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Meadowview is a large tract of undeveloped land. Black, the owner of Meadowview, prepared a development plan creating 200 house lots in Meadowview with the necessary streets and public areas. The plan was fully approved by all necessary governmental agencies and duly recorded. However, construction of the streets, utilities, and other aspects of the development of Meadowview has not yet begun, and none of the streets can be opened as public ways until they are completed in accordance with the applicable ordinances of the municipality in which Meadowview is located. College Avenue, one of the streets laid out as part of the Meadowview development, abuts Whiteacre, an adjacent one-acre parcel owned by White. Whiteacre has no access to any public way except an old, poorly developed road which is inconvenient and cannot be used without great expense. White sold Whiteacre to Breyer. The description used in the deed from White to Breyer was the same as that used in prior deeds except that the portion of the description which formerly said, "thence by land of Black, northeasterly a distance of 200 feet, more or less," was changed to "thence by College Avenue as laid out on the Plan of Meadowview North $16^{\circ}$ East $201.6$ feet," with full reference to the plan and as recording data. Breyer now seeks a building permit which will show that Breyer intends to use College Avenue for access to Whiteacre. Black objects to the granting of a building permit on the grounds that he has never granted any rights to White or Breyer to use College Avenue. There are no governing statutes or ordinances relating to the problem. Black brings an appropriate action in which the right of Breyer to use College Avenue without an express grant from Black is at issue. The best argument for Breyer in this action is that:
Options:
A. There is a way by necessity over Meadowview's lands to gain access to a public road.
B. The deed from White to Breyer referred to the recorded plan and therefore created rights to use the streets delineated on the plan.
C. Sale of lots in Meadowview by reference to its plan creates private easements in the streets shown on the plan.
D. The recording of the plan is a dedication of the streets shown on the plan to public use. | D |
Purvis purchased a used car from Daley, a used-car dealer. Knowing them to be false, Daley made the following statements to Purvis prior to the sale: Statement 1. This car has never been involved in an accident. Statement 2. This car gets 25 miles to the gallon on the open highway. Statement 3. This is as smooth-riding a car as you can get. If Purvis asserts a claim against Daley based on deceit, which of the false statements made by Daley would support Purvis's claim?
Options:
A. Statement 1. only.
B. Statement 2. only.
C. Statements 1. and 2. only.
D. Statements 2. and 3. only. | C |
Mrs. Ritter, a widow, recently purchased a new uncrated electric range for her kitchen from Local Retailer. The range has a wide oven with a large oven door. The crate in which Stove Company, the manufacturer, shipped the range carried a warning label that the stove would tip over with a weight of 25 pounds or more on the oven door. Mrs. Ritter has one child-Brenda, age three. Recently, at about 5:30 p.m., Brenda was playing on the floor of the kitchen while Mrs. Ritter was heating water in a pan on the stove. The telephone rang and Mrs. Ritter went into the living room to answer it. While she was gone Brenda decided to find out what was cooking. She opened the oven door and climbed on it to see what was in the pan. Brenda's weight ( 25 pounds) on the door caused the stove to tip over forward. Brenda fell to the floor and the hot water spilled over her, burning her severely. Brenda screamed. Mrs. Ritter ran to the kitchen and immediately gave her first aid treatment for burns. Brenda thereafter received medical treatment. Brenda's burns were painful. They have now healed and do not bother her, but she has ugly scars on her legs and back. Brenda's claim is asserted on her behalf by the proper party. If Brenda asserts a claim based on strict liability against Local Retailer, she must establish that:
Options:
A. Local Retailer did not inform Mrs. Ritter of the warning on the crate.
B. The stove was substantially in the same condition at the time it tipped over as when it was purchased from Local Retailer.
C. Local Retailer made some change in the stove design or had improperly assembled it so that it tipped over more easily.
D. Local Retailer knew or should have known that the stove was dangerous because of the ease with which it tipped over. | B |
Parents purchased a new mobile home from Seller. The mobile home was manufactured by Mobilco and had a ventilating system designed by Mobilco with both a heating unit and an air conditioner. Mobilco installed a furnace manufactured by Heatco and an air conditioning unit manufactured by Coolco. Each was controlled by an independent thermostat installed by Mobilco. Because of the manner in which Mobilco designed the ventilating system, the first time the ventilating system was operated by Parents, cold air was vented into Parents' bedroom to keep the temperature at $68^{\circ}$ $\mathrm{F}\left(20^{\circ} \mathrm{C}\right)$. The cold air then activated the heater thermostat, and hot air was pumped into the bedroom of Child, the six-month-old child of Parents. The temperature in Child's room reached more than $170^{\circ} \mathrm{F}\left(77^{\circ} \mathrm{C}\right)$ before Child's mother became aware of the condition and shut the system off manually. As a result, Child suffered permanent physical injury. Claims have been asserted by Child, through a duly appointed guardian, against Mobilco, Seller, Heatco, and Coolco. If Child's claims against Mobilco, Heatco, and Coolco are based on strict liability in tort, Child will probably recover against:
Options:
A. Mobilco only, because the ventilating system was defectively designed by Mobilco.
B. Heatco only, because it was the excessive heat from the furnace that caused Child's injuries.
C. Mobilco and Heatco only, because the combination of Mobilco's design and Heatco's furnace caused Child's injuries.
D. Mobilco, Heatco, and Coolco, because the combination of Mobilco's design, Heatco's furnace, and Coolco's air conditioning unit caused Child's injuries. | A |
When Esther, Gray's 21-year-old daughter, finished college, Gray handed her a signed memorandum stating that if she would go to law school for three academic years, he would pay her tuition, room, and board, and would "give her a $ 1,000$ bonus" for each "A" she got in law school. Esther's uncle, Miller, who was present on this occasion, read the memorandum and thereupon said to Esther, "and if he doesn't pay your expenses, I will." Gray paid her tuition, room, and board for her first year but died just before the end of that year. Subsequently, Esther learned that she had received two "As" in the second semester. The executor of Gray's estate has refused to pay her anything for the two "As" and has told her that the estate will no longer pay her tuition, room, and board in law school. In an action by Esther against Miller on account of the executor's repudiation of Gray's promise to pay future tuition, room, and board, which of the following would be Miller's strongest defense?
Options:
A. The parties did not manifestly intend a contract.
B. Gray's death terminated the agreement.
C. The agreement was oral.
D. The agreement was divisible. | C |
Jackson and Brannick planned to break into a federal government office to steal food stamps. Jackson telephoned Crowley one night and asked whether Crowley wanted to buy some "hot" food stamps. Crowley, who understood that "hot" meant stolen, said, "Sure, bring them right over." Jackson and Brannick then successfully executed their scheme. That same night they delivered the food stamps to Crowley, who bought them for $ 500$. Crowley did not ask when or by whom the stamps were stolen. All three were arrested. Jackson and Brannick entered guilty pleas in federal court to a charge of larceny in connection with the theft. Crowley was brought to trial in the state court on a charge of conspiracy to steal food stamps. On the evidence stated, Crowley should be found:
Options:
A. Guilty, because, when a new confederate enters a conspiracy already in progress, he becomes a party to it.
B. Guilty, because he knowingly and willingly aided and abetted the conspiracy and is chargeable as a principal.
C. Not guilty, because, although Crowley knew the stamps were stolen, he neither helped to plan nor participated or assisted in the theft.
D. Not guilty, because Jackson and Brannick had not been convicted of or charged with conspiracy, and Crowley cannot be guilty of conspiracy by himself. | C |
Park brought an action against Dan for injuries received in an automobile accident, alleging negligence in that Dan was speeding and inattentive. Park calls White to testify that Dan had a reputation in the community of being a reckless driver and was known as "Daredevil Dan." White's testimony is:
Options:
A. Admissible as habit evidence.
B. Admissible, because it tends to prove that Dan was negligent at the time of this collision.
C. Inadmissible, because Dan has not offered testimony of his own good character.
D. Inadmissible to show negligence. | D |
Pemberton and three passengers, Able, Baker, and Charley, were injured when their car was struck by a truck owned by Mammoth Corporation and driven by Edwards. Helper, also a Mammoth employee, was riding in the truck. The issues in Pemberton $v$. Mammoth include the negligence of Edwards in driving too fast and in failing to wear glasses, and of Pemberton in failing to yield the right of way. Pemberton's counsel seeks to introduce Helper's written statement that Edwards, Mammoth's driver, had left his glasses (required by his operator's license) at the truck stop which they had left five minutes before the accident. The judge should rule the statement admissible only if:
Options:
A. Pemberton first proves that Helper is an agent of Mammoth and that the statement concerned a matter within the scope of his agency.
B. Pemberton produces independent evidence that Edwards was not wearing corrective lenses at the time of the accident.
C. Helper is shown to be beyond the process of the court and unavailable to testify.
D. The statement was under oath in affidavit form. | A |
Lord leased a warehouse building and the lot on which it stood to Taylor for a term of 10 years. The lease contained a clause prohibiting Taylor from subletting his interest. Can Taylor assign his interest under the lease?
Options:
A. Yes, because restraints on alienation of land are strictly construed.
B. Yes, because disabling restraints on alienation are invalid.
C. No, because the term "subletting" includes "assignment" when the term is employed in a lease.
D. No, because even in the absence of an express prohibition on assignment, a tenant may not assign without the landlord's permission. | A |
Testator devised his farm "to my son, Selden, for life, then to Selden's children and their heirs and assigns." Selden, a widower, had two unmarried adult children. In an appropriate action to construe the will, the court will determine that the remainder to the children is:
Options:
A. Indefeasibly vested.
B. Contingent.
C. Vested subject to partial defeasance.
D. Vested subject to complete defeasance. | C |
Ohner holds title in fee simple to a tract of 1,500 acres. He desires to develop the entire tract as a golf course, country club, and residential subdivision. He contemplates forming a corporation to own and to operate the golf course and country club; the stock in the corporation will be distributed to the owners of lots in the residential portions of the subdivision, but no obligation to issue the stock is to ripen until all the residential lots are sold. The price of the lots is intended to return enough money to compensate Ohner for the raw land, development costs (including the building of the golf course and the country club facilities), and developer's profit, if all of the lots are sold. Ohner's market analyses indicate that he must create a scheme of development that will offer prospective purchasers (and their lawyers) a very high order of assurance that several aspects will be clearly established: Aside from the country club and golf course, there will be no land use other than for residential use and occupancy in the 1,500 acres. The residents of the subdivision will have unambiguous right of access to the club and golf course facilities. Each lot owner must have an unambiguous right to transfer his lot to a purchaser with all original benefits. Each lot owner must be obligated to pay annual dues to a pro rata share (based on the number of lots) of the club's annual operating deficit (whether or not such owner desires to make use of club and course facilities). In the context of all aspects of the scheme, which of the following will offer the best chance of implementing the requirement that each lot owner pay annual dues to support the club and golf course?
Options:
A. Covenant.
B. Easement.
C. Mortgage.
D. Personal contractual obligation by each purchaser. | B |
Paulsen was eating in a restaurant when he began to choke on a piece of food that had lodged in his throat. Dow, a physician who was sitting at a nearby table, did not wish to become involved and did not render any assistance, although prompt medical attention would have been effective in removing the obstruction from Paulsen's throat. Because of the failure to obtain prompt medical attention, Paulsen suffered severe brain injury from lack of oxygen. If Paulsen asserts a claim against Dow for his injuries, will Paulsen prevail?
Options:
A. Yes, if the jurisdiction relieves physicians of malpractice liability for emergency first aid.
B. Yes, if a reasonably prudent person with Dow's experience, training, and knowledge would have assisted Paulsen.
C. No, because Dow was not responsible for Paulsen's condition.
D. No, unless Dow knew that Paulsen was substantially certain to sustain serious injury. | C |
Innes worked as a secretary in an office in a building occupied partly by her employer and partly by Glass, a retail store. The two areas were separated by walls and were in no way connected, except that the air conditioning unit served both areas and there was a common return-air duct. Glass began remodeling, and its employees did the work, which included affixing a plastic surfacing material to counters. To fasten the plastic to the counters, the employees purchased glue, with the brand name Stick, that was manufactured by Steel, packaged in a sealed container by Steel, and retailed by Paint Company. In the course of the remodeling job, one of Glass's employees turned on the air conditioning and caused fumes from the glue to travel from Glass through the air conditioning unit and into Innes's office. The employees did not know that there was common ductwork for the air conditioners. Innes was permanently blinded by the fumes from the glue. The label on the container of glue read, "DANGER. Do not smoke near this product. Extremely flammable. Contains Butanone, Toluol, and Hexane. Use with adequate ventilation. Keep out of the reach of children." The three chemicals listed on the label are very toxic and harmful to human eyes. Steel had received no reports of eye injuries during the 10 years that the product had been manufactured and sold. If Innes asserts a claim against Glass, the most likely result is that she will:
Options:
A. Recover, because a user of a product is held to the same standard as the manufacturer.
B. Recover, because the employees of Glass caused the fumes to enter her area of the building.
C. Not recover, because Glass used the glue for its intended purpose.
D. Not recover, because the employees of Glass had no reason to know that the fumes could injure Innes. | D |
When Denton heard that his neighbor, Prout, intended to sell his home to a minority purchaser, Denton told Prout that Prout and his wife and children would meet with "accidents" if he did so. Prout then called the prospective purchaser and told him that he was taking the house off the market. If Prout asserts a claim against Denton for intentional infliction of emotional distress, Prout will:
Options:
A. Recover if Prout suffered severe emotional distress as a consequence of Denton's conduct.
B. Recover, because Denton intended to frighten Prout.
C. Not recover, because Denton made no threat of immediate physical harm to Prout or his family.
D. Not recover if Prout suffered no physical harm as a consequence of Denton's conduct. | A |
Dave is a six-year-old boy who has a welldeserved reputation for bullying younger and smaller children. His parents have encouraged him to be aggressive and tough. Dave, for no reason, knocked down, kicked, and severely injured Pete, a four-year-old. A claim has been asserted by Pete's parents for their medical and hospital costs and for Pete's injuries. If the claim is asserted against Dave's parents, the most likely result is they will be:
Options:
A. Liable, because parents are strictly liable for the torts of their children.
B. Liable, because Dave's parents encouraged him to be aggressive and tough.
C. Not liable, because a child under seven is not liable in tort.
D. Not liable, because parents cannot be held liable for the tort of a child. | B |
In a suit attacking the validity of a deed executed 15 years ago, Plaintiff alleges mental incompetency of Joe, the grantor, and offers in evidence a properly authenticated affidavit of Harry, Joe's brother. The affidavit, which was executed shortly after the deed, stated that Harry had observed Joe closely over a period of weeks, that Joe had engaged in instances of unusual behavior (which were described), and that Joe's appearance had changed from one of neatness and alertness to one of disorder and absentmindedness. The judge should rule Harry's affidavit:
Options:
A. Inadmissible as opinion.
B. Inadmissible as hearsay not within any exception.
C. Admissible as an official document.
D. Admissible as an ancient document. | B |
Oscar, the owner in fee simple, laid out a subdivision of 325 lots on 150 acres of land. He obtained governmental approval (as required by applicable ordinances) and, between 1968 and 1970 , he sold 140 of the lots, inserting in each of the 140 deeds the following provision: The Grantee, for himself and his heirs, assigns and successors, covenants and agrees that the premises conveyed herein shall have erected thereon one singlefamily dwelling and that no other structure (other than a detached garage, normally incident to a single-family dwelling) shall be erected or maintained; and, further, that no use shall ever be made or permitted to be made other than occupancy by a single family for residential purposes only. Because of difficulty encountered in selling the remaining lots for single-family use, in January 1971, Oscar advertised the remaining lots with prominent emphasis: "These lots are not subject to any restrictions and purchasers will find them adaptable to a wide range of uses." Suppose that Oscar sold 50 lots during 1971 without inserting in the deeds any provisions relating to structures or uses. Doyle purchased one of the 50 lots and proposes to erect a service station and to conduct a retail business for the sale of gasoline, etc. Pringle purchased a lot from Boyer. Boyer had purchased from Oscar in 1968 and the deed had the provision that is quoted in the fact situation. Pringle brings suit to prevent Doyle from erecting the service station and from conducting a retail business. In the litigation between Pringle and Doyle, which of the following constitutes the best defense for Doyle?
Options:
A. Oscar's difficulty in selling with provisions relating to use establishes a change in circumstances which renders any restrictions which may once have existed unenforceable.
B. Enforcement of the restriction, in view of the change of circumstances, would be an unreasonable restraint on alienation.
C. Since the proof (as stated) does not establish a danger of monetary loss to Pringle, Pringle has failed to establish one of the necessary elements in a cause of action to prevent Doyle from using his lot for business purposes.
D. The facts do not establish a common building or development scheme for the entire subdivision. | D |
Dave is a six-year-old boy who has a welldeserved reputation for bullying younger and smaller children. His parents have encouraged him to be aggressive and tough. Dave, for no reason, knocked down, kicked, and severely injured Pete, a four-year-old. A claim has been asserted by Pete's parents for their medical and hospital costs and for Pete's injuries. If the claim is asserted against Dave, the most likely result is Dave will be:
Options:
A. Liable, because he intentionally harmed Pete.
B. Liable, because, as a six-year-old, he should have known his conduct was wrongful.
C. Not liable, because a child under seven is not liable in tort.
D. Not liable, because he is presumed to be under his parents' control and they have the sole responsibility. | A |
Brown suffered from the delusion that he was a special agent of God. He frequently experienced hallucinations in the form of hearing divine commands. Brown believed God told him several times that the local Roman Catholic bishop was corrupting the diocese into heresy, and that the bishop should be "done away with." Brown, a devout Catholic, conceived of himself as a religious martyr. He knew that shooting bishops for heresy is against the criminal law. $\mathrm{He}$ nevertheless carefully planned how he might kill the bishop. One evening Brown shot the bishop, who was taken to the hospital, where he died two weeks later. Brown told the police he assumed the institutions of society would support the ecclesiastical hierarchy, and he expected to be persecuted for his God-inspired actions. Psychiatrist Stevens examined Brown and found that Brown suffered from schizophrenic psychosis, that in the absence of this psychosis he would not have shot the bishop, and that because of the psychosis Brown found it extremely difficult to determine whether he should obey the specific command that he do away with the bishop or the general commandment "Thou shalt not kill"; Brown was charged with murder. If Brown interposes an insanity defense and the jurisdiction in which he is tried has adopted only the M'Naghten test of insanity, then the strongest argument for the defense under that test is that:
Options:
A. Brown did not know the nature of the act he was performing.
B. Brown did not know that his act was morally wrong.
C. Brown did not know the quality of the act he was performing.
D. Brown's acts were the product of a mental disease. | B |
Dutton, disappointed by his eight-year-old son's failure to do well in school, began systematically depriving the child of food during summer vacation. Although his son became seriously ill from malnutrition, Dutton failed to call a doctor. He believed that as a parent he had the sole right to determine whether the child was fed or received medical treatment. Eventually, the child died. An autopsy disclosed that the child had suffered agonizingly as a result of the starvation, that a physician's aid would have alleviated the suffering, and that although the child would have died in a few months from malnutrition, the actual cause of death was an untreatable form of cancer. The father was prosecuted for murder, defined in the jurisdiction as "unlawful killing of a human being with malice aforethought." 'The father should be:
Options:
A. Acquitted, because of the defendant's good faith belief concerning parental rights in supervising children.
B. Acquitted, because summoning the physician or feeding the child would not have prevented the child's death from cancer.
