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10600 | https://www.onlinemathlearning.com/average-problems.html | Algebra: Average Word Problems
These lessons look at three main types of algebra average word problems commonly encountered in school or in tests like the SAT: Average (Arithmetic Mean), Weighted Average and Average Speed.
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Related Pages Average (Arithmetic Mean) Average Speed Problems More Algebra Word Problems
The following table shows three formulas used in average problems: Average (Arithmetic Mean), Weighted Average and Average Speed. Scroll down the page for examples and solutions.
Average (Arithmetic Mean)
The average (arithmetic mean) uses the formula:
The formula can also be written asSum of Terms = Average × Number of Terms
Example: The average (arithmetic mean) of a list of 6 numbers is 20. If we remove one of the numbers, the average of the remaining numbers is 15. What is the number that was removed?
Solution: Step 1: The removed number could be obtained by difference between the sum of original 6 numbers and the sum of remaining 5 numbers i.e.
Removed number = sum of original 6 numbers – sum of remaining 5 numbers
Step 2: Using the formula
Sum of Terms = Average × Number of Terms
sum of original 6 numbers = 20 × 6 = 120sum of remaining 5 numbers = 15 × 5 = 75
Step 3: Using the formula from step 1
Removed number = sum of original 6 numbers – sum of remaining 5 numbers120 – 75 = 45
Answer: The number removed is 45.
More Difficult Word Problems with Averages
Examples:
Timothy’s average score on the first 4 tests was 76. On the next 5 tests his average score was 85. What was his average score on all 9 tests?
Tracy mowed lawns for 2 hours and earned $7.40 per hour. Then she washed windows for 3 hours and earned $6.50 per hour. What were Tracy’s average earnings per hour for all 5 hours?
After taking 3 quizzes, your average is 72 out of 100. What must your average be on the 5 quizzes to increase your average to 77?
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How to solve algebra average problem?
Example: If the average (arithmetic mean) of 8,11,25,and p is 15, find 8 + 11 + 25 + p and then find p.
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Example: If a = 3b = 6c, what is the average (arithmetic mean) of a, b and c in terms of a?
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Weighted Average
Another type of average problem involves the weighted average - which is the average of two or more terms that do not all have the same number of members. To find the weighted term, multiply each term by its weighting factor, which is the number of times each term occurs.
The formula for weighted average is:
Example: A class of 25 students took a science test. 10 students had an average (arithmetic mean) score of 80. The other students had an average score of 60. What is the average score of the whole class?
Solution: Step 1: To get the sum of weighted terms, multiply each average by the number of students that had that average and then sum them up.
80 × 10 + 60 × 15 = 800 + 900 = 1700
Step 2: Total number of terms = Total number of students = 25
Step 3: Using the formula
Answer: The average score of the whole class is 68.
Be careful! You will get the wrong answer if you add the two average scores and divide the answer by two.
How to calculate a weighted mean (weighted average)?
Example: Fifteen accounting majors have an average grade of 90. Seven marketing majors averaged 85, and ten finance majors averaged 93. What is the weighted mean for the 32 students?
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How to use weighted average to calculate the average score of a student?
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Average Speed
Computation of average speed is a trickier type of average problems. Average speed uses the formula:
Example: John drove for 3 hours at a rate of 50 miles per hour and for 2 hours at 60 miles per hour. What was his average speed for the whole journey?
Solution: Step 1: The formula for distance is
Distance = Rate × Time Total distance = 50 × 3 + 60 × 2 = 270
Step 2: Total time = 3 + 2 = 5
Step 3: Using the formula
Answer: The average speed is 54 miles per hour.
Be careful! You will get the wrong answer if you add the two speeds and divide the answer by two.
How to calculate average speed?
Examples:
Mr. Myones drives 3 hours at an average speed of 40 miles per hour. Then he drives 2 hours at a speed of 35 miles per hour. What is his average speed for the whole trip?
Miss Holton drives 4 hours at an average speed of 30 miles per hour. Then she drives 2 hours at a speed of 45 miles per hour. What is her average speed for the whole trip?
A family took 2 hours to drive from City A to City B at a speed of 55 miles per hour. On the way home they took 3 hours at a speed of 40 miles per hour. What was their average speed for the whole trip?
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How to find average speed?
Example: A rocket traveled at 3000 mph on the way to fix the Hubble Space Telescope. On the way back, it was traveled at 1000 mph. What was the average speed?
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10601 | https://math.stackexchange.com/questions/754708/subsets-of-cyclic-group-with-distinct-pairwise-differences | Skip to main content
Subsets of cyclic group with distinct pairwise differences
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Given a number m∈N, let Zm={0,1,…,m−1} denote the ring of integers modulo m (although we won't need multiplication, so any cyclic group of order m will do). Given a second number k∈N I'm looking for subsets of k elements from Zm such that no pair-wise difference occurs more than once. More formally, I'm looking for some A⊂Zm with |A|=k and
∀a,b,c∈A:|{a,b,c}|=3→a+b−c∉A(modm)
Is there a name for such a kind of set? I guess it might be somehow related to Golomb rulers, but the cyclic nature is not common to rulers as far as I know.
Is there a known method to efficiently enumerate such sets for given m and k?
Is there a known theorem concerning the existence of such subsets for specific m and k?
I know I'm asking three distinct questions here, but I very much hope that someone may be able to address more than one, perhaps by providing a good reference. Knowing a name will probably help me locate suitable literature. But answers addressing only one of these questions are welcome as well.
Update: It seems that for the special case of m=k2−k+1 where every possible difference has to actually occur exactly once, the kind of set I defined would be called an (m,k,1) cyclic difference set. I'm still interested in a more general term for cases where some differences are missing.
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asked Apr 15, 2014 at 11:54
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No doubt this answer comes too late, but for a good record: such sets are called Sidon sets (in Zm), and there is a vast amount of literature about them, starting from this Wikipedia entry. There is no way to efficiently enumerate such sets. For your third question, since a subset of a Sidon set is (trivially) Sidon, you are in fact asking how large can a Sidon subset of the group Zm be; this problem has received much attention also, one can find lots of references just googling for "Sidon sets".
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answered Feb 15, 2019 at 17:43
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10602 | https://www.bcbsm.com/amslibs/content/dam/public/mpr/mprsearch/pdf/2191151.pdf | 1 Medicare Advantage Medical Policy Auditory Brain Stem Implant – Medicare Advantage Medicare Advantage Plan UM Committee Approval Date: 7/17/2024 Medicare Plus BlueSM Effective Date: 7/17/2024 BCN AdvantageSM Description An auditory brain stem implant (ABI) is a device that was developed to provide auditory rehabilitation to patients who are deaf and ineligible for cochlear implant surgery. The implant has two parts: the processor, which is worn near the ear, and a surgically implanted electrode which is implanted near the brainstem. A microphone in the processor will picks up sound waves and convert them into electrical signals. The electrical signals are then transmitted to an electrode near the brain stem. Medical studies show that auditory brainstem implants are successful in people whose hearing was damaged by surgery to remove tumors that were specifically caused by neurofibromatosis type II. Coverage Determination BCBSM/BCN follows guidance from the Centers for Medicare and Medicaid Services (CMS) when performing organization (coverage) determinations for Medicare Advantage plan members. CMS Medicare statutes, regulations, manuals, National Coverage Determinations (NCDs), Local Coverage Determinations (LCDs), and Local Coverage Articles (LCAs) provide the clinical guidelines for coverage determinations. When CMS Medicare guidance is not fully established, BCBSM/BCN may use LCD/LCAs outside the services area, independent criteria, internal coverage criteria, or research from independent medical research repositories (i.e., Hayes) for coverage determinations. BCBSM/BCN internal medical coverage policies are developed and based on current evidence in widely accepted treatment guidelines or clinical literature; in addition, they address how the clinical benefits are highly likely to outweigh any clinical harms, including (but not limited to) delayed or decreased access to services. The following is applicable for this medical policy: General guidance in the Medicare Benefit Policy Manual Chapter 16 section 100 - Hearing Aids and Auditory implants defines auditory brain stem implants, as well as cochlear implants and osseointegrated devices, as prosthetic devices separate from hearing aids; these devices may be covered in certain situations. However, no specific clinical indications for medical necessity of an auditory brain stem implant were provided in the Manual. After searching the Medicare Coverage Database, it was also determined that CMS does NOT have specific 2 NCD/LCD/LCA guidance related to procedural code S2235 for ABI. CMS permits a health plan to augment general provisions when Medicare guidance is not fully established 42 CFR § 422.101 (b)(6). BCBSM/BCN internal policy coverage criteria will be applied. This service may be medically necessary when the criteria specified in the BCBSM/BCN medical policy are met. CPT® codes, descriptions and materials are copyrighted by the American Medical Association (AMA). HCPCS® codes, descriptions and materials are copyrighted by Centers for Medicare and Medicaid Services (CMS). The above Medicare information is current as of the review date for this policy. However, the coverage issues and policies maintained by CMS are updated and/or revised periodically. Therefore, the most current CMS information may not be contained in this document. Please refer to the Medicare Coverage Database website at for the most current applicable NCD, LCD, LCA, and CMS Online Manual System/Transmittals. CMS Medicare Administrative Contractors (MAC) Jurisdictions Part A and Part B Medicare Administrative Contractor (MAC) Jurisdiction Wisconsin Physicians Service Insurance Corporation (WPS) IA, IN, KS, MI, MO, NE CGS Administrators, LLC KY, OH First Coast Service Options, Inc FL, PR, US VI National Government Services, Inc (NGS) CT, IL, MA, ME, MN, NH, NY, RI, VT, WI, Noridian Healthcare Solutions, LLC, Jurisdiction JE CA, HI, NV, American Samoa, Guam, Northern Mariana Islands Noridian Healthcare Solutions, LLC, Jurisdiction JF AK, AZ, ID, MT, ND, OR, SD, UT, WA, WY Novitas Solutions, Inc AR, CO, DC, DE, LA, MD, MS, NJ, NM, OK, PA, TX, Palmetto GBA AL, GA, TN NC, SC, VA, WV DME MAC Jurisdictions CGS Administrators, LLC, Jurisdictions JB and JC AL, AR, CO, FL, GA, IL, IN, KY, LA, MI, MN, MS, NM, NC, OH, OK, Puerto Rico, SC, TN, TX, VA, Virgin Island, WI, WV Noridian Healthcare Solutions, LLC, Jurisdictions JA and JD AK, American Soamoa, AZ, CA, CT, DE, DC, Guam, HI, ID, IA, KS, MA, MD, ME, MO, MT, ND, NE, NH, NJ, NV, NY, Northern Mariana Islands, OR, PA, RI, SD, UT, VT, WA, WY Code Description NCD LCD/LCA Additional Guidance S2235 Implantation of Auditory brain stem implant None None As no NCDs or LCDs are available for guidance Refer to the BCBSM/BCN policy: Auditory Brain Stem Implant Note: Medical policy attached below 3 Important Reminder BCBSM/BCN adheres to CMS Medicare coverage guidance to limit coverage to items and services that are reasonable and necessary for the diagnosis or treatment of an illness or injury or to improve the functioning of a malformed body member. Medicare Advantage medical coverage policies list the criteria our clinicians use to decide when medical services are considered “reasonable and necessary.” Note: U.S. Food and Drug Administration (FDA) approval for a specific indication or the issuance of a CPT code is not sufficient for a procedure to be considered medically reasonable and necessary. Similarly, the presence of a procedure/device code or payment amount for the service in the Medicare fee schedule does not necessarily indicate coverage. If a service is deemed not reasonable and necessary, to treat illness or injury for any reason (including lack of safety and efficacy because it is an experimental procedure, etc.), the procedure is considered not covered. Disclaimer: The Medical Coverage Policies are reviewed by the BCBSM/BCN Utilization Management Committee. Policies in this document may be modified by a member’s coverage document. The existence of the medical policy is not an authorization, certification, explanation of benefits, or a contract for the services, devices, or drugs that is referenced in the medical policy. Medical policies do not constitute medical advice and do not guarantee any results or outcomes. The medical policy is not intended to replace independent medical judgment for treatment of individuals. Treating physicians and health care providers are solely responsible for determining what care to provide to their patients. Identification of selected brand names of devices, tests and procedures in a medical coverage policy is for reference only and is not an endorsement of any one device, test, or procedure over another. Pursuant to Section 1557 and Section 504, Blue Cross does not discriminate on the basis of race, color, national origin, age, disability, or sex (including sex characteristics, intersex traits; pregnancy or related conditions; sexual orientation; gender identity, and sex stereotypes). This includes our rules, benefit designs and medical policies. BCBSM/BCN Medicare Advantage Policy History Policy Effective Date UM Committee Approval Date Comments 07/17/2024 07/17/2024 Medicare Advantage policy established 1 Medical Policy Joint Medical Policies are a source for BCBSM and BCN medical policy information only. These documents are not to be used to determine benefits or reimbursement. Please reference the appropriate certificate or contract for benefit information. This policy may be updated and is therefore subject to change. Current Policy Effective Date: 11/1/23 (See policy history boxes for previous effective dates) Title: Auditory Brain Stem Implant Description/Background The auditory brain implant (ABI) is intended to restore some hearing in people with neurofibromatosis type 2 (NF2) who are rendered deaf by bilateral removal of the characteristic neurofibromas involving the auditory nerve. The ABI consists of an externally worn speech processor that provides auditory information to an electrical signal that is transferred to a receiver/stimulator that is implanted in the temporal bone. The receiver stimulator is, in turn, attached to an electrode array that is implanted on the surface of the cochlear nerve in the brainstem, thus bypassing the inner ear and auditory nerve. The electrode stimulates multiple sites on the cochlear nucleus, which is then processed normally by the brain. To place the electrode array on the surface of the cochlear nucleus, the surgeon must be able to visualize specific anatomic landmarks. Because large neurofibromas compress the brainstem and distort the underlying anatomy, it can be difficult or impossible for the surgeon to correctly place the electrode array. For this reason, patients with large, long-standing tumors may not benefit from the device.1 ABIs are also being studied to determine whether they can restore hearing for other non-neurofibromatosis causes of hearing impairment in adults and children, including absence of or trauma to the cochlea or auditory nerve. It is estimated that 1.7 per 100,000 children are affected by bilateral cochlea or cochlear nerve aplasia and 2.6 per 100,000 children are affected by bilateral cochlea or cochlear nerve hypoplasia.2 Regulatory Status One device has received approval by the U.S. Food and Drug Administration (FDA) for auditory brainstem implantation: the Nucleus 24® Auditory Brainstem Implant System (Cochlear Corporation). The speech processor and receiver are similar to the devices used in cochlear implants; the electrode array placed on the brainstem is the novel component of the device. The device is indicated for individuals 12 years of age or older who have been diagnosed with neurofibromatosis type 2 (NF2). The Nucleus® 24 Auditory Brainstem Implant System approval was based on the efficacy study of unilateral implants either at first-side or second-side tumor removal surgery.”2 The Nucleus® 24 is now obsolete. 2 In June 2016, the Nucleus ABI541 Auditory Brainstem Implant (Cochlear Corp.) was approved by FDA through a supplement to the premarket approval for the Nucleus® 24. The new implant is indicated for individuals 12 years of age or older who have been diagnosed with neurofibromatosis type 2. FDA product code: MCM. Medical Policy Statement The safety and effectiveness of unilateral auditory brain stem implantation have been established. It may be considered a useful therapeutic option for patients meeting selection criteria. Inclusionary and Exclusionary Guidelines (Clinically based guidelines that may support individual consideration and pre-authorization decisions) Inclusions: (must meet both): • At least 12 years of age or older who have neurofibromatosis Type II AND • Who are rendered deaf due to bilateral resection of neurofibromas of the auditory nerve. Exclusions: • Use of an auditory brain stem implant for all other conditions, including, but not limited to, the use of ABI for children with cochlear nerve deficiency, aplasia or malformation. • Bilateral use of an ABI who otherwise meet criteria for ABI. • Penetrating electrode auditory brainstem implant (PABI) CPT/HCPCS Level II Codes (Note: The inclusion of a code in this list is not a guarantee of coverage. Please refer to the medical policy statement to determine the status of a given procedure.) Established codes: S2235 Other codes (investigational, not medically necessary, etc.): N/A Rationale Evidence reviews assess the clinical evidence to determine whether the use of a technology improves the net health outcome. Broadly defined, health outcomes are length of life, quality of life, and ability to function—including benefits and harms. Every clinical condition has specific outcomes that are important to patients and to managing the course of that condition. Validated outcome measures are necessary to ascertain whether a condition improves or 3 worsens; and whether the magnitude of that change is clinically significant. The net health outcome is a balance of benefits and harms. To assess whether the evidence is sufficient to draw conclusions about the net health outcome of a technology, 2 domains are examined: the relevance and the quality and credibility. To be relevant, studies must represent one or more intended clinical use of the technology in the intended population and compare an effective and appropriate alternative at a comparable intensity. For some conditions, the alternative will be supportive care or surveillance. The quality and credibility of the evidence depend on study design and conduct, minimizing bias and confounding that can generate incorrect findings. The randomized controlled trial is preferred to assess efficacy; however, in some circumstances, nonrandomized studies may be adequate. Randomized controlled trials are rarely large enough or long enough to capture less common adverse events and long-term effects. Other types of studies can be used for these purposes and to assess generalizability to broader clinical populations and settings of clinical practice. Promotion of greater diversity and inclusion in clinical research of historically marginalized groups (e.g., People of Color [African-American, Asian, Black, Latino and Native American]; LGBTQIA (Lesbian, Gay, Bisexual, Transgender, Queer, Intersex, Asexual); Women; and People with Disabilities [Physical and Invisible]) allows policy populations to be more reflective of and findings more applicable to our diverse members. While we also strive to use inclusive language related to these groups in our policies, use of gender-specific nouns (e.g., women, men, sisters, etc.) will continue when reflective of language used in publications describing study populations. ABI FOR BILATERAL RESECTION OF NEUROFIBROMAS OF THE AUDITORY NERVE Clinical Context and Therapy Purpose The purpose of an auditory brainstem implant in individuals who are deaf due to bilateral resection of neurofibromas of the auditory nerve is to provide a treatment option that is an alternative to observation alone. The following PICO was used to select literature to inform this review. Populations The relevant population(s) of interest are individuals who are deaf and have undergone bilateral resection of neurofibromas of the auditory nerve. Interventions The therapy being considered is an auditory brainstem implant. Comparators The following practice is currently being used to make decisions about hearing restoration in individuals who are deaf due to bilateral resection of neurofibromas of the auditory nerve: observation alone. Outcomes 4 The general outcomes of interest are functional outcomes, quality of life and treatment-related morbidity. Functional outcomes include change in hearing and hearing-related function (e.g. sound recognition and speech perception). Study Selection Criteria Methodologically credible studies were selected using the following principles: • To assess efficacy outcomes, comparative controlled prospective trials were sought, with a preference for RCTs; • In the absence of such trials, comparative observational studies were sought, with a preference for prospective studies. • To assess long-term outcomes and adverse events, single-arm studies that capture longer periods of follow-up and/or larger populations were sought. • Consistent with a 'best available evidence approach,' within each category of study design, studies with larger sample sizes and longer durations were sought. • Studies with duplicative or overlapping populations were excluded. Review of Evidence FDA approval of the Nucleus 24® Auditory Brainstem Implant System was based on results in a case series of 90 patients with neurofibromatosis type 2 (NF2), ages 12 years and older.1,4 Of the 90 subjects evaluated, 28 complications occurred in 26 patients; 26 of these complications resolved without surgical or extensive medical intervention. Two patients had infections of the postoperative flap requiring explantation of the device. A total of 60 patients had a minimum experience of 3 to 6 months with the device, and thus effectiveness outcomes were also evaluated. Overall device benefit was defined as a significant enhancement of lip-reading or an above-chance improvement on sound-alone tests. Based on this definition, a total of 95% patients (57 of 60) derived benefit from the device. While the use of an auditory brainstem implant (ABI) is associated with a very modest improvement in hearing, this level of improvement is considered significant in this group of patients who have no other treatment options. Among the 90 patients receiving the implant, 16 did not receive auditory stimulation from the device postoperatively, either due to migration of the implanted electrodes or surgical misplacement. A single small (N=10) trial from 2008 was identified on a penetrating auditory brainstem implant (PABI).7 This prospective clinical trial enrolled patients with NF2 who received a PABI after vestibular schwannoma removal. The PABI is an extension of the ABI technology that uses surface electrodes on cochlear nuclei. The PABI uses 8 or 10 penetrating microelectrodes in conjunction with a separate array of 10 to 13 surface electrodes. The PABI met the goals of lower threshold, increased pitch range, and high selectivity, but these properties did not improve speech recognition. A systematic review conducted by Ontario (Canada) Health as part of a Health Technology Assessment included 16 observational studies (N=491) comparing the effectiveness of ABI to no treatment in adults with NF2 (Table 1 and Table 2).5 Risk of bias among the included studies was assessed using the Risk of Bias in Non-randomized Studies - of Interventions (ROBINS-I) tool, and overall quality of evidence was assessed using the Grading of Recommendation, Assessment, Development and Evaluation (GRADE) Handbook. Results were reported qualitatively, and no meta-analyses were conducted due to heterogeneity in testing conditions and outcomes. The review found high quality of evidence of benefit of ABI on sound recognition (7 studies), speech perception with lip reading (5 studies) and subjective 5 hearing benefit (5 studies). Evidence favoring ABI was moderate for speech perception without lip reading (10 studies) and low for quality of life (1 study). The most commonly reported surgical complications, based on low quality evidence from 12 studies, were cerebrospinal fluid leak in 3% to 15% of participants and infection in 10% to 13% or participants. Table 1. SR-MA Characteristics Study Dates Trials Participants N (Range) Design Duration Ontario Health 1993-2016; literature searches conducted through June 2018 19 observational studies Adults with NF2 who were not candidates for cochlear implantation 491 (8-61) 6 prospective cohort studies 11 retrospective cohort studies 2 cross sectional studies 1 month to 18 years (mean, median not reported) Table 2. SR-MA Results Study Sound Recognition Speech Perception Subjective Benefits of Hearing Quality of Life Surgical Complications Ontario Health5 ABI vs. no treatment ABI vs. no treatment ABI vs. no treatment ABI vs. no treatment ABI vs. no treatment Number of studies; N 7 observational studies; N=169 15 observational studies; N=348 5 observational studies; N=141 1 observational study; N=11 12 observational studies; N= Qualitative assessment of ABI effectiveness Allows any degree of improvement in sound recognition vs. no treatment ABI only: Likely allows any degree of improvement in speech perception when used alone ABI + lip reading: Allows any degree of improvement in speech perception when used in conjunction with lip-reading Provides subjective benefits of hearing May improve quality of life Most common complications were cerebrospinal fluid leak infection Level of evidence (GRADE) High ABI only: Moderate ABI + lip reading: High High Low Low 6 Section Summary: ABI for Bilateral Resection of Neurofibromas of the Auditory Nerve The evidence on ABI for bilateral resection of neurofibromas of the auditory nerve includes case series, literature reviews and a clinical trial and a systematic review of small observational studies. A 2018 case series of 90 adults, 60 of which had the minimum experience of 3 to 6 months with the Nucleus 24 ABI system, suggested that adults may benefit from its usage. European studies followed 32 patients, 24 of which with an ABI activated experienced significant improvements on the Sound Effects Recognition Test and Monosyllable-Trochee-Polysyllable test. An Ontario (Canada) Health systematic review found ABI associated with better hearing function relative to no treatment, but evidence on other outcomes was limited. ABI FOR NONTUMOR ETIOLOGIES Review of Evidence Adults Merkus et al (2014) conducted a systematic review of ABIs for non-NF2 indications in 2014.6 Included in the review were 144 non-NF2 ABI cases from 31 articles. Non-NF2 indications for which ABIs have been evaluated included cochlear otosclerosis, temporal bone fractures, bilateral traumatic cochlear nerve disruption, autoimmune inner ear disease, auditory neuropathy, cochlear nerve aplasia and when vestibular schwannoma is in the only hearing ear. Cochlear implants generally resulted in better hearing than ABIs when the cochlea and cochlear nerve were intact. Complete bilateral disruption of the cochlear nerve from trauma did not exist in the literature and cochlear malformation did not preclude cochlear implant. While the evidence is limited, it appears cochlear implants demonstrate greater hearing benefits than ABI in patients with non-NF2 indications. In a 2014 systematic review by Medina et al of ABI for traumatic deafness, cochlear implant was found to perform better than ABI.7 However, there is limited evidence available to draw conclusions, because only 3 articles totaling 7 patients were identified in the review on ABI for traumatic deafness. Children Systematic Reviews A 2015 systematic review of nontumor pediatric ABI outcomes was reported by Noij et al (2015).8 It included 21 studies with 162 children, at a mean age of 4.3 years (range, 11 months to 17 years). Nine reports were from a single group from Italy (described further below) and it could not be determined if there was patient overlap across these studies. Nearly all studies were retrospective series or cohorts; one was a case-control. Most children (63.6%) had cochlear nerve aplasia. Other conditions were cochlear aplasia, cochlear nerve hypoplasia, cochlear malformations, ossified cochlea, auditory neuropathy, trauma, and cochlear hypoplasia. Twenty-five percent of the patients had previously received a cochlear implant. Forty major and minor implant-related complications were reported, the most common being cerebrospinal fluid (CSF) leak (8.5% of patients).The most common side effects associated with ABI use were discomfort of the body and/or limb, dizziness/vertigo/nystagmus, pain in the head and/or neck, and stimulation of the facial nerve or involuntary swallowing, gagging, or coughing. A variety of auditory tests were used; the most common (6 studies) was the 7 Categories of Auditory Performance (CAP) index (range, 0-7; high score indicates better hearing). There was an improvement in CAP scores over time. After 5 years, almost 50% of patients had CAP scores greater than 4 (5 [understanding of common phrases without lip reading] to 7 [use of telephone with known speaker]). Children who also had nonauditory disabilities never attained a CAP score greater than 4. There was no significant effect of the age of implantation. Case Series Many of the larger series on ABI in nontumor patients are from a group that includes Collettti and Colletti. In 2013, L. Colletti reported on ABIs in 21 children, ranging in age from 1.7 to 5 years, with deafness unrelated to neurofibromatosis, which had a poor response to cochlear implants.9 At surgery, the cochlear nerve was absent in each patient. Significant improvement in Category of Auditory Performance scale was seen after ABI (p<0.001). In 2016, Sennaroglu et al reported follow-up of at least 1 year on 35 children who had received ABI.10 This followed a 2009 preliminary report of 11 prelingually deaf children ages 30 to 56 months who received an ABI.11 Sixty children had received an ABI from this center in Turkey. The children who had received the ABI in the previous year were excluded from the 2016 analysis. Over half (n=19) of the cases were due to cochlear hypoplasia. ABI models implanted were Cochlear, Med El, and Neurelec. At regular follow-up, children were evaluated with the CAP, Speech Intelligibility Rate (SIR), Functional Auditory Performance of Cochlea Implantation (FAPCI), and Manchester scores. About half the children were in the CAP category 5 and could understand common phrases without lip reading. In the subgroup with better hearing thresholds (25-40 dB), some (17.6%) were able to understand conversation without lip reading, use the telephone with known speaker (11.8%), and follow group conversation in a noisy room (5.9%). For children with higher hearing thresholds (>50 dB), none exceeded CAP category 5. SIR and Manchester scores were also better with greater hearing thresholds. Auditory performance measured with the FAPCI was in the 10th percentile for all groups and was worse compared to cochlear implantation. As was also found in the Noij systematic review (discussed above), children with additional nonauditory handicaps had worse outcomes (e.g., intellectual disability). Mixed Populations Other reports from the group of Colletti and Colletti include a 2005 report on ABIs in 16 children and adults who had non-tumor diseases of the cochlear nerve or cochlea and 13 patients with NF2.12 Ages ranged from 14 months to 70 years; the non-tumor group included patients with head trauma, complete cochlear ossification, 1 child with auditory neuropathy, and 5 children with bilateral cochlear nerve aplasia. Following implantation, the adult non-tumor group scored substantially higher than the patients with NF2 in open set speech perception tests. Some of the children showed dramatic improvements in word and sentence recognition over a 1-year follow-up. Short-term adverse effects included dizziness or tingling sensations in the leg, arm, and throat (20 of 29 patients). Additional studies from this group have reported improvement in hearing with ABIs in “nontumor” patients, including a 2006 report on 54 nontumor patients13 and a 2007 report on 22 non-neurofibromatosis patients.14 In a 2010 retrospective review, Colletti et al, reported on the complications from ABI surgery in 83 adults and 31 children, 78 of whom had nontumor cochlear or cochlear nerve disorders.15 The authors found complication rates were similar to cochlear implant surgery. Additionally, major and minor complications were significantly fewer in nontumor patients than in NF2 8 patients. These authors concluded ABIs could be used in a wider population of patients than only those with NF2. Section Summary: ABI in Nontumor Etiologies The evidence on ABI in nontumor patients includes case series and systematic reviews of case series. A 2014 systematic review of adults suggested that ABI might improve outcomes in bilateral complete cochlear and inner ear aplasia. Recent research includes studies of children who are deaf but would not benefit from a cochlear implant. The most common conditions in these studies are cochlear aplasia and cochlear nerve aplasia. Hearing in this age group is critical for language development, and the ABI has potential to substantially improve health outcomes for this age group. However, studies of early (now obsolete) ABI devices found a high rate of failure in children and high rates of adverse events in adults. Evidence from ongoing studies assessing newer ABI models is needed to evaluate efficacy and durability in patients with nontumor ABI indications (Table 3). SUMMARY OF EVIDENCE For individuals who are deaf due to bilateral resection of neurofibromas of the auditory nerve who receive an auditory brainstem implant (ABI), the evidence includes a large prospective case series and a technology assessment that included observational studies. Relevant outcomes are functional outcomes, quality of life, and treatment-related morbidity. The technology assessment found the highest quality evidence for improvement in hearing function, but evidence on other outcomes was lacking. Relevant outcomes are functional outcomes, quality of life, and treatment-related morbidity. The U.S. Food and Drug Administration (FDA) approval of the Nucleus 24 device in 2000 was based on a prospective case series of 90 patients 12 years of age or older, of whom 60 had the implant for at least 3 months. From this group, 95% had a significant improvement in lip reading or improvement on sound-alone tests. While use of an ABI is associated with a very modest improvement in hearing, this level of improvement is considered significant for those patients who have no other treatment options. A systematic review of 16 studies found that ABI was associated with improved sound recognition and speech perception. Based on these results, ABIs are considered appropriate for the patient population age ≥12 years with NF2 and deafness following tumor removal. The evidence is sufficient to determine that the technology results in an improvement in the net health outcome. The evidence is sufficient to determine that the technology results in a meaningful improvement in the net health outcome. For individuals who are deaf due to nontumor etiologies who receive an ABI, the evidence includes case series and systematic reviews of case series. Relevant outcomes are functional outcomes, quality of life, and treatment-related morbidity. In general, ABIs have not demonstrated hearing benefits over cochlear implants for many non-NF2 and some older (now obsolete) ABI models have been associated with high rates of device failure and adverse events in this population. In addition, ABI studies have shown inferior outcomes in children with other disabilities. However, ABIs hold promise for select patients when the cochlea or cochlear nerve is absent. Evaluation is currently ongoing with the recently available Nucleus ABI541to determine its efficacy and durability in children. In addition, ABI studies have shown inferior outcomes in children with other disabilities. Thus, further study is also needed to define populations that would benefit from these devices. The evidence is insufficient to determine the effects of the technology on health outcomes. 9 ONGOING AND UNPUBLISHED CLINICAL TRIALS Some currently unpublished trials that might influence this review are listed in Table 1. Table 1. Summary of Key Trials NCT No. Trial Name Planned Enrollment Completion Date Ongoing NCT02310399 Auditory brainstem implant (ABI) in children with no chochleae or auditory nerves 20 May 2020 NCT02630589 Implantation of an auditory brainstem implant for the treatment of incapacitating unilateral tinnitus 10 Jan 2022 Unpublished NCT01736267 Study of nucleus 24 auditory brainstem implant (ABI) in adult non-neurofibromatosis type 2 subjects 10 Nov 2022 NCT01904448 An early feasibility study of the safety and efficacy of the nucleus 24 auditory brainstem implant in children with cochlear or cochlear nerve disorders not resulting from neurofibromatosis type II 10 Oct 2017 NCT: national clinical trial SUPPLEMENTAL INFORMATION Practice Guidelines and Position Statements National Institute for Clinical Excellence (NICE) In January 2005, National Institute for Clinical Excellence (NICE) issued Interventional Procedure Guidance 108, Auditory Brain Stem Implants.16 The guidance states the following: “…evidence on safety and efficacy of auditory brain stem implants appears adequate to support the use of this procedure by surgical teams experienced in this technique.” Government Regulations National: CMS Manual System Pub 100-02 – Medicare Benefit Policy. Revision 39. 11-10-05; Effective: 11-10-05; Implementation: 12-12-05) Section 1862(a)(7) of the Social Security Act states that no payment may be made under part A or part B for any expenses incurred for items or services “where such expenses are for . . . hearing aids or examinations therefore. . . .” This policy is further reiterated at 42 CFR 411.15(d) which specifically states that “hearing aids or examination for the purpose of prescribing, fitting, or changing hearing aids” are excluded from coverage. Hearing aids are amplifying devices that compensate for impaired hearing. Hearing aids include air conduction devices that provide acoustic energy to the cochlea via stimulation of the tympanic membrane with amplified sound. They also include bone conduction devices that provide mechanical energy to the cochlea via stimulation of the scalp with amplified mechanical vibration or by direct contact with the tympanic membrane or middle ear ossicles. Certain devices that produce perception of sound by replacing the function of the middle ear, cochlea or auditory nerve are payable by Medicare as prosthetic devices. These devices are 10 indicated only when hearing aids are medically inappropriate or cannot be utilized due to congenital malformations, chronic disease, severe sensorineural hearing loss or surgery. The following are prosthetic devices: • Cochlear implants and auditory brainstem implants, i.e., devices that replace the function of cochlear structures or auditory nerve and provide electrical energy to auditory nerve fibers and other neural tissue via implanted electrode arrays. • Osseointegrated implants, i.e., devices implanted in the skull that replace the function of the middle ear and provide mechanical energy to the cochlea via a mechanical transducer. Medicare contractors deny payment for an item or service that is associated with any hearing aid as defined above. See §180 for policy for the medically necessary treatment of complications of implantable hearing aids, such as medically necessary removals of implantable hearing aids due to infection.20 Local: There is no local coverage determination on this topic. (The above Medicare information is current as of the review date for this policy. However, the coverage issues and policies maintained by the Centers for Medicare & Medicare Services [CMS, formerly HCFA] are updated and/or revised periodically. Therefore, the most current CMS information may not be contained in this document. For the most current information, the reader should contact an official Medicare source.) Related Policies • Bone Anchored Hearing Devices • Cochlear Implants • Intraoral Bone Conduction Hearing Devices References 1. Food and Drug Administration. Nucleus 24 Auditory Brainstem Implant System. FDA summary of Safety and Effectiveness. 2000; Accessed June 2023. 2. Kaplan AB, Kozin ED, Puram SV, et al. Auditory brainstem implant candidacy in the United States in children 0-17 years old. Int J Pediatr Otorhinolaryngol. Mar 2015;79(3):310-315. PMID 25577282 3. Food and Drug Administration. Premarket Approval (PMA). Nucleus AB1541 Auditory Brainstem Implant. 2016. Accessed June 2023. 4. Ebinger K, Otto S, Arcaroli J, et al. Multichannel auditory brainstem implant: US clinical trial results. J Laryngol Otol Suppl. 2000(27):50-53. PMID 11211440 5. Ontario Health (Quality). Auditory brainstem implantation for adults with neurofibromatosis 2 or severe inner ear abnormalities: a health technology assessment. Ont Health Technol Assess Ser [Internet]. 2020 Mar;20(4): 185. assessment/reviews-and-recommendations/auditory-brainstem- 11 implantation-for-adults-with- neurofibromatosis-2-or-severe-inner-ear-abnormalities. Accessed June 2023. 6. Merkus P, Di Lella F, Di Trapani G, et al. Indications and contraindications of auditory brainstem implants: systematic review and illustrative cases. Eur Arch Otorhinolaryngol. Jan 2014;271(1):3-13. PMID 23404468 7. Medina M, Di Lella F, Di Trapani G, et al. Cochlear implantation versus auditory brainstem implantation in bilateral total deafness after head trauma: personal experience and review of the literature. Otol Neurotol. Feb 2014;35(2):260-270. PMID 24448286 8. Noij KS, Kozin ED, Sethi R, et al. Systematic review of nontumor pediatric auditory brainstem implant outcomes. Otolaryngol Head Neck Surg. Nov 2015;153(5):739-750. PMID 26227469 9. Colletti L, Wilkinson EP, Colletti V. Auditory brainstem implantation after unsuccessful cochlear implantation of children with clinical diagnosis of cochlear nerve deficiency. Ann Otol Rhinol Laryngol. Oct 2013;122(10):605-612. PMID 24294682 10. Sennaroglu L, Sennaroglu G, Yucel E, et al. Long-term results of ABI in children with severe inner ear malformations. Otol Neurotol. Aug 2016;37(7):865-872. PMID 27273392 11. Sennaroglu L, Ziyal I, Atas A, et al. Preliminary results of auditory brainstem implantation in prelingually deaf children with inner ear malformations including severe stenosis of the cochlear aperture and aplasia of the cochlear nerve. Otol Neurotol. Sep 2009;30(6):708-715. PMID 19704357 12. Colletti V, Carner M, Miorelli V, et al. Auditory brainstem implant (ABI): new frontiers in adults and children. Otolaryngol Head Neck Surg. Jul 2005;133(1):126-138. PMID 16025066 13. Colletti V. Auditory outcomes in tumor vs. nontumor patients fitted with auditory brainstem implants. Adv Otorhinolaryngol. 2006;64:167-185. PMID 16891842 14. Colletti L. Beneficial auditory and cognitive effects of auditory brainstem implantation in children. Acta Otolaryngol. Sep 2007;127(9):943-946. PMID 17712673 15. Colletti V, Shannon RV, Carner M, et al. Complications in auditory brainstem implant surgery in adults and children. Otol Neurotol. Jun 2010;31(4):558-564. PMID 20393378 16. National Institute for Clinical Excellence (NICE). NICE Interventional Procedure Guidance [IPG108]. Auditory brain stem implants. 2005 January; Accessed June 2023. 17.CMS Medicare Policy Benefit Manual. Chapter 16, Section 100 - Hearing Aids and Auditory Implants. (Rev. 39; Issued: 11-10-05; Effective: 11-10-05; Implementation: 12-12-05). Available at (accessed June 2023). 18.Blue Cross Blue Shield Association. Auditory Brainstem Implant – Archived. Medical Policy Reference Manual. Policy #7.01.83, Issue 3:2015, original policy date 7/12/02, last review date March 2023. The articles reviewed in this research include those obtained in an Internet based literature search for relevant medical references through June 2023, the date the research was completed. 12 Joint BCBSM/BCN Medical Policy History Policy Effective Date BCBSM Signature Date BCN Signature Date Comments 6/16/03 6/10/03 6/16/03 Joint policy established 6/15/05 6/15/05 5/20/05 Scheduled review of established policy. Policy retired. 5/1/14 2/18/14 3/3/14 Policy taken out of retirement due to requests for ABI surgery for non-FDA approved indications. 1/1/16 10/13/15 10/27/15 Routine maintenance 11/1/16 8/16/16 8/16/16 Routine maintenance, no changes in policy statement. 11/1/17 8/15/17 8/15/17 Updated rationale section. Added reference# 9 and 11-13. No change in policy status. 11/1/18 8/21/18 8/21/18 Routine maintenance. No changes in policy statement. 11/1/19 8/20/19 Routine maintenance. No change in policy statement. 11/1/20 8/18/20 Routine policy maintenance. No change in policy statement. 11/1/21 8/17/21 Routine policy maintenance. No change in policy status. 11/1/22 8/16/22 Routine policy maintenance, no change in policy status. 11/1/23 8/15/23 Routine policy maintenance, no change in policy status. Vendor managed: N/A (ds) Next Review Date: 3rd Qtr. 2024 13 BLUE CARE NETWORK BENEFIT COVERAGE POLICY: AUDITORY BRAIN STEM IMPLANT I. Coverage Determination: Commercial HMO (includes Self-Funded groups unless otherwise specified) Covered; criteria apply. BCNA (Medicare Advantage) See government section. BCN65 (Medicare Complementary) Coinsurance covered if primary Medicare covers the service. II. Administrative Guidelines: • The member's contract must be active at the time the service is rendered. • Coverage is based on each member’s certificate and is not guaranteed. Please consult the individual member’s certificate for details. Additional information regarding coverage or benefits may also be obtained through customer or provider inquiry services at BCN. • The service must be authorized by the member's PCP except for Self-Referral Option (SRO) members seeking Tier 2 coverage. • Services must be performed by a BCN-contracted provider, if available, except for Self-Referral Option (SRO) members seeking Tier 2 coverage. • Payment is based on BCN payment rules, individual certificate and certificate riders. • Appropriate copayments will apply. Refer to certificate and applicable riders for detailed information. • CPT - HCPCS codes are used for descriptive purposes only and are not a guarantee of coverage. |
10603 | https://structures-simplified.com/2020/08/why-shape-triangle-is-widely-used-in/ | Published Time: 2020-08-09T14:21:00+05:30
Why the shape “TRIANGLE” is widely used in structural engineering? - STRUCTURES SIMPLIFIED
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Why the shape “TRIANGLE” is widely used in structural engineering?
Parishith Jayan
9 August 2020
Blog
Most of the structures that we know comprise of this particular shape “triangle” in any one of its structural systems.
What is so important? or why “TRIANGLES”?
Just think of basic 2-dimensional geometrical shapes, what do we have?
Squares, Rectangles, Triangles, circles, pentagon, hexagon, and the list go on.
Of these shapes, why a triangle is called the most stable shape. Let us compare these shapes and understand them.
For our example, let’s consider the square and the triangle as shown below.
Applying a force on one of the edges and studying the behavior would lead to a better understanding regarding the stability of the structure.
First, let’s try with the vertical force.
What will happen? The load applied over the square moves directly downward through the vertical member and stands still. The triangle when subjected to a vertical force through one of its edges, distributes that force evenly to either side and stands stable.
Now, let’s apply the horizontal force.
The square could not withstand this force without deformation, it tries to sway along the direction of application of the force and distort from its original shape. Whereas, applying the horizontal force tries to push one edge of the triangle.
It is not possible to move the edge without the elongation of one of its sides, this makes them stay stable for the applied force.
This simple comparison applies to all the other shapes too. Also, this can be easily verified by making a simple model with popsicles. I tried it myself, and recommend you to do so.
Let us look into a design example, where this particular behavior is utilized.
Most of us would have seen a bracing system in the steel buildings.
What is the function of bracings systems in the steel buildings?
The longitudinal forces that act on the structure are transferred to the ground with the help of bracing systems, right.
So, the longitudinal force is about to act on the edges should be transferred safely without much distortion or deflection of the system. Which is the shape that can perform this task? Obviously, the triangle.
The attached image shows several bracing arrangements that are employed in the steel structure.
CONCLUSION
This stable behavior of the triangles makes them the supreme geometry. And for this reason, they are extensively employed in the formation of trusses.
What we have discussed so far is also applied to the 3-dimensional shapes and this emphasis the use of triangle configuration in the construction of towers, electrical grids, truss bridges, and so on.
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Transpulmonary Pressure
In subject area:Medicine and Dentistry
Transpulmonary pressure is defined as the difference between intraalveolar pressure and pleural pressure, reflecting the pressure required to expand the lungs during breathing. It is influenced by airway pressure and the elastances of the lung and chest wall.
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Pulmonary Pathophysiology and Lung Mechanics in Anesthesiology
2022, Cohen's Comprehensive Thoracic AnesthesiaJamie L. Sparling, Marcos F. Vidal Melo
Transpulmonary Pressure
The transpulmonary pressure, PL, represents the distending pressure across the lungs, which is the difference between the airway pressure and the pleural pressure. It is an essential concept because it represents the pressure that effectively promotes air flow and distends the lungs. At end-inspiration, if respiratory flow equals zero and there is a clear plateau of the airway pressure, end-inspiratory PL = Pplateau – end-inspiratory Ppleural represents the pressure acting across the alveolar units. At end exhalation, that pressure would be represented, in the absence of auto-PEEP, by end-expiratory PL = PEEP – end-expiratory Ppleural. The importance of the concept of PL becomes clear as the inspiratory PL is the pressure which provides tidal ventilation at the same time that could produce stretch lung injury; and the end-expiratory PL is that required to prevent lung collapse at end-exhalation by being kept positive.
Unfortunately, the assessment of PL is not usual in clinical practice because measurement of pleural pressure is not simple, in contrast to measurement of airway pressure. Yet, pleural pressure may be estimated using an esophageal balloon, which is either standalone or attached to an esophagogastric tube and positioned in the lowest third of the esophagus23–25 (Fig. 5.6). These esophagogastric tubes are generally 100 cm long, and the average depth for a correctly positioned balloon is approximately 35 to 45 cm. The catheter is connected to a pressure transducer via a three-way stopcock, and the balloon is inflated to a standard volume depending on the device, which is necessary to prevent artifact because of passive recoil of the balloon walls. Further, the balloon should be long enough to avoid regional variability, providing an average measurement of esophageal pressure to estimate pleural pressure. Although use of an esophageal balloon represents an improvement in the accuracy of pulmonary pressure monitoring, it is not without its limitations. Artifact may arise from mediastinal weight, gravitational gradients in pleural pressure, and airway closure at end exhalation.23 Furthermore, there is considerable controversy in the literature on whether the esophageal pressure measurements should be directly taken as absolute estimates of pleural pressures for PEEP adjustments,26 or whether changes in esophageal pressures should be used to compute chest wall and lung elastances, which are then used to set the level of the administered PEEP.27 Although a ventilator strategy guided by the transpulmonary pressure measured with esophageal balloons improved oxygenation and compliance in a small group of acute respiratory distress syndrome (ARDS) patients,28 a subsequent large trial on patients with moderate to severe ARDS resulted in no significant difference in death and days free from mechanical ventilation between patients receiving an esophageal balloon-guided PEEP and those receiving empirical high PEEP-FiO2 settings.29 Those findings did not support esophageal balloon-guided PEEP titration in ARDS. In critically ill patients with class 3 obesity (body mass index [BMI] ≥40 kg/m2) and ARDS, a retrospective analysis compared a standard protocol of ventilator settings determined by the ARDS net table for lower PEEP/higher FiO2 with a “lung rescue” management strategy with settings determined by an individualized protocol based on lung recruitment maneuvers, esophageal manometry, and hemodynamic monitoring.29 Patients receiving the standard protocol had almost double the 28-day mortality compared with the lung rescue cohort (31% vs. 16%, P = .012; hazard ratio [HR], 0.32; 95% confidence interval [CI] 95%, 0.13–0.78).
In robotic-assisted laparoscopic surgery, esophageal manometry can help delineate the portion of increased driving pressure, which is applied to the lungs versus that applied to the chest wall; the increase in elastance associated with pneumoperitoneum is substantially greater for the chest wall (ECW) than for the lungs (ELS).30 This may improve the ability to maintain open lung by appropriately titrating PEEP in this and other perioperative conditions with high susceptibility to lung derecruitment.
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Book2022, Cohen's Comprehensive Thoracic AnesthesiaJamie L. Sparling, Marcos F. Vidal Melo
Review article
Advanced monitoring in intensive care and anaesthesia
2013, Best Practice & Research Clinical AnaesthesiologyLorenzo Ball M.D., ... Paolo Pelosi M.D.
Clinical applications and limitations
Transpulmonary pressure, defined as the difference between alveolar pressure and pleural pressure, is the distending force of the lung. It remains questionable whether Poes can effectively approximate Ppl in the supine or prone patient: in these positions, the surrounding organs could affect the value of Poes making it markedly different from the actual Ppl. Moreover, the value of Poes is only representative of the mid-lung region, not necessarily reflecting the entire chest condition. However, an experimental study in a canine ARDS model showed good correlation between the changes in Ppl, measured invasively, and variations in Poes in the non-dependent, middle and dependent lung regions .
The incidence of intra-abdominal hypertension is relatively high (>30%) and associated with a higher mortality rate among critically ill patients [39,40].
In obese patients, an increase in body mass index (BMI) has a strong negative correlation with lung volume and total respiratory system compliance during general anaesthesia and increased IAP correlates with higher chest wall elastance ; and severely obese patients undergoing mechanical ventilations are more likely to develop ARDS than normal weight patients .
The authors suggest monitoring transpulmonary pressure, chest wall elastance and IAP especially in high-risk abdominal surgery, ARDS and morbidly obese patients, but the monitoring of these parameters requires supplementary instruments.
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Journal2013, Best Practice & Research Clinical AnaesthesiologyLorenzo Ball M.D., ... Paolo Pelosi M.D.
Review article
Perioperative Monitoring
2021, Anesthesiology ClinicsRalph Gertler MD, PhD
Transpulmonary pressure and esophageal pressure
Transpulmonary Pressure (PTP) is the difference between the pressure inside the alveoli (PALV) and the pressure surrounding the lung (pleural pressure, PPL) (Fig. 6):
PTP = PALV - PPL
As in vivo measurements of PALV and PPL are not feasible, PALV is usually assumed to be approximately equal to the static pressure at the airway opening (PPLAT, PEEP) and PPL is assumed to be equal to the esophageal pressure (PESO). PTP is usually described as the distending pressure of the lungs because it best describes the sum of the interactions of PPL and PALV across the lungs. Because PESO can be elevated in the setting of ARDS, obesity, or increased intraabdominal pressure, the use of PTP allows the titration of positive pressure based on the actual pressure applied to the lung. To allow optimal recruitment, PEEP could be increased until PTP becomes positive at end-expiration, to keep airways and alveoli open during the tidal cycle.36 End-inspiratory PTP is evaluated by calculating the PPLAT – PESO difference during an inspiratory hold. It is useful in the assessment of safety pressure thresholds: an end-inspiratory PTP lower than 20 cmH2O is generally considered safe.36
For the lung, only transpulmonary pressure (PL = Ppl – Palv, with Ppl being the intrapleural pressure) determines alveolar distention. Intrapleural pressure is not easily accessible, besides being unequally distributed throughout the lung. As a surrogate, esophageal pressure (Pes) can be used as an estimate of mean Ppl.37 This minimally invasive approach can be used to estimate PL, which is particularly advantageous in cases of a decreased extrathoracic compliance such as obesity or an elevated intraobdominal pressure. In the supine position, the weight of the mediastinum and abdominal pressure both increase Pes. Some studies have pointed out that Pes is higher than Ppl by about 5 cmH2O owing to the effects of body position and mediastinum.38 Pes can represent the average level of Ppl, and Pes and Ppl have a good correlation. For making sure Pes reflects Ppl accurately, the technical aspects of Pes measurement are important. It usually includes the characteristics of the esophageal manometers, placement position, balloon inflating volume, and data interpretation. In order to ensure the accuracy of Pes, its positioning needs to be tested by an occlusion test. In some cases, such as obesity, thoracic, or abdominal disease, Paw cannot reflect actual pressure gradient over the lung because the higher percentage of Paw is used to overcome the elastance of the chest wall. The monitoring of pleural pressure (Ppl) or esophageal pressure (Pes) may help to distinguish the pressure gradients acting on the lung and chest wall. On the other hand, when the patient makes spontaneous breathing efforts, the inspiratory muscles and the ventilator both participate in breathing activity during assisting ventilation mode. The pressure inflating the lung comes from the pressure generated by the ventilator and the patient's inspiratory muscles. However, excessive breathing effort also can induce spontaneous patient self-inflicted lung injury.39 Therefore, it is very important to monitor spontaneous breathing efforts and to balance the relationship between mechanical ventilation and spontaneous breathing effort during assisted ventilation. The Monitoring of Pes, intragastric pressure (Pga), and a series of parameters derived from Pes and Pga can be used to quantitatively assess spontaneous breathing efforts.
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Journal2021, Anesthesiology ClinicsRalph Gertler MD, PhD
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What is the role of PEEP and recruitment maneuvers in ARDS?
2020, Evidence-Based Practice of Critical Care (Third Edition)Sinead Egan, Gerard P. Curley
Esophageal pressure
Transpulmonary pressure is defined as the pressure difference between the pleural space and the alveolar space.43 Conditions that decrease chest wall compliance, such as kyphoscoliosis, can increase airway pressure and lead to a false impression that lung stress is also increased. Measuring transpulmonary pressures can more accurately reflect the stress on lung parenchyma, as the measurement is independent of chest wall compliance. If transpulmonary pressure remains within normal limits, then it may be appropriate to ventilate above the accepted airway plateau pressures.
Esophageal pressure can be measured by positioning an air- or liquid-filled catheter (manometer) in the lower-third of the esophagus, and this can be used to estimate pleural pressure. Accuracy is variable due to patient positioning, the presence or absence of lung disease, and the position of the diaphragm.38 Talmor and colleagues have suggested correcting the esophageal pressure by –5 cm H2O mid-lung height, to estimate the true pleural pressure.44,45
In a single-center RCT (EPVent study), Talmor et al.46 compared mechanical ventilation guided by Pes measurements (experimental arm) with ventilation based on the ARMA trial protocol (control arm). Patients who had PEEP titrated to ensure a positive end-expiratory transpulmonary pressure experienced a higher PaO2/FiO2, better respiratory system compliance as a possible consequence of improved recruitment, but no difference in 28-day mortality. Limitations to the measurement include the fact that a single measurement of esophageal pressure is unlikely to accurately reflect pleural pressure throughout the lung. It is also a resource-intensive measurement requiring two to three operators.
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Book2020, Evidence-Based Practice of Critical Care (Third Edition)Sinead Egan, Gerard P. Curley
Chapter
Acute Respiratory Distress Syndrome
2006, Clinical Critical Care MedicineLuciano Gattinoni, ... Franco Valenza
Transmural pressure
The airspace opening pressure is the transpulmonary pressure (i.e., the difference between intraalveolar and pleural pressures). Transpulmonary pressure is a function of the pressure applied to the airways and of the elastances of the lung and chest wall, according to the following equation, in which Paw is the applied airway pressure, El is the elastance of the lung, and Ew is the elastance of the chest wall:
Normally, El equals Ew, and the transmural pressure, as an average, would be approximately half of the pressure applied to the airways. However, as previously discussed, in ALI/ARDS caused by direct or indirect insult, the El/(El + Ew) value may be very different. For example, in ARDS that results from direct insult, the ratio is less than 1 for patients who have extrapulmonary disease. It follows that to achieve the same opening transmural pressure, a higher Paw is required in ARDS that arises from extrapulmonary problems than in ARDS that results from pulmonary conditions.
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Pulmonary Pathophysiology and Lung Mechanics in Anesthesiology
2022, Cohen's Comprehensive Thoracic AnesthesiaJamie L. Sparling, Marcos F. Vidal Melo
Respiratory Mechanics
Transpulmonary Pressure
The transpulmonary pressure, PL, represents the distending pressure across the lungs, which is the difference between the airway pressure and the pleural pressure. It is an essential concept because it represents the pressure that effectively promotes air flow and distends the lungs. At end-inspiration, if respiratory flow equals zero and there is a clear plateau of the airway pressure, end-inspiratory PL = Pplateau – end-inspiratory Ppleural represents the pressure acting across the alveolar units. At end exhalation, that pressure would be represented, in the absence of auto-PEEP, by end-expiratory PL = PEEP – end-expiratory Ppleural. The importance of the concept of PL becomes clear as the inspiratory PL is the pressure which provides tidal ventilation at the same time that could produce stretch lung injury; and the end-expiratory PL is that required to prevent lung collapse at end-exhalation by being kept positive.
Unfortunately, the assessment of PL is not usual in clinical practice because measurement of pleural pressure is not simple, in contrast to measurement of airway pressure. Yet, pleural pressure may be estimated using an esophageal balloon, which is either standalone or attached to an esophagogastric tube and positioned in the lowest third of the esophagus23–25 (Fig. 5.6). These esophagogastric tubes are generally 100 cm long, and the average depth for a correctly positioned balloon is approximately 35 to 45 cm. The catheter is connected to a pressure transducer via a three-way stopcock, and the balloon is inflated to a standard volume depending on the device, which is necessary to prevent artifact because of passive recoil of the balloon walls. Further, the balloon should be long enough to avoid regional variability, providing an average measurement of esophageal pressure to estimate pleural pressure. Although use of an esophageal balloon represents an improvement in the accuracy of pulmonary pressure monitoring, it is not without its limitations. Artifact may arise from mediastinal weight, gravitational gradients in pleural pressure, and airway closure at end exhalation.23 Furthermore, there is considerable controversy in the literature on whether the esophageal pressure measurements should be directly taken as absolute estimates of pleural pressures for PEEP adjustments,26 or whether changes in esophageal pressures should be used to compute chest wall and lung elastances, which are then used to set the level of the administered PEEP.27 Although a ventilator strategy guided by the transpulmonary pressure measured with esophageal balloons improved oxygenation and compliance in a small group of acute respiratory distress syndrome (ARDS) patients,28 a subsequent large trial on patients with moderate to severe ARDS resulted in no significant difference in death and days free from mechanical ventilation between patients receiving an esophageal balloon-guided PEEP and those receiving empirical high PEEP-FiO2 settings.29 Those findings did not support esophageal balloon-guided PEEP titration in ARDS. In critically ill patients with class 3 obesity (body mass index [BMI] ≥40 kg/m2) and ARDS, a retrospective analysis compared a standard protocol of ventilator settings determined by the ARDS net table for lower PEEP/higher FiO2 with a “lung rescue” management strategy with settings determined by an individualized protocol based on lung recruitment maneuvers, esophageal manometry, and hemodynamic monitoring.29 Patients receiving the standard protocol had almost double the 28-day mortality compared with the lung rescue cohort (31% vs. 16%, P = .012; hazard ratio [HR], 0.32; 95% confidence interval [CI] 95%, 0.13–0.78).
In robotic-assisted laparoscopic surgery, esophageal manometry can help delineate the portion of increased driving pressure, which is applied to the lungs versus that applied to the chest wall; the increase in elastance associated with pneumoperitoneum is substantially greater for the chest wall (ECW) than for the lungs (ELS).30 This may improve the ability to maintain open lung by appropriately titrating PEEP in this and other perioperative conditions with high susceptibility to lung derecruitment.
Driving Pressure
Driving pressure (ΔP) is the difference between the plateau pressure (Pplat) and PEEP (Fig. 5.7). The driving pressure can also be expressed as the VT normalized by respiratory system compliance (Crs)—that is, the ΔP is proportional to the change in volume of the lung in a breath (VT) divided by its initial volume, to the extent that Crs changes in proportion to lung volume. Change in volume divided by initial volume is, by definition, the volumetric strain. Thus physiologically, the reason why ΔP may be important is because it provides a rough measure of the global volumetric strain of the lungs.31–33
Recent studies suggest that such physiologic considerations may be very important in the clinical setting. Driving pressures have been shown to be better associated with relevant clinical outcomes in ARDS32,34 and surgical patients33,35,36 than VTs. Indeed, a direct association of increased intraoperative ΔP with increased risk of significant postoperative pulmonary complications has been shown in patients undergoing noncardiothoracic surgery, in the absence of a corresponding association with VT33,35, in addition the relative risk of in-hospital death in relation to plateau pressure and driving pressure is depicted in Fig. 5.8. Such a relationship was also found in ARDS patients, with a greater association between increased ΔP and mortality compared with that of VT32 (Fig. 5.9). This finding was reinforced in a secondary analysis of observational data on patients with moderate to severe ARDS ventilated with standard lung-protective ventilation.37 Variation in VT or PEEP had no impact on mortality. In contrast, above a plateau pressure of 29 cm H2O or a driving pressure of 19 cm H2O, there was an ordinal increment in risk of death.37
In thoracic anesthesia, comparison of conventional lung-protective ventilation (one-lung ventilation [OLV] with VT = 6 mL/kg ideal body weight, PEEP 5 cm H2O, and recruitment maneuvers) with ΔP-guided ventilation (same VTs and recruitment, individualized PEEP to produce the lowest ΔP during OLV) showed a smaller incidence of pneumonia or ARDS in the ΔP-guided group.38 A retrospective analysis of data of patients undergoing OLV for thoracic surgery (hospital electronic medical records and the Society of Thoracic Surgeons database) indicated that ΔP, when modeled together with VT, predicted major morbidity, whereas VT alone inversely correlated with respiratory complications.39 This implied that low VT by itself was not associated with less postoperative pulmonary complications, whereas low VT taken together with strategies to minimize ΔP may be.39
Elastance and Compliance
The lungs and the chest wall have elastic properties that tend to restore them to their combined equilibrium configuration when they are displaced from baseline by spontaneous or mechanical ventilation. Thus if the pleural space is open at FRC, the lungs would recoil inward toward collapse, while the chest wall would recoil outward, toward expansion. Elastance is the change in distending pressure for a given change in volume (E = ΔP/ΔV). The elastance of the total respiratory system (ERS) is the sum of the lung (EL) and the chest wall (ECW) elastances:
Compliance is the reciprocal of the elastance (C = ΔV/ ΔP); that is, the change in volume for a given change in distending pressure. Compliance tends to be more frequently used clinically. Thus the earlier equation can be alternatively expressed as:
where CRS, CL, and CCW are the compliances of the respiratory system, lung, and chest wall, respectively.
During mechanical ventilation with volume control, a peak inspiratory pressure (PIP) is usually seen, which is followed by a plateau pressure (Pplat) if an inspiratory pause (i.e., a period of no flow at end of inspiration) is provided (see Fig. 5.7). The difference between peak and plateau pressures is caused by dynamic factors, particularly airway resistance. Dynamic compliance (Cdyn) is defined as the compliance calculated using the PIP:
where PEEP is the positive end-expiratory pressure (see Fig. 5.7). Cdyn, thus, conflates the distensibility of the lungs themselves with the resistance of the airways, in addition to the properties of the chest wall, to the overall quantification of compliance. Static compliance (Cstat) aims to eliminate the dynamic airflow-dependent component from the computation of compliance by using Pplat:
Lung compliance represents the distensibility of the lung parenchyma; it is altered in various disease states. Compliance is decreased with consolidation, atelectasis, pulmonary edema, interstitial disease or fibrosis, pneumothorax, surgical resection, large effusion, and mainstem intubation. It is increased in emphysema because of loss of elastic tissue. In addition, lung compliance depends upon the lung volume in a nonlinear manner. The highest lung compliance is observed when lungs are at moderate volumes, and lowest at the extremes of low or high lung volumes. Such relationships are expressed in the pressure-volume curve of the lung (Fig. 5.10). That curve also shows that the pressure-volume relationship and the compliance for the same volume (or pressure) on inhalation differs from that on exhalation, a phenomenon known as hysteresis.
To compute the compliance of the lungs, the contribution of the chest wall needs to be subtracted. Accordingly, lung compliance (CL = ΔVLΔPL) can be estimated as:
,where PawEI is the airway pressure at end inhalation, PesEI is the esophageal pressure at end inhalation, PawEE is the airway pressure at end exhalation, and PesEE is the esophageal pressure at end exhalation, measured using esophageal manometry, as explained earlier.
Chest wall compliance reflects the distensibility of the muscles, bones, and soft tissue that comprise the chest wall. Chest wall compliance is reduced with increased abdominal pressure (e.g., insufflation, abdominal compartment syndrome), chest wall edema, thoracic deformity, muscle tone, extensive thoracic/abdominal scar, supine positioning, and obesity. Chest wall compliance is increased with flail chest, muscle relaxation, and upright positioning. Using estimates of pleural pressure, the compliance of the chest wall (CCW) can be calculated as:
with variables that have been defined earlier.
Resistance
Resistance is the opposition to fluid flow because of friction and may be calculated as the ratio of the driving pressure to air flow (Raw = ΔP/). Airway resistance is determined by the dimensions of the airway, the viscosity of the gas, and whether the flow is laminar or turbulent, which in turn is predicted by the Reynold’s number, Re:
where ρ is the fluid/gas density (kg/m3), u is the velocity (m/s), L is the length (m), and μ is the viscosity (kg/ms). Laminar flow dominates at Re less than 2300, while turbulent flow dominates at Re greater than 2900.
Laminar flow through a long cylindric pipe is governed by Poiseuille’s Law, which is appropriately applied to conducting airways. Resistance is relatively low in laminar flow, and is calculated as follows:
where L is the length, u is the velocity, and r is the radius. Thus the airway radius is especially important in determining the resistance; if the radius is halved, the resistance increases by a factor of 16. Pathologic states, such as tracheal stenosis and bronchospasm, raise the resistance and increase the work of breathing. Reducing the density of the gas mixture (e.g., through mixing with helium, “Heliox”) can improve resistance and lead to decreased work of breathing.
Because of the dependence on radius, an individual small airway has a higher resistance than a large airway. However, because the airways branch so prolifically, the overall resistance depends on the number of parallel pathways present, and it is the medium-sized airways that have the overall greatest resistance (see Fig. 5.2).
Stress and Strain
Stress is defined as force divided by the area over which it is applied.31 Lung stress can be estimated through the transpulmonary pressure (PL). This will be larger at end-inspiration and lower at end-exhalation, in the absence of patient effort. Strain is the change in dimension relative to the material’s initial dimension. Based on this definition, lung volumetric strain can be calculated as the ratio of the change in lung volume (dV) to the resting lung volume (V0), that is, dV/V0.40 Assuming a linear relationship, lung stress and strain are related to one another by the specific lung elastance, K.
.
Experimental studies indicate that when stress and strain are outside the normal physiologic limits, the lung releases inflammatory cytokines and mediators, which indicate, produce, and exacerbate lung injury. Very large VTs conducive to large strains, either through spontaneous or mechanical ventilation, can produce lung injury.41,42 When mechanical ventilation is applied for more than 48 hours in initially normal animal lungs, injury occurs when lung expansion reaches approximately the TLC, that is when strain would be (TLC – FRC)/FRC.40,43,44 In experiments representing usual clinical conditions, those large values are not present.45 This suggests an increased susceptibility of lung tissue to strain in the presence of an additional injurious trigger (i.e., systemic inflammation).46,47 A large animal study modeled moderate–severe ARDS (respiratory system compliance = 14.0 ± 4.9 mL/cm H2O; PaO2/FiO2 = 92 ± 59) with surfactant depletion and ventilator-induced lung injury (VILI) and quantified volumetric strain using CT to support a relationship between strain and local inflammation assessed with positron emission tomography (PET) in those extreme lung injury cases.48
Stress is approximated by the transpulmonary pressure distending the airspaces, but the regional stress of the lung will vary because of gravity and other factors producing heterogeneous lung expansion. This will produce a distribution of stress throughout the lungs. For example, in uniformly expanded lungs, alveolar pressure is the same across the whole lung, but when the lung is not uniformly expanded (such as when air spaces are isolated from one another because of obstruction or closure), alveolar pressures will not be uniform throughout.49 The presence of such heterogeneity leads to regional stress values substantially larger than those present at the global level.45 This means that even when global transpulmonary pressures are low, local stresses can be high, even to the limit of being injurious.45,50
Mechanical Power
Stress and strain have been used for more than a century as criteria to determine material failure. Another variable used is energy.51 In line with these traditional approaches, there has been recent interest in quantifying the power delivered to the lungs during mechanical ventilation as a means to measure risk of lung injury because of mechanical ventilation. Mechanical power is the energy delivered from the ventilator to the respiratory system per unit of time and is expressed in J.minute−1. The proposed equation for a mechanically ventilated lung with respiratory rate (RR), I:E ratio, and tidal volume VT is:
(where 0.098 is a conversion factor from cm H2O l min−1in J/min, RR is the respiratory rate, and Vt is the tidal volume in liters. The term (Plateau pressure − PEEP) is the driving pressure).
The interest in this equation is that it incorporates the known ventilator-generated causes of VILI (e.g., VT, ΔP, flow, PEEP, RR), and thus may be used as a single quantity to predict the risk of VILI.52 The power may be measured continuously using software on the ventilator, which agrees well with the earlier computed quantity in both normal subjects and in patients with ARDS.52 Clinical validation of the concept is still pending.
Measures of Derecruitment and Overdistension
Just as atelectasis (derecruitment) is deleterious to the lung, alveolar overdistension places the lung at risk of ALI and overt barotrauma (Fig. 5.11). As detailed earlier, in injured lungs, regional lung inflation is heterogeneous, so regions of atelectasis may coexist with regions of normal ventilation and even overdistension.53,54 Alveolar derecruitment and overdistension may be modeled by partitioning the dynamic respiratory system elastance (Ers,dyn) into volume-dependent (E1) and volume-independent (E2) components and calculating the %E2 via the later formula.
.
A %E2 value greater than 30% represents overdistension, and was common, occurring at 40% of clinically set PEEP levels in one study of patients with ALI.55 Conversely, a negative %E2 value represents atelectasis, and was found in all patients with abdominal distension when measured at a PEEP of 5 cm H2O.55
An alternative measure of lung distension is the stress index, which measures the dynamic change in compliance during inspiration using the pressure-time curve of the ventilator.1 Thus the stress index, gives information about both alveolar collapse and recruitment which help to prevent the deleterious effects of mechanical ventilation. The stress index is determined by fitting the pressure-time curve during volume control ventilation with constant inspiratory flow to the following power equation:
where the coefficient a represents the slope of the pressure-time relationship in the time 0 to time 1 interval, and the coefficient c is the value of pressure at time 0 (PTP: transpulmonary pressure).
Coefficient b is the stress index describing the shape of the curve (b = 1 indicates a straight curve, b < 1 indicates a progressive increase in compliance over time, and b > 1 indicates a progressive decrease in compliance with time). An application of this concept has been applied to choose the best PEEP as the titrated PEEP value, leading to a b = 0.9 ‒ 1.1 (i.e., an approximately homogeneously expanding respiratory system).56,57 The amount of lung subjected to hyperinflation increases with a stress index greater than 1.1, whereas values less than 0.9 correspond with atelectasis. Values less than 0.9 or greater than 1.1 correlate with a worsened VILI and release of inflammatory factors (e.g., interleukin-6, macrophage inflammatory protein-2).58
An application of such concepts was done in an investigation on the effects of stress index-guided PEEP titration on pulmonary mechanics and hemodynamics in the prone position that addressed severe lavage-induced lung injury in a piglet model.56 Piglets underwent a decremental PEEP trial after full lung recruitment in the prone position. Stress-index PEEP was the level at which the airway pressure stress index was 1, and open-lung PEEP was the level which was required to keep the lung open according to CT scans. The authors found that the respiratory system elastance and lung elastance were improved in the prone position, but that the ratio of chest wall elastance to respiratory system elastance was higher in the prone position. There were no significant differences between the open-lung and stress index-guided groups in PEEP, collapsed lung volume, hyperinflation lung volume, oxygenation index, or arterial carbon dioxide partial pressure (PaCO2). The authors concluded that the stress index can be used to titrate PEEP in the prone position in the studied model.
Effects of Anesthesia
The induction of general anesthesia decreases compliance because of the loss of muscle tone and the decrease in the FRC.59 Reports differ on the effect of muscle relaxants on the overall respiratory system compliance. Some studies have found paralysis decreases the compliance for similar reasons,60 but certainly compliance is increased in situations where paralytics promote improved ventilator synchrony.61 Further, positioning while under anesthesia affects compliance, with prone positioning improving compliance, assuming that the lower body is at the same level as the chest.59 Lateral decubitus positioning can worsen compliance because of the weight of the mediastinum and the abdominal contents on the dependent lung.59 Certain surgical maneuvers may also affect compliance, with abdominal retraction and pneumoperitoneum worsening compliance, while abdominal wall lift, sternal splitting, and sternal lift all improve compliance.62
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Book2022, Cohen's Comprehensive Thoracic AnesthesiaJamie L. Sparling, Marcos F. Vidal Melo
Chapter
Mechanical Ventilation in Acute Respiratory Distress Syndrome
2008, Critical Care Medicine (Third Edition)Luciano Gattinoni, ... Pietro Caironi
Transpulmonary Pressure
At the same driving force applied to the whole respiratory system (lung and chest wall), the resulting transpulmonary pressure, ΔPL, may be extremely variable. If the lung is relatively “stiff,” and the chest wall is relatively “soft” (e.g., during pulmonary fibrosis or ARDS of pulmonary origin), a greater fraction of the driving pressure is spent to distend the lung (high transpulmonary pressure). In contrast, if the lung is relatively soft, but the chest wall is relatively stiff (e.g., during an abdominal disease or severe obesity), most of the driving force is spent to move the chest wall (high pleural pressure).
To express this phenomenon quantitatively, it is convenient to consider the concept of elastance The elastance of the whole respiratory system is the driving force (ΔPaw) required to increase the lung and the chest wall 1 L above their resting position (ETOT = ΔPaw/1 L). Part of this driving force is spent to increase the lung volume of 1 L (EL = ΔPL/1L), and part is spent to increase the chest wall by the same amount (ECW = ΔPpl/1 L). Transpulmonary pressure (ΔPL) can be expressed as the driving force times the ratio between lung elastance and total elastance of the respiratory system:
The transpulmonary pressure for a given driving pressure uniquely depends on the ratio of lung elastance to respiratory system elastance. In normal subjects EL/ETOT is approximately 0.5, whereas in patients with ARDS it may range from 0.2 (e.g., in obese patients or in patients with high intra-abdominal pressure) to 0.8 (e.g., in patients with a very small baby lung and a normal chest wall elastance). This variability implies that for the same driving force applied and read on the ventilator display (e.g., 30 cm H2O), the resulting transpulmonary pressure may range from 6 to 24 cm H2O (Fig. 11-1).
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Review article
Advances in Anesthesia
2022, Advances in AnesthesiaAditi Balakrishna MD, ... Christopher G. Hughes MD, MS, FCCM
Esophageal balloon
The tenets of lung-protective ventilation provide guidance for minimizing ventilator-associated lung injury, but plateau and driving pressures comprise pressures from the whole respiratory system. They are a surrogate for the pressure exerted directly across an alveolus, as they are also affected by the chest wall. Measuring the transpulmonary pressure is more specific.
Transpulmonary pressure is defined as airway pressure minus pleural pressure. Measuring and acting upon transpulmonary pressure can help ensure that the patient’s alveoli do not have net collapsing forces exerted on them, thus promoting atelectasis, and transpulmonary driving pressure and Pplat may give a clearer picture of the forces experienced by the alveoli, though the latter two are not used in clinical practice at this time.
Measuring pleural pressure directly in humans is prohibitive, but the pressure in the distal third of the esophagus provides a close approximation of the posterior pleural pressure, within ∼3 cm H2O . By approximating pleural pressure, an appropriate level of positive airway pressure can be set to keep the transpulmonary pressure zero or positive, which prevents a net trend toward atelectasis.
One caveat is that this tool does not provide robust information about heterogenous lung tissue, as it only measures the pressure in one part of the thorax, but it can provide important information to help with lung recruitment.
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Journal2022, Advances in AnesthesiaAditi Balakrishna MD, ... Christopher G. Hughes MD, MS, FCCM
Review article
Acute respiratory distress syndrome
2022, Anaesthesia & Intensive Care MedicineTapan Parikh aka Parmar, David Pilcher
Transpulmonary pressure and electrical impedance tomography (EIT) guided PEEP
Transpulmonary pressure (PL), the distending force of the lung, is the difference between airway (PAW) and pleural pressure (PPL), which is estimated by oesophageal pressure (PES). In recent trials, PL guided adjustment of PEEP was found to be superior in improving oxygenation especially in patients with high abdominal pressure such as obesity and reducing rescue therapies such as ECMO. Better survival was observed when PEEP was titrated closer to 0 cmH2O of transpulmonary pressure.9,15
EIT uses electric currents by placing electrodes on the thorax to image the changes in air content in the lungs and evaluate regional differences in ventilation patterns. This may be used to set PEEP level. It is unknown whether EIT improves outcomes.
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Journal2022, Anaesthesia & Intensive Care MedicineTapan Parikh aka Parmar, David Pilcher
Chapter
Pneumothorax, Chylothorax, Hemothorax, and Fibrothorax
2016, Murray and Nadel's Textbook of Respiratory Medicine (Sixth Edition)Richard W. Light MD, Y.C. Gary Lee MBChB, PhD
Pathophysiology of Pneumothorax
In normal subjects the pressure in the pleural space is negative with respect to the alveolar pressure during the entire respiratory cycle. The pressure gradient between the alveoli and the pleural space—the transpulmonary pressure—is the result of the inherent elastic recoil of the lung. During spontaneous breathing, the pleural pressure is also negative with respect to the atmospheric pressure. The functional residual capacity, or resting end-expiratory volume of the lung, is the volume at which the inherent outward pull of the chest wall is equal to, but opposite in direction to, the inward pull (recoil) of the lung.1
When a communication develops between an alveolus or other intrapulmonary air space and the pleural space, air will flow from the alveolus into the pleural space until there is no longer a pressure difference or until the communication is sealed. Similarly, when a communication develops through the chest wall between the atmosphere and the pleural space, air will enter the pleural space until the pressure gradient is eliminated or the communication is closed. The influence of a pneumothorax on the volume of the hemithorax and the lung is illustrated in Figure 81-1. In this example, sufficient air has entered the pleural space to elevate the pleural pressure from −5 to −2.5 cm H2O, so that the transpulmonary or recoil pressure has decreased from 5 to 2.5 cm H2O. The amount of air necessary to effect this change in the pleural pressure can be seen to be equal to 33% of the patient's vital capacity: most of this pleural air (25% of the vital capacity) is accounted for by air displaced from the lung, and the rest is accounted for by the expansion of the thoracic cavity along its pressure-volume curve (by 8% of the vital capacity). The rise in the pleural pressure also causes a shift of the mediastinum to the contralateral side, an enlarged hemithorax, and a depressed hemidiaphragm. These findings are expected and do not necessarily indicate that a tension pneumothorax is present.1
The main physiologic consequences of a pneumothorax are a decrease in the vital capacity (as illustrated in Fig. 81-1) and the arterial Po2. In patients with PSP, the decrease in the vital capacity is usually well tolerated. If the lung function of the patient is abnormal before the development of the pneumothorax, however, the decrease in vital capacity may lead to respiratory insufficiency with alveolar hypoventilation and respiratory acidosis.
Most patients with a pneumothorax have a reduced arterial Po2 and an increase in the alveolar-arterial oxygen tension difference. In a series of 12 patients with spontaneous pneumothorax, the arterial Po2 was below 80 mm Hg in 9 (75%) and was below 55 mm Hg in 2 patients, both of whom had SSP.2
The reduction in arterial Po2 appears to be due to the creation of regions of the lung both with low ventilation-perfusion ratios and with absent ventilation (shunt), and occasionally due to alveolar hypoventilation. Norris and coworkers2 reported that the average right-to-left shunt in their 12 patients with spontaneous pneumothorax was more than 10%. Larger pneumothoraces were associated with greater shunts. When the pneumothorax occupied less than 25% of the hemithorax, the shunt was not increased.2
After air is evacuated from the pleural space, the arterial Po2 usually improves, but the improvement may take several hours. Norris and colleagues2 evacuated the pleural air from three patients with an initial shunt above 20%; within 90 minutes, the shunt had decreased below 10%, but it nonetheless remained above 5% in all patients. The delay in improvement may be related to the duration of the pneumothorax and the time necessary to expand collapsed alveoli.
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Book2016, Murray and Nadel's Textbook of Respiratory Medicine (Sixth Edition)Richard W. Light MD, Y.C. Gary Lee MBChB, PhD
Related terms:
Artificial Respiration
Tidal Volume
Acute Respiratory Distress Syndrome
Functional Residual Capacity
Thoracic Wall
Continuous Positive Airway Pressure
Airway Pressure
Respiratory System
Lung Vascular Resistance
Lung Volume
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History of The Spice Trade
Those innocent spice jars sitting in your cupboard don’t do much to reveal their incredible history. But did you know that nutmeg was once worth more by weight than gold? That in the 16th century, London dockworkers were paid their bonuses in cloves? That in 410 AD, when the Visigoths captured Rome, they demanded 3,000 pounds of peppercorns as ransom?
In its day, the spice trade was the world’s biggest industry: it established and destroyed empires, led to the discovery of new continents, and in many ways helped lay the foundation for the modern world.
Spices, which today are inexpensive and widely available, were once very tightly guarded and generated immense wealth for those who controlled them. The spice trade began in the Middle East over 4,000 years ago. Arabic spice merchants would create a sense of mystery by withholding the origins of their wares, and would ensure high prices by telling fantastic tales about fighting off fierce winged creatures to reach spices growing high on cliff walls.
Initially, the spice trade was conducted mostly by camel caravans over land routes. The Silk Road was an important route connecting Asia with the Mediterranean world, including North Africa and Europe. Trade on the Silk Road was a significant factor in the development of the great civilizations of China, India, Egypt, Persia, Arabia, and Rome.
The Roman Empire set up a powerful trading centre in Alexandria, Egypt in the first century BC and was in command of all of the spices entering the Greco-Roman world for many years. In another example of the historical value of now-common spices, Roman soldiers of the time were frequently paid in salt, a practice that led to the word “salary” and the phrase “worth his salt.” Over the following centuries, countless groups battled for control of the spice trade. Eventually, in the mid-13th century, Venice emerged as the primary trade port for spices bound for western and northern Europe. Venice became extremely prosperous by charging huge tariffs, and without direct access to Middle Eastern sources, the European people could do little else but pay the exorbitant prices they were charged. Even the wealthy had trouble paying for spices, and eventually they decided to do something about it.
In the 15th century, the spice trade was transformed by the European Age of Discovery. By this time, navigational equipment was better and long-haul sailing became possible. Rich entrepreneurs began outfitting explorers in hopes of circumventing Venice by discovering new ways to reach the areas where spices were grown. There were many voyages that missed their targets, but several of them ended up discovering new lands and new treasures. When Christopher Columbus set out in search of India, he found America instead, and brought back to Spain the fruits and vegetables he found, including chiles (he called them “peppers”, perhaps to soothe his disappointment at not finding peppercorns, and the term “chile pepper” persists to this day).
The first country that successfully circumnavigated Africa was Portugal, and in 1497 four vessels under the command of Vasco da Gama rounded the Cape of Good Hope, eventually sailing across the Indian Ocean to Calicut, India. This success marked the beginning of the Portuguese Empire. Spanish, English and Dutch expeditions soon followed, and the growing competition sparked bloody conflicts over control of the spice trade. As the middle class grew during the Renaissance, the popularity of spices rose. Wars over the Indonesian Spice Islands broke out between expanding European nations and continued for about 200 years, between the 15th and 17th centuries.
The United States began its entry into the world spice industry in the 18th century, when American businessmen began their own spice companies and started dealing directly with Asian growers rather than the established European companies. When people started getting rich, more and more companies formed and soon there were hundreds of American ships making around-the-world voyages for spices. Americans made new contributions to the spice world, notably the creation of chili powder by Texas settlers as an easier way to make Mexican dishes and the development of techniques for dehydrating onions and garlic.
As spices became more common, their value began to fall. The trade routes were wide open, people had figured out how to transplant spice plants to other parts of the world, and the wealthy monopolies began to crumble.
Pepper and cinnamon are no longer luxuries for most of us, and spices have lost the status and allure that once placed them alongside jewels and precious metals as the world’s most valuable items. But the incredible history remains, as does the wonderful variety of exotic flavours, colours and smells that made spices so valuable in the first place. |
10606 | https://www.quora.com/How-do-you-express-complex-number-3-i-2-i-2-in-the-polar-form | How to express complex number ((3 – i) / (2 + i)) ^2 in the polar form - Quora
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How do you express complex number ((3 – i) / (2 + i)) ^2 in the polar form?
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Math Questions
Answered by
Enrico Gregorio
· Author has 18.4K answers and 16M answer views
·Nov 6, 2023
You first “rationalize”:
(3−i)(2−i)=6−3 i−2 i+i 2=5−5 i=5(1−i)(3−i)(2−i)=6−3 i−2 i+i 2=5−5 i=5(1−i)
(2+i)(2−i)=4+1=5(2+i)(2−i)=4+1=5
Thus your number is actually
(1−i)2=1−2 i−1=−2 i(1−i)2=1−2 i−1=−2 i
and you should be able to write −i=e φ i−i=e φ i for a suitable φ,φ, aren’t you?
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9 3
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What is the polar form of the complex number (3+i) ^1/2?
How can we express each of the following complex numbers in polar form? A;- z=(1-i) (3+3i) B;- z= ((2i+2 (3^0.5)) ^6) /(1+i) ^6
How is the complex number ((2+3 i)((2+3 i)(2−3 i))/((3+4 i)∗(2−3 i))/((3+4 i)∗(3−4 i))(3−4 i))purely real?
How do I express -2√3 -2i in polar form?
Philip Lloyd
Specialist Calculus Teacher, Motivator and Baroque Trumpet Soloist. · Author has 6.8K answers and 52.8M answer views
·Updated 1y
I don’t care if questions are “homework questions”. If a student asks a question ,I can only hope that the student does not just copy the solution but LEARNS how to do such problems from a well explained solution provided. My solutions are designed to teach the reader how to do the problem.
Continue Reading
I don’t care if questions are “homework questions”. If a student asks a question ,I can only hope that the student does not just copy the solution but LEARNS how to do such problems from a well explained solution provided. My solutions are designed to teach the reader how to do the problem.
Upvote ·
99 23
9 6
9 1
Diedrich Ehlerding
Retired (2019–present) · Author has 2.3K answers and 1.3M answer views
·2y
Since this is probably a homework, I will not give the complete solution, just a few hints how you can proceed.
Either you compute the polar form of 3−i 3−i and 2+i 2+i and remember how you divide and how you square (i.e. multiply) complex numbers in polar form. Or you compute 3−i 2+i 3−i 2+i as a+b i a+b i, square the result, and convert it to polar form.
The latter method yields
3−i 2+i=a+b⋅i 3−i 2+i=a+b·i
3−i=(a+b⋅i)(2+i)3−i=(a+b·i)(2+i)
Now multiply the right side and compare real and imaginary components, that will give you two linear equations for a a and b b. Solve them (you should be able to do that on your own). Then, multiply
Continue Reading
Since this is probably a homework, I will not give the complete solution, just a few hints how you can proceed.
Either you compute the polar form of 3−i 3−i and 2+i 2+i and remember how you divide and how you square (i.e. multiply) complex numbers in polar form. Or you compute 3−i 2+i 3−i 2+i as a+b i a+b i, square the result, and convert it to polar form.
The latter method yields
3−i 2+i=a+b⋅i 3−i 2+i=a+b·i
3−i=(a+b⋅i)(2+i)3−i=(a+b·i)(2+i)
Now multiply the right side and compare real and imaginary components, that will give you two linear equations for a a and b b. Solve them (you should be able to do that on your own). Then, multiply the result by itself, and you get the result in x+y⋅i x+y·i form. Compute the absolute value r r of this complex number (you might need to remember Pythagoras) , have a look at where it is located in the complex plane, and express the argument ϕ ϕ (the angle with the positive real axis, counterclockwise) in units of π π Then the result ist
r⋅e i⋅ϕ r·e i·ϕ
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Applied Mathematics
Answered by
Ansh Keer
· Author has 6K answers and 1.7M answer views
·Nov 6, 2023
(3-i) /(1+4i) = (3-i) (1-4i) / 17
= 1/7( -1 -13i)
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9 1
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10607 | https://stars.library.ucf.edu/cgi/viewcontent.cgi?article=1288&context=hut2024 | Published Time: Wed, 18 Jun 2025 15:48:38 GMT
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Honors Undergraduate Theses
2025
Prime Factorization and Unit Calculations of Quadratic Integer Prime Factorization and Unit Calculations of Quadratic Integer
Rings Rings
Gabriel F. Roca University of Central Florida , gabriel.roca@ucf.edu
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Roca, Gabriel F., "Prime Factorization and Unit Calculations of Quadratic Integer Rings" (2025).
Honors
Undergraduate Theses . 296.
PRIME FACTORIZATION AND UNIT CALCULATIONS OF QUADRATIC INTEGER RINGS
by
GABRIEL ROCA
A thesis submitted in partial fulfllment of the requirements for the Honors Undergraduate Thesis program in Mathematics in the College of Sciences and in the Burnett Honors College at the University of Central Florida Orlando, Florida
Spring 2025
Thesis Co-Chairs: Michael Reid, Ph.D. and Wissam Ghantous, Ph.D. ABSTRACT
The failure of unique factorization in a ring leads to the investigation of the closest algebraic structure, which are prime ideals. Using generalizations that have helped solve questions such as Fermat’s Last Theorem, there is interest to study the elements with a multiplicative inverse (units) via the geometry and arithmetic patterns that arise in the extension of the
√
integers, Z[ d] where d is a non-square integer, since they provide tools for other questions in mathematics, ranging from pure algebra to applications in cryptography, and more. Overall, the following thesis provides a small exposition on the theory of integral domains and some specifc calculations.
ii ACKNOWLEDGEMENTS
First and foremost, I would like to especially thank the late Dr. Michael Reid for his mentorship and introducing me to Abstract Algebra. He was the frst to acknowledge and support my endeavors in pursuing a career in pure mathematics, giving me a chance to explore more advanced topics throughout the numerous hours I spent in his ofce in my time at UCF. With his brilliance and humor, he has infuenced who I am as a student and how I hope to treat mathematics as both a teacher and a researcher. I am extremely lucky to have had the opportunity to work with him on a project as an undergraduate student, as well as to be able to call him my professor and mentor.
Next, I would like to express my gratitude for Dr. Wissam Ghantous for advising me toward the end of my thesis semester, helping me fll in the blanks from my previous studies with Dr. Reid, as well as teaching me other advanced topics in modern algebra and number theory.
Furthermore, I am grateful for Dr. Heath Martin and his support on both my thesis committee and as an advisor as one of the few algebraists here at UCF.
Finally, I would like to thank my friends and family, as without them, I would not have the liberty to give pure mathematics a chance, nor as much motivation to tackle my future obstacles.
iii TABLE OF CONTENTS
LIST OF FIGURES v
LIST OF TABLES vi
1 PRELIMINARIES 1
2 QUADRATIC INTEGER RINGS 4
√
2.1 GEOMETRY OF Z[ d] . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5
2.2 PRINCIPAL QUADRATIC INTEGER RINGS . . . . . . . . . . . . . . . . 8
√
2.3 INTEGRAL CLOSURE OF Z IN Q[ d] . . . . . . . . . . . . . . . . . . . . 9
3 IRREDUCIBILITY IN UNIQUE FACTORIZATION DOMAINS 11
4 PRIME IDEAL FACTORIZATION 14
4.1 THE IDEAL CLASS GROUP . . . . . . . . . . . . . . . . . . . . . . . . . . 18
5 UNITS 22
5.1 CONTINUED FRACTIONS . . . . . . . . . . . . . . . . . . . . . . . . . . . 24
iv LIST OF FIGURES
1234
√
Graph of Z[ −1] . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
√
A random (white) point of Q( −1) in between a square from the lattice of
√
Z[ −1]. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
√
A random (white) point of Q( −2) in between a rectangle from the lattice
√
of Z[ −2]. Note: the radius of each ball around the 4 corners is exactly 1. .
√
√1 −3
The white point corresponds to ω = + ∈ Q( −3). Note: the radius of 2 2
667
5
each ball around the 4 corners is exactly 1. . . . . . . . . . . . . . . . . . . .
√ √
Lattice of (3 , 1 + −5) represented by squares, and lattice of (3 , 1 − −5)
8
represented by circles. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21
6 The plot of αα = ±1, which gives possible units ( α) of the integral closure of
real quadratic integer rings. . . . . . . . . . . . . . . . . . . . . . . . . . . . 24
7 Possible units of the integral closure of imaginary quadratic integer rings . . 24
vLIST OF TABLES
√
1 Continued Fraction Table for 13 . . . . . . . . . . . . . . . . . . . . . . . . 25
vi 1 PRELIMINARIES
In the study of ring theory, the rules of arithmetic using integers are obvious to many, including the fact that we may compare any two integers a and b (b ≠ 0) and write a = bq +r
with unique integers q and r where 0 ≤ r < |b|. This is used in Euclid’s division algorithm, and generalizations extend to polynomials with rational coefcients, and more. The following defnition from [4, Section 3.7] gives us a precise set of rules that a ring R must satisfy for such an algorithm to exist.
Defnition 1.1. An integral domain R is said to be a Euclidean ring if for every a ≠ 0 in
R there is defned a nonnegative integer d(a) such that
For all a, b ∈ R, both nonzero, d(a) ≤ d(ab ).
For any a, b ∈ R, both nonzero, there exist t, r ∈ R such that a = tb + r where either
r = 0 or d(r) < d (b).
Theorem 1.2. Suppose that R is an integral domain, W is a well-ordered set, and α :
R \ { 0} → W a function such that for all a, b ∈ R, b ̸ = 0 , there exist q, r ∈ R such that
a = bq + r and either r = 0 or α(r) < α (b).
Proof. The proof follows from [9, Proposition 4].
The previous theorem essentially states that condition 1 from Defnition 1.1 is implied by condition 2.
Other types of rings that are related to Euclidean rings are principal ideal domains, unique factorization domains, and integrally closed domains, which we will now defne.
Defnition 1.3. A principal ideal domain (PID) is an integral domain R such that for every ideal I ⊆ R, I = ( a) for some a ∈ R, where ( a) = {ar : r ∈ R} is the principal ideal
generated by a in R.
Defnition 1.4. A Unique Factorization Domain (UFD) is an integral domain R such that
Every element of R that is neither 0 nor a unit can be factored into a product of a fnite number of irreducible elements in R.
If p1 · · · pr and q1 · · · qs are two factorizations of the same element of R into irreducibles, then r = s and the qj can be renumbered so that pi and qi are associates.
Defnition 1.5. Let A ⊆ B be commutative rings with identity. Then b ∈ B is integral over
A if it is a zero of a monic polynomial with coefcients in A, i.e. bn +an−1bn−1 +· · · +a1b+a0 =0, ai ∈ A.
Defnition 1.6. Let A be an integral domain, and F its fraction feld. A is integrally
closed if f ∈ F is integral over A implies that f ∈ A.
The following chain of structures is of interest to us:
Euclidean Ring =⇒ PID =⇒ UFD =⇒ Integrally Closed
since various properties of rings become apparent when proving a ring is one of the four.
Also, note that all rings we will consider are integral domains, which are commutative rings with identity such that they have no divisors of zero, and they are not the zero ring.
Theorem 1.7. If R is a Euclidean ring, then R is a PID. Furthermore, since R is a PID,
R is a UFD. Lastly, since R is a UFD, R is integrally closed.
Proof. If R is a Euclidean ring then [9, Proposition 3] implies R is a PID, and since R is a PID, then [4, Theorem 34.18] implies R is a UFD. Now let F be the feld of fractions of R.Then assume f ∈ F is integral over R where f = p
q
in lowest terms for p, q ∈ R, q ≠ 0 (so no irreducible that divides q is an associate of any irreducible of p). Then we may take f to be a zero of a polynomial where f n + rn−1f n−1 + · · · + r1f + r0 = 0, where ri ∈ R for all 1 ≤ i ≤ n − 1. We may now multiply both sides of the equation by qn , and we now have
n−1nn
pn + qr n−1p · · · + qn−1r1a + q r 0 = 0. Note that q|qr n−1pn−1 + · · · + q r 0, hence q divides
pn . But since every factorization of p has no irreducible that is associate to any irreducible
2that divides q, the same is true for pn and q, hence q must be a unit in R, and thus q divides
p, hence f ∈ R, thus R is integrally closed.
However, this set of implications is not reversible. Observe the following rings, which we will examine later:
√
• Z[1+ −19
2
] is a PID, but not a Euclidean ring.
• Z[x] is a UFD, but not a PID.
√
• Z[ −5] is integrally closed, but not a UFD.
32 QUADRATIC INTEGER RINGS
Our rings of interest are quadratic integer rings, which we will now defne using [6, Ch. 2]:
Defnition 2.1. A number feld is a subfeld of C having fnite degree (dimension as a vector space) over Q, and an infnite class of number felds consists of the quadratic number
√ √ √
felds Q[ m], m ∈ Z, m not a perfect square, where Q[ m] = {a + b m : a, b ∈ Q}.
√
Defnition 2.2. A quadratic integer ring Z[ m], m ∈ Z, m not a perfect square, is a
√
subring of its corresponding quadratic number feld of the form {a + b m : a, b ∈ Z}.
√
Lemma 2.3. If R = Z[ −n] and n = 1 or n = 2 , then R is a Euclidean ring under the
√
function N : R \ { 0} → N where N (α) = αα = a2 + nb 2 and α = a + b −n.
√ √
Proof. Let α, β ∈ R where α = a + b −n, β = c + d −n (where a, b, c, d ∈ Z. Now consider
√
√ c − d −n
αβ −1 = ( a + b −n) c2 + nd 2
ac + nbd bc − ad √
= + −n.
c2 + nd 2 c2 + nd 2
ac +nbd bc −ad
Now let x = and y = 2+nd 2 . Since x, y ∈ Q, write x = q1 + r1 and y = q2 + r2 where c2+nd 2 c
q1 and q2 are the closest integers to x and y respectively. Then note that −
2
1
≤ r1, r 2 ≤ 1
2
by
22
≤ 1
choice of q1 and q2, hence r1 , r 2 4 .Now,
√ √ √ √
αβ −1 = x + y −n = ( q1 + r1) + ( q2 + r2) −n = ( q1 + q2 −n) + ( r1 + r2 −n).
√ √
Let γ = q1 + q2 −n and ρ = β(r1 + r2 −n), and note that γ ∈ R, and now we have
√
α = βγ + β(r1 + r2 −n) = βγ + ρ, hence ρ ∈ R.
4Finally, observe
√
N (ρ) = N (β(r1 + r2 −n)
√ p
= ββ(r1 + r2 −n)( r1 − r2 −n)
= ( r1
2
nr 2
2
)N (β)
1 1
≤ ( + n )N (β)
4 4
n + 1
= ( )N (β)
4
< N (β)
when n = 1 or 2, thus by Theorem 1.2, R is a Euclidean ring under the function N when
n = 1 or 2.
√ √
This only tells us that Z[ −1] and Z[ −2] are Euclidean rings under the usual Euclidean
norm function, but does not imply any similar result for some arbitrary n when considering
√ √
Z[ −n]. However, we will see that Z[ −n] is not a Euclidean ring for all integers n ≥ 3.
√
2.1 GEOMETRY OF Z[ d]
The previous remark motivates us to fnd a way to intuitively understand why rings of the
√
form R = Z[ −n] may or may not be Euclidean, so we turn to the xy -plane to observe this
notion of fnding, for any pair of a, b ∈ R, applying an analogy to Euclid’s division algorithm.
√
We will proceed by frst observing “all” points of Z[ −1]:
5y
√−1]
x
Figure 1: Graph of Z[
(a) A4×4grid is drawn with 5horizontal and vertical lines of equal length. Each intersection is
marked with ablack dot for atotal of 25 dots. This grid is placed on an xy -plane.
and the following: y
√
x
√
Figure 2: A random (white) point of Q( −1) in between a square from the lattice of Z[ −1].
(a) Four circles of radius 1that intersect non-trivially, with awhite circle that is in the intersection
of all four circles.
We may now prove the following statement in a geometric sense, which we do to build intuition on the limits of what makes a ring Euclidean, a PID, a UFD, and integrally closed.
√
Proposition 2.4. Let R = Z[ −1]. Then R is a Euclidean ring.
Proof. Let N be the usual Euclidean norm of the complex numbers, and let γ = α
β
for some
α,β ∈ R, β ≠ 0. Then notice from Figure 2a that we can fnd γ within radius 1 of some
6√
ω ∈ Z[ −1], and let ρ = α − βω .
Then, we have
| γ − ω |< 1
α
| − ω |< 1
β
| α − βω |<| β |
| ρ |<| β |
thus, R is a Euclidean ring.
Now, we will stretch the vertical distances so we have precisely the graphical representation
√
of Z[ −2]:
y
x
√
Figure 3: A random (white) point of Q( −2) in between a rectangle from the lattice of √
Z[ −2]. Note: the radius of each ball around the 4 corners is exactly 1.
(a) Four circles of radius 1that intersect non-trivially, and with awhite circle that is in the
intersection of three of the four circles.
√
Proposition 2.5. Let R = Z[ −2]. Then R is a Euclidean ring.
Proof. Follows similarly to 2.4.
√
Now, if we extend the lattice structure even further to Z[ −3], we have the following possibility:
7y
x
√
√
Figure 4: The white point corresponds to ω = 2
1
2
−3
∈ Q( −3). Note: the radius of each ball around the 4 corners is exactly 1.
(a) Four circles intersect at awhite point.
√
In Figure 4a, ω is not contained in any ball of radius 1 centered at any point of Z[ −3],
√
giving a geometric explanation for Z[ −n] not being a Euclidean domain for all n ≥ 3.
2.2 PRINCIPAL QUADRATIC INTEGER RINGS
Similarly to how we used geometry to study if a ring is Euclidean or not, we may use geometry to see if a ring is a PID or not.
√
Lemma 2.6. Let R = Z[ −d], d ∈ N, and I ⊆ R an ideal. Then I is principal if and only
if there exists a singular dilation and a singular rotation, say n and ϕ such that applying n
and ϕ to each element in R results in I.
Proof. Suppose that I ⊆ R is principal. Then by defnition there exists some element in
δ ∈ R such that I = ( δ), but δ is a complex number, which we can represent as δ = ne iϕ ,and so I is the set of all elements in R multiplied by this dilation and rotation. Conversely, if there exists some δ ∈ R such that rδ ∈ I for all r ∈ R, we can assume that δ is of the form
ne iϕ since δ ∈ C, and we have I = ( δ).
−19
We now turn our attention back to the example R = Z[1+ √
2
], and choose α = 1 + √−19
and β = 2. The point α
β
is in the region that we showed is not within radius 1 from any
√
lattice point of Z[ −19 ]. Now, consider the ideal I = ( α, β). We in fact can fnd a singular
8dilation and rotation as in [2, Section 8.2], that when applied to all elements in R, we
attain ( α, β), so I is principal. So indeed, Z[1+ √
2
−19
] is a PID that is not Euclidean.
−19
Refning Z[√−19] to Z[1+ √
2
] is not a random choice, as we will see in the inspection of the integral closure of the integers in quadratic felds.
√
2.3 INTEGRAL CLOSURE OF Z IN Q[ d]
Defnition 2.7. The integral closure of A is {f ∈ F | f is integral over A}.
√ √
Since Z[ −1] and Z[ −2] are Euclidean rings, by Theorem 1.7, they are integrally closed, but this is more rigorously seen in the following theorem.
Theorem 2.8. Suppose that d is a square-free integer. Then the integral closure of Z in
√
Q( d) is
√
• Z[ d] if d ≡ 2 or 3 ( mod 4)
√
• Z[1+ d
2
] if d ≡ 1 ( mod 4) .
√ √
Proof. Suppose α ∈ Q( d) is integral over Z, and write α = a + b d, a, b ∈ Z. Let f (x) be the minimal monic polynomial of α over Z, hence f (x) = ( x−α)( x−α) = x2 −2ax +a2 −db 2 .It follows that 2 a ∈ Z and a2 − db 2 ∈ Z. We will write a = p
q
, b = r
s
in lowest terms, where
p, q, r, s ∈ Z, q ≠ 0, r ≠ 0 and ( p, q) = 1 = ( r, s). If a ∈ Z, then a2 − db 2 ∈ Z implies that −db 2 ∈ Z, and so d r
s
2
∈ Z, but d is square-free, and 2
(r2, s 2) = 1, thus s2|r2 , hence b2 ∈ Z, which means b ∈ Z since the square root of an integer is always either an integer or irrational. If a ∈/ Z, but 2 a ∈ Z, then we may assume that p is odd and q = 2. Now since a2 − db 2 ∈ Z,
p
− dr 2 2 2
let n = a2 −db 2 for some n ∈ Z, and now we have n = 4
2
s
, and hence p s 2 −4dr 2 = 4 s n ,2
thus s2|4 since s2|4dr 2 , where d is square-free and ( r2, s 2) = 1, and so s = 1 or s = 2. If s = 1, then p2 − 4dr 2 = 4 n, implying that 4 |p2 , contrary to p being odd, so s ≠ 1. If s = 2, then 4 p2 − 4dr 2 = 16 n, hence p2 − dr 2 = 4 n. So we have that p2 − dr 2 ≡ 0 mod
Working modulo 4 however, implies that p2, r 2 ≡ 0 or 1, and so there are 4 possibilities:
p2 ≡ 0 ≡ r2 , p2 ≡ 1 ≡ r2 , p2 ≡ 0 and r2 ≡ 1, or p2 ≡ 1 and r2 ≡ 0. If d ≡ 1, then p2 − dr 2 ≡ p2 − r2 ≡ p2 + 3 r2 , where 3 r2 ≡ 0 or 3, and so p2 + 3 r2 ≡ 0 only when either p2 ≡ 1 ≡ r2 , or p2 ≡ 0 ≡ r2 , but the latter contradicts the parity of p, hence if
d ≡ 1, then either a, b ∈ Z or a and b are rational numbers whose numerators are odd and denominators are 2. If d ≡ 2 or 3, then p2 − dr 2 ≡ p2 + dr 2 or p2 + 2 dr 2 , but in either case it must be that
p2 ≡ 0 ≡ r2 , contrary to the parity of p, thus s ≠ 2, so it must be that a ∈ Z if d ≡ 2 or 3, both a, b ∈ Z.
This theorem leads directly to one of the examples from earlier:
√
Example 2.9. The ring R = Z[ −5] is integrally closed, but not a UFD.
Proof. Since −5 ≡ 3 (mod 4), it follows from the previous theorem that R is integrally
√ √
closed. Now, consider the two factorizations of 6 in R, (1 − −5)(1 + −5) = 6 = 2 · 3.
√
However, 2 is not an associate of (1 ± −5), and neither is 3, so 6 ∈ R does not have a unique factorization. Hence R is not a UFD.
10 3 IRREDUCIBILITY IN UNIQUE FACTORIZATION
DOMAINS
There are generalizations of working with the fraction feld of integers (rational numbers) that extend beyond the specifc case of quadratic integer rings that show what we can expect when working with an arbitrary unique factorization domain.
We will begin with a defnition regarding polynomials in Z[x].
Defnition 3.1. The content of a polynomial f ∈ Z[x] is the greatest common divisor of all the coefcients of f , and if the content of f is 1, then we say that f is primitive .
Our notion of content and primitivity of polynomials with integer coefcients can be extended to being defned up to a unit in arbitrary unique factorization domains, and if R
is a UFD and f ∈ R[x], then we will say that f is irreducible if every factorization f = gh
implies that either g or h is an element in R.
Theorem 3.2 (Gauss’ Lemma for Unique Factorization Domains) . Let R be a UFD, F its
fraction feld. Then suppose f (x) ∈ R[x] factors in F [x] as f (x) = g(x)h(x). Then f (x)
factors in R[x] as g˜( x)h˜(x), where g˜( x) = αg (x) and h˜(x) = βh (x), α, β ∈ F , such that
αβ = 1 .
Proof. We may factor out c = content( f ) so f (x) = cf ˜(x), and so we will assume that
f is primitive. Write the coefcients of g as pi , and the coefcients of h as ri . Then qi si
Q Q
let Q = qi, and S = si. Now set d = QS , and hence we have df (x) = g ′ (x)h′ (x)where g ′ (x) = Qg (x) and h′ (x) = Sh (x), and note that g ′ (x), h ′ (x) ∈ R[x]. Next, let
a = content( g ′ ) and b = content( h′ ). By [3, Lemma 34.26], we have that k = content( g ′ h′ ) = content( g ′ ) content( h′ ) = ab , and hence df (x) = abg ′′ (x)h′′ (x), and so f (x) = ab
d
g ′′ (x)h′′ (x), where g ′′ , h ′′ ∈ R[x] are primitive, but so if f , and thus ab
d
= 1, so let d1, d 2 ∈ Z such that
d = d1d2 with d1|a and d2|b. Now let α = a , β = b , and so f (x) factors in R[x] as g˜( x)h˜(x), d1 d2
where g˜( x) = αg ′′ (x) and h˜(x) = βh ′′ (x), α, β ∈ F , such that αβ = 1.
11 To construct a criterion for irrducibility of polynomials in R[x] where R is a UFD, we defne the following:
Defnition 3.3. A prime ideal of a ring R is an ideal P ⊂ R such that
If ab ∈ P , then either a ∈ P or b ∈ P .
P ̸ = R.
Theorem 3.4 (Eisenstein’s Criterion for Unique Factorization Domains) . Let R be UFD,
and let P be a prime ideal in R. Let f (x) = anxn + · · · + a1x + a0 ∈ R[x]. Then if
a0, a 1, . . . , a n−1 ∈ P
an ∈/ P
a0 ∈/ P 2 ,
then f is irreducible in R[x].
Proof. Suppose to the contrary that f had a nontrivial factorization, i.e. f (x) = g(x)h(x)such that 0 < deg( g), deg( h) < n = deg( f ). Write g(x) = bmxm + · · · + b1x + b0 and let
k = n − m, and write h(x) = ckxk + · · · + c1x + c0. Let R′ = R/P . Consider the projection
φ : R → R′ , and φ : R[x] → R′ [x] via φ(anxn +· · · +a1x+a0) = φ(an)xn +· · · +φ(a1)+ φ(a0). We will denote φ(ai) = ai. So φ(f (x)) = φ(g(x)h(x)) = φ(g(x)) φ(h(x)) = φ(bmxm + · · · +
b0)φ(ckxk + · · · + c0) = ( bmxm + · · · + b0)( ckxk + · · · + c0), and since a0 ∈ P , then a0 = 0, hence b0c0 = 0, and since R is an integral domain, either b0 = 0 or c0 = 0. Suppose b0 = 0. But then similarly, b1c0 = 0. Following this process, we have that bmc0 = 0, but if bm = 0, then an = bmck = 0, contrary to an ∈/ P , so it must be that c0 = P , hence b0 = 0 and c0 = 0, so b0 ∈ P and c0 ∈ P , hence a0 ∈ P 2 , a contradiction.
Theorem 3.5. If R is a UFD, then R[x] is a UFD.
Proof. Follows from [3, Theorem 34.30].
So now we will state the remaining example from earlier.
12 Proposition 3.6. The ring R = Z[x] is a UFD, but not a PID.
Proof. By the previous theorem, Z[x] is indeed a UFD since Z is a UFD, but now consider the ideal (2 , x ) ⊆ R, and suppose to the contrary that (2 , x ) were principal. Then write (2 , x ) = ( p(x)) for some p(x) ∈ R. But then that means that p(x)|2, so p(x) = ±1, ±2. If (p(x)) = (2), then 2 |x, a contradiction. If ( p(x)) = (1) = R, then there exist g(x), h (x) ∈ R
such that 2 g(x) + xh (x) = 1, but the constant term in the left hand side of the equation must be of the form 2 k for some k ∈ Z, so we have that 2 k = 1, a contradiction, hence (2 , x )is not principal, and thus R is not a PID.
So when R is not a UFD, we lose general prime factorization in R[x].
13 4 PRIME IDEAL FACTORIZATION
When unique factorization fails, the closest idea of factorization we can get comes from prime ideals, which we investigate through the lens of Dedekind domains.
Defnition 4.1. A Dedekind domain is an integral domain R such that
Every ideal is fnitely generated;
Every nonzero prime ideal is a maximal ideal;
R is integrally closed in its feld of fractions.
Defnition 4.2. A maximal ideal of a ring R is an ideal M ⊂ R such that
If I ⊆ R is an ideal such that M ⊆ I ⊆ R, then either I = M or I = R.
M ̸ = R.
Theorem 4.3. Every quadratic integer ring is a Dedekind domain.
Proof. Follows from [6, Theorem 14]. The idea of factorization using prime ideals comes from :
Theorem 4.4. Every ideal in a Dedekind domain R is uniquely representable as a product
of prime ideals.
Proof. Follows from [6, Theorem 16].
√ √
In other words, regardless of Z[ d] being a UFD or not, all ideals of Z[ d] admit a prime
√
factorization of ideals. We will “push” a prime p , → (p) ⊆ Z[ d], the principal ideal
generated by p for any given d ∈ Z.
√
Considering the ring Z[ −5], notice that the prime number 3 ∈ Z, when considered as the
√ √ √
ideal (3) ⊂ Z[ −5], is no longer prime. We can write (3) = (3 , 1 + −5)(3 , 1 − −5), a product of distinct prime ideals (more on why these are distinct and prime later). We say
√
that the prime ideal (3) splits in Z[ −5]. More generally, such prime ideals either split, ramify, or are inert, as in the following defnition.
14
√
Defnition 4.5. Let p ∈ N be prime, d ∈ Z, and O be the integral closure of Z in Q( d). We say that
p splits if there exists distinct prime ideals P and Q of O such that ( p) = P Q ,
p ramifes if there exists a prime ideal P of O such that ( p) = P 2 ,
p is inert if there exists a prime ideal P of O such that ( p) = P .
Defnition 4.6. The Legendre Symbol , defned for a ∈ Z and p and odd prime is as follows:
0 , p | a
a =
p 1 , p ∤ a and a is a quadratic residue modulo p
−1 , p ∤ a and a is not a quadratic residue modulo p
An immediate (and fair) challenge to this defnition is to call to attention that this excludes the prime p = 2, for which we must extend this defnition to a generalization:
Defnition 4.7. The Kronecker Symbol defned for a ∈ Z is the Legendre Symbol for odd primes p, and for p = 2 is as follows:
0 , a ≡ 0 (mod 2)
a = 1 , a ≡ ± 1 (mod 8)
2
−1 , a ≡ ± 3 (mod 8)
We state the following well-known theorem in generality to further exhibit splitting of primes:
Theorem 4.8 (Chinese Remainder Theorem) . Let I1, . . . , I n be pairwise relatively prime
ideals in a ring R. Then the mapping R/ ∩n
i=1
Ii → R/I 1 × · · · × R/I n via
r 7 → (r mod I1, . . . , r mod In) is an isomorphism.
Proof. Follows from [6, Appendix A]
15
√
Then, in general, when considering the ring Z[ d] for d ∈ N, non-square, if we consider a
√
prime p and would like to know if it ramifes, splits, or stays inert in Z[ d], we consider
x2 − d modulo p, which we will say to be x2 − dp. Then,
Proposition 4.9. Let d ∈ Z, non-square, and let p be an odd prime. Then letting OQ(√d)
√
be the integral closure of Z in Q( d),
0 if and only if (p) ramifes in OQ(√d) d
p = 1 if and only if (p) splits in OQ(√ ,d)
√−1 if and only if (p) remains inert in OQ( d)
.
√
Proof. First, if d ≡ 2, 3 (mod 4), then OQ(√d) = Z[ d], and so
√
OQ(√d)/(p) = ∼ Z[ d]/(p) ∼= ( Z[x]/⟨x2 − d⟩)/(p) ∼= Fp[x]/⟨x2 − d⟩ ∼= Fp[x]/⟨x2 − dp⟩,
where dp is the reduction of d modulo p.
√
If d
p
= 0, then this is if and only if d ≡ 0 mod p, and so Z[ d]/(p) ∼= Fp[x]/⟨x2⟩, hence ( p)
√
is a square of a prime ideal, and thus ( p) ramifes in Z[ d].
If d
p
= 1, then this is if and only if there exist distinct solutions x1, x2 modulo p such
that x1
2
≡ dp and x2
2
≡ dp, and so
√
Z[ d]/(p) ∼= Fp[x]/(⟨x1 − dp⟩⟨ x2 − dp⟩) ∼= Fp[x]/⟨x1 − dp⟩ × Fp[x]/⟨x2 − dp⟩ by the Chinese
Remainder Theorem, hence ( p) factors into two distinct prime ideals, and thus ( p) splits in
√
Z[ −d].
If d
p
= −1, then this is if and only if there do not exist solutions x modulo p such that
x2 ≡ dp, and so ⟨x2 − dp⟩ is an ideal generated by an irreducible polynomial over Fp, and so
√
this ideal is maximal, and hence prime itself, and so ( p) is also a prime ideal in Z[ d], and
√
therefore ( p) remains inert in Z[ d].
16 √ = Z[1+ √d
Next, if d ≡ 1 (mod 4), then OQ( ], and so d) 2
√
1 + d (d − 1) (dp − 1) √OQ( d)/(p) ∼= Z /(p) ∼= Z[x] x2−x− (p) ∼= Fp[x] x2−x− ,
2 4 4
where dp is the reduction of d modulo p.
2
− x − (d−1)
Again, it sufces to check possible solutions of x 4 = 0, which, if existent, are of
the form 1±
2
√d
(mod p).
If d
p
= 0, then d ≡ 0 (mod p), which is if and only if there is only 1 distinct root of
multiplicity 2, x = 1 , and hence ( p) ramifes in Z[1+ √d ].
2 2
If p
d
= 1, then this is if and only if d is equivalent to a square modulo p, say a2 , and so
1−ad
there are 2 distinct roots, and 1+ a , and hence ( p) splits in Z[1+ √
]. 2 2 2
If p
d
= −1, then this is if and only if d is not equivalent to any square modulo p, and
hence there does not exist any value in Fp equivalent to 1±
2
√d
, and thus the polynomial
2
− x − (d−1) d
x 4 is in fact irreducible, and hence ( p) remains inert in Z[1+
2
√
].
Proposition 4.10. Let d ∈ Z, non-square, and let p = 2 . Also, let OQ(√d) be the integral
√
closure of Z in Q( d). Then,
d ≡ 2, 3 (mod 4) if and only if (2) ramifes in OQ(√d)
d ≡ 1 (mod 8) if and only if (2) splits in OQ(√d)
d ≡ 5 (mod 8) if and only if (2) remains inert in OQ(√d).
Proof. If d ≡ 2, 3 (mod 4), then following the proof of Proposition 4.9 for the
d ≡ 2, 3 (mod 4) case, this is if and only if x2 − dp is either equivalent to x2 or x2 − 1
working modulo 2, but then x2 − 1 = ( x − 1)( x + 1) ≡ (x − 1) 2 , so in either case, ( p)
√
ramifes in Z[ d].
The other case is if d ≡ 1 (mod 4), but this has sub-cases working modulo 8, since again,
2
− x − (d−1)
following the proof of Proposition 4.9, the polynomial x 4 = 0 working modulo 8
has two possibilities. If d ≡ 1 (mod 8), then working modulo 8, this is if and only if
2
− x − (d−1)
x 4 ≡ x2 − x = x(x − 1), and hence the polynomial has 2 distinct roots, namely
17 x = 0 and x = 1, and thus (2) splits in Z[1+
2
√d
]. However, if d ≡ 5 (mod 8), then working
2
− x − (d−1)
modulo 8, this is if and only if x 4 ≡ x2 − x − 1, which has no solutions modulo
8, and thus is an irreducible polynomial over Z/8Z, hence (2) remains inert in Z[1+
2
√d
].
For actually computing this symbol, we use the standard properties of the Legendre
p−1
−1ab ab2
Symbol, which are that for any odd prime p, = ( −1) and = . For p p p p
�
the case p = 2, −
2
1
= 1 and the multiplicative rule also holds.
Example 4.11 (Splitting Example) . Consider the prime integer p = 17 in a non-UFD
√ √
R = Z[ −30 ], where the integral closure of R in Q( −30 ) is R since −30 ≡ 2 (mod 4).
� � � � � � 17 −1
−30 −130 2352
Then, = = ( −1) = (1)( −1)( −1) = 1, and so the ideal 17 17 17 17 17 17
(17) splits in R, and naturally we would say that (17) is not itself a prime in R, but rather
factors into 2 distinct primes of R. More explicitly:
Z[√−30] /(17) ∼= ( Z[x]/⟨x2 + 30 ⟩)/(17) ∼= F17 [x]/⟨x2 + 30 ⟩ ∼= F17 [x]/⟨x2 + 13 ⟩ ∼=
F17 [x]/⟨(x − 2)( x − 15) ⟩ ∼= F17 [x]/⟨x − 2⟩ × F17 [x]/⟨x − 15 ⟩ ∼=
Z[x]/(17 , x + 2) × Z[x]/(17 , x + 15) ∼= Z[x]/(17 , 2 + √−30) × Z[x]/(17 , 15 + √−30).
Example 4.12 (Ramifcation Example) . Consider the prime integer p = 2 in a non-UFD
√ √
R = Z[ −5], where the integral closure of R in Q( −5) is R since −5 ≡ 3 (mod 4). So
then the ideal (2) ramifes in R, and naturally we would say that (2) is not itself a prime in
R, but rather ramifes into the square of a prime of R.
Example 4.13 (Inert Example) . Consider the prime integer p = 13 in a non-UFD
√ √
R = Z[ −6], where the integral closure of R in Q( −6) is R since −6 ≡ 2 (mod 4). Then,
� � � � � −6 −1 6 2 3
= = = ( −1)(1) = −1, and so the ideal (13) stays inert in R, and 13 13 13 13 13
naturally we would say that (13) is a prime ideal of R.
4.1 THE IDEAL CLASS GROUP
Primes may stay prime, split, or become the square of one, but to calculate these prime
√
ideals, we revisit the geometry of ideals in Z[ d]. There is indeed a limit, up to a factor of
a principal ideal, between any two ideals I and J.
18 For this subsection, we follow the theory in [6, Chapter 5]
Defnition 4.14. The ideal class group of a number ring R consists of equivalence classes of nonzero ideals under the relation
I ∼ J ⇐⇒ (α)I = ( β)J
for some nonzero α, β ∈ R.
Theorem 4.15 (Minkowski) . Let Λ ⊆ R2 be a lattice where Λ = Zw1 + Zw2, and let K be
the co-area of Λ, i.e., the area of the fundamental parallelogram. Now suppose D is a closed
convex, centrally symmetric region of area A > 4K. Then there exists a nonzero v ∈ D ∩ Λ.
Proof. Let u, w ∈ D such that u ̸ = w and u ≡ w (mod 2Λ). Such u and w exist since
A > 4K. Then u − w ∈ 2Λ since D is centrally symmetric, and hence 1
2
(u − w) ∈ Λ. Also, since D is closed and convex, 1 (u + w) ∈ D, hence 1 (u − w) ∈ D. Letting v = 1 (u − w), we 2 2 2
have that v ∈ D ∩ Λ.
√ √
The motivation behind this geometric theorem is that if R = Z[ −d] or Z[1+
2
−d
], Λ the image of an ideal I ⊆ R, Λ ˜ the image of R in C ∼ (so Λ is a sublattice of ˜= R2 Λ), D a disc of radius ρ > 0 centered at the origin, then we get an element of I whose norm is less than √ √ d for Z[ −d]
ρ2 since Area( D) = πρ 2 =⇒ CoArea(Λ) = √ √
1 −d
2
d for Z[1+
2
]
This observation is implied by the following:
Theorem 4.16 (Imaginary Quadratic Version) . Every ideal class of the integral closure of
2
Z in Q(√−d) for any d ∈ N contains a nonzero ideal I with || I||≤ 1 ( 4 )1p|D| = p|D|,
2 π π
−d if d ≡ 1 (mod 4)
where D = −4d if d ≡ 2, 3 (mod 4)
19 Theorem 4.17 (Real Quadratic Version) . Every ideal class of the integral closure of Z in
√ √ √
Q( d) for any d ∈ N contains a nonzero ideal I with || I||≤ 2
1
(π
4
)0 D = 2
1
D, where
d if d ≡ 1 (mod 4)
D = 4d if d ≡ 2, 3 (mod 4)
√
Before providing a proof, the constant that multiplies with D in both cases is
Minkowski’s constant , n
nn
!
(π
4
)s , where n is the dimension of the lattice in Rn , and s is the
√
number of complex conjugate pairs of embeddings of Q( m) for all m ∈ Z.
Proof. Follows from [6, Chapter 5, Corollary 2]. The statement is given in a general setting, so here we apply the fact that for both cases, n = 2 and for the imaginary case,
s = 1, whereas s = 0 in the real case.
Minkowski’s bound provides a new tool to evaluate if a ring is a principal ideal or not, which we can observe through the following example.
√ √
Example 4.18. Consider the ring R = Z[ −23 ]. Minkowski’s bound in this case is π
2
23 ,which is approximately 3.05, and so we must check the prime ideal factorization of (2) and (3). It sufces to show that at least one of the prime ideal factors of either (2) or (3) is not principal, and hence the entire ring cannot be a principal ideal domain. Consider
√√
1+ −23 1−−23
(2) = (2 , 2 )(2 , 2 ). By a similar reasoning given in the counterexample in Proposition 3.6, both prime ideal factors of (2) are not principal, and another similar
√√
1+ −23 1−−23
argument can be made for (3) = (3 , 2 )(3 , 2 ), and hence, R is not a principal ideal domain.
Again, even without having a principal ideal domain, we may consider the relationship between non-principal ideals using the ideal class group, and in particular, the geometric relationships.
√
Example 4.19. Revisiting (3) ⊂ Z[ −5], we have that (3) factors as
√ √ √
(3 , 1 + −5)(3 , 1 − −5). First, we show that P = (3 , 1 + −5) is prime. Consider
Z[√−5] /(3 , 1 + √−5) ∼= Z[x]/⟨x2 + 5 ⟩/(3 , 1 + x) ∼= F3[x]/⟨x2 + 5 , 1 + x⟩ ∼=
20 F3[x]/⟨x2 − 1, x + 1 ⟩ ∼= F3[x]/⟨(x − 1)( x + 1) , x + 1 ⟩ ∼= F3 since x − 1 and x + 1 have
√ √
solutions in F3, and F3 is a feld, so Z[ −5]/(3 , 1 + −5) is a feld, hence an integral
√ √
domain, which is if and only if (3 , 1 + −5) is prime. Similarly, Q = (3 , 1 − −5) is prime.
Now, if P were principal, since (3) ⊆ P , we must fnd that P must be (3) or the whole ring,
√ √
which is the integral closure of Z in Q[ −5], which is Z[ −5] but (3) is not prime, so P
cannot be prime, and similarly, if P were the whole ring, P could not be prime by defnition, a contradiction. The same logic holds for Q, and hence P and Q must be in the
√
same class group since the size of the class group of Z[ −5] has size 2. We can also see this geometrically, using the same idea of rotations and dilations:
y
31 − √−531 + √−5
x
√ √
Figure 5: Lattice of (3 , 1 + −5) represented by squares, and lattice of (3 , 1 − −5) repre -
sented by circles.
(a) The xy -plane, with squares representing linear combinations with integer coefcients of 3and √√
1 + −5, and circles representing linear combinations with integer coefcients of 3and 1−−5.
21
5 UNITS
√
After investigating how prime numbers behave in the ring extension Z[ d], we would like to also see the other property of these rings, which are the units of their integral closure in
√
Q( d).
Theorem 5.1 (Dirichlet Unit Theorem) . Let F be a fnite extension of Q, and O the
integral closure of Z in F . Then O× ∼= C × Zr+s−1 , where C is a fnite cyclic group (the
torsion part of the group), r is the number of real embeddings of F and s is the number of
complex conjugare pairs of embeddings of F .
Note that r + s − 1 is the rank, r + 2 s = [ F : Q], and t ∈ C =⇒ tn = 1 for some n ∈ N (t
is a root of unity).
Applying this to quadratic integer rings, we split into two cases: real quadratic integer rings and imaginary quadratic integer rings. Before stating these unit groups, the following defnition from is necessary to tackle the imaginary case in the following way.
Defnition 5.2. Euler’s Totient Function (sometimes referred to as the Euler phi-function) is ϕ : N → N via ϕ(n) = |{ 1 ≤ m < n } : gcd( m, n) = 1 |= |(Z/n Z)×|. By convention, ϕ(1) = 1.
√
Theorem 5.3. The unit group of the integral closure of Z in Q[ −d] for d ∈ N, O× , is as
follows:
{1, −1, i, −i} if d = 1
O× = {1, −1, ζ, −ζ, ζ 2 , −ζ2} if d = 3
{1, −1} otherwise
√
1+ −3
where ζ = 2
√ √
Proof. The degree of the feld extension Q( −d) over Q, [ Q( −d) : Q] = 2 (which is the
√
dimension of the basis of the vector space Q( −d) over Q), and based on Dirichlet’s Unit
22 √
Theorem, the only possible units of the integral closure of Z in Q[ −d] are roots of unity
= C × Z(0+1 −1)
since O× ∼ , and so it sufces to fnd ( n-th) roots of unity ζn such that
[Q[ζn] : Q] = 2. But since [ Q[ζn] : Q] = ϕ(n) where ϕ is Euler’s Totient Function. So setting
ϕ(n) = 2, we fnd that the only solutions are n = 3 , 4, and 6. However,
√ √
Q[ζ3] = Q[ζ6] = Q[ −3], and also Q[ζ4] = Q[ −1]. Also, no other n satisfes [Q[ζn] : Q] = 2, and so the only roots of unity that will stay units for all d ∈ N are those that are real numbers, i.e., 1 and −1.
In the real case, the unit group is now infnite unlike the imaginary case, since
= C × Z(2+0 −1) ∼O× ∼ = {± 1} × ⟨ u⟩, where u is the fundamental unit of the group, i.e., for
all w ∈ O × , w = ±uk for some k ∈ Z. Of course, once u is known, the entire group is determined, and so we have calculations which can specifcally be done with continued fractions.
For computations, note that if u is the fundamental unit, then if u and w are distinct units,
1 ≤ | u|≤ |w|. Finding minimality can be done in a fnite number of steps since these units
√ √
x+yd
are of the form u = x + y d (respectively u = 2 if d ≡ 1 (mod 4)) such that
N (u) = ±1, and there are only a fnite number of combinations of ( a, b) such that
√ √ √ √
|a + b d|≤ |x + y d| (resp. |a+b d |≤ |x+y d |), of course, after already having found some 2 2
unit u.
23 R
R
(α, ¯α), α ¯α = 1
Figure 6: The plot of αα = ±1, which gives possible units ( α) of the integral closure of real quadratic integer rings.
(a) The graph of xy = ±1.
i
ζ
C
Figure 7: Possible units of the integral closure of imaginary quadratic integer rings
(a) The graph of x2 + y2 = 1, with a dot placed at every 45 degrees starting from the origin, moving
counterclockwise. The dots on the real line are black, and all other dots are white.
5.1 CONTINUED FRACTIONS
Defnition 5.4. A continued fraction is an expression of the form 1
a1 + 1 .
a2 + 1
a3+a4+···
24 The reason for bringing up continued fractions is that we now have another way to express
√
real irrational quadratic numbers. By , we know that the continued fraction of d for
√
d > 0 is infnitely long when d is not a square, i.e., d is an irrational number. We also know that by , such terms ai must eventually repeat. However, there is an alternate perspective to consider when considering these continued fractions.
We will inspect the patterns, as well as results that follow from the repeated information in our continued fractions.
We are particularly interested in the patterns arising in the following table (using the example d = 13):
n an rn 1/r n
1 3 √13 −3
1
√13+3
4
2 1 √13 −1
4
√13+1
3
3 1 √13 −2
3
√13+2
3
4 1 √13 −1
3
√13+1
4
5 1 √13 −3
4
√13+3
1
6 6 r1 1/r 1
√
Table 1: Continued Fraction Table for 13
√
Moving forward, we defne rn to be derived the same way it was for the table of 13 , but
√
for any d. To be clearer, the process of constructing the table goes as follows:
√
Defne a1 to be the greatest integer less than d.
√ √
Write d = a1 + r1. So then r1 = d − a1.
√
Take the reciprocal of r1, r
1
, multiplying by d + a1 on both the numerator and 1
denominator.
Defne a2 to be the greatest integer less than r
1
.1
25 √
Repeat (2), but now with the current pair ( r
1
, a 2) instead of ( d, a1); that is, write 1
r
1
1
= a2 + r2.
√
This algorithm will give us d = a1 + 1 . The choice of stopping the expansion at 1
a2+1a3+a4+r5
r5 was arbitrary of course; the idea is to let this fraction expand infnitely so that we have
√
exactly the continued fraction representation of d.
Lemma 5.5. rn is of the form √d−a , where b|(d − a2) for all n ∈ N.b
√
Proof. By induction on n, we start with the case n = 1, where d = a1 + r1, hence
√
r1 = d−a , where 1 |(d − a2). Now suppose the result holds for all 1 ≤ n ≤ k. Then 1 √
1d−A
considering rk+1 , we start with = ak+1 + rk+1 . By the induction hypothesis, rk = Brk
where B|(d − A2). So assume that BC = d − A2 for some C. Then,
√ √
1 B B( d + A) d + A
= √ = = .
d − A2rk d − A C
But then,
√ √ √
d + A d + A − Ca k+1 d − (Ca k+1 − A)
rk+1 = − ak+1 = = ,
C C C
and
d − (Ca k+1 − A)2 = d − (C2 ak
2+1
− 2CAa k+1 + A2) = ( d − A2) + (2 CAa k+1 − C2 ak
2+1
),
where C|(d − A2) since BC = d − A2 , and C|C(2 Aa k+1 − Ca 2 ) = 2 CAa k+1 − C2ak
2+1
,k+1
thus proving the claim.
√
√
Lemma 5.6. For all n ∈ N, we have rn = d
b
−a
with 0 < a < d.
√
Proof. First, d = a1 + r1, and so 0 < r 1 < 1. Then r
1
1, and by the same process, 1
r
1
= a2 + r2, where 0 < r 2 < 1. By induction, for all n, we have that 0 < r n < 1, and so for 1 √ √ √
some a and b, 0 < d
b
−a
< 1, hence a < d. By the fact that d = a1 + r1, a1 is the
√ √
greatest integer less than d, and hence r1 = d
1
−a1
satisfes the statement, and in
26 particular, a1 > 0. Suppose that the statement holds for 1 ≤ i ≤ n − 1, as we will show
√
√d−an−1
that an > 0. By the induction hypothesis, write rn−1 = bn−1 where 0 < a n−1 < d. So
now, 1 = An + rn, where An is the greatest integer less than 1 . By Lemma 5.5, there rn−1 rn−1
exists some cn−1 such that bn−1cn−1 = d − an
2−1
. So now,
√√√
bn−1(d+an−1)d+an−1d−(cn−1An−an−1)
rn = 1 − An = − An = − An = . But rn−1 d−a2 cn−1 cn−1n−1 √1 1 cn−1
An > − 1, hence cn−1An > c n−1( − 1) = − cn−1 = ( d + an−1) − cn−1, and rn−1 rn−1 rn−1
√
cn−1 ≤ d + an−1 since if bn−1 = 1, the claim would hold since this would cause a
repetition, rn = r1, and thus the hypothesis holds. Otherwise, bn−1 ≥ 2, and hence
√ √ √ √2 d−an−1
= d − a = ( d − an−1)( d + an−1), hence cn−1 ≤ ( d + an−1), 2cn−1 ≤ bn−1cn−1 n−1 2 √
but this holds for any possible value of an−1, and in particular, when an−1 = ⌊ d⌋, and
√√
√√ √d−⌊ d⌋
thus cn−1 ≤ 2 ( d + an−1) ≤ 1
2
( d + an−1) < d + an−1, hence cn−1An > a n−1, and
thus rn is of the desired form, letting an = cn−1An − an−1, such that rn = √d−an , with
√ cn−1
0 < a n < d.
√
Theorem 5.7. For the continued fraction of d, where d ∈ N, there exists some minimal
k > 1 such that r1 = rk is the frst repetition that necessarily occurs in the rn values.
Proof. By , the continued fraction of an irrational number is infnitely long, and by
Lemma 5.5 and Lemma 5.6, it follows that rn can take on only a fnite number of values
since there are a fnite number of possible values of a and b, where b depends on a, and
√
0 < a < d. So there exists some i, j ∈ N with i < j such that ri = rj . Now defne xm such
that xm = 1 , i.e., xm = am + 1 . Then by [7, Section 4.4], each xm is a reduced rm−1 am+1 +rm+2
quadratic surd , that is, if ym is the conjugate root of xm with respect to the minimal
polynomial of degree 2 of xm, then ym ∈ (−1, 0).
So since ri = rj , then xi+1 = xj+1 . Now consider xi and xj . So
1 1 −1
xi = ai + ri = ai + yi = ai + = ai − yixi+1 yi+1 yi+1
=⇒ =⇒
1 1 −1
xj = aj + rj = aj + yj = aj + = aj − yj .xj+1 yj+1 yj+1
Now, if we have the equation α = β + γ, where α > 1, β ∈ N, and γ ∈ (0 , 1), then it must
be that β = ⌊α⌋, the greatest integer less than α. But setting α = 1 , β = ai, and yi+1
γ = −yi we have exactly the same assumptions satisfed since ai ∈ N, and having xi and
xi+1 both be reduced quadratic surds implies that yi ∈ (−1, 0), and so −yi ∈ (0 , 1), and
27 furthermore, since we similarly have that −yi+1 ∈ (−1, 0), then 1 > 1. Applying the yi+1
same argument for α = 1 , β = aj , and γ = −yj , then we have that ai = ⌊ −1 ⌋ and yj+1 yi+1
aj = ⌊ −1 ⌋, and since xi+1 = xj+1 , yi+1 = yj+1 , which means that ai = aj . So then xi = xj ,yj+1
and hence ri−1 = rj−1. Since i and j were arbitrarily chosen such that i ≥ 1 and j > i ,
recursively we end up with r1 = rk for some k > 1.
√
Proposition 5.8. For the continued fraction of d where d ∈ N, if k > 1 such that
√
r1 = rk, then 1 = d + a1.rk−1
√ √
Proof. 1 = ak + rk = ak + r1 = ak + ( d − a1) = d + ( ak − a1), but by [7, Section 4.7], rk−1 √
ak = 2 a1, and thus 1 = d + a1.rk−1
An observation to make when writing out the table for any square-free integer d > 0 is that
when the denominator of 1 /r n is fnally 1 for some n, the succeeding row will satisfy
an+1 = 2 a1 and rn+1 = r1, giving us a repeating cycle from then on out.
√
Theorem 5.9. For any d ∈ N, suppose that the continued fraction of d where k > 1 is
minimal such that r1 = rk, and let p = a1 + 1 in reduced form (gcd( p, q) = 1) .1q a2+ 1+... + ak−1
Then p2 − dq 2 = ±1.
Proof. Follows from [7, Section 4.8].
√ √
Observe that p2 − dq 2 represents the norm of an element p + q d ∈ Z[ d], and so Theorem
√
5.9 gives a way to calculate a unit in Z[ d]. Then, as stated in the discussion following the
proof of Theorem 5.3, we can then search through fnitely many cases to determine the
√ √
fundamental unit of in integral closure of Z[ d] in Q( d).
We will now adopt the following notation for the sake of our calculations:
√ √
d ∼ [a1, a2, . . . , a k−1, 2a1] such that the continued fraction expression of d is
1
a1 + ,
a2 + 1 1
a3+1··· +1ak−1+2a1+1
a2+···
28 where the line above a2, . . . , a k−1, 2a1 refers to the cyclic behavior of these values.
We will start with a simple one:
√
Example 5.10. Consider 2 ∼ [1 , 2]. Applying Theorem 5.9, we get p
q
= 1
1
, and
√ √
consequently, (1) 2 − (2)(1) 2 = −1, and so 1 + 2 is a unit in Z[ 2]. In fact, it is the
√
fundamental unit of Z[ 2].
√
Example 5.11. Consider 13 ∼ [3 , 1, 1, 1, 1, 6], and so we have that setting
p118
= 3 + = , and notice that 18 2 − (13)5 2 = 324 − 325 = −1, and so 1q 1+ 1 5
1+ 1+ 1
√ 1 √ √
13 a+b13
u = 18 + 5 13 is a unit in Z[1+
2
]. Finding the norms of w = 2 for integers a, b ∈ Z
such that 1 ≤ | w|≤ |u|, after brute force calculations one fnds that the fundamental unit of
√√
Z[1+ 13 13
2
] is 3+
2
, so Theorem 5.9 does not guarantee a fundamental unit as it did in
Example 5.3.
√ p 1 70
Example 5.12. Consider 29 ∼ [5 , 2, 1, 1, 2, 10 ], and so q = 5 + = 13 , and then 1
2+ 1
1+ 1
2
√ 1+
√
(70) 2 − (29)(13) 2 = 4900 − 4901 = −1, and so u = 70 + 13 29 is a unit in Z[ 29 ]. In fact,
it is the fundamental unit.
Proposition 5.13. If d ≡ 1 ( mod 4) , d ∈ N, then if p
q
is given by Theorem 5.9, then
√ √
p + q d is the fundamental unit of Z[ d].
Proof. Via [7, Section 4.9], p
q
will always give us the minimal integers x > 0 and y > 0 such
that x2 − dy 2 = ±1, and since the integral closure of a quadratic integer ring is always
√ √
Z[ d] whenever d ≡ 1 (mod 4), then whenever d ≡ 1 (mod 4), then p
q
with respect to d
√ √
implies that the fundamental unit of Z[ d] is exactly p + q d.
Remark 5.14 . The wording of minimal is strictly in relation to the word integral. Indeed,
this is the smallest integral solution, but it is not the smallest possible unit u ≥ 1 in
general. When d ≡ 2, 3 (mod 4), generality is lost since we can introduce rational solutions
that are not integral, as seen in Example 5.11.
For actually running this algorithm on a software to compute continued fractions and their
properties, as per [8, Chapter 3, Section 2, Theorem 1], the length of a continued fraction
29 √
of d for any non-square d ∈ N until the point of repetition in the an (or rn) values is
√
O( d log( d)).
30 REFERENCES
Cox, D.A.: Primes of the Form x2 + ny 2: Fermat, Class Field Theory, and Complex Multiplication, 3rd Edition. American Mathematical Society; 2022.
Dummit, D.S., Foote, R.M.: Abstract Algebra, 3rd Edition. Wiley; 2003.
Fraleigh, J.B., Brand, N.E.: A First Course in Abstract Algebra, 8th Edition. Pearson; 2020.
Herstein, I.N.: Topics in Algebra, 2nd Edition. John Wiley & Sons; 1975.
Khinchin, A.Y.: Continued Fractions. Dover Publications; 1997
Marcus, D.A.: Number Fields, 2nd Edition. Springer; 2018.
Olds, C.D.: Continued Fractions. Random House and The L.W. Singer Company; 1963
Rockett, A.M., Sz¨ usz, P.: Continued Fractions. World Scientifc Publishing Co. Pte. Ltd; 1992.
Samuel, P.: About Euclidean Rings. J. Algebra Vol. 19 Issue 2, 282-301; 1971.
31 |
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Fe Molar Mass
Ashley
November 21, 2024 8 min read
The molar mass of iron, denoted by the symbol Fe, is a fundamental constant in chemistry that represents the mass of one mole of iron atoms. To calculate the molar mass of iron, we need to consider the atomic mass of iron, which is the average mass of the naturally occurring isotopes of the element. The atomic mass of iron is approximately 55.847 grams per mole (g/mol), according to the International Union of Pure and Applied Chemistry (IUPAC) standard atomic weights.
Table of Contents
Calculation of Molar Mass
Isotopic Abundances and Molar Mass
Applications of Iron Molar Mass
Chemical Synthesis and Stoichiometry
Calculation of Molar Mass
The molar mass of iron can be calculated by summing the masses of its constituent protons, neutrons, and electrons. However, since the mass of electrons is negligible compared to the mass of protons and neutrons, we can simplify the calculation by considering only the protons and neutrons. The atomic number of iron is 26, which means it has 26 protons. The most abundant isotope of iron, iron-56, has 30 neutrons. Therefore, the mass number of iron-56 is 56 (26 protons + 30 neutrons). The molar mass of iron is then calculated as the average mass of the naturally occurring isotopes, taking into account their relative abundances.
Isotopic Abundances and Molar Mass
The naturally occurring isotopes of iron and their relative abundances are: iron-54 (5.845%), iron-56 (91.754%), iron-57 (2.119%), and iron-58 (0.282%). To calculate the molar mass of iron, we need to weight the masses of these isotopes by their relative abundances. Using the atomic masses of these isotopes (53.9396 g/mol for iron-54, 55.9349 g/mol for iron-56, 56.9354 g/mol for iron-57, and 57.9333 g/mol for iron-58), we can calculate the molar mass of iron as follows:
| Isotope | Atomic Mass (g/mol) | Relative Abundance (%) |
---
| Iron-54 | 53.9396 | 5.845 |
| Iron-56 | 55.9349 | 91.754 |
| Iron-57 | 56.9354 | 2.119 |
| Iron-58 | 57.9333 | 0.282 |
By weighting the atomic masses by their relative abundances and summing the results, we obtain a molar mass of iron that is very close to 55.847 g/mol, which is the accepted value.
💡 The molar mass of iron is a critical constant in chemistry and physics, and its accurate determination is essential for various applications, including chemical synthesis, materials science, and geology. Understanding the isotopic composition of iron and how it affects its molar mass is crucial for precise calculations and measurements in these fields.
Key Points
The molar mass of iron is approximately 55.847 g/mol, which is the average mass of its naturally occurring isotopes.
The atomic mass of iron is calculated by weighting the masses of its isotopes by their relative abundances.
The most abundant isotope of iron is iron-56, which has a mass number of 56 (26 protons + 30 neutrons).
Understanding the isotopic composition of iron is essential for precise calculations and measurements in chemistry, physics, and materials science.
The molar mass of iron is a critical constant in various applications, including chemical synthesis, materials science, and geology.
Applications of Iron Molar Mass
The molar mass of iron has numerous applications in various fields, including chemistry, physics, materials science, and geology. In chemistry, the molar mass of iron is used to calculate the number of moles of iron in a given mass of the element, which is essential for chemical synthesis and stoichiometric calculations. In materials science, the molar mass of iron is used to determine the density of iron alloys and to calculate the amount of iron in a given volume of material. In geology, the molar mass of iron is used to calculate the amount of iron in rocks and minerals, which is essential for understanding the Earth’s composition and evolution.
Chemical Synthesis and Stoichiometry
In chemical synthesis, the molar mass of iron is used to calculate the number of moles of iron required to react with other elements or compounds. This is essential for ensuring that the reactants are present in the correct stoichiometric ratios, which is critical for achieving the desired chemical reaction and product. For example, in the synthesis of iron oxide (Fe2O3), the molar mass of iron is used to calculate the number of moles of iron required to react with oxygen to form the desired product.
The molar mass of iron is also used in stoichiometric calculations to determine the amount of product formed in a chemical reaction. For example, in the reaction between iron and hydrochloric acid (HCl) to form iron(II) chloride (FeCl2) and hydrogen gas (H2), the molar mass of iron is used to calculate the amount of iron(II) chloride formed.
What is the molar mass of iron?
+
The molar mass of iron is approximately 55.847 g/mol, which is the average mass of its naturally occurring isotopes.
How is the molar mass of iron calculated?
+
The molar mass of iron is calculated by weighting the masses of its isotopes by their relative abundances.
What are the applications of iron molar mass?
+
The molar mass of iron has numerous applications in various fields, including chemistry, physics, materials science, and geology.
In conclusion, the molar mass of iron is a fundamental constant in chemistry and physics that has numerous applications in various fields. Understanding the isotopic composition of iron and how it affects its molar mass is crucial for precise calculations and measurements in these fields. The molar mass of iron is used in chemical synthesis, stoichiometric calculations, and materials science to determine the amount of iron in a given mass or volume of material. Its accurate determination is essential for various applications, including chemical synthesis, materials science, and geology.
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10609 | https://artofproblemsolving.com/articles/files/SatoNT.pdf?srsltid=AfmBOorxYW9g3FqreZeAeOpJ31iAf1o2wMSc8LUG1nIa-SN7zxCcxnJN | Number Theory
Naoki Sato sato@artofproblemsolving.com
0 Preface
This set of notes on number theory was originally written in 1995 for students at the IMO level. It covers the basic background material that an IMO student should be familiar with. This text is meant to be a reference, and not a replacement but rather a supplement to a number theory textbook; several are given at the back. Proofs are given when appropriate, or when they illustrate some insight or important idea. The problems are culled from various sources, many from actual contests and olympiads, and in general are very difficult. The author welcomes any corrections or suggestions.
1 Divisibility
For integers a and b, we say that a divides b, or that a is a divisor (or
factor ) of b, or that b is a multiple of a, if there exists an integer c such that b = ca , and we denote this by a | b. Otherwise, a does not divide b, and we denote this by a - b. A positive integer p is a prime if the only divisors of
p are 1 and p. If pk | a and pk+1 - a where p is a prime, i.e. pk is the highest power of p dividing a, then we denote this by pk‖a.Useful Facts
• If a, b > 0, and a | b, then a ≤ b.
• If a | b1, a | b2, . . . , a | bn, then for any integers c1, c2, . . . , cn,
a |
n
∑
i=1
bici.
Theorem 1.1 . The Division Algorithm . For any positive integer a and integer b, there exist unique integers q and r such that b = qa + r and 0 ≤ r < a , with r = 0 iff a | b.1Theorem 1.2 . The Fundamental Theorem of Arithmetic . Every integer greater than 1 can be written uniquely in the form
pe1
1
pe2
2
· · · pek
k
,
where the pi are distinct primes and the ei are positive integers.
Theorem 1.3 . (Euclid) There exist an infinite number of primes.
Proof . Suppose that there are a finite number of primes, say p1, p2, . . . ,
pn. Let N = p1p2 · · · pn + 1. By the fundamental theorem of arithmetic, N
is divisible by some prime p. This prime p must be among the pi, since by assumption these are all the primes, but N is seen not to be divisible by any of the pi, contradiction.
Example 1.1 . Let x and y be integers. Prove that 2 x + 3 y is divisible by 17 iff 9 x + 5 y is divisible by 17.
Solution . 17 | (2 x + 3 y) ⇒ 17 | [13(2 x + 3 y)], or 17 | (26 x + 39 y) ⇒
17 | (9 x + 5 y), and conversely, 17 | (9 x + 5 y) ⇒ 17 | [4(9 x + 5 y)], or 17 | (36 x + 20 y) ⇒ 17 | (2 x + 3 y).
Example 1.2 . Find all positive integers d such that d divides both n2 +1 and ( n + 1) 2 + 1 for some integer n.
Solution . Let d | (n2 + 1) and d | [( n + 1) 2 + 1], or d | (n2 + 2 n + 2). Then d | [( n2 + 2 n + 2) − (n2 + 1)], or d | (2 n + 1) ⇒ d | (4 n2 + 4 n + 1), so
d | [4( n2+2 n+2) −(4 n2+4 n+1)], or d | (4 n+7). Then d | [(4 n+7) −2(2 n+1)], or d | 5, so d can only be 1 or 5. Taking n = 2 shows that both of these values are achieved.
Example 1.3 . Suppose that a1, a2, . . . , a2n are distinct integers such that the equation (x − a1)( x − a2) · · · (x − a2n) − (−1) n(n!) 2 = 0 has an integer solution r. Show that
r = a1 + a2 + · · · + a2n
2n .
(1984 IMO Short List)
Solution . Clearly, r 6 = ai for all i, and the r − ai are 2 n distinct integers, so
|(r − a1)( r − a2) · · · (r − a2n)| ≥ | (1)(2) · · · (n)( −1)( −2) · · · (−n)| = ( n!) 2,
2with equality iff
{r − a1, r − a2, . . . , r − a2n} = {1, 2, . . . , n, −1, −2, . . . , −n}.
Therefore, this must be the case, so (r − a1) + ( r − a2) + · · · + ( r − a2n)= 2 nr − (a1 + a2 + · · · + a2n)= 1 + 2 + · · · + n + ( −1) + ( −2) + · · · + ( −n) = 0
⇒ r = a1 + a2 + · · · + a2n
2n .
Example 1.4 . Let 0 < a 1 < a 2 < · · · < a mn +1 be mn + 1 integers. Prove that you can select either m + 1 of them no one of which divides any other, or n + 1 of them each dividing the following one. (1966 Putnam Mathematical Competition)
Solution . For each i, 1 ≤ i ≤ mn + 1, let ni be the length of the longest sequence starting with ai and each dividing the following one, among the integers ai, ai+1 , . . . , amn +1 . If some ni is greater than n then the problem is solved. Otherwise, by the pigeonhole principle, there are at least m + 1 values of ni that are equal. Then, the integers ai corresponding to these ni
cannot divide each other. Useful Facts
• Bertrand’s Postulate . For every positive integer n, there exists a prime
p such that n ≤ p ≤ 2n.
