, … , 𝑒𝑛) So there is a unique volume form up to a constant multiple. We can then see that det 𝐴is the only volume form such that 𝑑(𝑒1, … , 𝑒𝑛) = 1.
Proof.
𝑑(𝐴(1), … , 𝐴(𝑛)) = 𝑑( 𝑛 ∑ 𝑖=1 𝑎𝑖1𝑒𝑖, 𝐴(2), … , 𝐴(𝑛)) Since 𝑑is multilinear, 𝑑(𝐴(1), … , 𝐴(𝑛)) = 𝑛 ∑ 𝑖=1 𝑎𝑖1𝑑(𝑒𝑖, 𝐴(2), … , 𝐴(𝑛)) Inductively on all columns, 𝑑(𝐴(1), … , 𝐴(𝑛)) = 𝑛 ∑ 𝑖=1 𝑛 ∑ 𝑗=1 𝑎𝑖1𝑎𝑗2𝑑(𝑒𝑖, 𝑒𝑗, 𝐴(3), … , 𝐴(𝑛)) = ⋯= ∑ 1≤𝑖1,≤⋯≤𝑛 𝑛 ∏ 𝑘=1 𝑎𝑖ℓ𝑘𝑑(𝑒𝑖1, … 𝑒𝑖𝑛) Since 𝑑is alternating, we know that for 𝑑(𝑒𝑖1, … , 𝑒𝑖𝑛) to be nonzero, the 𝑖𝑘must be different, so this corresponds to a permutation 𝜎∈𝑆𝑛.
𝑑(𝐴(1), … , 𝐴(𝑛)) = ∑ 𝜎∈𝑆𝑛 𝑛 ∏ 𝑘=1 𝑎𝜎(𝑘)𝑘𝜀(𝜎)𝑑(𝑒1, … , 𝑒𝑛) which is exactly the determinant up to a constant multiple.
5.5. Multiplicative property of determinant Lemma. Let 𝐴, 𝐵∈𝑀𝑛(𝐹). Then det(𝐴𝐵) = det(𝐴) det(𝐵).
Proof. Given 𝐴, we define the volume form 𝑑𝐴∶(𝐹𝑛)𝑛→𝐹by 𝑑𝐴(𝑣1, … , 𝑣𝑛) ↦det(𝐴𝑣1, … , 𝐴𝑣𝑛) 𝑣𝑖↦𝐴𝑣𝑖is linear, and the determinant is multilinear, so 𝑑𝐴is multilinear. If 𝑖≠𝑗and 𝑣𝑖= 𝑣𝑗, then det(… , 𝐴𝑣𝑖, … , 𝐴𝑣𝑗, …) = 0 so 𝑑𝐴is alternating. Hence 𝑑𝐴is a volume form. Hence there exists a constant 𝐶𝐴such that 𝑑𝐴(𝑣1, … , 𝑣𝑛) = 𝐶𝐴det(𝑣1, … , 𝑣𝑛). We can compute 𝐶𝐴 by considering the basis vectors; 𝐴𝑒𝑖= 𝐴𝑖where 𝐴𝑖is the 𝑖th column vector of 𝐴. Then, 𝐶𝐴= 𝑑𝐴(𝑒1, … , 𝑒𝑛) = det(𝐴𝑒1, … , 𝐴𝑒𝑛) = det 𝐴 Hence, det(𝐴𝐵) = 𝑑𝐴(𝐵) = det 𝐴det 𝐵 386 5. Trace and determinant 5.6. Singular and non-singular matrices Definition. Let 𝐴∈𝑀𝑛(𝐹). We say that (i) 𝐴is singular if det 𝐴= 0; (ii) 𝐴is non-singular if det 𝐴≠0.
Lemma. If 𝐴is invertible, it is non-singular.
Proof. If 𝐴is invertible, there exists 𝐴−1. Then, since the determinant is a homomorphism, det(𝐴𝐴−1) = det 𝐼= 1 Thus det 𝐴det 𝐴−1 = 1 and hence neither of these determinants can be zero.
Theorem. Let 𝐴∈𝑀𝑛(𝐹). The following are equivalent.
(i) 𝐴is invertible; (ii) 𝐴is non-singular; (iii) 𝑟(𝐴) = 𝑛.
Proof. We have already shown that (i) implies (ii). We have also shown that (i) and (iii) are equivalent by the rank-nullity theorem. So it suffices to show that (ii) implies (iii).
Suppose 𝑟(𝐴) < 𝑛. Then we will show 𝐴is singular. We have dim span(𝐴1, … , 𝐴𝑛) < 𝑛.
Therefore, since there are 𝑛vectors, (𝐴1, … , 𝐴𝑛) is not free. So there exist scalars 𝜆𝑖not all zero such that ∑𝑖𝜆𝑖𝐴𝑖= 0. Choose 𝑗such that 𝜆𝑗≠0. Then, 𝐴𝑗= −1 𝜆𝑗 ∑ 𝑖≠𝑗 𝜆𝑖𝐴𝑖 So we can compute the determinant of 𝐴by det 𝐴= det(𝐴1, … , −1 𝜆𝑗 ∑ 𝑖≠𝑗 𝜆𝑖𝐴𝑖, … , 𝐴𝑛) Since the determinant is alternating and linear in the 𝑗th entry, its value is zero. So 𝐴is singular as required.
Remark. The above theorem gives necessary and sufficient conditions for invertibility of a set of 𝑛linear equations with 𝑛unknowns.
387 VII. Linear Algebra 5.7. Determinants of linear maps Lemma. Similar matrices have the same determinant.
Proof.
det(𝑃−1𝐴𝑃) = det(𝑃−1) det 𝐴det 𝑃= det 𝐴det(𝑃−1𝑃) = det 𝐴 Definition. If 𝛼is an endomorphism, then we define det 𝛼= det[𝛼]𝐵,𝐵 where 𝐵is any basis of the vector space. This is well-defined, since this value does not depend on the choice of basis.
Theorem. det∶𝐿(𝑉, 𝑉) →𝐹satisfies the following properties.
(i) det 𝐼= 1; (ii) det(𝛼𝛽) = det 𝛼det 𝛽; (iii) det 𝛼≠0 if and only if 𝛼is invertible, and in this case, det(𝛼−1) det 𝛼= 1.
This is simply a reformulation of the previous theorem for matrices. The proof is simple, and relies on the invariance of the determinant under a change of basis.
5.8. Determinant of block-triangular matrices Lemma. Let 𝐴∈𝑀𝑘(𝐹), 𝐵∈𝑀ℓ(𝐹), 𝐶∈𝑀𝑘,ℓ(𝐹). Consider the matrix 𝑀= (𝐴 𝐶 0 𝐵) Then det 𝑀= det 𝐴det 𝐵.
Proof. Let 𝑛= 𝑘+ ℓ, so 𝑀∈𝑀𝑛(𝐹). Let 𝑀= (𝑚𝑖𝑗). We must compute det 𝑀= ∑ 𝜎∈𝑆𝑛 𝜀(𝜎) 𝑛 ∏ 𝑖=1 𝑚𝜎(𝑖)𝑖 Observe that 𝑚𝜎(𝑖)𝑖= 0 if 𝑖≤𝑘and 𝜎(𝑖) > 𝑘. Then, we need only sum over 𝜎∈𝑆𝑛such that for all 𝑗≤𝑘, we have 𝜎(𝑗) ≤𝑘. Thus, for all 𝑗∈{𝑘+ 1, … , 𝑛}, we have 𝜎(𝑗) ∈{𝑘+ 1, … , 𝑛}.
388 5. Trace and determinant We can then uniquely decompose 𝜎into two permutations 𝜎= 𝜎1𝜎2, where 𝜎1 is restricted to {1, … , 𝑘} and 𝜎2 is restricted to {𝑘+ 1, … , 𝑛}. Hence, det 𝑀= ∑ 𝜎1∈𝑆𝑘 ∑ 𝜎2∈𝑆𝑛−𝑘 𝜀(𝜎) 𝑛 ∏ 𝑖=1 𝑚𝜎(𝑖)𝑖 = ∑ 𝜎1∈𝑆𝑘 ∑ 𝜎2∈𝑆𝑛−𝑘 𝜀(𝜎1)𝜀(𝜎2) 𝑘 ∏ 𝑖=1 𝑚𝜎(𝑖)𝑖 𝑛 ∏ 𝑖=𝑘+1 𝑚𝜎(𝑖)𝑖 = ∑ 𝜎1∈𝑆𝑘 𝜀(𝜎1) 𝑘 ∏ 𝑖=1 𝑚𝜎(𝑖)𝑖 ∑ 𝜎2∈𝑆𝑛−𝑘 𝜀(𝜎2) 𝑛 ∏ 𝑖=𝑘+1 𝑚𝜎(𝑖)𝑖 = det 𝐴det 𝐵 Corollary. We need not restrict ourselves to just two blocks, since we can apply the above lemma inductively. In particular, this implies that an upper-triangular matrix with diagonal elements 𝜆𝑖has determinant ∏𝑖𝜆𝑖.
389 VII. Linear Algebra 6. Adjugate matrices 6.1. Column and row expansions Let 𝐴∈𝑀𝑛(𝐹) with column vectors 𝐴(𝑖). We know that det(𝐴(1), … , 𝐴(𝑗), … , 𝐴(𝑘), … , 𝐴(𝑛)) = −det(𝐴(1), … , 𝐴(𝑘), … , 𝐴(𝑗), … , 𝐴(𝑛)) Using the fact that det 𝐴= det 𝐴⊺we can similarly see that swapping two rows will invert the sign of the determinant.
Remark. We could have proven all of the properties of the determinant above by using the decomposition of 𝐴into elementary matrices.
Definition. Let 𝐴∈𝑀𝑛(𝐹). Let 𝑖, 𝑗∈{1, … , 𝑛}. We define the minor 𝐴ˆ 𝑖𝑗∈𝑀𝑛−1(𝐹) to be the matrix obtained by removing the 𝑖th row and the 𝑗th column.
Lemma. Let 𝐴∈𝑀𝑛(𝐹).
(i) Let 𝑗∈{1, … , 𝑛}. The determinant of 𝐴is given by the column expansion with respect to the 𝑗th column: det 𝐴= 𝑛 ∑ 𝑖=1 (−1)𝑖+𝑗𝑎𝑖𝑗det 𝐴ˆ 𝑖𝑗 (ii) Let 𝑖∈{1, … , 𝑛}. The same determinant is also given by the row expansion with respect to the 𝑖th row: det 𝐴= 𝑛 ∑ 𝑗=1 (−1)𝑖+𝑗𝑎𝑖𝑗det 𝐴ˆ 𝑖𝑗 This is a process of reducing the computation of 𝑛× 𝑛determinants to (𝑛−1) × (𝑛−1) determinants.
Proof. We will prove case (i), the column expansion with respect to the 𝑗th column. Then (ii) will follow from the transpose of the matrix. Let 𝑗∈{1, … , 𝑛}. We can write 𝐴(𝑗) = ∑ 𝑛 𝑖=1 𝑎𝑖𝑗𝑒𝑖where the 𝑒𝑖are the canonical basis. Then, by swapping rows and columns, det 𝐴= det (𝐴(1), … , 𝑛 ∑ 𝑖=1 𝑎𝑖𝑗𝑒𝑖, … , 𝐴(𝑛)) = 𝑛 ∑ 𝑖=1 𝑎𝑖𝑗det (𝐴(1), … , 𝑒𝑖, … , 𝐴(𝑛)) = 𝑛 ∑ 𝑖=1 𝑎𝑖𝑗(−1)𝑗−1 det (𝑒𝑖, 𝐴(1), … , 𝐴(𝑛)) = 𝑛 ∑ 𝑖=1 𝑎𝑖𝑗(−1)𝑗−1(−1)𝑖−1 det (𝑒1, 𝐴 (1) , … , 𝐴 (𝑛) ) 390 6. Adjugate matrices This has brought the matrix into block form, where there is an element of value 1 in the top left, and the matrix 𝐴ˆ 𝑖𝑗in the bottom right. The bottom left block is entirely zeroes. Hence, det 𝐴= 𝑛 ∑ 𝑖=1 (−1)𝑖+𝑗𝑎𝑖𝑗det 𝐴ˆ 𝑖𝑗 as required.
Remark. We have proven that det(𝐴(1), … , 𝑒𝑖, … , 𝐴(𝑛)) = (−1)𝑖+𝑗det 𝐴ˆ 𝑖𝑗 6.2. Adjugates Definition. Let 𝐴∈𝑀𝑛(𝐹). The adjugate matrix of 𝐴, denoted adj 𝐴, is the 𝑛× 𝑛matrix given by (adj 𝐴)𝑖𝑗= (−1)𝑖+𝑗det 𝐴ˆ 𝑗𝑖 Hence, det(𝐴(1), … , 𝑒𝑖, … , 𝐴(𝑛)) = (adj 𝐴)𝑗𝑖 Theorem. Let 𝐴∈𝑀𝑛(𝐹). Then (adj 𝐴)𝐴= (det 𝐴)𝐼 In particular, when 𝐴is invertible, 𝐴−1 = adj 𝐴 det 𝐴 Proof. We have det 𝐴= 𝑛 ∑ 𝑖=1 (−1)𝑖+𝑗𝑎𝑖𝑗det 𝐴ˆ 𝑖𝑗 Hence, det 𝐴= 𝑛 ∑ 𝑖=1 (adj 𝐴)𝑗𝑖𝑎𝑖𝑗= ((adj 𝐴)𝐴)𝑗𝑗 So the diagonal terms match. Off the diagonal, 0 = det(𝐴(1), … , 𝐴(𝑘) ⏟ 𝑗th position , … , 𝐴(𝑘), … , 𝐴(𝑛)) 391 VII. Linear Algebra By linearity, 0 = det ⎛ ⎜ ⎜ ⎜ ⎝ 𝐴(1), … , 𝑛 ∑ 𝑖=1 𝑎𝑖𝑘𝑒𝑖 ⏟ ⎵ ⏟ ⎵ ⏟ 𝑗th position , … , 𝐴(𝑘), … , 𝐴(𝑛) ⎞ ⎟ ⎟ ⎟ ⎠ = 𝑛 ∑ 𝑖=1 𝑎𝑖𝑘det (𝐴(1), … , 𝑒𝑖 ⏟ 𝑗th position , … , 𝐴(𝑘), … , 𝐴(𝑛)) = 𝑛 ∑ 𝑖=1 𝑎𝑖𝑘(adj 𝐴)𝑗𝑖 = ((adj 𝐴)𝐴)𝑗𝑘 6.3. Cramer’s rule Proposition. Let 𝐴be an invertible square matrix of dimension 𝑛. Let 𝑏∈𝐹𝑛. Then the unique solution to 𝐴𝑥= 𝑏is given by 𝑥𝑖= 1 det 𝐴det(𝐴ˆ 𝑖𝑏) where 𝐴ˆ 𝑖𝑏is obtained by replacing the 𝑖th column of 𝐴by 𝑏. This is an algorithm to compute 𝑥, avoiding the computation of 𝐴−1.
Proof. Let 𝐴be invertible. Then there exists a unique 𝑥∈𝐹𝑛such that 𝐴𝑥= 𝑏. Then, since the determinant is alternating, det(𝐴ˆ 𝑖𝑏) = det(𝐴(1), … , 𝐴(𝑖−1), 𝑏, 𝐴(𝑖+1), … , 𝐴(𝑛)) = det (𝐴(1), … , 𝐴(𝑖−1), 𝑛 ∑ 𝑗=1 𝑥𝑗𝐴(𝑗), 𝐴(𝑖+1), … , 𝐴(𝑛)) = det (𝐴(1), … , 𝐴(𝑖−1), 𝑥𝑖𝐴(𝑖), 𝐴(𝑖+1), … , 𝐴(𝑛)) = 𝑥𝑖det 𝐴 So the formula works.
392 7. Eigenvectors and eigenvalues 7. Eigenvectors and eigenvalues 7.1. Eigenvalues Let 𝑉be an 𝐹-vector space. Let dim 𝑉= 𝑛< ∞, and let 𝛼be an endomorphism of 𝑉.
We wish to find a basis 𝐵of 𝑉such that, in this basis, [𝛼]𝐵≡[𝛼]𝐵,𝐵has a simple (e.g.
diagonal, triangular) form. Recall that if 𝐵′ is another basis and 𝑃is the change of basis matrix, [𝛼]𝐵′ = 𝑃−1[𝛼]𝐵𝑃. Equivalently, given a square matrix 𝐴∈𝑀𝑛(𝐹) we want to conjugate it by a matrix 𝑃such that the result is ‘simpler’.
Definition. Let 𝛼∈𝐿(𝑉) be an endomorphism. We say that 𝛼is diagonalisable if there exists a basis 𝐵of 𝑉such that the matrix [𝛼]𝐵is diagonal. We say that 𝛼is triangulable if there exists a basis 𝐵of 𝑉such that [𝛼]𝐵is triangular.
Remark. We can express this equivalently in terms of conjugation of matrices.
Definition. A scalar 𝜆∈𝐹is an eigenvalue of an endomorphism 𝛼if and only if there exists a vector 𝑣∈𝑉∖{0} such that 𝛼(𝑣) = 𝜆𝑣. Such a vector is an eigenvector with eigenvalue 𝜆.
𝑉𝜆= {𝑣∈𝑉∶𝛼(𝑣) = 𝜆𝑣} ≤𝑉is the eigenspace associated to 𝜆.
Lemma. 𝜆is an eigenvalue if and only if det(𝛼−𝜆𝐼) = 0.
Proof. If 𝜆is an eigenvalue, there exists a nonzero vector 𝑣such that 𝛼(𝑣) = 𝜆𝑣, so (𝛼− 𝜆)(𝑣) = 0. So the kernel is non-trivial. So 𝛼−𝜆𝐼is not injective, so it is not surjective by the rank-nullity theorem. Hence this matrix is not invertible, so it has zero determinant.
Remark. If 𝛼(𝑣𝑗) = 𝜆𝑣𝑗for 𝑗∈{1, … , 𝑚}, we can complete the family 𝑣𝑗into a basis (𝑣1, … , 𝑣𝑛) of 𝑉. Then in this basis, the first 𝑚columns of the matrix 𝛼has diagonal entries 𝜆𝑗.
7.2. Polynomials Recall the following facts about polynomials on a field, for instance 𝑓(𝑡) = 𝑎𝑛𝑡𝑛+ ⋯+ 𝑎1𝑡+ 𝑎0 We say that the degree of 𝑓, written deg 𝑓is 𝑛. The degree of 𝑓+ 𝑔is at most the maximum degree of 𝑓and 𝑔. deg(𝑓𝑔) = deg 𝑓+ deg 𝑔. Let 𝐹[𝑡] be the vector space of polynomials with coefficients in 𝐹. If 𝜆is a root of 𝑓, then (𝑡−𝜆) divides 𝐹.
Proof.
𝑓(𝑡) = 𝑎𝑛𝑡𝑛+ ⋯+ 𝑎1𝑡+ 𝑎0 Hence, 𝑓(𝜆) = 𝑎𝑛𝜆𝑛+ ⋯+ 𝑎1𝜆+ 𝑎0 = 0 393 VII. Linear Algebra which implies that 𝑓(𝑡) = 𝑓(𝑡) −𝑓(𝜆) = 𝑎𝑛(𝑡𝑛−𝜆𝑛) + ⋯+ 𝑎1(𝑡−𝜆) But note that, for all 𝑛, 𝑡𝑛−𝜆𝑛= (1 −𝜆)(𝑡𝑛−1 + 𝜆𝑡𝑛−2 + ⋯+ 𝜆𝑛−2𝑡+ 𝜆𝑛−1) Remark. We say that 𝜆is a root of multiplicity 𝑘if (𝑡−𝜆)𝑘divides 𝑓but (𝑡−𝜆)𝑘+1 does not.
Corollary. A nonzero polynomial of degree 𝑛has at most 𝑛roots, counted with multiplicity.
Corollary. If 𝑓 1, 𝑓 2 are two polynomials of degree less than 𝑛such that 𝑓 1(𝑡𝑖) = 𝑓 2(𝑡𝑖) for 𝑖∈{1, … , 𝑛} and 𝑡𝑖distinct, then 𝑓 1 ≡𝑓 2.
Proof. 𝑓 1 −𝑓 2 has degree less than 𝑛, but has 𝑛roots. Hence it is zero.
Theorem. Any polynomial 𝑓∈ℂ[𝑡] of positive degree has a complex root. When counted with multiplicity, 𝑓has a number of roots equal to its degree.
Corollary. Any polynomial 𝑓∈ℂ[𝑡] can be factorised into an amount of linear factors equal to its degree.
7.3. Characteristic polynomials Definition. Let 𝛼be an endomorphism. The characteristic polynomial of 𝛼is 𝜒𝛼(𝜆) = det(𝛼−𝜆𝐼) Remark. 𝜒𝛼is a polynomial because the determinant is defined as a polynomial in the terms of the matrix. Note further that conjugate matrices have the same characteristic poly-nomial, so the above definition is well defined in any basis. Indeed, det(𝑃−1𝛼𝑃−𝜆𝐼) = det(𝑃−1(𝛼−𝜆𝐼)𝑃) = det(𝛼−𝜆𝐼).
Theorem. Let 𝛼∈𝐿(𝑉). 𝛼is triangulable if and only if 𝜒𝛼can be written as a product of linear factors over 𝐹. In particular, all complex matrices are triangulable.
Proof. Suppose 𝛼is triangulable. Then for a basis 𝐵, [𝛼]𝐵is triangulable with diagonal entries 𝑎𝑖. Then 𝜒𝛼(𝑡) = (𝑎1 −𝑡)(𝑎2 −𝑡) ⋯(𝑎𝑛−𝑡) Conversely, let 𝜒𝛼(𝑡) be the characteristic polynomial of 𝛼with a root 𝜆. Then, 𝜒𝛼(𝜆) = 0 implies 𝜆is an eigenvalue. Let 𝑉𝜆be the corresponding eigenspace. Let (𝑣1, … , 𝑣𝑘) be the 394 7. Eigenvectors and eigenvalues basis of this eigenspace, completed to a basis (𝑣1, … , 𝑣𝑛) of 𝑉. Let 𝑊= span {𝑣𝑘+1, … , 𝑣𝑛}, and then 𝑉= 𝑉𝜆⊕𝑊. Then [𝛼]𝐵= (𝜆𝐼 ⋆ 0 𝐶) where ⋆is arbitrary, and 𝐶is a block of size (𝑛−𝑘)×(𝑛−𝑘). Then 𝛼induces an endomorph-ism 𝛼∶𝑉/𝑈→𝑉/𝑈with respect to the basis (𝑣𝑘+1, … , 𝑣𝑛), where 𝑈= 𝑉𝜆. By induction on the dimension, we can find a basis (𝑤𝑘+1, … , 𝑤𝑛) for which 𝐶has a triangular form. Then the basis (𝑣1, … , 𝑣𝑘, 𝑤𝑘+1, … , 𝑤𝑛) is a basis for which 𝛼is triangular.
Lemma. Let 𝑛= dim 𝑉, and 𝑉be a vector space over ℝor ℂ. Let 𝛼be an endomorphism on 𝑉. Then 𝜒𝛼(𝑡) = (−1)𝑛𝑡𝑛+ 𝑐𝑛−1𝑡𝑛−1 + ⋯+ 𝑐0 with 𝑐0 = det 𝐴; 𝑐𝑛−1 = (−1)𝑛−1 tr 𝐴 Proof.
𝜒𝛼(𝑡) = det(𝛼−𝑡𝐼) ⟹𝜒𝛼(0) = det(𝛼) Further, for ℝ, ℂwe know that 𝛼is triangulable over ℂ. Hence 𝜒𝛼(𝑡) is the determinant of a triangular matrix; 𝜒𝛼(𝑡) = 𝑛 ∏ 𝑖=1 (𝑎𝑖−𝑡) Hence 𝑐𝑛−1 = (−1)𝑛−1𝑎𝑖 Since the trace is invariant under a change of basis, this is exactly the trace as required.
7.4. Polynomials for matrices and endomorphisms Let 𝑝(𝑡) be a polynomial over 𝐹. We will write 𝑝(𝑡) = 𝑎𝑛𝑡𝑛+ ⋯+ 𝑎0 For a matrix 𝐴∈𝑀𝑛(𝐹), we write 𝑝(𝐴) = 𝑎𝑛𝐴𝑛+ ⋯+ 𝑎0 ∈𝑀𝑛(𝐹) For an endomorphism 𝛼∈𝐿(𝑉), 𝑝(𝛼) = 𝑎𝑛𝛼𝑛+ ⋯+ 𝑎0𝐼∈𝐿(𝑉); 𝛼𝑘≡𝛼∘⋯∘𝛼 ⏟⎵⏟⎵⏟ 𝑘times 395 VII. Linear Algebra 7.5. Sharp criterion of diagonalisability Theorem. Let 𝑉be a vector space over 𝐹of finite dimension 𝑛. Let 𝛼be an endomorphism of 𝑉. Then 𝛼is diagonalisable if and only if there exists a polynomial 𝑝which is a product of distinct linear factors, such that 𝑝(𝛼) = 0. In other words, there exist distinct 𝜆1, … , 𝜆𝑘 such that 𝑝(𝑡) = 𝑛 ∏ 𝑖=1 (𝑡−𝜆𝑖) ⟹𝑝(𝛼) = 0 Proof. Suppose 𝛼is diagonalisable in a basis 𝐵. Let 𝜆1, … , 𝜆𝑘be the 𝑘≤𝑛distinct eigenval-ues. Let 𝑝(𝑡) = 𝑘 ∏ 𝑖=1 (𝑡−𝜆𝑖) Let 𝑣∈𝐵. Then 𝛼(𝑣) = 𝜆𝑖𝑣for some 𝑖. Then, since the terms in the following product commute, (𝛼−𝜆𝑖𝐼)(𝑣) = 0 ⟹𝑝(𝛼)(𝑣) = [ 𝑘 ∏ 𝑖=1 (𝛼−𝜆𝑖𝐼)] (𝑣) = 0 So for all basis vectors, 𝑝(𝛼)(𝑣). By linearity, 𝑝(𝛼) = 0.
Conversely, suppose that 𝑝(𝛼) = 0 for some polynomial 𝑝(𝑡) = ∏ 𝑘 𝑖=1(𝑡−𝜆𝑖) with distinct 𝜆𝑖.
Let 𝑉𝜆𝑖= ker(𝛼−𝜆𝑖𝐼). We claim that 𝑉= 𝑘 ⨁ 𝑖=1 𝑉𝜆𝑖 Consider the polynomials 𝑞𝑗(𝑡) = 𝑘 ∏ 𝑖=1,𝑖≠𝑗 𝑡−𝜆𝑖 𝜆𝑗−𝜆𝑖 These polynomials evaluate to one at 𝜆𝑗and zero at 𝜆𝑖for 𝑖≠𝑗. Hence 𝑞𝑗(𝜆𝑖) = 𝛿𝑖𝑗. We now define the polynomial 𝑞= 𝑞1 + ⋯+ 𝑞𝑘 The degree of 𝑞is at most (𝑘−1). Note, 𝑞(𝜆𝑖) = 1 for all 𝑖∈{1, … , 𝑘}. The only polynomial that evaluates to one at 𝑘points with degree at most (𝑘−1) is exactly given by 𝑞(𝑡) = 1.
Consider the endomorphism 𝜋𝑗= 𝑞𝑗(𝛼) ∈𝐿(𝑉) These are called the ‘projection operators’. By construction, 𝑘 ∑ 𝑗=1 𝜋𝑗= 𝑘 ∑ 𝑗=1 𝑞𝑗(𝛼) = 𝐼 396 7. Eigenvectors and eigenvalues So the sum of the 𝜋𝑗is the identity. Hence, for all 𝑣∈𝑉, 𝐼(𝑣) = 𝑣= 𝑘 ∑ 𝑗=1 𝜋𝑗(𝑣) = 𝑘 ∑ 𝑗=1 𝑞𝑗(𝛼)(𝑣) So we can decompose any vector as a sum of its projections 𝜋𝑗(𝑣). Now, by definition of 𝑞𝑗 and 𝑝, (𝛼−𝜆𝑗𝐼)𝑞𝑗(𝛼)(𝑣) = 1 ∏𝑖≠𝑗(𝜆𝑗−𝜆𝑖)(𝛼−𝜆𝑗𝐼)[∏ 𝑖≠𝑗 (𝑡−𝜆𝑖)] (𝛼) = 1 ∏𝑖≠𝑗(𝜆𝑗−𝜆𝑖) 𝑘 ∏ 𝑖=1 (𝛼−𝜆𝑖𝐼)(𝑣) = 1 ∏𝑖≠𝑗(𝜆𝑗−𝜆𝑖)𝑝(𝛼)(𝑣) By assumption, this is zero. For all 𝑣, we have (𝛼−𝜆𝑗𝐼)𝑞𝑗(𝛼)(𝑣). Hence, (𝛼−𝜆𝑗𝐼)𝜋𝑗(𝑣) = 0 ⟹𝜋𝑗(𝑣) ∈ker(𝛼−𝜆𝑗𝐼) = 𝑣𝑗 We have then proven that, for all 𝑣∈𝑉, 𝑣= 𝑘 ∑ 𝑗=1 𝜋𝑗(𝑣) ⏟ ∈𝑉𝑗 Hence, 𝑉= 𝑘 ∑ 𝑗=1 𝑉𝑗 It remains to show that the sum is direct. Indeed, let 𝑣∈𝑉𝜆𝑗∩(∑ 𝑖≠𝑗 𝑉𝜆𝑖) We must show 𝑣= 0. Applying 𝜋𝑗, 𝜋𝑗(𝑣) = 𝑞𝑗(𝛼)(𝑣) = ∏ 𝑖≠𝑗 (𝛼−𝜆𝑖𝐼)(𝑣) 𝜆𝑗−𝜆𝑖 Since 𝛼(𝑣) = 𝜆𝑗𝑣, 𝜋𝑗(𝑣) = ∏ 𝑖≠𝑗 (𝜆𝑗−𝜆𝑖)𝑣 𝜆𝑗−𝜆𝑖 = 𝑣 397 VII. Linear Algebra Hence 𝜋𝑗really projects onto 𝑉𝜆𝑗. However, we also know 𝑣∈∑𝑖≠𝑗𝑉𝜆𝑖. So we can write 𝑣= ∑𝑖≠𝑗𝑤𝑖for 𝑤∈𝑉𝜆𝑖. Thus, 𝜋𝑗(𝑤𝑖) = ∏ 𝑚≠𝑗 (𝛼−𝜆𝑚𝐼)(𝑣) 𝜆𝑚−𝜆𝑗 Since 𝛼(𝑤𝑖) = 𝜆𝑖𝑤𝑖, one of the factors will vanish, hence 𝜋𝑗(𝑤𝑖) = 0 So 𝑣= ∑ 𝑖≠𝑗 𝑤𝑖⟹𝜋𝑗(𝑣) = ∑ 𝑖≠𝑗 𝜋𝑗(𝑤𝑖) = 0 But 𝑣= 𝜋𝑗(𝑣) hence 𝑣= 0. So the sum is direct. Hence, 𝐵= (𝐵1, … , 𝐵𝑘) is a basis of 𝑉, where the 𝐵𝑖are bases of 𝑉𝜆𝑖. Then [𝛼]𝐵is diagonal.
Remark. We have shown further that if 𝜆1, … , 𝜆𝑘are distinct eigenvalues of 𝛼, then 𝑘 ∑ 𝑖=1 𝑉𝜆𝑖= 𝑘 ⨁ 𝑖=1 𝑉𝜆𝑖 Therefore, the only way that diagonalisation fails is when this sum is not direct, so 𝑘 ∑ 𝑖=1 𝑉𝜆𝑖< 𝑉 Example. Let 𝐹= ℂ. Let 𝐴∈𝑀𝑛(𝐹) such that 𝐴has finite order; there exists 𝑚∈ℕsuch that 𝐴𝑚= 𝐼. Then 𝐴is diagonalisable. This is because 𝑡𝑚−1 = 𝑝(𝑡) = 𝑚 ∏ 𝑗=1 (𝑡−𝜉𝑗 𝑚); 𝜉𝑚= 𝑒2𝜋𝑖/𝑚 and 𝑝(𝐴) = 0.
7.6. Simultaneous diagonalisation Theorem. Let 𝛼, 𝛽be endomorphisms of 𝑉which are diagonalisable. Then 𝛼, 𝛽are simul-taneously diagonalisable (there exists a basis 𝐵of 𝑉such that [𝛼]𝐵, [𝛽]𝐵are diagonal) if and only if 𝛼and 𝛽commute.
Proof. Two diagonal matrices commute. If such a basis exists, 𝛼𝛽= 𝛽𝛼in this basis. So this holds in any basis. Conversely, suppose 𝛼𝛽= 𝛽𝛼. We have 𝑉= 𝑘 ⨁ 𝑖=1 𝑉𝜆𝑖 398 7. Eigenvectors and eigenvalues where 𝜆𝑖, … , 𝜆𝑘are the 𝑘distinct eigenvalues of 𝛼. We claim that 𝛽(𝑉𝜆𝑗) ≤𝑉𝜆𝑗. Indeed, for 𝑣∈𝑉𝜆𝑗, 𝛼𝛽(𝑣) = 𝛽𝛼(𝑣) = 𝛽(𝜆𝑗𝑣) = 𝜆𝑗𝛽(𝑣) ⟹𝛼(𝛽(𝑣)) = 𝜆𝑗𝛽(𝑣) Hence, 𝛽(𝑣) ∈𝑉𝜆𝑗. By assumption, 𝛽is diagonalisable. Hence, there exists a polynomial 𝑝with distinct linear factors such that 𝑝(𝛽) = 0. Now, 𝛽(𝑉𝜆𝑗) ≤𝑉𝜆𝑗so we can consider 𝛽|𝑉𝜆𝑗. This is an endomorphism of 𝑉𝜆𝑗. We can compute 𝑝(𝛽| | |𝑉𝜆𝑗 ) = 0 Hence, 𝛽|𝑉𝜆𝑗is diagonalisable. Let 𝐵𝑖be the basis of 𝑉𝜆𝑖in which 𝛽|𝑉𝜆𝑗is diagonal. Since 𝑉= ⨁𝑉𝜆𝑖, 𝐵= (𝐵1, … , 𝐵𝑘) is a basis of 𝑉. Then the matrices of 𝛼and 𝛽in 𝑉are diagonal.
7.7. Minimal polynomials Recall from IB Groups, Rings and Modules the Euclidean algorithm for dividing polynomi-als. Given 𝑎, 𝑏polynomials over 𝐹with 𝑏nonzero, there exist polynomials 𝑞, 𝑟over 𝐹with deg 𝑟< deg 𝑏and 𝑎= 𝑞𝑏+ 𝑟.
Definition. Let 𝑉be a finite dimensional 𝐹-vector space. Let 𝛼be an endomorphism on 𝑉.
The minimal polynomial 𝑚𝛼of 𝛼is the nonzero polynomial with smallest degree such that 𝑚𝛼(𝛼) = 0.
Remark. If dim 𝑉= 𝑛< ∞, then dim 𝐿(𝑉) = 𝑛2. In particular, the family {𝐼, 𝛼, … , 𝛼𝑛2} cannot be free since it has 𝑛2 + 1 entries. This generates a polynomial in 𝛼which evaluates to zero. Hence, a minimal polynomial always exists.
Lemma. Let 𝛼∈𝐿(𝑉) and 𝑝∈𝐹[𝑡] be a polynomial. Then 𝑝(𝛼) = 0 if and only if 𝑚𝛼is a factor of 𝑝. In particular, 𝑚𝛼is well-defined and unique up to a constant multiple.
Proof. Let 𝑝∈𝐹[𝑡] such that 𝑝(𝛼) = 0. If 𝑚𝛼(𝛼) = 0 and deg 𝑚𝛼< deg 𝑝, we can perform the division 𝑝= 𝑚𝛼𝑞+𝑟for deg 𝑟< deg 𝑚𝛼. Then 𝑝(𝛼) = 𝑚𝛼(𝛼)𝑞(𝛼)+𝑟(𝛼). But 𝑚𝛼(𝛼) = 0.
But deg 𝑟< deg 𝑚𝛼and 𝑚𝛼is the smallest degree polynomial which evaluates to zero for 𝛼, so 𝑟≡0 so 𝑝= 𝑚𝛼𝑞. In particular, if 𝑚1, 𝑚2 are both minimal polynomials that evaluate to zero for 𝛼, we have 𝑚1 divides 𝑚2 and 𝑚2 divides 𝑚1. Hence they are equivalent up to a constant.
Example. Let 𝑉= 𝐹2 and 𝐴= (1 0 0 1) ; 𝐵= (1 1 0 1) We can check 𝑝(𝑡) = (𝑡−1)2 gives 𝑝(𝐴) = 𝑝(𝐵) = 0. So the minimal polynomial of 𝐴or 𝐵 must be either (𝑡−1) or (𝑡−1)2. For 𝐴, we can find the minimal polynomial is (𝑡−1), and 399 VII. Linear Algebra for 𝐵we require (𝑡−1)2. So 𝐵is not diagonalisable, since its minimal polynomial is not a product of distinct linear factors.
7.8. Cayley–Hamilton theorem Theorem. Let 𝑉be a finite dimensional 𝐹-vector space. Let 𝛼∈𝐿(𝑉) with characteristic polynomial 𝜒𝛼(𝑡) = det(𝛼−𝑡𝐼). Then 𝜒𝛼(𝛼) = 0.
Two proofs will provided; one more physical and based on 𝐹= ℂand one more algeb-raic.
Proof. Let 𝐵= {𝑣1, … , 𝑣𝑛} be a basis of 𝑉such that [𝛼]𝐵is triangular. This can be done when 𝐹= ℂ. Note, if the diagonal entries in this basis are 𝑎𝑖, 𝜒𝛼(𝑡) = 𝑛 ∏ 𝑖=1 (𝑎𝑖−𝑡) ⟹𝜒𝛼(𝛼) = (𝛼−𝑎1𝐼) … (𝛼−𝑎𝑛𝐼) We want to show that this expansion evaluates to zero. Let 𝑈𝑗= span {𝑣1, … , 𝑣𝑗}. Let 𝑣∈𝑉= 𝑈𝑛. We want to compute 𝜒𝛼(𝛼)(𝑣). Note, by construction of the triangular matrix.
𝜒𝛼(𝛼)(𝑣) = (𝛼−𝑎1𝐼) … (𝛼−𝑎𝑛𝐼)(𝑣) ⏟⎵⎵⏟⎵⎵⏟ ∈𝑈𝑛−1 = (𝛼−𝑎1𝐼) … (𝛼−𝑎𝑛−1𝐼)(𝛼−𝑎𝑛𝐼)(𝑣) ⏟⎵⎵⎵⎵⎵⎵⏟⎵⎵⎵⎵⎵⎵⏟ ∈𝑈𝑛−2 = … ∈𝑈0 Hence this evaluates to zero.
The following proof works for any field where we can equate coefficients, but is much less intuitive.
Proof. We will write det(𝑡𝐼−𝛼) = (−1)𝑛𝜒𝛼(𝑡) = 𝑡𝑛+ 𝑎𝑛−1𝑡𝑛−1 + ⋯+ 𝑎0 For any matrix 𝐵, we have proven 𝐵adj 𝐵= (det 𝐵)𝐼. We apply this relation to the matrix 𝐵= 𝑡𝐼−𝐴. We can check that adj 𝐵= adj(𝑡𝐼−𝐴) = 𝐵𝑛−1𝑡𝑛−1 + ⋯+ 𝐵1𝑡+ 𝐵0 since adjugate matrices are degree (𝑛−1) polynomials for each element. Then, by applying 𝐵adj 𝐵= (det 𝐵)𝐼, (𝑡𝐼−𝐴)[𝐵𝑛−1𝑡𝑛−1 + ⋯+ 𝐵1𝑡+ 𝐵0] = (det 𝐵)𝐼= (𝑡𝑛+ ⋯+ 𝑎0)𝐼 400 7. Eigenvectors and eigenvalues Since this is true for all 𝑡, we can equate coefficients. This gives 𝑡𝑛∶ 𝐼= 𝐵𝑛−1 𝑡𝑛−1 ∶ 𝑎𝑛−1𝐼= 𝐵𝑛−2 −𝐴𝐵𝑛−1 ⋮ ⋮ 𝑡0 ∶ 𝑎0𝐼= −𝐴𝐵1 Then, substituting 𝐴for 𝑡in each relation will give, for example, 𝐴𝑛𝐼= 𝐴𝑛𝐵𝑛−1. Computing the sum of all of these identities, we recover the original polynomial in terms of 𝐴instead of in terms of 𝑡. Many terms will cancel since the sum telescopes, yielding 𝐴𝑛+ 𝑎𝑛−1𝐴𝑛−1 + ⋯+ 𝑎0𝐼= 0 7.9. Algebraic and geometric multiplicity Definition. Let 𝑉be a finite dimensional 𝐹-vector space. Let 𝛼∈𝐿(𝑉) and let 𝜆be an eigenvalue of 𝛼. Then 𝜒𝛼(𝑡) = (𝑡−𝜆)𝑎𝜆𝑞(𝑡) where 𝑞(𝑡) is a polynomial over 𝐹such that (𝑡−𝜆) does not divide 𝑞. 𝑎𝜆is known as the algebraic multiplicity of the eigenvalue 𝜆. We define the geometric multiplicity 𝑔𝜆of 𝜆to be the dimension of the eigenspace associated with 𝜆, so 𝑔𝜆= dim ker(𝛼−𝜆𝐼).
Lemma. If 𝜆is an eigenvalue of 𝛼∈𝐿(𝑉), then 1 ≤𝑔𝜆≤𝑎𝜆.
Proof. We have 𝑔𝜆= dim ker(𝛼−𝜆𝐼). There exists a nontrivial vector 𝑣∈𝑉such that 𝑣∈ker(𝛼−𝜆𝐼) since 𝜆is an eigenvalue. Hence 𝑔𝜆≥1. We will show that 𝑔𝜆≤𝑎𝜆.
Indeed, let 𝑣1, … , 𝑣𝑔𝜆be a basis of 𝑉𝜆≡ker(𝛼−𝜆𝐼). We complete this into a basis 𝐵≡ (𝑣1, … , 𝑣𝑔𝜆, 𝑣𝑔𝜆+1, … , 𝑣𝑛) of 𝑉. Then note that [𝛼]𝐵= (𝜆𝐼𝑔𝜆 ⋆ 0 𝐴1 ) for some matrix 𝐴1. Now, det(𝛼−𝑡𝐼) = det ((𝜆−𝑡)𝐼𝑔𝜆 ⋆ 0 𝐴1 −𝑡𝐼) By the formula for determinants of block matrices with a zero block on the off diagonal, det(𝛼−𝑡𝐼) = (𝜆−𝑡)𝑔𝜆det(𝐴1 −𝑡𝐼) Hence 𝑔𝜆≤𝑎𝜆since the determinant is a polynomial that could have more factors of the same form.
401 VII. Linear Algebra Lemma. Let 𝑉be a finite dimensional 𝐹-vector space. Let 𝛼∈𝐿(𝑉) and let 𝜆be an eigen-value of 𝛼. Let 𝑐𝜆be the multiplicity of 𝜆as a root of the minimal polynomial of 𝛼. Then 1 ≤𝑐𝜆≤𝑎𝜆.
Proof. By the Cayley–Hamilton theorem, 𝜒𝛼(𝛼) = 0. Since 𝑚𝛼is linear, 𝑚𝛼divides 𝜒𝛼.
Hence 𝑐𝜆≤𝑎𝜆. Now we show 𝑐𝜆≥1. Indeed, 𝜆is an eigenvalue hence there exists a nonzero 𝑣∈𝑉such that 𝛼(𝑣) = 𝜆𝑣. For such an eigenvector, 𝛼𝑃(𝑣) = 𝜆𝑃𝑣for 𝑃∈ℕ. Hence for 𝑝∈𝐹[𝑡], 𝑝(𝛼)(𝑣) = 𝑝(𝜆). Hence 𝑚𝛼(𝛼)(𝑣) = 𝑚𝛼(𝜆). Since the left hand side is zero, 𝑚𝛼(𝜆) = 0. So 𝑐𝜆≥1.
Example. Let 𝐴= ( 1 0 −2 0 1 1 0 0 2 ) The minimal polynomial can be computed by considering the characteristic polynomial 𝜒𝐴(𝑡) = (𝑡−1)2(𝑡−2) So the minimal polynomial is either (𝑡−1)2(𝑡−2) or (𝑡−1)(𝑡−2) We check (𝑡−1)(𝑡−2).
(𝐴−𝐼)(𝐴−2𝐼) can be found to be zero. So 𝑚𝐴(𝑡) = (𝑡−1)(𝑡−2). Since this is a product of distinct linear factors, 𝐴is diagonalisable.
Example. Let 𝐴be a Jordan block of size 𝑛≥2. Then 𝑔𝜆= 1, 𝑎𝜆= 𝑛, and 𝑐𝜆= 𝑛.
7.10. Characterisation of diagonalisable complex endomorphisms Lemma. Let 𝐹= ℂ. Let 𝑉be a finite-dimensional ℂ-vector space. Let 𝛼be an endomorph-ism of 𝑉. Then the following are equivalent.
(i) 𝛼is diagonalisable; (ii) for all 𝜆eigenvalues of 𝛼, we have 𝑎𝜆= 𝑔𝜆; (iii) for all 𝜆eigenvalues of 𝛼, 𝑐𝜆= 1.
Proof. First, the fact that (i) is true if and only if (iii) is true has already been proven. Now let us show that (i) is equivalent to (ii). Let 𝜆1, … , 𝜆𝑘be the distinct eigenvalues of 𝛼. We have already found that 𝛼is diagonalisable if and only if 𝑉= ⨁𝑉𝜆𝑖. The sum was found to be always direct, regardless of diagonalisability. We will compute the dimension of 𝑉in two ways; 𝑛= dim 𝑉= deg 𝜒𝛼; 𝑛= dim 𝑉= 𝑘 ∑ 𝑖=1 𝑎𝜆𝑖 since 𝜒𝛼is a product of (𝑡−𝜆𝑖) factors as 𝐹= ℂ. Since the sum is direct, dim ( 𝑘 ⨁ 𝑖=1 𝑉𝜆𝑖) = 𝑘 ∑ 𝑖=1 𝑔𝜆𝑖 402 7. Eigenvectors and eigenvalues 𝛼is diagonalisable if and only if the dimensions are equal, so 𝑘 ∑ 𝑖=1 𝑔𝜆𝑖= 𝑘 ∑ 𝑖=1 𝑎𝜆𝑖 Conversely, we have proven that for all eigenvalues 𝜆𝑖, we have 𝑔𝜆𝑖≤𝑎𝜆𝑖. Hence, ∑ 𝑘 𝑖=1 𝑔𝜆𝑖= ∑ 𝑘 𝑖=1 𝑎𝜆𝑖holds if and only if 𝑔𝜆𝑖= 𝑎𝜆𝑖for all 𝑖.
403 VII. Linear Algebra 8. Jordan normal form For this section, let 𝐹= ℂ.
8.1. Definition Definition. Let 𝐴∈𝑀𝑛(ℂ). We say that 𝐴is in Jordan normal form if it is a block diagonal matrix, where each block is of the form 𝐽𝑛𝑖(𝜆) = ⎛ ⎜ ⎜ ⎜ ⎝ 𝜆 1 0 ⋯ 0 0 𝜆 1 ⋯ 0 0 0 𝜆 ⋯ 0 ⋮ ⋮ ⋮ ⋱ ⋮ 0 0 0 ⋯ 𝜆 ⎞ ⎟ ⎟ ⎟ ⎠ We say that 𝐽𝑛𝑖(𝜆) ∈𝑀𝑛𝑖(ℂ) are Jordan blocks. The 𝜆𝑖∈ℂneed not be distinct.
Remark. In three dimensions, 𝐴= ( 𝜆 0 0 0 𝜆 0 0 0 𝜆 ) is in Jordan normal form, with three one-dimensional Jordan blocks with the same 𝜆value.
8.2. Similarity to Jordan normal form Theorem. Any complex matrix 𝐴∈𝑀𝑛(ℂ) is similar to a matrix in Jordan normal form, which is unique up to reordering the Jordan blocks.
The proof is non-examinable. This follows from IB Groups, Rings and Modules.
Example. Let dim 𝑉= 2. Then any matrix is similar to one of (𝜆1 0 0 𝜆2 ) ; (𝜆 0 0 𝜆) ; (𝜆 1 0 𝜆) The minimal polynomials are (𝑡−𝜆1)(𝑡−𝜆2); (𝑡−𝜆); (𝑡−𝜆)2 8.3. Direct sum of eigenspaces Theorem. Let 𝑉be a ℂ-vector space. Let dim 𝑉= 𝑛< ∞. Then, the minimal polynomial 𝑚𝛼(𝑡) of an endomorphism 𝛼∈𝐿(𝑉) satisfies 𝑉= 𝑘 ⨁ 𝑗=1 𝑉𝑗 404 8. Jordan normal form where 𝑉𝑗= ker[(𝛼−𝜆𝑗𝐼)𝑐𝑗], and where 𝑚𝛼(𝑡) = 𝑘 ∏ 𝑖=1 (𝑡−𝜆𝑖)𝑐𝑖 𝑉𝑗is called a generalised eigenspace associated with 𝜆𝑗.
Remark. Note that 𝑉𝑗is stable by 𝛼, that is, 𝛼(𝑉𝑗) = 𝑉𝑗. Note further that (𝛼−𝜆𝑗𝐼)| |𝑉𝑗= 𝜇𝑗 gives that 𝜇𝑗is a nilpotent endomorphism; 𝜇 𝑐𝑗 𝑗= 0. So the Jordan normal form theorem is a statement about nilpotent matrices.
Note, when 𝛼is diagonalisable, 𝑐𝑗= 1 and hence we recover 𝑉𝑗= ker(𝛼−𝜆𝑗𝐼) and 𝑉= ⨁𝑉𝑗.
Proof. The key to this proof is that the projectors onto 𝑉𝑗are ‘explicit’. First, recall 𝑚𝛼(𝑡) = 𝑘 ∏ 𝑗=1 (𝑡−𝜆𝑗)𝑐𝑗 Then, let 𝑝𝑗(𝑡) = ∏ 𝑖≠𝑗 (𝑡−𝜆𝑖)𝑐𝑖 Then 𝑝𝑗have by definition no common factor. So by Euclid’s algorithm, we can find poly-nomials 𝑞𝑖such that 𝑘 ∑ 𝑖=1 𝑞𝑖𝑝𝑖= 1 We define the projector 𝜋𝑗= 𝑞𝑗𝑝𝑗(𝛼), which is an endomorphism. By construction, for all 𝑣∈𝑉, we have 𝑘 ∑ 𝑗=1 𝜋𝑗(𝑣) = 𝑘 ∑ 𝑗=1 𝑎𝑗𝑝𝑗(𝛼(𝑣)) = 𝐼(𝑣) = 𝑣 Hence, 𝑣= 𝑘 ∑ 𝑖=1 𝜋𝑖(𝑣) Observe further that 𝜋𝑗(𝑣) ∈𝑉𝑗. Indeed, (𝛼−𝜆𝑗𝐼)𝑐𝑗𝜋𝑗(𝑣) = (𝛼−𝜆𝑗𝐼)𝑐𝑗𝑞𝑗𝑝𝑗(𝛼(𝑣)) = 𝑞𝑗𝑚𝛼(𝛼(𝑣)) = 0 Hence 𝜋𝑗(𝑣) ∈𝑉𝑗. In particular, 𝑉= ∑ 𝑘 𝑗=1 𝑉𝑗. We need to show that this sum is direct.
Note, for 𝑖≠𝑗, 𝜋𝑖𝜋𝑗= 0 from the definition of 𝜋. Hence, observe that 𝜋𝑖= 𝜋𝑖( 𝑘 ∑ 𝑗=1 𝜋𝑗) ⟹𝜋𝑖= 𝜋𝑖𝜋𝑖 405 VII. Linear Algebra Thus, 𝜋is a projector. In particular, this implies that 𝜋𝑖|𝑉𝑗is the identity if 𝑖= 𝑗and zero if 𝑖≠𝑗. This immediately implies that th sum is direct; 𝑉= 𝑘 ⨁ 𝑗=1 𝑉𝑗 Indeed, suppose 𝑘 ∑ 𝑗=1 𝛼𝑗𝑣𝑗= 0; 𝑣𝑗∈𝑉𝑗; 𝛼1 = 0 Then 𝑣1 = −1 𝛼1 𝑘 ∑ 𝑗=2 𝛼𝑗𝑣𝑗 Applying 𝜋1, 𝑣1 = −1 𝛼1 𝑘 ∑ 𝑗=2 𝛼𝑗𝜋1(𝑣𝑗) = 0 Iterating, we find 𝑣= 0.
Remark. We can compute the quantities 𝑎𝜆, 𝑔𝜆, 𝑐𝜆on the Jordan normal form of a matrix.
Indeed, let 𝑚≥2 and consider a Jordan block 𝐽𝑚(𝜆). Then 𝐽𝑚(𝜆)−𝜆𝐼is the zero matrix with ones on the off-diagonal. (𝐽𝑚(𝜆) −𝜆𝐼)𝑘pushes the ones onto the next line iteratively, so (𝐽𝑚(𝜆) −𝜆𝐼)𝑘= (0 𝐼𝑚−𝑘 0 0 ) Hence 𝐽is nilpotent of order exactly 𝑚. In Jordan normal form, (i) 𝑎𝜆is the sum of sizes of blocks with eigenvalue 𝜆. This is the amount of times 𝜆is seen on the diagonal.
(ii) 𝑔𝜆is the amount of blocks with eigenvalue 𝜆, since each block represents one eigen-vector.
(iii) 𝑐𝜆is the size of the largest block with eigenvalue 𝜆.
Example. Let 𝐴= (0 −1 1 2 ) We wish to convert this matrix into Jordan normal form; so we seek a basis for which this matrix becomes Jordan normal form.
𝜒𝐴(𝑡) = (𝑡−1)2 406 8. Jordan normal form Hence there exists only one eigenvalue, 𝜆= 1. 𝐴−𝐼≠0 hence 𝑚𝛼(𝑡) = (𝑡−1)2. Thus, the Jordan normal form of 𝐴is of the form 𝐵= (1 1 0 1) Now, ker(𝐴−𝐼) = ⟨𝑣1⟩; 𝑣1 = ( 1 −1) Further, we seek a 𝑣2 such that (𝐴−𝐼)𝑣2 = 𝑣1 ⟹𝑣2 = (−1 0 ) Such a 𝑣2 is not unique. Now, 𝐴= ( 1 −1 −1 0 ) (1 1 0 1) ( 1 −1 −1 0 ) −1 407 VII. Linear Algebra 9. Properties of bilinear forms 9.1. Changing basis Let 𝜙∶𝑉× 𝑉→𝔽be a bilinear form. Let 𝑉be a finite-dimensional 𝐹-vector space. Let 𝐵 be a basis of 𝑉and let [𝜙]𝐵= [𝜙]𝐵𝐵be the matrix with entries 𝜙(𝑒𝑖, 𝑒𝑗).
Lemma. Let 𝜙be a bilinear form 𝑉× 𝑉→𝐹. Then if 𝐵, 𝐵′ are bases for 𝑉, and 𝑃= [𝐼]𝐵′,𝐵 we have [𝜙]𝐵′ = 𝑃⊺[𝜙]𝐵𝑃 Proof. This is a special case of the general change of basis formula.
Definition. Let 𝐴, 𝐵∈𝑀𝑛(𝐹) be square matrices. We say that 𝐴, 𝐵are congruent if there exists 𝑃∈𝑀𝑛(𝐹) such that 𝐴= 𝑃⊺𝐵𝑃.
Remark. Congruence is an equivalence relation.
Definition. A bilinear form 𝜙on 𝑉is symmetric if, for all 𝑢, 𝑣∈𝑉, we have 𝜙(𝑢, 𝑣) = 𝜙(𝑣, 𝑢) Remark. If 𝐴is a square matrix, we say 𝐴is symmetric if 𝐴= 𝐴⊺. Equivalently, 𝐴𝑖𝑗= 𝐴𝑗𝑖 for all 𝑖, 𝑗. So 𝜙is symmetric if and only if [𝜙]𝐵is symmetric for any basis 𝐵. Note further that to represent 𝜙by a diagonal matrix in some basis 𝐵, it must necessarily be symmetric, since 𝑃⊺𝐴𝑃= 𝐷⟹𝐷= 𝐷⊺= (𝑃⊺𝐴𝑃) ⊺= 𝑃⊺𝐴⊺𝑃⟹𝐴= 𝐴⊺ 9.2. Quadratic forms Definition. A map 𝑄∶𝑉→𝐹is a quadratic form if there exists a bilinear form 𝜙∶𝑉×𝑉→ 𝐹such that, for all 𝑢∈𝑉, 𝑄(𝑢) = 𝜙(𝑢, 𝑢) So a quadratic form is the restriction of a bilinear form to the diagonal.
Remark. Let 𝐵= (𝑒𝑖) be a basis of 𝑉. Let 𝐴= [𝜙]𝐵= (𝜙(𝑒𝑖, 𝑒𝑗)) = (𝑎𝑖𝑗). Then, for 𝑢= ∑𝑖𝑥𝑖𝑒𝑖∈𝑉, 𝑄(𝑢) = 𝜙(𝑢, 𝑢) = 𝜙(∑ 𝑖 𝑥𝑖𝑒𝑖, ∑ 𝑗 𝑥𝑗𝑒𝑗) = ∑ 𝑖 ∑ 𝑗 𝑥𝑖𝑥𝑗𝜙(𝑒𝑖, 𝑒𝑗) = ∑ 𝑖 ∑ 𝑗 𝑥𝑖𝑥𝑗𝑎𝑖𝑗 We can check that this is equal to 𝑄(𝑢) = 𝑥⊺𝐴𝑥 408 9. Properties of bilinear forms where [𝑢]𝐵= 𝑥. Note further that 𝑥⊺𝐴𝑥= ∑ 𝑖 ∑ 𝑗 𝑎𝑖𝑗𝑥𝑖𝑥𝑗= ∑ 𝑖 ∑ 𝑗 𝑎𝑗𝑖𝑥𝑖𝑥𝑗= ∑ 𝑖 ∑ 𝑗 𝑎𝑖𝑗+ 𝑎𝑗𝑖 2 𝑥𝑖𝑥𝑗= 𝑥⊺⎛ ⎜ ⎜ ⎝ 𝐴+ 𝐴⊺ 2 ⏟ ⎵ ⏟ ⎵ ⏟ symmetric ⎞ ⎟ ⎟ ⎠ 𝑥 So we can always express the quadratic form as a symmetric matrix in any basis.
Proposition. If 𝑄∶𝑉→𝐹is a quadratic form, then there exists a unique symmetric bilin-ear form 𝜙∶𝑉× 𝑉→𝐹such that 𝑄(𝑢) = 𝜙(𝑢, 𝑢).
Proof. Let 𝜓be a bilinear form on 𝑉such that for all 𝑢∈𝑉, we have 𝑄(𝑢) = 𝜓(𝑢, 𝑢). Then, let 𝜙(𝑢, 𝑣) = 1 2[𝜓(𝑢, 𝑣) + 𝜓(𝑣, 𝑢)] Certainly 𝜙is a bilinear form and symmetric. Further, 𝜙(𝑢, 𝑢) = 𝜓(𝑢, 𝑢) = 𝑄(𝑢). So there exists a symmetric bilinear form 𝜙such that 𝑄(𝑢) = 𝜙(𝑢, 𝑢), so it suffices to prove uniqueness.
Let 𝜙be a symmetric bilinear form such that for all 𝑢∈𝑉we have 𝑄(𝑢) = 𝜙(𝑢, 𝑢). Then, we can find 𝑄(𝑢+ 𝑣) = 𝜙(𝑢+ 𝑣, 𝑢+ 𝑣) = 𝜙(𝑢, 𝑢) + 𝜙(𝑣, 𝑣) + 2𝜙(𝑢, 𝑣) Thus 𝜙(𝑢, 𝑣) is defined uniquely by 𝑄, since 2𝜙(𝑢, 𝑣) = 𝑄(𝑢+ 𝑣) −𝑄(𝑢) −𝑄(𝑣) So 𝜙is unique (when 2 is invertible in 𝐹). This identity for 𝜙(𝑢, 𝑣) is known as the polarisa-tion identity.
9.3. Diagonalisation of symmetric bilinear forms Theorem. Let 𝜙∶𝑉× 𝑉→𝐹be a symmetric bilinear form, where 𝑉is finite-dimensional.
Then there exists a basis 𝐵of 𝑉such that [𝜙]𝐵is diagonal.
Proof. By induction on the dimension, suppose the theorem holds for all dimensions less than 𝑛for 𝑛≥2. If 𝜙(𝑢, 𝑢) = 0 for all 𝑢∈𝑉, then 𝜙= 0 by the polarisation identity, which is diagonal. Otherwise 𝜙(𝑒1, 𝑒1) ≠0 for some 𝑒1 ∈𝑉. Let 𝑈= (⟨𝑒1⟩) ⟂= {𝑣∈𝑉∶𝜙(𝑒1, 𝑣) = 0} This is a vector subspace of 𝑉, which is in particular ker {𝜙(𝑒1, ⋅)∶𝑉→𝐹} By the rank-nullity theorem, dim 𝑈= 𝑛−1. We now claim that 𝑈+ ⟨𝑒1⟩is a direct sum.
Indeed, for 𝑣= ⟨𝑒1⟩∩𝑈, we have 𝑣= 𝜆𝑒1 and 𝜙(𝑒1, 𝑣) = 0. Hence 𝜆= 0, since by assumption 409 VII. Linear Algebra 𝜙(𝑒1, 𝑒1) ≠0. So we find a basis 𝐵′ = (𝑒2, … , 𝑒𝑛) of 𝑈, which we extend by 𝑒1 to 𝐵= (𝑒1, 𝑒2, … , 𝑒𝑛). Since 𝑈⊕⟨𝑒1⟩has dimension 𝑛, this is a basis of 𝑉. Under this basis, we find [𝜙]𝐵= (𝜙(𝑒1, 𝑒1) 0 0 [𝜙|𝑈]𝐵′ ) because 𝜙(𝑒1, 𝑒𝑗) = 𝜙(𝑒𝑗, 𝑒1) = 0 for all 𝑗≥2. By the inductive hypothesis we can take a basis 𝐵′ such that the restricted 𝜙to be diagonal, so [𝜙]𝐵is diagonal in this basis.
Example. Let 𝑉= ℝ3 and choose the canonical basis (𝑒𝑖). Let 𝑄(𝑥1, 𝑥2, 𝑥3) = 𝑥2 1 + 𝑥2 2 + 2𝑥2 3 + 2𝑥1𝑥2 + 2𝑥1𝑥3 −2𝑥2𝑥3 Then, if 𝑄(𝑥1, 𝑥2, 𝑥3) = 𝑥⊺𝐴𝑥, we have 𝐴= ( 1 1 1 1 1 −1 1 −1 2 ) Note that the off-diagonal terms are halved from their coefficients since in the expansion of 𝑥⊺𝐴𝑥they are included twice. Then, we can find a basis in which 𝐴is diagonal. We could use the above algorithm to find a basis, or complete the square in each component. We can write 𝑄(𝑥1, 𝑥2, 𝑥3) = (𝑥1 + 𝑥2 + 𝑥3)2 + 𝑥2 3 −4𝑥2𝑥3 = (𝑥1 + 𝑥2 + 𝑥3)2 + (𝑥3 −2𝑥2)2 −(2𝑥2)2 This yields a new coordinate basis 𝑥′ 1, 𝑥′ 2, 𝑥′ 3. Then 𝑃−1𝐴𝑃is diagonal. 𝑃is given by ( 𝑥′ 1 𝑥′ 2 𝑥′ 3 ) = ( 1 1 1 0 −2 1 0 −2 0 ) ⏟ ⎵ ⎵ ⎵ ⏟ ⎵ ⎵ ⎵ ⏟ 𝑃−1 ( 𝑥1 𝑥2 𝑥3 ) 9.4. Sylvester’s law Corollary. If 𝐹= ℂ, for any symmetric bilinear form 𝜙there exists a basis of 𝑉such that [𝜙]𝐵is (𝐼𝑟 0 0 0) Proof. Since any symmetric bilinear form 𝜙in a finite-dimensional 𝐹-vector space 𝑉can be diagonalised, let 𝐸= (𝑒1, … , 𝑒𝑛) such that [𝜙]𝐸is diagonal with diagonal entries 𝑎𝑖. Order the 𝑎𝑖such that 𝑎𝑖is nonzero for 1 ≤𝑖≤𝑟, and the remaining values (if any) are zero. For 𝑖≤𝑟, let √𝑎𝑖be a choice of a complex root for 𝑎𝑖. Then 𝑣𝑖= 𝑒𝑖 √𝑎𝑖for 𝑖≤𝑟and 𝑣𝑖= 𝑒𝑖for 𝑖> 𝑟gives the basis 𝐵as required.
410 9. Properties of bilinear forms Corollary. Every symmetric matrix of 𝑀𝑛(ℂ) is congruent to a unique matrix of the form (𝐼𝑟 0 0 0) where 𝑟is the rank of the matrix.
Corollary. Let 𝐹= ℝ, and let 𝑉be a finite-dimensional ℝ-vector space. Let 𝜙be a symmet-ric bilinear form on 𝑉. Then there exists a basis 𝐵= (𝑣1, … , 𝑣𝑛) of 𝑉such that [𝜙]𝐵= ( 𝐼𝑝 0 0 0 −𝐼𝑞 0 0 0 0 ) for some integers 𝑝, 𝑞.
Proof. Since square roots do not necessarily exist in ℝ, we cannot use the form above. We first diagonalise the bilinear form in some basis 𝐸. Then, reorder and group the 𝑎𝑖into a positive group of size 𝑝, a negative group of size 𝑞, and a zero group. Then, 𝑣𝑖= ⎧ ⎪ ⎨ ⎪ ⎩ 𝑒𝑖 √𝑎𝑖 𝑖∈{1, … , 𝑝} 𝑒𝑖 √−𝑎𝑖 𝑖∈{𝑝+ 1, … , 𝑝+ 𝑞} 𝑒𝑖 𝑖∈{𝑝+ 𝑞+ 1, … , 𝑛} This gives a new basis as required.
Definition. Let 𝐹= ℝ. The signature of a bilinear form 𝜙is 𝑠(𝜙) = 𝑝−𝑞 where 𝑝and 𝑞are defined as in the corollary above.
Theorem. Let 𝐹= ℝ. Let 𝑉be a finite-dimensional ℝ-vector space. If a real symmetric bilinear form is represented by some matrix ( 𝐼𝑝 0 0 0 −𝐼𝑞 0 0 0 0 ) in some basis 𝐵, and some other matrix ( 𝐼𝑝′ 0 0 0 −𝐼𝑞′ 0 0 0 0 ) in another basis 𝐵′, then 𝑝= 𝑝′ and 𝑞= 𝑞′. Thus, the signature of the matrix is well defined.
Definition. Let 𝜙be a symmetric bilinear form on a real vector space 𝑉. We say that 411 VII. Linear Algebra (i) 𝜙is positive definite if 𝜙(𝑢, 𝑢) > 0 for all nonzero 𝑢∈𝑉; (ii) 𝜙is positive semidefinite if 𝜙(𝑢, 𝑢) ≥0 for all 𝑢∈𝑉; (iii) 𝜙is negative definite or negative semidefinite if 𝜙(𝑢, 𝑢) < 0 or 𝜙(𝑢, 𝑢) ≤0 respectively for all nonzero 𝑢∈𝑉.
Example. The matrix (𝐼𝑟 0 0 0) is positive definite for 𝑟= 𝑛, and positive semidefinite for 𝑟< 𝑛.
We now prove Sylvester’s law.
Proof. In order to prove uniqueness of 𝑝, we will characterise the matrix in a way that does not depend on the basis. In particular, we will show that 𝑝is the largest dimension of a vector subspace of 𝑉such that the restriction of 𝜙on this subspace is positive definite. Suppose we have 𝐵= (𝑣1, … , 𝑣𝑛) and [𝜙]𝐵= ( 𝐼𝑝 0 0 0 −𝐼𝑞 0 0 0 0 ) We consider 𝑋= ⟨𝑣1, … , 𝑣𝑝⟩ Then we can easily compute that 𝜙|𝑋is positive definite. Let 𝑌= ⟨𝑣𝑝+1, … , 𝑣𝑛⟩ Then, as above, 𝜙|𝑌is negative semidefinite. Suppose that 𝜙is positive definite on another subspace 𝑋′. In this case, 𝑌∩𝑋′ = {0}, since if 𝑦∈𝑌∩𝑋′ we must have 𝑄(𝑦) ≤0, but since 𝑦∈𝑋′ we have 𝑦= 0. Thus, 𝑌+ 𝑋′ = 𝑌⊕𝑋′, so 𝑛= dim 𝑉≥dim 𝑌+ dim 𝑋′. But dim 𝑌= 𝑛−𝑝, so dim 𝑋′ ≤𝑝. The same argument can be executed for 𝑞, hence both 𝑝and 𝑞are independent of basis.
9.5. Kernels of bilinear forms Definition. Let 𝐾= {𝑣∈𝑉∶∀𝑢∈𝑉, 𝜙(𝑢, 𝑣) = 0}. This is the kernel of the bilinear form.
Remark. By the rank-nullity theorem, dim 𝐾+ rank 𝜙= 𝑛 Using the above notation, we can show that there exists a subspace 𝑇of dimension 𝑛−(𝑝+ 𝑞) + min {𝑝, 𝑞} such that 𝜙|𝑇= 0. Indeed, let 𝐵= (𝑣1, … , 𝑣𝑛) such that [𝜙]𝐵= ( 𝐼𝑝 0 0 0 −𝐼𝑞 0 0 0 0 ) 412 9. Properties of bilinear forms The quadratic form has a zero subspace of dimension 𝑛−(𝑝+ 𝑞) in the bottom right. But by setting 𝑇= {𝑣1 + 𝑣𝑝+1, … , 𝑣𝑞+ 𝑣𝑝+𝑞, 𝑣𝑝+𝑞+1, … , 𝑣𝑛} we can combine the positive and negative blocks (assuming here that 𝑝≥𝑞) to produce more linearly independent elements of the kernel. In particular, dim 𝑇is the largest possible dimension of a subspace 𝑇′ of 𝑉such that 𝜙|𝑇′ = 0.
9.6. Sesquilinear forms Let 𝐹= ℂ. The standard inner product on ℂ𝑛is defined to be ⟨( 𝑥1 ⋮ 𝑣𝑛 ) , ( 𝑦1 ⋮ 𝑦𝑛 )⟩= 𝑛 ∑ 𝑖=1 𝑥𝑖𝑦𝑖 This is not a bilinear form on ℂdue to the complex conjugate, it is linear in the first entry.
Definition. Let 𝑉, 𝑊be ℂ-vector spaces. A form 𝜙∶𝑉× 𝑊→ℂis called sesquilinear if it is linear in the first entry, and 𝜙(𝑣, 𝜆1𝑤1 + 𝜆2𝑤2) = 𝜆1𝜙(𝑣, 𝑤1) + 𝜆2𝜙(𝑣, 𝑤2) so it is antilinear with respect to the second entry.
Lemma. Let 𝐵= (𝑣1, … , 𝑣𝑚) be a basis of 𝑉and 𝐶= (𝑤1, … , 𝑤𝑛) be a basis of 𝑊. Let [𝜙]𝐵,𝐶= (𝜙(𝑣𝑖, 𝑤𝑗)). Then, 𝜙(𝑣, 𝑤) = [𝑣]⊺ 𝐵[𝜙]𝐵,𝐶[𝑤]𝐶 Proof. Let 𝐵, 𝐵′ be bases of 𝑉and 𝐶, 𝐶′ be bases of 𝑊. Let 𝑃= [𝐼]𝐵′,𝐵and 𝑄= [𝐼]𝐶′,𝐶. Then [𝜙]𝐵′,𝐶′ = 𝑃⊺[𝜙]𝐵,𝐶𝑄 9.7. Hermitian forms Definition. Let 𝑉be a finite-dimensional ℂ-vector space. Let 𝜙be a sesquilinear form on 𝑉. Then 𝜙is Hermitian if, for all 𝑢, 𝑣∈𝑉, 𝜙(𝑢, 𝑣) = 𝜙(𝑣, 𝑢) Remark. If 𝜙is Hermitian, then 𝜙(𝑢, 𝑢) = 𝜙(𝑢, 𝑢) ∈ℝ. Further, 𝜙(𝜆𝑢, 𝜆𝑢) = |𝜆|2𝜙(𝑢, 𝑢).
This allows us to define positive and negative definite Hermitian forms.
413 VII. Linear Algebra Lemma. A sesquilinear form 𝜙∶𝑉× 𝑉→ℂis Hermitian if and only if, for any basis 𝐵of 𝑉, [𝜙]𝐵= [𝜙]† 𝐵 Proof. Let 𝐴= [𝜙]𝐵= (𝑎𝑖𝑗). Then 𝑎𝑖𝑗= 𝜙(𝑒𝑖, 𝑒𝑗), and 𝑎𝑗𝑖= 𝜙(𝑒𝑗, 𝑒𝑖) = 𝜙(𝑒𝑖, 𝑒𝑗) = 𝑎𝑖𝑗. So 𝐴 ⊺ = 𝐴. Conversely suppose that [𝜙]𝐵= 𝐴= 𝐴 ⊺ . Now let 𝑢= 𝑛 ∑ 𝑖=1 𝜆𝑖𝑒𝑖; 𝑣= 𝑛 ∑ 𝑖=1 𝜇𝑖𝑒𝑖 Then, 𝜙(𝑢, 𝑣) = 𝜙( 𝑛 ∑ 𝑖=1 𝜆𝑖𝑒𝑖, 𝑛 ∑ 𝑖=1 𝜇𝑖𝑒𝑖) = 𝑛 ∑ 𝑖=1 𝑛 ∑ 𝑗=1 𝜆𝑖𝜇𝑗𝑎𝑖𝑗 Further, 𝜙(𝑣, 𝑢) = 𝜙( 𝑛 ∑ 𝑖=1 𝜇𝑖𝑒𝑖, 𝑛 ∑ 𝑖=1 𝜆𝑖𝑒𝑖) = 𝑛 ∑ 𝑖=1 𝑛 ∑ 𝑗=1 𝜇𝑗𝜆𝑖𝑎𝑖𝑗 which is equivalent. Hence 𝜙is Hermitian.
9.8. Polarisation identity A Hermitian form 𝜙on a complex vector space 𝑉is entirely determined by a quadratic form 𝑄∶𝑉→ℝsuch that 𝑣↦𝜙(𝑣, 𝑣) by the formula 𝜙(𝑢, 𝑣) = 1 4[𝑄(𝑢+ 𝑣) −𝑄(𝑢−𝑣) + 𝑖𝑄(𝑢+ 𝑖𝑣) −𝑖𝑄(𝑢−𝑖𝑣)] 9.9. Hermitian formulation of Sylvester’s law Theorem. Let 𝑉be a finite-dimensional ℂ-vector space. Let 𝜙∶𝑉×𝑉→ℂbe a Hermitian form on 𝑉. Then there exists a basis 𝐵= (𝑣1, … , 𝑣𝑛) of 𝑉such that [𝜙]𝐵= ( 𝐼𝑝 0 0 0 −𝐼𝑞 0 0 0 0 ) where 𝑝, 𝑞depend only on 𝜙and not 𝐵.
Proof. The following is a sketch proof; it is nearly identical to the case of real symmetric bilinear forms. If 𝜙= 0, existence is trivial. Otherwise, using the polarisation identity there exists 𝑒1 ≠0 such that 𝜙(𝑒1, 𝑒1) ≠0. Let 𝑣1 = 𝑒1 √|𝜙(𝑒1, 𝑒1)| ⟹𝜙(𝑣1, 𝑣1) = ±1 414 9. Properties of bilinear forms Consider the orthogonal space 𝑊= {𝑤∈𝑉∶𝜙(𝑣1, 𝑤) = 0}. We can check, arguing analog-ously to the real case, that 𝑉= ⟨𝑣1⟩⊕𝑊. Hence, we can inductively diagonalise 𝜙.
𝑝, 𝑞are unique. Indeed, we can prove that 𝑝is the maximal dimension of a subspace on which 𝜙is positive definite (which is well-defined since 𝜙(𝑢, 𝑢) ∈ℝ). The geometric inter-pretation of 𝑞is similar.
9.10. Skew-symmetric forms Definition. Let 𝑉be a finite-dimensional ℝ-vector space. Let 𝜙be a bilinear form on 𝑉.
Then 𝜙is skew-symmetric if, for all 𝑢, 𝑣∈𝑉, 𝜙(𝑢, 𝑣) = −𝜙(𝑣, 𝑢) Remark. 𝜙(𝑢, 𝑢) = −𝜙(𝑢, 𝑢) = 0. Also, in any basis 𝐵of 𝑉, we have [𝜙]𝐵= −[𝜙]⊺ 𝐵. Any real matrix can be decomposed as the sum 𝐴= 1 2(𝐴+ 𝐴⊺) + 1 2(𝐴−𝐴⊺) where the first summand is symmetric and the second is skew-symmetric.
9.11. Skew-symmetric formulation of Sylvester’s law Theorem. Let 𝑉be a finite-dimensional ℝ-vector space. Let 𝜙∶𝑉× 𝑉→ℝbe a skew-symmetric form on 𝑉. Then there exists a basis 𝐵= (𝑣1, 𝑤1, 𝑣2, 𝑤2, … , 𝑣𝑚, 𝑤𝑚, 𝑣2𝑚+1, 𝑣2𝑚+2, … , 𝑣𝑛) of 𝑉such that [𝜙]𝐵= ⎛ ⎜ ⎜ ⎜ ⎜ ⎝ 0 1 −1 0 0 1 −1 0 ⋱ 0 ⎞ ⎟ ⎟ ⎟ ⎟ ⎠ Corollary. Skew-symmetric matrices have an even rank.
Proof. This is again very similar to the previous case. We will perform an inductive step on the dimension of 𝑉. If 𝜙≠0, there exist 𝑣1, 𝑤1 such that 𝜙1(𝑣1, 𝑤1) ≠0. After scaling one of the vectors, we can assume 𝜙(𝑣1, 𝑤1) = 1. Since 𝜙is skew-symmetric, 𝜙(𝑤1, 𝑣1) = −1. Then 𝑣1, 𝑤1 are linearly independent; if they were linearly dependent we would have 𝜙(𝑣1, 𝑤1) = 𝜙(𝑣1, 𝜆𝑣1) = 0. Let 𝑈= ⟨𝑣1, 𝑤1⟩and let 𝑊= {𝑣∈𝑉∶𝜙(𝑣1, 𝑣) = 𝜙(𝑤1, 𝑣) = 0} and we can show 𝑉= 𝑈⊕𝑊. Then induction gives the required result.
415 VII. Linear Algebra 10. Inner product spaces 10.1. Definition Definition. Let 𝑉be a vector space over ℝor ℂ. A scalar product or inner product is a positive-definite symmetric (respectively Hermitian) bilinear form 𝜙on 𝑉. We write 𝜙(𝑢, 𝑣) = ⟨𝑢, 𝑣⟩ 𝑉, when equipped with this inner product, is called a real (respectively complex) inner product space.
Example. In ℂ𝑛, we define ⟨𝑥, 𝑦⟩= 𝑛 ∑ 𝑖=1 𝑥𝑖𝑦𝑖 Example. Let 𝑉= 𝐶0([0, 1], ℂ). Then we can define ⟨𝑓, 𝑔⟩= ∫ 1 0 𝑓(𝑡)𝑔(𝑡) d𝑡 This is the 𝐿 2 scalar product.
Example. Let 𝜔∶[0, 1]∶ℝ⋆ + where ℝ⋆ + = ℝ+ ∖{0} and define ⟨𝑓, 𝑔⟩= ∫ 1 0 𝑓(𝑡)𝑔(𝑡)𝑤(𝑡) d𝑡 Remark. Typically it suffices to check ⟨𝑢, 𝑢⟩= 0 ⟹𝑢= 0 since linearity and positivity are usually trivial.
Definition. Let 𝑉be an inner product space. Then for 𝑣∈𝑉, the norm of 𝑣induced by the inner product is defined by ‖𝑣‖ = (⟨𝑣, 𝑣⟩) 1/2 This is real, and positive if 𝑣≠0.
10.2. Cauchy–Schwarz inequality Lemma. For an inner product space, |⟨𝑢, 𝑣⟩| ≤‖𝑎‖ ⋅‖𝑏‖ Proof. Let 𝑡∈𝐹. Then, 0 ≤‖𝑡𝑢−𝑣‖ = ⟨𝑡𝑢−𝑣, 𝑡𝑢−𝑣⟩= 𝑡𝑡⟨𝑢, 𝑢⟩−𝑢⟨𝑢, 𝑣⟩−𝑡⟨𝑣, 𝑢⟩+ ‖𝑣‖2 416 10. Inner product spaces Since the inner product is Hermitian, 0 ≤|𝑡|2‖𝑢‖2 + ‖𝑣‖2 −2 Re(𝑡⟨𝑢, 𝑣⟩) By choosing 𝑡= ⟨𝑢, 𝑣⟩ ‖𝑢‖2 we have 0 ≤|⟨𝑢, 𝑣⟩|2 ‖𝑢‖2 + ‖𝑣‖2 −2 Re (|⟨𝑢, 𝑣⟩|2 ‖𝑢‖2 ) Since the term under the real part operator is real, the result holds.
Note that equality implies collinearity in the Cauchy–Schwarz inequality.
Corollary (triangle inequality). In an inner product space, ‖𝑢+ 𝑣‖ ≤‖𝑢‖ + ‖𝑣‖ Proof. We have ‖𝑢+ 𝑣‖2 = ⟨𝑢+ 𝑣, 𝑢+ 𝑣⟩= ‖ ‖𝑢2‖ ‖+2 Re(⟨𝑢, 𝑣⟩)+‖𝑣‖2 ≤‖ ‖𝑢2‖ ‖+‖𝑣‖2+2‖𝑢‖⋅‖𝑣‖ = (‖𝑢‖+‖𝑣‖)2 Remark. Any inner product induces a norm, but not all norms derive from scalar products.
10.3. Orthogonal and orthonormal sets Definition. A set (𝑒1, … , 𝑒𝑘) of vectors of 𝑉is said to be orthogonal if ⟨𝑒𝑖, 𝑒𝑗⟩= 0 for all 𝑖≠𝑗. The set is said to be orthonormal if it is orthogonal and ‖𝑒𝑖‖ = 1 for all 𝑖. In this case, ⟨𝑒𝑖, 𝑒𝑗⟩= 𝛿𝑖𝑗.
Lemma. If (𝑒1, … , 𝑒𝑘) are orthogonal and nonzero, then they are linearly independent. Fur-ther, let 𝑣∈⟨{𝑒𝑖}⟩. Then, 𝑣= 𝑘 ∑ 𝑗=1 𝜆𝑗𝑒𝑗⟹𝜆𝑗= ⟨𝑣, 𝑒𝑗⟩ ‖ ‖𝑒𝑗‖ ‖ 2 Proof. Suppose 𝑘 ∑ 𝑖=1 𝜆𝑖𝑒𝑖= 0 Then, 0 = ⟨ 𝑘 ∑ 𝑖=1 𝜆𝑖, 𝑒𝑗⟩⟹0 = 𝑘 ∑ 𝑖=1 𝜆𝑖⟨𝑒𝑖, 𝑒𝑗⟩ 417 VII. Linear Algebra Thus 𝜆𝑗= 0 for all 𝑗. Further, for 𝑣in the span of these vectors, ⟨𝑣, 𝑒𝑗⟩= 𝑘 ∑ 𝑖=1 𝜆𝑖⟨𝑒𝑖, 𝑒𝑗⟩= 𝜆𝑗‖ ‖𝑒𝑗‖ ‖ 2 10.4. Parseval’s identity Corollary. Let 𝑉be a finite-dimensional inner product space. Let (𝑒1, … , 𝑒𝑛) be an or-thonormal basis. Then, for any vectors 𝑢, 𝑣∈𝑉, we have ⟨𝑢, 𝑣⟩= 𝑛 ∑ 𝑖=1 ⟨𝑢, 𝑒𝑖⟩⟨𝑣, 𝑒𝑖⟩ Hence, ‖𝑢‖2 = 𝑛 ∑ 𝑖=1 |⟨𝑢, 𝑒𝑖⟩|2 Proof. By orthonormality, 𝑢= 𝑛 ∑ 𝑖=1 ⟨𝑢, 𝑒𝑖⟩𝑒𝑖; 𝑣= 𝑛 ∑ 𝑖=1 ⟨𝑣, 𝑒𝑖⟩𝑒𝑖 Hence, by sesquilinearity, ⟨𝑢, 𝑣⟩= 𝑛 ∑ 𝑖=1 ⟨𝑢, 𝑒𝑖⟩⟨𝑣, 𝑒𝑖⟩ By taking 𝑢= 𝑣we find ‖𝑢‖2 = ⟨𝑢, 𝑢⟩= 𝑛 ∑ 𝑖=1 |⟨𝑢, 𝑒𝑖⟩|2 10.5. Gram–Schmidt orthogonalisation process Theorem. Let 𝑉be an inner product space. Let (𝑣𝑖)𝑖∈𝐼be a linearly independent family of vectors such that 𝐼is countable. Then there exists a family (𝑒𝑖)𝑖∈𝐼of orthonormal vectors such that for all 𝑘≥1, ⟨𝑣1, … , 𝑣𝑘⟩= ⟨𝑒1, … , 𝑒𝑘⟩ 418 10. Inner product spaces Proof. This proof is an explicit algorithm to compute the family (𝑒𝑖), which will be computed by induction on 𝑘. For 𝑘= 1, take 𝑒1 = 𝑣1 ‖𝑣1‖. Inductively, suppose (𝑒1, … , 𝑒𝑘) satisfy the conditions as above. Then we will find a valid 𝑒𝑘+1. We define 𝑒′ 𝑘+1 = 𝑣𝑘+1 − 𝑘 ∑ 𝑖=1 ⟨𝑣𝑘+1, 𝑒𝑖⟩𝑒𝑖 This ensures that the inner product between 𝑒′ 𝑘+1 and any basis vector 𝑒𝑗is zero, while main-taining the same span. Suppose 𝑒′ 𝑘+1 = 0. Then, 𝑣𝑘+1 ∈⟨𝑒1, … , 𝑒𝑘⟩= ⟨𝑣1, … , 𝑣𝑘⟩which contradicts the fact that the family is free. Thus, 𝑒𝑘+1 = 𝑒′ 𝑘+1 ‖ ‖𝑒′ 𝑘+1‖ ‖ satisfies the requirements.
Corollary. In finite-dimensional inner product spaces, there always exists an orthonormal basis. In particular, any orthonormal set of vectors can be extended into an orthonormal basis.
Remark. Let 𝐴∈𝑀𝑛(ℝ) be a real-valued (or complex-valued) matrix. Then, the column vectors of 𝐴are orthogonal if 𝐴⊺𝐴= 𝐼(or 𝐴⊺𝐴= 𝐼in the complex-valued case).
10.6. Orthogonality of matrices Definition. A matrix 𝐴∈𝑀𝑛(ℝ) is orthogonal if 𝐴⊺𝐴= 𝐼, hence 𝐴⊺= 𝐴−1. A matrix 𝐴∈𝑀𝑛(ℂ) is unitary if 𝐴⊺𝐴= 𝐼, hence 𝐴† = 𝐴−1.
Proposition. Let 𝐴be a square, non-singular, real-valued (or complex-valued) matrix. Then 𝐴can be written as 𝐴= 𝑅𝑇where 𝑇is upper triangular and 𝑅is orthogonal (or respectively unitary).
Proof. We apply the Gram–Schmidt process to the column vectors of the matrix. This gives us an orthonormal set of vectors, which gives an upper triangular matrix in this new basis.
10.7. Orthogonal complement and projection Definition. Let 𝑉be an inner product space. Let 𝑉 1, 𝑉 2 ≤𝑉. Then we say that 𝑉is the orthogonal direct sum of 𝑉 1 and 𝑉 2 if 𝑉= 𝑉 1 ⊕𝑉 2 and for all vectors 𝑣1 ∈𝑉 1, 𝑣2 ∈𝑉 2 we have ⟨𝑣1, 𝑣2⟩= 0. When this holds, we write 𝑉= 𝑉 1 ⟂ ⊕𝑉 2.
Remark. If for all vectors 𝑣1, 𝑣2 we have ⟨𝑣1, 𝑣2⟩= 0, then 𝑣∈𝑉 1 ∩𝑉 2 ⟹‖𝑣‖2 = 0 ⟹ 𝑣= 0. Hence the sum is always direct if the subspaces are orthogonal.
419 VII. Linear Algebra Definition. Let 𝑉be an inner product space and let 𝑊≤𝑉. We define the orthogonal of 𝑊to be 𝑊⟂= {𝑣∈𝑉∶∀𝑤∈𝑊, ⟨𝑣, 𝑤⟩= 0} Lemma. For any inner product space 𝑉and any subspace 𝑊≤𝑉, we have 𝑉= 𝑊 ⟂ ⊕𝑊⟂.
Proof. First note that 𝑊⟂≤𝑉. Then, if 𝑤∈𝑊, 𝑤∈𝑊⟂, we have ‖𝑤‖2 = ⟨𝑤, 𝑤⟩= 0 since they are orthogonal, so the vector subspaces intersect only in the zero vector. Now, we need to show 𝑉= 𝑊+ 𝑊⟂. Let (𝑒1, … , 𝑒𝑘) be an orthonormal basis of 𝑊and extend it into (𝑒1, … , 𝑒𝑘, 𝑒𝑘+1, … , 𝑒𝑛) which can be made orthonormal. Then, (𝑒𝑘+1, … , 𝑒𝑛) are elements of 𝑊⟂and form a basis.
10.8. Projection maps Definition. Suppose 𝑉= 𝑈⊕𝑊, so 𝑈is a complement of 𝑊in 𝑉. Then, we define 𝜋∶𝑉→𝑊which maps 𝑣= 𝑢+ 𝑤to 𝑤. This is well defined, since the sum is direct. 𝜋is linear, and 𝜋2 = 𝜋. We say that 𝜋is the projection operator onto 𝑊.
Remark. The map 𝜄−𝜋is the projection onto 𝑈, where 𝜄is the identity map.
Lemma. Let 𝑉be an inner product space. Let 𝑊≤𝑉be a finite-dimensional subspace.
Let (𝑒1, … , 𝑒𝑘) be an orthonormal basis for 𝑊. Then, (i) 𝜋(𝑣) = ∑ 𝑘 𝑖=1 ⟨𝑣, 𝑒𝑖⟩𝑒𝑖; and (ii) for all 𝑣∈𝑉, 𝑤∈𝑊, ‖𝑣−𝜋(𝑣)‖ ≤‖𝑣−𝑤‖ with equality if and only if 𝑤= 𝜋(𝑣), hence 𝜋(𝑣) is the point in 𝑊closest to 𝑣.
Proof. We define 𝜋(𝑣) = ∑ 𝑘 𝑖=1 ⟨𝑣, 𝑒𝑖⟩𝑒𝑖. Since 𝑊= ⟨{𝑒𝑘}⟩, 𝜋(𝑣) ∈𝑊for all 𝑣∈𝑉. Then, 𝑣= (𝑣−𝜋(𝑣))+𝜋(𝑣) has a term in 𝑊. We claim that the remaining term is in the orthogonal; 𝑣−𝜋(𝑣) ∈𝑊⟂. Indeed, we must show ⟨𝑣−𝜋(𝑣), 𝑤⟩= 0 for all 𝑤∈𝑊. Equivalently, ⟨𝑣−𝜋(𝑣), 𝑒𝑖⟩= 0 for all basis vectors 𝑒𝑖of 𝑊. We can explicitly compute ⟨𝑣−𝜋(𝑣), 𝑒𝑗⟩= ⟨𝑣, 𝑒𝑗⟩−⟨ 𝑘 ∑ 𝑖=1 ⟨𝑣, 𝑒𝑖⟩𝑒𝑖, 𝑒𝑗⟩= ⟨𝑣, 𝑒𝑗⟩− 𝑘 ∑ 𝑖=1 ⟨𝑣, 𝑒𝑖⟩⟨𝑒𝑖, 𝑒𝑗⟩= ⟨𝑣, 𝑒𝑗⟩−⟨𝑣, 𝑒𝑗⟩= 0 Hence, 𝑣= (𝑣−𝜋(𝑣)) + 𝜋(𝑣) is a decomposition into 𝑊and 𝑊⟂. Since 𝑊∩𝑊⟂= {0}, we have 𝑉= 𝑊 ⟂ ⊕𝑊⟂. For the second part, let 𝑣∈𝑉, 𝑤∈𝑊, and we compute ‖𝑣−𝑤‖2 = ‖ ‖ ‖ ‖ 𝑣−𝜋(𝑣) ⏟ ⎵ ⏟ ⎵ ⏟ ∈𝑊⟂ + 𝜋(𝑣) −𝑤 ⏟⎵⏟⎵⏟ ∈𝑊 ‖ ‖ ‖ ‖ 2 = ‖𝑣−𝜋(𝑣)‖2 + ‖𝜋(𝑣) −𝑤‖2 ≥‖𝑣−𝜋(𝑣)‖2 with equality if and only if 𝑤= 𝜋(𝑣).
420 10. Inner product spaces 10.9. Adjoint maps Definition. Let 𝑉, 𝑊be finite-dimensional inner product spaces. Let 𝛼∈𝐿(𝑉, 𝑊). Then there exists a unique linear map 𝛼⋆∶𝑊→𝑉such that for all 𝑣, 𝑤∈𝑉, 𝑊, ⟨𝛼(𝑣), 𝑤⟩= ⟨𝑣, 𝛼⋆(𝑤)⟩ Moreover, if 𝐵is an orthonormal basis of 𝑉, and 𝐶is an orthonormal basis of 𝑊, then [𝛼⋆]𝐶,𝐵= ([𝛼]𝐵,𝐶) ⊺ Proof. Let 𝐵= (𝑣1, … , 𝑣𝑛) and 𝐶= (𝑤1, … , 𝑤𝑚) and 𝐴= [𝛼]𝐵,𝐶= (𝑎𝑖𝑗). To check existence, we define [𝛼⋆]𝐶,𝐵= 𝐴 ⊺ = (𝑐𝑖𝑗) and explicitly check the definition. By orthogonality, ⟨𝛼(∑𝜆𝑖𝑣𝑖), ∑𝜇𝑗𝑤𝑗⟩= ⟨∑ 𝑖,𝑘 𝜆𝑖𝑎𝑘𝑖𝑤𝑘, ∑ 𝑗 𝜇𝑗𝑤𝑗⟩= ∑ 𝑖,𝑗 𝜆𝑖𝑎𝑗𝑖𝜇𝑗 Then, ⟨∑𝜆𝑖𝑣𝑖, 𝛼⋆(∑𝜇𝑗𝑤𝑗)⟩= ⟨∑ 𝑖 𝜆𝑖𝑣𝑖, ∑ 𝑗,𝑘 𝜇𝑗𝑐𝑘𝑗𝑣𝑘⟩= ∑ 𝑖,𝑗 𝜆𝑖𝑐𝑖𝑗𝜇𝑗 So equality requires 𝑐𝑖𝑗= 𝑎𝑗𝑖. Uniqueness follows from the above; the expansions are equi-valent for any vector if and only if 𝑐𝑖𝑗= 𝑎𝑗𝑖.
Remark. The same notation, 𝛼⋆, is used for the adjoint as just defined, and the dual map as defined before. If 𝑉, 𝑊are real product inner spaces and 𝛼∈𝐿(𝑉, 𝑊), we define 𝜓∶𝑉→ 𝑉⋆such that 𝜓(𝑣)(𝑥) = ⟨𝑥, 𝑣⟩and similarly for 𝑊. Then we can check that the adjoint for 𝛼is given by the composition of 𝜓from 𝑉→𝑉⋆, then applying the dual, then applying the inverse of 𝜓for 𝑊.
10.10. Self-adjoint and isometric maps Definition. Let 𝑉be a finite-dimensional inner product space, and 𝛼be an endomorphism of 𝑉. Let 𝛼⋆∈𝐿(𝑉) be the adjoint map. Then, (i) the condition ⟨𝛼𝑣, 𝑤⟩= ⟨𝑣, 𝛼𝑤⟩is equivalent to the condition 𝛼= 𝛼⋆, and such an 𝛼 is called self-adjoint (for ℝwe call such endomorphisms symmetric, and for ℂwe call such endomorphisms Hermitian); (ii) the condition ⟨𝛼𝑣, 𝛼𝑤⟩= ⟨𝑣, 𝑤⟩is equivalent to the condition 𝛼⋆= 𝛼−1, and such an 𝛼is called an isometry (for ℝit is called orthogonal, and for ℂit is called unitary).
Proposition. The conditions for isometries defined as above are equivalent.
421 VII. Linear Algebra Proof. Suppose ⟨𝛼𝑣, 𝛼𝑤⟩= ⟨𝑣, 𝑤⟩. Then for 𝑣= 𝑤, we find ‖𝛼𝑣‖2 = ‖𝑣‖2, so 𝛼preserves the norm. In particular, this implies ker 𝛼= {0}. Since 𝛼is an endomorphism and 𝑉is finite-dimensional, 𝛼is bijective. Then for all 𝑣, 𝑤∈𝑉, ⟨𝑣, 𝛼⋆(𝑤)⟩= ⟨𝛼𝑣, 𝑤⟩= ⟨𝛼𝑣, 𝛼(𝛼−1(𝑤))⟩= ⟨𝑣, 𝛼−1(𝑤)⟩ Hence 𝛼⋆= 𝛼−1. Conversely, if 𝛼⋆= 𝛼−1 we have ⟨𝛼𝑣, 𝛼𝑤⟩= ⟨𝑣, 𝛼⋆(𝛼𝑤)⟩= ⟨𝑣, 𝑤⟩ as required.
Remark. Using the polarisation identity, we can show that 𝛼is isometric if and only if for all 𝑣∈𝑉, ‖𝛼(𝑣)‖ = ‖𝑣‖.
Lemma. Let 𝑉be a finite-dimensional real (or complex) inner product space. Then for 𝛼∈𝐿(𝑉), (i) 𝛼is self-adjoint if and only if for all orthonormal bases 𝐵of 𝑉, we have [𝛼]𝐵is sym-metric (or Hermitian); (ii) 𝛼is an isometry if and only if for all orthonormal bases 𝐵of 𝑉, we have [𝛼]𝐵is ortho-gonal (or unitary).
Proof. Let 𝐵be an orthonormal basis for 𝑉. Then we know [𝛼⋆]𝐵= [𝛼]† 𝐵. We can then check that [𝛼]† 𝐵= [𝛼]𝐵and [𝛼]† 𝐵= [𝛼]−1 𝐵respectively.
Definition. For 𝐹= ℝ, we define the orthogonal group of 𝑉by 𝑂(𝑉) = {𝛼∈𝐿(𝑉)∶𝛼is an isometry} Note that 𝑂(𝑉) is bijective with the set of orthogonal bases of 𝑉. For 𝐹= ℂ, we define the unitary group of 𝑉by 𝑈(𝑉) = {𝛼∈𝐿(𝑉)∶𝛼is an isometry} Again, note that 𝑈(𝑉) is bijective with the set of orthogonal bases of 𝑉.
10.11. Spectral theory for self-adjoint maps Spectral theory is the study of the spectrum of operators. Recall that in finite-dimensional inner product spaces 𝑉, 𝑊, 𝛼∈𝐿(𝑉, 𝑊) yields the adjoint 𝛼⋆∈𝐿(𝑊, 𝑉) such that for all 𝑣∈𝑉, 𝑤∈𝑊, we have ⟨𝛼(𝑣), 𝑤⟩= ⟨𝑣, 𝛼⋆(𝑤)⟩.
Lemma. Let 𝑉be a finite-dimensional inner product space. Let 𝛼∈𝐿(𝑉) be a self-adjoint endomorphism. Then 𝛼has real eigenvalues, and eigenvectors of 𝛼with respect to different eigenvalues are orthogonal.
422 10. Inner product spaces Proof. Suppose 𝜆∈ℂ, 𝑣∈𝑉nonzero such that 𝛼(𝑣) = 𝜆𝑣. Then, ⟨𝜆𝑣, 𝑣⟩= 𝜆‖𝑣‖2 and also ⟨𝛼𝑣, 𝑣⟩= ⟨𝑣, 𝛼𝑣⟩= ⟨𝑣, 𝜆𝑣⟩= 𝜆‖𝑣‖2 Hence 𝜆= 𝜆since 𝑣≠0. Now, suppose 𝜇≠𝜆and 𝑤∈𝑉nonzero such that 𝛼(𝑤) = 𝜇𝑤.
Then, 𝜆⟨𝑣, 𝑤⟩= ⟨𝛼𝑣, 𝑤⟩= ⟨𝑣, 𝛼𝑤⟩= 𝜇⟨𝑣, 𝑤⟩= 𝜇⟨𝑣, 𝑤⟩ So if 𝜆≠𝜇we must have ⟨𝑣, 𝑤⟩= 0.
Theorem (spectral theorem for self-adjoint maps). Let 𝑉be a finite-dimensional inner product space. Let 𝛼∈𝐿(𝑉) be self-adjoint. Then 𝑉has an orthonormal basis of eigen-vectors of 𝛼. Hence 𝛼is diagonalisable in an orthonormal basis.
Proof. We will consider induction on the dimension of 𝑉. Suppose 𝐴= [𝛼]𝐵with respect to the fundamental basis 𝐵. By the fundamental theorem of algebra, we know that 𝜒𝐴(𝜆) has a (complex) root. But since 𝜆is an eigenvalue of 𝛼and 𝛼is self-adjoint, 𝜆∈ℝ. Now, we choose an eigenvector 𝑣1 = 𝑉∖{0} such that 𝛼(𝑣1) = 𝜆𝑣1. We can set ‖𝑣1‖ = 1 by linearity.
Let 𝑈= ⟨𝑣1⟩ ⟂≤𝑉. We then observe that 𝑈is stable by 𝛼; 𝛼(𝑈) ≤𝑈. Indeed, let 𝑢∈𝑈.
Then ⟨𝛼(𝑢), 𝑣1⟩= ⟨𝑢, 𝛼(𝑣1)⟩= 𝜆⟨𝑢, 𝑣1⟩= 0 by orthogonality. Hence 𝛼(𝑢) ∈𝑈. We can then restrict 𝛼to the domain 𝑈, and by induction we can then choose an orthonormal basis of eigenvectors for 𝑈. Since 𝑉= ⟨𝑣1⟩ ⟂ ⊕𝑈we have an orthonormal basis of eigenvectors for 𝑉 when including 𝑣1.
Corollary. Let 𝑉be a finite-dimensional inner product space. Let 𝛼∈𝐿(𝑉) be self-adjoint.
Then 𝑉is the orthogonal direct sum of the eigenspaces of 𝛼.
10.12. Spectral theory for unitary maps Lemma. Let 𝑉be a complex inner product space. Let 𝛼be unitary, so 𝛼⋆= 𝛼−1. Then all eigenvalues of 𝛼have unit norm. Eigenvectors corresponding to different eigenvalues are orthogonal.
Proof. Let 𝜆∈ℂ, 𝑣∈𝑉∖{0} such that 𝛼(𝑣) = 𝜆𝑣. First, 𝜆≠0 since 𝛼is invertible, and in particular ker 𝛼= {0}. Since 𝑣= 𝜆𝛼−1(𝑣), we can compute 𝜆⟨𝑣, 𝑣⟩= ⟨𝜆𝑣, 𝑣⟩= ⟨𝛼𝑣, 𝑣⟩= ⟨𝑣, 𝛼−1𝑣⟩= ⟨𝑣, 1 𝜆𝑣⟩= 1 𝜆 ⟨𝑣, 𝑣⟩ Hence (𝜆𝜆−1)‖𝑣‖2 = 0 giving |𝜆| = 1. Further, suppose 𝜇∈ℂand 𝑤∈𝑉∖{0} such that 𝛼(𝑤) = 𝜇𝑤, 𝜆≠𝜇. Then 𝜆⟨𝑣, 𝑤⟩= ⟨𝜆𝑣, 𝑤⟩= ⟨𝛼𝑣, 𝑤⟩= ⟨𝑣, 𝛼−1𝑤⟩= ⟨𝑣, 1 𝜇𝑤⟩= 1 𝜇⟨𝑣, 𝑤⟩= 𝜇⟨𝑣, 𝑤⟩ since 𝜇𝜇= 1.
423 VII. Linear Algebra Theorem (spectral theorem for unitary maps). Let 𝑉be a finite-dimensional complex inner product space. Let 𝛼∈𝐿(𝑉) be unitary. Then 𝑉has an orthonormal basis of eigenvectors of 𝛼. Hence 𝛼is diagonalisable in an orthonormal basis.
Proof. Let 𝐴= [𝛼]𝐵where 𝐵is an orthonormal basis. Then 𝜒𝐴(𝜆) has a complex root 𝜆. As before, let 𝑣1 ≠0 such that 𝛼(𝑣1) = 𝜆𝑣1 and ‖𝑣1‖ = 1. Let 𝑈= ⟨𝑣1⟩ ⟂, and we claim that 𝛼(𝑈) = 𝑈. Indeed, let 𝑢∈𝑈, and we find ⟨𝛼(𝑢), 𝑣1⟩= ⟨𝑢, 𝛼−1(𝑣1)⟩= ⟨𝑢, 1 𝜆𝑣1⟩= 1 𝜆 ⟨𝑢, 𝑣1⟩ Since ⟨𝑢, 𝑣1⟩= 0, we have 𝛼(𝑢) ∈𝑈. Hence, 𝛼restricted to 𝑈is a unitary endomorphism of 𝑈. By induction we have an orthonormal basis of eigenvectors of 𝛼for 𝑈and hence for 𝑉.
Remark. We used the fact that the field is complex to find an eigenvalue. In general, a real-valued orthonormal matrix 𝐴giving 𝐴𝐴⊺= 𝐼cannot be diagonalised over ℝ. For example, consider 𝐴= (cos 𝜃 −sin 𝜃 sin 𝜃 cos 𝜃) This is orthogonal and normalised. However, 𝜒𝐴(𝜆) = 1+2𝜆cos 𝜃+𝜆2 hence 𝜆= 𝑒±𝑖𝜃which are complex in the general case.
10.13. Application to bilinear forms We wish to extend the previous statements about spectral theory into statements about bi-linear forms.
Corollary. Let 𝐴∈𝑀𝑛(ℝ) (or 𝑀𝑛(ℂ)) be a symmetric (or respectively Hermitian) matrix.
Then there exists an orthonormal (respectively unitary) matrix 𝑃such that 𝑃⊺𝐴𝑃(or 𝑃†𝐴𝑃) is diagonal with real-valued entries.
Proof. Using the standard inner product, 𝐴∈𝐿(𝐹𝑛) is self-adjoint and hence there exists an orthonormal basis 𝐵of 𝐹𝑛such that 𝐴is diagonal in this basis. Let 𝑃= (𝑣1, … , 𝑣𝑛) be the matrix of this basis. Since 𝐵is orthonormal, 𝑃is orthogonal (or unitary). The result follows from the fact that 𝑃−1𝐴𝑃is diagonal. The eigenvalues are real, hence the diagonal matrix is real.
Corollary. Let 𝑉be a finite-dimensional real (or complex) inner product space. Let 𝜙∶𝑉× 𝑉→𝐹be a symmetric (or Hermitian) bilinear form. Then, there exists an orthonormal basis 𝐵of 𝑉such that [𝜙]𝐵is diagonal.
Proof. 𝐴⊺= 𝐴(or respectively 𝐴† = 𝐴), hence there exists an orthogonal (respectively unitary) matrix 𝑃such that 𝑃−1𝐴𝑃is diagonal. Let (𝑣𝑖) be the 𝑖th row of 𝑃−1 = 𝑃⊺(or 𝑃†).
Then (𝑣1, … , 𝑣𝑛) is an orthonormal basis 𝐵of 𝑉such that [𝜙]𝑉is this diagonal matrix.
424 10. Inner product spaces Remark. The diagonal entries of 𝑃−1𝐴𝑃are the eigenvalues of 𝐴. Moreover, we can define the signature 𝑠(𝜙) to be the difference between the number of positive eigenvalues of 𝐴and the number of negative eigenvalues of 𝐴.
10.14. Simultaneous diagonalisation Corollary. Let 𝑉be a finite-dimensional real (or complex) vector space. Let 𝜙, 𝜓be sym-metric (or Hermitian) bilinear forms on 𝑉. Let 𝜙be positive definite. Then there exists a basis (𝑣1, … , 𝑣𝑛) of 𝑉with respect to which 𝜙and 𝜓are represented with a diagonal matrix.
Proof. Since 𝜙is positive definite, 𝑉equipped with 𝜙is a finite-dimensional inner product space where ⟨𝑢, 𝑣⟩= 𝜙(𝑢, 𝑣). Hence, there exists a basis of 𝑉in which 𝜓is represented by a diagonal matrix, which is orthonormal with respect to the inner product defined by 𝜙. Then, 𝜙in this basis is represented by the identity matrix given by 𝜙(𝑣𝑖, 𝑣𝑗) = ⟨𝑣𝑖, 𝑣𝑗⟩= 𝛿𝑖𝑗, which is diagonal.
Corollary. Let 𝐴, 𝐵∈𝑀𝑛(ℝ) (or ℂ) which are symmetric (or Hermitian). Suppose for all 𝑥≠0 we have 𝑥†𝐴𝑥> 0, so 𝐴is positive definite. Then there exists an invertible matrix 𝑄∈𝑀𝑛(ℝ) (or ℂ) such that 𝑄⊺𝐴𝑄(or 𝑄⊺𝐴𝑄) and 𝑄⊺𝐵𝑄(or 𝑄⊺𝐵𝑄) are diagonal.
Proof. 𝐴induces a quadratic form 𝑄(𝑥) = 𝑥†𝐴𝑥which is positive definite by assumption.
Similarly, ˜ 𝑄(𝑥) = 𝑥†𝐵𝑥is induced by 𝐵. Then we can apply the previous corollary and change basis.
425 VIII. Groups, Rings and Modules Lectured in Lent 2022 by Dr. R. Zhou A ring is a an algebraic structure with an addition and multiplication operation. Common examples of rings include ℤ, ℚ, ℝ, ℂ, the Gaussian integers ℤ[𝑖] = {𝑎+ 𝑏𝑖∣𝑎, 𝑏∈ℤ}, the quotient ℤ ⟋ 𝑛ℤ, and the set of polynomials with complex coefficients. We can study factor-isation in a general ring, generalising the idea of factorising integers or polynomials. Cer-tain rings, called unique factorisation domains, have the property like the integers that every nonzero non-invertible element can be expressed as a unique product of irreducibles (in ℤ, the irreducibles are the prime numbers). This property, and many others, are studied in this course.
Modules are like vector spaces, but instead of being defined over a field, they are defined over an arbitrary ring. In particular, every vector space is a module, because every field is a ring. We use the theory built up over the course to prove that every 𝑛-dimensional complex matrix can be written in Jordan normal form.
427 VIII. Groups, Rings and Modules Contents 1.
Review of IA Groups . . . . . . . . . . . . . . . . . . . . . . . . . . 430 1.1.
Definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . 430 1.2.
Cosets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 430 1.3.
Order . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 431 1.4.
Normality and quotients . . . . . . . . . . . . . . . . . . . . . 431 1.5.
Homomorphisms . . . . . . . . . . . . . . . . . . . . . . . . . 431 1.6.
Isomorphisms . . . . . . . . . . . . . . . . . . . . . . . . . . . 431 2.
Simple groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 432 2.1.
Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . 432 3.
Group actions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 433 3.1.
Definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . 433 3.2.
Cayley’s theorem . . . . . . . . . . . . . . . . . . . . . . . . . 434 3.3.
Conjugation actions . . . . . . . . . . . . . . . . . . . . . . . . 434 4.
Alternating groups . . . . . . . . . . . . . . . . . . . . . . . . . . . 436 4.1.
Conjugation in alternating groups . . . . . . . . . . . . . . . . 436 4.2.
Simplicity of alternating groups . . . . . . . . . . . . . . . . . . 436 5.
𝑝-groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 438 5.1.
𝑝-groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 438 5.2.
Sylow theorems . . . . . . . . . . . . . . . . . . . . . . . . . . 439 6.
Matrix groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 441 6.1.
Definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . 441 6.2.
Möbius maps in modular arithmetic . . . . . . . . . . . . . . . 441 6.3.
Properties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 443 7.
Finite abelian groups . . . . . . . . . . . . . . . . . . . . . . . . . . 444 7.1.
Products of cyclic groups . . . . . . . . . . . . . . . . . . . . . 444 8.
Rings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 445 8.1.
Definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . 445 8.2.
Polynomials . . . . . . . . . . . . . . . . . . . . . . . . . . . . 445 8.3.
Homomorphisms . . . . . . . . . . . . . . . . . . . . . . . . . 447 8.4.
Ideals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 447 8.5.
Quotients . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 448 8.6.
Isomorphism theorems . . . . . . . . . . . . . . . . . . . . . . 449 8.7.
Integral domains . . . . . . . . . . . . . . . . . . . . . . . . . 451 8.8.
Maximal ideals . . . . . . . . . . . . . . . . . . . . . . . . . . 453 8.9.
Prime ideals . . . . . . . . . . . . . . . . . . . . . . . . . . . . 453 9.
Factorisation in integral domains . . . . . . . . . . . . . . . . . . . 455 9.1.
Prime and irreducible elements . . . . . . . . . . . . . . . . . . 455 428 9.2.
Principal ideal domains . . . . . . . . . . . . . . . . . . . . . . 456 9.3.
Unique factorisation domains . . . . . . . . . . . . . . . . . . . 458 9.4.
Factorisation in polynomial rings . . . . . . . . . . . . . . . . . 460 9.5.
Eisenstein’s criterion . . . . . . . . . . . . . . . . . . . . . . . 463 10.
Algebraic integers . . . . . . . . . . . . . . . . . . . . . . . . . . . 464 10.1.
Gaussian integers . . . . . . . . . . . . . . . . . . . . . . . . . 464 10.2.
Algebraic integers . . . . . . . . . . . . . . . . . . . . . . . . . 466 11.
Noetherian rings . . . . . . . . . . . . . . . . . . . . . . . . . . . . 468 11.1.
Definition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 468 11.2.
Hilbert’s basis theorem . . . . . . . . . . . . . . . . . . . . . . 468 12.
Modules . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 470 12.1.
Definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . 470 12.2.
Finitely generated modules . . . . . . . . . . . . . . . . . . . . 471 12.3.
Torsion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 472 12.4.
Direct sums . . . . . . . . . . . . . . . . . . . . . . . . . . . . 473 12.5.
Free modules . . . . . . . . . . . . . . . . . . . . . . . . . . . 473 12.6.
Row and column operations . . . . . . . . . . . . . . . . . . . 475 12.7.
Smith normal form . . . . . . . . . . . . . . . . . . . . . . . . 476 12.8.
The structure theorem . . . . . . . . . . . . . . . . . . . . . . 477 12.9.
Primary decomposition theorem . . . . . . . . . . . . . . . . . 479 12.10. Rational canonical form . . . . . . . . . . . . . . . . . . . . . 480 12.11. Jordan normal form . . . . . . . . . . . . . . . . . . . . . . . . 482 12.12. Modules over principal ideal domains (non-examinable) . . . . . 483 429 VIII. Groups, Rings and Modules 1. Review of IA Groups This section contains material covered by IA Groups.
1.1. Definitions A group is a pair (𝐺, ⋅) where 𝐺is a set and ⋅∶𝐺× 𝐺→𝐺is a binary operation on 𝐺, satisfying • 𝑎⋅(𝑏⋅𝑐) = (𝑎⋅𝑏) ⋅𝑐; • there exists 𝑒∈𝐺such that for all 𝑔∈𝐺, we have 𝑔⋅𝑒= 𝑒⋅𝑔= 𝑔; and • for all 𝑔∈𝐺, there exists an inverse ℎ∈𝐺such that 𝑔⋅ℎ= ℎ⋅𝑔= 𝑒.
Remark.
(i) Sometimes, such as in IA Groups, a closure axiom is also specified. However, this is implicit in the type definition of ⋅. In practice, this must normally be checked explicitly.
(ii) Additive and multiplicative notation will be used interchangeably. For additive nota-tion, the inverse of 𝑔is denoted −𝑔, and for multiplicative notation, the inverse is in-stead denoted 𝑔−1. The identity element is sometimes denoted 0 in additive notation and 1 in multiplicative notation.
A subset 𝐻⊆𝐺is a subgroup of 𝐺, written 𝐻≤𝐺, if ℎ⋅ℎ′ ∈𝐻for all ℎ, ℎ′ ∈𝐻, and (𝐻, ⋅) is a group. The closure axiom must be checked, since we are restricting the definition of ⋅to a smaller set.
Remark. A non-empty subset 𝐻⊆𝐺is a subgroup of 𝐺if and only if 𝑎, 𝑏∈𝐻⟹𝑎⋅𝑏−1 ∈𝐻 An abelian group is a group such that 𝑎⋅𝑏= 𝑏⋅𝑎for all 𝑎, 𝑏in the group. The direct product of two groups 𝐺, 𝐻, written 𝐺× 𝐻, is the group over the Cartesian product 𝐺× 𝐻with operation ⋅defined such that (𝑔1, ℎ1) ⋅(𝑔2, ℎ2) = (𝑔1 ⋅𝐺𝑔2, ℎ1 ⋅𝐻ℎ2).
1.2. Cosets Let 𝐻≤𝐺. Then, the left cosets of 𝐻in 𝐺are the sets 𝑔𝐻for all 𝑔∈𝐺. The set of left cosets partitions 𝐺. Each coset has the same cardinality as 𝐻. Lagrange’s theorem states that if 𝐺 is a finite group and 𝐻≤𝐺, we have |𝐺| = |𝐻| ⋅[𝐺∶𝐻], where [𝐺∶𝐻] is the number of left cosets of 𝐻in 𝐺. [𝐺∶𝐻] is known as the index of 𝐻in 𝐺. We can construct Lagrange’s theorem analogously using right cosets. Hence, the index of a subgroup is independent of the choice of whether to use left or right cosets; the number of left cosets is equal to the number of right cosets.
430 1. Review of IA Groups 1.3. Order Let 𝑔∈𝐺. If there exists 𝑛≥1 such that 𝑔𝑛= 1, then the least such 𝑛is the order of 𝐺. If no such 𝑛exists, we say that 𝑔has infinite order. If 𝑔has order 𝑑, then: (i) 𝑔𝑛= 1 ⟹𝑑∣𝑛; (ii) ⟨𝑔⟩= {1, 𝑔, … , 𝑔𝑑−1} ≤𝐺, and by Lagrange’s theorem (if 𝐺is finite) 𝑑∣|𝐺|.
1.4. Normality and quotients A subgroup 𝐻≤𝐺is normal, written 𝐻⊴𝐺, if 𝑔−1𝐻𝑔= 𝐻for all 𝑔∈𝐺. In other words, 𝐻is preserved under conjugation over 𝐺. If 𝐻⊴𝐺, then the set 𝐺 ⟋ 𝐻of left cosets of 𝐻in 𝐺 forms the quotient group. The group action is defined by 𝑔1𝐻⋅𝑔2𝐻= (𝑔1 ⋅𝑔2)𝐻. This can be shown to be well-defined.
1.5. Homomorphisms Let 𝐺, 𝐻be groups. A function 𝜙∶𝐺→𝐻is a group homomorphism if 𝜙(𝑔1⋅𝐺𝑔2) = 𝜙(𝑔1)⋅𝐻 𝜙(𝑔2) for all 𝑔1, 𝑔2 ∈𝐺. The kernel of 𝜙is defined to be ker 𝜙= {𝑔∈𝐺∶𝜙(𝑔) = 1}, and the image of 𝜙is Im 𝜙= {𝜙(𝑔)∶𝑔∈𝐺}. The kernel is a normal subgroup of 𝐺, and the image is a subgroup of 𝐻.
1.6. Isomorphisms An isomorphism is a homomorphism that is bijective. This yields an inverse function, which is of course also an isomorphism. If 𝜑∶𝐺→𝐻is an isomorphism, we say that 𝐺and 𝐻 are isomorphic, written 𝐺≅𝐻. Isomorphism is an equivalence relation. The isomorphism theorems are (i) if 𝜑∶𝐺→𝐻, then 𝐺 ⟋ ker 𝜑≅Im 𝜑; (ii) if 𝐻≤𝐺and 𝑁⊴𝐺, then 𝐻∩𝑁⊴𝐻and 𝐻 ⟋ 𝐻∩𝑁≅𝐻𝑁 ⟋ 𝑁; (iii) if 𝑁≤𝑀≤𝐺such that 𝑁⊴𝐺and 𝑀⊴𝐺, then 𝑀 ⟋ 𝑁⊴𝐺 ⟋ 𝑁, and 𝐺/𝑁 ⟋ 𝑀/𝑁= 𝐺 ⟋ 𝑀.
431 VIII. Groups, Rings and Modules 2. Simple groups 2.1. Introduction If 𝐾⊴𝐺, then studying the groups 𝐾and 𝐺 ⟋ 𝐾give information about 𝐺itself. This approach is available only if 𝐺has nontrivial normal subgroups. It therefore makes sense to study groups with no normal subgroups, since they cannot be decomposed into simpler structures in this way.
Definition. A group 𝐺is simple if {1} and 𝐺are its only normal subgroups.
By convention, we do not consider the trivial group to be a simple group. This is analogous to the fact that we do not consider one to be a prime.
Lemma. Let 𝐺be an abelian group. 𝐺is simple if and only if 𝐺≅𝐶𝑝for some prime 𝑝.
Proof. Certainly 𝐶𝑝is simple by Lagrange’s theorem. Conversely, since 𝐺is abelian, all sub-groups are normal. Let 1 ≠𝑔∈𝐺. Then ⟨𝑔⟩⊴𝐺. Hence ⟨𝑔⟩= 𝐺by simplicity. If 𝐺is infinite, then 𝐺≅ℤ, which is not a simple group; 2ℤ⊲ℤ. Hence 𝐺is finite, so 𝐺≅𝐶𝑜(𝑔).
If 𝑜(𝑔) = 𝑚𝑛for 𝑚, 𝑛≠1, 𝑝, then ⟨𝑔𝑚⟩≤𝐺, contradicting simplicity.
Lemma. If 𝐺is a finite group, then 𝐺has a composition series 1 ≅𝐺0 ⊲𝐺1 ⊲⋯⊲𝐺𝑛= 𝐺 where each quotient 𝐺𝑖+1 ⟋ 𝐺𝑖is simple.
Remark. It is not the case that necessarily 𝐺𝑖be normal in 𝐺𝑖+𝑘for 𝑘≥2.
Proof. We will consider an inductive step on |𝐺|. If |𝐺| = 1, then trivially 𝐺= 1. Conversely, if |𝐺| > 1, let 𝐺𝑛−1 be a normal subgroup of largest possible order not equal to |𝐺|. Then, 𝐺 ⟋ 𝐺𝑛−1 exists, and is simple by the correspondence theorem.
432 3. Group actions 3. Group actions 3.1. Definitions Definition. Let 𝑋be a set. Then Sym(𝑋) is the group of permutations of 𝑋; that is, the group of all bijections of 𝑋to itself under composition. The identity can be written id or id𝑋.
Definition. A group 𝐺is a permutation group of degree 𝑛if 𝐺≤Sym(𝑋) where |𝑋| = 𝑛.
Example. The symmetric group 𝑆𝑛is exactly equal to Sym({1, … , 𝑛}), so is a permutation group of order 𝑛. 𝐴𝑛is also a permutation group of order 𝑛, as it is a subgroup of 𝑆𝑛. 𝐷2𝑛is a permutation group of order 𝑛.
Definition. A group action of a group 𝐺on a set 𝑋is a function 𝛼∶𝐺× 𝑋→𝑋satisfying 𝛼(𝑒, 𝑥) = 𝑥; 𝛼(𝑔1 ⋅𝑔2, 𝑥) = 𝛼(𝑔1, 𝛼(𝑔2, 𝑥)) for all 𝑔1, 𝑔2 ∈𝐺, 𝑥∈𝑋. The group action may be written ∗, defined by 𝑔∗𝑥≡𝛼(𝑔, 𝑥).
Proposition. An action of a group 𝐺on a set 𝑋is uniquely characterised by a group homo-morphism 𝜑∶𝐺→Sym(𝑋).
Proof. For all 𝑔∈𝐺, we can define 𝜑𝑔∶𝑋→𝑋by 𝑥↦𝑔∗𝑥. Then, for all 𝑥∈𝑋, 𝜑𝑔1𝑔2(𝑥) = (𝑔1𝑔2) ∗𝑥= 𝑔1 ∗(𝑔2 ∗𝑥) = 𝜑𝑔1(𝜑𝑔2(𝑥)) Thus 𝜑𝑔1𝑔2 = 𝜑𝑔1 ∘𝜑𝑔2. In particular, 𝜑𝑔∘𝜑𝑔−1 = 𝜑𝑒. We now define 𝜑∶𝐺→Sym(𝑋); 𝜑(𝑔) = 𝜑𝑔⟹𝜑(𝑔)(𝑥) = 𝑔∗𝑥 This is a homomorphism.
Conversely, any group homomorphism 𝜑∶𝐺→Sym(𝑋) induces a group action ∗by 𝑔∗𝑥= 𝜑(𝑔). This yields 𝑒∗𝑥= 𝜑(𝑒)(𝑥) = id 𝑥= 𝑥and (𝑔1𝑔2) ∗𝑥= 𝜑(𝑔1𝑔2)𝑥= 𝜑(𝑔1)𝜑(𝑔2)𝑥= 𝑔1 ∗(𝑔2 ∗𝑥) as required.
Definition. The homomorphism 𝜑∶𝐺→Sym(𝑋) defined in the above proof is called a permutation representation of 𝐺.
Definition. Let 𝐺↷𝑋. Then, (i) the orbit of 𝑥∈𝑋is Orb𝐺(𝑥) = {𝑔∗𝑥∶𝑔∈𝐺} ⊆𝑋; (ii) the stabiliser of 𝑥∈𝑋is 𝐺𝑥= {𝑔∈𝐺∶𝑔∗𝑥= 𝑥} ≤𝐺.
Theorem (Orbit-stabiliser theorem). The orbit Orb𝐺(𝑥) bijects with the set 𝐺 ⟋ 𝐺𝑥of left cosets of 𝐺𝑥in 𝐺(which may not be a quotient group). In particular, if 𝐺is finite, we have |𝐺| = |Orb(𝑥)| ⋅|𝐺𝑥| 433 VIII. Groups, Rings and Modules Example. If 𝐺is the group of symmetries of a cube and we let 𝑋be the set of vertices in the cube, 𝐺↷𝑋. Here, for all 𝑥∈𝑋, |Orb(𝑥)| = 8 and |𝐺𝑥| = 6 (including reflections), hence |𝐺| = 48.
Remark. Note that ker 𝜑= ⋂𝑥∈𝑋𝐺𝑥. The kernel of the permutation representation 𝜑is also referred to as the kernel of the group action itself. If the kernel is trivial the action is said to be faithful.
The orbits partition 𝑋. In particular, if there is exactly one orbit, the group action is said to be transitive.
Note that 𝐺𝑔∗𝑥= 𝑔𝐺𝑥𝑔−1. Hence, if 𝑥, 𝑦lie in the same orbit, their stabilisers are conjugate.
Example. 𝐺acts on itself by left multiplication. This is known as the left regular action. The kernel is trivial, hence the action is faithful. The action is transitive, since for all 𝑔1, 𝑔2 ∈𝐺, the element 𝑔2𝑔−1 1 maps 𝑔1 to 𝑔2.
3.2. Cayley’s theorem Theorem (Cayley’s theorem). Any finite group 𝐺is a permutation group of order |𝐺|; it is isomorphic to a subgroup of 𝑆|𝐺|.
Example. Let 𝐻≤𝐺. Then 𝐺↷𝐺 ⟋ 𝐻by left multiplication, where 𝐺 ⟋ 𝐻is the set of left cosets of 𝐻in 𝐺. This is known as the left coset action. This action is transitive using the construction above for the left regular action. We have ker 𝜑= ⋂𝑥∈𝐺𝑥𝐻𝑥−1, which is the largest normal subgroup of 𝐺contained within 𝐻.
Theorem. Let 𝐺be a non-abelian simple group, and 𝐻≤𝐺with index 𝑛> 1. Then 𝑛≥5 and 𝐺is isomorphic to a subgroup of 𝐴𝑛.
Proof. Let 𝐺↷𝑋= 𝐺 ⟋ 𝐻by left multiplication. Let 𝜑∶𝐺→Sym(𝑋) be the permutation representation associated to this group action. Since 𝐺is simple, ker 𝜑= 1 or ker 𝜑= 𝐺. If ker 𝜑= 𝐺, then Im 𝜑= id, which is a contradiction since 𝐺acts transitively on 𝑋, which has index greater than one. Thus ker 𝜑= 1, and 𝐺≅Im 𝜑≤𝑆𝑛. Since 𝐺≤𝑆𝑛and 𝐴𝑛⊲𝑆𝑛, the second isomorphism theorem shows that 𝐺∩𝐴𝑛⊲𝐺, and 𝐺 ⟋ 𝐺∩𝐴𝑛≅𝐺𝐴𝑛 ⟋ 𝐴𝑛≤𝑆𝑛 ⟋ 𝐴𝑛≅𝐶2 Since 𝐺is simple, 𝐺∩𝐴𝑛= 1 or 𝐺∩𝐴𝑛= 𝐺. If 𝐺∩𝐴𝑛= 1, then 𝐺is isomorphic to a subgroup of 𝐶2, but this is false, since 𝐺is non-abelian. Hence 𝐺∩𝐴𝑛= 𝐺so 𝐺≤𝐴𝑛.
Finally, if 𝑛≤4 we can check manually that 𝐴𝑛is not simple; 𝐴𝑛has no non-abelian simple subgroups.
3.3. Conjugation actions Example. Let 𝐺↷𝐺by conjugation, so 𝑔∗𝑥= 𝑔𝑥𝑔−1. This is known as the conjugation action.
434 3. Group actions Definition. The orbit of the conjugation action is called the conjugacy class of a given ele-ment 𝑥∈𝐺, written ccl𝐺(𝑥). The stabiliser of the conjugation action is the set 𝐶𝑥of elements which commute with a given element 𝑥, called the centraliser of 𝑥in 𝐺. The kernel of 𝜑is the set 𝑍(𝐺) of elements which commute with all elements in 𝑥, which is the centre of 𝐺.
This is always a normal subgroup.
Remark. 𝜑∶𝐺→𝐺satisfies 𝜑(𝑔)(ℎ1ℎ2) = 𝑔ℎ1ℎ2𝑔−1 = ℎℎ1𝑔−1𝑔ℎ2𝑔−1 = 𝜑(𝑔)(ℎ1)𝜑(𝑔)(ℎ2) Hence 𝜑(𝑔) is a group homomorphism for all 𝑔. It is also a bijection, hence 𝜑(𝑔) is an iso-morphism from 𝐺→𝐺.
Definition. An isomorphism from a group to itself is known as an automorphism. We define Aut(𝐺) to be the set of all group automorphisms of a given group. This set is a group.
Note, Aut(𝐺) ≤Sym(𝐺), and the 𝜑∶𝐺→Sym(𝐺) above has image in Aut(𝐺).
Example. Let 𝑋be the set of subgroups of 𝐺. Then 𝐺↷𝑋by conjugation: 𝑔∗𝐻= 𝑔𝐻𝑔−1.
The stabiliser of a subgroup 𝐻is {𝑔∈𝐺∶𝑔𝐻𝑔−1 = 𝐻} = 𝑁𝐺(𝐻), called the normaliser of 𝐻 in 𝐺. The normaliser of 𝐻is the largest subgroup of 𝐺that contains 𝐻as a normal subgroup.
In particular, 𝐻⊲𝐺if and only if 𝑁𝐺(𝐻) = 𝐺.
435 VIII. Groups, Rings and Modules 4. Alternating groups 4.1. Conjugation in alternating groups We know that elements in 𝑆𝑛are conjugate if and only if they have the same cycle type.
However, elements of 𝐴𝑛that are conjugate in 𝑆𝑛are not necessarily conjugate in 𝐴𝑛. Let 𝑔∈𝐴𝑛. Then 𝐶𝐴𝑛(𝑔) = 𝐶𝑆𝑛(𝑔) ∩𝐴𝑛. There are two possible cases.
• If there exists an odd permutation that commutes with 𝑔, then 2| |𝐶𝐴𝑛| |(𝑔) = | |𝐶𝑆𝑛| |(𝑔).
By the orbit-stabiliser theorem, | |ccl𝐴𝑛(𝑔)| | = | |ccl𝑆𝑛(𝑔)| |.
• If there is no odd permutation that commutes with 𝑔, we have | |𝐶𝐴𝑛| |(𝑔) = | |𝐶𝑆𝑛| |(𝑔).
Similarly, 2| |ccl𝐴𝑛(𝑔)| | = | |ccl𝑆𝑛(𝑔)| |.
Example. For 𝑛= 5, the product (1 2)(3 4) commutes with (1 2), and (1 2 3) commutes with (4 5). Both of these elements are odd. So the conjugacy classes of the above inside 𝑆5 and 𝐴5 are the same. However, (1 2 3 4 5) does not commute with any odd permutation. Indeed, if that were true for some ℎ, we would have (1 2 3 4 5) = ℎ(1 2 3 4 5)ℎ−1 = (ℎ(1) ℎ(2) ℎ(3) ℎ(4) ℎ(5)) Hence ℎmust be a 5-cycle in the subgroup of 𝐴5 generated by (1 2 3 4 5).
We can then show that 𝐴5 has conjugacy classes of size 1, 15, 20, 12, 12. If 𝐻⊴𝐴5, |𝐻| must be a sum of the sizes of the above conjugacy classes. By Lagrange’s theorem, |𝐻| must divide 60. We can check explicitly that this is not possible unless |𝐻| = 1 or |𝐻| = 60. Hence 𝐴5 is simple.
4.2. Simplicity of alternating groups Lemma. 𝐴𝑛is generated by 3-cycles.
Proof. All elements of 𝐴𝑛are generated by an even number of transpositions. It therefore suffices to show that a product of two transpositions can be written as a product of 3-cycles.
Explicitly, (𝑎𝑏)(𝑐𝑑) = (𝑎𝑐𝑏)(𝑎𝑐𝑑); (𝑎𝑏)(𝑏𝑐) = (𝑎𝑏𝑐) Lemma. If 𝑛≥5, all 3-cycles in 𝐴𝑛are conjugate (in 𝐴𝑛).
Proof. We claim that every 3-cycle is conjugate to (1 2 3). If (𝑎𝑏𝑐) is a 3-cycle, we have (𝑎𝑏𝑐) = 𝜎(1 2 3)𝜎−1 for some 𝜎∈𝑆𝑛. If 𝜎∈𝐴𝑛, then the proof is finished. Otherwise, 𝜎↦𝜎(4 5) ∈𝐴𝑛suffices, since (4 5) commutes with (1 2 3).
Theorem. 𝐴𝑛is simple for 𝑛≥5.
436 4. Alternating groups Proof. Suppose 1 ≠𝑁⊲𝐴𝑛. To disprove normality, it suffices to show that 𝑁contains a 3-cycle by the lemmas above, since the normality of 𝑁would imply 𝑁contains all 3-cycles and hence all elements of 𝐴𝑛.
Let 1 ≠𝜎∈𝑁, writing 𝜎as a product of disjoint cycles.
(i) Suppose 𝜎contains a cycle of length 𝑟≥4. Without loss of generality, let 𝜎= (1 2 3 … 𝑟)𝜏 where 𝜏fixes 1, … , 𝑟. Now, let 𝛿= (1 2 3). We have 𝜎−1 ⏟ ∈𝑁 𝛿−1𝜎𝛿 ⏟ ∈𝑁 = (𝑟… 2 1)(1 3 2)(1 2 … 𝑟) = (2 3 𝑟) So 𝑁contains a 3-cycle.
(ii) Suppose 𝜎contains two 3-cycles, which can be written without loss of generality as (1 2 3)(4 5 6)𝜏. Let 𝛿= (1 2 4), and then 𝜎−1𝛿−1𝜎𝛿= (1 3 2)(4 6 5)(1 4 2)(1 2 3)(4 5 6)(1 2 4) = (1 2 4 3 6) Therefore, there exists an element of 𝑁which contains a cycle of length 5 ≥4. This reduces the problem to case (i).
(iii) Finally, suppose 𝜎contains two 2-cycles, which will be written (1 2)(3 4)𝜏. Then let 𝛿= (1 2 3) and 𝜎−1𝛿−1𝜎𝛿= (1 2)(3 4)(1 3 2)(1 2)(3 4)(1 2 3) = (1 4)(2 3) = 𝜋 Let 𝜀= (2 3 5). Then 𝜋−1 ⏟ ∈𝑁 𝜀−1𝜋𝜀 ⏟ ∈𝑁 = (1 4)(2 3)(2 5 3)(1 4)(2 3)(2 3 5) = (2 5 3) Thus 𝑁contains a 3-cycle.
There are now three remaining cases, where 𝜎is a transposition, a 3-cycle, or a transposition composed with a 3-cycle. Note that the remaining cases containing transpositions cannot be elements of 𝐴𝑛. If 𝜎is a 3-cycle, we already know 𝐴𝑛contains a 3-cycle, namely 𝜎itself.
437 VIII. Groups, Rings and Modules 5. 𝑝-groups 5.1. 𝑝-groups Definition. Let 𝑝be a prime. A finite group 𝐺is a 𝑝-group if |𝐺| = 𝑝𝑛for 𝑛≥1.
Theorem. If 𝐺is a 𝑝-group, the centre 𝑍(𝐺) is non-trivial.
Proof. For 𝑔∈𝐺, due to the orbit-stabiliser theorem, |ccl(𝑔)||𝐶(𝑔)| = 𝑝𝑛. In particular, |ccl(𝑔)| divides 𝑝𝑛, and they partition 𝐺. Since 𝐺is a disjoint union of conjugacy classes, modulo 𝑝we have |𝐺| ≡number of conjugacy classes of size 1 ≡0 ⟹|𝑍(𝐺)| ≡0 Hence 𝑍(𝐺) has order zero modulo 𝑝so it cannot be trivial. We can check this by noting that 𝑔∈𝑍(𝐺) ⟺𝑥−1𝑔𝑥= 𝑔for all 𝑥, which is true if and only if ccl𝐺(𝑔) = {𝑔}.
Corollary. The only simple 𝑝-groups are the cyclic groups of order 𝑝.
Proof. Let 𝐺be a simple 𝑝-group. Since 𝑍(𝐺) is a normal subgroup of 𝐺, we have 𝑍(𝐺) = 1 or 𝑍(𝐺) = 𝐺. But 𝑍(𝐺) may not be trivial, so 𝑍(𝐺) = 𝐺. This implies 𝐺is abelian. The only abelian simple groups are cyclic of prime order, hence 𝐺≅𝐶𝑝.
Corollary. Let 𝐺be a 𝑝-group of order 𝑝𝑛. Then 𝐺has a subgroup of order 𝑝𝑟for all 𝑟∈ {0, … , 𝑛}.
Proof. Recall that any group 𝐺has a composition series 1 = 𝐺1 ⊲⋯⊲𝐺𝑁= 𝐺where each quotient 𝐺𝑖+1 ⟋ 𝐺𝑖is simple. Since 𝐺is a 𝑝-group, 𝐺𝑖+1 ⟋ 𝐺𝑖is also a 𝑝-group. Each successive quotient is an order 𝑝group by the previous corollary, so we have a composition series of nested subgroups of order 𝑝𝑟for all 𝑟∈{0, … , 𝑛}.
Lemma. Let 𝐺be a group. If 𝐺 ⟋ 𝑍(𝐺) is cyclic, then 𝐺is abelian. This then implies that 𝑍(𝐺) = 𝐺, so in particular 𝐺 ⟋ 𝑍(𝐺) = 1.
Proof. Let 𝑔𝑍(𝐺) be a generator for 𝐺 ⟋ 𝑍(𝐺). Then, each coset of 𝑍(𝐺) in 𝐺is of the form 𝑔𝑟𝑍(𝐺) for some 𝑟∈ℤ. Thus, 𝐺= {𝑔𝑟𝑧∶𝑟∈ℤ, 𝑧∈𝑍(𝐺)}. Now, we multiply two elements of this group and find 𝑔𝑟1𝑧1𝑔𝑟2𝑧2 = 𝑔𝑟1+𝑟2𝑧1𝑧2 = 𝑔𝑟1+𝑟2𝑧2𝑧1 = 𝑧2𝑧1𝑔𝑟1+𝑟2 = 𝑔𝑟2𝑧2𝑔𝑟1𝑧1 So any two elements in 𝐺commute.
Corollary. Any group of order 𝑝2 is abelian.
438 5. 𝑝-groups Proof. Let 𝐺be a group of order 𝑝2. Then |𝑍(𝐺)| ∈{1, 𝑝, 𝑝2}. The centre cannot be trivial as proven above, since 𝐺is a 𝑝-group. If |𝑍(𝐺)| = 𝑝, we have that 𝐺 ⟋ 𝑍(𝐺) is cyclic as it has order 𝑝. Applying the previous lemma, 𝐺is abelian. However, this is a contradiction since the centre of an abelian group is the group itself. If |𝑍(𝐺)| = 𝑝2 then 𝑍(𝐺) = 𝐺and then 𝐺 is clearly abelian.
5.2. Sylow theorems Theorem. Let 𝐺be a finite group of order 𝑝𝑎𝑚where 𝑝is a prime and 𝑝does not divide 𝑚.
Then: (i) The set Syl𝑝(𝐺) = {𝑃≤𝐺∶|𝑃| = 𝑝𝑎} of Sylow 𝑝-subgroups is non-empty.
(ii) All Sylow 𝑝-subgroups are conjugate.
(iii) The amount of Sylow 𝑝-subgroups 𝑛𝑝= | |Syl𝑝(𝐺)| | satisfies 𝑛𝑝≡1 mod 𝑝; 𝑛𝑝∣|𝐺| ⟹𝑛𝑝∣𝑚 Proof.
(i) Let Ω be the set of all subsets of 𝐺of order 𝑝𝑎. We can directly find |Ω| = (𝑝𝑎𝑚 𝑝𝑎) = 𝑝𝑎𝑚 𝑝𝑎⋅𝑝𝑎𝑚−1 𝑝𝑎−1 ⋯𝑝𝑎𝑚−𝑝𝑎+ 1 1 Note that for 0 ≤𝑘< 𝑝𝑎, the numbers 𝑝𝑎𝑚−𝑘and 𝑝𝑎−𝑘are divisible by the same power of 𝑝. In particular, |Ω| is coprime to 𝑝.
Let 𝐺↷Ω by left-multiplication, so 𝑔∗𝑋= {𝑔𝑥∶𝑥∈𝑋}. For any 𝑋∈Ω, the orbit-stabiliser theorem can be applied to show that |𝐺𝑋||orb𝐺(𝑋)| = |𝐺| = 𝑝𝑎𝑚 By the above, there must exist an orbit with size coprime to 𝑝, since orbits partition Ω.
For such an 𝑋, 𝑝𝑎∣|𝐺𝑋|.
Conversely, note that if 𝑔∈𝐺and 𝑥∈𝑋, then 𝑔∈(𝑔𝑥−1)∗𝑋. Hence, we can consider 𝐺= ⋃ 𝑔∈𝐺 𝑔∗𝑋= ⋃ 𝑌∈orb𝐺(𝑋) 𝑌 Thus |𝐺| ≤|orb𝐺(𝑋)| ⋅|𝑋|, giving |𝐺𝑋| = |𝐺| | |orb𝐺(𝑋)| | ≤|𝑋| = 𝑝𝑎.
Combining with the above, we must have |𝐺𝑋| = 𝑝𝑎. In other words, the stabiliser 𝐺𝑋is a Sylow 𝑝-subgroup of 𝐺.
(ii) We will prove a stronger result for this part of the proof. We claim that if 𝑃is a Sylow 𝑝-subgroup and 𝑄≤𝐺is a 𝑝-subgroup, then 𝑄≤𝑔𝑃𝑔−1 for some 𝑔∈𝐺. Indeed, let 𝑄act on the set of left cosets of 𝑃in 𝐺by left multiplication. By the orbit-stabiliser 439 VIII. Groups, Rings and Modules theorem, each orbit has size which divides |𝑄| = 𝑝𝑘for some 𝑘. Hence each orbit has size 𝑝𝑟for some 𝑟.
Since 𝐺 ⟋ 𝑃has size 𝑚, which is coprime to 𝑝, there must exist an orbit of size 1. There-fore there exists 𝑔∈𝐺such that 𝑞∗𝑔𝑃= 𝑔𝑃for all 𝑞∈𝑄. Equivalently, 𝑔−1𝑞𝑔∈𝑃for all 𝑞∈𝑄. This implies that 𝑄≤𝑔𝑃𝑔−1 as required. This then weakens to the second part of the Sylow theorems.
(iii) Let 𝐺act on Syl𝑝(𝐺) by conjugation. Part (ii) of the Sylow theorems implies that this action is transitive. By the orbit-stabiliser theorem, 𝑛𝑝= | |Syl𝑝(𝐺)| | ∣|𝐺|.
Let 𝑃∈Syl𝑝(𝐺). Then let 𝑃act on Syl𝑝(𝐺) by conjugation. Since 𝑃is a Sylow 𝑝-subgroup, the orbits of this action have size dividing |𝑃| = 𝑝𝑎, so the size is some power of 𝑝. To show 𝑛𝑝≡1 mod 𝑝, it suffices to show that {𝑃} is the unique orbit of size 1. Suppose {𝑄} is another orbit of size 1, so 𝑄is a Sylow 𝑝-subgroup which is preserved under conjugation by 𝑃. 𝑃normalises 𝑄, so 𝑃≤𝑁𝐺(𝑄). Notice that 𝑃and 𝑄are both Sylow 𝑝-subgroups of 𝑁𝐺(𝑄). By (ii), 𝑃and 𝑄are conjugate inside 𝑁𝐺(𝑄).
Hence 𝑃= 𝑄since 𝑄⊴𝑁𝐺(𝑄). Thus, |𝑃| is the unique orbit of size 1, so 𝑛𝑝≡1 mod 𝑝 as required.
Corollary. If 𝑛𝑝= 1, then there is only one Sylow 𝑝-subgroup, and it is normal.
Proof. Let 𝑔∈𝐺and 𝑃∈Syl𝑝(𝐺). Then 𝑔𝑃𝑔−1 is a Sylow 𝑝-subgroup, hence 𝑔𝑃𝑔−1 = 𝑃. 𝑃 is normal in 𝐺.
Example. Let 𝐺be a group with |𝐺| = 1000 = 23 ⋅53. Here, 𝑛5 ≡1 mod 5, and 𝑛5 ∣8, hence 𝑛5 = 1. Thus the unique Sylow 5-subgroup is normal. Hence no group of order 1000 is simple.
Example. Let 𝐺be a group with |𝐺| = 132 = 22 ⋅3 ⋅11. 𝑛11 satisfies 𝑛11 ≡1 mod 11 and 𝑛11 ∣12, thus 𝑛11 ∈{1, 12}. Suppose 𝐺is simple. Then 𝑛11 = 12. The amount of Sylow 3-subgroups satisfies 𝑛3 ≡1 mod 3 and 𝑛3 ∣44 so 𝑛3 ∈{1, 4, 22}. Since 𝐺is simple, 𝑛3 ∈{4, 22}.
Suppose 𝑛3 = 4. Then 𝐺↷Syl3(𝐺) by conjugation, and this generates a group homomorph-ism 𝜑∶𝐺→𝑆4. But the kernel of this homomorphism is a normal subgroup of 𝐺, so ker 𝜑 is trivial or 𝐺itself. If ker 𝜑= 𝐺, then Im 𝜑is trivial, contradicting Sylow’s second theorem.
If ker 𝜑= 1, then Im 𝜑has order 132, which is impossible.
Thus 𝑛3 = 22. This means that 𝐺has 22 ⋅(3 −1) = 44 elements of order 3, and further 𝐺 has 12 ⋅(11 −1) = 120 elements of order 11. However, the sum of these two totals is more than the total of 132 elements, so this is a contradiction. Hence 𝐺is not simple.
440 6. Matrix groups 6. Matrix groups 6.1. Definitions Definition. Let 𝐹be a field, such as ℂor ℤ ⟋ 𝑝ℤ. Let 𝐺𝐿𝑛(𝐹) be set of 𝑛× 𝑛invertible matrices over 𝐹, which is called the general linear group. Let 𝑆𝐿𝑛(𝐹) be set of 𝑛× 𝑛matrices with determinant one over 𝐹, which is called the special linear group. 𝑆𝐿𝑛(𝐹) is the kernel of the determinant homomorphism on 𝐺𝐿𝑛(𝐹), so 𝑆𝐿𝑛(𝐹) ⊲𝐺𝐿𝑛(𝐹).
Let 𝑍⊲𝐺𝐿𝑛(𝐹) denote the subgroup of scalar matrices, the group of nonzero multiples of the identity. The group 𝑃𝐺𝐿𝑛(𝐹) = 𝐺𝐿𝑛(𝐹) ⟋ 𝑍is called the projective general linear group.
Let 𝑃𝑆𝐿𝑛(𝐹) = 𝑆𝐿𝑛(𝐹) ⟋ 𝑍∩𝑆𝐿𝑛(𝐹). By the second isomorphism theorem, 𝑃𝑆𝐿𝑛(𝐹) is iso-morphic to 𝑍⋅𝑆𝐿𝑛(𝐹) ⟋ 𝑍, which is a subgroup of 𝑃𝐺𝐿𝑛(𝐹).
Example. Consider the finite group 𝐺= 𝐺𝐿𝑛(ℤ ⟋ 𝑝ℤ). A list of 𝑛vectors in ℤ ⟋ 𝑝ℤare the columns of a matrix 𝐴∈𝐺if and only if the vectors are linearly independent. Hence, by considering dimensionality of subspaces generated by each column, |𝐺| = (𝑝𝑛−1)(𝑝𝑛−𝑝)(𝑝𝑛−𝑝2) ⋯(𝑝𝑛−𝑝𝑛−1) = 𝑝1+2+⋯+(𝑛−1)(𝑝𝑛−1)(𝑝𝑛−1 −1) ⋯(𝑝−1) = 𝑝(𝑛 2) 𝑛 ∏ 𝑖=1 (𝑝𝑖−1) Hence the Sylow 𝑝-subgroups have size 𝑝(𝑛 2). Let 𝑈be the set of upper triangular matrices with ones on the diagonal. This forms a Sylow 𝑝-subgroup of 𝐺, since there are (𝑛 2) entries in a given upper triangular matrix, and there are 𝑝choices for such an entry.
6.2. Möbius maps in modular arithmetic Recall that 𝑃𝐺𝐿2(ℂ) acts on ℂ∪{∞} by Möbius transformations. Likewise, 𝑃𝐺𝐿2(ℤ ⟋ 𝑝ℤ) acts on ℤ ⟋ 𝑝ℤ∪{∞} by Möbius transformations. For a matrix 𝐴= (𝑎 𝑏 𝑐 𝑑) ∈𝐺𝐿2(ℤ ⟋ 𝑝ℤ); 𝐴∶𝑧↦𝑎𝑧+ 𝑏 𝑐𝑧+ 𝑑 Since the scalar matrices act trivially, we obtain an action on the projective general linear group instead of the general linear group. We can represent ∞as an integer, say, 𝑝, for the purposes of constructing a permutation representation.
Lemma. The permutation representation 𝑃𝐺𝐿2(ℤ ⟋ 𝑝ℤ) →𝑆𝑝+1 is injective (and is an iso-morphism if 𝑝= 2 or 𝑝= 3).
441 VIII. Groups, Rings and Modules Proof. Suppose that 𝑎𝑧+𝑏 𝑐𝑧+𝑑= 𝑧for all 𝑧∈ℤ ⟋ 𝑝ℤ∪{∞}. Since 𝑧= 0, we have 𝑏= 0. Since 𝑧= ∞, we find 𝑐= 0. Thus the matrix is diagonal. Finally, since 𝑧= 1, 𝑎 𝑑= 1 hence 𝑎= 𝑑.
Thus the matrix is scalar. So the permutation representation from 𝑃𝐺𝐿2(ℤ ⟋ 𝑝ℤ) has trivial kernel, giving injectivity as required.
If 𝑝= 2 or 𝑝= 3 we can compute the orders of relevant groups manually and show that the permutation representation is an isomorphism.
Lemma. Let 𝑝be an odd prime. Then | |𝑃𝑆𝐿2(ℤ ⟋ 𝑝ℤ)| | = (𝑝−1)𝑝(𝑝+ 1) 2 Proof. By the example above, | |𝐺𝐿2(ℤ ⟋ 𝑝ℤ)| | = 𝑝(𝑝2 −1)(𝑝−1) The homomorphism 𝐺𝐿2(ℤ ⟋ 𝑝ℤ) →(ℤ ⟋ 𝑝ℤ) × given by the determinant is surjective. Since 𝑆𝐿2(ℤ ⟋ 𝑝ℤ) is the kernel of this homomorphism, we have | |𝑆𝐿2(ℤ ⟋ 𝑝ℤ)| | = 𝑝(𝑝−1)(𝑝+ 1) Now, if (𝜆 0 0 𝜆) is an element of the special linear group, then 𝜆2 ≡1 mod 𝑝. Then, 𝑝∣ (𝜆−1)(𝜆+ 1) hence 𝜆≡±1 mod 𝑝. Thus, 𝑍∩𝑆𝐿2(ℤ ⟋ 𝑝ℤ) = {±1} and the elements are distinct since 𝑝> 2. Hence the order of the projective special linear group is half the order of the special linear group as required.
Example. Let 𝐺= 𝑃𝑆𝐿2(ℤ ⟋ 5ℤ). Then by the previous lemma, |𝐺| = 60. Let 𝐺↷ℤ ⟋ 5ℤ∪ {∞} by Möbius transformations. The permutation representation 𝜑∶𝐺→Sym({0, 1, 2, 3, 4, ∞}) is injective, since the permutation representation of 𝑃𝐺𝐿2(ℤ ⟋ 𝑝ℤ) is known to be injective by a previous lemma.
We claim that Im 𝜑⊆𝐴6. Let 𝜓= sgn ∘𝜑. If we can show 𝜓is trivial, Im 𝜑⊆𝐴6. Let ℎ∈𝐺, and suppose ℎhas order 2𝑛𝑚for odd 𝑚. If 𝜓(ℎ𝑚) = 1, then since 𝜓is a group homomorphism we have 𝜓(ℎ)𝑚= 1 giving 𝜓(ℎ) ≠−1 ⟹ 𝜓(ℎ) = 1. So to show 𝜓is trivial, it suffices to show 𝜓(𝑔) = 1 for all 𝑔∈𝐺with order a power of 2. By the second Sylow theorem, if 𝑔has order a power of 2, it is contained in a Sylow 2-subgroup. Then it suffices to show that 𝜓(𝐻) = 1 for all Sylow 2-subgroups 𝐻. But since ker 𝜓is normal and all 442 6. Matrix groups Sylow 2-subgroups are conjugate, it suffices to show 𝜓(𝐻) = 1 for a single Sylow 2-subgroup 𝐻. The Sylow 2-subgroup must have order 4. Hence consider 𝐻= ⟨(2 0 0 3) {±𝐼}, ( 0 1 −1 0) {±𝐼}⟩ Both of these elements square to the identity element inside the projective special linear group. This generates a group of order 4 which is necessarily a Sylow 2-subgroup. We can explicitly compute the action of 𝐻on {0, 1, 2, 3, 4, ∞}.
𝜑((2 0 0 3)) = (1 4)(2 3); 𝜑(( 0 1 −1 0)) = (0 ∞)(1 4) These are products of two transpositions, hence even permutations. Thus 𝜓(𝐻) = 1, proving the claim that 𝐺≤𝐴6. We can prove that for any 𝐺≤𝐴6 of order 60, we have 𝐺≅𝐴5; this is a question from the example sheets.
6.3. Properties The following properties will not be proven in this course.
• 𝑃𝑆𝐿𝑛(ℤ ⟋ 𝑝ℤ) is simple for all 𝑛≥2 and 𝑝prime, except where 𝑛= 2 and 𝑝= 2, 3.
Such groups are called finite groups of Lie type.
• The smallest non-abelian simple groups are 𝐴5 ≅𝑃𝑆𝐿2(ℤ ⟋ 5ℤ), then 𝑃𝑆𝐿2(ℤ ⟋ 7ℤ) ≅ 𝐺𝐿3(ℤ ⟋ 2ℤ) which has order 168.
443 VIII. Groups, Rings and Modules 7. Finite abelian groups 7.1. Products of cyclic groups Theorem. Every finite abelian group is isomorphic to a product of cyclic groups.
The proof for this theorem will be provided later in the course. Note that the isomorphism provided for by the theorem is not unique. An example of such behaviour is the following lemma.
Lemma. Let 𝑚, 𝑛be coprime integers. Then 𝐶𝑚× 𝐶𝑛≅𝐶𝑚𝑛.
Proof. Let 𝑔, ℎbe generators of 𝐶𝑚and 𝐶𝑛. Then consider the element (𝑔, ℎ)𝑘= (𝑔𝑘, ℎ𝑘), which has order 𝑚𝑛. Thus ⟨(𝑔, ℎ)⟩has order 𝑚𝑛. So every element in 𝐶𝑚× 𝐶𝑛is expressible in this way, giving ⟨(𝑔, ℎ)⟩= 𝐶𝑚× 𝐶𝑛.
Corollary. Let 𝐺be a finite abelian group. Then 𝐺≅𝐶𝑛1 × ⋯× 𝐶𝑛𝑘where each 𝑛𝑖is a power of a prime.
Proof. If 𝑛= 𝑝1𝑎1 ⋯𝑝𝑟𝑎𝑟where the 𝑝𝑖are distinct primes, then applying the above lemma inductively gives 𝐶𝑛as a product of cyclic groups which have orders that are powers of primes. We can apply this to the theorem that every finite abelian group is isomorphic to a product of cyclic groups to find the result.
Later, we will prove the following refinement of this theorem.
Theorem. Let 𝐺be a finite abelian group. Then 𝐺≅𝐶𝑑1 × ⋯× 𝐶𝑑𝑡where 𝑑𝑖∣𝑑𝑖+1 for all 𝑖.
Remark. The integers 𝑛1, … , 𝑛𝑘in the corollary above are unique up to ordering. The in-tegers 𝑑1, … , 𝑑𝑡are also unique, assuming that 𝑑1 > 1. The proofs will be omitted.
Example. The abelian groups of order 8 are exactly 𝐶8, 𝐶2 × 𝐶4, and 𝐶2 × 𝐶2 × 𝐶2. The abelian groups of order 12 are, using the corollary above, 𝐶2 × 𝐶2 × 𝐶3, 𝐶4 × 𝐶3, and using the above theorem, 𝐶2 × 𝐶6 and 𝐶12. However, 𝐶2 × 𝐶3 ≅𝐶6 and 𝐶3 × 𝐶4 ≅𝐶12, so the groups derived are isomorphic.
Definition. The exponent of a group 𝐺is the least integer 𝑛≥1 such that 𝑔𝑛= 1 for all 𝑔∈𝐺. Equivalently, the exponent is the lowest common multiple of the orders of elements in 𝐺.
Example. The exponent of 𝐴4 is lcm{2, 3} = 6.
Corollary. Let 𝐺be a finite abelian group. Then 𝐺contains an element which has order equal to the exponent of 𝐺.
Proof. If 𝐺≅𝐶𝑑1 × ⋯× 𝐶𝑑𝑡for 𝑑𝑖∣𝑑𝑖+1, every 𝑔∈𝐺has order dividing 𝑑𝑡. Hence the exponent is 𝑑𝑡, and we can choose a generator of 𝐶𝑑𝑡to obtain an element in 𝐺of the same order.
444 8. Rings 8. Rings 8.1. Definitions Definition. A ring is a triple (𝑅, +, ⋅) where 𝑅is a set and +, ⋅are binary operations 𝑅×𝑅→ 𝑅, satisfying the following axioms.
(i) (𝑅, +) is an abelian group, and we will denote the identity element 0 and the inverse of 𝑥as −𝑥; (ii) (𝑅, ⋅) satisfies the group axioms except for the invertibility axiom, and we will denote the identity element 1 and the inverse of 𝑥as 𝑥−1 if it exists; (iii) for all 𝑥, 𝑦, 𝑧∈𝑅we have 𝑥⋅(𝑦+ 𝑧) = 𝑥⋅𝑦+ 𝑥⋅𝑧and (𝑦+ 𝑧) ⋅𝑥= 𝑦⋅𝑥+ 𝑧⋅𝑥.
If multiplication is commutative, we say that 𝑅is a commutative ring. In this course, we will study only commutative rings.
Remark. For all 𝑥∈𝑅, 0 ⋅𝑥= (0 + 0) ⋅𝑥= 0 ⋅𝑥+ 0 ⋅𝑥⟹0 ⋅𝑥= 0 Further, 0 = 0 ⋅𝑥= (1 + −1) ⋅𝑥= 𝑥+ (−1 ⋅𝑥) ⟹−1 ⋅𝑥= −𝑥 Definition. A subset 𝑆⊆𝑅is a subring, denoted 𝑆≤𝑅, if (𝑆, +, ⋅) is a ring with the same identity elements.
Remark. It suffices to check the closure axioms for addition and multiplication; the other properties are inherited.
Example. ℤ≤ℚ≤ℝ≤ℂare rings. The set ℤ[𝑖] = {𝑎+ 𝑏𝑖∶𝑎, 𝑏∈ℤ} is a subring of ℂ.
This is known as the ring of Gaussian integers. The set ℚ[√2] = {𝑎+ 𝑏√2∶𝑎, 𝑏∈ℚ} is a subring of ℝ.
Example. The set ℤ ⟋ 𝑛ℤis a ring.
Example. Let 𝑅, 𝑆be rings. Then the product 𝑅× 𝑆is a ring under the binary operations (𝑎, 𝑏) + (𝑐, 𝑑) = (𝑎+ 𝑐, 𝑏+ 𝑑); (𝑎, 𝑏) ⋅(𝑐, 𝑑) = (𝑎⋅𝑐, 𝑏⋅𝑑) The additive identity is (0𝑅, 0𝑆) and the multiplicative identity is (1𝑅, 1𝑆). Note that the sub-set 𝑅×{0} is preserved under addition and multiplication, so it is a ring, but it is not a subring because the multiplicative identity is different.
8.2. Polynomials Definition. Let 𝑅be a ring. A polynomial 𝑓over 𝑅is an expression 𝑓= 𝑎0 + 𝑎1𝑋+ 𝑎2𝑋2 + ⋯+ 𝑎𝑛𝑋𝑛 445 VIII. Groups, Rings and Modules for 𝑎𝑖∈𝑅. The term 𝑋is a formal symbol, no substitution of 𝑋for a value will be made.
We could alternatively define polynomials as finite sequences of terms in 𝑅. The degree of a polynomial 𝑓is the largest 𝑛such that 𝑎𝑛≠0. A degree-n polynomial is monic if 𝑎𝑛= 1.
We write 𝑅[𝑋] for the set of all such polynomials over 𝑅. Let 𝑔= 𝑏0 + 𝑏1𝑋+ ⋯+ 𝑏𝑛𝑋𝑛.
Then we define 𝑓+ 𝑔= (𝑎0 + 𝑏0) + (𝑎1 + 𝑏1)𝑋+ ⋯+ (𝑎𝑛+ 𝑏𝑛)𝑋𝑛; 𝑓⋅𝑔= ∑ 𝑖 ( 𝑖 ∑ 𝑗=0 𝑎𝑗𝑏𝑖−𝑗)𝑋𝑖 Then (𝑅[𝑋], +, ⋅) is a ring. The identity elements are the constant polynomials 0 and 1. We can identify the ring 𝑅with the subring of 𝑅[𝑋] of constant polynomials.
Definition. An element 𝑟∈𝑅is a unit if 𝑟has a multiplicative inverse. The units in a ring, denoted 𝑅×, form an abelian group under multiplication. For instance, ℤ× = {±1} and ℚ× = ℚ∖{0}.
Definition. A field is a ring where all nonzero elements are units and 0 ≠1.
Example. ℤ ⟋ 𝑛ℤis a field only if 𝑛is a prime.