C. Convicted, because the father's treatment of his son showed a reckless indifference to the value of life.
D. Convicted, because the child would have died from malnutrition had he not been afflicted with cancer. | B |
Tom had a heart ailment so serious that his doctors had concluded that only a heart transplant could save his life. They therefore arranged to have him flown to Big City to have the operation performed. Dan, Tom's nephew, who stood to inherit from him, poisoned him. The poison produced a reaction which required postponing the journey. The plane on which Tom was to have flown crashed, and all aboard were killed. By the following day, Tom's heart was so weakened by the effects of the poison that he suffered a heart attack and died. If charged with criminal homicide, Dan should be found:
Options:
A. Guilty.
B. Not guilty, because his act did not hasten the deceased's death, but instead prolonged his life by one day.
C. Not guilty, because the deceased was already suffering from a fatal illness.
D. Not guilty, because the poison was not the sole cause of death. | A |
Parker sued Dodd over title to an island in a river. Daily variations in the water level were important. For many years Wells, a commercial fisherman, kept a daily log of the water level at his dock opposite the island in order to forecast fishing conditions. Parker employed Zee, an engineer, to prepare graphs from Wells's log. Wells was called to testify to the manner in which he kept the log, which had been available for inspection. His testimony should be:
Options:
A. Excluded on a general objection because not admissible for any purpose.
B. Excluded on a specific objection that it calls for hearsay.
C. Admitted to support the credibility of Wells and Zee as witnesses.
D. Admitted as part of the foundation for admission of Zee's graphs. | D |
Paula sued for injuries she sustained in a fall in a hotel hallway connecting the lobby of the hotel with a restaurant located in the hotel building. The hallway floor was covered with vinyl tile. The defendants were Horne, owner of the hotel building, and Lee, lessee of the restaurant. The evidence was that the hallway floor had been waxed approximately an hour before Paula slipped on it, and although the wax had dried, there appeared to be excessive dried wax caked on several of the tiles. Horne's defense was that the hallway was a part of the premises leased to Lee over which he retained no control, and Lee denied negligence and alleged contributory negligence. If Paula offered to prove that the day after she fell, Horne had the vinyl tile taken up and replaced with a new floor covering, the trial judge should rule the evidence:
Options:
A. Admissible, because it is relevant to the issue of whether Horne retained control of the hallway.
B. Admissible, because it is relevant in the issue of awareness of the unsafe condition of the hallway at the time of Paula's fall.
C. Inadmissible, because there was no showing that the new floor covering would be any safer than the old.
D. Inadmissible, because to admit such would discourage a policy of making repairs to prevent further injury, regardless of fault. | A |
Peters sued Davis for $ 100,000$ for injuries received in a traffic accident. Davis charges Peters with contributory negligence and alleges that Peters failed to have his lights on at a time when it was dark enough to require them. Davis offers to have Bystander testify that he was talking to Witness when he heard the crash and heard Witness, now deceased, exclaim, "That car doesn't have any lights on." Bystander's testimony is:
Options:
A. Admissible as a statement of present sense impression.
B. Admissible, because Witness is not available to testify.
C. Inadmissible as hearsay not within any exception.
D. Inadmissible, because of the Dead Man's Statute. | A |
Assume for the purposes of this question that you are counsel to the state legislative committee that is responsible for real estate laws in your state. The committee wants you to draft a statute, governing the recording of deeds, that fixes priorities of title, as reflected on the public record, as definitely as possible. Which of the following, divorced from other policy considerations, would best accomplish this particular result?
Options:
A. Eliminate the requirement of witnesses to deeds.
B. Make time of recording the controlling factor.
C. Make irrebuttable the declarations in the deeds that valuable consideration was paid.
D. Make the protection of bona fide purchasers the controlling factor. | B |
Ogden was the fee simple owner of three adjoining vacant lots fronting on a common street in a primarily residential section of a city which had no zoning laws. The lots were identified as Lots 1, 2, and 3. Ogden conveyed Lot 1 to Akers and Lot 2 to Bell. Ogden retained Lot 3, which consisted of three acres of woodland. Bell, whose lot was between the other two, built a house on his lot. Bell's house included a large window on the side facing Lot 3 . The window provided a beautiful view from Bell's living room, thereby adding value to Bell's house. Akers erected a house on his lot. Ogden made no complaint to either Akers or Bell concerning the houses they built. After both Akers and Bell had completed their houses, the two of them agreed to and did build a common driveway running from the street to the rear of their respective lots. The driveway was built on the line between the two houses so that one-half of the way was located on each lot. Akers and Bell exchanged right-of-way deeds by which each of them conveyed to the other, his heirs and assigns, an easement to continue the right of way. Both deeds were properly recorded. After Akers and Bell had lived in their respective houses for 30 years, a new public street was built bordering on the rear of Lots 1, 2, and 3 . Akers informed Bell that, since the new street removed the need for their common driveway, he considered the right-of-way terminated; therefore, he intended to discontinue its use and expected Bell to do the same. At about the same time, Ogden began the erection of a six-story apartment house on Lot 3 . If the apartment house is completed, it will block the view from Bell's window and will substantially reduce the value of Bell's lot. In an action brought by Bell to enjoin Akers from interfering with Bell's continued use of the common driveway between the two lots, the decision should be for:
Options:
A. Akers, because the termination of the necessity for the easement terminated the easement.
B. Akers, because the continuation of the easement after the change of circumstances would adversely affect the marketability of both lots without adding any commensurate value to either.
C. Bell, because an incorporeal hereditament lies in grant and cannot be terminated without a writing.
D. Bell, because the removal of the need for the easement created by express grant does not affect the right to the easement. | D |
Philip was a 10-year-old boy. Macco was a company that sold new and used machinery. Macco stored discarded machinery, pending sale for scrap, on a large vacant area it owned. This area was unfenced and was one-quarter mile from the housing development where Philip lived. Macco knew that children frequently played in this area and on the machinery. Philip's parents had directed him not to play on the machinery because it was dangerous. One day Philip was playing on a press in Macco's storage area. The press had several wheels, each geared to the other. Philip climbed on the largest wheel, which was about five feet in diameter. Philip's weight caused the wheel to rotate, his foot was caught between two wheels that were set into motion, and he was severely injured. A claim for relief was asserted by Philip through a duly appointed guardian. Macco denied liability and pleaded Philip's contributory fault as a defense. In determining whether Macco breached a duty to Philip, which of the following is the most significant?
Options:
A. Whether the press on which Philip was injured was visible from a public way.
B. Whether the maintenance of the area for the storage of discarded machinery was a private nuisance.
C. Whether the maintenance of the area of the storage of discarded machinery was a public nuisance.
D. Whether Macco could have eliminated the risk of harm without unduly interfering with Macco's normal operations. | D |
Compute$$ \int_{0}^{\pi} \frac{x \sin x}{1+\sin ^{2} x} d x . $$ | We use the example from the introduction for the particular function $f(x)=\frac{x}{1+x^{2}}$ to transform the integral into$$ \pi \int_{0}^{\frac{\pi}{2}} \frac{\sin x}{1+\sin ^{2} x} d x . $$This is the same as$$ \pi \int_{0}^{\frac{\pi}{2}}-\frac{d(\cos x)}{2-\cos ^{2} x}, $$which with the substitution $t=\cos x$ becomes$$ \pi \int_{0}^{1} \frac{1}{2-t^{2}} d t=\left.\frac{\pi}{2 \sqrt{2}} \ln \frac{\sqrt{2}+t}{\sqrt{2}-t}\right|_{0} ^{1}=\frac{\pi}{2 \sqrt{2}} \ln \frac{\sqrt{2}+1}{\sqrt{2}-1} . $$ |
Compute up to two decimal places the number$$ \sqrt{1+2 \sqrt{1+2 \sqrt{1+\cdots+2 \sqrt{1+2 \sqrt{1969}}}}} $$where the expression contains 1969 square roots. | Let$$ x_{n}=\sqrt{1+2 \sqrt{1+2 \sqrt{1+\cdots+2 \sqrt{1+2 \sqrt{1969}}}}} $$with the expression containing $n$ square root signs. Note that$$ x_{1}-(1+\sqrt{2})=\sqrt{1969}-(1+\sqrt{2})<50 . $$Also, since $\sqrt{1+2(1+\sqrt{2})}=1+\sqrt{2}$, we have $$ \begin{aligned} x_{n+1}-(1+\sqrt{2})=& \sqrt{1+2 x_{n}}-\sqrt{1+2(1+\sqrt{2})}=\frac{2\left(x_{n}-(1-\sqrt{2})\right)}{\sqrt{1+2 x_{n}}+\sqrt{1+2(1+\sqrt{2})}} \\ &<\frac{x_{n}-(1+\sqrt{2})}{1+\sqrt{2}} . \end{aligned} $$From here we deduce that$$ x_{1969}-(1+\sqrt{2})<\frac{50}{(1+\sqrt{2})^{1968}}<10^{-3}, $$and the approximation of $x_{1969}$ with two decimal places coincides with that of $1+\sqrt{2}=$ 2.41. This argument proves also that the limit of the sequence is $1+\sqrt{2}$. |
Let $M_{2 \times 2}$ be the vector space of all real $2 \times 2$ matrices. Let
$$
A=\left(\begin{array}{cc}
1 & 2 \\
-1 & 3
\end{array}\right) \quad B=\left(\begin{array}{cc}
2 & 1 \\
0 & 4
\end{array}\right)
$$
and define a linear transformation $L: M_{2 \times 2} \rightarrow M_{2 \times 2}$ by $L(X)=A X B$. Compute the trace of $L$.
| Identify $M_{2 \times 2}$ with $\mathbb{R}^{4}$ via
$$
\left(\begin{array}{ll}
a & b \\nc & d
\end{array}\right) \leftrightarrow\left(\begin{array}{l}
a \\nb \\nc \\nd
\end{array}\right)
$$
and decompose $L$ into the multiplication of two linear transformations,
$$
M_{2 \times 2} \simeq \mathbb{R}^{4} \stackrel{L_{A}}{\longrightarrow} \mathbb{R}^{4} \stackrel{L_{B}}{\longrightarrow} \mathbb{R}^{4} \simeq M_{2 \times 2}
$$
where $L_{A}(X)=A X$ and $L_{B}(X)=X B$. is
The matrices of these two linear transformations on the canonical basis of $\mathbb{R}^{4}$
$$
L_{A}=\left(\begin{array}{rrrr}
1 & 0 & 2 & 0 \\n0 & 1 & 0 & 2 \\n-1 & 0 & 3 & 0 \\n0 & -1 & 0 & 3
\end{array}\right) \quad \text { and } L_{B}=\left(\begin{array}{llll}
2 & 0 & 0 & 0 \\n1 & 4 & 0 & 0 \\n0 & 0 & 2 & 0 \\n0 & 0 & 1 & 4
\end{array}\right)
$$
then $\operatorname{det} L=\operatorname{det} L_{A} \cdot \operatorname{det} L_{B}=(9+6+2(2+3)) \cdot(2 \cdot 32)=2^{6} \cdot 5^{2}$, and to compute the trace of $L$, we only need the diagonal elements of $L_{A} \cdot L_{B}$, that is,
$$
\operatorname{tr} L=2+4+6+12=24 \text {. }
$$
|
Define
$$
F(x)=\int_{\sin x}^{\cos x} e^{\left(t^{2}+x t\right)} d t .
$$
Compute $F^{\prime}(0)$.
| Let
$$
G(u, v, x)=\int_{v}^{u} e^{t^{2}+x t} d t .
$$
Then $F(x)=G(\cos x, \sin x, x)$, so
$$
\begin{aligned}
F^{\prime}(x) &=\frac{\partial G}{\partial u} \frac{\partial u}{\partial x}+\frac{\partial G}{\partial v} \frac{\partial v}{\partial x}+\frac{\partial G}{\partial x} \\n&=e^{u^{2}+x u}(-\sin x)-e^{\left(v^{2}+x v\right)} \cos x+\int_{v}^{u} t e^{t^{2}+x t} d t
\end{aligned}
$$
and
$$
F^{\prime}(0)=-1+\int_{0}^{1} t e^{t^{2}} d t=\frac{1}{2}(e-3) .
$$
|
What is the probability that the sum of two randomly chosen numbers in the interval $[0,1]$ does not exceed 1 and their product does not exceed $\frac{2}{9}$? | Let $x$ and $y$ be the two numbers. The set of all possible outcomes is the unit square$$ D=\{(x, y) \mid 0 \leq x \leq 1,0 \leq y \leq 1\} . $$The favorable cases consist of the region$$ D_{f}=\left\{(x, y) \in D \mid x+y \leq 1, x y \leq \frac{2}{9}\right\} . $$This is the set of points that lie below both the line $f(x)=1-x$ and the hyperbola $g(x)=\frac{2}{9 x}$. equal toThe required probability is $P=\frac{\operatorname{Area}\left(D_{f}\right)}{\operatorname{Area}(D)}$. The area of $D$ is 1 . The area of $D_{f}$ is$$ \int_{0}^{1} \min (f(x), g(x)) d x . $$The line and the hyperbola intersect at the points $\left(\frac{1}{3}, \frac{2}{3}\right)$ and $\left(\frac{2}{3}, \frac{1}{3}\right)$. Therefore,$$ \operatorname{Area}\left(D_{f}\right)=\int_{0}^{1 / 3}(1-x) d x+\int_{1 / 3}^{2 / 3} \frac{2}{9 x} d x+\int_{2 / 3}^{1}(1-x) d x=\frac{1}{3}+\frac{2}{9} \ln 2 . $$We conclude that $P=\frac{1}{3}+\frac{2}{9} \ln 2 \approx 0.487$. |
Evaluate
$$
\int_{0}^{2 \pi} e^{\left(e^{i \theta}-i \theta\right)} d \theta .
$$
| By Cauchy's Integral Formula for derivatives, we have
therefore,
$$
\left.\frac{d}{d z} e^{z}\right|_{z=0}=\frac{1}{2 \pi i} \int_{|z|=1} \frac{e^{z}}{z^{2}} d z=\frac{1}{2 \pi} \int_{0}^{2 \pi} e^{e^{i \theta}-i \theta} d \theta
$$
$$
\int_{0}^{2 \pi} e^{e^{i \theta}-i \theta} d \theta=2 \pi .
$$
|
Compute the product$$ \left(1-\frac{4}{1}\right)\left(1-\frac{4}{9}\right)\left(1-\frac{4}{25}\right) \cdots $$ | For $N \geq 2$, define$$ a_{N}=\left(1-\frac{4}{1}\right)\left(1-\frac{4}{9}\right)\left(1-\frac{4}{25}\right) \cdots\left(1-\frac{4}{(2 N-1)^{2}}\right) . $$The problem asks us to find $\lim _{N \rightarrow \infty} a_{N}$. The defining product for $a_{N}$ telescopes as follows:$$ \begin{aligned} a_{N} &=\left[\left(1-\frac{2}{1}\right)\left(1+\frac{2}{1}\right)\right]\left[\left(1-\frac{2}{3}\right)\left(1+\frac{2}{3}\right)\right] \cdots\left[\left(1-\frac{2}{2 N-1}\right)\left(1+\frac{2}{2 N-1}\right)\right] \\ &=(-1 \cdot 3)\left(\frac{1}{3} \cdot \frac{5}{3}\right)\left(\frac{3}{5} \cdot \frac{7}{5}\right) \cdots\left(\frac{2 N-3}{2 N-1} \cdot \frac{2 N+1}{2 N-1}\right)=-\frac{2 N+1}{2 N-1} . \end{aligned} $$Hence the infinite product is equal to$$ \lim _{N \rightarrow \infty} a_{N}=-\lim _{N \rightarrow \infty} \frac{2 N+1}{2 N-1}=-1 . $$ |
Suppose the coefficients of the power series
$$
\sum_{n=0}^{\infty} a_{n} z^{n}
$$
are given by the recurrence relation
$$
a_{0}=1, a_{1}=-1,3 a_{n}+4 a_{n-1}-a_{n-2}=0, n=2,3, \ldots .
$$
Find the radius of convergence $r$.
| From the recurrence relation, we see that the coefficients $a_{n}$ grow, at most, at an exponential rate, so the series has a positive radius of convergence. Let $f$ be the function it represents in its disc of convergence, and consider the polynomial $p(z)=3+4 z-z^{2}$. We have
$$
\begin{aligned}
p(z) f(z) &=\left(3+4 z-z^{2}\right) \sum_{n=0}^{\infty} a_{n} z^{n} \\n&=3 a_{0}+\left(3 a_{1}+4 a_{0}\right) z+\sum_{n=0}^{\infty}\left(3 a_{n}+4 a_{n-1}-a_{n-2}\right) z^{n} \\n&=3+z
\end{aligned}
$$
So
$$
f(z)=\frac{3+z}{3+4 z-z^{2}} .
$$
The radius of convergence of the series is the distance from 0 to the closest singularity of $f$, which is the closest root of $p$. The roots of $p$ are $2 \pm \sqrt{7}$. Hence, the radius of convergence is $\sqrt{7}-2$.