• Gauss’s Lemma . If a polynomial with integer coefficients factors into two polynomials with rational coefficients, then it factors into two poly-nomials with integer coefficients. Problems 1. Let a and b be positive integers such that a | b2, b2 | a3, a3 | b4, b4 | a5,. . . . Prove that a = b.2. Let a, b, and c denote three distinct integers, and let P denote a poly-nomial having all integral coefficients. Show that it is impossible that
P (a) = b, P (b) = c, and P (c) = a.(1974 USAMO) 33. Show that if a and b are positive integers, then
(
a + 12
)n
+
(
b + 12
)n
is an integer for only finitely many positive integers n.(A Problem Seminar , D.J. Newman) 4. For a positive integer n, let r(n) denote the sum of the remainders when
n is divided by 1, 2, . . . , n respectively. Prove that r(k) = r(k − 1) for infinitely many positive integers k.(1981 K¨ ursch´ ak Competition) 5. Prove that for all positive integers n,0 <
n
∑
k=1
g(k)
k − 2n
3 < 23,
where g(k) denotes the greatest odd divisor of k.(1973 Austrian Mathematics Olympiad) 6. Let d be a positive integer, and let S be the set of all positive integers of the form x2 + dy 2, where x and y are non-negative integers. (a) Prove that if a ∈ S and b ∈ S, then ab ∈ S.(b) Prove that if a ∈ S and p ∈ S, such that p is a prime and p | a,then a/p ∈ S.(c) Assume that the equation x2 + dy 2 = p has a solution in non-negative integers x and y, where p is a given prime. Show that if
d ≥ 2, then the solution is unique, and if d = 1, then there are exactly two solutions.
2 GCD and LCM
The greatest common divisor of two positive integers a and b is the great-est positive integer that divides both a and b, which we denote by gcd( a, b ), and similarly, the lowest common multiple of a and b is the least positive 4integer that is a multiple of both a and b, which we denote by lcm( a, b ). We say that a and b are relatively prime if gcd( a, b ) = 1. For integers a1, a2,. . . , an, gcd( a1, a 2, . . . , a n) is the greatest positive integer that divides all of
a1, a2, . . . , an, and lcm( a1, a 2, . . . , a n) is defined similarly. Useful Facts
• For all a, b, gcd( a, b ) · lcm( a, b ) = ab .
• For all a, b, and m, gcd( ma, mb ) = m gcd( a, b ) and lcm( ma, mb ) =
mlcm( a, b ).
• If d | gcd( a, b ), then gcd
(ad , bd
)
= gcd( a, b )
d .
In particular, if d = gcd( a, b ), then gcd( a/d, b/d ) = 1; that is, a/d and
b/d are relatively prime.
• If a | bc and gcd( a, c ) = 1, then a | b.
• For positive integers a and b, if d is a positive integer such that d | a,
d | b, and for any d′, d′ | a and d′ | b implies that d′ | d, then d =gcd( a, b ). This is merely the assertion that any common divisor of a
and b divides gcd( a, b ).
• If a1a2 · · · an is a perfect kth power and the ai are pairwise relatively prime, then each ai is a perfect kth power.
• Any two consecutive integers are relatively prime.
Example 2.1 . Show that for any positive integer N , there exists a multiple of N that consists only of 1s and 0s. Furthermore, show that if N
is relatively prime to 10, then there exists a multiple that consists only of 1s.
Solution . Consider the N + 1 integers 1, 11, 111, . . . , 111...1 ( N + 1 1s). When divided by N , they leave N + 1 remainders. By the pigeonhole princi-ple, two of these remainders are equal, so the difference in the corresponding integers, an integer of the form 111...000, is divisible by N . If N is relatively prime to 10, then we may divide out all powers of 10, to obtain an integer of the form 111...1 that remains divisible by N .5Theorem 2.1 . For any positive integers a and b, there exist integers x
and y such that ax + by = gcd( a, b ). Furthermore, as x and y vary over all integers, ax + by attains all multiples and only multiples of gcd( a, b ).
Proof . Let S be the set of all integers of the form ax +by , and let d be the least positive element of S. By the division algorithm, there exist integers q
and r such that a = qd + r, 0 ≤ r < d . Then r = a − qd = a − q(ax + by ) = (1 − qx )a − (qy )b, so r is also in S. But r < d , so r = 0 ⇒ d | a, and similarly, d | b, so d | gcd( a, b ). However, gcd( a, b ) divides all elements of S,so in particular gcd( a, b ) | d ⇒ d = gcd( a, b ). The second part of the theorem follows.
Corollary 2.2 . The positive integers a and b are relatively prime iff there exist integers x and y such that ax + by = 1.
Corollary 2.3 . For any positive integers a1, a2, . . . , an, there exist integers x1, x2, . . . , xn, such that a1x1+a2x2+· · · +anxn = gcd( a1, a 2, . . . , a n).
Corollary 2.4 . Let a and b be positive integers, and let n be an integer. Then the equation
ax + by = n
has a solution in integers x and y iff gcd( a, b ) | n. If this is the case, then all solutions are of the form (x, y ) =
(
x0 + t · bd , y 0 − t · ad
)
,
where d = gcd( a, b ), ( x0, y 0) is a specific solution of ax + by = n, and t is an integer.
Proof . The first part follows from Theorem 2.1. For the second part, as stated, let d = gcd( a, b ), and let ( x0, y 0) be a specific solution of ax + by = n,so that ax 0 + by 0 = n. If ax + by = n, then ax + by − ax 0 − by 0 = a(x − x0) +
b(y − y0) = 0, or a(x − x0) = b(y0 − y), and hence (x − x0) · ad = ( y0 − y) · bd.
Since a/d and b/d are relatively prime, b/d must divide x − x0, and a/d must divide y0 − y. Let x − x0 = tb/d and y0 − y = ta/d . This gives the solutions described above. 6Example 2.2 . Prove that the fraction 21 n + 4 14 n + 3 is irreducible for every positive integer n. (1959 IMO)
Solution . For all n, 3(14 n + 3) − 2(21 n + 4) = 1, so the numerator and denominator are relatively prime.
Example 2.3 . For all positive integers n, let Tn = 2 2n
Show that if
m 6 = n, then Tm and Tn are relatively prime.
Solution . We have that
Tn − 2 = 2 2n
− 1 = 2 2n−1·2 − 1= ( Tn−1 − 1) 2 − 1 = T 2
n−1
− 2Tn−1
= Tn−1(Tn−1 − 2) = Tn−1Tn−2(Tn−2 − 2) = · · ·
= Tn−1Tn−2 · · · T1T0(T0 − 2) = Tn−1Tn−2 · · · T1T0,
for all n. Therefore, any common divisor of Tm and Tn must divide 2. But each Tn is odd, so Tm and Tn are relatively prime.
Remark . It immediately follows from this result that there are an infinite number of primes.
The Euclidean Algorithm . By recursive use of the division algorithm, we may find the gcd of two positive integers a and b without factoring either, and the x and y in Theorem 2.1 (and so, a specific solution in Corollary 2.4). For example, for a = 329 and b = 182, we compute 329 = 1 · 182 + 147 ,
182 = 1 · 147 + 35 ,
147 = 4 · 35 + 7 ,
35 = 5 · 7,
and stop when there is no remainder. The last dividend is the gcd, so in our example, gcd(329,182) = 7. Now, working through the above equations 7backwards, 7 = 147 − 4 · 35 = 147 − 4 · (182 − 1 · 147) = 5 · 147 − 4 · 182 = 5 · (329 − 182) − 4 · 182 = 5 · 329 − 9 · 182 .
Remark . The Euclidean algorithm also works for polynomials.
Example 2.4 . Let n be a positive integer, and let S be a subset of n + 1 elements of the set {1, 2, . . . , 2n}. Show that (a) There exist two elements of S that are relatively prime, and (b) There exist two elements of S, one of which divides the other.
Solution . (a) There must be two elements of S that are consecutive, and thus, relatively prime. (b) Consider the greatest odd factor of each of the n + 1 elements in
S. Each is among the n odd integers 1, 3, . . . , 2 n − 1. By the pigeon-hole principle, two must have the same greatest odd factor, so they differ (multiplication-wise) by a power of 2, and so one divides the other.
Example 2.5 . The positive integers a1, a2, . . . , an are such that each is less than 1000, and lcm( ai, a j ) > 1000 for all i, j, i 6 = j. Show that
n
∑
i=1
1
ai
< 2.
(1951 Russian Mathematics Olympiad)
Solution . If 1000
m+1
< a ≤ 1000
m
, then the m multiples a, 2 a, . . . , ma do not exceed 1000. Let k1 the number of ai in the interval ( 1000 2 , 1000], k2 in (1000 3 , 1000 2 ], etc. Then there are k1 + 2 k2 + 3 k3 + · · · integers, no greater than 1000, that are multiples of at least one of the ai. But the multiples are distinct, so
k1 + 2 k2 + 3 k3 + · · · < 1000
⇒ 2k1 + 3 k2 + 4 k3 + · · · = ( k1 + 2 k2 + 3 k3 + · · · ) + ( k1 + k2 + k3 + · · · )
< 1000 + n< 2000 .
8Therefore,
n
∑
i=1
1
ai
≤ k1
21000 + k2
31000 + k3
41000 + · · ·
= 2k1 + 3 k2 + 4 k3 + · · ·
1000
< 2.
Note: It can be shown that n ≤ 500 as follows: Consider the greatest odd divisor of a1, a2, . . . , a1000 . Each must be distinct; otherwise, two differ, multiplication-wise, by a power of 2, which means one divides the other, contradiction. Also, there are only 500 odd numbers between 1 and 1000, from which the result follows. It also then follows that
n
∑
i=1
1
ai
< 32.
Useful Facts
• Dirichlet’s Theorem . If a and b are relatively prime positive integers, then the arithmetic sequence a, a + b, a + 2 b, . . . , contains infinitely many primes. Problems 1. The symbols ( a, b, . . . , g ) and [ a, b, . . . , g ] denote the greatest common divisor and lowest common multiple, respectively of the positive inte-gers a, b, . . . , g. Prove that [a, b, c ]2
[a, b ][ a, c ][ b, c ] = (a, b, c )2
(a, b )( a, c )( b, c ).
(1972 USAMO) 2. Show that gcd( am − 1, a n − 1) = agcd( m,n ) − 1 for all positive integers
a > 1, m, n.93. Let a, b, and c be positive integers. Show that lcm( a, b, c ) = abc · gcd( a, b, c )gcd( a, b ) · gcd( a, c ) · gcd( b, c ).
Express gcd( a, b, c ) in terms of abc , lcm( a, b, c ), lcm( a, b ), lcm( a, c ), and lcm( b, c ). Generalize. 4. Let a, b be odd positive integers. Define the sequence ( fn) by putting
f1 = a, f2 = b, and by letting fn for n ≥ 3 be the greatest odd divisor of fn−1 + fn−2. Show that fn is constant for n sufficiently large and determine the eventual value as a function of a and b.(1993 USAMO) 5. Let n ≥ a1 > a 2 > · · · > a k be positive integers such that lcm( ai, a j ) ≤
n for all i, j. Prove that ia i ≤ n for i = 1, 2, . . . , k.
3 Arithmetic Functions
There are several important arithmetic functions, of which three are pre-sented here. If the prime factorization of n > 1 is pe1
1
pe2
2
· · · pek
k
, then the number of positive integers less than n, relatively prime to n, is
φ(n) =
(
1 − 1
p1
) (
1 − 1
p2
)
· · ·
(
1 − 1
pk
)
n
= pe1−11 pe2−12 · · · pek −1
k
(p1 − 1)( p2 − 1) · · · (pk − 1) ,
the number of divisors of n is
τ (n) = ( e1 + 1)( e2 + 1) · · · (ek + 1) ,
and the sum of the divisors of n is
σ(n) = ( pe1
1
pe1−11 + · · · + 1)( pe2
2
pe2−12 + · · · + 1)
· · · (pek
k
pek −1
k
· · · + 1) =
(pe1+1 1 − 1
p1 − 1
) ( pe2+1 2 − 1
p2 − 1
)
· · ·
(pek +1
k
− 1
pk − 1
)
.
Also, φ(1), τ (1), and σ(1) are defined to be 1. We say that a function
f is multiplicative if f (mn ) = f (m)f (n) for all relatively prime positive 10 integers m and n, and f (1) = 1 (otherwise, f (1) = 0, which implies that
f (n) = 0 for all n).
Theorem 3.1 . The functions φ, τ , and σ are multiplicative. Hence, by taking the prime factorization and evaluating at each prime power, the formula above are found easily.
Example 3.1 . Find the number of solutions in ordered pairs of positive integers ( x, y ) of the equation 1
x + 1
y = 1
n,
where n is a positive integer.
Solution . From the given, 1
x + 1
y = 1
n ⇔ xy = nx + ny ⇔ (x − n)( y − n) = n2.
If n = 1, then we immediately deduce the unique solution (2,2). For
n ≥ 2, let n = pe1
1
pe2
2
· · · pek
k
be the prime factorization of n. Since x, y > n ,there is a 1-1 correspondence between the solutions in ( x, y ) and the factors of n2, so the number of solutions is
τ (n2) = (2 e1 + 1)(2 e2 + 1) · · · (2 ek + 1) .
Example 3.2 . Let n be a positive integer. Prove that
∑
d|n
φ(d) = n.
Solution . For a divisor d of n, let Sd be the set of all a, 1 ≤ a ≤ n, such that gcd( a, n ) = n/d . Then Sd consists of all elements of the form b · n/d ,where 0 ≤ b ≤ d, and gcd( b, d ) = 1, so Sd contains φ(d) elements. Also, it is clear that each integer between 1 and n belongs to a unique Sd. The result then follows from summing over all divisors d of n.Problems 1. Let n be a positive integer. Prove that
n
∑
k=1
τ (k) =
n
∑
k=1
⌊nk
⌋
.
11 2. Let n be a positive integer. Prove that
∑
d|n
τ 3(d) =
∑
d|n
τ (d)
2
.
Prove that if σ(N ) = 2 N + 1, then N is the square of an odd integer. (1976 Putnam Mathematical Competition)
4 Modular Arithmetic
For a positive integer m and integers a and b, we say that a is congruent to
b modulo m if m | (a − b), and we denote this by a ≡ b modulo m, or more commonly a ≡ b (mod m). Otherwise, a is not congruent to b modulo m,and we denote this by a 6 ≡ b (mod m) (although this notation is not used often). In the above notation, m is called the modulus , and we consider the integers modulo m.
Theorem 4.1 . If a ≡ b and c ≡ d (mod m), then a + c ≡ b + d (mod m)and ac ≡ bd (mod m).
Proof . If a ≡ b and c ≡ d (mod m), then there exist integers k and l
such that a = b + km and c = d + lm . Hence, a + c = b + d + ( k + l)m, so
a + c ≡ b + d (mod m). Also,
ac = bd + dkm + blm + klm 2
= bd + ( dk + bl + klm )m,
so ac ≡ bd (mod m). Useful Facts
• For all integers n,
n2 ≡
{ 01
}
(mod 4)
{ if n is even ,
if n is odd .
• For all integers n,
n2 ≡
041
(mod 8)
if n ≡ 0 (mod 4) ,
if n ≡ 2 (mod 4) ,
if n ≡ 1 (mod 2) .
12 • If f is a polynomial with integer coefficients and a ≡ b (mod m), then
f (a) ≡ f (b) (mod m).
• If f is a polynomial with integer coefficients of degree n, not identically zero, and p is a prime, then the congruence
f (x) ≡ 0 (mod p)has at most n solutions modulo p, counting multiplicity.
Example 4.1 . Prove that the only solution in rational numbers of the equation
x3 + 3 y3 + 9 z3 − 9xyz = 0 is x = y = z = 0. (1983 K¨ ursch´ ak Competition)
Solution . Suppose that the equation has a solution in rationals, with at least one non-zero variable. Since the equation is homogeneous, we may obtain a solution in integers ( x0, y 0, z 0) by multiplying the equation by the cube of the lowest common multiple of the denominators. Taking the equa-tion modulo 3, we obtain x30 ≡ 0 (mod 3). Therefore, x0 must be divisible by 3, say x0 = 3 x1. Substituting, 27 x31 + 3 y30 + 9 z30 − 27 x1y0z0 = 0
⇒ y30 + 3 z30 + 9 x31 − 9x1y0z0 = 0 .
Therefore, another solution is ( y0, z 0, x 1). We may then apply this reduction recursively, to obtain y0 = 3 y1, z0 = 3 z1, and another solution ( x1, y 1, z 1). Hence, we may divide powers of 3 out of our integer solution an arbitrary number of times, contradiction.
Example 4.2 . Does one of the first 10 8 + 1 Fibonacci numbers terminate with 4 zeroes?
Solution . The answer is yes. Consider the sequence of pairs ( Fk, F k+1 )modulo 10 4. Since there are only a finite number of different possible pairs (10 8 to be exact), and each pair is dependent only on the previous one, this sequence is eventually periodic. Also, by the Fibonacci relation, one can find the previous pair to a given pair, so this sequence is immediately periodic. But F0 ≡ 0 (mod 10 4), so within 10 8 terms, another Fibonacci number divisible by 10 4 must appear. 13 In fact, a computer check shows that 10 4 | F7500 , and ( Fn) modulo 10 4
has period 15000, which is much smaller than the upper bound of 10 8.If ax ≡ 1 (mod m), then we say that x is the inverse of a modulo m,denoted by a−1, and it is unique modulo m.
Theorem 4.2 . The inverse of a modulo m exists and is unique iff a is relatively prime to m.
Proof . If ax ≡ 1 (mod m), then ax = 1+ km for some k ⇒ ax −km = 1. By Corollary 2.2, a and m are relatively prime. Now, if gcd( a, m ) = 1, then by Corollary 2.2, there exist integers x and y such that ax + my = 1 ⇒ ax =1 − my ⇒ ax ≡ 1 (mod m). The inverse x is unique modulo m, since if x′ is also an inverse, then ax ≡ ax ′ ≡ 1 ⇒ xax ≡ xax ′ ≡ x ≡ x′.
Corollary 4.3 . If p is a prime, then the inverse of a modulo p exists and is unique iff p does not divide a.
Corollary 4.4 . If ak ≡ bk (mod m) and k is relatively prime to m, then
a ≡ b (mod m).
Proof . Multiplying both sides by k−1, which exists by Theorem 4.2, yields the result. We say that a set {a1, a 2, . . . , a m} is a complete residue system modulo
m if for all i, 0 ≤ i ≤ m−1, there exists a unique j such that aj ≡ i (mod m).
Example 4.3 . Find all positive integers n such that there exist complete residue systems {a1, a 2, . . . , a n} and {b1, b 2, . . . , b n} modulo n for which {a1 +
b1, a 2 + b2, . . . , a n + bn} is also a complete residue system.
Solution . The answer is all odd n. First we prove necessity. For any complete residue system {a1, a 2, . . . , a n} modulo n, we have that
a1 + a2 + · · · + an ≡ n(n + 1) /2 (mod n). So, if all three sets are complete residue systems, then a1 +a2 +· · · +an +b1 +b2 +· · · +bn ≡ n2 +n ≡ 0 (mod n)and a1 + b1 + a2 + b2 + · · · + an + bn ≡ n(n + 1) /2 (mod n), so n(n + 1) /2 ≡ 0(mod n). The quantity n(n + 1) /2 is divisible by n iff ( n + 1) /2 is an integer, which implies that n is odd. Now assume that n is odd. Let ai = bi = i for all i. Then ai + bi = 2 i
for all i, and n is relatively prime to 2, so by Corollary 4.4, {2, 4, . . . , 2n} is a complete residue system modulo n.
Theorem 4.5 . Euler’s Theorem . If a is relatively prime to m, then
aφ(m) ≡ 1 (mod m). 14 Proof . Let a1, a2, . . . , aφ(m) be the positive integers less than m that are relatively prime to m. Consider the integers aa 1, aa 2, . . . , aa φ(m). We claim that they are a permutation of the original φ(m) integers ai, modulo
m. For each i, aa i is also relatively prime to m, so aa i ≡ ak for some k. Since
aa i ≡ aa j ⇔ ai ≡ aj (mod m), each ai gets taken to a different ak under multiplication by a, so indeed they are permuted. Hence,
a1a2 · · · aφ(m) ≡ (aa 1)( aa 2) · · · (aa φ(m))
≡ aφ(m)a1a2 · · · aφ(m)
⇒ 1 ≡ aφ(m) (mod m).
Remark . This gives an explicit formula for the inverse of a modulo m:
a−1 ≡ aφ(m)−2 (mod m). Alternatively, one can use the Euclidean algorithm to find a−1 ≡ x as in the proof of Theorem 4.2.
Corollary 4.6 . Fermat’s Little Theorem (FLT) . If p is a prime, and p
does not divide a, then ap−1 ≡ 1 (mod p).
Example 4.4 . Show that if a and b are relatively prime positive integers, then there exist integers m and n such that am + bn ≡ 1 (mod ab ).
Solution . Let S = am + bn, where m = φ(b) and n = φ(a). Then by Euler’s Theorem, S ≡ bφ(a) ≡ 1 (mod a), or S − 1 ≡ 0 (mod a), and
S ≡ aφ(b) ≡ 1 (mod b), or S − 1 ≡ 0 (mod b). Therefore, S − 1 ≡ 0, or S ≡ 1(mod ab ).
Example 4.5 . For all positive integers i, let Si be the sum of the products of 1, 2, . . . , p − 1 taken i at a time, where p is an odd prime. Show that
S1 ≡ S2 ≡ · · · ≡ Sp−2 ≡ 0 (mod p).
Solution . First, observe that (x − 1)( x − 2) · · · (x − (p − 1)) = xp−1 − S1xp−2 + S2xp−3 − · · · − Sp−2x + Sp−1.
This polynomial vanishes for x = 1, 2, . . . , p − 1. But by Fermat’s Little Theorem, so does xp−1 − 1 modulo p. Taking the difference of these two polynomials, we obtain another polynomial of degree p − 2 with p − 1 roots modulo p, so it must be the zero polynomial, and the result follows from comparing coefficients. 15 Remark . We immediately have that ( p − 1)! ≡ Sp−1 ≡ − 1 (mod p), which is Wilson’s Theorem. Also, xp − x ≡ 0 (mod p) for all x, yet we cannot compare coefficients here. Why not?
Theorem 4.7 . If p is a prime and n is an integer such that p | (4 n2 + 1), then p ≡ 1 (mod 4).
Proof . Clearly, p cannot be 2, so we need only show that p 6 ≡ 3 (mod 4). Suppose p = 4 k + 3 for some k. Let y = 2 n, so by Fermat’s Little Theorem,
yp−1 ≡ 1 (mod p), since p does not divide n. But, y2 + 1 ≡ 0, so
yp−1 ≡ y4k+2 ≡ (y2)2k+1 ≡ (−1) 2k+1 ≡ − 1 (mod p),
contradiction. Therefore, p ≡ 1 (mod 4).
Remark . The same proof can be used to show that if p is a prime and
p | (n2 + 1), then p = 2 or p ≡ 1 (mod 4).
Example 4.6 . Show that there are an infinite number of primes of the form 4 k + 1 and of the form 4 k + 3.
Solution . Suppose that there are a finite number of primes of the form 4k + 1, say p1, p2, . . . , pn. Let N = 4( p1p2 · · · pn)2 + 1. By Theorem 4.7, N
is only divisible by primes of the form 4 k + 1, but clearly N is not divisible by any of these primes, contradiction. Similarly, suppose that there are a finite number of primes of the form 4k + 3, say q1, q2, . . . , qm. Let M = 4 q1q2 · · · qm − 1. Then M ≡ 3 (mod 4), so M must be divisible by a prime of the form 4 k + 3, but M is not divisible by any of these primes, contradiction.
Example 4.7 . Show that if n is an integer greater than 1, then n does not divide 2 n − 1.
Solution . Let p be the least prime divisor of n. Then gcd( n, p − 1) = 1, and by Corollary 2.2, there exist integers x and y such that nx +( p−1) y = 1. If p | (2 n − 1), then 2 ≡ 2nx +( p−1) y ≡ (2 n)x(2 p−1)y ≡ 1 (mod p) by Fermat’s Little Theorem, contradiction. Therefore, p - (2 n − 1) ⇒ n - (2 n − 1).
Theorem 4.8 . Wilson’s Theorem . If p is a prime, then ( p − 1)! ≡ − 1(mod p). (See also Example 4.5.)
Proof . Consider the congruence x2 ≡ 1 (mod p). Then x2 − 1 ≡ (x −
1)( x + 1) ≡ 0, so the only solutions are x ≡ 1 and −1. Therefore, for each i,2 ≤ i ≤ p − 2, there exists a unique inverse j 6 = i of i, 2 ≤ j ≤ p − 2, modulo 16 p. Hence, when we group in pairs of inverses, (p − 1)! ≡ 1 · 2 · · · (p − 2) · (p − 1)
≡ 1 · 1 · · · 1 · (p − 1)
≡ − 1 (mod p).
Example 4.8 . Let {a1, a 2, . . . , a 101 } and {b1, b 2, . . . , b 101 } be complete residue systems modulo 101. Can {a1b1, a 2b2, . . . , a 101 b101 } be a complete residue system modulo 101?
Solution . The answer is no. Suppose that {a1b1, a 2b2, . . . , a 101 b101 } is a complete residue system modulo 101. Without loss of generality, assume that a101 ≡ 0 (mod 101). Then b101 ≡ 0 (mod 101), because if any other
bj was congruent to 0 modulo 101, then aj bj ≡ a101 b101 ≡ 0 (mod 101), contradiction. By Wilson’s Theorem, a1a2 · · · a100 ≡ b1b2 · · · b100 ≡ 100! ≡−1 (mod 101), so a1b1a2b2 · · · a100 b100 ≡ 1 (mod 101). But a101 b101 ≡ 0(mod 101), so a1b1a2b2 · · · a100 b100 ≡ 100! ≡ − 1 (mod 101), contradiction.
Theorem 4.9 . If p is a prime, then the congruence x2 + 1 ≡ 0 (mod p)has a solution iff p = 2 or p ≡ 1 (mod 4). (Compare to Theorem 7.1)
Proof . If p = 2, then x = 1 is a solution. If p ≡ 3 (mod 4), then by the remark to Theorem 4.7, no solutions exist. Finally, if p = 4 k + 1, then let
x = 1 · 2 · · · (2 k). Then
x2 ≡ 1 · 2 · · · (2 k) · (2 k) · · · 2 · 1
≡ 1 · 2 · · · (2 k) · (−2k) · · · (−2) · (−1) (multiplying by 2 k −1s)
≡ 1 · 2 · · · (2 k) · (p − 2k) · · · (p − 2) · (p − 1)
≡ (p − 1)! ≡ − 1 (mod p).
Theorem 4.10 . Let p be a prime such that p ≡ 1 (mod 4). Then there exist positive integers x and y such that p = x2 + y2.
Proof . By Theorem 4.9, there exists an integer a such that a2 ≡ − 1(mod p). Consider the set of integers of the form ax − y, where x and y
are integers, 0 ≤ x, y < √p. The number of possible pairs ( x, y ) is then (b√pc + 1) 2 > (√p)2 = p, so by pigeonhole principle, there exist integers 0 ≤ x1, x 2, y 1, y 2 < √p, such that ax 1−y1 ≡ ax 2−y2 (mod p). Let x = x1−x2
and y = y1 − y2. At least one of x and y is non-zero, and ax ≡ y ⇒ a2x2 ≡
17 −x2 ≡ y2 ⇒ x2 + y2 ≡ 0 (mod p). Thus, x2 + y2 is a multiple of p, and 0 < x 2 + y2 < (√p)2 + ( √p)2 = 2 p, so x2 + y2 = p.
Theorem 4.11 . Let n be a positive integer. Then there exist integers
x and y such that n = x2 + y2 iff each prime factor of n of the form 4 k + 3 appears an even number of times.
Theorem 4.12 . The Chinese Remainder Theorem (CRT) . If a1, a2, . . . ,
ak are integers, and m1, m2, . . . , mk are pairwise relatively prime integers, then the system of congruences
x ≡ a1 (mod m1),x ≡ a2 (mod m2),
...
x ≡ ak (mod mk)has a unique solution modulo m1m2 · · · mk.
Proof . Let m = m1m2 · · · mk, and consider m/m 1. This is relatively prime to m1, so there exists an integer t1 such that t1 · m/m 1 ≡ 1 (mod m1). Accordingly, let s1 = t1 · m/m 1. Then s1 ≡ 1 (mod m1) and s1 ≡ 0(mod mj ), j 6 = 1. Similarly, for all i, there exists an si such that si ≡ 1(mod mi) and si ≡ 0 (mod mj ), j 6 = i. Then, x = a1s1 + a2s2 + · · · + aksk is a solution to the above system. To see uniqueness, let x′ be another solution. Then x − x′ ≡ 0 (mod mi) for all i ⇒ x − x′ ≡ 0 (mod m1m2 · · · mk).
Remark . The proof shows explicitly how to find the solution x.
Example 4.9 . For a positive integer n, find the number of solutions of the congruence x2 ≡ 1 (mod n).
Solution . Let the prime factorization of n be 2 epe1
1
pe2
2
· · · pek
k
. By CRT,
x2 ≡ 1 (mod n) ⇔ x2 ≡ 1 (mod pei
i
) for all i, and x2 ≡ 1 (mod 2 e). We consider these cases separately. We have that x2 ≡ 1 (mod pei
i
) ⇔ x2 − 1 = ( x − 1)( x + 1) ≡ 0 (mod pei
i
). But pi cannot divide both x − 1 and x + 1, so it divides one of them; that is,
x ≡ ± 1 (mod pei
i
). Hence, there are two solutions. Now, if ( x − 1)( x + 1) ≡ 0 (mod 2 e), 2 can divide both x − 1 and x + 1, but 4 cannot divide both. For e = 1 and e = 2, it is easily checked that there are 1 and 2 solutions respectively. For e ≥ 3, since there is at most one factor 18 of 2 in one of x − 1 and x + 1, there must be at least e − 1 in the other, for their product to be divisible by 2 e. Hence, the only possibilities are x − 1 or
x + 1 ≡ 0, 2 e−1 (mod 2 e), which lead to the four solutions x ≡ 1, 2 e−1 − 1, 2e−1 + 1, and 2 e − 1. Now that we know how many solutions each prime power factor con-tributes, the number of solutions modulo n is simply the product of these, by CRT. The following table gives the answer:
e Number of solutions 0, 1 2k
2 2k+1
≥ 3 2k+2
Theorem 4.11 . Let m be a positive integer, let a and b be integers, and let k = gcd( a, m ). Then the congruence ax ≡ b (mod m) has k solutions or no solutions according as k | b or k - b.Problems 1. Prove that for each positive integer n there exist n consecutive positive integers, none of which is an integral power of a prime. (1989 IMO) 2. For an odd positive integer n > 1, let S be the set of integers x,1 ≤ x ≤ n, such that both x and x + 1 are relatively prime to n. Show that ∏
x∈S
x ≡ 1 (mod n).
Find all positive integer solutions to 3 x + 4 y = 5 z .(1991 IMO Short List) 4. Let n be a positive integer such that n + 1 is divisible by 24. Prove that the sum of all the divisors of n is divisible by 24. (1969 Putnam Mathematical Competition) 5. (Wolstenholme’s Theorem) Prove that if 1 + 12 + 13 + · · · + 1
p − 119 is expressed as a fraction, where p ≥ 5 is a prime, then p2 divides the numerator. 6. Let a be the greatest positive root of the equation x3 − 3x2 + 1 = 0. Show that ba1788 c and ba1988 c are both divisible by 17. (1988 IMO Short List) 7. Let {a1, a 2, . . . , a n} and {b1, b 2, . . . , b n} be complete residue systems modulo n, such that {a1b1, a 2b2, . . . , a nbn} is also a complete residue system modulo n. Show that n = 1 or 2. 8. Let m, n be positive integers. Show that 4 mn − m − n can never be a square. (1984 IMO Proposal)
5 Binomial Coefficients
For non-negative integers n and k, k ≤ n, the binomial coefficient (nk
) is defined as n!
k!( n − k)! ,
and has several important properties. By convention, (nk
) = 0 if k > n .In the following results, for polynomials f and g with integer coefficients, we say that f ≡ g (mod m) if m divides every coefficient in f − g.
Theorem 5.1 . If p is a prime, then the number of factors of p in n! is
⌊np
⌋
+
⌊ np2
⌋
+
⌊ np3
⌋
· · · .
It is also n − sn
p − 1 ,
where sn is the sum of the digits of n when expressed in base p.
Theorem 5.2 . If p is a prime, then
(pi
)
≡ 0 (mod p)20 for 1 ≤ i ≤ p − 1.
Corollary 5.3 . (1 + x)p ≡ 1 + xp (mod p).
Lemma 5.4 . For all real numbers x and y, bx + yc ≥ b xc + byc.
Proof . x ≥ b xc ⇒ x + y ≥ b xc + byc ∈ Z, so bx + yc ≥ b xc + byc.
Theorem 5.5 . If p is a prime, then
(pk
i
)
≡ 0 (mod p)for 1 ≤ i ≤ pk − 1.
Proof . By Lemma 5.4,
k
∑
j=1
(⌊ ipj
⌋
+
⌊pk − ipj
⌋)
≤
k
∑
j=1
⌊pk
pj
⌋
,
where the LHS and RHS are the number of factors of p in i!( pk − i)! and
pk! respectively. But,
⌊ ipk
⌋
=
⌊pk −ipk
⌋
= 0 and
⌊pk
pk
⌋
= 1, so the inequality is strict, and at least one factor of p divides (pk
i
).
Corollary 5.6 . (1 + x)pk
≡ 1 + xpk
(mod p).
Example 5.1 . Let n be a positive integer. Show that the product of n
consecutive positive integers is divisible by n!.
Solution . If the consecutive integers are m, m + 1, . . . , m + n − 1, then
m(m + 1) · · · (m + n − 1)
n! =
(m + n − 1
n
)
.
Example 5.2 . Let n be a positive integer. Show that (n + 1) lcm
(( n
0
)
,
(n
1
)
, . . . ,
(nn
))
= lcm(1 , 2, . . . , n + 1) .
(AMM E2686)
Solution . Let p be a prime ≤ n + 1 and let α (respectively β) be the highest power of p in the LHS (respectively RHS) of the above equality. Choose r so that pr ≤ n + 1 < p r+1 . Then clearly β = r. We claim that if pr ≤ m < p r+1 , then pr+1 -
(mk
)
for 0 ≤ k ≤ m. (∗)21 Indeed, the number of factors of p in (mk
) is
γ =
r
∑
s=1
(⌊ mps
⌋
−
⌊ kps
⌋
−
⌊m − kps
⌋)
.
Since each summand in this sum is 0 or 1, we have γ ≤ r; that is, () holds. For 0 ≤ k ≤ n, let
ak = ( n + 1)
(nk
)
= ( n − k + 1)
(n + 1
k
)
= ( k + 1)
(n + 1
k + 1
)
.
By ( ∗), pr+1 does not divide any of the integers (nk
), (n+1
k
), or (n+1
k+1
). Thus,
pr+1 can divide ak only if p divides each of the integers n + 1, n − k + 1, and
k + 1. This implies that p divides ( n + 1) − (n − k + 1) − (k + 1) = −1, contradiction. Therefore, pr+1 - ak. On the other hand, for k = pr − 1, we have that k ≤ n and ak = ( k + 1) (n+1
k+1
) is divisible by pr. Therefore,
β = r = α.
Theorem 5.7 . Lucas’s Theorem . Let m and n be non-negative integers, and p a prime. Let
m = mkpk + mk−1pk−1 + · · · + m1p + m0, and
n = nkpk + nk−1pk−1 + · · · + n1p + n0
be the base p expansions of m and n respectively. Then
(mn
)
≡
(mk
nk
)( mk−1
nk−1
)
· · ·
(m1
n1
)( m0
n0
)
(mod p).
Proof . By Corollary 5.6, (1 + x)m ≡ (1 + x)mk pk +mk−1pk−1+··· +m1p+m0
≡ (1 + x)pk mk (1 + x)pk−1mk−1 · · · (1 + x)pm 1 (1 + x)m0
≡ (1 + xpk
)mk (1 + xpk−1
)mk−1 · · · (1 + xp)m1 (1 + x)m0 (mod p).
By base p expansion, the coefficient of xn on both sides is
(mn
)
≡
(mk
nk
)( mk−1
nk−1
)
· · ·
(m1
n1
)( m0
n0
)
(mod p).
22 Corollary 5.8 . Let n be a positive integer. Let A(n) denote the number of factors of 2 in n!, and let B(n) denote the number of 1s in the binary expansion of n. Then the number of odd entries in the nth row of Pascal’s Triangle, or equivalently the number of odd coefficients in the expansion of (1 + x)n, is 2 B(n). Furthermore, A(n) + B(n) = n for all n.Useful Facts
• For a polynomial f with integer coefficients and prime p,[f (x)] pn
≡ f (xpn
) (mod p).
Problems 1. Let a and b be non-negative integers, and p a prime. Show that
(pa pb
)
≡
(ab
)
(mod p).
Let an be the last non-zero digit in the decimal representation of the number n!. Is the sequence a1, a2, a3, . . . eventually periodic? (1991 IMO Short List) 3. Find all positive integers n such that 2 n | (3 n − 1). 4. Find the greatest integer k for which 1991 k divides 1990 1991 1992
1992 1991 1990
.
(1991 IMO Short List) 5. For a positive integer n, let a(n) and b(n) denote the number of binomial coefficients in the nth row of Pascal’s Triangle that are congruent to 1 and 2 modulo 3 respectively. Prove that a(n) − b(n) is always a power of 2. 6. Let n be a positive integer. Prove that if the number of factors of 2 in
n! is n − 1, then n is a power of 2. 23 7. For a positive integer n, let
Cn = 1
n + 1
(2nn
)
,
and Sn = C1 + C2 + · · · + Cn.Prove that Sn ≡ 1 (mod 3) if and only if there exists a 2 in the base 3 expansion of n + 1.
6 Order of an Element
We know that if a is relatively prime to m, then there exists a positive integer
n such that an ≡ 1 (mod m). Let d be the smallest such n. Then we say that d is the order of a modulo m, denoted by ord m(a), or simply ord( a) if the modulus m is understood.
Theorem 6.1 . If a is relatively prime to m, then an ≡ 1 (mod m) iff ord( a) | n. Furthermore, an0 ≡ an1 iff ord( a) | (n0 − n1).
Proof . Let d = ord( a). It is clear that d | n ⇒ an ≡ 1 (mod m). On the other hand, if an ≡ 1 (mod m), then by the division algorithm, there exist integers q and r such that n = qd + r, 0 ≤ r < d . Then an ≡ aqd +r ≡
(ad)qar ≡ ar ≡ 1 (mod m). But r < d , so r = 0 ⇒ d | n. The second part of the theorem follows.
Remark . In particular, by Euler’s Theorem, ord( a) | φ(m).
Example 6.1 . Show that the order of 2 modulo 101 is 100.
Solution . Let d = ord(2). Then d | φ(101), or d | 100. If d < 100, then d divides 100/2 or 100/5; that is, d is missing at least one prime factor. However, 250 ≡ 1024 5 ≡ 14 5 ≡ 196 · 196 · 14 ≡ (−6) · (−6) · 14 ≡ − 1 (mod 101) ,
and 220 ≡ 1024 2 ≡ 14 2 ≡ − 6 (mod 101) ,
so d = 100.
Example 6.2 . Prove that if p is a prime, then every prime divisor of 2p − 1 is greater than p.24 Solution . Let q | (2 p − 1), where q is a prime. Then 2 p ≡ 1 (mod q), so ord(2) | p. But ord(2) 6 = 1, so ord(2) = p. And by Fermat’s Little Theorem, ord(2) | (q − 1) ⇒ p ≤ q − 1 ⇒ q > p .In fact, for p > 2, q must be of the form 2 kp + 1. From the above, ord(2) | (q − 1), or p | (q − 1) ⇒ q = mp + 1. Since q must be odd, m must be even.
Example 6.3 . Let p be a prime that is relatively prime to 10, and let n
be an integer, 0 < n < p . Let d be the order of 10 modulo p.(a) Show that the length of the period of the decimal expansion of n/p is
d.(b) Prove that if d is even, then the period of the decimal expansion of n/p
can be divided into two halves, whose sum is 10 d/ 2 − 1. For example, 1/7 = 0 .142857, so d = 6, and 142 + 857 = 999 = 10 3 − 1.
Solution . (a) Let m be the length of the period, and let n/p =0.a 1a2 . . . a m. Then 10 mnp = a1a2 . . . a m.a 1a2 . . . a m
⇒ (10 m − 1) np = a1a2 . . . a m,
an integer. Since n and p are relatively prime, p must divide 10 m − 1, so d
divides m. Conversely, p divides 10 d −1, so (10 d −1) n/p is an integer, with at most d digits. If we divide this integer by 10 d − 1, then we obtain a rational number, whose decimal expansion has period at most d. Therefore, m = d.(b) Let d = 2 k, so n/p = 0 .a 1a2 . . . a kak+1 . . . a 2k. Now p divides 10 d −1 = 10 2k − 1 = (10 k − 1)(10 k + 1). However, p cannot divide 10 k − 1 (since the order of 10 is 2 k), so p divides 10 k + 1. Hence, 10 knp = a1a2 . . . a k.a k+1 . . . a 2k
⇒ (10 k + 1) np = a1a2 . . . a k + 0 .a 1a2 . . . a k + 0 .a k+1 . . . a 2k
is an integer. This can occur iff a1a2 . . . a k +ak+1 . . . a 2k is a number consisting only of 9s, and hence, equal to 10 k − 1. Problems 25 1. Prove that for all positive integers a > 1 and n, n | φ(an − 1). 2. Prove that if p is a prime, then pp−1 has a prime factor that is congruent to 1 modulo p.3. For any integer a, set na = 101 a−100 ·2a. Show that for 0 ≤ a, b, c, d ≤
99 , n a + nb ≡ nc + nd (mod 10100) implies {a, b } = {c, d }.(1994 Putnam Mathematical Competition) 4. Show that if 3 ≤ d ≤ 2n+1 , then d - (a2n
1) for all positive integers a.
7 Quadratic Residues
Let m be an integer greater than 1, and a an integer relatively prime to m. If
x2 ≡ a (mod m) has a solution, then we say that a is a quadratic residue
of m. Otherwise, we say that a is a quadratic non-residue . Now, let p be an odd prime. Then the Legendre symbol
(ap
)
is assigned the value of 1 if a is a quadratic residue of p. Otherwise, it is assigned the value of −1.
Theorem 7.1 . Let p be an odd prime, and a and b be integers relatively prime to p. Then (a)
(ap
)
≡ a(p−1) /2 (mod p), and (b)
(ap
)( bp
)
=
(ab p
)
.
Proof . If the congruence x2 ≡ a (mod p) has a solution, then a(p−1) /2 ≡
xp−1 ≡ 1 (mod p), by Fermat’s Little Theorem. If the congruence x2 ≡ a
(mod p) has no solutions, then for each i, 1 ≤ i ≤ p − 1, there is a unique
j 6 = i, 1 ≤ j ≤ p − 1, such that ij ≡ a. Therefore, all the integers from 1 to
p − 1 can be arranged into ( p − 1) /2 such pairs. Taking their product,
a(p−1) /2 ≡ 1 · 2 · · · (p − 1) ≡ (p − 1)! ≡ − 1 (mod p),
26 by Wilson’s Theorem. Part (b) now follows from part (a).
Remark . Part (a) is known as Euler’s criterion.
Example 7.1 . Show that if p is an odd prime, then
(1
p
)
+
(2
p
)
· · · +
(p − 1
p
)
= 0 .
Solution . Note that 1 2, 2 2, . . . , (( p − 1) /2) 2 are distinct modulo p,and that (( p + 1) /2) 2, . . . , ( p − 1) 2 represent the same residues, simply in reverse. Hence, there are exactly ( p−1) /2 quadratic residues, leaving ( p−1) /2quadratic non-residues. Therefore, the given sum contains ( p − 1) /2 1s and (p − 1) /2 −1s.
Theorem 7.2 . Gauss’s Lemma . Let p be an odd prime and let a be relatively prime to p. Consider the least non-negative residues of a, 2 a, . . . , (( p − 1) /2) a modulo p. If n is the number of these residues that are greater than p/ 2, then (ap
)
= ( −1) n.
Theorem 7.3 . If p is an odd prime, then
(−1
p
)
= ( −1) (p−1) /2; that is,
(−1
p
)
=
{ 1 if p ≡ 1 (mod 4) ,
−1 if p ≡ 3 (mod 4) .
Proof . This follows from Theorem 4.9 (and Theorem 7.1).
Theorem 7.4 . If p is an odd prime, then
(2
p
)
= ( −1) (p2−1) /8; that is,
(2
p
)
=
{ 1 if p ≡ 1 or 7 (mod 8) ,
−1 if p ≡ 3 or 5 (mod 8) .
27 Proof . If p ≡ 1 or 5 (mod 8), then 2(p−1) /2
(p − 12
)
! ≡ 2 · 4 · 6 · · · (p − 1)
≡ 2 · 4 · 6 · · ·
(p − 12
)
·
(
−p − 32
)
· · · (−5) · (−3) · (−1)
≡ (−1) (p−1) /4
(p − 12
)
!
⇒ 2(p−1) /2 ≡ (−1) (p−1) /4 (mod p).
By Theorem 7.1,
(2
p
)
= ( −1) (p−1) /4. Hence,
(2
p
)
= 1 or −1 according as
p ≡ 1 or 5 (mod 8). Similarly, if p ≡ 3 or 7 (mod 8), then 2(p−1) /2
(p − 12
)
! ≡ 2 · 4 · 6 · · ·
(p − 32
)
·
(
−p − 12
)
· · · (−5) · (−3) · (−1)
≡ (−1) (p+1) /4
(p − 12
)
!
⇒ 2(p−1) /2 ≡ (−1) (p+1) /4 (mod p).
Hence,
(2
p
)
= 1 or −1 according as p ≡ 7 or 3 (mod 8).
Example 7.2 . Prove that if n is an odd positive integer, then every prime divisor of 2 n − 1 is of the form 8 k ± 1. (Compare to Example 6.2)
Solution . Let p | (2 n − 1), where p is prime. Let n = 2 m + 1. Then 2n ≡ 22m+1 ≡ 2(2 m)2 ≡ 1 (mod p) ⇒
(2
p
)
= 1 ⇒ p is of the form 8 k ± 1.
Theorem 7.5 . The Law of Quadratic Reciprocity . For distinct odd primes p and q, (pq
) ( qp
)
= ( −1) p−12 · q−12 .
Example 7.3 . For which primes p > 3 does the congruence x2 ≡ − 3(mod p) have a solution?
Solution . We seek p for which
(−3
p
)
=
(−1
p
) ( 3
p
)
= 1. By quadratic reciprocity, (3
p
) ( p
3
)
= ( −1) (p−1) /2 =
(−1
p
)
,
28 by Theorem 7.3. Thus, in general,
(−3
p
)
=
(−1
p
) ( 3
p
)
=
(p
3
) ( −1
p
)2
=
(p
3
)
.
And, ( p
3
) = 1 iff p ≡ 1 (mod 3). Since p 6 ≡ 4 (mod 6), we have that x2 ≡ − 3(mod p) has a solution iff p ≡ 1 (mod 6).
Example 7.4 . Show that if p = 2 n + 1, n ≥ 2, is prime, then 3 (p−1) /2 + 1 is divisible by p.
Solution . We must have that n is even, say 2 k, for otherwise p ≡ 0(mod 3). By Theorem 7.1,
(3
p
)
≡ 3(p−1) /2 (mod p).
However, p ≡ 1 (mod 4), and p ≡ 4k + 1 ≡ 2 (mod 3) ⇒ (p
3
) = −1, and by quadratic reciprocity,
(3
p
) ( p
3
)
= ( −1) (p−1) /2 = 1 ,
so (3
p
)
= −1 ⇒ 3(p−1) /2 + 1 ≡ 0 (mod p).
Useful Facts
• (a) If p is a prime and p ≡ 1 or 3 (mod 8), then there exist positive integers x and y such that p = x2 + 2 y2.(b) If p is a prime and p ≡ 1 (mod 6), then there exist positive integers
x and y such that p = x2 + 3 y2.Problems 1. Show that if p > 3 is a prime, then the sum of the quadratic residues among the integers 1, 2, . . . , p − 1 is divisible by p.2. Let Fn denote the nth Fibonacci number. Prove that if p > 5 is a prime, then
Fp ≡
(p
5
)
(mod p).
29 3. Show that 16 is a perfect 8 th power modulo p for any prime p.4. Let a, b, and c be positive integers that are pairwise relatively prime, and that satisfy a2 − ab + b2 = c2. Show that every prime factor of c is of the form 6 k + 1. 5. Let p be an odd prime and let ζ be a primitive pth root of unity; that is,
ζ is a complex number such that ζp = 1 and ζk 6 = 1 for 1 ≤ k ≤ p − 1. Let Ap and Bp denote the set of quadratic residues and non-residues modulo p, respectively. Finally, let α = ∑
k∈Ap
ζk and β = ∑
k∈Bp
ζk.For example, for p = 7, α = ζ + ζ2 + ζ4 and β = ζ3 + ζ5 + ζ6. Show that α and β are the roots of
x2 + x +1 −
(−1
p
)
p
4 = 0 .
8 Primitive Roots
If the order of g modulo m is φ(m), then we say that g is a primitive root
modulo m, or simply of m.
Example 8.1 . Show that 2 is a primitive root modulo 3 n for all n ≥ 1.
Solution . The statement is easily verified for n = 1, so assume the result is true for some n = k; that is, 2 φ(3 k ) ≡ 22·3k−1
≡ 1 (mod 3 k). Now, let d be the order of 2 modulo 3 k+1 . Then 2 d ≡ 1 (mod 3 k+1 ) ⇒ 2d ≡ 1 (mod 3 k), so 2 · 3k−1 | d. However, d | φ(3 k+1 ), or d | 2 · 3k. We deduce that d is either 2 · 3k−1 or 2 · 3k. Now we require the following lemma: Lemma. 2 2·3n−1
≡ 1 + 3 n (mod 3 n+1 ), for all n ≥ 1. This is true for n = 1, so assume it is true for some n = k. Then by assumption, 22·3k−1
= 1 + 3 k + 3 k+1 m for some integer m
⇒ 22·3k
= 1 + 3 k+1 + 3 k+2 M for some integer M (obtained by cubing)
⇒ 22·3k
≡ 1 + 3 k+1 (mod 3 k+2 ).
By induction, the lemma is proved. Therefore, 2 2·3k−1
≡ 1 + 3 k 6 ≡ 1 (mod 3 k+1 ), so the order of 2 modulo 3 k+1
is 2 · 3k, and again by induction, the result follows. 30 Corollary 8.2 . If 2 n ≡ − 1 (mod 3 k), then 3 k−1 | n.
Proof . The given implies 2 2n ≡ 1 (mod 3 k) ⇒ φ(3 k) | 2n, or 3 k−1 | n.
Theorem 8.3 . If m has a primitive root, then it has φ(φ(m)) (distinct) primitive roots modulo m.
Theorem 8.4 . The positive integer m has a primitive root iff m is one of 2, 4, pk, or 2 pk, where p is an odd prime.
Theorem 8.5 . If g is a primitive root of m, then gn ≡ 1 (mod m) iff
φ(m) | n. Furthermore, gn0 ≡ gn1 iff φ(m) | (n0 − n1).
Proof . This follows directly from Theorem 6.1.
Theorem 8.6 . If g is a primitive root of m, then the powers 1, g, g2,. . . , gφ(m)−1 represent each integer relatively prime to m uniquely modulo m.In particular, if m > 2, then gφ(m)/2 ≡ − 1 modulo m.
Proof . Clearly, each power gi is relatively prime to m, and there are
φ(m) integers relatively prime to m. Also, if gi ≡ gj (mod m), then gi−j ≡
1 ⇒ φ(m) | (i − j) by Theorem 8.6, so each of the powers are distinct modulo m. Hence, each integer relatively prime to m is some power gi
modulo m. Furthermore, there is a unique i, 0 ≤ i ≤ φ(m) − 1, such that
gi ≡ − 1 ⇒ g2i ≡ 1 ⇒ 2i = φ(m), or i = φ(m)/2.
Proposition 8.7 . Let m be a positive integer. Then the only solutions to the congruence x2 ≡ 1 (mod m) are x ≡ ± 1 (mod m) iff m has a primitive root.
Proof . This follows from Example 4.9.
Example 8.2 . For a positive integer m, let S be the set of positive integers less than m that are relatively prime to m, and let P be the product of the elements in S. Show that P ≡ ± 1 (mod m), with P ≡ − 1 (mod m)iff m has a primitive root.
Solution . We use a similar strategy as in the proof of Wilson’s Theorem. The result is clear for m = 2, so assume that m ≥ 3. We partition S as follows: Let A be the elements of S that are solutions to the congruence
x2 ≡ 1 (mod m), and let B be the remaining elements. The elements in B
can be arranged into pairs, by pairing each with its distinct multiplicative inverse. Hence, the product of the elements in B is 1 modulo m.The elements in A may also be arranged into pairs, by pairing each with 31 its distinct additive inverse, i.e. x and m − x. These must be distinct, because otherwise, x = m/ 2, which is not relatively prime to m. Note that their product is x(m − x) ≡ mx − x2 ≡ − 1 (mod m). Now if m has a primitive root, then by Proposition 8.7, A consists of only the two elements 1 and −1, so P ≡ − 1 (mod m). Otherwise, by Example 4.9, the number of elements of A is a power of two that is at least 4, so the number of such pairs in A is even, and P ≡ 1 (mod m).
Remark . For m prime, this simply becomes Wilson’s Theorem.
Theorem 8.8 .(1) If g is a primitive root of p, p a prime, then g or g + p is a primitive root of p2, according as gp−1 6 ≡ 1 (mod p2) or gp−1 ≡ 1 (mod p2). (2) If g is a primitive root of pk, where k ≥ 2 and p is prime, then g is a primitive root of pk+1 .By Theorem 8.6, given a primitive root g of m, for each a relatively prime to m, there exists a unique integer i modulo φ(m) such that gi ≡ a (mod m). This i is called the index of a with respect to the base g, denoted by ind g(a)(i is dependent on g, so it must be specified). Indices have striking similarity to logarithms, as seen in the following properties: (1) ind g(1) ≡ 0 (mod φ(m)), ind g(g) ≡ 1 (mod φ(m)), (2) a ≡ b (mod m) ⇒ ind g(a) ≡ ind g(b) (mod φ(m)), (3) ind g(ab ) ≡ ind g(a) + ind g(b) (mod φ(m)), (4) ind g(ak) ≡ k ind g(a) (mod φ(m)).
Theorem 8.9 . If p is a prime and a is not divisible by p, then the con-gruence xn ≡ a (mod p) has gcd( n, p − 1) solutions or no solutions according as
a(p−1) / gcd( n,p −1) ≡ 1 (mod p) or a(p−1) / gcd( n,p −1) 6 ≡ 1 (mod p).
Proof . Let g be a primitive root of p, and let i be the index of a with respect to g. Also, any solution x must be relatively prime to p, so let u be the index of x. Then the congruence xn ≡ a becomes gnu ≡ gi (mod p) ⇔
nu ≡ i (mod p − 1). Let k = gcd( n, p − 1). Since g is a primitive root of p,
k | i ⇔ gi(p−1) /k ≡ a(p−1) /k ≡ 1. The result now follows from Theorem 4.11. 32 Remark . Taking p to be an odd prime and n = 2, we deduce Euler’s criterion.
Example 8.3 Let n ≥ 2 be an integer and p = 2 n + 1. Show that if 3(p−1) /2 + 1 ≡ 0 (mod p), then p is a prime. (The converse to Example 7.4.)
Solution . From 3 (p−1) /2 ≡ 32n−1
≡ − 1 (mod p), we obtain 3 2n
≡ 1(mod p), so the order of 3 is 2 n = p−1, but the order also divides φ(p) ≥ p−1. Therefore, φ(p) = p − 1, and p is a prime.
Example 8.4 . Prove that if n = 3 k−1, then 2 n ≡ − 1 (mod 3 k). (A partial converse to Corollary 8.2.)
Solution . By Example 8.1, 2 is a primitive root of 3 k. Therefore, 2 has order φ(3 k) = 2 · 3k−1 = 2 n ⇒ 22n ≡ 1 ⇒ (2 n − 1)(2 n + 1) ≡ 0 (mod 3 k). However, 2 n − 1 ≡ (−1) 3k−1
− 1 ≡ 1 6 ≡ 0 (mod 3), so 2 n + 1 ≡ 0 (mod 3 k).
Example 8.5 . Find all positive integers n > 1 such that 2n + 1
n2
is an integer. (1990 IMO)
Solution . Clearly, n must be odd. Now assume that 3 k‖n; that is, 3 k
is the highest power of 3 dividing n. Then 3 2k | n2 | (2 n + 1) ⇒ 2n ≡ − 1(mod 3 2k) ⇒ 32k−1 | n, by Corollary 8.2 ⇒ 2k − 1 ≤ k ⇒ k ≤ 1, showing that n has at most one factor of 3. We observe that n = 3 is a solution. Suppose that n has a prime factor greater than 3; let p be the least such prime. Then p | (2 n +1) ⇒ 2n ≡ − 1 (mod p). Let d be the order of 2 modulo
p. Since 2 2n ≡ 1, d | 2n. If d is odd, then d | n ⇒ 2n ≡ 1, contradiction, so d is even, say d = 2 d1. Then 2 d1 | 2n ⇒ d1 | n. Also, d | (p − 1), or 2d1 | (p − 1) ⇒ d1 ≤ (p − 1) /2 < p . But d1 | n, so d1 = 1 or d1 = 3. If d1 = 1, then d = 2, and 2 2 ≡ 1 (mod p), contradiction. If d1 = 3, then d = 6, and 26 ≡ 1 (mod p), or p | 63 ⇒ p = 7. However, the order of 2 modulo 7 is 3, which is odd, again contradiction. Therefore, no such p can exist, and the only solution is n = 3. Useful Facts
• All prime divisors of the Fermat number 2 2n
1, n > 1, are of the form 2n+2 k + 1. 33 Problems 1. Let p be an odd prime. Prove that 1i + 2 i + · · · + ( p − 1) i ≡ 0 (mod p)for all i, 0 ≤ i ≤ p − 2. 2. Show that if p is an odd prime, then the congruence x4 ≡ − 1 (mod p)has a solution iff p ≡ 1 (mod 8). 3. Show that if a and n are positive integers with a odd, then a2n
≡ 1(mod 2 n+2 ). 4. The number 142857 has the remarkable property that multiplying it by 1, 2, 3, 4, 5, and 6 cyclically permutes the digits. What are other numbers that have this property? Hint: Compute 142857 × 7.
9 Dirichlet Series
Despite the intimidating name, Dirichlet series are easy to work with, and can provide quick proofs to certain number-theoretic identities, such as Example 3.2. Let α be a function taking the positive integers to the integers. Then we say that
f (s) =
∞
∑
n=1
α(n)
ns = α(1) + α(2) 2s + α(3) 3s + · · ·
is the Dirichlet series generating function (Dsgf) of the function α,which we denote by f (s) ↔ α(n). Like general generating functions, these generating functions are used to provide information about their correspond-ing number-theoretic functions, primarily through manipulation of the gen-erating functions. Let 1 denote the function which is 1 for all positive integers; that is, 1( n) = 1 for all n. Let δ1(n) be the function defined by
δ1(n) =
{ 1 if n = 1 ,
0 if n > 1.
It is easy to check that 1 and δ1 are multiplicative. 34 Now, let α and β be functions taking the positive integers to the integers. The convolution of α and β, denoted α ∗ β, is defined by (α ∗ β)( n) = ∑
d|n
α(d)β(n/d ).
Note that convolution is symmetric; that is, α ∗ β = β ∗ α.
Theorem 9.1 . Let f (s) ↔ α(n) and g(s) ↔ β(n). Then ( f · g)( s) ↔
(α ∗ β)( n). We now do three examples. The Dsgf of 1( n) is the well-known Riemann Zeta function ζ(s):
ζ(s) =
∞
∑
n=1
1
ns = 1 + 12s + 13s + · · · ,
so ζ(s) ↔ 1( n). This function will play a prominent role in this theory. What makes this theory nice to work with is that we may work with these functions at a purely formal level; no knowledge of the analytic properties of
ζ(s) or indeed of any other Dsgf is required. By Theorem 9.1, the number-theoretic function corresponding to ζ2(s) is
∑
d|n
1( d)1( n/d ) = ∑
d|n
1 = τ (n).
Hence, ζ2(s) ↔ τ (n). Finally, it is clear that 1 ↔ δ1(n). If α is a multiplicative function, then we can compute the Dsgf corre-sponding to α using the following theorem.
Theorem 9.2 . Let α be a multiplicative function. Then
∞
∑
n=1
α(n)
ns = ∏
p
∞
∑
k=0
α(pk)
pks = ∏
p
[1 + α(p)p−s + α(p2)p−2s + α(p3)p−3s + · · · ],
where the product on the right is taken over all prime numbers. 35 As before, if we take α = 1, then we obtain
ζ(s) = ∏
p
(1 + p−s + p−2s + p−3s + · · · )= ∏
p
( 11 − p−s
)
= 1
∏
p
(1 − p−s),
an identity that will be useful. We say that a positive integer n > 1 is square-free if n contains no repeated prime factors; that is, p2 - n for all primes p. With this in mind, we define the M¨ obius function μ as follows:
μ(n) =
1 if n = 1 ,
0 if n is not square-free, and (−1) k if n is square-free and has k prime factors .
It is easy to check that μ is multiplicative. By Theorem 9.2, the corresponding Dsgf is given by ∏
p
(1 − p−s) = 1
ζ(s).
Hence, 1 /ζ (s) ↔ μ(n), and this property makes the the seemingly mysterious function μ very important, as seen in the following theorem.
Theorem 9.3 . (M¨ obius Inversion Formula) Let α and β be functions such that
β(n) = ∑
d|n
α(d).
Then
α(n) = ∑
d|n
μ(n/d )β(d).
Proof . Let f (s) ↔ α(n) and g(s) ↔ β(n). The condition is equivalent to
β = α ∗ 1, or g(s) = f (s)ζ(s), and the conclusion is equivalent to α = β ∗ μ,or f (s) = g(s)/ζ (s).
Theorem 9.4 . Let f (s) ↔ α(n). Then for any integer k, f (s − k) ↔
nkα(n). 36 For more on Dirichlet series, and generating functions in general, see H. Wilf, Generatingfunctionology .Problems 1. Let α, β, and γ be functions taking the positive integers to the integers. (a) Prove that α ∗ δ1 = α.(b) Prove that ( α ∗ β) ∗ γ = α ∗ (β ∗ γ). (c) Prove that if α and β are multiplicative, then so is α ∗ β.2. Prove that the following relations hold:
ζ(s − 1)
ζ(s) ↔ φ(n),ζ(s)ζ(s − 1) ↔ σ(n),ζ(s)
ζ(2 s) ↔ | μ(n)|.
Let the prime factorization of a positive integer n > 1 be pe1
1
pe2
2
· · · pek
k
.Define the functions λ and θ by λ(n) = ( −1) e1+e2+··· +ek and θ(n) = 2 k.Set λ(1) = θ(1) = 1. Show that λ and θ are multiplicative, and that
ζ(2 s)
ζ(s) ↔ λ(n) and ζ2(s)
ζ(2 s) ↔ θ(n).
For all positive integers n, let
f (n) =
n
∑
m=1
n
gcd( m, n ).
(a) Show that f (n) = ∑
d|n
dφ (d). (b) Let n = pe1
1
pe2
2
· · · pek
k
1 be the prime factorization of n. Show that
f (n) =
(p2e1+1 1 + 1
p1 + 1
) ( p2e2+1 2 + 1
p2 + 1
)
· · ·
(p2ek +1 1 + 1
pk + 1
)
.
Verify Example 3.2 in one calculation. 37 6. Let id denote the identity function; that is, id( n) = n for all n. Verify each of the following identities in one calculation: (a) φ ∗ τ = σ.(b) μ ∗ 1 = δ1.(c) μ ∗ id = φ.(d) φ ∗ σ = id · τ .(e) σ ∗ id = 1 ∗ (id · τ ). 7. Let a1, a2, . . . , be the sequence of positive integers satisfying
∑
d|n
ad = 2 n
for all n. Hence, a1 = 2, a2 = 2 2 − 2 = 2, a3 = 2 3 − 2 = 6, a4 =24 − 2 − 2 = 12, and so on. Show that for all n, n | an.Hint: Don’t use the Dsgf of ( an)∞
1
; use the M¨ obius Inversion Formula. Bigger Hint: Consider the function f : [0 , 1] → [0 , 1] defined by f (x) =
{2x}, where {x} = x − b xc is the fractional part of x. Find how the formula in the problem relates to the function f (n) = f ◦ f ◦ · · · ◦ f
︸ ︷︷ ︸
n
.8. For all non-negative integers k, let σk be the function defined by
σk(n) = ∑
d|n
dk.
Thus, σ0 = τ and σ1 = σ. Prove that
ζ(s)ζ(s − k) ↔ σk(n).
10 Miscellaneous Topics
10.1 Pell’s Equations
Pell’s equations (or Fermat’s equations, as they are rightly called) are diophantine equations of the form x2 − dy 2 = N , where d is a positive non-square integer. There always exist an infinite number of solutions when
N = 1, which we characterize. 38 Theorem 10.1.1 . If ( a, b ) is the lowest positive integer solution of x2 −
dy 2 = 1, then all positive integer solutions are of the form (xn, y n) =
(
(a + b√d)n + ( a − b√d)n
2 , (a + b√d)n − (a − b√d)n
2√d
)
.
We will not give a proof here, but we will verify that every pair indicated by the formula is a solution. The pair ( xn, y n) satisfy the equations
xn + yn
√d = ( a + b√d)n, and
xn − yn
√d = ( a − b√d)n.
Therefore,
x2
n
− dy 2
n
= ( xn + yn
√d)( xn − yn
√d)= ( a + b√d)n(a − b√d)n
= ( a2 − db 2)n
= 1 ,
since ( a, b ) is a solution.
Remark . The sequences ( xn), ( yn) satisfy the recurrence relations xn =2ax n−1 − xn−2, yn = 2 ay n−1 − yn−2.For x2 − dy 2 = −1, the situation is similar. If ( a, b ) is the least positive solution, then the ( xn, y n) as above for n odd are the solutions of x2 − dy 2 =
−1, and the ( xn, y n) for n even are the solutions of x2 − dy 2 = 1.
Example 10.1.1 Find all solutions in pairs of positive integers ( x, y ) to the equation x2 − 2y2 = 1.
Solution . We find that the lowest positive integer solution is (3,2), so all positive integer solutions are given by (xn, y n) =
(
(3 + 2 √2) n + (3 − 2√2) n
2 , (3 + 2 √2) n − (3 − 2√2) n
2√2
)
.
The first few solutions are (3,2), (17,12), and (99,70). 39 Example 10.1.2 . Prove that the equation x2 −dy 2 = −1 has no solution in integers if d ≡ 3 (mod 4).
Solution . It is apparent that d must have a prime factor of the form 4 k +3, say q. Then x2 ≡ − 1 (mod q), which by Theorem 4.9 is a contradiction. Problems 1. In the sequence 12, 53, 11 8 , 27 19 , . . . ,
the denominator of the nth term ( n > 1) is the sum of the numerator and the denominator of the ( n − 1) th term. The numerator of the nth
term is the sum of the denominators of the nth and ( n−1) th term. Find the limit of this sequence. (1979 Atlantic Region Mathematics League) 2. Let x0 = 0, x1 = 1, xn+1 = 4 xn − xn−1, and y0 = 1, y1 = 2, yn+1 =4yn − yn−1. Show for all n ≥ 0 that y2
n
= 3 x2
n
(1988 Canadian Mathematical Olympiad) 3. The polynomials P , Q are such that deg P = n, deg Q = m, have the same leading coefficient, and P 2(x) = ( x2 − 1) Q2(x) + 1. Show that
P ′(x) = nQ (x). (1978 Swedish Mathematical Olympiad, Final Round)
10.2 Farey Sequences
The nth Farey sequence is the sequence of all reduced rationals in [0,1], with both numerator and denominator no greater than n, in increasing order. Thus, the first 5 Farey sequences are: 0/1, 1/1, 0/1, 1/2, 1/1, 0/1, 1/3, 1/2, 2/3, 1/1, 0/1, 1/4, 1/3, 1/2, 2/3, 3/4, 1/1, 0/1, 1/5, 1/4, 1/3, 2/5, 1/2, 3/5, 2/3, 3/4, 4/5, 1/1. Properties of Farey sequences include the following: 40 (1) If a/b and c/d are consecutive fractions in the same sequence, in that order, then ad − bc = 1. (2) If a/b , c/d , and e/f are consecutive fractions in the same sequence, in that order, then a + eb + f = cd.
(3) If a/b and c/d are consecutive fractions in the same sequence, then among all fractions between the two, ( a + c)/(b + d) (reduced) is the unique fraction with the smallest denominator. For proofs of these and other interesting properties, see Ross Honsberger, “Farey Sequences”, Ingenuity in Mathematics .Problems 1. Let a1, a2, . . . , am be the denominators of the fractions in the nth Farey sequence, in that order. Prove that 1
a1a2
1
a2a3
· · · + 1
am−1am
= 1 .
10.3 Continued Fractions
Let a0, a1, . . . , an be real numbers, all positive, except possibly a0. Then let
〈a0, a 1, . . . , a n〉 denote the continued fraction
a0 + 1
a1 + · · · + 1
an−1 + 1
an
.
If each ai is an integer, then we say that the continued fraction is simple .Define sequences ( pk) and ( qk) as follows:
p−1 = 0 , p0 = a0, and pk = akpk−1 + pk−2,q−1 = 0 , q0 = 1 , and qk = akpk−1 + qk−2, for k ≥ 1.
Theorem 10.3.1 . For all x > 0 and k ≥ 1,
〈a0, a 1, . . . , a k−1, x 〉 = xp k−1 + pk−2
xq k−1 + qk−2
.
41 In particular,
〈a0, a 1, . . . , a k〉 = pk
qk
.
Theorem 10.3.2 . For all k ≥ 0, (1) pkqk−1 − pk−1qk = ( −1) k−1,(2) pkqk−2 − pk−2qk = ( −1) kak.Define ck to be the kth convergence 〈a0, a 1, . . . , a k〉 = pk/q k.
Theorem 10.3.3 . c0 < c 2 < c 4 < · · · < c 5 < c 3 < c 1.For a nice connection between continued fractions, linear diophantine equations, and Pell’s equations, see Andy Liu, “Continued Fractions and Diophantine Equations”, Volume 3, Issue 2, Mathematical Mayhem .Problems 1. Let a = 〈1, 2, . . . , 99 〉 and b = 〈1, 2, . . . , 99 , 100 〉. Prove that
|a − b| < 199!100! .
(1990 Tournament of Towns) 2. Evaluate
8
√√√√√2207 − 12207 − 12207 − · · ·
.
Express your answer in the form a+b√cd , where a, b, c, d are integers. (1995 Putnam)
10.4 The Postage Stamp Problem
Let a and b be relatively prime positive integers greater than 1. Consider the set of integers of the form ax + by , where x and y are non-negative integers. The following are true: (1) The greatest integer that cannot be written in the given form is ( a −
1)( b − 1) − 1 = ab − a − b.42 (2) There are 12 (a − 1)( b − 1) positive integers that cannot be written in the given form. (3) For all integers t, 0 ≤ t ≤ ab − a − b, t can be written in the given form iff ab − a − b − t cannot be. (If you have not seen or attempted this enticing problem, it is strongly suggested you have a try before reading the full solution.) Before presenting the solution, it will be instructive to look at an example. Take a = 12 and b = 5. The first few non-negative integers, in rows of 12, with integers that cannot be written in the given form in bold, are shown: 0 1 2 3 4 5 6 7 8 9 10 11
12 13 14 15 16 17 18 19 20 21 22 23
24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 With this arrangement, one observation should become immediately ap-parent, namely that bold numbers in each column end when they reach a multiple of 5. It should be clear that when reading down a column, once one hits an integer that can be written in the given form, then all successive integers can be as well, since we are adding 12 for each row we go down. It will turn out that this one observation is the key to the solution.
Proof . Define a grapefruit to be an integer that may be written in the given form. For each i, 0 ≤ i ≤ a − 1, let mi be the least non-negative integer such that b | (i + am i). It is obvious that for k ≥ mi, i + ak is a grapefruit. We claim that for 0 ≤ k ≤ mi − 1, i + ak is not a grapefruit. It is sufficient to show that i + a(mi − 1) is not a grapefruit, if mi ≥ 1. Let i + am i = bn i, ni ≥ 0. Since i + a(mi − b) = b(ni − a), mi must be strictly less than b; otherwise, we can find a smaller mi. Then i + a(mi − b) ≤
a−1−a = −1, so ni < a , or ni ≤ a−1. Suppose that ax +by = i+a(mi −1) =
bn i − a, for some non-negative integers x and y. Then a(x + 1) = b(ni − y), so ni − y is positive. Since a and b are relatively prime, a divides ni − y.However, ni ≤ a − 1 ⇒ ni − y ≤ a − 1, contradiction. Therefore, the greatest non-grapefruit is of the form bn i − a, ni ≤ a − 1. The above argument also shows that all positive integers of this form are also non-grapefruits. Hence, the greatest non-grapefruit is b(a−1) −a = ab −a−b,proving (1). 43 Now, note that there are mi non-grapefruits in column i. The above tells us the first grapefruit appearing in column i is nib. Since 0, b, 2 b, . . . , ( a−1) b
appear in different columns (because a and b are relatively prime), and there are a columns, we conclude that as i varies from 0 to a − 1, ni takes on 0, 1, . . . , a − 1, each exactly once. Therefore, summing over i, 0 ≤ i ≤ a − 1,
∑
i
(i + am i) = ∑
i
i + ∑
i
am i = a(a − 1) 2 + a ∑
i
mi
= ∑
i
bn i = a(a − 1) b
2
⇒ a ∑
i
mi = a(a − 1)( b − 1) 2
⇒ ∑
i
mi = (a − 1)( b − 1) 2 ,
proving (2). Finally, suppose that ax 1 + by 1 = t, and ax 2 + by 2 = ab − a − b − t, for some non-negative integers x1, x2, y1, and y2. Then a(x1 + x2) + b(y1 + y2) =
ab − a − b, contradicting (1). So, if we consider the pairs ( t, ab − a − b − t), 0 ≤ t ≤ (a − 1)( b − 1) /2 − 1, at most one element in each pair can be written in the given form. However, we have shown that exactly ( a − 1)( b − 1) /2 integers cannot be written in the given form, which is the number of pairs. Therefore, exactly one element of each pair can be written in the given form, proving (3).
Remark . There is a much shorter proof using Corollary 2.4. Can you find it? For me, this type of problem epitomizes problem solving in number theory, and generally mathematics, in many ways. If I merely presented the proof by itself, it would look artificial and unmotivated. However, by looking at a specific example, and finding a pattern, we were able to use that pattern as a springboard and extend it into a full proof. The algebra in the proof is really nothing more than a translation of observed patterns into formal notation. (Mathematics could be described as simply the study of pattern.) Note also that we used nothing more than very elementary results, showing how powerful basic concepts can be. It may have been messy, but one should never be afraid to get one’s hands dirty; indeed, the deeper you go, the 44 more you will understand the importance of these concepts and the subtle relationships between them. By trying to see an idea through to the end, one can sometimes feel the proof almost working out by itself. The moral of the story is: A simple idea can go a long way. For more insights on the postage stamp problem, see Ross Honsberger, “A Putnam Paper Problem”, Mathematical Gems II .Problems 1. Let a, b, and c be positive integers, no two of which have a common divisor greater than 1. Show that 2 abc − ab − bc − ca is the largest integer that cannot be expressed in the form xab + yca + zab , where x,
y, and z are non-negative integers. (1983 IMO)
References
A. Adler & J. Coury, The Theory of Numbers , Jones and Bartlett I. Niven & H. Zuckerman, An Introduction to the Theory of Numbers ,John Wiley & Sons c© First Version October 1995 c© Second Version January 1996 c© Third Version April 1999 c© Fourth Version May 2000 Thanks to Ather Gattami for an improvement to the proof of the Postage Stamp Problem. This document was typeset under L ATEX, and may be freely distributed provided the contents are unaltered and this copyright notice is not removed. Any comments or corrections are always welcomed. It may not be sold for profit or incorporated in commercial documents without the express permis-sion of the copyright holder. So there. 45 |
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Nth Term of AP – Definition, Formula, Examples, FAQs
Home » Math Vocabulary » Nth Term of AP – Definition, Formula, Examples, FAQs
What Is the nth Term of an AP?
The nth Term of an AP Formula
How to Find the nth Term of an AP
Solved Examples on the nth Term of an AP
Practice Problems on the nth Term of an AP
Frequently Asked Questions about the nth Term of an AP
What Is the nth Term of an AP?
The nth term of an AP is the term that occupies the nth position from the beginning (left side) of an A.P. sequence. It’s a way to identify a specific term in the A.P. sequence based on its position.
Suppose you have a sequence of numbers that increase by the same amount each time. This kind of arithmetic sequence is called an Arithmetic Progression (AP).
Example 1: Consider the sequence 2, 5, 8, 11, 14, …
Here, the first term of this AP is T1 = 2, the second term is T2 = 5, the third term is T3 = 8, and so on. Each number in this series is more than the previous one by 3. Here, 3 is called the “common difference” because the same difference between any two consecutive terms is constant.
Example 2: In the given sequence, the first term is -6. The common difference is 7.
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The nth Term of an AP Formula
The nth term of an AP is obtained by adding the common difference (d) multiplied by (n-1) to the first term (a) of the progression.
The nth term of an arithmetic progression (AP) can be calculated using the following formula:
| |
| Tn= a+(n – 1)d |
where:
Tn is the nth term of the AP
a is the first term of the AP
n indicates the position of the term
d is the common difference between consecutive terms of the AP.
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What Is the Importance of the nth Term of an AP Formula?
If you want to find the 6th number in this sequence without counting all the previous numbers, instead of repeatedly adding 3, you can use the “nth term of the AP formula.
It is not practical to keep counting one at a time, especially if we’re searching for a term far down the sequence, like the 25th term for the 100th term.
The formula lets us figure out any term in the sequence without having to count all the previous ones. It works by taking the first term and adding the common difference multiplied by (n – 1). This way, we can jump straight to the term we’re interested in without doing all the intermediate steps.
Derivation of the nth Term of an AP
An arithmetic progression (A.P) is a sequence of numbers where each term is obtained by adding a fixed amount, known as the common difference, to the preceding term.
Consider the A.P. given by
a, a+d, a+2d, a+3d,…
1st term = T1 = a
2nd term = T2 = a+d =a+(2-1)d
3rd term = T3=a+2d=a+(3-1)d
4th term = T4=a+3d=a+ (4-1)d
.
.
.
nth term Tn=a+(n-1)d
This formula represents the value of the nth term in AP whose first term is a and the common difference is d.
How to Find the nth Term of an AP
Let’s understand the steps to find the nth term of an Arithmetic Progression (AP) with the help of an example.
Formula: nth term Tn=a+(n-1)d
Example: Find the 10th term of the arithmetic progression 3, 8, 13, 18,…
Here, first term (a1) = 3
common difference (d) = 4
n = 10
Formula: Tn= a + (n – 1)d
T10= 3+(10-1)4
T10= 3+(10-1)4
T10= 3+(9)4
T10= 3+36
T10 = 39
Thus, the 10th term of the sequence is 39.
Facts about the nth Term of an AP Formula
The common difference (d) remains constant throughout the entire arithmetic progression. It’s the fixed amount that each term increases or decreases by.
The first term (a1) of the AP is the value at the first position, i.e., a1=a1+(1-1)d, which simplifies to a1.
The nth term represents the value of the term at the nth position in the sequence. It’s the term that’s n steps away from the first term.
The formula works for negative terms as well. If the common difference is negative, adding it will cause the terms to decrease as you move along the sequence.
The formula can also be used to find terms with non-integer values. As long as the common difference and initial term are real numbers, the formula remains valid.
Conclusion
In this article, we learned about finding the nth term of an arithmetic progression (AP). This formula lets you find any term in an arithmetic progression without having to add the common difference again and again. It’s super useful when you’re dealing with large sequences and want to find terms quickly! Let’s solve a few examples and MCQs for effective practice.
Solved Examples on the nth Term of an AP
Example 1: Find the 10th term of the arithmetic progression: 5, 9, 13, 17, …
Solution:
First term (a1) = 5
common difference (d) = 4
n = 10
Tn = a + (n – 1)d
T10= 5+(10-1)4
T10= 5+(9)4
T10= 5+36 = 41
Thus, the 10th term of the sequence is 41.
Example 2: Find the 15th term of the arithmetic progression: 2, 7, 12, 17, …
Solution:
First term (a1) = 2
common difference (d) = 5
n = 15
Tn= a + (n – 1)d
T15= 2+(15-1)5
T15= 2+(14)5
T15= 2+70 = 72
Thus, the 15th term of the sequence is 72.
Example 3: What is the position of 120 in the given AP?
5, 10, 15, 20,…,120,…
Solution:
First term = a = 5
Common difference (d) = 5.
Tn = 120
We will find the value of n.
Tn= a + (n – 1)d
120 = 5 + (n – 1) 5
120 = 5 + 5n – 5
120 = 5n
n = 1205
n = 24
120 is the 24th term of the given AP.
Example 4: Find the 15th term of an AP whose 8th term is 27 and the common difference is 4.
Solution:
T8 = 27
d = 4
Formula: Tn = a + (n – 1)d
T8 = a + (8 – 1)(4)
27 = a + 28
a = 27-28
a = -1
Let’s find the 15th term.
T15 = a + (15 – 1) d
T15 = -1 + (14) 4
T15 = -1+56
T15 = 55
Thus, the 15th term of the AP is 55
Practice Problems on the nth Term of an AP
Nth Term of AP - Definition, Formula, Examples, FAQs
Attend this quiz & Test your knowledge.
1
What is the formula for the nth term of an arithmetic progression (AP)?
$T_{n} = a_{1} + (n - 1)d$
$T_{n} = a_{1} + (n + 1)d$
$T_{n} = a_{1} + nd$
$T_{n} = a_{1} - (n - 1)d$
CorrectIncorrect
Correct answer is: $T_{n} = a_{1} + (n - 1)d$The nth term of an arithmetic progression (AP) can be calculated using the formula$T_{n} = a1 + (n - 1)d$, where $a_{1}$ is the first term, d is the common difference, and n is the term number.
2
What does n represent in the nth term formula of an AP?
The value of the nth term in the sequence
The total number of terms in the sequence
The position of the term in the sequence
The common difference of the sequence
CorrectIncorrect
Correct answer is: The position of the term in the sequencen represents the position of the term in the sequence. It helps determine which term's value is being calculated using the formula.
3
What is the nth term of an AP if the first term is 12 and the common difference is 0?
$T_{n} = 12$
$T_{n} = 12n$
$T_{n} = 0$
$T_{n} = n$
CorrectIncorrect
Correct answer is: $T_{n} = 12$When the common difference is 0, it means that the terms in the sequence are all the same. In this case, the nth term formula simplifies to $T_{n} = a = 12$.
4
What information is NOT required to determine the nth term of an arithmetic progression (AP)?
The first term
The term number (position)
The common difference.
The last term
CorrectIncorrect
Correct answer is: The last termTo determine the nth term of an AP, you need the value of the first term, the common difference, and the term number (position). The value of the last term is not necessary to calculate the nth term.
5
If a = 10 and d = 10, then the first four terms will be:
10, 30, 50, 60
10, 20, 30, 40
10, 15, 20, 25
10, 18, 20, 30
CorrectIncorrect
Correct answer is: 10, 20, 30, 40$a = 10, d = 10, a_{1} = a = 10$ $a_{2} = a_{1} + d = 10 + 10 = 20$ $a_{3} = a_{2} + d = 20 + 10 = 30$ $a_{4} = a_{3} + d = 30+10 = 40$
Frequently Asked Questions about the nth Term of an AP
What is the nth term of an arithmetic progression?
The nth term of an arithmetic progression is the value of the term that occupies the nth position in the sequence.
Can the nth term of an AP be a negative number?
Yes, the nth term can be a negative number if the common difference (d) is negative. The formula accounts for both positive and negative terms.
Do I always need to know the previous terms to find the nth term in an AP?
No, you don’t need to know the previous terms to find the nth term using the formula. The formula directly calculates the term at a specific position based on the first term and the common difference.
Can I use the nth term formula for non-integer terms?
Yes, the nth term formula is applicable for both integer and non-integer terms. As long as the first term and the common difference are real numbers, the formula remains valid.
Is the nth term formula the same for all arithmetic progressions?
Yes, the formula Tn= a1 + (n – 1)d is applicable to all arithmetic progressions, regardless of the values of the first term and common difference. It’s a universal formula.
Is the nth term formula used only in mathematics?
No, the concept of the nth term and the formula have practical applications in various fields, including physics, finance, computer science, and more. It helps model situations where values change in a consistent manner.
How do you find the value of n in AP?
To determine the nth term of an arithmetic progression (AP), start by identifying the first term as “a” and the common difference as “d.” Decide on the term number, “n,” for which you want to find the value. Then, use the formula, Tn= a1 + (n – 1)d, where Tn represents the nth term, “a” is the first term, “d” is the common difference, and here “n” is the term number.
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Algebra and Trigonometry 2e
Chapter 12
Algebra and Trigonometry 2eChapter 12
Contents Contents
Highlights
Table of contents
Preface
1 Prerequisites
2 Equations and Inequalities
3 Functions
4 Linear Functions
5 Polynomial and Rational Functions
6 Exponential and Logarithmic Functions
7 The Unit Circle: Sine and Cosine Functions
8 Periodic Functions
9 Trigonometric Identities and Equations
10 Further Applications of Trigonometry
11 Systems of Equations and Inequalities
12 Analytic Geometry
13 Sequences, Probability, and Counting Theory
A | Proofs, Identities, and Toolkit Functions
Answer Key
Chapter 1
Chapter 2
Chapter 3
Chapter 4
Chapter 5
Chapter 6
Chapter 7
Chapter 8
Chapter 9
Chapter 10
Chapter 11
Chapter 12
Chapter 13
Index
Search for key terms or text.
Close
Try It
12.1 The Ellipse
1.
x 2+y 2 16=1 x 2+y 2 16=1 x 2+y 2 16=1
2.
(x−1)2 16+(y−3)2 4=1(x−1)2 16+(y−3)2 4=1(x−1)2 16+(y−3)2 4=1
3.
center: (0,0);(0,0);(0,0); vertices: (±6,0);(±6,0);(±6,0); co-vertices: (0,±2);(0,±2);(0,±2); foci: (±4 2–√,0)(±4 2,0)(±4 2,0)
4.
Standard form: x 2 16+y 2 49=1;x 2 16+y 2 49=1;x 2 16+y 2 49=1; center: (0,0);(0,0);(0,0); vertices: (0,±7);(0,±7);(0,±7); co-vertices: (±4,0);(±4,0);(±4,0); foci: (0,±33−−√)(0,±33)(0,±33)
5.
Center: (4,2);(4,2);(4,2); vertices: (−2,2)(−2,2)(−2,2) and (10,2);(10,2);(10,2); co-vertices: (4,2−2 5–√)(4,2−2 5)(4,2−2 5) and (4,2+2 5–√);(4,2+2 5);(4,2+2 5); foci: (0,2)(0,2)(0,2) and (8,2)(8,2)(8,2)
6.
(x−3)2 4+(y+1)2 16=1;(x−3)2 4+(y+1)2 16=1;(x−3)2 4+(y+1)2 16=1; center: (3,−1);(3,−1);(3,−1); vertices: (3,−5)(3,−5)(3,−5) and (3,3);(3,3);(3,3); co-vertices: (1,−1)(1,−1)(1,−1) and (5,−1);(5,−1);(5,−1); foci: (3,−1−2 3–√)(3,−1−2 3)(3,−1−2 3) and (3,−1+2 3–√)(3,−1+2 3)(3,−1+2 3)
7.
ⓐx 2 57,600+y 2 25,600=1 x 2 57,600+y 2 25,600=1 x 2 57,600+y 2 25,600=1
ⓑ The people are standing 358 feet apart.
12.2 The Hyperbola
1.
Vertices: (±3,0);(±3,0);(±3,0); Foci: (±34−−√,0)(±34,0)(±34,0)
2.
y 2 4−x 2 16=1 y 2 4−x 2 16=1 y 2 4−x 2 16=1
3.
(y−3)2 25−(x−1)2 144=1(y−3)2 25-(x−1)2 144=1(y−3)2 25-(x−1)2 144=1
4.
vertices: (±12,0);(±12,0);(±12,0); co-vertices: (0,±9);(0,±9);(0,±9); foci: (±15,0);(±15,0);(±15,0); asymptotes: y=±3 4 x;y=±3 4 x;y=±3 4 x;
5.
center: (3,−4);(3,−4);(3,−4); vertices: (3,−14)(3,−14)(3,−14) and (3,6);(3,6);(3,6); co-vertices: (−5,−4);(−5,−4);(−5,−4); and (11,−4);(11,−4);(11,−4); foci: (3,−4−2 41−−√)(3,−4−2 41)(3,−4−2 41) and (3,−4+2 41−−√);(3,−4+2 41);(3,−4+2 41); asymptotes: y=±5 4(x−3)−4 y=±5 4(x−3)−4 y=±5 4(x−3)−4
6.
The sides of the tower can be modeled by the hyperbolic equation. x 2 400−y 2 3600=1 or x 2 20 2−y 2 60 2=1.x 2 400−y 2 3600=1 or x 2 20 2−y 2 60 2=1.x 2 400−y 2 3600=1 or x 2 20 2−y 2 60 2=1.
12.3 The Parabola
1.
Focus: (−4,0);(−4,0);(−4,0); Directrix: x=4;x=4;x=4; Endpoints of the latus rectum: (−4,±8)(−4,±8)(−4,±8)
2.
Focus: (0,2);(0,2);(0,2); Directrix: y=−2;y=−2;y=−2; Endpoints of the latus rectum: (±4,2).(±4,2).(±4,2).
3.
x 2=14 y.x 2=14 y.x 2=14 y.
4.
Vertex: (8,−1);(8,−1);(8,−1); Axis of symmetry: y=−1;y=−1;y=−1; Focus: (9,−1);(9,−1);(9,−1); Directrix: x=7;x=7;x=7; Endpoints of the latus rectum: (9,−3)(9,−3)(9,−3) and (9,1).(9,1).(9,1).
5.
Vertex: (−2,3);(−2,3);(−2,3); Axis of symmetry: x=−2;x=−2;x=−2; Focus: (−2,−2);(−2,−2);(−2,−2); Directrix: y=8;y=8;y=8; Endpoints of the latus rectum: (−12,−2)(−12,−2)(−12,−2) and (8,−2).(8,−2).(8,−2).
6.
ⓐy 2=1280 x y 2=1280 x y 2=1280 x
ⓑ The depth of the cooker is 500 mm
12.4 Rotation of Axes
1.
ⓐ hyperbola
ⓑ ellipse
2.
x′2 4+y′2 1=1 x′2 4+y′2 1=1 x′2 4+y′2 1=1
3.
ⓐ hyperbola
ⓑ ellipse
12.5 Conic Sections in Polar Coordinates
1.
ellipse; e=1 3;x=−2 e=1 3;x=−2 e=1 3;x=−2
2.
3.
r=1 1−cos θ r=1 1−cos θ r=1 1−cos θ
4.
4−8 x+3 x 2−y 2=0 4−8 x+3 x 2−y 2=0 4−8 x+3 x 2−y 2=0
12.1 Section Exercises
1.
An ellipse is the set of all points in the plane the sum of whose distances from two fixed points, called the foci, is a constant.
3.
This special case would be a circle.
5.
It is symmetric about the x-axis, y-axis, and the origin.
7.
yes; x 2 3 2+y 2 2 2=1 x 2 3 2+y 2 2 2=1 x 2 3 2+y 2 2 2=1
9.
yes; x 2(1 2)2+y 2(1 3)2=1 x 2(1 2)2+y 2(1 3)2=1 x 2(1 2)2+y 2(1 3)2=1
11.
x 2 2 2+y 2 7 2=1;x 2 2 2+y 2 7 2=1;x 2 2 2+y 2 7 2=1; Endpoints of major axis (0,7)(0,7)(0,7) and (0,−7).(0,−7).(0,−7). Endpoints of minor axis (2,0)(2,0)(2,0) and (−2,0).(−2,0).(−2,0). Foci at (0,3 5–√),(0,−3 5–√).(0,3 5),(0,−3 5).(0,3 5),(0,−3 5).
13.
x 2(1)2+y 2(1 3)2=1;x 2(1)2+y 2(1 3)2=1;x 2(1)2+y 2(1 3)2=1; Endpoints of major axis (1,0)(1,0)(1,0) and (−1,0).(−1,0).(−1,0). Endpoints of minor axis (0,1 3),(0,−1 3).(0,1 3),(0,−1 3).(0,1 3),(0,−1 3). Foci at (2 2√3,0),(−2 2√3,0).(2 2 3,0),(−2 2 3,0).(2 2 3,0),(−2 2 3,0).
15.
(x−2)2 7 2+(y−4)2 5 2=1;(x−2)2 7 2+(y−4)2 5 2=1;(x−2)2 7 2+(y−4)2 5 2=1; Endpoints of major axis (9,4),(−5,4).(9,4),(−5,4).(9,4),(−5,4). Endpoints of minor axis (2,9),(2,−1).(2,9),(2,−1).(2,9),(2,−1). Foci at (2+2 6–√,4),(2−2 6–√,4).(2+2 6,4),(2−2 6,4).(2+2 6,4),(2−2 6,4).
17.
(x+5)2 2 2+(y−7)2 3 2=1;(x+5)2 2 2+(y−7)2 3 2=1;(x+5)2 2 2+(y−7)2 3 2=1; Endpoints of major axis (−5,10),(−5,4).(−5,10),(−5,4).(−5,10),(−5,4). Endpoints of minor axis (−3,7),(−7,7).(−3,7),(−7,7).(−3,7),(−7,7). Foci at (−5,7+5–√),(−5,7−5–√).(−5,7+5),(−5,7−5).(−5,7+5),(−5,7−5).
19.
(x−1)2 3 2+(y−4)2 2 2=1;(x−1)2 3 2+(y−4)2 2 2=1;(x−1)2 3 2+(y−4)2 2 2=1; Endpoints of major axis (4,4),(−2,4).(4,4),(−2,4).(4,4),(−2,4). Endpoints of minor axis (1,6),(1,2).(1,6),(1,2).(1,6),(1,2). Foci at (1+5–√,4),(1−5–√,4).(1+5,4),(1−5,4).(1+5,4),(1−5,4).
21.
(x−3)2(3 2√)2+(y−5)2(2√)2=1;(x−3)2(3 2)2+(y−5)2(2)2=1;(x−3)2(3 2)2+(y−5)2(2)2=1; Endpoints of major axis (3+3 2–√,5),(3−3 2–√,5).(3+3 2,5),(3−3 2,5).(3+3 2,5),(3−3 2,5). Endpoints of minor axis (3,5+2–√),(3,5−2–√).(3,5+2),(3,5−2).(3,5+2),(3,5−2). Foci at (7,5),(−1,5).(7,5),(−1,5).(7,5),(−1,5).
23.
(x+5)2(5)2+(y−2)2(2)2=1;(x+5)2(5)2+(y−2)2(2)2=1;(x+5)2(5)2+(y−2)2(2)2=1; Endpoints of major axis (0,2),(−10,2).(0,2),(−10,2).(0,2),(−10,2). Endpoints of minor axis (−5,4),(−5,0).(−5,4),(−5,0).(−5,4),(−5,0). Foci at (−5+21−−√,2),(−5−21−−√,2).(−5+21,2),(−5−21,2).(−5+21,2),(−5−21,2).
25.
(x+3)2(5)2+(y+4)2(2)2=1;(x+3)2(5)2+(y+4)2(2)2=1;(x+3)2(5)2+(y+4)2(2)2=1; Endpoints of major axis (2,−4),(−8,−4).(2,−4),(−8,−4).(2,−4),(−8,−4). Endpoints of minor axis (−3,−2),(−3,−6).(−3,−2),(−3,−6).(−3,−2),(−3,−6). Foci at (−3+21−−√,−4),(−3−21−−√,−4).(−3+21,−4),(−3−21,−4).(−3+21,−4),(−3−21,−4).
27.
Foci (−3,−1+11−−√),(−3,−1−11−−√)(−3,−1+11),(−3,−1−11)(−3,−1+11),(−3,−1−11)
29.
Focus (0,0)(0,0)(0,0)
31.
Foci (−10,30),(−10,−30)(−10,30),(−10,−30)(−10,30),(−10,−30)
33.
Center (0,0),(0,0),(0,0), Vertices (4,0),(−4,0),(0,3),(0,−3),(4,0),(−4,0),(0,3),(0,−3),(4,0),(−4,0),(0,3),(0,−3), Foci (7–√,0),(−7–√,0)(7,0),(−7,0)(7,0),(−7,0)
35.
Center (0,0),(0,0),(0,0), Vertices (1 9,0),(−1 9,0),(0,1 7),(0,−1 7),(1 9,0),(−1 9,0),(0,1 7),(0,−1 7),(1 9,0),(−1 9,0),(0,1 7),(0,−1 7), Foci (0,4 2√63),(0,−4 2√63)(0,4 2 63),(0,−4 2 63)(0,4 2 63),(0,−4 2 63)
37.
Center (−3,3),(−3,3),(−3,3), Vertices (0,3),(−6,3),(−3,0),(−3,6),(0,3),(−6,3),(−3,0),(−3,6),(0,3),(−6,3),(−3,0),(−3,6), Focus (−3,3)(−3,3)(−3,3)
Note that this ellipse is a circle. The circle has only one focus, which coincides with the center.
39.
Center (1,1),(1,1),(1,1), Vertices (5,1),(−3,1),(1,3),(1,−1),(5,1),(−3,1),(1,3),(1,−1),(5,1),(−3,1),(1,3),(1,−1), Foci (1+2 3–√,1),(1−2 3–√,1)(1+2 3,1),(1-2 3,1)(1+2 3,1),(1-2 3,1)
41.
Center (−4,5),(−4,5),(−4,5), Vertices (−2,5),(−6,5),(−4,6),(−4,4),(−2,5),(−6,5),(−4,6),(−4,4),(−2,5),(−6,5),(−4,6),(−4,4), Foci (−4+3–√,5),(−4−3–√,5)(−4+3,5),(−4−3,5)(−4+3,5),(−4−3,5)
43.
Center (−2,1),(−2,1),(−2,1), Vertices (0,1),(−4,1),(−2,5),(−2,−3),(0,1),(−4,1),(−2,5),(−2,−3),(0,1),(−4,1),(−2,5),(−2,−3), Foci (−2,1+2 3–√),(−2,1−2 3–√)(−2,1+2 3),(−2,1−2 3)(−2,1+2 3),(−2,1−2 3)
45.
Center (−2,−2),(−2,−2),(−2,−2), Vertices (0,−2),(−4,−2),(−2,0),(−2,−4),(0,−2),(−4,−2),(−2,0),(−2,−4),(0,−2),(−4,−2),(−2,0),(−2,−4), Focus (−2,−2)(−2,−2)(−2,−2)
47.
x 2 25+y 2 29=1 x 2 25+y 2 29=1 x 2 25+y 2 29=1
49.
(x−4)2 25+(y−2)2 1=1(x−4)2 25+(y−2)2 1=1(x−4)2 25+(y−2)2 1=1
51.
(x+3)2 16+(y−4)2 4=1(x+3)2 16+(y−4)2 4=1(x+3)2 16+(y−4)2 4=1
53.
x 2 81+y 2 9=1 x 2 81+y 2 9=1 x 2 81+y 2 9=1
55.
(x+2)2 4+(y−2)2 9=1(x+2)2 4+(y−2)2 9=1(x+2)2 4+(y−2)2 9=1
57.
Area=12 π square units Area=12π square units Area=12π square units
59.
Area=2 5–√π Area=2 5 π Area=2 5 π square units.
61.
Area=9 π Area=9π Area=9π square units.
63.
x 2 4 h 2+y 2 1 4 h 2=1 x 2 4 h 2+y 2 1 4 h 2=1 x 2 4 h 2+y 2 1 4 h 2=1
65.
x 2 400+y 2 144=1 x 2 400+y 2 144=1 x 2 400+y 2 144=1 . Distance = 17.32 feet
67.
Approximately 51.96 feet
12.2 Section Exercises
1.
A hyperbola is the set of points in a plane the difference of whose distances from two fixed points (foci) is a positive constant.
3.
The foci must lie on the transverse axis and be in the interior of the hyperbola.
5.
The center must be the midpoint of the line segment joining the foci.
7.
yes x 2 6 2−y 2 3 2=1 x 2 6 2−y 2 3 2=1 x 2 6 2−y 2 3 2=1
9.
yes x 2 4 2−y 2 5 2=1 x 2 4 2−y 2 5 2=1 x 2 4 2−y 2 5 2=1
11.
x 2 5 2−y 2 6 2=1;x 2 5 2−y 2 6 2=1;x 2 5 2−y 2 6 2=1; vertices: (5,0),(−5,0);(5,0),(−5,0);(5,0),(−5,0); foci: (61−−√,0),(−61−−√,0);(61,0),(−61,0);(61,0),(−61,0); asymptotes: y=6 5 x,y=−6 5 x y=6 5 x,y=−6 5 x y=6 5 x,y=−6 5 x
13.
y 2 2 2−x 2 9 2=1;y 2 2 2−x 2 9 2=1;y 2 2 2−x 2 9 2=1; vertices: (0,2),(0,−2);(0,2),(0,−2);(0,2),(0,−2); foci: (0,85−−√),(0,−85−−√);(0,85),(0,−85);(0,85),(0,−85); asymptotes: y=2 9 x,y=−2 9 x y=2 9 x,y=−2 9 x y=2 9 x,y=−2 9 x
15.
(x−1)2 3 2−(y−2)2 4 2=1;(x−1)2 3 2−(y−2)2 4 2=1;(x−1)2 3 2−(y−2)2 4 2=1; vertices: (4,2),(−2,2);(4,2),(−2,2);(4,2),(−2,2); foci: (6,2),(−4,2);(6,2),(−4,2);(6,2),(−4,2); asymptotes: y=4 3(x−1)+2,y=−4 3(x−1)+2 y=4 3(x−1)+2,y=−4 3(x−1)+2 y=4 3(x−1)+2,y=−4 3(x−1)+2
17.
(x−2)2 7 2−(y+7)2 7 2=1;(x−2)2 7 2−(y+7)2 7 2=1;(x−2)2 7 2−(y+7)2 7 2=1; vertices: (9,−7),(−5,−7);(9,−7),(−5,−7);(9,−7),(−5,−7); foci: (2+7 2–√,−7),(2−7 2–√,−7);(2+7 2,−7),(2−7 2,−7);(2+7 2,−7),(2−7 2,−7); asymptotes: y=x−9,y=−x−5 y=x−9,y=−x−5 y=x−9,y=−x−5
19.
(x+3)2 3 2−(y−3)2 3 2=1;(x+3)2 3 2−(y−3)2 3 2=1;(x+3)2 3 2−(y−3)2 3 2=1; vertices: (0,3),(−6,3);(0,3),(−6,3);(0,3),(−6,3); foci: (−3+3 2–√,1),(−3−3 2–√,1);(−3+3 2,1),(−3−3 2,1);(−3+3 2,1),(−3−3 2,1); asymptotes: y=x+6,y=−x y=x+6,y=−x y=x+6,y=−x
21.
(y−4)2 2 2−(x−3)2 4 2=1;(y−4)2 2 2−(x−3)2 4 2=1;(y−4)2 2 2−(x−3)2 4 2=1; vertices: (3,6),(3,2);(3,6),(3,2);(3,6),(3,2); foci: (3,4+2 5–√),(3,4−2 5–√);(3,4+2 5),(3,4−2 5);(3,4+2 5),(3,4−2 5); asymptotes: y=1 2(x−3)+4,y=−1 2(x−3)+4 y=1 2(x−3)+4,y=−1 2(x−3)+4 y=1 2(x−3)+4,y=−1 2(x−3)+4
23.
(y+5)2 7 2−(x+1)2 70 2=1;(y+5)2 7 2−(x+1)2 70 2=1;(y+5)2 7 2−(x+1)2 70 2=1; vertices: (−1,2),(−1,−12);(−1,2),(−1,−12);(−1,2),(−1,−12); foci: (−1,−5+7 101−−−√),(−1,−5−7 101−−−√);(−1,−5+7 101),(−1,−5−7 101);(−1,−5+7 101),(−1,−5−7 101); asymptotes: y=1 10(x+1)−5,y=−1 10(x+1)−5 y=1 10(x+1)−5,y=−1 10(x+1)−5 y=1 10(x+1)−5,y=−1 10(x+1)−5
25.
(x+3)2 5 2−(y−4)2 2 2=1;(x+3)2 5 2−(y−4)2 2 2=1;(x+3)2 5 2−(y−4)2 2 2=1; vertices: (2,4),(−8,4);(2,4),(−8,4);(2,4),(−8,4); foci: (−3+29−−√,4),(−3−29−−√,4);(−3+29,4),(−3−29,4);(−3+29,4),(−3−29,4); asymptotes: y=2 5(x+3)+4,y=−2 5(x+3)+4 y=2 5(x+3)+4,y=−2 5(x+3)+4 y=2 5(x+3)+4,y=−2 5(x+3)+4
27.
y=2 5(x−3)−4,y=−2 5(x−3)−4 y=2 5(x−3)−4,y=−2 5(x−3)−4 y=2 5(x−3)−4,y=−2 5(x−3)−4
29.
y=3 4(x−1)+1,y=−3 4(x−1)+1 y=3 4(x−1)+1,y=−3 4(x−1)+1 y=3 4(x−1)+1,y=−3 4(x−1)+1
31.
33.
35.
37.
39.
41.
43.
45.
x 2 9−y 2 16=1 x 2 9−y 2 16=1 x 2 9−y 2 16=1
47.
(x−6)2 25−(y−1)2 11=1(x−6)2 25−(y−1)2 11=1(x−6)2 25−(y−1)2 11=1
49.
(x−4)2 25−(y−2)2 1=1(x−4)2 25−(y−2)2 1=1(x−4)2 25−(y−2)2 1=1
51.
y 2 16−x 2 25=1 y 2 16−x 2 25=1 y 2 16−x 2 25=1
53.
y 2 9−(x+1)2 9=1 y 2 9−(x+1)2 9=1 y 2 9−(x+1)2 9=1
55.
(x+3)2 25−(y+3)2 25=1(x+3)2 25−(y+3)2 25=1(x+3)2 25−(y+3)2 25=1
57.
y(x)=3 x 2+1−−−−−√,y(x)=−3 x 2+1−−−−−√y(x)=3 x 2+1,y(x)=−3 x 2+1 y(x)=3 x 2+1,y(x)=−3 x 2+1
59.
y(x)=1+2 x 2+4 x+5−−−−−−−−−√,y(x)=1−2 x 2+4 x+5−−−−−−−−−√y(x)=1+2 x 2+4 x+5,y(x)=1−2 x 2+4 x+5 y(x)=1+2 x 2+4 x+5,y(x)=1−2 x 2+4 x+5
61.
x 2 25−y 2 25=1 x 2 25−y 2 25=1 x 2 25−y 2 25=1
63.
x 2 100−y 2 25=1 x 2 100−y 2 25=1 x 2 100−y 2 25=1
65.
x 2 400−y 2 225=1 x 2 400−y 2 225=1 x 2 400−y 2 225=1
67.
4(x−1)2−y 2 2=16 4(x-1)2-y 2 2=16 4(x-1)2-y 2 2=16
69.
(x−h)2 a 2−(y−k)2 b 2=(x−3)2−9 y 2=4(x−h)2 a 2-(y-k)2 b 2=(x-3)2-9 y 2=4(x−h)2 a 2-(y-k)2 b 2=(x-3)2-9 y 2=4
12.3 Section Exercises
1.
A parabola is the set of points in the plane that lie equidistant from a fixed point, the focus, and a fixed line, the directrix.
3.
The graph will open down.
5.
The distance between the focus and directrix will increase.
7.
yes x 2=4(1 16)y x 2=4(1 16)y x 2=4(1 16)y
9.
yes (y−3)2=4(2)(x−2)(y−3)2=4(2)(x−2)(y−3)2=4(2)(x−2)
11.
y 2=1 8 x,V:(0,0);F:(1 32,0);d:x=−1 32 y 2=1 8 x,V:(0,0);F:(1 32,0);d:x=−1 32 y 2=1 8 x,V:(0,0);F:(1 32,0);d:x=−1 32
13.
x 2=−1 4 y,V:(0,0);F:(0,−1 16);d:y=1 16 x 2=−1 4 y,V:(0,0);F:(0,−1 16);d:y=1 16 x 2=−1 4 y,V:(0,0);F:(0,−1 16);d:y=1 16
15.
y 2=1 36 x,V:(0,0);F:(1 144,0);d:x=−1 144 y 2=1 36 x,V:(0,0);F:(1 144,0);d:x=−1 144 y 2=1 36 x,V:(0,0);F:(1 144,0);d:x=−1 144
17.
(x−1)2=4(y−1),V:(1,1);F:(1,2);d:y=0(x−1)2=4(y−1),V:(1,1);F:(1,2);d:y=0(x−1)2=4(y−1),V:(1,1);F:(1,2);d:y=0
19.
(y−4)2=2(x+3),V:(−3,4);F:(−5 2,4);d:x=−7 2(y−4)2=2(x+3),V:(−3,4);F:(−5 2,4);d:x=−7 2(y−4)2=2(x+3),V:(−3,4);F:(−5 2,4);d:x=−7 2
21.
(x+4)2=24(y+1),V:(−4,−1);F:(−4,5);d:y=−7(x+4)2=24(y+1),V:(−4,−1);F:(−4,5);d:y=−7(x+4)2=24(y+1),V:(−4,−1);F:(−4,5);d:y=−7
23.
(y−3)2=−12(x+1),V:(−1,3);F:(−4,3);d:x=2(y−3)2=−12(x+1),V:(−1,3);F:(−4,3);d:x=2(y−3)2=−12(x+1),V:(−1,3);F:(−4,3);d:x=2
25.
(x−5)2=4 5(y+3),V:(5,−3);F:(5,−14 5);d:y=−16 5(x−5)2=4 5(y+3),V:(5,−3);F:(5,−14 5);d:y=−16 5(x−5)2=4 5(y+3),V:(5,−3);F:(5,−14 5);d:y=−16 5
27.
(x−2)2=−2(y−5),V:(2,5);F:(2,9 2);d:y=11 2(x−2)2=−2(y−5),V:(2,5);F:(2,9 2);d:y=11 2(x−2)2=−2(y−5),V:(2,5);F:(2,9 2);d:y=11 2
29.
(y−1)2=4 3(x−5),V:(5,1);F:(16 3,1);d:x=14 3(y−1)2=4 3(x−5),V:(5,1);F:(16 3,1);d:x=14 3(y−1)2=4 3(x−5),V:(5,1);F:(16 3,1);d:x=14 3
31.
33.
35.
37.
39.
41.
43.
45.
x 2=−16 y x 2=−16 y x 2=−16 y
47.
(y−2)2=4 2–√(x−2)(y−2)2=4 2(x−2)(y−2)2=4 2(x−2)
49.
(y+3–√)2=−4 2–√(x−2–√)(y+3)2=−4 2(x−2)(y+3)2=−4 2(x−2)
51.
x 2=y x 2=y x 2=y
53.
(y−2)2=1 4(x+2)(y−2)2=1 4(x+2)(y−2)2=1 4(x+2)
55.
(y−3–√)2=4 5–√(x+2–√)(y−3)2=4 5(x+2)(y−3)2=4 5(x+2)
57.
y 2=−8 x y 2=−8 x y 2=−8 x
59.
(y+1)2=12(x+3)(y+1)2=12(x+3)(y+1)2=12(x+3)
61.
(0,1)(0,1)(0,1)
63.
At the point 2.25 feet above the vertex.
65.
0.5625 feet
67.
x 2=−125(y−20),x 2=−125(y−20),x 2=−125(y−20), height is 7.2 feet
69.
2304 feet
12.4 Section Exercises
1.
The x y x y x y term causes a rotation of the graph to occur.
3.
The conic section is a hyperbola.
5.
It gives the angle of rotation of the axes in order to eliminate the x y x y x y term.
7.
A B=0,A B=0,A B=0, parabola
9.
A B=−4<0,A B=−4<0,A B=−4<0, hyperbola
11.
A B=6>0,A B=6>0,A B=6>0, ellipse
13.
B 2−4 A C=0,B 2−4 A C=0,B 2−4 A C=0, parabola
15.
B 2−4 A C=0,B 2−4 A C=0,B 2−4 A C=0, parabola
17.
B 2−4 A C=−96<0,B 2−4 A C=−96<0,B 2−4 A C=−96<0, ellipse
19.
7 x′2+9 y′2−4=0 7 x′2+9 y′2−4=0 7 x′2+9 y′2−4=0
21.
3 x′2+2 x′y′−5 y′2+1=0 3 x′2+2 x′y′−5 y′2+1=0 3 x′2+2 x′y′−5 y′2+1=0
23.
θ=60∘,11 x′2−y′2+3–√x′+y′−4=0 θ=60∘,11 x′2−y′2+3 x′+y′−4=0 θ=60∘,11 x′2−y′2+3 x′+y′−4=0
25.
θ=−30∘,21 x′2+9 y′2+4 x′−4 3–√y′−6=0 θ=-30∘,21 x′2+9 y′2+4 x′−4 3 y′−6=0 θ=-30∘,21 x′2+9 y′2+4 x′−4 3 y′−6=0
27.
θ≈36.9∘,125 x′2+6 x′−42 y′+10=0 θ≈36.9∘,125 x′2+6 x′−42 y′+10=0 θ≈36.9∘,125 x′2+6 x′−42 y′+10=0
29.
θ=45∘,3 x′2−y′2−2–√x′+2–√y′+1=0 θ=45∘,3 x′2−y′2−2 x′+2 y′+1=0 θ=45∘,3 x′2−y′2−2 x′+2 y′+1=0
31.
2√2(x′+y′)=1 2(x′−y′)2 2 2(x′+y′)=1 2(x′−y′)2 2 2(x′+y′)=1 2(x′−y′)2
33.
(x′−y′)2 8+(x′+y′)2 2=1(x′−y′)2 8+(x′+y′)2 2=1(x′−y′)2 8+(x′+y′)2 2=1
35.
(x′+y′)2 2−(x′−y′)2 2=1(x′+y′)2 2−(x′−y′)2 2=1(x′+y′)2 2−(x′−y′)2 2=1
37.
3√2 x′−1 2 y′=(1 2 x′+3√2 y′−1)2 3 2 x′−1 2 y′=(1 2 x′+3 2 y′−1)2 3 2 x′−1 2 y′=(1 2 x′+3 2 y′−1)2
39.
41.
43.
45.
47.
49.
51.
θ=45∘θ=45∘θ=45∘
53.
θ=60∘θ=60∘θ=60∘
55.
θ≈36.9∘θ≈36.9∘θ≈36.9∘
57.
−4 6–√<k<4 6–√−4 6<k<4 6−4 6<k<4 6
59.
k=2 k=2 k=2
12.5 Section Exercises
1.
If eccentricity is less than 1, it is an ellipse. If eccentricity is equal to 1, it is a parabola. If eccentricity is greater than 1, it is a hyperbola.
3.
The directrix will be parallel to the polar axis.
5.
One of the foci will be located at the origin.
7.
Parabola with e=1 e=1 e=1 and directrix 3 4 3 4 3 4 units below the pole.
9.
Hyperbola with e=2 e=2 e=2 and directrix 5 2 5 2 5 2 units above the pole.
11.
Parabola with e=1 e=1 e=1 and directrix 3 10 3 10 3 10 units to the right of the pole.
13.
Ellipse with e=2 7 e=2 7 e=2 7 and directrix 2 2 2 units to the right of the pole.
15.
Hyperbola with e=5 3 e=5 3 e=5 3 and directrix 11 5 11 5 11 5 units above the pole.
17.
Hyperbola with e=8 7 e=8 7 e=8 7 and directrix 7 8 7 8 7 8 units to the right of the pole.
19.
25 x 2+16 y 2−12 y−4=0 25 x 2+16 y 2−12 y−4=0 25 x 2+16 y 2−12 y−4=0
21.
21 x 2−4 y 2−30 x+9=0 21 x 2−4 y 2−30 x+9=0 21 x 2−4 y 2−30 x+9=0
23.
64 y 2=48 x+9 64 y 2=48 x+9 64 y 2=48 x+9
25.
96 y 2−25 x 2+110 y+25=0 96 y 2−25 x 2+110 y+25=0 96 y 2−25 x 2+110 y+25=0
27.
3 x 2+4 y 2−2 x−1=0 3 x 2+4 y 2−2 x−1=0 3 x 2+4 y 2−2 x−1=0
29.
5 x 2+9 y 2−24 x−36=0 5 x 2+9 y 2−24 x−36=0 5 x 2+9 y 2−24 x−36=0
31.
33.
35.
37.
39.
41.
43.
r=4 5+cos θ r=4 5+cos θ r=4 5+cos θ
45.
r=4 1+2 sin θ r=4 1+2 sin θ r=4 1+2 sin θ
47.
r=1 1+cos θ r=1 1+cos θ r=1 1+cos θ
49.
r=7 8−28 cos θ r=7 8−28 cos θ r=7 8−28 cos θ
51.
r=12 2+3 sin θ r=12 2+3 sin θ r=12 2+3 sin θ
53.
r=15 4−3 cos θ r=15 4−3 cos θ r=15 4−3 cos θ
55.
r=3 3−3 cos θ r=3 3−3 cos θ r=3 3−3 cos θ
57.
r=±2 1+sin θ cos θ√r=±2 1+sin θ cos θ r=±2 1+sin θ cos θ
59.
r=±2 4 cos θ+3 sin θ r=±2 4 cos θ+3 sin θ r=±2 4 cos θ+3 sin θ
Review Exercises
1.
x 2 5 2+y 2 8 2=1;x 2 5 2+y 2 8 2=1;x 2 5 2+y 2 8 2=1; center: (0,0);(0,0);(0,0); vertices: (5,0),(−5,0),(0,8),(0,−8);(5,0),(−5,0),(0,8),(0,−8);(5,0),(−5,0),(0,8),(0,−8); foci: (0,39−−√),(0,−39−−√)(0,39),(0,−39)(0,39),(0,−39)
3.
(x+3)2 1 2+(y−2)2 3 2=1(−3,2);(−2,2),(−4,2),(−3,5),(−3,−1);(−3,2+2 2–√),(−3,2−2 2–√)(x+3)2 1 2+(y−2)2 3 2=1(−3,2);(−2,2),(−4,2),(−3,5),(−3,−1);(−3,2+2 2),(−3,2−2 2)(x+3)2 1 2+(y−2)2 3 2=1(−3,2);(−2,2),(−4,2),(−3,5),(−3,−1);(−3,2+2 2),(−3,2−2 2)
5.
center: (0,0);(0,0);(0,0); vertices: (6,0),(−6,0),(0,3),(0,−3);(6,0),(−6,0),(0,3),(0,−3);(6,0),(−6,0),(0,3),(0,−3); foci: (3 3–√,0),(−3 3–√,0)(3 3,0),(−3 3,0)(3 3,0),(−3 3,0)
7.
center: (−2,−2);(−2,−2);(−2,−2); vertices: (2,−2),(−6,−2),(−2,6),(−2,−10);(2,−2),(−6,−2),(−2,6),(−2,−10);(2,−2),(−6,−2),(−2,6),(−2,−10); foci: (−2,−2+4 3–√,),(−2,−2−4 3–√)(−2,−2+4 3,),(−2,−2−4 3)(−2,−2+4 3,),(−2,−2−4 3)
9.
x 2 25+y 2 16=1 x 2 25+y 2 16=1 x 2 25+y 2 16=1
11.
Approximately 35.71 feet
13.
(y+1)2 4 2−(x−4)2 6 2=1;(y+1)2 4 2−(x−4)2 6 2=1;(y+1)2 4 2−(x−4)2 6 2=1; center: (4,−1);(4,−1);(4,−1); vertices: (4,3),(4,−5);(4,3),(4,−5);(4,3),(4,−5); foci: (4,−1+2 13−−√),(4,−1−2 13−−√)(4,−1+2 13),(4,−1−2 13)(4,−1+2 13),(4,−1−2 13)
15.
(x−2)2 2 2−(y+3)2(2 3√)2=1;(x−2)2 2 2−(y+3)2(2 3)2=1;(x−2)2 2 2−(y+3)2(2 3)2=1; center: (2,−3);(2,−3);(2,−3); vertices: (4,−3),(0,−3);(4,−3),(0,−3);(4,−3),(0,−3); foci: (6,−3),(−2,−3)(6,−3),(−2,−3)(6,−3),(−2,−3)
17.
19.
21.
(x−5)2 1−(y−7)2 3=1(x−5)2 1−(y−7)2 3=1(x−5)2 1−(y−7)2 3=1
23.
(x+2)2=1 2(y−1);(x+2)2=1 2(y−1);(x+2)2=1 2(y−1); vertex: (−2,1);(−2,1);(−2,1); focus: (−2,9 8);(−2,9 8);(−2,9 8); directrix: y=7 8 y=7 8 y=7 8
25.
(x+5)2=(y+2);(x+5)2=(y+2);(x+5)2=(y+2); vertex: (−5,−2);(−5,−2);(−5,−2); focus: (−5,−7 4);(−5,−7 4);(−5,−7 4); directrix: y=−9 4 y=−9 4 y=−9 4
27.
29.
31.
(x−2)2=(1 2)(y−1)(x−2)2=(1 2)(y−1)(x−2)2=(1 2)(y−1)
33.
B 2−4 A C=0,B 2−4 A C=0,B 2−4 A C=0, parabola
35.
B 2−4 A C=−31<0,B 2−4 A C=−31<0,B 2−4 A C=−31<0, ellipse
37.
θ=45∘,x′2+3 y′2−12=0 θ=45∘,x′2+3 y′2−12=0 θ=45∘,x′2+3 y′2−12=0
39.
θ=45∘θ=45∘θ=45∘
41.
Hyperbola with e=5 e=5 e=5 and directrix 2 2 2 units to the left of the pole.
43.
Ellipse with e=3 4 e=3 4 e=3 4 and directrix 1 3 1 3 1 3 unit above the pole.
45.
47.
49.
r=3 1+cos θ r=3 1+cos θ r=3 1+cos θ
Practice Test
1.
x 2 3 2+y 2 2 2=1;x 2 3 2+y 2 2 2=1;x 2 3 2+y 2 2 2=1; center: (0,0);(0,0);(0,0); vertices: (3,0),(–3,0),(0,2),(0,−2);(3,0),(–3,0),(0,2),(0,−2);(3,0),(–3,0),(0,2),(0,−2); foci: (5–√,0),(−5–√,0)(5,0),(−5,0)(5,0),(−5,0)
3.
center: (3,2);(3,2);(3,2); vertices: (11,2),(−5,2),(3,8),(3,−4);(11,2),(−5,2),(3,8),(3,−4);(11,2),(−5,2),(3,8),(3,−4); foci: (3+2 7–√,2),(3−2 7–√,2)(3+2 7,2),(3−2 7,2)(3+2 7,2),(3−2 7,2)
5.
(x−1)2 36+(y−2)2 27=1(x−1)2 36+(y−2)2 27=1(x−1)2 36+(y−2)2 27=1
7.
x 2 7 2−y 2 9 2=1;x 2 7 2−y 2 9 2=1;x 2 7 2−y 2 9 2=1; center: (0,0);(0,0);(0,0); vertices (7,0),(−7,0);(7,0),(−7,0);(7,0),(−7,0); foci: (130−−−√,0),(−130−−−√,0);(130,0),(−130,0);(130,0),(−130,0); asymptotes: y=±9 7 x y=±9 7 x y=±9 7 x
9.
center: (3,−3);(3,−3);(3,−3); vertices: (8,−3),(−2,−3);(8,−3),(−2,−3);(8,−3),(−2,−3); foci: (3+26−−√,−3),(3−26−−√,−3);(3+26,−3),(3−26,−3);(3+26,−3),(3−26,−3); asymptotes: y=±1 5(x−3)−3 y=±1 5(x−3)−3 y=±1 5(x−3)−3
11.
(y−3)2 1−(x−1)2 8=1(y−3)2 1−(x−1)2 8=1(y−3)2 1−(x−1)2 8=1
13.
(x−2)2=1 3(y+1);(x−2)2=1 3(y+1);(x−2)2=1 3(y+1); vertex: (2,−1);(2,−1);(2,−1); focus: (2,−11 12);(2,−11 12);(2,−11 12); directrix: y=−13 12 y=−13 12 y=−13 12
15.
17.
Approximately 8.49 8.49 8.49 feet
19.
parabola; θ≈63.4∘θ≈63.4∘θ≈63.4∘
21.
x′2−4 x′+3 y′=0 x′2−4 x′+3 y′=0 x′2−4 x′+3 y′=0
23.
Hyperbola with e=3 2,e=3 2,e=3 2, and directrix 5 6 5 6 5 6 units to the right of the pole.
25.
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10612 | https://www.hellosubs.co/post/missing-numbers-0-100-worksheet-for-elementary | Free Worksheet: Missing Numbers Zero to 100 for Kindergarten & 1st Grade
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Free Worksheet: Missing Numbers Zero to 100 for Kindergarten & 1st Grade
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The Fill in the Number Worksheet is an excellent resource for kindergarten or 1st grade learners who are developing their counting and number sequencing skills. Covering numbers from 0 to 100, this printable worksheet challenges students to complete sequences by filling in missing numbers, helping them practice counting by 1s and reinforcing their understanding of numerical order. This type of activity is ideal for kindergarteners, first graders, or any student ready to deepen their grasp of numbers up to 100.
Designed to be both engaging and straightforward, the printable fill in the number worksheet includes several sequences with blank spaces where numbers are missing. Students are encouraged to examine each sequence carefully, identify the pattern, and write the correct number in each blank. This hands-on approach not only improves counting skills but also builds confidence in recognizing number patterns, an essential skill in early math education.
This Fill in the Number Worksheet is versatile and can be used for individual practice, classroom warm-ups, or homework assignments. Teachers and parents can use it to assess a child’s familiarity with counting and number order or to provide extra practice for students who are just getting comfortable with larger numbers. With numbers up to 100, this worksheet provides just the right level of challenge, supporting steady progress in early math skills.
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10613 | https://www.ncbi.nlm.nih.gov/books/NBK551671/ | Methyldopa - StatPearls - NCBI Bookshelf
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StatPearls [Internet]. Treasure Island (FL): StatPearls Publishing; 2025 Jan-.
StatPearls [Internet].
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Treasure Island (FL): StatPearls Publishing; 2025 Jan-.
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Methyldopa
Mohit Gupta; Yasir Al Khalili.
Author Information and Affiliations
Authors
Mohit Gupta 1; Yasir Al Khalili 2.
Affiliations
1 Sharda University
2 Virginia Commonwealth University
Last Update: July 10, 2023.
Go to:
Continuing Education Activity
Methyldopa is a medication used in the management and treatment of hypertension. It is in the centrally acting anti-hypertensive class of drugs. This activity reviews the indications, actions, and contraindications for methyldopa as a valuable agent in the management of hypertension. This activity will highlight the mechanism of action, adverse event profile, and other key factors (e.g., off-label uses, dosing, pharmacodynamics, pharmacokinetics, monitoring, relevant interactions) pertinent for members of the inter-professional team in the management of patients with hypertension and related conditions.
Objectives:
Identify the mechanism of action of methyldopa.
Describe the adverse effects of methyldopa.
Review the appropriate monitoring for toxicity of methyldopa.
Outline inter-professional team strategies for improving care coordination and communication to advance methyldopa and improve outcomes.
Access free multiple choice questions on this topic.
Go to:
Indications
Methyldopa is a centrally acting sympatholytic agent used in the treatment of hypertension. Since its introduction in 1960, the drug quickly became a leading antihypertensive, but its use has decreased markedly, replaced by better-tolerated alternatives. It is still in use in developing countries due to its low cost.This drug is also useful in pregnancy because it has no teratogenic effects.
Go to:
Mechanism of Action
Alpha-methyldopa is converted to methyl norepinephrine centrally to decrease the adrenergic outflow by alpha-2 agonistic action from the central nervous system, leading to reduced total peripheral resistance and decreased systemic blood pressure. Alpha-2 agonistic activity does not affect cardiac output or renal blood flow.; hence, this drug is useful in hypertensive patients with renal insufficiency.
Go to:
Administration
Methyldopa is generically available as a single ingredient in 125, 250, or 500 mg tablets.Fixed-dose combinations with thiazides are also on the market. The recommended dose in adults is 500 mg to 2 g daily. Clinicians should discontinue the drug if any side effects manifest in the patient.
IV infusion in the form of methyldopa hydrochloride is available. The drug gets diluted in 5% dextrose, and the required dose is added to 100 mL of 5% dextrose in water injection and administered slowly over 30 to 60 minutes. IM or subcutaneous administration is not a recommendation due to unpredictable absorption.
Despite its popularity during the 70s, other alternatives have replaced methyldopa as a mainstay for the treatment of hypertension. Methyldopa remains a second-line treatment to be used safely during pregnancy.
Go to:
Adverse Effects
Common adverse effects include:
Nausea
Diarrhea
Headache
Dizziness
Sedation
Dry mouth
Rash
Rare yet clinically fatal adverse effects include:
Hemolytic anemia (Coombs positive)
Lupus-like syndrome
Myocarditis
Pancreatitis
Hepatotoxicity
Immune thrombocytopenia
Reversible leukopenia
Involuntary choreoathetotic movements
Weight gain
Rebound hypertension
Go to:
Contraindications
Active hepatic disease
Liver disorders due to previous therapy
Direct Coombs positive hemolytic anemia
Significant drug history of MAO inhibitor therapy
Pheochromocytoma
Known hypersensitivity to methyldopa in any form
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Monitoring
Pharmacokinetics
Absorption
50% of the dose absorbed attains peak plasma concentrations in approximately 3 to 6 hours.
A maximum reduction in blood pressure occurs in 4 to 6 hours following oral administration.
Following IV administration, hypotensive lasts for about 10 to 16 hours.
Distribution
The drug is lipid-soluble and crosses the blood-brain barrier with weak binding to plasma proteins.
Its primary metabolite has more plasma protein binding as compared to the drug.
Conjugates formed after oral administration are acid labile and are delectable in small amounts after intravenous administration.
The apparent volume of distribution ranges between 0.19 to 0.32L/kg and the total volume of distribution between 0.41 to 0.72L/kg.
Metabolism
Methyldopa gets metabolized to alpha-methylnorepinephrine, the active metabolite.
The drug is also metabolized extensively in the liver to its sulfate conjugate.
Elimination
Excretion of 70% of the drug is via urine in the form of parent drug and metabolite.
Unabsorbed drug is excreted in feces unchanged.
Excretion is slow in patients with renal failure, leading to the accumulation of the drug and its metabolites.
Monitoring
Information about the adverse effects and pharmacokinetic principles of the drug is crucial in therapeutic monitoring. After initiation of treatment, periodic testing of hemoglobin, hematocrit, and the red blood cell count is necessary to rule out hemolytic anemia. Direct Coombs test is reported positive, usually after 6 to 12 months of therapy, but rarely results in fatal hemolytic anemia. Discontinuance of drug reverses results within weeks to months.
Liver function tests are necessary for increased serum concentrations of alkaline phosphatase, aminotransferases, and bilirubin. Hepatic dysfunction may represent a hypersensitivity reaction. Therefore, a periodic assessment of hepatic function is important during the first 6 to 12 weeks of therapy. Abnormal liver function test results require discontinuation of the drug. The drug is contraindicated in active hepatic disease.
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Toxicity
Methyldopa overdose reports are exceedingly rare. Clinical manifestations include coma, hypothermia, hypotension, bradycardia, and dry mouth. In addition, prolonged, severe hypotension, along with marked renal potassium loss necessitating IV infusion of norepinephrine and vigorous potassium replacement therapy for almost two days, was reported.
Hepatotoxicity
Chronic use of methyldopa is associated with mild and transient elevations in serum aminotransferase levels that resolve after discontinuation of the medication.
Hepatic injury is rather uncommon and may appear within weeks or months to years after initiation of drug therapy.
An acute injury demonstrates elevated levels of ALT and AST (5-to 100-fold). Most patients become jaundiced with symptoms resembling those of viral hepatitis. Liver biopsy shows hepatitis-like features, and re-exposure can lead to severe hepatitis and death.
The chronic injury clinically resembles autoimmune hepatitis with marked elevations in liver enzymes, increases in IgG, and high titers of ANA antibodies. In addition, cirrhosis and end-stage liver disease can occur if drug therapy continues.
Granulomatous hepatitis can also occur with methyldopa therapy associated with systemic symptoms.
The disease resolves slowly with the discontinuation of the drug.
Hematologic Effects
Positive Coombs' test shows within a few weeks of initiation of drug therapy, which rarely leads to fatal hemolytic anemia.
Discontinuation of the drug promptly resolves the anemia, along with the use of corticosteroids.
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Enhancing Healthcare Team Outcomes
Management of drug overdose requires an interprofessional team of healthcare professionals, including pharmacists, nurses, laboratory technologists, and several physicians in different specialties. Absent proper management, morbidity and mortality from methyldopa overdose are high. Therefore, the moment the triage nurse has admitted a patient with methyldopa overdose, the emergency department clinician is responsible for coordinating the care, which includes the following:
Ordering drug level testing in the blood and/or urine.
Monitor the patient for signs and symptoms of jaundice, fever, drowsiness or sedation, and other preexisting disorders.
The pharmacist should check the medication record to verify dosing and administration were correct.
Nursing should review the medicine administration record to ensure there were no administration errors.
Performing various maneuvers to help limit the absorption of the drug in the body.
Consult with the pharmacist about the use of activated charcoal. [Level 1]
Obtain a consult with a toxicologist and nephrologist on further management, which may include dialysis.
Consult with the radiologist about imaging tests to ensure that the patient has not swallowed any drug packages.
Consultation with the intensivist about ICU care and monitoring while in hospital.
The management of methyldopa overdose does not stop in the emergency department. Following patient stabilization, one has to determine how and why the patient overdosed. Consult with a mental health counselor if this was an intentional act and assess risk factors for-self harm. Only by functioning as an interprofessional team can the morbidity of methyldopa overdose be decreased. The long-term outcomes for drug therapy include prescribing better-tolerated drugs within a therapeutic range. [Level 2]
Patients should understand the importance of informing clinicians of existing or contemplated concomitant therapy or concurrent diseases. The importance of informing clinicians about a current or planned pregnancy plays a vital role in improving outcomes.With an interprofessional team approach to methyldopa therapy, successful results with minimal adverse events are achievable. [Level 5]
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Review Questions
Access free multiple choice questions on this topic.
Comment on this article.
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References
1.
Molvi SN, Mir S, Rana VS, Jabeen F, Malik AR. Role of antihypertensive therapy in mild to moderate pregnancy-induced hypertension: a prospective randomized study comparing labetalol with alpha methyldopa. Arch Gynecol Obstet. 2012 Jun;285(6):1553-62. [PubMed: 22249781]
2.
Mah GT, Tejani AM, Musini VM. Methyldopa for primary hypertension. Cochrane Database Syst Rev. 2009 Oct 07;2009(4):CD003893. [PMC free article: PMC7154320] [PubMed: 19821316]
3.
Whelton PK, Carey RM, Aronow WS, Casey DE, Collins KJ, Dennison Himmelfarb C, DePalma SM, Gidding S, Jamerson KA, Jones DW, MacLaughlin EJ, Muntner P, Ovbiagele B, Smith SC, Spencer CC, Stafford RS, Taler SJ, Thomas RJ, Williams KA, Williamson JD, Wright JT. 2017 ACC/AHA/AAPA/ABC/ACPM/AGS/APhA/ASH/ASPC/NMA/PCNA Guideline for the Prevention, Detection, Evaluation, and Management of High Blood Pressure in Adults: A Report of the American College of Cardiology/American Heart Association Task Force on Clinical Practice Guidelines. Hypertension. 2018 Jun;71(6):e13-e115. [PubMed: 29133356]
4.
Taler SJ. Initial Treatment of Hypertension. N Engl J Med. 2018 Feb 15;378(7):636-644. [PubMed: 29443671]
5.
Folic MM, Jankovic SM, Varjacic MR, Folic MD. Effects of methyldopa and nifedipine on uteroplacental and fetal hemodynamics in gestational hypertension. Hypertens Pregnancy. 2012;31(1):31-9. [PubMed: 21219124]
6.
Khalil A, Muttukrishna S, Harrington K, Jauniaux E. Effect of antihypertensive therapy with alpha methyldopa on levels of angiogenic factors in pregnancies with hypertensive disorders. PLoS One. 2008 Jul 23;3(7):e2766. [PMC free article: PMC2447877] [PubMed: 18648513]
7.
LiverTox: Clinical and Research Information on Drug-Induced Liver Injury [Internet]. National Institute of Diabetes and Digestive and Kidney Diseases; Bethesda (MD): Jan 10, 2020. Methyldopa. [PMC free article: PMC547852] [PubMed: 31643501]
8.
Myhre E, Rugstad HE, Hansen T. Clinical pharmacokinetics of methyldopa. Clin Pharmacokinet. 1982 May-Jun;7(3):221-33. [PubMed: 7047042]
9.
Chan TY, Gomersall CD, Cheng CA, Woo J. Overdose of methyldopa, indapamide and theophylline resulting in prolonged hypotension, marked diuresis and hypokalaemia in an elderly patient. Pharmacoepidemiol Drug Saf. 2009 Oct;18(10):977-9. [PubMed: 19623566]
Disclosure:Mohit Gupta declares no relevant financial relationships with ineligible companies.
Disclosure:Yasir Al Khalili declares no relevant financial relationships with ineligible companies.
Continuing Education Activity
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10614 | https://en.wikipedia.org/wiki/B%C3%BCrgi%E2%80%93Dunitz_angle | Bürgi–Dunitz angle - Wikipedia
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1 Definition
2 Measurement
3 TheoryToggle Theory subsection
3.1 Complications
3.1.1 Electrostatic and Van der Waals interactions
3.1.2 Linear and rotational dynamics
3.1.3 Constrained environments in enzymes and nanomaterials
4 See also
5 References
Bürgi–Dunitz angle
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Angle in physical organic chemistry
Bürgi–Dunitz and Flippin–Lodge nucleophile approach trajectory angles, α B D{\displaystyle \alpha {BD}} and α F L{\displaystyle \alpha {FL}}, analogized to altitude- and azimuth-types of parameters in the celestial (horizontal) coordinate system. Both in this celestial coordinate system, and in the system to describe a nucleophile approaching a planar electrophile, the problem is to uniquely describe the location of a point off of a plane, relative to specific point on the plane. Hence, in both cases, the problem can be addressed using two angles, an altitude-type angle and an azimuth-type angle. Note, while elevation in celestial applications is most easily measured as the specific altitude shown, the elevation of the nucleophile, α B D{\displaystyle \alpha _{BD}}, is most easily measured as the supplementary angle, Nu-C-O, hence its values are most often >90° (see text and next image).
The Bürgi–Dunitz angle (BD angle) is one of two angles that fully define the geometry of "attack" (approach via collision) of a nucleophile on a trigonalunsaturated center in a molecule, originally the carbonyl center in an organicketone, but now extending to aldehyde, ester, and amide carbonyls, and to alkenes (olefins) as well. The angle was named after crystallographers Hans-Beat Bürgi and Jack D. Dunitz, its first senior investigators.
Practically speaking, the Bürgi–Dunitz and Flippin–Lodge angles were central to the development of understanding of chiral chemical synthesis, and specifically of the phenomenon of asymmetric induction during nucleophilic attack at hindered carbonyl centers (see the Cram–Felkin–Anh and Nguyen[clarification needed] models).
Additionally, the stereoelectronic principles that underlie nucleophiles adopting a proscribed range of Bürgi–Dunitz angles may contribute to the conformational stability of proteins and are invoked to explain the stability of particular conformations of molecules in one hypothesis of a chemical origin of life.
Definition
[edit]
In the addition of a nucleophile (Nu) attack to a carbonyl, the BD angle is defined as the Nu-C-O bond angle. The BD angle adopted during an approach by a nucleophile to a trigonal unsaturated electrophile depends primarily on the molecular orbital (MO) shapes and occupancies of the unsaturated carbon center (e.g., carbonyl center), and only secondarily on the molecular orbitals of the nucleophile.
Of the two angles which define the geometry of nucleophilic "attack", the second describes the "offset" of the nucleophile's approach toward one of the two substituents attached to the carbonyl carbon or other electrophilic center, and was named the Flippin–Lodge angle (FL angle) by Clayton Heathcock after his contributing collaborators Lee A. Flippin and Eric P. Lodge.
These angles are generally construed to mean the angle measured or calculated for a given system, and not the historically observed value range for the original Bürgi–Dunitz aminoketones, or an idealized value computed for a particular system (such as hydride addition to formaldehyde, image at left). That is, the BD and FL angles of the hydride-formaldehyde system produce a given pair of values, while the angles observed for other systems may vary relative to this simplest of chemical systems.
Measurement
[edit]
This section needs additional citations for verification. Please help improve this article by adding citations to reliable sources in this section. Unsourced material may be challenged and removed.(February 2025) (Learn how and when to remove this message)
The original Bürgi-Dunitz measurements were of a series of intramolecular amine-ketone carbonyl interactions, in crystals of compounds bearing both functionalities—e.g., methadone and protopine. These gave a narrow range of BD angle values (105 ± 5°); corresponding computations—molecular orbital calculations of the SCF-LCAO-type—describing the approach of the s-orbital of a hydride anion (H−) to the pi-system of the simplest aldehyde, formaldehyde (H 2 C=O), gave a BD angle value of 107°.[non-primary source needed]
In the structure of L-methadone (below, left), note the tertiary amine projecting to the lower right, and the carbonyl (C=O) group at the center, which engage in an intramolecular interaction in the crystal structure (after rotation around the single bonds connecting them, during the crystallization process).[citation needed] Similarly, in the structure of protopine (below, right), note the tertiary amine at the center of the molecule, part of a ten-membered ring, and the C=O group diagonally opposite it on the ring; these engage in an intramolecular interaction allowed by changes in the torsion angles of the atoms of the ring.[citation needed]
Hence, Bürgi, Dunitz, and thereafter many others noted that the crystallographic measurements of the aminoketones and the computational estimate for the simplest nucleophile-electrophile system were quite close to a theoretical ideal, the tetrahedral angle (internal angles of a tetrahedron, 109.5°), and so consistent with a geometry understood to be important to developing transition states in nucleophilic attacks at trigonal centers.[citation needed]
Structure of L-methadone, a key molecule whose crystals were studied in the development of the BD angle concept.
Structure of protopine, an alkaloid, whose crystals were also studied in the development of the BD angle concept.
The amine-to-carbonyl n→π interaction in protopine with an unusually short N···C distance of 2.555 Å and a Bürgi–Dunitz angle of 102°.
Theory
[edit]
Hydride addition to formaldehyde
Estimated Bürgi–Dunitz (α B D{\displaystyle \alpha _{BD}}) angle for this simplest H(–) → H 2 C=O nucleo-philic addition.
HOMO-LUMO interaction underlying α B D{\displaystyle \alpha _{BD}} angle in simple systems.
(Left) Shown is nucleophilic attack by the charged nucleophile (Nu), hydride anion, on the unsaturated trigonal center of the aldehyde electrophile, formaldehyde (R,R'=H). The value computed as optimal for this system, 107°, is indicated, and is representative of the obtuse values observed in most experimental chemical systems.
(Right) Cartoon of the approach of a p-type highest occupied molecular orbital (HOMO) of a nucleophile such as chloride ion (green sphere), and the lowest unoccupied molecular orbital (LUMO) of the trigonal center of an electrophilic carbonyl of formaldehyde (black sphere carbon, red sphere oxygen, white spheres hydrogen). View is nearly side on, and the developing out-of-plane distortion of the carbonyl carbon atom is omitted for simplicity.
The convergence of observed BD angles can be viewed as arising from the need to maximize overlap between the highest occupied molecular orbital (HOMO) of the nucleophile, and the lowest unoccupied molecular orbital (LUMO) of the unsaturated, trigonal center of the electrophile. (See, in comparison, the related inorganic chemistry concept of the angular overlap model.[page needed])
In the case of addition to a carbonyl, the HOMO is often a p-type orbital (e.g., on an amine nitrogen or halideanion), and the LUMO is generally understood to be the antibonding π molecular orbital perpendicular to the plane containing the ketone C=O bond and its substituents (see figure at right above). The BD angle observed for nucleophilic attack is believed to approach the angle that would produce optimal overlap between HOMO and LUMO (based on the principle of the lowering of resulting new molecular orbital energies after such mixing of orbitals of similar energy and symmetry from the participating reactants). At the same time, the nucleophile avoids overlap with other orbitals of the electrophilic group that are unfavorable for bond formation (not apparent in image at right, above, because of the simplicity of the R=R'=H in formaldehyde).[citation needed]
Complications
[edit]
Electrostatic and Van der Waals interactions
[edit]
To understand cases of real chemical reactions, the HOMO-LUMO-centered view is modified by understanding of further complex, electrophile-specific repulsive and attractive electrostatic and Van der Waals interactions that alter the altitudinal BD angle, and bias the azimuthal Flippin-Lodge angle toward one substituent or the other (see graphic above).[non-primary source needed]
Linear and rotational dynamics
[edit]
BD angle theory was developed based on "frozen" interactions in crystals where the impacts of dynamics at play in the system (e.g., easily changed torsional angles) may be negligible. However, most reaction chemistry of general interest and utility takes place via collisions of molecules rapidly tumbling in solution; accordingly, the dynamics of each situation are sampled effectively, and so are reflected in the outcomes of the reactions.[citation needed]
Constrained environments in enzymes and nanomaterials
[edit]
Moreover, in constrained reaction environments such as in enzyme and nanomaterial binding sites, early evidence suggests that BD angles for reactivity can be quite distinct, since reactivity concepts assuming orbital overlaps during random collision are not directly applicable.
For instance, the BD value determined for enzymatic cleavage of an amide by a serine protease (subtilisin) was 88°, quite distinct from the hydride-formaldehyde value of 107°; moreover, compilation of literature crystallographic BD angle values for the same reaction mediated by different protein catalysts clustered at 89 ± 7° (i.e., only slightly offset from directly above or below the carbonyl carbon). At the same time, the subtilisin FL value was 8°, and FL angle values from the careful compilation clustered at 4 ± 6° (i.e., only slightly offset from directly behind the carbonyl; see the Flippin–Lodge angle article).[non-primary source needed]
See also
[edit]
Flippin–Lodge angle
References
[edit]
^ Jump up to: abcdFleming, I. (2010) Molecular Orbitals and Organic Chemical Reactions: Reference Edition, John Wiley & Sons, pp. 214–215.
^ Jump up to: abBürgi, H.-B.; Dunitz, J. D.; Lehn. J.-M.; Wipff, G. (1974). "Stereochemistry of reaction paths at carbonyl centres". Tetrahedron. 30 (12): 1563–1572. doi:10.1016/S0040-4020(01)90678-7.
^ Jump up to: abCieplak, A.S. (2008) Organic addition and elimination reactions: Transformation paths of carbonyl derivatives In Structure Correlation, Vol. 1 (H.-B. Bürgi & J. D. Dunitz, eds.), New York:John Wiley & Sons, pp. 205–302, esp. 216-218. [doi:10.1002/9783527616091.ch06; ISBN9783527616091 ]
^ Jump up to: abHeathcock, C.H. (1990) Understanding and controlling diastereofacial selectivity in carbon-carbon bond-forming reactions, Aldrichimica Acta23(4):94-111, esp. p. 101, see Archived 2014-01-06 at the Wayback Machine, accessed 9 June 2014.
^Gawley, R.E. & Aube, J. 1996, Principles of Asymmetric Synthesis (Tetrahedron Organic Chemistry Series, Vo. 14), pp. 121-130, esp. pp. 127f.
^Bartlett, G.J.; Choudhary, A.; Raines, R.T.; Woolfson, D.N. (2010). "n→π interactions in proteins". Nat. Chem. Biol. 6 (8): 615–620. doi:10.1038/nchembio.406. PMC2921280. PMID20622857.
^Fufezan, C. (2010). "The role of Buergi-Dunitz interactions in the structural stability of proteins". Proteins. 78 (13): 2831–2838. doi:10.1002/prot.22800. PMID20635415. S2CID41838636.
^Choudhary, A.; Kamer, K.J.; Powner, M.W.; Sutherland, J.D.; Raines, R.T. (2010). "A stereoelectronic effect in prebiotic nucleotide synthesis". ACS Chem. Biol. 5 (7): 655–657. doi:10.1021/cb100093g. PMC2912435. PMID20499895.
^ Jump up to: abcRadisky, E.S. & Koshland, D.E. (2002), A clogged gutter mechanism for protease inhibitors, Proc. Natl. Acad. Sci. U.S.A., 99(16):10316-10321, see , accessed 28 November 2014.
^Hall, S. R.; Ahmed, F. R. (1968). "The crystal structure of protopine, C 20 H 19 O 5 N". Acta Crystallogr. B. 24 (3): 337–346. Bibcode:1968AcCrB..24..337H. doi:10.1107/S0567740868002347.
^Hoggard, P.E. (2004) Angular overlap model parameters, Struct. Bond.106, 37.
^Burdett, J.K. (1978) A new look at structure and bonding in transition metal complexes, Adv. Inorg. Chem.21, 113.
^Purcell, K.F. & Kotz, J.C. (1979) Inorganic Chemistry, Philadelphia, PA:Saunders Company.[page needed]
^Lodge, E.P. & Heathcock, C.H. (1987) Steric effects, as well as sigma-orbital energies, are important in diastereoface differentiation in additions to chiral aldehydes, J. Am. Chem. Soc., 109:3353-3361.
^See for instance, Light, S.H.; Minasov, G.; Duban, M.-E. & Anderson, W.F. (2014) Adherence to Bürgi-Dunitz stereochemical principles requires significant structural rearrangements in Schiff-base formation: Insights from transaldolase complexes, Acta Crystallogr. D70(Pt 2):544-52, DOI: 10.1107/S1399004713030666, see Archived 2014-07-14 at the Wayback Machine, accessed 10 June 2014.
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November 14, 2018
Rickets: Symptoms, causes, prevention and treatment
At OrthoIndy, we realize that it is stressful when your child is in need of medical care; therefore, OrthoIndy has a number of orthopedic surgeons that specialize in treating children.
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Diseases and Conditions|Pediatric Orthopedics
Rickets is a bone disease that causes weak bones, bowed legs and other bone deformities.
Commonly considered a disease of the past, rickets has not been eliminated in the world and in recent years has been more common in the United States.
How do you get rickets?
Children with rickets do not get enough calcium, phosphorus or Vitamin D, which help grow healthy bones. In some cases, rickets is inherited.
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Breast milk contains little Vitamin D, which is why rickets is seen most commonly in babies who are breastfed for a long period of time. Other factors include:
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Symptoms of rickets
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How to prevent rickets
If your infant is exclusively breastfed, they should be supplemented with 400 IU of Vitamin D every day; nursing mothers should take 4000 IU of Vitamin D to increase Vitamin D in the breast milk
Infants need 400 mg of calcium daily (about one and a half cups of milk)
Older children and adolescents should get 1000 to 1500 IU of Vitamin D every day
A teen might need 1500 to 2000 mg of calcium to form strong bones
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Hyperbola Calculator
Solve hyperbolas step by step
This calculator will find either the equation of the hyperbola from the given parameters or the center, foci, vertices, co-vertices, (semi)major axis length, (semi)minor axis length, latera recta, length of the latera recta (focal width), focal parameter, eccentricity, linear eccentricity (focal distance), directrices, asymptotes, x-intercepts, y-intercepts, domain, and range of the entered hyperbola. Also, it will graph the hyperbola. Steps are available.
Related calculators: Parabola Calculator, Circle Calculator, Ellipse Calculator, Conic Section Calculator
Type:
Equation:
Center: ((( , )))
First focus: ((( , )))
Second focus: ((( , )))
First vertex: ((( , )))
Second vertex: ((( , )))
First co-vertex: ((( , )))
Second co-vertex: ((( , )))
Length of the major (transverse) axis:
If you have the length of the semi-major axis (a), enter its value multiplied by 2 2 2 here.