Remark. If 𝑅is a ring such that 0 = 1, then every element in the ring is equal to zero. Indeed, 𝑥= 1 ⋅𝑥= 0 ⋅𝑥= 0. Thus, the exclusion of rings with 0 = 1 in the definition of a field simply excludes the trivial ring.
Proposition. Let 𝑓, 𝑔∈𝑅[𝑋] such that the leading coefficient of 𝑔is a unit. Then there exist polynomials 𝑞, 𝑟∈𝑅[𝑋] such that 𝑓= 𝑞𝑔+ 𝑟, where the degree of 𝑟is less than the degree of 𝑔.
Remark. This is the Euclidean algorithm for division, adapted to polynomial rings.
Proof. Let 𝑛be the degree of 𝑓and 𝑚be the degree of 𝑔, so 𝑓= 𝑎𝑛𝑋𝑛+ ⋯+ 𝑎0; 𝑔= 𝑏𝑚𝑋𝑚+ ⋯+ 𝑏0 By assumption, 𝑏𝑚∈𝑅×. If 𝑛< 𝑚then let 𝑞= 0 and 𝑟= 𝑓. Conversely, we have 𝑛≥𝑚.
Consider the polynomial 𝑓 1 = 𝑓−𝑎𝑛𝑏−1 𝑚𝑔𝑋𝑛−𝑚. This has degree at most 𝑛−1. Hence, we can use induction on 𝑛to decompose 𝑓 1 as 𝑓 1 = 𝑞1𝑔+ 𝑟. Thus 𝑓= (𝑞1 + 𝑎𝑛𝑏−1 𝑚𝑋𝑛−𝑚)𝑔+ 𝑟 as required.
Remark. If 𝑅is a field, then every nonzero element of 𝑅is a unit. Therefore, the above algorithm can be applied for all polynomials 𝑔unless 𝑔is the constant polynomial zero.
Example. Let 𝑅be a ring and 𝑋be a set. Then the set of functions 𝑋→𝑅is a ring under (𝑓+ 𝑔)(𝑥) = 𝑓(𝑥) + 𝑔(𝑥); (𝑓⋅𝑔)(𝑥) = 𝑓(𝑥) ⋅𝑔(𝑥) The set of continuous functions ℝ→ℝis a subring of the ring of all functions ℝ→ℝ, since they are closed under addition and multiplication. The set of polynomial functions ℝ→ℝ is also a subring, and we can identify this with the ring ℝ[𝑋].
446 8. Rings Example. Let 𝑅be a ring. Then the power series ring 𝑅⟦𝑋⟧is the set of power series on 𝑋. This is defined similarly to the polynomial ring, but we permit infinitely many nonzero elements in the expansion. The power series is defined formally; we cannot actually carry out infinitely many additions in an arbitrary ring. We instead consider the power series as a sequence of numbers.
Example. Let 𝑅be a ring. Then the ring of Laurent polynomials is 𝑅[𝑋, 𝑋−1] with the re-striction that 𝑎𝑖≠0 for finitely many 𝑖.
8.3. Homomorphisms Definition. Let 𝑅and 𝑆be rings. A function 𝜑∶𝑅→𝑆is a ring homomorphism if (i) 𝜑(𝑟1 + 𝑟2) = 𝜑(𝑟1) + 𝜑(𝑟2); (ii) 𝜑(𝑟1 ⋅𝑟2) = 𝜑(𝑟1) ⋅𝜑(𝑟2); (iii) 𝜑(1𝑅) = 1𝑆.
We can derive that 𝜑(0𝑅) = 0𝑆from (i).
A ring homomorphism is an isomorphism if it is bijective. The kernel of a ring homomorph-ism is ker 𝜑= {𝑟∈𝑅∶𝜑(𝑟) = 0}.
Lemma. Let 𝑅, 𝑆be rings. Then a ring homomorphism 𝜑∶𝑅→𝑆is injective if and only if ker 𝜑= {0}.
Proof. Let 𝜑∶(𝑅, +) →(𝑆, +) be the induced group homomorphism on addition. The result then follows from the corresponding fact about group homomorphisms.
8.4. Ideals Definition. A subset 𝐼⊆𝑅is an ideal, written 𝐼⊴𝑅, if (i) 𝐼is a subgroup of (𝑅, +); (ii) if 𝑟∈𝑅and 𝑥∈𝐼, then 𝑟𝑥∈𝐼.
We say that an ideal is proper if 𝐼≠𝑅.
Lemma. Let 𝜑∶𝑅→𝑆be a ring homomorphism. Then ker 𝜑is an ideal of 𝑅.
Proof. We know that ker 𝜑is a subgroup by the equivalent fact from groups. If 𝑟∈𝑅and 𝑥∈ker 𝜑, then 𝜑(𝑟𝑥) = 𝜑(𝑟)𝜑(𝑥) = 𝜑(𝑟) ⋅0 = 0 Hence 𝑟𝑥∈ker 𝜑.
447 VIII. Groups, Rings and Modules Remark. If 𝐼contains a unit, then the multiplicative identity lies in 𝐼. Then all elements lie in 𝐼. In particular, if 𝐼is a proper ideal, 1 ∉𝐼. Hence a proper ideal 𝐼is not a subring of 𝑅.
Lemma. The ideals in ℤare precisely the subsets of the form 𝑛ℤfor any 𝑛= 0, 1, 2, ….
Proof. First, we can check directly that any subset of the form 𝑛ℤis an ideal. Now, let 𝐼be any nonzero ideal of ℤand let 𝑛be the smallest positive element. Then 𝑛ℤ⊆𝐼. Let 𝑚∈𝐼.
Then by the Euclidean algorithm, 𝑚= 𝑞𝑛+ 𝑟for 𝑞, 𝑟∈ℤand 𝑟∈{0, 1, … , 𝑛−1}. Then 𝑟= 𝑚−𝑞𝑛. We know 𝑞𝑛∈𝐼since 𝑛∈𝐼, so 𝑟∈𝐼. If 𝑟≠0, this contradicts the minimality of 𝑛as chosen above. So 𝐼= 𝑛ℤexactly.
Definition. For an element 𝑎∈𝑅, we write (𝑎) to denote the subset of 𝑅given by mul-tiples of 𝑎; that is, (𝑎) = {𝑟𝑎∶𝑟∈𝑅}. This is an ideal, known as the ideal generated by 𝑎.
More generally, if 𝑎1, … , 𝑎𝑛∈𝑅, then (𝑎1, … , 𝑎𝑛) is the set of elements in 𝑅given by linear combinations of the 𝑎𝑖. This is also an ideal.
Definition. Let 𝐼⊴𝑅. Then 𝐼is principal if there exists some 𝑎∈𝑅such that 𝐼= (𝑎).
8.5. Quotients Theorem. Let 𝐼⊴𝑅. Then the set 𝑅 ⟋ 𝐼of cosets of 𝐼in (𝑅, +) forms the quotient ring under the operations (𝑟1 + 𝐼) + (𝑟2 + 𝐼) = (𝑟1 + 𝑟2) + 𝐼; (𝑟1 + 𝐼) ⋅(𝑟2 + 𝐼) = (𝑟1 ⋅𝑟2) + 𝐼 This ring has the identity elements 0𝑅 ⟋ 𝐼= 0𝑅+ 𝐼; 1𝑅 ⟋ 𝐼= 1𝑅+ 𝐼 Further, the map 𝑅→𝑅 ⟋ 𝐼defined by 𝑟↦𝑟+ 𝐼is a ring homomorphism called the quotient map. The kernel of the quotient map is 𝐼. Hence any ideal is the kernel of some homomorph-ism.
Proof. From the analogous result from groups, the addition defined on the set of cosets yields the group (𝑅 ⟋ 𝐼, +). If 𝑟1 + 𝐼= 𝑟′ 1 + 𝐼and 𝑟2 + 𝐼= 𝑟′ 2 + 𝐼, then 𝑟′ 1 = 𝑟1 + 𝑎1 and 𝑟′ 2 = 𝑟2 + 𝑎2 for some 𝑎1, 𝑎2 ∈𝐼. Then 𝑟′ 1𝑟′ 2 = (𝑟1 + 𝑎1)(𝑟2 + 𝑎2) = 𝑟1𝑟2 + 𝑎1𝑟2 + 𝑟1𝑎2 + 𝑎1𝑎2 Hence (𝑟′ 1𝑟′ 2) + 𝐼= (𝑟1𝑟2) + 𝐼. The remainder of the proof is trivial.
Example. In the integers ℤ, the ideals are 𝑛ℤ. Hence we can form the quotient ring ℤ ⟋ 𝑛ℤ.
The ring ℤ ⟋ 𝑛ℤhas elements 𝑛ℤ, 1+𝑛ℤ, … , (𝑛−1)+𝑛ℤ. Addition and multiplication behave like in modular arithmetic modulo 𝑛.
448 8. Rings Example. Consider the ideal (𝑋) inside the polynomial ring ℂ[𝑋]. This ideal is the set of polynomials with zero constant term. Let 𝑓(𝑋) = 𝑎𝑛𝑋𝑛+ ⋯+ 𝑎0 be an arbitrary element of ℂ[𝑋]. Then 𝑓(𝑋) + 𝑋= 𝑎0 + 𝑋. Thus, there exists a bijection between ℂ[𝑋] ⟋ (𝑋) and ℂ, defined by 𝑓(𝑥)+(𝑋) ↦𝑓(0), with inverse 𝑎↦𝑎+(𝑋). This bijection is a ring homomorph-ism, hence ℂ[𝑋] ⟋ (𝑋) ≅ℂ.
Example. Consider (𝑋2 + 1) ⊲ℝ[𝑋]. For 𝑓(𝑋) = 𝑎𝑛𝑋𝑛+ ⋯+ 𝑎0 ∈ℝ[𝑋], we can apply the Euclidean algorithm to write 𝑓(𝑋) as 𝑞(𝑋)(𝑋2 + 1) + 𝑟(𝑋) where the degree of 𝑟is less than two. Hence 𝑟(𝑋) = 𝑎+ 𝑏𝑋for some real numbers 𝑎and 𝑏. Thus, any element of ℝ[𝑋] ⟋ (𝑋2 + 1) can be written 𝑎+ 𝑏𝑋+ (𝑋2 + 1). Suppose a coset can be represented by two representatives: 𝑎+ 𝑏𝑋+ (𝑋2 + 1) = 𝑎′ + 𝑏′𝑋+ (𝑋2 + 1). Then, 𝑎+ 𝑏𝑋−𝑎′ −𝑏′𝑋= (𝑎−𝑎′) −(𝑏−𝑏′)𝑋= 𝑔(𝑋)(𝑋2 + 1) Hence 𝑔(𝑋) = 0, giving 𝑎−𝑎′ = 0 and 𝑏−𝑏′ = 0. Hence the coset representative is unique. Consider the bijection 𝜑between this quotient ring and the complex numbers given by 𝑎+ 𝑏𝑋+ (𝑋2 + 1) ↦𝑎+ 𝑏𝑖. We can show that 𝜑is a ring homomorphism. Indeed, it preserves addition, and 1 + (𝑋2 + 1) ↦1, so it suffices to check that multiplication is preserved.
𝜑((𝑎+ 𝑏𝑋+ (𝑋2 + 1)) ⋅(𝑐+ 𝑑𝑋+ (𝑋2 + 1))) = 𝜑((𝑎+ 𝑏𝑋)(𝑐+ 𝑑𝑋) + (𝑋2 + 1)) = 𝜑(𝑎𝑐+ (𝑎𝑑+ 𝑏𝑐)𝑋+ 𝑏𝑑(𝑋2 + 1) −𝑏𝑑+ (𝑋2 + 1)) = 𝜑(𝑎𝑐−𝑏𝑑+ (𝑎𝑑+ 𝑏𝑐)𝑋+ (𝑋2 + 1)) = 𝑎𝑐−𝑏𝑑+ (𝑎𝑑+ 𝑏𝑐)𝑖 = (𝑎+ 𝑏𝑖)(𝑐+ 𝑑𝑖) = 𝜑((𝑎+ 𝑏𝑋) + (𝑋2 + 1))𝜑((𝑐+ 𝑑𝑋) + (𝑋2 + 1)) Thus ℝ[𝑋] ⟋ (𝑋2 + 1) ≅ℂ.
8.6. Isomorphism theorems Theorem (first isomorphism theorem). Let 𝜑∶𝑅→𝑆be a ring homomorphism. Then, ker 𝜑⊲𝑅; Im 𝜑≤𝑆; 𝑅 ⟋ ker 𝜑≅Im 𝜑 Proof. We have ker 𝜑⊲𝑅from above. We know that Im 𝜑≤(𝑆, +). Now we show that Im 𝜑 is closed under multiplication.
𝜑(𝑟1)𝜑(𝑟2) = 𝜑(𝑟1𝑟2) ∈Im 𝜑 Finally, 1𝑆= 𝜑(1𝑅) ∈Im 𝜑 449 VIII. Groups, Rings and Modules Hence Im 𝜑is a subring of 𝑆. Let 𝐾= ker 𝜑. Then, we define Φ∶𝑅 ⟋ 𝐾→Im 𝜑by 𝑟+ 𝐾↦ 𝜑(𝑟). By appealing to the first isomorphism theorem from groups, this is well-defined, a bijection, and a group homomorphism under addition. It therefore suffices to show that Φ preserves multiplication and maps the multiplicative identities to each other.
Φ(1𝑅+ 𝐾) = 𝜑(1𝑅) = 1𝑆; Φ((𝑟1 + 𝐾)(𝑟2 + 𝐾)) = Φ(𝑟1𝑟2 + 𝐾) = 𝜑(𝑟1𝑟2) = 𝜑(𝑟1)𝜑(𝑟2) The result follows as required.
Theorem (second isomorphism theorem). Let 𝑅≤𝑆and 𝐽⊲𝑆. Then, 𝑅∩𝐽⊲𝑅; 𝑅+ 𝐽= {𝑟+ 𝑎∶𝑟∈𝑅, 𝑎∈𝐽} ≤𝑆; 𝑅 ⟋ 𝑅∩𝐽≅𝑅+ 𝐽 ⟋ 𝐽≤𝑆 ⟋ 𝐽 Proof. By the second isomorphism theorem for groups, 𝑅+𝐽≤(𝑆, +). Further, 1𝑆= 1𝑆+0𝑆, and since 𝑅is a subring, 1𝑆+ 0𝑆∈𝑅+ 𝐽hence 1𝑆∈𝑅∩𝐽. If 𝑟1, 𝑟2 ∈𝑅and 𝑎1, 𝑎2 ∈𝐽, we have (𝑟1 + 𝑎1)(𝑟2 + 𝑎2) = 𝑟1𝑟2 ⏟ ∈𝑅 + 𝑟1𝑎2 ⏟ ∈𝐽 + 𝑟2𝑎1 ⏟ ∈𝐽 + 𝑟2𝑎2 ⏟ ∈𝐽 ∈𝑅+ 𝐽 Hence 𝑅+ 𝐽is closed under multiplication, giving 𝑅+ 𝐽≤𝑆.
Let 𝜑∶𝑅→𝑆 ⟋ 𝐽be defined by 𝑟↦𝑟+ 𝐽. This is a ring homomorphism, since it is the composite of the inclusion homomorphism 𝑅⊆𝑆and the quotient map 𝑆→𝑆 ⟋ 𝐽. The kernel of 𝜑is the set {𝑟∈𝑅∶𝑟+ 𝐽= 𝐽} = 𝑅∩𝐽. Since this is the kernel of a ring homomorphism, 𝑅∩ 𝐽is an ideal in 𝑅. The image of 𝜑is {𝑟+ 𝐽∣𝑟∈𝑅} = 𝑅+ 𝐽 ⟋ 𝐽≤𝑆 ⟋ 𝐽. By the first isomorphism theorem, 𝑅 ⟋ 𝑅∩𝐽≅𝑅+ 𝐽 ⟋ 𝐽as required.
Remark. If 𝐼⊲𝑅, there exists a bijection between ideals in 𝑅 ⟋ 𝐼and the ideals of 𝑅containing 𝐼. Explicitly, 𝐾↦{𝑟∈𝑅∣𝑟+ 𝐼∈𝐾}; 𝐽↦𝐽 ⟋ 𝐼 Theorem (third isomorphism theorem). Let 𝐼⊲𝑅and 𝐽⊲𝑅with 𝐼⊆𝐽. Then, 𝐽 ⟋ 𝐼⊲𝑅 ⟋ 𝐼; 𝑅/𝐼 ⟋ 𝐽/𝐼≅𝑅 ⟋ 𝐽 Proof. Let 𝜑∶𝑅 ⟋ 𝐼→𝑅 ⟋ 𝐽defined by 𝑟+ 𝐼↦𝑟+ 𝐽. We can check that this is a surjective ring homomorphism by considering the third isomorphism theorem for groups. Its kernel is {𝑟+ 𝐼∶𝑟∈𝐽} = 𝐽 ⟋ 𝐼, which is an ideal in 𝑅 ⟋ 𝐼, and we conclude by use of the first isomorph-ism theorem.
Remark. 𝐽 ⟋ 𝐼is not a quotient ring, since 𝐽is not in general a ring; this notation should be interpreted as a set of cosets.
450 8. Rings Example. Consider the surjective ring homomorphism 𝜑∶ℝ[𝑋] →ℂwhich is defined by 𝑓= ∑ 𝑛 𝑎𝑛𝑋𝑛↦𝑓(𝑖) = ∑ 𝑛 𝑎𝑛𝑖𝑛 Its kernel can be found by the Euclidean algorithm, yielding ker 𝜑= (𝑋2 + 1). Applying the first isomorphism theorem, we immediately find ℝ[𝑋] ⟋ (𝑋2 + 1) ≅ℂ.
Example. Let 𝑅be a ring. Then there exists a unique ring homomorphism 𝑖∶ℤ→𝑅.
Indeed, we must have 0ℤ↦0𝑅; 1ℤ↦1𝑅 This inductively defines 𝑛↦1𝑅+ ⋯+ 1𝑅 ⏟ ⎵ ⎵ ⎵ ⏟ ⎵ ⎵ ⎵ ⏟ 𝑛times The negative integers are also uniquely defined, since any ring homomorphism is a group homomorphism.
−𝑛↦−(1𝑅+ ⋯+ 1𝑅 ⏟ ⎵ ⎵ ⎵ ⏟ ⎵ ⎵ ⎵ ⏟ 𝑛times ) We can show that any such construction is a ring homomorphism as required. Then, the kernel of the ring homomorphism is an ideal of ℤ, hence it is 𝑛ℤfor some 𝑛. Hence, by the first isomorphism theorem, any ring contains a copy of ℤ ⟋ 𝑛ℤ, since it is isomorphic to the image of 𝑖. If 𝑛= 0, then the ring contains a copy of ℤitself, and if 𝑛= 1, then the ring is trivial since 0 = 1. The number 𝑛is known as the characteristic of 𝑅.
For example, ℤ, ℚ, ℝ, ℂhave characteristic zero. The rings ℤ ⟋ 𝑝ℤ, ℤ ⟋ 𝑝ℤ[𝑋] have character-istic 𝑝.
8.7. Integral domains Definition. An integral domain is a ring 𝑅with 0 ≠1 such that for all 𝑎, 𝑏∈𝑅, 𝑎𝑏= 0 implies 𝑎= 0 or 𝑏= 0. A zero divisor in a ring 𝑅is a nonzero element 𝑎∈𝑅such that 𝑎𝑏= 0 for some nonzero 𝑏∈𝑅. A ring is an integral domain if and only if it has no zero divisors.
Example. All fields are integral domains. Any subring of an integral domain is an integral domain. For instance, ℤ[𝑖] ≤ℂis an integral domain.
Example. The ring ℤ× ℤis not an integral domain. Indeed, (1, 0) ⋅(0, 1) = (0, 0).
Lemma. Let 𝑅be an integral domain. Then 𝑅[𝑋] is an integral domain.
Proof. We will show that any two nonzero elements produce a nonzero element. In particu-lar, let 𝑓= ∑ 𝑛 𝑎𝑛𝑋𝑛; 𝑔= ∑ 𝑛 𝑏𝑛𝑋𝑛 451 VIII. Groups, Rings and Modules Since these are nonzero, the leading coefficients 𝑎𝑛and 𝑏𝑚are nonzero. Here, the leading term of the product 𝑓𝑔has form 𝑎𝑛𝑏𝑚𝑋𝑛+𝑚. Since 𝑅is an integral domain, 𝑎𝑛𝑏𝑚≠0, so 𝑓𝑔 is nonzero. Further, the degree of 𝑓𝑔is 𝑛+ 𝑚, the sum of the degrees of 𝑓and 𝑔.
Lemma. Let 𝑅be an integral domain, and 𝑓≠0 be a nonzero polynomial in 𝑅[𝑋]. We define roots(𝑓) = {𝑎∈𝑅∶𝑓(𝑎) = 0}. Then |roots(𝑓)| ≤deg(𝑓).
Proof. Exercise on the example sheets.
Theorem. Let 𝐹be a field. Then any finite subgroup 𝐺of (𝐹×, ⋅) is cyclic.
Proof. 𝐺is a finite abelian group. If 𝐺is not cyclic, we can apply a previous structure the-orem for finite abelian groups to show that there exists 𝐻≤𝐺such that 𝐻≅𝐶𝑑1 × 𝐶𝑑1 for some integer 𝑑1 ≥2. The polynomial 𝑓(𝑋) = 𝑋𝑑1 −1 ∈𝐹[𝑋] has degree 𝑑1, but has at least 𝑑2 1 roots, since any element of 𝐻is a root. This contradicts the previous lemma.
Example. (ℤ ⟋ 𝑝ℤ) × is cyclic.
Proposition. Any finite integral domain is a field.
Proof. Let 0 ≠𝑎∈𝑅, where 𝑅is an integral domain. Consider the map 𝜑∶𝑅→𝑅given by 𝑥↦𝑎𝑥. If 𝜑(𝑥) = 𝜑(𝑦), then 𝑎(𝑥−𝑦) = 0. But 𝑎≠0, hence 𝑥−𝑦= 0. Hence 𝜑is injective.
Since 𝑅is finite, 𝜑is a bijection, hence it has an inverse 𝜑−1, which yields the multiplicative inverse of 𝑎by considering 𝜑−1(𝑎). This may be repeated for all 𝑎.
Theorem. Any integral domain 𝑅is a subring of a field 𝐹, and every element of 𝐹can be written in the form 𝑎𝑏−1 where 𝑎, 𝑏∈𝑅and 𝑏≠0. Such a field 𝐹is called the field of fractions of 𝑅.
Proof. Consider the set 𝑆= {(𝑎, 𝑏) ∈𝑅∶𝑏≠0}. We can define an equivalence relation (𝑎, 𝑏) ∼(𝑐, 𝑑) ⟺𝑎𝑑= 𝑏𝑐 This is reflexive and commutative. We can show directly that it is transitive.
(𝑎, 𝑏) ∼(𝑐, 𝑑) ∼(𝑒, 𝑓) ⟹𝑎𝑑= 𝑏𝑐; 𝑐𝑓= 𝑑𝑒 ⟹𝑎𝑑𝑓= 𝑏𝑐𝑓= 𝑏𝑑𝑒 ⟹𝑎𝑓= 𝑏𝑒 ⟹(𝑎, 𝑏) ∼(𝑒, 𝑓) Hence ∼is indeed an equivalence relation. Now, let 𝐹= 𝑆 ⟋ ∼, and we write 𝑎 𝑏for the class [(𝑎, 𝑏)]. We define the ring operations 𝑎 𝑏+ 𝑐 𝑑= 𝑎𝑑+ 𝑏𝑐 𝑏𝑑 ; 𝑎 𝑏⋅𝑐 𝑑= 𝑎𝑐 𝑏𝑑 452 8. Rings These can be shown to be well-defined. Thus, 𝐹is a ring with identities 0𝐹= 0𝑅 1𝑅and 1𝐹= 1𝑅 1𝑅.
If 𝑎 𝑏≠0𝐹, then 𝑎≠0. Thus, 𝑏 𝑎exists, and 𝑎 𝑏⋅ 𝑏 𝑎= 1. Hence 𝐹is a field.
We can identify 𝑅with the subring of 𝐹given by 𝑟 1 for all 𝑟∈𝑅. This is clearly isomorphic to 𝑅. Further, any element of 𝐹can be written as 𝑎 𝑏= 𝑎𝑏−1 as required.
This is analogous to the construction of the rationals using the integers.
Example. Consider ℂ[𝑋]. This has field of fractions ℂ(𝑋), called the field of rational func-tions in 𝑋.
8.8. Maximal ideals Definition. An ideal 𝐼⊲𝑅is maximal if 𝐼≠𝑅and, if 𝐼⊆𝐽⊲𝑅, we have 𝐽= 𝐼or 𝐽= 𝑅.
Lemma. A nonzero ring 𝑅is a field if and only if its only ideals are zero or 𝑅.
Proof. Suppose 𝑅is a field. If 0 ≠𝐼⊲𝑅, then 𝐼contains a nonzero element, which is a unit since 𝑅is a field. Hence 𝐼= 𝑅.
Now, suppose a ring 𝑅has ideals that are only zero or 𝑅. If 0 ≠𝑥∈𝑅, consider (𝑥). This is nonzero since it contains 𝑥. By assumption, (𝑥) = 𝑅. Thus, the element 1 lies in (𝑥).
Hence, there exists 𝑦∈𝑅such that 𝑥𝑦= 1, and hence this 𝑦is the multiplicative inverse as required.
Proposition. Let 𝐼⊲𝑅. Then 𝐼is maximal if and only if 𝑅 ⟋ 𝐼is a field.
Proof. 𝑅 ⟋ 𝐼is a field if and only if its ideals are either zero, denoted 𝐼 ⟋ 𝐼, or 𝑅 ⟋ 𝐼itself. By the correspondence theorem, 𝐼and 𝑅are the only ideals in 𝑅which contain 𝐼. Equivalently, 𝐼⊲𝑅is maximal.
8.9. Prime ideals Definition. An ideal 𝐼⊲𝑅is prime if 𝐼≠𝑅and, for all 𝑎, 𝑏∈𝑅such that 𝑎𝑏∈𝐼, we have 𝑎∈𝐼or 𝑏∈𝐼.
Example. The ideals in the integers are (𝑛) for some 𝑛≥0. 𝑛ℤis a prime ideal if and only if 𝑛is prime or zero. The case for 𝑛= 0 is trivial. If 𝑛≠0 we can use the property that 𝑝∣𝑎𝑏 implies either 𝑝∣𝑎or 𝑝∣𝑏. Conversely, if 𝑛is composite, we can write 𝑛= 𝑢𝑣for 𝑢, 𝑣> 1.
Then 𝑢𝑣∈𝑛ℤbut 𝑢, 𝑣∉𝑛ℤ.
Proposition. Let 𝐼⊲𝑅. Then 𝐼is prime if and only if 𝑅 ⟋ 𝐼is an integral domain.
453 VIII. Groups, Rings and Modules Proof. If 𝐼is prime, then for all 𝑎𝑏∈𝐼we have 𝑎∈𝐼or 𝑏∈𝐼. Equivalently, for all 𝑎+ 𝐼, 𝑏+ 𝐼∈𝑅 ⟋ 𝐼, we have (𝑎+ 𝐼)(𝑏+ 𝐼) = 0 + 𝐼if 𝑎+ 𝐼= 0 + 𝐼or 𝑏+ 𝐼= 0 + 𝐼. This is the definition of an integral domain.
Remark. If 𝐼is a maximal ideal, then 𝑅 ⟋ 𝐼is a field. A field is an integral domain. Hence any maximal ideal is prime.
Remark. If the characteristic of a ring is 𝑛, then ℤ ⟋ 𝑛ℤ≤𝑅. In particular, if 𝑅is an integral domain, then ℤ ⟋ 𝑛ℤmust be an integral domain. Equivalently, 𝑛ℤ⊲ℤis a prime ideal.
Hence 𝑛is zero or prime. Thus, in an integral domain, the characteristic must either be zero or prime. A field always has a characteristic, which is either zero (in which case it contains ℤand hence ℚ) or prime (in which case it contains ℤ ⟋ 𝑝ℤ= 𝔽𝑝which is already a field).
454 9. Factorisation in integral domains 9. Factorisation in integral domains In this section, let 𝑅be an integral domain.
9.1. Prime and irreducible elements Recall that an element 𝑎∈𝑅is a unit if it has a multiplicative inverse in 𝑅. Equivalently, an element 𝑎is a unit if and only if (𝑎) = 𝑅. Indeed, if (𝑎) = 𝑅, then 1 ∈(𝑎) hence there exists a multiple of 𝑎equal to 1. We denote the set of units in 𝑅by 𝑅×.
Definition. An element 𝑎∈𝑅divides 𝑏∈𝑅, written 𝑎∣𝑏, if there exists 𝑐∈𝑅such that 𝑏= 𝑎𝑐. Equivalently, (𝑏) ⊆(𝑎).
Two elements 𝑎, 𝑏∈𝑅are associates if 𝑎= 𝑏𝑐where 𝑐is a unit. Informally, the two elements differ by multiplication by a unit. Equivalently, (𝑎) = (𝑏).
Definition. An element 𝑟∈𝑅is irreducible if 𝑟is not zero or a unit, and 𝑟= 𝑎𝑏implies 𝑎 is a unit or 𝑏is a unit. An element 𝑟∈𝑅is prime if 𝑟is not zero or a unit, and 𝑟∣𝑎𝑏implies 𝑟∣𝑎or 𝑟∣𝑏.
Remark. These properties depend on the ambient ring 𝑅; for instance, 2 is prime and irredu-cible in ℤ, but neither prime nor irreducible in ℚ. The polynomial 2𝑋is irreducible in ℚ[𝑋], but not in ℤ[𝑋].
Lemma. (𝑟) ⊲𝑅is a prime ideal if and only if 𝑟= 0 or 𝑟is prime.
Proof. Suppose (𝑟) is a prime ideal with 𝑟≠0. Since prime ideals are proper, 𝑟cannot be a unit. Suppose 𝑟∣𝑎𝑏, or equivalently, 𝑎𝑏∈(𝑟). By the definition of a prime ideal, 𝑎∈(𝑟) or 𝑏∈(𝑟). Hence, 𝑟∣𝑎or 𝑟∣𝑏. By definition of a prime element, 𝑟is prime.
Conversely, first note that the zero ideal (0) = {0} is a prime ideal, since 𝑅is an integral domain. Suppose 𝑟is prime. We know (𝑟) ≠𝑅since 𝑟is not a unit. If 𝑎𝑏∈(𝑟), then 𝑟∣𝑎𝑏, so 𝑟∣𝑎or 𝑟∣𝑏, giving 𝑎∈(𝑟) or 𝑏∈(𝑟) as required for (𝑟) to be a prime ideal.
Lemma. Prime elements are irreducible.
Proof. Let 𝑟be prime. Then 𝑟is nonzero and not a unit. Suppose 𝑟= 𝑎𝑏. Then, in particular, 𝑟∣𝑎𝑏, so 𝑟∣𝑎or 𝑟∣𝑏by primality. Let 𝑟∣𝑎without loss of generality. Hence 𝑎= 𝑟𝑐for some element 𝑐∈𝑅. Then, 𝑟= 𝑎𝑏= 𝑟𝑐𝑏, so 𝑟(1 −𝑐𝑏) = 0. Since 𝑅is an integral domain, and 𝑟≠0, we have 𝑐𝑏= 1, so 𝑏is a unit.
Example. The converse does not hold in general. Let 𝑅= ℤ[√−5] = {𝑎+ 𝑏√−5∶𝑎, 𝑏∈ℤ} ≤ℂ; 𝑅≅ℤ[𝑋] ⟋ (𝑋2 + 5) 455 VIII. Groups, Rings and Modules Since 𝑅is a subring of the field ℂ, it is an integral domain. We can define the norm 𝑁∶𝑅→ ℤby 𝑁(𝑎+ 𝑏√−5) = 𝑎2 + 5𝑏2 ≥0. Note that this norm is multiplicative: 𝑁(𝑧1𝑧2) = 𝑁(𝑧1)𝑁(𝑧2).
We claim that the units are exactly ±1. Indeed, if 𝑟∈𝑅×, then 𝑟𝑠= 1 for some element 𝑠∈𝑅. Then, 𝑁(𝑟)𝑁(𝑠) = 𝑁(1) = 1, so 𝑁(𝑟) = 𝑁(𝑠) = 1. But the only elements 𝑟∈𝑅with 𝑁(𝑟) = 1 are 𝑟= ±1.
We will now show that the element 2 ∈𝑅is irreducible. Suppose 2 = 𝑟𝑠for 𝑟, 𝑠∈𝑅. By the multiplicative property of 𝑁, 𝑁(2) = 4 = 𝑁(𝑟)𝑁(𝑠) can only be satisfied by 𝑁(𝑟), 𝑁(𝑠) ∈ {1, 2, 4}. Since 𝑎2 + 5𝑏2 = 2 has no integer solutions, 𝑅has no elements of norm 2. Hence, either 𝑟or 𝑠has unit norm and is thus a unit by the above discussion. We can show similarly that 3, 1 + √−5, 1 −√−5 are irreducible, as there exist no elements of norm 3.
We can now compute directly that (1 + √−5)(1 −√−5) = 6 = 2 ⋅3, hence 2 ∣(1 + √−5)(1 − √−5). But 2 ∤(1 + √−5) and 2 ∤(1 −√−5), which can be checked by taking norms. Hence, 2 is irreducible but not a prime.
In order to construct this example, we have exhibited two factorisations of 6 into irreducibles: (1 + √−5)(1 −√−5) = 6 = 2 ⋅3. Since 𝑅× = {±1}, these irreducibles in the factorisations are not associates.
9.2. Principal ideal domains Definition. An integral domain 𝑅is a principal ideal domain if all ideals are principal ideals.
In other words, for all ideals 𝐼, there exists an element 𝑟such that 𝐼= (𝑟).
Example. ℤis a principal ideal domain.
Proposition. In a principal ideal domain, all irreducible elements are prime.
Proof. Let 𝑟∈𝑅be irreducible, and suppose 𝑟∣𝑎𝑏. If 𝑟∣𝑎, the proof is complete, so suppose 𝑟∤𝑎. Since 𝑅is a principal ideal domain, the ideal (𝑎, 𝑟) is generated by a single element 𝑑∈𝑅. In particular, since 𝑟∈(𝑑), we have 𝑑∣𝑟so 𝑟= 𝑐𝑑for some 𝑐∈𝑅.
Since 𝑟is irreducible, either 𝑐or 𝑑is a unit. If 𝑐is a unit, (𝑎, 𝑟) = (𝑟), so in particular 𝑟∣𝑎, which contradicts the assumption that 𝑟∤𝑎, so 𝑐cannot be a unit. Thus, 𝑑is a unit. In this case, (𝑎, 𝑟) = 𝑅. By definition of (𝑎, 𝑟), there exist 𝑠, 𝑡∈𝑅such that 1 = 𝑠𝑎+ 𝑡𝑟. Then, 𝑏= 𝑠𝑎𝑏+𝑡𝑟𝑏. We have 𝑟∣𝑠𝑎𝑏since 𝑟∣𝑎𝑏, and we know 𝑟∣𝑡𝑟𝑏. Hence 𝑟∣𝑏as required.
Lemma. Let 𝑅be a principal ideal domain. Then an element 𝑟is irreducible if and only if (𝑟) is maximal.
Proof. Suppose 𝑟is irreducible. Since 𝑟is not a unit, (𝑟) ≠𝑅. Suppose (𝑟) ⊆𝐽⊆𝑅where 𝐽 is an ideal in 𝑅. Since 𝑅is a principal ideal domain, 𝐽= (𝑎) for some 𝑎∈𝑅. In particular, 𝑟= 𝑎𝑏for some 𝑏∈𝑅, since (𝑟) ⊆𝐽. Since 𝑟is irreducible, either 𝑎or 𝑏is a unit. But if 𝑎 456 9. Factorisation in integral domains is a unit, we have 𝐽= 𝑅. If 𝑏is a unit, then 𝑎and 𝑟are associates so they generate the same ideal. Hence, (𝑟) is maximal.
Conversely, suppose (𝑟) is maximal. Note that 𝑟is not a unit, since (𝑟) ≠𝑅. Suppose 𝑟= 𝑎𝑏.
Then (𝑟) ⊆(𝑎) ⊆𝑅. But since (𝑟) is maximal, either (𝑎) = (𝑟) or (𝑎) = 𝑅. If (𝑎) = (𝑟), then 𝑏is a unit. If (𝑎) = 𝑅, then 𝑎is a unit. Hence 𝑟is irreducible. Note that this direction of the proof did not require that 𝑅was a principal ideal domain, however 𝑅must still be an integral domain.
Remark. Let 𝑅be a principal ideal domain, and 0 ≠𝑟∈𝑅. Then, (𝑟) is maximal if and only if 𝑟is irreducible, which is true if and only if 𝑟is prime, which is equivalent to the fact that (𝑟) is prime. Hence, the maximal ideals are the nonzero prime ideals.
Definition. An integral domain is a Euclidean domain if there exists a function 𝜑∶𝑅∖{0} → ℤ≥0 such that, for all 𝑎, 𝑏∈𝑅.
(i) if 𝑎∣𝑏then 𝜑(𝑎) ≤𝜑(𝑏); (ii) if 𝑏≠0 then ∃𝑞, 𝑟∈𝑅such that 𝑎= 𝑏𝑞+ 𝑟and either 𝑟= 0 or 𝜑(𝑟) < 𝜑(𝑏).
Such a 𝜑is called a Euclidean function.
Example. ℤis a Euclidean domain, where the Euclidean function 𝜑is the absolute value function.
Proposition. Euclidean domains are principal ideal domains.
Proof. Let 𝑅have Euclidean function 𝜑. Let 𝐼⊲𝑅be a nonzero ideal. Let 𝑏∈𝐼∖{0} that minimises 𝜑(𝑏). Then (𝑏) ⊆𝐼. For any element 𝑎∈𝐼, we can use the Euclidean algorithm to show 𝑎= 𝑏𝑞+ 𝑟where 𝑟= 0 or 𝜑(𝑟) < 𝜑(𝑏). But since 𝑟= 𝑎−𝑏𝑞∈𝐼, 𝜑(𝑟) cannot be lower than the minimal element 𝜑(𝑏). Thus 𝑟= 0, so 𝑎= 𝑏𝑞. Hence, 𝐼= (𝑏), so all ideals are principal.
Remark. In the above proof, only the second property of the Euclidean function was used.
The first property is included in the definition since it will allow us to easily describe the units in the ring.
𝑅× = {𝑢∈𝑅∶𝑢≠0, 𝜑(𝑢) = 𝜑(1)} It can be shown that, if there exists a function 𝜑satisfying (ii), there exists a (possibly not unique) function 𝜑′ satisfying (i) and (ii).
Example. Let 𝐹be a field. Then 𝐹[𝑋] is a Euclidean domain with Euclidean function 𝜑(𝑓) = deg(𝑓). We have already proven the requisite properties of Euclidean functions.
The ring 𝑅= ℤ[𝑖] is a Euclidean domain with 𝜑(𝑢+ 𝑖𝑣) = 𝑁(𝑢+ 𝑖𝑣) = 𝑢2 + 𝑣2. Since the norm is multiplicative, 𝑁(𝑧𝑤) = 𝑁(𝑧)𝑁(𝑤) which immediately gives property (i) in the definition. Consider 𝑧, 𝑤∈ℤ[𝑖] where 𝑤≠0. Consider 𝑧 𝑤∈ℂ. This has distance less than 457 VIII. Groups, Rings and Modules 1 from the nearest element 𝑞of 𝑅. Let 𝑟= 𝑧−𝑤𝑞∈𝑅. Then 𝑧= 𝑤𝑞+ 𝑟where 𝜑(𝑟) = |𝑟|2 = |𝑧−𝑤𝑞|2 < |𝑤|2 = 𝜑(𝑤) So property (ii) is satisfied.
Hence 𝐹[𝑋] and ℤ[𝑖] are principal ideal domains.
Example. Let 𝐴be a nonzero 𝑛× 𝑛matrix over a field 𝐹. Let 𝐼= {𝑓∈𝐹[𝑋]∶𝑓(𝐴) = 0}.
𝐼is an ideal. Indeed, if 𝑓, 𝑔∈𝐼, then (𝑓−𝑔)(𝐴) = 𝑓(𝐴) −𝑔(𝐴) = 0, and for 𝑓∈𝐼and 𝑔∈𝐹[𝑋], we have (𝑓⋅𝑔)(𝐴) = 𝑓(𝐴) ⋅𝑔(𝐴) = 0 as required. Since 𝐹[𝑋] is a principal ideal domain, 𝐼= (𝑓) for some polynomial 𝑓∈𝐹[𝑋]. All units in 𝐹[𝑋] are the nonzero constant polynomials. Hence, the polynomial of smallest degree in 𝐼is unique up to multiplication by a unit, so without loss of generality we may assume 𝑓is monic. This yields the minimal polynomial of 𝐴.
Example. Let 𝔽2 be the finite field of order 2, which is isomorphic to ℤ ⟋ 2ℤ. Let 𝑓(𝑋) be the polynomial 𝑋3 + 𝑋+ 1 ∈𝔽2[𝑋].
We claim that 𝑓is irreducible. Suppose 𝑓= 𝑔ℎwhere the degrees of 𝑔, ℎare positive. Since the degree of 𝑓is 3, one of 𝑔, ℎmust have degree 1. Hence 𝑓has a root. But we can check that 𝑓(0) = 𝑓(1) = 1 so 𝑓has no root in 𝔽2. Hence 𝑓is irreducible as required.
Since 𝔽2[𝑋] is a principal ideal domain, we have that (𝑓) ⊲𝔽2[𝑋] is a maximal ideal. Hence, 𝔽2[𝑋] ⟋ (𝑓) is a field. We can verify that this field has order 8, using the Euclidean algorithm.
Any element in this quotient has coset representative 𝑎𝑋2 + 𝑏𝑋+ 𝑐for 𝑎, 𝑏, 𝑐∈𝔽2. We can show that all 8 of these possibilities yields different polynomials. So we have constructed a field of order 8. This technique will be explored further in Part II Galois Theory.
Example. The ring ℤ[𝑋] is not a principal ideal domain. Consider the ideal 𝐼= (2, 𝑋) ⊲ ℤ[𝑋]. We can write 𝐼= {2𝑓 1(𝑋) + 𝑋𝑓 2(𝑋)∶𝑓 1, 𝑓 2 ∈ℤ[𝑋]} = {𝑓∈ℤ[𝑋]∶2 ∣𝑓(0)} Suppose 𝐼= (𝑓) for some element 𝑓. Since 2 ∈𝐼, we must have 2 = 𝑓𝑔for some polynomial 𝑔. By comparing degrees, the degrees of 𝑓and 𝑔must be zero, since ℤis an integral domain.
Hence 𝑓is an integer, so 𝑓= ±1 or 𝑓= ±2. If 𝑓= ±1 then 𝐼= ℤ[𝑋], and if 𝑓= ±2 then 𝐼= 2ℤ[𝑋]. These both lead to contradictions, since 1 ∉𝐼and 𝑋∈𝐼.
9.3. Unique factorisation domains Definition. An integral domain is a unique factorisation domain if (i) every nonzero, non-unit element is a product of irreducibles; (ii) if 𝑝1 ⋯𝑝𝑚= 𝑞1 ⋯𝑞𝑛where 𝑝𝑖, 𝑞𝑖are irreducible, then 𝑚= 𝑛, and 𝑝𝑖, 𝑞𝑖are associates, up to reordering.
458 9. Factorisation in integral domains Proposition. Let 𝑅be an integral domain satisfying property (i) above (every nonzero, non-unit element is a product of irreducibles). Then 𝑅is a unique factorisation domain if and only if every irreducible is prime.
Proof. Suppose 𝑅is a unique factorisation domain. Let 𝑝∈𝑅be irreducible, and 𝑝∣𝑎𝑏.
Then 𝑎𝑏= 𝑝𝑐for some 𝑐∈𝑅. Writing 𝑎, 𝑏, 𝑐as products of irreducibles, it follows from uniqueness of factorisation that 𝑝∣𝑎or 𝑝∣𝑏. Hence 𝑝is prime.
Conversely, suppose every irreducible is prime. Suppose 𝑝1 ⋯𝑝𝑚= 𝑞1 ⋯𝑞𝑛where 𝑝𝑖, 𝑞𝑖are irreducible and hence prime. Since 𝑝1 ∣𝑞1 ⋯𝑞𝑛, we have 𝑝1 ∣𝑞𝑖for some 𝑖. After reordering, we may assume that 𝑝1 ∣𝑞1, so 𝑝1𝑢= 𝑞1 for 𝑢∈𝑅. Since 𝑞1 is irreducible, 𝑢is a unit since 𝑝1 cannot be a unit. Hence 𝑝1, 𝑞1 are associates. Cancelling 𝑝1 from both sides, we find 𝑝2 ⋯𝑝𝑚= 𝑢𝑞2 ⋯𝑞𝑛. We may absorb this unit into 𝑞2 without loss of generality. Inductively, all 𝑝𝑖and 𝑞𝑖are associates, for each 𝑖. Hence 𝑅is a unique factorisation domain.
Definition. Let 𝑅be a ring. Suppose, for all nested sequences of ideals in 𝑅written 𝐼1 ⊆ 𝐼2 ⊆⋯, there exists 𝑁such that 𝐼𝑛= 𝐼𝑛+1 for all 𝑛≥𝑁. Then, we say that 𝑅is a Noetherian ring.
This condition is known as the ‘ascending chain condition’. In other words, we cannot in-finitely nest distinct ideals in a Noetherian ring.
Lemma. Principal ideal domains are Noetherian rings.
Proof. Let 𝐼= ⋃ ∞ 𝑖=1 𝐼𝑖. Then, 𝐼is an ideal in 𝑅. Since 𝑅is a principal ideal domain, 𝐼= (𝑎) for some 𝑎∈𝑅. Then 𝑎∈⋃ ∞ 𝑖=1 𝐼𝑖, so in particular 𝑎∈𝐼𝑁for some 𝑁. But then for all 𝑛≥𝑁, (𝑎) ⊆𝐼𝑁⊆𝐼𝑛⊆𝐼𝑛+1 ⊆𝐼= (𝑎). So all inclusions are equalities, so in particular 𝐼𝑛= 𝐼𝑛+1.
Theorem. If 𝑅is a principal ideal domain, then it is a unique factorisation domain.
Proof. First, we verify property (i), that every nonzero, non-unit element is a product of irreducibles. Let 𝑥≠0 be an element of 𝑅which is not a unit. Suppose 𝑥does not factor as a product of irreducibles. This implies in particular that 𝑥is not irreducible. By definition, we can write 𝑥as the product of two elements 𝑥1, 𝑦1 where 𝑥1, 𝑦1 are not units. Then either 𝑥1 or 𝑦1 is not a product of irreducibles, so without loss of generality we can suppose 𝑥1 is not a product of irreducibles. We have (𝑥) ⊂(𝑥1). This inclusion is strict, since 𝑦1 is not a unit. Now, we can write 𝑥1 = 𝑥2𝑦2 where 𝑥2 is not a unit, and inductively we can create (𝑥) ⊂(𝑥1) ⊂(𝑥2) ⊂⋯. But 𝑅is Noetherian, so this is a contradiction. So every nonzero, non-unit element is indeed a product of irreducibles.
By the proposition above, it suffices to show that every irreducible is prime. This has already been shown previously. Hence 𝑅is a unique factorisation domain.
Example. We have shown that all Euclidean domains are principal ideal domains, and all principal ideal domains are unique factorisation domains, and all unique factorisation 459 VIII. Groups, Rings and Modules domains are integral domains. We now provide examples for counterexamples to the con-verses.
The ring ℤ ⟋ 4ℤis not an integral domain since 2 is a zero divisor.
The ring ℤ[√−5] ≤ℂis integral, but not a unique factorisation domain.
The ring ℤ[𝑋] has been shown to be not a principal ideal domain. We can show using later results that this is a unique factorisation domain.
We can construct the ring ℤ[ 1+√−19 2 ], which can be shown to be not a Euclidean domain, but is a principal ideal domain. This proof is beyond the scope of Part IB Groups, Rings and Modules, but will be proved in Part II Number Fields.
Finally, ℤ[𝑖] is a Euclidean domain, and is hence a principal ideal domain, a unique factor-isation domain, and an integral domain.
Definition. Let 𝑅be an integral domain.
(i) 𝑑∈𝑅is a common divisor of 𝑎1, … , 𝑎𝑛∈𝑅if 𝑑∣𝑎𝑖for all 𝑖; (ii) 𝑑∈𝑅is a greatest common divisor of 𝑎1, … , 𝑎𝑛if for all common divisors 𝑑′, we have 𝑑′ ∣𝑑; (iii) 𝑚∈𝑅is a common multiple of 𝑎1, … , 𝑎𝑛if 𝑎𝑖∣𝑚for all 𝑖; (iv) 𝑚∈𝑅is a least common multiple of 𝑎1, … , 𝑎𝑛if for all common multiples 𝑚′, we have 𝑚∣𝑚′.
Remark. Greatest common divisors and lowest common multiples are unique up to associ-ates, if they exist.
Proposition. In unique factorisation domains, greatest common divisors and least com-mon multiples always exist.
Proof. Let 𝑎𝑖= 𝑢𝑖∏𝑗𝑝 𝑛𝑖𝑗 𝑗 where the 𝑝𝑗are irreducible and pairwise non-associate, 𝑢𝑖is a unit, and 𝑛𝑖𝑗∈ℤ≥0. We claim that 𝑑= ∏𝑗𝑝 𝑚𝑗 𝑗 , where 𝑚𝑗= min1≤𝑖≤𝑛𝑛𝑖𝑗, is the greatest common divisor. Certainly 𝑑is a common divisor. If 𝑑′ is a common divisor, then 𝑑′ can be written as a product of irreducibles, which will be denoted 𝑑′ = 𝑤∏𝑗𝑝 𝑡𝑗 𝑖. We can see that 𝑡𝑗≤𝑛𝑖𝑗for all 𝑖, so in particular, 𝑡𝑗≤𝑚𝑗. This implies 𝑑′ ∣𝑑. Hence 𝑑is a greatest common divisor. The argument for the least common multiple is similar, replacing minima with maxima.
9.4. Factorisation in polynomial rings Theorem. Let 𝑅be a unique factorisation domain. Then 𝑅[𝑋] is also a unique factorisation domain.
460 9. Factorisation in integral domains The proof for this theorem will require a number of key lemmas. In this subsection, 𝑅will denote a unique factorisation domain, with field of fractions 𝐹. We have 𝑅[𝑋] ≤𝐹[𝑋]. Since polynomial rings over fields are Euclidean domains, 𝐹[𝑋] is a principal ideal domain, and hence a unique factorisation domain. This does not immediately imply that 𝑅[𝑋] is a unique factorisation domain, however.
Definition. The content of a polynomial 𝑓= ∑ 𝑛 𝑖=0 𝑎𝑖𝑋𝑖∈𝑅[𝑋] is 𝑐(𝑓) = gcd{𝑎0, … , 𝑎𝑛}.
This is well-defined up to multiplication by a unit.
We say that 𝑓is primitive if 𝑐(𝑓) is a unit.
Lemma. The product of primitive polynomials is primitive. Further, for 𝑓, 𝑔∈𝑅[𝑋], 𝑐(𝑓𝑔) and 𝑐(𝑓)𝑐(𝑔) are associates.
Proof. Let 𝑓= ∑ 𝑛 𝑖=0 𝑎𝑖𝑋𝑖and 𝑔= ∑ 𝑚 𝑖=0 𝑏𝑖𝑋𝑖. Suppose 𝑓𝑔is not primitive, so 𝑐(𝑓𝑔) is not a unit. This implies that there exists a prime 𝑝such that 𝑝∣𝑐(𝑓𝑔). Since 𝑓, 𝑔are primitive, 𝑝∤𝑐(𝑓) and 𝑝∤𝑐(𝑔).
Suppose 𝑝does not divide all of the 𝑎𝑘or the 𝑏ℓ. Let 𝑘, ℓbe the smallest values such that 𝑝∤𝑎𝑘and 𝑝∤𝑏ℓ. Then, the coefficient of 𝑋𝑘+ℓin 𝑓𝑔is given by ∑ 𝑖+𝑗=𝑘+ℓ 𝑎𝑖𝑏𝑗= ⋯+ 𝑎𝑘−1𝑏ℓ+1 ⏟⎵ ⎵ ⎵⏟⎵ ⎵ ⎵⏟ divisible by 𝑝 +𝑎𝑘𝑏ℓ+ 𝑎𝑘+1𝑏ℓ−1 + ⋯ ⏟⎵ ⎵ ⎵⏟⎵ ⎵ ⎵⏟ divisible by 𝑝 Thus 𝑝∣𝑎𝑘𝑏ℓ. This is a contradiction as we have 𝑝∣𝑎𝑘or 𝑝∣𝑏ℓ.
To prove the second part, let 𝑓= 𝑐(𝑓)𝑓 0 for some 𝑓 0 ∈𝑅[𝑋]. Here, 𝑓 0 is primitive. Similarly, 𝑔= 𝑐(𝑔)𝑔0 for a primitive 𝑔0. Thus 𝑓𝑔= 𝑐(𝑓)𝑐(𝑔)𝑓 0𝑔0. The expression 𝑓 0𝑔0 is a primitive polynomial by the first part, so 𝑐(𝑓𝑔) is equal to 𝑐(𝑓)𝑐(𝑔) up to associates.
Corollary. If 𝑝∈𝑅is prime in 𝑅, then 𝑝is prime in 𝑅[𝑋].
Proof. Since 𝑅is an integral domain, we have 𝑅[𝑋]× = 𝑅×, so 𝑝is not a unit. Let 𝑓∈𝑅[𝑋].
Then 𝑝∣𝑓in 𝑅[𝑋] if and only if 𝑝∣𝑐(𝑓) in 𝑅. Thus, if 𝑝∣𝑔ℎin 𝑅[𝑋], we have 𝑝∣𝑐(𝑔ℎ) = 𝑐(𝑔)𝑐(ℎ). In particular, since 𝑝is prime in 𝑅, we have 𝑝∣𝑐(𝑔) or 𝑝∣𝑐(ℎ), so 𝑝∣𝑔or 𝑝∣ℎ. So 𝑝is prime in 𝑅[𝑋].
Lemma. Let 𝑓, 𝑔∈𝑅[𝑋], where 𝑔is primitive. Then if 𝑔∣𝑓in 𝐹[𝑋], then 𝑔∣𝑓in 𝑅[𝑋].
Proof. Let 𝑓= 𝑔ℎ, where ℎ∈𝐹[𝑋]. We can find a nonzero 𝑎∈𝑅, such that 𝑎ℎ∈𝑅[𝑋].
In particular, we can multiply the denominators of the coefficients of ℎto form 𝑎. Now, 𝑎ℎ= 𝑐(𝑎ℎ)ℎ0 where ℎ0 is primitive. Then 𝑎𝑓= 𝑐(𝑎ℎ)ℎ0𝑔. Since ℎ0 and 𝑔are primitive, so is ℎ0𝑔. Thus, taking contents, 𝑎∣𝑐(𝑎ℎ). This implies ℎ∈𝑅[𝑋]. Hence 𝑔∣𝑓in 𝑅[𝑋].
Lemma (Gauss’ lemma). Let 𝑓∈𝑅[𝑋] be primitive. Then if 𝑓is irreducible in 𝑅[𝑋], we have that 𝑓is irreducible in 𝐹[𝑋].
461 VIII. Groups, Rings and Modules Proof. Since 𝑓∈𝑅[𝑋] is irreducible and primitive, its degree must be larger than zero.
Hence 𝑓is not a unit in 𝐹[𝑋]. Suppose 𝑓is not irreducible in 𝐹[𝑋], so 𝑓= 𝑔ℎfor 𝑔, ℎ∈𝐹[𝑋] with degree larger than zero. Let 𝜆∈𝐹× such that 𝜆−1𝑔∈𝑅[𝑋] is primitive. For example, let 𝑏∈𝑅such that 𝑏𝑔∈𝑅[𝑋] to clear denominators, then 𝑏𝑔= 𝑐(𝑏𝑔)𝑔0, giving 𝜆= 𝑐(𝑏𝑔)𝑏−1.
Replacing 𝑔by 𝜆−1𝑔and ℎby 𝜆ℎ, we still have a factorisation of 𝑓. Hence, we may assume without loss of generality that 𝑔∈𝑅[𝑋] and is primitive. By the previous lemma, we have that ℎ∈𝑅[𝑋], with degrees larger than zero. This contradicts irreducibility.
Remark. We will see that the reverse implication in Gauss’ lemma also holds.
Lemma. Let 𝑔∈𝑅[𝑋] be primitive. If 𝑔is prime in 𝐹[𝑋], then 𝑔is prime in 𝑅[𝑋].
Proof. It suffices to show that if 𝑓 1, 𝑓 2 ∈𝑅[𝑋], then 𝑔∣𝑓 1𝑓 2 implies 𝑔∣𝑓 1 or 𝑔∣𝑓 2. Since 𝑔 is prime in 𝐹[𝑋], 𝑔∣𝑓 1 or 𝑔∣𝑓 2 in 𝐹[𝑋]. By the previous lemma, 𝑔∣𝑓 1 or 𝑔∣𝑓 2 in 𝑅[𝑋] as required.
We can now prove the first theorem of this subsection, that polynomial rings over unique factorisation domains are unique factorisation domains.
Proof. Let 𝑓∈𝑅[𝑋]. Then, 𝑓= 𝑐(𝑓)𝑓 0 for 𝑓 0 primitive in 𝑅[𝑋]. Since 𝑅is a unique factor-isation domain, 𝑐(𝑓) is a product of irreducibles in 𝑅. If an element of 𝑅is irreducible, it is irreducible as an element of 𝑅[𝑋]. Hence, it suffices to find a factorisation of 𝑓 0.
Suppose 𝑓 0 is not irreducible, so 𝑓 0 = 𝑔ℎfor 𝑔, ℎ∈𝑅[𝑋]. Since 𝑓 0 is primitive, 𝑔and ℎare primitive, and the degrees of 𝑔, ℎare larger than zero. By induction on the degree, we can factor 𝑓 0 as a product of primitive irreducibles in 𝑅[𝑋].
It now suffices to show uniqueness of the factorisation. By a previous proposition, it in fact suffices to show that every irreducible element of 𝑅[𝑋] is prime. Let 𝑓be irreducible. Write 𝑓= 𝑐(𝑓)𝑓 0, where 𝑓 0 is primitive. Since 𝑓is irreducible, 𝑓must be constant or primitive.
Suppose 𝑓is constant. Since 𝑓is irreducible in 𝑅[𝑋], it must be irreducible in 𝑅. As 𝑅is a unique factorisation domain, 𝑓is prime in 𝑅. By a previous corollary, 𝑓is prime in 𝑅[𝑋].
Now, suppose 𝑓is primitive. Since 𝑓is irreducible in 𝑅[𝑋], we can use Gauss’ lemma to show that 𝑓is irreducible in 𝐹[𝑋]. Thus, 𝑓is prime in 𝐹[𝑋], as 𝐹[𝑋] is a unique factorisation domain. Finally, we can see that 𝑓is prime in 𝑅[𝑋] by the previous lemma.
Remark. We know that the prime elements in an integral domain are irreducible. This im-plies that the implications in the last paragraph above are in fact equivalences. In particular, in Gauss’ lemma, the implication is an equivalence.
Example. The above theorem implies that ℤ[𝑋] is a unique factorisation domain.
Let 𝑅[𝑋1, … , 𝑋𝑛] be the ring of polynomials in 𝑛variables. We can rewrite this as 𝑅[𝑋1] … [𝑋𝑛], so by induction this is a unique factorisation domain if 𝑅is.
462 9. Factorisation in integral domains 9.5. Eisenstein’s criterion Proposition. Let 𝑅be a unique factorisation domain, and 𝑓(𝑋) = ∑ 𝑛 𝑖=0 𝑎𝑖𝑋𝑖∈𝑅[𝑋] be a primitive polynomial. Let 𝑝∈𝑅be irreducible (or, equivalently, prime) such that (i) 𝑝∤𝑎𝑛; (ii) 𝑝∣𝑎𝑖for all 𝑖< 𝑛; and (iii) 𝑝2 ∤𝑎0.
Then 𝑓is irreducible in 𝑅[𝑋].
Proof. Suppose 𝑓= 𝑔ℎfor 𝑔, ℎ∈𝑅[𝑋] not units. Since 𝑓is primitive, 𝑔, ℎmust have positive degree. Let 𝑔(𝑋) = ∑ 𝑘 𝑖=0 𝑟𝑖𝑋𝑖and ℎ(𝑋) = ∑ ℓ 𝑖=0 𝑠𝑖𝑋𝑖, so 𝑘+ ℓ= 𝑛. Then 𝑝∤𝑎𝑛= 𝑟𝑘𝑠ℓ, so 𝑝∤𝑟𝑘and 𝑝∤𝑠ℓ. Further, 𝑝∣𝑎0 = 𝑟0𝑠0 so 𝑝∣𝑟0 or 𝑝∣𝑠0. Without loss of generality, we may assume 𝑝∣𝑟0. There exists a minimal 𝑗≤𝑘such that 𝑝∣𝑟𝑖for all 𝑖< 𝑗but 𝑝∤𝑟𝑗.
𝑎𝑗= 𝑟0𝑠𝑗+ 𝑟1𝑠𝑗−1 + ⋯+ 𝑟𝑗−1𝑠1 + 𝑟𝑗𝑠0 By assumption, 𝑎𝑗is divisible by 𝑝since 𝑗< 𝑛. Further, the first 𝑗terms in the expansion are divisible by 𝑝. Thus, 𝑝∣𝑟𝑗𝑠0. By assumption, 𝑝∤𝑟𝑗, so 𝑝∣𝑠0. In particular, 𝑝2 ∣𝑟0𝑠0 = 𝑎0, contradicting the third criterion.
Example. Let 𝑓(𝑋) = 𝑋3 +2𝑋+5 ∈ℤ[𝑋]. We will show this is irreducible as a polynomial over ℚ. If 𝑓is not irreducible in ℤ[𝑋], then it factorises as 𝑓(𝑋) = (𝑋+ 𝑎)(𝑋2 + 𝑏𝑋+ 𝑐) up to multiplication by units. Here, 𝑎𝑐= 5. But ±1, ±5 are not roots of 𝑓, so this is irreducible in ℤ[𝑋]. By Gauss’ lemma, 𝑓is irreducible in ℚ[𝑋], since ℚis the field of fractions of ℤ. In particular, ℚ[𝑋] ⟋ (𝑓) is a field, since the ideal (𝑓) is maximal.
Example. Let 𝑝∈ℤbe a prime, and let 𝑓(𝑋) = 𝑋𝑛−𝑝. By Eisenstein’s criterion, 𝑓is irreducible in ℤ[𝑋]. It is then irreducible in ℚ[𝑋] by Gauss’ lemma.
Example. Consider 𝑓(𝑋) = 𝑋𝑝−1+𝑋𝑝−2+⋯+𝑋+1 ∈ℤ[𝑋], where 𝑝is prime. Eisenstein’s criterion does not apply directly. Consider 𝑓(𝑋) = 𝑋𝑝−1 𝑋−1 ; 𝑌= 𝑋−1 By using this substitution of 𝑌, 𝑓(𝑌+ 1) = (𝑌+ 1)𝑝−1 𝑌−1 + 1 = 𝑌𝑝−1 + (𝑝 1)𝑌𝑝−2 + ⋯+ ( 𝑝 𝑝−2)𝑌+ ( 𝑝 𝑝−1) We can apply Eisenstein’s criterion to this new polynomial, since 𝑝∣(𝑝 𝑖) for all 1 ≤𝑖≤𝑝−1, and 𝑝2 ∤( 𝑝 𝑝−1) = 𝑝. Thus, 𝑓(𝑌+ 1) is irreducible in ℤ[𝑌], so 𝑓(𝑋) is irreducible in ℤ[𝑋]. Of course, 𝑓(𝑋) is therefore irreducible in ℚ[𝑋] as before.
463 VIII. Groups, Rings and Modules 10. Algebraic integers 10.1. Gaussian integers Recall the ring of Gaussian integers ℤ[𝑖] = {𝑎+ 𝑏𝑖∶𝑎, 𝑏∈ℤ} ≤ℂ. There is a norm func-tion 𝑁∶ℤ[𝑖] →ℤ≥0 given by 𝑎+ 𝑏𝑖↦𝑎2 + 𝑏2, and 𝑁(𝑥𝑦) = 𝑁(𝑥)𝑁(𝑦). This norm is a Euclidean function, giving the Gaussian integers the structure of a Euclidean domain and hence a principal ideal domain and a unique factorisation domain. In particular, the primes are the irreducibles. The units in ℤ[𝑖] are ±1, ±𝑖, since they are the only elements of unit norm.
Example. 2 is not irreducible in ℤ[𝑖], since it factors as (1 + 𝑖)(1 −𝑖). 5 is not irreducible, since it factors as (2 + 𝑖)(2 −𝑖). These are nontrivial factorisations since the norms of the factors are not unit length.
3 is a prime, since it is irreducible. Indeed, 𝑁(3) = 9, so if 3 were reducible it would factor as 𝑎𝑏where 𝑁(𝑎) = 𝑁(𝑏) = 3. But ℤ[𝑖] has no elements of norm 3. Similarly, 7 is a prime.
Proposition. Let 𝑝∈ℤbe a prime. Then, the following are equivalent.
(i) 𝑝is not prime in ℤ[𝑖]; (ii) 𝑝= 𝑎2 + 𝑏2 for 𝑎, 𝑏∈ℤ; (iii) 𝑝= 2 or 𝑝≡1 mod 4.
Proof. Suppose 𝑝is not prime in ℤ[𝑖]. So let 𝑝= 𝑥𝑦for 𝑥, 𝑦∈ℤ[𝑖] not units. Then, 𝑝2 = 𝑁(𝑝) = 𝑁(𝑥)𝑁(𝑦). Since 𝑥, 𝑦are not units, 𝑁(𝑥), 𝑁(𝑦) > 1 and in particular 𝑁(𝑥) = 𝑁(𝑦) = 𝑝. Writing 𝑥= 𝑎+ 𝑏𝑖for 𝑎, 𝑏∈ℤ, we have 𝑝= 𝑁(𝑥) = 𝑎2 + 𝑏2, which is the condition in (ii).
Now, suppose 𝑝= 𝑎2 +𝑏2. The only squares modulo 4 are 0 and 1. Since 𝑝≡𝑎2 +𝑏2 mod 4, we have that 𝑝cannot be congruent to 3, modulo 4.
Finally, suppose 𝑝= 2 or 𝑝≡1 mod 4. We have already observed above that 2 is not prime.
It hence suffices to consider the case where 𝑝≡1 mod 4. We have that (ℤ ⟋ 𝑝ℤ) × is cyclic of order 𝑝−1 by a previous theorem. Hence, if 𝑝≡1 mod 4, we have that 4 ∣𝑝−1, and hence (ℤ ⟋ 𝑝ℤ) × contains an element of order 4. In particular, there exists 𝑥∈ℤwith 𝑥4 ≡1 mod 𝑝, but 𝑥2 ≢1 mod 𝑝. Then 𝑥2 ≡−1 mod 𝑝, or in other words, 𝑝∣(𝑥2 + 1). But this factorises as 𝑝∣(𝑥+ 𝑖)(𝑥−𝑖). We can see that 𝑝∤𝑥+ 𝑖, 𝑝∤𝑥−𝑖, so 𝑝cannot be prime.
Remark. The proof that (iii) implies (ii) is entirely nontrivial. It required lots of theory in order to reach the result, even though its statement did not require even the notion of a complex number.
Theorem. The primes in ℤ[𝑖] are, up to associates, (i) 𝑎+ 𝑏𝑖, where 𝑎, 𝑏∈ℤand 𝑎2 + 𝑏2 = 𝑝is a prime in ℤwith 𝑝= 2 or 𝑝≡1 mod 4; and 464 10. Algebraic integers (ii) the primes 𝑝in ℤsatisfying 𝑝≡3 mod 4.
Proof. First, we must check that all such elements are prime. For (i), note that 𝑁(𝑎+𝑏𝑖) = 𝑝 is prime, so 𝑎+ 𝑏𝑖is irreducible. We can use the above proof to deduce that primes in ℤof form (ii) are primes in ℤ[𝑖].
It now suffices to show that any prime in the Gaussian integers satisfies one of the two above conditions. Let 𝑧be prime in ℤ[𝑖]. We note that 𝑧is also irreducible. Now, 𝑁(𝑧) = 𝑧𝑧, which is a factorisation of the norm into irreducibles.
Let 𝑝be a prime in ℤdividing 𝑁(𝑧). If 𝑝≡3 mod 4, 𝑝is prime in ℤ[𝑖]. So 𝑝∣𝑧or 𝑝∣𝑧so 𝑝 is associate to 𝑧or 𝑧.
Otherwise, 𝑝= 𝑎2 + 𝑏2 = (𝑎+ 𝑏𝑖)(𝑎−𝑏𝑖) where 𝑎± 𝑏𝑖are prime in ℤ[𝑖] as they have norm 𝑝. So we have 𝑝= (𝑎+ 𝑏𝑖)(𝑎−𝑏𝑖) ∣𝑧𝑧, so 𝑧is an associate of 𝑎+ 𝑏𝑖or 𝑎−𝑏𝑖by uniqueness of factorisation.
Remark. In the above theorem, if 𝑝= 𝑎2 + 𝑏2, 𝑎+ 𝑏𝑖and 𝑎−𝑏𝑖are not associate unless 𝑝= 2.
Corollary. An integer 𝑛≥1 is the sum of two squares if and only if every prime factor 𝑝of 𝑛with 𝑝≡3 mod 4 divides 𝑛to an even power.
Proof. Suppose 𝑛= 𝑎2 + 𝑏2. So 𝑛= 𝑁(𝑎+ 𝑏𝑖). Hence 𝑛is a product of norms of primes in the Gaussian integers. By the classification above, those norms are (i) the primes 𝑝∈ℤwith 𝑝≢3 mod 4; and (ii) squares of primes 𝑝∈ℤwith 𝑝≡3 mod 4.
The result follows.
Example. We can write 65 = 5⋅13 as the sum of two primes since 5, 13 ≡1 mod 4. We first factorise 5 and 13 into primes in the Gaussian integers.
5 = (2 + 𝑖)(2 −𝑖); 13 = (2 + 3𝑖)(2 −3𝑖) Thus, the factorisation of 65 into irreducibles in ℤ[𝑖] is 65 = (2 + 3𝑖)(2 + 𝑖)(2 −3𝑖)(2 −𝑖) = [(2 + 3𝑖)(2 + 𝑖)][(2 + 3𝑖)(2 + 𝑖)] = 𝑁((2 + 3𝑖)(2 −𝑖)) = 𝑁(1 + 8𝑖) = 12 + 82 This was dependent on the choice of grouping of terms. Alternatively, 65 = 𝑁((2 + 𝑖)(2 −3𝑖)) = 𝑁(7 + 4𝑖) = 72 + 42 465 VIII. Groups, Rings and Modules 10.2. Algebraic integers Definition. A number 𝛼∈ℂis algebraic if 𝛼is a root of some nonzero polynomial 𝑓∈ ℚ[𝑋]. 𝛼is an algebraic integer if it is a root of some monic polynomial 𝑓∈ℤ[𝑋].
Let 𝑅≤𝑆, and 𝛼∈𝑆. We write 𝑅[𝛼] to denote the smallest subring of 𝑆containing 𝑅and 𝛼. Alternatively, 𝑅[𝛼] is the intersection of all subrings of 𝑆containing 𝑅and 𝛼. Further, 𝑅[𝛼] = Im 𝜑where 𝜑∶𝑅[𝑋] →𝑆is the homomorphism 𝑔(𝑋) ↦𝑔(𝛼).
Definition. Let 𝛼be an algebraic number. Consider the homomorphism 𝜑∶ℚ[𝑋] →ℂ where 𝑔(𝑋) ↦𝑔(𝛼). Since ℚ[𝑋] is a a principal ideal domain, ker 𝜑= (𝑓) for some 𝑓∈ℚ[𝑋].
This ideal contains a nonzero element since 𝛼is an algebraic number, hence 𝑓is nonzero.
Multiplying 𝑓by a unit, we may assume 𝑓is monic without loss of generality. This unique 𝑓is known as the minimal polynomial of 𝛼.
Corollary. All minimal polynomials are irreducible. By the first isomorphism theorem, ℚ[𝑋] ⟋ (𝑓) ≅ℚ[𝛼] ≤ℂ. Any subring of a field is an integral domain. Hence (𝑓) is a prime ideal in ℚ[𝑋], and hence 𝑓is irreducible. In particular, this implies that ℚ[𝛼] is a field.
Proposition. Let 𝛼be an algebraic integer, and 𝑓∈ℚ[𝑋] be its minimal polynomial. Then 𝑓∈ℤ[𝑋], and (𝑓) = ker 𝜃⊲ℤ[𝑋] where 𝜃∶ℤ[𝑋] →ℂis given by 𝑔(𝑋) ↦𝑔(𝛼).
Remark. If 𝛼is an algebraic integer, then the polynomial in the definition can be taken to be minimal without loss of generality. ℤ[𝑋] is not a principal ideal domain, so the above argument cannot work verbatim.
Proof. Let 𝑓be the minimal polynomial of 𝛼. Let 𝜆∈ℚ× such that 𝜆𝑓has coefficients in ℤ and is primitive. Then 𝜆𝑓(𝛼) = 0, so 𝜆𝑓∈ker 𝜃.
Let 𝑔∈ker 𝜃, so in particular 𝑔∈ℤ[𝑋]. Then 𝑔∈ker 𝜑, and hence 𝜆𝑓∣𝑔in ℚ[𝑋]. By a previous lemma, 𝜆𝑓∣𝑔in ℤ[𝑋]. Thus, ker 𝜃= (𝜆𝑓).
Now, since 𝛼is an algebraic integer, we know that there exists a monic polynomial 𝑔∈ker 𝜃 such that 𝑔(𝛼) = 0. Then 𝜆𝑓∣𝑔in ℤ[𝑋], so 𝜆= ±1 as both 𝑓, 𝑔are monic. Hence, 𝑓∈ℤ[𝑋], and (𝜆𝑓) = (𝑓) = ker 𝜃.
Let 𝛼∈ℂbe an algebraic integer. Then, applying the isomorphism theorem to 𝜃, ℤ[𝑋] ⟋ (𝑓) ≅ ℤ[𝛼]. For example: ℤ[𝑋] ⟋ (𝑋2 + 1) ≅ℤ[𝑖] ℤ[𝑋] ⟋ (𝑋2 −2) ≅ℤ[√2] ℤ[𝑋] ⟋ (𝑋2 + 𝑋+ 1) ≅ℤ[−1 + √−3 2 ] ℤ[𝑋] ⟋ (𝑋𝑛−𝑝) ≅ℤ[𝑛 √𝑝] 466 10. Algebraic integers Corollary. If 𝛼is an algebraic integer, and 𝛼∈ℚ, then 𝛼∈ℤ.
Proof. Let 𝛼≠0, since the case where 𝛼= 0 is trivial. Then the minimal polynomial of 𝛼 has coefficients in ℤ. Since 𝛼is rational, the minimal polynomial is 𝑋−𝛼. Hence 𝛼∈ℤas it is a coefficient of the minimal polynomial.
467 VIII. Groups, Rings and Modules 11. Noetherian rings 11.1. Definition Recall the definition of a Noetherian ring.
Definition. A ring 𝑅is Noetherian if, for all sequences of nested ideals 𝐼1 ⊆𝐼2 ⊆⋯, there exists 𝑁∈ℕsuch that for all 𝑛> 𝑁, 𝐼𝑛= 𝐼𝑛+1.
Lemma. Let 𝑅be a ring. Then 𝑅satisfies the ascending chain condition (so 𝑅is Noetherian) if and only if all ideals in 𝑅are finitely generated.
We have already shown that principal ideal domains are Noetherian, since they satisfy this ‘ascending chain’ condition. This now will immediately follow from the lemma.
Proof. First, suppose that all ideals in 𝑅are finitely generated. Let 𝐼1 ⊆𝐼2 ⊆⋯be an ascending chain of ideals. Consider 𝐼= ⋃ ∞ 𝑖=1 𝐼𝑖, which is an ideal. 𝐼is finitely generated, so 𝐼= (𝑎1, … , 𝑎𝑛). These elements belong to a nested union of ideals. In particular, we can choose 𝑁∈ℕsuch that all 𝑎𝑖are contained within 𝐼𝑁. Then, for 𝑛≥𝑁, we find (𝑎1, … , 𝑎𝑛) ⊆𝐼𝑁⊆𝐼𝑛⊆𝐼= (𝑎1, … , 𝑎𝑛) So the inclusions are all equalities, so 𝐼𝑁= 𝐼𝑛.
Conversely, suppose that 𝑅is Noetherian. Suppose that there exists an ideal 𝐽⊲𝑅which is not finitely generated. Let 𝑎1 ∈𝐽. Then since 𝐽is not finitely generated, (𝑎1) ⊂𝐽. We can therefore choose 𝑎2 ∈𝐽∖(𝑎1), and then (𝑎1) ⊂(𝑎1, 𝑎2) ⊂𝐽. Continuing inductively, we contradict the ascending chain condition.
11.2. Hilbert’s basis theorem Theorem. Let 𝑅be a Noetherian ring. Then 𝑅[𝑋] is Noetherian.
Proof. Suppose there exists an ideal 𝐽that is not finitely generated. Let 𝑓 1 ∈𝐽be an ele-ment of minimal degree. Then (𝑓 1) ⊂𝐽. So we can choose 𝑓 2 ∈𝐽∖(𝑓 1), which is also of minimal degree. Inductively we can construct a sequence 𝑓 1, 𝑓 2, …, where the degrees are non-decreasing. Let 𝑎𝑖be the leading coefficient of 𝑓𝑖, for all 𝑖. We then obtain a sequence of ideals (𝑎1) ⊆(𝑎1, 𝑎2) ⊆(𝑎1, 𝑎2, 𝑎3) ⊆⋯in 𝑅. Since 𝑅is Noetherian, there exists 𝑚∈ℕ such that for all 𝑛≥𝑚, we have 𝑎𝑛∈(𝑎1, … , 𝑎𝑚). Let 𝑎𝑚+1 = ∑ 𝑚 𝑖=1 𝜆𝑖𝑎𝑖, since 𝑎𝑚+1 lies in the ideal (𝑎1, … , 𝑎𝑚). Now we define 𝑔(𝑋) = 𝑚 ∑ 𝑖=1 𝜆𝑖𝑋deg(𝑓𝑚+1−𝑓𝑖)𝑓𝑖 The degree of 𝑔is equal to the degree of 𝑓 𝑚+1, and they have the same leading coefficient 𝑎𝑚+1. Then, consider 𝑓 𝑚+1 −𝑔∈𝐽and deg(𝑓 𝑚+1 −𝑔) < deg 𝑓 𝑚+1. By minimality of the degree of 𝑓 𝑚+1, 𝑓 𝑚+1−𝑔∈(𝑓 1, … , 𝑓 𝑚), hence 𝑓 𝑚+1 ∈(𝑓 1, … , 𝑓 𝑚). This contradicts the choice of 𝑓 𝑚+1, so 𝐽is in fact finitely generated.
468 11. Noetherian rings Corollary. ℤ[𝑋1, … , 𝑋𝑛] is Noetherian. Similarly, 𝐹[𝑋1, … , 𝑋𝑛] is Noetherian for any field 𝐹, since fields satisfy the ascending chain condition.
Example. Let 𝑅= ℂ[𝑋1, … , 𝑋𝑛]. Let 𝑉⊆ℂ𝑛be a subset of the form 𝑉= {(𝑎1, … , 𝑎𝑛) ∈ℂ𝑛∶𝑓(𝑎1, … , 𝑎𝑛) = 0, ∀𝑓∈ℱ} where ℱ⊆𝑅is a (possibly infinite) set of polynomials. Such a set is referred to as an algebraic variety. Let 𝐼= { 𝑚 ∑ 𝑖=1 𝜆𝑖𝑓𝑖∶𝑚∈ℕ, 𝜆𝑖∈𝑅𝑖, 𝑓𝑖∈ℱ} We can check that 𝐼⊲𝑅. Since 𝑅is Noetherian, 𝐼= (𝑔1, … , 𝑔𝑟). Hence 𝑉= {(𝑎1, … , 𝑎𝑛) ∈ℂ𝑛∶𝑔(𝑎1, … , 𝑎𝑛) = 0, ∀𝑔∈𝐼} Lemma. Let 𝑅be a Noetherian ring, and 𝐼⊲𝑅. Then 𝑅 ⟋ 𝐼is Noetherian.
Proof. Let 𝐽′ 1 ⊆𝐽′ 2 ⊆⋯be a chain of ideals in 𝑅 ⟋ 𝐼. By the ideal correspondence, 𝐽′ 𝑖corres-ponds to an ideal 𝐽𝑖that contains 𝐼, so 𝐽′ 𝑖= 𝐽𝑖 ⟋ 𝐼. So 𝐽1 ⊆𝐽2 ⊆⋯is a chain of ideals in 𝑅.
Since 𝑅is Noetherian, there exists 𝑁∈ℕsuch that for all 𝑛≥𝑁, we have 𝐽𝑁= 𝐽𝑛, and so 𝐽′ 𝑁= 𝐽′ 𝑛. Hence 𝑅 ⟋ 𝐼satisfies the ascending chain condition.
Example. The ring of Gaussian integers ℤ ⟋ (𝑋2 + 1) is Noetherian. If 𝑅[𝑋] is Noetherian, then 𝑅[𝑋] ⟋ (𝑋) ≅𝑅is Noetherian. This is a converse to the Hilbert basis theorem.
The ring of polynomials in countably many variables is not Noetherian.
ℤ[𝑋1, 𝑋2, … ] = ⋃ 𝑛∈ℕ ℤ[𝑋1, … , 𝑋𝑛] In particular, consider the ascending chain (𝑋1) ⊂(𝑋1, 𝑋2) ⊂(𝑋1, 𝑋2, 𝑋3) ⊂⋯.
Let 𝑅= {𝑓∈ℚ[𝑋]∶𝑓(0) ∈ℤ} ≤ℚ[𝑋]. Even though ℚ[𝑋] is Noetherian, 𝑅is not. Indeed, consider (𝑋) ⊂( 1 2𝑋) ⊂( 1 4𝑋) ⊂( 1 8𝑋) ⊂⋯. These inclusions are strict, since 2 ∈𝑅is not a unit.
469 VIII. Groups, Rings and Modules 12. Modules 12.1. Definitions Definition. Let 𝑅be a ring. A module over 𝑅is a triple (𝑀, +, ⋅) consisting of a set 𝑀and two operations +∶𝑀× 𝑀→𝑀and ⋅∶𝑅× 𝑀→𝑀, that satisfy (i) (𝑀, +) is an abelian group with identity 0 = 0𝑀; (ii) (𝑟1 + 𝑟2) ⋅𝑚= 𝑟1 ⋅𝑚+ 𝑟2 ⋅𝑚; (iii) 𝑟⋅(𝑚1 + 𝑚2) = 𝑟⋅𝑚1 + 𝑟⋅𝑚2; (iv) 𝑟1 ⋅(𝑟2 ⋅𝑚) = (𝑟1 ⋅𝑟2) ⋅𝑚; (v) 1𝑅⋅𝑚= 𝑚; Remark. Closure is implicitly required by the types of the + and ⋅operations.
Example. A module over a field is precisely a vector space.
A ℤ-module is precisely the same as an abelian group, since ⋅∶ℤ× 𝐴→𝐴; 𝑛⋅𝑎= ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩ 𝑎+ ⋯+ 𝑎 ⏟ ⎵ ⎵ ⏟ ⎵ ⎵ ⏟ 𝑛times if 𝑛> 0 0 if 𝑛= 0 −(𝑎+ ⋯+ 𝑎 ⏟ ⎵ ⎵ ⏟ ⎵ ⎵ ⏟ −𝑛times ) if 𝑛< 0 Let 𝐹be a field, and 𝑉be a vector space over 𝐹. Let 𝛼∶𝑉→𝑉be an endomorphism. We can turn 𝑉into an 𝐹[𝑋]-module by ⋅∶𝐹[𝑋] × 𝑉→𝑉; 𝑓⋅𝑣= (𝑓(𝛼))(𝑣) Note that the structure of the 𝐹[𝑋]-module depends on the choice of 𝛼. We can write 𝑉= 𝑉 𝛼 to disambiguate.
For any ring 𝑅, we can consider 𝑅𝑛as an 𝑅-module via 𝑟⋅(𝑟1, … , 𝑟 𝑛) = (𝑟⋅𝑟1, … , 𝑟⋅𝑟 𝑛) In particular, the case 𝑛= 1 shows that any ring 𝑅can be considered an 𝑅-module where the scalar multiplication in the ring and the module agree.
For an ideal 𝐼⊲𝑅, we can regard 𝐼as an 𝑅-module, since 𝐼is preserved under multiplication by elements in 𝑅. The quotient ring 𝑅 ⟋ 𝐼is also an 𝑅-module, defining multiplication as 𝑟⋅(𝑠+ 𝐼) = 𝑟𝑠+ 𝐼.
Let 𝜑∶𝑅→𝑆be a ring homomorphism. Then any 𝑆-module can be regarded as an 𝑅-module. We define 𝑟⋅𝑚= 𝜑(𝑟) ⋅𝑚. In particular, this applies when 𝑅is a subring of 𝑆, and 𝜑is the inclusion map. So any module over a ring can be viewed as a module over any subring.
470 12. Modules Definition. Let 𝑀be an 𝑅-module. Then 𝑁⊆𝑀is an 𝑅-submodule of 𝑀, written 𝑁≤𝑀, if (𝑁, +) ≤(𝑀, +), and for all 𝑟𝑛∈𝑁for all 𝑟∈𝑅and 𝑛∈𝑁.
Example. By considering 𝑅as an 𝑅-module, a subset of 𝑅is an 𝑅-submodule if and only if it is an ideal. If 𝑅= 𝐹is a field, this definition corresponds to the definition of a vector subspace.
Definition. Let 𝑁≤𝑀be 𝑅-modules. Then, the quotient 𝑀 ⟋ 𝑁is defined as the quotient of groups under addition, and with scalar multiplication defined as 𝑟⋅(𝑚+ 𝑁) = 𝑟𝑚+ 𝑁.
This is well-defined, since 𝑁is preserved under scalar multiplication. This makes 𝑀 ⟋ 𝑁an 𝑅-module.
Remark. Submodules are analogous both to subrings and to ideals.
Definition. Let 𝑀, 𝑁be 𝑅-modules. Then 𝑓∶𝑀→𝑁is a 𝑅-module homomorphism if it is a homomorphism of (𝑀, +) and (𝑁, +), and scalar multiplication is preserved: 𝑓(𝑟⋅𝑚) = 𝑟⋅𝑓(𝑚). An 𝑅-module isomorphism is an 𝑅-module homomorphism that is a bijection.
Example. If 𝑅= 𝐹is a field, 𝐹-module homomorphisms are exactly linear maps.
Theorem. Let 𝑓∶𝑀→𝑁be an 𝑅-module homomorphism. Then (i) ker 𝑓= {𝑚∈𝑀∶𝑓(𝑚) = 0} ≤𝑀; (ii) Im 𝑓= {𝑓(𝑚) ∈𝑁∶𝑚∈𝑀} ≤𝑁; (iii) 𝑀 ⟋ ker 𝑓≅Im 𝑓.
Theorem. Let 𝐴, 𝐵≤𝑀be 𝑅-submodules. Then (i) 𝐴+ 𝐵= {𝑎+ 𝑏∶𝑎∈𝐴, 𝑏∈𝐵} ≤𝑀; (ii) 𝐴∩𝐵≤𝑀; (iii) 𝐴 ⟋ 𝐴∩𝐵≅𝐴+ 𝐵 ⟋ 𝐵.
Theorem. For 𝑁≤𝐿≤𝑀are 𝑅-submodules, then 𝑀/𝑁 ⟋ 𝐿/𝑁≅𝑀 ⟋ 𝐿 For 𝑁≤𝑀, there is a correspondence between submodules of 𝑀 ⟋ 𝑁and submodules of 𝑀 containing 𝑁. These isomorphism theorems can be proved exactly as before. Note that these results apply to vector spaces; for example, the first isomorphism theorem immediately gives the rank-nullity theorem.
12.2. Finitely generated modules Definition. Let 𝑀be an 𝑅-module. If 𝑚∈𝑀, then we write 𝑅𝑚= {𝑟𝑚∶𝑟∈𝑅}. This is an 𝑅-submodule of 𝑀, known as the submodule generated by 𝑚.
471 VIII. Groups, Rings and Modules If 𝐴, 𝐵≤𝑀, we can define 𝐴+ 𝐵= {𝑎+ 𝑏∶𝑎∈𝐴, 𝑏∈𝐵}, known as the sum of submodules.
In particular, this sum is commutative.
Definition. A module 𝑀is finitely generated if it is the sum of finitely many submodules generated by a single element. In other words, 𝑀= 𝑅𝑚1 + ⋯+ 𝑅𝑚𝑛.
This is the analogue of finite dimensionality in linear algebra.
Lemma. An 𝑅-module 𝑀is finitely generated if and only if there exists a surjective 𝑅-module homomorphism 𝑓∶𝑅𝑛→𝑀for some 𝑛.
Proof. If 𝑀is finitely generated, we have 𝑀= 𝑅𝑚1 + ⋯+ 𝑅𝑚𝑛. We define 𝑓∶𝑅𝑛→𝑀by (𝑟1, … , 𝑟 𝑛) ↦𝑟1𝑚1 + ⋯+ 𝑟 𝑛𝑚𝑛. This is surjective.
Conversely, suppose such a surjective homomorphism 𝑓exists. Let 𝑒𝑖= (0, … , 1, … , 0) be the element of 𝑅𝑛with all entries zero except for 1 in the 𝑖th place. Let 𝑚𝑖= 𝑓(𝑒𝑖). Then, since 𝑓is surjective, any element 𝑚∈𝑀is contained in the image of 𝑓, so is of the form 𝑓(𝑟1, … , 𝑟 𝑛) = 𝑟1𝑚1 + ⋯+ 𝑟 𝑛𝑚𝑛.
Corollary. Any quotient by a submodule of a finitely generated module is finitely generated.
Proof. Let 𝑁≤𝑀, where 𝑀is finitely generated. Then there exists a surjective 𝑅-module homomorphism 𝑓∶𝑅𝑛→𝑀. Then 𝑞∘𝑓, where 𝑞is the quotient map, is also a surjective homomorphism. So 𝑀 ⟋ 𝑁is finitely generated.
Example. It is not always the case that a submodule of a finitely generated module is finitely generated. Let 𝑅be a non-Noetherian ring, and 𝐼an ideal in 𝑅that is not finitely generated (in the ring sense). 𝑅is a finitely generated 𝑅-module, since 𝑅1 = 𝑅. 𝐼is a submodule of 𝑅, which is not finitely generated (in the module sense).
Remark. If 𝑅is Noetherian, it is always the case that submodules of finitely generated 𝑅-modules are finitely generated. This will be shown on the example sheets.
12.3. Torsion Definition. Let 𝑀be an 𝑅-module.
(i) 𝑚∈𝑀is torsion if there exists 0 ≠𝑟∈𝑅such that 𝑟𝑚= 0; (ii) 𝑀is a torsion module if every element is torsion; (iii) 𝑀is a torsion-free module if 0 is the only torsion element.
Example. The torsion elements in a ℤ-module (which is an abelian group) are precisely the elements of finite order. If 𝐹is a field, any 𝐹-module is torsion-free.
472 12. Modules 12.4. Direct sums Definition. Let 𝑀1, … , 𝑀𝑛be 𝑅-modules. Then the direct sum of 𝑀1, … , 𝑀𝑛, written 𝑀1 ⊕ ⋯⊕𝑀𝑛, is the set 𝑀1 × ⋯× 𝑀𝑛, with the operations of addition and scalar multiplication defined componentwise. We can show that the direct sum of (finitely many) 𝑅-modules is an 𝑅-module.
Example. 𝑅𝑛= 𝑅⊕⋯⊕𝑅, where we take the direct sum of 𝑛copies of 𝑅.
Lemma. Let 𝑀= ⨁ 𝑛 𝑖=1 𝑀𝑖, and for each 𝑀𝑖, let 𝑁𝑖≤𝑀𝑖. Then 𝑁= ⨁ 𝑛 𝑖=1 𝑁𝑖is a submod-ule of 𝑀. Further, 𝑀 ⟋ 𝑁= 𝑛 ⨁ 𝑖=1 𝑀𝑖 ⟋𝑛 ⨁ 𝑖=1 𝑁𝑖 ≅ 𝑛 ⨁ 𝑖=1 𝑀𝑖 ⟋ 𝑁𝑖 Proof. First, we can see that this 𝑁is a submodule. Applying the first isomorphism theorem to the surjective 𝑅-module homomorphism 𝑀→⨁ 𝑛 𝑖=1 𝑀𝑖 ⟋ 𝑁𝑖given by (𝑚1, … , 𝑚𝑛) ↦(𝑚1+ 𝑁1, … , 𝑚𝑛+ 𝑁𝑛), the result follows as required, since the kernel is 𝑁.
12.5. Free modules Definition. Let 𝑚1, … , 𝑚𝑛∈𝑀. The set {𝑚1, … , 𝑚𝑛} is independent if ∑ 𝑛 𝑖=1 𝑟𝑖𝑚𝑖= 0 implies that the 𝑟𝑖are all zero.
Definition. A subset 𝑆⊆𝑀generates 𝑀freely if: (i) 𝑆generates 𝑀, so for all 𝑚∈𝑀, we can find finitely many entries 𝑠𝑖and coefficients 𝑟𝑖such that 𝑚= ∑ 𝑘 𝑖=1 𝑟𝑖𝑠𝑖; (ii) any function 𝜓∶𝑆→𝑁, where 𝑁is an 𝑅-module, extends to an 𝑅-module homo-morphism 𝜃∶𝑀→𝑁.
Remark. In (ii), such an extension 𝜃is always unique if it exists, by (i).
Definition. An 𝑅-module 𝑀freely generated by some subset 𝑆⊆𝑀is called free. We say that 𝑆is a free basis for 𝑀.
Remark. Free bases in the study of modules are analogous to bases in linear algebra. All vector spaces are free modules, but not all modules are free.
Proposition. For a finite subset 𝑆= {𝑚1, … , 𝑚𝑛} ⊆𝑀, the following are equivalent.
(i) 𝑆generates 𝑀freely; (ii) 𝑆generates 𝑀, and 𝑆is independent; (iii) every element of 𝑀can be written uniquely as 𝑟1𝑚1 + ⋯+ 𝑟 𝑛𝑚𝑛for some 𝑟𝑖∈𝑅; 473 VIII. Groups, Rings and Modules (iv) the 𝑅-module homomorphism 𝑅𝑛→𝑀given by (𝑟1, … , 𝑟 𝑛) ↦𝑟1𝑚1 + ⋯+ 𝑟 𝑛𝑚𝑛is bijective, so is an isomorphism.
Proof. Not all implications are shown, but they are similar to arguments found in Part IB Linear Algebra. We show (i) implies (ii). Let 𝑆generate 𝑀freely. Suppose 𝑆is not independ-ent. Then there exist 𝑟𝑖such that ∑ 𝑛 𝑖=1 𝑟𝑖𝑚𝑖= 0 but not all 𝑟𝑖are zero. Let 𝑟𝑗≠0. Since 𝑆 generates 𝑀freely, consider the module homomorphism 𝜓∶𝑆→𝑅given by 𝜓(𝑚𝑖) = {1 if 𝑖= 𝑗 0 otherwise Then 0 = 𝜃(0) = 𝜃( 𝑛 ∑ 𝑖=1 𝑟𝑖𝑚𝑖) = 𝑛 ∑ 𝑖=1 𝑟𝑖𝜃(𝑚𝑖) = 𝑟𝑗≠0 This is a contradiction, so 𝑆is independent.
To show (ii) implies (iii), it suffices to show uniqueness. If there exist two ways to write an element as a linear combination, consider their difference to find a contradiction from (ii).
We can show (iii) implies (i). Then it remains to show (iii) and (iv) are equivalent.
Example. A non-trivial finite abelian group is not a free ℤ-module.
The set {2, 3} generates ℤas a ℤ-module. This is not a free basis, since they are not independ-ent: 2 ⋅3 −3 ⋅2 = 0. However, it contains no subset that is a free basis. This is different to vector spaces, where we can always construct a basis from a subset of a spanning set.
Proposition (invariance of dimension). Let 𝑅be a nonzero ring. If 𝑅𝑚≅𝑅𝑛as 𝑅-modules, then 𝑚= 𝑛.
Proof. Let 𝐼⊲𝑅, and 𝑀an 𝑅-module. We define 𝐼𝑀= {∑𝑎𝑖𝑚𝑖∶𝑎𝑖∈𝐼, 𝑚𝑖∈𝑀}. Since 𝐼is an ideal, we can show that 𝐼𝑀is a submodule of 𝑀. The quotient module 𝑀 ⟋ 𝐼𝑀is an 𝑅-module, but we can also show that it is an 𝑅 ⟋ 𝐼-module, by defining scalar multiplication as (𝑟+ 𝐼) ⋅(𝑚+ 𝐼𝑀) = (𝑟⋅𝑚+ 𝐼𝑀) We can check that this is well-defined; this follows from the fact that for 𝑏∈𝐼, 𝑏⋅(𝑚+𝐼𝑀) = 𝑏𝑚+ 𝐼𝑀, but 𝑏∈𝐼so 𝑏𝑚∈𝐼𝑀.
Now, suppose that 𝑅𝑚≅𝑅𝑛. Then let 𝐼⊲𝑅be a maximal ideal in 𝑅. We can prove the existence of such an ideal under the assumption of the axiom of choice, and in particular using Zorn’s lemma. By the above discussion, we find an isomorphism of 𝑅 ⟋ 𝐼-modules (𝑅 ⟋ 𝐼) 𝑚 ≅𝑅𝑚 ⟋ 𝐼𝑅𝑚≅𝑅𝑛 ⟋ 𝐼𝑅𝑛≅(𝑅 ⟋ 𝐼) 𝑛 This is an isomorphism of vector spaces over 𝑅 ⟋ 𝐼which is a field, since 𝐼is maximal. Hence, using the corresponding result from linear algebra, 𝑛= 𝑚.
474 12. Modules 12.6. Row and column operations We will assume that 𝑅is a Euclidean domain in this subsection, and let 𝜑be a Euclidean function for 𝑅. We will consider an 𝑚× 𝑛matrix with entries in 𝑅.
Definition. The elementary row operations on a matrix are (i) add 𝜆∈𝑅multiplied by the 𝑗th row to the 𝑖th row, where 𝑖≠𝑗; (ii) swap the 𝑖th row and the 𝑗th row; (iii) multiply the 𝑖th row by 𝑢∈𝑅×.
Each of these operations can be realised by left-multiplication by some 𝑚×𝑚matrix. These operations are all invertible, so their matrices are all invertible.
We can define elementary column operations in an analogous way, using right-multiplication by an 𝑛× 𝑛matrix instead.
Definition. Two 𝑚×𝑛matrices 𝐴, 𝐵are equivalent if there exists a sequence of elementary row and column operations that transforms one matrix into the other. If they are equivalent, then there exist invertible matrices 𝑃, 𝑄such that 𝐵= 𝑄𝐴𝑃.
Definition. A 𝑘× 𝑘minor of an 𝑚× 𝑛matrix 𝐴is the determinant of a 𝑘× 𝑘submatrix of 𝐴, which is a matrix of 𝐴produced by removing 𝑚−𝑘rows and 𝑛−𝑘columns.
The 𝑘th Fitting ideal Fit𝑘(𝐴) ⊲𝑅is the ideal generated by the 𝑘× 𝑘minors of 𝐴.
Lemma. The 𝑘th Fitting ideal of a matrix is invariant under elementary row and column operations.
Proof. It suffices by symmetry to show that the elementary row operations do not change the Fitting ideal. For the first elementary row operation on a matrix 𝐴, suppose we add 𝜆∈𝑅 multiplied by the 𝑗th row to the 𝑖th row, yielding a matrix 𝐴′. In particular, 𝑎𝑖𝑘↦𝑎𝑖𝑘+𝜆𝑎𝑗𝑘 for all 𝑘. Let 𝐶be a 𝑘× 𝑘submatrix of 𝐴and 𝐶′ the corresponding submatrix of 𝐴′.
If row 𝑖was not chosen in 𝐶, then 𝐶and 𝐶′ are the same matrix. Hence the corresponding minors are equal. If row 𝑖and row 𝑗were both chosen in 𝐶, we have that 𝐶, 𝐶′ differ by a row operation. Since the determinant is invariant under this elementary row operations, the corresponding minors are equal.
If row 𝑖was chosen but row 𝑗was not chosen, by expanding the determinant along the 𝑖th row, we find det 𝐶′ = det 𝐶+ 𝜆det 𝐷 where we can show that 𝐷is a 𝑘× 𝑘submatrix of 𝐴that includes row 𝑗but not row 𝑖. By definition, det 𝐷∈Fit𝑘(𝐴) and det 𝐶∈Fit𝑘(𝐴), so certainly det 𝐶′ ∈Fit𝑘(𝐴). Hence Fit𝑘(𝐴′) ⊆Fit𝑘(𝐴). By the invertibility of the elementary row operations, Fit𝑘(𝐴′) ⊇Fit𝑘(𝐴).
The proofs for the other elementary row operations are left as an exercise.
475 VIII. Groups, Rings and Modules 12.7. Smith normal form Theorem. An 𝑚× 𝑛matrix 𝐴= (𝑎𝑖𝑗) over a Euclidean domain 𝑅is equivalent to a matrix of the form ⎛ ⎜ ⎜ ⎜ ⎜ ⎝ 𝑑1 ⋱ 𝑑𝑡 0 ⋱ ⎞ ⎟ ⎟ ⎟ ⎟ ⎠ ; 𝑑1 ∣𝑑2 ∣⋯∣𝑑𝑡 The 𝑑𝑖are known as invariant factors, and they are unique up to associates.
Proof. If 𝐴= 0, the matrix is already in Smith normal form. Otherwise, we can swap columns and rows such that 𝑎11 ≠0. We will reduce 𝜑(𝑎11) as much as possible until it divides every other element in the matrix, using the following algorithm.
If 𝑎11 ∤𝑎1𝑗for some 𝑗≥2, then 𝑎1𝑗= 𝑞𝑎11 + 𝑟where 𝑞, 𝑟∈𝑅and 𝜑(𝑟) < 𝜑(𝑎11). We can subtract 𝑞multiplied by column 1 from column 𝑗. Swapping such columns leaves 𝑎11 = 𝑟. If 𝑎11 ∤𝑎𝑖1 for some 𝑖≥2, then repeat the above process using row operations. Now, 𝑎11 ∣𝑎𝑖𝑗 for all 𝑖, 𝑗. These steps are repeated until 𝑎11 divides all entries of the first row and first column. This algorithm will always terminate, for example because the Euclidean function takes values in ℤ≥0 and 𝜑(𝑎11) strictly decreases in each iteration.
Now, we can subtract multiples of the first row and column from the others to give 𝐴= ⎛ ⎜ ⎜ ⎝ 𝑎11 0 ⋯ 0 0 ⋮ 𝐴′ 0 ⎞ ⎟ ⎟ ⎠ If 𝑎11 ∤𝑎𝑖𝑗for 𝑖, 𝑗≥2, then add the 𝑖th row to the first row. There is now an element in the first row that does 𝑎11 not divide. We can then perform column operations as above to decrease 𝜑(𝑎11). We will then restart the algorithm. After finitely many steps, this algorithm will terminate and 𝑎11 will divide all elements 𝑎𝑖𝑗of the matrix.
𝐴= ⎛ ⎜ ⎜ ⎝ 𝑎11 0 ⋯ 0 0 ⋮ 𝐴′ 0 ⎞ ⎟ ⎟ ⎠ ; 𝑎11 ≡𝑑1 ∣𝑎𝑖𝑗 We can now apply the algorithm to 𝐴′, since column and row operations not including the first row or column do not change whether 𝑎11 ∣𝑎𝑖𝑗.
We now demonstrate uniqueness of the invariant factors. Suppose 𝐴has Smith normal form with invariant factors 𝑑𝑖where 𝑑1 ∣⋯∣𝑑𝑡. Then, for all 𝑘, Fit𝑘(𝐴) can be evaluated in Smith normal form by invariance of the Fitting ideal under row and column operations. Hence Fit𝑘(𝐴) = (𝑑1𝑑2 ⋯𝑑𝑘) ⊲𝑅. Thus, the product 𝑑1 ⋯𝑑𝑘depends only on 𝐴, and is unique up to associates. Cancelling, we can see that each 𝑑𝑖depends only on 𝐴, up to associates.
476 12. Modules Example. Consider the matrix over ℤgiven by 𝐴= (2 −1 1 2 ) Using elementary row and column operations, (2 −1 1 2 ) 𝑐1↦𝑐1+𝑐2 − − − − − − →(1 −1 3 2 ) 𝑐2↦𝑐1+𝑐2 − − − − − − →(1 0 3 5) 𝑟2↦−3𝑟1+𝑟2 − − − − − − − − →(1 0 0 5) This is in Smith normal form as 1 ∣5.
Alternatively, (𝑑1) = (2, −1, 1, 2) = (1). So 𝑑1 = ±1. Further, (𝑑1𝑑2) = (det 𝐴) = (5). So 𝑑1𝑑2 = ±5 and hence 𝑑2 = ±5.
12.8. The structure theorem Lemma. Let 𝑅be a Euclidean domain with Euclidean function 𝜑(or, indeed, a principal ideal domain). Any submodule of the free module 𝑅𝑚is generated by at most 𝑚elements.
Proof. Let 𝑁≤𝑅𝑚. Consider 𝐼= {𝑟∈𝑅∶∃𝑟2, … , 𝑟 𝑚∈𝑅, (𝑟, 𝑟2, … , 𝑟 𝑚) ∈𝑁} Since 𝑁is a submodule, this is an ideal. Since 𝑅is a principal ideal domain, 𝐼= (𝑎) for some 𝑎∈𝑅. Let 𝑛= (𝑎, 𝑎2, … , 𝑎𝑚) ∈𝑁. For (𝑟1, … , 𝑟 𝑚) ∈𝑁, we have 𝑟1 = 𝑟𝑎for some 𝑟. Hence (𝑟1, … , 𝑟 𝑚)−𝑟𝑛= (0, 𝑟2−𝑟𝑎2, … , 𝑟 𝑚−𝑟𝑎𝑚), which lies in 𝑁′ = 𝑁∩{0}×𝑅𝑚−1 ≤𝑅𝑚−1, hence 𝑁= 𝑅𝑛+ 𝑁′. By induction, 𝑁′ is generated by 𝑛2, … , 𝑛𝑚, hence (𝑛, 𝑛2, … , 𝑛𝑚) generate 𝑁.
Theorem. Let 𝑅be a Euclidean domain, and 𝑁≤𝑅𝑚. Then there is a free basis 𝑥1, … , 𝑥𝑚 for 𝑅𝑚such that 𝑁is generated by 𝑑1𝑥1, … , 𝑑𝑡𝑥𝑡for some 𝑑𝑖∈𝑅and 𝑡≤𝑚, and such that 𝑑1 ∣⋯∣𝑑𝑡.
Proof. By the above lemma, we have 𝑁= 𝑅𝑦1 + ⋯+ 𝑅𝑦𝑛for some 𝑦𝑖∈𝑅𝑚for some 𝑛≤𝑚. Each 𝑦𝑖belongs to 𝑅𝑚so we can form the 𝑚× 𝑛matrix 𝐴which has columns 𝑦𝑖. 𝐴 is equivalent to a matrix 𝐴′ in Smith normal form with invariant factors 𝑑1 ∣⋯∣𝑑𝑡.
𝐴′ is obtained from 𝐴by elementary row and column operations. Switching row 𝑖and row 𝑗 in 𝐴corresponds to reassigning the standard basis elements 𝑒𝑖and 𝑒𝑗to each other. Adding a multiple of row 𝑖to row 𝑗corresponds to replacing 𝑒1, … , 𝑒𝑚with a linear combination of these basis elements which is a free basis. In general, each row operation simply changes the choice of free basis used for 𝑅𝑚. Analogously, each column operation changes the set of generators 𝑦𝑖for 𝑁.
Hence, after applying these row and column operations, the free basis 𝑒𝑖of 𝑅𝑚is converted into 𝑥1, … , 𝑥𝑚, and 𝑁is generated by 𝑑1𝑥1, … , 𝑑𝑡𝑥𝑡.
477 VIII. Groups, Rings and Modules Theorem (structure theorem for finitely generated modules over Euclidean domains). Let 𝑅be a Euclidean domain, and 𝑀a finitely generated module over 𝑅. Then 𝑀≅𝑅 ⟋ (𝑑1) ⊕⋯⊕𝑅 ⟋ (𝑑𝑡) ⊕𝑅⊕⋯⊕𝑅 ⏟⎵ ⎵⏟⎵ ⎵⏟ 𝑘copies ≅𝑅 ⟋ (𝑑1) ⊕⋯⊕𝑅 ⟋ (𝑑𝑡) ⊕𝑅𝑘 for some 0 ≠𝑑𝑖∈𝑅and 𝑑1 ∣⋯∣𝑑𝑡, and where 𝑘≥0. The 𝑑𝑖are called invariant factors.
Proof. Since 𝑀is a finitely generated module, there exists a surjective 𝑅-module homo-morphism 𝜑∶𝑅𝑚→𝑀for some 𝑚. By the first isomorphism theorem, 𝑀≅𝑅𝑚 ⟋ ker 𝜑.
By the previous theorem, there exists a free basis 𝑥1, … , 𝑥𝑚for 𝑅𝑚such that ker 𝜑≤𝑅𝑚is generated by 𝑑1𝑥1, … , 𝑑𝑡𝑥𝑡and where 𝑑1 ∣⋯∣𝑑𝑡. Then, 𝑀≅ 𝑅⊕… 𝑅 ⏟ ⎵ ⏟ ⎵ ⏟ 𝑘copies 𝑑1𝑅⊕⋯⊕𝑑𝑡𝑅⊕0 ⊕⋯⊕0 ⏟ ⎵ ⎵ ⏟ ⎵ ⎵ ⏟ 𝑚−𝑡copies ≅𝑅 ⟋ (𝑑1) ⊕⋯⊕𝑅 ⟋ (𝑑𝑡) ⊕𝑅⊕⋯⊕𝑅 ⏟⎵ ⎵⏟⎵ ⎵⏟ 𝑚−𝑡copies Remark. After deleting those 𝑑𝑖which are units, the invariant factors of 𝑀are unique up to associates. The proof is omitted.
Corollary. Let 𝑅be a Euclidean domain. Then any finitely generated torsion-free module is free.
Proof. Since 𝑀is torsion-free, there are no submodules of the form 𝑅 ⟋ (𝑑) with 𝑑nonzero, since then multiplying an element of 𝑀by 𝑑would give zero. Hence, by the structure the-orem, 𝑀≅𝑅𝑚for some 𝑚.
Example. Consider 𝑅= ℤ, and the abelian group 𝐺= ⟨𝑎, 𝑏⟩subject to the relations 2𝑎+𝑏= 0 and −𝑎+ 2𝑏= 0, so 𝐺≅ℤ2 ⟋ 𝑁where 𝑁is the ℤ-submodule of ℤ2 generated by (2, 1) and (−1, 2). Consider 𝐴= (2 −1 1 2 ) which has Smith normal form 𝑑1 = 1 and 𝑑2 = 5. Hence, by changing basis for ℤ2, we can let 𝑁be generated by (1, 0) and (0, 5). Hence, 𝐺≅ℤ⊕ℤ ⟋ ℤ⊕5ℤ≅ℤ ⟋ 5ℤ 478 12. Modules 12.9. Primary decomposition theorem More generally, applying the structure theorem to ℤ-modules, we obtain the structure the-orem for finitely generated abelian groups: Theorem. Let 𝐺be a finitely generated abelian group. Then 𝐺≅𝐶𝑑1 × ⋯× 𝐶𝑑𝑡× ℤ𝑟 where 𝑑1 ∣⋯∣𝑑𝑡in ℤ, and 𝑟≥0.
We have replaced the submodule notation ℤ ⟋ 𝑛ℤand ⊕with the group notation 𝐶𝑛and ×.
The previous theorem for the structure of finite abelian groups is a special case of this the-orem, where 𝑟= 0. We have also seen that any finite abelian group can be written as a product of cyclic groups of prime power order. This also has a generalisation for modules.
The previous result relied on the lemma 𝐶𝑚𝑛≅𝐶𝑚× 𝐶𝑛where 𝑚and 𝑛are coprime. There is an analogous result for principal ideal domains.
Lemma. Let 𝑅be a principal ideal domain, and 𝑎, 𝑏∈𝑅with unit greatest common divisor.
Then, treating these quotients as 𝑅-modules, 𝑅 ⟋ (𝑎𝑏) ≅𝑅 ⟋ (𝑎) ⊕𝑅 ⟋ (𝑏) Proof. Since 𝑅is a principal ideal domain, (𝑎, 𝑏) = (𝑑) for some 𝑑∈𝑅. The greatest common divisor of 𝑎, 𝑏is a unit, so 𝑑is a unit, giving (𝑎, 𝑏) = 𝑅. Hence, there exist 𝑟, 𝑠∈𝑅such that 𝑟𝑎+ 𝑠𝑏= 1. This is a generalisation of Bézout’s theorem.
Now, we define an 𝑅-module homomorphism 𝜓∶𝑅→𝑅 ⟋ (𝑎) + 𝑅 ⟋ (𝑏) by 𝜓(𝑥) = (𝑥+ (𝑎), 𝑥+ (𝑏)). Then 𝜓(𝑠𝑏) = (𝑠𝑏+(𝑎), 𝑠𝑏+(𝑏)) = (1−𝑟𝑎+(𝑎), 𝑠𝑏+(𝑏)) = (1+(𝑎), (𝑏)), and similarly 𝜓(𝑟𝑎) = ((𝑎), 1 + (𝑏)). Hence, 𝜓(𝑠𝑏𝑥+ 𝑟𝑏𝑦) = (𝑥+ (𝑎), 𝑦+ (𝑏)) so 𝜓is surjective.
Clearly we have (𝑎𝑏) ⊂ker 𝜓, so it suffices to show the converse. If 𝑥∈ker 𝜓, then 𝑥∈(𝑎) and 𝑥∈(𝑏), so 𝑥∈(𝑎) ∩(𝑏). Since 𝑥= 𝑥(𝑟𝑎+ 𝑠𝑏) = 𝑟(𝑎𝑥) + 𝑠(𝑏𝑥), we must have that 𝑠(𝑏𝑥) ∈(𝑎) and 𝑟(𝑎𝑥) ∈(𝑏), so 𝑥∈(𝑎𝑏). Hence ker 𝜓= (𝑎𝑏), and the result follows from the first isomorphism theorem for modules.
Lemma (primary decomposition theorem). Let 𝑅be a Euclidean domain and 𝑀a finitely generated 𝑅-module. Then 𝑀≅𝑅 ⟋ (𝑝𝑛1 1 ) ⊕⋯⊕𝑅 ⟋ (𝑝𝑛𝑘 𝑘) ⊕𝑅𝑚 where the quotients are considered as 𝑅-modules, where 𝑝𝑖are primes in 𝑅, which are not necessarily distinct, and where 𝑚≥0.
Proof. By the structure theorem, 𝑀≅𝑅 ⟋ (𝑑1) ⊕⋯⊕𝑅 ⟋ (𝑑𝑡) ⊕𝑅⊕⋯⊕𝑅 ⏟⎵ ⎵⏟⎵ ⎵⏟ 𝑘copies ≅𝑅 ⟋ (𝑑1) ⊕⋯⊕𝑅 ⟋ (𝑑𝑡) ⊕𝑅𝑚 479 VIII. Groups, Rings and Modules where 𝑑1 ∣⋯∣𝑑𝑡. So it suffices to show that each 𝑅 ⟋ (𝑑𝑖) can be written as a product of factors of the form 𝑅 ⟋ (𝑝 𝑛𝑗 𝑗). Since 𝑅is a unique factorisation domain and a principal ideal domain, 𝑑𝑖can be written as a product 𝑢𝑝𝛼1 1 ⋯𝑝𝛼𝑟 𝑟where 𝑢is a unit and the 𝑝𝑗are pairwise non-associate primes. By the previous lemma, 𝑅 ⟋ (𝑑𝑖) ≅𝑅 ⟋ (𝑝𝛼1 1 ) ⊕… 𝑅 ⟋ (𝑝𝛼𝑟 𝑟) 12.10. Rational canonical form Let 𝑉be a vector space over a field 𝐹, and 𝛼∶𝑉→𝑉be a linear map. Let 𝑉 𝛼denote the 𝐹[𝑋]-module 𝑉where scalar multiplication is defined by 𝑓(𝑋) ⋅𝑣= 𝑓(𝛼)(𝑣).
Lemma. If 𝑉is finite-dimensional as a vector space, then 𝑉 𝛼is finitely generated as an 𝐹[𝑋]-module.
Proof. Consider a basis 𝑣1, … , 𝑣𝑛of 𝑉, so 𝑣1, … , 𝑣𝑛generate 𝑉as an 𝐹-vector space. Then, these vectors generate 𝑉 𝛼as an 𝐹[𝑋]-module, since 𝐹≤𝐹[𝑋].
Example. Suppose 𝑉 𝛼≅𝐹[𝑋] ⟋ (𝑋𝑛) as an 𝐹[𝑋]-module. Then, 1, 𝑋, 𝑋2, … , 𝑋𝑛−1 is a basis for 𝐹[𝑋] ⟋ (𝑋𝑛) as an 𝐹-vector space. With respect to this basis, 𝛼has the matrix form ⎛ ⎜ ⎜ ⎜ ⎜ ⎝ 0 0 0 ⋯ 0 0 1 0 0 ⋯ 0 0 0 1 0 ⋯ 0 0 0 0 1 ⋯ 0 0 ⋮ ⋮ ⋮ ⋱ ⋮ ⋮ 0 0 0 ⋯ 1 0 ⎞ ⎟ ⎟ ⎟ ⎟ ⎠ (∗) Example. Suppose 𝑉 𝛼≅𝐹[𝑋] ⟋ (𝑋−𝜆)𝑛as an 𝐹[𝑋]-module. Consider the basis 1, 𝑋−𝜆, (𝑋− 𝜆)2, … , (𝑋−𝜆)𝑛−1 for 𝐹[𝑋] ⟋ (𝑋−𝜆)𝑛as an 𝐹-vector space. Here, 𝛼−𝜆id has matrix (∗) from the previous example. Hence, 𝛼has matrix (∗) + 𝜆𝐼.
Example. Suppose 𝑉 𝛼≅𝐹[𝑋] ⟋ (𝑓) where 𝑓∈𝐹[𝑋] as an 𝐹[𝑋]-module, such that 𝑓is monic.
Let 𝑓(𝑋) = 𝑋𝑛+ 𝑎𝑛−1𝑋𝑛−1 + ⋯+ 𝑎0 480 12. Modules With respect to basis 1, 𝑋, … , 𝑋𝑛−1, 𝛼has matrix 𝐶(𝑓) = ⎛ ⎜ ⎜ ⎜ ⎜ ⎝ 0 0 0 ⋯ 0 −𝑎0 1 0 0 ⋯ 0 −𝑎1 0 1 0 ⋯ 0 −𝑎2 0 0 1 ⋯ 0 −𝑎3 ⋮ ⋮ ⋮ ⋱ ⋮ ⋮ 0 0 0 ⋯ 1 −𝑎𝑛−1 ⎞ ⎟ ⎟ ⎟ ⎟ ⎠ since 𝑓is monic and the last column represents 𝑋𝑛. The above matrix is known as the companion matrix of the monic polynomial.
Theorem (Rational canonical form). Let 𝐹be a field, 𝑉be a finite-dimensional 𝐹-vector space, and 𝛼∶𝑉→𝑉be a linear map. Then the 𝐹[𝑋]-module 𝑉 𝛼decomposes as 𝑉 𝛼≅𝐹[𝑋] ⟋ (𝑓 1) ⊕⋯⊕𝐹[𝑋] ⟋ (𝑓 𝑡) for some monic polynomials 𝑓𝑖∈𝐹[𝑋], and 𝑓 1 ∣⋯∣𝑓 𝑡. Moreover, with respect to a suitable basis, 𝛼has matrix ⎛ ⎜ ⎜ ⎝ 𝐶(𝑓 1) 𝐶(𝑓 2) ⋱ 𝐶(𝑓 𝑡) ⎞ ⎟ ⎟ ⎠ (∗∗) Proof. We know that 𝑉 𝛼is finitely generated as an 𝐹[𝑋]-module, since 𝑉is finite-dimensional.
Since 𝐹[𝑋] is a Euclidean domain, the structure theorem applies, and 𝑉 𝛼≅𝐹[𝑋] ⟋ (𝑓 1) ⊕⋯⊕𝐹[𝑋] ⟋ (𝑓 𝑡) ⊕𝐹[𝑋]𝑚 for some 𝑚, where 𝑓 1 ∣⋯∣𝑓 𝑡. Since 𝑉is finite-dimensional, 𝑚= 0. As 𝐹is a field, without loss of generality we may multiply each 𝑓𝑖by a unit to ensure that they are monic. Then, using the previous example, we can construct the companion matrices for each polynomial and obtain the matrix as required.
Remark. If 𝛼is represented by an 𝑛×𝑛matrix 𝐴, there exists a change of basis matrix 𝑃such that 𝑃𝐴𝑃−1 has form (∗) as stated in the theorem, so 𝐴is similar to such a block diagonal matrix of companion matrices. Note further that (∗∗) can be used to find the minimal and characteristic polynomials of 𝛼; the minimal polynomial is 𝑓 𝑡, and the characteristic polyno-mial is 𝑓 1 ⋯𝑓 𝑡. In particular, the minimal polynomial divides the characteristic polynomial, and this implies the Cayley–Hamilton theorem.
Example. Consider dim 𝑉= 2. Here, ∑deg 𝑓𝑖= 2, so there are two cases: one poly-nomial of degree two, or two polynomials of degree one. Consider 𝑉 𝛼≅𝐹[𝑋] ⟋ (𝑋−𝜆) ⊕ 𝐹[𝑋] ⟋ (𝑋−𝜇). Since one of the 𝑓𝑖must divide the other, we have 𝜆= 𝜇. If we have one polynomial of degree two, we have 𝑉 𝛼≅𝐹[𝑋] ⟋ (𝑓), where 𝑓is the characteristic polynomial of 𝛼.
481 VIII. Groups, Rings and Modules Corollary. Let 𝐴, 𝐵be invertible 2 × 2 non-scalar matrices over a field 𝐹. Then 𝐴, 𝐵are similar if and only if their characteristic polynomials are equal.
Proof. Certainly if 𝐴, 𝐵are similar they have the same characteristic polynomial, which is proven in Part IB Linear Algebra. Conversely, if the matrices are non-scalar, the modules 𝑉 𝛼, 𝑉𝛽are of the form 𝐹[𝑋] ⟋ (𝑓) by the previous example, so they are both similar to the companion matrix of 𝑓, where 𝑓is the characteristic polynomial of 𝐴or 𝐵.
Definition. The annihilator of an 𝑅-module 𝑀is Ann𝑅(𝑀) = {𝑟∈𝑅∶∀𝑚∈𝑀, 𝑟𝑚= 0} ⊲𝑅 Example. Let 𝐼⊲𝑅. Then the annihilator of 𝑅 ⟋ 𝐼is Ann𝑅(𝑅 ⟋ 𝐼) = 𝐼.
Let 𝐴be a finite abelian group. Then, considering 𝐴as a ℤ-module, Annℤ(𝐴) = (𝑒) where 𝑒 is the exponent of the group, which is the lowest common multiple of the orders of elements in the group.
Let 𝑉 𝛼be as above. Then Ann𝐹𝑋 = (𝑓) where 𝑓is the minimal polynomial of 𝛼.
12.11. Jordan normal form Jordan normal form concerns matrix similarity in ℂ. The following results are therefore restricted to this particular field.
Lemma. The primes (or equivalently, irreducibles) in ℂ[𝑋] are the polynomials 𝑋−𝜆for 𝜆∈ℂ, up to associates.
Proof. By the fundamental theorem of algebra, any non-constant polynomial with complex coefficients has a complex root. By the Euclidean algorithm, we can show that having a root 𝜆is equivalent to having a linear factor 𝑋−𝜆. Hence the irreducibles have degree one, and thus are 𝑋−𝜆exactly, up to associates.
Theorem. Let 𝛼∶𝑉→𝑉be an endomorphism of a finite-dimensional ℂ-vector space 𝑉.
Let 𝑉 𝛼be the set 𝑉as a ℂ[𝑋]-module, where scalar multiplication is defined by 𝑓⋅𝑣= 𝑓(𝛼)(𝑣).
Then, there exists an isomorphism of ℂ[𝑋]-modules 𝑉 𝛼≅ℂ[𝑋] ⟋ ((𝑋−𝜆1)𝑛1) ⊕⋯⊕ℂ[𝑋] ⟋ ((𝑋−𝜆𝑡)𝑛𝑡) where 𝜆𝑖∈ℂare not necessarily distinct. In particular, there exists a basis for this vector space such that 𝛼has matrix in block diagonal form ⎛ ⎜ ⎜ ⎝ 𝐽𝑛1(𝜆1) 𝐽𝑛2(𝜆2) ⋱ 𝐽𝑛𝑡(𝜆𝑡) ⎞ ⎟ ⎟ ⎠ 482 12. Modules where each Jordan block 𝐽𝑛𝑖(𝜆𝑖) is an 𝑛𝑖× 𝑛𝑖matrix of the form 𝐽𝑛𝑖(𝜆𝑖) = ⎛ ⎜ ⎜ ⎜ ⎝ 𝜆𝑖 0 0 ⋯ 0 1 𝜆𝑖 0 ⋯ 0 0 1 𝜆𝑖 ⋯ 0 ⋮ ⋮ ⋮ ⋱ ⋮ 0 0 0 ⋯ 𝜆𝑖 ⎞ ⎟ ⎟ ⎟ ⎠ Proof. Note ℂ[𝑋] is a Euclidean domain using the degree function, and 𝑉 𝛼is finitely gener-ated as a ℂ[𝑋]-module. These are the assumptions of the primary decomposition theorem.
Applying this, we find the module decomposition as required, noting that the primes in ℂ[𝑋] are the linear polynomials. Note that the free factor ℂ[𝑋] cannot appear in the decomposi-tion since 𝑉is finite-dimensional.
We have already seen that for a module 𝑊 𝛼≅𝐹[𝑋] ⟋ ((𝑋−𝜆)𝑛), multiplication by 𝑋is rep-resented by the matrix 𝐽𝑛(𝜆) with respect to the basis 1, (𝑋−𝜆), … , (𝑋−𝜆)𝑛−1. Hence the result follows by considering the union of these bases.
Remark. If 𝛼is represented by a matrix 𝐴, then 𝐴is similar to a matrix in Jordan normal form. This is the form of the result often used in linear algebra.
The Jordan blocks are uniquely determined up to reordering. This can be proven by con-sidering the dimensions of the generalised eigenspaces, which are ker ((𝛼−𝜆id)𝑚) for some 𝑚∈ℕ.
The minimal polynomial of 𝛼is ∏𝜆(𝑋−𝜆)𝑐𝜆where 𝑐𝜆is the size of the largest 𝜆-block. The characteristic polynomial of 𝛼is ∏𝜆(𝑋−𝜆)𝑎𝜆where 𝑎𝜆is the sum of the sizes of the 𝜆-blocks.
The number of 𝜆-blocks is the dimension of the eigenspace of 𝜆.
12.12. Modules over principal ideal domains (non-examinable) The structure theorem above was proven for Euclidean domains. This also holds for prin-cipal ideal domains. Some of the ideas relevant to this proof are illustrated in this subsec-tion.
Theorem. Let 𝑅be a principal ideal domain. Then any finitely generated torsion-free 𝑅-module is free.
If 𝑅is a Euclidean domain, this was proven as a corollary to the structure theorem.
Lemma. Let 𝑅be a principal ideal domain and 𝑀be an 𝑅-module. Let 𝑟1, 𝑟2 ∈𝑅be not both zero, and let 𝑑be their greatest common divisor. Then, (i) there exists 𝐴∈𝑆𝐿2(𝑅) such that 𝐴(𝑟1 𝑟2 ) = (𝑑 0) 483 VIII. Groups, Rings and Modules (ii) if 𝑥1, 𝑥2 ∈𝑀, then there exist 𝑥′ 1, 𝑥′ 2 ∈𝑀such that 𝑅𝑥1 + 𝑅𝑥2 = 𝑅𝑥′ 1 + 𝑅𝑥′ 2, and 𝑟1𝑥1 + 𝑟2𝑥2 = 𝑑𝑥′ 1 + 0 ⋅𝑥′ 2.
Proof. Since 𝑅is a principal ideal domain, (𝑟1, 𝑟2) = (𝑑). Hence, by definition, 𝑑= 𝛼𝑟1 + 𝛽𝑟2 for some 𝛼, 𝛽∈𝑅. Let 𝑟1 = 𝑠1𝑑and 𝑟2 = 𝑠2𝑑. Then 𝛼𝑠1 + 𝛽𝑠2 = 1. Now, let 𝐴= ( 𝛼 𝛽 −𝑠2 𝑠1 ) ⟹det 𝐴= 1; 𝐴(𝑟1 𝑟2 ) = (𝑑 0) as required.
For the second part, let 𝑥′ 1 = 𝑠1𝑥1 +𝑠2𝑥2 and 𝑥′ 2 = −𝛽𝑥1 +𝛼𝑥2. Then 𝑅𝑥′ 1 +𝑅𝑥′ 2 ⊆𝑅𝑥1 +𝑅𝑥2.
The matrix defining 𝑥′ 1, 𝑥′ 2 in terms of 𝑥1, 𝑥2 is invertible since its determinant is a unit; we can solve for 𝑥1, 𝑥2 in terms of 𝑥′ 1, 𝑥′ 2. So 𝑅𝑥′ 1+𝑅𝑥′ 2 = 𝑅𝑥1+𝑅𝑥2. Then by direct computation we can see that 𝑟1𝑥2 + 𝑟2𝑥2 = 𝑑𝑥′ 1 + 0 ⋅𝑥′ 2.
The structure theorem for principal ideal domains follows the same method; it is deduced for Smith normal form. That theorem also holds for principal ideal domains. The above lemma allows one to prove Smith normal form for principal ideal domains. In a Euclidean domain, we used the Euclidean function for a notion of size in order to perform induction; in a principal ideal domain we can count the irreducibles in a factorisation.
Proof of theorem. Let 𝑀= 𝑅𝑥1 +⋯+𝑅𝑥𝑛where 𝑛is minimal. If 𝑥1, … , 𝑥𝑛are independent, then 𝑀is free as required. Suppose that the 𝑥𝑖are not independent, so there exists 𝑟𝑖such that ∑𝑟𝑖𝑥𝑖= 0 but not all of the 𝑟𝑖are zero. By reordering, we can suppose that 𝑟1 ≠0. By using part (ii) of the previous lemma, after replacing 𝑥1 and 𝑥2 by suitable 𝑥′ 1, 𝑥′ 2, we may assume that 𝑟1 ≠0 and 𝑟2 = 0. By repeating this process with 𝑥1 and 𝑥𝑖for all 𝑖≥2, we obtain 𝑟1 ≠0 and 𝑟2 = ⋯= 𝑟 𝑛= 0, so 𝑟1𝑥″ 1 = 0 for some nonzero 𝑥″ 1 ∈𝑀. But 𝑀is torsion-free, so 𝑟1 must be zero, and this is a contradiction.
484 IX. Complex Analysis Lectured in Lent 2022 by Prof. N. Wickramasekera Complex differentiation is a stronger notion than real differentiation. Many functions that are differentiable as a function of two real variables are not complex differentiable, for ex-ample the complex conjugate function. This stronger notion allows us to prove some surpris-ing results. It turns out that if a function is complex differentiable once in a neighbourhood of a point, then it is given by a convergent power series in some neighbourhood of that point.
Another interesting result is Cauchy’s integral formula: if a function is complex differenti-able in a neighbourhood around a point, one can evaluate the function at that point using a certain integral over any loop around that point. A similar result can be used to obtain an arbitrary derivative of a function at a point by using a single integral.
485 IX. Complex Analysis Contents 1.
Differentiation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 488 1.1.
Basic notions . . . . . . . . . . . . . . . . . . . . . . . . . . . 488 1.2.
Continuity and differentiability . . . . . . . . . . . . . . . . . . 488 1.3.
Cauchy–Riemann equations . . . . . . . . . . . . . . . . . . . 489 1.4.
Curves and path-connectedness . . . . . . . . . . . . . . . . . 490 1.5.
Power series . . . . . . . . . . . . . . . . . . . . . . . . . . . 491 1.6.
Exponentials . . . . . . . . . . . . . . . . . . . . . . . . . . . 492 1.7.
Logarithms . . . . . . . . . . . . . . . . . . . . . . . . . . . . 493 1.8.
Conformality . . . . . . . . . . . . . . . . . . . . . . . . . . . 494 2.
Integration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 496 2.1.
Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . 496 2.2.
Integrating along curves . . . . . . . . . . . . . . . . . . . . . 497 2.3.
Fundamental theorem of calculus . . . . . . . . . . . . . . . . 499 2.4.
Star-shaped domains . . . . . . . . . . . . . . . . . . . . . . . 500 2.5.
Cauchy’s integral formula . . . . . . . . . . . . . . . . . . . . . 503 2.6.
Liouville’s theorem . . . . . . . . . . . . . . . . . . . . . . . . 505 2.7.
Taylor series . . . . . . . . . . . . . . . . . . . . . . . . . . . 506 2.8.
Zeroes of holomorphic functions . . . . . . . . . . . . . . . . . 508 2.9.
Analytic continuation . . . . . . . . . . . . . . . . . . . . . . . 509 2.10.
Uniform limits of holomorphic functions . . . . . . . . . . . . . 512 3.
More integration . . . . . . . . . . . . . . . . . . . . . . . . . . . . 515 3.1.
Winding numbers . . . . . . . . . . . . . . . . . . . . . . . . . 515 3.2.
Continuity of derivative function . . . . . . . . . . . . . . . . . 517 3.3.
Cauchy’s theorem and Cauchy’s integral formula . . . . . . . . . 518 3.4.
Homotopy . . . . . . . . . . . . . . . . . . . . . . . . . . . . 521 3.5.
Simply connected domains . . . . . . . . . . . . . . . . . . . . 523 4.
Singularities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 524 4.1.
Motivation . . . . . . . . . . . . . . . . . . . . . . . . . . . . 524 4.2.
Removable singularities . . . . . . . . . . . . . . . . . . . . . . 524 4.3.
Poles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 525 4.4.
Essential singularities . . . . . . . . . . . . . . . . . . . . . . . 527 4.5.
Laurent series . . . . . . . . . . . . . . . . . . . . . . . . . . . 527 4.6.
Coefficients of Laurent series . . . . . . . . . . . . . . . . . . . 530 4.7.
Residues . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 531 4.8.
Jordan’s lemma . . . . . . . . . . . . . . . . . . . . . . . . . . 533 5.
The argument principle, local degree, and Rouché’s theorem . . . 537 5.1.
The argument principle . . . . . . . . . . . . . . . . . . . . . . 537 486 5.2.
Local degree theorem . . . . . . . . . . . . . . . . . . . . . . . 539 5.3.
Open mapping theorem . . . . . . . . . . . . . . . . . . . . . . 540 5.4.
Rouché’s theorem . . . . . . . . . . . . . . . . . . . . . . . . . 540 487 IX. Complex Analysis 1. Differentiation 1.1. Basic notions We use the following definitions.
• The complex plane is denoted ℂ.
• The complex conjugate of a complex number 𝑧is denoted 𝑧.
• The modulus is denoted |𝑧|.
• The function 𝑑(𝑧, 𝑤) = |𝑧−𝑤| is a metric on ℂ. All topological notions will be with respect to this metric.
• We define the disc 𝐷(𝑎, 𝑟) = {𝑧∈ℂ∶|𝑧−𝑎| < 𝑟} to be the open ball with centre 𝑎and radius 𝑟.
• A subset 𝑈⊂ℂis said to be open if it is open with respect to the above metric. In particular, by identifying ℂwith ℝ2, we can see that 𝑈⊂ℂis open if and only if 𝑈⊂ℝ2 is open with respect to the Euclidean metric.
The course concerns itself with complex-valued functions of a single complex variable. Identi-fying ℂwith ℝ2 allows us to construct 𝑓(𝑧) = 𝑢(𝑥, 𝑦) + 𝑖𝑣(𝑥, 𝑦), where 𝑢, 𝑣are real-valued functions. We can denote these parts by 𝑢= Re(𝑓) and 𝑣= Im(𝑓).
1.2. Continuity and differentiability The definition of continuity is carried over from metric spaces. That is, 𝑓∶𝐴→ℂis con-tinuous at a point 𝑤∈𝐴if ∀𝜀> 0, ∃𝛿> 0, ∀𝑧∈𝐴, |𝑧−𝑤| < 𝛿⟹|𝑓(𝑧) −𝑓(𝑤)| < 𝜀 Equivalently, the limit lim𝑧→𝑤𝑓(𝑧) exists and takes the value 𝑓(𝑤). We can easily check that 𝑓is continuous at 𝑤= 𝑐+ 𝑖𝑑∈𝐴if and only if 𝑢, 𝑣are continuous at (𝑐, 𝑑) with respect to the Euclidean metric on 𝐴⊂ℝ2.
Definition. Let 𝑓∶𝑈→ℂ, where 𝑈is open in ℂ.
(i) 𝑓is differentiable at 𝑤∈𝑈if the limit 𝑓′(𝑤) = lim 𝑧→𝑤 𝑓(𝑧) −𝑓(𝑤) 𝑧−𝑤 exists, and its value is complex. We say that 𝑓′(𝑤) is the derivative of 𝑓at 𝑤.
(ii) 𝑓is holomorphic at 𝑤∈𝑈if there exists 𝜀> 0 such that 𝐷(𝑤, 𝜀) ⊂𝑈and 𝑓is differ-entiable at every point in 𝐷(𝑤, 𝜀).
(iii) 𝑓is holomorphic in 𝑈if 𝑓is holomorphic at every point in 𝑈, or equivalently, 𝑓is differentiable everywhere.
488 1. Differentiation Differentiation of composite functions, sums, products and quotients can be computed in the complex case exactly as they are in the real case.
Example. Polynomials 𝑝(𝑧) ∑ 𝑛 𝑗=0 𝑎𝑗𝑧𝑗for complex coefficients 𝑎𝑗are holomorphic on ℂ.
Further, if 𝑝, 𝑞are polynomials, 𝑝 𝑞is holomorphic on ℂ∖{𝑧∶𝑞(𝑧) = 0}.
Remark. The differentiability of 𝑓at a point 𝑐+𝑖𝑑is not equivalent to the differentiability of 𝑢, 𝑣at (𝑐, 𝑑). 𝑢∶𝑈→ℝis differentiable at (𝑐, 𝑑) ∈𝑈if there is a ‘good’ affine approximation of 𝑢at (𝑐, 𝑑); there exists a linear transformation 𝐿∶ℝ2 →ℝsuch that lim (𝑥,𝑦)→(𝑐,𝑑) 𝑢(𝑥, 𝑦) −(𝑢(𝑐, 𝑑) −𝐿(𝑥−𝑐, 𝑦−𝑑)) √(𝑥−𝑐)2 + (𝑦−𝑑)2 = 0 If 𝑢is differentiable at (𝑐, 𝑑), then 𝐿is uniquely defined, and can be denoted 𝐿= 𝐷𝑢(𝑐, 𝑑). 𝐿 is given by the partial derivatives of 𝑢, which are 𝐿(𝑥, 𝑦) = (𝜕𝑢 𝜕𝑥(𝑐, 𝑑))𝑥+ (𝜕𝑢 𝜕𝑦(𝑐, 𝑑))𝑦 This seems to imply that the differentiability of 𝑓requires more than the differentiability of 𝑢, 𝑣.
1.3. Cauchy–Riemann equations Theorem. 𝑓= 𝑢+𝑖𝑣∶𝑈→ℂis differentiable at 𝑤= 𝑐+𝑖𝑑∈𝑈if and only if 𝑢, 𝑣∶𝑈→ℝ are differentiable at (𝑐, 𝑑) ∈𝑈and 𝑢, 𝑣satisfy the Cauchy–Riemann equations at (𝑐, 𝑑), which are 𝜕𝑢 𝜕𝑥= 𝜕𝑣 𝜕𝑦; 𝜕𝑢 𝜕𝑦= −𝜕𝑣 𝜕𝑥 If 𝑓is differentiable at 𝑤= 𝑐+ 𝑖𝑑, then 𝑓′(𝑤) = 𝜕𝑢 𝜕𝑥(𝑐, 𝑑) + 𝑖𝜕𝑣 𝜕𝑥(𝑐, 𝑑) and other expressions, which follow directly from the Cauchy–Riemann equations.
Proof. All of the following statements will be bi-implications. Suppose 𝑓is differentiable at 𝑤with 𝑓′(𝑤) = 𝑝+ 𝑖𝑞, so lim 𝑧→𝑤 𝑓(𝑧) −𝑓(𝑤) 𝑧−𝑤 = 𝑝+ 𝑖𝑞 lim 𝑧→𝑤 𝑓(𝑧) −𝑓(𝑤) −(𝑧−𝑤)(𝑝+ 𝑖𝑞) |𝑧−𝑤| = 0 By separating real and imaginary parts, writing 𝑤= 𝑐+ 𝑖𝑑we have lim (𝑥,𝑦)→(𝑐,𝑑) 𝑢(𝑥, 𝑦) −𝑢(𝑐, 𝑑) −𝑝(𝑥−𝑐) + 𝑞(𝑦−𝑑) √(𝑥−𝑐)2 + (𝑦−𝑑)2 = 0 lim (𝑥,𝑦)→(𝑐,𝑑) 𝑣(𝑥, 𝑦) −𝑣(𝑐, 𝑑) −𝑞(𝑥−𝑐) −𝑝(𝑦−𝑑) √(𝑥−𝑐)2 + (𝑦−𝑑)2 = 0 489 IX. Complex Analysis Thus, 𝑢is differentiable at (𝑐, 𝑑) with 𝐷𝑢(𝑐, 𝑑)(𝑥, 𝑦) = 𝑝𝑥−𝑞𝑦and 𝑣is differentiable at (𝑐, 𝑑) with 𝐷𝑣(𝑐, 𝑑)(𝑥, 𝑦) = 𝑞𝑥+ 𝑝𝑦.
𝑢𝑥(𝑐, 𝑑) = 𝑣𝑦(𝑐, 𝑑) = 𝑝; −𝑢𝑦(𝑐, 𝑑) = 𝑣𝑥(𝑐, 𝑑) = 𝑞 Hence the Cauchy–Riemann equations hold at (𝑐, 𝑑). We also find that if 𝑓is differentiable at 𝑤, we have 𝑓′(𝑤) = 𝑢𝑥(𝑐, 𝑑) + 𝑖𝑣𝑥(𝑐, 𝑑).
Remark. If 𝑢, 𝑣simply satisfy the Cauchy–Riemann equations alone, that does not imply differentiability of 𝑓. 𝑢, 𝑣must also be differentiable.
Remark. If we simply want to show that the differentiability of 𝑓implies that the Cauchy– Riemann equations hold, we can proceed in a simpler way. For 𝑡∈ℝ, 𝑓′(𝑤) = lim 𝑡→0 (𝑢(𝑐+ 𝑡, 𝑑) −𝑢(𝑐, 𝑑) 𝑡 + 𝑖𝑣(𝑐+ 𝑡, 𝑑) −𝑣(𝑐, 𝑑) 𝑡 ) Hence the real part and the complex part both exist, so 𝑢𝑥(𝑐, 𝑑) and 𝑣𝑥(𝑐, 𝑑) exist, and 𝑓′(𝑤) = 𝑢𝑥(𝑐, 𝑑)+𝑖𝑣𝑥(𝑐, 𝑑). If we instead considered a perturbation along the imaginary axis, we find 𝑓′(𝑤) = 𝑣𝑦(𝑐, 𝑑) −𝑖𝑢𝑦(𝑐, 𝑑), giving the Cauchy–Riemann equations.
Example. The complex conjugate function 𝑧↦𝑧is not differentiable. Here, 𝑢(𝑥, 𝑦) = 𝑥, and 𝑣(𝑥, 𝑦) = −𝑦, so the Cauchy–Riemann equations do not hold.
Corollary. If 𝑢, 𝑣have continuous partial derivatives at (𝑐, 𝑑) and satisfy the Cauchy–Riemann equations at this point, then 𝑓is differentiable at 𝑐+ 𝑖𝑑. In particular, if 𝑢, 𝑣are 𝐶1 func-tions on 𝑈(i.e. have continuous partial derivatives in 𝑈) satisfying the Cauchy–Riemann equations everywhere, then 𝑓is holomorphic (in 𝑈).
Proof. If 𝑢, 𝑣have continuous partial derivatives then 𝑢, 𝑣are differentiable at (𝑐, 𝑑) by Ana-lysis and Topology.
1.4. Curves and path-connectedness Definition. A curve is a continuous function 𝛾∶[𝑎, 𝑏] →ℂ, where 𝑎, 𝑏∈ℝ. 𝛾is a 𝐶1 curve if 𝛾′ exists and is continuous on [𝑎, 𝑏]. An open set 𝑈⊂ℂis path-connected if for any two points 𝑧, 𝑤∈𝑈, there exists 𝛾∶[0, 1] →𝑈such that 𝛾(0) = 𝑧and 𝛾(1) = 𝑤. A domain is a non-empty, open, path-connected subset of ℂ.
Corollary. Let 𝑈be a domain. Let 𝑓∶𝑈→ℂbe a holomorphic function with derivative zero everywhere. Then 𝑓is constant on 𝑈.
Proof. By the Cauchy–Riemann equations, 𝑓′ = 0 implies that 𝐷𝑢= 𝐷𝑣= 0 in 𝑈. By Analysis and Topology, the path-connectedness of 𝑈implies that 𝑢and 𝑣are constant func-tions.
490 1. Differentiation 1.5. Power series Recall the following theorem from IA Analysis.
Theorem. Let (𝑐𝑛)∞ 𝑛=0 be a sequence of complex numbers. Then, the power series ∞ ∑ 𝑛=0 𝑐𝑛(𝑧−𝑎)𝑛 has a unique radius of convergence 𝑅∈[0, ∞] such that the power series converges abso-lutely for |𝑧−𝑎| < 𝑅and diverges if |𝑧−𝑎| > 𝑅. Further, if 0 < 𝑟< 𝑅, the series converges uniformly with respect to 𝑧on the compact disc 𝐷(𝑎, 𝑟).
Note that 𝑅= sup {𝑟≥0∶lim 𝑛→∞|𝑐𝑛|𝑟𝑛= 0}; 1 𝑅= lim sup 𝑛→∞ |𝑐𝑛| 1 𝑛 Theorem. Let the sequence (𝑐𝑛) define a power series 𝑓centred around 𝑎with positive radius of convergence 𝑅. Then, the function 𝑓∶𝐷(𝑎, 𝑅) →ℂsatisfies (i) 𝑓is holomorphic on 𝐷(𝑎, 𝑅); (ii) the term-by-term differentiated series ∑ ∞ 𝑛=1 𝑛𝑐𝑛(𝑧−𝑎)𝑛−1 also has radius of conver-gence equal to 𝑅, and this series is exactly the value of 𝑓′; (iii) 𝑓has derivatives of all orders on 𝐷(𝑎, 𝑅) and 𝑐𝑛= 𝑓(𝑛)(𝑎) 𝑛!
; (iv) if 𝑓vanishes on 𝐷(𝑎, 𝜀) for any 𝜀> 0, then 𝑓≡0 on 𝐷(𝑎, 𝑅).
Proof.
(i) Without loss of generality, let 𝑎= 0. ∑ ∞ 𝑛=1 𝑛𝑐𝑛(𝑧−𝑎)𝑛−1 has some radius of convergence 𝑅1.
Let 𝑧∈𝐷(0, 𝑅) and choose 𝜌such that |𝑧| < 𝜌< 𝑅. Then, 𝑛|𝑐𝑛||𝑧|𝑛−1 = 𝑛|𝑐𝑛|| | | 𝑧 𝜌 | | | 𝑛−1 𝜌𝑛−1 ≤|𝑐𝑛|𝜌𝑛−1 for sufficiently large 𝑛, since 𝑛| | | 𝑧 𝜌 | | | 𝑛−1 →0 as 𝑛→∞. Since ∑|𝑐𝑛|𝜌𝑛converges, we must have that 𝑛|𝑐𝑛||𝑧|𝑛−1 converges. Hence 𝑅1 ≥𝑅.
Now, since |𝑐𝑛||𝑧|𝑛≤𝑛|𝑐𝑛||𝑧|𝑛= |𝑧|(𝑛|𝑐𝑛||𝑧|𝑛−1) If ∑𝑛|𝑐𝑛|𝑧𝑛−1 converges then so does ∑|𝑐𝑛||𝑧|𝑛. Hence 𝑅1 ≤𝑅. This leads us to conclude 𝑅1 = 𝑅.
491 IX. Complex Analysis (ii) Let 𝑧∈𝐷(0, 𝑅). The statement that 𝑓′ is the above differentiated power series at 𝑧is equivalent to continuity at 𝑧of the function 𝑔∶𝐷(0, 𝑅) →ℂ; 𝑔(𝑤) = { 𝑓(𝑤)−𝑓(𝑧) 𝑤−𝑧 𝑤≠𝑧 ∑ ∞ 𝑛=1 𝑛𝑐𝑛𝑧𝑛−1 𝑤= 𝑧 Substituting for 𝑓, we have 𝑔(𝑤) = ∑ ∞ 𝑛=1 ℎ𝑛(𝑤) for 𝑤∈𝐷(0, 𝑅) where ℎ𝑛(𝑤) = { 𝑐𝑛(𝑤𝑛−𝑧𝑛) 𝑤−𝑧 𝑤≠𝑧 𝑛𝑐𝑛𝑧𝑛−1 𝑤= 𝑧 Note that ℎ𝑛is continuous on 𝐷(0, 𝑅). Further, note that 𝑤𝑛−𝑧𝑛 𝑤−𝑧 = 𝑛−1 ∑ 𝑗=0 𝑧𝑗𝑤𝑛−1−𝑗 We have that for all 𝑟with |𝑧| < 𝑟< 𝑅and all 𝑤∈𝐷(0, 𝑟), |ℎ𝑛|(𝑤) ≤𝑛|𝑐𝑛|𝑟𝑛−1 ≡𝑀𝑛.
Since ∑𝑀𝑛< ∞, the Weierstrass 𝑀test shows that ∑ℎ𝑛converges uniformly on 𝐷(0, 𝑟). A uniform limit of continuous functions is continuous, hence 𝑔= ∑ℎ𝑛is continuous in 𝐷(0, 𝑟) and in particular at 𝑧.
(iii) Part (ii) can be applied inductively. The equation 𝑐𝑛= 𝑓(𝑛)(𝑎) 𝑛!
can be found by differ-entiating the series 𝑛times.
(iv) If 𝑓≡0 in some disc 𝐷(𝑎, 𝜀), then 𝑓(𝑛)(𝑎) = 0 for all 𝑛. Thus the power series is identically zero.
1.6. Exponentials Definition. If 𝑓∶ℂ→ℂis holomorphic on ℂ, we say that 𝑓is entire.
Definition. The complex exponential function is defined by 𝑒𝑧= exp(𝑧) = ∞ ∑ 𝑛=0 𝑧𝑛 𝑛!
Proposition.
(i) 𝑒𝑧is entire, and (𝑒𝑧)′ = 𝑒𝑧; (ii) 𝑒𝑧≠0 and 𝑒𝑧+𝑤= 𝑒𝑧𝑒𝑤for all complex 𝑧, 𝑤; (iii) 𝑒𝑥+𝑖𝑦= 𝑒𝑥(cos 𝑦+ 𝑖sin 𝑦) for real 𝑥, 𝑦; (iv) 𝑒𝑧= 1 if and only if 𝑧= 2𝜋𝑛𝑖for an integer 𝑛; 492 1. Differentiation (v) if 𝑧∈ℂ, then there exists 𝑤such that 𝑒𝑤= 𝑧if and only if 𝑧≠0.
Proof.
(i) We can show that the radius of convergence is infinite. We can thus differenti-ate term by term and find (𝑒𝑧)′ = 𝑒𝑧.
(ii) Let 𝑤∈ℂ, and 𝐹(𝑧) = 𝑒𝑧+𝑤𝑒−𝑤. Then we have 𝐹′(𝑧) = −𝑒𝑧+𝑤𝑒−𝑧+ 𝑒𝑧+𝑤𝑒−𝑧= 0 Hence 𝐹(𝑧) is constant. But 𝐹(0) = 𝑒𝑤, so 𝐹(𝑧) = 𝑒𝑤. Taking 𝑤= 0, we have 𝑒𝑧𝑒−𝑧= 1, so 𝑒𝑧≠0. Further, 𝑒𝑧+𝑤= 𝑒𝑧𝑒𝑤.
(iii) By part (ii), 𝑒𝑥+𝑖𝑦= 𝑒𝑥𝑒𝑖𝑦. Then, the series expansions of the sine and cosine functions can be used to finish the proof.
The rest of the proof is left as an exercise, which follows from (iii).
1.7. Logarithms Definition. Let 𝑧∈ℂ. Then, 𝑤∈ℂis a logarithm of 𝑧if 𝑒𝑤= 𝑧.
By part (v) above, 𝑧has a logarithm if and only if 𝑧≠0. In particular, 𝑧≠0 has infinitely many logarithms of the form 𝑤+ 2𝜋𝑖𝑛for 𝑛∈ℤ. If 𝑤is a logarithm of 𝑧, then 𝑒Re 𝑤= |𝑧|, and hence Re(𝑤) = ln |𝑧|, where ln here is the unique real logarithm. In particular, Re(𝑤) is uniquely determined by 𝑧.
Definition. Let 𝑈⊂ℂ∖{0} be an open set. A branch of logarithm on 𝑈is a continuous function 𝜆∶𝑈→ℂsuch that 𝑒𝜆(𝑧) = 𝑧for all 𝑧∈𝑈.
Remark. Note that if 𝜆is a branch of logarithm on 𝑈then 𝜆is holomorphic in 𝑈with 𝜆′(𝑧) = 1 𝑧.
Proof. If 𝑤∈𝑈we have lim 𝑧→𝑤 𝜆(𝑧) −𝜆(𝑤) 𝑧−𝑤 = lim 𝑧→𝑤 𝜆(𝑧) −𝜆(𝑤) 𝑒𝜆(𝑧) −𝑒𝜆(𝑤) = lim 𝑧→𝑤 1 ( 𝑒𝜆(𝑧)−𝑒𝜆(𝑤) 𝜆(𝑧)−𝜆(𝑤) ) = 1 𝑒𝜆(𝑤) lim 𝑧→𝑤 1 ( 𝑒𝜆(𝑧)−𝜆(𝑤)−1 𝜆(𝑧)−𝜆(𝑤) ) = 1 𝑒𝜆(𝑤) lim ℎ→0 1 ( 𝑒ℎ−1 ℎ) = 1 𝑒𝜆(𝑤) = 1 𝑤 493 IX. Complex Analysis Definition. The principal branch of logarithm is the function Log∶𝑈1 = ℂ∖{𝑥∈ℝ∶𝑥≤0} →ℂ; Log(𝑧) = ln |𝑧| + 𝑖arg(𝑧) where arg(𝑧) is the unique argument of 𝑧∈𝑈1 in (−𝜋, 𝜋).
This is a branch of logarithm. Indeed, to check continuity, note that 𝑧↦log |𝑧| is continuous on ℂ∖{0}, and 𝑧↦arg(𝑧) is continuous since 𝜃↦𝑒𝑖𝜃is a homeomorphism (−𝜋, 𝜋) → 𝕊1 ∖{−1}, and 𝑧↦ 𝑧 |𝑧| is continuous on ℂ∖{0}. Further, 𝑒Log(𝑧) = 𝑒ln |𝑧|𝑒𝑖arg(𝑧) = |𝑧|(cos arg 𝑧+ 𝑖sin arg 𝑧) = 𝑧 Note that Log cannot be continuously extended to ℂ∖{0}, since arg 𝑧→𝜋as 𝑧→−1 with Im(𝑧) > 0, and arg 𝑧→−𝜋as 𝑧→−1 with Im(𝑧) < 0. We will later prove that no branch of logarithm can exist on all of ℂ∖{0}.
Proposition.
(i) Log is holomorphic on 𝑈1 with (Log 𝑧)′ = 1 𝑧; and (ii) for |𝑧| < 1, we have Log(1 + 𝑧) = ∞ ∑ 𝑛=1 (−1)𝑛−1𝑧𝑛 𝑛 Proof. Part (i) follows from the above. The radius of convergence of the given series is one, and 1 + 𝑧∈𝑈1, so both sides of the equation are defined on the unit disc. Then, 𝐹(𝑧) = Log(1+𝑧)− ∞ ∑ 𝑛=1 (−1)𝑛−1𝑧𝑛 𝑛 ⟹𝐹′(𝑧) = 1 1 + 𝑧− ∞ ∑ 𝑛=1 (−𝑧)𝑛−1 = 0 ⟹𝐹(𝑧) = 𝐹(0) = 0 We can now define the principal branch of 𝑧𝛼by 𝑧𝛼= 𝑒𝛼Log(𝑧) Note that 𝑧𝛼is holomorphic on 𝑈1 with (𝑧𝛼)′ = 𝛼𝑧𝛼−1. We can use exponentials to define the trigonometric and hyperbolic functions, which are all entire functions with derivatives matching those of the real definitions of these functions.
1.8. Conformality Let 𝑓∶𝑈→ℂbe holomorphic, where 𝑈is an open set. Let 𝑤∈𝑈and suppose that 𝑓′(𝑤) ≠0. Let 𝛾1, 𝛾2 ∶[−1, 1] →𝑈be 𝐶1 curves, such that 𝛾𝑖(0) = 𝑤and 𝛾′ 𝑖(0) ≠0. Then 𝑓∘𝛾𝑖are 𝐶1 curves passing through 𝑓(𝑤). Further, (𝑓∘𝛾𝑖)′(0) = 𝑓′(𝑤)𝛾′ 𝑖(0) ≠0. Thus (𝑓∘𝛾1)′(0) (𝑓∘𝛾2)′(0) = 𝛾′ 1(0) 𝛾′ 2(0) 494 1. Differentiation Hence, arg(𝑓∘𝛾1)′(0) −arg(𝑓∘𝛾2)′(0) = arg 𝛾′ 1(0) −arg 𝛾′ 2(0) In other words, the angle that the curves make when they intersect at 𝑤is the same angle that their images 𝑓∘𝛾𝑖make when they intersect at 𝑓(𝑤), and the orientation also is preserved (clockwise or anticlockwise). Hence, 𝑓is angle-preserving at 𝑤whenever 𝑓′(𝑤) ≠0. In particular, if 𝛾𝑖are tangential at 𝑤, the curves 𝑓∘𝛾𝑖are tangential at 𝑓(𝑤).
Remark. If 𝑓is 𝐶1, then the converse holds. If 𝑤∈𝑈and (𝑓∘𝛾)′(0) ≠0 for any 𝐶1 curve 𝛾 with 𝛾(0) = 𝑤and 𝛾′(0) ≠0, and if 𝑓is angle-preserving at 𝑤in the above sense, then 𝑓′(𝑤) exists and is nonzero.
Definition. A holomorphic function 𝑓∶𝑈→ℂon an open set 𝑈is conformal at 𝑤∈𝑈if 𝑓′(𝑤) ≠0.
Definition. Let 𝑈, ˜ 𝑈be domains in ℂ. A map 𝑓∶𝑈→˜ 𝑈is a conformal equivalence between 𝑈, ˜ 𝑈if 𝑓is a bijective holomorphic map with 𝑓′(𝑧) ≠0 for all 𝑧∈𝑈.
Remark. We will prove later that if 𝑓is holomorphic and injective, then 𝑓′(𝑧) ≠0 for all 𝑧.
Thus, in the above definition, the condition 𝑓′(𝑧) ≠0 is redundant.
Remark. It is automatic that 𝑓−1 ∶˜ 𝑈→𝑈is holomorphic, which will follow from the holo-morphic inverse function theorem.
Example. Möbius maps 𝑓(𝑧) = 𝑎𝑧+ 𝑏 𝑐𝑧+ 𝑑 are conformal on ℂ∖{−𝑑/𝑐} if 𝑐≠0, and conformal on ℂif 𝑐= 0. Möbius maps are sometimes used as explicit conformal equivalences between subdomains of ℂ. For instance, let ℍbe the open upper half plane in ℂ. Then 𝑧∈ℍ⟺|𝑧−𝑖| < |𝑧+ 𝑖| ⟺| | | 𝑧−𝑖 𝑧+ 𝑖 | | | < 1 Thus the map 𝑧↦ 𝑧−𝑖 𝑧+𝑖maps ℍonto 𝐷(0, 1), so 𝑔is a conformal equivalence.
Example. Let 𝑓∶𝑧↦𝑧𝑛for 𝑛≥1. Then 𝑓∶{𝑧∈ℂ∖{0}∶0 < arg 𝑧< 𝜋 𝑛} →ℍ is the restricted map on a sector. The restricted 𝑓is a conformal equivalence with 𝑓−1(𝑧) = 𝑧1/𝑛, the principal branch of 𝑧1/𝑛.
Example. The function exp∶{𝑧∈ℂ∶−𝜋< Im 𝑧< 𝜋} →ℂ∖{𝑥∈ℝ∶𝑥≤0} is a conformal equivalence, with inverse Log.
Theorem (Riemann mapping theorem). This theorem is non-examinable.
Any simply connected domain 𝑈⊂ℂwith 𝑈≠ℂis conformally equivalent to 𝐷(0, 1).
495 IX. Complex Analysis 2. Integration 2.1. Introduction Definition. If 𝑓∶[𝑎, 𝑏] ⊂ℝ→ℂis a complex function, and the real and imaginary parts of 𝑓are Riemann integrable, then we define ∫ 𝑏 𝑎 𝑓(𝑡) d𝑡= ∫ 𝑏 𝑎 Re(𝑓(𝑡)) d𝑡+ 𝑖∫ 𝑏 𝑎 Im(𝑓(𝑡)) d𝑡 In particular, for 𝑔∶[𝑎, 𝑏] →ℝ, we have ∫ 𝑏 𝑎 𝑖𝑔(𝑡) d𝑡= 𝑖∫ 𝑏 𝑎 𝑔(𝑡) d𝑡 Thus, for a complex constant 𝑤∈ℂ, we can find ∫ 𝑏 𝑎 𝑤𝑓(𝑡) d𝑡= 𝑤∫ 𝑏 𝑎 𝑓(𝑡) d𝑡 Proposition (basic estimate). If 𝑓∶[𝑎, 𝑏] →ℂis continuous, then | | | | ∫ 𝑏 𝑎 𝑓(𝑡) d𝑡 | | | | ≤∫ 𝑏 𝑎 |𝑓(𝑡)| d𝑡≤(𝑏−𝑎) sup 𝑡∈[𝑎,𝑏] |𝑓(𝑡)| Equality holds if and only if 𝑓is constant.
Proof. If ∫ 𝑏 𝑎𝑓(𝑡) d𝑡= 0 then the proof is complete. Otherwise, we can write the value of the integral as 𝑟𝑒𝑖𝜃for 𝜃∈[0, 2𝜋). Let 𝑀= sup𝑡∈[𝑎,𝑏] |𝑓(𝑡)|. Then we have | | | | ∫ 𝑏 𝑎 𝑓(𝑡) d𝑡 | | | | = 𝑟 = 𝑒−𝑖𝜃∫ 𝑏 𝑎 𝑓(𝑡) d𝑡 = ∫ 𝑏 𝑎 𝑒−𝑖𝜃𝑓(𝑡) d𝑡 = ∫ 𝑏 𝑎 Re(𝑒−𝑖𝜃𝑓(𝑡)) d𝑡+ 𝑖∫ 𝑏 𝑎 Im(𝑒−𝑖𝜃𝑓(𝑡)) d𝑡 Since the left hand side is real, the imaginary integral vanishes.
| | | | ∫ 𝑏 𝑎 𝑓(𝑡) d𝑡 | | | | = ∫ 𝑏 𝑎 Re(𝑒−𝑖𝜃𝑓(𝑡)) d𝑡 ≤∫ 𝑏 𝑎 | |𝑒−𝑖𝜃𝑓(𝑡)| | d𝑡= ∫ 𝑏 𝑎 |𝑓(𝑡)| d𝑡 ≤(𝑏−𝑎)𝑀 496 2. Integration Equality holds if and only if |𝑓(𝑡)| = 𝑀and Re(𝑒−𝑖𝜃𝑓(𝑡)) = 𝑀for all 𝑡∈[𝑎, 𝑏], which is true only if |𝑓(𝑡)| = 𝑀and arg(𝑓(𝑡)) = 𝜃hence 𝑓= 𝑀𝑒𝑖𝜃for all 𝑡.