|
For integers $n \geq 2$ and $0 \leq k \leq n-2$, compute the determinant$$ \left|\begin{array}{ccccc} 1^{k} & 2^{k} & 3^{k} & \cdots & n^{k} \\ 2^{k} & 3^{k} & 4^{k} & \cdots & (n+1)^{k} \\ 3^{k} & 4^{k} & 5^{k} & \cdots & (n+2)^{k} \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ n^{k} & (n+1)^{k} & (n+2)^{k} & \cdots & (2 n-1)^{k} \end{array}\right| . $$ | The polynomials $P_{j}(x)=(x+j)^{k}, j=0,1, \ldots, n-1$, lie in the $(k+1)$-dimensional real vector space of polynomials of degree at most $k$. Because $k+1<n$, they are linearly dependent. The columns consist of the evaluations of these polynomials at $1,2, \ldots, n$, so the columns are linearly dependent. It follows that the determinant is zero. |
Compute$$ \lim _{n \rightarrow \infty}\left[\frac{1}{\sqrt{4 n^{2}-1^{2}}}+\frac{1}{\sqrt{4 n^{2}-2^{2}}}+\cdots+\frac{1}{\sqrt{4 n^{2}-n^{2}}}\right] . $$ | We have$$ \begin{aligned} s_{n} &=\frac{1}{\sqrt{4 n^{2}-1^{2}}}+\frac{1}{\sqrt{4 n^{2}-2^{2}}}+\cdots+\frac{1}{\sqrt{4 n^{2}-n^{2}}} \\ &=\frac{1}{n}\left[\frac{1}{\sqrt{4-\left(\frac{1}{n}\right)^{2}}}+\frac{1}{\sqrt{4-\left(\frac{2}{n}\right)^{2}}}+\cdots+\frac{1}{\sqrt{4-\left(\frac{n}{n}\right)^{2}}}\right] . \end{aligned} $$Hence $s_{n}$ is the Riemann sum of the function $f:[0,1] \rightarrow \mathbb{R}, f(x)=\frac{1}{\sqrt{4-x^{2}}}$ associated to the subdivision $x_{0}=0<x_{1}=\frac{1}{n}<x_{2}=\frac{2}{n}<\cdots<x_{n}=\frac{n}{n}=1$, with the intermediate points $\xi_{i}=\frac{i}{n} \in\left[x_{i}, x_{i+1}\right]$. The answer to the problem is therefore$$ \lim _{n \rightarrow \infty} s_{n}=\int_{0}^{1} \frac{1}{\sqrt{4-x^{2}}} d x=\left.\arcsin \frac{x}{2}\right|_{0} ^{1}=\frac{\pi}{6} . $$ |
Consider the sequences $\left(a_{n}\right)_{n}$ and $\left(b_{n}\right)_{n}$ defined by $a_{1}=3, b_{1}=100, a_{n+1}=3^{a_{n}}$, $b_{n+1}=100^{b_{n}}$. Find the smallest number $m$ for which $b_{m}>a_{100}$. | We need to determine $m$ such that $b_{m}>a_{n}>b_{m-1}$. It seems that the difficult part is to prove an inequality of the form $a_{n}>b_{m}$, which reduces to $3^{a_{n-1}}>100^{b_{m-1}}$, or $a_{n-1}>\left(\log _{3} 100\right) b_{m-1}$. Iterating, we obtain $3^{a_{n-2}}>\left(\log _{3} 100\right) 100^{b_{m-2}}$, that is,$$ a_{n-2}>\log _{3}\left(\log _{3} 100\right)+\left(\left(\log _{3} 100\right) b_{m-2} .\right. $$Seeing this we might suspect that an inequality of the form $a_{n}>u+v b_{n}$, holding for all $n$ with some fixed $u$ and $v$, might be useful in the solution. From such an inequality we would derive $a_{n+1}=3^{a_{n}}>3^{u}\left(3^{v}\right)^{b_{m}}$. If $3^{v}>100$, then $a_{n+1}>3^{u} b_{m+1}$, and if $3^{u}>u+v$, then we would obtain $a_{n+1}>u+v b_{m+1}$, the same inequality as the one we started with, but with $m+1$ and $n+1$ instead of $m$ and $n$.The inequality $3^{v}>100$ holds for $v=5$, and $3^{u}>u+5$ holds for $u=2$. Thus $a_{n}>2+5 b_{m}$ implies $a_{n+1}>2+5 b_{m+1}$. We have $b_{1}=100, a_{1}=3, a_{2}=27, a_{3}=3^{27}$, and $2+5 b_{1}=502<729=3^{6}$, so $a_{3}>2+5 b_{1}$. We find that $a_{n}>2+5 b_{n-2}$ for all $n \geq 3$. In particular, $a_{n} \geq b_{n-2}$.On the other hand, $a_{n}<b_{m}$ implies $a_{n+1}=3^{a_{n}}<100^{b_{m}}<b_{m+1}$, which combined with $a_{2}<b_{1}$ yields $a_{n}<b_{n-1}$ for all $n \geq 2$. Hence $b_{n-2}<a_{n}<b_{n-1}$, which implies that $m=n-1$, and for $n=100, m=99$. |
Let the sequence $a_{0}, a_{1}, \ldots$ be defined by the equation
$$
1-x^{2}+x^{4}-x^{6}+\cdots=\sum_{n=0}^{\infty} a_{n}(x-3)^{n} \quad(0<x<1) .
$$
Find
$$
\limsup _{n \rightarrow \infty}\left(\left|a_{n}\right|^{\frac{1}{n}}\right) .
$$
| As
$$
1-x^{2}+x^{4}-x^{6}+\cdots=\frac{1}{1+x^{2}}
$$
which has singularities at $\pm i$, the radius of convergence of
$$
\sum_{n=0}^{\infty} a_{n}(x-3)^{n}
$$
is the distance from $3$ to $\pm i, and |3 \mp i|=\sqrt{10}$. We then have
$$
\limsup _{n \rightarrow \infty}\left(\left|a_{n}\right|^{\frac{1}{n}}\right)=\frac{1}{\sqrt{10}} \cdot
$$
|
Find the radius of convergence $R$ of the Taylor series about $z=1$ of the function $f(z)=1 /\left(1+z^{2}+z^{4}+z^{6}+z^{8}+z^{10}\right)$.
| The rational function
$$
f(z)=\frac{1-z^{2}}{1-z^{12}}
$$
has poles at all nonreal twelfth roots of unity (the singularities at $z^{2}=1$ are removable). Thus, the radius of convergence is the distance from 1 to the nearest singularity:
$$
R=|\exp (\pi i / 6)-1|=\sqrt{(\cos (\pi / 6)-1)^{2}+\sin ^{2}(\pi / 6)}=\sqrt{2-\sqrt{3}} .
$$
|
Find the number of permutations $a_{1}, a_{2}, a_{3}, a_{4}, a_{5}, a_{6}$ of the numbers $1,2,3,4,5,6$ that can be transformed into $1,2,3,4,5,6$ through exactly four transpositions (and not fewer). | The condition from the statement implies that any such permutation has exactly two disjoint cycles, say $\left(a_{i_{1}}, \ldots, a_{i_{r}}\right)$ and $\left(a_{i_{r+1}}, \ldots, a_{i_{6}}\right)$. This follows from the fact that in order to transform a cycle of length $r$ into the identity $r-1$, transpositions are needed. Moreover, we can only have $r=5,4$, or 3 .When $r=5$, there are $\left(\begin{array}{l}6 \\ 1\end{array}\right)$ choices for the number that stays unpermuted. There are $(5-1) !$ possible cycles, so in this case we have $6 \times 4 !=144$ possibilities. When $r=4$, there are $\left(\begin{array}{l}6 \\ 4\end{array}\right)$ ways to split the numbers into the two cycles (two cycles are needed and not just one). One cycle is a transposition. There are $(4-1) !=6$ choices for the other. Hence in this case the number is 90 . Note that here exactly four transpositions are needed.Finally, when $r=3$, then there are $\left(\begin{array}{l}6 \\ 3\end{array}\right) \times(3-1) ! \times(3-1) !=40$ cases. Therefore, the answer to the problem is $144+90+40=274$. |
Given the fact that $\int_{-\infty}^{\infty} e^{-x^{2}} d x=\sqrt{\pi}$, evaluate the integral
$$
I=\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} e^{-\left(x^{2}+(y-x)^{2}+y^{2}\right)} d x d y .
$$ | We write
$$
\begin{aligned}
I &=\int_{-\infty}^{\infty} e^{-3 y^{2} / 2}\left(\int_{-\infty}^{\infty} e^{-2 x^{2}+2 x y-y^{2} / 2} d x\right) d y \\n&=\int_{-\infty}^{\infty} e^{-3 y^{2} / 2}\left(\int_{-\infty}^{\infty} e^{-2\left(x-\frac{y}{2}\right)^{2}} d x\right) d y
\end{aligned}
$$
making the substitution $t=\sqrt{2}\left(x-\frac{y}{2}\right), d t=\sqrt{2} d x$ on the inner integral, we get
$$
\begin{aligned}
I &=\frac{1}{\sqrt{2}} \int_{-\infty}^{\infty} e^{-3 y^{2} / 2}\left(\int_{-\infty}^{\infty} e^{-t^{2}} d t\right) d y \\n&=\sqrt{\frac{\pi}{2}} \int_{-\infty}^{\infty} e^{-3 y^{2} / 2} d y
\end{aligned}
$$
now making the substitution: $s=\sqrt{\frac{3}{2}} y, d s=\sqrt{\frac{3}{2}} d y$ we obtain
$$
\begin{aligned}
I &=\sqrt{\frac{\pi}{3}} \int_{-\infty}^{\infty} e^{-s^{2}} d s \\n&=\frac{\pi}{\sqrt{3}} .
\end{aligned}
$$
|
The zeros of the polynomial $P(x)=x^{3}-10 x+11$ are $u$, $v$, and $w$. Determine the value of $\arctan u+\arctan v+\arctan w$. | First solution: Let $\alpha=\arctan u, \beta=\arctan v$, and $\arctan w$. We are required to determine the sum $\alpha+\beta+\gamma$. The addition formula for the tangent of three angles, $$ \tan (\alpha+\beta+\gamma)=\frac{\tan \alpha+\tan \beta+\tan \gamma-\tan \alpha \tan \beta \tan \gamma}{1-(\tan \alpha \tan \beta+\tan \beta \tan \gamma+\tan \alpha \tan \gamma)}, $$implies$$ \tan (\alpha+\beta+\gamma)=\frac{u+v+w-u v w}{1-(u v+v w+u v)} . $$Using Viète's relations,$$ u+v+w=0, \quad u v+v w+u w=-10, \quad u v w=-11, $$we further transform this into $\tan (\alpha+\beta+\gamma)=\frac{11}{1+10}=1$. Therefore, $\alpha+\beta+\gamma=\frac{\pi}{4}+k \pi$, where $k$ is an integer that remains to be determined.From Viète's relations we can see the product of the zeros of the polynomial is negative, so the number of negative zeros is odd. And since the sum of the zeros is 0 , two of them are positive and one is negative. Therefore, one of $\alpha, \beta, \gamma$ lies in the interval $\left(-\frac{\pi}{2}, 0\right)$ and two of them lie in $\left(0, \frac{\pi}{2}\right)$. Hence $k$ must be equal to 0 , and $\arctan u+$ $\arctan v+\arctan w=\frac{\pi}{4}$.Second solution: Because$$ \operatorname{Im} \ln (1+i x)=\arctan x, $$we see that$$ \begin{aligned} \arctan u+\arctan v+\arctan w &=\operatorname{Im} \ln (i P(i))=\operatorname{Im} \ln (11+11 i) \\ &=\arctan 1=\frac{\pi}{4} . \end{aligned} $$ |
Find the maximum of $x_{1}^{3}+\ldots+x_{10}^{3}$ for $x_{1}, \ldots, x_{10} \in[-1,2]$ such that $$ x_{1}+\ldots+x_{10}=10.$$ | Look at the behavior of expression $x_{1}^{3}+x_{2}^{3}$ when we move the variables $x_{1} \leq x_{2}$ together, that is, replace them with $x_{1}+\varepsilon$ and $x_{2}-\varepsilon, 0<\varepsilon \leq \frac{x_{2}-x_{1}}{2}$, and when we move them apart, that is, replace them with $x_{1}-\varepsilon$ and $x_{2}+\varepsilon, \varepsilon>0$. After moving together, the product $x_{1} x_{2}$ is not decreasing, and after moving apart, it is not increasing. Hence the sum of cubes of two variables $$ x_{1}^{3}+x_{2}^{3}=\left(x_{1}+x_{2}\right)\left(\left(x_{1}+x_{2}\right)^{2}-3 x_{1} x_{2}\right) $$ is not decreasing as two variables move apart, if $x_{1}+x_{2} \geq 0$, and as they move together, if $x_{1}+x_{2} \leq 0$. We will move apart couples of variables $x_{i}, x_{j}$, for which $x_{i}+x_{j} \geq 0$, until one of the variables coincides with an end of the segment $[-1 ; 2]$. That can be done while at least one couple of variables with nonnegative sum remains in the interval $(-1 ; 2)$. After that we get one of the next two situations: (1) some variables are equal to 2 or $-1$, and the rest of the variables have negative pairwise sums (and moreover each of the variables in the interval $(-1 ; 2)$ has negative sum with -1), or (2) some variables are equal to 2 or $-1$, and a single variable in the interval $(-1 ; 2)$ has nonnegative sum with $-1$. In the first case, let $x$ be the mean value of all the variables in $[-1 ; 2)$. Then $x<0$ because all pairwise sums are negative. We will move together couples of variables from the interval $[-1 ; 2)$ as described above, until one of the variables in the couple equals $x$. Thus, we can start with an arbitrary collection of points $x_{1}, \ldots, x_{10}$ and use the described above moving together and apart (which preserve the sum of elements and not decrease the sum of cubes) to reach one of the following collections: (1) $k$ variables are at a point $x \in[-1 ; 0]$ and $10-k$ variables are at the point 2 $(k=0, \ldots, 10)$ (2) $k$ variables are at the point $-1$, a single variable equals $x \in[1 ; 2]$, and $9-k$ variables are at the point $2(k=0, \ldots, 9)$. From the conditions $x_{1}+\ldots+x_{10}=10$ and $x \in[-1 ; 0]$ (or $x \in[1 ; 2]$ ), we obtain that either $k=4$ or $k=5$ for collections of the first type and $k=3$ for collections of the second type. It remains to examine the following collections: $$ \begin{gathered} \left(-\frac{1}{2},-\frac{1}{2},-\frac{1}{2},-\frac{1}{2}, 2,2,2,2,2,2\right), \quad(0,0,0,0,0,2,2,2,2,2), \\ (-1,-1,-1,1,2,2,2,2,2,2) . \end{gathered} $$ The maximal value of the sum of cubes is equal to $47.5$ and attained at the first collection. Answer: $47 |
Two airplanes are supposed to park at the same gate of a concourse. The arrival times of the airplanes are independent and randomly distributed throughout the 24 hours of the day. What is the probability that both can park at the gate, provided that the first to arrive will stay for a period of two hours, while the second can wait behind it for a period of one hour? | The set of possible events is modeled by the square $[0, 24] \times[0,24]$. It is, however, better to identify the 0th and the 24th hours, thus obtaining a square with opposite sides identified, an object that in mathematics is called a torus (which is, in fact, the Cartesian product of two circles. The favorable region is outside a band of fixed thickness along the curve $x=y$ on the torus as depicted in Figure 110. On the square model this region is obtained by removing the points $(x, y)$ with $|x-y| \leq 1$ together with those for which $|x-y-1| \leq 1$ and $|x-y+1| \leq 1$. The required probability is the ratio of the area of the favorable region to the area of the square, and is$$ P=\frac{24^{2}-2 \cdot 24}{24^{2}}=\frac{11}{12} \approx 0.917 . $$ |
Find a limit $$ \lim _{n \rightarrow \infty}\left(\int_{0}^{1} e^{x^{2} / n} d x\right)^{n}. $$ | Fix an arbitrary $\varepsilon>0$. Since the function $y=e^{t}$ is convex, for $t \in[0, \varepsilon]$ its graph lies above the tangent $y=1+t$ but under the secant $y=1+a_{\varepsilon} t$, where $a_{\varepsilon}=\frac{1}{\varepsilon}\left(e^{\varepsilon}-1\right)$. Hence for $\frac{1}{n} \leq \varepsilon$ it holds $$ \left(\int_{0}^{1} e^{x^{2} / n} d x\right)^{n} \geq\left(\int_{0}^{1}\left(1+\frac{x^{2}}{n}\right) d x\right)^{n}=\left(1+\frac{1}{3 n}\right)^{n} \rightarrow e^{1 / 3}, \text { as } n \rightarrow \infty, $$ and $$ \left(\int_{0}^{1} e^{x^{2} / n} d x\right)^{n} \leq\left(\int_{0}^{1}\left(1+\frac{a_{\varepsilon} x^{2}}{n}\right) d x\right)^{n}=\left(1+\frac{a_{\varepsilon}}{3 n}\right)^{n} \rightarrow e^{a_{\varepsilon} / 3}, \text { as } n \rightarrow \infty $$ Since $a_{\varepsilon}=\frac{1}{\varepsilon}\left(e^{\varepsilon}-1\right) \rightarrow 1$, as $\varepsilon \rightarrow 0+$, we get $\lim _{n \rightarrow \infty}\left(\int_{0}^{1} e^{x^{2} / n} d x\right)^{n}=e^{1 / 3}$. Answer: $e^{1 / 3}$ |
Let $x_{0}=1$ and
$$
x_{n+1}=\frac{3+2 x_{n}}{3+x_{n}}, \quad n \geqslant 0 .
$$
Find the limit $x_{\infty}=\lim _{n \rightarrow \infty} x_{n}$.
| Obviously, $x_{n} \geqslant 1$ for all $n$; so, if the limit exists, it is $\geqslant 1$, and we can pass to the limit in the recurrence relation to get
$$
x_{\infty}=\frac{3+2 x_{\infty}}{3+x_{\infty}} \text {; }
$$
in other words, $x_{\infty}^{2}+x_{\infty}-3=0$. So $x_{\infty}$ is the positive solution of this quadratic equation, that is, $x_{\infty}=\frac{1}{2}(-1+\sqrt{13})$.