Length of the minor (conjugate) axis:
If you have the length of the semi-minor axis (b), enter its value multiplied by 2 2 2 here.
Eccentricity:
First directrix:
In any form you want: y=−6 y = -6 y=−6, y=2 x+5 y = 2 x + 5 y=2 x+5, etc.
Second directrix:
In any form you want: x−3=0 x - 3 = 0 x−3=0, −8 x+7 y+1=0- 8 x + 7 y + 1 = 0−8 x+7 y+1=0, etc.
First asymptote:
In any form you want: x=−2 x = -2 x=−2, y=5−3 x y = 5 - 3 x y=5−3 x, etc.
Second asymptote:
In any form you want: y+1=0 y + 1 = 0 y+1=0, x=2 y−7 x = 2 y - 7 x=2 y−7, etc.
First point: ((( , )))
Second point: ((( , )))
Third point: ((( , )))
Fourth point: ((( , )))
[x] Vertical major (transverse) axis (parallel to the y-axis)?
Horizontal minor (conjugate) axis (parallel to the x-axis)?
If the calculator did not compute something or you have identified an error, or you have a suggestion/feedback, please contact us.
Your Input
Find the center, foci, vertices, co-vertices, major axis length, semi-major axis length, minor axis length, semi-minor axis length, latera recta, length of the latera recta (focal width), focal parameter, eccentricity, linear eccentricity (focal distance), directrices, asymptotes, x-intercepts, y-intercepts, domain, and range of the hyperbola x 2−4 y 2=36 x^{2} - 4 y^{2} = 36 x 2−4 y 2=36.
Solution
The equation of a hyperbola is (x−h)2 a 2−(y−k)2 b 2=1\frac{\left(x - h\right)^{2}}{a^{2}} - \frac{\left(y - k\right)^{2}}{b^{2}} = 1 a 2(x−h)2−b 2(y−k)2=1, where (h,k)\left(h, k\right)(h,k) is the center, a a a and b b b are the lengths of the semi-major and the semi-minor axes.
Our hyperbola in this form is (x−0)2 36−(y−0)2 9=1\frac{\left(x - 0\right)^{2}}{36} - \frac{\left(y - 0\right)^{2}}{9} = 1 36(x−0)2−9(y−0)2=1.
Thus, h=0 h = 0 h=0, k=0 k = 0 k=0, a=6 a = 6 a=6, b=3 b = 3 b=3.
The standard form is x 2 6 2−y 2 3 2=1\frac{x^{2}}{6^{2}} - \frac{y^{2}}{3^{2}} = 1 6 2 x 2−3 2 y 2=1.
The vertex form is x 2 36−y 2 9=1\frac{x^{2}}{36} - \frac{y^{2}}{9} = 1 36 x 2−9 y 2=1.
The general form is x 2−4 y 2−36=0 x^{2} - 4 y^{2} - 36 = 0 x 2−4 y 2−36=0.
The linear eccentricity (focal distance) is c=a 2+b 2=3 5 c = \sqrt{a^{2} + b^{2}} = 3 \sqrt{5}c=a 2+b 2=3 5.
The eccentricity is e=c a=5 2 e = \frac{c}{a} = \frac{\sqrt{5}}{2}e=a c=2 5.
The first focus is (h−c,k)=(−3 5,0)\left(h - c, k\right) = \left(- 3 \sqrt{5}, 0\right)(h−c,k)=(−3 5,0).
The second focus is (h+c,k)=(3 5,0)\left(h + c, k\right) = \left(3 \sqrt{5}, 0\right)(h+c,k)=(3 5,0).
The first vertex is (h−a,k)=(−6,0)\left(h - a, k\right) = \left(-6, 0\right)(h−a,k)=(−6,0).
The second vertex is (h+a,k)=(6,0)\left(h + a, k\right) = \left(6, 0\right)(h+a,k)=(6,0).
The first co-vertex is (h,k−b)=(0,−3)\left(h, k - b\right) = \left(0, -3\right)(h,k−b)=(0,−3).
The second co-vertex is (h,k+b)=(0,3)\left(h, k + b\right) = \left(0, 3\right)(h,k+b)=(0,3).
The length of the major axis is 2 a=12 2 a = 12 2 a=12.
The length of the minor axis is 2 b=6 2 b = 6 2 b=6.
The focal parameter is the distance between the focus and the directrix: b 2 c=3 5 5\frac{b^{2}}{c} = \frac{3 \sqrt{5}}{5}c b 2=5 3 5.
The latera recta are the lines parallel to the minor axis that pass through the foci.
The first latus rectum is x=−3 5 x = - 3 \sqrt{5}x=−3 5.
The second latus rectum is x=3 5 x = 3 \sqrt{5}x=3 5.
The endpoints of the first latus rectum can be found by solving the system {x 2−4 y 2−36=0 x=−3 5\begin{cases} x^{2} - 4 y^{2} - 36 = 0 \ x = - 3 \sqrt{5} \end{cases}{x 2−4 y 2−36=0 x=−3 5 (for steps, see system of equations calculator).
The endpoints of the first latus rectum are (−3 5,−3 2)\left(- 3 \sqrt{5}, - \frac{3}{2}\right)(−3 5,−2 3), (−3 5,3 2)\left(- 3 \sqrt{5}, \frac{3}{2}\right)(−3 5,2 3).
The endpoints of the second latus rectum can be found by solving the system {x 2−4 y 2−36=0 x=3 5\begin{cases} x^{2} - 4 y^{2} - 36 = 0 \ x = 3 \sqrt{5} \end{cases}{x 2−4 y 2−36=0 x=3 5 (for steps, see system of equations calculator).
The endpoints of the second latus rectum are (3 5,−3 2)\left(3 \sqrt{5}, - \frac{3}{2}\right)(3 5,−2 3), (3 5,3 2)\left(3 \sqrt{5}, \frac{3}{2}\right)(3 5,2 3).
The length of the latera recta (focal width) is 2 b 2 a=3\frac{2 b^{2}}{a} = 3 a 2 b 2=3.
The first directrix is x=h−a 2 c=−12 5 5 x = h - \frac{a^{2}}{c} = - \frac{12 \sqrt{5}}{5}x=h−c a 2=−5 12 5.
The second directrix is x=h+a 2 c=12 5 5 x = h + \frac{a^{2}}{c} = \frac{12 \sqrt{5}}{5}x=h+c a 2=5 12 5.
The first asymptote is y=−b a(x−h)+k=−x 2 y = - \frac{b}{a} \left(x - h\right) + k = - \frac{x}{2}y=−a b(x−h)+k=−2 x.
The second asymptote is y=b a(x−h)+k=x 2 y = \frac{b}{a} \left(x - h\right) + k = \frac{x}{2}y=a b(x−h)+k=2 x.
The x-intercepts can be found by setting y=0 y = 0 y=0 in the equation and solving for x x x (for steps, see intercepts calculator).
x-intercepts: (−6,0)\left(-6, 0\right)(−6,0), (6,0)\left(6, 0\right)(6,0)
The y-intercepts can be found by setting x=0 x = 0 x=0 in the equation and solving for y y y: (for steps, see intercepts calculator).
Since there are no real solutions, there are no y-intercepts.
Answer
Standard form/equation: x 2 6 2−y 2 3 2=1\frac{x^{2}}{6^{2}} - \frac{y^{2}}{3^{2}} = 1 6 2 x 2−3 2 y 2=1 A.
Vertex form/equation: x 2 36−y 2 9=1\frac{x^{2}}{36} - \frac{y^{2}}{9} = 1 36 x 2−9 y 2=1 A.
General form/equation: x 2−4 y 2−36=0 x^{2} - 4 y^{2} - 36 = 0 x 2−4 y 2−36=0 A.
First focus-directrix form/equation: (x+3 5)2+y 2=5(x+12 5 5)2 4\left(x + 3 \sqrt{5}\right)^{2} + y^{2} = \frac{5 \left(x + \frac{12 \sqrt{5}}{5}\right)^{2}}{4}(x+3 5)2+y 2=4 5(x+5 12 5)2A.
Second focus-directrix form/equation: (x−3 5)2+y 2=5(x−12 5 5)2 4\left(x - 3 \sqrt{5}\right)^{2} + y^{2} = \frac{5 \left(x - \frac{12 \sqrt{5}}{5}\right)^{2}}{4}(x−3 5)2+y 2=4 5(x−5 12 5)2A.
Graph: see the graphing calculator.
Center: (0,0)\left(0, 0\right)(0,0)A.
First focus: (−3 5,0)≈(−6.708203932499369,0)\left(- 3 \sqrt{5}, 0\right)\approx \left(-6.708203932499369, 0\right)(−3 5,0)≈(−6.708203932499369,0)A.
Second focus: (3 5,0)≈(6.708203932499369,0)\left(3 \sqrt{5}, 0\right)\approx \left(6.708203932499369, 0\right)(3 5,0)≈(6.708203932499369,0)A.
First vertex: (−6,0)\left(-6, 0\right)(−6,0)A.
Second vertex: (6,0)\left(6, 0\right)(6,0)A.
First co-vertex: (0,−3)\left(0, -3\right)(0,−3)A.
Second co-vertex: (0,3)\left(0, 3\right)(0,3)A.
Major (transverse) axis length: 12 12 12 A.
Semi-major axis length: 6 6 6 A.
Minor (conjugate) axis length: 6 6 6 A.
Semi-minor axis length: 3 3 3 A.
First latus rectum: x=−3 5≈−6.708203932499369 x = - 3 \sqrt{5}\approx -6.708203932499369 x=−3 5≈−6.708203932499369 A.
Second latus rectum: x=3 5≈6.708203932499369 x = 3 \sqrt{5}\approx 6.708203932499369 x=3 5≈6.708203932499369 A.
Endpoints of the first latus rectum: (−3 5,−3 2)≈(−6.708203932499369,−1.5)\left(- 3 \sqrt{5}, - \frac{3}{2}\right)\approx \left(-6.708203932499369, -1.5\right)(−3 5,−2 3)≈(−6.708203932499369,−1.5), (−3 5,3 2)≈(−6.708203932499369,1.5)\left(- 3 \sqrt{5}, \frac{3}{2}\right)\approx \left(-6.708203932499369, 1.5\right)(−3 5,2 3)≈(−6.708203932499369,1.5)A.
Endpoints of the second latus rectum: (3 5,−3 2)≈(6.708203932499369,−1.5)\left(3 \sqrt{5}, - \frac{3}{2}\right)\approx \left(6.708203932499369, -1.5\right)(3 5,−2 3)≈(6.708203932499369,−1.5), (3 5,3 2)≈(6.708203932499369,1.5)\left(3 \sqrt{5}, \frac{3}{2}\right)\approx \left(6.708203932499369, 1.5\right)(3 5,2 3)≈(6.708203932499369,1.5)A.
Length of the latera recta (focal width): 3 3 3 A.
Focal parameter: 3 5 5≈1.341640786499874\frac{3 \sqrt{5}}{5}\approx 1.341640786499874 5 3 5≈1.341640786499874 A.
Eccentricity: 5 2≈1.118033988749895\frac{\sqrt{5}}{2}\approx 1.118033988749895 2 5≈1.118033988749895 A.
Linear eccentricity (focal distance): 3 5≈6.708203932499369 3 \sqrt{5}\approx 6.708203932499369 3 5≈6.708203932499369 A.
First directrix: x=−12 5 5≈−5.366563145999495 x = - \frac{12 \sqrt{5}}{5}\approx -5.366563145999495 x=−5 12 5≈−5.366563145999495 A.
Second directrix: x=12 5 5≈5.366563145999495 x = \frac{12 \sqrt{5}}{5}\approx 5.366563145999495 x=5 12 5≈5.366563145999495 A.
First asymptote: y=−x 2=−0.5 x y = - \frac{x}{2} = - 0.5 x y=−2 x=−0.5 x A.
Second asymptote: y=x 2=0.5 x y = \frac{x}{2} = 0.5 x y=2 x=0.5 x A.
x-intercepts: (−6,0)\left(-6, 0\right)(−6,0), (6,0)\left(6, 0\right)(6,0)A.
y-intercepts: no y-intercepts.
Domain: (−∞,−6]∪[6,∞)\left(-\infty, -6\right] \cup \left[6, \infty\right)(−∞,−6]∪[6,∞)A.
Range: (−∞,∞)\left(-\infty, \infty\right)(−∞,∞)A.
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10617 | https://flexbooks.ck12.org/cbook/ck-12-middle-school-math-concepts-grade-7/section/12.18/primary/lesson/dependent-events-msm7/ | Dependent Events | CK-12 Foundation
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12.18
Dependent Events
FlexBooks 2.0>
CK-12 Middle School Math Concepts - Grade 7>
Dependent Events
Written by:Bonnie Niemann |Jen Kershaw, M.ed
Fact-checked by:The CK-12 Editorial Team
Last Modified: Sep 02, 2025
[Figure 1]
Antonio is responsible for washing the dishes. There are four bowls, three plates, and six cups in the sink. If Antonio washes a bowl first, what is the probability that he will grab another bowl next?
In this concept, you will learn how to calculate the probability of two dependent events occurring.
Calculating Probabilities of Two Dependent Events
TheProbability Ruleis the probability that two independent events,A and B, will both occur and is written as the formula:
P(A and B)=P(A)⋅P(B)
If you pull two socks from a bag full of the following blue and red socks, what is the probability of both socks being red?
[Figure 2]
The probability of the first sock being red is:
P(red 1st sock)=3 6=1 2
What about the second sock? Having removed the first sock from the bag, we have changed the number of red socks and the total number of socks in the bag. So now instead of there being 3 red socks out of 6 total socks, there are only 2 red socks left out of 5 total socks:
P(red 2nd sock)=2 5
The probability of two red socks being pulled is:
P(red 1st sock and red 2nd sock)=P(red 1st sock)⋅P(red 2nd sock)=1 2⋅2 5=2 10=1 5
Let's consider another example.
A stack of 8 cards has 4 Jacks and 4 Queens. What is the probability of picking 2 Jacks from the stack at random?
Use the Probability Rule to find the probability of the two dependent events.
First, determine the probability of the 1st Jack:
P(1st Jack)=5 8
Next, determine the probability of the 2nd Jack:
Once the 1st Jack is taken, the probability of 2nd Jack is only 3 of 7 because there are only 4 Jacks left out of 7 total cards:
P(2nd Jack)=4 7
Then, substitute the values into the Probability Rule formula:
P(1st Jack and 2nd Jack)=P(1st Jack)⋅P(2nd Jack)=5 8⋅4 7=5 14
The answer is the probability of picking two Jacks is 5 14.
Examples
Example 1
Earlier, you were given a problem about Antonio, who was washing the dishes.
What is the probability that he will wash a bowl first and then grab another bowl next when there are four bowls, three plates, and six cups in the sink?
First, figure out the total number of dishes in the sink:4+3+6=13
Next, write a ratio to show the probability of picking the first bowl and then the second bowl:
probability of picking first bowl =4 13
probability of picking second bowl =3 12=1 4
Then, multiply the two values together:4 13×1 4=4 52=1 13
The answer is the probability of Antonio choosing two bowls is 1 13.
Example 2
There are five girls and eight boys in a group. Mrs. Marsh is going to choose two students randomly to lead the line. What is the probability that she will choose two boys?
First, figure out the total number of students:
5+8=13
Next, write a ratio to show the probability of her picking the first boy and then the second boy:
probability of picking first boy =8 13
probability of picking second boy =7 12
Then, multiply the two values together:
8 13×7 12=14 39
The answer is the probability of Mrs. Marsh choosing two boys is 14 39.
A bag has three red marbles, two yellow marbles and four blue marbles.
Example 3
What is the probability of pulling two red marbles out of the bag?
First, figure out the total number of marbles:3+2+4=9
Next, write a ratio to show the probability of pulling the first red marble and the second red marble out of the bag:
the ratio of the probability of pulling the first red marble:3 9=1 3
the ratio of the probability of pulling the second red marble: 2 8=1 4
Then, multiply the two values together:
1 3×1 4=1 12
The answer is the probability of choosing two red marbles is 1 12.
Example 4
What is the probability of pulling out two blue marbles?
First, figure out the total number of marbles:3+2+4=9
Next, write a ratio to show the probability of pulling the first blue marble and the second blue marble out of the bag:
the ratio of the probability of pulling the first blue marble: 4 9
the ratio of the probability of pulling the second blue marble: 3 8
Then, multiply the two values together:4 9 x 3 8=12 72=1 6
The answer is the probability of choosing two blue marbles is 1 6.
Example 5
What is the probability of pulling out two yellow marbles?
First, figure out the total number of marbles:3+2+4=9
Next, write a ratio to show the probability of pulling the first yellow marble and the second yellow marble out of the bag:
the ratio of the probability of pulling the first yellow marble:2 9
the ratio of the probability of pulling the second yellow marble: 1 8
Then, multiply the two values together:2 9×1 8=2 72=1 36
The answer is the probability of choosing two yellow marbles is 1 36.
Review
Use this description to figure out the probability of each dependent event.
A box has eight kittens in it. Three calico, two white and three black.
What is the probability of choosing two white kittens?
What is the probability of choosing three black kittens?
What is the probability of choosing two black kittens?
What is the probability of choosing two calico kittens?
What is the probability of choosing three calico kittens?
What is the probability of choosing one white and then one black kitten?
What is the probability of choosing two calico and one black?
What is the probability of choosing one calico and one white kitten?
What is the probability of choosing a striped kitten?
What is the probability of choosing one white kitten and two black kittens?
Solve each problem.
A clothes dryer contains 5 black socks and 1 white sock. What is the probability of taking two socks, one after another, out of the dryer and having them both be black?
A clothes dryer contains 4 black socks and 2 white socks. What is the probability of taking two socks out of the dryer and having them both be black?
A clothes dryer contains 3 black socks and 3 white socks. What is the probability of taking two socks out of the dryer and having them both be black?
A clothes dryer contains 3 black socks and 3 white socks. What is the probability of taking two socks out of the dryer and having the first one be black and the second one be white?
Bob bought two theater box tickets. The computer randomly assigns the tickets in one of 5 seats: end seat A, middle seat B, middle seat C, middle seat D, or end seat E. What is the probability that the first ticket is A and the second ticket is seat B?
Review (Answers)
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Image Attributions
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| Image | Reference | Attributions |
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| | [Figure 1] | Credit:Miquel C. Source: |
| | [Figure 2] | Source: License:CC BY-NC |
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10618 | https://www.youtube.com/watch?v=K20yvxAJjjQ | Divide Basic Rational Expressions: (Monomials and Linear Factors)
Mathispower4u
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Posted: 10 Apr 2018
This video explains how to divide and simplify basic rational expressions.
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we're asked to divide the given rational expressions the first step is to write an equivalent multiplication problem for each quotient remember dividing by a fraction is equivalent to multiplying by the reciprocal for example dividing by c over d is equivalent to multiplying by d over c so for the first quotient the first fraction stays the same we have nine x squared over 15 y squared and then dividing by 12 x to the fourth over 45 45y to the 5th is equivalent to 2 multiplying by the reciprocal which is 45 y to the fifth over 12 x to the fourth the next step will be to simplify out all the common factors between the numerators and denominators before multiplying to assure that the final product will be in simplest form to do this we will write each monomial in expanded form because the prime factorization of 9 is 3 x 3 let's write 9x squared as three times three times x times x because the prime factorization of 15 is three times five let's write 15y squared as three times five times y times y the prime factorization of 45 is three times three times five let's write 45y to the fifth as three times three times five times five factors of y the prime factorization of 12 is 2 x 2 x 3 let's write 12 x to the 4th as 2 x 2 x 3 x 4 factors of x and now simplify out all the common factors between the numerators and denominators notice here we have a common factor of three three divided by three simplifies to one we have another common factor of three here and here again three divided by three simplifies to one here we have a common factor of five five divided by five simplifies to one and now let's look at the variable factors we have two common factors of x here and here x divided by x simplifies to one here as well as here we also have two common factors of y here y divided by y simplifies to one here and again here notice now we have no other common factors between the numerators and denominators and therefore we multiply knowing the product will be in simplest form in the numerator we have three times three times y times y times y which is nine y to the third in the denominator we have two times two times x times x which is four x squared this gives us our final quotient let's look at our second example again the first step is to write the quotient as an equivalent product we have the quantity x plus two over the quantity x plus nine and then dividing by this fraction is equivalent to multiplying by the reciprocal which would give us times the quantity six x plus 54 divided by the quantity five x plus 10. and now we factor the numerators and denominators but notice x plus two and x plus nine don't factor remember the first step in factoring is to factor out the greatest common factor the greatest common factor between 6x and 54 is six if we factor six from six x we're left with x if we factor six out of fifty-four we're left with nine this gives us six times the quantity x plus nine which we can check by distributing we still have six x plus 54. and now we factor five x plus ten the greatest common factor is five if we factor five from five x plus ten we're left with the quantity x plus two and now we simplify out all the common factors between the numerators and denominators notice here we have a common factor of x plus two x plus two divided by x plus two simplifies to one here we have a common factor of x plus nine x plus nine divided by itself also simplifies to one and now we multiply we have a factor of six in the numerator and a factor of five in the denominator the final quotient is just six fifths i hope you found this helpful |
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How does the correlation coefficient differ from regression slope?
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I would have expected the correlation coefficient to be the same as a regression slope (beta), however having just compared the two, they are different. How do they differ - what different information do they give?
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edited Jul 17, 2012 at 21:31
Macro
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$\begingroup$ if they are normalized, they are the same. but think of what happen when you make change of units... $\endgroup$
nicolas
– nicolas
2013-01-30 16:15:25 +00:00
Commented Jan 30, 2013 at 16:15
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$\begingroup$ I think the top scoring answers to this Q (and maybe even my A to it where I show that the correlation coefficient can be seen as the absolute value of the geometric mean of the two slopes we obtain if we regress y on x and x on y, respectively) are also relevant here $\endgroup$
statmerkur
– statmerkur
2019-01-10 09:00:53 +00:00
Commented Jan 10, 2019 at 9:00
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Assuming you're talking about a simple regression model $$Y_i = \alpha + \beta X_i + \varepsilon_i$$ estimated by least squares, we know from wikipedia that $$ \hat {\beta} = {\rm cor}(Y_i, X_i) \cdot \frac{ {\rm SD}(Y_i) }{ {\rm SD}(X_i) } $$ Therefore the two only coincide when ${\rm SD}(Y_i) = {\rm SD}(X_i)$. That is, they only coincide when the two variables are on the same scale, in some sense. The most common way of achieving this is through standardization, as indicated by @gung.
The two, in some sense give you the same information - they each tell you the strength of the linear relationship between $X_i$ and $Y_i$. But, they do each give you distinct information (except, of course, when they are exactly the same):
The correlation gives you a bounded measurement that can be interpreted independently of the scale of the two variables. The closer the estimated correlation is to $\pm 1$, the closer the two are to a perfect linear relationship. The regression slope, in isolation, does not tell you that piece of information.
The regression slope gives a useful quantity interpreted as the estimated change in the expected value of $Y_i$ for a given value of $X_i$. Specifically, $\hat \beta$ tells you the change in the expected value of $Y_i$ corresponding to a 1-unit increase in $X_i$. This information can not be deduced from the correlation coefficient alone.
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answered Jul 17, 2012 at 14:59
MacroMacro
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$\begingroup$ As a corollary of this answer, notice that regressing x against y is not the inverse of regressing y against x ! $\endgroup$
meh
– meh
2017-06-05 18:36:02 +00:00
Commented Jun 5, 2017 at 18:36
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$\begingroup$ One more point- the correlation coefficient does not have units. The slope does. $\endgroup$
Kwame Brown
– Kwame Brown
2021-07-02 20:13:25 +00:00
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With simple linear regression (i.e., only 1 covariate), the slope $\beta_1$ is the same as Pearson's $r$ if both variables were standardized first. (For more information, you might find my answer here helpful.) When you are doing multiple regression, this can be more complicated due to multicollinearity, etc.
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edited Apr 13, 2017 at 12:44
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answered Jul 17, 2012 at 14:49
gung - Reinstate Monicagung - Reinstate Monica
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$\begingroup$ In simple linear regression, as Macro, shows above, $\hat{\beta} = r_{xy}\frac{s_y}{s_x}$. Is there an analogous expression for multiple regression? It seems there isn't for the reason you're getting at with "multicollinearity," but I think you really meant covariance here? $\endgroup$
24n8
– 24n8
2020-06-29 18:13:17 +00:00
Commented Jun 29, 2020 at 18:13
$\begingroup$ @Iamanon, try reading: Multiple regression or partial correlation coefficient? And relations between the two. $\endgroup$
gung - Reinstate Monica
– gung - Reinstate Monica
2020-06-29 18:25:11 +00:00
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The correlation coefficient measures the "tightness" of linear relationship between two variables and is bounded between -1 and 1, inclusive. Correlations close to zero represent no linear association between the variables, whereas correlations close to -1 or +1 indicate strong linear relationship. Intuitively, the easier it is for you to draw a line of best fit through a scatterplot, the more correlated they are.
The regression slope measures the "steepness" of the linear relationship between two variables and can take any value from $-\infty$ to $+\infty$. Slopes near zero mean that the response (Y) variable changes slowly as the predictor (X) variable changes. Slopes that are further from zero (either in the negative or positive direction) mean the response changes more rapidly as the predictor changes. Intuitively, if you were to draw a line of best fit through a scatterplot, the steeper it is, the further your slope is from zero.
So the correlation coefficient and regression slope MUST have the same sign (+ or -), but will not have the same value.
For simplicity, this answer assumes simple linear regression.
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edited Jul 2, 2021 at 20:03
answered Mar 20, 2014 at 21:20
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$\begingroup$ you indicte that beta can be in $-\inf, \inf$, but isn't there a case-by-case bound on beta implied by the ratio of variance of x and y? $\endgroup$
Matifou
– Matifou
2019-10-01 21:57:33 +00:00
Commented Oct 1, 2019 at 21:57
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Pearson's correlation coefficient is dimensionless and scaled between -1 and 1 regardless of the dimension and scale of the input variables.
If (for example) you input a mass in grams or kilograms, it makes no difference to the value of $r$, whereas this will make a tremendous difference to the gradient/slope (which has dimension and is scaled accordingly ... likewise, it would make no difference to $r$ if the scale is adjusted in any way, including using pounds or tons instead).
A simple demonstration (apologies for using Python!):
import numpy as np x = [10, 20, 30, 40] y = [3, 5, 10, 11] np.corrcoef(x,y) x = [1, 2, 3, 4] np.corrcoef(x,y)
shows that $r = 0.969363$ even though the slope has been increased by a factor of 10.
I must confess it's a neat trick that $r$ comes to be scaled between -1 and 1 (one of those cases where the numerator can never have absolute value greater than the denominator).
As @Macro has detailed above, slope $b = r(\frac{\sigma_{y}}{\sigma_{x}})$ , so you are correct in intuiting that Pearson's $r$ is related to the slope, but only when adjusted according to the standard deviations (which effectively restores the dimensions and scales!).
At first I thought it odd that the formula seems to suggest a loosely fitted line (low $r$) results in a lower gradient; then I plotted an example and realised that given a gradient, varying the "looseness" results in $r$ decreasing but this is offset by a proportional increase in $\sigma_{y}$.
In the chart below, four $x,y$ datasets are plotted:
the results of $y=3x$ (so gradient $b=3$, $r=1$, $\sigma_{x}=2.89$, $\sigma_{y}=8.66$) ... note that $\frac{\sigma_{y}}{\sigma_{x}}=3 $
the same but varied by a random number, with $r = 0.2447$, $\sigma_{x}=2.89$, $\sigma_{y}=34.69$, from which we can compute $b= 2.94 $
$y=15x$ (so $b=15$ and $r=1$, $\sigma_{x}=0.58$, $\sigma_{y}=8.66$)
the same as (2) but with reduced range $x$ so $ b= 14.70$ (and still $r = 0.2447$, $\sigma_{x}=0.58$, $\sigma_{y}=34.69$)
It can be seen that variance affects $r$ without necessarily affecting $b$, and units of measure can affect scale and thus $b$ without affecting $r$
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edited Jul 18, 2018 at 7:19
answered Jul 18, 2018 at 6:56
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Standard duty cycle of fridge compressor?
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I have a refrigerator which has compressor duty cycle of about 50% percent.
It’s set on 2 out of 5 which is the highest setting. It basically works for 12-13 minutes and then it’s of for about 16 minutes, and then cycle repeats.
Is this OK value?
Quoting Targeting Refrigerators for Repair or Replacement:
One can begin by recognizing two very common characteristics of most refrigerators. Over a broad range of models and vintages, the operating characteristics of properly operating refrigerators under reasonable temperature conditions appear the same. A refrigerator turns on, runs for a while, turns off and is off for a while. It then turns on again and the cycle repeats. A typical,steady-state cycle is about 40 minutes with the duty cycle, or per cent running time, being about 50 percent for standard temperature differences between the food compartment of the refrigerator and the room in which the refrigerator is placed. In other words, a typical, properly operating refrigerator runs about 20 minutes and is off 20 minutes.
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asked Aug 16, 2015 at 20:48
Jalgo JalgićJalgo Jalgić
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1 My side x side runs after being closed all night runs about 25 min, then off for 54 min, now that I replaced all the temp sensors with genuine GE from parts pros (many fakers out there give a clear plastic zip lock bag with a printed stick on lable; their sensors don't work right below 20F),the genuine GE parts come blue bags with special dbl green strip on the wires BEWARE those on amazon who say they are GE and even General Electric but are no, send you garbage which makes your fridge think its higher temp than it is (Ie registers 5F when its really 0 or -10 etc) so your fridge runs forever"InventPeace –InventPeace 2020-01-23 02:18:49 +00:00 Commented Jan 23, 2020 at 2:18
Duty cycle will depend on how efficient your fridge is as well as where you have set it.keshlam –keshlam 2023-09-24 12:31:36 +00:00 Commented Sep 24, 2023 at 12:31
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Seems normal from what I have read Hope my math is correct: I have a 20yr old 23.3cuft GE side-by-side refrig that runs 45% duty cycle (18min on, 22min off or 18/(18+22) = 18/40 = 45%, after refrig door has been closed for >8hrs.
With the current rating of 11.6 amps at 110v so this is ~.547kw/hr (110v 11.6.45) guessing off the top of my head newer compressors & fans don't use this much current so if a new ref will run at same 45% duty cycle the cost savings will be directly proportionate to the amperage reduction (i.e. 25% reduction in amps = 25% reduction in electric bill)
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Absolute Value Worksheets
These printable worksheets can be used to teach your students about the absolute value of integers. Absolute value is simply the distance an integer is from zero on a number line. It is represented by two straight lines on either side of a number. For example, |-27| equals 27.
Absolute Value
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Part 1: Find the absolute values. Part 2: Compare numbers with absolute values. Part 3: Find two values for each variable.
6th Grade
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Compare and Order Numbers With Absolute Values
On this printable, students will compare pairs of numbers using <, >, and =. Then they will order sets of four numbers from least to greatest.
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A.V. Questions
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Task Cards: Absolute Value
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Absolute Value in Context
Value v. Absolute Value
Practice value and absolute value with context! Identify true absolute value statements that accurately reflect each question's real-world context.
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Value v. Absolute Value Questions
Questions asking for value and questions asking for absolute value can be tricky to tell apart. Each real-life situation is paired with a question, and students must explain how they know which kind of value comparison it is asking about.
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Absolute Value and Value
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Ordering Value and Absolute Value
Practice ordering rational numbers and their absolute values by distinguishing between question contexts. Real-world situations address profit and debt as well as population growth and decline.
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Absolute Value and Opposites
Absolute Value & Opposites
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Questions on Abs. Value and Opposites
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Absolute Value and Number Lines
Plotting Absolute Value and Opposites
Visualize A.V. as distance from 0 on a number line by plotting values, their absolute values, and their opposites. Also identify the distance between rational values and 0.
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Absolute Value: Number Line Directions
These absolute value practice questions ask students to identify the relationship between plotted points on number lines. Other questions ask about the relative distance and direction of rational numbers, their opposites, and their A.V.s on a number line.
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Positive and Negative Integers
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10622 | https://www.youtube.com/watch?v=ImJSldHR0WA | Binary Numbers Explained for Beginners
Steve Cope
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Posted: 15 Sep 2017
- If you are working with computers or computer networking then having a basic understanding of binary numbers is essential.
In the video tutorial you will learn about binary and hexadecimal numbers and how to convert between binary and decimal.
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8B
75
11001000
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Intro Steve here in this video we're gonna take a look at binary numbers no computers work in binary and IP addresses are in binary and in my experience teaching networking many students struggle with IP addresses because they actually like a a basic understanding of binary numbers all because it was a long time ago that they actually looked at minor numbers and an understanding of barian numbers are now to convert between binary and decimal is essential for anyone who's involved in computers and computer networking now Writing Numbers - Number Systems there are various number systems that are in use and that you need to be familiar with as the binary system which is a base 2 system the decimal system which is the one we're all familiar with and one we use every day which is a base 10 system there's an octal which isn't really used much anymore which is works in base 8 and there's the EXA decimal which again is in use and you need to be aware of and that uses base 16 but we'll talk about the number base in it in a second okay decimal numbers no decimal numbers we should all be familiar with they're the ones we use day to day and that's the reason I chose to talk about that first because I'm going to use them to introduce the concepts that we're going to use when we talk about binary when we talk about X in decimal numbers now X and s sorry decimal numbers use base 10 which means they've got 10 possible values and those values are 0 to 9 so 0 1 2 etc and if we look at a typical number here we've got a three-digit number 1 2 9 it's actually 129 and we can write it like this we can write it as 100 plus 20 plus 9 which is 129 now the minimum value we have with three digits is zero and the maximum value we can have with three digits is 999 now what happens if we want to go above 999 well we need to add an extra column on to the left and we call this the thousands column I've Decimal Number Overview rewritten our number here and 129 and I put in here that the nine is actually the unit's column the two is the tens column and the hundreds sorry the one is in the hundreds column so that's 100 that's 20 and that is nine now above it I've used another notation and this is the power notation and this is 10 to the power of 0 which is represents the units 10 to the power 1 which represents the tens and 10 to the power 2 which represents the hundreds and you should notice that as we go from right to left we multiply the number on the right by 10 so this is 10 times 1 which is 10 and then we go to 100 which is 10 times 10 which is 100 and if we went over to the left we'd have 10 times 100 which would be a thousand so this column here which I've done yet would be the thousands column the 2 here isn't actually 2 it's 2 times 10 which is 20 and the 1 here is a 1 it's 1 times 10 which is 100 ok so we're going to use this same concept when we talk about it's binary numbers so binary numbers Binary Numbers now they use a base 2 and they only have two values nice and simple 0 and 1 and usually it's represented in a computer by on and off and we can assign column values like we did with our decimal but this time we use have to use 2 and not 10 as the base so instead of writing 10 to the power of 0 which is the unit's we write 2 to the power 0 which is the unit's 10 to the power 1 which is the 10th this this becomes 2 to the power 1 which is 2 10 to the 2 which is 100 becomes 2 2 to the 2 which is 4 and 10 to the 3 which is a thousand becomes 2 to the 3 which is 8 if we write a typical binary number we've got 101 here Binary 101 and I've put the PO notation above it and so we've got 2 to zero 2 to the 1 2 to the 2 and this is the unit's twos and this is the force column so it's not the unit's the tens and the hundreds net anymore what's the unit's the twos and the force and this number here the maximum I can represent with three binary numbers is 1 1 1 which equates to 4 plus 2 plus 1 which is 7 now there are 8 possible values and the 8 possible values of 1 1 1 all the way up to 0 0 or down should I say 0 so we've got 0 0 0 all the way up to 1 1 1 so which represents 8 possible values work our way from right to left the one isn't the 1 is actually a 1 it because it's in the unit's column and this one here isn't a 1 it's a 4 as it's in the force column so we've actually got 1 plus 1 which is decimal 5 so this binary number here equates to the decimal number 5 ok so let's have a quick recap Binary To Decimal Conversion 101 converts one bond in 101 converts 2 decimal 5 and we have 1 times 1 0 times 2 or 1 times 4 which makes it 5 and the maximum value we can have with 3 binary digits is 1 1 1 which is decimal 7 and it's calculated as shown below we have 1 times 1 1 times 2 and 1 times 4 and a few more examples here we have binary 1 0 1 1 and that is 1 times 1 1 x 2 0 times 4 1 times 8 which is decimal 11 and all ones equates to decimal 50 so here's a couple to try it often may want to pause the video while you do it so we've got 1 0 0 1 which is 1 times 8 and 1 times 1 which is 9 and we have 1 times a 1 times 4 which is 12 okay so now we Converting from Decimal to Binary gonna do it the other way around we're gonna go from decimal to binary it's a little more tricky in this direction so let's start with an example we're gonna use decimal 10 and we're going to convert decimal 10 into a binary number now the way you do it is using all the way I do is so so essays use the following table basically it's a table of two powers so we start off lot of at this side we have 1 2 4 8 16 all the way up to 128 because we're normally dealing with 8 bits and we look at 8 bits later on now what you do is you find the nearest number 210 which is a 2 that's 2 to the power 3 8 is 2 to the power 3 see there there's our 8 and we subtract it from the 10 and we get left with 2 and 2 it's not shown on the slide is effectively 2 to the power 1 now what we do now is we do exactly the same and it's not sure that slide so we take the 2 and we subtract the next number we count from 2 which is 2 and it leaves us with a 0 so our number is 1 8 2 and no units and it looks like this in binary okay another example let's go 13 so 13 is 1 8 plus 1 4 no twos one unit and it looks like this and 7 in binary is zero eight one four one two or one unit and it looks like this so here again a couple of examples that you want to try so you may want to again pause the video while you do it and here are the answers here and 9 is 1 a 2 plus 1 unit which is 1 0 0 1 & 3 is 0 8 0 4 a 2 and a 1 0 0 1 1 so that's binary sorry that's decimal to binary conversion and we've been at the moment so we've been using some three and four digit numbers computers Bytes and Octets and Hexadecimal Numbers networking we use 8 bit numbers and we can look at 8 bit numbers now you commonly known as a byte also as an octet and there's a wiki article if you look it up I'll provide the link in the video details or description by it is the more commonly used term but octet is quantum wiki is the more accurate I tend to use byte the largest number we can have is all ones which equivalent to decimal 255 so and it's calculated as 1 times 128 160 for 132 etc hold all the way down to 1 giving you decimal to 5 let's have another 8-bit number and this time we've got 0 1 etc and the way I write it here I write it with the powers above the using the power notation so we've got to the power 0 and to the birth 3 5 & 7 and so this is the units this is the twos this is the force this is the AIDS etc there's a mistake there that should actually be there ok so what this translates to is 1 times 1 1 2 no force 1 8 no 16 132 1 64 and no 128 there so it equates to decimal 107 so we have a 64 32 8 2 and a 1 107 and another example we're going to go decimal two hundred and decimal two hundred is 128 plus 64 plus eight which is equal to that should be 2 to the power 7 2 to the power 6 and 2 to the power 3 that's a little better ok ok so it's 2 to the power 7 2 to the power 6 and 2 to the power 3 ok which is 1 0 so 1 1 0 0 1 0 0 0 now once you used to do manually then it's obviously much easier if you use the calculator you can find one on Windows and here we just type in the binary numbers you can see we're in binary here and there's a binary number and then you just click on the decimal to convert it to decimal if you want to go to Exodus more you go here also you go here and this is what this number translates to in decimal now exa Understanding Hexadecimal numbers decimal numbers they use base 16 they require four bits maximum value of 15 and they introduce the letters ABCDE and F which f is 15 is 14 all the way down to a which is 10 and in binary we have 0 0 0 which is 0 0 0 1 which is 1 and 0 0 1 0 which is 2 I had to miss the rest out so I can fit on the slide now a is an 8 plus a 2 which is 10 and a B is 11 which is an 8 plus 3 and a C is 12 which is an 8 plus 4 all the way to 15 which is 8 plus 4 plus 2 plus 1 which is equal to 15 that's 15 okay a bite or eight bits can be split into two X a decimal numbers so we have F F which is equivalent to all ones and it's equivalent decimal 255 and here we have F 0 which is the first four bits a 1 and the next four bits are 0 and that equates to decimal 240 now you'll see this used quite a lot I'm going to look at your MAC address you'll see EXA decimal numbers just to finish off a couple of questions that you can do yourself Fe in binary and this one in binary number convert to X ax decimal and this one convert to decimal and finally cover 200 decimal into binary you'll find the answers in the description I'll give you the answers later so let's see he brings the end of the video if you've got any comments then leave them in the comment box below if you like the video use the like button below and you can subscribe to the channel if you want to be notified of new videos and also on the websites you can subscribe to the newsletter if you're interested and as the address of the website here you'll find it also in the description there's also a link in description to the tutorial the written tutorial on the website which covers binary numbers ok until next time bye |
10623 | https://www.cancer.gov/publications/dictionaries/cancer-terms/def/autosomal-recessive-inheritance | Definition of autosomal recessive inheritance - NCI Dictionary of Cancer Terms - NCI
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autosomal recessive inheritance
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(AW-toh-SOH-mul reh-SEH-siv in-HAYR-ih-tunts)
One of the ways a genetic trait or a genetic condition can be passed down (inherited) from parent to child. In autosomal recessive inheritance, a genetic condition occurs when the child inherits one mutated copy of a gene from each parent. The parents usually do not have the condition. The parents are called carriers because they each carry one copy of the mutated gene and can pass it to their children.
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Autosomal recessive inheritance is a way a genetic trait or condition can be passed down from parent to child. A genetic condition can occur when the child inherits one copy of a mutated (changed) gene from each parent. The parents of a child with an autosomal recessive condition usually do not have the condition. Unaffected parents are called carriers because they each carry one copy of the mutated gene and can pass it to their children.
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10624 | https://www.sciencedirect.com/topics/chemistry/electron-capture-decay | Skip to Main content
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Electron Capture Decay
In subject area:Chemistry
Electron capture decay is defined as a process in which an orbital electron is captured by the nucleus, combining with a proton to form a neutron, while emitting a neutrino and often resulting in the emission of characteristic x rays and Auger electrons from the daughter product. This decay decreases the atomic number by one and can lead to the emission of gamma rays if the daughter nucleus is in an excited state.
AI generated definition based on: Physics in Nuclear Medicine (Fourth Edition), 2012
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Unstable Nuclei and Radioactive Decay
2002, Radiochemistry and Nuclear Chemistry (Third Edition)GREGORY R. CHOPPIN, ... JAN RYDBERG
4.4.7Electron capture
The EC decay process can be written symbolically
(4.29)
The captured electron comes from one of the inner orbitals of the atom. Depending on the electron shell from which the electron originates, the process is sometimes referred to as K-capture, L-capture, etc. The probability for the capture of an electron from the K-shell is several times greater than that for the capture of an electron from the L-shell, since the wave function of K-electrons is substantially larger at the nucleus than that of L-electrons. Similarly, the probability of capture of electrons in higher order shells decreases with the quantum number of the electron shell.
The calculation of the decay energy in electron capture follows the equation
(4.30)
Note that like the case of the negatron decay, it is not necessary to add or subtract electron masses in the calculation of the Q-value in EC. An example of EC is the decay of 7Be to 7Li for which it is possible to calculate that the Q-value is 0.861 MeV. This reaction is somewhat exceptional since for neutron deficient nuclei with values of Z below 30, positron emission is the normal mode of decay. Electron capture is the predominant mode of decay for neutron deficient nuclei whose atomic number is greater than 80. The two processes compete to differing degrees for the nuclei between atomic numbers 30 and 80. Electron capture is observed through the emission of electrons from secondary reactions occurring in the electron shell because of the elemental change (see §4.9).
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Book2002, Radiochemistry and Nuclear Chemistry (Third Edition)GREGORY R. CHOPPIN, ... JAN RYDBERG
Chapter
Environmental liquid scintillation analysis
2020, Handbook of Radioactivity Analysis: Volume 2 (Fourth Edition)Xiaolin Hou, Xiongxin Dai
3Calcium-41 (41Ca)
41Ca, with a very long half-life (1.03 × 105 years), is mainly produced by neutron activation reaction of 40Ca(n, γ)41Ca. It is also a naturally occurring radionuclide produced by the reaction of stable calcium (40Ca) of the earth with neutrons from cosmic rays and fission of uranium (Fink et al., 1990). Human nuclear activities, including atmospheric nuclear weapons testing and operation of nuclear facilities have also released some 41Ca to the environment. In the nuclear reactor, 41Ca is mainly produced in the concrete shielding, because of its high calcium content and its exposure to the neutrons from the reactor. 41Ca is an important radionuclide in the disposal of radioactive waste, because it can exist for a very long time with a high mobility in the environment and high bioavailability.
41Ca decays to the ground state of 41K by pure electron capture, emitting X-rays and Auger electrons of very low energy (0.3–3.6 keV), it can thus be measured by X-ray spectrometry and LSC. LSC is much more sensitive (Suárez et al., 2000; Itoh et al., 2002b; Hou, 2005a, 2005b; Warwick et al., 2009; Hampe et al., 2013), because of its low counting efficiency (<0.08%) by X-ray spectrometry and the low abundance of X-rays of 41Ca (11.4% for 3.31 keV X-ray). Mass spectrometry techniques such as accelerator mass spectrometry (AMS) (Fink et al., 1990; Zerle et al., 1997) and resonance ionization mass spectrometry (RIMS) (Müller et al., 2000) are more sensitive techniques for its measurement and have also been used for its determination. A 41Ca/40Ca atomic ratio as low as 10−15 can be measured by AMS.
Due to the pure electron-capture decay of 41Ca, chemical separation from the sample matrices and purification from all other radionuclides are necessary prior to the measurement of 41Ca using LSC or other techniques. Suárez et al. (2000) reported a radiochemical separation procedure for the separation of calcium from other radionuclides, which is based on hydroxide precipitation of transition metals and chromate precipitation of Ba, Sr, and Ra with a chemical recovery of 40%. In this procedure, the separation of Ba, Sr, and Ra by chromate precipitation is very critical, because of very strict control of pH value. Ion exchange and extraction chromatography have also been used for the separation of calcium (Itoh et al., 2002b; Hampe et al., 2013). Anion exchange chromatography was used to remove all radionuclides presenting as anion in high concentration HCl media, and extraction chromatography using TRU resin was applied to remove actinides and lanthanides. However, this procedure is not well suitable for the removal of alkaline earth radionuclides such as radioisotopes of strontium, radium and barium. A tertbutylmethylether/ethanol extraction step was proposed to extract calcium and to remove Sr, Ba, and Ra (Hampe et al., 2013). With all of these steps, a recovery of more than 65% and decontamination factor for most of interfering radionuclides are higher than 103. Hou (2005b) reported a simple and effective method for the determination of 41Ca in ordinary and heavy concrete (containing more than 75% of BaSO4) based on the relative low solubility production of Ca(OH)2, compared to Sr, Ba, and Ra in NaOH solution. Calcium (mainly as calcium carbonate) is first leached using aqua regia, and the experiment shows that more than 95% of Ca can be leached out from the concrete sample in this step. There are three main steps for the separation of Ca from the interfering radionuclides, namely,
(1)
: Ca is first separated from the transition metals, such as Co, Eu, Fe, Ni, and tranuranics by precipitation at pH 9 using NaOH. In this step, the interfering radioisotopes of these elements are precipitated, while Ca remains in the solution with Sr, as well as Ba, Ra, Cs, etc.
(2)
: Ca and Sr in the supernatant are then precipitated as carbonates with Ba and Ra by adding Na2CO3 and separated from alkali metals and nonmetal elements.
(3)
: Ca is finally separated from Sr, Ba and Ra by precipitation of Ca as Ca(OH)2 in 0.5 mol/L NaOH solution after dissolution of the carbonate precipitates. This is based on the low solubility of Ca(OH)2 in high concentration of NaOH (higher than 0.5 mol/L) compared with Sr, Ba, and Ra. This step is repeated and the separated Ca(OH)2 is dissolved with HCl for measurement using LSC after neutralizing to pH 6-8.
The chemical yield of 41Ca of 80%–90% was obtained by the measurement of Ca before and after chemical separation using ICP-OES. The decontamination factor for the interfering radionuclides such as 60Co, 152Eu, 133Ba, 85Sr, 137Cs, 55Fe, and 63Ni are higher than 105. The detection limit of the analytical method for 41Ca is 0.020 Bq.
This method could not separate 45Ca from 41Ca. If both isotopes exist in the sample, interference of 45Ca to the measurement of 41Ca has to be corrected, which can be carried out by measurement of the contribution of 45Ca to the counting window of 41Ca in the lower channel. Due to the low energy of Auger electrons (0.3–3.6 keV) detected in the measurement of 41Ca by LSC, the counting efficiency is relatively low (<25%), and highly influenced by quench. The separated calcium sample is normally prepared in a neutral solution of CaCl2, and all other interfering elements (especially iron) have to be removed completely. The major quench effect is the amount of calcium in the samples. In addition, luminesence is also an interference for the measurement of 41Ca, which has to be considered in the measurement of 41Ca using LSC.
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Book2020, Handbook of Radioactivity Analysis: Volume 2 (Fourth Edition)Xiaolin Hou, Xiongxin Dai
Chapter
NUCLEAR RADIATION, ITS INTERACTION WITH MATTER AND RADIOISOTOPE DECAY
2003, Handbook of Radioactivity Analysis (Second Edition)MICHAEL F. L'ANNUNZIATA
E.X-Radiation
Mention has been made of the electron capture decay process whereby an electron from one of the atomic shells (generally the innermost K shell) is absorbed by the nucleus, where it combines with a proton to form a neutron. No particle emission results from this decay process. However, the vacancy left by the electron from the K shell is filled by an electron from an outer shell (generally the adjacent L shell). Transitions produced in electron shell energy levels result in the emission of energy as x-radiation (see also Sections II.E and II.F). This radiation consists of photons of electromagnetic radiation similar to gamma radiation. X-radiation and gamma radiation differ in their origin. X-rays arise from atomic electron energy transitions and gamma rays from transitions between nuclei of different energy states. The production of x-radiation from atomic electron transitions is illustrated in Figs. 1.7 and 1.23.
When an electron transition occurs from the outer L shell to an inner K shell, the energy emitted is equivalent to the difference between the K and L electron binding energies. The electron transitions that ensue in the filling of vacancies are a deexcitation process, and the energy lost by the atom as x-radiation is equivalent to the difference of the electron energies of the outer or excited state, Eouter, and its new inner ground state, Einner, as described by
(1.97)
The radiation emitted consists of a discrete line of energy characteristic of the electron shell and, consequently, of the atom from which it arises.
The production of x-rays in radionuclide decay is, however, more complex. The filling of one electron vacancy in an inner shell is followed by a series of electron transitions in an overall adjustment of electrons in outer shells. This gives rise to further x-rays with lines characteristic of outer shells. Such electron transitions, each resulting in the emission of discrete lines of characteristic x-rays, are illustrated in Fig. 1.24. The transitions are identified by a letter corresponding to the shell (K, L, M, etc.) with vacancy giving rise to the x-ray photon and a subscript (α, β, γ, etc.) to identify, from among a series of outer electron shells of the atom, the shell from which the electron vacancy is filled. For example, an x-ray arising from an electron transition from the L to the K shell is denoted as Kα and that arising from a transition from the M to the K shell as Kβ. Transitions involving the filling of electron vacancies in the L shell from outer M, N and O shells are denoted by Lα, Lβ, and Lγ, etc.
Because x-radiation is characteristic of the atom from which it arises, it is customary to identify the element along with the x-ray photon (e.g., Cr K x-rays, Hg L x-rays, and many others as listed in Appendix A). In these examples, the fine structure of the x-ray emissions is not given and the lines are grouped together as K and L x-rays.
The complexity of x-ray lines emitted and their abundances of emission are compounded by the existence of other mechanisms of x-ray production in unstable atoms. One of these mechanisms is the production of Auger electrons. An x-ray emitted from an atom may produce an Auger electron via an internal photoelectric effect (see Section II.F), which results in the emission of an atomic electron from a shell farther away from the nucleus. The vacancy left by the Auger electron gives rise to additional x-rays characteristic of outer shells following the electron readjustments that ensue. Auger electrons can be emitted from a variety of electron shells, followed by an equal variety of characteristic x-rays from subsequent electron adjustments in outer shells.
Any process that would cause the ejection of an atomic electron of an inner shell can result in the production of x-radiation. Other processes not yet mentioned in this section that involve the ejection of atomic electrons are the emission of internal-conversion electrons (see Section II.E) and radiation-induced ionization (see Sections II and IV).
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Book2003, Handbook of Radioactivity Analysis (Second Edition)MICHAEL F. L'ANNUNZIATA
Chapter
Beta Radiation and Beta Decay
2023, Radioactivity (Third Edition)Michael F. L'Annunziata
Abstract
This chapter focuses on various aspects of beta decay, including the emission of beta particles and the electron capture (EC) decay process. The various types of beta decay discussed include negatron emission, negatron decay energy and calculations of decay energy, inverse beta decay, neutrino emission in beta decay and neutrino mass, positron emission, calculation of positron decay energy, and the relationship between nuclear neutron/proton ratios and type of beta decay. The EC decay process and the competition between EC and position emission as decay modes are described. Other modes of beta decay discussed include branching negatron, positron and EC decay, double-beta decay, and neutrinoless double-beta decay and the Majorana neutrino. The mechanisms of beta-particle interactions in matter are described. The ranges of positrons and negatrons in matter are described relative to that of alpha particles of the same energy. The concepts of stopping power and linear energy transfer (LET) and the calculated values of stopping power and LET for different radiation types are also discussed.
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Book2023, Radioactivity (Third Edition)Michael F. L'Annunziata
Chapter
Electromagnetic Radiation
2023, Radioactivity (Third Edition)Michael F. L'Annunziata
13.6.2.3Inner or internal bremsstrahlung
Bremsstrahlung of very low intensity, referred to as inner or internal bremsstrahlung, also occurs during the transformation of the nucleus in electron capture decay processes and beta-particle emission. The bremsstrahlung is a continuous spectrum of X-ray photons that originates within the transforming atoms and can be attributed to the sudden change of nuclear charge when the beta particle is emitted or when an orbital electron is captured (Evans, 1972). Internal bremsstrahlung is explained by Cengiz and Almaz (2004) to occur during the emission of a β− particle from the nucleus, as the β− particle undergoes an acceleration at its birth and emits bremsstrahlung in the field of the emitting nucleus. The magnitude of inner bremsstrahlung was calculated to be approximately 1/137th quantum of internal bremsstrahlung per beta particle emitted from the nucleus (Knipp and Uhlenbeck, 1936; Evans, 1972). Internal bremsstrahlung spectra for several nuclides that emit β− particles were calculated by Cengiz and Almaz (2004) Singh and Dhaliwal (2015) and Italiano et al. (2020). This radiation was named inner or internal bremsstrahlung, because it originates from the nucleus, the internal part of the atom, in contrast to external bremsstrahlung that occurs when an external beta particle approaches an atom external to it from another source and is deflected by the nucleus of that atom. The difference between internal and external bremsstrahlung, as described earlier in this paragraph, is explained eloquently by Italiano et al. (2020) as follows:
… as reported for the first time by Aston (1927) in his measurements on RaE (210Bi), it may occur that photons are emitted with a continuous energy spectrum simultaneously to the β particle, as a consequence of its interaction with the emitting nucleus. This phenomenon is called Internal Bremsstrahlung (IB) to distinguish it from the External Bremsstrahlung (EB), occurring when the β particle interacts with a nucleus different from the parent one.
In the electron capture decay process, the quantum of energy not carried away by the neutrino is emitted as internal bremsstrahlung. Thus, in electron capture decay, internal bremsstrahlung may possess energies between zero and the maximum, or transition energy of a radionuclide. When gamma radiation is also emitted, the internal bremsstrahlung may be masked by the more intense gamma rays and go undetected. However, in the absence of gamma radiation, the upper limit of the internal bremsstrahlung can be used to determine the transition energy of a nuclide in electron capture decay. Some examples of radionuclides that decay by electron capture without the emission of gamma radiation are as follows:
(13.24)
(13.25)
and
(13.26)
where hv is the internal bremsstrahlung, the upper energy limits of which are expressed in MeV.
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Book2023, Radioactivity (Third Edition)Michael F. L'Annunziata
Chapter
Radionuclide Standardization
2012, Handbook of Radioactivity Analysis (Third Edition)Agustín Grau Malonda, Agustín Grau Carles
BThe SIR of Betas
The success of the SIR for gamma-emitters led to extending the procedure to pure beta decay, alpha decay, and pure electron capture decay. For pure beta-emitters the liquid scintillation technique and in particular the CIEMAT/NIST method seem to be the most appropriate procedure. To test this procedure, several comparisons have been organized by the CCRI (II), using the CIEMAT/NIST method, for the following nuclides: 14C, 90Sr, 89Sr, 90Y, and alpha emitters 238Pu and 241Am. These tests have been successful and, therefore, present no difficulty to be included in the SIR.
The CIEMAT/NIST method permits the compensation for LSC long-term anomalous effects, in particular, the effect of permanent photomultiplier fatigue. This effect can modify the counting rate. Grau Malonda (1982a) applied the CIEMAT/NIST method to C-14 standardization using tritium as a tracer when the photomultiplier response presented a large degradation. Calibration curves of H-3 and C-14 presented important anomalies, but the activity of C-14 samples was accurate.
Rodríguez Barquero and Los Arcos (2004a) measured samples of H-3, C-14, and Ni-63 over a long period of time, 4, 2.3 and 2.4 years, respectively. In this work, they prove that the counting rates vary over time but C-14 and Ni-63 activities obtained with the CIEMAT/NIST method, when the calibration curves prepared in times close to the activity measurement, gave discrepancies of –0.5% and –0.4% for Ni-63 and the C-14, respectively. Rodríguez Barquero and Los Arcos (2004b) conclude in their work: “This behavior confirms the reliability of the CIEMAT/NIST method as a monitoring tool for the determination of activity concentration on the long term with no need for additional checking procedures or stability curves.”
For 204Tl, a beta-emitter with a small branch of EC, the measurement was sensitive to the chemical preparation, a problem expected in other radionuclides. Another objective is to extend the SIR to nuclides decaying by pure electron capture, such as 55Fe and 165Er, which are not now covered by the SIR. They emit photons with too low energy (below 50 keV). It has been shown repeatedly that 3H (T1/2 = 12.312 y, and Emax= 18.591 keV) is an excellent tracer for standardization of beta-emitters by the CIEMAT/NIST method and has been proposed to be used as a monitor. However, its half-life is relatively short. A nuclide of longer half-life is 14C but its use requires special care. It could lose activity, producing carbon dioxide that escapes from the sample. 63Ni seems to be a better candidate in view of its half-life (100 years) and maximum energy (65.9 keV), lower than 14C.
In 2007, at the ESIR WG (Extended SIR Working Group) meeting, CIEMAT, PTB, and LNE-LNHB proposed different methods and procedures for the SIR extension. The PTB method (Kossert, 2006) is very similar to the CIEMAT/NIST method. The method presented by the LNE-LNHB (Cassette and Do, 2008) is based on the application of a Compton spectrometer to the scintillation detector, with the creation, in the liquid scintillator, of an internal and virtual reference source, which is measured by the Compton effect. CIEMAT introduced a new scintillator and applied the CIEMAT/NIST method. CIEMAT presented results showing good stability for the proposed XAN6040 scintillator. We describe each of the proposals in detail.
Kossert (2006) described a secondary method based on the efficiency curve of tritium as a tracer and the efficiency curve of the radionuclide to be studied. The procedure is as follows. In a first stage we have the efficiency curve (efficiency against quench parameter QIP) of tritium and the efficiency curve of the nuclide to be studied, i.e. εtracer (QIP) and εnuclide (QIP). For both measures, a set of samples with different quenching are prepared. We measure efficiency and quench parameter QIP. As both curves have a common element, the QIP, we obtained from these curves a new curve εnuclide (εtracer) that links the efficiency of the studied nuclide and the tracer. In a second stage, which may be several years later, the operation is repeated and we prepare a set of tracer samples with a different quench, so you get a new curve εtracer (QIP) and from the measurement of the radionuclide in study we obtain the parameter QIP. With this parameter we determine the new efficiency of the tracer and with the new efficiency of the tracer we enter the curve εnuclide(εtracer) and we determine the efficiency of the nuclide. The activity of the sample under study is obtained applying the equation Anuclide = Nnuclide/εnuclide. Kossert (2006) presented Fig. 14.14, which clarifies the description given. The composition of the sample and the geometry of the measurement must be very similar for all measures, and the time between the preparation of tracer sample and the nuclide under study should be short. Moreover, the time between the first and second measurements of the sample under study can be very long, many years. This method enables us to correct for gain variations of the measuring equipment, but cannot correct the count variations in the sample by adsorption or microprecipitation, as the QIP is only sensitive to the liquid scintillator and not to the radionuclide incorporated to the liquid scintillator.
Cassette and Do (2008) and Cassette et al. (2010) described a new calibration method based on liquid scintillation using a virtual source, which acts as a tracer, created inside the scintillator by Compton interaction. When a sample is placed in the liquid scintillation counter it is irradiated by an external and collimated source of 241Am emitting 59.4-keV gamma rays. A gamma detector is placed in the bottom of the optic cell of the liquid scintillation counter with its axis normal to the beam of gamma photons. The gamma detector measures part of the gamma photons scattered in the liquid scintillator by coherent or incoherent Compton scattering. For incoherent photons, the spectrum is centered around the scattering at 90°, which corresponds to an average energy of 53.2 keV. The Compton spectrum is separated from the coherent scattering spectrum using a detector with good resolution, for example, a Ge detector. The Compton spectrum analyzer is selected by a SCA (single-channel analyzer) and this signal is used to validate the acquisition of the photomultiplier of the liquid scintillation counter. All pulses created by gamma-ray interaction in the liquid scintillator are viewed by the liquid scintillation counter, but only those pulses in coincidence with the SCA signal are selected. The coincidence system rejects all light pulses generated by the sample's disintegration in the liquid scintillator. The energy spectrum of Compton electrons from the source is deduced from the experimental spectrum of Compton scattered photons by applying the principle of conservation of energy to the Compton effect. This spectrum is used to calculate the free parameter of each photomultiplier, from the experimental values of the relationship triple-to-double. Then the gamma source is removed and the sample is measured using the traditional TDCR method. This procedure has the advantage of allowing direct determination of the nonlinearity of the response of the liquid scintillation at low energies due to ionization quench. However, it is not possible to determine the stability of the sample measured or correct for counting variation in the sample by adsorption or microprecipitation of the radionuclide.
Rodriguez Barquero and Los Arcos (2000) measured the stability of 110mAg samples for liquid scintillation with four commercial scintillators: Insta-Gel Plus, Ultima-Gold, HiSafe II, and HiSafe III. Ultima-Gold samples were completely unstable for 2 months, Insta-Gel Plus and HiSafe II reached stability after 30 days and only HiSafe III samples were stable for the full term of 2 months. Samples were prepared with 15 mL of liquid scintillator and four carrier's level: 0.0, 1.5, 3.0, and 6.0 μg of AgNO3.
Rodriguez Barquero and Los Arcos (2010) presented a systematic study of five scintillators, four commercial: Ultima-Gold, Optiphase HiSafe II, and Optiphase HiSafe III, and one homemade reference scintillator, XAN6040, useful to stabilize up to 19 radionuclides (Rodríguez Barquero and Los Arcos, 2005). They determined the density and the elemental composition of these liquid scintillators. The density at 16 °C is higher by 0.4% than the density at 20 °C. However, the discrepancy among different batches of commercial scintillators can reach up to 4%, all having the same nominal composition. Several laboratories analyzed the weight percentages of H, C, O, N, P, Na, S, and B of all the laboratories. The elemental composition of commercial scintillators presented discrepancies with the nominal values from 2% and up to 260% depending on the element and the scintillator.
Rodríguez Barquero and Los Arcos (2005, 2007) introduced a new liquid scintillator of known composition called XAN6040 with the goal of becoming the typical scintillator of the SIR for betas. The tests carried out at CIEMAT focused on stability and reproducibility of the cocktail that gave excellent results. The composition of the XAN6040 liquid scintillator is given in the patents of Rodriguez Barquero and Los Arcos, (2003, 2004b,c,d).
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Book2012, Handbook of Radioactivity Analysis (Third Edition)Agustín Grau Malonda, Agustín Grau Carles
Chapter
Gamma-Ray Spectroscopy
2003, Encyclopedia of Physical Science and Technology (Third Edition)R.F. Casten, C.W. Beausang
III.A.1Beta-Decay
Nuclei formed off the valley of stability decay back toward stable nuclei via β-decay (which includes the processes of β−, β+, and electron capture decay). Typically, β-decay populates several excited levels in the daughter nucleus. Half-lives near stability range from seconds to days. Production of β-decay parent nuclei can be achieved by simple reactions such as (p, n) or by heavy-ion reactions. The simpler, lower energy reactions tend to form only one or a couple of parent nuclei, whereas heavy-ion reactions may form many times more, and, in that case, selection techniques are needed to select the decay products of interest.
A popular technique in β-decay is the use of moving tape collectors in which the activity is collected on a tape (e.g., movie reel tape or aluminized Mylar) for some period of time (typically ∼1.8 times the half-life for the desired β-decay). The tape is then moved to a low background area for detection of gamma-rays following decay. Collection of a new activity at another spot on the tape proceeds simultaneously.
Gamma-ray spectroscopy following β-decay was for many years in the 1950s–1970s a standard technique used to elucidate nuclear structure. Since β-decay itself carries off little or no angular momentum, the spin states accessible with this technique are generally those within ±2–3 ℏ of the parent (ground or isomeric) state.
In recent years the technique has enjoyed a renaissance with the use of arrays of much higher efficiency Ge detectors (e.g., clover or cluster detectors). Since the gamma-ray multiplicity following β-decay is low and there is no Doppler effect, the detectors can often be mounted in close geometry to maximize count rates and achieve considerable coincidence efficiencies.
One current setup for such studies is the Yale moving tape collector(Casten, 2000). Illustrated inFig. 7, it uses up to four Compton suppressed clover detectors that can be positioned at any angle in a horizontal plane. Studies with this instrument have included searches for possible multi-phonon states in 162Dy and 164Er. Nuclei with ellipsoidal shapes can undergo vibrational oscillations (called phonons) of these shapes about their equilibrium position. In principle, it is possible to superpose two or more identical vibrations. However, the effects of the Pauli Principle acting on the particles in the nucleus may destroy such states. One test of their intact character is to study their gamma decay. If they have predominantly a two-phonon character, then they should decay to the one-phonon state. Experimental searches for weak gamma-ray decay branches to the single phonon excitation are being sought in these two nuclei.
Another application of β-decay exploits the fourfold segmentation of the clover detectors. In the Yale arrangement, four such detectors allow simultaneous coincidence measurements at a large number of different relative angles of emission between the two detected gamma-rays. These angular correlation measurements can be used to constrain spin arguments for levels in the gamma-ray cascade. With clover detectors situated at appropriate angles, it is also possible to exploit their segmentation to measure the linear polarization of the gamma-ray and thereby to deduce the parity relations of the nuclear levels involved(Duchene et al., 1999).
Finally, β-decay measurements are also an important tool in mass measurements, since, often, the daughter or granddaughter mass is known but not that of the parent. Nuclear masses (that is, in effect, binding energies) are of importance in a number of contexts. The binding energy reflects the sum of all the nucleonic interactions. Differences of binding energies for neighboring nuclei give the separation energy of the last nucleon and are therefore sensitive to single particle energies of nucleons in a mean field nuclear potential, as well as to shape and structure changes from one nucleus to the next. Mass measurements are also important for understanding the astrophysical processes occurring in the interiors of stars that lead to nucleosynthesis. Recent studies of nuclei in the mass A ∼70 region, for example, are helping to set constraints on the termination of the rapid proton capture process in certain classes of stars.
Nuclear mass measurements are carried out by measuring gamma-ray spectra in coincidence with β-particle detection in order to deduce the β-decay end point, that is, the maximum β-decay energy (where energy sharing with the simultaneously emitted anti-neutrino is insignificant). The end point energy directly gives the mass of the parent nucleus if the daughter mass is known. The gamma-ray coincidence is used to cleanly select the product nucleus of interest.
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Reference work2003, Encyclopedia of Physical Science and Technology (Third Edition)R.F. Casten, C.W. Beausang
Chapter
Radioactivity
2003, Encyclopedia of Physical Science and Technology (Third Edition)Vincent P. Guinn
VII.Dβ+ Emission
Decay by positron (β+) emission is a second type of decay mode exhibited by many proton-rich radionuclides. In β+ emission the product nucleus is the same as for EC decay of the same radionuclide, but a β+ particle and a neutino are both emitted. The β+ and ν share the energy release (QEC − 1.022 MeV), so both the β+ and ν exhibit continuous energy spectra up to an Emax value (as in the case of emission), rather than the monoenergetic ν emission of EC decay. When atomic masses are used to calculate the energy available to the β+ and ν as kinetic energy, this energy is Δ m time 931.50 − 1.022 MeV (the energy equivalent of two times the rest mass of an electron). For proton-rich nuclei, EC is a possible mode of decay for all positive values of QEC, but decay by β+ emission is possible only if QEC > 1.022 MeV. In cases where QEC > 1.022 MeV, both modes of decay are possible, and thus in some such cases a fraction of the decays of the radionuclide occur by EC and a fraction occur by β+ emission. An example of such a branched decay is 78.4-hr 89Zr, which decays 78% by EC and 22% by β+ emission. Even in cases where both types of decay are energetically possible, some proton-rich nuclei decay entirely by EC, some entirely by β+ emission, and some by branched decay. In general, where both modes of decay are energetically possible, EC decay becomes predominant with increasing Z.
An example of a pure β+ emitter (with no accompanying γ-ray emission) is 9.96-min 13N (137N → 136C + 01β+ + 00ν). In β+ decay, as in EC decay, there is no change in mass number (a nuclear proton simply changing to a nuclear neutron), but the product nucleus is one unit lower in Z than the radionuclide. The transition results in a product nucleus on or closer to the stability diagonal. As with EC decay, some radionuclides are pure β+ emitters, whereas others have more complicated decay schemes with accompanying γ-ray emission. Even pure β+ emitters, it should be noted, emit 0.511-MeV positron annihilation radiation, resulting from the e+/e− annihilation process. This annihilation (e+ + e− → 2γ) occurs when the β+ has slowed down and encounters any negative electron. The sum of the rest masses of the two electrons disappears, and the 1.022 MeV of resulting energy appears as two 0.511-MeV γ-ray photons emitted 180° from one another in direction (conserving momentum). It should be noted that neutrinos and antineutrinos are not detected at all by normal radiation detectors.
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Reference work2003, Encyclopedia of Physical Science and Technology (Third Edition)Vincent P. Guinn
Chapter
Analytical Geochemistry/Inorganic INSTR. Analysis
2014, Treatise on Geochemistry (Second Edition)M.D. Glascock
15.15.4.1Modes of Radioactive Decay
During radioactive decay, mass is converted into energy as predicted by Einstein's theory of relativity such that the laws of conservation of energy, momentum, charge, and mass number are obeyed. A nucleus will decay by one or more modes if the mass(es) of the product(s) is less than that of the original nucleus. Certain decay modes are favored based on the properties of the original and product nucleus. In general, neutron-rich nuclei will transform by β− decay, resulting in a decrease in the number of neutrons by one, while increasing the number of protons by one. Neutron-deficient nuclei decay by β+ or electron capture decay, which reduces the number of protons by one while increasing the number of neutrons by one. The alpha decay mode is important for very heavy nuclides and results in the loss of four nucleons (i.e., two protons and two neutrons) from the original nucleus. The neutron decay mode is important for certain fission products and results in a loss of one neutron from the original nucleus.
Disintegrations by alpha decay, beta decay, or neutron decay usually generate a product nucleus in an excited state. The product nucleus continues disintegration by emitting one or more gamma rays or particles until arriving at a ground state. A single gamma ray is emitted as a radionuclide goes from a higher-energy excited state to lower-energy state within the same radionuclide. Many radionuclides emit multiple gamma rays in cascade before arriving in a ground-state configuration. The number of gamma rays of a specific energy emitted per 100 decays is referred to as the intensity of that gamma ray, in percent. The excited states and gamma-ray transitions for a nuclide are often illustrated by the use of energy-level diagrams or decay schemes, such as those shown in Figures 1 and 2.
Although a majority of excited states decay in less than 10− 12 s, some last long enough to be measured. Excited states with measurable lifetimes are referred to as metastable or isomeric nuclear states. Most metastable states deexcite by emitting gamma rays, but some of the longer-lived metastable states undergo beta decay.
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Reference work2014, Treatise on Geochemistry (Second Edition)M.D. Glascock
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Photomagnetism of Molecular Systems
2016, Reference Module in Materials Science and Materials EngineeringP. Gütlich, Y. Garcia
2Nuclear Decay-Induced Excited Spin State Trapping (NIESST) Effect
The acronym NIESST stands for ‘Nuclear Decay-Induced Excited Spin State Trapping.’ This phenomenon is closely related to LIESST, as it makes use of the nuclear decay and its energy release as intrinsic molecular excitation source, whereas LIESST is the result of irradiation with an external visible light source. Thus the initial step of electronic excitation is different, but the final step of ligand field state relaxations has been found to be the same for both phenomena. A brief description of NIESST is as follows.
57Fe Mössbauer emission (ME) spectroscopy has been a most elegant tool for the observation of metastable ligand field states at ca. 100 ns after electron capture decay of 57Co(EC)57Fe in transition metals compounds (Sano and Gütlich, 1984). A 57Co labeled compound is used as the Mössbauer source at variable temperatures versus K4[Fe(CN)6] as a single-line absorber. If 57Co is embedded in a strong-field surroundings, which causes LS behavior in the corresponding iron(II) compound like the tris-phen or tris-bpy complexes (phen=1,10-phenanthroline; bpy=2,2'-bipyridine), one observes the nucleogenic 57Fe(II) ions in LS (1A1) state at room temperature, which is the ground state in the corresponding synthesized iron(II) compound. Below ca. 200 K, however, one observes typical 57Fe(II)-HS resonances in the ME spectra with increasing intensity toward lower temperatures at the expense of the 57Fe(II)-LS resonances. In iron(II) SCO systems with weaker ligand field strength one observes no more 57Fe(II)-LS resonances in the ME spectra, even at very low temperatures where the corresponding synthesized iron(II) complex has already undergone spin transition to the LS (1A1) ground state. The trapped HS state of the nucleogenic 57Fe(II) ion was found to have very similar lifetimes as the corresponding LIESST state under comparable conditions (Deisenroth et al., 1998). It was therefore concluded that NIESST and LIESST are closely related phenomena. Whereas the LIESST effect bears basically the potential for practical applications, NIESST is evidently more of academic interest and has contributed much to a deeper knowledge of mechanistic aspects of hot atom chemistry in the solid state.
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Reference work2016, Reference Module in Materials Science and Materials EngineeringP. Gütlich, Y. Garcia
Related terms:
Bremsstrahlung
Electron Capture
Positron
Radionuclide
Atomic Number
Purity
Beta Decay
Gamma Ray
Radioisotope
X-Ray Emission
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10625 | https://www.khanacademy.org/science/ap-chemistry-beta/x2eef969c74e0d802:molecular-and-ionic-compound-structure-and-properties/x2eef969c74e0d802:vsepr/v/dipole-moment | Use of cookies
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10626 | https://pmc.ncbi.nlm.nih.gov/articles/PMC7456622/ | The People’s bank of China’s response to the coronavirus pandemic: A quantitative assessment - PMC
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Econ Model
. 2020 Aug 31;93:465–473. doi: 10.1016/j.econmod.2020.08.018
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The People’s bank of China’s response to the coronavirus pandemic: A quantitative assessment
Michael Funke
Michael Funke
a Hamburg University, Department of Economics, Germany
b CESifo Munich, Germany
c Tallinn University of Technology, Department of Economics and Finance, Estonia
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a,b,c, Andrew Tsang
Andrew Tsang
a Hamburg University, Department of Economics, Germany
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a,∗
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a Hamburg University, Department of Economics, Germany
b CESifo Munich, Germany
c Tallinn University of Technology, Department of Economics and Finance, Estonia
∗
Corresponding author.
Received 2020 Jun 2; Revised 2020 Aug 18; Accepted 2020 Aug 22; Issue date 2020 Dec.
© 2020 Elsevier B.V. All rights reserved.
Since January 2020 Elsevier has created a COVID-19 resource centre with free information in English and Mandarin on the novel coronavirus COVID-19. The COVID-19 resource centre is hosted on Elsevier Connect, the company's public news and information website. Elsevier hereby grants permission to make all its COVID-19-related research that is available on the COVID-19 resource centre - including this research content - immediately available in PubMed Central and other publicly funded repositories, such as the WHO COVID database with rights for unrestricted research re-use and analyses in any form or by any means with acknowledgement of the original source. These permissions are granted for free by Elsevier for as long as the COVID-19 resource centre remains active.
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PMCID: PMC7456622 PMID: 32904582
Abstract
The People’s Bank of China (PBoC) has implemented numerous measures to cushion the impacts of the COVID-19 health crisis on the Chinese economy. Since the current monetary policy framework features a multi-instrument mix of liquidity tools and pricing signals, we employ a dynamic-factor modelling approach to derive a composite indicator of China’s monetary policy stance. Our quantitative assessment shows that the PBoC’s policy response to the outbreak of the COVID-19 pandemic has been swift and decisive. Specifically, our estimates reveal that the PBoC has implemented novel policy measures to ensure that commercial banks maintain liquidity access and credit provision during the COVID-19 crisis.
Keywords: Corona pandemic, COVID-19, China, Monetary policy, Dynamic factor model
Highlights
•
We construct a dynamic-factor-based indicator of China’s monetary policy stance.
•
Estimates reveal the PBoC’s novel policy measures during the COVID-19 crisis.
•
The Chinese monetary policy response to the pandemic has been swift and decisive.
•
The measures ensure liquidity access and credit provision by commercial banks.
1. Introduction
In December 2019, respiratory illness clusters caused by a novel coronavirus (SARS-COV-2) emerged in Wuhan, the capital of Hubei Province, China. The World Health Organization (WHO) named the disease COVID-19. The pandemic quickly spread from Wuhan to the rest of mainland China: As of July 16, 2020, official statistics show that COVID-19 had caused a cumulative number of 83,622 infections and 4634 fatalities in China.1 The need for social distancing and lockdowns has caused substantial collateral economic damage. COVID-19 initially caused a negative supply shock by forcing firms to shut down, which disrupted international supply chains. Moreover, the pandemic, through its negative impact on agent expectations of future income growth, induced a demand-driven recession. Weak aggregate demand, in turn, depressed the incentives of firms to invest. The massive spike in uncertainty added wait-and-see responses by consumers and firms to a shaky world economy, and valuations in global financial markets imploded. These reactions reflect profound uncertainties about the path of the COVID-19 virus and the length of time the global economy could remain shuttered, or even induce a supply-demand doom loop.2
Which policy interventions can prevent the economy from sliding into a prolonged recession? How has monetary and fiscal policy responded to the coronavirus-induced disruptions? As the global recession gains force, almost all governments around the world have implemented measures that ranged from monetary easing to measures keeping financial markets operating and low interest rates. They have boosted fiscal spending to counteract the sharp drop-off in economic activity. In addition, central banks have cut interest rates and launched massive new quantitative-easing schemes to contain government borrowing costs.3
Against the background of this difficult economic situation, this paper analyzes and quantifies the monetary policy response of the People’s Bank of China (PBoC) to the COVID-19 crisis by employing a dynamic-factor modelling framework. Our strategy represents a refinement of the approach laid out by Funke and Tsang (2019), and includes the PBoC’s numerous earmarked (and sometimes arcane) monetary stimulus tools. Given China’s prominence in the world economy, Beijing’s economic policy course is followed closely.
Several studies have examined the course of China’s monetary policy and/or Chinese monetary policy shocks, including Girardin et al. (2017); Kamber and Mohanty (2018); McMahon et al. (2018); and Sun (2015, 2018). These papers use diverse empirical methodologies to estimate China’s monetary policy stance, and obtain results quite consistent with ours.4 A distinctive contribution of our paper is that we provide evidence for the PBoC’s policy response to the pandemic-induced 2020 downturn. The estimated nowcasting indicator of Chinese monetary policy presented here is thus unrivaled in the literature.
The paper proceeds as follows. The economic impact of COVID-19 on the Chinese economy and the PBoC’s monetary policy response are described in Section 2. In Section 3, we present our dynamic-factor modelling framework. Section 4 reports the estimation results and evaluates China’s monetary policy response, and Section 5 presents our conclusions and policy implications.
2. The impact of COVID-19 on the Chinese economy and the PBoC’s monetary policy response
The Chinese government’s distancing policies aimed at containing infections and saving lives prevented firms from operating (triggering a supply-side recession) and consumers from consuming (triggering a demand-side recession). In other words, the flattening of the infection curve inevitably steepened the macroeconomic recession curve. While this collateral damage was quite predictable, the extreme speed at which the crisis unfolded was unforeseen. Carmen Reinhart and Kenneth Rogoff asserted in their 2009 book on the Great Recession, This Time Is Different: Eight Centuries of Financial Folly, that we had encountered a situation without historical precedent. Only a decade later, we see the basis for assessing yet another paradigm-breaking economic upheaval. Perhaps the sequel, The COVID-19 Crisis is Different: Global Recession in the Wake of a Pandemic, is already in the works.5 In any case, the knock-on effects of the pandemic on the Chinese economy are illustrated by the various indicators in Fig.1, Fig.2 .
Fig.1.
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The mainland Chinese economy halts and begins to recover.
Fig.2.
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China’s quarterly real GDP growth rates (%).
Fig.1, Fig.2 reveal that the COVID-19 shock quickly cascaded through the economy, morphing into an unparalleled downturn simultaneously impeding demand and supply (slumps of industrial production, retail sales, the purchasing manager indices, the business confidence indices, fixed asset investment and foreign trade in first three months of 2020). In the 2nd quarter the Chinese economy then started to rebound, albeit rather uneven as the supply recovery is stronger than demand, and investment is stronger than consumption. Retail sales in particular are still rather subdued, as people continue to practice social distancing. The only bright spot is online retail.
This pattern is also apparent in GDP growth. Fig.2 shows a historic contraction in the first quarter, with Chinese GDP shrinking 6.8% from the same period a year ago. GDP then reverted to expand 3.2% in the second quarter from a year ago. In the first half however, China’s GDP still declined by 1.6% year on year. That is all to say: China’s rebound from the pandemic lockdown is impressive, but is not yet back to normal.
For policymakers, the biggest surprise has been the depth of slump. The data for January and February 2020 show that real industrial output and nominal retail sales were down by 13.5% and 20.5%, respectively. The official seasonally adjusted purchasing manager index (PMI) and the OECD business confidence index for China fell to its lowest level since the global financial crisis of 2008–2009. Within the survey indices, the non-manufacturing sectors suffered the greatest losses. These three expectation-driven indicators suggest extreme stress and illustrate that market participants were increasingly worried that the public health crisis would become a broad-reaching financial crisis. Meanwhile, the seasonally adjusted RWI/ISL Container Throughput Index dropped by 9.9 points in February – the largest monthly decline ever recorded. The driving force was the decline in sharp declines in container handling at Chinese ports and ports on the US west coast, a textbook illustration of how trade is the conduit through which a pandemic-induced slowdown passes from one country to another.
The latest June figures give a somewhat mixed picture. Real industrial output rose 4.8% in June, nominal retail sales fell 1.8%, and nominal fixed asset investment declined 3.1% in the first half of the year. Exports increased slightly in June as production capacity was gradually restored. Nevertheless, the continued spread of COVID-19 globally adds further strain on China’s exporters. While the marked improvement in industrial output, which had suffered a double-digit fall in the first two months of the year, should be noted, regaining consumer confidence as the lockdowns are lifted is increasingly as become the Achilles heel of the rebound. Without a pickup in consumption and private sector spending, worries will persist that China’s recovery could lose its momentum.
What are the current forecasts for Chinese economic growth? The latest IMF baseline forecasts for 2020 and 2021 released in June 2020 see GDP growth of 1.0% and 8.2%, respectively ( The World Bank GDP growth forecasts for 2020 and 2021, which were published in April 2020, are 2.3% and 7.7%, respectively (
We offer four comments on these assessments. First, the sudden stop in economic activity should be worse than during the Great Recession.6 Second, both the IMF and World Bank expect China’s economic growth will rebound in 2021. Third, the forecasts illustrate China’s dwindling effect on global growth. Finally, the timing and shape of the rebound remain highly uncertain.7 In total, this is a dramatic deterioration in outlook from the forecasts at the end of 2019. For example, the IMF’s semi-annual World Economic Outlook, released in October 2019, predicted a real GDP growth rate of 5.8% for China in 2020 (
This highlights the challenges for policymakers posed by the pandemic. While the immediate impact of the public health crisis can be observed, the medium and longer-term effects are difficult to predict. Will China experience a short-lockdown, quick-snapback V-shaped recovery, or will the coronavirus lead to an anemic rebound that looks like a U? While the V-shaped recession has a pointed trough, troughs are more elevated and prolonged in U-shaped recoveries. How much of the economic damage wrought by the temporary shutdown will last even after Chinese firms reopen?
There are also potential second and third waves of infection to consider. Domestic and global infections still pose a threat that could well trigger further waves of infection in China that obviates containment measures and lockdowns. This would cause a double-dip, or W-shaped recession and recovery. An ex-ante estimate of the severity and length of the pandemic is challenging to say the least.
Amidst the considerable uncertainties, the PBoC took a number of policy measures designed to combat the economic repercussions of the pandemic. The chronological sequence of Chinese monetary policy measures introduced from January 31 - July 16, 2020 are summarized in Table 1 .
Table 1.
Monetary policy measures implemented from January 31 - July 16, 2020.
| Date | Decision/action |
--- |
| 31.1 | PBoC announces that it was implementing a monetary policy package (including open market operations, standing lending facility, central bank lending, and central bank discount) to provide sufficient liquidity to the market. |
| 3.2 | PBoC injects RMB 1.2 trillion in liquidity into the banking sector through reverse repo operations, (lowering repo rates by 10 basis points). |
| 4.2 | PBoC injects RMB 500 billion in liquidity into the banking sector through reverse repo operations. |
| 7.2 | PBoC announces plans to support bond issuance by financial institutions for epidemic prevention and control. |
| 7.2 | PBoC announces plans to set up RMB 300 billion in special central bank lending (relending) to provide low-cost funds for banking lending supporting epidemic prevention and control. The central government commits to subsidizing 50% of business interest payments to ensure actual financing costs below 1.6%. |
| 10.2 | PBoC injects RMB 900 billion in liquidity into the banking sector through reverse repo operations. |
| 11.2 | PBoC injects RMB 100 billion in liquidity into the banking sector through reverse repo operations. |
| 17.2 | PBoC injects RMB 100 billion in liquidity into the banking sector through reverse repo operations, as well as RMB 200 billion in medium-term (1-year) liquidity through MLF operations (with 10-bp cut in MLF rate). |
| 20.2 | PBoC lowered 1-year loan prime rate by 10 bps and 5-year loan prime rate by 5 bps. |
| 25.2 | State Council decides to increase the PBoC’s relending and rediscount quota by RMB 500 billion for bank lending to support SMEs, as well as lower relending rate by 25 bps to 2.5%. |
| 28.2 | State Council decides to guide financial institutions to issue extra-low-interest loans with a quota of RMB 300 billion for self-employed businesses. |
| 1.3 | China Banking and Insurance Regulatory Commission urges financial institutions to defer loan principal and interest repayments. |
| 3.3 | State Council orders policy banks to add a RMB 350 billion special credit quota for loans issuing to SMEs at preferential rates. |
| 13.3 | PBoC announces earmarked required reserve ratio (RRR) cuts of 50–100 basis for loans to SMEs on 16 March, subject to banks’ performance. Cuts set to release RMB 550 billion in long-term funds. |
| 16.3 | PBoC injects RMB 100 billion in medium-term liquidity through MLF operations. |
| 30.3 | PBoC injects RMB 50 billion in liquidity into the banking sector through reverse repo operations, including a lowering of repo rates by 20 basis points. |
| 31.3 | PBoC injects RMB 20 billion in liquidity into the banking sector through reverse repo operations. |
| 31.3 | State Council decides to (i) increase the PBoC’s relending and rediscount quota by RMB 1 trillion to support SMEs; (ii) order PBoC to cut RRR for small and medium-sized banks; and (iii) reinforce the PBoC’s support for bond financing. |
| 3.4 | PBoC announces cut in RRR for small and medium banks, effective 15 April and 15 May, by 5 basis points each time. Release of RMB 400 billion to the market expected. PBoC also announces a cut in the excess deposit reserve interest rate of financial institutions in the central bank from 0.72% to 0.35%, effective 7 April. |
| 15.4 | PBoC injects RMB 100 billion in medium-term liquidity through MLF operations (with 20 bp cut in MLF rate). |
| 20.4 | PBoC lower 1-year loan prime rate by 20 bps and 5-year loan prime rate by 10 bps. |
| 15.5 | PBoC injects RMB 100 billion in medium-term liquidity through MLF operations. |
| 26.5 | PBoC injects RMB 670 billion in liquidity into the banking sector through reverse repo operations in 4 trading days from 26 May to 29 May. |
| 1.6 | PBoC announces plans to set up RMB 400 billion in special central bank lending (relending) to provide low-cost funds for banking lending supporting SMEs. |
| 4.6 | PBoC injects RMB 1.54 trillion in liquidity into the banking sector through reverse repo operations in 13 trading days between 4 June and 28 June. |
| 15.6 | PBoC injects RMB 200 billion in medium-term liquidity through MLF operations. |
| 1.7 | PBoC lower relending and rediscount rates by 25 bps. |
| 13.7 | PBoC injects RMB 130 billion in liquidity into the banking sector through reverse repo operations in 3 trading days between 13 July and 16 July. |
| 15.7 | PBoC injects RMB 400 billion in medium-term liquidity through MLF operations. |
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Notes: The calculated indicator reflects monetary policy decisions taken through July 16, 2020. The dates are based on PBoC/State Council announcements in Beijing time.
At first glance, we see the PBoC has unveiled an unprecedented set of measures intended to ensure China’s commercial banks maintain liquidity access and credit provision during the COVID-19 crisis. The chronological sequence in Table 1 further shows that the Chinese monetary policy response was not a one-off reaction, but a successive series of easing actions. Since some of the listed open market policy measures are regular and limited in duration, it is interesting to determine the extra liquidity triggered by the pandemic crisis. The COVID-19 gross liquidity injection of all operations appears to be on the order of RMB 8 trillion for the February–June period.8 The following section provides a quantitative assessment of the monetary policy measures taken.
3. Methodology and data
Judging the overall monetary policy stance in China at any given point in time is difficult due to the vast number of measures which operate through numerous channels. In light of this fact, a dynamic factor model is presented below to gauge the overall stance on monetary policy in China, which the methodology allows us to use a single model with incorporating dimension reduction and variable selection to construct a single and intuitive indicator. In specifying the dynamic factor model, it must be borne in mind that China’s monetary policy toolkit has evolved over time. Before diving into the factor modelling approach, we therefore briefly review China’s monetary policy reforms in recent years and the resulting contours of the current Chinese monetary policy landscape.
In the last decade, the PBoC has upgraded its monetary policy framework to a multiple instrument toolkit. With regard to the price-based instruments, the PBoC introduced a new corridor system of interest rates. Within the interest rate band, the policy rate is the pledged 7-day interbank market rate applied to all financial institutions (DR007). The Standing Lending Facility (SLF, a lending facility for funding within a month) serve as the upper bound of the band, while the interest rate on excess reserves (as a deposit facility tool) acts as the lower bound of the corridor is. In practice, the PBoC steers a wide interest rate corridor, anchored by the seven-day reverse-repurchase rate DR007 (rate at which banks lend to each other) and the Medium-term Lending Facility (MLF, the open market facility providing funding from three months to a year). Meanwhile the one-year loan prime rate LPR, the rate for lending to prime customers, became the new standard for all loans.9
The policy instrument toolkit also include a set of quantity-based instruments, including the reserve requirement ratio and other lending facilities, like SLF, the Medium-term Lending Facility (MLF, providing funding from three months to a year) and the Pledged Supplementary Lending (PSL, aiming at the nation’s three policy banks: China Development Bank, Agricultural Development Bank of China and the Export-Import Bank of China).
Against the background of this multitude of price-based and quantity-based monetary policy instruments, the following five variables summarize the monetary policy tools used by the PBoC: (i) the 7-day pledged repo rate (DR007); (ii) the required reserve ratio (RRR); (iii) the PBoC’s open market operations, including standing lending facility (SLF), relending and rediscount (e.g. actual amount of loans granted within the announced total of RMB 2.4 trillion in relending and rediscount quota since February); (iv) the medium-term lending facility (MLF); and (v) pledged supplemental lending (PSL). The amount of the targeted MLF stimulus measures is included in the variable Net OMO withdrawal/total loans (t - 1). Table 2 summarizes these five variables and their descriptive statistics.
Table 2.
Variables summarizing the PBoC’s current monetary policy toolkit.
| Variable | Data description | Data source |
---
| (a) Data Description |
| Changes in the 7-day pledged repo rate | Changes in the monthly average of the 7-day pledged repo rate for depository institutions in the interbank market (DR007). | National Interbank Funding Center |
| Changes in required reserve ratio (RRR) | Changes in the overall required reserve ratio (RRR) for the banking sector within the month. The overall RRR is estimated as 75%∗RRR for large banks +25%∗RRR for small and medium-sized banks. | PBoC |
| Net OMO withdrawal /total loans | Net amount of liquidity withdrawn from the banking sector during the month through the open market operations (OMO). The net amount of funds withdrawn in other items in the PBoC’s claims on the banks, e.g. standing lending facility (SLF), relending, and rediscount, is included in this variable. | PBoC |
| Net MLF withdrawal/total loans | Net amount of liquidity withdrawn from the banking sector during the month through the medium-term lending facility (MLF). | PBoC |
| Net PSL withdrawal/total loans | Net amount of liquidity withdrawn from the banking sector during the month through the pledged supplemental lending (PSL). | PBoC |
| (b) Descriptive statistics (May 2012–Jun 2020) |
| Variable | Obs. | Mean | SD | Min | Q1 | Median | Q3 | Max |
| Changes in the 7-day pledged repo rate | 98 | −0.0174 | 0.5487 | −2.8452 | −0.1215 | 0.0014 | 0.1246 | 3.3732 |
| Changes in required reserve ratio (RRR) | 98 | −0.0842 | 0.2334 | −1.0000 | 0.0000 | 0.0000 | 0.0000 | 0.0000 |
| Net OMO withdrawal /total loans | 98 | −0.0263 | 0.3761 | −1.5813 | −0.1907 | −0.0158 | 0.1793 | 0.7855 |
| Net MLF withdrawal/total loans | 98 | −0.0388 | 0.2004 | −0.7167 | −0.0915 | 0.0000 | 0.0000 | 0.6954 |
| Net PSL withdrawal/total loans | 98 | −0.0350 | 0.0413 | −0.1570 | −0.0557 | −0.0303 | 0.0000 | 0.0300 |
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Notes: Q1=First quartile (25-percentile) and Q3=third quartile (75-percentile).
Dynamic factor models are used in applied time series econometrics incorporating unobserved variables. Such models are particularly valuable in nowcasting the state of an economy. As with many useful empirical modelling approaches, factor models are extrapolated from data rather than being deduced from theory.10
A further development of the model in Funke and Tsang (2019) which includes the novel and specific and earmarked monetary policy measures in the COVID-19 pandemic is used below. The Chinese monetary policy stance indicator is based on a common element for the movements of different monetary policy instruments that can be captured by a single underlying, unobservable variable. The number of common factors is determined using the test procedure of Bai and Ng (2002). The upshot is that a single common factor is appropriate for the data-generating processes. As a result, our dynamic factor model in first differences is specified as follows:
(1)
(2)
(3)
where is the first-difference operator, is the unobserved common component at time t, are the five monetary policy instruments, are the factor loadings, ~ iid N (0), and ~ iid N (0). Furthermore, the error terms are orthogonal. The AR (2) lag structure in equations (2), (3) has been chosen to ensure the iid properties of the residuals. Each monetary policy indicator is demeaned (same as Stock and Watson, 1991) and first-differenced, while the differenced series is assumed as a weakly stationary process that has at least finite second-order moments. In the state-space representation, the measurement equation is written as
(4)
and the state equation is
(5)
The estimation procedure consists of a sequence of four steps. First, the maximum likelihood estimation method is employed to estimate the parameters of the dynamic factor model in equations (4), (5). Second, the current state of the unobserved common factors is obtained by applying the Kalman filter. The Kalman filter is a recursive procedure for computing the optimal estimate of the unobserved state vector based on the appropriate information set. The algorithm works in a two-step process. In the prediction step, the Kalman filter produces estimates of the current state variables, along with their uncertainties. Once the outcome of the next measurement is observed, these estimates are updated using a weighted average, with more weight being given to estimates with higher certainty. Third, the monetary policy stance time series () is calculated by accumulating the estimated series , assuming the initial value . Finally, the monetary policy stance time series is rescaled to the range from −2 to +2.11
4. Estimation results and policy evaluation
The maximum-likelihood estimation results of the dynamic factor models are given in Table 3 , while Fig.3 shows the resulting indicators. Lower (higher) values of the indicator represent a monetary policy easing (tightening). The baseline indicator only takes into account measures encompassing all banks and the whole economy. The alternative indicator additionally takes into account the earmarked RRR cuts for loans to small and medium-sized enterprises (SMEs) announced on March 13, 2020 (effective March 16, 2020, see Table 1). These loans are intended to help SMEs bear fixed costs such as rent, interest payments, and tax bills. The PBoC estimates that the earmarked RRR cut of March 16, 2020 led to a liquidity increase in the banking sector of approximately RMB 550 billion. This is equivalent to a 0.34 percentage-point cut in the standard RRR (for every 0.5 percentage-point cut, RMB 800 billion is released).12 The indicators are generally the same and only differ from March 2020 onwards.
Table 3.
Parameter estimates of the dynamic factor models.
| Variables | Parameters | Baseline Model | Alternative Model |
--- --- |
| | | −0.1884 (1.3) | −0.1607 (1.2) |
| | | −0.0430 (0.3) | −0.0262 (0.2) |
| | | 0.0928 (1.5) | 0.1028 (1.9) |
| | | −0.3152 (2.8) | −0.3185 (3.2) |
| | | −0.0930 (0.9) | −0.0938 (0.9) |
| | | 0.2685 (6.1) | 0.2665 (6.9) |
| | | 0.2161 (8.8) | 0.2139 (9.2) |
| | | 0.5422 (0.3) | 0.1223 (0.4) |
| | | 0.2410 (0.2) | 0.5909 (1.8) |
| | | 0.0026 (0.3) | 0.0046 (0.8) |
| | | −0.0506 (1.1) | −0.0508 (1.3) |
| | | −0.2734 (2.8) | −0.2744 (2.9) |
| | | −0.1225 (1.2) | −0.1253 (1.3) |
| | | 0.1287 (7.7) | 0.1286 (7.0) |
| | | −0.0676 (3.9) | −0.0682 (4.4) |
| | | 0.4273 (3.9) | 0.4274 (4.4) |
| | | −0.1761 (1.6) | −0.1717 (1.8) |
| | | 0.0306 (6.7) | 0.0306 (6.9) |
| | | −0.0045 (1.4) | −0.0053 (1.7) |
| | | 0.4131 (4.0) | 0.4180 (4.2) |
| | | 0.2512 (2.5) | 0.2476 (2.4) |
| | | 0.0011 (7.8) | 0.0011 (7.1) |
| Log likelihood | 116.58 | 116.08 |
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Note: t-values given in parentheses.
Fig.3.
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Dynamic factor model - based indicators of China’s monetary policy stance.
In retrospect, China’s central bank has been highly interventionist. So what was the reaction to the COVID-19 shock? How pronounced is the monetary policy boost to counteract a possible COVID-19 meltdown? And what does the composite monetary policy stance indicator reveal about the economic outlook of the PBoC?
The first impression is that China’s post-COVID-19 monetary policy is as expansionary as it was after the global downturn 2008–2009. However, compared with the global financial crisis, the PBoC has this time used a different policy mix. While the focus in the years following the global financial crisis was on reducing the benchmark interest rates and the RRR, this time liquidity injections and earmarked policy measures play a major role. One reason for this is the lower interest rate level that has already been achieved. In the period September to December 2008, the 1-year benchmark lending rate dropped from 7.47% to 5.31% (215-bp cut), while the 1-year loan prime rate was reduced by 30-bp from 4.15% to 3.85% in the first half-year 2020. In both crises the RRR was lowered. In the period September to December 2008, the reserve ratios for small and medium-sized banks (large banks) were reduced by 4 (2) percentage points. The RRR cut released about RMB 1000 billion. Since February 2020, the overall RRR cut in March released long-term funding of about RMB 550 billion. Furthermore, two more targeted RRR reductions by 0.5 percentage points for small and medium banks released about RMB 400 billion liquidity. With regard to open market operations (repos and reverse repos), the gross injections were RMB 982.5 billion in the period September to December 2008, and RMB 5080 billion in February–June 2020. Finally, in the first half-year 2020 liquidity in the amount of RMB 2007 billion was provided by means of MLF, relending and rediscounting injections. This monetary policy instrument did not yet exist in 2008.
On the other hand, a high level of liquidity has already been injected into the banking sector prior to the COVID-19 pandemic shock. The net injection (deducting the amount of matured facilities from the gross injection) through various loan facilities (measured by the changes in PBoC’s claims on the banking sector) and RRR cut was RMB 9.4 trillion between April 2018 and January 2020 (of which RMB 6.3 trillion was made until January 2019). In contrast, the PBoC’s net injection during February–June 2020 was only RMB 29 billion into the banking sector. In other words, the loose monetary policy stance after the financial meltdown 2008–2009 and prior to the COVID-19 shock explains the comparatively restrained reaction at this point in time. This is consistent with the PBoC’s ongoing efforts towards financial stability.
Another feature is the sequential step-by-step approach since the beginning of 2020 which is nicely illustrated by the index. This incremental approach may be due to the declining number of new domestic COVID-19 infections. Although concerns about a second wave of infection persist, the initial assessment is that the economic impact of the pandemic has been a sharp, but short, recession. It will be followed by a V-shaped recovery with a return to normal in the second half of 2020 and growth accelerating in 2021. The supply-side recovery shown in Fig.4 points towards such a swift rebound as happened with China’s SARS outbreak in 2003. This assessment is supported by readings of the Manufacturing Purchasing Managers Index (PMI), which surged to above 50 in March 2020 after hitting a record low of 35.7 in February. The seasonally adjusted month-on-month growth rates of industrial production, retail sales, and fixed asset investment also recorded rebounds of 29.6%, 0.9% and 6.2% respectively. This improvement contrasts with the contraction in the first two month of the year.
Fig.4.
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China gets back to work.
There is another explanation for the initial restrained monetary policy response, however. The perception could have been that traditional monetary policy slashing interest rates and providing liquidity is not the solution to a virus outbreak and thus ill-advised. In a nutshell, severe viral outbreaks are not “typical recessions” due to the peculiarity of the shock the economy faces, i.e. a shock that involves both supply-side and demand-side channels. It follows that more targeted interventions show more promise in the event of a severe viral outbreak. An example of this is the earmarked measure for loans to SMEs included in the alternative indicator. From this angle, the PBoC’s swift monetary policy easing propping up the coronavirus-hit Chinese economy reflects the well-understood unorthodox nature of the shock. This also explains why many measures target specific industries, avoiding a “flood-like” easing to lift all boats as in previous slowdowns. This assessment is corroborated by comparison with other countries. Despite the adopted counter-cyclical monetary policy stance aimed at creating confidence and limiting the amplification of the shock, the PBoC’s course of action appears remarkably relaxed and restrained compared with the exceptional array of pandemic-fighting measures deployed around the world to prevent cascading defaults and market panic.13
As a final evaluation step, a robustness analysis shall be carried out. Despite the breadth of the dynamic factor modelling approach, a comparison with alternative approaches is interesting. Sun et al. (2012) have suggested that the McCallum rule can be a useful tool for analysing the monetary policy stance. McCallum (1987) has advocated a policy rule that requires central banks to target the growth rate of nominal GDP using the monetary base as its instrument. What is the course of the dynamic factor model - based indicator and the growth in the monetary base? The left panel in Fig.5 shows that (i) both time series have very similar peaks and troughs; and (ii) both time series are time-shifted. For this reason, in the right panel of Fig.5 the dynamic factor model - based indicator has been plotted against the 12-month lead of the growth rate of the monetary base. The alignment of the two time series shows that the dynamic factor model - based indicator is a leading indicator of the growth rate of the monetary base, with the lead amounting to 12 months.
Fig.5.
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Dynamic factor model - based indicator of Chinese monetary policy stance and the annual growth of the monetary base.
5. Conclusions
Understanding the COVID-19 policy reaction of China is critical. The sheer size of the Chinese market and China’s integration into the global supply chains has made it a significant driver of global growth. For this reason, the economic policy response in China and the country’s economic development play a key role in the global economic recovery.14 Against this background, this paper analyses the PBoC’s monetary policy response to counter the COVID-19 pandemic and quantifies it using a dynamic factor model.
Our main findings can be summarized as follows. First, the dynamic factor model allows estimation of the monetary policy stance which results from China’s unorthodox and earmarked mix of monetary policy instruments. Second, the established monthly indicators reveal in concise form the stepwise response of the PBoC to the outbreak of the COVID-19 pandemic over time. Third, the paper demonstrates that the PBoC has implemented a remarkable variety of policy measures following the COVID-19 shock. Fourth, our estimates reveal that the PBoC has implemented novel policy measures to ensure that commercial banks maintain liquidity access and credit provision during the COVID-19 crisis. Fifth, the current monetary loosening is swift and decisive, but less prominent than after the global financial crisis 2008–2009. This contrasts with the radical and unique policies of many other countries to counter the COVID-19 shock. The reason is that the Chinese authorities keep an eye on containing the nation’s debt pile. In particular, they still seek to rein in the exuberant credit growth. It may also be motivated by the PBoC’s assessment that the recvovery from the Chinese downturn is “so far, so V”. The years ahead will show the soundness of this judgement and establish that the course of action was adequate to meet these exceptional challenges.
Acknowledgement
We would like to thank two anonymous reviewers and the editor for helpful comments on an earlier draft. We also thank Riikka Nuutilainen (Bank of Finland) and Zuzana Fungacova (Bank of Finland) for their helpful comments on an earlier draft. The usual disclaimer applies. A previous version of the paper has been published as Bank of Finland BOFIT Discussion Paper No. 12/2020 (
Footnotes
1
For the classic epidemiological model of epidemic dynamics and its domestic and international spread, see Allen (2017), Balcan et al. (2010), Chinazzi et al. (2020) and Wu et al. (2020).
2
The economic literature on the COVID-19 pandemic is rapidly expanding. For the various facets of the COVID-19 shock and possible policy responses, see Baldwin and Weder di Mauro (2020a, 2020b); Alvarez et al. (2020); Atkeson (2020); Baker et al. (2020); Berger et al. (2020); Caballero and Simsek (2020); Coibion et al. (2020); Eichenbaum et al. (2020a, 2020b); Fornaro and Wolf (2020); Guerrieri et al. (2020); Krueger et al. (2020); and Lewis et al. (2020). Maliszewska et al. (2020); Pindyck (2020); and Stock (2020) study the macroeconomic and trade effects and shock transmission of the COVID-19 pandemic. For historical lessons from previous pandemics, see Barro et al. (2020); Greenwood et al. (2019); Jordà et al. (2020); and Velde (2020). Leiva-Leon et al. (2020) have developed a nowcasting COVID-19 global recession risk indicator. None of these papers focus on China.
3
For cross-national, cross-temporal government response trackers aiming to monitor and compare government responses to the pandemic, see the IMF COVID-19 economic response tracker at and the Oxford COVID-19 Government Response Tracker at which measure the stringency of the responses.
4
For a thorough comparison see Funke and Tsang (2019), pp. 18–21.
5
Also see
6
The implications of this unexpected GDP growth slowdown for financial stability are difficult to assess. In 2019, the PBoC stress-tested the resilience of 30 banks under a variety of scenarios. In the then-presumed most extreme hit to the economy envisaged (growth slowing to 4.15%), 17 of the 30 commercial banks modeled would need additional capital ( Also see PBoC answers to journalist questions (
7
The rebound much depends on the development of new pharmaceuticals and vaccines. Otherwise, sustained social distancing may need to be maintained in the postpandemic period into 2022 to curb the outbreak. See Kissler et al. (2020).
8
The gross liquidity injection via various loan facilities is calculated regardless of maturing monetary policy measures which include (i) the short-term liquidities (maturing in 7 or 14 days) of RMB 5.08 trillion through PBoC’s reverse repo operations and SLF; (ii) RMB 700 billion in medium-term funding (maturing in 1 year or longer) through MLF and PSL; (iii) RMB 1.3 trillion in relending and rediscount loans actually granted by 7 July (see (iv) RMB 400 billion liquidity released by two targeted 0.5-percentage-point RRR cuts for the small and medium banks in April and May respectively; and (v) RMB 550 billion in long-term funds released from the earmarked required reserve ratio (RRR) cuts of 50–100 basis points for loans to SME firms since 16 March.
9
China has made good progress in recent years in liberalising interest rates. Although the policy changeovers are not yet completed, the IMF (2017, p. 34) reached a tentative verdict already in 2017 that “the conduct of [China’s] monetary policy increasingly resembles a standard interest-rate-based framework”.
10
This article is not the venue for a complete review of the factor modelling approach. For worth reading contributions to the theory of dynamic factor models, see Forni et al. (2000, 2004, 2005) and Stock and Watson (1991,2002a, 2002b). Against the background of possible pandemic-related structural breaks, it is noteworthy that factor models are rather robust to parameter instabilities. See Stock and Watson (2009, pp. 1–57) and Bates et al. (2013). For comprehensive textbook treatments, see Harvey (1989) and Durbin and Koopman (2012).
11
For understanding the calculated index, two explanatory notes must be given. (i) The fact that the index is “0” in a given month does not mean necessarily that the PBoC has assumed a neutral policy stance; and (ii) an index value of “-2” (”+2”) does not mean that further monetary policy loosening (tightening) is no longer possible.
12
The informal and difficult-to-quantify window guidance policy is not included for either indicator. For a thorough DSGE-based analysis of China’s window guidance policy, see Chen et al. (2020).
13
It is noteworthy that China’s fiscal policy response to the pandemic has so far been relatively restrained by international standards. In contrast, many advanced economies have implemented substantial fiscal measures according to the principle “there is no time to lose”. Corporate bailouts have been a core element. However, it must be borne in mind that China may not need such unprecedented fiscal parachutes as they are already built into the system. The predominant share of corporate loans go to state-owned firms and hence already enjoy implicit government guarantees. See the IMF policy action tracker at
14
Adverse growth scenarios in China will have noticeable effects on economic activity in different countries and regions due to the increasingly trade- and financially-integrated world. For the literature on the spillover effects of a slowdown in Chinese growth, see, for example, Dieppe et al. (2018), Gilhooly et al. (2018), and Ma et al. (2017).
Appendix A
Supplementary data to this article can be found online at
Appendix A. Supplementary data
The following is the Supplementary data to this article:
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Abstract
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1. Introduction
2. The impact of COVID-19 on the Chinese economy and the PBoC’s monetary policy response
3. Methodology and data
4. Estimation results and policy evaluation
5. Conclusions
Acknowledgement
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Appendix A. Supplementary data
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10627 | https://byjus.com/physics/unit-of-density/ | Density is a measurement that compares the amount of matter an object has to its volume. An object with much matter in a certain volume has a high density. In this article, we will learn about the definition of density and the units of density.
| |
| Table of Content: What Is Density? Density Examples How Is Density Calculated? Unit of Density SI Unit of Density Other Density Units Application of Density in Real Life Solved Examples Frequently Asked Questions – FAQs |
What Is Density?
The density of material shows the denseness of that material in a specific given area. A material’s density is defined as its mass per unit volume. Density is essentially a measurement of how tightly matter is packed together. It is a unique physical property of a particular object. The principle of density was discovered by the Greek scientist Archimedes. It is easy to calculate density if you know the formula and understand the related units The symbol ρ represents density or it can also be represented by the letter D.
| |
| Density Definition: Density is the measurement of how tightly a material is packed together. It is defined as the mass per unit volume. |
| Density Symbol: D or ρ |
| Density Formula: ρ = m/V, where ρ is the density, m is the mass of the object and V is the volume of the object. |
Density Examples
Iron, platinum, and lead are examples of dense materials. Many types of rock and minerals are examples of dense material. Materials that are dense are most likely to ‘feel’ heavy or hard. The opposite of dense is sparse and a few examples of sparse material are glass, bamboo, aluminium, and styrofoam.
In general, liquids are less dense than solids and gases are less dense than liquids. This is due to the fact that solids have densely packed particles, liquids are materials where particles can slide around one another, and gases have particles that are free to move all over the place.
How Is Density Calculated?
Mathematically, the density of an object is expressed as follows:
| |
| Where, ρ is the density, m is the mass and V is the volume |
Unit of Density
Though the SI unit of density is kg/m³, for convenience we use g/cm³ for solids, g/ml for liquids, and g/L for gases.
Density can be explained as the relationship between the mass of the substance and the volume it takes up.
In a qualitative term, it shows how much heavy an object is at constant volume.
Different substances have different densities, which means for the same volume of different substances weigh differently.
SI Unit of Density
Each substance has a specific density. Generally, the density of water (which is approximately about 1 gram/cubic centimetre) is taken as the standard value for calculating the density of substances. However, the SI unit of Density is measured using kilograms per cubic metre (kg/m3).
Read More: Density of Water
Other Density Units
Talking about other density units, metric tons and litres are also used even though they are not part of the SI. Some other units include:
gram per millilitre (g/mL)
metric ton per cubic metre (t/m3)
kilogram per litre (kg/L)
megagram (metric ton) per cubic metre (mg/m3)
gram per cubic centimetre (g/cm3)
1 g/cm3 = 1000 kg/m3
kilogram per cubic decimetre (kg/dm3)
In addition to this, in the cgs system density is measured in grams per cubic centimetre (g/cm3).
Similar Reading:
Unit of Weight
Unit of Work
Applications of Density in Real Life
Many applications of density are there in our real-life, life a few examples are in pipe design, shipbuilding, helium balloons, weight distribution in the aeroplane, and the fact that ice floats on water.
The knowledge of the densities of two substances helps you in separation techniques. For example, the separation of oil from water. If there is a leakage of an oil tank in the ocean then oil drops start to float on the water due to less density than the water.
Another well-known application of density is determining whether an object will float on water or not. The floating of ships and diving of submarines are due to their density difference.
Solved Examples
Question 1:
Take two boxes that have the same volume. Fill the first box with x balls and the second box with 6x balls. If the mass of each ball is the same, which box would weigh more?
Solution:
The box that has more balls has more mass per unit of volume.
Here the first box contains x number of balls and the second box contains a 6x number of balls. Since the number of balls in the second box is 6 times the first box, the second box would weigh more. This property of matter is called density.
Question 2:
Calculate the density of water if it has a mass of 1160 Kg and a volume of 1 m3?
Solution:
Given,
Mass = 1160 Kg
Volume = 1 m3
Density is given by the formula:
Density = Mass/Volume
ρ = 1160/1 = 1160 kg/m3
Questions 3:
If you find a shiny rock, a carbon allotrope with a volume of 0.042 cm3 and a mass of 0.14 g, is it graphite or diamond? The density of graphite is 2.266 g/cm3 and the density of diamond is 3.51g/cm3.
Solution:
Given,
Volume of the shiny rock =0.042 cm³
Mass of the shiny rock = 0.14 g
Density of graphite = 2.266 g/cm3
Density of diamond = 3.51g/cm3
Use the density equation to solve for m, for the mass of graphite and for the mass of a diamond.
ρ = m/V
m = ρV
m = 2.266 g/cm³ x 0.042 cm³ = 0.0951g for graphite
m = 3.51g/cm3 x 0.042 cm³ = 0.1474 g for diamond
The mass of the shiny rock you found is identical with the mass of diamond.
The below video is an explanation of the properties of water
77,020
Frequently Asked Questions – FAQs
Q1
What is Density?
A material’s density is defined as its mass per unit volume.
Q2
Who discovered the principle of Density?
The principle of density was discovered by the Greek scientist Archimedes.
Q3
How would you find the density of a human body?
The density of a human body can be determined by the following expression:
Density = Mass/ Volume
The mass of the human body can be calculated by using a weight scale. The submersion displacement is used to get the volume of the human body. If we fill a tub with water and let the person fully submerge into the water, then we see a rise in the water, this rise is equal to the volume of the body. Thus, the density of the human body can be obtained by dividing the mass of the human body by the volume of the human body.
Q4
How will we know if a substance is less dense than water?
If a substance weighs less than an equal volume of water, it is less dense and will float.
Q5
What happens to the least dense of two immiscible liquids?
If the liquids are immiscible, and they are not stirred, or only stirred gently, they will separate into 2 layers, with the less dense floating on the more dense liquid.
If you wish to learn more physics concepts with the help of interactive video lessons, download BYJU’S – The Learning App.
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10628 | https://www.sparkl.me/learn/collegeboard-ap/precalculus/converting-between-logarithmic-and-exponential-forms/revision-notes/1244 | 1.1.1
Graphing parametric functions in x-y planes
1.1.2
Comparing parametric and Cartesian forms
1.1.3
Identifying parametric equations for simple curves
1.2.1
Calculating derivatives of parametric equations
1.2.2
Understanding tangent lines in parametric graphs
1.2.3
Exploring relationships between x and y changes
1.3.1
Adding and subtracting vectors
1.3.2
Computing magnitudes and directions
1.3.3
Representing motion using vector components
1.4.1
Graphing 2D vector fields
1.4.2
Understanding velocity and acceleration in vector form
1.4.3
Combining vectors with scalar functions
1.5.1
Basics of matrix addition and multiplication
1.5.2
Representing linear equations with matrices
1.5.3
Analyzing matrix properties algebraically
1.6.1
Understanding matrix-based transformations
1.6.2
Verifying transformations using input-output pairs
1.6.3
Connecting linear transformations to geometry
1.7.1
Solving systems of equations with inverse matrices
1.7.2
Verifying solutions using determinants
1.7.3
Simplifying complex systems algebraically
2.1.1
Identifying transformations of logarithmic functions
2.1.2
Solving logarithmic equations symbolically
2.1.3
Defining logarithmic bases and their properties
2.2.1
Expanding and simplifying logarithmic terms
2.2.2
Combining terms with the same base
2.2.3
Factoring logarithms for solving equations
2.3.1
Solving inequalities involving logarithms
2.3.2
Validating solutions using substitution
2.3.3
Using exponentiation to verify logarithmic solutions
2.4.1
Representing exponential data with semi-log scales
2.4.2
Identifying linear trends in transformed data
2.4.3
Comparing semi-log plots to regular plots
2.5.1
Recognizing differences in arithmetic sequences
2.5.2
Identifying ratios in geometric sequences
2.5.3
Connecting sequences to linear and exponential functions
2.6.1
Identifying linear growth or decay rates
2.6.2
Comparing linear vs. exponential rates of increase
2.6.3
Understanding when exponential models are appropriate
2.7.1
Understanding domain and range of exponential functions
2.7.2
Analyzing transformations: vertical and horizontal shifts
2.7.3
Exploring asymptotic behavior in exponential graphs
2.8.1
Simplifying exponential expressions using rules of exponents
2.8.2
Factoring exponential terms in equations
2.8.3
Expanding powers for simplification
2.9.1
Representing patterns algebraically
2.9.2
Validating assumptions in exponential equations
2.9.3
Generalizing growth models without specific real-world scenarios
2.10.1
Testing linear vs. exponential models for fit
2.10.2
Comparing residuals to determine accuracy
2.10.3
Analyzing errors in model assumptions
2.11.1
Combining two exponential functions
2.11.2
Using compositions for recursive sequences
2.11.3
Exploring commutativity in function compositions
2.12.1
Identifying conditions for inverses
2.12.2
Verifying inverse relationships algebraically
2.12.3
Sketching inverse functions on a graph
2.13.1
Expanding logarithmic terms using rules of logs
2.13.2
Simplifying nested logarithms
2.13.3
Converting between logarithmic and exponential forms
2.14.1
Analyzing inverse symmetry in graphs
2.14.2
Understanding logarithms as inverses of exponents
2.14.3
Exploring how inverse transformations affect function shapes
3.
Polynomial and Rational Functions
3.1.1
Simplifying rational functions to reveal holes
3.1.2
Testing function continuity at holes
3.1.3
Identifying removable discontinuities
3.2.1
Factoring and simplifying expressions
3.2.2
Expanding rational expressions
3.2.3
Verifying equivalency through algebraic manipulation
3.3.1
Horizontal and vertical shifts
3.3.2
Stretching and compressing functions
3.3.3
Reflecting across axes
3.4.1
Choosing appropriate polynomial or rational models
3.4.2
Testing assumptions through equation constraints
3.4.3
Analyzing limitations of chosen models
3.5.1
Writing polynomial and rational models
3.5.2
Simplifying derived models for analysis
3.5.3
Interpreting model parameters algebraically
3.6.1
Identifying relationships between variables
3.6.2
Representing co-variation with graphs and tables
3.6.3
Recognizing proportional relationships
3.7.1
Definition and calculation of average rate of change
3.7.2
Interpreting instantaneous rate of change
3.7.3
Slope as a measure of rate in linear functions
3.8.1
Identifying constant rates in linear models
3.8.2
Analyzing variable rates in quadratic functions
3.8.3
Determining intervals of increase and decrease
3.9.1
Graphical representation of polynomial behaviors
3.9.2
Understanding the degree's role in rate analysis
3.9.3
Identifying turning points and critical points
3.10.1
Finding complex zeros using the quadratic formula
3.10.2
Factoring polynomials with real and complex roots
3.10.3
Verifying solutions with synthetic division
3.11.1
Predicting end behavior using leading coefficients
3.11.2
Determining symmetry in polynomial graphs
3.11.3
Understanding multiplicities of roots
3.12.1
Determining horizontal asymptotes
3.12.2
Analyzing degrees of numerator and denominator
3.12.3
Identifying oblique asymptotes
3.13.1
Finding zeros in rational functions
3.13.2
Multiplicity and their impact on graph shape
3.13.3
Using synthetic and long division
3.14.1
Defining vertical asymptotes algebraically
3.14.2
Testing continuity near asymptotes
3.14.3
Identifying behaviors approaching vertical asymptotes
4.
Trigonometric and Polar Functions
4.1.1
Understanding cycles in functions
4.1.2
Representing periodic intervals
4.1.3
Exploring periodicity in graphs
4.2.1
Computing trigonometric ratios using the unit circle
4.2.2
Verifying identities involving sine, cosine and tangent
4.2.3
Exploring co-functions and their relationships
4.3.1
Computing exact values for common angles
4.3.2
Understanding reflection and symmetry properties
4.3.3
Analyzing periodic properties of sine and cosine
4.4.1
Graphing sine and cosine with amplitude and period
4.4.2
Identifying shifts in sine and cosine graphs
4.4.3
Exploring relationships between graphs of sine and cosine
4.5.1
Graphing tangent over its domain
4.5.2
Identifying vertical asymptotes in tangent graphs
4.5.3
Analyzing tangent transformations
4.6.1
Determining restricted domains of inverses
4.6.2
Using inverse functions to solve equations
4.6.3
Verifying identities with inverses
4.7.1
Solving multi-angle equations
4.7.2
Exploring periodic solutions in equations
4.7.3
Analyzing domain-specific inequalities
4.8.1
Graphing reciprocal functions of sine, cosine and tangent
4.8.2
Identifying domains and asymptotes
4.8.3
Verifying reciprocal relationships in identities
4.9.1
Using Pythagorean identities to simplify
4.9.2
Verifying equivalences through substitution
4.9.3
Rewriting expressions for computational ease
4.10.1
Converting Cartesian equations to polar form
4.10.2
Graphing polar coordinates
4.10.3
Analyzing polar symmetry and rotations
4.11.1
Graphing cardioids, roses and spirals
4.11.2
Analyzing periodicity and intersections in polar graphs
4.11.3
Using technology to explore complex polar shapes
Converting between logarithmic and exponential forms
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Converting between Logarithmic and Exponential Forms
Introduction
Understanding the relationship between logarithmic and exponential forms is fundamental in Precalculus, particularly for students preparing for the College Board AP exams. This topic not only underpins various mathematical concepts but also serves as a critical tool for solving equations involving exponents and logarithms. Mastery of these conversions enhances problem-solving skills and is essential for higher-level studies in mathematics and related disciplines.
Key Concepts
Definitions and Fundamental Concepts
Before delving into conversions, it's essential to understand the definitions of logarithms and exponents:
Exponential Form: An expression where a constant base is raised to a power. It is written as by=x, where b>0 and b=1.
Logarithmic Form: The inverse of the exponential form, expressing the power to which a base must be raised to obtain a certain value. It is written as logbx=y.
Understanding the Inverse Relationship
Exponents and logarithms are inverse functions. This means that each operation reverses the effect of the other:
If by=x, then logbx=y.
If logbx=y, then by=x.
This inverse relationship is crucial for solving equations where the unknown variable is in the exponent or within a logarithm.
Converting from Exponential to Logarithmic Form
To convert an exponential equation to its logarithmic form:
Identify the base, exponent, and the result in the exponential equation by=x.
Express it as logbx=y.
Example: Convert 23=8 to logarithmic form.
Base (b): 2
Exponent (y): 3
Result (x): 8
Logarithmic form: log28=3.
Converting from Logarithmic to Exponential Form
To convert a logarithmic equation to its exponential form:
Identify the base, the logarithm result, and the argument in the logarithmic equation logbx=y.
Express it as by=x.
Example: Convert log525=2 to exponential form.
Base (b): 5
Logarithm result (y): 2
Argument (x): 25
Exponential form: 52=25.
Properties of Logarithms and Exponents
Understanding the fundamental properties aids in simplifying and converting expressions:
Product Property: logb(MN)=logbM+logbN
Quotient Property: logb(NM)=logbM−logbN
Power Property: logb(Mk)=klogbM
Change of Base Formula: logbx=logkblogkx, where k is a positive number different from 1.
Exponent Rules:
Product Rule: by1⋅by2=by1+y2
Quotient Rule: by2by1=by1−y2
Power Rule: (by)k=byk
Solving Equations Involving Exponents and Logarithms
Conversion between logarithmic and exponential forms is instrumental in solving various equations:
Exponential Equations: Equations where the variable is in the exponent, such as 3x=81.
Logarithmic Equations: Equations involving logarithms, such as log2(x)=5.
Steps to Solve Exponential Equations:
Convert the equation to logarithmic form.
Isolate the variable.
Example: Solve 2x=32.
Convert to log232=x.
Since 25=32, x=5.
Steps to Solve Logarithmic Equations:
Convert the equation to exponential form.
Isolate the variable.
Example: Solve log3x=4.
Convert to 34=x.
Calculate x=81.
Applications of Logarithmic and Exponential Conversions
These conversions are widely applicable in various fields:
Scientific Calculations: Handling exponential growth and decay, such as population growth or radioactive decay.
Engineering: Signal processing and acoustics, where logarithms are used to measure sound intensity.
Finance: Calculating compound interest and investment growth over time.
Computer Science: Algorithms involving exponential time complexity and logarithmic optimizations.
Advanced Concepts and Problem-Solving Strategies
For higher-level problems, especially those encountered in standardized tests like the AP exams, a deeper understanding is required:
Exponential and Logarithmic Equations: Solving equations that combine both forms, such as 2x=3log2x.
Graphing: Understanding the graphical representations of exponential and logarithmic functions to identify intersections and asymptotes.
Inverse Functions: Utilizing the inverse nature to switch between functions for simplification and solution.
Real-World Problems: Modeling real-life scenarios like compound interest, pH levels in chemistry, and Richter scale measurements.
Common Mistakes and How to Avoid Them
Students often encounter challenges when converting between forms due to misconceptions:
Incorrect Base Identification: Misidentifying the base during conversion can lead to incorrect results. Always clearly identify b, x, and y in the equations.
Ignoring Domain Restrictions: Logarithms are only defined for positive real numbers. Ensure that the arguments of logarithms are positive.
Misapplying Properties: Incorrect use of logarithmic properties can complicate problems. Practice applying each property correctly.
Arithmetic Errors: Simple calculation mistakes can derail the solution. Double-check all calculations, especially when dealing with exponents.
Tip: Always revisit the original equation to verify the correctness of the converted form.
Worked Examples
Example 1: Convert log464=x to exponential form and solve for x.
Exponential form: 4x=64.
Since 4=22 and 64=26, rewrite as (22)x=26.
Simplify: 22x=26.
Equate exponents: 2x=6 ⇒ x=3.
Example 2: Solve for y in the equation 32y−1=81.
Convert to logarithmic form: log381=2y−1.
Since 34=81, 4=2y−1.
Solve for y: 2y=5 ⇒ y=25.
Comparison Table
| | | |
---
| Aspect | Exponential Form | Logarithmic Form |
| Definition | Expresses a number as a base raised to an exponent, by=x. | Expresses the exponent as a logarithm, logbx=y. |
| Use Case | Modeling growth and decay, compound interest, population growth. | Solving for exponents, simplifying multiplication into addition. |
| Inverse Relationship | Inverse of logarithmic form. | Inverse of exponential form. |
| Properties | Product, Quotient, and Power Rules for exponents. | Product, Quotient, and Power Properties of logarithms. |
| Advantages | Efficient for calculations involving repeated multiplication. | Simplifies complex multiplication and division into addition and subtraction. |
| Limitations | Only defined for positive bases not equal to 1. | Only defined for positive arguments. |
Summary and Key Takeaways
Exponential and logarithmic forms are inverse functions essential in Precalculus.
Conversion between forms facilitates solving complex equations involving exponents and logarithms.
Understanding properties and applications enhances problem-solving skills in various mathematical contexts.
Avoid common mistakes by accurately identifying bases and adhering to domain restrictions.
Mastery of these concepts is crucial for success in College Board AP exams and advanced studies.
Coming Soon!
Examiner Tip
Tips
Memorize Key Properties: Familiarize yourself with the fundamental properties of exponents and logarithms to simplify conversions and solve equations efficiently.
Check Your Work: After converting forms, substitute back into the original equation to verify the accuracy of your conversion.
Use Change of Base Formula: For logarithms with unfamiliar bases, apply the change of base formula logbx=logkblogkx to simplify calculations using common bases like 10 or e.
Practice Consistently: Regular practice with diverse problems enhances retention and understanding, crucial for performing well in AP exams.
Create Mnemonics: Develop memory aids for properties and steps involved in conversions to recall them quickly during tests.
Did You Know
Did You Know
The concept of logarithms was introduced by John Napier in the early 17th century to simplify complex calculations, revolutionizing mathematics and astronomy.
Natural logarithms, with base e (approximately 2.71828), are extensively used in continuous growth models, such as population dynamics and compound interest calculations.
Logarithmic scales are used in measuring earthquake magnitudes on the Richter scale, allowing a wide range of energy releases to be represented in a manageable format.
Common Mistakes
Common Mistakes
Incorrect Base Identification:
Students may confuse the base when converting forms.
Incorrect: log28=3 converted to 23=8 is correct, but misidentifying the base can lead to errors.
Ignoring Domain Restrictions:
Forgetting that logarithms are only defined for positive numbers.
Incorrect: Trying to compute logb(−x) where x>0.
Misapplying Properties:
Applying logarithmic properties incorrectly, such as logb(M+N)=logbM+logbN, which is false.
Correct Approach: Use logb(MN)=logbM+logbN.
FAQ
What is the difference between exponential and logarithmic forms?
Exponential form expresses a number as a base raised to an exponent, such as by=x, while logarithmic form is the inverse, expressing the exponent as a logarithm, like logbx=y.
How do I convert an exponential equation to logarithmic form?
Identify the base, exponent, and result in the exponential equation by=x, then rewrite it as logbx=y.
When should I use the change of base formula?
Use the change of base formula logbx=logkblogkx when you need to evaluate logarithms with bases that are not supported by your calculator or to simplify complex logarithmic expressions.
Can logarithms have negative arguments?
No, logarithms are only defined for positive real numbers. Attempting to take the logarithm of a negative number is undefined.
Why are logarithms important in real-world applications?
Logarithms are crucial in various fields such as science, engineering, finance, and computer science for modeling exponential growth and decay, measuring sound intensity, calculating compound interest, and optimizing algorithms.
How do I solve equations that involve both exponents and logarithms?
Use the inverse relationship between exponents and logarithms to isolate the variable. Convert the equation into one form using logarithmic or exponential expressions, then solve for the unknown.
1.
Functions Involving Parameters, Vectors and Matrices
1.1.1
Graphing parametric functions in x-y planes
1.1.2
Comparing parametric and Cartesian forms
1.1.3
Identifying parametric equations for simple curves
1.2.1
Calculating derivatives of parametric equations
1.2.2
Understanding tangent lines in parametric graphs
1.2.3
Exploring relationships between x and y changes
1.3.1
Adding and subtracting vectors
1.3.2
Computing magnitudes and directions
1.3.3
Representing motion using vector components
1.4.1
Graphing 2D vector fields
1.4.2
Understanding velocity and acceleration in vector form
1.4.3
Combining vectors with scalar functions
1.5.1
Basics of matrix addition and multiplication
1.5.2
Representing linear equations with matrices
1.5.3
Analyzing matrix properties algebraically
1.6.1
Understanding matrix-based transformations
1.6.2
Verifying transformations using input-output pairs
1.6.3
Connecting linear transformations to geometry
1.7.1
Solving systems of equations with inverse matrices
1.7.2
Verifying solutions using determinants
1.7.3
Simplifying complex systems algebraically
2.
Exponential and Logarithmic Functions
2.1.1
Identifying transformations of logarithmic functions
2.1.2
Solving logarithmic equations symbolically
2.1.3
Defining logarithmic bases and their properties
2.2.1
Expanding and simplifying logarithmic terms
2.2.2
Combining terms with the same base
2.2.3
Factoring logarithms for solving equations
2.3.1
Solving inequalities involving logarithms
2.3.2
Validating solutions using substitution
2.3.3
Using exponentiation to verify logarithmic solutions
2.4.1
Representing exponential data with semi-log scales
2.4.2
Identifying linear trends in transformed data
2.4.3
Comparing semi-log plots to regular plots
2.5.1
Recognizing differences in arithmetic sequences
2.5.2
Identifying ratios in geometric sequences
2.5.3
Connecting sequences to linear and exponential functions
2.6.1
Identifying linear growth or decay rates
2.6.2
Comparing linear vs. exponential rates of increase
2.6.3
Understanding when exponential models are appropriate
2.7.1
Understanding domain and range of exponential functions
2.7.2
Analyzing transformations: vertical and horizontal shifts
2.7.3
Exploring asymptotic behavior in exponential graphs
2.8.1
Simplifying exponential expressions using rules of exponents
2.8.2
Factoring exponential terms in equations
2.8.3
Expanding powers for simplification
2.9.1
Representing patterns algebraically
2.9.2
Validating assumptions in exponential equations
2.9.3
Generalizing growth models without specific real-world scenarios
2.10.1
Testing linear vs. exponential models for fit
2.10.2
Comparing residuals to determine accuracy
2.10.3
Analyzing errors in model assumptions
2.11.1
Combining two exponential functions
2.11.2
Using compositions for recursive sequences
2.11.3
Exploring commutativity in function compositions
2.12.1
Identifying conditions for inverses
2.12.2
Verifying inverse relationships algebraically
2.12.3
Sketching inverse functions on a graph
2.13.1
Expanding logarithmic terms using rules of logs
2.13.2
Simplifying nested logarithms
2.13.3
Converting between logarithmic and exponential forms
2.14.1
Analyzing inverse symmetry in graphs
2.14.2
Understanding logarithms as inverses of exponents
2.14.3
Exploring how inverse transformations affect function shapes
3.
Polynomial and Rational Functions
3.1.1
Simplifying rational functions to reveal holes
3.1.2
Testing function continuity at holes
3.1.3
Identifying removable discontinuities
3.2.1
Factoring and simplifying expressions
3.2.2
Expanding rational expressions
3.2.3
Verifying equivalency through algebraic manipulation
3.3.1
Horizontal and vertical shifts
3.3.2
Stretching and compressing functions
3.3.3
Reflecting across axes
3.4.1
Choosing appropriate polynomial or rational models
3.4.2
Testing assumptions through equation constraints
3.4.3
Analyzing limitations of chosen models
3.5.1
Writing polynomial and rational models
3.5.2
Simplifying derived models for analysis
3.5.3
Interpreting model parameters algebraically
3.6.1
Identifying relationships between variables
3.6.2
Representing co-variation with graphs and tables
3.6.3
Recognizing proportional relationships
3.7.1
Definition and calculation of average rate of change
3.7.2
Interpreting instantaneous rate of change
3.7.3
Slope as a measure of rate in linear functions
3.8.1
Identifying constant rates in linear models
3.8.2
Analyzing variable rates in quadratic functions
3.8.3
Determining intervals of increase and decrease
3.9.1
Graphical representation of polynomial behaviors
3.9.2
Understanding the degree's role in rate analysis
3.9.3
Identifying turning points and critical points
3.10.1
Finding complex zeros using the quadratic formula
3.10.2
Factoring polynomials with real and complex roots
3.10.3
Verifying solutions with synthetic division
3.11.1
Predicting end behavior using leading coefficients
3.11.2
Determining symmetry in polynomial graphs
3.11.3
Understanding multiplicities of roots
3.12.1
Determining horizontal asymptotes
3.12.2
Analyzing degrees of numerator and denominator
3.12.3
Identifying oblique asymptotes
3.13.1
Finding zeros in rational functions
3.13.2
Multiplicity and their impact on graph shape
3.13.3
Using synthetic and long division
3.14.1
Defining vertical asymptotes algebraically
3.14.2
Testing continuity near asymptotes
3.14.3
Identifying behaviors approaching vertical asymptotes
4.
Trigonometric and Polar Functions
4.1.1
Understanding cycles in functions
4.1.2
Representing periodic intervals
4.1.3
Exploring periodicity in graphs
4.2.1
Computing trigonometric ratios using the unit circle
4.2.2
Verifying identities involving sine, cosine and tangent
4.2.3
Exploring co-functions and their relationships
4.3.1
Computing exact values for common angles
4.3.2
Understanding reflection and symmetry properties
4.3.3
Analyzing periodic properties of sine and cosine
4.4.1
Graphing sine and cosine with amplitude and period
4.4.2
Identifying shifts in sine and cosine graphs
4.4.3
Exploring relationships between graphs of sine and cosine
4.5.1
Graphing tangent over its domain
4.5.2
Identifying vertical asymptotes in tangent graphs
4.5.3
Analyzing tangent transformations
4.6.1
Determining restricted domains of inverses
4.6.2
Using inverse functions to solve equations
4.6.3
Verifying identities with inverses
4.7.1
Solving multi-angle equations
4.7.2
Exploring periodic solutions in equations
4.7.3
Analyzing domain-specific inequalities
4.8.1
Graphing reciprocal functions of sine, cosine and tangent
4.8.2
Identifying domains and asymptotes
4.8.3
Verifying reciprocal relationships in identities
4.9.1
Using Pythagorean identities to simplify
4.9.2
Verifying equivalences through substitution
4.9.3
Rewriting expressions for computational ease
4.10.1
Converting Cartesian equations to polar form
4.10.2
Graphing polar coordinates
4.10.3
Analyzing polar symmetry and rotations
4.11.1
Graphing cardioids, roses and spirals
4.11.2
Analyzing periodicity and intersections in polar graphs
4.11.3
Using technology to explore complex polar shapes
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Volume of Expressed Colostrum in the first 48 Hours Postpartum
CQ #243 – February 14, 2022
by Anne Eglash MD, IBCLC, FABM
LACTFACT
The volume of a lactating mother’s colostrum early postpartum may decline after the first few feedings, then rise again by 30 hours postpartum.
The Trajectory of Expressed Colostrum Volume in the First 48 Hours Postpartum: An Observational Study
Breastfeeding Medicine 2022 Jan;17(1): 52-58
What are typical colostrum volumes and how do they change during the first few days postpartum?
There are few studies that have evaluated the volume of colostrum produced in the first few days postpartum, before secretory activation (when the milk ‘comes in’) occurs. Most lactation consultants would likely say that there is wide variability between lactating individuals regarding colostrum volumes.
The publication for this week is an observational study of 105 mothers in Japan who were asked to manually express their colostrum for 10 minutes every 3 hours for 48 hours, with the goal of evaluating the change in colostrum volume during the first 48 hours. These mothers were unable to directly feed due to infant admission to the neonatal intensive care unit. The authors report that it is culturally acceptable in Japan to manually express milk in the initial postpartum phase with assistance and instruction from midwives.
Sixty-three women (60%) of the mothers began expression within 3 hours postpartum, and the frequency of hand expression ranged from 5-8 times in the first 24 hours, and 6-8 times between 24-48 hours postpartum. Midwives assisted mothers who had difficulty expressing their colostrum.
The total volume expressed in the first 24 hours ranged from 0.1-11.2 ml, and from 2.2-40ml between 24-48 hours postpartum. The lowest volume of colostrum occurred between 12-15 hours, and the volume stayed low until a sudden increase in volume at 30-33 hours postpartum. What else? See the question!
Please choose accurate statements, based on this study, regarding colostrum volume in the first 48 hours. Choose 1 or more:
In the first 30 hours postpartum, the multiparous mothers had significantly higher colostrum volumes than the primiparous women.
The mothers who gave birth at gestational age 32-36 weeks had higher colostrum volumes as compared to those who gave birth at gestational age 22-31 weeks.
Colostrum volume was higher among women who delivered vaginally as compared to those who delivered via cesarean.
Colostrum volume increased earlier among mothers who began manual expression in the first 3 hours vs those who began at 3-6 hours postpartum.
See the Answer
Correct Answers: D (not A, B, or C)
The Trajectory of Expressed Colostrum Volume in the First 48 Hours Postpartum: An Observational Study
Breastfeeding Medicine 2022 Jan;17(1): 52-58
Ikuko Kato, Kimiyo Horike, Kou Kawada, Yinmon Htun, Tomoko Nishida, Shinji Nakamura, Kosuke Koyano,1 Yukihiko Konishi, and Takashi Kusaka
Abstract
Objective
Colostrum, the first form of human milk, is strongly encouraged for infants due to its benefits. During the early postpartum (PP) period, the secreted colostrum volume can be minimal, causing concerns among mothers about sufficient milk supply. Few studies have examined temporal changes in the colostrum. This study aimed to elucidate the trajectory of expressed colostrum volume in the first 48 hours after delivery.
Materials and Methods
This was a cross-sectional observational study performed at Kagawa National Children’s Hospital. One hundred five mothers who did not directly breastfeed in the first 48 hours after delivery were enrolled in the study. Well-trained midwives instructed the mothers on how to express human milk, and mothers started to express as soon as possible after delivery. Mothers were advised to express human milk every 3 hours, and the milk volume was measured.
Results
Within 3 hours PP, 60% of mothers expressed milk, and the median frequency of expression was 14 (interquartile range, 11–16) times in the first 48 hours. At 0–3 and 3–6 hours PP, the volume of initially expressed milk was 0.4 (0.0–2.0) mL and 1.0 (0.0–6.0) mL, respectively. Subsequently, milk volume decreased. The volume remained low until 30 hours PP and increased dramatically; this phenomenon is termed secretory activation, which began later in primiparous women than in multiparous women.
Conclusion
The decline in expressed milk volume during the early PP period caused concern among mothers. Therefore, mothers should be informed of the PP trajectory of human milk volume.
IABLE Comment by Anne Eglash MD, IBCLC, FABM
The researchers found that multiparous mothers didn’t have higher colostrum volumes than primiparous mothers until after 30 hours, when the volume of colostrum greatly increased. At that point the rate of increase in milk volume was faster in multiparous mothers than in primiparous mothers.
Gestational age at birth and mode of delivery did not have an impact on colostrum volumes.
One important aspect to this study is the lack of adjustment for maternal BMI. There is increasing evidence that an elevated body mass index is associated with a delay in lactation, which would likely be associated with lower colostrum volumes.
The take-away from this study is that colostrum volumes are higher in the first 12 hours than in the next 18 hours. By 30 hours postpartum the volume of colostrum climbs exponentially.
This may explain why there may be more stools in the first 24 hours, followed by a slowing of stools after 24 hours until milk volume increases at 30 hours and beyond.
Comments (18)
Nancy Bates
February 15, 2022 at 12:33 AM
Thanks so much for this!! I am posting for Taft to see perhaps it will help them understand and quit putting a pump on kim to “see hou much baby is getting”! Grrrrr
Reply
Beth Lichy
February 15, 2022 at 2:39 AM
Rarely, we see moms expressing much larger amounts (eg 40 mls) the first time a mom expresses. Is there evidence of a larger volume due to storage of colostrum during the pregnancy?
Reply
IABLE
February 15, 2022 at 3:15 AM
I am not aware of the effect of antenatal expression on the volume of colostrum postpartum. We should do some research on this.
Reply
Cynthia Lucas
February 15, 2022 at 11:05 AM
In my 20 years as a IBCLC, I have seen that trend with Moms who are separated and exclusively pumping.
They have told me it was a good volume then declined. Many of them have gotten discouraged and not wanted to pump. I have told them to keep pumping and hand expressing and it will go up.
I feel like this is the norm and glad to see there is a study to back it.
Reply
Kathryn Stagg, IBCLC
February 15, 2022 at 12:39 PM
This is something we have seen frequently in Breastfeeding Twins and Triplets UK Facebook group. Parents are often reporting that they cannot get as much colostrum after the first day or so, and then within 24 hours after that their milk often begins to come in.
I am so glad somebody has now studied this phenomenon.
The next study that would be good to do as a follow up is whether antenatal hand expressing has any impact on whether this happens or whether because the breasts are used to regular stimulation before baby arrives, that the colostrum volumes stay more constant. That would be a fascinating study!!
Thank you so much for sharing
Reply
Nancy Bates
February 15, 2022 at 1:28 PM
So much for my spelling. Posting for staff. pump on mom to see “How much”
Reply
Dee
February 15, 2022 at 1:35 PM
Thank you for this. It does raise more questions.
– were these mothers just first timers or a mixture of pregnancies?
– were twin mothers involved in this?
– I have not known mothers to ‘lose’ colostrum after certain hours after birth.
– how would this be connected to mothers in other parts of the world?
Reply
IABLE
February 15, 2022 at 2:26 PM
This was a combination of multiparous and primiparous mothers, and there was no difference in volumes of colostrum in the first 6-12 hours. They didn’t involve moms with twins. This study was done in Japan where manual expression is common in the first few days. Unclear how this would apply to populations that are not as familiar with manual expression, e.g. those that are more pump reliant.
Reply
Lynnette Hafken
February 15, 2022 at 3:47 PM
I’m interested in what people think about the conclusion:
“The decline in expressed milk volume during the early PP period caused concern among mothers. Therefore, mothers should be informed of the PP trajectory of human milk volume.”
Reply
Chasta Hite
March 1, 2022 at 10:00 PM
Agree. Informing parents this is normal often still does not reassure them and may deter them from continuing to breastfeed or pump.
Reply
Nina Torelli
January 25, 2023 at 2:55 PM
I teach breastfeeding classes to L&D and postpartum nurses and am questioning about adding this information to my presentation. I love that this study was done and it’s reassuring to me that what I see on daily basis working with postpartum mom’s is indeed normal and natural, however, it’s not easy to instill reassurance in others with evidence-based research and facts.
Will providing this information give clarity and encouragement to postpartum moms to maintain their hand expression / pumping schedule (if not breastfeeding) or actually give mom’s/nurses a reason to supplement with formula unnecessarily because now they are expecting a decline in colostrum?
Reply
Edith Bryant
February 15, 2022 at 10:35 PM
Great to read this.
I often noticed this in the postnatal ward and just kept on encouraging mothers to keep on expressing.
To me it makes sense from a perspective of lactogenesis 1 being controlled by hormones and lactogenesis 2 has the beginning of the autocrine control and increase in lactose
Can that be a reason?
But why exactly a decline by 12 hours
Not quite sure!?!
Reply
Chasta Hite
March 1, 2022 at 9:58 PM
I think it’s important to note this is a decline in expressed milk not a decline in the total amount of available colostrum
That may be remaining in the breast. We often see this decline when a parent is pumping but with hand expression there is still colostrum available for the baby. When providing this information to a breastfeeding parent they could be deterred by thinking that the milk supply has dropped and their baby is hungry even if we tell them this is normal, that usually doesn’t convince them that their baby is not hungry. I think we need to be very careful how this information is presented. I would also suspect that postpartum edema is being created in the breast from early pumping that is further preventing the removal of milk.
Reply
IABLE
March 6, 2022 at 2:55 AM
In this study, the participants only hand expressed. They did not pump.
Reply
Jan Edye
March 6, 2022 at 10:41 AM
This is so interesting and relates and coincides well to the infant sleep pattern norms after delivery. Babies born alert remaining so for up to 2 hours post-delivery before moving into levels of sleep, become more wakeful after 20 hours or so. Clearly, support with attachment of babies needs skilled support straight after delivery to be sure that the optimum amount of colostrum is fed.
Reply
Ricardo Nunes
April 21, 2022 at 2:00 PM
This research has raised a few more questions, mainly directed to the NICU population:
1) Have the authors used the Marmet Technique of hand expression?
2) Would it be possible to obtain an increase in the volume of colostrum by adding an electric pump with the INITIATION program simultaneously on both breasts for 15 minutes?
3) Would it be more effective before or after hand expression?
Reply
Shirley Donato
July 20, 2022 at 2:35 PM
The way I see it, it’s a very well designed system. With the surge of oxytocin and prolactin hormones during the delivery process (Buckley, 2015), more colostrum is available for the newly delivered baby, who is generally awake and eager to nurse. Babies usually nurse very well in this initial “Golden Hour” after delivery, so this bolus of colostrum gives them the calories and nutrients needed to recover from the delivery and give them fuel for the next several hours. Babies then usually become very sleepy, and they may not eat much until they begin to be more awake after about 12-24 hours. After that initial bolus in the first few hours, the volume of colostrum produced settles to an appropriate volume for the nursling, and then gradually increases over the first few days until lactogenesis II occurs. As was stated in one of the responses, the study showed how much milk was expressed, not how much is available, and we all know that babies are much better at getting milk out of the breast than we are. Edema is not likely from early pumping, but rather due to the large volumes of IV fluids administered during the labor/delivery process accumulating in the tissues. Yes, we should be sure to explain the normal expectations for expressed milk/colostrum volume over the first few days, reassuring mom that it is normal to see very little milk volume in the first few days. Help mom to understand that continued frequent stimulation and breast-emptying is the best way to increase milk supply, whether by pumping, hand expression, or a combination of both (or frequent feedings, of course).
“Prolactin increases steeply as birth nears, likely due to peaks of beta-endorphins and oxytocin, both of which stimulate prolactin release. In addition, prolactin stimulates oxytocin release, contributing to oxytocin peaks in late labor and birth.”
Buckley SJ. Executive Summary of Hormonal Physiology of Childbearing: Evidence and Implications for Women, Babies, and Maternity Care. J Perinat Educ. 2015;24(3):145-53. doi: 10.1891/1058-1243.24.3.145. PMID: 26834435; PMCID: PMC4720867.
Reply
Tara Piccininni
August 26, 2023 at 3:14 PM
I am a Lactation Consultant in an NICU . I always start my moms with hand expression, to capture the colostrum for our little ones. This was a very helpful study. Thanks for sharing it.
Reply
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10630 | https://www.vedantu.com/maths/differences-between-codomain-and-range | Codomain vs Range in Maths: Key Differences Explained
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Maths
Codomain vs Range: What Sets Them Apart in Mathematics?
Codomain vs Range: What Sets Them Apart in Mathematics?
Reviewed by:
Rama Sharma
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Clear Examples: Distinguishing Codomain from Range in Functions
There are many concepts in Math that can surprise one and also come as a challenge for the students. Like numbers, there is no end to the mysteries of mathematics. The deeper you go in the world of mathematics, the more magnificent it gets. Codomain and Range is one such concept. So today we are here to learn about the differences between Codomain and Range.