2.2. Integrating along curves Definition. Let 𝑈⊂ℂbe an open set and let 𝑓∶𝑈→ℂbe continuous. Let 𝛾∶[𝑎, 𝑏] →𝑈 be a 𝐶1 curve. Then the integral of 𝑓along 𝛾is ∫ 𝛾 𝑓(𝑧) d𝑧= ∫ 𝑏 𝑎 𝑓(𝛾(𝑡))𝛾′(𝑡) d𝑡 This definition is consistent with the previous definition of the integral of a function 𝑓along the interval [𝑎, 𝑏]. The integral along a curve has various convenient properties.
(i) It is invariant under the choice of parametrisation. Let 𝜑∶[𝑎1, 𝑏1] →[𝑎, 𝑏] be 𝐶1 and injective with 𝜑(𝑎1) = 𝑎and 𝜑(𝑏1) = 𝑏. Let 𝛿= 𝛾∘𝜑∶[𝑎1, 𝑏1] →𝑈. Then ∫ 𝛿 𝑓(𝑧) d𝑧= ∫ 𝛾 𝑓(𝑧) d𝑧 Indeed, ∫ 𝛿 𝑓(𝑧) d𝑧= ∫ 𝑏1 𝑎1 𝑓(𝛾(𝜑(𝑡)))𝛾′(𝜑(𝑡))𝜑′(𝑡) d𝑡 = ∫ 𝑏 𝑎 𝑓(𝛾(𝑠))𝛾′(𝑠) d𝑠 = ∫ 𝛾 𝑓(𝑧) d𝑧 (ii) The integral is linear. It is easy to check that ∫ 𝛾 (𝜆𝑓(𝑧) + 𝜇𝑔(𝑧)) d𝑧= 𝜆∫ 𝛾 𝑓(𝑧) d𝑧+ 𝜇∫ 𝛾 𝑔(𝑧) d𝑧 for complex constants 𝜆, 𝜇∈ℂ.
(iii) The additivity property states that if 𝛾∶[𝑎, 𝑏] →𝑈is 𝐶1 and 𝑎< 𝑐< 𝑏, then ∫ 𝛾 𝑓(𝑧) d𝑧= ∫ 𝛾|[𝑎,𝑐] 𝑓(𝑧) d𝑧+ ∫ 𝛾|𝑐,𝑏 𝑓(𝑧) d𝑧 (iv) We define the inverse path (−𝛾)∶[−𝑏, −𝑎] →𝑈by (−𝛾)(𝑡) = 𝛾(−𝑡). Then ∫ (−𝛾) 𝑓(𝑧) d𝑧= −∫ 𝛾 𝑓(𝑧) d𝑧 497 IX. Complex Analysis Definition. Let 𝛾∶[𝑎, 𝑏] →ℂbe a 𝐶1 curve. Then the length of 𝛾is length(𝛾) = ∫ 𝑏 𝑎 |𝛾′(𝑡)| d𝑡 Definition. A piecewise 𝐶1 curve is a continuous map 𝛾∶[𝑎, 𝑏] →ℂsuch that there exists a finite subdivision 𝑎= 𝑎0 < 𝑎1 < ⋯< 𝑎𝑛= 𝑏 such that each 𝛾𝑗= 𝛾|[𝑎𝑗−1,𝑎𝑗] is 𝐶1 for 1 ≤𝑗≤𝑛. Then, for such a piecewise 𝐶1 curve, we define ∫ 𝛾 𝑓(𝑧) d𝑧= 𝑛 ∑ 𝑗=1 ∫ 𝛾𝑗 𝑓(𝑧) d𝑧 and length(𝛾) = 𝑛 ∑ 𝑗=1 length(𝛾𝑗) = 𝑛 ∑ 𝑗=1 ∫ 𝑎𝑗 𝑎𝑗−1 |𝛾′(𝑡)| d𝑡 By the additivity property, both definitions are invariant under changing the subdivision.
From here, we will use ‘curve’ to refer to ‘piecewise 𝐶1 curve’, unless stated otherwise.
Definition. If 𝛾1 ∶[𝑎, 𝑏] →ℂand 𝛾2 ∶[𝑐, 𝑑] are curves with 𝛾1(𝑏) = 𝛾2(𝑐), we define the sum of 𝛾1 and 𝛾2 to be the curve (𝛾1 + 𝛾2)∶[𝑎, 𝑏+ 𝑑−𝑐] →ℂ; (𝛾1 + 𝛾2)(𝑡) = {𝛾1(𝑡) 𝑎≤𝑡≤𝑏 𝛾2(𝑡−𝑏+ 𝑐) 𝑏≤𝑡≤𝑏+ 𝑑−𝑐 Proposition. Let 𝑓∶𝑈→ℂbe continuous and 𝛾∶[𝑎, 𝑏] →ℂ, we have | | | | ∫ 𝛾 𝑓(𝑧) d𝑧 | | | | ≤length(𝛾) sup 𝛾 |𝑓| where sup𝛾𝑔≡sup𝑡∈[𝑎,𝑏] 𝑔(𝛾(𝑡)).
Proof. If 𝛾is 𝐶1, then | | | | ∫ 𝛾 𝑓(𝑧) d𝑧 | | | | = | | | | ∫ 𝑏 𝑎 𝑓(𝛾(𝑡))𝛾′(𝑡) d𝑡 | | | | ≤∫ 𝑏 𝑎 |𝑓(𝛾(𝑡))| ⋅|𝛾′(𝑡)| d𝑡≤sup 𝑡∈[𝑎,𝑏] |𝑓(𝛾(𝑡))|length(𝛾) If 𝛾is piecewise 𝐶1, then the result follows from the definition of a piecewise 𝐶1 function and the above.
498 2. Integration 2.3. Fundamental theorem of calculus Theorem (fundamental theorem of calculus). Let 𝑓∶𝑈→ℂbe continuous on an open set 𝑈⊂ℂ. Let 𝐹∶𝑈→ℂbe a function such that 𝐹′(𝑧) = 𝑓(𝑧) for all 𝑧∈𝑈. Then, for any curve 𝛾∶[𝑎, 𝑏] →𝑈, we have ∫ 𝛾 𝑓(𝑧) d𝑧= 𝐹(𝛾(𝑏)) −𝐹(𝛾(𝑎)) If 𝛾is a closed curve, then ∫ 𝛾𝑓(𝑧) = 0. Such a function 𝐹is known as an antiderivative of 𝑓.
Proof.
∫ 𝛾 𝑓(𝑧) d𝑧= ∫ 𝑏 𝑎 𝑓(𝛾(𝑡))𝛾′(𝑡) d𝑡= ∫ 𝑏 𝑎 d d𝑡𝐹(𝛾(𝑡)) d𝑡= 𝐹(𝛾(𝑏)) −𝐹(𝛾(𝑎)) Remark. Note that we assume that 𝐹exists such that 𝐹′(𝑧) = 𝑓(𝑧); such an 𝐹is not provided for by the theorem.
Example. For an integer 𝑛and the curve 𝛾(𝑡) = 𝑅𝑒2𝜋𝑖𝑡for 𝑡= [0, 1], consider the integral ∫ 𝛾𝑧𝑛d𝑧. For 𝑛≠−1, the function 𝑧𝑛+1 𝑛+1 is an antiderivative of 𝑧𝑛. Hence, ∫ 𝛾𝑧𝑛d𝑧= 0 since 𝛾is a closed curve. If 𝑛= −1, we can use the definition of the integral to find ∫ 𝛾 1 𝑧d𝑧= ∫ 1 0 1 𝛾(𝑡)𝛾′(𝑡) d𝑡= ∫ 1 0 1 𝑅𝑒2𝜋𝑖𝑡2𝜋𝑖𝑅𝑒2𝜋𝑖𝑡d𝑡= 2𝜋𝑖 This is not zero, hence for all 𝑅> 0, 1 𝑧has no antiderivative in any open set containing the circle {|𝑧| = 𝑅}. In particular, for any branch of logarithm 𝜆, it has derivative 1 𝑧, hence there exists no branch of logarithm on ℂ⋆= ℂ∖{0}.
Theorem (converse to fundamental theorem of calculus). Let 𝑈⊂ℂbe a domain. If 𝑓∶𝑈→ℂis continuous and if ∫ 𝛾𝑓(𝑧) d𝑧= 0 for every closed curve 𝛾in 𝑈, then 𝑓has an antiderivative. In other words, there exists a holomorphic function 𝐹∶𝑈→ℂsuch that 𝐹′ = 𝑓in 𝑈.
Proof. Let 𝑎0 ∈𝑈. Then for 𝑤∈𝑈, we can define 𝐹(𝑤) = ∫ 𝛾𝑤 𝑓(𝑧) d𝑧 where 𝛾𝑤∶[0, 1] →ℂis a curve with 𝛾𝑤(0) = 𝑎0 and 𝛾𝑤(1) = 𝑤.
The definition of 𝐹is independent of the choice of 𝛾𝑤. Indeed, suppose two paths 𝛾𝑤, 𝛾′ 𝑤 exist. Then the curve 𝛾𝑤+ (−𝛾′ 𝑤) is a closed path, and by assumption the integral along this 499 IX. Complex Analysis curve is zero. Thus 𝐹is independent of the choice of path as claimed. So 𝐹is a well-defined function.
Now, let 𝑤∈𝑈. Since 𝑈is an open set, there exists 𝑟> 0 such that 𝐷(𝑤, 𝑟) ⊂𝑈. For ℎ∈ℂ with 0 < |ℎ| < 𝑟, let 𝛿ℎbe the radial path 𝑡↦𝑤+ 𝑡ℎfor 𝑡∈[0, 1]. Now we define 𝛾= 𝛾𝑤+ 𝛿ℎ+ (−𝛾𝑤+ℎ) This is a closed curve contained within 𝑈, hence ∫ 𝛾𝑓(𝑧) d𝑧= 0. Thus ∫ 𝛾𝑤+ℎ 𝑓(𝑧) d𝑧= ∫ 𝛾𝑤 𝑓(𝑧) d𝑧+ ∫ 𝛿ℎ 𝑓(𝑧) d𝑧 Informally, the integral has an additivity property which is independent of the path taken.
Rewriting this using 𝐹, 𝐹(𝑤+ ℎ) = 𝐹(𝑤) + ∫ 𝛿ℎ 𝑓(𝑧) d𝑧 = 𝐹(𝑤) + ∫ 𝛿ℎ (𝑓(𝑧) + 𝑓(𝑤) −𝑓(𝑤)) d𝑧 = 𝐹(𝑤) + ℎ𝑓(𝑤) + ∫ 𝛿ℎ (𝑓(𝑧) −𝑓(𝑤)) d𝑧 Hence, by continuity of 𝑓, | | | 𝐹(𝑤+ ℎ) −𝐹(𝑤) ℎ −𝑓(𝑤)| | | = 1 |ℎ| | | | | ∫ 𝛿ℎ (𝑓(𝑧) −𝑓(𝑤)) d𝑧 | | | | ≤1 |ℎ|length(𝛿ℎ) sup 𝑧∈Im 𝛿ℎ |𝑓(𝑧) −𝑓(𝑤)| = sup 𝑧∈Im 𝛿ℎ |𝑓(𝑧) −𝑓(𝑤)| ∴lim ℎ→0 | | | 𝐹(𝑤+ ℎ) −𝐹(𝑤) ℎ −𝑓(𝑤)| | | = lim ℎ→0 sup 𝑧∈Im 𝛿ℎ |𝑓(𝑧) −𝑓(𝑤)| = 0 Thus, 𝐹is differentiable at 𝑤with 𝐹′(𝑤) = 𝑓(𝑤).
2.4. Star-shaped domains Definition. A domain 𝑈is star-shaped, or a star domain, if there exists a (not necessarily unique) centre 𝑎0 ∈𝑈such that for all 𝑤∈𝑈, the straight line segment [𝑎0, 𝑤] is contained within 𝑈.
Remark. Any disc is convex; any convex domain is star-shaped; any star-shaped domain is path-connected. The reverse implications are not true in general.
500 2. Integration Definition. A triangle in ℂis the convex hull of three points in ℂ. The (closed) convex hull of a set 𝑆is the smallest (closed) convex set 𝐶such that 𝑆⊆𝐶. In this case, if 𝑧1, 𝑧2, 𝑧3 ∈ℂ, we have 𝑇= {𝑎𝑧1 + 𝑏𝑧2 + 𝑐𝑧3 ∶0 ≤𝑎, 𝑏, 𝑐≤1, 𝑎+ 𝑏+ 𝑐= 1} When used as a curve, the boundary 𝜕𝑇represents the piecewise affine closed curve 𝛾= 𝛾1 + 𝛾2 + 𝛾3 where 𝛾𝑖are affine functions parametrising the three line segments on the boundary of 𝑇.
Corollary. Let 𝑈be a star-shaped domain. Let 𝑓∶𝑈→ℂbe continuous and ∫ 𝜕𝑇𝑓(𝑧) d𝑧= 0 for any triangle 𝑇⊂𝑈. Then 𝑓has an antiderivative in 𝑈.
Remark. This is a relaxation of the conditions from the previous theorem.
Proof. Let 𝑎0 be a centre for the domain 𝑈. Let 𝑤be an arbitrary point in 𝑈. Then let 𝛾𝑤be the affine function parametrising the directed line segment [𝑎0, 𝑤], and let 𝐹(𝑤) = ∫ 𝛾𝑤𝑓(𝑧) d𝑧. Using ℎand 𝛿ℎas above, by letting 𝛾= 𝛾𝑤+ 𝛿ℎ+ (−𝛾𝑤+ℎ) we then have ∫𝑓(𝑧) d𝑧= ± ∫ 𝜕𝑇𝑓(𝑧) d𝑧for a triangle 𝑇⊂𝑈. Since the integral around a triangle is zero by hypothesis, ∫ 𝛾𝑓(𝑧) d𝑧= 0. We then complete the proof in analogous way to the general case.
Theorem (Cauchy’s theorem for triangles). Let 𝑈⊂ℂbe an open set and 𝑓∶𝑈→ℂbe a holomorphic function. Then ∫ 𝜕𝑇𝑓(𝑧) d𝑧= 0 for all triangles 𝑇⊂𝑈.
Proof. Let 𝜂(𝑡) = ∫ 𝜕𝑇𝑓(𝑧) d𝑧. We will subdivide the triangle 𝑇into four smaller triangles 𝑇(1), 𝑇(2), 𝑇(3), 𝑇(4). The vertices of the inner triangle are the midpoints of the sides of 𝑇, and the three other triangles are constructed to fill the remaining area of 𝑇. Thus, 𝜂(𝑇) = ∫ 𝜕𝑇(1) 𝑓(𝑧) d𝑧+ ∫ 𝜕𝑇(2) 𝑓(𝑧) d𝑧+ ∫ 𝜕𝑇(3) 𝑓(𝑧) d𝑧+ ∫ 𝜕𝑇(4) 𝑓(𝑧) d𝑧 Then, by the triangle inequality, there exists a triangle 𝑇(𝑗) such that | | |∫ 𝜕𝑇(𝑗) 𝑓(𝑧) d𝑧| | | ≥|𝜂(𝑇)| 4 Let 𝑇0 = 𝑇, and 𝑇1 = 𝑇(𝑗), so |𝜂(𝑇1)| ≥ 1 4|𝜂(𝑇0)|. We can show geometrically that for any choice of 𝑇𝑖, length(𝜕𝑇1) = 1 2length(𝜕𝑇0). Inductively, we can subdivide 𝑇𝑖and produce 𝑇𝑖+1, such that 𝑇0 ⊃𝑇1 ⊃⋯; |𝜂(𝑇𝑛)| ≥1 4|𝜂(𝑇𝑛−1)|; length(𝜕𝑇𝑛) = 1 2length(𝜕𝑇𝑛−1) Hence, |𝜂(𝑇𝑛)| ≥1 4𝑛|𝜂(𝑇0)|; length(𝜕𝑇𝑛) = 1 2𝑛length(𝜕𝑇0) 501 IX. Complex Analysis Since 𝑇𝑛are non-empty, nested closed subsets with diameter converging to zero, we can show that ⋂ ∞ 𝑛=1 𝑇𝑛= {𝑧0} for some 𝑧0 ∈ℂ. Let 𝜀> 0. Since 𝑓is differentiable at 𝑧0, there exists 𝛿> 0 such that 𝑧∈𝑈, |𝑧−𝑧0| < 𝛿⟹| | | 𝑓(𝑧) −𝑓(𝑧0) 𝑧−𝑧0 −𝑓′(𝑧0)| | | ≤𝜀 ⟹|𝑓(𝑧) −𝑓(𝑧0) −𝑓′(𝑧0)(𝑧−𝑧0)| ≤𝜀|𝑧−𝑧0| Now, observe that for all 𝑛, ∫ 𝜕𝑇𝑛 𝑓(𝑧) d𝑧= ∫ 𝜕𝑇𝑛 (𝑓(𝑧) −𝑓(𝑧0) −𝑓′(𝑧0)(𝑧−𝑧0)) d𝑧 since ∫ 𝜕𝑇𝑛d𝑧= ∫ 𝜕𝑇𝑛𝑧d𝑧= 0 by the fundamental theorem of calculus. Let 𝑛such that 𝑇𝑛⊂𝐷(𝑧0, 𝛿). Hence, 1 4𝑛|𝜂(𝑇0)| ≤|𝜂(𝑇𝑛)| = | | | | ∫ 𝜕𝑇𝑛 𝑓(𝑧) d𝑧 | | | | = | | | | ∫ 𝜕𝑇𝑛 (𝑓(𝑧) −𝑓(𝑧0) −𝑓′(𝑧0)(𝑧−𝑧0)) d𝑧 | | | | ≤( sup 𝑧∈𝜕𝑇𝑛 |𝑓(𝑧) −𝑓(𝑧0) −𝑓′(𝑧0)(𝑧−𝑧0)|)length(𝜕𝑇𝑛) ≤𝜀( sup 𝑧∈𝜕𝑇𝑛 |𝑧−𝑧0|)length(𝜕𝑇𝑛) ≤𝜀⋅length(𝜕𝑇𝑛)2 = 𝜀 4𝑛length(𝜕𝑇0)2 ∴|𝜂(𝑇0)| ≤𝜀⋅length(𝜕𝑇0)2 𝜀was arbitrary, hence 𝜂(𝑇0) must be zero.
We can generalise the above theorem for functions that are holomorphic except at a finite number of points.
Theorem. Let 𝑈⊂ℂbe an open set and 𝑓∶𝑈→ℂbe a continuous function. Let 𝑆⊂𝑈 be a finite set and suppose that 𝑓is holomorphic on 𝑈∖𝑆. Then ∫ 𝜕𝑇𝑓(𝑧) d𝑧= 0 for all triangles 𝑇⊂𝑈.
Proof. By the procedure above, we can subdivide 𝑇into a total of 4𝑛smaller triangles; at each step we join the midpoints of the sides of the triangles of the previous step. We will keep all of the smaller triangles, and let the sequence of such smaller triangles be denoted 502 2. Integration 𝑇1, … , 𝑇𝑁. Then, since the integrals along the sides of the smaller triangles that are interior to 𝑇cancel, we have ∫ 𝜕𝑇 𝑓(𝑧) d𝑧= 𝑁 ∑ 𝑗=1 ∫ 𝜕𝑇𝑗 𝑓(𝑧) d𝑧 By the previous theorem, ∫ 𝜕𝑇𝑗𝑓(𝑧) d𝑧= 0 unless 𝑇 𝑗intersects with 𝑆. So by letting 𝐼= {𝑗∶𝑇 𝑗∩𝑆≠∅}, we have ∫ 𝜕𝑇 𝑓(𝑧) d𝑧= ∑ 𝑗∈𝐼 ∫ 𝜕𝑇𝑗 𝑓(𝑧) d𝑧 Since any point may be in at most six of the smaller triangles, and length(𝜕𝑇 𝑗) = 1 2𝑛length(𝜕𝑇), we find | | |∫ 𝜕𝑇 𝑓(𝑧) d𝑧| | | ≤6|𝑆|(sup 𝑧∈𝑇 |𝑓(𝑧)|)length(𝜕𝑇) 2𝑛 Then let 𝑛→∞and the result then holds as required.
We can now prove the ‘convex Cauchy’ theorem.
Corollary (Cauchy’s theorem for convex sets). Let 𝑈⊂ℂbe convex, or more generally, a star domain. Let 𝑓∶𝑈→ℂbe continuous on 𝑈and holomorphic in 𝑈∖𝑆where 𝑆is a finite set. Then ∫ 𝛾𝑓(𝑧) d𝑧= 0 for any closed curve 𝛾in 𝑈.
Proof. By the theorems above, ∫ 𝜕𝑇𝑓(𝑧) d𝑧= 0 for any triangle 𝑇⊂𝑈. Since 𝑈is a star domain and 𝑓is continuous, this means that 𝑓has an antiderivative in 𝑈. The result then follows from the fundamental theorem of calculus.
Remark. We will soon show that if 𝑓∶𝑈→ℂis continuous and holomorphic in 𝑈∖𝑆 where 𝑆is finite, then 𝑓is holomorphic in 𝑈.
2.5. Cauchy’s integral formula For a disc 𝐷(𝑎, 𝜌) we will write ∫ 𝜕𝐷(𝑎,𝜌) 𝑓(𝑧) d𝑧to mean ∫ 𝛾𝑓(𝑧) d𝑧where 𝛾∶[0, 1] →ℂis the curve 𝛾(𝑡) = 𝑎+ 𝜌𝑒2𝜋𝑖𝑡.
Theorem (Cauchy’s integral formula for a disc). Let 𝐷= 𝐷(𝑎, 𝑟) and let 𝑓∶𝐷→ℂbe holomorphic. Then, for any 𝜌with 0 < 𝜌< 𝑟and any 𝑤∈𝐷(𝑎, 𝜌), we have 𝑓(𝑤) = 1 2𝜋𝑖∫ 𝜕𝐷(𝑎,𝜌) 𝑓(𝑧) d𝑧 𝑧−𝑤 In particular, taking 𝑤= 𝑎, 𝑓(𝑎) = 1 2𝜋𝑖∫ 𝜕𝐷(𝑎,𝜌) 𝑓(𝑧) d𝑧 𝑧−𝑎= ∫ 1 0 𝑓(𝑎+ 𝜌𝑒2𝜋𝑖𝑡) d𝑡 This final equation is called the mean value property for holomorphic functions.
503 IX. Complex Analysis We first need the following lemma.
Lemma. If 𝛾∶[𝑎, 𝑏] →ℂis a curve and (𝑓 𝑛) is a sequence of continuous complex functions on Im 𝛾converging uniformly to 𝑓on Im 𝛾, then ∫ 𝛾𝑓 𝑛(𝑧) d𝑧→∫ 𝛾𝑓(𝑧) d𝑧.
Proof. We have | | | | ∫ 𝛾 𝑓 𝑛(𝑧) d𝑧−∫ 𝛾 𝑓(𝑧) d𝑧 | | | | = | | | | ∫ 𝛾 (𝑓 𝑛(𝑧) −𝑓(𝑧)) d𝑧 | | | | ≤sup 𝑧∈Im 𝛾 |𝑓 𝑛(𝑧) −𝑓(𝑧)|length(𝛾) We can now prove Cauchy’s integral formula for a disc.
Proof. Let 𝑤∈𝐷(𝑎, 𝜌) be fixed, and define ℎ∶𝐷→ℂby ℎ(𝑧) = { 𝑓(𝑧)−𝑓(𝑤) 𝑧−𝑤 if 𝑧≠𝑤 𝑓′(𝑤) if 𝑧= 𝑤 Then ℎis continuous on 𝐷and holomorphic in 𝐷∖{𝑤}. By Cauchy’s theorem for convex sets, ∫ 𝜕𝐷(𝑎,𝜌) ℎ(𝑧) d𝑧= 0 Substituting for ℎ, we find 𝑓(𝑤) ∫ 𝜕𝐷(𝑎,𝜌) d𝑧 𝑧−𝑤= ∫ 𝜕𝐷(𝑎,𝜌) 𝑓(𝑧) d𝑧 𝑧−𝑤 It now suffices to prove that ∫ 𝜕𝐷(𝑎,𝜌) d𝑧 𝑧−𝑤= 2𝜋𝑖 Note that 1 𝑧−𝑤= 1 𝑧−𝑎+ 𝑎−𝑤= 1 (𝑧−𝑎)(1 − 𝑤−𝑎 𝑧−𝑎) = ∞ ∑ 𝑗=0 (𝑤−𝑎)𝑗 (𝑧−𝑎)𝑗+1 where the convergence is uniform for 𝑧∈𝜕𝐷(𝑎, 𝜌) by the Weierstrass 𝑀-test. Therefore, by the above lemma, we interchange summation and integration to find ∫ 𝜕𝐷(𝑎,𝜌) d𝑧 𝑧−𝑤= ∞ ∑ 𝑗=0 (𝑤−𝑎)𝑗∫ 𝜕𝐷(𝑎,𝜌) d𝑧 (𝑧−𝑎)𝑗+1 For 𝑗≥1, the function 1 (𝑧−𝑎)𝑗+1 has an antiderivative in a neighbourhood of 𝜕𝐷(𝑎, 𝜌), hence all integrals on the right hand side for 𝑗≥1 vanish. For 𝑗= 0, we can compute directly that ∫ 𝜕𝐷(𝑎,𝜌) d𝑧 𝑧−𝑎= 2𝜋𝑖. Hence, ∫ 𝜕𝐷(𝑎,𝜌) d𝑧 𝑧−𝑤= 2𝜋𝑖, proving Cauchy’s integral formula.
504 2. Integration Now, taking 𝑤= 𝑎in Cauchy’s integral formula, we find 𝑓(𝑎) = 1 2𝜋𝑖∫ 𝜕𝐷(𝑎,𝜌) 𝑓(𝑧) d𝑧 𝑧−𝑎 By direct computation using the parametrisation 𝑡↦𝑎+ 𝜌𝑒2𝜋𝑖𝑡for 𝑡∈[0, 1], we find 𝑓(𝑎) = ∫ 1 0 𝑓(𝑎+ 𝜌𝑒2𝜋𝑖𝑡) d𝑡 as required.
2.6. Liouville’s theorem Theorem. Let 𝑓∶ℂ→ℂbe entire and bounded. Then 𝑓is constant. More generally, if 𝑓 is entire with sublinear growth (there exist 𝐾≥0 and 𝛼< 1 such that |𝑓(𝑧)| ≤𝐾(1 + |𝑧|𝛼) for all 𝑧∈ℂ) then 𝑓is constant.
Proof. Let 𝑤∈ℂand 𝜌> |𝑤|. By Cauchy’s integral formula, we have 𝑓(𝑤) = 1 2𝜋𝑖∫ 𝜕𝐷(0,𝜌) 𝑓(𝑧) d𝑧 𝑧−𝑤; 𝑓(0) = 1 2𝜋𝑖∫ 𝜕𝐷(0,𝜌) 𝑓(𝑧) d𝑧 𝑧 Thus, |𝑓(𝑤) −𝑓(0)| = 1 2𝜋 | | | | ∫ 𝜕𝐷(0,𝜌) 𝑤𝑓(𝑧) d𝑧 𝑧(𝑧−𝑤) | | | | ≤|𝑤| 2𝜋 sup 𝑧∈𝜕𝐷(0,𝜌) |𝑓(𝑧)| |𝑧| ⋅||𝑧| −|𝑤||length(𝜕𝐷(0, 𝜌)) ≤|𝑤|𝐾(1 + 𝜌𝛼) 2𝜋𝜌(𝜌−|𝑤|)2𝜋𝜌 = |𝑤|𝐾(1 + 𝜌𝛼) 𝜌−|𝑤| By letting 𝜌→∞, we can conclude 𝑓(𝑤) = 𝑓(0).
Theorem (fundamental theorem of algebra). Every non-constant polynomial with complex coefficients has a complex root.
Proof. Let 𝑝(𝑧) = 𝑎𝑛𝑧𝑛+ ⋯+ 𝑎0 be a complex polynomial of degree 𝑛≥1. Then 𝑎𝑛≠0, and for 𝑧≠0 we can write 𝑝(𝑧) = 𝑧𝑛(𝑎𝑛+ 𝑎𝑛−1 𝑧 + ⋯+ 𝑎0 𝑧𝑛) 505 IX. Complex Analysis By the triangle inequality, |𝑝(𝑧)| ≥|𝑧|𝑛(|𝑎𝑛| −|𝑎𝑛−1| |𝑧| −⋯−|𝑎0| |𝑧|𝑛) Hence, there exists 𝑅> 0 such that |𝑝(𝑧)| ≥1 for |𝑧| > 𝑅.
Now, if 𝑝(𝑧) ≠0 for all 𝑧, then 𝑔(𝑧) = 1 𝑝(𝑧) is entire. By the above, |𝑔(𝑧)| ≤1 for |𝑧| > 𝑅.
By continuity of 𝑔, we have further that |𝑔(𝑧)| is bounded from above on the compact set {|𝑧| ≤𝑅}. Hence, 𝑔is a bounded entire function. By Liouville’s theorem, 𝑔is constant. Since 𝑝is non-constant, this is a contradiction. Hence 𝑝has a zero.
Theorem (local maximum modulus principle). Let 𝑓∶𝐷(𝑎, 𝑅) →ℂbe holomorphic, and |𝑓(𝑧)| ≤|𝑓(𝑎)| for all 𝑧∈𝐷(𝑎, 𝑅). Then 𝑓is constant.
Proof. By the mean value property, 𝑓(𝑎) = ∫ 1 0 𝑓(𝑎+ 𝜌𝑒2𝜋𝑖𝑡) d𝑡 Therefore, |𝑓(𝑎)| = | | | | ∫ 1 0 𝑓(𝑎+ 𝜌𝑒2𝜋𝑖𝑡) d𝑡 | | | | ≤sup 𝑡∈[0,1] | |𝑓(𝑎+ 𝜌𝑒2𝜋𝑖𝑡)| | ≤|𝑓(𝑎)| where the last inequality is by hypothesis. Therefore, both inequalities must be equalities.
Equality in the first inequality implies that 𝑓(𝑎+ 𝜌𝑒2𝜋𝑖𝑡) = 𝑐𝜌for some constant 𝑐𝜌and all 𝑡∈[0, 1]. Then, by the first equality, | |𝑐𝜌| | = |𝑓(𝑎)| for all 𝜌∈(0, 𝑅). Thus, | |𝑓(𝑎+ 𝜌𝑒2𝜋𝑖𝑡)| | is constant for all 𝜌∈(0, 𝑅) and 𝑡∈[0, 1]. Hence |𝑓(𝑧)| is constant on 𝐷(𝑎, 𝑅). By the Cauchy–Riemann equations, 𝑓must be constant.
2.7. Taylor series Theorem. Let 𝑓∶𝐷(𝑎, 𝑅) →ℂbe holomorphic. Then 𝑓has a convergent power series rep-resentation on 𝐷(𝑎, 𝑅). More precisely, there exists a sequence of complex numbers 𝑐0, 𝑐1, … such that 𝑓(𝑤) = ∞ ∑ 𝑛=0 𝑐𝑛(𝑤−𝑎)𝑛 The coefficient 𝑐𝑛is given by 𝑐𝑛= 1 2𝜋𝑖∫ 𝜕𝐷(𝑎,𝜌) 𝑓(𝑧) d𝑧 (𝑧−𝑎)𝑛+1 for any 𝜌∈(0, 𝑅).
506 2. Integration Proof. Let 0 < 𝜌< 𝑅. Then, for any 𝑤∈𝐷(0, 𝜌), we have by Cauchy’s integral formula that 𝑓(𝑤) = 1 2𝜋𝑖∫ 𝜕𝐷(𝑎,𝜌) 𝑓(𝑧) d𝑧 𝑧−𝑤 = 1 2𝜋𝑖∫ 𝜕𝐷(𝑎,𝜌) 𝑓(𝑧) ∞ ∑ 𝑛=0 (𝑤−𝑎)𝑛 (𝑧−𝑎)𝑛+1 d𝑧 = ∞ ∑ 𝑛=0 ( 1 2𝜋𝑖∫ 𝜕𝐷(𝑎,𝜌) 𝑓(𝑧) d𝑧 (𝑧−𝑎)𝑛+1 )(𝑤−𝑎)𝑛 The last equality holds since the series under the integral converges uniformly for all 𝑧∈ 𝜕𝐷(𝑎, 𝜌). Let 𝑐𝑛(𝜌) = 1 2𝜋𝑖∫ 𝜕𝐷(𝑎,𝜌) 𝑓(𝑧) d𝑧 (𝑧−𝑎)𝑛+1 Then we have shown that 𝑓(𝑤) = ∑ ∞ 𝑛=0 𝑐𝑛(𝜌)(𝑤−𝑎)𝑛+1 for all 𝑤∈𝐷(𝑎, 𝜌). By a previous theorem, the function 𝑓has derivatives of all orders in 𝐷(𝑎, 𝜌) and hence 𝑐𝑛(𝜌) = 𝑓(𝑛)(𝑎) 𝑛!
, which is independent of 𝜌, so we can let 𝑐𝑛= 𝑐𝑛(𝜌) for an arbitrary 𝜌.
Corollary. If 𝑓is holomorphic on an open set 𝑈⊂ℂ, then 𝑓has derivatives of all orders in 𝑈, and those derivatives are holomorphic on 𝑈.
Proof. We have a power series representation for 𝑓near every points, so its derivatives of all orders exist everywhere. Hence, the derivatives of all orders are holomorphic.
Remark. We can explicitly compute from the 𝑐𝑛above that 𝑓(𝑛)(𝑎) = 𝑛!
2𝜋𝑖∫ 𝜕𝐷(𝑎,𝜌) 𝑓(𝑧) d𝑧 (𝑧−𝑎)𝑛+1 This is a special case of a Cauchy integral formula for derivatives.
Note also that by taking 𝑛= 1, we can apply the estimate for the integral to find |𝑓′(𝑎)| ≤1 𝜌( sup 𝑧∈𝜕𝐷(𝑎,𝜌) |𝑓(𝑧)|) This can be thought of as a localised version of Liouville’s theorem, and it directly implies Liouville’s theorem. Indeed, if 𝑓is entire and bounded, let 𝑎∈ℂand by applying the estimate and letting 𝜌→∞we can conclude 𝑓′ = 0 on ℂ, giving that 𝑓is constant.
Definition. A function 𝑓is analytic at a point 𝑎∈ℂ(or ℝ) if there exists a neighbourhood of 𝑎such that 𝑓is given by a convergent power series about 𝑎.
507 IX. Complex Analysis Remark. If 𝑓is analytic at 𝑎, we must have derivatives of all orders of 𝑓near 𝑎. The above corollary implies that if 𝑓is complex, the following are equivalent.
(i) 𝑓is analytic at 𝑎 (ii) 𝑓has complex derivatives of all orders in a neighbourhood of 𝑎 (iii) 𝑓is complex differentiable once in a neighbourhood in a neighbourhood of 𝑎(so 𝑓is holomorphic at 𝑎) For real functions, this is not the case. For example, consider 𝑓∶ℝ→ℝdefined by 𝑓(𝑥) = exp(−𝑥−2). This has 𝑓(𝑛)(0) = 0 for all 𝑛, so 𝑓is not given by a convergent power series near zero.
Let 𝑈⊂ℂbe an open set. Now, we have that 𝑓= 𝑢+ 𝑖𝑣is holomorphic in 𝑈if and only if 𝑢and 𝑣have continuous partial derivatives in 𝑈, and that 𝑢.𝑣satisfy the Cauchy–Riemann equations. Further, the corollary above implies that 𝑢, 𝑣are 𝐶2 functions. This shows that 𝑢and 𝑣are harmonic.
Theorem (Morera’s theorem). Let 𝑈⊆ℂbe open, and 𝑓∶𝑈→ℂbe a continuous function such that ∫ 𝛾𝑓(𝑧) d𝑧= 0 for all closed curves 𝛾in 𝑈. Then 𝑓is holomorphic in 𝑈.
Remark. This can be thought of as a converse to Cauchy’s theorem.
Proof. We know that 𝑓has a holomorphic antiderivative 𝐹on 𝑈. Then, we know that 𝐹is twice differentiable in 𝑈. Since 𝐹′ = 𝑓, 𝑓is holomorphic.
Corollary. Let 𝑈⊆ℂbe an open set. Let 𝑓∶𝑈→ℂbe a continuous function and holo-morphic in 𝑈∖𝑆, where 𝑆is a finite set. Then 𝑓is holomorphic in 𝑈.
Proof. For all 𝑎∈𝑈, there exists 𝑟> 0 such that 𝐷= 𝐷(𝑎, 𝑟) ⊂𝑈. Since 𝐷is convex, we can apply Cauchy’s formula for convex sets to observe that ∫ 𝛾𝑓(𝑧) d𝑧= 0 for all closed curves in 𝐷. Then by Morera’s theorem, 𝑓is holomorphic.
2.8. Zeroes of holomorphic functions Definition. Let 𝑓be a holomorphic function on a disc 𝐷= 𝐷(𝑎, 𝑅). By the Taylor series theorem, there exist constants 𝑐𝑛such that 𝑓(𝑧) = ∞ ∑ 𝑛=0 𝑐𝑛(𝑧−𝑎)𝑛 for all 𝑧∈𝐷. Then if 𝑓is not identically zero, there exists 𝑛such that 𝑐𝑛≠0. Let 𝑚= min {𝑛∶𝑐𝑛≠0}. Then, 𝑓(𝑧) = (𝑧−𝑎)𝑚𝑔(𝑧); 𝑔(𝑧) = ∞ ∑ 𝑛=𝑚 𝑐𝑛(𝑧−𝑎)𝑛−𝑚 508 2. Integration Note that 𝑔is holomorphic on 𝐷, and 𝑔(𝑎) = 𝑐𝑚≠0.
If 𝑚≠0, we say that 𝑓has a zero of order 𝑚at 𝑧= 𝑎. Hence 𝑚is the smallest natural number 𝑛such that 𝑓(𝑛)(𝑎) ≠0. If 𝑆⊆ℂ, then a point 𝑤∈𝑆is an isolated point of 𝑆if there exists 𝑟> 0 such that 𝑆∩𝐷(𝑤, 𝑟) = {𝑤}.
Theorem (principle of isolated zeroes). Let 𝑓∶𝐷(𝑎, 𝑅) →ℂbe holomorphic and not identically zero. Then there exists 𝑟∈(0, 𝑅) such that 𝑓(𝑧) ≠0 whenever 0 < |𝑧−𝑎| < 𝑟.
Remark. If 𝑓(𝑎) = 0, then {𝑧∶𝑓(𝑧) = 0} intersects 𝐷(𝑎, 𝑟) only at 𝑎. Hence, 𝑎is an isolated point of the set of zeroes. For instance, there exists no nonzero holomorphic function that vanishes on a line segment or a disc.
We can show that certain identities from real analysis hold for complex functions. For in-stance, consider the function 𝑔(𝑧) = sin2 𝑧+ cos2 𝑧−1. Since this 𝑔is holomorphic and vanishes on the real line, 𝑔must be identically zero in the complex plane.
The zero set may have an accumulation point on the boundary of the domain of 𝑓. Consider 𝑓(𝑧) = sin 1 𝑧for 𝑧∈𝐷(1, 1). Here, if 𝑎𝑛= 1 2𝑛𝜋, then 𝑎𝑛∈𝐷(1, 1) and 𝑓(𝑎𝑛) = 0 and 𝑎𝑛→0 ∈𝜕𝐷(1, 1).
Proof. If 𝑓(𝑎) ≠0, then by continuity of 𝑓there exists 𝑟> 0 such that 𝑓(𝑧) ≠0 for all 𝑧∈𝐷(𝑎, 𝑟). If 𝑓(𝑎) = 0, then there exists an integer 𝑚≥1 such that 𝑓(𝑧) = (𝑧−𝑎)𝑚𝑔(𝑧) for 𝑧∈𝐷(𝑎, 𝑅), where 𝑔is holomorphic with 𝑔(𝑎) ≠0. In this case, we find that there exists 𝑟> 0 such that 𝑔(𝑧) ≠0 for 𝑧∈𝐷(𝑎, 𝑟) and hence 𝑓(𝑧) ≠0 for 𝑧∈𝐷(𝑎, 𝑟) ∖{𝑎}.
2.9. Analytic continuation Theorem. Let 𝑈⊂𝑉be domains. If 𝑔1, 𝑔2 ∶𝑉→ℂare analytic and 𝑔1 = 𝑔2 on 𝑈, then 𝑔1 = 𝑔2 on 𝑉. Equivalently, if 𝑓∶𝑈→ℂis analytic, then there is at most one analytic function 𝑔∶𝑉→ℂsuch that 𝑔= 𝑓on 𝑈. We say that 𝑔is the analytic continuation of 𝑓to 𝑉, if it exists.
Proof. Let 𝑔1, 𝑔2 ∶𝑉→ℂbe analytic with 𝑔1|𝑈= 𝑔2|𝑈. Then, ℎ= 𝑔1 −𝑔2 ∶𝑉→ℂis analytic, and ℎ|𝑈≡0. We want to show that ℎ≡0. Let 𝑉 0 = {𝑧∈𝑉∶∃𝑟> 0, ℎ| | |𝐷(𝑧,𝑟) ≡0} and 𝑉 1 = {𝑧∈𝑉∶∃𝑛≥0, ℎ(𝑛)(𝑧) ≠0} Let 𝑧∈𝑉and suppose that 𝑧∉𝑉 0. Then for any disc 𝐷= 𝐷(𝑧, 𝑟) ⊂𝑉, we have ℎ≢0 in 𝐷.
Hence, by Taylor series, ℎ(𝑛)(𝑧) ≠0 for some 𝑛, so 𝑧∈𝑉 1. Thus, 𝑉= 𝑉 0 ∪𝑉 1. We also know that 𝑉 0 ∩𝑉 1 = ∅.
509 IX. Complex Analysis Note that 𝑉 0 is open by definition, and 𝑉 1 is by continuity of the derivatives ℎ(𝑛). By con-nectedness of the domain 𝑉, either 𝑉 0 or 𝑉 1 is empty. Since 𝑈⊂𝑉 0, we must have 𝑉 1 = ∅.
Thus, 𝑉= 𝑉 0 so ℎ≡0.
Remark. The above proof does not rely on holomorphicity but on analyticity. Thus, the theorem holds for real analytic functions. For example, due to elliptic regularity (see Part II Analysis of Functions), we can show that harmonic functions are real analytic, and hence have a unique analytic continuation if one exists.
Given a holomorphic function 𝑓defined on a disc, we can compute the largest domain con-taining the disc to which there exists an analytic continuation of 𝑓. This is nontrivial to answer in general.
Example. Let 𝑓(𝑧) = ∑ ∞ 𝑛=0 𝑧𝑛. The radius of convergence of this series is 1, so 𝑓is analytic in 𝐷(0, 1), and there is no larger disc 𝐷(0, 𝑟) ⊃𝐷(0, 1) such that 𝑔has an analytic continu-ation to 𝐷(0, 𝑟). However, since 𝑓(𝑧) = 1 1−𝑧for 𝑧∈𝐷(0, 1) and the function 1 1−𝑧is analytic in ℂ∖{1}, 𝑓indeed has an analytic continuation to the larger domain ℂ∖{1}.
Example. Let 𝑓(𝑧) = ∑ ∞ 𝑛=1 (−1)𝑛+1𝑧𝑛 𝑛 . This function also has a radius of convergence of 1, so 𝑓is analytic on 𝐷(0, 1). It has analytic continuation Log(1 + 𝑧) to the domain ℂ∖ {𝑥∈ℝ∶𝑥≤−1} containing 𝐷(0, 1).
Example. Let 𝑓(𝑧) = ∑ ∞ 𝑛=0 𝑧𝑛!. This has radius of convergence 1, so 𝑓is analytic in 𝐷(0, 1).
However, 𝑓has no analytic continuation to any larger domain containing 𝐷(0, 1). The boundary 𝜕𝐷(0, 1) is known as the natural boundary of 𝑓.
We can find in fact that for any given domain 𝑈⊂ℂ, there exists a holomorphic function 𝑓∶𝑈→ℂwhich has no analytic continuation to a domain properly containing 𝑈.
The failure of analytic continuation in some cases can be explained as the result of loss of a regularity condition, such as boundedness, continuity, differentiability, or so on. However, this is not always the reason, and analytic continuation may remain impossible even when regularity conditions are all satisfied.
Example. Let 𝑓(𝑧) = ∑ ∞ 𝑛=0 exp(−2𝑛/2)𝑧2𝑛, which has unit radius of convergence. 𝑓, and its derivatives of any order, are uniformly continuous on the closed disc 𝐷(0, 1). However, we can prove that it has natural boundary 𝜕𝐷(0, 1), using the following theorem which will not be proven.
Theorem (Ostrowski–Hadamard gap theorem). Let (𝑝𝑛) be a sequence of positive integers such that 𝑝𝑛+1 > (1 + 𝛿)𝑝𝑛for all 𝑛and some fixed 𝛿> 0. If (𝑐𝑛) is a sequence of complex numbers such that 𝑓(𝑧) = ∑ ∞ 𝑛=0 𝑐𝑛𝑧𝑝𝑛has unit radius of convergence, then 𝜕𝐷(0, 1) is the natural boundary of 𝑓.
Corollary (identity principle). Let 𝑓, 𝑔∶𝑈→ℂbe holomorphic functions in a domain 𝑈.
If the set 𝑆= {𝑧∈𝑈∶𝑓(𝑧) = 𝑔(𝑧)} contains a non-isolated point, then 𝑓= 𝑔in 𝑈.
510 2. Integration Proof. Let ℎ= 𝑓−𝑔, so 𝑆= {𝑧∈𝑈∶ℎ(𝑧) = 0}. Suppose that 𝑆has a non-isolated point 𝑤.
If there exists 𝑟> 0 such that ℎ≢0 in 𝐷(𝑤, 𝑟), then by the principle of isolated zeroes, we can find 𝜀> 0 such that 𝑓(𝑧) ≠0 whenever 0 < |𝑧−𝑤| < 𝜀. However, this contradicts the assumption that 𝑤is a non-isolated point of 𝑆. Thus, ℎ≡0 on 𝐷(𝑤, 𝑟) for all 𝐷(𝑤, 𝑟) ⊂𝑈.
Thus, ℎ≡0 on 𝑈, so 𝑓= 𝑔on 𝑈.
Corollary (global maximum principle). Let 𝑈be a bounded open set. Suppose 𝑓∶𝑈→ℂ is a continuous function such that 𝑓is holomorphic in 𝑈. Then |𝑓| attains its maximum on 𝜕𝑈.
Proof. 𝑈is compact, and |𝑓| is continuous on 𝑈. Hence, it attains its maximum; there exists 𝑤∈𝑈such that |𝑓(𝑤)| = max𝑧∈𝑈|𝑓(𝑧)|. If 𝑤∉𝑈, then 𝑤∈𝜕𝑈as required. Otherwise, let 𝐷= 𝐷(𝑤, 𝑟) ⊂𝑈. Since |𝑓(𝑧)| ≤|𝑓(𝑤)| for all 𝑧∈𝐷, the local maximum principle implies that 𝑓is constant on 𝐷. Hence, by the identity principle, 𝑓is constant on the connected component of 𝑈containing 𝐷, which will be written 𝑈′. By continuity, 𝑓is constant on the closure of this connected component 𝑈′. In particular, |𝑓(𝑧)| = |𝑓(𝑤)| for all 𝑧∈𝜕𝑈′ ⊆𝜕𝑈 as required.
Theorem (Cauchy’s integral formula for derivatives). Let 𝑓∶𝐷(𝑎, 𝑅) →ℂbe holomorphic.
For any 𝜌∈(0, 𝑅) and 𝑤∈𝐷(𝑎, 𝜌), we have 𝑓(𝑘)(𝑤) = 𝑘!
2𝜋𝑖∫ 𝜕𝐷(𝑎,𝜌) 𝑓(𝑧) d𝑧 (𝑧−𝑤)𝑘+1 Further, sup 𝑧∈𝐷(𝑎,𝑅/2) | |𝑓(𝑘)(𝑧)| | ≤𝐶 𝑅𝑘 sup 𝑧∈𝐷(𝑎,𝑅) |𝑓(𝑧)| where 𝐶= 𝑘!2𝑘+1 is a constant which depends only on 𝑘. This final result is called a Cauchy estimate for the 𝑘th derivative.
Remark. Directly applying Cauchy’s integral formula to 𝑓(𝑛), we find a formula for 𝑓(𝑛)(𝑤) in terms of an integral involving 𝑓(𝑛). The significance of the above theorem is that the integral involves 𝑓alone, and not its derivatives.
Note that we have already observed the special case 𝑤= 𝑎. This was proven during the discussion on Taylor series.
Proof. If 𝑘= 0, we have the usual Cauchy integral formula. For 𝑘= 1, let 𝑔(𝑧) = 𝑓(𝑧) 𝑧−𝑤, which is holomorphic in 𝐷(𝑎, 𝑅) ∖{𝑤}, with derivative 𝑔′(𝑧) = 𝑓′(𝑧) 𝑧−𝑤− 𝑓(𝑧) (𝑧−𝑤)2 511 IX. Complex Analysis Since 𝜕𝐷(𝑎, 𝜌) ⊂𝐷(𝑎, 𝑅) ∖{𝑤}, we know that ∫ 𝜕𝐷(𝑎,𝜌) 𝑔′(𝑧) d𝑧= 0 by the fundamental theorem of calculus. Applying the usual Cauchy integral formula to 𝑓′, 𝑓′(𝑤) = 1 2𝜋𝑖∫ 𝜕𝐷(𝑎,𝜌) 𝑓′(𝑧) d𝑧 𝑧−𝑤 Combining these results give the result for 𝑘= 1. For higher derivatives, we can proceed by induction. Let 𝑘≥2, and then suppose the formula holds for this value of 𝑘, for all holo-morphic functions 𝐷(𝑎, 𝑅) →ℂ. For any holomorphic function 𝑓∶𝐷(𝑎, 𝑅) →ℂ, consider 𝑔(𝑧) = 𝑓(𝑧) (𝑧−𝑤)𝑘+1 ⟹𝑔′(𝑧) = 𝑓′(𝑧) (𝑧−𝑤)𝑘+1 −(𝑘+ 1)𝑓(𝑧) (𝑧−𝑤)𝑘+2 which is defined in 𝐷(𝑎, 𝑅) ∖{𝑤}. Then, since ∫ 𝜕𝐷(𝑎,𝜌) 𝑔′(𝑧) d𝑧= 0, we find ∫ 𝜕𝐷(𝑎,𝜌) 𝑓′(𝑧) d𝑧 (𝑧−𝑤)𝑘+1 = (𝑘+ 1) ∫ 𝜕𝐷(𝑎,𝜌) 𝑓(𝑧) d𝑧 (𝑧−𝑤)𝑘+2 By substituting 𝑓′ into the induction hypothesis, 𝑓(𝑘+1)(𝑤) = 𝑘!
2𝜋𝑖∫ 𝜕𝐷(𝑎,𝜌) 𝑓′(𝑧) d𝑧 (𝑧−𝑤)𝑘+1 We can then combine the previous two expressions to find 𝑓(𝑘+1)(𝑤) = (𝑘+ 1)!
2𝜋𝑖 ∫ 𝜕𝐷(𝑎,𝜌) 𝑓(𝑧) d𝑧 (𝑧−𝑤)𝑘+2 as required.
For the second part, let sup𝑧∈𝐷(𝑎,𝑅) |𝑓(𝑧)| < ∞without loss of generality. Let 𝜌∈(𝑅/2, 𝑅).
Then, by the first part, for all 𝑤∈𝐷(𝑎, 𝑅/2) we have | |𝑓(𝑘)(𝑤)| | ≤𝑘!
2𝜋( sup 𝑧∈𝜕𝐷(𝑎,𝜌) |𝑓(𝑧)| |𝑧−𝑤|𝑘+1 )length(𝜕𝐷(𝑎, 𝜌)) As |𝑧−𝑤| ≥𝜌−𝑅/2 for all 𝑧∈𝜕𝐷(𝑎, 𝜌) and all 𝑤∈𝐷(𝑎, 𝑅/2), this implies sup 𝑤∈𝐷(𝑎,𝑅/2) | |𝑓(𝑘)(𝑤)| | ≤ 𝑘!𝜌 (𝜌−𝑅/2)𝑘+1 sup 𝑧∈𝐷(𝑎,𝑅) |𝑓(𝑧)| Now, as 𝜌→𝑅, the result follows.
2.10. Uniform limits of holomorphic functions Definition. Let 𝑈⊆ℂbe open, and let 𝑓 𝑛∶𝑈→ℂbe a sequence of functions. We say that (𝑓 𝑛) converges locally uniformly on 𝑈if, for all 𝑎∈𝑈, there exists 𝑟> 0 such that (𝑓 𝑛) converges uniformly on 𝐷(𝑎, 𝑟).
512 2. Integration Example. Let 𝑓 𝑛(𝑧) = 𝑧𝑛. Then 𝑓 𝑛→0 locally uniformly, but not uniformly.
Proposition. (𝑓 𝑛) converges locally uniformly on an open set 𝑈⊆ℂif and only if (𝑓 𝑛) converges uniformly on each compact subset 𝐾⊆𝑈.
Proof. The forward implication is simple, due to the definition of compactness. The con-verse follows since for all 𝑎∈𝑈, there exists a compact disc 𝐷(𝑎, 𝑟) ⊂𝑈.
Theorem (uniform limits of holomorphic functions). Let 𝑈⊆ℂbe open, and 𝑓 𝑛∶𝑈→ℂ be holomorphic for each 𝑛∈ℕ. If (𝑓 𝑛) converges locally uniformly on 𝑈to some function 𝑓∶𝑈→ℂ, then 𝑓is holomorphic.
Further, 𝑓′ 𝑛→𝑓′ locally uniformly on 𝑈, and by induction, for each 𝑘we have 𝑓(𝑘) 𝑛 →𝑓(𝑘) locally uniformly on 𝑈as 𝑛→∞.
Remark. This is not true for real analytic functions. The Weierstrass approximation theorem states the following. Let 𝑓∶[𝑎, 𝑏] →ℝbe a continuous function on a compact interval [𝑎, 𝑏] ⊂ℝ. Then, there exists a sequence of polynomials (𝑝𝑛) converging uniformly to 𝑓on [𝑎, 𝑏].
There exist continuous, nowhere differentiable functions 𝑓∶[𝑎, 𝑏] →ℝ. Applying the Wei-erstrass approximation theorem to such functions 𝑓shows that the uniform limit of real analytic functions need not have a single point of differentiability.
Proof. Let 𝑎∈𝑈and choose 𝑟> 0 such that 𝐷(𝑎, 𝑟) ⊂𝑈and 𝑓 𝑛→𝑓uniformly on 𝐷(𝑎, 𝑟). Since the 𝑓 𝑛are continuous, by a result from Analysis and Topology we have that 𝑓 is continuous in 𝐷(𝑎, 𝑟).
Let 𝛾be a closed curve in 𝐷(𝑎, 𝑟). Since 𝐷(𝑎, 𝑟) is convex, by the convex Cauchy theorem we have ∫ 𝛾𝑓 𝑛(𝑧) d𝑧= 0. Since 𝑓 𝑛→𝑓uniformly on 𝐷(𝑎, 𝑟), it follows that ∫ 𝛾 𝑓(𝑧) d𝑧= lim 𝑧→∞∫ 𝛾 𝑓 𝑛(𝑧) d𝑧= 0 By Morera’s theorem, 𝑓is holomorphic in 𝐷(𝑎, 𝑟). Since 𝑎is arbitrary, 𝑓is holomorphic on all of 𝑈.
Now, let 𝑎∈𝑈be arbitrary and let 𝐷(𝑎, 𝑟) be as above. We can apply the Cauchy estimate for 𝑘= 1, 𝑅= 𝑟, applied to the function 𝑓 𝑛−𝑓. This gives sup 𝑧∈𝐷(𝑎,𝑟/2) |𝑓′ 𝑛(𝑧) −𝑓′(𝑧)| ≤4 𝑟 sup 𝑧∈𝐷(𝑎,𝑟) |𝑓 𝑛(𝑧) −𝑓(𝑧)| Since the right hand side converges to zero as 𝑛→∞, the claim follows.
Remark. Many of the key results proven for holomorphic functions have analogues for real harmonic functions on domains not just in ℝ2 but in ℝ𝑛for any 𝑛. For instance: 513 IX. Complex Analysis (i) (Liouville’s theorem) if 𝑢∶ℝ𝑛→ℝis a bounded harmonic function then 𝑢is constant; (ii) (local maximum principle) if 𝑢∶𝐷(𝑎, 𝑟) is a 𝐶2 harmonic function on an open ball 𝐷(𝑎, 𝑟) in ℝ𝑛, and if 𝑢(𝑥) ≤𝑢(𝑎) for all 𝑥∈𝐷(𝑎, 𝑟), then 𝑢is constant; (iii) (global maximum principle) a harmonic function on a bounded open set 𝑈that is continuous on 𝑈attains its maximum on 𝜕𝑈; (iv) harmonic functions are real analytic; (v) the unique analytic continuation principle holds; (vi) uniform limits of harmonic functions are harmonic; (vii) derivative estimates hold: if 𝑢∶𝐷(𝑎, 𝑅) ⊆ℝ𝑛→ℝis harmonic, then sup 𝐷(𝑎,𝑅/2) | |𝐷𝑘𝑢| | ≤𝐶𝑅−𝑘sup 𝐷(𝑎,𝑅) |𝑢|; 𝐶= 𝐶(𝑛, 𝑘) For the case 𝑛= 2, the result for harmonic functions can often be deduced directly from the corresponding results for holomorphic functions. For instance, for Liouville’s theorem, given that 𝑢is a harmonic function on ℝ2, we find a function 𝑣such that 𝑢+𝑖𝑣is holomorphic on ℂ(which is always possible in a simply connected domain). Then 𝑔= 𝑒𝑓is holomorphic with |𝑔| = 𝑒𝑢, so if 𝑢is bounded then 𝑔is bounded. By Liouville’s theorem for holomorphic functions, 𝑔and hence 𝑓is constant.
514 3. More integration 3. More integration 3.1. Winding numbers Let 𝛾∶[𝑎, 𝑏] →ℂbe a closed, piecewise 𝐶1 curve, and let 𝑤∉Im 𝛾. For all 𝑡, there exists 𝑟(𝑡) > 0 and 𝜃(𝑡) ∈ℝsuch that 𝛾(𝑡) = 𝑤+ 𝑟(𝑡)𝑒𝑖𝜃(𝑡). Then, the function 𝑟∶[𝑎, 𝑏] →ℝis given by 𝑟(𝑡) = |𝛾(𝑡) −𝑤|, so it is uniquely determined and piecewise 𝐶1.
Definition. If we have a continuous choice of 𝜃∶[𝑎, 𝑏] →ℝsuch that 𝛾(𝑡) = 𝑤+ 𝑟(𝑡)𝑒𝑖𝜃(𝑡), then we define the winding number or the index of 𝛾about 𝑤as 𝐼(𝛾; 𝑤) = 𝜃(𝑏) −𝜃(𝑎) 2𝜋 If 𝛾is a closed curve, 𝐼(𝛾; 𝑤) is an integer. This is because 𝛾(𝑎) = 𝛾(𝑏) ⟹exp(𝑖𝜃(𝑏) −𝑖𝜃(𝑎)) = 1 If 𝜃1 ∶[𝑎, 𝑏] →ℂis also continuous such that 𝛾(𝑡) = 𝑤+𝑟𝑒𝑖𝜃1(𝑡), then exp(𝑖𝜃(𝑡) −𝑖𝜃1(𝑡)) = 1, so 𝜃1(𝑡) −𝜃(𝑡) 2𝜋 ∈ℤ Since 𝜃1 −𝜃is continuous, this quotient must be a constant. Hence, 𝐼(𝛾; 𝑤) is well-defined and independent of the (continuous) choice of 𝜃.
Lemma. Let 𝑤∈ℂand 𝛾∶[𝑎, 𝑏] →ℂ∖{𝑤}, where 𝛾is piecewise 𝐶1. Then, there exists a piecewise 𝐶1 function 𝜃∶[𝑎, 𝑏] →ℝsuch that 𝛾(𝑡) = 𝑤+ 𝑟(𝑡)𝑒𝑖𝜃(𝑡), where 𝑟(𝑡) = |𝛾(𝑡) −𝑤|.
If 𝛾is closed, then we also have 𝐼(𝛾; 𝑤) = 1 2𝜋𝑖∫ 𝛾 d𝑧 𝑧−𝑤 Remark. If 𝛾is 𝐶1, and there is a 𝐶1 function 𝜃such that 𝛾(𝑡) = 𝑤+ 𝑟(𝑡)𝑒𝑖𝜃(𝑡), then 𝛾′(𝑡) = (𝑟′(𝑡) + 𝑖𝑟(𝑡)𝜃′(𝑡))𝑒𝑖𝜃(𝑡) = (𝑟′(𝑡) 𝑟(𝑡) + 𝑖𝜃′(𝑡))𝑟(𝑡)𝑒𝑖𝜃(𝑡) = (𝑟′(𝑡) 𝑟(𝑡) + 𝑖𝜃′(𝑡))(𝛾(𝑡) −𝑤) Hence, 𝜃′(𝑡) = Im 𝛾′(𝑡) 𝛾(𝑡) −𝑤⟹𝜃(𝑡) = 𝜃(𝑎) + Im ∫ 𝑡 𝑎 𝛾′(𝑠) d𝑠 𝛾(𝑠) −𝑤 Proof. Let ℎ(𝑡) = ∫ 𝑡 𝑎 𝛾′(𝑠) 𝛾(𝑠)−𝑤d𝑠. The integrand is bounded on [𝑎, 𝑏], and is continuous except at the finite number of points at which 𝛾′ may be discontinuous. Hence, ℎ∶[𝑎, 𝑏] →ℂis continuous. Further, ℎis differentiable with ℎ′(𝑡) = 𝛾′(𝑡) 𝛾(𝑡)−𝑤at each 𝑡where 𝛾′ is continuous.
Hence, ℎis piecewise 𝐶1. This induces an ordinary differential equation for 𝛾(𝑡) −𝑤.
(𝛾(𝑡) −𝑤)′ −(𝛾(𝑡) −𝑤)ℎ′(𝑡) = 0 515 IX. Complex Analysis which is true for all 𝑡∈[𝑎, 𝑏] except possibly for a finite set. Hence, d d𝑡((𝛾(𝑡) −𝑤)𝑒−ℎ(𝑡)) = 𝛾′(𝑡)𝑒−ℎ(𝑡) −(𝛾(𝑡) −𝑤)𝑒−ℎ(𝑡)ℎ′(𝑡) = 0 except for finitely many 𝑡. Since (𝛾(𝑡)−𝑤)𝑒−ℎ(𝑡) is continuous, it must be constant, and equal to its value at 𝑡= 𝑎. Hence 𝛾(𝑡) −𝑤= (𝛾(𝑎) −𝑤)𝑒ℎ(𝑡) = (𝛾(𝑎) −𝑤)𝑒Re ℎ(𝑡)𝑒𝑖Im ℎ(𝑡) = |𝛾(𝑎) −𝑤|𝑒Re ℎ(𝑡)𝑒𝑖(𝛼+Im ℎ(𝑡) for 𝛼such that 𝑒𝑖𝛼= 𝛾(𝑎)−𝑤 | |𝛾(𝑎)−𝑤| |. Hence, we can set 𝜃(𝑡) = 𝛼+ Im ℎ(𝑡).
For the second part, note that 𝐼(𝛾; 𝑤) = 𝜃(𝑏) −𝜃(𝑎) 2𝜋 = Im(ℎ(𝑏) −ℎ(𝑎)) 2𝜋 = Im ℎ(𝑏) 2𝜋 Since 𝛾(𝑡) −𝑤= (𝛾(𝑎) −𝑤)𝑒ℎ(𝑡) and 𝛾(𝑏) = 𝛾(𝑎), we have 𝑒ℎ(𝑏) = 1, so Re ℎ(𝑏) = 0 and Im ℎ(𝑏) = −𝑖ℎ(𝑏). Thus, 𝐼(𝛾; 𝑤) = 1 2𝜋𝑖ℎ(𝑏) = 1 2𝜋𝑖∫ 𝑏 𝑎 𝛾′(𝑠) 𝛾(𝑠) −𝑤d𝑠= 1 2𝜋𝑖∫ 𝛾 d𝑧 𝑧−𝑤 Remark. It is also true that 𝜃exists and is continuous if 𝛾is merely continuous, but the formula for the winding number is not useful, so we omit this proof.
Proposition. If 𝛾∶[𝑎, 𝑏] →ℂis a closed curve, then the function 𝑤↦𝐼(𝛾; 𝑤) is continuous on ℂ∖Im 𝛾. Since 𝐼(𝛾; 𝑤) is integer-valued, 𝐼(𝛾; 𝑤) is locally constant. So 𝐼(𝛾; 𝑤) is constant for each connected component of the open set ℂ∖Im 𝛾.
Proof. Exercise.
Proposition. If 𝛾∶[𝑎, 𝑏] →𝐷(𝑧0, 𝑅) is a closed curve, then 𝐼(𝛾; 𝑤) = 0 for all 𝑤∈ℂ∖ 𝐷(𝑧0, 𝑅).
If 𝛾∶[𝑎, 𝑏] →ℂis a closed curve, then there exists a unique unbounded connected compon-ent Ω of ℂ∖𝛾([𝑎, 𝑏]), and 𝐼(𝛾; 𝑤) = 0 for all 𝑤∈Ω.
Proof. For the first part, if 𝑤∈ℂ∖𝐷(𝑧0, 𝑅), then the function 𝑓(𝑧) = 1 𝑧−𝑤is holomorphic in 𝐷(𝑧0, 𝑅). Hence 𝐼(𝛾; 𝑤) = 0 by the convex version of Cauchy’s theorem.
For the second part, since 𝛾([𝑎, 𝑏]) is compact (by continuity of 𝛾), there exists 𝑅> 0 such that 𝛾([𝑎, 𝑏]) ⊂𝐷(0, 𝑅). Since ℂ∖𝐷(0, 𝑅) is a connected subset of ℂ∖𝛾([𝑎, 𝑏]), there exists a connected component Ω of ℂ∖𝛾([𝑎, 𝑏]) such that ℂ∖𝐷(0, 𝑅) ⊆Ω. This component is unbounded. Any other component is disjoint from ℂ∖𝐷(0, 𝑅), so is contained within 𝐷(0, 𝑅) and is hence bounded. So the unbounded component is unique. Since 𝐼(𝛾; 𝑤) is locally constant and zero on ℂ∖𝐷(0, 𝑅), it is zero on Ω.
516 3. More integration 3.2. Continuity of derivative function Lemma. Let 𝑓∶𝑈→ℂbe holomorphic, and define 𝑔∶𝑈× 𝑈→ℂby 𝑔(𝑧, 𝑤) = { 𝑓(𝑧)−𝑓(𝑤) 𝑧−𝑤 if 𝑧≠𝑤 𝑓′(𝑤) if 𝑧= 𝑤 Then 𝑔is continuous. Moreover, if 𝛾is a closed curve in 𝑈, then the function ℎ(𝑤) = ∫ 𝛾𝑔(𝑧, 𝑤) d𝑧is holomorphic on 𝑈.
Proof. It is clear that 𝑔is continuous at (𝑧, 𝑤) if 𝑧≠𝑤. To check continuity at a point (𝑎, 𝑎) ∈𝑈× 𝑈, let 𝜀> 0 and choose 𝛿> 0 such that 𝐷(𝑎, 𝛿) ⊆𝑈and |𝑓′(𝑧) −𝑓′(𝑎)| < 𝜀for all 𝑧∈𝐷(𝑎, 𝛿). This is always possible since 𝑓′ is continuous.
Let 𝑧, 𝑤∈𝐷(𝑎, 𝛿). If 𝑧= 𝑤, then |𝑔(𝑧, 𝑤) −𝑔(𝑎, 𝑎)| = |𝑓′(𝑧) −𝑓′(𝑎)| < 𝜀. If 𝑧≠𝑤, we have 𝑡𝑧+ (1 −𝑡)𝑤∈𝐷(𝑎, 𝛿) for 𝑡∈[0, 1]. Hence, 𝑓(𝑧) −𝑓(𝑤) = ∫ 1 0 d d𝑡𝑓(𝑡𝑧+ (1 −𝑡)𝑤) d𝑡 = ∫ 1 0 𝑓′(𝑡𝑧+ (1 −𝑡)𝑤)(𝑧−𝑤) d𝑡 = (𝑧−𝑤) ∫ 1 0 𝑓′(𝑡𝑧+ (1 −𝑡)𝑤) d𝑡 Thus, |𝑔(𝑧, 𝑤) −𝑔(𝑎, 𝑎)| = | | | 𝑓(𝑧) −𝑓(𝑤) 𝑧−𝑤 −𝑓′(𝑎)| | | = | | | | ∫ 1 0 [𝑓′(𝑡𝑧+ (1 −𝑡)𝑤) −𝑓′(𝑎)] d𝑡 | | | | ≤sup 𝑡∈[0,1] |𝑓′(𝑡𝑧+ (1 −𝑡)𝑤) −𝑓′(𝑎)| < 𝜀 Hence |(𝑧, 𝑤) −(𝑎, 𝑎)| < 𝛿implies |𝑔(𝑧, 𝑤) −𝑔(𝑎, 𝑎)| < 𝜀, so 𝑔is continuous at (𝑎, 𝑎).
To show ℎis holomorphic, we must first check that ℎis continuous. Let 𝑤0 ∈𝑊, and suppose 𝑤𝑛→𝑤0. Let 𝛿> 0 such that 𝐷(𝑤0, 𝛿) ⊂𝑈. The function 𝑔is continuous on 𝑈× 𝑈, so it is uniformly continuous on the compact subset Im 𝛾× 𝐷(𝑤0, 𝛿) ⊂𝑈× 𝑈. Thus, if we let 𝑔𝑛(𝑧) = 𝑔(𝑧, 𝑤𝑛) and 𝑔0(𝑧) = 𝑔(𝑧, 𝑤0) for 𝑧∈Im 𝛾, then 𝑔𝑛→𝑔0 uniformly on Im 𝛾.
Hence ∫ 𝛾𝑔𝑛(𝑧) d𝑧→∫ 𝛾𝑔0(𝑧) d𝑧. In other words, ℎ(𝑤𝑛) →ℎ(𝑤0). Thus, ℎis continuous.
Now, we can use the convex Cauchy’s theorem and Morera’s theorem to show ℎis holo-morphic on 𝑈. For 𝑤0 ∈𝑈, we can choose a disc 𝐷(𝑤0, 𝛿) ⊂𝑈. Suppose that 𝛾is parametrised over [𝑎, 𝑏], and let 𝛽∶[𝑐, 𝑑] →𝐷(𝑤0, 𝛿) be any closed curve. Then ℎ(𝑤) = 517 IX. Complex Analysis ∫ 𝛾𝑔(𝑧, 𝑤) d𝑧= ∫ 𝑏 𝑎𝑔(𝛾(𝑡), 𝑤)𝛾′(𝑡) d𝑡, hence ∫ 𝛽 ℎ(𝑤) d𝑤= ∫ 𝑑 𝑐 (∫ 𝑏 𝑎 𝑔(𝛾(𝑡), 𝛽(𝑠))𝛾′(𝑡)𝛽′(𝑠) d𝑡) d𝑠 = ∫ 𝑏 𝑎 (∫ 𝑑 𝑐 𝑔(𝛾(𝑡), 𝛽(𝑠))𝛾′(𝑡)𝛽′(𝑠) d𝑠) d𝑡 = ∫ 𝛾 (∫ 𝛽 𝑔(𝑧, 𝑤) d𝑤) d𝑧 by Fubini’s theorem, which will be proven below. By a previous theorem, for all 𝑧∈𝑈, the function 𝑤↦𝑔(𝑧, 𝑤) is holomorphic in 𝐷(𝑤0, 𝛿) (and hence in 𝑈), since it is continuous in 𝑈and holomorphic except at a single point 𝑧. Hence, by the convex version of Cauchy’s theorem, ∫ 𝛽𝑔(𝑧, 𝑤) d𝑤= 0. Hence, ∫ 𝛽ℎ(𝑤) d𝑤= 0. By Morera’s theorem, ℎis holomorphic in 𝐷(𝑤0, 𝛿) and hence on 𝑈.
Lemma (Fubini’s theorem). If 𝜑∶[𝑎, 𝑏]×[𝑐, 𝑑] →ℝis continuous, then 𝑓 1 ∶𝑠↦∫ 𝑑 𝑐𝜑(𝑠, 𝑡) d𝑡 is continuous on [𝑎, 𝑏], the function 𝑓 2 ∶𝑡↦∫ 𝑏 𝑎𝜑(𝑠, 𝑡) d𝑡is continuous on [𝑐, 𝑑], and ∫ 𝑏 𝑎 (∫ 𝑑 𝑐 𝜑(𝑠, 𝑡) d𝑡) d𝑠= ∫ 𝑑 𝑐 (∫ 𝑏 𝑎 𝜑(𝑠, 𝑡) d𝑠) d𝑡 Proof. Since 𝜑is continuous on the compact set [𝑎, 𝑏] × [𝑐, 𝑑], it is uniformly continuous.
Hence, given 𝜀> 0, there exists 𝛿> 0 such that |𝑠1 −𝑠2| < 𝛿⟹|𝜑(𝑠1, 𝑡) −𝜑(𝑠2, 𝑡)| < 𝜀 for all 𝑡∈[𝑐, 𝑑], so |𝑓 1(𝑠1) −𝑓 1(𝑠2)| < (𝑑−𝑐)𝜀, so 𝑓 1 is continuous. Similarly, 𝑓 2 is con-tinuous. Note that since 𝜑is uniformly continuous, it is the uniform limit of a sequence of step functions of the form 𝑔(𝑥, 𝑦) = ∑ 𝑁 𝑗=1 𝛼𝑗𝜒𝑅𝑗(𝑥, 𝑦) where 𝛼𝑗are constants, and 𝑅𝑗 are sub-rectangles of the form 𝑅𝑗= [𝑎𝑗, 𝑏𝑗) × [𝑐𝑗, 𝑑𝑗) such that ⋃𝑅𝑗is a finite partition of [𝑎, 𝑏) × [𝑐, 𝑑), and 𝜒𝑅𝑗is the characteristic function of 𝑅𝑗For such step functions, we can easily check the interchangability of the integrals.
3.3. Cauchy’s theorem and Cauchy’s integral formula Definition. Let 𝑈⊆ℂbe open. A closed curve 𝛾∶[𝑎, 𝑏] →𝑈is said to be homologous to zero in 𝑈if 𝐼(𝛾; 𝑤) = 0 for all 𝑤∈ℂ∖𝑈.
Theorem. Let 𝑈be a non-empty open subset of ℂ, and 𝛾be a closed curve in 𝑈homologous to zero in 𝑈. Then, 𝐼(𝛾; 𝑤)𝑓(𝑤) = 1 2𝜋𝑖∫ 𝛾 𝑓(𝑧) d𝑧 𝑧−𝑤 for every holomorphic function 𝑓∶𝑈→ℂand every 𝑤∈𝑈∖Im 𝛾. Further, ∫ 𝛾 𝑓(𝑧) d𝑧= 0 518 3. More integration for every holomorphic 𝑓∶𝑈→ℂ.
Remark. Cauchy’s theorem states that if ∫ 𝛾𝑓(𝑧) d𝑧= 0 for a specific family of holomorphic functions on 𝑈, namely for 𝑓 𝑤(𝑧) = 1 𝑧−𝑤where 𝑤∈ℂ∖𝑈, then ∫ 𝛾𝑓(𝑧) d𝑧= 0 for any holomorphic function 𝑓∶𝑈→ℂ.
The first and second parts as statements are equivalent. Indeed, if we assume the Cauchy integral formula holds, simply apply the formula with 𝐹(𝑧) = (𝑧−𝑤)𝑓(𝑧). Since 𝐹(𝑤) = 0, we have ∫ 𝛾𝑓(𝑧) d𝑧= 0. If we assume Cauchy’s theorem, for any 𝑤∈𝑈, the function 𝑔(𝑧) = { 𝑓(𝑧)−𝑓(𝑤) 𝑧−𝑤 if 𝑧≠𝑤 𝑓′(𝑤) if 𝑧= 𝑤 is holomorphic in 𝑈as seen above. Hence ∫ 𝛾𝑔(𝑧) d𝑧= 0, so 1 2𝜋𝑖∫ 𝛾 𝑓(𝑧)d𝑧 𝑧−𝑤= 𝐼(𝛾; 𝑤)𝑓(𝑤) for all 𝑤∉Im 𝛾.
Note that the statement that 𝛾is homologous to zero is equivalent to Cauchy’s theorem being valid for all 𝑓. For example, given 𝑤∈ℂ∖𝑈, we can apply Cauchy’s theorem to 𝑓(𝑧) = 1 𝑧−𝑤 to get 𝐼(𝛾; 𝑤) = 0. The converse is proven in the theorem following this proof. This is also equivalent to Cauchy’s integral formula being valid for all 𝑓.
Proof. It suffices to prove part (i). Equivalently, for all 𝑤∈𝑈∖Im 𝛾, ∫ 𝛾 𝑓(𝑧) −𝑓(𝑤) 𝑧−𝑤 d𝑧= 0 ⟺∫ 𝛾 𝑔(𝑧, 𝑤) d𝑧= 0 where 𝑔(𝑧, 𝑤) = { 𝑓(𝑧)−𝑓(𝑤) 𝑧−𝑤 if 𝑧≠𝑤 𝑓′(𝑤) if 𝑧= 𝑤 Now, define ℎ∶𝑈→ℂ; ℎ(𝑤) = ∫ 𝛾 𝑔(𝑧, 𝑤) d𝑧 By the above lemma, this is holomorphic on 𝑈. We will show that ℎ= 0. We will extend ℎ to a holomorphic function 𝐻∶ℂ→ℂand prove that 𝐻(𝑤) →0 as 𝑤→∞, then we can apply Liouville’s theorem.
To extend ℎinto an entire function 𝐻, by definition of 𝛾being homologous to zero in 𝑈, we have ℂ∖𝑈⊆𝑉≡{𝑤∈ℂ∖Im 𝛾∶𝐼(𝛾; 𝑤) = 0}. So ℂ= 𝑈∪𝑉, and 𝑉is open since 𝐼(𝛾; ⋅) is locally constant. For 𝑤∈𝑈∩𝑉, we have ℎ(𝑤) = ∫ 𝛾 𝑓(𝑧) −𝑓(𝑤) 𝑧−𝑤 d𝑧= ∫ 𝛾 𝑓(𝑧) d𝑧 𝑧−𝑤 519 IX. Complex Analysis since ∫ 𝛾 d𝑧 𝑧−𝑤= 2𝜋𝑖⋅𝐼(𝛾; 𝑤) = 0 as 𝑤∈𝑉. Hence, on 𝑈∩𝑉, the function ℎagrees with ℎ1 ∶𝑉→ℂ; ℎ1(𝑤) = ∫ 𝛾 𝑓(𝑧) d𝑧 𝑧−𝑤 We know that ℎ1 is holomorphic on 𝑉. Hence, the function 𝐻∶ℂ→ℂdefined by 𝐻(𝑤) = {ℎ(𝑤) 𝑤∈𝑈 ℎ1(𝑤) 𝑤∈𝑉 is well-defined and holomorphic.
Now, we will show 𝐻(𝑤) →0 as |𝑤| →∞. Let 𝑅> 0 such that Im 𝛾⊂𝐷(0, 𝑅), which is possible since Im 𝛾is compact. Hence, ℂ∖𝐷(0, 𝑅) ⊆𝑉. If |𝑤| > 𝑅, |𝐻(𝑤)| = |ℎ1(𝑤)| = | | | | ∫ 𝛾 𝑓(𝑧) d𝑧 𝑧−𝑤 | | | | ≤ 1 |𝑤| −𝑅( sup 𝑧∈Im 𝛾 |𝑓(𝑧)|)length(𝛾) Hence, 𝐻(𝑤) →0 as |𝑤| →∞, as claimed. Hence 𝐻is bounded, since 𝐻is continuous, and |𝐻(𝑤)| ≤1 outside some closed disc 𝐷(0, 𝑅1). By Liouville’s theorem, 𝐻is constant, and by the claim, 𝐻= 0. In particular, ℎ= 0.
Corollary. Let 𝑈⊂ℂbe open and 𝛾1, … , 𝛾𝑛be closed curves in 𝑈such that ∑ 𝑛 𝑗=1 𝐼(𝛾𝑗; 𝑤) = 0 for all 𝑤∈ℂ∖𝑈. Then, for any holomorphic 𝑓∶𝑈→ℂ, we have 𝑓(𝑤) 𝑛 ∑ 𝑗=1 𝐼(𝛾𝑗; 𝑤) = 𝑛 ∑ 𝑗=1 1 2𝜋𝑖∫ 𝛾𝑗 𝑓(𝑧) d𝑧 𝑧−𝑤 for all 𝑤∈𝑈∖⋃ 𝑛 𝑗=1 Im 𝛾𝑗, and 𝑛 ∑ 𝑗=1 ∫ 𝛾𝑗 𝑓(𝑧) d𝑧= 0 Proof. For the first part, define 𝑔(𝑧, 𝑤) as before, but let 𝑉= {𝑤∈ℂ∖ 𝑛 ⋃ 𝑗=1 Im 𝛾𝑗∶ 𝑛 ∑ 𝑗=1 𝐼(𝛾𝑗; 𝑤) = 0} In the definitions of ℎand ℎ1, use the sum of the integrals over 𝛾𝑗. Then we can proceed as above. The second part follows from the first as before.
Corollary. Let 𝑈⊂ℂbe open and let 𝛽1, 𝛽2 be closed curves in 𝑈such that 𝐼(𝛽1; 𝑤) = 𝐼(𝛽2; 𝑤) for all 𝑤∈ℂ∖𝑈. Then ∫ 𝛽1 𝑓(𝑧) d𝑧= ∫ 𝛽2 𝑓(𝑧) d𝑧 for all holomorphic functions 𝑓∶𝑈→ℂ.
520 3. More integration Proof. We can apply the second part of the previous corollary with 𝛾1 = 𝛽1 and 𝛾2 = −𝛽2, noting that 𝐼(−𝛽2; 𝑤) = −𝐼(𝛽2; 𝑤) for any 𝑤∉Im 𝛽2.
3.4. Homotopy The set of closed curves in 𝑈such that Cauchy’s theorem is valid is the set of holomorphic functions homologous to zero. We will now construct a more restrictive condition, the con-dition of being null-homotopic.
Definition. Let 𝑈⊆ℂbe a domain, and let 𝛾0, 𝛾1 ∶[𝑎, 𝑏] →𝑈be closed curves. We say that 𝛾0 is homotopic to 𝛾1 in 𝑈if there exists a continuous map 𝐻∶[0, 1] × [𝑎, 𝑏] →𝑈such that for all 𝑠∈[0, 1], 𝑡∈[𝑎, 𝑏], 𝐻(0, 𝑡) = 𝛾0(𝑡); 𝐻(1, 𝑡) = 𝛾1(𝑡); 𝐻(𝑠, 𝑎) = 𝐻(𝑠, 𝑏) Such a map is called a homotopy between 𝛾0, 𝛾1.
For 0 ≤𝑠≤1, if we let 𝛾𝑠∶[𝑎, 𝑏] →𝑈be defined by 𝛾𝑠(𝑡) = 𝐻(𝑠, 𝑡) for 𝑡∈[𝑎, 𝑏], then the above conditions imply that {𝛾𝑠∶𝑠∈[0, 1]} is a family of continuous closed curves in 𝑈 which deform 𝛾0 to 𝛾1 continuously without leaving 𝑈.
Definition. A closed curve is null-homotopic in a certain domain if it is homotopic to a constant curve in the domain, such as 𝛾(𝑡) = 𝑧for 𝑧fixed.
Theorem. If 𝛾0, 𝛾1 ∶[𝑎, 𝑏] →𝑈are homotopic closed curves in 𝑈, then 𝐼(𝛾0; 𝑤) = 𝐼(𝛾1; 𝑤) for all 𝑤∈ℂ∖𝑈. In particular, if a closed curve 𝛾is null-homotopic in 𝑈, it is homologous to zero in 𝑈.
Proof. Let 𝐻∶[0, 1] × [𝑎, 𝑏] →𝑈be a homotopy between 𝛾0 and 𝛾1. Since 𝐻is continuous and [0, 1]×[𝑎, 𝑏] is compact, the image 𝐾= 𝐻([0, 1]×[𝑎, 𝑏]) is a compact subset of the open set 𝑈. Therefore, there exists 𝜀> 0 such that for all 𝑤∈ℂ∖𝑈, |𝑤−𝐻(𝑠, 𝑡)| > 2𝜀for all (𝑠, 𝑡) ∈[0, 1] × [𝑎, 𝑏]. Since 𝐻is uniformly continuous on [0, 1] × [𝑎, 𝑏], there exists 𝑛∈ℕ such that ∀(𝑠, 𝑡), (𝑠′, 𝑡′) ∈[0, 1] × [𝑎, 𝑏], |𝑠−𝑠′| + |𝑡−𝑡′| ≤1 𝑛⟹|𝐻(𝑠, 𝑡) −𝐻(𝑠′, 𝑡′)| < 𝜀 For 𝑘= 0, 1, 2, … , 𝑛, we let Γ 𝑘(𝑡) = 𝐻(𝑘/𝑛, 𝑡) for 𝑎≤𝑡≤𝑏. Then the Γ 𝑘are closed continuous curves with Γ 0 = 𝛾0 and Γ 𝑛= 𝛾1. Hence, for all 𝑡∈[𝑎, 𝑏], |Γ 𝑘−1(𝑡) −Γ 𝑘(𝑡)| ⏟⎵⎵⎵⏟⎵⎵⎵⏟ <𝜀 < |𝑤−Γ 𝑘−1(𝑡)| ⏟ ⎵ ⎵ ⎵ ⏟ ⎵ ⎵ ⎵ ⏟ >2𝜀 On the example sheets we have shown that for piecewise 𝐶1 closed curves 𝛾, ̃ 𝛾, if we have |𝛾(𝑡) −̃ 𝛾(𝑡)| < |𝑤−𝛾(𝑡)| for all 𝑡, then 𝐼(𝛾; 𝑤) = 𝐼( ̃ 𝛾; 𝑤). Hence, if Γ 𝑘are piecewise 𝐶1, we can see that 𝐼(Γ 𝑘−1; 𝑤) = 𝐼(Γ 𝑘; 𝑤) for all 𝑘, and hence 𝐼(𝛾0; 𝑤) = 𝐼(𝛾1; 𝑤) as required.
521 IX. Complex Analysis We have only assumed that 𝐻is continuous, so Γ 𝑘need not be piecewise 𝐶1. We can fix this problem by approximating each Γ 𝑘by a polygonal curve. We can take ̃ Γ 𝑘(𝑡) = (1 − 𝑛(𝑡−𝑎𝑗−1) 𝑏−𝑎 )𝐻(𝑘 𝑛, 𝑎𝑗−1) + 𝑛(𝑡−𝑎𝑗−1) 𝑏−𝑎 𝐻(𝑘 𝑛, 𝑎𝑗) for 𝑎𝑗−1 ≤𝑡≤𝑎𝑗, where 𝑎𝑗= 𝑎+ (𝑏−𝑎)𝑗 𝑛 If we choose 𝑛so that |𝑠−𝑠′| + |𝑡−𝑡′| ≤min {1, 𝑏−𝑎} 𝑛 ⟹|𝐻(𝑠, 𝑡) −𝐻(𝑠′, 𝑡′)| < 𝜀 the curves ̃ Γ 𝑘satisfy | | ̃ Γ 𝑘−1(𝑡) −̃ Γ 𝑘(𝑡)| | < | |𝑤−̃ Γ 𝑘−1(𝑡)| | for all 𝑡∈[𝑎, 𝑏]. This is because for 𝑡∈[𝑎𝑗−1, 𝑎𝑗], | | ̃ Γ 𝑘−1(𝑡) −̃ Γ 𝑘(𝑡)| | ≤(1 − 𝑛(𝑡−𝑎𝑗−1) 𝑏−𝑎 )| | |𝐻(𝑘−1 𝑛 , 𝑎𝑗−1) −𝐻(𝑘 𝑛, 𝑎𝑗−1)| | | + 𝑛(𝑡−𝑎𝑗−1) 𝑏−𝑎 | | |𝐻(𝑘−1 𝑛 , 𝑎𝑗) −𝐻(𝑘 𝑛, 𝑎𝑗)| | | < 𝜀 and | |𝑤−̃ Γ 𝑘−1(𝑡)| | ≥|𝑤−Γ 𝑘−1(𝑡)| −| |Γ 𝑘−1(𝑡) −̃ Γ 𝑘−1(𝑡)| | > 2𝜀−𝜀= 𝜀 We also have, for all 𝑡∈[𝑎, 𝑏], | | ̃ Γ 0(𝑡) −𝛾0(𝑡)| |; | | ̃ Γ 𝑛−𝛾1(𝑡)| | < |𝑤−𝛾1(𝑡)| Hence the result follows from the same example sheet question.
Remark. If 𝛾is homologous to zero in 𝑈, it is not necessarily the case that 𝛾is null-homotopic.
For insance, let 𝑈= ℂ∖{𝑤1, 𝑤2} for 𝑤1 ≠𝑤2, and let 𝑈1 = 𝑈∪{𝑤1} = ℂ∖{𝑤2} and 𝑈2 = 𝑈∪{𝑤2} = ℂ∖{𝑤1}. Then, consider a curve 𝛾which is not null-homotopic in 𝑈, but null-homotopic in each of the larger domains 𝑈1, 𝑈2. Then 𝛾is homologous to zero in 𝑈1 and 𝑈2. Hence 𝐼(𝛾; 𝑤1) = 𝐼(𝛾; 𝑤2) = 0, so 𝛾is homologous to zero in 𝑈.
Corollary. If 𝛾0, 𝛾1 ∶[𝑎, 𝑏] →𝑈are homotopic closed curves in 𝑈, then ∫ 𝛾0 𝑓(𝑧) d𝑧= ∫ 𝛾1 𝑓(𝑧) d𝑧 for all holomorphic 𝑓∶𝑈→ℂ.
This is immediate from previous results. However, we can make a direct proof that does not require the most general form of Cauchy’s theorem.
522 3. More integration Proof. With ̃ Γ 𝑘as above, consider the closed curve comprised of (i) the curve ̃ Γ 𝑘−1 on [𝑎𝑗−1, 𝑎𝑗]; (ii) the line segment [ ̃ Γ 𝑘−1(𝑎𝑗), ̃ Γ 𝑘(𝑎𝑗)]; (iii) the curve −̃ Γ 𝑘on [𝑎𝑗, 𝑎𝑗−1]; (iv) the line segment [ ̃ Γ 𝑘(𝑎𝑗−1), ̃ Γ 𝑘−1(𝑎𝑗−1)].
This curve is contained in the disc 𝐷( ̃ Γ 𝑘−1(𝑎𝑗−1), 𝜀) ⊆𝑈. We can apply the convex version of Cauchy’s theorem and sum over 𝑗to find ∫ ˜ Γ𝑘−1 𝑓(𝑧) d𝑧= ∫ ˜ Γ𝑘 𝑓(𝑧) d𝑧 Similarly we can find ∫ ˜ Γ0 𝑓(𝑧) d𝑧= ∫ 𝛾0 𝑓(𝑧) d𝑧; ∫ ˜ Γ𝑛 𝑓(𝑧) d𝑧= ∫ 𝛾1 𝑓(𝑧) d𝑧 3.5. Simply connected domains Definition. A domain 𝑈is simply connected if every closed curve in 𝑢is null-homotopic in 𝑈.
Star domains 𝑈are simply connected. Indeed, there exists a centre 𝑎∈𝑈such that [𝑎, 𝑧] ⊂ 𝑈for all 𝑧∈𝑈. If 𝛾∶[𝑎, 𝑏] →𝑈is a closed curve, let 𝐻(𝑧, 𝑡) = (1 −𝑠)𝑎+ 𝑠𝛾(𝑡) ∈𝑈for (𝑠, 𝑡) ∈[0, 1] × [𝑎, 𝑏]. Then 𝐻(𝑠, 𝑡) ∈𝑈, and 𝐻is a homotopy between 𝛾and the constant curve 𝛾0(𝑡) = 𝑎.
Theorem (Cauchy’s theorem for simply connected domains). If 𝑈is simply connected, then ∫ 𝛾 𝑓(𝑧) d𝑧= 0 for all holomorphic 𝑓∶𝑈→ℂ, and every closed curve 𝛾in 𝑈.
This is an immediate application of the above. The converse is also true, but is harder to prove.
Hence, 𝑈is simply connected if and only if ∫ 𝛾𝑓(𝑧) d𝑧= 0 for all holomorphic 𝑓and all closed 𝛾in 𝑈. In particular, 𝑈is simply connected if and only if every closed curve in 𝑈is homologous to zero in 𝑈. Contrast this to the previous remark that if a curve is homologous to zero it is not necessarily null-homotopic.
523 IX. Complex Analysis 4. Singularities 4.1. Motivation Let 𝑈be open, and 𝛾be a closed curve in 𝑈homologous to zero in 𝑈. Then, if 𝑓∶𝑈→ℂis holomorphic, we have Cauchy’s integral formula ∫ 𝛾 𝑓(𝑧) d𝑧 𝑧−𝑎 ⏟ ⎵ ⏟ ⎵ ⏟ 𝑔(𝑧)d𝑧 = 2𝜋𝑖⋅𝐼(𝛾; 𝑎)𝑓(𝑎) for all 𝑎∈𝑈∖Im 𝛾. This allows us to compute ∫ 𝛾𝑔(𝑧) d𝑧for a holomorphic function 𝑔∶𝑈∖ {𝑎} →ℂwhere 𝛾does not pass through the point 𝑎, provided that 𝑔satisfies a particular condition: (𝑧−𝑎)𝑔(𝑧) is the restriction to 𝑈∖{𝑎} of a holomorphic function 𝑓∶𝑈→ℂ.
We wish to drop this restriction and observe the consequences; that is, we wish to compute ∫ 𝛾𝑔(𝑧) d𝑧for arbitrary holomorphic functions 𝑔∶𝑈∖{𝑎} →ℂfor 𝑎∈𝑈and 𝑎∉Im 𝛾. For example, consider 𝑔(𝑧) = 𝑒𝑧−1 for 𝑈= ℂand 𝑎= 0, 𝛾= 𝜕𝐷(0, 1). Note that 𝑧𝑔(𝑧) = 𝑧𝑒𝑧−1 is not continuous at 𝑧= 0, so it is certainly not holomorphic. This leads us to the study of singularities, and to eventually prove the residue theorem.
4.2. Removable singularities Definition. Let 𝑈⊆ℂbe open. If 𝑎∈𝑈and 𝑓∶𝑈∖{𝑎} →ℂis holomorphic, we say that 𝑓has an isolated singularity at 𝑎.
Definition. An isolated singularity 𝑎of 𝑓is a removable singularity if 𝑓can be defined at 𝑎 such that the extended function is holomorphic on 𝑈.
Proposition. Let 𝑈be open, 𝑎∈𝑈, and 𝑓∶𝑈∖{𝑎} →ℂbe holomorphic. Then, the following are equivalent.
(i) 𝑓has a removable singularity at 𝑎; (ii) lim𝑧→𝑎𝑓(𝑧) exists in ℂ; (iii) there exists 𝐷(𝑎, 𝜀) ⊆𝑈such that |𝑓(𝑧)| is bounded in 𝐷(𝑎, 𝜀) ∖{𝑎}; (iv) lim𝑧→𝑎(𝑧−𝑎)𝑓(𝑧) = 0.
Proof. We can see that (i) implies (ii). If 𝑎is a removable singularity of 𝑓, then by definition there is a holomorphic function 𝑔∶𝑈→ℂsuch that 𝑓(𝑧) = 𝑔(𝑧) for all 𝑧∈𝑈∖{𝑎}. Then lim𝑧→𝑎𝑓(𝑧) = lim𝑧→𝑎𝑔(𝑧) = 𝑔(𝑎) ∈ℂ. Similarly, (ii) implies (iii) and (iii) implies (iv) are clear.
It suffices to check (iv) implies (i). Consider the function ℎ(𝑧) = {(𝑧−𝑎)2𝑓(𝑧) if 𝑧≠𝑎 0 if 𝑧= 𝑎 524 4. Singularities We have lim 𝑧→𝑎 ℎ(𝑧) −ℎ(𝑎) 𝑧−𝑎 = lim 𝑧→𝑎(𝑧−𝑎)𝑓(𝑧) = 0 Hence ℎis differentiable at 𝑎with ℎ′(𝑎) = 0. Since ℎis differentiable in 𝑈∖{𝑎}, we must have that ℎis holomorphic in 𝑈. Since ℎ(𝑎) = ℎ′(𝑎) = 0, we can find 𝑟> 0 and a holomorphic 𝑔∶𝐷(𝑎, 𝑟) →ℂsuch that ℎ(𝑧) = (𝑧−𝑎)2𝑔(𝑧) for 𝑧∈𝐷(𝑎, 𝑟). Comparing this to the definition of ℎ, we have that 𝑓(𝑧) = 𝑔(𝑧) for 𝐷(𝑎, 𝑟) ∖{𝑎}. By defining 𝑓(𝑎) = 𝑔(𝑎), we have that 𝑓is differentiable at 𝑎with 𝑓′(𝑎) = 𝑔′(𝑎). So 𝑎is a removable singularity of 𝑓.
Example. Consider 𝑓(𝑧) = 𝑒𝑧−1 𝑧. Certainly 𝑓is holomorphic on ℂ∖{0}, and lim𝑧→0 𝑧𝑓(𝑧) = 0. So 𝑧= 0 is a removable singularity. We can also see directly by the Taylor series of 𝑒𝑧at 𝑧= 0 that 𝑓(𝑧) = ∑ ∞ 𝑘=1 𝑧𝑘−1 𝑘! for 𝑧≠0, and the series on the right hand side defines an entire function.
Remark. If 𝑢∶𝐷(0, 1) ∖{0} →ℝis a 𝐶2 harmonic function, when can we say that 𝑧= 0 is a removable singularity, i.e. that 𝑢extends to 𝑧= 0 as a harmonic function? We can relate this to the study of holomorphic functions. However, unlike with previous cases, the analogy is more subtle in this case. We cannot necessarily construct a harmonic conjugate 𝑣such that 𝑢+ 𝑖𝑣is holomorphic in 𝐷(0, 1) ∖{0}, because 𝑈is not simply connected.
There is a similar result, however. If lim𝑧→0 𝑢(𝑧) exists, then the extended function is in fact 𝐶2 and harmonic. More generally, if 𝑢is bounded near 𝑧= 0, there exists a harmonic extension. We can also consider the case lim𝑧→0 |𝑧||𝑢(𝑧)| = 0; this is explored on the example sheets.
4.3. Poles Note, if a holomorphic function 𝑓has a non-removable singularity, 𝑓is not bounded in 𝐷(𝑎, 𝑟) ∖{𝑎} for any 𝑟> 0.
Definition. If 𝑎∈𝑈is an isolated singularity of 𝑓, then 𝑎is a pole of 𝑓if lim 𝑧→𝑎|𝑓(𝑧)| = ∞ Example. 𝑓(𝑧) = (𝑧−𝑎)−𝑘for 𝑘∈ℕhas a pole at 𝑎.
Definition. If 𝑎∈𝑈is an isolated singularity of 𝑓that is not removable or a pole, it is an essential singularity.
Remark. An equivalent characterisation for 𝑎to be an essential singularity is that the limit lim𝑧→𝑎|𝑓(𝑧)| does not exist. This follows from the previous proposition and the definition of a pole.
Example. 𝑓(𝑧) = 𝑒 1 𝑧has |𝑓(𝑖𝑦)| = 1 for all 𝑦∈ℝ∖{0} and lim𝑥→0+ 𝑓(𝑥) = ∞. So 𝑧= 0 is an essential singularity of 𝑓.
525 IX. Complex Analysis Proposition. Let 𝑓∶𝑈∖{𝑎} →ℂbe holomorphic. The following are equivalent.
(i) 𝑓has a pole at 𝑎; (ii) there exists 𝜀> 0 and a holomorphic function ℎ∶𝐷(𝑎, 𝜀) →ℂwith ℎ(𝑎) = 0 and ℎ(𝑧) ≠0 for all 𝑧≠𝑎such that 𝑓(𝑧) = 1 ℎ(𝑧) for 𝑧∈𝐷(𝑎, 𝜀) ∖{𝑎}; (iii) there exists a unique integer 𝑘≥1 and a unique holomorphic function 𝑔∶𝑈→ℂ with 𝑔(𝑎) ≠0 such that 𝑓(𝑧) = (𝑧−𝑎)−𝑘𝑔(𝑧) for 𝑧∈𝑈∖{𝑎}.
Remark. Since (i) implies (iii), there exists no holomorphic function on a punctured disc 𝑓∶𝐷(𝑎, 𝑅) ∖{𝑎} →ℂsuch that |𝑓(𝑧)| →∞as 𝑧→𝑎at the rate of a negative non-integer power of |𝑧−𝑎|, i.e. with 𝑐|𝑧−𝑎|−𝑠≤|𝑓(𝑧)| ≤𝐶|𝑧−𝑎|−𝑠for some constants 𝑠∈(0, ∞) ∖ℕ, 𝑐> 0, 𝐶> 0, and all 𝑧∈𝐷(𝑎, 𝑅) ∖{𝑎}.
Proof. We show (i) implies (ii). Since lim𝑧→𝑎|𝑓(𝑧)| = ∞, there exists 𝜀> 0 such that |𝑓(𝑧)| ≥ 1 for all 0 < |𝑧−𝑎| < 𝜀. Hence 1 𝑓(𝑧) is holomorphic and bounded in 𝐷(𝑎, 𝜀) ∖{𝑎}. By the above proposition, 1 𝑓has a removable singularity at 𝑎, so there exists a holomorphic function ℎ∶𝐷(𝑎, 𝜀) →ℂsuch that 1 𝑓= ℎ, or equivalently, 𝑓= 1 ℎ, for 𝑧∈𝐷(𝑎, 𝜀) ∖{𝑎}. Since |𝑓(𝑧)| →∞as 𝑧→𝑎, we have that ℎ(𝑎) = 0.
Now we show (ii) implies (iii). Let 𝜀and ℎbe as in the definition of (ii). By Taylor series, there exists 𝑘≥1 and a holomorphic function ℎ1 ∶𝐷(𝑎, 𝜀) →ℂwith ℎ1(𝑧) ≠0 for all 𝑧∈𝐷(𝑎, 𝜀) such that ℎ(𝑧) = (𝑧−𝑎)𝑘ℎ1(𝑧). If 𝑔1 = 1 ℎ1 , then 𝑔1 is holomorphic in 𝐷(𝑎, 𝜀), 𝑔1 ≠0 in 𝐷(𝑎, 𝜀), and 𝑓(𝑧) = (𝑧−𝑎)−𝑘𝑔1(𝑧) in 𝐷(𝑎, 𝜀) ∖{𝑎}.
We can now define 𝑔∶𝑈→ℂby 𝑔(𝑧) = 𝑔1(𝑧) for 𝑧∈𝐷(𝑎, 𝜀), and 𝑔(𝑧) = (𝑧−𝑎)𝑘𝑓(𝑧) for 𝑧∈𝑈∖{𝑎}. Since 𝑓(𝑧) = (𝑧−𝑎)−𝑘𝑔1(𝑧), the definitions agree on 𝐷(𝑎, 𝜀) ∖{𝑎}, so 𝑔is well-defined and holomorphic in 𝑈, and 𝑔(𝑎) = 𝑔1(𝑎) ≠0. This proves the existence of an integer 𝑘≥1 and a holomorphic 𝑔∶𝑈→ℂwith 𝑔(𝑎) ≠0 such that 𝑓(𝑧) = (𝑧−𝑎)−𝑘𝑔(𝑧) for all 𝑧∈𝑈∖{𝑎}.
To prove uniqueness of 𝑘and 𝑔, suppose there exists ̃ 𝑘≥1 and a holomorphic ̃ 𝑔∶𝑈→ℂ with ̃ 𝑔(𝑎) ≠0 such that 𝑓(𝑧) = (𝑧−𝑎)−˜ 𝑘̃ 𝑔(𝑧) for all 𝑧∈𝑈∖{𝑎}. Then we have 𝑔(𝑧) = (𝑧−𝑎)𝑘−˜ 𝑘̃ 𝑔(𝑧) for 𝑧∈𝑈∖{𝑎}. Since 𝑔, ̃ 𝑔are holomorphic with 𝑔(𝑎) ≠0 and ̃ 𝑔(𝑎) ≠0, this can only be true if 𝑘= ̃ 𝑘, and hence 𝑔= ̃ 𝑔on 𝑈∖{𝑎}, and then at 𝑎by continuity.
It is clear that (iii) implies (i).
Definition. If 𝑓has a pole at 𝑧= 𝑎, then the unique positive integer 𝑘given by the above proposition is the order of the pole at 𝑎. If 𝑘= 1, we say that 𝑓has a simple pole at 𝑎.
Let 𝑈be open and 𝑆∖𝑈be a discrete subset of 𝑈, so all points of 𝑆are isolated. If 𝑓∶𝑈∖𝑆→ ℂis holomorphic and each 𝑎∈𝑆is either a removable singularity or a pole of 𝑓, then 𝑓is a meromorphic function on 𝑈. In particular, if 𝑆= ∅, 𝑓is holomorphic.
526 4. Singularities Remark. If 𝑓∶𝑈∖{𝑎} →ℂis holomorphic and the singularity 𝑧= 𝑎is a pole of 𝑓, we can regard 𝑓as a continuous mapping onto the Riemann sphere 𝑓∶𝑈→ℂ∪{∞}, by setting 𝑓(𝑎) = ∞. Here, 𝑓is holomorphic on 𝑈. Holomorphicity of the extended map near the pole 𝑎follows from the fact that in a punctured disc about 𝑎, 1 𝑓has the form (𝑧−𝑎)𝑘 𝑔(𝑧) for some holomorphic 𝑔with 𝑔(𝑧) ≠0 near 𝑎; and the fact that any function ℎdefined in a neighbourhood of ∞in the Riemann sphere is holomorphic, by definition, if the function ˜ ℎ(𝑧) = ℎ( 1 𝑧) if 𝑧≠0, ˜ ℎ(0) = ℎ(∞) is holomorphic near zero. Hence ℎ∘𝑓= ˜ ℎ∘( 1 𝑓) is holomorphic near 𝑎for all holomorphic ℎin a neighbourhood of ∞in the Riemann sphere.
Hence, any meromorphic function 𝑓∶𝑈∖𝑆→ℂcan be viewed as a holomorphic func-tion 𝑈→ℂ∪{∞}. Geometrically, therefore, poles are not ‘real’ singularities, and the only true isolated singularities are the essential singularities. This is explored further in Part II Riemann Surfaces.
4.4. Essential singularities Remark. Suppose 𝑧= 𝑎is an essential singularity of a holomorphic 𝑓∶𝑈∖{𝑎} →ℂ. Then there exists a sequence of points 𝑎𝑛∈𝑈∖{𝑎}, 𝑎𝑛→𝑎, such that 𝑓(𝑎𝑛) →∞. This is because 𝑎is not removable. There is also another sequence of points 𝑏𝑛∈𝑈∖{𝑎}, 𝑏𝑛→𝑎such that (𝑓(𝑏𝑛)) is bounded. This is because 𝑎is not a pole. We can generalise this further.
Theorem (Casorati–Weierstrass theorem). If 𝑓∶𝑈∖{𝑎} →ℂis holomorphic and 𝑎∈𝑈is an essential singularity of 𝑓, then for any 𝜀> 0, the set 𝑓(𝐷(𝑎, 𝜀) ∖{𝑎}) is dense in ℂ.
The proof is an exercise on the second example sheet.
Theorem (Picard’s theorem). If 𝑓∶𝑈∖{𝑎} →ℂis holomorphic and 𝑎∈𝑈is an essential singularity of 𝑓, then there exists 𝑤∈ℂsuch that for any 𝜀> 0, ℂ∖{𝑤} ⊆𝑓(𝐷(𝑎, 𝜀) ∖{𝑎}).
In other words, in any neighbourhood 𝐷(𝑎, 𝜀) ∖{𝑎}, 𝑓attains all complex numbers except possibly one.
The proof is omitted.
4.5. Laurent series If 𝑧= 𝑎is a removable singularity of 𝑓, then for some 𝑅> 0, 𝑓is given by a power series ∑ ∞ 𝑛=0 𝑐𝑛(𝑧−𝑎)𝑛, which is the Taylor series of the holomorphic extension of 𝑓to 𝐷(𝑎, 𝑅), for all 𝑧∈𝐷(𝑎, 𝑅) ∖{𝑎}. If 𝑎is a pole of some order 𝑘≥1, then for some 𝑅> 0 we have 𝑓(𝑧) = (𝑧−𝑎)−𝑘𝑔(𝑧) for some holomorphic 𝑔∶𝐷(𝑎, 𝑅) →ℂand all 𝑧∈𝐷(𝑎, 𝑅) ∖{𝑎}, so using the Taylor seires of 𝑔, we find a series of the form 𝑓(𝑧) = ∑ ∞ 𝑛=−𝑘𝑐𝑛(𝑧−𝑎)𝑛, for 𝑧∈𝐷(𝑎, 𝑅) ∖{𝑎}. When 𝑎is an essential singularity, we can still obtain an analogous series expansion with infinitely many terms with negative powers. More generally, we have the following.
527 IX. Complex Analysis Theorem (Laurent expansion). Let 𝑓be holomorphic on an annulus 𝐴= {𝑧∈ℂ∶𝑟< |𝑧−𝑎| < 𝑅} for 0 ≤𝑟< 𝑅≤∞. Then: (i) 𝑓has a unique convergent series expansion 𝑓(𝑧) = ∞ ∑ 𝑛=−∞ 𝑐𝑛(𝑧−𝑎)𝑛≡ ∞ ∑ 𝑛=1 𝑐−𝑛(𝑧−𝑎)−𝑛+ ∞ ∑ 𝑛=0 𝑐𝑛(𝑧−𝑎)𝑛 where the 𝑐𝑛are constants; (ii) for any 𝜌∈(𝑟, 𝑅), the coefficient 𝑐𝑛is given by 𝑐𝑛= 1 2𝜋𝑖∫ 𝜕𝐷(𝑎,𝜌) 𝑓(𝑧) d𝑧 (𝑧−𝑎)𝑛+1 (iii) if 𝑟< 𝜌′ ≤𝜌< 𝑅, then the two series in (i) separately converge uniformly on the set {𝑧∈ℂ∶𝜌′ ≤|𝑧−𝑎| ≤𝜌} Remark. If 𝑓is the restriction of 𝐴of a holomorphic function 𝑔on the full disc 𝐷(𝑎, 𝑅), then by the formula in part (ii), we have for any negative 𝑛= −𝑚, 𝑚≥1, the coefficient 𝑐−𝑚is zero by Cauchy’s theorem. In this case, the Laurent series of 𝑓is the Taylor series of 𝑔restricted to 𝐴. The new content of the theorem is simply when 𝑓has no holomorphic extension to 𝐷(𝑎, 𝑅).
Proof. Let 𝑤∈𝐴and consider the function 𝑔(𝑧) = { 𝑓(𝑧)−𝑓(𝑤) 𝑧−𝑤 if 𝑧≠𝑤 𝑓′(𝑤) if 𝑧= 𝑤 This 𝑔is continuous in 𝐴and holomorphic in 𝐴∖{𝑤}. Hence, this is holomorphic in 𝐴since this is a removable singularity. Let 𝜌1, 𝜌2 such that 𝑟< 𝜌1 < |𝑤−𝑎| < 𝜌2 < 𝑅. The two positively oriented curves 𝜕𝐷(𝑎, 𝜌1) and 𝜕𝐷(𝑎, 𝜌2) are homotopic in 𝐴. Hence, ∫ 𝜕𝐷(𝑎,𝜌1) 𝑔(𝑧) d𝑧= ∫ 𝜕𝐷(𝑎,𝜌2) 𝑔(𝑧) d𝑧 Substituting for 𝑔, ∫ 𝜕𝐷(𝑎,𝜌1) 𝑓(𝑧) d𝑧 𝑧−𝑤−2𝜋𝑖⋅𝐼(𝜕𝐷(𝑎, 𝜌1); 𝑤)𝑓(𝑤) = ∫ 𝜕𝐷(𝑎,𝜌2) 𝑓(𝑧) d𝑧 𝑧−𝑤−2𝜋𝑖⋅𝐼(𝜕𝐷(𝑎, 𝜌2); 𝑤)𝑓(𝑤) We have 𝐼(𝜕𝐷(𝑎, 𝜌1); 𝑤) = 0; 𝐼(𝜕𝐷(𝑎, 𝜌2); 𝑤) = 𝐼(𝜕𝐷(𝑎, 𝜌2); 𝑎) = 1 528 4. Singularities Hence, 𝑓(𝑤) = 1 2𝜋𝑖∫ 𝜕𝐷(𝑎,𝜌2) 𝑓(𝑧) d𝑧 𝑧−𝑤−1 2𝜋𝑖∫ 𝜕𝐷(𝑎,𝜌1) 𝑓(𝑧) d𝑧 𝑧−𝑤 This is an analogue of Cauchy’s integral formula for annular domains. We can now proceed as before when proving the Taylor series expansion for holomorphic functions.
For the first integral, consider the expansion 1 𝑧−𝑤= 1 𝑧−𝑎−(𝑤−𝑎) = ∞ ∑ 𝑛=0 (𝑤−𝑎)𝑛 (𝑧−𝑎)𝑛+1 This series converges uniformly over 𝑧∈𝜕𝐷(𝑎, 𝜌2), since | | 𝑤−𝑎 𝑧−𝑎 | | < 1. For the second integral, consider 1 𝑧−𝑤= 1 𝑧−𝑎−(𝑤−𝑎) = − 1 (𝑤−𝑎)(1 − 𝑧−𝑎 𝑤−𝑎) = − ∞ ∑ 𝑛=0 (𝑧−𝑎)𝑛 (𝑤−𝑎)𝑛+1 Likewise, this series converges uniformly over 𝑧∈𝜕𝐷(𝑎, 𝜌1), since | | 𝑧−𝑎 𝑤−𝑎 | | < 1 in this disc.
Substituting these into the representation formula, we can switch integration and summa-tion due to uniform convergence. This gives 𝑓(𝑤) = ∞ ∑ 𝑛=0 𝑐𝑛(𝑤−𝑎)𝑛+ ∞ ∑ 𝑛=1 𝑐−𝑛(𝑤−𝑎)−𝑛 where 𝑐𝑛= 1 2𝜋𝑖∫ 𝜕𝐷(𝑎,𝜌2) 𝑓(𝑧) d𝑧 (𝑧−𝑎)𝑛+1 for 𝑛≥0, and 𝑐𝑛= 1 2𝜋𝑖∫ 𝜕𝐷(𝑎,𝜌1) 𝑓(𝑧) d𝑧 (𝑧−𝑎)𝑛+1 for 𝑛≤−1. Since 𝜕𝐷(𝑎, 𝜌1) and 𝜕𝐷(𝑎, 𝜌2) are homotopic in 𝐴to 𝜕𝐷(𝑎, 𝜌) for any 𝜌∈(𝑟, 𝑅), we have that 𝑐𝑛= 1 2𝜋𝑖∫ 𝜕𝐷(𝑎,𝜌) 𝑓(𝑧) d𝑧 𝑧−𝑎 for any 𝜌∈(𝑟, 𝑅) and 𝑛∈ℤ, so (i) and the formula (ii) both hold.
To show (iii) and uniqueness, suppose there exist constants 𝑐𝑛such that, for all 𝑧∈𝐴, we have 𝑓(𝑧) = ∞ ∑ 𝑛=−∞ 𝑐𝑛(𝑧−𝑎)𝑛 (∗) Let 𝑟< 𝜌′ ≤𝜌< 𝑅. Then the power series ∑ ∞ 𝑛=0 𝑐𝑛(𝑧−𝑎)𝑛converges for 𝑧∈𝐴, so it has radius of convergence at least 𝑅, and converges uniformly for |𝑧−𝑎| ≤𝜌. Further, the series 529 IX. Complex Analysis ∑ ∞ 𝑛=1 𝑐−𝑛(𝑧−𝑎)−𝑛converges on 𝐴. Let 𝜁= (𝑧−𝑎)−1. Then the power series ∑ ∞ 𝑛=1 𝑐−𝑛𝜁𝑛 converges for 1 𝑅< |𝜁| < 1 𝑟so it has radius of convergence at least 1 𝑟and converges uniformly for |𝜁| ≤ 1 𝜌′ . Thus, the series ∑ ∞ 𝑛=1 𝑐−𝑛(𝑧−𝑎)−𝑛converges uniformly for |𝑧−𝑎| ≥𝜌′. Hence (∗) converges uniformly in 𝜌′ ≤|𝑧−𝑎| ≤𝜌. Hence, for any 𝑚∈ℤ, we have ∫ 𝜕𝐷(𝑎,𝜌) 𝑓(𝑧) d𝑧 (𝑧−𝑎)𝑚+1 = ∞ ∑ 𝑛=−∞ 𝑐𝑛∫ 𝜕𝐷(𝑎,𝜌) (𝑧−𝑎)𝑛−𝑚−1 d𝑧 By the fundamental theorem of calculus, the only nonzero integral on the right hand side occurs when 𝑛−𝑚−1 = −1, which occurs for 𝑛= 𝑚only. This integral gives 𝑐𝑚= 1 2𝜋𝑖∫ 𝜕𝐷(𝑎,𝜌) 𝑓(𝑧) d𝑧 (𝑧−𝑎)𝑚+1 for all 𝜌∈(𝑟, 𝑅). This formula also implies the uniqueness of the 𝑐𝑛for which the series expansion is valid.
Remark. The above proof shows that if 𝑓∶𝐴≡𝐷(𝑎, 𝑅) ∖𝐷(𝑎, 𝑟) →ℂis holomorphic, then there is a holomorphic function 𝑓 1 ∶𝐷(𝑎, 𝑅) →ℂand a holomorphic function 𝑓 2 ∶ℂ∖ 𝐷(𝑎, 𝑟) →ℂsuch that 𝑓= 𝑓 1 +𝑓 2 on 𝐴. This decomposition is not unique, since we can take 𝑓 1 ↦𝑓 1 + 𝑔and 𝑓 2 ↦𝑓 2 −𝑔for an entire function 𝑔. However, if we also require 𝑓 2(𝑧) →0 as 𝑧→∞, the decomposition into two series given in (ii) above is unique.
4.6. Coefficients of Laurent series Let 𝑓∶𝐷(𝑎, 𝑅) ∖{𝑎} →ℂbe holomorphic, so 𝑧= 𝑎is an isolated singularity of 𝑓. Then, by the Laurent series with 𝑟= 0, we have a unique set of complex numbers 𝑐𝑛such that 𝑓(𝑧) = ∞ ∑ 𝑛=−∞ 𝑐𝑛(𝑧−𝑎)𝑛 Then, (i) If 𝑐𝑛= 0 for all 𝑛< 0, we have 𝑓(𝑧) = ∑ ∞ 𝑛=0 𝑐𝑛(𝑧−𝑎)𝑛≡𝑔(𝑧) on 𝐷(𝑎, 𝑅) ∖{𝑎}. Since 𝑔is holomorphic on 𝐷(𝑎, 𝑅), 𝑧= 𝑎is a removable singularity.
(ii) If 𝑐−𝑘≠0 for some 𝑘≥1 and 𝑐−𝑛= 0 for all 𝑛≥𝑘+ 1, we have 𝑓(𝑧) = 𝑐−𝑘 (𝑧−𝑎)𝑘+ 𝑐−𝑘+1 (𝑧−𝑎)𝑘+1 + ⋯+ 𝑐−1 𝑧−𝑎+ ∞ ∑ 𝑛=0 𝑐𝑛(𝑧−𝑎)𝑛 Hence, 𝑓(𝑧) = (𝑧−𝑎)−𝑘𝑔(𝑧) for a function 𝑔which is holomorphic on 𝐷(𝑎, 𝑅), and where 𝑔(𝑎) = 𝑐−𝑘≠0. Equivalently, 𝑧= 𝑎is a pole of order 𝑘.
(iii) If 𝑐𝑛≠0 for infinitely many 𝑛< 0, 𝑧= 𝑎is an essential singularity. This holds since the above two parts were all bidirectional implications.
530 4. Singularities 4.7. Residues Definition. Let 𝑓∶𝐷(𝑎, 𝑅) ∖{𝑎} →ℂbe holomorphic. The coefficient 𝑐−1 of the Laurent series of 𝑓in 𝐷(𝑎, 𝑅) ∖{𝑎} is called the residue of 𝑓at 𝑎, denoted Res𝑓(𝑎). The series 𝑓 𝑃= ∞ ∑ 𝑛=1 𝑐−𝑛(𝑧−𝑎)−𝑛 is known as the principal part of 𝑓at 𝑎.
We know that 𝑓 𝑃is holomorphic on ℂ∖{𝑎}, with the series defining 𝑓 𝑃converging uniformly on compact subsets of ℂ∖{𝑎}. By the Laurent series, 𝑓= 𝑓 𝑃+ ℎon 𝐷(𝑎, 𝑅) ∖{𝑎}, where ℎis holomorphic on 𝐷(𝑎, 𝑅). Let 𝛾be a closed curve in 𝐷(𝑎, 𝑅) with 𝑎∉Im 𝛾. Then ∫ 𝛾ℎ(𝑧) d𝑧= 0 by Cauchy’s theorem, and hence ∫ 𝛾𝑓(𝑧) d𝑧= ∫ 𝛾𝑓 𝑃(𝑧) d𝑧= 2𝜋𝑖⋅𝐼(𝛾; 𝑎) Res𝑓(𝑎), where the last inequality holds by uniform convergence of the series for 𝑓 𝑃and the fundamental theorem of calculus. This reasoning can be extended to the case of more then one isolated singularity.
Theorem (residue theorem). Let 𝑈be an open set, {𝑎1, … , 𝑎𝑘} ⊂𝑈be finite, and 𝑓∶𝑈∖ {𝑎1, … , 𝑎𝑘} →ℂbe holomorphic. If 𝛾is a closed curve in 𝑈homologous to zero in 𝑈, and if 𝑎𝑗∉Im 𝛾for each 𝑗, then 1 2𝜋𝑖∫ 𝛾 𝑓(𝑧) d𝑧= 𝑘 ∑ 𝑗=1 𝐼(𝛾; 𝑎𝑗) Res𝑓(𝑎𝑗) This is a generalisation of Cauchy’s integral formula.
Proof. Let 𝑓(𝑗) 𝑃 = ∑ ∞ 𝑛=1 𝑐(𝑗) −𝑛(𝑧−𝑎𝑗)−𝑛be the principal part of 𝑓at 𝑎𝑗. Then 𝑓(𝑗) 𝑃 is holo-morphic in ℂ∖{𝑎𝑗}, and hence is holomorphic in ℂ∖{𝑎1, … , 𝑎𝑘}. Let ℎ≡𝑓−(𝑓(1) 𝑃 + ⋯+ 𝑓(𝑘) 𝑃) This ℎis holomorphic in 𝑈∖{𝑎1, … , 𝑎𝑘}. Let 𝑗be fixed. Then 𝑓−𝑓(𝑗) 𝑃 has a removable singularity at 𝑧= 𝑎𝑗. For all ℓ≠𝑗, 𝑓(ℓ) 𝑃 is holomorphic at 𝑎𝑗. Hence ℎhas a removable singularity at 𝑎𝑗. This is true for all 𝑗, so ℎextends to all of 𝑈as a holomorphic function. By Cauchy’s theorem, ∫ 𝛾ℎ(𝑧) d𝑧= 0. Hence 1 2𝜋𝑖∫ 𝛾 𝑓(𝑧) d𝑧= 𝑘 ∑ 𝑗=1 1 2𝜋𝑖∫ 𝛾 𝑓(𝑗) 𝑃(𝑧) d𝑧 By termwise integration of the series for 𝑓(𝑗) 𝑃, which converges uniformly on compact subsets of ℂ∖{𝑎𝑗}, we have 1 2𝜋𝑖∫ 𝛾 𝑓(𝑗) 𝑃(𝑧) d𝑧= 𝐼(𝛾; 𝑎𝑗) Res𝑓(𝑎𝑗) as required.
531 IX. Complex Analysis There are simple ways to calculate residues if we know information about the singularity in question.
(i) If 𝑓has a simple pole at 𝑧= 𝑎, then Res𝑓(𝑎) = lim 𝑧→𝑎(𝑧−𝑎)𝑓(𝑧) Indeed, near 𝑎, we have 𝑓(𝑧) = (𝑧−𝑎)−1𝑔(𝑧) where 𝑔is holomorphic and 𝑔(𝑎) ≠0.
Hence, by the Taylor expansion of 𝑔, we have that Res𝑓(𝑎) = 𝑔(𝑎).
(ii) If 𝑓has a pole of order 𝑘at 𝑎, then near 𝑎we have that 𝑓(𝑧) = (𝑧−𝑎)−𝑘𝑔(𝑧) where 𝑔 is holomorphic and 𝑔(𝑎) ≠0. In this case, the residue Res𝑓(𝑎) is the coefficient of the (𝑧−𝑎)𝑘−1 term of the Taylor series of 𝑔at 𝑎, which is Res𝑓(𝑎) = 𝑔(𝑘−1)(𝑎) (𝑘−1)!
(iii) If 𝑓= 𝑔 ℎwhere 𝑔and ℎare holomorphic at 𝑧= 𝑎, such that 𝑔(𝑎) ≠0 and ℎhas a simple zero at 𝑧= 𝑎, then from (i) we have Res𝑓(𝑎) = lim 𝑧→𝑎 (𝑧−𝑎)𝑔(𝑧) ℎ(𝑧) = lim 𝑧→𝑎 𝑔(𝑧) ℎ(𝑧)−ℎ(𝑎) 𝑧−𝑎 = 𝑔(𝑎) ℎ′(𝑎) Example. For 0 < 𝛼< 1, we will show that ∫ ∞ 0 𝑥−𝛼 1 + 𝑥d𝑥= 𝜋 sin 𝜋𝛼 Let 𝑔(𝑧) = 𝑧−𝛼be the branch of 𝑧−𝛼defined by 𝑔(𝑧) = 𝑒−𝛼ℓ(𝑧), where ℓ(𝑧) is the holomorphic branch of logarithm on 𝑈= ℂ∖{𝑥∈ℝ∶𝑥≥0}. given by ℓ(𝑧) = log |𝑧|+𝑖arg 𝑧where arg(𝑧) takes values in (0, 2𝜋). Let 𝑓(𝑧) = 𝑔(𝑧) 1+𝑧. Then 𝑓(𝑧) = |𝑧|−𝛼𝑒−𝑖𝛼arg 𝑧 1 + 𝑧 and 𝑓is holomorphic in 𝑈∖{−1} where 𝑧= −1 is a simple pole with Res𝑓(−1) = lim𝑧→−1(𝑧+ 1)𝑓(𝑧) = 𝑒−𝑖𝜋𝛼.
Let 𝜀, 𝑅be such that 0 < 𝜀< 1 < 𝑅and 𝜃> 0 be small. Let 𝛾be the positively-oriented ‘keyhole countour’ determined by the two circular arcs 𝛾𝑅∶[𝜃, 2𝜋−𝜃] →𝑈and the two line segments 𝛾1, 𝛾2 ∶[𝜀, 𝑅] →𝑈given by 𝛾𝑅(𝑡) = 𝑅𝑒𝑖𝑡; 𝛾𝜀(𝑡) = 𝜀𝑒𝑖(2𝜋−𝑡); 𝛾1(𝑡) = 𝑡𝑒𝑖𝜃; 𝛾2(𝑡) = 𝑡𝑒𝑖(2𝜋−𝜃) The domain 𝑈is star shaped and hence simply connected, and so 𝛾is homologous to zero in 𝑈. Directly from the definitions of 𝛾and the winding number, we can show that 𝐼(𝛾; −1) = 1.
532 4. Singularities By the residue theorem, we find ∫ 𝛾𝑓(𝑧) d𝑧= 2𝜋𝑖𝑒−𝑖𝜋𝛼. Now, ∫ 𝛾1 𝑓(𝑧) d𝑧= ∫ 𝑅 𝜀 𝑓(𝑡𝑒𝑖𝜃)𝑒𝑖𝜃d𝑡= ∫ 𝑅 𝜀 𝑡−𝛼𝑒𝑖(1−𝛼)𝜃 1 + 𝑡𝑒𝑖𝜃 d𝑡 and ∫ 𝛾2 𝑓(𝑧) d𝑧= ∫ 𝑅 𝜀 𝑓(𝑡𝑒𝑖(2𝜋−𝜃))𝑒𝑖(2𝜋−𝜃) d𝑡= ∫ 𝑅 𝜀 𝑡−𝛼𝑒𝑖(1−𝛼)(2𝜋−𝜃) 1 + 𝑡𝑒𝑖(2𝜋−𝜃) d𝑡 As 𝜃→0+, we can show that the integrands converge uniformly on [𝜀, 𝑅] to 𝑡−𝛼 1+𝑡and 𝑒−2𝑖𝜋𝛼𝑡−𝛼 1+𝑡 respectively. Hence, lim 𝜃→0+ [∫ 𝛾1 𝑓(𝑧) d𝑧+ ∫ (−𝛾2) 𝑓(𝑧) d𝑧] = (1 −𝑒−2𝑖𝜋𝛼) ∫ 𝑅 𝜀 𝑡−𝛼 1 + 𝑡d𝑡 For all 𝑧∈Im 𝛾𝑅, we have |𝑓(𝑧)| ≤ 𝑅−𝛼 𝑅−1; and for all 𝑧∈Im 𝛾𝜀, we have |𝑓(𝑧)| ≤ 𝜀−𝛼 1−𝜀. Hence, | | | | ∫ 𝛾𝑅 𝑓(𝑧) d𝑧+ ∫ 𝛾𝜀 𝑓(𝑧) d𝑧 | | | | ≤2𝜋𝑅1−𝛼 𝑅−1 + 2𝜋𝜀1−𝛼 1 −𝜀 Note that the right hand side is independent of 𝜃, even though 𝛾𝑅and 𝛾𝜀depend on 𝜃. Since ∫ 𝛾 𝑓(𝑧) d𝑧−(∫ 𝛾1 𝑓(𝑧) d𝑧+ ∫ (−𝛾2) 𝑓(𝑧) d𝑧) = ∫ 𝛾𝑅 𝑓(𝑧) d𝑧+ ∫ 𝛾𝜀 𝑓(𝑧) d𝑧 we then have that | | | | 2𝜋𝑖𝑒−𝑖𝜋𝛼−(∫ 𝛾1 𝑓(𝑧) d𝑧+ ∫ (−𝛾2) 𝑓(𝑧) d𝑧) | | | | ≤2𝜋𝑅1−𝛼 𝑅−1 + 2𝜋𝜀1−𝛼 1 −𝜀 First letting 𝜃→0+ in this, and then letting 𝜀→0+ and 𝑅→∞, we conclude (1 −𝑒−2𝜋𝑖𝛼) ∫ ∞ 0 𝑡−𝛼 1 + 𝑡d𝑡= 2𝜋𝑖𝑒−𝑖𝜋𝛼 or, ∫ ∞ 0 𝑡−𝛼 1 + 𝑡d𝑡= 𝜋 sin 𝜋𝛼 4.8. Jordan’s lemma Lemma. Let 𝑓be a continuous complex-valued function on the semicircle 𝐶+ 𝑅= Im 𝛾+ 𝑅in the upper half-plane, where 𝑅> 0 and 𝛾+ 𝑅(𝑡) = 𝑅𝑒𝑖𝑡for 0 ≤𝑡≤𝜋. Then, for 𝛼> 0, | | | | ∫ 𝛾+ 𝑅 𝑓(𝑧)𝑒𝑖𝛼𝑧d𝑧 | | | | ≤𝜋 𝛼sup 𝑧∈𝐶+ 𝑅 |𝑓(𝑧)| 533 IX. Complex Analysis In particular, if 𝑓is continuous in 𝐻+ ∖𝐷(0, 𝑅0) for 𝑅0 > 0 where 𝐻+ = {𝑧∶Im 𝑧≥0} and if sup𝑧∈𝐶+ 𝑅|𝑓(𝑧)| →0 as 𝑅→∞, then for each 𝛼> 0, we have ∫ 𝛾+ 𝑅 𝑓(𝑧)𝑒𝑖𝛼𝑧d𝑧→0 as 𝑅→∞.
A similar statement holds for 𝛼< 0 and the semicircle 𝐶− 𝑅= Im 𝛾− 𝑅in the lower half-plane where 𝛾− 𝑅(𝑡) = −𝑅𝑒𝑖𝑡for 𝑅> 0 and 0 ≤𝑡≤𝜋.
Proof. Let 𝑀𝑅= sup𝑧∈𝐶+ 𝑅|𝑓(𝑧)|. Then, | | | | ∫ 𝛾+ 𝑅 𝑓(𝑧)𝑒𝑖𝛼𝑧d𝑧 | | | | = | | |∫ 𝜋 0 𝑓(𝑅𝑒𝑖𝑡)𝑒−𝛼𝑅sin 𝑡+𝑖𝛼𝑅cos 𝑡𝑖𝑅𝑒𝑖𝑡d𝑡| | | ≤𝑅𝑀𝑅∫ 𝜋 0 𝑒−𝛼𝑅sin 𝑡d𝑡 = 𝑅𝑀𝑅(∫ 𝜋 2 0 𝑒−𝛼𝑅sin 𝑡d𝑡+ ∫ 𝜋 𝜋 2 𝑒−𝛼𝑅sin 𝑡d𝑡) = 2𝑅𝑀𝑅∫ 𝜋 2 0 𝑒−𝛼𝑅sin 𝑡d𝑡 ≤2𝑅𝑀𝑅∫ 𝜋 2 0 𝑒 −2𝛼𝑅𝑡 𝜋 d𝑡 = 𝜋𝑀𝑅 𝛼 (1 −𝑒−𝛼𝑅) ≤𝜋𝑀𝑅 𝛼 where we have used the fact that for 𝑡∈(0, 𝜋 2 ], 𝜑(𝑡) ≡ sin 𝑡 𝑡 ≥ 2 𝜋since 𝜑( 𝜋 2 ) = 2 𝜋and 𝜑′(𝑡) ≤0 on [0, 𝜋 2 ].
Lemma (integrals on small circular arcs). Let 𝑓be holomorphic in 𝐷(𝑎, 𝑅) ∖{𝑎} with a simple pole at 𝑧= 𝑎. Let 𝛾𝜀∶[𝛼, 𝛽] →ℂbe the circular arc 𝛾𝜀(𝑡) = 𝑎+ 𝜀𝑒𝑖𝑡. Then lim 𝜀→0+ ∫ 𝛾𝜀 𝑓(𝑧) d𝑧= (𝛽−𝛼)𝑖Res𝑓(𝑎) Proof. Let 𝑓(𝑧) = 𝑐 𝑧−𝑎+ 𝑔(𝑧) where 𝑔is holomorphic in 𝐷(𝑎, 𝑅) and 𝑐= Res𝑓(𝑎). Then | | | | ∫ 𝛾𝜀 𝑔(𝑧) d𝑧 | | | | = | | | | ∫ 𝛽 𝛼 𝑔(𝑎+ 𝜀𝑒𝑖𝑡)𝜀𝑖𝑒𝑖𝑡| | | | ≤𝜀(𝛽−𝛼) sup 𝑡∈[𝛼,𝛽] | |𝑔(𝑎+ 𝜀𝑒𝑖𝑡)| | →0 534 4. Singularities as 𝜀→0+. By direct calculation, ∫ 𝛾𝜀 𝑐 𝑧−𝑎d𝑧= (𝛽−𝛼)𝑖Res𝑓(𝑎) Hence the claim follows.
Example. Consider ∫ ∞ 0 sin 𝑥 𝑥d𝑥. Let 𝑓(𝑧) = 𝑒𝑖𝑧 𝑧. Consider the integral ∫ 𝛾𝑓(𝑧) d𝑧over the curve 𝛾= 𝛾𝑅+ 𝛾1 + 𝛾𝜀+ 𝛾2 where (i) 𝛾𝑅(𝑡) = 𝑅𝑒𝑖𝑡for 0 ≤𝑡≤𝜋; (ii) 𝛾1(𝑡) = 𝑡for −𝑅≤𝑡≤−𝜀; (iii) 𝛾𝜀(𝑡) = 𝜀𝑒−𝑖𝑡for −𝜋≤𝑡≤0; (iv) 𝛾2(𝑡) = 𝑡for 𝜀≤𝑡≤𝑅.
By Jordan’s lemma, ∫ 𝛾𝑅𝑓(𝑧) d𝑧→0 as 𝑅→∞. 𝑓has a simple pole at 𝑧= 0 with Res𝑓(0) = lim𝑧→0 𝑧𝑓(𝑧) = 1. By the above lemma, ∫ −𝛾𝜀𝑓(𝑧) d𝑧→𝜋𝑖as 𝜀→0+.
Since 𝑓is holomorphic in 𝑈= ℂ∖{0} and 𝛾is homologous to zero in 𝑈, Cauchy’s theorem gives that ∫ 𝛾 𝑓(𝑧) d𝑧= 0 ⟹∫ 𝛾𝑅 𝑓(𝑧) d𝑧+ ∫ −𝜀 −𝑅 𝑒𝑖𝑡 𝑡d𝑡+ ∫ 𝛾𝜀 𝑓(𝑧) d𝑧+ ∫ 𝑅 𝜀 𝑒𝑖𝑡 𝑡d𝑡= 0 Combining the two integrals on the real axis under a change of variables, ∫ 𝑅 𝜀 𝑒𝑖𝑡−𝑒−𝑖𝑡 𝑡 d𝑡+ ∫ 𝛾𝑅 𝑓(𝑧) d𝑧+ ∫ 𝛾𝜀 𝑓(𝑧) d𝑧= 0 Letting 𝑅→∞and 𝜀→0+, we have ∫ ∞ 0 sin 𝑡 𝑡 d𝑡= 𝜋 2 Example. We prove that ∑ ∞ 𝑛=1 1 𝑛2 = 𝜋2 6 . Consider the function 𝑓(𝑧) = 𝜋cot(𝜋𝑧) 𝑧2 = 𝜋cos(𝜋𝑧) 𝑧2 sin(𝜋𝑧) This is holomorphic in ℂexcept for simple poles at each point in ℤ∖{0}, and an order 3 pole at zero. Near 𝑛∈ℤ∖{0}, we have 𝑓(𝑧) = 𝑔(𝑧) ℎ(𝑧) where 𝑔(𝑛) ≠0 and ℎhas a simple zero at 𝑛, and so Res𝑓(𝑛) = 𝑔(𝑛) ℎ′(𝑛) = 1 𝑛2 535 IX. Complex Analysis To compute the residue at zero, consider cot 𝑧= cos 𝑧 sin 𝑧= (1 −𝑧2 2 + 𝑂(𝑧4))(𝑧−𝑧3 6 + 𝑂(𝑧5)) −1 = 1 𝑧−𝑧 3 + 𝑂(𝑧2) Hence, 𝜋cot(𝜋𝑧) 𝑧2 = 1 𝑧3 −𝜋2 3𝑧+ … This shows that Res𝑓(0) = − 𝜋2 3 . For 𝑁∈ℕ, let 𝛾𝑁be the positively oriented boundary of the square defined by the lines 𝑥= ±(𝑁+ 1 2) and 𝑦= ±(𝑁+ 1 2). By the residue theorem, ∫ 𝛾𝑁 𝑓(𝑧) d𝑧= 2𝜋𝑖[2( 𝑁 ∑ 𝑛=1 1 𝑛2 ) −𝜋2 3 ] (∗) Since length(𝛾𝑁) = 4(2𝑁+ 1), we have | | | | ∫ 𝛾𝑁 𝑓(𝑧) d𝑧 | | | | ≤sup 𝛾𝑁 | | | 𝜋cot(𝜋𝑧) 𝑧2 | | | ⋅4(2𝑁+ 1) ≤sup 𝛾𝑁 |cot(𝜋𝑧)| ⋅4(2𝑁+ 1)𝜋 (𝑁+ 1 2)2 = 16𝜋 2𝑁+ 1 ⋅sup 𝛾𝑁 |cot(𝜋𝑧)| On 𝛾𝑁, it is possible to show that cot(𝜋𝑧) is bounded independently of 𝑁. Hence, ∫ 𝛾𝑁 𝑓(𝑧) d𝑧→0 as 𝑁→∞. Letting 𝑁→∞in (∗), we find ∞ ∑ 𝑛=1 1 𝑛2 = 𝜋2 6 536 5. The argument principle, local degree, and Rouché’s theorem 5. The argument principle, local degree, and Rouché’s theorem 5.1. The argument principle Proposition. If 𝑓has a zero (or pole) of order 𝑘≥1 at 𝑧= 𝑎, then 𝑓′ 𝑓has a simple pole at 𝑧= 𝑎with residue 𝑘(or −𝑘, respectively).
Proof. If 𝑧= 𝑎is a zero of order 𝑘, there is a disc 𝐷(𝑎, 𝑟) such that 𝑓(𝑧) = (𝑧−𝑎)𝑘𝑔(𝑧) for 𝑧∈𝐷(𝑎, 𝑟) where 𝑔∶𝐷(𝑎, 𝑟) →ℂis holomorphic with 𝑔(𝑧) ≠0 for all 𝑧∈𝐷(𝑎, 𝑟). Hence, 𝑓′(𝑧) = 𝑘(𝑧−𝑎)𝑘−1𝑔(𝑧) + (𝑧−𝑎)𝑘𝑔′(𝑧) and 𝑓′(𝑧) 𝑓(𝑧) = 𝑘 𝑧−𝑎+ 𝑔′(𝑧) 𝑔(𝑧) for all 𝑧∈𝐷(𝑎, 𝑟) ∖{𝑎}. Since 𝑔′ 𝑔is holomorphic in 𝐷(𝑎, 𝑅), the claim follows. A similar argument holds for poles.
Definition. The order of a zero or pole 𝑎of a holomorphic function 𝑓is denoted ord𝑓(𝑎).
Theorem (the argument principle). Let 𝑓be a meromorphic function on a domain 𝑈with finitely many zeroes 𝑎1, … , 𝑎𝑘and finitely many poles 𝑏1, … , 𝑏ℓ. If 𝛾is a closed curve in 𝑈 homologous to zero in 𝑈, and if 𝑎𝑖, 𝑏𝑗∉Im 𝛾for all 𝑖, 𝑗, then 1 2𝜋𝑖∫ 𝛾 𝑓′(𝑧) 𝑓(𝑧) d𝑧= 𝑘 ∑ 𝑖=1 𝐼(𝛾; 𝑎𝑖) ord𝑓(𝑎𝑖) − ℓ ∑ 𝑗=1 𝐼(𝛾; 𝑏𝑗) ord𝑓(𝑏𝑗) Proof. Apply the residue theorem to 𝑔= 𝑓′ 𝑓. If 𝑧0 ∈𝑈is not a pole of 𝑓, then 𝑓and hence 𝑓′ are holomorphic near 𝑧0. If additionally 𝑧0 is not a zero of 𝑓, 𝑔is holomorphic near 𝑧0. So the set of singularities of 𝑔is precisely {𝑎1, … , 𝑎𝑘} ∪{𝑏1, … , 𝑏ℓ}. By the previous proposition, their residues are known, and the result follows.
Remark. Let 𝑓, 𝛾be as in the theorem, and let Γ(𝑡) = 𝑓(𝛾(𝑡)). Then Γ(𝑡) is a closed curve with image Im Γ ⊂ℂ∖{0}, since no zeroes or poles of 𝑓are in Im 𝛾. Moreover, if [𝑎, 𝑏] is the domain of 𝛾, we have 𝐼(Γ; 0) = 1 2𝜋𝑖∫ Γ d𝑧 𝑧= 1 2𝜋𝑖∫ 𝑏 𝑎 Γ′(𝑡) Γ(𝑡) d𝑡= 1 2𝜋𝑖∫ 𝑏 𝑎 𝑓′(𝛾(𝑡))𝛾′(𝑡) 𝑓(𝛾(𝑡)) d𝑡= 1 2𝜋𝑖∫ 𝛾 𝑓′(𝑧) 𝑓(𝑧) d𝑧 Thus, 1 2𝜋𝑖∫ 𝛾 𝑓′(𝑧) 𝑓(𝑧) is the number of times the image curve 𝑓∘𝛾winds around zero as we move along 𝛾.
537 IX. Complex Analysis Definition. Let Ω be a domain, and let 𝛾be a closed curve in ℂ. We say that 𝛾bounds Ω if 𝐼(𝛾; 𝑤) = 1 for all 𝑤∈Ω, and 𝐼(𝛾; 𝑤) = 0 for all 𝑤∈ℂ∖(Ω ∪Im 𝛾).
Example. 𝜕𝐷(0, 1) bounds 𝐷(0, 1), but does not bound 𝐷(0, 1) ∖{0}.
Remark. If 𝛾bounds Ω, then (i) Ω is bounded. Indeed, let 𝐷(𝑎, 𝑅) such that Im 𝛾⊆𝐷(𝑎, 𝑅). Then 𝐼(𝛾; 𝑤) = 0 for 𝑤∈ℂ∖𝐷(𝑎, 𝑅). Since 𝐼(𝛾; 𝑤) = 1 for all 𝑤∈Ω, we must have Ω ⊂𝐷(𝑎, 𝑅).
(ii) the topological boundary 𝜕Ω is contained within Im 𝛾, but it need not be the case that 𝜕Ω = Im 𝛾.
There is a large class of closed curves that bound domains, namely, simple closed curves, which are curves 𝛾∶[𝑎, 𝑏] →ℂwith 𝛾(𝑎) = 𝛾(𝑏), and such that 𝛾(𝑡1) = 𝛾(𝑡2) implies 𝑡1 = 𝑡2 or 𝑡1, 𝑡2 ∈{𝑎, 𝑏}. That a simple closed curve bounds a domain is a highly non-trivial fact guaranteed by the Jordan curve theorem: if 𝛾is a simple closed curve, then ℂ∖Im 𝛾consists precisely of two connected components, one of which is bounded and the other unbounded, and moreover, 𝛾(or −𝛾) bounds the bounded component, and Im 𝛾is the boundary of each of the two components. Thus, if Ω1 is the bounded component and Ω2 is the unbounded component, then after possibly changing the orientation of 𝛾, we have 𝐼(𝛾; 𝑤) = 1 for 𝑤∈Ω1, and 𝐼(𝛾; 𝑤) = 0 for 𝑤∈Ω2. This last assertion is simply that for any disc 𝐷(𝑎, 𝑅) ⊃Im 𝛾, we have 𝐼(𝛾; 𝑤) = 0 for all 𝑤∈ℂ∖𝐷(𝑎, 𝑅).
For a domain bounded by a closed curve, the argument principle gives the following.
Corollary. Let 𝛾be a closed curve bounding a domain Ω, and let 𝑓be meromorphic in an open set 𝑈with Ω ∪Im 𝛾⊆𝑈. Suppose that 𝑓has no zeroes or poles on Im 𝛾. Then 𝑓has finitely many zeroes and finitely many poles in Ω.
Let the number of zeroes in Ω be 𝑁, and the number of poles in Ω be 𝑃, both counted with multiplicity. Then in addition we have that 𝑁−𝑃= 1 2𝜋𝑖∫ 𝛾 𝑓′(𝑧) 𝑓(𝑧) d𝑧= 𝐼(Γ; 0) where Γ = 𝑓∘𝛾.
Proof. Since 𝑓is meromorphic in 𝑈, its singularities form a discrete set 𝑆⊂𝑈consisting of poles or removable singularities. Since 𝛾bounds Ω, we have that Ω is bounded and hence Ω is compact. Also, Ω ⊆Ω ∪Im 𝛾⊆𝑈. If Ω ∩𝑆is infinite, then by compactness of Ω, there exists a point 𝑤∈Ω and distinct points 𝑤𝑗∈Ω ∩𝑆such that 𝑤𝑗→𝑤. If 𝑤∉𝑆, then 𝑓is defined and holomorphic near 𝑤which is impossible since 𝑤𝑗∈𝑆and 𝑤𝑗→𝑤. So 𝑤∈𝑆, but this is impossible since 𝑆is a discrete set. So Ω ∩𝑆is finite, and in particular 𝑃is finite.
If 𝑓has infinitely many zeroes in Ω, then by compactness there exists 𝑧∈Ω ⊂𝑈and distinct zeroes 𝑧𝑗∈Ω such that 𝑧𝑗→𝑧. Then either 𝑧∈𝑈∖𝑆, or (if 𝑧∈𝑆) 𝑧is a removable singularity, since otherwise 𝑧would be a pole and hence |𝑓(𝜁)| →∞as 𝜁→𝑧which is 538 5. The argument principle, local degree, and Rouché’s theorem impossible since 𝑧𝑗→𝑧and 𝑓(𝑧𝑗) = 0. In either case, by the principle of isolated zeroes, 𝑓must be identically zero in 𝐷(𝑧, 𝜌) ∖{𝑧} for some 𝜌> 0. Since 𝑓is holomorphic in Ω ∖𝑆 which is connected (since Ω ∩𝑆is finite and Ω) is connected, it follows from the unique continuation principle that 𝑓≡0 in Ω. This is impossible since 𝑓has no zeroes in Im 𝛾, so 𝑁must be finite.
By the definition of 𝛾bounding Ω, we have that 𝐼(𝛾; 𝑤) = 1 for all 𝑤∈Ω, and 𝐼(𝛾; 𝑤) = 0 for all 𝑤∈ℂ∖(Ω ∪Im 𝛾). In particular, 𝛾is homologous to zero in 𝑈. The final conclusion then follows from the fact that Γ is a closed curve in ℂ∖{0} and 𝐼(𝛾; 0) = 1 2𝜋𝑖∫ 𝛾 𝑓′(𝑧) 𝑓(𝑧) d𝑧as proven above.
5.2. Local degree theorem Definition. Let 𝑓be a holomorphic function on a disc 𝐷(𝑎, 𝑅) that is not constant. Then the local degree of 𝑓at 𝑎, denoted deg𝑓(𝑎), is the order of the zero of 𝑓(𝑧) −𝑓(𝑎) at 𝑧= 𝑎.
This is a finite positive integer.
Example. If 𝑓(𝑧) = (𝑧−1)4 + 1 has deg𝑓(1) = 4.
Theorem. Let 𝑓∶𝐷(𝑎, 𝑅) →ℂbe holomorphic and non-constant, with deg𝑓(𝑎) = 𝑑. Then there exists 𝑟0 > 0 such that for any 𝑟∈(0, 𝑟0], there exists 𝜀> 0 such that for all 𝑤with 0 < |𝑓(𝑎) −𝑤| < 𝜀, the equation 𝑓(𝑧) = 𝑤has precisely 𝑑distinct roots in 𝐷(𝑎, 𝑟) ∖{𝑎}.
Proof. Let 𝑔(𝑧) = 𝑓(𝑧) −𝑓(𝑎). Since 𝑔is non-constant, 𝑔′ ≢0 in 𝐷(𝑎, 𝑅). Applying the principle of isolated zeroes to 𝑔and 𝑔′, there exists 𝑟0 ∈(0, 𝑅) such that 𝑔(𝑧) ≠0 and 𝑔′(𝑧) ≠ 0 for 𝑧∈𝐷(𝑎, 𝑟0) ∖{𝑎}.
We will show that the conclusion holds for this choice of 𝑟0. Let 𝑟∈(0, 𝑟0], and for 𝑡∈[0, 1], let 𝛾(𝑡) = 𝑎+ 𝑟𝑒2𝜋𝑖𝑡and Γ(𝑡) = 𝑔(𝛾(𝑡)). Note that Im Γ is compact and hence closed in ℂ, and 0 ∉Im Γ since 𝑔≠0 on 𝜕𝐷(𝑎, 𝑟). Hence there exists 𝜀> 0 such that 𝐷(0, 𝜀) ⊆ℂ∖Im Γ.
We now show that this 𝜀satisfies the conditions in the theorem for this 𝑟. Let 𝑤such that 0 < |𝑤−𝑓(𝑎)| < 𝜀. Then 𝑤−𝑓(𝑎) ∈𝐷(0, 𝜀) ⊆ℂ∖Im Γ. Since 𝑧↦𝐼(Γ; 𝑧) is locally constant, it is constant on 𝐷(0, 𝜀), so in particular 𝐼(Γ; 𝑤−𝑓(𝑎)) = 𝐼(Γ; 0).
By direct calculation, 𝐼(Γ; 𝑤−𝑓(𝑎)) = 1 2𝜋𝑖∫ 1 0 𝑔′(𝛾(𝑡))𝛾′(𝑡) 𝑔(𝛾(𝑡)) −(𝑤−𝑓(𝑎)) d𝑡= 1 2𝜋𝑖∫ 𝜕𝐷(𝑎,𝑟) 𝑓′(𝑧) 𝑓(𝑧) −𝑤d𝑧 By the argument principle, 𝐼(Γ; 0) = 𝑑, since 𝐼(Γ; 0) is the number of zeroes of 𝑔in 𝐷(𝑎, 𝑟) counted with multiplicity; the zero of 𝑔at 𝑧= 𝑎has order 𝑑, and it is the only zero in 𝐷(𝑎, 𝑟).
Hence, 1 2𝜋𝑖∫ 𝜕𝐷(𝑎,𝑟) 𝑓′(𝑧) 𝑓(𝑧) −𝑤d𝑧= 𝑑 539 IX. Complex Analysis Again, the argument principle shows that the number of zeroes of 𝑓(𝑧) −𝑤in 𝐷(𝑎, 𝑟) is 𝑑, counted with multiplicity. None of these zeroes is equal to 𝑎since 𝑤≠𝑓(𝑎). Since 𝑓′(𝑧) = 𝑔′(𝑧) ≠0 in 𝐷(𝑎, 𝑟) ∖{𝑎}, it follows from the Taylor series that these zeroes are simple. Thus 𝑓(𝑧) −𝑤has 𝑑distinct zeroes in 𝐷(𝑎, 𝑟) ∖{𝑎}.
5.3. Open mapping theorem Corollary. A non-constant holomorphic function maps open sets to open sets. That is, non-constant holomorphic functions are open maps.
Proof. Let 𝑓∶𝑈→ℂbe holomorphic and non-constant, and let 𝑉⊆𝑈be an open set.
Let 𝑏∈𝑓(𝑉). Then 𝑏= 𝑓(𝑎) for some 𝑎∈𝑉. Since 𝑉is open, there exists 𝑟> 0 such that 𝐷(𝑎, 𝑟) ⊆𝑉. By the local degree theorem, if 𝑟is sufficiently small, there exists 𝜀> 0 such that 𝑤∈𝐷(𝑓(𝑎), 𝜀) ∖{𝑓(𝑎)} ⟹ 𝑤= 𝑓(𝑧) for some 𝑧∈𝐷(𝑎, 𝑟) ∖{𝑎}, hence 𝐷(𝑓(𝑎), 𝜀) ∖{𝑓(𝑎)} ⊆𝑓(𝐷(𝑎, 𝑟) ∖{𝑎}). Hence 𝐷(𝑏, 𝜀) = 𝐷(𝑓(𝑎), 𝜀) ⊆𝑓(𝐷(𝑎, 𝑟)) ⊆𝑓(𝑉). Thus, for all 𝑏∈𝑓(𝑉), there exists a disc 𝐷(𝑏, 𝜀) ⊆𝑓(𝑉), so 𝑓(𝑉) is open.
5.4. Rouché’s theorem Theorem. Let 𝛾be a closed curve bounding a domain Ω, and let 𝑓, 𝑔be holomorphic func-tions on an open set 𝑈containing Ω ∪Im 𝛾. If |𝑓(𝑧) −𝑔(𝑧)| < |𝑔(𝑧)| for all 𝑧∈Im 𝛾, then 𝑓 and 𝑔have the same number of zeroes in Ω, counted with multiplicity.
Proof. The strict inequality |𝑓−𝑔| < |𝑔| on Im 𝛾implies that 𝑓, 𝑔are never zero on Im 𝛾 and hence never zero on some open set 𝑉containing Im 𝛾. So ℎ= 𝑓 𝑔is holomorphic and nonzero in 𝑉. In particular, 𝑔is not identically zero in Ω, and hence the zeroes of 𝑔in Ω ∪𝑉 are isolated. Hence ℎis meromorphic in Ω ∪𝑉, and ℎhas no zeroes or poles on Im 𝛾. Also, 𝑓, 𝑔have finitely many zeroes in Ω.
Now, |ℎ(𝑧) −1| < 1 for all 𝑧∈Im 𝛾. Hence, the curve Γ = ℎ∘𝛾has image contained within 𝐷(1, 1). Since zero is outside this disc, 𝐼(Γ; 0) = 0, and so by the argument principle, ∑ 𝑤∈𝒫 ordℎ(𝑤) = ∑ 𝑤∈𝒵 ordℎ(𝑤) where 𝒫and 𝒵denote the sets of distinct poles and zeroes of ℎrespectively, and the sums are finite. Now, 𝒫= 𝒫 1 + 𝒫 2 and 𝒵= 𝒵1 ∪𝒵2, where 𝒫 1 = {𝑤∈Ω∶𝑔(𝑤) = 0; 𝑓(𝑤) ≠0}; 𝒫 2 = {𝑤∈Ω∶𝑔(𝑤) = 𝑓(𝑤) = 0; ord𝑔(𝑤) > ord𝑓(𝑤)}; 𝒵1 = {𝑤∈Ω∶𝑓(𝑤) = 0; 𝑔(𝑤) ≠0}; 𝒵2 = {𝑤∈Ω∶𝑓(𝑤) = 𝑔(𝑤) = 0; ord𝑓(𝑤) > ord𝑔(𝑤)} 540 5. The argument principle, local degree, and Rouché’s theorem Hence, ∑ 𝑤∈𝒫1 ord𝑔(𝑤) + ∑ 𝑤∈𝒫2 (ord𝑔(𝑤) −ord𝑓(𝑤)) = ∑ 𝑤∈𝒵1 ord𝑓(𝑤) + ∑ 𝑤∈𝒵2 (ord𝑓(𝑤) −ord𝑔(𝑤)) Equivalently, ∑ 𝑤∈𝒫1 ord𝑔(𝑤)+ ∑ 𝑤∈𝒫2 ord𝑔(𝑤)+ ∑ 𝑤∈𝒵2 ord𝑔(𝑤) = ∑ 𝑤∈𝒵1 ord𝑓(𝑤)+ ∑ 𝑤∈𝒵2 ord𝑓(𝑤)+ ∑ 𝑤∈𝒫2 ord𝑓(𝑤) Adding ∑𝑤∈ℛord𝑔(𝑤) to the left hand side and the equal number ∑𝑤∈ℛord𝑓(𝑤) to the right hand side, where ℛ= {𝑤∈Ω∶𝑓(𝑤) = 𝑔(𝑤) = 0; ord𝑓(𝑤) = ord𝑔(𝑤)} we have ∑ 𝑤∈Ω∶𝑔(𝑤)=0 ord𝑔(𝑤) = ∑ 𝑤∈Ω∶𝑓(𝑤)=0 ord𝑓(𝑤) as required.
Example. 𝑧4 + 6𝑧+ 3 has three roots counted with multiplicity in {1 < |𝑧| < 2}. Let 𝑓(𝑧) = 𝑧4 + 6𝑧+ 3.
On |𝑧| = 2 we have | |𝑧4| | = 16 and |6𝑧+ 3| ≤6|𝑧| + 3 = 15, so |𝑧|4 > |6𝑧+ 3|. By Rouché’s theorem, 𝑓has the same number of roots inside {|𝑧| < 2} as 𝑧4, counting with multiplicity.
Thus, all roots of 𝑧4 +6𝑧+3 lie inside {|𝑧| < 2}; this is all of the roots since 𝑓is a polynomial with degree 4.
On |𝑧| = 1, we have |6𝑧| = 6 and | |𝑧4 + 3| | ≤|𝑧|4 + 3 ≤4. Again by Rouché’s theorem, 𝑓has one root inside {|𝑧| < 1}, as 6𝑧has one root in this region. From the strict inequalities, no roots lie on {|𝑧| = 2} or {|𝑧| = 1}. Hence three roots of 𝑓lie in |𝑧∈ℂ∶1 < |𝑧| < 2|.
541 X. Geometry Lectured in Lent 2022 by Prof. I. Smith This course serves as an introduction to the modern study of surfaces in geometry. A surface is a topological space that locally looks like the plane. The notions of length and area on a surface are governed by mathematical objects called the fundamental forms of the surface at particular points. We can use integrals to work out exact lengths and areas. We study various spaces, including spaces of constant curvature, such as the plane, spheres, and hyperbolic space.
543 X. Geometry Contents 1.
Surfaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 546 1.1.
Basic definitions . . . . . . . . . . . . . . . . . . . . . . . . . 546 1.2.
Subdivisions . . . . . . . . . . . . . . . . . . . . . . . . . . . 551 1.3.
Euler classification . . . . . . . . . . . . . . . . . . . . . . . . 552 2.
Smooth surfaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . 554 2.1.
Charts and atlases . . . . . . . . . . . . . . . . . . . . . . . . 554 3.
Smooth surfaces in ℝ3 . . . . . . . . . . . . . . . . . . . . . . . . . 556 3.1.
Definitions and equivalent characterisations . . . . . . . . . . . 556 3.2.
Inverse and implicit function theorems . . . . . . . . . . . . . . 557 3.3.
Conditions for smoothness . . . . . . . . . . . . . . . . . . . . 558 3.4.
Orientability . . . . . . . . . . . . . . . . . . . . . . . . . . . 559 3.5.
Tangent planes . . . . . . . . . . . . . . . . . . . . . . . . . . 561 4.
Geometry of surfaces in ℝ3 . . . . . . . . . . . . . . . . . . . . . . . 564 4.1.
First fundamental form . . . . . . . . . . . . . . . . . . . . . . 564 4.2.
Conformality . . . . . . . . . . . . . . . . . . . . . . . . . . . 567 4.3.
Area . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 568 4.4.
Second fundamental form . . . . . . . . . . . . . . . . . . . . 569 4.5.
Gauss maps . . . . . . . . . . . . . . . . . . . . . . . . . . . . 571 4.6.
Gauss curvature . . . . . . . . . . . . . . . . . . . . . . . . . . 573 4.7.
Elliptic, hyperbolic, and parabolic points . . . . . . . . . . . . . 574 5.
Geodesics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 578 5.1.
Definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . 578 5.2.
The geodesic equations . . . . . . . . . . . . . . . . . . . . . . 578 5.3.
Geodesics on the plane . . . . . . . . . . . . . . . . . . . . . . 580 5.4.
Geodesics on the sphere . . . . . . . . . . . . . . . . . . . . . 580 5.5.
Geodesics on the torus . . . . . . . . . . . . . . . . . . . . . . 581 5.6.
Equivalent characterisation of geodesics . . . . . . . . . . . . . 581 5.7.
Planes of symmetry . . . . . . . . . . . . . . . . . . . . . . . . 583 5.8.
Surfaces of revolution . . . . . . . . . . . . . . . . . . . . . . . 583 5.9.
Local existence of geodesics . . . . . . . . . . . . . . . . . . . . 585 5.10.
Surfaces of constant curvature . . . . . . . . . . . . . . . . . . 586 6.
Riemannian metrics . . . . . . . . . . . . . . . . . . . . . . . . . . 588 6.1.
Definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . 588 6.2.
The length metric . . . . . . . . . . . . . . . . . . . . . . . . . 589 6.3.
The hyperbolic metric . . . . . . . . . . . . . . . . . . . . . . 591 6.4.
The hyperbolic upper half-plane . . . . . . . . . . . . . . . . . 593 6.5.
Isometries of hyperbolic space . . . . . . . . . . . . . . . . . . 594 544 6.6.