To prove that the limit exists, we use the recurrence relation to get
$$
\begin{aligned}
x_{n+1}-x_{n} &=\frac{3+2 x_{n}}{3+x_{n}}-\frac{3+2 x_{n-1}}{3+x_{n-1}} \\n&=\frac{3\left(x_{n}-x_{n-1}\right)}{\left(3+x_{n}\right)\left(3+x_{n+1}\right)}
\end{aligned}
$$
Hence, $\left|x_{n+1}-x_{n}\right| \leqslant \frac{1}{3}\left|x_{n}-x_{n-1}\right|$. Iteration gives
$$
\left|x_{n+1}-x_{n}\right| \leqslant 3^{-n}\left|x_{1}-x_{0}\right|=\frac{1}{3^{n} \cdot 4} .
$$
The series $\sum_{n=1}^{\infty}\left(x_{n+1}-x_{n}\right)$, of positive terms, is dominated by the convergent series $\frac{1}{4} \sum_{n=1}^{\infty} 3^{-n}$ and so converges. We have $\sum_{n=1}^{\infty}\left(x_{n+1}-x_{n}\right)=\lim x_{n}-x_{1}$ and we are done. |
Consider a sequence $x_{n}=x_{n-1}-x_{n-1}^{2}, n \geq 2, x_{1} \in(0,1)$. Calculate $$ \lim _{n \rightarrow \infty} \frac{n^{2} x_{n}-n}{\ln n} . $$ | By the monotone convergence theorem, it is easy to show that $x_{n} \rightarrow 0$, as $n \rightarrow \infty$. Next, by the Stolz-Cesaro theorem it holds $$ \begin{aligned} n x_{n}=\frac{n}{\frac{1}{x_{n}}} \sim \frac{1}{\frac{1}{x_{n}}-\frac{1}{x_{n-1}}}=\frac{x_{n-1} x_{n}}{x_{n-1}-x_{n}}=\\ \quad=\frac{x_{n-1}\left(x_{n-1}-x_{n-1}^{2}\right)}{x_{n-1}^{2}}=1-x_{n-1} \rightarrow 1, \text { as } n \rightarrow \infty . \end{aligned} $$ Transform the expression in question and use the Stolz-Cesaro theorem once again: $$ \begin{aligned} \frac{n^{2} x_{n}-n}{\ln n}=\frac{n x_{n} \cdot\left(n-\frac{1}{x_{n}}\right)}{\ln n} \sim \frac{n-\frac{1}{x_{n}}}{\ln n} \sim \frac{1-\frac{1}{x_{n}}+\frac{1}{x_{n-1}}}{\ln n-\ln (n-1)}=\frac{1-\frac{1}{x_{n}}+\frac{1}{x_{n-1}}}{-\ln \left(1-\frac{1}{n}\right)} \sim \\ \sim n\left(1-\frac{1}{x_{n-1}-x_{n-1}^{2}}+\frac{1}{x_{n-1}}\right)=-\frac{n x_{n-1}}{1-x_{n-1}} \rightarrow-1, \text { as } n \rightarrow \infty \end{aligned} $$ Answer: $-1 |
Let $\varphi(x, y)$ be a function with continuous second order partial derivatives such that
1. $\varphi_{x x}+\varphi_{y y}+\varphi_{x}=0$ in the punctured plane $\mathbb{R}^{2} \backslash\{0\}$,
2. $r \varphi_{x} \rightarrow \frac{x}{2 \pi r}$ and $r \varphi_{y} \rightarrow \frac{y}{2 \pi r}$ as $r=\sqrt{x^{2}+y^{2}} \rightarrow 0$.
Let $C_{R}$ be the circle $x^{2}+y^{2}=R^{2}$. Evaluate the line integral
$$
\int_{C_{R}} e^{x}\left(-\varphi_{y} d x+\varphi_{x} d y\right)
$$
| Consider the annular region $\mathcal{A}$ between the circles of radius $r$ and $R$, then by Green's Theorem
$$
\begin{aligned}
&\int_{R} e^{x}\left(-\varphi_{y} d x+\varphi_{x} d y\right)-\int_{r} e^{y}\left(-\varphi_{y} d x+\varphi_{x} d y\right)= \\n&=\int_{\partial \mathcal{A}} e^{x}\left(-\varphi_{x} d x+\varphi_{x} d y\right) \\n&=\iint_{\mathcal{A}} d\left(e^{x}\left(-\varphi_{y} d x+\varphi_{x} d y\right)\right) \\n&=\iint_{\mathcal{A}}-e^{x} \varphi_{y y} d y \wedge d x+\left(e^{x} \varphi_{x}+e^{x} \varphi_{x x}\right) d x \wedge d y \\n&=\iint_{\mathcal{A}} e^{x}\left(\varphi_{x x}+e^{x} \varphi_{x x}+\varphi_{x}\right) d x \wedge d y=0
\end{aligned}
$$
showing that the integral does not depend on the radius $r$. Now, parametrizing the circle of radius $r$
$$
\begin{aligned}
&\int_{C_{r}} e^{x}\left(-\varphi_{y} d x+\varphi_{x} d y\right)= \\n&=\int_{0}^{2 \pi} e^{r \cos \theta}\left(\varphi_{y}(r \cos \theta, r \sin \theta) \sin \theta+\varphi_{x}(r \cos \theta, r \sin \theta) \cos \theta\right) r d \theta
\end{aligned}
$$
but when $r \rightarrow 0$ the integrand converges uniformly to
$$
\frac{\sin \theta}{2 \pi} \cdot \sin \theta+\frac{\cos \theta}{2 \pi} \cdot \cos \theta=\frac{1}{2 \pi}
$$
so the integral approaches $1$ when $r \rightarrow 0$ and that is the value of the integral.
|
Compute$$ \lim _{n \rightarrow \infty} \sum_{k=1}^{n}\left(\frac{k}{n^{2}}\right)^{\frac{k}{n^{2}}+1} . $$ | We use the fact that$$ \lim _{x \rightarrow 0^{+}} x^{x}=1 . $$As a consequence, we have$$ \lim _{x \rightarrow 0^{+}} \frac{x^{x+1}}{x}=1 . $$For our problem, let $\epsilon>0$ be a fixed small positive number. There exists $n(\epsilon)$ such that for any integer $n \geq n(\epsilon)$,$$ 1-\epsilon<\frac{\left(\frac{k}{n^{2}}\right)^{\frac{k}{n^{2}}+1}}{\frac{k}{n^{2}}}<1+\epsilon, \quad k=1,2, \ldots, n . $$From this, using properties of ratios, we obtain$$ 1-\epsilon<\frac{\sum_{k=1}^{n}\left(\frac{k}{n^{2}}\right)^{\frac{k}{n^{2}}+1}}{\sum_{k=1}^{n} \frac{k}{n^{2}}}<1+\epsilon, \quad \text { for } n \geq n(\epsilon) . $$Knowing that $\sum_{k=1}^{n} k=\frac{n(n+1)}{2}$, this implies$$ (1-\epsilon) \frac{n+1}{2 n}<\sum_{k=1}^{n}\left(\frac{k}{n^{2}}\right)^{\frac{k}{n^{2}}+1}<(1+\epsilon) \frac{n+1}{2 n}, \quad \text { for } n \geq n(\epsilon) . $$It follows that$$ \lim _{n \rightarrow \infty} \sum_{k-1}^{n}\left(\frac{k}{n^{2}}\right)^{\frac{k}{n^{2}}+1}=\frac{1}{2} . $$ |
A sequence $\left\{x_{n}, n \geq 1\right\}$ satisfies $x_{n+1}=x_{n}+e^{-x_{n}}, n \geq 1$, and $x_{1}=1$. Find $\lim _{n \rightarrow \infty} \frac{x_{n}}{\ln n}$. | The sequence $\left\{x_{n}\right\}$ is increasing, hence it has a finite or infinite limit. If $\lim _{n \rightarrow \infty} x_{n}=$ $x<+\infty$, then $x=x+e^{-x}$, a contradiction. Therefore, $\lim _{n \rightarrow \infty} x_{n}=+\infty$. We use twice the Stolz-Cesaro theorem and obtain $$ \lim _{n \rightarrow \infty} \frac{x_{n}}{\ln n}=\lim _{n \rightarrow \infty} \frac{x_{n+1}-x_{n}}{\ln (n+1)-\ln n}=\lim _{n \rightarrow \infty} \frac{e^{-x_{n}}}{\ln \left(1+\frac{1}{n}\right)}=\lim _{n \rightarrow \infty} \frac{n}{e^{x_{n}}}=\lim _{n \rightarrow \infty} \frac{1}{e^{x_{n+1}}-e^{x_{n}}} . $$ It remains to notice that $$ \lim _{n \rightarrow \infty}\left(e^{x_{n+1}}-e^{x_{n}}\right)=\lim _{n \rightarrow \infty}\left(e^{x_{n}+e^{-x_{n}}}-e^{x_{n}}\right)=\lim _{n \rightarrow \infty} \frac{e^{e^{-x_{n}}}-1}{e^{-x_{n}}}=\lim _{t \rightarrow 0} \frac{e^{t}-1}{t}=1 . $$ Answer: 1 |
Find the limit $$ \lim _{N \rightarrow \infty} \sqrt{N}\left(1-\max _{1 \leq n \leq N}\{\sqrt{n}\}\right), $$ where $\{x\}$ denotes the fractional part of $x$. | Estimate $\max _{1 \leq n \leq N}\{\sqrt{n}\}$. For $k^{2} \leq n<(k+1)^{2}$ it holds $$ \{\sqrt{n}\}=\sqrt{n}-k \leq \sqrt{(k+1)^{2}-1}-k, $$ where equality is attained for $n=(k+1)^{2}-1$. If $k<l$ then $$ \begin{aligned} 1 &-\left\{\sqrt{(k+1)^{2}-1}\right\}=k+1-\sqrt{(k+1)^{2}-1}=\\ &=\frac{1}{k+1+\sqrt{(k+1)^{2}-1}}>\frac{1}{l+1+\sqrt{(l+1)^{2}-1}}=1-\left\{\sqrt{(l+1)^{2}-1}\right\}, \end{aligned} $$ hence $\left\{\sqrt{(k+1)^{2}-1}\right\}$ is increasing in $k$. Thus, for $N=m^{2}-1$ we get $$ \max _{1 \leq n \leq N}\{\sqrt{n}\}=\left\{\sqrt{m^{2}-1}\right\}=\sqrt{m^{2}-1}-m+1 . $$ Now consider $m^{2} \leq N \leq(m+1)^{2}-1$. Since $\max _{1 \leq n \leq N}\{\sqrt{n}\}$ is increasing in $N$, it holds $$ \sqrt{m^{2}-1}-m+1 \leq \max _{1 \leq n \leq N}\{\sqrt{n}\} \leq \sqrt{(m+1)^{2}-1}-m . $$ Therefore, $$ 1+m-\sqrt{(m+1)^{2}-1} \leq 1-\max _{1 \leq n \leq N}\{\sqrt{n}\} \leq m-\sqrt{m^{2}-1} $$ Also we have $m \leq \sqrt{N}<m+1$, so $$ \begin{aligned} &\frac{m}{m+1} \cdot(m+1)\left(m+1-\sqrt{(m+1)^{2}-1}\right) \leq \\ &\quad \leq \sqrt{N}\left(1-\max _{1 \leq n \leq N}\{\sqrt{n}\}\right)<\frac{m+1}{m} \cdot m\left(m-\sqrt{m^{2}-1}\right) . \end{aligned} $$ Since $m\left(m-\sqrt{m^{2}-1}\right)=\frac{m}{m+\sqrt{m^{2}-1}} \rightarrow \frac{1}{2}$ and $\frac{m}{m+1} \rightarrow 1$, as $m \rightarrow \infty$, we have $\lim _{N \rightarrow \infty} \sqrt{N}\left(1-\max _{1 \leq n \leq N}\{\sqrt{n}\}\right)=\frac{1}{2}$ Answer: $\frac{1}{2} |
Compute
$$
L=\lim _{n \rightarrow \infty}\left(\frac{n^{n}}{n !}\right)^{1 / n} .
$$
| Let $p_{1}=1, p_{2}=(2 / 1)^{2}, p_{3}=(3 / 2)^{3}, \ldots, p_{n}=(n /(n-1))^{n}$. Then
$$
\frac{p_{1} p_{2} \cdots p_{n}}{n}=\frac{n^{n}}{n !},
$$
and since $p_{n} \rightarrow e$, we have $\lim \left(n^{n} / n !\right)^{1 / n}=e$ as well (using the fact that $\left.\lim n^{1 / n}=1\right)$.
|
Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be the function of period $2 \pi$ such that $f(x)=x^{3}$ for $-\pi \leqslant x<\pi$.
The Fourier series for $f$ has the form $\sum_{1}^{\infty} b_{n} \sin n x$. Find
$$
\sum_{n=1}^{\infty} b_{n}^{2}.
$$
| Using Parseval's Identity
$$
\frac{1}{2} a_{0}^{2}+\sum_{n=1}^{\infty}\left(a_{n}^{2}+b_{n}^{2}\right)=\frac{1}{\pi} \int_{-\pi}^{\pi} f^{2}(x) d x
$$
and the fact that all $a_{n}=0$,
$$
\sum_{n=1}^{\infty} b^{2}=\frac{1}{\pi} \int_{-\pi}^{\pi} x^{6} d x=\frac{2}{7} \pi^{6} .
$$
|
For a real $2 \times 2$ matrix
$$
X=\left(\begin{array}{ll}
x & y \\
z & t
\end{array}\right),
$$
let $\|X\|=x^{2}+y^{2}+z^{2}+t^{2}$, and define a metric by $d(X, Y)=\|X-Y\|$. Let $\Sigma=\{X \mid \operatorname{det}(X)=0\}$. Let
$$
A=\left(\begin{array}{ll}
1 & 0 \\
0 & 2
\end{array}\right) \text {. }
$$
Find the minimum distance from $A$ to $\Sigma$. | For $X \in \Sigma$, we have
$$
\begin{aligned}
\|A-X\|^{2} &=(1-x)^{2}+y^{2}+z^{2}+(2-t)^{2} \\n&=y^{2}+z^{2}+1-2 x+x^{2}+(2-t)^{2} \\n& \geqslant \pm 2 y z+1-2 x+2|x|(2-t) \\n&=4|x|-2 x+2(\pm y z-|x| t)+1
\end{aligned}
$$
We can choose the sign, so $\pm y z-|x| t=0$ because $\operatorname{det} X=0$. As $4|x|-2 x \geqslant 0$, we have $\|A-X\| \geqslant 1$ with equality when $4|x|-2 x=0,|x|=2-t, y=\pm z$,
|
Find the probability that in the process of repeatedly flipping an unbiased coin, one will encounter a run of 5 heads before one encounters a run of 2 tails. | We call a successful string a sequence of $H$ 's and $T$ 's in which $H H H H H$ appears before $T T$ does. Each successful string must belong to one of the following three types:(i) those that begin with $T$, followed by a successful string that begins with $H$;(ii) those that begin with $H, H H, H H H$, or $H H H H$, followed by a successful string that begins with $T$;(iii) the string $H H H H H$.Let $P_{H}$ denote the probability of obtaining a successful string that begins with $H$, and let $P_{T}$ denote the probability of obtaining a successful string that begins with $T$. Then$$ \begin{aligned} P_{T} &=\frac{1}{2} P_{H}, \\ P_{H} &=\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}\right) P_{T}+\frac{1}{32} . \end{aligned} $$Solving these equations simultaneously, we find that$$ P_{H}=\frac{1}{17} \quad \text { and } \quad P_{T}=\frac{1}{34} . $$Hence the probability of obtaining five heads before obtaining two tails is $\frac{3}{34}$. |
Compute the integral $\iint_{D} x d x d y$, where$$ D=\left\{(x, y) \in \mathbb{R}^{2} \mid x \geq 0,1 \leq x y \leq 2,1 \leq \frac{y}{x} \leq 2\right\} . $$ | The domain is bounded by the hyperbolas $x y=1, x y=2$ and the lines $y=x$ and $y=2 x$. This domain can mapped into a rectangle by the transformation$$ T: \quad u=x y, \quad v=\frac{y}{x} . $$Thus it is natural to consider the change of coordinates$$ T^{-1}: \quad x=\sqrt{\frac{u}{v}}, \quad y=\sqrt{u v} . $$The domain becomes the rectangle $D^{*}=\left\{(u, v) \in \mathbb{R}^{2} \mid 1 \leq u \leq 2,1 \leq v \leq 2\right\}$. The Jacobian of $T^{-1}$ is $\frac{1}{2 v} \neq 0$. The integral becomes$$ \int_{1}^{2} \int_{1}^{2} \sqrt{\frac{u}{v}} \frac{1}{2 v} d u d v=\frac{1}{2} \int_{1}^{2} u^{1 / 2} d u \int_{1}^{2} v^{-3 / 2} d v=\frac{1}{3}(5 \sqrt{2}-6) . $$ |
A husband and wife agree to meet at a street corner between 4 and 5 o'clock to go shopping together. The one who arrives first will await the other for 15 minutes, and then leave. What is the probability that the two meet within the given time interval, assuming that they can arrive at any time with the same probability? | Denote by $x$, respectively, $y$, the fraction of the hour when the husband, respectively, wife, arrive. The configuration space is the square $$ D=\{(x, y) \mid 0 \leq x \leq 1,0 \leq y \leq 1\} . $$In order for the two people to meet, their arrival time must lie inside the region$$ D_{f}=\left\{(x, y)|| x-y \mid \leq \frac{1}{4}\right\} . $$The desired probability is the ratio of the area of this region to the area of the square.The complement of the region consists of two isosceles right triangles with legs equal to $\frac{3}{4}$, and hence of areas $\frac{1}{2}\left(\frac{3}{4}\right)^{2}$. We obtain for the desired probability$$ 1-2 \cdot \frac{1}{2} \cdot\left(\frac{3}{4}\right)^{2}=\frac{7}{16} \approx 0.44 . $$ |
Evaluate the product$$ \left(1-\cot 1^{\circ}\right)\left(1-\cot 2^{\circ}\right) \cdots\left(1-\cot 44^{\circ}\right) \text {. } $$ | We have$$ \begin{aligned} \left(1-\cot 1^{\circ}\right)\left(1-\cot 2^{\circ}\right) \cdots\left(1-\cot 44^{\circ}\right) \\ &=\left(1-\frac{\cos 1^{\circ}}{\sin 1^{\circ}}\right)\left(1-\frac{\cos 2^{\circ}}{\sin 2^{\circ}}\right) \cdots\left(1-\frac{\cos 44^{\circ}}{\sin 44^{\circ}}\right) \\ &=\frac{\left(\sin 1^{\circ}-\cos 1^{\circ}\right)\left(\sin 2^{\circ}-\cos 2^{\circ}\right) \cdots\left(\sin 44^{\circ}-\cos 44^{\circ}\right)}{\sin 1^{\circ} \sin 2^{\circ} \cdots \sin 44^{\circ}} \end{aligned} $$Using the identity $\sin a-\cos a=\sqrt{2} \sin \left(a-45^{\circ}\right)$ in the numerators, we transform this further into$$ \begin{gathered} \frac{\sqrt{2} \sin \left(1^{\circ}-45^{\circ}\right) \cdot \sqrt{2} \sin \left(2^{\circ}-45^{\circ}\right) \cdots \sqrt{2} \sin \left(44^{\circ}-45^{\circ}\right)}{\sin 1^{\circ} \sin 2^{\circ} \cdots \sin 44^{\circ}} \\ =\frac{(\sqrt{2})^{44}(-1)^{44} \sin 44^{\circ} \sin 43^{\circ} \cdots \sin 1^{\circ}}{\sin 44^{\circ} \sin 43^{\circ} \cdots \sin 1^{\circ}} . \end{gathered} $$After cancellations, we obtain $2^{22}$. |
Find the number of roots of
$$
z^{7}-4 z^{3}-11=0
$$
which lie between the two circles $|z|=1$ and $|z|=2$.