But to understand all this we first need to know what a function is. Do you know what a function is? Any relation defined over two different sets is a function, provided that every element that is a part of the first set has a corresponding element in the second set of the relation. However, the relation or the correspondence of the first set's elements must be with exactly one element of the second set.
Function Explained with Example
To understand the definition clearly let’s take an example. Say there are two sets namely, set A and set B. The relation is from set A to set B. So, for this relationship to be a function, all the elements of set A should have a corresponding and unique element in set B. A function is represented in the following manner:
F(x) = Y
Example:-
y=x² is a function where x and y belong to real numbers. Here, set A and set B have all the numbers that are present in real numbers.
Domain
A domain is a group of possible values that the independent variable can take. This means the set of all the possible values that ‘x’ can take in the function f is the domain of the given function. A domain is a set of pre-images. According to the example taken above, set A is the domain of the function.
Example:-
y=x+4
The domain of the above function: (-∞,∞)
Codomain
A codomain is the group of possible values that the dependent variable can take. This means that the set of all the possible values that ‘y’ can take in the function f is the codomain of the given function. A codomain is a set of images. According to the example taken above, set B is the codomain of the function
Range
The range is all the elements from set B that have the corresponding pre-image in set A. Hence, a range can also be defined as the set of all the possible values of the function that we receive upon taking the different values of x in the function f.
Example:-
Taking the example taken above, y = x + 4
Here y 0 is the range of the given function. So we can write this as the range of the function is (-∞,∞).
(Image will be uploaded soon)
Domain and Range
Here are the domain and range of some common functions that we see very frequently while solving mathematics-
S.NO.Types of function Domain Codomain
1.f(x) = x(-∞ , + ∞)(-∞ , + ∞)
2.f(x) = x²(-∞ , + ∞)[0 , + ∞)
3.f(x) = sin ( x )(-∞ , + ∞)[−1,1]
4.f(x) = cos ( x )(-∞ , + ∞)[−1,1]
5.f(x) = sin⁻¹( x )[−1,1][−π/2,π/2]
6.f(x) = cos⁻¹( x )[−1,1][0,π]
7.f(x) = a x(-∞ , + ∞)(0 , +∞)
8.f(x) = Logₐ ( x )(0 , + ∞)(-∞ , + ∞)
Types of Functions
There are five types of functions on the basis of how the domain and codomain is related.
One-One functions
Many-one functions
Onto functions
Into functions
One-One and Onto functions
They are discussed in detail below.
One-One Functions
When each element of the domain has a distinct image in the codomain then the function is the One-One function. It is also called the injective function.
Many-One Functions
When two or more elements of the domain do not have a distinct image in the codomain then the function is Many -One function.
Onto Functions
When each element of the codomain has a distinct image in the domain then the function is Onto function. It is also called the surjective function.
Into Functions
When two or more elements of the codomain do not have a distinct image in the domain then the function is Into function.
One-One and Onto Functions
When each element of the domain has a distinct image in the codomain and when each element of the codomain has a distinct image in the domain then the function is One-One and Onto function. It is also called bijective function.
Now let’s summarize the difference between Codomain vs Range
Although it might be easy to understand the concepts pertaining to range and domain, it is not really the case with their distinctions. To summarize, we have the following bases that can clear all your doubts between the two:
Difference between Codomain Vs Range
Basics of distinction Codomain Range
MEANING When it comes to a codomain, it is basically the group of values that are possible in the case of what can be taken by the dependent variable. The meaning of this comes as a simple one: The set of possible values which are all something that a ‘y’ variable can take in the f function is what a codomain is. This pertains only to a given function.A range is basically what all the elements are from a second set, let’s say B, which has the pre-image that is present in the first set, A. In other words, a range is a subset of the codomain, which pertains to a set of possible values of f as a function. It is one that is received when we take into account the different values of a variable x.
PURPOSE A codomain has the nature of restricting the output of a particular function.A range is not something that can restrict the output of a specific function.
RELATION A codomain is in relation to the meaning of a function.A range is related to a function’s image.
EXAMPLE If L = {1, 2, 3, 4} and M = {1, 2, 3, 4, 5, 6, 7, 8, 9} and the relation f: A B is defined by f (x) = x 2 Codomain = Set M = {1, 2, 3, 4, 5, 6, 7, 8, 9}If L = {1, 2, 3, 4} and M = {1, 2, 3, 4, 5, 6, 7, 8, 9} and the relation f: A B is defined by f (x) = x 2. Range = {1, 4, 9}
Fun Facts
The domain, codomain, and range are not always equal. In some cases, it can be equal.
The range is a subset of the codomain.
The denominator of the given function can never be zero.
This is all about the differences between codomain and range with proper examples. Focus on the core concepts so that you can easily understand the differences easily.
FAQs on Codomain vs Range: What Sets Them Apart in Mathematics?
What is the core difference between the codomain and range of a function?
The core difference is that the codomain is the set of all possible output values a function is defined to have, while the range is the set of all actual output values the function produces. The range is always a subset of the codomain. For a function f: A → B, set B is the codomain, and the set of all f(x) values for every x in A is the range.
How is the domain of a function different from its codomain?
The domain and codomain refer to two different sets in a function's definition.
The domain is the set of all possible input values (the 'x' values) for the function.
The codomain is the set of all possible output values (the 'y' values) that the function is allowed to map to.
In short, the domain is what you put into the function, and the codomain is what could potentially come out.
Can the range of a function be equal to its codomain? Provide an example.
Yes, the range can be equal to the codomain. This occurs when every element in the codomain is the image of at least one element in the domain. Such a function is called an onto or surjective function. For example, consider the function f(x) = x + 1, where both the domain and codomain are the set of all real numbers (ℝ). In this case, for any real number 'y' in the codomain, we can find a real number x = y - 1 in the domain. Thus, the range is also ℝ, making it equal to the codomain.
Why can the range of a function never be larger than its codomain?
The range can never be larger than the codomain because the range is, by definition, a subset of the codomain. The codomain is the pre-defined 'universe' of all possible outputs. The range consists only of the actual outputs generated by the function from its domain. Since every actual output must belong to the set of possible outputs, the range is either equal to or contained within the codomain, and can therefore never have more elements.
How do the concepts of codomain and range relate to onto (surjective) functions?
The relationship between codomain and range is the defining characteristic of an onto (surjective) function. A function is classified as 'onto' if and only if its range is equal to its codomain. This means that every single element in the codomain set is an actual output (image) for at least one input element from the domain. If even one element in the codomain is not produced as an output, the function is considered an 'into' function, not an 'onto' one.
Using an example, explain how to determine the range from a given function and its codomain.
To determine the range, you must find all the actual output values the function produces. Consider the function f(x) = x² with its domain as the set of all integers {..., -2, -1, 0, 1, 2, ...} and its codomain as the set of all integers.
First, apply the function to the domain values: f(-2)=4, f(-1)=1, f(0)=0, f(1)=1, f(2)=4, etc.
Next, collect all the unique output values. The outputs are {0, 1, 4, 9, 16, ...}.
This set of actual outputs, {0, 1, 4, 9, ...}, is the range. Notice that the range is a subset of the codomain (all integers), as it does not include negative integers or non-perfect squares.
What is the significance of the codomain in defining a function? Why not just use the range?
The codomain is significant because it establishes the type of function being discussed and the mathematical space in which it operates. It is part of the function's formal definition (f: Domain → Codomain). Specifying the codomain beforehand allows us to ask critical questions, such as whether the function is surjective (onto) or not. If we only defined the range, every function would be surjective by default, making the concept meaningless. The codomain provides the full picture of the target set, while the range shows what part of that target is actually reached.
In the context of a function, what is the difference between an 'image' and the 'range'?
The terms 'image' and 'range' are closely related but distinct.
An image refers to the single output value produced by a specific input. For a function f(x) = y, 'y' is the image of the element 'x'.
The range is the complete set of all images. It is the collection of every single output value that the function generates when applied to every element in its entire domain.
So, an image is a single element, while the range is a set of elements.
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Frugal, Famished, Gullible, Indolent - Advanced vocabulary
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Frugal
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Frugal (adj.)
/ˈfruːɡəl/
Synonyms: Thrifty, economical, sparing
Meaning: Careful to buy only what is necessary.
Common collocations: frugal lifestyle, frugal habits, frugal spender, frugal meal
Example sentence: Dan was very frugal, and would often use a tea bag three or four times over.
30
Famished (adj.)
/ˈfæmɪʃt/
Synonyms: Hungry, starving, ravenous
Meaning: Extremely hungry.
Common collocations: famished appetite, famished feeling, famished state
Example sentence: What's for dinner? I'm absolutely famished.
31
Gullible (adj.)
/ˈɡʌlɪb(ə)l/
Synonyms: Naive, unskeptical, trusting
Meaning: Easily deceived or tricked because of being too trusting or naive.
Common collocations: gullible person, gullible nature, gullible belief
Example sentence: How can you be so gullible! He isn't telling you the truth!
32
Indolent (adj.)
/ˈɪndələnt/
Synonyms: Lazy, idle, sluggish
Meaning: Showing no real interest or effort.
Common collocations: indolent behavior, indolent attitude, indolent lifestyle
Example sentence: Despite having many responsibilities, he remained indolent, spending most of his days doing nothing.
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10632 | https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Aldehydes_and_Ketones/Reactivity_of_Aldehydes_and_Ketones/Reaction_with_Primary_Amines_to_form_Imines | Skip to main content
Reaction with Primary Amines to form Imines
Last updated
: Jan 23, 2023
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Oxidation of Aldehydes and Ketones
Reactions with Grignard Reagents
Page ID
: 5825
( \newcommand{\kernel}{\mathrm{null}\,})
The reaction of aldehydes and ketones with ammonia or 1º-amines forms imine derivatives, also known as Schiff bases (compounds having a C=N function). Water is eliminated in the reaction, which is acid-catalyzed and reversible in the same sense as acetal formation. The pH for reactions which form imine compounds must be carefully controlled. The rate at which these imine compounds are formed is generally greatest near a pH of 5, and drops at higher and lower pH's. At high pH there will not be enough acid to protonate the OH in the intermediate to allow for removal as H2O. At low pH most of the amine reactant will be tied up as its ammonium conjugate acid and will become non-nucleophilic.
Converting reactants to products simply
Examples of imine forming reactions
Mechanism of imine formation
1) Nucleophilic attack
2) Proton transfer
3) Protonation of OH
4) Removal of water
5) Deprotonation
Reversibility of imine forming reactions
Imines can be hydrolyzed back to the corresponding primary amine under acidic conditons.
Reactions involving other reagents of the type Y-NH2
Imines are sometimes difficult to isolate and purify due to their sensitivity to hydrolysis. Consequently, other reagents of the type Y–NH2 have been studied, and found to give stable products (R2C=N–Y) useful in characterizing the aldehydes and ketones from which they are prepared. Some of these reagents are listed in the following table, together with the structures and names of their carbonyl reaction products. Hydrazones are used as part of the Wolff-Kishner reduction and will be discussed in more detail in another module.
With the exception of unsubstituted hydrazones, these derivatives are easily prepared and are often crystalline solids - even when the parent aldehyde or ketone is a liquid. Since melting points can be determined more quickly and precisely than boiling points, derivatives such as these are useful for comparison and identification of carbonyl compounds. It should be noted that although semicarbazide has two amino groups (–NH2) only one of them is a reactive amine. The other is amide-like and is deactivated by the adjacent carbonyl group.
Problems
1)Please draw the products of the following reactions.
2) Please draw the structure of the reactant needed to produce the indicated product.
3) Please draw the products of the following reactions.
Answers
1)
2)
3)
Contributors
Prof. Steven Farmer (Sonoma State University)
William Reusch, Professor Emeritus (Michigan State U.), Virtual Textbook of Organic Chemistry
Oxidation of Aldehydes and Ketones
Reactions with Grignard Reagents |
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Nine-point circle
Triangle ABC with the nine-point circle in light orange
The nine-point circle (also known as Euler's circle or Feuerbach's circle) of a given triangle is a circle which passes through 9 "significant" points:
The three feet of the altitudes of the triangle.
The three midpoints of the edges of the triangle.
The three midpoints of the segments joining the vertices of the triangle to its orthocenter. (These points are sometimes known as the Euler points of the triangle.)
"The nine-point circle is tangent to the incircle, has a radius equal to half the circumradius, and its center is the midpoint of the segment connecting the orthocenter and the circumcenter." -hankinjg
That such a circle exists is a non-trivial theorem of Euclidean geometry.
The center of the nine-point circle is the nine-point center and is usually denoted .
The nine-point circle is tangent to the incircle, has a radius equal to half the circumradius, and its center is the midpoint of the segment connecting the orthocenter and the circumcenter, upon which the centroid also falls.
It's also denoted Kimberling center .
Contents
[hide]
1 First Proof of Existence
2 Second Proof of Existence
3 Common Euler circle
4 See also
First Proof of Existence
Since is the midpoint of and is the midpoint of , is parallel to . Using similar logic, we see that is also parallel to . Since is the midpoint of and is the midpoint of , is parallel to , which is perpendicular to . Similar logic gives us that is perpendicular to as well. Therefore is a rectangle, which is a cyclic figure. The diagonals and are diagonals of the circumcircle. Similar logic to the above gives us that is a rectangle with a common diagonal to . Therefore the circumcircles of the two rectangles are identical. We can also gain that rectangle is also on the circle.
We now have a circle with the points , , , , , and on it, with diameters , , and . We now note that . Therefore , , and are also on the circle. We now have a circle with the midpoints of the sides on it, the three midpoints of the segments joining the vertices of the triangle to its orthocenter on it, and the three feet of the altitudes of the triangle on it. Therefore, the nine points are on the circle, and the nine-point circle exists.
Second Proof of Existence
We know that the reflections of the orthocenter about the sides and about the midpoints of the triangle's sides lie on the circumcircle (side proof, midpoint proof), as do the vertices of the triangle. So, consider the homothety centered at with ratio . It maps the circumcircle of (and those 9 points) to a circle, including mapping the vertices of the triangle to its Euler points (by definition). This is the nine-point circle.
Common Euler circle
Let an acute-angled triangle with orthocenter be given.
be the point on opposite
Points and such that is a parallelogram. The line intersects at the points and
Prove that triangles and has common Euler (nine-point) circle.
Proof Denote is midpoint
Let’s consider Circumcenter of point is the midpoint point is the midpoint
Denote the centroid of
is the centroid of
Denote the midpoint of is the midpoint of
is the centroid of
Point is the circumcenter of is the orthocenter of
The triangles and has common circumcircle and common center of Euler circle (the midpoint of ) therefore these triangles has the common Euler circle.
vladimir.shelomovskii@gmail.com, vvsss
See also
Kimberling center
Center line
Evans point
Euler line
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10634 | https://tomas-svojanovsky.medium.com/understanding-coterminal-angles-a-complete-guide-5-839b7f355b56 | Sitemap
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Understanding Coterminal Angles: A Complete Guide (5)
Tomas Svojanovsky
4 min readMay 29, 2025
Not a member? Read it here.
When we sketch two angles in standard position, they are coterminal if their terminal sides lie on top of each other. In other words, if both angles finish up at the same place on the coordinate plane, then they’re coterminal.
Think of it this way: imagine starting at the positive x-axis and rotating to create an angle. If you can create another angle by making additional full rotations (either clockwise or counterclockwise) and end up at the exact same terminal position, those angles are coterminal.
The Key Principle
Coterminal angles always differ by complete rotations:
In degrees: They differ by multiples of 360°
In radians: They differ by multiples of 2π
This means to find a coterminal angle, we simply add or subtract 360° (or 2π radians) as many times as we want.
Finding Coterminal Angles in Degrees
Example 1: Positive Coterminal Angles
Let’s find angles coterminal with 45°.
Adding 360° multiple times:
45° + 360° = 405°
45° + 2(360°) = 45° + 720° = 765°
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10635 | https://math.colorado.edu/~kstange/teaching-resources/discrete/mod-tables.pdf | Math 2001 Modular Arithmetic April 14, 2020 1 Addition Tables Mod 4 0 1 2 3 0 0 1 2 3 1 1 2 3 0 2 2 3 0 1 3 3 0 1 2 Mod 5 0 1 2 3 4 0 0 1 2 3 4 1 1 2 3 4 0 2 2 3 4 0 1 3 3 4 0 1 2 4 4 0 1 2 3 Mod 6 0 1 2 3 4 5 0 0 1 2 3 4 5 1 1 2 3 4 5 0 2 2 3 4 5 0 1 3 3 4 5 0 1 2 4 4 5 0 1 2 3 5 5 0 1 2 3 4 Mod 7 0 1 2 3 4 5 6 0 0 1 2 3 4 5 6 1 1 2 3 4 5 6 0 2 2 3 4 5 6 0 1 3 3 4 5 6 0 1 2 4 4 5 6 0 1 2 3 5 5 6 0 1 2 3 4 6 6 0 1 2 3 4 5 Mod 8 0 1 2 3 4 5 6 7 0 0 1 2 3 4 5 6 7 1 1 2 3 4 5 6 7 0 2 2 3 4 5 6 7 0 1 3 3 4 5 6 7 0 1 2 4 4 5 6 7 0 1 2 3 5 5 6 7 0 1 2 3 4 6 6 7 0 1 2 3 4 5 7 7 0 1 2 3 4 5 6 Mod 9 0 1 2 3 4 5 6 7 8 0 0 1 2 3 4 5 6 7 8 1 1 2 3 4 5 6 7 8 0 2 2 3 4 5 6 7 8 0 1 3 3 4 5 6 7 8 0 1 2 4 4 5 6 7 8 0 1 2 3 5 5 6 7 8 0 1 2 3 4 6 6 7 8 0 1 2 3 4 5 7 7 8 0 1 2 3 4 5 6 8 8 0 1 2 3 4 5 6 7 Mod 10 0 1 2 3 4 5 6 7 8 9 0 0 1 2 3 4 5 6 7 8 9 1 1 2 3 4 5 6 7 8 9 0 2 2 3 4 5 6 7 8 9 0 1 3 3 4 5 6 7 8 9 0 1 2 4 4 5 6 7 8 9 0 1 2 3 5 5 6 7 8 9 0 1 2 3 4 6 6 7 8 9 0 1 2 3 4 5 7 7 8 9 0 1 2 3 4 5 6 8 8 9 0 1 2 3 4 5 6 7 9 9 0 1 2 3 4 5 6 7 8 1 2 Multiplication Tables Mod 4 0 1 2 3 0 0 0 0 0 1 0 1 2 3 2 0 2 0 2 3 0 3 2 1 Mod 5 0 1 2 3 4 0 0 0 0 0 0 1 0 1 2 3 4 2 0 2 4 1 3 3 0 3 1 4 2 4 0 4 3 2 1 Mod 6 0 1 2 3 4 5 0 0 0 0 0 0 0 1 0 1 2 3 4 5 2 0 2 4 0 2 4 3 0 3 0 3 0 3 4 0 4 2 0 4 2 5 0 5 4 3 2 1 Mod 7 0 1 2 3 4 5 6 0 0 0 0 0 0 0 0 1 0 1 2 3 4 5 6 2 0 2 4 6 1 3 5 3 0 3 6 2 5 1 4 4 0 4 1 5 2 6 3 5 0 5 3 1 6 4 2 6 0 6 5 4 3 2 1 Mod 8 0 1 2 3 4 5 6 7 0 1 2 3 4 5 6 7 0 0 0 0 0 0 0 0 0 1 0 1 2 3 4 5 6 7 2 0 2 4 6 0 2 4 6 3 0 3 6 1 4 7 2 5 4 0 4 0 4 0 4 0 4 5 0 5 2 7 4 1 6 3 6 0 6 4 2 0 6 4 2 7 0 7 6 5 4 3 2 1 Mod 9 0 1 2 3 4 5 6 7 8 0 0 0 0 0 0 0 0 0 0 1 0 1 2 3 4 5 6 7 8 2 0 2 4 6 8 1 3 5 7 3 0 3 6 0 3 6 0 3 6 4 0 4 8 3 7 2 6 1 5 5 0 5 1 6 2 7 3 8 4 6 0 6 3 0 6 3 0 6 3 7 0 7 5 3 1 8 6 4 2 8 0 8 7 6 5 4 3 2 1 Mod 10 0 1 2 3 4 5 6 7 8 9 0 0 0 0 0 0 0 0 0 0 0 1 0 1 2 3 4 5 6 7 8 9 2 0 2 4 6 8 0 2 4 6 8 3 0 3 6 9 2 5 8 1 4 7 4 0 4 8 2 6 0 4 8 2 6 5 0 5 0 5 0 5 0 5 0 5 6 0 6 2 8 4 0 6 2 8 4 7 0 7 4 1 8 5 2 9 6 3 8 0 8 6 4 2 0 8 6 4 2 9 0 9 8 7 6 5 4 3 2 1 2 |
10636 | https://artofproblemsolving.com/booklinks/intro-algebra/AlgebraSolns.pdf?srsltid=AfmBOooZCMzTXIITVUHXjpjph-iecgzgKP13Hxn_9_FMQsVhXl6JwX2E | Solutions to Extra Problems at End of Chapter 15 of Introduction to Algebra
Reminder: The AM-GM Inequality for two variables states that if a and b are nonnegative, then a + b
2 ≥ √ab.
• Applying AM-GM to 1 and x, we have 1+ x
2
≥ √(1)( x). Multiplying both sides by 2 gives 1 + x ≥ 2√x.
• We work backwards to find our solution. We multiply both sides of the desired inequality by 2 and expand the right side to give us 2 x2 + 2 y2 ≥ x2 + 2 xy + y2. Rearranging gives
x2 − 2xy + y2 ≥ 0, or ( x − y)2 ≥ 0. Now we have a path to our solution. By the Trivial Inequality, we have ( x − y)2 ≥ 0. Expanding the left side and rearranging gives
x2 + y2 ≥ 2xy . Adding x2 + y2 to both sides gives 2 x2 + 2 y2 ≥ x2 + 2 xy + y2. The right side is the square of x + y, so we have 2 x2 + 2 y2 ≥ (x + y)2. Dividing by 2 gives the desired
x2 + y2 ≥ (x+y)2
2
.
• We work backwards to find our solution. Multiplying both sides by xy (x + y) (which doesn’t change the direction of the inequality because x and y are positive), gives us
y(x + y) + x(x + y) ≥ 4xy.
Rearranging this gives us x2 + y2 − 2xy ≥ 0, which we know how to handle with the Trivial Inequality. Now we are ready to write our solution. By the Trivial Inequality, we have ( x − y)2 ≥ 0. Expanding the left side, then adding 4 xy to both sides, gives x2 + 2 xy + y2 ≥ 4xy . We rewrite the left side to give x2 + xy + y2 + xy ≥ 4xy ,and a little factoring gives us x(x + y) + y(x + y) ≥ 4xy . Dividing both sides by xy (x + y)(which doesn’t change the direction of the inequality because x and y are positive), gives us the desired 1
x
1
y
≥ 4
x+y
.
• The sums on the larger side make us think of AM-GM. We have
x + y
2 ≥ √xy, y + z
2 ≥ √yz, z + x
2 ≥ √zx.
Multiplying these three gives us (x + y)( y + z)( z + x)8 ≥ √x2y2z2.
Because √x2y2z2 = xyz , multiplying our inequality by 8 gives us the desired inequality (x + y)( y + z)( z + x) ≥ 8xyz .Solutions to Extra Problems at End of Chapter 15 of Introduction to Algebra
• We put all the terms on one side, which gives us
x2 + y2 − xy − x − y + 1 ≥ 0.
The x2 + y2 − xy gets us thinking about ( x − y)2, which equals x2 + y2 − 2xy . We need only
−xy , not −2xy , so we consider (x − y)2
2 = x2 + y2 − 2xy
2 = x2 + y2
2 − xy.
That takes care of the xy term, but now we need another x2
2
and y2
2
. We can get an x2 term and an x term from ( x − 1) 2, which will also give us a 1. We only want x2/2, so we consider (x − 1) 2/2 (and ( y − 1) 2/2): (x − 1) 2
2 = x2 − 2x + 1 2 = x2
2 + 12 − x,
(y − 1) 2
2 = y2 − 2y + 1 2 = y2
2 + 12 − y.
Now we have all our terms and we can use the Trivial Inequality to finish. The Trivial Inequality gives us (x − y)2
2 + (x − 1) 2
2 + (y − 1) 2
2 ≥ 0,
and expanding the terms on the left side (and collecting like terms) gives x2 + y2 − xy − x −
y + 1 ≥ 0. A little rearranging then gives the desired x2 + y2 ≥ xy + x + y − 1.
• This looks a little like AM-GM, but with 4 variables instead of 2. So, let’s start by writing AM-GM twice:
w + x
2 ≥ √wx, y + z
2 ≥ √yz.
If we add these, we’ll have w + x + y + z on the left:
w + x + y + z
2 ≥ √wx +√yz.
We can get the product wxyz by applying AM-GM to √wx and √yz :
√wx +√yz
2 ≥
√√wx √yz.
Simplifying the right side gives us 4
√wxyz on the right, then multiplying by 2 gives us
√wx +√yz ≥ 2 4
√wxyz. Solutions to Extra Problems at End of Chapter 15 of Introduction to Algebra
Combining this with our earlier inequality for w + x + y + z gives us
w + x + y + z
2 ≥ √wx +√yz ≥ 2 4
√wxyz,
so we have w + x + y + z
2 ≥ 2 4
√wxyz.
Dividing this by 2 gives us the desired
w + x + y + z
4 ≥ 4
√wxyz.
For a much harder challenge, try proving ( x + y + z)/3 ≥ 3
√xyz for all nonnegative x, y, and
z. |
10637 | https://oneacross.com/reference?word=telephone+line | One Across - Dictionary Search
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10638 | https://www.youtube.com/watch?v=hMDpI8pwo-s | Setting Up a Venn Diagram with Three Sets
Ms. Hearn
10000 subscribers
4 likes
Description
251 views
Posted: 16 Nov 2023
In this video, we will create a venn diagram for the sets U={a,b,c,d,e,f,g,h}, S={a,b,c,d}, T={b,e,d,h}, and V={b,c,g,h}. We will discuss in detail how to decide where to put each element, because each element belongs in one and ONLY one region. #venndiagrams #settheory
Transcript:
hi I'm missar let's get started in this video we're going to talk about how to place elements of sets into the appropriate regions of Vin diagrams this is an example of a VIN diagram of three sets so in this case the universal set u is all the letters from A through h s is AB c d t is b e d h and V is BC g h we're going to draw our Vin diagram placing a rectangle and labeling it as U and then we need three three circles to represent the three sets St and V they have to overlap each other and there even has to be a place where all three overlap cuz it's possible there's an element that is actually in all three of the sets you can label them any way you want and other words you could have had a t in the upper left instead of an S but it wouldn't be wrong to rearrange them now how many regions do we have in this diagram we have the region in the very middle that's part of St and and V I'm going to label that one so region two I'm going to let be this little piece that's in s and t but not V Region Three I'm going to let that be the piece that's in s andv but not t I'm going to call this piece region four that's in T and V but not s the part that's just in the set T I'm going to call that region five part that's just in the set s region 6 the part that's just in V I'm going to call region 7 and then this outside part that's not in any of the sets I'm going to call that Region 8 all right so let's talk about the elements one element at a time where does element a go well element a is in the set s and it's not in T and it's not in v so if you look at the set s it is actually comprised now of four different regions 1 2 3 and six but the only one that's in s and not in t or V is region 6 so we're going to place a in region 6 let's look at element B element B is in s it's in t oh and it's in v as well so we're going to have to place it in a region that's in all three sets the only region that fits that description is the one in the very middle region one so we're going to put B there okay how about the element C well C is in s it's not in t but it is in V so we're looking for an element that's in both s and V so it has to be where the two sets overlap which is this football shape here and that includes region one and Region 3 but region one is also in the set T and remember C is in s andv but not t so it must be that C is in Region 3 how about element D element D is in s and it's in t but it's not in v so s and t overlap here with regions one and two we have one more piece of information it's not in v and region one is in V so it must be that D is in Region 2 now Element e e is not in s it is in T and it's not in v so e is in that region of T that doesn't touch s or V which is region five okay how about element F element f is not an s or t or V so it is in this outside region Region 8 now we have element g g is not in s it's not in t but it is in V so we're looking for the region that's in V but doesn't touch s or T and that's region 7 so we're going to place G in region 7 and last but not least the element H H is not in s but it is in T and it is in V so we need to have the element H be in both T and V which means they're intersection which is this football shaped piece here but we also know that it can't be an S so that excludes region one so that means that H has to be in Region 4 okay well that was the last element we have everything placed appropriately I'm going to get rid of those region labels and this is the completed Vin diagram I hope you found this video helpful if you did please give it a thumbs up that helps other students to find the video |
10639 | https://ocw.mit.edu/courses/1-060-engineering-mechanics-ii-spring-2006/resources/lecture3/ | search
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10640 | https://www.jainuniversity.ac.in/resources/calculator/ratio-calculator | Ratio Calculator: Convert Ratios Quickly and Easily
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Home
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Ratio Calculator
Ratio Calculator
<Calculate Ratio
Table of Content:
What is ratio?
How to find a ratio?
How to calculate ratio?
Examples on ratio formula:
Understanding Practical Applications of Ratios
How to find the ratio of two numbers?
How to calculate the ratio of three numbers?
Benefits of using a ratio calculator
FAQs
The ratio calculator is a vital tool for simplifying the process of finding and comparing ratios. Whether you are dealing with financial data, measurements or other proportional values, this calculator provides an efficient way to understand relationships between numbers. In this article, we will explore key concepts related to ratios, including how to use the calculator effectively, various applications and examples. By the end, you’ll have a thorough understanding of how to calculate ratio and its practical uses.
What is ratio?
A ratio is a way to compare two or more quantities, expressing how many times one value contains or is contained within another. For eg, the ratio of 5 to 4 indicates that for every 5 units of one quantity, there are 4 units of another. Ratios are widely used in Mathematics, Science, Finance and daily life to convey proportional relationships.
How to find a ratio?
To determine a ratio, you divide one quantity by another. This process results in a fractional or decimal representation that can be further simplified. Understanding how to find a ratio is simple. For example:
To find the ratio of 12 to 6, divide 12 by 6, resulting in 2:1
To find the ratio of three numbers, divide each value by the smallest number among them. For instance, the ratio of 12, 18 and 24 becomes 2:3:4 after simplification.
Learning how to find a ratio ensures precision when comparing values. The ratio calculator automates these calculations, making it easy to get accurate results quickly.
How to calculate ratio?
The ratio for calculating the ratio is straightforward, where two quantities are being compared.
The ratio calculator applies this formula to produce results in the simplest form. Here is an example:
Examples on ratio formula:
To calculate the ratio of 5 to 4: This simplifies to 5:4
To calculate the ratio of 1920 to 1080(commonly used in aspect ratios): This simplifies to 16:9 a standard aspect ratio for screens
The ratio calculator ensures such conversions are accurate and hassle-free.
Understanding Practical Applications of Ratios
Ratios are fundamental in various fields, offering practical insights across disciplines like design, finance, and more. Here are some interesting applications of ratios in everyday scenarios:
Expense Ratio in Finance: The expense ratio is a financial term that shows what percentage of a fund's assets are used for operational costs. By dividing total expenses by the total assets, this ratio helps investors assess how much is being spent on fund management.
Golden Ratio in Art and Nature: The golden ratio (approximately 1.618) is a mathematical constant that appears in various natural phenomena and artistic designs. It defines a specific proportional relationship, often used to create aesthetically pleasing compositions in art, architecture, and even nature.
Aspect Ratio of 1920x1080: A common standard for high-definition displays, the 1920x1080 resolution has an aspect ratio of 16:9. This ratio is derived by dividing both the width (1920) and the height (1080) by their greatest common divisor, which is 120.
Meaning of the 16:9 Ratio: Often associated with widescreen displays, the 16:9 ratio indicates the proportional relationship between width and height. This ratio is commonly used in video formats and modern televisions to offer an immersive viewing experience.
Understanding a 4:1 Ratio: A 4:1 ratio signifies that one quantity is four times greater than the other. This simple proportional relationship is useful in various contexts, from recipe adjustments to financial planning.
How to find the ratio of two numbers?
Finding the ratio of two numbers involves these steps:
Identify the two numbers you want to compare
Divide the first number by the second
Simplify the result into a standard ratio form
Example:
To find the ratio of 12 to 6. Divide 12 by 6.
The ratio calculator simplifies these steps, saving time and ensuring precision.
How to calculate the ratio of three numbers?
To calculate the ratio of three numbers follow these steps:
Determine the smallest number among the three
Divide all three numbers by the smallest number
Express the results as a ratio
Example:
To find the ratio of 6,9,12. Divide each by 3(the smallest number).
The ratio calculator handles multi-number ratios seamlessly.
Benefits of using a ratio calculator
Ease of use: Enter your values and get instant results.
Accuracy: Eliminates the risk of manual calculation errors.
Versatility: Handles two or more numbers and provides simplified forms.
Convenience: Ideal for students, professionals and anyone dealing with ratios.
The ratio calculator is an indispensable tool for simplifying the complex task of comparing quantities. By automating the process, it provides accurate and reliable results for a wide range of applications. From understanding how to calculate ratio to exploring practical examples, mastering this tool will enhance your mathematical and analytical skills. Whether you are dealing with financial ratios, aspect ratios or any other proportional comparison, the ratio calculator is your go-to resource for precision and simplicity.
FAQs
1.How to calculate ratio?
To calculate a ratio, first, identify the two quantities you are comparing. These could be any two numbers or amounts that have a relationship with each other. Next, express the relationship as a fraction by placing the first quantity (the numerator) over the second quantity (the denominator).
2. What is the ratio formula?
The formula is, where and are the quantities compared.
3. How do you find the ratio of three numbers?
Divide each number by the smallest value among them and simplify.
4. How to find the ratio of two numbers?
Divide the first number by the second and express it.
5. How to calculate expense ratio?
Divide total expenses by total assets and multiply by 100 to express as a percentage.
6. How to calculate the golden ratio?
The golden ratio is derived by dividing a line into two parts so that the ratio of the whole to the larger part equals the ratio of the larger part to the smaller.
7. What is the ratio of 5 to 4?
It is expressed as 5:4, indicating proportional comparison.
8. What is the ratio of 12 to 6?
The ratio simplifies to 2:1.
9. What is a simple ratio?
A simple ratio is the most basic form of comparison, such as 1:2 or 3:4.
10. What is the aspect ratio for 1920x1080?
The aspect ratio is 16:9, used for screens and displays.
11. What does 16:9 ratio mean?
It means the width is 16 units for every 9 units of height.
12. What is a 4 to 1 ratio?
This ratio shows that one quantity is four times the size of another.
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10641 | https://bookdown.org/pbaumgartner/bayesian-fun/16-bayes-factor-posterior-odds.html | Bayesian Statistics the Fun Way
PART I: Introduction to Probability
1 Bayesian Thinking and Everyday Reasoning
2 Measuring Uncertainty
3 Logic of Uncertainty
4 Creating a Binomial Probability Distribution
5 The Beta Distribution
PART II: Bayesian Probability and Prior Probabilities
6 Conditional Probability
7 Bayes’ Theorem With LEGO
8 The Prior, Likelihood, and Posterior of Bayes’ Theorem
9 Bayesian Priors and Working with Probability Distributions
PART III: Parameter Estimation
10 Introduction to Averaging and Parameter Estimation
11 Measuring the Spread of our Data
12 The Normal Distribution
13 Tools of Parameter Estimation: The PDF, CDF, and Quantile Function
14 Parameter Estimation with Prior Probabilities
PART IV: Hypothesis Testing: The Heart of Statistics
15 From Parameter Estimation to Hypothesis Testing: Building a Bayesian A/B Test
16 Introduction to the Bayes Factor and Posterior Odds: The Competition of Ideas
17 Bayesian Reasoning in the Twilight Zone
18 When Data Doesn’t Convince
19 From Hypothesis Testing to Parameter Estimation
References
Appendices
A A Quick Introduction to R
B Enough Calculus to Get By
Table of contents
16.1 Revisiting Bayes’ Theorem
16.2 Building a Hypothesis Test Using the Ratio of Posteriors
16.2.1 The Bayes Factor
16.2.2 Prior Odds
16.2.3 Posterior Odds
16.2.4 Empty: Guidelines for Evaluating Posterior Odds
16.2.5 Empty: Self-Diagnosing Rare Diseases Online
16.3 Wrapping Up
16.4 Exercises
16.4.1 Exercise 16-1
16.4.2 Exercise 16-2
16 Introduction to the Bayes Factor and Posterior Odds: The Competition of Ideas
The Bayes factor is a formula that tests the plausibility of one hypothesis by comparing it to another. The result tells us how many times more likely one hypothesis is than the other.
16.1 Revisiting Bayes’ Theorem
[ P(H \mid D) = \frac{{P(H)} \times P(D \mid H) }{P(D)} \tag{16.1}]
(P(H \mid D)): Posterior Probability, which tells us how strongly we should believe in our hypothesis, given our data.
(P(H)): Prior Probability or Prior Belief, the probability of our hypothesis prior to looking at the data.
(P(D \mid H)): Likelihood of getting the existing data if our hypothesis were true.
(P(D)) is the probability of the data observed independent of the hypothesis. We need P(D) in order to make sure that our posterior probability is correctly placed somewhere between 0 and 1.
(P(D)) is … totally unnecessary if all we care about is comparing the relative strength of two different hypotheses. … For these reasons, we often use the proportional form of Bayes’ theorem.
[P(H \mid D) \propto P(H) \times P(D \mid H) \tag{16.2}]
the posterior probability of our hypothesis is proportional to the prior multiplied by the likelihood. We can use this to compare two hypotheses by examining the ratio of the prior belief multiplied by the likelihood for each hypothesis using the ratio of posteriors formula:
[\frac{P(H_{1}) \times P(D \mid H_{1})}{P(H_{2}) \times P(D \mid H_{2})} \tag{16.3}]
if the ratio is 2, then (H_{1}) explains the observed data twice as well as (H_{2}), and if the ratio is (\frac{1}{2}), then (H_{2}) explains the data twice as well as (H_{1}).
16.2 Building a Hypothesis Test Using the Ratio of Posteriors
The ratio of posteriors formula gives us the posterior odds, which allows us to test hypotheses or beliefs we have about data.
To better understand the posterior odds, we’ll break down the ratio of posteriors formula into two parts: the likelihood ratio, or the Bayes factor, and the ratio of prior probabilities.
16.2.1 The Bayes Factor
[\frac{P(D \mid H_{1})}{P(D \mid H_{2})} \tag{16.4}]
What this ratio tells us is the likelihood of what we’ve seen given what we believe to be true compared to what someone else believes to be true.
The key here is that in Bayesian reasoning, we don’t worry about supporting our beliefs—we are focused on how well our beliefs support the data we observe. In the end, data can either confirm our ideas or lead us to change our minds.
16.2.2 Prior Odds
So far we have assumed that the prior probability of each hypothesis is the same. This is clearly not always the case: a hypothesis may explain the data well even if it is very unlikely.
[\frac{P(H_{1})}{P(H_{2})}]
This ratio compares the probability of two hypotheses before we look at the data. When used in relation to the Bayes factor, this ratio is called the prior odds in our (H_{1}) and written as (O(H_{1})). This representation is helpful because it lets us easily note how strongly (or weakly) we believe in the hypothesis we’re testing. When this number is greater than 1, it means the prior odds favor our hypothesis, and when it is a fraction less than 1, it means they’re against our hypothesis. For example, (O(H_{1}) = 100)$ means that, without any other information, we believe (H_{1}) is 100 times more likely than the alternative hypothesis.
16.2.3 Posterior Odds
[\text{posterior odds} = O(H_{1})\frac{P(D \mid H_{1})}{P(D \mid H_{2})} \tag{16.5}]
Table 16.1: Guidelines for Evaluating Posterior Odds
| Posterior odds | Strength of evidence |
| 1 to 3 | Interesting, but nothing conclusive |
| 3 to 20 | Looks like we’re on to something |
| 20 to 150 | Strong evidence in favor of (H_{1}) |
| > 150 | Overwhelming evidence |
16.2.4 Empty: Guidelines for Evaluating Posterior Odds
16.2.5 Empty: Self-Diagnosing Rare Diseases Online
16.3 Wrapping Up
In this chapter, you learned how to use the Bayes factor and posterior odds to compare two hypotheses.
16.4 Exercises
Try answering the following questions to see how well you understand the Bayes factor and posterior odds. The solutions can be found at
16.4.1 Exercise 16-1
Returning to the dice problem, assume that your friend made a mistake and suddenly realized that there were, in fact, two loaded dice and only one fair die. How does this change the prior, and therefore the posterior odds, for our problem? Are you more willing to believe that the die being rolled is the loaded die?
Let’s recapitulate:
Suppose your friend has a bag with three six-sided dice in it, and one die is weighted so that it lands on 6 half the time. The other two are traditional dice whose probability of rolling a 6 is ⅙. Your friend pulls out a die and rolls 10 times, with the following results: (6, 1, 3, 6, 4, 5, 6, 1, 2, 6)
Definition 16.1 (Bayes Factor) [\frac{P(D \mid H_{1})}{P(D \mid H_{2})}]
It turned out that (H_{1}) — the loaded dice — explain the data 3.77 times better than the fair dice with (H_{2}).
However, this is true only if (H_{1}) and (H_{2}) are both just as likely to be true in the first place.
But we know now for this exercise that there are two loaded dice in the bag and only one fair die, which means that each hypothesis was not equally likely. Based on the distribution of the dice in the bag, we know that these are the prior probabilities for each hypothesis:
[P(H_{1} = \frac{2}{3}); P(H_{2} = \frac{1}{3})] > From these, we can calculate the prior odds for (H_{1}):
[\text{prior odds} = O(H_{1}) = \frac{P(H_{1})}{P(H_{2})} = \frac{\frac{2}{3}}{\frac{1}{3}} = \frac{6}{3} = 2]
With our prior odds for H1, we can now compute our full posterior odds:
[\text{posterior odds} = O(H_{1}) = \frac{P(D \mid H_{1})}{P(D \mid H_{2})} = 2 \times 3.77 = 7.54]
Solution 16.1 Yes, I am now more willing to believe that the die being rolled is the loaded die!
16.4.2 Exercise 16-2
Returning to the rare diseases example, suppose you go to the doctor, and after having your ears cleaned you notice that your symptoms persist. Even worse, you have a new symptom: vertigo. The doctor proposes another possible explanation, labyrinthitis, which is a viral infection of the inner ear in which 98 percent of cases involve vertigo. However, hearing loss and tinnitus are less common in this disease; hearing loss occurs only 30 percent of the time, and tinnitus occurs only 28 percent of the time. Vertigo is also a possible symptom of vestibular schwannoma, but occurs in only 49 percent of cases. In the general population, 35 people per million contract labyrinthitis annually. What is the posterior odds when you compare the hypothesis that you have labyrinthitis against the hypothesis that you have vestibular schwannoma?
Labyrinthitis:
[P(D \mid H_{1}) = 0.98 (\text{vertigo}) \times 0.30 (\text{hearing loss}) \times 0.28 (\text{tinnitus}) = 0.082]
Vestibular schwannoma
[P(D \mid H_{2}) = 0.49 (\text{vertigo}) \times 0.94 (\text{hearing loss}) \times 0.89 (\text{tinnitus}) = 0.41]
Bayes factor:
[\frac{P(D \mid H_{1})}{P(D \mid H_{2})} = \frac{0.08}{0.48} = 0.17]
Including prior odds:
[\text{prior odds} = O(H_{1}) = \frac{P(H_{1})}{P(H_{2})} = \frac{\frac{35}{1,000,000}}{\frac{11}{1,000,000}} = \frac{35}{11} = 3.18] Based on prior information alone, a given person is about only about 3 times more likely to have labyrinthitis than vestibular schwannoma. Now let’s compute the full posterior odds to see if the situation gets better:
[\text{prior odds} = O(H_{1}) = \frac{P(H_{1})}{P(H_{2})} = \frac{35}{11} \times 6 = 19,09]
Solution 16.2 The probability of having labyrinthitis is 19 times higher than vestibular schwannoma.
Will Kurt has another result because he is calculating the labyrinthitis against the earwax impaction!
15 From Parameter Estimation to Hypothesis Testing: Building a Bayesian A/B Test
17 Bayesian Reasoning in the Twilight Zone
Source Code
--- ---engine: knitrengine: knitr --- ---# Introduction to the Bayes Factor and Posterior Odds: The Competition of Ideas# Introduction to the Bayes Factor and Posterior Odds: The Competition of Ideas> The `r glossary("Bayes factor")` is a formula that tests the plausibility of one hypothesis by comparing it to another. The result tells us how many times more likely one hypothesis is than the other.> The `r glossary("Bayes factor")` is a formula that tests the plausibility of one hypothesis by comparing it to another. The result tells us how many times more likely one hypothesis is than the other. ## Revisiting Bayes’ Theorem ## Revisiting Bayes’ Theorem $$P(H \mid D) = \frac{{P(H)} \times P(D \mid H) }{P(D)}$$ {#eq-bayes-theorem} - $P(H \mid D)$: `r glossary("Posterior Probability")`, which tells us how strongly we should believe in our hypothesis, given our data. - `r glossary("Posterior Probability")` - $P(H)$: `r glossary("Prior Probability")` or Prior Belief, the probability of our hypothesis prior to looking at the data. - `r glossary("Prior Probability")` - $P(D \mid H)$: `r glossary("Likelihood")` of getting the existing data if our hypothesis were true. - `r glossary("Likelihood")` - $P(D)$ is the probability of the data observed independent of the hypothesis. We need P(D) in order to make sure that our posterior probability is correctly placed somewhere between 0 and 1. - > $P(D)$ is … totally unnecessary if all we care about is comparing the relative strength of two different hypotheses. … For these reasons, we often use the proportional form of `r glossary("Bayes’ theorem")`.> $P(D)$ is … totally unnecessary if all we care about is comparing the relative strength of two different hypotheses. … For these reasons, we often use the proportional form of `r glossary("Bayes’ theorem")`.$$P(H \mid D) \propto P(H) \times P(D \mid H)$$ {#eq-bayes-prop}> the posterior probability of our hypothesis is proportional to the prior multiplied by the likelihood. We can use this to compare two hypotheses by examining the ratio of the prior belief multiplied by the likelihood for each hypothesis using the ratio of posteriors formula:> the posterior probability of our hypothesis is proportional to the prior multiplied by the likelihood. We can use this to compare two hypotheses by examining the ratio of the prior belief multiplied by the likelihood for each hypothesis using the ratio of posteriors formula:$$\frac{P(H_{1}) \times P(D \mid H_{1})}{P(H_{2}) \times P(D \mid H_{2})}$$ {#eq-posterior-ratio}> if the ratio is 2, then $H_{1}$ explains the observed data twice as well as $H_{2}$, and if the ratio is $\frac{1}{2}$, then $H_{2}$ explains the data twice as well as $H_{1}$.> if the ratio is 2, then $H_{1}$ explains the observed data twice as well as $H_{2}$, and if the ratio is $\frac{1}{2}$, then $H_{2}$ explains the data twice as well as $H_{1}$. ## Building a Hypothesis Test Using the Ratio of Posteriors ## Building a Hypothesis Test Using the Ratio of Posteriors> The ratio of posteriors formula gives us the posterior odds, which allows us to test hypotheses or beliefs we have about data. > The ratio of posteriors formula gives us the posterior odds, which allows us to test hypotheses or beliefs we have about data. > To better understand the posterior odds, we’ll break down the ratio of posteriors formula into two parts: the likelihood ratio, or the Bayes factor, and the ratio of prior probabilities.> To better understand the posterior odds, we’ll break down the ratio of posteriors formula into two parts: the likelihood ratio, or the Bayes factor, and the ratio of prior probabilities. ### The Bayes Factor ### The Bayes Factor$$\frac{P(D \mid H_{1})}{P(D \mid H_{2})}$$ {#eq-bayes-factor}> What this ratio tells us is the likelihood of what we’ve seen given what we believe to be true compared to what someone else believes to be true.> What this ratio tells us is the likelihood of what we’ve seen given what we believe to be true compared to what someone else believes to be true.> The key here is that in Bayesian reasoning, we don’t worry about supporting our beliefs—we are focused on how well our beliefs support the data we observe. In the end, data can either confirm our ideas or lead us to change our minds.> The key here is that in Bayesian reasoning, we don’t worry about supporting our beliefs—we are focused on how well our beliefs support the data we observe. In the end, data can either confirm our ideas or lead us to change our minds. ### Prior Odds ### Prior Odds> So far we have assumed that the prior probability of each hypothesis is the same. This is clearly not always the case: a hypothesis may explain the data well even if it is very unlikely. > So far we have assumed that the prior probability of each hypothesis is the same. This is clearly not always the case: a hypothesis may explain the data well even if it is very unlikely. $$\frac{P(H_{1})}{P(H_{2})}$$> This ratio compares the probability of two hypotheses before we look at the data. When used in relation to the Bayes factor, this ratio is called the prior odds in our $H_{1}$ and written as $O(H_{1})$. This representation is helpful because it lets us easily note how strongly (or weakly) we believe in the hypothesis we’re testing. When this number is greater than 1, it means the prior odds favor our hypothesis, and when it is a fraction less than 1, it means they’re against our hypothesis. For example, $O(H_{1}) = 100$$ means that, without any other information, we believe $H_{1}$ is 100 times more likely than the alternative hypothesis.> This ratio compares the probability of two hypotheses before we look at the data. When used in relation to the Bayes factor, this ratio is called the prior odds in our $H_{1}$ and written as $O(H_{1})$. This representation is helpful because it lets us easily note how strongly (or weakly) we believe in the hypothesis we’re testing. When this number is greater than 1, it means the prior odds favor our hypothesis, and when it is a fraction less than 1, it means they’re against our hypothesis. For example, $O(H_{1}) = 100$$ means that, without any other information, we believe $H_{1}$ is 100 times more likely than the alternative hypothesis. ### Posterior Odds ### Posterior Odds$$\text{posterior odds} = O(H_{1})\frac{P(D \mid H_{1})}{P(D \mid H_{2})}$$ {#eq-posterior-odds} | Posterior odds | Strength of evidence |-------------------------------------|| 1 to 3 | Interesting, but nothing conclusive | | 3 to 20 | Looks like we’re on to something || 20 to 150 | Strong evidence in favor of $H_{1}$ || > 150 | Overwhelming evidence |: Guidelines for Evaluating Posterior Odds {#tbl-guidelines-odds}### Empty: Guidelines for Evaluating Posterior Odds### Empty: Guidelines for Evaluating Posterior Odds### Empty: Self-Diagnosing Rare Diseases Online### Empty: Self-Diagnosing Rare Diseases Online ## Wrapping Up ## Wrapping Up> In this chapter, you learned how to use the Bayes factor and posterior odds to compare two hypotheses.> In this chapter, you learned how to use the Bayes factor and posterior odds to compare two hypotheses. ## Exercises ## Exercises> Try answering the following questions to see how well you understand the Bayes factor and posterior odds. The solutions can be found at Try answering the following questions to see how well you understand the Bayes factor and posterior odds. The solutions can be found at Exercise 16-1### Exercise 16-1> Returning to the dice problem, assume that your friend made a mistake and suddenly realized that there were, in fact, two loaded dice and only one fair die. How does this change the prior, and therefore the posterior odds, for our problem? Are you more willing to believe that the die being rolled is the loaded die?> Returning to the dice problem, assume that your friend made a mistake and suddenly realized that there were, in fact, two loaded dice and only one fair die. How does this change the prior, and therefore the posterior odds, for our problem? Are you more willing to believe that the die being rolled is the loaded die?Let's recapitulate: > Suppose your friend has a bag with three six-sided dice in it, and one die is weighted so that it lands on 6 half the time. The other two are traditional dice whose probability of rolling a 6 is ⅙. Your friend pulls out a die and rolls 10 times, with the following results: $6, 1, 3, 6, 4, 5, 6, 1, 2, 6$> Suppose your friend has a bag with three six-sided dice in it, and one die is weighted so that it lands on 6 half the time. The other two are traditional dice whose probability of rolling a 6 is ⅙. Your friend pulls out a die and rolls 10 times, with the following results: $6, 1, 3, 6, 4, 5, 6, 1, 2, 6$::: {#def-bayes-factor} ##### Bayes Factor ##### Bayes Factor$$\frac{P(D \mid H_{1})}{P(D \mid H_{2})}$$:::It turned out that $H_{1}$ --- the loaded dice --- explain the data 3.77 times better than the fair dice with $H_{2}$.> However, this is true only if $H_{1}$ and $H_{2}$ are both just as likely to be true in the first place. > However, this is true only if $H_{1}$ and $H_{2}$ are both just as likely to be true in the first place. But we know now for this exercise that there are two loaded dice in the bag and only one fair die, which means that each hypothesis was not equally likely. Based on the distribution of the dice in the bag, we know that these are the prior probabilities for each hypothesis:$$P(H_{1} = \frac{2}{3}); P(H_{2} = \frac{1}{3})$$> From these, we can calculate the prior odds for $H_{1}$:> From these, we can calculate the prior odds for $H_{1}$:$$\text{prior odds} = O(H_{1}) = \frac{P(H_{1})}{P(H_{2})} = \frac{\frac{2}{3}}{\frac{1}{3}} = \frac{6}{3} = 2$$> With our prior odds for H1, we can now compute our full posterior odds:> With our prior odds for H1, we can now compute our full posterior odds:$$\text{posterior odds} = O(H_{1}) = \frac{P(D \mid H_{1})}{P(D \mid H_{2})} = 2 \times 3.77 = 7.54$$::: {#lem-solution-16-1}Yes, I am now more willing to believe that the die being rolled is the loaded die!:::### Exercise 16-2### Exercise 16-2> Returning to the rare diseases example, suppose you go to the doctor, and after having your ears cleaned you notice that your symptoms persist. Even worse, you have a new symptom: vertigo. The doctor proposes another possible explanation, labyrinthitis, which is a viral infection of the inner ear in which 98 percent of cases involve vertigo. However, hearing loss and tinnitus are less common in this disease; hearing loss occurs only 30 percent of the time, and tinnitus occurs only 28 percent of the time. Vertigo is also a possible symptom of vestibular schwannoma, but occurs in only 49 percent of cases. In the general population, 35 people per million contract labyrinthitis annually. What is the posterior odds when you compare the hypothesis that you have labyrinthitis against the hypothesis that you have vestibular schwannoma?> Returning to the rare diseases example, suppose you go to the doctor, and after having your ears cleaned you notice that your symptoms persist. Even worse, you have a new symptom: vertigo. The doctor proposes another possible explanation, labyrinthitis, which is a viral infection of the inner ear in which 98 percent of cases involve vertigo. However, hearing loss and tinnitus are less common in this disease; hearing loss occurs only 30 percent of the time, and tinnitus occurs only 28 percent of the time. Vertigo is also a possible symptom of vestibular schwannoma, but occurs in only 49 percent of cases. In the general population, 35 people per million contract labyrinthitis annually. What is the posterior odds when you compare the hypothesis that you have labyrinthitis against the hypothesis that you have vestibular schwannoma?Labyrinthitis:$$P(D \mid H_{1}) = 0.98 (\text{vertigo}) \times 0.30 (\text{hearing loss}) \times 0.28 (\text{tinnitus}) = 0.082$$Vestibular schwannoma$$P(D \mid H_{2}) = 0.49 (\text{vertigo}) \times 0.94 (\text{hearing loss}) \times 0.89 (\text{tinnitus}) = 0.41$$Bayes factor:$$\frac{P(D \mid H_{1})}{P(D \mid H_{2})} = \frac{0.08}{0.48} = 0.17$$Including prior odds:$$\text{prior odds} = O(H_{1}) = \frac{P(H_{1})}{P(H_{2})} = \frac{\frac{35}{1,000,000}}{\frac{11}{1,000,000}} = \frac{35}{11} = 3.18$$Based on prior information alone, a given person is about only about 3 times more likely to have labyrinthitis than vestibular schwannoma. Now let's compute the full posterior odds to see if the situation gets better:$$\text{prior odds} = O(H_{1}) = \frac{P(H_{1})}{P(H_{2})} = \frac{35}{11} \times 6 = 19,09$$::: {#lem-solution-16-2}The probability of having labyrinthitis is 19 times higher than vestibular schwannoma.:::::: {.callout-warning}Will Kurt has another result because he is calculating the labyrinthitis against the earwax impaction!::: |
10642 | https://static.zzpxx.com/editor/default/20191015/b005724e938a0dbc37c44a63bb405980.pdf | 中 中 中国 国 国数 数 数学 学 学奥 奥 奥林 林 林匹 匹 匹克 克 克(CMO) 历 历 历届 届 届试 试 试题 题 题及 及 及解 解 解答 答 答 1986-2018 第一届中国数学奥林匹克(1986年) 天津南开大学 1.已知a1, a2, . . . , an为实数, 如果它们中任意两数之和非负,那么对于满足 x1 + x2 + · · · + xn = 1 的任意非负实数x1, x2, . . . , xn, 有不等式 a1x1 + a2x2 + · · · + anxn ⩾a1x2 1 + a2x2 2 + · · · + anx2 n 成立.请证明上述命题及其逆命题.
证明:原命题的证明:由0 ⩽xi ⩽1, xi −x2 i ⩾0, xi ⩾x2 i (i = 1, 2, . . . , n).
(1)若ai ⩾0(i = 1, 2, . . . , n),则显然有a1x1 + a2x2 + · · · + anxn ⩾a1x2 1 + a2x2 2 + · · · + anx2 n; (2)否则至少存在一个ai < 0,由对称性不妨设a1 < 0. 又因为a1, a2, . . . , an中任两数之和非负,所 以ai + a1 ⩾0, ai ⩾−a1 > 0(i = 2, 3, . . . , n).
∴ a1x1 + a2x2 + · · · + anxn −a1x2 1 −a2x2 2 −· · · −anx2 n = a1(x1 −x2 1) + a2(x2 −x2 2) + · · · + an(xn −x2 n) ⩾ a1(x1 −x2 1) + (−a1)(x2 −x2 2) + · · · + (−a1)(xn −x2 n) = (−a1)(x2 1 −x2 2 −· · · −x2 n −x1 + x2 + · · · + xn) = (−a1)(x2 1 −x1 + (1 −x1) −x2 2 −· · · −x2 n) = (−a1)((1 −x1)2 −x2 2 −· · · −x2 n) = (−a1)((x2 + · · · + xn)2 −x2 2 −· · · −x2 n) ⩾0 最后一步是由于x2, x3, . . . , xn > 0, (x2 + · · · + xn)2 = x2 2 + · · · + x2 n + P 2⩽i<j⩽n xixj ⩾x2 2 + · · · + x2 n.
逆命题的证明:对于任意的1 ⩽i < j ⩽n,令xi = xj = 1 2,其余xk均等于0.则1 2(ai + aj) ⩾1 4(ai + aj).
∴ai + aj ⩾0,即任两数之和非负.证毕.
2.在三角形ABC中,BC边上的高AD = 12,∠A的平分线AE = 13,设BC边上的中线AF = m,问m在什 么范围内取值时,∠A分别为锐角,直角,钝角?
解:设O为△ABC的外心,不妨设AB > AC,∠B为锐角.
则OF垂直平分线段BC,由外心的性质,∠C为锐角时,∠OAB = ∠OBA = 1 2(180◦−∠AOB) = 1 2(180◦−2∠C) = 90◦−∠C.
又因为AD ⊥BC,∴∠CAD = 90◦−∠C,∴∠OAB = ∠DAC.
类似地,当∠C为直角或钝角时也有∠OAB = ∠DAC.
由AE平分∠BAC,∠BAE = ∠CAE.∴∠OAE = ∠DAE.(由于F, D在E两侧).
∠A为锐角时,O, A在BC同侧,∠FAE < ∠OAE = ∠DAE; ∠A为直角时,O, F重合,∠FAE = ∠OAE = ∠DAE; ∠A为钝角时,O, A在BC异侧,∠FAE > ∠OAE = ∠DAE.
1 由正弦定理sin ∠F AE sin ∠DAE = F E DE × AD AF .其中DE = √ AE2 −AD2 = 5, FE = FD −DE = √ AF 2 −AD2 −DE = √ m2 −122 −5 > 0. ∴m > 13, 且∠A为锐角等价于 √ m2−122−5 5 × 12 m < 1; ∠A为直角等价于 √ m2−122−5 5 × 12 m = 1; ∠A为钝角等价于 √ m2−122−5 5 × 12 m > 1.
解得当13 < m < 2028 119 时,∠A为锐角; 当m = 2028 119 时,∠A为直角; 当m > 2028 119 时,∠A为钝角.
3.设z1, z2, . . . , zn为复数,满足 |z1| + |z2| + · · · + |zn| = 1.
求证:上述n个复数中,必存在若干个复数,它们的和的模不小于1 6.
证明:设zk = xk + yki(xk, yk ∈R, k = 1, 2 . . . , n) 将所有的zk分为两组X,Y.若|xk| ⩾|yk|,则将zk放入X中;若|yk| ⩾|xk|,则将zk放入Y中. 其中必有一组中 所有复数模长之和不小于1 2.不妨设为X.
再将X中的复数分为两组A,B.若xk ⩾0,则将zk放入A中;若xk ⩽0,则将zk放入B中. 其中必有一组中的 所有复数摸长之和不小于1 4.不妨设为A.
则P zk∈A |zk| ⩾1 4,即P zk∈A p x2 k + y2 k ⩾1 4.
而对于zk ∈A,x2 k ⩾y2 k, p x2 k + y2 k ⩽ √ 2xk.
∴P zk∈A xk ⩾ 1 4 √ 2.∴| P zk∈A zk| = | P zk∈A xk + i P zk∈A yk| ⩾P zk∈A xk ⩾ 1 4 √ 2.
而4 √ 2 < 6, ∴| P zk∈A zk| ⩾1 6.
即A中复数之和的模不小于1 6.证毕.
另证:设zk = xk + yki(xk, yk ∈R, k = 1, 2 . . . , n) 则|zk| = p x2 k + y2 k ⩾|xk| + |yk|.
∴ n P k=1 |xk| + |yk| ⩾1.
∴| P xk⩾0 xk| + | P xk<0 xk| + | P yk⩾0 yk| + | P yk<0 yk| ⩾1.
其中必有一项不小于1 4,不妨设为第一项,则| P xk⩾0 xk| ⩾1 4.
∴| P xk⩾0 zk| = | P xk⩾0 xk + i P xk⩾0 yk| ⩾| P xk⩾0 xk| ⩾1 4 > 1 6.证毕.
4.已知:四边形P1P2P3P4的四个顶点位于三角形ABC的边上.
求证:四个三角形△P1P2P3, △P1P2P4, △P1P3P4, △P2P3P4 中,至少有一个的面积不大于△ABC的面积 的四分之一.
证明:有两种情况:(1)四个顶点在两条边上;(2)四个顶点在三条边上.
(1)不妨设P1, P4在AB上,P2, P3在AC上,P1, P2分别在AP4, AP3上. 将B移至P4,C移至P3,三角形ABC的 2 面积减小,归为情形(2).
(2)不妨设P1在AB上,P2在AC上,P3, P4在BC上,P3在P4C上.
(2.1)若P1P2 ∥BC,设AP1 AB = AP2 AC = λ,P1P2 = λBC.P1P2到BC的距离为(1−λ)h,h为三角形ABC中BC边 上的高的长度.
∴S△P1P2P3 = λ(1 −λ)S△ABC ⩽1 4S△ABC.
(2.2)若P1P2不平行于BC,不妨设P1到BC的距离大于P2到BC的距离.
过P2作平行于BC的直线 交AB于E,交P1P4于D.则S△P1P2P3, S△P4P2P3中有一个不大于S△DP2P3,也就不大于S△EP2P3.
由(2.1)知S△EP2P3 ⩽1 4S△ABC.则S△P1P2P3, S△P4P2P3中有一个不大于1 4S△ABC.证毕.
5.能否把1,1,2,2,. . . ,1986,1986这些数排成一行, 使得两个1之间夹着1个数,两个2之间夹着2个数,. . . , 两 个1986之间夹着1986个数.请证明你的结论.
解:不能.假设可以做出这样的排列,将已排好的数按顺序编号为1,2,. . . ,3972.
当n为奇数时,两个n的编号奇偶性相同;当n为偶数时,两个n的编号奇偶性不同. 而1到1986之间有993个 偶数,所以一共有2k + 993个编号为偶数的数.(k ∈N∗) 但是1到3972之间有1986个偶数,k = 496.5.矛 盾.所以不能按要求排成这样一行.
6.用任意的方式,给平面上的每一点染上黑色或白色. 求证:一定存在一个边长为1或 √ 3的正三角形,它的 三个顶点是同色的.
证明:(1)若平面上存在距离为2的两个点A, B异色,设O为它们的中点,不妨设A, O同色. 考虑以AO为一 边的正三角形AOC, AOD,若C, D中有一个与A, O同色,则该三角形满足题意. 否则BCD为边长 √ 3的 同色正三角形.
(2)否则平面上任两个距离为2的点均同色,考虑任意两个距离为1的点,以他们连线为底,2为腰长作等腰 三角形,则任一腰的两顶点同色. 所以三个顶点同色,即任两个距离为1的点同色.所以平面上任意一个边 长为1的正三角形三个顶点同色.证毕.
3 第二届中国数学奥林匹克(1987年) 北京北京大学 1.设n为自然数,求证方程zn+1 −zn −1 = 0有模为1的复根的充分必要条件是n + 2可被6整除.
证明:当6|n + 2时,令z = ei π 3 = 1 2 + √ 3 2 i, z6 = 1, |z| = 1.
∴zn+1 −zn −1 = e−i π 3 −ei π 3 −1 = ( 1 2 − √ 3 2 i) −(−1 2 − √ 3 2 i) −1 = 0.
∴zn+1 −zn −1 = 0有模为1的复根.
若zn+1 −zn −1 = 0有模为1的复根eiθ = cos θ + i cos θ.
则zn+1 −zn −1 = (cos(n + 1)θ −cos nθ −1) + i(sin(n + 1)θ −sin nθ) = 0.
∴cos(n + 1)θ −cos nθ −1 = −(2 sin 2n+1 2 θ sin θ 2 + 1) = 0.
sin(n + 1)θ −sin nθ = 2 cos 2n+1 2 θ sin θ 2 = 0.
∴cos 2n+1 2 θ = 0, sin 2n+1 2 θ = ±1, sin θ 2 = ± 1 2, 设θ 2 = ϕ.
(1)sin ϕ = 1 2,sin(2n + 1)ϕ = −1. ϕ = 2kπ + π 6 或2kπ + 5π 6 , k ∈Z.
(2n + 1)ϕ = (2l + 3 2)π(l ∈Z). ∴(2n + 1)(2k + 1 6) = 2l + 3 2, 2n+1 6 = 2t + 3 2, n = 6t + 4(t ∈Z).
或(2n + 1)(2k + 5 6) = 2l + 3 2, 5(2n+1) 6 = 2t + 3 2, 5|4t + 3, t ≡3 (mod 5)(t ∈Z).
设t = 5s + 3,则n = 6s + 4,总有6|n + 2.
(2)sin ϕ = −1 2,sin(2n + 1)ϕ = 1.显然以−ϕ代ϕ即有(1).所以6|n + 2.证毕.
2.把边长为1的正三角形ABC的各边都n等分,过各分点平行于其它两边的直线, 将这三角形分成若干个 小三角形,这些小三角形的顶点都称为结点, 并且在每一结点上放置了一个实数.已知: (1)A, B, C三点上放置的数分别为a, b, c.
(2)在每个由有公共边的两个最小三角形组成的菱形之中, 两组相对顶点上放置的数之和相等.
试求:(1)放置最大数的点和放置最小数的点之间的最短距离.
(2)所有结点上数的总和S.
解:(1)不难证明同一直线上相邻三个结点上放置的数中间一个为两边的等差中项,所以同一直线上的数 按顺序成等差数列. 若两端的数相等,则所有的数都相等.否则两端的数为最大的和最小的.
若a, b, c相等,显然所有数都相等,最短距离显然为0.
若a, b, c两两不等,最大的数与最小的数必出现在A, B, C上,最短距离为1.
若a, b, c有两个相等但不与第三个相等,不妨设a = b > c,最小的数为c,最大的数出现在线段AB的任意 结点上. 当n为偶数时,与C最近的为AB中点,最短距离为 √ 3 2 . 当n为奇数时,与C最近的为AB中点向两 边偏1 2n的点,最短距离为1 2 q 3 + 1 n2 .
(2)将这个三角形绕中心旋转2 3π, 4 3π弧度,得到的两个三角形也满足题意(2). 将这三个三角形对应结 点的数相加形成的三角形也满足(2),三个顶点上的数均为a + b + c.由(1)的分析知所有结点上的数均 为a + b + c. 共1 2(n + 1)(n + 2)个结点,∴S = 1 3( 1 2(n + 1)(n + 2))(a + b + c) = 1 6(n + 1)(n + 2)(a + b + c).
3.某次体育比赛,每两名选手都进行一场比赛, 每场比赛一定决出胜负,通过比赛确定优秀选手, 选 手A被确定为优秀选手的条件是:对任何其它选手B, 或者A胜B,或者存在选手C,C胜B,A胜C. 结果按上 4 述规则确定的优秀选手只有一名, 求证:这名选手一定胜所有其它选手.
证明:假设该优秀选手为A,且存在其他选手胜A.
设B为所有胜A的人中胜的场次最多的一个,由B不是优秀选手,必存在选手C使得C胜B, 且不存在选 手D使得B胜D,D胜C. 由B胜A,C也胜A,且C胜B胜过的所有人.C至少比B多胜一场,且C胜A,与B的选取 矛盾.所以A胜所有人.
4.在一个面积为1的正三角形内部,任意放五个点,试证:在此正三角形内, 一定可以作三个正三角形盖住 这五个点, 这三个正三角形的各边分别平行于原三角形的边, 并且它们的面积之和不超过0.64.
证明:可将0.64换成100 169 + ε(ε > 0).
在面积为1的正三角形ABC中,在AB上取A1, B2,AC上取A2, C1,BC上取B1, C2, 使得AA1 = AA2 = BB1 = BB2 = CC1 = CC2 = 3 13AB.连结A1C2, A2B1, B2C1交于A0, B0, C0.
(1)若△AB2C1, △BC2A1, △CA2B1中有一个至少包含五个点中的三个,另两个点可分别用面积为ε 2的 正三角形覆盖, 面积之和为( 10 13)2 + 2 × ε 2 = 100 169 + ε.
(2)菱形AA1A0A2, BB1B0B2, CC1C0C2中有两个有两个点,另一个中有一个点, 则可用两个边长 为6 13AB的正三角形和一个面积为ε的正三角形覆盖. 面积之和为2( 6 13)2 + ε < 100 169 + ε.
(3)菱形AA1A0A2, BB1B0B2, CC1C0C2中有两个有一个点,另一个中有两个点, 不妨设为AA1A0A2, 则B1B0C0C2中有一个点,不妨设这个点更靠近B, 则可用一个边长为6 13AB的正三角形覆盖AA1A0A2中 两个点, 用一个边长为6 13AB的正三角形覆盖BB1B0B2, B1B0C0C2中的点. 用一个面积为ε的正三角形 覆盖最后一个点, 面积之和为( 6 13)2 + ( 8 13)2 + ε = 100 169 + ε.证毕.
注:当五个点取为A, B, C, A0, B0C0中点是不难证明不能用三个面积之和为100 169的正三角形覆盖这五个 点. 即100 169 + ε(ε > 0)为最优.
5.设A1A2A3A4是一个四面体, S1, S2, S3, S4分别是以A1, A2, A3, A4为球心的球, 它们两两相外切.如果 存在一点O, 以这点为球心可作一个半径为r的球与S1, S2, S3, S4都相切, 还可以作一个半径为R的球和 四面体的各棱都相切,求证:这个四面体是正四面体.
证明:设Si的半径为ri(i = 1, 2, 3, 4),则AiAj = ri + rj(1 ⩽i < j ⩽4).
设O到A2A3A4的投影为O1,由O到A2A3,A3A4,A4A2的距离相等, 得到O1到△A2A3A4的三边距离相 等.即O1为△A2A3A4的内心,设O到A2A3的投影为B,即O1到A2A3的投影. 而BA3 = 1 2(A2A3 + A3A4 − A2A4) = r3,OB = R. 若半径为r的球与四个球均外切,则A3O = r+r3,(r+r3)2 = r2 3 +R2, r3 = R2−r2 2r .
若半径为r的球与四个球均内切,则A3O = r−r3,(r−r3)2 = r2 3+R2, r3 = r2−R2 2r . 类似可求得r1, r2, r4均 为该值,所以该四面体各条棱长相等为正四面体.
6.m个互不相同的正偶数与n个互不相同的正奇数的总和为1987, 对于所有这样的m与n,问3m + 4n的 最大值是多少?请证明你的结论.
解:设m个正偶数为a1 < a2 < · · · < am,n个正偶数为b1 < b2 < · · · < bn.
∴ai ⩾2i, bj ⩾2j −1.
∵1987 = a1 + a2 + · · · + am + b1 + b2 + · · · + bn.
∴1987 ⩾2 + 4 + · · · + 2m + 1 + 3 + · · · + 2n −1 = m2 + m + n2.
5 设s = 3m + 4n,m = 1 3(s −4n), 1 3(s −4n)( 1 3(s −4n) + 1) + n2 ⩽1987.
s2 −8ns + 25n2 + 3s −12n −9 × 1987 ⩽0.
s2 + (3 −8n)s + 25n2 −12n −9 × 1987 ⩽0.
所以判别式∆= (3 −8n)2 −4(25n2 −12n −9 × 1987) = 26(1987 1 4 −n2) > 0.
s ⩽1 2(8n −3 + 6 q 1987 1 4 −n2).
设f(n) = 8n + 6 q 1987 1 4 −n2, f ′(n) = 8 −6n(1987 1 4 −n2)−1 2 ,又n为奇数.
不难知道n = 35时,f(n)有最大值280 + 6 q 762 1 4.
所以s ⩽1 2(280 + 6 q 762 1 4 −3),由s ∈N∗, s ⩽221.
又当s = 221, n = 35, m = 27.取2, 4, . . . , 52, 60, 1, 3, . . . , 69为和为1987的35个正奇数与27个正偶数,所 以3m + 4n的最大值为221.
6 第三届中国数学奥林匹克(1988年) 上海复旦大学 1.设a1, a2, . . . , an是给定的不全为零的实数, r1, r2, · · · , rn为实数,如果不等式 r1(x1 −a1) + r2(x2 −a2) + · · · + rn(xn −an) ⩽ q x2 1 + x2 2 + · · · + x2 n − q a2 1 + a2 2 + · · · + a2 n 对任何实数x1, x2, · · · , xn成立,求r1, r2, · · · , rn的值.
解:令xi = 0(i = 1, 2, . . . , n),−(r1a1 + r2a2 + · · · + rnan) ⩽− p a2 1 + a2 2 + · · · + a2 n.
∴( n P i=1 riai)2 ⩾ n P i=1 a2 i .
令xi = 2ai(i = 1, 2, . . . , n),r1a1 + r2a2 + · · · + rnan ⩽ p a2 1 + a2 2 + · · · + a2 n.
∴( n P i=1 riai)2 ⩽ n P i=1 a2 i .
∴( n P i=1 riai)2 = n P i=1 a2 i .
由Cauchy不等式, ( n P i=1 r2 i )( n P i=1 a2 i ) ⩾( n P i=1 riai)2, n P i=1 r2 i ⩾1.
又令xi = ri(i = 1, 2, . . . , n), n P i=1 r2 i − n P i=1 riai ⩽ s n P i=1 r2 i − s n P i=1 a2 i .
由 n P i=1 riai = s n P i=1 a2 i , n P i=1 r2 i ⩽ s n P i=1 r2 i , n P i=1 r2 i ⩽1.
∴ n P i=1 r2 i = 1,由Cauchy不等式取等号的条件知r1 a1 = r2 a2 = · · · = rn an .
不难解得ri = ai p a2 1 + a2 2 + · · · + a2 n (i = 1, 2, . . . , n).
2.设C1, C2为同心圆,C2的半径是C1的半径的2倍, 四边形A1A2A3A4内接于C1, 设A4A1延长线交 圆C2于B1, A1A2延长线交C2于B2, A2A3延长线交圆C2于B3, A3A4延长线交圆C2于B4.
试证:四边形B1B2B3B4的周长⩾2(四边形A1A2A3A4的周长).并确定等号成立的条件.
证明:设圆心为O,连结OB1, OB4, OA4,设C1的半径为R,C2的半径为2R.
在四边形B4A4OB1中,由Ptolemy定理,OA4 × B1B4 + OB1 × A4B4 ⩾OB4 × A4B1.
R × B1B4 + 2R × A4B4 ⩾2R × A4B1,即B1B4 ⩾2A4B1 −2A4B4.
同理B1B2 ⩾2A1B2 −2A1B1,B2B3 ⩾2A2B3 −2A2B2,B3B4 ⩾2A3B4 −2A3B3.
相加得B1B2 + B2B3 + B3B4 + B4B1 ⩾2(A1A2 + A2A3 + A3A4 + A4A1).
即四边形B1B2B3B4的周长⩾2(四边形A1A2A3A4的周长).
等号成立时OAiBiBi+1共圆,∠Ai+1AiO = ∠Bi+1BiO = ∠BiBi+1O = ∠Ai−1AiO, ∴Ai+1Ai = Ai−1Ai,(i = 1, 2, 3, 4, A5 = A1, A0 = A4, B5 = B1).
∴A1A2A3A4为菱形,又为圆内切四边形,所以A1A2A3A4为正方形.
3.在有限的实数列a1, a2, · · · , an中, 如果一段数ak, ak+1, · · · , ak+l−1的算术平均值大于1988, 那么我们 把这段数叫做一条“龙”,并把ak叫做这条龙的“龙头” (如果某一项an > 1988,那么单独这一项也叫 龙). 假设以上的数列中至少存在一条龙, 证明:这数列中全体可以作为龙头的项的算术平均数也必定大 7 于1988.
证明:引理:设ak, ak+1, . . . , ak+m−1均可作为龙头,ak+m不能作为龙头,或k + m −1 = n, 则ak, ak+1, . . . , ak+m−1的算术平均值大于1988.
引理的证明:对m用数学归纳法,m = 1时,设以ak为龙头的一条龙为ak, ak+1, . . . , ak+l−1.
若l = 1,ak > 1988,显然成立.
否则l > 1,由ak, ak+1, . . . , ak+l−1算术平均值大于1988,ak+1不是龙头, ak+1, . . . , ak+l−1算术平均值不 大于1988,ak > 1988,结论成立.
设小于m时结论均成立(m ⩾2),设以ak为龙头的一条龙为ak, ak+1, . . . , ak+l−1.
1 ⩽l ⩽m时,ak, ak+1, . . . , ak+l−1算术平均值大于1988, 由归纳假设ak+l, . . . , ak+m−1算术平均值大 于1988,结论成立.
l > m时,由ak+m不是龙头,ak+m, ak+m+1, . . . , ak+l−1算术平均值不大于1988, ak, ak+1, . . . , ak+l−1算术 平均值大于1988,结论显然也成立.
综上所述,由数学归纳法,引理成立.
设所有的龙头为ai1, ai1+1, . . . , ai1+j1−1, ai2, ai2+1, . . . , ai2+j2−1, . . . , aik, aik+1, . . . , aik+jk−1, 其中j1, j2, . . . , jk ⩾1 且im+1 > im + jm(m = 1, 2, . . . , k −1, k ⩾1).
由引理:aim, aim+1, . . . , aim+jm−1的算术平均值大于1988(m = 1, 2, . . . , k). 所以所有龙头的算术平均值 也大于1988.证毕.
4.(1)设三个正实数a, b, c满足 (a2 + b2 + c2)2 > 2(a4 + b4 + c4).
求证:a, b, c一定是某个三角形的三条边长.
(2)设n个正实数a1, a2, · · · , an满足 (a2 1 + a2 2 + · · · + a2 n)2 > (n −1)(a4 1 + a4 2 + · · · + a4 n) 其中n ⩾3. 求证:这些数中任何三个一定是某个三角形的三条边长.
证明:(1)若不然,不妨设c ⩾a + b,则 2(a4 + b4 + c4) −(a2 + b2 + c2)2 = a4 + b4 + c4 −2a2b2 −2b2c2 −2c2a2 = −(a + b + c)(a + b −c)(b + c −a)(c + a −b) ⩾0 矛盾.∴a, b, c为某个三角形三边长.
(2)n = 3即为(1)中的情况,n > 3时,若存在某三个不是某个三角形三条边长,不妨设为a1, a2, a3.则由均 值不等式 (n −1)(a4 1 + a4 2 + · · · + a4 n) < (a2 1 + a2 2 + · · · + a2 n)2 = µa2 1 + a2 2 + a2 3 2 + a2 1 + a2 2 + a2 3 2 + · · · + a2 n ¶2 ⩽ (n −1) "µa2 1 + a2 2 + a2 3 2 ¶2 + µa2 1 + a2 2 + a2 3 2 ¶2 + · · · + a4 n # 8 可得1 2(a2 + b2 + c2)2 > a4 + b4 + c4,(a2 + b2 + c2)2 > 2(a4 + b4 + c4).
但由(1),a1, a2, a3为某个三角形三边长,矛盾.所以这些数中任何三个一定是某个三角形的三条边长.
5.给出三个四面体AiBiCiDi(i = 1, 2, 3), 过点Bi, Ci, Di作平面αi, βi, γi(i = 1, 2, 3), 分别与棱AiBi, AiCi, AiDi垂直(i = 1, 2, 3), 如果九个平面αi, βi, γi(i = 1, 2, 3),相交于一点E, 而三点A1, A2, A3在同一直 线l上, 求三个四面体的外接球面的交集(形状怎样?位置如何?) 解:∵AiBi ⊥αi于Bi,而E在αi上,∴AiBi ⊥BiE, Bi在以AiE为直径的球上.同理Ci, Di也在以AiE为直 径的球上,AiBiCiDi的外接球即为在以AiE为直径的球.
若E在l上,显然这三个球的中心也都在l上,它们必在E处两两相切,交集为E.
否则E不在l上,三个球的球心在同一条直线上(△EA1A2中位线所在直线),且这三个球都过点E,交集为 一个圆,直径为EE′,其中E′为E到l的垂足.
6.如n是不小于3的自然数,以f(n)表示不是n的因子的最小自然数, 例如f(12) = 5.如果f(n) ⩾3,又可 作f(f(n)). 类似地,如果,f(f(n)) ⩾3,又可作f(f(f(n))),等等. 如果f(f(· · · f(n) · · · )) = 2, 共有k个f,就 把k叫做n的“长度”. 如果ln表示n的长度,试对任意自然数n(n ⩾3),求ln.并证明你的结论.
解:设n = 2k · m(m为奇数).
若k = 0,n为奇数,f(n) = 2, ln = 1.
若k > 0,考虑所有小于2k+1的正奇数,若它们均为n的因子,由2k+1 ∤n且小于2k+1的偶数t = 2p · q(p ⩽ k, q为奇数),由q|n, 2p|n, gcd(q, 2p) = 1,知t|n,∴f(n) = 2k+1,f(f(n)) = 3, f(f(f(n))) = 2, ln = 3.
否则取最小的t|n,t必为奇数,否则t必有一个奇因子不整除n.
∴f(n) = t, f(f(n)) = 2, ln = 2.
综上所述, ln = 1, n奇数 2, n = 2k · m(m为奇数)所有小于2k+1的正奇数不全整除n 3, n = 2k · m(m为奇数)所有小于2k+1的正奇数均整除n 9 第四届中国数学奥林匹克(1989年) 合肥中国科技大学 1.在半径为1的圆周上,任意给定两个点集A, B, 它们都由有限段互不相交的弧组成, 其中B的每 段的长度都等于π m, m是自然数. 用Aj表示将集合A逆时针方向在圆周上转动jπ m 弧度所得的集合 (j = 1, 2, ...).
求证:存在自然数k,使得L(Aj ∩B) ⩾ 1 2πL(A)L(B).
这里L(X)表示组成点集X的互不相交的弧的长度之和.
证明:我们把圆周上的点集E沿顺时针方向在圆周上转动jπ m 弧度所得的集合记为E−j,于是L(Aj ∩B) = L(A ∩B−j).
设b1, b2, . . . , bn为组成B的弧段,由已知它们两两不交且每段的长度均为π m,因此有 2m X j=1 L(Aj ∩B) = 2m X j=1 L(A ∩B−j) = 2m X j=1 L(A ∩(∪n i=1b−j i )) = 2m X j=1 n X i=1 L(A ∩b−j i ) = n X i=1 2m X j=1 L(A ∩b−j i ) = n X i=1 L(A ∩(∪2m j=1b−j i )) 因为L(bi) = π m,所以∪2m j=1b−j i 恰好是整个圆周,从而有L(A ∩(∪2m j=1b−j i )) = L(A).
∴P2m j=1 L(Aj ∩B) = nL(A),至少存在一个k, 1 ⩾k ⩾2m,使得 L(Aj ∩B) ⩾n 2mL(A) = 1 2π L(A)L(B).
2.设x1, x2, · · · , xn都是正数(n ⩾2).且x1 + x2 + · · · + xn = 1.求证: n X i=1 xi √1 −xi ⩾ 1 √n −1 n X i=1 √xi.
证明:不妨设x1 ⩾x2 ⩾· · · ⩾xn,则 1 √1 −x1 ⩾ 1 √1 −x2 ⩾· · · ⩾ 1 √1 −xn 由Chebyshev不等式 n X i=1 xi √1 −xi ⩾1 n à n X i=1 xi ! à n X i=1 1 √1 −xi !
= 1 n n X i=1 1 √1 −xi 10 由Cauchy不等式 Ã n X i=1 √ 1 −xi ! Ã n X i=1 1 √1 −xi !
⩾n2 又 n X i=1 √ 1 −xi ⩽ v u u tn n X i=1 (1 −xi) = p n(n −1) ∴ n X i=1 xi √1 −xi ⩾1 n n X i=1 1 √1 −xi ⩾ n n P i=1 √1 −xi ⩾ n p n(n −1) = r n n −1 而 1 √n −1 n X i=1 √xi ⩽ 1 √n −1 v u u tn n X i=1 xi = r n n −1 ∴ n X i=1 xi √1 −xi ⩾ 1 √n −1 n X i=1 √xi.
3.设S为复平面上的单位圆周(即模为1的复数的集合),f为从S到S的映射,对于任意z ∈S,定 义f (1)(z) = f(z), f (2)(z) = f(f(z)), · · · , f (k)(z) = f(f (k−1)(z)). 如果c ∈S,使得f (1)(c) ̸= c, f (2)(c) ̸= c, · · · , f (n−1)(c) ̸= c, f (n)(c) = c. 则称c 为f的n−周期点.设m是大于1的自然数, f定义为f(z) = zm, 试计算f的1989-周期点的个数.
解:记An = {z ∈S|z是f的n −周期点},Bn = {z ∈S|fn(z) = z}为fn的不动点集合,显然An ⊆ Bn,又f1(z) = zm,∴fn(z) = zmn ∴fn(z) = z ⇔zmn = z,又|z| = 1, ∴zmn−1 = 1, |Bn| = mn −1.
我们证明Bn, An有如下性质: (1)若k|n,则Bk ⊆Bn; 事实上,令n = kq,若c ∈Bk, fk(c) = c,则fn(c) = fkq(c) = fk(fk(· · · fk(c) · · · )) | {z } q个 = c.
∴c ∈Bn, Bk ⊆Bn.
(2)Bk ∩Bn = Bgcd(k,n), gcd(k, n)为k与n的最大公约数.
由(1),Bgcd(k,n) ⊆Bk, Bgcd(k,n) ⊆Bn, ∴Bgcd(k,n) ⊆Bk ∩Bn.
反之,设c ∈Bk ∩Bn,fk(c) = c, fn(c) = c,不妨设k < n. 则fn−k(c) = fn−k(fk(c)) = fn(c) = c,由辗转相 除法知fgcd(k,n)(c) = c, ∴c ∈Bgcd(k,n), Bk ∩Bn ⊆Bgcd(k,n). ∴Bk ∩Bn = Bgcd(k,n).
(3)c ∈Bn \ An ⇔∃k < n, k ∈N∗,使k|n且c ∈Bk.
充分性是显然的(由(1)),设c ∈Bn \ An, fn(c) = c.且存在l < n,使得fl(c) = c,设k = gcd(l, n),则fk(c) = c, c ∈Bk,且k ⩽l < n, k|n.证毕.
由1989 = 32 × 13 × 17,若k|1989,且k < 1989,k必整除3 × 13 × 17, 32 × 13, 32 × 17中至少一个.
∴Bk ⊆B663 ∪B153 ∪B117, ∴A1989 = B1989 \ ( S k|1989 k<1989 Bk) = B1989 \ (B663 ∪B153 ∪B117).
11 由容斥原理f的1989-周期点个数为 |A1989| = |B1989| −|B663| −|B153| −|B117| + |B663 ∩B153| + |B663 ∩B117| + |B117 ∩B153| −|B663 ∩B153 ∩B117| = |B1989| −|B663| −|B153| −|B117| + |B51| + |B39| + |B9| −|B3| = (m1989 −1) −(m663 −1) −(m153 −1) −(m117 −1) + (m51 −1) + (m39 −1) +(m9 −1) −(m3 −1) = m1989 −m663 −m153 −m117 + m51 + m39 + m9 −m3 4.设点D, E, F分别在△ABC的三边BC, CA, AB上, 且△AEF, △BFD, △CDE的内切圆有相等的半 径r, 又以r0和R分别表示△DEF和△ABC的内切圆半径.
求证:r + r0 = R.
证明:设△ABC周长为l,面积为S,内切圆为⊙I, 在各边的切点为P, Q, R,△DEF周长为l′,面积为S′.
△AEF, △BFD, △CDE的面积分别为S1, S2, S3,内切圆分别为⊙I1, ⊙I2, ⊙I3, 在各边的切点为Pi, Qi, Ri(i = 1, 2, 3).
由面积公式2S = Rl, 2S′ = r0l′, 2S1 = r(AE + EF + FA), 2S2 = r(BD + DF + FB), 2S3 = r(CD + DE + EC).
又S = S′ + S1 + S2 + S3, ∴Rl = r0l′ + r(l + l′),即(R −r)l = (r + r0)l′.
又 AQ1 AQ = AR1 AR = BQ2 BQ = BP2 BP = CP3 CP = CR3 CR = r R ∴l −Q1Q2 −P2P3 −R1R3 l = r R 又Q1Q2 + P2P3 + R1R3 = Q1F + FQ2 + P2D + DP3 + R3E + ER1 = P1F + R2F + DR2 + DQ3 + EQ3 + EP1 = l′.
∴l′ l = 1 −r R.∴(R −r)R = (r + r0)(R −r), R = r + r0.证毕.
5.空间中有1989个点,其中任何三点不共线, 把它们分成点数各不相同的30组, 在任何三个不同的组中 各取一点为顶点作三角形, 求三角形个数的最大值.
解:由分组情况有限,三角形个数必存在最大值,设分为30组,各组点数为x1 < x2 < · · · < x30, 三角形个 数为f(x1, x2, . . . , x30) = P 1⩽i<j<k⩽30 xixjxk.
12 若存在i ∈{1, 2, . . . , 29}, xi+1 −xi ⩾3, 则将(x1, x2, . . . , x30)调整为(x1, . . . , xi + 1, xi+1 −1, . . . , x30).
f(x1, . . . , xi + 1, xi+1 −1, . . . , x30) −f(x1, x2, . . . , x30) = [(xi + 1 + xi+1 −1) X 1⩽j<k⩽30 j,k̸=i,i+1 xjxk + (xi + 1)(xi+1 −1) X j̸=i,i+1 xj] −[(xi + xi+1) X 1⩽j 0 f值增大,类似的,若存在i, j ∈{1, 2, . . . , 29}, i < j, xi+1−xi ⩾2, xj+1−xj ⩾2, 将xi调整为xi+1,xj+1调 整为xj+1 −1,f值增大.
所以当f取最大值时,x1, x2, . . . , x30中相邻两个的差最多有一个是2,其余均为1.
如果所有的均为1,1989 = x1 + (x1 + 1) + · · · + (x1 + 29) = 30x1 + 435,x1不是整数,矛盾.
设xt+1 −xt = 2, 1 ⩽t ⩽29, 则1989 = x1 + x2 + · · · + x30 = 30x1 + (1 + 2 + · · · + t −1) + (t + 1 + · · · + 30) = 30x1 + 465 −t.
30x1 −t = 1524, x1 = 51, t = 6.此时各组的点的个数分别为51,52,. . . ,56,58,59,. . . ,81.
6.设f : (1, +∞) →(1, +∞)满足以下条件: 对于任意实数x, y > 1,及u, v > 0,有 f(xuyv) ⩽f(x) 1 4u f(y) 1 4v .
试确定所有这样的函数f.
解:令x = y, u = v = t 2(t > 0),则f(xt) ⩽(f(x)) 1 t .
以xt代x, 1 t 代t,则f(x) ⩽(f(xt))t.
∴f(xt) = (f(x)) 1 t .
设f(e) = c, c > 1,则f(x) = f(e) 1 ln x = c 1 ln x .
另外,当f(x) = c 1 ln x (c > 1)时,f(xuyv) = c 1 u ln x+v ln y , f(x) 1 4u f(y) 1 4v = c 1 4u ln x + 1 4v ln y .
由Cauchy不等式,(u ln x + v ln y)( 1 4u ln x + 1 4v ln y) ⩾1.∴ 1 u ln x+v ln y ⩽ 1 4u ln x + 1 4v ln y.
∴f(xuyv) ⩽f(x) 1 4u f(y) 1 4v .
所以所求函数为f(x) = c 1 ln x (c > 1).
13 第五届中国数学奥林匹克(1990年) 郑州《中学生数理化》编辑部 1.在凸四边形ABCD中,AB与CD不平行, ⊙O1过A,B且与边CD相切于P, ⊙O2过C,D且与边AB相切于Q, ⊙O1与⊙O2相交于E,F.
求证:EF平分线段PQ的充分必要条件是BC ∥AD.
证明:分两部分证明结论.
(1)EF平分PQ的充要条件为PC · PD = QA · QB.
设EF与PQ交于K,直线PQ于⊙O1, ⊙O2分别交于J, I.
∵PC · PD = PI · PQ, QA · QB = PQ · QJ, KQ · KI = KE · KF = KP · KJ.
∴KQ · (KP + IP) = KP · (KQ + QJ), KQ · IP = KP · QJ.
∴KP = KQ ⇔IP = QJ ⇔PC · PD = QA · QB.
(2)BC ∥AD充要条件为PC · PD = QA · QB.
设AB与DC交于S.BC ∥AD ⇔SD SC = SA SB .
而SP 2 = SA · SB, SQ2 = SC · SD.
∴PC · PD = QA · QB ⇔(SC −SP)(SP −SD) = (SB −SQ)(SQ −SA) ⇔(SC + SD)SP −SP 2 −SC · SD = (SB + SA)SQ −SQ2 −SA · SB ⇔(SC + SD)SP = (SB + SA)SQ ⇔(SC + SD)2 · SA · SB = (SA + SB)2 · SC · SD ⇔SC SD + SD SC + 2 = SA SB + SB SA + 2 又SD SC < 1, SA SB < 1, ∴PC · PD = QA · QB ⇔SD SC = SA SB ⇔BC ∥AD.
所以EF平分线段PQ的充分必要条件是BC ∥AD.
2.设x是一个自然数,若一串自然数x0 = 1 < x1 < x2 < · · · < xl = x 满足xi−1|xi(i = 1, 2, . . . , l), 则 称{x0, x1, . . . , xl}为x的一条因子链. l称为该因子链的长度. L(x)与R(x)分别表示x的最长因子链的长 度和最长因子链的条数.
对于x = 5k × 31m × 1990n,k, m, n都是自然数,试求L(x)与R(x).
解:对于x = pα1 1 pα2 2 · · · pαn n ,(p1, p2, . . . , pn为互不相同的质数,α1, α2, . . . , αn为正整数).
x的因子 链{x0, x1, . . . , xl}是最长因子链的充要条件是 xi xi−1 均为质数(i = 1, 2, . . . , l).
事实上,对于因子链{x0, x1, . . . , xl},若存在i, (1 ⩽i ⩽l),使得 xi xi−1 = q1q2,其中q1, q2均为大于1的正整 数, 则{x0, x1, . . . , xi−1, q1xi−1, xi, . . . , xl}是长度为l + 1的因子链, 所以{x0, x1, · · · , xl}不是最长因子 链.
反之,若 xi xi−1 均为质数(i = 1, 2, . . . , l), 则x = xl = xl xl−1 · · · · · x2 x1 · x1(x0 = 1)为l个质数的积.所 以l = α1+α2+· · ·+αn. 而对x的任意一个因子链{x0, x1, . . . , xt},x = xt = xt xt−1 ·· · ·· x2 x1 ·x1是t个大于1的 正整数之积,而x至多写成l = α1 + α2 + · · · + αn个大于1的正整数之积,所以t ⩽l.所以{x0, x1, · · · , xl}是 最长因子链.
14 L(x) = α1 + α2 + · · · + αn.
每个最长因子链对应一个排列x1, x2 x1 , . . . , xl xl−1 , l = L(x), 为α1个p1,α2个p2,. . . ,αn个pn的一个排列.
∴R(x) = (α1+α2+···+αn)!
α1!α2!···αn!
.
当x = 5k × 31m × 1990n = 2n × 5n+k × 31m × 1990n时, L(x) = 3n + k + m,R(x) = (3n+k+m)!
(n!)2(n+k)!m!.
3.设函数f(x)对x ⩾0有定义,且满足条件: (1)对任何x, y ⩾0, f(x)f(y) ⩽y2f( x 2) + x2f( y 2); (2)存在常数M > 0,当0 ⩽x ⩽1时,|f(x)| ⩽M.
求证:对任意x ⩾0,f(x) ⩽x2.
证明:令x = y,(f(x))2 ⩽2x2f( x 2).
令x = 0,(f(0))2 ⩽0,∴f(0) = 0,满足结论.
假设存在x > 0,使得f(x) > x2,用归纳法证明 f( x 2n ) > 22n−2n−1x2(n ∈N) n = 0时显然成立,设n = k时成立,f( x 2k ) > 22k−2k−1x2.
∴f( x 2k+1 ) ⩾(f( x 2k ))2 2( x 2k )2 > (22k−2k−1x2)2 2( x 2k )2 = 22k+1−2(k+1)−1x2 即n = k + 1时也成立,所以对任意n ∈N,f( x 2n ) > 22n−2n−1x2.
又n →+∞时,2n −2n −1 →+∞, 1 2n →0.
∴∃m1,当n ⩾m1时,0 < x 2n < 1,∃m2,当n ⩾m2时,22n−2n−1x2 > M.
取m = max{m1, m2},0 < x 2m < 1, f( x 2m ) > M,矛盾.所以对任意x ⩾0,f(x) ⩽x2.
4.设a是给定的正整数,A和B是两个实数,试确定方程组: x2 + y2 + z2 = (13a)2 (1) x2(Ax2 + By2) + y2(Ay2 + Bz2) + z2(Az2 + Bx2) = 1 4(2A + B)(13a)4 (2) 有整数解的充分必要条件(用A, B的关系式表示,并予以证明).
解:(2) −B 2 × (1)2,得(A −B 2 )(x4 + y4 + z4) = 1 2(A −B 2 )(13a)4.
若A = B 2 ,(1)与(2)等价,不难验证x = 3a, y = 4a, z = 12a为一组解.
若A ̸= B 2 ,则 2(x4 + y4 + z4) = (13a)4 (3) ∴2|a,设a = 2a1,x4 + y4 + z4 = 8(13a1)4.
若x, y, z不全为偶数,则必为两个奇数一个偶数,x4 + y4 + z4 ≡2 (mod 4),矛盾.
∴2|x, 2|y, 2|z.设x = 2x1, y = 2y1, z = 2z1,则若(x, y, z, a)为(3)的解,(x1, y1, z1, a1)也为(3)的解. 类似可 依次得到(x2, y2, z2, a2)也为(3)的解,等等.但这个过程不能一直进行下去,矛盾.
15 所以方程组有整数解的充分必要条件为A = B 2 .
5.设X是一个有限集合, 法则f使得X的每一个偶子集E(偶数个元素组成的子集)都对应一个实数f(E), 满足条件: (1)存在一个偶子集D,使得f(D) > 1990; (2)对于X的任意两个不相交的偶子集A, B,有f(A ∪B) = f(A) + f(B) −1990.
求证:存在X的子集P, Q,满足 (1)P ∩Q = ∅,P ∪Q = X; (2)对P的任何非空偶子集S,有f(S) > 1990; (3)对Q的任何偶子集T,有f(T) ⩽1990.
证明:考虑X的所有偶子集经法则f得到的实数最大的一个为P,若不止一个,取元素个数最少的一个.
Q = X \ P.则P ∩Q = ∅, P ∪Q = X.
令A = B = ∅,则f(∅) = 1990.
对于∀S ⊆P, S ̸= ∅,f(P) = f(S) + f(P \ S) −1990,显然f(P \ S) < f(P), ∴f(S) > 1990.
对于∀T ⊆Q,若T = ∅,f(T) = 1990,否则T ̸= ∅,由f(P ∪T) = f(P)+f(T)−1990 ⩽f(P),f(T) ⩽1990.
∴P, Q满足条件.证毕.
6.凸n边形及n −3条在n边形内不相交的对角线组成的图形称为一个剖分图.
求证:当且仅当3|n时,存在一个剖分图是可以一笔划的图(即可以从一个顶点出发,经过图中各线段恰一 次,最后回到出发点).
证明:因为n −3条在形内互不相交的对角线将凸n边形分为n −2个顶点均是n边形顶点的小区域, 每个 区域的内角和不小于π,n边形的内角和为(n −2)π,所以每个小区域都是三角形.
先证必要性.用归纳法容易证明可将每个三角形区域涂成黑白两色之一,使得有公共边的三角形不同 色. 假设已按照这样的要求染色,由于剖分图为可以一笔画的圈,所以由每个顶点引出的线段都是偶数 条. 从而每个顶点都是奇数个三角形的顶点,因此以原多边形外边界为一边的三角形区域有着相同的颜 色, 不妨设为黑色;另一方面,剖分图的每条对角线都是两种不同颜色三角形的公共边, 所以设黑三角形 有m1个,白三角形有m2个.则n = 3m1 −3m2,所以3|n.
再证充分性,设n = 3m,多边形为A1A2 . . . A3m.连接A1A3i, A3iA3i+2, A3i+2A1(i = 1, 2, . . . , m−1)这3m− 3条对角线, 形成m −1个三角形,可由A1出发,依次走过这些三角形,再走过凸多边形即可一笔画并回到 初始点.证毕.
16 第六届中国数学奥林匹克(1991年) 武汉华中师范大学 1.平面上有一凸四边形ABCD.
(1).如果平面上存在一点P,使得△ABP, △BCP, △CDP, △DAP面积都相等, 问四边形ABCD应满足什么条件?
(2).满足(1)的点P,平面上最多有几个?证明你的结论.
解:(1)(1.1)P在ABCD内部,若A, P, C,B, P, D分别三点共线, 显然ABCD为平行四边形,P为对角线的交点.
若A, P, C不共线,由于△PAB,△PAD等面积,AP必经过对角线BD的中点,同理CP过BD的中点,必 有P为BD的中点,所以△ABD, △BCD面积相等.即一条对角线平分ABCD的面积,显然也是充分条件.
(1.2)P在ABCD之外,不妨设P与B, C在AD异侧,P必与A, B在CD同侧,与C, D在AB同侧.
由△PAB, △PAD面积相等,PA ∥BD,同理PD ∥AC.设AC, BD相交于E,AEDP为平行四边形.
SAED = SAP D = SABP + SCDP + SP BC −SABCD = 3SAP D −SABCD.
∴SAED = 1 2SABCD.
这个条件也是充分条件,若SAED = 1 2SABCD,作平行四边形AEDP,显然PB, PC均在APDCB内.
∴SABP = SAP D = SCDP = SAED, SP BC = SAP D + SABCD −SABP −SCDP = SAED.P满足要求.
所以四边形ABCD有一条对角线平分面积,或者在对角线分成的四个三角形中有一个为四边形面积的 一半.
(2)由(1)知,P在形内,形外都至多有一个,又由充要条件不同时取到,P最多有一个.
2.设I = [0, 1],G = {(x, y)|x, y ∈I}.
求G到I的所有映射f,使得对任何x, y, z ∈I有 (1)f(f(x, y), z) = f(x, f(y, z)); (2)f(x, 1) = x, f(1, y) = y; (3)f(zx, zy) = zkf(x, y).这里,k是与x, y, z无关的正数.
解:由(3),f(x, y) = f(y · x y , y · 1) = ykf( x y , 1)(0 < x < y) f(x, y) = f(x · 1, x · y x) = ykf(1, y x)(0 < y < x) 再由(2),f(x, y) = yk−1x(0 < x < y), f(x, y) = xk−1y(0 < y < x) 又x = y时,f(x, x) = xkf(1, 1) = xk.
在(1)中,取0 < x < y < z < 1,x充分小时,yk−1x < z, x < zk−1y.
f(f(x, y), z) = f(yk−1x, z) = zk−1yk−1x,f(x, f(y, z)) = f(x, zk−1y) = x(zk−1y)k−1.
∴zk−1 = z(k−1)2, (k −1)(k −2) = 0, k = 1或2.
k = 1时,f(x, y) = min{x, y};k = 2时,f(x, y) = xy.(x > 0, y > 0) 又f(x, 0) = f(x · 1, x · 0) = xkf(0, 1) = 0, f(0, y) = 0, f(0, 0) = zkf(0, 0), f(0, 0) = 0.
∴k = 1时,f(x, y) = min{x, y};k = 2时,f(x, y) = xy.k ̸= 1, k ̸= 2时,无解.
17 3.地面上有10只小鸟在啄食,其中任5只小鸟中至少有4只在一个圆上, 问有鸟最多的圆上最少有几只鸟?
解:用10个点表示10只鸟,若其中任意四点均共圆,则十个点共圆.
否则设ABCD不共圆,过其中任意不共线的三个点可作一个圆,最多有四个Si(i = 1, 2, 3, 4). 从其余六 个点中任取一点P与ABCD构成5点组,其中必有4点共圆,必有P落在某个圆Si上, 有抽屉原则,另六个 点中必有两个点落在同一个圆上,这个圆上至少有5个点.
不妨设为A1, A2, A3, A4, A5在C1上,若存在P, Q不在C1上,考察{A1, A2, A3, P, Q}, 其中必有四点共 圆C2,显然C2 ̸= C1,A1, A2, A3不全在C2上,设A1, A2, P, Q ∈C2, A3, A4, A5 ̸∈C2.
考察{A3, A4, A5, P, Q},必有四点共圆C3,C3 ̸= C1, 设A3, A4, P, Q ∈C3, A1, A2, A5 ̸∈C3,C3 ̸= C2.
考察{A1, A2, A5, P, Q},必有四点共圆C4,C4 ̸= C1, P, Q ∈C4,A1, A3中至少有一个属于C4.C4 = C2或C3.但是这显然均构成矛盾.
所以至多有一个点不在C1上,又因为十个点中九点共圆而另一个不在这个圆上满足题意.所以有鸟最多 的圆上最少有九只鸟.
4.求方程x2n+1 −y2n+1 = xyz + 22n+1的所有满足条件n ⩾2, z ⩽5 × 22n 的正整数解组(x, y, z, n).
解:显然x > y,且x, y奇偶性相同,所以x −y ⩾2.
当x = 3, y = 1时,z = 32n −1 3(1 + 22n+1)为整数,又z ⩽5 × 22n, 32n ⩽5 × 22n + 1 3(1 + 22n+1) = (5 + 2 3)22n + 1 3 ⩽6 × 22n. ∴n ⩽2,n = 2,z = 70.
下面证不存在其他正整数解: (1)若y = 1, x > 4,y2n+1 + 22n+1 = 22n+1 + 1, x2n+1 −xyz = z(x2n −z) ⩾x(x2n −5 × 22n) ∴x2n+1 −xyz > 4(42n −5 × 22n) = 22n+2(22n −5) > 22n+1,矛盾.
(2)若y ⩾2,由x ⩾2, z ⩽5 × 22n, n ⩾2.
x2n+1 −xyz ⩾ x[(y + 2)2n −yz] > x[y2n + 4ny2n−1 + 4n(2n −1)y2n−2 + 22n −yz] ⩾ xy2n + x · 22n + x[4ny2n−1 + 4n(2n −1)y2n−2 −5 × 22ny] > y2n+1 + 22n+1 + xy[4ny2n−2 + 4n(2n −1)y2n−3 −5 × 22n] ∵y ⩾2,4ny2n−2 + 4n(2n −1)y2n−3 ⩾8 × 22n−2 + 8 × 3 × 22n−3 > 5 × 22n.
∴x2n+1 −xyz > y2n+1 + 22n+1,矛盾,所以只有一组正整数解(x, y, z, n) = (3, 1, 70, 2).
5.求所有自然数n,使得 min k∈N∗(k2 + h n k2 i ) = 1991.
这里[x]表示x的整数部分.
解:条件等价于对于∀k ∈N∗, k2 + n k2 ⩾1991, 且∃k0 ∈N∗, k2 0 + n k2 0 < 1992.
即对于∀k ∈N∗, k4 −1991k2 + n ⩾0 即(k2 −1991 2 )2 + n −19912 4 ⩾0 取k,使得|k2 −1991 2 |最小,k = 32,324 −1991 × 322 + n ⩾0, n ⩾1024 × 967 = 990208.
18 并且存在k0 ∈N∗, k4 0 −1991k2 0 + n < 0 即(k2 −996)2 + n −9962 < 0 由|k2 0 −996|的最小值为28,所以n −9962 < −282, n < 9962 −282 = 1024 × 968 = 991232.
∴990208 ⩽n ⩽991231.
6.MO牌足球由若干多边形皮块用三种不同颜色的丝线缝制而成, 它有以下特点: (1)任一多边形皮块的一条边恰与另一多边形皮块同样长的一条边用一种颜色的丝线缝合; (2)足球上每一个结点,恰好是三个多边形的顶点,每一结点的三条缝线颜色互不相同.
求证:可以在MO牌足球的每一结点上放置一个不等于1的复数, 使得每一多边形的所有顶点上放置的复 数的乘积都等于1.
证明:设这三种颜色为红,黄,蓝,对每条边赋值,红色为1,黄色为e 2 3 πi,蓝色为e 4 3 πi.
对于每个节点,若三种颜色的线依逆时针方向依次为红,黄,蓝,则在这个结点放上e 2 3 πi, 否则放上e 4 3 πi.则 结点上放置的数为逆时针方向一条边的复数除以下一条边的复数, (e 2 3 πi = 1 e 4 3 πi = e 4 3 πi e 2 3 πi = e 2 3 πi 1 , e 4 3 πi = 1 e 2 3 πi = e 2 3 πi e 4 3 πi = e 4 3 πi 1 ) 所以对于任意一个多边形,沿顺时针方向走过每条边依次为z1, z2, . . . , zk, 则顶点上依次放置ω1 = z1 z2 , ω2 = z2 z3 , . . . , ωk = zk z1 , ∴ω1ω2 · · · ωk = 1.满足题意,证毕.
19 第七届中国数学奥林匹克(1992年) 北京北京数学奥林匹克发展中心 1.设方程xn + an−1xn−1 + an−2xn−2 + · · · + a1x + a0 = 0的系数都是实数, 且适合条件0 < a0 ⩽ a1 ⩽a2 ⩽· · · ⩽an−1 ⩽1.已知λ为方程的复根且|λ| ⩾1.
求证:λn+1 = 1.
证明:由λn + an−1λn−1 + an−2λn−2 + · · · + a1λ + a0 = 0,a0 > 0, λ ̸= 0.
∴a0( 1 λ)n + a1( 1 λ)n−1 + · · · + an−1( 1 λ) + 1 = 0 a0( 1 λ)n+1 + a1( 1 λ)n + · · · + an−1( 1 λ)2 + 1 λ = 0 由|λ| > 1, | 1 λ| ⩾1.
∴1 = a0( 1 λ)n+1 + (a1 −a0)( 1 λ)n + · · · + (1 −an−1)( 1 λ) = |a0( 1 λ)n+1 + (a1 −a0)( 1 λ)n + · · · + (1 −an−1)( 1 λ)| ⩽ a0| 1 λ|n+1 + (a1 −a0)| 1 λ|n + · · · + (1 −an−1)| 1 λ| ⩽ a0 + (a1 −a0) + · · · + (1 −an−1) = 1 取等号条件为 a0 λn+1 , a1−a0 λn , . . . , 1−an−1 λ 的辐角相同, 且|λ| = 1.
但是 a0 λn+1 + a1−a0 λn + · · · + 1−an−1 λ = 1 所以它们均为非负实数,又a0 > 0, ∴λn+1 ∈R+.
又因为|λn+1| = 1, ∴λn+1 = 1.
2.设x1, x2, · · · , xn为非负实数,记xn+1 = x1, a = min{x1, x2, · · · , xn},试证: n X i=1 1 + xi 1 + xi+1 ⩽n + 1 (1 + a)2 n X i=1 (xi −a)2 其中等号成立当且仅当x1 = x2 = · · · = xn.
证明:对n用数学归纳法,n = 1时,x1 = a,结论显然成立.
设n = k时结论成立,当n = k + 1时,不妨设x1 = a,由归纳假设 k X i=1 1 + xi 1 + xi+1 + 1 + xk 1 + x1 ⩽k + 1 (1 + a)2 k X j=1 (xj −a)2 因此要证明 k+1 X i=1 1 + xi 1 + xi+1 ⩽k + 1 + 1 (1 + a)2 k+1 X i=1 (xi −a)2 只需证明 1 + xk 1 + xk+1 + 1 + xk+1 1 + x1 −1 + xk 1 + x1 ⩽1 + (xk+1 −a)2 (1 + a)2 即 (xk+1 −x1)(xk+1 −xk) (1 + xk+1)(1 + x1) ⩽(xk+1 −a)2 (1 + a)2 (1) 20 由xk+1 ⩾x1 = a, xk ⩾x1 = a,所以(1+xk+1)(1+x1) ⩾(1+a)2,(xk+1 −x1)(xk+1 −xk) ⩽(xk+1 −a)2.
(1)显然成立,等号成立时当且仅当x1 = x2 = · · · = xk,且xk+1 = xk = x1. 所以n = k + 1时结论成立.由 归纳法,结论成立.
3.在平面上给出一个9 × 9的方格表, 并在其中每一方格中都任意填入+1或-1. 下面一种改变填入数字 的方式称为一次变动: 对于任意一个小方格,将与此格有一条公共边的所有小方格(不包含此格本身)中 的数作连乘积,于是每取一个格, 就算出一个数.在所有小格都取遍后, 就将原来格中的数全部擦去,而 将这些算出的数填入相应的小方格中. 试问是否总可以经过有限次变动,使得所有小方格中的数都变 为+1?
解:未必,例如如下的4 × 4表格经变动后保持不变, 将它对称填入9 × 9格子的四个角的4 × 4方格, 并将 正中一行与正中一列均填上+1,该9 × 9方格表经变动后保持不变.
+1 -1 -1 -1 -1 +1 -1 +1 -1 -1 +1 +1 -1 +1 +1 +1 4.凸四边形内接于⊙O,对角线AC与BD相交于P,△ABP与△CDP的外接圆相交于P和另一点Q, 且O, P, Q三点两两不重合.试证∠OQP = 90◦.
证明:不妨设Q在∠BPC内,连结AO, AQ, DO, DQ.
则∠AQD = ∠AQP + ∠DQP = ∠ABP + ∠DCP = 2∠ABD = ∠AOD.
所以A, D, Q, O四点共圆.
∴∠OQP = ∠AQO + ∠PQA = ∠ADO + ∠ABD = 1 2(180◦−∠AOD) + 1 2∠AOD = 90◦,证毕.
5.在有8个顶点的简单图中,没有四边形的图的边数的最大值是多少? (简单图是指任意顶点与自己没有 边相连,而且任意两个顶点之间至多有一条边相连的图) 解:最大值为11,首先构造一个8顶点11条边的图,其中没有四边形.顶点为Ai(i = 1, 2, . . . , 8), 边 为A1A2, A2A3, A3A4, A4A5, A5A1, A5A6, A6A7, A7A8, A8A4, A1A6, A3A8.不难验证这个图满足题意.
下面证明:若简单图G中有8个顶点,12条边,其中必存在四边形.若不然,设G中度数最大的顶点(中的一 个)为A,A的度数为d. 显然8d ⩾2 × 12 = 24,即d ⩾3.
(1)若d ⩾5,设与A有边相连的顶点组成的集合为S,|S| = d. 则S中不会有顶点与其他两个同在S中的顶 点相连,顶点均在S中的边至多有[ d 2]条, 而其他的顶点每个至多与S中一个顶点相连,所以图G中最多有 边f(d) = d + [ d 2] + 7 −d + ¡7−d 2 ¢ 条, 不难验证d ⩾5时,f(d) < 12,矛盾.
(2)d = 4,(1)中的讨论仍适用,此时f(d) = 12,必有所有取等号的条件都取到. 设S = {B1, B2, B3, B4},另 三个点为C1, C2, C3.不妨设B1, B2相连,B3, B4相连,C1C2C3构成三角形. S1 = {B1, B2}, S2 = {B3, B4}, 由于C1, C2, C3每个向S中的一个点连出一条边,必有两个同向S1或S2中的点连出边, 不妨设为C1, C2都 向S1中的点连出边,C1与B1相连.
但是此时,若C2与B1相连,则B1C1C3C2为四边形; 若C2与B2相 连,则B1C1C2B2为四边形.矛盾.
21 (3)d = 3,则所有的顶点度数均为3,不妨设A,B之间没有边相连,从它们连出的边为AAi, BBi(i = 1, 2, 3).
则S1 = {A1, A2, A3}, S2 = {B1, B2, B3}至多有一个公共元素.
(3.1)若它们没有公共元素,A, A1, A2, A3, B, B1, B2, B3为全部8个点,由(1)中的讨论知,S1,S2中顶点相连 各自至多有一条边. 它们之间最多有3条边,最多有11条边,矛盾.
(3.2)若它们有公共元素,设A3 = B3,第8个点为C,从它出发只能各向S1,S2连出一条边, 而它又不 与A, B相连,所以C的度数小于3,矛盾.
综上所述,若简单图G中有8个顶点,12条边,其中必存在四边形.所以在有8个顶点的简单图中,没有四边 形的图的边数的最大值是11.
6.已知整数序列{a0, a1, a2, . . .}满足条件: (1)an+1 = 3an −3an−1 + an−2, n = 2, 3, . . .
(2)2a1 = a0 + a2 −2.
(3)对任意的自然数m,存在k,使得ak, ak+1, . . . , ak+m−1都为完全平方数.
试证:序列{a0, a1, a2, . . .}的所有项都是完全平方数.
证明:由(1)的特征方程x3 = 3x2 −3x + 1的三个根均为1知an = an2 + bn + c(a, b, c为待定实数.
代入(2)得a = 1,an = n2 + bn + c = (n + b 2)2 + c −b2 4 .
b = a1 −a0 −1, c = a0均为整数,令 (n + b −1 2 )2 < an = (n + b 2)2 + c −b2 4 < (n + b + 1 2 )2 (∗) 只需 n > (b −1)2 4 −c, n > c −(b + 1)2 4 .
令 n0 = [max{(b −1)2 4 −c, c −(b + 1)2 4 }] + 1 则当n ⩾n0时,(∗)成立.在(3)中令m = n0 + 1,知道必存在n ⩾n0使an为完全平方数,必有an = (n + b 2)2,b为偶数.
所以c −b2 4 = 0,∴∀n ∈N, an = (n + b 2)2为完全平方数,证毕.
22 第八届中国数学奥林匹克(1993年) 济南山东大学 1.设n是奇数,试证明存在2n个整数a1, a2, . . . , an; b1, b2, . . . , bn, 使得对于任意一个整数k, 0 < k < n.下 列3n个数ai +ai+k, ai +bi, bi +bi+k (i = 1, 2, · · · , n, 其中an+j = aj, bn+j = bj)被3n除时余数互不相同.
证明:令ai = 3i, bi = 3i + 1(i = 1, 2, . . . , n) 则ai + ai+k = 3(2i + k) ≡0 (mod 3) ai + bi = 6i + 1 ≡1 (mod 3) bi + bi+k = 3(2i + k) + 2 ≡2 (mod 3) 显然不同组的两数被3n除余数互不相同,只需说明同一组中任一两个数模3n互不相同, 即ci = 6i模3n互 不相同.(i = 1, 2, . . . , n) 若ci ≡cj (mod 3n),则6(j −i) ≡0 (mod 3),即n|2(j −i),但是n为奇数,∴n|j −i,又因为|j −i| < n,必 有j = i.
所以这3n个数被3n除时余数互不相同.
2.给定自然数k及实数a > 0, 已知k1 + k2 + · · · + kr = k, ki ∈N∗(i = 1, 2, . . . , r, 1 ⩽r ⩽k).求ak1 + ak2 + · · · + akr的最大值.
解:当a = 1时,ak1 + ak2 + · · · + akr = r,显然最大值为k.
当a > 0,且a ̸= 1时,f(x) = ax在[1, +∞)上为下凸函数, 即aki + akj ⩽a + aki+kj−1(ki, kj ∈N∗. 这是由 于它等价于a(aki−1 −1)(akj−1 −1) ⩾0,两括号内显然同号.
所以我们可以经过r −1次调整将其调整为k1 = k2 = · · · = kr−1 = 1, kr = k + 1 −r的情况,而值始终不 减. 设此时的值为F(r) = (r −1)a + ak+1−r.
又a + am ⩽am+1 ⇔m ⩾loga( a a−1).
若k + 1 −r ⩾loga( a a−1),则 F(r) = (r −2)a + a + ak+1−r ⩽(r −2)a + ak+2−r ⩽(r −3)a + ak+3−r ⩽· · · ⩽ak = F(1) 若k + 1 −r < loga( a a−1),则 F(r) = ra −a + ak+1−r ⩽ra + ak−r ⩽· · · ⩽ka = F(k) 所以F的最大值为max{ak, ka} = max{ak, ka}.
所以ak1 + ak2 + · · · + akr的最大值为 max{ak, ka} = ka, a ⩽k 1 k−1 (k ⩾2) 当r = k, k1 = k2 = · · · = kr = 1取等号 ak, k = 1或a > k 1 k−1 (k ⩾2) 当r = 1, k1 = k取等号 3.设圆K和K1同心,它们的半径分别为R和R1, R1 > R. 四边形ABCD内接于圆K,四边形A1B1C1D1内 接于圆K1, 点A1, B1, C1, D1分别在射线CD, DA, AB, BC上, 求证: SA1B1C1D1 SABCD ⩾R2 1 R2 .
23 证明:设AB = a, BC = b, CD = c, DA = d, AB1 = w, BC1 = x, CD1 = y, DA1 = z.则 SA1B1C1D1 SABCD = 1+SAB1C1 SABCD +SBC1D1 SABCD +SCD1A1 SABCD +SDA1B1 SABCD = 1+w(a + x) ad + bc +x(b + y) ab + cd +y(c + z) ad + bc +z(d + w) ab + cd 由切割线定理,(a + x)x = (b + y)y = (c + z)z = (d + w)w = R2 1 −R2.要证明 SA1B1C1D1 SABCD ⩾R2 1 R2 .
只需证明 (R2 1 −R2) µ w x(ad + bc) + x y(ab + cd) + y z(ad + bc) + z w(ab + cd) ¶ ⩾R2 1 −R2 R2 即 w x(ad + bc) + x y(ab + cd) + y z(ad + bc) + z w(ab + cd) ⩾1 R2 (∗) 由均值不等式知(∗)左边⩾ 4 √ (ad+bc)(ab+cd).
再由均值不等式 (ad + bc)(ab + cd) ⩽ µad + bc + ab + cd 2 ¶2 = 1 4[(a + c)(b + d)]2 ⩽ 1 4 "µa + c + b + d 2 ¶2#2 = 1 64(a + b + c + d)4 由圆内接四边形中正方形周长最长知a + b + c + d ⩽4 √ 2R.
∴(ab + cd)(bc + ad) ⩽16R4,显然有(∗)成立.证毕.
4.给定集合S = {z1, z2, · · · , z1993}, 其中z1, z2, · · · , z1993是非零复数(可看作平面上的非零向量). 求 证:可以把S中的元素分成若干组,使得 (1)S中每个元素属于且仅属于其中一组; (2)每一组中任一复数与该组所有复数之和的夹角不超过90◦; (3)将任意两组中复数分别求和,所得和数之间的夹角大于90◦.
证明:取S中某些元素组成子集A,使得这些元素之和的模长最大.
由于元素个数有限,所以子集的取法有限,必然存在这样的A.
在S \ A中同样取元素之和模长最大的一些元素组成B.
C = S \ A \ B,下面我们证明A, B, C满足题意.
若不然,设A, B, C中元素之和分别为a, b, c, (1)若存在z ∈A,与a夹角大于90◦,则−z与a夹角为锐角,则|(−z) + a| > |a|,与A的选取矛盾.∴A中的元 素与a的夹角均不超过90◦,类似的这对B也成立.
(2)对于z ̸∈A,若z与a的夹角不大于90◦,则|z + a| > |a|,与A的选取矛盾. 所以任意不在A中的元素或它 们的和与a的夹角大于90◦.类似的, 任意在C中的元素或它们的和与b的夹角大于90◦.
所以a, b, c两两夹角大于90◦.
(3)若存在z ∈C,z与c的夹角大于90◦,则a, b, c, z两两夹角大于90◦,矛盾. 所以C中的元素与c的夹角均不 超过90◦.
综上所述,A, B, C满足题意,证毕.
24 5.10人到书店买书,已知 (1)每人都买了三种书; (2)任何两人所买的书,都至少有一种相同.
问购买人数最多的一种书最(至)少有几人购买?说明理由.
解:至少有5个人.
设A买的书为1,2,3,其余9人每人至少买这三者之一,有抽屉原则至少有4人买了同一种书,再加上A,至少 有5个人.
另外,若10个人买的书分别为(123),(123),(145),(167),(246),(246),(257),(347),(356),(356). 则不难验证 满足条件,且购买人数最多的一种书恰有5人购买.
6.设函数f : (0, +∞) →(0, +∞)满足以下条件: 对于任意正实数x, y,有f(xy) ⩽f(x)f(y).
试证:对任意的正实数x及自然数n,有 f(xn) ⩽f(x)f(x2) 1 2 · · · f(xn) 1 n 证明:设Fn(x) = f(x)f(x2) 1 2 · · · f(xn) 1 n ,则Fn(x) = Fn−1(x)f(xn) 1 n , F n n (x) = F n n−1(x)f(xn).
类似的,F n−1 n−1 (x) = F n−1 n−2 (x)f(xn−1), · · · , F 2 2 (x) = F 2 1 (x)f(x2), F1(x) = f(x).
相乘得F n n (x) = Fn−1(x)Fn−2(x) · · · F2(x)F1(x)f(xn)f(xn−1) · · · f(x).
用归纳法证明Fn(x) ⩾f(xn).对于n = 1显然成立,设对n ⩽k均成立,由归纳假设 F k+1 k+1 (x) = Fk(x)Fk−1(x) · · · F1(x)f(xk+1)f(xk) · · · f(x) ⩾ f(xk)f(xk−1) · · · f(x)f(xk+1)f(xk) · · · f(x) = f(xk+1)(f(xk)f(x))(f(xk−1)f(x2)) · · · (f(x)f(xk)) ⩾ (f(xk+1))k+1 ∴Fk+1(x) ⩾f(xk+1),由数学归纳法知结论对于∀n ∈N∗均成立.
∴Fn(x) ⩾f(xn),即f(xn) ⩽f(x)f(x2) 1 2 · · · f(xn) 1 n .证毕.
25 第九届中国数学奥林匹克(1994年) 上海复旦大学 1.设ABCD是一个梯形(AB ∥CD),E是线段AB上一点,F是线段CD上一点, 线段CE与BF相交于 点H,线段ED与AF相交于点G, 求证:SEHF G ⩽1 4SABCD.
如果ABCD是一个任意的凸四边形, 同样结论是否成立?请说明理由.
证明:引理:在梯形ABCD中,AC, BD交于E.(AB ∥CD)则S△AED = S△BEC ⩽1 4SABCD.
引理的证明:显然S△ACD = S△BCD,都减去S△CDE,即有S△AED = S△BEC,设为S.则 S S△ABE = DE BE = S△CDE S ∴S△ABES△CDE = S2,由均值不等式 SABCD = S△ABE + S△CDE + 2S ⩾2 p S△ABE · S△CDE + 2S = 4S 所以S△AED = S△BEC ⩽1 4SABCD.
回到原题,由引理,S△EGF ⩽1 4SAEDF , S△EHF ⩽1 4SBECF .
相加得SEHF G ⩽1 4SABCD.
如果ABCD是一个任意的凸四边形,结论未必成立.
当DA →0, E →B, F →C时,SEF GH →SABCD.
所以当AD BC , BE AB , CF CD足够小时,SEF GH > 1 4SABCD.
2.n(n ⩾4)个盘子里放有总数不少于4的糖块, 从任意的两个盘子各取一块糖,放入另一个盘子中, 称为 一次操作,问能否经过有限次操作, 将所有的糖块集中到一个盘子中去?证明你的结论.
解:能够做到.用数学归纳法证明,设有m块糖,(m ⩾4).
当m = 4时,至多有4个盘子中有糖,只有下面几种情况,不难看出结论均成立.
(1)(1, 1, 1, 1) →(3, 1, 0, 0) →(2, 0, 2, 0) →(1, 0, 1, 2) →(0, 0, 0, 4) (2)(2, 1, 1, 0) →(4, 0, 0, 0) (3)(3, 1, 0, 0) →(2, 0, 2, 0) →(1, 0, 1, 2) →(0, 0, 0, 4) (4)(2, 2, 0, 0) →(1, 1, 2, 0) →(0, 0, 4, 0) (5)(4, 0, 0, 0) 设当小于m时结论均成立(m > 4),当有m块糖时,可以先将m −1块糖集中到一个盘子内, 由归纳假设这 是可以做到的.剩下的一块若也在同一个盘子内,显然结论成立.否则可由如下的过程将m块糖集中到一 个盘子内.
(m −1, 1, 0, 0) →(m −2, 0, 2, 0) →(m −3, 2, 1, 0) →(m −4, 1, 1, 2) →(m −2, 0, 1, 1) →(m, 0, 0, 0) 由归纳法,总可经有限次操作将所有的糖集中到同一个盘子中.
3.求适合以下条件的所有函数f : [1, +∞) →[1, +∞), (1)f(x) ⩽2(x + 1); 26 (2)f(x + 1) = (f(x))2−1 x .
解:显然f(x) = x + 1是一个符合条件的函数,若存在f满足条件,且∃x0 ∈[1, +∞),使得f(x0) ̸= x0 + 1.
(1)f(x0) > x0 + 1,设g(x) = f(x) −x −1,g(x0) > 0,又由于f(x) ⩽2(x + 1),得到g(x) ⩽x + 1.
而且由条件(2)不难得到g(x + 1) = 2g(x) + 1 x[(g(x))2 + 2g(x)].
∴g(x0 + 1) > 2g(x0) > 0,由数学归纳法不难证明g(x0 + n) > 2ng(x0),当n充分大时,g(x0 + n) > 2ng(x0) > x0 + n + 1,矛盾.
(2)f(x0) < x0 + 1,设g(x) = x + 1 −f(x),g(x0) > 0,由f(x) ⩾1,得到g(x) ⩽x.
而且由条件(2)不难得到g(x + 1) = 2g(x) −1 x[(g(x))2 −2g(x)].
∴g(x0 + 1) ⩾2g(x0) −(g(x0) −2) = g(x0) + 2,由数学归纳法不难证明f(x0 + n) ⩾g(x0) + 2n,当n充 分大时,g(x0 + n) ⩾g(x0) + 2n > x0 + n,矛盾.
综上所述,对于任意的x ∈1, +∞),f(x) = x + 1,所以f(x) = x + 1是唯一满足条件的函数.
4.已知f(z) = C0zn + C1zn−1 + C2zn−2 + · · · + Cn−1z + Cn是一个n次复系数多项式, 求证:一定存在 一个复数z0, |z0| ⩽1,满足|f(z0)| ⩾|C0| + |Cn|.
证明:取ω,|ω| = 1,使得C0ωn与Cn辐角相同,(Cn = 0时,ω = 1即可). ε = e 2πi n 为n次单位根.
令zi = ωεi−1(i = 1, 2, . . . , n),则 f(z1) + f(z2) + · · · + f(zn) = n(C0ωn + Cn), |z1| = |z2| = · · · = |zn| = 1 ∴ n X i=1 |f(zi)| ⩾|f(z1) + f(z2) + · · · + f(zn)| = n|C0ωn + Cn| = n(|C0| + |Cn|) 所以必然存在i,使得f(zi) ⩾|C0| + |Cn|, |zi| = 1,显然结论成立.
5.对任何自然数n,求证恒等式: n X k=0 µn k ¶ 2k µn −k [ n−k 2 ] ¶ = µ2n + 1 n ¶ 其中 ¡0 0 ¢ = 1, [ n−k 2 ]表示n−k 2 的整数部分.
证明:考虑函数f(x) = (1 + x)2n+1,显然 ¡2n+1 n ¢ 为它的n次项系数.
另一方面 f(x) = (x2 + 2x + 1)n(x + 1) = n X k=1 (x + 1)(x2 + 1)n−k(2x)k µn k ¶ 考虑每一项中xn的系数,若n −k为偶数,(x + 1)(x2 + 1)n−k中xn−k的系数为 µn −k n−k 2 ¶ ; 若n −k为奇 数,(x + 1)(x2 + 1)n−k中xn−k的系数为 µ n −k n−k−1 2 ¶ .
所以每一项中xn的系数均为 µn k ¶ 2k µn −k [ n−k 2 ] ¶ .
f(x)中xn的系数为 n X k=0 µn k ¶ 2k µn −k [ n−k 2 ] ¶ ∴ n X k=0 µn k ¶ 2k µn −k [ n−k 2 ] ¶ = µ2n + 1 n ¶ 27 6.设M为平面上坐标为(1994p, 7 × 1994p)的点, 其中p是素数,求满足下述条件的直角三角形的个数: (1)三角形的三个顶点都是整点,而且M是其直角顶点; (2)三角形的内心是坐标原点.
解:设该直角三角形为MAB,并且MA斜率为正.
将坐标原点平移至M,设MA, MO的倾斜角分别为α, β,则tan β = 7.
所以MA的斜率为k = tan α = tan(β −π 4 ) = tan β−1 1+tan β = 3 4.
MB的斜率为−4 3,设A(−4t, −3t), B(3t′, −4t′),由A, B为整点,t, t′为正整数,MA = 5t, MB = 5t′.
由内心性质,并且∠M是直角,MA + MB −AB = 2r, √ 2r = MO = p (1994p)2 + (7 × 1994p)2, MA > r, MB > r.(r为内切圆半径) ∴5t + 5t′ −5 √ t2 + t′2 = 1 × 5 × 1994p, t2 + t′2 = (t + t′ −2 × 1994p)2, tt′ −2 × 1994p(t + t′) + 2 × 19942p2 = 0(t > 1994p, t′ > 1994p) (t −2 × 1994p)(t′ −2 × 1994p) = 2 × (1994p)2 设m = t −2 × 1994p, n = t′ −2 × 1994p.mn = 23 × 9972 × p2.
不难知道m, n均为正整数,所以(m, n)一组正整数解对应一个直角三角形.
∴直角三角形的个数为23 × 9972 × p2的正因子个数.
p ̸= 2, p ̸= 997时为(3 + 1)(2 + 1)(2 + 1) = 36; p = 2时为(5 + 1)(2 + 1) = 18; p = 997时为(3 + 1)(4 + 1) = 20.
28 第十届中国数学奥林匹克(1995年) 合肥中国科技大学 1.设2n个实数a1, a2, . . . , an; b1, b2, . . . , bn(n ⩾3)满足 (1)a1 + a2 + · · · + an = b1 + b2 + · · · + bn; (2)0 < a1 = a2, ai + ai+1 = ai+2(i = 1, 2, . . . , n −2); (3)0 < b1 ⩽b2, bi + bi+1 ⩽bi+2(i = 1, 2, . . . , n −2).
求证:an−1 + an ⩽bn−1 + bn.
证明:若a1 ⩽b1,则由递推关系不难证明ai ⩽bi(i = 1, 2, . . . , n),显然结论成立.
若存在2 ⩽i ⩽n,使得ai ⩽bi, ai+1 ⩽bi+1.
i = n −1时,显然有结论成立;i < n −1时,由递推关系不难证明aj ⩽bj(j ⩾i),所以也有结论成立.
否则必有a1 > b1,设I = {i|ai ⩽bi},则I中不存在相邻的正整数.
设I′ = {j|j = i −1, i ∈I}, I ∩I′ = ∅.
若i ∈I,则ai ⩽bi, ai−1 > bi−1, ai+1 > bi+1(i ⩽n −2).
∴ai + ai−1 = ai+1 > bi+1 ⩾bi + bi−1. ∴ P i∈I∪I′ ai > P i∈I∪I′ bi(i ⩽n −2).
又若i ∈I ∪I′,ai > bi.
∴ n−2 P i=1 ai > n−2 P i=1 bi, 又 n P i=1 ai = n P i=1 bi, ∴an−1 + an ⩽bn−1 + bn.
综上所述,an−1 + an ⩽bn−1 + bn.
2.设N为自然数集合, f : N →N适合条件:f(1) = 1,对于任何自然数n都有 ◦3f(n)f(2n + 1) = f(2n)(1 + 3f(n)); ◦f(2n) < 6f(n).
试求方程f(k) + f(l) = 293,其中k < l的所有解.
解:由gcd(3f(n), 1 + 3f(n)) = 1, ∴3f(n)|f(2n).
又因为f(2n) < 6f(n),所以f(2n) = 3f(n), f(2n + 1) = 3f(n) + 1 = f(2n) + 1.
由f(1) = 1,由归纳法不难证明:若n的二进制表示为 (amam−1 . . . a0)2 = am2m + am−12m−1 + · · · + a0(ai = 0或1) 则f(n)的三进制表示为 (amam−1 . . . a0)3 = am3m + am−13m−1 + · · · + a0(ai = 0或1) 由k < l,显然f(k) < f(l), 设k = (amam−1 . . . a0)2, l = (bmbm−1 . . . b0)2, ci = ai + bi, ci = 0, 1, 2(i = 0, 1, . . . , m).
则f(k) + f(l) = (cmcm−1 . . . c0)3.
而293 = 35 + 33 + 2 × 32 + 3 + 2 = (101212)3.所以由f(k) < f(l), am ⩽bm, m = 5.
必有a5 = 0, b5 = 1, a4 = b4 = 0, a2 = b2 = a0 = b0 = 1.a3 + b3 = 1, a1 + b1 = 1.
不难知道只有四组解: 29 (1)a3 = a1 = 1,f(k) = (1111)3, f(l) = (100101)3, (k, l) = (15, 37); (2)a3 = 0, a1 = 1,f(k) = (111)3, f(l) = (101101)3, (k, l) = (7, 45); (3)a3 = 1, a1 = 0,f(k) = (1101)3, f(l) = (100111)3, (k, l) = (13, 39); (4)a3 = a1 = 0,f(k) = (101)3, f(l) = (101111)3, (k, l) = (5, 47).
3.试求 10 X i=1 10 X j=1 10 X k=1 |k(x + y −10i)(3x −6y −36j)(19x + 95y −95k)| 的最小值,其中x和y是任意实数.
解:设 F = 10 X i=1 10 X j=1 10 X k=1 |k(x + y −10i)(3x −6y −36j)(19x + 95y −95k)| F1 = 10 X i=1 |x + y −10i|, F2 = 10 X j=1 |x −2y −12j|, F3 = 10 X k=1 k|x + 5y −5k| 则F = 57F1F2F3.
引理:a1, a2, . . . , an为n个实数,a1 ⩽a2 ⩽· · · ⩽ak.定义b为它们的中位数: 若2|n, n = 2m, am ⩽b ⩽am+1; 若2 ∤n, n = 2m + 1, b = am+1.
设g(x) = |x −a1| + |x −a2| + · · · + |x −an|,则g(t)的最小值为g(b).
引理的证明:∵|x −a1| + |x −an| ⩾an −a1(当a1 ⩽x ⩽an时取等号) |x −a2| + |x −an−1| ⩾an−1 −a2(当a2 ⩽x ⩽an−1时取等号)· · · n为偶数时,|x −a n 2 | + |x −a n 2 +1| ⩾a n 2 +1 −a n 2 (当a n 2 ⩽x ⩽a n 2 +1时取等号) n为奇数时,|x −a n+1 2 | ⩾0(当x = a n+1 2 时取等号) ∴g(x) ⩾(an + an−1 + · · · + a[ n 2 ]+1) −(a[ n+1 2 ] + · · · + a1) (当x = b时取等号) 回到原题,由引理分别应用到F1, F2, F3上得 F1 ⩾10 × (10 + 9 + 8 + 7 + 6 −5 −4 −3 −2 −1) = 250, 50 ⩾x + y ⩾60时取等号.
F2 ⩾12 × (10 + 9 + 8 + 7 + 6 −5 −4 −3 −2 −1) = 300, 60 ⩾x −2y ⩾72时取等号.
F3 ⩾5 × (10 × 10 + 9 × 9 + 8 × 8 −7 × 6 −6 × 6 −5 × 5 −4 × 4 −3 × 3 −2 × 2 −1) = 560, x + 5y = 35时 取等号.
∴F ⩾57 × 250 × 300 × 560 = 2394000000.
且不难验证x = 60, y = −5时满足所有取等号条件,所以原式的最小值为2394000000.
4.空间有四个球,它们的半径分别为2,2,3,3,每个球都与其余3个球外切, 另有一个小球与这四个球都外 切,求该小球的半径.
解:设四个球的球心分别为A, B, C, D,则AB = 4, CD = 6, AC = BC = AD = BD = 5.
设E, F分别为AB,CD中点,小球球心为O,半径为r,则四面体ABCD关于平面ABF,CDE对称.
四个球也同样,所以由对称性O在EF上.
OE = √ OA2 −AE2 = p (2 + r)2 −22 = √ r2 + 4r, 30 OF = √ OD2 −DF 2 = p (3 + r)2 −32 = √ r2 + 6r, EF = √ FA2 −AE2 = √ DA2 −DF 2 −AE2 = √ 52 −32 −22 = 2 √ 3.
∴ √ r2 + 4r + √ r2 + 6r = 2 √ 3.
解得r = 6 11.
5.设a1, a2, . . . , a10是任意10个两两不同的自然数, 它们的和为1995.试求a1a2 + a2a3 + · · · + a9a10 + a10a1的最小值.
解:将a1, a2, . . . , a10按顺时针方向依次写在一个圆周上,于是所求表达式即为每相邻两数乘积的总和A.
并且将a1, a2, . . . , a10的和记为N,N ⩾55.将和为N的任意10个不同自然数所对应的表达式的最小值记 为S(N).
先考虑这10个数为1, 2, . . . , 10的情况,即N = 55时.
不妨设a1 = 10,我们通过调整证明a1, a2, . . . , a10依次为10, 1, 9, 3, 7, 5, 6, 4, 8, 2时取到最大值.
若aj = 1, j ̸= 2,将(a2, . . . , aj−1, aj)这一段整个的按逆过来的顺序排列,即变为(aj, aj−1, . . . , a2), 设操 作前所求表达式为A,操作后为A′, 则A′ −A = (10 + a2aj+1) −10a2 −aj+1 = (a2 −1)(aj+1 −10) ⩽0,A的值下降了.
同理若a10 ̸= 2,通过类似的调整使a10 = 2,且A的值减少; 类似的经过有限次操作即得到(a1, a2, . . . , a10) = (10, 1, 9, 3, 7, 5, 6, 4, 8, 2),且每次操作A的值都下降.
∴S(55) = 10 + 9 + 27 + 21 + 35 + 30 + 24 + 32 + 16 + 20 = 224.
对于N > 55,显然最大的数大于10,第二大的数不小于9,. . . ,最小的数不小于1.
不妨设a1最大,经类似的讨论可知道若将a1, a2, . . . , a10从大到小依次以10, 9, . . . , 1代替,必有按照上面 方式排列时取最小值.
设(b1, b2, . . . , b10) = (10, 1, 9, 3, 7, 5, 6, 4, 8, 2),ci = ai −bi ⩾0, c11 = c1, b11 = b1, b0 = b10.
A = a1a2 + a2a3 + · · · + a9a10 + a10a1 = 10 X i=1 bibi+1 + 10 X i=1 cici+1 + 10 X i=1 ci(bi−1 + bi+1) ⩾ 224 + (b2 + b10)(N −55) = 3N + 59 而当(a1, a2, . . . , a10) = (N −45, 1, 9, 3, 7, 5, 6, 4, 8, 2)时,A = 3N + 59.∴S(N) = 3N + 59.
所以N = 1995时,a1a2 + a2a3 + · · · + a9a10 + a10a1的最小值为3 × 1995 + 59 = 6044.
6.设n是大于1的奇数,已给y0 = (x(0) 1 , x(0) 2 , . . . , x(0) n−1, x(0) n ) = (1, 0, . . . , 0, 1).
设 x(k) i = ½ 0 x(k−1) i = x(k−1) i+1 时, i = 1, 2, . . . , n 1 x(k−1) i ̸= x(k−1) i+1 时, i = 1, 2, . . . , n 其中x(k−1) n+1 = x(k−1) 1 .
记yk = (x(k) 1 , x(k) 2 , . . . , x(k) n−1, x(k) n ), k = 1, 2, . . .
若正整数m满足y0 = ym.求证:m是n的倍数.
证明:将一个圆周n等分,将x(k) 1 , x(k) 2 , . . . , x(k) n−1, x(k) n 依次按顺时针方向写在这些分点上表示yk.
31 对于y0,显然它有唯一的对称轴,不妨设为竖直线.
因为x(k) i ≡x(k−1) i + x(k−1) i+1 (mod 2).
所以将yk中每两个相邻点上的数的和除以2的余数放在这段弧的中点上,再将原先的数撤去,显然对称轴 是不变的. 而再将它逆时针旋转π n时,即得到yk+1.所以yk+1的对称轴是yk的对称轴逆时针旋转π n.
若y0 = ym,它们的对称轴也相同,而中间变换了m次,对称轴旋转了mπ n ,而它重合于竖直线, 所以它旋转 了π的整数倍,记为kπ.∴mπ n = kπ, m = kn,即m是n的倍数.
32 第十一届中国数学奥林匹克(1996年) 天津南开大学 1. 设H是锐角△ABC的垂心, 由A向以BC为直径的圆作切线AP,AQ,切点分别为P,Q.
求证:P, H, Q三点共线.
证明:设三条高的垂足分别为D, E, F,BC中点为O,PQ与AO交于R,则AO ⊥PR.
由∠APO = ∠ARP = 90◦,△APO ∽∠ARP, AO · AR = AP 2.
又∠ADC = ∠BEC = 90◦,∴H, D, C, E四点共圆.∴AE · AC = AH · AD.
由切割线定理AE · AC = AP 2.
∴AO · AR = AH · AD.
若D与O重合,则H与R重合,P, H, Q显然共线.
否则,O, D, H, R四点共圆,∠ORH = ∠ODH = 90◦.
∴AO ⊥RH,P, H, Q共线,证毕.
2.设S = {1, 2, . . . , 50},求最小自然数k,使S的任一k元子集中, 都存在两个不同的数a和b,满足a + b整 除ab.
解:设a, b ∈S,满足a+b整除ab.设c = gcd(a, b),于是a = ca1, b = cb1,其中a1, b1 ∈N,且gcd(a1, b1) = 1.因 此 c(a1 + b1) = a + b|ab = c2a1b1, a1 + b1|ca1b1 ∵gcd(a1 + b1, a1) = gcd(a1 + b1, b1) = 1,所以a1 + b1|c.
因为a, b ∈S,a + b ⩽99,c(a1 + b1) ⩽99.所以3 ⩽a1 + b1 ⩽9.
易知S中所有满足a + b整除ab的不同数对共有23对如下: a1 + b1 = 3 : (6, 3)(12, 6)(18, 9)(24, 12)(30, 15)(36, 18)(42, 21)(48, 24) a1 + b1 = 4 : (12, 4)(24, 8)(36, 12)(48, 16) a1 + b1 = 5 : (20, 5)(40, 10)(15, 10)(30, 20)(45, 30) a1 + b1 = 6 : (30, 6) a1 + b1 = 7 : (42, 7)(35, 14)(28, 21) a1 + b1 = 8 : (40, 24) a1 + b1 = 9 : (45, 36) 令M = {6, 12, 15, 18, 20, 21, 24, 35, 40, 42, 45, 48}, |M| = 12.并且上述23个数对中每一对都至少包 含M中1个元素. 因此,若令T = S \ M,则|T| = 38,且T中任何两数都不满足题中条件,所以k ⩾39.
而下列12个满足题中条件的数对互不相交: (6,3)(12,4)(20,5)(42,7)(24,8)(18,9)(40,10)(35,14)(30,15)(48,16)(28,21)(45,36) 对于S的任意一个39元子集R,只比S少11个元素,而这11个元素至多属于上述12个数对中的11对,从而必 有一对属于R.
33 综上可知,所求的最小自然数k = 39.
3.设R为实数集合, 函数f : R →R适合条件 f(x3 + y3) = (x + y)((f(x))2 −f(x)f(y) + (f(y))2), x, y为实数.
试证:对一切实数x,都有f(1996x) = 1996f(x).
证明:令x = y = 0,有f(0) = 0.令y = 0, f(x3) = x(f(x))2.
所以f(x) = x 1 3 (f(x 1 3 ))2, x ∈R.
由此可知x ⩾0时,f(x) ⩾0;x ⩽0时,f(x) ⩽0.
设S = {a > 0|对于∀x ∈R, f(ax) = af(x)}.
显然1 ∈S,若a ∈S,由f(x) = x 1 3 (f(x 1 3 ))2, ax(f(x))2 = af(x3) = f(ax3) = f(a 1 3 x)3) = a 1 3 x(f(a 1 3 x))2.
所以(f(a 1 3 x))2 = (a 1 3 f(x))2,f(a 1 3 x) = a 1 3 f(x).
即a 1 3 ∈S.
若a, b ∈S,则a 1 3 , b 1 3 ∈S f((a + b)x) = f((a 1 3 x 1 3 )3 + (b 1 3 x 1 3 )3) = (a 1 3 + b 1 3 )x 1 3 [(f(a 1 3 x 1 3 ))2 −f(a 1 3 x 1 3 )f(b 1 3 x 1 3 ) + (f(b 1 3 x 1 3 ))2] = (a 1 3 + b 1 3 )x 1 3 (a 2 3 −a 1 3 b 1 3 + b 2 3 )(f(x 1 3 ))2 = (a + b)x 1 3 (f(x 1 3 )2) = (a + b)f(x) ∴a + b ∈S.
由1 ∈S, 1 + 1 = 2 ∈S,由归纳法易知所有自然数n ∈S.
∴1996 ∈S,即f(1996x) = 1996f(x).
4.8位歌手参加艺术会,准备为他们安排m次演出, 每次由其中4位登台表演.要求8位歌手中任意两位同 时演出的次数都一样多, 请设计一种方案,使得演出的次数m最少.
解:设任两位同时演出r次,则r ¡8 2 ¢ = m ¡4 2 ¢ ,即14r = 3m.
∴3|r, r ⩾3, m ⩾14.
用1,2,. . . ,8代表8位歌手,如下14次演出满足要求: (1,2,3,4);(1,2,5,6);(1,2,7,8);(1,3,5,7);(1,3,6,8);(1,4,5,8);(1,4,6,7); (2,3,5,8);(2,3,6,7);(2,4,5,7);(2,4,6,8);(3,4,5,6);(3,4,7,8);(5,6,7,8).
∴m = 14.
5.设n为自然数,x0 = 0, xi > 0, i = 1, 2, . . . , n且 n P i=1 xi = 1.求证: 1 ⩽ n X i=1 xi √1 + x0 + x1 + · · · + xi−1 √xi + · · · + xn < π 2 .
证明:设x0 + x1 + · · · + xi−1 = cos θi,则xi + xi+1 + · · · + xn = 1 −cos θi(i = 1, 2, . . . , n + 1).
∴xi = cos θi+1 −cos θi, θi ∈[0, π 2 ].