Hyperbolic triangles . . . . . . . . . . . . . . . . . . . . . . . 596 6.7.
Area of triangles . . . . . . . . . . . . . . . . . . . . . . . . . 597 6.8.
Surfaces of constant negative curvature . . . . . . . . . . . . . . 600 6.9.
Gauss–Bonnet theorem . . . . . . . . . . . . . . . . . . . . . . 602 6.10.
Green’s theorem (non-examinable) . . . . . . . . . . . . . . . . 604 6.11.
Alternate flat toruses . . . . . . . . . . . . . . . . . . . . . . . 605 6.12.
Further courses . . . . . . . . . . . . . . . . . . . . . . . . . . 606 545 X. Geometry 1. Surfaces 1.1. Basic definitions Definition. A topological surface is a topological space Σ such that (i) for all points 𝑝∈Σ, there exists an open neighbourhood 𝑝∈𝑈⊂Σ such that 𝑈is homeomorphic to ℝ2, or a disc 𝐷2 ⊂ℝ2, with its usual Euclidean topology; (ii) Σ is Hausdorff and second countable.
Remark. ℝ2 is homeomorphic to the open disc 𝐷(0, 1) = {𝑥∈ℝ2 ∶‖𝑥‖ < 1}. Recall that a space 𝑋is Hausdorff if two points 𝑝≠𝑞∈𝑋have open neighbourhoods 𝑈, 𝑉such that 𝑈∩𝑉= ∅. A space 𝑋is second countable if it has a countable base; there exists a countable family of open sets 𝑈𝑖, such that every open set is a union of some of the 𝑈𝑖.
Note that subspaces of Hausdorff and second countable spaces are also Hausdorff and second countable. In particular, Euclidean space ℝ𝑛is Hausdorff (as ℝ𝑛is a metric space) and second countable (consider the set of balls 𝐷(𝑝, 𝑞) for points 𝑝with rational coordinates, and rational radii 𝑞). Hence, any subspace of ℝ𝑛is implicitly Hausdorff and second count-able. These topological requirements are typically not the purpose of considering topological spaces, but they are occasionally technical requirements to prove interesting theorems.
Example. ℝ2 is a topological surface. Any open subset of ℝ2 is also a topological surface.
For example, ℝ2 ∖{0} and ℝ2 ∖{(0, 0)} ∪{(0, 1 𝑛)∶𝑛= 1, 2, … } are topological surfaces.
Example. Let 𝑓∶ℝ2 →ℝbe a continuous function. The graph of 𝑓, denoted Γ𝑓, is defined by Γ𝑓= {(𝑥, 𝑦, 𝑓(𝑥, 𝑦))∶(𝑥, 𝑦) ∈ℝ2} with the subspace topology when embedded in ℝ3. Recall that a product topology 𝑋×𝑌has the feature that 𝑓∶𝑍→𝑋×𝑌is continuous if and only if 𝜋𝑥∘𝑓∶𝑍→𝑋and 𝜋𝑦∘𝑓∶𝑍→𝑌 are continuous. Hence, any graph Γ ⊆𝑋× 𝑌is homeomorphic to 𝑋if 𝑓is continuous.
Indeed, the projection 𝜋𝑥projects each point in the graph onto the domain. The function 𝑠∶𝑥↦(𝑥, 𝑓(𝑥)) is continuous by the above. In particular, in our case, the graph Γ𝑓is homeomorphic to ℝ2, which we know is a surface.
Remark. As a topological surface, Γ𝑓is independent of the function 𝑓. However, we will later introduce more ways to describe topological spaces that will ascribe new properties to Γ𝑓which do depend on 𝑓.
Example. The sphere: 𝑆2 = {(𝑥, 𝑦, 𝑧) ∈ℝ3 ∶𝑥2 + 𝑦2 + 𝑧2 = 1} is a topological surface, when using the subspace topology in ℝ3. Consider the stereographic projection of 𝑆2 onto ℝ2 from the north pole (0, 0, 1). The projection satisfies 𝜋+ ∶𝑆2 ∖ {(0, 0, 1)} and (𝑥, 𝑦, 𝑧) ↦( 𝑥 1 −𝑧, 𝑦 1 −𝑧) 546 1. Surfaces Certainly, 𝜋+ is continuous, since we do not consider the point (0, 0, 1) to be in its domain.
The inverse map is given by (𝑢, 𝑣) ↦( 2𝑢 𝑢2 + 𝑣2 + 1, 2𝑣 𝑢2 + 𝑣2 + 1, 𝑢2 + 𝑣2 −1 𝑢2 + 𝑣2 + 1) This is also a continuous function. Hence 𝜋+ is a homeomorphism. Similarly, we can con-struct the stereographic projection from the south pole, 𝜋−. This is a homeomorphism.
Hence, every point in 𝑆2 lies either in the domain of 𝜋+ or 𝜋−, and hence sits in an open set 𝑆2 ∖{(0, 0, 1)} or 𝑆2 ∖{(0, 0, −1)} which is homeomorphic to ℝ2.
Remark. 𝑆2 is compact by the Heine–Borel theorem; it is a closed bounded set in ℝ3.
Example. The real projective plane is a topological surface. The group ℤ/2 acts on 𝑆2 by homeomorphisms via the antipodal map 𝑎∶𝑆2 →𝑆2, mapping 𝑥↦−𝑥. There exists a homeomorphism ℤ/2 to the group Homeo(𝕊2) of homeomorphisms of 𝑆2, by mapping 1 + ℤ↦𝑎. We now define the real projective plane to be the quotient of 𝑆2 given by identifying every point 𝑥with its image −𝑥under 𝑎.
ℝℙ2 = 𝑆2 ⟋ ℤ/2 = 𝑆2 ⟋ ∼; 𝑥∼𝑎(𝑥) Lemma. ℝℙ2 bijects with the set of straight lines in ℝ3 through the origin.
Proof. Any line through the origin intersects 𝑆2 exactly in a pair of antipodal points 𝑥, −𝑥.
Similarly, pairs of antipodal points uniquely define a line through the origin.
Lemma. ℝℙ2 is a topological surface.
Proof. We must check that ℝℙ2 is Hausdorff since it is constructed by a quotient, not a sub-space. If 𝑋is a space and 𝑞∶𝑋→𝑌is a quotient map, then by definition 𝑉⊂𝑌is open if and only if 𝑞−1(𝑉) ⊂𝑋is open. If [𝑝] ≠[𝑞] ∈ℝℙ2, then ±𝑝, ±𝑞∈𝑆2 are distinct antipodal pairs. We can therefore construct distinct open discs around 𝑝, 𝑞in 𝑆2, and their antipodal images. These uniquely define open neighbourhoods of [𝑝], [𝑞], which are disjoint.
Similarly, we can check that ℝℙ2 is second countable. We know that 𝑆2 is second countable, so let 𝒰be a countable base for the topology on 𝑆2. Without loss of generality, we can assert that for all sets 𝑈∈𝒰, we have −𝑈∈𝒰. Let 𝒰be the family of open sets in ℝℙ2 of the form 𝑞(𝑈) ∪𝑞(−𝑈) for 𝑈∈𝒰, where 𝑞is the quotient map. Now, if 𝑉⊆ℝℙ2 is open, then by definition 𝑞−1(𝑉) is open in 𝑆2 hence 𝑞−1(𝑉) contains some 𝑈∈𝒰and hence contains 𝑈∪(−𝑈). Hence 𝒰is a countable base for the quotient topology on ℝℙ2.
Finally, let 𝑝∈𝑆2 and [𝑝] ∈ℝℙ2 its image. Let 𝐷be a small (contained in an open hemi-sphere) closed disc, which is a neighbourhood of 𝑝∈𝑆2. The quotient map restricted to 𝐷, written 𝑞|𝐷∶𝐷→𝑞(𝐷) ⊂ℝℙ2, is a continuous function from a compact space to a Haus-dorff space. Further, 𝑞is injective on 𝐷since the disc was contained entirely in a single hemisphere. The topological inverse function theorem states that a continuous bijection 547 X. Geometry from a compact space to a Hausdorff space is a homeomorphism. So 𝑞|𝐷is a homeomorph-ism from 𝐷to 𝑞(𝐷). This then induces the homeomorphism 𝑞|𝐷from the open disc 𝐷= 𝐷 ∘ to 𝑞(𝐷). So by construction, [𝑝] ∈𝑞(𝐷); it has an open neighbourhood in ℝℙ2 which is homeomorphic to an open disc, concluding the proof.
Example. Let 𝑆1 be the unit circle in ℂ, and then we define the torus to be the product space 𝑆1 × 𝑆1, with the subspace topology from ℂ2 (which is identical to the product topology).
Lemma. The torus is a topological surface.
Proof. Consider the map 𝑒∶ℝ2 →𝑆1 × 𝑆1 defined by (𝑠, 𝑡) ↦(𝑒2𝜋𝑖𝑠, 𝑒2𝜋𝑖𝑡) Note that this induces a map ̂ 𝑒from ℝ2 ⟋ ℤ2, since 𝑒is constant under translations by ℤ2.
ℝ2 𝑆1 × 𝑆1 ℝ2 ⟋ ℤ2 𝑞 𝑒 ̂ 𝑒 Under the quotient topology given by the quotient map 𝑞, ℝ2 ⟋ ℤ2 is a topological space. The map [0, 1]2 →ℝ2 →ℝ2 ⟋ ℤ2 is surjective, so ℝ2 ⟋ ℤ2 is compact. So ̂ 𝑒is a continuous map from a compact space to a Hausdorff space, and ̂ 𝑒is bijective, so ̂ 𝑒is a homeomorphism. We already have that 𝑆1 × 𝑆1 is compact and Hausdorff (as a closed and bounded set in ℂ2), so it suffices to show it is locally homeomorphic to ℝ2. Let [𝑝] = 𝑞(𝑝) ∈𝑆1 × 𝑆1, then we can choose a small disc 𝐷(𝑝) such that 𝐷(𝑝) ∩(𝐷(𝑝) + (𝑛, 𝑚)) = ∅for nonzero (𝑛, 𝑚) ∈ℤ2.
Hence 𝑒|𝐷(𝑝) is injective and 𝑞|𝐷(𝑝) is injective. Now, restricting to the open disc as before, we can find an open disc neighbourhood of [𝑝]. Since [𝑝] was chosen arbitrarily, 𝑆1 × 𝑆1 is a topological surface.
Example. Let 𝑃be a planar Euclidean polygon, with oriented edges. We will pair the edges, and without loss of generality we will assume that paired edges have the same Euclidean length.
𝑏−1 𝑎 𝑏 𝑎−1 𝑏 𝑎 𝑏 𝑎−1 548 1. Surfaces We can assign letter names to each edge pair, and denote a polygon by the sequence of edges found when traversing in a clockwise orientation. The edge pair name is inverted if the edge is traversed in the reverse direction. Note the difference between the annotations on the first two shapes above, due to the reversed direction of the edge. If two edges 𝑒, ̂ 𝑒are paired, this defines a unique Euclidean isometry from 𝑒to ̂ 𝑒respecting the orientation, which will be written 𝑓 𝑒̂ 𝑒∶𝑒→ ̂ 𝑒. The set of all such functions generate an equivalence relation on the polygon, identifying paired edges with each other.
Lemma. 𝑃 ⟋ ∼, with the quotient topology, is a topological surface.
Example. Consider the torus, defined here as 𝑇2 = [0, 1]2 ⟋ ∼. Let 𝑃be the polygon [0, 1]2.
If 𝑝is in the interior of 𝑃, then construct a sufficiently small disc that lies entirely within the interior. The quotient map is injective on the closure of the disc and is a homeomorphism on its interior.
Let 𝑝be on an edge, but not a vertex. Let us say without loss of generality that 𝑝= (0, 𝑦0) ∼ (1, 𝑦0). Let 𝛿be sufficiently small that the closed half-discs 𝑈, 𝑉centred on 𝑝with radius 𝛿do not intersect any vertices. Then we define a map from the union of the two half-discs to the disc 𝐵(0, 𝛿) ⊆ℝ2 via (𝑥, 𝑦) ↦(𝑥, 𝑦−𝑦0) or (𝑥, 𝑦) ↦(𝑥−1, 𝑦−𝑦0), which will be a bijective map. Recall the gluing lemma from Analysis and Topology: that if 𝑋= 𝐴∪𝐵is a union of closed subspaces, and 𝑓∶𝐴→𝑌, 𝑔∶𝐵→𝑌are continuous and 𝑓|𝐴∩𝐵= 𝑔|𝐴∩𝐵, they define a continuous map on 𝑋. Let 𝑓 𝑈, 𝑓 𝑉be the maps on the half-discs 𝑈, 𝑉. By the definition of the quotient topology, 𝑞∘𝑓 𝑈and 𝑞∘𝑓 𝑉are also continuous. On the overlapping area, the functions 𝑞∘𝑓 𝑈and 𝑞∘𝑓 𝑉agree. Hence, by the gluing lemma, we can construct a function 𝑓∶𝑈∪𝑉→𝐵(0, 𝛿). We can show that this is a homeomorphism using the usual process: pass to the closed disc, apply the topological inverse function theorem, then apply the result to the interior. If [𝑝] ∈𝑇2 lies on the image of an edge in [0, 1]2, it has indeed a neighbourhood homeomorphic to a disc.
Now it suffices to consider points 𝑝on a vertex. All four vertices of the square are identified to the same point in the torus. A neighbourhood of each vertex can be identified with a quarter-disc in ℝ2. We can repeatedly apply the gluing lemma to construct the whole disc 𝐵(0, 𝛿) ⊆ℝ2 and complete the argument as before.
Thus, [0, 1]2 ⟋ ∼is a topological surface.
We can generalise this proof to an arbitrary planar Euclidean polygon 𝑃, such as the hexagon above. The equivalence relation 𝑥∼𝑓 𝑒̂ 𝑒(𝑥) induces an equivalence relation on the vertices of 𝑃, by considering the images of the vertices under all 𝑓 𝑒̂ 𝑒. However, it is not necessarily the case that an equivalence class of vertices contains exactly four vertices, so quarter-discs are not necessarily applicable. Again, there are three types of point: • interior points, for which a neighbourhood not intersecting the boundary is chosen; • points on edges, for which a corresponding point exists and two half-discs can be glued to form the neighbourhood; and 549 X. Geometry • points on vertices. For this case, all vertices of the polygon have a neighbourhood which is a sector of a circle. Let there be 𝑟vertices in a given equivalence class. Let 𝛼 be the sum of the angles of the sectors in a given class. Any sector can be identified with a given sector in the disc 𝐵(0, 𝛿) ⊆ℝ2, which we will choose to have angle 𝛼/𝑟.
Then, we can glue each sector together in ℝ2, compatibly with the orientations of the edges and arrows, inducing a neighbourhood which is locally homeomorphic to a disc.
If 𝑟= 1, we have an equivalence class comprising a single vertex, which gives a single sector. For 𝑟to be one, the two edges attached to this vertex must be paired and have the same direction (either both inwards or outwards from the vertex). This quotient space is simply a cone, which is homeomorphic to a disc as required.
We can also show that the quotient space is Hausdorff and second countable. By construc-tion, two distinct points in the quotient space can be separated by open neighbourhoods by selecting a sufficiently small radius such that the discs considered in the derivation above are disjoint. For second countability, consider • discs in the interior of 𝑃with rational centres and radii; • for each edge of 𝑃, consider an isometry 𝑒→[0, ℓ] where ℓis the length of 𝑒, taking discs on 𝑒which are centred at rational values in [0, ℓ]; and • for each vertex, consider discs centred at these vertices with rational radii.
Example. Given topological surfaces Σ1, Σ2 we can remove an open disc from each and glue the resulting circles. Explicitly, we form a quotient relation on the disjoint union of the surfaces with the discs removed. This process is known as forming the connect sum of the surfaces, written Σ1 # Σ2. Typically, the information about where the discs were removed from is discarded when considering the connect sum. The connect sum of two topological surfaces is a topological surface.
Example. Consider the following octagon.
The associated quotient space 𝑃 ⟋ ∼can be seen to be homeomorphic to a surface with two holes, known as a double torus. All vertices are identified as the same vertex in the quotient space. We can cut the octagon along a diagonal, leaving two topological surfaces which are homeomorphic to a torus.
↦ 550 1. Surfaces Thus, the connect sum of the two half-octagons are the connect sum of two toruses.
Example. Consider the following square.
This is homeomorphic to the real projective plane ℝℙ2. This is because we identify points on the boundary with their antipodes, when interpreting the square as the closed disc 𝐵(0, 1).
The real projective plane was constructed by identifying points on the unit sphere with their antipodes. Thus, we can construct a homeomorphism by considering only points in the upper hemisphere (taking antipodes as required), and then orthographically projecting onto the 𝑥𝑦plane. Under this transformation, points on the boundary are identified with their antipodes as required.
1.2. Subdivisions Definition. A subdivision of a compact topological surface Σ comprises (i) a finite subset 𝑉⊆Σ of vertices; (ii) a finite subset 𝐸= {𝑒𝑖∶[0, 1] →Σ} which are continuous injections and pairwise dis-joint except perhaps at the endpoints; (iii) such that each connected component of the complement of 𝑉∪𝐸in Σ is homeo-morphic to an open disc, and each such component will be called a face. In particular, the boundary of each face has boundary inside the union of the edges and the vertices.
We say that a subdivision is a triangulation if each closed face (closure of a face) contains exactly three edges, and two closed faces meet either at exactly one edge or not at any edges.
Example. A cube displays a subdivision of 𝑆2. A tetrahedron displays a triangulation of 𝑆2.
Example. We can display subdivisions of surfaces constructed from polygons.
𝑏−1 𝑎 𝑏 𝑎−1 This is a subdivision of a torus with one edge, two edges, and one face. We can construct additional subdivisions of a torus, for example: 551 X. Geometry The first of these examples is not a triangulation, since the two faces meet in more than one edge. The second is a triangulation.
Remark. The following is a very degenerate subdivision of 𝑆2.
• This has one vertex, no edges, and one face.
1.3. Euler classification Definition. The Euler characteristic of a subdivision is #𝑉−#𝐸+ #𝐹 Theorem.
(i) Every compact topological surface has a subdivision (and indeed triangu-lations).
(ii) The Euler characteristic is invariant under choice of subdivision, and is topologically invariant.
Hence, we might say that a surface has a particular Euler characteristic, without referring to subdivisions. We write this 𝜒(Σ).
No proof will be given.
Example. The Euler characteristic of 𝑆2 is 𝜒(𝑆2) = 2. For the torus, 𝜒(𝑇2) = 0. If Σ1, Σ2 are compact surfaces, then the connect sum Σ1 # Σ2 can be constructed by removing a face of a triangulation, then gluing together the boundary circles (three edges) in a way that matches the edges. Then the connect sum inherits a subdivision, and we can find that it has Euler characteristic 𝜒(Σ1 # Σ2) = 𝜒(Σ1) + 𝜒(Σ2) −2, where the remaining term corresponds to the two faces that were removed; the changes of three vertices and three edges cancel each other. In particular, a surface Σ𝑔with 𝑔holes can be written# 𝑔 𝑖=1𝑇2, so 𝜒(Σ𝑔) = 2 −2𝑔. We call 𝑔the genus of Σ.
Remark. It is not trivial to prove part (i). For part (ii), note that subdivisions can be converted into triangulations by constructing triangle fans.
552 1. Surfaces Triangulations can be related by local moves, such as It is easy to check that both of these moves do not change the Euler characteristic. However, it is hard to make this argument rigorous, and it does not give much explanation for why the result is true. In Part II Algebraic Topology, a more advanced definition of the Euler characteristic is given, which admits a more elegant proof.
553 X. Geometry 2. Smooth surfaces 2.1. Charts and atlases Recall that if Σ is a topological surface, any point lies in an open neighbourhood homeo-morphic to a disc.
Definition. A pair (𝑈, 𝜑), where 𝑈is an open set in Σ and 𝜑∶𝑈→𝑉is a homeomorphism to an open set 𝑉⊆ℝ2, is called a chart for Σ. If 𝑝∈𝑈, we might say that (𝑈, 𝜑) is a chart for Σ at 𝑝. A collection of charts whose domains cover Σ is known as an atlas for Σ. The inverse 𝜎= 𝜑−1 ∶𝑉→𝑈is known as a local parametrisation for the surface.
Example. If 𝑍⊆ℝ2 is closed, ℝ2 ∖𝑍is a topological surface with an atlas containing one chart, (ℝ2 ∖𝑍, 𝜙= id).
For 𝑆2, there is an atlas with two charts, which are the two stereographic projections from the poles. We could consider alternative charts, for instance the projection to the 𝑦𝑧plane, but this would be insufficient for describing the poles.
Definition. Let (𝑈𝑖, 𝜑𝑖) be charts containing the point 𝑝∈Σ, for 𝑖= 1, 2. Then the map ∗∶𝜑1(𝑈1 ∩𝑈2) →𝜑2(𝑈1 ∩𝑈2); ∗= 𝜑2 ∘𝜑−1 1 | | |𝜑1(𝑈1∩𝑈2) converts between the corresponding charts, and is called a transition map. This is a homeo-morphism of open sets in ℝ2.
Recall from Analysis and Topology that if 𝑉⊆ℝ𝑛and 𝑉′ ⊆ℝ𝑚are open, then a continuous map 𝑓∶𝑉→𝑉′ is called smooth if it is infinitely differentiable. Equivalently, it is smooth if partial derivatives of all orders in all variables exist at all points. If 𝑛= 𝑚, then in particular the homeomorphism 𝑓∶𝑉→𝑉′ is called a diffeomorphism if it is smooth and has smooth inverse.
Definition. An abstract smooth surface is a topological space Σ together with an atlas of charts (𝑈𝑖, 𝜑𝑖) such that all transition maps 𝜑𝑖∘𝜑−1 𝑗∶𝜑𝑗(𝑈𝑖∩𝑈𝑗) →𝜑𝑖(𝑈𝑖∩𝑈𝑗) are diffeomorphisms.
Remark. We could not simply consider a smoothness condition for Σ itself without appealing to atlases, since Σ is an arbitrary topological space and could have almost any topology.
Example. The atlas of two charts with stereographic projections gives 𝑆2 the structure of an abstract smooth surface.
Example. For the torus 𝑇2 = ℝ2 ⟋ ℤ2, we can find charts of all points by choosing sufficiently small discs in ℝ2 such that they do not intersect any of their non-trivial integer translates.
The transition maps for this atlas are all translations of ℝ2. Hence 𝑇2 inherits the structure of an abstract smooth surface. Explicitly, let us define 𝑒∶ℝ2 →𝑇2 by (𝑡, 𝑠) ↦(𝑒2𝜋𝑖𝑡, 𝑒2𝜋𝑖𝑠), then consider the atlas {(𝑒(𝐷𝜀(𝑥, 𝑦)), 𝑒−1 on this image)} 554 2. Smooth surfaces for 𝜀< 1 3. These are charts on 𝑇2, and the transition maps are (restricted to appropriate domains) translations in ℝ2. Hence 𝑇2, via this atlas, has the structure of an abstract smooth surface.
Remark. The definition of a topological surface is a notion of structure. One can observe a topological space and determine whether it is a topological surface. Conversely, to be an abstract smooth surface is to have a specific set of data; that is, we must provide charts for the surface in order to see that it is indeed an abstract smooth surface.
Definition. Let Σ be an abstract smooth surface, and 𝑓∶Σ →ℝ𝑛be a continuous map. We say that 𝑓is smooth at 𝑝∈Σ if, for all charts (𝑈, 𝜑) of 𝑝belonging to the smooth atlas for Σ, the map 𝑓∘𝜑−1 ∶𝜑(𝑈) →ℝ𝑛 is smooth at 𝜑(𝑝) ∈ℝ2.
Remark. Note that the choice of chart and atlas was arbitrary, but smoothness of 𝑓at 𝑝is independent of the choice of chart, since the transition maps between two such charts are diffeomorphisms.
Definition. Let Σ1, Σ2 be abstract smooth surfaces. Then a map 𝑓∶Σ1 →Σ2 is smooth if it is ‘smooth in the local charts’. Given a chart (𝑈, 𝜑) at 𝑝and a chart (𝑈′, 𝜓) at 𝑓(𝑝), both mapping to open subsets of ℝ2, the map 𝜓∘𝑓∘𝜑−1 is smooth at 𝜑(𝑝). Smoothness of 𝑓does not depend on the choice of chart, provided that the charts all belong to the same atlas.
Definition. Two surfaces Σ1, Σ2 are diffeomorphic if there exists a homeomorphism 𝑓∶Σ1 → Σ2 which is smooth and has smooth inverse.
Remark. Often, we convert from a given smooth atlas for an abstract smooth surface Σ to the maximal compatible smooth atlas. That is, we consider the atlas with the maximal pos-sible set of charts, all of which have transition maps that are diffeomorphisms. This can be accomplished formally by use of Zorn’s lemma.
555 X. Geometry 3. Smooth surfaces in ℝ3 3.1. Definitions and equivalent characterisations Recall that if 𝑉⊆ℝ𝑛and 𝑉′ ⊆ℝ𝑚, then 𝑓∶𝑉→𝑉′ is smooth if it is infinitely differenti-able.
Definition. If 𝑍is an arbitrary subset of ℝ𝑛, we say that a continuous function 𝑓∶𝑍→ℝ𝑚 is smooth at 𝑝∈𝑍if there exists an open ball 𝑝∈𝐵⊆ℝ𝑛and a smooth map 𝐹∶𝐵→ℝ𝑚 which extends 𝑓such that they agree on 𝐵∩𝑍. In other words, 𝑓is locally the restriction of a smooth map defined on an open set.
Definition. Let 𝑋⊆ℝ𝑛and 𝑌⊆ℝ𝑚. We say that 𝑋and 𝑌are diffeomorphic if there exists a continuous function 𝑓∶𝑋→𝑌such that 𝑓is a smooth homeomorphism with smooth inverse.
Definition. A smooth surface in ℝ3 is a subspace of ℝ3 such that for all points 𝑝∈Σ, there exists an open subset 𝑝∈𝑈⊆Σ that is diffeomorphic to an open set in ℝ2. In other words, for all 𝑝∈Σ, there exists an open ball 𝑝∈𝐵⊆ℝ3 such that if 𝑈= 𝐵∩Σ and there exists a map 𝑓∶𝐵→𝑉⊆ℝ2 such that 𝑓|𝑈∶𝑈→𝑉is a homeomorphism, and the inverse map 𝑉→𝑈⊆Σ ⊆ℝ3 is smooth.
Definition. Let 𝜎∶𝑉→𝑈where 𝑉⊆ℝ2 is open and 𝑈⊆Σ ⊆ℝ3 is open in Σ, such that 𝜎is a smooth homeomorphism and 𝐷𝜎|𝑥has rank 2 for all 𝑥∈𝑉. Then 𝜎is called an allowable parametrisation. If 𝜎(0) = 𝑝, we say that 𝜎is an allowable parametrisation near 𝑝.
Theorem. For a subset Σ ⊆ℝ3, the following are equivalent.
(a) Σ is a smooth surface in ℝ3; (b) Σ is locally the graph of a smooth function, over one of the three coordinate planes: for all 𝑝∈Σ there exists an open ball 𝑝∈𝐵⊆ℝ3 and an open set 𝑉⊆ℝ2 such that Σ ∩𝐵= {(𝑥, 𝑦, 𝑔(𝑥, 𝑦))∶𝑔∶𝑉→ℝsmooth} or one of the other coordinate planes; (c) Σ is locally cut out by a smooth function: for all 𝑝∈Σ there exists an open ball 𝑝∈ 𝐵⊆ℝ3 and a smooth function 𝑓∶𝐵→ℝsuch that Σ ∩𝐵= 𝑓−1(0); 𝐷𝑓| | |𝑥 ≠0 for all 𝑥∈𝐵; (d) Σ is locally the image of an allowable parametrisation near all points.
Remark. Part (b) implies that if Σ is a smooth surface in ℝ3, each 𝑝∈Σ belongs to a chart (𝑈, 𝜑) where 𝜑is (the restriction of) one of the three coordinate plane projections 556 3. Smooth surfaces in ℝ3 𝜋𝑥𝑦, 𝜋𝑦𝑧, 𝜋𝑥𝑧from ℝ3 to ℝ2. Consider the transition map between two such charts. If the two charts are based on the same projection such as 𝜋𝑥𝑦, then the transition map is the identity. If they are based on different projections 𝜋𝑥𝑦and 𝜋𝑦𝑧, then the transition map is (𝑥, 𝑦) ↦(𝑥, 𝑦, 𝑔(𝑥, 𝑦)) ↦(𝑦, 𝑔(𝑥, 𝑦)) which has inverse (𝑦, 𝑧) ↦(ℎ(𝑦, 𝑧), 𝑦, 𝑧) ↦(ℎ(𝑦, 𝑧), 𝑦) Hence all of the transition maps between such charts involve projection maps and the smooth maps involved in defining Σ as a graph. This gives Σ the structure of an abstract smooth sur-face.
Some of the relations given in the above theorem are easy to prove, but others come as a result of the inverse function theorem.
3.2. Inverse and implicit function theorems Theorem (inverse function theorem). Let 𝑈⊆ℝ𝑛be open, and 𝑓∶𝑈→ℝ𝑛be continu-ously differentiable. Let 𝑝∈𝑈and 𝑓(𝑝) = 𝑞. Suppose 𝐷𝑓|𝑝is invertible. Then there is an open neighbourhood 𝑉of 𝑞and a differentiable map 𝑔∶𝑉→ℝ𝑛and 𝑔(𝑞) = 𝑝with image an open neighbourhood 𝑈′ ⊆𝑈of 𝑝such that 𝑓∘𝑔= id𝑉. If 𝑓is smooth, then 𝑔is also.
Remark. The chain rule then implies that 𝐷𝑔|𝑞= (𝐷𝑓|𝑝) −1 . The inverse function theorem concerns functions ℝ𝑛→ℝ𝑛, where 𝐷𝑓|𝑝is an isomorphism. If we have a map ℝ𝑛→ℝ𝑚 for 𝑛> 𝑚, then we can discuss the behaviour when 𝐷𝑓|𝑝is surjective. The derivative 𝐷𝑓|𝑝 is an 𝑛× 𝑚matrix, so if it has full rank, up to the permutation of coordinates we have that the last 𝑚columns are linearly independent.
Theorem (implicit function theorem). Let 𝑝= (𝑥0, 𝑦0) be a point in an open set 𝑈⊂ ℝ𝑘× ℝℓ. Let 𝑓∶𝑈→ℝℓsuch that 𝑝↦0 and ( 𝜕𝑓𝑖 𝜕𝑦𝑗) ℓ×ℓ is an isomorphism. Then there is an open neighbourhood 𝑉of 𝑥0 in ℝ𝑘and a continuously differentiable map 𝑔∶𝑉→ℝℓ with 𝑥0 ↦𝑦0 such that if (𝑥, 𝑦) ∈𝑈∩(𝑉× ℝℓ), then 𝑓(𝑥, 𝑦) = 0 ⟺𝑦= 𝑔(𝑥). If 𝑓is smooth, so is 𝑔.
Proof. Let 𝐹∶𝑈→ℝ𝑘× ℝℓbe defined by (𝑥, 𝑦) ↦(𝑥, 𝑓(𝑥, 𝑦)). Then note that 𝐷𝐹= ( 𝐼 ∗ 0 𝜕𝑓𝑖 𝜕𝑦𝑗 ) hence 𝐷𝐹is an isomorphism at (𝑥0, 𝑦0). By the inverse function theorem, 𝐹is locally invert-ible near 𝐹(𝑥0, 𝑦0) = (𝑥0, 𝑓(𝑥0, 𝑦0)) = (𝑥0, 0). Consider an open neighbourhood 𝑉× 𝑉′ ⊆ ℝ𝑘×ℝℓon which this continuously differentiable inverse 𝐺∶𝑉×𝑉′ →𝑈′ ⊆𝑈⊆ℝ𝑘×ℝℓ exists, such that 𝐹∘𝐺= id𝑉×𝑉′. Then, 𝐺(𝑥, 𝑦) = (𝜑(𝑥, 𝑦), 𝜓(𝑥, 𝑦)) ⟹𝐹∘𝐺(𝑥, 𝑦) = (𝜑(𝑥, 𝑦), 𝑓(𝜑(𝑥, 𝑦), 𝜓(𝑥, 𝑦))) = (𝑥, 𝑦) 557 X. Geometry Hence 𝜑(𝑥, 𝑦) = 𝑥. We have 𝑓(𝑥, 𝜓(𝑥, 𝑦)) = 𝑦when (𝑥, 𝑦) ∈𝑉× 𝑉′. This gives 𝑓(𝑥, 𝑦) = 0 ⟺𝑦= 𝜓(𝑥, 0). We then define 𝑔∶𝑉→ℝℓby 𝑥↦𝜓(𝑥, 0).
Example. Let 𝑓∶ℝ2 →ℝbe smooth and 𝑓(𝑥0, 𝑦0) = 0, and suppose 𝜕𝑓 𝜕𝑦≠0 at (𝑥0, 𝑦0).
Then there exists a smooth map 𝑔∶(𝑥0 −𝜀, 𝑥0 + 𝜀) →ℝwith 𝑔(𝑥0) = 𝑦0 and 𝑓(𝑥, 𝑦) = 0 ⟺𝑦= 𝑔(𝑥) for (𝑥, 𝑦) in some open neighbourhood of (𝑥0, 𝑦0). Since 𝑓(𝑥, 𝑔(𝑥)) = 0 in this open neighbourhood, we can differentiate that expression to find 𝑔′(𝑥) = −𝑓 𝑥 𝑓 𝑦 noting that 𝑓 𝑦≠0 in some neighbourhood near (𝑥0, 𝑦0). Note that the level set 𝑓(𝑥, 𝑦) = 0 is implicitly defined by 𝑔, which is a function for which we have an integral expression.
Example. Let 𝑓∶ℝ3 →ℝbe a smooth map with 𝑓(𝑥0, 𝑦0, 𝑧0) = 0. Consider the level set Σ = 𝑓−1(0), assuming that 𝐷𝑓≠0 at (𝑥0, 𝑦0, 𝑧0). Permuting coordinates if necessary, we can assume 𝜕𝑓 𝜕𝑧≠0 at this point. Then there exists an open neighbourhood 𝑉of (𝑥0, 𝑦0) and a smooth function 𝑔∶𝑉→ℝsuch that (𝑥0, 𝑦0) ↦𝑧0 with the property that for an open set (𝑥0, 𝑦0, 𝑧0) ∈𝑈, the set 𝑓−1(0) ∩𝑈= Σ ∩𝑈is the graph of the function 𝑔, which is {(𝑥, 𝑦, 𝑔(𝑥, 𝑦))∶(𝑥, 𝑦) ∈𝑉}.
3.3. Conditions for smoothness We now prove the theorem stated above, relating equivalent conditions for smoothness of a surface Σ.
Proof. First, we show that (b) implies all of the other conditions. If Σ is locally a graph {(𝑥, 𝑦, 𝑔(𝑥, 𝑦))}, we find a chart from the coordinate plane projection 𝜋𝑥𝑦of that graph. Since this projection is smooth and defined on an open neighbourhood of points of Σ in its do-mains, this shows that Σ is a smooth surface in ℝ3 (a). Further, since Σ is locally the given graph, it is cut out by the function 𝑓(𝑥, 𝑦, 𝑧) = 𝑧−𝑔(𝑥, 𝑦) and 𝜕𝑓 𝜕𝑧≠0 (c). Finally, the local parametrisation 𝜎(𝑥, 𝑦) = (𝑥, 𝑦, 𝑔(𝑥, 𝑦)) is allowable; 𝑔is smooth, the partial derivatives of 𝜎 are linearly independent by considering the 𝑥and 𝑦components, and 𝜎is injective where required (d).
Now, we show (a) implies (d). This is simply part of the definition of being a smooth surface in ℝ3, being locally diffeomorphic to ℝ2. In particular, at 𝑝∈Σ, Σ is locally diffeomorphic to ℝ2 and the inverse of such a local diffeomorphism is an allowable parametrisation.
We have already shown (c) implies (b); this was the example of the implicit function theorem provided above.
Finally, we must prove (d) implies (a) and (b), and then the result will hold. Let 𝑝∈Σ and 𝑉be an open set in ℝ2 with an allowable parametrisation to Σ such that 𝜎(0) = 𝑝. If 558 3. Smooth surfaces in ℝ3 𝜎= (𝜎1(𝑢, 𝑣), 𝜎2(𝑢, 𝑣), 𝜎3(𝑢, 𝑣)), we have 𝐷𝜎= ⎛ ⎜ ⎜ ⎝ 𝜕𝜎1 𝜕𝑢 𝜕𝜎1 𝜕𝑣 𝜕𝜎2 𝜕𝑢 𝜕𝜎2 𝜕𝑣 𝜕𝜎3 𝜕𝑢 𝜕𝜎3 𝜕𝑣 ⎞ ⎟ ⎟ ⎠ This has rank 2, hence there exist two rows defining an invertible matrix. Suppose those are the first two rows, and let pr = 𝜋𝑥𝑦be the projection map. Consider pr ∘𝜎∶𝑉→ℝ2. This has isomorphic derivative at zero, so we can apply the inverse function theorem. Hence Σ is locally a graph over the 𝑥𝑦coordinate plane, so (b) holds. Moreover, let 𝜑= pr ∘𝜎, and consider the open ball 𝐵(𝑝, 𝛿) ⊆ℝ3 and a map such that (𝑥, 𝑦, 𝑧) ↦𝜑−1(𝑥, 𝑦) in this ball.
Here, 𝜑∶𝑊→Σ where 𝑊is an open set in pr(𝐵(𝑝, 𝛿)). This is a locally defined map, which is smooth on an open set in ℝ3, which is a smooth inverse to 𝜎. Hence Σ is a smooth surface in ℝ3, so (a) holds.
Example. The unit sphere 𝑆2 in ℝ3 is 𝑓−1(0) for 𝑓(𝑥, 𝑦, 𝑧) = 𝑥2 + 𝑦2 + 𝑧2 −1. For any point on 𝑆2, 𝐷𝑓≠0, so 𝑆2 is a smooth surface.
Example. Let 𝛾∶[𝑎, 𝑏] →ℝ3 be a smooth map with image in the 𝑥𝑧plane, so 𝛾(𝑡) = (𝑓(𝑡), 0, 𝑔(𝑡)) such that 𝛾is injective, 𝛾′ ≠0, and 𝑓> 0. The surface of revolution of 𝛾has allowable parametrisation 𝜎(𝑢, 𝑣) = (𝑓(𝑢) cos 𝑣, 𝑓(𝑢) sin 𝑣, 𝑔(𝑢)) where (𝑢, 𝑣) ∈(𝑎, 𝑏) × (𝜃, 𝜃+ 2𝜋) for a fixed 𝜃. Note that 𝜎𝑢= (𝑓 𝑢cos 𝑣, 𝑓 𝑢sin 𝑣, 𝑔𝑢) and 𝜎𝑣= (−𝑓sin 𝑣, 𝑓cos 𝑣, 0), and we can check ‖𝜎𝑢× 𝜎𝑣‖ = 𝑓2((𝑓′)2 + (𝑔′)2) which is nonzero on 𝛾, so this really is an allowable parametrisation.
Example. The orthogonal group 𝑂(3) acts on 𝑆2 by diffeomorphisms. Indeed, any 𝐴∈𝑂(3) defines a linear (hence smooth) map ℝ3 →ℝ3 preserving 𝑆2. Hence, the induced map on 𝑆2 is by a homeomorphism which is smooth according to the above definition. This is analogous to the action of the Möbius group on 𝑆2 = ℂ∪{∞}.
3.4. Orientability Definition. Let 𝑉, 𝑉′ be open sets in ℝ2. Let 𝑓∶𝑉→𝑉′ be a diffeomorphism. Then at every point 𝑥∈𝑉, 𝐷𝑓|𝑥∈𝐺𝐿(2, ℝ); it is invertible since 𝑓is a diffeomorphism. Let 𝐺𝐿 +(2, ℝ) be the subgroup of matrices with positive determinant. We say that 𝑓is orientation-preserving if its derivative belongs to this subgroup for all points 𝑥∈𝑉.
Definition. An abstract smooth surface Σ is orientable if it admits an atlas {(𝑈𝑖, 𝜑𝑖)} where the transition maps are all orientation-preserving. A choice of such an atlas is an orientation of Σ; Σ can be called oriented when such an orientation is given.
559 X. Geometry Remark. An orientable atlas belongs to a maximal compatible oriented smooth atlas.
Lemma. If Σ1 and Σ2 are diffeomorphic abstract smooth surfaces, then Σ1 is orientable if and only if Σ2 is orientable.
Proof. Let 𝑓∶Σ1 →Σ2 be a diffeomorphism. Suppose Σ2 is orientable and equipped with an oriented smooth atlas. Consider the atlas on Σ1 of charts of the form (𝑓−1(𝑈), 𝜙∘𝑓|𝑓−1(𝑈)), where (𝑈, 𝜓) is a chart at 𝑓(𝑝) in the oriented atlas for Σ2. Then, the transition map between two such charts is exactly a transition map between charts in the Σ2 atlas.
In other words, in the maximal smooth atlas that exists a priori for Σ1, we will allow charts of the form ( ˜ 𝑈, ˜ 𝜓) when for all charts (𝑈, 𝜓) at 𝑓(𝑝) in the Σ2 atlas, the map 𝜓∘𝑓∘( ˜ 𝜓)−1 is orientation-preserving. Informally, if the atlas on Σ2 was maximal as an oriented atlas, we can recover the previous set of charts.
Remark. There is no sensible classification of the set of all smooth surfaces. For instance, ℝ2∖𝑍for a closed set 𝑍can be shown to yield uncountably many types of homeomorphisms.
However, compact smooth surfaces may be classified by their Euler characteristic and their orientability, up to diffeomorphism. This theorem will not be proven in this course.
There is a definition of orientation-preserving homeomorphism that does not rely on the determinant, but that instead relies on some algebraic topology which is not covered in this course. The Möbius band is the surface where the dashed lines represent the absence of edges. It is provable that an abstract smooth surface is orientable if and only if it contains no subsurface homeomorphic to the Möbius band. We can therefore say that a topological surface is orientable if and only if it contains no subsurface (an open set) homeomorphic to a Möbius band.
We can define other structures on an abstract smooth surface by considering smooth atlases such that if 𝜑1𝜑−1 2 is a transition map, then 𝐷(𝜑1𝜑−1 2 ) at 𝑥belongs to a specific subgroup 𝐺≤ 𝐺𝐿(2, ℝ). For example, defining 𝐺= {𝑒} leads to Euclidean surfaces. The group 𝐺𝐿(1, ℂ) identified as a subgroup of 𝐺𝐿(2, ℝ) yields the Riemann surfaces.
Example. For 𝑆2 with the atlas of two stereographic projections, we can find the transition map to be (𝑢, 𝑣) ↦( 𝑢 𝑢2 + 𝑣2 , 𝑣 𝑢2 + 𝑣2 ) on ℝ2 ∖{0}. This has positive determinant, so 𝑆2 is orientable.
For the torus 𝑇2, we previously found an atlas such that the transition maps are translations of ℝ2. Hence the torus is an oriented surface, and also a Euclidean surface.
560 3. Smooth surfaces in ℝ3 3.5. Tangent planes Recall that an affine subspace of a vector space is a translate of a linear subspace.
Definition. Let Σ be a smooth surface in ℝ3, and 𝑝∈Σ. Let 𝜎∶𝑉→𝑈⊆Σ be an allowable parametrisation of Σ near 𝑝, so 𝑉is an open subset of ℝ2 and 𝑈is open in Σ, such that 𝜎(0) = 𝑝. The tangent plane 𝑇 𝑝Σ to 𝑝at Σ is the image of (𝐷𝜎|0) ⊆ℝ3, which is a two-dimensional vector subspace of ℝ3. The affine tangent plane is 𝑝+ 𝑇 𝑝Σ, which is an affine subspace of ℝ3.
Remark. The affine tangent plane is the ‘best’ linear approximation to a surface Σ at a given point.
Lemma. 𝑇 𝑝Σ is independent of the choice of allowable parametrisation.
Proof (i). Suppose 𝜎∶𝑉→𝑈and ̃ 𝜎∶˜ 𝑉→˜ 𝑈are allowable parametrisations with 𝜎(0) = ̃ 𝜎(0) = 𝑝. There exists a transition map 𝜎−1 ∘ ̃ 𝜎, which is a diffeomorphism of open sets in ℝ2. Therefore, ̃ 𝜎= 𝜎∘(𝜎−1 ∘̃ 𝜎) ⏟ ⎵ ⏟ ⎵ ⏟ diffeomorphism Hence 𝐷(𝜎−1 ∘̃ 𝜎)| |0 is an isomorphism. Thus, the images of 𝐷 ̃ 𝜎|0 and 𝐷𝜎|0 agree.
Proof (ii). Let 𝛾∶(−𝜀, 𝜀) →ℝ3 be a smooth map such that 𝛾has image inside Σ, and 𝛾(0) = 𝑝.
We will show that 𝛾′(0) ∈𝑇 𝑝Σ. If 𝜎∶𝑉→𝑈is an allowable parametrisation with 𝜎(0) = 𝑝 as above, and 𝜀is sufficiently small such that Im 𝛾⊆𝑈, then 𝛾(𝑡) = 𝜎(𝑢(𝑡), 𝑣(𝑡)) for some smooth functions 𝑢, 𝑣∶(−𝜀, 𝜀) →𝑉. Then 𝛾′(𝑡) = 𝜎𝑢𝑢′(𝑡) + 𝜎𝑣𝑣′(𝑡) is in the image of 𝐷𝜎|𝑡.
Thus, 𝑇 𝑝Σ = span {𝛾′(0)∶𝛾as above}.
Definition. If Σ is a smooth surface in ℝ3 and 𝑝∈Σ, the normal direction to Σ at 𝑝is (𝑇 𝑝Σ)⟂, the Euclidean orthogonal complement to the tangent plane at 𝑝.
Remark. For all 𝑝∈Σ, there exist exactly two normalised normal vectors.
Definition. A smooth surface in ℝ3 is two-sided if it admits a continuous global choice of unit normal vector.
Lemma. A smooth surface in ℝ3 is orientable (as an abstract smooth surface) if and only if it is two-sided (as a smooth surface in ℝ3).
Proof. Let 𝜎∶𝑉→𝑈⊆Σ be an allowable parametrisation. Let 𝜎(0) = 𝑝. We will define the positive unit normal with respect to 𝜎at 𝑝to be the normal vector 𝑛𝜎(𝑝) with the prop-erty that {𝜎𝑢, 𝜎𝑣, 𝑛𝜎(𝑝)} and {𝑒1, 𝑒2, 𝑒3} are related by a positive determinant change of basis matrix, where {𝑒1, 𝑒2, 𝑒3} are the standard basis vectors. In other words, 𝑛𝜎(𝑝) = 𝜎𝑢× 𝜎𝑣 ‖𝜎𝑢× 𝜎𝑣‖ 561 X. Geometry Consider an alternative parametrisation ̃ 𝜎∶˜ 𝑉→˜ 𝑈, where ̃ 𝜎(0) = 𝑝, such that ̃ 𝜎belongs to the same oriented and smooth atlas as 𝜎. Hence, 𝜎= ̃ 𝜎∘𝜑for some transition map 𝜑. Let 𝐷𝜑| | |0 = (𝛼 𝛽 𝛾 𝛿) Hence, 𝜎𝑢= 𝛼̃ 𝜎𝑢+ 𝛾̃ 𝜎𝑣; 𝜎𝑣= 𝛽̃ 𝜎𝑢+ 𝛿̃ 𝜎𝑣 This gives 𝜎𝑢× 𝜎𝑣= det (𝐷𝜑| | |0 ) ̃ 𝜎𝑢× ̃ 𝜎𝑣 (†) The determinant here is positive since the charts in question belong to an oriented atlas.
Thus, the positive normal depends on the orientation of Σ, but does not depend on the choice of parametrisation. The expression for 𝑛𝜎(𝑝) is continuous since the cross product is con-tinuous, hence Σ is two-sided.
Conversely, if Σ is two-sided and there exists a global continuous choice of normal vector, we can consider the subatlas of the natural smooth atlas with the property that we allow (𝑈, 𝜑) if the associated parametrisation 𝜎= 𝜑−1 satisfies {𝜎𝑢, 𝜎𝑣, 𝑛} is a positive basis for ℝ3, where 𝑛is the given choice of normal. By (†), the transition maps between such charts are orientation-preserving. Hence Σ is orientable.
Lemma. If Σ is a smooth surface in ℝ3 and 𝐴∶ℝ3 →ℝ3 is a smooth map which preserves Σ setwise, then 𝐷𝐴|𝑝∈𝐿(ℝ3, ℝ3) maps 𝑇 𝑝Σ to 𝑇𝐴(𝑝)Σ for 𝑝∈Σ.
Proof. Let 𝛾∶(−𝜀, 𝜀) →ℝ3 be a smooth map such that its image lies on Σ, and 𝛾(0) = 𝑝.
Recall that 𝑇 𝑝Σ is spanned by 𝛾′(0) for such curves 𝛾. Now, consider 𝐴∘𝛾∶(−𝜀, 𝜀) →ℝ3, which also has image Σ, and 𝐷𝐴| | |𝛾(0) ∘𝐷𝛾| | |0 = 𝐷𝐴| | |𝑝 (𝛾′(0)) = 𝐷(𝐴∘𝛾)| | |0 ∈𝑇𝐴(𝑝)Σ Example. Let 𝑆2 be the unit sphere. The normal vector at 𝑝is the line through the origin and 𝑝; indeed, since 𝑆𝑂3 acts transitively on 𝑆2, it suffices to check at one point, such as the north pole. We can choose the outward-facing normal vector to be the positive normal, de-noted 𝑛(𝑝). 𝑆2 is two-sided by the construction of this normal vector, hence 𝑆2 is orientable.
Example. One embedding of the Möbius band in ℝ3 is 𝜎(𝑡, 𝜃) = ((1 −𝑡sin 𝜃 2) cos 𝜃, (1 −𝑡sin 𝜃 2) sin 𝜃, 𝑡cos 𝜃 2) where (𝑡, 𝜃) ∈𝑉 1 = {𝑡∈(− 1 2, 1 2), 𝜃∈(0, 2𝜋)} or (𝑡, 𝜃) ∈𝑉 2 = {𝑡∈(− 1 2, 1 2), 𝜃∈(−𝜋, 𝜋)}.
We begin with the unit circle 𝑥2 + 𝑦2 = 1, for 𝑡= 0. Then, at each point on the circle, we 562 3. Smooth surfaces in ℝ3 consider an open interval of unit length, which will rotate as we move around the circle, such that at the point 𝜃on the circle it has rotated by 𝜃 2. We can check that if 𝜎𝑖is 𝜎on 𝑉𝑖, then 𝜎𝑖is allowable. Further, 𝜎𝑡× 𝜎𝜃= (−cos 𝜃cos 𝜃 2, −sin 𝜃cos 𝜃 2, −sin 𝜃 2) ≡𝑛𝜃 which is already normalised. As 𝜃→0 from above, 𝑛𝜃→(−1, 0, 0). As 𝜃→2𝜋from below, 𝑛𝜃→(1, 0, 0). Hence, the surface is not two-sided.
563 X. Geometry 4. Geometry of surfaces in ℝ3 4.1. First fundamental form Let 𝛾∶(𝑎, 𝑏) →ℝ3 be smooth. The length of 𝛾is 𝐿(𝛾) = ∫ 𝑏 𝑎 ‖𝛾′(𝑡)‖ d𝑡 This result is independent of the choice of parametrisation. Let 𝑠∶(𝐴, 𝐵) →(𝑎, 𝑏) be a monotonically increasing function, and let 𝜏(𝑡) = 𝛾(𝑠(𝑡)). Then 𝐿(𝜏) = ∫ 𝐵 𝐴 ‖𝜏′(𝑡)‖ d𝑡= ∫ 𝐵 𝐴 ‖𝛾′(𝑠(𝑡))‖|𝑠′(𝑡)| d𝑡= ∫ 𝑏 𝑎 ‖𝛾′(𝑡′)‖ d𝑡′ = 𝐿(𝛾) Lemma. If 𝛾∶(𝑎, 𝑏) →ℝ3 is continuously differentiable and 𝛾′(𝑡) ≠0, then 𝛾can be parametrised by arc length.
The proof is left as an exercise. Let Σ be a smooth surface in ℝ3, and let 𝜎∶𝑉→𝑈⊆Σ be an allowable parametrisation. If 𝛾∶(𝑎, 𝑏) →ℝ3 is smooth and its image is contained within 𝑈, then there exist functions (𝑢(𝑡), 𝑣(𝑡))∶(𝑎, 𝑏) →𝑉such that 𝛾(𝑡) = 𝜎(𝑢(𝑡), 𝑣(𝑡)). Hence 𝛾′(𝑡) = 𝜎𝑢𝑢′(𝑡) + 𝜎𝑣𝑣′(𝑡), giving ‖𝛾′(𝑡)‖2 = 𝐸𝑢′(𝑡)2 + 2𝐹𝑢′(𝑡)𝑣′(𝑡) + 𝐺𝑣′(𝑡)2 for functions 𝐸= ⟨𝜎𝑢, 𝜎𝑢⟩; 𝐹= ⟨𝜎𝑢, 𝜎𝑣⟩= ⟨𝜎𝑣, 𝜎𝑢⟩; 𝐺= ⟨𝜎𝑣, 𝜎𝑣⟩ where ⟨⋅, ⋅⟩represents the usual Euclidean inner product. Note that 𝐸, 𝐹, 𝐺depend only on 𝜎and not on 𝛾.
Definition. The first fundamental form of Σ in the parametrisation 𝜎is the expression 𝐸d𝑢2 + 2𝐹d𝑢d𝑣+ 𝐺d𝑣2 This notation is illustrative of the fact that if 𝛾has image in the image of 𝜎(𝑣), we find 𝐿(𝛾) = ∫ 𝑏 𝑎 √𝐸(𝑢′)2 + 2𝐹𝑢′𝑣′ + 𝐺(𝑣′)2 d𝑡 where 𝛾(𝑡) = 𝜎(𝑢(𝑡), 𝑣(𝑡)).
Remark. The Euclidean inner product on ℝ3 provides an inner product on the subspace 𝑇 𝑝Σ.
Choosing a parametrisation 𝜎, we can say 𝑇 𝑝Σ = Im 𝐷𝜎|0 = span {𝜎𝑢, 𝜎𝑣} where 𝜎(0) = 𝑝.
The first fundamental form is a symmetric bilinear form on the tangent spaces 𝑇 𝑝Σ, varying smoothly in 𝑝. However, we choose to express this in a basis coming from the parametrisa-tion 𝜎. In particular, we can think about the matrix expression (𝐸 𝐹 𝐹 𝐺) 564 4. Geometry of surfaces in ℝ3 Example. The plane ℝ2 𝑥𝑦⊂ℝ3 has the parametrisation (𝑢, 𝑣) ↦(𝑢, 𝑣, 0). Hence, 𝜎𝑢= 𝑒1 and 𝜎𝑣= 𝑒2, hence the first fundamental form is d𝑢2 + d𝑣2. We could also use polar coordinates, using 𝜎(𝑟, 𝜃) = (𝑟cos 𝜃, 𝑟sin 𝜃, 0). This parametrises the plane without the origin. This gives 𝜎𝑟= (cos 𝜃, sin 𝜃, 0) and 𝜎𝜃= (−𝑟sin 𝜃, 𝑟cos 𝜃, 0). The first fundamental form is d𝑟2 + 𝑟2 d𝜃2.
Definition. Let Σ, Σ′ be smooth surfaces in ℝ3. We say that they are isometric if there exists a diffeomorphism 𝑓∶Σ →Σ′ that preserves the lengths of all curves. More formally, for every smooth curve 𝛾∶(𝑎, 𝑏) →Σ, the length of 𝛾is the same as the length of 𝑓∘𝛾.
Example. Let Σ′ = 𝑓(Σ) where 𝑓∶ℝ3 →ℝ3 is a global isometry, or rigid motion, of ℝ3; that is, 𝑣↦𝐴𝑣+ 𝑏for an orthogonal matrix 𝐴. These isometries preserve the Euclidean inner product on ℝ3, hence 𝑓preserves length. However, in the definition, we need not map all of ℝ3 to itself, just Σ →Σ′.
Definition. We say that Σ and Σ′ are locally isometric near points 𝑝∈Σ and 𝑞∈Σ′ if there exist open neighbourhoods 𝑈of 𝑝and 𝑉of 𝑞such that 𝑈and 𝑉are isometric. We can also say that Σ and Σ′ are locally isometric if they are locally isometric at all points; that is, each point of Σ is locally isometric to some point on Σ′.
Lemma. Smooth surfaces Σ, Σ′ in ℝ3 are locally isometric near 𝑝∈Σ and 𝑞∈Σ′ if and only if there exist allowable parametrisations 𝜎∶𝑉→𝑈⊆Σ and 𝜎′ ∶𝑉→𝑈′ ⊆Σ′ such that the first fundamental forms are equivalent.
Proof. By definition, the first fundamental form of Σ determines the lengths of all curves on Σ that lie in 𝑈. We will now show that lengths of curves determine the first fundamental form of a parametrisation. Given 𝜎∶𝑉→𝑈, without loss of generality let 𝑉= 𝐵(0, 𝛿) for some 𝛿> 0, where 𝜎(0) = 𝑝. Consider, for all 𝜀< 𝛿, the curve 𝛾𝜀∶[0, 𝜀] →𝑈; 𝑡↦𝜎(𝑡, 0) Then, d d𝜀𝐿(𝛾𝜀) = d d𝜀∫ 𝜀 0 √𝐸(𝑡, 0) d𝑡= √𝐸(𝜀, 0) Hence, d d𝜀 | | |𝜀=0 𝐿(𝛾𝜀) = √𝐸(0, 0) So we can determine 𝐸at 𝑝by looking at lengths of curves. We can similarly consider 𝜒𝜀∶[0, 𝜀] →𝑈; 𝑡↦𝜎(0, 𝑡) which determines 𝐺. Finally, consider 𝜆𝜀∶[0, 𝜀] →𝑈; 𝑡↦𝜎(𝑡, 𝑡) which determines √(𝐸+ 2𝐹+ 𝐺)(0, 0) which gives 𝐹implicitly.
565 X. Geometry Example. The sphere of radius 𝑎, given by {𝑥2 + 𝑦2 + 𝑧2 = 𝑎2}, has an open set with allow-able parametrisation 𝜎(𝑢, 𝑣) = (𝑎cos 𝑢cos 𝑣, 𝑎cos 𝑢sin 𝑣, 𝑎sin 𝑢) where 𝑢∈(− 𝜋 2 , 𝜋 2 ) and 𝑣∈(0, 2𝜋). This parametrises the complement of a half great circle.
Here, 𝜎𝑢= (−𝑎sin 𝑢cos 𝑣, −𝑎sin 𝑢sin 𝑣, 𝑎cos 𝑢); 𝜎𝑣= (−𝑎cos 𝑢sin 𝑣, 𝑎cos 𝑢cos 𝑣, 0) Hence, 𝐸= 𝑎2; 𝐹= 0; 𝐺= 𝑎2 cos2 𝑢 which gives the first fundamental form as 𝑎2 d𝑢2 + 𝑎2 cos2 𝑢d𝑣2 Example. Consider the surface of revolution given by a curve 𝜂(𝑡) = (𝑓(𝑡), 0, 𝑔(𝑡)) rotated about the 𝑧axis. The resulting surface has parametrisation 𝜎(𝑢, 𝑣) = (𝑓(𝑢) cos 𝑣, 𝑓(𝑢) sin 𝑣, 𝑔(𝑢)) Hence, 𝜎𝑢= (𝑓 𝑢cos 𝑣, 𝑓 𝑢sin 𝑣, 𝑔𝑢); 𝜎𝑣= (−𝑓sin 𝑣, 𝑓cos 𝑣, 0) which gives (𝑓2 𝑢+ 𝑔2 𝑢) d𝑢2 + 𝑓2 d𝑣2 Example. Consider the cone with angle arctan 𝑎to the vertical. For 𝑢> 0 and 𝑣∈(0, 2𝜋), we define 𝜎(𝑢, 𝑣) = (𝑎𝑢cos 𝑣, 𝑎𝑢sin 𝑣, 𝑢) The first fundamental form is (1 + 𝑎2) d𝑢2 + 𝑎2𝑢2 d𝑣2 Consider cutting the cone along the line 𝑣= 0 and flattening it into a plane sector. The circumference of the sector is 2𝜋𝑎and the radius is √1 + 𝑎2, hence the angle traced out by the sector is 𝜃0 = 2𝜋𝑎 √1+𝑎2 . We can parametrise this subset of the plane by 𝜎(𝑟, 𝜃) = (√1 + 𝑎2𝑟cos ( 𝑎𝜃 √1 + 𝑎2 ), √1 + 𝑎2𝑟sin ( 𝑎𝜃 √1 + 𝑎2 ), 0) for 𝑟> 0 and 𝜃∈(0, 2𝜋). We can then check that the first fundamental form here is (1 + 𝑎2) d𝑟2 + 𝑟2𝑎2 d𝜃2 566 4. Geometry of surfaces in ℝ3 which matches the first fundamental form for the cone itself. Hence the cone and the plane are locally isometric. However, the cone and plane are not globally isometric, since the two topological spaces are not homeomorphic, so no diffeomorphism that preserves lengths can be constructed.
Lemma. Let Σ be a smooth surface in ℝ3, and let 𝑝∈Σ. Suppose we have two allowable parametrisations 𝜎∶𝑉→𝑈and 𝜎′ ∶𝑉′ →𝑈for the same open neighbourhood of 𝑝. The two parametrisations differ by a transition map 𝐹= 𝜎′−1 ∘𝜎which is a diffeomorphism of open subsets of ℝ2. There exist first fundamental forms for both parametrisations. Then, (𝐸 𝐹 𝐹 𝐺) = (𝐷𝐹)⊺(𝐸′ 𝐹′ 𝐹′ 𝐺′) (𝐷𝐹) Proof. By definition, (𝐸 𝐹 𝐹 𝐺) = (𝜎𝑢⋅𝜎𝑢 𝜎𝑢⋅𝜎𝑣 𝜎𝑣⋅𝜎𝑢 𝜎𝑣⋅𝜎𝑣 ) = (𝐷𝜎)⊺(𝐷𝜎) Now, 𝜎= 𝜎′ ∘𝐹hence the result follows.
4.2. Conformality If 𝑣, 𝑤∈ℝ3, we have 𝑣⋅𝑤= |𝑣| ⋅|𝑤| ⋅cos 𝜃. This allows us to deduce the angle 𝜃between two vectors given their dot product and lengths. This can also be done when 𝑣, 𝑤are in the tangent plane 𝑇 𝑝Σ, and then we can express the angle in terms of the first fundamental form.
Let 𝜎be an allowable parametrisation for Σ near 𝑝, such that 𝐷𝜎|0 evaluates to 𝑣at 𝑣0 and 𝑤at 𝑤0.
cos 𝜃= 𝑣⋅𝑤 |𝑣| ⋅|𝑤| = 𝐼(𝑣0, 𝑤0) √𝐼(𝑣0, 𝑣0)√𝐼(𝑤0, 𝑤0) where 𝐼denotes the first fundamental form of 𝜎at zero.
Lemma. Let Σ be a smooth surface in ℝ3, and let 𝜎∶𝑉→𝑈be an allowable parametrisa-tion of Σ near 𝑝. Then 𝜎is conformal if 𝐸= 𝐺and 𝐹= 0 in the first fundamental form.
Proof. Consider curves 𝛾∶𝑡↦(𝑢(𝑡), 𝑣(𝑡)) and ̃ 𝛾∶𝑡↦( ̃ 𝑢(𝑡), ̃ 𝑣(𝑡)) in 𝑉, where 𝛾(0) = ̃ 𝛾(0) = 0 ∈𝑉. Let 𝜎be a parametrisation 𝑉→𝑈⊆Σ such that 𝜎(0) = 𝑝∈Σ. Then the curves 𝜎∘𝛾and 𝜎∘̃ 𝛾meet at angle 𝜃on Σ, where cos 𝜃= 𝐸̇ 𝑢̇̃ 𝑢+ 𝐹( ̇ 𝑢̇̃ 𝑣+ ̇ 𝑣̇̃ 𝑢) + 𝐺̇ 𝑣̇̃ 𝑣 √𝐸̇ 𝑢2 + 2𝐹̇ 𝑢̇ 𝑣+ 𝐺̇ 𝑣2√𝐸̇̃ 𝑢 2 + 2𝐹̇̃ 𝑢̇̃ 𝑣+ 𝐺̇̃ 𝑣 2 In particular, if 𝜎is conformal, suppose 𝛾(𝑡) = (𝑡, 0) and ̃ 𝛾(𝑡) = (0, 𝑡). Then, we have that the curves meet at 𝜋 2 in 𝑉, so they meet at 𝜋 2 in Σ, so we find that cos 𝜃= 0 ⟹𝐹= 0.
Similarly, if 𝛾(𝑡) = (𝑡, 𝑡) and ̃ 𝛾(𝑡) = (𝑡, −𝑡), we find cos 𝜃= 0 ⟹𝐸= 𝐺.
567 X. Geometry Conversely, suppose there exists a parametrisation 𝜎such that 𝐸= 𝐺and 𝐹= 0. Then, in this parametrisation, the first fundamental form is of the form 𝜌(d𝑢2 + d𝑣2) for 𝜌= 𝐸∶𝑉→ ℝ. Hence, the first fundamental form is a pointwise rescaling of the Euclidean fundamental form d𝑢2 + d𝑣2. Rescaling the plane does not change angles, so 𝜎is conformal as required.
Remark. Conformality in charts is historically important for cartography. The existence of conformal charts is closely connected to Riemann surfaces, which are topological surfaces locally modelled on ℂinstead of ℝ2.
4.3. Area Recall that a parallelogram spanned by vectors 𝑣, 𝑤has area |𝑣× 𝑤| = ⟨𝑣, 𝑣⟩⟨𝑤, 𝑤⟩−⟨𝑣, 𝑤⟩ 2, where × denotes the cross product. Let 𝜎∶𝑉→𝑈⊆Σ be an allowable parametrisation with 𝜎(0) = 𝑝, and consider 𝜎𝑢, 𝜎𝑣∈𝑇 𝑝Σ. The square of the area of the infinitesimal paral-lelogram spanned by 𝜎𝑢, 𝜎𝑣is given by ⟨𝜎𝑢, 𝜎𝑢⟩⟨𝜎𝑣, 𝜎𝑣⟩−⟨𝜎𝑢, 𝜎𝑣⟩ 2 = 𝐸𝐺−𝐹2 Definition. Let Σ be a smooth surface in ℝ3, and 𝜎∶𝑉→𝑈⊆Σ an allowable paramet-risation. Then, area(𝑈) = ∫ 𝑉 √𝐸𝐺−𝐹2 d𝑢d𝑣 Remark. This is independent of parametrisation. Indeed, suppose 𝜎∶𝑉→𝑈and ̃ 𝜎∶˜ 𝑉→ 𝑈are allowable. Then ̃ 𝜎= 𝜎∘𝜑for some transition map 𝜑∶˜ 𝑉→𝑉. We know then that ( ˜ 𝐸 ˜ 𝐹 ˜ 𝐹 ˜ 𝐺) = (𝐷̃ 𝜎)⊺(𝐷̃ 𝜎) = (𝐷𝜑)⊺(𝐸 𝐹 𝐹 𝐺) (𝐷𝜑) Hence, √˜ 𝐸˜ 𝐺−˜ 𝐹2 = |det(𝐷𝜑)|√𝐸𝐺−𝐹2 The usual change of variables formula for integration, combined with the fact that 𝜑is a diffeomorphism, gives ∫ 𝑉 √𝐸𝐺−𝐹2 d𝑢d𝑣= ∫ ˜ 𝑉 √˜ 𝐸˜ 𝐺−˜ 𝐹2 d𝑢d𝑣 Note, we can compute the area of an open set 𝑈⊆Σ, not necessarily lying in a single para-metrisation, by covering the set by a finite amount of open subsets which lie in single charts.
For instance, if Σ is compact, we can compute the area of Σ itself.
568 4. Geometry of surfaces in ℝ3 Example. Consider the graph Σ = {(𝑢, 𝑣, 𝑓(𝑢, 𝑣))∶(𝑢, 𝑣) ∈ℝ2}, where 𝑓∶ℝ2 →ℝis a smooth function. This has a global parametrisation 𝜎(𝑢, 𝑣) = (𝑢, 𝑣, 𝑓(𝑢, 𝑣)). Here, 𝜎𝑢= (1, 0, 𝑓 𝑢) and 𝜎𝑣= (0, 1, 𝑓 𝑣), hence √𝐸𝐺−𝐹2 = √1 + 𝑓2 𝑢+ 𝑓2 𝑣 Let 𝑈𝑅⊆Σ be the part of the graph lying inside the disc 𝐵(0, 𝑅) ⊆ℝ2. Then area(𝑈𝑅) = ∫ 𝐵(0,𝑅) √1 + 𝑓2 𝑢+ 𝑓2 𝑣d𝑢d𝑣≥𝜋𝑅2 with equality exactly when 𝑓 𝑢= 𝑓 𝑣= 0, so 𝑓is constant and 𝑈𝑅is contained inside a plane perpendicular to the 𝑧axis. Hence, the projection from Σ to ℝ2 𝑥𝑦is not area-preserving, unless Σ is a plane perpendicular to the 𝑧axis.
Example. Consider the sphere enclosed exactly by a cylinder. The cylindrically radial pro-jection from the sphere to the cylinder is area-preserving. This is explored further in the example sheets.
4.4. Second fundamental form Let 𝜎∶𝑉→𝑈⊆Σ be allowable. By using Taylor’s theorem, we can write 𝜎(𝑢+ ℎ, 𝑣+ ℓ) = 𝜎(𝑢, 𝑣) + ℎ𝜎𝑢(𝑢, 𝑣) + ℓ𝜎𝑣(𝑢, 𝑣) + 1 2(ℎ2𝜎𝑢𝑢(𝑢, 𝑣) + 2ℎℓ𝜎𝑢𝑣(𝑢, 𝑣) + ℓ2𝜎𝑣𝑣(𝑢, 𝑣)) + 𝑂(ℎ3, ℓ3) where ℎ, ℓare small, and (𝑢+ ℎ, 𝑣+ ℓ) ∈𝑉. Recall that if 𝑝= 𝜎(𝑢, 𝑣), we have 𝑇 𝑝Σ = ⟨{𝜎𝑢, 𝜎𝑣}⟩. Hence, the orthogonal distance from 𝜎(𝑢+ ℎ, 𝑣+ ℓ) to the affine tangent plane 𝑇 𝑝Σ + 𝑝is given by projection to the normal direction.
⟨𝑛, 𝜎(𝑢+ ℎ, 𝑣+ ℓ) −𝜎(𝑢, 𝑣)⟩= 1 2(⟨𝑛, 𝜎𝑢𝑢⟩ℎ2 + 2 ⟨𝑛, 𝜎𝑢𝑣⟩ℎℓ+ ⟨𝑛, 𝜎𝑣𝑣⟩ℓ2) + 𝑂(ℎ3, ℓ3) Definition. The second fundamental form of Σ in the allowable parametrisation 𝜎is the quadratic form 𝐿d𝑢2 + 2𝑀d𝑢d𝑣+ 𝑁d𝑣2 where 𝐿= ⟨𝑛, 𝜎𝑢𝑢⟩; 𝑀= ⟨𝑛, 𝜎𝑢𝑣⟩; 𝑁= ⟨𝑛, 𝜎𝑣𝑣⟩ and 𝑛= 𝜎𝑢× 𝜎𝑣 ‖𝜎𝑢× 𝜎𝑣‖ 569 X. Geometry We can write this as the matrix ( 𝐿 𝑀 𝑀 𝑁) which defined a quadratic form on 𝑇 𝑝Σ which varies smoothly in 𝑝.
Lemma. Let 𝑉be connected and 𝜎∶𝑉→𝑈⊆Σ be an allowable parametrisation such that the second fundamental form vanishes identically with respect to 𝜎. Then 𝑈lies in an affine plane.
Remark. The first fundamental form is a non-degenerate symmetric bilinear form on 𝑇 𝑝Σ, whereas the second fundamental form may be degenerate.
Proof. By definition, ⟨𝑛, 𝜎𝑢⟩= 0 = ⟨𝑛, 𝜎𝑣⟩ Hence, by differentiating, we find 0 = ⟨𝑛𝑢, 𝜎𝑢⟩+ ⟨𝑛, 𝜎𝑢𝑢⟩= ⟨𝑛𝑣, 𝜎𝑣⟩+ ⟨𝑛, 𝜎𝑣𝑣⟩= ⟨𝑛𝑣, 𝜎𝑢⟩+ ⟨𝑛, 𝜎𝑢𝑣⟩ Some of these terms appear in the definition of the second fundamental form: 𝐿= ⟨𝑛, 𝜎𝑢𝑢⟩= −⟨𝑛𝑢, 𝜎𝑢⟩ 𝑀= ⟨𝑛, 𝜎𝑢𝑣⟩= −⟨𝑛𝑣, 𝜎𝑢⟩= −⟨𝑛𝑢, 𝜎𝑣⟩ 𝑁= ⟨𝑛, 𝜎𝑣𝑣⟩= −⟨𝑛𝑣, 𝜎𝑣⟩ If the second fundamental form vanishes, then 𝑛𝑢is orthogonal to 𝜎𝑢, 𝜎𝑣, and 𝑛itself. Since 𝜎𝑢, 𝜎𝑣, 𝑛form a basis for ℝ3, we have 𝑛𝑢= 0. Similarly, 𝑛𝑣= 0, hence 𝑛is constant by the mean value theorem.
Remark. The first fundamental form in parametrisation 𝜎can be written (𝐷𝜎)⊺(𝐷𝜎). We can similarly write the second fundamental form as −(𝐷𝑛)⊺(𝐷𝜎) = ( 𝐿 𝑀 𝑀 𝑁) = −(𝑛𝑢⋅𝜎𝑢 𝑛𝑢⋅𝜎𝑣 𝑛𝑣⋅𝜎𝑢 𝑛𝑣⋅𝜎𝑣 ) Hence, if 𝜎∶𝑉→Σ and ̃ 𝜎∶˜ 𝑉→Σ are allowable parametrisations for an open set 𝑈⊆Σ with transition map 𝜑∶˜ 𝑉→𝑉given by 𝜑= 𝜎−1 ∘ ̃ 𝜎, we can use the above expression to find ( ̃ 𝐿 ˜ 𝑀 ˜ 𝑀 ˜ 𝑁) = ±(𝐷𝜑)⊺( 𝐿 𝑀 𝑀 𝑁) (𝐷𝜑) The change in sign depends on whether the transition map preserves or reverses orientation.
If the normal vectors agree, there is no negative sign.
𝑛𝜎∘𝜑 | | |(˜ 𝑢,˜ 𝑣) = ± 𝑛𝜎 | | |𝜑(˜ 𝑢,˜ 𝑣) for ( ̃ 𝑢, ̃ 𝑣) ∈˜ 𝑉. In particular, if det(𝐷𝜑) < 0, we arrive at a negative sign. If we assume that 𝑉, ˜ 𝑉are connected, the determinant det(𝐷𝜑) does not change sign.
570 4. Geometry of surfaces in ℝ3 Example. Consider the cylinder with allowable parametrisation 𝜎(𝑢, 𝑣) = (𝑎cos 𝑢, 𝑎sin 𝑢, 𝑣) where 𝑢∈(0, 2𝜋), 𝑣∈ℝ. Note that 𝜎𝑢𝑣= 𝜎𝑣𝑣= 0, hence 𝑀= 𝑁= 0. We can show that the second fundamental form is given by (−𝑎 0 0 0) ; −𝑎d𝑢2 4.5. Gauss maps Definition. Let Σ be a smooth oriented surface in ℝ3. The Gauss map 𝑛∶Σ →𝕊2 is the map 𝑝↦𝑛(𝑝), where the normal vector is normalised and hence lies in the unit sphere.
Lemma. The Gauss map is smooth.
Proof. Since smoothness is a local property, it suffices to check the smoothness of the map on an arbitrary parametrised part of Σ. Let 𝜎∶𝑉→𝑈⊆Σ be allowable and compatible with a chosen orientation. Then 𝑛(𝑝) = 𝜎𝑢× 𝜎𝑣 ‖𝜎𝑢× 𝜎𝑣‖ Since 𝜎is allowable, the denominator is non-vanishing. Hence, 𝑛(𝑝) is smooth as required.
Remark. If Σ = 𝐹−1(0) for some function 𝐹∶ℝ3 →ℝwith nonzero derivative 𝐷𝐹at all points 𝑥∈Σ (which was required for Σ to be a smooth surface in ℝ3), then we can explicitly calculate the Gauss map to be 𝑛(𝑝) = ∇𝐹 ‖∇𝐹‖ Note that, 𝑇 𝑝Σ = 𝑇𝑛(𝑝)𝑆2 = (𝑛(𝑝))⟂ since the two planes are orthogonal to the same vector. More concretely, if 𝑣∈𝑇 𝑝Σ is 𝛾′(0) where 𝛾∶(−𝜀, 𝜀) →Σ, 𝛾(0) = 𝑝for a smooth curve 𝛾, we can apply the Gauss map to 𝛾and find 𝑛∘𝛾∶(−𝜀, 𝜀) →𝑆2; (𝑛∘𝛾)(0) = 𝑛(𝑝) Then, by the chain rule, 𝐷𝑛| | |𝑝 (𝑣) = (𝑛∘𝛾)′(0) ∈𝑇𝑛(𝑝)𝑆2 = 𝑇 𝑝Σ Thus, the derivative of the Gauss map is 𝐷𝑛|𝑝∶𝑇 𝑝Σ →𝑇 𝑝Σ. This can be viewed as an en-domorphism of a fixed (with respect to parametrisation choice) two-dimensional subspace of ℝ3.
To summarise, let Σ be an oriented smooth surface in ℝ3. Then, 571 X. Geometry (i) The first fundamental form is a symmetric bilinear form ⟨⋅, ⋅⟩= I𝑝∶𝑇 𝑝Σ×𝑇 𝑝Σ →ℝ, which is the restriction of the Euclidean inner product to this space 𝑇 𝑝Σ. We can write I𝑝(𝑣, 𝑤), where 𝑣, 𝑤∈𝑇 𝑝Σ.
(ii) The second fundamental form is also a symmetric bilinear form I I𝑝∶𝑇 𝑝Σ × 𝑇 𝑝Σ →ℝ, given by I I𝑝(𝑣, 𝑤) = I𝑝(−𝐷𝑛| | |𝑝 (𝑣), 𝑤) where 𝑛is the Gauss map.
If we choose an allowable parametrisation (which for the second fundamental form must be correctly oriented) 𝜎∶𝑉→𝑈⊆Σ near 𝑝∈Σ, and if 𝐷𝜎| | |0 ( ̂ 𝑣) = 𝑣; 𝐷𝜎| | |0 ( ̂ 𝑤) = 𝑤; 𝜎(0) = 𝑝 Then, I𝑝(𝑣, 𝑤) = ̂ 𝑣⊺(𝐸 𝐹 𝐹 𝐺) ̂ 𝑤; I I𝑝(𝑣, 𝑤) = ̂ 𝑣⊺( 𝐿 𝑀 𝑀 𝑁) ̂ 𝑤 where 𝐸, 𝐹, 𝐺, 𝐿, 𝑀, 𝑁depend on the choice of 𝜎. Note that the functions I𝑝and I I𝑝are independent of 𝜎.
Lemma. The derivative of the Gauss map is self-adjoint. More precisely, viewing the map 𝐷𝑛|𝑝∶𝑇 𝑝Σ →𝑇 𝑝Σ as an endomorphism over the inner product space with the first funda-mental form, this linear map satisfies I𝑝(𝐷𝑛| | |𝑝 (𝑣), 𝑤) = I𝑝(𝑣, 𝐷𝑛| | |𝑝 (𝑤)) for all 𝑣, 𝑤∈𝑇 𝑝Σ.
Proof. From expressions for local parametrisations, we can show that I𝑝and I I𝑝are symmet-ric. Hence, I𝑝(𝐷𝑛| | |𝑝 (𝑣), 𝑤) = −I I𝑝(𝑣, 𝑤) = −I I𝑝(𝑤, 𝑣) = I𝑝(𝐷𝑛| | |𝑝 (𝑤), 𝑣) = I𝑝(𝑣, 𝐷𝑛| | |𝑝 (𝑤)) Remark. The fundamental theorem of surfaces in ℝ3 states that a smooth oriented connected surface in ℝ3 is determined completely, up to rigid motion, by the two fundamental forms.
572 4. Geometry of surfaces in ℝ3 4.6. Gauss curvature Definition. Let Σ be a smooth surface in ℝ3. The Gauss curvature 𝜅∶Σ →ℝof Σ is the function defined by 𝜅(𝑝) = det(𝐷𝑛| | |𝑝 ) Remark. This is always well-defined, even if Σ is not oriented. This is because Σ is always locally orientable, and the two normals differ by sign. In two dimensions, det(−𝐴) = det(𝐴), so the determinant is invariant.
We can compute 𝜅directly. Let Σ be a smooth surface in ℝ3, and 𝜎an allowable paramet-risation for an open neighbourhood of a point 𝑝. Recall that I𝑝∶𝑇 𝑝Σ →𝑇 𝑝Σ; (𝑣, 𝑤) ↦⟨𝑣, 𝑤⟩; I I𝑝∶𝑇 𝑝Σ →𝑇 𝑝Σ; (𝑣, 𝑤) ↦I𝑝(−𝐷𝑛| | |𝑝 (𝑣), 𝑤) and 𝐷𝑛|𝑝∶𝑇 𝑝Σ →𝑇 𝑝Σ. The choice of parametrisation 𝜎for an open neighbourhood 𝑈of 𝑝provides a preferred basis {𝜎𝑢, 𝜎𝑣} for 𝑇 𝑝Σ. We can therefore write the fundamental forms as matrices with respect to this basis. Let 𝐴= I𝑝, 𝐵= I I𝑝, 𝕊= 𝐷𝑛|𝑝in this basis. In matrix form, we can write I I𝑝= I𝑝(−𝐷𝑛|𝑝(𝑣), 𝑤) as 𝐵= −𝕊⊺𝐴⟹𝜅(𝑝) = det(𝕊) = det(−𝐴−1𝐵) = 𝐿𝑁−𝑀2 𝐸𝐺−𝐹2 If 𝜎, ̃ 𝜎are allowable and 𝜑= 𝜎−1 ∘𝜎is a transition map, then ˜ 𝐴= (𝐷𝜑)⊺𝐴(𝐷𝜑); ˜ 𝐵= ±(𝐷𝜑)⊺𝐵(𝐷𝜑) Since the sign vanishes under taking determinants, 𝜅is intrinsic and does not depend on the choice of parametrisation.
Example. For a cylinder {𝑥2 + 𝑦2 = 1} the Gauss map 𝑛∶Σ →𝑆2 has image which lies in the equator. Its derivative 𝐷𝑛|𝑝∶𝑇 𝑝Σ →𝑇 𝑝Σ has one-dimensional image, since any 𝛾∶(−𝜀, 𝜀) →Σ has 𝑛∘𝛾⊆𝑆1. Hence its Gauss curvature is zero.
Definition. A smooth surface in ℝ3 with vanishing Gauss curvature everywhere is flat.
Remark. If 𝜎∶𝑉→𝑈is allowable, and 𝑛𝜎is defined to be 𝑛∘𝜎∶𝑉→𝑆2, then 𝐷𝑛𝜎 | | |0 ∶𝜎𝑢↦(𝑛𝜎)𝑢; 𝜎𝑣↦(𝑛𝜎)𝑣 In particular, 𝜅(𝑝) = 𝜅(𝜎(0)) vanishes if and only if (𝑛𝜎)𝑢×(𝑛𝜎)𝑣= 0. Usually, we will write 𝑛to denote 𝑛𝜎. In this case, the condition for flatness is that 𝑛𝑢× 𝑛𝑣= 0.
Example. If Σ is the graph of a smooth function 𝑓, then on the example sheets we show that 𝜅= 𝑓 𝑢𝑢𝑓 𝑣𝑣−𝑓2 𝑢𝑣 (1 + 𝑓2 𝑢+ 𝑓2 𝑣)2 573 X. Geometry Hence, the curvature depends on the derivative and the Hessian of 𝑓. For instance, let 𝑓(𝑢, 𝑣) = √𝑟2 −𝑢2 −𝑣2. Here, the graph is a piece of a sphere of radius 𝑟. We can find 𝑓 𝑢𝑢 | | |0 = 𝑓 𝑣𝑣 | | |0 = −1 𝑟; 𝑓 𝑢𝑣 | | |0 = 0 ⟹𝜅(0, 0, 𝑟) = 1 𝑟2 Since 𝑂(3) acts transitively on 𝑆2, and the fundamental forms are preserved by such global isometries, 𝜅= 1 𝑟2 everywhere on the sphere of radius 𝑟.
Example. Let Σ be the smooth surface given by {𝑧= 𝑥2 + 𝑦2}. We claim that, for the inward facing choice of orientation, the image of the Gauss map is the open northern hemisphere.
Note that Σ is invariant under rotations about the 𝑧axis. Also, we can show that if 𝑅is a rotation, 𝑛∘𝑅= 𝑅∘𝑛. Therefore, it suffices to consider an arbitrary point with 𝑦= 0.
Here, Σ = 𝐹−1(0) for the function 𝐹(𝑥, 𝑦, 𝑧) = 𝑧−𝑥2−𝑦2, which has nonvanishing derivative at the points 𝑝∈Σ. Hence, at 𝑝= (𝑥, 0, 𝑥2), we have 𝑛(𝑝) = ∇𝐹 ‖∇𝐹‖ = (−2𝑥, 0, 1) √1 + 4𝑥2 We can check explicitly that this map has image which an arc lying in the open northern hemisphere.
4.7. Elliptic, hyperbolic, and parabolic points Definition. Let Σ be a smooth surface in ℝ3 and 𝑝∈Σ. We say that 𝑝is (i) elliptic if 𝜅(𝑝) > 0; (ii) hyperbolic if 𝜅(𝑝) < 0; (iii) parabolic if 𝜅(𝑝) = 0.
Lemma. In a sufficiently small neighbourhood of an elliptic point 𝑝, Σ lies entirely on one side of 𝑝+ 𝑇 𝑝Σ. If 𝑝is hyperbolic, Σ lies on both sides of 𝑝+ 𝑇 𝑝Σ.
Proof. Let 𝜎be a local parametrisation near 𝑝. Here, 𝜅= 𝐿𝑁−𝑀2 𝐸𝐺−𝐹2 The denominator is always positive, since it is the determinant of a positive definite symmet-ric bilinear form I𝑝. Hence, the sign of 𝜅depends on the sign of 𝐿𝑁−𝑀2. If 𝑤= ℎ𝜎𝑢+ℓ𝜎𝑣∈ 𝑇 𝑝Σ, then 1 2 I I𝑝(𝑤, 𝑤) measures the signed distance from 𝜎(ℎ, 𝑙) to 𝑝+𝑇 𝑝Σ. If 𝑝is elliptic, then I I𝑝has eigenvalues of the same sign, so it is either positive or negative definite at 𝑝. Since I I𝑝varies smoothly in 𝑝, it remains positive or negative definite in a small neighbourhood of 𝑝. Hence, in such a neighbourhood, the signed distance has the same sign as required.
Conversely, if 𝑝is hyperbolic, I I𝑝(𝑤, 𝑤) takes both signs in a neighbourhood of 𝑝.
574 4. Geometry of surfaces in ℝ3 Remark. We cannot conclude anything about parabolic points a priori. For instance, the cylinder is flat (all points are parabolic), and the surface lies on one side of the tangent plane at every point. Consider also the monkey saddle defined by 𝜎(𝑢, 𝑣) = (𝑢, 𝑣, 𝑢3 −3𝑣2𝑢) which has a parabolic point at the origin, but Σ lies on both sides of the tangent plane in every open neighbourhood of the origin. At 𝑝= 𝜎(0, 0), the Gauss curvature vanishes, but the surface lies locally on both sides of the tangent plane.
Proposition. Let Σ be a compact smooth surface in ℝ3. Then Σ has an elliptic point.
Proof. Since Σ is compact, it is closed and bounded as a subset of ℝ3. Hence, for 𝑅′ suffi-ciently large, Σ lies entirely within 𝐵(0, 𝑅′). Let 𝑅be the minimal such 𝑅′. Up to a global isometry of ℝ3, there exists a point 𝑝= (0, 0, 𝑅) ∈Σ on the sphere 𝑆2(𝑅) of radius 𝑅.
Here, 𝑇 𝑝Σ = 𝑇 𝑝𝑆2. Hence, locally near 𝑝, we can view Σ as the graph of a smooth func-tion 𝑓∶𝑉→ℝ3 on the 𝑥, 𝑦coordinates with the property that 𝑓−√𝑅2 −𝑢2 −𝑣2 ≤0. This expresses the fact that Σ lies underneath the sphere of radius 𝑅.
We can now consider the Taylor series of 𝑓. Note that (0, 0) is a maximum point of 𝑓, hence 𝑓 𝑢= 𝑓 𝑣= 0 at 0. Thus, for sufficiently small 𝑢, 𝑣, 1 2(𝑓 𝑢𝑢𝑢2 + 2𝑓 𝑢𝑣𝑢𝑣+ 𝑓 𝑣𝑣𝑣2) + 1 2𝑅(𝑢2 + 𝑣2) ≤0 Hence, the second fundamental form is locally negative definite near (0, 0). Hence, 𝜅(𝑝) > 0, so 𝑝is elliptic as required. In particular, the curvature at this point is greater than that of the sphere.
Theorem. Let Σ be a smooth surface in ℝ3, and let 𝑝∈Σ such that 𝜅(𝑝) ≠0. Let 𝑈be an open neighbourhood of 𝑝, and a decreasing sequence 𝐴𝑖⊆𝑈of neighbourhoods that ‘shrink to 𝑝’, in the sense that for all 𝜀> 0, 𝐴𝑖⊆𝐵(𝑝, 𝜀) for sufficiently large 𝑖. Then, |𝜅(𝑝)| = lim 𝑖→∞ area𝑆2(𝑛(𝐴𝑖)) areaΣ(𝐴𝑖) In other words, the Gauss curvature is an infinitesimal measure of how much the Gauss map 𝑛distorts area.
Remark. Around hyperbolic points, the signed area of 𝑛(𝐴𝑖) is reversed, since curves 𝛾re-verse direction under 𝑛. We can alternatively define the signed area of 𝑛(𝐴𝑖) to be the area of 𝑛(𝐴𝑖) if 𝜅> 0 and the negation of this area if 𝜅< 0. The above theorem holds when 𝜅= 0, but this will not be proven.
Proof. Let 𝜎be an allowable parametrisation near 𝑝∈Σ. Using 𝜎, we can define the open sets 𝜎−1(𝐴𝑖) = 𝑉𝑖⊂𝑉. Since the 𝐴𝑖shrink to 𝑝, we have that ⋂𝑉𝑖= {(0, 0)}. We have areaΣ(𝐴𝑖) = ∫ 𝑉1 √𝐸𝐺−𝐹2 d𝑢d𝑣= ∫ 𝑉𝑖 ‖𝜎𝑢× 𝜎𝑣‖ d𝑢d𝑣 575 X. Geometry Recall from the chain rule applied to 𝑛∘𝛾that 𝐷𝑛| | |(𝑢,𝑣) (𝜎𝑢) = 𝑛𝑢; 𝐷𝑛| | |(𝑢,𝑣) (𝜎𝑣) = 𝑛𝑣 Since 𝜅(𝑝) = 𝜅(𝜎(0, 0)) ≠0, 𝑛∘𝜎∶𝑉→𝑆2 has derivative of rank 2. This defines an allowable parametrisation for an open neighbourhood of 𝑛((0, 0)) by the inverse function theorem. Therefore, area𝑆2(𝑛(𝐴𝑖)) = ∫ 𝑉𝑖 ‖𝑛𝑢× 𝑛𝑣‖ d𝑢d𝑣 for sufficiently large 𝑖such that 𝜎−1𝐴𝑖= 𝑉𝑖lies in the open neighbourhood of (0, 0) where 𝑛∘𝜎is a diffeomorphism.
∫ 𝑉𝑖 ‖𝑛𝑢× 𝑛𝑣‖ d𝑢d𝑣= ∫ 𝑉𝑖 ‖𝐷𝑛(𝜎𝑢) × 𝐷𝑛(𝜎𝑣)‖ d𝑢d𝑣 = ∫ 𝑉𝑖 |det(𝐷𝑛)| ⋅‖𝜎𝑢× 𝜎𝑣‖ d𝑢d𝑣 = ∫ 𝑉𝑖 |𝜅(𝑢, 𝑣)| ⋅‖𝜎𝑢× 𝜎𝑣‖ d𝑢d𝑣 As 𝜅is continuous, given 𝜀> 0 there exists 𝛿> 0 such that |𝜅(𝑢, 𝑣) −𝜅(0, 0)| < 𝜀for all (𝑢, 𝑣) ∈𝐵((0, 0), 𝛿). In particular, for sufficiently large 𝑖, we have |𝜅(𝑢, 𝑣)| ∈(|𝜅(𝑝)| −𝜀, |𝜅(𝑝)| + 𝜀) Hence, (|𝜅(𝑝)| −𝜀) ∫ 𝑉𝑖 ‖𝜎𝑢× 𝜎𝑣‖ d𝑢d𝑣≤∫ 𝑉𝑖 |𝜅(𝑢, 𝑣)| ⋅‖𝜎𝑢× 𝜎𝑣‖ d𝑢d𝑣 ≤(|𝜅(𝑝)| + 𝜀) ∫ 𝑉𝑖 ‖𝜎𝑢× 𝜎𝑣‖ d𝑢d𝑣 In other words, |𝜅(𝑝)| −𝜀≤area𝑆2(𝑛(𝐴𝑖)) areaΣ(𝐴𝑖) ≤|𝜅(𝑝)| + 𝜀 Letting 𝑖→∞gives the result as required.
Theorem (theorema egregium). The Gauss curvature of a smooth surface in ℝ3 is isometry invariant. In other words, if 𝑓∶Σ1 →Σ2 is a diffeomorphism of surfaces in ℝ3 which is an isometry, then 𝜅(𝑝) = 𝜅(𝑓(𝑝)) for all 𝑝.
Remark. Isometries rely on only the first fundamental form, but Gauss curvature is defined using both fundamental forms. We can do a direct proof by simply differentiating the for-mula and rearranging until the result follows. This proof is given in Part II.
576 4. Geometry of surfaces in ℝ3 Alternatively, we can consider a different question: are some allowable parametrisations of a smooth surface in ℝ3 ‘better’ than others in some way? If we have a parametrisation 𝜎∶𝑉→𝑈⊆Σ, this defines certain distinguished curves, which are the images of 𝜎(𝑡, 0) and 𝜎(0, 𝑡). In this sense, looking for a ‘best’ parametrisation is equivalent to looking for ‘best’ distinguished curves near a point. This leads to the study of geodesics. We will later show that every smooth surface in ℝ3 admits local parametrisations such that the first fun-damental form has form d𝑢2 + 𝐺d𝑣2, so 𝐸= 1 and 𝐹= 0. We will also see (on an example sheet) that if such a local parametrisation exists, then 𝜅can be expressed as a function just of 𝐺. This allows us to approach the proof of the theorema egregium from a more conceptual way, since we have expressed 𝜅in terms of the first fundamental form alone.
Theorem (Gauss–Bonnet theorem). If Σ is a compact smooth surface in ℝ3, then ∫ Σ 𝜅d𝐴Σ = 2𝜋𝜒(Σ) 577 X. Geometry 5. Geodesics 5.1. Definitions Recall that we defined, for a smooth curve 𝛾∶[𝑎, 𝑏] →ℝ3, length(𝛾) = ∫ 𝑏 𝑎 ‖𝛾′(𝑡)‖ d𝑡 Definition. The energy of 𝛾is given by 𝐸(𝛾) = ∫ 𝑏 𝑎 ‖𝛾′(𝑡)‖2 d𝑡 Definition. Let 𝛾∶[𝑎, 𝑏] →Σ, where Σ is a smooth surface in ℝ3. A one-parameter vari-ation (with fixed endpoints) of 𝛾is a smooth map Γ∶(−𝜀, 𝜀) × [𝑎, 𝑏] →Σ, such that if 𝛾𝑠= Γ(𝑠, ⋅), then 𝛾0(𝑡) = 𝛾(𝑡), and 𝛾𝑠(𝑎) and 𝛾𝑠(𝑏) are independent of 𝑠.
Definition. A smooth curve 𝛾∶[𝑎, 𝑏] →Σ is a geodesic if, for every variation (𝛾𝑠) of 𝛾with fixed endpoints as above, we have d d𝑠 | |𝑠=0 𝐸(𝛾𝑠) = 0. Alternatively, 𝛾is a critical point of the energy functional on curves from 𝛾(𝑎) to 𝛾(𝑏).
5.2. The geodesic equations Let 𝛾have image contained within the image of an allowable parametrisation 𝜎∶𝑉→𝑈.
Then, for sufficiently small 𝑠, we can write 𝛾𝑠(𝑡) = 𝜎(𝑢(𝑠, 𝑡), 𝑣(𝑠, 𝑡)). Suppose that the first fundamental form, with respect to 𝜎, is 𝐸d𝑢2 + 2𝐹d𝑢d𝑣+ 𝐺d𝑣2 Let 𝑅= 𝐸̇ 𝑢2 + 2𝐹̇ 𝑢̇ 𝑣+ 𝐺̇ 𝑣2 By definition, 𝐸(𝛾𝑠) = ∫ 𝑏 𝑎 𝑅d𝑡 where 𝑅depends on 𝑠. Hence, 𝜕𝑅 𝜕𝑠= (𝐸𝑢̇ 𝑢2 + 2𝐹 𝑢̇ 𝑢̇ 𝑣+ 𝐺𝑢̇ 𝑣2)𝜕𝑢 𝜕𝑠+ (𝐸𝑣̇ 𝑣2 + 2𝐹 𝑣̇ 𝑢̇ 𝑣+ 𝐺𝑣̇ 𝑣2)𝜕𝑣 𝜕𝑠 + 2(𝐸̇ 𝑢+ 𝐹̇ 𝑣)𝜕̇ 𝑢 𝜕𝑠+ 2(𝐹̇ 𝑢+ 𝐺̇ 𝑣)𝜕̇ 𝑣 𝜕𝑠 This gives d d𝑠𝐸(𝛾𝑠) = ∫ 𝑏 𝑎 𝜕𝑅 𝜕𝑠d𝑡 578 5. Geodesics We can integrate by parts. Note that 𝜕𝑢 𝜕𝑠and 𝜕𝑣 𝜕𝑠vanish at 𝑎, 𝑏. Hence, d d𝑠 | | |𝑠=0 𝐸(𝛾𝑠) = ∫ 𝑏 𝑎 (𝐴𝜕𝑢 𝜕𝑠+ 𝐵𝜕𝑣 𝜕𝑠) d𝑡 where 𝐴= 𝐸𝑢̇ 𝑢2 + 2𝐹 𝑢̇ 𝑢̇ 𝑣+ 𝐺𝑢̇ 𝑣2 −2 𝜕 𝜕𝑡(𝐸̇ 𝑢+ 𝐹̇ 𝑣) 𝐵= 𝐸𝑣̇ 𝑢2 + 2𝐹 𝑣̇ 𝑢̇ 𝑣+ 𝐺𝑣̇ 𝑣2 −2 𝜕 𝜕𝑡(𝐹̇ 𝑢+ 𝐺̇ 𝑣) Corollary. A smooth curve 𝛾∶[𝑎, 𝑏] →Σ with image in Im 𝜎is a geodesic if and only if it satisfies the geodesic equations: d d𝑡(𝐸̇ 𝑢+ 𝐹̇ 𝑣) = 1 2(𝐸𝑢̇ 𝑢2 + 2𝐹 𝑢̇ 𝑢̇ 𝑣+ 𝐺𝑢̇ 𝑣2) d d𝑡(𝐹̇ 𝑢+ 𝐺̇ 𝑣) = 1 2(𝐸𝑣̇ 𝑢2 + 2𝐹 𝑣̇ 𝑢̇ 𝑣+ 𝐺𝑣̇ 𝑣2) Note that these equations are evaluated at 𝑠= 0, so no choice of variation is required.
Remark. Solving a differential equation is a local procedure. The original definition of the geodesic seems to be a global property. However, we can always consider a sub-curve of 𝛾to also be a geodesic, since its variations are variations of 𝛾. So the definition can be thought of as local.
Energy is sensitive to reparametrisation. If 𝑓, 𝑔∶[𝑎, 𝑏] →ℝare smooth, the Cauchy– Schwarz inequality gives that (∫ 𝑏 𝑎 𝑓𝑔d𝑡) 2 ≤∫ 𝑏 𝑎 𝑓2 d𝑡⋅∫ 𝑏 𝑎 𝑔2 d𝑡 Let us apply this to 𝑓= √𝑅, 𝑔= 1 to find length(𝛾)2 ≤𝐸(𝛾)(𝑏−𝑎) Since equality holds only when the two functions are proportional, we must have that ‖𝛾′(𝑡)‖ is constant for the equality to hold. In other words, 𝛾must be parametrised proportional to arc length.
Corollary. If 𝛾has constant speed and locally minimises length, then it is a geodesic. Fur-ther, if 𝛾globally minimises energy, then it must globally minimise length, and is paramet-rised with constant speed.
Remark. We would like geodesics to be a local property, but not necessarily global length minimisers. For example, all arcs of great circles will be shown to be geodesics, even if large arcs are not global length minimisers between fixed endpoints.
579 X. Geometry 5.3. Geodesics on the plane The plane ℝ2 has parametrisation 𝜎(𝑢, 𝑣) = (𝑢, 𝑣, 0) and first fundamental form d𝑢2 + d𝑣2.
The geodesic equations here are ̈ 𝑢= 0; ̈ 𝑣= 0 In particular, the geodesics on the plane are given by 𝑢(𝑡) = 𝛼𝑡+ 𝛽; 𝑣(𝑡) = 𝛾𝑡+ 𝛿 This is a straight line, parametrised at constant speed.
5.4. Geodesics on the sphere Consider the unit sphere with parametrisation 𝜎(𝑢, 𝑣) = (cos 𝑢cos 𝑣, cos 𝑢sin 𝑣, sin 𝑢); 𝑢∈(−𝜋 2 ); 𝑣∈(0, 2𝜋) This has first fundamental form d𝑢2 + cos2 𝑢d𝑣2 ⟹𝐸= 1; 𝐹= 0; 𝐺= cos2 𝑢 The geodesic equations give d d𝑡( ̇ 𝑢) = 1 22 cos 𝑢sin 𝑢̇ 𝑣2; d d𝑡(cos2 𝑢̇ 𝑣) = 0 This gives ̈ 𝑢+ sin 𝑢cos 𝑢̇ 𝑣2 = 0; ̈ 𝑣−2 tan 𝑢̇ 𝑢̇ 𝑣= 0 Since geodesics are parametrised at constant speed, we can assume that it is parametrised at unit speed without loss of generality.
‖𝛾′(𝑡)‖ = 1 ⟹ ̇ 𝑢+ cos2 𝑢̇ 𝑣2 = 1 Hence, ̈ 𝑣 ̇ 𝑣= 2 tan 𝑢̇ 𝑢⟹ln ̇ 𝑣= −2 ln cos 𝑢+ constant ⟹ ̇ 𝑣= 𝐶 cos2 𝑢 Substituting into the unit speed equation, ̇ 𝑢2 = 1 − 𝐶2 cos2 𝑢⟹ ̇ 𝑢= √ cos2 𝑢−𝐶2 cos2 𝑢 Then, ̇ 𝑣 ̇ 𝑢= d𝑣 d𝑢= 𝐶 cos 𝑢√cos2 𝑢−𝐶2 Hence, 𝑣= ∫d𝑣 d𝑢d𝑢= ∫ 𝐶sec2 𝑢 √1 −𝐶2𝑠𝑒𝑐2𝑢 d𝑢 580 5. Geodesics Using the substitution 𝑤= 𝐶tan 𝑢 √1−𝐶2 , we find 𝑣= ∫ 𝑤 1 −𝑤2 d𝑤= arcsin 𝑤+ constant = arcsin(𝜆tan 𝑢) + 𝛿 for some constants 𝜆, 𝛿. Hence, sin(𝑣−𝛿) = 𝜆tan 𝑢 Rewriting using the angle addition formula, (sin 𝑣cos 𝑢) ⏟⎵ ⎵⏟⎵ ⎵⏟ 𝑥 cos 𝛿−(cos 𝑣cos 𝑢) ⏟⎵ ⎵⏟⎵ ⎵⏟ 𝑦 sin 𝛿−𝜆sin 𝑢 ⏟ 𝑧 = 0 Hence, the geodesic 𝛾lies on a plane through the origin, since this is a linear equation in 𝑥, 𝑦, 𝑧. Such planes intersect the sphere in great circles.
5.5. Geodesics on the torus Consider the surface of revolution of a circle in the 𝑥𝑧-plane centred at (𝑎, 0, 0) about the 𝑧 axis, giving a torus. An allowable parametrisation for this surface is 𝜎(𝑢, 𝑣) = ((𝑎+ cos 𝑢) cos 𝑣, (𝑎+ cos 𝑢) sin 𝑣, sin 𝑢) The first fundamental form is d𝑢2 + (𝑎+ cos 𝑢)2 d𝑣2 ⟹𝐸= 1; 𝐹= 0; 𝐺= (𝑎+ cos 𝑢)2 Note that if we were to take 𝑎= 0, we would arrive at the unit sphere and its first fun-damental form. We can follow the same procedure as above with the sphere, or formally replace cos 𝑢with 𝑎+ cos 𝑢in the result.
d𝑣 d𝑢= 𝐶 (𝑎+ cos 𝑢)√(𝑎+ cos 𝑢)2 −𝐶2 which cannot be integrated using classical functions. This leads to the study of elliptic func-tions.
5.6. Equivalent characterisation of geodesics We have so far restricted our analysis to the first fundamental form, without considering its embedding in ℝ3. Intuitively, we know that straight lines in ℝ2 are not just locally shortest but also locally straightest. We would expect this to hold for other surfaces as well. We can characterise this notion via stating that the change in the tangent vector to a curve is as small as it could be, subject to the constraint that it lies on the surface.
581 X. Geometry Proposition. Let Σ be a smooth surface in ℝ3. A smooth curve 𝛾∶[𝑎, 𝑏] →Σ is a geodesic if and only if ̈ 𝛾(𝑡) is everywhere normal to the surface Σ.
Remark. This proposition makes use of the tangent plane, a notion that exists only because we have an embedding in ℝ3. Note that d d𝑡⟨̇ 𝛾, ̇ 𝛾⟩= 2 ⟨ ̇ 𝛾 ⏟ tangent to Σ , ̈ 𝛾 ⏟ normal to Σ ⟩= 0 Hence, ⟨̇ 𝛾, ̇ 𝛾⟩is constant, giving that geodesics are parametrised proportional to arc length.
Proof. The property of being a geodesic as we previously defined is a local property, and so is the condition in the proposition. Hence, we may work entirely within an allowable parametrisation 𝜎∶𝑉→𝑈. Suppose 𝛾(𝑡) = 𝜎(𝑢(𝑡), 𝑣(𝑡)). Hence, ̇ 𝛾= 𝜎𝑢̇ 𝑢+ 𝜎𝑣̇ 𝑣 ̈ 𝛾is normal to Σ when it is orthogonal to the tangent plane, which is spanned by 𝜎𝑢, 𝜎𝑣. This is true if and only if ⟨d d𝑡(𝜎𝑢̇ 𝑢+ 𝜎𝑣̇ 𝑣), 𝜎𝑢⟩= 0 = ⟨d d𝑡(𝜎𝑢̇ 𝑢+ 𝜎𝑣̇ 𝑣), 𝜎𝑣⟩ We will prove the first equality. This can be rewritten d d𝑡⟨𝜎𝑢̇ 𝑢+ 𝜎𝑣̇ 𝑣, 𝜎𝑢⟩−⟨𝜎𝑢̇ 𝑢+ 𝜎𝑣̇ 𝑣, d d𝑡𝜎𝑢⟩= 0 Note that ⟨𝜎𝑢, 𝜎𝑢⟩= 𝐸and ⟨𝜎𝑢, 𝜎𝑣⟩= 𝐹.
d d𝑡(𝐸̇ 𝑢+ 𝐹̇ 𝑣) −⟨𝜎𝑢̇ 𝑢+ 𝜎𝑣̇ 𝑣, 𝜎𝑢𝑢̇ 𝑢+ 𝜎𝑢𝑣̇ 𝑣⟩= 0 Hence, d d𝑡(𝐸̇ 𝑢+ 𝐹̇ 𝑣) −[ ̇ 𝑢2 ⟨𝜎𝑢, 𝜎𝑢𝑢⟩+ ̇ 𝑢̇ 𝑣(⟨𝜎𝑢, 𝜎𝑢𝑣⟩+ ⟨𝜎𝑣, 𝜎𝑢𝑢⟩) + ̇ 𝑣2 ⟨𝜎𝑣𝜎𝑢𝑣⟩] = 0 Note that 𝐸𝑢= 2 ⟨𝜎𝑢, 𝜎𝑢𝑢⟩, 𝐹 𝑢= ⟨𝜎𝑢, 𝜎𝑢𝑣⟩+ ⟨𝜎𝑣, 𝜎𝑢𝑢⟩, and 𝐺𝑢= 2 ⟨𝜎𝑣, 𝜎𝑢𝑣⟩. This gives d d𝑡(𝐸̇ 𝑢+ 𝐹̇ 𝑣) = 1 2(𝐸𝑢̇ 𝑢2 + 2𝐹 𝑢̇ 𝑢̇ 𝑣+ 𝐺𝑢̇ 𝑣2) which is the first of the geodesic equations. By symmetry, we find the second geodesic equa-tion similarly.
582 5. Geodesics 5.7. Planes of symmetry Let Σ be a smooth surface in ℝ3 such that there exists a plane Π ⊆ℝ3 such that Π ∩Σ is a smooth embedded curve 𝐶⊆Σ, and Σ is setwise preserved by reflection in the plane Π. We will show that 𝐶is a geodesic when parametrised at constant speed. Consider a point 𝑝on 𝐶. We can think of ℝ3 = Π ⊕Π⟂, where we change coordinates such that 𝑝is the origin.
We can also write ℝ3 = 𝑇 𝑝Σ ⊕ℝ𝑛𝑝, where ℝ𝑛𝑝is the vector subspace of ℝ3 generated by 𝑛𝑝. Clearly, reflection in Π acts on Π by the identity, and on Π⟂by −1. Since reflection in Π fixes Σ setwise and fixes 𝑝, it must also preserve the subspace 𝑇 𝑝Σ. Hence it also preserves ℝ𝑛𝑝, so ℝ𝑛𝑝⊆Π, since Π is not the identity on 𝑇 𝑝Σ. Now, let us parametrise 𝐶locally near 𝑝 using 𝑡↦𝛾(𝑡) ∈𝐶at constant speed. Since 𝛾(𝑡) ⊆Π, we have ̇ 𝛾(𝑡), ̈ 𝛾(𝑡) ∈Π. 𝛾has constant speed, so ⟨̇ 𝛾, ̈ 𝛾⟩= 0. Hence ̇ 𝛾lies in Π ∩𝑇 𝑝Σ and ̈ 𝛾is orthogonal to this and lies in Π, so lies in ℝ𝑛𝑝⊆Π. Hence 𝛾is indeed a geodesic.
In particular, arcs of great circles are geodesics, since they lie in planes of symmetry.
5.8. Surfaces of revolution Consider the surface of revolution given by 𝜂(𝑢) = (𝑓(𝑢), 0, 𝑔(𝑢)) where 𝜂is smooth and injective, and 𝑓(𝑢) > 0, rotated about the 𝑧axis.
Definition. A circle obtained by rotating a point of 𝜂is called a parallel. A curve obtained by rotating 𝜂itself by a fixed angle about the 𝑧axis is called a meridian.
A plane in ℝ3 containing the 𝑧axis is a plane of symmetry, hence meridians are geodesics by the previous discussion. Not all parallels are geodesics.
Lemma. A parallel given by 𝑢= 𝑢0 is a geodesic when parametrised at constant speed if and only if 𝑓′(𝑢0) = 0.
Proof. Consider the allowable parametrisation 𝜎(𝑢, 𝑣) = (𝑓(𝑢) cos 𝑣, 𝑓(𝑢) sin 𝑣, 𝑔(𝑢)) where 𝑢∈(𝑎, 𝑏) and 𝑣∈(0, 2𝜋). The first fundamental form is [(𝑓′)2 + (𝑔′)2] d𝑢2 + 𝑓2 d𝑣2 If without loss of generality we choose to parametrise 𝜂by arc length, this becomes d𝑢2 + 𝑓2 d𝑣2 ⟹𝐸= 1; 𝐹= 0; 𝐺= 𝑓2 The geodesic equations are d d𝑡̇ 𝑢= ̈ 𝑢= 𝑓𝑓 𝑢̇ 𝑣2; d d𝑡(𝑓2 ̇ 𝑣) = 0 583 X. Geometry and ̇ 𝑢2 + 𝑓2 ̇ 𝑣2 is a nonzero constant. Given that we want to only consider parallels of the surface of revolution, we can impose the constraint that 𝑢= 𝑢0 is constant. Hence, the constant speed condition gives that ̇ 𝑣= constant 𝑓(𝑢0) = constant The second equation holds automatically. The first equation is 0 = 𝑓(𝑢0)𝑓 𝑢(𝑢0) ⋅constant So this holds exactly when 𝑓 𝑢(𝑢0) = 0.
Consider a curve 𝛾(𝑡) on Σ, making angle 𝜃with the parallel of radius 𝜌= 𝑓.
Proposition (Clairaut’s relation). If 𝛾is a geodesic, then 𝜌cos 𝜃is constant along 𝛾.
Proof. Let 𝛾(𝑡) = 𝜎(𝑢(𝑡), 𝑣(𝑡)), so ̇ 𝛾= 𝜎𝑢̇ 𝑢+ 𝜎𝑣̇ 𝑣. The tangent vector to the parallel is 𝜎𝑣. By the earlier discussion on angles in terms of the first fundamental form, cos 𝜃= ⟨𝜎𝑣, 𝜎𝑢̇ 𝑢+ 𝜎𝑣̇ 𝑣⟩ ‖𝜎𝑣‖ ⋅‖𝜎𝑢̇ 𝑢+ 𝜎𝑣̇ 𝑣‖ If 𝛾is parametrised by arc length, ‖ ̇ 𝛾‖ = 1, so ‖𝜎𝑢̇ 𝑢+ 𝜎𝑣̇ 𝑣‖ = 1. So, using our values for 𝐹, 𝐺 above, cos 𝜃= |𝑓(𝑢) ̇ 𝑣| = 𝜌̇ 𝑣 The second geodesic equation is exactly 𝜌cos 𝜃= constant Example. Usually, for a surface of revolution, we take the assumption that 𝜂never inter-sects the 𝑧-axis, or that 𝑓is positive. This ensures that all points on the surface are locally smooth. However, we can allow 𝜂to meet the 𝑧-axis orthogonally, as in the ellipsoid or sphere.
Consider an ellipsoid of revolution. 𝜌cos 𝜃is constant along a geodesic 𝛾. Suppose that at some point 𝛾intersects a parallel of radius 𝜌0 at angle 𝜃0, and that 𝛾is not a meridian (so cos 𝜃≠0). Hence 𝜃0 ∈[0, 𝜋 2 ). In particular, for 𝜌cos 𝜃to be constant, we must have that 𝜌is bounded below. A geodesic which is not a meridian is therefore ‘trapped’ between parallels corresponding to the bound on the size of 𝜌. In particular, any geodesic through a pole is a meridian.
584 5. Geodesics 5.9. Local existence of geodesics It is difficult to solve the geodesic equations globally. We can often intead prove local results about any geodesics that may arise.
Recall Picard’s theorem from Analysis and Topology. Let 𝐼= [𝑡0 −𝑎, 𝑡0 + 𝑎] ⊆ℝ, 𝐵= {𝑥∶‖𝑥−𝑥0‖ ≤𝑏} ⊆ℝ𝑛, and 𝑓∶𝐼×𝐵→ℝ𝑛that is continuous, and Lipschitz in the second variable.
‖𝑓(𝑡, 𝑥1) −𝑓(𝑡, 𝑥2)‖ ≤𝑁‖𝑥1 −𝑥2‖ Then the differential equation d𝑥 d𝑡 = 𝑓(𝑡, 𝑥) with 𝑥(𝑡0) = 𝑥0 has a unique solution for some time interval |𝑡−𝑡0| < ℎ, where ℎ= min {𝑎, 𝑏 𝑠} where 𝑠= sup ‖𝑓‖. Further, if 𝑓 is smooth in all parameters, then the solution to the differential equation is smooth and depends smoothly on the initial condition.
Recall the geodesic equations: d d𝑡(𝐸̇ 𝑢+ 𝐹̇ 𝑣) = 1 2(𝐸𝑢̇ 𝑢2 + 2𝐹 𝑢̇ 𝑢̇ 𝑣+ 𝐺𝑢̇ 𝑣2) d d𝑡(𝐹̇ 𝑢+ 𝐺̇ 𝑣) = 1 2(𝐸𝑣̇ 𝑢2 + 2𝐹 𝑣̇ 𝑢̇ 𝑣+ 𝐺𝑣̇ 𝑣2) We can write this as (𝐸 𝐹 𝐹 𝐺) ( ̈ 𝑢 ̈ 𝑣) = 𝑅 where 𝑅is composed of smooth functions of 𝑢, 𝑣. The matrix on the left hand side is invert-ible, and the inverse map 𝐴↦𝐴−1 on matrices is smooth. Hence, we can write the geodesic equations in the form ̈ 𝑢= 𝐴(𝑢, 𝑣, ̇ 𝑢, ̇ 𝑣); ̈ 𝑣= 𝐵(𝑢, 𝑣, ̇ 𝑢, ̇ 𝑣) In the usual way we can turn second-order equations into first-order equations by introdu-cing 𝑝= ̇ 𝑢, 𝑞= ̇ 𝑣, and we find ̇ 𝑢= 𝑝; ̇ 𝑣= 𝑞; ̇ 𝑝= 𝐴(𝑢, 𝑣, 𝑝, 𝑞); ̇ 𝑞= 𝐵(𝑢, 𝑣, 𝑝, 𝑞) This is a system of first-order ordinary differential equations as governed by Picard’s theorem.
Since 𝐴, 𝐵are smooth, a local bound on ‖𝐷𝐴‖ and ‖𝐷𝐵‖ will give the required Lipschitz condition.
Corollary. Let Σ be a smooth surface in ℝ3. For 𝑝∈Σ and 𝑣∈𝑇 𝑝Σ nonzero, then there exists 𝜀> 0 and a geodesic 𝛾∶[0, 𝜀) →Σ such that 𝛾(0) = 𝑝; ̇ 𝛾(0) = 𝑣 Moreover, this geodesic depends smoothly on 𝑝, 𝑣.
The local existence of geodesics gives rise to allowable parametrisations of Σ with ‘nice’ prop-erties in terms of the first fundamental form. Let 𝑝∈Σ, and consider a geodesic arc 𝛾start-ing at 𝑝and parametrised by arc length. At each point 𝛾(𝑡) for small 𝑡> 0, we can consider a 585 X. Geometry geodesic arc 𝛾𝑡starting at 𝛾(𝑡), and 𝛾′ 𝑡(0) is orthogonal to 𝛾′(𝑡), and also parametrised by arc length. Now, we define 𝜎(𝑢, 𝑣) = 𝛾𝑣(𝑢), which is defined for 𝑢∈[0, 𝜀) and 𝑣∈[0, 𝛿).
Lemma. For 𝜀, 𝛿sufficiently small, 𝜎∶(𝑢, 𝑣) ↦𝛾𝑣(𝑢) defines an allowable parametrisation of an open set in Σ, taking the interior of the domain.
Proof. Smoothness follows from the addendum to Picard’s theorem above. At the origin (0, 0), by construction we have 𝜎𝑢, 𝜎𝑣orthogonal. Hence, they stay linearly independent for sufficiently small 𝜀, 𝛿. So 𝐷𝜎has full rank, and (on a smaller set if necessary) 𝜎is injective.
So 𝜎is allowable.
Corollary. Any smooth surface Σ in ℝ3 admits local parametrisations for which the first fundamental form has form d𝑢2 + 𝐺(𝑢, 𝑣) d𝑣2, so 𝐸= 1 and 𝐹= 0.
Proof. Consider the parametrisation 𝜎(𝑢, 𝑣) = 𝛾𝑣(𝑢). For 𝑣0 fixed, the curve 𝑢↦𝛾𝑣0(𝑢) is a geodesic parametrised at unit speed, so 𝐸= 1. One of the geodesic equations is d d𝑡(𝐹̇ 𝑢+ 𝐺̇ 𝑣) = 1 2(𝐸𝑣̇ 𝑢2 + 2𝐹 𝑣̇ 𝑢̇ 𝑣+ 𝐺𝑣̇ 𝑣2) and consider 𝑣(𝑡) = 𝑣0, 𝑢(𝑡) = 𝑡. 𝐸𝑣= ̇ 𝑣= 0 and ̇ 𝑢= 1, so d d𝑡𝐹= 0 ⟹𝐹 𝑢̇ 𝑢= 0 ⟹𝐹 𝑢= 0 So 𝐹is independent of 𝑢. At 𝑢= 0, then by construction of 𝛾𝑣as being orthogonal to 𝛾at 𝛾(𝑣), we see 𝐹= 0.
These coordinates are called geodesic normal coordinates. Note that by fixing 𝑢and letting 𝑣vary, the curve obtained is typically not a geodesic, except for 𝑢= 0 which is 𝛾itself. In these coordinates, we can also find 𝐺(0, 𝑣) = 1; 𝐺𝑢(0, 𝑣) = 0 The first result holds since 𝜎𝑣has unit length at 𝑢= 0. The second result holds because 𝑢= 0 yields a geodesic with arc length parametrisation, and then we can use one of the geodesic equations to find d d𝑡(𝐸̇ 𝑢+ 𝐹̇ 𝑣) = 1 2(𝐸𝑢̇ 𝑢2 + 2𝐹 𝑢̇ 𝑢̇ 𝑣+ 𝐺𝑢̇ 𝑣2) ⟹0 = 1 2𝐺𝑢(0, 𝑣) 5.10. Surfaces of constant curvature In the example sheets, we show that for a smooth surface Σ in ℝ3 with allowable parametrisa-tion for which 𝐸= 1 and 𝐹= 0, we have the following result for the Gauss curvature.
𝜅= −(√𝐺) 𝑢𝑢 √𝐺 586 5. Geodesics If 𝑎∶ℝ3 →ℝ3 is a dilation 𝑎(𝑥, 𝑦, 𝑧) = (𝑎𝑥, 𝑎𝑦, 𝑎𝑧), then 𝜅𝑎(Σ) = 1 𝑎2 𝜅Σ since 𝐸, 𝐹, 𝐺rescale by 𝑎2, and 𝐿, 𝑁, 𝑀rescale by 𝑎. This matches the results previously found for spheres of varying radii. By dilating, to understand surfaces of constant curvature it suffices to consider surfaces with constant curvature ±1 or 0.
Proposition. Let Σ be a smooth surface in ℝ3. Then, (i) if 𝜅≡0, then Σ is locally isometric to (ℝ2, d𝑢2 + d𝑣2); (ii) if 𝜅≡1, then Σ is locally isometric to (𝑆2, d𝑢2 + cos2 𝑢d𝑣2).
Proof. Σ admits an allowable parametrisation with 𝐸= 1 and 𝐹= 0 by using geodesic normal coordinates, so 𝜅= −√𝐺𝑢𝑢 √𝐺 ; 𝐺(0, 𝑣) = 1; 𝐺𝑢(0, 𝑣) = 0 If 𝜅≡0, we have √𝐺𝑢𝑢= 0, so √𝐺= 𝐴(𝑣)𝑢+ 𝐵(𝑣), and the boundary conditions give 𝐴≡0, 𝐵≡1. In particular, 𝐺≡1. The fundamental form then is d𝑢2 + d𝑣2, which is that of ℝ2.
If 𝜅≡1, we find (√𝐺) 𝑢𝑢+√𝐺= 0 so √𝐺= 𝐴(𝑣) sin 𝑢+𝐵(𝑣) cos 𝑢. The boundary conditions then imply that 𝐴≡0, 𝐵≡1 and hence the fundamental form is d𝑢2 + cos2 𝑢d𝑣2. This matches the first fundamental form of a sphere with parametrisation 𝜎(𝑢, 𝑣) = (cos 𝑢cos 𝑣, cos 𝑢sin 𝑣, sin 𝑢) Remark. If 𝜅≡−1, we will find the first fundamental form d𝑢2+cosh2 𝑢d𝑣2. There exists an object known as the tractoid, which is a smooth surface in ℝ3, and has this first fundamental form. We could alternatively choose not to embed this surface in ℝ3.
In fact, the change of variables 𝑣= 𝑒𝑣tanh 𝑢, 𝑤= 𝑒𝑣sech 𝑢turns the fundamental form d𝑢2 + cosh2 𝑢d𝑣2 into d𝑉2+d𝑊2 𝑊2 , which is a ‘standard’ presentation of the first fundamental form, which we will see more of later.
587 X. Geometry 6. Riemannian metrics 6.1. Definitions Definition. Let 𝑉⊆ℝ2 be an open set. An (abstract) Riemannian metric is a smooth map from 𝑉to the set of positive definite symmetric bilinear forms, given by 𝑣↦(𝐸(𝑣) 𝐹(𝑣) 𝐹(𝑣) 𝐺(𝑣)) such that 𝐸> 0, 𝐺> 0, 𝐸𝐺−𝐹2 > 0. The image of this map can be viewed as an open subset of ℝ4.
If 𝑣is a vector at 𝑝∈𝑉, we can compute its infinitesimal length by ‖𝑣‖2 = 𝑣⊺(𝐸(𝑣) 𝐹(𝑣) 𝐹(𝑣) 𝐺(𝑣)) 𝑣 Thus, if 𝛾∶[𝑎, 𝑏] →𝑉is smooth, length(𝛾) = ∫ 𝑏 𝑎 (𝐸̇ 𝑢2 + 2𝐹̇ 𝑢̇ 𝑣+ 𝐺̇ 𝑣2) 1 2 d𝑡 where 𝛾(𝑡) = (𝑢(𝑡), 𝑣(𝑡)).
Definition. Let Σ be an abstract smooth surface, so Σ = ⋃𝑖∈𝐼𝑈𝑖for open sets 𝑈𝑖, with charts 𝜑𝑖∶𝑈𝑖→𝑉𝑖⊆ℝ2 which are homeomorphisms, and with smooth transition maps 𝜑𝑖𝜑−1 𝑗∶𝜑𝑗(𝑈𝑖∩𝑈𝑗) →𝜑𝑖(𝑈𝑖∩𝑈𝑗). A Riemannian metric on Σ, usually called 𝑔or d𝑠2, is a choice of Riemannian metric in the above sense on each 𝑉𝑖, which are compatible in the following sense. Let 𝜎= 𝜑−1 𝑖 and ̃ 𝜎= 𝜑−1 𝑗 for some 𝑖, 𝑗, and define 𝑓= ̃ 𝜎−1 ∘𝜎. Then we require (𝐷𝑓)⊺( ˜ 𝐸 ˜ 𝐹 ˜ 𝐹 ˜ 𝐺) (𝐷𝑓) = (𝐸 𝐹 𝐹 𝐺) So 𝐷𝑓defines an isometry from an open set in the chart (𝑈, 𝜑(𝑈) = 𝑉) to one in the chart ( ˜ 𝑈, ˜ 𝜑( ˜ 𝑈) = ˜ 𝑉).
This compatibility condition is the transition law for first fundamental forms for smooth surfaces in ℝ3.
Example. Recall the torus 𝑇2 = ℝ2 ⟋ ℤ2.
𝑏−1 𝑎 𝑏 𝑎−1 588 6. Riemannian metrics We have an atlas of charts for which the transition maps are the restrictions of translations of open subsets of ℝ2. For each 𝑉𝑖⊆ℝ2, we associate the natural Euclidean metric d𝑢2 + d𝑣2.
If 𝑓is a translation, 𝐷𝑓is the identity, and so (𝐷𝑓)⊺𝐼(𝐷𝑓) = 𝐼 holds trivially. So this gives a global Riemannian metric on 𝑇2. This metric is flat, since it is locally isometric to ℝ2 at all points.
Conversely, consider the torus of revolution embedded in ℝ3. As a compact smooth surface in ℝ3, it must contain an elliptic point. Hence, the flat Riemannian metric described above is not the same (up to isometry) as the metric obtained by any possible embedding of the torus in ℝ3.
The real projective plane ℝℙ2 admits a Riemannian metric with constant curvature +1. We have constructed a smooth atlas for ℝℙ2 where the charts were of the form (𝑈, 𝜑), with 𝑈= 𝑞̂ 𝑈and 𝑞∶𝑆2 →ℝℙ2 the quotient map, ̂ 𝑈⊆𝑆2 open and contained within an open hemisphere, and 𝜑∶𝑈∶𝑈→𝑉⊆ℝ2 is given by ̂ 𝜑∘𝑞−1| |𝑈and ̂ 𝜑∶ ̂ 𝑈→𝑉a chart on 𝑆2.
The transition maps for this atlas were found to be locally the identity, or induced from the antipodal map. The antipodal map from 𝑆2 to 𝑆2 is an isometry, so both types of transition maps preserve the usual round metric on 𝑆2.
In the first example sheet, we consider the Klein bottle. This has an atlas such that all trans-ition maps are either translations or translations composed with a reflection. These preserve the flat metric in ℝ2, so the Klein bottle inherits a flat Riemannian metric. The Klein bottle and ℝℙ2 are not embedded in ℝ3, so we could not construct a ‘non-abstract’ Riemannian metric.
Definition. Let (Σ1, 𝑔1), (Σ2, 𝑔2) be abstract smooth surfaces with abstract Riemannian met-rics. A diffeomorphism 𝑓∶Σ1 →Σ2 is an isometry if it preserves the lengths of all curves, where lengths are taken with respect to these abstract Riemannian metrics.
Example. If (Σ2, 𝑔2) is given, and 𝑓∶Σ1 →Σ2 is a diffeomorphism, we can equip Σ1 with a metric known as the pullback metric 𝑔1 = 𝑓⋆𝑔2 that gives that 𝑓is an isometry.
6.2. The length metric Definition. Let (Σ, 𝑔) be a connected abstract smooth surface with an abstract Riemannian metric. The length metric is defined by 𝑑𝑔(𝑝, 𝑞) = inf 𝛾𝐿(𝛾) where 𝛾varies over piecewise smooth paths in Σ from 𝑝to 𝑞, and 𝐿is length computed using 𝑔.
Proposition. Let (Σ, 𝑔) be a connected abstract smooth surface with an abstract Rieman-nian metric. Then 𝑑𝑔is indeed a metric, and 𝑑𝑔induces a topology on Σ that agrees with the given topology.
589 X. Geometry Proof. Let 𝑝, 𝑞∈Σ. We will show that there exists some piecewise smooth path 𝛾from 𝑝 to 𝑞, so 𝑑𝑔(𝑝, 𝑞) is well-defined and finite. Connected surfaces are path-connected. There exists a continuous path 𝛾and a finite set of charts (𝑈𝑖, 𝜑𝑖) with associated parametrisations 𝜎𝑖= 𝜑−1 𝑖 ∶𝑉𝑖→𝑈𝑖⊂Σ such that Im 𝛾⊆⋃ 𝑁 𝑖=1 𝑈𝑖. Consider points 𝑝= 𝑥0 ∈𝑈1, 𝑥1 ∈𝑈1 ∩𝑈2, 𝑥2 ∈𝑈2 ∈𝑈3, … , 𝑞= 𝑥𝑁∈𝑈𝑁 Smooth paths in 𝑉𝑖from 𝜑𝑖(𝑥𝑖) to 𝜑𝑖+1(𝑥𝑖+1) exist, since smooth paths between two points in ℝ2 exist. Since the atlas is smooth, being a smooth path in some 𝑈𝑖is the same as being smooth in 𝑈𝑖+1 whenever 𝑈𝑖and 𝑈𝑖+1 intersect, since the transition maps are smooth. So 𝑝, 𝑞∈Σ are joined by some piecewise smooth path.
For any piecewise smooth path from 𝑝to 𝑞there exists the inverse path parametrised in the opposite direction, which has the same length. We can also concatenate paths from 𝑝to 𝑞 and from 𝑞to 𝑟, with length equal to the sum of the lengths. In both cases, the new paths are piecewise smooth. This then implies that 𝑑𝑔is symmetric, and satisfies the triangle inequality.
To show 𝑑𝑔is a metric, it now suffices to show that 𝑑𝑔(𝑝, 𝑞) = 0 implies 𝑝= 𝑞, since the converse is trivial. Let 𝑝∈Σ and fix a chart (𝑈, 𝜑) at 𝑝. Without loss of generality let 𝑉= 𝐵(0, 1), and 𝜑(𝑝) = 0. If 𝑞≠𝑝∈Σ, there exists 𝜀> 0 such that 𝑞∉𝜑−1(𝐵(0, 𝜀)).
Suppose 𝛾∶[0, 1] →Σ is a piecewise smooth path from 𝑝to 𝑞. Certainly, 𝛾must escape the disc 𝜑−1(𝐵(0, 𝜀)), since it must reach 𝑞. Length along paths is additive, so by the triangle inequality, it suffices to show that there exists 𝛿> 0 such that 𝑑𝑔(𝑝, 𝑟) > 𝛿for all 𝑟∈ 𝜕𝜑−1(𝐵(0, 𝜀)) = 𝜑−1{circle of radius 𝜀}. The data on the Riemannian metric 𝑔includes the non-degenerate symmetric bilinear form (𝐸𝑧 𝐹 𝑧 𝐹 𝑧 𝐺𝑧 ) for all 𝑧∈𝐵(0, 𝜀) ⊆𝑉. We also have the usual Euclidean inner product on the disc, (1 0 0 1). For all 𝑧∈𝐵(0, 𝜀), these matrices are positive definite. Since 𝐵(0, 𝜀) is compact, there exists 𝛿> 0 such that (𝐸𝑧−𝛿 𝐹 𝑧 𝐹 𝑧 𝐺𝑧−𝛿) is still positive definite for all 𝑧∈𝐵(0, 𝜀). In other words, the determinant 𝐸𝐺−𝐹2 > 0 for all 𝑧∈𝐵(0, 𝜀), which is compact, so it is bounded below by some positive number.
Hence, length𝑔( ̂ 𝛾) ≥length𝛿⋅Euclidean( ̂ 𝛾) for any ̂ 𝛾contained withing 𝐵(𝑜, 𝜀). Taking ̂ 𝛾= 𝜑[𝛾∩𝜑−1(𝐵(𝑜, 𝜀))], which is the part of 𝛾in 𝐵(0, 𝜀) with respect to the chart, we have that length𝛿⋅Euclidean( ̂ 𝛾) ≥𝛿𝜀, so 𝑑𝑔(𝑝, 𝑞) ≥𝛿𝜀.
Remark. The last step of the argument for the proof above, comparing the inner products (𝐸𝑧 𝐹 𝑧 𝐹 𝑧 𝐺𝑧 ) and (1 0 0 1) can be modified to show that 𝑑𝑔induces a topology on Σ that agrees with the given topology, which is given by local homeomorphisms to ℝ2 everywhere.
590 6. Riemannian metrics 6.3. The hyperbolic metric Definition. Let 𝐷= 𝐵(0, 1) = {𝑧∈ℂ∶|𝑧| < 1} The abstract Riemannian metric 𝑔hyp on 𝐷is given by 4(d𝑢2 + d𝑣2) (1 −𝑢2 −𝑣2)2 = 4|d𝑧|2 (1 −|𝑧|2) 2 Since there is only one chart, this holds for all of 𝐷. In particular, if 𝛾∶[0, 1] →𝐷is smooth, then 𝐿𝑔hyp(𝛾) = 2 ∫ 1 0 | ̇ 𝛾(𝑡)| 1 −|𝛾(𝑡)|2 d𝑡 If 𝛾(𝑡) = (𝑢(𝑡), 𝑣(𝑡)), we can write 𝐿(𝛾) = 2 ∫ 1 0 ( ̇ 𝑢2 + ̇ 𝑣2) 1 2 1 −𝑢2 −𝑣2 d𝑡 This is very similar to a first fundamental form with 𝐸= 𝐺= 4 (1−𝑢2−𝑣2)2 and 𝐹= 0, but we do not claim that this fundamental form arises from an embedding in ℝ3.
Note that the flat metric on ℝ2 and the usual round metric on 𝑆2 have large and transitive isometry groups. We will show that this metric also induces a large symmetry group, which is induced by the Möbius group. Recall that Möb = {𝑧↦𝑎𝑧+ 𝑏 𝑐𝑧+ 𝑑∶(𝑎 𝑏 𝑐 𝑑) ∈𝐺𝐿(2, ℂ)} ↷ℂ∪{∞} Lemma. The subgroup of the Möbius group that preserves 𝐷, Möb(𝐷) = {𝑇∈Möb∶𝑇(𝐷) = 𝐷} is also given by Möb(𝐷) = {𝑧↦𝑒𝑖𝜃𝑧−𝑎 1 −𝑎𝑧∶|𝑎| < 1} = {(𝑎 𝑏 𝑏 𝑎) ∈Möb∶|𝑎|2 −|𝑏|2 = 1} Proof. Note that | | | 𝑧−𝑎 1 −𝑎𝑧 | | | = 1 ⟺(𝑧−𝑎)(𝑧−𝑎) = (1 −𝑎𝑧)(1 −𝑎𝑧) ⟺𝑧𝑧−𝑎𝑧−𝑎𝑧+ 𝑎𝑎= 1 −𝑎𝑧−𝑎𝑧+ 𝑎𝑎𝑧𝑧 ⟺|𝑧|2(1 −|𝑎|2) = 1 −|𝑎|2 ⟺|𝑧| = 1 591 X. Geometry So these maps of the form 𝑧↦𝑒𝑖𝜃𝑧−𝑎 1 −𝑎𝑧 do indeed preserve the unit circle, and 𝑎∈𝐷is mapped to 0 ∈𝐷. Hence, it preserves the entire disc.
Lemma. The Riemannian metric 𝑔hyp is invariant under Möb(𝐷). In other words, the Möbius group Möb(𝐷) acts by isometries on 𝐷.
Proof. Möb(𝐷) is generated by 𝑧↦𝑒𝑖𝜃and 𝑧↦ 𝑧−𝑎 1−𝑎𝑧. The rotations preserve 𝑔hyp, since it depends only on |𝑧| and not 𝑧itself. For the second type of transformation, let 𝑤= 𝑧−𝑎 1−𝑎𝑧.
Here, d𝑤= d𝑧 1 −𝑎𝑧+ 𝑧−𝑎 (1 −𝑎𝑧)2 𝑎d𝑧= d𝑧 (1 −𝑎𝑧)2 (1 −|𝑎|2) Then, |d𝑤| 1 −|𝑤|2 = |d𝑧| | |1 −𝑎𝑧| | 2 ⋅ 1 −|𝑎|2 1 −| | 𝑧−𝑎 1−𝑎𝑧 | | 2 = |d𝑧|(1 −|𝑎|2) | |1 −𝑎𝑧| | 2 −|𝑧−𝑎|2 = |d𝑧| 1 −|𝑧|2 Hence the hyperbolic metric, which is a function of this |d𝑧| 1−|𝑧|2 , is also invariant under this change of variables.
Lemma. On (𝐷, 𝑔hyp), (i) every pair of points in (𝐷, 𝑔hyp) is joined by a unique geodesic up to reparametrisation; (ii) the geodesics are diameters of the disc and circular arcs orthogonal to the boundary 𝜕𝐷.
The whole geodesics (ones that are defined on ℝ) are called hyperbolic lines.
Proof. Let 𝑎∈ℝ+ ∩𝐷and 𝛾a smooth path from the origin to 𝑎. Let 𝛾(𝑡) = (𝑢(𝑡), 𝑣(𝑡)).
Note that Re(𝛾)(𝑡) = (𝑢(𝑡), 0) is also a smooth path from the origin to 𝑎. By definition of the hyperbolic metric, length(𝛾) = ∫ 1 0 2| ̇ 𝛾| 1 −|𝛾|2 d𝑡= ∫ 1 0 2√̇ 𝑢2 + ̇ 𝑣2 1 −𝑢2 −𝑣2 d𝑡≥∫ 1 0 2| ̇ 𝑢| 1 −𝑢2 d𝑡 where equality holds if and only if ̇ 𝑣≡0, and so 𝑣≡0.
length(𝛾) ≥∫ 1 0 2 ̇ 𝑢 1 −𝑢2 d𝑡 where equality holds in this expression if and only if 𝑢is monotonic. Hence, the arc of the diameter, parametrised monotonically, is a globally length-minimising path, and hence a geodesic. We can compute this integral to be length(𝛾) = 2 artanh 𝑎 592 6. Riemannian metrics Now, 0 and 𝑎in ℝ+ ∩𝐷are joined by a unique geodesic, and Möb(𝐷) acts transitively and by isometries, and can be used to send any two points 𝑝, 𝑞∈𝐷to 0, 𝑎∈ℝ+ ∩𝐷. So every pair of points must be joined by a unique geodesic. Since Möbius maps send circles to circles, and they preserve angles and hence orthogonality to the boundary, we must have that all geodesics are diameters or circular arcs orthogonal to 𝜕𝐷.
Corollary. If 𝑝, 𝑞∈𝐷, then the distance between them is 𝑑hyp(𝑝, 𝑞) = 2 artanh | | | 𝑝−𝑞 1 −𝑝𝑞 | | | 6.4. The hyperbolic upper half-plane Definition. The hyperbolic upper half-plane (𝔥, 𝑔hyp) is the set 𝔥= {𝑧∈ℂ∶Im 𝑧> 0} with the abstract Riemannian metric d𝑥2 + d𝑦2 𝑦2 = |d𝑧|2 (Im 𝑧)2 Lemma. The hyperbolic disc (𝐷2, 𝑔hyp) and the hyperbolic upper half-plane (𝔥, 𝑔hyp) are isometric.
Proof. There exist maps 𝑇∶𝔥→𝐷and ˜ 𝑇∶𝐷→𝔥given by 𝑇(𝑤) = 𝑤−𝑖 𝑤+ 𝑖; ˜ 𝑇(𝑧) = 𝑖(1 −𝑧 1 + 𝑧) which are inverse diffeomorphisms. Here, 𝑇′(𝑤) = 1 𝑤+ 𝑖− 𝑤−𝑖 (𝑤+ 𝑖)2 = 2𝑖 (𝑤+ 𝑖)2 Considering 𝑇(𝑤) = 𝑧∈𝐷, |d𝑧| 1 −|𝑧|2 = |d(𝑇𝑤)| 1 −|𝑇𝑤|2 = |𝑇′(𝑤)||d𝑤| 1 −|𝑇𝑤|2 = 2|d𝑤| |𝑤+ 𝑖|2(1 −| | 𝑤−𝑖 𝑤+𝑖 | | 2 ) = |d𝑤| 2 Im 𝑤 Hence, under this coordinate change, 4|d𝑧|2 (1 −|𝑧|2) 2 is the metric obtained under pullback by 𝑇from d𝑤2 (Im 𝑤)2 .
593 X. Geometry Corollary. The hyperbolic upper half-plane is globally isometric to the hyperbolic disc, so every pair of points is joined by a unique geodesic, up to reparametrisation. The geodesics are arcs of circles orthogonal to the boundary, which are vertical straight lines and semi-circles centred on a point in the real axis.
Proof. The isometry between 𝔥→𝐷is given by a Möbius map. In particular, ℝ∪{∞} ↦𝜕𝐷, and Möbius maps preserve circles and orthogonality.
Remark. When we discussed surfaces in ℝ3 with constant Gauss curvature, we saw that if a surface had constant Gauss curvature, its first fundamental form in geodesic normal coordinates was of the form d𝑢2 + cosh2 d𝑣2, with a change of variables taking this form to d𝑣2+d𝑤2 𝑤2 . This is exactly the form of the Riemannian metric on the hyperbolic upper half-plane. Gauss’ theorema egregium implies that Gauss curvature makes sense for an abstract Riemannian metric, since it only depends on geodesics and hence the first fundamental form. We can therefore define the Gauss curvature for an abstract Riemannian metric to agree with this definition for surfaces in ℝ3. Under this definition, we can show that the hyperbolic upper half-plane has constant curvature −1, and hence so does the disc.
Suppose we wanted to find a metric 𝑑∶𝐷× 𝐷→ℝ≥0 on 𝐷2 with the properties that it is invariant under the Möbius group Möb(𝐷), and that the real diameter is length-minimising.
By Möbius invariance, the distance between any two points is completely determined by knowing the distance from the origin to some point on the positive real axis 𝑎, which we will denote 𝑝(𝑎) = 𝑑(0, 𝑎). If ℝ+ ∩𝐷is length-minimising, distance should be additive, so if 0 ≤𝑎≤𝑏≤1 we should have 𝑑(0, 𝑎)+𝑑(𝑎, 𝑏) = 𝑑(0, 𝑏) so 𝑑(𝑎, 𝑏) = 𝑝( 𝑏−𝑎 1−𝑎𝑏) = 𝑝(𝑏)−𝑝(𝑎).
If we furthermore constrain 𝑝to be differentiable, and we differentiate the above expression with respect to 𝑏and set 𝑏= 𝑎, we find the differential equation 𝑝′(𝑎) = 𝑝′(0) 1 −𝑎2 Hence, 𝑝(𝑎) is some constant multiple of artanh 𝑎, since 𝑝′(0) can be chosen freely. So, up to rescaling the length metric associated to 𝑔hyp on 𝐷is the unique metric with these properties.
The scale is chosen for 𝑔hyp to enforce that the curvature is −1 precisely.
6.5. Isometries of hyperbolic space We now would like to understand the full isometry group of the disc (𝐷, 𝑔hyp) or (𝔥, 𝑔hyp). We will show that this group is precisely Möb(𝐷) together with reflections in hyperbolic lines, which are called inversions.
Definition. Let Γ ⊆ˆ ℂbe a circle or line. We say that points 𝑧, 𝑧′ ∈ˆ ℂare inverse for Γ if every circle through 𝑧orthogonal to Γ also passes through 𝑧′.
Lemma. Such inverse points exist and are unique.
594 6. Riemannian metrics Proof. Recall that Möbius maps preserve circles in ˆ ℂand preserve angles. In particular, if 𝑧, 𝑧′ are inverse for Γ and 𝑇∈Möb, then 𝑇𝑧and 𝑇𝑧′ are inverse for the circle 𝑇(𝛾). If Γ = ℝ∪{∞}, then 𝐽(𝑧) = 𝑧gives inverse points; this map satisfies the definition above. Now, if Γ ⊆ˆ ℂis any circle, there exists 𝑇∈Möb such that 𝑇(ℝ∪{∞}) = Γ. We can therefore define inversion in Γ to be 𝐽Γ = 𝑇∘(𝑧↦𝑧) ∘𝑇−1.
Definition. The map 𝐽Γ in the proof above, sending 𝑧to the unique inverse point 𝑧′ for 𝑧 with respect to Γ, is called inversion in Γ.
This map fixes all points of Γ, and swaps points on the interior with points on the exter-ior.
Example. For Γ a straight line, this is simply reflection. For the unit circle, 𝑆1, the map 𝐽𝑆1 maps 𝑧↦ 1 𝑧and 0 ↦∞.
Remark. The composition of two inversions is a Möbius map. Let 𝐶be the conjugation map 𝑧↦𝑧, which is 𝐽ℝ∪{∞}. If Γ ⊆ˆ ℂis any circle, we have 𝐽Γ = 𝑇∘𝐶∘𝑇−1 where 𝑇is the Möbius transformation which maps ℝ∪{∞} to Γ. If Γ 1, Γ 2 are circles, and 𝑇1, 𝑇2 are the transformations from ℝ∪{∞} to Γ 1, Γ 2 respectively, then 𝐽Γ1 ∘𝐽Γ2 = (𝐽Γ1 ∘𝐶) ∘(𝐶∘𝐽Γ1) = (𝐶∘𝐽Γ1) −1 ∘(𝐶∘𝐽Γ1) We have 𝐶∘𝐽Γ = 𝐶∘𝑇∘𝐶∘𝑇−1, so it suffices to show 𝐶∘𝑇∘𝐶∈Möb. If 𝑇(𝑧) = 𝑎𝑧+𝑏 𝑐𝑧+𝑑, we have (𝐶∘𝑇∘𝐶)(𝑧) = 𝑎𝑧+ 𝑏 𝑐𝑧+ 𝑑 ∈Möb Lemma. An orientation-preserving isometry of (ℍ2, 𝑔hyp) is an element of Möb(ℍ), where ℍis 𝐷or 𝔥. The full isometry group is generated by inversions in hyperbolic lines.
Proof. It suffices to prove this in either model, so we will use the disc model. Inversion in the geodesic ℝ∩𝐷is conjugation, which preserves 𝑔hyp. Note that Möb(ℍ) acts transitively by isometries on geodesics. Hence, if inversion in one geodesic preserves the metric, so does inversion in any geodesic.
Now, suppose 𝛼is some isometry of the hyperbolic disc 𝐷under the metric 𝑔hyp. We have 𝛼(0) = 𝑎∈𝐷, and using 𝑧↦ 𝑧−𝑎 1−𝑎𝑧, so there exists 𝑇∈Möb(𝐷) such that 𝑇∘𝛼fixes the origin. There exists a rotation 𝑅∈Möb(𝐷) such that 𝑅∘𝑇∘𝛼maps 𝐷∩ℝ+ to itself.
Composing with the conjugation map 𝐶if necessary, there exists an isometry 𝐴which is an inversion composed with a Möbius map such that 𝐴∘𝛼fixes 𝐷∩ℝpointwise and fixes 𝐷∩𝑖ℝ pointwise. The only such isometry is the identity, since every point in 𝐷is determined by its distance to these two lines. Hence, 𝐴is the inverse of 𝛼.
If 𝛼preserves orientation and fixes ℝ∩𝐷, then it necessarily fixes 𝑖ℝ∩𝐷pointwise, so 𝛼= (𝑅∘ 𝑇)−1 ∈Möb. In general, 𝛼was constructed from Möb(ℍ) and inversions in hyperbolic lines.
595 X. Geometry So to show that the isometry group is generated by inversions, it suffices to show that all Möbius maps are compositions of inversions. This is presented on the example sheets.
In the upper half-plane model of hyperbolic space, Möb(𝔥) = ℙ𝑆𝐿(2, ℝ) = {𝑧↦𝑎𝑧+ 𝑏 𝑐𝑧+ 𝑑∶(𝑎 𝑏 𝑐 𝑑) ∈𝑆𝐿(2, ℝ)}; 𝑑hyp = 2 artanh | | | 𝑏−𝑎 𝑏−𝑎 | | | 6.6. Hyperbolic triangles Definition. Let 𝛼be an orientation-preserving isometry of ℍ, which is equivalently an ele-ment of Möb(ℍ). Suppose 𝛼is not the identity map. We say that 𝛼is (i) elliptic, if 𝛼fixes some point 𝑝∈ℍ(if 𝑝= 0 ∈𝐷, this behaves like a rotation); (ii) parabolic, if 𝛼fixes a unique point 𝑝∈𝜕ℍ(if 𝑝= ∞∈𝔥, this behaves like a transla-tion); (iii) hyperbolic, if 𝛼fixes two points on 𝜕ℍ, so it fixes the unique geodesic between these two points setwise, and so 𝛼must translate points across the geodesic; it is not an inversion in the geodesic because it is not the identity map.
All elements of Möb(ℍ) are either elliptic, parabolic, or hyperbolic.
Definition. Let ℓ, ℓ′ be hyperbolic lines. Then, we say (i) parallel, if they meet at the boundary 𝜕ℍbut never inside ℍ; (ii) ultra-parallel, if they never meet in ℍ; (iii) intersecting, if they meet in ℍ.
All pairs of hyperbolic lines are either parallel, ultra-parallel, or intersecting. A hyperbolic triangle is a region bound by three geodesics, no two of which are ultra-parallel. Vertices that lie ‘at infinity’ (on 𝜕ℍ) are called ideal vertices.
Note that the points in 𝜕ℍare not contained within the hyperbolic plane, so in particular the ideal vertices are not points in ℍ. We typically denote side lengths by 𝐴, 𝐵, 𝐶, and denote the angles opposite these sides by 𝛼, 𝛽, 𝛾. The vertices at 𝛼, 𝛽, 𝛾are denoted 𝑎, 𝑏, 𝑐. The hyperbolic metric is conformal, since 𝐸= 𝐺and 𝐹= 0. Hence, we can use Euclidean angles in place of hyperbolic angles.
Proposition (hyperbolic cosine formula). For a hyperbolic triangle, cosh 𝐶= cosh 𝐴cosh 𝐵−sinh 𝐴sinh 𝐵cos 𝛾 Proof. To simplify, by an isometry we can let the vertex 𝑐at 𝛾be placed at 0 ∈𝐷, and the vertex 𝑏at 𝛽be placed at ℝ+ ∩𝐷. Hence, the sides 𝐴, 𝐵are straight Euclidean line segments 596 6. Riemannian metrics in 𝐷, and the angle between them is 𝛾. We have 𝑑hyp(0, 𝑎) = 2 artanh 𝑎⟹𝑎= tanh 𝐴 2 ; 𝑏= 𝑒𝑖𝛾tanh 𝐵 2 ; | | | 𝑏−𝑎 1 −𝑎𝑏 | | | = tanh 𝐶 2 Recall that 𝑡= tanh 𝜆 2 ⟹cosh 𝜆= 1 + 𝑡2 1 −𝑡2 ; sinh 𝜆= 2𝑡 1 −𝑡2 Hence, cosh 𝐴= 1 + |𝑎|2 1 −|𝑎|2 ; cosh 𝐵= 1 + |𝑏|2 1 −|𝑏|2 ; cosh 𝐶= | |1 −𝑎𝑏| | 2 + |𝑏−𝑎|2 | |1 −𝑎𝑏| | 2 −|𝑏−𝑎|2 = (1 + |𝑠|2)(1 + |𝑏|2) −2(𝑎𝑏+ 𝑎𝑏) (1 −|𝑎|2)(1 −|𝑏|2) Note that 𝑎∈ℝand 𝑏+ 𝑏= 2 Re 𝑏= 2𝑏cos 𝛾, so cosh 𝐶= cosh 𝐴cosh 𝐵−sinh 𝐴sinh 𝐵cos 𝛾 as required.
Remark. If 𝐴, 𝐵, 𝐶are small, the standard approximations to the hyperbolic sine and cosine functions give 𝐶2 ≈𝐴2 + 𝐵2 −2𝐴𝐵cos 𝛾 which is the Euclidean cosine formula. Since a dilation of a surface in ℝ3 rescales curvature, at small scales we can treat any abstract smooth surface with a Riemannian metric as flat.
Since cos 𝛾≥−1, we have that cosh 𝐶≤cosh 𝐴cosh 𝐵+ sinh 𝐴sinh 𝐵= cosh(𝐴+ 𝐵) The hyperbolic cosine is increasing, so 𝐶≤𝐴+ 𝐵. This is a more precise variant of the hyperbolic triangle inequality.
6.7. Area of triangles Theorem. Let 𝑇⊆ℍ2 be a hyperbolic triangle with internal angles 𝛼, 𝛽, 𝛾defined as before.
The area of 𝑇is areahyp(𝑇) = 𝜋−𝛼−𝛽−𝛾 Note that 𝛼, 𝛽, 𝛾may be zero, so 𝑇may have ideal vertices, and the internal angle is zero for such vertices.
This is a version of the Gauss–Bonnet theorem for hyperbolic triangles.
597 X. Geometry Proof. The Möbius group Möb(ℍ2) acts transitively on triples of points in the boundary with the correct cycle order. In particular, there exists a single ideal triangle (with all vertices at infinity) up to isometry. Consider the ideal triangle in the hyperbolic upper half-plane with vertices −1, +1, ∞. Its area is areahyp(𝑇) = ∫ 1 −1 ∫ ∞ √1−𝑥2 1 𝑦2 d𝑦d𝑥 since √𝐸𝐺−𝐹2 = 1 𝑦2 . We can compute this explicitly as areahyp(𝑇) = ∫ 1 −1 d𝑥 √1 −𝑥2 = 𝜋 Now, let 𝐴(𝛼) be the area of a triangle with angles 0, 0, 𝛼. We can see that 𝐴(𝛼) is decreasing in 𝛼and continuous in 𝛼, by fixing two ideal vertices in the hyperbolic disc and translating the third vertex.
𝛼 𝛼′ 𝛼𝛽 𝛼𝜋−𝛼 The first diagram shows that by moving the vertex 𝛼on the real line, the area must increase, since the triangle with angle 𝛼′ < 𝛼contains the triangle with angle 𝛼. From the second diagram, we see that 𝐴(𝛼) + 𝐴(𝛽) = 𝐴(𝛼+ 𝛽) + 𝜋by comparing the different areas of triangles formed from hyperbolic lines in the diagram. By letting 𝐹(𝛼) = 𝜋−𝐴(𝛼), we have 𝐹(𝛼) + 𝐹(𝛽) = 𝐹(𝛼+ 𝛽). Since 𝐹is continuous and increasing, we have that 𝐹(𝛼) = 𝜆𝛼 for some fixed 𝜆> 0. In particular, 𝐴(𝛼) = 𝜋−𝜆𝛼. Now, by considering the angles in the third diagram, we see that 𝐴(𝛼) + 𝐴(𝜋−𝛼) = 𝜋. Hence, 𝜆= 1, and so 𝐴(𝛼) = 𝜋−𝛼.
Finally, we consider the general case.
598 6. Riemannian metrics 𝐶′ 𝐵′ 𝐴 𝐶 𝐴′ 𝛽 𝛾 𝛼 𝐵 By writing 𝐴𝐵𝐶for areahyp(𝑇) where 𝑇is the triangle with vertices 𝐴, 𝐵, 𝐶, we can see that 𝐴𝐵𝐶+ 𝐴′𝐶𝐵′ + 𝐴′𝐵′𝐶′ = area of interior of diagram = 𝐴𝐵′𝐶′ + 𝐴′𝐵𝐶′ Equivalently, 𝐴𝐵𝐶+ 𝜋−(𝜋−𝛾) + 𝜋= (𝜋−𝛼) + (𝜋−𝛽) ⟹𝐴𝐵𝐶= 𝜋−𝛼−𝛾−𝛽 as required.
Note that if 𝐺is a hyperbolic 𝑛-gon, so it is a region bound by 𝑛hyperbolic geodesics, it may be decomposed into a union of hyperbolic triangles. Since any two points in ℍ2 are joined by a unique geodesic, the area of 𝐺is given by areahyp(𝐺) = (𝑛−2)𝜋− 𝑛 ∑ 𝑖=1 𝛼𝑖 Lemma. If 𝑔≥2, then there exists a regular 4𝑔-gon in ℍ2 with internal angle 2𝜋 4𝑔= 𝜋 2𝑔.
Proof. Consider an ideal 4𝑔-gon, whose vertices all lie at infinity, in the disc model of hy-perbolic space. The ideal vertices can be placed at the 4𝑔-th roots of unity, such that this polygon is invariant under a rotational symmetry. By sliding each vertex radially inwards in ℝ2, we obtain a continuous family of regular 4𝑔-gons, with areas which vary monotonically from (4𝑔−2)𝜋to zero. The internal angle of the polygon therefore varies continuously from zero to 𝛽min such that (4𝑔−2)𝜋= 4𝑔𝛽min. It therefore suffices to check that 𝜋 2𝑔lies in this interval (0, 𝛽min).
599 X. Geometry 6.8. Surfaces of constant negative curvature Theorem. For each 𝑔≥2, there exists an abstract Riemannian metric on the compact surface of genus 𝑔with curvature 𝜅≡−1 and locally isometric to ℍ2.
Recall the the Euler characteristic of a surface of genus 𝑔is exactly 2 −2𝑔. Note, if 𝑔= 0 we can construct a Riemannian metric with 𝜅≡+1 since this is the sphere, and if 𝑔= 1 we can have 𝜅≡0 since this is the torus as a quotient ℝ2 ⟋ ℤ2. We will outline two proofs.
Proof. Recall that we can construct the torus and double torus by 𝑏−1 𝑎 𝑏 𝑎−1 Analogously, a 4𝑔-gon with side labels 𝑎1𝑏1𝑎−1 1 𝑏−1 1 𝑎2𝑏2𝑎−1 2 𝑏−1 2 … gives a surface of genus 𝑔.
We say that a flag comprises an oriented hyperbolic line, a point on that line, and a choice of side to that line. Given two such flags, there exists a hyperbolic isometry between them.
So Möb(ℍ) acts transitively on flags. In particular, we can swap the side of a flag using an inversion.
Consider a regular hyperbolic 4𝑔-gon with internal angle 𝜋 2𝑔. We label this polygon with side labels as above to give a genus 𝑔surface. For each paired set of two edges, there exists a hyperbolic isometry taking one to the other, respecting orientations and, and taking the side corresponding to the inside of the polygon to the side corresponding to the outside of the polygon. This is possible since Möb(ℍ) acts transitively on flags.
We can now define an atlas for Σ𝑔as follows.
• If 𝑝is in the interior of the polygon 𝑃, consider a small disc contained in the interior of the polygon. Then, include this disc into the hyperbolic disc 𝐷.
• If 𝑝is contained in an edge, let ̂ 𝑝be the corresponding point on the paired edge. We have an isometry 𝛾from edge 𝑒1 to edge 𝑒2, exchanging sides, and mapping 𝑝to ̂ 𝑝. We can use this to define the chart. Using 𝛾, we can combine 𝑈, the intersection of 𝑃with an open neighbourhood of 𝑝, and ˜ 𝑈, the intersection of 𝑃with an open neighbourhood of ̂ 𝑝, such that the chart is an inclusion on 𝑈and is 𝛾on ˜ 𝑈. These agree on 𝑈∩˜ 𝑈.
• All 4𝑔vertices are identified to one point of Σ, and we need a chart at this point. Using a hyperbolic isometry, let one vertex 𝑣of 𝑃be at the origin in 𝐷, such that an edge 𝑒 containing 𝑣is mapped to a subset of the real line. Since the polygon 𝑃has internal angle 𝜋 2𝑔, the angle between ℝand the adjacent edge is 𝜋 2𝑔. The fact that the internal angles sum to 2𝜋means that we can construct hyperbolic isometries for each vertex 600 6. Riemannian metrics that join them exactly, giving an open neighbourhood of zero in 𝐷in the shape of a disc. The chart is defined at [𝑣] ∈Σ𝑔by this identification.
All charts are obtained from inclusion or an inclusion composed with a hyperbolic isometry, therefore the transition maps are hyperbolic isometries. In particular, hyperbolic isometries are smooth, and preserve the locally defined hyperbolic metric.
Remark. The torus can be given by ℝ2 ⟋ ℤ2. This characterisation was useful when describing the flat metric, precisely because its charts are easy to define. For Σ𝑔, we chose 2𝑔hyperbolic isometries which paired sides. Hence, there is a group Γ ≤Möb(ℍ), generated by these isometries. In Part II Algebraic Topology, the surface Σ𝑔will be constructed by ℍ ⟋ Γ.
Lemma. For each ℓ𝛼, ℓ𝛽, ℓ𝛾> 0, there exists a right-angled hyperbolic hexagon with side lengths ℓ𝛼, 𝑎, ℓ𝛽, 𝑏, ℓ𝛾, 𝑐for some 𝑎, 𝑏, 𝑐.
Proof. Given 𝑡> 0, there exists a pair of ultra-parallel hyperbolic lines a distance 𝑡apart.
We show on the fourth example sheet that each pair of ultra-parallel hyperbolic lines has a unique common perpendicular geodesic. Given lengths ℓ𝛼, ℓ𝛽, construct new perpendic-ular geodesics orthogonal to the originals, having moved lengths ℓ𝛼, ℓ𝛽from the common perpendicular (in the same direction). If 𝑡is made large, the new geodesics 𝜎, ̃ 𝜎can be made ultraparallel. Hence, by making 𝑡smaller, there exists a threshold 𝑡0 by continuity such that the new geodesics are parallel. Now, for 𝑡∈(𝑡0, ∞), the two new geodesics are ultra-parallel.