| Let $p(z)=z^{7}-4 z^{3}-11$. For $z$ in the unit circle, we have
$$
|p(z)-11|=\left|z^{7}-4 z^{3}\right| \leqslant 5<11
$$
so, by Rouch\u00e9's Theorem, the given polynomial has no zeros in the unit disc. For $|z|=2$,
$$
\left|p(z)-z^{7}\right|=\left|4 z^{3}+11\right| \leqslant 43<128=\left|z^{7}\right|
$$
so there are seven zeros inside the disc $\{z|| z \mid<2\}$ and they are all between the two given circles.
|
Let $\mathcal{S}=\left\{(x, y, z) \in \mathbb{R}^{3} \mid x^{2}+y^{2}+z^{2}=1\right\}$ denote the unit sphere in $\mathbb{R}^{3}$. Evaluate the surface integral over $\mathcal{S}$ :
$$
\iint_{\mathcal{S}}\left(x^{2}+y+z\right) d A .
$$
| Using the change of variables
$$
\begin{cases}x=\sin \varphi \cos \theta & 0<\theta<2 \pi \\ y=\sin \varphi \sin \theta & 0<\varphi<\pi \\ z=\cos \varphi & \end{cases}
$$
we have
$$
d A=\sin \varphi d \theta d \varphi
$$
and
$$
\iint_{\mathcal{S}}\left(x^{2}+y+z\right) d A=\int_{0}^{\pi} \int_{0}^{2 \pi}\left(\sin ^{2} \varphi \cos ^{2} \theta+\sin \varphi \sin \theta+\cos \varphi\right) \sin \varphi d \theta d \varphi .
$$
Breaking the integral in three terms, we get
$$
\begin{aligned}
\int_{0}^{\pi} \int_{0}^{2 \pi} \sin \varphi \cos \varphi d \theta d \varphi=2 \pi \cdot \frac{1}{2} \int_{0}^{\pi} \sin 2 \varphi d \varphi=0 \\n\int_{0}^{\pi} \int_{0}^{2 \pi} \sin ^{2} \varphi \sin \theta d \theta d \varphi=\left(\int_{0}^{\pi} \sin ^{2} \varphi d \varphi\right) \int_{0}^{2 \pi} \sin \theta d \theta=0 \\n\int_{0}^{\pi} \int_{0}^{2 \pi} \sin ^{3} \varphi \cos ^{2} \theta d \theta d \varphi &=\left(\int_{0}^{\pi} \sin ^{3} \varphi d \varphi\right)\left(\int_{0}^{2 \pi} \cos ^{2} \theta d \theta\right) \\n&=\int_{0}^{\pi} \frac{1}{4}\left(3 \sin \varphi-\sin ^{3} \varphi\right) d \varphi \int_{0}^{2 \pi} \cos ^{2} \theta d \theta \\n&=\frac{1}{4}\left(-3 \cos \varphi+\left.\frac{1}{3} \cos ^{3} \varphi\right|_{0} ^{\pi}\right)\left(\int_{0}^{2 \pi} \frac{1+\cos 2 \theta}{2} d \theta\right) \\n&=\frac{1}{4}\left(-3(-2)+\frac{1}{3}(-2)\right) \pi \\n&=\frac{1}{4}\left(6-\frac{2}{3}\right) \pi=\frac{4}{3} \pi .
\end{aligned}
$$
Therefore,
$$
\iint_{\mathcal{S}}\left(x^{2}+y+z\right) d A=\frac{4}{3} \pi
$$
|
Find the sum of the series $$ \sum_{n=0}^{\infty} \frac{1}{n ! 2^{n}} \cos \frac{\pi n-1}{2} . $$ | For all $x \in \mathbb{R}$, it holds $$ \begin{aligned} &\sum_{n=0}^{\infty} \frac{\cos \left(\frac{\pi n}{2}-x\right)}{n !} x^{n}=\cos x \sum_{n=0}^{\infty} \frac{\cos \frac{\pi n}{2}}{n !} x^{n}+\sin x \sum_{n=0}^{\infty} \frac{\sin \frac{\pi n}{2}}{n !} x^{n}= \\ &=\cos x \sum_{k=0}^{\infty} \frac{(-1)^{k}}{(2 k) !} x^{2 k}+\sin x \sum_{k=0}^{\infty} \frac{(-1)^{k}}{(2 k+1) !} x^{2 k+1}=\cos ^{2} x+\sin ^{2} x=1 . \end{aligned} $$ In particular for $x=\frac{1}{2}$ we obtain the sum of the series in question. Answer: 1 |
What is the probability that a uniformly random permutation of the first $n$ positive integers has the numbers 1 and 2 within the same cycle? | The total number of permutations is of course $n$ !. We will count instead the number of permutations for which 1 and 2 lie in different cycles.If the cycle that contains 1 has length $k$, we can choose the other $k-1$ elements in $\left(\begin{array}{c}n-2 \\ k-1\end{array}\right)$ ways from the set $\{3,4, \ldots, n\}$. There exist $(k-1)$ ! circular permutations of these elements, and $(n-k)$ ! permutations of the remaining $n-k$ elements. Hence the total number of permutations for which 1 and 2 belong to different cycles is equal to$$ \sum_{k=1}^{n-1}\left(\begin{array}{l} n-2 \\ k-1 \end{array}\right)(k-1) !(n-k) !=(n-2) ! \sum_{k=1}^{n-1}(n-k)=(n-2) ! \frac{n(n-1)}{2}=\frac{n !}{2} . $$It follows that exactly half of all permutations contain 1 and 2 in different cycles, and so half contain 1 and 2 in the same cycle. The probability is $\frac{1}{2}$. |
Find$$ \int_{0}^{1} \frac{\ln (1+x)}{1+x^{2}} d x .$$ | With the substitution $\arctan x=t$ the integral takes the form$$ I=\int_{0}^{\frac{\pi}{4}} \ln (1+\tan t) d t . $$This we already computed in the previous problem. ("Happiness is longing for repetition," says M. Kundera.) So the answer to the problem is $\frac{\pi}{8} \ln 2$. |
Let $C$ be the unit circle $x^{2}+y^{2}=1$. A point $p$ is chosen randomly on the circumference of $C$ and another point $q$ is chosen randomly from the interior of $C$ (these points are chosen independently and uniformly over their domains). Let $R$ be the rectangle with sides parallel to the $x$-and $y$-axes with diagonal $p q$. What is the probability that no point of $R$ lies outside of $C$? | The pair $(p, q)$ is chosen randomly from the three-dimensional domain $C \times$ int $C$, which has a total volume of $2 \pi^{2}$ (here int $C$ denotes the interior of $C$ ). For a fixed $p$, the locus of points $q$ for which $R$ does not have points outside of $C$ is the rectangle whose diagonal is the diameter through $p$ and whose sides are parallel to the coordinate axes (Figure 112). If the coordinates of $p$ are $(\cos \theta, \sin \theta)$, then the area of the rectangle is $2|\sin 2 \theta|$Figure 112The volume of the favorable region is therefore$$ V=\int_{0}^{2 \pi} 2|\sin 2 \theta| d \theta=4 \int_{0}^{\pi / 2} 2 \sin 2 \theta d \theta=8 . $$Hence the probability is$$ P=\frac{8}{2 \pi^{2}}=\frac{4}{\pi^{2}} \approx 0.405 . $$ |
Let $M$ be a subset of $\{1,2,3, \ldots, 15\}$ such that the product of any three distinct elements of $M$ is not a square. Determine the maximum number of elements in $M$. | Note that the product of the three elements in each of the sets $\{1,4,9\},\{2,6,12\}$, $\{3,5,15\}$, and $\{7,8,14\}$ is a square. Hence none of these sets is a subset of $M$. Because they are disjoint, it follows that $M$ has at most $15-4=11$ elements.Since 10 is not an element of the aforementioned sets, if $10 \notin M$, then $M$ has at most 10 elements. Suppose $10 \in M$. Then none of $\{2,5\},\{6,15\},\{1,4,9\}$, and $\{7,8,14\}$ is a subset of $M$. If $\{3,12\} \not \subset M$, it follows again that $M$ has at most 10 elements. If $\{3,12\} \subset M$, then none of $\{1\},\{4\},\{9\},\{2,6\},\{5,15\}$, and $\{7,8,14\}$ is a subset of $M$, and then $M$ has at most 9 elements. We conclude that $M$ has at most 10 elements in any case.Finally, it is easy to verify that the subset$$ M=\{1,4,5,6,7,10,11,12,13,14\} $$has the desired property. Hence the maximum number of elements in $M$ is 10 . |
Compute the limit
$$
\lim _{z \rightarrow 0}\left((\tan z)^{-2}-z^{-2}\right)
$$
where $z$ is a complex variable.
| As
$$
(\tan z)^{-2}-z^{-2}=\frac{z^{2}-(\tan z)^{2}}{z^{2}(\tan z)^{2}}
$$
the Maclaurin expansion of the numerator has no terms of degree up to 3 , whereas the expansion of the denominator starts with $z^{4}$, therefore, the limit is finite.
As
$$
\tan z=z+\frac{1}{3} z^{3}+o\left(z^{4}\right) \quad(z \rightarrow 0)
$$
we have
$$
(\tan z)^{-2}-z^{-2}=\frac{z^{2}-z^{2}-\frac{2}{3} z^{4}+o\left(z^{4}\right)}{z^{4}+o\left(z^{4}\right)} \quad(z \rightarrow 0)
$$
so the limit at 0 is $-2 / 3$.
|
Let $f(x)=\frac{1}{4}+x-x^{2}$. For any real number $x$, define a sequence $\left(x_{n}\right)$ by $x_{0}=x$ and $x_{n+1}=f\left(x_{n}\right)$. If the sequence converges, let $x_{\infty}$ denote the limit. For $x=0$, find $x_{\infty}=\lambda$.
| We have
$$
f(x)=\frac{1}{2}-\left(x-\frac{1}{2}\right)^{2}
$$
so $x_{n}$ is bounded by $1 / 2$ and, by the Induction Principle, nondecreasing. Let $\lambda$ be its limit. Then
$$
\lambda=\frac{1}{2}-\left(\lambda-\frac{1}{2}\right)^{2}
$$
and, as the sequence takes only positive values,
$$
\lambda=\frac{1}{2} \text {. }
$$
|
The clock-face is a disk of radius 1 . The hour-hand is a disk of radius $1 / 2$ is internally tangent to the circle of the clock-face, and the minute-hand is a line segment of length 1 . Find the area of the figure formed by all intersections of the hands in 12 hours (i.e., in one full turn of the hour-hand). | Let the minute hand makes an angle $\varphi \in[0 ; 2 \pi)$ with the vertical position. At that moment, the hour hand makes one of the angles $$ \psi_{k}=\frac{\varphi+2 \pi k}{12}(\bmod 2 \pi), k \in \mathbb{Z}, $$ with the vertical position. The hands intersect along a segment of length $$ \max \left(0, \cos \left(\psi_{k}-\varphi\right)\right), $$ hence the area in question equals $\frac{1}{2} \int_{0}^{2 \pi} \rho^{2}(\varphi) d \varphi$, where $$ \rho(\varphi)=\max \left(0, \max _{k} \cos \left(\psi_{k}-\varphi\right)\right), \varphi \in \mathbb{R} . $$ Notice that $$ \rho(\varphi)=\cos \left(\min _{k}\left|\frac{\varphi+2 \pi k}{12}-\varphi\right|\right)=\cos \left(\min _{k}\left|\frac{11 \varphi}{12}-\frac{\pi k}{6}\right|\right) . $$ This function is periodic with period of $\frac{2 \pi}{11}$. Since for $\varphi \in\left[-\frac{\pi}{11}, \frac{\pi}{11}\right]$ it holds $\rho(\varphi)=$ $\cos \left(\frac{11 \varphi}{12}\right)$, the area of the figure formed by intersections of hands equals $$ \begin{gathered} \frac{1}{2} \int_{0}^{2 \pi} \rho^{2}(\varphi) d \varphi=\frac{11}{2} \int_{-\frac{\pi}{11}}^{\frac{\pi}{11}} \rho^{2}(\varphi) d \varphi=\frac{11}{2} \int_{-\frac{\pi}{11}}^{\frac{\pi}{11}} \cos ^{2}\left(\frac{11 \varphi}{12}\right) d \varphi= \\ =\frac{11}{4} \int_{-\frac{\pi}{11}}^{\frac{\pi}{11}}\left(1+\cos \left(\frac{11 \varphi}{6}\right)\right) d \varphi=\left.\left(\frac{11 \varphi}{4}+\frac{3}{2} \sin \left(\frac{11 \varphi}{6}\right)\right)\right|_{-\frac{\pi}{11}} ^{\frac{\pi}{11}}=\frac{\pi+3}{2} . \end{gathered} $$ Answer: $\frac{\pi+3}{2} |
Evaluate
$$
\iint_{\mathcal{A}} e^{-x^{2}-y^{2}} d x d y,
$$
where $\mathcal{A}=\left\{(x, y) \in \mathbb{R}^{2} \mid x^{2}+y^{2} \leqslant 1\right\}$.
| Using polar coordinates, we have
$$
\begin{aligned}
\iint_{\mathcal{A}} e^{-x^{2}-y^{2}} d x d y &=\int_{0}^{2 \pi} \int_{0}^{1} \rho e^{-\rho^{2}} d \rho d \theta \\n&=-\frac{1}{2} \int_{0}^{2 \pi} \int_{0}^{1}-2 \rho e^{-\rho^{2}} d \rho d \theta \\n&=-\frac{1}{2} \int_{0}^{2 \pi}\left(e^{-1}-1\right) \\n&=\pi\left(e^{-1}-1\right) .
\end{aligned}
$$
|
Let $S_{9}$ denote the group of permutations of nine objects and let $A_{9}$ be the subgroup consisting of all even permutations. Denote by $1 \in S_{9}$ the identity permutation. Determine the minimum of all positive integers $m$ such that every $\sigma \in A_{9}$ satisfies $\sigma^{m}=1$.
| The order of a $k$-cycle is $k$, so the smallest $m$ which simultaneously annihilates all 9-cycles, 8-cycles, 7-cycles, and 5-cycles is $2^{3} \cdot 3^{2} \cdot 5 \cdot 7= 2520$. Any $n$-cycle, $n \leqslant 9$, raised to this power is annihilated, so $n=2520$.
To compute $n$ for $A 9$, note that an 8-cycle is an odd permutation, so no 8-cycles are in $A_{9}$. Therefore, $n$ need only annihilate 4-cycles (since a 4-cycle times a transposition is in A9), 9-cycles, 7-cycles, and 5-cycles. Thus, $n=2520 / 2= 1260$.
|
A bag contains 1993 red balls and 1993 black balls. We remove two uniformly random balls at a time repeatedly and
(i) discard them if they are of the same color,
(ii) discard the black ball and return to the bag the red ball if they are of different colors.