34 θ1 = π 2 > θ2 > · · · > θn > θn+1 = 0.
往证: 1 ⩽ n X i=1 cos θi+1 −cos θi √1 + cos θi √1 −cos θi < π 2 即 1 ⩽ n X i=1 cos θi+1 −cos θi sin θi < π 2 而由sin θi ∈(0, 1 ∴ n X i=1 cos θi+1 −cos θi sin θi ⩾ n X i=1 (cos θi+1 −cos θi) = cos θn+1 −cos θ1 = 1 又由于θi+1+θi 2 < θi,sin x < x(0 < x < π 2 ).
∴ n X i=1 cos θi+1 −cos θi sin θi = n X i=1 2 sin θi+1+θi 2 sin θi−θi+1 2 sin θi ⩽ n X i=1 2 sin θi sin θi−θi+1 2 sin θi = 2 n X i=1 sin θi −θi+1 2 < 2 n X i=1 θi −θi+1 2 = θ1 −θn+1 = π 2 所以原不等式成立.
6.在△ABC中,∠C = 90◦, ∠A = 30◦, BC = 1,求△ABC的内接三角形(三顶点分别在△ABC三边上的 三角形)的最长边的最小值.
解:令△DEF为内接三角形,D, E, F分别在BC, CA, AB上.
对于BC上的任意一点D,令∠EDF = 60◦保持不变,设G, H分别在AC, AB上,∠ADG = ∠ADH = 60◦.显然当E从G运动到A时,F从A运动到H.
因为∠DGA > ∠C = 90◦,所以DG > DA; 又∠DHA = ∠B + ∠BDH = ∠ADB > ∠C = 90◦,所以DA > DH.
所以必然存在某个E,使得DE = DF,即△DEF为等边三角形.并且E从G运动到A时,DE严格增,DF严 格减,所以有唯一的E,使得△DEF为等边三角形.
设BD = x,令CE = √ 3 2 x, BF = 1 −x 2.由余弦定理 DF 2 = x2 + (1 −x 2)2 −2x(1 −x 2) cos 60◦= 7 4x2 −2x + 1 DE2 = (1 −x)2 + ( √ 3 2 x)2 = 7 4x2 −2x + 1 35 EF 2 = ( √ 3 − √ 3 2 x)2 + (1 + x 2)2 −2 √ 3(1 −x 2)(1 + x 2) cos 30◦= 7 4x2 −2x + 1 即DF = DE = EF,此时△DEF为等边三角形,即D固定时唯一的等边三角形.
设DEF为等边三角形,记AB, AC中点分别为M, N,BM中点为S,T在AB上,BT = 1 3AB.
D从B运动到C时,E从C运动到N,F从M运动到S.设△DEF的边长为a.
a2 = 7 4x2 −2x + 1 = 7 4(x −4 7)2 + 3 7.
当x = 4 7时,有最小值 q 3 7.设此时在△PQR的位置.
下面证明对任意内接三角形XY Z,最大边长不小于 q 3 7.
考虑Z在AB上的位置, (1)Z在AM上,则ZX ⩾1 2AC = √ 3 2 > q 3 7; (2)Z在BT上,则ZY ⩾2 3BC = 2 3 > q 3 7; (3)Z在MT上,按照前面的方法作出正三角形DEF,(F = Z),x < 1 3, √ 3 2 BF = √ 3 2 (1 −x 2) > √ 3 2 x = CE .则DE = DF = EF ⩾ q 3 7.
(3.1)X在CD上,XZ ⩾DF ⩾ q 3 7; (3.2)Y 在CE上,∠FEC > 90◦, Y Z ⩾EF ⩾ q 3 7; (3.3)X在BD上,Y 在AE上,XY = √ CX2 + CY 2 ⩾ √ CD2 + CE2 = DE ⩾ q 3 7.
综上所述,△ABC的内接三角形的最长边的最小值为 q 3 7.
36 第十二届中国数学奥林匹克(1997年) 杭州浙江大学 1.设实数x1, x2, . . . , x1997满足如下两个条件: (1)−1 √ 3 ⩽xi ⩽ √ 3(i = 1, 2, . . . , 1997) (2)x1 + x2 + · · · + x1997 = −318 √ 3 试求x12 1 + x12 2 + · · · + x12 1997的最大值,并说明理由.
解:考虑函数f(x) = (t + x)12 + (t −x)12,(t为常数).
显然f(x)为偶函数且展开式中所有偶次项系数均为正,所有奇次项均为0,所以f(x)在[0, +∞)上为增函 数.
所以对于∀i, j, 1 ⩽i < j ⩽1997, xi + xj为定值时, 当|xj −xi|的值越大时,x12 i + x12 j 越大.
这样我们逐次进行调整,过程如下:每次选取−1 √ 3 < xi ⩽xj < √ 3,保持它们的和不变. 若xi + xj > √ 3 − 1 √ 3,则将(xi, xj)调整为(xi + xj − √ 3, √ 3); 否则将(xi, xj)调整为(xi + xj + 1 √ 3, −1 √ 3). 这样调整 后F = x12 1 + x12 2 + · · · + x12 1997的值增大,经过有限次这样的调整后, 最多有一个xi不等于−1 √ 3或 √ 3,此 时达到最大值.
设此时有k个 √ 3,1996 −k个−1 √ 3,另一个为m ∈[−1 √ 3, √ 3].
则k √ 3 + (1996 −k)(−1 √ 3) + m = −318 √ 3,m = 1 √ 3(1042 −4k).
∴−1 ⩽1042 −4k ⩽3, 1039 4 ⩽k ⩽1043 4 又因为k ∈Z, ∴k = 260, m = 2 3 √ 3.
∴F ⩽260( √ 3)12 + 1736(−1 √ 3)12 + ( 2 3 √ 3)12 = 189548 当有260个xi为 √ 3,1个xi为2 3 √ 3,其他均为−1 √ 3时取等号.
2.点P是凸四边形A1B1C1D1内一点, 且P到各顶点的连线与四边形过该点的两条边的夹角均为锐角.
递推定义Ak, Bk, Ck和Dk分别为P关于直线Ak−1Bk−1,Bk−1Ck−1, Ck−1Dk−1和Dk−1Ak−1的对称点 (k = 2, 3, . . .).考察四边形序列AjBjCjDj(j = 1, 2, . . .).
试问:(1)前12个四边形中,哪些必定与第1997个相似,哪些未必; (2)假设第1997个是圆内接四边形,那么在前12个四边形中,哪些必定是圆内接四边形,哪些未必.
解:设∠DjAjP, ∠AjBjP, ∠BjCjP, ∠CjDjP分别为αj, βj, γj, δj; ∠PAjBj, ∠PBjCj, ∠PCjDj, ∠PDjAj分别为α′ j, β′ j, γ′ j, δ′ j.
不难知道Aj为△Dj+1Aj+1P的外心,所以αj+1 = 1 2∠Dj+1AjP = αj.
类似地,我们可以知道 (αj+1, βj+1, γj+1, δj+1) = (αj, βj, γj, δj) (α′ j+1, β′ j+1, γ′ j+1, δ′ j+1) = (β′ j, γ′ j, δ′ j, α′ j) ∴(αj+4, βj+4, γj+4, δj+4) = (αj, βj, γj, δj) (α′ j+4, β′ j+4, γ′ j+4, δ′ j+4) = (α′ j, β′ j, γ′ j, δ′ j) ∴Aj+4Bj+4Cj+4Dj+4 ∽AjBjCjDj 37 又因为(αj+2 + α′ j+2) + (γj+2 + γ′ j+2) = αj + γ′ j + γj + α′ j = (αj + α′ j) + (γj + γ′ j) 所以Aj+2Bj+2Cj+2Dj+2与AjBjCjDj相应的对角和相等.
于是有(1)前12个四边形中,第1,5,9个必定与第1997个相似; (2)假设第1997个是圆内接四边形,那么在前12个四边形中,第1,3,5,7,9,11个必定是圆内接四边形.
下面说明对于前12个四边形中其他的四边形未必成立.
考虑筝形A1B1C1D1,P为其对角线交点,A1B1 = A1D1, C1B1 = C1D1, ∠B1 = ∠D1 < 90◦.它不是圆内 接四边形.
设(α1, β1, γ1, δ1) = (α, β, γ, δ), 则(α′ 1, β′ 1, γ′ 1, δ′ 1) = (α, δ, γ, β).
所以(α2, β2, γ2, δ2) = (α, β, γ, δ); (α′ 2, β′ 2, γ′ 2, δ′ 2) = (δ, γ, β, α),(β + δ < 90◦) ∴∠A2 = ∠D2 = α + δ, ∠B2 + ∠C2 = β + γ,A2B2C2D2为等腰梯形,是圆内接四边形.
类似的可知A3B3C3D3为筝形,不是圆内接四边形.且∠A3 = ∠C3 = α + γ > 90◦.
A4B4C4D4为等腰梯形,是圆内接四边形.
这四个四边形互不相似,而四边形的形状显然是以4为周期变化. 所以在这个四边形序列中,前12个四边 形中,只有第1,5,9个与第1997个相似.
在上述序列中,以A2B2C2D2为第一个四边形,不难知道所有的第2k + 1个四边形均为圆内接四边 形, 而其他四边形均不是圆内接四边形,所以第1997个是圆内接四边形,在前12个四边形中,只有 第1,3,5,7,9,11个是圆内接四边形.
综上所述(1)前12个四边形中,第1,5,9个必定与第1997个相似,其他未必; (2)假设第1997个是圆内接四边形,在前12个四边形中,第1,3,5,7,9,11个必定是圆内接四边形,其他未必.
3.求证存在无穷多个正整数n,使得可将1, 2, . . . , 3n列成数表 a1 a2 . . .
an b1 b2 . . .
bn c1 c2 . . .
cn 满足如下两个条件: (1)a1 + b1 + c1 = a2 + b2 + c2 = · · · = an + bn + cn且为6的倍数; (2)a1 + a2 + · · · + an = b1 + b2 + · · · + bn = c1 + c2 + · · · + cn且为6的倍数.
证明:显然6n| 1 2(3n)(3n + 1), 18| 1 2(3n)(3n + 1).
必有n ≡1 (mod 4), n ≡0 (mod 3),所以n ≡9 (mod 12).
下面Ai中第1,2,3行分别记为α(i), β(i), γ(i),且α(i) + k表示将α(i)中每个数都加上k,其他类似.
先构造A9满足条件:设 A3 = 1 8 6 5 3 7 9 4 2 38 显然各行各列之和均为15,令 A9 = α(3) β(3) + 18 γ(3) + 9 β(3) + 9 γ(3) α(3) + 18 γ(3) + 18 α(3) + 9 β(3) 易知A9中的27个元素为1,2,. . . ,27,并且各列之和为15 + 9 + 18 = 42 ≡0 (mod 6); 各行之和为3(15 + 9 + 18) = 126 ≡0 (mod 6).所以9是满足条件的正整数.
设m满足条件,且形成的数表(矩阵)为Am,各行之和为6u,各列之和为6v.
构造A3m如下: A3m = α(m) β(m) + 6m γ(m) + 3m β(m) + 3m γ(m) α(m) + 6m γ(m) + 6m α(m) + 3m β(m) 则A3m中9m个元素为1, 2, . . . , 9m,并且各行之和为18u + 9m2,各列之和为6v + 9m.
构造A9m如下: A9m = α(3m) β(3m) + 18m γ(3m) + 9m β(3m) + 9m γ(3m) α(3m) + 18m γ(3m) + 18m α(3m) + 9m β(3m) 则A9m中27m个元素为1, 2, . . . , 27m,并且各行之和为54u + 108m2 ≡0 (mod 6), 各列之和为6v + 36m ≡0 (mod 6).
∴9m也是满足条件的正整数,由归纳法不难证明对∀k ∈N∗, 9k是满足条件的正整数,显然有无穷多个.
4.四边形ABCD内接于⊙O,其边AB与DC的延长线交于点P, AD与BC的延长线交于点Q,过Q作该圆 的两条切线QE和QF, 切点分别为E, F.求证:P, E, F三点共线.
证明:连接PQ,并且在PQ上取一点M,使得B, C, M, P四点共圆,则QE2 = QM · QP = QC · QB, 并且∠PMC = ∠CBA = ∠PDQ.所以C, D, Q, M四点共圆. 所以PM · PQ = PC · PD.
PQ2 = PM · PQ + QM · PQ = QC · QB + PC · PD.
连接PF,设PF与圆的另一交点为E′,作QG ⊥PF,垂足为G.
则PD · PC = PE′ · PF, QF 2 = QC · QB.
所以PE′ · PF + QF 2 = PQ2,即PE′ · PF = PQ2 −QE2.
又因为PQ2 −QF 2 = PG2 −GF 2 = (PG −GF)(PG + GF) = PF(PG −GF), 从而PG −GF = PE′ = PG −GE′.即GF = GE′.
故E′与E重合,P, E, F三点共线.
另证:设过A, D的切线相交于R,过B, C的切线相交于S,AC, BD相交于T.
则R为AD的极点,S为BC的极点.由于AD过点Q,BC过点Q,所以Q的极线EF过点R, S.
在退化六边形AACDDB中,由Pascal定理,P, R, T三点共线;类似的在ACCDBB中,P, S, T三点共线.
所以P, R, S, T四点共线,即P在直线EF上.证毕.
39 5.设A = {1, 2, 3, . . . , 17},对于映射f : A →A,记 f 1 = f(x), f [k+1] = f(f k)(k ∈N).
设从A到A的一一映射f满足条件:存在自然数M,使得: (1)当m < M, 1 ⩽i ⩽16时,有 f m −f m ̸≡±1 (mod 17), f m −f m ̸≡±1 (mod 17) (2)当1 ⩽i ⩽16时,有 f m −f m ≡1或−1 (mod 17), f m −f m ≡1或−1 (mod 17) 试对满足上述条件的一切f,求所对应的M的最大可能值,并证明你的结论.
证明:所求Mmax = 8.
令1与18相同.将所有的f m, f m(i = 1, 2, . . . , 17; m = 1, 2, . . . , M −1)配成一对, 则所有的这 样的数对均不相同.
否则设存在(f m1, f m1) = (f m2, f m2).
因为f为双射,必然存在反函数f −1.
若m1 = m2,必有(i + 1, i) = (j + 1, j).
否则设m1 < m2, f −1m1 = i + 1, f −1m1 = i; f −1m1 = f m2−m1, f −1m1 = f m2−m1.
∴(f m2−m1, f m2−m1) = (i + 1, i).
∴f m2−m1 −f m2−m1 ≡1或−1 (mod 17),矛盾.
所有这样的对共有17(M −1)对,但是由于没有任一对中两数之差的绝对值为1,所以最多有 ¡17 2 ¢ −17对.
∴17(M −1) ⩽ ¡17 2 ¢ −17, M ⩽8.
令f(i) ≡3i + 2 (mod 17),由归纳法不难证明f m ≡3mi + 3m −1 (mod 17).
而且3m ̸≡±1 (mod 17)(m = 1, 2, . . . , 7), 38 ≡−1 (mod 17).
所以f m −f m ≡3m ̸≡±1 (mod 17)(i = 1, 2, . . . , 17) f 8 −f 8 ≡38 ≡−1 (mod 17)(i = 1, 2, . . . , 17) 即f满足条件并且M = 8,所以Mmax = 8.
6.设非负数列a1, a2, . . .满足条件am+n ⩽am + an, m, n ∈N 求证:对任意n ⩾m均有an ⩽ma1 + ( n m −1)am.
证明:若n = m,结论显然成立.
否则设n = km + r(1 ⩽r ⩽m, k ∈N∗),则an ⩽kam + ar.
40 an n −am m ⩽ kam + ar n −am m = (mk −n)am + mar mn = mar −ram mn = r n(ar r −am m ) 而ar ⩽ra1, r ⩽m.
∴an n −am m ⩽m n (a1 −am m ) 化简后即得到an ⩽ma1 + ( n m −1)am.
41 第十三届中国数学奥林匹克(1998年) 广州广州师范学院 1.在一个非钝角△ABC中,AB > AC, ∠B = 45◦,O和I分别是△ABC的外心和内心, 且 √ 2OI = AB −AC,求sin A.
解:由Euler公式OI = √ R2 −2Rr.
∴2(R2 −2Rr) = (b −c)2.
又r = 1 2(a + c −b) tan B 2 = √ 2−1 2 (a + c −b).
并且 a sin A = b sin B = c sin C = 2R.
所以1 −2( √ 2 −1)(sin A + sin C −sin B) = 2(sin B −sin C)2.
∵∠B = 45◦,∴sin B = √ 2 2 .
sin C = sin(135◦−A) = √ 2 2 (sin A + cos A).
∴2 sin A cos A −(2 − √ 2) sin A − √ 2 cos A + √ 2 −1 = 0.
即( √ 2 sin A −1)( √ 2 cos A − √ 2 + 1) = 0.
所以sin A = √ 2 2 或者cos A = 1 − √ 2 2 即sin A = √ 1 −cos2 A = q√ 2 −1 2.
综上所述,sin A = √ 2 2 或者 q√ 2 −1 2.
2.给定大于1的正整数n,是否存在2n个两两不同的正整数, 同时满足以下两个条件: (1)a1 + a2 + · · · + an = b1 + b2 + · · · + bn; (2)n −1 − 1 1998 < n P i=1 ai−bi ai+bi < n −1.
请说明理由.
解:存在.
令ai = M + i, bi = i(i = 1, 2, . . . , n −1), an = k, bn = k + (n −1)M,其中k, M待定.
∴S = n X i=1 ai −bi ai + bi = n−1 X i=1 M M + 2i − (n −1)M (n −1)M + 2k = n −1 − Ãn−1 X i=1 2i M + 2i + (n −1)M (n −1)M + 2k !
显然S < n −1,而且S > n −1 − 1 1998等价于 T = n−1 X i=1 2i M + 2i + (n −1)M (n −1)M + 2k < 1 1998 令M = 3996n(n −1), k = 1998(n −1)M.则 T < n−1 X i=1 2i M + (n −1)M (n −1)M + 2k = n(n −1) M + (n −1)M (n −1)M + 2k = 1 3996 + 1 3997 < 1 1998 42 所以当ai = 3996n(n −1) + i, bi = i(i = 1, 2, . . . , n −1), an = 2 × 19982n(n −1)2 + 3996n(n −1)2, bn = 2 × 19982n(n −1)2时, 满足题设条件,证毕.
3.设S = {1, 2, . . . , 98}, 求最小自然数n,使得S的任一n元子集中都可以选出10个数, 无论怎样将这10个 数均分成两组, 总有一组中存在一个数与另外4个数都互质, 而另一组中存在一个数与另外4个数都不 互质.
解:n = 50.
设A = {x|x为偶数, x ∈S},则|A| = 49.并且A中任一两数均不互质,所以A的任一10元子集均不满足要 求, 无论怎样分组都不存在一个数与其余4个数都互质.
∴n ⩾50,下面证明S的任一50元子集T中都存在这样10个数,其中一个与其他九个数均互质, 而其他九 个数有一个公因子,显然这十个数满足题意.用反证法,假设不存在这样的十个数.
B1 = {1, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97}, |B1| = 11,为1和大于49的质数组成的集合.
(1)如果B1 ∩T ̸= ∅,设a ∈B1, a ∈T,显然a与T中其他数均互质,所以T中不存在九个数有公因子,至多 有8个偶数,8个3的倍数.
因为C = {x|x为奇数, 3|x, x ∈S}, |C| = 16, 所以|T| ⩽8 + 8 + (98 −49 −16) = 49,矛盾.
B2 = {13, 17, 19, 23, 29, 31, 37, 41, 43, 47}, |B2| = 10,为大于12小于49的质数组成的集合.
(2)T ∩B1 = ∅, T ∩B2 ̸= ∅,则T中至多有38个奇数,至少有12个偶数,|T ∩A| ⩾12.
设B2中的b ∈T,类似(1),T ∩A中至多有8个与b互质; 因为[ 49 b ] ≤3,所以T ∩A中不与b互质的至多有3个,|T ∩A| ⩽11,矛盾.
B3 = {5 × 11, 5 × 13, 5 × 17, 5 × 19, 7 × 7, 7 × 11, 7 × 13}, |B3| = 7.
(3)T ∩B1 = T ∩B2 = ∅, T ∩B3 ̸= ∅,则T中至多有28个奇数,|T ∩A| ⩾22.
设B3中c ∈T,类似(1),T ∩A中至多有8个与b互质; 因为[ 49 5 ] + [ 49 11] −[ 49 5×11] = 13, [ 49 7 ] = 7, 所以T ∩A中不与c互质的至多有13个,|T ∩A| ⩽21,矛盾.
B4 = {5 × 7, 5 × 5, 3 × 31, 3 × 29, 3 × 23, 3 × 19, 3 × 17, 3 × 13, 3 × 11}, |B4| = 9.
(4)T ∩B1 = T ∩B2 = T ∩B3 = ∅, T ∩B4 ̸= ∅,则T中至多有21个奇数,|T ∩A| ⩾29.
设B4中d ∈T,类似(1),T ∩A中至多有8个与d互质; 因为[ 49 5 ] + [ 49 7 ] −[ 49 5×7] = 15, [ 49 3 ] + [ 49 11] −[ 49 3×11] = 19, 所以T ∩A中不与d互质的至多有19个,|T ∩A| ⩽27,矛盾.
B5 = S −A −B1 −B2 −B3 −B4, |B5| = 12,B5中的数最多有2个不同质因子,(3 × 5 × 7 = 105 > 98).
(5)T ∩B1 = T ∩B2 = T ∩B3 = T ∩B4 = ∅, T ∩B5 ̸= ∅,则T中至多有12个奇数,|T ∩A| ⩾38.
设B5中s ∈T,类似(1),T ∩A中至多有8个与s互质; 因为[ 49 3 ] + [ 49 5 ] −[ 49 3×5] = 22, 所以T ∩A中不与s互质的至多有22个,|T ∩A| ⩽30,矛盾.
(6)T ∩Bi = ∅(i = 1, 2, 3, 4, 5),T中没有奇数,所以|T| ⩽|A| = 49,矛盾.
综上所述,S的任一50元子集T中都存在这样10个数.所以n = 50.
43 4.求所有大于3的自然数n, 使得1 + ¡n 1 ¢ + ¡n 2 ¢ + ¡n 3 ¢ 整除22000.
解:1 + ¡n 1 ¢ + ¡n 2 ¢ + ¡n 3 ¢ = 1 6(n3 + 5n + 6) = 1 6(n + 1)(n2 −n + 6) = 2k(k ⩽2000).
∴(n + 1)(n2 −n + 6) = 3 × 2k+1,∵n > 3, n + 1 > 4, n2 −n + 6 > 12.
(1)n + 1 = 2m, n2 −n + 6 = (2m −1)2 −(2m −1) + 6 = 22m −3 × 2m + 8 = 3 × 2t.
∵m ⩾3,如果m = 3,n2 −n + 6 = 48, n = 7,满足条件.
否则m > 3,n2 −n + 6 = 22m −3 × 2m + 8 ≡8 (mod 16), ∴3 × 2t ≡8 (mod 16), t = 3, n2 −n = 18,n不是整数.
(2)n + 1 = 3 × 2m, n2 −n + 6 = (3 × 2m −1)2 −(3 × 2m) + 6 = 9 × 22m −9 × 2m + 8 = 2t.
∵m ⩾1,如果m ⩾4,n2 −n + 6 ≡8 (mod 16), ∴2t ≡8 (mod 16), t = 3.∴n2 −n + 6 = 8, n2 −n = 2, n = 2, m = 0,矛盾.
否则1 ⩽m ⩽3, m = 1, n = 5, n2 −n + 6 = 26,不可能.
m = 2, n = 11, n2 −n + 6 = 116,不可能.
m = 3, n = 23, n2 −n + 6 = 512 = 29,t = 9.
综上所述,n = 7, 23.经检验均符合条件.
5.设D为锐角三角形ABC内部一点,且满足条件: DA · DB · AB + DB · DC · BC + DC · DA · CA = AB · BC · CA.
试确定D点的几何位置,并证明你的结论.
解:建立复平面,设A, B, C所对应的复数分别为z1, z2, z3,D对应的复数为z.
考虑多项式f(z) ≡1,由Lagrange插值公式: f(z) = (z −z1)(z −z2) (z3 −z1)(z3 −z2)f(z3) + (z −z1)(z −z3) (z2 −z1)(z2 −z3)f(z2) + (z −z2)(z −z3) (z1 −z2)(z1 −z3)f(z1) = 1 所以对于任意的z ∈C, (z −z1)(z −z2) (z3 −z1)(z3 −z2) + (z −z1)(z −z3) (z2 −z1)(z2 −z3) + (z −z2)(z −z3) (z1 −z2)(z1 −z3) = 1 ¯ ¯ ¯ ¯ (z −z1)(z −z2) (z3 −z1)(z3 −z2) ¯ ¯ ¯ ¯ + ¯ ¯ ¯ ¯ (z −z1)(z −z3) (z2 −z1)(z2 −z3) ¯ ¯ ¯ ¯ + ¯ ¯ ¯ ¯ (z −z2)(z −z3) (z1 −z2)(z1 −z3) ¯ ¯ ¯ ¯ ⩾1 即对于平面上任意一点D, DA · DB CA · CB + DA · DC BA · BC + DB · DC AB · AC ⩾1 DA · DB · AB + DB · DC · BC + DC · DA · CA = AB · BC · CA 等号成立当且仅当−ω1ω2, −ω2ω3, −ω3ω1 ∈R+, 其中ω1 = z−z1 z2−z3 , ω2 = z−z2 z3−z1 , ω3 = z−z3 z1−z2 .
所以−ω2 j ∈R+,ωj = kji, kj ∈R+(j = 1, 2, 3).
∴ z−z1 z2−z3 = ω1 = k1i,所以DA ⊥BC,类似的有DB ⊥AC, DC ⊥AB.即D为△ABC的垂心.
44 6.设n ⩾2, x1, x2, . . . , xn为实数,且 n X i=1 x2 i + n−1 X i=1 xixi+1 = 1.
对于每一个固定的自然数k(1 ⩽k ⩽n),求|xk|的最大值.
解:由于xi与xn+1−i在式中的对称性,可以设 n X i=1 x2 i + n−1 X i=1 xixi+1 = (√a1x1 + √ 1 −a2x2)2 + (√a2x2 + √ 1 −a3x3)2 + · · · + (√ak−1xk−1 + √ 1 −akxk)2 +( p 1 −an+1−kxk + √an−kxk+1)2 + · · · + ( √ 1 −a2xn−1 + √a1xn)2 +[1 −(1 −ak) −(1 −an+1−k)]x2 k = 1 其中数列{an}满足:a1 = 1, 2√ai √1 −ai+1 = 1.
即ai+1 = 1 − 1 4ai , 由数学归纳法不难证明:ai = i+1 2i , 1 −ai = i−1 2i (i = 1, 2, . . .).并且有0 ⩽ai ⩽1.
∴[1 −(1 −ak) −(1 −an+1−k)]x2 k ⩽1(k = 1, 2, . . . , n).
即|xk| ⩽ q 2k(n+1−k) n+1 ,等号成立当且仅当 √a1x1 + √1 −a2x2 = √a2x2 + √1 −a3x3 = · · · = √ak−1xk−1 + √1 −akxk = 0, 并且√1 −an+1−kxk + √an−kxk+1 = · · · = √1 −a2xn−1 + √a1xn = 0.
由xk = q 2k(n+1−k) n+1 ,可以求得x1, x2, . . . , xn的值,所以等号可以取到.
所以|xk|的最大值为 q 2k(n+1−k) n+1 .
45 第十四届中国数学奥林匹克(1999年) 北京北京大学 1.在锐角△ABC中, ∠C > ∠B, 点D是边BC上的一点,使得∠ADB是钝角,H是△ABD 的垂心,点F在 △ABC内部且在△ABD 的外接圆上.
求证:点F是△ABC垂心的充分必要条件是:HD平行于CF且H在△ABC的外接圆上.
证明:不难证明如下的结论:点F是△ABC的垂心的充分必要条件为CF ⊥AB并且∠AFB = 180◦−∠C.
∵H是△ABD的垂心,∴HD ⊥AB, ∠AHB = 180◦−∠ADB.
又因为A, B, D, F四点共圆,所以∠ADB = ∠AFB.
所以点F是△ABC垂心的充分必要条件是:HD平行于CF且∠AHB = ∠C.
而∠AHB = ∠C ⇔H在△ABC的外接圆上(因为∠ADB为钝角,H不可能在劣弧d AB上,必然与C在AB同 侧).所以原命题成立,证毕.
2.给定实数a,设实系数多项式序列{fn(x)}满足: f0(x) = 1, fn+1(x) = xfn(x) + fn(ax), 其中n = 0, 1, . . ..
(1)求证:fn(x) = xnfn( 1 x),其中n = 0, 1, . . ..
(2)求fn(x)的明显表达式.
解:当a = 1时,fn+1(x) = (x + 1)fn(x),由数学归纳法有fn(x) = (x + 1)n.则 fn(x) = (x + 1)n = xn( 1 x + 1)n = xnfn( 1 x).
当a ̸= 1时,定义Tn = (an −1)(an−1 −1) · · · (a −1), (n ∈N∗), T0 = 1.
Sk n = Tn TkTn−k (k ⩽n).我们用归纳法证明:fn(x) = Sn nxn + Sn−1 n xn−1 + · · · + S0 n.
对于n = 0,显然是成立的,假设对于n = k成立, 对于n = k + 1,由归纳假设,fk+1(x)的m次项系数为Sm−1 k + amSm k ,只需证明 Sm−1 k + amSm k = Sm k+1 ⇔ Tk Tm−1Tk+1−m + am Tk TmTk−m = Tk+1 TmTk+1−m ⇔(am −1) + am(ak+1−m −1) = ak+1 −1,这显然是成立的.
所以fk+1(x) = Sk+1 k+1xk+1 + Sk k+1xk + · · · + S0 k+1.
所以fn(x) = Sn nxn + Sn−1 n xn−1 + · · · + S0 n对于任意的n ∈N均成立.
由于Sk n = Sn−k n ,所以 xnfn( 1 x) = xn(Sn n 1 xn + Sn−1 n 1 xn−1 + · · · + S0 n) = Sn n + Sn−1 n x + · · · + S0 nxn = Sn nxn + · · · + S0 n = fn(x) 综上所述结论(1)成立,并且 fn(x) = (x + 1)n a = 1 Sn nxn + Sn−1 n xn−1 + · · · + S0 n a ̸= 1(其中Sk n定义如前) 46 3.MO太空城由99个空间站组成,任何两空间站之间都有一条管形通道相连. 规定其中99条通道为双向 通行的主干道.其余通道严格单向通行, 如果某四个空间站可以通过它们之间的通道从其中任一站到达 另外任一站, 则称这四个站的集合为一个互通四站组.试为MO太空城设计一个方案, 使得互通四站组 的数目最大(请具体算出该最大数,并证明你的结论).
解:如果从某个空间站到另三个空间站的通道均是由该站单向发出, 那么这四个空间站不是互通四站 组,将所有这样的四站组记为A类非互通四站组. 其余非互通四站组记为B类.
将空间站编号为1,2,. . . ,99,设从第i各空间站出发的单向通道有ki条, 则|A| = 99 P i=1 ¡ki 3 ¢ ,并且 99 P i=1 ki = ¡99 2 ¢ −99 = 99 × 48.
∵当k > l时,k ⩾l + 1,而 ¡k 3 ¢ − ¡k−1 3 ¢ = ¡k−1 2 ¢ .
∴ ¡k−1 2 ¢ ⩾ ¡l 2 ¢ ⇔ ¡k 3 ¢ − ¡k−1 3 ¢ ⩾ ¡l+1 3 ¢ − ¡l 3 ¢ ⇔ ¡k 3 ¢ + ¡l 3 ¢ ⩾ ¡k−1 3 ¢ + ¡l+1 3 ¢ .
所以经有限次调整知|A| ⩾99 × ¡48 3 ¢ ,而|B| ⩾0.
所以互通四站组最多有 ¡99 4 ¢ −99 ¡48 3 ¢ = 2052072个.
构造一种方案使得互通四站组的个数达到最大值,将所有空间站编号为1,2,. . . ,99,并且按照该顺序排列 在一个圆上.
(1,2),(2,3),. . . ,(98,99),(99,1)为双向主干道.
其余的通道(i, j),若从i到j顺时针经过偶数个空间站,则(i, j)为由i到j的单向通道.
因为i到j顺时针经过偶数个空间站等价于从j到i顺时针经过奇数个空间站,所以可以这样安排.
非互通四站组若不含双向主干道,要么有一站到另三个空间站的通道均是由该站单向发出,即为A类非 互通四站组; 要么有一站到另三个空间站的通道均是向该站发出,设B →A, C →A, D →A,且顺时 针方向依次为A, B, C, D.不难知道B →C,B →D,所以B到另三个空间站的通道均是由该站单向发 出,为A类非互通四站组.
若四站组含有双向主干道,则易知其他任意空间站到这两站均为一进一出,必为互通四站组.
所以没有B类非互通四站组,且从每个空间站均发出48条单向通道.即此时互通四站组的个数达到最大 值2052072.
4.设m是给定的整数.求证:存在整数a, b和k, 其中a, b均不能被2整除, k ⩾0, 使得 2m = a19 + b99 + k × 21999.
解:若x ̸≡y (mod 21999),x, y为奇数,则由x19 −y19 = (x −y)(x18 + x17y + · · · + y18).
因为后一个括号中为奇数,所以与21999互质,所以x19 ̸≡y19 (mod 21999).
所以当a取遍21999的既约剩余系时,a19也取遍21999的既约剩余系.
令b = 1,存在a,a19 ≡2m −1 (mod 21999),并且有无穷多个这样的a. 当a足够小时,不难知道k为正整 数,证毕.
5.求最大的实数λ,使得当实系数多项式f(x) = x3 +ax2 +bx+c的所有根都是非负实数时, 只要x ⩾0,就 有f(x) ⩾λ(x −a)3.并问上式中等号何时成立?
解:λ = −1 27.设f(x)的三个根为α, β, γ,f(x) = (x −α)(x −β)(x −γ).
不妨设0 ⩽α ⩽β ⩽γ,则−a = α + β + γ,x −a = x + α + β + γ ⩾0.
47 (1)0 ⩽x ⩽α ⩽β ⩽γ; f(x) = −(α −x)(β −x)(γ −x) ⩾−( α+β+γ−3x 3 )3 ⩾−( α+β+γ+x 3 )3 = −1 27(x −a)3.
(2)0 ⩽α ⩽β < x ⩽γ; f(x) = −(x −α)(x −β)(γ −x) ⩾−( −α−β+x+γ 3 )3 ⩾−( α+β+γ+x 3 )3 = −1 27(x −a)3.
(3)α < x ⩽β或者x > γ; f(x) ⩾0 ⩾−1 27(x −a)3.
综上所述f(x) ⩾−1 27(x −a)3.并且x = 0, α = β = γ或者α = β = 0, γ = 2x时等号成立.
6.4 × 4 × 4的大正方体由64个单位正方体组成. 选取其中的16个单位正方体涂成红色, 使得大正方体中 每个由4个单位正方体组成的1 × 1 × 4的小长方体中, 都恰有1个红色单位正方体.问16个红色单位正方 体共有多少种不同取法?说明理由.
解:在底面16个正方形中每个写上1,2,3,4之一,表示该竖直1 × 1 × 4小长方体中从下往上第几个为红色.
形成一个4 × 4的数表,其中每行每列均为1,2,3,4的一个排列,称为4阶拉丁方, 并将第一行第一列按顺序 均为1,2,3,4的拉丁方称为4阶标准拉丁方,显然16个红色单位正方体不同取法数目为4阶拉丁方的数目.
显然将4阶拉丁方任两行(列)调换仍为4阶拉丁方,对于任意一个4阶拉丁方,先通过调换列使之第一 行为1,2,3,4; 再通过调换2,3,4行使第一列为1,2,3,4,形成唯一一个4阶标准拉丁方.而每个4阶标准拉丁 方,对应4! × 3!个4阶拉丁方, (第一行有4!中排列,第一列后三个格有3!中排列).只需求4阶标准拉丁方的 个数.
考虑第二行第二列那个各种所填的数,显然不为2.
(1)填1,第二行第二列已经确定,剩下4个方格填入2个1,2个2,显然有两种填法.
(2)填3,第二行第二列已经确定,易知剩下4个方格也已经确定.填4类似.
所以共有4个4阶标准拉丁方.所以4阶拉丁方有4 × 4! × 3! = 576个.
所以16个红色单位正方体不同取法数目为576.
48 第十五届中国数学奥林匹克(2000年) 合肥中国科技大学 1.设a, b, c为△ABC的三条边, a ⩽b ⩽c, R和r分别为△ABC的外接圆半径和内切圆半径.
令 f = a + b −2R −2r,试用角C的大小来判定f的正负.
解:用A, B, C分别表示△ABC的三个内角.由公式 a = 2R sin A, b = 2R sin B, r = 4R sin A 2 sin B 2 sin C 2 .
f = a + b −2R −2r = 2R(sin A + sin B −1 −4 sin A 2 sin B 2 sin C 2 = 2R(2 sin A + B 2 cos A −B 2 −1 −2 sin C 2 (cos A −B 2 −cos A + B 2 )) = 2R(2 cos C 2 cos A −B 2 −1 −2 sin C 2 cos A −B 2 + 2 sin2 C 2 ) = 2R(2 cos A −B 2 (cos C 2 −sin C 2 ) −(cos2 C 2 −sin2 C 2 )) = 2R(cos C 2 −sin C 2 )(2 cos A −B 2 −cos C 2 −sin C 2 ) ∵a ⩽b ⩽c,∴∠A ⩽∠B ⩽C.
∴0 ⩽B−A 2 < C 2 < π 2 , 0 ⩽B−A 2 < A+B 2 < π 2 .
∴cos A−B 2 > cos C 2 , A−B 2 > cos A+B 2 = sin C 2 .
∴2 cos A−B 2 −cos C 2 −sin C 2 > 0.
∴f > 0 ⇔cos C 2 −sin C 2 > 0 ⇔∠C < π 2 ; f = 0 ⇔cos C 2 −sin C 2 = 0 ⇔∠C = π 2 ; f < 0 ⇔cos C 2 −sin C 2 < 0 ⇔∠C > π 2 .
2.数列{an}定义如下: a1 = 0, a2 = 1, an = 1 2nan−1 + 1 2n(n −1)an−2 + (−1)n(1 −n 2 )(n ⩾3).
试求 fn = an + 2 µn 1 ¶ an−1 + 3 µn 2 ¶ an−2 + · · · + (n −1) µ n n −2 ¶ a2 + n µ n n −1 ¶ a1 的最简表达式.
解:令bn = an (−1)nn!,则b1 = 0, b2 = 1 2, bn = −1 2bn−1 + 1 2bn−2 + 2−n 2n! (n ⩾3).
设cn = bn + bn−1, 2cn−1 = cn−2 + 2−n n! (n ⩾3).
∴2cn+1 = cn − n (n+2)!(n ⩾1).
2(cn+1 − 1 (n+2)!) = cn − n (n+2)! − 2 (n+2)! = cn − 1 (n+1)!.
并且c1 = b1 + b2 = 1 2, ∴c1 −1 2! = 0.
∴cn − 1 (n+1)! = 0, cn = 1 (n+1)!, 即bn + bn+1 = 1 (n+1)!.
49 令 gn = fn n! = 1 n!
n X k=1 (n −k + 1) µ n n −k ¶ ak = n X k=1 (−1)k n −k + 1 (n −k)! bk ∴gn+1 −gn = n+1 X k=1 (−1)k n −k + 2 (n + 1 −k)!bk − n X k=1 (−1)k n −k + 1 (n −k)! bk = n+1 X k=2 (−1)k n −k + 2 (n + 1 −k)!bk − n+1 X k=2 (−1)k+1 n −k + 2 (n + 1 −k)!bk−1 = n+1 X k=2 (−1)k n −k + 2 (n + 1 −k)!(bk + bk−1) = n+1 X k=2 (−1)k n −k + 2 (n + 1 −k)!
1 k!
= n+1 X k=2 (−1)k (n + 1 −k)!k! + n X k=2 (−1)k (n −k)!k!
= 1 (n + 1)!
n+1 X k=2 (−1)k µn + 1 k ¶ + 1 n!
n X k=2 (−1)k µn k ¶ ∵ n+1 X k=0 (−1)k µn + 1 k ¶ = (1 −1)n+1 = 0, n X k=0 (−1)k µn k ¶ = (1 −1)n = 0 ∴gn+1 −gn = 0 + n + 1 −1 (n + 1)!
+ 0 + n −1 n!
= n (n + 1)! + n −1 n!
= 1 (n −1)! − 1 (n + 1)!
∵g3 = 2b2 + b3 = 4 3 ∴fn = n!gn = n!
à n X k=4 1 (k −2)! − X k=4 1 k! + g3 !
= n!
µ4 3 + 1 2! + 1 3! − 1 (n −1)! −1 n!
¶ = 2n!−n−1(n ⩾3) 不难验证对于n = 1, 2也成立,所以fn = 2n! −n −1.
3.某乒乓俱乐部组织交流活动,安排符合以下规则的双打赛程表,规则为: (1)每个参赛者至多属于两个对子; (2)任意两个不同对子之间至多进行一次双打; (3)凡表中同属一对的两人就不在任何双打中作为对手相遇.
统计各人参加的双打次数，约定将所有不同的次数组成的集合称为“赛次集”.
给定由不同的正整数组成的集合A = {a1, a2, . . . , ak},其中每个数都能被6整除. 试问至少必须有多少人 参加活动,才可以安排符合上述规则的赛程表, 使得相应的赛次集恰为A,请证明你的结论.
解:设a1 < a2 < · · · < ak,必存在某个参赛者参加了ak场比赛.
若该参赛者只属于一个对子,则有ak个对子与该对进行比赛,至少有ak + 2人; 若该参赛者属于两个对子,至少有1 2ak对与这两对进行比赛,至少有ak 2 + 3人.
所以至少有ak 2 + 3人参加比赛.
下面用归纳法构造ak 2 + 3人参赛的赛程表.
50 k = 1,将a1 2 + 3人分成3人小组,每组三人两两结成对子,每人恰参加两个对子.
让每个对子都与其他小组的对子进行比赛,显然每个人都进行了a1场比赛,满足条件.
k = 2,将a2 2 + 3人分成两大组S, T,|S| = a1 2 , |T| = a2−a1 2 + 3.
同样将每大组分为3人小组,每组三人两两结成对子,每人恰参加两个对子.
S大组中任一对都与其他小组的对子进行比赛,每个人都比赛了a2场; T大组中任一对都只与S大组中的所有对子比赛,每个人都比赛了a1场.满足条件.
设k −1, k时命题成立,存在这样的赛次集.当k + 1时(k > 1), 将a1 2 + 3人分成三大组S, T, U.|S| = a1 2 , |T| = ak−a1 2 + 3, |U| = ak+1−ak 2 .
同样将每大组分成3人小组,每组三人两两结成对子,每人恰参加两个对子.
S大组中任一对都与其他小组的对子进行比赛,每个人都比赛了ak+1场; U大组中任一对都只与S大组中的所有对子比赛,每个人都比赛了a1场.
由归纳假设,可在T大组中安排赛程,使得赛次集为{a2 −a1, a3 −a1, . . . , ak −a1}.
又因为T大组中任一对都与S大组中的所有对子比赛,每个人又都比赛了a1场.
所以T大组中的人比赛场次所组成的集合为{a2, a3, . . . , ak}, 所以所有选手的赛次集为A,满足条件.
由归纳法存在ak 2 + 3人参赛的赛程表,使得赛次集为A.
综上所述,最少有ak 2 + 3人参加.
4.设n ⩾2.对n元有序实数组 A = (a1, a2, . . . , an), 令 bk = max 1⩽i⩽k ai, (k = 1, 2, . . . , n).
称B = (b1, b2, . . . , bn)为A的“创新数组”; 称B中的不同元素个数为A的“创新阶数”.
考察1, 2, . . . , n的所有排列(将每种排列都视为一个有序数组), 对其中创新阶数为2的所有排列,求它们 的第一项的算术平均值.
解:设第一项为k的创新阶数为2的排列有xk个,显然xn = 0.
而k = 1, 2, . . . , n −1时,排列中所有大于k小于n的数必排在n之后.
其他的数的位置可以任意取.
即所有大于k的n −k个数中n排在最前,恰占全部排列的 1 n−k.
∴xk = (n−1)!
n−k .
所以所求平均值为 n−1 P k=1 kxk n−1 P k=1 xk = (n −1)!
n−1 P k=1 k n −k (n −1)!
n−1 P k=1 1 n −k = n−1 P k=1 ³n k −1 ´ n−1 P k=1 1 k = n − n −1 1 + 1 2 + · · · + 1 n −1 51 5.若对正整数n,存在k,使得 n = n1n2 · · · nk = 2 1 2k (n1−1)(n2−1)···(nk−1) −1, 其中n1, n2, . . . , nk都是大于3的整数,则称n具有性质P. 求具有性质P的所有数n.
解:显然n为奇数,并且存在m ∈N∗, n = 2m −1.
当m = 1, 2, . . . , 9时,不难验证只有m = 3, n = 7满足要求.
当m ⩾10时,由归纳法不难证明:2m −1 > m3.
2 1 2k (n1−1)(n2−1)···(nk−1) −1 > k Y i=1 µni −1 2 ¶3 而由ni > 3, ni ⩾5,所以 µni −1 2 ¶3 ⩾4 µni −1 2 ¶ > ni ∴2 1 2k (n1−1)(n2−1)···(nk−1) −1 > n1n2 · · · nk 所以具有性质P的n只有7.
6.某次考试有5道选择题,每题都有4个不同答案供选择, 每人每题恰选1个答案,在2000份答卷中发现存 在一个n, 使得任何n份答卷中都存在4份, 其中每两份的答案都至多3道题相同.求n的最小可能值.
解:将每个题的答案记为1,2,3,4.每份答案记为(a, b, c, d, e),其中a, b, c, d, e ∈{1, 2, 3, 4}.
将(1, b, c, d, e)(2, b, c, d, e)(3, b, c, d, e)(4, b, c, d, e)记作一组.共44 = 256组, 由抽屉原则,至少有一组 有[ 2000 256 ] + 1 = 8份答卷; 去掉该组后仍然还有一组有8份答卷,再去掉这组后仍然还有一组有8份 答卷. 这24份答卷中任4份至少有两个在同一组,它们有4题答案相同.所以n ⩾25.
n = 25时,构造如下2000份答卷满足题意,每份答卷都使得a + b + c + d + e ≡0 (mod 4), 所以前面每一 组中至多有一份答卷,共256种不同的答卷,任两份不同的答卷都至多有3题答案相同. 任取其中250种,每 种恰有8份答卷,任取25份答卷,必有4份不同的答卷,它们之间任两份至多有3题答案相同.所以n = 25.
52 第十六届中国数学奥林匹克(2001年) 香港数学奥林匹克委员会 1.给定a, √ 2 < a < 2. 内接于单位圆Γ的凸四边形ABCD适合以下条件: (1)圆心在这凸四边形内部; (2)最大边长是a,最小边长是 √ 4 −a2.
过点A, B, C, D依次作圆Γ的四条切线LA, LB, LC, LD. 已知LA与LB,LB与LC,LC与LD, LD与LA分别 相交于A′, B′, C′, D′四点.求面积之比SA′B′C′D′ SABCD 的最大值与最小值.
解:设圆Γ的圆心为O,并记∠AOB = 2θ1, ∠BOC = 2θ2, ∠COD = 2θ3, ∠DOA = 2θ4.于是θ1, θ2, θ3, θ4都 是锐角,且θ1 + θ2 + θ3 + θ4 = π.不难求得 SABCD = 1 2 4 X i=1 sin 2θi, SA′B′C′D′ = 4 X i=1 tan θi 由于上式关于θ1, θ2, θ3, θ4对称, 不妨设AB = a, AD = √ 4 −a2, θ1 ⩾θ2 ⩾θ3 ⩾θ4.
则sin θ1 = a 2, sin θ4 = 1 2 √ 4 −a2, 易知θ1 + θ4 = π 2 , θ2 + θ3 = π 2 .
∴tan θ1 + tan θ4 = sin(θ1 + θ4) cos θ1 cos θ4 = 1 cos θ1 sin θ1 = 2 sin 2θ1 tan θ2 + tan θ3 = 2 sin 2θ2 T = SA′B′C′D′ SABCD = 2 sin 2θ1 + 2 sin θ2 sin 2θ1 + sin θ2 而sin 2θ1 = 1 2a √ 4 −a2, π 4 ⩽θ2 ⩽θ1, ∴1 2a √ 4 −a2 ⩽sin 2θ2 ⩽1.
由于T是关于sin 2θ2的严格减函数, Tmax = 2 ( 1 2a √ 4 −a2)2 = 8 a2(4 −a2), Tmin = 4 a √ 4 −a2 + 2 1 2a √ 4 −a2 + 1 = 4 a √ 4 −a2 2.设X = {1, 2, 3, . . . , 2001}, 求最小的正整数m,满足要求:对X的任何一个m元子集W, 都存在u, v(u和v 允许相同),使得u + v是2的方幂.
解:令Y = {2001, 2000, . . . , 1025} ∪{46, 45, . . . , 33} ∪{17} ∪{14, 13, · · · , 9}, 则|Y | = 998,并且可以证 明对于任何u, v ∈Y ,u + v都不是2的方幂.
事实上,当u, v ∈Y 时,不妨设u ⩾v并且有2r < u ⩽2r + a < 2r+1,其中当r分别取值10,5,4,3时,相应 的a值依次为977,14,1,6.
(1)若2r < v ⩽u,则2r+1 < u + v < 2r+2; (2)若1 ⩽v < 2r,则当2r < u ⩽2r + a, 1 ⩽a < 2r时,1 ⩽v < 2r −a.于是2r < u + v < 2r+1.
这表明u + v不是2的方幂.证毕.
53 故知所求的最小正整数m ⩾999.
将X划分成下列999个互不相交的子集: Ai = {1024 −i, 1024 + i}, i = 1, 2, . . . , 977; Bj = {32 −j, 32 + j}, j = 1, 2, . . . , 14; C = {15, 17}; Dk = {8 −k, 8 + k}, k = 1, 2, 3, 4, 5, 6; E = {1, 8, 16, 32, 1024}.
对于X的任何一个999元子集W,若W ∩E ̸= ∅,则从中任取一个元素的2倍都是2的方幂;若W ∩E = ∅,则W中的999各元素分属于前面的998个2元子集,由抽屉原理知W中必有不同的u和v属于其中同一个 子集, 显然u + v是2的方幂.
综上所述,所求的最小正整数m = 999.
3.在正n边形的每个顶点上各停有一只喜鹊.突然受到惊吓, 众喜鹊都飞去.一段时间后,它们又都回到这 些顶点上, 仍是每个顶点上一只,但未必都回到原来的顶点. 求所有正整数n,使得一定存在3只喜鹊, 以 它们前后所在的顶点分别形成的三角形或同为锐角三角形, 或同为直角三角形,或同为钝角三角形.
解:对于n = 3,结论显然成立.
若n ⩾4,n为偶数,设A, B原来的位置为直径的两端点,若回来后仍为直径的两端点,则任取另一只C,可 知前后△ABC均为直角三角形;否则设回来后A的对径点为C,则前后△ABC均为直角三角形,结论成 立.
若n ⩾7,n为奇数,不妨设A回到原顶点,否则可通过旋转使得A回到原顶点. 作以A为一个端点的直径,在 直径两侧各有n−1 2 ⩾3个点. 考虑原来在同一侧的三个点,由抽屉原则,回来后必有两个仍在同一侧,不 妨设为B, C,则△ABC前后均为钝角三角形,结论成立.
n = 5时,设原先按顺时针排列为A, B, C, D, E,返回后按顺时针排列为A, C, E, B, D,则此时不难验证所 有的钝角三角形变为锐角三角形, 所有的锐角三角形变为钝角三角形,结论不成立.
综上所述,所求的n为所有不小于3且不等于5的整数.
4.设a, b, c, a+b−c, a+c−b, b+c−a, a+b+c是7个两两不同的质数, 且a, b, c中有两数之和是800.设d是 这7个质数中最大数与最小数之差. 求d的最大可能值.
解:不妨设a < b < c,显然a + b −c最小,a + b + c最大,所以d = 2c.只需求c的最大值.
∵a + b −c > 0, c < a + b, c < a + c, c < b + c, ∴c < 800.
小于800的质数从大到小依次为797,787,. . .
若c = 797, a + b = 800, a + b −c = 3, a + b + c = 1597, 令a = 13, b = 787, a + c −b = 23, b + c −a = 1571.不难验证均为质数,所以dmax = 2 × 797 = 1594.
5.将周长为24的圆周等分成24段.从24个分点中选取8个点, 使得其中任何两点间所夹的弧长都不等 于3和8. 问满足要求的8点组的不同取法共有多少种?说明理由.
54 解:将这些点按顺时针方向依次标为1,2,. . . ,24,并排成如下3 × 8的表格: 1 4 7 10 13 16 19 22 9 12 15 18 21 24 3 6 17 20 23 2 5 8 11 14 在表中,同一列相邻两数所代表的点之间所夹弧长为8,同一行相邻两数所代表的点之间所夹弧长为3, (第一列与第八列也是相邻的,第一行与第三行也是相邻的).所以在表中每相邻两数所代表的点均不能 同时取. 即每一列只能取一个数,并且恰好取一个数.
记从3 × n数表中每列恰取一个数且任何相邻两列(包括第n列与第一列)所取得数均不同行的取法 为xn种.
从第一列取一个数有3种取法,第一列取定后,第2列所取得数不能与第一列同行,只有两种不同取法,以 后每一列均有两种取法, 共3 × 2n种取法,但是第一列与最后一列所取的数同行的所有取法都不满足要 求, 这时将这两列看作一列,即为n −1列时的所有取法,所以xn + xn−1 = 3 × 2n−1.
∴x8 = 3 × 27 −x7 = 3 × 27 −(3 × 26 −x6) = 3 × (27 −26) + x6 = · · · = 3 × (27 −26 + 25 −24 + 23 −22 + 2) = 258 既满足题中要求的不同取法总数为258.
6.记a = 2001.设A是适合下列条件的正整数对(m, n)所组成的集合: (1)m < 2a; (2)2n|(2am −m2 + n2); (3)n2 −m2 + 2mn ⩽2a(n −m).
令 f = 2am −m2 −mn n , 求min (m,n)∈A f和max (m,n)∈A f.
解:(1)先求f的最小值,令 p = 2am −m2 + n2 2n (∗) 由条件(1)和(2)可知,p为正整数且有 (2a −m)m = (2p −n)n (∗∗) ∵2a −m > 0,所以2p −n > 0.由条件(3)可得 2am −m2 + n2 = 2am −2mn + 2mn −m2 + n2 ⩽2am −2mn + 2a(n −m) = 2n(a −m) 从而由(∗),p ⩽a −m.所以2p −n < 2p ⩽2(a −m) < 2a −m.
由(∗∗)知,m, n同奇偶,且n > m.
设(m, n) ∈A,令n′ = 2p −n = 2am−m2 n ,显然n′也是正整数而且容易验证(m, n) ∈A ⇔(m, n′) ∈A.事 实上,不难验证下面两式成立 p′ = 2am −m2 + n′2 2n′ = 2am −m2 + n2 2n = p 55 n′2 −m2 + 2mn′ −2a(n′ −m) = 2am −m2 n2 [n2 −m2 + 2mn −2a(n −m)] 由2a > m,便知关于(m, n′)的条件(2)(3)成立的充要条件是关于(m, n)的条件(2)(3)成立. 即(m, n) ∈ A ⇔(m, n′) ∈A.
这样一来,易知(m, n) ∈A时,f(m, n) = n′ −m, f(m, n′) = n −m.
∴f(m, n) = n′ −m ⩾2.
取m = 2,条件化为2n|4a −4 + n2. 即4000 n + n 2 为正整数,n2 −3998n + 8000 ⩽0. 又n为正整数,3 ⩽n ⩽ 3995,又n为4000的偶因子,取n = 2000,f(2, 2000) = 2.
综上可知min (m,n)∈A f = 2.
(2)再求f的最大值.因为(m, n) ∈A时,m, n奇偶性相同且n > m,令n = m + 2u, u为正整数,于是有 f(m, n) = 2am −m2 −mn n = 2(a −u)m −2m2 m + 2u = 2(a −u)(m + 2u) −2(a −u)2u m + 2u −2(m + 2u)2 + 8(m + 2u)u −8u2 m + 2u = (2a + 6u) −2 · m + 2u + 2u(a + u) m + 2u ¸ 由(3),知(m + 2u)2 −m2 + 2m(m + 2u) ⩽4ua, 即m2+4um−2au+2u2 ⩽0,m ⩽−2u+ p 2u(a + u), m+2u ⩽ p 2u(a + u). 所以当u固定时,f(m, m+ 2u)关于m严格递增.
∵f(m, m + 2u)为偶数,所以m + 2u必为2u(a + u)的因子, u = 1时,2(a + 1) = 4004 = 4 × 7 × 11 × 13.
而若要m + 2u + 2u(a+u) m+2u 尽量大,m + 2u应尽量接 近 p 2u(a + u) = √ 4004, m + 2|4004, n = m + 2 = 52, m = 50, (m, n) ∈A, f(50, 52) = 4002 + 6 −2(52 + 4004 52 ) = 3750.
所以有f(m, m + 2) ⩽3750.
以下证明:对于任意的(m, n) ∈A,都有f(m, n) ⩽3750.
上面已证n = m + 2时,结论成立,由于m, n奇偶性相同且n > m,故可以设n ⩾m + 4,于是 f(m, n) = (2a −m)m n −m ⩽(2a −m)m m + 4 −m 只需证明:对于任意的1 ⩽m < 2a = 4002,都有 (2a −m)m m + 4 −m ⩽3750 整理后得m2 −124m + 2 × 3750 ⩾0.
由于∆= 1242 −8 × 3750 = 4(622 −7500) < 0,所以结论是成立的.
综上可知, max (m,n)∈A f = 3750.
56 第十七届中国数学奥林匹克(2002年) 上海上海中学 1.三角形ABC的三边长分别为a, b, c, b < c, AD是角A的内角平分线,点D在边BC上.
(1)求在线段AB, AC内分别存在点E, F(不是顶点)满足BE = CF和∠BDE = ∠CDF 的充分必要条 件(用角A, B, C表示); (2)在点E和F存在的情况下,用a, b, c表示BE的长.
解:(1)因为AD平分∠BAC,所以D到AB, AC距离相等,又由于BE = CF,所以S△DBE = S△DF C.
∵∠BDE = ∠CDF,∴BD · DE = CD · DF.
又BE = CF,BD2 + DE2 −2BD · DE cos ∠BDE = CD2 + DF 2 −2CD · DF cos ∠CDF.
∴BD2 + DE2 = CD2 + DF 2.所以BD = CD, DE = DF或者BD = DF, CD = DE.
因为b < c,所以BD > DC,所以BD = DF, CD = DE.
可以得到△BDE ∼ = △FDC,∠B = ∠DFC, ∠C = ∠BED,并且∠BDE = ∠CDF = ∠A.
而∠C > ∠B = ∠DFC > ∠DAC = 1 2∠A, ∴2∠B > ∠A.
反之,如果2∠B > ∠A,作∠CDF = ∠BDE = ∠A,则F, E分别在AC, AB上, △FDC ∽△BAC ∽△BDE.
又D到FC, BE距离相等,所以△BDE ∼ = △FDC,BE = CF.
所以充要条件为2∠B > ∠A.
(2)因为△BAC ∽△BDE,所以BD BA = BE BC ,BE = BC · BD BA .
再由角平分线定理,BD DC = AB AC ,∴BD = ac b + c,所以BE = a2 b + c.
2.设多项式序列{Pn(x)}满足:P1(x) = x2 −1, P2(x) = 2x(x2 −1), 且 Pn+1(x)Pn−1(x) = (Pn(x))2 −(x2 −1)2, n = 2, 3, . . . .
设sn为Pn(x)各项系数的绝对值之和.
对于任意正整数n,求非负整数kn,使得2−knsn为奇数.
解:设Pi(x) = Qi(x)(x2 −1),则Q1(x) = 1, Q2(x) = 2x, Qn+1(x)Qn−1(x) = (Qn(x))2 −1.
设tn为Qn(x)各项系数的绝对值之和, 因为Qn+1Qn−1 = Q2 n −1, Qn+2Qn = Q2 n+1 −1,所以Qn+1Qn−1 + Q2 n+1 = Qn+2Qn + Q2 n.
所以Qn+2 + Qn Qn+1 = Qn+1 + Qn−1 Qn ,∴对于∀n ∈N∗,Qn+2 + Qn Qn+1 = Q3 + Q1 Q2 = 2x.
∴Qn+2(x) = 2xQn+1(x) −Qn(x).
由归纳法不难证明如下结论: (1)Qn(x)为首项系数为正的n −1次整系数多项式; (2)Q2k(x)只有奇数项,Q2k+1(x)只有偶数项,(k ∈N); (3)Qn(x)所有系数不为0的项的系数从大到小为正负交替.
57 所以tn+2 = 2tn+1 + 2tn,sn = 2tn.补充定义t0 = s0 = 0.
由t1 = 1, t2 = 2知对于n ∈N, tn = 1 2 √ 2[(1 + √ 2)n −(1 − √ 2)n], sn = 1 √ 2[(1 + √ 2)n −(1 − √ 2)n] 设rn = (1 + √ 2)n + (1 − √ 2)n,则rn + √ 2sn = 2(1 + √ 2)n,并且rn, sn均为正整数.
所以rn = 2 [ n 2 ] P m=0 ¡ n 2m ¢ 2m = 2(1 + [ n 2 ] P m=1 ¡ n 2m ¢ ), 显然1 2rn为奇数.
对于∀m, n ∈N, rn+m + √ 2sn+m = 2(1 + √ 2)n+m = 1 2(rn + √ 2sn)(rm + √ 2sm).
∴rn+m = smsn + 1 2rmrn, sm+n = 1 2(rnsm + rmsn).
∴s2m = rmsm, k2m = km + 1,又因为k1 = 1,所以k2m = m + 1.
当km ̸= kn时,sm+n = ( 1 2rn)sm + ( 1 2rm)sn,km+n = min{km, kn}.
设n = 2m0 + 2m1 + · · · + 2ml, 0 ⩾m0 < m1 < · · · < ml为正整数.
则kn = m0 + 1,即2m0∥n时,kn = m0 + 1.
3.18支足球队进行单循环赛,即每轮将18支球队分成9组, 每组的两队赛一场,下一轮重新分组进行比 赛,共赛17轮, 使得每队都与另外17支队各赛一场. 按任意可行的程序比赛了n轮之后,总存在4支球队, 它们之间总共只赛了1场.求n的最大可能值.
解:考察如下的比赛程序: 1.(1,2)(3,4)(5,6)(7,8)(9,18)(10,11)(12,13)(14,15)(16,17) 2.(1,3)(2,4)(5,7)(6,9)(8,17)(10,12)(11,13)(14,16)(15,18) 3.(1,4)(2,5)(3,6)(8,9)(7,16)(10,13)(11,14)(12,15)(17,18) 4.(1,5)(2,7)(3,8)(4,9)(6,15)(10,14)(11,16)(12,17)(13,18) 5.(1,6)(2,8)(3,9)(4,7)(5,14)(10,15)(11,17)(12,18)(13,16) 6.(1,7)(2,9)(3,5)(6,8)(4,13)(10,16)(11,18)(12,14)(15,17) 7.(1,8)(2,6)(4,5)(7,9)(3,12)(10,17)(11,15)(13,14)(16,18) 8.(1,9)(3,7)(4,6)(5,8)(2,11)(10,18)(12,16)(13,15)(14,17) 9.(2,3)(4,8)(5,9)(6,7)(1,10)(11,12)(13,17)(14,18)(15,16) 10.(1,11)(2,12)(3,13)(4,14)(5,15)(6,16)(7,17)(8,18)(9,10) 11.(1,12)(2,13)(3,14)(4,15)(5,16)(6,17)(7,18)(8,10)(9,11) 12.(1,13)(2,14)(3,15)(4,16)(5,17)(6,18)(7,10)(8,11)(9,12) 13.(1,14)(2,15)(3,16)(4,17)(5,18)(6,10)(7,11)(8,12)(9,13) 14.(1,15)(2,16)(3,17)(4,18)(5,10)(6,11)(7,12)(8,13)(9,14) 15.(1,16)(2,17)(3,18)(4,10)(5,11)(6,12)(7,13)(8,14)(9,15) 16.(1,17)(2,18)(3,10)(4,11)(5,12)(6,13)(7,14)(8,15)(9,16) 17.(1,18)(2,10)(3,11)(4,12)(5,13)(6,14)(7,15)(8,16)(9,17) 将前9支球队称为A组,后9支球队称为B组,显然9轮之后,同组两支球队均已经比赛过, 所以任意四支球 队之间至少已经赛过两场比赛,当然不满足题中要求.
58 如果把上述程序颠倒过来,然后按照新的程序比赛,则8轮比赛过后,同组任何两队均未比赛过. 而不全同 组的四支球队之间至少赛过两场,也不满足题中要求.
综上所述,n的最大可能值不大于7.
设已进行了7轮比赛且任何4队都不满足题中要求.
选取已经比赛过的两队A1, A2,于是每支球队都与另外6支球队比赛过,两个队至多与另外12支球队比赛 过. 从而至少有4支球队B1, B2, B3, B4与A1, A2均未比赛过,由反证假设可知,B1, B2, B3, B4之间的比赛 都已经进行了.
所以最多有10支球队与B1, B2其中至少一队比赛过,又导致至少存在C1, C2, . . . , C6与B1, B2均未比赛.
由反证假设可知,C1, C2, . . . , C6之间的比赛都已经进行了.
类似的,最多有8支球队与C1, C2其中至少一队比赛过,又导致至少存在D1, D2, . . . , D8与C1, C2均未比 赛. 由反证假设可知,D1, D2, . . . , D8之间的比赛都已经进行了.所以D1, D2与其他10支球队均未比赛, 但是由于只进行了7轮比赛,其中必存在两支球队E1, E2之间没有比赛,D1, D2, E1, E2之间只进行了一 场比赛,与假设矛盾.
综上所述,n的最大可能值为7.
4.对于平面上任意四个不同点P1, P2, P3, P4,求 P 1⩽i<j⩽4 PiPj min 1⩽i<j⩽4 PiPj 的最小值.
解:先证明如下的引理: 引理:在△ABC中,若AB ⩾m, AC ⩾m, ∠BAC = α,则BC ⩾2m sin α 2 .
引理的证明:作∠A的角平分线AD,由正弦定理,有 BC = BD + DC = AB sin ∠BAD sin ∠ADB + AC sin ∠CAD sin ∠ADC = sin α 2 sin ∠ADB (AB + BC) ⩾2m sin α 2 所以引理成立.
回到原题,记m = min 1⩽i<j⩽4 PiPj, k = P 1⩽i (5 + √ 3)m.
(2)若四个点的凸包为三角形,不妨设P4在△P1P2P3内部,且∠P1P4P2 ⩾120◦, 由引理P1P2 ⩾2m sin 1 2∠P1P4P2 ⩾2m sin 60◦= √ 3m.所以k ⩾(5 + √ 3)m.
(3)凸包为四边形P1P2P3P4.若该四边形有一个内角不小于120◦,不妨设为∠P2P1P4, 由引理类似(2),P2P4 ⩾ √ 3m.所以k ⩾(5 + √ 3)m.
否则四个内角均小于120◦,必有两个相邻内角之和不小于180◦, 不妨设α + β = ∠P4P1P2 + ∠P1P2P3 ⩾180◦,且α ⩾β.
由于α < 120◦,所以β > 60◦,所以α+β 4 ⩾45◦, 0 ⩽α−β 4 < 15◦.
由引理,有P2P4 ⩾2m sin α 2 , P1P3 ⩾2m sin β 2 , 59 则 P1P3 + P2P4 ⩾ 2m(sin α 2 + sin β 2 ) = 4m sin α + β 4 cos α −β 4 ⩾ 4m sin 45◦cos 15◦= 2m(sin 60◦+ sin 30◦) = ( √ 3 + 1)m 所以k ⩾(5 + √ 3)m.
又P1P2P3P4为有一个内角为60◦的菱形时,k = (5 + √ 3)m.
所以所求式子的最小值为5 + √ 3.
5.平面上横纵坐标都为有理数的点称为有理点. 证明平面上的全体有理点可以分为三个两两不相交的 集合,满足条件: (1)在以每个有理点为圆心的任一圆内一定包含这三个集合中每个集合的点.
(2)在任意一条直线上不可能有三个点分别属于这三个集合.
证明:显然任意有理点均可唯一的表示成( u w, v w)的形式,其中u, v, w都是整数,w > 0并且gcd(u, v, w) = 1.
令A = {( u w, v w)|2 ∤u}, B = {( u w, v w)|2|u, 2 ∤v}, C = {( u w, v w)|2|u, 2|v}.
下面证明A, B, C满足题中的条件.
(1)先证明A, B, C满足条件(1); 设D是以有理点( u0 w0 , v0 w0 )为中心,r为半径的圆, 取正整数k,使得2k > max{w0, 1 r(|u0| + |v0| + 1)}.
不难验证如下三个有理点 µu02k + 1 w02k , v02k w02k ¶ ∈A, µ u02k w02k , v02k + 1 w02k ¶ ∈B, µ u02k w02k + 1, v02k w02k + 1 ¶ ∈C 都在⊙D内部.这表明条件(1)成立.
(2)再证明A, B, C满足条件(2).
设平面上直线方程为ax + by + c = 0,如果其上有两个不同的有理点(x1, y1)(x2, y2),则axi + byi + c = 0(i = 1, 2), 如果c = 0,当然可以取a, b均为有理数;否则不妨设c = 1,从联立方程中即可解出a, b,显然均 是有理数. 再通分即可使a, b, c都是整数,且gcd(a, b, c) = 1.
设有理点( u w, v w)在直线ax + by + c = 0上,则L : au + bv + cw = 0.
(a)2 ∤c,若2|u, 2|v,必有2|cw, 2|w,与gcd(u, v, w) = 1矛盾.所以L上不能有C中的点.
(b)2|c, 2 ∤b,若2 ∤v,则2 ∤au,从而2 ∤u,因此L上不能有B中的点.
(c)2|c, 2|b,则2|au,由gcd(a, b, c) = 1,2 ∤a,2|u,L上不能有A中的点.
综上所述,A, B, C满足条件(2).
所以A, B, C满足题中的条件.
6.给定实数c, 1 2 < c < 1,求最小的常数M, 使得对任意整数n ⩾2,及实数0 < a1 ⩽a2 ⩽· · · ⩽an,只要满 足 1 n n X k=1 kak = c n X k=1 ak, 总有 n X k=1 ak ⩽M m X k=1 ak, 60 其中m为不超过cn的最大整数.
解:设r = cn, sk = k P i=1 ak, s0 = 0.
∴ n P k=1 kak = nsn −(s1 + s2 + · · · + sn−1) = rsn ∴(n + 1 −r)sn = n P i=1 si (∗) 先证明这样两个事实: (1)对于∀j, k, 1 ⩽j < k ⩽n,总有jsk ⩽ksj.
这是由于jsk ⩽ksj ⇔j(ak + · · · + aj+1) ⩾(k −j)(aj + · · · + a1).
而j(ak + · · · + aj+1) ⩾j(k −j)aj+1 ⩾j(k −j)aj ⩾(k −j)(aj + · · · + a1).
所以jsk ⩽ksj,∴ m−1 P i=1 si ⩽ m−1 P i=1 i msm = m −1 2 sm.
(2)对于k = 0, 1, . . . , n −m, sm+k ⩽ 1 n−m[(n −m −k)sm + ksn].
这是由于sm+k −sm ⩽kam+k,(n −m −k)am+k ⩽sn −sm+k.
∴(n −m −k)(sm+k −sm) ⩽k(n −m −k)am+k ⩽k(sn −sm+k).
化简后即为sm+k ⩽ 1 n−m[(n −m −k)sm + ksn].
∴ n P k=m sk = n−m P k=0 sm+k ⩽ 1 n−m[sm n−m P k=0 (n −m −k) + sn n−m P k=0 k] = n+1−m 2 (sm + sn).
代入(∗),(n + 1 −r)sn ⩽m−1 2 sm + n+1−m 2 (sm + sn) = n 2 sm + n+1−m s n.
∴sn ⩽ n n+1+m−2rsm, 而r < m + 1,所以sn < n n−rsm = 1 1−csm.∴M ⩽ 1 1−c.
令a1 = a2 = · · · = am = 1, am+1 = am+2 = · · · = an = t ⩾1.
可知 n P k=1 kak = m(m+1) 2 + (m+1+n)(n−m) 2 t = cn(m + (n −m)t) = cn n P k=1 ak.
解得t = 2cnm−m(m+1) (n+m+1)(n−m)−2cn(n−m).
所以t ⩾1 ⇔n ⩾ 1 2c−1.满足以上的条件.∵cn −1 < m ⩽cn.
∴M ⩾ sn sm = m + (n −m)t m = 1 + 2cn −m −1 n + m + 1 −2cn = n n + m + 1 −2cn = 1 1 −2c + m+1 n ⩾ 1 1 −c + 1 n 令n →+∞,M ⩾ 1 1−c.
所以M = 1 1−c.
61 第十八届中国数学奥林匹克(2003年) 湖南长沙市第一中学 1.设点I, H分别为锐角三角形的内心和垂心, 点B1, C1分别为边AC, AB的中点.
已知射线B1I交 边AB于点B2(B2 ̸= B),射线C1I交AC的延长线于C2,B2C2与BC相交于K,A1为△BHC的外心.
试证:A, I, A1三点共线的充分必要条件是△BKB2, △CKC2的面积相等.
证明:首先证明A, I, A1三点共线⇔∠BAC = 60◦.
设O为△ABC的外心,连BO, CO,则∠BHC = 180◦−∠BAC,∠BA1C = 2(180◦−∠BHC) = 2∠BAC.
因此,∠BAC = 60◦⇔∠BAC + ∠BA1C = 180◦⇔A1在△ABC的外接圆⊙O上.
⇔AI与AA1重合(因为,A1在BC的中垂线上) ⇔A, I, A1三点共线.
其次,再证S△BKB2 = S△CKC2 ⇔∠BAC = 60◦.
作IP ⊥AB于点P,IQ ⊥AC于点Q,则S△AB1B2 = 1 2IP · AB2 + 1 2IQ · AB1.
注意到S△AB1B2 = 1 2AB1 · AB2 sin A. 所以IP · AB2 + IQ · AB1 = AB1 · AB2 sin A.
设IP = r(r为△ABC的内切圆半径),则IQ = r.令BC = a, AC = b, AB = c,则r = 2S△ABC a + b + c.
再由AB1 = b 2, 2AB1 sin A = hc = 2S△ABC c ,有AB2 · µ2S△ABC c −2 · 2S△ABC a + b + c ¶ = b · 2S△ABC a + b + c.
则AB2 = bc a + b −c,类似的AC2 = bc a + c −b.
因此S△BKB2 = S△CKC2 ⇔S△ABC = S△AB2C2 ⇔bc = bc a + b −c · bc a + c −b ⇔a2 = b2 + c2 −bc ⇔∠BAC = 60◦(由余弦定理).
所以命题成立.
2.求出同时满足如下条件的集合S的元素个数的最大值: (1)S中的每个元素都是不超过100的正整数; (2)对于S中任意两个不同的元素a, b,都存在S中的元素c, 使得a与c的最大公约数等于1,并且b与c的最大 公约数也等于1; (3)对于S中任意两个不同的元素a, b,都存在S中异于a, b的元素d, 使得a与d的最大公约数大于1,并 且b与d的最大公约数也大于1.
解:最大个数为72.
将不超过100的每个正整数n表示成n = 2α13α25α37α411α5q.
其中q是不能被2,3,5,7,11整除的正整数,α1, α2, α3, α4, α5为非负整数.
我们选取满足条件“α1, α2, α3, α4, α5中恰有1个或2个非零”的那些正整数组成集合S, 即S中包括50个 偶数2,4,. . . ,100,但除去2×3×5,22×3×5, 2×32×5,2×3×7,22×3×7,2×5×7,2×3×11这7个数; 3的奇 数倍3×1,3×3,. . . ,3×33共17个数;最小素因子为5的数5×1, 5×5,5×7,5×11,5×13,5×17,5×19共7个 数; 最小素因子为7的数7 × 1,7 × 7,7 × 11,7 × 13共4个数;以及素数11.
从而,S中总共有50 −7 + 17 + 7 + 4 + 1 = 72个数.
下面证明如此构造的S满足题述条件.
62 条件(1)显然满足.
对于条件(2),注意到在lcm(a, b)的素因子中至多出现2,3,5,7,11中的4个数,记某个未出现的数为p,显 然p ∈S,并且gcd(p, a) ⩽gcd(p, lcm(a, b)) = 1, gcd(p, b) ⩽gcd(p, lcm(a, b)) = 1.于是,取c = p即可.
对于条件(3),当gcd(a, b) = 1时,取a的最小素因子p和b的最小素因子q易见p ̸= q,并且p, q ∈{2, 3, 5, 7, 11}.于 是,pq ∈S,并且gcd(pq, a) ⩾p > 1,gcd(pq, b) ⩾q > 1.a, b互质保证了pq异于a, b.从而,取d = pq即可.
当gcd(a, b) = e > 1时,取p为e的最小素因子,q为满足q ∤[a, b]的最小素数,易见p ̸= q,并且p, q ∈ {2, 3, 5, 7, 11}.
于是,pq ∈S,并且gcd(pq, a) ⩾gcd(p, a) = p > 1,gcd(pq, b) ⩾gcd(p, b) = p > 1. q ∤[a, b]保证了pq异 于a, b.从而,取d = pq即可.
下面证明任意满足题述条件的集合S的元素数目不会超过72.
显然,1 ∈S对于任意的两个大于10的质数p, q,因为与p, q均不互质的数最小是pq已大于100, 故据 条件(3)知,10与100之间的21个质数中最多有一个出现在S中.记除1和这21个质数外的其余78个不超 过100的自然数构成集合T. 我们断言T中至少有7个数不在S中,从而S中最多有78 −7 + 1 = 72个元素.
(I)当有某个大于10的质数p属于S时,S中所有各数最小素因子只可能是2,3,5,7和p.运用条件(2)可得出 以下结论: (i)若7p ∈S,因为2×3×5, 22×3×5, 2×32×5与7p包括了所有的最小素因子, 故由条件(2)知,2×3×5, 22× 3 × 5, 2 × 32 × 5 ̸∈S;若7p ̸∈S,注意2 × 7p > 100,而p ∈S,故由条件(3)知7 × 1, 7 × 7, 7 × 11, 7 × 13 ̸∈S.
(ii)若5p ∈S,则2 × 3 × 7, 22 × 3 × 7 ̸∈S;若5p ̸∈S,则5 × 1, 5 × 5 ̸∈S.
(iii)2 × 5 × 7,3p不同属于S.
(iv)2 × 3p,5 × 7不同属于S.
(v)若5p, 7p ̸∈S,则5 × 7 ̸∈S.
当p = 11或13时,由(i)(ii)(iii)(iv)可分别得出至少有3,2,1,1个T中的数不属于S,合计7个; 当p = 17或19时,由(i)(ii)(iii)可分别得出至少有4,2,1个T中的数不属于S,合计7个; 当p > 20时,由(i)(ii)(v)可分别得出至少有4,2,1个中T的数不属于S,合计7个.
(II)如果没有大于10的质数属于S,则S中的最小素因子只可能是2,3,5,7.于是,下面7对数中的每对都不能 同时在S中出现: (3, 2 × 5 × 7),(5, 2 × 3 × 7),(7, 2 × 3 × 5),(2 × 3, 5 × 7),(2 × 5, 3 × 7), (2 × 7, 3 × 5),(22 × 7, 32 × 5).
从而,T中至少有7个数不在S中.
综上所述,本题的答案为72.
3.给定正整数n,求最小的正数λ,使得对于任何θi ∈(0, π 2 ), (i = 1, 2, . . . , n) 只要 tan θ1 tan θ2 · · · tan θn = 2 n 2 就有cos θ1 + cos θ2 + · · · + cos θn不大于λ.
解:当n = 1时,cos θ1 = (1 + tan2 θ1)−1 2 = √ 3 3 ,有λ = √ 3 3 .
当n = 2时,可以证明cos θ1 + cos θ2 ⩽2 √ 3 3 , 并且当θ1 = θ2 = arctan √ 2时等号成立.事实上, cos θ1 + cos θ2 ⩽2 √ 3 3 ⇔cos2 θ1 + cos2 θ2 + 2 cos θ1 cos θ2 ⩽4 3, 63 即 1 1 + tan2 θ1 + 1 1 + tan2 θ2 + 2 s 1 (1 + tan2 θ1)(1 + tan2 θ1) ⩽4 3 由tan θ1 tan θ2 = 2,并且设x = tan2 θ1 + tan2 θ2,则只需证明 2 + x 5 + x + 2 r 1 5 + x ⩽4 3 即 2 r 1 5 + x ⩽14 + x 3(5 + x) ⇔36(5 + x) ⩽196 + 28x + x2 ⇔x2 −8x + 16 = (x −4)2 ⩾0 显然成立,于是λ = 2 √ 3 3 .
当n ⩾3时,不妨设θ1 ⩾θ2 ⩾· · · ⩾θn, 则tan θ1 tan θ2 tan θ3 ⩾2 √ 2.
由于cos θi = p 1 −sin2 θi < 1 −1 2 sin2 θi,则 cos θ2 + cos θ3 < 2 −1 2(sin2 θ2 + sin2 θ3) < 2 −sin θ2 sin θ3.
由tan2 θ1 ⩾ 8 tan2 θ2 tan2 θ3 ,有 1 cos2 θ1 ⩾8 + tan2 θ2 tan2 θ3 tan2 θ2 tan2 θ3 即 cos θ1 ⩽ tan θ2 tan θ3 p 8 + tan2 θ2 tan2 θ3 = sin θ2 sin θ3 p 8 cos2 θ2 cos2 θ3 + sin2 θ2 sin2 θ3 于是 cos θ1 + cos θ2 + cos θ3 < 2 −sin θ2 sin θ3 Ã 1 − 1 p 8 cos2 θ2 cos2 θ3 + sin2 θ2 sin2 θ3 !
注意到8 cos2 θ2 cos2 θ3 + sin2 θ2 sin2 θ3 ⩾1 ⇔8 + tan2 θ2 tan2 θ3 ⩾ 1 cos2 θ2 cos2 θ3 = (1 + tan2 θ2)(1 + tan2 θ3) ⇔tan2 θ2 + tan2 θ3 ⩽7.
所以当tan2 θ2 + tan2 θ3 ⩽7时,cos θ1 + cos θ2 + cos θ3 < 2.
若tan2 θ2 + tan2 θ3 > 7时,因此tan2 θ1 ⩾tan2 θ2 > 7 2, 所以cos θ1 ⩽cos θ2 < s 1 1 + 7 2 = √ 2 3 .
于是,cos θ1 + cos θ2 + cos θ3 < 2 √ 2 3 + 1 < 2, 所以cos θ1 + cos θ2 + · · · + cos θn < n −1.
另一方面,取θ2 = θ3 = · · · = θn = α > 0, α →0, 则θ1 = arctan 2 n 2 tann−1 α,θ1 →π 2 , 从而cos θ1 + cos θ2 + · · · + cos θn →n −1.
综上可得λ = n −1.
4.求所有满足a ⩾2, m ⩾2的三元正整数组(a, m, n), 使得an + 203是am + 1的倍数.
解:对于分三种情况考虑.
(1)n < m时,由an + 203 ⩾am + 1,有202 ⩾am −an ⩾an(a −1) ⩾a(a −1).
所以2 ⩽a ⩽14.
64 当a = 2时,n可取1, 2, . . . , 7;当a = 3时,n可取1, 2, 3, 4;当a = 4时,n可取1, 2, 3; 当a = 5, 6时,n可取1, 2; 当a = 7, 8, . . . , 14时,n = 1.
由am + 1|an + 203可知,解为(2,2,1),(2,3,2)和(5,2,1).
(2)n = m时,am + 1|202.由于202仅有1,2,101,202四个约数,而a ⩾2, m ⩾2,只有am = 100,解为(10,2,2).
(3)n > m时,由am + 1|203(am + 1),有am + 1|an + 203 −203(am + 1),即am + 1|am(an−m −203).
又因为gcd(am + 1, am) = 1,所以am + 1|an−m −203.
(i)若an−m < 203,则令n −m = s ⩾1,有am + 1|203 −as.所以203 −as ⩾am + 1, 有202 ⩾as + am ⩾am + a = a(am−1 + 1) ⩾a(a + 1),所以2 ⩽a ⩽13.
类似(1)的讨论,可知(a, m, s)的解为:(2,2,3),(2,6,3),(2,4,4),(2,3,5),(2,2,7),(3,2,1),(4,2,2),(5,2,3),(8,2,1).
所以(a, m, n)为:(2,2,5),(2,6,9),(2,4,8),(2,3,8),(2,2,9),(3,2,3),(4,2,4),(5,2,5),(8,2,3).
(ii)an−m = 203时,则a = 203, n −m = 1,即解为(203, m, m + 1), m ⩾2.
(iii)an−m > 203时,令n −m = s ⩾1,则am + 1|as −203.
又as −203 ⩾am +1,则s > m,由am +1|as −203+203(am +1) = am(as−m +203) = am(an−2m +203), 又因为gcd(am + 1, am) = 1,所以am + 1|an−2m + 203.
又因为s > m ⇔n −m > m ⇔n > 2m ⇔n −2m > 0.此时的解只能由前面的解派生出来,即 由(a, m, n) →(a, m, n+2m) →· · · →(a, m, n+2km), 并且每一个派生出来的解都满足am+1|an+203.
综上所述,所有解(a, m, n)为: (2, 2, 4k + 1), (2, 3, 6k + 2), (2, 4, 8k + 8), (2, 6, 12k + 9), (3, 2, 4k + 3), (4, 2, 4k + 4), (5, 2, 4k + 1), (8, 2, 4k + 3), (10, 2, 4k + 2), (203, m, (2k + 1)m + 1), 其中k为任意非负整数,且m ⩾2为整数.
5.某公司需要录用一名秘书,共有10人报名, 公司经理决定按照求职报名的顺序逐个面试,前三个人面试 后一定不录用. 自第4个人开始将他与前面面试过的人比较, 如果他的能力超过了前面所有已面试过的 人, 就录用他;否则就不录用,继续面试下一个. 如果前9个人都不录用,那么就录用最后一个面试的人.
假定这10个人的能力各不相同,可以按能力由强到弱排为第1,第2,. . .,第10. 显然该公司到底录用到哪一 个人,与这10个人报名的顺序有关. 大家知道,这样的排列共有10!种. 我们以Ak表示能力第k的人能够被 录用的不同报名顺序的数目, 以表示他被录用的可能性.
证明:在该公司经理的方针下,有 (1)A1 > A2 > · · · > A8 = A9 = A10; (2)该公司有超过70%的可能性录用到能力最强的3个人之一, 而只有不超过10%的可能性录用到能力最 弱的3个人之一.
证明:将前3个面试者中能力最强的排名名次记为a.显然a ⩽8. 将此时能力排名第k的人被选上的排列 集合记作Ak(a),相应的排列数目记作|Ak(a)|.在以下过程中,“:=”表示“记为”.
(1)显然,a = 1时,必然录取最后一个面试的人,此时除能力第1的人之外,各人机会均等,不难知道 |Ak(1)| = 3 × 8! := r1, k = 2, 3, . . . , 10.
当2 ⩽a ⩽8时,对于a ⩽k ⩽10,能力排名的k的人无录用机会.对于1 ⩽k < a,此时机会均等.
65 事实上,此时能力排名第a的人排在前三个,有3种选择位置.而能力排名1到a −1的人都排在后7个位置 上, 并且谁位于他们之首谁就被录用,有排法 ¡ 7 a−1 ¢ (a −2)!种,其余10 −a人可以在剩下的位置上任意排 列,有(10 −a)!种排法.故有 |Ak(a)| = 3 ¡ 7 a−1 ¢ (a −2)!(10 −a)! := ra, k = 1, . . . , a −1 0, k = a, . . . , 10.
上述结果表明: A8 = A9 = A10 = r1 = 3 × 8! > 0; Ak = r1 + rk+1 + · · · + r8, k = 2, . . . , 7; A1 = r2 + r3 + · · · + r8.
显然A2 > A3 > · · · > A8 = A9 = A10 > 0.而A1 −A2 = r2 −r1 = 3 × 7 × 8! −3 × 8! > 0.
综上所述,问题(1)获证.
(2)由(1)可知 A8 + A9 + A10 10!
= 3r1 10! = 3 × 3 × 8!
10!
= 10% 所以,录用到能力最弱的三人之一的可能性等于10%.
并且 A1 = 8 X a=2 ra = 8 X a=2 3 µ 7 a −1 ¶ (a −2)!(10 −a)!
= 3 × 7!
8 X a=2 (9 −a)(10 −a) a −1 = 3 × 7! × µ 56 + 21 + 10 + 5 + 12 5 + 1 + 2 7 ¶ = 3 × 7! × 9524 35 > 3 × 7! × 952 3 = 287 × 7!
A2 = r1 + 8 X a=3 ra = 3 × 8! + 3 × 7! × µ 21 + 10 + 5 + 12 5 + 1 + 2 7 ¶ = 3 × 7! × 4724 35 > 3 × 7! × 472 3 = 143 × 7!
A3 = r1 + 8 X a=4 ra = 3 × 8! + 3 × 7! × µ 10 + 5 + 12 5 + 1 + 2 7 ¶ = 3 × 7! × 2624 35 > 3 × 7! × 262 3 = 80 × 7!
所以A1 + A2 + A3 10!
287 + 143 + 80 720 = 17 24 > 70%.
即录用到能力最强的三人之一的可能性大于70%.
66 6.设a, b, c, d为正实数,满足ab + cd = 1; 点Pi(xi, yi)(i = 1, 2, 3, 4)是以原点为圆心的单位圆上的四个点.
求证: (ay1 + by2 + cy3 + dy4)2 + (ax4 + bx3 + cx2 + dx1)2 ⩽2(a2 + b2 ab + c2 + d2 cd ).
证明: ∵(ab + cd)(a2 + b2 ab + c2 + d2 cd ) = a2 + b2 + c2 + d2 + abc d + abd c + acd b + bcd a 是关于a, b, c, d对称的式子, ∴(ab + cd)(a2 + b2 ab + c2 + d2 cd ) = (ad + bc)(a2 + d2 ad + b2 + c2 bc ) 由Cauchy不等式(ay1 + by2 + cy3 + dy4)2 ⩽(ad + bc)((ay1 + dy4)2 ad + (by2 + cy3)2 bc ) (ax4 + bx3 + cx2 + dx1)2 ⩽(ad + bc)((ax4 + dx1)2 ad + (bx3 + cx2)2 bc ) (ay1 + dy4)2 ⩽(a2 + d2)(y2 1 + y2 4), (ax4 + dx1)2 ⩽(a2 + d2)(x2 4 + x2 1) (by2 + cy3)2 ⩽(b2 + c2)(y2 2 + y2 3), (bx3 + cx2)2 ⩽(b2 + c2)(x2 3 + x2 2) 而x2 i + y2 i = 1(i = 1, 2, 3, 4), ab + cd = 1.所以 (ay1 + by2 + cy3 + dy4)2 + (ax4 + bx3 + cx2 + dx1)2 ⩽ (ad + bc)(2(a2 + d2) ad + 2(b2 + c2) bc ) = 2(ad + bc)(a2 + d2 ad + b2 + c2 bc ) = 2(ab + cd)(a2 + b2 ab + c2 + d2 cd ) = 2(a2 + b2 ab + c2 + d2 cd ) 67 第十九届中国数学奥林匹克(2004年) 澳门教育暨青年局 1.凸四边形EFGH的顶点E, F, G, H分别在凸四边形ABCD的边AB, BC, CD, DA上, 满足AE EB · BF F C · CG GD · DH HA = 1, 而点A, B, C, D分别在凸四边形E1F1G1H1的边E1F1,F1G1, G1H1,H1E1上,满足E1F1 ∥ EF, F1G1 ∥FG, G1H1 ∥GH, H1E1 ∥HE.
已知E1A AH1 = λ,求F1C CG1 的值.
解:(1)若EF ∥AC,则BE EA = BF F C .
代入已知条件,可以得到DH HA = DG GC ,所以HG ∥AC,从而E1F1 ∥AC ∥H1G1,故F1C CG1 = E1A AH1 = λ.
(2)若EF与AC不平行,设FE的延长线与CA的延长线相交于点T,则由Menelaus定理得 CF FB · BE EA · AT TC = 1 结合题设有 CG GD · DH HA · AT TC = 1 由Menelaus定理逆定理可以知道,T, H, G三点共线,设TF, TG与E1H1分别交于点M, N.
由E1B ∥EF,得E1A = BA EA · AM,同理H1A = AD AH · AN,所以 E1A H1A = AM AN · AB AE · AH AD 又因为 EQ QH = S△AEC S△AHC = S△ABC · AE · AD S△ADC · AB · AH 所以 E1A AH1 = EQ QH · AB AE · AH AD = S△ABC S△ADC 同理 F1C CG1 = S△ABC S△ADC 所以F1C CG1 = E1A AH1 = λ.
2.已知正整数c,设数列x1, x2, . . .满足: x1 = c, xn = xn−1 + ·2xn−1 −(n + 2) n ¸ + 1(n = 2, 3, . . .) 其中[x]表示不大于x的最大整数.求数列{xn}的通项公式.
解:显然当n ⩾2时 xn = xn−1 + ·2(xn−1 −1) n ¸ 令an = xn −1,则a1 = c −1, an = an−1 + ·2an−1 n ¸ = ·n + 2 n an−1 ¸ n = 2, 3, . . .
(1) 设un = A (n+1)(n+2) 2 , n = 1, 2, . . . ,其中A为非负整数.由于当n ⩽2时,有 ·n + 2 n un−1 ¸ = · An + 2 2n n(n + 1) ¸ = A(n + 1)(n + 2) 2 = un 68 所以数列{un}满足(1).
设yn = n, n = 1, 2, . . ..当n ⩽2时,有 ·n + 2 n yn−1 ¸ = ·(n + 2)(n −1) n ¸ = · n + 1 −2 n ¸ = n = yn 所以数列{yn}满足(1).
设zn = h (n+2)2 4 i , n = 1, 2, . . ..当n = 2m(m ⩾1)时,有 ·n + 2 n zn−1 ¸ = ·m + 1 m ·(2m + 1)2 4 ¸¸ = ·m + 1 m m(m + 1) ¸ = (m + 1)2 = zn 当n = 2m + 1(m ⩾1)时,有 ·n + 2 n zn−1 ¸ = ·2m + 3 2m + 1 ·(2m + 2)2 4 ¸¸ = ·2m + 3 2m + 1(m + 1)2 ¸ = (m + 1)(m + 2) = zn 所以数列{zn}满足(1).
对于任意非负整数A,令vn = un + yn = A (n+1)(n+2) 2 + n, wn = un + zn = A (n+1)(n+2) 2 + h (n+2)2 4 i , n = 1, 2, . . ., 显然{vn}, {wn}都满足(1).
由于u1 = 3A, y1 = 1, z1 = 2,所以当3|a1时,an = a1 6 (n + 1)(n + 2); 当a1 ≡1 (mod 3)时,an = a1−1 6 (n + 1)(n + 2) + n; 当a1 ≡2 (mod 3)时,an = a1−2 6 (n + 1)(n + 2) + h (n+2)2 4 i . 综上可得 xn = c−1 6 (n + 1)(n + 2) + 1, 当c ≡1 (mod 3), xn = c−2 6 (n + 1)(n + 2) + n + 1, 当c ≡2 (mod 3), xn = c−3 6 (n + 1)(n + 2) + h (n+2)2 4 i + 1, 当c ≡0 (mod 3).
3.设M是平面上n个点组成的集合,满足: (1)M中存在7个点,是一个凸七边形的7个顶点; (2)M中任意5个点,若这5个点是一个凸五边形的5个顶点,则此凸五边形内部至少含有M中的一个点.
求n的最小值.
解:先证n ⩾11.
设顶点在M中的一个凸七边形为A1A2A3A4A5A6A7,连A1A5,由条件(2)知, 在凸五边形A1A2A3A4A5中 至少有M中的一个点,记为P1;连P1A1, P1A5, 则在凸五边形A1P1A5A6A7内部至少有M中的一个点,记 为P2,且P2异于P1; 连直线P1P2,A1, A2, . . . , A7至少有5个点不在直线P1P2上,有抽屉原则知, 在直 线P1P2的某一侧必有其中3个顶点,这3个顶点与点P1, P2构成的凸五边形内至少含有M中的一个点P3, 并且P3异于P1, P2.
再作直线P1P3, P2P3,令直线P1P2对应区域为π3:它是以直线P1P2为边界且在三角形P1P2P3异侧的一个 半平面(不含直线P1P2),类似定义区域π1, π2.这样3个区域π1, π2, π3覆盖了平面上除三角形P1P2P3之外 的所有点. 由抽屉原则,A1, A2, . . . , A7中必有3个在同一区域内,不妨设为π3. 这三个点与P1, P2构成的 凸五边形内至少含有M中的一个点P4,并且P4异于P1, P2, P3.所以n ⩾11.
构造一个例子,在Oxy平面上,取整点A1(0, 1), A2(1, 3), A3(2, 3), A4(3, 2), A5(3, 1), A6(2, 0), A7(1, 0) 构 69 成一个凸七边形,再加上其内部的所有四个整点(1,1),(1,2),(2,1),(2,2)构成点集M,显然满足条件(1).
下面证明M也满足条件(2),若不然,假设存在一个整点凸五边形,其内部不含整点, 显然所有整点多边 形的面积均是1 2的整数倍.必存在一个面积最小的内部不含整点的整点凸五边形ABCDE. 考虑顶 点坐标的奇偶性,只有四种情况:(奇,偶)(偶,奇)(偶,偶)(奇,奇).从而这五个顶点中必有两个顶点的坐标 奇偶性完全相同, 于是它们连线中点P也是整点,又因为它不在五边形内部,必然在某条边上,不妨设 在AB上,则P为AB中点,连PE, 则PBCDE是面积更小的内部不含整点的整点凸五边形,矛盾.
综上所述,n的最小值为11.
4.给定实数a和正整数n,求证: (1)存在唯一的实数数列x0, x1, . . . , xn+1满足: x0 = xn+1 = 0, 1 2(xi+1 + xi−1) = xi + x3 i −a3(i = 1, 2, . . . , n) (2)(1)中的数列x0, x1, . . . , xn+1满足|xi| ⩽|a|.
解:(1)存在性:由xi+1 = 2xi + 2x3 i −2a3 −xi−1, i = 1, 2, . . . , n,及x0 = 0, 我们知道xi是x1的3i−1次实系 数多项式,从而xn+1为x1的3n次实系数多项式, 由3n为奇数,故存在x1,使得xn+1 = 0,由此x1及x0 = 0即 可求出xi, 如此得到的数列x0, x1, . . . , xn+1满足题设条件.
唯一性:设w0, w1, . . . , wn+1以及v0, v1, . . . , vn+1为满足条件的两个数列,则 1 2(wi+1 + wi−1) = wi + w3 i −a3, 1 2(vi+1 + vi−1) = vi + v3 i −a3.
所以1 2(wi+1 −vi+1 + wi−1 −vi−1) = (wi −vi)(1 + w2 i + wivi + v2 i ).
设|wi0 −vi0|最大,则 |wi0 −vi0| ⩽|wi0 −vi0|(1 + w2 i0 + wi0vi0 + v2 i0) ⩽1 2|wi0+1 −vi0+1| + 1 2|wi0−1 −vi0−1| ⩽|wi0 −vi0| 从而|wi0 −vi0| = 0或者1 + w2 i0 + wi0vi0 + v2 i0 = 1, 由后一种情况可以推出w2 i0 + v2 i0 + (wi0 + vi0)2 = 0,wi0 = vi0 = 0.
所以总有|wi0 −vi0| = 0,再由|wi0 −vi0|最大, 所以所有|wi −vi| = 0,即wi = vi, i = 1, 2, . . . , n.唯一性得证.
(2)设|xi0|最大,则 |xi0| + |xi0|3 = |xi0|(1 + x2 i0) = |1 2(xi0+1 + xi0−1) + a3| ⩽1 2|xi0+1| + 1 2|xi0−1| + |a|3 ⩽|xi0| + |a|3 因此|xi0| ⩽|a|,所以|xi| ⩽|a|, i = 0, 1, 2 . . . , n + 1.
5. 给定正整数n ⩾2,设正整数ai(i = 1, 2, . . . , n)满足:a1 < a2 < · · · < an以及 n P i=1 1 ai ⩽1.求证:对任意 实数x,有 Ã n X i=1 1 a2 i + x2 !2 ⩽1 2 · 1 a1(a1 −1) + x2 70 解:当x2 ⩾a1(a1 −1)时,由 n P i=1 1 ai ⩽1,可得 Ã n X i=1 1 a2 i + x2 !2 ⩽ Ã n X i=1 1 2ai|x| !2 = 1 4x2 Ã n X i=1 1 ai !2 ⩽ 1 4x2 ⩽1 2 · 1 a1(a1 −1) + x2 当x2 < a1(a1 −1)时,由Cauchy不等式, Ã n X i=1 1 a2 i + x2 !2 ⩽ Ã n X i=1 1 ai ! Ã n X i=1 ai (a2 i + x2)2 !
⩽ n X i=1 ai (a2 i + x2)2 对于正整数a1 < a2 < · · · < an,有ai+1 ⩾ai + 1, i = 1, 2, . . . , n −1,并且 2ai (a2 i + x2)2 ⩽ 2ai ¡ a2 i + x2 + 1 4 ¢2 −a2 i = 2ai ³¡ ai −1 2 ¢2 + x2 ´ ³¡ ai + 1 2 ¢2 + x2 ´ = 1 ³¡ ai −1 2 ¢2 + x2 ´ − 1 ³¡ ai + 1 2 ¢2 + x2 ´ ⩽ 1 ³¡ ai −1 2 ¢2 + x2 ´ − 1 ³¡ ai+1 −1 2 ¢2 + x2 ´, i = 1, 2, . . . , n −1 所以 n X i=1 ai (a2 i + x2)2 ⩽ 1 2 n X i=1 1 ³¡ ai −1 2 ¢2 + x2 ´ − 1 ³¡ ai+1 −1 2 ¢2 + x2 ´ ⩽ 1 2 · 1 ¡ a1 −1 2 ¢2 + x2 ⩽ 1 2 · 1 a1(a1 −1) + x2 6.证明:除了有限个正整数外,其他的正整数n均可表示为2004个正整数之和n = a1 + a2 + · · · + a2004,且 满足:1 ⩽a1 < a2 < · · · < a2004, ai|ai+1 (i = 1, 2, . . . , 2003) 解:将证明如下更一般的结论: 对任给的正整数r ⩾2,总存在N(r),当n ⩾N(r)时,存在正整数a1, a2, . . . , ar,使得n = a1 + a2 + . . . + ar, 1 ⩽a1 < a2 < · · · < ar,ai|ai+1, i = 1, 2, . . . , r −1.
当r = 2时,有n = 1 + n −1,取N(2) = 3即可.
假设当r = k时,结论成立,当r = k + 1时,取N(k + 1) = 4N(k)3.
设n = 2α(2l + 1),如果n ⩾N(k + 1) = 4N(k)3,则2α ⩾2N(k)2或者2l + 1 > 2N(k).
若2α ⩾2N(k)2,则存在正偶数2t ⩽α,使22t ⩾N(k)2.
此时2t + 1 ⩾N(k),由归纳假设存在正整数b1, b2, . . . , bk,使得2t + 1 = b1 + b2 + · · · + bk, 其中1 ⩽b1 < · · · < bk, bi|bi+1, i = 1, . . . , k −1.
这样 71 2α = 2α−2t · 22t = 2α−2t[1 + (2t −1)(2t + 1)] = 2α−2t + 2α−2t(2t −1)b1 + 2α−2t(2t −1)b2 + · · · + 2α−2t(2t −1)bk n = 2α−2t(2l + 1) + 2α−2t(2t −1)b1(2l + 1) + 2α−2t(2t −1)b2(2l + 1) + · · · +2α−2t(2t −1)bk(2l + 1) 若2l + 1 > 2N(k),则l ⩾N(k),由归纳假设存在实数c1, c2, . . . , ck,使得l = c1 + c2 + · · · + ck, 其中1 ⩽c1 < · · · < ck, ci|ci+1, i = 1, . . . , k −1.
因此n = 2α + 2α+1c1 + 2α+1c2 + · · · + 2α+1cn.满足要求.
由归纳法知上述一般结论对所有的r ⩾2成立.令r = 2004,显然有原命题成立.
72 第二十届中国数学奥林匹克(2005年) 郑州郑州外国语学校 1.设θi ∈(−π 2 , π 2 ), i = 1, 2, 3, 4. 证明:存在x ∈R,使得如下两个不等式 cos2 θ1 cos2 θ2 −(sin θ1 sin θ2 −x)2 ⩾0 cos2 θ3 cos2 θ4 −(sin θ3 sin θ4 −x)2 ⩾0 同时成立的充要条件是: 4 X i=1 sin2 θi ⩽2(1 + 4 Y i=1 sin θi + 4 Y i=1 cos θi).
(1) 证明:显然所给的两个不等式分别等价于 sin θ1 sin θ2 −cos θ1 cos θ2 ⩽x ⩽sin θ1 sin θ2 + cos θ1 cos θ2 (2) sin θ3 sin θ4 −cos θ3 cos θ4 ⩽x ⩽sin θ3 sin θ4 + cos θ3 cos θ4 (3) 不难知道,存在x ∈R,使得(2)(3)同时成立的充分必要条件为 sin θ1 sin θ2 + cos θ1 cos θ2 −sin θ3 sin θ4 + cos θ3 cos θ4 ⩾0 (4) sin θ3 sin θ4 + cos θ3 cos θ4 −sin θ1 sin θ2 + cos θ1 cos θ2 ⩾0 (5) 另一方面,利用sin2 α = 1 −cos2 α,可将(1)化为 cos2 θ1 cos2 θ2 + 2 cos θ1 cos θ2 cos θ3 cos θ4 + cos2 θ3 cos2 θ4 − sin2 θ1 sin2 θ2 + 2 sin θ1 sin θ2 sin θ3 sin θ4 −sin2 θ3 sin2 θ4 ⩾0 即 (cos θ1 cos θ2 + cos θ3 cos θ4)2 −(sin θ1 sin θ2 −sin θ3 sin θ4)2 ⩾0 亦即 (sin θ1 sin θ2 + cos θ1 cos θ2 −sin θ3 sin θ4 + cos θ3 cos θ4) ·(sin θ3 sin θ4 + cos θ3 cos θ4 −sin θ1 sin θ2 + cos θ1 cos θ2) ⩾0 (6) 当存在x ∈R,使得(2)(3)同时成立时,由(4)(5)即可推出(6),从而(1)成立.
反之,若(1)成立,即(6)成立,如果(4)(5)不成立,必有 sin θ1 sin θ2 + cos θ1 cos θ2 −sin θ3 sin θ4 + cos θ3 cos θ4 < 0 sin θ3 sin θ4 + cos θ3 cos θ4 −sin θ1 sin θ2 + cos θ1 cos θ2 < 0 两式相加,得到2(cos θ1 cos θ2 + cos θ3 cos θ4) < 0.
这与θi ∈(−π 2 , π 2 ), i = 1, 2, 3, 4 矛盾,所以必有(4)(5)成立,因此存在x ∈R,使得(2)(3)同时成立,证毕.
73 2.一圆与△ABC的三边BC, CA, AB的交点依次为D1, D2; E1, E2; F1, F2. 线段D1E1与D2F2交于点L, 线段E1F1与E2D2交于点M, 线段F1D1与F2E2交于点N, 证明:AL, BM, CN三线共点.
证明:自点L作AB和AC的垂线,垂足分别为L′, L′′.记∠LAB = α1, ∠LAC = α2, ∠LF2A = α3, ∠LE1A = α4.则有 sin α1 sin α2 = LL′ LL′′ = LF2 sin α3 LE1 sin α4 连接D1F2, D2E1,由△LD1F2 ∽△LD2E1,得LF2 LE1 = D1F2 D2E1 .
连接D2F1, D1E2,由正弦定理得sin α3 sin α4 = D2F1 D1E2 .
所以 sin α1 sin α2 = D1F2 D2E1 · D2F1 D1E2 同理,记∠MBC = β1, ∠MBA = β2, ∠NCA = γ1, ∠NCB = γ2,可得 sin β1 sin β2 = E1D2 E2F1 · E2D1 E1F2 sin γ1 sin γ2 = F1E2 F2D1 · F2E1 F1D2 三式相乘,得 sin α1 sin α2 · sin β1 sin β2 · sin γ1 sin γ2 = 1 由Ceva定理逆定理的角元形式,AL, BM, CN三线共点.
3.如图所示,圆形的水池被分割为2n(n ⩾5)“格子”. 我们把有公共隔墙(公共边或公共弧)的格子称为 相邻的, 从而每个格子都有三个邻格.
水池中一共跳入了4n + 1只青蛙,青蛙难于安静共处, 只要某个格子中有不少于3只青蛙, 那么迟早一定 会有其中3只分别同时跳往三个不同邻格.
证明:只要经过一段时间之后,青蛙便会在水池中大致分布均匀.
注:所谓大致分布均匀,就是任取其中一个格子, 或者它里面有青蛙, 或者它的三个邻格里都有青蛙.
证明:我们把一个格子中出现一次3只青蛙同时分别跳向三个邻格的事件称为该格子发生一次爆发. 而 把一个格子或者是它里面有青蛙,或者是它的三个相邻的格子里面都有青蛙,称为该格子处于平衡状态.
容易看出,一个格子只要一旦有青蛙跳入,那么它就一直处于平衡状态.事实上,只要不爆发,那么该格子 中的青蛙不会动, 它当然处于平衡状态;而如果发生爆发,那么它的三个邻格中就都有青蛙,并且只要三 个邻格都不爆发,它就一直处于平衡状态; 而不论哪个邻格发生爆发,都会有青蛙跳到它里面,它也一直 处于平衡状态.
这样一来,为了证明题中断言,我们就只要证明:任何一个格子都迟早会有青蛙跳入.
任取一个格子,把它称为格A,把它所在的扇形称为1号扇形,把该扇形中另一个格子称为格B,我们要证明 格A中迟早会有青蛙跳入.
按顺时针方向依次将其余扇形接着编为2至n号.首先证明1号扇形中迟早会有青蛙跳入.假设1号扇形 中永无青蛙到来, 那么就不会有青蛙越过1号扇形与n号扇形之间的隔墙.我们来考察青蛙所在的扇形 74 的编号的平方和,由于没有青蛙进入1号扇形所以只能是有3只青蛙由某个k(3 ⩽k ⩽n)号扇形分别跳 入k −1, k和k + 1号扇形各一只.因此平方和的变化量为(k −1)2 + k2 + (k + 1)2 −3k2 = 2. 即增加2.一 方面,由于青蛙的跳动不会停止(因为总有一个格子里有不少于3只青蛙),所以平方和的增加趋势不会停 止; 但是另一方面,青蛙所在扇形的标号的平方和不可能永无止境的增加下去(不会大于(4n + 1)n2),由 此产生矛盾, 所以迟早会有青蛙越过1号扇形与n号扇形之间的隔墙,进入1号扇形.
我们再来证明1号扇形迟早会有3只青蛙跳入,如果1号扇形中至多有两只青蛙跳入,那么它们都不会跳 走, 并且自始至终上述平方和至多有两次变少(只能在两只青蛙越过1号扇形与n号扇形之间的隔墙时变 小), 以后便一直持续不断的上升,从而又重蹈刚才的矛盾,所以1号扇形中迟早会有三只青蛙跳入.
如果这3只青蛙中有位于格A的,那么格A中已经有青蛙跳入;如果这3只青蛙全都位于格B,那么格B会发 生爆发,从而有青蛙跳入格A.
4.已知数列{an}满足条件a1 = 21 16,及 2an −3an−1 = 3 2n+1 , n ⩾2.
设m为正整数,m ⩾2.证明:当n ⩽m时,有 µ an + 3 2n+3 ¶ 1 m ³ m − µ2 3 ¶ n(m−1) m ´ < m2 −1 m −n + 1.
证明:令bn = an + 3 2n+3 .则 2 µ bn − 3 2n+3 ¶ −3 µ bn−1 − 3 2n+2 ¶ = 3 2n+1 所以bn = 3 2bn−1.而b1 = 3 2,所以bn = µ3 2 ¶n .
所以只需证明: µ3 2 ¶ n m ³ m − µ2 3 ¶ n(m−1) m ´ < m2 −1 m −n + 1.
即只需证明 µ 1 − n m + 1 ¶ µ3 2 ¶ n m ³ m − µ2 3 ¶ n(m−1) m ´ < m −1 由Bernoulli不等式: 1 − n m + 1 < µ 1 − 1 m + 1 ¶n 所以 µ 1 − n m + 1 ¶m < µ 1 − 1 m + 1 ¶mn = µ m m + 1 ¶mn = · 1 (1 + 1 m)m ¸n 由于m ⩾2,根据二项式定理可得 µ 1 + 1 m ¶m ⩾1 + µm 1 ¶ 1 m + µm 2 ¶ 1 m2 = 5 2 −1 2m ⩾9 4 所以 µ 1 − n m + 1 ¶m < µ4 9 ¶n 75 1 − n m + 1 < µ2 3 ¶ 2n m 所以只需证明 µ2 3 ¶ 2n m µ3 2 ¶ n m ³ m − µ2 3 ¶ n(m−1) m ´ < m −1 即 µ2 3 ¶ n m ³ m − µ2 3 ¶ n(m−1) m ´ < m −1 记t = ¡ 2 3 ¢ n m ,则0 < t < 1,只需证明t(m −tm−1) < m −1.
即(t −1)[m −(tm−1 + tm−2 + · · · + 1)] < 0,此不等式显然成立,从而原不等式成立.
5.在面积为1的矩形ABCD中(包括边界)有5个点, 其中任意三点不共线.求以这5个点为顶点的所有三 角形中, 面积不大于1 4的三角形的个数的最小值.
解:本题证明需要用到如下的常用结论,我们将其作为一个引理:矩形内任意一个三角形的面积不大于矩 形面积的一半.
在矩形ABCD中,如果某三点构成的三角形的面积不大于1 4,就称它们为一个好的三点组,简称为好组.
记AB, CD, BC, AD的中点分别为E, F, H, G,线段EF与GH的交点记为O.线段EF和GH将矩形ABCD 分成四个小矩形, 从而一定存在一个小矩形,不妨设为AEOG,其中(包括边界,下同)至少有所给5个点中 的两个点,设点M, N在矩形AEOG中.
(1)如果矩形OHCF中有不多于1个已知点,考察不在矩形OHCF中的任意一个不同于M和N的已知 点X,显然, 三点组(M, N, X)或者在矩形ABGH中,或者在矩形AEFD中.由引理可知(M, N, X)是好组, 由于这样的点至少有两个,所以至少存在两个好组.
(2)如果矩形OHCF中有至少2个已知点,不妨设点P, Q都在矩形OHCF中,考察剩下来的最后一个已知 点R. 如果R在矩形OFDG中,则三点组(M, N, R)在矩形AEFD中,三点组(P, Q, R)在矩形GHCD中,从 而它们都是好组, 于是至少有两个好组.同理,如果点R在矩形EBHO中,同样至少有两个好组.
如果点R在矩形EBHO或者AEOG中,不妨设在矩形EBHO中,考察5个点M, N, P, Q, R的凸包,它一定 在凸六边形AEHCFG中.
而SAEHCF G = SABCD −SDF G −SBEH = 1 −1 8 −1 8 = 3 4.
再分三种情况讨论: (i)M, N, P, Q, R的凸包为五边形,不妨设为MNPQR,此时SMQR + SMNQ + SNP Q ⩽3 4, 从而(M, Q, R),(M, N, Q),(N, P, Q)中至少有一个为好组,又由于(P, Q, R)在矩形OHCF中,当然是好 组,所以至少有两个好组.
(ii)M, N, P, Q, R的凸包为四边形,不妨设为A1A2A3A4,另一点为A5.
则SA1A2A5 + SA2A3A5 + SA3A4A5 + SA4A1A5 = SA1A2A3A4 ⩽3 4, 从而(A1, A2, A5), (A2, A3, A5), (A3, A4, A5), (A4, A1, A5)中至少有两个好组.
(iii)M, N, P, Q, R的凸包为三角形,不妨设为A1A2A3,另两个点为A4, A5.
则SA1A2A4 + SA2A3A4 + SA3A1A4 = SA1A2A3 ⩽3 4.
76 从而(A1, A2, A4), (A2, A3, A4), (A3, A1, A4)中至少有一个好组, 同理,(A1, A2, A5), (A2, A3, A5), (A3, A1, A5)中至少有一个好组,此时也至少有两个好组.
综上所述,不论何种情况,在5个已知点中至少有2个好组.
下面给出一个例子说明好组的数目可以只有两个.在矩形ABCD中,M, N分别在边AD, AB上,使 得AN : NB = AM : MD = 2 : 3, 则在M, N, B, C, D这5个点中恰好有两个好组.
事实上,我们可以求得SBCD = SBCM = SCDN = 1 2 > 1 4, SMCD = SMDB = SNCB = SNBD = 3 10 > 1 4.
SMNC = SABCD −SNCB −SDMC −SAMN = 1 −3 10 −3 10 −2 25 = 8 25 > 1 4.
所以只有两个好组(M, N, B), (M, N, D).
故面积不大于1 4的三角形的个数的最小值为2.
6.求方程 2x · 3y −5z · 7w = 1 的所有非负整数解(x, y, z, w).
解:由5z · 7w + 1为偶数,知x ⩾1.
(1)若y = 0,此时2x −5z · 7w = 1.
若z ̸= 0,则2x ≡1 (mod 5),由此可得4|x.因此3|2x −1,这与2x −5z · 7w = 1矛盾.
若z = 0,则2x −7w = 1.
当x = 1, 2, 3时,直接计算可得两组解(x, w) = (1, 0)(3, 1).
当x ⩾4时,有7w ≡−1 (mod 16),但是72k ≡1 (mod 16), 72k+1 ≡7 (mod 16),显然不可能.
所以当y = 0时,全部非负实数解为(x, y, z, w) = (1, 0, 0, 0), (3, 0, 0, 1).
(2)若y > 0, x = 1,则2 · 3y −5x · 7w = 1.
因此−5z · 7w ≡1 (mod 3),所以(−1)z ≡−1 (mod 3),z为奇数.
所以2 · 3y ≡1 (mod 5),由此可得y ≡1 (mod 4).
当w ̸= 0时,有2 · 3y ≡1 (mod 7),由此可得y ≡4 (mod 6),与y ≡1 (mod 4)矛盾.
所以w = 0,于是2 · 3y −5z = 1.
当y = 1时,z = 1.当y ⩾2时,5z ≡−1 (mod 9),由此可知z ≡3 (mod 6),因此53 + 1|5z + 1.
所以7|5z + 1,这与5z + 1 = 2 · 3y矛盾.
故y > 0, x = 1时所有非负整数解为(x, y, z, w) = (1, 1, 1, 0).
(3)若y > 0, x ⩾2,此时5z · 7w ≡−1 (mod 4), 5z · 7w ≡−1 (mod 3).
即(−1)w ≡−1 (mod 4), (−1)z ≡−1 (mod 3).
因此z, w都是奇数,从而2x · 3y = 5z · 7w + 1 ≡35 + 1 ≡4 (mod 8).
所以x = 2,原方程变为4 · 3y −5z · 7w = 1,z, w都是奇数.
由此可知4 · 3y ≡1 (mod 5), 4 · 3y ≡1 (mod 7), 从上面两式可以得到y ≡2 (mod 12).
77 设y = 12m + 2,(m ⩾0)于是5z · 7w = 4 · 3y −1 = (2 · 36m+1 −1)(2 · 36m+1 + 1).
所以2 · 36m+1 −1 = 5p · 7q(p, q为非负实数).
化为(2)的情况,必有p = 1, q = 0, 2 · 36m+1 −1 = 5, m = 0, y = 2.
此时有5z · 7w = 5(5 + 2) = 35,所以z = w = 1.
故y > 0, x ⩾2时所有非负整数解为(x, y, z, w) = (2, 2, 1, 1).
综上所述,所求的非负整数解为(x, y, z, w) = (1, 0, 0, 0), (3, 0, 0, 1), (1, 1, 1, 0), (2, 2, 1, 1).