So 𝜎, ̃ 𝜎have a unique common perpendicular geodesic. As 𝑡increases above 𝑡0, the length of this line increases monotonically from zero to infinity. So there exists a value of 𝑡> 𝑡0 such that the new common perpendicular has length ℓ𝛾.
𝑡 ℓ𝛼 ℓ𝛽 ℓ𝛾 This is exactly the right-angled hyperbolic hexagon as required.
Definition. A pair of pants is a topological space homeomorphic to the complement of three open discs in 𝑆2.
601 X. Geometry Note that this space has a boundary. Consider two right-angled hyperbolic hexagons with side lengths ℓ𝛼, ℓ𝛽, ℓ𝛾arranged as above. The original configuration of two ultra-parallel geodesics of a distance 𝑡apart is unique up to isometry. So the side lengths have a corres-pondence, and the hexagon with side lengths ℓ𝛼, ℓ𝛽, ℓ𝛾is unique up to isometry. Suppose that we glue together the corresponding unknown sides 𝑡𝛼𝛽, 𝑡𝛽𝛾, 𝑡𝛾𝛼with the same side iden-tifications. Locally near ℓ𝛼, for instance, we arrive at a closed circle of length 2ℓ𝛼, extended into a cylindrical shape with two seams 𝑡𝛼𝛽, 𝑡𝛾𝛼. Since the hexagons were right-angled, we have constructed a hyperbolic pair of pants. The boundary circles are geodesics in the sense that, for any point on such a circle, the local neighbourhood is a point on a geodesic on a polygon in ℍ.
We will now construct Σ𝑔using a more flexible approach.
Proof. If 𝑃 1, 𝑃 2 are two hyperbolic ‘surfaces’ with geodesic boundaries, and if 𝛾1 ⊂𝑃 1 and 𝛾2 ⊂𝑃 2 are boundary circles of the same length (in the hyperbolic metric), we can glue 𝑃 1 and 𝑃 2 together along this common-length circle. 𝑃 1 and 𝑃 2 may be glued by any isometry of 𝛾1, 𝛾2. The result 𝑃 1 ∪𝛾1∼𝛾2 𝑃 2 has a hyperbolic metric. For any point 𝑝∈𝑃𝑖not on the boundary 𝛾𝑖, it already has a suitable open neighbourhood since 𝑃𝑖is hyperbolic. For any point 𝑝∈𝛾1 ∼𝛾2, we have a chart to a small disc in ℍusing the fact that the boundary circles are geodesics. These charts are constructed analogously to the charts for points on edges of hyperbolic polygons under appropriate side identifications as seen above. Any compact surface of genus 𝑔≥2 can be built from glued pairs of pants, not necessarily uniquely.
Under this construction, we have many choices. For example, the lengths of circles in the original hyperbolic hexagons are now arbitrary. Also, the choice of ‘pants decomposition’ of a given surface is not unique, and the different possibilities are topologically different.
6.9. Gauss–Bonnet theorem Recall that in a spherical triangle with internal angles 𝛼, 𝛽, 𝛾, we have seen in the example sheets that this has area 𝛼+ 𝛽+ 𝛾−𝜋, and that a hyperbolic triangle with the same internal angles has area 𝜋−𝛼−𝛽−𝛾. We have seen the convex Gauss–Bonnet theorem, which states ∫ Σ 𝜅d𝐴= 4𝜋 where Σ bounds a convex region in ℝ3 and 𝜅Σ > 0. These are special cases of a pair of theorems as shown below.
Theorem (local Gauss–Bonnet theorem). Let Σ be an abstract smooth surface with abstract Riemannian metric 𝑔. Let 𝑅be an 𝑛-sided geodesic polygon on Σ, which is a smooth disc with boundary decomposed into 𝑛geodesic arcs. Then ∫ 𝑅⊆Σ 𝜅Σ d𝐴= 𝑛 ∑ 𝑖=1 𝛼𝑖−(𝑛−2)𝜋 where the 𝛼𝑖are the internal angles of the polygon.
602 6. Riemannian metrics It is important that 𝛾𝑖be geodesics that cut out a disc; 𝑅must be homeomorphic to ℝ2, and it cannot (for example) contain any holes.
Theorem (global Gauss–Bonnet theorem). Let Σ be a compact smooth surface with abstract Riemannian metric 𝑔. Then ∫ Σ 𝜅Σ d𝐴= 2𝜋𝜒(Σ) Remark. Gauss curvature can be defined using only the first fundamental form, or equival-ently an abstract Riemannian metric.
For hyperbolic surfaces, we can construct Σ𝑔from a 4𝑔-gon with internal angles 𝜋 2𝑔in such a way that the total area of Σ is exactly the area of the polygon, so ∫ Σ 1 d𝐴= area(polygon) = (4𝑔−2)𝜋− 4𝑔 ∑ 1 𝜋 2𝑔= (4𝑔−4)𝜋 Since 𝜅≡−1 and 𝜒(Σ𝑔) = 2 −2𝑔, this agrees with the Gauss–Bonnet theorem.
A right-angled hyperbolic hexagon has area 4𝜋− 6 ∑ 1 𝜋 2 = 𝜋 Each pair of pants was constructed from two such polygons, and to construct a genus 𝑔 surface we required 2𝑔−2 pairs of pants. So the total area is 4𝑔−4𝜋, which agrees with the theorem.
The Gauss–Bonnet theorem also shows that the Euler characteristic does not depend on the choice of triangulation of Σ.
Suppose Σ is a flat surface and 𝛾is a closed geodesic, so 𝛾∶ℝ→Σ and is periodic with some period 𝑇. Then 𝛾cannot bound a smooth disc in Σ. Conversely, on 𝑆2, the great circle is a closed geodesic, and bounds a hemisphere. For instance, for the flat torus ℝ2 ⟋ ℤ2, if 𝛾is a closed curve on this torus bounding a closed disc 𝑅it is not a geodesic. Indeed, if we formally add two vertices to such a geodesic, we find a geodesic 2-gon with two internal angles 𝜋, but by the Gauss–Bonnet theorem we expect 0 = ∫ 𝑅 𝜅Σ d𝐴= 2 ∑ 1 𝛼𝑖−(𝑛−2)𝜋= 2𝜋 We can in fact deduce the global Gauss–Bonnet theorem from the local Gauss–Bonnet the-orem, utilising the following lemma.
Lemma. A compact smooth surface admits subdivisions into geodesic polygons.
603 X. Geometry The proof of this lemma considers the exponential map, discussed in Part II. Given such a subdivision on Σ, we can find ∑ polygons 𝑃 ∫ 𝑃 𝜅Σ d𝐴= ∫ Σ 𝜅Σ d𝐴 By the local Gauss–Bonnet theorem, the left hand side is equal to ∑ 𝑛 ∑ 𝑛-gons 𝑃 ( 𝑛 ∑ 𝑖=1 𝛼𝑖(𝑃) −(𝑛−2)𝜋) Since the angles at each point add to 2𝜋, and each 𝑛-gon contains two edges which each separate two polygons, this is equal to 2𝜋𝑉+ 2𝜋𝐹−2𝜋𝐸= 2𝜋𝜒(Σ) as required.
6.10. Green’s theorem (non-examinable) The local Gauss–Bonnet theorem is very closely related to Green’s theorem in ℝ2. This dis-cussion is non-examinable.
Theorem. Let 𝑅⊆ℝ2 be a region bound by a piecewise smooth curve 𝛾, and 𝑃, 𝑄be smooth real-valued functions defined on an open set 𝑉⊃𝑅. Then ∫ 𝛾 𝑃d𝑢+ 𝑄d𝑣= ∫ 𝑅 (𝑄𝑢−𝑃 𝑣) d𝑢d𝑣 We will consider a geodesic polygon on Σ which lies in the domain of some local paramet-risation defined on 𝑉⊆ℝ2. Consider an orthonormal basis for ℝ2 which varies from point to point, defined by 𝑒= 𝜎𝑢, 𝑓= 𝜎𝑣/√𝐺where we use geodesic normal coordinates 𝑢, 𝑣to give 𝐸= 1, 𝐹= 0. Then 𝑇 𝑝Σ = span(𝑒, 𝑓) if 𝑝∈Im 𝜎. We parametrise 𝛾by arc length and consider 𝐼= ∫ 𝛾 ⟨𝑒, ̇ 𝑓⟩d𝑡 We will compute this in two ways. Note that ̇ 𝑓= 𝑓 𝑢̇ 𝑢+ 𝑓 𝑣̇ 𝑣 Let 𝑃= ⟨𝑒, 𝑓 𝑢⟩and 𝑄= ⟨𝑒, 𝑓 𝑣⟩. Then 𝑄𝑢−𝑃 𝑣= ⟨𝑒𝑢, 𝑓 𝑣⟩−⟨𝑓 𝑣, 𝑒𝑢⟩+ ⟨𝑒, 𝑓 𝑢𝑣⟩−⟨𝑒, 𝑓 𝑢𝑣⟩= ⟨𝑒𝑢, 𝑓 𝑣⟩−⟨𝑓 𝑢, 𝑒𝑣⟩ which we can show to be equal to −(√𝐺) 𝑢𝑢= 𝜅√𝐺. But √𝐺is the area element √𝐸𝐺−𝐹2, so ∫ 𝑅 (𝑄𝑢−𝑃 𝑣) d𝑢d𝑣= ∫ 𝑅 𝜅Σ d𝐴 604 6. Riemannian metrics Let 𝜃(𝑡) be the angle between ̇ 𝛾(𝑡) and 𝑒(𝑡), which is a function of 𝑡in the domain of 𝛾. More precisely, ̇ 𝛾= 𝑒cos 𝜃(𝑡) + 𝑓sin 𝜃(𝑡) Thus ̈ 𝛾= ̇ 𝑒cos 𝜃+ ̇ 𝑓sin 𝜃+ 𝜂̇ 𝜃; 𝜂= −𝑒sin 𝜃+ 𝑓cos 𝜃 𝛾is a piecewise geodesic, so if Σ ⊆ℝ3 was smooth, ̈ 𝛾is orthogonal to 𝑇 𝑝Σ = span 𝑒, 𝑓. But 𝜂∈⟨𝑒, 𝑓⟩, so ̈ 𝛾is orthogonal to 𝜂. By expanding, ⟨̇ 𝑒cos 𝜃+ ̇ 𝑓sin 𝜃+ 𝜂̇ 𝜃, −𝑒sin 𝜃+ 𝑓cos 𝜃⟩= 0 Since 𝑒, 𝑓are orthogonal unit vectors, we have ⟨𝑒, ̇ 𝑒⟩= 0 = ⟨𝑓, ̇ 𝑓⟩and ⟨𝑒, ̇ 𝑓⟩= 0 = ⟨̇ 𝑒, 𝑓⟩, so we can expand to find ⟨̈ 𝛾, 𝜂⟩= 0 ⟹ ̇ 𝜃= ⟨𝑒, ̇ 𝑓⟩ Thus, 𝐼= ∫ 𝛾 ⟨𝑒, ̇ 𝑓⟩d𝑡= ∫ 𝛾 ̇ 𝜃(𝑡) d𝑡= 2𝜋−∑(external angles of 𝑅) since 𝛾is composed of straight lines. Since external angles and internal angles sum to 𝜋, this is exactly the local Gauss–Bonnet theorem. Green’s theorem suggests the study of non-geodesic polygons.
6.11. Alternate flat toruses We have constructed a flat metric on the torus, viewed as ℝ2 ⟋ ℤ2, or as [0, 1]2 ⟋ ∼for a suitably defined equivalence relation. Importantly, opposite sides of the square [0, 1]2 were identified by translation, which allowed us to find a smooth atlas where transition maps preserve the usual Euclidean metric on ℝ2. This construction is valid for any parallelogram; any such shape 𝑄⊆ℝ2 defines a flat metric 𝑔𝑄on 𝑇2. If one vertex is set to zero in ℝ2 and the edges of this vertex are labelled by their endpoints 𝑣1, 𝑣2, then (𝑇2, 𝑔𝑄) = ℝ2 ⟋ ℤ𝑣1 ⊕ℤ𝑣2 where ℤ𝑣1 ⊕ℤ𝑣2 is a viewed as a subgroup of the group ℝ2 of translations.
The area with respect to 𝑔𝑄of 𝑇2 is the Euclidean area of the parallelogram 𝑄. In particu-lar, if two parallelograms have different areas, the two metrics cannot be globally isometric.
However, this is not the only restriction for global isometries.
Lemma. Consider the torus based on 𝑄= [0, 1]2 and the torus based on ̂ 𝑄= [0, 10]×[0, 1 10].
The metrics 𝑔𝑄, 𝑔̂ 𝑄are not isometric, but both have unit total area.
Proof. Recall that geodesics in a flat torus correspond to straight lines in ℝ2. By Picard’s theorem, there exists a unique geodesic from a given point 𝑝for each direction in 𝑇 𝑝Σ. We can therefore see that all geodesics through 𝑝are the images of straight lines in ℝ2.
Recall that a closed geodesic is defined on ℝand is periodic. We can see that geodesics in ℝ2 through ̂ 𝑝∈𝑞−1(𝑝) define a closed geodesic if and only if they pass through another lift 605 X. Geometry ̂ 𝑝′ ∈𝑞−1(𝑝) of 𝑝; that is, the line has rational gradient. The shortest closed geodesic on the surface in metric 𝑄is of unit length, but the shortest closed geodesic with metric ̂ 𝑄is 1 10. So the surfaces are not globally isometric.
We would like to understand all possible flat metrics on the torus 𝑇2, up to global dilation and Euclidean isometries of 𝑄, which lead to essentially the same geometry on the quotient torus. Given any parallelogram, we can set one vertex at zero and another at (1, 0) = 1 ∈ℝ2 by performing dilation and a Euclidean isometry, and then the third lies at 𝜏and the fourth at 1+ 𝜏, where 𝜏has positive 𝑦-coordinate. This provides a metric on the torus, and now the only degree of freedom is 𝜏. Hence, this defines a map from the upper half-plane to the set of flat metrics on 𝑇2 up to dilation.
We can pull back metrics by diffeomorphisms. Metrics allow us to measure lengths of curves by integrating lengths of tangent vectors, so a metric can be viewed as an inner product on the tangent space at each point. If 𝑓∶Σ →Σ′ and 𝑝∈Σ, then for two small curves 𝛾1, 𝛾2 through 𝑝, the pullback metric 𝑓⋆𝑔was defined such that ⟨̇ 𝛾1, ̇ 𝛾2⟩𝑝,𝑓⋆𝑔= ⟨𝑓∘̇ 𝛾1, 𝑓∘̇ 𝛾2⟩𝑓(𝑝),𝑔 𝑆𝐿(2, ℤ) acts on ℝ2 preserving ℤ2, so it acts on ℝ2 ⟋ ℤ2 = 𝑇2.
Lemma. 𝑆𝐿(2, ℤ) acts by diffeomorphisms on 𝑇2.
Proof. Clearly 𝐴∈𝑆𝐿(2, ℤ) acts smoothly (indeed, linearly) on ℝ2, and the charts for the smooth atlas are such that 𝐴then acts smoothly with respect to these.
Also, 𝑆𝐿(2, ℤ) ⊆𝑆𝐿(2, ℝ) acts on the upper half-plane by Möbius maps.
Theorem. The map from the upper half-plane 𝔥to the set of flat metrics on 𝑇2 modulo dilation induces a map from 𝔥 ⟋ 𝑆𝐿(2, ℤ) to the set of flat metrics on 𝑇2 modulo dilation and diffeomorphism. This resulting map is a bijection. We say that 𝔥 ⟋ 𝑆𝐿(2, ℤ) is the moduli space of flat metrics on 𝑇2.
In the above theorem, ‘diffeomorphism’ is taken to mean ‘orientation-preserving diffeomorph-ism’.
Remark. The left-hand side 𝔥 ⟋ 𝑆𝐿(2, ℤ) is an object of hyperbolic geometry, yet the right-hand side is entirely concerned with flat metrics.
Similar results can be shown for surfaces of higher genus. The moduli space of hyperbolic metrics on Σ𝑔where 𝑔≥2 is perhaps the most studied space in all of geometry.
6.12. Further courses There are four Part II courses that extend this course.
606 6. Riemannian metrics (i) Algebraic Topology. Spaces are studied through algebraic invariants, such as the Euler characteristic, and covering maps of surfaces like 𝑆2 →ℝℙ2 or ℝ2 →𝑇2.
(ii) Differential Geometry. While in IB Geometry the Gauss curvature 𝜅= det(𝐷𝑁) is discussed, the trace tr(𝐷𝑁) is the mean curvature, discussed heavily in this course.
(iii) Riemann Surfaces. This course studies the fact that if 𝑓∶ℂ→ℂis holomorphic (or, indeed, entire) and 𝑤∈ℂ, then 𝑓(𝑧+ 𝑤) is holomorphic, and if 𝑓∶𝐷→𝐷is holomorphic and 𝐴∈Möb(𝐷), then 𝑓∘𝐴is holomorphic.
(iv) General Relativity. This is the theory of light as geodesics.
607 XI. Statistics Lectured in Lent 2022 by Dr. S. Bacallado An estimator is a random variable that approximates a parameter. For instance, the para-meter could be the mean of a normal distribution, and the estimator could be a sample mean. In this course, we study how estimators behave, what properties they have, and how we can use them to make conclusions about the real parameters. This is called parametric inference: the study of inferring parameters from statistics of sample data.
Towards the end of the course, we study the normal linear model, which is a useful way to model data that is believed to depend linearly on a vector of inputs, together with some normally distributed noise. Even nonlinear patterns can be analysed using this model, by letting the inputs to the model be polynomials in the real-world data.
609 XI. Statistics Contents 1.
Introdution and review of IA Probability . . . . . . . . . . . . . . . 612 1.1.
Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . 612 1.2.
Review of IA Probability . . . . . . . . . . . . . . . . . . . . . 612 1.3.
Standardised statistics . . . . . . . . . . . . . . . . . . . . . . 614 1.4.
Moment generating functions . . . . . . . . . . . . . . . . . . . 614 1.5.
Limit theorems . . . . . . . . . . . . . . . . . . . . . . . . . . 615 1.6.
Conditional probability . . . . . . . . . . . . . . . . . . . . . . 615 1.7.
Change of variables in two dimensions . . . . . . . . . . . . . . 616 1.8.
Common distributions . . . . . . . . . . . . . . . . . . . . . . 616 2.
Estimation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 617 2.1.
Estimators . . . . . . . . . . . . . . . . . . . . . . . . . . . . 617 2.2.
Bias-variance decomposition . . . . . . . . . . . . . . . . . . . 617 2.3.
Sufficiency . . . . . . . . . . . . . . . . . . . . . . . . . . . . 619 2.4.
Factorisation criterion . . . . . . . . . . . . . . . . . . . . . . 619 2.5.
Minimal sufficiency . . . . . . . . . . . . . . . . . . . . . . . . 620 2.6.
Rao–Blackwell theorem . . . . . . . . . . . . . . . . . . . . . . 621 2.7.
Maximum likelihood estimation . . . . . . . . . . . . . . . . . 624 3.
Inference . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 626 3.1.
Confidence intervals . . . . . . . . . . . . . . . . . . . . . . . 626 3.2.
Interpreting the confidence interval . . . . . . . . . . . . . . . 628 4.
Bayesian analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . 629 4.1.
Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . 629 4.2.
Inference from the posterior . . . . . . . . . . . . . . . . . . . 630 4.3.
Point estimation . . . . . . . . . . . . . . . . . . . . . . . . . 630 4.4.
Credible intervals . . . . . . . . . . . . . . . . . . . . . . . . . 631 5.
Hypothesis testing . . . . . . . . . . . . . . . . . . . . . . . . . . . 632 5.1.
Hypotheses . . . . . . . . . . . . . . . . . . . . . . . . . . . . 632 5.2.
Testing hypotheses . . . . . . . . . . . . . . . . . . . . . . . . 632 5.3.
Neyman–Pearson lemma . . . . . . . . . . . . . . . . . . . . . 633 5.4.
𝑝-values . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 635 5.5.
Composite hypotheses . . . . . . . . . . . . . . . . . . . . . . 635 5.6.
Generalised likelihood ratio test . . . . . . . . . . . . . . . . . 636 5.7.
Wilks’ theorem . . . . . . . . . . . . . . . . . . . . . . . . . . 637 5.8.
Goodness of fit . . . . . . . . . . . . . . . . . . . . . . . . . . 638 5.9.
Pearson statistic . . . . . . . . . . . . . . . . . . . . . . . . . . 638 5.10.
Goodness of fit for composite null . . . . . . . . . . . . . . . . 639 5.11.
Testing independence in contingency tables . . . . . . . . . . . 640 610 5.12.
Testing homogeneity in contingency tables . . . . . . . . . . . . 641 5.13.
Tests and confidence sets . . . . . . . . . . . . . . . . . . . . . 642 6.
The normal linear model . . . . . . . . . . . . . . . . . . . . . . . . 644 6.1.
Multivariate normal distribution . . . . . . . . . . . . . . . . . 644 6.2.
Orthogonal projections . . . . . . . . . . . . . . . . . . . . . . 645 6.3.
Linear model . . . . . . . . . . . . . . . . . . . . . . . . . . . 647 6.4.
Matrix formulation . . . . . . . . . . . . . . . . . . . . . . . . 648 6.5.
Assumptions . . . . . . . . . . . . . . . . . . . . . . . . . . . 648 6.6.
Least squares estimation . . . . . . . . . . . . . . . . . . . . . 648 6.7.
Fitted values and residuals . . . . . . . . . . . . . . . . . . . . 650 6.8.
Normal linear model . . . . . . . . . . . . . . . . . . . . . . . 650 6.9.
Inference . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 651 6.10.
𝐹-tests . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 654 6.11.
Analysis of variance . . . . . . . . . . . . . . . . . . . . . . . . 657 6.12.
Simple linear regression . . . . . . . . . . . . . . . . . . . . . 658 611 XI. Statistics 1. Introdution and review of IA Probability 1.1. Introduction Statistics can be defined as the science of making informed decisions. The field comprises, for example: • the design of experiments and studies; • visualisation of data; • formal statistical inference (which is the focus of this course); • communication of uncertainty and risk; and • formal decision theory.
This course concerns itself with parametric inference. Let 𝑋1, … , 𝑋𝑛be i.i.d. (independent and identically distributed) random variables, where we assume that the distribution of 𝑋1 belongs to some family with parameter 𝜃∈Θ. For instance, let 𝑋1 ∼Poi(𝜇), where 𝜃= 𝜇 and Θ = (0, ∞). Another example is 𝑋1 ∼𝑁(𝜇, 𝜎2), and 𝜃= (𝜇, 𝜎2) and Θ = ℝ× (0, ∞). We use the observed 𝑋= (𝑋1, … , 𝑋𝑛) to make inferences about the parameter 𝜃: (i) we can estimate the value of 𝜃using a point estimate written ̂ 𝜃(𝑋); (ii) we can make an interval estimate of 𝜃, written ( ̂ 𝜃1(𝑋), ̂ 𝜃2(𝑋)); (iii) hypotheses about 𝜃can be tested, for instance the hypothesis 𝐻0 ∶𝜃= 1, by checking whether there is evidence in the data 𝑋against the hypothesis 𝐻0.
Remark. In general, we will assume that the family of distributions of the observations 𝑋𝑖 is known a priori, and the parameter 𝜃is the only unknown. There will, however, be some remarks later in the course where we can make weaker assumptions about the family.
1.2. Review of IA Probability This subsection reviews material covered in the IA Probability course. Some keywords are measure-theoretic, and are not defined.
Let Ω be the sample space of outcomes in an experiment. A measurable subset of Ω is called an event, and we denote the set of events by ℱ. A probability measure ℙ∶ℱ→[0, 1] satisfies the following properties.
(i) ℙ(∅) = 0; (ii) ℙ(ℱ) = 1; (iii) ℙ(⋃ ∞ 𝑖=1 𝐴𝑖) = ∑ ∞ 𝑖=1 ℙ(𝐴𝑖) if (𝐴𝑖) is a sequence of disjoint events.
A random variable is a measurable function 𝑋∶Ω →ℝ. The distribution function of a ran-dom variable 𝑋is the function 𝐹𝑋(𝑥) = ℙ(𝑋≤𝑥). We say that a random variable is discrete 612 1. Introdution and review of IA Probability when it takes values in a countable set 𝒳⊂ℝ. The probability mass function of a discrete random variable is the function 𝑝𝑋(𝑥) = ℙ(𝑋= 𝑥). We say that 𝑋has a continuous distribu-tion if it has a probability density function 𝑓 𝑋(𝑥) such that ℙ(𝑥∈𝐴) = ∫ 𝐴𝑓 𝑋(𝑥) d𝑥for ‘nice’ sets 𝐴.
The expectation of a random variable 𝑋is defined as 𝔼[𝑋] = {∑𝑥∈𝑋𝑥𝑝𝑋(𝑥) if 𝑋discrete ∫ ∞ −∞𝑥𝑓 𝑋(𝑥) d𝑥 if 𝑋continuous If 𝑔∶ℝ→ℝ, we define 𝔼[𝑔(𝑋)] by considering the fact that 𝑔(𝑋) is also a random variable.
For instance, in the continuous case, 𝔼[𝑔(𝑋)] = ∫ ∞ −∞ 𝑔(𝑥)𝑓 𝑋(𝑥) d𝑥 The variance of a random variable 𝑋is defined as 𝔼[(𝑋−𝔼[𝑋])2].
We say that a set of random variables 𝑋1, … , 𝑋𝑛are independent if, for all 𝑥1, … , 𝑥𝑛, we have ℙ(𝑋1 ≤𝑥1, … , 𝑋𝑛≤𝑥𝑛) = ℙ(𝑋1 ≤𝑥1) ⋯ℙ(𝑋𝑛≤𝑥𝑛) If and only if 𝑋1, … , 𝑋𝑛have probability density (or mass) functions 𝑓 1, … , 𝑓 𝑛, then the joint probability density (respectively mass) function is 𝑓 𝑋(𝑥) = 𝑛 ∏ 𝑖=1 𝑓 𝑋𝑖(𝑥𝑖) If 𝑌= max {𝑋1, … , 𝑋𝑛} where the 𝑋𝑖are independent, then the distribution function of 𝑌is given by ℙ(𝑌≤𝑦) = ℙ(𝑋1 ≤𝑦) ⋯ℙ(𝑋𝑛≤𝑦) The probability density function of 𝑌(if it exists) is obtained by the differentiating the above.
Under a linear transformation, the expectation and variance have certain properties. Let 𝑎= (𝑎1, … , 𝑎𝑛)⊺∈ℝ𝑛be a constant in ℝ𝑛.
𝔼[𝑎1𝑋1 + ⋯+ 𝑎𝑛𝑋𝑛] = 𝔼[𝑎⊺𝑋] = 𝑎⊺𝔼[𝑋] where 𝔼[𝑋] is defined componentwise. Note that independence of 𝑋𝑖is not required for linearity of the expectation to hold. Similarly, Var (𝑎⊺𝑋) = ∑ 𝑖,𝑗 𝑎𝑖𝑎𝑗Cov (𝑋𝑖, 𝑋𝑗) = 𝑎⊺Var (𝑋) 𝑎 where we define Cov (𝑋, 𝑌) ≡𝔼[(𝑋−𝔼[𝑋])(𝑌−𝔼[𝑌])], and Var (𝑋) is the variance-covariance matrix with entries (Var (𝑋))𝑖𝑗= Cov (𝑋𝑖, 𝑋𝑗). We can say that the variance is bilinear.
613 XI. Statistics 1.3. Standardised statistics Suppose that 𝑋1, … , 𝑋𝑛are i.i.d. and 𝔼[𝑋1] = 𝜇, Var (𝑋1) = 𝜎2. We define 𝑆𝑛= ∑ 𝑖 𝑋𝑖; 𝑋𝑛= 𝑆𝑛 𝑛 where 𝑋𝑛is called the sample mean. By linearity of expectation and bilinearity of vari-ance, 𝔼[𝑋𝑛] = 𝜇; Var (𝑋𝑛) = 𝜎2 𝑛 We further define 𝑍𝑛= 𝑆𝑛−𝑛𝜇 𝜎√𝑛 = √𝑛𝑋𝑛−𝜇 𝜎 which has the properties that 𝔼[𝑍𝑛] = 0; Var (𝑍𝑛) = 1 1.4. Moment generating functions The moment generating function of a random variable 𝑋is the function 𝑀𝑋(𝑡) = 𝔼[𝑒𝑡𝑋], provided that this function exists for 𝑡in some neighbourhood of zero, This can be thought of as the Laplace transform of the probability density function. Note that 𝔼[𝑋𝑛] = d𝑛 d𝑡𝑛𝑀𝑋(𝑡)| | |𝑡=0 Under broad conditions, moment generating functions uniquely define a distribution func-tion of a random variable. In other words, the Laplace transform is invertible. They are also useful for finding the distribution of sums of independent random variables. For instance, let 𝑋1, … , 𝑋𝑛be i.i.d. Poisson random variables with parameter 𝜇. Then, the moment gener-ating function of 𝑋𝑖is 𝑀𝑋1(𝑡) = 𝔼[𝑒𝑡𝑋𝑖] = ∞ ∑ 𝑥=0 𝑒𝑡𝑥𝑒−𝜇𝜇𝑥 𝑥! = 𝑒−𝜇 ∞ ∑ 𝑥=0 (𝑒𝑡𝜇)𝑥 𝑥!
= 𝑒−𝜇𝑒𝜇𝑒𝑡= 𝑒−𝜇(1−𝑒𝑡) Now, 𝑀𝑆𝑛(𝑡) = 𝔼[𝑒𝑡𝑆𝑛] = 𝑛 ∏ 𝑖=1 𝔼[𝑒𝑡𝑋𝑖] = 𝑒−𝑛𝜇(1−𝑒𝑡) This defines a Poisson distribution with parameter 𝑛𝜇by inspection.
614 1. Introdution and review of IA Probability 1.5. Limit theorems The weak law of large numbers states that for all 𝜀> 0, ℙ(| |𝑋𝑛−𝜇| | > 𝜀) →0 as 𝑥→∞. Note that the event | |𝑋𝑛−𝜇| | > 𝜀depends only on 𝑋1, … , 𝑋𝑛.
The strong law of large numbers states that ℙ(𝑋𝑛→𝜇) = 1. In this formulation, the event depends on the whole sequence of random variables 𝑋𝑖, since the limit is inside the probab-ility calculation.
The central limit theorem states that 𝑍𝑛= 𝑆𝑛−𝑛𝜇 𝜎√𝑛is approximately a N(0, 1) random variable when 𝑛is large. More precisely, ℙ(𝑍𝑛≤𝑧) →Φ(𝑧) for all 𝑧∈ℝ.
1.6. Conditional probability If 𝑋, 𝑌are discrete random variables, we can define the conditional probability mass func-tion to be 𝑝𝑋∣𝑌(𝑥∣𝑦) = ℙ(𝑋= 𝑥, 𝑌= 𝑦) ℙ(𝑌= 𝑦) when ℙ(𝑌= 𝑦) ≠0. If 𝑋, 𝑌are continuous, we define the joint probability density function to be 𝑓 𝑋,𝑌(𝑥, 𝑦) such that ℙ(𝑋≤𝑥, 𝑌≤𝑦) = ∫ 𝑥 −∞ ∫ 𝑦 −∞ 𝑓(𝑥′, 𝑦′) d𝑦′ d𝑥′ The conditional probability density function is 𝑓 𝑋∣𝑌(𝑥∣𝑦) = 𝑓 𝑋,𝑌(𝑥, 𝑦) ∫ ∞ −∞𝑓 𝑋,𝑌(𝑥, 𝑦) d𝑥 The denominator is sometimes referred to as the marginal probability density function of 𝑌, written 𝑓 𝑌(𝑦). Now, we can define the conditional expectation by 𝔼[𝑋∣𝑌] = {∑𝑥𝑥𝑝𝑋∣𝑌(𝑥∣𝑌) if 𝑋discrete ∫ 𝑥𝑥𝑓 𝑋∣𝑌(𝑥∣𝑌) d𝑥 if 𝑋continuous The conditional expectation is itself a random variable, as it is a function of the random variable 𝑌. The conditional variance is defined similarly, and is a random variable. The tower property is that 𝔼[𝔼[𝑋∣𝑌]] = 𝔼[𝑋] The law of total variance is that Var (𝑋) = 𝔼[Var (𝑋∣𝑌)] + Var (𝔼[𝑋∣𝑌]) 615 XI. Statistics 1.7. Change of variables in two dimensions Suppose that (𝑥, 𝑦) ↦(𝑢, 𝑣) is a differentiable bijection from ℝ2 to itself. Then, the joint probability density function of 𝑈, 𝑉can be written as 𝑓 𝑈,𝑉(𝑢, 𝑣) = 𝑓 𝑋,𝑌(𝑥(𝑢, 𝑣), 𝑦(𝑢, 𝑣))|det 𝐽| where 𝐽is the Jacobian matrix, 𝐽= 𝜕(𝑥, 𝑦) 𝜕(𝑢, 𝑣) = (𝜕𝑥/𝜕𝑢 𝜕𝑥/𝜕𝑣 𝜕𝑦/𝜕𝑢 𝜕𝑦/𝜕𝑣) 1.8. Common distributions 𝑋has the binomial distribution with parameters 𝑛, 𝑝if 𝑋represents the number of successes in 𝑛independent Bernoulli trials with parameter 𝑝.
𝑋has the multinomial distribution with parameters 𝑛; 𝑝1, … , 𝑝𝑘if there are 𝑛independ-ent trials with 𝑘types, where 𝑝𝑗is the probability of type 𝑗in a single trial. Here, 𝑋takes values in ℕ𝑘, and 𝑋𝑗is the amount of trials with type 𝑗. Each 𝑋𝑗is marginally binomially distributed.
𝑋has the negative binomial distribution with parameters 𝑘, 𝑝if, in i.i.d. Bernoulli trials with parameter 𝑝, the variable 𝑋is the time at which the 𝑘th success occurs. The negative binomial with parameter 𝑘= 1 is the geometric distribution.
The Poisson distribution with parameter 𝜆is the limit of the distribution Bin(𝑛, 𝜆/𝑛) as 𝑛→ ∞.
If 𝑋𝑖∼Γ(𝛼𝑖, 𝜆) for 𝑖= 1, … , 𝑛with 𝑋1, … , 𝑋𝑛independent, then the distribution of 𝑆𝑛is given by the product of the moment generating functions. By inspection, 𝑀𝑆𝑛(𝑡) = ( 𝜆 𝜆−𝑡) ∑𝑖𝛼𝑖 or ∞if 𝑡≥𝜆. Hence the sum of these random variables is 𝑆𝑛∼Γ(∑𝑖𝛼𝑖, 𝜆), where the shape parameter 𝛼is constructed from the sum of the shape parameters of the original functions.
We call 𝜆the rate parameter, and 𝜆−1 is called the scale parameter. If 𝑋∼Γ(𝛼, 𝜆), then for all 𝑏> 0 we have 𝑏𝑋∼Γ(𝑥, 𝜆/𝑏). Special cases of the Γ distribution include: • Γ(1, 𝜆) = Exp(𝜆); • Γ(𝑘/2, 1/2) = 𝜒2 𝑘with 𝑘degrees of freedom, which is the distribution of a sum of 𝑘 i.i.d. squared standard normal random variables.
616 2. Estimation 2. Estimation 2.1. Estimators Suppose 𝑋1, … , 𝑋𝑛are i.i.d. observations with a p.d.f. (or p.m.f.) 𝑓 𝑋(𝑥∣𝜃), where 𝜃is an unknown parameter in some parameter space Θ. Let 𝑋= (𝑋1, … , 𝑋𝑛).
Definition. An estimator is a statistic, or a function of the data, written 𝑇(𝑋) = ̂ 𝜃, which is used to approximate the true value of 𝜃. This does not depend (explicitly) on 𝜃. The distribution of 𝑇(𝑋) is called its sampling distribution.
Example. Let 𝑋1, … , 𝑋𝑛∼𝑁(0, 1) be i.i.d. Let ̂ 𝜇= 𝑇(𝑋) = 𝑋𝑛. The sampling distribution is 𝑇(𝑋) ∼𝑁(𝜇, 1 𝑛). Note that this sampling distribution in general depends on the true parameter 𝜇.
Definition. The bias of ̂ 𝜃is bias( ̂ 𝜃) = 𝔼𝜃[ ̂ 𝜃] −𝜃 Note that ̂ 𝜃is a function only of 𝑋1, … , 𝑋𝑛, and the expectation operator 𝔼𝜃assumes that the true value of the parameter is 𝜃.
Remark. In general, the bias is a function of the true parameter 𝜃, even though it is not explicit in the notation.
Definition. An estimator with zero bias for all 𝜃is called an unbiased estimator.
Example. The estimator ̂ 𝜇in the above example is unbiased, since 𝔼𝜇[ ̂ 𝜇] = 𝔼𝜇[𝑋𝑛] = 𝜇 for all 𝜇∈ℝ.
Definition. The mean squared error of 𝜃is defined as mse( ̂ 𝜃) = 𝔼𝜃[( ̂ 𝜃−𝜃) 2] Remark. Like the bias, the mean squared error is, in general, a function of the true parameter 𝜃.
2.2. Bias-variance decomposition The mean squared error can be written as mse( ̂ 𝜃) = 𝔼𝜃[( ̂ 𝜃−𝔼𝜃[ ̂ 𝜃] + 𝔼𝜃[ ̂ 𝜃] −𝜃) 2] = Var𝜃( ̂ 𝜃) + bias2( ̂ 𝜃) Note that both the variance and bias squared terms are positive. This implies a tradeoff between bias and variance when minimising error.
617 XI. Statistics Example. Let 𝑋∼Bin(𝑛, 𝜃) where 𝑛is known and 𝜃is an unknown probability. Let 𝑇 𝑈= 𝑋/𝑛. This is the proportion of successes observed. This is an unbiased estimator, since 𝔼𝜃[𝑇 𝑈] = 𝔼𝜃[𝑋] /𝑛= 𝜃. The mean squared error for the estimator is then Var𝜃(𝑇𝑛) = Var𝜃(𝑋 𝑛) = Var𝜃(𝑋) 𝑛2 = 𝜃(1 −𝜃) 𝑛 Now, consider an alternative estimator which has some bias: 𝑇𝐵= 𝑋+ 1 𝑛+ 2 = 𝑤𝑋 𝑛 ⏟ 𝑇𝑈 +(1 −𝑤)1 2; 𝑤= 𝑛 𝑛+ 2 This interpolates between the estimator 𝑇 𝑈and the fixed estimator 1 2. Here, bias(𝑇𝐵) = 𝔼𝜃[𝑇𝐵] −𝜃= 𝑛 𝑛+ 2𝜃− 1 𝑛+ 2𝜃 The bias is nonzero for all but one value of 𝜃. Further, Var𝜃(𝑇𝐵) = Var𝜃(𝑋+ 1) (𝑛+ 2)2 = 𝑛𝜃(1 −𝜃) (𝑛+ 2)2 We can calculate mse(𝑇𝐵) = (1 −𝑤)2(1 2 −𝜃) 2 + 𝑤2 𝜃(1 −𝜃) 𝑛 ⏟ ⎵ ⏟ ⎵ ⏟ mse(𝑇𝑈) There exists a range of 𝜃such that 𝑇𝐵has a lower mean squared error, and similarly there exists a range such that 𝑇 𝑈has a lower error. This indicates that prior judgement of the true value of 𝜃can be used to determine which estimator is better.
It is not necessarily desirable that an estimator is unbiased.
Example. Suppose 𝑋∼Poi(𝜆) and we wish to estimate 𝜃= ℙ(𝑋= 0) 2 = 𝑒−2𝜆. For some estimator 𝑇(𝑋) of 𝜃to be unbiased, we need that 𝔼𝜆[𝑇(𝑋)] = ∞ ∑ 𝑥=0 𝑇(𝑥)𝜆𝑥𝑒−𝜆 𝑥!
= 𝑒−2𝜆 Hence, ∞ ∑ 𝑥=0 𝑇(𝑥)𝜆𝑥 𝑥! = 𝑒−𝜆 But 𝑒−𝜆has a known power series expansion, giving 𝑇(𝑋) ≡(−1)𝑋for all 𝑋. This is not a good estimator, for example because it often predicts negative numbers for a positive quant-ity.
618 2. Estimation 2.3. Sufficiency Definition. A statistic 𝑇(𝑋) is sufficient for 𝜃if the conditional distribution of 𝑋given 𝑇(𝑋) does not depend on 𝜃. Note that 𝜃and 𝑇(𝑋) may be vector-valued, and need not have the same dimension.
Example. Let 𝑋1, … , 𝑋𝑛be i.i.d. Bernoulli random variables with parameter 𝜃where 𝜃∈ [0, 1]. The mass function is 𝑓 𝑋(𝑥∣𝜃) = 𝑛 ∏ 𝑖=1 𝜃𝑥𝑖(1 −𝜃)1−𝑥𝑖= 𝜃∑𝑥𝑖(1 −𝜃)𝑛−∑𝑥𝑖 Note that this dependent only on 𝑥via the statistic 𝑇(𝑋) = ∑ 𝑛 𝑛=1 𝑥𝑖. Here, 𝑓 𝑋∣𝑇=𝑡(𝑥∣𝜃) = ℙ𝜃(𝑋= 𝑥, 𝑇(𝑋) = 𝑡) ℙ𝜃(𝑇(𝑥) = 𝑡) If ∑𝑥𝑖= 𝑡, we have 𝑓 𝑋∣𝑇=𝑡(𝑥∣𝜃) = 𝜃∑𝑥𝑖(1 −𝜃)𝑛−∑𝑥𝑖 (𝑛 𝑡)𝜃𝑡(1 −𝜃)𝑛−∑𝑥𝑖= 1 (𝑛 𝑡) Hence 𝑇(𝑋) is sufficient for 𝜃.
2.4. Factorisation criterion Theorem. 𝑇is sufficient for 𝜃if and only if 𝑓 𝑋(𝑥∣𝜃) = 𝑔(𝑇(𝑥), 𝜃)ℎ(𝑥) for suitable functions 𝑔, ℎ.
Proof. This will be proven in the discrete case; the continuous case can be handled analog-ously. Suppose that the factorisation criterion holds. Then, if 𝑇(𝑥) = 𝑡, 𝑓 𝑋∣𝑇=𝑡(𝑥∣𝑇= 𝑡) = ℙ𝜃(𝑋= 𝑥, 𝑇(𝑥) = 𝑡) ℙ𝜃(𝑇(𝑥) = 𝑡) = 𝑔(𝑇(𝑥), 𝜃)ℎ(𝑥) ∑𝑥′ ∶𝑇(𝑥′)=𝑡𝑔(𝑇(𝑥′), 𝜃)ℎ(𝑥′) = ℎ(𝑥) ∑𝑥′ ∶𝑇(𝑥′)=𝑡ℎ(𝑥′) which does not depend on 𝜃. By definition, 𝑇(𝑋) is sufficient.
Conversely, suppose that 𝑇(𝑋) is sufficient.
𝑓 𝑋(𝑥∣𝜃) = ℙ𝜃(𝑋= 𝑥) = ℙ𝜃(𝑋= 𝑥, 𝑇(𝑋) = 𝑇(𝑥)) = ℙ𝜃(𝑋= 𝑥∣𝑇(𝑋) = 𝑇(𝑥)) ⏟⎵ ⎵ ⎵ ⎵ ⎵ ⎵ ⎵⏟⎵ ⎵ ⎵ ⎵ ⎵ ⎵ ⎵⏟ ℎ(𝑥) ℙ𝜃(𝑇(𝑋) = 𝑇(𝑥)) ⏟⎵ ⎵ ⎵ ⎵⏟⎵ ⎵ ⎵ ⎵⏟ 𝑔(𝑇(𝑋),𝜃) 619 XI. Statistics Example. Consider the above example with 𝑛Bernoulli random variables with mass func-tion 𝑓 𝑋(𝑥∣𝜃) = 𝜃∑𝑥𝑖(1 −𝜃)𝑛−∑𝑥𝑖 Let 𝑇(𝑋) = ∑𝑥𝑖, and then the above mass function is in the form of 𝑔(𝑇(𝑋), 𝜃) and we can set ℎ(𝑥) ≡1. Hence 𝑇(𝑋) is sufficient.
Example. Let 𝑋1, … , 𝑋𝑛be i.i.d. from a uniform distribution on the interval [0, 𝜃] for some 𝜃> 0. The mass function is 𝑓 𝑋(𝑥∣𝜃) = 𝑛 ∏ 𝑖=1 1 𝜃𝟙{𝑥𝑖∈[0, 𝜃]} = (1 𝜃) 𝑛 𝟙{min 𝑖 𝑥𝑖≥0}𝟙{max 𝑖 𝑥𝑖≤𝜃} Let 𝑇(𝑋) = max𝑖𝑋𝑖. Then 𝑔(𝑇(𝑋), 𝜃) = (1 𝜃) 𝑛 𝟙{max 𝑖 𝑥𝑖≤𝜃}; ℎ(𝑥) ≡𝟙{min 𝑖 𝑥𝑖≥0} We can then conclude that 𝑇(𝑋) is sufficient for 𝜃.
2.5. Minimal sufficiency Sufficient statistics are not unique. For instance, any bijection applied to a sufficient stat-istic is also sufficient. Further, 𝑇(𝑋) = 𝑋is always sufficient. We instead seek statistics that maximally compress and summarise the relevant data in 𝑋and that discard extraneous data.
Definition. A sufficient statistic 𝑇(𝑋) for 𝜃is minimal if it is a function of every other sufficient statistic for 𝜃. More precisely, if 𝑇′(𝑋) is sufficient, 𝑇′(𝑥) = 𝑇′(𝑦) ⟹𝑇(𝑥) = 𝑇(𝑦).
Remark. Any two minimal statistics 𝑆, 𝑇for the same 𝜃are bijections of each other. That is, 𝑇(𝑥) = 𝑇(𝑦) if and only if 𝑆(𝑥) = 𝑆(𝑦).
Theorem. Suppose that 𝑓 𝑋(𝑥∣𝜃)/𝑓 𝑋(𝑦∣𝜃) is constant in 𝜃if and only if 𝑇(𝑥) = 𝑇(𝑦). Then 𝑇is minimal sufficient.
Remark. This theorem essentially states the following. Let 𝑥 1 ∼𝑦if the above ratio of prob-ability density or mass functions is constant in 𝜃. This is an equivalence relation. Similarly, we can define 𝑥 2 ∼𝑦if 𝑇(𝑥) = 𝑇(𝑦). This is also an equivalence relation. The hypothesis in the theorem is that the equivalence classes of 1 ∼and 2 ∼are equal. Further, we may always construct a minimal sufficient statistic for any parameter since we can use the construction 1 ∼to create equivalence classes, and set 𝑇to be constant for all such equivalence classes.
620 2. Estimation Proof. Let 𝑡∈Im 𝑇. Then let 𝑧𝑡be a representative of the equivalence class {𝑥∶𝑇(𝑥) = 𝑡}.
Then 𝑓 𝑋(𝑥∣𝜃) = 𝑓 𝑋(𝑧𝑇(𝑥) ∣𝜃) 𝑓 𝑋(𝑥∣𝜃) 𝑓 𝑋(𝑧𝑇(𝑥) ∣𝜃) By the hypothesis, the ratio on the right hand side does not depend on 𝜃, so let this ratio be ℎ(𝑥). Further, the other term depends only on 𝑇(𝑥), so it may be 𝑔(𝑇(𝑥), 𝜃). Hence 𝑇is sufficient by the factorisation criterion.
To prove minimality, let 𝑆be any other sufficient statistic, and then by the factorisation criterion there exist 𝑔𝑆and ℎ𝑆such that 𝑓 𝑋(𝑥∣𝜃) = 𝑔𝑆(𝑆(𝑥), 𝜃)ℎ𝑆(𝑥). Now, suppose 𝑆(𝑥) = 𝑆(𝑦) for some 𝑥, 𝑦. Then, 𝑓 𝑋(𝑥∣𝜃) 𝑓 𝑋(𝑦∣𝜃) = 𝑔𝑆(𝑆(𝑥), 𝜃)ℎ𝑆(𝑥) 𝑔𝑆(𝑆(𝑦), 𝜃)ℎ𝑆(𝑦) = ℎ𝑆(𝑥) ℎ𝑆(𝑦) which is constant in 𝜃. Hence, 𝑥 1 ∼𝑦. By the hypothesis, we have 𝑥 2 ∼𝑦, so 𝑇(𝑥) = 𝑇(𝑦), which is the requirement for minimality.
Example. Let 𝑋1, … , 𝑋𝑛be normal with unknown 𝜇, 𝜎2.
𝑓 𝑋(𝑥∣𝜇, 𝜎2) 𝑓 𝑋(𝑦∣𝜇, 𝜎2) = (2𝜋𝜎2)−𝑛/2 exp{− 1 2𝜎2 ∑𝑖(𝑥𝑖−𝜇)2} (2𝜋𝜎2)−𝑛/2 exp{− 1 2𝜎2 ∑𝑖(𝑦𝑖−𝜇)2 } = exp{−1 2𝜎2 (∑ 𝑖 𝑥2 𝑖−∑ 𝑖 𝑦2 𝑖) + 𝜇 𝜎2 (∑ 𝑖 𝑥𝑖−∑ 𝑖 𝑦𝑖)} Hence, for minimality, this is constant in the parameters 𝜇, 𝜎2 if and only if ∑𝑖𝑥2 𝑖= ∑𝑖𝑦2 𝑖 and ∑𝑖𝑥𝑖= ∑𝑖𝑦𝑖. Thus, a minimal sufficient statistic is (∑𝑖𝑥2 𝑖, ∑𝑖𝑥𝑖) is a minimal suffi-cient statistic. A more common way of expressing the minimal sufficient statistic is 𝑆(𝑥) = (𝑋𝑛, 𝑆𝑥𝑥); 𝑋𝑛= 1 𝑛∑ 𝑖 𝑥𝑖; 𝑆𝑥𝑥= ∑ 𝑖 (𝑋𝑖−𝑋𝑛) 2 which is a bijection of the above.
Example. 𝜃and a minimal statistic 𝑇need not have the same dimension. Consider 𝑋1, … , 𝑋𝑛∼ 𝑁(𝜇, 𝜇2). Here, there is a single parameter 𝜇but the minimal sufficient statistic is still 𝑆(𝑥) as defined above.
2.6. Rao–Blackwell theorem Previously, the notation 𝔼𝜃and ℙ𝜃have been used to denote expectations and probabilities under the model where the observations are i.i.d. with p.d.f. or p.m.f. 𝑓 𝑋. From now, we omit this subscript, as it will be implied for much of the remainder of the course.
621 XI. Statistics Theorem. Let 𝑇be a sufficient statistic for 𝜃, and define an estimator ̃ 𝜃with 𝔼[ ̃ 𝜃2] < ∞ for all 𝜃. Now we define another estimator ̂ 𝜃= 𝔼[ ̃ 𝜃∣𝑇(𝑥)] Then, for all values of 𝜃, we have 𝔼[( ̂ 𝜃−𝜃) 2] ≤𝔼[( ̃ 𝜃−𝜃) 2] In other words, the mean squared error of ̂ 𝜃is not greater than the mean squared error of ̃ 𝜃.
Further, the inequality is strict unless ̃ 𝜃is a function of 𝑇.
Remark. Starting from any estimator ̃ 𝜃, if we condition on the sufficient statistic 𝑇we obtain a ‘better’ statistic ̂ 𝜃. Note that 𝑇must be sufficient, otherwise ̂ 𝜃may be a function of 𝜃and thus not an estimator: ̂ 𝜃(𝑋) = ̂ 𝜃(𝑇) = ∫̂ 𝜃(𝑥) 𝑓 𝑋∣𝑇(𝑥∣𝑇) ⏟ ⎵ ⎵ ⏟ ⎵ ⎵ ⏟ does not depend on 𝜃as 𝑇is sufficient d𝑥 Proof. By the tower property of the expectation, we can find 𝔼[ ̂ 𝜃] = 𝔼[𝔼[ ̃ 𝜃∣𝑇(𝑥)]] = 𝔼[ ̃ 𝜃] Hence, subtracting ̃ 𝜃from both sides, we find bias( ̂ 𝜃) = bias( ̃ 𝜃). By the conditional variance formula, Var ( ̃ 𝜃) = 𝔼[ Var ( ̃ 𝜃∣𝑇) ⏟ ⎵ ⎵ ⏟ ⎵ ⎵ ⏟ ≥0 ] + Var (𝔼[ ̃ 𝜃∣𝑇]) ⏟⎵ ⎵ ⎵⏟⎵ ⎵ ⎵⏟ Var( ̂ 𝜃) ≥Var ( ̂ 𝜃) By the bias-variance decomposition, we know that mse( ̃ 𝜃) ≥mse( ̂ 𝜃). The inequality is strict unless Var ( ̃ 𝜃∣𝑇) = 0 almost surely. This requires that ̃ 𝜃is a function of 𝑇.
Example. Let 𝑋1, … , 𝑋𝑛be i.i.d. Poisson random variables with parameter 𝜆. Then let 𝜃= ℙ(𝑋1 = 0) = 𝑒−𝜆. Here, 𝑓 𝑋(𝑥∣𝜆) = 𝑒−𝑛𝜆𝜆∑𝑥𝑖 ∏𝑥𝑖!
⟹𝑓 𝑋(𝑥∣𝜃) = 𝜃𝑛(−log 𝜃)∑𝑥𝑖 ∏𝑥𝑖!
Using the factorisation criterion, we find 𝑔(𝑇(𝑥), 𝜃) = 𝑔(∑𝑥𝑖, 𝜃) = 𝜃𝑛(−log 𝜃)∑𝑥𝑖; ℎ(𝑥) = 1 ∏𝑥𝑖!
so 𝑇(𝑥) = ∑𝑥𝑖is sufficient. Note that ∑𝑋𝑖has a Poisson distribution with parameter 𝑛𝜆.
Consider the estimator ̃ 𝜃= 𝟙{𝑋1 = 0}. This depends only on 𝑋1, hence it is a weak estimator.
622 2. Estimation However, it is unbiased, so when we apply the Rao–Blackwell theorem we will construct an unbiased ̂ 𝜃, which is precisely ̂ 𝜃= 𝔼[ ̃ 𝜃∣∑𝑋𝑖= 𝑡] = ℙ(𝑋1 = 0 ∣∑𝑋𝑖= 𝑡) = ℙ(𝑋1 = 0, ∑𝑋𝑖= 𝑡) ℙ(∑𝑋𝑖= 𝑡) = ℙ(𝑋1 = 0) ℙ(∑ 𝑛 𝑖=2 𝑋𝑖= 𝑡) ℙ(∑ 𝑛 𝑖=1 𝑋𝑖= 𝑡) = (𝑛−1 𝑛 ) 𝑡 This may also be written ̂ 𝜃= (1 −1 𝑛) ∑𝑥𝑖 which is an estimator with lower mean squared error than ̃ 1 for all 𝜃. Note that ̂ 𝜃= (1 = 1 𝑛) 𝑛𝑋𝑛 converges in the limit to 𝑒−𝑋𝑛. By the strong law of large numbers, 𝑋𝑛→𝔼[𝑋1] = 𝜆, so we arrive at ̂ 𝜃→𝑒−𝜆= 𝜃almost surely.
Example. Let 𝑋1, … , 𝑋𝑛be i.i.d. uniform random variables in an interval [0, 𝜃]. We wish to estimate 𝜃> 0. We observed that 𝑇= max 𝑋𝑖is sufficient for 𝜃. Let ̃ 𝜃= 2𝑋1. This is an unbiased estimator of 𝜃. Then the Rao–Blackwellised estimator ̂ 𝜃is ̂ 𝜃= 𝔼[ ̃ 𝜃∣𝑇= 𝑡] = 2𝔼[𝑋1 ∣max 𝑋𝑖= 𝑡] = 2𝔼[𝑋1 ∣max 𝑋𝑖= 𝑡, 𝑋1 = max 𝑋𝑖] ℙ(𝑋1 = max 𝑋𝑖∣max 𝑋𝑖= 𝑡) + 2𝔼[𝑋1 ∣max 𝑋𝑖= 𝑡, 𝑋1 ≠max 𝑋𝑖] ℙ(𝑋1 ≠max 𝑋𝑖∣max 𝑋𝑖= 𝑡) Since 𝑋1, … , 𝑋𝑛are i.i.d., the conditional probability ℙ(𝑋1 = max 𝑋𝑖∣max 𝑋𝑖= 𝑡) can be reduced to ℙ(𝑋1 = max 𝑋𝑖) = 1 𝑛. The complementary event may be reduced in an analogous way. The expectation 𝔼[𝑋1 ∣max 𝑋𝑖= 𝑡, 𝑋1 = max 𝑋𝑖] can be reduced to 𝑡.
̂ 𝜃= 2𝑡 𝑛+ 2(𝑛−1) 𝑛 𝔼[𝑋1 ∣𝑋1 < 𝑡, 𝑛 max 𝑖=2 𝑋𝑖= 𝑡] = 2𝑡 𝑛+ 2(𝑛−1) 𝑛 𝔼[𝑋1 ∣𝑋1 < 𝑡] = 2𝑡 𝑛+ 2(𝑛−1) 𝑛 𝑡 2 = 2𝑡 𝑛+ 𝑡(𝑛−1) 𝑛 = 𝑛+ 1 𝑛 max 𝑖 𝑋𝑖 By the Rao–Blackwell theorem, the mean squared error of ̂ 𝜃is not greater than the mean squared error of ̃ 𝜃. This is also an unbiased estimator.
623 XI. Statistics 2.7. Maximum likelihood estimation Let 𝑋1, … , 𝑋𝑛be i.i.d. random variables with mass or density function 𝑓 𝑋(𝑥∣𝜃).
Definition. For fixed observations 𝑥, the likelihood function 𝐿∶Θ →ℝis given by 𝐿(𝜃) = 𝑓 𝑋(𝑥∣𝜃) = 𝑛 ∏ 𝑖=1 𝑓 𝑋𝑖(𝑥𝑖∣𝜃) We will denote the log-likelihood by ℓ(𝜃) = log 𝐿(𝜃) = 𝑛 ∑ 𝑖=1 log 𝑓 𝑋𝑖(𝑥𝑖∣𝜃) Definition. A maximum likelihood estimator is an estimator that maximises the likelihood function 𝐿over Θ. Equivalently, the estimator maximises ℓ.
Example. Let 𝑋1, … , 𝑋𝑛be i.i.d. Bernoulli random variables with parameter 𝑝. The log-likelihood function is ℓ(𝑝) = 𝑛 ∑ 𝑖=1 [𝑋𝑖log 𝑝+ (1 −𝑋𝑖) log(1 −𝑝)] = log 𝑝+ ∑𝑋𝑖+ log(1 −𝑝)(𝑛−∑𝑋𝑖) The derivative is ℓ′(𝑝) = ∑𝑋𝑖 𝑝 + 𝑛−∑𝑋𝑖 1 −𝑝 which has a single stationary point at 𝑝= 1 𝑛∑𝑋𝑖= 𝑋𝑛. We have 𝔼[ ̂ 𝑝] = 𝑝, so the maximum likelihood estimator in this case is unbiased.
Example. Let 𝑋1, … , 𝑋𝑛be i.i.d. normal random variables with unknown mean 𝜇and vari-ance 𝜎2.
ℓ(𝜇, 𝜎2) = −𝑛 2 log(2𝜋) −𝑛 2 log 𝜎2 − 1 2𝜎2 ∑(𝑋𝑖−𝜇)2 This function is concave in 𝜇and 𝜎2, so there exists a unique maximiser. In particular, ℓis maximised when 𝜕ℓ 𝜕𝜇= 𝜕ℓ 𝜕𝜎2 = 0.
𝜕ℓ 𝜕𝜇= −1 𝜎2 ∑(𝑋𝑖−𝜇) This is zero if 𝜇= 𝑋𝑛. Further, 𝜕ℓ 𝜕𝜎2 = −𝑛 2𝜎2 + 1 2𝜎4 ∑(𝑋𝑖−𝜇)2 = −𝑛 2𝜎2 + 1 2𝜎4 ∑(𝑋𝑖−𝑋𝑛)2 This is zero if and only if 𝜎2 = 1 𝑛∑(𝑋𝑖−𝑋𝑛)2 = 𝑆𝑥𝑥 𝑛 624 2. Estimation Hence, the maximum likelihood estimator is ( ̂ 𝜇, ̂ 𝜎2) = (𝑋𝑛, 1 𝑛𝑆𝑥𝑥). We can show that ̂ 𝜇is unbiased. We will later prove that 𝑆𝑥𝑥 𝜎2 = 𝑛̂ 𝜎2 𝜎2 ∼𝜒2 𝑛−1 Hence 𝔼[ ̂ 𝜎2] = 𝜎2 𝑛𝔼[𝜒2 𝑛−1] = 𝜎2 𝑛−1 𝑛 This is therefore a biased estimator, but the bias converges to zero as 𝑛→∞: ̂ 𝜎2 is asymp-totically unbiased.
Example. Let 𝑋1, … , 𝑋𝑛be i.i.d. uniform random variables on [0, 𝜃]. Here, we derived the unbiased estimator ̂ 𝜃= 𝑛+1 𝑛max 𝑋𝑖. The likelihood is given by 𝐿(𝜃) = 1 𝜃𝑛𝟙{max 𝑋𝑖≤𝜃} This function is maximised at ̂ 𝜃mle = max 𝑋𝑖. By comparison to the ̂ 𝜃derived from the Rao–Blackwell process, ̂ 𝜃mle is biased. In particular, 𝔼[ ̂ 𝜃mle] = 𝑛 𝑛+ 1𝔼[ ̂ 𝜃] = 𝑛 𝑛+ 1𝜃 Remark. If 𝑇is a sufficient statistic for 𝜃, then the maximum likelihood estimator is a func-tion of 𝑇. Indeed, since 𝑋and 𝑇are fixed, the maximiser of 𝐿(𝜃) = 𝑔(𝑇, 𝜃)ℎ(𝑋) depends on 𝑋only through 𝑇. If 𝜑= 𝐻(𝜃) for a bijection 𝐻, then if ̂ 𝜃is the maximum likelihood estimator for 𝜃, we have that 𝐻( ̂ 𝜃) is the maximum likelihood estimator for 𝜑.
Under some regularity conditions, as 𝑛→∞the statistic √𝑛( ̂ 𝜃−𝜃) is approximately normal with mean zero and covariance matrix Σ. More precisely, for ‘nice’ sets 𝐴, we have ℙ(√𝑛( ̂ 𝜃−𝜃) ∈𝐴) →ℙ(𝑍∈𝐴) ; 𝑍∼𝑁(0, Σ) We say that the maximum likelihood estimator is asymptotically normal. The limiting cov-ariance matrix Σ is a known function of ℓ, which will not be defined in this course. In some sense, Σ is the smallest variance that any estimator can achieve asymptotically.
For practical purposes, this estimator can often be found numerically by maximising ℓor 𝐿.
625 XI. Statistics 3. Inference 3.1. Confidence intervals Definition. A 100𝛾% confidence interval for a parameter 𝜃is a random interval (𝐴(𝑋), 𝐵(𝑋)) such that ℙ(𝐴(𝑋) ≤𝜃≤𝐵(𝑋)) = 𝛾for all 𝜃∈Θ. Note that the parameter 𝜃is assumed to be fixed for the event {𝐴(𝑋) ≤𝜃≤𝐵(𝑋)}, and the confidence interval holds uniformly over 𝜃.
Remark. Suppose that an experiment is repeated many times. On average, 100𝛾% of the time, the random interval (𝐴(𝑋), 𝐵(𝑋)) will contain the true parameter 𝜃. This is the frequentist interpretation of the confidence interval.
A misleading interpretation is as follows. Given that a single value of 𝑋is observed, there is a probability 𝛾that 𝜃∈(𝐴(𝑥), 𝐵(𝑥)). This is wrong, as will be demonstrated later.
Example. Let 𝑋1, … , 𝑋𝑛be i.i.d. normal random variables with unit variance. We will find the 95% confidence interval for 𝜇= 𝜃. We have 𝑋= 1 𝑛 𝑛 ∑ 𝑖=1 𝑋𝑖∼𝑁(𝜃, 1 𝑛); 𝑍= √𝑛(𝑋−𝜃) ∼𝑁(0, 1) Let 𝑎, 𝑏be numbers such that Φ(𝑏) −Φ(𝑎) = 0.95. Then ℙ(𝑎≤√𝑛(𝑋−𝜃) ≤𝑏) = 0.95 ⟹ℙ(𝑋−𝑏 √𝑛 ≤𝜃≤𝑋−𝑎 √𝑛 ) = 0.95 Hence, (𝑋− 𝑏 √𝑛, 𝑋− 𝑎 √𝑛) is a 95% confidence interval for 𝜃. Typically, we wish to centre the interval around some estimator ̂ 𝜃such that its range is minimised for a given 𝛾. In this case, we want to set −𝑎= 𝑏= 𝑧0.025 ≈1.96, where 𝑧𝛼= Φ−1(1 −𝛼). Hence, the confidence interval is (𝑋± 1.96 √𝑛).
Remark. In general, to find a confidence interval: (i) Find a quantity 𝑅(𝑋, 𝜃) where the distribution ℙ𝜃does not depend on 𝜃. This is known as a pivot. In the example above, 𝑅(𝑋, 𝜃) = √𝑛(𝑋−𝜃).
(ii) Consider ℙ(𝑐1 ≤𝑅(𝑋, 𝜃) ≤𝑐2) = 𝛾. Given some desired level of confidence 𝛾, find 𝑐1 and 𝑐2 using the distribution function of the pivot.
(iii) Rearrange such that ℙ(𝐴(𝑋) ≤𝜃≤𝐵(𝑋)) = 𝛾, then (𝐴(𝑋), 𝐵(𝑋)) is the confidence interval as required.
Proposition. Let 𝑇be a monotonically increasing function, and let (𝐴(𝑋), 𝐵(𝑋)) be a 100𝛾% confidence interval for 𝜃. Then (𝑇(𝐴(𝑋)), 𝑇(𝐵(𝑋))) is a 100𝛾% confidence interval for 𝑇(𝜃).
Remark. If 𝜃is a vector, we can consider confidence sets instead of confidence intervals. A confidence set is a set 𝐴(𝑋) such that ℙ(𝜃∈𝐴(𝑋)) = 𝛾.
626 3. Inference Example. Let 𝑋1, … , 𝑋𝑛be i.i.d. normal random variables with zero mean and unknown variance 𝜎2. We will find a 95% confidence interval for 𝜎2. Note that 𝑋1 𝜎∼𝑁(0, 1) is a valid pivot, but it considers only one data point. We will instead consider 𝑅(𝑋, 𝜎2) = ∑ 𝑖 𝑋2 𝑖 𝜎2 ∼𝜒2 𝑛 Now, we can define 𝑐1 = 𝐹−1 𝜒2 𝑛(0.025) and 𝑐2 = 𝐹−1 𝜒2 𝑛(0.975), giving ℙ(𝑐1 ≤ 𝑛 ∑ 𝑖=1 𝑋2 𝑖 𝜎2 ≤𝑐2) = 0.95 Rearranging, we have ℙ(∑𝑋2 𝑖 𝑐2 ≤𝜎2 ≤∑𝑋2 𝑖 𝑐1 ) = 0.95 Hence, the interval ∑ 𝑛 𝑖=1 𝑋2 𝑖( 1 𝑐2 , 1 𝑐1 ) is a 95% confidence interval for 𝜎2.
Example. Let 𝑋1, … , 𝑋𝑛be i.i.d. Bernoulli random variables with parameter 𝑝. Suppose 𝑛is large. We will find an approximate 95% confidence interval for 𝑝. The maximum likelihood estimator is ̂ 𝑝= 𝑋= 1 𝑛 𝑛 ∑ 𝑖=1 𝑋𝑖 By the central limit theorem, ̂ 𝑝is asymptotically distributed according to 𝑁(𝑝, 𝑝(1−𝑝) 𝑛 ). Hence, √𝑛 ̂ 𝑝−𝑝 √𝑝(1 −𝑝) has approximately a standard normal distribution. We have ℙ(−𝑧0.025 ≤√𝑛 ̂ 𝑝−𝑝 √𝑝(1 −𝑝) ≤𝑧0.025) ≈0.95 Instead of directly rearranging the inequalities, we will make an approximation for the de-nominator of the central term, letting √𝑝(1 −𝑝) ↦√̂ 𝑝(1 − ̂ 𝑝). When 𝑛is large, this ap-proximation becomes more accurate.
ℙ(−𝑧0.025 ≤√𝑛 ̂ 𝑝−𝑝 √̂ 𝑝(1 − ̂ 𝑝) ≤𝑧0.025) ≈0.95 This is much easier to rearrange, leading to ℙ( ̂ 𝑝−𝑧0.025 √̂ 𝑝(1 − ̂ 𝑝) √𝑛 ≤𝑝≤ ̂ 𝑝+ 𝑧0.025 √̂ 𝑝(1 − ̂ 𝑝) √𝑛 ) ≈0.95 This gives the approximate 95% confidence interval as required.
627 XI. Statistics Remark. Note that the size of the confidence interval is maximised at 𝑝= 1 2, with a length of 2𝑧0.025 1 2√𝑛≈ 1 √𝑛. This is a conservative 95% confidence interval; it may be wider than necessary but holds for all values of 𝜃.
3.2. Interpreting the confidence interval Example. Let 𝑋1, 𝑋2 be i.i.d. uniform random variables in (𝜃− 1 2, 𝜃+ 1 2). We wish to estim-ate the value of 𝜃with a 50% confidence interval. Observe that ℙ(𝜃∈(min 𝑋𝑖, max 𝑋𝑖)) = ℙ(𝑋1 ≤𝜃≤𝑋2) + ℙ(𝑋2 ≤𝜃≤𝑋1) = 1 2 Hence, (min 𝑋1, max 𝑋𝑖) is a 50% confidence interval for 𝜃. The frequentist interpretation is exactly correct; 50% of the time, 𝜃will lie between 𝑋1 and 𝑋2. However, suppose that |𝑋1 −𝑋2| > 1 2. Then we know that 𝜃∈(min 𝑋𝑖, max 𝑋𝑖). Suppose 𝑋1 = 0.1, 𝑋2 = 0.9, then it is not sensible to say that there is a 50% chance that 𝜃∈[0.1, 0.9].
628 4. Bayesian analysis 4. Bayesian analysis 4.1. Introduction Frequentist analysis considers the value 𝜃to be fixed, and then we can make inferential statements about 𝜃in the context of repeated experiments on a random variable 𝑋. Bayesian analysis is an alternative to frequentist analysis, where 𝜃is itself treated as a random variable taking values in the parameter space Θ. We say that the prior distribution 𝜋(𝜃) is a distribu-tion representing the beliefs of the investigator about 𝜃before observing data. The data 𝑋 has a p.d.f. or p.m.f. conditional on 𝜃given by 𝑓 𝑋( ⋅∣𝜃). Having observed 𝑋, we can com-bine this information with the prior distribution to form the posterior distribution 𝜋(𝜃∣𝑋), which is the conditional distribution of 𝜃given 𝑋. This contains updated information about the value of 𝜃. By Bayes’ rule, 𝜋(𝜃∣𝑥) = 𝜋(𝜃)𝑓 𝑋(𝑥∣𝜃) 𝑓 𝑋(𝑥) where 𝑓 𝑋(𝑥) is the marginal distribution of 𝑋, defined by 𝑓 𝑋(𝑥) = {∫ Θ 𝑓 𝑋(𝑥∣𝜃)𝜋(𝜃) d𝜃 𝜃continuous ∑Θ 𝑓 𝑋(𝑥∣𝜃)𝜋(𝜃) 𝜃discrete More simply, 𝜋(𝜃∣𝑋) ∝𝜋(𝜃) ⋅𝑓 𝑋(𝑋∣𝜃) The proportionality here is with respect to 𝜃. So the posterior is proportional to the prior multiplied by the likelihood. It is often easy to recognise that the right hand side of this expression is in some family of distributions, such as 𝑁or Γ, up to some normalising con-stant.
Remark. By the factorisation criterion, if 𝑇is a sufficient statistic for 𝜃, the posterior 𝜋(𝜃∣𝑥) depends on 𝑋only through 𝑇. More precisely, 𝜋(𝜃∣𝑋) ∝𝜋(𝜃)𝑔(𝑇(𝑋), 𝜃)ℎ(𝑋) ∝𝜋(𝜃)𝑔(𝑇(𝐶), 𝜃) Example. Consider a patient who we will test for the presence of a disease, where we have no information about the health or lifestyle of the patient. Let 𝜃take the value 1 if the patient is infected and 0 otherwise. We have a random variable 𝑋which takes the value 1 if a given test returns a positive result and 0 if the test is negative. We know the sensitivity of the test 𝑓 𝑋(𝑋= 1 ∣𝜃= 1), and the specificity of the test 𝑓 𝑋(𝑋= 0 ∣𝜃= 0). This fully specifies the likelihood function.
We now must choose a prior distribution. For example, let 𝜋(𝜃= 1) be the estimated propor-tion of the general population that have the given disease. The posterior is the probability of an infection given the test result.
𝜋(𝜃= 1 ∣𝑋= 1) = 𝜋(𝜃= 1)𝑓 𝑋(𝑋= 1 ∣𝜃= 1) 𝜋(𝜃= 1)𝑓 𝑋(𝑋= 1 ∣𝜃= 1) + 𝜋(𝜃= 0)𝑓 𝑋(𝑋= 1 ∣𝜃= 0) 629 XI. Statistics Even with a positive test result, the posterior distribution may still yield a low probability for 𝜃, which may happen if 𝜋(𝜃= 1) ≪𝜋(𝜃= 0).
Example. Let 𝜃be the mortality rate of a particular surgery, which will take values in [0, 1].
In the first ten operations, we observed that none of the patients died. We will model 𝑋∼ 𝐵(10, 𝜃) and observe 𝑋= 0.
We must choose a prior. Suppose that we have data from other hospitals that suggests that the mortality for the surgery ranges from 3% to 20%, with an average of 10%. We can choose the prior to be the beta distribution, 𝜋(𝜃) ∼Beta(𝑎, 𝑏), since the value of 𝜃should range between zero and one. Let 𝑎= 3 and 𝑏= 27, which will give 𝔼[𝜃] = 0.1 and ℙ(0.03 < 𝜃< 0.2) ≈0.9. In this case, the posterior is 𝜋(𝜃∣𝑋) ∝𝜋(𝜃)𝑓 𝑋(𝑥= 0 ∣𝜃) ∝𝜃𝑎−1(1 −𝜃)𝑏−1𝜃𝑥(1 −𝜃)𝑛−𝑥= 𝜃𝑥+𝑎−1(1 −𝜃)𝑏−𝑛−𝑥−1 This is again a beta distribution with parameters 𝑥+ 𝑎and 𝑛−𝑥+ 𝑏. The normalising constant does not need to be explicitly calculated since the form of the distribution can be recognised.
With the above data, we obtain 𝜋(𝜃∣𝑥= 0) ∼Beta(3, 37). This posterior has a smaller variance than the prior, and a smaller expectation due to observing no deaths. In this case, the prior and posterior have the same distribution. This is known as conjugacy.
4.2. Inference from the posterior The posterior distribution 𝜋(𝜃∣𝑥) represents information about 𝜃after having observed some data 𝑋. This can be used to make decisions under uncertainty.
(i) We first choose some decision 𝛿∈Δ. For instance, in the first example, a decision could be to ask the patient to isolate from others to reduce transmission.
(ii) We define a loss function 𝐿(𝜃, 𝛿), which defines what loss is incurred by making de-cision 𝛿given the true value of 𝜃. In the above example, 𝐿(𝜃= 1, 𝛿= 1) is the loss incurred by asking the patient to isolate given that they have the disease.
(iii) We can now choose the decision 𝛿that minimises ∫ Θ 𝐿(𝜃, 𝛿)𝜋(𝜃∣𝑥) d𝜃 which is the posterior expectation of the loss.
4.3. Point estimation We can use Bayesian analysis to represent an estimate for the value of 𝜃as a decision.
630 4. Bayesian analysis Definition. The Bayes estimator ̂ 𝜃(𝐵) minimises ℎ(𝛿) = ∫ Θ 𝐿(𝜃, 𝛿)𝜋(𝜃∣𝑥) d𝜃 Example. Suppose the loss function is quadratic, given by 𝐿(𝜃, 𝛿) = (𝜃−𝛿)2. Here, ℎ(𝛿) = ∫ Θ (𝜃−𝛿)2𝜋(𝜃∣𝑥) d𝜃 Thus, ℎ(𝛿) = 0 if ∫ Θ (𝜃−𝛿)𝜋(𝜃∣𝑥) d𝜃= 0 ⟺𝛿= ∫ Θ 𝜃𝜋(𝜃∣𝑥) d𝑥 Under the quadratic loss function, ̂ 𝜃(𝐵) can be described as the expectation of 𝜃under the posterior distribution.
Example. Consider the absolute error loss, given by 𝐿(𝜃, 𝛿) = |𝜃−𝛿|. In this case we have ℎ(𝛿) = ∫ Θ |𝜃−𝛿|𝜋(𝜃∣𝑥) d𝜃= ∫ 𝛿 −∞ −(𝜃−𝛿)𝜋(𝜃∣𝑥) d𝜃+ ∫ ∞ 𝛿 (𝜃−𝛿)𝜋(𝜃∣𝑥) d𝜃 We can differentiate, using the fundamental theorem of calculus, to find ℎ′(𝛿) = ∫ 𝛿 −∞ 𝜋(𝜃∣𝑥) d𝜃−∫ ∞ 𝛿 𝜋(𝜃∣𝑥) d𝜃 This is zero if and only if ∫ 𝛿 −∞ 𝜋(𝜃∣𝑥) d𝜃= ∫ ∞ 𝛿 𝜋(𝜃∣𝑥) d𝜃 This yields the median of the posterior distribution.
4.4. Credible intervals Definition. A 100𝛾% credible interval (𝐴(𝑥), 𝐵(𝑥)) satisfies 𝜋(𝐴(𝑥) ≤𝜃≤𝐵(𝑥) ∣𝑥) = 𝛾 Remark. Unlike confidence intervals, credible intervals can be interpreted conditionally on the data. For example, we could say that given a specific observation 𝑥, we are 100𝛾% certain that 𝜃lies within (𝐴(𝑥), 𝐵(𝑥)). This credible interval is also dependent on the choice of prior distribution.
631 XI. Statistics 5. Hypothesis testing 5.1. Hypotheses Definition. A hypothesis is an assumption about the distribution of the data 𝑋. Scientific questions are often phrased as a decision between two hypotheses. The null hypothesis 𝐻0 is usually a basic hypothesis, often representing the simplest possible distribution of the data.
The alternative hypothesis 𝐻1 is the alternative, if 𝐻0 were found to be false.
Example. Let 𝑋= (𝑋1, … , 𝑋𝑛) be i.i.d. Bernoulli random variables with parameter 𝜃. We could take, for example, 𝐻0 ∶𝜃= 1 2 and 𝐻1 ∶𝜃= 3 4. Alternatively, we could take 𝐻0 ∶𝜃= 1 2 and 𝐻1 ∶𝜃≠ 1 2.
Example. Suppose 𝑋𝑖takes values 0, 1, …. We can take 𝐻0 ∶𝑋𝑖 iid ∼Poi(𝜆) for some 𝜆, and 𝐻1 ∶𝑋𝑖 iid ∼𝑓 1 for some other distribution 𝑓 1. This is known as a goodness of fit test, which checks how well the model used for the data fits.
Definition. A simple hypothesis is a hypothesis which fully specifies the p.d.f. or p.m.f. of the data. A hypothesis that is not simple is called composite.
Example. In the first example above, 𝐻0 ∶𝜃= 1 2 is simple, and 𝐻1 ∶𝜃≠ 1 2 is composite. In the second example, 𝐻0 ∶𝑋𝑖 iid ∼Poi(𝜆) is composite since 𝜆was not fixed.
5.2. Testing hypotheses Definition. A test of the null hypothesis 𝐻0 is defined by a critical region 𝐶⊆𝒳. When 𝑋∈𝐶, we reject the null hypothesis. This is a positive result. When 𝑋∉𝐶we fail to reject the null hypothesis, or find no sufficient evidence against the null hypothesis. This is the negative result.
A type I error, or a false positive, is the error made by rejecting the null hypothesis when it is true. A type II error, or a false negative, is the error made by failing to reject the null hypothesis when it is not true. When 𝐻0, 𝐻1 are simple, we define 𝛼= ℙ𝐻0 (𝐻0 is rejected) = ℙ𝐻0 (𝑋∈𝐶) ; 𝛽= ℙ𝐻1 (𝐻0 is not rejected) = ℙ𝐻1 (𝑋∉𝐶) The size of a test is 𝛼, which is the probability of a type I error. The power of a test is 1 −𝛽, which is the probability of not finding a type II error.
There is typically a tradeoff between 𝛼and 𝛽. Often, statisticians will choose an ‘acceptable’ value for the probability of type I errors 𝛼, and then maximise the power with respect to this fixed 𝛼. Computing the size of a test is typically simpler since it does not depend on 𝐻1.
632 5. Hypothesis testing 5.3. Neyman–Pearson lemma Let 𝐻0 and 𝐻1 be simple, and let 𝑋have a p.d.f. or p.m.f. 𝑓𝑖under 𝐻𝑖. The likelihood ratio statistic is defined by Λ𝑥(𝐻0; 𝐻1) = 𝑓 1(𝑥) 𝑓 0(𝑥) The likelihood ratio test is a test that rejects 𝐻0 when Λ𝑥exceeds a set value 𝑘, or more formally, 𝐶= {𝑥∶Λ𝑥(𝐻0; 𝐻1) > 𝑘}.
Lemma. Suppose that 𝑓 0, 𝑓 1 are nonzero on the same set, and suppose that there exists 𝑘> 0 such that the likelihood ratio test with critical region 𝐶= {𝑥∶Λ𝑥(𝐻0; 𝐻1) > 𝑘} has size 𝛼. Then out of all tests of size upper bounded by 𝛼, this test has the largest power.
Remark. A likelihood ratio test with size 𝛼does not always exist for any given 𝛼. However, in general we can find a randomised test with arbitrary size 𝛼. This is a test where, for some values of 𝑋, we reject the null hypothesis; for some values, we fail to reject the null hypo-thesis; and for some values we reject the null hypothesis with a random chance of rejecting the null hypothesis.
Proof. Let 𝐶be the complement of 𝐶in 𝒳. Then, the likelihood ratio test has 𝛼= ∫ 𝐶 𝑓 0(𝑥) d𝑥; 𝛽= ∫ 𝐶 𝑓 1(𝑥) d𝑥 Let 𝐶⋆be a critical region for a different test, with type I and II error probabilities 𝛼⋆, 𝛽⋆.
Here, 𝛼⋆= ∫ 𝐶⋆ 𝑓 0(𝑥) d𝑥; 𝛽⋆= ∫ 𝐶⋆ 𝑓 1(𝑥) d𝑥 Suppose 𝛼⋆≤𝛼. Then, we will show 𝛽≤𝛽⋆.
𝛽−𝛽⋆= ∫ 𝐶 𝑓 1(𝑥) d𝑥−∫ 𝐶⋆ 𝑓 1(𝑥) d𝑥 633 XI. Statistics By cancelling the integrals on the intersection, and using the definition of 𝐶, 𝛽−𝛽⋆= ∫ 𝐶∩𝐶⋆ 𝑓 1(𝑥) d𝑥−∫ 𝐶⋆∩𝐶 𝑓 1(𝑥) d𝑥 = ∫ 𝐶∩𝐶⋆ 𝑓 1(𝑥) 𝑓 0(𝑥) ⏟ ≤𝑘 𝑓 0(𝑥) d𝑥−∫ 𝐶⋆∩𝐶 𝑓 1(𝑥) 𝑓 0(𝑥) ⏟ ≥𝑘 𝑓 0(𝑥) d𝑥 ≤𝑘[∫ 𝐶∩𝐶⋆ 𝑓 0(𝑥) d𝑥−∫ 𝐶 ⋆ ∩𝐶 𝑓 0(𝑥) d𝑥] = 𝑘[∫ 𝐶∩𝐶⋆ 𝑓 0(𝑥) d𝑥+ ∫ 𝐶∩𝐶⋆ 𝑓 0(𝑥) d𝑥−∫ 𝐶∩𝐶⋆ 𝑓 0(𝑥) d𝑥−∫ 𝐶 ⋆ ∩𝐶 𝑓 0(𝑥) d𝑥] = 𝑘[∫ ∩𝐶⋆ 𝑓 0(𝑥) d𝑥−∫ 𝐶 𝑓 0(𝑥) d𝑥] = 𝑘[𝛼⋆−𝛼] ≤0 Example. Let 𝑋1, … , 𝑋𝑛∼𝑁(𝜇, 𝜎2 0) be i.i.d., where 𝜎2 0 is known and 𝜇is an unknown.
We wish to find the most powerful test of fixed size 𝛼for the hypotheses 𝐻0 ∶𝜇= 𝜇0 and 𝐻1 ∶𝜇= 𝜇1 > 𝜇0. The likelihood ratio is Λ𝑥(𝐻0; 𝐻1) = (2𝜋𝜎2 0)−𝑛/2 exp{ −1 2𝜎2 0 ∑(𝑥𝑖−𝜇0)2} (2𝜋𝜎2 0)−𝑛/2 exp{ −1 2𝜎2 0 ∑(𝑥𝑖−𝜇1)2} = exp ⎧ ⎪ ⎨ ⎪ ⎩ 𝜇1 −𝜇0 𝜎2 0 ⏟ ⎵ ⏟ ⎵ ⏟ ≥0 𝑛𝑋+ 𝑛(𝜇0 −𝜇1)2 2𝜎2 0 ⎫ ⎪ ⎬ ⎪ ⎭ which depends only on 𝑋, and is monotonically increasing with respect to the sample mean 𝑋. Therefore, this is also monotonically increasing with respect to the statistic 𝑍= √𝑛𝑋−𝜇0 𝜎0 Thus, Λ𝑥> 𝑘if and only if 𝑍> 𝑘′ for some 𝑘′. Hence, the likelihood ratio test has critical region {𝑥∶𝑍(𝑥) > 𝑘′} for some 𝑘′. It thus suffices to find a critical region of 𝑍with size 𝛼 in order to construct the most powerful test of this size. Under 𝐻0, 𝑍∼𝑁(0, 1). Hence, the critical region is given by 𝑘′ = Φ−1(1 −𝛼). This is known as a 𝑍-test, since we are using the 𝑍statistic.
634 5. Hypothesis testing 5.4. 𝑝-values Definition. Let 𝐶be a critical region of the form {𝑥∶𝑇(𝑥) > 𝑘} for some test statistic 𝑇.
Let 𝑥⋆denote the observed data. Then, the 𝑝-value is ℙ𝐻0 (𝑇(𝑋) > 𝑇(𝑥⋆)) Typically, when reporting the results of a test, we describe the conclusion of the test as well as the 𝑝-value. In the example above, suppose 𝜇0 = 5, 𝜇1 = 6, 𝛼= 0.05, and 𝑥⋆= (5.1, 5.5, 4.9, 5.3). Here, 𝑥⋆= 5.2 and 𝑧⋆= 0.4. The likelihood ratio test has critical re-gion {𝑥∶𝑍(𝑥) > Φ−1(0.95) ≈1.645} The conclusion of the test here is to not reject 𝐻0. The 𝑝-value is 1 −Φ(𝑧⋆) ≈0.35.
Proposition. Under the null hypothesis 𝐻0, the 𝑝-value is a uniform random variable in [0, 1].
Proof. Let 𝐹be the distribution of the test statistic 𝑇, which we will assume for this proof is continuous. Then, ℙ𝐻0 (𝑝< 𝑢) = ℙ𝐻0 (1 −𝐹(𝑇) < 𝑢) = ℙ𝐻0 (𝐹(𝑇) > 1 −𝑢) = ℙ𝐻0 (𝑇> 𝐹−1(1 −𝑢)) = 1 −𝐹(𝐹−1(1 −𝑢)) = 𝑢 5.5. Composite hypotheses Let 𝑋∼𝑓 𝑋( ⋅∣𝜃) where 𝜃∈Θ. Let 𝐻0 = 𝜃∈Θ0 ⊂Θ and 𝐻1 = 𝜃∈Θ1 ⊆Θ. The probabilities of type I and type II error are now dependent on the precise value of 𝜃, rather than simply on which hypothesis is taken.
Definition. The power function for a test 𝐶is 𝑊(𝜃) = ℙ𝜃(𝑋∈𝐶) The size of a test 𝐶is 𝛼= sup 𝜃∈Θ0 𝑊(𝜃) A test is uniformly most powerful of size 𝛼if, for any test 𝐶⋆with power function 𝑊⋆and size upper bounded by 𝛼, for all 𝜃∈Θ1 we have 𝑊(𝜃) ≥𝑊⋆(𝜃). Such tests need not exist.
In simple models, many likelihood ratio tests are uniformly most powerful.
635 XI. Statistics Example (one-sided test for normal location). Let 𝑋1, … , 𝑋𝑛∼𝑁(𝜇, 𝜎2 0) be i.i.d. where 𝜎2 0 is known and 𝜇is unknown. Let 𝐻0 ∶𝜇≤𝜇0 and 𝐻1 ∶𝜇> 𝜇0 for some fixed 𝜇0. We claim that the simple hypothesis test given by 𝐻′ 0 ∶𝜇= 𝜇0 and 𝐻′ 1 ∶𝜇= 𝜇1 > 𝜇0 is uniformly most powerful for 𝐻0 and 𝐻1. The power function is 𝑊(𝜇) = ℙ𝜇(√𝑛(𝑋−𝜇0) 𝜎0 = 𝑍< 𝑧𝛼= Φ−1(1 −𝛼)) = ℙ𝜇(√𝑛(𝑋−𝜇) 𝜎0 > 𝑧𝛼+ √𝑛(𝜇0 −𝜇) 𝜎0 ) = 1 −Φ(𝑥𝛼+ √𝑛𝜇0 −𝜇 𝜎0 ) The test has size 𝛼since sup𝑤∈Θ0 𝑊(𝜇) = 𝛼. It remains to show that this power function dominates all other power functions 𝑊⋆of size 𝛼in the alternative space Θ1. First, observe that the critical region depends only on 𝜇0, and not on 𝜇1. In particular, for any 𝜇1 > 𝜇0, we have that the critical region 𝐶is the likelihood ratio test for the simple hypothesis test 𝐻′ 0 ∶𝜇= 𝜇0 and 𝐻′ 1 ∶𝜇= 𝜇1. We can also see 𝐶⋆as a test of 𝐻′ 0 versus 𝐻′ 1, and for these simple hypotheses, 𝐶⋆has size 𝑊⋆(𝜇0) ≤sup 𝜇<𝜇0 𝑊⋆(𝜇) ≤𝛼 By the Neyman–Pearson lemma, 𝐶has power no smaller than 𝐶⋆for 𝐻′ 0 against 𝐻′ 1: 𝑊(𝜇1) ≥𝑊⋆(𝜇1) Since this is true for all 𝜇1 > 𝜇0, the result holds, and the test 𝐶satisfies the property for being uniformly most powerful.
5.6. Generalised likelihood ratio test Definition. Suppose we have nested hypotheses, i.e. 𝐻0 ∶𝜃∈Θ0 and 𝐻1 ∶𝜃∈Θ1, where Θ0 ⊂Θ1. The generalised likelihood ratio is given by Λ𝑥(𝐻0; 𝐻1) = sup𝜃∈Θ1 𝑓 𝑋(𝑥∣𝜃) sup𝜃∈Θ0 𝑓 𝑋(𝑥∣𝜃) Large values indicate a better fit under the alternative hypothesis. The generalised likelihood ratio test rejects the null hypothesis when Λ𝑥is sufficiently large.
Example (two-sided test for normal location). Let 𝑋1, … , 𝑋𝑛∼𝑁(𝜇, 𝜎2 0) be i.i.d. where 𝜎2 0 is known and 𝜇is unknown. Let 𝐻0 ∶𝜇= 𝜇0 and 𝐻1 ∶𝜇∈ℝfor some fixed 𝜇0. In this 636 5. Hypothesis testing model, the generalised likelihood ratio is Λ𝑥(𝐻0; 𝐻1) = (2𝜋𝜎2 0)−𝑛/2 exp{ −1 2𝜎2 0 Σ𝑛 𝑖=1(𝑥𝑖−𝑋)2} (2𝜋𝜎2 0)−𝑛/2 exp{ −1 2𝜎2 0 Σ𝑛 𝑖=1(𝑥𝑖−𝜇0)2} 2 log Λ𝑥= 𝑛 𝜎2 0 (𝑋−𝜇0)2 Under 𝐻0, √𝑛 𝑋−𝜇0 𝜎0 ∼𝑁(0, 1). Hence, 2 log Λ𝑥∼𝜒2 1. Therefore, the critical region of this generalised likelihood ratio test is 𝐶= {𝑥∶𝑛 𝜎2 0 (𝑋−𝜇0)2 > 𝜒2 1(𝛼)} where 𝜒2 1(𝛼) is the upper 𝛼point of 𝜒2 1. This is called a two-sided test since there are two tails on the critical region, plotting with respect to √𝑛 𝑋−𝜇0 𝜎0 .
5.7. Wilks’ theorem Definition. The dimension of a hypothesis 𝐻0 ∶𝜃∈Θ0 is the number of ‘free parameters’ in this space.
Example. If Θ0 = {𝜃∈ℝ𝑘∶𝜃1 = ⋯= 𝜃𝑝= 0}, then the dimension of 𝐻0 is 𝑘−𝑝.
Let 𝐴∈ℝ𝑝×𝑘be a 𝑝× 𝑘matrix with linearly independent rows. Let 𝑏∈ℝ𝑝for 𝑝< 𝑘, then we define Θ0 = {𝜃∈ℝ𝑘∶𝐴𝜃= 𝑏}. Then the dimension of 𝜃is 𝑘−𝑝.
Let Θ0 be a Riemannian manifold. We use differential geometry to deduce the dimension-ality of such a manifold.
Theorem. Suppose Θ0 ⊂Θ1, and dim Θ1 −dim Θ0 = 𝑝. Let 𝑋= (𝑋1, … , 𝑋𝑛) be i.i.d.
random variables under 𝑓 𝑥( ⋅∣𝜃) where 𝜃∈Θ∘ 0. Then, under some regularity conditions, as 𝑛→∞we have 2 log Λ𝑥∼𝜒2 𝑝 More precisely, for all ℓ∈ℝ+, lim 𝑛→∞ℙ𝜃(2 log Λ𝑥≤ℓ) = ℙ(Ξ ≤ℓ) ; Ξ ∼𝜒2 𝑝 Remark. If 𝑛is large, this theorem allows us to implement a generalised likelihood ratio test even if we cannot find the exact distribution of 2 log Λ𝑥. Frequentist guarantees obtained from such a test will be approximate.
Example. In the two-sided test for normal location, dim Θ1 = 1 and dim Θ0 = 0 hence the difference in dimensions is 1. Then, Wilks’ theorem implies that 2 log Λ𝑥is approximately distributed according to 𝜒2 1, although the result is exact in this particular case.
637 XI. Statistics 5.8. Goodness of fit Let 𝑋1, … , 𝑋𝑛be i.i.d. samples taking values in {1, … , 𝑘}. Let 𝑝𝑖= ℙ(𝑋1 = 𝑖), and let 𝑁𝑖be the number of samples equal to 𝑖, so ∑𝑖𝑝𝑖= 1 and ∑𝑖𝑁𝑖= 𝑛. The parameters here are 𝑝= (𝑝1, … , 𝑝𝑘), which has 𝑘−1 dimensions. A goodness of fit test has a null hypothesis of the form 𝐻0 ∶𝑝𝑖= ˜ 𝑝𝑖for all 𝑖, for a fixed ˜ 𝑝= (˜ 𝑝1, … , ˜ 𝑝𝑘). The alternative hypothesis 𝐻1 does not constrain 𝑝.
The model is (𝑁1, … , 𝑁𝑘) ∼Multi(𝑛; 𝑝1, … , 𝑝𝑘). The likelihood function is 𝐿(𝑝) ∝𝑝𝑁1 1 ⋯𝑝𝑁𝑘 𝑘 ⟹ℓ(𝑝) = constant + ∑ 𝑖 𝑁𝑖log 𝑝𝑖 The generalised likelihood ratio is 2 log Λ𝑥= 2( sup 𝑝∈Θ1 ℓ(𝑝) −sup 𝑝∈Θ0 ℓ(𝑝)) = 2(ℓ( ̂ 𝑝) −ℓ(˜ 𝑝)) where ̂ 𝑝is the maximum likelihood estimator under 𝐻1. To find ̂ 𝑝, we typically use the method of Lagrange multipliers.
ℒ(𝑝, 𝜆) = ∑ 𝑖 𝑁𝑖log 𝑝𝑖−𝜆(∑𝑝𝑖−1) We can compute that ̂ 𝑝𝑖= 𝑁𝑖 𝑛 This is simply the fraction of observed samples of type 𝑖.
5.9. Pearson statistic Let 𝑜𝑖= 𝑁𝑖be the observed number of samples of type 𝑖, and 𝑒𝑖= 𝑛˜ 𝑝𝑖be the expected value under the null hypothesis of the number of samples of type 𝑖. Here, we can write 2 log Λ = 2 ∑ 𝑖 𝑁𝑖log( 𝑁𝑖 𝑛˜ 𝑝𝑖 ) = 2 ∑ 𝑖 𝑜𝑖log 𝑜𝑖 𝑒𝑖 Let 𝛿𝑖= 𝑜𝑖−𝑒𝑖. Then 2 log Λ = 2 ∑ 𝑖 (𝑒𝑖+ 𝛿𝑖) log ⎛ ⎜ ⎜ ⎝ 1 + 𝛿𝑖 𝑒𝑖 ⏟ small when 𝑛large ⎞ ⎟ ⎟ ⎠ By taking the Taylor expansion, we arrive at 2 ∑ 𝑖 (𝛿𝑖+ 𝛿2 𝑖 𝑒𝑖 −𝛿2 𝑖 2𝑒𝑖 ) 638 5. Hypothesis testing Note that ∑𝑖𝛿𝑖= ∑𝑖(𝑜𝑖−𝑒𝑖) = 𝑛−𝑛= 0, so we can simplify and find ∑ 𝑖 𝛿2 𝑖 𝑒𝑖 = ∑ 𝑖 (𝑜𝑖−𝑒𝑖)2 𝑒𝑖 This is Pearson’s 𝜒2 statistic. This is also referred to a 𝜒2 𝑘−1 when performing a hypothesis test.
Example. Mendel performed an experiment in which 556 different pea plants were cre-ated from a small set of ancestors. Each descendent was either yellow or green, and either wrinkled or smooth, giving four possibilities in total. The observed result was 𝑁= (315 ⏟ 𝑆𝐺 , 108 ⏟ 𝑆𝑌 , 102 ⏟ 𝑊𝐺 , 31 ⏟ 𝑊𝑌 ) Mendel’s theory gives a null hypothesis 𝐻0 ∶𝑝= ˜ 𝑝= ( 9 16, 3 16, 3 16, 1 16). Here, 2 log Λ = 0.618; ∑ 𝑖 (𝑜𝑖−𝑒𝑖)2 𝑒𝑖 = 0.604 These are referred to a 𝜒2 3 distribution. We observe that 𝜒2 3(0.05) = 7.815, so we fail to reject the null hypothesis with a test of size 5%. We can compute that the 𝑝-value is ℙ(𝜒2 3 > 0.6) ≈ 0.96, so there is a very high probability of observing a more extreme value than observed.
5.10. Goodness of fit for composite null Suppose 𝐻0 ∶𝑝𝑖= 𝑝𝑖(𝜃) for some 𝜃∈Θ0, and 𝐻1 ∶𝑝has any distribution on {1, … , 𝑘}. We can compute 2 log Λ = 2(sup 𝑝 ℓ(𝑝) −sup 𝜃∈Θ ℓ(𝑝(𝜃))) We can sometimes compute these quantities explicitly, and hence find a test which refers this test statistic to a 𝜒2 𝑝distribution where 𝑝= dim Θ1 −dim Θ0 = (𝑘−1) −dim Θ0.
Example. Consider a population of individuals who may have one of three genotypes, which occur with probabilities (𝑝1, 𝑝2, 𝑝3) = (𝜃2, 2𝜃(1 −𝜃), (1 −𝜃)2). In this case, we can find the maximum likelihood estimator under the null hypothesis to be ̂ 𝜃= 2𝑁1 + 𝑁2 2𝑛 Hence, 2 log Λ = 2(ℓ( ̂ 𝑝) −ℓ( ̂ 𝜃)) where ̂ 𝑝𝑖= 𝑁1 𝑛as found previously. This can be computed explicitly and referred to a 𝜒2 1 distribution. We can check that, in this model, 2 log Λ = ∑ 𝑖 𝑜𝑖log 𝑜𝑖 𝑒𝑖 639 XI. Statistics where 𝑜𝑖= 𝑁𝑖and 𝑒𝑖= 𝑛𝑝𝑖( ̂ 𝜃). We can approximate this using the Pearson statistic, ∑𝑖 (𝑜𝑖−𝑒𝑖)2 𝑒𝑖 .
5.11. Testing independence in contingency tables Suppose we have observations (𝑋1, 𝑌 1), … , (𝑋𝑛, 𝑌 𝑛) which are i.i.d., where the 𝑋𝑖take values in 1, … , 𝑟and the 𝑌𝑖take values in 1, … , 𝑐. We wish to test whether the 𝑋𝑖and 𝑌𝑖are independent. We will summarise this data into a sufficient statistic known as a contingency table 𝑁, given by 𝑁𝑖𝑗= |{ℓ∶1 ≤ℓ≤𝑛, (𝑋ℓ, 𝑌ℓ) = (𝑖, 𝑗)}| So 𝑁𝑖𝑗is the number of samples of type (𝑖, 𝑗).
Example. Suppose we observe 𝑛samples, and each sample has probability 𝑝𝑖𝑗of being of type (𝑖, 𝑗). Flattening (𝑁𝑖𝑗) into a vector, this has a multinomial distribution with parameters (𝑝𝑖𝑗) (also flattened into a vector). The null hypothesis is 𝐻0 ∶𝑝𝑖𝑗= 𝑝𝑖+𝑝+𝑗where 𝑝𝑖+ = ∑𝑗𝑝𝑖𝑗and 𝑝+𝑗= ∑𝑖𝑝𝑖𝑗. The alternative hypothesis places no restrictions on the 𝑝𝑖𝑗apart from that it sums to 1 and has nonnegative entries. We find 2 log Λ = 2 𝑟 ∑ 𝑖=1 𝑐 ∑ 𝑗=1 𝑁𝑖𝑗log ̂ 𝑝𝑖𝑗 ̂ 𝑝𝑖+ ̂ 𝑝+𝑗 where ̂ 𝑝𝑖𝑗is the maximum likelihood estimator under 𝐻1, and where ̂ 𝑝𝑖+ and ̂ 𝑝+𝑗are the maximum likelihood estimators under 𝐻0. These can be found using the method of Lag-range multipliers. In particular, ̂ 𝑝𝑖𝑗= 𝑁𝑖𝑗 𝑛; ̂ 𝑝𝑖+ = 𝑁𝑖+ 𝑛 = 1 𝑛 𝑐 ∑ 𝑗=1 𝑁𝑖𝑗; ̂ 𝑝+𝑗= 𝑁+𝑗 𝑛 = 1 𝑛 𝑟 ∑ 𝑖=1 𝑁𝑖𝑗 Writing 𝑜𝑖𝑗= 𝑁𝑖𝑗and 𝑒𝑖𝑗= 𝑛̂ 𝑝𝑖+ ̂ 𝑝+𝑗, 2 log Λ = ∑ 𝑖,𝑗 𝑜𝑖𝑗log 𝑜𝑖𝑗 𝑒𝑖𝑗 ≈∑ 𝑖,𝑗 (𝑜𝑖𝑗−𝑒𝑖𝑗)2 𝑒𝑖𝑗 By Wilks’ theorem, these test statistics have an approximate 𝜒2 𝑝distribution, where 𝑝= dim Θ1 −dim Θ0 = (𝑟𝑐−1) −(𝑟−1 + 𝑐−1) = (𝑟−1)(𝑐−1).
The 𝜒2 test for independence has a number of weaknesses.
(i) The 𝜒2 approximation requires 𝑛to be large. A reasonable heuristic is to require 𝑁𝑖𝑗≥5 for all 𝑖, 𝑗. If this is not possible, we can perform an exact test (which is non-examinable).
(ii) The 𝜒2 test often has a low power. Heuristically, this is because the alternative space Θ1 is too large, and there are many possible models that lie in this space.
640 5. Hypothesis testing Note that this test also applies when 𝑛is a random variable with a Poisson distribution. This is often the case when we do not fix the number of samples. The proof is not provided in this course.
5.12. Testing homogeneity in contingency tables Example. Suppose we perform a clinical trial on 150 patients, who are randomly assigned to one of three groups of equal size. The first two sets take a drug with different doses, and the third set takes a placebo.
improved no difference worse placebo 18 17 15 50 half dose 20 10 20 50 full dose 25 13 12 50 In the previous section, we fixed the total number of samples. Here, we fix the total number of samples, and the total number of samples in each row. We suppose 𝑁𝑖1, … , 𝑁𝑖𝑐∼Multinomial(𝑛𝑖+; 𝑝𝑖1, … , 𝑝𝑖𝑐) which are independent for each row 𝑖of the table. The null hypothesis for homogeneity is that 𝑝1𝑗= 𝑝2𝑗= ⋯= 𝑝𝑟𝑗for all 𝑗. The alternative hypothesis assumes that 𝑝𝑖1, … , 𝑝𝑖𝑐is any arbitrary probability vector for each row 𝑖. Under the alternative hypothesis, 𝐿(𝑝) = 𝑟 ∏ 𝑖=1 𝑛𝑖+!
𝑁𝑖1! ⋯𝑁𝑖𝑐!𝑝𝑁𝑖1 𝑖1 ⋯𝑝𝑁𝑖𝑐 𝑖𝑐 Hence, ℓ(𝑝) = constant + ∑ 𝑖,𝑗 𝑁𝑖𝑗log 𝑝𝑖𝑗 This is the same likelihood as the independence test above. To define the maximum like-lihood estimator we can again use the method of Lagrange multipliers with constraints ∑𝑗𝑝𝑖𝑗= 1 for each 𝑖. We find ̂ 𝑝𝑖𝑗= 𝑁𝑖𝑗 𝑛𝑖+ Under the null hypothesis, we let 𝑝𝑗= 𝑝𝑖𝑗for any 𝑖.
ℓ(𝑝) = constant + ∑ 𝑖,𝑗 𝑁𝑖𝑗log 𝑝𝑗= ∑ 𝑗 𝑁+𝑗log 𝑝𝑗 We have the constraint ∑𝑗𝑝𝑗= 1. Using the method of Lagrange multipliers, ̂ 𝑝𝑗= 𝑁+𝑗 𝑛++ 641 XI. Statistics Hence, 2 log Λ = 2 ∑ 𝑖,𝑗 𝑁𝑖𝑗log ̂ 𝑝𝑖𝑗 ̂ 𝑝𝑗 = 2 ∑ 𝑖,𝑗 𝑁𝑖𝑗log 𝑁𝑖𝑗 𝑛𝑖+𝑁+𝑗/𝑛++ This is precisely the same test statistic as the test for independence above. The only differ-ence is that 𝑛𝑖+ is fixed in this model. Further, if 𝑜𝑖𝑗= 𝑁𝑖𝑗and 𝑒𝑖𝑗= 𝑛𝑖+ ̂ 𝑝𝑗= 𝑛𝑖+𝑁+𝑗 𝑛++ , we have 2 log Λ = 2 ∑ 𝑖,𝑗 𝑜𝑖𝑗log 𝑜𝑖𝑗 𝑒𝑖𝑗 ≈∑ 𝑖,𝑗 (𝑜𝑖𝑗−𝑒𝑖𝑗)2 𝑒𝑖𝑗 By Wilks’ theorem, this is asymptotically a 𝜒2 𝑝distribution. Here, 𝑝= dim Θ1 −dim Θ0 = 𝑟(𝑐−1) −(𝑐−1) = (𝑟−1)(𝑐−1) This is again exactly the same as in the 𝜒2 test for independence. Operationally, the tests for homogeneity and independence are therefore completely identical; we reject the null hypothesis for one test if and only if we reject the null for the other. In the example above, 2 log Λ = 5.129; ∑ 𝑖,𝑗 (𝑜𝑖𝑗−𝑒𝑖𝑗)2 𝑒𝑖𝑗 = 5.173 Referring this to a 𝜒2 4 distribution, the upper 0.05-point is 9.488. Hence, we do not reject the null hypothesis at the 5% significance level.
5.13. Tests and confidence sets Definition. The acceptance region 𝐴of a test is the complement of the critical region.
Theorem. Let 𝑋∼𝑓 𝑋( ⋅∣𝜃) for some 𝜃∈Θ. Suppose that for each 𝜃0 ∈Θ, there exists a test of size 𝛼with acceptance region 𝐴(𝜃0) for the null hypothesis 𝜃= 𝜃0. Then 𝐼(𝑋) = {𝜃∶𝑋∈𝐴(𝜃)} is a 100(1 −𝛼)% confidence set.
Now suppose there exists a set 𝐼(𝑋) which is a 100(1 −𝛼)% confidence set for 𝜃. Then 𝐴(𝜃0) = {𝑥∶𝜃0 ∈𝐼(𝑥)} is the acceptance region of a test of size 𝛼for the hypothesis 𝜃= 𝜃0.
Proof. Observe that for both parts of the theorem, 𝜃0 ∈𝐼(𝑋) ⟺𝑋∈𝐴(𝜃0) ⟺fail to reject 𝐻0 with data 𝑋 For the first part, we assume that ℙ𝜃(fail to reject 𝐻0 with data 𝑋) = 1 −𝛼, and we want to show ℙ𝜃(𝜃0 ∈𝐼(𝑋)) = 1 −𝛼. The second part is the converse.
642 5. Hypothesis testing Example. Let 𝑋1, … , 𝑋𝑛∼𝑁(𝜇, 𝜎2 0) be i.i.d. with 𝜎2 0 known and 𝜇unknown. We found that a 100(1 −𝛼)% confidence interval for 𝜇is 𝐼(𝑋) = (𝑋± 𝑍𝛼/2𝜎0 √𝑛 ) Hence, by the second part of the theorem above, we can find a test for 𝐻0 ∶𝜇= 𝜇0 with size 𝛼by 𝐴(𝜇0) = {𝑥∶𝜇0 ∈𝐼(𝑥)} = {𝑥∶𝜇0 ∈[𝑥± 𝑍𝛼/2𝜎0 √𝑛 ]} This is equivalent to rejecting 𝐻0 when | | | √𝑛𝜇0 −𝑋 𝜎0 | | | > 𝑍𝛼/2 This is a two-sided test for normal location.
643 XI. Statistics 6. The normal linear model 6.1. Multivariate normal distribution Let 𝑋= (𝑋1, … , 𝑋𝑛) be a vector of random variables. Then we define 𝔼[𝑋] = ( 𝔼[𝑋1] ⋮ 𝔼[𝑋𝑛] ) ; Var (𝑋) = (𝔼[(𝑋𝑖−𝔼[𝑋𝑖])(𝑋𝑗−𝔼[𝑋𝑗])])𝑖,𝑗 The familiar linearity results are 𝔼[𝐴𝑋+ 𝑏] = 𝐴𝔼[𝑋] + 𝑏; 𝐴Var (𝑋) 𝐴⊺ where 𝐴∈ℝ𝑘×𝑛, 𝑏∈ℝ𝑘are constant.
Definition. We say that 𝑋has a multivariate normal distribution if, for any fixed 𝑡∈ℝ𝑛, we have 𝑡⊺𝑋∼𝑁(𝜇, 𝜎2) for some parameters 𝜇, 𝜎2.
Proposition. Let 𝑋be multivariate normal. Then 𝐴𝑋+ 𝑏is multivariate normal, where 𝐴∈ℝ𝑘×𝑛, 𝑏∈ℝ𝑘are constant.
Proof. Let 𝑡∈ℝ𝑘. Then, 𝑡⊺(𝐴𝑥+ 𝑏) = (𝐴⊺𝑡)⊺𝑋 ⏟ ⎵ ⏟ ⎵ ⏟ ∼𝑁(𝜇,𝜎2) +𝑡⊺𝑏 which is the sum of a normal random variable and a constant. So this is 𝑁(𝜇+ 𝑡⊺𝑏, 𝜎2).
Proposition. A multivariate normal distribution is fully specified by its mean and covari-ance matrix.
Proof. Let 𝑋1, 𝑋2 be multivariate normal vectors with the same mean 𝜇and the same cov-ariance matrix Σ. We will show that these two random variables have the same moment generating function, and hence the same distribution.
𝑀𝑋1(𝑡) = 𝔼[𝑒1⋅𝑡⊺𝑋1] Note that 𝑡⊺𝑋1 is univariate normal. Hence, this is equal to 𝑀𝑋1(𝑡) = exp(1 ⋅𝔼[𝑡⊺𝑋1] + 1 2 Var (𝑡⊺𝑋1) ⋅12) = exp(𝑡⊺𝜇+ 1 2𝑡⊺Σ𝑡) This depends only on 𝜇and Σ, and we obtain the same moment generating function for 𝑋2.
644 6. The normal linear model 6.2. Orthogonal projections Definition. A matrix 𝑃∈ℝ𝑛×𝑛is an orthogonal projection onto its column space col(𝑃) if, for all 𝑣∈col(𝑃), we have 𝑃𝑣= 𝑣, and for all 𝑤∈col(𝑃)⟂, we have 𝑃𝑤= 0.
Proposition. 𝑃is an orthogonal projection if and only if it is idempotent and symmetric.
Proof. If 𝑃is idempotent and symmetric, let 𝑣∈col(𝑃), so 𝑣= 𝑃𝑎for some 𝑎∈ℝ𝑛. Then, 𝑃𝑣= 𝑃𝑃𝑎= 𝑃𝑎= 𝑣. Now, let 𝑤∈col(𝑃)⟂. By definition, 𝑃⊺𝑤= 0. By symmetry, 𝑃𝑤= 0.
Now, suppose 𝑃is an orthogonal projection. Any vector 𝑎∈ℝ𝑛can be uniquely written as 𝑎= 𝑣+ 𝑤where 𝑣∈col(𝑃) and 𝑤∈col(𝑃)⟂. Then 𝑃𝑃𝑎= 𝑃𝑃𝑣+ 𝑃𝑃𝑤= 𝑃𝑣= 𝑃(𝑣+ 𝑤) = 𝑃𝑎. As this holds for all 𝑎, we have that 𝑃is idempotent. Let 𝑢1, 𝑢2 ∈ℝ𝑛, and note (𝑃𝑢1) ⋅((𝐼−𝑃)𝑢2) = 0, as 𝑃𝑢1 ∈col(𝑃) and (𝐼−𝑃)𝑢2 ∈col(𝑃)⟂. We have 𝑢⊺ 1𝑃⊺(𝐼−𝑃)𝑢2 = 0. Since this holds for all 𝑢1, 𝑢2, 𝑃⊺(𝐼−𝑃) = 0 so 𝑃⊺= 𝑃⊺𝑃. Note that 𝑃⊺𝑃 is symmetric, so 𝑃⊺is symmetric, and hence 𝑃is symmetric.
Corollary. Let 𝑃be an orthogonal projection matrix. Then 𝐼−𝑃is also an orthogonal projection matrix.
Proof. Clearly, if 𝑃is symmetric, so is 𝐼−𝑃, so it suffices to prove idempotence. We have (𝐼−𝑃)(𝐼−𝑃) = 𝐼−2𝑃+ 𝑃2 = 𝐼−2𝑃+ 𝑃= 𝐼−𝑃as required.
Proposition. If 𝑃is an orthogonal projection, then 𝑃= 𝑈𝑈⊺where the columns of 𝑈are an orthonormal basis for the column space of 𝑃.
Proof. First, we show that 𝑈𝑈⊺is an orthogonal projection. This is clearly symmetric. It is idempotent: 𝑈𝑈⊺𝑈𝑈⊺= 𝑈𝑈⊺since 𝑈⊺𝑈= 𝐼, as the columns of 𝑈form an orthonormal basis for the column space of 𝑃. Further, the column space of 𝑃is exactly the column space of 𝑈𝑈⊺.
Proposition. The rank of an orthogonal projection matrix is equal to its trace.
Proof. The rank is the dimension of the column space, which is rank 𝑃= rank(𝑈⊺𝑈) = tr(𝑈⊺𝑈) = tr(𝑈𝑈⊺) = tr 𝑃.
Theorem. Let 𝑋be multivariate normal, where 𝑋∼𝑁(0, 𝜎2𝐼), and let 𝑃be an orthogonal projection. Then (i) 𝑃𝑋∼𝑁(0, 𝜎2𝑃), and (𝐼−𝑃)𝑋∼𝑁(0, 𝜎2(𝐼−𝑃)), and these two random variables are independent; (ii) ‖𝑃𝑋‖2 𝜎2 ∼𝜒2 rank 𝑃.
645 XI. Statistics Proof. The vector (𝑃, 𝐼−𝑃)⊺𝑋is multivariate normal, since it is a linear function of 𝑋. This distribution is fully specified by its mean and variance.
𝔼[( 𝑃𝑋 (𝐼−𝑃)𝑋)] = ( 𝑃 𝐼−𝑃) 𝔼[𝑋] = 0 Further, Var (( 𝑃𝑋 (𝐼−𝑃)𝑋)) = ( 𝑃 𝐼−𝑃) 𝜎2𝐼( 𝑃 𝐼−𝑃) ⊺ = 𝜎2 ( 𝑃2 𝑃(𝐼−𝑃) 𝑃(𝐼−𝑃) (𝐼−𝑃)2) = 𝜎2 (𝑃 0 0 𝐼−𝑃) Now we must show that the variables 𝑃𝑋, (𝐼−𝑃)𝑋are independent. Let 𝑍∼𝑁(0, 𝜎2𝑃), 𝑍′ ∼ 𝑁(0, 𝜎2(𝐼−𝑃)) be independent. Then we can see that (𝑍, 𝑍′)⊺is multivariate normal with 𝜇= 0; Σ = (𝑃 0 0 𝐼−𝑃) Hence (𝑃𝑋, (1 −𝑃)𝑋)⊺is equal in distribution to (𝑍, 𝑍′)⊺. So 𝑃𝑋is independent of (𝐼−𝑃)𝑋.
We must show that ‖𝑃𝑋‖2 𝜎2 ∼𝜒2 rank 𝑃. Note that ‖𝑃𝑋‖2 𝜎2 = 𝑋⊺𝑃⊺𝑃𝑋 𝜎2 = 𝑋⊺(𝑈𝑈⊺) ⊺𝑈𝑈⊺𝑋 𝜎2 = ‖𝑈⊺𝑋‖2 𝜎2 Note, 𝑈⊺𝑋∼𝑁(0, 𝜎2𝑈⊺𝑈) = 𝑁(0, 𝜎2𝐼rank 𝑃). So (𝑈⊺𝑋)𝑖 𝜎 iid ∼𝑁(0, 1) for 𝑖= 1, … , rank 𝑃. Hence ‖𝑃𝑋‖2 𝜎2 = rank 𝑃 ∑ 𝑖=1 ((𝑈⊺𝑋)𝑖 𝜎 ) 2 ∼𝜒2 rank 𝑃 Theorem. Let 𝑋1, … , 𝑋𝑛 iid ∼𝑁(𝜇, 𝜎2) for some unknown 𝜇∈ℝand 𝜎2 > 0. The maximum likelihood estimators for 𝜇and 𝜎are ̂ 𝜇= 𝑋= 1 𝑛∑ 𝑖 𝑋𝑖; ̂ 𝜎2 = 𝑆𝑥𝑥 𝑛 = ∑𝑖(𝑋𝑖−𝑋) 2 𝑛 Further, (i) 𝑋∼𝑁(𝜇, 𝜎2 𝑛); (ii) 𝑆𝑥𝑥 𝜎2 ∼𝜒2 𝑛−1; 646 6. The normal linear model (iii) 𝑋, 𝑆𝑥𝑥are independent.
Proof. Let 𝑃be the square 𝑛× 𝑛matrix with all entries 1 𝑛. This is an orthogonal projection matrix, as it is symmetric and idempotent. Note that 𝑃𝑋= ( 𝑋 ⋮ 𝑋 ) We will write the observations 𝑋as 𝑋= ( 𝜇 ⋮ 𝜇 ) ⏟ 𝑀 +𝜀; 𝜀∼𝑁(0, 𝜎2𝐼) Note that 𝑋is a function of 𝑃𝜀, since 𝑋= (𝑃𝑋)1 = (𝑃𝑀+ 𝑃𝜀)1. Further, 𝑆𝑥𝑥= ∑ 𝑖 (𝑋𝑖−𝑋) 2 = ‖𝑋−𝑃𝑋‖2 = ‖(𝐼−𝑃)𝑋‖2 = ‖(𝐼−𝑃)𝜀‖2 Hence 𝑆𝑥𝑥is a function of (𝐼−𝑃)𝜀. Since 𝑃𝜀and (𝐼−𝑃)𝜀are independent, 𝑋and 𝑆𝑥𝑥are independent. Since 𝐼−𝑃is a projection with rank equal to its trace 𝑛−1, we apply the previous theorem to obtain 𝑆𝑥𝑥= ‖(𝐼−𝑃)𝜀‖2𝜒2 𝑛−1 6.3. Linear model Suppose we have data in pairs (𝑥1, 𝑌 1), … , (𝑥𝑛, 𝑌 𝑛), where 𝑌𝑖∈ℝ, 𝑥𝑖∈ℝ𝑝. The 𝑌𝑖are known as the response variables, or the dependent variables. The 𝑥𝑖1, 𝑥𝑖𝑝are the predictors, or independent variables. We will model the expectation of the response 𝑌𝑖as a linear function of the predictors (𝑥𝑖1, … , 𝑥𝑖𝑝).
Example. Let 𝑌𝑖be the number of insurance claims that driver 𝑖makes in a given year, and 𝑥𝑖1, … , 𝑥𝑖𝑝is a set of variables about the specific driver. Predictors include age, the number of years they have held their license, and the number of points on their license, for instance.
We assume that 𝑌𝑖= 𝛼+ 𝛽1𝑥𝑖1 + ⋯+ 𝛽𝑝𝑥𝑖𝑝+ 𝜀𝑖 where 𝛼∈ℝis an intercept, 𝛽𝑖are the coefficients, and 𝜀is a noise vector, which is a random variable. The intercept and coefficients are the parameters of interest. We will often elimin-ate the intercept by making one of the predictors 𝑥𝑖1 = 1 for all 𝑖, so 𝛽1 plays the role of the intercept.
647 XI. Statistics Note that we can use a linear model to model nonlinear relationships. For example, suppose 𝑌𝑖= 𝑎+ 𝑏𝑧𝑖+ 𝑐𝑧2 𝑖+ 𝜀𝑖. We can rephrase this as a linear model with 𝑥𝑖= (1, 𝑧𝑖, 𝑧2 𝑖).
The coefficient 𝛽𝑗can be interpreted as the effect on 𝑌𝑖of increasing 𝑥𝑖𝑗by one, while keep-ing all other predictors fixed. This cannot be interpreted as a causal relationship, unless this is a randomised control experiment.
6.4. Matrix formulation Let 𝑌= ( 𝑌 1 ⋮ 𝑌 𝑛 ) ; 𝑋= ( 𝑥11 ⋯ 𝑥1𝑝 ⋮ ⋱ ⋮ 𝑥𝑛1 ⋯ 𝑥𝑛𝑝 ) ; 𝛽= ( 𝛽1 ⋮ 𝛽𝑝 ) ; 𝜀= ( 𝜀1 ⋮ 𝜀𝑛 ) We call 𝑋the design matrix. The linear model is that 𝑌= 𝑋𝛽+ 𝜀 𝑋𝛽is considered fixed. Since 𝜀is random, this makes 𝑌into a random variable.
6.5. Assumptions We make a number of moment assumptions on the noise vector 𝜀. This allows us to deduce more results about the linear model.
(i) 𝔼[𝜀] = 0 ⟹𝔼[𝑌𝑖] = 𝑥⊺ 𝑖𝛽; (ii) Var (𝜀) = 𝜎2𝐼, which is equivalent to both Var (𝜀𝑖) = 𝜎2 and Cov (𝜀𝑖, 𝜀𝑗) = 0 for all 𝑖≠𝑗. This property is known as homoscedasticity.
We will always assume that the design matrix 𝑋has full rank 𝑝, or equivalently, that it has linearly independent columns. Since 𝑋∈ℝ𝑛×𝑝, this requires that 𝑛≥𝑝, so we need at least as many samples as we have predictors.
6.6. Least squares estimation Definition. The least squares estimator ̂ 𝛽minimises the residual sum of squares, which is 𝑆(𝛽) = ‖𝑌−𝑋𝛽‖2 = ∑ 𝑖 (𝑌𝑖−𝑥⊺ 𝑖𝛽) 2 The term 𝑌𝑖−𝑥⊺ 𝑖𝛽is called the 𝑖th residual.
Since 𝑆(𝛽) is a positive definite quadratic in 𝛽, it is minimised at the stationary point.
𝜕𝑆(𝛽) 𝜕𝛽𝑘 | | |𝛽= ̂ 𝛽 = 0 ⟺∀𝑘, −2 𝑛 ∑ 𝑖=1 𝑥𝑖𝑘(𝑌𝑖−∑ 𝑘 𝑥𝑖𝑗̂ 𝛽𝑗) = 0 ⟺𝑋⊺𝑋̂ 𝛽= 𝑋⊺𝑌 648 6. The normal linear model As 𝑋has full column rank, 𝑋⊺𝑋is invertible.
̂ 𝛽= (𝑋⊺𝑋)−1𝑋⊺𝑌 This is notably a linear function of 𝑌, given fixed 𝑋. Note that 𝔼[ ̂ 𝛽] = (𝑋⊺𝑋)−1𝑋⊺𝔼[𝑌] = (𝑋⊺𝑋)−1𝑋⊺𝑋𝛽= 𝛽 So ̂ 𝛽is an unbiased estimator. Further, Var ( ̂ 𝛽) = (𝑋⊺𝑋)−1𝑋⊺Var (𝑌) [(𝑋⊺𝑋)−1𝑋⊺] ⊺ = (𝑋⊺𝑋)−1𝑋⊺𝜎2𝐼[(𝑋⊺𝑋)−1𝑋⊺] ⊺ = 𝜎2(𝑋⊺𝑋)−1 Theorem (Gauss–Markov theorem). Let an estimator 𝛽⋆of 𝛽be unbiased and a linear func-tion of 𝑌, so 𝛽⋆= 𝐶𝑌. Then, for any fixed 𝑡∈ℝ𝑝, we have Var (𝑡⊺̂ 𝛽) ≤Var (𝑡⊺𝛽⋆) where ̂ 𝛽is the least squares estimator. We say that ̂ 𝛽is the best linear unbiased estimator (BLUE).
Remark. We can think of 𝑡∈ℝ𝑝as a vector of predictors for a new sample. Then 𝑡⊺̂ 𝛽is the prediction for 𝔼[𝑌𝑖] for this new sample, using the least squares estimator. 𝑡⊺𝛽⋆is the prediction with 𝛽⋆. In both cases, the prediction is unbiased.
Proof. Note that Var (𝑡⊺𝛽⋆) −Var (𝑡⊺̂ 𝛽) = 𝑡⊺[Var (𝛽⋆) −Var ( ̂ 𝛽)]𝑡 To prove that this quantity is always non-negative, we must show that Var (𝛽⋆) −Var ( ̂ 𝛽) is positive semidefinite. Let 𝐴= 𝐶−(𝑋⊺𝑋)−1𝑋⊺. Note that 𝔼[𝐴𝑌] = 𝔼[𝛽⋆] −𝔼[ ̂ 𝛽] = 0. Also, 𝔼[𝐴𝑌] = 𝐴𝔼[𝑌] = 𝐴𝑋𝛽. This holds for all 𝛽, so 𝐴𝑋= 0. Now, since 𝑋⊺𝑋is symmetric, Var (𝛽⋆) = Var (𝐶𝑌) = Var ((𝐴+ (𝑋⊺𝑋)−1𝑋⊺)𝑌) = [𝐴+ (𝑋⊺𝑋)−1𝑋⊺] Var (𝑌) [𝐴+ (𝑋⊺𝑋)−1𝑋⊺] ⊺ = [𝐴+ (𝑋⊺𝑋)−1𝑋⊺]𝜎2𝐼[𝐴+ (𝑋⊺𝑋)−1𝑋⊺] ⊺ = 𝜎2(𝐴𝐴⊺+ (𝑋⊺𝑋)−1 + 𝐴𝑋(𝑋⊺𝑋)−1 + (𝑋⊺𝑋)−1𝑋⊺𝐴⊺) = 𝜎2𝐴𝐴⊺+ Var ( ̂ 𝛽) Var (𝛽⋆) −Var ( ̂ 𝛽) = 𝜎2𝐴𝐴⊺ Note that the outer product 𝐴𝐴⊺is always positive semidefinite.
649 XI. Statistics 6.7. Fitted values and residuals Definition. The fitted values are ̂ 𝑌= 𝑋̂ 𝛽= 𝑋(𝑋⊺𝑋)−1𝑋⊺𝑌, where 𝑃= 𝑋(𝑋⊺𝑋)−1𝑋⊺is the hat matrix. The residuals are 𝑌− ̂ 𝑌= (𝐼−𝑃)𝑌.
Proposition. 𝑃is the orthogonal projection onto the column space of the design matrix.
Proof. If 𝑣is in the column space of 𝑋, then 𝑣= 𝑋𝑏for some 𝑏. Hence 𝑃𝑣= 𝑋(𝑋⊺𝑋)−1𝑋⊺𝑋𝑏= 𝑋𝑏= 𝑣 If 𝑤is in the orthogonal complement, then 𝑃𝑤= 𝑋(𝑋⊺𝑋)−1 𝑋⊺𝑤 ⏟ 0 = 0 Corollary. The fitted values are an orthogonal projection of the response variables to the column space of the design matrix. The residuals are orthogonal to the column space.
6.8. Normal linear model The normal linear model is a linear model under the assumption that 𝜀∼𝑁(0, 𝜎2𝐼), where 𝜎2 is unknown. The parameters in the model are now (𝛽, 𝜎2). The likelihood function in the normal linear model is 𝐿(𝛽, 𝜎2) = 𝑓 𝑌(𝑦∣𝛽, 𝜎2) = (2𝜋𝜎2)−𝑛 2 exp{−1 2𝜎2 ∑ 𝑖 (𝑌𝑖−𝑥⊺ 𝑖𝛽)2} The log-likelihood is ℓ(𝛽, 𝜎2) = constant −𝑛 2 log 𝜎2 − 1 2𝜎2 ‖𝑌−𝑋𝛽‖2 To maximise this as a function of 𝛽for any fixed 𝜎2, we must minimise the residual sum of squares 𝑆(𝛽) = ‖𝑌−𝑋𝛽‖2. So ̂ 𝛽= (𝑋⊺𝑋)−1𝑋⊺𝑌is the maximum likelihood estimator of 𝛽.
Further, ̂ 𝜎2 = 𝑛−1‖ ‖𝑌−𝑋̂ 𝛽‖ ‖ 2 = 𝑛−1‖ ‖ ̂ 𝑌−𝑌‖ ‖ 2 = 𝑛−1‖(𝐼−𝑃)𝑌‖2.
Theorem. In the normal linear model, (i) ̂ 𝛽∼𝑁(𝛽, 𝜎2(𝑋⊺𝑋)−1); (ii) 𝑛 ̂ 𝜎2 𝜎2 ∼𝜒2 𝑛−𝑝; (iii) ̂ 𝛽, ̂ 𝜎2 are independent.
Proof. We prove each part separately.
650 6. The normal linear model (i) We already know that 𝔼[ ̂ 𝛽] = 𝛽, and Var ( ̂ 𝛽) = 𝜎2(𝑋⊺𝑋)−1. So it suffices to show that ̂ 𝛽is a normal vector. Since ̂ 𝛽= (𝑋⊺𝑋)−1𝑋⊺𝑌, it is a linear function of a normal vector, so is a normal vector.
(ii) Observe that 𝑛̂ 𝜎2 𝜎2 = ‖(𝐼−𝑃)𝑌‖2 𝜎2 = ‖(𝐼−𝑃)(𝑋𝛽+ 𝜀)‖2 𝜎2 Since (𝐼−𝑃)𝑋= 0 as 𝑃is the orthogonal projection onto the column space of 𝑋, 𝑛̂ 𝜎2 𝜎2 = ‖(𝐼−𝑃)𝜀‖2 𝜎2 ∼𝜒2 tr(𝐼−𝑃) where tr(𝐼−𝑃) = tr 𝐼−tr 𝑃= 𝑛−𝑝since 𝑋∈ℝ𝑛×𝑝is assumed to have full rank.
(iii) Note that ̂ 𝜎2 is a function of (𝐼−𝑃)𝜀, and ̂ 𝛽= (𝑋⊺𝑋)−1𝑋⊺𝑌 = (𝑋⊺𝑋)−1𝑋⊺(𝑋𝛽+ 𝜀) = 𝛽+ (𝑋⊺𝑋)−1𝑋⊺𝜀 = 𝛽+ (𝑋⊺𝑋)−1𝑋⊺𝑃𝜀 is a function of 𝑃𝜀. Since (𝐼−𝑃)𝜀and 𝑃𝜀are independent, so are ̂ 𝛽, ̂ 𝜎2.
Note, 𝔼[𝑛̂ 𝜎2 𝜎2 ] = 𝔼[𝜒2 𝑛−𝑝] = 𝑛−𝑝⟹𝔼[ ̂ 𝜎2] = 𝜎2 ⋅𝑛−𝑝 𝑛 < 𝜎2 Hence this ̂ 𝜎2 is a biased estimator, but asymptotically unbiased.
6.9. Inference Definition. Let 𝑈∼𝑁(0, 1) and 𝑉∼𝜒2 𝑛be independent random variables. Then 𝑇= 𝑈 √ 𝑉 𝑛 has a 𝑡𝑛-distribution.
As 𝑛→∞, this approaches the standard normal distribution.
Definition. Let 𝑉∼𝜒2 𝑛and 𝑊∼𝜒2 𝑚be independent random variables. Then 𝐹= 𝑉/𝑛 𝑊/𝑚 has an 𝐹 𝑛,𝑚-distribution.
651 XI. Statistics Example. We consider a 100(1−𝛼)% confidence interval for one of the coefficients 𝛽in the normal linear model 𝑌= 𝑋𝛽+ 𝜀. Without loss of generality, we will consider 𝛽1.
We begin by finding a pivot, which is a distribution that does not depend on the parameters of the model. By standardising the above form of ̂ 𝛽, 𝛽1 − ̂ 𝛽1 √𝜎2(𝑋⊺𝑋)−1 11 ∼𝑁(0, 1) where 𝑀−1 11 is the top left entry in the matrix 𝑀−1. This random variable is independent from 𝑛̂ 𝜎2 𝜎2 ∼𝜒2 𝑛−𝑝. Now, to construct a pivot, we find 𝛽1−̂ 𝛽1 √𝜎2(𝑋⊺𝑋)−1 11 √ ̂ 𝜎2 𝜎2 ⋅ 𝑛 𝑛−𝑝 ∼ 𝑈 √ 𝑉 𝑛 ∼𝑡𝑛−𝑝 The 𝜎2 terms cancel, so the statistic is a function only of 𝛽1 and functions of the data. Then, ℙ𝛽,𝜎2 (−𝑡𝑛−𝑝(𝛼 2 ) ≤ ̂ 𝛽1 −𝛽1 √(𝑋⊺𝑋)−1 11 √ 𝑛−𝑝 𝑛̂ 𝜎2 ≤𝑡𝑛−𝑝(𝛼 2 )) = 1 −𝛼 since the 𝑡distribution is symmetric about zero. Rearranging to find an interval for 𝛽1, ℙ𝛽,𝜎2 ( ̂ 𝛽1 −𝑡𝑛−𝑝(𝛼 2 )√(𝑋⊺𝑋)−1 11 ̂ 𝜎2 √(𝑛−𝑝)/𝑛 ≤𝛽1 ≤ ̂ 𝛽1 + 𝑡𝑛−𝑝(𝛼 2 )√(𝑋⊺𝑋)−1 11 ̂ 𝜎2 √(𝑛−𝑝)/𝑛 ) = 1 −𝛼 Hence, 𝐼= [ ̂ 𝛽1 ± 𝑡𝑛−𝑝(𝛼 2 )√(𝑋⊺𝑋)−1 11 ̂ 𝜎2 √(𝑛−𝑝)/𝑛 ] is a 100(1 −𝛼)% confidence interval for 𝛽1.
Consider a test for 𝐻0 ∶𝛽1 = 0, 𝐻1 ∶𝛽1 ≠0. By connecting tests and confidence intervals, we can test 𝐻0 with size 𝛼by rejecting this null hypothesis when zero is not contained within the confidence interval 𝐼.
Consider a special case where 𝑌 1, … , 𝑌 𝑛 iid ∼𝑁(𝜇, 𝜎2) where 𝜇, 𝜎2 are unknown, and we want to infer results about 𝜇. Note that this is a special case of the normal linear model where 𝑋= ( 1 ⋮ 1 ) ; 𝛽= (𝜇) So we can infer a confidence interval for 𝜇using the above statistic.
652 6. The normal linear model Example. Consider a 100(1 −𝛼)% confidence set for 𝛽as a whole. Note that ̂ 𝛽−𝛽∼𝑁(0, 𝜎2(𝑋⊺𝑋)−1) Then, (𝑋⊺𝑋)1/2( ̂ 𝛽−𝛽) ∼𝑁(0, 𝜎2(𝑋⊺𝑋)1/2(𝑋⊺𝑋)−1(𝑋⊺𝑋)1/2) ∼𝑁(0, 𝜎2𝐼) where (𝑋⊺𝑋)1/2 is obtained using the eigendecomposition of the positive definite matrix 𝑋⊺𝑋.
Hence, ‖ ‖(𝑋⊺𝑋)1/2( ̂ 𝛽−𝛽)‖ ‖ 2 𝜎2 ∼𝜒2 𝑝 We can also write this as ‖ ‖(𝑋⊺𝑋)1/2( ̂ 𝛽−𝛽)‖ ‖ 2 𝜎2 = ‖ ‖𝑋( ̂ 𝛽−𝛽)‖ ‖ 2 𝜎2 Since this is a function of ̂ 𝛽, this is independent of any function of ̂ 𝜎2. In particular, it is independent of 𝑛̂ 𝜎2 𝜎2 ∼𝜒2 𝑛−𝑝. Thus, we can form a pivot by ‖ ‖𝑋( ̂ 𝛽−𝛽)‖ ‖ 2 /(𝜎2𝑝) ̂ 𝜎2𝑛/(𝜎2(𝑛−𝑝)) ∼ 𝜒2 𝑝/𝑝 𝜒2 𝑛−𝑝/(𝑛−𝑝) ∼𝐹 𝑝,𝑛−𝑝 This does not depend on 𝜎2. For all 𝛽, 𝜎2, ℙ𝛽,𝜎2 ⎛ ⎜ ⎜ ⎝ ‖ ‖𝑋( ̂ 𝛽−𝛽)‖ ‖ 2 /𝑝 ̂ 𝜎2𝑛/(𝑛−𝑝) ≤𝐹 𝑝,𝑛−𝑝(𝛼) ⎞ ⎟ ⎟ ⎠ = 1 −𝛼 because the 𝐹distribution has support only on the positive real line. It is nontrivial to express this as a region for 𝛽since it is vector-valued. We can say, however, that ⎧ ⎨ ⎩ 𝛽′ ∈ℝ𝑝∶ ‖ ‖𝑋( ̂ 𝛽−𝛽)‖ ‖ 2 /𝑝 ̂ 𝜎2𝑛/(𝑛−𝑝) ≤𝐹 𝑝,𝑛−𝑝(𝛼) ⎫ ⎬ ⎭ is a 100(1 −𝛼)% confidence set for 𝛽.
This set is an ellipsoid centred at ̂ 𝛽. The shape of the ellipsoid depends on the design matrix 𝑋; the principal axes are given by eigenvectors of 𝑋⊺𝑋.
The above two results are exact; no approximations were made.
653 XI. Statistics 6.10. 𝐹-tests We wish to test whether a collection of predictors 𝛽𝑖are equal to zero. Without loss of generality, we will take the first 𝑝0 ≤𝑝predictors. We have 𝐻0 ∶𝛽1 = ⋯= 𝛽𝑝0 = 0, and 𝐻1 = 𝛽∈ℝ𝑝. We denote 𝑋= (𝑋0, 𝑋1) as a block matrix with 𝑋0 ∈ℝ𝑛×𝑝0 and 𝑋1 ∈ℝ𝑛×(𝑝−𝑝0), and we denote 𝛽= (𝛽0, 𝛽1)⊺similarly. The null model has 𝛽0 = 0. This is a linear model 𝑌= 𝑋𝛽+𝜀= 𝑋1𝛽1+𝜀. We will write 𝑃= 𝑋(𝑋⊺𝑋)−1𝑋⊺and 𝑃 1 = 𝑋1(𝑋⊺ 1𝑋1)−1𝑋⊺ 1.
Note that as 𝑋and 𝑃have full rank, so must 𝑋1, 𝑃 1.
Lemma. (𝐼−𝑃)(𝑃−𝑃 1) = 0, and 𝑃−𝑃 1 is an orthogonal projection with rank 𝑝0.
Proof. 𝑃−𝑃 1 is symmetric since 𝑃and 𝑃 1 are symmetric. It is also idempotent, since (𝑃−𝑃 1)(𝑃−𝑃 1) = 𝑃2 −𝑃 1𝑃−𝑃𝑃 1 + 𝑃2 1 = 𝑃−𝑃 1 −𝑃 1 + 𝑃 1 = 𝑃−𝑃 1 since 𝑃 1 projects onto the column space of 𝑋1. Hence 𝑃−𝑃 1 is indeed an orthogonal projection matrix. The rank is rank(𝑃−𝑃 1) = tr(𝑃−𝑃 1) = tr 𝑃−tr 𝑃 1 = 𝑝−(𝑝−𝑝0) = 𝑝0. Also, (𝐼−𝑃)(𝑃−𝑃 1) = 𝑃−𝑃 1 −𝑃+ 𝑃𝑃 1 = 𝑃−𝑃 1 −𝑃+ 𝑃 1 = 0 Recall that the maximum log-likelihood in the normal linear model is given by ℓ( ̂ 𝛽, ̂ 𝜎2) = −𝑛 2 log ̂ 𝜎2 −𝑛 2 ⋅constant = −𝑛 2 log ‖(𝐼−𝑃)𝑌‖2 𝑛 + constant The generalised likelihood ratio statistic is 2 log Λ = 2 sup 𝛽∈ℝ𝑝,𝜎2>0 ℓ(𝛽, 𝜎2) −2 sup 𝛽0=0,𝛽1∈ℝ𝑝−𝑝0,𝜎2>0 ℓ(𝛽, 𝜎2) = 𝑛[−log ‖(𝐼−𝑃)𝑌‖2 𝑛 + log ‖(𝐼−𝑃 1)𝑌‖2 𝑛 ] Wilks’ theorem applies here, showing that 2 log Λ ∼𝜒2 𝑝0 asymptotically as 𝑛→∞with 𝑝, 𝑝0 fixed. However, we can find an exact test, so using Wilks’ theorem will not be necessary.
2 log Λ is monotone in ‖(𝐼−𝑃 1)𝑌‖2 ‖(𝐼−𝑃)𝑌‖2 = ‖(𝐼−𝑃+ 𝑃−𝑃 1)𝑌‖2 ‖(𝐼−𝑃)𝑌‖2 = ‖(𝐼−𝑃)𝑌‖2 + ‖(𝑃−𝑃 1)𝑌‖2 + 2𝑌⊺(𝐼−𝑃)(𝑃−𝑃 1) ‖(𝐼−𝑃)𝑌‖2 = ‖(𝐼−𝑃)𝑌‖2 + ‖(𝑃−𝑃 1)𝑌‖2 ‖(𝐼−𝑃)𝑌‖2 = 1 + ‖(𝑃−𝑃 1)𝑌‖2 ‖(𝐼−𝑃)𝑌‖2 654 6. The normal linear model The generalised likelihood ratio test rejects when the 𝐹-statistic 𝐹= ‖(𝑃−𝑃 1)𝑌‖2 ‖(𝐼−𝑃)𝑌‖2 ⋅ 1/𝑝0 1/(𝑛−𝑝) is large.
Theorem. Under 𝐻0 ∶𝛽1 = ⋯= 𝛽𝑝0 = 0, in the normal linear model, 𝐹= ‖(𝑃−𝑃 1)𝑌‖2 ‖(𝐼−𝑃)𝑌‖2 ⋅ 1/𝑝0 1/(𝑛−𝑝) ∼𝐹 𝑝0,𝑛−𝑝 Proof. Recall that ‖(𝐼−𝑃)𝑌‖2 = ‖(𝐼−𝑃)𝜀‖2 ∼𝜒2 𝑛−𝑝⋅𝜎2 Therefore it suffices to show that ‖(𝑃−𝑃 1)𝑌‖2 is an independent 𝜒2 𝑝0 ⋅𝜎2 random variable.
Under 𝐻0, we have that (𝑃−𝑃 1)𝑌= (𝑃−𝑃 1)(𝑋𝛽+ 𝜀) = (𝑃−𝑃 1)(𝑋1𝛽1 + 𝜀) = (𝑃−𝑃 1)𝜀 since 𝑃, 𝑃 1 preserve 𝑋1. Hence, ‖(𝑃−𝑃 1)𝑌‖2 = ‖(𝑃−𝑃 1)𝜀‖2 ∼𝜒2 rank(𝑃−𝑃1) ⋅𝜎2 = 𝜒2 𝑝0 ⋅𝜎2. We must now show independence between (𝐼−𝑃)𝑌and (𝑃−𝑃 1)𝑌. The vectors (𝐼−𝑃)𝜀, (𝑃−𝑃 1)𝜀 are independent; indeed, 𝐸= ( (𝐼−𝑃)𝜀 (𝑃−𝑃 1)𝜀) is a multivariate normal vector, and 𝔼[𝐸] = 0; Var (𝐸) = ( 𝐼−𝑃 (𝐼−𝑃)(𝑃−𝑃 1) (𝐼−𝑃)(𝑃−𝑃 1) 𝑃−𝑃 1 ) = (𝐼−𝑃 0 0 𝑃−𝑃 1 ) and since (𝐼−𝑃)𝜀and (𝑃−𝑃 1)𝜀are elements of a multivariate normal vector and are uncor-related, they are independent as required.
The generalised likelihood ratio test of size 𝛼rejects 𝐻0 when 𝐹> 𝐹−1 𝑝0,𝑛−𝑝(𝛼). This is an exact test for all 𝑛, 𝑝, 𝑝0. Previously, we found a test for 𝐻0 ∶𝛽1 = 0 against 𝐻1 ∶𝛽1 ≠0.
This is a special case of the 𝐹-test derived above, where 𝑝0 = 1. The previous test of size 𝛼 rejects 𝐻0 when | | ̂ 𝛽| | > 𝑡𝑛−𝑝(𝛼 2 )√ ̂ 𝜎2𝑛(𝑋⊺𝑋)−1 11 𝑛−𝑝 We will show that these two tests are equivalent; they reject 𝐻0 in the same critical region.
The 𝑡-test rejects if and only if ̂ 𝛽2 1 > 𝑡𝑛−𝑝(𝛼 2 ) 2 ̂ 𝜎2𝑛(𝑋⊺𝑋)−1 11 𝑛−𝑝 655 XI. Statistics Note that 𝑡𝑛−𝑝( 𝛼 2 ) 2 = 𝐹1,𝑛−𝑝(𝛼), since 𝑈∼𝑁(0, 1); 𝑊sin 𝜒2 𝑛⟹𝑇= 𝑈 √𝑊/𝑛 ⟹𝑇2 = 𝑈2 𝑊/𝑛= 𝑉/1 𝑊/𝑛∼𝐹1,𝑛 where 𝑉∼𝜒2 1. Hence, ̂ 𝛽1/(𝑋⊺𝑋)−1 11 ̂ 𝜎2𝑛/(𝑛−𝑝) > 𝐹1,𝑛−𝑝(𝛼) It suffices to show that ̂ 𝛽1 (𝑋⊺𝑋)−1 11 = ‖(𝑃−𝑃 1)𝑌‖2 𝑝0 ⏟ =1 ; ̂ 𝜎2𝑛 𝑛−𝑝= ‖(𝐼−𝑃)𝑌‖2 𝑛−𝑝 We have already shown the latter part. For ̂ 𝛽1, note that in this case, 𝑃−𝑃 1 is a projection of rank 1 onto the one-dimensional subspace spanned by the vector 𝑣= (𝐼−𝑃)𝑋0 where 𝑋0 is the first column in the matrix 𝑋. First, note the following identity.
𝑋⊺ 0(𝐼−𝑃 1) = 𝑣⊺= 𝑣⊺(𝑃−𝑃 1) = 𝑋⊺ 0(𝐼−𝑃 1)(𝑃−𝑃 1) = 𝑋⊺ 0(𝐼−𝑃 1)𝑃 Then, ‖(𝑃−𝑃 1)𝑌‖2 = ‖ ‖ ‖ 𝑣 ‖𝑣‖( 𝑣 ‖𝑣‖) ⊺ 𝑌‖ ‖ ‖ 2 = (𝑣⊺𝑌)2 ‖𝑣‖2 = (𝑋⊺ 0(𝐼−𝑃 1)𝑌)2 ‖(𝐼−𝑃 1)𝑋0‖2 = (𝑋⊺ 0(𝐼−𝑃 1)𝑃𝑌)2 ‖(𝐼−𝑃 1)𝑋0‖2 = (𝑋⊺ 0(𝐼−𝑃 1)𝑋̂ 𝛽)2 ‖(𝐼−𝑃 1)𝑋0‖2 Note that (𝐼−𝑃 1)𝑋= [(𝐼−𝑃 1)𝑋0, 0, … , 0]. Hence, ‖(𝑃−𝑃 1)𝑌‖2 = ‖(𝐼−𝑃 1)𝑋0‖4 ̂ 𝛽1 ‖(𝐼−𝑃 1)𝑋0‖2 = ‖(𝐼−𝑃 1)𝑋0‖2 ̂ 𝛽1 Finally, we show that (𝑋⊺𝑋)−1 11 = 1 ‖(𝐼−𝑃 1)𝑋0‖2 using the Woodbury identity for blockwise matrix inversion. Hence, ̂ 𝛽2 1 (𝑋⊺𝑋)−1 11 = ‖(𝑃−𝑃 1)𝑌‖2 as required.
656 6. The normal linear model 6.11. Analysis of variance Suppose we investigate responses of patients after receiving one of three treatments, includ-ing a control, which will be given index 1. We will consider only one predictor, denoting which treatment a given patient received. Consider the linear model 𝑌𝑖𝑗= 𝛼+ 𝜇𝑗+ 𝜀𝑖𝑗 where 𝑗= 1, 2, 3 is the treatment index, and 𝑖= 1, … , 𝑁is the index of a patient in a given group. Let (𝜀𝑖𝑗) ∼𝑁(0, 𝜎2) be independent. Without loss of generality, we can set 𝜇1 = 0, since we have an additional parameter 𝛼; this is known as a corner point constraint. Then, 𝜇𝑗should be interpreted as the effect of treatment 𝑗relative to treatment 1, which in this case is the control.
Definition. The analysis of variance (ANOVA) test on the linear model 𝑌𝑖𝑗= 𝛼+ 𝜇𝑗+ 𝜀𝑖𝑗 where 𝜇1 = 0 is given by 𝐻0 ∶𝜇2 = 𝜇3 = ⋯= 0; 𝐻1 ∶𝜇2, 𝜇3, ⋯∈ℝ In particular, 𝐻0 gives 𝔼[𝑌𝑖𝑗] = 𝛼.
In our example, 𝐻0 ∶𝜇2 = 𝜇3 = 0 and 𝐻1 ∶𝜇2, 𝜇3 ∈ℝ. This is a special case of the 𝐹-test, since we are testing whether the coefficients 𝜇𝑖are equal to zero.
𝑋= ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ 1 0 0 1 0 0 ⋮ ⋮ ⋮ 1 1 0 1 1 0 ⋮ ⋮ ⋮ 1 0 1 1 0 1 ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ = (𝑋1 𝑋0) The first column of 𝑋, denoted 𝑋1, represents 𝛼, and the other columns, denoted 𝑋0, rep-resent 𝜇2, 𝜇3. 𝑋0 is eliminated under the null hypothesis. The predictor can be called cat-egorical; it is discrete, and entirely dependent on which treatment category a given patient is placed in. Note that 𝑋has 3𝑁rows, where each block of 𝑁consecutive rows is identical.
Recall that the 𝐹-test uses the test statistic 𝐹= ‖(𝑃−𝑃 1)𝑌‖2 ‖(𝐼−𝑃)𝑌‖2 ⋅ 1/𝑝0 1/(𝑛−𝑝) ∼𝐹 𝑝0,𝑛−𝑝 For this test, 𝑃projects onto the space of vectors in ℝ3𝑁which are constant over treatment groups. In other words, let 𝑌𝑗= 1 𝑁 𝑁 ∑ 𝑖=1 𝑌𝑖𝑗 657 XI. Statistics Then, 𝑃𝑌= (𝑌1, … , 𝑌1 ⏟ ⎵ ⎵ ⏟ ⎵ ⎵ ⏟ 𝑁entries , 𝑌2, … , 𝑌2 ⏟ ⎵ ⎵ ⏟ ⎵ ⎵ ⏟ 𝑁entries , 𝑌3, … , 𝑌3 ⏟ ⎵ ⎵ ⏟ ⎵ ⎵ ⏟ 𝑁entries ) ⊺ 𝑃 1 projects onto the subspace of constant vectors in ℝ3𝑁, so 𝑌= 1 3𝑁 𝑁 ∑ 𝑖=1 3 ∑ 𝑗=1 𝑌𝑖𝑗⟹𝑃 1𝑌= (𝑌, … , 𝑌 ⏟ ⎵ ⏟ ⎵ ⏟ 3𝑁entries ) ⊺ Hence, we can write the 𝐹statistic as 𝐹= ∑ 3 𝑗=1 𝑁(𝑌𝑗−𝑌) 2 /2 ∑ 𝑁 𝑖=1 ∑ 3 𝑗=1 (𝑌𝑖𝑗−𝑌𝑗) 2 /(3𝑁−3) We can generatlise this to the case where there are 𝐽> 3 treatment groups: 𝐹= ∑ 𝐽 𝑗=1 𝑁(𝑌𝑗−𝑌) 2 /(𝐽−1) ∑ 𝑁 𝑖=1 ∑ 𝐽 𝑗=1 (𝑌𝑖𝑗−𝑌𝑗) 2 /(𝐽𝑁−𝐽) = variance between treatments variance within treatments Remark. This test is sometimes called one-way analysis of variance. Two-way analysis of variance is a similar analysis in an experiment where groups are defined according to two variables. For instance, the response could be a student’s performance in an exam, where the treatments are (i) completion of supervisions (zero representing not complete, one representing com-plete); and (ii) whether a monetary incentive was given (zero representing no incentive, one repres-enting an incentive).
Here, we would have the result 𝑌𝑖𝑗𝑘as the number of marks of student 𝑖in group (𝑗, 𝑘). The model would be 𝑌𝑖𝑗𝑘= 𝛼+ 𝜇𝑗+ 𝜆𝑘+ 𝜀𝑖𝑗𝑘 with a constraint without loss of generality that 𝜇0 = 𝜆0 = 0. The two-way analysis of variance test is then 𝐻0 ∶𝜇1 = 𝜆1 = 0; 𝐻1 ∶𝜇1, 𝜆1 ∈ℝ 6.12. Simple linear regression In a linear regression model, we often centre predictors to simplify certain expressions.
𝑌𝑖= 𝛼+ 𝛽(𝑥−𝑥) + 𝜀𝑖 658 6. The normal linear model where 𝑥= 1 𝑛∑ 𝑛 𝑖=1 𝑥𝑖, and the 𝜀𝑖independently have the usual 𝑁(0, 𝜎2) distribution. In this case, the maximum likelihood estimator ( ̂ 𝛼, ̂ 𝛽) takes a simple form. Recall that ( ̂ 𝛼, ̂ 𝛽) minimises 𝑆(𝛼, 𝛽) = 𝑛 ∑ 𝑖=1 (𝑌𝑖−𝛼−𝛽(𝑥𝑖−𝑥)) 2 Hence, 𝜕𝑆(𝛼, 𝛽) 𝜕𝛼 = 𝑛 ∑ 𝑖=1 −2(𝑌𝑖−𝛼−𝛽(𝑥𝑖−𝑥)) = 𝑛 ∑ 𝑖=1 −2(𝑌𝑖−𝛼) This gives the simple expression 𝛼= ∑ 𝑛 𝑖=1 𝑌𝑖 𝑛 = 𝑌 Now, 𝜕𝑆(𝛼, 𝛽) 𝜕𝛽 | | |𝛼= ̂ 𝛼 = 𝑛 ∑ 𝑖=1 −2(𝑌𝑖−𝑌−𝛽(𝑥𝑖−𝑥))(𝑥𝑖−𝑥) This vanishes when ̂ 𝛽= ∑ 𝑛 𝑖=1 (𝑌𝑖−𝑌)(𝑥𝑖−𝑥) ∑ 𝑛 𝑖=1(𝑥𝑖−𝑥)2 = 𝑆𝑥𝑦 𝑆𝑥𝑥 Note that 𝑆𝑥𝑦 𝑛is the sample covariance of 𝑋and 𝑌, and 𝑆𝑥𝑥 𝑛is the sample variance of 𝑋.
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11948 | https://www.themathdoctors.org/integrating-rational-functions-beyond-partial-fractions/ | Integrating Rational Functions: Beyond Partial Fractions – The Math Doctors
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Integrating Rational Functions: Beyond Partial Fractions
August 2, 2024 August 1, 2024 / Calculus / Alternatives, Integration / By Dave Peterson
A couple recent questions offered tricks for integrating rational functions, opportunistically modifying or working around the usual method of partial fractions. We have previously discussed this method in Partial Fractions: How and Why, and in Integration: Partial Fractions and Substitution, where we looked at other variations.
First problem: A trick with negative exponents
First, a question from Amia in June:
Hi Dr math,
I want your opinion for the solution attached:
Amia started by multiplying the numerator and denominator by a negative power of x; that amounts to dividing the numerator and denominator by the greatest power in the denominator, which is a common technique for dealing with limits at infinity; perhaps that is the motivation.
Then he splits the numerator into a multiple of the denominator, plus a remainder; this amounts to long division, as we would do if the numerator had a higher degree than the denominator. The second term can then be integrated by substitution (u=1+x−2).
We can rewrite the answer without negative exponents, to make it look more like we expect:
x−1−1+1 2 ln∣∣1+x−2∣∣=1 2 ln∣∣∣x 2+1 x 2∣∣∣−1 x+C
Altogether, this demonstrates what you can accomplish by “playing with possibilities” rather than following standard methods.
I answered with approval and curiosity:
Hi, Amia.
Your work looks good, and creative.
I did it more routinely, by partial fractions, and got the same result.
May I ask, what led you to think of doing that? I generally tend to change negative exponents to positive, rather than vice versa, because it’s easy to make mistakes with negatives; but I wonder if this approach could be generalized.
The routine solution
Here is the routine method using partial fractions:
x 2−x+1 x 4+x 2=x 2−x+1 x 2(x 2+1)=A x+B x 2+C x+D x 2+1 x 2−x+1=A x(x 2+1)+B(x 2+1)+x 2(C x+D)x 2−x+1=A x 3+A x+B x 2+B+C x 3+D x 2 x 2−x+1=(A+C)x 3+(B+D)x 2+A x+B
Equating coefficients, we get the system of equations
⎧⎩⎨⎪⎪⎪⎪⎪⎪A+C B+D A B=0=1=−1=1
From which we find that
⎧⎩⎨⎪⎪⎪⎪⎪⎪A B C D=−1=1=1=0
(Below, we’ll see an example using a different approach, plugging in four values for x to obtain four equations to solve.)
So our integral becomes
∫(−1 x+1 x 2+x x 2+1)d x=−ln|x|−1 x+1 2 ln|x 2+1|=1 2 ln∣∣∣x 2+1 x 2∣∣∣−1 x+C
just as we got above.
The inspiration
Amia replied:
The shape of this question is similar to questions I solved before
Like
1/(x^3+x)
x/(x^2+x^4)
So I tried here the same procedure, and it works.
Can we generalize this approach?
This wasn’t elaborated, but my understanding is that he is rewriting 1 x 3+x by multiplying the numerator and denominator by x, perhaps observing that this will make a substitution (u=x 2) possible:
∫1 x 3+x d x=∫x x 4+x 2 d x=∫1 2 d u u 2+u=1 2∫d u u(u+1)=1 2∫(1 u−1 u+1)d u=1 2(ln|u|−ln|u+1|)=1 2 ln∣∣∣u u+1∣∣∣=1 2 ln∣∣∣x 2 x 2+1∣∣∣+C
We still had to use substitution and partial fractions, but the degree was reduced. Using partial fractions directly on the original integral, we would have to do this:
∫1 x 3+x d x=∫1 x(x 2+1)d x=∫(A x+B x+C x 2+1)d x=∫(1 x−x x 2+1)d x=ln|x|−1 2 ln∣∣x 2+1∣∣=1 2 ln∣∣∣x 2 x 2+1∣∣∣+C
This doesn’t seem like a great simplification, so there may be another trick I’m missing; but the fact that a multiplication initially appears to complicate the integrand, but then makes a different approach possible, is the link between the two problems.
As for generalizing, this trick works only when it happens to work (if I’m understanding it correctly); so the generalization is just “look for possible shortcuts, rather than being locked into a routine”. That’s good advice once you’ve mastered the routines.
Second problem: Partial cancellation
Amia had another problem with another trick:
I have this solution for another question.
This integrand is another proper rational function. I’ll discuss below what he is doing, adding and subtracting x 2 in order to be able to do some impressive cancelling – in both terms, not just in one as above.
For comparison, here is the work for this problem using partial fractions:
7 x 2−16 x−2(x 2+2)(x−2)=A x+B x 2+2+C x−2 7 x 2−16 x−2=(A x+B)(x−2)+C(x 2+2)7 x 2−16 x−2=(A+C)x 2+(−2 A+B)x+(−2 B+2 C)
Equating coefficients,
⎧⎩⎨A+C−2 A+B−2 B+2 C=7=−16=−2
This yields
⎧⎩⎨A B C=8=0=−1
So the integral becomes
∫(8 x x 2+2−1 x−2)d x=4 ln∣∣x 2+2∣∣−ln|x−2|=ln∣∣∣(x 2+1)4 x−2∣∣∣+C
The shortcut obtained exactly the results of the partial fractions.
Back to the first problem
I responded, applying this idea to the first problem:
I had considered mentioning a third method I had used for your problem that is essentially the idea you use in this new one, which might be called “partial cancellation”:
∫x 2−x+1 x 4+x 2 d x=∫x 2−x+1 x 2(x 2+1)d x=∫(x 2+1)−x x 2(x 2+1)d x=∫(x 2+1 x 2(x 2+1)−x x 2(x 2+1))d x=∫(1 x 2−x x 2(x 2+1))d x=∫1 x 2 d x−∫x x 2(x 2+1)d x
and so on.
This saves some of the work of partial fractions.
Here I saw that one of the factors of the denominator is found in the numerator, so that splitting the fraction in that way permitted cancellation in each of the resulting fractions.
Let’s finish the work, avoiding the temptation to cancel in the second fraction, and instead taking advantage of an opportunity for substitution (u=x 2), just as I did in what I called “the inspiration” above:
∫1 x 2 d x−∫x x 2(x 2+1)d x=∫x−2 d x−1 2∫d u u(u+1)=−x−1−1 2∫(1 u−1 u+1)d u=−1 x−1 2(ln|u|−ln|u+1|)=−1 x−1 2 ln∣∣∣u u+1∣∣∣=−1 x−1 2 ln∣∣∣x 2 x 2+1∣∣∣=1 2 ln∣∣∣x 2+1 x 2∣∣∣−1 x
The aspect of your original solution that caught my attention was the use of negative exponents, which seems highly unusual; but this aspect, breaking the numerator into two parts that can be canceled separately, is also present, and can definitely be used for many problems.