The process ends when there are less than two balls in the bag. What is the probability that this process will terminate with one red ball in the bag? | First, observe that since at least one ball is removed during each stage, the process will eventually terminate, leaving no ball or one ball in the bag. Because red balls are removed 2 at a time and since we start with an odd number of red balls, the number of red balls in the bag at any time is odd. Hence the process will always leave red balls in the bag, and so it must terminate with exactly one red ball. The probability we are computing is therefore 1 . |
Let $T_{1}, T_{2}, T_{3}$ be points on a parabola, and $t_{1}, t_{2}, t_{3}$ the tangents to the parabola at these points. Compute the ratio of the area of triangle $T_{1} T_{2} T_{3}$ to the area of the triangle determined by the tangents. | Choose a Cartesian system of coordinates such that the equation of the parabola is $y^{2}=4 p x$. The coordinates of the three points are $T_{i}\left(4 p \alpha_{i}^{2}, 4 p \alpha_{i}\right)$, for appropriately chosen $\alpha_{i}, i=1,2,3$. Recall that the equation of the tangent to the parabola at a point $\left(x_{0}, y_{0}\right)$ is $y y_{0}=2 p\left(x+x_{0}\right)$. In our situation the three tangents are given by$$ 2 \alpha_{i} y=x+4 p \alpha_{i}^{2}, \quad i=1,2,3 . $$If $P_{i j}$ is the intersection of $t_{i}$ and $t_{j}$, then its coordinates are $\left(4 p \alpha_{i} \alpha_{j}, 2 p\left(\alpha_{i}+\alpha_{j}\right)\right)$. The area of triangle $T_{1} T_{2} T_{3}$ is given by a Vandermonde determinant: $$ \pm \frac{1}{2}\left|\begin{array}{lll} 4 p \alpha_{1}^{2} & 4 p \alpha_{1} & 1 \\ 4 p \alpha_{2}^{2} & 4 p \alpha_{2} & 1 \\ 4 p \alpha_{3}^{2} & 4 p \alpha_{3} & 1 \end{array}\right|=\pm 8 p^{2}\left|\begin{array}{lll} \alpha_{1}^{2} & \alpha_{1} & 1 \\ \alpha_{2}^{2} & \alpha_{2} & 1 \\ \alpha_{3}^{2} & \alpha_{3} & 1 \end{array}\right|=8 p^{2}\left|\left(\alpha_{1}-\alpha_{2}\right)\left(\alpha_{1}-\alpha_{3}\right)\left(\alpha_{2}-\alpha_{3}\right)\right| . $$The area of the triangle $P_{12} P_{23} P_{31}$ is given by$$ \begin{aligned} & \pm \frac{1}{2}\left|\begin{array}{lll}4 p \alpha_{1} \alpha_{2} & 2 p\left(\alpha_{1}+\alpha_{2}\right) & 1 \\4 p \alpha_{2} \alpha_{3} & 2 p\left(\alpha_{2}+\alpha_{3}\right) & 1 \\4 p \alpha_{3} \alpha_{1} & 2 p\left(\alpha_{3}+\alpha_{1}\right) & 1\end{array}\right| \\ & =\pm 4 p^{2}\left|\begin{array}{ll}\alpha_{1} \alpha_{2}\left(\alpha_{1}+\alpha_{2}\right) & 1 \\\alpha_{2} \alpha_{3}\left(\alpha_{2}+\alpha_{3}\right) & 1 \\\alpha_{3} \alpha_{1}\left(\alpha_{3}+\alpha_{1}\right) & 1\end{array}\right|=\pm 4 p^{2}\left|\begin{array}{rr}\left(\alpha_{1}-\alpha_{3}\right) \alpha_{2}\left(\alpha_{1}-\alpha_{3}\right) & 0 \\\left(\alpha_{2}-\alpha_{1}\right) \alpha_{3}\left(\alpha_{2}-\alpha_{1}\right) & 0 \\\alpha_{3} \alpha_{1}\left(\alpha_{3}+\alpha_{1}\right) & 1\end{array}\right| \\ & =4 p^{2}\left|\left(\alpha_{1}-\alpha_{3}\right)\left(\alpha_{1}-\alpha_{2}\right)\left(\alpha_{2}-\alpha_{3}\right)\right| \text {. } \end{aligned} $$We conclude that the ratio of the two areas is 2 , regardless of the location of the three points or the shape of the parabola. |
For each continuous function $f:[0,1] \rightarrow \mathbb{R}$, we define $I(f)=\int_{0}^{1} x^{2} f(x) d x$ and $J(f)=\int_{0}^{1} x(f(x))^{2} d x$. Find the maximum value of $I(f)-J(f)$ over all such functions $f$. | We change this into a minimum problem, and then relate the latter to an inequality of the form $x \geq 0$. Completing the square, we see that$$ \left.x(f(x))^{2}-x^{2} f(x)=\sqrt{x} f(x)\right)^{2}-2 \sqrt{x} f(x) \frac{x^{\frac{3}{2}}}{2}=\left(\sqrt{x} f(x)-\frac{x^{\frac{3}{2}}}{2}\right)^{2}-\frac{x^{3}}{4} $$Hence, indeed,$$ J(f)-I(f)=\int_{0}^{1}\left(\sqrt{x} f(x)-\frac{x^{\frac{3}{2}}}{2}\right)^{2} d x-\int_{0}^{1} \frac{x^{3}}{4} d x \geq-\frac{1}{16} $$It follows that $I(f)-J(f) \leq \frac{1}{16}$ for all $f$. The equality holds, for example, for $f:[0,1] \rightarrow \mathbb{R}, f(x)=\frac{x}{2}$. We conclude that$$ \max _{f \in \mathcal{C}^{0}([0,1])}(I(f)-J(f))=\frac{1}{16} . $$ |
Compute the integral$$ \iint_{D} \frac{d x d y}{\left(x^{2}+y^{2}\right)^{2}}, $$where $D$ is the domain bounded by the circles$$ \begin{array}{ll} x^{2}+y^{2}-2 x=0, & x^{2}+y^{2}-4 x=0, \\ x^{2}+y^{2}-2 y=0, & x^{2}+y^{2}-6 y=0 . \end{array} $$ | The domain $D$ is depicted in Figure 71 . We transform it into the rectangle $D_{1}=$ $\left[\frac{1}{4}, \frac{1}{2}\right] \times\left[\frac{1}{6}, \frac{1}{2}\right]$ by the change of coordinates$$ x=\frac{u}{u^{2}+v^{2}}, \quad y=\frac{v}{u^{2}+v^{2}} . $$The Jacobian is Figure 71$$ J=-\frac{1}{\left(u^{2}+v^{2}\right)^{2}} . $$Therefore,$$ \iint_{D} \frac{d x d y}{\left(x^{2}+y^{2}\right)^{2}}=\iint_{D_{1}} d u d v=\frac{1}{12} . $$ |
Let $v$ and $w$ be distinct, randomly chosen roots of the equation $z^{1997}-1=0$. Find the probability that $\sqrt{2+\sqrt{3}} \leq|v+w|$. | Because the 1997 roots of the equation are symmetrically distributed in the complex plane, there is no loss of generality to assume that $v=1$. We are required to find the probability that$$ |1+w|^{2}=|(1+\cos \theta)+i \sin \theta|^{2}=2+2 \cos \theta \geq 2+\sqrt{3} . $$This is equivalent to $\cos \theta \geq \frac{1}{2} \sqrt{3}$, or $|\theta| \leq \frac{\pi}{6}$. Because $w \neq 1, \theta$ is of the form $\pm \frac{2 k \pi}{1997} k$, $1 \leq k \leq\left\lfloor\frac{1997}{12}\right\rfloor$. There are $2 \cdot 166=332$ such angles, and hence the probability is $\frac{332}{1996}=\frac{83}{499} \approx 0.166$ |
The numbers $1,2,3,4,5,6,7$, and 8 are written on the faces of a regular octahedron so that each face contains a different number. Find the probability that no two consecutive numbers are written on faces that share an edge, where 8 and 1 are considered consecutive. | Consider the dual cube to the octahedron. The vertices $A, B, C, D, E, F, G$, $H$ of this cube are the centers of the faces of the octahedron (here $A B C D$ is a face of the cube and $(A, G),(B, H),(C, E),(D, F)$ are pairs of diagonally opposite vertices). Each assignment of the numbers $1,2,3,4,5,6,7$, and 8 to the faces of the octahedron corresponds to a permutation of $A B C D E F G H$, and thus to an octagonal circuit of these vertices. The cube has 16 diagonal segments that join nonadjacent vertices. The problem requires us to count octagonal circuits that can be formed by eight of these diagonals.Six of these diagonals are edges of the tetrahedron $A C F H$, six are edges of the tetrahedron $D B E G$, and four are long diagonals, joining opposite vertices of the cube. Notice that each vertex belongs to exactly one long diagonal. It follows that an octagonal circuit must contain either 2 long diagonals separated by 3 tetrahedron edges (Figure 108a), or 4 long diagonals (Figure 108b) alternating with tetrahedron edges.
Figure 108When forming a (skew) octagon with 4 long diagonals, the four tetrahedron edges need to be disjoint; hence two are opposite edges of $A C F H$ and two are opposite edges of $D B E G$. For each of the three ways to choose a pair of opposite edges from the tetrahedron $A C F H$, there are two possible ways to choose a pair of opposite edges from tetrahedron $D B E G$. There are $3 \cdot 2=6$ octagons of this type, and for each of them, a circuit can start at 8 possible vertices and can be traced in two different ways, making a total of $6 \cdot 8 \cdot 2=96$ permutations.An octagon that contains exactly two long diagonals must also contain a three-edge path along the tetrahedron $A C F H$ and a three-edge path along tetrahedron the $D B E G$. A three-edge path along the tetrahedron the $A C F H$ can be chosen in $4 !=24$ ways. The corresponding three-edge path along the tetrahedron $D B E G$ has predetermined initial and terminal vertices; it thus can be chosen in only 2 ways. Since this counting method treats each path as different from its reverse, there are $8 \cdot 24 \cdot 2=384$ permutations of this type.In all, there are $96+384=480$ permutations that correspond to octagonal circuits formed exclusively from cube diagonals. The probability of randomly choosing such a permutation is $\frac{480}{8 !}=\frac{1}{84}$. |
Let $V$ be the vector space of all real $3 \times 3$ matrices and let $A$ be the diagonal matrix
$$
\left(\begin{array}{lll}
1 & 0 & 0 \\
0 & 2 & 0 \\
0 & 0 & 1
\end{array}\right) .
$$
Calculate the determinant of the linear transformation $T$ on $V$ defined by $T(X)=\frac{1}{2}(A X+X A)$.
| Let $X=\left(x_{i j}\right)$ be any element of $M_{3}(\mathbb{R})$. A calculation gives
$$
T(X)=\left(\begin{array}{ccc}
x_{11} & 3 x_{12} / 2 & x_{13} \\n3 x_{21} / 2 & 2 x_{22} & 3 x_{23} / 2 \\nx_{31} & 3 x_{32} / 2 & x_{33}
\end{array}\right) .
$$
It follows that the basis matrices $M_{i j}$ are eigenvectors of $T$. Taking the product of their associated eigenvalues, we get $\operatorname{det} T=2(3 / 2)^{4}=81 / 8$.
|
An old woman went to the market and a horse stepped on her basket and smashed her eggs. The rider offered to pay for the eggs and asked her how many there were. She did not remember the exact number, but when she had taken them two at a time there was one egg left, and the same happened when she took three, four, five, and six at a time. But when she took them seven at a time, they came out even. What is the smallest number of eggs she could have had? | We are to find the smallest positive solution to the system of congruences$$ \begin{aligned} &x \equiv 1(\bmod 60), \\ &x \equiv 0(\bmod 7) . \end{aligned} $$The general solution is $7 b_{1}+420 t$, where $b_{1}$ is the inverse of 7 modulo 60 and $t$ is an integer. Since $b_{1}$ is a solution to the Diophantine equation $7 b_{1}+60 y=1$, we find it using Euclid's algorithm. Here is how to do it: $60=8 \cdot 7+4,7=1 \cdot 4+3,4=1 \cdot 3+1$. Then$$ \begin{aligned} 1 &=4-1 \cdot 3=4-1 \cdot(7-1 \cdot 4)=2 \cdot 4-7=2 \cdot(60-8 \cdot 7)-7 \\ &=2 \cdot 60-17 \cdot 7 \end{aligned} $$Hence $b_{1}=-17$, and the smallest positive number of the form $7 b_{1}+420 t$ is $-7 \cdot 17+$ $420 \cdot 1=301$. |
Consider a spinless particle represented by the wave function
$$
\psi=K(x+y+2 z) e^{-\alpha r}
$$
where $r=\sqrt{x^{2}+y^{2}+z^{2}}$, and $K$ and $\alpha$ are real constants.
If the $z$-component of angular momentum, $L_{z}$, were measured, what is the probability that the result would be $L_{z}=+\hbar$ ?
You may find the following expressions for the first few spherical harmonics useful:
$$
\begin{aligned}
&Y_{0}^{0}=\sqrt{\frac{1}{4 \pi}}, \quad Y_{1}^{\pm 1}=\mp \sqrt{\frac{3}{8 \pi}} \sin \theta e^{\pm i \phi}, \\
&Y_{1}^{0}=\sqrt{\frac{3}{4 \pi}} \cos \theta, \quad Y_{2}^{\pm 1}=\mp \sqrt{\frac{15}{8 \pi}} \sin \theta \cos \theta e^{\pm i \phi} .
\end{aligned}
$$ | The wave function may be rewritten in spherical coordinates as
$$
\psi=K r(\cos \phi \sin \theta+\sin \phi \sin \theta+2 \cos \theta) e^{-\alpha r}
$$
its angular part being
$$
\psi(\theta, \phi)=K^{\prime}(\cos \phi \sin \theta+\sin \phi \sin \theta+2 \cos \theta),
$$
where $K^{\prime}$ is the normalization constant such that
$$
K^{\prime 2} \int_{\theta}^{\pi} d \theta \int_{0}^{2 \pi} \sin \theta(\cos \phi \sin \theta+\sin \phi \sin \theta+2 \cos \theta)^{2} d \phi=1 \text {. }
$$
Since
$$
\cos \phi=\frac{1}{2}\left(e^{i \phi}+e^{-i \phi}\right), \quad \sin \phi=\frac{1}{2 i}\left(e^{i \phi}-e^{-i \phi}\right)
$$
we have
$$
\begin{aligned}
\psi(\theta, \phi) &=K^{\prime}\left[\frac{1}{2}\left(e^{i \phi}+e^{-i \phi}\right) \sin \theta+\frac{1}{2 i}\left(e^{i \phi}-e^{-i \phi}\right) \sin \theta+\cos 2 \theta\right], \\
&=K^{\prime}\left[-\frac{1}{2}(1-i) \sqrt{\frac{8 \pi}{3}} Y_{1}^{1}+\frac{1}{2}(1+i) \sqrt{\frac{8 \pi}{3}} Y_{1}^{-1}+2 \sqrt{\frac{4 \pi}{3}} Y_{1}^{0}\right] .
\end{aligned}
$$
The normalization condition and the orthonormality of $Y_{l}^{m}$ then give
$$
K^{\prime 2}\left[\frac{1}{2} \cdot \frac{8 \pi}{3}+\frac{1}{2} \cdot \frac{8 \pi}{3}+4 \cdot \frac{4 \pi}{3}\right]=1
$$
or
$$
K^{\prime}=\sqrt{\frac{1}{8 \pi}},
$$
and thus
$$
\begin{aligned}
\psi(\theta, \phi)=& \sqrt{\frac{1}{8 \pi}}\left[-\frac{1}{2}(1-i) \sqrt{\frac{8 \pi}{3}} Y_{1}^{1}\right.\\
&\left.+\frac{1}{2}(1+i) \sqrt{\frac{8 \pi}{3}} Y_{1}^{-1}+2 \sqrt{\frac{4 \pi}{3}} Y_{1}^{0}\right] .
\end{aligned}
$$
The total angular momentum of the particle is
$$
\sqrt{\left\langle\mathbf{L}^{2}\right\rangle}=\sqrt{l(l+1)} \hbar=\sqrt{2} \hbar .
$$
as the wave function corresponds to $l=1$.The $z$-component of the angular momentum is
$$
\begin{aligned}
\left\langle\psi^{*}\left|L_{z}\right| \psi\right\rangle=& K^{\prime 2}\left[\frac{1}{2} \cdot \frac{8 \pi}{3} \cdot \hbar\left(Y_{1}^{1}\right)^{2}+\frac{1}{2} \cdot \frac{8 \pi}{3}(-\hbar)\left(Y_{1}^{-1}\right)^{2}\right.\\
&\left.+4 \cdot \frac{4 \pi}{3}(0)\left(Y_{1}^{0}\right)^{2}\right] \\
=& \frac{1}{8 \pi}\left[\frac{1}{2} \cdot \frac{8 \pi}{3}(+\hbar)+\frac{1}{2} \cdot \frac{8 \pi}{3}(-\hbar)\right]=0 .
\end{aligned}
$$The probability of finding $L_{z}=+\hbar$ is
$$
\begin{aligned}
P &=\left|\left\langle L_{z}=+\hbar \mid \psi(\theta, \phi)\right\rangle\right|^{2} \\
&=\frac{1}{8 \pi} \cdot \frac{1}{2} \cdot \frac{8 \pi}{3}=\frac{1}{6} .
\end{aligned}
$$ |
A student near a railroad track hears a train's whistle when the train is coming directly toward him and then when it is going directly away. The two observed frequencies are $250$ and $200 \mathrm{~Hz}$. Assume the speed of sound in air to be $360 \mathrm{~m} / \mathrm{s}$. What is the train's speed? | Let $\nu_{0}, \nu_{1}, \nu_{2}$ be respectively the frequency of the whistle emitted by the train, and the frequencies heard by the student when the train is coming and when it is moving away. The Doppler effect has it that
$$
\begin{aligned}
&\nu_{1}=\left(\frac{c}{c-v}\right) \nu_{0}, \\
&\nu_{2}=\left(\frac{c}{c+v}\right) \nu_{0},
\end{aligned}
$$
where $c$ is the speed of sound and $v$ is the speed of the train, and thus
$$
\frac{\nu_{1}}{\nu_{2}}=\frac{c+v}{c-v} .
$$
Putting in the data, we have
$$
1.25=\frac{360+v}{360-v},
$$
or
$$
\frac{2.25}{0.25}=\frac{720}{2 v},
$$
and thus
$$
v=\frac{360}{9}=40 \mathrm{~m} / \mathrm{s} .
$$ |
An opaque sheet has a hole in it of $0.5 \mathrm{~mm}$ radius. If plane waves of light $(\lambda=5000 \AA)$ fall on the sheet, find the maximum distance from this sheet at which a screen must be placed so that the light will be focused to a bright spot. | The maximum distance $r$ of the screen from the hole for which the light will be focused to a bright spot is that for which the area of the hole corresponds to the first Fresnel zone only, and is given by
$$
\rho_{1}^{2}=\lambda r,
$$
where $\rho$ is the radius of the hole. Hence
$$
r=\frac{\rho_{1}^{2}}{\lambda}=\frac{0.5^{2}}{5000 \times 10^{-7}}=500 \mathrm{~mm} .
$$
Without the opaque screen, the number of the Fresnel zones will approach infinity. The total amplitude is then
hence
$$
\begin{aligned}
A &=A_{1}-A_{2}+A_{3}-\ldots \ldots \\
&=\frac{A_{1}}{2}+\left(\frac{A_{1}}{2}-A_{2}+\frac{A_{3}}{2}\right)+\left(\frac{A_{3}}{2}-A_{4}+\frac{A_{5}}{2}\right)+\ldots \\
& \approx \frac{A_{1}}{2}
\end{aligned}
$$
$$
I^{\prime} \sim A_{1}^{2} / 4 \text {, i.e., } I^{\prime}=I_{1} / 4
$$
Thus the intensity is now $1 / 4$ that before the opaque sheet is removed. |
Assume a visible photon of $3 \mathrm{eV}$ energy is absorbed in one of the cones (light sensors) in your eye and stimulates an action potential that produces a $0.07$ volt potential on an optic nerve of $10^{-9} \mathrm{~F}$ capacitance.
Calculate the charge needed. | $Q=V C=0.07 \times 10^{-9}=7 \times 10^{-11}$ Coulomb. |
Imagine the universe to be a spherical cavity, with a radius of $10^{28} \mathrm{~cm}$ and impenetrable walls.
If the temperature were $0 \mathrm{~K}$, and the universe contained $10^{80}$ electrons in a Fermi distribution, calculate the Fermi momentum of the electrons. | The number of photons in the angular frequency range from $\omega$ to $\omega+d \omega$ is
$$
d N=\frac{V}{\pi^{2} c^{3}} \frac{\omega^{2} d \omega}{e^{\beta \hbar \omega}-1}, \quad \beta=\frac{1}{k T} .
$$
The total number of photons is
$$
\begin{aligned}
N &=\frac{V}{\pi^{2} c^{3}} \int_{0}^{\infty} \frac{\omega^{2} d \omega}{e^{\beta \omega h / 2 \pi}-1}=\frac{V}{\pi^{2} c^{3}} \frac{1}{(\beta \hbar)^{3}} \int_{0}^{\infty} \frac{x^{2} d x}{e^{x}-1} \\
&=\frac{V}{\pi^{2}}\left(\frac{k T}{\hbar c}\right)^{3} \cdot 2 \sum_{n=1}^{\infty} \frac{1}{n^{3}} \approx \frac{2 \times 1.2}{\pi^{2}} \cdot V\left(\frac{k T}{\hbar c}\right)^{3} \\
&=\frac{2.4}{\pi^{2}} \cdot \frac{4}{3} \pi \cdot\left(10^{28}\right)^{3} \cdot\left(\frac{1.38 \times 10^{-16} \times 3}{1.05 \times 10^{-27} \times 3 \times 10^{10}}\right)^{3} \\
& \approx 2.5 \times 10^{87} .