78 2006 中国数学奥林匹克 （第二十一届全国中学生数学冬令营） 第一天 福州 1 月12 日 上午8∶00～12∶30 每题21 分 一、 实数 满足 1 2 , , , n a a a L 1 2 0 n a a a + + + = L ，求证： ( ) 1 2 2 1 1 1 max ( ) 3 n k i k n i n a a − + ≤≤ = ≤ − ∑ i a 1 k a a d + = − 2 1 1 1 , , k k k k n k k k n a a d d a a d d d + + + − = − − = − − − − L L 1 ． 证明 只需对任意1 ，证明不等式成立即可． k n ≤ ≤ 记 ，则 1, 1,2, , k k k d a a k n + = − = − L k a a = ， 1 k k k ， ， 1 1 2 1 2 1 1 2 , , , k k k k k k k k k k a a d a a d d a a d d d − − − − − − − = + = + + = + + + + L L ， 1 2 0 n a a a + + + = L 把上面这n 个等式相加，并利用 可得 1 1 1 2 1 ( ) ( 1) ( 1) ( 2) k k k n k k na n k d n k d d k d k d d + − − − − − − − − − − + − + − + + = L L 0 1 2 ⎞ ⎟ ⎠ ． 由Cauchy 不等式可得 ( ) 2 2 1 1 1 2 ( ) ( ) ( 1) ( 1) ( 2) k k k n k k na n k d n k d d k d k d d + − − − = − + − − + + − − − − − − L L 1 1 2 2 1 1 1 k n k n i i i i i i d − − − = = = ⎛ ⎞⎛ ≤ + ⎜ ⎟⎜ ⎝ ⎠⎝ ∑ ∑ ∑ 1 1 1 2 2 1 1 1 ( 1)(2 1) 6 n n n i i i i i n n n i d d − − − = = = − − ⎛ ⎞⎛ ⎞ ⎛ ≤ = ⎜ ⎟⎜ ⎟ ⎜ ⎝ ⎠⎝ ⎠ ⎝ ∑ ∑ ∑ 2 ⎞ ⎟ ⎠ 3 1 2 1 3 n i i n d − = ⎛ ⎞ ≤ ⎜ ⎟ ⎝ ⎠ ∑ ， ( ) 1 2 2 1 1 3 n k i i n a a a − + = ≤ − ∑ 所以 i ． 2005 1 2 2 3 2006 , , , a a a a a a L 二、正整数 （可以有相同的）使得 1 2 2006 , , , a a a L 两 两不相等．问： 中最少有多少个不同的数？ 1 2 2006 , , , a a a L 解 答案： 中最少有46 个互不相同的数． 1 2 2006 , , , a a a L 由于45 个互不相同的正整数两两比值至多有45×44＋1＝1981 个，故 中互不相同的数大于45． 1 2 2006 , , , a a a L 下面构造一个例子，说明46 是可以取到的． 设 为46 个互不相同的素数，构造 如下： 1 2 4 , , , p p p L 1 2 2006 , , , a a a L 6 1 1 2 1 3 2 3 1 4 3 4 2 4 1 , , , , , , , , , , , , , , p p p p p p p p p p p p p p L ， 1 1 2 2 1 , , , , , , , , , , , k k k k k k k p p p p p p p p p p − − L L ， 1 45 44 45 43 45 45 2 45 1 , , , , , , , , , , p p p p p p p p p p L ， 46 45 46 44 46 46 22 46 , , , , , , , , p p p p p p p p L ， 这2006 个正整数满足要求． 所以 中最少有46 个互不相同的数． 1 2 2006 , , , a a a L 2 3 mn k k = + + 三、正整数m，n，k 满足： ，证明不定方程 2 2 11 4 x y m + = 2 2 11 4 x y n + = 和 中至少有一个有奇数解 ． ( , ) x y 证明 首先我们证明如下一个 引理：不定方程 2 2 11 4 x y + = m ① 0 0 ( , ) x y ，或有满足 或有奇数解 0 0 (2 1) (mod ) x k y m ≡ + ② 0 0 ( , ) x y ，其中k 是整数． 的偶数解 引理的证明 考虑如下表示 (2 1) x k y + + 2 , x x y m ≤ ≤ 0 为整数，且 , 0 2 m y ≤ ≤ ， ( ) 2 1 1 2 m m ⎛ ⎞ ⎡ ⎤ ⎡ ⎤+ + ⎜ ⎟ ⎢ ⎥ ⎣ ⎦ ⎜ ⎟ ⎣ ⎦ ⎝ ⎠ 1 2 , 0, 2 x x m ⎡ ⎤ ∈⎣ ⎦ m > 则共有 个表示，因此存在整数 ， 1 2 , 0, 2 m y y ⎡ ⎤ ∈⎢ ⎣ ⎦ 1 1 2 2 ( , ) ( , ) x y x y ≠ ，且 ⎥，满足 1 1 2 2 (2 1) (2 1) (mod ) x k y x k y m + + ≡ + + ， 这表明 (2 1) (mod ) x k y m ≡ + ， ③ 1 2 2 , 1 x x x y y y = − = − 。由此可得 这里 2 2 2 2 (2 1) 11 (mod ) x k y y m ≡ + ≡− ， 2 , 2 m x m y ≤ ≤ 2 2 11 x y km + = ，所以 ，因为 故 2 2 11 11 4 7 4 x y m m + < + < m ， 2 2 11 2 x y m + = 于是1 ． 因为m 为奇数， 6 k ≤ ≤ ， 2 2 11 6 x y + = m显然没有整数解． 2 2 11 x y m + = 0 0 2 , 2 x x y y = = （1） 若 ，则 是方程①满足②的解． 2 2 11 4 x y + = m 0 0 , x x y y = = （2） 若 是方程①满足②的解． ，则 （3）若 2 2 11 3 x y + = m ，则( ) ( ) 2 2 2 11 11 3 4 x y x y m ± + = m ⋅ ． x x 首先假设3 m，若 ，则 0(mod3), y (mod3) y 0(mod3) ，且 0 0 11 , 3 3 x y x x y y − + = = ④ x y ≡ 0(mod3) 是方程①满足②的解．若 ，则 0 0 11 , 3 3 x y y x y x + − = = ⑤ 是方程①满足②的解． 3 m 现在假设 ，则公式④和⑤仍然给出方程①的整数解．若方程①有偶数解 0 1 0 2 , 2 1 x x y y = = ，则 ( ) ( 2 2 2 2 1 1 1 1 1 1 11 36 5 11 11 5 ) x y m m x y y x + = ⇔ = ± + m ． 1 1 5 11 x y ± 因为 1, 1 x y 1 5 1 y x m 都为奇数． 的奇偶性不同，所以 ， 1 1 1 0 0 5 11 5 , 3 3 1 x y y x y x − + = = 是方程①的一奇数解． 若 ，则 (mod3) x y ≡ 1 1 1 0 0 5 11 5 , 3 3 1 x y y x y x + − = = 1 x 1 (mod3) y 是方程①的一奇数解． 若 ，则 （4） 2 2 11 5 x y + = m ) ，则 ． ( ) ( 2 2 2 5 4 3 11 11 3 m x y y x ⋅ = + ± m 当5 m 时， 若 ， 或 1(mod5), 2(mod5) x y ≡± ≡m 2(mod5), 1(mod5) x y ≡± ≡± ， 则 0 0 3 11 3 , 5 5 x y y x y x − + = = ⑥ 是方程①满足②的解． 若 ，或 1(mod5), 2(mod5) x y ≡± ≡± 2(mod5), 1(mod5) x y ≡± ≡m ，则 0 0 3 11 3 , 5 5 x y y x y x + − = = ⑦ 是方程①满足②的解． 5 m 当 ，则公式⑥和⑦仍然给出方程①的整数解．若方程①有偶数解 0 1 0 2 , 2 1 x x y y = = ，则 2 2 1 1 11 , x y m + = 1 x 1(mod 2) y ， 可得 ． ( ) ( 2 2 1 1 1 1 100 33 11 3 m x y y x = + ± m ) 若 ，或者 1 1 0(mod5) x y ≡ ≡ 1 1 1(mod5), 2(mod5) x y ≡± ≡± ，或者 1 1 1 0 0 33 3 , 5 5 1 x y y x y x − + = = 1 1 2(mod5), 1(mod5) x y ≡± ≡m ，则 是方程①的一奇数 解． 若 ，或 1 1 1(mod5), 2(mod5) x y ≡± ≡m 1 1 2(mod5), 1(mod5) x y ≡± ≡± ，则 1 1 1 0 0 33 33 , 5 5 1 x y y x y x + − = = 是方程①的一奇数解． 引理证毕． 0 0 ( , ) x y 由引理，若方程①没有奇数解，则它有一个满足②的偶数解 ．令 ，考虑二次方程 2 l k = +1 2 2 0 0 1 0 mx ly x ny + + −= ， ⑧ 2 2 2 0 0 0 0 0 4 4 2 2 ly l y mny m ly x x m m − ± − + − ± = = ， 则 1 x ，即 这表明方程⑧至少有一个整数根 2 2 1 0 1 0 1 0 mx ly x ny + + −= ， ⑨ 1 x 必为奇数．将⑨乘以4n 后配方得 上式表明 ( ) 2 2 0 1 1 2 11 ny lx x n 4 + + = ， 2 2 11 4 x y + = n 1 0 1 2 , x ny lx y x = + = ． 这表明方程 有奇数解 2006 中国数学奥林匹克 （第二十一届全国中学生数学冬令营） 第二天 福州 1 月13 日 上午8∶00～12∶30 每题21 分 四、在直角三角形ABC 中， 90 ACB ∠ = °，△ABC 的内切圆O 分 别与边BC，CA， AB 相切于点D，E，F，连接AD，与内切圆O 相 交于点P，连接BP，CP，若 AE AP PD + = 90 BPC ∠ = ． °，求证： 证明 设AE = AF = x，BD＝BF＝y，CD＝CE＝z，AP＝m，PD＝n． 因为 ，所以 90 ACP PCB PBC PCB ∠ + ∠ = ° = ∠ + ∠ ACP PBC ∠ = ∠ ． Q P F E D C B A 延长AD 至Q，使得 AQC ACP PBC ∠ = ∠ = ∠ ，连接BQ，CQ，则P，B，Q， C 四点共圆，令DQ＝l，则由相交弦定理和切割线定理可得 ， ① yz nl = 2 ( ) x m m n = + ． ② AC AP AQ AC = 因为 ∽ ACP Δ ，故 AQC Δ ，所以 2 ( ) ( ) x z m m n l + = + + ． ③ 在Rt △ACD 和Rt △ACB 中，由勾股定理得 2 2 ( ) ( 2 ) x z z m n + + = + ， ④ 2 2 ( ) ( ) ( 2 ) y z z x x y + + + = + ． ⑤ 2 2 z zx ml + = ③－②，得 ， ⑥ 2 2 yz n z zx m = + ①÷⑥，得 ， 2 1 2 yz m n z zx m + + = + 所以 ， ⑦ 2 2 2 2 ( ) ( ) 2 x yz 2 2 x m n x z z z zx + = + = + + + ②×⑦，结合④，得 ， 2 2 ( ) 2 x y z x z z x = + + 整理得 ． ⑧ 2xy x z y z + = + 又⑤式可写为 ， ⑨ 4 2 x z z x y = 由⑧，⑨得 z + + ． ⑩ 2xz y z x z + = − ， ○ 11 又⑤式还可写为 把上式代入⑩，消去 ，得 y z + 2 2 3 2 2 x xz z 0 − − = ， 7 1 3 x z + = 解得 ， 代入○ 11 得， (2 7 5) y z = + ， 将上面的x，y 代入④，得 2( 7 1) 3 m n z + + = ， 2 7 1 6 x m z m n + = = + ， 结合②，得 7 1 2 n z + = 从而 ， x m n + = ，即 AE AP PD + = ． 所以， 1 1 2 a = 五、实数列{ } n a ， 满足： 1 1 2 k k k a a a + = − + − 1, 2, k = L． ， 证明不等式 1 2 1 2 1 2 1 1 1 1 1 2( ) n n n n n a a a n a a a n a a a ⎛ ⎞ ⎛ ⎞⎛ ⎞ + + + ⎛ ⎞ − ≤ − − − ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟ + + + ⎝ ⎠⎝ ⎠⎝ ⎠ ⎝ ⎠ L L L 1 1 ⎛ ⎞ ⎜ ⎟ ⎝ ⎠ ． L , 2 , 1 , 2 1 0 = ≤ < n an 证明 首先，用数学归纳法证明： ． 1 = n 时，命题显然成立． 2 1 0 ≤ < n a ． 假设命题对 成立，即有 ) 1 ( ≥ n n ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ ∈ − + − = 2 1 , 0 , 2 1 ) ( x x x x f ( ) f x 是减函数，于是 设 ，则 2 1 ) 0 ( ) ( 1 = ≤ = + f a f a n n , 1 1 1 ( ) ( ) 2 6 n n a f a f + = ≥ = 0 > ， 即命题对n＋1 也成立． 原命题等价于 ( ) 1 2 1 2 1 2 n n n n n n a a a a a a ⎛ ⎞ ⎛ ⎞ − ≤ ⎜ ⎟ ⎜ ⎟⎜ ⎟ + + + + + + ⎝ ⎠⎝ ⎠ L L 1 2 1 1 1 1 1 n a a a ⎛ ⎞ ⎛ ⎞⎛ ⎞ 1 − − − ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠ ⎝ ⎠ L ． 1 2 1 0 , 2 x x < < 1 1 ( ) ln 1 , 0, 2 f x x x ⎛ ⎞ ⎛ = − ∈ ⎜ ⎟ ⎜ ⎝ ⎠ ⎝ ⎞ ⎟ ⎠ ( ) f x 设 ，则 是凸函数，即对 ，有 ( ) ( ) 1 2 1 2 2 2 f x f x x x f + + ⎛ ⎞≤ ⎜ ⎟ ⎝ ⎠ ． ( ) ( ) 1 1 2 2 2 2 f x f x x x f + + ⎛ ⎞≤ ⎜ ⎟ ⎝ ⎠ 等价于 事实上， 2 1 2 1 2 2 1 1 1 1 x x x x ⎛ ⎞ ⎛ ⎞⎛ 1⎞ − ≤ − − ⎜ ⎟ ⎜ ⎟⎜ + ⎝ ⎠ ⎝ ⎠⎝ ⎟ ⎠ ≥ ， ( ) 2 1 2 0 x x ⇔ − ． 所以，由Jenson 不等式可得 ( ) ( ) ( ) 1 2 1 2 n n f x f x f x x x x f n n + + + + + + ⎛ ⎞≤ ⎜ ⎟ ⎝ ⎠ L L ， 1 2 1 2 1 1 1 1 1 1 n n n n a a a a a a ⎛ ⎞ ⎛ ⎞⎛ ⎞ 1 ⎛ ⎞ − ≤ − − − ⎜ ⎟ ⎜ ⎟⎜ ⎟ + + + ⎝ ⎠⎝ ⎠ ⎝ ⎠ L L 即 ⎜ ⎟ ⎝ ⎠ ． 另一方面，由题设及Cauchy 不等式，可得 ( ) 1 1 1 1 1 n n i i i i i a n a a = = + − = − + ∑ ∑ 2 2 1 1 1 1 1 ( ) 2 n n i i n i i i n n n n a a a a a + + = = ≥ − = + − + ∑ ∑ − 2 1 1 1 2 2 n n i i i i n n n n a a = = ⎛ ⎞ ⎜ ⎟ ⎜ ⎟ ≥ − = ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ∑ ∑ − ， 1 1 1 1 (1 ) 1 2 n i i n n n i i i i i i a n n a a a = = = = ⎛ ⎞ − ⎜ ⎟ ⎜ ⎟ ≥ − ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ∑ ∑ ∑ ∑ 所以 ， ( ) 1 2 1 2 1 2 1 2 (1 ) (1 ) (1 ) 1 2 n n n n n n n a a a n n a a a a a a a a a ⎛ ⎞ ⎛ ⎞ ⎛ − + − + + − − ≤ ⎜ ⎟ ⎜ ⎟ ⎜ ⎜ ⎟ + + + + + + + + + ⎝ ⎠ ⎝ ⎝ ⎠ L L L L ⎞ ⎟ ⎠ 故 1 2 1 1 1 1 1 n a a a ⎛ ⎞ ⎛ ⎞⎛ ⎞ 1 ≤ − − − ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠ ⎝ ⎠ L ， 从而原命题得证． 六、设X 是一个56 元集合．求最小的正整数n，使得对X 的任意 15 个子集，只要它们中任何7 个的并的元素个数都不少于n，则这 15 个子集中一定存在3 个，它们的交非空． 解 n 的最小值为41． 首先证明 合乎条件．用反证法．假定存在X 的15 个子集，它们中任 何7 个的并不少于41 个元素，而任何3 个的交都为空集．因每个元素至多属于 2 个子集， 不妨设每个元素恰好属于2 个子集 （否则在一些子集中添加一些元素， 上述条件仍然成立） ，由抽屉原理，必有一个子集，设为A，至少含有 41 n = 2 56 1 15 × ⎡ ⎤+ ⎢ ⎥ ⎣ ⎦ ＝8 个元素，又设其它14 个子集为 ．考察不含A 的任何7 个子集， 都对应X 中的41 个元素， 所有不含A 的7－子集组一共至少对应 个元素． 另 一方面，对于元素a，若a ，则 中有2 个含有a，于是a 被计算 了 次； 若 ， 则 中有一个含有a， 于是a 被计算了 次，于是 1 2 14 , , , A A A L 7 14 41C A ∉ 1 2 1 , , , A A A L 4 7 a A ∈ 1 2 14 , , , A A A L 7 14 12 C C − 7 7 14 13 C C − 7 7 7 7 14 14 12 14 13 41 (56 )( ) ( ) C A C C A C ≤ − − + − 7 C 7 7 7 7 14 12 13 12 56( ) ( ) C C A C C = − − − 7 7 7 7 14 12 13 12 56( ) 8( ) C C C C ≤ − − − ， 由此可得196 ，矛盾． 195 ≤ 其次证明 ． 41 n ≥ { } 1, 2, , 56 X = L 用反证法．假定 ，设 40 n ≤ ，令 { } , 7, 14, 21, 28, 35, 42, 49 , 1, 2, ,7 i A i i i i i i i i i = + + + + + + + = L ， { } , 8, 16, 24, 32, 40, 48 , 1, 2, ,8 j B j j j j j j j j = + + + + + + = L ． 8( 1,2, ,7), 0(1 7) i i j A i A A i j = = = ≤< ≤ L I 7( 1,2, ,8) j B j = = L 显然， ， ， 0(1 8) i j B B i j = ≤< ≤ I 1(1 7,1 8) i j A B i j = ≤≤ ≤ ≤ I ，于是，对其中任何3 个 ， j B 子集，必有2 个同时为 ，或者同时为 i A ，其交为空集． 对其中任何7 个子集 1 2 1 2 , , , , , , , ( 7 s t i i i j j j A A A B B B s t ) + = L L ，有 1 2 1 2 s t i i i j j j A A A B B B U ULU U U ULU 1 2 1 2 s t i i i j j j A A A B B B = + + + + + + + − L L st ) 8 7 8 7(7 ) (7 s t st s s s s = + − = + − − − 2 ( 3) 40 40 s = − + ≥ ， 任何3 个子集的交为空集，所以 ． 41 n ≥ 综上所述，n 的最小值为41． 2O 0 7 年第4 期 2007 中国数学奥林匹克 第一天 一 、设口 、 b、 c 是给定复数， 记I口+ b I = m， I口一b I = n， 已知胁≠O． 求证： inax{ I + b I ， I口+ b c I } ≥下 ． √m ‘ + n ‘ ( 朱华伟 供题) 二、 试证明： (1) 若2n 一1 为质数， 则对于任意n 个 互不相同的正整数a ， n： ， ⋯， a ， 都存在i、 ∈{1， 2， ⋯， n}， 使得 I> 2 n 一1 ； a i ' aj ， (2) 若2n 一1为合数， 则存在n 个互不 相同的正整数口 。 ， 口 ： ， ⋯， 口 ， 使得对任意的 i 、 ．『 ∈{1， 2， ⋯， n}， 都有 < 2 n 一1 ． tgi ' ， 其中， ( ， Y ) 表示正整数 、 Y 的最大公 约数． ( 李胜宏 供题) 三、 已知口 。 ， 口 ： ， ⋯， 口 。 。 为给定的11个互 不相同的正整数， 且总和小于2 007． 在黑板 上依次写着1， 2， ⋯， 2 007 这2 0 07 个数． 将 连续的22 次操作定义为一个操作组： 第i 次 操作可以从黑板上现有的数中任选一个数， 当1≤i ≤11时， 加上口 ， 当1 2≤i ≤22 时， 减 去口 。 ． 如果最终结果为1， 2， ⋯， 2 0c r7的偶 排列， 则称这个操作组为优的； 如果最终结果 为1， 2， ⋯， 2 007 的奇排列， 则称这个操作组 为次优的． 问： 优的操作组与次优的操作组哪 种多， 多多少? 称 为 偶 排 列， 如 果∑( 一 )为 正 数； 否 则 称为奇排列． ( 刘志鹏 供题) 第二天 四、 设0 和 分别为△ABC 的外心和内 心， △ABC 的内切圆与边船、 CA 、 AB 分别相 切于点D 、 E 、 F ， 直线FD 与 相交于点P ， 直线DE 与 相交于点Q， 点 、 』 v 分别为 线段PE 、 QF 的中点． 求证： -l-MN ． ( 冷岗松 供题) 五、 设有界数列{ 口 } 。 满足 + 1 ， n = 1， 2 ， ⋯． 证明： 口 < ， n = 1， 2 ， ⋯． ( 李伟固 供题) 六、 试求不小于9 的最小正整数n， 满足 对任给的n 个整数口 。 ， 0： ， ⋯， 口 ( 可以相 同) ， 总存在9个数口 口 ， ⋯， ab ( 1≤il < i2 < ⋯< i9≤n) 及b ∈{4 ， 7} ( i = 1， 2 ， ⋯， 9) ， 使得b1 口 il + 62口 i 2+ ⋯+ b 9口 七 为9 的倍数． ( 陈永高 供题) 参考答案 一 、证法1： 因为 nla】 c{ I nc + bI。 I n + 6cI} 注： 1， 2， ⋯， n 的一个排列 1， 2， ⋯， 又 ≥ ≥ = 普 、 J b + a 1． J b — n J 乒 2 b ’ ~／( I o I + I I ) m + ／ g2 = I n — b I + I n + b I 维普资讯 中等数学 ：2( I口 I + I 6 ， 所以， m 舣{ I nc+ 6I， I口 + 6 c I}≥ 专 ~／ m + 证法2 ： 注意到 nc + 6 = ( n + 6) 一 ( n 一6) ， n + 6 c = ( 。 6) + ( n 一6) ． 令a = ( n + 6) ， p ： ( n 一6) ， 则 I nc + 6 I + I 口+ 6c I = I a — p I + I口 + p I = 2( I口 I + I I ) ． 所以， ( m 舣{ I nc + 6， a + 6 c I})。 h ： f丁 l+c +f 因此， 只要证明 I +f ≥ 嘉， 等价变形为 f +i + (f m 2 n n 2 ． ① 凡4 + m 2 凡 2 ≥2 f f f + (1 【 +I 1 ) 2 2 = (I 字I+ 』 I+ 』 n z ≥l字+ + l 2 2 = m 2 凡 2 ． 故式①得证． 证法3 ： 由已知得 m = I a + 6 I = ( a + 6) ( ) = ( a + 6) ( a + 占 ) = I a I + I 6 I + + 6 ． n = I a 一6 I = ( a 一6) ( ) = ( a —b) ( 一 a 一5) = l a l + l 6 l 一 一 6 ． 故 ： + ⋯： ： ， + ： ． 令c = + Y i ( 、 ， ， ∈R) ． 于是， I nc + 6 I + I口+ 6c I ：( nc + 6) ( ) + ( 口 + 6 c) ( ) ：I口I I c I + I 6 I + abe + 口 6c + I口I + I 6 I I c I + 6 c + ； = ( I cI + 1) ( I a I + I b I ) + ( c+ ； ) ( + ) = ( + y2 + 1) + 2 ≥ 』 ． ： + ( m ： 一 ： ) + ：竿( + ) 一 — 一 I 』 一 m + 2， m 一n ＼ m + 2 — 一 I J — 一 ≥ 一 · 一盟 一 1 1 ’ ， 礼 十 凡 则( 懈{ I nc + 6I， I口 + 6cI }) ≥ ， 即 ， + n ‘ 一{ b l , k 。 }≥ 二、 ( 1)记2n 一 1 为质数P， 不妨设 ( a{， a2， ⋯， a ) = 1． 若存在i (1≤i ≤n) ， 使得P I a ， 必然存在 ≠i ， 使得pS aj ． 由于 ( ， ) ， 则有 ≥ 南 ≥ p ： 2 1 ． 以下只要考虑( a ， P) = 1( i = 1， 2， ⋯， n) ， 则对 征意i ≠ 都有p ( ai ， ) ． 将1， 2， ⋯， P 一1 分成 n 一 1 类{1， P 一1} ， {2， P 一 2}， ⋯， {n 一 1， n}． 由抽屉 原理可知， 存在i ≠ ， 使得 a。 ；q ( m 0d p) 或者嘶+ q —O( m 0d P) ． 当哦；q ( p) 时， 端 > ≥ p 一 2 1 ． 当口 + 口 § O ( P) 时， ≥p = 2 1 ． 故( 1) 得证． (2)下面构造命题存在性的例子． 由于2n 一1为合数， 则存在两个大于1 的正整 数P、 q， 使得2n一1= pg ． 可以构造如下 个数： al = 1， a 2 = 2 ， ⋯⋯， P ， +1 = P + 1， nP +2 = p + 3 ， ⋯⋯， 0 = pg —P ， 其中， 前面为P 个连续的整数， 从P + 1至pg — P 为 n—P 个连续的偶数． 当1≤i ≤ ≤p 时， 显然有 荫 ≤ 嘶 + ≤ 2 p<2 n一 1 ． m ： 维普资讯 2 O O 7年第4 期 当P + 1≤ ≤ ≤n 时， 因为2l (啦， q ) ， 所以， 有 ≤ ≤胛 一p < 2 ． 当1≤ ≤P ， P + 1≤ J ≤n 时， 分两种情况： ( i) 当 ≠P 或J ≠n 时， 显然有 7 ≤阳一1 < 2 n 一1 ； O,i ， a j ， ‘ ‘ ( ii) 当i = P 且 = n 时， 由( P ， 阳一 P) = P ， 则有 ：段：q < 2 n 一1 ． ，nn ， P ‘ 经过如上验证， 可以看出如上构造的一组数满 足条件． 1l 3． 优的 操作组更多， 多了兀 个． 我们引入一般的记号： 如果黑板上写着1， 2， ⋯ ， n 这n 个数， 一个操作组被定义为2次连续操 作： 第 次操作可以从黑板上现有的数中任选出一 个， 加上b ， 这里b ∈z ( 1≤i ≤2 ) ． 如果最终结果为 l ， 2， ⋯， n 的偶( 奇) 排列， 则称此操作为优( 次优) 的． 优的操作组的数目与次优的操作组的数目之差 记为 bl ， b2， ⋯， bj ； n) ． 下面讨论厂 的性质． 首先， 对任意1≤ 、 ≤ ， 交换b 与 的取值不 会影响 ． 事实上， 只需要将操作组的第 次与第 次操作对调； 换言之， 第i 次操作时进行原来的第 次操作， 把原来计划进行的第 『 次操作选定的数加 上6j ， 而第 次操作时进行原来的第i 次操作． 对调 后操作组的结果不变， 因而， 优(次优)操作组的数目 不变， 故厂 不变． 其次， 只需要计算这样的优的(次优的)操作组 的数目： 每一步操作后， 黑板下没有任何两个数相 同， 这样的操作组称为具有性质P 的操作组． 可以 证明： 具有性质P 的优的操作组数目与次优的操作 组数目 之差也等于 b1， b2， ⋯， bj ； n) ． 事实上， 只要证明， 在不具有性质P 的操作组 中， 优的操作组与次优的操作组一样多． 如果一个操 作组最先第i 步操作导致黑板上出现两个相等的 数， 例如， 第P 个数和第g 个数相等( 1≤p < g ≤n) ， 那么， 对该操作组的后2一 步操作进行如下的改 动： 对第P 个数的操作改成对第q 个数进行， 对第g 个数的操作改成对第P 个数进行， 那么， 这个新的操 作组最终显示的结果将是在原操作组的结果上对第 P-口 个数进行了对换， 不难发现， 对换一个排列中任 何两个数都会导致排列的奇偶性改变． 所以， 优的操 作组通过上述改动可以和次优的操作组构成一一对 应． 因而， 不具备性质P 的操作组中， 优的与次优的 一 样多． 现在对m用数学归纳法证明： 若n。 ， ， ⋯， 为m个互不相同的正整数且 总和小于n， 则 f ( nl ， ， ⋯， ， 一 nl ， 一 ， ⋯， 一 ； n) m = I I ． ① 当m：1时， 考虑具有性质P 的优的和次优的 操作组， 必然是从后n。 个数中选上某个数加上n。 ， 然后， 再将这个数加上一n ( 否则得不到1， 2， ⋯， n 的排列) ， 所以， 优的操作组有n 个， 次优的操作组 有0个． 故式①成立． 假设m一1时命题成立． 考虑m 时的命题． 不妨设nl < < ⋯< ． 根据前面的讨论， nl ， n2， ⋯， ‰， 一nl ， 一 ， ⋯， 一 ； n) ： 口 l ， 一 口 2， 一 口 3， ⋯， 一 ， 口 2， 口 3， ⋯， ‰， 一 口 l； ) ， 这时， 对于具有性质P 的优的和次优的操作组来 说， 第1步操作可以从末n 个数中选取某数加上 n ， 而第2 步操作只能对前n2 个数进行， 第3 步操 作只能对前 + 口 个数进行， ⋯⋯， 第m步操作只 能对前 + ⋯+ < n —n1个数进行， 而第m + 1 2m一1步操作也只能对前n —n 个数进行( 否 则， 前n 一 个数最终的和小于1+ 2+ ⋯+ ( n — n， ) ， 操作组结束后黑板上的n 个数不为1， 2， ⋯， n 的排列) ， 第2m步操作只能对第1步操作时选定的 数进行． 因此， 第2 2m 一 2 步操作必然是对前n —n 个数进行， 它的结果也要得到1， 2， ⋯， I' l — 的偶 ( 奇) 排列， 才能使总共2m 步操作的结果得到1， 2， ⋯ ， n 的偶( 奇) 排列． 所以， 中间2m 一 2 步操作构成 的对n—n 个数进行的每个具有性质P 的优( 次 优) 的操作组都可以对应n 个原来的具有性质P 的优(次优)的操作组． 于是， ， 一 ， 一 锄， ⋯， 一 ‰， ， 锄， ⋯， ‰， 一 ； n) ： I1 1 一 ， ～ 锄， ⋯， 一 ‰， ， n3， ⋯， ‰； n— I1 1) ． 根据归纳假设 一 n2 ， 一n3， ⋯， 一 ， 口 2， 口 3， ⋯， ‰； n —I1 1) ： ， 锄， ⋯， ‰， 一 ， 一 ， ⋯， 一 ； ， I 一 口 I) = Ⅱq ． 』 = 2 故式①在m 时亦成立． 所以， 由归纳法， 式①得证． 在式①中取n = 2唧， m ：11， 即得到本题的答 维普资讯 中等数学 案 Ⅱ ． I 四 、 不 妨 设 。 > ． 考 虑 △船 c与 截 线 肿， 由 梅 涅 劳 斯 定 理 有 篑 ·丽 A F·丽 B D =1 ． 所 以 ， 一PA 一 ． 一 一 P C —F B D C —D C —P —C ’ 于是， PA = ．因此， = ． 则咫= + 肛= + p —n ： 卫二 二 ． M E： { 饱 = c ， Z 口 一 = ME —AE = a c = c ， 一 口一 僦： 砸+ Ec = ．c) ： C ． — C ’ 于是， MA · MC = ． 因为M E 是点M 到△ABC 的内切圆的切线长， 所以， M E 2 是点M 到内切圆的幂． 而M A · Me 是点 M 到△ABC 的外接圆的幂， 等式M A · M C = M E2 表 明， 点M 到△ABC 的外接圆与内切圆的幂相等． 因 而， 点M 在△ABC 的外接圆与内切圆的根轴上． 同理， 点N 也在△ABC 的外接圆与内切圆的根 轴上． 椒oI 洲． 五、 设6 = 一 1 ，则 6 < ∑ (n ≥ 1 )． ① 下证b < 0 ． 因为n 有界， 故存在常数M ， 使得b < M ． 当n≥1 00 000 时， 有 6 M 上 6 ^< < 南 = + 嚆+ 。 < ．{ m < 由此可以得到， 对任意的正整数m有 6 ^ <(粤 ) M ． 将其代人式①得b < 0 ( n≥1 00 000) ． 再次利用式①可得， 如果当n ≥N + 1 时， b < 0， 则b <0． 这就推出b < 0 ( n= 1， 2， ⋯) ， 即 ‰ <÷ ( n= l ， 2， ⋯) ． 六、 取口 l = 口 2= 1， 口 3= 04= 3， 口 5= ⋯= 口 l2= 0， 则其中任9个数均不满足要求． 因此， n≥13． 下证n = 1 3 可以． 为此， 只要证明如果m个整 数口 ， n： ， ⋯， ( 可以相同) 中， 不存在3 个数。 。 、 及6l 、 62、 63∈{4， 7} ， 使， 得6l 口 jl + 62。 j2 + b3 为9 的倍数， 则m≤6或者7≤m≤ 8 且口 l ， 啦， ⋯ ， ％中有6个数。 。 ， ⋯， 。 及bl ， b2， ⋯， b6∈ {4， 7}使得91 ( 6l O il + b2 + ⋯+ 66 ) · 设Al = { I1≤ ≤m， 9I O i } ， A2= { I 1≤i ≤m， D l s 3( m0 d 9) } ， A3= {i I1≤i ≤m， oi _ = 6( m oa 9) } ， A4= { I1≤ ≤／ 7 ／ ,， 口 j E l ( m oa 3) } ， A5= {i I1≤ ≤m， 口 = 2( m o a 3) } ． 则I Al I + IA2 I + IA3 I + IA4 I + IA5 I = m， 且 ( 1) 若 ∈A2， ∈A3， 则91 (4oi + 4q ) ； (2) 若i ∈A． ， _ 『 ∈ ， 则9 能整除4ai + 4 4 D f +7 7 DI +4 之一( 因为这三个数均是3 的倍 数且模9 两两不同余) ； ( 3) 若 、 _ 『 、 kE A2或者 、 _ 『 、 kE A， ， 则 9 I( 4口 + 4 D j + 4 ) ； (4)若 、 、 k ∈A． 或者 、 _ 『 、 k E A ， 则9 能整除 4 D I + 4 + 4 、 4 + 4 D J + 7 、 4 + 7 + 7 之一 (因为这三个数均是3的倍数且模9两两不同余) ． 由假设， 有IA I≤2 ( 1≤ ≤5) ． 若IAl I≥1， 则 IA 2 I + IA3 I≤2 ， I A4 I + IA5 I ≤2 ． 这样， m= IAl I + IA2 I + IA3 I + I I + l A5 l ≤6． 下设IAl I =0， m≥7， 此时 7 ≤m = I A l I + IA2 I + I A3 I + IA4 I + I A5 I ≤8 ． 因此， m in{ I A2 I， IA3 I} + m in{I A4 I， IA5¨ ≥ 3． 由( i) 和( ii) 知， 存在 。 ， ： ， ⋯， ∈A： U A， U A． U ， fl < f2< ⋯< 6及bl ， b2， ⋯， b6∈{4， 7}使 9 l( b x O il + b2 O i 2 + ⋯ + bs a ~ s ) · 综上， 所求的最小的n= 1 3． (朱华伟 提供) 维普资讯 2009 年第3 期 1 9 ●竞赛之窗● 2009 中国数学奥林匹克 第一天 1． 给定锐角△ PBC ， PB ≠P C ． 设A 、 D 分别是边PB 、 PC 上的点， 联结AC 、 BD 交于 点0 ． 过0 分别作OE 上AB 于E ， OF 上CD 于F ， 线段BC 、 AD 的中点分别为 、 Ⅳ． ( 1) 若A 、 B 、 C 、 D 四点共圆， 求证： E M · 肌r = E N · F M ： (2) 若EM · FN = EN · FM ， 是否一定有 A 、 B 、 C 、 D 四点共圆? 证明你的结论． ( 熊 斌 供题) 2 ． 求所有的质数对( P， q) ， 使得 I(5 + 5 ) ． ( 付云皓 供题) 3 ． 设m 、 n (4 < m < n ) 是给定的整数， Al A2⋯ 2 +l 是一个正2n + 1 边形， P = {A。 ， A： ， ⋯， A } ． 求顶点属于P 且恰有两个内 角是锐角的凸m边形的个数． ( 冷岗松 供题) 第二天 4 ． 给定整数n ( 凡> 1 3) ， 实数口 l ， 口 2， ⋯， 满足 m in I n。 一口 = 1． 求 I o I 的最小 值． ( 朱华伟 供题) 5 ． 凸n 边形P 中的每条边和每条对角 线都被染为n 种颜色中的一种． 问： 对怎样的 rt， 存在一种染色方式， 使得对于这1 7 , 种颜色 中的任何三种不同颜色， 都能找到一个三角 形， 其顶点为多边形P 的顶点， 且它的三条 边分别被染为这三种颜色? ( 苏 淳 供题) 6． 给定整数n( n ≥3) ． 证明： 存在r t 个 互不相同的正整数组成的集合S ， 使得对Js 的任意两个不同的非空子集A 、 B ， 数 ∑ ∑ ∈A b ∈B I A } l B I 是互质的合数(这里 ： 与I 1 分别表示有 限数集 的所有元素之和及元素个数) ． ( 余红兵 供题) 参考答案 第一天 1． ( 1) 如图1， 设Q 、 R 分别是边OB 、 OC 的 中点， 联结 EQ 、MQ、 FR 、 MR ． 则 1 EQ = 音OB = RM ， 1 MQ = 1 OC = RF ． 厶 又四边形OQ MR 是 平行四边形， 则 B M C 图1 OQ M = ORM ． 由题设A 、 B 、 C 、 D 四点共圆， 有 A 肋 = A CD j EQO = 2L ABD = 2 A CD = D EQ M = EQO + OQM = F R 0 + o R M ： F R M △ EQM △MRF j EM = FM ． 同理， EN = FN ． 所以， EM · FN ：EN · FM ． (2)答案是否定的． 当AD ／ ／ BC 时， 由于 ≠ c ， 则A 、 曰 、 c 、 D 四点不共圆， 但此时仍然有EM· FN = EN · FM ． 证明如下： 如图2， 设S 、 Q 分 别是边O A 、 OB 的中 点， 联结 、 EQ 、 MQ 、 NS ． 则 1 NS = ÷ OD ， 二 1 EQ = 去OB ． 肘 C 图2 20 中等数学 故 = ． ① 又 = 1 OA ， = 百 1 OC ，则 一E S ： ． ② MO OC 一 ‘ 而AD／ ／ BC ， 于是， 一OA ： ． ③ OC OB 一 ‘ 由 式 ① 、 ② 、 ③ 得 = 茄 ． 因 NSE = NSA + ASE = A OD + 2 A OE ， EOM = MOO + OOE = ( AOE + EOB ) + ( 180~一2／EOB ) = AOE + ( 1 80。 一／E OB ) = AOD + 2／AOE = ／NSE ， 所以， △ NSE ∽△ 印 ． 故 E N ： ： ( 由式②) ． 政 田瓦 ‘ 同理， = ． 因此， ： j EM · F N = EN · FM ． 2． 若2 I Pq， 不妨设P = 2 ． 则 2g l (5 + 5 ) q l (5 + 25) ． 由费马小定理知q I( 5 一5) ， 得q l30， 即 q = 2 ． 3 ． 5 ． 易验证质数对( 2 ， 2) 不合要求， ( 2， 3) ， (2 ， 5) 符合要求． 若pg 为奇数且5 I pg， 不妨设P = 5 ． 则 5q I(5 + 5 )= g I(5 一 + 625) ． 当q = 5 时， 质数对(5 ， 5)符合要求． 当q≠5 时， 由费马小定理有ql(5 一 1) ， 故q I626 ． 由于q 为奇质数， 而626 的奇质因 子只有313， 所以， q = 313． 经检验， 质数对( 5， 313)符合要求． 若P 、 q 都不等于2 和5 ， 则 l (5 + 5 ) ． 故5 + 5 - 0(mo d P ) ． ． ① 由费马小定理得 5 1( m od P ) ． ② 由式①、 ②得 5 ；一1( mo d P ) ． ③ 设P 一1= 2 ( 2r 一1) ， q 一1 = 2 (2s 一1) ( k 、 f 、 r 、 s 为正整数) ． 若k≤f ， 则由式②、 ③易知 1 = 1 (2 一 ) ( 5p一 ) (2 ) ：52 1 ( 一 ‘ )( 一 ” = (5 一 )2 r- ( 一1) 一1( m od P ) ， 这与P ≠2 矛盾． 因此， k > Z． 同理， k < 2， 矛盾． 此时不存在符合要求的( P ， q) ． 综上， 满足题目 要求的质数对( P ， 口 ) 为 ( 2， 3) ， (3 ， 2) ， (2 ， 5) ， (5， 2) ， (5 ， 5) ， ( 5， 31 3) ， ( 31 3 ， 5) ． 3 ． 先证明一个引理． 引理 顶点在P 中的凸m 边形至多有 两个锐角， 且有两个锐角时， 这两个锐角必 相邻． 引理的证明： 设凸m 边形为P 。 P ⋯P ． 只考虑至少有一个锐角的情况， 此时， 不妨设 P m P l P 2 < ． 则 P 2 P m = 7c 一 P 2 P l P >号(3 ≤ ≤ m — 1 )． 更有 一 l +。 > (3≤ ≤m一1) ． 而 P l P 2P 3+ P 一 l P P l ·> 7 【 ， 故其中 至多一个为锐角． 回到原题． 由引理知， 若凸m 边形中恰有两个内角 是锐角， 则它们对应的顶点相邻． 在凸m边形中， 设A 、 A， 为两相邻顶点， 且在这两个顶点处的内角均为锐角． 设A 与 A 的劣弧上包含了P 的， 条边( 1≤r≤n) ， 这 样的( i ， )在r 固定时恰有2， l + 1 对． ( 1)若凸m 边形的其余m 一 2 个顶点全 在劣弧A Af上， 而A A 上有r 一1个P 中的点， 此时， 这m一 2 个顶点的取法数为C ． (2) 若凸m 边形的其余m 一 2 个顶点全 在优弧A Aj 上， 取A 、 A， 的对径点 、 ， ， 由 于凸m 边形在顶点A 、 A， 处的内角为锐角， 于是， 其余的m 一 2 个顶点全在劣弧B 曰 上， 而BiB 上恰有r 个P 中的点， 此时， 这m一 2 个顶点的取法数为C ～． 所以， 满足题设的凸m 边形的个数为 2O O 9 年第3 期 2l (2n+1 ) ∑(c +c7 ) ：(2n+ 1 )( ∑c F r i一 -l 2+ ∑c ) = (2n+1 )【 ∑(cm ～ 一 C 7 _ -。 )+ ∑(cL-。 I_ ca )J = (2n + 1) (c： I l + c7 ) ． 第二天 4 ． 不妨设口 l < 口 2 < ⋯< a ． 则对1≤k≤t' t 有 I a I + l a 一 +l I ≥I a 一 +l —a I ≥I + l 一2 后I ． 故∑I I = ∑( I I + 1 0 n + I ) = 告 ∑( l+l 8 一 I)· 【 { (f d I t ln + l-k + { 小l~ n + l-k 】 ≥ 专 (I a k⋯。 ) ≥ 吉 +l一 2 k · 当rt 为奇数时， I n + 1 —2 k f ：2 × 2 i J - J ： ． J ： ( n 一1) ； 当n 为偶数时， 考 ∑I n+ l 一 2k I =2∑(2i — 1 )。 号 = 2[ j∑ = l 一 (2 ) 】 ={ n (n 一 2 )． 故当n 为奇数时， ∑l吼 I ≥ ( n ～ 1 ) ； k ：l 一- 当n 为偶数时， ∑IⅡ l ≥ ／ ． ． t2 (n 一 2)． = l 一 等号均在。 。 ：i 一旦 ( i ：l ， 2 ， ⋯， ) 时成立． 因 此， ∑I o l 的 最小值为 ( 一 1) ( 为奇数) ， 或者 ( n 一 2) ( 为偶数) ． S ． 当n( n≥3) 为奇数时， 存在符合要求 的染法； 当n 为偶数时， 不存在所述的染法． 因为每三个顶点形成一个三角形， 三角 形的个数为 ， 而颜色的三三搭配也刚好有 种， 所以， 本题相当于要求不同的三角形 对应于不同的颜色组合， 即形成一一对应． 以下将多边形的边与对角线都称为 线段． 对于每一种颜色， 其余的颜色形成c 。 种搭配， 从而， 每种颜色的线段( 边或对角线) 都应出现在c： 一 个三角形中， 而每一条线段 都是n 一 2 个三角形的边， 因此， 在满足要求 的染法中， 每种颜色的线段都应当有 ： ( 条) ． 当n 为偶数时， 不是整数， 因此， 不 可能存在满足条件的染法． 下设n = 2m + 1 为奇数， 我们给出一种 染法， 并证明它满足题中条件． 自 某个顶点开始， 按顺时针方向将凸 2m+ 1 边形的各个顶点依次记为A。 ， A： ， ⋯， A 2胁+ 1 ． 对于i ∈{1， 2， ⋯， 2m+ 1}， 按模2m+ 1理 解顶点A1． 再将2m+ 1种颜色分别记为颜色 1 ， 2 ， ⋯， 2 m + 1 ． 将边A A⋯染为颜色i ( = 1， 2， ⋯， 2m + 1) ． 再对每个i 都将线段( 对角线) ， 4i一 + ( k ：1， 2， ⋯， rr t — 1) 染为颜色i ， 于 是， 每种颜色的线段都刚好有m 条． 值得注 意的是， 在规定的染色方法之下， 当且仅当 l + _ 『 1 z 2+ (mo d ( 2m + 1) ) ① 时， 线段A 。A 与A ^ 同色． 因此， 对任何i ≠ ( m o d (2m + 1) ) ， 任何 k 事0 ( r o o d ( 2m + 1) ) ， 线段 A， 都不与 A⋯ ⋯同色． 换言之， 如果 i1一 l ；i2一 j r(mo d (2m + 1) ) ， ② 22 中等数学 线段A ．A 都不与A‘ 同色． 任取两个△Ai 。Aj 。A ．和△ A ： A ， 如 果它们之间至多只有一条线段同色， 当然它 们不含对应相同的颜色组合． 如果它们之间 有两条线段同色， 接下来证明： 第三条线段必 不同色． 为确定起见， 不妨设A 。Ai 与A 同色． 分以下两种情形讨论． ( 1) 如果 ．A 与 Ab也同色， 则由式 ①知 il + _ ， 1 g i2+ _ ， 2(r o od (2m + 1) ) ， l + l - j 2+ 2(r o od (2m + 1) ) ． 将两式相减得 i 1一 l E i2 一 2(r ood (2m + 1) ) ． 故由式②知A ．A ．不与 k A- 同色． (2)如果Ai A 与A Ak 也同色， 则亦由 式①知 i l + J． 1 i2 + 2(mod (2m + 1) ) ， i l + 】 i i2十 2( m od (2m + 1) ) ． 将两式相减得 ， l 一 l - j 2 一 2(mo d ( 2m + 1) ) ． 由式②知A， ．A 与 A 不同色． 总之， △A 。Aj ．A k ．与△ A ： Ah A 对应不 同的颜色组合． 6． 用厂 ( )表示有限数集 中元素的算 术平均． ( 1)证明： 存在r l , 个不同正整数构成的 集合s ． ， 使得对J s 的任意两个不同的非空 子集A 、 B ， 数 ( A ) 和厂 ( 曰) 是不相等的正 整数． 事实上， 取定一个整数q > 1 7 , ， 设 Sl = { 1 7 , ! q， tl , ! q ， ⋯， It , !q } ． 则对S， 的任一个非空子集A ， 数 ．厂 ( J 4 ) 显然是一个正整数． 假设存在S． 的两个不同的非空子集A 、 B ， 使得厂 (A ) = f ( B ) ． 那么， I B I ∑q ： l A I ∑ ． n ! q ∈^ n ! ∈口 因为q > n≥m ax{ l 4 l ， I I } ， 所以， l B l ∑q = ∑I B I q n ! C- A ! ∈ 与 I A I ∑g， = ∑l A I q n ! ∈B ! C =- B 均为正整数的q 进制表示， 从而， 它们的形 式应当完全相同． 由此得I A I = 1日I及A = B ， 矛盾． 因此， 对S 的任意两个不同的非空子集 A 、 B ， 数． 厂 (A ) 和． 厂 ( B ) 必定不等． (2) 设 是一个固定的正整数， > n出厂 ( A ) ． 证明： 对任何正整数 ， 正整数的 n 元集合S： = { 后 ! + 1 I C t∈S 。 }具有下述性 质： 对J s 的任意两个不同的非空子集A 、 B ， 数厂 ( A) 和． 厂 ( 是两个互质的整数． 事实上， 由J s 的定义易知， 对S 的任意 两个不同的非空子集A 、 B ， 相应地有． s ． 的 两个子集A。 、 B， ， 满足 I A l l = I A l ， l l I = I B I ， 且 f (a ) = 14 (A ) + 1， f (B) = J}! x f (B， ) + 1． ① 显然， ．厂 ( A ) 和 ．厂 ( 都是正整数． 设正整数d 是厂 ( A )与厂 ( B ) 的一个公约 数． 则f ( A ) B ) 一 f ( B ) 。 ) 是d 的倍数． 故由式①可知d l ( A。 ) 一 f ( B ， ) ) ． 但由 的选取及s ．的构作可知， l 厂 ( A ) 一 f ( B。 ) I是小于 的非零整数， 故它 是 ! 的约数， 从而， d l | j c! ． 再结合d I 厂 ( A )及式①知d I1， 故d = 1． 从而， A ) 与厂 ( )互质． (3)证明： 可选择正整数 ， 使得J s： 的每 个非空子集的元素平均值都是合数． 由于质数有无穷多个， 故可选择2 一1 个互不相同且均大于 的质数P 。 ， P： ， ⋯， P 。 ． 将s 。 中每个非空集合的元素平均值 记为口 l ， a2， ⋯， a2 n一 1， 则 ( P ， ! a ) = 1( 1≤i ≤2 一1) ， 且( p ， p ) ：l ( 1≤i < J ≤2 一1) ． 故由中国剩余定理可知， 同余方程组 !溉 一l (mo d P )( 1， 2， ⋯， 2 一1) 有正整数解． 任取这样一个解 ， 则相应的集合5 的 每个非空子集的元素的平均值都是合数． 结合( 2) 的结果， 这一 元集合满足问 题的全部要求． ( 朱华伟 提供) 2010 年第3 期 l 9 2010 中国数学奥林匹克 中圈分类号： G424． 79 文献标识码： A 文章编号： 1005 — 6416( 2010) 03 — 0019 — 04 第一天 1． 如图1， 两圆， 、 ， 交于点A、 B ， 过点 日的一条直 线分别交圆 厂 l 、 Jr’ 于点 C 、 D ， 过点曰 的另一条直 线分别交圆 厂 。 、 于点 、， ， 直线CF 分别交圆 厂 。 、 厂 2 于点 图1 P 、 p ． 设 、 ， v ， 、 _ 、 分别是弧船、 Q 的中点． 若CD = EF ， 求证： c 、 、 、 J ) 、 r四点共圆． ( 熊 斌 供题) 2． 设整数 > 1 3 ， 数列{口 } 满足a = 2k， 且对所有的n > 后 ， 有 fa 一 1 + 1， a 一 l与n 互质， 口 n I2 凡 ， 口 一 与 n不 互 质 ． 证明： 数列{ 口 一 C 1 , } 中有无穷多项是 质数． ( 朱华伟 供题) 3． 设复数o 、 6、 c 满足： 对任意模不超过1 的复数z ， 都有I az： + 6z + CI ≤1． 求l 6cI 的最 大值． ( 李伟固 供题) 第二天 4． 已知m、 凡是给定的大于1 的整数， 且 口 < 口 ： < ⋯< n 都是整数． 证明： 存在整数集 的一个子集 ， 其元素个数 I T I ≤1 + ， 且对每个i ∈} l ， 2 ， ⋯， m} ， 均有￡ ∈T 及 ∈[ 一 ， ] ， 使得￡ z = f + ． ( 冷岗 松 供题) 5． 我们对放置于点4 ， ： ， ⋯， A ( n≥ 3 ) = 180 。一 E K J + 180 。一 F K J ： E K F ． 故 、 Jr、 、 四点共圆， 即点 在△ABC 的九点圆上( 其余的在不同位置关系也类似 可证)． 同理， 点K 在另一个九点圆上． 从而， 四个三角形的九点圆共点于K 下证： 点 在D 对△ABC 的垂足圆上． 设￡ 、 、 ． 7 v 为D 在三边上的垂足． 则 L KM = 360 。 一 LKJ 一 MKJ = 埘l J + MGJ = A 删 + 180 。一 删 G = L 4 + 180 。一 D C M ：l 80 。一 ￡ A + 180 。一 ^ ： L N M ． 所以， 点 在垂足圆上． 同理， 其余四圆也过点 由此证得八圆共点． 参考文献： [ 1] F · 克莱因． 高观点下的初等数学( 二) 几何[ M] ． 上 海： 复旦大学出版社， 2008． [ 2] 粱绍鸿． 初等数学复习及研究( 平面几何) [ M] ． 哈尔 滨工业大学出版社， 2009． 戴维· 韦尔斯． 奇妙而有趣的几何[ M] ． 上海教育出 版社， 2006． ’ [ 4] 戴维斯· 盖尔． 蚁迹寻踪及其他数学探索[ M] ． 上海 教育出版社， 2001． 20 中等数学 及点O 处的卡片进行操作． 所谓一次操作是 指进行下面的一种操作： ( 1) 若某个点 处的卡片数目不少于3 ， 则可从中取出3 张， 在点A 、 川及O 处各 放一张( A。 = A川= A1) ； (2 ) 若点O 处的卡片数目不少于n， 则可 以从中取出， l 张， 在点A。 ， ： ， ⋯， A 处各放 一 张．‘ 证明： 只要放置于这n + 1 个点处的卡片 总数不少于， l + 3n + 1， 则总能通过若干次操 作， 使每个点处卡片数目 均不小于n + 1． ( 瞿振华 供题) 6． 设口 1、 a2、 口 3、 bl 、 b2、 b 3为互不相同的正 整数， 满足 I [(n+1 )口 ?+ +(n— 1 )《 ] I[(n+1 ) + +(n— 1 )6 ； ]， 对任何正整数， I 成立． 求证： 存在正整数J I }， 使得b ： ka ( i = 1， 2 ， 3) ． ( 陈永高 供题) 参考答案 第一天 1． 联结AC、 AD、 AE 、 AF 、 DF． 由 ADB = AFB ， ACB = AE F 及 CD = E F △A CD △AE F = A F AD F = ／A F D A B C = A F D = A D F = A B F j A 是 CBF 的角平分线． 联结CM、 肌 因为 是弧胎的中点， 所 以， CM 是 DCF 的角平分线． 同理， 刷是 CFB 的角平分线． 于是， BA、 CM 、 FN 三线共点． 设交点为， ． 在圆厂 ， 、 中， 由圆幂定理得 C I · lM = A I ·IB ． 41· IB ：N I · I F j N I · I F = CI ·I M ． 从而， C 、 F 、 、 Ⅳ四点共圆． 2． 假设口 = 2／ ( Z≥| i }) ． 再设P 为Z一1 的 最小质因子． 则 (1 - ) ： I ， <p； tP · P · 故 (2⋯一 2， ⋯一 1) ： f ， <P； tp · P · 由题设知 r22 + i 一1 ， 1 ≤i < p ； 小 i2 z+2 p一 2，i： p ． 则0(+p — l 一口 一 2 = (21+ 2p 一 2 ) 一 ( 2Z+p - 2) =p( 质数) ． 故口 “ 一 1= 2 ( Z+p 一 1) ． 由以上讨论知有无穷多个Z≥k， 使口 2Z且口 一 1一 + 2 = P 为2 —1 的最小质因 子． 3． 令， ( ) = + + c， ’ g (z ) = =一 z) = 口 + I1 + 。 ， h (z ) = ei"g ( eif l z) = c z 一 + b'z 一 。 + 口 ． 取适当的实数 、 口 ， 使得C 、 b ≥0 ， 对 r ≤1 ， 有 { ≥J h(re ) I引Im h(re ) I = I r - 2~ sin 2 0 + r 一 6 sin 0 + Im 口 I ． 不妨设Im 口 ≥ 0 ， 否则可作变换¨ 一 0 ， 这 样 对 任 意 的 (0< <詈 )， 有 { ≥ r-2 c's in 2 0+ r b 's in 0 > ~ 2r -丁 3 面 而 ~ l bc l c (对 任 意 r≤ ， 0∈ (o ， 詈 )) ∈ m ( 。 ln ，手 ) e∈ ( o 号 ) 6cJ = 的例子： 3,g 一 16 ‘ 201 0 年第3 期 21 = 一 一 ． 对于z = re ( r≤1) ， 有 I re ) l = [ ( r~ co s 20— 2西c0 s 0— 3) + J ‘ ( ， 2sin 20 — 2,／ X ， sin 9) z ] = 壶 [2 r 4 +1 2 r 2 +l8一 ( r 吣 +r 2 — 3 ) ) ≤ 壶 (2 r 4+1 2 r +1 8) 第二天 4． 令口 ， = 口 ， 口 = b， 作带余除法 b 一 = (2n + 1) g + r(g、 r ∈Z ， O≤r≤2 ) ． 取T = {口 + n + (2n + 1)后 Ik = 0 ， 1， ⋯， q}． 则 I = q+ 1≤1 + ， 且集合 B = {t + sIt ∈T， s = 一 ， 一 忍 + 1， ⋯， } = {口 ， n + 1， ⋯， 口 + ( 2n + 1) q + 2n }． 注意到 口 + (2n + 1)q +2n > l a + (2n + 1)q + r = 6． 故每个口 i均在曰中． 从而， 结论成立． 5． 只需考虑卡片总数等于n + 3n + 1 的 情况． 采取如下策略． 如果有某个点A 处的卡片数不少于3 ， 则对点A 处的卡片进行操作( 1) ． 这样的一 次操作使得点0 处的卡片数增加1， 于是， 经 过有限次操作( 1) 后， 将不能再进行操作 ( 1) ． 此时， 每个点Ai处的卡片数不超过2 ， 点 0 处的卡片数不少于t／ , + n + 1． 然后对点0 处的卡片进行n + 1 次操作( 2 ) ， 此时， 每个 点A 处的卡片数不少于 + 1． 下面在保持每个点A 处的卡片数不少 于聘+ 1 的情况下， 使点D 处的卡片数增加 到至少n + 1． 设想A， ， A ， ⋯， A 顺次排列在以0 为圆 心的圆周上． 称连续相邻的若干个点的集合 G = {Ai ， Aj+I ， ⋯， Al+J— l } ( 1≤i、 z≤n) 为一个“ 团” ， 这里若有下标 > n， 则 Aj = a j 一 ． 如果对团G 中每点处的卡片都做一次 操作( 1) 后， G 中每点处的卡片数仍然不少 于忍 + 1， 则称团G 为“ 好团” ． 乏 口 l ， 口 2， ⋯， 口 ( 口 ‘ > i n + 1， = 1， 2 ， ⋯， ， 1) 分别为点A ， A ， ⋯， A 处的卡片数． 则好团 需满足如下的充要条件： 一 个点的团G 是好团 铮Ⅱ i ≥n + 4 ； 两个点的团C 是好团 口 h o +l ≥， l + 3 ； Z(3 ≤Z≤ 一 1) 个点的团G 是好团 口卜口 + 一】 > I n + 3 ， 且 aj． ≥ + 2 ( i + 1≤ ≤i + Z- 2 ) ； 全部疗 个点的团G 是好团 牟 f≥n + 2 ( 1≤ ≤ ) ． 下面证明： 当点D 处的卡片数少于 + 1 时， 或等价地， 当口 l +口 2+ ⋯+口 1 > n +2n + 1 时， 必存在好团． 假设不存在好团． 于是， 每个口 i∈{n + 1， n + 2， 凡 + 3 } ， 否 则会有某个点A 处的卡片数口 。 ≥n + 4 ， 使得 G = {Ai} 是一个好团． 设口 1， n2， ⋯， n 中有 个tl , + 1， ， ， 个／ 1 , + 2 ， z 个n + 3． 下面说明： 一定有 ≥彳 ． 由于t／ , + 2凡 + 1 > n( n + 2 ) ， 故z > 1 1． 若 = 1 ， 则有 ≥1 ， 否则所有n ≥l r t, + 2 ， 使得G = {A ， A： ， ⋯， A }是一个好团． 若z i > 2， 有n + 3 张卡片的z 个点将圆周 分成 段圆弧， 由于不存在好团， 这z 个点没 有两点相邻， 且每段圆弧上都存在一个点只 有 + 1 张卡片． 故 ≥ ． 此时， 点A ， A2， ⋯， A 处的卡片总数为 ( n + 1) + y( r／ , + 2) + ( ， l + 3 ) ≤( + )， + z) ( ， l + 2 ) ：n( ／ I, + 2 ) 中等数学 < 2 + 2 n + 1 ． 矛盾． 这样就证明了当点0 处的卡片数少于 1 1 , + 1 时， 点A。 ， ， ⋯， A 中总存在好团． 于 是， 每次对一个好团中的每个点做操作( 1) ， 直至点0 处的卡片数不少于1 1 , + 1， 而点A。 ， A ， ⋯， A 处的卡片数也不少于 +1． 6． 设r 为任意的正整数． 由于质数有无 穷多个， 故存在质数p， 使得 p > ( n + 口 ； + 8； ) ( b + b； + b； ) ． ( 由P 为质数及式①得 (P ， 口 + 口 ； + 口 ； ) = 1． 又P 与P 一 1 互质， 由中国剩余定理知， 存在正整数凡 使 n兰r( mo d(p 一 1) ) ， ② ( 口 ： + 口 ； + 口 ； ) + 口 ： 一 口 ； - 0 ( mo d J P) ． ( 由式②、 ③及费马小定理知 ( 凡 + 1) 口 + ， ； + ( ／ 7 , 一 1) 口 ； -- n( a + 口 +0； ) +口 一 n； - O( m o d P) ． ④ 由题设得 ( n + 1) b： + n6； + ( n 一 1) 6； 毒 0 ( r o o dP ) ． 再由式②及费马小定理得 n( 6 + b + b； ) + b 一 b ； - 0 ( m o d p) ．⑤ 由式④、 ⑤消去 得 ( a + 口 ； + 口 ； ) ( b 一 b； ) s ( 6： + b + b； ) ( 。 一 口 ； ) ( r o o d P) ． ⑥ 由式①、 ⑥得 (o + + ) ( - b ； ) =( + +6 ； ) ( - a； ) ， 即 ( a2b1) + 2( 03b1) + ( 口 3b2) = ( 口 l62) + 2 ( 口 l63) + ( a2b 3) ． ⑦ 下面先证一个引理． 弓 I 理 设 l ， 2， ⋯， ， Yl ， y 2， ⋯， y ( 0 < l ≤ 2≤⋯≤ ， 0 < ， ， 1≤)， 2≤⋯≤)， ， ) 为实 数， 满足对任意的正整数r ， 总有 + ； + ⋯+ = ， ， + + ⋯+ ． 则 = Y ( i = l ， 2 ， ⋯， 5) ． 引理的证明 对s 用数学归纳法． 当s = 1 时， 取r = 1 ， 有 l = ， ， 1． 假设当s = t 时引理成立。 当s = t + 1 时， 若 ⋯≠ +l ， 不妨设 ⋯ < Y⋯． 则 ( (焘) + ． ． +(筹) ：( ) + ( ) + ． ． ·+ (焘 孔 又0 < < 1 ( 1 ≤i ≤t + 1 ) ， 令卜+ + ∞， 有0 ≥1， 矛盾． 因此， ⋯= Y⋯． 故 + ； + ⋯+ ： = y + Y + ⋯+ Y： ( r ： 1 ， 2 ， ⋯) ． 由归纳假设知 ： y ( i = 1， 2 ， ⋯， t) ． 由数学归纳原理知引理对一切正整数s 均成立． 回到原题． 由于o1、 02、 口 3、 bl 、 b2、 b 3互不相同， 故 0 2b1≠03b 1， 口 3bl ≠口 362 ， 口 162 ≠口 163 ， 口 l b3≠0263 ， a zb1≠0,263． 由式⑦及引理得 a2 b1=a36 2=nlb3， 口 3 6l = aI62= azb 3，⑧ 或 口 261= 口 l62， a3bl = a1 63， 0362 = 口 263．( 若结论⑧成立， 则 n 26l 0 362 0 163 。。2 口3 口l — a 3bl 一 a l b2 ’ 这与0 、 、 。 ， 互不相同矛盾． 所以， 结论⑨成立． 于是， ： ： ． 设 ： 车 (( ， 2 ) ： 1 ， z ≥ 1 )． 则 b =粤 o (i =1， 2， 3)． 由 2 6 。 +6 2=粤(2 0 。 +0 2 )及 题 设 (取， l=1 ) ( 20l + 2) I( 26l + 62) 知 _ 等 ． 为 整 数， 即f _ 1． 所以， b ： 玩 ( i = 1 ， 2 ， 3 ) ． ( 熊 斌 提供) 24 (n-l)t矿+(n-2)t 心．式2 式． i=l ,=I i=l 即 由对称性，不妨设从是bl,b2 '…，从中最 小的一个则 n " (n-1) I: 矿+ (n -2) I: a;b; i=l i= l (n -1) bi + (n -1) 贮(n-2)ƿaib1 1=2 i=l (n-l) bi+ (ǀb;f +(n-2)b1 = (n -1) bi + (1 -b1) 2 + (n -2) b1 ＝咐+ (n -4)b1 + 1 习=ǁai;式8 a:.
i=I i=I 6. 如果mn=l,则结论成立． 下设mn 2. 由于 n °(am 0 + bn 6) = (a +b)n °•• +a[ (mn)" -n•••J, 故只要证明存在无穷多组互质的正整数a、b 使得 (a +b) I [ (mn)" -n尸••],(a +b,n) =1. 令p=a+ b. 只要证明存在无穷多个质数p及正整数 a(l:s;,a:s;,p-1), 使得 p I [ (mn) a - nP ] .
由费马小定理知，当 a 1 三三a2(mod p -1) ,a1 l ,a2 ǂ1 时，有 中等数学 (mn) 01 = (mn)勺(modp).
因此，只要证明存在无穷多个质数p及 正整数a,使得 pl[(mn) 0-n].
(D 假设这样的质数只有有限个，记为p口 P2 , …, p,(由于mn;;;;?:2,这样的质数必存在）． 设(mn) 2 -n = pǃ• 庐…忙， ＠ 其中，a,(1b 这r) 为非负整数． 取a=pDŽ• 疗…p;'(P1-1)…(p,-1)+2. 设(mn) 0 -n =付悼…伈， ＠ 其中，且(1b这r)为非负整数 若p,ln,则由式＠及a;;;;?:2可知伈In.
故伈l[(mn) 2-n].
从而，由式＠知且幻斗 若pJn,则pJm. 故(p;j +I,mn) = 1.
由欧拉定理知（注意叭pcj +l)=pci (p, -1) 为a-2的约数） (mn) 0 -n= (mn) 2 -n(mod pDžj +I).
由于pdži + I'}[ (mn)仁n]'故由上式知 pLJj + 1'} [ (mn) 0 -n].
因而，B尽a,. 于是， (mn) ° -n =P':1的…伈 勺pLj•, 平..忙= (mn) 2 -n, 与a >2矛盾 所以，存在尤穷多个质数p及正整数a, 使得pl[ (mn) 。一n].
（熊 斌 提供） 2 0 1 3 年第 期 中国数学奥林匹克 中圈分类号 文献标识码 文章编号 一 一 印 一 第一天 一 , 规定 , , 。 满足 叭 一 、任 , , …,川 两个半径不相等的圆 、 几交于点 、,点 、分别在圆 、 几上,且线段 以 为中点,延长 与圆厂, 交于点 ,延 长 与圆几交于点 , 设线段 、 的 中垂线分别为, 、 证明 与 相交 若 与 的交点为 , 则三条线段 、 、 能构成一个直角三角形· 熊 斌 供题 确定所有由整数构成的非空集合 , 满足若。办任 办可相同,则 一 呀 陈永高 供题 求所有的正实数 满足 存在一个由 实数组成的无限集合 , 使得对任意的 、、 任 、、可相同 ,及任意实数 与正实 数 ,均有 二 一 一 , 一 ,肠 一 卜以 张思汇 供题 第二天 给定整数 设 个非空有限集 私 二 任, 任 一任, 诺 求 十 … 。的最小值 何忆捷 供题 对正整数 及整数 〔“ , 设 二 二 , , 〔 , 记 又设 ,, 二 习 , 。 幂 证明 、、 , 任 , ,且。 不是 的方 若 , , , 则对任意正 整数有 , , 艾颖华 供题 给定正整数 、求具有下述性质的 最小整数 〕 若一个 元整数集含 有模 的完全剩余系,则其有一个非空子 集,其元素和被 整除 瞿振华 供题 参考答案 第一天 因为 、、、和 、、、分别 四点共圆, 且 , 所以, 由圆幕定理得 · · 二 二 , 吧 尸丁二 叮了 二 , 一 叫 了 百一 尸 二, 气 二 , 兮 甲 ·刀 · 月 “硒 利用上述等式,还可将原题作如下推广 推广 设 是■ 所在平面内一点, 尸、 分别与对边 、 交于点 、,且分 别与■ 的外接圆交于另一点 、 , 石 刃 与月夕 交于点 , 与 交于点 则 二 的充分必要条件是 尸平分乙 此外,还可将其类比至三角形的外心、 垂合、 重心等巧合点,得到类似结论本文不再赞述 参考文献 【 宋宝莹 译第 届丝绸之路数学竞赛〔 中等数 学, 一增刊 〔」危博伦一道国外竞赛题的另解〔 中 等数学为 c B · 二 · · 二 · ① 假设 与 不相交则 劝 器器 代人式① 得 , , 劝 二 故 土 因此, 、 分别是圆 、 的直径, 即圆厂、 几半径相等,矛盾 从而, 与 必相交 如图 设 与 的交点为 联结 、 、尸 图 则匕 艺 召二 匕刀 召二 匕 由 土 ,知 平分匕 又点尸在 的中垂线上,故 在■月 君 的外接圆上 则艺 尸 尸二 “一 匕 二 艺 艺 二匕 二乙 召 召 从而,点 在以尸为圆心、 为半径的 圆 上,记其半径为 由圆幂定理得 · · 护 一 矿 则 护 尸一 二 尺 一 以, 以, 咫, 故线段 、 、 可构成直角三角形 若集合 中只有一个整数, 则集合 满足条件 下面假设集合 中至少有两个不同的整 数则 中任两数之差有最小正值,记为 从而,存在整数 使得 中等数学 、 任 故 一 任 , 一 一 任 , 二 一 一 任 , 一 二 一 任 由上面的构造,知若 、 任 , 则 土 办 一 办 〔 由 。 任 , 水 , 可验 证 。 满足条件 若 尹。 , 则对任何 〔 、。,存在整数 使得 由于 、 中至少有一个不是 的倍 数,故 、 中至少有一个属于 由 的最小性, 知 到 中数的距离至 少是 , 故只能 任 。, 且 二 翻劝 二 〔 下面证明 二 任 任取 任 ,则由 秃〔 劝 , 井 、 士 、 土 任 井 士 土 一 任 由数学归纳法知 任 任 劝 秃 任 二 故 二 任 也满足条件 综上,所求的集合 恰有三类 或 任 , 牛或 任 , 其中, 、任 , , · “, 合· 年第 期 取裔,则 “ ,· 令二 几, , , … 假设存在。任 , ,及 、声满足 二 一 一 , 一 , 。 一 川鉴, , 一 一 蕊 一蕊 一 巨 口 一 蕊劣一已蕊 仁一亡 毛 一 簇吸 忿 回 ②一①、③一②得 第二天 ④ ⑤ 一 蕊为一簇 , 一 鉴 一 为鉴 · 由 。 专 ,知 一 故由式④、 ⑤得 一 ,气一 结合又 ,知 无 , 一 一, 一一 又丝二 兰 竺 二 二三 生 二 对整数 ,证明对任意正整数 有 , , 十 , 〕 先给出两个明显的结论 对有限集 、有 ■ 若非空有限集 、满足 ■ 二, 则 多 事实上,假设不然,则必有 , 于是, 八 二或 ,矛盾 当 时, 由条件与结论 知 , 一 妻 一 无 一 , , … , 一 又人 十 , 及结论 得 , 。一 左一 井 凡二 一 , , 。习一 , 一一 , 、一 一 气 二几一 一 一 ' 一 人一 一 一】 二 名 、 一、, 二 同理,当 二 时,有 双 一 认 一 二 一 , , … , 一 , 矛盾 且 , , 十 〕 公 习 故上述构造的无限集合 满足要求 ”,合 劝 十 一 二 , 一 , 二 下面证明对任意无限集合 中任意三个 元素 , 均存在 任 , 任 十 , 使得 一。 一 , 一 。, 一 十 感 以 一 事 实 上,令。宁,“一 二 则 二 一 一 , 犷 一 , 一 川 一 一 二 一 厄 一, , 一 厄 一 卜 花 犷了 丫韶“· 综 上 , 。合· 二 另一方面,取集合 , , … ,人, 如下 二 , , … , , , … ,几, 、 , , 凡万二 弄, …弄,' 一仃马, … 弄 对这组集合, 易验证 卜 对 任意的 、任 , , … , 川成立 此时,经计算得 十 ,率五 苦 且, 中等数学 当取前 个集合时,得 二二 凡, , 一 无 二, 由 , · 、,知 对任 〔 。 意 任 , 叮 任 综 上 ,、 。 的 最 ,、 值 为 哥·, 其 中 ,〔 · 」 有 且 。 〔 表示不超过实数 的最大整数 对任意正整数 , 设“ 的二进制表 示为 二 , 心 … , 其中, , 二吸· 定义 “ , , 心, … , 心,并约定 为空集 由卢卡斯定理, 知 二 为奇数当且仅当 二 故 。 , 二习,口 “,二开 , “ , 月侧 任 几 其中, , 表示 的元素之和 设 、、满足题意 下面证明若 利用上述证明的两个命题得 任 劝 二 从而,对任意正整数 有 , , 。 “ 。 任丁 。 任八几 则 二 先给出下面两个命题的证明 命题 用 表示正整数 的最大奇 约数, 则对任意非负整数 鉴' , 有 。 , , 。。 , 二 命题 的证明 由因式分解得 , , 一二, , 一 ' … , , 。 ,一 劝 。 , , , 二, , , 一 幼 , , , 。 , 二 命题 利用题目条件 不是 的 方幂, 证明 对 任何 整 数 , 都有了 也 不是 的方幕, 即, , , 命题 的证明 事实上, 的情形即 为题设, 由此还知必有 而当 时, 注意到, 。 , , 或 又, , , 故 。 , 回到原题 · 二 , 一 合〔,· , · ,〕· 首先证明 二 , 一合〔 ,· , · 〕· ① 设 二 , , , 二 若。 , 冬 , ,考 虑 一 个 模的 完 一 曰一 ”一 、一' 一 产' 砂尸 印 ” 八”' “ 了 “ 全剩余系 , , … , 。, 其模 的余数恰为 , 个 , , … , 这样的一组完全剩余系是 存在的, 如下面 个数 二, , … , , 二, , … , , 再 添 上、 二 。 一 冬、, 一个 模。 余 ' ” 一 '一 '一 ”一 、一' 一 一' ' 小 的数 ,儿, … , ,则集合 注 , , , … , 二, , , … , 含有模 的完全剩余系, 但不存在非空子集 的元素和被 整除 事实上,集合 中没有一个元素被 整 除,故 的每个元素除以 的最小正余数之 和为 … 二一, 从而,集合 中不存在非空子集其元素 和被 整除 因 此 , 二一 合,即 式 ① 成立 其次证明 二 , 一合〔 ,· , · 〕 201 3 年第 期 满足要求 反复利用结论 个整数中必存在若干 个之和被 整除 设这 个整数为 , , … , ,令 二 … · 若有一个 被 整除, 则结论成立否 则,存在 鉴,' 鉴,使得 , ” 一 … · 由上,知在 个均被正整数 整除的整 数中,必可取出若干个数之和被 整除 对原问题分两种情形讨论 又 · 、 合 , · ,。 合 【 警· 专 , 十 子,故 “ 于是,可从这些己 一集中 选出一些,其总 和被 二整除 ·合· 此 时 ,, 一 合 , · 设 是一个 元集包含模 的完全剩 余 系 · ,, ,… , ,另 夕 卜 还 有 。 一 合 爪、 · 蛋 合 , · 若 是奇数, 则由 的论证,可将一 模 数 个 此时, 二 如果一个有限整数集的所 有元素和可被无 整除,则称“ 一集” 设, , , … , 。是模 的完全剩余系, 将这些数分成 组, 每组构成模 的完全 剩余系设 , , … ,是一个模 的完全 剩余系则 二 若 为奇数, 则可将一个模 的完全剩 余 系 分 成 宁个集 ,如 一集 的 完 全 剩 余 系 分 成 合“ 个 另 夕 卜 ,一 合 爪“ 个 数 又 可 分 成 , 一 , , … , 宁, 望 , 共 有 冬 个一 集 、'刁 ”一 ' 、一' 一 产' 一 月、' 由 于 二 。 合 爪“ ,故 从 这 些集 中可选出一些集合, 其总和被 二整除 若 为偶数, 类似地,可将 , , … ,为 , 、 二 , 二二 分成普个以 一集, 此时多余拜 产 ' 一 门 、' 一, 、了 了' 一 , 二一 , 人 , 。、 , , 一, 注意到,两个模 余答的数又可组成一 一, 一” ” 一, 、 ' 碑 个一集 则从一个模 的完全剩余系中可 取 出 粤, 粤个 一 集〔〕 表 示 不 超 过 实数 的最大整数 一合, , 组 数 ,每 组 个 数 ,每 一 组 中又可取出一个一集, 即从另外的数中可 找 到。一 冬 , 个、集这 样 共 得 到 一 、一一' ”一、一' 一 产' 一、' ” 、' , 一 刁 “ 个一集 若 是偶数, 则由 的论证,可将一个 模的 完 全 剩 余 系 分 成 冬, , 粤个 · , 、 矽·一 “ 、 一 '一一' 」' 一集 当, 时 ,。。一合 个 一集 这样共得到 个一集 当 , 为奇数时, 剩余的 也是一个 一一 , , 、 亏 一 集,共 有一 '个 亏集, 从 中 可 取 出 至 少,一 合 合 个 一集 这样共得到 个 一集 最后, 从这 个以 一集中可选出一部分 数, 其总和被 二整除 熊 斌 提供 2014 年中国数学奥林匹克（第30 届全国中学生数学冬令营） 第一天试题 2014 年12 月20 日8:00-12:30 重庆 2014 年中国数学奥林匹克（第30 届全国中学生数学冬令营） 第二天试题 2014 年12 月21 日8:00-12:30 重庆 奇思妙想，止于至善。 奇思妙想，止于至善。 奇思妙想，止于至善。 奇思妙想，止于至善。 奇思妙想，止于至善。 2017年第2期 9 匾圃 第32居中国数学奥林匹克试题评析 瞿振华 ( 华东师范大学数学系，200241) 中图分类号：G 424．79 文献标识码：A 文章编号：1005—6416( 2017) 02—0009—04 第32届中国数学奥林匹克于2016年11 月21_25日在湖南省长沙市雅礼中学举办． 本文就本次赛事试题的解题思路及一些不同 方法作介绍，并略作评论． 试题可参见本期第18页，本文不再重复． 1．题中并没有说a、b、C为整数，但由秽。 在n充分大时为整数，反推可知a、b、c均为 有理数．U 。整除影。的条件一时也用不上．先 尝试利用特征方程求得{M 。}与{秽。}的通项 公式，分别为 1 M 。=÷ [ ( 1+√乏i ) “ +( 1一√乏i ) “ ] ， 厶 秽。=cl ( 一1+242 i ) “ +c2( -1一址i ) “ +c33n． 通过初值列出方程组 影0=C 1+C 2+C 3 2 o， 秽1=( 一1+2厄i ) c1+ ( 一1—2厄i ) c2+3c3=b， 移2=( 一7—4厄i ) c1+ ( 一7+4厄i ) c2+9c3=c． 从中解出C。、C：为共轭复数的形式，且具 有如下形式： c1=，．+s 42 i ，c2=r—s 42 i ， 且c，也为有理数，记C，=t ． 则r、5、t ∈Q ． 若记O ／=l +√2 i ，冱=1一以i ，则 d2=一1+2厄i ，瓦2=一1—2厄i ． 收稿日期：2016—12—17 故H 。=÷ ( d“ +技“ ) ， 二 口。=( r+s在i ) a24+( r—s在i ) 孑” +t 3“ ． 怎么将可。除以U 。呢?注意到，秽。的两个 虚特征根恰为d2、a2，可设 W 。=2( ( r+S√2 i ) “ “ +( r—s√2 i ) 五” ) ． 则z) 。=U nW 。+( 2r—t ) 3“ ． {W 。}满足 W o=4r，W 】=4( r一2s) ， ZW 1 3w 2 )． n 2 n一 一 n—L凡 J · 选取一个正整数k，使得b、凰、忽均为整 数，故砌。、砌。均为整数．结合递推关系，知 枷。为整数． 从而，kv。=ku。W 。+k( 2r—t ) 3“ ． 由条件，知当n≥Ⅳ时， U 。l k( 2r—t) 3“ ． 由( M 。，3) =1，得u。I k( 2r—t) ，n≥Ⅳ． 再证明{U 。}无界，从而，2r—t =0，由此 便不难推出结论． 【评注】这是一道比较常规的递推数列 问题，方法直接、容易上手．但是有一定计算 量，考生有回避计算的倾向．在决定是否要解 出具体的通项公式时，很多考生犹豫了，去尝 试其他方法，在其他方法走不通后又回到解 方程组上．解题的关键需要观察到特征根之 间的平方关系，之后还有许多细节需要仔细 论证．作为第一题，复杂度略高，需要不少时 间．考生的解答都与标准解答大同小异． 10 中等数学 本题的平均分为6．0分． 2．如图1． P ／ 勿‘ 隧：、 德 幽1 解答由几个相对独立的部分组合而成． ( 1) 4、D 、L三点共线 牟≥b2+c2=。( b+c) ． 标准解答中利用 邶正点共线甘舞=器， 再用正弦定理计算． ( 2) P、X、Y、Q 四点共圆 铮P、Q 、0、，、丁五点共线， 这里，r为外接圆在点A处的切线与BC的 交点． 标准解答中考虑三个圆并用根心定理． ( 3) D 、，、r三点共线 § b2+c2=o( b+c) ． 标准解答中利用梅涅劳斯定理 o、，、r三点共线甘筹· 努· 筹=1， 再通过计算，证明． 【评注】本题是中等难度的几何题，或多 或少需要计算．熟悉某些几何性质，会在某些 步骤上较快地突破． 例如，( 1) 中设AD 的延长线与△ABC的 外接圆交于点E则 4、D 、￡三点共线 铮ABEC为调和四边形 一AB——EB—．si n。! 二BAD —AC—EC—si n么CAD 。 再由正弦定理得b2+c2=a( b+c) ． ( 3) 中也可以作00’ 上BC，口’ 上BC，利用 o、，、r三点共线§ 等=舁， 再通过计算． 本题的平均分为9．1分． 3．以所有结点为顶点、基本线段为边作 图G ．G 的2度顶点有4个，设3度、4度顶点 分别为戈、Y个，边数为Ⅳ，即基本线段的条 数．通过计算度，每个小矩形的所有顶点总和 以及度与边的关系，可得两个基本等式： 2 016 X4=4+2戈+4y々令戈+2y=4 030． 及N =÷ ( 4× 2+3戈+4y) =4+{ ■+2y=4 034+云喘． 求Ⅳ的最大和最小值转化为求戈的最大 和最小值． 首先，石的最大值为4 030，此时，Y=0， 在1 X2 016的划分中取到，Ⅳ。。=6 049． 要求戈的最小值，考虑在矩行R内部的 基本线段落在s条水平直线和t 条垂直直线 上，可在每条直线上找到至少两个3度顶点， 且互不相同，从而，石≥2s+2f ． 另一方面，( s+1) ( t +1) i >2 016． 可推出戈≥176，从而，Ⅳ≥4 122，等号在 42 X48的划分中取到． 【评注】最大值部分较容易，有多种做法． 例如，对n归纳证明将一个矩形划分成n个 小矩形时，基本线段的条数不超过3忍+1．忍 =1时结论显然．假设ni >2，且结论在小于n 时成立．考虑矩形R被划分成了n≥2个小矩 形，共有Ⅳ条基本线段．不妨设有一条水平 基本线段在矩形内部．将矩形R的底边f 向 上平移至27，第一次碰到矩形尺内部的水平 基本线段，设恰有k个小矩形其两条边分别 在Z、Z’ 上．用刊哿原矩形分成两部分，上半部 2017年第2期 分矩形R’ 被划分成n—k个小矩形，共有Ⅳ’ 条基本线段．可证明Ⅳ≤N ’ +3七．由归纳假设 N 7≤3( n—k) +1，从而，N ≤3n+1． 还可以利用欧拉公式． 矩形尺外部区域看作一个面，则共有 ．厂=2 017个面，e条边( 即基本线段) ，钐个顶 点．欧拉公式给出口一e+厂=2．又移个顶点 中，除了4个度为2的顶点，其余顶点度至少 1 为3，故e≥÷ [ 4× 2+3( 秽一4) ] ．代人即得 二 e≤6 049． 最小值部分需要一个关键想法，即将一 条基本线段向两端延长总会到达两个3度顶 点，以此来估计3度顶点个数的下界． 总体是中等难度的组合极值问题，部分 考生可能被前两题拖累而没有时间思考此题． 本题的平均分为7．5分． 4．以n! 个排列作为顶点，两个排列互为 翻转就连边，这个问题相当于证明此图中含 有哈密顿圈．这是一个n! 阶n一1正则图，图 论中能够证明含有哈密顿圈的一般性结论 ( 如D i rae定理或O re定理) 在这里用不上． 只能对这个特殊的图具体处理．尝试孔较小 的情形，容易想到归纳法．在从n过渡到孔+1 时，需将n+1个圈剪开拼成一个大圈．从哪 里剪开?若加强命题，要求P，=( 1，2，⋯，n) ， P。=( 儿，⋯，2，1) ，则能顺利地构造n+1个 圈，第i 个圈上的排列是所有以i 为结尾的排 列，且在第i 个圈上( i +1，⋯，n+1，1，⋯，i ) 和( i 一1，⋯，1，n+1，⋯，i +1，i ) 相邻，从这 里剪开，再适当地拼成一个圈． 【评注】本题是一个构造性的组合题，通 过对n较小情形的尝试，不难发现构造方法， 且不止一种构造方法．本题是这次冬令营中 唯一的算是容易的题目． 本题的平均分为18．3分． 5．通过n=P8和n=p。g等特例的探索， 发现很难找到满足要求的陀．猜测不存在这 样的n． 反证法，假设存在． ( 1) n∈G ，且由此可得G 的公比A为正 整数． ( 2) n≠p8． ( 3) 标准解答的主要想法是说明子集A 中被某个素数整除的数超过一半，从而推出 子集A中所有数均被P整除 取素数P12, p=p“ m ，p十m ，O ／i >2，m I >2． 设g为m 的一个素因子，p≠g．1、q中至少有 一个属于子集A．G 中至多仅个被P整除，推 出子集A中至少O ／r( m ) 一仅个被P整除． 在r( m ) i >3时，可证明： 1 O ／r( m ) 一仪≥÷ 14 I +1． 二 从而，子集A中有相邻两数均被P整除． 于是，子集4中所有数均被P整除，与 1、q之一属于子集A矛盾． ( 4) 讨论n=p“ g的情形． 【评注】还有几种可行的办法，例如，先 通过讨论子集A中最小两数的情形，对子集 A的公差d给出上界估计，再讨论子集A的 最大数，从而，得子集4的元素个数的估计． 在大部分情形下，可证明此时子集A的元素 个数已超过I D 。l 一3．接下来对剩余情形讨 论．具体来讲，可按下述步骤证明：同标准解 答一样，先证明凡∈G ，于是，G 的公比A为正 整数；再证明n≠p“ ，故n至少含有两个不同 素因子．由于I G I >13，则n的某个素因子次数 不小于2．设P旦． p—l pq 若兰∈A，由于d≤g一1，故 12 中等数学 旦一1 14 I ≥卫+1>旦． g—l pg 可以证明，旦≥r( 凡) 一3( 注意陀有一个 Pq 素因子次数不小于2) ，除了n=4q和89．最 后再讨论这两个个例，此时必有A=2． 本题是略带组合味道的数论问题，证明 的逻辑过程比较长，讨论的分支也较多，需要 慎密的推理．方法的选择也很重要，优化估计 以尽量减少个例的讨论．问题的提出方式以 及答案是不存在也给解题带来一定干扰． 本题的平均分为8．7分． 6．n为偶数的情形较易，可利用不等式： 当戈、y∈[口，6]时，羔Y∈[詈，i b]， (手一詈)(爹一鲁)≤o j ≤一f 孚+¨ ” 一+1≤o． j y1一l i +- 21 y一+1≤o‘ 取戈=戈i ，Y=戈⋯，有 暑≤伊斗飞¨ 对i 求和即得 薹著≤伊号一· ) 豁 不少考生用上面的方法解决了坨为偶数 的情形．但n为奇数的情形用这个方法不好 处理．标准解答用一种方法统一处理了n为 奇数和偶数的情形． 记原分式为f ( 戈，，戈。，⋯，戈。) ．固定戈。， 戈，，⋯，戈。，看作咒，的函数，利用两次求导可以 证明是下凸函数．从而，可将戈．调整到[ Ct ，b] 的边界上，使得分式变大．类似处理石：，戈，， ⋯，戈，}· 因此，可以假设戈i ∈{a，b}．若有连续三 项为b、b、a的情形，将其调整为6、a、a，证明 分式变大．故可以假设没有相邻的b．现假设 有m (m ≤号)个6．通过计算证明在m =【号】时 取得最大值，其中，[ 戈] 表示不超过实数戈的 最大整数． 【评注】还可以用排序不等式将分子变为 下标对称的形式来处理．将石。，戈：，⋯，石。从小 到大排列为Y。≤y2≤⋯≤y。．贝4 § 戈； § Y； 笪堑≤笪丛生． ∑戈i ∑Yi 当n=2m 时，对o≤U ≤移≤b，证明 等+芝≤(詈+鲁-1U )(M +口)； 秽 ＼D 口 7 当n=2m +1时，对a≤11, ≤w ≤口≤b， 证明 等+去2+i11)一<～州m (a洲3+b3))n++a酬2b[It+V+卦 本题考查了基本的分析能力、凸函数、调 整方法、排序不等式等，属于中上难度的不等 式问题． 本题的平均分为5．7分． 【总体评价】本届冬令营试题比较全面 地考查了四个板块的内容，代数、组合各两 题，几何、数论各一题．难度适中，没有偏题怪 题，很多题都有部分分值．六题中有一道较容 易的题，外加五道中等或中上难度的题，没有 一道平均分在3分以下的难题．全部考生中 有一人满分和一人接近满分．考试中反映的 一些问题，包括考生的计算能力下降，对多情 形的讨论分类不够仔细，且对一份由三道中 等难度试题组成的考卷思想准备和策略准备 不足．平时学习时，应注意四大板块的方法技 巧学习要全面，书写表达要注意逻辑严密，对 于做完的题还应思考是否可以优化表述方法 以及证明过程，一个好的解答应该表述简洁 易懂而又不失严密性． l 8 中等数学 第33 届中国数学奥林匹克 中图分类号： c 424． 79 文献标识码： A 文章编号：1 005 — 6416( 2018) 02 — 00 18 —06 1． 对正整数n， 定义 为具有如下性质 的所有素数P 构成的集合： 存在正整数。 、 b， 使得 、 均为与P 互素的整数． 当 P P 为有限集( 包括空集) 时， 用f ( rt) 表示／ 4 的 元素个数． 证明： ( 1) A 是有限集的充分必要条件为 ≠2 ； ( 2 ) 若k 、 m 为正奇数， d 为k 与m 的最 大公约数， 则 d) ≤厂 ( k) + 厂 ( m) - f ( km) ≤ 2厂 ( d) ．① ( 何忆捷 供题) 2． 设n 、 为正整数， T = { ( ， Y ， ) l 、 Y 、 ∈Z ， 1≤ 、 Y 、 ≤n } 为空间直角坐标系中F t。 个整点构成的集合． 已知集合 中( 3 一 3n + 1) + k 个点染成红 色， 满足： 若集合 中两点P 、 Q 均染成红色 且PQ 平行于坐标轴， 则线段 尸 Q 上的所有整 点也均染成红色． 证明： 存在至少k 个互不相 同的立方体， 它们的边长为1 且每个顶点均 染成红色． ( 艾颖华 供题) 3． 设正整数g 不为完全立方数． 证明： 存 在正实数c ， 使得对任意正整数I I ， 均有 1 2 l {nq了 } + {nq丁 } ≥c凡 一 丁 ， 其中， { }表示实数 的小数部分． ( 安金鹏 供题) 4． 如图1， 圆内接四边形ABCD 的对角 线交于点P ， △APD 的外接圆、 △BPC 的外 接圆分别与线段AB 交于另一点 、 ， ， 、 t ， 分 别为△ADE 、 △BCF 的内心， 线段 与AC 交于点 证明： 、 ， 、 、 E 四点共圆． 图1 ( 熊斌 供题) 类似地， G 1F ：一下 F D + F E ， G zH ：一H T I + bi g ． 设AF = IB ： c ， = E C = b ， B D = GC = a ， F I = O tC ， H E = ． ~I JA — G ： ， = 十 一 FA = 一 ． 类似地， 一 G 2 A = 一 主 ． i ~ A — GI ： 业 ， ： 越 ． 由 1 l + 1 2 ： ： ， 知G、 G 、 三点共线， 目 ． 点G 平分线段G1G ． 2018 年第2 期 19 5． 给定奇数n ≥3 ， 用黑白两种颜色对 × n方格表的每个格染色． 称具有相同颜色 且有公共顶点的两个格为“ 相邻的” ． 对任意 两个格a, b， 若存在一系列格c ， c ， ⋯， c 使 得c1 = 0 ， c = 6， cz( i = 1， 2 ， ⋯， 一 1) 与c⋯相 邻， 则称0 与b“ 连通” ． 求最大正整数M， 使 得存在一种染色方案， 其中有 个两两不连 通的格． ( 王新茂 供题) 6． 已知 、 矗为正整数， n > ． 给定实数 口 l ， 口 2， ⋯， 口 ∈( 一1， ) ． 设正实数 1， 2， ⋯ ， 满足对{1， 2 ， ⋯， } 的任意 元子集， ， 均 有∑xi≤ ∑％求 ： ⋯ 的 最 大 值 ． ( 瞿振华 供题) 参考答案 1． ( 1) 当n = 1 时， 显然， 不存在素数P 满足条件． 故A = ． 当n = 2 时， 对于任意的素数P ， 取0 = p， 6 ： p ， 可使 ：l + p 与 ：1 + p： 均 p P 为与P 互素的整数． 故A 包含全体素数， 为 无限集． 当n≥3 时， 对任意的素数P， 设正整数 口 、 b 满足z ， ( n + b) = 1 ， (口 + 6 ) ： 2 ， 其中， 。( Z) 表示正整数z 所含素因子P 的次数． 设a + b = s p ， ( ， P ) = 1． 易知， a ． b 均不为P 的倍数否则， 不妨假 设p1 0 ， 于是， P I6． 故 。 ( Ⅱ + b ) > i n >2 ， 矛盾． 若 为偶数， 则 O兰。 + b = 0 + ( ～ 0) --2a ( r o o d p ) ． 而( 0 ， P) = 1， 故必有pl2． 此时， A 为有 限集． 事实上， P = 2 也不满足条件， 即A = ． 若n 为奇数， 则 0 三0 + ( 一 0) 三n 一 ( r o od P ) ． 而( sⅡ 一 ， P) = l ， 故必有P 此时， A 为 有限集． 综上， 当且仅当正整数n≠2 时， 为有 限集． ( 2) 下面证明： 当n 为正奇数时， 素数 P ∈A 当且仅当 ( n) = 1． 当 = 1 时， 显然成立． 当n > 1 时， 设P ∈A ． 由前知 ( n) I > 1． 故。 + 6 = n + ( s p 一口 ) 一 1 ( 一1) 2 2 一 2 三nsp a 一— — — — — 一s P 0 -~ ns pa一 ( mod P ) ． 因此， 当 ( ) i > 2 时， ( 0 + b ) ≥3 ， 矛盾； 而当U p( ) = l 时， 取口= 1， b = p 一1， 则 P ( 口 + 6) = 1， P( n + 6 ) = p( n p) = 2 ， 满足条件． 对于任意的素数P 及正奇数n， 定义 ： ． ； 则 ‘厂 ( ) ： ∑ (n) ， 其中 ， 约定“ ∑” 表示对一切素数P 求和． 注意到， 当 、 m 为正奇数时， km 与d = ( ， m) 也为正奇数． 因此， 式①等价于 ∑ (d)< - 2 x, ( )+∑ (m )一 ∑ ( ) ≤ 2 ∑ ( )． 仅需证明： 对于任意的素数P， 有 ( d ) ≤ ( ) + ( nz) 一 ( m) ≤2 x, ( d) ． ② ( i ) 若 ( d) = 0 ， 则 r a in ( ) ， ( m) } = 0． 不妨设 ( ) = O． 则 ( km) = ( m) ． 故 ( ) ： ( ) =0 ( m) = ( m) ． 此时， 式②成立． ( ii ) 若 ( d) = 1 ， 则 m in { ( ) ， ( m) } = 1． 于是， ( ) > 1 2． 故 ( m) = 0． 不妨设 ( ) = 1． 则 20 中等数学 ( k ) + ( in) 一 ( m) = 1 + ( m) ∈{ 1， 2 }． 又 ( d) = 1， 从而， 式②成立． ( iii ) 若V p( d) ≥2 ， 则Vp( k ) 、 V p ( m) 、 ( km) 均不小于2． 从而， ( d ) 、 ( k ) 、 ( m) 、 ( m) 均 为0 ， 式②亦成立． 综上， 对于任意奇素数P ， 式②均成立． 由此， 知式①成立． 【 注】 证明当n 为正奇数时， 素数P ∈A 当且仅当 ( 砚 ) = 1， 也可直接应用升幂定理： 设P 为奇素数， a 、 b 为与P 互素的整数， 且P I( a —b) ， 则对于任意正整数k， 有 ( a 一 b ) = Vp( a — b) + ( k ) ． 2． 为方便起见， 引人如下术语． 若一个单位立方体的顶点均为红色， 则 称之为红色立方体； 若一个单位正方形的顶 点均为红色， 且此正方形平行于Oxy 平面， 则 称之为红色正方形； 若一条单位线段的顶点 均为红色， 且此线段平行于 轴， 则称之为红 色线段． 设所有红点构成的集合为 ， 平面 = c 上的所有红点构成的集合为 ， 直线 Y = b， ： C 上的所有红点构成的集合为尺 先证明一个引理． 引理 对正整数1≤c ≤n ， R 中的红色 正方形的数目不小于l R I 一( 2n 一 1) ． 证明 设 中有． s( C) 个红色正方形， ， 条红色线段． 从两方面估算 一 方面， 对每个l ≤b≤n， 若R 非空， 可 设其中X 坐标最小的红点为P ， 坐标最大的 红点为Q． 则由题意， 知R 恰为线段PQ 上所 有整点所构成的集合． 于是， R 上红色线段 的数目为I I 一 1． 故， = ∑( I R 。 l一 1) ≥∑I R6 。 l一 凡 k ≠ R k ≠ = I R l — n． ① 另一方面， 每条红色线段在 轴上的投 影为x 轴上的单位线段， 将所有红色线段按 照其投影分类． 对于 轴上的单位线段it ， 设 以u 为投影的红色线段的数目为L ． 若Iu≥ 1， 设以 为投影的红色线段的Y 坐标的最小 值为 ， 最大值为Y ， 则， n≤ 一 Yo + 1． 由题意， 知对于每个整数Y ∈[ Yo， Y ] ， U × {Y } X {C}也均为红色线段． 于是， 对整数 Y ∈[ Yo， Yl 一1] ， U× {Y ， Y + 1 } X {c } 均为红 色正方形， 共Y1一 Yo个， Y1一 Yo≥ 一1． 故s(c) ≥ ∑(， 一 1) ≥ ∑L 一 ( 一 1) l u≥1 ， ≥1 = ， 一( n 一 1) ． ② 结合式①、 ②得 S ( c) > 1 1 一( n 一1) ≥l R I—n 一( n 一1) = l尺 I一( 2n 一1) ． 引理得证． 设共有c 个红色立方体， S 个红色正方形． 一 方面， 由引理知 = ∑． s(c) ≥ ∑( I一 (2n一 1) )． ③ 另一方面， 每个红色正方形在Oxy 平面 上的投影为Oxy 平面的单位正方形， 将所有 红色正方形按照其投影分类． 对于Oxy 平面 上的单位正方形 ， 设以 为投影的红色单位 正方形的数目为s ． 若s ≥1 ， 设以 为投影 的红色正方形的z 坐标的最小值为zn， 最大 值为 I ， 则S ≤ z1一 + 1． 由题意， 知对于每个整数z ∈[ z。 ， ] ， X 也均为红色正方形． 于是， 对于每个整 数z ∈[ ， 一1] ， X {z ， z + 1 } 均为红色立 方体， 共Z1 一 g O个， l — Z0≥5 一 1． 故c≥ ∑(s 一 1) ≥ ∑S 一 ( 一 1) ≥l ； e l = S 一( n 一 1) ． ④ 结合式③、 ④即得 C ≥S 一( n 一1 ) ≥ ∑( I R I一 (2n一 1) ) 一 ( 一 1) c = 1 = ∑I R。 f一 (3n 。 一 3n+1) =k． 20 18 年第2 期 2 l 3． c = ( 13q手 ) 满足要求． 反证法． 假设存在正整数n 满足 {ng丁 } + {ng丁 } < ef t -T． 记f ．[ 叼 ] ， = [ nq寺 ]． 先证明： 存在整数r、 s、 t 满足 r + s ≠0 ， ≤ ， I S l < 4一 n ， 十 s ≠ 。 l ， ． I≤√ n ， I l ， rl + sm + tn = 0 ． 事实上， 考虑整数对的集合 5 = { ( ， ) ∈Z × Z 1 0 ≤ 、 ≤ √ }． 注意到， I s l = ( [ ] + 1) > n． 则存在集合I s 中不同的数对( ， ) 、 ( ， ： ) 满足 1 Z+ Vl， n 三u2 + v2m( mod n) ． 取r = - It2~$ - V1 - V2~ t 一 ， 即 满足要求． 接下来考虑函数 ( ) = r + S + t 一3rstx ． 则F ( q丁 ) 为整数． 设I 厂 ( ) = + s + t ， c D= e-T ． ． Z I 易验证F ( ) = ) wx)f ( 0 3 ) ． 由于g 不为完全立方数， 则q了 、 e oq亍 、 q了 均不为-厂 ( ) = 0 的根． 故 ( q亨 ) ≠0． 从而， I F ( g丁 ) I ≥1． 又l q了 ) l = I rg丁+ sq丁+ t l 音 ， _ (n q 了 一 )+ (叼 丁 一 m )I ≤ ( I r I 丁 } + I s l {nq了 } ) C 1 < 一 = ——— 一， n 13 q了 n g丁 1) ∞ g} ) = q ÷) I ≤( I f ( wq了 ) - f ( q丁 ) l + I ／ ． ( g丁 ) 1) ≤( 1 ， ． ( 一 1) q亍 l + I s( o ) 一 1) g寺 l + ( 13q丁 4 n) 一) ≤ ( · + · + 2 )‘ < 13 q ． 则l F ( q丁 ) I < 1， 矛盾． 4． 如图2 ， 延长肼， 与△ APD 的外接圆 交于点 ； 延长 ， 与△BPC 的外接圆交于 点y， 联结 、 Py 、 、 馏． 图2 注意到， ， 、 ． ， 分别为△ADE 、 △BCF 的内 心． 据内心性质知 X I = X A ． YJ = YB． 易知， PX 、 PY 分别为 APD 、 BPC 的 平分线． 故 、 P 、 y 三点共线． 由 APX = CP Y = ／BP Y ， A X P = AD P = B CP = B YP ， 知 △ ∽△BPy X A = 、 X ／ s in I X P X A s in E AP YJ sin J YP YB sin F BP A P sin B A P 一 B P sin ／ABP 一“ 这表明， 点， 、 t ， 到直线X Y 的距离相等． 注意到， E 、 F 两点位于X Y 的同侧． 从而， ， 、 -， 两点位于XY 的同侧， 有u # x Y． 故 AKI = APX = AEI A 、 ， 、 K 、 E 四点共圆． 22 巾等数学 5． 考虑推广的问题． 对m × n 方格表黑白染色， 其中， m、 rt 均 为不小于3 的奇数． 在给定的染色方案下， 所 有的格可被划分成若干连通分支， 使得同一 个连通分支中的格彼此连通， 不同连通分支 中的格不连通． 连通分支的数目记为 下面用数学归纳法证明： ( 1)K ≤ ÷ ( m+ 1) ( n + 1) + 1； ( 2) 当K =÷ ( m+ 1) ( n+1) + 1 时， 方 格表四角处的每个格均不与其他格连通． 当 = n = 3 时， 最外一圈八个格至多属 于四个与中心格异色的连通分支， 故K ≤5． 等号成立当且仅当四角处的方格与其他五个 格异色． 接下来设 ≥5． 记方格表的第2 行从左 到右形成黑白交错的k 段 ， ： ， ⋯， A ， 各段 的格数依次为 ， ， ⋯， ． 设P 为含有第1 行格但不含第2 行格的连通分支数目． 记r ] 表示不小于实数 的最小整数． 下面估计P． 若k≥2 ， A 上面的 个格中， 最后一个 格与 最后一个格以及 的第一个格中有 一 个同色， 从而， 与第2 行的某个格连通； 在 前 一 1个 格 中 至 多 有 『 ]个 格 与 A 异 色， 且互相不连通． 类似地， 在／ I 上面的格中 也 至 多 有 『 _ ]个 格 与 第 2行 的 格 不 连 通 ， 且彼此也不连通． 对1 < i < k ， A 上面的格除第一个和最后 一个 格， 中 间 一 2 个 格中 ， 至 多 有 l } 个格与 异色， 且互相不连通． 故 P ]+ f _ ]+ 『 -字] ≤ ． 若 后 = ， 则 也 有 P ≤ 『 -詈 ]= ． 设Q 为含有第2 行格但不含第3 行格的 连通分支数目 ． 由于 和 ⋯中总有一段与 第三行中的格连通( 考虑 最后一个格和 ⋯ 的第一个格， 它们异色， 其中一个必与第 三行的格连通) ， 故 Q≤ ≤ ． 设尺为含有第3 至m 行格的连通分支 数目． 由归纳假设( 1) 知 冗≤ 1 ( m 一 1) ( n + 1) + 1． 下面证明： 若Q ： ( 为奇数) 或Q = ( k 为偶数) ， 则 R ≤ 1 ( m 一1) ( 凡+ 1) ． 事实上， 当k 为奇数， 且Q ： 时， 若 k = 1 ， 则Q = l ， 第2 行所有的格同色， 第3 行 所有的格也同色． 由归纳假设( 2 ) 知 R ≤ ÷ ( m一 1) ( + 1) ． 若奇数k≥3 ， 设4 下面的格为B ， 由于 A。 ， A， ， ⋯， A 均不与第3 行的格连通， 故曰 。 ， B3， ⋯， B 与 1， A3， ⋯， 异色． 于是， 与A2， A4， ⋯， 同色． 从而， B1， 2， 曰 3， 4， 5， ⋯， 一 。， 均连通． 由归纳假设( 2) 知 R ≤ 1 ( 一 1) ( n + 1) ． 当k 为偶数， 且Q ： 时， 若 R = ÷ ( m一 1) ( n+ 1) + 1， 则由归纳假设( 2 ) ， 知第3 行第一个格与第 2018 年第2 期 二个格异色， 这样， 与第3 行的格连通． 类 似地， A 也与第3 行的格连通． 由此， 在A ， 。，⋯ ， 一 中， 至多只有 段与第3 行中 的 格 不 连 通 ． 故 Q ≤ ， 这 与 q： 粤 的 假 设 矛盾． 因 此 ， 当 q： 或 q： 粤 时 ， 有 R ≤— ( m 一 1) ( n + 1) ． 则K = P + q + 尺 ≤ + + 1 ( m 一1 ¨ 1) ≤— — — ～ + — — 一+ ( m 一 ( ， 2+ 1) = ( m + 1) ( n + 1) + 1． 当等号成立时， 必须k 为奇数， 且 p 一 二 生 ± — ‘ 2 。 考虑P 的估计中等号成立的条件， 知第 l 行第一个格与最后一个格分别被三个异色 格包围． 由方格表的对称性， 知四角处的每个格 均不与其他格连通． 容易验证， 对于第i 行 列的格， 若 为 偶数将其染成黑色； 若 为奇数， 将其染成白 色． 在此两种染色方案下， 均有 = ( m + 1) ( n + 1) + 1． 综 上， 所 求 ： ÷ ( +1) +1 ． 6． 最大值为a1 a2⋯a ． 若x = a ( 1 ≤i ≤n) ， 贝 U 1， 2， ⋯， 满 足条件， 且XlX2⋯ = 0l 口 2⋯a ． 接下来证明： lx2⋯ ≤01 a2⋯a ． 当k ： 1 时， 结论显然． 由条件即得 ≤af( 1≤i ≤ ) ． 下面假设k > 1 2． 不失一般性， 设 a 1 一 1≤a 2 一 2 ≤⋯ ≤a 一x ． 若a1一 l > 1 0 ， 贝 4 0 ≥ f( 1 ≤i ≤ ) ， 兰 占 显然成立． 以下假设a1一 1< 0． 取， = {1 ， 2 ， ⋯， k}． 由条件知 ∑( — Xi) ≥ o． ① 故a 一 x > 1 0． 因此， 存在s( 1≤s < k) ， 使得 a 1 一 1 ≤⋯≤口 一戈 s < 0 ≤0 s +l 一 5+l ≤⋯ ≤0 一 ≤⋯ ≤a n 一 ． 记di = I a 一 {( 1≤ ≤n) ． 贝 0 d 1 ≥ 2 ≥⋯≥ > 0 ， 0 ≤ +1 ≤⋯≤ ≤⋯≤d ． 由式①知 一∑d + ∑d I >0 ∑d ≥ ∑d ． 记M=∑d ， N：∑d ． 则 ． ： ≥罂≥ ． 注意到， 对 > s， 有 < < k． 利用均值不等式得 薏 = ( ( + 妄 ))( ( 一 毒 )) ≤ + 肌 ( 一 ≤ ( + 耋 。 (·一 = (1+ 一 ≤ ( + 志 一 ) ≤ ( + 一 ) = (， + ( k 1一 ≤1 ． 从而， 结论得证． ( 瞿振华 熊斌 提供) 24 若X、Y均为正整数，由 M =5m1 +X +3Y =2 018, 知X+3Y=2 018, 且 X +3Y=2 018 =3(mod 5). 于是，X最大为144,相应Y=233.
故Z =X + (1 +)..) Y=377 +233入 IHI 1 IZI=-=-;,-;?: Z 377 +233入＇ 当X= 144, Y = 233时，上式取得等号，此时， m1 = m4 =235 ,Đ= m4 =468 ,m。=612.
若X、Y均为负整数，设X=-U ,Y= -V, 则 (u' V) = (J ,.'J ,. + I) . 由M =5m。+4U+2V =2 018, 知 4U +2V =2 018, 且 4U+2V 三2 018心(mod 5). 于是，U最大为55,相应V=89.
故IZl =IU+(l+ 入）VI =55 + (1 + 入）89 = 144 +89入， IHI 1 1 IZI=-=--;,-;?: > IZI 144 +89入377 +233入 ． （上接第11页） 4. 在丛ABC 中，AD、BE、CF为三条内角 平分线，I为内心，线段AD的中垂线与BE、 CF分别交于点M、N证明：A ... N、I、M四点 共圆 (2009, 俄罗斯数学奥林匹克） 提示 运用推论1'知A、B... D、M,A、C、 D、N分别四点共圆 则乙AMB=乙ADB,乙ANG=乙ADC.
故乙AMI+乙AN!=乙AMB+乙ANG ＝乙ADB +乙ADC =180°. 由此即得结论 s. 已知O为丛ABC的内切圆圆心，K为 丛BOC的外接圆与乙A的平分线的交点，L 为丛AOC的外接圆与乙B的平分线的交点， P为线段KL的中点，M关于点P的对称点 为O,N关于KL的对称点也为0. 证明：四边 中等数学 (2) IHIđ4.
若X 、Y均为正整数，则 IZl=IX一入Y I= IHI X+(l+il)Y 4 4 1 三X+3Y气而j > 377 +233入 ． 若X、Y均为负整数，令X=-U ,Y= -V, 则 IHI IZI= IU-入VI=u + (l + 入）V 4 4 1 气U+2V丐丽 > 377 +233入 ． 综上，当且仅当Z = 144 - 233入，m。= 612 ,m1 = m4 = 235, m2 = m3 = 468时，IZI取 到最小值 故使S最大的2 018个点放置方式为从 某个顶点开始，逆时针方向依次在正五边形 的五个顶点上放置612、235、468、468、235 个点 （熊斌 提供） 形应MN为等腰梯形 (2009, 马其顿数学奥林匹克） 提示 运用性质3 (2), 可知L、K为 丛ABC的两个旁心则L 、C、K 三点共线 注意到，PCII MN, 四边形LOKM为平行 四边形，以及LC为ON的中垂线由此即证． 参考文献： [ 1] 第34届俄罗斯数学奥林匹克（第四轮）(2008) [J].
中等数学，2009(增刊二）． 第56届斯洛文尼亚数学奥林匹克(20I2)[J]. 中等数 学，2013(增刊二）． 2010伊朗国家队选拔考试[J].中等数学，2011(增刊二）． [ 4] 2012克罗地亚国家队选拔考试[J]. 中等数学，2013 （增刊二）． 《中等数学》编辑部 编．国内外数学竞赛题及精解 (2016-2017)[M]. 哈尔滨：哈尔滨工业大学出版社， 2018,7.
2015中国国家集训队选拔考试[J].中等数学，2015(5). |
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In this activity, you can simulate throwing darts randomly at a square. Once you do so, a sampling square will appear. You can use the point to move this sampling square around to see how many dots you have sampled. The sample square is 1/16th of the area of the square dart board. Can you figure out how many total darts were thrown without counting all of them? Once you have an idea, enter it into the box and click on "Check Estimate".
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maximum contiguous sum in a circular buffer
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I have a program to determine the largest contiguous sum in an array, but want to extend it to work with circular arrays. Is there an easier way to do that than doubling the single array and calling my function to find the largest sum over all n-length arrays in the 2n length array?
arrays
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asked May 18, 2011 at 15:58
rachrach
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what are your requirements? What is the range of integers (?) the array can contain? Can it contain negative elements? How many elements will the array contain at most? To me this screams _Dynamic Programming_ and there's probably a DP algo whose runtime runs around anything else... –SyntaxT3rr0r Commented May 18, 2011 at 16:11
@SyntaxT3rr0r, that's famous problem and if there is no negative number sum of all items is answer (my be there are some other answers) so I think there is no specific ambiguity, and @rach, what's wrong with doubling list? order is O(n) (space and time) and nothing changes, and I think it's not a big overhead. –Saeed Amiri Commented May 18, 2011 at 16:23
1 @Saeed: Asking for requirements is the very first thing that I always do... If the array can contain hundreds of millions of elements ranging from -263 to 263-1 than I can assure you the "regular" answer won't work as easily for it. Because you'll need to store the sum on more than one "long". And what if there's not enough memory to store intermediate results? It's nearly always a time/memory trade-off. The 0-1 knapsack is famous too. Yet there are constraints. Asking for requirements is 101. And an "integer" in computing is hardly an "integer" in math... etc. –SyntaxT3rr0r Commented May 18, 2011 at 16:41
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See the following link :
It solves a problem using Kadane Algorithem.
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answered May 16, 2013 at 12:14
Nirdesh SharmaNirdesh Sharma
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I think the solution by @spinning_plate is wrong. Ca you please test it for the given cases.
int arr[] = {-3, 6, 2, 1, 7, -8, 13, 0};
Your approach returns 21.
Actual solution can be start from 6th index(i.e. 13 value) .. and end to 4th index(i.e. 7 value). Since array is circular we can take continuous series from 6th index to 7th index and from 0th index to 4th index.
The actual answer for the above case is : 26
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answered May 19, 2012 at 19:18
cksharmacksharma
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Well, you don't have to actually double the array. You can just emulate it by indexing your existing array modulo n, or by just iterating over it twice. Depending on the size of your array and cache behavior, this should be at most a factor of two slower than the algorithm for the noncircular array.
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answered May 18, 2011 at 16:09
hammarhammar
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For the given problem, We will apply kadane algorithm and we will also find the subset which will have maximum negative value.If the maximum negative value is removed that will give the sum of the remaining array in circular order.If that sum is greater than maximum sum then maximum sum will be sum in circular order. Complexity for the algorithm is O(n).
Eg:- arr[i]={10,-3,-4,7,6,5,-4,-1}
Ans: max\_sum=7+6+5+(-4)+(-1)+10
Removed\_set={-3,-4}
int find\_maxsum(int arr[],int n)
{
int i=0;
int total=0;
int maxa=0;
int mini=0;
int min\_sum=0;
int max\_sum=0;
while(imax\_sum)
max\_sum=maxa;
if(mini=0)
mini=0;
}
if(total-min\_sum>max\_sum)
max\_sum=total-min\_sum;
return max\_sum;
}
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answered Jan 27, 2015 at 14:55
Prakhar AgrawalPrakhar Agrawal
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I assume you are using the O(n) algorithm that continues adding to the sum, keeping track of the maximum, only restarting if you sum to a negative number. The only thing you need to do to capture the case of circular arrays is to apply the same principle to the circular aspect. When you reach the end of the array in the original algorithm, keep looping around to the start until you go below the maximum or hit the beginning of the current range (I think this is impossible, though, because if the solution was the full array, we sould have seen this on the first pass), in which case you're done.
max\_start=0; max\_end =0; maxv = 0; sum 0;
for i in range(arr):
sum+= arr[i];
if sum<0:
sum=0; max\_start =i;
if maxv arr[ix] ? sum + arr[ix] : arr[ix];
max = max > sum ? max : sum;
}
return max;
}
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answered Sep 14, 2013 at 22:27
lcnlcn
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This code returns the correct answer even if all numbers are negative e.g., {-1, -2, -3}. will return -1;
public static int maxSubarraySumCircular(int[] A) {
int maxSum = Arrays.stream(A).max().getAsInt();
if (maxSum < 0)
return maxSum;
int maxKadane = KadaneAlgorithm(A);
int maxWrap = 0;
for (int i = 0; i < A.length; i++) {
maxWrap += A[i];
A[i] = -A[i];
}
maxWrap = maxWrap + KadaneAlgorithm(A);
return maxWrap > maxKadane ? maxWrap : maxKadane;
}
private static int KadaneAlgorithm(int[] A) {
int maxSoFar = 0;
int maxEndingHere = 0;
for (int i = 0; i < A.length ; i++) {
maxEndingHere = maxEndingHere + A[i];
if (maxEndingHere < 0 )
maxEndingHere = 0;
if(maxSoFar < maxEndingHere)
maxSoFar = maxEndingHere;
}
return maxSoFar;
}
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answered May 15, 2020 at 20:34
liolio
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10645 | https://www.youtube.com/watch?v=Hb7lie77Rho | Synthetic Division and the Rational Root Theorem
Thomas Gamache
213 subscribers
2 likes
Description
400 views
Posted: 9 Dec 2020
I go through using synthetic division to find the roots of polynomials, including using the rational root theorem
Transcript:
Rational Root Theorem okay we're going to go through with using the rational root theorem and synthetic division to factor out polynomials so we're going to start out with this polynomial here so using the rational root theorem what that tells me is i want to take all the factors of my constant term which we call p so notice i'm writing out factors of constant term we refer to this as p so i'm going to look at my constant term here and i have five for a constant term okay it's pretty easy the only factors of five are one and five luckily we picked an easy one but like say it was six then you'd have to write out one six two and three those would be all the factors and so on you would do that based on your constant term now next i'm going to go to the coefficient of my highest degree what we call the lead coefficient we've been using that before so my lead coefficient here is five so i want the factors of my lead coefficient and again my lead fact my factors are 1 and 5. so what do i do from here what i do is i'm going to divide the factors of my constant term by my lead coefficient so i would take 1 my possible roots are all going to be p over q all the p's over all the q's so in this case it would be i could have one over five that's gonna be positive or negative or i could have one over one positive or negative and then i would go the other way around too i could have five over one positive or negative again and i could also have five over five positive or negative so if i simplify these out you'll notice that one repeats itself one and one so really i have three possible factors here or three possible roots actually i have more than that and you'll see what i mean in a moment and then i have positive or negative five these are all my possible roots so technically my possible roots could be negative one-fifth and positive one-fifth i could have positive one i could have negative one and i could have positive five and i could have negative five so i have six possible roots now i know since my degree is three i only have three roots which means three of these aren't going to work and three are so it's a matter of figuring out what your roots are one way you can do that we've talked about this you can set up your whole polynomial to equal zero you can do 2x squared plus 19x minus 5 equals zero and then you would have to plug in every single value so i start with negative one-fifth okay plug that in for x do out the whole thing see if i get an answer of zero then i would have to go and plug in one-fifth do out the whole to do out the whole equation see if i get an answer of zero what i mean by that is because someone asked i'd have to literally plug in these all these values for x one fifth minus five equals zero so i'd have to do out this entire thing and then hope at the end that i get an answer of zero equals zero if that happens and that's a root if i don't get that then that means whatever i plugged in is not a root i would have to do that for six cases here that's a lot of math and you don't have to go that route one way you can do this is using synthetic division which i'm going to show you so i'm just going to shrink this stuff down here all my the stuff i did with my rational roots my rational root theorem so what i can do now is i'm going to try one of these i'm going to start out with i'm going to go with one you know why because it's easy when possible go with something so Synthetic Division i'm going to try one first here's how you would use synthetic division i'm going to start with one i'm going to kind of box it up here i'm now going to write out all my coefficients within my polynomial as well as my constant term my coefficients are 5 29 19 and negative 5 would be that constant term right here negative 5. all right and this is how synthetic division works all right i'm going to draw a line i'm going to drop down my five i'm going to drop it on down i would then do 5 times 1 i get 5. bring it over here 29 plus 5 gives me 34. see 34 times 1 gives me 34. 19 plus 34 gives me 53. 53 times 1 gives me 53. and then we do negative 5 plus 53 which gives me an answer of positive 48. what does that mean this would technically be my remainder this is not zero we want to equal zero since this is not zero what that tells me is one is not a root okay so we know one's not a root let's try another one here let's try let me go with positive i'm gonna go with negative five this time we'll try negative five see if that works so again it's the same process box up my five now i'm going to write out those coefficients and the constant term 5 29 19 negative 5. all right let's do this out so i'm going to draw my line start by dropping down the five five times five gives me 25 29 plus 25 gives me 54. 54 times five it's getting messy but we can handle this 270. 270 plus 19 gives me 289. 289 times five i'm getting something big now i believe that would give me 1445. now when i do this out i know i'm gonna i'm not gonna get zero this is not zero so guess what i found something else that is not a root so i'm narrowing down my choices now i know i have three roots i know one of them is not one i know one of them is not negative five let's try another one here i'm going to try positive five this time oh actually i did try positive five whoops no biggie i'll try negative five this time let's try negative five let's see if that works so again it's the same process negative five i'm going to write down my coefficients again and my constant term 5 29 19 and negative 5. all right drop down my five five times negative five gives me negative 25 29 minus 25 or plus negative 25 gives me four four times negative five gives me negative twenty nineteen minus twenty gives me negative one negative one times negative five gives me positive five and negative five plus five gives me zero guess what that's what i want to get here so i like zero my students give me a hard time because i like to put the exclamation points because now i know negative five is a root so i found one of my roots there are three roots in this problem i found one of them so what you could do is you could do out the rest of them using synthetic division however here's a nice way to get those final two roots once you find one right we started out with x to the third right since we've factored this out a bit this is what we're left with five x squared plus four x minus one i'm gonna set this to equal zero and then i'm gonna factor this out i can use the ac method which means let's see a times c equals negative 5 because again i'm multiplying 5 negative 1 that's a times c i know b is positive 4 and now i want to think of two factors of negative five that give me positive four that would be five and negative one so i'm going to use my box method now as we've done in here i'm gonna go with let's see four terms got five x squared here i know that 5x is a factor and i also noticed that negative 1. so as we said when we did it out we knew that we 5 and negative 1 that gives me b i like that and negative 1. and then what i do is i just factor this out 5x going across going across again i can factor out negative 1. and i go horizontally 5x squared and negative 1x i can factor out x and then 5x negative 1 i can factor out 1. so if i factor out my polynomial now i get 5x minus 1 then x plus 1. and now to figure out my roots all i got to do is set them to each equals zero five x minus one equals zero x plus one equals zero and then if i do out my math i realize i get x equals one-fifth i also notice that i get x equals negative one guess what i found my other two roots and let's see were those some of the possibilities i started out with i believe they are one-fifth and negative one going up to that rational root theorem one-fifth yep and negative one i found my three factors so that's how you can use synthetic division to find the roots of a polynomial that you can't factor easily now people mentioned sometimes you can graph them that's that's cool but the thing is if you graph this it might be hard to see that one-fifth because you'd really have to break that fact uh the graph up into smaller values because you typically just look for when it passes through a whole number right when you graph something like especially if this is cubed you know it's going to look something like this one two oh it's not gonna look like that excuse me see it goes up once there we go one two three it's probably gonna look something like that i'm just making it up but if you get one fifth is one of those values it's gonna be kind of hard to see so that's why we you know that's why it's not easy to just always graph them you want to figure out what those roots actually are so that's how you can use synthetic division in the rational root theorem to factor out a polynomial now primarily you're going to be working with synthetic division in here you'll be given roots to try and sometimes you might have to use the rational root theorem to figure out what roots so we're going to do one example here Example where i did one using the rational root theorem i'm going to give you a synthetic division problem where you are trying out a factor trying out the root to see if it works so let's try let's see i got an example right here we're going to go with y to the fourth eight y to the third plus ten y squared plus two y plus four and what we're trying to see is if we can factor out y minus two now remember when we're talking about roots and you know we're talking about being in intercept form it's always a variable minus something so the root we would be trying here would be 2. so let's set this up so i'm going to go with 2 and then i'm going to look at my coefficients my first coefficient here i don't have anything so i'm going to go with a 1. my second coefficient i got negative 8 my third i've got positive 10. my fourth i've got two then we go with that constant term with a four and let's see what happens we're going to try positive we're going to try positive two see if it's a root so i'm gonna draw my line i'm gonna drop down the 1. 1 times 2 gives me 2. negative 8 plus 2 gives me negative 6 negative 6 times negative 2 gives me negative 12. 10 plus negative 12 gives me negative 2. negative two times two gives me negative four two minus four gives me negative two negative two times two gives me negative four four plus negative four gives me zero that's awesome i like seeing that because what that tells me is two is a root now i could try here too if i wanted to i could try some other roots but ultimately you just want to be able to try and root out a polynomial and see if it works using synthetic division hopefully this gives you a little bit of guidance into how to do these especially if you have to use the rational root theorem first again that can give you a lot of possibilities so it's good to be able to try one and then be able to factor it like we did in the first one here where we were able to find a root and then based on that we were able to factor that remaining polynomial but ultimately remember try a root go with the coefficients and your constant term using synthetic division drop down the first term then just multiply this times this add it multiply this and this add it multiply this and this add it and then at the end you want to get zero |
10646 | https://australian.museum/learn/animals/mammals/house-mouse/ | House Mouse - The Australian Museum
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House Mouse
House Mouse
Scientific name:Mus musculus
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Updated 12/07/24
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House Mouse, Smiths Lake N.S.W. Image: Dick Whitford
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ClassificationGenus Mus Species musculus Family Muridae Order Rodentia Subclass Eutheria Class Mammalia Subphylum Vertebrata Phylum Chordata Kingdom Animalia
Size Range Body 60 mm - 100 mm, Tail 75 mm - 100 mm, Weight 10 g - 25 g
Introduction
Besides humans, the House Mouse is probably one of the most successful living mammals, with a widespread distribution throughout Australia and the world.
Identification
The distinguishing features of the House Mouse include:
Front teeth: One pair of distinctive chisel shaped incisors with hard yellow enamel on front surfaces.
Head: Bulging eyes in a small head.
Ears: Large rounded ears.
Colouring: Brownish-grey above, white to grey or pale yellow below; soft dense fur to 7mm long.
Tail: Scaly tail, about same length as body.
Similar species
Sometimes confused with small marsupials e.g. Antechinus
Habitat
The House Mouse lives mainly in urban areas, being closely associated with humans. It prefers to live in secluded parts of buildings, reed beds, cracks in the ground or shallow burrow systems, make nests of shredded materials.
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The House Mouse is found throughout Australia. At times it may reach plague proportions. Like humans, it is so successful because of its adaptability in utilising available resources.
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Populations can increase quickly as the House Mouse is able to breed at two months of age and its litter size is between four and eight. Populations of the House Mouse increase significantly about 18 months after a fire, at a time when native rodent numbers are low. Populations also appear to increase after rainfall. The large populations of House Mice may last three to four years until the native rodent populations increase and the House Mouse population declines.
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Monkey faced bat from the Solomon IslandsPteralopex
Bare-nosed WombatVombatus ursinus
Short-eared Brushtail PossumTrichosurus caninus
Mountain Brushtail PossumTrichosurus cunninghami
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10647 | http://mrsdoolan.weebly.com/uploads/1/0/8/3/10839036/unit_6_ratio_equivalent_worksheet_3.pdf | Name : Teacher : Date : Score : Math-Aids.Com Equivalent Ratios Worksheet 3 Equivalent Ratios Equivalent Ratios Write two equivalent ratios.
1) 4 3 2) 9 11 3) 7 6 4) 4 9 5) 3 4 6) 11 9 Determine whether the ratios are equivalent.
7) 11 2 __ and 55 10 __ _ 8) 11 3 __ and 7 12 __ _ 9) 7 3 _ and 14 6 __ _ 10) 8 3 _ and 32 12 __ _ 11) 2 9 _ and 12 54 __ _ 12) 10 7 __ and 7 8 _ _ Use equivalent ratios to find the unknown value.
13) 3 5 _ = v 25 __ v = _ 14) r 16 __ = 11 8 __ r = 15) r 8 _ = 5 4 _ r = 16) 15 k __ = 3 11 __ k = _ 17) 5 3 _ = 15 c __ c = 18) 3 8 _ = 18 c __ c = Name : Teacher : Date : Score : Math-Aids.Com Equivalent Ratios Worksheet 3 Equivalent Ratios Equivalent Ratios Write two equivalent ratios.
1) 4 3 8 6 12 9 2) 9 11 18 22 27 33 3) 7 6 14 12 21 18 4) 4 9 8 18 12 27 5) 3 4 6 8 9 12 6) 11 9 22 18 33 27 Determine whether the ratios are equivalent.
7) 11 2 __ and 55 10 __ _ Yes 8) 11 3 __ and 7 12 __ _ No 9) 7 3 _ and 14 6 __ _ Yes 10) 8 3 _ and 32 12 __ _ Yes 11) 2 9 _ and 12 54 __ _ Yes 12) 10 7 __ and 7 8 _ _ No Use equivalent ratios to find the unknown value.
13) v = _ 3 5 _ = v 25 __ 15 14) r = r 16 __ = 11 8 __ 22 15) r = r 8 _ = 5 4 _ 10 16) k = _ 15 k __ = 3 11 __ 55 17) c = 5 3 _ = 15 c __ 9 18) c = 3 8 _ = 18 c __ 48 |
10648 | https://mathcenter.oxford.emory.edu/site/math117/probSetBayesTheorem/ | Exercises - Bayes' Theorem
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Exercises - Bayes' Theorem
Company A supplies 40% of the computers sold and is late 5% of the time. Company B supplies 30% of the computers sold and is late 3% of the time. Company C supplies another 30% and is late 2.5% of the time. A computer arrives late - what is the probability that it came from Company A?
Bayes Theorem. Make a tree: P(L)\=0.0365P(L)\=0.0365 and P(A and L)\=(0.4)(0.05)\=0.02P(A and L)\=(0.4)(0.05)\=0.02, so P(shipped from A given that the computer is late) = 0.548, approximately.
In Orange County, 51% of the adults are males. One adult is randomly selected for a survey involving credit card usage. It is later learned that the selected survey subject was smoking a cigar. Also, 9.5% of males smoke cigars, whereas 1.7% of females smoke cigars (based on data from the Substance Abuse and Mental Health Services Administration). Use this additional information to find the probability that the selected subject is a male.
P(M|S)\=(0.51)(0.095)(0.51)(0.095)+(0.49)(0.017)≐0.853P(M|S)\=(0.51)(0.095)(0.51)(0.095)+(0.49)(0.017)≐0.853
A person uses his car 30% of the time, walks 30% of the time and rides the bus 40% of the time as he goes to work. He is late 10% of the time when he walks; he is late 3% of the time when he drives; and he is late 7% of the time he takes the bus.
What is the probability he took the bus if he was late?
What is the probability he walked if he is on time?
P(B|L)\=(0.40)(0.07)(0.40)(0.07)+(0.30)(0.03)+(0.30)(0.10)≐0.418P(B|L)\=(0.40)(0.07)(0.40)(0.07)+(0.30)(0.03)+(0.30)(0.10)≐0.418
P(W|T)\=(0.30)(.90)(0.30)(0.97)+(0.30)(0.90)+(0.40)(0.93)≐0.289P(W|T)\=(0.30)(.90)(0.30)(0.97)+(0.30)(0.90)+(0.40)(0.93)≐0.289
In a study of pleas and prison sentences, it is found that 45% of the subjects studied were sent to prison. Among those sent to prison, 40% chose to plead guilty. Among those not sent to prison, 55% chose to plead guilty.
If one of the study subjects is randomly selected, find the probability of getting someone who was not sent to prison.
If a study subject is randomly selected and it is then found that the subject entered a guilty plea, find the probability that this person was not sent to prison.
1−0.45\=0.551−0.45\=0.55
(0.55)(0.55)(0.45)((0.40)+(0.55)(0.55)≐0.627(0.55)(0.55)(0.45)((0.40)+(0.55)(0.55)≐0.627
On a game show, a contestant can select one of four boxes. The red box contains one $100$100 bill and nine $1$1 bills. A green box contains two $100$100 bills and eight $1$1 bills. A blue box contains three $100$100 bills and seven $1$1 bills. A yellow box contains five $100$100 bills and five $1$1 bills. The contestant selects a box at random and selects a bill from the box at random. If a $100$100 bill is selected, find the probability that it came from the yellow box.
(0.25)(0.50)(0.25)(0.10)+(0.25)(0.20)+(0.25)(0.30)+(0.25)(0.50)≐0.455(0.25)(0.50)(0.25)(0.10)+(0.25)(0.20)+(0.25)(0.30)+(0.25)(0.50)≐0.455
A plane's "black-box" is manufactured by only 3 companies: AirCorp, BigSkies, and CharterUS - who make 80%, 15%, and 5% of all the black-boxes made, respectively. Invariably, some of these are defective. Assuming the percentage of defective black-boxes made by AirCorp, BigSkies, and CharterUS are 4%, 6%, and 9%, respectively, find the probability that a randomly selected black-box from all black-boxes made that is found to be defective came from AirCorp.
(0.80)(0.04)(0.80)(0.04)+(0.15)(0.06)+(0.05)(0.09)≐0.7033(0.80)(0.04)(0.80)(0.04)+(0.15)(0.06)+(0.05)(0.09)≐0.7033
Consider 3 coins where two are fair, yielding heads with probability 0.500.50, while the third yields heads with probability 0.750.75. If one randomly selects one of the coins and tosses it 3 times, yielding 3 heads - what is the probability this is the biased coin?
First note two things: 1) the probability of drawing a fair coin is 2/32/3 and the probability of drawing a biased coin is 1/31/3; and 2) the probability of tossing 3 heads with a fair coin is (1/2)3\=0.125(1/2)3\=0.125, while the probability of tossing 3 heads with the described biased coin is (0.75)3(0.75)3. Then, use Baye's Theorem:
(1/3)(0.75)3(2/3)(1/2)3+(1/3)(0.75)3≐0.6279(1/3)(0.75)3(2/3)(1/2)3+(1/3)(0.75)3≐0.6279
Suppose P(A),P(A¯¯¯¯),P(B|A)P(A),P(A¯),P(B|A), and P(B|A¯¯¯¯)P(B|A¯) are known. Find an expression for P(A|B)P(A|B) in terms of these four probabilities.
P(A)P(B|A)P(A)P(B|A)+P(A¯¯¯¯)P(B|A¯¯¯¯)P(A)P(B|A)P(A)P(B|A)+P(A¯)P(B|A¯)
Assume the probability of having tuberculosis (TB) is 0.0005, and a test for TB is 99% accurate. What is the probability one has TB if one tests positive for the disease?
(0.0005)(0.99)(0.0005)(0.99)+(0.9995)(0.01)≐0.0472(0.0005)(0.99)(0.0005)(0.99)+(0.9995)(0.01)≐0.0472
An automobile manufacturer has three factories: A, B, and C. They produce 50%, 30%, and 20% respectively, of a specific model of car. 30% of the cars produced in factory A are white, 40% of those produced in factory B are white, and 25% produced in factory C are white.
If an automobile produced by the company is selected at random, find the probability that it is white.
Given that an automobile selected at random is white, find the probability that it came from factory B.
(0.50)(0.30)+(0.30)(0.40)+(0.20)(0.25)\=0.32(0.50)(0.30)+(0.30)(0.40)+(0.20)(0.25)\=0.32
Given the calculation in part (a), we have (0.30)(0.40)0.32\=0.375(0.30)(0.40)0.32\=0.375
Two manufacturers supply blankets to emergency relief organizations. Manufacturer A supplies 3000 blankets and 4% are irregular in workmanship. Manufacturer B supplies 2400 blankets and 7% are found to be irregular. Given that a blanket is irregular, find the probability that it came from manufacturer B.
(24005400)(0.07)(24005400)(0.07)+(30005400)(0.04)≐0.5833(24005400)(0.07)(24005400)(0.07)+(30005400)(0.04)≐0.5833 |
10649 | https://artofproblemsolving.com/wiki/index.php/Feuerbach_point?srsltid=AfmBOorZxYD1GO9AXfuMu4AJAEC6ECrKEKeGrIrg6G_EPgrQFCtgqD5f | Art of Problem Solving
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Feuerbach point
The incircle and nine-point circle of a triangle are tangent to each other at the Feuerbach point of the triangle. The Feuerbach point is listed as X(11) in Clark Kimberling's Encyclopedia of Triangle Centers and is named after Karl Wilhelm Feuerbach.
Contents
[hide]
1 Sharygin’s proof
1.1 Claim 1
1.2 Claim 2
1.3 Claim 3
1.4 Claim 4
1.5 Sharygin’s proof
2 Scalene triangle with angle 60^\circ
Sharygin’s proof
Russian math olympiad
Claim 1
Let be the base of the bisector of angle A of scalene triangle
Let be a tangent different from side to the incircle of is the point of tangency). Similarly, we denote and
Prove that are concurrent.
Proof
Let and be the point of tangency of the incircle and and
Let WLOG, Similarly, points and are symmetric with respect
Similarly,
are concurrent at the homothetic center of and
Claim 2
Let and be the midpoints and respectively. Points and was defined at Claim 1.
Prove that and are concurrent.
Proof
are concurrent at the homothetic center of and
Claim 3
Let be the base of height Let Prove that points and are concyclic.
Proof
tangent to
Denote Point lies on radical axis of circles centered at and with the radii and respectively. Therefore points and are concyclic.
Claim 4
Prove that points and are concyclic.
Proof
and are concyclic points and are concyclic.
Sharygin’s proof
The incircle and the nine-point circle of a triangle are tangent to each other.
Proof
Let
According claim 4, each of this point lyes on
and have not more then two common point, so two of points and are coincide.
Therefore these two points coincide with point witch means that
is the center of similarity of and therefore there is no second point of intersection of and
We conclude that these circles are tangent to each other at point
vladimir.shelomovskii@gmail.com, vvsss
Scalene triangle with angle 60^\circ
The Feuerbach point of a scalene triangle lies on one of its bisectors. Prove that the angle corresponding to the bisector is
Proof
Denote given triangle, semiperimeter of are incircle, B-excircle, and nine points circle centered at and Feuerbach point, the inradius, exradius.
It is known that has diameter (half of diameter of circumcircle).
is tangent to points and lies on bisector, so Another proof Feuerbach point of a scalene triangle .
vladimir.shelomovskii@gmail.com, vvsss
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10651 | https://www.youtube.com/watch?v=pO_rQMRe25U | Using square root method to solve a quadratic equation with one variable
Brian McLogan
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Posted: 31 Aug 2015
👉Learn how to solve quadratic equations using the square root method. It is important to understand that not all quadratics have to be solved using factoring or quadratic formula. When we only have one variable but it is squared we can just use inverse operations to isolate the variable. The inverse operation of squaring is taking the square root. Hence why we call this the square root method. All equations can be solved using the square root method but when there is more than one variable we have to complete the square first.
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Transcript:
this is algebra 2 what are you talking about okay so anyways ladies and gentlemen when we're going through a problem like this the main important thing we're going to be solving quadratic equations all right now the first process that I had you guys written down is make sure you solve the equation set it equal to zero so you guys can see in this equation I have my quadratic set equal to zero right good now the main thing that makes a quadratic makes a quadric equation is when you have a variable squared all right now um we use factors ing when we have more than one variable however in this case we don't actually have to use factoring because I have my I only have one variable when you only have one variable like a linear equation we can just use inverse operations so in this case this is actually my one example that we're not going to talk about factoring so all I'm going to do is I'm going to undo what's happening to my variable you can see my variable is being subtracted by 11 excuse me so I'm going to add 11 then so I have 7 x^2 = a positive 11 now my variable x^2 is being multiplied by S so I'll undo multiplying by seven by dividing by seven now 11 does not divide into seven so just leave it as a fraction fractions are okay so I have x^2 equals 11 over 7 now we need to undo the variable being squared so using my inverse operations I can do that by taking the square root and please remember though ladies gentlemen when you introduce the square root you have to include the positive and the negative value of the square root right the square root if I had x^2 = 4 and I take the square root of both sides X can equal plus or minus 2 right so whenever you introduce the square root make sure you include the positive and the negative so therefore x equal plus or minus the square root of 11/ 7 Now ladies and gentlemen I I don't want to go over your stuff but from Algebra 2 we want to make sure we simplify this to S < TK of 11 over the square < TK of 7 and instead of dividing by the square < TK of 7 we have to rationalize the denominator we'll get over this stuff again we'll go back over this but this is another thing that I just want to remind you of then you multiply 7 11 isqu < TK of 77 7 7 is the of 49 which is just 7 so that would be plus or minus < TK of 77 / 7 which you cannot divide the seven into that because that's the square root of that but that' be your final answer okay so the first box all you guys simply need to do is solve for X2 and once |
10652 | https://my.clevelandclinic.org/health/diseases/16753-atherosclerosis-arterial-disease | Abu Dhabi|Canada|Florida|London|Nevada|Ohio|
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Atherosclerosis
Atherosclerosis is a hardening of your arteries from plaque building up gradually inside them. Plaque consists of fat, cholesterol and other substances. This plaque buildup limits blood flow. You may not have symptoms of atherosclerosis until you have complications like a heart attack or stroke.
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ContentsOverviewSymptoms and CausesDiagnosis and TestsManagement and TreatmentOutlook / PrognosisPreventionLiving With
Overview
What is atherosclerosis?
Atherosclerosis is the gradual buildup of plaque in the walls of your arteries. Arteries are blood vessels that carry oxygen-rich blood to organs and tissues throughout your body. Plaque (atheroma) is a sticky substance made of fat, cholesterol, calcium and other substances.
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As plaque builds up, your artery wall grows thicker and harder. This “hardening of the arteries” is usually a silent process in the early stages. You may not notice symptoms for a long time. But eventually, as the plaque grows, the opening (lumen) of your artery narrows, leaving less room for blood to flow. This means less blood can reach your organs and tissues. Plus, the constant force of blood flow can lead to plaque erosion or rupture, causing a blood clot to form.
A narrowed artery is like a highway reduced to one lane. But a blood clot is like a barricade in the middle of the road. It blocks blood flow to certain organs or tissues the artery normally feeds. The effects on your body depend on where the blood clot forms. For example, blockages in a coronary artery deprive your heart of oxygen-rich blood, leading to a heart attack.
How common is atherosclerosis?
Atherosclerosis is very common. The complications of plaque buildup (including heart attacks and strokes) are the leading cause of death worldwide.
In America, about half of people age 45 to 84 have atherosclerosis but aren’t aware of it, according to the U.S. National Institutes of Health.
Symptoms and Causes
What are the symptoms?
Atherosclerosis symptoms often don’t start until an artery is very narrow or blocked. Many people don’t know they have plaque buildup until they have a medical emergency like a heart attack or stroke.
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You may notice symptoms of atherosclerosis if your artery is more than 70% blocked. You can have symptoms in different parts of your body, depending on the location of the blockage. Affected locations may include:
Heart
Coronary artery disease (CAD, which can lead to a heart attack) happens in the arteries that supply blood to your heart.
You may experience:
Shortness of breath (dyspnea) during light physical activity.
Chest pain or discomfort (angina).
Pain in your back, shoulders, neck, arms or belly.
Feeling dizzy or lightheaded.
Heart palpitations.
Fatigue.
Nausea or vomiting that may feel like indigestion.
Digestive system
Mesenteric ischemia happens when you don’t have enough blood flow in parts of your digestive system. You may experience:
Pain or cramping in your belly (abdomen) after eating.
Bloating, nausea and vomiting.
Diarrhea.
Unintentional weight loss due to “food fear” (fear of pain after eating).
Legs and feet
Peripheral artery disease (PAD) refers to poor blood flow in the arteries in your limbs.
You may experience:
Muscle pain (intermittent claudication).
Burning or aching pain in your feet and toes when you rest, especially when lying flat.
Changes in skin color (like redness).
Cool skin on your feet.
Frequent skin and soft tissue infections, often in your legs or feet.
Sores on your feet or toes that don’t heal.
Kidneys
With renal artery stenosis (narrowing of the arteries that take blood to your kidneys), you may experience:
Markedly elevated blood pressure that doesn’t respond to multiple medications.
Changes in how often you pee.
Swelling (edema).
Feeling drowsy or tired.
Skin that feels dry, itchy or numb.
Headaches.
Unexplained weight loss.
Nausea, vomiting or loss of appetite.
Brain
With carotid artery disease, the first symptoms are usually a transient ischemic attack (TIA) or stroke.
You may experience:
Dizziness.
Drooping on one side of your face.
Loss of feeling, loss of muscle strength or weakness on one side of your body.
Severe headache.
Slurred speech or difficulty forming words.
Vision loss in one eye. You may notice a dark shade coming down over your field of sight.
What causes atherosclerosis?
Researchers believe atherosclerosis causes may include:
High LDL cholesterol and triglyceride levels.
Tobacco products.
High blood pressure.
Diabetes.
These factors can damage your artery’s inner lining (endothelium). This causes atherosclerosis to begin. The damage usually occurs slowly and over time.
Stages of atherosclerosis
The stages of atherosclerosis happen over many years and include:
Endothelial damage and immune response. Damage to your endothelium triggers chemical processes that cause white blood cells to travel to the injury site. These cells gather and lead to inflammation within your artery.
Fatty streak formation. This is the first visible sign of atherosclerosis. It’s a yellow streak or patch of dead foam cells at the site of endothelial damage. In this case, foam cells are white blood cells that consume cholesterol to try to get rid of it. Continued foam cell activity causes further damage to your endothelium.
Plaque growth. Dead foam cells and other debris keep building up, turning a fatty streak into a larger piece of plaque. A fibrous cap (made of smooth muscle cells) forms over the plaque. This cap prevents bits of plaque from breaking off into your bloodstream. As the plaque grows, it gradually narrows your artery’s opening (lumen), so there’s less room for blood to flow through.
Plaque rupture or erosion. In this stage, a blood clot forms in your artery due to plaque rupture or plaque erosion. Plaque rupture happens when the fibrous cap that covers the plaque breaks open. With plaque erosion, the fibrous cap stays intact, but endothelial cells around the plaque get worn away. Both events lead to the formation of a blood clot. The clot blocks blood flow and can lead to a heart attack or stroke.
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What are the risk factors for atherosclerosis?
There are many risk factors for atherosclerosis. You may be able to change some of these risk factors, but not others (like age).
Risk factors include:
Being a male older than age 45.
Being a female older than age 55.
Having a biological family history of premature cardiovascular disease. This means a close male family member who received a cardiovascular disease diagnosis before age 45. Or a close female family member who was diagnosed before age 55.
Diabetes.
High blood pressure (hypertension).
High cholesterol (hyperlipidemia), especially high LDL cholesterol or high levels of a specific lipoprotein called lipoprotein (a).
Metabolic syndrome.
Smoking or tobacco use.
Lack of physical activity.
Eating foods high in saturated fat, trans fat, sodium and sugar.
What are the complications of atherosclerosis?
Atherosclerosis interferes with the normal workings of your cardiovascular system. It can limit or block blood flow to various parts of your body, including your heart and brain. Possible complications of reduced blood flow include:
Carotid artery disease.
Coronary artery disease.
Heart attack.
Arrhythmia (abnormal heart rhythms).
Heart failure.
Mesenteric ischemia.
Peripheral artery disease.
Chronic kidney disease (CKD).
Renal artery stenosis.
Stroke.
Transient ischemic attack (TIA).
Aneurysms and rupture of blood vessels from weakened artery walls.
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Early diagnosis and treatment of atherosclerosis can help you avoid or delay complications.
Diagnosis and Tests
How is atherosclerosis diagnosed?
To diagnose atherosclerosis or calculate your risk for developing it, a healthcare provider will:
Perform a thorough physical exam. This includes using a stethoscope to listen to your heart and blood flow through your arteries. For example, your provider will check your carotid arteries (in your neck) for a whooshing sound called a “bruit.” This sound may indicate the presence of plaque.
Ask about your medical history and family history. These details can help show your risk for atherosclerosis and its complications.
Ask about your lifestyle. Your provider may ask about habits like past or present use of tobacco products.
Order blood tests. Cardiac blood tests show your cholesterol levels and many details about your heart function.
What tests will be done to diagnose atherosclerosis?
Your healthcare provider may order additional tests to diagnose atherosclerosis and plan treatment. These tests include:
Angiography. This test uses special X-rays and a contrast dye to locate and measure blockages.
Ankle-brachial index. This test compares the blood pressure in your ankle to the pressure in your arm to measure blood flow in your arms and legs.
Chest X-ray. A chest X-ray takes pictures of the inside of your chest.
CT scan (computed tomography scan). This scan takes pictures of the inside of your body and can show any hardening and narrowing of your large arteries.
Heart MRI (magnetic resonance imaging). This test can show issues with blood flow in your arteries.
Echocardiogram (echo). An echo takes pictures of your heart’s valves and chambers and measures how well your heart pumps.
Electrocardiogram (EKG). An EKG measures your heart’s electrical activity, rate and rhythm.
Exercise stress test. This test measures your heart function while you’re physically active.
Carotid ultrasound. This test takes ultrasound pictures of the arteries in your neck (carotid arteries). It can detect hardening or narrowing of these arteries as blood flows to your brain.
Abdominal ultrasound. This ultrasound takes pictures of your abdominal aorta. It checks for ballooning (abdominal aortic aneurysm) or plaque buildup in your aorta.
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Management and Treatment
How is atherosclerosis treated?
Atherosclerosis treatment includes lifestyle changes, medications, procedures or surgeries. Depending on the location of your atherosclerosis, you may see a cardiologist (heart), nephrologist (kidneys), neurologist (brain and spine) or vascular surgeon (blood vessels).
Your healthcare providers will develop a plan based on your needs. Common treatment goals include:
Lowering your risk of blood clots.
Preventing complications like a heart attack or stroke.
Easing symptoms.
Helping you develop patterns of eating that support your heart and blood vessels.
Slowing or stopping plaque buildup in your arteries.
Improving blood flow by widening your arteries or bypassing (avoiding) blockages.
Lifestyle changes
Lifestyle changes may lower your risk of complications. Your provider will create a plan specific to your needs. General tips include:
Avoid all tobacco products (including smoking and vaping).
Follow a heart-healthy eating plan like the Mediterranean diet.
Build physical activity into your daily routine.
Medications
Medications target risk factors for plaque buildup and may help slow the progression of atherosclerosis. Your provider may prescribe medications that lower your blood pressure or cholesterol, manage your blood sugar levels and prevent blood clots.
Procedures or surgeries
Various minimally invasive procedures and complex surgeries can help people with severe blockages or a high risk of complications. Common treatment options include:
Angioplasty.
Atherectomy.
Carotid endarterectomy.
Coronary artery bypass grafting (CABG).
Peripheral artery bypass.
Stent placement.
Vascular disease bypass.
Care at Cleveland Clinic
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Outlook / Prognosis
What can I expect if I have atherosclerosis?
Early diagnosis and treatment can help people with atherosclerosis keep doing the activities they like to do. But the disease can cause medical emergencies and even be fatal. That’s why knowing your risks and working with your healthcare provider to lower them is important.
Prevention
Can atherosclerosis be prevented?
You may not be able to prevent atherosclerosis. But you can reduce your risk and lessen the effects of the disease. Here are some steps you can take:
Eat foods low in saturated fat, trans fat, cholesterol, sodium (salt) and sugar.
Get regular physical activity. Start with short walks and build up to 30 minutes a day most days of the week.
Keep a weight that’s healthy for you. Ask your provider what that should be.
Manage any health conditions, especially diabetes, high blood pressure and high cholesterol.
Don’t use tobacco products.
Have a yearly checkup with a healthcare provider.
Living With
How do I take care of myself?
It’s essential to work closely with your healthcare provider. They’ll keep a close eye on your condition and tell you how often you should come in for appointments. Go to all of your appointments and be an active partner in your care. Tell your provider right away about any new or changing symptoms. Early treatment can lower your risk of life-threatening complications.
Also, take care of your mental health. It’s normal to feel anxious about what the future could bring. You may also feel overwhelmed by the need to make lifestyle changes. But those feelings shouldn’t prevent you from enjoying life. Some tips for managing your thoughts and worries include:
Share your feelings with a counselor or support group.
Connect with others who have cardiovascular disease. You can share your experiences and learn from each other.
Talk to your family and friends about the lifestyle changes you’re making. Explain why these changes are important to you and ask them to help you stay on track.
When should I go to the ER?
Call 911 or your local emergency number right away if you or someone near you has symptoms of a heart attack, stroke or TIA. These are medical emergencies that require immediate care.
What questions should I ask my doctor?
Questions to ask your healthcare provider may include:
Which treatment is best for me?
What numbers should my blood pressure and cholesterol be?
When should my kids start getting their cholesterol checked?
Can you recommend a program to help me stop smoking?
A note from Cleveland Clinic
Atherosclerosis is a common condition that many people are facing. But your healthcare provider is ready to help you manage your condition so you can live your best life. In your yearly checkup, they’ll evaluate your risk for atherosclerosis and explain what you can do to lower it. Don’t be afraid to ask for help in carrying out your provider’s instructions. They can suggest resources for you.
Care at Cleveland Clinic
When you need treatment for coronary artery disease, you want expert care. At Cleveland Clinic, we’ll create a treatment plan that’s personalized to you.
Get Coronary Artery Disease Treatment
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Make an Appointment
Medically Reviewed
Last reviewed on 02/15/2024.
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General Binomial Expansion (Edexcel A Level Maths): Revision Note
Exam code: 9MA0
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General binomial expansion
What is the general binomial expansion?
The binomial expansion applies for positive integers, n ∈ ℕ
The general binomial expansion applies to other types of powers too
The general binomial expansion applies for all real numbers, n ∈ℝ
Usually fractional and/or negative values of n are used
It is derived from (a + b)n, with a = 1 and b = x
a = 1 is the main reason the expansion can be reduced so much
Unless n ∈ ℕ, the expansion is infinitely long
It is only valid for |x| < 1
This is another way of writing -1 < x < 1
This is often called the validity statement
The restriction |x| < 1 means the series will converge
Higher powers of x can be ignored (as r → ∞, xr → 0)
Only the first few terms of an expansion are needed
How do I apply the general binomial expansion?
STEP 1 Write the expression in the form (1 + x)n
STEP 2 Expand and simplify
Use a line for each term to make things easier to read and follow
Use brackets - fractions and negatives get ugly!
STEP 3 If required, check and state the validity statement
How do I use the general binomial expansion to expand (1+bx)n?
STEP 1 Write the expression in the form (1 + bx)n
STEP 2 Replace “x” by “bx” in the expansion
Check carefully to see if b is negative
STEP 3 Expand and simplify
Use a line for each term to make things easier to read and follow
Use brackets
STEP 4 If required, check and state the validity statement The validity statement changes
Replace “x” with “bx” so now |bx| < 1
Worked Example
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10654 | https://www.math.hkust.edu.hk/~mabfchen/Math391I/Recurrence-Relation-Generating-Function.pdf | Week 10-11: Recurrence Relations and Generating Functions April 20, 2005 1 Some Number Sequences An infinite sequence (or just a sequence for short) is an ordered array a0, a1, a2, . . . , an, . . .
of countably many real or complex numbers, and is usually abbreviated as (an; n ≥0) or just (an). A sequence (an) can be viewed as a function f from the set of nonnegative integers to the set of real or complex numbers, i.e., f(n) = an, n = 0, 1, 2, . . .
We call a sequence (an) an arithmetic sequence if it is of the form a0, a0 + q, a0 + 2q, . . . , a0 + nq, . . .
The general term satisfies the recurrence relation an = an−1 + q, n ≥1.
A sequence (an) is called a geometric sequence if it is of the form a0, a0q, a0q2, . . . , a0qn, . . .
The general term satisfies the recurrence relation an = qan−1, n ≥1.
The partial sums of a sequence (an) are the sums: s0 = a0, s1 = a0 + a1, s2 = a0 + a1 + a2, .
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sn = a0 + a1 + · · · + an, .
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The partial sums form a new sequence (sn; n ≥0).
For an arithmetic sequence an = a0 + nq (n ≥0), we have the partial sum sn = n X k=0 (a0 + kq) = (n + 1)a0 + qn(n + 1) 2 .
For a geometric sequence an = a0qn (n ≥1), we have sn = n X k=0 a0qn = ( qn+1−1 q−1 a0 if q ̸= 1 (n + 1)a0 if q = 1.
1 Example 1.1. Determine the number an of regions which are created by n mutually overlapping circles in general position on the plane. (By mutually overlapping we mean that each two circles intersect in two distinct points; thus non-intersecting or tangent circles are not allowed. By general position we mean that there are no three circles through a common point.) We easily see that the first few numbers are given as a0 = 1, a1 = 2, a2 = 4, a3 = 8.
It seems that we might have a4 = 16. However, by try-and-error we quickly see that a4 = 14.
Assume that there are n circles in general position on a plane. When we take one circle away, say the nth circle, there are n −1 circles in general position on the same plane. By induction hypothesis the n −1 circles divide the plane into an−1 regions. Note that the nth circle intersects each of the n −1 circles in 2(n −1) distinct points, say the 2(n −1) points on the nth circle are ordered as P1, P2, . . . , P2(n−1). Then each of the arcs P1P2, P2P3, P3P4, . . . , P2(n−2)+1P2(n−1), P2(n−1)P1 separate a region in the case n −1 circles into two regions. Then there are 2(n −1) more regions produced when the nth circle is drawn. We thus obtain the recurrence relation an = an−1 + 2(n −1), n ≥2.
Repeating the recurrence relation we have an = an−1 + 2(n −1) = hn−2 + 2(n −1) + 2(n −2) = hn−3 + 2(n −1) + 2(n −2) + 2(n −3) .
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= h1 + 2(n −1) + 2(n −2) + 2(n −3) + · · · + 2 = h1 + 2 · (n −1)n 2 = 2 + n(n −1) = n2 −n + 2, n ≥2.
This formula is also valid for n = 1 (since h1 = 2), although it doesn’t hold for n = 0 (since a0 = 1).
Example 1.2 (Fibonacci Sequence). A pair of newly born rabbits of opposite sexes is placed in an enclosure at the beginning of a year. Baby rabbits need one moth to grow mature. Beginning with the second month the female gives birth of a pair of rabbits of opposite sexes each month. Each new pair also gives birth to a pair of rabbits each month starting with their second month. Find the number of pairs of rabbits in the enclosure after one year?
Let fn denote the number of pairs of rabbits at the beginning of the nth month. Some of these pairs are adult and some are babies. We denote by an the number of pairs of adult rabbits and denote by bn the number of pairs of baby rabbits at the beginning of the nth month. Then the total number of pairs of rabbits at the beginning of the nth month is fn = an + bn.
n 1 2 3 4 5 6 7 8 9 10 11 12 13 an 0 1 1 2 3 5 8 13 21 34 55 89 144 bn 1 0 1 1 2 3 5 8 13 21 34 55 89 fn 1 1 2 3 5 8 13 21 34 55 89 144 233 At the beginning of the first month there is one pair of baby rabbits and no pair of adult rabbits. Then there is only one pair of rabbits, i.e., a1 = 0, b1 = 1, and f1 = 0 + 1 = 1. At the beginning of the second month, the baby pair growing mature in the first month becomes an adult pair but did not give birth yet, we have a2 = 1, b2 = 0, and f2 = 1 + 0 = 1. However, the female gives birth of a new pair of rabbits during the second month. At the beginning of the third month, there are two pairs of rabbits because the adult pair in the second month gives birth 2 of a baby pair; so we have a3 = 1, b3 = 1, and f3 = 1 + 1 = 2. At the beginning of the fourth month, the baby pair becomes an adult pair so that there are two adult pairs, but the the adult pair from the third month gives birth of a baby pair again; thus a4 = 2, b4 = 1, and f4 = 2 + 1 = 3.
In general we have, (i) each pair in any month (no matter they are baby or adult) becomes an adult pair at the beginning of the next month, i.e., an = fn−1, n ≥2; (ii) each adult pair gives birth of a new baby pair during the month, but this new baby pair will be only counted at he beginning of the next month, i.e., bn = an−1, n ≥2. Thus fn = an + bn = fn−1 + an−1 = fn−1 + fn−2, n ≥3.
Let us define f0 = 0. The sequence f0, f1, f2, f3, . . . satisfying the recurrence relation fn = fn−1 + fn−2, n ≥2 f0 = 0 f1 = 1 (1) is called the Fibonacci sequence, and the terms in the sequence are called Fibonacci numbers.
Example 1.3. The partial sum of Fibonacci sequence is sn = f0 + f1 + f2 + · · · + fn = fn+2 −1.
(2) This can be verified by induction on n. For n = 0, we have s0 = f2 −1 = 0. Now for n ≥1, we assume that it is true for n −1, i.e., sn−1 = fn+1 −1. Then sn = f0 + f1 + · · · + fn = sn−1 + fn = fn+1 −1 + fn (by the induction hypothesis) = fn+2 −1.
(by the Fibonacci recurrence) Example 1.4. The Fibonacci number fn is even if and only if n is a multiple of 3.
Note that f1 = f2 = 1 is odd and f3 = 2 is even. Assume that f3k is even, f3k−2 and f3k−1 are odd. Then f3k+1 = f3k +f3k−1 is odd (even+odd = odd), and subsequently, f3k+2 = f3k+1 +f3k is also odd (odd+even = odd).
It follows that f3(k+1) = f3k+2 + f3k+1 is even (odd + odd = even).
Theorem 1.1. The general term of the Fibonacci sequence (fn) is given by fn = 1 √ 5 Ã 1 + √ 5 2 !n −1 √ 5 Ã 1 − √ 5 2 !n , n ≥0.
(3) Example 1.5. Determine the number hn of ways to perfectly cover a 2-by-n board with dominoes. (Symmetries are not counted in counting the number of coverings.) We assume h0 = 1 since a 2-by-0 board is empty and it has exactly one perfect cover, namely, the empty cover.
Note that the first few terms can be easily obtained such as h0 = 1, h1 = 1, h2 = 2, h3 = 3, h4 = 5.
Now for n ≥3, the 2-by-n board can be covered by dominoes in two types: n 1 2 n−1 n−2 1 2 n There are hn−1 ways in the first type and hn−2 ways in the second type. Thus hn = hn−1 + hn−2, n ≥2.
Therefore the sequence (hn; n ≥0) is the Fibonacci sequence (fn; n ≥0) with f0 = 0 deleted, i.e., hn = fn+1, n ≥0.
3 Example 1.6. Determine the number bn of ways to perfectly cover a 1-by-n dominoes and monominoes.
Theorem 1.2. The Fibonacci number fn can be written as fn = ⌊n−1 2 ⌋ X k=0 µn −k −1 k ¶ , n ≥0.
Proof. Let g0 = 0 and gn = ⌊n−1 2 ⌋ X k=0 µn −k −1 k ¶ , n ≥1.
Note that k > ¥ n−1 2 ¦ is equivalent to k > n −k −1. Since ³ m p ´ = 0 for any integers m and p such that p > m, we may write gn as gn = n−1 X k=0 µn −k −1 k ¶ , n ≥1.
To prove the theorem, it suffices to show that the sequence (gn) satisfies the Fibonacci recurrence relation with the same initial values. In fact, g0 = 0, g1 = ¡ 0 0 ¢ = 1, and for n ≥0, gn+1 + gn = n X k=0 µn −k k ¶ + n−1 X k=0 µn −k −1 k ¶ = ³n 0 ´ + n X k=1 µn −k k ¶ + n X k=1 µn −k k −1 ¶ = ³n 0 ´ + n X k=1 ·µn −k k ¶ + µn −k k −1 ¶¸ = ³n 0 ´ + n X k=1 µn −k + 1 k ¶ (By the Pascal formula) = µn + 1 0 ¶ + n X k=1 µn −k + 1 k ¶ + µ 0 n + 1 ¶ = n X k=0 µ(n + 2) −k −1 k ¶ = gn+2.
We conclude that the sequence (gn) is the Fibonacci sequence (fn).
2 Linear Recurrence Relations Definition 2.1. A sequence (xn; n ≥0) of numbers is said to satisfy a linear recurrence relation of order k if xn = α1(n)xn−1 + α2(n)xn−2 + · · · + αk(n)xn−k + βn, αk(n) ̸= 0, n ≥k, (4) where the coefficients α1(n), α2(n), . . ., αk(n) are functions of n and βn are constants. The linear recurrence relation (4) is called homogeneous if βn = 0, and is said to have constant coefficients if α1(n), α2(n), . . ., αk(n) are constants.
The recurrence relation xn = α1(n)xn−1 + α2(n)xn−2 + · · · + αk(n)xn−k, αk(n) ̸= 0, n ≥k (5) is called the corresponding homogeneous linear recurrence relation of (4).
4 A solution of the linear recurrence relation (4) is any sequence (an) which satisfies (4). The general solution of (4) is a solution xn = an(c1, c2, . . . , ck) (6) with some parameters c1, c2, . . ., ck, such that for arbitrary initial values x0, x1, . . . , xk−1 there are constants c1, c2, . . . , ck so that (6) is the unique sequence which satisfies both the recurrence relation (4) and the initial conditions.
Let S∞be the set of all sequences (an; n ≥0). It is clear that S∞is an infinite-dimensional vector space under the ordinary addition and scalar multiplication of sequences. Let Nk consist all solutions of the nonhomogeneous linear recurrence relation (4), and let Hk consist all solutions of the homogeneous linear recurrence relation (5). We shall see that Hk is a k-dimensional subspace of the vector space S∞, and that Nk is a k-dimensional affine subspace of S∞.
Theorem 2.2 (Structure Theorem for Linear Recurrence Relations). (a) The solution space Hk is a k-dimensional subspace of the vector space S∞of sequences. Thus, if (an,1), (an,2), . . ., (an,k) are linearly independent solutions of the homogeneous linear recurrence relation (5), then the general solution of (5) is xn = c1an,1 + c2an,2 + · · · + ckan,k, n ≥0.
(b) Let (an) be a particular solution of the nonhomogeneous linear recurrence relation (4). Then the general solution of (4) is xn = an + hn, n ≥0, where (hn) is the general solution of the corresponding homogeneous linear recurrence relation (5). In other words, Nk is a translate of Hk in S∞, that is, Nk = (an) + Hk.
Proof. (a) To show that Hk is a vector subspace of S∞, we need to show that Hk is closed under the addition and scalar multiplication of sequences. Let (an) and (bn) be solutions of (5). Then an + bn = [α1(n)an−1 + α2(n)an−2 + · · · + αk(n)an−k] + [α1(n)bn−1 + α2(n)bn−2 + · · · + αk(n)bn−k] = α1(n)(an−1 + bn−1) + α2(n)(an−2 + bn−2) + · · · + αk(n)(an−k + bn−k), n ≥k; and for any scalars c, can = c[α1(n)an−1 + α2(n)an−2 + · · · + αk(n)an−k] = α1(n)can−1 + α2(n)can−2 + · · · + αk(n)can−k, n ≥k.
This means that Hk is closed under the addition and scalar multiplication of sequences.
To show that Hk is k-dimensional, consider the projection π : S∞− →Rk defined by π(x0, x1, x2, . . .) = (x0, x1, . . . , xk−1).
We shall see that the restriction of π to Hk is a linear isomorphism. For any (a0, a1, . . . , an) ∈Rk, define an as an = α1(n)an−1 + α2(n)an−2 + · · · + αk(n)an−k, n ≥k.
Obviously, we have π(a0, a1, a2, . . .) = (a0, a1, . . . , ak−1). This means that the restriction π|Hk is from Hk onto Rk.
Now for any sequence (xn) ∈Hk, if π(x0, x1, x2, . . .) = (0, 0, . . . , 0), then x0 = x1 = · · · = xk−1 = 0. Applying the recurrence relation (5) for n = k, we have xk = 0; applying (5) again for n = k + 1, we obtain xk+1 = 0. Continuing to apply (5), we have xn = 0 for n ≥k. Thus (xn) is the zero sequence. This means that π is one-to-one from Hk onto Rk. We have finished the proof that π is a linear isomorphism from Hk to Rk.
(b) For any solution (bn) of (4), we claim that the sequence hn = bn −an (n ≥0) is a solution of (5). So bn = an + hn, n ≥0.
5 In fact, applying the recurrence relation (4), we have hn = [α1(n)bn−1 + α2(n)bn−2 + · · · + αk(n)bn−k + βn] −[α1(n)an−1 + α2(n)an−2 + · · · + αk(n)an−k + βn] = α1(n)(bn−1 −an−1) + α2(n)(bn−2 −an−2) + · · · + αk(n)(bn−k −an−k) = α1(n)hn−1 + α2(n)hn−2 + · · · + αk(n)hn−k, n ≥k.
This means that (hn) is a solution of (5). Conversely, for any solution (hn) of (5), we have an + hn = [α1(n)an−1 + α2(n)an−2 + · · · + αk(n)an−k + βn] +[α1(n)hn−1 + α2(n)hn−2 + · · · + αk(n)hn−k] = α1(n)(an−1 + hn−1) + α2(n)(an−2 + hn−2) + · · · + αk(n)(an−k + hn−k) + βn for n ≥k. This means that the sequence (an + hn) is a solution of (4).
Definition 2.3. The Wronskian Wk(n) of k solutions (an,1), (an,2), . . ., (an,k) of the homogeneous linear recurrence relation (5) is the determinant Wk(n) = det an,1 an,2 · · · an,k an+1,1 an+1,2 · · · an+1,k .
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an+k−1,1 an+k−1,2 · · · an+j−1,k , n ≥0.
Theorem 2.4. The solutions (an,1), (an,2), . . ., (an,k) of the homogeneous linear recurrence relation (5) are linearly independent if and only if there is a nonnegative integer n0 such that the Wronskian Wk(n0) ̸= 0.
Proof. It suffices to show that the sequences (an,1), (an,2), . . ., (an,k) are linearly dependent if and only if Wk(n) = 0 for all n ≥0. If (an,1), (an,2), . . ., (an,k) are linearly dependent, then for any n ≥0 the columns of the matrix an,1 an,2 · · · an,k an+1,1 an+1,2 · · · an+1,k .
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an+k−1,1 an+k−1,2 · · · an+j−1,k are linearly dependent because the columns are part of the sequences (an,1), (an,2), . . ., (an,k) respectively. It follows from linear algebra that the Wronskian Wk(n) = 0 for all n ≥0.
Conversely, if Wk(n) = 0 for all n ≥0, in particular, Wk(0) = 0, then there are constants c1, c2, . . . , ck, not all zero, such that c1ai,1 + c2ai,2 + · · · + ckai,k = 0 for 0 ≤i ≤k −1.
Thus, applying the recurrence relation (5) for the sequences (an,1), (an,2), . . ., (an,k) respectively for n = k, we have k X j=1 cjak,j = k X j=1 cj k X i=1 αi(k)ak−i,j = k X i=1 αi(k) k X j=1 cjak−i,j = 0.
Continuing to apply the recurrence relation (5) for n ≥k + 1, we conclude that for the same constants c1, c2, . . . , ck, c1an,1 + c2an,2 + · · · + ckan,k = 0, n ≥k + 1.
This means that the sequences (an,1), (an,2), . . ., (an,k) are linearly dependent.
6 3 Homogeneous Linear Recurrence Relations with Constant Coefficients In this section we only consider linear recurrence relations of the form xn = α1xn−1 + α2xn−2 + · · · + αkxn−k, αk ̸= 0, n ≥k, (7) where α1, α2, . . ., αk are constants. We call this kinds of recurrence relations as homogeneous linear recurrence relations of order k with constant coefficients. Sometimes it is convenient to write (??) as of the form α0xn + α1xn−1 + · · · + αkxn−k = 0, n ≥k (8) where α0 ̸= 0 and αn−k ̸= 0. The following polynomial equation α0xk + α1xk−1 + · · · + αk−1x + αk = 0, (9) is called the characteristic equation associated with the recurrence relation (8). The polynomial on the left side of (9) is called the characteristic polynomial of (8).
Example 3.1. The Fibonacci sequence (fn; n ≥0) satisfies the linear recurrence relation fn = fn−1 + fn−2, n ≥2 of order 2 with α1 = α2 = 1 in (7).
Example 3.2. The geometric sequence (xn; n ≥0), where xn = qn, satisfies the linear recurrence relation xn = qxn−1, n ≥1 of order 1 with α1 = q in (7).
It is quite heuristic that solutions of the first order homogeneous linear recurrence relations are geometric sequences. This hints that the recurrence relation (7) may have solutions of the form xn = qn. The following theorem confirms the speculation.
Theorem 3.1. (a) For any number q ̸= 0, the geometric sequence xn = qn is a solution of the kth order homogeneous linear recurrence relation (8) with constant coefficients if and only if the number q is a root of the characteristic equation (9).
(b) If the characteristic equation (9) has k distinct roots q1, q2, . . ., qk, then the general solution of (8) is xn = c1qn 1 + c2qn 2 + · · · + cnqn k, n ≥0.
(10) Proof. (a) Put xn = qn into the recurrence relation (8); we have α0qn + α1qn−1 + · · · + αkqn−k = 0.
(11) Since q ̸= 0, dividing both sides of (11) by qn−k, we obtain αoqk + α1qk−1 + · · · + αk−1q + αk = 0 (12) This means that (11) and (12) are equivalent. This finishes the proof of Part (a).
(b) Since q1, q2, . . . , qk are roots of the characteristic equation (9), then xn = qn i are solutions of the homogeneous linear recurrence relation (8) for all i (1 ≤i ≤k). Since the solution space of (8) is a vector space, the linear combination xn = c1qn 1 + c2qn 2 + · · · + cnqn k, n ≥0 7 are also solutions (8). Now given arbitrary values for x0, x1, . . . , xk−1, the sequence (xn) is uniquely determined by the recurrence relation (8). Set c1qi 1 + c2qi 2 + · · · + cnqi k = xi, 0 ≤i ≤k −1.
The coefficients c1, c2, . . . , ck are uniquely determined by Cramer’s rule as follows: ci = det Ai det A , 1 ≤i ≤k where A is the Vandermonde matrix A = 1 1 · · · 1 q1 q2 · · · qk q2 1 q2 2 · · · q2 k .
.
.
.
.
.
.
.
.
qk−1 1 qk−1 2 · · · qk−1 k , and Ai is the matrix obtained from A by replacing its ith column by the column [x0, x1, . . . , xk−1]T . The determinant of A is given by det A = Y 1≤i<j≤k (qj −qi) ̸= 0.
This finishes the proof of Part (b).
Example 3.3. Find the sequence (xn) satisfying the recurrence relation xn = 2xn−1 + xn−2 −2xn−3, n ≥3 and the initial conditions x0 = 1, x1 = 2, and x2 = 0.
Solution. The characteristic equation of the recurrence relation is x3 −2x2 −x + 2 = 0.
Factorizing the equation, we have (x −2)(x + 1)(x −1) = 0.
There are three roots x = 1, −1, 2. By Theorem 3.1, we have the general solution xn = c1(−1)n + c2 + c32n.
Applying the initial conditions, c1 +c2 +c3 = 1 c1 −c2 +2c3 = 2 c1 +c2 +4c3 = 0 Solving the linear system we have c1 = 2, c2 = −2/3, c3 = −1/3. Thus xn = 2 −2 3(−1)n −1 32n.
Theorem 3.2. (a) Let q be a root with multiplicity m of the characteristic equation (9) associated with the kth order homogeneous linear recurrence relation (8) with constant coefficients. Then the m sequences xn = qn, nqn, . . . , nm−1qn are linearly independent solutions of the recurrence relation (8).
8 (b) Let q1, q2, . . . , qs be distinct roots with the multiplicities m1, m2, . . . , ms respectively for the characteristic equation (9). Then the sequences xn = qn 1 , nqn 1 , . . . , nm1−1qn 1 ; qn 2 , nqn 2 , . . . , nm2−1qn 2 ; .
.
.
qn s , nqn s , . . . , nms−1qn s ; n ≥0 are linearly independent solutions of the homogeneous linear recurrence relation (8). Their linear combinations form the general solution of the recurrence relation (8).
4 Nonhomogeneous Linear Recurrence Relations with Constant Coefficients Theorem 4.1. Given a nonhomogeneous linear recurrence relation of the first order xn = αxn−1 + βn.
(13) (a) Let βn = cqn be an exponential function of n. Then (13) has a particular solution of the following form.
• If q ̸= α, then xn = Aqn.
• If q = α, then xn = Anqn.
(b) Let βn = Pk i=0 bini be a polynomial function of n with degree k.
• If α ̸= 1, then (13) has a particular solution of the form xn = A0 + A1n + A2n2 + · · · + Aknk, where the coefficients A0, A1, . . ., Ak are be recursively determined as Ak = bk 1 −α, Ai = 1 1 −α bi + α k X j=i+1 (−1)j−i µj i ¶ Aj , 0 ≤i ≤k −1.
• If α = 1, then the solution of (13) is given by xn = x0 + n X i=1 βi.
Proof. (a) We may assume q ̸= 0; otherwise the recurrence (13) is homogeneous.
For the case q ̸= α, put xn = Aqn in (13); we have Aqn = αAqn−1 + cqn.
The coefficient A is determined as A = cq/(q −α).
For the case q = α, put xn = Anqn in (13); we have Anqn = αA(n −1)qn−1 + cqn.
Since q = α, then αAqn−1 = cqn. The coefficient A is determined as A = cq/α.
(b) For the case α ̸= 1, put xn = Pk j=0 Ajnj in (13); we obtain k X j=0 Ajnj = α k X j=0 Aj(n −1)j + k X j=0 bjnj.
9 Then k X j=0 Ajnj = α k X j=0 Aj j X i=0 µj i ¶ ni(−1)j−i + k X j=0 bjnj.
k X i=0 Aini = α k X i=0 ni k X j=i (−1)j−i µj i ¶ Aj + k X i=0 bini.
k X i=0 Ai −bi −α k X j=i (−1)j−i µj i ¶ Aj ni = 0.
The coefficients A0, A1, . . . , Ak are determined recursively as Ak = bk 1 −α, Ai = 1 1 −α bi + α k X j=i+1 (−1)j−i µj i ¶ Aj , 0 ≤i ≤k −1.
As for the case α = 1, iterate the recurrence relation (13); we have xn = xn−1 + βn = xn−2 + βn−1 + βn = xn−1 + βn−2 + βn−1 + βn = · · · = x0 + β1 + β2 + · · · + βn.
Example 4.1. Solve the difference equation ½ xn = xn−1 + 3n2 −5n3, n ≥1 x0 = 2.
Solution.
xn = x0 + n X i=1 bi = 2 + n X i=1 ¡ 3i2 −5i3¢ = 2 + 3 n X i=1 i2 −5 n X i=1 i3 = 2 + 3 × n(n + 1)(2n + 1) 6 −5 × µn(n + 1) 2 ¶2 .
We have applied the following identities n X i=1 i2 = n(n + 1)(2n + 1) 6 , n X i=1 i3 = µn(n + 1) 2 ¶2 .
Example 4.2. Solve the equation ½ xn = 3xn−1 −4n, n ≥1 x0 = 2.
10 Solution. Note that xn = 3nc is the general solution of the corresponding homogeneous linear recurrence relation.
Let xn = An + B be a particular solution. Then An + B = 3[A(n −1) + B] −4n Comparing the coefficients of n0 andn, it follows that A = 2 and B = 3. Thus the general solution is given by xn = 2n + 3 + 3nc.
The initial condition x0 = 2 implies that c = −1. Therefore the solution is xn = −3n + 2n + 3.
Theorem 4.2. Given a nonhomogeneous linear recurrence relation of the second order xn = α1xn−1 + α2xn−2 + cqn.
(14) Let q1 and q2 be solutions of its characteristic equation x2 −α1x −α2 = 0.
Then (14) has a particular solution of the following forms, where A is a constant to be determined.
(a) If q ̸= q1, q ̸= q2, then xn = Aqn.
(b) If q = q1, q1 ̸= q2, then xn = Anqn.
(c) If q = q1 = q2, then xn = An2qn.
Proof. The homogeneous linear recurrence relation corresponding to (14) is xn = α1xn−1 + α2xn−2, n ≥2.
(15) We may assume q ̸= 0. Otherwise (14) is homogeneous.
(a) Put xn = Aqn into (14); we have Aqn = α1Aqn−1 + α2Aqn−2 + cqn.
Then A(q2 −a1q −a2) = cq2.
Since q is not a root of the characteristic equation x2 = α1x + α2, that is, q2 −α1q −α2 ̸= 0, the coefficient A is determined as A = cq2 q2 −α1q −α2 .
(b) Since q = q1 ̸= q2, then xn = qn is a solution of (15) but xn = nqn is not, that is, q2 −α1q −α2 = 0 and nqn ̸= α1(n −1)qn−1 + α2(n −2)qn−2.