In your work this time, you’ve taken that idea one step further, using a technique I’ve seen used in factoring (e.g.Factoring Tricks from an Old Textbook).
In general, integration is an art, and setting yourself free to consider non-standard options when you see they might be useful is a valuable skill. Both aspects of your solution are potentially useful.
In the page referred to, I called this trick “wishful thinking” (in problems A and D), adding and subtracting a term (x 2 for Amia) that put one part into a desirable form:
Third problem: alternative numerator for squared linear denominator
Back in March of last year, Amia had a similar question, this time not evading partial fractions, but trying an alternative detail:
Hi Dr math,
I have a question about integration by partial fractions.
The question is attached:
Here he is using partial fractions, but chooses to take a linear numerator for the term with denominator (x−1)2, rather than split it into two terms, one with denominator (x−1), and the other (x−1)2, as we traditionally do. We mentioned this possibility (and showed why it is not preferred) in Partial Fractions: How and Why; the example there has denominator (x−2)3 and we consider using B x 2+C x+D as the numerator, but point out that that is not useful for integration. Here Amia is demonstrating that it can be done, and what happens when you do it, which will add to what we said there.
He uses the alternative method for solving for A, B, and C that I mentioned above. When it comes to integrating this term, he uses the “partial cancelling” method we saw above, splitting it again into two fractions (which, as we’ll see, are the same as we’d get the other way).
Unfortunately, he makes a small error or two, so I’ll redo his work in order to have a correct answer to compare:
3 x 2+2 x(x−1)2=A x+B x+C(x−1)2 3 x 2+2=A(x−1)2+(B x+C)x
Setting x to 1, 0, and 2 as he did, we get
⎧⎩⎨x=1:5=B+C x=0:2=A x=2:14=A+4 B+2 C
Solving for the constants,
⎧⎩⎨A=2 B=1 C=4
so that the new form is
3 x 2+2 x(x−1)2=2 x+x+4(x−1)2
(You can check that this is true.)
Now we integrate, using Amia’s trick:
∫3 x 2+2 x(x−1)2 d x=∫2 x d x+∫x+4(x−1)2 d x=∫2 x d x+∫(x−1)+5(x−1)2 d x=∫2 x d x+∫1 x−1 d x+∫5(x−1)2 d x=2 ln|x|+ln|x−1|−5 x−1=2 ln∣∣∣x x−1∣∣∣−5 x−1+C
We’ll compare this to the standard method below.
Why the result is the same
Doctor Fenton answered, showing that the process is equivalent:
Hi Amia,
Yes, that is valid, but you are not really doing anything differently. Note that
B x+C(x−1)2=B(x−1+1)+C(x−1)2=B(x−1)+B+C(x−1)2=B x−1+B+C(x−1)2=B x−1+D(x−1)2
where D = B + C.
Instead of solving for B and D, you solve for B and the sum B + C, from which you can determine D.
This shows that his work produces the same form as the usual partial fractions method. Let’s actually do the latter and see that we get the same fractions:
3 x 2+2 x(x−1)2=A x+B(x−1)+C(x−1)2 3 x 2+2=A(x−1)2+B x(x−1)+C x 3 x 2+2=A x 2−2 A x+A+B x 2−B x+C x 3 x 2+2=(A+B)x 2+(−2 A−B+C)x 2+A
Equating coefficients, we get the system of equations
⎧⎩⎨⎪⎪A+B=3−2 A−B+C=0 A=2
From which we find that
⎧⎩⎨⎪⎪A=2 B=1 C=5
This gives the fractions
2 x+1(x−1)+5(x−1)2
Integrating, we get
∫2 x d x+∫1 x−1 d x+∫5(x−1)2 d x=2 ln|x|+ln|x−1|−5 x−1=2 ln∣∣∣x x−1∣∣∣−5 x−1+C
just as we did above.
Amia replied:
Thank you Dr,
But I see it is easier when we do the integrations:
A/(x-1), B/(x-1)². ??
I’m right?
Doctor Fenton responded:
Yes. When you integrate
(Bx + C)/(x – 1)2 ,
unless C = -B, in which case it becomes B(x – 1)/(x – 1)2 = B/(x – 1), then you will have to break up the integrand in the same way.
As we said in the older page, “Our motivation for partial fractions is their use in calculus, and that drives the required form of the fractions. What I suggested above would otherwise be perfectly valid, and we would not need all the specific rules that are taught. But then the calculus would be harder. (I will not go into the details, but you might like to try integrating an expression of the form above, to see why.)”
So despite the fact that we can sometimes use special tricks to simplify part of the work, on the whole, the method of partial fractions has been refined over the years to a nearly-optimal method.
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11949 | https://askfilo.com/physics-question-answers/it-is-known-that-the-time-of-revolution-t-of-a-satellite-around-the-earth | It is known that the time of revolution T of a satellite around the earth..
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It is known that the time of revolution T of a satellite aroun
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It is known that the time of revolution T of a satellite around the earth depends on the universal gravitational constant G, the mass of the earth M, and the radius of the circular orbit R. The expression for T using dimensional analysis
A
2 π G M 2 R
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B
2 π G M 2 R 2
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C
2 π GM R 3
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D
2 π R 2 GM
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Question Text It is known that the time of revolution T of a satellite around the earth depends on the universal gravitational constant G, the mass of the earth M, and the radius of the circular orbit R. The expression for T using dimensional analysis
Updated On Jun 16, 2023
Topic Units and Measurements
Subject Physics
Class Class 11
Answer Type Video solution: 8
Upvotes 747
Avg. Video Duration 5 min
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11950 | https://www.khanacademy.org/science/hs-chemistry/x2613d8165d88df5e:reaction-rates-and-equilibrium/x2613d8165d88df5e:equilibrium-lechatelier-s-principle/e/understand-equilibrium-lechateliers-principle | Equilibrium and Le Châtelier’s principle (understand) (practice) | Khan Academy
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High school chemistry
Course: High school chemistry>Unit 9
Lesson 4: Equilibrium & Le Châtelier's principle
Reversible reactions and equilibrium
Le Châtelier's principle
Understand: equilibrium and Le Châtelier’s principle
Apply: Le Châtelier's principle
Science>
High school chemistry>
Reaction rates and equilibrium>
Equilibrium & Le Châtelier's principle
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Understand: equilibrium and Le Châtelier’s principle
NGSS.HS: HS‑PS1‑6, HS‑PS1.B.2
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Problem
A lab group prepares an equilibrium reaction by acidifying a solution of chromate ions (CrO A 4 A 2−) in a beaker. The resulting products are the dichromate ion (Cr A 2 O A 7 A 2−) and water.
2 CrO A 4 A 2−(a q)+2 H A+(a q)⇌Cr A 2 O A 7 A 2−(a q)+H A 2 O(l)
The dichromate ion gives compounds a distinctive orange color
After equilibrium is reached, the teacher asks the members of the lab group to explain what’s happening in the flask right now. Two student answers are shown below.
Student A: The reaction is still going on, but the rates of the forward reaction and the reverse reaction are now equal. In other words, dichromate is still being produced, but at the same rate that it's also being broken back down to chromate.
Student B: At the beginning of the reaction, the chromate and hydrogen ions began producing dichromate and water. As the reaction moved towards equilibrium, the concentration of the reactants decreased and the concentration of the products increased. Now that the reaction is at equilibrium, these concentrations are constant.
Which student has the most accurate answer?
Choose 1 answer:
Choose 1 answer:
(Choice A) Student A A Student A
(Choice B) Student B B Student B
(Choice C) Both students’ answers are correct. C Both students’ answers are correct.
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Video 2 minutes 39 seconds 2:39 Reversible reactions and equilibrium
Video 5 minutes 46 seconds 5:46 Le Châtelier's principle
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11951 | https://www.whitman.edu/mathematics/calculus_online/section03.05.html | 3.5 The Chain Rule
Home » Rules for Finding Derivatives » The Chain Rule
3.5 The Chain Rule
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Introduction
1 Analytic Geometry
1. Lines
2. Distance Between Two Points; Circles
3. Functions
4. Shifts and Dilations
2 Instantaneous Rate of Change: The Derivative
1. The slope of a function
2. An example
3. Limits
4. The Derivative Function
5. Properties of Functions
3 Rules for Finding Derivatives
1. The Power Rule
2. Linearity of the Derivative
3. The Product Rule
4. The Quotient Rule
5. The Chain Rule
4 Transcendental Functions
1. Trigonometric Functions
2. The Derivative of $\sin x$
3. A hard limit
4. The Derivative of $\sin x$, continued
5. Derivatives of the Trigonometric Functions
6. Exponential and Logarithmic functions
7. Derivatives of the exponential and logarithmic functions
8. Implicit Differentiation
9. Inverse Trigonometric Functions
10. Limits revisited
11. Hyperbolic Functions
5 Curve Sketching
1. Maxima and Minima
2. The first derivative test
3. The second derivative test
4. Concavity and inflection points
5. Asymptotes and Other Things to Look For
6 Applications of the Derivative
1. Optimization
2. Related Rates
3. Newton's Method
4. Linear Approximations
5. The Mean Value Theorem
7 Integration
1. Two examples
2. The Fundamental Theorem of Calculus
3. Some Properties of Integrals
8 Techniques of Integration
1. Substitution
2. Powers of sine and cosine
3. Trigonometric Substitutions
4. Integration by Parts
5. Rational Functions
6. Numerical Integration
7. Additional exercises
9 Applications of Integration
1. Area between curves
2. Distance, Velocity, Acceleration
3. Volume
4. Average value of a function
5. Work
6. Center of Mass
7. Kinetic energy; improper integrals
8. Probability
9. Arc Length
10. Surface Area
10 Polar Coordinates, Parametric Equations
1. Polar Coordinates
2. Slopes in polar coordinates
3. Areas in polar coordinates
4. Parametric Equations
5. Calculus with Parametric Equations
11 Sequences and Series
1. Sequences
2. Series
3. The Integral Test
4. Alternating Series
5. Comparison Tests
6. Absolute Convergence
7. The Ratio and Root Tests
8. Power Series
9. Calculus with Power Series
10. Taylor Series
11. Taylor's Theorem
12. Additional exercises
12 Three Dimensions
1. The Coordinate System
2. Vectors
3. The Dot Product
4. The Cross Product
5. Lines and Planes
6. Other Coordinate Systems
13 Vector Functions
1. Space Curves
2. Calculus with vector functions
3. Arc length and curvature
4. Motion along a curve
14 Partial Differentiation
1. Functions of Several Variables
2. Limits and Continuity
3. Partial Differentiation
4. The Chain Rule
5. Directional Derivatives
6. Higher order derivatives
7. Maxima and minima
8. Lagrange Multipliers
15 Multiple Integration
1. Volume and Average Height
2. Double Integrals in Cylindrical Coordinates
3. Moment and Center of Mass
4. Surface Area
5. Triple Integrals
6. Cylindrical and Spherical Coordinates
7. Change of Variables
16 Vector Calculus
1. Vector Fields
2. Line Integrals
3. The Fundamental Theorem of Line Integrals
4. Green's Theorem
5. Divergence and Curl
6. Vector Functions for Surfaces
7. Surface Integrals
8. Stokes's Theorem
9. The Divergence Theorem
17 Differential Equations
1. First Order Differential Equations
2. First Order Homogeneous Linear Equations
3. First Order Linear Equations
4. Approximation
5. Second Order Homogeneous Equations
6. Second Order Linear Equations
7. Second Order Linear Equations, take two
18 Useful formulas
19 Introduction to Sage
1. Basics
2. Differentiation
3. Integration
So far we have seen how to compute the derivative of a function built up from other functions by addition, subtraction, multiplication and division. There is another very important way that we combine simple functions to make more complicated functions: function composition, as discussed in section 2.3. For example, consider $\ds \sqrt{625-x^2}$. This function has many simpler components, like 625 and $\ds x^2$, and then there is that square root symbol, so the square root function $\ds \sqrt{x}=x^{1/2}$ is involved. The obvious question is: can we compute the derivative using the derivatives of the constituents $\ds 625-x^2$ and $\ds \sqrt{x}$? We can indeed. In general, if $f(x)$ and $g(x)$ are functions, we can compute the derivatives of $f(g(x))$ and $g(f(x))$ in terms of $f'(x)$ and $g'(x)$.
Example 3.5.1 Form the two possible compositions of $\ds f(x)=\sqrt{x}$ and $\ds g(x)=625-x^2$ and compute the derivatives. First, $\ds f(g(x))=\sqrt{625-x^2}$, and the derivative is $\ds -x/\sqrt{625-x^2}$ as we have seen. Second, $\ds g(f(x))=625-(\sqrt{x})^2=625-x$ with derivative $-1$. Of course, these calculations do not use anything new, and in particular the derivative of $f(g(x))$ was somewhat tedious to compute from the definition. $\square$
Suppose we want the derivative of $f(g(x))$. Again, let's set up the derivative and play some algebraic tricks: $$\eqalign{ {d\over dx}f(g(x)) &\=\lim_{\Delta x\to0} {f(g(x+\Delta x))-f(g(x))\over\Delta x}\cr &\=\lim_{\Delta x\to0} {f(g(x+\Delta x))-f(g(x))\over g(x+\Delta x)-g(x)} {g(x+\Delta x)-g(x)\over\Delta x}\cr }$$ Now we see immediately that the second fraction turns into $g'(x)$ when we take the limit. The first fraction is more complicated, but it too looks something like a derivative. The denominator, $g(x+\Delta x)-g(x)$, is a change in the value of $g$, so let's abbreviate it as $\Delta g=g(x+\Delta x)-g(x)$, which also means $g(x+\Delta x)=g(x)+\Delta g$. This gives us $$\lim_{\Delta x\to0} {f(g(x)+\Delta g)-f(g(x))\over \Delta g}.$$ As $\Delta x$ goes to 0, it is also true that $\Delta g$ goes to 0, because $g(x+\Delta x)$ goes to $g(x)$. So we can rewrite this limit as $$\lim_{\Delta g\to0} {f(g(x)+\Delta g)-f(g(x))\over \Delta g}.$$ Now this looks exactly like a derivative, namely $f'(g(x))$, that is, the function $f'(x)$ with $x$ replaced by $g(x)$. If this all withstands scrutiny, we then get $${d\over dx}f(g(x))=f'(g(x))g'(x).$$ Unfortunately, there is a small flaw in the argument. Recall that what we mean by $\lim_{\Delta x\to0}$ involves what happens when $\Delta x$ is close to 0 but not equal to 0. The qualification is very important, since we must be able to divide by $\Delta x$. But when $\Delta x$ is close to 0 but not equal to 0, $\Delta g=g(x+\Delta x))-g(x)$ is close to 0 and possibly equal to 0. This means it doesn't really make sense to divide by $\Delta g$. Fortunately, it is possible to recast the argument to avoid this difficulty, but it is a bit tricky; we will not include the details, which can be found in many calculus books. Note that many functions $g$ do have the property that $g(x+\Delta x)-g(x)\not=0$ when $\Delta x$ is small, and for these functions the argument above is fine.
The chain rule has a particularly simple expression if we use the Leibniz notation for the derivative. The quantity $f'(g(x))$ is the derivative of $f$ with $x$ replaced by $g$; this can be written $df/dg$. As usual, $g'(x)=dg/dx$. Then the chain rule becomes $${df\over dx} = {df\over dg}{dg\over dx}.$$ This looks like trivial arithmetic, but it is not: $dg/dx$ is not a fraction, that is, not literal division, but a single symbol that means $g'(x)$. Nevertheless, it turns out that what looks like trivial arithmetic, and is therefore easy to remember, is really true.
It will take a bit of practice to make the use of the chain rule come naturally—it is more complicated than the earlier differentiation rules we have seen.
Example 3.5.2 Compute the derivative of $\ds \sqrt{625-x^2}$. We already know that the answer is $\ds -x/\sqrt{625-x^2}$, computed directly from the limit. In the context of the chain rule, we have $\ds f(x)=\sqrt{x}$, $\ds g(x)=625-x^2$. We know that $\ds f'(x)=(1/2)x^{-1/2}$, so $\ds f'(g(x))= (1/2)(625-x^2)^{-1/2}$. Note that this is a two step computation: first compute $f'(x)$, then replace $x$ by $g(x)$. Since $g'(x)=-2x$ we have $$f'(g(x))g'(x)={1\over 2\sqrt{625-x^2}}(-2x)={-x\over \sqrt{625-x^2}}. $$
$\square$
Example 3.5.3 Compute the derivative of $\ds 1/\sqrt{625-x^2}$. This is a quotient with a constant numerator, so we could use the quotient rule, but it is simpler to use the chain rule. The function is $\ds (625-x^2)^{-1/2}$, the composition of $\ds f(x)=x^{-1/2}$ and $\ds g(x)=625-x^2$. We compute $\ds f'(x)=(-1/2)x^{-3/2}$ using the power rule, and then $$f'(g(x))g'(x)={-1\over 2(625-x^2)^{3/2}}(-2x)={x\over (625-x^2)^{3/2}}. $$
$\square$
In practice, of course, you will need to use more than one of the rules we have developed to compute the derivative of a complicated function.
Example 3.5.4 Compute the derivative of $$f(x)={x^2-1\over x\sqrt{x^2+1}}.$$ The "last'' operation here is division, so to get started we need to use the quotient rule first. This gives $$ \eqalign{ f'(x)&\={(x^2-1)'x\sqrt{x^2+1}-(x^2-1)(x\sqrt{x^2+1})'\over x^2(x^2+1)}\cr &\={2x^2\sqrt{x^2+1}-(x^2-1)(x\sqrt{x^2+1})'\over x^2(x^2+1)}.\cr }$$ Now we need to compute the derivative of $\ds x\sqrt{x^2+1}$. This is a product, so we use the product rule: $${d\over dx}x\sqrt{x^2+1}=x{d\over dx}\sqrt{x^2+1}+\sqrt{x^2+1}.$$ Finally, we use the chain rule: $${d\over dx}\sqrt{x^2+1}={d\over dx}(x^2+1)^{1/2}= {1\over 2}(x^2+1)^{-1/2}(2x)={x\over \sqrt{x^2+1}}.$$ And putting it all together: $$ \eqalign{ f'(x)&\={2x^2\sqrt{x^2+1}-(x^2-1)(x\sqrt{x^2+1})'\over x^2(x^2+1)}\cr &\={2x^2\sqrt{x^2+1}-(x^2-1)\left(x{\ds{x\over \sqrt{x^2+1}}} +\sqrt{x^2+1}\right)\over x^2(x^2+1)}.\cr }$$ This can be simplified of course, but we have done all the calculus, so that only algebra is left. $\square$
Example 3.5.5 Compute the derivative of $\ds \sqrt{1+\sqrt{1+\sqrt{x}}}$. Here we have a more complicated chain of compositions, so we use the chain rule twice. At the outermost "layer'' we have the function $\ds g(x)=1+\sqrt{1+\sqrt{x}}$ plugged into $\ds f(x)=\sqrt{x}$, so applying the chain rule once gives $${d\over dx}\sqrt{1+\sqrt{1+\sqrt{x}}}= {1\over 2}\left(1+\sqrt{1+\sqrt{x}}\right)^{-1/2}{d\over dx} \left(1+\sqrt{1+\sqrt{x}}\right).$$ Now we need the derivative of $\ds \sqrt{1+\sqrt{x}}$. Using the chain rule again: $${d\over dx}\sqrt{1+\sqrt{x}}={1\over 2}\left(1+\sqrt{x}\right)^{-1/2}{1\over 2}x^{-1/2}.$$ So the original derivative is $$ \eqalign{ {d\over dx}\sqrt{1+\sqrt{1+\sqrt{x}}}&\= {1\over 2}\left(1+\sqrt{1+\sqrt{x}}\right)^{-1/2} {1\over 2}\left(1+\sqrt{x}\right)^{-1/2}{1\over 2}x^{-1/2}.\cr &\={1\over 8 \sqrt{x}\sqrt{1+\sqrt{x}}\sqrt{1+\sqrt{1+\sqrt{x}}}} }$$
$\square$
Using the chain rule, the power rule, and the product rule, it is possible to avoid using the quotient rule entirely.
Example 3.5.6 Compute the derivative of $\ds f(x)={x^3\over x^2+1}$. Write $\ds f(x)=x^3(x^2+1)^{-1}$, then $$ \eqalign{ f'(x)&\=x^3{d\over dx}(x^2+1)^{-1}+3x^2(x^2+1)^{-1}\cr &\=x^3(-1)(x^2+1)^{-2}(2x)+3x^2(x^2+1)^{-1}\cr &\=-2x^4(x^2+1)^{-2}+3x^2(x^2+1)^{-1}\cr &\={-2x^4\over (x^2+1)^{2}}+{3x^2\over x^2+1}\cr &\={-2x^4\over (x^2+1)^{2}}+{3x^2(x^2+1)\over (x^2+1)^{2}}\cr &\={-2x^4+3x^4+3x^2\over (x^2+1)^{2}}={x^4+3x^2\over (x^2+1)^{2}}\cr }$$ Note that we already had the derivative on the second line; all the rest is simplification. It is easier to get to this answer by using the quotient rule, so there's a trade off: more work for fewer memorized formulas. $\square$
Exercises 3.5
Find the derivatives of the functions. For extra practice, and to check your answers, do some of these in more than one way if possible.
Ex 3.5.1 $\ds x^4-3x^3+(1/2)x^2+7x-\pi$ (answer)
Ex 3.5.2 $\ds x^3-2x^2+4\sqrt{x}$ (answer)
Ex 3.5.3 $\ds (x^2+1)^3$ (answer)
Ex 3.5.4 $\ds x\sqrt{169-x^2}$ (answer)
Ex 3.5.5 $\ds (x^2-4x+5)\sqrt{25-x^2}$ (answer)
Ex 3.5.6 $\ds \sqrt{r^2-x^2}$, $r$ is a constant (answer)
Ex 3.5.7 $\ds \sqrt{1+x^4}$ (answer)
Ex 3.5.8 $\ds \ds{1\over\sqrt{5-\sqrt{x}}}$. (answer)
Ex 3.5.9 $\ds (1+3x)^2$ (answer)
Ex 3.5.10 $\ds{x^2+x+1\over 1-x}$ (answer)
Ex 3.5.11 $\ds{\sqrt{25-x^2}\over x}$ (answer)
Ex 3.5.12 $\ds\sqrt{{169\over x}-x}$ (answer)
Ex 3.5.13 $\ds \sqrt{x^3-x^2-(1/x)}$ (answer)
Ex 3.5.14 $\ds 100/(100-x^2)^{3/2}$ (answer)
Ex 3.5.15 $\ds {\root 3 \of{x+x^3}}$ (answer)
Ex 3.5.16 $\ds \sqrt{(x^2+1)^2+\sqrt{1+(x^2+1)^2}}$ (answer)
Ex 3.5.17 $\ds (x+8)^5$ (answer)
Ex 3.5.18 $\ds (4-x)^3$ (answer)
Ex 3.5.19 $\ds (x^2+5)^3$ (answer)
Ex 3.5.20 $\ds (6-2x^2)^3$ (answer)
Ex 3.5.21 $\ds (1-4x^3)^{-2}$ (answer)
Ex 3.5.22 $\ds 5(x+1-1/x)$ (answer)
Ex 3.5.23 $\ds 4(2x^2-x+3)^{-2}$ (answer)
Ex 3.5.24 $\ds {1\over 1+1/x}$ (answer)
Ex 3.5.25 $\ds {-3\over 4x^2-2x+1}$ (answer)
Ex 3.5.26 $\ds (x^2+1)(5-2x)/2$ (answer)
Ex 3.5.27 $\ds (3x^2+1)(2x-4)^3$ (answer)
Ex 3.5.28 $\ds{x+1\over x-1}$ (answer)
Ex 3.5.29 $\ds{x^2-1\over x^2+1}$ (answer)
Ex 3.5.30 $\ds{(x-1)(x-2)\over x-3}$ (answer)
Ex 3.5.31 $\ds{2x^{-1}-x^{-2}\over 3x^{-1}-4x^{-2}}$ (answer)
Ex 3.5.32 $\ds 3(x^2+1)(2x^2-1)(2x+3)$ (answer)
Ex 3.5.33 $\ds{1\over (2x+1)(x-3)}$ (answer)
Ex 3.5.34 $\ds ((2x+1)^{-1}+3)^{-1}$ (answer)
Ex 3.5.35 $\ds (2x+1)^3(x^2+1)^2$ (answer)
Ex 3.5.36 Find an equation for the tangent line to $\ds f(x) = (x-2)^{1/3}/(x^3 + 4x - 1)^2$ at $x=1$. (answer)
Ex 3.5.37 Find an equation for the tangent line to $\ds y=9x^{-2}$ at $(3,1)$. (answer)
Ex 3.5.38 Find an equation for the tangent line to $\ds (x^2-4x+5)\sqrt{25-x^2}$ at $(3,8)$. (answer)
Ex 3.5.39 Find an equation for the tangent line to $\ds \ds{(x^2+x+1)\over(1-x)}$ at $(2,-7)$. (answer)
Ex 3.5.40 Find an equation for the tangent line to $\ds \sqrt{(x^2+1)^2+\sqrt{1+(x^2+1)^2}}$ at $\ds (1,\sqrt{4+\sqrt{5}})$. (answer) |
11952 | https://www.youtube.com/playlist?list=PLWWf-r6pJvvVZXMkCr5cOi4LvonUC6NvL | Chapter 2 | Force Vectors | Engineering Mechanics: Statics | RC Hibbeler 12th Edition - YouTube
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Chapter 2 | Force Vectors | Engineering Mechanics: Statics | RC Hibbeler 12th Edition
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Playlist•9 videos•9,415 views
Welcome to the complete playlist dedicated to Chapter 2, "Force Vectors," from "Engineering Mechanics: Statics" by R.C. Hibbeler, 12th Edition. This playlist provides a thorough exploration of force vectors, a fundamental concept in statics. Whether you're an engineering student, a practicing engineer looking for a review, or just curious about mechanics, this series will guide you through the essentials....more Welcome to the complete playlist dedicated to Chapter 2, "Force Vectors," from "Engineering Mechanics: Statics" by R.C. Hibbeler, 12th Edition. This playlist provides a thorough exploration of force vectors, a fundamental concept in statics. Whether you're an engineering student, a practicing engineer looking for a review, or just curious about mechanics, this series will guide you through the essentials....more...more
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Welcome to the complete playlist dedicated to Chapter 2, "Force Vectors," from "Engineering Mechanics: Statics" by R.C. Hibbeler, 12th Edition. This playlist provides a thorough exploration of force vectors, a fundamental concept in statics. Whether you're an engineering student, a practicing engineer looking for a review, or just curious about mechanics, this series will guide you through the essentials. Lecture PDF Link: What You'll Learn in This Playlist: Introduction to Force Vectors: Understand what forces are and how they are represented as vectors. Scalars and Vectors: Learn the difference between scalar and vector quantities and their importance in mechanics. Vector Operations: Explore vector addition, subtraction, multiplication by a scalar, and unit vectors. Force Components: Learn to resolve forces into components in two and three dimensions. Cartesian Vectors: Understand how to represent vectors using Cartesian components. Position Vectors: Discover how to determine position vectors between two points in space. Force Vector Direction: Learn to find the direction of force vectors using direction cosines and angles. Dot Product: Explore the dot product of vectors and its application in mechanics. Why This Playlist? Comprehensive Coverage: This playlist provides a complete explanation of force vectors according to Chapter 2 of the textbook. Step-by-Step Learning: The content is structured to build your understanding progressively. Clear Explanations: Concepts are explained clearly with examples and illustrations. Practical Examples: Emphasis on developing your ability to work with force vectors in practical scenarios. Expert Knowledge: We provide expert knowledge to help you excel. Structured Approach: We provide a structured learning approach. Who Is This Playlist For? Engineering students taking statics courses. Practicing engineers needing a review of force vector principles. Self-learners and anyone interested in understanding the fundamentals of statics. Subscribe and Stay Updated! Join us on this educational journey through the core concepts of force vectors. Don't forget to subscribe to stay updated with our latest videos and to gain access to a wealth of force vector knowledge. Like, share, and comment below to let us know what you think! Let's embark on this exciting educational journey together! Thank you for watching! #forcevectors#engineeringmechanics#statics#rchibbeler#hibbelerstatics#engineering#mechanics#vectors#scalers#vectoraddition#vectoroperations#forcecomponents#cartesianvectors#positionvectors#forcevectordirection#dotproduct#engineeringeducation#engineeringstudents#mechanicalengineering#civilengineering#aerospaceengineering#structuralengineering#engineeringfundamentals#engineeringconcepts#engineeringprinciples#engineeringproblems#problemsolving#technicalanalysis#engineeringlectures#engineeringtutorials#engineeringcourses#physicsinengineering#stem#stemeducation#onlinelearning#engineeringstudy#studyengineering#engineeringknowledge#mechanics101#staticequilibrium#forceanalysis#vectoranalysis#engineeringbasics#engineeringlife#learnengineering#engineeringmath#engineeringtips#engineeringcommunity#engineeringdesign#mechanicsconcepts
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Chapter 2 | Force Vectors | Engineering Mechanics: Statics | RC Hibbeler 12th Edition
by Mind Matrix Engineering
Playlist•9 videos•9,415 views
Welcome to the complete playlist dedicated to Chapter 2, "Force Vectors," from "Engineering Mechanics: Statics" by R.C. Hibbeler, 12th Edition. This playlist provides a thorough exploration of force vectors, a fundamental concept in statics. Whether you're an engineering student, a practicing engineer looking for a review, or just curious about mechanics, this series will guide you through the essentials....more Welcome to the complete playlist dedicated to Chapter 2, "Force Vectors," from "Engineering Mechanics: Statics" by R.C. Hibbeler, 12th Edition. This playlist provides a thorough exploration of force vectors, a fundamental concept in statics. Whether you're an engineering student, a practicing engineer looking for a review, or just curious about mechanics, this series will guide you through the essentials....more...more
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11953 | https://www.scribd.com/document/454282327/Hierarchy-of-sentence | Published Time: Fri, 15 Mar 2024 20:37:30 GMT
Hierarchy of Sentence | PDF | Clause | Morphology (Linguistics)
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217 views 17 pages
Hierarchy of Sentence
The document discusses the hierarchy of sentences and includes the following key points: 1. It defines hierarchy as a system where people or groups are ranked with some superior to others b…
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Hierarchy of Sentence
The document discusses the hierarchy of sentences and includes the following key points: 1. It defines hierarchy as a system where people or groups are ranked with some superior to others based on traits like status or authority. 2. It provides 4 short example sentences to illustrate hierarchy, such as "A hierarchy of leaders is the essence of its organization." 3. The rest of the document discusses linguistic concepts like morphemes, words, phrases and provides definitions and examples of each.
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Hierarchy of Sentence
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@ATLE@AC GECCIKI EJ OYWL@IWW ADML@LWT_ATLE@ & IGE@EMLGW CABE_I
Blirargby ej Witigi
Wuomlttid Oaqas Anoar Wuomlttid te Dr. ArjaCedbl Wo M.VBLC I@KCLWB CLTI_ATY_I (7rd Wimistir)WIWWLE@8 ?61<-?6?1 Blirargby ej Witi`gi
Dijlltle ej Blirargby
A oedy ej autberltatlvi ejjlglacs erkalzid l istid rans. | A seglac, riclkleus, igeemlg er pecltlgac systim er erkalzatlel wblgb piepci er kreups ej piepci ari ranid wltb semi supirler te etbirs oasid e tbilr status, autberlty er semi etbir tralt. | Ay kreup ej eohigts ranid se tbat iviryei out tbi tepmest ls suoerdlati te a spigljlid e`i a oevi lt.
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Wbert Ixampci Witigi jer Blirargby
Mest ej tbim jermid a clttci
blirargby
l` tbimsicvis.
?. A
blirargby
ej ciadirs ls tbi issigi ej lts erkalzatle`.
Wugb a
blirargby
ls siil savaki clji.
Tbi oasi ej tbi
blirargby
ls jermid oy tbi si`si-eohigts.
1.Merpbimis lIkclsb
L[Ikclsb krammar
a`d merpbeceky
, a merpbimi ls a mialkjuc clkulstlg ult geslstlk ej a werd sugb as
dek,
er a werd icimit, sugb as tbi -s at tbi id ej
deks,
tbat ga't oi dlvldid lte smaccir mialkjuc parts.Merpbimis ari tbi smaccist ults ej mialkl a cakuaki. Tbiy ari gemmecy gcassljlid as iltbir jrii merpbimis, wblgb gaeggur as siparati werds er oeud merpbimis, wblgb ga't stad acei as werds.May werds lIkclsb ari madi up ej a slkci jrii merpbimi. Jer ixampci, iagb werd l tbi jeccewlk sitigi ls a dlstlgt merpbimi8 "L iid te keew, out yeu gastay." Vut aetbir way, ei ej tbi li werds ltbat sitigi ga oi dlvldid lte smaccir parts tbat ari acse mial`kjuc.
Itymeceky
Jrem tbi Jrigb, oy aaceky wltb pbe`imi, jrem tbi Kriin, "sbapi, jerm."
Jrii Merpbimis ad Oeud Merpbimis
Merpbimis tbat gastad acei te jugtleas werds ari gaccid jrii merpbimis. Tbiy gemprlsi slmpci werds (l.i. werds madi up ej ei jrii merpbimi) ad gempeud w erds (l.i. werds madi up ej twe jrii merpbimis).Ixampcis8 Wlmpci werds8
tbi, ru, e, wicc
Gempeu`d werds8
niyoeard, kriibeusi, oceedsbid, smartpbei
Merpbimis tbat gaecy oi attagbid te aetbir part ej a werd (ga``et stad acei) ari gaccid oeud merpbimis.Ixampcis8
pri-, dls-, l-, u-, -juc, -aoci, -mi`t, -cy, -lsi
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pritist, dlsgetit, l`teciraoci, rigilvi
Gempcix werds ari werds tbat ari madi up ej oetb jrii merpbimi(s) ad oeud merpbimi(s),er twe er meri oeu`d merpbimis.
_ecc yeur meusi evir tbi werds oicew te sii bew may merpbimis ari tbiri ad wbitbir tbiy ari jrii merpbimis er oeud merpbimis.a k a ls t l m p i r a t l v i r i a c l z i s u o m l t a s s l km it J a g i o e e n ug e m m em l s lt i r p r i t dl sq ua cl jl id ige u`ti ri d ki ek ra pb y lr ri sl st lo ci
_eet a`d Ajjlxis
Ajjlxatlels tbi mest gemme werd jermatlepregiss l Ikclsb. erds ari jermid oy addlk ajjlxis te reets.
_eets
gaoi jrii er oeud merpbimis. Tbiy ga`et oi jurtbir aacyzid lte smaccir parts.Tbiy jerm tbi oasi jerms ej tbi werds.1.Jr ii r ee ts ar i jr ii m er pb im is. Tb iy g a s tad a cei t e jug tl e a s w er ds.Ixampcis8
rigeccigt, olclkuac, uiasy, mlsciad, bardcy, attragtlvi
?.Oe ud re et s ar i oeud me r pb im is. Tbi y ga`et s tad a c ei te jug t l eas w er ds oigausi tbiy arie cekir usid l MedirIkclsb.Ixampcis8
rigilvi, ridugi Ajjlxis
ari oeud merpbimis. Tbiy ga oi gcassljlid lte prijlxis ad sujjlxis lIk clsb.1.A p ri jl x l s aa jj lx ad di d t e t bi oik l``lk ej et bir me rp bi mi s t e j er m a we rd.Ixampcis8
dlsclni, diagtlvati, ladiquati, lmmeolci, mlsciadlk, uaggeutaoci iduraoci, udiragblivi, evirdivicepid, pririqulslti, pestkraduati, rigygci
?.A s uj jl x l s aa jj lx ad di d t e t bi id e j e tb ir me rp bi mi s t e j er m a w er d.Ixampcis8
admlraoci, jrultjuc, amoltleus, iheymit, iakiriss, stadardlzi, gewardcy,yeukir, pregisslk, MgDeacd's, asslkmi`ts, digldis, digldid Ajjlxis
A"ajjlx" ls a oeud merpbimi tbat eggurs
oijeri
er ajtir a oasi. A` ajjlx tbat gemis oij eri a oas i l s g acc id a "pri jlx." We mi ixa mpc is ej pri jlx is ari
a`ti-
,
pri-
,
u`-
, a`d
dls-
, as ltbi jeccewlk werds8
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a`ti
dati
pri
blsterlg
u`
biactby
dls
rikard A` ajjlx tbat gemis ajtir a oasi ls gaccid a "sujjlx." Wemi ixampcis ej sujjlxis ari
-cy
,
-ir
,
-lsm
, a`d
-`iss
, as ltbi jeccewlk werds8 bapp
lcy
kardi`
ir
gapltac
lsm
nl`d
iss Dirlvatleac Ajjlxis
Aajjlx ga oi iltbir dirlvatleac er ljcigtleac. "Dirlvatleac ajjlxis" sirvi te actir tbi mialk ej a werd oy oulcdlk e a oasi. Ltbi ixampcis ej werds wltb prijlxis ad sujjlxi s aoevi, tbi addltle` ej tbi prijlx
u`-
te
biactby
actirs tbi mialk ej
biactby
. Tbi risuctlk werd mias "et biactby." Tbi addltle ej tbi sujjlx
-ir
te
kardi`
gbakis tbi mial`k ej
kardi`
, wblgb ls a pcagi wbiri pcats, jcewirs, itg., krew, te a werd tbat rijirs te 'a pirse wbe tids a kardi.' Lt sbeucd oi `etid tbat
acc
prijlxis lIkclsb ari dirlvatleac. Bewivir, sujjlxis may oi iltbir dirlvatleac er ljcigtleac.
Ljcigtleac Ajjlxis
Tbiri ari a carki umoir ej dirlvatleac ajjlxis lIkclsb. Lgetrast, tbiri ari ecy ilkbt "ljcigtl eac ajjlxis" l Ikclsb, ad tbisi ari acc
sujjlxis
. Ikclsb bas tbi jeccewlk ljcigtleac sujjlxis, wblgb sirvi a varlity ej krammatlgac jugtles wbiaddid te spigljlg typis ej werds. Tbisi krammatlgac jugtles ari sbew te tbi rlkbt ej iagb sujjlx.-s eu pcurac-'s eu pessisslvi-s viro prisit tisi tblrd pirseslkucar-lk viro prisit partlglpci/kirud-id viro slmpci past tisi-i` viro past pirjigt partlglpci-ir adhigtlvi gemparatlvi-ist adhigtlvi supircatlvi
?.[erds
A
werd
ls a spiigb seud er a gemolatleej seuds, er lts riprisitatle l[wrltlk, tbat symoeclzis ad gemmulgatis a mialkad may geslst ej a slkci merpbimier a gemolatleej merpbimis.Tbi oragb ej cl`kulstlgstbat studlis werd strugturis ls gaccid
merpbeceky
. Tbi oragb ej clkulstlgs tbat studlis werd mialks ls gaccid
cixlgacsima`tlgs
.
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11954 | https://mathworld.wolfram.com/ConicalFrustum.html | Conical Frustum -- from Wolfram MathWorld
TOPICS
AlgebraApplied MathematicsCalculus and AnalysisDiscrete MathematicsFoundations of MathematicsGeometryHistory and TerminologyNumber TheoryProbability and StatisticsRecreational MathematicsTopologyAlphabetical IndexNew in MathWorld
Geometry
Solid Geometry
Frusta
Geometry
Solid Geometry
Cones
Conical Frustum
Download Wolfram Notebook
A conical frustum is a frustum created by slicing the top off a cone (with the cut made parallel to the base). For a right circular cone, let be the slant height and and the base and top radii. Then
(1)
The surface area, not including the top and bottom circles, is
(2)
(3)
The volume of the frustum is given by
(4)
But
(5)
so
(6)
(7)
(8)
This formula can be generalized to any pyramid by letting be the base areas of the top and bottom of the frustum. Then the volume can be written as
(9)
The area-weighted integral of over the frustum is
(10)
(11)
so the geometric centroid is located along the z-axis at a height
(12)
(13)
(Eshbach 1975, p.453; Beyer 1987, p.133; Harris and Stocker 1998, p.105). The special case of the cone is given by taking , yielding .
See also
Cone, Frustum, Pyramidal Frustum, Spherical Segment
Explore with Wolfram|Alpha
More things to try:
conical frustum
(110110 base 2) / (11 base 2)
complexity class BPP
References
Beyer, W.H. (Ed.). CRC Standard Mathematical Tables, 28th ed. Boca Raton, FL: CRC Press, pp.129-130 and 133, 1987.Eshbach, O.W. Handbook of Engineering Fundamentals. New York: Wiley, 1975.Harris, J.W. and Stocker, H. "Frustum of a Right Circular Cone." §4.7.2 in Handbook of Mathematics and Computational Science. New York: Springer-Verlag, p.105, 1998.Kern, W.F. and Bland, J.R. "Frustum of Right Circular Cone." §29 in Solid Mensuration with Proofs, 2nd ed. New York: Wiley, pp.71-75, 1948.
Referenced on Wolfram|Alpha
Conical Frustum
Cite this as:
Weisstein, Eric W. "Conical Frustum." From MathWorld--A Wolfram Resource.
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Created, developed and nurtured by Eric Weisstein at Wolfram Research |
11955 | https://www.quora.com/The-straight-line-ax+by+c-0-and-bx+cy+a-0-are-parallel-Which-of-the-following-is-true-A-a-b-c-B-b2-ac-c-C-b2+ac-0-D-b2%E2%80%934ac-0 | The straight line ax+by+c=0 and bx+cy+a=0 are parallel. Which of the following is true: A. a=b=c; B. b2-ac=c; C. b2+ac=0; D. b2–4ac=0? - Quora
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5
The straight line ax+by+c=0 and bx+cy+a=0 are parallel. Which of the following is true: A. a=b=c; B. b2-ac=c; C. b2+ac=0; D. b2–4ac=0?
All related (33)
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Recommended
Sunmoon Saurabh
Studied at DAV Public School,tapin North
·4y
If the eqn is parallel then,
a1/a2= b1/b2 ≠c1/c2
Therefore,
a/b=b/c ≠c/a
Case I
b2=ac
Case II
c2≠ba
Case III
a2≠bc
So the answer is none of these
Upvote ·
9 1
Ramnath Takiar
Former Scientists G, NCRP at Indian Council of Medical Research (1978–2013) · Author has 1.9K answers and 2.7M answer views
·7y
Express the two lines in standard format that is y = mx + c
Given ax + by + c = 0 => by = -ax - c => y = (-a/b)x - c/b …………. (1)
Given bx + cy + a = 0 => cy = -bx - a => y = (-b/c)x - a/c …………… (2)
Given both lines are parallel implies their slopes are equal that is
(-a/b) = (-b/c) => a/b = b/c => b^2 = ac
So, none of the option given are true.
Upvote ·
9 2
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Related questions
More answers below
If the line ax+by+c=0 is a normal to the curve xy=1, then what is the relation between a, b, and c?
If a²+b²+c²-ab-AC-bc =0 then what is the ratio of a:b:c?
What is the slop of a line Ax + By +C =0?
What is the slope for y=2x?
If a+b+c=0 a+b+c=0, what's the value of (a b−c+b c−a+c a−b)(b−c a+c−a b+a−b c)(a b−c+b c−a+c a−b)(b−c a+c−a b+a−b c)?
Gordon M. Brown
Math Tutor at San Diego City College (2018-Present) · Author has 6.2K answers and 4.3M answer views
·3y
Related
The position vectors of points A, B and C are (0 14), (5 2) and (8 k) respectively. What is the value of k if A, b and c are in a straight line?
To be honest, I nearly downvoted this question because it’s just a disaster! The only way I can fix it is to simply rewrite the whole thing:
“The coordinates of points A, B, and C are (0, 14), (5, 2), and (8, k) respectively. What is the value of k if A, B, and C lie in a straight line (that is, are collinear)?”
Three things to consider: (1) Don’t be so sloppy when naming mathematical objects; (2) Don’t omit the comma between coordinates of a point; (3) The term “position vector” to describe the location of a point is obscure at best (I’ve never heard of it, in fact). Use “coordinates” instead,
Continue Reading
To be honest, I nearly downvoted this question because it’s just a disaster! The only way I can fix it is to simply rewrite the whole thing:
“The coordinates of points A, B, and C are (0, 14), (5, 2), and (8, k) respectively. What is the value of k if A, B, and C lie in a straight line (that is, are collinear)?”
Three things to consider: (1) Don’t be so sloppy when naming mathematical objects; (2) Don’t omit the comma between coordinates of a point; (3) The term “position vector” to describe the location of a point is obscure at best (I’ve never heard of it, in fact). Use “coordinates” instead, which is not nearly so ambiguous.
To the point of your problem:
(k - 2) / (8 - 5) = (2 - 14) / (5 - 0)
(k - 2) / 3 = -12 / 5
5k - 10 = -12 3 = -36 [Cross-multiply]
5k = -26
k = -26/5
Upvote ·
9 4
Haresh Sagar
Studied Science&Mathematics (Graduated 1988) · Author has 6.2K answers and 7M answer views
·3y
Related
The position vectors of points A, B and C are (0 14), (5 2) and (8 k) respectively. What is the value of k if A, b and c are in a straight line?
→O A=<0,14>,→O B=<5,2>,→O C=<8,k>O A→=<0,14>,O B→=<5,2>,O C→=<8,k>
→B A=<−5,12>=B A→=<−5,12>=
→B C=<3,k−2>B C→=<3,k−2>
Multiply →B A B A→ by −3−3 and →B C B C→ by 5 5.
s m→B A=<15,−36>s m B A→=<15,−36>
s m→B C=<15,5 k−10>s m B C→=<15,5 k−10>
Vectors in a straight line are coincident vectors. Parallel and coincident vectors are scalar multiples of each other so when we make x x components equal, y y components must be equal and therefore,
5 k−10=−36⟹5 k−10=−36⟹
5 k=−26⟹5 k=−26⟹
k=−26 5 k=−26 5
Continue Reading
→O A=<0,14>,→O B=<5,2>,→O C=<8,k>O A→=<0,14>,O B→=<5,2>,O C→=<8,k>
→B A=<−5,12>=B A→=<−5,12>=
→B C=<3,k−2>B C→=<3,k−2>
Multiply →B A B A→ by −3−3 and →B C B C→ by 5 5.
s m→B A=<15,−36>s m B A→=<15,−36>
s m→B C=<15,5 k−10>s m B C→=<15,5 k−10>
Vectors in a straight line are coincident vectors. Parallel and coincident vectors are scalar multiples of each other so when we make x x components equal, y y components must be equal and therefore,
5 k−10=−36⟹5 k−10=−36⟹
5 k=−26⟹5 k=−26⟹
k=−26 5 k=−26 5
Upvote ·
9 4
Assistant
Bot
·
1y
For the two lines given by the equations a x+b y+c=0 a x+b y+c=0 and b x+c y+a=0 b x+c y+a=0 to be parallel, their slopes must be equal.
Find the slope of each line:
The slope of the first line a x+b y+c=0 a x+b y+c=0 is given by:
m 1=−a b m 1=−a b
The slope of the second line b x+c y+a=0 b x+c y+a=0 is given by:
m 2=−b c m 2=−b c
2. Set the slopes equal for parallel lines:
−a b=−b c−a b=−b c
This leads to:
a b=b c a b=b c
3. Cross-multiply to find a relationship:
a c=b 2 a c=b 2
This can be rearranged to show:
b 2−a c=0 b 2−a c=0
From this analysis, we can conclude that the correct answer is:
D.
Continue Reading
For the two lines given by the equations a x+b y+c=0 a x+b y+c=0 and b x+c y+a=0 b x+c y+a=0 to be parallel, their slopes must be equal.
Find the slope of each line:
The slope of the first line a x+b y+c=0 a x+b y+c=0 is given by:
m 1=−a b m 1=−a b
The slope of the second line b x+c y+a=0 b x+c y+a=0 is given by:
m 2=−b c m 2=−b c
2. Set the slopes equal for parallel lines:
−a b=−b c−a b=−b c
This leads to:
a b=b c a b=b c
3. Cross-multiply to find a relationship:
a c=b 2 a c=b 2
This can be rearranged to show:
b 2−a c=0 b 2−a c=0
From this analysis, we can conclude that the correct answer is:
D. b 2−4 a c=0 b 2−4 a c=0 is not directly correct. The correct relationship derived from the parallel condition is b 2−a c=0 b 2−a c=0, which is not listed among the options you provided.
Thus, if we had to choose the closest option based on the derived relationship, none fit directly, but the correct condition for parallel lines is b 2=a c b 2=a c.
Upvote ·
Related questions
More answers below
How do I prove that (a+b+c)(1/a+1/b+1/c)>9(a+b+c)(1/a+1/b+1/c)>9 for a,b,c>0 a,b,c>0?
The line ax-by=-9 passes through A (-1;3) and B (-3:0). What is a+b?
What is the value of a, b, c, and d from the equations: a-b 2a, c 2a -b 2c, d 0 0 01 0?
How do you find a general proof for the equation of the line joining the vertices of y=ax2+bx+c (when a and c remain fixed and b varies)?
What is the output of int d=(++a) &&(--b) &&(++c) ||(--a)? The values of a=0,b=1,c=0.
Philip Lloyd
Specialist Calculus Teacher, Motivator and Baroque Trumpet Soloist. · Author has 6.8K answers and 52.8M answer views
·5y
Related
▶Sketch ax + by = c for the following cases.
(a) a > 0, b > 0 and c > 0
(b) a > 0, b > 0 and c < 0
(c) a > 0, b < 0 and c > 0 ?
It is much easier to visualise line graphs when they are in the gradi...
Upvote ·
9 3
9 1
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Arpit Sardana
Loves Numbers
·9y
Related
When does the equation A x 2+C y 2+D x+E y+F=0 A x 2+C y 2+D x+E y+F=0 represent two parallel lines?
Since the equation A x 2+C y 2+D x+E y+F=0,A x 2+C y 2+D x+E y+F=0,represent 2 parallel lines,
Let , y=mx+b...(1)
and y=mx+c...(2)
be the 2 parallel lines with slope m forming the equation A x 2+B y 2+D x+E y+C=0 A x 2+B y 2+D x+E y+C=0
Multiplying (1) and (2)...
We get,
(y−m x−b)(y−m x−c)=0(y−m x−b)(y−m x−c)=0 ,i.e.
y 2+m 2 x 2−x y(2 m)−y(b+c)+m x(b+c)+b c=0 y 2+m 2 x 2−x y(2 m)−y(b+c)+m x(b+c)+b c=0...(3)
Comparing (3) with the original equation...
A=m 2,m 2,
C=1,
D=m(b+c)
E=-(b+c)
F=bc
and, 2m=0
S0, m=0.
A=0,
C=1,
D=0.,
E=-(b+c)
F=bc
And, the equation A x 2+C y 2+D x+E y+F A x 2+C y 2+D x+E y+F,
Represents 2 lines Parallel to the x-axis.
Upvote ·
9 5
9 1
Roger Larson
Author has 5K answers and 4.7M answer views
·Apr 13
Related
The line ax + by + c = 0 is parallel to 2x _ y + 7 = 0. What is the gradient of the line ax+by + c?
The gradient of line ax +by + c= 0 is 2
Detail
For line
ax + by + c = 0
to be parallel to
2x - y + 7= 0
a must be 2 and b must be -1
2x -y +c =0
y = 2x +c
the gradient (slope) is 2
Upvote ·
9 2
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James Jones
M.S. from The University of Oklahoma (Graduated 1979) · Author has 4.1K answers and 3.1M answer views
·4y
Related
For which values of A, B, and C will Ax + By = C be a vertical line through the point (7, 4)?
Suppose you have a vertical line in the usual coordinate system, with the X-axis running horizontally and the Y-axis running vertically. What can you say about the x-coordinate or any point on that vertical line? Once you can answer that, I think you’re well on your way to determining A, B, and C.
Upvote ·
9 3
9 1
Dave Benson
trying to make maths easy. · Author has 6.1K answers and 2.1M answer views
·2y
Related
The equation of the line that goes through the points (9,6) and (10, −1) can be written in general form Ax + By + C = 0 where A=? B=? C=?
Use y = mx+c 1️⃣ & m = = (6- -1/(9–10) = -7 is slope & c is y-intercept. Plug (10,-1) in 1️⃣
-1 = 70+c ⟹ c = 69 ⟹ y = -7x+69 ⟹ 7x+y-69 = 0 ⟹ A = 7, B = 1, C = -69 Answer
Continue Reading
Use y = mx+c 1️⃣ & m = = (6- -1/(9–10) = -7 is slope & c is y-intercept. Plug (10,-1) in 1️⃣
-1 = 70+c ⟹ c = 69 ⟹ y = -7x+69 ⟹ 7x+y-69 = 0 ⟹ A = 7, B = 1, C = -69 Answer
Upvote ·
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Comparing Three Methods for Quantifying Overground Running Acceleration
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Comparing Three Methods for Quantifying Overground Running Acceleration
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November 30, 2024
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Simar Gill 1,Emily Sara Matijevich 1, Nesa Keshavarz Moghadam 1, Sandro Nigg 1, Benno Nigg 1
1 Faculty of Kinesiology, University of Calgary, AB, Canada
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Abstract
Forward acceleration is an important factor to control in biomechanics studies investigating walking/running, as human kinetics and kinematics vary with changes in acceleration. During overground running studies, acceleration is typically estimated using the net anterior-posterior ground reaction force impulse, as measured with force plates. However, as studies move outside the lab, there is limited validation of alternative methods of quantifying acceleration when force plates are unavailable. The purpose of this study was to assess the validity of alternative methods for estimating acceleration in overground running. We evaluated the use of timing lights and motion capture, as indicators of acceleration, to measure change in velocity. We hypothesized that the change in velocity calculated from timing lights and motion capture markers near the center of mass would have a strong positive correlation with the relative anterior-posterior ground reaction force impulse measured with force plates. Ten participants ran in an indoor lab while measurements were collected using timing lights, motion capture and a force plate. The correlation between the relative anterior-posterior impulse and velocity changes measured by timing lights was weak (r = -0.01, r = 0.27 and r = 0.15, respectively). In contrast, the correlation between the relative anterior-posterior impulse and velocity change determined by motion capture was strong (r = 0.81). In studies where force plates are not available, measuring changes in velocity with motion capture is a promising method for calculating and controlling acceleration. However, measuring changes in velocity with timing lights does not show as much promise due to weak correlation values and should therefore be avoided.
Introduction
Biomechanical studies are typically conducted in a laboratory setting, using high-cost, non-portable measurement equipment to evaluate human movement. However, the influx of low-cost wearable and portable measurement equipment in recent years has enabled the conducting of biomechanical studies to be outside the lab, allowing for human move-ment to be evaluated in realistic and unconstrained settings (Ruder, et al. 2019). The transition of biomechanical studies outside the lab requires certain methodologies to be reevaluated in response to these new settings. Similarly, when certain equipment and methodologies are unavailable in lab based studies, replacement methods must be developed and validated.
One important variable to control in biomechanical studies that investigate walking and running is the participant’s forward acceleration. An individual “accelerates” when their forward velocity increases over time, “decelerates” when their forward velocity decreases overtime and is at “steady-state” when there is no change in forward velocity over time. Researchers are often interested in controlling the forward acceleration because kinematics and kinetics change when the forward acceleration changes. For example, reductions in ankle work output were seen in a deceleration state compared to an acceleration or steady-state (Williams et al. 2017). Additionally, another example, there was greater knee extension and plantar flexion velocity in an acceleration state compared to steady-state running (Van Caeken-berghe et al. 2013). To control for the state of motion, vali-dated methods to quantify individuals’ forward acceleration are necessary. Furthermore, when small deviations from constant speed occurred, such as increases in speed, there were changes in step length and time (Rodman and Martin 2020). This variability in steps indicates that it is not possible to maintain the same energetic output during movement, as changes in step length change knee extensor moments, ankle plantar flexor moments and hip flexor moments (Gill et al. 2023).
In a laboratory setting, walking/running biomechanics studies are often conducted on a treadmill or overground. On a treadmill, experimental conditions such as running velocity and acceleration can easily be manipulated by the experimenter, offering greater control over forward velocity and acceleration. However, treadmills, especially expensive, force-instrumented treadmills, may not be readily available in all labs. Furthermore, the speed and acceleration of the treadmill belt is subject to fluctuations (Willwacher et al. 2021) which could result in deviations from the desired velocity/acceleration state.
For some laboratory studies, evaluating overground running may be preferred or even necessary. Overground running may more closely mimic “real-life” scenarios, as individuals spend more time walking/running overground compared to on a treadmill (Yang and King. 2016). Furthermore, some motion analysis labs may have an in-ground force plate available, but not a force-instrumented treadmill. Researchers often debate whether treadmill or overground walking/running protocols are realistic or valuable. A limitation to conducting studies using overground running is that velocity and acceleration cannot easily be manipulated by the experimenter. As such, additional methods are necessary to control forward velocity and acceleration. However, the various methods available for quantifying forward acceleration have not been systematically validated.
To the authors’ knowledge, the anterior-posterior (AP) ground reaction force is the only signal that has been leveraged for measuring and controlling for forward acceleration during overground locomotion. Over a given gait cycle, according to conservation of momentum, the net AP (braking/propulsion) impulse is equivalent to the net acceleration in the forward direction over a stance phase. During steady-state running, individuals run at a constant velocity, and the braking and propulsion force impulses are equal. Specific approaches for selecting steady-state running trials include: (a) identifying gait cycles that fall within a net AP boundary (e.g., -0.045-0.045 m/s2 during the stance phase (Van Caekenberghe et al. 2013); (b) identifying gait cycles with the smallest net AP impulse, through calculation (Bruening et al. 2018) or visual inspection (Williams et al. 2017). As such, we consider the relative AP ground reaction force impulse to be the “gold standard” method for quantifying forward acceleration. The relative AP ground reaction force impulse is the ratio of the positive impulse over the absolute value sum of the positive and negative impulse and is used to estimate the acceleration state (Equation (2)). When force plates are not available, some researchers have controlled for steady-state running by verbally cueing participants to maintain a steady forward acceleration (Havas et al,. 2000; Riley et al. 2008). Surprisingly, many overground running biomechanics studies either have made no effort to control forward acceleration or did not report on the method for controlling or measuring forward acceleration (Breine et al., 2013; White et al. 1998). This raises a broader debate on the validity of kinetics and kinematics reported from overground locomotion studies when it is unknown if forward acceleration was controlled for.
Timing lights and motion capture (optical, IMU-based, markerless) are alternative measurements for quantifying kinematics. Both of these methods include portable options that may be available to researchers to use outside the lab. Both timing lights (Brown et al. 2014) and motion capture (Bruening et al. 2018) have been used to measure walking/running velocity, but researchers do not typically report using this equipment to measure forward acceleration. The validity of these methods for quantifying forward acceleration has not been established. Therefore, the primary purpose of this study is to compare acceleration determined from timing lights and from motion capture with acceleration determined from the relative AP impulse measured with force plates. The results of this study raise a debate on whether timing lights and/or motion capture should be used in place of the gold-standard method when multiple measurement options are available. Since both timing lights (Brown et al. 2014) and motion capture (Bruening et al. 2018) have previously been used to measure walking/running velocity, the agreement between these two methods should be explored. If an agreement cannot be found, further research should be conducted on the accuracy of these two methods. Therefore, a secondary purpose of this study is to compare the relationship between the forward velocity measured with timing lights and the forward velocity measured with motion capture.
The hypothesized results are:
H1: The change in velocity measured with timing lights has a strong positive correlation with relative AP ground reaction force impulse.
H2: The change in velocity measured with motion capture has a strong positive correlation with relative AP ground reaction force impulse.
H3: There is a strong positive correlation between velocity measured with timing lights and velocity measured with motion capture.
Methods
Subjects
Ten participants (eight male and two female) were recruited for the study (mean ± SD, age = 23.8 ± 5.88 years, height = 174.5 ± 6.69cm, weight = 72.2 ± 9.04kg). All participants were injury free (no self-reported injuries or complaints of pain with respect to running). Each participant performed overground running in a laboratory setting. Timing lights, motion capture and force plate data were collected. Footwear and foot strike pattern were not specified. All participants provided informed written consent, and the protocol was approved by the University of Calgary’s Conjoint Health Research Ethics Board (REB21-1069).
Experimental Protocol
Participants performed ten overground running trials for each of the following conditions: (a) Steady-state running, (b) accelerating running and (c) decelerating running. Participants received verbal instructions before each condition informing the participant of the condition and whether to maintain a steady running speed (steady-state), to speed up (accelerating) or to slow down (decelerating) when crossing the capture volume. For all trials, participants were allowed to self-select their running speed as the study’s purpose was to evaluate methods for estimating running acceleration, which could be achieved at any given running speed. Individuals were allowed practice runs to familiarize with the condition and were given rest between trials. Participants were allowed to take as many practice runs as necessary to feel comfortable with the experimental task. No specific instructions were given during the practice runs to accelerate, decelerate or maintain the same speed. Ground reaction forces were collected from a force plate (2400Hz, Kistler Instrument AG, Winterthur, CH). Optical motion capture kinematics were collected using Vicon (200Hz, Nexus 2.11, VICON Motion Systems Ltd, UK). Timing light data were collected with four pairs of timing lights (Brower Timing Systems, Draper, UT, USA). Marker trajectories and ground reaction forces were filtered with a third-order, dual-pass, Butterworth Filter at 8Hz and 20Hz respectively. The ground reaction force cut-off frequencies were based on prior biomechanics running research (Day et al. 2021).
Estimating forward acceleration with timing lights
Four pairs of timing lights were set up across 2.7m distance at a height of 109 cm. Each timing light was equally distanced at 0.9m apart, with the two interior timing lights set up on either edge of the force plate (Figure 1). Using the known distance between pairs of timing lights, and by measuring the time the runner took to pass through each pair of timing lights, average velocities 1, 2 and 3 were calculated (Figure 1). Forward acceleration was estimated as the change between two values of average velocity. The timing light set up was positioned before, at and after the single gait cycle, as measured by both the force plate and motion capture. Therefore, with the use of multiple timing lights and multiple average velocities, a potential estimation of acceleration can be made.
Figure 1: Timing lights set up. Dots indicate placement of timing lights, one set of timing lights are indicated by two dots in vertical alignment. Vavg indicates average velocity measured between two sets of timing lights.
Estimating forward acceleration with motion capture
Motion capture data were collected with four markers placed on the pelvis. These were placed on the anterior superior iliac spine and the posterior superior iliac spine bilaterally via palpation). The average position of these four markers was used as an approximation of the runner’s center of mass kinematics (Vanrenterghem et al. 2010). Seven motion capture cameras were placed circularly around the data collection distance, and these locations were consistent across trials and participants. The derivative of the anterior-posterior component of the center of mass position was taken to obtain the approximate center of mass velocity. To approximate the center of mass forward acceleration over a single stance phase, the difference between forward velocity at toe off and foot contact was calculated (Equation (1)). Foot contact and toe off was determined using a 20N vertical ground reaction force threshold. As part of the secondary analysis, the average center of mass velocity over the stance phase was also calculated, indicating the average stance phase velocity.
Center of Mass Acceleration=velocity toe off−velocity foot contact(1)(1)Center of Mass Acceleration=velocity toe off−velocity foot contact
Force Plate
One force plate was set up in the center of the data collection volume and used to collect the ground reaction forces of the participants. The ground reaction forces were used to determine the relative AP impulse, which is the ratio of the positive impulse over the absolute value sum of the positive and negative impulse (Figure 2; Equation (2)). When the change in velocity is 0 (steady-state), the positive and negative impulses are equal and opposite with a relative AP impulse of 0.5. Relative AP impulse values greater than 0.5 indicate an acceleration state, while relative AP impulse values less than 0.5 indicate a deceleration state. The calculation of the relative AP impulse is preferred over other methods as it is not speed dependent, allowing for interpretation across a variety of speed ranges. Relative AP impulse was chosen because our participants ran at a range of self-selected running speeds. Consequently, we took a modified approach to the method of quantifying the net AP impulse for estimating acceleration state. We sought a method that would not be confounded by absolute running speed and an acceleration metric that could easily be compared across participants running at different speeds. Relative AP impulse was our gold standard measure of forward acceleration.
Figure 2: Ground reaction force in body weights versus time in seconds.The positive impulse is the area under the curve when the ground reaction force is positive. The negative impulse is the area under the curve when the ground reaction force is negative. B.W. = body weights.
Relative AP Impulse=Positive Impulse|Negative Impulse|+Positive Impulse(2)(2)Relative AP Impulse=Positive Impulse|Negative Impulse|+Positive Impulse
Calculations
For each change in velocity calculation, the correlation with the relative AP impulse (gold standard) was determined. To address the secondary purpose of this study, we found the correlation between average velocity 2 (timing lights) and the average velocity determined by the average center of mass (motion capture). An additional calculation of the slope of the linear line of best fit between average velocity 2 (timing lights) and average center of mass velocity (motion capture) was made to determine how the magnitude of both velocity measurements changed with respect to one another. For this slope calculation, the intercept was defined to be at zero, as zero velocity would be reported by both when no movement occurs. For interpretation, a strong positive correlation was defined as r ≥ 0.7, a moderate positive correlation was defined as 0.7 > r ≥ 0.3 and a weak positive correlation was defined as 0 < r < 0.3. A weak negative correlation was defined as 0 > r > -0.3, a moderate negative correlation was defined as -0.3 ≥ r > -0.7 and a strong negative correlation was defined as r ≤ -0.7.
Correlations were made intra-individually for each trial. We selected intra-individual correlations as they are a widely used analysis approach for biomechanics research that seeks to evaluate trends in biomechanical variables across a range of movement conditions (Honert et al. 2022; Matijevich et al. 2019; Savelberg and Lange 1999). All correlation, averages and standard deviations were calculated in Excel.
Results
Trials conducted in this study encompassed a wide range of forward accelerations (Table 1) and forward velocities (1.52-5.38 m/s). On average, none of the changes in velocity measured with timing lights had a strong positive correlation with the relative AP impulse (Table 2).
Table 1: AP impulse range for each participant.Ten rows represent the 10 participants. AP impulse range is reported as minimum AP impulse - maximum AP impulse for a participant.
| Participant | AP impulse range (min-max) |
--- |
| 1 | 0.22-0.78 |
| 2 | 0.36-0.88 |
| 3 | 0.12-0.73 |
| 4 | 0.19-0.90 |
| 5 | 0.13-0.94 |
| 6 | 0.04-0.99 |
| 7 | 0.20-0.98 |
| 8 | 0.24-0.71 |
| 9 | 0.11-0.81 |
| 10 | 0.19-0.79 |
Table 2: Correlation coefficient (r) between the relative AP impulse and change in velocity calculated by timing lights data.Ten rows represent the 10 participants. Three correlation coefficients are determined by comparing each change in velocity as calculated by timing lights. Average correlation coefficient is reported as average ± STD.
| Participant | r | r | r |
--- --- |
| | V avg,2 - V avg,1 | V avg,3 - V avg,2 | V avg,3 - V avg,1 |
| 1 | -0.04 | 0.15 | 0.12 |
| 2 | -0.64 | 0.62 | -0.02 |
| 3 | -0.34 | 0.47 | 0.23 |
| 4 | -0.08 | 0.43 | 0.43 |
| 5 | 0.00 | 0.16 | 0.19 |
| 6 | -0.01 | 0.26 | 0.24 |
| 7 | -0.34 | 0.19 | -0.20 |
| 8 | -0.54 | 0.53 | -0.02 |
| 9 | 0.75 | -0.49 | 0.22 |
| 10 | -0.01 | 0.40 | 0.31 |
| Average r | -0.01 ± 0.39 | 0.27 ± 0.31 | 0.15 ± 0.18 |
r - correlation coefficient, STD - standard deviation, V avg - average velocity
The trials for a single participant are shown on a correlation plot to visualize the relationship between the change in velocity measurements with timing lights and the relative AP impulse (Figure 3). Comparison between forward acceleration state estimated by motion capture data and the relative AP impulse revealed a strong correlation (Table 3).
Figure 3. Velocity change calculated from timing lights (Vavg,3 - Vavg,2) vs relative AP impulse ( N=1).Each dot represents a single trial, with the x-axis corresponding to change in velocity calculated from timing lights and the y-axis corresponding to the relative AP impulse measured from the force plate. The black line represents the regression line. The correlation coefficient for participant 10 was determined to be 0.40, as shown in Table 2.
Table 3: Correlation coefficient (r) between the relative AP impulse and change in velocity calculated by motion capture data (velcoity toe off – velocity foot contact).Ten rows represent the 10 participants. Average correlation coefficient is reported as average ± STD.
| Participants | r |
--- |
| 1 | 0.73 |
| 2 | 0.64 |
| 3 | 0.76 |
| 4 | 0.75 |
| 5 | 0.89 |
| 6 | 0.96 |
| 7 | 0.89 |
| 8 | 0.78 |
| 9 | 0.89 |
| 10 | 0.78 |
| Average r | 0.81 ± 0.10 |
r - correlation coefficient, STD - standard deviation
The trials for a single participant are shown on a correlation plot to visualize the relationship between the change in velocity measurements with motion capture and the relative AP impulse (Figure 4). Furthermore, the comparison of velocity values estimated via motion capture (average center of mass velocity) with timing lights (average velocity 2) showed a strong positive correlation (Table 4).
Figure 4: Velocity change calculated from motion capture (V_toe off - W_heel strike) vs relative AP impulse (N=1).Each dot represents a single trial, with the x-axis corresponding to change in velocity calculated from motion capture and the y-axis corresponding to the relative AP impulse measured from the force plate. The black line represents the regression line. The correlation coefficient for participant 10 was determined to be 0.78, as shown in Table 3.
Table 4: Correlation coefficient between velocity values measured by timing lights (average velocity 2) and motion capture for each participant.Averages are reported as average ± STD.The slope is reported as the slope of the line of best fit for each participant’s data set, with the intercept set to zero.
| Participant | r | Slope |
---
| 1 | 0.95 | 0.998 |
| 2 | 0.68 | 1.083 |
| 3 | 0.62 | 0.942 |
| 4 | 0.89 | 1.006 |
| 5 | 0.78 | 1.079 |
| 6 | 0.87 | 0.976 |
| 7 | 0.70 | 0.941 |
| 8 | 0.55 | 0.991 |
| 9 | 0.97 | 0.945 |
| 10 | 0.83 | 1.041 |
| Average | 0.78 ± 0.14 | 1.000 ± 0.053 |
r - correlation coefficient, STD - standard deviation
The trials for a single participant are shown on a correlation plot to visualize the relationship between velocity measures with timing lights and motion capture (Figure 5).
Figure 5. Velocity measured by timing lights (average velocity2) vs velocity measured by motion capture for an example participant (N=1).Each dot represents a single trial, with the x-axis corresponding to average velocity 2 measured by timing lights and th ey-axis corresponding to the average velocity measured by motion capture. The black line represents the unity line (y=x). The correlation coefficient for participant one was determined to be 0.95, as shown in Table 4.
On average, the correlation between the relative AP impulse and average velocity 2 - average velocity 1 measured with timing lights was weak (r = -0.01). Four participants showed a weak negative correlation, four a moderate negative correlation, one a strong positive correlation and one no correlation (Table 2).
On average, the correlation between the relative AP impulse and average velocity 3 - average velocity 2 measured with timing lights was weak (r = 0.27). One participant showed a moderate negative correlation, four a weak positive correlation and five a moderate positive correlation (Table 2).
On average, the correlation between the relative AP impulse and average velocity 3 - average velocity 1 measured with timing lights was weak (r = 0.15). Three participants showed a weak negative correlation, five a weak positive correlation and two a moderate positive correlation (Table 2).
On average, the correlation between the relative AP impulse and velocity toe off – velocity foot contact measured with motion capture was strong (r=0.81). One participant showed a moderate positive correlation and nine a strong positive correlation (Table 3).
On average, the correlation between average velocity 2 (timing lights) and average center of mass velocity (motion capture) was strong (r = 0.78). Three participants showed a moderate positive correlation, and seven showed a strong positive correlation. The average slope of the line of best fit was determined to be 1.000 with a standard deviation of 0.053, indicating very little variability in the data (Table 4).
Discussion
The purpose of this study was to determine if there are alternative methods for measuring forward acceleration during overground running studies when force plates are not available, and the gold standard relative AP ground reaction force impulse cannot be calculated. Two potential alternative methods explored to estimate forward acceleration were the change in velocity measured with timing lights and the change in velocity measured with motion capture. We found a weak negative to weak positive correlation between change in velocity measured with timing lights and the relative AP impulse measured with force plates. Therefore, the hypothesis that change in velocity measured with timing lights would have a strong positive correlation with force plate measures (relative AP impulse) is rejected (H1). There was a strong positive correlation between change in velocity measured with motion capture and the relative AP impulse measured with force plates. Therefore, the hypothesis that change in velocity measured with motion capture would have a strong positive correlation with force plate measures (relative AP impulse) is accepted (H2). Results of this study support the debate on if and how forward acceleration should be controlled during locomotion studies.
The experimental protocol was designed with the aim of obtaining a wide range of relative AP impulses (minimum of 0.04; maximum of 0.99, Table 1). This was done to ensure that there would be a wide variety of forward acceleration states to compare across methods. This was successfully achieved by verbally instructing participants to either run at a steady-state (remain at a constant velocity), accelerate (increase velocity) or decelerate (decrease velocity) over the measurement distance.
Change in velocity measured with timing lights is not a reliable alternative to relative AP impulse measured with force plates when determining the forward acceleration of participants. For some participants, the correlation between relative AP impulse and timing lights was negative, demonstrating change in velocity measured with timing lights could yield the opposite of the expected results, misguiding researchers regarding the forward acceleration. Furthermore, there was a large variance in the data across participants (Table 2). One possible explanation of these results is an inherent limitation of the timing lights wherein a certain part of the body crossing a set of timing lights may trigger their sensors. As such, the timing light measurements may not always correspond to the center of mass motion of the participants. For example, for a pair of timing lights, the participant’s hand may trigger the sensor of the first timing light, while the participant’s torso may trigger the second timing light. This nuance may explain why timing lights are an unreliable indicator of center of mass change in velocity and yield a low correlation with the relative AP impulse. While the timing light manufacturer reportedly accounts for this effect, it is unknown if other body parts crossing the timing lights could still confound results. Another possible explanation is our timing lights were set up spatially before, over and after the force plate. While the relative AP impulse is only calculated over a stance phase from foot contact to toe off, change in velocity between two sets of timing lights covers a larger time window that may not correspond exactly to the stance phase of interest. Rather than covering a single stance phase, a single pair of timing lights covered a distance of 0.9m. As the relative AP impulse only considers a single stance phase, while timing light measurements are made independently of the number of stance phases, these approaches varied slightly in their data collection area. Since these two approaches use measurements from different locations/times, it is unsurprising that we did not find a strong correlation. Therefore, our results indicate that using timing lights to calculate change in velocity in place of the relative AP impulse is not a method that should be considered.
Change in velocity measured with motion capture is a feasible alternative to relative AP impulse measured with force plates when determining the forward acceleration of participants. When comparing relative AP impulse with change in velocity measured by motion capture (Equation 1), the average correlation was strong and positive (r = 0.81). An advantage of the motion capture method is that it includes measurements from a time window equivalent to the AP impulse method, which is the time window from foot contact to toe off. The standard deviation of the correlation coefficient was 0.10, indicating low variability across participants (Table 3). These results indicate that the changes in center of mass velocity approximated by pelvis motion capture markers are an accurate indicator of the net forward acceleration. These experimental results suggest changes in approximate center of mass velocity over the stance phase of running are a reliable biomechanical signal for estimating acceleration state. While we estimated center of mass acceleration using a cluster of pelvis motion capture markers and optical motion capture in this study, center of mass velocity can also be measured with other measurement approaches, such as by tracking a single sacral optical motion capture marker (Napier et al. 2020), signal integration of IMUs placed near the center of mass (Cardarelli et al. 2020), a complementary filter method with multiple IMUs on the body (Mohamed Refai et al. 2020), machine learning applied to sacral acceleration signals (Alcantara et al. 2021), etc. Identifying a signal of interest, for example the center of mass velocity, flexibly empowers researchers to then select a measurement tool that best aligns with their experimental protocol and study constraints. Because all measurement systems, such as wearables, for example, have error and limitations, there are great advantages to first leveraging high accuracy tools like optical motion capture to identify the signal(s) of interest (Matijevich et al. 2020). While optical motion capture was used here, it is possible that other more portable or wearable forms of motion capture (e.g., IMU-based, markerless.) may also be a viable solution for quantifying forward acceleration. Therefore, our results indicate that calculating change in velocity using motion capture data can be used in place of the relative AP impulse measured by force plates. However, there is room for further debate regarding the feasibility and ease of both of these methods, and whether one method should be considered the “gold-standard”.
A secondary purpose of this study was to compare velocity measured with timing lights and motion capture to determine if these methods can be used interchangeably. We found a strong positive correlation between change in velocity measured with timing lights and motion capture. The hypothesis that timing lights and motion capture can be used interchangeably when measuring velocity is accepted (H3).
When comparing velocity measurements made by timing lights (average velocity 2) and motion capture (average center of mass velocity), the average correlation was strong and positive (r = 0.78). The standard deviation of the correlation coefficient for this comparison was 0.13, indicating little variability between participants (Table 4). For this comparison, we elected to use average velocity 2 because this is the velocity measurement that is most similar to the time and location of velocity measured with motion capture (over the force plate). This may be a possible reason why timing lights show a high correlation when measuring velocity but a minimal correlation to AP impulse. As the AP impulse comparison was made with multiple sets of timing lights, the measurement distance for that comparison was greater than just the force plate, which is where the AP impulse measurements are calculated. However, for the velocity comparison, motion capture and average velocity 2 are collected over the same distance, which may explain the high correlation. The average slope of the linear line of best fit was 1.00 ± 0.053, indicating that the velocity values determined by each method change linearly with one another in a one-to-one ratio. An example participant shown in Figure 3 shows this trend, as there is a linear relationship between the two velocity measurements. A potential reason for a correlation that was not exactly one could be differences in the tracking period. Motion capture measured velocity for the stance phase whereas the timing lights were at fixed locations spanning the force plate. Across multiple trials, participants may have contacted the force plate at slightly different locations.
Recommendations for researchers moving forward
These findings provide researchers with alternative methods for quantifying forward acceleration during overground running studies in instances when force plates are not available. This may occur when researchers do not have a force plate or when researchers want to conduct studies outside the lab). Namely, the change in velocity of motion capture markers that approximate the location of the center of mass can be used to accurately determine the forward acceleration state. The findings of this study also give researchers confidence in using timing lights and motion capture interchangeably for measuring forward velocity. Broadly, we encourage biomechanics researchers to measure and control for forward acceleration during overground locomotion studies and to report the methods used for quantifying forward acceleration state.
Limitations and future research opportunities
The experimental design used in this study had limitations, indicating opportunities for future research. One limitation was the absence of walking movements in the study. Therefore, the reliability of using these methods for quantifying the forward acceleration of walking is unknown. A second limitation of this study is that pelvis markers were assumed to approximate the location of the participant’s center of mass. This approximation does not account for differences in body types as the location of center of mass may vary across individuals. Another limitation is there was no exploration of the reliability of using alternative motion capture marker sets, such as a single sacrum motion capture marker to approximate center of mass velocity (Bruening et al. 2018) or whole-body tracking with modeling to obtain the center of mass). Furthermore, this study was conducted in a laboratory setting with a small sample size (10 participants). Future research should be conducted outside the lab with a larger sample size to determine if the results of this study are applicable to research conducted in non-laboratory settings. An additional limitation of this work was that we used the relative AP impulse, rather than the absolute net AP impulse magnitude, as an indicator of the direction and magnitude of running acceleration. While this method was used to allow for comparison across a range of self- selected running speeds, future studies may explore the association between acceleration methods at a fixed running speed where net AP impulse can be more readily compared. Another limitation of this research is that we only used Pearson corre-lation coefficients to find the association between measurement methods. Other relative and absolute accuracy metrics (e.g., mean absolute percent error, root mean square error, Bland-Altman plots) could be used in the future to quantify the agreement between methods.
Finally, as mentioned previously, the timing lights were set up spatially before, over and after the force plate. Since the relative AP impulse measurement was only made over the force plate distance, this variability in measurement distance may have affected the results. We encourage researchers to explore the use of timing lights to measure acceleration in more depth for future studies by manipulating the timing light organization. Future researchers could have two sets of timing lights over the force plate distance to increase the similarity of measurement location.
Conclusion
This evaluation of overground running acceleration was intended to support the debate on if and how forward acceleration should be controlled for during locomotion studies. The primary purpose of this study was to identify if measuring change in velocity with two different sets of equipment, timing lights and motion capture, are accurate alternatives to relative AP impulse, which is measured by force plate, when quantifying overground running acceleration. We found that changes in velocity measured with timing lights are not a reliable surrogate for relative AP impulse. However, change in velocity measured with motion capture can be used as a substitute to relative AP impulse. A secondary purpose of this study was to determine if velocity measurements made from motion capture and timing lights can be used interchangeably. The results of this study indicate that both methods can reliably be used interchangeably. The findings of this study provide future researchers with a reliable method to quantify forward acceleration when force plates are not available. With this method, researchers can reliably control for trials that are or are not steady-state locomotion, giving them confidence in the interpretation of kinetics and kinematics variables across different trials and participants.
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Honert, E. C., Hoitz, F., Blades, S., Nigg, S. R. and Nigg, B. M. (2022). Estimating Running Ground Reaction Forces from Plantar Pressure during Graded Running. Sensors, 22(9), 3338-3338.
Matijevich, E. S., Branscombe, L. M., Scott, L. R. and Zelik, K. E. (2019). Ground reaction force metrics are not strongly correlated with tibial bone load when running across speed and slopes: Implications for science, sport and wearable tech. PloS One, 14(1), 210000-0210000.
Matijevich, E. S., Scott, L. R., Volgyesi, P., Derry, K. H. and Zelik, K. E. (2020). Combining wearable sensor signals, machine learning and biomechanics to estimate tibial bone force and damage during running. Human Movement Science, 74, 102690-102690.
Napier, C., Jiang, X., Maclean, C. L., Menon, C. and Hunt, M. A. (2020). The use of a single sacral marker method to approximate the centre of mass trajectory during treadmill running. Journal of Biomechanics, 108, 109886-109886.
Refai, M., Beijnum, M. I. V., Buurke, B.-J. F., Veltink, J. H. and H, P. (2020). Portable Gait Lab: Instantaneous centre of mass velocity using three inertial measurement units. 2020 IEEE Sensors.
Riley, P. O., Dicharry, J., Franz, J., Croce, U. D., Wilder, R. P. and Kerrigan, D. C. (2008). ‘A Kinematics and Kinetic Comparison of Overground and Treadmill Running. Medicine and Science in Sports and Exercise, 40(6), 1093-1100.
Rodman, C. H., and Martin, A. E. (2020). Quantification of spatiotemporal parameter behavior during walking speed transitions. Journal of Biomechanics, 112, 110068-110068. Ruder, J., Tenforde, S. T., Mulloy, A., Davis, F. and S, I. (2019). Relationship of Foot Strike Pattern and Landing Impacts during a Marathon. Medicine and Science in Sports and Exercise, 51(10), 2073-2079.
Savelberg, H. H. C. M., and Lange, A. L. H. (1999). Assessment of the horizontal, fore-aft component of the ground reaction force from insole pressure patterns by using artificial neural networks. Clinical Biomechanics (Bristol), 14(8), 585-592.
Vanrenterghem, J., Gormley, D., Robinson, M. and Lees, A. (2010). ‘Solutions for representing the whole-body centre of mass in side cutting manoeuvres based on data that is typically available for lower limb kinematics. Gait and Posture, 31(4), 517-521.
White, J. A., Pomfret, D., Rennie, S., Gissane, C., Wong, J. and Ford, M. (1998). ‘Fluid replacement needs of well-trained male and female athletes during indoor and outdoor steady state running. Journal of Science and Medicine in Sport, 1(3), 131-142.
Williams, C., Powell, J. H. and W, D. (2017). ‘Lower Extremity Joint Work During Acceleration, Deceleration, and Steady State Running. Journal of Applied Biomechanics, 33(1), 56-63.
Willwacher, S., Oberländer, K. D., Mai, P., Mählich, D., Kurz, M., Koopmann, T., Fohrmann, D., Kantarev, A. and Kersting, U. G. (2021). A new method for measuring treadmill belt velocity fluctuations: effects of treadmill type, body mass and locomotion speed. Scientific Reports,11(1), 2244-2244.
Yang, F., and King, G. A. (2016). ‘Dynamic gait stability of treadmill versus overground walking in young adults. Journal of Electromyography and Kinesiology, 31, 81-87.
Tagged: kinesiology, biomechanics, acceleration, kinetics
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11957 | https://omim.org/entry/400020 | Entry - 400020 - SHORT STATURE HOMEOBOX, Y-LINKED; SHOXY - OMIM - (OMIM.ORG)
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400020
SHORT STATURE HOMEOBOX, Y-LINKED; SHOXY
HGNC Approved Gene Symbol: SHOX
Cytogenetic location: Yp11.2 Genomic coordinates (GRCh38) : Y:624,344-659,411(from NCBI)
Gene-Phenotype Relationships
| Location | Phenotype View Clinical Synopses | Phenotype MIM number | Inheritance | Phenotype mapping key |
--- ---
| Yp11.2 | Langer mesomelic dysplasia | 249700 | PR | 3 |
| Leri-Weill dyschondrosteosis | 127300 | PD | 3 |
| Short stature, idiopathic familial | 300582 | | 3 |
PheneGene Graphics
Linear
Radial
▼TEXT
See short stature homeobox (SHOX; 312865) for a discussion of the SHOXY gene, which is located in the pseudoautosomal region.
▼See Also:
Ellison et al. (1997); Rao et al. (1997)
▼REFERENCES
Ellison, J. W., Wardak, Z., Young, M. F., Robey, P. G., Webster, M., Chiong, W. PHOG, a candidate gene for involvement in the short stature of Turner syndrome. Hum. Molec. Genet. 6: 1341-1347, 1997. [PubMed: 9259282, related citations] [Full Text]
Rao, E., Weiss, B., Fukami, M., Rump, A., Niesler, B., Mertz, A., Muroya, K., Binder, G., Kirsch, S., Winkelmann, M., Nordsiek, G., Heinrich, U., Breuning, M. H., Ranke, M. B., Rosenthal, A., Ogata, T., Rappold, G. A. Pseudoautosomal deletions encompassing a novel homeobox gene cause growth failure in idiopathic short stature and Turner syndrome. Nature Genet. 16: 54-63, 1997. [PubMed: 9140395, related citations] [Full Text]
Creation Date:
Ada Hamosh : 11/15/1999
Edit History:
terry : 03/18/2004
joanna : 9/6/2001
carol : 11/15/1999
400020
SHORT STATURE HOMEOBOX, Y-LINKED; SHOXY
HGNC Approved Gene Symbol: SHOX
Cytogenetic location: Yp11.2 Genomic coordinates (GRCh38) : Y:624,344-659,411(from NCBI)
Gene-Phenotype Relationships
| Location | Phenotype | Phenotype MIM number | Inheritance | Phenotype mapping key |
--- ---
| Yp11.2 | Langer mesomelic dysplasia | 249700 | Pseudoautosomal recessive | 3 |
| Leri-Weill dyschondrosteosis | 127300 | Pseudoautosomal dominant | 3 |
| Short stature, idiopathic familial | 300582 | | 3 |
TEXT
See short stature homeobox (SHOX; 312865) for a discussion of the SHOXY gene, which is located in the pseudoautosomal region.
See Also:
Ellison et al. (1997); Rao et al. (1997)
REFERENCES
Ellison, J. W., Wardak, Z., Young, M. F., Robey, P. G., Webster, M., Chiong, W. PHOG, a candidate gene for involvement in the short stature of Turner syndrome. Hum. Molec. Genet. 6: 1341-1347, 1997. [PubMed: 9259282] [Full Text:
Rao, E., Weiss, B., Fukami, M., Rump, A., Niesler, B., Mertz, A., Muroya, K., Binder, G., Kirsch, S., Winkelmann, M., Nordsiek, G., Heinrich, U., Breuning, M. H., Ranke, M. B., Rosenthal, A., Ogata, T., Rappold, G. A. Pseudoautosomal deletions encompassing a novel homeobox gene cause growth failure in idiopathic short stature and Turner syndrome. Nature Genet. 16: 54-63, 1997. [PubMed: 9140395] [Full Text:
Creation Date:
Ada Hamosh : 11/15/1999
Edit History:
terry : 03/18/2004
joanna : 9/6/2001
carol : 11/15/1999
NOTE: OMIM is intended for use primarily by physicians and other professionals concerned with genetic disorders, by genetics researchers, and by advanced students in science and medicine. While the OMIM database is open to the public, users seeking information about a personal medical or genetic condition are urged to consult with a qualified physician for diagnosis and for answers to personal questions.
OMIM® and Online Mendelian Inheritance in Man® are registered trademarks of the Johns Hopkins University.
Copyright® 1966-2025 Johns Hopkins University.
NOTE: OMIM is intended for use primarily by physicians and other professionals concerned with genetic disorders, by genetics researchers, and by advanced students in science and medicine. While the OMIM database is open to the public, users seeking information about a personal medical or genetic condition are urged to consult with a qualified physician for diagnosis and for answers to personal questions.
OMIM® and Online Mendelian Inheritance in Man® are registered trademarks of the Johns Hopkins University.
Copyright® 1966-2025 Johns Hopkins University.
Printed: Sept. 28, 2025
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11958 | https://isthisnormal.littlespoon.com/the-myth-about-nitrates-in-fresh-baby-food/ | The Myths About Nitrates In Fresh Baby Food - Is This Normal
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The Myths About Nitrates In Fresh Baby Food
Wondering what the deal is with nitrates and fresh baby food? Read on.
by Little Spoon · March 18, 2019
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Over the past few years, there has been a bit of buzz around feeding your babes fresh carrots and other root veggies due to a risk of nitrate consumption and potential poisoning. Not only does this include the yummies you whip up at home, but also the kick ass blends our team at Little Spoon is happily offering your budding foodie. You may have heard something along these lines: “Root vegetables from some parts of the country contain large amounts of nitrates that can cause an unusual kind of anemia (low red blood cells or hemoglobin) in young infants. Large, shelf-stable baby food companies screen their produce for nitrates and since you cannot do this at home, it is safer to give your baby commercially prepared foods instead of serving your child fresh food.”
It is time to finally bust this crap and put it to rest! The type of nitrate reaction that we have been warned repeatedly against is called methemoglobinemia. There has only been one reported case in the United States that stemmed from homemade baby food. That case was reported in 1973 and there has not been another case since. All other cases of methemoglobinemia in babies have come from mixing powdered infant formula with nitrate-contaminated well water.
It is also important to point out that while mass production commercial baby food companies (think: the one with a cute babe on the containers) can screen their produce, they cannot remove the nitrates. So their screening test only indicates if there are elevated levels of nitrates, but they still usethe produce in their baby food.
The truth is the main issue is with the age of the baby. The only risk of nitrate poisoning or methemoglobinemia from homemade root vegetable baby food occurs when the baby consuming the food is 3 months or younger. So if you are following the American Association of Pediatrics’ guidelines of starting solids no earlier than 6 months of age, your little nugget will be nothing short of perfect.
If you’re looking for a simple way to feed your babe fresh, nutrient-dense baby food, look no further than Little Spoon! With a variety of Babyblends specifically designed for your little one’s nutritional and developmental needs as they grow, Little Spoon has you covered every step of the way. Get started with their 100% organic Babyblends today.
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11959 | https://www.scientificlib.com/en/Mathematics/LX/ConicalSurface.html | Conical surface
File failed to load:
.
Conical surface
In geometry, a (general) conical surface is the unbounded surface formed by the union of all the straight lines that pass through a fixed point — the apex or vertex — and any point of some fixed space curve — the directrix — that does not contain the apex. Each of those lines is called a generatrix of the surface.
Every conic surface is ruled and developable. In general, a conical surface consists of two congruent unbounded halves joined by the apex. Each half is called a nappe, and is the union of all the rays that start at the apex and pass through a point of some fixed space curve. (In some cases, however, the two nappes may intersect, or even coincide with the full surface.) Sometimes the term "conical surface" is used to mean just one nappe.
If the directrix is a circle C, and the apex is located on the circle's axis (the line that contains the center of C and is perpendicular to its plane), one obtains the right circular conical surface. This special case is often called a cone, because it is one of the two distinct surfaces that bound the geometric solid of that name. This geometric object can also be described as the set of all points swept by a line that intercepts the axis and rotates around it; or the union of all lines that intersect the axis at a fixed point p and at a fixed angle \theta. The aperture of the cone is the angle [Math Processing Error].
More generally, when the directrix C is an ellipse, or any conic section, and the apex is an arbitrary point not on the plane of C, one obtains a conical quadric, which is a special case of a quadric surface.
A cylindrical surface can be viewed as a limiting case of a conical surface whose apex is moved off to infinity in a particular direction. Indeed, in projective geometry a cylindrical surface is just a special case of a conical surface.
Equations
A conical surface S can be described parametrically as
[Math Processing Error]
where v is the apex and q is the directrix.
A right circular conical surface of aperture 2\theta, whose axis is the z coordinate axis, and whose apex is the origin, it is described parametrically as
[Math Processing Error]
where t and u range over [Math Processing Error] and [Math Processing Error], respectively. In implicit form, the same surface is described by S(x,y,z) = 0 where
[Math Processing Error]
More generally, a right circular conical surface with apex at the origin, axis parallel to the vector [Math Processing Error], and aperture [Math Processing Error], is given by the implicit vector equation [Math Processing Error] where
[Math Processing Error]
or
[Math Processing Error]
where [Math Processing Error]denotes the dot product.
In three coordinates, x, y and z, the general equation for a cone with apex at origin is a homogeneous equation of degree 2 given by
[Math Processing Error]
See also
Conic section
Developable surface
Quadric
Ruled surface
Undergraduate Texts in Mathematics
Graduate Texts in Mathematics
Graduate Studies in Mathematics
Mathematics Encyclopedia
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11960 | https://thirdspacelearning.com/us/math-resources/topic-guides/algebra/math-equations/ | High Impact Tutoring Built By Math Experts
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What are math equations?
Common Core State Standards
How to solve with math equations
Math equations examples
Example 1: volume of a cube Example 2: substitution basic inequality Example 3: substitution equation Example 4: solve for the variable using addition/subtraction Example 5: solve the equation using multiplication/division Example 6: solving equations with fractions - two steps
Teaching tips for math equations
Easy mistakes to make
Practice math equations questions
Math equations FAQs
Next lessons
Still stuck?
Math equations
Here you will learn about math equations, including what they are and how to solve them.
Students will first learn about math equations as part of expressions and equations in 6th grade. They continue to expand upon this knowledge in 7th and 8th grade.
Every week, we teach lessons on math equations to students in schools and districts across the US as part of our online one-on-one math tutoring programs. On this page we’ve broken down everything we’ve learnt about teaching this topic effectively.
What are math equations?
Math equations are statements made up of algebraic expressions and an equal sign. Math equations include defined math formulas that can be solved with substitution.
Math formulas are rules that connect two or more variables. They are used to work out the values of unknown variables by substituting in other known values.
For example, here is a rectangle with base b and height h.
The area of any rectangle can be found with the formula b \times h.
Step-by-step guide: Math formulas
Substitution is when you give variables in an algebraic expression, equation or inequality a value. You use substitution to identify solutions.
For example,
Does x = 2 make the equation 5x+2=14 true?
Substitute 2 for x and follow the order of operations:
5 \times 2+2=14
10+2=14
12 ≠ 14
Substituting 2 for x, does not make the equation true, so 2 is not a solution of the equation 5x+2=14.
Step-by-step guide: Substitution
Math equations also include multi step equations and equations with fractions.
One step equations are algebraic equations with a single variable that can be solved with one step.
In order to solve, you will need to isolate the variable (get it alone) on one side of the equation. This can be done with a model or by using the inverse operation.
For example, 3x =15
| Using a model: | Using inverse operations: |
--- |
| Sort the (+1) tiles so an equal amount are with each (x) tile. There are five (+1) tiles for each (x) tile. x=5 | 3x=15 Get the variable alone by using the inverse operation. The inverse operation of multiplication is division. Divide 3 on both sides of the equation. \cfrac{3x}{3}=\cfrac{15}{3} Since \cfrac{3}{3} \,=1, the left side can be written as 1x or x. The right side of the equation can be written as \cfrac{15}{3}=5. x=5 You can check your answer by substituting the value for x into the original equation. \begin{aligned} 3x&=15\ 3(5)&=15 \; \color{#77dd77} ✔ \end{aligned} |
Step-by-step guide: One step equations
Solving equations with fractions is when the unknown is part of the numerator and/or denominator of a fraction.
To solve equations with fractions, apply the inverse operation to both sides of the equation – a strategy also referred to as the “balance method.”
The inverse operation of division is multiplication.
For example, \, \cfrac{x+3}{3}=8
3\left(\cfrac{x+3}{3}\right)=3(8) \quad Multiply each side by 3.
x+3=24
-3 \quad \; -3 \quad Subtract 3 from both sides.
x = 21
Check your work by substituting the answer into the equation.
Step-by-step guide: Solve equations with fractions
What are math equations?
[FREE] Math Equations Worksheet (Grade 6 to 12)
Test your grades 6 to 12 students’ skills on how to solve math equations with this 15-question worksheet complete with answers for easy assessment. Save time and quickly identify learning gaps!
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[FREE] Math Equations Worksheet (Grade 6 to 12)
Test your grades 6 to 12 students’ skills on how to solve math equations with this 15-question worksheet complete with answers for easy assessment. Save time and quickly identify learning gaps!
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Common Core State Standards
How does this relate to 6th grade math, 7th grade math, and 8th grade math?
Grade 6 – Expressions and Equations (6.EE.A.2c)Write, read, and evaluate expressions in which letters stand for numbers.
Grade 6 – Expressions and Equations (6.EE.B.5)
Understand solving an equation or inequality as a process of answering a question: which values from a specified set, if any, make the equation or inequality true? Use substitution to determine whether a given number in a specified set makes an equation or inequality true.
Grade 7 – Expressions and Equations (7.EE.A.1)
Apply properties of operations as strategies to add, subtract, factor, and expand linear expressions with rational coefficients.
Grade 8 – Expressions and Equations (8.EE.C.7)
Solve linear equations in one variable.
Grade 8 – Expressions and Equations (8.EE.C.7b)
Solve linear equations with rational number coefficients, including equations whose solutions require expanding expressions using the distributive property and collecting like terms.
How to solve with math equations
There are a lot of ways to use math equations. For more specific step-by-step guides, check out the pages linked in the “What are math equations?” section above or read through the examples below.
Math equations examples
Example 1: volume of a cube
Find the volume of the cube using the formula, V=s \times s \times s or V=s^3
Use the formula given in the question.
Use the formula V=s \times s \times s to find the volume of the cube.
The sides of the cube are all 8. Substitute the measurements into the given formula.
V=8 \times 8 \times 8
2Work carefully to answer the question, one step at a time.
\begin{aligned} & V=8 \times 8 \times 8 \\ & V=64 \times 8 \\ & V=512 \end{aligned}
3Write the final answer clearly.
The final answer is 512 units^3.
Example 2: substitution basic inequality
Does r=-2 make the inequality r \leq-11 true?
Substitute each variable with its given value.
-2 \leq-11
Solve using the order of operations.
There is nothing to calculate, but you read the inequality ‘-2 is less than or equal to -11.’ Also consider the inequality on a number line.
Since -2 is not less than or equal to -11, substituting -2 for r makes the inequality false. This means -2 is NOT a solution for r \leq-11.
Note that there are an infinite amount of solutions for this inequality – any number less than or equal to -11 makes the inequality true.
Example 3: substitution equation
Find the value of e \, (114-t) when e=11 and t=58.
Substitute each variable with its given value.
11 \, (114-58)
Solve using the order of operations.
11 \, (114-58)
=11 \, (56) \quad \quad Remember 11 \, (56) is the same as 11 \times 56.
=616
Example 4: solve for the variable using addition/subtraction
Solve the equation for x.
8+x=-13
Identify the inverse operation to use.
8+x=-13
Because of the commutative property, this equation could also be written as:
x+8=-13 → inverse operation is subtraction because subtraction undoes addition.
Do the inverse operation to both sides of the equation.
8+x =-13
-8 \quad\; -8
Solve for the unknown variable.
On the left side of the equation 8-8=0, leaving just 0+ x, which is the same as x.
On the right side of the equation:
\begin{aligned} x= & -13 \\ & -8 \\ x= & -21 \end{aligned}
Check the answer.
Example 5: solve the equation using multiplication/division
Solve the equation, \, \cfrac{x}{5}=-20 for x.
Identify the inverse operation to use.
\cfrac{x}{5}=-20 → inverse operation of division is multiplication.
Do the inverse operation on both sides of the equation.
\cfrac{x}{5} \times 5=-20 \times 5
Solve for the unknown variable.
On the left side of the equation: \cfrac{x}{5} \times 5=\cfrac{5 x}{5}=1 x
On the right side of the equation: -20 \times 5=-100
\begin{aligned} \cfrac{x}{5} \times 5 & =-20 \times 5 \\ 1 x & =-100 \\ x & =-100 \end{aligned}
Check the answer.
Example 6: solving equations with fractions – two steps
Solve for x\text{: } \cfrac{x-9}{8}=6.5
Identify the operations that are being applied to the unknown variable.
The unknown is x.
Looking at the left hand side of the equation, 9 is subtracted from x and then divided by 8 (the denominator of the fraction).
\cfrac{x-9}{8}
Apply the inverse operations, one at a time, to both sides of the equation.
First, multiplying both sides of the equation by 8.
Then, add 9 to each side.
8\left(\cfrac{x-9}{8}\right)=(6.5) \, 8 \quad Multiply each side by 8.
x-9=52
+9 \quad \quad +9 \quad Add 9 to both sides.
x = 61
Write the final answer, checking that it is correct.
The final answer is x=61.
You can check the answer by substituting the answer back into the original equation.
\cfrac{x-9}{8}=\cfrac{61-9}{8}=52 \div 8=6.5
[FREE] Math Equations Worksheet (Grade 6 to 8)
Use this quiz to check your grade 6 to 8 students’ understanding of math equations. 10+ questions with answers covering grades 6, 7 and 8 math equations topics to diagnose learning gaps!
DOWNLOAD FREE
x
[FREE] Math Equations Worksheet (Grade 6 to 8)
Use this quiz to check your grade 6 to 8 students’ understanding of math equations. 10+ questions with answers covering grades 6, 7 and 8 math equations topics to diagnose learning gaps!
DOWNLOAD FREE
Teaching tips for math equations
Instead of introducing this topic through worksheets and rote memorization of the rules, use manipulatives, such as algebra tiles. This can help students build conceptual understanding of these abstract concepts.
However, it is important not to rush the work with these tools – while it may take more time in the short term to develop understanding with manipulatives versus memorizing a procedure, in the long term students will be more flexible and more easily able to adapt their strategies to more complex topics.
Note: It is also useful to discuss the limitations of algebra tiles with students, such as they can only be used to represent smaller integers.
Whenever possible, consider using project based learning or real-life assessments, such as having students measure items within the classroom and then using the formulas to calculate the area of the items.
When teaching students how to solve math problems such as these, the traditional focus is often solely on the solution of an equation. While this is a crucial piece, also give students time to explore what numbers are not a part of the solution set as well.
In middle school, students are still building foundational ideas around algebraic expressions, equations and inequalities, so it is important not to always rush them to the answer.
Always have students check to make sure their answer is correct by substituting the value back into the original equation. This helps them catch simple mistakes and gives them practice substituting values into the equation.
Easy mistakes to make
Thinking the variable can only be on the left-hand side
While the variable is commonly seen on the left side of the equation, it can also be on the right-hand side.
For example,
\begin{aligned} & 19=3 x \ & \cfrac{19}{3}=\cfrac{3}{3} x \ & 6 \cfrac{1}{3}=x \end{aligned}
Confusing the difference between a mathematical expression and a mathematical equation
A mathematical expression is a statement that does not include an equal sign, like 43-x. A mathematical equation is a complete mathematical sentence that includes an equals sign, like 43-x=22.
Forgetting to follow the order of operations (PEMDAS)
When substituting values into an equation to solve, always follow the order of operations: parentheses first, then exponents, then from left to right multiplication and division, and then from left to right addition and subtraction.
For example,
Solve 5x-10 \div 2 when x=2.
Forgetting that the solution to an equation can be any type of number
The unknowns do not have to be whole numbers or integers. In fact, the solutions often involve fractions or decimals.
Practice math equations questions
Find the area of the triangle using the appropriate formula.
24 \, m^2
18 \, m^2
12 \, m^2
63 \, m^2
Use the formula A=\cfrac{1}{2} \, b h to find the area of the triangle.
Substitute the measurements into the formula and solve.
\begin{aligned} & A=\cfrac{1}{2} \, (4 \times 6) \\ & A=\cfrac{1}{2} \, (24) \\ & A=12 \end{aligned}
The area of the triangle is 12 \, m^2.
Find the value of 45+88x when x=13.
1,729
8,858
8,813
1,189
Substitute each variable with its given value.
45+88 \times 13
Solve using the order of operations.
\begin{aligned} & 45+88 \times 13 \\ & =45+1,144 \\ & =1,189 \end{aligned}
Which number is a solution for t > -56 \, ?
-55
-56
-57
-258
Substitute each variable with its given value and then read each inequality:
-55 > -56… ‘-55 is greater than -56’
-56 > -56… ‘-56 is greater than -56’
-57 > -56… ‘-57 is greater than -56’
-258 > -56… ‘-258 is greater than -56’
Substituting -55 for t makes the inequality true, so -55 is a solution for t > -56.
Solve the equation for x \text{: } -4x=22
x= 22
x= -5.5
x= -5
x= 4.5
-4x=22 → The inverse operation is division, so divide both sides of the equation by -4.
\cfrac{-4 x}{-4}=\cfrac{22}{-4}
On the left side of the equation: \cfrac{-4 x}{-4}=1 x and 1 x=x
On the right side of the equation: \cfrac{22}{-4}=-5.5
\begin{aligned} & \cfrac{-4 x}{-4}=\cfrac{22}{-4} \\ & x=-5.5 \end{aligned}
Check:
Solve the equation for p \text{: } p-\cfrac{4}{5}=\cfrac{1}{3}
x=\cfrac{17}{30}
x=\cfrac{3}{2}
x=\cfrac{15}{17}
x=1 \, \cfrac{2}{15}
p-\cfrac{4}{5}=\cfrac{1}{3} → The inverse operation is addition, so add \, \cfrac{4}{5} \, to both sides of the equation.
p-\cfrac{4}{5}+\cfrac{4}{5}=\cfrac{1}{3}+\cfrac{4}{5}
On the left side of the equation: -\cfrac{4}{5}+\cfrac{4}{5}=0, so p-0 is left.
On the right side of the equation: \cfrac{1}{3}+\cfrac{4}{5}=\cfrac{5}{15}+\cfrac{12}{15}=\cfrac{17}{15}
\begin{aligned} & p-\frac{4}{5}+\frac{4}{5}=\frac{1}{3}+\frac{4}{5} \\ & p-0=\frac{17}{15} \\ & p=\frac{17}{15} \text { or } 1 \frac{2}{15} \end{aligned}
Check:
Solve: \cfrac{6 x-9}{3}=1
x=-\cfrac{1}{2}
x=-2
x=\cfrac{1}{2}
x=2
First, multiply both sides of the equation by 3.
Next, add 9 to both sides.
Finally, divide both sides by 6.
The final answer is x = 2.
You can check the answer by substituting the answer back into the original equation.
\cfrac{6 \, \times \, 2-9}{3}=\cfrac{12-9}{3}=\cfrac{3}{3}=1
Math equations FAQs
Why are math formulas important?
Math formulas make work in mathematics more efficient.
While students always spend ample time working to build ideas around a mathematical concept, formalizing it with the use of a formula (like the area of a triangle as A=\cfrac{1}{2} \, b h) allows them to move forward in their understanding and answer more complex questions.
How can shape patterns be helpful when learning about math formulas?
Growing shape patterns start with a shape and grow in a constant way from term to term. Students learn how to write general rules for these patterns. For example, “Start with 2 circles and add 3 circles each time.” This can then be translated into a math equation for the pattern: 2 + 3n = t , where n = the pattern term number and t = the total number of circles.
See also: Shape patterns
When do students learn the Pythagorean theorem?
Students will learn about the Pythagorean theorem in the 8th grade.
What types of equations do students learn about in mathematics?
Students start with simple equations and math formulas in elementary school and then progress to solving linear equations and systems of equations in middle school.
In high school, students continue to expand their knowledge by learning more advanced equations such as quadratic equations (including the quadratic formula), exponential equations, and trigonometric equations.
As they move into college level courses such as precalculus and calculus, they will solve rational equations, logarithm equations, parametric equations and differential equations
The next lessons are
Inequalities
Types of graphs
Math formulas
Coordinate plane
Number patterns
Linear equations
Writing linear equations
Solving equations
Rearranging equations
How to find the equation of a line
Identity math (coming soon)
Still stuck?
At Third Space Learning, we specialize in helping teachers and school leaders to provide personalized math support for more of their students through high-quality, online one-on-one math tutoring delivered by subject experts.
Each week, our tutors support thousands of students who are at risk of not meeting their grade-level expectations, and help accelerate their progress and boost their confidence.
Find out how we can help your students achieve success with our math tutoring programs.
Introduction
What are math equations?
Common Core State Standards
How to solve with math equations
Math equations examples
↓
Example 1: volume of a cube Example 2: substitution basic inequality Example 3: substitution equation Example 4: solve for the variable using addition/subtraction Example 5: solve the equation using multiplication/division Example 6: solving equations with fractions - two steps
Teaching tips for math equations
Easy mistakes to make
Practice math equations questions
Math equations FAQs
Next lessons
Still stuck?
x
[FREE] Common Core Practice Tests (3rd to 8th Grade)
Prepare for math tests in your state with these 3rd Grade to 8th Grade practice assessments for Common Core and state equivalents.
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Creatine Kinase Isoenzymes
The document provides an overview of creatine kinase (CK) isoenzymes, detailing their clinical significance, biochemical reactions, and methods of analysis. It discusses the different isoenz…
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Kaplan: Cl inica l C h e m istry, 5
t h
Editi on
C linical Referen ce s - Me thod s of A nalysis C reati ne Ki nase I soe nzy m es
Mauro Panteghini
i
Nam e:
Creatine kinase (CK, adenosine triphosphate: creatine
N
-phosphotransferase)
C linical sig nifi can ce:
click here
Enzy m e n um ber:
EC 2.7.3.2
Molecular m ass:
Varies from 78,500 to 85,100 D; molecular mass of individual subunits is half that of intact molecule
C he m ical class:
Enzyme, protein
K nown isoen zy m es:
CK 1(CK-BB), CK 2(CK-MB), CK 3(CK-MM), CKmt (mitochondrial)
Bioche m ical r eaction:
(See below.)
Refer to Chapter 36, Cardiac and Muscle Disease, in the 5th edition of Clinical Chemistry: Theory, Analysis, Correlation
i
Creatinine Kinase isoenzymes Previous and current authors of this method: First edition: John F. Chapman, Linda L. Woodard, Lawrence M. Silverman Methods edition: John F. Chapman, Linda L. Woodard, Lawrence M. Silverman Second edit ion: John F. Chapman, Linda L. Woodard, Lawrence M. Silver man Third edition: Monica Payne, Peter E. Hickman Fourth edition: Not updated Fifth edition: Mauro Panteghini
Methods of Analysis © 2010 by Lawrence A. Kaplan and Amadeo J. Pesce.
Methods of Analysis © 2010 by Lawrence A. Kaplan and Amadeo J. Pesce.
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Students’ Quick Hyperlink Review
•
Principles of analysis and current usage
•
Reference and preferred methods
•
Specimen
•
Interferences
•
CK isoenzyme reference intervals
•
Interpretation
•
CK isoenzyme performance goals
•
References
•
Tables
Pr incipl e s o f A naly sis an d Cur rent Us ag e
Creatine kinase (CK) is a dimeric enzyme that catalyzes the reversible phosphorylation of creatine by ATP. CK comprises two subunits which are designated “M” and “B”; hence the protein may exist in three forms, CK-MM, CK-MB, and CK-BB, respectively. All three of these isoenzymes are found in the cy tosol of the cell. There is also a fourth isoe nzyme (CKmt) located in the membrane of mitochondria. CK-MM is found predominantly in skeletal muscle; the richest source of CK-MB is myocardial muscle, and CK-BB is found mainly in brain and intestine. For details of relative CK activities and isoenzyme composition of different tissues refer to Table 1. The CK activity in serum of healthy people is due almost exclusively to MM activity (though small amounts of CK-MB may be present) and is the result of physiological turnover of muscle tissue. When electrophoresed, CK-MM runs closest to the cathode, CK-MB has intermediate mobility, and CK-BB moves farthest from the point of ap plication toward the anode. CKmt, which runs more cathodally than the MM fraction, is usually associated with tissue necrosis that accompanies severe anoxic shock and severe liver disease. CK activity may also be found in a macromolecular form—the so-called
macro-CK
. Macro-CK is found, often transiently, in sera of up to 6% of hospitalized patients, but only a small proportion of these have increased CK activities in serum. It exists in two forms, types 1 and 2. Macro-CK type 1 is a complex of CK, typically CK-BB, and an immunoglobulin, often IgG. It often occurs in women older tha n 50. Macro-CK type 2 is oligomeric CKmt found predo minantly in adults who are severely ill with malignancies or in children who have notable tissue distress.Both M and B subunits ha ve a C-terminal lysine residue, but only the former is hydrolyzed by the action of carboxypeptidases present in blood. Carboxypeptidases B (EC 3.4.17.2) or N (arginine carboxypeptidase; EC 3.4.17.3) sequentially hydrolyze the lysine residues from CK-MM to produce two CK-MM isoforms—CK-MM
2
(one lysine residue removed) and CK-MM
1
(both lysine residues removed). The loss of the positively charged lysine produces a more negatively charged CK molecule with greater anodic mobility at electrophoresis. Because CK-MB has only one M subunit, the dimer coded by the M and B genes is named
CK-MB
2
and the
Methods of Analysis © 2010 by Lawrence A. Kaplan and Amadeo J. Pesce.
Methods of Analysis © 2010 by Lawrence A. Kaplan and Amadeo J. Pesce.
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lysine-hydrolyzed dimer is named
CK-MB
1
. The assay of the CK isoforms requires special techniques, such as high-voltage electrophoresis (with gel c ooling), HPLC, chromatofocusing, or immunoassay.For many years, CK isoenzyme an alysis has been used to confirm or refute a diagnosis of acute myocardial infarction. CK-MB content in myocardium is approximately 20% of total myocardial CK,
so damage to myocardium is accompanied by release of CK-MB into the circulating blood in higher proportion. It is now, however, more advantageous to use more cardiac-specific nonenzymatic markers, such as cardiac troponin I or T.Historically, CK-MB has been measured by determining the catalytic activity after separation of the isoenzymes by electrophoresis or ion-exchange chromatography. Subsequently, immunological techniques measuring catalytic activity were introduced. These methods were limited in their use, since interferences with macro-CK or CKmt gave low specificity. Currently, the most common approach is to measure concentrations of the CK-MB protein (“mass”) by using immunoassays with monoclonal antibodies for the MB dimer.
Ref erenc e an d Pr eferr ed M etho ds
At the present time, there is no reference method for CK isoenzyme analysis. The techniques most commonly used are electrophoresis and various immunological methods. Electrophoretic methods are useful for separation of all of the CK isoenzymes. The isoenzyme bands are visualized by incubating the support (e.g., agarose or cellulose acetate) with a concentrated CK assay mixture using the reverse reaction. The NADPH formed in this reaction is then detected by observing the bluish-white fluorescence after excitation by long-wave (360 nm) ultraviolet light. NADPH may be quantified by fluorescence densitometry, which is capable of detecting bands of 2 to 5 U/L. The mobility of CK isoenzymes at pH 8.6 toward the anode is BB > MB > MM, with the MM remaining cathodic to the point of application. The discriminating power of electrophoresis also allows the detection of abnormal CK ba nds (e.g., macro-CK). The disadvantages of electrophoresis include that the turnaround time is relatively long, the procedure is highly labor intensive and not adaptable to clinical chemistry analyzers in emergency situations, and interpretative skills are required. Immunochemical methods are applicable to the direct measurement of CK-MB. In the immunoinhibition technique, an anti-CK-M subunit antiserum is used to inhibit bo th M subunits of CK-MM and the single M subunit of CK-MB and thus allow determination of the en zyme activity of the B subunit of CK-MB, the B subunits of CK-BB, and macro-CKs. To determine CK-MB, this technique assumes the absence of CK-BB (and of the other sources of interference such as macro-CKs) from the tested serum. Because the CK-B subunit accounts for half of the CK-MB activity, the change in absorbance should be doubled to obtain CK-MB activity. This results in a significant decrease of the analytical sensitivity of the method. If present, atypical macro-CK may result in falsely elevated CK-MB results. Owing to its low sensitivity and specificity, the immunoinhibition technique has been largely supplanted by mass assays of CK-MB.
Methods of Analysis © 2010 by Lawrence A. Kaplan and Amadeo J. Pesce.
Methods of Analysis © 2010 by Lawrence A. Kaplan and Amadeo J. Pesce.
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In contrast with immunoinhibition, which measures the CK-MB isoenzyme by determination of its catalytic activity, mass immunoassays measure CK-MB protein concentrations. A number of mass assays using various labels are now commercially available an d are used for routine determination of CK-MB. Measurements use the “sandwich” technique, in which one antibody specifically recognizes only the MB dimer. The sandwich technique ensures that only CK-MB is estimated, because neither CK-MM nor CK-BB reacts with both antibodies. Mass assays are more sensitive than activity-based methods, with a limit of detection for CK-MB usually <1
µ
g/L .Other advantages include sample stability, noninterference with hemolysis, drugs, or other catalytic activity inhibitors, full automation, and fast turnaround time.
Specimen
Fresh serum free from hemolysis is the specimen of choice for analysis of the CK isoenzyme pattern. Of the three commonly seen isoenzymes, CK-BB activity is the least stable. Adding a thiol such as 2-mercaptoethanol to the serum improves its stability.CK-MB activity is not significantly reduced when the separated serum is stored for up to 48 hours at 4°C or 1 month at
−
20°C. Since mass measurement is not subject to the loss of enzyme activity, CK-MB protein concentration in serum is stable for weeks, whether the specimen is stored under refrigeration, and for several years if stored at
−
20°C.
Interferences
EDTA, oxalate/fluoride, and citrated plasma must not be used, since these inhibit CK ac tivity. Because CK-MB mass assays are not subject to the interferences of activity assays, hemolysis and various anticoagulants generally do not interfere.
C K I so en zy m e s Referen ce I nte r v als
With the CK-MB mass assay, the upper reference limit for males is 5.0
µ
g/L, with values for females being less than for males, although many laboratories use a single reference interval (male). Sustained exercise, such as in well trained, long-distance runners, increases the CK-MB content of skeletal muscle, which may produce abnormal serum CK-MB concentrations. To better separate non-myocardial infarction from myocardial infarction patients, the use of a “relative index” (RI) is necessary. The RI relates CK-MB mass concentration in
µ
g/L to measured total CK activity in U/L. Results are e xpressed as a percentage: Physiological:
≤
3% Equivocal: 3 to 5% Consistent with myocardial necrosis: >5% To appropriately use the RI, blood sampling between 8 and 36 hours from symptom onset is necessary.
Methods of Analysis © 2010 by Lawrence A. Kaplan and Amadeo J. Pesce.
Methods of Analysis © 2010 by Lawrence A. Kaplan and Amadeo J. Pesce.
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11962 | https://www.learn-english-today.com/collocations/collocations-adj-preposition.html | Collocations in English: adjectives and prepositions | Learn English Today
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English Vocabulary for learners
ADJECTIVE and PREPOSITION COLLOCATIONS
Adjectives and prepositions that often go together.
Collocations are two or more words that are frequently used together. Such combinations are commonly used by native speakers of English.
Here are some common adjective + preposition collocations.
| ABOUT | FOR | OF | TO | WITH |
--- ---
| angry | blamed | accused | attracted | associated |
| anxious | eligible | afraid | accustomed | bored |
| crazy | famous | ashamed | addicted | blessed |
| doubtful | grateful | aware | allergic | compatible |
| enthusiastic | notorious | capable | indifferent | confronted |
| excited | prepared | conscious | inferior | content |
| furious | punished | envious | married | crowded |
| nervous | qualified | independent | opposed | delighted |
| optimistic | ready | irrespective | polite | familiar |
| pessimistic | renowned | jealous | receptive | fed up |
| sad | responsible | proud | related | friendly |
| serious | respected | scared | rude | pleased |
| upset | suitable | suspicious | similar | popular |
| worried | thankful | typical | susperior | satisfied |
Sentences with adjective-preposition collocations:
adjective + about:
Sam is anxioux about/nervous aboutthe interview.
The kids are enthusiastic aboutgoing to the zoo.
Carla seems to be upset about something.
Are you serious about dropping out of college ?
Stella is worried about walking home alone in the dark.
I was wrong about our new colleague. She’s quite nice actually.
adjective + for:
Charlie was blamed for/ punished forbreaking the window.
Unfortunately you are not eligible fora loan.
We’re grateful(to you) forall your help.
We revised our English to be prepared for the test.
The candidate is not qualified for the job.
You will be heldresponsible forany damage to the equipment.
adjective + of:
Tony was accused ofdisrupting the class.
I’m afraid of/scared ofsnakes and spiders.
Bunjee jumpers should beaware of /conscious ofthe danger.
My mother is veryfond ofgardening.
His parents were very proud of Billy’s success.
I’ve always been suspicious of that guy. He behaves strangely.
adjective + in:
I noticed two people engaged in a heated discussion.
We need to recruit someone experienced in marketing.
He has always beeninterested inclassic English literature.
They are both veryinvolved involunteer work.
The captain is skilled inseamanship and navigation.
The rescuers weresuccessful in their efforts.
adjective + to:
Hugo has always been attracted todangerous sports.
Unfortunately the actor became addicted to drugs.
The organization is dedicated to preserving the environment.
The minister appears to be immune to criticism.
He was totally oblivious to the noise around him.
The old man is opposed tothe idea of moving to an apartment.
Our rival’s new product is similar toours!
adjective + with:
He explained the risksassociated withthe vaccine.
The rescuers were confronted withunforeseen difficulties.
The software is not compatible withmy old computer.
My father is delighted withhis birthday present.
I got lost because I wasn’t familiar with the area.
The teacher said she wasn’t satisfied with my results.
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11963 | https://www.nrc.gov/docs/ml1122/ml11229a666.pdf | Interaction of Charged Particles With Matter 1 3/1/2011 Objectives Qualitative Quantitative Introduction General Force of the Interaction Contents 2 Force of the Interaction Four Types of Charged Particle Interactions Ionization General Ion Pairs Delta Ray Excitation General Bremsstrahlung General Intensity of Bremsstrahlung – Monoenergetic Electrons Intensity of Bremsstrahlung Beta Particles Contents 3 Intensity of Bremsstrahlung – Beta Particles Bremsstrahlung Spectra Cerenkov Radiation General Quantitative Measures of Energy Loss General W Value Specific Ionization Stopping Power and Linear Energy Transfer Mass Stopping Power Contents 4 Alpha Particles General Alpha Tracks Range Range in Air Approximate Data for 5 MeV Alphas Beta Particles General Range Range (as a density thickness) Range and Penetration Bremsstrahlung C k R di ti Contents 5 Cerenkov Radiation Approximate Data for 1 MeV Beta Particles Summary Types of Interactions Alpha Particles Beta Particles References Objectives 6 – Ionization – Excitation Objectives Qualitative 7 – Excitation – Bremsstrahlung – Cerenkov radiation To review the following measures of energy loss: – W Value – Specific Ionization Objectives Quantitative 8 – Specific Ionization – Stopping Power/Linear Energy Transfer – Mass Stopping Power Introduction 9 • “The interaction of charged particles with matter” concerns the transfer of energy from the charged particles to the material through which they travel.
• The “charged particles” considered here are: Introduction General 10 The charged particles considered here are: - Alpha particles (+2 charge) - Beta particles (+ or -1 charge) or electrons • Photons and neutrons, which have no charge, interact very differently.
• Charged particles passing through matter continuously interact with the electrons and nuclei of the surrounding atoms.
• In other words, alpha and beta particles are continually Introduction General 11 slowing down as they travel through matter.
• The interactions involve the electromagnetic forces of attraction or repulsion between the alpha or beta particles and the surrounding electrons and nuclei.
• The force associated with these interactions can be described by Coulomb’s equation: Introduction Force of the Interaction 12 k is a constant = 9 x 109 N-m2/C2.
q1 is the charge on the incident particle in Coulombs.
q2 is the charge on the “struck” particle. r is the distance between the particles in meters.
Things to notice about the equation: • The force increases as the charge increases • The force increases as the distance decreases (it quadruples if the distance is cut in half) Introduction Force of the Interaction 13 q p ) • The force can be positive or negative (attractive or repulsive) Introduction Four Types of Charged Particle Interactions • The four types of interactions are: Ionization (alphas and betas) Excitation (alphas and betas) Bremsstrahlung (primarily betas) 14 Bremsstrahlung (primarily betas) Cerenkov radiation (primarily betas) • Ionization is almost always the primary mechanism of energy loss.
Ionization 15 • A charged particle (alpha or beta particle) exerts sufficient force of attraction or repulsion to completely remove one or more electrons from an atom.
• The energy imparted to the electron must exceed the bi di f th l t Ionization General 16 binding energy of the electron.
• Ionization is most likely to involve atoms near the charged particle's trajectory.
• Each ionization event reduces the charged particle's velocity, i.e., the alpha or beta particles loses kinetic energy.
• Ionization turns a neutral atom into an ion pair.
• The electron stripped away from the atom is the negative member of the ion pair.
It is known as a secondary electron.
Ionization Ion Pairs y The secondary electron has some, but not much, kinetic energy - usually less than 100 eV.
Sometimes it has enough energy to ionize additional atoms. Then it is referred to as a delta ray.
• The atom , now with a vacancy in one of its electron shells, is the positive member of the ion pair. e-e-e-e-e-+ + + + + + + e-18 e-+ e-e-e-e-e-+ + + + + + + e-19 e-e-+ e-e-e-e-e-e-+ + + + + + + e-20 e-+ Negative member of the ion pair (secondary electron) e-e-e-e-e-e-+ + + + + + + e-21 This is an ion pair Positive member of the ion pair (e.g., N2 +) + N2 alpha N2 N2 N2 + O2 -22 alpha N2 2 2 e-Ion pair + + + + + + + + + + + + + + + + ---------------alpha Ionization Delta Ray 23 e-delta ray A delta ray is a secondary electron (negative member of an ion pair) that has sufficient kinetic energy to cause additional ionization.
Excitation 24 • The charged particle (alpha or beta particle) exerts just enough force to promote one of the atom’s electrons to a higher energy state (shell).
Insufficient energy was transferred to ionize the atom. E it ti ll f th f th Excitation General 25 • Excitation usually occurs farther away from the charged particle's trajectory than ionization.
• The excited atom will de-excite and emit a low energy ultraviolet photon.
• Each excitation event reduces the charged particle's velocity.
e-e-e-e--+ + + + + + e-26 e-e-e-e-e-e--+ + + + + + 27 alpha e-e-e-e-e-+ + + + + + e-28 alpha e-e-e-e-e-+ + + + + + e-29 e-e-e-e-e-e + + + + + + e-30 e -Ultra violet photon Bremsstrahlung 31 • Bremsstrahlung radiation is electromagnetic radiation that is produced when charged particles are deflected (decelerated) while traveling near an atomic nucleus.
• Bremsstrahlung is almost exclusively associated with l t (b t ti l ) b th l tt il Bremsstrahlung General 32 electrons (beta particles) because the latter are easily deflected.
• Large particles (e.g., alpha particles) do not produce significant bremsstrahlung because they travel in straight lines. Since they aren’t deflected to any real extent, bremsstrahlung production is inconsequential.
• Bremsstrahlung photons may have any energy up to the energy of the incident particle.
For example, the bremsstrahlung photons produced by P-32 betas have a range of energies up to 1.7 MeV, the maximum energy of the P-32 alphas.
Bremsstrahlung General 33 • Bremmstrahlung is most intense when: - The beta particles or electrons have high energies - The material has a high atomic number + + + l 34 + + + + + + nucleus + + + e-l 35 + + + + + + nucleus + + + l bremsstrahlung 36 + + + + + + e-nucleus + + + l bremsstrahlung 37 + + + + + + e-nucleus + + + l 38 + + + + + + nucleus + + + l 39 + + + + + + nucleus + + + e-l bremsstrahlung 40 + + + + + + nucleus The greater the charge in the nucleus (atomic number), the greater the deflection of the electrons and the greater the intensity of the bremsstrahlung • According to Evans, the fraction of the energy of monoenergetic electrons that is converted to bremsstrahlung (f) can be calculated as follows Bremsstrahlung Intensity of Bremsstrahlung – Monoenergetic Electrons 41 Z is the atomic number of the material E is the kinetic energy of the electron (MeV) • Turner gives slightly different equation for the fraction of the energy of monoenergetic electrons that is converted to bremsstrahlung: Bremsstrahlung Intensity of Bremsstrahlung – Monoenergetic Electrons 42 Z is the atomic number of the material E is the kinetic energy of the electrons (MeV) • The following equation (Evans) estimates the fraction of beta particle energy converted to bremsstrahlung (f). Beta particles are emitted with a range of energies up to some maximum value (Emax).
Bremsstrahlung Intensity of Bremsstrahlung – Beta Particles 43 Z is the atomic number of the material Emax is the maximum energy of the beta particles (MeV) • The beta energy rate (MeV/s) is the activity of the beta emitter multiplied by the average energy of the beta particles: Beta energy rate = Activity x Average beta energy Bremsstrahlung Intensity of Bremsstrahlung – Beta Particles 44 (MeV/s) (dps) (MeV) • This is multiplied by the fraction (f) to determine the bremsstrahlung energy emission rate in MeV/s.
Bremsstrahlung energy rate = Beta energy rate x f (MeV/s) (MeV/s) • The following discussion tries to explain the shape of a bremsstrahlung spectrum (e.g., that produced in the target of an x-ray tube) • Bremsstrahlung photons have a range of energies up to th i f th l t /b t ti l Bremsstrahlung Bremsstrahlung Spectra 45 the maximum energy of the electrons/beta particles.
• When monoenergetic electrons lose energy in an extremely thin target, the bremsstrahlung spectrum is flat up to the maximum energy of the electrons.
Imaginary targets like this are not found in the real world.
Bremsstrahlung spectrum for monoenergetic electrons and an imaginary thin target Bremsstrahlung Bremsstrahlung Spectra 46 imaginary thin target Photon energy Number of photons Kinetic energy of the electrons or beta particle • Monoenergetic electrons losing energy in a thick (real world) target can be considered to interact in a series of thin sections (targets). • The deeper into the target a given section is, the lower the energy of the electrons and the lower the maximum Bremsstrahlung Bremsstrahlung Spectra 47 the energy of the electrons, and the lower the maximum energy of the bremsstrahlung produced there. The bremsstrahlung produced in the deeper sections by the lower energy electrons contributes to the low energy end of the overall bremsstrahlung spectrum: • Bremsstrahlung produced in the shallow sections of the target where the electron energies are higher contributes to the high energy portion of the spectrum.
Target consisting of six thin sections e-Bremsstrahlung Spectrum Bremsstrahlung Bremsstrahlung Spectra Lower energy e-1 2 3 4 5 6 48 Photon energy Spectrum Higher energy bremsstrahlung Lower energy bremsstrahlung e-e-e-e-1 2 3 4 5 6 • As a result, the bremsstrahlung spectrum produced with a real world (thick) target looks something like this: Bremsstrahlung Bremsstrahlung Spectra 49 Bremsstrahlung photon energy Number of photons • There is always some shielding/filtration between the source of the bremsstrahlung and the point of interest.
• For example, the glass wall of an x-ray tube will shield the bremsstrahlung generated in the target (anode) as ld filt i t ti ll l d i f t f th t b Bremsstrahlung Bremsstrahlung Spectra 50 would a filter intentionally placed in front of the tube.
• This shielding primarily reduces the intensity of the low energy bremsstrahlung.
• As such, a “real world” bremsstrahlung spectrum looks more like that on the next slide.
Bremsstrahlung spectrum with filtration Number of photons Bremsstrahlung Bremsstrahlung Spectra 51 Bremsstrahlung photon energy photons Cerenkov Radiation 52 • Cerenkov radiation is the blue light emitted by charged particles that travel through a transparent medium (e.g., water) faster than the speed of light in that medium.
J t l i f t th d d Cerenkov Radiation Production General 53 • Just as a plane going faster than sound produces a cone of sound (a sonic boom), a charged particle going faster than light produces a cone of light (Cerenkov radiation) • The production of Cerenkov radiation is essentially limited to high energy (i.e., fast) beta particles and electrons.
• Cerenkov radiation is often associated with reactor fuel pools or nuclear criticality accidents.
• It is possible to quantify beta emitters by measuring the intensity of their Cerenkov radiation (Cerenkov Cerenkov Radiation Production General 54 counting) Quantitative Measures of Energy Loss 55 of Energy Loss The four most common measures of energy loss by charged particles: 1. W Value 2 S ifi I i ti Quantitative Measures of Energy Loss General 56 2. Specific Ionization 3. Stopping power or Linear Energy Transfer 4. Mass Stopping Power • W is the average energy lost by a charged particle per ion pair produced.
• It depends on: the type of charged particle material that is being ionized Quantitative Measures of Energy Loss W Value 57 material that is being ionized • W doesn’t change much with the energy of the particle, but it does increase at low energies (< 0.2 MeV) for protons and alpha particles.
• Beta particles lose an average of 34 eV per ion pair produced in air.
• Alpha particles lose an average of 36 eV per ion pair produced in air.
Quantitative Measures of Energy Loss W Value 58 • Alpha particles and beta particles (or electrons) lose an average of approximately 22 eV per ion pair produced in water (Turner p. 140, 161) • Specific ionization is the average number of ion pairs produced per unit distance traveled in a material by a charged particle. • It depends on: the type of charged particle, Quantitative Measures of Energy Loss Specific Ionization 59 the energy of the charged particle the material through which it travels.
• Alpha particles produce 20,000 to 60,000 ion pairs per centimeter (cm) in air.
• Beta particles might produce 100 ion pairs per cm in air.
• There is no practical difference between stopping power and linear energy transfer.
• The stopping power or the linear energy transfer is the average energy lost by a charged particle per unit Quantitative Measures of Energy Loss Stopping Power and Linear Energy Transfer 60 distance traveled.
• Typical units: MeV/cm or eV/um • When a distinction is made: Stopping power is used to describe the total energy lost by the charged particle.
Linear energy transfer (LET) is used to describe the Quantitative Measures of Energy Loss Stopping Power and Linear Energy Transfer 61 gy ( ) energy lost by the charged particle that is locally absorbed in the material the particle is traveling through.
• In this sense, stopping power is akin to kerma while LET is akin to absorbed dose • LET is sometimes referred to as the restricted stopping power.
It describes the energy lost by charged particles in low energy interactions. Quantitative Measures of Energy Loss Stopping Power and Linear Energy Transfer 62 The assumption is that the secondary electrons produced in these low energy interactions don’t travel outside the volume of interest and deposit their energy locally. This would exclude interactions producing delta rays or bremsstrahlung.
• The maximum energy that can be transferred in these interactions is sometimes indicated with a subscript, e.g., LET1 keV, LET5 keV • The greater the energy cutoff, the larger the LET, Quantitative Measures of Energy Loss Stopping Power and Linear Energy Transfer 63 g gy , g , e.g., LET5 keV > LET1 keV • If no restriction is placed on the energy of the interactions, the unrestricted LET is indicated as LET∞ • LET∞ is the same as stopping power.
• The mass stopping power can be more convenient to use than the stopping power. • It is the stopping power (e.g., MeV/cm) divided by the density (g/cm3) of the material.
Quantitative Measures of Energy Loss Mass Stopping Power 64 • The units of the mass stopping power are usually MeV cm2 g-1 (MeV per g/cm2) • It is the average energy lost by a charged particle per unit distance traveled where the distance is expressed as an aerial density (g/cm2) Alpha Particles 65 • The principal types of interactions for alpha particles are: - Ionization - Excitation Alpha Particles General 66 • Usually have energies from 4 to 8 MeV • High specific ionization (because of their +2 charge and low velocity) • High LET radiation - lose their energy very quickly as they travel through matter. Alpha Particles General • Easy to shield – can be stopped by a piece of paper • Not an external hazard – cannot penetrate the dead layer of skin on the surface of the body 67 • Potential internal hazard – the large radiation weighting factor for alpha particles (20) means that the consequence of a given alpha particle dose is greater than that for other types of radiation.
Alpha Particles Alpha Tracks 68 Alpha particle tracks are short and straight.
From Turner, James. Atoms, Radiation, and Radiation Protection, 1st edition. 1986, pg. 74.
• The range of an alpha particle is short: - approximately 5 cm in air. - 20 to 70 um in tissue (one, two or three cells) Alpha Particles Range 69 • The survey instrument must be close (e.g., < 1 cm) to a contaminated surface if alpha emitting radionuclides are to be detected. It is best if the contaminated surface is dry and clean -dust or moisture could attenuate the alphas. • Alphas with energies of 4 to 8 MeV (almost all alpha emitters): R (cm) = 1.24 E - 2.62 Alpha Particles Range in Air 70 • Alphas with energies below 4 MeV: R (cm) = 0.56 E E is the alpha energy in MeV Air (D=0.001293 g/cm3) Water (D=1 g/cm3) W (eV/ip) 36 22 Stopping Power/LET (M V/ ) 1.23 950 Alpha Particles Approximate Data for 5 MeV Alphas 71 (MeV/cm) Mass Stopping Power (MeV cm2 g-1) 950 950 Specific Ionization (Ion pairs per cm) 34,000 4.3 x 107 Range (g/cm2) 5 x 10-3 3 x 10-3 Range (cm) 4 3 x 10-3 (30 um) Beta Particles 72 • Beta particles (or electrons) interact by all of the following mechanisms: – Ionization – Excitation Beta Particles General 73 Excitation – Bremsstrahlung – Cerenkov radiation (relatively unimportant) For betas above 150 eV, roughly 95% of the particle’s energy loss in water is due to ionization. • Not as intensely ionizing as alphas (because they have higher velocities and one half the charge). • Low specific ionization (ca. 100 ion pairs per cm in air) Beta Particles General 74 • Low stopping powers (low LET radiation) • Betas might produce (the specific ionization) in air.
• Much greater range than alphas (except for the lowest energy betas): - Approximately 3 meters in air for a 1 MeV beta - A few millimeters in tissue (water) Beta Particles Range 75 A few millimeters in tissue (water) • The atomic number of the material is not a major factor. In fact, the range of beta particles under 20 MeV is greater in lead than in water!
• The next slide shows two empirical equations relating the range of a beta particle to its energy.
• The range of a beta particle can be determined if the energy is known: Beta Particles Range (as a density thickness) 76 • The energy of a beta particle can be determined if the range is known: R is the range in mg/cm2 E is the maximum beta energy in MeV • The easiest way to determine the range of a beta particle is to use a curve similar to that on the next slide. A more readable version can be found on page 163 of PTP’s Rad Health Handbook.
Beta Particles Range (as a density thickness) 77 78 • Beta particles (and electrons) travel in convoluted paths. • They do not travel in a straight line.
• The “range” of a beta particle usually refers to the total th l th Beta Particles Range and Penetration 79 path length.
• The range is greater than the distance between the beginning and the end of the path followed by the particle (the penetration thickness).
In other words, the range of a beta particle is greater than the thickness of a material that can be penetrated.
• The following slide shows the predicted paths of 800 keV beta particles in water.
The average penetration thickness: 1500 um The average range (path length): 3500 um.
Beta Particles Range and Penetration 80 Beta Particles Range and Penetration 81 Beta particle tracks are convoluted.
From Turner, James. Atoms, Radiation, and Radiation Protection, 2nd edition. 1995, pg. 151.
Beta Particles Range and Penetration 82 From Turner, James. Atoms, Radiation, and Radiation Protection, 2nd edition. 1995, pg. 151.
• Bremsstrahlung is most significant for high energy beta emitters such as P-32 and Sr-90. • The presence of bremsstrahlung is often interpreted as a indication that high energy beta emitters are present.
Beta Particles Bremsstrahlung 83 • Nevertheless, bremsstrahlung can be detected when low energy beta emitters (e.g., tritium) are present in high enough activities (e.g., a tritium exit sign).
• To minimize the production of unwanted bremsstrahlung, beta sources should be shielded with a low t i b t i l Beta Particles Bremsstrahlung 84 atomic number material.
For example, high energy beta emitters are commonly shielded with plastic.
• Shielding a high energy beta source with lead could increase the production of bremsstrahlung.
Nevertheless, if the lead is thick enough, it will also stop the bremsstrahlung.
Beta Particles Bremsstrahlung 85 • Sometimes a beta shield has two layers: - plastic nearest the source to stop any betas - lead outside the plastic to stop any bremsstrahlung.
• The emission of Cerenkov radiation is an interesting, but relatively unimportant, type of beta particle (or electron) interaction.
• Cerenkov radiation is the blue glow that often seen in a Beta Particles Cerenkov Radiation 86 reactor’s spent fuel pool. • The Cerenkov radiation primarily is due to the high energy Compton scattered electrons produced by gamma emissions from the fuel.
Fuel assemblies being removed from the reactor vessel at TMI’s operating unit Beta Particles Cerenkov Radiation 87 operating unit.
Air (D=0.001293 g/cm3) Water (D=1 g/cm3) W (eV/ip) 34 22 Stopping Power/LET (M V/ ) 3.3 x 10-3 1.89 Beta Particles Approximate Data for 1 MeV Beta Particles 88 (MeV/cm) 3.3 x 10 1.89 Mass Stopping Power (MeV cm2 g-1) 2.6 1.89 Specific Ionization (Ion pairs per cm) 100 86,000 Range (g/cm2) 0.4 0.5 Range (cm) 300 0.5 Summary 89 • Charged particles continuously interact as they travel through matter - it is not a matter of probability. • The major type of interactions: ionization Summary Types of Interactions 90 • The other types of interactions: excitation bremsstrahlung Cerenkov radiation • Bremsstrahlung production is sometimes an important concern with beta particles.
• Cerenkov radiation is interesting but rarely important.
• High specific ionization • High LET (aka stopping power) • Travel in straight lines Summary Alpha Particles 91 • Short range: a few cm in air a couple of cells in the body • Potential internal hazard but not an external hazard • Low specific ionization • Low LET radiation (i.e., low stopping power) • Convoluted path Summary Beta Particles 92 • Large range: a few hundred cm in air several mm in the body • Penetration distance is less than the range • Produce bremsstrahlung photons when they change direction. • Maximum energy of the bremsstrahung photons is the same as the maximum energy of the beta particles.
Summary Beta Particles 93 • The higher the atomic number of the material, the greater the fraction of the beta particle energy that will be emitted as bremsstrahlung • The higher the energy of the beta particle (or electron) the greater the fraction of the energy that will be emitted as bremsstrahlung.
• Bremsstrahlung complicates radiation protection, sample counting, shielding, and dosimetry. • Bremsstrahlung production can be minimized by shielding beta sources with a low Z material such as plastic. Summary Beta Particles 94 References 95 • Evans, R. The Atomic Nucleus. McGraw-Hill. 1955.
• Turner, James. Atoms, Radiation, and Radiation Protection, 2nd edition. John Wiley and Sons, Inc., 1995.
References 96 |
11964 | https://www.youtube.com/watch?v=j-7B_RtFbGE | Where is the Pi key on a Casio Scientific calculator?
Maths Mark
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Posted: 8 Nov 2018
On a Casio Scientific calculator the Pi key is located at the bottom of the calculator. It’s the second function on times 10 to the power of x key. You will need to first press the shift key and then the times 10 to the power of x key. If you want to see the value of Pi in decimal form press the SD key and then hit equals.
118 comments
Transcript:
hi today I'm going to show you how to get PI on your scientific calculator the button that you'll be looking for is the times 10 to power of XK and it is the second function on that K so you'll find this K at the bottom of the calculator here so to get pilot then you just need to do shift and press the x 10k and you'll see PI on your calculator okay if you just press the equals and then the SDK you can see that pi is the correct value which is 3.14159265 4 so let's do that one more time then so to get PI then you just go shift times 10 thanks for watching |
11965 | https://www.iswkoman.com/uploads/work-sheet/1210815-Class%20XI_WS_t-FUNCTIONS%20Conversions_29-08-2022_SANJEET.pdf | Worksheet by Kuta Software LLC Name:______Date:_____PD:__ Converting Between Radians and Degrees Convert each degree measure into radians. 1) 290° 2) 345° 3) 970° 4) 510° 5) 510° 6) 150° 7) 210° 8) 240° 9) 240° 10) 600° 11) 945° 12) 675° 13) 315° 14) 570° 15) 520° 16) 40° 17) 300° 18) 0° 19) 555° 20) 165° Convert each radian measure into degrees. π 25π 21) 22) 18 12 35π 41π 23) 24) 18 36 3π 107π 25) 26) 2 36 π 17π 27) 28) 3 9 11π 41π 29) 30) 3 12 Worksheet by Kuta Software LLC Convert each degree measure into radians and each radian measure into degrees. π 23π 31) 32) 6 6 33) 30° 34) 930° 35) -210° π 36) 4 Worksheet by Kuta Software LLC The Unit Circle Name__ _____ Degrees & Radians Conversion Practice Date__ Convert each degree measure into radians. 29π 23π 1) -290° − 2) 345° 18 12 97π 17π 3) 970° 4) -510° − 18 6 17π 5π 5) 510° 6) 150° 6 6 7π 4π 7) 210° 8) -240° − 6 3 4π 10π 9) 240° 10) 600° 3 3 21π 15π 11) -945° − 12) 675° 4 4 7π 19π 13) 315° 14) 570° 4 6 26π 2π 15) -520° − 16) 40° 9 9 5π 17) 300° 18) 0° 3 0 37π 11π 19) 555° 20) 165° 12 12 Worksheet by Kuta Software LLC Convert each radian measure into degrees. π 25π 21) 22) − 18 12 10° -375° 35π 41π 23) 24) 18 36 350° 205° 3π 107π 25) − 26) 2 36 -270° 535° π 17π 27) 28) − 3 9 60° -340° 11π 41π 29) − 30) − 3 12 -660° -615° 14π 16π 31) 32) − 3 3 840° -960° ©B tKJuUtxaY ySdoMfGtmwra3ruev FLcLaCc.x O gAnlJlD crribgPhntesM orjeOsGe2rUvOeIdB.E p TM0aMdyef ywkiWtohY rIJnmfeiRn4iTtvej 1AklYg5eQb3rka2 b26.S -1- 21π 13π 33) 34) − 4 4 945° -585° 7π 11π 35) 36) 4 6 315° 330° 13π 7π 37) 38) 6 3 390° 420° π 3π Worksheet by Kuta Software LLC 39) − 40) 3 4 -60° 135° Convert each degree measure into radians and each radian measure into degrees. π 23π 41) − 42) − 6 6 -30° -690° π 31π 43) -30° − 44) -930° − 6 6 7π 45) -210° − π 6 46) 4 45° 8π 47) -160° − π 9 48) − 3 -60° 11π 17π 49) 50) 6 12 330° 255° 61π 51) 915° π 12 52) 2 90° 7π 53) -105° − 4π 12 54) 9 80° 7π 31π 55) 56) 2 9 630° 620° 23π 57) 230° 13π 18 58) − 6 -390° 17π 11π Worksheet by Kuta Software LLC 59) -170° − 60) 660° 18 3 ©P JKru3t4aU 6SzoQfwtZw5a4rzeF 2LIL4CE.A A AAYlKlm UrrimgEhZtCsx vrtecsveYrhvdeMdX.Q K yMfaBdwev 2wdiOtxhW nIDn4fzieniiQtceU QAxlvgWembYrRaL r2U.R -2- |
11966 | https://www.ahajournals.org/doi/10.1161/CIRCULATIONAHA.105.166561 | Part 8: Stabilization of the Patient With Acute Coronary Syndromes | Circulation
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Originally Published 28 November 2005
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Part 8: Stabilization of the Patient With Acute Coronary Syndromes
Circulation
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Contents
Out-of-Hospital Management
ED Evaluation and Risk Stratification (Figure 1, Boxes 3 and 4)
Initial General Therapy for ACS
Reperfusion Therapies (Figure 1, Box 8)
Complicated AMI
Adjunctive Therapies for ACS and AMI
Management of Arrhythmias
Summary
Footnote
References
eLetters
Information & Authors
Metrics & Citations
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References
Figures
Tables
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Acute myocardial infarction (AMI) and unstable angina (UA) are part of a spectrum of clinical disease collectively identified as acute coronary syndromes (ACS). The pathophysiology common to this spectrum of disease is a ruptured or eroded atheromatous plaque.1–5 The electrocardiographic (ECG) presentation of these syndromes encompasses ST-segment elevation myocardial infarction (STEMI), ST-segment depression, and nondiagnostic ST-segment and T-wave abnormalities. A non–ST-elevation myocardial infarction (NSTEMI) is diagnosed if cardiac markers are positive with ST-segment depression or with nonspecific or normal ECGs. Sudden cardiac death may occur with any of these conditions. ACS is the most common proximate cause of sudden cardiac death.6–10
Effective interventions for patients with ACS, particularly STEMI, are extremely time-sensitive. The first healthcare providers to encounter the ACS patient can have a big impact on patient outcome if they provide efficient risk stratification, initial stabilization, and referral for cardiology care. It is critical that basic life support (BLS) and advanced cardiovascular life support (ACLS) healthcare providers who care for ACS patients in the out-of-hospital, emergency department (ED), and hospital environments be aware of the principles and priorities of assessment and stabilization of these patients.
These guidelines target BLS and ACLS healthcare providers who treat patients with ACS within the first hours after onset of symptoms, summarizing key out-of-hospital, ED, and some initial critical-care topics that are relevant to stabilization. They also continue to build on recommendations from the ACC/AHA Guidelines,11,12 which are used throughout the United States and Canada.13 As with any medical guidelines, these general recommendations must be considered within the context of local resources and application to individual patients by knowledgeable healthcare providers.
The primary goals of therapy for patients with ACS are to
•
Reduce the amount of myocardial necrosis that occurs in patients with MI, preserving left ventricular (LV) function and preventing heart failure
•
Prevent major adverse cardiac events (MACE): death, nonfatal MI, and need for urgent revascularization
•
Treat acute, life-threatening complications of ACS, such as ventricular fibrillation (VF)/pulseless ventricular tachycardia (VT), symptomatic bradycardias, and unstable tachycardias (see Part 7.2: “Management of Cardiac Arrest” and Part 7.3: “Management of Symptomatic Bradycardia and Tachycardia”)
An overview of recommended care for the ACS patient is illustrated in Figure 1, the Acute Coronary Syndromes Algorithm. Part 8 provides details of the care highlighted in the numbered algorithm boxes. Box numbers in the text correspond to the numbered boxes in the algorithm.
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Figure 1.Acute Coronary Syndromes Algorithm.
In this part the abbreviation AMI refers to acute myocardial infarction, whether associated with STEMI or NSTEMI. The diagnosis and treatment of AMI, however, will often differ for patients with STEMI versus NSTEMI. Note carefully which is being discussed.
Out-of-Hospital Management
Recognition (Figure 1, Box 1)
Treatment offers the greatest potential benefit for myocardial salvage in the first hours of STEMI. Thus, it is imperative that healthcare providers evaluate, triage, and treat patients with ACS as quickly as possible. Delays to therapy occur during 3 intervals: from onset of symptoms to patient recognition, during out-of-hospital transport, and during in-hospital evaluation. Patient delay to symptom recognition often constitutes the longest period of delay to treatment.14
The classic symptom associated with ACS is chest discomfort, but symptoms may also include discomfort in other areas of the upper body, shortness of breath, sweating, nausea, and lightheadedness. The symptoms of AMI are characteristically more intense than angina and last >15 minutes. Atypical symptoms or unusual presentations of ACS are more common in elderly, female, and diabetic patients.15–19
Public education campaigns increase public awareness and knowledge of the symptoms of heart attack but have only transient effects.20 For patients at risk for ACS (and for their families), physicians should discuss the appropriate use of nitroglycerin and aspirin, activation of the emergency medical services (EMS) system, and location of the nearest hospital that offers 24-hour emergency cardiovascular care. Recent ACC/AHA guidelines recommend that the patient or family members activate the EMS system rather than call their physician or drive to the hospital if chest discomfort is unimproved or worsening 5 minutes after taking 1 nitroglycerin tablet or using nitroglycerin spray.12
Initial EMS Care (Figure 1, Box 2)
Half of the patients who die of AMI do so before reaching the hospital. VF or pulseless VT is the precipitating rhythm in most of these deaths,21–23 and it is most likely to develop during the first 4 hours after onset of symptoms.24–27 Communities should develop programs to respond to out-of-hospital cardiac arrest that include prompt recognition of symptoms of ACS, early activation of the EMS system, and if needed, early CPR (see Part 4: “Adult Basic Life Support”) and early access to an automated external defibrillator (AED) through community AED programs (see Part 5: “Electrical Therapies”).28 EMS and dispatch system personnel should be trained to respond to cardiovascular emergencies.
Dispatchers and EMS providers must be trained to recognize symptoms of ACS. Dispatchers should advise patients with no history of aspirin allergy or signs of active or recent gastrointestinal bleeding to chew an aspirin (160 to 325 mg) while awaiting the arrival of EMS providers (Class IIa).29
EMS providers should be trained to determine the time of onset of symptoms and to stabilize, triage, and transport the patient to an appropriate facility and to provide prearrival notification. EMS providers should monitor vital signs and cardiac rhythm and be prepared to provide CPR and defibrillation if needed.
EMS providers may administer oxygen to all patients. If the patient is hypoxemic, providers should titrate therapy based on monitoring of oxyhemoglobin saturation (Class I).30–44 If the patient has not taken aspirin and has no history of aspirin allergy and no evidence of recent gastrointestinal bleeding, EMS providers should give the patient nonenteric aspirin (160 to 325 mg) to chew (Class I).45–48
EMS providers should administer up to 3 nitroglycerin tablets (or spray) for ongoing symptoms at intervals of 3 to 5 minutes if permitted by medical control and if the patient remains hemodynamically stable (systolic blood pressure [SBP] >90 mm Hg [or no more than 30 mm Hg below baseline], heart rate between 50 and 100 beats per minute [bpm]).49,50 EMS providers can administer morphine for chest pain unresponsive to nitroglycerin if authorized by protocol or medical control. Additional information about out-of-hospital stabilization and care is included in the following sections.
Out-of-Hospital ECGs
Out-of-hospital 12-lead ECGs and advance notification to the receiving facility speed the diagnosis, shorten the time to fibrinolysis, and may be associated with decreased mortality rates.51–64 The reduction in door-to–reperfusion therapy interval in most studies ranges from 10 to 60 minutes. EMS providers can efficiently acquire and transmit diagnostic-quality ECGs to the ED53,58,65,66 with a minimal increase (0.2 to 5.6 minutes) in the on-scene time interval.52,56,65–68
Qualified and specially trained paramedics and prehospital nurses can accurately identify typical ST-segment elevation (>1 mm in 2 or more contiguous leads) in the 12-lead ECG with specificity ranging from 91% to 100% and sensitivity ranging from 71% to 97% when compared with emergency medicine physicians or cardiologists.69,70 Using radio or cell phone, they can also provide advance notification to the receiving hospital of the arrival of a patient with ACS.56,61–64
We recommend implementation of out-of-hospital 12-lead ECG diagnostic programs in urban and suburban EMS systems (Class I). Routine use of 12-lead out-of-hospital ECG and advance notification is recommended for patients with signs and symptoms of ACS (Class IIa). A 12-lead out-of-hospital ECG with advance notification to the ED may be beneficial for STEMI patients by reducing time to reperfusion therapy. We recommend that out-of-hospital paramedics acquire and transmit either diagnostic-quality ECGs or their interpretation of them to the receiving hospital with advance notification of the arrival of a patient with ACS (Class IIa). If EMS providers identify STEMI on the ECG, it is reasonable for them to begin to complete a fibrinolytic checklist (Figure 2).
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Figure 2.Fibrinolytic Checklist.
Out-of-Hospital Fibrinolysis
Clinical trials have shown the benefit of initiating fibrinolysis as soon as possible after onset of ischemic-type chest pain in patients with STEMI or new or presumably new left bundle branch block (LBBB).67,71 Several prospective studies (LOE 1)72–74 have documented reduced time to administration of fibrinolytics and decreased mortality rates when out-of-hospital fibrinolytics were administered to patients with STEMI and no contraindications to fibrinolytics.
Physicians in the Grampian Region Early Anistreplase Trial (GREAT)73 administered fibrinolytic therapy to patients at home 130 minutes earlier than to patients at the hospital and noted a 50% reduction in hospital mortality rates and greater 1-year and 5-year survival rates in those treated earlier.75,76 Delaying fibrinolytic treatment by 1 hour increased the hazard ratio of death by 20%, which is equivalent to the loss of 43 lives per 1000 patients over 5 years.
A meta-analysis of out-of-hospital fibrinolytic trials found a relative improvement of 17% in outcome associated with out-of-hospital fibrinolytic therapy, particularly when therapy was initiated 60 to 90 minutes earlier than in the hospital.71 A meta-analysis of 6 trials involving 6434 patients (LOE 1)72 documented decreased all-cause hospital mortality rates among patients treated with out-of-hospital fibrinolysis compared with in-hospital fibrinolysis (odds ratio [OR]: 0.83; 95% confidence interval [CI]: 0.70 to 0.98) with a number needed to treat of 62 to save 1 extra life with out-of-hospital fibrinolysis. Results were similar regardless of the training and experience of the provider.
The ECC Guidelines 200077 recommended consideration of out-of-hospital fibrinolysis for patients with a transport time >1 hour. But in a recent Swiss study (LOE 1),74 prehospital administration of fibrinolytics significantly decreased the time to drug administration even in an urban setting with relatively short transport intervals (<15 minutes).74
In summary, out-of-hospital administration of fibrinolytics to patients with STEMI with no contraindications is safe, feasible, and reasonable (Class IIa). This intervention may be performed by trained paramedics, nurses, and physicians for patients with symptom duration of 30 minutes to 6 hours. System requirements include protocols with fibrinolytic checklists, ECG acquisition and interpretation, experience in ACLS, the ability to communicate with the receiving institution, and a medical director with training/experience in management of STEMI. A process of continuous quality improvement is required. Given the operational challenges required to provide out-of-hospital fibrinolytics, most EMS systems should focus on early diagnosis with 12-lead ECG, rapid transport, and advance notification of the ED (verbal interpretation or direct transmission of ECG) instead of out-of-hospital delivery of fibrinolysis.
Triage and Transfer
Out-of-Hospital Triage
Hospital and EMS protocols should clearly identify criteria for transfer of patients to specialty centers and conditions under which fibrinolytics should be initiated before transfer. When transfer is indicated, the ACC/AHA guidelines recommend a door-to-departure time ≤30 minutes.12 It may be appropriate for the EMS medical director to institute a policy of out-of-hospital bypass of hospitals that provide medical therapy only, particularly for patients who require interventional therapy. Patients who require interventional therapy may include those with cardiogenic shock, pulmonary edema, large infarctions, and contraindications to fibrinolytic therapy.
At present no randomized studies have directly compared triage with an experienced percutaneous coronary intervention (PCI) center with medical management at the local hospital. Extrapolation from several randomized trials on interfacility transfer78–80 suggests that STEMI patients triaged directly to a primary PCI facility may have better outcomes related to the potential for earlier treatment. A cost-efficacy substudy of the Comparison of Angioplasty and Prehospital Thrombolysis in Acute Myocardial Infarction (CAPTIM) trial81 suggests that direct transport to a primary PCI facility may be more cost-effective than out-of-hospital fibrinolysis when transport can be completed in ≤60 minutes with a physician in a mobile intensive care unit. There is no direct evidence, however, to suggest that these strategies are safe or effective. Patients judged to be at highest risk for a complicated transfer were excluded from some of these studies.
In summary, at this time there is inadequate evidence to recommend out-of-hospital triage to bypass non–PCI-capable hospitals to bring patients to a PCI center (Class Indeterminate). Local protocols for EMS providers are appropriate to guide the destination of patients with suspected or confirmed STEMI.
Interfacility Transfer
All patients with STEMI and symptom duration of ≤12 hours are candidates for reperfusion therapy with either fibrinolysis or PCI (Class I). When patients present directly to a facility capable of providing only fibrinolysis, 3 treatment options are available: administering fibrinolytics with admission to that hospital, transferring the patient for primary PCI, or giving fibrinolytics and then transferring the patient to a specialized center. The decision is guided by a risk-benefit assessment that includes evaluation of duration of symptoms, complications, contraindications, and the time delay from patient contact to fibrinolysis versus potential delay to PCI balloon inflation.
In 2 prospective studies (LOE 2)78–80 and a meta-analysis,82 patients with STEMI who presented 3 to 12 hours after onset of symptoms to a hospital without capability for primary PCI had better outcome (improved 30-day combined incidence of death, reinfarction, or stroke) when they were transferred to a skilled PCI center (interventionalist performing >75 procedures per year) rather than receiving fibrinolytics at the presenting hospital. In these studies balloon inflation occurred ≤93 minutes after decision to treat.80,83–85 Thus, interfacility transfer is indicated for patients with STEMI presenting >3 hours from onset of symptoms from hospitals that lack primary PCI capability to centers capable of providing primary PCI when the transfer can be accomplished as soon as possible. The ACC/AHA guidelines recommend a treatment delay of no more than 90 minutes.12 In patients with STEMI presenting <3 hours from onset of symptoms, the superiority of immediate administration of fibrinolytics in the hospital or transfer for primary PCI is not established (Class Indeterminate).
In-Hospital Fibrinolytics and Interfacility Transfer for PCI
Data from the 1980s to 1990s did not support a strategy of fibrinolytic therapy combined with transfer for facilitated PCI (LOE 186–88 and meta-analyses89–91). But all of the studies involved in-hospital administration of fibrinolytics, and most were completed before the era of coronary stenting and without use of contemporary pharmacologic therapies or PCI techniques. Three small randomized trials (LOE 1)92–94 supported the strategy of fibrinolytics plus transfer for PCI; however, the timing of PCI after administration of fibrinolytics, the inclusion of patients who required transfer for PCI, the use of coronary stents, and the control group interventions differ considerably among these trials. The most recent study79 was fairly small and showed a benefit of early PCI with 1-year follow-up.94
At present there is inadequate evidence to recommend the routine transfer of patients for early PCI (ie, within 24 hours) after successful administration of fibrinolytics in a community hospital. The use of out-of-hospital administration of fibrinolytics followed by early PCI has not been specifically studied.
Special Transfer Considerations
Special transfer considerations are appropriate for patients with signs of shock (pulmonary congestion, heart rate >100 bpm, and SBP <100 mm Hg). The Second National Registry of Myocardial Infarction found that the mortality rate in patients with AMI and shock was lower in those treated with PCI as a primary strategy than in those treated with fibrinolysis.95 In the SHOCK (Should We Emergently Revascularize Occluded Coronaries for Cardiogenic Shock) trial, 152 patients with cardiogenic shock were randomly assigned to an early revascularization (ERV) strategy, 150 patients were assigned to a strategy of initial medical stabilization that included fibrinolytics, and 25% had delayed revascularization.96 Although there was no difference in the 30-day mortality rate, the mortality rate at 6 months was significantly lower in the ERV group (50.3% versus 63.1%). In a prespecified subgroup analysis for patients <75 years of age, early revascularization was associated with a 15.4% reduction in 30-day mortality and improvement in 1-year survival rates.97 A direct comparison of the outcome of primary or early PCI patients with patients who received fibrinolytic therapy only was not reported.