\end{aligned}
$$
The total energy is
$$
\begin{aligned}
E &=\frac{V \hbar}{\pi^{2} c^{3}} \int_{0}^{\infty} \frac{\omega^{3} d \omega}{e^{\beta \omega h / 2 \pi}-1}=\frac{\pi^{2} k^{4}}{15(h c / 2 \pi)^{3}} V T^{4} \\
& \approx 2.6 \times 10^{72} \mathrm{ergs} .
\end{aligned}
$$
The Fermi momentum of the electrons is
$$
p_{F}=\hbar\left(3 \pi^{2} \frac{N}{V}\right)^{1 / 3}=2 \times 10^{-26} \mathrm{~g} \cdot \mathrm{cm} / \mathrm{s} .
$$ |
In a simplified model of a relativistic nucleus-nucleus collision, a nucleus of rest mass $m_{1}$ and speed $\beta_{1}$ collides head-on with a target nucleus of mass $m_{2}$ at rest. The composite system recoils at speed $\beta_{0}$ and with center of mass energy $\varepsilon_{0}$. Assume no new particles are created.
Calculate $\beta_{0}$ for a ${ }^{40} \mathrm{Ar}$ nucleus impinging at $\beta_{1}=0.8$ on a ${ }^{238} \mathrm{U}$ nucleus. | As implied by the question the velocity of light is taken to be one for convenience.
For a system, $E^{2}-p^{2}$ is invariant under Lorentz transformation. In the laboratory frame $\Sigma$, writing $\gamma_{1}=\frac{1}{\sqrt{1-\beta_{1}^{2}}}$,
$$
E^{2}-p^{2}=\left(m_{1} \gamma_{1}+m_{2}\right)^{2}-\left(m_{1} \gamma_{1} \beta_{1}\right)^{2} .
$$
In the center of mass frame $\Sigma^{\prime}, E^{\prime 2}-p^{\prime 2}=\varepsilon_{0}^{2}$. Hence
$$
\begin{aligned}
\varepsilon_{0}^{2} &=\left(m_{1} \gamma_{1}+m_{2}\right)^{2}-\left(m_{1} \gamma_{1} \beta_{1}\right)^{2} \\
&=m_{1}^{2} \gamma_{1}^{2}\left(1-\beta_{1}^{2}\right)+2 m_{1} m_{2} \gamma_{1}+m_{2}^{2} \\
&=m_{1}^{2}+m_{2}^{2}+2 m_{1} m_{2} \gamma_{1},
\end{aligned}
$$
or
$$
\varepsilon_{0}=\sqrt{m_{1}^{2}+m_{2}^{2}+\frac{2 m_{1} m_{2}}{\sqrt{1-\beta_{1}^{2}}}} .
$$
In the laboratory, the system of $m_{1}, m_{2}$ has total momentum $m_{1} \gamma_{1} \beta_{1}$ and total energy $m_{1} \gamma_{1}+m_{2}$. These are conserved quantities so that after the collision the composite system will move with velocity
$$
\beta_{0}=\frac{m_{1} \gamma_{1} \beta_{1}}{m_{1} \gamma_{1}+m_{2}}=\frac{m_{1} \beta_{1}}{m_{1}+m_{2} \sqrt{1-\beta_{1}^{2}}} .
$$
The masses are approximately
$$
\begin{aligned}
&m_{1}=40 \times 0.94=37.6 \mathrm{GeV}, \\
&m_{2}=238 \times 0.94=223.7 \mathrm{GeV} .
\end{aligned}
$$
Then
$$
\begin{aligned}
\varepsilon_{0} &=\sqrt{37.6^{2}+223.7^{2}+\frac{2 \times 37.6 \times 223.7}{\sqrt{1-0.64}}} \\
&=282 \mathrm{GeV}=2.82 \times 10^{5} \mathrm{MeV}, \\
\beta_{0} &=\frac{37.6 \times 0.8}{37.6+223.7 \times \sqrt{1-0.64}}=0.175
\end{aligned}
$$ |
In a region of empty space, the magnetic field (in Gaussian units) is described by
$$
\mathrm{B}=B_{0} e^{a x} \hat{\mathbf{e}}_{z} \sin w
$$
where $w=k y-\omega t$.
Find the speed of propagation $v$ of this field. | Express $B$ as $\operatorname{Im}\left(B_{0} e^{a x} e^{i w}\right) \hat{\mathbf{e}}_{z}$.
Using Maxwell's equation
$$
\nabla \times \mathbf{B}=\frac{1}{c} \frac{\partial \mathrm{E}}{\partial t}
$$
and the definition $k=\frac{\omega}{c}$ for empty space, we obtain
$$
\mathbf{E}=\frac{i c}{\omega} \nabla \times \mathbf{B}=\frac{i}{k}\left|\begin{array}{ccc}
\mathbf{e}_{x} & \mathbf{e}_{y} & \mathbf{e}_{z} \\
\frac{\partial}{\partial x} & \frac{\partial}{\partial y} & 0 \\
0 & 0 & B_{z}
\end{array}\right|
$$
where $\frac{\partial}{\partial z}=0$ as $\mathbf{B}$ does not depend on $z$.
Hence
$$
\begin{gathered}
E_{x}=\operatorname{Im}\left(\frac{i}{k} B_{0} e^{a x} i k e^{i w}\right)=-B_{0} e^{a x} \sin w \\
E_{y}=\operatorname{Im}\left(-\frac{i}{k} B_{0} a e^{a x} e^{i w}\right)=-\frac{a c}{\omega} B_{0} e^{a x} \cos w \\
E_{z}=0
\end{gathered}
$$If the wave form remains unchanged during propagation, we have
$$
d w=k d y-w d t=0,
$$
or $\frac{d y}{d t}=\frac{w}{k}=c$. Hence the wave propagates along the $y$ direction with a speed $v=c$. |
A meteorite of mass $1.6 \times 10^{3} \mathrm{~kg}$ moves about the earth in a circular orbit at an altitude of $4.2 \times 10^{6} \mathrm{~m}$ above the surface. It suddenly makes a head-on collision with another meteorite that is much lighter, and loses $2.0 \%$ of its kinetic energy without changing its direction of motion or its total mass.
Find the meteorite's distance of closest approach to the earth after the collision. | The laws of conservation of mechanical energy and conservation of angular momentum apply to the motion of the heavy meteorite after its collision.
For the initial circular motion, $E<0$, so after the collision we still have $E<0$. After it loses $2.0 \%$ of its kinetic energy, the heavy meteorite will move in an elliptic orbit.
From
$$
\frac{m v^{2}}{r}=\frac{G m M}{r^{2}},
$$
we obtain the meteorite's kinetic energy before collision:
$$
\begin{aligned}
\frac{1}{2} m v^{2} &=\frac{G m M}{2 r}=\frac{m g R^{2}}{2 r} \\
&=\frac{m \times 9.8 \times 10^{3} \times 6400^{2}}{2(6400+4200)}=1.89 \times 10^{7} \mathrm{~m} \text { Joules },
\end{aligned}
$$
where $m$ is the mass of the meteorite in kg. The potential energy of the meteorite before collision is
$$
-\frac{G m M}{r}=-m v^{2}=-3.78 \times 10^{7} m \text { Joules } .
$$
During the collision, the heavy meteorite's potential energy remains constant, while its kinetic energy is suddenly reduced to
$$
1.89 \times 10^{7} \mathrm{~m} \times 98 \%=1.85 \times 10^{7} \mathrm{~m} \text { Joules. }
$$
Hence the total mechanical energy of the meteorite after the collision is
$$
E=(1.85-3.78) \times 10^{7} \mathrm{~m}=-1.93 \times 10^{7} \mathrm{~m} \text { Joules } .
$$
From
$$
E=\frac{-G m M}{2 a}=\frac{-m R^{2} g}{2 a},
$$
we obtain the major axis of the ellipse as
$$
\begin{aligned}
2 a &=\frac{R^{2} g}{1.93 \times 10^{7}}=\frac{\left(6400 \times 10^{3}\right)^{2} \times 9.8}{1.93 \times 10^{7}} \\
&=2.08 \times 10^{7} \mathrm{~m}=2.08 \times 10^{4} \mathrm{~km} .
\end{aligned}
$$
As after the collision, the velocity of the heavy meteorite is still perpendicular to the radius vector from the center of the earth, the meteorite is at the apogee of the elliptic orbit. Then the distance of the apogee from the center of the earth is $6400+4200=10600 \mathrm{~km}$ and the distance of the perigee from the center of the earth is
$$
r_{\min }=20800-10600=10200 \mathrm{~km} .
$$
Thus the meteorite's distance of closest approach to the earth after the collision is $10200-6400=3800 \mathrm{~km}$.
From the above calculations, we see that it is unnecessary to know the mass of the meteorite. Whatever the mass of the meteorite, the answer is the same as long as the conditions remain unchanged. |
X-rays are reflected from a crystal by Bragg reflection. If the density of the crystal which is of an accurately known structure is measured with an rms error of 3 parts in $10^{4}$, and if the angle the incident and reflected rays make with the crystal plane is $6^{\circ}$ and is measured with an rms error of $3.4$ minutes of arc, then what is the rms error in the determination of the X-ray wavelength? | For simplicity consider a crystal whose primitive cell is simple cubic with edge $d$ (to be multiplied by a factor of about one for the primitive cells of other crystal structures). For first order reflection, $n=1$ and Bragg's law gives
$$
2 d \sin \theta=\lambda .
$$
Differentiating, we have
$$
\left|\frac{\Delta \lambda}{\lambda}\right|=\left|\frac{\Delta d}{d}\right|+\cot \theta|\Delta \theta| .
$$
The volume of a unit cell is
$$
d^{3}=\frac{M}{\rho N_{0}},
$$
where $M$ is the molar weight and $\rho$ the mass density of the crystal, and $N_{0}$ is Avogadro's number. This differentiates to give
$$
\left|\frac{\Delta d}{d}\right|=\frac{1}{3}\left|\frac{\Delta \rho}{\rho}\right| .
$$
Thus
$$
\left|\frac{\Delta \lambda}{\lambda}\right|=\frac{1}{3}\left|\frac{\Delta \rho}{\rho}\right|+\cot \theta \cdot|\Delta \theta|,
$$
and, in terms of rms errors,
$$
\frac{\sigma_{\lambda}}{\lambda}=\left[\left(\frac{1}{3} \frac{\sigma_{\rho}}{\rho}\right)^{2}+\left(\sigma_{\theta} \cot \theta\right)^{2}\right]^{1 / 2} \text {. }
$$
As $\frac{\sigma_{\rho}}{\rho}=3 \times 10^{-4}, \sigma_{\theta} \cot \theta=\frac{\sigma_{\theta} \cos \theta}{\sin \theta} \approx \frac{\sigma_{\theta}}{\theta}=\frac{3.4}{60 \times 6}=9.4 \times 10^{-3}$,
$$
\frac{\sigma_{\lambda}}{\lambda}=\sqrt{10^{-8}+\left(9.4 \times 10^{-3}\right)^{2}}=9.4 \times 10^{-3} .
$$ |
Find the angular separation in seconds of arc of the closest two stars resolvable by the following reflecting telescope: $8 \mathrm{~cm}$ objective, $1.5$ meter focal length, $80$X eyepiece. Assume a wavelength of $6000 \AA$. ($1 \AA=10^{-8}$ cm). | The angular resolving power of the telescope is
$$
\Delta \theta_{1}=1.22 \frac{\lambda}{D}=1.22 \times \frac{6000 \times 10^{-8}}{8} \approx 2^{\prime \prime} .
$$
The resolving power of the human eye is about 1 minute of arc. Using an eyepiece of magnification $80 \mathrm{X}$, the eye's resolving power is $\Delta \theta_{2}=1^{\prime} / 80<$ $2^{\prime \prime}$. Therefore, the resolvable angular separation of stars using the telescope is $\Delta \theta=\max \left(\Delta \theta_{1}, \Delta \theta_{2}\right)=2^{\prime \prime}$. |
A self-luminous object of height $h$ is $40 \mathrm{~cm}$ to the left of a converging lens with a focal length of $10 \mathrm{~cm}$. A second converging lens with a focal length of $20 \mathrm{~cm}$ is $30 \mathrm{~cm}$ to the right of the first lens.
Calculate the ratio of the height of the final image to the height $h$ of the object. | From $\frac{1}{f_{1}}=\frac{1}{u_{1}}+\frac{1}{v_{1}}$, where $f_{1}=10 \mathrm{~cm}, u_{1}=40 \mathrm{~cm}$, we obtain $v_{1}=13 \frac{1}{3} \mathrm{~cm}$.
From $\frac{1}{f_{2}}=\frac{1}{u_{2}}+\frac{1}{v_{2}}$, where $f_{2}=20 \mathrm{~cm}, u_{2}=\left(30-\frac{10}{3}\right) \mathrm{cm}$, we obtain $v_{2}=-100 \mathrm{~cm}$, i.e., the final image is $100 \mathrm{~cm}$ to the left of the second lens.The ratio of the final image height to the height $h$ of the object can be calculated from
$$
m=m_{2} \times m_{1}=\left(\frac{v_{2}}{u_{2}}\right)\left(\frac{v_{1}}{u_{1}}\right)=-2 .
$$
The minus sign signifies an inverted image. |
A currently important technique for precisely measuring the lifetimes of states of multiply ionized atoms consists of exciting a beam of the desired ions with a laser tuned to a resonance wavelength of the ion under study, and measuring the emission intensity from the ion beam as a function of down- stream distance. When a particular ion beam with a resonance wavelength at $4885 \AA$ is excited with a $4880 \AA$ argon-ion laser output, the intensity is found to drop by a factor of two, $10 \mathrm{~mm}$ down-stream from the excitation point. If ion beam velocity is $v / c=5 \times 10^{-3}$, calculate the angle the laser beam must make with the normal to the ion beam to achieve resonance and the lifetime of the excited state of the ion.
| Take two coordinate frames $\Sigma, \Sigma^{\prime} . \Sigma$ is the laboratory frame in which the laser source $L$ is at rest and the ions move with velocity $\beta c$ along the $x$-axis as shown in Fig. 1.46. $\Sigma^{\prime}$ is the rest frame of an ion in the beam and moves with velocity $\beta c$ relative to $\Sigma$ along the $x$ direction. Consider the transformation of the 4 -vector $\left(\mathbf{k}, \frac{\omega}{c}\right)$. We have
$$
\frac{\omega^{\prime}}{c}=\gamma\left(\frac{\omega}{c}-\beta k \sin \theta\right),
$$
where $\gamma=\left(1-\beta^{2}\right)^{-\frac{1}{2}}, k=\frac{\omega}{c}$. Hence
$$
\begin{aligned}
\sin \theta &=\frac{1}{\beta}\left(1-\frac{\lambda}{\lambda^{\prime}} \sqrt{1-\beta^{2}}\right) \\
& \approx \frac{1}{\beta}\left[1-\frac{\lambda}{\lambda^{\prime}}\left(1-\frac{\beta^{2}}{2}\right)\right]
\end{aligned}
$$
With $\lambda=4885 \AA, \lambda^{\prime}=4880 \AA, \beta=5 \times 10^{-3}$, we find
$$
\sin \theta=0.2074, \quad \text { or } \quad \theta=11.97^{\circ} \text {. }
$$
As the excited ions move with velocity $\beta c$, on account of time dilation the laboratory lifetime of an ion is $\gamma \tau$, where $\tau$ is the proper lifetime. Hence the excited ion beam will decay according to
$$
I=I_{0} \exp \left(-\frac{t}{\gamma \tau}\right)
$$
giving
$$
\begin{aligned}
\tau &=\frac{t}{\gamma \ln \left(\frac{I_{0}}{I}\right)}=\frac{x}{\gamma \beta c \ln \left(\frac{I_{0}}{I}\right)} \\
& \approx\left(1-\frac{\beta^{2}}{2}\right) \frac{x}{\beta c \ln \left(\frac{I_{0}}{I}\right)}=9.6 \times 10^{-9} \mathrm{~s} .
\end{aligned}
$$ |
A one-dimensional square well of infinite depth and $1 \AA$ width contains 3 electrons. The potential well is described by $V=0$ for $0 \leq x \leq 1 \AA$ and $V=+\infty$ for $x<0$ and $x>1 \AA$. For a temperature of $T=0 \mathrm{~K}$, the average energy of the 3 electrons is $E=12.4 \mathrm{eV}$ in the approximation that one neglects the Coulomb interaction between electrons. In the same approximation and for $T=0 \mathrm{~K}$, what is the average energy for 4 electrons in this potential well? | For a one-dimensional potential well the energy levels are given by
$$
E_{n}=E_{1} n^{2},
$$
where $E_{1}$ is the ground state energy and $n=1,2, \ldots$. Pauli's exclusion principle and the lowest energy principle require that two of the three electrons are in the energy level $E_{1}$ and the third one is in the energy level $E_{2}$. Thus $12 \cdot 4 \times 3=2 E_{1}+4 E_{1}$, giving $E_{1}=6 \cdot 2 \mathrm{eV}$. For the case of four electrons, two are in $E_{1}$ and the other two in $E_{2}$, and so the average energy is
(Note: the correct value of $E_{1}$ is
$$
E=\frac{1}{4}\left(2 E_{1}+2 E_{2}\right)=\frac{5}{2} E_{1}=15.5 \mathrm{eV} .
$$
$$
\begin{aligned}
\frac{\pi^{2} \hbar^{2}}{2 m a^{2}} &=\frac{1}{2 m c^{2}}\left(\frac{\pi \hbar c}{a}\right)^{2}=\frac{1}{1.02 \times 10^{6}}\left(\frac{\pi \times 6.58 \times 10^{-16} \times 3 \times 10^{10}}{10^{-8}}\right)^{2} \\
&=37.7 \mathrm{eV} .)