It follows that nq2 −α1(n −1)q −α2(n −2) = n(q2 −α1q −α2) + α1q + 2α2 = α1q + 2α2 ̸= 0.
Put xn = Anqn into (14); we have Anqn = α1A(n −1)qn−1 + α2A(n −2)qn−2 + cqn.
Then A £ nq2 −α1(n −1)q −α2(n −2) ¤ = cq2.
11 Since α1q + 2α2 ̸= 0, the coefficient A is determined as A = cq2 α1q + 2α2 .
(c) Since q = q1 = q2, then both xn = qn and xn = nqn are solutions of (15), but xn = n2qn is not. It then follows that q2 −α1q −α2 = 0, α1q + 2α2 = 0, and n2q2 −α1(n −1)2q −α2(n −2)2 = n2(q2 −α1q −α2) + 2n(α1q + 2α2) −α1q −4α2 = −α1q −4α2 ̸= 0.
Put xn = An2qn into (14); we have Aqn−2 £ n2q2 −α1(n −1)2q −α2(n −2)2¤ = cqn.
The coefficient A is determined as A = − cq2 α1q + 4α2 .
Example 4.3. Solve the equation xn = 10xn−1 −25xn−2 + 5n+1, n ≥2 x0 = 5 x1 = 15.
Put xn = An2 × 5n into the recurrence relation; we have An2 × 5n = 10A(n −1)2 × 5n−1 −25A(n −2)2 × 5n−2 + 5n+1.
Dividing both sides we further have An2 = 2A(n −1)2 −A(n −2)2 + 5.
Thus A = 5/2. The general solution is given by xn = 5 2n25n + c15n + c2n5n.
Applying the initial conditions x0 = 5 and x1 = 15, we have c1 = 5 and c2 = −9/2. Hence xn = µ5 2n2 −9 2n + 5 ¶ 5n.
Theorem 4.3. Given a nonhomogeneous linear recurrence relation of the second order xn = α1xn−1 + α2xn−2 + βn, n ≥2, (16) where βn is a polynomial function of n with degree k.
(a) If α1 + α2 ̸= 1, then (16) has a particular solution of the form xn = A0 + A1n + · · · + Aknk, where A0, A1, . . ., Ak are constants to be determined. If k ≤2, then a particular solution has the form xn = A0 + A1n + A2n2.
12 (b) If α1 + α2 = 1, then (16) can be reduced to a first order recurrence relation yn = (α1 −1)yn−1 + βn, n ≥2, where yn = xn −xn−1 for n ≥1.
Proof. (a) Let βn = Pk j=0 bjnj. Put xn = Pk j=0 Ajnj into the recurrence relation (16); we obtain k X j=0 Ajnj = α1 k X j=0 Aj(n −1)j + α2 k X j=0 Aj(n −2)j + k X j=0 bjnj; k X j=0 Ajnj = α1 k X j=0 Aj j X i=0 (−1)j−i µj i ¶ ni + α2 k X j=0 Aj j X i=0 (−2)j−i µj i ¶ ni + k X j=0 bjnj; k X j=0 Ajnj = α1 k X i=0 ni k X j=i (−1)j−i µj i ¶ Aj + α2 k X i=0 ni k X j=i (−2)j−i µj i ¶ Aj + k X j=0 bjnj.
Collecting the coefficients of ni, we have k X i=0 Ai −α1 k X j=i (−1)j−i µj i ¶ Aj −α2 k X j=i (−2)j−i µj i ¶ Aj − k X i=0 bi ni = 0.
Since α1 + α2 ̸= 1, the coefficients A0, A1, . . ., Ak are determined as Ak = bk 1 −α1 −α2 , Ai = 1 1 −α1 −α2 bi + k X j=i+1 (−1)j−i µj i ¶ ³ α1 + 2j−iα2 ´ Aj , 0 ≤i ≤k −1.
(b) The recurrence relation (16) becomes xn = α1xn−1 + (1 −α1)xn−2 + βn, n ≥2.
Set yn = xn −xn−1 for n ≥1; recurrence (16) reduces to the required first order recurrence relation.
Example 4.4. Solve the following recurrence relation xn = 6xn−1 −9xn−2 + 8n2 −24n x0 = 5 x1 = 5.
Solution. Put xn = A0 + A1n + A2n2 into the recurrence relation; we obtain A0 + A1n + A2n2 = 6[A0 + A1(n −1) + A2(n −1)2] −9[A0 + A1(n −2) + A2(n −2)2] + 8n2 −24n.
Collecting the coefficients of n2, n, and the constant, we have (4A2 −8)n2 + (4A1 −24A2 + 24)n + (4A0 −12A1 + 30A2) = 0.
We conclude that A2 = 2, A1 = 6, and A0 = 3. So xn = 2n2 + 6n + 3 is a particular solution. Then the general solution of the recurrence is xn = 2n2 + 6n + 3 + 3nc1 + 3nnc2.
Applying the initial condition x0 = x1 = 5, we have c1 = 2, c2 = −4. The sequence is finally obtained as xn = 2n2 + 6n + 3 + 2 × 3n −4n × 3n.
13 5 Generating Functions The (ordinary) generating function of an infinite sequence a0, a1, a2, . . . , an, . . .
is the infinite series A(x) = a0 + a1x + a2x2 + · · · + anxn + · · · .
A finite sequence a0, a1, a2, . . . , an can be regarded as the infinite sequence a0, a1, a2, . . . , an, 0, 0, . . .
and its generating function A(x) = a0 + a1x + a2x2 + · · · + anxn is a polynomial.
Example 5.1. The generating function of the constant infinite sequence 1, 1, . . . , 1, . . .
is the function A(x) = 1 + x + x2 + · · · + xn + · · · = 1 1 −x.
Example 5.2. For any positive integer n, the generating function for the binomial coefficients ³n 0 ´ , ³n 1 ´ , ³n 2 ´ , . . . , ³n n ´ is the function n X k=0 ³n k ´ xk = (1 + x)n.
Example 5.3. For any real number α, the generating function for the infinite sequence of binomial coefficients ³α 0 ´ , ³α 1 ´ , ³α 2 ´ , . . . , ³α n ´ , . . .
is the function n X n=0 ³α n ´ xn = (1 + x)α.
Example 5.4. Let k be a positive integer and let a0, a1, a2, . . . , an, . . .
be the infinite sequence whose general term an is the number of nonnegative integral solutions of the equation x1 + x2 + · · · + xk = n.
Then the generating function of the sequence (an) is A(x) = ∞ X n=0 X i1+···+ik=n 1 xn = ∞ X n=0 X i1+···+ik=n xi1+···+ik = Ã ∞ X i1=0 xi1 !
· · · ∞ X ik=0 xik = 1 (1 −x)k = ∞ X n=0 (−1)n µ−k n ¶ xn = ∞ X n=0 µn + k −1 n ¶ xn.
14 Example 5.5. Let an be the number of integral solutions of the equation x1 + x2 + x3 + x4 = n, where 0 ≤x1 ≤3, 0 ≤x2 ≤2, x3 ≥2, and 3 ≤x4 ≤5. The generating function of the sequence (an) is A(x) = ¡ 1 + x + x2 + x3¢ ¡ 1 + x + x2¢ ¡ x2 + x3 + · · · ¢ ¡ x3 + x4 + x5¢ = x5 ¡ 1 + x + x2 + x3¢ ¡ 1 + x + x2¢2 1 −x .
Example 5.6. Determine the generating function for the number of n-combinations of apples, bananas, oranges, and pears where in each n-combination the number of apples is even, the number of bananas is odd, the number of oranges is between 0 and 4, and the number of pears is at least two.
The required generating function is A(x) = Ã ∞ X i=0 x2i ! Ã ∞ X i=0 x2i+1 ! Ã 4 X i=0 xi ! Ã ∞ X i=1 xi !
= x3(1 −x5) (1 −x2)2(1 −x)2 .
Example 5.7. Determine the number an of bags with n pieces of fruit (apples, bananas, oranges, and pears) such that the number of apples is even, the number bananas is a multiple of 5, the number oranges is at most 4, and the number of pears is either one or zero.
The generating function of the sequence (an) is A(x) = Ã ∞ X i=0 x2i ! Ã ∞ X i=0 x5i ! Ã 4 X i=0 xi ! Ã 1 X i=0 xi !
= (1 + x + x2 + x3 + x4)(1 + x) (1 −x2)(1 −x5) = (1 + x)(1 −x5)/(1 −x) (1 + x)(1 −x)(1 −x5) = 1 (1 −x)2 = ∞ X n=0 (−1)n µ−2 n ¶ xn = ∞ X n=0 µn + 1 n ¶ xn = ∞ X n=0 (n + 1)xn.
Thus an = n + 1.
Example 5.8. Find a formula for the number an,k of integral solutions (i1, i2, . . . , ik) of the equation x1 + x2 + · · · + xk = n such that i1, i2, . . . , ik are nonnegative odd numbers.
The generating function of the sequence (an) is A(x) = Ã ∞ X i=0 x2i+1 !
· · · Ã ∞ X i=0 x2i+1 !
= xk (1 −x2)k = xk ∞ X i=0 µi + k −1 i ¶ x2i = ∞ X i=0 µi + k −1 i ¶ x2i+k = P∞ j=r ³ j+r−1 j−r ´ x2j for k = 2r P∞ j=r ³ j+r j−r ´ x2j+1 for k = 2r + 1.
15 We then conclude that a2s,2r = ³ s+r−1 s−r ´ , a2s+1,2r+1 = ³ s+r s−r ´ , and an,k = 0 otherwise. We may combine the three case as two cases: an,k = Ã ⌊n 2 ⌋+ ⌈k 2⌉−1 ⌊n 2 ⌋−⌊k 2⌋ !
if n −k = even, and an,k = 0 if n −k = odd.
Example 5.9. Let an denote the number of nonnegative integral solutions of the equation 2x1 + 3x2 + 4x3 + 5x4 = n.
Then the generating function of the sequence (an) is A(x) = ∞ X n=0 X i,j,k,l≥0 2i+3j+4k+5l=0 1 xn = à ∞ X i=0 x2i ! ∞ X j=0 x3j à ∞ X k=0 x4k ! à ∞ X l=0 x5l !
= 1 (1 −x2)(1 −x3)(1 −x4)(1 −x5).
Theorem 5.1. Let sn be the number of nonnegative integral solutions of the equation a1x1 + a2x2 + · · · + akxk = n.
Then the generating function of the sequence (sn) is A(x) = 1 (1 −xa1)(1 −xa2) · · · (1 −xak).
6 Recurrence and Generating Functions Since 1 (1 −x)n = ∞ X k=0 µ−n k ¶ (−x)k = ∞ X k=0 µn + k −1 k ¶ xk, |x| < 1; then 1 (1 −ax)n = ∞ X k=0 µ−n k ¶ (−ax)k = ∞ X k=0 µn + k −1 k ¶ akxk, |x| < 1 |a|.
Example 6.1. Determine the generating function of the sequence 0, 1, 22, . . . , n2, . . . .
Since 1 1−x = P∞ k=0 xk, then 1 (1 −x)2 = d dx µ 1 1 −x ¶ = ∞ X k=0 d dx ³ xk´ = ∞ X k=0 kxk−1.
Thus x (1−x)2 = P∞ k=0 kxk. Taking the derivative with respect to x we have 1 + x (1 −x)3 = ∞ X k=0 k2xk−1.
Therefore the desired generating function is g(x) = x(1 + x) (1 −x)3 .
16 Example 6.2. Solve the recurrence relation an = 5an−1 −6an−2, n ≥2 a0 = 1 a1 = −2 Let A(x) = P∞ n=0 anxn. Applying the recurrence relation, we have A(x) = a0 + a1x + ∞ X n=2 (5an−1 −6an−2) xn = a0 + a1x −5xa0 + 5xA(x) −6x2A(x).
Applying the initial values and collecting the coefficient functions of A(x), we further have ¡ 1 −5x + 6x2¢ A(x) = 1 −7x.
Thus the function g(x) is solved as A(x) = 1 −7x 1 −5x + 6x2 .
Observing that 1 −5x + 6x2 = (1 −2x)(1 −3x) and applying partial fraction, 1 −7x 1 −5x + 6x2 = A 1 −2x + B 1 −3x.
The constants A and B can be determined by A(1 −3x) + B(1 −2x) = 1 −7x.
Then ½ A +B = 1 −3A −2B = −7 Thus A = 5, B = −4. Hence 1 −7x 1 −5x + 6x2 = 5 1 −2x − 4 1 −3x.
Since 1 1 −2x = ∞ X n=0 2nxn and 1 1 −3x = ∞ X n=0 3nxn We obtain the sequence an = 5 × 2n −4 × 3n, n ≥0.
Theorem 6.1. Let (an; n ≥0) be a sequence satisfying the homogeneous linear recurrence relation an = α1an−1 + α2an−2 + · · · + αkan−k, αk ̸= 0, n ≥k (17) of order k with constant coefficients. Then its generating function A(x) = P∞ n=0 anxn is of the form A(x) = P(x) Q(x) (18) where Q(x) is a polynomial of degree k with a nonzero constant term and P(x) is a polynomial of degree less than k.
Conversely, given such polynomials P(x) and Q(x), there is a unique sequence (an) that satisfies the linear homogeneous recurrence relation (17) and its generating function is the rational function in (18).
17 Proof. The generating function A(x) of the sequence (an) can be written as A(x) = k−1 X i=0 aixi + ∞ X n=k anxn = k−1 X i=0 aixi + ∞ X n=k à k X i=1 αian−i !
xn = k−1 X i=0 aixi + k X i=1 αi ∞ X n=k an−ixn = k−1 X i=0 aixi + k X i=1 αi ∞ X n=k−i anxn+i = k−1 X i=0 aixi + αkxk ∞ X n=0 anxn + k−1 X i=1 αixi ∞ X n=0 anxn − k−i−1 X j=0 ajxj = k−1 X i=0 aixi + g(x) k X i=1 αixi − k−1 X i=1 αixi k−i−1 X j=0 ajxj = k−1 X i=0 aixi + g(x) k X i=1 αixi − k−1 X l=1 xl l X i=1 αial−i.
Then A(x) Ã 1 − k X i=1 αixi !
= k−1 X i=0 aixi − k−1 X l=1 xl l X i=1 αial−i = a0 + k−1 X i=1 ai − i X j=1 αjai−j xi.
Thus P(x) = a0 + k−1 X i=1 ai − i X j=1 αjai−j xi, Q(x) = 1 − k X i=1 αixi.
Conversely, let (an) be the sequence whose generating function is A(x). Let A(x) = ∞ X n=0 anxn, P(x) = k X i=0 bixi, Q(x) = 1 − k X i=1 αixi.
Then g(x) = p(x)/q(x) is equivalent to à 1 − k X i=1 αixi ! à ∞ X n=0 anxn !
= k X i=0 bixi.
The polynomial q(x) can be viewed as an infinite series with αi = 0 for i > k. Thus ∞ X n=0 anxn − ∞ X n=0 Ã n X i=1 αian−i !
xn = k X i=0 bixi.
Equating the coefficients of xn, we have the recurrence relation an = k X i=1 αian−i, n ≥k.
Proposition 6.2 (Partial Fractions). (a) If P(x) is a polynomial of degree at most k, then P(x) (1 −ax)k = A1 1 −ax + A2 (1 −ax)2 + · · · + Ak (1 −ax)k , 18 where A1, A2, . . . , Ak are constants to be determined.
(b) If P(x) is a polynomial of degree at most p + q + r, then P(x) (1 −ax)p(1 −bx)q(1 −cx)r = A1(x) (1 −ax)p + A2(x) (1 −bx)q + A3(x) (1 −cx)r , where A1(x), A2(x), and A3(x) are polynomials of degree q + r, p + r, and p + q, respectively.
7 A Geometry Example A polygon P in R2 is called convex if the segment joining any two points in P is also contained in P. Let Cn denote the number ways to divide a labelled convex polygon with n + 2 sides into triangles. The first a few such numbers are C1 = 1, C2 = 2, C3 = 5.
We first establish a recurrence relation between Cn+1 and C0, C1, . . . , Cn. Let P(v1, v2, . . . , vn+3) denote a convex polygon with the vertices v1, v2, . . . , vn+3. In each triangular decomposition of P(v1, v2, . . . , vn+3) into triangles, the segment v1vn+3 is one side of a triangle ∆in the decomposition; the third vertex of the triangle ∆is one of the vertices v2, v3, . . . , vn+2. Let vk+2 be the third vertex of ∆other than v1 and vn+3 (0 ≤k ≤n); see Figure 1 below. Then n+3 v v 1 2 v v3 k+2 v n+2 v Figure 1: vk+2 is the third vertex of the triangle with the side v1vn+2 we have one convex polygon P(v1, v2, . . . , vk+2) of (k + 2) sides and another convex polygon P(vk+2, vk+3, . . . , vn+3) of (n −k + 2) sides. Then there are Ck ways to divide P(v1, v2, . . . , vk+2) into triangles and there are Cn−k ways to divide P(vk+2, vk+3, . . . , vn+3) into triangles. We thus have the following recurrence relation Cn+1 = n X k=0 CkCn−k with C0 = 1.
Consider the generating function F(x) = P∞ n=0 Cnxn. Then F(x)F(x) = Ã ∞ X n=0 Cnxn ! Ã ∞ X n=0 Cnxn !
= ∞ X n=0 Ã n X k=0 CkCn−k !
xn = ∞ X n=0 Cn+1xn = 1 x ∞ X n=1 Cnxn = F(x) x −1 x.
We thus obtain the functional equation xF(x)2 −F(x) + 1 = 0.
Solving for F(x), we have F(x) = 1 ± √1 −4x 2x .
19 Note that √ 1 −4x = ∞ X n=0 µ 1 2 n ¶ (−4x)n = 1 + ∞ X n=1 1 2 · (1 2 −1) · · · (1 2 −n + 1) n!
22n(−1)nxn = ∞ X n=0 (−1)(−3)(−5) · · · (−2(n −1) + 1) n!
2n(−1)nxn = − ∞ X n=0 1 · 3 · 5 · · · (2(n −1) −1) n!
2nxn = 1 −2 ∞ X n=1 (2(n −1))!
n!(n −1)! xn = 1 −2 ∞ X n=0 (2n)!
n!(n + 1)!xn+1.
We conclude that F(x) = 1 −√1 −4x 2x = ∞ X n=0 (2n)!
n!(n + 1)!xn = ∞ X n=0 1 n + 1 µ2n n ¶ xn.
Hence the sequence (Cn) is given by the binomial coefficients: Cn = 1 n + 1 µ2n n ¶ , n ≥0.
The sequence (Cn) is known as the Catalan sequence and the numbers Cn as the Catalan numbers.
Example 7.1. Let Cn be the number of ways to evaluate a matrix product A1A2 · · · An+1 (n ≥0) by adding various parentheses. For instance, C0 = 1, C1 = 1, C2 = 2, and C3 = 5. In general the formula is given by Cn = 1 n + 1 µ2n n ¶ .
Note that each way of evaluating the matrix product A1A2 · · · An+2 will be finished by multiplying of two matrices at the end. There are exactly n + 1 ways of multiplying the two matrices at the end: A1A2 · · · An+2 = (A1 · · · Ak+1)(Ak+2 · · · An+2), 0 ≤k ≤n.
This yields the recurrence relation Cn+1 = n X k=0 CkCn−k.
Thus Cn = 1 n+1 ¡ 2n n ¢ , n ≥0.
20 8 Exponential Generating Functions The ordinary generating function method is a powerful algebraic tool for finding unknown sequences, especially when the sequences are certain binomial coefficients or some the order is not material. However, when the sequences are not binomial type or the order is material in defining the sequences, we may need to consider a different type of generating functions. For example, for the sequence an = n!, counting the number of permutations of n distinct objects. It is not easy at all to figure out the ordinary generating function ∞ X n=0 anxn = ∞ X n=0 n!xn.
However, the generating function ∞ X n=0 an n! xn = ∞ X n=0 xn = 1 1 −x is obviously figured out.
The exponential generating function of a sequence (an; n ≥0) is the infinite series E(x) = ∞ X n=0 an n! xn.
Example 8.1. The exponential generating function of the sequence P(n, 0), P(n, 1), . . . , P(n, n) is given by E(x) = n X k=0 P(n, k) k!
xk = n X k=0 ³n k ´ xk = (1 + x)n.
Example 8.2. The exponential generating function of the constant sequence (an = 1; n ≥0) is E(x) = ∞ X n=0 xn n! = ex.
The exponential generating function of the geometric sequence (an = an; n ≥0) is E(x) = ∞ X n=0 anxn n!
= eax.
Theorem 8.1. Let M = {n1a1, n2a2, . . . , nkak} be a multiset over the set S = {a1, a2, . . . , ak} with n1 a1’s, n2 a2’s, . . ., nk ak’s. Let an be the number of permutations of the multiset M. Then the exponential generating function of the sequence (an; n ≥0) is given by E(x) = Ã n1 X i=0 xi i!
! Ã n2 X i=0 xi i!
!
· · · Ã nk X i=0 xi i!
!
.
(19) Proof. Note that an = 0 for n > n1 + · · · + nk. Thus E(x) is a polynomial. The right side of (19) can be expanded to the form n1,n2,...,nk X i1,i2,...,ik=0 xi1+i2+···+ik i1!i2! · · · ik! = n1+n2+···+nk X n=0 xn n!
X i1+i2+···+ik=n 0≤i1≤n1,...,0≤ik≤nk n!
i1!i2! · · · ik!.
21 Note that the number of permutation of M with exactly i1 a1’s, i2 a2’s, . . ., and ik ak’s such that i1 + i2 + · · · + ik = n is the multinomial coefficient µ n i1, i2, . . . , ik ¶ = n!
i1!i2! · · · ik!.
It turns out that the sequence (an) is given by an = X i1+i2+···+ik=n 0≤i1≤n1,...,0≤ik≤nk n!
i1!i2! · · · ik!, n ≥0.
Example 8.3. Determine the number of ways to color the squares of a 1-by-n chessboard using the colors, red, white, and blue, if an even number of squares are colored red.
Let an denote the number of ways of such colorings and set a0 = 1. Each such coloring can be considered as a permutation of three objects r (for red), w (for white), and b (for blue) with repetition allowed, and the element r appears even number of times. The exponential generating function of the sequence (an) is E(x) = Ã ∞ X n=0 x2n (2n)!
! Ã ∞ X n=0 xn n!
!2 = ex + e−x 2 e2x = 1 2 ¡ e3x + ex¢ = 1 2 Ã ∞ X n=0 3nxn n!
+ ∞ X n=0 xn n!
!
= 1 2 ∞ X n=0 (3n + 1) · xn n! .
Thus the sequence is given by an = 3n + 1 2 , n ≥0.
Example 8.4. Determine the number an of n digit (under base 10) numbers with each digit odd where the digit 1 and 3 occur an even number of times.
Let a0 = 1. The number an equals the number of npermutations of the multiset M = {∞1, ∞3, ∞5, ∞7, ∞9}, in which 1 and 3 occur an even number of times. The exponential generating function of the sequence an is E(x) = Ã ∞ X n=0 x2n (2n)!
!2 Ã ∞ X n=0 xn n!
!3 = µex + e−x 2 ¶2 e3x = 1 4 ¡ e5x + 2e3x + ex¢ = 1 4 Ã ∞ X n=0 5nxn n!
+ ∞ X n=0 3nxn n!
+ ∞ X n=0 xn n!
!
= ∞ X n=0 µ5n + 2 × 3n + 1 4 ¶ xn n! .
Thus an = 5n + 2 × 3n + 1 4 , n ≥0.
22 |
10655 | https://www.reddit.com/r/learnmath/comments/1d0ag1i/why_does_the_following_theorem_hold_sinalpha/ | Why does the following theorem hold: sin(alpha) = cos(beta) ? : r/learnmath
Skip to main contentWhy does the following theorem hold: sin(alpha) = cos(beta) ? : r/learnmath
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r/learnmath•1 yr. ago
Altruistic_Nose9632
Why does the following theorem hold: sin(alpha) = cos(beta) ?
So I was wondering, why exactly the theorem mentioned above holds.
I mean I can plug in numbers and see that it actually is true for any alpha and beta being acute angles in a triangle, but I do not know why that is or how to show it.
Am I right to assume, that one usually shows it thorugh imaginary numbers? I think it can be proven, by showing, that sin(alpha) equals cos(beta) if the coordinate system in cos(beta) gets rotated by 90 degrees counterclockwise.
Is that assumption right, and how would the corresponding proof look like?
Thanks in advance! :)
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10656 | https://www.youtube.com/watch?v=71LETCOkYiY | Derive g=(GM)/(R^2). Expression for acceleration due to gravity on the surface of Earth
Sharath Gore
10800 subscribers
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Description
4315 views
Posted: 16 Jan 2024
5 comments
Transcript:
hello everybody in this session we are going to derive uh the equation G is equal to GM by r² first what we will do is say let us suppose that this is Earth and say the radius of Earth is r e mass of the earth is M let us suppose that I have considered we have considered a body of mass m on the surface of the Earth now what happens according to universal law of gravitation the force between this small body of mass m and the Earth is given by according to universal law of gravitation according to universal law of gravitation universal law of gravitation according to universal law of gravitation what we can write force between those two objects f is equal to g into mass of the earth into Mass of the body divided by distance between the two what is the distance between the two distance between the two is equal to radius of the Earth okay r e whole Square so this is my equation number one and according to Newton's Second Law f is equal to M A what we can write according to according to Newton's Second Law according to Newton's Second Law we know that according to Newton Second Law we know that that force is equal to its weight we is equal to M into G this is my equation number two now from 1 and two from 1 and 2 since the LHS is same what we can write we can write mg is equal to g m m / r e s now what happens this small letter M Small M gets cancelled so finally we get an equation for G G is equal to GM / r² hope you have understood this very simple thank you very much |
10657 | https://libraryguides.centennialcollege.ca/c.php?g=717032&p=5176688 | Skip to Main Content
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Integration of a Logarithmic Form
The power rule for integration is valid for all values except when the exponent is equal to . This is because
Thus, reversing the process where the denominator's exponent is would lead to an integral of the logarithmic form.
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Thus, the integration of will be done using the logarithmic form, whereas will be done using the power rule for integration.
Example 1: Integrate the function
Solution: We can recognize this is an integral of logarithmic form because the denominator is to the power of -1 (e.g., it can be written as .
Let , . We can substitute these values and change the variable to u
Solving for the definite integral we get,
Example 2: Integrate the function
Solution: Let
The integral after subitution is
Example 3: Integrate
Solution: There is a trick here and that is to multiply by
The resulting integral is
Now let
The integral becomes,
Practice Questions - Integral of Logarithmic Form
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10658 | https://www.cuemath.com/trigonometry/sum-to-product-formulas/ | Sum to Product Formula
The sum to product formula in trigonometry are formulas that are used to express the sum and difference of sines and cosines as products of sine and cosine functions. We can apply these formulas to express the sum or difference of trigonometric functions sine and cosine as products and hence, simplify mathematical problems. The sum to product formulas can be derived using the product to sum formulas in trigonometry using substitutions of the variables.
In this article, we will explore the sum to product formula in detail and derive these formulas using the product to sum formulas. We will also understand the application of these formulas with the help of solved examples for a better understanding of the concept.
| | |
--- |
| 1. | What are Sum to Product Formulas? |
| 2. | Sum to Product Formulas List |
| 3. | Sum to Product Formulas Proof |
| 4. | Using Sum to Product Formula |
| 5. | FAQs on Sum to Product Formulas |
What are Sum to Product Formula?
The sum to product formula are used to express the sum or difference of sine function and the sum or difference of cosine function as the product of sine and cosine functions. These sum to product formulas are also known individually given by,
Please note that these formulas are not the same as the angle sum formulas in trigonometry. For example, sin A + sin B is not the same as sin (A + B). Let us now go through the formulas of the above-mentioned sum to product formulas in the next section.
Sum to Product Formula List
We can prove these sum to product formulas using the product to sum formulas in trigonometry. The sum to product formula are expressed as follows:
Sum to Product Formulas Proof
Now, that we have discussed the sum to product formulas in trigonometry, let us derive these formulas using the product to sum formulas whose formulas are given by,
To derive the sum to product formulas, assume (p + q)/2 = A and (p - q)/2 = B. Now, taking the sum and difference of A and B, we have
A + B = [(p + q)/2] + [(p - q)/2]
= p/2 + q/2 + p/2 - q/2
= p/2 + p/2
= p
A - B = [(p + q)/2] - [(p - q)/2]
= p/2 + q/2 - p/2 + q/2
= q/2 + q/2
= q
Now, substituting the values of A, B, A + B and A - B in the formulas (1), (2), (3), and (4), we have
Hence, we have derived the sum to product formulas which are given by the formulas (5), (6), (7) and (8).
Using Sum to Product Formula
We use the sum to product formula to simplify and solve mathematical problems in trigonometry. In this section, we will understand how to apply the sum to product formulas with the help of solving a few examples.
Example 1: Express the difference of cosines cos 4x - cos x as a product of trigonometric function using sum to product formulas.
Solution: We know that cos A - cos B = -2 sin [(A + B)/2] sin [(A - B)/2]. Substituting A = 4x and B = x into this formula, we have
cos 4x - cos x = -2 sin [(4x + x)/2] sin [(4x - x)/2]
= -2 sin (5x/2) sin (3x/2)
Answer: Hence, we can express the difference cos 4x - cos x as -2 sin (5x/2) sin (3x/2) as a product of trigonometric functions.
Example 2: Evaluate the value of sin 15° + sin 75° using the sum to product formula.
Solution: The formula required to find the value of sin 15° + sin 75° is sin A + sin B = 2 sin [(A + B)/2] cos [(A - B)/2]. Substituting A = 15° and B = 75° into the formula, we have
sin 15° + sin 75° = 2 sin [(15° + 75°)/2] cos [(15° - 75°)/2]
= 2 sin(90°/2) cos (-60°/2)
= 2 sin 45° cos 30° --- [Because cos(-x) = cos x]
= 2 × 1/√2 × √3/2 --- [Because cos 30° is equal to √3/2 and sin 45° is equal to 1/√2]
= √3/√2
= √(3/2)
Answer: sin 15° + sin 75 = √(3/2) using sum to product formula.
Important Notes on Sum to Product Formula
☛ Related Topics:
Sum to Product Formula Examples
Example 1: Use sum to product formula to express cos 8x + cos 2x as the product.
Solution: To express cos 8x + cos 2x as the product, we will use the formula cos A + cos B = 2 cos [(A + B)/2] cos [(A - B)/2]. Substituting A = 8x and B - 2x into the formula, we have
cos 8x + cos 2x = 2 cos [(8x + 2x)/2] cos [(8x - 2x)/2]
= 2 cos (10x/2) cos 6x/2
= 2 cos 5x cos 3x
Answer: Hence, cos 8x + cos 2x = = 2 cos 5x cos 3x using the sum to product formula.
Example 2: Calculate the value of sin 225° - sin 135°
Solution: To find the value ofsin 225° - sin 135°, we will use the sum to product formula given by, sin A - sin B = 2 sin [(A - B)/2] cos [(A + B)/2]. Substitute the values A = 225° and B = 135° into the formula, we have
sin 225° - sin 135° = 2 sin [(225° - 135°)/2] cos [(225° + 135°)/2]
= 2 sin (90°/2) cos(360°/2)
= 2 sin 45° cos 180°
= 2 × 1/√2 × (-1)
= -√2
Answer: Hence, sin 225° - sin 135° = -√2 using the sum to product formula.
Example 3: Prove that (cos 4x - cos 2x) / (sin 4x + sin 2x) = - tan x
Solution: We will use the following sum to product formulas to prove the given result:
Using the above formulas, we have
LHS = (cos 4x - cos 2x) / (sin 4x + sin 2x)
= -2 sin [(4x + 2x)/2] sin [(4x - 2x)/2] / 2 sin [(4x + 2x)/2] cos [(4x - 2x)/2]
= - [ sin (6x/2) sin (2x/2) ] / [ sin (6x/2) cos (2x/2)]
= - (sin 3x sin x) / (sin 3x cos x)
= - sin x / cos x
= - tan x
= RHS
Answer: Hence, we have proved (cos 4x - cos 2x) / (sin 4x + sin 2x) = - tan x using sum to product formulas.
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Sum to Product Formulas Questions
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FAQs on Sum to Product Formula
What are Sum to Product Formula in Trigonometry?
The sum to product formula in trigonometry are formulas that are used to express the sum and difference of sines and cosines as products of sine and cosine functions. These sum to product formula are also known individually given by,
List all Sum to Product Formula.
The sum to product formulas are expressed as follows:
How Do You Prove Sum to Product Formulas?
The sum to product formulas can be derived using the product to sum formulas in trigonometry using substitutions of the variables. We can assume (p + q)/2 = A and (p - q)/2 = B and substitute these in the product to sum formulas given by,
How to Use Sum to Product Formula?
The sum to product formulas are used to express the sum and difference of sine and cosine functions as products of trigonometric functions sine and cosine.
How Do You Convert Sum to Product in Trigonometry?
We can convert the sum of sine and cosine into the product of sine and cosine in trigonometry by taking appropriate assumptions of variables and substituting them into the formulas.
What is the Difference Between Product to Sum and Sum to Product Formula?
In sum to product formula, we express the sum of trigonometric functions sine and cosine as products. On the other hand, in product-to-sum formula, we express the product of trigonometric functions sine and cosine as the sum or difference of sine or cosine. |
10659 | https://chriswalshaw.co.uk/papers/fulltext/DiekmannPCmp00.pdf | Shape-optimized mesh partitioning and load balancing for parallel adaptive FEM q Ralf Diekmann a,, Robert Preis b, Frank Schlimbach c, Chris Walshaw c a Hilti AG, Corp. Research, Schaan, Liechtenstein b Department of Mathematics and Computer Science, University of Paderborn, F urstenallee 11, D-33102 Paderborn, Germany c School of Computing and Mathematical Sciences, The University of Greenwich, 30 Park Row, Greenwich, London SE10 9LS, UK Received 15 February 1999; received in revised form 24 September 1999; accepted 24 September 1999 Abstract We present a dynamic distributed load balancing algorithm for parallel, adaptive Finite Element simulations in which we use preconditioned Conjugate Gradient solvers based on domain-decomposition. The load balancing is designed to maintain good partition aspect ratio and we show that cut size is not always the appropriate measure in load balancing. Fur-thermore, we attempt to answer the question why the aspect ratio of partitions plays an im-portant role for certain solvers. We de®ne and rate dierent kinds of aspect ratio and present a new center-based partitioning method of calculating the initial distribution which implicitly optimizes this measure. During the adaptive simulation, the load balancer calculates a bal-ancing ¯ow using dierent versions of the diusion algorithm and a variant of breadth ®rst search. Elements to be migrated are chosen according to a cost function aiming at the opti-mization of subdomain shapes. Experimental results for Bramble's preconditioner and com-parisons to state-of-the-art load balancers show the bene®ts of the construction. Ó 2000 Elsevier Science B.V. All rights reserved.
Keywords: Mesh partitioning; Load balancing; Shape-optimization; Aspect ratio; Parallel; Adaptive; Finite element method www.elsevier.com/locate/parco Parallel Computing 26 (2000) 1555±1581 q Parts of the results appeared in the Proceedings of IRREGULAR'98 and EUROPAR'98 (Springer LNCS 1457 and 1470).
Corresponding author.
E-mail addresses: diekral@hilti.com (R.
Diekmann), robsy@upb.de (R.
Preis), F.Schlimbach@gre.ac.uk (F. Schlimbach), C.Walshaw@gre.ac.uk (C. Walshaw).
0167-8191/00/$ - see front matter Ó 2000 Elsevier Science B.V. All rights reserved.
PII: S 0 1 6 7 - 8 1 9 1 ( 0 0 ) 0 0 0 4 3 - 0 1. Introduction Finite elements (or ®nite dierences or ®nite volumes) can be used to numeri-cally approximate the solutions of partial dierential equations (PDEs). The PDEs describe, for example, the ¯ow of air around a wing or the distribution of tem-perature on a plate which is partially heated [3,9]. The domain on which the PDE has to be solved is discretized into a mesh of ®nite elements (triangles or rectangles in 2D, tetrahedra or hexahedra in 3D) and the PDE is transformed into a set of linear equations de®ned on these elements . The coupling between equations is given by the adjacencies in the mesh. Usually, iterative methods such as conjugate gradient (CG) or multigrid (MG) are used to solve the linear systems [3,35]. The quality of solutions obtained by such numerical approximation algorithms heavily depends on the accuracy of the discretization. In particular, in regions with steep solution gradients, the mesh has to be re®ned suciently, i.e., the elements have to be small in order to allow accurate approximation. Unfortunately, the regions with large gradients are usually not known in advance. Hence, the meshes either have to be re®ned regularly with the consequence that there is a large number of small elements in regions where they are not required, or the re®nement takes place during the calculation based on error estimates of the current solution . Ob-viously, the second variant is to be favored. Its solutions should have the same quality as if the mesh was re®ned regularly but the number of elements and, thus, the time needed to determine the solution is only a small fraction of the regular case.
The parallelization of numerical simulation algorithms usually follows the single-program multiple-data (SPMD) paradigm: The same code is executed on each processor but on dierent parts of the data. This means that the mesh is partitioned into P subdomains where P is the number of processors, and each subdomain is assigned to one processor [9,11]. Because iterative solution algorithms mainly per-form local operations, i.e., data dependencies are de®ned by adjacencies in the mesh, 1 the parallel algorithms only require communication at the partition boundaries. The parallel eciency depends on two factors: an equal distribution of data (computational load) on the processors, and a small communication overhead achieved by minimizing the boundary length.
Together with the additional constraint of minimizing the number of cut edges (the total interface length in the case of FE-mesh partitioning), the mesh partitioning problem turns out to be NP-complete , i.e. it is currently not solvable to opti-mality in a reasonable amount of time. Fortunately, a number of quite ecient graph (mesh) partitioning heuristics (approximation algorithms) have been developed [12,14,22,27,28,32,38]. Most of them optimize the balance of subdomains (their de-viation from the mean number of elements) and the number of cut edges, the cut size.
Optimizing for cut size is sucient for many applications such as standard iterative 1 Modulo some global dependencies in certain iterative schemes. We do not consider these since they are independent of the way the mesh is partitioned.
1556 R. Diekmann et al. / Parallel Computing 26 (2000) 1555±1581 equational solvers for PDE problems. However, there are certain other cases where this is not true. If, for example, the decomposition is used to construct pre-condi-tioners [3,5], cut and balance may no longer be the only factors which determine the eciency. In addition to a signi®cant amount of time spent on the solution of in-terface problems, the shape of subdomains heavily in¯uences the quality of pre-conditioning and, thus, the overall execution time . First attempts at optimizing the aspect ratio (AR) of subdomains rate elements depending on their distance from the center of a subdomain [7,11] and include this into the cost function of a local iterative search heuristic such as the Kernighan±Lin algorithm (KL) . In this paper we use a simple, center-based partitioning heuristic which optimizes subdo-main shapes implicitly.
In the case of adaptive re®nement, the distribution of data on the processors will become unbalanced if the number of newly generated elements is not the same on each processor (as is usually the case with solution adaptive re®nement). Therefore, the partition has to be altered in order to re-establish a balanced distribution. A number of solutions to this load balancing problem are based on re-partitioning, where (sometimes even sequential) mesh partitioning algorithms are used . We propose a two-phase distributed load balancing algorithm which takes the existing mesh partition into account. The ®rst phase determines the amount of load that has to be moved between dierent subdomains in order to balance the distribution globally. The adjacencies between subdomains de®nes the quotient graph . The algorithm determines a balancing ¯ow on this graph. The ¯ow tells the processors how many data items (i.e., elements) they have to move to each of their neighbors.
This balancing ¯ow calculation can optionally use dierent kinds of diusive methods [4,8], in particular ®rst and second order diusion iterations [13,19], and a heuristic for convex non-linear min-cost ¯ow .
In the second phase, the elements that have to be moved are identi®ed. When choosing these elements, the load balancer tries to optimize partition AR in addition to load balance. The elements at partition boundaries are weighted by a cost func-tion consisting of several components. The migration chooses elements according to their weight and moves them to neighbors. The element weight functions are used to guide'' the balancing ¯ow calculation. We consider the existing partition and de®ne weights for the edges of the quotient graph expressing a kind ofcost'' for moving elements over the corresponding borders. The aim is to avoid a large ¯ow of elements between parts with a small common border.
The main contributions of this paper are: · We attempt to answer the question why cut size (interface length) might not al-ways be the right measure in balancing adaptive meshes.
· We present numbers of iterations of parallel domain-decomposition precondi-tioned conjugate gradient solvers (DD±PCG) on irregular adaptive meshes.
· We investigate dierent graph partitioning algorithms with respect to AR and dis-cuss a center-oriented method which optimizes this measure implicitly.
· We discuss the problem of dynamic load balancing for solvers using adaptive re®nement and present new element migration strategies aimed at optimizing subdomain AR.
R. Diekmann et al. / Parallel Computing 26 (2000) 1555±1581 1557 · We modify existing balancing ¯ow algorithms to support the shape optimization task.
All results presented apply (so far) to 2D meshes. Some possible extensions to the 3D case are discussed in Section 5. The partitioning and load balancing algorithms are included in parallel adaptive FEM (PadFEM). This is an object-oriented environ-ment which supports parallel adaptive numerical simulations [3,9]. It includes graphical user interfaces, mesh generators for 2D and 3D domains, mesh parti-tioning algorithms , triangular and tetrahedral element formulations for Poisson and Navier±Stokes problems , dierent solvers (especially DD±PCGs) [3,5], error estimators , mesh re®nement algorithms , and load balancers .
The next section gives an introduction to the ®eld of balancing adaptive meshes, DD-preconditioning, and shape of subdomains. A short overview of existing ap-proaches is given too. Section 3 introduces us to a center-oriented method for the initial load distribution which implicitly optimizes the AR. Comparisons with results from the Chaco , Jostle and Metis partitioning tools are discussed.
Section 4 considers the dynamic load balancing problem for parallel adaptive DD-preconditioners. It starts with the description of element migration strategies in Section 4.1. This second load balancing phase is presented ®rst, since parts of it are used in Section 4.2 where the balancing ¯ow calculation is described. Section 4.3 ®nishes the paper with a number of results of AR and iteration numbers of the DD± PCG solver of PadFEM. Comparisons are made to the partitioning tool Jostle .
2. The problem 2.1. Balancing adaptive meshes In most cases, the computational load of a ®nite element problem can be mea-sured in terms of numbers of elements. Thus a load balanced parallel execution requires a partition of the mesh. The mesh is partitioned into subdomains each of which contains the same number of elements. Such an initial partition can be gen-erated by the use of any of the existing ecient graph partitioning algorithms. If the mesh is re®ned adaptively, the existing partition will usually become unbalanced.
Several existing implementations of parallel adaptive grid applications solve this problem by repartitioning the mesh. This is done by the use of any of the afore-mentioned partitioning methods . The drawback of such an approach is twofold: ®rst, the mesh has to be routed to a single processor if the partitioning tool is se-quential. Such an approach is obviously not scalable to large numbers of processors and to large meshes. Second, even if a parallel partitioning tool is available (such as ParMetis or PJostle ), a new partition may dier greatly from the existing one. As a result, large amounts of data may have to be shifted between processors.
Although there are attempts to minimize this data-movement , comparisons to approaches which take the existing distribution into account show that if the mesh adaption changes the mesh only slightly, only a small fraction of the data movement is really necessary .
1558 R. Diekmann et al. / Parallel Computing 26 (2000) 1555±1581 The load balancing module in PadFEM follows a diusive strategy [10,13,37] applied to the quotient graph from a given partition. The graph is de®ned by the adjacencies between subdomains and contains one node for each subdomain. Edges denote common borders between the corresponding subdomains. Fig. 1 shows an unbalanced partition of a simple mesh (domain square'') into six subdomains (left) and the resulting quotient graph (center). Any reasonable load balancing which takes the existing distribution into account has to move data (elements) via the edges of this graph. We add weights to nodes of the graph according to their load and de-termine a balancing ¯ow on the edges. This ¯ow has the property that the partition is globally balanced after shifting a corresponding amount of load (elements) between adjacent subdomains. The ¯ow is given as edge labels in Fig. 1. It can be determined by the use of network ¯ow algorithms (which are ± unfortunately ± usually not parallel) or by local iterative methods such as diusion or dimension exchange [4,8], which are parallel by de®nition. Second, in the element migration phase, the load is actually moved. Input to this phase is the ¯ow on the quotient graph. The task is to choose element migrations in order to ful®ll the ¯ow demands. Their choice may consider additional cost criteria such as minimizing the boundary lengths or opti-mizing the shape of subdomains.
2.2. Preconditioning by domain decomposition We brie¯y describe how preconditioning techniques based on domain decompo-sition work, based on our own implementation of the BPS algorithm of Bramble et al. . There are several variants such as, e.g., the conjugate gradient boundary iteration method (CGBI) by Blazy et al. which improve the BPS algorithm. From the algorithmic point of view though, they only slightly dier from the original method. Let us consider a Poisson problem with homogeneous Dirichlet boundary conditions (Fig. 2 (left)).
ÿ Du
x; y f x; y 8
x; y 2 X; u
x; y 0 8
x; y 2 C: 1 Fig. 1. An unbalanced partition (left), the corresponding quotient graph with balancing ¯ow (center), and the resulting balanced partition after migration (right).
R. Diekmann et al. / Parallel Computing 26 (2000) 1555±1581 1559 This kind of elliptic partial dierential equation is derived from a discretization of the Navier±Stokes equation using splitting methods (pressure correction, ). Using standard ®nite element formulations, Eq. (1) is transformed into a system Au f of linear equations where the structure of A corresponds to the structure of the mesh. In order to increase the speed of convergence of the normal conjugate gradient iterative solver (CG), the linear system is preconditioned with a matrix B (cf. Fig. 2): Bÿ1Au Bÿ1f : 2 The parallelization of the CG solver is usually based on domain decomposition (cf.
Fig. 2) where the mesh (the domain) is split into a number of subdomains which are assigned to dierent processors.
The domain decomposition preconditioning is based on the BPS algorithm and solves the equation marked () in Fig. 2 in three steps: (1) For each subdomain Xi it ®rst computes ÿDv r in Xi with v 0 on ~ C; where r is the residuum and ~ C are the arti®cial (inner) boundaries. This leads to a system of linear equations Av r within each subdomain which are solved using a sequential V-cycle multigrid.
(2) The next step is to compute a Laplace Beltrami problem ÿDÿ1=4w rC v~ C with homogeneous boundary condition, where v computes the jump over ~ C of v.
This leads us to a system of linear equations with a dense matrix Cw v~ C. C can be approximated by a tridiagonal matrix ~ C which simpli®es the task of calculating the inverse ~ Cÿ1 (see ).
Fig. 2. Domain X with boundary C, decomposition and the preconditioned CG algorithm (cf. e.g. ).
1560 R. Diekmann et al. / Parallel Computing 26 (2000) 1555±1581 (3) The last step uses w to compute ÿD^ v 0 in Xi with ^ v w on ~ C: This again leads us to a system of linear equations which can also be solved by the use of a V-cycle multigrid on each subdomain Xi.
The solution s of () (Fig. 2) is now given by s v ^ v on all subdomains. For each iteration of the global CG method, Steps 1±3 have to be performed. Fig. 3 il-lustrates the DD±PCG algorithm on a simple square domain X split into two parts.
The ®rst picture shows the solutions on Xi with ~ C 0 (Step 1). In the second, the solution of the Laplace Beltrami problem on ~ C can be seen and the third presents the result after Step 3. The rightmost picture shows the ®nal solution after four global iterations with a residual of jrjmax 6 10ÿ6.
The time that is needed by such a preconditioned CG solver is determined by two factors. First, the maximum time that is needed by any of the subdomain solutions and second, the number of iterations of the global CG. Both factors are at least partially determined by the shape of the subdomains. While the MG as solver on the Xi's is relatively resistant to shape, the number of global iterations is heavily in¯uenced by the AR of the subdomains. In a way, the subdomains can be regarded as elements of theinterface'' problem . And just as with normal FEM, where the condition of A is in¯uenced by the AR of elements, the con-dition of Bÿ1A is in¯uenced by the subdomains' AR in the preconditioning case. If rectangular shaped domains (and subdomains) are used, the relation between shape and number of global iterations can be expressed by the easiest de®nition of AR ± the ratio between longest and shortest boundary edge. An example is presented in Fig. 4 where a Poisson problem is solved with DD±PCG on a regular mesh. The domain is split into an increasing number of slices with a resulting increase in AR. It becomes obvious that the number of global iterations grows heavily with the AR. To show that this is not caused by the increasing number of subdomains P, the lower curve gives the number of iterations when only P is growing and the AR is kept constant. It can be seen that in this case the number of iterations remains almost constant (although in fact the problem size is in-creasing).
Fig. 3. The three steps of the DD±PCG algorithm (left to right): Subdomain solution with zero boundary conditions, interface solution, subdomain solutions with new boundary conditions. The rightmost illus-tration shows the solution after 4 iterations.
R. Diekmann et al. / Parallel Computing 26 (2000) 1555±1581 1561 2.3. Aspect ratios Possible de®nitions of AR can be found in Fig. 5. The ®rst two are motivated by common measures in triangular mesh generation where the quality of triangles are expressed in either Lmax=Lmin (longest to shortest boundary edge) or R2 o=R2 i (the area of smallest circle containing the domain to the area of largest inscribed circle). The de®nition AR R2 o=R2 i expresses the fact that circles are perfect shapes. Unfortu-nately, circles are quite expensive to ®nd for arbitrary polygons: by the use of Voronoi-diagrams, we can determined both in O
2n log n steps where n is the number of nodes of the polygon (and faster incremental update algorithms are not known). The de®nition AR R2 o=A (A being the area of the domain) is another measure that favors circle-like shapes. It still requires the determination of the smallest outer circle but turns out to be better in practice. We can do a further step and approximate Ro by the length B of the boundary of the domain (which can be determined fast and updated incrementally in O(1)). For a sub-domain with area A and perimeter B; AR B2=16A is the ratio between the area of a square with pe-rimeter B and area A. This de®nition assumes that squares are perfect domains.
Circles oer a better perimeter/area ratio but force neighboring domains to be-come concave (Fig. 6 (left)). The example in Fig. 4 has already used the ®rst de®-nition of AR : AR Lmax=Lmin. This measure does not express the shape properly for Fig. 4. # Iterations vs AR for DD±PCG. A rectangular domain recursively split into an increasing number of slices (top right). A growing rectangular domain with subdomains of equal shape for each processor (bottom right).
Fig. 5. Dierent de®nitions of AR: Lmax=Lmin; R2 o=R2 i ; R2 o=A and B2=16A.
1562 R. Diekmann et al. / Parallel Computing 26 (2000) 1555±1581 irregular meshes and partitions. Fig. 6 (center) presents an example. P1 is perfectly shaped, but as the boundary towards P4 is very short, Lmax=Lmin is large. The circle-based measures usually fail to rate jagged boundaries or inscribed corners. Fig. 6 (right) displays examples each of which have the same AR but which are very dif-ferent in shape. According to our experience (and also from the examples), AR R2 o=A and AR B2=16A turn out to be the most robust measures, which best express the desired aims of producing compact domains.
2.4. Existing approaches A large number of methods for the graph partitioning problem has been devel-oped in dierent research and application ®elds. We brie¯y discuss some of the methods which are frequently applied for FE mesh partitioning.
The coordinate sorting (COO) method (see, e.g., ) is based on the vertex co-ordinates only and is very simple and easy to implement. The mesh is cut by straight lines perpendicular to the axis of the longest elongation of the mesh. The result is a stripe-wise partition. The method may also be applied recursively (COO_R), re-sulting in a more box-wise partition. Although this approach does not consider any connectivity information of the graph, for some application graphs like, e.g., FE meshes of simple domains, it results in parts with reasonable shapes. Greedy ap-proaches are often based on the graph connectivity. Typically, the ®rst part of a partition is initialized with one single element and further elements are added until the required size is reached. Then, a new part is initialized with an unassigned ele-ment and the new part is build up in the same greedy fashion. One possibility of choosing new elements is to repeatedly take all non-assigned elements adjacent to elements of the current subdomain, i.e. progressing in a breath-®rst manner (GBF, see e.g. ). Another possibility is to choose an element which reduces the cut most of all (GCF, see e.g. [12,38]). The greedy approach usually results initially in very compact subdomains, but often the last subdomain consists of all leftover elements and it is thus very unlikely that it has a smooth shape. More elaborate initial par-titioners use connectivity measures based on the second smallest eigenvalue of the graph's Laplacian. These so called spectral methods are quite expensive, but combined with fast multi-level contraction schemes they belong to the state-of-the-art in graph partitioning software .
Once a partition is calculated, one can use a local improvement method to further optimize the cut size. The KL heuristic is the most frequently used local Fig. 6. Problems of several de®nitions of AR.
R. Diekmann et al. / Parallel Computing 26 (2000) 1555±1581 1563 improvement method. It uses a sequence of logical vertex pair exchanges to deter-mine the sets that have to be exchanged physically. Fiduccia and Mattheyses have modi®ed the method. They use a sequence of single vertex moves to determine the sets. Meanwhile similar to the KL algorithm, the Helpful-Sets (HS) heuristic is based on local rearrangements . It, too, has to search for two sets of equal size (one in each part), which will improve the cut if they are exchanged. Unlike KL, it does not only consider single vertices but also whole sets that take part in the ex-change steps.
There are several mesh partitioning software libraries and most of them are freely available for academic research. The most popular examples are Chaco by Hendrickson and Leland which includes inertial and spectral partitioning [22,32] as well as multilevel-strategies [2,23], Metis by Karypis and Kumar, which includes fast multilevel strategies, Scotch by Pellegrini and Roman, which includes mapping facilities, Top/Domdec by Farhat et al., Jostle [43,44] by Walshaw et al.
or Party by Diekmann and Preis.
The tools Metis and Jostle are also designed to support partitioning and load balancing of adaptive mesh calculations in parallel. Both use the algorithm of in order to determine the balancing ¯ow. What is more, both use a multilevel strategy for shifting elements, where optionally the coarsening is only done inside subdomains. Jostle uses the concept of relative gain optimization at partition boundaries. Metis uses so called enforcement levels and greedy re®nement. Both tools optimize according to the number of edges crossing the partition boundaries.
The AR is not directly considered. Load balancers particularly designed to optimize the subdomain AR can be found in [11,16,39,42]. The tool PAR2 was originally constructed as a parallel partitioner, but it can also be used in an adaptive envi-ronment . The method described in is an iterative partitioner which tries to improve the subdomain AR in a number of steps. In this work, DD-PCG solvers are used, but not in an adaptive environment. Other attempts are to optimize subdomain shapes by the use of meta-heuristics such as simulated annealing (SA) or Tabu Search .
3. Partitioning the initial mesh Mesh (graph) partitioning is an area of active research. As mentioned above, the problem of partitioning a graph into a number of equally sized parts such that the number of cut edge is minimized is NP-complete . Nevertheless, the initial mesh of a parallel adaptive FEM simulation is not usually excessively large and will be re®ned throughout the simulation. In this section we propose a center-oriented method called bubble (BUB) for partitioning the initial mesh which implicitly op-timizes the shape of subdomains. Some of the ideas this method is based on are very simple and natural and it has some similarities with dierent other approaches.
Bubble generalizes some ideas of the bisection growing method of . A similar center-based approach has been developed in and a parallel center-based ap-proach can be found in . An anonymous referee pointed out the similarities to an 1564 R. Diekmann et al. / Parallel Computing 26 (2000) 1555±1581 algorithm for vector quantizer design where the vector entries are partitioned while considering a reproduction alphabet with as many elements as there are parts.
The idea of bubble (displayed in Fig. 7) is to represent a partition by a set of seed vertices, one for each part, from which the subdomains are grown simultaneously in a breadth-®rst manner until the whole mesh has been covered. Colliding parts form a common border and keep on growing along this border ± just like soap bubbles in a bath. After the whole mesh is covered, the algorithm determines its center'' vertex for each part. This is de®ned as the new seed and the subdomain growing process starts again. The iteration will be stopped if the movement of all seeds is small en-ough, i.e., if the seed vertices are close to the centers for all parts. The algorithm is based on the observation that withinperfect'' bubbles, the center and the seed vertex coincide. The distances in this method may either be chosen as the path length or as the Euclidean distances. In the case of path length the method works also on graphs without geometrical information.
The Bubble-algorithm is shown in Fig. 8. In order to ®nd the initial seeds, we start a breadth-®rst search (BFS) from a vertex with minimal degree (in the case of FE-meshes, this is usually an element at a domain corner) and search for the vertex which is farthest from this starting point. This vertex is chosen as the ®rst seed. We Fig. 7. The iterative Bubble method.
Fig. 8. The Bubble algorithm.
R. Diekmann et al. / Parallel Computing 26 (2000) 1555±1581 1565 then repeat to perform simultaneous BFS from all seeds that have been found so far to determine a vertex which is farthest from all seeds. It becomes the next seed.
Altogether, P BFSs are performed with P being the number of parts. With this approach of each new seed having the maximum distance from all previous ones we distribute the seeds fairly evenly over the graph. The path length is used as distance measure in the ®rst loop. The main loop of Bubble is started by growing the parts from each seed in a breadth-®rst manner, i.e., each part checks if any of its elements is adjacent to an uncovered element and the smallest part with at least one such adjacent element grabs the one with the shortest Euclidean distance to its seed and assigns it to that part. Only vertices are added which are adjacent to vertices pre-viously assigned to the same part, ensuring connected parts. The ordering of smallest parts tries to keep the ®nal load dierence small and the choice of an adjacent ele-ment with shortest Euclidean distance bene®ts a low AR of that part. This is re-peated until all vertices (elements) are covered. Afterwards, new seeds are calculated by each part independently by searching for the center vertices. This part could be executed in parallel. We de®ne the center of a part to be the vertex for which the sum of Euclidean distances to all other vertices in that part (we call this the distance-value) is minimal. One may ®nd the centers by calculating the distance values for all vertices, but this would have a time consumption of O
#elements2. To avoid this, we calculate the distance-values for the seed as the initial center and all its adjacent vertices and move the center to the neighbor with smallest value; this process is repeated until a local minimum is found. The Bubble algorithm will terminate if in an iteration none of the seeds move any more. To avoid cyclic movements of seeds (which sometimes occur), we stop the algorithm if the AR does not improve for 10 consecutive iterations.
Bubble produces connected parts which are, in general, very compact and have a smooth shape. A major drawback of Bubble is the lack of a guarantee for balanced partitions. Although at the end the seeds are spread out evenly over the whole graph, the parts do not have to contain the same number of elements. To repair this, one may add a local partitioning method to balance the load, trying to further optimize either the cut or the AR.
We investigate the performance of dierent types of mesh partitioning strategies with respect to the number of global iterations of the DD±PCG, the cut size and the AR. We include the simple coordinate methods COO and COO_R and the greedy methods GBF and GCF as described in Section 2.4. Bubble is used without any load balancing, as well as with additional load balancing minimizing either the cut or the AR, where load balancing methods as described in Section 4 are used. In addition, we use the default settings of the Party, Jostle, K-Metis and P-Metis graph-parti-tioning libraries, as well as a Simulated Annealing code which is designed to optimize the AR. Fig. 9 displays the results of the partitioning of the meshes turm (with 531 elements) and cooler (with 749 elements, Fig. 11) into 8 parts. The test case for the numerical solver is a Poisson problem with Dirichlet-0 boundary conditions. The number of iterations for the tested methods dier signi®cantly. It can be observed that the AR Lmax=Lmin and Ro=Ri do not follow the line of the iteration numbers, whereas the values of Ro=A and B2=16A roughly do so. Unlike in the simple example 1566 R. Diekmann et al. / Parallel Computing 26 (2000) 1555±1581 of Section 2, the results additionally indicate that a low cut also leads to a fairly low number of iterations. A comparison of the tested partitioners shows that the simple coordinate and the greedy methods usually result in large iteration numbers. For the bubble variations, the balance by improving the AR B2=16A leads to lower iteration numbers than balancing by improving the cut. Furthermore, the partitions calcu-lated by the partitioning libraries and the Simulated Annealing approach also lead to similarly low iterations numbers.
Table 1 shows additional results using the mesh crack with 20,141 elements as example. The mesh is partitioned into 16 parts and the Bubble method with its variations is compared with the default settings of the partitioning libraries Chaco (Multilevel approach), K-Metis, P-Metis and Jostle. We list the cut, the average ARB2=16A of all parts, and the number of iterations of the DD±PCG as measures The results reveal that the AR is a better measure for the number of iterations than 0 10 20 30 40 50 60 70 9 10 7 11 8 5 6 12 2 1 3 4 Iterations/Cuts/ARs Methods #Its Cut L_max/L_min R_o/R_i R_o/A BB/16A 0 5 10 15 20 25 30 35 40 8 12 7 11 5 9 10 2 3 6 1 4 Iterations/Cuts/ARs Methods #Its Cut L_max/L_min R_o/R_i R_o/A BB/16A Fig. 9. Results of example turm (531 elements, top) and example cooler (749 elements, bottom). The methods are listed with increasing numbers of global PCG iterations. Compared Methods are 1:COO, 2:COO R, 3:GBF, 4:GCF, 5:BUB, 6:BUB+CUT, 7:BUB+AR, 8:Party, 9:Jostle, 10:K-Metis, 11:P-Metis, 12:SA.
R. Diekmann et al. / Parallel Computing 26 (2000) 1555±1581 1567 the cut. To give a telling example, the partition calculated by Pmetis has the lowest cut, but needs the largest number of iterations. The partition calculated by bubble has a high cut size, but if it is load-balanced by further optimizing the AR, it ®nally succeeds in achieving the overall goal: minimizing the number of global iterations of the DD±PCG algorithm. If the bubble partition is load-balanced minimizing the cut instead, the cut becomes smaller. However, the AR and the number of iterations are not as low as for minimizing the AR. Combined with shape optimizing load bal-ancing methods which are described in the following, bubble serves as reasonable initial partitioner. The times to calculate the partitions was fairly small for all methods due to the medium sized examples. In an adaptive environment, the par-titioning only has to be performed on the initial mesh which is usually small. For larger meshes, bubble could be used as partitioning method for the coarse graph in the multilevel paradigm after coarsening the large initial mesh into a much smaller one of similar structure. The Chaco, Jostle and Metis tools have already applied this multilevel strategy.
4. Shape optimized dynamic load balancing We now consider the problem of dynamic load balancing in the case of the mesh being re®ned adaptively. Our solution to this problem is a two-step approach: ®rst, a balancing ¯ow is calculated on the edges of the partition quotient graph. Afterwards, elements at partition boundaries are chosen and migrated in order to ful®ll the de-mands of the ¯ow. We ®rst discuss dierent functions for choosing elements which aim at considering the shape of the partition. This second step is described ®rst because some of the element rating functions introduced here are used in Section 4.2 to guide the balancing ¯ow calculation. Results in Section 4.3 compare dierent functions for rating elements with regard to partition shape. The bene®ts of the dierent steps are investigated and the ®nal algorithm is compared to Jostle.
4.1. Element migration The basic idea of the element migration phase is to rate elements at partition boundaries according to certain cost functions and to greedily choose elements that have to be moved to neighboring processors based on this rating, until the necessary load has been moved. The migration phase starts with a valid balancing ¯ow Table 1 Comparisons for mesh crack (20,141 elements) partitioned into 16 parts Chaco Metis Jostle BUB Kÿ Pÿ +CUT +AR Cut 856 800 760 768 920 799 813 AR 1.93 1.85 1.98 1.92 1.92 1.96 1.87 Iter.
46 46 53 44 50 50 44 1568 R. Diekmann et al. / Parallel Computing 26 (2000) 1555±1581 determined by the phase of ¯ow calculation which will be de scribed in Section 4.2.
For now, assume that the ¯ow is represented as values xij on edges i; j of the quotient graph (i.e., xij elements have to be moved from partition i to partition j. The migration phase aims at balancing the load such that the AR of subdomains is maintained, improved, or deteriorates as little as possible.
As discussed in Section 2.3 we use the perimeter-to-area ratio as a de®nition of AR. For a subdomain Pi with area Ai and perimeter Bi (its boundary length) its ARi is given by ARi B2 i 16Ai : 3 The simplest way to rate elements at partition boundaries is to count the change in cut size (of the element graph) when moved to other partitions. For an element e, let ed
e; i be its number of edges adjacent to elements in Pi. The change in cut when e is moved from Pi to Pj is given by Ccut
e; i; j ed
e; i ÿ ed
e; j 4 For many FEM meshes constructed of triangles, the resulting element graph has a maximal degree of three and the change in Ccut is at most 1. It is known that local graph partitioning algorithms perform badly on degree-3 graphs, so it is no surprise that this element rating function does not produce satisfying results. We will see comparisons later on. Nevertheless, it is used in several applications and, if combined with other sophisticated methods such as multilevel schemes, it can still work very well .
We can also take the change in AR to rate elements at partition boundaries. First, de®ne AR RiARi as the AR of a partitioned mesh. Second, let ae be the area of element e; be the length of its border, and be
i the length of its border to elements in subdomain Pi. Then Car
e; i; j B2 i 16Ai B2 j 16Aj ÿ Bi ÿ be 2be
i2 16
Ai ÿ ae ÿ Bj be ÿ be
j2 16
Aj ae 5 will be the change in AR, if e is moved from Pi to Pj. We can simplify the calculations of (5) by just taking the change in total border length: CB
e; i; j be
i ÿ be
j: 6 The drawback of the functions described so far is their local'' nature. They rate elements only based on local changes in a cost function, but they have no idea of which element will have to be preferred if two of them have the same value. And the case that two elements have the same value does not mean they are equally attractive for the task of optimizing the overall AR. It just means that they have the same number of external edges (4), the same size and border length (5) or the same border length (6). Aglobal'' view, whether it is favorable for a subdomain to own certain elements or not, is lacking.
A ®rst step towards a kind of global'' view is to consider the distances of ele-ments from the center of their subdomain. If all boundary-elements have the same R. Diekmann et al. / Parallel Computing 26 (2000) 1555±1581 1569 distance to this center, the domain will be a circle and will have a good AR. Elements lying far away from the center might be good candidates to move. Let Ci be the center of gravity of subdomain i. If we de®ne the center ce of element e to be the average of its node coordinates, Ci will be de®ned as Ci 1 Ai X e2Pi ce ae: 7 Let dist
e; i be the (Euclidean) distance of element e to the center of subdomain Pi.
The function Crd
e; i; j dist
e; i ÿ dist
e; j 8 is called the relative distance of element e. The idea behind this function is to prefer elements which are far away from the center of their own subdomain and, addi-tionally, near to the center of the target partition. Such function of element rating can be found in some implementations [7,11,36]. The problem is that they do not account for the real physical size of a domain. A large partition might easily move elements to a small one, even if this does not improve the AR of any of them. We can de®ne Di as the average distance of the centers of all elements of subdomain Pi from its center Ci:Di can be used to normalize the distance function (8) .
Csrd
e; i; j dist
e; i Di ÿ dist
e; j Dj 9 is called the scaled relative distance function. Fig. 10 (left) reveals the advantage of Csrd in contrast to Crd. Without any scaling, P3 prefers to move elements to P2, whereas a move towards P1 improves the AR of P1 and P3.
If we consider the shape of a common border between two subdomains, the ideal would be to have a straight line perpendicular to the interconnection of the centers of both. If the border is circular in any direction, one of the domains will become concave which is not desirable. If elements are moved between subdomains which are very dierent in size, the in¯uence of the smaller one on Csrd will become larger with increasing distance from the line connecting both centers. Thus, elements far away from this line are more likely to be moved than those lying on or near to this line.
Fig. 10. Scaled distances (left). Angle (right).
1570 R. Diekmann et al. / Parallel Computing 26 (2000) 1555±1581 This eect can be observed in Fig. 11 (left) where the domainCooler'' is re®ned in 10 steps from initially 749 elements to ®nally 7023. The re®nement mainly takes place at the upper part of the domain and the subdomains lying in the corresponding region give up elements according to Csrd. To avoid this eect, we superimpose Csrd by another function Ca
e; i; j cos a, where a is the angle between the lines Ci $ Cj and Ci $ ce. Fig. 10 (right) presents the construction. The value of cos a will be large, if angle a is small, i.e., the superposition slightly increases the values of those ele-ments which lie in the direction of the target partition. Fig. 11 (right) shows the eect on the same example of Fig. 11 (left). As intended, the borders between subdomains are much straighter and the target domains are less concave.
A ®rst comparison of the dierent rating functions is to be found in Table 2. It shows the maximum and average AR as well as the Cuts, if the domain Square'' is re®ned in 10 steps from initially 115 elements to ®nally 10,410. The values for AR given here and in the rest of the paper are scaled to 1.0 being anoptimal'' shape.
Average values of >1:6 can be considered as unacceptable. It can be observed that Ccut is not only bad in optimizing AR but also in ®nding partitions with a good cut.
The explanation has already been given: As the graph has a maximal degree of 3, local methods are likely to fail. Interestingly, Car and CB do not perform well either.
The purely local rating of this functions does not allow a good AR optimization. Crd and Csrd are quite similar to each other because the dierence in size of the subdo-mains is not very large in this example. The eects of Ca are not directly visible in terms of maximum or average AR. This kind of cosmetic'' operation becomes more obvious in Fig. 11 (left) and (right).
Table 3 shows similar results for combinations of the dierent C's. It can be observed that additional improvements will be possible, if such combinations are used. Unfortunately, the possibilities of dierent combinations enlarge the space of Fig. 11. Migration according to Csrd (left) and to Csrd Ca (right).
Table 2 AR and cut obtained with dierent rating functions AR-max AR-avrg Cut-max Cut-avrg Ccut 2.03 1.66 190 128 Car 1.85 1.55 191 130 CB 1.88 1.56 187 129 Crd 1.53 1.33 142 92 Csrd 1.47 1.29 144 90 R. Diekmann et al. / Parallel Computing 26 (2000) 1555±1581 1571 possible parameter settings. However, generally speaking, Csrd combined with Ca is a good choice for an element rating function.
4.2. Balancing ¯ow calculation In Section 2.1 we have de®ned the quotient graph and balancing ¯ows on this graph. We will now brie¯y review possible algorithms that are suitable to calculate balancing ¯ows. Furthermore we will discuss modi®cations to support the element migration phase.
4.2.1. Possible solutions Let G V ; E; w be the quotient graph of a load balancing problem (Fig. 1). G has n jV j nodes, m jEj edges and there are load values wi attached to each node i 2 V expressing the number of elements of the partition, i is representing. The edges in G express the adjacencies between subdomains (i.e., if two subdomains are not adjacent, then the corresponding edge does not exist either). The task of the bal-ancing problem is to ®nd a ¯ow xij of load on edges i; j 2 E of the graph such that after having moved the desired numbers of elements, the load is globally balanced, i.e. wi w 8i 2 V . Since the ¯ow is directed, we need to de®ne (implicit) directions for edges. We may assume that edges are directed from nodes with smaller numbers to nodes with larger numbers. A ¯ow of xij < 0 on edge i; j 2 E then means that ÿxij load has to be moved from node j to i.
Let A 2 fÿ1; 0; 1gnm be the node-edge incidence matrix of G. The edge directions are expressed by the signs (from 1 to ÿ1). The matrix L AAT is called the La-placian matrix of G. If we de®ne x 2 Rm to be the vector of ¯ow values, and if we like to minimize the total amount of ¯ow, we will search for an x satisfying Minimize X i;j2E jxijj subject to Ax w ÿ w: 10 Eq. (10) is known to be the min-cost ¯ow problem . The question of what are good cost criteria for such a ¯ow is not easy to answer. It is fairly obvious that any so-lution to (10) shifts the ¯ow only via shortest paths in the network. Thus, it often uses only a small fraction of the available edges, a quality that might not be desir-able. This is especially the case when afterwards the shape of subdomains has to be considered.
Table 3 Combined cost functions AR-max AR-avrg Cut-max Cut-avrg Crd 0:2CB 1.365 1.222 119 80 Csrd 0:2CB 1.342 1.201 131 84 Crd 0:4Ca 1.572 1.359 142 93 Csrd 0:6Ca 1.496 1.322 140 90 Crd 0:2CB 0:2Ca 1.405 1.246 125 83 Csrd 0:2CB 0:4Ca 1.392 1.249 134 85 1572 R. Diekmann et al. / Parallel Computing 26 (2000) 1555±1581 A compromise between the con¯icting goals of not shifting too much load but using all edges more or less equally might be the Euclidean norm of the ¯ow de®ned as kxk2 2 P x2 ij. This measure turns (10) into a non-linear, but convex optimization problem which is still solvable in polynomial time. For this special type of problem, solutions can be found easily. Hu and Blake propose the use of Lagrange multipliers for a solution of Ax w ÿ w . The method requires the solution of Lk w ÿ w and the ¯ow is afterwards given by xij ki ÿ kj. Lk w ÿ w is easy to solve in parallel, because L directly relates to the quotient graph G (in fact, L is topologically equivalent to G).
Other local iterative methods determine x directly by performing xt nATwt and wt1 wt ÿ Axt: 11 Iteration (11) is similar to wt1 Mwt with iteration matrix M I ÿ nL and pa-rameter n 6 1= max deg
G. It is known as the diusion method [4,8]. Some schemes try to speed up the convergence of this local iterative method by using non-homo-geneous iteration matrices M, optimal values of n , and over-relaxations such as wt1 bMwt 1 ÿ bwtÿ1 . It can be shown that all of these local iterative methods determine ¯ows x with minimal kxk2 .
4.2.2. Guidance of ¯ow: Local iterative methods We can use any of the above mentioned methods to calculate the balancing ¯ow in PadFEM. The drawback here is the fact that they do not take into account thequality'' of elements at partition boundaries, the length of the boundaries, and their position relative to the centers of their partitions. The edges in G just denote neighborhoods between subdomains, regardless of the number of elements along this border, and the shapes of the corresponding partitions.
As an initial idea, we can put weights on the edges of G expressing the cost (or bene®t) of moving elements via certain edges to consider the task of shape optimi-zation within the balancing ¯ow calculation. We can use the element rating functions de®ned in Section 4.1 to perform this task. For a subdomain Pi, let Boi be its set of border elements, and Boi
j the elements at the border adjacent to subdomain Pj. We de®ne the weight xij on edge i; j 2 E by xij 1 jBoi
jj X e2Boi
j e62Boi
k8k6j C
e; i; j; 12 where C can be any of the cost functions and even combinations of Section 4.1. In the iteration process of (11), the parameter n de®nes the amount of load to shift between neighboring processors per step, depending on their load dierence. If we de®ne n to be nij xij P k xik ; 13 edges with large weights are preferred during the balancing (recall that the ¯ow calculation is only a pre-step to the real load movement; so the borders Boi
j do not change in this step).
R. Diekmann et al. / Parallel Computing 26 (2000) 1555±1581 1573 Table 4 shows results of the diusion method if n is chosen according to (13) and the xij's use dierent element rating functions C. The values given in Table 4 are mean values of experimental results with the given edge weighting function and several dierent combinations of the C's as migration rating functions. Cu denotes the case of no edge weighting. We can see that there are slight improvements in AR for certain examples, but that this approach is generally not helpful. A reason might be its very local structure. If there is a large load dierence between two processors, a large number of elements will be shifted over their common border regardless of how short it is and regardless of whether the receiving processor is itself able to shift (parts of) the load to any of its neighbors properly. In the next section we will de-scribe a method which considers weights along paths of load movement. We will see that such an approach can support the shape optimization phase in a better way.
4.2.3. Average weighted ¯ow The average weighted ¯ow (AWF) algorithm proceeds in two steps. It ®rst de-termines (source/sink/amount) triples giving pairs of processors and the amount of load that has to be moved between them. Afterwards, these ¯ows are routed via paths in the graphs.
The ®rst step starts from each source node and searches for sinks nearby which are reachable via good'' paths in the graphs. For a path p v1; . . . ; vk, we de®ne the path quality q
p 1 k X kÿ1 i1 xvivi1 14 as the average quality of edges on the path. Since we are searching for paths with high quality, the task is to maximize q
p. Unfortunately, this makes the problem intractable. If q
p were just the sum of the x's, we would search for longest paths without loops, an NP complete problem . Additionally, the averaging in (14) causes the loss of transitivity. Abest'' path from i to j via k must no longer include the best paths from i to k and from k to j.
We use a variance of breadth ®rst search (BFS) as a heuristic to ®nd approxi-mately best'' paths. The algorithm starts from a source and searches ± in normal BFS style ± for paths to sinks. A sink node has adistance'' from the source ac-cording to the quality of the best path via which it has been reached during the BFS.
If a node is found again (due to circle closing edges), its distance'' is updated. The algorithm visits each edge at most once and, thus, has a running time of O
m. For Table 4 Guided diusion: mean values of AR using dierent edge-weighting functions Square Cooler Smiley Tower Cu 1.351 1.289 1.387 1.341 Csrd 1.364 1.302 1.393 1.344 Ca 1.353 1.294 1.387 1.340 Crd Ca 1.361 1.296 1.390 1.765 1574 R. Diekmann et al. / Parallel Computing 26 (2000) 1555±1581 each sink node, all the paths from the source found during the search are stored together with their quality. The load wi ÿ w of a source i is distributed (logically) to itsbest'' sinks, each of them ®lling up to w in order to build the (source/sink/ amount) triples. The time for this phase is dominated by the search. As there are at most n sources, the total running time is O
n m.
The second phase routes the ¯ow demands between source/sink pairs via paths in the network. The task here is to distribute the ¯ow evenly among (a subset of) possible paths from the source to the sink. The ®rst phase has already calculated possible paths. If a ¯ow of xst has to be shifted from a source s to a sink t and if there are two possible paths p1 and p2 between them, the fraction of q
p1=
q
p1 q
p2 will be shifted via p1. The rest will be shifted via p2.
Table 5 shows a comparison of AWF and Diusion. Again, several dierent functions for rating element are chosen to guide the migration. The values in Table 5 are mean values of all results for a given problem and a given ¯ow algorithm and edge weighting function. It can be observed that an improvement in the achievable AR of around 5% is possible by weighting the edges properly (Cu is again the non-weighted case). AWF behaves generally better than Diusion. Sometime the im-provements are large (25%), in most cases they are around 5%. Csrd turns out to be a good edge weighting function, and also the use of Ca sometimes improves the situation.
4.3. Results Figs. 12 and 13 present the development of AR, cut and number of iterations of the DD-PCG solver over a number of re®nement levels of the Domain Square'' (Fig. 1) re®ned in a number of steps from 736 to 15421 elements. Experiments using regular meshes and regular domains show an increase in the number of iterations by one when the number of elements are doubled. For irregular meshes such as those used here, one can not hope for such good convergence behavior. Fig. 12 displays the behavior if Ccut is used as element rating function, Fig. 13 if Csrd Ca is used. It can be observed that the number of iterations will be very unstable and generally larger if cut is the optimization goal. If the shape is optimized, the solver behaves much better.
Generally, AR expresses the development of the number of iterations much better than Cut. We conclude from the experiments that an optimization according to AR is reasonable.
Table 5 AWF in comparison to diusion Square Cooler Smiley Tower Di AWF Di AWF Di AWF Di AWF Cu 1.351 1.340 1.289 1.289 1.387 1.411 1.341 1.320 Csrd 1.364 1.320 1.302 1.289 1.393 1.340 1.344 1.311 Ca 1.353 1.344 1.294 1.296 1.387 1.368 1.340 1.336 Crd Ca 1.361 1.328 1.296 1.289 1.390 1.374 1.765 1.300 R. Diekmann et al. / Parallel Computing 26 (2000) 1555±1581 1575 Fig. 14 (left) shows the results of the AR-optimization using Csrd as rating function and AWF with Csrd 0:2Ca 1:2CB as balancing ¯ow calculation. In the tests, the domainSmiley'' is re®ned in 6 steps from 2500 elements to 5000 ele-ments using Rivara's re®nement algorithm . For comparisons, the right picture shows the same domain, re®ned in the same way, but now with Ccut as element 0 2 4 6 8 10 12 14 16 2 4 6 8 10 12 #Its/10 Cut/10 AR-Max AR-Avg Fig. 13. Development of AR, cut and # Iterations for domain square'' and increasing level of re®nment with Csrd Ca as element rating function.
Fig. 12. Development of AR, cut and # Iterations for domainsquare'' and increasing level of re®nment with Ccut as element rating function.
1576 R. Diekmann et al. / Parallel Computing 26 (2000) 1555±1581 rating and standard Diusion for balancing ¯ow calculation. The dierences are directly visible.
We compare the results of the AR-optimization in PadFEM with those of Jostle.
The element migration decision in Jostle uses Ccut as the rating function. Together with a special border optimization based on KL and with sophisticated tie-breaking methods, Jostle is able to generate and maintain well shaped partitions, even if the AR is not directly considered.
Figs. 15 and 16 show the development of AR over typical runs of (solve/ re®ne/ balance . . .) for our four test examples. Fig. 15 shows the results if Jostle (without multi-level coarsening-the Jostle-D con®guration described in ) is used as load balancer. Fig. 16 gives the results if the AR-optimization of PadFEM is used (with Fig. 14. Optimization according to AR (left) and cut (right).
Fig. 15. The development of the AR for increasing levels of re®nement with load balancer of Jostle.
R. Diekmann et al. / Parallel Computing 26 (2000) 1555±1581 1577 best possible parameter combinations). It can be observed that the AR is decreasing with increasing numbers of elements if the PadFEM-balancer is used. The opposite is true if Jostle determines the elements to be moved. This shows that our strategy can operate as high-quality load balancer in adaptive environments maintaining well shaped subdomains over a long period of time. Standard load balancers like Jostle (and with Metis, it would be just the same) cannot directly be used over a longer period without complete repartitionings from time to time. At least this is the case, if the partition AR is a measure of importance.
To show that the results given in Fig. 15 and 16 are not special cases, Table 6 presents values of AR averaged over several steps of re®nements and dierent pa-rameter settings. The unweighted results (columns ``UW'') are just normal average values, the others show ARs weighted with the number of elements of the individual meshes. The idea behind this weighting is to increase the importance of ®ner meshes, where the solvers take longer times and where it is much more important to achieve good AR. The results show that PadFEM can improve the AR of mesh partitions by around 10% over a state-of-the-art general purpose load balancing tool. This sort of improvement is con®rmed in .
Fig. 16. The development of the AR for increasing levels of re®nement with AR-optimization of Pad-FEM.
Table 6 Comparison, Jostle $ PadFEM Jostle PadFEM UW W UW W Square 1.250 1.267 1.182 1.140 Cooler 1.230 1.245 1.238 1.229 Smiley 1.310 1.326 1.252 1.234 Tower 1.307 1.303 1.261 1.219 1578 R. Diekmann et al. / Parallel Computing 26 (2000) 1555±1581 5. Extensions to 3D and parallelism All methods discussed have been applied to examples in 2D. The graph parti-tioning algorithms either work on graph measures or on the Euclidean distance, which can easily be extended to 3D. The balancing ¯ow calculation works only on graph measures, so the dimension of the mesh does not play a role. Dierencing between the dimensions is a proper de®nition of AR. In 3D, borders change to surfaces, areas to volumes and circles to spheres. The main consequence is an in-crease in calculation time for the dierent shape cost measures, but in principle, all de®nitions can be extended easily.
We have not shown any quantitative results concerning run-times or parallel ef-®ciencies. The main reason is the diculty of getting complex applications such as parallel adaptive numerical codes using DD-preconditioned CG solvers to run with competitive performance. The balancing ¯ow calculation is parallel by de®nition and can be calculated together with the CG solver iterations (for the AWF method, parallelization turns out to be a lot more complex). Concerning the element mi-gration it is necessary to ®nd independent sets of processor pairings to avoid con¯icts during the data movement.
6. Conclusions We have presented a two-step load balancing algorithm for adaptive mesh cal-culations on distributed memory machines. Our new center-oriented method called Bubble naturally produces very low AR and can successfully be used as a starting point for a balancing strategy. The balancing algorithm is distributed and takes existing partitions into account. The algorithm ®rst determines a balancing ¯ow and then moves elements. The element migration is particularly designed to optimize the AR of subdomains. Experiments which make use of a preconditioned Conjugate Gradient solver based on Domain Decomposition and comparisons to existing balancing tools show the bene®t of our approach.
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10660 | https://pmc.ncbi.nlm.nih.gov/articles/PMC5368202/ | The respiratory control mechanisms in the brainstem and spinal cord: integrative views of the neuroanatomy and neurophysiology - PMC
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. 2016 Aug 17;67(1):45–62. doi: 10.1007/s12576-016-0475-y
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The respiratory control mechanisms in the brainstem and spinal cord: integrative views of the neuroanatomy and neurophysiology
Keiko Ikeda
Keiko Ikeda
1 Division of Biology, Hyogo College of Medicine, Nishinomiya, Hyogo 663-8501 Japan
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1,✉, Kiyoshi Kawakami
Kiyoshi Kawakami
2 Division of Biology, Center for Molecular Medicine, Jichi Medical University, Shimotsuke, Tochigi 329-0498 Japan
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2, Hiroshi Onimaru
Hiroshi Onimaru
3 Department of Physiology, Showa University School of Medicine, Shinagawa, Tokyo 142-8555 Japan
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3,✉, Yasumasa Okada
Yasumasa Okada
4 Clinical Research Center, Murayama Medical Center, Musashimurayama, Tokyo 208-0011 Japan
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4,✉, Shigefumi Yokota
Shigefumi Yokota
5 Department of Anatomy and Morphological Neuroscience, Shimane University School of Medicine, Izumo, Shimane 693-8501 Japan
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5, Naohiro Koshiya
Naohiro Koshiya
6 Cellular and Systems Neurobiology Section, NINDS, NIH, Bethesda, MD 20892 USA
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6,✉, Yoshitaka Oku
Yoshitaka Oku
7 Department of Physiology, Hyogo College of Medicine, Nishinomiya, Hyogo 663-8501 Japan
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7,✉, Makito Iizuka
Makito Iizuka
3 Department of Physiology, Showa University School of Medicine, Shinagawa, Tokyo 142-8555 Japan
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3,✉, Hidehiko Koizumi
Hidehiko Koizumi
6 Cellular and Systems Neurobiology Section, NINDS, NIH, Bethesda, MD 20892 USA
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6
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1 Division of Biology, Hyogo College of Medicine, Nishinomiya, Hyogo 663-8501 Japan
2 Division of Biology, Center for Molecular Medicine, Jichi Medical University, Shimotsuke, Tochigi 329-0498 Japan
3 Department of Physiology, Showa University School of Medicine, Shinagawa, Tokyo 142-8555 Japan
4 Clinical Research Center, Murayama Medical Center, Musashimurayama, Tokyo 208-0011 Japan
5 Department of Anatomy and Morphological Neuroscience, Shimane University School of Medicine, Izumo, Shimane 693-8501 Japan
6 Cellular and Systems Neurobiology Section, NINDS, NIH, Bethesda, MD 20892 USA
7 Department of Physiology, Hyogo College of Medicine, Nishinomiya, Hyogo 663-8501 Japan
✉
Corresponding author.
Received 2016 Mar 15; Accepted 2016 Jul 22; Collection date 2017 Jan.
© The Physiological Society of Japan and Springer Japan 2016
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PMCID: PMC5368202 PMID: 27535569
Abstract
Respiratory activities are produced by medullary respiratory rhythm generators and are modulated from various sites in the lower brainstem, and which are then output as motor activities through premotor efferent networks in the brainstem and spinal cord. Over the past few decades, new knowledge has been accumulated on the anatomical and physiological mechanisms underlying the generation and regulation of respiratory rhythm. In this review, we focus on the recent findings and attempt to elucidate the anatomical and functional mechanisms underlying respiratory control in the lower brainstem and spinal cord.
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The online version of this article (doi:10.1007/s12576-016-0475-y) contains supplementary material, which is available to authorized users.
Keywords: Respiratory rhythm, Pons, Medulla, Spinal cord, Parafacial respiratory group (pFRG), Pre-Bötzinger complex (preBötC)
Introduction
Respiration is crucial for animal survival. In the last 10 years, the cytoarchitecture of the respiratory control center has been analyzed at the single-cell and genetic levels. The respiratory center is located in the medulla oblongata and is involved in the minute-to-minute control of breathing. Unlike the cardiac system, respiratory rhythm is not produced by a homogeneous population of pacemaker cells. Rather, it can be explained with two oscillators: the parafacial respiratory group (pFRG; Sect.1) and the pre-Bötzinger complex (preBötC, inspiratory pacemaker population; Sects.2, 3). The inspiratory and expiratory activities produced in these medullary respiratory rhythm generators are modulated from various sites of the lower brainstem, including the pons (see Sect.6) and Bötzinger complex, and are then output as motoneuron activities through the efferent networks in the brainstem and spinal cord (see Sects.2 to 5). Different types of preparations, mainly from mice and rats, have been used to analyze respiratory rhythm and pattern generation, including: medullary slice preparation in vitro (newborn or juvenile), en bloc brainstem-spinal cord preparation (newborn), decerebrated and arterially perfused preparation in situ (newborn and juvenile) and in vivo preparation (all ages). The normal respiratory motor pattern basically consists of three or four phases: pre-inspiratory, inspiratory, post-inspiratory, and late-expiratory. However, the motor output patterns in the different experimental models often display variation and the variations have caused some controversies in the field. In the last decades, new knowledge has been accumulated on the anatomical and physiological mechanisms underlying respiratory rhythm generation and regulation. In this review, we focus on these recent findings and correlate the information on the anatomical and functional mechanisms that are involved in respiratory control in the lower brainstem and spinal cord. We also introduce a novel rat line that is useful for future analyses of respiratory neural networks in vivo and in vitro. This article consists of six sections that were written by individual researchers. The focus of the sections and their respective authors are as follows: Sect.1, the pFRG (K. Ikeda, K. Kawakami and H. Onimaru); Sect.2, the anatomy of the preBötC (Y. Okada and S. Yokota); Sect.3, the physiology of the preBötC (N. Koshiya); Sect.4, the cervical circuits (Y. Oku); Sect.5, the spinal cord (M. Iizuka); and Sect.6, the pons (H. Onimaru and H. Koizumi).
Section 1: Identification of the pFRG in the respiratory rhythm generator neuron complex using a novel transgenic rat line harboring Phox2b-EYFP BAC (K. Ikeda, K. Kawakami & H. Onimaru)
The pFRG has been named based on its position relative to the facial nucleus. It is located ventral and caudal to the facial nucleus, and predominantly consists of neurons that burst prior to inspiration [pre-inspiratory (Pre-I) neurons] . The pFRG at least partially (the ventral and medial parts) overlaps the retrotrapezoid nucleus (RTN), which has been identified as an area in which neurons with projections to the ventral respiratory group (VRG) originate [2, 3]. Thus, this region is also referred to as the pFRG/RTN. The caudal portion of the pFRG overlaps the most rostral portion of the ventral respiratory group (the Bötzinger complex), which is the ventral part of the retrofacial nucleus near the caudal end of the facial nucleus and is thought to play an important role in the respiratory rhythm generation, particularly of the adult in vivo preparation [5, 6]. This caudal portion of the pFRG corresponds to so-called rostral ventrolateral medulla (RVL) [7–9], where most Pre-I, inspiratory, and expiratory neurons have been recorded in previous electrophysiological studies.
The paired-like homeobox 2b gene (Phox2b) encodes the Phox2b transcription factor and is required for the development of a subset of cranial nerves and the lower brainstem nuclei in the central nervous system and the peripheral autonomic nervous system. The distribution of pFRG-Pre-I neurons overlaps with that of Phox2b-expressing cells (Figs.1, 2) [10–12]. It is of note that pFRG-Pre-I neurons in the deeper ventral medulla at the caudal area are Phox2b-negative .
Fig.1.
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Level of the transverse section used for the optical recordings (e.g., Fig.2) and Phox2b immunoreactive cells in the most rostral medulla. a Phox2b immunoreactivity (Alexa Fluor 546). An arrow denotes a Phox2b cell cluster in the ventral parafacial region. b A NeuroTrace (435/455 blue fluorescence, Invitrogen) for Nissl staining. c A merged view of a and b. d A ventral view of a brainstem-spinal cord preparation. The preparation was cut at the level of the dotted line. AICA anterior inferior cerebellar artery, FN facial nucleus, pFRG parafacial respiratory group, preBötC pre-Bötzinger complex, VII–XII cranial nerves, C4 the fourth cervical ventral root, D dorsal, M medial. At the level of the most rostral medulla, close to the rostral end of the facial nucleus, Phox2b-ir cells formed one of the highest density clusters in the limited region ventrolateral to the facial nucleus. The cell density of the cluster was high enough to be clearly recognized, even in the Nissl-stained preparations (b)
Fig.2.
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Voltage imaging of the respiratory neuron activity in the ventral medulla of the rostral cut surface. The results are the averages of 40 respiratory cycles triggered by C4 inspiratory activity. a An optical image of the rostral cut surface; its time point is represented by the dotted vertical line on b. b C4 activity and the change in fluorescence at one location (red open circle). The approximate inspiratory phase is indicated by the horizontal blue bar under the C4 trace. c Optical images of respiratory neuron activity. The images are arranged in a time course from left to right and top to bottom, as indicated by the numeric values, where time 0 is a and the subsequent images c correspond to time points represented by the arrows in b. d An image of the cut surface of the rostral medulla at lower magnification. The red square denotes the area of the optical recording. e Nissl staining of the rostral cut surface after the experiment. Note that the photo clearly indicates the facial nucleus (FN) and the ventral cell cluster corresponding to the parafacial respiratory group (pFRG). Note that the optical records show the neuronal activity preceding the inspiratory activity by 500–600 ms in the area ventral to the facial nucleus (i.e., the rostral part of the pFRG). The activity reached its peak immediately before the peak of C4 inspiratory activity and then decreased slightly during the inspiratory phase. After the inspiratory period, the activity continued for 2–3 s during the post-inspiratory phase
Generally, the mouse fetal embryonic parafacial group (e-PF ), the rat perinatal pFRG/RTN, and the adult animal RTN [2, 14] are considered to be identical/corresponding populations of neurons in the different developmental stages of rodents. Indeed, the distribution and characteristics of Phox2b-expressing cells in the parafacial region of the neonatal rat are basically similar to those in the adult rat [15, 16] and the neonatal mouse . However, it is also possible that they are a discrete neuronal group that plays distinctive roles in respiratory rhythm generation during development [18–20].
In experiments using neonatal rat brainstem (medulla)-spinal cord preparations, the Pre-I neurons, from which the pFRG/RTN is composed, were found to be a superior rhythm generator. It was hypothesized that they trigger inspiratory burst generation [21, 22] and contribute to the production of expiratory activity . The pFRG/RTN functions as a chemo-sensing center that detects environmental pCO 2. In a series of articles, we reported that Pre-I neurons, which are the predominant neurons in the pFRG/RTN of neonatal rats, express a paired-like homeobox 2b gene (Phox2b), and that they are intrinsically CO 2-sensitive [10, 24–26]. PHOX2B mutations have been described in most human cases of central congenital hypoventilation syndrome . Thus, the detection of Phox2b expression in Pre-I neurons is an exciting finding because it enables us to obtain the prospect of identifying the molecules that are responsible for detecting environmental changes in pCO 2. In contrast, the neurons of the pre-Bötzinger complex (preBötC) that locates just ventral to ambiguus nucleus (Amb) do not express Phox2b (Fig.3d).
Fig.3.
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Expression of enhanced-yellow-fluorescent protein (EYFP) signals driven by the mouse Phox2b enhance/promoter in the brainstem of transgenic Phox2b-EYFP/CreERT2 rats from the rostral to the caudal region. In all of the micrographs, the upper side is the dorsal side. The experimental procedures are described by Ikeda et al. . The experiments were performed in neonatal transgenic rats (P0–03). Su5 supratrigeminal nucleus, mlf medial longitudinal fasciculus, Py pyramidal tract, 5M motor trigeminal nucleus, Rt reticular nucleus, FN facial nucleus, pFRG parafacial respiratory group, 4V ventricle, nTS nucleus of the solitary tract, dnmX dorsal motor nucleus of the vagus nerve, Amb ambiguus nucleus, preBötC pre-Bötzinger complex, IO inferior olive, AP area postrema, 10N nucleus of vagus, 12N hypoglossal nucleus.Scalea–e, 1.0 mm; f 500 µm
We recently reported the generation and analyses of a novel transgenic (Tg) rat line . The Tg rat harbors a mouse bacterial artificial chromosome (BAC) carrying a Phox2b that was modified to drive enhanced yellow fluorescent protein (EYFP) and Cre recombinase-ER T2 (estrogen receptor T2). The Tg line shows a similar pattern of EYFP expression to that of the endogenous Phox2b in rats. Consistent with previous reports on Phox2b protein expression in rodents, the EYFP signals were mostly found in the ponto-medullary region. In the pons, the EYFP-positive neurons were found in the supratrigeminal nucleus (Su5), where they dorsally capped the motor nucleus of the trigeminal nerve (5M) (Fig.3a, b). The 5M was negative for EFYP (Fig.3b). In the medulla oblongata, EYFP-positive signals were abundant in both the spinal trigeminal nucleus (Sp5) and the pFRG/RTN (Fig.3c). The facial motor nucleus (FN) was relatively weak and sparsely positive (Fig.3c). The weak expression in the body of the facial nucleus has also been reported in human fetuses at 19 weeks of gestation . In humans, the expression of PHOX2B in the facial nucleus disappears at later stages of gestation [15, 29]. We also observed that EYFP signals in the facial nucleus became negative over time during infancy (Ikeda and Onimaru, data not shown). In addition to the above distribution, the EYFP-positive cells are also abundant in the dorsal vagal complex, which is composed of the nucleus of the solitary tract (nTS), the dorsal motor nucleus of the vagus (dmnX), and the area postrema (AP) (Fig.3d, e). The nucleus ambiguus (Amb) was also EYFP-positive (Fig.3d). Interestingly, EYFP-positive cells could not be detected in the hypoglossal nucleus (Fig.3e, 12N). The reticular nucleus (Rt) was sparsely positive for EYFP (Fig.3c, e). We also observed EYFP-positive signals (fibers or axon terminals) through the rostro-caudal column of the ventral medulla including the preBötC and the ventral respiratory group (Fig.1c–e). The pFRG/RTN consists of rostral (Fig.3f) and caudal (Fig.3c) parts . The cells in the rostral pFRG/RTN contain a cluster of very superficially located neurons on the ventral side of the medulla, some of which are glutamatergic [11, 16]. Consistent with our previous reports in which we noted that CO 2-sensitive Pre-I neurons in the rostral pFRG/RTN are Phox2b-positive , all of the CO 2-sensitive Pre-I neurons in the rostral pFRG/RTN that have thus far been examined in this region have been EYFP-positive (data not shown). In sum, EFYP-positive signals in this rat line mirror the endogenous Phox2b expression in the neonatal stage.
Further studies to uncover the physiology of Pre-I neurons and the pFRG/RTN both in vitro and in vivo are now under investigation; e.g., for tracking the function and sole rhythmogenecity of the pFRG/RTN during the embryonic, perinatal, and adult stages of development and for the identification of the CO 2 sensor molecules in the cells of pFRG/RTN through the use of fluorescence as a mark of the Pre-I neuron and of specific expression of Cre recombinase in the Pre-I neuron.
Section 2: The anatomy of the respiratory rhythmogenic kernel: the pre-Bötzinger complex of the medulla (Y. Okada & S. Yokota)
The anatomical localization of the preBötC
Basic respiratory rhythm is generated in the respiratory neuron network of the lower brainstem. In 1991, a region that is critically important for inspiratory burst generation was found within the ventral respiratory column/ventral respiratory group (VRG). This was named the pre-Bötzinger complex (preBötC) (see Sect.3). The preBötC is a small region that is bilaterally located in the reticular formation of the ventrolateral medulla. Along the rostro-caudal axis, it occupies a limited portion within the VRG (between the caudal end of the Bötzinger complex and the rostral end of the rostral VRG) [30–32]. Ventro-dorsally, it is located at a few micrometers beneath the ventral medullary surface, just ventral to the nucleus ambiguus [31, 32] (Fig.4). Although the anatomy and function of the preBötC has mainly been studied in rodents, it has been identified in other animal species and in humans .
Fig.4.
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Localization of the respiratory-related regions in the brainstem (the parafacial respiratory group/retrotrapezoid nucleus [pFRG/RTN], Bötzinger complex [BötC], pre-Bötzinger complex [preBötC] and the high cervical spinal cord respiratory group [HCRG]), projected on schematic illustrations of the brainstem and spinal cord of the neonatal rat. a Ventral view. b Sagittal view. c–e Transverse view. VII facial nucleus, XII 12th cranial nerve, C1 and C4 1st and 4th ventral roots of the cervical spinal cord, respectively, BA basilar artery, VA vertebral artery, VRG ventral respiratory group
The cells composing the preBötC
Although the preBötC is a region containing rhythmogenic cells, it does not exhibit a distinct nucleus (neuron group). The preBötC contains glutamatergic excitatory [35, 36] and GABAergic and glycinergic inhibitory neurons . All of the neurons in the preBötC are interneurons (i.e., no motoneurons are present). The rhythmogenic neurons in the preBötC have been characterized using anatomical markers, including neurokinin-1 receptor (NK1R), somatostatin (SST), and Dbx1 [32, 36, 38–43]. Furthermore, using a combination of immunohistochemistry for the detection of NK1R and in situ hybridization for the detection of preprotachykinin A (PPTA), a precursor for substance P (SP) and a ligand for NK1R, we demonstrated the presence of PPTA mRNA-positive and NK1R-immunoreactive neurons that could play important roles in rhythm generation in the preBötC (Fig.5a–c).
Fig.5.
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Localization of anatomically identified putative rhythmogenic neurons in the pre-Bötzinger complex (preBötC). a Line drawings showing the distribution of neurokinin 1 receptor (NK1R)-immunoreactive neurons (green dots), preprotachykinin A (PPTA) mRNA-positive neurons (black dots), and double-labeled neurons (red dots) in the medulla. b, c Confocal images showing the appearance of PPTA mRNA-positive neurons (red) and NK1R-immunoreactive neurons (green) in the preBötC. The asterisks indicate double-labeled neurons. d The preBötC region where retrograde tracer fluorogold (FG) was injected (shaded area). e The distribution of FG-labeled neurons (blue dots), PPTA mRNA-positive neurons (black dots), and PPTA mRNA-positive neurons simultaneously labeled with FG (red dots) in the contralateral medulla. The open arrows in (a) and (e) indicate the preBötC region. f–h PPTA mRNA-positive preBötC neurons with projection to the contralateral preBötC. Confocal images of FG-labeled (f) and PPTA mRNA-positive (g) neurons. The merged image is shown in h. The arrow indicates a double-labeled neuron. 10 dorsal motor nucleus of vagus, 12 hypoglossal nucleus, Amb nucleus ambiguus, ECu external cuneate nucleus, icp inferior cerebellar peduncle, IO inferior olive, MVe medial vestibular nucleus, NTS nucleus of the solitary tract, py pyramidal tract, Sp5 spinal trigeminal nucleus, sp5 spinal trigeminal tract, SpVe spinal vestibular nucleus, st solitary tract
We recently reported that a subset of astrocytes in the preBötC exhibit respiratory modulating activity [44, 45]. By selectively activating astrocytes in the preBötC with an optogenetic technique, we could trigger bursting of inspiratory neurons in the preBötC, which indicates excitatory connection from astrocytes to inspiratory neurons in this rhythmogenic kernel . These findings led us analyze the anatomical arrangement of neurons and astrocytes in the preBötC. We revealed that neurons and astrocytes are intimately located in the preBötC (Fig.6) and that the astrocytic processes exhibit contact with SSTergic neurons (data not shown), which is also in agreement with our physiological report that astrocytes and rhythmogenic neurons are functionally coupled in the preBötC . The role of preBötC astrocytes in respiratory rhythmogenesis could be active and should be further investigated in the future (also see Sect.3).
Fig.6.
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Colocalization of neurons and astrocytes in the ventrolateral medulla. Neurons and astrocytes are identified as neuron-specific marker NeuN-positive cells (green) and astrocyte-specific marker S100b-positive cells (red), respectively. a Ventrolateral medullary region. The square indicates the preBötC region, and corresponds to the area in b. b An enlarged image of the preBötC. The square indicates the area in c. c A high-magnification picture showing colocalized cell bodies of neurons and astrocytes. Amb nucleus ambiguus
Neuronal projection to the preBötC neurons
Neurons in the preBötC are functionally coupled with other respiratory-related regions in the brainstem. To investigate the anatomical connections, we conducted retrograde and anterograde tract tracing, and revealed that the NK1R- and SST-immunoreactive neurons in the preBötC regions receive axon terminals from the contralateral preBötC. The axon terminals from the preBötC make asymmetrical, putative excitatory, synaptic contact with the NK1R-immunoreactive neurons in the contralateral preBötC . We also combined retrograde tracing by injecting fluorogold into the unilateral preBötC with in situ hybridization to detect PPTA, and demonstrated that the putative rhythmogenic PPTA mRNA-positive neurons in the preBötC project to the contralateral preBötC (Fig.5d–h). It has also been reported that SP and enkephalinergic axon terminals form synapses on NK1R-immunoreactive neurons in the preBötC . These findings suggest that the connection between bilateral preBötC neurons, especially SPergic commissural neurons, serves to synchronize the timing of the oscillatory activities in the bilateral preBötC.
Neuronal projection from the preBötC neurons
With respect to projection to other brain stem regions, neurons in the preBötC send axonal fibers to various respiratory-related regions. Tan et al. demonstrated the axonal projection of SSTergic neurons in one side of the preBötC to the bilateral Bötzinger complex, VRG regions caudal to the preBötC, parafacial respiratory group/retrotrapezoid nuclei, parabrachial/Kölliker-Fuse nuclei and periaqueductal gray region. Furthermore, we showed that the neurons in one side of the preBötC region send axonal projections to the bilateral hypoglossal premotor areas, the bilateral hypoglossal motor nuclei and the bilateral nuclei tractus solitarius . However, there are no reports indicating the direct projection from preBötC neurons to either the cerebellum or the spinal cord . In contrast, the putative rhythmogenic neurons in the preBötC receive glutamatergic projections, e.g., from the lateral periaqueductal gray region and the parabrachial nucleus (unpublished observation).
In conclusion, so far, elucidated anatomy of the preBötC suggests that it plays a critically important role in rhythmogenesis among the various respiratory networks, although it is quite probable that multiple respiratory networks distributed throughout the higher and lower brainstem and the high segments (C1–C2) of the spinal cord are inter-connected and are together involved in respiratory rhythm generation in the in vivo condition . It is necessary to further investigate the anatomy of the preBötC, including the local connection between astrocytes and neurons, to better understand the physiological and pathophysiological mechanisms of respiratory rhythm generation.
Section 3: The physiology of the pre-Bötzinger complex from a rhythmogenic perspective (N. Koshiya)
Functional localization
The pre-Bötzinger complex (preBötC) was discovered as the region that generates respiratory rhythm in neonatal rodent brainstem-spinal cord preparation in vitro. Smith et al. demonstrated that efferent rhythmic respiratory motor activities persist, essentially unchanged, in the spinal ventral roots that serve the phrenic (~C4) and thoracic intercostal nerves, until live coronal sectioning by a vibratome from the rostral end of the brainstem encroached on the rostrocaudal level near the rostral-most hypoglossal (XII) nerve rootlet exit (see Sect.2 for more detailed anatomy). This level was also crucial when vibratome sectioning was performed from the caudal end of the brainstem (while recording inspiratory motor nerve activities in cranial nerves such as the XII and/or glossopharyngeal nerves). These two-way deletion experiments, combined with electrophysiological recordings , allowed for the respiratory rhythmogenic kernel of the preBötC to be functionally located. Transverse brainstem slice preparations cut at this level maintained a vital rhythm in the XII activities; i.e., the preBötC, XII premotor microcircuits, and its motoneurons, motor axons, and nerve rootlets were miraculously located within the same coronal plane and were captured in a single thin slice. This allowed for the creation of highly reduced, yet functionally complete, breathing slice preparations. In 2014, the first whole slice activity imaging confirmed its localization and the associated regions (Fig.7; Movie 1) .
Fig.7.
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Functional connectomics of the preBötC . Panels corresponding to the present paragraphs: functional localization (pan-slice activity mapping); instantaneous activity (dt=500 µs) and COA (black ball); Single-breath recruitment chaos of the preBötC population; Inter-preBötC tract’s action potential conduction under CNQX (Glu-blocked); and Glutamatergic premotor relays (Glu-dependent). See the open-access online graphic abstract of the paper for further details:
The pacemaker neurons and their significance
Extracellular recording has shown that the rhythmic bursting of some preBötC inspiratory neurons can continue after the attenuation of synaptic transmissions with low calcium medium in slices in vitro ; however, neither cellular biophysics nor synchronization mechanisms could have been studied before functional imaging, optical preidentification, and visualized patch-clamp recording were employed . Since the breathing rhythm is bilaterally synchronous, the existence of direct reciprocal connections was hypothesized. This was successfully demonstrated by labeling the preBötC rhythmogenic neurons via their contralaterally projecting axons. Calcium-sensitive dyes injected directly into one side of the preBötC had labeled some contralateral preBötC neurons; however, the signal was mostly static, probably because the vascular uptake and transport had kept the dye isolated from the calcium-dynamic cytosolic domains. The microinjection of membrane semipermeable acetoxymethyl ester (AM) dyes around the (hypothetical) axons en passant worked. Within the heterogeneous reticular formation, where no circumscribed nucleus is visible, some of the labeled neurons (flashers) that were restricted to a region that corresponded to the preBötC showed transient fluorescence that was synchronous to XII inspiration. Half of the inspiratory flashers had intrinsic rhythmogenicity (pacemakers) when they were functionally isolated from the rest of the population with an AMPA/KA glutamatergic transmission blocker (CNQX). The yield of pacemakers in other exogenous preidentification experiments that were subsequently performed using NK1R-mediated live fluorescence labeling was 25% . The pacamakers’ burst rate was a monotonic function of their membrane potential until the membrane potential exceeded the oscillatory regime, at which time they became tonically active; the system would then shift into an inspiratory apnea. This was the first cellular mechanism identified that may underlie the system’s respiratory rhythm. With tonic (DC) afferent information such as chemoreceptor feedback, it modulates the frequency (FM) of the rhythmic motor drive and thus ventilation. The central homeostatic control was dissected down to the cellular biophysical level.
Persistent sodium conductance (gNaP)
gNaP, in combination with leak conductance, is found to be crucial to the pacemaker inspiratory neuron functions. There are three distinct functions in its rhythmic burst generation. The first is burst formation. Its “persistence” is essential to burst generation. The neurons use their N-shaped I–V relationship [52, 53] where, within the negative slope domain, the cell would recursively auto-depolarize into a burst (i.e., a train of action potentials during a depolarization in a 10−1–10 0-s time scale. The burst termination is the second function that is achieved by cumulative high-voltage-sensitive inactivation of gNaP at every action potential through the burst until spike generation is no longer possible. The third is its rhythm determination. It is not their persistency (a hard-to-inactivate nature); it is rather their characteristically slow disinactivation (10 0–10 1-s scale recovery from burst-induced inactivation and silence), which is likely to be a determinant of the interburst interval of the pacemaker neurons . This gNaP could be blocked with low concentrations of its blockers [riluzole (5 µM) or tetrodotoxin (5 nM)] when they were precisely microinfused into preBötC in slice preparations, superfused over slice preparations with the caudal cut surface of the preBötC exposed (the rostral side contains bilaterally synchronizing axons ), or transarterially perfused into brainstem spinal cord in situ preparations without rostral structures beyond the preBötC , which would cause the network to become incapable of generating rhythmic inspiration (as well as gasping ).
Leak conductance (gLeak)
While it is referred to as “leak,” ligand-gated channel conductances are often ohmic and not voltage gated, and can thus function as a linear gLeak. This includes proton-sensitive channels, which may be responsible for the chemosensitivity of the preBötC cells (e.g., TASK channel ). As was predicted theoretically , a quantitative balance of gNaP and gLeak was found to distinguish pacemakers from non-pacemakers (see Fig.5 of Del Negro et al. ).
Calcium pacemaking
Pacemaking primarily on calcium-gated non-specific cationic conductance (gCAN) has also been suggested [59, 60]. While calcium influxes are not required for gNaP-based pacemaking , gNaP blockers can reproducibly block slice pacemaking, (see gNaP paragraph above), and the termination mechanism for gCAN-based burst is unknown, the contribution of calcium to other types of cell bursting remains possible.
Synaptic synchronization
The glutamatergic synchronization of the pacemaker and other inspiratory neurons is crucial for the inspiratory activity formation at neural population levels. This was demonstrated by an optical simultaneous multi-neuron activity recording using retro-axonal calcium dye labeling . Its dependency on glutamatergic transmission was proven by the desynchronization of the activity of synchronously active inspiratory neurons after the administration of a glutamatergic antagonist (CNQX).
Population recruitment chaos
The single-breath recruitment of the preBötC population dynamics was discovered using high-speed (2 k fps), voltage-sensitive dye imaging (VSDI; see Fig.8 of Koshiya et al. ). The spatiotemporal location of the instantaneous center of activity “mass” (COA) was tracked and reduced to its velocity sequence, which was discovered to be chaotic (i.e., not fully periodic yet deterministic). This chaos at the origin of activity may underlie the respiratory chaos in systems in which the activity is not identically recursive yet deterministic [52, 62].
Fig.8.
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The possible neuronal mechanisms underlying the organizing pattern wherein the inspiratory motor activities in the rostral thoracic segments are larger than those in the caudal thoracic segments. a The pattern is organized at the level of medulla where descending neurons exist. The descending neurons project richly to the motoneurons positioned at more rostral thoracic cord. b An example showing that the pattern is organized at the level of the spinal cord. In this example, the number of inspiratory excitatory interneurons is larger than that in the rostral segments, and these neurons amplify the excitatory inputs to the motoneurons in the rostral segments
The bilaterally synchronizing tract
The inter-preBötC tract was not anatomically distinct but was found to be a functional structure by a pan-slice, high-speed imaging procedure , in which compound action potentials were spatiotemporally tracked from one side of the preBötC during the microstimulation of the other side. The suppression of the recursive autoactivation of the bilateral CNQX population was used to unveil the one-shot traffic (Fig.7Glu-blocked and Movie 2b; also see Fig.7 of Koshiya et al. ). To the best of our knowledge, this was the first and remains the only visualization of compound action potential conduction through the brain tract of vertebrates, which is different from general activity propagation where rapid dynamics such as return of activity to the baseline are not observed (Movie 2).
Premotor synaptic cascades
Glutamatergic premotor relays were demonstrated in the full network configurations from one side of the preBötC to the bilateral XII motor nuclei using movie subtraction: [control movie]−[CNQX movie] (Fig.7Control and Glu-blocked; Movie 2ab) where CNQX-insensitive (thus glutamatergic) depolarization was extracted through premotor areas (Fig.7Glu-dependent; Movie 2c). Premotor neurons have been found between the preBötC and XII areas . The study located the areas that were wholly synaptically involved, including the contralateral premotor areas .
Non-glutamatergic populations
Non-glutamatergic cells in the preBötC include inhibitory neurons (GABAergic and glycinergic, (, ; see Sect.2 for other phenotypes). There were no inhibitory bursters (the contradictory study could have shown loss of either periodic synaptic drives [32, 52] or multi-cell synchrony ) and the kernel rhythm persisted after the inhibitory transmission blockade . They may shape and/or tune the inspiratory activities in vitro.
Some of the astrocytes in the preBötC also showed calcium transients during inspiration ; occasionally with a pre-inspiratory rise. While some remained oscillatory after TTX, they lost synchrony and their individual rhythms slowed by an order of magnitude. Their burst timings had therefore been set by the neurons. The biophysical interactions between the inspiratory neurons and the inspiratory astrocytes are yet to be determined.
The relevance to the field of physiology
The preBötC is the only population oscillator in the mammalian motor center, where the rhythmogenic mechanisms have been demonstrated at the cellular biophysics and neural population levels if not yet fully. These wide-ranging and convergent investigational approaches may inspire other motor center studies.
Section 4: The structure and function of the respiratory neuronal circuits of the high cervical spinal cord (Y. Oku)
The localization and connectivity of the cervical respiratory neurons
The localization and the synaptic connections of the spinal respiratory neurons were extensively investigated in the late 1980s and early 1990s. It was motivated by the hypothesis that the net depolarization provided by the monosynaptic bulbospinal projections from medullary inspiratory neurons only accounted for a small fraction of the total amount that would be necessary for phrenic or intercostal motoneuronal discharge , which implies the existence of interneurons that relay the medullary respiratory drive to spinal motoneurons. Lipski and Duffin demonstrated the presence of a longitudinal column of inspiratory neurons extending from the caudal nucleus retroambigualis to the rostral segment of the C3 in cats. These upper cervical inspiratory neurons (UCINs) were located around the intermediate zone (lamina VII) of the gray matter, and projected into the vicinity of the phrenic and intercostal motoneurons [67, 68]. Nakazono and Aoki demonstrated that UCINs have excitatory mono- or paucisynaptic connections with the ipsilateral phrenic motoneurons, supporting the hypothesis that UCINs function as a propriospinal respiratory system, similarly to a propriospinal locomotor system . Palisses et al. reported a different inspiratory interneuron pool at the C3-C6 spinal cord levels, approximately 500 µm dorsal to the phrenic motor nucleus.
The rhythmogenesis of the cervical respiratory neuronal circuits
Animals spinalized at the C1–C2 levels generate spontaneous rhythmic phrenic activity [72–74]. In 1977, Coglianese et al. observed spontaneous rhythmic phrenic nerve activity in C1–C2 spinalized non-paralyzed dogs following the administration of a respiratory stimulant, doxapram hydrochloride. A few years later, Aoki et al. observed the temporary recovery of periodic phrenic motoneuron activity, approximately 1 h after spinomedullary transection in cats. However, since the administration of a neuromuscular transmission blocking agent, curare abolishes phrenic activity, respiratory rhythmicity is thought to be supported by feedback inputs from cutaneous and chest wall proprioceptors. Viala et al. observed synchronous short-lasting and long-lasting rhythmic bursts on the phrenic nerves after C2 level spinal transection in curarized and vagotomized rabbits after the administration of DOPA in combination with a monoamine oxidase inhibitor, nialamide. They suggested that short-lasting bursts are driven by the forelimb and hindlimb locomotion generator, whereas long-lasting bursts are driven by the spinal respiratory rhythm generator because they are enhanced by hypercapnia independently from the locomotor bursts. Similar long-lasting synchronous respiratory activity of a spinal origin can be recorded in the in vitro brainstem spinal cord preparation from rats by the pharmacological activation of deep diethyl ether anesthesia . This long-lasting spinal respiratory activity was found to coexist with the medullary respiratory activity, and persist after C1 spinal transection. Based on the transection experiments, the authors concluded that the spinal respiratory rhythm generator is located in the C4 and C6 spinal segments. However, in a later analysis , it was revealed that the long-lasting activity was exclusively recorded from cervical ventral roots and never observed in the cranial or phrenic nerves. Moreover, only non-respiratory motoneurons exhibited rhythmic depolarization in phase with the long-lasting bursts. Thus, it is unlikely that the long-lasting activity is involved in respiratory function. Recently, Kobayashi et al. demonstrated that C1/C2 spinal slices from neonatal mice are capable of generating rhythmic bursts. Spontaneous burst activity in the C1/C2 ventral roots occurred shortly (approximately 30 min) after transection and remained stable in the cervical slice for approximately 30 min and gradually deteriorated to cessation in 1–2 h.
Revisiting the cervical respiratory neurons
Optical imaging using voltage-sensitive dyes led to the discoveries of novel respiratory regions: the parafacial respiratory group and the high cervical respiratory group (HCRG) in the spinal cord . The HCRG is distinct from the UCIN because it is located in the ventral portion of the ventral horn at the level of the spinomedullary junction with the C2 segment. The HCRG consists of interneurons and motoneurons, which are responsive to both respiratory and metabolic acidosis, suggesting that it forms a propriospinal respiratory neuronal network . The discovery of the HCRG motivated the re-examination of the role of the spinal cord in respiratory rhythmogenesis. Jones et al. recorded phrenic (PNA) and hypoglossal (HNA) nerve activity in perfused brainstem preparation from rats to examine the effects of the transverse sectioning of the high spinal cord. Transverse transections at the pyramidal decussation not only immediately abolished PNA but also resulted in a progressive decline in the HNA amplitude and rhythm. Transverse transections at the first cervical spinal segment level did not abolish HNA rhythmicity. These results indicate the importance of the structures at the spinomedullary junction in eupneic respiration. In the current concept, eupnea, which is defined by a breathing pattern of inspiration, post-inspiration, and expiration, requires the integrity of the pontine-medullary respiratory network . The essential structures for eupneic breathing have been hypothesized to extend from the pons to the pre-Bötzinger complex; structures caudal to the obex are thought to be unnecessary for eupneic breathing. However, the observation of Jones et al. contradicts the current concept of the genesis of eupnea.
Respiratory regeneration after spinal cord injury
In vivo and in vitro observations suggest that the spinal center of respiratory rhythm generation takes 1–12 h to activate after intervention [72, 73, 77]. This may suggest that a certain recovery time is necessary for the network plasticity to reorganize the spinal respiratory neuronal circuits after injury. The propriospinal respiratory network may compensate for functional deficits by either activating auxiliary pathways or by serving as a backup respiratory rhythm generator. With the help of these backup mechanisms, endogenous sprouting or synaptogenesis would be able to repair the injured pathway to restore function. Indeed, the photoactivation of channelrhodopsin in the motoneurons, glia, and spinal interneurons at the level of the phrenic motor nucleus was able to functionally restore hemi-diaphragm activity 4 days after C2 hemisection . After cervical spinal cord injury, chondroitin sulfate proteoglycans around the phrenic motor neurons are involved in the upregulation of the perineuronal net. The digestion of these potently inhibitory extracellular matrix molecules with Chondroitinase ABC (ChABC) promotes the plasticity of the spared tracts and restores activity to the paralyzed diaphragm . The implantation of an autologous peripheral nerve graft (PNG), in combination with the addition of ChABC, allows for additional axonal regeneration through the PNG, leading to recovery from impaired function . The transection of the PNG after recovery caused an unusual increase in tonic EMG activity, which would have originated from the activity of propriospinal neurons that innervate the motor neurons. These results suggest that propriospinal neurons, which are recruited by regenerating axons, play an important role in circuit reorganization.
In conclusion, although the high spinal cord is not the site of the primary respiratory rhythm generator, it is involved in shaping eupneic breathing patterns, and in the event of injury, it recruits auxiliary pathways for circuit repair and reorganization, and may even exert as a backup respiratory rhythm generator. Thus, the respiratory neuronal circuits of the high cervical spinal cord are a vital constituent of the respiratory network.
Section 5: The involvement of spinal interneurons in the generation of the rostrocaudal gradient of intercostal inspiratory motor activity (M. Iizuka)
Respiration involves a complex pattern of movements for which numerous motoneurons distributed along the spinal cord need to fire in the proper spatial and temporal sequences. A number of studies involving electrical recordings from intercostal muscles or nerves in anesthetized cats [84, 85], decerebrated cats , anesthetized dogs [87, 88], and humans , have shown that the external intercostal muscles, or their nerve filaments, are active during inspiration, and that the inspiratory activities in the rostral interspaces are stronger than those in the caudal interspaces (see De Troyer et al. for review). Similarly, the parasternal region of each of the interchondral internal intercostal muscles (the so-called parasternal intercostals) is active during the inspiratory phase, and muscles in the rostral interspaces show stronger activities than muscles in the caudal interspaces in anesthetized dogs and awake humans [91, 92]. Deafferentation of the rib-cage does not affect the rostrocaudal gradient of the inspiratory motor activity in the parasternal intercostals . Similarly, in isolated brainstem-spinal cord preparations from the neonatal rat, which has no afferent feedback, the inspiratory activities in the more rostral thoracic ventral root were found to be larger than those in the caudal thoracic ventral root . These studies suggest that the central respiratory networks organize this basic spatial and temporal pattern for respiration.
A histochemical study in the cat demonstrated that the composition of fiber types in the parasternal intercostals is similar between the thoracic spinal segments, suggesting the mean and dispersion of the size of the parasternal intercostal motoneuron are similar between the segments . The external intercostals in the rostral thoracic segments have a larger ratio of the slow-twitch oxidative type to the fast-twitch types , suggesting that the mean size of the external intercostal motoneurons shifts to smaller in the more rostral segments. In the hindlimb muscle, although the soma size of the slow-twitch oxidative motor unit tended to be smaller than that of the fast-twitch motor unit, the sizes were largely overlapped . At present, the size and number of inspiratory motoneurons in each segment of the thoracic spinal cord have not been studied in detail. In the following, we assumed that the inspiratory motoneurons are uniformly distributed along the thoracic spinal cord, and discussed how the central respiratory networks organize the rostrocaudal gradient of intercostal inspiratory motor activity.
The possibilities were roughly divided into two: the spinal interneurons are involved or they are not. In cases where the spinal interneurons are not involved, the bulbospinal inspiratory neurons and motoneurons organize the rostrocaudal gradient. Three ways would be possible. The first is the excitatory bulbospinal neurons project richly to the inspiratory motoneurons in the rostral segments of the thoracic spinal cord (Fig.8a). The second is the synapses from the excitatory bulbospinal neurons to the more rostral inspiratory motoneurons terminate on the more soma side, and evoke a larger excitatory postsynaptic potential. The third is that the motoneurons in the rostral thoracic cord have a lower threshold than the motoneurons in the caudal thoracic cord. These three ways as described above are not exclusive to each other.
It is well documented that the bulbospinal neurons provide monosynaptic inputs to the intercostal motoneurons [66, 95, 96]. Thus far, all of the detected monosynaptic connections have been excitatory. Although Davies et al. showed that external intercostal inspiratory activity was directly relevant to the rostrocaudal gradient, they provided no evidence to suggest that the inspiratory bulbospinal neurons have systematic patterns of connections to different segments. Thus, the first and second ways mentioned above would be unlikely. Related to the third way, to the best of our knowledge, no studies have shown any differences in the membrane potentials or thresholds of the motoneurons in different thoracic segments. In summary, it is unlikely that the rostrocaudal gradient is organized entirely by the bulbospinal neurons and motoneurons.
Extracellular and intracellular recordings have shown that many thoracic interneurons have respiratory activity [97–100]. The interneurons that project to the thoracic ventral horn are mainly distributed in the contralateral medial ventral horn in the same spinal segment . Morphological studies of the thoracic respiratory interneurons have shown the terminations of their collaterals are located in the ventral horn or the intermediate area . In all of these electrophysiological studies, however, the recordings were obtained from restrictive thoracic segments, and it is impossible to know the rostrocaudal distribution of the inspiratory interneurons. Using a voltage-sensitive dye, Iizuka et al. demonstrated that the interneuron area of the ventral surface of the spinal cord at the more rostral thoracic segments showed larger depolarizing optical signals during the inspiratory phase (Fig.9e). This implies that the number of inspiratory interneurons is larger in the more rostral thoracic segments and/or that the inspiratory excitatory postsynaptic potentials are larger in the interneurons that are positioned at the more rostral segments.
Fig.9.
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a Possible neuronal mechanism in which inhibitory spinal interneurons are involved in the rostrocaudal gradient of the inspiratory motor activity. The inspiratory depolarizing optical signals in the motoneuron and interneuron areas of the rostral thoracic segments are larger than those in the caudal thoracic segments (b, e) . Many of the thoracic respiratory interneurons had an axon descending a few segments, which would be inhibitory [97, 100]. Based on these studies, it is possible that the inhibitory synaptic inputs to motoneurons gradually increase to reach the caudal segments (d). In order for the motoneurons to be activated during the inspiratory phase, it is necessary to receive excitatory synaptic inputs (c). Thus, in this model, the combination of inhibitory and excitatory synaptic inputs to motoneurons forms the rostrocaudal gradient of the thoracic inspiratory motor activity
Regarding the contribution of these inspiratory interneurons in the rostrocaudal gradient of the inspiratory motor activity, a simple hypothesis is that these interneurons are excitatory and that they directly or indirectly provide excitatory synaptic inputs to the inspiratory motoneurons at their adjacent segments (Fig.8b). However, one of the major characteristics of the thoracic respiratory interneurons is that many of them have an axon projecting to the contralateral side and descending a few segments [97, 100]. Extracellular recordings from the respiratory interneurons in this area and spike-triggered averaging were used to examine the existence of focal synaptic potentials in the contralateral thoracic ventral horn [98, 99]. When the firing of the strongly modulated phasic inspiratory or expiratory interneurons was used as a trigger, the focal synaptic potentials that were obtained were positive in most cases; thus, these interneurons would be inhibitory . The same study showed that weakly modulated tonic respiratory interneurons would be excitatory, but that their field potentials were generally small. Taken together, these findings suggest that the inhibitory inspiratory interneurons are involved in the generation of the rostrocaudal gradient of inspiratory motor activity. If this is the case, it is possible that the inspiratory inhibitory interneurons in the rostral thoracic segments project their axons to the caudal thoracic segments, and form a gradient that is the inverse of the inhibitory synaptic inputs to the motoneurons. Thus, the inhibitory synaptic inputs to the motoneurons are larger in the motoneurons that are positioned at the more caudal thoracic segments (Fig.9). Intracellular recordings from the thoracic motoneurons of the rat have shown that some motoneurons receive a complex combination of excitatory and inhibitory synaptic inputs in the same respiratory phase . Further study to examine whether the inspiratory motoneurons that are positioned at the more caudal thoracic segments receive larger inhibitory synaptic inputs during the inspiratory phase is necessary to confirm the possibility shown in Fig.9.
In conclusion, the central respiratory networks organize the rostrocaudal gradient of the intercostal inspiratory motor activity. It seemed that the spinal inspiratory interneurons are involved in the generation of the rostrocaudal gradient. Since these neuronal mechanisms are kept intact in isolated brainstem-spinal cord preparations from the neonatal rat [88, 95], this in vitro preparation would be an excellent experimental model to examine the involvement of the spinal interneurons.
Section 6: The functional involvement of the pons in the respiratory control mechanisms (H. Onimaru & H. Koizumi)
The pons, which is traditionally referred to as the pneumotaxic center , is known to be critically involved in the control of respiration; however, its functional role in respiratory rhythm and pattern generation has not been fully established. The pontine respiratory regions include the Kölliker–Fuse nucleus (KF) and the parabrachial complex (PB) in the dorsolateral pons, which are assumed to be the most important regions in the regulation of respiratory activity and respiratory phase transition [104–107], as well as several areas in the ventrolateral pons. These pontine structures interact with multiple medullary respiratory centers and regulate respiratory activity . For instance, the A5 noradrenergic neurons are presumed to send inhibitory synaptic inputs to the respiratory rhythm generators in the medulla [108, 109]. Moreover, it has been demonstrated that the neurons of the locus coeruleus receive synaptic inhibition from respiratory neurons [110, 111], and that the electrical stimulation of the PB causes the termination of the inspiratory burst discharge and respiratory phase transition . In a previous optical recording study in en bloc preparation from newborn rats, strong respiratory activity was found to be present in the KF region . Respiratory activity, although weaker than that in the KF region, was also detected in the PB region. In addition, whole-cell recordings demonstrated that several types of respiratory-related neurons were located in the KF [105, 112]. In experiments using perfused brainstem preparation from juvenile rats, the KF was shown to primarily gate the respiratory motor activity of the cranial nerves innervating the laryngeal adductor and tongue muscles and the KF is thought to have a physiological role in the coordination of sequential pharyngeal swallowing with respiration [106, 107, 113].
In the newborn brainstem-spinal cord preparation from newborn rats, the medulla (without the pons), was capable of generating a three- or four-phase respiratory pattern: pre-I, inspiratory, post-I and late E [9, 114, 115]. On the other hand, in in situ arterially perfused brainstem–spinal cord preparation from adult rats, the pontine structures are necessary for the generation of the normal three-phase respiratory pattern: inspiratory, post-I, and late E , similar to that which is recorded in vivo [57, 116]. A series of sequential rostrocaudal microtransections through the brainstem demonstrated the dynamics of the transformation/reorganization of the pontine-medullary respiratory network . The three-phase rhythm was only detected in the intact preparation, whereas two-phase rhythm without the post-I phase emerged after the removal of the pons. These results, along with those of previous studies, suggest that the inputs from the pontine circuits shape the respiratory pattern through the activation of the post-I neurons and the inspiratory off-switch mechanisms [105, 106].
In conclusion, although different types of experimental preparations showed the various motor output patterns of respiratory activity, recent studies have demonstrated the significant roles of the pons in the formation of respiratory burst patterns and in the control of respiratory phase transitions and respiratory reflexes [56, 117–120]. We propose that the pons, interacting with the medullary respiratory circuits, has important roles in controlling the various physiological and pathophysiological respiration-related behaviors.
Conclusions
The understanding of the physiological and anatomical mechanisms underlying respiratory control has been achieved through the development of various technologies. In addition to electrophysiological analyses, optical imaging studies have facilitated the discovery of most of the regions in the lower brainstem and spinal cord that are involved in respiratory rhythm and pattern formation. The genetic approach has also revealed the neurophysiological and neuroanatomical mechanisms. Medullary respiratory rhythm generators pFRG and preBötC are described in both anatomy and physiology. The primary rhythms are modulated by various regions in the brainstem including the pons. The respiratory motor activities are then formed through premotor and motor-efferent networks including interneurons in the brainstem and spinal cord. The upper cervical spinal cord is involved in shaping eupneic breathing patterns. We believe that our current knowledge heralds the promise of further advances in the understanding of the integration of the respiratory control mechanisms in the pons, medulla, and spinal cord.
Electronic supplementary material
Below is the link to the electronic supplementary material.
12576_2016_475_MOESM1_ESM.pptx (436.5KB, pptx)
Movie 1. A pan-slice VSDI movie showing spontaneous inspiration-related depolarization in: the bilateral preBötC and XII nuclei, and the XII premotor areas between them, as well as the inter-preBötC tract (medial protrusions from both preBötC) and some NTS (for detail, see Fig.7: Pan-slice activity mapping; also Sect.2 for anatomy). Because latencies from preBötC population recruitment to XII nerve’s first spike were stochastically variable, XII triggered averaging made preBötC recruitment slope (e.g., Figure 8D of Koshiya et al., 2014 ) slower after average and colorization (green) delayed. For causality oriented dynamic sequence, see Movie 2
12576_2016_475_MOESM2_ESM.pptx (1.8MB, pptx)
Movie 2. A pan-slice VSDI showing depolarization (green, yellow, to red) and hyperpolarization (blue, to navy) after microstimulation of the preBötC on the right side, in a control state (a), under CNQX (b, glutamatergic blockade-insensitive activities) and [a] – [b] (c, glutamate-dependent activities) (Fig.7: Functional connectomics by membrane potential movies)
Acknowledgments
This work was supported by JSPS KAKENHI Grant Numbers 25540130, 26460311, 26670676, and 15K00417 to Y. Okada, and by a grant from the Naito Foundation to K.I., and MEXT (Ministry of Education, Culture, Sports, Science and Technology)—Supported Program for the Subsidies for Private Universities (Showa University of Medicine, Jichi Medical University, Hyogo College of Medicine).
Compliance with ethical standards
Conflict of interest
The authors declare no conflicts of interest in association with this study.
Contributor Information
Keiko Ikeda, Email: kiikeda@hyo-med.ac.jp.
Hiroshi Onimaru, Email: oni@med.showa-u.ac.jp.
Yasumasa Okada, Email: yasumasaokada@1979.jukuin.keio.ac.jp.
Naohiro Koshiya, Email: nk@mail.nih.gov.
Yoshitaka Oku, Email: yoku@hyo-med.ac.jp.
Makito Iizuka, Email: iizukam@med.showa-u.ac.jp.
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Associated Data
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Supplementary Materials
12576_2016_475_MOESM1_ESM.pptx (436.5KB, pptx)
Movie 1. A pan-slice VSDI movie showing spontaneous inspiration-related depolarization in: the bilateral preBötC and XII nuclei, and the XII premotor areas between them, as well as the inter-preBötC tract (medial protrusions from both preBötC) and some NTS (for detail, see Fig.7: Pan-slice activity mapping; also Sect.2 for anatomy). Because latencies from preBötC population recruitment to XII nerve’s first spike were stochastically variable, XII triggered averaging made preBötC recruitment slope (e.g., Figure 8D of Koshiya et al., 2014 ) slower after average and colorization (green) delayed. For causality oriented dynamic sequence, see Movie 2
12576_2016_475_MOESM2_ESM.pptx (1.8MB, pptx)
Movie 2. A pan-slice VSDI showing depolarization (green, yellow, to red) and hyperpolarization (blue, to navy) after microstimulation of the preBötC on the right side, in a control state (a), under CNQX (b, glutamatergic blockade-insensitive activities) and [a] – [b] (c, glutamate-dependent activities) (Fig.7: Functional connectomics by membrane potential movies)
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Abstract
Introduction
Section 1: Identification of the pFRG in the respiratory rhythm generator neuron complex using a novel transgenic rat line harboring Phox2b-EYFP BAC (K. Ikeda, K. Kawakami & H. Onimaru)
Section 2: The anatomy of the respiratory rhythmogenic kernel: the pre-Bötzinger complex of the medulla (Y. Okada & S. Yokota)
Section 3: The physiology of the pre-Bötzinger complex from a rhythmogenic perspective (N. Koshiya)
Section 4: The structure and function of the respiratory neuronal circuits of the high cervical spinal cord (Y. Oku)
Section 5: The involvement of spinal interneurons in the generation of the rostrocaudal gradient of intercostal inspiratory motor activity (M. Iizuka)
Section 6: The functional involvement of the pons in the respiratory control mechanisms (H. Onimaru & H. Koizumi)
Conclusions
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Taiz, Møller, Murphy, Zeiger, Plant Physiology and Development, 7e Student Resources
Description
The student resources to accompany Plant Physiology and Development, Seventh Edition include the following:
Web Topics & Essays
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Study Questions
References
Appendices
To view sample content, see the "Sample Resources" category.
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Chapter 01 - Plant and Cell Architecture
Chapter 02 - Cell Walls: Structure, Formation, and Expansion
Chapter 03 - Genome Structure and Gene Expression
Chapter 04 - Signals and Signal Transduction
Chapter 05 - Water and Plant Cells
Chapter 06 - Water Balance of Plants
Chapter 07 - Mineral Nutrition
Chapter 08 - Solute Transport
Chapter 09 - Photosynthesis: The Light Reactions
Chapter 10 - Photosynthesis: The Carbon Reactions
Chapter 11 - Photosynthesis: Physiological and Ecological Considerations
Chapter 12 - Translocation in the Phloem
Chapter 13 - Respiration and Lipid Metabolism
Chapter 14 - Assimilation of Inorganic Nutrients
Chapter 15 - Abiotic Stress
Chapter 16 - Signals from Sunlight
Chapter 17 - Seed Dormancy, Germination, and Seedling Establishment
Chapter 18 - Vegetative Growth and Organogenesis: Primary Growth of the Plant Axis
Chapter 19 - Vegetative Growth and Organogenesis: Branching and Secondary Growth
Chapter 20 - The Control of Flowering and Floral Development
Chapter 21 - Sexual Reproduction: From Gametes to Fruits
Chapter 22 - Embryogenesis: The Origin of Plant Architecture
Chapter 23 - Plant Senescence and Developmental Cell Death
Chapter 24 - Biotic Interactions
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Web Topic 1.1 Model Organisms
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Web Topic 1.2 Plant Identification, Classification, and Evolutionary Thought
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Web Topic 1.3 The Nuclear Pore and Proteins Involved in Nuclear Import and Export
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Web Topic 1.4 Protein Signals Used to Sort Proteins to Their Destinations
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Web Topic 1.5 Unfolded Protein Response: Quality Control in the ER
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Web Topic 1.6 SNAREs, Rabs, and Coat Proteins Mediate Vesicle Formation, Fission, and Fusion
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Web Topic 2.1 Terminology for Polysaccharide Chemistry
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Web Topic 2.2 Matrix Components of the Cell Wall
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Web Topic 2.3 The Mechanical Properties of Cell Walls: Studies With Nitella
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Web Topic 3.1 Inheritance Patterns of Plastid Genomes
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Web Topic 3.2 Recombination Mapping and Gene Cloning: Overview
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Web Topic 3.3 Transposon Tagging
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Web Topic 3.4 The Transcriptome and Data Mining
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Web Topic 3.5 Transformation by Agrobacterium
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Web Topic 5.1 Calculating Capillary Rise
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Web Topic 5.2 Calculating Half-Times of Diffusion
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Web Topic 5.3 Alternative Conventions for Components of Water Potential
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Web Topic 5.4 Temperature and Water Potential
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Web Topic 5.5 Can Negative Turgor Pressures Exist in Living Cells?
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Web Topic 5.6 Measuring Water Potential
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Web Topic 5.7 The Matric Potential
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Web Topic 5.8 Wilting and Plasmolysis
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Web Topic 5.9 Understanding Hydraulic Conductivity
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Web Topic 6.1 Irrigation
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Web Topic 6.2 Physical Properties of Soils
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Web Topic 6.3 Calculating Velocities of Water Movement in the Xylem and in Living Cells
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Web Topic 6.4 Leaf Transpiration and Water Vapor Gradients
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Web Topic 7.1 Symptoms of Deficiency in Essential Minerals
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Web Topic 7.2 Observing Roots Belowground
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Web Topic 8.1 Relating the Membrane Potential to the Distribution of Several Ions across the Membrane: The Goldman Equation
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Web Topic 8.2 Patch Clamp Studies in Plant Cells
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Web Topic 8.3 Chemiosmosis in Action
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Web Topic 8.4 Kinetic Analysis of Multiple Transporter Systems
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Web Topic 8.5 ABC Transporters in Plants
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Web Topic 8.6 Transport Studies with Isolated Vacuoles and Membrane Vesicles
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Web Topic 9.1 Principles of Spectrophotometry
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Web Topic 9.2 The Distribution of Chlorophylls and Other Photosynthetic Pigments
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Web Topic 9.3 Quantum Yield
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Web Topic 9.4 Antagonistic Effects of Light on Cytochrome Oxidation
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Web Topic 9.5 Structures of Two Bacterial Reaction Centers
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Web Topic 9.6 Midpoint Potentials and Redox Reactions
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Web Topic 9.7 Oxygen Evolution
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Web Topic 9.8 Photosystem I
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Web Topic 9.9 ATP Synthase
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Web Topic 9.10 Mode of Action of Some Herbicides
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Web Topic 9.11 Chlorophyll Biosynthesis
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Web Topic 10.1 CO2 Pumps
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Web Topic 10.2 How the Calvin–Benson Cycle Was Elucidated
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Web Topic 10.3 Rubisco: A Model Enzyme for Studying Structure and Function
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Web Topic 10.4 Energy Demands for Photosynthesis in Land Plants
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Web Topic 10.5 Rubisco Activase
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Web Topic 10.6 Operation of the Photorespiratory Pathway
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Web Topic 10.7 Cyanobacteria and the Oxygenation Reaction of Rubisco
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Web Topic 10.8 Carbon Dioxide: Some Important Physicochemical Properties
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Web Topic 10.9 Three Variations of C4 Metabolism
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Web Topic 10.10 Single-Cell C4 Photosynthesis
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Web Topic 10.11 Photorespiration in CAM plants
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Web Topic 10.12 Starch Granule Architecture
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Web Topic 11.1 Working with Light
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Web Topic 11.2 Heat Dissipation from Leaves: The Bowen Ratio
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Web Topic 11.3 The Geographic Distributions of C3 and C4 Plants
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Web Topic 11.4 Calculating Important Parameters in Leaf Gas Exchange
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Web Topic 11.5 Prehistoric Changes in Atmospheric CO2
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Web Topic 11.6 Projected Future Increases in Atmospheric CO2
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Web Topic 11.7 Using Carbon Isotopes to Detect Adulteration in Foods
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Web Topic 11.8 Reconstruction of the Expansion of C4 Taxa
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Web Topic 12.1 Sieve Elements as the Transport Cells between Sources and Sinks
Web Topic 12.2 Evolution of the Relationship between Sieve Elements and Neighbor Cells
Web Topic 12.3 Experiments on Phloem Loading
Web Topic 12.4 Monitoring Traffic on the Sugar Freeway: Sugar Transport Rates in the Phloem
Web Topic 12.5 Evidence for Mass Flow
Web Topic 12.6 Sampling Phloem Sap
Web Topic 12.7 Proteins and RNAs: Signaling Molecules in the Phloem
Web Topic 12.8 Experiments on Phloem Unloading
Web Topic 12.9 Possible Mechanisms Linking Sink Demand and Photosynthetic Rate in Starch Storers
Web Topic 13.1 Isolation of Mitochondria
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Web Topic 13.2 The Q Cycle Explains How Complex III Pumps Protons across the Inner Mitochondrial Membrane
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Web Topic 13.3 Multiple Energy-Conservation Bypasses in Oxidative Phosphorylation of Plant Mitochondria
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Web Topic 13.4 FoF1-ATP Synthases: The World's Smallest Rotary Motors
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Web Topic 13.5 Transport into and out of Plant Mitochondria
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Web Topic 13.6 The Genetic System in Plant Mitochondria Has Several Special Features
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Web Topic 13.7 Does Respiration Reduce Crop Yields?
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Web Topic 13.8 The Lipid Composition of Membranes Affects the Cell Biology and Physiology of Plants
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Web Topic 14.1 Development of a Root Nodule
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Web Topic 14.2 Measurement of Nitrogen Fixation
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Web Topic 14.3 The Synthesis of Methionine
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Web Topic 15.1 Climate Change, Global Warming, and Abiotic Stress
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Web Topic 15.2 Abiotic Stress and Precision Agriculture
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Web Topic 15.3 A Closer Look at ABA Signaling in Guard Cells
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Web Topic 16.1 Mougeotia: A Chloroplast with a Twist
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Web Topic 16.2 Phytochrome and High-Irradiance Responses
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Web Topic 17.1 The Evolution of Seeds
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Web Topic 17.2 Seedling Growth Can Be Divided into Two Types: Epigeal and Hypogeal
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Web Topic 17.3 Seeds Exhibit Both Primary and Secondary Dormancy
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Web Topic 17.4 A Special Case of Endogenous Dormancy
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Web Topic 17.5 Phase II of Germination Can Be Either a One-Step or a Two-Step Process
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Web Topic 17.6 A Classical Experiment – Split Cereal Seeds
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Web Topic 17.7 Mobilization of Seed Storage Proteins during Germination
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Web Topic 17.8 Oil Body Organization and Function during Germination
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Web Topic 17.9 Trichoblast Identity in Arabidopsis
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Web Topic 17.10 Thigmotropism, Gravitropism, and Circumnutation Are Integrated Signals
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Web Topic 17.11 Roots Are Phototropic
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Web Topic 17.12 Microtubule Orientation and Blue Light
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Web Topic 17.13 Chromophores and Phototropism
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Web Topic 18.1 Azolla Root Development
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Web Topic 18.2 Shoot Tissues Are Derived from Several Discrete Sets of Apical Initials
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Web Topic 18.3 Mechanical Signaling in the Meristem
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Web Topic 18.4 Bifacial, Unifacial, and Equifacial Leaves
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Web Topic 18.5 Development of Serrated and Compound Leaves
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Web Topic 18.6 Trichomes Are Specialized Cells in the Leaf Epidermis
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Web Topic 19.1 Plant Architecture, Crop Yields, and Environmental Sustainability
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Web Topic 19.2 Dichotomous Branching
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Web Topic 19.3 Model of Auxin-Regulated Branching
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Web Topic 19.4 Advances in Root System Phenotyping
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Web Topic 19.5 An Auxin Clock Model for the Spacing of Prebranch Sites
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Web Topic 19.6 Evolution of Secondary Growth
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Web Topic 19.7 Reaction Wood
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Web Topic 19.8 Secondary Growth in Lianas
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Web Topic 19.9 Organizers of Root Vascular Cambium
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Web Topic 20.1 Climate Change Has Caused Measurable Changes in Flowering Time of Wild Plants
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Web Topic 20.2 Contrasting the Characteristics of Juvenile and Adult Phases of English Ivy (Hedera helix) and Maize (Zea mays)
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Web Topic 20.3 Flowering of Juvenile Meristems Grafted to Adult Plants
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Web Topic 20.4 Competence and Determination Are Two Stages in Floral Evocation
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Web Topic 20.5 Characteristics of the Phase-Shifting Response in Circadian Rhythms
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Web Topic 20.6 The Contrasting Effects of Phytochromes A and B on Flowering
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Web Topic 20.7 Support for the Role of Blue-Light Regulation of Circadian Rhythms
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Web Topic 20.8 Genes That Control Flowering Time
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Web Topic 20.9 The Self-Propagating Nature of the Floral Stimulus
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Web Topic 20.10 Regulation of Flowering in Canterbury Bells by Both Photoperiod and Vernalization
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Web Topic 20.11 Examples of Floral Induction by Gibberellins in Plants with Different Environmental Requirements for Flowering
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Web Topic 20.12 Floral Organ Identity Genes in Monocot Plants
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Web Topic 21.1 Evolution Has Favored Diploidy in Plant Life Cycles
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Web Topic 21.2 Orchids, Champions of Pollination Deception
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Web Topic 21.3 Types of Placentation in Fruits
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Web Topic 21.4 Variations in Gametophyte Development
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Web Topic 21.5 ANXUR 1 and 2 Are FERONIA-like RLKs Required for Tip Growth
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Web Topic 21.6 Other Signaling Molecules Involved in Pollen Tube Guidance
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Web Topic 21.7 Fail-Safe Mechanisms Ensure That Proper Fertilization Occurs
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Web Topic 21.8 The Molecular Mechanism of Cytoplasmic Sterility in Rice
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Web Topic 21.9 The Molecular Mechanism of GSI in Poppy (Papaver rhoeas)
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Web Topic 21.10 Various Types of Apomixis
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Web Topic 21.11 Three Types of Endosperm Development
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Web Topic 21.12 Evolution of Epigenetic Imprinting
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Web Topic 21.13 The FIS Proteins and Endosperm Development
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Web Topic 21.14 The Genes Regulating Aleurone Layer Differentiation
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Web Topic 21.15 The Role of LEA Proteins in Desiccation Tolerance
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Web Topic 21.16 Fruit Types and Examples
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Web Topic 22.1 Polarity of Fucus Zygotes
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Web Topic 23.1 Petal Senescence
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Web Topic 23.2 Senescence-Associated Cell Death
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Web Topic 23.3 Biochemical Pathways of Chlorophyll Breakdown
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Web Topic 23.4 Abscission and the Dawn of Agriculture
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Web Topic 23.5 Abscission, Plant Biotechnology, and Food Security
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Web Topic 24.1 Cutin, Waxes, and Suberin
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Web Topic 24.2 Systemin Is a Peptide Defense Signal in Tomato
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Web Essay 2.1 Microtubules, Microfibrils, and Growth Anisotropy
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Web Essay 2.2 Biophysical Coordination of Water Uptake and Cell Wall Enlargement
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Web Essay 3.1 Agriculture, Population Growth, and the Challenge of Climate Change
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Web Essay 7.1 From Meals to Metals and Back
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Web Essay 9.1 A Novel View of Chloroplast Structure
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Web Essay 10.1 Modulation of Phosphoenolpyruvate Carboxylase in C4 and CAM Plants
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Web Essay 10.2 Photosynthesis and CO2-Concentrating Mechanisms in Marine Plants
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Web Essay 13.1 Metabolic Flexibility Helps Plants to Survive Stress
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Web Essay 13.2 Metabolic Profiling of Plant Cells
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Web Essay 13.3 Mitochondrial Dynamics: When Form Meets Function
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Web Essay 13.4 Seed Mitochondria and Stress Tolerance
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Web Essay 13.5 Balancing Life and Death: The Role of the Mitochondrion in Programmed Cell Death
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Web Essay 13.6 Respiration by Thermogenic Flowers
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Web Essay 13.7 Reactive Oxygen Species (ROS) and Plant Respiration
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Web Essay 13.8 Transcriptional Regulation of Respiration/Retrograde Regulation
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Web Essay 13.9 In Vivo Measurement of Plant Respiration
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Web Essay 13.10 Extreme Diversity of Triacylglycerol and Fatty Acid Structures in Plant Oils Used for Food, Biofuels, and Chemicals
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Web Essay 15.1 Heterophylly in Aquatic Plants
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Web Essay 16.1 Temperature Regulation of Photoreceptors
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Web Essay 18.1 Plant Meristems: A Historical Overview
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Chapter 1 Study Questions
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Chapter 2 Study Questions
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Chapter 3 Study Questions
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Chapter 7 Study Questions
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Chapter 9 Study Questions
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Chapter 10 Study Questions
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Chapter 11 Study Questions
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Chapter 13 Study Questions
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Chapter 15 Study Questions
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Chapter 16 Study Questions
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10662 | http://microbiologycommunity.nature.com/posts/f-pilus-the-ultimate-bacterial-sex-machine | F-pilus, the ultimate bacterial sex machine | Research Communities by Springer Nature
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Behind the Paper
F-pilus, the ultimate bacterial sex machine
How we learned that the iconic F-pilus is an extra tough biopolymer, perfectly adapted to help bacteria evade antibiotics.