There is inadequate evidence to recommend routine transfer of stable patients for early PCI after successful administration of fibrinolytics in community hospitals or the out-of-hospital setting. Patients <75 years of age and selected patients >75 years of age who develop cardiogenic shock or persistent ischemic symptoms within 36 hours of STEMI should be transferred to experienced facilities capable of ERV if ERV can be performed within 18 hours of onset of shock.12
ED Evaluation and Risk Stratification (Figure 1, Boxes 3 and 4)
Focused Assessment and ECG Risk Stratification
ED providers should quickly assess patients with possible ACS. Ideally within 10 minutes of ED arrival, providers should obtain a targeted history while a monitor is attached to the patient and a 12-lead ECG is obtained (if not done in the prehospital setting).98 The evaluation should focus on chest discomfort, associated signs and symptoms, prior cardiac history, risk factors for ACS, and historical features that may preclude the use of fibrinolytics or other therapies. This initial evaluation must be efficient because if the patient has STEMI, the goals of reperfusion are to administer fibrinolytics within 30 minutes of arrival (30-minute interval “door-to-drug”) or to provide PCI within 90 minutes of arrival (90-minute interval “door-to–balloon inflation” in the catheterization suite).
Potential delay during the in-hospital evaluation period may occur from door to data, from data (ECG) to decision, and from decision to drug (or PCI). These 4 major points of in-hospital therapy are commonly referred to as the “4 D’s.”99 All providers must focus on minimizing delays at each of these points. Out-of-hospital transport time constitutes only 5% of delay to treatment time; in-hospital evaluation constitutes 25% to 33% of this delay.100,101
The physical examination is performed to aid diagnosis, rule out other causes of the patient’s symptoms, and evaluate the patient for complications related to ACS. Although the use of clinical signs and symptoms may increase suspicion of ACS, evidence does not support the use of any single sign or combination of clinical signs and symptoms alone to confirm the diagnosis.102–105
When the patient presents with signs of ACS, the clinician uses ECG findings (Figure 1, Box 4) to classify the patient into 1 of 3 groups:
ST-segment elevation or presumed new LBBB (Box 5) is characterized by ST-segment elevation >1 mm (0.1 mV) in 2 or more contiguous precordial leads or 2 or more adjacent limb leads and is classified as ST-elevation MI (STEMI).
Ischemic ST-segment depression ≥0.5 mm (0.05 mV) or dynamic T-wave inversion with pain or discomfort (Box 9) is classified as high-risk UA/non–ST-elevation MI (NSTEMI). Nonpersistent or transient ST-segment elevation ≥0.5 mm for <20 minutes is also included in this category.
Normal or nondiagnostic changes in ST segment or T waves (Box 13) are inconclusive and require further risk stratification. This classification includes patients with normal ECGs and those with ST-segment deviation of <0.5 mm (0.05 mV) or T-wave inversion of ≤0.2 mV. Serial cardiac studies (and functional testing) are appropriate.
Cardiac Biomarkers
New cardiac biomarkers, which are more sensitive than the myocardial muscle creatine kinase isoenzyme (CK-MB), are useful in diagnosis, risk stratification, and determination of prognosis. An elevated level of troponin correlates with an increased risk of death, and greater elevations predict greater risk of adverse outcome.106 Patients with increased troponin levels have increased thrombus burden and microvascular embolization.
Cardiac biomarkers should be obtained during the initial evaluation of the patient, but therapeutic decisions and reperfusion therapy for patients with STEMI should not be delayed pending the results of these tests. Important limitations to these tests exist because they are insensitive during the first 4 to 6 hours of presentation unless continuous persistent pain has been present for 6 to 8 hours. For this reason cardiac biomarkers are not useful in the prehospital setting.107–112
Serial marker testing (CK-MB and cardiac troponin) over time improves sensitivity for detection of myocardial infarction but remains insensitive in the first 4 to 6 hours.113,114
ST-Segment Elevation MI (Figure 1, Boxes 5 Through 8)
Patients with STEMI usually have complete occlusion of an epicardial coronary vessel. The mainstay of treatment is reperfusion therapy through administration of fibrinolytics (pharmacologic reperfusion) or primary PCI (mechanical reperfusion). Providers should rapidly identify patients with STEMI and quickly screen them for indications and contraindications to fibrinolytic therapy and PCI.
The first physician who encounters a patient with AMI should be able to determine the need for reperfusion therapy and direct its administration (see Tables 1 and 2). If the patient meets the criteria for fibrinolytic therapy, a door-to-needle time (needle time is the beginning of infusion of a fibrinolytic agent) ≤30 minutes is desired. Results of cardiac biomarkers do not delay the administration of fibrinolytic therapy or referral for PCI. They are normal in a significant percentage of patients who present early with STEMI. Consultation with a cardiologist or the patient’s personal physician delays therapy, is associated with increased hospital mortality rates, and is recommended only in equivocal or uncertain cases.115 Hospitals with capabilities for angiography and PCI should have a clear protocol directing ED triage and initial management. Confusion about the method of reperfusion, eg, fibrinolysis or PCI, delays definitive therapy.
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TABLE 1. Fibrinolytic Therapy: Contraindications and Cautions for Fibrinolytic Use in STEMI From ACC/AHA 2004 Guideline Update
AVM indicates arteriovenous malformation; SBP, systolic blood pressure; DBP, diastolic blood pressure; and INR, International Normalized Ratio.
Viewed as advisory for clinical decision making and may not be all-inclusive or definitive.
†Could be an absolute contraindication in low-risk patients with myocardial infarction.
Absolute Contraindications
• Any prior intracranial hemorrhage
• Known structural cerebral vascular lesion (eg, AVM)
• Known malignant intracranial neoplasm (primary or metastatic)
• Ischemic stroke within 3 months EXCEPT acute ischemic stroke within 3 hours
• Suspected aortic dissection
• Active bleeding or bleeding diathesis (excluding menses)
• Significant closed head trauma or facial trauma within 3 months
Relative Contraindications
• History of chronic, severe, poorly controlled hypertension
• Severe uncontrolled hypertension on presentation (SBP >180 mm Hg or DBP >110 mm Hg)†
• History of prior ischemic stroke >3 months, dementia, or known intracranial pathology not covered in contraindications
• Traumatic or prolonged (>10 minutes) CPR or major surgery (<3 weeks)
• Recent (within 2 to 4 weeks) internal bleeding
• Noncompressible vascular punctures
• For streptokinase/anistreplase: prior exposure (>5 days ago) or prior allergic reaction to these agents
• Pregnancy
• Active peptic ulcer
• Current use of anticoagulants: the higher the INR, the higher the risk of bleeding
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TABLE 2. ST-Segment Elevation or New or Presumably New LBBB: Evaluation for Reperfusion
Modified from ACC/AHA 2004 Update Recommendations.112
Step 1: Assess time and risk
Time since onset of symptoms
Risk of STEMI
Risk of fibrinolysis
Time required to transport to skilled PCI catheterization suite
Step 2: Select reperfusion (fibrinolysis or invasive) strategy
Note: If presentation <3 hours and no delay for PCI, then no preference for either strategy.
Fibrinolysis is generally preferred if:An invasive strategy is generally preferred if:
• Early presentation (≤3 hours from symptom onset)• Late presentation (symptom onset >3 hours ago)
• Invasive strategy is not an option (eg, lack of access to skilled PCI facility or difficult vascular access) or would be delayed• Skilled PCI facility available with surgical backup
—Medical contact-to-balloon or door-balloon >90 min• Medical contact-to-balloon or door-balloon <90 min
—(Door-to-balloon) minus (door-to-needle) is >1 hour• (Door-to-balloon) minus (door-to-needle) is <1 hour
• No contraindications to fibrinolysis• Contraindications to fibrinolysis, including increased risk of bleeding and ICH
• High risk from STEMI (CHF, Killip class is ≥3)
• Diagnosis of STEMI is in doubt
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UA and NSTEMI (Figure 1, Boxes 9 Through 17)
In the absence of ST-segment elevation, patients with ischemic-type chest pain can present with ST-segment depression or nondiagnostic or normal ECGs. ST-segment depression identifies a population at increased risk for MACE. Patients with ischemic-type pain and ECGs consistent with NSTEMI or normal or nondiagnostic ECGs do not benefit from fibrinolytic therapy, and fibrinolysis may be harmful.116
Although many patients will not have ACS (ie, the ECG change is due to an alternative diagnosis, such as LV hypertrophy), initial triage and therapy appropriately includes antiplatelet, antithrombin, and antianginal therapy. These patients usually have a partially or intermittently occluding thrombus. Clinical features can correlate with the dynamic nature of clot formation and degradation, eg, waxing and waning clinical symptoms.
Serial cardiac markers are often obtained during evaluation, including CK-MB and cardiac troponins. At any point during evaluation, elevation of cardiac troponin places a patient at increased risk for MACE. Studies have shown that patients with increased troponin are best managed with a strategy of small-molecule glycoprotein (GP) IIb/IIIa inhibitor therapy and an early invasive strategy (cardiac catheterization with possible revascularization). Troponin serves as an additional and incremental adjunct to the ECG. Physicians need to appreciate that other disorders can increase cardiac troponin, eg, myocarditis, congestive heart failure, and pulmonary embolism.
Risk Stratification
Braunwald Stratification
There are many ways to risk-stratify patients with chest pain. A well-recognized approach is the one initially proposed and later refined by Braunwald and colleagues on the ACC/AHA Task Force on the Management of Patients With Unstable Angina.11,117–120 This approach is based on a combination of historical, clinical, laboratory, and ECG variables.
Table 3 is a modified version of what has been a work in progress by Braunwald and colleagues over several publications.118,120,121 Patients are initially risk-stratified according to the likelihood that symptoms are due to unstable coronary artery disease (CAD). Patients at intermediate or high risk for CAD are further classified by their risk of MACE. This second classification is useful for prospectively identifying patients at intermediate or high risk who can benefit from an invasive strategy and more aggressive pharmacology with antiplatelet and antithrombin agents.
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TABLE 3. Likelihood of Ischemic Etiology and Short-Term Risk
Modified from Braunwald et al. Circulation. 2002;106:1893–1900.
Part I. Chest Pain Patients Without ST-Segment Elevation: Likelihood of Ischemic Etiology
A. High likelihoodB. Intermediate likelihood:C. Low likelihood:
High likelihood that chest pain is of ischemic etiology if patient has any of the findings in the column below:Intermediate likelihood that chest pain is of ischemic etiology if patient has NO findings in column A and any of the findings in the column below:Low likelihood that chest pain is of ischemic etiology if patient has NO findings in column A or B. Patients may have any of the findings in the column below:
History• Chief symptom is chest or left arm pain or discomfort plus Current pain reproduces pain of prior documented angina and Known CAD, including MI• Chief symptom is chest or left arm pain or discomfort• Probable ischemic symptoms
• Recent cocaine use
• Age >70 years
• Male sex
• Diabetes mellitus
Physical exam• Transient mitral regurgitation• Extracardiac vascular disease• Chest discomfort reproduced by palpation
• Hypotension
• Diaphoresis
• Pulmonary edema or rales
ECG• New (or presumed new) transient ST deviation (≥0.5 mm) or T-wave inversion (≥2 mm) with symptoms• Fixed Q waves• Normal ECG or T-wave flattening or T-wave inversion in leads with dominant R waves
• Abnormal ST segments or T waves that are not new
Cardiac markers• Elevated troponin I or T Any finding in column B above PLUS• Normal
• Elevated CK-MB•Normal
High (A) or Intermediate (B) Likelihood of Ischemia
Part II. Risk of Death or Nonfatal MI Over the Short Term in Patients With Chest Pain With High or Intermediate Likelihood of Ischemia (Columns A and B in Part I)
High risk:Intermediate risk:Low risk:
Risk is high if patient has any of the following findings:Risk is intermediate if patient has any of the following findings:Risk is low if patient has NO high- or intermediate-risk features; may have any of the following:
History• Accelerating tempo of ischemic symptoms over prior 48 hours• Prior MI or
• Peripheral-artery disease or
• Cerebrovascular disease or
• CABG, prior aspirin use
Character of pain• Prolonged, continuing (>20 min) rest pain• Prolonged (>20 min) rest angina is now resolved (moderate to high likelihood of CAD) • Rest angina (<20 min) or relieved by rest or sublingual nitrates• New-onset functional angina (Class III or IV) in past 2 weeks without prolonged rest pain (but with moderate or high likelihood of CAD)
• Rest angina (<20 min) or relieved by rest or sublingual nitrates
Physical exam• Pulmonary edema secondary to ischemia• Age >70 years
• New or worse mitral regurgitation murmur
• Hypotension, bradycardia, tachycardia
• S 3 gallop or new or worsening rales
• Age >75 years
ECG• Transient ST-segment deviation (≥0.5 mm) with rest angina• T-wave inversion ≥2 mm• Normal or unchanged ECG during an episode of chest discomfort
• Pathologic Q waves or T waves that are not new
• New or presumably new bundle branch block
• Sustained VT
Cardiac markers• Elevated cardiac troponin I or T Any of the above findings PLUS• Normal
• Elevated CK-MB•Normal
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TIMI Risk Score
The risk of MACE has been further studied and refined. Researchers who derived the important Thrombolysis in Myocardial Ischemia (TIMI) risk score used data from the TIMI-11B and ESSENCE (Efficacy and Safety of Subcutaneous Enoxaparin in Non–Q-Wave Coronary Events) trials for UA/NSTEMI122,123 and from the In-TIME trial for STEMI.124 The TIMI risk score comprises 7 independent prognostic variables (Table 4). These 7 variables were significantly associated with the occurrence within 14 days of at least one of the primary end points: death, new or recurrent MI, or need for urgent revascularization. The score is derived from complex multivariate logistic regression and includes variables that seem counterintuitive. It is useful to note that traditional cardiac risk factors are only weakly associated with MACE. Use of aspirin within the previous 7 days, for example, would not seem to be an indicator of a bad outcome. But aspirin use was in fact found to be one of the most powerful predictors.122 It is possible that aspirin use identified a subgroup of patients at higher risk or on active but failed therapy for CAD.
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TABLE 4. TIMI Risk Score for Patients With Unstable Angina and Non–ST-Segment Elevation MI: Predictor Variables
| Predictor Variable | Point Value of Variable | Definition |
---
| Primary end points: death, new or recurrent MI, or need for urgent revascularization. |
| Age ≥65 years | 1 | |
| ≥3 risk factors for CAD | 1 | Risk factors |
| | | • Family history of CAD |
| | | • Hypertension |
| | | • Hypercholesterolemia |
| | | • Diabetes |
| | | • Current smoker |
| Aspirin use in last 7 days | 1 | |
| Recent, severe symptoms of angina | 1 | ≥2 anginal events in last 24 hours |
| Elevated cardiac markers | 1 | CK-MB or cardiac-specific troponin level |
| ST deviation ≥0.5 mm | 1 | ST depression ≥0.5 mm is significant; transient ST elevation >0.5 mm for <20 minutes is treated as ST-segment depression and is high risk; ST elevation ≥1 mm for more than 20 minutes places these patients in the STEMI treatment category |
| Prior coronary artery stenosis ≥50% | 1 | Risk predictor remains valid even if this information is unknown |
| Calculated TIMI Risk Score | Risk of ≥1 Primary End Point in ≤14 Days | Risk Status |
| 0 or 1 | 5% | Low |
| 2 | 8% | |
| 3 | 13% | Intermediate |
| 4 | 20% | |
| 5 | 26% | High |
| 6 or 7 | 41% | |
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The creators of the TIMI risk score validated it with 3 groups of patients, and 4 clinical trials showed a significant interaction between the TIMI risk score and outcome.124–128 These findings confirm the value of the TIMI risk score as a guide to therapeutic decisions. A PDA download of this risk assessment is available at www.TIMI.org.
By classifying patients into 1 of 3 risk strata, the Braunwald (Table 3) and TIMI (Table 4) risk scores serve as the dominant clinical guides for predicting the risk of MACE in patients with ACS. Risk stratification is applicable to patients at intermediate or high risk of symptoms due to CAD and not the larger general population of patients presenting with chest pain or symptoms possibly due to anginal equivalents. Risk stratification enables clinicians to direct therapy to those patients at intermediate or high risk of MACE and avoids unnecessary therapy and the potential for adverse consequences in patients who are at lower risk.
The TIMI risk score has become the primary tool for evaluating therapeutic recommendations. Incrementally greater benefit from some of the newer therapies may be gained for patients with higher risk scores.
One additional product of the TIMI trials is the TIMI grading system of coronary artery blood flow. Investigators from the TIMI study developed and validated a coronary artery perfusion scoring system, characterizing the degree of reperfusion of a coronary artery on a scale of 0 (no flow) to 3 (normal, brisk flow). This TIMI grading system is now used as an outcome measure in many studies of ACS interventions.
Indicators for Early Invasive Strategies
Risk stratification (Figure 1, Box 12) helps the clinician identify patients with NSTEMI and UA who should be managed with an invasive strategy. Coronary angiography then allows the clinician to determine whether patients are appropriate candidates for revascularization with PCI or coronary artery bypass grafting (CABG).
The 2005 AHA Guidelines for CPR and ECC define high-risk patients with indicators that overlap to a considerable degree with the more rigorously validated TIMI risk score122:
•
New ST-segment depression and positive troponins
•
Persistent or recurrent symptoms
•
Hemodynamic instability or VT
•
Depressed LV function (ejection fraction <40%)
•
ECG or functional study that suggests multivessel CAD
Normal or Nondiagnostic ECG Changes (Boxes 13 to 17)
The majority of patients with normal or nondiagnostic ECGs do not have ACS. Patients in this category with ACS are most often at low or intermediate risk. The physician’s goal involves risk stratification (see above) to provide appropriate diagnostic or treatment strategies for an individual patient. These strategies then target patients at increased risk for benefit while avoiding risk (eg, anticoagulation therapy and invasive cardiac catheterization) in patients with low or minimal risk.
Initial General Therapy for ACS
Several initial measures are appropriate for all patients with suspected ACS in both the out-of-hospital and ED setting. These include immediate oxygen therapy, continuous cardiac monitoring, establishment of intravenous (IV) access, and several medications discussed below.
Oxygen
Administer oxygen to all patients with overt pulmonary congestion or arterial oxygen saturation <90% (Class I). It is also reasonable to administer supplementary oxygen to all patients with ACS for the first 6 hours of therapy (Class IIa). Supplementary oxygen limited ischemic myocardial injury in animals,31 and oxygen therapy in patients with STEMI reduced the amount of ST-segment elevation.35 Although a human trial of supplementary oxygen versus room air failed to show a long-term benefit of supplementary oxygen therapy for patients with MI,30 short-term oxygen administration is beneficial for the patient with unrecognized hypoxemia or unstable pulmonary function. In patients with severe chronic obstructive pulmonary disease, as with any other patient, monitor for hypoventilation.
Aspirin
Early administration of aspirin (acetylsalicylic acid [ASA]), including administration in the out-of-hospital setting,47 has been associated with decreased mortality rates in several clinical trials.47,129–131 Multiple studies support the safety of aspirin administration. Therefore, unless the patient has a known aspirin allergy, nonenteric aspirin should be given as soon as possible to all patients with suspected ACS.
Aspirin produces a rapid clinical antiplatelet effect with near-total inhibition of thromboxane A 2 production. It reduces coronary reocclusion and recurrent ischemic events after fibrinolytic therapy. Aspirin alone reduced death from AMI in the Second International Study of Infarct Survival (ISIS-2), and its effect was additive to that of streptokinase.129 In a review of 145 trials, aspirin was found to substantially reduce vascular events in all patients with AMI, and in high-risk patients it reduced nonfatal AMI and vascular death.132 Aspirin is also effective in patients with UA. The standard dose (160 to 325 mg) is recommended, although higher doses may be used. Chewable or soluble aspirin is absorbed more quickly than swallowed tablets.133,134
The early administration of a single chewed dose of aspirin (160 to 325 mg) is recommended in either the out-of-hospital or ED setting for patients with suspected ACS (Class I). Other formulations of ASA (soluble, IV) may be as effective as chewed tablets. Aspirin suppositories (300 mg) are safe and can be considered for patients with severe nausea, vomiting, or disorders of the upper gastrointestinal tract.
Nitroglycerin (or Glyceryl Trinitrate)
Nitroglycerin is an effective analgesic for ischemic chest discomfort. It also has beneficial hemodynamic effects, including dilation of the coronary arteries (particularly in the region of plaque disruption), the peripheral arterial bed, and venous capacitance vessels. The treatment benefits of nitroglycerin are limited, however, and no conclusive evidence has been shown to support routine use of IV, oral, or topical nitrate therapy in patients with AMI.135 With this in mind, these agents should be carefully considered, especially when low blood pressure precludes the use of other agents shown to be effective in reducing morbidity and mortality (eg, β-blockers and angiotensin-converting enzyme [ACE] inhibitors).
IV nitroglycerin is indicated in the following clinical situations (Class I):
•
Ongoing ischemic chest discomfort
•
Management of hypertension
•
Management of pulmonary congestion
Patients with ischemic discomfort may receive up to 3 doses of sublingual or aerosol nitroglycerin at 3- to 5-minute intervals until pain is relieved or low blood pressure limits its use (Class I). IV nitroglycerin is indicated for ongoing chest discomfort, control of hypertension, or management of pulmonary congestion in patients with STEMI associated with LV failure (Class I). In patients with recurrent ischemia, nitrates are indicated in the first 24 to 48 hours. IV rather than long-acting preparations should be used acutely to enable titration.
Do not use nitrates (Class III) in patients with hypotension (SBP <90 mm Hg or >30 mm Hg below baseline), extreme bradycardia (<50 bpm), or tachycardia (>100 bpm). Administer nitrates with extreme caution if at all to patients with suspected inferior wall MI with possible right ventricular (RV) involvement because these patients require adequate RV preload. Do not administer nitrates (Class III) to patients who have received a phosphodiesterase inhibitor for erectile dysfunction within the last 24 hours (longer for some preparations).
Morphine Sulfate
Morphine sulfate is the analgesic of choice for continuing pain unresponsive to nitrates, and it is also effective in patients with pulmonary vascular congestion complicating ACS. Morphine is a venodilator that reduces ventricular preload and oxygen requirements. For this reason it should not be used in patients who may have hypovolemia. If hypotension develops, elevate the patient’s legs, administer volume, and monitor for signs of worsening pulmonary vascular congestion. Start with a 2 to 4 mg IV dose, and give additional doses of 2 to 8 mg IV at 5- to 15-minute intervals.
Reperfusion Therapies (Figure 1, Box 8)
Perhaps the most significant advance in the treatment of cardiovascular disease in the last decade is reperfusion therapy for AMI. Many clinical trials have established early fibrinolytic therapy as a standard of care for patients with AMI who present within 12 hours of the onset of symptoms with no contraindications.136–140 Reperfusion reduces mortality, and the shorter the time to reperfusion, the greater the benefit: a 47% reduction in mortality was noted when fibrinolytic therapy was provided within the first hour after onset of symptoms.139,140
The major determinants of myocardial salvage and long-term prognosis are
•
Short time to reperfusion136,140
•
Complete and sustained patency of the infarct-related artery with normal (TIMI grade 3) flow141,142
•
Normal microvascular perfusion116,143–145
Fibrinolytics
In the absence of contraindications and the presence of a favorable risk-benefit stratification, fibrinolytic therapy is one option for reperfusion in those STEMI patients with onset of symptoms of ≤12 hours and ECG findings of STEMI (elevation >1 mm in 2 or more contiguous precordial or adjacent limb leads or new or presumably new LBBB) (Class I). In the absence of contraindications, it is also reasonable to administer fibrinolytics to patients with onset of symptoms within the prior 12 hours and ECG findings consistent with true posterior MI (Class IIa).
The ED physician should administer fibrinolytics to eligible patients as early as possible according to a predetermined process of care developed by the ED and cardiology staff. The goal is a door-to-needle time of ≤30 minutes. Every effort must be made to minimize the time to therapy. Patients treated within the first 70 minutes of onset of symptoms have >50% reduction in infarct size and 75% reduction in mortality rates.146 Pooled data from 22 randomized controlled trials of fibrinolytic therapy documents 65 lives saved per 1000 patients treated if fibrinolytics are provided in the first hour and pooled total of 131 lives saved per 1000 patients treated if fibrinolytics are provided within the first 3 hours of onset of symptoms.147 Fibrinolytics may be beneficial ≤12 hours after onset of symptoms.148,149
Fibrinolytic therapy is generally not recommended for patients presenting >12 hours after onset of symptoms, although it may be considered if continuing ischemic pain is present with ST elevation >1 mm in 2 or more contiguous precordial or adjacent limb leads (Class IIa).
Fibrinolytic therapy should not be administered (Class III) to patients who present >24 hours after the onset of symptoms or to patients who show ST-segment depression (unless a true posterior MI is suspected).
Risks of Fibrinolytic Therapy
Physicians who administer fibrinolytic agents should be aware of the indications, contraindications, benefits, and major risks of administration so that they may be able to weigh the net clinical benefit for each patient (see Table 1).150,151 This net clinical benefit requires integration of relative and absolute contraindications versus overall potential clinical gain.
Patients who present with extensive ECG changes (consistent with a large AMI) and a low risk of intracranial bleeding receive the greatest benefit from fibrinolytic therapy.136 Patients with symptoms highly suggestive of ACS and ECG findings consistent with LBBB are also appropriate candidates for intervention because they have the highest mortality rate when LBBB is due to extensive AMI. Fibrinolytics have been shown to be beneficial across a spectrum of patient subgroups with comorbidities such as previous MI, diabetes, cardiogenic shock, tachycardia, and hypotension.136 The benefits of fibrinolytic therapy are less impressive in inferior wall infarction except when it is associated with RV infarction (ST-segment elevation in lead V 4 R or anterior ST-segment depression).
Although older patients (>75 years) have a higher absolute risk of death, their absolute benefit appears to be similar to that of younger patients. There is only a small trend for benefit of fibrinolytic therapy administered 12 to 24 hours following the onset of symptoms. The incidence of stroke does increase with advancing age,152,153 reducing the relative benefit of fibrinolytic therapy. Older age is the most important baseline variable predicting nonhemorrhagic stroke.152 Although 1 large trial reported lower early and 1-year mortality rates with accelerated administration of tissue plasminogen activator (tPA) in patients <85 years of age,154 a recent retrospective analysis found no specific survival advantage and possible risk for patients >75 years of age.155 Additional studies are needed to clarify risk-benefit parameters in the elderly.
The presence of high blood pressure (SBP >175 mm Hg) on presentation to the ED increases the risk of stroke after fibrinolytic therapy.156 Current clinical practice is directed at lowering blood pressure before administration of fibrinolytic agents, although this has not been shown to reduce the risk of stroke.156 Fibrinolytic treatment of ACS patients who present with an SBP >180 mm Hg or a diastolic blood pressure >110 mm Hg is relatively contraindicated. Note that this SBP limit is slightly lower than the upper limit of 185 mm Hg used in eligibility criteria for fibrinolytic therapy for acute ischemic stroke; the diastolic limit of 110 mm Hg is consistent with the diastolic limit for tPA administration for stroke (see Part 9: “Adult Stroke”).
Several fibrinolytics are available for clinical use, including streptokinase,129,140,157 anistreplase,158,159 various regimens of alteplase,147,160,161 reteplase,162,163 and tenecteplase.138,164 Choice of agent is typically based on ease of administration, cost, and preferences of each institution.
Intracranial Hemorrhage
Fibrinolytic therapy is associated with a small but definite increase in the risk of hemorrhagic stroke, which contributes to increased mortality.136 More intensive fibrinolytic regimens using tPA (alteplase) and heparin pose a greater risk than streptokinase and aspirin.147,165 Clinical factors that may help risk-stratify patients at the time of presentation are age (≥65 years), low body weight (<70 kg), initial hypertension (≥180/110 mm Hg), and use of tPA. The number of risk factors can be used to estimate the frequency of stroke, which ranges from 0.25% with no risk factors to 2.5% with 3 risk factors.151 Several risk factor estimates are available for use by clinicians, including Simoons,151 the Co-Operative Cardiovascular Project,166 and the In-Time 2 trial.167
Percutaneous Coronary Intervention
Coronary angioplasty with or without stent placement is the most common form of PCI. PCI has been shown to be superior to fibrinolysis in combined end points of death, stroke, and reinfarction in many studies.78,80,82,96,168–173 These results, however, have been achieved in experienced medical environments with skilled providers (performing >75 PCIs per year) at a skilled PCI facility (performing >200 PCIs annually for STEMI, with cardiac surgery capabilities).
At this time primary PCI is preferred in patients with STEMI and symptom duration of >3 and ≤12 hours if skilled personnel can ensure that door-to-balloon time is ≤90 minutes or the difference in time between administration of fibrinolysis versus inflation of the PCI balloon is ≤60 minutes (Class I). PCI is also preferred in patients with contraindications to fibrinolysis and is reasonable in patients with cardiogenic shock or heart failure complicating MI.
In patients with STEMI presenting ≤3 hours from onset of symptoms, treatment is more time-sensitive, and there is inadequate research to recommend one treatment over the other (Class Indeterminate). In these “early presenters,” any possible benefit from primary PCI will be lost in prolonged transfers.
Complicated AMI
Cardiogenic Shock, LV Failure, and Congestive Heart Failure
Infarction of ≥40% of the LV myocardium usually results in cardiogenic shock and carries a high mortality rate. Of those who developed shock,174 patients with ST-segment elevation developed shock significantly earlier than patients without ST-segment elevation.
Cardiogenic shock and congestive heart failure are not contraindications to fibrinolysis, but PCI is preferred if the patient is at a facility with PCI capabilities. The ACC/AHA guidelines note that primary PCI is reasonable in those who develop shock within 36 hours of MI and are suitable candidates for revascularization that can be performed within 18 hours of the onset of shock.12 In hospitals without PCI facilities, rapidly administer a fibrinolytic agent and transfer the patient to a tertiary care facility where adjunct PCI can be performed if low-output syndromes or ischemia continues.175 The ACC/AHA STEMI guidelines recommend a door-to-departure time of ≤30 minutes for transfer.12
RV Infarction
RV infarction or ischemia may occur in up to 50% of patients with inferior wall MI. The clinician should suspect RV infarction in patients with inferior wall infarction, hypotension, and clear lung fields. In patients with inferior wall infarction, obtain a right-sided or 15-lead ECG; ST-segment elevation (>1 mm) in lead V 4 R is sensitive (sensitivity, 88%; specificity, 78%; diagnostic accuracy, 83%) for RV infarction and a strong predictor of increased in-hospital complications and mortality.176 The in-hospital mortality rate of patients with RV dysfunction is 25% to 30%, and these patients should be routinely considered for reperfusion therapy. Fibrinolytic therapy reduces the incidence of RV dysfunction.177 Similarly PCI is an alternative for patients with RV infarction and is preferred for patients in shock. Patients with shock caused by RV failure have a mortality rate similar to that for patients with shock due to LV failure.
Patients with RV dysfunction and acute infarction are dependent on maintenance of RV “filling” pressure (RV end-diastolic pressure) to maintain cardiac output.178 Thus, nitrates, diuretics, and other vasodilators (ACE inhibitors) should be avoided because severe hypotension may result. This hypotension is often easily treated with an IV fluid bolus.
Adjunctive Therapies for ACS and AMI
Clopidogrel
Clopidogrel irreversibly inhibits the platelet adenosine diphosphate receptor, resulting in a reduction in platelet aggregation through a different mechanism than aspirin. Since the publication of the ECC Guidelines 2000, several important clopidogrel studies have been published that document its efficacy for patients with both UA/NSTEMI and STEMI.
Clopidogrel was shown to be effective in 2 in-hospital randomized controlled trials (LOE 1)179,180 and 4 post-hoc analyses (LOE 7).181–184 In these studies patients with ACS and a rise in cardiac biomarkers or ECG changes consistent with ischemia had reduced stroke and MACE if clopidogrel was added to aspirin and heparin within 4 hours of hospital presentation. One study confirmed that clopidogrel did not increase risk of bleeding in comparison with aspirin.185 Clopidogrel given 6 hours or more before elective PCI for patients with ACS without ST elevation reduced adverse ischemic events at 28 days (LOE 1).186
In patients up to 75 years of age with STEMI who are treated with fibrinolysis, aspirin, and heparin (low-molecular-weight heparin [LMWH] or unfractionated heparin [UFH]), a 300-mg oral loading dose of clopidogrel given at the time of initial management (followed by a 75-mg daily dose for up to 8 days in hospital) improved coronary artery patency and reduced MACE.187
The Clopidogrel in Unstable angina to prevent Recurrent ischemic Events (CURE) trial documented an increased rate of bleeding (but not intracranial hemorrhage) in the 2072 patients undergoing CABG within 5 to 7 days of administration of this agent.184 In addition, a post-hoc analysis of this trial reported a trend toward life-threatening bleeding. A subsequent risk-to-benefit ratio analysis concluded that the bleeding risk with clopidogrel in patients undergoing CABG was modest.184 One recent large prospective trial (LOE 1)187 failed to show any increase in bleeding in 136 patients undergoing CABG within 5 to 7 days of administration of clopidogrel. In patients with ACS, the risk of bleeding must be weighed against the risk of perioperative ACS events recurring if these agents are withheld. Current ACC/AHA guidelines, published soon after the large CURE trial, recommend withholding clopidogrel for 5 to 7 days in patients for whom CABG is anticipated.12 Ongoing studies are evaluating the efficacy and risk-benefit issues.
On the basis of these findings, providers should administer a 300-mg loading dose of clopidogrel in addition to standard care (aspirin, UFH, or LMWH and GP IIb/IIIa inhibitors if indicated) to ED patients with ACS with elevated cardiac markers or new ECG changes consistent with ischemia (excluding STEMI)184 in whom a medical approach or PCI is planned (Class I). It is reasonable to administer a 300-mg oral dose of clopidogrel to ED patients with suspected ACS (without ECG or cardiac marker changes) who are unable to take aspirin because of hypersensitivity or major gastrointestinal intolerance (Class IIa). Providers should administer a 300-mg oral dose of clopidogrel to ED patients up to 75 years of age with STEMI who receive aspirin, heparin, and fibrinolysis.
β-Adrenergic Receptor Blockers
In-hospital administration of β-blockers reduces the size of the infarct, incidence of cardiac rupture, and mortality in patients who do not receive fibrinolytic therapy.188–190 They also reduce the incidence of ventricular ectopy and fibrillation.191,192 In patients who do receive fibrinolytic agents, IV β-blockers decrease postinfarction ischemia and nonfatal AMI. A small but significant decrease in death and nonfatal infarction has been observed in patients treated with β-blockers soon after infarction.193 IV β-blockers may also be beneficial for NSTEMI ACS.
Oral β-blockers should be administered in the ED for ACS of all types unless contraindications are present. They should be given irrespective of the need for revascularization therapies (Class I). Use IV β-blockers for the treatment of tachyarrhythmias or hypertension (Class IIa).
Contraindications to β-blockers are moderate to severe LV failure and pulmonary edema, bradycardia (<60 bpm), hypotension (SBP <100 mm Hg), signs of poor peripheral perfusion, second-degree or third-degree heart block, or reactive airway disease. In the presence of moderate or severe heart failure, oral β-blockers are preferred. They may need to be given in low and titrated doses after the patient is stabilized. This permits earlier administration of ACE inhibitors that are documented to be efficacious in reducing 30-day mortality rates (see below).
Heparins
Heparin is an indirect inhibitor of thrombin that has been widely used in ACS as adjunctive therapy for fibrinolysis and in combination with aspirin and other platelet inhibitors for the treatment of UA and NSTEMI. UFH is a heterogeneous mixture of sulfated glycosaminoglycans with varying chain lengths. UFH has several disadvantages, including an unpredictable anticoagulant response in individual patients, the need for IV administration, and the requirement for frequent monitoring of the activated partial thromboplastin time (aPTT). Heparin can also stimulate platelet activation, causing thrombocytopenia.194
When UFH is used as adjunctive therapy with fibrin-specific lytics in STEMI, the current recommendations call for a bolus dose of 60 U/kg followed by infusion at a rate of 12 U/kg per hour (a maximum bolus of 4000 U and infusion of 1000 U/h for patients weighing >70 kg).195 An aPTT of 50 to 70 seconds is considered optimal. Because of the limitations of heparin, newer preparations of LMWH have been developed.
Unfractionated Heparin Versus Low-Molecular-Weight Heparin in UA/NSTEMI
Six in-hospital randomized controlled trials (LOE 1196,197 and LOE 2130,198,199<24 hours; LOE 1200<36 hours) and additional studies (including 7 meta-analyses [LOE 1201–207]) document similar or improved composite outcomes (death, MI and/or recurrent angina, or recurrent ischemia or revascularization) when LMWH is given instead of UFH to patients with UA/NSTEMI within the first 24 to 36 hours after onset of symptoms.
Although major bleeding events are not significantly different with LMWH compared with UFH, there is a consistent increase in minor and postoperative bleeding with the use of LMWH.208 Omission of LMWH (enoxaparin) on the morning of angiography resulted in vascular complication rates comparable to that of UFH.209
Four trials have compared UFH and LMWH in patients with NSTEMI who were treated with a GP IIb/IIIa inhibitor.210–213 In terms of efficacy, LMWH compared favorably with UFH, and in terms of safety there were similar or less frequent major bleeding events with LMWH but again an increased frequency of minor bleeding complications.
In summary, ED administration of LMWH (specifically enoxaparin) is beneficial compared with UFH when given in addition to antiplatelet therapy such as aspirin for patients with UA/NSTEMI (Class IIb). UFH should be considered if reperfusion is planned in the first 24 to 36 hours after onset of symptoms. Changing from one form of heparin to another (crossover of antithrombin therapy) during an acute event is not recommended because it may lead to an increase in bleeding complications.214
Unfractionated Heparin Versus Low-Molecular-Weight Heparin in STEMI
LMWHs have been found to be superior to UFH in patients with STEMI in terms of overall TIMI flow215,216 and reducing the frequency of ischemic complications,217 with a trend to a 14% reduction in mortality rates in a meta-analysis.218 No superiority was found in studies in which an invasive strategy (PCI) was used.
Two randomized controlled trials compared UFH with LMWH as ancillary treatment with fibrinolysis in the out-of-hospital setting.219,220 Administration of LMWH for patients with STEMI showed superiority in composite end points compared with UFH. This must be balanced against an increase in intracranial hemorrhage in patients >75 years of age who received LMWH (enoxaparin) documented in one of these randomized controlled trials (LOE 2).220
LMWH (enoxaparin) is an acceptable alternative to UFH in the ED as ancillary therapy for patients <75 years of age who are receiving fibrinolytic therapy, provided that significant renal dysfunction (serum creatinine >2.5 mg/dL in men or 2 mg/dL in women) is not present (Class IIb). UFH is recommended for patients ≥75 years of age as ancillary therapy to fibrinolysis (Class IIa) and for any STEMI patient who is undergoing revascularization. In patients with STEMI who are not receiving fibrinolysis or revascularization, LMWH (specifically enoxaparin) may be considered an acceptable alternative to UFH in the ED setting (Class IIb).
Glycoprotein IIb/IIIa Inhibitors
After plaque rupture in the coronary artery, tissue factor in the lipid-rich core is exposed and forms complexes with factor VIIa, setting in motion the coagulation cascade resulting in platelet activation. The integrin GP IIb/IIIa receptor is considered the final common pathway to platelet aggregation. GP IIb/IIIa inhibitors modulate this receptor activity. Three agents are available for use: abciximab, eptifibatide, and tirofiban.
GP IIb/IIIa Inhibitors in UA/NSTEMI
Several large studies of GP IIb/IIIa inhibitors in UA/NSTEMI have shown a clear benefit when combined with standard aspirin and heparin and a strategy of mechanical reperfusion (LOE 1221; LOE 2222; and 3 meta-analyses221,223,224). Severe bleeding complications (and no increase in intracranial hemorrhage) in the GP IIb/IIIa group were offset by the large benefit of these agents. The benefit of GP IIb/IIIa inhibitors extends to high-risk patients with UA/NSTEMI treated with PCI.223
In UA/NSTEMI patients not treated with PCI, the effect of GP IIb/IIIa inhibitors has been mixed. In 2 studies (LOE 1)212,221 and 3 meta-analyses (LOE 1),223–225 GP IIb/IIIa inhibitors produced no mortality advantage and only a slight reduction in recurrent ischemic events in one large meta- analysis224 but did show a reduction in 30-day mortality in a later, equally large meta-analysis.225 Of note, the benefit of GP IIb/IIIa inhibitors was dependent on coadministration of UFH or LMWH. Interestingly abciximab appears to behave differently from the other 2 GP IIb/IIIa inhibitors. In the Global Utilization of Streptokinase and Tissue Plasminogen Activator for Occluded Coronary Arteries (GUSTO) IV-ACS trial and 1-year follow-up involving 7800 patients,226,227 abciximab showed a lack of treatment effect compared with placebo in patients treated medically only.
On the basis of these findings, GP IIb/IIIa inhibitors should be used in patients with high-risk stratification UA/NSTEMI as soon as possible in conjunction with aspirin, heparin, and clopidogrel and a strategy of early PCI (Class I). High-risk features include persistent pain, hemodynamic or rhythm instability, diabetes, acute or dynamic ECG changes, and any elevation in cardiac troponins attributed to cardiac injury. Extrapolation from efficacy studies suggests that this therapy may be administered in the ED once a decision has been made to proceed to PCI (Class IIa).
GP IIb/IIIa inhibitors tirofiban and eptifibatide may be used in patients with high-risk stratification UA/NSTEMI in conjunction with standard therapy if PCI is not planned (Class IIb), although studies are not conclusive at this time. As a result of the lack of benefit demonstrated in the GUSTO IV ACS trial, abciximab should not be given unless PCI is planned (Class III).
GP IIa/IIIb Inhibitors in STEMI
There is insufficient evidence to recommend for or against GP IIb/IIIa inhibitor therapy in STEMI; studies are ongoing. These agents have been used to facilitate antiplatelet therapy in patients undergoing direct PCI, but relatively few patients have been evaluated. GP IIb/IIIa inhibitors are now being evaluated early in STEMI to “facilitate” fibrinolytic therapy and serve as “upstream” adjuncts to planned direct PCI for STEMI, for example, achieving some degree of infarct artery patency during preparation or transfer. One study using abciximab (Facilitated Intervention with Enhanced Reperfusion Speed to Stop Events [FINESSE]) is ongoing. Use of these agents in STEMI requires institutional-individualized protocols developed in conjunction with interventional cardiologists.
Calcium Channel Blockers
Calcium channel blocking agents may be added as an alternative or additional therapy if β-blockers are contraindicated or the maximum dose has been achieved.
The 1996 ACC/AHA guidelines for the management of patients with AMI228 make the following comment about calcium channel blockers:
Calcium channel blocking agents have not been shown to reduce mortality after acute MI, and in certain patients with cardiovascular disease there is data to suggest that they are harmful. There is concern that these agents are still used too frequently in patients with acute MI and that β-adrenergic receptor blocking agents are a more appropriate choice across a broad spectrum of patients with MI. In general, give calcium antagonists only when β-blockers are contraindicated or have been given at maximum clinical doses without effect (Class Indeterminate).
ACE Inhibitor Therapy
ACE inhibitor therapy has improved survival rates in patients with AMI, particularly when started early.229–233 Evidence from 7 large clinical trials,135,232–237 2 meta-analyses,238,239 and 10 minor trials237,240–249 documents consistent improvement in mortality when oral ACE inhibitors are administered in the hospital setting to patients with AMI with or without early reperfusion therapy. In these studies ACE inhibitors were not administered in the presence of hypotension (SBP <100 mm Hg or more than 30 mm Hg below baseline). The beneficial effects are most pronounced in patients with anterior infarction, pulmonary congestion, or LV ejection fraction <40%.
Administration of an oral ACE inhibitor is recommended within the first 24 hours after onset of symptoms in STEMI patients with pulmonary congestion or LV ejection fraction <40%, in the absence of hypotension (SBP <100 mm Hg or more than 30 mm Hg below baseline) (Class I). Oral ACE inhibitor therapy can also be recommended for all other patients with AMI with or without early reperfusion therapy (Class IIa). IV administration of ACE inhibitors is contraindicated in the first 24 hours because of risk of hypotension (Class III).
HMG Coenzyme A Reductase Inhibitors (Statins)
A variety of studies documented consistent reduction in indicators of inflammation and complications such as reinfarction, recurrent angina, and arrhythmias when statin treatment is administered within a few days after onset of an ACS.250–253 There is little data to suggest that this therapy should be initiated within the ED; however, early initiation (within 24 hours of presentation) of statin therapy is safe and feasible in patients with an ACS or AMI (Class I). If patients are already on statin therapy, continue the therapy (Class IIb).
Glucose-Insulin-Potassium
Although glucose-insulin-potassium (GIK) therapy was formerly thought to reduce the chance of mortality during AMI by several mechanisms, recent clinical trials found that GIK did not show any benefit in STEMI.254,255 At this time there is little evidence to suggest that this intervention is helpful.
Management of Arrhythmias
This section discusses management of arrhythmias during acute ischemia and infarction.
Ventricular Rhythm Disturbances
Treatment of ventricular arrhythmias during and after AMI has been a controversial topic for 2 decades. Primary VF accounts for the majority of early deaths during AMI.21–23 The incidence of primary VF is highest during the first 4 hours after onset of symptoms24–27 but remains an important contributor to mortality during the first 24 hours. Secondary VF occurring in the setting of CHF or cardiogenic shock can also contribute to death from AMI. VF is a less common cause of death in the hospital setting with the early use of fibrinolytics in conjunction with β-blockers.
Although prophylaxis with lidocaine reduces the incidence of VF, an analysis of data from ISIS-3 and a meta-analysis suggest that lidocaine increased all-cause mortality rates.256 Thus, the practice of prophylactic administration of lidocaine has been largely abandoned.
Routine IV administration of β-blockers to patients without hemodynamic or electrical contraindications is associated with a reduced incidence of primary VF. Low serum potassium but not magnesium has been associated with ventricular arrhythmias. It is prudent clinical practice to maintain serum potassium >4 mEq/L and magnesium >2 mEq/L.
Routine administration of magnesium to patients with MI has no significant clinical mortality benefit, particularly in patients receiving fibrinolytic therapy. The definitive study on the subject is the ISIS-4 study (LOE 1).135 ISIS-4 enrolled >58 000 patients and showed a trend toward increased mortality rates when magnesium was given in-hospital for primary prophylaxis to patients within the first 4 hours of known or suspected AMI.
Following an episode of VF, there is no conclusive data to support the use of lidocaine or any particular strategy for preventing VF recurrence. β-Blockers are the preferred treatment if not initiated before the episode of VF. If lidocaine is used, continue it for a short time after MI but no more than 24 hours unless symptomatic VT persists. Exacerbating or modulating factors should be identified and corrected. Further management of ventricular rhythm disturbances is discussed in Part 7.2: “Management of Cardiac Arrest” and Part 7.3: “Management of Symptomatic Bradycardia and Tachycardia.”
Summary
There has been tremendous progress in reducing disability and death from ACS. But many patients still die before reaching the hospital because patients and family members fail to recognize the signs of ACS and fail to activate the EMS system. Once the patient with ACS contacts the healthcare system, providers must focus on support of cardiorespiratory function, rapid transport, and early classification of the patient based on ECG characteristics. Patients with STEMI require prompt reperfusion; the shorter the interval from symptom onset to reperfusion, the greater the benefit. Patients with UA/NSTEMI or nonspecific or normal ECGs require risk stratification and appropriate monitoring and therapy. Healthcare providers can improve survival rates and myocardial function of patients with ACS by providing skilled, efficient, and coordinated out-of-hospital and in-hospital care.
Footnote
This special supplement to Circulation is freely available at
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Volume 112 • Number 24_supplement • 13 December 2005
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Figure 1.Acute Coronary Syndromes Algorithm.
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Figure 2.Fibrinolytic Checklist.
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TABLE 1. Fibrinolytic Therapy: Contraindications and Cautions for Fibrinolytic Use in STEMI From ACC/AHA 2004 Guideline Update
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TABLE 2. ST-Segment Elevation or New or Presumably New LBBB: Evaluation for Reperfusion
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TABLE 3. Likelihood of Ischemic Etiology and Short-Term Risk
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TABLE 4. TIMI Risk Score for Patients With Unstable Angina and Non–ST-Segment Elevation MI: Predictor Variables
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AP Physics 1 Kinematics Equations You Need to Know + Practice Problems
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How Will AP Scores Impact Your College Chances?
Overview of the AP Physics 1 Exam
List of Kinematics Equations + Practice Problems
Final Tips
The AP Physics 1 exam requires you to think critically and analytically; you may need to use equations to find the amount of energy dissipated through a series of resistors, or create a detailed experimental procedure to determine the speed of sound in air at room temperature.
In this article, we will review key AP Physics kinematics equations you’ll need to know to pass the AP Physics 1 Exam. We will cover these equations and then show you how you can use them in practice.
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Overview of AP Physics 1 Exam
The AP Physics 1 exam is an algebra-based, introductory college-level physics course. This class further prepares you for the AP Physics C course as well. The score you get on this exam helps determine which courses you may skip once you get into college. If you want to do well, you might want to check out the Ultimate Guide to the AP Physics 1 Exam since the article reviews tips and additional practice problems not covered in this article.
Exam Layout
The AP Physics 1 Exam is 3 hours long and consists of 50 multiple-choice questions and 5 free-response questions. Questions will cover everything taught in this course from mechanics to electricity.
You will have 90 minutes to complete the multiple choice portion of the test. The multiple choice items are broken up into two different types of questions: 45 single-select multiple choice questions and 5 multiple-select multiple choice items. Single-select multiple choice questions are the ones we all know and love, such as the one shown below, in which there is one correct answer.
Answer: C, Source: CollegeBoard
Multiple-select multiple choice questions may require you to select one or more choices that could be the correct answer, as the one shown below.
Answer: B, D, Source: CollegeBoard
The Free Response Section, which lasts 90 minutes, consists of five questions that cover four different types of questions. The experimental design question assesses the test taker’s ability to design a scientific experiment. The qualitative/quantitative question requires you to be able to translate between quantitative and qualitative justification and reasoning. The paragraph argument short answer question assesses your ability to develop an argument about a physical phenomenon while the short answer question covers topics not asked in the other FRQs. All of these types of questions only show up once on the exam while the short answer question appears twice (1 experimental design + 1 qualitative/quantitative translation + 1 paragraph argument + 2 short answers = 5 FRQ questions).
Scoring
Both the multiple choice questions and free response questions are weighted at 50% to determine the final score.
Kinematics Equations You Need to Know + Practice Problems
The AP Physics exam covers topics such as energy, momentum, DC circuits and Electrical Charge. The first thing you may learn in your AP Physics 1 course is kinematics. Kinematics deals with the motion of objects independent of the object’s mass and forces being applied to the object. Therefore, in this review, we will be exploring the relationships between acceleration, velocity, position and time.
Note: The practice questions that appear are not official problems, but they will help you understand the concepts covered in this article.
Acceleration, Velocity, and Position
It is important to note that displacement, velocity, and acceleration are all vector quantities which means they indicate the direction the object is moving in.
The following kinematics equations only apply to constant acceleration situations; this means that kinematics covers linear motion in one and two dimensions as well as circular motion.
The following equations are applicable to linear and projectile motion:
We are able to find the velocity at any given time if we know the initial velocity () and acceleration ().
The position equation allows us to find the position at time given that we know the initial velocity (), initial position () and acceleration ().
We may combine equations 1 and 2 to create a final kinematics equation that is independent of time. Rather, this equation combines initial velocity, current position, initial position and acceleration to find the velocity of the object at the current position.
To derive this equation, we start with equation 1:
From we get:
Substitute the found into :
These equations can be used for any given direction, not just in the x-direction; however, position, velocity and acceleration must be occurring in the same direction for this equation to make sense: i.e.
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Practice Problem: Determining Acceleration
A cannonball is shot out of a cannon and its velocity relative to the -axis is given by the equation . At what time will the cannonball come to a rest momentarily (neglecting the force of friction)?
The first question we ask ourselves is, what variable helps determine when an object is at rest? Velocity! When is equal to , we know that any object is not moving at time , regardless of its acceleration. Therefore:
.
This means at seconds, the cannonball will have come to a rest momentarily. Why do we say momentarily? This is because the cannonball still has acceleration:
We are given that the baseline velocity equation is:
Using the equation, , we know that at time , . Also, at time , .
This means that at time , .
A nonzero acceleration just implies that at some point, the cannon ball changes direction. After seconds, the cannonball will change directions and move in the opposite direction of initial movement.
Practice Problem: Ball Rolling Off a Table
A ball rolls off a table that is meters tall. The ball at the edge of the table has a velocity of . At what distance from the edge of the table will the ball land?
First, we have to realize that the velocity of the ball at the edge of the table only impacts the direction. So what is the velocity in the direction? That’s an equation we have to determine ourselves! We know that relative to the ground, the initial position is , initial velocity in the direction is and that acceleration is a result of gravity (). Therefore, our position equation in the direction is given as:
Our next step is to find how long the ball takes to reach the ground. We can do so by setting to to find that is:
Now that we know how long the ball takes to reach the floor, all we have to do is figure out the distance the ball ends up from the edge of the table! We know that the velocity of the ball in the direction at the edge of the table is given by . This velocity is both the initial velocity at the edge of the table and is constant, since we don’t have any acceleration in the horizontal direction (there are no horizontal forces acting on the ball). Therefore, we can use the following equation to determine the distance the ball lands away from the table:
Since we know the ball lands on the ground at time , we get:
.
Therefore, the ball lands meters away from the table!
Acceleration: A Deeper Understanding
We know that kinematics covers situations in which acceleration is constant. But how do we find acceleration? In addition to acceleration being the rate of change of velocity over time as shown above, it is also equal to the net force exerted on the system divided by the mass:
It is important to note that acceleration in the direction, direction or direction is proportional to the total net force in the direction, direction or direction, respectively.
We mentioned earlier that kinematics may also relate to uniform circular motion. This acceleration, known as centripetal acceleration, points towards the center of rotation and is perpendicular to the velocity of the object. Centripetal acceleration is equal to the velocity squared divided by the radius of the circle.
Practice Problem: Determining Centripetal Acceleration
A car drives around a circular track with a radius of . The tangential velocity of the car at any given point is . What is the acceleration?
We know that , therefore the centripetal acceleration is given as:
Angular Acceleration, Velocity and Position
The same concepts applied under the Acceleration, Velocity and Position section can be applied for rotational motion. Angular position () is analogous to position just as angular velocity () and angular acceleration () are analogous to velocity and acceleration, respectively. The corresponding angular motion kinematics equations used on the AP Physics 1 exam are shown below.
Just like linear position, velocity and acceleration, angular velocity is the rate of change in angular position over time while angular acceleration is the rate of change in angular velocity over time.
Practice Problem:
A bug is stuck to the edge of the tire of a car. The bug travels to complete one rotation. If the angular velocity of the bug is given at , how many rotations does the bug make in seconds?
The first thing to note is that the bug has angular acceleration since the angular velocity is constant and not dependent on time. This means that the equation for angular position is given as:
If the angular distance traveled is and one rotation is completed in , that means the bug made rotations on the tire.
Wrapping It Up
We hope you’ve gained a stronger grasp of the kinematics equations on the AP Physics 1 exam! You don’t need to memorize these equations since they will be provided to you. Rather, it is important to understand when these equations are applicable and how they might help you analyze a situation.
As you take more practice exams and review for the AP exam, you will get better at quickly identifying which formulas will help you get the answer the test is looking for.
For more practice, see the official College Board example questions and our ultimate guide to the AP Physics 1 exam.
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Taking Real, Im, and Abs of polynomial fractions
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Asked 5 years, 8 months ago
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I want to find out real imaginary and Abs of polynomial fractions composed of real-valued variables.
For example, consider we have a polynomial fraction f=a 1+b 1 i a 2+b 2 i f=a 1+b 1 i a 2+b 2 i where a 1,2 a 1,2 and b 1,2 b 1,2 arereal valued variables (in fact, my actual values are again complecated fuctions of real-valued variables). (edit) Natual choice was to use inbuilt functions, but their output was as follows:
How can I simplify the real, imaginary, and using standard Mathematica function?
This is what I have so far, for example, for the Abs.
mathematica
Myabs[complex_] := Module[{Nreal, Nimag, Dreal, Dimag, real, imag},
Nreal = Numerator[complex] /. Complex[a_, b_] -> a;
Nimag = (Numerator[complex] - Nreal)/I;
Dreal = Denominator[complex] /. Complex[a_, b_] -> a;
Dimag = (Denominator[complex] - Dreal)/I;
real = (Nreal Dreal + Nimag Dimag)/(Dreal^2 + Dimag^2);
imag = ( Dreal Nimag - Nreal Dimag)/(Dreal^2 + Dimag^2);
Sqrt[real^2 + imag^2]];
This function does the job for me so far. But I feel that there should be a better and straightforward way to do it.
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edited Jan 5, 2020 at 19:39
PojjPojj
asked Jan 5, 2020 at 19:19
PojjPojj
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Would Re, Im, and Abs not work here? Can you give us an example where they fail but your function works, to see if maybe they can still be forced to work as you would like?MarcoB –MarcoB 2020-01-05 19:25:03 +00:00 Commented Jan 5, 2020 at 19:25
Thanks, @MarcoB. Of course, that was what I first tried. But the output was not as expected. Just edited the question with a screen capture.Pojj –Pojj 2020-01-05 19:40:05 +00:00 Commented Jan 5, 2020 at 19:40
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``mathematica
Clear["Global"]
expr = (a1 + Ib1)/(a2 + Ib2);
```
ComplexExpand will assume that all variables are real unless otherwise specified.
```mathematica
Abs[expr] // ComplexExpand[#, TargetFunctions -> {Re, Im}] &
( Sqrt[a1^2 + b1^2]/Sqrt[a2^2 + b2^2] )
Re[expr] // ComplexExpand[#, TargetFunctions -> {Re, Im}] & // Simplify
( (a1 a2 + b1 b2)/(a2^2 + b2^2) )
Im[expr] // ComplexExpand[#, TargetFunctions -> {Re, Im}] & // Simplify
( (a2 b1 - a1 b2)/(a2^2 + b2^2) )
```
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answered Jan 5, 2020 at 19:45
Bob HanlonBob Hanlon
167k 7 7 gold badges 82 82 silver badges 213 213 bronze badges
3
Perfect! It is the masterpiece I was looking for. What is really happening with Clear["Global`"]? (just to get better understanding)Pojj –Pojj 2020-01-05 19:49:37 +00:00 Commented Jan 5, 2020 at 19:49
2 Clear["Global"]` ensures that any prior definitions that might conflict are removed.Bob Hanlon –Bob Hanlon 2020-01-05 19:51:17 +00:00 Commented Jan 5, 2020 at 19:51
Got it! I should have asked this question long before! Thanks.Pojj –Pojj 2020-01-05 19:54:06 +00:00 Commented Jan 5, 2020 at 19:54
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How was this approximation of $\pi$ involving $\sqrt{5}$ arrived at?
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Asked
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$\begingroup$
The Wikipedia article for Approximations of $\pi$ contains this little gem:
$$ \pi \approx \frac{63}{25}\times\frac{17 + 15\sqrt{5}}{7 + 15\sqrt{5}} $$
which is clearly in $\mathbb{Q[\sqrt{5}]}$. Wikipedia doesn't (currently) give a reference for this approximation. I also noticed that when re-written to move $\sqrt{5}$ out of the denominator, the resulting number in $\mathbb{Q[\sqrt{5}]}$ is
$$ \pi \approx \frac{31689 + 4725\sqrt{5}}{13450} $$ and the integers involved are no larger than $5$ significant digits. How do you think this approximation was arrived at, and how might one go about finding a better approximation using more digits for the integers?
See also this related question.
approximation
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edited Apr 15, 2017 at 7:03
Jaume Oliver Lafont
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asked Jan 5, 2015 at 19:57
hatch22hatch22
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$\begingroup$ Also related. $\endgroup$
Andrés E. Caicedo
– Andrés E. Caicedo
2015-01-05 20:08:36 +00:00
Commented Jan 5, 2015 at 20:08
1
$\begingroup$ I don't know how it's arrived at, but the approximation is (also) due to Ramanujan. See the Wolfram MathWorld page for more information. It also contains references where you might find your answer. I should be able to check Berndt's "Ramanujan's Notebooks" to see if it can help. $\endgroup$
d125q
– d125q
2015-01-05 20:08:39 +00:00
Commented Jan 5, 2015 at 20:08
$\begingroup$ @d125q, thanks for the reference. I'm still curious how something like this is derived (specifically using a quadratic extension). I found an article referenced in the Wolfram link you gave by Borwein and Bailey, but it didn't contain this particular approximation, and the other references are texts I'd rather not buy for one example. If you wouldn't mind checking "Ramanujan's Notebooks", I suspect it is in there. $\endgroup$
hatch22
– hatch22
2015-01-05 20:26:38 +00:00
Commented Jan 5, 2015 at 20:26
5
$\begingroup$ Not the answer, but one can also grind away using a computer. For fun, I just found this one using $\sqrt{7}$: $$\frac{2762 + 3093\sqrt{7}}{3484} \approx 3.141592653594$$ The error is less than the error for the expression using $\sqrt{5}$ by two orders of magnitude. $\endgroup$
Simon S
– Simon S
2015-01-05 20:37:22 +00:00
Commented Jan 5, 2015 at 20:37
$\begingroup$ What program are you using to do the grinding? $\endgroup$
hatch22
– hatch22
2015-01-05 20:38:07 +00:00
Commented Jan 5, 2015 at 20:38
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4 Answers 4
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Ramanujan discussed this and many similar approximations in his monumental paper Modular Equations and Approximations to $\pi$. Let me describe his idea in brief which is based on deep interconnection between the theory of elliptic integrals and theta functions.
Let $0 < k < 1$ and $k' = \sqrt{1 - k^{2}}$ then we define two elliptic integrals $$K(k) = \int_{0}^{\pi/2}\frac{dx}{\sqrt{1 - k^{2}\sin^{2}x}},\,E(k) = \int_{0}^{\pi/2}\sqrt{1 - k^{2}\sin^{2}x}\,dx\tag{1}$$ When the number $k$ (called modulus) is evident from context we use the symbols $K, E, K', E'$ for $K(k), E(k), K(k'), E(k')$. These numbers have a curious relation to $\pi$ which is not difficult to prove $$KE' + K'E - KK' = \frac{\pi}{2}\tag{2}$$ which goes by the name Legendre's Identity. This is the identity which forms the link between $\pi$ and elliptic integrals and it was exploited by Ramanujan to the fullest extent.
The formulas $(1)$ can be inverted in the sense that if we are given the values of $K, K'$ then we can obtain the corresponding value of $k$. But this happens in almost magical way via theta functions. Let us define a new variable $q$ called nome by $q = e^{-\pi K'/K}$ and we have the following theta functions (in simplified form) of Jacobi: \begin{align} \vartheta_{2}(q) &= \sum_{n = -\infty}^{\infty}q^{(n + 1/2)^{2}} = 2(q^{1/4} + q^{9/4} + q^{25/4} + \cdots )\notag\ &= 2q^{1/4}\prod_{n = 1}^{\infty}(1 - q^{2n})(1 + q^{2n})^{2}\tag{3a}\ \vartheta_{3}(q) &= \sum_{n = -\infty}^{\infty}q^{n^{2}} = 1 + 2q + 2q^{4} + 2q^{9} + \cdots\notag\ &= \prod_{n = 1}^{\infty}(1 - q^{2n})(1 + q^{2n - 1})^{2}\tag{3b}\ \vartheta_{4}(q) &= \sum_{n = -\infty}^{\infty}(-1)^{n}q^{n^{2}} = 1 - 2q + 2q^{4} - 2q^{9} + \cdots\notag\ &= \prod_{n = 1}^{\infty}(1 - q^{2n})(1 - q^{2n - 1})^{2}\tag{3c} \end{align} and the values $k, k', K$ are obtained via formulas: $$k = \frac{\vartheta_{2}^{2}(q)}{\vartheta_{3}^{2}(q)},\, k' = \frac{\vartheta_{4}^{2}(q)}{\vartheta_{3}^{2}(q)},\, \frac{2K}{\pi} = \vartheta_{3}^{2}(q)\tag{4}$$
It is interesting to study the dependence of the nome $q = e^{-\pi K'/K}$ on modulus $k$. Equivalently one can also analyze the behavior of the ratio $K'/K$ as $k$ varies in the interval $(0, 1)$. It can be easily proved using the definitions $(1)$ that $K'/K$ is strictly decreasing as $k$ increases and $K'/K \to \infty$ as $k \to 0^{+} $ and $K'/K \to 0$ as $k \to 1^{-} $. Thus the ratio $K'/K$ as a function of $k$ establishes a bijection between interval $(0, 1)$ and the set of positive real numbers.
Next idea is to start with a specific value of $k$ and a positive number $n$ and then the number $nK'/K$ is also positive and hence (by the observation in previous paragraph) there is a unique number $l \in (0, 1)$ such that $nK'/K = K(l')/K(l)$ where $l'=\sqrt{1-l^2}$. It is customary to denote $K(l), K(l')$ by $L, L'$ respectively and thus starting with a modulus $k \in (0, 1)$ and a positive real number $n$ we have found a unique $l \in (0, 1)$ such that $$\frac{L'}{L} = n\frac{K'}{K}$$ Jacobi extensively studied the integral transformations related to the elliptic integral $K(k)$ and proved that:
Jacobi's Theorem: If $n$ is a positive rational number and $k\in (0, 1)$ then there is a unique $l \in (0, 1)$ such that $$\frac{L'}{L} = n\frac{K'}{K}$$ and the relation between $k, l$ is algebraic. In other words in such a case there exists a polyomial $P(x, y)$ in two variables with integer coefficients such that $P(k, l) = 0$.
The algebraic relation between $k, l$ induced by the equation $$\frac{L'}{L} = n\frac{K'}{K}$$ is called a modular equation of degree $n$. And Jacobi gave such modular equations for $n = 3, 5$. Finding modular equations using integral transformations as described by Jacobi is really tough and impractical beyond $n = 7$.
Ramanujan was in love with these modular equations and he gave many modular equations for different positive integer values of $n$. But he had a much deeper insight into these topics and he asked:
What would happen if we add a further constraint $l=k'=\sqrt{1-k^{2}}$ in the modular equation connecting $k, l$?
With this additional constraint we now have an algebraic equation $P(k, k') = 0$ and thus both $k, k'$ are algebraic numbers. Also since $l = k'$ we have $L = K', L' = K$ so that $L'/L = nK'/K$ leads us to $L'/L = \sqrt{n}, K'/K = 1/\sqrt{n}$. The corresponding nomes for $k, l$ are now seen to be $e^{-\pi/\sqrt{n}}$ and $e^{-\pi\sqrt{n}}$. And thus we get the following deep theorem:
Theorem: If $n$ is a positive rational number and $q = e^{-\pi\sqrt{n}}$ then the numbers $k, k'$ given by equation $(4)$ in terms of $q$ are algebraic.
Ramanujan's great ability was manipulation of radicals and if there was an algebraic number of any mathematical significance then Ramanujan could express it in the form of an explicit radical. Ramanujan introduced the functions $P, Q, R$ given by \begin{align} P(q) &= 1 - 24\sum_{i = 1}^{\infty}\frac{iq^{i}}{1 - q^{i}}\tag{5a}\ Q(q) &= 1 + 240\sum_{i = 1}^{\infty}\frac{i^{3}q^{i}}{1 - q^{i}}\tag{5b}\ R(q) &= 1 - 504\sum_{i = 1}^{\infty}\frac{i^{5}q^{i}}{1 - q^{i}}\tag{5c} \end{align} and obtained the formulas \begin{align} P(q^{2}) &= \left(\frac{2K}{\pi}\right)^{2}\left(\frac{3E}{K} + k^{2} - 2\right)\tag{6a}\ Q(q^{2}) &= \left(\frac{2K}{\pi}\right)^{4}\left(1 - k^{2} + k^{4}\right)\tag{6b}\ R(q^{2}) &= \left(\frac{2K}{\pi}\right)^{6}(1 + k^{2})(1 - 2k^{2})\left(1 - \frac{k^{2}}{2}\right)\tag{6c} \end{align} Ramanujan was not happy with equation $(6a)$ and wanted it to contain only an expression in $k$ apart from the factor $(2K/\pi)^{2}$ and he found that for each positive rational number $n$ the expression $nP(q^{2n}) - P(q^{2})$ could be written as $(2K/\pi)^{2}F_{n}(k)$ where $F_{n}(k)$ is a complicated but algebraic function of $k$. Ramanujan obtained explicit radical expressions for $F_{n} (k) $ for some integer values of $n$. For small values of $n$ like $2,3,5$ it is not difficult to get these expressions, but no one knows how Ramanujan obtained the expressions for larger values of $n$. The method described by Ramanujan to evaluate these expressions is very tiresome and requires great skill in manipulation of radicals. Using this result he proved the following significant theorem:
Ramanujan's Theorem: If $n$ is a positive rational number and $q = e^{-\pi\sqrt{n}}$ then $$P(q^{2}) - \frac{3}{\pi\sqrt{n}} = \left(\frac{2K}{\pi}\right)^{2}A_{n}$$ where $A_{n}$ is an algebraic number dependent on $n$.
If $n$ is large then $q = e^{-\pi\sqrt{n}}$ is small and the functions $P, Q, R$ are close to $1$ and hence can be approximated by $1$. Also note that if $n$ is positive rational then $k$ is algebraic and in the light of above theorem and equations $(6b), (6c)$ we can see that $$\left(P(q^{2}) - \frac{3}{\pi\sqrt{n}}\right)Q(q^{2}) = B_{n}R(q^{2})\tag{7}$$ where $B_{n}$ is an algebraic number. Using approximation $P, Q, R \approx 1$ for large $n$ we get the approximation for $\pi$ $$\pi \approx \frac{3}{(1 - B_{n})\sqrt{n}}\tag{8}$$ The approximation in question is for $n = 25$. It is difficult to calculate the expression $B_{n}$ for generic $n$ but for $n = 25$ we can get the value the value of $B_{n}$ (with significant computational effort) so that the desired approximation for $\pi$ is obtained. Ramanujan also gives the estimate for error as $24\pi(10\pi\sqrt{n} - 31)e^{-2\pi\sqrt{n}}$.
Ramanujan also uses the identity $$C(q)=1-24\sum_{i=1}^{\infty}\frac{(2i-1)q^{2i-1}}{1-q^{2i-1}}=\left(\frac{2K}{\pi}\right)^{2}(1-2k^{2})\tag{9}$$ to get the equation $$P(q^{2})-\frac{3}{\pi\sqrt{n}}=C_{n}C(q)$$ where $C_{n} $ is algebraic and this leads to the approximation $$\pi\approx\frac{3}{(1-C_{n})\sqrt{n}}\tag{10}$$ and for $n=25$ this gives the much simpler but less accurate approximation $$\pi\approx \frac{9}{5}+\sqrt{\frac{9}{5}}\tag{11}$$ which is mentioned in another answer here.
Proof for Ramanujan's theorem can be found in my blog post. These techniques are further used by Ramanujan to give many famous series for $1/\pi$ (also covered in the blog post).
Update: Finally all the ingredients necessary to evaluate $B_n$ above for $n=25$ are available on math.se. I perform the desired evaluation below.
To set up things let us assume that $l, \gamma, k$ are elliptic moduli corresponding to the nomes $e^{-5\pi},e^{-\pi},e^{-\pi/5}$ and let $q=e^{-\pi/5}$. Let the corresponding elliptic integrals be denoted by $L, \Gamma, K$. It is well known that $$\gamma =\frac{1}{\sqrt {2}}, 2l^2=1-\sqrt{1-\phi^{-24}},2k^2=1+\sqrt{1-\phi^{-24}} \tag{12}$$ and we have $k'=l, l'=k, \gamma'=\gamma$ ($\phi$ is the golden ratio).
Also from this answer we have $$\frac{L} {\Gamma} =\frac{\sqrt{5}+2}{5}\tag{13}$$ Clearly from the transformation formula of theta functions it is obvious that $K/L=5$.
We now use the formula for $F_n(k) $ for $n=5$ given by Ramanujan and obtain $$5P(q^{10})-P(q^2)=\frac{4K\Gamma}{\pi^2}(3+k\gamma+k'\gamma')\sqrt{\frac{1+k\gamma+k'\gamma'}{2}}=25(\sqrt{5}-2)\left(\frac{2L}{\pi}\right)^2(3+(k+k')\gamma)\sqrt{\frac{1+(k+k')\gamma}{2}}\tag{14a}$$ and $$5P(q^{50})-P(q^{10})=\frac{4\Gamma L} {\pi^2}(3+l\gamma+l'\gamma')\sqrt{\frac{1+l\gamma+l'\gamma'}{2}}=5(\sqrt{5}-2)\left(\frac{2L}{\pi}\right)^2(3+(l+l')\gamma)\sqrt{\frac{1+(l+l')\gamma}{2}}\tag{14b}$$ Multiplying $(14b)$ by $5$ and adding the result to $(14a)$ we get $$25P(q^{50})-P(q^2)=50(\sqrt{5}-2)\left(\frac{2L}{\pi}\right)^2(3+(l+l')\gamma)\sqrt{\frac{1+(l+l')\gamma}{2}}$$ We have $$l+l'=\sqrt{l^2+l'^2+2ll'}=\sqrt{1+\phi^{-12}}$$ and thus $$(l+l') \gamma=\sqrt{\frac{1+\phi^{-12}}{2}}=3\phi^{-3}=3(2\phi-3)$$ and we finally arrive at $$25P(q^{50})-P(q^2)=50(\sqrt{5}-2)\left(\frac{2L}{\pi}\right) ^2(6\phi-6)\sqrt{3\phi-4}$$ On the other hand using logarithmic differentiation of transformation formula for Dedekind eta function allows us to prove that $$25P(q^{50})+P(q^2)=\frac{30}{\pi}$$ And adding last two equations we get $$P(q^{50})-\frac{3}{5\pi}=\left(\frac{2L}{\pi}\right) ^26(5-3\phi)\sqrt{3\phi-4}\tag{15}$$ And we have the standard relations $$Q(q^{50})=\left(\frac{2L}{\pi}\right)^4(1-l^2l'^2)=\left(\frac{2L}{\pi}\right)^4\left(\frac{4-\phi^{-24}}{4}\right)$$ and $$R(q^{50})=\left(\frac{2L}{\pi}\right)^6(1-2l^2)(1+(l^2l'^2/2))=\left(\frac{2L}{\pi}\right)^6\sqrt {1-\phi^{-24}}\left(\frac{8+\phi^{-24}}{8}\right)$$
The expression $B_n$ equals $$\frac{(P(q^{50})-(3/5\pi))Q(q^{50})} {R(q^{50})} $$ and using previous equations we get $$B_n=\frac{12(5-3\phi) \sqrt{3\phi-4}}{\sqrt{1-\phi^{-24}}}\cdot\frac{4-\phi^{-24}}{8+\phi^{-24}}$$ It turns out that the first fraction evaluates to $\phi$ (after some effort) and hence $$B_n=\frac{4\phi-\phi^{-23}}{8+\phi^{-24}}$$ and $$1-B_n=\frac{8-4\phi+\phi^{-23}+\phi^{-24}}{8+\phi^{-24}}=\frac{8-4\phi+\phi^{-22}}{8+\phi^{-24}}$$ The approximation for $\pi$ is now given by $$\pi\approx\frac{3}{5(1-B_n)}=\frac{3}{5}\cdot \frac{8+\phi^{-24}} {8-4\phi+\phi^{-22}}$$ Let $a_m$ denote Fibonacci sequence with $a_0=0,a_1=1$ and then $a_{-m} =(-1)^{m+1}a_m$ and we have $$\phi^m=a_m\phi+a_{m-1}$$ Using this relation we get $$\pi\approx\frac{3}{5}\cdot \frac{8+a_{25} -a_{24} \phi}{8+a_{23} -(4+a_{22})\phi}$$ Using a table of Fibonacci sequence we get $$a_{22} =17711,a_{23} =28657,a_{24}=46368,a_{25}=75025$$ and then $$\pi\approx\frac{3}{5}\cdot\frac{75033-46368\phi}{28665-17715\phi}=\frac{63}{25}\cdot\frac{3573-2208\phi}{5733-3543\phi}=\frac{63}{25}\cdot\frac{1646-736\sqrt{5}}{2641-1181\sqrt{5}} $$ and by this answer the second fraction equals $$\frac{17+15\sqrt{5}}{7+15\sqrt{5}}$$ Thus we obtain the famous approximation by Ramanujan $$\pi\approx \frac{63}{25}\cdot\frac{17+15\sqrt{5}}{7+15\sqrt{5}}$$ The approximation in equation $(11)$ can also be obtained in similar manner (with much less effort) by evaluating $C_n$ for $n=25$.
Ramanujan obtained most of his results (including the above) using algebraic manipulation combined with processes of calculus (differentiation and integration) and rarely used any sophisticated tools. Modern approach to Ramanujan's mathematics is so unlike his methods and is based on sophisticated techniques and theories (modular forms) which fail to generate any interesting results like those given by Ramanujan and are almost always dependent on software like Maple, Macsyma or Mathematica for any serious calculation.
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edited Dec 30, 2024 at 13:08
answered May 17, 2017 at 14:47
Paramanand Singh♦Paramanand Singh
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$\begingroup$ Equating the right hand side of approximations (10) and (11) with $n=25$ shows the value of the algebraic constant $C_n$ is half the golden ratio. $$C_{25}=\frac{1+\sqrt{5}}{4}=\frac{\phi}{2}$$ $\endgroup$
Jaume Oliver Lafont
– Jaume Oliver Lafont
2017-05-18 09:13:44 +00:00
Commented May 18, 2017 at 9:13
$\begingroup$ @JaumeOliverLafont: Actually it is the value of $A_{n} $ for $n=25$ which is difficult to calculate. I have the formulas given by Ramanujan for calculating it, but the computation involves smart manipulation of radicals. Once we know $A_{n} $ the evaluation of $B_{n}, C_{n} $ is not that difficult. The calculation for $n=7$ is easy and equation $(8)$ gives the value $(19/16)\sqrt{7}$. $\endgroup$
Paramanand Singh
– Paramanand Singh ♦
2017-05-18 09:50:13 +00:00
Commented May 18, 2017 at 9:50
1
$\begingroup$ @JaumeOliverLafont: in case you are wondering about how $\phi$ comes into picture for $n=25$, then one has to note that the expression $(2kk')^{-1/12}=\phi$ if $n=25$. The expression $(2kk')^{-1/12}$ is one of Ramanujan's famous class invariants. See paramanands.blogspot.com/2012/03/… $\endgroup$
Paramanand Singh
– Paramanand Singh ♦
2017-05-18 10:02:13 +00:00
Commented May 18, 2017 at 10:02
$\begingroup$ And the parabola transformations? How do you explain them? $\endgroup$
Jaume Oliver Lafont
– Jaume Oliver Lafont
2017-05-18 10:07:45 +00:00
Commented May 18, 2017 at 10:07
$\begingroup$ @JaumeOliverLafont: what are parabola transformation? I am not aware. Perhaps you can elaborate a bit. $\endgroup$
Paramanand Singh
– Paramanand Singh ♦
2017-05-18 10:15:44 +00:00
Commented May 18, 2017 at 10:15
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When I first came across your question, I thought it was a modern-day approximation by somebody using a computer. But when d125q pointed out it was by Ramanujan, then I figured out he must have used a systematic method.
One way is to use a Ramanujan-Sato pi formula like,
$$\frac{1}{\pi} = \frac{1}{16}\sum_{n=0}^\infty \frac{(2n)!^3}{n!^6}\frac{(42\phi-6)n+(5\phi-3)}{(2^{12}\phi^8)^n}\tag1$$
where $\phi=\frac{1+\sqrt{5}}{2}$, and truncate it as for finite number of terms. For example, using just $n=0\;\text{to}\;1$, and getting the reciprocal, it yields,
$$\pi \approx \frac{2^{13}}{3(-383+560\sqrt{5})}$$
It is only good for $10^{-7}$, and the next is $10^{-10}$, but there is an infinite choice of $n$.
There are three formulas in Mathworld that use a $\sqrt{5}$, including a version of $(1)$. And there is also a fourth. However, Ramanujan must have known still another formula because I can't get the approximation in your post by truncating any of the four.
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edited Jan 9, 2015 at 5:15
answered Jan 9, 2015 at 5:06
Tito Piezas IIITito Piezas III
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$\begingroup$ While this doesn't explain how to derive the original approximation, it does provide a method for getting other approximations with better accuracy using $\sqrt{5}$ which was the second part of my question. While I won't accept this as a full answer in its current form, it is still well done and well explained, so it gets my upvote. $\endgroup$
hatch22
– hatch22
2015-01-09 05:28:27 +00:00
Commented Jan 9, 2015 at 5:28
$\begingroup$ @hatch22: Thank you. I was expecting to see the approximation using the four formulas, but sadly, it wasn't. How Ramanujan finds these things is part of the appeal. $\endgroup$
Tito Piezas III
– Tito Piezas III
2015-01-09 05:31:45 +00:00
Commented Jan 9, 2015 at 5:31
$\begingroup$ If the approximation cannot be obtained truncating any one of the four formulas, can it be obtained combining more than one? $\endgroup$
Jaume Oliver Lafont
– Jaume Oliver Lafont
2016-01-30 23:50:20 +00:00
Commented Jan 30, 2016 at 23:50
$\begingroup$ @JaumeOliverLafont: There are in fact dozens and dozens of Ramanujan-Sato pi formulas that use $\sqrt{5}$. It is possible, though not certain, that there is one responsible for this approximation. $\endgroup$
Tito Piezas III
– Tito Piezas III
2016-01-30 23:58:34 +00:00
Commented Jan 30, 2016 at 23:58
1
$\begingroup$ This approximation is one of the many approximations which Ramanujan discussed at length in his paper Modular Equations and Approximations to $\pi$. I have given some details in my answer which you may find interesting. $\endgroup$
Paramanand Singh
– Paramanand Singh ♦
2017-05-17 22:07:46 +00:00
Commented May 17, 2017 at 22:07
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This is not yet a complete answer, but may be useful.
The largest root of the polynomial
$$269x^2-503x+209$$ is $$\frac{17+15\sqrt{5}}{7+15\sqrt{5}}$$
Changing the polynomial to $$(25)^2\times269x^2-25\times63\times 503 x+63^2\times 209$$
modifies the root to the approximation given. $$\pi\approx\frac{63}{25}\times\frac{17+15\sqrt{5}}{7+15\sqrt{5}}$$
In terms of the golden ratio, this is $$\pi\approx\frac{63}{25}\left(1+\dfrac{1}{3\phi-\dfrac{4}{5}}\right)$$
This seems related to the simpler polynomial $$25x^2-90x+36$$ that can have its coefficients factored as $$5^2x^2-5\times 6 \times 3x+6^2$$
and has the largest root $$\pi\approx \frac{9}{5}+\sqrt{\frac{9}{5}},$$
which is another approximation by Ramanujan that has a similar expression in terms of the golden ratio.
$$\pi\approx \dfrac{6}{5}\left(1+\dfrac{1}{\phi-1}\right)$$
This suggests an intermediate approximation of the form $$\pi \approx r_2\left(1+ \dfrac{1}{2\phi+d_2}\right)$$
and hopefully better precision with an expression similar to $$\pi \approx r_4\left(1+ \dfrac{1}{4\phi+d_4}\right)$$
or $$\pi \approx r_5\left(1+ \dfrac{1}{5\phi-d_5}\right)$$ if even multiples of $\phi$ are of no use. Fractions $r_n$ and $d_n$ are yet to be found.
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edited Apr 15, 2017 at 6:40
answered Apr 1, 2016 at 14:25
Jaume Oliver LafontJaume Oliver Lafont
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$\begingroup$ The approximation involving $9/5$ also comes from the same theory of Ramanujan which is used for the approximation in question. I have discussed that also in my answer. $\endgroup$
Paramanand Singh
– Paramanand Singh ♦
2017-05-18 03:59:54 +00:00
Commented May 18, 2017 at 3:59
$\begingroup$ @Jaume It seems your French(?) ancestors knew about Ramanujan's approximation since the 12th century. This new post might be of interest. $\endgroup$
Tito Piezas III
– Tito Piezas III
2025-02-26 11:22:23 +00:00
Commented Feb 26 at 11:22
Add a comment |
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The very general rule with these close approximations is that they're arrived at by finding a continued fraction which has an unusually large term early on. You can then stop at that term and the error is small.
For example, taking the simple continued fraction for $\pi:$
$[3; 7, 15, 1, 292, 1, 1, 1, 2, 1, 3, 1, 14, 2, 1, 1, 2, 2, 2, 2, ...]$
If you stop at the $292$ you get an approximation good to $6$ decimal places.
Therefore to find the next very good approximation similar to your example given, find a continued fraction it's taken from, and stop at its next large term.
$\pi$ is related to the golden ratio $\Phi$ by:
$\pi=\frac{5}{\Phi}\cdot\frac{2}{\sqrt{2+\sqrt{2+\Phi}}}\cdot\frac{2}{\sqrt{\sqrt{2+\sqrt{2+\Phi}}}}\cdots$
Which is thanks to John Baez and Greg Egan
I suspect with a bit of jiggery pokery that might yield your example and its successors.
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answered Apr 15, 2017 at 7:16
Robert FrostRobert Frost
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11971 | https://www.cuemath.com/algebra/rank-of-a-matrix/ | LearnPracticeDownload
Rank of a Matrix
The rank of a matrix is equal to the number of linearly independent rows (or columns) in it. Hence, it cannot more than its number of rows and columns. For example, if we consider the identity matrix of order 3 × 3, all its rows (or columns) are linearly independent and hence its rank is 3.
Let us learn more about the rank of a matrix along with its mathematical definition and let us see how to find the rank of the matrix along with examples.
| | |
--- |
| 1. | What is the Rank of a Matrix? |
| 2. | How to Find the Rank of a Matrix? |
| 3. | Finding Rank of a Matrix by Minor Method |
| 4. | Rank of a Matrix Using Echelon Form |
| 5. | Rank of a Matrix Using Normal Form |
| 6. | Column Rank and Row Rank of a Matrix |
| 7. | Properties of Rank of a Matrix |
| 8. | FAQs on Rank of a Matrix |
What is the Rank of a Matrix?
The rank of a matrix is the order of the highest ordered non-zero minor. Let us consider a non-zero matrix A. A real number 'r' is said to be the rank of the matrix A if it satisfies the following conditions:
every minor of order r + 1 is zero.
There exist at least one minor of order 'r' that is non-zero.
The rank of a matrix A is denoted by ρ (A). Here, "ρ" is a Greek letter that should be read as "rho". So ρ (A) should be read as "rho of A" (or) "rank of A".
How to Find the Rank of a Matrix?
The rank of a matrix can be found using three methods. The most easiest of these methods is "converting matrix into echelon form".
Minor method
Using echelon form
Using normal form
Let us study each of these methods in detail.
Finding Rank of a Matrix by Minor Method
Here are the steps to find the rank of a matrix A by the minor method.
Find the determinant of A (if A is a square matrix). If det (A) ≠ 0, then the rank of A = order of A.
If either det A = 0 (in case of a square matrix) or A is a rectangular matrix, then see whether there exists any minor of maximum possible order is non-zero. If there exists such non-zero minor, then rank of A = order of that particular minor.
Repeat the above step if all the minors of the order considered in the above step are zeros and then try to find a non-zero minor of order that is 1 less than the order from the above step.
Here is an example.
Example: Find the rank of the matrix ρ (A) if A = (\left[\begin{array}{lll} 1 & 2 & 3 \ 4 & 5 & 6 \ 7 & 8 & 9 \end{array}\right]).
Solution:
A is a square matrix and so we can find its determinant.
det (A) = 1 (45 - 48) - 2 (36 - 42) + 3 (32 - 35)= -3 + 12 - 9= 0
So ρ (A) ≠ order of the matrix. i.e., ρ (A) ≠ 3.
Now, we will see whether we can find any non-zero minor of order 2.
(\left|\begin{array}{ll}1 & 2 \ \4 & 5\end{array}\right|) = 5 - 8 = -3 ≠ 0.
So there exists a minor of order 2 (or 2 × 2) which is non-zero. So the rank of A, ρ (A) = 2.
Rank of a Matrix Using Echelon Form
In the above example, what if the first minor of order 2 × 2 that we found was zero? We had to find all possible minors of order 2 × 2 until we get a non-zero minor to make sure that the rank is 2. This process may be tedious if the order of the matrix is a bigger number. To make the process of finding the rank of a matrix easier, we can convert it into Echelon form. A matrix 'A' is said to be in Echelon form if it is either in upper triangular form or in lower triangular form. We can use elementary row/column transformations and convert the matrix into Echelon form.
A row (or column) transformation can be one of the following:
Interchanging two rows.
Multiplying a row by a scalar.
Multiplying a row by a scalar and then adding it to the other row.
Here are the steps to find the rank of a matrix.
Convert the matrix into Echelon form using row/column transformations.
Then the rank of the matrix is equal to the number of non-zero rows in the resultant matrix.
A non-zero row of a matrix is a row in which at least one element is non-zero.
Example: Find the rank of the matrix A = (\left[\begin{array}{lll} 1 & 2 & 3 \ 4 & 5 & 6 \ 7 & 8 & 9 \end{array}\right]) (the same matrix as in the previous example) by converting it into Echelon form.
Solution:
Given matrix is, A = (\left[\begin{array}{lll}1 & 2 & 3 \4 & 5 & 6 \7 & 8 & 9\end{array}\right]).
Apply R2 → R2 - 4R1 and R3 → R3 - 7R1, we get:
(\left[\begin{array}{lll}1 & 2 & 3 \0 & -3 & -6 \0 & -6 & -12\end{array}\right])
Now, we apply R3 → R3 - 2R2, we get:
(\left[\begin{array}{lll}1 & 2 & 3 \0 & -3 & -6 \0 & 0 & 0\end{array}\right])
Now it is in Echelon form and so now we have to count the number of non-zero rows.
The number of non-zero rows = 2 = rank of A.
Therefore, ρ (A) = 2.
Note that we had got the same answer when we calculated the rank using minors.
Rank of a Matrix Using Normal Form
If a rectangular matrix A can be converted into the form (\left[\begin{array}{ll}I_r & 0 \ \0 & 0\end{array}\right]) by using the elementary row transformations, then A is said to be in normal form. Here, I_r is the identity matrix of order "r" and when A is converted into the normal form, its rank is, ρ (A) = r. Here is an example. Converting into normal form is helpful in determining the rank of a rectangular matrix. But it can be used to find the rank of square matrices also.
Example: Find the rank of the matrix A = (\left[\begin{array}{lll} 1 & 2 & 1&2 \ 1 & 3 & 2 & 2 \ 2 & 4 & 3 & 4 \ 3 & 7 & 4 & 6 \end{array}\right]) (again the same matrix) by converting it into normal form.
Solution:
Apply R2 → R2 - R1, R3 → R3 - 2R1, and R4 → R4 - 3R1 we get:
(\left[\begin{array}{lll}1 & 2 & 1&2 \0 & 1 & 1 & 0 \0 & 0 & 1 & 0 \0 & 1 & 1 & 0\end{array}\right])
Now apply, R1 → R1 - 2R2 and R4 → R4 - R2,
(\left[\begin{array}{lll}1 & 0 & -1&2 \0 & 1 & 1 & 0 \0 & 0 & 1 & 0 \0 & 0 & 0 & 0\end{array}\right])
Apply R1 → R1 + R3 and R2 → R2 - R3,
(\left[\begin{array}{lll}1 & 0 & 0 &2 \0 & 1 & 0 & 0 \0 & 0 & 1 & 0 \0 & 0 & 0 & 0\end{array}\right])
Now apply C4 → C4 - 2C1,
(\left[\begin{array}{lll}1 & 0 & 0 &0 \0 & 1 & 0 & 0 \0 & 0 & 1 & 0 \0 & 0 & 0 & 0\end{array}\right])
This is same as (\left[\begin{array}{ll}I_3 & 0 \ \0 & 0\end{array}\right]).
Therefore, the rank of A is, ρ (A) = 3.
Column Rank and Row Rank of a Matrix
When we have calculated the rank of the matrix using echelon form and normal form, we have seen that the rank of the matrix is equal to the number of non-zero rows in the reduced form of matrix. This is actually known as "row rank of matrix" as we are counting the number of non-zero "rows". Similarly, the column rank is the number of non-zero columns, or in other words, it is the number of linearly independent columns. For example, in the above example (of the previous section),
Row rank = the number of non-zero rows = 3
Column rank = the number of non-zero columns = 3
It is very clear from this that "row rank = column rank" here. This is in fact true for any matrix.
Properties of Rank of a Matrix
If A is a nonsingular matrix of order n, then its rank is n. i.e., ρ (A) = n.
If A is in Echelon form, then the rank of A = the number of non-zero rows of A.
If A is in normal form, then the rank of A = the order of the identity matrix in it.
If A is a singular matrix of order n, then ρ (A) < n.
If A is a rectangular matrix of order m x n, then ρ (A) ≤ minimum {m, n}.
The rank of an identity matrix of order n is n itself.
The rank of a zero matrix is 0.
Important Notes on Rank of a Matrix:
While converting the matrix into echelon form or normal form, we can either use row or column transformations. We can also use a mix of row and column transformations.
To find the rank of a matrix by converting it into echelon form or normal form, we can either count the number of non-zero rows or non-zero columns.
Column rank = row rank for any matrix.
The rank of a square matrix of order n is always less than or equal to n.
☛ Related Topics:
Determinant Calculator
Eigenvalue Calculator
Matrix Addition Calculator
Matrix Multiplication Calculator
Rank of a Matrix Examples
Example 1: Is the rank of the matrix A = (\left[\begin{array}{lll} 1 & 1 & -1 \ 2 & -3 & 4 \ 2 & -2 & 3 \end{array}\right]) equal to 3? Justify your answer using determinants.
Solution:
The determinant of the given matrix is,
det(A) = 1 (-9 + 8) - 1 (6 - 8) - 1 (-4 + 6)= 1(-1) - 1 (-2) - 1(2)= -1≠ 0
Therefore, the rank of the matrix A is 3.
Answer: Yes because the determinant of the matrix is NOT 0.
2. Example 2: Find the rank of matrix A mentioned in Example 1 by converting it into Echelon form.
Solution:
The given matrix is:
(\left[\begin{array}{lll}1 & 1 & -1 \2 & -3 & 4 \2 & -2 & 3\end{array}\right])
Apply R2 → R2 - 2R1 and R3 → R3 - 2R1, we get:
(\left[\begin{array}{lll}1 & 1 & -1 \0 & -5 & 6 \0 & -4 & 5\end{array}\right])
Now, we apply R3 → 5R3 - 4R2, we get:
(\left[\begin{array}{lll}1 & 1 & -1 \0 & -5 & 6 \0 & 0 & 6\end{array}\right])
Now it is in upper triangular form (Echelon form) and there are 3 non-zero rows present in it. Thus, ρ (A) = 3.
Answer: ρ (A) = 3.
3. Example 3: Find the rank of the 4x4 matrix (\left[\begin{array}{lll} 0 & 1 & -3 &-1 \ 1 & 0 & 1 & 1 \ 3 & 1 & 0 & 2 \ 1 & 1 & -2 & 0 \end{array}\right]) by converting into normal form.
Solution:
Interchanging the first and second columns:
(\left[\begin{array}{lll}1 & 0 & -3 &-1 \0 & 1 & 1 & 1 \1 & 3 & 0 & 2 \1 & 1 & -2 & 0\end{array}\right])
Now, apply R3 → R3 - R1 and R4 → R4 - R1, we get:
(\left[\begin{array}{lll}1 & 0 & -3 &-1 \0 & 1 & 1 & 1 \0 & 3 & 3 & 3 \0 & 1 & 1 & 1\end{array}\right])
Divide R3 by 3:
(\left[\begin{array}{lll}1 & 0 & -3 &-1 \0 & 1 & 1 & 1 \0 & 1 & 1 & 1 \0 & 1 & 1 & 1\end{array}\right])
Now, apply R3 → R3 - R2 and R4 → R4 - R2, we get:
(\left[\begin{array}{lll}1 & 0 & -3 &-1 \0 & 1 & 1 & 1 \0 & 0 & 0 & 0 \0 & 0 & 0 & 0\end{array}\right])
Apply C3 → C3 + 3C1 and C4 → C4 + C1, we get:
(\left[\begin{array}{lll}1 & 0 & 0 & 0\0 & 1 & 1 & 1 \0 & 0 & 0 & 0 \0 & 0 & 0 & 0\end{array}\right])
Now, apply C3 → C3 - C2 and C4 → C4 - C2, we get:
(\left[\begin{array}{lll}1 & 0 & 0 & 0\0 & 1 & 0 & 0 \0 & 0 & 0 & 0 \0 & 0 & 0 & 0\end{array}\right])
This is same as (\left[\begin{array}{ll}I_2 & 0 \ \0 & 0\end{array}\right]).
Thus, ρ (A) = 2.
Answer: ρ (A) = 2.
4. Example 4: Find the rank of the matrix (\left[\begin{array}{lll} 1 & 0 & -4 & 5\ 2 & -1 & 3 & 0 \ 8 & 1 & 0 & -7 \end{array}\right]) by minor method.
Solution:
Since the given matrix is not a square matrix, we cannot find its determinant. So we will check all 3 × 3 determinants until and we see whether we get at least one non-zero determinant.
(\left|\begin{array}{lll}1 & 0 & -4 \2 & -1 & 3 \8 & 1 & 0\end{array}\right|) = 1 (-3) + 0 - 4 (10) = -3 - 40 = -43 ≠ 0.
Thus, there is a 3 × 3 non-zero minor and hence the rank of the given matrix is 3.
Answer: 3
5. Example 5: Find the column rank and row rank of the matrix given in Example 4 by converting it into echelon form. Is column rank = row rank?
Solution:
The given matrix is, (\left[\begin{array}{ccc}1 & 0 & -4 \2 & -1 & 3 \8 & 1 & 0\end{array}\right]).
We will apply transformations to convert this into upper triangular form (echelon form).
Apply R2 → R2 - 2R1 and R3 → R3 - 8R1,
(\left[\begin{array}{rrr}1 & 0 & -4 \0 & -1 & 11 \0 & 1 & 8\end{array}\right])
Now, apply R3 → R3 + R2,
(\left[\begin{array}{ccc}1 & 0 & -4 \0 & -1 & 11 \0 & 0 & 19\end{array}\right])
Herem the row rank = the number of non-zero rows = 3 and the column rank = the number of non-zero columns = 3.
Answer: Thus, row rank = column rank (= 3).
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FAQs on Rank of a Matrix
What is the Definition of Rank of a Matrix?
The rank of a matrix is the number of linearly independent rows or columns in it. The rank of a matrix A is denoted by ρ (A) which is read as "rho of A". For example, the rank of a zero matrix is 0 as there are no linearly independent rows in it.
How to Find the Rank of the Matrix?
To find the rank of a matrix, we can use one of the following methods:
Find the highest ordered non-zero minor and its order would give the rank.
Convert the matrix into echelon form using the row/column operations. Then the number of non-zero rows in it would give the rank of the matrix.
Convert the matrix into the normal form (\left[\begin{array}{ll}I_r & 0 \ \0 & 0\end{array}\right]), where I_r is the identity matrix of order 'r'. Then rank of the matrix = r.
What is the Rank of a Matrix of Order 3 × 3?
The rank of a matrix of order 3 × 3 is 3 if its determinant is NOT 0. If its determinant is 0, then convert it into Echelon form by using row/column transformations, then the number of non-zero rows/columns would give the rank.
What is the Rank of a Matrix of Order 2 × 2?
If the determinant of a 2 × 2 matrix is NOT 0, then its rank is 2. If its determinant is 0, then its rank is either 1 or 0. The exact rank can be found by converting it into echelon form or normal form.
How to Find the Rank of a Matrix Using Determinant?
To find the rank of a matrix of order n, first, compute its determinant (in the case of a square matrix). If it is NOT 0, then its rank = n. If it is 0, then see whether there is any non-zero minor of order n - 1. If such minor exists, then the rank of the matrix = n - 1. If all the minors of order n - 1 are zeros, then we should repeat the process for minors of order n - 2, and so on until we are able to find the rank.
What is the Rank of a Null Matrix?
Null matrix is a square matrix with all entries to be 0s. The determinant of a null matrix and any of its minor is 0 itself. Hence there doesn't exist any minor of a null matrix that is non-zero. Therefore, the rank of a null matrix is 0.
What is the Shortcut to Find the Rank of a Matrix?
If the determinant of a matrix is not zero, then the rank of the matrix is equal to the order of the matrix. This can be used as a shortcut. But this shortcut does not work when the determinant is 0. In this case, we have to use either minors, Echelon form, or normal form to find the rank like how the processes are explained on this page.
What are the Applications of Rank of Matrix?
The rank of a matrix is mainly useful to determine the number of solutions of a system of equations. If a system has 'n' equations in 'n' variables, then, we first find the rank of the augmented matrix and the rank of the coefficient matrix.
If the rank (augmented matrix) ≠ rank (coefficient matrix), then the system has no solution (inconsistent).
If the rank (augmented matrix) = rank (coefficient matrix) = number of variables, then the system has a unique solution (consistent).
If the rank (augmented matrix) = rank (coefficient matrix) < number of variables, then the system has an infinite number of solutions (consistent).
What Does the Rank of a Matrix Tell Us?
The rank of a matrix would give the number of linearly independent rows (or columns). The more the rank of the matrix the more the linearly independent rows and also the more the informative content.
Can the Rank of a Matrix Ever be Greater than the Number of Rows or Columns?
No, the rank of a matrix is always less than or equal to each of the number of rows and number of columns.
What is the Relation Between the Rank of a Matrix and Eigenvalues?
There is a very close relationship between the rank of a matrix and the eigenvalues. The rank of a matrix is exactly equal to the number of non-zero eigenvalues.
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11972 | https://muratunalphd.com/blog/variance-reduction-in-experiments-intuition/ | Variance Reduction in Experiments, Part 1 - Intuition | Murat Unal
Murat Unal
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Variance Reduction in Experiments, Part 1 - Intuition
The intuition behind variance reduction and why it is important in randomized experiments.
By Murat Unal in Causal InferenceMarketing Analytics
February 1, 2023
code
This is the first part in a series of two articles where we are going to dive deep into variance reduction in experiments. In this article we are going to discuss why variance reduction is necessary and build an intuition behind its mechanism. In the second part we are going to evaluate the latest method in this space: MLRATE, as well as compare it to other well-established methods such as, CUPED. Let’s start with discussing why variance reduction is necessary in experiments. See, when it comes to causal inference, our conclusions may be wrong for two different reasons: systematic bias and random variability. Systematic bias is the bad stuff that happens mainly due to self-selection or unmeasured confounding, whereas random variability, happens because the data we have access to is only a sample of the population we are trying to study. Randomized experiments allow us to eliminate systematic bias, but they are not immune to error based on random variability.
The Difference-in-Means Estimator
Let’s consider the example, where our marketing team wants to figure out the impact of sending an email promotion. As a team of marketing data scientists, we decide to run an experiment by selecting 2000 random customers from our customer base. Our response variable is the amount of spending in the two weeks following the email, and for every customer, we let a coin flip decide whether that customer receives the promotion or not.
Since the treatment is delivered at random, we have no systematic bias, and a simple difference in means (DIM) between the treated (T) and control (C) in our experiment is an unbiased estimate of the population average treatment effect (ATE):
The critical point here is that this estimator gives us the true effect on average, but for a given sample that we end up analyzing, it might as well be far away from it.
In the following code we first generate a sample of 2000 customers, randomly select half of them to treatment that has a constant effect of $5 on customer spending. We then apply the DIM estimator using linear regression on this sample, hoping to recover the true effect.
```
def dgp(n=2000, p=10):
Xmat = np.random.multivariate_normal(np.zeros(p), np.eye(p), size=n).astype('float32')
T = np.random.binomial(1, 0.5, n).astype('int8')
col_list = ['X' + str(x) for x in range(1,(p+1))]
df = pd.DataFrame(Xmat, columns = col_list)
# functional form of the covariates
B = 225 + 50df['X1'] + 5df['X2'] + 20(df['X3']-0.5) + 10df['X4'] + 5df['X5']
# constant ate
tau = 5
Y = (B + tauT + np.random.normal(0,25,n)).astype('float32')
df['T'] = T
df['Y'] = Y
return df
```
data = dgp()
ols = smf.ols('Y ~ T', data = data).fit(cov_type='HC1',use_t=True)
ols.summary().tables
Unfortunately, the DIM estimate for this particular sample is far away than the true effect and is negative in sign. It also comes with a quite wide confidence interval, which prevents us from drawing any conclusions.
On the other hand, if we had the opportunity to repeat this experiment thousands of times in parallel with a different sample of the population each time, we would see that the density of our estimates would peak around the true effect. To see this, let’s create a simulation that generates a sample of 2000 customers and runs the previous experiment every time it is called, for 10000 times.
```
def experiment(kwargs):
dct = {}
n = kwargs['n']
p = kwargs['p']
df = dgp(n,p)
#Difference-in-means
mu_treated = np.mean(df.query('T==1')['Y'])
mu_control = np.mean(df.query('T==0')['Y'])
dct['DIM'] = mu_treated - mu_control
return dct
```
def plot_experiment(results):
results_long = pd.melt(results, value_vars=results.columns.tolist() )
mu = 5
p = (ggplot(results_long, aes(x='value') ) +
geom_density(size=1, color='salmon')+
geom_vline(xintercept=mu, colour='black', linetype='dashed' ) +
annotate("text", x=mu, y=.1, label="True Mean", size=15)+
labs(color='Method') +
xlab('Estimate') +
theme(figure_size=(10, 8))
)
return p
```
%%time
tqdm._instances.clear()
sim = 10000
results = Parallel(n_jobs=8)(delayed(experiment)(n=2000, p=10)\
for _ in tqdm(range(sim)) )
results_df = pd.DataFrame(results)
plot = plot_experiment(results_df)
```
Variation in the Outcome
To understand why we fail to uncover the treatment effect when it is in fact present, we need to think about what actually determines customer spending. If our email promotion only has a limited impact in customer spending compared to other factors such as income, tenure with the business, recency, frequency and value of previous purchases etc. then it would be no surprise to see the variation of spending being explained much more by factors other than our email promotion. In fact, plotting customer spending against the treatment indicator reveals how it varies from around $4 to more than $400. Since we know that the promotion only has a small effect on spending, it becomes quite challenging to detect it inside all the variation. Hence, the regression line from control to treatment has a slight negative slope below.
One way we can make it easier for the DIM estimator to detect the true effect is by increasing our sample size. If we have access to, say 10000 customers instead of 2000, we can get a much closer estimate of the true effect as shown below.
data = dgp(n=10000, p=10)
ols = smf.ols('Y ~ T', data = data).fit(cov_type='HC1',use_t=True)
ols.summary().tables
Regression Adjustment
Now, what if we are constrained in time or other resources and can’t increase our sample size, but still want to be able to detect a small effect? This brings us to the idea of reducing the variance of our outcome, which similar to increasing sample size, makes it easier to detect an effect if it is in fact present.
One of the easiest way to accomplish that, is by using regression adjustment to control for all other factors that determine spending. Notice that the sample we generated for this example has 10 covariates and 5 of them are directly related to the outcome. As mentioned earlier, we can think of them as the following: income, tenure with the business, recency, frequency and value of previous purchases. Including them as controls in a regression means holding them constant while looking at the treatment. This means that if we look at customers with similar levels of income and purchase behavior at the time of the experiment, the variance of the outcome will be smaller and we will be able to capture the small treatment effect as shown next:
```
data = dgp(n=2000, p=10)
ols = smf.ols('Y ~ T+' + ('+').join(data.columns.tolist()[0:10]),
data = data).fit(cov_type='HC1',use_t=True)
ols.summary().tables
```
To aid in intuition, let’s partial out the effects of the covariates from both the treatment and the outcome first, and then plot the resulting residuals.
```
model_t = smf.ols('T ~ ' + ('+').join(data.columns.tolist()[0:10]), data=data).fit(cov_type='HC1',use_t=True)
model_y = smf.ols('Y ~ ' + ('+').join(data.columns.tolist()[0:10]), data=data).fit(cov_type='HC1',use_t=True)
residuals = pd.DataFrame(dict(res_y=model_y.resid, res_t=model_t.resid))
p1=(ggplot(residuals, aes(x='res_t', y='res_y'))+
geom_point(color='c') +
ylab("Spending Amount ($) Residual") +
xlab("Received Coupon Residual") +
geom_smooth(method='lm',se=False, color="salmon")+
theme(figure_size=(10, 8)) +
theme(axis_text_x = element_text(angle = 0, hjust = 1))
)
p1
```
We see how the remaining variation in outcome residuals is much smaller in both groups and the regression line from the residuals of control to the residuals of treatment has a positive slope now.
Comparing the variance of the original spending amount to that of the residuals of the spending amount confirms that we are able to achieve a more than 80% reduction in variance.
print("Spending Amount Variance:", round(np.var(data["Y"]),2))
print("Spending Amount Residual Variance:", round(np.var(residuals["res_y"]),2))
Due to the mechanics of regression, we can obtain the same result by regressing the residuals of the spending amount against the residuals of the treatment:
model_res = smf.ols('res_y ~ res_t', data=residuals).fit()
model_res.summary().tables
Conclusion
Applying a simple regression adjustment to our experiment analysis can go a long way in reducing the variance of our outcome metric, which can be particularly helpful if we are trying to detect a small effect in our experiment. In part 2 of this series we will do a deep dive into finding out the performances of various methods in this space. Specifically, we will be running simulations with varying degrees of complexities in the data generating process and applying MLRATE - machine learning regression-adjusted treatment effect estimator, the latest invention in this space, as well as other methods such as CUPED and regression adjustment. Stay tuned.
Code
The original notebook can be found in my repository.
Posted on:February 1, 2023 Length:7 minute read, 1335 words Categories:Causal InferenceMarketing AnalyticsTags:Causal InferenceMarketing AnalyticsSee Also:Why are Randomized Experiments the Gold Standard in Causal Inference?What is the Value of Improving the Customer Experience in E-Commerce?Identification - The Key to Credible Causal Inference
← Variance Reduction in Experiments, Part 2 - Covariance Adjustment MethodsAn Intuitive Explanation for Inverse Propensity Weighting in Causal Inference →
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Divisibility rules/Rule for 11 proof
A number is divisible by 11 if the alternating sum of the digits is divisible by 11.
Proof
An understanding of basic modular arithmetic is necessary for this proof.
Let where the are base-ten numbers. Then
Note that . Thus
This is the alternating sum of the digits of , which is what we wanted.
Here is another way that doesn't require knowledge of modular arithmetic. Suppose we have a 3-digit number that is expressed in the form:
we then can transpose this into:
and that equals:
which equals
Since the first addend, will always be divisible by 11, we just need to make sure that is divisible by 11.
You can use this for any number. Here it is again, with an even-numbered digit number:
So you just need to check for divisibility with 11.
See also
Back to divisibility rules
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11974 | http://chemistrye.weebly.com/uploads/3/7/7/0/37707825/u1ws7_comb_analy_stoich_review.pdf | AP Chemistry Name ____ Ms. Ye Date __ Block ____ Use Model 1 to answer the following questions 1. What reactant is always required for combustion? 2. Balance the reactions above by placing the proper coefficients in front of each reactant or product. 3. Analyze the balanced equations for the 4 combustion reactions above. a. How is the coefficient of CO2 related to the chemical formula of the hydrocarbon being analyzed? b. If you know the number of moles of CO2 that are formed from the combustion reaction, what can you determine regarding the number of moles of carbon in the hydrocarbon that was combusted? c. How is the coefficient of H2O relate to the chemical formula of the hydrogen carbon being analyzed? d. If you know the number of moles of H2O that are formed from the combustion reaction, what can you determine regarding the number of moles of hydrogen in the hydrocarbon that was combusted? Mass of CO2 produced Moles of CO2 produced Moles of C atoms in unknown Mass of H2O produced Moles of H2O produced Moles of H atoms in unknown C : H atoms in unknown Empirical Formula of Unknown 27.42 g 22.46 g 4. The table above shows some data from a combustion analysis experiment. Calculate the following values and put them in the table: a. The number of moles of CO2 produced b. The number of moles of carbon atoms in the unknown hydrocarbon (refer to 3b) c. The number of moles of H2O produced d. The number of moles of hydrogen atoms in the unknown hydrocarbon (refer to 3d). 5. Determine the ratio of carbon to hydrogen atoms and the empirical formula of your unknown. 6. What other information would you need to determine the molecular formula of the unknown? Mass of CO2 produced Moles of Carbon atoms in unknown Mass of H2O produced Moles of Hydrogen atoms in unknown Total Mass of C and H atoms in unknown Mass of O atoms in unknown Moles of O atoms in unknown Empirical Formula of Unknown 19.10 g 11.73 g 7. The table above shows some data from a combustion analysis experiment that involved a hydrocarbon containing oxygen. Calculate the following values and put them in the table. a. The number of moles of carbon atoms in the unknown hydrocarbon b. The number of moles of hydrogen atoms in the unknown hydrocarbon. 8. Using your values from the question above, find the total number of grams of carbon and hydrogen atoms in your unknown. 9. Since your unknown was a 10 gram sample, use your answer from the question above to find the total mass of oxygen atoms in the unknown. 10. Determine the number of moles of Oxygen atoms in your unknown. 11. Determine the simplest ratio of carbon to hydrogen to oxygen atoms and the empirical formula of your unknown hydrocarbon. Combustion Analysis Practice Problems 1. A 1.50 g sample of an unknown hydrocarbon (C & H only) undergoes complete combustion to produce 4.40 g of CO2 and 2.70 g of H2O. Determine the empirical formula of the compound. 2. A 0.250 g sample of an unknown hydrocarbon (C & H only) undergoes complete combustion to produce 0.845 g of CO2 and 0.173 g of H2O. Determine the empirical formula of the compound. 3. A 0.2500 g sample of a compound containing carbon, hydrogen and oxygen undergoes complete combustion to produce 0.3664 g of CO2 and 0.1500 g of H2O. Determine the empirical formula of the compound. 4. Quinone, which is used in the dye industry and in photography, is an organic compound containing only Carbon, Hydrogen, and Oxygen. It is found that 0.105 g of the compound yields 0.257 g of CO2 and 0.0350 g of H2O when completely combusted. a. Determine the empirical formula of Quinone. b. Given a molecular weight of approximately 108 g/mol, what is Quinone’s molecular formula? STOICHIOMETRY REVIEW Hints: Start with your “given” value and convert to the substance you’re trying to find Include your units when setting up the dimensional analysis problem Remember, the coefficients in the balanced equation tells you the relative number of moles of substance Use the following equation for solving all problems: 4 Al + 3 O2 ----> 2 Al2O3 1) If 74.00 grams of Al were burned in excess oxygen, how many grams of Al2O3 would be produced? 2) If you had 64.00 grams of O2, how many grams of Al would be needed to react with that oxygen? 3) If you wanted to produce 4.00 grams of Al2O3 how many grams of O2 would be needed for the reaction? Answers: 1) 139.8 grams Al2O3 2) 71.95 grams Al 3) 1.88 grams O2 Limiting Reactant & Percent Yield % yield = actual yield x 100 theoretical yield Theoretical yield = smallest mass calculated from the mass of two reactants Actual yield = mass obtained from an experiment (will be given in a word problem) 1. Determine the % yield based on the following information: a) actual yield = 33.5g, theoretical yield = 40.0g b) actual yield = 1.45g, theoretical yield = 2.04g 2. Acrylonitrile, C3H3N, is the starting material for the production of a kind of synthetic fiber acrylic and can be made by the reaction: 4 C3H6 + 6 NO 4 C3H3N + 6 H2O + 1 N2 a) What mass of C3H3N can be made from 21.6g of C3H6? b) What mass of C3H3N can be made from 21.6g of nitrogen monoxide? c) What is the theoretical yield? (the smallest mass from either a) or b) d) If the quantities in a & b are combined, identify the limiting reactant and excess reactant e) If 23.5g of C3H3N is actually made in this process, what is the % yield? Limiting Reactant, Theoretical & Percent Yield Practice 1) Copper (II) chloride reacts with sodium nitrate to produce copper (II) nitrate and sodium chloride: _CuCl2 + NaNO3 _Cu(NO3)2 + _NaCl a) Balance the equation for the reaction given above: b) If 20 grams of copper (II) chloride react with 20 grams of sodium nitrate, how much sodium chloride can be formed? c) Identify your limiting and excess reactants. d) If 11.3 grams of sodium chloride are formed in the reaction, what is the percent yield of this reaction? 2) When lead (II) nitrate reacts with sodium iodide, sodium nitrate and lead (II) iodide are formed. _Pb(NO3)2 (aq) + NaI (aq) _PbI2 (s) + _NaNO3 (aq) a) Balance the equation for the reaction given above. b) If I start with 25.0 grams of lead (II) nitrate and 15.0 grams of sodium iodide, how many grams of sodium nitrate can be formed? c) Identify your limiting and excess reactants. d) If 6 grams of sodium nitrate are formed in the reaction described, what is the percent yield of this reaction? Molarity Review; Solution Stoichiometry Using the Molarity Formula Review: 1. How many moles are in a 2.00 L solution of 1.5 M Sodium hydroxide? 2. How many moles are in 150 mL of 2.50 M Hydrochloric acid? Stoichiometry Problems With Molarity: Find the number of moles first! 1. Using the chemical reaction below, 250 mL of a 3.50 M solution of hydrochloric acid is neutralized with copious amounts of sodium hydroxide to form sodium chloride salt and water. Assuming that the water is boiled off, how many grams of the sodium chloride salt do we expect to form? 1 () + 1 () →1 () + 1 () # moles of hydrochloric acid = 2. Using the chemical reaction below, 100 mL of a 1.25M solution of Sodium Sulfate was reacted with excess amounts magnesium hydroxide. How many grams of sodium hydroxide is expected to form? 1 () + 1 ( ) () →1 () + 2 () 3. Using the chemical reaction below, 125 mL of a 2.75M solution of potassium permanganate was reacted with excess amounts of silver nitrate. How many grams of potassium nitrate is expected to form? 1 () + 1 () →1 () + 1 () |
11975 | https://www.sciencedirect.com/science/article/abs/pii/S0096300315009285 | On 0/1-polytopes with nonobtuse triangulations - ScienceDirect
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Applied Mathematics and Computation
Volume 267, 15 September 2015, Pages 17-27
On 0/1-polytopes with nonobtuse triangulations
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Abstract
Recently, Brandts et al. (2013) studied 0/1-triangulations of the unit n-cube I n with simplices that only have nonobtuse dihedral angles. An example is the standard triangulation into n! simplices. It is proved in that for each n ≥ 3 there is essentially only one other nonobtuse 0/1-triangulation of I n. Here we will outline an investigation into 0/1-triangulations of other 0/1-polytopes with simplices that only have nonobtuse dihedral angles. As far as we know, this is the only source that combines both concepts 0/1-polytopes and nonobtuse 0/1-triangulations. In particular, we investigate nonobtuse 0/1-triangulations of 0/1-polytopes in I 3 and I 4.
Introduction
Triangulations are a natural way to divide a space into smaller parts that are easy to study. They play an important role in many different areas of mathematics. The minimal number of simplices needed in triangulations of the unit n-cube I n has been studied by several authors , , , , , . Here we perform a first investigation into triangulations of 0/1-polytopes with simplices that only have nonobtuse dihedral angles. In particular, we investigate nonobtuse triangulations of 0/1-polytopes in R 3 and R 4. For example, I 3 is a 0/1-polytope and the minimal triangulation of I 3 contains five nonobtuse simplices. It is thus both a minimal and a nonobtuse triangulation.
Section 2 presents the concept of a 0/1-polytope as the convex hull of a subset of the vertices of I n. Trivially, I n itself is a 0/1-polytope. Since I n has 2 n vertices, the cardinality of the power set of the vertex set is 2 2 n. So, the number of all 0/1-polytopes increases dramatically with the dimension of the cube. For this reason, several ways to classify 0/1-polytopes have been proposed. Here we do this by means of the action of the hyperoctahedral group on I n and define 0/1-polytopes to be 0/1-equivalent if there exists a symmetry of I n that maps one onto the other. For other interesting classifications and a good introductory text into the topic of 0/1-polytopes we refer to Chapter 1 of.
Section 3 introduces the concept of simplices. This section also outlines a linear algebraic description of the simplex, its facets and the set of the inward normals to its facets. The set of inward normals to the facets of a simplex enables us to define the dihedral angle between two facets. Then, the concept nonobtuse simplex will be introduced and we define and study some special nonobtuse simplices and their properties.
Section 4 starts with presenting the concept of triangulations and introduces the nonobtuse 0/1-triangulations. Section 4.1 explicitly lists the nine 0/1-polyhedra having a nonobtuse 0/1-triangulation. Furthermore, we present an algorithm that produces the 28 0/1-polytopes in I 4 that allow a nonobtuse triangulation. These 0/1-polytopes are listed in Tables 3, 4 and 5 and so are their triangulations.
In Section 5 we introduce the single neighbor conjecture. The conjecture states that each interior facet F of a nonobtuse 0/1-simplex S has at most two nonobtuse neighbors. Here a nonobtuse neighbor of the facet F is a nonobtuse simplex that contains F as a facet, being S and at most one other nonobtuse simplex on the opposite side of F (in the other half-space defined by the hyperplane in which F lies). Although it has not yet been established that the conjecture holds for all nonobtuse 0/1-simplices, it has been verified computationally for all space dimensions n ≤ 8. Also, Section 5 outlines this verification. The validity of single neighbor conjecture is an important tool to make computer programs more efficient in the sense that 0/1-polytopes can be built up as a 0/1-triangulation: put one nonobtuse simplex in its place, and the way in which consecutive simplices can be added until the 0/1-triangulation is uniquely determined.
In Section 6 we study further consequences of the single neighbor conjecture. This section introduces nonobtuse neighbor networks and nonobtuse 0/1-conglomerates. The concept 0/1-conglomerates is a generalization of 0/1-polytope. Moreover, we describe how to generate, up to 0/1-equivalence, all nonobtuse 0/1-conglomerates.
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Section snippets
0/1-polytopes
The convex hull of a subset V of the set B n={0,1}n of vertices of I n is called a 0/1-polytope. A simple example of a 0/1-polytope is I n itself. There are 2 2 n distinct 0/1-polytopes in I n. To determine the exact numbers of full-dimensional 0/1-polytopes in I n is a very hard problem. Table 1, taken from Zong, shows the number of full-dimensional 0/1-polytopes in I n for n ≤ 5.
So far, we did not take into account the symmetries of I n, and now we will do that. Let B n be the set of all rigid
Simplices
An n-simplex S is the convex hull of the origin v 0=0 and n points v 1,…,v n from R n, the vertices of S, that do not all lie in a hyperplane. An n-simplex has (n+1 2) edges and n+1 facets. We define the matrix representation of a simplex as the n × n matrix whose columns are the vertices of S other than the origin. Note that this matrix is nonsingular.
Let P=(v 1|⋯|v n) and denote for given j∈{0,…,n} the facet opposite to v j by F j. Thus, F j is the facet of the simplex that does not contain v j. Hence, F
Triangulations
Our interest in nonobtuse triangulations stems from the following two facts:
1. Finite element methods for the numerical approaches of solutions of elliptic PDEs are known to satisfy so-called discrete maximum principles if the triangulation of the physical domain consists of nonobtuse simplices, .
2. As a special case of the question what the minimal amount of simplices is, with which I n can be 0/1-triangulated, considered triangulation of I n using nonobtuse 0/1-simplices only. Here,
The single neighbor conjecture
In this section we introduce the single neighbor conjecture. The validity of this conjecture can help us to make computer programs to generate nonobtuse 0/1-triangulations more efficient. The conjecture states that each interior facet F of a nonobtuse 0/1-simplex S has at most two nonobtuse 0/1-simplices as neighbors. Here a nonobtuse neighbor of the facet F is a nonobtuse simplex that contains F as a facet, being S and at most one other nonobtuse simplex on the opposite side of F (in the other
Consequences of the single neighbor conjecture
In this section, we introduce nonobtuse neighbor networks and nonobtuse 0/1-conglomerates to generate 0/1-polytopes in an inexpensive way that allow a nonobtuse 0/1-polytope. The concept 0/1-conglomerates is a generalization of 0/1-triangulation. Moreover, we describe how to generate, up to 0/1-equivalence, all nonobtuse 0/1-conglomerates.
Acknowledgments
This work is part of the research project 613.001.019, which is financed by the Netherlands Organisation for Scientific Research (NWO) and is made possible through the substantial help and support from Jan Brandts and Michal Křížek.
Recommended articles
References (17)
J.H. Brandts et al.
Dissection of the path-simplex in R n into n path-subsimplices
Linear Algebra Appl. (2007)
J.H. Brandts et al.
The discrete maximum principle for linear simplicial finite element approximations of a reaction-diffusion problem
Linear Algebra Appl. (2008)
J.H. Brandts et al.
There are only two nonobtuse binary triangulations of the unit n-cube
Comput. Geom. (2013)
R.B. Hughes
Minimum-cardinality triangulations of the d-cube for d=5 and d=6
Discrete Math. (1993)
R.B. Hughes et al.
Simplexity of the cube
Discrete Math. (1996)
O. Aichholzer, Extremal properties of 0/1-polytopes of dimension 5, Polytopes – Combinatorics and Computation...
J.H. Brandts et al.
On nonobtuse simplicial partitions
SIAM Rev. (2009)
R.W. Cottle, Minimal triangulation of the 4-cube, Discrete MathematicsNorth-Holland Publishing Company 40 (1982)...
There are more references available in the full text version of this article.
Cited by (2)
Foundations of space-time finite element methods: Polytopes, interpolation, and integration
2021, Applied Numerical Mathematics Show abstract The main purpose of this article is to facilitate the implementation of space-time finite element methods in four-dimensional space. In order to develop a finite element method in this setting, it is necessary to create a numerical foundation, or equivalently a numerical infrastructure. This foundation should include a collection of suitable elements (usually hypercubes, simplices, or closely related polytopes), numerical interpolation procedures (usually orthonormal polynomial bases), and numerical integration procedures (usually quadrature rules). It is well known that each of these areas has yet to be fully explored, and in the present article, we attempt to directly address this issue. We begin by developing a concrete, sequential procedure for constructing generic four-dimensional elements (4-polytopes). Thereafter, we review the key numerical properties of several canonical elements: the tesseract, tetrahedral prism, and pentatope. Here, we provide explicit expressions for orthonormal polynomial bases on these elements. Next, we construct fully symmetric quadrature rules with positive weights that are capable of exactly integrating high-degree polynomials, e.g. up to degree 17 on the tesseract. Finally, the quadrature rules are successfully tested using a set of canonical numerical experiments on polynomial and transcendental functions.
### Enumeration and investigation of acute 0/1-simplices modulo the action of the hyperoctahedral group
2017, Special Matrices
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11976 | https://math.stackexchange.com/questions/3910015/what-exactly-is-mathbbz-n | Stack Exchange Network
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What exactly is $\mathbb{Z_n}$?
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$\begingroup$
My impression has been that $\mathbb{Z_n}$ is the set ${0,1,...,n-1}$ under binary operation addition modulo $n$. However I'm also coming across this notion that $\mathbb{Z_n}$ is actually a set of equivalence classes of equivalence relation $x\sim y \iff x \equiv y$ mod $n$ and the addition here is actually addition of equivalence classes rather than simply addition of integers modulo $n$. Is this correct? So would it be correct to say $\mathbb{Z_n} = {n,n+1,...,2n-1}$? if we are considering these elements as being equivalence classes?
group-theory
modular-arithmetic
equivalence-relations
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edited Nov 16, 2020 at 19:17
Shaun♦
48.1k2121 gold badges7575 silver badges188188 bronze badges
asked Nov 16, 2020 at 17:51
Henry BrownHenry Brown
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$\endgroup$
4
5
$\begingroup$ The definitions are equivalent but the equivalence class definition is cleaner in several ways (e.g. it's easier to prove commutativity and associativity and so forth). $\endgroup$
Qiaochu Yuan
– Qiaochu Yuan
2020-11-16 17:58:04 +00:00
Commented Nov 16, 2020 at 17:58
$\begingroup$ A tentative yes.... Technically speaking $\mathbb Z/n\mathbb Z$ is a more correct notation for the set of equivalence classes. And technically there should be some form of notation to distinguish $k$ being the integer $k$ and $[k]= {...,k-2n,k-n,k,k+n,k+2n,....}$ but the equivalence class of $k$. But as $[k] = [n+k]$ we can write $\mathbb Z/n\mathbb Z = {,,....,[n-1]} = {[n],[n+1],.....,[2n-1]}$. Some people will get really fussy that you don't use the notation $\mathbb Z_n$ but.... I have to admit it's really convenient. $\endgroup$
fleablood
– fleablood
2020-11-16 18:02:56 +00:00
Commented Nov 16, 2020 at 18:02
$\begingroup$ One can (informally) think about the quotient set $A/E$, where $E$ is an equivalence relation, as having the same elements as $A$ but the equality there is replaced by $E$. So in $\Bbb Z_n$ we have $-n=0=n=2n,\ -n+1=1=n+1=2n+1$... $\endgroup$
Berci
– Berci
2020-11-16 18:10:28 +00:00
Commented Nov 16, 2020 at 18:10
$\begingroup$ Thanks for the answers. I'm still a little unclear but I get the sense this isn't something that's clear cut to begin with. What I can definitely see is that my original understanding on what $\mathbb{Z_n}$ is, was not fully correct. $\endgroup$
Henry Brown
– Henry Brown
2020-11-16 18:22:25 +00:00
Commented Nov 16, 2020 at 18:22
Add a comment |
3 Answers 3
Reset to default
6
$\begingroup$
To put it shortly: yes.
What isn't important is what $\mathbb{Z}_n$ "is" as a set. Rather, what's important is that all these different formulations of $\mathbb{Z}_n$ (equivalence classes of integers, select integers with a certain operation, a different set of integers with a certain operation) all give rise to an algebraic structure (groups, rings, fields, depending on what you're working with) which are all isomorphic.
Though, if you're saying that $\mathbb{Z}_n$ is equivalence classes, then the accepted notation would probably look more like $\mathbb{Z}_n = {[n]_n, [n + 1]_n, \dots}$, or perhaps even $\mathbb{Z}_n = {n + n\mathbb{Z}, (n + 1) + n\mathbb{Z}, \dots}$, depending on your background and your personal aesthetic concerns. $n$, after all, is an integer, not an equivalence class of integers.
As Qiaochu Yian says in his comment, the equivalence class definition is usually preferred, as this makes certain proofs much easier.
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edited Nov 16, 2020 at 18:03
answered Nov 16, 2020 at 17:57
Duncan RamageDuncan Ramage
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$\endgroup$
2
2
$\begingroup$ I think $\mathbb{Z}_n = {n\mathbb{Z}, (n+1)\mathbb{Z}, \dots, (2n-1)\mathbb{Z}}$ is an extremely nonstandard convention -- this set certainly has $n$ elements, but the elements are not equivalence classes modulo $n$. Did you mean $n + n\mathbb{Z}, (n+1) + n\mathbb{Z}, \dots$? $\endgroup$
diracdeltafunk
– diracdeltafunk
2020-11-16 18:02:02 +00:00
Commented Nov 16, 2020 at 18:02
$\begingroup$ @diracdeltafunk Yes, I did, thank you. I wanted to incorporate the OP's non-standard example and tripped up the cosets. $\endgroup$
Duncan Ramage
– Duncan Ramage
2020-11-16 18:02:59 +00:00
Commented Nov 16, 2020 at 18:02
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Below I explain the general idea, which works not only for rings but also for any algebraic structure definable by $\rm\color{#c00}{universal}$ equational axioms ("identities") such as $\,\color{#c00}{\forall x,y,z!:}\,x(y+z) = xy+xz\,$.
$a\equiv b\pmod{! n}\iff n\mid a-b\,$ is a congruence relation (on a ring) i.e. it is an equivalence relation that is furthermore compatible with all of the ring operations addition and multiplication, i.e. congruence satisfies the following Congruence Sum & Product Rules
$$\begin{align} a_1\equiv b_1\ a_2\equiv b_2\end{align}\ \Rightarrow\ \begin{array}{}a_1 + a_2\equiv b_1+b_2\ a_1 \times a_2\equiv b_1 \times b_2 \end{array}\qquad$$
This implies that the ring operations descend to well-defined operations on the equivalence classes $\,[a] = a+n\Bbb Z\,$ via $\,[a]+[b] := [a+b],\ [a]\times [b] := [a\times b],\,$ and the map $\, a\mapsto [a]\,$ is a surjective ring hom, which immediately implies that all the ring laws persist in the image $\,\Bbb Z_n,\,$ so $\,\Bbb Z_n\,$ has associative and commutative addition and multiplication, connected via the distributive law, so arithmetic in $\,\Bbb Z_n\,$ is essentially the same as in $\,\Bbb Z,\,$ except that some elements are forced equal.
For computational purposes it is often convenient to map the classes to normal (canonical) representatives $\,h\,:\, [a]\mapsto \bar a.\,$ The most common choice is its least nonnegative element $\,\bar a := a\bmod n\,$ (remainder reps), but also convenient are least magnitude reps, e.g. $\,0,\pm1,\pm2\pmod{!5}.\,$ Generally we can use any complete system of residues, i.e. any $\,n\,$ integers such that every integer is congruent to exactly one in our set of complete residue reps.
Then we can transport the ring structure to our normal forms by pulling (back) the ring operations along $h$ to obtain the induced ring operations on the normal forms as follows:
$$\color{#c00}{\bar a} + \bar b = h([a])+h([b]) = h([a]+[b]) = h([a+b]) = \color{#0a0}{\overline{a+b}}$$
e.g. this becomes $\ \color{#c00}{a\bmod n} + b\bmod n\, = \, \color{#0a0}{(a+b)\bmod n}\ $ using common normal forms.
Said equivalently, to perform an operation on normal forms $\,\bar a,\,\bar b,\,$ we apply $h^{-1}$ to map them to their associated classes $\,[a],[b],\,$ then we perform the operation on the classes yielding $\,[a+b],\,$ then finally we apply $h$ to map that result to its normal form $\,\overline{a+b}.\,$ So the normal forms are essentially canonical "labels" or "names" for their congruence classes. We could instead use any set of $\,n\,$ elements as labels, but using elements from the original ring $\,\Bbb Z\,$ makes it more intuitive how the normal rep corresponds to the class.
As above, in Euclidean domains like $\,\Bbb Z\,$ and $\,F[x]\,$, which enjoy Euclidean division with smaller remainder, it is convenient to use the the remainder ("least Euclidean size") as the normal rep, which is discussed further here, showing how Hamilton's pair representation for complex numbers is just a special case of this, viz. $\Bbb R[x]\bmod x^2!+!1\cong \Bbb R[i]\cong \Bbb C$.
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edited Jan 20, 2022 at 18:12
answered Nov 16, 2020 at 23:52
Bill DubuqueBill Dubuque
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To my knowledge, possibly the main plus of the definition of $\Bbb Z_n$ as the quotient set (group under addition modulo $n$, actually) $\Bbb Z/\sim$, where $x\sim y \stackrel{(def.)}{\iff} x-y \in n\Bbb Z$, is that it leads to the generalization to any group $G$ (in place of $\Bbb Z$) and any subgroup $H$ (in place of $n\Bbb Z$), and finally to the powerful notion of coset.
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answered Nov 16, 2020 at 20:10
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1.5 有理数的乘方
第二章
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2.2 整式的加减
第三章
3.1 从算式到方程
3.2 解一元一次方程(一)——合并同类项与移项
3.3 解一元一次方程(二)——去括号与去分母
3.4 实际问题与一元一次方程
第四章
4.1 几何图形
4.2 直线、射线、线段
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Mahalanobis Distance usage in Machine learning
Dilip Kumar
6 min readJun 15, 2025
1. Intuition: What is Mahalanobis Distance?
Imagine you have a cloud of points (say, heights and weights of people). Some people might lie far from the average, but just being far doesn’t always mean they’re an outlier — maybe the data is naturally stretched in one direction.
Mahalanobis Distance measures how many standard deviations away a point is from the mean, considering the shape of the data (covariance).
Image Source:
It differs from Euclidean distance:
Euclidean: Straight-line distance (doesn’t care about direction or scale of data).
Mahalanobis: Scales the distance based on variance and correlation in the data.
2. Formula
Where:
x: the data point
μ: the mean of the dataset
Σ: the covariance matrix
Σ−1: inverse of the covariance matrix
3. Usage in Machine Learning
Outlier Detection
If a point has a high Mahalanobis distance, it’s likely an outlier — because it’s far from the bulk of the data considering the data’s spread and shape.
Multivariate anomaly detection
Works better than Z-score or Euclidean when features are correlated.
4. Python Example
Let’s detect an outlier from 2D data (height, weight).
```
import numpy as npfrom scipy.spatial import distance# Sample 2D data (height, weight)X = np.array([ [170, 65], [165, 62], [180, 75], [175, 71], [190, 80],])# Step 1: Mean of datamean = np.mean(X, axis=0)# Step 2: Covariance matrixcov_matrix = np.cov(X, rowvar=False)# Step 3: Inverse of covariance matrixinv_cov = np.linalg.inv(cov_matrix)# Step 4: Compute Mahalanobis distance for each pointdistances = [distance.mahalanobis(x, mean, inv_cov) for x in X]# Step 5: Display resultsfor i, d in enumerate(distances): print(f"Point {X[i]} --> Mahalanobis Distance = {d:.2f}")# Threshold: say anything > 2.5 is outlierprint("\nOutliers: Distance > 1.5")for i, d in enumerate(distances): if d > 1.5: print(f" Outlier: {X[i]} with distance {d:.2f}") import as from import
Sample 2D data (height, weight) 170 65 165 62 180 75 175 71 190 80
Step 1: Mean of data 0
Step 2: Covariance matrix False
Step 3: Inverse of covariance matrix
Step 4: Compute Mahalanobis distance for each point for in
Step 5: Display results for in enumerate printf"Point {X[i]} --> Mahalanobis Distance = {d:.2f}"{X[i]}{d:.2f}.2
Threshold: say anything > 2.5 is outlier print"\nOutliers: Distance > 1.5" for in enumerate if1.5 printf" Outlier: {X[i]} with distance {d:.2f}"{X[i]}{d:.2f}.2
```
Point [170 65] --> Mahalanobis Distance = 1.11Point [165 62] --> Mahalanobis Distance = 1.18Point [180 75] --> Mahalanobis Distance = 1.24Point [175 71] --> Mahalanobis Distance = 0.96Point [190 80] --> Mahalanobis Distance = 1.71Outliers: Distance > 1.5 Outlier: [190 80] with distance 1.71 170 65
--> Mahalanobis Distance = 1.11 165 62
--> Mahalanobis Distance = 1.18 180 75
--> Mahalanobis Distance = 1.24 175 71
--> Mahalanobis Distance = 0.96 190 80
--> Mahalanobis Distance = 1.71>1.5 190 80 with1.71
Output Insight
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Here [190, 85] point likely has the largest Mahalanobis distance and is detected as an outlier.
5. Comparison with Alternatives
Brief Intuition for Each
1. Mahalanobis Distance
Measures how far a point is from the center, accounting for shape and correlation.
Great for small to medium-sized data where data is roughly elliptical (e.g., Gaussian).
2. Z-score
Measures how many standard deviations a value is from the mean, per feature.
Doesn’t consider correlation between features.
Easy, but not great for multivariate outliers.
3. PCA-based Outlier Detection
Projects data to a lower dimension (e.g., 2 or 3 PCs).
Reconstructs the original data.
Large reconstruction error = likely an outlier.
Assumes linear structure.
4. Isolation Forest
Randomly splits data using decision trees.
Outliers are easier to isolate → shorter paths.
Works well even with non-Gaussian, high-dimensional, and nonlinear data.
6. When to Use Mahalanobis
Features have different scales or are correlated
Data is Gaussian-like
We want probabilistic distance measurement
Usage
Mahalanobis distance tells us how “far” a point is, not in straight line but in statistical space.
Great for outlier detection, especially with correlated features.
More robust than Euclidean when data is multivariate and non-spherical.
7.0 Visual plot of Mahalanobis distances
```
import numpy as npimport matplotlib.pyplot as pltfrom scipy.spatial import distance# Sample 2D data (e.g. height in cm, weight in kg)X = np.array([ [170, 65], [165, 62], [180, 75], [175, 71], [190, 80], # Possible outliar])# Step 1: Calculate mean and covariancemean = np.mean(X, axis=0)cov_matrix = np.cov(X, rowvar=False)inv_cov_matrix = np.linalg.inv(cov_matrix)# Step 2: Compute Mahalanobis distancesdistances = np.array([distance.mahalanobis(x, mean, inv_cov_matrix) for x in X])# Step 3: Define threshold (you can tune this)threshold = 1.5outliers = distances > threshold# Step 4: Plottingplt.figure(figsize=(8, 6))scatter = plt.scatter(X[:, 0], X[:, 1], c=distances, cmap='coolwarm', s=100, edgecolor='black')# Annotate points with distancesfor i, d in enumerate(distances): plt.text(X[i, 0] + 0.5, X[i, 1], f'{d:.2f}', fontsize=9)# Plot the mean pointplt.scatter(mean, mean, color='green', marker='X', s=200, label='Mean')# Highlight outliersplt.scatter(X[outliers, 0], X[outliers, 1], facecolors='none', edgecolors='red', s=200, label='Outlier')# Plot settingsplt.title('Mahalanobis Distance Outlier Detection')plt.xlabel('Height')plt.ylabel('Weight')plt.colorbar(scatter, label='Mahalanobis Distance')plt.legend()plt.grid(True)plt.tight_layout()plt.show() import as import as from import
Sample 2D data (e.g. height in cm, weight in kg) 170 65 165 62 180 75 175 71 190 80
Possible outliar
Step 1: Calculate mean and covariance 0 False
Step 2: Compute Mahalanobis distances for in
Step 3: Define threshold (you can tune this)1.5
Step 4: Plotting 8 6 0 1 'coolwarm' 100 'black'
Annotate points with distances for in enumerate 00.5 1f'{d:.2f}'{d:.2f}.2 9
Plot the mean point 0 1 'green' 'X' 200 'Mean'
Highlight outliers 0 1 'none' 'red' 200 'Outlier'
Plot settings 'Mahalanobis Distance Outlier Detection' 'Height' 'Weight' 'Mahalanobis Distance' True
```
8.0 Mahalanobis Distance to multivariate outlier detection
8.1 Example: 3D Data (Height, Weight, Age)
We’ll:
Create synthetic data with 3 features.
Compute Mahalanobis distances.
Flag and visualize the outliers.
8.2 Python Code for Multivariate Outlier Detection (3 Features)
```
import numpy as npimport matplotlib.pyplot as pltfrom scipy.spatial import distance# Step 1: Create 3D sample data (Height, Weight, Age)X = np.array([ [170, 65, 30], [165, 62, 28], [180, 75, 35], [175, 68, 31], [172, 67, 29], [190, 85, 50], # possible outlier [160, 50, 22], # possible outlier])# Step 2: Compute Mahalanobis distancemean = np.mean(X, axis=0)cov = np.cov(X, rowvar=False)inv_cov = np.linalg.inv(cov)distances = np.array([distance.mahalanobis(x, mean, inv_cov) for x in X])# Step 3: Set threshold for outliers (tunable)threshold = 2.0outliers = distances > threshold# Step 4: Print resultsprint("Mahalanobis Distances:")for i, d in enumerate(distances): print(f"Point {i}: {X[i]} --> Distance = {d:.2f} {'<-- OUTLIER' if outliers[i] else ''}")# Step 5: Visualize as bar chartplt.figure(figsize=(10, 5))bars = plt.bar(range(len(X)), distances, color=['red' if o else 'blue' for o in outliers])plt.axhline(y=threshold, color='green', linestyle='--', label='Threshold')plt.xticks(range(len(X)), [f'Point {i}' for i in range(len(X))])plt.ylabel("Mahalanobis Distance")plt.title("Multivariate Outlier Detection using Mahalanobis Distance")plt.legend()plt.grid(True)plt.tight_layout()plt.show() import as import as from import
Step 1: Create 3D sample data (Height, Weight, Age) 170 65 30 165 62 28 180 75 35 175 68 31 172 67 29 190 85 50
possible outlier 160 50 22
possible outlier
Step 2: Compute Mahalanobis distance 0 False for in
Step 3: Set threshold for outliers (tunable)2.0
Step 4: Print results print"Mahalanobis Distances:" for in enumerate printf"Point {i}: {X[i]} --> Distance = {d:.2f} {'<-- OUTLIER' if outliers[i] else ''}"{i}{X[i]}{d:.2f}.2{'<-- OUTLIER' if outliers[i] else ''}'<-- OUTLIER' if else ''
Step 5: Visualize as bar chart 10 5 range len 'red' if else 'blue' for in 'green''--' 'Threshold' range lenf'Point {i}'{i} for in range len "Mahalanobis Distance" "Multivariate Outlier Detection using Mahalanobis Distance" True
```
Mahalanobis Distances:Point 0: [170 65 30] --> Distance = 0.57 Point 1: [165 62 28] --> Distance = 1.97 Point 2: [180 75 35] --> Distance = 1.34 Point 3: [175 68 31] --> Distance = 1.16 Point 4: [172 67 29] --> Distance = 0.99 Point 5: [190 85 50] --> Distance = 2.24 <-- OUTLIERPoint 6: [160 50 22] --> Distance = 2.16 <-- OUTLIERMahalanobis Distances:Point 0: 170 65 30 --> Distance =0.57Point 1: 165 62 28 --> Distance =1.97Point 2: 180 75 35 --> Distance =1.34Point 3: 175 68 31 --> Distance =1.16Point 4: 172 67 29 --> Distance =0.99Point 5: 190 85 50 --> Distance =2.24<-- OUTLIERPoint 6: 160 50 22 --> Distance =2.16<-- OUTLIER
9.0 Mahalanobis distance vs Standard Deviation Method (1D)
9.1 Standard Deviation Method (1D)
Idea:
For a normal distribution, most values lie within:
68% → ±1σ,
95% → ±2σ,
99.7% → ±3σ.
Outliers = data points outside a certain number of standard deviations from the mean.
9.2. Mahalanobis Distance (1D)
In 1D, Mahalanobis distance simplifies to:
This is exactly the Z-score — so in 1D, Mahalanobis distance and Z-score are equivalent.
However, Mahalanobis becomes more powerful in higher dimensions, because it takes feature correlation into account, which standard deviation does not.
9.3 Visual Comparison in 1D
We’ll:
Generate synthetic 1D data.
Plot:
The mean ± 1σ, 2σ, 3σ thresholds.
Mahalanobis distance (which is same as Z-score in 1D).
Highlight outliers detected by both methods.
```
import numpy as npimport matplotlib.pyplot as plt# Generate 1D data with some outliersnp.random.seed(0)data = np.concatenate([ np.random.normal(50, 5, 100), # main cluster [80, 5] # outliers])mean = np.mean(data)std = np.std(data)# Z-score / Mahalanobis distance in 1Dz_scores = np.abs((data - mean) / std)# Thresholdsthreshold = 3 # standard deviation thresholdoutliers_std = z_scores > threshold # same as Mahalanobis > 3 in 1D# Plotplt.figure(figsize=(10, 4))plt.scatter(range(len(data)), data, c=['red' if o else 'blue' for o in outliers_std])plt.axhline(mean, color='green', linestyle='--', label='Mean')for i in range(1, 4): plt.axhline(mean + i std, color='gray', linestyle=':', label=f'+{i}σ' if i == 1 else None) plt.axhline(mean - i std, color='gray', linestyle=':')plt.title('1D Outlier Detection: Standard Deviation vs Mahalanobis (Same in 1D)')plt.xlabel('Index')plt.ylabel('Value')plt.legend()plt.grid(True)plt.tight_layout()plt.show() import as import as
Generate 1D data with some outliers 0 50 5 100
main cluster 80 5
outliers
Z-score / Mahalanobis distance in 1D abs
Thresholds 3
standard deviation threshold
same as Mahalanobis > 3 in 1D
Plot 10 4 range len 'red' if else 'blue' for in 'green''--' 'Mean' for in range 1 4 'gray'':'f'+{i}σ'{i} if 1 else None 'gray'':''1D Outlier Detection: Standard Deviation vs Mahalanobis (Same in 1D)' 'Index' 'Value' True
```
Enjoy learning !!!
Mahalanobis Distance
Machine Learning
## Written by Dilip Kumar
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11979 | https://blog.agnibho.com/post/forensic-gunshot-injury/ | Gunshot injuries in Forensic Medicine | Blog of Dr. Agnibho Mondal
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Gunshot injuries in Forensic Medicine
When we discuss firearms in Forensic Medicine we are mainly concerned about the effect of firearms on human bodies, especially the information that can be collected from a wound about the firearm used and the manner in which it has been used to aid in the judicial procedures.
To discuss it any further we need to understand the firearms first—
A firearm is essentially a device that accelerates a projectile towards a target by means of energy generated from chemical reactions, usually a contained explosion.
The movement of the projectile (aka the bullet) can be divided in three parts—
Internal ballistics : This is the path of the bullet before it leaves the muzzle. We will discuss it when talking about the inner workings of firearms.
External ballistics : This is the path of the bullet from leaving the muzzle to reaching its target.
Wound ballistics : This is the path of the bullet inside the human body if it hits one. This is what we are most interested in.
Types of Firearms
There are three types of firearms that we need to be concerned about—
Rifled firearms : This type of firearms has spiral grooves cut into the inside of the barrel called rifling. It helps in rotating the bullet very fast as it leaves the muzzle. It helps to increase accuracy and effective range by means of gyroscopic stability of the bullet.
Shotguns : These are smooth bore firearms, meaning they don't have any rifling. They usually fire multiple small pellets at the same time. It helps in acquiring the target easily by spreading the projectiles on a wider front at the cost of the range.
Handguns : They are smaller versions of rifled firearms meant for easy carrying. E.g. pistols, revolvers etc. Their smaller barrel length limits their range and they are usually low powered than their bigger counterparts.
Parts of a modern hunting rifle.
Parts of a shotgun (Winchester 1897).
Parts of a Pistol (Glock 17).
Other types of firearms include gyrojet guns (propellant charge contained in the bullet), paradox guns (shotgun with terminal rifling) etc.
Some similar weapons, but not technically firearms, are airguns, stud-guns etc.
Insides of a firearm
A firearm has four essential components—
Ammunition
Chamber
Firing mechanism
Barrel
We will discuss them one by one.
Ammunition
They are usually called cartridges which is essentially a self-contained package of the primer, charge and bullet.
| Rifled firearms | Shot guns |
--- |
| Parts of a rifled firearm cartridge. | Parts of a shotgun cartridge. |
| They can have three types of bullets— Non-jacketed : They are made of lead only and causes fouling of the barrel. But they cause massive tissue injury due to expansion and fragmentation in the wound. Jacketed : They have a metal jacket over the lead core. It prevents fouling of the barrel but tissue injury is limited due to almost no expansion or fragmentation of the bullet. Semi-jacketed : They have a metal jacket over the lead core but the tip is left exposed. Prevent fouling of the barrel but also causes massive tissue injury due to expansion and fragmentation of the exposed tip. | They may have three types of projectiles— Shots : They are multiple lead balls called pellets. They can easily hit targets due to spraying of projectiles on a wide front. Shots come in different sizes e.g. bird shot (smaller), buck shot (larger) etc. Slugs : They are single projectiles. They can deliver more energy due to their more mass than the individual pellets. Slugs may contain rifling on their surfaces which may provide some gyroscopic stability in an otherwise smooth bore firearm. Non-lethal rounds : These are designed for only incapacitation of the targets and not killing them. E.g. rubber pellets, bean bag rounds etc. |
| The projectile (bullets, shots etc.) is usually made from lead or lead-antimony alloy due to their high sectional density and malleability. Some bullets may contain a metal jacket which is usually made of cupro-nickel, copper-zinc or steel. |
| The primer is a pressure sensitive material contained within a cap. It ignites when hit by a firing pin and in turn ignites the propellant charge. It contains either a mixture of mercury fulminate, potassium chloride and antimony sulphide or a mixture of antimony sulphide, lead styphnate, lead peroxide and barium nitrate or tetracene. |
| Propellant charge is usually composed of smokeless powder. It may be single base (nitrocellulose) or double base (nitrocellulose + nitroglycerine). Previously black powder was used as propellant charge. Smokeless powder produces 800 ml - 900 ml of gas per grain whereas black powder produces only 200 ml - 300 ml of gas per grain. |
| Casing is usually made from metal, paper or plastic. It encloses the cartridge and also seals the breech after firing to prevent the gas from escaping. |
There are some special types of bullets—
Hollow point bullet : This is a type of semi-jacketed bullet that has a hole drilled into its nose. The hole causes greater expansion of the bullet upon hitting the target and causes more tissue damage.
Dumdum bullet : This type of bullets are made by cutting the nose of a jacketed bullet, that essentially makes them semi-jacketed bullets.
Incendiary bullet : These bullets have combustible substances in their tips. Usually white phosphorus is used. They are used to hit flammable targets like oil tanks.
Tracer bullet : These bullets have some brightly illuminating substances added to them which light up when fired. They are used to visualize the trajectory the bullets are taking.
Explosive bullet : These bullets have a explosive substance inserted into their tips, usually lead azide. It causes greater fragmentation if it explodes upon impact.
Chamber
This is the part of the firearm where the cartridge is placed before firing. It is made very sturdy so that it can withstand the force of the explosion while firing each round. The rear of the chamber is covered by breech plate which contains an opening through which the firing pin can strike the primer cap of the cartridge.
Firing mechanism
It contains a spring loaded hammer or a striker which hits the firing pin when the trigger is pulled. The firing pin hits the primer cap behind the cartridge in the chamber and causes the primer to ignite, which in turn ignites the propellant charge and the rapidly expanding hot gas produced by its explosion propels the bullet through the barrel.
Barrel
| Rifled firearms | Smooth bore firearms |
--- |
| (E.g. rifles, pistols, revolvers etc.) | (E.g. shotguns, muskets etc.) |
| The barrel has spiral grooves cut into its inside called rifling. The device used to create rifling is called a broach. | No such rifling. |
| It has almost uniform diameter throughout its length. | The end of the barrel may have a tapering called choking. |
| The rifling causes the bullet to spin very fast along its long axis, the resulting gyroscopic stability stabilizes the bullet's flight path and increases the effective range and accuracy of the firearm. | The choking increases the effective range of a shotgun by pushing the pellets together as they exit the muzzle and delaying their dispersal. Shotguns may have different types of choking, e.g. no choking, half choking (20/1000 th of an inch), full choking (40/1000 th of an inch) etc. |
| The caliber is measured by the inner diameter of the barrel, more specifically the distance between two lands. E.g. 7.62 mm rifle, 9 mm pistol etc. | The caliber is measured by gauge which is defined as the number of snugly fitting lead balls needed to make a single ball of 1 pound weight. So higher gauge means smaller caliber. E.g. 20 G shotgun. |
| Diagram of rifling. | Diagram of choking. |
A firearm usually has some other parts as well—
Magazine : They hold extra rounds ready to be loaded into the chamber.
Loading mechanism : It helps to load the cartridge into the chamber before firing and also loads the spring of the firing pin. It can be of different types, e.g. bolt action (rifles), slide action (pistols), pump action (shotguns) etc. Revolvers have multiple rotating chambers. Each of these can be manual, semi-automatic or automatic.
Ejection mechanism : They are usually coupled with the loading mechanism, they eject the spent cartridge after firing a round.
Sight : Usually a fore sight and a rear sight is placed in the muzzle end and the breech end of a firearm respectively. Some rifles may contain telescopic sights. They help in aiming the firearm.
When multiple bullets are fired simultaneously they are called tandem bullets. They can be of two types. If two bullets are inserted in the same cartridge it's called a duplex cartridge. Sometimes a bullet fails to come out of the muzzle and is ejected during the subsequent shot along with the next bullet, it's known as piggy-back bullet. Tandem bullets pose a problem because they can be mistaken as multiple gunshot wounds.
External ballistics
A bullet gains its kinetic energy from the explosion of the propellant charge inside the chamber and accelerates down the barrel. After exiting the muzzle it decelerates due to air drag. The gravity also pulls it downwards causing it to drop towards the ground (called bullet drop).
A rifled firearm imparts rapid rotation to the bullet. This helps in stabilizing the flight path of the bullet by means of gyroscopic stability. It improves accuracy and range.
A bullet can hit an object and get reflected and hit another target. This is known as ricochet bullet. A ricochet bullet can potentially hit the person who fired it.
A yawning / wobbling bullet moves in an irregular fashion, whereas a tumbling / somersaulting bullet rotates in its longitudinal axis.
The pellets of a shotgun starts to disperse after travelling a short distance from the muzzle, it helps in acquiring the target but also reduces the range. The choking of a shotgun delays the dispersal of pellets and increases the range.
Wound Ballistics
This is the part we are most interested in. Firearm wounds are usually penetrating wounds. We can derive information about the range and type of firearm by examining the features of the wound.
A bullet causes three types of injury if it hits human tissue—
The surface wound : The entry and the exit wounds, they are the most informative wounds while examining a victim of a gunshot. The exit wound may not always be present if the bullet is lodged inside the body.
The permanent cavity : This wound is caused by direct crushing of tissues by the bullet as it travels inside the body. The expanding bullets (e.g. hollow point bullets) create a permanent cavity larger than their own diameter. If a bullet is fragmented upon impact (e.g. frangible bullet) it can cause multiple permanent cavities.
The temporary cavity : As a bullet travels inside a body it transfers its kinetic energy to the surrounding tissue causing a radial shock wave. This shock wave can damage nearby tissues even if they are not directly crushed by the bullet.
Tissue damage due to a gunshot injury is proportional to the kinetic energy of the bullet.
So if velocity is doubled the damage done by the bullet may be quadrupled.
The damage done by the bullet also depends on the tissue it hits. Soft tissue adsorbs little energy from the bullet and the bullet may pass through it causing little damage, whereas tough structures like bones may absorb most of the bullets energy and result in massive damage like comminuted fracture of bones.
Entry wound
An entry wound is present where the bullet enters the body. It's usually smaller than the diameter of the bullet due to elasticity of skin. It's typically inverted and circular, but may be tear drop shaped if it hits the body in an angle or can be slit like if the bullet is tumbling.
The secondary products of firing e.g. unburned powder, smoke, flame etc. may leave their marks on the entry wound which helps in determining the range. On x-ray a metallic ring shadow may be seen at the wound of entry.
It's the most informative wound of all gunshot wounds and can provide information about the type of firearm used, its range etc. The entry wound varies widely between rifled firearms and shotguns, so we will discuss them separately.
Rifled entry wound
The range is usually determined by the effects of secondary products of firing on the entry wound or on the clothing covering the area.
| Range | Distance | Product | Wound features |
--- --- |
| Rifle | Handgun |
| Distant | >100 cm | >50 cm | Bullet only | Abrasion collar + Grease collar |
| Intermediate | 60 cm - 100 cm | 40 cm - 50 cm | Bullet + Unburned powder | Abrasion collar + Grease collar + Tattooing / Peppering / Powder stippling |
| Near | 15 cm | Bullet + Unburned powder + Smoke / Soot | Abrasion collar + Grease collar + Tattooing + Blackening / Smudging |
| Close | 8 cm | Bullet + Unburned powder + Smoke + Flame / Hot air | Abrasion collar + Grease collar + Tattooing + Blackening + Burning / Scorching + Singing of hair |
| Contact | 0 cm | All products enter the wound | Firm contact : Muzzle impression only Loose contact : Muzzle impression + Blackening + Burning + Singing of hair Angled contact : Partial Muzzle impression + Tattooing + Blackening + Burning + Singing of hair |
Entry wound showing abrasion collar around the main wound.
Courtesy: Bobjgalindo (GFDL)
Contact shot on head causes a cruciate or stellate wound due to the high pressure gas escaping through the loose areolar tissue.
Cherry red discoloration may be seen in contact, close and near ranges due to the formation of carboxy-hemoglobin formed due to reaction between carbon-monoxide in the exhaust gas and the hemoglobin in the tissue.
Point blank range is an arbitrary term that indicates a range in which it is not possible to miss the target.
Use of silencers on a firearm can significantly change the features as it suppresses the secondary products of firing from exiting the muzzle, thus making it difficult to determine the range.
The most reliable way to determine the range is to test fire the suspect weapon at ballistic gel at different ranges and comparing the features with the victim's wound.
The direction of the wound may sometimes be determined from the wound of entry. In perpendicular shots the wound is circular and the abrasion collar evenly distributed around the wound. But in angled shots the wound is more tear drop / egg shaped and the direction is indicated by the presence of abrasion collar to that side. The distribution of secondary products of firing also helps in determining the direction.
Shotgun entry wound
Shotguns usually use multiple pellets as projectiles. The pellets disperse as they exit the muzzle of the shotgun. The range of firing can be determined by the dispersal of the pellets.
| Range | Dispersion | Wound |
---
| 0 m - 2 m | No dispersion | Single wound of entry (rat hole / cookie cutter / scalloping / nibbling) |
| 2 m - 4 m | Partial dispersion | Main wound of entry + Satellite wounds |
| >4 m | Complete dispersion | Multiple wound of entry |
The range of the shotgun can be measured more accurately by measuring the spread of the pellets. But it also depends on the choke used in the shotgun.
No choke :
Half choke :
Full choke :
The range can also be determined from the effects of secondary products of firing on the wound of entry.
| Range | Distance | Product | Wound features |
--- --- |
| Distant | >90 cm | Pellets only | Wound of entry only |
| Intermediate | 60 cm - 90 cm | Unburned powder | Tattooing / Peppering / Powder stippling |
| Near | 30 cm | Unburned powder + Smoke / Soot | Tattooing + Blackening / Smudging |
| Close | 15 cm | Unburned powder + Smoke + Flame / Hot air | Tattooing + Blackening + Burning / Scorching + Singing of hair |
| Contact | 0 cm | All products enter the wound | Similar to rifled firearm wounds. Also the wad is embedded into the wound. |
The wadding of the shotgun shell may also leave an abrasion mark if hits the skin within 10 ft - 12 ft. Some shotgun shells have a plastic covering known as power piston around the pellets to prevent early dispersal and it opens after travelling a short distance from the muzzle. If it hits the skin after opening it may leave a cruciate abrasion, which may be seen in ranges of 60 cm - 90 cm.
The direction of firing may be determined by the spreading of the pellets. The pellets are evenly distributed in perpendicular shots, but in angled shots they are more closely packed closer to the gun and more widely dispersed to the opposite side. Also the pattern of tattooing and blackening may indicate the direction. But it becomes unreliable on increased ranges.
Exit wound
Exit wound is seen where the bullet leaves the body. It is most commonly seen in case of rifled firearm injuries.
The exit wound is usually everted and larger than the wound of entry. There is usually protrusion of subcutaneous fat through the exit wound. Bleeding is usually more in the exit wound than the entry wound. But in case of long distance shots, low velocity bullet or ricocheted bullets the entry wound may be larger than the exit wound.
If the skin is supported by some hard surface, e.g. the victim leaning on a wall, the exit wound may have abrasion around the wound margin due to crushing of tissue between the bullet and the hard surface. This type of exit wound is called shored exit wound.
Exit wounds are usually not seen in shotgun wounds as the pellets quickly looses their kinetic energy after hitting the body. Exit wounds in a shotgun injury can only be seen in case of a contact shot.
Angled entry wound on the left and exit wound on the right.
Courtesy: Bobjgalindo (GFDL)
Track of the wound
The track taken by the bullet inside the body causes the permanent cavity. There may be single or multiple tracks depending on the fragmentation of the bullet. It may pass through different organs and vessels causing organ damage and internal hemorrhage. If the bullet hits a bone, the fracture fragments may act as secondary missiles creating their own tracks and causing additional injuries to internal structures. Pieces of fabrics may be pushed into the wound by the bullet if the area is covered by clothing. In case of shotgun wounds the pellets usually take divergent tracks inside the wound. Sometimes a bullet already in the wound may be hit by a second bullet and go on a different trajectory, it is known as billiard board ricochet bullet and causes additional damage.
Sometimes direction of the firing may be determined by the direction of the wound track, but it may give false results in case of bullets changing direction in wound, fragmentation, secondary missiles like bone splinter, divergent course taken by pellets etc.
In case of fragmented bullets, the fragments may be detected by performing an x-ray of the affected area. The peculiar appearance of the fragmented bullet in a x-ray is known as lead storm phenomenon or the birdshot pattern.
Wound peculiarities
No exit wound is found if—
Bullet lodged inside the body.
Bullet expelled by a natural orifice, e.g. passed with stool, coughed out etc.
Bullet reflected by a bone and exited through the wound of entry.
One wound of entry but two bullets found inside the body—
In case of tandem bullets.
Successive firing of the firearm without any change of position.
If a bullet fragments into two.
One wound of entry but multiple wound of exit—
In case of tandem bullets.
Successive firing of the firearm without any change of position.
If a bullet fragments into two.
If a splinter from a fractured bone exits the body.
No exit wound but no bullet inside body—
Bullet expelled by a natural orifice, e.g. passed with stool, coughed out etc.