\end{aligned}
$$ |
Evaluate
$$
I=\int_{0}^{\infty} \frac{d x}{4+x^{4}} .
$$ | As
$$
y=1+2 x+3 x^{2}+4 x^{3}+\ldots+n x^{n-1}+\ldots
$$
we have
$$
x y=x+2 x^{2}+3 x^{3}+4 x^{4}+\ldots+n x^{n}+\ldots
$$
and thus
$$
y-x y=1+x+x^{2}+x^{3}+\ldots+x^{n-1}+\ldots=\frac{1}{1-x}
$$
since $|x|<1$. Hence
$$
y=\frac{1}{(1-x)^{2}} .
$$
The mean value of $x$ is
$$
\bar{x}=\frac{\int_{0}^{\infty} x f(x) d x}{\int_{0}^{\infty} f(x) d x}=\frac{\int_{0}^{\infty} x^{2} e^{-x / \lambda} d x}{\int_{0}^{\infty} x e^{-x / \lambda} d x}=\frac{2 \lambda \int_{0}^{\infty} x e^{-x / \lambda} d x}{\int_{0}^{\infty} x e^{-x / \lambda} d x}=2 \lambda,
$$
where integration by parts has been used for the numerator and $\lambda>0$ has been assumed . The most probable value of $x, x_{m}$, is given by
$$
\left(\frac{d t}{d x}\right)_{x_{m}}=0,
$$
whence $x_{m}=\lambda$.
Note that if $\lambda<0, f(x)$ is infinite at $\infty$ and the mean and most probable values are also infinite.
Fig. $3.14$Write
$$
\begin{aligned}
I &=\frac{1}{2} \int_{0}^{\infty} \frac{d x}{x^{4}+4}+\frac{1}{2} \int_{0}^{\infty} \frac{d x}{x^{4}+4} \\
&=\frac{1}{2} \int_{-\infty}^{0} \frac{d x}{x^{4}+4}+\frac{1}{2} \int_{0}^{\infty} \frac{d x}{x^{4}+4}=\frac{1}{2} \int_{\infty}^{\infty} \frac{d x}{x^{4}+4}
\end{aligned}
$$
and consider
$$
\oint_{C} \frac{d z}{z^{4}+4}
$$
where $C$ is the closed contour of Fig. $3.14$ consisting of the line from $-R$ to $R$ and the semicircle $\Gamma$, traversed in anticlockwise sense, in the complex plane. As
$$
\frac{1}{z^{4}+4}=\frac{1}{\left(z-z_{1}\right)\left(z-z_{2}\right)\left(z-z_{3}\right)\left(z-z_{4}\right)},
$$
where
$$
z_{1}=1+i, \quad z_{2}=-1+i, \quad z_{3}=-1-i, \quad z_{4}=1-i
$$
the residue theorem gives
$$
\begin{aligned}
&\oint_{C} \frac{d z}{z^{4}+4} \\
&=2 \pi i\left[\frac{1}{\left(z_{1}-z_{2}\right)\left(z_{1}-z_{3}\right)\left(z_{1}-z_{4}\right)}+\frac{1}{\left(z_{2}-z_{1}\right)\left(z_{2}-z_{3}\right)\left(z_{2}-z_{4}\right)}\right] \\
&=2 \pi i\left[\frac{1}{2(2+2 i) \cdot 2 i}-\frac{1}{2 \cdot 2 i(-2+2 i)}\right]=\frac{\pi}{4} .
\end{aligned}
$$
But
$$
\oint_{C} \frac{d z}{z^{4}+4}=\int_{-R}^{R} \frac{d x}{x^{4}+4}+\int_{\Gamma} \frac{d z}{z^{4}+4} \rightarrow \int_{-\infty}^{\infty} \frac{d x}{x^{4}+4}
$$
as $R \rightarrow \infty$. Hence
$$
I=\frac{1}{2} \cdot \frac{\pi}{4}=\frac{\pi}{8} .
$$ |
The deuteron is a bound state of a proton and a neutron of total angular momentum $J=1$. It is known to be principally an $S(L=0)$ state with a small admixture of a $D(L=2)$ state.
Calculate the magnetic moment of the pure $D$ state $n-p$ system with $J=1$. Assume that the $n$ and $p$ spins are to be coupled to make the total spin $S$ which is then coupled to the orbital angular momentum $L$ to give the total angular momentum $J$. Express your result in nuclear magnetons. The proton and neutron magnetic moments are $2.79$ and $-1.91$ nuclear magnetons respectively.
| The parities of the $S$ and $D$ states are positive, while the parity of the $P$ state is negative. Because of the conservation of parity in strong interaction, a quantum state that is initially an $S$ state cannot have a $P$ state component at any later moment.
The possible spin values for a system composed of a proton and a neutron are 1 and 0 . We are given $\mathbf{J}=\mathbf{L}+\mathbf{S}$ and $J=1$. If $S=0, L=1$, the system would be in a $P$ state, which must be excluded as we have seen in (a). The allowed values are then $S=1, L=2,1,0$. Therefore a $G$ state $(L=4)$ cannot contribute.
The total spin is $\mathbf{S}=\mathbf{s}_{p}+\mathbf{s}_{n}$. For a pure $D$ state with $J=1$, the orbital angular momentum (relative to the center of mass of the $n$ and $p$ ) is $L=2$ and the total spin must be $S=1$. The total magnetic moment arises from the coupling of the magnetic moment of the total spin, $\mu$, with that of the orbital angular momentum, $\boldsymbol{\mu}_{L}$, where $\boldsymbol{\mu}=\boldsymbol{\mu}_{p}+\boldsymbol{\mu}_{n}, \boldsymbol{\mu}_{p}, \boldsymbol{\mu}_{n}$ being the spin magnetic moments of $p$ and $n$ respectively.
The average value of the component of $\mu$ in the direction of the total spin $\mathbf{S}$ is
$$
\mu_{s}=\frac{\left(g_{p} \mu_{N} \mathbf{s}_{p}+g_{n} \mu_{N} \mathbf{s}_{n}\right) \cdot \mathbf{S}}{\mathbf{S}^{2}} \mathbf{S}=\frac{1}{2}\left(g_{p}+g_{n}\right) \mu_{N} \mathbf{S}
$$
where
$$
\mu_{N}=\frac{e \hbar}{2 m_{p} c}, \quad g_{p}=5.58, \quad g_{n}=-3.82,
$$
as $\mathbf{s}_{p}=\mathbf{s}_{n}=\frac{1}{2} \mathbf{S}$.
The motion of the proton relative to the center of mass gives rise to a magnetic moment, while the motion of the neutron does not as it is uncharged. Thus
$$
\mu_{L}=\mu_{N} \mathbf{L}_{p},
$$
where $\mathbf{L}_{p}$ is the angular momentum of the proton relative to the center of mass. As $\mathbf{L}_{p}+\mathbf{L}_{n}=\mathbf{L}$ and we may assume $\mathbf{L}_{p}=\mathbf{L}_{n}$, we have $\mathbf{L}_{p}=\mathbf{L} / 2$ (the center of mass is at the mid-point of the connecting line, taking $m_{p} \approx m_{n}$ ). Consequently, $\mu_{L}=\mu_{N} \mathbf{L} / 2$.
The total coupled magnetic moment along the direction of $\mathbf{J}$ is then
$$
\mu_{T}=\frac{\left[\frac{1}{2} \mu_{N} \mathbf{L} \cdot \mathbf{J}+\frac{1}{2}\left(g_{p}+g_{n}\right) \mu_{N} \mathbf{S} \cdot \mathbf{J}\right] \mathbf{J}}{J(J+1)} .
$$
Since $\mathbf{J}=\mathbf{L}+\mathbf{S}, \mathbf{S} \cdot \mathbf{L}=\frac{1}{2}\left(J^{2}-L^{2}-S^{2}\right)$. With $J=1, L=2, S=1$ and so $J^{2}=2, L^{2}=6, S^{2}=2$, we have $\mathbf{S} \cdot \mathbf{L}=-3$ and thus $\mathbf{L} \cdot \mathbf{J}=3$, $\mathbf{S} \cdot \mathbf{J}=-1$. Hence
$$
\begin{aligned}
\boldsymbol{\mu}_{T} &=\left[\frac{1}{2} \mu_{N} \cdot 3+\frac{1}{2}\left(g_{p}+g_{n}\right) \mu_{N}(-1)\right] \mathbf{J} / 2 \\
&=\left[1.5-\frac{1}{2}\left(g_{p}+g_{n}\right)\right] \frac{1}{2} \mu_{N} \mathbf{J}=0.31 \mu_{N} \mathbf{J} .
\end{aligned}
$$
Taking the direction of $\mathbf{J}$ as the $z$-axis and letting $J_{z}$ take the maximum value $J_{z}=1$, we have $\mu_{T}=0.31 \mu_{N}$. |
At room temperature, $k_{\mathrm{B}} T / e=26 \mathrm{mV}$. A sample of cadmium sulfide displays a mobile carrier density of $10^{16} \mathrm{~cm}^{-3}$ and a mobility coefficient $\mu=10^{2} \mathrm{~cm}^{2} /$ volt sec.
The carriers are continuously trapped into immobile sites and then being thermally reionized into mobile states. If the average free lifetime in a mobile state is $10^{-5}$ second, what is the rms distance a carrier diffuses between successive trappings? | The electrical conductivity is given by $\sigma=n e \mu$. With $n=10^{22} \mathrm{~m}^{-3}$, $e=1.6 \times 10^{-19} \mathrm{C}, \mu=10^{-2} \mathrm{~m}^{2} \mathrm{~V}^{-1} \mathrm{~s}^{-1}$, we have for the material $\sigma=$ $16 \Omega^{-1} \mathrm{~m}^{-1}$.The law of equipartition of energy
$$
\frac{1}{2} m \bar{v}_{x}^{2}=\frac{1}{2} m \bar{v}_{y}^{2}=\frac{1}{2} m \bar{v}_{z}^{2}=\frac{1}{2} k_{\mathrm{B}} T
$$
gives
$$
\frac{1}{2} m \bar{v}^{2}=\frac{3}{2} k_{\mathrm{B}} T,
$$
or
$$
\bar{v}^{2}=\sqrt{\frac{3 k_{\mathrm{B}} T}{m}} .
$$
The rms distance $l$ between successive trappings is given by
$$
l^{2}=\bar{v}^{2} t^{2} .
$$
Hence
$$
l=\sqrt{\frac{3 k_{\mathrm{B}} T}{m}} t=\sqrt{3\left(\frac{k_{\mathrm{B}} T}{e}\right) \frac{e}{m}} t .
$$
With $\frac{k_{\mathrm{B}} T}{e}=26 \times 10^{-3} \mathrm{~V}, \frac{e}{m}=1.76 \times 10^{11} \mathrm{C} \mathrm{kg}^{-1}, t=10^{-5} \mathrm{~s}$, we have $l=1.17 \mathrm{~m}$. |
A closely wound search coil has an area of $4 \mathrm{~cm}^{2}, 160$ turns and a resistance of $50 \Omega$. It is connected to a ballistic galvanometer whose resistance is $30 \Omega$. When the coil rotates quickly from a position parallel to a uniform magnetic field to one perpendicular, the galvanometer indicates a charge of $4 \times 10^{-5} \mathrm{C}$. What is the flux density of the magnetic field? | Suppose the coil rotates from a position parallel to the uniform magnetic field to one perpendicular in time $\Delta t$. Since $\Delta t$ is very short, we have
$$
\varepsilon=\frac{\Delta \phi}{\Delta t}=i(R+r)
$$
As $q=i \Delta t$, the increase of the magnetic flux is
$$
\Delta \phi=q(R+r)=B A N,
$$
since the coil is now perpendicular to the field. Hence the magnetic flux density is
$$
\begin{aligned}
B &=\frac{(R+r) q}{A N}=\frac{(50+30) \times\left(4 \times 10^{-5}\right)}{4 \times 10^{-4} \times 160} \\
&=0.05 \mathrm{~T}=50 \mathrm{Gs} .
\end{aligned}
$$ |
A solenoid has an air core of length $0.5 \mathrm{~m}$, cross section $1 \mathrm{~cm}^{2}$, and 1000 turns. A secondary winding wrapped around the center of the solenoid has 100 turns. A constant current of $1 \mathrm{~A}$ flows in the secondary winding and the solenoid is connected to a load of $10^{3}$ ohms. The constant current is suddenly stopped. How much charge flows through the resistance? (Neglect end effects.) | Let the current in the winding of the solenoid be $i$. The magnetic induction inside the solenoid is then $B=\mu_{0} n i$ with direction along the axis, $n$ being the number of turns per unit length of the winding.
The total magnetic flux linkage is
$$
\psi=N \phi=N B S=N^{2} \mu_{0} S i / l .
$$
Hence the self-inductance is
$$
\begin{aligned}
L &=\frac{\psi}{i}=N^{2} \mu_{0} S / l \\
& \approx \frac{1000^{2} \times 4 \pi \times 10^{-7} \times 10^{-4}}{1 / 2}=2.513 \times 10^{-4} \mathrm{H} .
\end{aligned}
$$
The total magnetic flux linkage in the secondary winding produced by the currrent $i$ is $\psi^{\prime}=N^{\prime} \phi$, giving the mutual inductance as
$$
M=\frac{\psi^{\prime}}{i}=\frac{N N^{\prime} \mu_{o} S}{l}=2.513 \times 10^{-5} \mathrm{H} .
$$
Because of the magnetic flux linkage $\psi^{\prime}=M I, I$ being the current in the secondary, an emf will be induced in the solenoid when the constant current $I$ in the secondary is suddenly stopped. Kirchhoff's law gives for the induced current $i$ in the solenoid
$$
-\frac{d \psi^{\prime}}{d t}=R i+L \frac{d i}{d t},
$$
or
$$
-d \psi^{\prime}=R i d t+L d i=R d q+L d i
$$
Integrating over $t$ from $t=0$ to $t=\infty$ gives $-\Delta \psi^{\prime}=R q$, since $i(0)=$ $i(\infty)=0$. Thus the total charge passing through the resistance is
$$
q=\frac{-\Delta \psi^{\prime}}{R}=\frac{M I}{R}=\frac{2.513 \times 10^{-5} \times 1}{10^{3}}=2.76 \times 10^{-7} \mathrm{C}
$$ |
In a region of empty space, the magnetic field (in Gaussian units) is described by
$$
\mathrm{B}=B_{0} e^{a x} \hat{\mathbf{e}}_{z} \sin w
$$
where $w=k y-\omega t$.
Find the speed of propagation $v$ of this field. | Express $B$ as $\operatorname{Im}\left(B_{0} e^{a x} e^{i w}\right) \hat{\mathbf{e}}_{z}$.
Using Maxwell's equation
$$
\nabla \times \mathbf{B}=\frac{1}{c} \frac{\partial \mathrm{E}}{\partial t}
$$
and the definition $k=\frac{\omega}{c}$ for empty space, we obtain
$$
\mathbf{E}=\frac{i c}{\omega} \nabla \times \mathbf{B}=\frac{i}{k}\left|\begin{array}{ccc}
\mathbf{e}_{x} & \mathbf{e}_{y} & \mathbf{e}_{z} \\
\frac{\partial}{\partial x} & \frac{\partial}{\partial y} & 0 \\
0 & 0 & B_{z}
\end{array}\right|
$$
where $\frac{\partial}{\partial z}=0$ as $\mathbf{B}$ does not depend on $z$.
Hence
$$
\begin{gathered}
E_{x}=\operatorname{Im}\left(\frac{i}{k} B_{0} e^{a x} i k e^{i w}\right)=-B_{0} e^{a x} \sin w \\
E_{y}=\operatorname{Im}\left(-\frac{i}{k} B_{0} a e^{a x} e^{i w}\right)=-\frac{a c}{\omega} B_{0} e^{a x} \cos w \\
E_{z}=0
\end{gathered}
$$If the wave form remains unchanged during propagation, we have
$$
d w=k d y-w d t=0,
$$
or $\frac{d y}{d t}=\frac{w}{k}=c$. Hence the wave propagates along the $y$ direction with a speed $v=c$. |
The electric field of an electromagnetic wave in vacuum is given by
$$
\begin{gathered}
E_{x}=0, \\
E_{y}=30 \cos \left(2 \pi \times 10^{8} t-\frac{2 \pi}{3} x\right), \\
E_{z}=0,
\end{gathered}
$$
where $E$ is in volts/meter, $t$ in seconds, and $x$ in meters. Determine the wavelength $\lambda$. | $$
k=\frac{2 \pi}{3} \mathrm{~m}^{-1}, \quad \omega=2 \pi \times 10^{8} \mathrm{~s}^{-1} .
$$
$f=\frac{\omega}{2 \pi}=10^{8} \mathrm{~Hz}$.$\lambda=\frac{2 \pi}{k}=3 \mathrm{~m}$. |
A common lecture demonstration is as follows: hold or clamp a onemeter long thin aluminium bar at the center, strike one end longitudinally (i.e. parallel to the axis of the bar) with a hammer, and the result is a sound wave of frequency $2500 \mathrm{~Hz}$.
From this experiment, calculate the speed of sound in aluminium. | The point where the bar is struck is an antinode and the point where it is held a node. With the bar held at the center and its one end struck, the wavelength $\lambda$ is related to its length $L$ by $\lambda=2 L$. Hence the speed of sound propagation in the aluminium bar is
$$
v_{\mathrm{Al}}=\nu \lambda=2 \nu L=2 \times 2500 \times 1=5000 \mathrm{~m} / \mathrm{s} \text {. }
$$
The speed of sound in a solid is
$$
v=\sqrt{\frac{Y}{\rho}},
$$
where $Y$ is the Young's modulus of its material and $\rho$ its density. The speed of sound in a fluid is
$$
v=\sqrt{\frac{M}{\rho}}
$$
where $M$ is its bulk modulus and $\rho$ its density. For adiabatic compression of a gas, $M=\gamma p$, where $p$ is its pressure and $\gamma$ the ratio of its principal specific heats; $\gamma=1.4$ for air, a diatomic gas. Hence
$$
\frac{v_{\text {air }}}{v_{\mathrm{Al}}}=\sqrt{\frac{1.4 p \rho_{\mathrm{Al}}}{Y \rho_{\mathrm{air}}}} .
$$
With
$$
\begin{aligned}
p &=1.013 \times 10^{6} \mathrm{dyn} / \mathrm{cm}^{2} \quad(\text { standard atmosphere) }\\
Y &=7.05 \times 10^{11} \mathrm{dyn} / \mathrm{cm}^{2} \\
\rho_{\mathrm{Al}} &=2.7 \mathrm{~g} / \mathrm{cm}^{3} \\
\rho_{\text {air }} &=1.165 \times 10^{-3} \mathrm{~g} / \mathrm{cm}^{3} \quad\left(\text { at } 30^{\circ} \mathrm{C}\right) \\
v_{\text {air }} &=6.83 \times 10^{-2} \times 5000=341 \mathrm{~m} / \mathrm{s} .
\end{aligned}
$$$v_{\mathrm{Al}}=5000 \mathrm{~m} / \mathrm{s}$. |