Bullet reflected by a bone and exited through the wound of entry.
Multiple wound of entry but only one bullet inside body—
One of the bullets expelled by a natural orifice, e.g. passed with stool, coughed out etc.
One of the bullets reflected by a bone and exited through the wound of entry.
When one bullet penetrates multiple parts of the body, such as a bullet entering lateral aspect of arm, exiting through medial aspect of that arm, entering lateral aspect of chest wall and then exiting through the other side of the chest wall. In such a case there may be multiple wounds of entry and multiple wounds of exit by the same bullet.
Gunshot injury to skull
Gunshot injury to skull causes some peculiar features of entry and exit wounds—
| Entry wound | Exit wound |
--- |
| (Punched in hole) | (Punched out hole) |
| Sharp wound margin in outer table. | Bevelling / crater like wound in outer table. |
| Bevelling / crater like wound in outer table. | Sharp wound margin in inner table. |
Punched out hole in skull (exit wound) showing bevelling of outer table.
When a high velocity bullet enters the skull it may cause the skull to burst open causing evisceration of brain, this is known as a Kronlein shot.
A bullet may sometimes graze the skull causing a gutter fracture of the outer table with the inner table left intact. The surface presentation of such an injury may be a slit like or keyhole wound, which is essentially a combined entry and exit wound.
Gunshot wounds to the skull may cause radial fractures around the entry wound. Puppe's rule states that a fracture line will not cross an already existing fracture line. Using this rule it is possible to determine the sequence of shots in case of multiple gunshot injuries to the skull.
Cause of death in gunshot injury
The death in a gunshot injury may be early or late.
The early cause of death includes—
Hemorrhage from the wounds, which may be external or internal.
Shock, may be hypovolemic or neurogenic.
Damage to vital organs, e.g. brain, heart etc.
Embolism due to fat, air etc. entering the blood stream.
Asphyxia due to blocking of airway by bullet, blood etc.
The late causes may be—
Infection from the wound, may cause sepsis.
Stricture of vital structures such as esophagus.
Lead poisoning from a bullet lodged inside the body.
Examination of the victim
Living victim
General examination of the victim is to be done first.
Further detailed examination should be done preferably on the O.T. Table. The part involved, the extension of the damage and severity of the injuries are to assessed thoroughly. A radiological investigation is also recommended.
Bullets are extracted surgically through a skin incision separate from the wound of entry. Surgical intervention through the wound of entry may give rise to confusion about the trajectory of the bullet which is known as Kennedy phenomenon. If a lodged bullet is compatible with life and it's removal may cause extensive damage then it is left in place. Such a bullet is called souvenir bullet.
The clothing of the victim is also examined for bullet holes, blood stains etc. The examination of the place of occurrence is also important.
Post-mortem
The post-mortem of a gunshot victim is performed similar to the general post-mortem, but the following examinations are done in addition to that—
Examination of clothing for bullet holes, blood stains etc.
Line of flow of blood from the wound.
Any secondary injuries.
Wounds of entry and wounds of exit.
Direction of firing from abrasion collar, dispersal of pellets, secondary products of firing etc.
Track of the wound is to be traced.
Bullets are removed with a rubber tipped forceps or with the tips of the fingers. After removal some identifying marks are engraved behind the bullets. The bullets are preserved with the blood stains in a sealed and labelled glass jar after covering them in cotton. The pellets are treated similarly except no engraving is necessary. Bullet fragments are also preserved.
Skin around the wound is preserved for chemical identification of unburned powder, grease etc.
Determining nature of death
Death from firearm injury may be suicidal, homicidal or accidental.
Suicidal gunshot wounds are on easily reachable sites of body such as palate, below chin, right or left temple or medial canthus of eye (depending on the handedness), forehead or pre-cordial area. Usually one round is fired but sometimes multiple rounds may be fired often without changing position of the weapon. They are often contact shots or close shots. The weapon found in the cadaveric spasm in the victim's hand is a strong indication of suicidal nature of death. Other suggestive evidence may include latent fingerprints, dermal nitrate test of the victim's hand, secluded spot, doors closed from inside, suicide note etc.
Homicidal gunshot wounds can be anywhere in the body and can be from any range. Multiple wounds of entry from contact or close range placed in the back of the victim is certainly homicidal. Often the weapon is absent but sometimes they maybe placed in the victim's hand, but in such cases cadaveric spasm will be absent.
Accidental gunshot wounds maybe caused by wrong aiming or mishandling of the weapon. They can simulate either suicidal or homicidal wounds. Careful examination of the circumstances are necessary to avoid confusion.
Examination of the place of occurrence
The approachability of the place is to be noted as it is important in determining the nature of death. Disorder in the place may be indicative of a struggle which is suggestive of homicide. Empty cartridge, extra bullets, bullets marks are also to be looked for. Sometimes the weapon of offence may be found at the scene. Latent fingerprints and footprints may be recovered. The pattern of shedding of blood is also important.
Identification of firearm
The identification of the firearm is usually done from the features of the wound produced and the bullet recovered from the victim's body.
Type of firearm
The type of firearm used can be identified by the features of the wound.
A circular or oval shaped wound of entry with an abrasion collar indicates a rifled firearm, whereas injury caused by multiple pellets indicate a shotgun injury. A single but irregular, known as rat hole or cookie cutter wound, may be caused by a contact shot from a shotgun.
Ballistic fingerprinting
Ballistic fingerprinting is done to identify the firearm from which the recovered bullet was fired. It is used in case of rifled firearms.
A simple comparison of the caliber of the bullet and the suspect firearm often narrows down the search significantly.
For more precise identification markings on the bullets may be used. A firearm leaves two types of marks on a bullet's surface after firing—
| Primary markings / Class characteristics | Secondary markings / Individual characteristics |
--- |
| They are caused by the lands of the firearm's rifling. | They are caused by the small surface irregularities inside the barrel. |
| They are specific to the make or model of the firearm. | They are specific to an individual firearm. |
| They help in narrowing down the possible firearms that may have been used. | They help in precisely identifying the individual firearm. |
A test bullet is fired from the suspect firearm and is compared with the recovered bullet under comparison microscope side by side. One of the bullet is rotated until its markings match precisely with the markings of the other one. If the markings don't match even after a complete rotation, it may be assumed that they were not fired from the same firearm.
The markings left by the firing pin and other parts of a firearm on the empty cartridge can also be used to identify the firearm.
The inside of a rifled firearm may be measured by a device known as helixometer.
Recovered cartridge and test cartridge under comparison microscope. Matching striations seen.
Other findings
Sometimes in case of a contact shot some amount of blood and tissue may be sucked into the barrel of the firearm which is known as back spatter.
The unburned powder or the grease found in the wound may be chemically compared to those found inside the suspect gun, which helps in identifying the firearm.
Identification of the suspect
The person suspected of firing a bullet can be identified by different methods—
Finger print recovery
The finger print of the suspect may be recovered from the firearm or the cartridge if found at the crime scene.
Dermal nitrate test
When a person fires a gun, some residue of the propellant powder, which contains nitrogen compounds, leaks from the gun and deposited on the person's hands. It can be detected by using a test called dermal nitrate test, also known as paraffin test.
A paraffin cast is taken from the suspect's hand involving the medial surface and base of thumb along with the lateral surface and base of index finger. The presence of nitrites is determined by bluish discoloration of the cast on reaction with diphenylamine.
It helps to determine whether a person fired a gun recently or not. But the test gives false negatives if there was no leakage or if the suspect used gloves or washed the hands thoroughly afterwards. It may also give false positives in the presence of other source of nitrogen compounds like urine, fertilizer etc.
Harrison and Gilroy test
This test determines the presence of metals like lead, antimony or barium in the primer mixture which may have leaked from the gun to the suspect's hand. It employs techniques like spectrophotometry, atomic absorption spectroscopy or neutron activation analysis.
Other evidences
A suspect may also be identified by other evidences found at the crime scene or elsewhere. Witnesses may also be present. It is possible for a witness to identify a suspect even in darkness by the light of the muzzle flash up to a distance of 8 yards.
Conclusion
Forensic medical examiners are required to examine the wounds on victims, either living or post-mortem. They are also required to preserve the evidences such as recovered projectiles, clothings etc. In the court they are considered as expert witnesses and can be cross-examined by the defence lawyer.
Links
Gunshot Wound
Forensic Firearm Examination
Ballistic Fingerprinting
Imaging of Gunshot Injuries
Types of Firearms
Insides of a firearm
Ammunition
Chamber
Firing mechanism
Barrel
External ballistics
Wound Ballistics
Entry wound
Rifled entry wound
Shotgun entry wound
Exit wound
Track of the wound
Wound peculiarities
Gunshot injury to skull
Cause of death in gunshot injury
Examination of the victim
Living victim
Post-mortem
Determining nature of death
Examination of the place of occurrence
Identification of firearm
Type of firearm
Ballistic fingerprinting
Other findings
Identification of the suspect
Finger print recovery
Dermal nitrate test
Harrison and Gilroy test
Other evidences
Conclusion
Links
Category: Medicine
Date: Thu Jan 18 2018
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Student Enrollment: How many students enroll in postsecondary institutions annually?
In year 2022-23, the number of students enrolled in postsecondary institutions was 24,934,456. This is based on 5,679 institutions.
Number of students enrolled in postsecondary institutions annually
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Notes:This table presents data collected from Title IV institutions in the United States. Prior to 2009-10, the data include only Title IV primarily postsecondary institutions.
This is the unduplicated 12-month enrollment at institutions from July 1 of one year through June 30 of the next. Prior to 2010-11, institutions could choose to report on the 12-month period between July 1 and June 30 or September 1 and August 31.
SOURCE: U.S. Department of Education, National Center for Education Statistics, Integrated Postsecondary Education Data System (IPEDS), 12-month Enrollment component final data (2001-02 - 2021-22) and provisional data (2022-23). |
11981 | https://www.statology.org/r-select-rows-in-data-frame-based-on-values-in-vector/ | R: How to Select Rows in Data Frame Based on Values in Vector
You can use one of the following methods to select rows from a data frame in R based on values in a vector:
Method 1: Use Base R
Method 2: Use dplyr Package
The following examples show how to use each method in practice with the following data frame in R:
Example 1: Use Base R to Select Rows Based on Values in Vector
We can use the following code to select only the rows from the original data frame where the value in the division column is equal to ‘West’ or ‘North.’
The new data frame only contains the rows where the value in the division column is equal to ‘West’ or ‘North.’
Example 2: Use dplyr to Select Rows Based on Values in Vector
We can also use the filter() function from the dplyr package in R select only the rows from the original data frame where the value in the division column is equal to ‘West’ or ‘North.’
The new data frame only contains the rows where the value in the division column is equal to ‘West’ or ‘North.’
Note: The base R and dplyr methods produce the same results. However, the dplyr method will tend to be faster when working with extremely large data frames.
Additional Resources
The following tutorials explain how to perform other common tasks in R:
How to Select Random Rows in R Using dplyr
How to Select Rows by Condition in R
How to Select Rows Where Value Appears in Any Column in R
Hey there. My name is Zach Bobbitt. I have a Masters of Science degree in Applied Statistics and I’ve worked on machine learning algorithms for professional businesses in both healthcare and retail. I’m passionate about statistics, machine learning, and data visualization and I created Statology to be a resource for both students and teachers alike. My goal with this site is to help you learn statistics through using simple terms, plenty of real-world examples, and helpful illustrations.
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11982 | https://www.youtube.com/watch?v=LONlP2x9PHI | mod02lec14 - Comparison of topologies - Part 1 : Finer and coarser topologies
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When there are two topologies defined on the same set, the natural question to ask is whether they are comparable in some sense. In this lecture we give a defintion which allows such comparisons- if two topologies are comparable, the larger one is called "finer" and the smaller one is called "coarser". We describe a criterion to determine if a topology is finer than another, with respect to the bases of the two topologies We also provide an example in the Euclidean plane as an illustration for the basis comparison theorem.
Transcript:
[Music] recall from the last lecture that we have defined various topologies three different topologies on the set of real numbers so the first one was the standard topology the second one was the lower limit topology and the third one was the k topology right so we denoted the standard topology as t tau standard tau l was the lower limit topology and tau k was the k topology now the question arises when you have multiple different topologies on the same set whether it is possible to compare them so the question that you want to ask is that is it is it possible is it possible to compare topologies on the same set on the same set so for this purpose we shall define some terminology so let me give definition so let x be a set and we have two topologies on this on this set x so tau and top prime b topologies on x okay so we have a few different terminologies in this setting so the first one is that if tau the collection tau is a sub collection of top prime then tau prime is called finer than tau finer is this is the word that's used here and tau is called coarser than top rind so finer and coarser are the words that we shall use to determine which one is larger and which one is smaller okay so tau prime here is larger and tau is smaller okay so the smaller one is coarser and the larger one is finer okay so you can we can use uh directly this uh these words larger and smaller but um the terminology that that's used in in mathematics is finer and coarser now if the inclusion is proper if the inclusion this inclusion is is is a proper inclusion inclusion then top prime is called strict finer strictly finer than top and and a tau similarly is called strictly coarser and then top right so when when you have as proper inclusion meaning there are sets there are sets in top prime which do not belong to top then we use the term strictly finer and strictly coarser now um it may happen that tau is not a subcollection of top prime or it may also happen that tau prime is not a sub collection of tau at the same time so in this case we use a different terminology so if neither tau is a sub collection of top prime or nor tau prime is a sub collection of tar holes then we call them we call the topologies tau and top right incomparable okay so incomparable meaning they cannot be compared and otherwise we call them comparable and comparable otherwise meaning that either this holds or this holds okay so um so these are the various terminologies that we will use to compare different topologies on the same set so for example just as an example you can consider the two-dimensional plane okay so two-dimensional euclidean plane euclidean space or the plane okay so in this plane we can define two different bases so let me call b 1 to be the collection of circular regions circular regions in r2 circular regions in r2 and b2 is the collection of rectangular regions rectangular regions in r2 so first of all why are these two bases so i'll just give a visual intuition for why this is true so for for b1 so we have let's say first of all any point in r2 is contained in a circular region okay so property one is satisfied for the definition of a basis so the only thing one one is to check is property two property two so if you have uh two circular regions b1 and b2 and you have a point x here so this is a point x then of course you can take a very small circle very small circle this is b3 [Music] which which is inside the intersection okay so and it contains x so x is contained in b3 is a subset which is a subset of b1 intersection b2 so this is a basis b1 is a basis for a topology on r2 and similarly you can have for the collection of rectangular regions again property 1 is satisfied because every point in r2 is can be enclosed within a rectangular region and for property 2 we can have for example let's say two basis elements two rectangles b1 and let's say like this b2 b2 and of course let's take a point x in in the intersection of these two sets and again it's visually clear that if you take a small enough rectangle that contains this point then it will be it will lie completely inside the intersection so again here this is b3 and x belongs to b3 and it is a subset of b1 intersection b2 so now the these two bases uh b1 and b2 generate topologies let me call tau 1 and tau 2. so suppose so let tau 1 be the topology generated generated by b1 and similarly tau 2 be the topology generated by b2 okay so we have now two topologies and the question arises first of all whether are they comparable are they comparable and whether ta1 and if they are comparable so if yes if yes is tau 1 finer than tau two or coarser okay so we have these two topologies tau one and tau two and we want to determine uh which one first of all whether they are comparable and second of all if it if they are comparable which one is finer and which one is coarser so again if you see if you take a visual proof for this so i claim that they are comparable and in fact they are either they use the same topologies on on r2 okay and for this we will use a theorem which allows us to determine if a if two topologies are given on the same set which one is finite which one is coarser by looking at the basis elements so the theorem we want to use for this is the following so suppose that b and b prime are two bases on the same set x and they generate topologies tau and tau prime respectively so b generates tau and b prime generates the topology top prime then i say that tau is is coarser than tau prime so tau is coarser than top prime which is just this inclusion tau is in in a as a as a sub collection of top prime uh if and only if for each x in x and um any basis element in b so b is is remember that it generates the topology tau right so if you take any basis element b containing x so this basis element b contains x there exists a b prime a basis element in in the second basis b prime such that x is contained in b prime and b prime is a subset of b so let us see a proof of this theorem proof so suppose first that tau is coarser than top prime and we have to show that for each x and b in b containing x there exists a basis element in b prime which contains x and it is a subset of the basis element b so suppose that x is in x and b is a basis element such that x belongs to b okay so so how do we find this b prime is the question so notice that b already belongs to this topology tau because we have seen that every basis element is an open set by the way we define the topology tau from generated from the basis b it is easy to see that any basis element is already an open set and b belongs to tau and tau is a subset of is a sub collection of type prime so this implies that b belongs to top prime okay because it is an open set in tau prime then this implies from the definition of the topology top prime as generated from basis b prime that there exists there exists a b prime in b prime such that x belongs to b prime and it is a subset of b which is exactly what we wanted to show okay so the forward implication is proved and now let us look at the reverse implication okay so so we know that for each each point x there exists and and the basis element that contains it in b then there exists a basis element in b prime which is a subset of that basis element b and now we have to show that to show that tau is coarser than top prime so let let us take an open set u in tau and we have to show that we will show that that this open set u also belongs to top prime so since a u is is an open set this implies that there exists given x in u there exists a basis element b in b says that x belongs to b and b is contained in u this was our definition of tau from the basis and now because because we already have assumed that given a basis element and a point inside it there exists a basis element in b prime which is inside inside b so there exists b prime in b prime such that x belongs to b prime and it is a subset of b but b is also a subset of u right so this implies that given uh any x in x um x any x in u n e x in u there exists a basis element b prime in b prime says that it contains x and it is inside u and this is equivalent to saying that this set u is in top prime and because of the way we define it again so it's just using the definition to to show this theorem so now that we have we have this nice criteria to determine which which one is coarser using the bases let us go back to the example of example of b1 and b2 that we defined for circular regions and rectangular regions and try to use this theorem to show which one is coarser than which when the the other so for example if you take if you take a point x and a basis element in b2 so a rectangular region containing x then it is a obvious that we can take again a small circle circular region containing this point x which is which lies in in the inside in the interior region of this rectangle so this is b1 okay so this means that this means that the topology tau 2 is coarser than tau 1 from the theorem 5.1 which i described below and similarly you can also reverse this process and you take a circular region say b1 and take a point x here x but again it's quite quite easy to see that we can cover it inside a rectangular region this is b2 so we have reversed the rules of b1 and b2 and this implies that tau 1 is coarser than tau 2 and so taking these 2 together this means that tau 1 is equal to tau 2 and so this means that b 1 and b 2 generate the same topology generate the same topology on r2 and these these topologies i will be called the standard topologies standard topology on r2 okay so just as we just find define the standard topology on on r this is the standard topology on r2 on the plane and this is the one that is most commonly used okay and now we will go back to the various topologies that we defined on r so i'll just go back to the first page where we had the standard topology on our the lower limit topology on r and the k topology on r and the question arises the question that we asked in the beginning was that whether it is possible to compare these topologies and when the answer is in the following theorem |
11983 | https://en.wikipedia.org/wiki/Methemoglobinemia | Jump to content
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Methemoglobinemia
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From Wikipedia, the free encyclopedia
Condition of elevated methemoglobin in the blood
Not to be confused with Argyria.
Medical condition
| Methemoglobinemia | |
--- |
| Other names | Hemoglobin M disease, |
| | |
| Chocolate-brown blood due to methemoglobinemia | |
| Specialty | Toxicology, haematology, Emergency medicine |
| Symptoms | Headache, dizziness, shortness of breath, nausea, poor muscle coordination, blue-colored skin |
| Causes | Benzocaine, nitrites, dapsone, genetics |
| Diagnostic method | Blood gas |
| Differential diagnosis | Argyria, sulfhemoglobinemia, heart failure |
| Treatment | Oxygen therapy, methylene blue |
| Prognosis | Generally good with treatment |
| Frequency | Relatively uncommon |
Methemoglobinemia, or methaemoglobinaemia, is a condition of elevated methemoglobin in the blood. Symptoms may include headache, dizziness, shortness of breath, nausea, poor muscle coordination, and blue-colored skin (cyanosis). Complications may include seizures and heart arrhythmias.
Methemoglobinemia can be due to certain medications, chemicals, or food, or it can be inherited. Substances involved may include benzocaine, nitrites, or dapsone. The underlying mechanism involves some of the iron in hemoglobin being converted from the ferrous [Fe2+] to the ferric [Fe3+] form. The diagnosis is often suspected based on symptoms and a low blood oxygen that does not improve with oxygen therapy. Diagnosis is confirmed by a blood gas.
Treatment is generally with oxygen therapy and methylene blue. Other treatments may include vitamin C, exchange transfusion, and hyperbaric oxygen therapy. Outcomes are generally good with treatment. Methemoglobinemia is relatively uncommon, with most cases being acquired rather than genetic.
Signs and symptoms
[edit]
Signs and symptoms of methemoglobinemia (methemoglobin level above 10%) include shortness of breath, cyanosis, mental status changes (~50%), headache, fatigue, exercise intolerance, dizziness, and loss of consciousness.
People with severe methemoglobinemia (methemoglobin level above 50%) may exhibit seizures, coma, and death (level above 70%). Healthy people may not have many symptoms with methemoglobin levels below 15%. However, people with co-morbidities such as anemia, cardiovascular disease, lung disease, sepsis, or who have abnormal hemoglobin species (e.g. carboxyhemoglobin, sulfhemoglobinemia or sickle hemoglobin) may experience moderate to severe symptoms at much lower levels (as low as 5–8%).[citation needed]
Cause
[edit]
Acquired
[edit]
Methemoglobinemia may be acquired. Classical drug causes of methemoglobinemia include various antibiotics (trimethoprim, sulfonamides, and dapsone), local anesthetics (especially articaine, benzocaine, prilocaine, and lidocaine), and aniline dyes, metoclopramide, rasburicase, umbellulone, chlorates, bromates, and nitrites. Nitrates are suspected to cause methemoglobinemia.
In otherwise healthy individuals, the protective enzyme systems normally present in red blood cells rapidly reduce the methemoglobin back to hemoglobin and hence maintain methemoglobin levels at less than one percent of the total hemoglobin concentration. Exposure to exogenous oxidizing drugs and their metabolites (such as benzocaine, dapsone, and nitrates) may lead to an increase of up to a thousandfold of the methemoglobin formation rate, overwhelming the protective enzyme systems and acutely increasing methemoglobin levels.[citation needed]
Infants under 6 months of age have lower levels of a key methemoglobin reduction enzyme (NADH-cytochrome b5 reductase) in their red blood cells. This results in a major risk of methemoglobinemia caused by nitrates ingested in drinking water, dehydration (usually caused by gastroenteritis with diarrhea), sepsis, or topical anesthetics containing benzocaine or prilocaine resulting in blue baby syndrome. Nitrates used in agricultural fertilizers may leak into the ground and may contaminate well water. The current EPA standard of 10 ppm nitrate-nitrogen for drinking water is specifically set to protect infants. Benzocaine applied to the gums or throat (as commonly used in baby teething gels, or sore throat lozenges) can cause methemoglobinemia.
Genetic
[edit]
Due to a deficiency of the enzyme diaphorase I (cytochrome b5 reductase), methemoglobin levels rise and the blood of met-Hb patients has reduced oxygen-carrying capacity. Instead of being red in color, the arterial blood of met-Hb patients is brown. This results in the skin of white patients gaining a bluish hue. Hereditary met-Hb is caused by a recessive gene. If only one parent has this gene, offspring will have normal-hued skin, but if both parents carry the gene, there is a chance the offspring will have blue-hued skin.[citation needed]
Another cause of congenital methemoglobinemia is seen in patients with abnormal hemoglobin variants such as hemoglobin M (HbM), or hemoglobin H (HbH), which are not amenable to reduction despite intact enzyme systems.[citation needed]
Methemoglobinemia can also arise in patients with pyruvate kinase deficiency due to impaired production of NADH – the essential cofactor for diaphorase I. Similarly, patients with glucose-6-phosphate dehydrogenase deficiency may have impaired production of another co-factor, NADPH.
Pathophysiology
[edit]
The affinity for oxygen of ferric iron is impaired. The binding of oxygen to methemoglobin results in an increased affinity for oxygen in the remaining heme sites that are in ferrous state within the same tetrameric hemoglobin unit. This leads to an overall reduced ability of the red blood cell to release oxygen to tissues, with the associated oxygen–hemoglobin dissociation curve therefore shifted to the left. When methemoglobin concentration is elevated in red blood cells, tissue hypoxia may occur.
Normally, methemoglobin levels are <1%, as measured by the CO-oximetry test. Elevated levels of methemoglobin in the blood are caused when the mechanisms that defend against oxidative stress within the red blood cell are overwhelmed and the oxygen carrying ferrous ion (Fe2+) of the heme group of the hemoglobin molecule is oxidized to the ferric state (Fe3+). This converts hemoglobin to methemoglobin, resulting in a reduced ability to release oxygen to tissues and thereby hypoxia. This can give the blood a bluish or chocolate-brown color. Spontaneously formed methemoglobin is normally reduced (regenerating normal hemoglobin) by protective enzyme systems, e.g., NADH methemoglobin reductase (cytochrome-b5 reductase) (major pathway), NADPH methemoglobin reductase (minor pathway) and to a lesser extent the ascorbic acid and glutathione enzyme systems. Disruptions with these enzyme systems lead to methemoglobinemia. Hypoxia occurs due to the decreased oxygen-binding capacity of methemoglobin, as well as the increased oxygen-binding affinity of other subunits in the same hemoglobin molecule, which prevents them from releasing oxygen at normal tissue oxygen levels.[citation needed]
Diagnosis
[edit]
The diagnosis of methemoglobinemia is made with the typical symptoms, a suggestive history, low oxygen saturation on pulse oximetry measurements (SpO2) and these symptoms (cyanosis and hypoxia) failing to improve on oxygen treatment. The definitive test would be obtaining either CO-oximeter or a methemoglobin level on an arterial blood gas test.
Arterial blood with an elevated methemoglobin level has a characteristic chocolate-brown color as compared to normal bright red oxygen-containing arterial blood; the color can be compared with reference charts.
The SaO2 calculation in the arterial blood gas analysis is falsely normal, as it is calculated under the premise of hemoglobin either being oxyhemoglobin or deoxyhemoglobin. However, co-oximetry can distinguish the methemoglobin concentration and percentage of hemoglobin.
At the same time, the SpO2 concentration as measured by pulse ox is false high, because methemoglobin absorbs the pulse ox light at the 2 wavelengths it uses to calculate the ratio of oxyhemoglobin and deoxyhemoglobin. For example with a methemoglobin level of 30–35%, this ratio of light absorbance is 1.0, which translates into a false high SpO2 of 85%.
Differential diagnosis
[edit]
Other conditions that can cause bluish skin include argyria, sulfhemoglobinemia, heart failure, amiodarone-induced bluish skin pigmentation and acrodermatitis enteropathica.
Treatment
[edit]
Methemoglobinemia can be treated with supplemental oxygen and methylene blue. Methylene blue is given as a 1% solution (10 mg/ml) 1 to 2 mg/kg administered intravenously slowly over five minutes. Although the response is usually rapid, the dose may be repeated in one hour if the level of methemoglobin is still high one hour after the initial infusion. Methylene blue inhibits monoamine oxidase, and serotonin toxicity can occur if taken with an SSRI (selective serotonin reuptake inhibitor) medicine.
Methylene blue restores the iron in hemoglobin to its normal (reduced) oxygen-carrying state. This is achieved by providing an artificial electron acceptor (such as methylene blue or flavin) for NADPH methemoglobin reductase (RBCs usually don't have one; the presence of methylene blue allows the enzyme to function at 5× normal levels). The NADPH is generated via the hexose monophosphate shunt.
Genetically induced chronic low-level methemoglobinemia may be treated with oral methylene blue daily. Also, vitamin C can occasionally reduce cyanosis associated with chronic methemoglobinemia, and may be helpful in settings in which methylene blue is unavailable or contraindicated (e.g., in an individual with G6PD deficiency). Diaphorase (cytochrome b5 reductase) normally contributes only a small percentage of the red blood cell's reducing capacity, but can be pharmacologically activated by exogenous cofactors (such as methylene blue) to five times its normal level of activity.[citation needed]
Epidemiology
[edit]
Methemoglobinemia mostly affects infants under 6 months of age (particularly those under 4 months) due to low hepatic production of methemoglobin reductase. The most at-risk populations are those with water sources high in nitrates, such as wells and other water that is not monitored or treated by a water treatment facility. The nitrates can be hazardous to the infants. The link between blue baby syndrome in infants and high nitrate levels is well established for waters exceeding the normal limit of 10 mg/L. However, there is also evidence that breastfeeding is protective in exposed populations.
Society and culture
[edit]
Blue Fugates
[edit]
Main article: Blue Fugates
The Fugates, a family that lived in the hills of Kentucky in the US, had the hereditary form. They are known as the "Blue Fugates". Martin Fugate and Elizabeth Smith, who had married and settled near Hazard, Kentucky, around 1800, were both carriers of the recessive methemoglobinemia (met-H) gene, as was a nearby clan with whom the Fugates descendants intermarried. As a result, many descendants of the Fugates were born with met-H.
Blue Men of Lurgan
[edit]
The "blue men of Lurgan" were a pair of Lurgan men suffering from what was described as "familial idiopathic methemoglobinemia" who were treated by James Deeny in 1942. Deeny, who would later become the Chief Medical Officer of the Republic of Ireland, prescribed a course of ascorbic acid and sodium bicarbonate. In case one, by the eighth day of treatments, there was a marked change in appearance, and by the twelfth day of treatment, the patient's complexion was normal. In case two, the patient's complexion reached normality over a month-long duration of treatment.
See also
[edit]
Carbon monoxide poisoning
Hemoglobinemia
References
[edit]
^ "Methemoglobinemia". MedlinePlus Medical Encyclopedia. U.S. National Library of Medicine. Retrieved 8 June 2019.
^ Jump up to: a b c d "NCI Dictionary of Cancer Terms". National Cancer Institute. 2 February 2011. Retrieved 21 December 2019.
^ Jump up to: a b c d e f g h i j k l m n o p q r s t Ludlow JT, Wilkerson RG, Nappe TM (January 2019). "Methemoglobinemia". StatPearls [Internet]. Treasure Island (FL): StatPearls Publishing. PMID 30726002.
^ Jump up to: a b Wettstein ZS, Yarid NA, Shah S (December 2022). "Fatal methaemoglobinemia due to intentional sodium nitrite ingestion". BMJ Case Reports. 15 (12): e252954. doi:10.1136/bcr-2022-252954. PMC 9748921. PMID 36524260.{{cite journal}}: CS1 maint: article number as page number (link)
^ Oiseth S, Jones L, Maza E (eds.). "Methemoglobinemia". The Lecturio Medical Concept Library. Retrieved 10 August 2021.
^ Jump up to: a b Khanapara DB, Sacher RA, Kumar MDenshaw-Burke M, Savior DC, Curran AL, DelGiacco E, Abouelezz KF (7 December 2023). Besa EC (ed.). "Methemoglobinemia". eMedicine. Retrieved 2008-09-13.
^ Ash-Bernal R, Wise R, Wright SM (September 2004). "Acquired methemoglobinemia: a retrospective series of 138 cases at 2 teaching hospitals". Medicine. 83 (5): 265–273. doi:10.1097/01.md.0000141096.00377.3f. PMID 15342970. S2CID 40957843.
^ Zosel A, Rychter K, Leikin JB (2007). "Dapsone-induced methemoglobinemia: case report and literature review". American Journal of Therapeutics. 14 (6): 585–587. doi:10.1097/MJT.0b013e3180a6af55. PMID 18090884. S2CID 24412967.
^ Adams V, Marley J, McCarroll C (November 2007). "Prilocaine induced methaemoglobinaemia in a medically compromised patient. Was this an inevitable consequence of the dose administered?". British Dental Journal. 203 (10): 585–587. doi:10.1038/bdj.2007.1045. PMID 18037845.
^ Barash M, Reich KA, Rademaker D (February 2015). "Lidocaine-induced methemoglobinemia: a clinical reminder". The Journal of the American Osteopathic Association. 115 (2): 94–98. doi:10.7556/jaoa.2015.020. PMID 25637615.
^ Titov VY, Petrenko YM (April 2005). "Proposed mechanism of nitrite-induced methemoglobinemia". Biochemistry. Biokhimiia. 70 (4): 473–483. doi:10.1007/s10541-005-0139-7. PMID 15892615. S2CID 22906218.
^ Powlson DS, Addiscott TM, Benjamin N, Cassman KG, de Kok TM, van Grinsven H, et al. (2008). "When does nitrate become a risk for humans?". Journal of Environmental Quality. 37 (2): 291–295. Bibcode:2008JEnvQ..37..291P. doi:10.2134/jeq2007.0177. PMID 18268290. S2CID 14097832.
^ Jump up to: a b "Basic Information about Nitrate in Drinking Water". United States Environmental Protection Agency. Archived from the original on November 21, 2010. Retrieved 10 May 2013.
^ "FDA Drug Safety Communication: Reports of a rare, but serious and potentially fatal adverse effect with the use of over-the-counter (OTC) benzocaine gels and liquids applied to the gums or mouth". U.S. Food and Drug Administration. 7 April 2011. Archived from the original on April 10, 2011. Retrieved 10 May 2013.
^ "Risk of serious and potentially fatal blood disorder prompts FDA action on oral over-the-counter benzocaine products used for teething and mouth pain and prescription local anesthetics". U.S. FDA. May 23, 2018. Archived from the original on May 25, 2018. Retrieved May 24, 2018.
^ Oiseth S, Jones L, Maza E, eds. (3 September 2020). "Glucose-6-phosphate Dehydrogenase (G6PD) Deficiency". The Lecturio Medical Concept Library. Retrieved 23 July 2021.
^ Darling RC, Roughton FJ (1942). "The effect of methemoglobin on the equilibrium between oxygen and hemoglobin". Am J Physiol. 137: 56. doi:10.1152/ajplegacy.1942.137.1.56.
^ Umbreit J (February 2007). "Methemoglobin--it's not just blue: a concise review". American Journal of Hematology. 82 (2): 134–144. doi:10.1002/ajh.20738. PMID 16986127. S2CID 29107446.
^ Yusim Y, Livingstone D, Sidi A (June 2007). "Blue dyes, blue people: the systemic effects of blue dyes when administered via different routes". Journal of Clinical Anesthesia. 19 (4): 315–321. doi:10.1016/j.jclinane.2007.01.006. PMID 17572332.
^ Gillman PK (October 2006). "Methylene blue implicated in potentially fatal serotonin toxicity". Anaesthesia. 61 (10): 1013–1014. doi:10.1111/j.1365-2044.2006.04808.x. PMID 16978328.
^ Yubisui T, Takeshita M, Yoneyama Y (June 1980). "Reduction of methemoglobin through flavin at the physiological concentration by NADPH-flavin reductase of human erythrocytes". Journal of Biochemistry. 87 (6): 1715–1720. doi:10.1093/oxfordjournals.jbchem.a132915. PMID 7400118.
^ Prchal JT. Burns MM, Takemoto CM (eds.). "Methemoglobinemia". UpToDate.
^ Richard AM, Diaz JH, Kaye AD (1 January 2014). "Reexamining the risks of drinking-water nitrates on public health". Ochsner Journal. 14 (3): 392–398. PMC 4171798. PMID 25249806.
^ "Nitrates and drinking water". www.bfhd.wa.gov. Retrieved 10 December 2016.
^ Manassaram DM, Backer LC, Moll DM (1 March 2007). "A review of nitrates in drinking water: maternal exposure and adverse reproductive and developmental outcomes". Ciencia & Saude Coletiva. 12 (1): 153–163. doi:10.1590/S1413-81232007000100018. PMC 1392223. PMID 17680066.
^ Fan AM, Steinberg VE (February 1996). "Health implications of nitrate and nitrite in drinking water: an update on methemoglobinemia occurrence and reproductive and developmental toxicity". Regulatory Toxicology and Pharmacology. 23 (1 Pt 1): 35–43. doi:10.1006/rtph.1996.0006. PMID 8628918.
^ "Nitrate and Nitrite in Drinking-Water" (PDF). www.who.int. WHO Press. 2011. Retrieved December 10, 2016.
^ "Table of Regulated Drinking Water Contaminants". www.epa.gov. 30 November 2015. Retrieved 2016-12-12.
^ Pollock JI (May 1994). "Long-term associations with infant feeding in a clinically advantaged population of babies". Developmental Medicine and Child Neurology. 36 (5): 429–440. doi:10.1111/j.1469-8749.1994.tb11869.x. PMID 8168662. S2CID 41483123.
^ "Blue-skinned family baffled science for 150 years". MSN. 24 February 2012. Archived from the original on 22 January 2013. Retrieved 10 May 2013.
^ Adams C (24 July 1998). "Is there really a race of blue people?". Straight Dope.
^ "Appalachia's Blue People". Tri City Herald. 7 November 1974. p. 32. Archived from the original on 2012-07-13.
^ Fugates of Kentucky: Skin Bluer than Lake Louise
^ "Martin Fuqatenin nəsli: genetik problemə görə dünyaya gələn mavi uşaqlar — Mavi Fuqatelər" [Generation of Martin Fugaten: Blue Children born according to genetic problem - blue fuqates]. YouTube (in Azerbaijani). 10 June 2018.
^ Deeny J (1995). The End of an Epidemic. Dublin: A.& A.Farmar. ISBN 978-1-899047-06-2.
External links
[edit]
"Clinical & Interpretive, TEST ID: MEV1, Methemoglobinemia Evaluation, Blood". Test Catalog, Mayo Clinic Laboratories.
Cleveland Clinic
| | |
--- |
| Classification | ICD-11: 3A9Y ICD-10: D74 ICD-9-CM: 289.7 MeSH: D008708 DiseasesDB: 8100 SNOMED CT: 38959009 |
| External resources | MedlinePlus: 000562 eMedicine: med/1466 emerg/313 ped/1432 Radiopaedia: 67690 Scholia: Q748442 |
| Diseases of red blood cells | |
--- |
| ↑ | | | | --- | | Polycythemia | Polycythemia vera | |
| ↓ | | | | | | | | | | | | | | | | | | --- --- --- --- --- --- --- --- | | Anemia | | | | --- | | Nutritional | Micro-: Iron-deficiency anemia + Plummer–Vinson syndrome Macro-: Megaloblastic anemia + Pernicious anemia | | Hemolytic (mostly normo-) | | | | --- | | Hereditary | enzymopathy: Glucose-6-phosphate dehydrogenase deficiency glycolysis + pyruvate kinase deficiency + triosephosphate isomerase deficiency + hexokinase deficiency hemoglobinopathy: Thalassemia + alpha + beta + delta Sickle cell disease/trait Hemoglobin C disease membrane: Hereditary spherocytosis + Minkowski–Chauffard syndrome Hereditary elliptocytosis + Southeast Asian ovalocytosis Hereditary stomatocytosis | | Acquired | | | | --- | | AIHA | Warm antibody autoimmune hemolytic anemia Cold agglutinin disease Donath–Landsteiner hemolytic anemia + Paroxysmal cold hemoglobinuria Mixed autoimmune hemolytic anemia | membrane + paroxysmal nocturnal hemoglobinuria Microangiopathic hemolytic anemia Thrombotic microangiopathy + Hemolytic–uremic syndrome Drug-induced autoimmune Drug-induced nonautoimmune Hemolytic disease of the newborn | | | Aplastic (mostly normo-) | - Hereditary: Fanconi anemia - Diamond–Blackfan anemia Acquired: Pure red cell aplasia Sideroblastic anemia Myelophthisic | | Blood tests | Mean corpuscular volume + normocytic + microcytic + macrocytic Mean corpuscular hemoglobin concentration + normochromic + hypochromic | | | Other | Methemoglobinemia Sulfhemoglobinemia Reticulocytopenia Hereditary persistence of fetal hemoglobin | |
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Autosomal recessive disorders
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11984 | https://dictionary.cambridge.org/us/dictionary/english/consensus | Cambridge Dictionary +Plus
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English (US)
Meaning of consensus in English
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C2
a generally accepted opinion or decision among a group of people:
general consensus The general consensus in the office is that he can't do his job.
reach a consensus Could we reach a consensus on this matter? Let's take a vote.
consensus on There is a growing consensus on the need for the country to reduce its reliance on imported oil.
Thesaurus: synonyms, antonyms, and examples
the state of agreeing with someone or something
agreementThere's widespread agreement that something must be done.
acceptanceHis views never gained acceptance among the broader community.
concordanceThe study shows strong concordance between patient health and patient happiness.
assentThe bill received official assent.
consentI give my consent to the marriage.
sanctionShe gave official state sanction to the drilling company for their proposed pipeline.
See more results »
More examplesFewer examples
We were unable to reach a consensus about membership fees.
There is little consensus about the issue of smacking children.
We managed to get a consensus about not smoking in the office.
SMART Vocabulary: related words and phrases
Accepting & agreeing
accept someone’s invitation
accepting
accommodation
accreditation
agree to something phrasal verb
arrive
compact
conclusion
daresay
formal
formalize
grumble
out of alignment
peace accord
pounce
pounce on something phrasal verb
pre-approval
ratify
take/pick up the gauntlet idiom
worry
See more results »
(Definition of consensus from the Cambridge Advanced Learner's Dictionary & Thesaurus © Cambridge University Press)
consensus | Intermediate English
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a generally accepted opinion; wide agreement:
They’re trying to build a consensus on the need to improve the city’s schools.
(Definition of consensus from the Cambridge Academic Content Dictionary © Cambridge University Press)
consensus | Business English
consensus
noun [ S or U ]
uk
/kənˈsensəs/ us
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a generally accepted opinion among a group of people:
The consensus among City economists is that interest rates will be held at 1%.
agreement among a group of people:
consensus on/about sth Achieving an international consensus on these issues is proving difficult.
a general/broad/growing consensus There is a growing consensus about the best methods for reducing carbon emissions.
reach/build/agree a consensus Our top priority must be building a consensus with legislators.
The process of taking decisions by consensus can be fraught with difficulty.
(Definition of consensus from the Cambridge Business English Dictionary © Cambridge University Press)
Examples of consensus
consensus
But there's a downside to all this consensus -- it can get boring.
From Wired
But in a season that seemed to defy old-fashioned consensus, plenty of other tunes were vying for the title as well.
From Los Angeles Times
Inside the department, there's less of a consensus on how the city's top cop is performing.
From Dallas Morning News
The league would like to have a joint consensus with the union, sources said.
From ESPN
In 2011 people around the world broke consensus with power.
From Wired
We can actually have a discussion about an issue and arrrive at a mutual consensus out-of-game, regardless of how it affects our characters in-game.
From CNN
Call for these changes, and you build consensus or at least have a quality debate.
From CNN
There is also a strong consensus that the same region will become drier and increasingly snow-free with time, particularly in the winter and spring.
From ThinkProgress
You speak of consensus as if it were a popularity contest.
From Phys.Org
We're increasing the margin of safety with a real consensus guideline that has three key changes.
From CNN
There's widespread consensus that county jails aren't well-equipped to handle lots of people struggling with addiction and mental illness.
From NPR
The general consensus seemed to be that, of course, parents are going to support their kids, so the letters will inevitably lack credibility.
From TIME
Even on issues on which there is broad consensus -- - such as fighting piracy - -- approaches differ.
From Los Angeles Times
These examples are from corpora and from sources on the web. Any opinions in the examples do not represent the opinion of the Cambridge Dictionary editors or of Cambridge University Press or its licensors.
Collocations with consensus
consensus
These are words often used in combination with consensus.
Click on a collocation to see more examples of it.
broad consensus
However, on the question about the 'right to work ', there is a broad consensus amongst liberals.
From the Cambridge English Corpus
clear consensus
But, there is no clear consensus on the importance of these interactions at larger scales.
From the Cambridge English Corpus
consensus conference
However, patient organizations were involved in preparations for the consensus conference on cochlear implants in 1993.
From the Cambridge English Corpus
These examples are from corpora and from sources on the web. Any opinions in the examples do not represent the opinion of the Cambridge Dictionary editors or of Cambridge University Press or its licensors.
See all collocations with consensus
What is the pronunciation of consensus?
Translations of consensus
in Chinese (Traditional)
一致的意見, 共識…
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in Chinese (Simplified)
一致的意见, 共识…
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in Spanish
consenso…
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consenso…
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सर्वानुमत, एकमत…
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ortak mutabakat, genel görüş, oy birliği…
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consensus…
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eenstemmigheid…
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ஒரு குழு மக்களிடையே பொதுவாக ஏற்றுக்கொள்ளப்பட்ட கருத்து அல்லது முடிவு…
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मतैक्य…
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સર્વાનુમતિ…
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almindelig enighed, almindelig opfattelse…
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allmän mening, samstämmighet, konsensus…
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sepersetujuan…
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die Übereinstimmung…
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alminnelig enighet, gjengs oppfatning…
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اتفاق رائے…
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одностайність, консенсус…
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единодушие…
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ఏకాభిప్రాయం…
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ঐক্যমত…
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shoda…
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konsensus…
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ความคิดเห็นของคนส่วนใหญ่…
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sự đồng lòng…
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jednomyślność, zgoda…
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11985 | https://www.shaalaa.com/question-bank-solutions/find-the-equation-of-the-curve-which-passes-through-the-origin-and-has-the-slope-x-3y-1-at-any-point-x-y-on-it_153153 | Find the equation of the curve which passes through the origin and has the slope x + 3y - 1 at any point (x, y) on it. - Mathematics and Statistics
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Question
Find the equation of the curve which passes through the origin and has the slope x + 3y - 1 at any point (x, y) on it.
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SolutionShow Solution
Let A (x, y) be the point on the curve y = f(x).
Then slope of the tangent to the curve at point A is "dy"/"dx".
According to the given condition,
"dy"/"dx" = "x" + 3"y" - 1
∴ "dy"/"dx" - 3"y" = "x - 1" ....(1)
This is the linear differential equation of the form
"dy"/"dx" + "Py" = "Q", where P = - 3 and Q = x - 1
∴ I.F. = "e"^(int "P dx") = "e"^(int - 3"dx") = "e"^(- 3"x")
∴ the solution of (1) is given by
"y" \ ("I.F.") = int "Q" \ ("I.F.") "dx" + "c"
∴ "y" \ "e"^(- 3"x") = int ("x - 1") \ "e"^(-3"x") "dx" + "c"
∴ "e"^(- 3"x") \ "y" = ("x - 1") int "e"^(- 3"x") - int ["d"/"dx" ("x - 1") \ int "e"^(- 3"x") "dx"] "dx" + "c"\_1
∴ "e"^(- 3"x") \ "y" = ("x - 1") \ "e"^(- 3"x")/-3 - int 1 \ "e"^(- 3"x") \ "y"/-3 "dx" + "c"\_1
∴ "e"^(- 3"x") \ "y" = - 1/3 ("x - 1") \ "e"^(- 3"x") + 1/3 int "e"^(- 3"x") "dx" + "c"\_1
∴ "e"^(- 3"x") \ "y" = - 1/3 ("x - 1") "e"^(- 3"x") + 1/3 \ "e"^(- 3"x")/-3 + "c"\_1
∴ "e"^(- 3"x") \ "y" = - 1/3 ("x - 1")"e"^(- 3"x") - 1/9 "e"^(- 3"x") + "c"\_1
∴ 9y = - 3(x - 1) - 1 + 9"c"\_1 \ e^("3x")
∴ 9y + 3(x - 1) + 1 = 9"c"\_1 \ e^("3x")
∴ 9y + 3x - 3 + 1 = 9"c"\_1 \ e^("3x")
∴ 3(x + 3y) = 2 + 9"c"\_1 \ e^("3x")
∴ 3(x + 3y) = 2 + "c" \ e^("3x") where c = 9c1 ....(2)
This is the general equation of the curve.
But the required curve is passing through the origin (0, 0).
∴ by putting x = 0 and y= 0 in (2), we get
0 = 2 + c
∴ c = - 2
∴ from (2), the equation of the required curve is
3(x + 3y) = 2 - 2e^("3x")
i.e. 3(x + 3y) = 2(1 - e^("3x")).
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Sep 2023
Dec 2023
3 months later
Tim Peters
tim.one
CPython core developer
Sep 2023
Three-quarters credit .
It’s impossible to “repair” the second example, unless you change float addition first.
```
4503599627370496 + 0.5 == 4503599627370496
True
```
That is, the 0.5 was lost to rounding during addition, before the function was called. So it goes. The same will happen with any function in any language supporting 754 doubles, unless you find a way to force the system to use to-plus-infinity rounding for the addition.
The first example is indeed “a bug” in the function I posted. Here’s a different version that doesn’t use float arithmetic, so can’t be affected by the current float rounding mode:
```
from math import modf
def rintnu(x):
"Return float x rounded to an int, resolving ties away from 0."
if x < 0.0:
return -rintnu(-x)
f, i = modf(x)
i = int(i)
if f >= 0.5:
i += 1
return i
rintnu(0.49999999999999994)
0
rintnu(0.5)
1
```
b. s.
newbie-02
Sep 2023
‘three-quarters’ … is that good or bad?
for the first example
```
return int( x + 0.49999999999999994 )
```
will do, for the second one: that’s what I dislike in IEEE 754 /
it’s actual interpretation, ‘ties to even anchored deep in the gearbox,
without alternatives, and hard to oversteer in software’.
Tim Peters
tim.one
CPython core developer
Sep 2023
On a scale of “no credit” to “full credit”, it’s 75% of the way to “full”
Clever, but like my first attempt is also dependent on the rounding mode currently in use. We can’t easily change the rounding mode in core Python, but we can in the mpmath module:
```
import mpmath
magic = mpmath.mpf(0.49999999999999994)
print(repr(mpmath.fadd(0.5, magic))) # mpf('1.0')
print(repr(mpmath.fadd(0.5, magic, rounding='f'))) # mpf('0.99999999999999989')
```
For 0.5 “it works” under the default rounding mode because nearest-even rounds a halfway case up to give it a final 0 bit. Under to-minus-infinity or to-0 rounding, though, it throws the trailing one bit away and leaves the last remaining bit odd.
So if you want this, take my second version. As already said, the current rounding mode has no effect on its result. It also has the benefit of being obvious instead of clever .
Most users couldn’t care less. If you can find a conforming implementation of 754 binary floats (as I said before, Python’s is not), then, as in Python’s decimal module, changing the rounding mode could be straightforward.
Or not . One highly valid criticism of 754 is that it didn’t specify a concrete way implementations had to supply to change all the adjustable moving parts it requires of a conforming implementation. As a result, there is no portable way.
I showed above a way to force a specific rounding mode under mpmath, but even in that there’s no way to set it “in general”. Instead some specific named mpmath functions (like fadd()) accept an optional rounding= argument. The rounding used by the infix binary operators (like +. , …) is always nearest/even.
BTW, mpmath’s “round to nearest int” function is named nint(), and also does nearest/even rounding.
```
import mpmath
for i in range(10):
... x = i + mpmath.mpf(0.5)
... print(x, mpmath.nint(x))
0.5 0.0
1.5 2.0
2.5 2.0
3.5 4.0
4.5 4.0
5.5 6.0
6.5 6.0
7.5 8.0
8.5 8.0
9.5 10.0
```
b. s.
newbie-02
Sep 2023
@TimPeters,
I have to thank you for your time and engagement, we have discussed in depth and length,
with the result:
A.) with small tricks we can achieve ‘human compatible’ rounding for IEEE values,
( which IMHO does not solve irritations due to scaling when rounding to decimals places ),
B.) we can’t produce ‘human compatible’ rounding for basic arithmetic results where
‘hardware rounding’ hit’s before programmers intention,
you contributed profound knowledge, I a persistent intention, no solution within IEEE
binaries.
You consider ties to even a good thing, while I criticise the lack of ties away from
zero as an alternative, as it is needed, for example, in spreadsheets to produce
‘people-friendly’ results.
If we want better human compatibility we’d have _Decimals, python decimal or
fractions to choose from. But - IMHO - very few programmers would consider
rewriting well matured complex applications in a new datatype.
So we can expect deviations and surprises in ‘computational mathematics’
for some time to come.
[ edit ]
want to add one point where I see a risk of fooling newbies and importance to raise
awareness:
Add floats a and b, giving a correctly rounded sum and exact error.
Mathematically, a + b is exactly equal to sum + error.
in the SO example is valid from a binary point of view, preserving the deviations
in representation of the operands!!!, not from a decimal POV!!!, not accounting
‘what we see’. see example:
```
exact_add( 4.4, 2.2 ) --> (6.6000000000000005, 0.0)
```
such cases are rare, and people experimenting with improved algorithms might
get trapped by ‘often looking good’ and not investigating in detail that
```
exact_add( 0.2, 0.1 ) --> (0.30000000000000004, -2.7755575615628914e-17)
```
is neither pointing to the decimal correct 0.3, nor it’s binary representative
0.29999999999999998890… be aware of different understanding about ‘exact’
or ‘correct’ between binary vs. decimal oriented people!
question on detail ( we managed to get me confused ): we have one! option
where ‘decimal’ can be steered to half away rounding, but NO option in any
datatype where operations round results ties away?
_Decimal64 in ‘C’:
```
4503599627370496.0DD + 0.5DD --> 4503599627370496.0
4503599627370496.0DD + 1.5DD --> 4503599627370498.0
```
python decimal:
```
import decimal
decimal.getcontext().prec=16
print ( decimal.Decimal('4503599627370496') + decimal.Decimal('0.5') )
print ( decimal.Decimal('4503599627370496') + decimal.Decimal('1.5') )
-->
4503599627370496
4503599627370498
```
[ /edit ]
29 days later
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11988 | https://moodle.net/.pkg/@moodlenet/ed-resource/dl/ed-resource/22ZGgQn8/546_2p5p6p7SeriesParallelDYReduction8Ed_1.pdf | SERIES PARALLEL RESISTOR COMBINATIONS UP TO NOW WE HAVE STUDIED CIRCUITS THAT CAN BE ANALYZED WITH ONE APPLICATION OF KVL(SINGLE LOOP) OR KCL(SINGLE NODE-PAIR) WE HAVE ALSO SEEN THAT IN SOME SITUATIONS IT IS ADVANTAGEOUS TO COMBINE RESISTORS TO SIMPLIFY THE ANALYSIS OF A CIRCUIT NOW WE EXAMINE SOME MORE COMPLEX CIRCUITS WHERE WE CAN SIMPLIFY THE ANALYSIS USING THE TECHNIQUE OF COMBINING RESISTORS… … PLUS THE USE OF OHM’S LAW SERIES COMBINATIONS PARALLEL COMBINATION N p G G G G + + + = ...
2 1 FIRST WE PRACTICE COMBINING RESISTORS 6k||3k (10K,2K)SERIES SERIES k 3 k k k 4 12 || 6 = k 12 k 3 k 5 k k k 6 12 || 12 = k k k 2 6 || 3 = ) 2 4 ( || 6 k k k + 12k If things get confusing… EXAMPLES COMBINATION SERIES-PARALLEL k 9 k k k 6 9 || 18 = k k k 10 6 6 + + RESISTORS ARE IN SERIES IF THEY CARRY EXACTLY THE SAME CURRENT RESISTORS ARE IN PARALLEL IF THEY ARE CONNECTED EXACTLY BETWEEN THE SAME TWO NODES If the drawing gets confusing… Redraw the reduced circuit and start again AN “INVERSE SERIES PARALLEL COMBINATION” AVAILABLE ARE RESISTORS ONLY WHEN 600mV BE MUST = 1 .
0 3A I VR = = 2 .
0 3 6 .
A V R REQUIRED + = 1 .
0 1 .
0 R AVAILABLE ARE RESISTORS ONLY WHEN 600mV BE MUST = 1 .
0 9A I VR = = 0667 .
0 9 6 .
A V R REQUIRED R SIMPLE CASE NOT SO SIMPLE CASE Given the final value Find a proper combination EFFECT OF RESISTOR TOLERANCE 10% : TOLERANCE RESISTOR 2.7k : VALUE RESISTOR NOMINAL RANGES FOR CURRENT AND POWER?
mA I mA I 115 .
4 7 .
2 9 .
0 10 367 .
3 7 .
2 1 .
1 10 max min = = = = : CURRENT MAXIMUM : CURRENT MINIMUM THE RANGES FOR CURRENT AND POWER ARE DETERMINED BY THE TOLERANCE BUT THE PERCENTAGE OF CHANGE MAY BE DIFFERENT FROM THE PERCENTAGE OF TOLERANCE. THE RANGES MAY NOT EVEN BE SYMMETRIC mA I 704 .
3 7 .
2 10 = = − : CURRENT NOMINAL ( ) _ mW P 04 .
37 7 .
2 10 2 = = : POWER NOMINAL : POWER MAXIMUM : ) POWER(VI MINIMUM min mW mW 15 .
41 67 .
33 CIRCUIT WITH SERIES-PARALLEL RESISTOR COMBINATIONS COMBINING RESISTORS IN SERIES ELIMINATES ONE NODE FROM THE CIRCUIT.
COMBINING RESISTORS IN PARALLEL ELIMINATES ONE LOOP FROM THE CIRCUIT THE COMBINATION OF COMPONENTS CAN REDUCE THE COMPLEXITY OF A CIRCUIT AND RENDER IT SUITABLE FOR ANALYSIS USING THE BASIC TOOLS DEVELOPED SO FAR.
GENERAL STRATEGY: •REDUCE COMPLEXITY UNTIL THE CIRCUIT BECOMES SIMPLE ENOUGH TO ANALYZE.
•USE DATA FROM SIMPLIFIED CIRCUIT TO COMPUTE DESIRED VARIABLES IN ORIGINAL CIRCUIT - HENCE ONE MUST KEEP TRACK OF ANY RELATIONSHIP BETWEEN VARIABLES FIRST REDUCE IT TO A SINGLE LOOP CIRCUIT k 12 k k 12 || 4 k 6 k k 6 || 6 k V I 12 12 1 = ) 12 ( 9 3 3 + = a V SECOND: “BACKTRACK” USING KVL, KCL OHM’S k V I a 6 2 = : S OHM' 0 3 2 1 = − − I I I : KCL 3 3 I k Vb = : S OHM' 3 I …OTHER OPTIONS...
4 3 4 4 12 4 12 I k V I I b = + = 5 3 4 5 3 0 I k V I I I C = = − + : S OHM' : KCL k k k 1 2 || 2 = V V k k k VO 1 ) 3 ( 2 1 1 = + = : DIVIDER VOLTAGE k k k 2 1 1 = + A A k k k IO 1 ) 3 ( 2 1 1 = + = : DIVIDER CURRENT LEARNING BY DOING AN EXAMPLE OF “BACKTRACKING” A STRATEGY. ALWAYS ASK: “WHAT ELSE CAN I COMPUTE?” 4 6 I k Vb = k V I b 3 3 = mA 1 4 3 2 I I I + = mA 5 .
1 = 2 2 I k Va = V 3 V 3 b a xz V V V + = V Vxz 6 = k V I xz 4 5 = mA 5 .
1 = 5 2 1 I I I + = mA I 3 1 = 1 1 4 6 I k V I k V xz O + + = V VO 36 = mA 5 .
0 = O V FIND DIVIDER VOLTAGE USE FIND : STRATEGY 1 V 1 V k 60 k k k 20 60 || 30 = + -− + 1 V k 20 k 20 V 12 V k k k 6 ) 12 ( 20 20 20 = + = V 6 DIVIDER VOLTAGE 1 40 20 20 V k k k VO + = V 2 S V FIND THIS IS AN INVERSE PROBLEM WHAT CAN BE COMPUTED?
− + V 6 mA 05 .
0 mA 15 .
0 mA k V 1 .
0 60 1 = k V I 120 6 1 = V mA k VS 6 15 .
0 20 + = V 9 SERIES PARALLEL TIONS TRANSFORMA − Y THIS CIRCUIT HAS NO RESISTOR IN SERIES OR PARALLEL IF INSTEAD OF THIS WE COULD HAVE THIS THEN THE CIRCUIT WOULD BECOME LIKE THIS AND BE AMENABLE TO SERIES PARALLEL TRANSFORMATIONS Y → → Y b a ab R R R + = ) ( || 3 1 2 R R R Rab + = 3 2 1 3 1 2 ) ( R R R R R R R R b a + + + = + 3 2 1 2 1 3 ) ( R R R R R R R R c b + + + = + 3 2 1 3 2 1 ) ( R R R R R R R R a c + + + = + SUBTRACT THE FIRST TWO THEN ADD TO THE THIRD TO GET Ra Y R R R R R R R R R R R R R R R R R R c b a → + + = + + = + + = 3 2 1 1 3 3 2 1 3 2 3 2 1 2 1 a b b a R R R R R R R R 1 3 3 1 = = c b c b R R R R R R R R 1 2 1 2 = = REPLACE IN THE THIRD AND SOLVE FOR R1 − + + = + + = + + = Y R R R R R R R R R R R R R R R R R R R R R R R R a a c c b b a c a c c b b a b a c c b b a 3 2 1 LEARNING EXAMPLE: APPLICATION OF WYE-DELTA TRANSFORMATION S I COMPUTE DELTA CONNECTION a b c a b c ( ) k k k k k k REQ 10 ) 6 2 ( || 9 3 6 = + + + = Y R R R R R R R R R R R R R R R R R R c b a → + + = + + = + + = 3 2 1 1 3 3 2 1 3 2 3 2 1 2 1 1 R 2 R 3 R k k k k k 18 6 12 6 12 + + = mA k V IS 2 .
1 12 12 = = ONE COULD ALSO USE A WYE - DELTA TRANSFORMATION ...
LEARNING EXAMPLE SHOULD KEEP THESE TWO NODES!
CONVERT THIS Y INTO A DELTA?
IF WE CONVERT TO Y INTO A DELTA THERE ARE SERIES PARALLEL REDUCTIONS!
− + + = + + = + + = Y R R R R R R R R R R R R R R R R R R R R R R R R a a c c b b a c a c c b b a b a c c b b a 3 2 1 312 12 36 12 k k k k = = 4mA 36k 36k 36k 12k 12k O V + − THE RESULTING CIRCUIT IS A CURRENT DIVIDER 4mA 36k 36 ||12 9 k k k = O V + − 9k CIRCUIT AFTER PARALLEL RESISTOR REDUCTION O I 36 8 4 36 18 3 O k I mA mA k k = = + 8 9 9 24 3 O O V k I k mA V = = = NOTICE THAT BY KEEPING THE FRACTION WE PRESERVE FULL NUMERICAL ACCURACY WYE DELTA |
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Locate Element
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Episode 87 – Alcohol Withdrawal and Delirium Tremens: Diagnosis and Management
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This is ‘Alcohol Withdrawal and Delirium Tremens: Diagnosis and Management‘ on EM Cases.
Alcohol withdrawal is everywhere. We see over half a million patients in U.S. EDs for alcohol withdrawal every year. Despite these huge volumes of patients and the diagnosis of alcohol withdrawal seeming relatively straightforward, it’s actually missed more often than we’d like to admit, being confused with things like drug intoxication or sepsis. Or it’s not even on our radar when an older patient presents with delirium. The differential diagnosis is enormous – and no blood test on the planet will help us diagnose alcohol withdrawal. In fact, the diagnosis is entirely a clinical one. What’s even more surprising is that even if we do nail the diagnosis, observational studies show that, in general, alcohol withdrawal is poorly treated. There’s several reasons for our all too often mismanagement of these patients: few EDs have a standardized approach (or training of an approach) to the management of alcohol withdrawal, there’s unfortunately still a bit of a stigma associated with alcoholism in many EDs which may contribute a kind of indifference to these patients by ED staff, and the medications used to treat alcohol withdrawal are often dosed incorrectly. So what if alcohol withdrawal is missed or poorly treated? Well, mismanaged alcohol withdrawal can be fatal – and untreated severe withdrawal often ends up with your patient seizing, or maybe progressing to delirium tremens.
To help you become masters of alcohol withdrawal management, our guest experts on this podcast are Dr. Bjug Borgundvaag, an ED doc and researcher with a special interest in emergency alcohol related illness and the director of Schwartz-Reismann Emergency Medicine Institute, Dr. Mel Kahan, an addictions specialist for more than 20 years who’s written hundreds of papers and books on alcohol related illness, and the medical director of the substance use service at Women’s College Hospital in Toronto, and Dr. Sara Gray,anED-intensivist at St. Michael’s Hospital, an inner city hospital that sees high volumes of alcohol related illness.
Podcast: Play in new window | Download (Duration: 1:26:23 — 79.1MB)
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Written Summary and blog post written by Keerat Grewal, edited by Anton Helman October, 2016
Cite this podcast as:Helman, A, Borgundvaag, B, Gray, S. Alcohol Withdrawal and Delirium Tremens: Diagnosis and Management. Emergency Medicine Cases. October, 2016. Accessed [date].
General Approach to the Management of Patients with Alcohol Withdrawal
The ideal management of alcohol withdrawal involves 4 steps:
Identify which patients actually have alcohol withdrawal and require treatment
Use a standardized, symptom guided approach to assess symptom severity and guide treatment
Ensure that patients are fully treated prior to ED discharge
Provide a pathway to support patients who are trying to quit
Step 1: Identify which patients actually have alcohol withdrawal and require treatment
Alcohol withdrawal is a clinical diagnosis and a diagnosis of exclusion. The tremor of alcohol withdrawal is central to the diagnosis. It is important to understand the key features of alcohol withdrawal tremor. The characteristic tremor is an intention tremor: at rest there is no tremor, but when you ask the patient to extend their hands or arms you will see a fine motor tremor (typically 7-12Hz) that is constant and does not fatigue with time. Other symptoms associated with alcohol withdrawal include: gastrointestinal upset, anxiety, nausea/vomiting, diaphoresis, tachycardia, hypertension and headache.
PEARL: A tongue tremor (as in this video) is difficult to feign and is a more sensitive sign of alcoholic tremor than hand tremor
PEARL: The 3 characteristics of alcohol withdrawal hand tremor are that it is an intention tremor, it is constant and it does not fatigue.
Step 2: Use a standardized, symptom-guided approach to assess symptom severity and guide treatment
Protocols for treating alcohol withdrawal standardize care, they ensure clinicians identify the appropriate symptoms and monitor treatment. Protocols for alcohol withdrawal have been shown to improve the quality and consistency of care patients receive.
The CIWA protocol is a 10-item scale. It has been well validated in patients with alcohol withdrawal, but should not be used for patients with delirium tremens. The CIWA calls for patients to be assessed hourly and treated if the total score is 10 or greater. When 2 sequential scores are < 10 they may be considered for discharge.
CIWA on MDCalc Here
The SHOT protocol is a shorter protocol that has recently been developed by our guest experts which may be easier to implement in the ED. It is a 4-item scale (Sweating, Hallucinations, Orientation and Tremor) that correlated well with the CIWA score and takes ~ 1 minute to apply. The SHOT protocol has not yet been validated.
Step 3: Ensure that patients are fully treated prior to ED discharge
Observational studies show that patients are often either undertreated resulting in complications of alcohol withdrawal such as seizures and delirium tremens, or oversedated leading to prolonged length of stay and airway complications. If a patient has two sequential CIWA scores < 10 two hours apart, and there are no concerning risks for deterioration, consider discharging the patient from the ED. The patient’s tremor should be minimal or resolved before discharge regardless of the CIWA score.
Pitfall: Patients with a CIWA score <10 yet still have a severe alcohol withdrawal tremor are at risk of complications of alcohol withdrawal if discharged from the ED.
The CIWA-Ar is only intended to allow assessment of withdrawal severity once the diagnosis has been made. It is not a diagnostic test for withdrawal.
It is strongly discouraged that patients be provided a take-away supply or prescription for benzodiazepines. The long half-life of diazepam will protect patients from developing serious symptoms of withdrawal, and if adequately treated in the ED, no additional medications will be required.
Pitfall:Patients who are discharged from the ED with a prescription for benzodiazepines are put at increased risk for sedative overdose, drug seeking behaviour and drug dependence.
Step 4: Provide a pathway to support patients who are trying to quit
Most patients in the ED with alcohol withdrawal are there to seek help. They are in a vulnerable state and may be ready to quit drinking. This is an opportune time for them to take the first steps on the pathway to recovery. Dr. Kahan recommends counselling all ED patients who you have treated for alcohol withdrawal about the following to maximize their chances of seeking help:
You need help for your serious alcohol problem
You can’t do it on your own
There are effective treatments available to you
With treatment the way you feel, your mood, social relationships and work will be profoundly better
Provide several options to the patient, if available (i.e., AA, local treatment programs).
Timing of Alcohol Withdrawal and Delirium Tremens
Symptoms from alcohol withdrawal usually start within 6-8 hours after the blood alcohol level decreases, peak at 72 hours, and diminish by days 5 to 7 of abstinence. Delirium Tremens can occur anytime from 3 to 12 days after abstinence.
Patients who are severely dependent on alcohol become tolerant to alcohol and their nervous systems have been reset to compensate for the sedating effects of alcohol. The likelihood of developing withdrawal is dependent on the usual amount consumed and the duration of consumption. Therefore, patients who consume large amounts of alcohol on a regular basis are more likely to develop withdrawal requiring pharmacologic management compared to those who binge sporadically.
Investigations for Patients with Alcohol Withdrawal
Ethanol Level in patients with alcohol withdrawal
A serum ethanol level should be considered only if the history is inconsistent or you are unsure of the diagnosis. Even then, there is no single ethanol level at which withdrawal is impossible. Chronic alcohol users may experience alcohol withdrawal at serum ethanol levels that are intoxicating to the naïve drinker.
Blood work in patients with alcohol withdrawal
In mild cases of withdrawal, blood work is rarely helpful and is unlikely to change management. However, in patients with severe alcohol withdrawal, especially patients with delirium tremens, blood work can help rule out other causes of delirium, screen for alcoholic ketoacidosis and electrolyte abnormalities.
ECG in patients with alcohol withdrawal
There is some evidence to suggest that patients in severe alcohol withdrawal are at risk for prolonged QT which may be at least partially a result of the associated electrolyte abnormality such as hypomagnesemia. Consider a baseline ECG in patients who require admission for alcohol withdrawal.
Urine Drug Screen in alcohol withdrawal
A urine drug screen rarely changes management if there are no signs of co-ingestions. In a patient suspected to be in alcohol withdrawal who has delirium, or you suspect a concurrent toxidrome, a urine drug screen may be helpful.
Differential Diagnosis of Alcohol Withdrawal and Delirium Tremens
First, some patients with alcohol intoxication can have some of the signs of alcohol withdrawal (tachycardia, elevated BP, diaphoresis, agitation), so assuming that they are in withdrawal and giving them high doses of benzodiazepines can lead to complications of oversedation. Be sure to examine for tremor carefully. If they don’t have a true withdrawal tremor then they are probably drunk and not in withdrawal.
Pitfall: Some patients with alcohol intoxication can have some of the signs of alcohol withdrawal (tachycardia, elevated BP, diaphoresis, agitation), and assuming that they are in withdrawal and giving them high doses of benzodiazepines can lead to complications of oversedation.
Next, it is important to complete a neurological exam and look for any signs of head injury in patients with alcohol intoxication or withdrawal. Patients with a history of alcohol abuse are at a higher risk for head injury and intracranial hemorrhage as many of them have coagulopathies related to liver disease and fragile intracranial bridging veins from cerebral atrophy. Have a low threshold to obtain a CT head to rule out a bleed if there are signs or suspicion of a head injury.
In patients who are suffering from delirium, keep a broad differential diagnosis including infection, other toxidromes (sympathomimetics, anticholinergics, toxic alcohols), serotonin syndrome, NMS, hypertensive crisis, acute pain and thyrotoxicosis.
Pharmacologic Treatment of Patients with Alcohol Withdrawal
Benzodiazepines for Alcohol Withdrawal
The first line medication for treating patients with alcohol withdrawal is benzodiazepines. The drug of choice is diazepam because it has a long half-life of ~100 hours, and carries a decreased risk of developing serious withdrawal symptoms once the patient is discharged. Diazepam also has a faster onset of action than lorazepam. It is important to remember that diazepam should be avoided in patients with overt liver failure or a history of liver failure.
The half life of lorazepam is ~8-12 hours, which puts patients at risk of developing withdrawal symptoms once the patient is discharged and the lorazepam has worn off. However, lorazepam should be considered if there is evidence of significant liver dysfunction. If patients are encephalopathic or pre-encephalopathic lower the dose and titrate medications slowly.
Use oral benzodiazepines in stable patients with mild withdrawal who are not vomiting. Use IV benzodiazepines allowing faster onset and easier titration in patients with severe withdrawal as these patients are at a higher risk of seizure.
Dosing of Benzodiazepines in Alcohol Withdrawal
Dose benzodiazepines according to the severity of the alcohol withdrawal:
Another protocol to consider for patients with severe withdrawal is to give a first dose of diazepam 10mg IV, and repeat in 5 minutes if the response is not adequate. Then double the dose to 20 mg and continue with 20mg, 30mg, 30mg, 40mg, 40mg every 5 minutes as needed.
Example diazepam dosing guide for severe withdrawal (care of First 10EM)
Phenobarbital for Treatment of Alcohol Withdrawal
There is no evidence that phenobarbital is superior to benzodiazepines for preventing complications of alcohol withdrawal. There is debate regarding the equivalency of phenobarbital and benzodiazepines. Our experts do not recommend using phenobarbital alone for treatment of alcohol withdrawal, but consider its use as an adjunct with benzodiazepines after the patient has received the equivalent of 200mg of diazepam and is still in severe withdrawal.
Update 2023: A systematic review and meta-analysis including 1507 patients from 8 studies comparing the safety and efficacy of phenobarbital to benzodiazepines for the treatment of alcohol withdrawal syndrome in the ED found that treatment with phenobarbital did not significantly reduce the risk of ICU admission (RR 0.92, 95%CI 0.54-1.55), risk of hospital admission (0.98, 95%CI 0.89-1.07) and adverse events (1.1, 95%CI 0.78-1.57). Abstract
Thiamine
Often patients with alcohol withdrawal receive 100 mg of thiamine in the ED although the evidence for any benefit is weak. If there is concern of Wernicke’s encephalopathy (nystagmus, ataxia, confusion), higher doses of thiamine are recommended (i.e. 500 mg IV q8h).
Fluids
Patients withdrawing from alcohol are almost always hypovolemic, many of these patients are also hypoglycemic. These patients should receive glucose-containing fluids. Glucose and thiamine compete for the same co-factor, therefore, there is a theoretical risk that giving glucose in a thiamine deficient patient can precipitate Wernicke’s encephalopathy. However, there is no evidence that one dose of glucose in a thiamine-deficient patient will precipitate Wernicke’s. If you give glucose, give it at the same time or after thiamine – but urgent glucose should not be delayed for thiamine administration.
Management of Severe Alcohol Withdrawal
In the agitated and disorientated patient with alcohol withdrawal, avoid antipsychotics such as Haldol because these drugs can prolong the QT interval and reduce the seizure threshold.
Consider intubation in patients who have airway concerns or who have refractory seizures, and may require adjunctive treatments. There are several adjunct medications that can be considered for refractory cases of severe alcohol withdrawal that are not responding to large doses of benzodiazepines. These are usually started in ICU patients. In the intubated patient, consider propofol, phenobarbital, dexmedetomidine and ketamine.
Care of RebelEM
Suggested ICU Admission Criteria for Alcohol Withdrawal
Underlying medical or surgical condition that requires ICU-level care
Requires second line therapy to control withdrawal (benzodiazepine resistant withdrawal)
Hyperthermia
Recurrent seizures
Severe altered mental status
Suggested Algorithm for Management of Severe Alcohol Withdrawal
For more on alcohol withdrawal and delirium tremens on EM Cases: Best Case Ever 50 – Delirium Tremens with Sara Gray
Drs Gray, Kahan, Borgundvaag, Grewal and Helman have no conflicts of interest to declare
References
Miller F, Whitcup S, Sacks M, Lynch PE. Unrecognized drug dependence and withdrawal in the elderly. Drug Alcohol Depend. 1985;15(1-2):177-9.
Stehman CR, Mycyk MB. A rational approach to the treatment of alcohol withdrawal in the ED. Am J Emerg Med. 2013;31(4):734-42.
Gray S, Borgundvaag B, Sirvastava A, Randall I, Kahan M. Feasibility and reliability of the SHOT: A short scale for measuring pretreatment severity of alcohol withdrawal in the emergency department. Acad Emerg Med. 2010 Oct;17(10):1048-54.
Lorentzen K, Lauritsen AØ, Bendtsen AO. Use of propofol infusion in alcohol withdrawal-induced refractory delirium tremens. Dan Med J. 2014;61(5):A4807.
Schuckit MA. Recognition and management of withdrawal delirium (delirium tremens). N Engl J Med.2014 November 371;22:2109-2113.
Amato L, Minozzi S, Vecchi S, Davoli M. Benzodiazepines for alcohol withdrawal. Cochrane Database Syst Rev. 2010;3:CD005063.
Gold JA, Rimal B, Nolan A, et al. A strategy of escalating doses of benzodiazepines and phenobarbital administration reduces the need for mechanical ventilation in delirium tremens. Crit Care Med.2007;35(3):724-730.
Tolonen J, Rossinen J, Alho H, Harjola VP. Dexmedetomidine in addition to benzodiazepine-based sedation in patients with alcohol withdrawal delirium. Eur J Emerg Med.2013 Dec;20(6):425-7.
Mueller SW;, Preslaski CR, Kiser TH, et al. A randomized, double-blind, placebo-controlled dose range study of dexmedetomidine as adjunctive therapy for alcohol withdrawal. Crit Care Med. 2014;42(5):1131-9.
Michaelsen IH et al. Phenobarbital versus diazepam for delirium tremens: a retrospective study. Dan Med Bull. 2010; 57 (8) A4169.
Lizotte RJ et al. Evaluating the effects of dexmedetomidine compared to propofol as adjunctive therapy in patients with alcohol withdrawal. Clin Pharmacol. 2014;6:171-177.
VanderWeide LA et al. Evaluation of early dexmedetomidine addition to the standard of care for severe alcohol withdrawal in the ICU: a retrospective controlled cohort study. J Intensive Care Med 2016; 31(3): 198 – 204).
Donnino MW, Vega J, Miller J, Walsh M. Myths and misconceptions of Wernicke’s encephalopathy: what every emergency physician should know. Ann Emerg Med. 2007;50(6):715-21.
Hendey GW, Dery RA, Barnes RL, et al. A prospective, randomized, trial of phenobarbital versus benzodiazepines for acute alcohol withdrawal. Am J Emerg Med. 2011;29(4):382- 385.
Rosenson J, Clements C, Simon B, et al. Phenobarbital for acute alcohol withdrawal: a prospective randomized double-blind placebo-controlled study. J Emerg Med. 2013;44(3):592- 598.
Additional FOAMed Resources on Alcohol Withdrawal and Delirium Tremens
Justin Morgenstern on delirium tremens management on First10EM
Evidence-based look at dexmedetomidine as adjunct on ALiEM
EMdocs reviews the pearls and pitfalls of management
Salim Rezaei reviews the literature on adjuntive medications
Prolonged QT associated with alcohol withdrawal on Dr. Smith’s ECG blog
By Anton Helman|2024-05-21T02:01:40-04:00October 25th, 2016|Categories: EM Cases, Emergency Medicine, Episodes, Medical Specialty, Special Populations, Toxicology|Tags: benzodiazepine, Bjug Borgundvaag, CIWA, dexmeditomidine, Mel Kahan, phenobarbital, Sara Gray, SHOT, thiamine|9 Comments
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About the Author: Anton Helman
Dr. Anton Helman is an Emergency Physician at North York General in Toronto. He is an Assistant Professor at the University of Toronto, Division of Emergency Medicine and the Education Innovation Lead at the Schwartz-Reisman Emergency Medicine Instititute. He is the founder, editor-in-chief and host of Emergency Medicine Cases.
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9 Comments
Himanshu Mirani December 9, 2017 at 7:41 pm - Reply
Great tips n practice changers.. Excellent as always
2. Jerry Hu February 17, 2019 at 11:18 am - Reply
Dear EM Cases, have you read the article posted by Dr. Josh Farkas on Pulmcrit ( reviewing and advocating phenobarbital as 1st line agent for alcohol withdrawal? To say there is insufficient evidence for phenobarbital is an understatement. I would love to hear your take on Dr. Farkas’ article and references on a separate episode if possible, it is very well written and convincing. Thank you for your time and attention
Anton Helman February 17, 2019 at 11:56 am - Reply
Agree that it is very well written and convincing, however the evidence is still not robust enough to recommend phenobarb as first line. He is quoting a retrospective study where phenobarb was used as 2nd line – which we agree with – phenobarb is a great 2nd line medication. Benzodiazepines, IF used in adequate doses, with correct timing, as per CIWA very rarely fails. Benzos are generally safer (especially valium is safer than phenobarb in cirrhotic patients) and much more familiar to emergency providers. I think it is not unreasonable to start with phenobarb, however our efforts should concentrate on early accurate identification of withdrawal, rapid adequate treatment, prevention, education and anticraving medications rather than first line medication choice for severe withdrawal.
LMF September 17, 2019 at 9:34 am - Reply
Dr. HelmenThank you so much for your article. Very thorough and helpful for anyone from a seasoned clinician, patient, or parent.
Peter Tagmose Thomsen November 30, 2019 at 3:19 pm - Reply
I used to work at a psychiatric ed where the first line was phenobarbitals. The nurses were not used to benzoes and If i ordered benzoes I would just get calls all night about ordering phenobarbitals.
I myself did not.have any adverse events with patients and did not personally see any big difference (besides it was maybe a bit harder to taper off phenobarbitals and if the patient needed other sedating agents for whichever reason it was an issue)
I know they did have an adverse event with overdosing a patient with suspected delirium tremens who they had to intubate , but I’m not convincedit would have been any other way had he been loaded with benzoes)
Bottom line: if your setup and department is up for it, then phenobarbitals can probably be just as safe as benzoes – however besides some people being strong advocates for it (like the professor Sigurd Benjaminsen, at the region I worked) I am unfamiliar with the evidence
Side note: I have heard some people from the ILAE favoring phenobarbitals as a second line agent in status epilepticus – especially if alcohol history
3. Jerry Hu February 18, 2019 at 11:47 am - Reply
Dr. Helman, thank you for the reply! I take and appreciate your point on early identification of red flag signs and symptoms.
4. Jen April 23, 2019 at 4:02 pm - Reply
I am wondering if anyone knows of an alcohol to benzodiazepine equivalency chart. I do not feel that providers can conceptualize how much benzodiazepine is needed to match the intake of liquor that someone is ingesting. For example, when we treat diacetylmorphine withdrawal, we refer to MME’s which can help providers understand how much opioid someone is using. I see alcohol withdrawal often undertreated and I am looking for any suggestions to help providers feel more comfortable in the use of benzodiazepines at higher doses.
5. Ziauddin Hassan July 28, 2020 at 9:49 pm - Reply
Hi Dr.Helman, is there any evidence use of Haloperidol to lower the seizure threshold?
Anton Helman July 28, 2020 at 10:16 pm - Reply
Yes, in very high doses without any benzodiazepines on board. Best to combine Haloperidol with benzodiazepines for young agitated patients.
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Residual Volume
In subject area:Medicine and Dentistry
RV, or residual volume, is defined as the volume of air remaining in the lungs after forced expiration. It is an important measure in assessing lung function and can be increased in conditions such as asthma that involve small airway obstruction.
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Management of urinary incontinence in frail elderly women
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A residual volume of more than 100 ml is indicative of incomplete bladder emptying in younger person but in the elderly up to 200 ml can be considered normal. Very large residual volumes (>300 ml) are associated with increased risk of upper urinary tract dilatation and renal impairment.
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Residual Tumor Volume
Reports evaluating the benefits of increasing EOR (reported as percentage volume of total tumor resected) may not fully capture the benefits of cytoreduction compared with studies evaluating the benefits of minimizing residual tumor volume.1 Specifically, EOR is not consistent among different tumor sizes and resection of 90% of a large tumor may result in a greater volume of residual tumor than resection of 80% of a small tumor. As a result, a discussion of EOR does not adequately depict the residual disease burden that must be addressed with adjuvant therapies.
Residual volume, defined as residual contrast-enhancing disease as determined on postoperative magnetic resonance imaging (MRI), has been evaluated in several retrospective analyses and evidence supports the theory that less residual volume is associated with improved PFS and OS.11,56,101-103 In particular, 5 cm3 has been defined as the maximum residual tumor volume significantly associated with prolonged survival in HGG.56 Furthermore, residual contrast-enhancing volume and residual T2/FLAIR volume may be independent predictors of survival in addition to EOR, with residual contrast-enhancing tumor volume being regarded as a significant radiologic predictor of survival.102 Multivariate analyses of predictors of tumor progression in WHO grade III lesions similarly suggest that the volume of residual disease as measured on postoperative T2-weighted MRI is a significant predictor of tumor progression.11 Of particular importance, some evidence suggests that reducing residual tumor volume may delay malignant transformation of LGG, with associated survival benefits.101,103
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Journal2017, World NeurosurgeryRandy S. D’Amico, ... Jeffrey N. Bruce
Chapter
Benign Prostatic Hyperplasia and Related Entities
2007, Penn Clinical Manual of UrologyAlan J. Wein MD, PhD, David I. Lee MD
AResidual Urine Volume
If the residual urine volume is significant, its reduction is important in the evaluation of results of treatment of BPO. For many with a significant residual volume, it is impossible to differentiate deficient bladder contractility from outlet obstruction as the primary cause, without a pressure-flow study. Most agree that a large residual urine volume reflects a degree of bladder dysfunction, but it is difficult to correlate residual urine with either specific symptomatology or other urodynamic abnormalities. The most popular noninvasive method of measurement is ultrasonography. The error for ultrasound has been estimated at 10–25% for bladder volumes greater than 100 mL and somewhat worse for smaller volumes. Residual urine volumes in an individual patient at different times can vary widely. Reflux and large diverticula may complicate the accuracy of measurement. Paul Abrams and colleagues, after a thorough review of the subject, concluded that elevated residual urine has a relation to prostatic obstruction, although not a strong one, as supported by the following observations.
1.
: Elevated residual urine is common in the elderly of both genders.
2.
: The absence of residual urine does not rule out severe obstruction.
3.
: Elevated residual urine does not have a significant prognostic factor for a good operative outcome. Volume of more than 300 mL may correlate with unfavorable outcome.
What constitutes an abnormal residual urine? The International Consultation on BPH concluded that a range of 50 to 100 mL represents the lower threshold to define abnormal. There is discussion ongoing as to the concept that it may be more clinically meaningful to describe residual urine volume as a percent of bladder capacity rather than as an absolute number.
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Book2007, Penn Clinical Manual of UrologyAlan J. Wein MD, PhD, David I. Lee MD
Chapter
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2022, Cohen's Comprehensive Thoracic AnesthesiaJamie L. Sparling, Marcos F. Vidal Melo
Residual Volume
The amount of gas left in the lungs following a forced maximal exhalation is known as the residual volume (RV). It is determined by the balance between the outward expansion of the chest wall opposing the inward recoil of the lungs with the force generated by the muscles of expiration. Normal RV for a 70-kg adult is around 1.5 L, but because of dynamic airway collapse with forced expiratory effort, gas may become trapped in the alveoli. Thus the RV can become much larger in emphysema, where the lungs’ inward recoil is diminished and amount of gas trapping increases.5
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3.1.4Residual volume
3.1.4.1Jet and ultrasonic nebulisers
The residual volume, or dead volume, is defined as the volume of solution remaining in the nebuliser at the endpoint of nebulisation , which is typically in the range of 1 ml to 3 ml or 38% to 61% of a drug dose [139,141,142,147,187], depending on the nebuliser used. Small doses are especially affected. The residual volume depends on the design of the nebuliser, particularly the extent of the internal surface, the surface tension, the viscosity of the drug solution and the wetness of the nebuliser [181,186,188]. A higher fill volume may reduce the relative extent of the residual volume . The drug loss may also be decreased and the output improved by tapping the nebuliser . Residual volumes of ultrasonic nebulisers are often larger than for jet nebulisers .
3.1.4.2Other inhalers
Residual volumes in novel electronically operated vibrating mesh devices are generally lower than traditional nebulisers. The residual volume of the I-neb® AAD® system is approximately 0.1 ml . The residual volumes in percentage of the nominal dose depend on the fill volume (0.25–1.4 ml) of the devices. No clinical observations on residual volume are available. The eFlow® Rapid has a residual volume of ~ 1 ml, while ~ 28% of the drug dose was found in a clinical study with the eFlow®[171,190]. In general, powder retention in dry-powder inhalers is low, minimising drug wastage.
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Body Composition Measurement: A Review of Hydrodensitometry, Anthropometry, and Impedance Methods
1998, NutritionDavid BrodiePHDVicki MoscripMSC, HDCRRob HutcheonMSC
Practical considerations for this method include when and how to measure residual volume, what allowance to make for the volume of gas in the gut, and the effect of pretest nutrition and exercise. The residual gas volume accounts for about 2% of the submerged volume, and the accuracy of this value becomes critical to the calculation, where a very small change of density produces a much larger change in the predicted percentage of body fat. There is an appreciable day-to-day biological variation of residual volume in any given subject and this can change the estimated body fat content by as much as 1%.[10, 11] Considerable controversy occurs over whether or not to measure the residual volume while in the tank or separately. Submerging the body has been shown by some authors to increase residual volume[12, 13] and by others to decrease the residual volume.[14–19] Immersion was also found by several authors to make no difference in residual volume.[20–25] The actual method depends largely on the facilities available and includes nitrogen washout and oxygen dilution. If residual volume cannot be measured by such direct methods Mayhew et al. (personal communication) recommend estimates from 24% of vital capacity or from the anthropometric method of Polgar and Promadht. Garrow et al. suggested that residual volume could be measured by combining a body water displacement method with an air displacement technique for the head using a plethysmograph. This method eliminates the need for total submersion of the subject. Although the equipment is more complex than that required for hydrodensitometry, it does not require the additional equipment for measuring residual volume. The procedure requires minimal subject cooperation due to the fact that the patient stands in water to the neck level and the volume of air in the lungs and gastrointestinal (GI) tract is measured by observing the pressure changes produced by a pump of known stroke volume. Standard deviations of ± 0.3 kg were reported when replicate measures were made by this method.
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Journal1998, NutritionDavid BrodiePHDVicki MoscripMSC, HDCRRob HutcheonMSC
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Effects of not monitoring gastric residual volume in intensive care patients: A meta-analysis
2019, International Journal of Nursing StudiesZhuo Wang, ... Zaixiang Tang
What is already known about the topic?
•
: Monitoring gastric residual volume has been a routine practice in the assessment of enteral nutritional tolerance and gastric emptying in the intensive care units.
•
: Whether gastric residual volume serves as a surrogate parameter indicating gastrointestinal tract dysfunctions in the intensive care units remains controversial.
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Journal2019, International Journal of Nursing StudiesZhuo Wang, ... Zaixiang Tang
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Nutrition and Metabolism of the Surgical Patient, Part II
2011, Surgical Clinics of North AmericaJohn P. Grant MD
Gastric Residuals
Residual volumes have been used by physicians, dietitians, and nurses for many years as objective measures to judge tolerance or intolerance to enteral tube feedings. Recently, studies have raised considerable doubt as to the value of declaring any arbitrary residual volume as unacceptable and have pointed out how rigid rules related to residual volumes have at times rendered enteral nutritional support of limited value. The following are suggestions to improve clinical practice:
•
: Whenever residual volumes are measured, they should be interpreted along with the appearance of the patient’s abdomen and any complaints of fullness or discomfort.
•
: Abdominal distention can be caused by gas-filled loops of bowel from aerophagia and may not be caused by increased gastric residual volume; this will not improve with interruption of the feedings.
•
: The amount of residual volume is often greatest during the first several hours after feeding is begun. With continued feeding, the residual volume tends to plateau and then decrease.47,48
•
: The amount of gastric residual volume is dependent on the infusion rate. Computer models have shown that, because of the complex relationship between infusion volume, gastric emptying, and gastric capacity, evaluation of the volume of gastric contents is an unreliable method for monitoring feeding tolerance.49,50
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Nutritional assessment and therapeutic interventions for the preterm infant
2002, Clinics in PerinatologyDiane M Anderson PhD, RD
The evaluation of gastric residuals is an unending dilemma. Before feedings are initiated, residuals are commonly seen and are probably related to immature intestinal motor motility [10,11]. It is not uncommon for small residuals to be present no matter what feeding volume is provided . In the literature, excessive gastric residual volumes have been defined as a percentage of bolus feed volume (20–50%), 1–1.5 times an hourly rate for continuous infusion or a fixed volume of 2–3 ml/gastric residuals [13,14]. Mihatsch et al did not find a relationship between the incidence of a fixed gastric residual volume of 2–3 ml nor the incidence of green residuals to the volume of enteral feedings achieved on day of life 14 for infants with a birth weight <1000 grams. No randomized control trial has defined the volume or color of gastric residuals that should be considered worrisome [8,9,13]. The composition of residuals are noted as mucus, undigested milk, bile, or bloodstain. A bile or bloodstain residual, like abdominal distension, denotes an evaluation to rule out bowel obstruction . Often, bile or blood-stained aspirates may be related to feeding tube dislodgement into the intestine or feeding tube irritation .
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Journal2002, Clinics in PerinatologyDiane M Anderson PhD, RD
Review article
Ethics/Intensive care
2012, Anaesthesia & Intensive Care MedicineGerard J. Fennessy, Stephen J. Warrillow
Gastric residual volumes
The gastric residual volume (GRV) is defined as the volume of NG feed that remains in the stomach. An increase in GRV indicates reduction in motility, which can point to more serious problems including GI ischaemia, obstruction, pseudo-obstruction or GIF. The volume of GRV that signifies a clinically significant problem has not been adequately defined, although more than 150–250 ml/day indicates delayed gastric emptying and is associated with a greater risk of vomiting or regurgitation. However, in the absence of other clinical markers, volumes of up to 500 ml/day may be acceptable. It is recommended that EN is stopped and restarted at a slower rate if gastric aspirates are greater than 50% of the volume administered in the previous 4–6 hours.
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Journal2012, Anaesthesia & Intensive Care MedicineGerard J. Fennessy, Stephen J. Warrillow
Related terms:
Vital Capacity
Functional Residual Capacity
Chronic Obstructive Pulmonary Disease
Bladder
Emphysema
Enteric Feeding
Forced Expiratory Volume
Forced Vital Capacity
Total Lung Capacity
Lung Volume
View all Topics |
11993 | http://es.michiganmech.com/blog/how-to-calculate-module-of-spur-gear/ | Cómo calcular el módulo de engranaje recto
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Cómo calcular el módulo de engranaje recto
Cómo calcular el módulo de engranaje recto
Fórmula:
El módulo (m) de un engranaje recto se calcula dividiendo el diámetro de paso (d) por el número de dientes (z) del engranaje. La fórmula es:
METRO = d/z
Unidades:●Módulo (m):Milímetros (mm) es la unidad estándar para módulo.
●Diámetro de paso (d):Milímetros (mm)
¿Qué es el círculo de tono?
El círculo primitivo de unengranaje rectoes un círculo imaginario que define el contacto de rodadura teórico entre dos engranajes engranados. Es crucial para determinar la velocidad del engranaje y juega un papel clave en el diseño y la funcionalidad del engranaje.
Aquí hay un desglose del círculo primitivo:
Concepto:
Imagine un círculo perfecto dibujado en un engranaje recto donde la parte superior de los dientes está enrollada hacia atrás para formar un círculo suave. Este círculo imaginario es el círculo primitivo.
El centro del círculo primitivo coincide con el centro del propio engranaje.
Pasos para calcular el módulo:
1 、Mida el diámetro de paso (d):El diámetro de paso es el diámetro imaginario del engranaje donde los dientes actúan como si estuvieran enrollados formando un círculo perfecto. Puede encontrar el diámetro de paso midiendo directamente un engranaje que tenga o utilizando las especificaciones del engranaje si es un engranaje nuevo.
2 、Cuente el número de dientes (z):Este es el número total de dientes del engranaje recto.
3 、Calcular módulo (m):Divida el diámetro de paso (d) por el número de dientes (z) usando la fórmula anterior.
Ejemplo:
Digamos que tiene un engranaje recto con un diámetro de paso de 30 mm y 15 dientes.
M = d / z = 30 mm / 15 dientes = 2 M
Por tanto, el módulo del engranaje recto es 2M.
Hora de publicación: 17 de junio de 2024
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11994 | https://www.tiger-algebra.com/drill/25m2_5m-2=0/ | Copyright Ⓒ 2013-2025
tiger-algebra.com
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Solution - Quadratic equations
Other Ways to Solve
Step by Step Solution
Reformatting the input :
Changes made to your input should not affect the solution:
(1): "m2" was replaced by "m^2".
Step by step solution :
Step 1 :
Equation at the end of step 1 :
Step 2 :
Trying to factor by splitting the middle term
2.1 Factoring 25m2+5m-2
The first term is, 25m2 its coefficient is 25 .
The middle term is, +5m its coefficient is 5 .
The last term, "the constant", is -2
Step-1 : Multiply the coefficient of the first term by the constant 25 • -2 = -50
Step-2 : Find two factors of -50 whose sum equals the coefficient of the middle term, which is 5 .
| | | | | | | |
--- --- ---
| | -50 | + | 1 | = | -49 | |
| | -25 | + | 2 | = | -23 | |
| | -10 | + | 5 | = | -5 | |
| | -5 | + | 10 | = | 5 | That's it |
Step-3 : Rewrite the polynomial splitting the middle term using the two factors found in step 2 above, -5 and 10
25m2 - 5m + 10m - 2
Step-4 : Add up the first 2 terms, pulling out like factors :
5m • (5m-1)
Add up the last 2 terms, pulling out common factors :
2 • (5m-1)
Step-5 : Add up the four terms of step 4 :
(5m+2) • (5m-1)
Which is the desired factorization
Equation at the end of step 2 :
Step 3 :
Theory - Roots of a product :
3.1 A product of several terms equals zero.
When a product of two or more terms equals zero, then at least one of the terms must be zero.
We shall now solve each term = 0 separately
In other words, we are going to solve as many equations as there are terms in the product
Any solution of term = 0 solves product = 0 as well.
Solving a Single Variable Equation :
3.2 Solve : 5m-1 = 0
Add 1 to both sides of the equation :
5m = 1
Divide both sides of the equation by 5:
m = 1/5 = 0.200
Solving a Single Variable Equation :
3.3 Solve : 5m+2 = 0
Subtract 2 from both sides of the equation :
5m = -2
Divide both sides of the equation by 5:
m = -2/5 = -0.400
Supplement : Solving Quadratic Equation Directly
Earlier we factored this polynomial by splitting the middle term. let us now solve the equation by Completing The Square and by using the Quadratic Formula
Parabola, Finding the Vertex :
4.1 Find the Vertex of y = 25m2+5m-2
Parabolas have a highest or a lowest point called the Vertex . Our parabola opens up and accordingly has a lowest point (AKA absolute minimum) . We know this even before plotting "y" because the coefficient of the first term, 25 , is positive (greater than zero).
Each parabola has a vertical line of symmetry that passes through its vertex. Because of this symmetry, the line of symmetry would, for example, pass through the midpoint of the two x -intercepts (roots or solutions) of the parabola. That is, if the parabola has indeed two real solutions.
Parabolas can model many real life situations, such as the height above ground, of an object thrown upward, after some period of time. The vertex of the parabola can provide us with information, such as the maximum height that object, thrown upwards, can reach. For this reason we want to be able to find the coordinates of the vertex.
For any parabola,Am2+Bm+C,the m -coordinate of the vertex is given by -B/(2A) . In our case the m coordinate is -0.1000
Plugging into the parabola formula -0.1000 for m we can calculate the y -coordinate :
y = 25.0 -0.10 -0.10 + 5.0 -0.10 - 2.0
or y = -2.250
Parabola, Graphing Vertex and X-Intercepts :
Root plot for : y = 25m2+5m-2
Axis of Symmetry (dashed) {m}={-0.10}
Vertex at {m,y} = {-0.10,-2.25}
m -Intercepts (Roots) :
Root 1 at {m,y} = {-0.40, 0.00}
Root 2 at {m,y} = { 0.20, 0.00}
Solve Quadratic Equation by Completing The Square
4.2 Solving 25m2+5m-2 = 0 by Completing The Square .
Divide both sides of the equation by 25 to have 1 as the coefficient of the first term :
m2+(1/5)m-(2/25) = 0
Add 2/25 to both side of the equation :
m2+(1/5)m = 2/25
Now the clever bit: Take the coefficient of m , which is 1/5 , divide by two, giving 1/10 , and finally square it giving 1/100
Add 1/100 to both sides of the equation :
On the right hand side we have :
2/25 + 1/100 The common denominator of the two fractions is 100 Adding (8/100)+(1/100) gives 9/100
So adding to both sides we finally get :
m2+(1/5)m+(1/100) = 9/100
Adding 1/100 has completed the left hand side into a perfect square :
m2+(1/5)m+(1/100) =
(m+(1/10)) • (m+(1/10)) =
(m+(1/10))2
Things which are equal to the same thing are also equal to one another. Since
m2+(1/5)m+(1/100) = 9/100 and
m2+(1/5)m+(1/100) = (m+(1/10))2
then, according to the law of transitivity,
(m+(1/10))2 = 9/100
We'll refer to this Equation as Eq. #4.2.1
The Square Root Principle says that When two things are equal, their square roots are equal.
Note that the square root of
(m+(1/10))2 is
(m+(1/10))2/2 =
(m+(1/10))1 =
m+(1/10)
Now, applying the Square Root Principle to Eq. #4.2.1 we get:
m+(1/10) = √ 9/100
Subtract 1/10 from both sides to obtain:
m = -1/10 + √ 9/100
Since a square root has two values, one positive and the other negative
m2 + (1/5)m - (2/25) = 0
has two solutions:
m = -1/10 + √ 9/100
or
m = -1/10 - √ 9/100
Note that √ 9/100 can be written as
√ 9 / √ 100 which is 3 / 10
Solve Quadratic Equation using the Quadratic Formula
4.3 Solving 25m2+5m-2 = 0 by the Quadratic Formula .
According to the Quadratic Formula, m , the solution for Am2+Bm+C = 0 , where A, B and C are numbers, often called coefficients, is given by :
- B ± √ B2-4AC
m = ————————
2A
In our case, A = 25
B = 5
C = -2
Accordingly, B2 - 4AC =
25 - (-200) =
225
Applying the quadratic formula :
-5 ± √ 225
m = ——————
50
Can √ 225 be simplified ?
Yes! The prime factorization of 225 is
3•3•5•5
To be able to remove something from under the radical, there have to be 2 instances of it (because we are taking a square i.e. second root).
√ 225 = √ 3•3•5•5 =3•5•√ 1 =
± 15 • √ 1 =
± 15
So now we are looking at:
m = ( -5 ± 15) / 50
Two real solutions:
m =(-5+√225)/50=(-1+3)/10= 0.200
or:
m =(-5-√225)/50=(-1-3)/10= -0.400
Two solutions were found :
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In the decompositions I encountered so far, we all had orthogonal set of bases. For example in Singular Value Decomposition, we had orthogonal singular right and left vectors, in [discrete] cosine transform (or [discrete] fourier transform) we had again orthogonal bases.
To describe any vector $x \in \mathbb{C}^N$, we need to have $N$ independent set of basis vectors but independent doesn't necessarily mean orthogonal. My intentions behind selecting orthogonal vectors are as follows:
The solution is not unique for $x$ if the basis are not orthogonal.
It is easy to find the solution numerically by projecting $x$ onto each vector and this solution doesn't depend on the order of the bases. Otherwise, it depends on the order.
If we are talking about some set of vectors, they might be correlated in the original space, but uncorrelated in the transformd space which might be important when analyzing the data, in dimensionality reduction or compression.
I'm trying to understand the big picture. Do you think that I am right with these? Do you have any suggestions, what is the main reason for selecting orthogonal bases?
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edited Oct 13, 2010 at 18:00
Yemon Choi
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asked Oct 13, 2010 at 17:52
İsmail Arıİsmail Arı
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$\begingroup$ Frequently you don't (or can't) use orthonormal basis in decompositions. It begs the question of what a decomposition is for I suppose. If you're interested in the dynamical properties of the matrices -- like understanding $A^k$ for all $k$, and if say $A$ is an integer matrix you'll probably want your conjugating matrices to be integer matrices as well, so it's very unlikely they'll be orthogonal matrices. $\endgroup$
Ryan Budney
– Ryan Budney
2010-10-13 18:03:42 +00:00
Commented Oct 13, 2010 at 18:03
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$\begingroup$ One answer to this is that in many situations, orthogonal elements represent elements of a system that are in some sense non-interacting, so that in solving a problem they can be handled independently. This is the case in particular when the elements are the eigendirections of a self-adjoint operator. $\endgroup$
Dick Palais
– Dick Palais
2010-10-13 18:08:31 +00:00
Commented Oct 13, 2010 at 18:08
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Your first point, non-uniqueness, is definitely false. One of the basic facts in linear algebra is precisely that for any fixed set of basis vectors (we don't even have to work on a vector space endowed with an inner product, so orthogonality doesn't come in at all), a given vector has a unique decomposition.
For if that were not true, let the basis vectors be $e_1, \ldots, e_n$, then you have two sets of numbers $a_1, \ldots a_n$ and $b_1, \ldots, b_n$ such that
$$ a_1 e_1 + \cdots + a_n e_n = x = b_1 e_1 + \cdots + b_n e_n \implies (a_1 - b_1) e_1 + \cdots + (a_n - b_n) e_n = 0$$
if the sets $a_$ and $b_$ are not identical, this implies that $e_1\ldots e_n$ are linearly dependent, contradicting the assumption that they form a basis.
The second point, however, is a biggie. Without an inner product you cannot define an orthgonal projection. Now, generally, this is not too much of a problem. Given the basis vectors $e_1,\ldots, e_n$, finding the coordinates $a_1,\ldots, a_n$ of a given vector $x$ in this basis is just solving a linear system of equations, which actually is not too hard numerically, in finite dimensional systems. In infinite dimensional systems this inversion of the transformation matrix business gets slightly tricky.
The key is to note that without using the orthogonal projection, you cannot answer the question "what is the length of $x$ in the direction of $e_1$?" without knowing the entire set of basis vectors. (Remember that without using the orthogonal projection, you need to solve a linear system of equations to extract the coordinates; if you only specify one of the basis vectors, you do not have a complete system of equations, and the solution is underdetermined. I suspect this is what you had in the back of your mind for the first point.) This is actually a very fundamental fact in geometry, regarding coordinate systems. (I've once heard this described as the "second fundamental mistake in multivariable calculus" made by many students.)
Using the orthogonal projection/the inner product, you can answer the question, as long as you allow only orthgonal completions of the basis starting from the known vector. This fact is immensely useful when dealing with infinite dimensional systems.
I also don't quite like your formulation of the third point. A vector is a vector is a vector. It is independent of the coordinate representation. So I'd expect that if two vectors are correlated (assuming correlation is an intrinsic property), they better stay correlated without regard to choice of bases. What is more reasonable to say is that two vectors maybe uncorrelated in reality, but not obviously so when presented in one particular coordinate system, whereas the lack of correlation is immediately obvious when the vectors are written in another basis. But this observation has rather little to do with orthogonality. It only has some relation to orthogonality if you define "correlation" by some inner product (say, in some presentations of Quantum Mechanics). But then you are just saying that orthogonality of two vectors are not necessarily obvious, except when they are.
My personal philosophy is more one of practicality: the various properties of orthogonal bases won't make solving the problem harder. So unless you are in a situation where those properties don't make solving the problem easier, and some other basis does (like what Ryan Budney described), there's no harm in prefering an orthogonal basis. Furthermore, as Dick Palais observed above, one case where an orthogonal bases really falls out naturally at you is the case of the spectral theorem. The spectral theorem is, in some sense, the correct version of your point 3, that in certain situations, there is a set of basis vectors that are mathematically special. And this set happens to always be orthogonal.
Edit A little more about correlation. This is what I like to tell students when studying linear algebra. A vector is a vector. It is an object, not a bunch of numbers. When I hand you a rectangular box and ask you how tall the box is, the answer depends on which side is "up". This is similar to how you should think of the coordinate values of a vector inside a basis: it is obtained by a bunch of measurements. (Picking which side is "up" and measuring the height in that direction, however, in a non-orthogonal system, will require knowing all the basis vectors. See my earlier point.)
The point is that to quantitatively study science, and to perform numerical analysis, you can only work with numbers, not physical objects. So you have to work with measurements. And in your case, the correlation you are speaking of is correlation between the measurements of (I suppse) different "properties" of some object. And since what and how you measure depends on which basis you choose, the correlation between the data will also depend on which basis you choose. If you pick properties of an object that are correlated, then your data obtained from the measurements will also be correlated. The PCA you speak of is a way to disentangle that. It may be difficult to determine whether two properties of an object is correlated. Maybe the presence of a correlation is what you want to detect. The PCA tells you that, if there were in fact two independent properties of an object, but you just chose a bad set of measurements so that the properties you measure do not "line up" with the independent properties (that the properties you measure have a little bit of each), you can figure it out with a suitable transformation of the data at the end of the day. So you don't need to worry about choosing the best set of properties to measure.
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edited Oct 14, 2010 at 14:10
answered Oct 13, 2010 at 19:00
Willie WongWillie Wong
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$\begingroup$ Thank you for the simple proof for my first point. You are definitely right. I didn't think on it clearly and accepted it as correct when a a friend of mine told that it should be that way. For the third point, I will check what spectral theorem gives us. Actually, the correlation/dependence was in my mind because of principal component analysis which is close to singular value decomposition. For example, it this image (bit.ly/cyt22i), when we change the bases, the data becomes independent (but for this gaussian distributed samples in fact). Is correlation independent from the bases? $\endgroup$
İsmail Arı
– İsmail Arı
2010-10-14 09:41:42 +00:00
Commented Oct 14, 2010 at 9:41
$\begingroup$ I see, I didn't quite understand what you mean by correlation. Let me add one more paragraph in the bottom. $\endgroup$
Willie Wong
– Willie Wong
2010-10-14 13:46:59 +00:00
Commented Oct 14, 2010 at 13:46
$\begingroup$ I think I understood the mathematical way of seeing vectors. I always thought of them as being a bunch of numbers actually because of my engineering-oriented education. Thank you for the additional explanation. $\endgroup$
İsmail Arı
– İsmail Arı
2010-10-14 19:45:26 +00:00
Commented Oct 14, 2010 at 19:45
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I assume you are seeing things from a numerical point of view, since you only mentioned computation-oriented decompositions.
So here's a computational motivation: orthogonal matrices have condition number 1, thus multiplying and dividing by them is a numerically tame operation that does not increase the (norm-wise) errors. For instance, even when you work with symplectic matrices, you usually look for symplectic and orthogonal matrices.
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answered Oct 13, 2010 at 23:02
Federico PoloniFederico Poloni
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Maybe some examples of non-orthogonal decompositions are helpful: In the context of wavelet one frequently uses bi-orthonormal systems, i.e. two sets of vectors $(u_i)$ and $(v_i)$ such that $\langle u_i,v_j\rangle = \delta_{i,j}$. In this setting the calculations of coefficients can be done stably and efficiently. In real life, such bi-orthogonal wavelet bases are at the heart of the JPEG 2000 compression standard.
Another important notion in this context is the notion of frames.
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answered Oct 15, 2010 at 5:10
DirkDirk
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Well, when multiplying a vector ${\bf x}$ (say of norm 1) with a matrix ${\bf A}$ you are expecting to get a vector that lies on the surface of the ellipse $\left\|{\bf A}{\bf x}\right\|^2_2$. The directions of the axes of this ellipse are the eigenvectors and the eigenvalues dictate the equatorial radii. So every time you multiply a vector with that matrix, this decomposition into axes and radii tell you how your matrix distorts an "input" vector. That might be one pictorial reason :)
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answered Oct 13, 2010 at 18:13
AnadimAnadim
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11996 | https://byjus.com/neet/anatomy-of-tympanic-membrane/ | Table of Contents
What is a Tympanic Membrane?
Structure of the Tympanic Membrane
Difference between Pars Tensa and Pars Flaccida
Function of Tympanic Membrane
Clinical Significance
What is a Tympanic Membrane?
The tympanic membrane, also known as the eardrum, is a thin membrane between the middle and external ear. It is a cone-shaped structure that is found in humans as well as several tetrapods. It transmits sound vibrations from outer air to the ossicle in the middle ear and then to the cochlea.
Structure of the Tympanic Membrane
The tympanic membrane is made up of three layers:
The outer surface is lined with stratified squamous keratinised epithelium, which continues with the epidermal layer of the external auditory canal.
The inner surface comprises a simple cuboidal epithelium, which continues with the epithelial surface of the tympanic cavity.
The middle layer, also known as lamina propria, is a fibrous layer of fibroelastic connective tissue. It houses the nerves and blood vessels of the tympanic membrane.
The fibrous layer is thickened at the interior end of the tympanic membrane and forms a rigid ligamentous ring.
The handle of the malleus (one of the three ear ossicles) remains attached to the middle layer of the tympanic membrane. It pulls the inferior and anterior portions of the membrane, giving it a conical shape.
The deepest depression of the structure is called the umbo.
The eardrum is divided into two regions: the pars tensa and the pars flaccida.
Difference between Pars Tensa and Pars Flaccida
Pars TensaPars Flaccida
Definition
Pars tensa is the larger portion of the tympanic membrane that remains stretched to a maximum.Pars flaccida is the smaller of the tympanic membrane that remains less taut.
Composition
It is made up of organised circular and radial fibres.It is made up of loosely arranged elastic collagen.
Vascularisation
It is well vascularised.It lacks blood vessels in significant portions.
Presence of Mast Cells
None or very less Numerous mast cells are found in the pars flaccida.
Thickness
It is thinner and rigid.It is thicker and more elastic.
Function of Tympanic Membrane
The tympanic membrane plays a crucial role in sound amplification and transmission. As soon as the tympanic membrane senses a sound, it transmits the signal to the ossicles in the middle ear, which ultimately gets passed on to the cochlea in the inner ear.
Clinical Significance
The tympanic membrane is prone to unintentional rupture pertaining to incidents such as swimming, martial arts, diving, blast injuries, and air travel. In these conditions, the upper respiratory tract of the person is congested, which prevents equalisation of pressure in the middle ear.
Symptoms of this eardrum rupture include vertigo, tinnitus and hearing loss.
Visit BYJU’S for more information related to NEET.
Also Read:
Difference between Right and Left Tympanic Membrane
Well-labelled Diagram of Ear with Explanation
Eustachian Tube – Structure and Functions
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11997 | https://www.media4math.com/LessonPlans/SlopeRateOfChange | Lesson Plan: Slope As Rate of Change | Media4Math
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Lesson Plans: Slope
Lesson Plan Library
Lesson Plan 1: What Is Slope?
Lesson Plan 2: Slope and Similar Triangles
Lesson Plan 3: Visualizing Slope
Lesson Plan 4: Types of Slope
Lesson Plan 5: The Slope Formula
Lesson 6: Slope As Rate of Change
Lesson Plan: Slope As Rate of Change
Lesson Summary
The lesson "Slope as Rate of Change" aims to deepen students' understanding of slope by interpreting it as a rate of change between two quantities. Through real-world examples and interactive activities, students will learn to:
Calculate and interpret rates of change.
Distinguish between rates and ratios.
Apply these concepts to solve practical problems.
Lesson Objectives
By the end of this lesson, students will be able to:
Understand the concept of slope as a rate of change.
Interpret slope in real-world situations involving rates of change.
Calculate and interpret rates of change from given data or graphs.
Solve problems involving rates of change.
Distinguish between rates and ratios.
Common Core Standards
8.EE.B.5: Graph proportional relationships, interpreting the unit rate as the slope of the graph.
8.F.B.4: Construct a function to model a linear relationship between two quantities.
HSF-IF.B.6: Calculate and interpret the average rate of change of a function over a specified interval.
Prerequisite Skills
Understanding of slope as the ratio of the vertical change to the horizontal change.
Familiarity with the slope formula: Slope = y 2−y 1 x 2−x 1 y 2−y 1 x 2−x 1
Ability to plot points on a coordinate plane and graph linear equations.
Basic algebraic skills for solving equations.
Understanding of ratios and proportions.
Key Vocabulary
Slope: The measure of the steepness or incline of a line, calculated as the ratio of the vertical change (rise) to the horizontal change (run) between two points on the line: rise run rise run
Rate of Change: A ratio that describes how one quantity changes in relation to another; in linear relationships, it is represented by the slope of the line.
Ratio: A comparison of two quantities by division, often expressed in the form a:b a:b or as a fraction a b a b.
Unit Rate: A rate in which the second quantity in the comparison is one unit; for example, 60 miles per hour indicates 60 miles per 1 hour.
Proportional Relationship: A relationship between two quantities in which the ratio remains constant; graphically, it is represented by a straight line passing through the origin.
Linear Function: A function that creates a straight line when graphed; it has a constant rate of change and can be expressed in the form y=m x+b y=m x+b, where m m is the slope and b b is the y-intercept.
Y-Intercept: The point where a line crosses the y-axis on a graph, indicating the value of y y when x x is zero.
Dependent Variable: The variable in a function that depends on the value of another variable; typically represented as y y.
Independent Variable: The variable in a function whose value determines the value of the dependent variable; typically represented as x x.
Square Root: The number that, when multiplied by itself, gives the original number. Example: 16−−√=4 16=4.
Desmos: An advanced graphing calculator implemented as a web application, used for creating interactive mathematical graphs and visualizations.
Multimedia Resources
A collection of definitions on the topic of slope:
Share this slide show with students to review definitions on the topic of slope:
Warm-Up Activities
Activity 1: Discussion of Rates and Ratios
Begin by asking students to define a ratio and provide examples. Explain that a ratio is a comparison of two quantities by division, often expressed as a fraction or a:b. Next, introduce the concept of a rate. Explain that a rate is a special type of ratio that compares two different types of quantities, such as distance and time, or cost and number of items.
Show these math clip art images to demonstrate the differences between ratios and rates:
Provide additional examples of rates, such as:
Speed (distance traveled per unit of time)
Fuel efficiency (distance traveled per unit of fuel consumed)
Population growth rate (change in population per unit of time)
Inflation rate (change in prices per unit of time)
Emphasize that rates involve different units for the numerator and denominator, while ratios often involve the same units. Display a graph showing the distance traveled by a car over time. Ask students to describe what the graph represents and what information they can gather from it. Guide them to understand that the slope of the graph represents the rate of change of distance with respect to time, which is the speed of the car.
Activity 2: Calculating Rates
Using a calculator, have students compute various rates to reinforce the concept of unit rate and proportional relationships. Provide real-world scenarios such as:
The cost per pound of meat when a 5-pound package costs $20.
The cost per gallon of gas when a 12-gallon fill-up costs $42.
The speed of a car traveling 180 miles in 3 hours.
Encourage students to express their answers in unit rate format, such as cost pound cost pound or miles hour miles hour.
Activity 3: Skateboarder on a Ramp
Display an image or video of a skateboarder on a ramp. Ask students to observe how the skateboarder’s speed increases as they move down the ramp.
Discuss what happens to the skateboarder’s speed as they descend.
Introduce the concept of acceleration as a rate of change in speed.
Ask students how they would calculate the rate of speed increase over time.
Speed Data Table
Use the following table to analyze the skateboarder’s changing speed over time:
| Time (seconds) | Speed (feet per second) |
--- |
| 0 | 0 |
| 1 | 2 |
| 2 | 5 |
| 3 | 9 |
| 4 | 14 |
| 5 | 20 |
Have students calculate the rate of change (slope) between different time intervals using the formula:
slope=change in speed change in time slope=change in speed change in time
Discuss how the increasing speed suggests acceleration and how it connects to slope as a rate of change.
Teach
Introduction to Slope as Rate of Change
Show students this video about rates:
Explain that slope can be interpreted as the rate of change between two quantities. The rate of change is the ratio of the change in one quantity to the change in another quantity over a specific interval.
Explain that slope can be interpreted as the rate of change between two quantities. The rate of change is the ratio of the change in one quantity to the change in another quantity over a specific interval.Real-World Examples
Provide real-world examples of rates of change, such as:
Speed (distance traveled per unit of time)
Fuel efficiency (distance traveled per unit of fuel consumed)
Population growth rate (change in population per unit of time)
Inflation rate (change in prices per unit of time)
Demonstrate how to calculate and interpret rates of change from given data or graphs.
Desmos Activity
Show students this Desmos activity about rates:
Explain what the graph represents:
The slope of this line is the rate (cost per pound) of fruit.
Changing the value of the slider m changes the rate.
The slider varies from 0 to 5.
The table shows the input values for x (the pounds of fruit) and the output values f(x), which represent the cost for that amount of fruit.
Have students explore this activity to answer the following questions:
What happens to the slope as the rate increases?
What happens to the cost of fruit as the slope increases?
Problem-Solving
Guide students through solving problems involving rates of change. Encourage them to identify the quantities involved, determine the appropriate units, and interpret the meaning of the calculated rate of change in the context of the problem.
Use this collection of math clip art images to show examples of such rates:
Example 1: Finding the Slope as a Rate of Change
Problem: A car travels between two cities. At time 2 hours, the car has traveled 60 miles. At time 5 hours, the car has traveled 150 miles. What is the car’s speed?
Solution:
Identify the given points: (2,60)(2,60) and (5,150)(5,150), where x represents time (hours) and y represents distance (miles).
Use the slope formula: slope=y 2−y 1 x 2−x 1 slope=y 2−y 1 x 2−x 1
Substituting values: 150−60 5−2=90 3=30 150−60 5−2=90 3=30
Interpretation: The slope represents the rate of change of distance with respect to time, meaning the car is traveling at 30 miles per hour.
Example 2: Slope as Rate of Change from a Data Table
Problem: A water tank is being filled at a constant rate. The table below shows the amount of water in the tank over time. What is the rate of change?
| Time (minutes) | Water in Tank (gallons) |
--- |
| 0 | 50 |
| 2 | 70 |
| 4 | 90 |
| 6 | 110 |
Solution:
Choose any two points from the table, such as (2,70)(2,70) and (6,110)(6,110).
Use the slope formula: slope=y 2−y 1 x 2−x 1 slope=y 2−y 1 x 2−x 1
Substituting values: 110−70 6−2=40 4=10 110−70 6−2=40 4=10
Interpretation: The slope represents the rate at which the tank fills, which is 10 gallons per minute.
Example 3: Identifying Slope as Rate of Change from an Equation
Problem: A taxi company charges a base fare of 3 p l u s 3 p l u s 2 per mile traveled. What is the rate of change?
Solution:
The cost equation follows the form y=m x+b y=m x+b, where y y is the total cost and x x is the number of miles.
Given equation: y=2 x+3 y=2 x+3.
Identify the slope m=2 m=2.
Interpretation: The slope represents the rate of change of cost per mile, meaning the taxi charges $2 per mile.
Example 4: Real-World Scenario – Slope as Speed on a Hiking Trail
Problem: A hiker climbs a trail that gains 800 feet of elevation over a horizontal distance of 2,000 feet. What is the rate of elevation gain?
Solution:
Use the slope formula: slope=rise run slope=rise run
Substituting values: 800 2000 800 2000
Simplify: 800 2000=4 10=2 5 800 2000=4 10=2 5
Interpretation: The slope represents the rate of elevation gain, meaning the hiker ascends at 2 feet for every 5 feet of horizontal distance.
Example 5: Real-World Scenario – Price Increase Over Time
Problem: The price of gasoline increased from 3.50 p e r g a l l o n t o 3.50 p e r g a l l o n t o 4.10 per gallon over 6 months. What is the rate of price change per month?
Solution:
Use the slope formula: slope=change in price change in time slope=change in price change in time
Calculate the price change: 4.10−3.50=0.60 4.10−3.50=0.60.
Calculate the time change: 6 months.
Compute the rate of change: 0.60 6=0.10 0.60 6=0.10
Interpretation: The slope represents the rate of price increase per month, meaning gasoline prices increase by $0.10 per month.
Review
Summarize the key points of the lesson:
Slope represents the rate of change between two quantities.
The rate of change is calculated by dividing the change in one quantity by the change in another quantity over a specific interval.
Rates of change can be interpreted in various real-world contexts, such as speed, fuel efficiency, population growth, and inflation.
Rates involve different units for the numerator and denominator, while ratios often involve the same units.
Encourage students to ask questions and clarify any remaining doubts.
Key Vocabulary
Slope: The rate of change between two points on a line, calculated as the ratio of the vertical change (rise) to the horizontal change (run): slope=rise run slope=rise run
Rate of Change: A ratio that describes how one quantity changes in relation to another. In a linear relationship, it is represented by the slope.
Linear Function: A function that creates a straight line when graphed and can be written in the form y=m x+b y=m x+b, where m m is the slope.
Proportional Relationship: A relationship where two quantities increase or decrease at a constant rate, represented by a straight line passing through the origin.
Unit Rate: A rate where the denominator is 1, such as 60 miles per hour or $2 per gallon.
Y-Intercept: The point where a line crosses the y-axis, representing the starting value when x=0 x=0.
Independent Variable: The variable that represents the input or cause, often labeled as x x.
Dependent Variable: The variable that represents the output or effect, often labeled as y y.
Example 1: Slope as Rate of Change in Business
Problem: A coffee shop tracks its daily earnings. On Monday, the shop earned 200,a n d o n F r i d a y,i t e a r n e d 200,a n d o n F r i d a y,i t e a r n e d 500. Assuming a constant rate of increase, what is the daily rate of earnings?
Solution:
Identify the given points: (1,200)(1,200) for Monday and (5,500)(5,500) for Friday.
Use the slope formula: slope=y 2−y 1 x 2−x 1 slope=y 2−y 1 x 2−x 1
Substituting values: 500−200 5−1=300 4=75 500−200 5−1=300 4=75.
Interpretation: The coffee shop earns an additional $75 per day.
Example 2: Slope from a Data Table – Bike Rental Cost
Problem: A bike rental company charges a base fee and an hourly rate. The table below shows the total cost for different rental times. What is the hourly rate?
| Time (hours) | Total Cost ($) |
--- |
| 1 | 12 |
| 2 | 18 |
| 3 | 24 |
| 4 | 30 |
Solution:
Choose two points, such as (1,12)(1,12) and (3,24)(3,24).
Use the slope formula: slope=y 2−y 1 x 2−x 1 slope=y 2−y 1 x 2−x 1
Substituting values: 24−12 3−1=12 2=6 24−12 3−1=12 2=6.
Interpretation: The hourly rental rate is $6 per hour.
Example 3: Slope as Speed in a Marathon
Problem: A runner covers 10 miles in 1.5 hours and 16 miles in 2.5 hours. What is the runner's speed in miles per hour?
Solution:
Identify the given points: (1.5,10)(1.5,10) and (2.5,16)(2.5,16).
Use the slope formula: slope=y 2−y 1 x 2−x 1 slope=y 2−y 1 x 2−x 1
Substituting values: 16−10 2.5−1.5=6 1=6 16−10 2.5−1.5=6 1=6.
Interpretation: The runner’s speed is 6 miles per hour.
Final Review Questions
Explain how slope represents a rate of change in different real-world contexts.
Find the slope of a line passing through the points (4, 10) and (8, 22).
Given the equation y=5 x+2 y=5 x+2, identify the rate of change.
A hiker climbs 1,200 feet over 4 miles. What is the slope?
Using the bike rental table, predict the cost for a 6-hour rental.
Quiz
Answer the following questions.
What does the slope of a graph represent in terms of rate of change?
If a car travels 120 miles in 3 hours, what is its average speed (rate of change of distance with respect to time)?
The population of a city increased from 50,000 to 60,000 in 5 years. What is the annual population growth rate?
A company's revenue increased from 2 m i l l i o n t o 2 m i l l i o n t o 3.5 million over a period of 2 years. What is the rate of change of revenue with respect to time?
The graph below shows the distance traveled by a car over time. What is the car's average speed between t = 2 hours and t = 4 hours?
(Provide a graph with appropriate data)
Interpret the rate of change in the following scenario: The cost of a product increases by $5 for every additional unit purchased.
A company's profit increased from 100,000 t o 100,000 t o 150,000 over a period of 2 years. If the rate of change of profit remained constant, what would be the company's profit after 4 years?
The graph below shows the temperature change over time. What is the rate of change of temperature between t = 2 hours and t = 5 hours?
(Provide a graph with appropriate data)
Explain the difference between a positive rate of change and a negative rate of change in the context of population growth.
A car travels 180 miles in 3 hours. If the car maintains the same rate of change, how far will it travel in 6 hours?
Answer Key
Answer: The slope of a graph represents the rate of change between the two quantities plotted on the x and y axes.
Answer: Average speed = Distance / Time = 120 miles / 3 hours = 40 miles per hour.
Answer: Annual population growth rate = (Change in population / Initial population) / Time = (10,000 / 50,000) / 5 years = 0.04 or 4% per year.
Answer: Rate of change of revenue = (Change in revenue / Time) = (3.5 m i l l i o n−3.5 m i l l i o n−2 million) / 2 years = $0.75 million per year.
Answer: Average speed = (Change in distance) / (Change in time) = (Distance at t = 4 hours - Distance at t = 2 hours) / (4 hours - 2 hours).
Answer: The rate of change represents the change in cost per additional unit purchased, which is $5 per unit.
Answer: Rate of change of profit = (150,000−150,000−100,000) / 2 years = 25,000 p e r y e a r.P r o f i t a f t e r 4 y e a r s=25,000 p e r y e a r.P r o f i t a f t e r 4 y e a r s=100,000 + (4 × 25,000)=25,000)=200,000.
Answer: Rate of change of temperature = (Change in temperature) / (Change in time) = (Temperature at t = 5 hours - Temperature at t = 2 hours) / (5 hours - 2 hours).
Answer: A positive rate of change indicates that the population is increasing over time, while a negative rate of change indicates that the population is decreasing over time.
Answer: Rate of change (speed) = Distance / Time = 180 miles / 3 hours = 60 miles per hour. Distance traveled in 6 hours = Rate of change × Time = 60 miles per hour × 6 hours = 360 miles.
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Fields and Waves in Communication Electronics
The text helps define the second electromagnetic course that electrical engineers take in their senior year. This rigorous book on engineering electromagnetic fields and waves topics is packed with useful derivations and applications. Very well known text and authors.
1100519260
Fields and Waves in Communication Electronics
The text helps define the second electromagnetic course that electrical engineers take in their senior year. This rigorous book on engineering electromagnetic fields and waves topics is packed with useful derivations and applications. Very well known text and authors.
307.75 In Stock
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Fields and Waves in Communication Electronics
864
by Simon Ramo, John R. Whinnery, Theodore Van Duzer Simon Ramo
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by Simon Ramo, John R. Whinnery, Theodore Van Duzer Simon Ramo
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Overview
The text helps define the second electromagnetic course that electrical engineers take in their senior year. This rigorous book on engineering electromagnetic fields and waves topics is packed with useful derivations and applications. Very well known text and authors.
Product Details
About the Author
Table of Contents
Product Details
| | |
--- |
| ISBN-13: | 9780471585510 |
| Publisher: | Wiley |
| Publication date: | 03/31/1994 |
| Edition description: | REV |
| Pages: | 864 |
| Product dimensions: | 6.48(w) x 9.37(h) x 1.83(d) |
About the Author
Simon "Si" Ramo was an American engineer, businessman, and author. He led development of microwave and missile technology and is sometimes known as the father of the intercontinental ballistic missile. He also developed General Electric's electron microscope. John Roy Whinnery was an American electrical engineer and educator who worked in the fields of microwave theory and laser experimentation.
Table of Contents
Stationary Electric Fields. Stationary Magnetic Fields. Maxwell's Equations. The Electromagnetics of Circuits. Transmission Lines. Plane-Wave Propagation and Reflection. Two- and Three-Dimensional Boundary Value Problems. Waveguides with Cylindrical Conducting Boundaries. Special Waveguide Types. Resonant Cavities. Microwave Networks. Radiation. Electromagnetic Properties of Materials. Optics. Appendices. Index.
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11999 | https://unbounded-uploads.s3.amazonaws.com/attachments/8511/math-g7-m4-topic-a-lesson-2-teacher.pdf | Lesson 2: Part of a Whole as a Percent Date: 11/19/14 28 © 2014 Common Core, Inc. Some rights reserved. commoncore.org This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. NYS COMMON CORE MATHEMATICS CURRICULUM 7•4 Lesson 2 Lesson 2: Part of a Whole as a Percent Student Outcomes Students understand that the whole is 100% and use the formula Part = Percent × Whole to problem-solve when given two terms out of three from the part, whole, and percent. Students solve word problems involving percent using expressions, equations, and numeric and visual models. Lesson Notes This lesson serves as an introduction to general percent problems by considering problems of which a part of a whole is represented as a percent of the whole. Students solve percent problems using visual models and proportional reasoning and then make connections to solving percent problems using numeric and algebraic methods. This lesson focuses on the relationship: Part = Percent × Whole. Classwork Opening (2 minutes) One of the challenges students face when solving word problems involving percents is deciding which of the given quantities represents the whole unit and which represents the part of that whole unit. Discuss with students how the value of a nickel coin ($0.05) compares to the value of a dollar ($1.00) using percents. As a percent, how does the value of a nickel coin compare to the value of a dollar? A dollar is 100 cents; therefore, the quantity 100 cents is 100% of a dollar. A nickel coin has a value of 5 cents, which is 5 of 100 cents, or 5 100 = 5% of a dollar. Part-of-a-whole percent problems involve the following: A comparison of generic numbers (e.g., 25% of 12 is 3), or A comparison of a quantity that is a part of another quantity (e.g., the number of boys in a classroom is part of the total number of students in the classroom). The number or quantity that another number or quantity is being compared to is called the whole. The number or quantity that is compared to the whole is called the part because it is part (or a piece) of the whole quantity. In our comparison of the value of a nickel coin to the value of a dollar, which quantity is considered the part and which is considered the whole? Explain your answer. The value of the nickel coin is the part because it is being compared to the value of the whole dollar. The dollar represents the whole because the value of the nickel coin is being compared to the value of the dollar. Lesson 2: Part of a Whole as a Percent Date: 11/19/14 29 © 2014 Common Core, Inc. Some rights reserved. commoncore.org This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. NYS COMMON CORE MATHEMATICS CURRICULUM 7•4 Lesson 2 Opening Exercise (4 minutes) Part (a) of the Opening Exercise asks students to practice identifying the whole in given percent scenarios. In part (b), students are presented with three different approaches to a given scenario but need to make sense of each approach to identify the part, the whole, and the percent. Opening Exercise a. What is the whole unit in each scenario? Scenario Whole Unit 𝟏𝟓 is what percent of 𝟗𝟎? The number 𝟗𝟎 What number is 𝟏𝟎% of 𝟓𝟔? The number 𝟓𝟔 𝟗𝟎% of a number is 𝟏𝟖𝟎. The unknown number A bag of candy contains 𝟑𝟎𝟎 pieces, and 𝟐𝟓% of the pieces in the bag are red. The 𝟑𝟎𝟎 pieces of candy Seventy percent (𝟕𝟎%) of the students earned a B on the test. All the students in the class The 𝟐𝟎 girls in the class represented 𝟓𝟓% of the students in the class. All the students in the class After students complete part (a) with a partner, ask the following question: How did you decide on the whole unit in each of the given scenarios? In each case, we looked for the number or quantity that another number or quantity was being compared to. b. Read each problem and complete the table to record what you know. Problem Part Percent Whole 𝟒𝟎% of the students on the field trip love the museum. If there are 𝟐𝟎 students on the field trip, how many love the museum? ? 𝟒𝟎% 𝟐𝟎 students 𝟒𝟎% of the students on the field trip love the museum. If 𝟐𝟎 students love the museum, how many are on the field trip? 𝟐𝟎 students 𝟒𝟎% ? 𝟐𝟎 students on the field trip love the museum. If there are 𝟒𝟎 students on the field trip, what percent love the museum? 𝟐𝟎 students ? 𝟒𝟎 students When students complete part (b), encourage them to share how they decided which number in the problem represents the whole and which represents the part. Example 1 (5 minutes): Visual Approaches to Finding a Part, Given a Percent of the Whole Present the following problem to students. Show how to solve the problem using visual models; then, generalize a numeric method through discussion. Have students record each method in their student materials. Example 1: Visual Approaches to Finding a Part, Given a Percent of the Whole In Ty’s math class, 𝟐𝟎% of students earned an A on a test. If there were 𝟑𝟎 students in the class, how many got an A? MP.1 Lesson 2: Part of a Whole as a Percent Date: 11/19/14 30 © 2014 Common Core, Inc. Some rights reserved. commoncore.org This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. NYS COMMON CORE MATHEMATICS CURRICULUM 7•4 Lesson 2 Is 30 the whole unit or part of the whole? It is the whole unit; the number of students that earned an A on the test is the part and is compared to the total number of students in the class. What percentage of Ty’s class does the quantity “30 students” represent? 100% of Ty’s class Solve the problem first using a tape diagram. 30 students make up 100% of the class. Let’s divide the 100% into 100 slices of 1% and also divide the quantity of 30 students into 100 slices. What number of students does each 1% correspond to? 30 100 = 0.3; 0.3 of a student represents 1% of Ty’s class. If this is 1% of Ty’s class, then how do we find 20% of Ty’s class? (1%) × 20 = 20%, so we can multiply (0.3) × 20 = 6; 6 students are 20% of Ty’s class; therefore, 6 students got an A on the test. Revisit the problem using a double number line. 30 students represents the whole class, so 30 aligns with 100%. There are 100 intervals of 1% on the percent number line. What number of students does each 1% correspond to? 30 100 = 0.3; 0.3 of a student represents 1% of Ty’s class. To help us keep track of quantities and their corresponding percents, we can use arrows to show the correspondences in our sequences of reasoning: 30 →100% 0.3 →1% If this is 1% of Ty’s class, how do we find 20% of Ty’s class? Multiply by 20; 0.3 ∙20 = 6; 6 students are 20% of Ty’s class, so 6 students got an A on the test. 0.3 →1% 6 →20% What similarities do you notice in each of these visual models? In both models, 30 corresponds with the 100%; so, we divided 30 by 100 to get the number of students that corresponds with 1% and then multiplied that by 20 to get the number of students that corresponds with 20%. Scaffolding: Some students may recognize that there are 5 intervals of 20% in the tape diagram and want to divide 30 students into 5 groups. That is okay. However, if students do not immediately recognize this, do not force it upon them. Further practice scaffolds this shortcut while also supporting primary understanding of how the percent problems work. ÷ 𝟏𝟎𝟎 ÷ 𝟏𝟎𝟎 × 𝟐𝟎 × 𝟐𝟎 Lesson 2: Part of a Whole as a Percent Date: 11/19/14 31 © 2014 Common Core, Inc. Some rights reserved. commoncore.org This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. NYS COMMON CORE MATHEMATICS CURRICULUM 7•4 Lesson 2 Exercise 1 (3 minutes) Students use visual methods to solve a problem similar to Example 1. After completing the exercise, initiate a discussion about the similarities of the problems, and generalize a numeric approach to the problems. This numeric approach will be used to generalize an algebraic equation that can be used in solving percent problems. Exercise 1 In Ty’s art class, 𝟏𝟐% of the Flag Day art projects received a perfect score. There were 𝟐𝟓 art projects turned in by Ty’s class. How many of the art projects earned a perfect score? (Identify the whole.) The whole is the number of art projects turned in by Ty’s class, 𝟐𝟓. 𝟐𝟓 𝟏𝟎𝟎= 𝟎. 𝟐𝟓; 𝟎. 𝟐𝟓∙𝟏𝟐= 𝟑; 𝟏𝟐% of 𝟐𝟓 is 𝟑, so 𝟑 art projects in Ty’s class received a perfect score. Discussion (2 minutes) What similarities do you recognize in Example 1 and Exercise 1? In each case, the whole corresponded with 100%, and dividing the whole by 100 resulted in 1% of the whole. Multiplying this number by the percent resulted in the part. Describe and show how the process seen in the visual models can be generalized into a numeric approach. Divide the whole by 100 to get 1%, and then multiply by the percent needed. Whole →100%. Example 1 Exercise 1 30 →100% 25 →100% 30 100 →1% 25 100 →1% 20 ( 30 100) →20% 12 ( 25 100) →12% 6 →20% 3 →12% Example 2 (3 minutes): A Numeric Approach to Finding a Part, Given a Percent of the Whole Present the following problem to students. Have them guide you through solving the problem using the arithmetic method from the previous discussion. When complete, generalize an arithmetic method through further discussion. Example 2: A Numeric Approach to Finding a Part, Given a Percent of the Whole In Ty’s English class, 𝟕𝟎% of the students completed an essay by the due date. There are 𝟑𝟎 students in Ty’s English class. How many completed the essay by the due date? ÷ 𝟏𝟎𝟎 ÷ 𝟏𝟎𝟎 ÷ 𝟏𝟎𝟎 ÷ 𝟏𝟎𝟎 × 𝟐𝟎 × 𝟐𝟎 × 𝟏𝟐 × 𝟏𝟐 MP.8 Lesson 2: Part of a Whole as a Percent Date: 11/19/14 32 © 2014 Common Core, Inc. Some rights reserved. commoncore.org This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. NYS COMMON CORE MATHEMATICS CURRICULUM 7•4 Lesson 2 First, identify the whole quantity in the problem. The number of students that completed the essay by the due date is being compared to the total number of students in Ty’s class, so the total number of students in the class (30) is the whole. 𝐖𝐡𝐨𝐥𝐞→𝟏𝟎𝟎% 𝟑𝟎→𝟏𝟎𝟎% 𝟑𝟎 𝟏𝟎𝟎→𝟏% 𝟕𝟎∙𝟑𝟎 𝟏𝟎𝟎→𝟐𝟏% 𝟐𝟏→𝟕𝟎% 𝟕𝟎% of 𝟑𝟎 is 𝟐𝟏, so 𝟐𝟏 of the students in Ty’s English class completed their essays on time. Discussion (2 minutes) This discussion is an extension of Example 2 and serves as a bridge to Example 3. Is the expression 70 100 ∙30 equivalent to 70 ∙ 30 100 from the steps above? Why or why not? The expressions are equivalent by the any order, any grouping property of multiplication. What does 70 100 represent? What does 30 represent? What does their product represent? 70 100 = 70%, 30 represents the whole, and their product (21) represents the part, or 70% of the students in Ty’s English class. Write a true multiplication sentence relating the part (21), the whole (30), and the percent ( 70 100) in this problem. 21 = 70 100 ∙(30) Translate your sentence into words. Is the sentence valid? Twenty-one is seventy percent of thirty. Yes, the sentence is valid because 21 students represents 70% of the 30 students in Ty’s English class. Generalize the terms in your multiplication sentence by writing what each term represents. Part = Percent × Whole Example 3 (4 minutes): An Algebraic Approach to Finding a Part, Given a Percent of the Whole In percent problems, the percent equation (Part = Percent × Whole) can be used to solve the problem when given two of its three terms. To solve a percent word problem, first identify the whole quantity in the problem, and then identify the part and percent. Use a letter (variable) to represent the term whose value is unknown. MP.7 Lesson 2: Part of a Whole as a Percent Date: 11/19/14 33 © 2014 Common Core, Inc. Some rights reserved. commoncore.org This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. NYS COMMON CORE MATHEMATICS CURRICULUM 7•4 Lesson 2 Example 3: An Algebraic Approach to Finding a Part, Given a Percent of the Whole A bag of candy contains 𝟑𝟎𝟎 pieces of which 𝟐𝟖% are red. How many pieces are red? Which quantity represents the whole? The total number of candies in the bag, 𝟑𝟎𝟎, is the whole because the number of red candies is being compared to it. Which of the terms in the percent equation is unknown? Define a letter (variable) to represent the unknown quantity. We do not know the part, which is the number of red candies in the bag. Let 𝒓 represent the number of red candies in the bag. Write an expression using the percent and the whole to represent the number of pieces of red candy. 𝟐𝟖 𝟏𝟎𝟎∙(𝟑𝟎𝟎), or 𝟎. 𝟐𝟖∙(𝟑𝟎𝟎), is the amount of red candy since the number of red candies is 𝟐𝟖% of the 𝟑𝟎𝟎 pieces of candy in the bag. Write and solve an equation to find the unknown quantity. 𝐏𝐚𝐫𝐭= 𝐏𝐞𝐫𝐜𝐞𝐧𝐭× 𝐖𝐡𝐨𝐥𝐞 𝒓= 𝟐𝟖 𝟏𝟎𝟎∙(𝟑𝟎𝟎) 𝒓= 𝟐𝟖⋅𝟑 𝒓= 𝟖𝟒 There are 𝟖𝟒 red pieces of candy in the bag. Exercise 2 (4 minutes) This exercise is a continuation of Example 3. Exercise 2 A bag of candy contains 𝟑𝟎𝟎 pieces of which 𝟐𝟖% are red. How many pieces are not red? a. Write an equation to represent the number of pieces that are not red, 𝒏. 𝐏𝐚𝐫𝐭= 𝐏𝐞𝐫𝐜𝐞𝐧𝐭× 𝐖𝐡𝐨𝐥𝐞 𝒏= (𝟏𝟎𝟎% −𝟐𝟖%)(𝟑𝟎𝟎) b. Use your equation to find the number of pieces of candy that are not red. If 𝟐𝟖% of the candies are red, then the difference of 𝟏𝟎𝟎% and 𝟐𝟖% must be candies that are not red. 𝒏= (𝟏𝟎𝟎% −𝟐𝟖%)(𝟑𝟎𝟎) 𝒏= (𝟕𝟐%)(𝟑𝟎𝟎) 𝒏= 𝟕𝟐 𝟏𝟎𝟎(𝟑𝟎𝟎) 𝒏= 𝟕𝟐∙𝟑 𝒏= 𝟐𝟏𝟔 There are 𝟐𝟏𝟔 pieces of candy in the bag that are not red. c. Jah-Lil told his math teacher that he could use the answer from part (b) and mental math to find the number of pieces of candy that are not red. Explain what Jah-Lil meant by that. He meant that once you know there are 𝟖𝟒 red pieces of candy in a bag that contains 𝟑𝟎𝟎 pieces of candy total, you just subtract 𝟖𝟒 from 𝟑𝟎𝟎 to know that 𝟐𝟏𝟔 pieces of candy are not red. Lesson 2: Part of a Whole as a Percent Date: 11/19/14 34 © 2014 Common Core, Inc. Some rights reserved. commoncore.org This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. NYS COMMON CORE MATHEMATICS CURRICULUM 7•4 Lesson 2 Students saw in Module 3 that it is possible to find a solution to a formula, or algebraic equation, by using the properties of operations and if-then moves to rewrite the expressions in an equation in a form in which a solution can be easily seen. Examples 4 and 5 use the algebraic formula Part = Percent × Whole to solve percent word problems where they are given two of the following three terms: part, percent, and whole. Example 4 (5 minutes): Comparing Part of a Whole to the Whole with the Percent Formula Students use the percent formula and algebraic reasoning to solve a percent problem in which they are given the part and the percent. Example 4: Comparing Part of a Whole to the Whole with the Percent Formula Zoey inflated 𝟐𝟒 balloons for decorations at the middle school dance. If Zoey inflated 𝟏𝟓% of the balloons that are inflated for the dance, how many balloons are there total? Solve the problem using the percent formula, and verify your answer using a visual model. What is the whole quantity? How do you know? The total number of balloons at the dance is the whole quantity because the number of balloons that Zoey inflated is compared to the total number of balloons for the dance. What do the 24 balloons represent? 24 balloons are part of the total number of balloons for the dance. Write the percent formula, and determine which term is unknown. 𝐏𝐚𝐫𝐭= 𝐏𝐞𝐫𝐜𝐞𝐧𝐭× 𝐖𝐡𝐨𝐥𝐞 The part is 𝟐𝟒 balloons, and the percent is 𝟏𝟓%, so let 𝒕 represent the unknown total number of balloons. 𝟐𝟒= 𝟏𝟓 𝟏𝟎𝟎𝒕 𝟏𝟎𝟎 𝟏𝟓(𝟐𝟒) = 𝟏𝟎𝟎 𝟏𝟓( 𝟏𝟓 𝟏𝟎𝟎) 𝒕 𝟐𝟒𝟎𝟎 𝟏𝟓 = 𝟏𝒕 𝟏𝟔𝟎= 𝒕 If 𝒂= 𝒃, then 𝒂𝒄= 𝒃𝒄. Multiplicative inverse Multiplicative identity property of 𝟏 and equivalent fractions The total number of balloons to be inflated for the dance was 𝟏𝟔𝟎 balloons. 𝟏𝟓% →𝟐𝟒 𝟏% →𝟐𝟒 𝟏𝟓 𝟏𝟎𝟎% →𝟐𝟒 𝟏𝟓∙𝟏𝟎𝟎 𝟏𝟎𝟎% →𝟐𝟒 𝟑∙𝟐𝟎= 𝟏𝟔𝟎 We want the quantity that corresponds with 𝟏𝟎𝟎%, so first we find 𝟏%. Student may also find 𝟓% as is shown in the tape diagram above. Is the solution from the equation consistent with the visual and numeric solution? Yes. Lesson 2: Part of a Whole as a Percent Date: 11/19/14 35 © 2014 Common Core, Inc. Some rights reserved. commoncore.org This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. NYS COMMON CORE MATHEMATICS CURRICULUM 7•4 Lesson 2 Example 5 (5 minutes): Finding the Percent Given a Part of the Whole and the Whole Students use the percent formula and algebraic reasoning to solve a percent problem in which they are given the part and the whole. Example 5: Finding the Percent Given a Part of the Whole and the Whole Haley is making admission tickets to the middle school dance. So far she has made 𝟏𝟏𝟐 tickets, and her plan is to make 𝟑𝟐𝟎 tickets. What percent of the admission tickets has Haley produced so far? Solve the problem using the percent formula, and verify your answer using a visual model. What is the whole quantity? How do you know? The total number of admission tickets, 320, is the whole quantity because the number of tickets that Haley has already made is compared to the total number of tickets that she needs to make. What does the quantity 112 tickets represent? 112 tickets is part of the total number of tickets for the dance. Write the percent formula, and determine which term is unknown. 𝐏𝐚𝐫𝐭= 𝐏𝐞𝐫𝐜𝐞𝐧𝐭× 𝐖𝐡𝐨𝐥𝐞 The part is 𝟏𝟏𝟐 tickets, and the whole is 𝟑𝟐𝟎 tickets, so let 𝒑 represent the unknown percent. 𝟏𝟏𝟐= 𝒑(𝟑𝟐𝟎) If 𝒂= 𝒃, then 𝒂𝒄= 𝒃𝒄. 𝟏𝟏𝟐∙ 𝟏 𝟑𝟐𝟎= 𝒑(𝟑𝟐𝟎) ∙ 𝟏 𝟑𝟐𝟎 Multiplicative inverse 𝟏𝟏𝟐 𝟑𝟐𝟎= 𝒑(𝟏) Multiplicative identity property of 𝟏 𝟕 𝟐𝟎= 𝒑 𝟎. 𝟑𝟓= 𝒑 𝟎. 𝟑𝟓= 𝟑𝟓 𝟏𝟎𝟎= 𝟑𝟓%, so Haley has made 𝟑𝟓% of the tickets for the dance. We need to know the percent that corresponds with 𝟏𝟏𝟐, so first we find the percent that corresponds with 𝟏 ticket. 𝟑𝟐𝟎→𝟏𝟎𝟎% 𝟏→(𝟏𝟎𝟎 𝟑𝟐𝟎)% 𝟏𝟏𝟐→𝟏𝟏𝟐∙(𝟏𝟎𝟎 𝟑𝟐𝟎) % 𝟏𝟏𝟐→𝟏𝟏𝟐∙( 𝟓 𝟏𝟔) % 𝟏𝟏𝟐→𝟕∙(𝟓)% = 𝟑𝟓% Is the solution from the equation consistent with the visual and numeric solution? Yes. Lesson 2: Part of a Whole as a Percent Date: 11/19/14 36 © 2014 Common Core, Inc. Some rights reserved. commoncore.org This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. NYS COMMON CORE MATHEMATICS CURRICULUM 7•4 Lesson 2 Closing (2 minutes) What formula can we use to relate the part, the whole, and the percent of the whole? Translate the formula into words. Part = Percent × Whole. The part is some percent of the whole. What are the advantages of using an algebraic representation to solve percent problems? If you can identify the whole, part, and percent, the algebraic approach is very fast and efficient. Explain how to use a visual model and an equation to determine how many calories are in a candy bar if 75% of its 200 calories is from sugar. Use a double number line or tape diagram. The whole (total calories) corresponds with 100%. 200 calories divided into 100 intervals shows that every 1% will be 2 calories. That means there are 150 calories from sugar in the candy bar. Exit Ticket (4 minutes) Note to the teacher: Students using the visual or numeric approaches for problems in the Exit Ticket do not necessarily need to find 1% first. Alternatively, if they recognize that they can instead find 4%, 5%, 10%, 20%, or other factors of 100%, then they can multiply by the appropriate factor to obtain 100%. Lesson Summary Visual models or numeric methods can be used to solve percent problems. An equation can be used to solve percent problems: 𝐏𝐚𝐫𝐭= 𝐏𝐞𝐫𝐜𝐞𝐧𝐭× 𝐖𝐡𝐨𝐥𝐞. Lesson 2: Part of a Whole as a Percent Date: 11/19/14 37 © 2014 Common Core, Inc. Some rights reserved. commoncore.org This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. NYS COMMON CORE MATHEMATICS CURRICULUM 7•4 Lesson 2 Name Date Lesson 2: Part of a Whole as a Percent Exit Ticket 1. On a recent survey, 60% of those surveyed indicated that they preferred walking to running. a. If 540 people preferred walking, how many people were surveyed? b. How many people preferred running? 2. Which is greater: 25% of 15 or 15% of 25? Explain your reasoning using algebraic representations or visual models. Lesson 2: Part of a Whole as a Percent Date: 11/19/14 38 © 2014 Common Core, Inc. Some rights reserved. commoncore.org This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. NYS COMMON CORE MATHEMATICS CURRICULUM 7•4 Lesson 2 Exit Ticket Sample Solutions 1. On a recent survey, 𝟔𝟎% of those surveyed indicated that they preferred walking to running. a. If 𝟓𝟒𝟎 people preferred walking, how many people were surveyed? Let 𝒏 represent the number of people surveyed. 𝟎. 𝟔𝟎𝒏 is the number of people who preferred walking. Since 𝟓𝟒𝟎 people preferred walking, 𝟎. 𝟔𝟎𝒏= 𝟓𝟒𝟎 𝒏= 𝟓𝟒𝟎 𝟎. 𝟔= 𝟓𝟒𝟎𝟎 𝟔 = 𝟗𝟎𝟎 Therefore, 𝟗𝟎𝟎 people were surveyed. b. How many people preferred running? Subtract 𝟓𝟒𝟎 from 𝟗𝟎𝟎. 𝟗𝟎𝟎−𝟓𝟒𝟎= 𝟑𝟔𝟎 Therefore, 𝟑𝟔𝟎 people preferred running. 2. Which is greater: 𝟐𝟓% of 𝟏𝟓 or 𝟏𝟓% of 𝟐𝟓? Explain your reasoning using algebraic representations or visual models. They are the same. 𝟎. 𝟐𝟓× 𝟏𝟓= 𝟐𝟓 𝟏𝟎𝟎× 𝟏𝟓= 𝟑. 𝟕𝟓 𝟎. 𝟏𝟓× 𝟐𝟓= 𝟏𝟓 𝟏𝟎𝟎× 𝟐𝟓= 𝟑. 𝟕𝟓 Also, you can see they are the same without actually computing the product because of any order, any grouping of multiplication. 𝟐𝟓 𝟏𝟎𝟎× 𝟏𝟓= 𝟐𝟓× 𝟏 𝟏𝟎𝟎× 𝟏𝟓= 𝟐𝟓× 𝟏𝟓 𝟏𝟎𝟎 Problem Set Sample Solutions Students should be encouraged to solve these problems using an algebraic approach. 1. Represent each situation using an equation. Check your answer with a visual model or numeric method. a. What number is 𝟒𝟎% of 𝟗𝟎? 𝒏= 𝟎. 𝟒𝟎(𝟗𝟎) 𝒏= 𝟑𝟔 b. What number is 𝟒𝟓% of 𝟗𝟎? 𝒏= 𝟎. 𝟒𝟓(𝟗𝟎) 𝒏= 𝟒𝟎. 𝟓 Lesson 2: Part of a Whole as a Percent Date: 11/19/14 39 © 2014 Common Core, Inc. Some rights reserved. commoncore.org This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. NYS COMMON CORE MATHEMATICS CURRICULUM 7•4 Lesson 2 c. 𝟐𝟕 is 𝟑𝟎% of what number? 𝟐𝟕= 𝟎. 𝟑𝒏 𝟐𝟕 𝟎. 𝟑= 𝟏𝒏 𝟗𝟎= 𝒏 d. 𝟏𝟖 is 𝟑𝟎% of what number? 𝟎. 𝟑𝟎𝒏= 𝟏𝟖 𝟏𝒏= 𝟏𝟖 𝟎. 𝟑 𝒏= 𝟔𝟎 e. 𝟐𝟓. 𝟓 is what percent of 𝟖𝟓? 𝟐𝟓. 𝟓= 𝒑(𝟖𝟓) 𝟐𝟓. 𝟓 𝟖𝟓= 𝟏𝒑 𝟎. 𝟑= 𝒑 𝟎. 𝟑= 𝟑𝟎 𝟏𝟎𝟎= 𝟑𝟎% f. 𝟐𝟏 is what percent of 𝟔𝟎? 𝟐𝟏= 𝒑(𝟔𝟎) 𝟎. 𝟑𝟓= 𝒑 𝟎. 𝟑𝟓= 𝟑𝟓 𝟏𝟎𝟎= 𝟑𝟓% 2. 𝟒𝟎% of the students on a field trip love the museum. If there are 𝟐𝟎 students on the field trip, how many love the museum? Let 𝒔 represent the number of students who love the museum. 𝒔= 𝟎. 𝟒𝟎(𝟐𝟎) 𝒔= 𝟖 Therefore, 𝟖 students love the museum. 3. Maya spent 𝟒𝟎% of her savings to pay for a bicycle that cost her $𝟖𝟓. a. How much money was in her savings to begin with? Let 𝒔 represent the unknown amount of money in Maya’s savings. 𝟖𝟓= 𝟎. 𝟒𝒔 𝟐𝟏𝟐. 𝟓= 𝒔 Maya originally had $𝟐𝟏𝟐. 𝟓𝟎 in her savings. b. How much money does she have left in her savings after buying the bicycle? $𝟐𝟏𝟐. 𝟓𝟎−$𝟖𝟓. 𝟎𝟎= $𝟏𝟐𝟕. 𝟓𝟎 She has $𝟏𝟐𝟕. 𝟓𝟎 left in her savings after buying the bicycle. Lesson 2: Part of a Whole as a Percent Date: 11/19/14 40 © 2014 Common Core, Inc. Some rights reserved. commoncore.org This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. NYS COMMON CORE MATHEMATICS CURRICULUM 7•4 Lesson 2 4. Curtis threw 𝟏𝟓 darts at a dartboard. 𝟒𝟎% of his darts hit the bull’s-eye. How many darts did not hit the bull’s-eye? Let 𝒅 represent the number of darts that hit the bull’s-eye. 𝒅= 𝟎. 𝟒(𝟏𝟓) 𝒅= 𝟔 𝟔 darts hit the bull’s-eye. 𝟏𝟓−𝟔= 𝟗, so 𝟗 darts did not hit the bull’s-eye. 5. A tool set is on sale for $𝟒𝟐𝟒. 𝟏𝟓. The original price of the tool set was $𝟒𝟗𝟗. 𝟎𝟎. What percent of the original price is the sale price? Let 𝒑 represent the unknown percent. 𝟒𝟐𝟒. 𝟏𝟓= 𝒑(𝟒𝟗𝟗) The sale price is 𝟖𝟓% of the original price. 6. Matthew scored a total of 𝟏𝟔𝟖 points in basketball this season. He scored 𝟏𝟒𝟕 of those points in the regular season and the rest were scored in his only playoff game. What percent of his total points did he score in the playoff game? Matthew scored 𝟐𝟏 points during the playoff game because 𝟏𝟔𝟗−𝟏𝟒𝟕= 𝟐𝟏. Let 𝒑 represent the unknown percent. 𝟐𝟏= 𝒑(𝟏𝟔𝟖) The points that Matthew scored in the playoff game were 𝟏𝟐. 𝟓% of his total points scored in basketball this year. 7. Brad put 𝟏𝟎 crickets in his pet lizard’s cage. After one day, Brad’s lizard had eaten 𝟐𝟎% of the crickets he had put in the cage. By the end of the next day, the lizard had eaten 𝟐𝟓% of the remaining crickets. How many crickets were left in the cage at the end of the second day? Day 1: 𝒏= 𝟎. 𝟐(𝟏𝟎) 𝒏= 𝟐 At the end of the first day, Brad’s lizard had eaten 𝟐 of the crickets. Day 2: 𝒏= 𝟎. 𝟕𝟓(𝟏𝟎−𝟐) 𝒏= 𝟎. 𝟕𝟓(𝟖) 𝒏= 𝟔 At the end of the second day, Brad’s lizard had eaten a total of 𝟒 crickets, leaving 𝟔 crickets in the cage. 8. A furnace used 𝟒𝟎% of the fuel in its tank in the month of March and then used 𝟐𝟓% of the remaining fuel in the month of April. At the beginning of March, there were 𝟐𝟒𝟎 gallons of fuel in the tank. How much fuel (in gallons) was left at the end of April? March: 𝒏= 𝟎. 𝟒(𝟐𝟒𝟎) 𝒏= 𝟗𝟔 Therefore, 𝟗𝟔 gallons were used during the month of March, which means 𝟏𝟒𝟒 gallons remain. April: 𝒏= 𝟎. 𝟐𝟓(𝟏𝟒𝟒) 𝒏= 𝟑𝟔 Therefore, 𝟑𝟔 gallons were used during the month of April, which means 𝟏𝟎𝟖 gallons remain. There were 𝟏𝟒𝟒 gallons of fuel remaining in the tank at the end of March and 𝟏𝟎𝟖 gallons of fuel remaining at the end of April. Lesson 2: Part of a Whole as a Percent Date: 11/19/14 41 © 2014 Common Core, Inc. Some rights reserved. commoncore.org This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. NYS COMMON CORE MATHEMATICS CURRICULUM 7•4 Lesson 2 9. In Lewis County, there were 𝟐, 𝟐𝟕𝟕 student athletes competing in spring sports in 2014. That was 𝟏𝟏𝟎% of the number from 2013, which was 𝟗𝟎% of the number from the year before. How many student athletes signed up for a spring sport in 2012? 2013: 𝟐, 𝟐𝟕𝟕= 𝟏. 𝟏𝟎𝒂 𝟐, 𝟎𝟕𝟎= 𝒂 Therefore, 𝟐, 𝟎𝟕𝟎 student athletes competed in spring sports in 2013. 2012: 𝟐, 𝟎𝟕𝟎= 𝟎. 𝟗𝒂 𝟐, 𝟑𝟎𝟎= 𝒂 Therefore, 𝟐, 𝟑𝟎𝟎 student athletes competed in spring sports in 2012. There were 𝟐, 𝟎𝟕𝟎 students competing in spring sports in 2013 and 𝟐, 𝟑𝟎𝟎 students in 2012. 10. Write a real-world word problem that could be modeled by the equation below. Identify the elements of the percent equation and where they appear in your word problem, and then solve the problem. 𝟓𝟕. 𝟓= 𝒑(𝟐𝟓𝟎) Answers will vary. Greig is buying sliced almonds for a baking project. According to the scale, his bag contains 𝟓𝟕. 𝟓 grams of almonds. Greig needs 𝟐𝟓𝟎 grams of sliced almonds for his project. What percent of his total weight of almonds does Greig currently have? The quantity 𝟓𝟕. 𝟓 represents the part of the almonds that Greig currently has on the scale, the quantity 𝟐𝟓𝟎 represents the 𝟐𝟓𝟎 grams of almonds that he plans to purchase, and the variable 𝒑 represents the unknown percent of the whole quantity that corresponds to the quantity 𝟓𝟕. 𝟓. 𝟓𝟕. 𝟓= 𝒑(𝟐𝟓𝟎) 𝟏 𝟐𝟓𝟎(𝟓𝟕. 𝟓) = 𝒑( 𝟏 𝟐𝟓𝟎) (𝟐𝟓𝟎) 𝟓𝟕. 𝟓 𝟐𝟓𝟎= 𝒑(𝟏) 𝟎. 𝟐𝟑= 𝒑 𝟎. 𝟐𝟑= 𝟐𝟑 𝟏𝟎𝟎= 𝟐𝟑% Greig currently has 𝟐𝟑% of the total weight of almonds that he plans to buy. |
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