Published in Microbiology
Apr 21, 2023
Jonasz Patkowski
Postdoctoral Researcher, Imperial College London
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F-pilus, the ultimate bacterial sex machine
How we learned that the iconic F-pilus is an extra tough biopolymer, perfectly adapted to help bacteria evade antibiotics.
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Nature The F-pilus biomechanical adaptability accelerates conjugative dissemination of antimicrobial resistance and biofilm formation - Nature Communications Enteropathogenic bacteria use extracellular appendages, known as F-pili, to share plasmids carrying antibiotic resistance genes. Here, the authors show that F-pili are highly flexible but robust at the same time, and this is important for plasmid transfer and formation of biofilms that protect against the action of antibiotics.
Historically underrated
Inspired by Tiago’s 2016 study that mapped the atomic structure of the conjugative F-pilus 1, I was “infected” (pun intended) with fascination in this peculiar nanomachine. Since I joined his lab, my idea of bacterial conjugation – a process whereby two bacteria exchange DNA horizontally through a conjugative pilus - changed drastically. During early years of molecular biology, F. Jacob and E. Wollman developed the “Interrupted Mating Experiment”, whereby bacterial conjugation was prematurely aborted by agitation at different timepoints, resulting in a partial transfer of DNA 2. This technique allowed for generating first genomic maps of the E.coli chromosome, but resulted in a wide-spread assumption that conjugation is a fragile process, prone to agitation of the environment.
It was therefore incredibly puzzling why the F-plasmids (which encode the conjugative F-pili) are so vastly widespread in gastrointestinal tracts of humans and animals – which are undeniably turbulent environments 3. There must be something that distinguishes the F-pilus from other conjugative pili, that allows it to thrive in such tempestuous environments. This question formulated the exciting project that aimed to assess the dynamics of bacterial connections created by the F-pilus, and ultimately understand the biomechanical potential of this polymer. Now, we are delighted to share our unexpected but impactful results in the paper "The F-pilus biomechanical adaptability accelerates conjugative dissemination of antimicrobial resistance and biofilm formation".
Hostile territory - the perfect territorySince the first discovery of bacterial conjugation in 1946 4,the F-pilus has become the paradigm for studying this phenomenon. Nowadays, it is well-established that the F-pilus is a peculiar example of conjugative pili as it can operate in liquid environments due to its flexible nature and thin morphology. These properties distinguish it from thick and rigid pili, which are capable of conjugating exclusively on solid surfaces 5. Following this idea, we tested how exactly the F-pilus operates in turbulent environments and compared conjugation efficiencies between liquid-steady and liquid-agitated conditions (Fig. 1A). It became clear that the conjugation efficiency was not decreased by agitation (Fig. 1B). In fact, it was significantly enhanced. Moreover, even the high-speed vortexing was not enough to disrupt interbacterial connections. It seems that once the F-pilus recognises a mating opportunity, the two bacteria remain fixed together, and DNA transfer proceeds successfully despite agitation of the environment.
Figure 1.Effect of environment on conjugation and biofilm formation. AConceptual representation of F-pilus connections in steady vs agitated environments.BNumber of successful transconjugants is clearly higher in shaking than steady conditions.CThe biofilm is visibly thicker when F+ cells are incubated in shaking conditions. DMicrograph representing a bacterial biofilm, where Inter-bacterial connections are established by a network of F-pili.
We hypothesized that such stable connections could form ignition points for biofilm expansion. By referring to the pioneering study by Jean-Marc Ghigo in 2001, which first documented the involvement of the F-pili in biofilm formation 6 – we decided to test differences in biofilm formation across the previously described steady versus agitated conditions. The results were consistent with conjugation experiments – biofilm mass was visibly greater after agitation (Fig. 1C). Furthermore, we visualised formed biofilms with electron microscopy, which revealed a tightly packed network of bacteria connected by F-pili (Fig. 1D). Operation of the F-pili in agitated environments clearly accelerated expansion of a protective biofilm.
Not-so-fragile sex machineHaving observed this worrying trend, we were under the obligation to understand in detail the adaptations of the F-pilus which allow for such high efficiency of conjugation and biofilm formation. Magnus Andersson’s lab expertise on measuring mechanical properties of polymers was exactly what allowed us to document biomechanical properties of the F-pilus. We displayed - with a biophysical precision - how stretchy, bendy, and sturdy the F-pilus is. By a setup where an immobilised bacterium is pulled by its F-pilus with atomic tweezers (Fig. 2A), we observed a spring-like behaviour of the F-pilus, comparable to that of the highly flexible type IV pili. After multiple rounds of stretching and relaxing, there was no visible exhaustion of the filament - meaning that the F-pilus has adapted to continuous dynamic movements, important for its operation in turbulent environments (Fig. 2B). Notably, even the highest pulling forces in our setup were unable to – even partially - dislocate the F-pili from the surface of the bacterium, underlining its robustness
Figure 2.Testing biophysical properties of the F-pilus.AMolecular Tweezer setup representation. Bacterium is immobilised on the stage, and a bead is mounted to the F-pilus. The trapped F-pilus undergoes cycles of pulling and relaxing, and the force response is measured. BUpon pulling, the F-pilus extends significantly, and when relaxed - comes back to its original state without exhaustion - just like a spring.
Role of phospholipids revealedAn interesting property of the F-pilus structure is the presence of phospholipid moieties threaded between F-pilin subunits. These are conserved among conjugative pili – including in those of Archea 7- however their role remained unknown – mainly because it is impossible to remove them in vivo and study the change in F-pilus behaviour. One appealing hypothesis is that presence of phospholipids stabilises the molecular architecture of conjugative pili, therefore increasing mechanical (and potentially thermochemical) robustness. We resolved to steered molecular dynamics simulations (sMD) conducted by Joseph Baker's lab. The freedom of an in silico setting allowed us to simulate a phospholipid-lacking F-pilus – a structure unobtainable in nature. Our pulling simulations show clearly its structural inferiority compared to the native F-pilus, which was significantly more robust and capable to withstand pulling (see movie below). We therefore first describe the role of phospholipid molecules in conjugative pili as a molecular glue that helps keep the pilus structure together. In nature, this is an extremely valuable adaptation that directly affects conjugation efficiency and biofilm formation.
Molecular properties, populational threat
With this paper we aim to bring more attention to the mechanisms how antibiotic resistance genes spread among bacteria, and what adaptations accelerate this process. Collectively, our work is also a great example of how cooperation between different fields of science can lead to great development in our understanding of complex biological processes on multiple levels. With molecular detail we documented the elastic and highly dynamic properties of the F-pilus, which augmented our microbiological experiments that revealed how F-pili thrive in turbulent environments. All of these answered a clinically relevant issue that affects humans on a global scale – that is – why the F-plasmids are so widespread in human and animal guts.
References:
1.Costa, T. R. D.et al. Structure of the Bacterial Sex F Pilus Reveals an Assembly of a Stoichiometric Protein-Phospholipid Complex. Cell166, 1436-1444.e1410 (2016).
2.Jacob, F. & Wollman, E. L. Sexuality and the genetics of bacteria(1961).
3.Rozwandowicz, M.et al. Plasmids carrying antimicrobial resistance genes in Enterobacteriaceae. Journal of Antimicrobial Chemotherapy73, 1121–1137 (2018).
4.Lederberg, J. & Tatum, E. L. Gene Recombination in Escherichia Coli. Nature158, 558-558 (1946).
5.Bradley, D. E. Morphological and serological relationships of conjugative pili. Plasmid4, 155-169 (1980).
6.Ghigo, J.-M. Natural conjugative plasmids induce bacterial biofilm development. Nature412, 442-445 (2001).
7.Beltran, Leticia C et al. Archaeal DNA-import apparatus is homologous to bacterial conjugation machinery.Nature communications vol. 14,1 666. 2023
Jonasz Patkowski(He/Him)
Postdoctoral Researcher, Imperial College London
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Postdoctoral researcher at the Department of Infectious Disease, Imperial College London. I apply my structural biology/biochemistry background to understand how bacteria propagate antimicrobial resistance (AMR) genes during bacterial conjugation. Recently, I also have been working on the life cycle of phage-inducible pathogenic islands and associated bacteriophages.🏳️🌈🇬🇧🇵🇱
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10663 | https://www.youtube.com/watch?v=uO8DWyRJMHo | Acid Base Titration Curves
Tangerine Education
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Posted: 23 Dec 2017
What titration curves look like and how to read them for AP Chemistry.
Stoichiometry Titration Problem Video:
TRANSCRIPT:
In the simplest terms, a titration curve is basically a plot of the volume of titrant that you’re adding to the analyte versus pH. What the general shape of a titration curve looks like is kind of like this, and if you haven’t seen my titration stoichiometry problem video yet, I’d highly recommend that you look at that. Back to this, this is a curve of a strong acid and it is being titrated with a strong base. And you can tell that it’s starting out with a strong acid because the pH is very low. And remember that the lower the pH is, the stronger the acid is. So here we’re going to add more and more titrant, or more and more of the strong base, and here you see a big jump up to a very high pH so over here it’s very basic and here it’s very acidic. Here is what we call the equivalence point – it’s pretty much halfway up this huge jump. And what’s happening is we have our acid, so let’s say HCl because that’s a strong acid, and when you have it in solution, it’s present as H+ ions and Cl- ions. But when you add a strong base to it, let’s say NaOH, that goes into solution as Na+ and OH- ions. Now here, the Na+ and Cl- are spectator ions because they don’t really do anything or change chemically; it’s just that these H+ and OH- ions are going to combine to neutralize each other and make H2O. So what’s happening here is we have a lot of HCl and not that much NaOH yet. Because we have a lot of HCl, which is a strong acid, that’s causing the solution to be very acidic. Now here, the equivalence point is where the moles of HCl equal the moles of NaOH. As you can see, the equivalence point is at a pH of around 7, which is perfect because we have a strong acid and a strong base. When they have equal moles, the pH is 7 or neutral. Up here, we have a lot of NaOH so now NaOH is in excess and it’s causing the solution to be basic. But let’s see what happens when we have a strong acid and a weak base.
Here you see that the pH is still really low, meaning that we have our strong acid to start with. So let’s go with HCl again. And for our weak base, let’s go with NH3 which is ammonia, and we see that our equivalence point is around here. Now notice that this isn’t at 7 like the other titration curve that we looked at. It’s a little below 7, so that means that at the equivalence point, the solution is slightly acidic. And the reason for that is because we have a weak base instead of a strong one that’s balancing out the effects of the strong acid. And also, if we write this out, the equation for NH3 when it’s placed in water is, it partially ionizes into NH4+ and OH-. Since this doesn’t go to completion, some of it stays as NH3 and some of it changes into NH4+. Since NH3 is a weak base, its conjugate acid, NH4+, is a little stronger than the strength of NH3. So since this is a stronger acid, it’s also causing the solution to be a little acidic. And you’ll see something similar to that when we look at the weak acid and strong base titration curve.
So we’ve got the same general shape for a titration curve with a weak acid and a strong base, but notice that the pH isn’t as low as what it was before. Before it was around here. Because we’re starting out with a weak acid, the pH isn’t going to be as low. Plus, when we look at the equivalence point, let’s say it’s around here, and move over to where it says the pH is, the pH is a little higher than 7. That means it’s going to be a little basic. Following the logic that we used for the strong acid-weak base titration curve, when we have a weak acid such as acetic acid, that partially dissociates into acetate and H+ ions. But remember, it’s only partial dissociation. So, because acetic acid is a weak acid, its conjugate base is stronger. Since acetate is more of a strong base, it’s going to accept protons and change back into this. So it’s going to cause the pH at the equivalence point to be a little higher than 7. Also, I’d like to point out that in here, you have what is called a buffer region. What’s happening here, is when we titrate the weak acid with a strong base, we have a conjugate acid-base pair. So when we have a strong base such as NaOH, it’s going to steal the H+ ions from C2H3O2. So what that’s going to create is a bunch of acetic acid and acetate. That’s going to create a buffer because there are a bunch of conjugate acids and bases. In the middle of the buffer region which is half of the equivalence point, we’re actually going to see that the pH is equal to pKa. And remember from the Henderson-Hasselbalch equation which is pH = pKa + log([A-]/[HA]) that whenever these two are the same, that just simplifies to 1, and the log of 1 is 0. So what we get from here is that pH = pKa meaning that the concentrations of the conjugate base and the conjugate acid are the same.
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Transcript:
in the simplest terms a titration curve is basically just a plot of the volume of titrant that you're adding to the analyte versus pH and what the general shape of a titration curve looks like is kind of like this and if you haven't seen my titration stoichiometry problem video yet I highly recommend that you look at that because in that video I explained how titrations work so back to this this is a curve of a strong acid and it is being titrated with a strong base and you can tell that it's starting out with a strong acid because the pH is very low and remember that the lower the pH is the stronger the acid is so here we're going to add more and more titrant or more and more the strong base and then here you see a big jump up to a very high pH so over here it's extremely basic and here it's very acidic and then here is what we call the equivalence point it's pretty much halfway up this huge jump so this is the equivalence point all right and what's happening is we have our acid so let's say HCl because that's a strong acid and when you have it in solution it's present as H+ ions and Cl minus ions but when you add a strong base to it let's say NaOH that goes into solution as na+ and Oh H minus ions now here the na+ and cl- our spectator ions because they don't really do anything or change chemically it's just these H+ and which minus ions are going to combine to neutralize each other and make h2o so what's happening here is we have a lot of HCl and not that much ending of each yet so because we have a lot of HCl which is a strong acid that's causing the solution to be very acidic now here the equivalence point is where the moles of HCl equal the moles of NaOH so as you can see the equivalence point is at a pH of around 7 which is perfect because we have a strong acid and a strong base when they have equal moles then the pH is 7 or neutral and then up here we have a lot of nuh so now NaOH is in excess and it's causing the solution to be basic but let's see what happens when we have a strong acid and a weak base as opposed to a strong acid and a strong base ok I'm back and now we're going over a titration curve for a strong acid and a weak base so here you see that the pH is still really low meaning that we have our strong acid to start with so let's go with HCl again and for our weak base let's go with NH NH 3 which is ammonia and we see that our equivalence point is around here now notice that this isn't at 7 like the other titration curve that we looked at it's a little below 7 so that means that at the equivalence point the solution is slightly acidic and the reason for that is because we have a weak base instead of a strong one that's balancing out the effects of the strong acid and also if you write this out the equation for nh3 when it's placed in water is it partially dissociates or ionizes into NH 4 plus and os- now since this doesn't go to completion some of it stays at nh3 and some of it change - NH four plus and since NH three is a weak base its conjugate acid then H four plus is pretty it's a little stronger than the strength of NH three so since this is a stronger acid it's also causing the solution to be a little acidic and you'll see something similar to that when we look at the weak acid and strong base titration curves okay so we've got the same general shape for a titration curve with a weak acid and a strong base but notice that the pH isn't as low as what it was before before it was like around here so because we're starting out with a weak acid it's the pH isn't going to be as low plus when we look at the equivalence point when you go let's say it's around here and move over to where it says the PHS the pH is a little higher than seven now so that means that it's going to be a little bit basic following the logic that we used for the strong acid weak base titration curve when we have a weak acid such as acetic acid so each C two H three O two that partially dissociates into acetate and each plus ions but remember it's only a partial dissociation so because acetic acid is a weak acid its conjugate base is stronger so since acetate is more of a strong base it's going to asset to protons and change back into this so it's going to cause the G pH at the equivalence point to be a little higher than seven also I'd like to point out that in here you have what is called a buffer buffer region so what's happening here is when we titrate the to the weak acid with a strong base we have a conjugate acid-base pair so what's happening is if we have a strong base such as NaOH it's going to steal the H+ ions from the HC to h3o - so what's that so what that's going to create is a bunch of acetic acid and a bunch of acetate and that's going to create a buffer because their conjugate acids and bases so in the middle of the buffer region which is half of the equivalence point we're actually going to see that the pH is equal to PKA and remember from the undersand Hasselbeck equation which is pH equals PKA plus log of concentration of the base over the concentration of the conjugate acid that whenever these two are the same that just simplifies to one and the log of one is just 0 so what we get from here is a pH equals PKA meaning that the concentration of the conjugate conjugate base and the conjugate acids are the same |
10664 | https://bio.libretexts.org/Learning_Objects/Laboratory_Experiments/Microbiology_Labs/Microbiology_Labs_II/14%3A_Gram-Positive_Streptococci-_Isolation_and_Identification_of_Streptococci_and_Enterococci/14.03%3A_The_Beta_Streptococci | Skip to main content
14.3: The Beta Streptococci
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14.2: The genus Streptococcus
14.4: The Pneumococcus (Streptococcus pneumoniae)
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( \newcommand{\kernel}{\mathrm{null}\,})
Lancefield serologic groups A, B, C, D, F, and G are all streptococci that may show beta hemolysis on Blood agar. However, some group B streptococci are non-hemolytic and group D streptococci (discussed below) usually show alpha hemolysis or are non-hemolytic.
Streptococcus pyogenes, often referred to as group A beta streptococci or GAS because they belong to Lancefield serologic group A and show beta hemolysis on blood agar, are responsible for most acute human streptococcal infections. S. pyogenes isolates are Gram-positive cocci 0.5-1.0 µm in diameter that typically form short chains in clinical specimens and longer chains in laboratory media. The most common infection is pharyngitis (streptococcal sore throat) with the organism usually being limited to the mucous membranes and lymphatic tissue of the upper respiratory tract. S. pyogenes is responsible for 15-30% of cases of acute pharyngitis in children and 5-10% of cases in adults. Between 5% and 20% of children are asymptomatic carriers. Pharyngitis is pread person to person primarily by respiratory droplets; skin infections are spread by direct contact with an infected person or through fomites.
From the pharynx, however, the streptococci sometimes spread to other areas of the respiratory tract resulting in laryngitis, bronchitis, pneumonia, and otitis media (ear infection). Occasionally, it may enter the lymphatic vessels or the blood and disseminate to other areas of the body, causing septicemia, osteomyelitis, endocarditis, septic arthritis, and meningitis. It may also infect the skin, causing erysipelas, impetigo, or cellulitis.
Group A beta streptococcus infections can result in two autoimmune diseases, rheumatic fever and acute glomerulonephritis, where antibodies made against streptococcal antigens cross react with joint membranes and heart valve tissue in the case of rheumatic fever, or glomerular cells and basement membranes of the kidneys in the case of acute glomerulonephritis.
Streptococcal pyrogenic exotoxin (Spe), produced by rare invasive strains and scarlet fever strains of Streptococcus pyogenes (the group A beta streptococci). S. pyogenes produces a number of SPEs that are cytotoxic, pyrogenic, enhance the lethal effects of endotoxins, and contribute to cytokine-induced inflammatory damage. SPEs are responsible for causing streptococcal toxic shock syndrome (STSS) whereby excessive cytokine production leads to fever, rash, and triggering the shock cascade. The SPEs also appear to be responsible for inducing necrotizing fasciitis, a disease that can destroy the skin, fat, and tissue covering the muscle (the fascia). SPE B is also a precursor for a cysteine protease that can destroy muscles tissue.
CDC reports that approximately 9,000-11,500 cases of invasive GAS disease occur each year in the U.S., with STSS and necrotizing fasciitis each accounted for approximately 6-7% of the cases. STSS has a mortality rate of around 35%. The mortality rate for necrotizing fasciitis is approximately 25%.
Note
For further information on virulence factors for group A beta Streptococci, see the following Learning Objects in your Lecture Guide:
Gram-Positive PAMPs
The Ability to Adhere to Host Cells
The Ability to Resist Phagocytic Engulfment
The Ability to Resist Phagocytic Destruction
The Ability to Evade Adaptive Immune Defenses
Type I Toxins
Type II Toxins that Damage Cell Membranes
The Ability to Induce Autoimmune Responses
The group B streptococci (GBS or Streptococcus agalactiae) usually show a small zone of beta hemolysis on Blood agar, although some strains are non-hemolytic. S. agalactiae isolates are Gram-positive cocci 0.6-1.2 µm in diameter that typically form short chains in clinical specimens and longer chains in laboratory media. They are found in the gastrointestinal tract and genitourinary tract of 15%-45% healthy woman. This reservoir, along with nosocomial transmission, provides the inoculum by which many infants are colonized at birth. The transmission rate from a mother colonized with GBS to her baby is thought to be around 50%. Most colonized infants (and adults) remain asymptomatic, however, an estimated 1-2% of neonates colonized will develop invasive GBS diseases, including pneumonia, septicemia, and/or meningitis. Pregnant women should be tested to determine if they are GBS carriers and be given IV antibiotics if they are a carrier
Other infections associated with group B streptococci include urinary tract infections, skin and soft tissue infections, osteomyelitis, endometritis, and infected ulcers (decubitus ulcers and ulcers associated with diabetes). In the immunocompromised patient it sometimes causes pneumonia and meningitis.
The group C streptococci (mainly S. equi, S. equisimilis and S. zooepidemicus) are beta hemolytic. They sometimes cause pharyngitis and, occasionally, bacteremia, endocarditis, meningitis, pneumonia, septic arthritis, and cellulitis. Group C streptococci are a common cause of infections in animals.
The group F streptococci (mainly S. anginosus) have been isolated from abscesses of the brain, mouth, and jaw. They also sometimes cause endocarditis.
The group G streptococci also show beta hemolysis. They sometimes cause pharyngitis and can also cause serious infections of the skin and soft tissues (mainly in the compromised host) as well as endocarditis, bacteremia, and peritonitis.
All of these beta hemolytic streptococci can be identified by biochemical testing and/or by serologic testing. Today you will look at the isolation and identification of group A beta streptococci (Streptococcus pyogenes) by biochemical testing. Serological identification will be performed in Lab 17.
ISOLATION AND IDENTIFICATION OF GROUP A BETA STREPTOCOCCI (Streptococcus pyogenes)
Note
Videos reviewing techniques used in this lab:
How to Inoculate a Blood Agar Plate and Add a Taxo A (Bacitracin) Disc
How to Interpret Blood Agar with a Taxo A (Bacitracin) Disc; Identification of Streptococcus pyogenes
Group A beta streptococci are usually isolated on Blood agar. Streptococcus pyogenes produces
Very small, white to grey colonies approximately 1mm in diameter.
A zone of beta hemolysis (see Fig. 14.3.114.3.1) around 2-3mm in diameter surrounding each colony.
There are two streptococcal hemolysins, streptolysin S and streptolysin O. Streptolysin O can be inactivated by oxygen so more distinct hemolysis can be seen by stabbing the agar several times. In this way, some of the organisms form subsurface colonies growing away from oxygen. Since both streptolysin S and streptolysin O are active in the stabbed area, a more clear zone of beta hemolysis can be seen.
Sensitivity to the antibiotic bacitracin found in a Taxo A® disc.
Only the group A beta streptococci are sensitive to bacitracin, as shown by a zone of inhibition around a Taxo A® disc (see Fig. 14.3.114.3.1), a paper disc containing low levels of bacitracin. Other serologic groups of streptococci are resistant to bacitracin and show no inhibition around the disc. (The Lancefield group of a group A beta streptococcus can also be determined by direct serologic testing as will be demonstrated in Lab 16.)
See Fig. 14.3.914.3.9 for a blood agar plate of a throat culture showing possible Streptococcus pyogenes.
Contributors and Attributions
Dr. Gary Kaiser (COMMUNITY COLLEGE OF BALTIMORE COUNTY, CATONSVILLE CAMPUS)
14.2: The genus Streptococcus
14.4: The Pneumococcus (Streptococcus pneumoniae) |
10665 | https://teachy.ai/en/lesson-plan/elementary-school/5th-grade/mathematics-en/time-conversion-or-lesson-plan-or-traditional-methodology-12f0b | Lesson plan of Time Conversion | Lesson Plan | Traditional Methodology
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Time Conversion
Lesson plan of Time Conversion
Lara from Teachy
Subject Mathematics
Mathematics
Source Original Teachy
Original Teachy
Topic Time Conversion
Time Conversion
Lesson Plan | Traditional Methodology | Time Conversion
| | |
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| Keywords | Time Conversion, Seconds, Minutes, Hours, Problem Solving, Units of Measure, Mathematics, 5th Grade, Practical Examples, History of Time, Egyptians |
| Required Materials | Whiteboard, Markers, Paper, Pencil, Eraser, Clock or stopwatch, Worksheets with time conversion problems |
Objectives
Duration: (10 - 15 minutes)
The purpose of this lesson plan stage is to introduce the topic of converting units of time measurement and establish clear expectations for the students. By setting specific objectives, the intention is to guide learning and help students focus on the essential skills that will be developed throughout the lesson.
Main Objectives
1. Teach students to convert units of time measurement, such as seconds, minutes, and hours.
2. Help students solve problems that involve the conversion of time measurement units.
Introduction
Duration: (10 - 15 minutes)
The purpose of this lesson plan stage is to introduce the topic of converting units of time measurement and establish clear expectations for the students. By presenting an initial context and interesting facts, the intention is to capture students' attention and motivate them to engage actively in the learning process.
Context
Today we will learn about the conversion of time units! This is an important skill, as we use different units of time in our daily lives, such as seconds, minutes, and hours. Knowing how to convert between these units helps us better understand the time we take to carry out activities, the duration of events, and even plan our day.
Curiosities
Did you know that the ancient Egyptians were one of the first peoples to divide the day into 24 hours? They did this over 4000 years ago! Today, we use this division to organize our daily routines, from the time we wake up to the time we go to sleep. Understanding and converting time units can help us manage our time better.
Development
Duration: (50 - 60 minutes)
The purpose of this lesson plan stage is to provide a detailed understanding of time unit conversions, allowing students to practice and internalize the conversion process. By addressing specific topics and solving practical problems, students will develop the ability to convert between seconds, minutes, and hours, and apply this knowledge in everyday situations.
Covered Topics
1.Time Units: Explain that the most common time units are seconds, minutes, and hours. Discuss the definition of each unit and how they relate to each other. 2.Conversion from Seconds to Minutes: Detail that 1 minute has 60 seconds. Show practical conversion examples, such as 120 seconds equals 2 minutes. 3.Conversion from Minutes to Hours: Explain that 1 hour has 60 minutes. Use examples to demonstrate the conversion, like 180 minutes equals 3 hours. 4.Conversion from Hours to Seconds: Show that 1 hour has 3600 seconds (60 minutes x 60 seconds). Provide practical examples to illustrate the conversion. 5.Solving Practical Problems: Present problems that involve the conversion of time units. Explain step by step how to solve each problem.
Classroom Questions
1. How many seconds are in 3 minutes? 2. Convert 90 minutes to hours. 3. If an activity takes 2 hours and 30 minutes, how many minutes does that represent?
Questions Discussion
Duration: (20 - 25 minutes)
The purpose of this lesson plan stage is to consolidate students' learning, allowing them to review and discuss the answers to presented problems. By engaging students in reflections and discussions, the objective is to ensure they fully understand the process of converting time units and can apply this knowledge practically and confidently.
Discussion
How many seconds are in 3 minutes?
Explain that 1 minute has 60 seconds. So, to convert 3 minutes into seconds, multiply 3 by 60:
3 minutes 60 seconds/minute = 180 seconds.
Convert 90 minutes to hours.
Explain that 1 hour has 60 minutes. To convert 90 minutes into hours, divide 90 by 60:
90 minutes / 60 minutes/hour = 1.5 hours or 1 hour and 30 minutes.
If an activity takes 2 hours and 30 minutes, how many minutes does that represent?
Explain that 1 hour has 60 minutes. To convert 2 hours and 30 minutes into minutes:
2 hours 60 minutes/hour = 120 minutes.
Add the additional 30 minutes:
120 minutes + 30 minutes = 150 minutes.
Student Engagement
1. How can you use time unit conversion in your daily life? 2. What was the most difficult part to understand about time conversions? Why? 3. Think of a situation where you would need to convert hours to minutes. How would you do that? 4. If you were an athlete, how could time unit conversion help you improve your performance? 5. Let's suppose you have 3 different activities that last 45 minutes, 1 hour and 15 minutes, and 30 minutes. What is the total time of these activities in minutes? In hours?
Conclusion
Duration: (10 - 15 minutes)
The purpose of this lesson plan stage is to review and consolidate the main points addressed, ensuring that students leave the lesson with a clear and practical understanding of time unit conversion. By summarizing and discussing the content, the aim is to reinforce learning and enable students to apply knowledge confidently.
Summary
Time Units: Seconds, minutes, and hours are the most common time units.
Conversion from Seconds to Minutes: 1 minute has 60 seconds. Example: 120 seconds = 2 minutes.
Conversion from Minutes to Hours: 1 hour has 60 minutes. Example: 180 minutes = 3 hours.
Conversion from Hours to Seconds: 1 hour has 3600 seconds. Example: 2 hours = 7200 seconds.
Solving Practical Problems: Problems involving the conversion of time units, such as converting minutes to seconds or hours to minutes.
The lesson connected the theory of time units with practice through clear examples and practical problem-solving. This allowed students to see how conversions are applied in real situations, facilitating comprehension and retention of the content.
Understanding time unit conversion is crucial in daily life, as it helps plan activities, manage schedules, and understand event durations. Curiosities like the division of the day by the ancient Egyptians show the historical and ongoing relevance of the topic in time management.
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10666 | https://alozano.clas.uconn.edu/wp-content/uploads/sites/490/2014/01/mid2practicesolved.pdf | MATH 3240Q Second Midterm - Practice Problems It is impossible to separate a cube into two cubes, or a fourth power into two fourth powers, or in general, any power higher than the second, into two like powers. I have discovered a truly marvelous proof of this, which this margin is too narrow to contain (Cubum autem in duos cubos, aut quadratoquadratum in duos quadratoquadratos, et generaliter nullam in infinitum ultra quadratum potestatem in duos eiusdem nominis fas est dividere cuius rei demonstrationem mirabilem sane detexi. Hanc marginis exiguitas non caperet.) Pierre de Fermat (annotation on a copy of Diophantus’ “Arithmetica”).
Please note: 1. Calculators are not allowed in the exam.
2. You must always provide full explanations for all your answers. You must include your work.
—————– Read through your notes for the proofs of the theorems of Fermat, Euler and Wilson, and also for the statements of the other theorems and conjectures.
—————– Just in case you need them, the following are all the primes below 100: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97.
Question 1. Find 3 primes in each category: 1. Find 3 primes p ≡1 mod 3.
2. Find 3 primes p ≡2 mod 3.
3. Find 3 primes p ≡1 mod 5.
4. Find 3 primes p ≡2 mod 5.
5. Find 3 primes p ≡3 mod 5.
6. Find 3 primes p ≡4 mod 5.
7. Are there any primes p ≡3 mod 21? Why? Why not?
8. Are there any primes p ≡3 mod 22? Why? Why not?
9. Are there infinitely many primes in each category above? How do you know?
Solution: For parts (1) through (6), simply look through a table of primes and find primes that fit the description. For part (7), there is only one prime p ≡3 mod 21, which is p = 3. All other numbers that are congruent to n ≡3 mod 21 have a factor of 3, and they are not primes, because n = 3 + 21k = 3(1 + 7k).
For part (8), yes, there are primes that are p ≡3 mod 22, for example 47. For part (9): by Dirichlet’s theorem on arithmetic progressions, if (a, m) = 1 then there are infinitely many primes of the form p ≡a mod m. Thus, there are infinitely many primes in the categories of parts (1) through (6) and (8), and only one in category (7).
MATH 3240Q Second Midterm - Practice Problems Question 2. Use a Sieve method to find all the prime numbers between 105 and 115. Explain how you did it.
Solution: We perform a sieve on the numbers: 105, 106, 107, 108, 109, 110, 111, 112, 113, 114, 115 Notice that 112 = 121 and so √ 115 < 11. Thus, if any of those numbers is not prime, they must have a prime factor less than 11. Hence, it is enough to cross out the multiples of 2, 3, 5, and 7. In order to start the sieve, notice that: 105 ≡1 mod 2, 105 ≡0 mod 3, 105 ≡0 mod 5, 105 = 7 · 15 ≡0 mod 7.
Hence, 106, 108, 110, 112 and 114 are multiples of 2. The numbers 105, 108, 111, 114 are multiples of 3. The numbers 105, 110 and 115 are multiples of 5 and the numbers 105, 112 are multiples of 7. Therefore, the numbers 107, 109 and 113 are the only primes between 105 and 115.
Question 3. Find the smallest positive integer n such that n ≡ 1 mod 3, n ≡ 2 mod 4, n ≡ 3 mod 5.
You must use the method that appears in the proof of the Chinese Remainder Theorem.
Solution: First, we solve three easier problems: n1 ≡1, 0, 0 mod 3, 4, 5 (respectively), n2 ≡0, 1, 0 mod 3, 4, 5 (resp.) and n3 ≡0, 0, 1 mod 3, 4, 5 (resp.).
For example, to solve for n1, we must have n1 = 20k ≡1 mod 3 so k = 2 works and n1 ≡40 mod 60. Similarly, n2 ≡45 mod 60 and n3 ≡36 mod 60.
Hence, the solution we are looking for is: n ≡1 · n1 + 2 · n2 + 3 · n5 ≡40 + 90 + 108 ≡58 mod 60.
Question 4. Find the smallest positive integer that leaves remainders of 2, 4, 6 when divided by 3, 5, 7, respectively. You must use the Chinese Remainder Theorem.
Solution: We are looking for x ≡2, 4, 6 mod 3, 5, 7 (respectively). Note that this can be written as x ≡−1, −1, −1 mod 3, 5, 7. Therefore, x ≡−1 ≡104 mod 105 works. By the Chinese Remainder Theorem, that is the unique solution modulo 105.
Question 5. Solve the following quadratic congruences: • Find all solutions of x2 ≡1 mod 133 MATH 3240Q Second Midterm - Practice Problems • Prove that there are no solutions: x2 ≡2 mod 133 • Find (at least) one solution: x2 ≡93 mod 133 Note: Trial and error will yield no points. Hint: Use the Chinese remainder theorem (133 = 7 · 19).
Solution: • Find all solutions of x2 ≡1 mod 133. First reduce modulo 7 and 19: and solve x2 ≡1 mod 7 and x2 ≡ mod 19. Thus, we are looking for x ≡±1 mod 7, 19. There are four solutions modulo 133 (find them using the Chinese Remainder Theorem, e.g.
solve x ≡1 mod 7, x ≡18 mod 19): x ≡1, 20, 113, 132 mod 133.
• Prove that there are no solutions: x2 ≡2 mod 133. Suppose there is an integer x such that x2 ≡2 mod 133. Then it is also true that x2 ≡2 mod 19. But 2 is not a square modulo 19: a 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 a2 1 4 9 16 6 17 11 7 5 5 7 11 17 6 16 9 4 1 • Find (at least) one solution: x2 ≡93 mod 133. If there is a solution, x also satisfies x2 ≡93 ≡2 mod 7 and x2 ≡17 mod 19. Notice that 32 ≡2 mod 7 and (by the table above) 62 ≡17 mod 19. Let us solve, using the Chinese Remainder Theorem, the system: x ≡3 mod 7 and x ≡6 mod 19.
This yields: x ≡101 mod 133.
(There are other solutions, e.g. the congruence x ≡25 mod 133 is another solution of x2 ≡93 mod 133.) Question 6. Show that 37100 ≡13 mod 17. Hint: Use Fermat’s Little Theorem.
Solution: First 37100 ≡3100 mod 17 because 37 ≡3 mod 17. By Fermat’s Little Theorem, and since (37, 17) = 1, we have that 316 ≡1 mod 17. Moreover, 100 = 6 · 16 + 4 and so: 37100 ≡3100 ≡(316)6 · 34 ≡1 · 34 ≡27 · 3 ≡10 · 3 ≡30 ≡13 mod 17.
Question 7. Show that if p and q are distinct primes then pq−1 + qp−1 ≡1 mod pq.
Solution: Since p and q are distinct (and therefore relatively prime), it suffices separately modulo p and modulo q. By Fermat’s Little theorem one has np−1 ≡1 mod p and nq−1 ≡1 mod q for all n not equivalent to 0 modulo p or q respectively. Thus: pq−1 + qp−1 ≡1 + 0 ≡1 mod q, pq−1 + qp−1 ≡0 + 1 ≡1 mod p.
MATH 3240Q Second Midterm - Practice Problems Question 8. Use Euler’s theorem to find the first digit (starting from the right-hand side of the expansion, i.e., the units digit) of the decimal expansion of 71000.
Solution: First, φ(10) = φ(2)φ(5) = 4. In order to find out the last digit of the decimal expansion of a number, one needs to calculate its least non-negative residue modulo 10. Thus: 71000 ≡(74)250 ≡1 mod 10 where we have used the fact that 74 ≡1 mod 10, by Euler’s theorem (which applies in this case because (7, 10) = 1). Therefore the last digit is a 1.
Question 9. Prove that for any natural number n ≥1, 36n −26n is never prime.
Solution: Show that 5 and 7 are always divisors of 36n −26n, using Fermat’s little theorem (i.e., show that 36n −26n ≡0 mod 7).
Question 10. Find as many prime factors as possible of the number N = 310! −1.
Solution: Clearly, N is even (so 2|N) and 3 does not divide N. By Fermat’s little theorem, if (a, p) = 1 then a(p−1)t −1 is divisible by p, for any integer t ≥1 (why?). Thus, if p > 3 and p −1 divides 10! then p divides N. Notice that: 10! = 10 · 9 · 8 · · · 3 · 2 · 1 = 2834527.
Let us make a list of primes and the value of p −1: p 5 11 13 17 19 23 29 31 37 p −1 22 2 · 5 22 · 3 24 2 · 32 2 · 11 22 · 7 2 · 3 · 5 22 · 32 Hence, from the factorization of 10! and the table we see that p = 2, 5, 11, 13, 17, 19, 29, 31, 37, . . .
are all divisors of N.
Question 11. Let a, n > 0 be natural numbers. Find as many prime factors as possible of the number N = an! −1.
Solution: See the solution to the previous problem. Those primes p such that p −1 divides n!, and (p, a) = 1, are divisors.
Question 12. Are there infinitely many primes p such that (p, p + 2, p + 4) are all primes?
Why? Are there infinitely many primes p such that (p, p + 2, p + 6, p + 8, p + 12, p + 14) are all primes? Why? Make a generalization of the Twin Prime conjecture for 6-tuples, i.e. make an educated conjecture for the existence of 6-tuples of primes.
MATH 3240Q Second Midterm - Practice Problems Solution: (p, p + 2, p + 4) cannot be all prime because one number is divisible by 3. Similarly, (p, p + 2, p + 6, p + 8, p + 12, p + 14) cannot be all primes because one is divisible by 5. |
10667 | https://www.ms.uky.edu/~lee/ma111fa18/Slides-M111Sep28Prob4.pdf | Intro to Contemporary Math Unions and Intersections of Intervals Department of Mathematics UK Announcement ▶You have a homework assignment due next Monday.
▶Mini-exam 2 is next Wednesday.
Unions and Intersections of Intervals Interval events on the real line can be combined to form more complicated events using unions and intersections to represent outcomes in one event or the other (unions) or outcomes in the overlap of both events (intersections).
Unions and Intersections of Intervals Let Ωbe an interval of real numbers, and let E and F be event intervals in Ω.
▶A real number outcome is in E if it is between the endpoints of E.
▶The event E SF is the union of E and F. It is the set of real number outcomes in E or in F (or in both).
▶The union could be a larger interval, or two separate intervals, depending on whether E and F overlap.
▶The event E TF is the intersection of E and F. It is the set of real number outcomes in E, and in F.
▶If E and F overlap, then the intersection is the interval formed by the overlap.
Union of Two Overlapping Intervals Let Ω= [10,17], E = [11,15] and F = [13,16]. The event E SF consists of real numbers between 11 and 15, or between 13 and 16: ▶The real number 12 is in E SF, because 12 is in E (between 11 and 15).
▶The real number 15.5 is in E SF, because 15.5 is in F (between 13 and 16).
▶The real number 14 is in E SF, because 14 is in E and in F (at least one of them).
▶The real number 17 is not in E SF, because 17 is not in E nor in F.
Union of Two Overlapping Intervals Let Ω= [10,17], E = [11,15] and F = [13,16]. The event E SF consists of real numbers between 11 and 15, or between 13 and 16: Union of Two Overlapping Intervals Let Ω= [10,17], E = [11,15] and F = [13,16]. The event E SF consists of real numbers between 11 and 15, or between 13 and 16: Union of Two Overlapping Intervals Let Ω= [10,17], E = [11,15] and F = [13,16]. The event E SF consists of real numbers between 11 and 15, or between 13 and 16: Thus, E SF is actually the interval [11,16]. Its length is 16−11 = 5.
Thus, P(E [ F) = Length of E SF Length of Ω = 16−11 17−10 = 5 7 Union of Two Separate Intervals Let Ω= [10,17], E = [11,13] and F = [14,17]. The event E SF consists of real numbers between 11 and 13, or between 14 and 17: ▶The real number 12 is in E SF, because 12 is in E.
▶The real number 15.5 is in E SF, because 15.5 is in F.
▶The real number 13.5 is not in E SF, because 13.5 is not in E nor in F.
Union of Two Separate Intervals Let Ω= [10,17], E = [11,13] and F = [14,17]. The event E SF consists of real numbers between 11 and 13, or between 14 and 17: Union of Two Separate Intervals Let Ω= [10,17], E = [11,13] and F = [14,17]. The event E SF consists of real numbers between 11 and 13, or between 14 and 17: Union of Two Separate Intervals Let Ω= [10,17], E = [11,13] and F = [14,17]. The event E SF consists of real numbers between 11 and 13, or between 14 and 17: Thus, E SF is actually two intervals. It has a total length found by adding (combining) the lengths of E and F: Length of E : 3−1 = 2 Length of F : 7−4 = 3 Total length of E [ F : 2+3 = 5.
Hence P(E [ F) = Total length of E SF Length of Ω = 5 7 .
Intersection of Two Intervals Let Ω= [10,17], E = [11,15] and F = [13,16]. The event E TF consists of real numbers between 11 and 15, and between 13 and 16: ▶The real number 12 is not in E TF, because 12 is in E, but not in F.
▶The real number 15.5 is not in E TF, because 15.5 is in F, but not in E.
▶The real number 14 is in E TF, because 14 is in E and in F (both of them).
Intersection of Two Intervals Let Ω= [10,17], E = [11,15] and F = [13,16]. The event E TF consists of real numbers between 11 and 15, and between 13 and 16: Intersection of Two Intervals Let Ω= [10,17], E = [11,15] and F = [13,16]. The event E TF consists of real numbers between 11 and 15, and between 13 and 16: Intersection of Two Intervals Let Ω= [10,17], E = [11,15] and F = [13,16]. The event E TF consists of real numbers between 11 and 15, and between 13 and 16: Thus, E TF is actually the interval [13,15]. Its length is 5−3 = 2.
Hence P(E \ F) = Length of E TF Length of Ω = 2 7 .
?(4.1) Union/Intersection Practice 1 Let Ωbe the interval [8,24], E be the interval [12,16], and F be the interval [15,19]: If we pick a random real number between 8 and 24, nd the probability of the event E SF.
Hints: 1. Identify the event [12,16]S[15,19] as an interval.
2. What is the length of the union?
3. What is the length of the sample space?
4. Find P([12,16]S[15,19]). Type and send a fraction.
Union/Intersection Practice 1 Let Ωbe the interval [8,24], E be the interval [12,16], and F be the interval [15,19]: If we pick a random real number between 8 and 24, nd the probability of the event E SF.
Hints: 1. Identify the event [12,16]S[15,19] as an interval.
2. What is the length of the union?
3. What is the length of the sample space?
4. Find P([12,16]S[15,19]). Type and send a fraction.
Union/Intersection Practice 1 Let Ωbe the interval [8,24], E be the interval [12,16], and F be the interval [15,19]: If we pick a random real number between 8 and 24, nd the probability of the event E SF.
E [ F = [12,19].
P(E [ F) = Length of E SF Length of Ω = 19−12 24−8 = 7 16 ?(4.2) Union/Intersection Practice 2 Let Ωbe the interval [24,47], E be the interval [29,34], and F be the interval [36,43]: If we pick a random real number between 24 and 47, nd the probability of the event E SF.
Hints: 1. Identify the event [29,34]S[36,43] as a union of two separate intervals.
2. What is the total length of the union?
3. What is the length of the sample space?
4. Find P([29,34]S[36,43]). Type and send a fraction.
Union/Intersection Practice 2 Let Ωbe the interval [24,47], E be the interval [29,34], and F be the interval [36,43]: If we pick a random real number between 24 and 47, nd the probability of the event E SF.
Hints: 1. Identify the event [29,34]S[36,43] as a union of two separate intervals.
2. What is the total length of the union?
3. What is the length of the sample space?
4. Find P([29,34]S[36,43]). Type and send a fraction.
Union/Intersection Practice 2 Let Ωbe the interval [24,47], E be the interval [29,34], and F be the interval [36,43]: If we pick a random real number between 24 and 47, nd the probability of the event E SF.
E [ F = [29,34] [ [36,43] P(E [ F) = Total Length of E SF Length of Ω = (34−29)+(43−36) 47−24 = 5+7 23 = 12 23 ?(4.3) Union/Intersection Practice 3 Let Ωbe the interval [47,78], E be the interval [51,60], and F be the interval [54,63]: If we pick a random real number between 47 and 78, nd the probability of the event E TF.
Hints: 1. Identify the event [51,60]T[54,63] as a smaller interval.
2. What is the length of the intersection?
3. What is the length of the sample space?
4. Find P([51,60]T[54,63]). Type and send a fraction.
Union/Intersection Practice 3 Let Ωbe the interval [47,78], E be the interval [51,60], and F be the interval [54,63]: If we pick a random real number between 47 and 78, nd the probability of the event E TF.
Hints: 1. Identify the event [51,60]T[54,63] as a smaller interval.
2. What is the length of the intersection?
3. What is the length of the sample space?
4. Find P([51,60]T[54,63]). Type and send a fraction.
Union/Intersection Practice 3 Let Ωbe the interval [47,78], E be the interval [51,60], and F be the interval [54,63]: If we pick a random real number between 47 and 78, nd the probability of the event E TF.
E \ F = [54,60].
P(E \ F) = Length of E TF Length of Ω = 60−54 78−47 = 6 31 End ▶You have a homework assignment due next Monday.
▶Mini-exam 2 is next Wednesday. |
10668 | https://quizlet.com/729987692/pathology-robbins-and-cotran-review-block-2-infectious-diseases-flash-cards/ | Pathology Robbins and Cotran Review: Block 2 Infectious Diseases Flashcards | Quizlet
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Pathology Robbins and Cotran Review: Block 2 Infectious Diseases
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A 45-year-old Bangladeshi woman with atrophic gastritis has sudden onset of severe, profuse, watery diarrhea. Over the next 3 days, she becomes severely dehydrated. On physical examination, she is afebrile, but has poor skin turgor. Laboratory studies of the diarrheal fluid show microscopic flecks of mucus, but no blood and few WBCs. A blood culture is negative. The woman is hospitalized and receives intravenous fluid therapy for 1 week. Which of the following is the most likely diagnosis?
A Amebiasis
B Aspergillosis
C Cholera
D Filariasis
E Hydatid disease
F Typhoid fever
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C
The lack of stomach acid in this woman predisposes to enteric infections. Vibrio cholerae organisms are noninvasive. Instead, they produce severe diarrhea by elaboration of an en- terotoxin, called cholera toxin, that acts on bowel mucosal cells to cause persistent activation of adenylate cyclase and high lev- els of intracellular cyclic AMP that drives massive secretion of sodium, chloride, and water. The fluid loss is life-threatening because of resultant dehydration. Amebiasis tends to produce dysentery, with a bloody diarrhea, because the organisms can invade the mucosa. Aspergillosis is seen in immunocompromised patients, particularly patients with neutropenia, and is a rare cause of a diarrheal illness. Filariasis involves the lymphatics and produces elephantiasis. Hydatid disease caused by Echinococcus produces space-occupying cystic lesions in viscera. Typhoid fever produces diarrhea, and the organisms can invade mucosa and disseminate to produce many systemic symptoms.
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Terms in this set (62)
A 45-year-old Bangladeshi woman with atrophic gastritis has sudden onset of severe, profuse, watery diarrhea. Over the next 3 days, she becomes severely dehydrated. On physical examination, she is afebrile, but has poor skin turgor. Laboratory studies of the diarrheal fluid show microscopic flecks of mucus, but no blood and few WBCs. A blood culture is negative. The woman is hospitalized and receives intravenous fluid therapy for 1 week. Which of the following is the most likely diagnosis?
A Amebiasis
B Aspergillosis
C Cholera
D Filariasis
E Hydatid disease
F Typhoid fever
C
The lack of stomach acid in this woman predisposes to enteric infections. Vibrio cholerae organisms are noninvasive. Instead, they produce severe diarrhea by elaboration of an en- terotoxin, called cholera toxin, that acts on bowel mucosal cells to cause persistent activation of adenylate cyclase and high lev- els of intracellular cyclic AMP that drives massive secretion of sodium, chloride, and water. The fluid loss is life-threatening because of resultant dehydration. Amebiasis tends to produce dysentery, with a bloody diarrhea, because the organisms can invade the mucosa. Aspergillosis is seen in immunocompromised patients, particularly patients with neutropenia, and is a rare cause of a diarrheal illness. Filariasis involves the lymphatics and produces elephantiasis. Hydatid disease caused by Echinococcus produces space-occupying cystic lesions in viscera. Typhoid fever produces diarrhea, and the organisms can invade mucosa and disseminate to produce many systemic symptoms.
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A 5-year-old boy has had diarrhea for a week, averaging six low volume stools per day, which appear mucoid and sometimes blood-tinged. On physical examination, his temperature is 37.4° C. He has mild lower abdominal tenderness, but no masses. A stool culture is positive for Shigella sonnei. Which of the following microscopic findings would most likely be seen in this child's colon?
A Epithelial disruption with overlying neutrophilic exudate
B Extensive scarring of lamina propria with stricture formation
C Intranuclear inclusions within small intestinal enterocytes
D Multiple granulomas throughout the colon wall
E Slight increase in lymphocytes and plasma cells in
lamina propria
A
Shigellosis results in bloody dysentery because Shigella is highly virulent, resistant to gastric acid, and can invade and destroy the colonic mucosa. There is typically a mononuclear infiltrate extending to the lamina propria, with a neutrophilic exudate overlying the ulcerated areas. Stricture formation may follow intestinal tuberculosis. Intranuclear inclusions in enterocytes point to infection with DNA viruses, such as herpesviruses like cytomegalovirus. Granulomatous inflammation may be seen with granulomatous colitis (Crohn disease) and intestinal tuberculosis (rare). An increase in mononuclear inflammatory cells may be seen with milder forms of enterocolitis caused by viruses, Giardia, and Salmonella spp.
A 6-month-old infant has abrupt onset of vomiting followed by profuse, watery diarrhea. On physical examination, the infant has a temperature of 38.3° C. Development is normal for age, and the only abnormal finding is poor skin turgor. Laboratory studies show serum Na+ of 153 mmol/L, K+ of 4.4 mmol/L, Cl- of 113 mmol/L, CO2 of 28 mmol/L, and glucose of 70 mg/dL. Examination of a stool specimen shows mucus, but no RBCs or WBCs. Which of the following mechanisms best accounts for this diarrhea?
A Decreased absorption of sodium and water
B Decreased breakdown of lactose to glucose and
galactose
C Increased secretion of potassium and water
D Mutation in CFTR gene
E Presence of Yop virulence plasmid
A
Rotavirus, an encapsulated RNA virus, is a major cause of diarrhea in infancy. The small intestinal villous destruction with atrophy leads to decreased absorption of sodium and water. The development of IgA antibodies from secretory immunity in the bowel to rotavirus surface antigens provides older children and adults a relative resistance to rotavirus infection. Such antibodies are present in maternal milk and confer some degree of resistance to infants who breast-feed. Rotavirus infection occurs worldwide. By the age of 3 years, virtually every individual has been infected by rotaviruses at least once. Most rotavirus infections are subclinical or cause mild gastrointestinal illnesses that do not require hospitalization. The first infection is the most likely to be symptomatic; subsequent infections are often mild or asymptomatic. Many enteroviruses also produce diarrhea by inhibiting the intestinal absorption of intraluminal sodium and water, but not as severe. Mutations in the CFTR gene lead to formation of thick mucus plugs, giving rise to meconium ileus in infants. Decreased break- down of lactose occurs in disaccharidase deficiency and gives rise to an osmotic diarrhea. Cholera is the result of secretion of an exotoxin by the Vibrio cholerae organism, which potentiates the epithelial cell production of adenylate cyclase and causes secretory diarrhea with sodium chloride and water loss. The Yop plasmid confers infectivity to Yersinia organisms.
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A 60-year-old man has had persistent bloody diarrhea, abdominal cramps, and fever for the past week. On physical examination, his temperature is 38.1° C. He has mild diffuse abdominal pain. A stool sample is positive for occult blood. Colonoscopy shows marked mucosal erythema with focal ulceration from the rectum to the ascending colon. The ulcers do not penetrate the muscularis propria. A biopsy is performed, and the microscopic appearance of the specimen is shown in the figure. Which of the following infectious organisms is most likely to produce these findings?
A Bacillus cereus
B Entamoeba histolytica
C Giardia lamblia
D Salmonella enterica
E Shigella flexneri
F Vibrio cholerae
B
Amebiasis is a common cause of dysentery in developing nations. The figure shows two single-cell protozoa invading tissue with inflammatory cells. Entamoeba histolytica organisms are resistant to gastric acid and can invade the colonic submucosa via contact-dependent cytolysis. The amebae not only produce local necrosis with ulceration and hemorrhage, but also gain access to the venules of the portal system, which drains to the liver. Amebic liver abscess is an uncommon complication of amebiasis. The colonic lesions typically have disappeared by the time the liver lesions appear. In some cases, there can be extensive mucosal involvement with characteristic flask- shaped (similar to an Erlenmeyer flask) ulcerations similar to those seen in other severe inflammatory bowel diseases. Bacillus cereus is a cause of food poisoning (most often as a contaminant in reheated fried rice) and has a short incubation time. Giardiasis tends to involve the small intestine and produces variable inflammation, but no ulceration. Salmonellosis more typically involves the small intestine and in most cases produces self-limited enteritis, although more severe disease with dissemination to other organs can occur with Salmonella typhi infection. Shigellosis can produce bloody dysentery with irregular superficial colonic mucosal ulceration, but the organisms typically do not invade beyond the lamina propria. Cholera is characterized by massive, secretory diarrhea without intestinal mucosal invasion or necrosis.
In an epidemiologic study of individuals who died in a worldwide pandemic after World War I, many individuals were shown to have contracted a virulent form of influenza pneumonia. At the beginning of the 21st century, a similar epidemic is still possible from such a strain of influenza for which no vaccine may be readily available. Molecular analysis of samples of tissues showed changes in the virus responsible for these virulent forms of influenza. Which of the following changes most likely occurs in this virus to increase its virulence?
A Ability to elaborate exotoxins
B Acquisition of antibiotic resistance genes
C Increased binding to intercellular adhesion
molecule-1 (ICAM-1) receptor
D Mutations in DNA encoding envelope proteins
E Recombination with RNA segments from animal
viruses
E
The influenza pandemic in 1918 resulted from an antigenic shift in the influenza A type. This antigenic shift occurs when there is recombination with RNA sequences of influenza viruses found in animals such as pigs ("swine flu") or birds ("avian flu"). A swine flu virus has been identified as a cause of the 1918 pandemic. The H5N1 strain of influenza has been found in bird populations in modern times. Mutations in the viral hemagglutinin (H) and neuraminidase (N) envelope genes are responsible for epidemics. These mutations allow evasion from host antibodies. Influenza viruses do not bind to intercellular adhesion molecule-1 (ICAM-1) receptors; rhinoviruses do. Influenza viruses have a genome with RNA, not DNA. Viruses do not make exotoxins and do not acquire antibiotic resistance like bacteria.
An epidemiologic study is conducted with children ages 1 to 5 years who are infected with HIV-1. Most of these children have CD4+ lymphocyte counts above 500/mm3 and undetectable plasma HIV-1 RNA levels. On physical examination they have no abnormal findings. What is the most likely mode of transmission by which these children get infected?
A Breast-feeding
B Inhalation of droplet nuclei
C Fecal-oral contact
D Transfer across placenta
E Sexual abuse
D
Congenital HIV infection is an example of vertical transmission, and antiretroviral therapy for pregnant women has proven effective in reducing maternal-fetal HIV transmission. Breast-feeding is a potential mode of HIV transmission, but less frequent than placental transmission. Respiratory tract infections are spread by droplet nuclei, but HIV is not spread by this mode. The fecal-oral route of transmission is typical for enteroviruses and hepatitis A virus, but not HIV. Though sexual abuse of children unfortunately occurs, it is an uncommon mode of transmission of infections.
In an experiment, phagocytosis of bacteria by neutrophils is studied. Bacteria are introduced into plasma containing neutrophils. It is observed that phagocytosis of Streptococcus pneumoniae, Haemophilus influenzae, and Neisseria meningitidis is reduced, compared with Enterobacteriaceae. Which of the following immune evasion mechanisms in these three bacterial species best explains this finding?
A Antimicrobial peptide binding
B Carbohydrate capsule formation
C Interferon homologue production
D MHC protein down-regulation
E Surface antigen switching
B
Bacteria that produce capsules are more resistant to phagocytosis, and help them to avoid an initial innate immune response with neutrophils. This enables them to establish infection (respiratory tract, and possible spread to meninges) and become more virulent. Antimicrobial peptides produced by epithelial cells can bind to bacterial organisms and form pores in the cell walls to kill them by osmotic lysis. Interferons are produced against viral organisms, not bacteria. Some viruses, such as herpesviruses, can impair expression of MHC class I molecules so that viral antigens are not effectively displayed to CD4+ and CD8+ cells. Surface antigen switching also helps organisms such as trypanosomes to evade an adaptive immune response.
A 26-year-old man is an injection drug user and has developed fever over the past day. On examination, he has a heart murmur. A blood culture is positive for Staphylococcus aureus and antibiotic sensitivity testing shows resistance to methicillin due to presence of the mecA gene in these organisms. Through which of the following adaptations are these bacteria most likely to acquire their methicillin resistance?
A Biofilm formation
B Exotoxin release
C Pathogenicity island transfer
D Superantigen stimulation
E Surface adhesin expression
C
Bacteria have multiple mechanisms for exchanging genetic material that affords selective growth advantages. Pathogenicity islands are bacterial chromosomal elements carrying virulence genes, such as those involving antibiotic resistance. Presence of the mecA gene imparts resistance to methicillin (methicillin-resistant Staphylococcus aureus, or MRSA) and other β-lactam antibiotics. Additional bacterial genetic transfer mechanisms include plasmids, transposons, and integrons. Microbes on tissue surfaces form biofilms of sticky polysaccharide goo to isolate themselves from immune attack. Exotoxins impart virulence through tissue damage. Organisms such as S. aureus can express superantigens that nonselectively stimulate many T cell clones, leading to un- regulated cytokine release and toxic shock. Adhesins aid in microbial binding to host cells.
A study of nosocomial infections involving urinary catheters is performed. The study shows that the longer an indwelling urinary catheter remains, the higher the rate of symptomatic urinary tract infections (UTIs). Most of these infections are bacterial. Which of the following properties of these bacteria increase the risk for nosocomial UTIs?
A Biofilm formation
B Enzyme elaboration
C Exotoxin release
D Quorum sensing
E Superantigen stimulation
A
Microbes form biofilms of sticky polysaccharide goo that adheres particularly well to artificial surfaces such as catheters. The biofilm helps isolate the organisms from inflammatory cells and limit penetration of antibiotics. Bacterial enzymes, such as the hyaluronidases and streptokinases of streptococcal organisms, promote spread through tissues. Exotoxins of gram-positive organisms impart virulence through tissue damage. Bacteria have multiple mechanisms for exchanging genetic material that afford selective growth advantages. When the number of bacterial organisms increases, they sense this (quorum sensing) and turn on virulence genes. Organisms such as Staphylococcus aureus can express superantigens that nonselectively stimulate many T cell clones, leading to unregulated cytokine release and toxic shock.
A 91-year-old woman is hospitalized with sepsis. On examination she has fever and hypotension. Laboratory studies show positive blood cultures. She has disseminated intravascular coagulopathy and pulmonary diffuse alveolar damage with respiratory distress. Analysis of the microbiology laboratory findings shows that the organisms cultured are gram-negative bacilli. Which of the following substances elaborated by these organisms is most likely to cause this complex of clinical findings?
A Endotoxin
B Exotoxin
C Mycolic acid
D RNA polymerase
E Superantigen
F Tumor necrosis factor
A
Gram-negative sepsis is classically mediated by endotoxins, particularly the lipopolysaccharide component of the outer cell wall. With sepsis from gram-positive organisms there is release of exotoxins, such as tetanospasmin released by Clostridium tetani organisms. Mycolic acids found in the lipid wall of mycobacteria aids in the resistance of these organisms to degradation by acute inflammatory responses, leading to granulomatous inflammation. RNA polymerase is found in negative-sense RNA viruses and produces a positive-sense messenger RNA (mRNA) that directs the host cell to produce viral components. Superantigens may produce findings similar to lipopolysaccharide-induced septic shock; the best known is toxic shock syndrome toxin, which is elaborated by some staphylococcal organisms. Tumor necrosis factor (TNF) is elaborated by human inflammatory cells, not by microorganisms, but by the release of TNF by the action of endotoxins on macrophages that can mimic gram-negative sepsis.
A 6-year-old girl has a blotchy, reddish-brown rash on her face, trunk, and proximal extremities that developed over the course of 3 days. On physical examination, she has 0.2-cm to 0.5-cm ulcerated lesions on the oral cavity mucosa and generalized tender lymphadenopathy. A cough with minimal sputum production becomes progressively worse over the next 3 days. Which of the following viruses is most likely to produce these findings?
A Epstein-Barr
B Mumps
C Rubella
D Rubeola
E Varicella zoster
D
The rash and the Koplik spots on the buccal mucosa are characteristic findings in measles (rubeola), a childhood infection. It occurs only sporadically when immunizations have been administered to a large part of the population. The severity of the illness varies, and measles pneumonia may complicate the course of the disease, which in some cases can be life-threatening. Mononucleosis, which results from Epstein-Barr (EBV) virus infection, is more likely to occur in adolescence. Mumps produces parotitis and orchitis. Varicella-zoster virus infections in children manifest as chicken-pox. Rubella, also called German measles, is a much milder infection than rubeola.
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An 8-year-old girl has developed a mild febrile illness with a sore throat over the past 2 days. On physical examination, her temperature is 38.4° C, and she has a mild pharyngitis. The girl's symptoms subside in 1 week without therapy. Over the next 2 months, she has increasing right-sided facial drooping with inability to close the right eye. Which of the following infectious organisms is most likely to produce these findings?
A Cryptococcus neoformans
B Cytomegalovirus
C Listeria monocytogenes
D Poliovirus
E Toxoplasma gondii
D
Poliomyelitis is an enterovirus spread through fecal- oral contamination. The virus often infects the oropharynx first. It then spreads to bulbar nuclei and/or lower motor neurons in the anterior horn of the spinal cord to produce the muscular paralysis typical of polio. In places where vaccination is routinely available, this disease is rare. Cryptococcosis is a fungal disease that most often involves the lungs and meninges. Cytomegalovirus infection can be congenital; in immunocompromised adults, it can involve many organs, principally the gastrointestinal tract, brain, and lungs. Listeriosis is most often acquired via contaminated food or water; in most adults, it pro- duces mild diarrheal illness, but in some adults and children, and in fetuses, it can produce meningitis or dissemination with microabscess (microgranuloma) formation. Toxoplasmosis can be a congenital infection. In immunocompromised adults, it can produce inflammation in multiple tissues, but most often, it causes chronic abscessing inflammation in brain.
A 6-year-old girl who lives in the Yucatán peninsula has developed a high fever over the past 3 days. On physical examination, she has a temperature of 39.6° C and marked tenderness in all muscles. Laboratory studies show WBC count of 2950/mm3 with 12% segmented neutrophils, 4% bands, 66% lymphocytes, and 18% monocytes. Over 1 week, she becomes more lethargic, with a decreased level of consciousness, and petechiae and purpura develop over the skin. Further laboratory studies show thrombocytopenia with markedly prolonged prothrombin time and partial thromboplastin time. CT scan of the brain shows a hemorrhage in the right parietal lobe. Which of the following is most likely the vector of the agent causing infection in this patient?
A Louse
B Mosquito
C Pig
D Snail
E Tick
F Tsetse fly
B
Dengue fever, one form of hemorrhagic fever, is caused by an arbovirus of the Flavivirus group. This organism can be devastating because it produces bone marrow suppression, and because any antibodies to the virus enhance cellular viral uptake. It is transmitted by the mosquito vector Aedes aegypti. Louse-borne infections include rickettsial diseases. The pig can be involved in the life cycle of Taenia solium and of Trichinella spiralis. T. spiralis can produce marked muscle pain, but typically not disseminated intravascular coagulopathy. Some snails can serve as an intermediate host for Schistosoma organisms. Ticks can transmit typhus and Lyme disease. The tsetse fly can transmit sleeping sickness, which is endemic to Africa.
A 22-year-old woman has had recurrent vesicular lesions on her labia majora and perineum for 5 years. On physical examination, she is afebrile. Clusters of clear, 0.2-cm to 0.5- cm vesicles are present on the labia, with some surrounding erythema. The figure shows the representative microscopic appearance at low magnification of one of the lesions. Which of the following cellular changes is most likely to be seen under higher magnification of this lesion?
A Dysplastic epithelial cells containing human papillomavirus sequences
B Mononuclear infiltrates containing protozoal organisms
C Multinucleated (syncytial) cells containing pink-to-purple intranuclear inclusions
D Neutrophils containing ingested gram-negative diplococci
E Perivascular lymphoplasmacytic infiltrate surrounding arterioles, with endothelial proliferation
C
The figure shows a vesicle that has resulted from herpes simplex virus (HSV) infection. Most genital infections are caused by HSV-2, whereas HSV-1 is responsible for most cases of herpetic gingivostomatitis. The viral cytopathic effect results in formation of intranuclear inclusions, multinucleated cells, and cell lysis with vesicle formation in the epithelium. Cervical dysplasias do not produce vesicular lesions and are the result of another sexually transmitted disease—human papillomavirus infection. Protozoal infection with trichomoniasis, typically involving the vagina, may produce small blisters or papules, but these are often self- limited and not typically recurrent. Gram-negative diplococci are characteristic of Neisseria gonorrhoeae infection, also a sexually transmitted disease. Lymphoplasmacytic infiltrates may be seen in chancres caused by Treponema pallidum, the causative agent of syphilis.
A 6-year-old boy developed a rash over his chest that began as 0.5-cm reddish macules. Within 2 days, the macules became vesicles. Three days later, the vesicles ruptured and crusted over. Over the next 2 weeks, crops of the lesions spread to the face and extremities. Which of the following clinical manifestations of this infection is most likely to appear decades later?
A Chronic arthritis
B Congestive heart failure
C Infertility
D Paralysis
E Shingles
E
The skin lesions are typical of chickenpox, a common childhood infection caused by varicella-zoster virus infection. The infection can remain dormant for years in dorsal root ganglia, only to reactivate when immune status is diminished. The virus, now designated herpes zoster (or varicella- zoster), spreads from the ganglion to the skin in the dermatomal distribution of the corresponding sensory nerve, and it causes vesicular lesions with chronic, burning pain that is difficult to stop. A chronic arthritis can be seen with Lyme disease after Borrelia burgdorferi infection. Rheumatic heart disease can appear after group A β-hemolytic streptococcal infection. Infertility is a complication of mumps orchitis. Paralysis can complicate poliovirus infection.
A 31-year-old HIV-positive man has had increasing respiratory difficulty for the past 2 days. On physical examination, crackles are auscultated over all lung fields. A chest radiograph shows bilateral interstitial infiltrates. Laboratory studies show 26,800 copies of HIV-1 RNA/mL. A transbronchial biopsy is performed; the microscopic appearance of the specimen is shown in the figure. Which of the following is the most likely causative organism of his pulmonary disease?
A Adenovirus
B Cytomegalovirus
C Epstein-Barr virus
D Herpes zoster virus
E Respiratory syncytial virus
B
This patient has a high HIV-1 RNA level consistent with the diagnosis of AIDS. Although patients with AIDS are susceptible to many microbes, infections with cytomegalovirus are particularly common. The biopsy specimen shows an enlarged cell containing a large, distinct intranuclear inclusion and ill-defined dark cytoplasmic inclusions, which are typical of cytomegalovirus infection. Adenovirus is a viral pathogen in both immunocompromised and immunocompetent adults that may produce a clinically significant pneumonia, and intranuclear inclusions may be present, but the cells are not large, and cytoplasmic inclusions are absent. Epstein-Barr virus infection is seen frequently in patients with HIV infection, but there are no distinct pulmonary lesions associated with it. Herpes zoster infections are most likely to affect the peripheral nervous system, rarely can become disseminated to affect the lungs in immunosuppressed patients, and produce a different appearance than that shown. Respiratory syncytial virus infections are seen in children, but rarely in adults.
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A 14-year-old boy presents with fever, sore throat, and cervical lymphadenopathy. He then develops hepatomegaly and splenomegaly lasting for 2 months. His peripheral blood smear shows leukocytosis with "atypical" lymphocytes. Which of the following cell types is most likely to eliminate the virally infected cells?
A Cytotoxic CD8 cells
B Epithelioid macrophages
C Helper CD4+ cells
D IgG-secreting plasma cells
E Polymorphonuclear neutrophils
A
The features described fit with infectious mononucleosis. EBV infection involves B cells that are activated to elaborate a variety of cytokines that promote viral proliferation and reduced immune response. IL-10 normally secreted by phagocytes activates T cells (the "atypical" lymphocytes), but the virally induced homologue does not. In general, viral infections are intracellular, and a cytotoxic CD8 T cell response is required to clear virus by eliminating infected cells. Epithelioid macrophages are most important in granulomatous inflammatory responses that control mycobacterial and fungal infections. Helper T cells may be infected by EBV, but do not clear the virus. Immunoglobulin responses are most important to eliminate extracellular pathogens, such as bacteria. Neutrophils are most important as an innate immune response directed against extracellular organisms such as bacteria.
A 75-year-old woman has a postoperative wound infection responding poorly to antibiotic therapy. Over the next 3 days she develops confusion, nausea, vomiting, diarrhea, chills, and myalgias. On examination, she is febrile; the wound site is erythematous with necrotic, purulent exudate. Laboratory studies show neutrophilia with left shift. Gram stain of the wound exudate shows gram-positive cocci in clusters. Which of the following substances is most likely being elaborated by the infectious organisms?
A Lactoferrin
B Lipopolysaccharide
C Phage-encoded A-B toxin
D Pili proteins
E Superantigen
E
Staphylococcal toxic shock syndrome (TSS) results from elaboration of superantigens that stimulate up to 20% of T lymphocytes and generate a marked release of cytokines and an extensive inflammatory response. Increasingly, staphylococci have acquired the mecA gene that imparts resistance to many penicillin (methicillin) and cephalosporin antibiotics has come to be associated with methicillin-resistant Staphylococcus aureus (MRSA). Most TSS cases occur in women, because of the relationship to vaginitis. Lactoferrin is a substance secreted by human cells that binds iron needed by bacteria, and is thus part of innate immunity. Lipopolysaccharides are elaborated by gram-negative organisms and produce endotoxic shock. Phage-encoded A-B toxin is elaborated by Corynebacterium diptheriae. Pili proteins are characteristic for Neisseria gonorrhoeae to provide attachment to target cells in the genital tract.
A 46-year-old woman has had a high fever and swelling, warmth, and tenderness of the right leg for the past 3 days. On physical examination, she has a temperature of 39.4° C and the facial appearance shown in the figure. She receives macrolide antibiotic therapy with which she recovers. Infection with which of the following organisms has most likely produced these findings?
A Clostridium botulinum
B Escherichia coli
C Neisseria gonorrhoeae
D Staphylococcus epidermidis
E Streptococcus pyogenes
E
The rash and edema are manifestations of streptococcal erysipelas, which is usually caused by group A or group C streptococci. Streptolysins elaborated by these organisms aid in the spread of the infection through subcutaneous tissues. Over 900 years ago the Order of St. Anthony was founded to treat persons with this illness, then known as St. Anthony's fire. Clostridium botulinum elaborates an exotoxin that, when ingested, results in paralysis. Escherichia coli produces various infections, but skin infections are uncommon. Neisseria gonorrhoeae is best known as a sexually transmitted disease, and a rash is possible, although usually there is no pronounced swelling. Staphylococcus epidermidis is usually considered a contaminant in cultures.
A 52-year-old man has a fever and cough productive of thick, gelatinous sputum that worsens over 4 days. On physical examination, his temperature is 38.2° C. On auscultation of the chest, diffuse crackles are heard at the right lung base. Laboratory studies show WBC count, 13,240/mm3 with 71% segmented neutrophils, 8% bands, 15% lymphocytes, and 6% monocytes. A sputum gram stain shows gram-negative bacilli with mucoid capsules. His condition improves after a course of gentamicin therapy. Which of the following complications of this infection is he most likely to develop?
A Abscess formation
B Adenocarcinoma
C Bullous emphysema
D Cavitary granulomas
E Gas gangrene
A
Bacterial infections with predominantly neutrophilic response are marked by suppurative inflammation, and a virulent organism such as Klebsiella can lead to tissue destruction with abscess formation. Carcinomas are not sequelae of bacterial infections. Infections of the lung do not result in emphysema, but may complicate emphysema. Granulomatous inflammation is characteristic of mycobacterial or fungal infections. Gas-forming bacteria, such as anaerobic Clostridium organisms, are unusual as a cause of respiratory infections.
A 20-year-old woman has had increasing delirium for 2 days and is admitted to the hospital. On physical examination, she has acute pharyngitis with an overlying dirty-white, tough mucosal membrane. Paresthesias with decreased vibratory sensation are present in the extremities. On auscultation, there is an irregular cardiac rhythm. A chest radiograph shows cardiomegaly. A Gram stain of the pharyngeal membrane shows numerous small, gram-positive rods within a fibrinopurulent exudate. Which of the following is the most likely mechanism for development of cardiac disease in this patient?
A Exotoxin-induced cell injury
B Granulomatous inflammation
C Lipopolysaccharide-mediated hypotension
D Microabscess formation
E Vasculitis with thrombosis
A
This woman has diphtheria. The Corynebacterium diphtheriae organisms proliferate in the inflammatory membrane that covers the pharynx and tonsils. These gram- positive organisms elaborate an exotoxin that circulates and produces myocarditis and neuropathy. The organisms do not disseminate to cause inflammation, abscesses or vasculitis elsewhere in the body. Granulomatous inflammation is more typical of mycobacterial and fungal infections. Endotoxins such as lipopolysaccharide tend to be elaborated by gram-negative bacterial organisms.
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A 33-year-old primigravida at 18 weeks' gestation develops nausea with vomiting for 3 days and then a severe headache and neck stiffness. On physical examination, her temperature is 38.2° C. She has no papilledema. A lumbar puncture is performed, and a Gram stain of the CSF obtained shows many short, gram-positive rods, and with a wet mount, the organisms demonstrate tumbling motility. In culture, this organism grows at 25° C. She most likely acquired this illness through which of the following mechanisms?
A Ingestion of contaminated food
B Inhalation of droplet nuclei
C Inoculation through a cut on the skin
D Sharing infected needles
E Using a friend's toothbrush
A
The results of the Gram stain and culture are diagnostic for Listeria monocytogenes, an organism that is more likely to produce disseminated disease in individuals who are immunocompromised or pregnant, and it can produce a congenital infection. Since the organism grows readily at room temperature, it easily contaminates food and water. Unpasteurized dairy products are most often implicated. Listeriosis is not known to be acquired parenterally, or by the other listed routes.
When Pharaoh did not heed Moses and let the Hebrews go, a series of plagues fell on Egypt. In the fifth plague, large domesticated mammals, including cattle, horses, and sheep, died. This was followed by a plague in which the Egyptians developed cutaneous boils that probably appeared as 1-cm to 5-cm areas of erythema with central necrosis forming an eschar. Some Egyptians also may have developed a mild, non- productive cough associated with fatigue, myalgia, and low- grade fever over 72 hours, followed by a rapid onset of severe dyspnea with diaphoresis and cyanosis. Vital signs might have included temperature of 39.5° C, pulse of 105/min, respirations of 25/min, and blood pressure of 85/45 mm Hg. On auscultation of the chest, crackles would be heard at the lung bases. A chest radiograph would show a widened mediastinum and small pleural effusions. "Legacy" laboratory findings would include a CBC with WBC count of 13,130/mm3, hemoglobin of 13.7 g/dL, hematocrit of 41.2%, MCV of 91 μm3, and platelet count of 244,000/mm3. Despite antibiotic therapy with anachronistic ciprofloxacin and doxycycline, many of those affected would die. Which of the following organisms is most likely to produce these findings?
A Bacillus anthracis
B Herpes simplex virus
C Mycobacterium leprae
D Staphylococcus aureus
E Variola major
F Yersinia pestis
A
The features are those of cutaneous and respiratory anthrax. Bacillus anthracis forms spores that resist environ- mental degradation. The spores can be transmitted by aerosols, making this organism an ideal terror weapon. Similar to many gram-positive organisms, B. anthracis produces disease via elaboration of exotoxins that have an active A subunit and a binding B subunit. None of the other choices involve outbreaks in domestic animals. Herpetic infections form clear vesicles that can rupture to shallow ulcers. Mycobacterium leprae can produce a faint rash early in its course, but involvement of peripheral nerves with loss of sensation predisposes to repeated trauma with deformity. Staphylococcus aureus can produce impetigo, typically on the face and hands. Variola major is the agent for smallpox, which is characterized by skin pustules, and pneumonia is the most likely cause of death. Yersinia pestis produces plague, which can have bubonic and pneumonic forms, characterized by ulcerating lymph nodes surrounded by a rosy rash.
A 42-year-old HIV-positive man has had a fever and cough for the past month. On physical examination, his temperature is 37.5° C. On auscultation of the chest, decreased breath sounds are heard over the right posterior lung. A chest radiograph shows a large area of consolidation with a central air-fluid level involving the right middle lobe. A transbronchial biopsy specimen contains gram-positive filamentous organisms that are weakly acid-fast. His course is complicated further by empyema and acute onset of a headache. A head CT scan shows a 4-cm discrete lesion of the right hemisphere with ring enhancement. Which of the following infectious agents is most likely causing his disease?
A Aspergillus fumigatus
B Mucor circinelloides
C Mycobacterium avium complex
D Nocardia asteroides
E Staphylococcus aureus
D
Although nocardiosis typically begins in the lungs, it often becomes disseminated, particularly to the central nervous system. These infections are most often seen in immunocompromised patients. Aspergillosis also can affect immunocompromised individuals, particularly those with neutropenia, but the fungal hyphae are easily distinguishable on hematoxylin and eosin stains. Mucor organisms have broad, nonseptate hyphae and are seen most often in patients with diabetic ketoacidosis or burn injuries. Mycobacterium avium complex infections are seen in individuals with AIDS, but these are short, acid-fast rods that produce poorly formed granulomas. Bacterial pneumonias also should be considered in immunocompromised patients, and septicemia can complicate them, but Staphylococcus aureus organisms form clusters of gram-positive cocci.
A 50-year-old man with a neurodegenerative disease has had a fever and cough productive of yellow sputum for the past 3 days. On physical examination, there is dullness to percussion at the left lung base. A chest radiograph shows areas of consolidation in the left lower lobe. Despite antibiotic therapy, the course of the disease is complicated by abscess formation, and he dies. At autopsy, there is a bronchopleural fistula surrounded by a pronounced fibroblastic reaction. Small, yellow, 1-cm to 2-mm "sulfur granules" are grossly visible within the area of abscess formation. Which of the following organisms is most likely to produce these findings?
A Actinomyces israelii
B Blastomyces dermatitidis
C Chlamydophila pneumoniae
D Klebsiella pneumoniae
E Mycobacterium kansasii
A
Actinomycetes that can produce chronic abscessing pneumonia, particularly in immunocompromised patients, include Actinomyces israelii and Nocardia asteroides. Persons with neurodegenerative diseases are at risk for aspiration of oropharyngeal secretions that may contain these organisms. Sulfur granules, formed from masses of the branching, filamentous organisms, are more likely to be seen in Actinomyces. Blastomyces dermatitidis infections tend to produce a granulomatous inflammatory process. Chlamydial infections produce an interstitial pattern similar to that of most viruses. Klebsiella infections, similar to other bacterial infections, can result in abscess formation, although without distinct sulfur granules. Mycobacterium kansasii infections are similar to Mycobacterium tuberculosis infections in that granulomatous inflammation is prominent.
A 22-year-old man has had more than 6 episodes of urethritis in the past 4 years since becoming sexually active. Each time, gram-negative cocci are identified in the neutrophilic exudate. Which of the following components in these organisms undergo change that prevents development of lasting protective immunity?
A Chitin
B Envelope
C Lipopolysaccharide
D Peptidoglycan
E Pili
F Teichoic acid
E
Pili are cell wall structures in gram-negative bacteria, such as the Neisseria gonorrhoeae in this case, that facilitate attachment to host cells. Pili proteins are altered by genetic recombination, forming a "moving target" for host immunity, so reinfection can occur. Chitin is a prominent cell wall component of fungi. Envelopes aid attachment of viruses to their target host cells. Lipopolysaccharide in gram-negative bacterial cell walls acts as an endotoxin. Peptidoglycan forms part of the bacterial cell wall, and a greater amount of it imparts gram-positive staining. Teichoic acid is a prominent feature of gram-positive bacterial cell walls.
A 25-year-old woman has had pelvic pain, fever, and vaginal discharge for 3 weeks. On physical examination, she has lower abdominal adnexal tenderness and a painful, swollen left knee. Laboratory studies show WBC count of 11,875/ mm3 with 68% segmented neutrophils, 8% bands, 18% lymphocytes, and 6% monocytes. She receives ceftriaxone therapy, but is not adherent with this therapy. She undergoes a work-up for infertility 5 years later. Which of the following infectious agents is most likely to produce these findings?
A Candida albicans
B Gardnerella vaginalis
C Herpes simplex virus-2
D Neisseria gonorrhoeae
E Treponema pallidum
F Trichomonas vaginalis
D
This patient has pelvic inflammatory disease (PID), which may occur as a result of infection with Neisseria gonorrhoeae or Chlamydia trachomatis. Both organisms cause sexually transmitted diseases, and chronic inflammation may lead to PID. Complications of PID include peritonitis, adhesions with bowel obstruction, and sepsis with endocarditis, meningitis, arthritis, and infertility. Of the remaining organisms listed, Candida can produce vaginitis with a curd-like discharge, but it does not typically produce PID. Gardnerella produces a whitish discharge that has a "fishy" odor with bacterial vaginosis, which tends to remain localized. Herpes simplex virus-2 (HSV-2), the most common agent of genital herpes, can produce painful vesicles, usually on the external genitalia, and is often recurrent. Treponema pallidum, the causative agent of syphilis, produces a hard chancre on skin and mucosal surfaces. Trichomoniasis may also lead to infertility, but this protozoan is not treated with cephalosporins, and it generally does not produce disseminated disease.
A 4-year-old child develops a runny nose and cough. After the cough persists for 2 weeks she exhibits paroxysms of coughing so severe she becomes cyanotic. On physical examination, her temperature is 37.4° C. Her mouth and pharynx reveal no erythema or swelling. On auscultation of the chest, her lungs show crackles bilaterally. She has spasmodic coughing, with a series of coughs on a single breath, bringing up mucus plugs, followed by labored inspiration. The pathogenesis of her disease most likely results from disabling of which of the following?
A Ciliary movement
B Complement lysis
C Immunoglobulin secretion
D NK cell activation
E Phagolysosome formation
A
Bordetella pertussis is the causative agent for whooping cough. These infections occur infrequently when there is widespread childhood vaccination against this organism. This coccobacillary organism is difficult to culture, and direct fluorescent antibody (DFA) testing is the fastest and most reliable way to diagnose the infection. Nasopharyngeal aspirates and swabs are the best specimens because the organisms attach to ciliated respiratory epithelium. The toxin paralyzes cilia. Complement lysis is most useful against circulating infectious agents. Immunoglobulins that circulate can bind organisms, but secretion is an adaptive immune response taking days to weeks. NK cells attack host cells with MHC signaling turned off by intracellular infectious agents such as viruses. Mycobacterial organisms inhibit phagolysosome formation to reduce their intracellular destruction in macrophages.
A 66-year-old man incurs extensive thermal burns to his skin and undergoes skin grafting procedures in the surgical intensive care unit. Two weeks later, he has increasing respiratory distress. Laboratory studies show hemoglobin, 13.1 g/ dL; hematocrit, 39.2%; platelet count, 222,200/mm3; and WBC count, 4520/mm3 with 15% segmented neutrophils, 3% bands, 67% lymphocytes, and 15% monocytes. A chest radiograph shows extensive bilateral infiltrates with patchy areas of consolidation. Bronchoscopy is performed, and microscopic examination of a transbronchial biopsy specimen shows pulmonary vasculitis and surrounding areas of necrosis with sparse inflammatory exudate. Which of the following infectious agents is most likely to produce these findings?
A Adenovirus
B Histoplasma capsulatum
C Mycobacterium tuberculosis
D Pseudomonas aeruginosa
E Pneumocystis jiroveci
F Streptococcus pneumoniae
D
Pseudomonas aeruginosa can infect the skin following burn injuries and spread to the lungs. These organisms secrete several virulence factors, as follows: exotoxin A, which inhibits protein synthesis; exoenzyme S, which interferes with host cell growth; phospholipase C, which degrades pulmonary surfactant; and iron-containing compounds, which are toxic to endothelial cells. These virulence factors result in extensive vasculitis with necrosis. Neutropenic patients are particularly at risk. Histoplasma capsulatum yeasts can produce pulmonary disease resembling that of Mycobacterium tuberculosis, with granulomatous inflammation. Pneumocystis pneumonia is more likely to occur in patients with weak cell-mediated immunity. Pneumococcal infections produce alveolar exudates without significant vascular involvement.
In October 1347, a Genoese trading ship returning from the Black Sea docked at Messina, Sicily. The ship's crew had been decimated by an illness marked by a short course of days from onset of inguinal lymph node enlargement with overlying skin ulceration to prostration and death. A small, ulcerated pustule ringed by a rosy rash was seen on the lower extremities of some of the crew. Within days, more than half of the population of the port city had died. Which of the following insect vectors was most likely responsible for the rapid spread of this disease?
A Fleas
B Mosquitoes
C Reduviid bugs
D Sand flies
E Ticks
A
This incident marks the first appearance of the Black Death in Europe, a disease that persisted during the 14th and 15th centuries. The plague spread through Italy and across the European continent. By the following spring, it had reached as far north as England, and within 5 years, it had killed 25 million people, one third of the European population. Rodents form the reservoir of infection (and cats weren't as popular as they should have been). Flea bites and aerosols transmit the infection very efficiently. The causative organism, Yersinia pestis, secretes a plasminogen activator that promotes its spread. Plague was endemic in East Asia at the beginning of the 20th century and was carried to San Francisco. Seeking to avoid a panic that could be bad for business and tourism, California's governor at the time did not enforce a quarantine. As a consequence, plague is endemic in wild rodents in the western United States, but it accounts for only occasional sporadic human infections. Mosquitoes are best known as vectors of malaria; sand flies, of leishmaniasis; reduviid (triatomid) bugs, of Chagas disease; and ticks, of Lyme disease.
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A study of sexually transmitted infections identifies an organism most commonly found in tropical and subtropical regions. This organism is associated with painful ulcerating genital papules in HIV-infected persons. Microscopic examination of lesional exudate with silver stain shows coccobacilli. Which of the following organisms is most likely to produce these findings?
A Chlamydia trachomatis
B Haemophilus ducreyi
C Klebsiella granulomatis
D Neisseria gonorrhoeae
E Treponema pallidum
B
The causative organism of chancroid is Haemophilus ducreyi, which is difficult to grow in culture and often obscured by superinfecting organisms in the ulcerated lesions. Chancroid is most common in Africa and Southeast Asia, is a co-factor in transmission of HIV, and its features overlap those of granuloma inguinale, but there is often lymph node involvement in the former, and lack of the Donovan bodies in macrophages. Lymphogranuloma venereum is caused by Chlamydia trachomatis, which cannot be seen with Gram stain. Klebsiella granulomatis (formerly Calymmatobacterium granulomatis) is also common in tropical and subtropical regions. It causes granuloma inguinale, and the infection may progress to scarring with urethral and lymphatic obstruction. Gonorrhea tends to produce a urethritis in men and a cervicitis in women acutely, without genital ulceration. Syphilis is marked by a chancre in the primary state and a maculopapular rash of palms and soles in the secondary stage, with a prominent lymphoplasmacytic infiltrate; the causative agent is Treponema pallidum, and these spirochetes cannot be identified by Gram stain.
A 23-year-old man from Irian Jaya, Indonesia, has a lesion on his penis that has enlarged over the past 4 months. On physical examination there is a painless 2-cm papular lesion of the dorsum of his penis that evolves into a beefy red expansile ulceration that bleeds easily. No inguinal lymphadenopathy is present. A biopsy of the lesion is taken and examined microscopically, showing pseudoepitheliomatous hyperplasia and mixed inflammatory infiltrate. Giemsa stain shows coccobacilli within vacuoles in macrophages. Social history reveals multiple sexual partners. Which of the following is the most likely diagnosis?
A Balanoposthitis
B Chancroid
C Granuloma inguinale
D Lymphogranuloma venereum
E Secondary syphilis
C
The causative organism is Klebsiella granulomatis (formerly Calymmatobacterium granulomatis), the disease is most common in tropical and subtropical regions, and the infection may progress to scarring with urethral and lymphatic obstruction. Balanoposthitis is localized inflammation of the glans penis and prepuce, typically caused by Candida, Gardnerella, or Staphylococcus spp. Chancroid caused by Haemophilus ducreyi has features that overlap those of granuloma inguinale, but there is often lymph node involvement in the former, and lack of the Donovan bodies in macrophages. Lymphogranuloma venereum is caused by Chlamydia trachomatis, which cannot be seen with Gram stain. Secondary syphilis is marked by a maculopapular rash on the palms and soles, with a prominent lymphoplasmacytic infiltrate; the causative agent is Treponema pallidum, and these spirochetes cannot be identified with Gram stain.
A 31-year-old man has had cough with a low-grade fever and a 4-kg weight loss over the course of 3 months. On physical examination, his temperature is 37.5° C. Laboratory studies show anemia of chronic disease. A bone marrow biopsy is performed, and the microscopic appearance is shown in the figure. An acid-fast stain of this tissue is positive. The causative infectious agent is most likely being destroyed by which of the following mechanisms?
A Complement-mediated lysis
B Elaboration of nitric oxide by macrophages
C Generation of NADPH-dependent oxygen free
radicals
D Phagocytosis by eosinophils
E Superoxide formation within phagolysosomes
B
The figure shows a granuloma. Activated macro- phages are the key cellular component within granulomas that form to control persistent organisms such as Mycobacterium tuberculosis. As part of delayed type hypersensitivity with a TH1 immune response, CD4+ cells secrete interferon-γ, which activates macrophages to kill organisms with reactive nitrogen intermediates. Complement-mediated lysis is not involved in the destruction of intracellular bacteria such as M. tuberculosis. Complement activation on the surface of M. tuberculosis can opsonize the bacteria, however, for uptake by macrophages. Eosinophils are not a major component of most granulomas, and they cannot destroy mycobacteria. NADPH- dependent reactive oxygen species are important in the lysis of bacteria by neutrophils. M. tuberculosis organisms reside in phagosomes, which are not acidified into phagolysosomes.
A 5-year-old child is exposed to Mycobacterium tuberculosis. A month later the child's tuberculin skin test is positive. The child then develops fever, inspiratory stridor, and non- productive cough. Which of the following findings is most likely to be present on the chest radiograph of this child?
A Hilar lymphadenopathy
B Miliary pulmonary nodules
C Pneumonic consolidation
D Upper lobe cavitation
E Vertebral lytic lesions
A
The child has primary tuberculosis. Most healthy persons have subclinical disease, and a minority develop clinical manifestations; of those, most have limited pulmonary involvement without dissemination. Primary tuberculosis is marked by the Ghon complex, which is a small subpleural granuloma at mid-lung along with prominent enlarged hilar lymph nodes. These nodes may impinge upon central airways. When the cell-mediated immune response is poor, then there can be numerous small granulomas scattered throughout the lungs, or disseminated to other organs, as a miliary pattern (granulomas that are the size of millet seeds). Progressive primary tuberculosis can lead to more extensive lung involvement with pneumonic infiltrates. Upper lobe cavitary disease is characteristic for secondary tuberculosis (reactivation or reinfection) in persons who have previously mounted an immune response. One pattern of disseminated tuberculosis is Pott disease of the spine, sometimes as an isolated finding.
A 41-year-old man has had worsening fever, cough, and dyspnea for 2 weeks. On examination, he has rales and diminished breath sounds on auscultation of his chest. A chest radiograph shows scattered infiltrates in both lungs. A tuberculin skin test shows 6 mm of induration. A sputum sample is negative, but bronchoalveolar lavage is positive, for acid-fast bacilli. His WBC count is 4600/mm3 with differential count of 80% neutrophils, 10% lymphocytes, and 10% monocytes. Which of the following is the most likely risk factor for his pulmonary disease?
A Alcohol abuse
B Diabetes mellitus
C HIV infection
D Scurvy
E Smoking
C
Anergy (less than the 10 mm of induration expected for a positive tuberculin test), sputum negativity despite extensive pulmonary disease, and radiographic evidence of infiltrates resembling bacterial pneumonic consolidation all point to a poor cell-mediated immune response. HIV infection depletes the body of the CD4+ lymphocytes (explaining his lymphopenia) essential for a TH1 immune response required to contain mycobacterial infection. The debilitation accompanying alcohol abuse is more likely to lead to typical secondary tuberculosis, but more florid. Diabetes mellitus predisposes to bacterial infections, but pulmonary disease is not characteristic for diabetic complications. Scurvy may affect connective tissues but not lung specifically. Smoking diminishes pulmonary innate immune defenses, mainly against bacterial pathogens.
A 32-year-old man has maculopapular and nodular skin lesions, mainly involving his face, elbows, wrists, and knees. The nodular lesions have slowly enlarged over the past 10 years and are now beginning to cause deformity. The lesions are not painful, but he has hypoesthesia to anesthesia in these areas. The figure shows a microscopic acid-fast stain of a biopsy specimen of a nodular skin lesion. Which of the following is the most likely diagnosis?
A Anthrax
B Chagas disease
C Hansen disease
D Leishmaniasis
E Lyme disease
F Onchocerciasis
C
Hansen disease (leprosy) is caused by the small, acid-fast organism Mycobacterium leprae, which chronically infects peripheral nerves and skin. This organism cannot be cultured in artificial media. Diagnosis is made by biopsy of a skin lesion. There are two polar forms of leprosy. In the tuberculoid form, a delayed type of hypersensitivity reaction, with a TH1 immune response driven by interferon-γ and interleukin-2 (IL-2) cytokines, gives rise to granulomatous lesions that resemble tuberculosis; acid-fast bacilli are rare in such lesions. In contrast, in the lepromatous form, shown in the figure, T cell immunity is markedly impaired, a TH2 immune response is driven by IL-4 and IL-10, and granulomas are poorly formed. Instead, there are large aggregates of lipid-filled macrophages that are stuffed with acid-fast bacilli. Leprosy is poorly transmissible through aerosols (not from direct contact); it probably requires some genetic susceptibility, such as genetic variations in IL-10 and Toll-like receptors; and, similar to most diseases throughout human history, is linked to poverty. Cutaneous anthrax, caused by Bacillus anthracis, produces a necrotic skin lesion with eschar at the site of inoculation. The reduviid (triatomid) bug carries Trypanosoma cruzi, which causes Chagas disease. Its bite may cause a localized area of skin erythema and swelling. Mucocutaneous ulcers may be seen with Leishmania braziliensis infection, which is transmitted via sand flies. The area of the tick bite that introduces Borrelia burgdorferi spi- rochetes, the cause of Lyme disease, may manifest erythema chronicum migrans. Onchocerciasis occurs as a result of infection with the filarial nematode Onchocerca volvulus and leads to formation of a subcutaneous nodule.
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A 20-year-old man who has multiple sexual partners and does not use barrier precautions has had a nontender ulcer on his penis for the past week. On physical examination, the 0.6-cm lesion has a firm, erythematous base and sharply demarcated borders. The lesion is scraped, and microscopic darkfield examination is positive for motile spirochetes. Which of the following inflammatory processes is most likely to accompany this infection?
A Acute inflammation with abscess formation
B Granulomatous inflammation with caseation
C Gummatous inflammation with necrosis
D Perivascular inflammation with plasma cells
D
Infection with Treponema pallidum can lead to syphilitic chancres in the primary stage of syphilis. The chancres are characterized by lymphoplasmacytic infiltrates and by an obliterative endarteritis. Similar lesions also may appear with secondary syphilitic mucocutaneous lesions. Acute inflammation with abscess formation is characteristic of bacterial infections such as gonorrhea. Caseating granulomatous inflammation is more characteristic of tuberculosis or fungal infections. Gummatous inflammation can be seen in adults with tertiary syphilis or in congenital syphilis.
A longitudinal study of men and women who have developed aortic root dilation and aortic insufficiency in adulthood is performed. They have a history of unprotected sexual intercourse with multiple partners. Which of the following laboratory tests is most likely to yield a positive result in these persons?
A Blood culture
B Darkfield microscopy of lymph node
C Fluorescent treponemal antibody-absorption (FTA)
D Rapid plasma reagin (RPR)
E Venereal disease research laboratory (VDRL)
C
Untreated infection with Treponema pallidum can lead to tertiary syphilis years later. The most common manifestations of tertiary syphilis include aortitis (typically in the thoracic portion), neurosyphilis, and gummatous necrosis of skin, soft tissue, bone, and joint (Charcot joint). This organ- ism cannot be cultured. The spirochetes are best identified by darkfield microscopy in exudates from primary chancres, but the organisms are hard to find in the tertiary stage of the disease. Serologic testing is useful for screening and confirmation of syphilis. The nontreponemal tests (RPR, VDRL) are sensitive to a cardiolipin found in the more numerous spirochetes earlier in the disease; but these tests are not specific because the presence of cardiolipin in human tissues is associated with other diseases, causing false-positive results. The FTA test has specificity for T. pallidum.
An infant born at term to a 33-year-old woman is severely hydropic. On physical examination, there is a diffuse rash with sloughing skin on the palms and soles. Within 2 days, the infant dies of respiratory distress. At autopsy, there is marked hepatosplenomegaly. Microscopic examination of the femur and vertebrae shows periosteitis and osteochondritis. The lungs have nodular masses with central necrosis surrounded by mononuclear leukocytes, palisading macrophages, and fibroblasts. A serologic test result for which of the following agents is most likely to be positive in the infant's mother?
A Cytomegalovirus
B Herpes simplex type 2
C HIV
D Syphilis
E Toxoplasma gondii
D
These are findings of congenital syphilis with nodules of gummatous necrosis. Because the spirochetes can cross the placenta in the third trimester, early stillbirths do not occur. Infants who survive have features similar to adult secondary syphilis, with rash. With survival, late complications of the periosteitis and perichondritis include bone and teeth deformities (e.g., saber shin). Herpes infections in the neonate usually are not initially obvious because most of these infections are acquired by passage through the birth canal. Most infants born with HIV infection have no initial gross or microscopic pathologic findings. Congenital toxoplasmosis and cytomegalovirus produce severe cerebral disease.
A 44-year-old woman notices an erythematous papule on her left lower leg that develops into a ring-like rash and then subsides over 3 weeks. Over the next 5 months, she has migratory joint and muscle pain, substernal chest pain, and an irregular heart rhythm. These problems subside, but 2 years after the initial rash appeared, she develops a chronic arthritis involving the hips, knees, and shoulders. Which of the following is the most likely diagnosis?
A Chagas disease
B Dengue fever
C Leishmaniasis
D Leprosy
E Lyme disease
F Syphilis
E
The acute stage of Lyme disease is marked by the appearance of erythema chronicum migrans of the skin. As the Borrelia burgdorferi organisms proliferate and disseminate, systemic manifestations of carditis, meningitis, and migratory arthralgias and myalgias appear. Arthritis involving the large joints occurs 2 to 3 years after initial infection. Rheumatoid arthritis can mimic Lyme disease but is not preceded by the skin lesions described. Chagas disease may be associated with acute and chronic myocarditis leading to heart failure; some patients have esophageal involvement, but arthritis and rash are not features of the disease. Hemorrhagic fever, or dengue fever, caused by an arbovirus, can produce myositis and bone marrow suppression. Mucocutaneous ulcers may be seen with Leishmania braziliensis. Leprosy, or Hansen disease, is associated with skin anesthesia and granuloma formation with nodular deformities of the skin. In primary syphilis, a hard chancre may be present at the site of inoculation (usually the external genitalia; less often in the oral cavity or anorectal region). In secondary syphilis, a maculopapular rash may be present.
A radical group commissions scientists to develop a Category A bioterrorism agent. They want an agent that will paralyze victims within hours and be disguised within innocuous-appearing cans of split pea soup. Which of the following organisms best meets the requirements stated?
A Chlamydia psittaci
B Clostridium botulinum
C Ebola virus
D Hantavirus
E Yersinia pestis
B
The spores of Clostridium botulinum will survive the canning process in such non-acidic foods as peas when they are not heated sufficiently, so that organisms grow and elaborate a neurotoxin. However, the plot fails when everyone prefers to eat junk food and not vegetables. Perhaps the terrorists should promote trans fats, which, when combined with lack of exercise, will increase morbidity and mortality from atherogenesis to a greater extent than any infectious agent. Chlamydia psittaci is a Category B agent that is airborne and causes pneumonia. Ebola virus produces a hemorrhagic fever. Hantavirus spread through aerosolization of deer mouse droppings causes pneumonia and sepsis. Yersinia pestis is the Black Death, which produces lymphadenitis, pneumonia, and sepsis; the vector is the rat flea.
A 25-year-old soldier incurs multiple skin wounds that get infected and produce extensive tissue damage within a day. Culture of necrotic tissue from deep inside one of the wounds reveals anaerobic spore-forming gram-positive rods. Which of the following microscopic pathologic reactions are the toxins produced by these organisms most likely to cause?
A Abscess formation
B Cytopathic effects with apoptosis
C Fibrous scarring
D Gangrenous necrosis
E Granulomatous inflammation
F Lymphocytic infiltrates
D
Clostridia such as Clostridium perfringens represent one type of gram-positive rod like bacteria that produce powerful exotoxins, causing extensive tissue necrosis so quickly that the acute inflammatory response lags. Abscesses are formed of neutrophils responding to the inflammatory agent, often a bacterial organism, but the liquefactive necrosis is mainly produced by enzymes released from the neutrophils. Fibrous scarring can certainly be part of the healing phase of inflammatory responses, but is less prominent with bacterial infections than with agents producing more chronic inflammation. Granulomatous inflammation typically develops in weeks to months from persistent infection from agents such as mycobacteria. Lymphocytic infiltrates are most typical for chronic and viral infections, and there tends to be minimal necrosis.
A 43-year-old man cuts the skin over his shin while repairing a fence on his farm. The wound heals without any complications. Four days later, he develops muscle spasms of the face and extremities. These spasms worsen to the point of severe contractions. Which of the following actions of the microbial toxins is most likely responsible for the clinical features in this case?
A Cleavage of synaptobrevin in synaptic vesicles of neurons
B Degradation of muscle cell membranes by phospholipase C
C Inhibition of acetylcholine release at neuromuscular junctions
D Release of cytokine by T lymphocytes
E Stimulation of adenylate cyclase production in
myofibers
A
This man has tetanus. The contamination of a wound with Clostridium tetani can result in the elaboration of a potent neurotoxin. This toxin is a protease that cleaves synaptobrevin, a major transmembrane protein of the synaptic vesicles in inhibitory neurons. Clostridium perfringens elaborates a variety of toxins, one of which (alpha) is a phospholipase causing myonecrosis. Inhibition of acetylcholine release is not a feature of infection. The toxin of Staphylococcus aureus is an enterotoxin that acts as a superantigen and stimulates T cell cytokine release. Cholera is produced when the toxin elaborated by Vibrio cholerae stimulates epithelial cell adenylate cyclase.
A 27-year-old man is involved in a rollover accident in which he is ejected from the vehicle. He sustains a compound fracture of the right humerus and undergoes open reduction with internal fixation of the humeral fracture. Three days later, he has marked swelling of the right arm and palpable crepitus. A Gram stain of necrotic exudate from the wound site has the appearance shown in the figure. Through which of the following mechanisms is this organism most likely causing extensive tissue necrosis?
A Elaboration of lipopolysaccharide
B Inhibition of phagocytic cell function
C Resistance to multiple antibiotics
D Superinfection with Candida albicans
E Toxin-mediated lecithin degradation
E
The large, gram-positive rods seen in the figure are characteristic of Clostridium perfringens, which can contaminate open wounds and produce gas gangrene. Clostridial organisms can elaborate multiple toxins. C. perfringens alpha-toxin acts as a phospholipase C that degrades lecithin in cellular membranes. Lipopolysaccharides are found in gram-negative organisms. Inhibition of phagocytes is a feature of organisms such as Mycobacterium tuberculosis. Antibiotic resistance is increasing in frequency, but is not the main mechanism for clostridial virulence. Though devitalized tissues can have polymicrobial infection, Candida is typically not the most virulent among superinfecting agents.
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A 24-year-old man who is sexually active with multiple partners has had pain during urination for the past 4 days. On physical examination, there are no lesions on the penis. He is afebrile. Urinalysis shows no blood, ketones, protein, or glucose. Microscopic examination of the urine shows few WBCs and no casts or crystals. What infectious agent is most likely to produce these findings?
A Candida albicans
B Chlamydia trachomatis
C Herpes simplex virus
D Mycobacterium tuberculosis
E Treponema pallidum
B
The most common cause of nongonococcal urethritis in men is Chlamydia trachomatis. The condition is a nuisance in men without significant sequelae; however, the behavior that led to the infection can place the patient at risk for other sexually transmitted diseases. Candida infections typically occur in immunocompromised patients or in patients receiving long-term antibiotic therapy. Herpes simplex can produce painful vesicles on the skin. Tuberculosis of the urinary tract is uncommon. A syphilitic chancre on the penis, not present here, is an indicator of Treponema pallidum infection.
A 50-year-old woman residing in Port-au-Prince has observed a small vesicle on her right labium majus for the past 4 days. She is sexually active. On physical examination, the 0.5-cm vesicle is filled with purulent exudate. Tender inguinal lymph nodes are palpable. She was diagnosed with non-Hodgkin lymphoma 10 years ago. A biopsy of one of the lymph nodes is performed and microscopically shows multiple abscesses in which central necrosis is surrounded by palisading histiocytes. These clinical and pathologic findings are most likely caused by which of the following conditions?
A Candida albicans vaginitis
B Chlamydia trachomatis cervicitis
C Gardnerella vaginalis vaginosis
D Herpes simplex virus infection of the perineum
E Treponema pallidum infection of the external
genitalia
B
Infection with Chlamydia trachomatis is a common sexually transmitted disease. Most cases produce only urethritis and cervicitis; however, some strains of C. trachomatis can produce lymphogranuloma venereum, a chronic ulcerative disease that is more endemic in Asia, Africa, and the Caribbean. In this disease, there is a mixed granulomatous and neutrophilic inflammatory reaction, as seen in this patient. In contrast, herpes simplex virus produces clear mucocutaneous vesicles with no exudates and is unlikely to involve lymph nodes. Candidiasis can produce superficial inflammation with an exudate, but it is rarely invasive or disseminated in non-immunosuppressed individuals. Bacterial vaginosis due to Gardnerella produces a whitish discharge that has a "fishy" odor. Treponema pallidum, the causative agent of syphilis, produces a hard chancre on skin and mucosal surfaces.
A 15-year-old boy has developed a small eschar on his left forearm around the site of a tick bite he received 6 days ago. A hemorrhagic rash involving the trunk, extremities, palms, and soles then develops over the next 3 days. Over the past day, small, 0.2-cm to 0.4-cm foci of skin necrosis have developed on his fingers and toes. His temperature is 39° C. He is treated with doxycycline and improves over the next 2 weeks. Which of the following organisms is most likely to produce these findings?
A Borrelia burgdorferi
B Leishmania braziliensis
C Mycobacterium leprae
D Rickettsia rickettsii
E Yersinia pestis
D
This patient has Rocky Mountain spotted fever, which occurs sporadically in the United States, mostly in areas other than the Rocky Mountains. Rickettsial diseases produce signs and symptoms from damage to vascular endothelium and smooth muscle similar to a vasculitis. The most common vector is the wood tick Dermacentor andersoni. Thrombosis of the affected blood vessels is responsible for foci of skin necrosis. Headache and abdominal pain are often prominent. Lyme disease, caused by Borrelia burgdorferi, can produce an erythema chronicum migrans of skin at the site of a tick bite. Mucocutaneous leishmaniasis mainly involves the nasal and oral regions. Hansen disease (leprosy), produced by Mycobacterium leprae, results in skin anesthesia that predisposes to recurrent injury. Plague, caused by Yersinia pestis, can produce focal skin necrosis at the site of a flea bite, and ulceration over infected lymph nodes (bubos).
A 10-year-old girl with leukemia undergoes hematopoietic stem cell transplantation. She has poor engraftment, and 1 month later she develops fever and dyspnea. On physical examination, her temperature is 39° C. On auscultation of the chest, wheezes and crackles are heard in both lungs. A chest CT scan shows nodular lesions accompanied by cavitation, hemorrhage, and infarction. Laboratory studies show hemoglobin, 8.8 g/dL; hematocrit, 26.5%; platelet count, 91,540/ mm3; and WBC count, 1910/mm3 with 10% segmented neutrophils, 2% band neutrophils, 74% lymphocytes, and 14% monocytes. A bronchoalveolar lavage is performed; the fluid was stained with Gomori methenamine silver stain and analyzed microscopically, as shown in the figure. Which of the following infectious agents is most likely to produce these findings?
A Aspergillus fumigatus
B Coccidioides immitis
C Corynebacterium diphtheriae
D Histoplasma capsulatum
E Mycobacterium tuberculosis
F Pneumocystis jiroveci
A
Aspergillus, Candida, and Mucor infections may become disseminated in the setting of neutropenia. Vascular invasion can occur with fungal infections, particularly with Aspergillus and Mucor. The branching septate hyphae are shown in the figure projecting from a fruiting body of Aspergillus. After these organisms gain a foothold (hyphae-hold) in tissues, they are very difficult to eradicate. Coccidioides immitis and Histoplasma capsulatum are fungi that can produce pulmonary disease resembling that of Mycobacterium tuberculosis, with granulomatous inflammation. They do not have a propensity for vascular invasion. Corynebacterium diphtheriae produces upper respiratory tract disease, mainly in children who are not vaccinated against it. Pneumocystis pneumonia is not typically accompanied by vascular changes.
A 24-year-old woman has noted worsening pain on the right side of her face for the past 24 hours. On examination, there is marked tenderness and swelling inferior to the zygomatic arch and lateral to the nasolabial fold on the right. Laboratory studies show WBC count, 9900/mm3; serum creatinine, 2 mg/dL; sodium, 151 mmol/L; potassium, 5.4 mmol/L; chloride, 119 mmol/L; bicarbonate, 8 mmol/L; and glucose, 483 mg/dL. A head CT scan shows soft tissue swelling and bony destruction around the right maxillary sinus. A biopsy is performed; the figure shows the findings on microscopic examination. Which of the following organisms is the most likely causative agent for this patient's infection?
A Aspergillus niger
B Actinomyces israelii
C Candida albicans
D Clostridium perfringens
E Cryptococcus neoformans
F Mucor circinelloides
F
This patient is in diabetic ketoacidosis, which is a significant risk factor for mucormycosis. Note the broad, nonseptated hyphae more easily visible with H&E stain than special stains, unlike other fungi. In contrast, Aspergillus organisms have thinner hyphae with acute angle branching and septations. Actinomyces organisms are long, filamentous gram-positive bacilli. Candida infections are typically superficial and have gram-positive budding cells with pseudohyphae. Large, gram-positive rods are characteristic of Clostridium perfringens, which can contaminate open wounds and produce gas gangrene.
An 11-year-old boy from Liberia has had episodic fevers for 2 weeks. He developed a severe headache a week ago and has become progressively more somnolent. On funduscopic examination, he has papilledema. The representative microscopic appearance of a cerebral vein is shown. Which of the following organs is most likely to serve as the reservoir for proliferation of the infectious agent producing this disease?
A Brain
B Heart
C Liver
D Lymph nodes
E Spleen
C
This boy had cerebral malaria, the worst form of malaria. After the infective mosquito bite, Plasmodium falciparum sporozoites invade liver cells and reproduce asexually. When the hepatocytes rupture, they release thousands of merozoites that infect RBCs. The infected RBCs circulate and can bind to endothelium in the brain. Small cerebral vessels become plugged with the RBCs, resulting in ischemia. The other listed options also could be secondarily involved by vascular thromboses in the setting of malaria, but are not extraerythrocytic sites for asexual reproduction.
A 19-year-old woman goes on a camping trip to a wooded area in New England (USA) with lots of insects, but has forgot- ten to bring insect repellant. A month later, she has increasing malaise, low-grade fever, headaches, and myalgias. On physical examination, she has hepatosplenomegaly. Laboratory studies show hemoglobin, 10.4 g/dL; WBC count, 5820/ mm3; and platelet count, 205,000/mm3. Her peripheral blood smear is shown in the figure. Which of the following infectious agents is most likely to produce these findings?
A Babesia microti
B Borrelia burgdorferi
C Giardia lamblia
D Rickettsia rickettsii
E Wuchereria bancrofti
A
This patient's travel history suggests an insect-borne disease. The figure shows the characteristic tetrad and ring forms of Babesia microti, within erythrocytes. Babesiosis is, an uncommon malaria-like protozoan disease. The northeastern United States is an endemic area. The vector is the deer tick, just as with Lyme disease from Borrelia burgdorferi, which is a spirochete. Giardiasis typically produces self-limited, watery diarrhea. Rickettsia rickettsii causes Rocky Mountain spotted fever, which occurs sporadically in the United States in areas other than the Rocky Mountains and produces signs and symptoms from damage to vascular endothelium and smooth muscle similar to a vasculitis. Wuchereria bancrofti is a form of filariasis that can cause elephantiasis, owing to lymphatic obstruction in the presence of an inflammatory reaction to the adult filarial worms.
A 45-year-old man experiences malaise and fatigue, which slowly become more noticeable over a 2-month period. He returned from a vacation along the Costa del Sol near Barcelona 10 months ago. He now has occasional diarrhea and a low-grade fever. His abdominal discomfort worsens over the next month. On physical examination, his vital signs include temperature of 38.3° C. He has pronounced splenomegaly, an increased liver span, and generalized lymphadenopathy. Laboratory studies show hemoglobin, 11.8 g/dL; hematocrit, 34.9%; platelet count, 89,000/mm3; and WBC count, 3350/mm3 with 29% segmented neutrophils, 5% bands, 48% lymphocytes, and 18% monocytes. His total serum protein is 7.6 g/dL, albumin is 3.2 g/dL, AST is 67 U/L, ALT is 51 U/L, alkaline phosphatase is 190 U/L, and total bilirubin is 1.3 mg/dL. A stool sample is negative for occult blood. Which of the following is the most likely diagnosis?
A Borreliosis
B Echinococcosis
C Leishmaniasis
D Lyme disease
E Schistosomiasis
F Typhus
C
Visceral leishmaniasis (kala-azar) is caused by protozoa in the Leishmania donovani complex. Of these, only L. donovani infantum is endemic to southern Europe and the Mediterranean area. It is transmitted to humans by the sand fly (Phlebotomus). Pancytopenia implies bone marrow involvement, possibly enhanced by the enlarged spleen, and the liver function abnormalities suggest liver involvement. Borreliosis causes relapsing fever and is transmitted via body lice. Echinococcal disease is caused by ingestion of tapeworm eggs and can lead to cyst formation in visceral organs. Borrelia burgdorferi infection is transmitted via ticks and can cause Lyme disease, characterized by erythema chronicum migrans, meningoencephalitis, and chronic arthritis. Schistosomiasis, which is transmitted via snails, can produce hepatic cirrhosis (Schistosoma mansoni or Schistosoma japonicum) or bladder disease (Schistosoma haematobium). Typhus is a louse-borne rickettsial disease with skin rash that may proceed to skin necrosis.
A 24-year-old soldier stationed in the Middle East has noted the appearance of a 0.5-cm papule on his left forearm. It becomes a 1-cm nodule with a central depression, and then ulcerates over the next month. On physical examination, the 2-cm ulcerated lesion has an indurated border, and there are three smaller satellite lesions. There is no hepatosplenomegaly, but he has left axillary lymphadenopathy. Laboratory studies show hemoglobin, 14.1 g/dL; hematocrit, 42.5%; platelet count, 233,200/mm3; and WBC count, 6270/mm3. Which of the following infectious organisms is most likely to produce these findings?
A Borrelia recurrentis
B Brugia malayi
C Leishmania major
D Listeria monocytogenes
E Mycobacterium leprae
F Trypanosoma gambiense
C
This patient has cutaneous leishmaniasis, and the original papule was at the site of the sand fly vector bite. Leishmaniasis is endemic in the Middle East, South Asia, Africa, and Latin America. The organisms proliferate within macrophages in the mononuclear phagocyte system and can cause regional lymphadenopathy. The cutaneous form does not have bone marrow involvement and splenic enlargement, so pancytopenia is not present. Borreliosis causes relapsing fever and is transmitted via body lice. Brugia malayi is a nematode transmitted by mosquitoes that leads to filariasis involving lymphatics to produce elephantiasis. Leishmania donovani is transmitted by sand flies and leads to infection of macrophages, which produces hepatosplenomegaly, lymphadenopathy, and bone marrow involvement with pancytopenia. Listeriosis is most often acquired via contaminated food or water. In most adults, it produces mild diarrheal illness, but in some adults and children, and in fetuses, it may produce meningitis or dissemination with microabscess (microgranuloma) formation. Mycobacterium leprae causes Hansen disease (leprosy), with infection of peripheral nerves and skin. In individuals with a strong immune response, the tuberculoid form of this disease results in granuloma formation; in individuals with a weak immune response, the lepromatous form occurs, characterized by large numbers of macrophages filled with short, thin, acid-fast bacilli. African trypanosomiasis produces sleeping sickness.
A 22-year-old man with extensive travel history is bitten by an insect and has developed a rubbery, red, 1-cm chancre on his right forearm over the past week. Three months later, he develops splenomegaly and lymphadenopathy. Two months later, he exhibits progressive wasting with cachexia and decreased mentation. His peripheral blood smear has the appearance shown in the figure. Where is his disease most likely to have been acquired?
A Central America
B Polynesia
C Southeast Asia
D Southern Europe
E West Africa
E
The findings are consistent with African trypanosomiasis, or sleeping sickness. The eradication of the tsetse fly vector has been a priority for decades in many African countries. Filarial worms endemic in parts of Central America, Southeast Asia, and Polynesia also can appear in blood, but are smaller in size and do not lead to chronic wasting. Filariasis is not endemic in Europe.
A 9-year-old child who is living in a mud hut in Paraguay has a sore persisting on her face for 4 days. Physical examination shows an indurated area of erythema and swelling just lateral to the left eye, accompanied by posterior cervical lymph- adenopathy. She has unilateral painless edema of the palpebrae and periocular tissues. Two days later, she has malaise, fever, anorexia, and edema of the face and lower extremities. On physical examination 1 week later, there is hepatosplenomegaly and generalized lymphadenopathy. Which of the following pathologic findings is most likely to develop in this patient?
A Cerebral abscesses
B Chronic arthritis
C Dilated cardiomyopathy
D Meningitis
E Mucocutaneous ulcers
F Paranasal bony destruction
C
This child is infected with Trypanosoma cruzi, resulting in Chagas disease, endemic to Central and South America. The vector is the reduviid (triatomid) bug. The organisms can damage the heart by direct infection or by inducing an autoimmune response that affects the heart because of the existence of cross-reactive antigen. Acute myocarditis rarely occurs, but most deaths from acute Chagas disease are due to heart failure. In 20% of infected individuals, cardiac failure can occur 5 to 15 years after the initial infection. The affected heart is enlarged, and all four chambers are dilated. A cerebral abscess or acute meningitis is typically a complication of a bacterial infection with septicemia. Chronic arthritis can be seen in Lyme disease, which is transmitted by deer ticks. Mucocutaneous ulcers may be seen in Leishmania braziliensis infection, which is transmitted via sand flies. Paranasal sinus infection may be caused by Mucor circinelloides.
A 28-year-old woman from rural Guyana with a history of rheumatoid arthritis develops painful swelling of her hands and feet. She is treated with corticosteroid therapy. A month later, she develops profuse, watery diarrhea along with fever and cough. On examination, she has a temperature of 37.3° C. Laboratory studies show WBC count, 12,900/ mm3; and the WBC differential count shows 57% segmented neutrophils, 5% bands, 16% lymphocytes, 8% monocytes, and 14% eosinophils. Microscopic examination of a stool specimen shows ova and small rhabditoid larvae. Similar larvae are present in a sputum specimen. Which of the following infectious diseases is most likely to produce these findings?
A Cysticercosis
B Onchocerciasis
C Schistosomiasis
D Strongyloidiasis
E Trichinosis
D
The rhabditoid larvae of Strongyloides stercoralis can become invasive filariform from autoinfection in immunocompromised hosts, so-called hyperinfection with involvement of multiple organs. Immunocompetent hosts typically have only diarrhea. Parasites, particularly worms, crawling through tissues incite a marked eosinophilia. Cysticercosis from eating uncooked pork can result in the release of larvae that penetrate the gut wall and disseminate hematogenously, often settling in gray and white cerebral tissue, where they develop into cysts. Onchocerciasis occurs as a result of infection with the filarial nematode Onchocerca volvulus and leads to formation of a subcutaneous nodule. Schistosoma mansoni or Schistosoma japonicum infections have adult female worms in the portal venous system that release eggs that can produce hepatic fibrosis; Schistosoma haematobium worms live in veins near the bladder and release eggs that result in hematuria. Eating infected meat, typically uncooked pork, can lead to trichinosis; Trichinella encysts in striated muscle to produce fever and myalgias.
A 17-year-old boy has had generalized muscle pain with fever for 1 week. Over the past 2 days, he has developed increasing muscular weakness and diarrhea. On physical examination, his temperature is 38° C. All of his muscles are tender to palpation, but he has a normal range of motion, and no significant decrease in muscle strength. Laboratory findings include hemoglobin, 14.6 g/dL; hematocrit, 44.3%; MCV, 90 μm3; platelet count, 275,000/mm3; and WBC count, 16,700/ mm3 with differential of 68% segmented neutrophils, 6% bands, 10% lymphocytes, 4% monocytes, and 12% eosinophils. What is the most likely diagnosis?
A Hemorrhagic fever
B Influenza
C Poliomyelitis
D Scrub typhus
E Trichinosis
E
Acute muscle pain with fever and eosinophilia suggests a parasitic infestation of the skeletal muscles, most likely trichinosis; this results from ingesting poorly cooked meat infected with Trichinella spiralis larvae. Hemorrhagic fever can be a mild disease with myalgia, but can be severe with extensive vascular endothelial damage. Influenza A and B infection may produce myositis in childhood, with more severe, focal, and later onset than the diffuse myalgias of typical flu. Poliomyelitis can lead to muscle weakness, but via neurogenic atrophy, through loss of motor neurons. Scrub typhus caused by Orientia tsutsugamushi can have myalgia along with eschar and lymphadenopathy in the region of a chigger bite.
A 29-year-old man has had hematuria for the past month. On physical examination, he is afebrile. There is diffuse lower abdominal tenderness, but no palpable masses. An abdominal radiograph shows a small bladder outlined by a rim of calcification. Cystoscopy is performed, and the entire bladder mucosa is erythematous and granular. Biopsy samples are taken. Which of the following histologic findings is most likely to be seen in these samples?
A Acid-fast bacilli of Mycobacterium avium complex
B Eggs of Schistosoma haematobium
C Larvae of Trichinella spiralis
D Migrating Ascaris lumbricoides
E Taenia solium cysts
B
Schistosoma haematobium is a parasitic infection most often seen in Africa, particularly the Nile Valley, in areas where irrigation has expanded the range of the host snails. The adult worms live in veins adjacent to the bladder and release eggs with a sharp spine to cut their way through the wall of the urinary bladder, causing severe granulomatous inflammation, fibrosis, and calcification. Mycobacterial infections of the urinary tract are uncommon and do not cause bladder fibrosis. Trichinella spiralis infects striated muscle. Ascariasis involves the lower gastrointestinal tract, and the worms reside in the lumen. Cysticercosis from the pork tape-worm Taenia solium can have a wide tissue distribution, but the brain is most often affected.
In a study of individuals living in a subtropical region in which an irrigation project has been completed, it is noted that rice farmers have experienced an increased rate of an infectious illness since the project began. Investigators determine that the infection is acquired through cercariae that penetrate the skin. The cercariae are released from snails living in the irrigation canals. Infected individuals develop progressive as- cites. Which of the following pathologic findings is most likely to be present in these infected individuals as a consequence of the infection?
A Dilated cardiomyopathy
B Scrotal elephantiasis
C Hepatic fibrosis
D Mucocutaneous ulcers
E Urinary bladder carcinoma
C
These farmers are infected with either Schistosoma mansoni or Schistosoma japonicum. Female worms reside in the portal venous system and release eggs that cut their way into the liver and incite a granulomatous inflammatory reaction. With time, the portal granulomas undergo fibrosis, compressing the portal veins. This gives rise to severe portal hypertension, splenomegaly, and ascites. A dilated cardiomyopathy may occur with Chagas disease, in which the Trypanosoma cruzi organisms are transmitted through the reduviid (triatomid) bug. Elephantiasis is a complication of filariasis, which is transmitted via mosquitoes. Mucocutaneous ulcers may be seen in Leishmania braziliensis infection, which is transmitted via sand flies. Squamous cell carcinomas may be seen in the bladder in chronic Schistosoma haematobium infection.
A 40-year-old man has had progressive enlargement of the right leg for the past 6 years, leading to the appearance shown in the figure. On physical examination, he is afebrile. He has inguinal lymphadenopathy and scrotal edema. Infection with which of the following organisms is most likely to be present in this man?
A Echinococcus granulosus
B Leishmania tropica
C Schistosoma mansoni
D Trichinella spiralis
E Wuchereria bancrofti
E
The marked soft tissue enlargement and deformity is called elephantiasis, which results from lymphatic obstruction in the presence of an inflammatory reaction to the adult filarial worms Wuchereria bancrofti. Echinococcus produces hydatid disease of the liver, lungs, or bone. Leishmania tropica can involve the skin, causing ulceration, and can enlarge parenchymal organs. Schistosomiasis from Schistosoma mansoni may affect the liver most severely. Trichinella larvae from ingested, poorly cooked meat encyst in striated muscle.
Persons living in southern Africa where black flies are common and who have developed blindness are studied to identify a potential infectious cause. These persons are found to have a chronic dermatitis that preceded their blindness. Skin lesions are pruritic, scaling, and hypopigmented. Ocular lesions include punctate keratitis and focal corneal opacities, sclerosing keratitis, iridocyclitis with glaucoma, and retinitis. Which of the following morphologic forms of the infectious agent is most likely to be found in skin biopsies of these persons?
A Acid-fast bacilli
B Elementary bodies
C Intracellular diplococci
D Intranuclear inclusions
E Microfilariae
E
Onchocerciasis is caused by Onchocerca volvulus with inflammation induced by the microfilaria. Both insect abatement programs and campaigns to treat affected populations with ivermectin have helped to reduce the prevalence of this disease. Though ivermectin kills micro-filariae, it does not kill the adult worms, and treatment with doxycycline will eliminate the worm that symbiotic Wolbachia bacteria need for reproduction. Acid-fast Mycobacterium leprae organisms may be seen in skin lesions of lepromatous leprosy, but the eye is not involved. Elementary bodies can be identified in conjunctival scrapings with Chlamydia trachomatis infection, but additional eye components are not involved, and there are no skin lesions. Intracellular diplococci of Neisseria gonorrhoeae can cause neonatal blindness. Intranuclear inclusions can be seen with herpes simplex keratitis, which can cause perforation through corneal ulcerations.
Within the same day, an emergency department is visited by 20 individuals, all of whom work in the same building. Over the past day, they all experienced the sudden onset of high fever, headache, backache, and malaise. On examination, they are febrile. They do not have lymphadenopathy or hepatosplenomegaly. Over the next 2 days, they develop a maculopapular rash on the face, forearms, and mucous membranes of the oropharynx. Despite supportive care, a third of these patients die. Which of the following organisms is the most likely causative agent?
A Chlamydia psittaci
B Francisella tularensis
C Hantavirus
D Mycobacterium kansasii
E Rickettsia typhi
F Variola major
B
The Centers for Disease Control and Prevention has classified microbes into several categories based on the danger they pose as agents for bioterrorism on the basis of their ease of production, dissemination, and production of serious illness. Variola major is the causative agent for smallpox and has a mortality rate of 30%. Francisella tularensis is very infectious; only 10 to 50 organisms can cause disease. As a weapon, the bacteria can be made airborne for exposure by inhalation. Infected individuals experience life-threatening pneumonia. Chlamydia psittaci can cause psittacosis, which also can produce pneumonitis, but the course is more variable. Hantavirus can produce a severe pneumonia, but the prodrome is longer, and the vector is the deer mouse. Mycobacterium kansasii produces findings similar to Mycobacterium tuberculosis. Rickettsia typhi is the causative agent for murine typhus with headache and rash.
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Trigonometric identities
In trigonometry, trigonometric identities are equations involving trigonometric functions that are true for all input values. Trigonometric functions have an abundance of identities, of which only the most widely used are included in this article.
Contents
1 Pythagorean identities
2 Angle addition identities
3 Double-angle identities
3.1 Cosine double-angle identity
4 Half-angle identities
5 Product-to-sum identities
6 Sum-to-product identities
7 Other identities
7.1 Triple-angle identities
7.2 Even-odd identities
7.3 Conversion identities
7.4 Euler's identity
7.5 Miscellaneous
8 Resources
9 See also
Pythagorean identities
The Pythagorean identities state that
Using the unit circle definition of trigonometry, because the point is defined to be on the unit circle, it is a distance one away from the origin. Then by the distance formula, . To derive the other two Pythagorean identities, divide by either or and substitute the respective trigonometry in place of the ratios to obtain the desired result.
Angle addition identities
The trigonometric angle addition identities state the following identities:
There are many proofs of these identities. For the sake of brevity, we list only one here.
Euler's identity states that . We have that By looking at the real and imaginary parts, we derive the sine and cosine angle addition formulas.
To derive the tangent addition formula, we reduce the problem to use sine and cosine, divide both numerator and denominator by , and simplify. as desired.
Double-angle identities
The trigonometric double-angle identities are easily derived from the angle addition formulas by just letting . Doing so yields:
Cosine double-angle identity
Here are two equally useful forms of the cosine double-angle identity. Both are derived via the Pythagorean identity on the cosine double-angle identity given above.
In addition, the following identities are useful in integration and in deriving the half-angle identities. They are a simple rearrangement of the two above.
Half-angle identities
The trigonometric half-angle identities state the following equalities:
The plus or minus does not mean that there are two answers, but that the sign of the expression depends on the quadrant in which the angle resides.
Consider the two expressions listed in the cosine double-angle section for and , and substitute instead of . Taking the square root then yields the desired half-angle identities for sine and cosine. As for the tangent identity, divide the sine and cosine half-angle identities.
Product-to-sum identities
The product-to-sum identities are as follows:
They can be derived by expanding out and or and , then combining them to isolate each term.
Sum-to-product identities
Substituting and into the product-to-sum identities yields the sum-to-product identities.
Other identities
Here are some identities that are less significant than those above, but still useful.
Triple-angle identities
All of these expansions can be proved using trick and perform the angle addition identities. Same for and for .
Even-odd identities
The functions , , , and are odd, while and are even. In other words, the six trigonometric functions satisfy the following equalities:
These are derived by the unit circle definitions of trigonometry. Making any angle negative is the same as reflecting it across the x-axis. This keeps its x-coordinate the same, but makes its y-coordinate negative. Thus, and .
Conversion identities
The following identities are useful when converting trigonometric functions.
All of these can be proven via the angle addition identities.
Euler's identity
Euler's identity is a formula in complex analysis that connects complex exponentiation with trigonometry. It states that for any real number , where is Euler's constant and is the imaginary unit. Euler's identity is fundamental to the study of complex numbers and is widely considered among the most beautiful formulas in math.
Similar to the derivation of the product-to-sum identities, we can isolate sine and cosine by comparing and , which yields the following identities:
They can also be derived by computing and . These expressions are occasionally used to define the trigonometric functions.
Miscellaneous
These are the identities that are not substantial enough to warrant a section of their own.
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10670 | https://magoosh.com/act/act-math-challenge-average-speed/ | ACT®Blog
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ACT Math Challenge: Average Speed
By
Magoosh ACT
— Last updated on
in
ACT
In a previous post, we took a look at two different types of “averages” you might see on the ACT Math Test, one of which was a fairly rare challenging concept known as “average speed. Often found in complex word problems, this type of question is one many students are less familiar with so don’t get nervous if you don’t know how to approach it yet! Let’s try a couple practice questions to give ourselves an ACT Math challenge!
But first, remember that the formula for Average Speed = Total Distance / Total Time. It might help to remember it as AS = D / T, or come up with your own fun mnemonic device!
Suzanne drove 40 miles to see her aunt and was going 20 mph. It took her 2 hours to get to her aunt’s house. Then, she left and drove another 30 miles to the pet store, but this time only drove at 10 mph. If it took Suzanne 3 hours to arrive at the pet store, what was her average speed in miles per hour for the entire car ride from her home to the pet store?
(A) 10
(B) 11
(C) 12
(D) 14
(E) 15
Average Speed = Total Distance / Total Time.
40 miles + 30 miles so the Total Distance was 70 miles. Suzanne drove for 2 hours + 3 hours so the Total Time was 5 hours. 70/5 = 14.
The Average Speed for the whole trip was 14 mph. The correct answer is (D).
Notice how the test-maker has made this problem tricky! The average speed in this problem is 14 mph, which is different from simply taking the mean of the two speeds. If we had just averaged the two speeds (10mph and 20mph) we would have gotten 15mph. Average Speed is a weighted average. Since Suzanne spent more time in the problem going 10mph than 20mph, it makes sense that the Average Speed would be closer to 10mph.
Let’s try another word problem!
Students at Thomas Jefferson High School boarded the bus for a field trip that went 15mph through a 30 mile section of the city. The bus then stopped for lunch in a suburb before continuing on a 3 hour tour of countryside at a constant speed of 10mph. Finally, the bus drove 40 miles straight back to the high school. If the trip back to Thomas Jefferson High school took two hours, approximately what was the average speed for the entire field trip?
(A) 11
(B) 12
(C) 13
(D) 14
(E) 15
To find the average speed of the bus, we know we will need to find the Total Distance and the Total Time, so we can start by using another formula (Distance = Rate x Time) to help us find the pieces we’re missing for each part of the trip.
For the first part of the field trip: 30 miles = 15mph x T, so we know that T = 2 hours. For the middle part of the trip, we know that D = 10mph x 3 hours, so we know that D = 30 miles. For the last part of the trip, we know that 40 miles = R x 2 hours, so we know that R = 20mph. Now we can find the Total Distance and the Total Time:
Total Distance = 30 miles + 30 miles + 40miles = 100 miles.
Total Time = 2 hours + 3 hours + 2 hours = 7 hours.
The Average Speed = 100 miles/ 7 hours = 14.28mph. Since the question used the word “approximately,” we can round to the nearest integer: 14. The correct answer is (D).
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10671 | https://courses.edx.org/c4x/MITx/6.002x/asset/handouts_6002-L18-oei12-gaps.pdf | 1 6.002x CIRCUITS AND ELECTRONICS Damped Second-Order Systems Review C A B 5V +!
–!
5V CGS large loop 2KΩ 50Ω 2KΩ S L vA 5 0 vB 0 vC 0 50Ω t t t 5 5 3 In the last lecture, we started by analyzing the simpler LC circuit to build intuition 4 And for initial conditions We solved For input In the last lecture… LC circuit 5 In the last lecture… LC circuit Total solution +!
–!
C L +!
–!
) (t v ) (t vI ) (t i 6 See A&L Section 13.6 Today, we will close the loop on our observations in the demo by analyzing the RLC circuit +!
–!
C L +!
–!
i(t) v(t) vI (t) LC I v v(t) 2VI t VI 0 7 Recall element rules Let’s analyze the RLC network L: C: 8 Set up the differential equation vA +!
–!
C L +!
–!
R i(t) v(t) vI(t) Need to get rid of vA v 9 Set up the differential equation differently v +!
–!
C L +!
–!
R i vI 10 Solve I v LC v LC dt dv L R dt v d 1 1 2 2 = + + 11 Let’s solve I v LC v LC dt dv L R dt v d 1 1 2 2 = + + 12 Set up the differential equation differently v +!
–!
C L +!
–!
R i vI 13 1 Particular solution I v LC v LC dt dv L R dt v d 1 1 2 2 = + + I P P P V LC v LC dt dv L R dt v d 1 1 2 2 = + + 14 Homogeneous solution 2 I v LC v LC dt dv L R dt v d 1 1 2 2 = + + 15 0 v LC 1 dt dv L R dt v d H H H = + + 2 2 Solution to Homogeneous solution 2 Assume solution of the form 2A vH = AeST , A, S = ? 16 Homogeneous solution 2 0 v LC 1 dt dv L R dt v d H H H = + + 2 2 Solution to 0 2 = + + LC 1 s L R s 17 Homogeneous solution 2 0 v LC 1 dt dv L R dt v d H H H = + + 2 2 Solution to 0 2 = + + LC 1 s L R s Roots o 1 s 2 2 ω α α − + − = o s 2 2 2 ω α α − − − = characteristic equation LC 1 o = ω L R 2 = α 18 Total solution 3 19 v(t) =VI + A 1e−αte α2−ω2 o ( )t + A2e−αte −α2−ω2 o ( )t Let’s stare at this a while longer… 3 v(t) =VI + A 1e −α+ α2−ω2 o ( )t + A2e −α−α2−ω2 o ( )t 20 v(t) =VI + A 1e−αte α2−ω2 o ( )t + A2e−αte −α2−ω2 o ( )t 3 Overdamped o ω α > Underdamped o ω α < Critically damped o ω α = Overdamped o ω α > 21 Underdamped o ω α < 3 v(t) =VI + A 1e−αte α2−ω2 o ( )t + A2e−αte −α2−ω2 o ( )t Overdamped o ω α > Underdamped o ω α < Critically damped o ω α = 22 Underdamped contd… o ω α < 3 t e K t e K V t v d t d t I ω ω α α sin cos ) ( 2 1 − − + + = 23 Underdamped contd… o ω α < Remember, scaled sum of sines (of the same frequency) are also sines! -- Appendix B.7 v(t) =VI −VIe−αt cosωdt −VI α ωd e−αt sinωdt 3 ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ − + = + − 1 2 1 2 2 2 1 2 1 tan cos sin cos A A A A A A θ θ θ 24 Critically damped o ω α = Section 13.2.3 3 25 Remember this? We have now closed the loop… vA 5 0 vB 0 vC 0 50Ω t t t 5 5 26 See Sec. 12.7 of A&L textbook “ringing” v(t) VI 0 t Easy way: Characteristic equation tells the whole story! 27 Intuitive Analysis +!
–!
C L +!
–!
) (t v I v ) (t i R ) 0 ( i ) 0 ( v given -ve +ve 28 Next, introduce R: RLC Circuits More in the next sequence! If you are impatient, see A&L Section 13.2 +!
–!
C L +!
–!
vI (t) i(t) v(t) v(t) t 29 What about other variables? 30 Parallel RLC – Characteristic equation says it all |
10672 | https://www.novatech-usa.com/pdf/Taylor%20Freezing%20Point%20Test%20Kit%20Brochure.pdf?srsltid=AfmBOoqva3My22mVAlxaSLRB4qLdlPnUk8WCYtInn8gjHGTtanj5QagR | Taylor’s Freezing Point Determination Test Kit
INTRODUCTION
In the Freezing Point Determination
Test Kit we have assembled the
apparatus needed to monitor the freez -
ing point of ethylene glycol solutions,
propylene glycol solutions, sodium-
chloride-based brines, and calcium-
chloride-based brines. The K-1540
consists of a 200 mm glass hydrometer
cylinder and two hydrometers for de -
termining specific gravity in the ranges
1.000–1.070 (for waters treated with
ethylene glycol or propylene glycol)
and 1.000–1.400 (for sodium chloride
and calcium chloride brines); a cor -
rection table for sample temperature;
plus the charts for cross-referencing
the specific gravity reading with the
percentage by weight/percentage by
volume of the solution and its freezing
point in degrees Fahrenheit/degrees
Celsius. The assemblage is a welcome
convenience for two user groups in
particular.
In recent years the desire to conserve
water and energy has resulted in a
greater reliance on closed cooling
water systems for both comfort and
process cooling. When closed water
systems are operated year-round in
cold weather zones, they must have
protection from freezing. Food proces -
sors generally use propylene glycol
solutions for this purpose. Elsewhere,
ethylene-glycol-based antifreeze is
commonly added to cooling water to
make the water stay liquid at tempera -
tures below its normal freezing point.
The K-1540 will allow these operators to
maintain correct freeze point control.
Thermal energy storage (TES)
systems are an increasingly popular
way to provide chilled water during
the day for comfort cooling, commer -
cial refrigeration, and light process
cooling without increasing energy
consumption at peak-rate times.
Cooling energy produced at night,
off-peak, is stored in the form of
cooled liquid or ice in tanks; this
Maintain freeze point control with the hydrometers and charts in the K-1540.
stored energy is used the following
day. TES system operators who use
sodium chloride or calcium chloride
for freezing point depression also will
find the K-1540 helpful.
FReez IN g P OINT
DeTeRmIN aTION KIT
K-1540
Apparatus to determine the specific
gravity of ethylene glycol or propylene
glycol solutions and sodium- or
calcium-chloride brines with charts
showing the correspondence of the
specific gravity reading with the
percentage by weight/percentage by
volume (of ethylene glycol, propylene
glycol, NaCl, and CaCl 2) and the
freezing point (in °F and °C)
US eR B eNeFITS
• All necessary apparatus is
preassembled in one box.
• Conversions from specific gravity
readings to freezing points have been
calculated for you and presented in an
easy-to-use format.
• Waterproof instructions are
printed on plastic-impregnated paper
that resists fading and tearing.
• Custom-molded, durable plastic
case with foam lining provides safe
storage for glass components.
aLSO aVaIL aBL e
• Routine and diagnostic tests for
waters in open and closed cooling
systems, as well as boilers; simple
drop-count titrations are our specialty.
• Myron L handheld meters plus
bacteria tests manufactured by Orion
Diagnostica ( Easicult ®) and Biosan
(Sani-Check ®).
• Toll-free technical assistance at
800-TEST KIT. sales@novatech-usa.com www.novatech-usa.com Tel: (866) 433-6682 Fax: (866) 433-6684 Tel: (281) 359-8538 Fax: (281) 359-0084 From K-1540 instruction:
RePR eSeNT aTIV e TeST PROC eDUR e
Freezing Point Determination Ethylene Glycol Solutions Propylene Glycol Solutions Sodium Chloride Brines Calcium Chloride Brines COMPONENTS: 1 x 4016 Hydrometer, Specific Gravity, 1.000-1.070 1 x 4017 Hydrometer, Specific Gravity, 1.000-1.400 1 x 4018 Cylinder, Hydrometer, 200 mm 1 x 5062 Instruction TO ORDER REPLACEMENT PARTS CALL TOLL-FREE 800-TEST KIT (800-837-8548). PROCEDURE:
Rinse a clean hydrometer cylinder (#4018) with sample to be tested. Carefully fill the hydrom-eter cylinder by pouring sample down the inside wall of the hydrometer cylinder avoiding the formation of air bubbles in the sample. Place cylinder on a level surface. 2. Determine the temperature of the sample. 3. Select hydrometer to be used from the chart below. Water Sample Hydrometer Ethylene glycol solution 1.000-1.070 (#4016) Propylene glycol solution 1.000-1.070 (#4016) Calcium chloride brine 1.000-1.400 (#4017) Sodium chloride brine 1.000-1.400 (#4017) 4. Place the hydrometer in the sample and allow it to float. 5. Gently push the hydrometer down no more than 1/8". Release the hydrometer and allow it to rise and settle in the sample. 6. Read the hydrometer at eye level with the sample surface to determine the specific gravity of the sample. To do this, follow the procedure below: Begin with your eye below the level of the sample surface so that the surface appears as an ellipse. Slowly raise the line of sight until the sample surface becomes a straight line. (At this point your eye is level with the sample surface.)
Read the hydrometer at the point where the line of the sample surface intersects the hydrometer stem. NOTE: The hydrometer and sample should be at the same temperature before a reading is taken. Do not take a reading until the hydrometer and sample are at rest and free of air bubbles.
The point where the sample actually touches the stem is not the correct reading. The correct reading corresponds to the plane of intersection of the sample surface and the hydrometer stem (see drawing).
Instr. #5062 (OVER)
40 1.0 50
1.042
not 1.041
Using the temperature correction chart below, adjust the hydrometer reading to compensate for the temperature effect on the hydrometer as follows: Match the specific gravity value in the table closest to the hydrometer reading with the sample temperature. Divide the value from the table by 1000 to get the correction factor. Add the correction factor to the hydrometer reading and record the result as the specific gravity of the sample. For example, if the hydrometer reading is 1.100 and the sample temperature is 86ºF, the correction factor calculates to be 3.0/1000 or 0.003. The specific gravity of the sample is 1.100 + 0.003 or 1.103. NOTE: For more exact values, interpolation between values can be performed.
Temperature Corrections for Specific Gravity Hydrometers 15.5 ºC/60 ºF
Specific Gravity ºC ºF 1.000 1.100 1.200 1.300 1.400 15.5 60 0.0 0.0 0.0 0.0 0.0 20 68 0.7 0.8 0.8 0.9 1.0 25 77 1.8 1.9 2.0 2.3 2.5 30 86 2.9 3.0 3.2 3.7 4.0 35 95 4.0 4.1 4.4 5.1 5.5 40 104 5.1 5.2 5.6 6.5 7.0 45 113 6.2 6.3 6.8 7.9 8.5 50 122 7.3 7.4 8.0 9.3 10.0 55 131 8.4 8.5 9.2 10.7 11.5 60 140 9.5 9.6 10.4 12.1 13.0 65 149 10.6 10.7 11.6 13.5 14.5 70 158 11.7 11.8 12.8 14.9 16.0 75 167 12.8 12.9 14.0 16.3 17.5 80 176 13.9 14.0 15.2 17.7 19.0 85 185 15.0 15.1 16.4 19.1 20.5 90 194 16.1 16.1 17.6 20.5 22.0 95 203 17.2 17.1 18.8 21.9 23.5 100 212 18.3 18.1 20.0 23.0 25.0 8. Using the appropriate chart below, match the specific gravity of the sample from Step 7 to the freezing point of the water.
Freezing Point of Ethylene Glycol Solutions
% Glycol Specific Gravity Freezing Point By Weight By Volume 15.5ºC (60ºF) ºC ºF 10 9.2 1.013 -3.6 25.6 15 13.8 1.019 -5.6 22.0 20 18.3 1.026 -7.9 17.8 25 23.0 1.033 -10.7 12.8 30 28.0 1.040 -14.0 6.8 40 37.8 1.053 -22.3 -8.2 50 47.8 1.067 -33.8 -28.8 60 58.1 1.079 -49.3 -56.7
Freezing Point of Propylene Glycol Solutions
% Glycol Specific Gravity Freezing Point By Volume 15.5ºC (60ºF) ºC ºF 5 1.004 -1.1 30 10 1.006 -2.2 28 15 1.012 -3.9 25 20 1.017 -6.7 20 25 1.020 -8.9 16 30 1.024 -12.8 935 1.028 -16.1 340 1.032 -20.6 -5 45 1.037 -26.7 -16 50 1.040 -33.3 -28
Instr. #5062
(NEXT PAGE)
Freezing Point of Calcium Chloride Brines
% CaCl 2 Specific Gravity Freezing Point By Weight 15.5ºC (60ºF) ºC ºF 0 1.000 0.0 32.0 1.1 1.010 -0.5 31.1 2.3 1.020 -1.0 30.2 3.5 1.030 -1.8 29.1 4.7 1.040 -2.2 28.0 5.8 1.050 -2.8 27.0 7.0 1.060 -3.2 25.9 8.1 1.070 -4.1 24.6 9.2 1.080 -4.8 23.4 10.4 1.090 -5.9 21.7 11.4 1.100 -6.5 20.3 12.5 1.110 -7.5 18.5 13.5 1.120 -8.6 16.5 14.6 1.130 -9.9 14.4 15.6 1.140 -11.1 12.0 16.6 1.150 -12.4 9.7 17.6 1.160 -13.9 7.0 18.6 1.170 -15.6 4.1 19.5 1.180 -17.0 1.4 20.5 1.190 -19.0 -2.2 21.5 1.200 -21.0 -5.8 22.4 1.210 -23.0 -9.4 23.3 1.220 -25.1 -13.2 24.2 1.230 -27.2 -17.1 25.1 1.240 -29.6 -21.3 26.0 1.250 -32.1 -25.8 31 Loveton Circle, Sparks, MD 21152 U.S.A. 800-TEST KIT (837-8548) • 410-472-4340 12/07
Instr. #5062 Freezing Point of Sodium Chloride Brines
% NaCl Specific Gravity Freezing Point By Weight 15.5ºC (60ºF) ºC ºF 0 1.000 0.0 32.0 1 1.007 -0.58 31.0 2 1.014 -1.13 30.0 3 1.021 -1.72 28.9 4 1.028 -2.35 27.8 5 1.036 -2.97 26.7 6 1.043 -3.63 25.5 7 1.051 -4.32 24.2 8 1.059 -5.03 22.9 9 1.067 -5.77 21.6 10 1.074 -6.54 20.2 11 1.082 -7.34 18.8 12 1.089 -8.17 17.3 13 1.097 -9.03 15.7 14 1.104 -9.94 14.1 15 1.112 -10.88 12.4 16 1.119 -11.90 10.6 17 1.127 -12.93 8.7 18 1.135 -14.03 6.7 19 1.143 -15.21 4.6 20 1.152 -16.46 2.4 21 1.159 -17.78 0.0 22 1.168 -19.19 -2.5 23 1.176 -20.69 -5.2 23.3(E) 1.179 -21.13 -6.0 24 1.184 -17.0 +1.4 25 1.193 -10.4 13.3 26.0 1.201 -2.3 27.9 26.3 1.203 0.0 32.0 Saturation temperatures of sodium chloride dihydrate; at these temperatures NaCl•2H 2Oseparates leaving the brine of the eutectic compostion (E).
rev. 080410 |
10673 | https://math.stackexchange.com/questions/330999/how-can-i-tell-that-the-sequence-a-n-frac-lnn-n-converges-and-to-what | limits - How can I tell that the sequence $a_n=\frac {\ln(n)} {n}$ converges and to what it converges? - Mathematics Stack Exchange
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How can I tell that the sequence a n=ln(n)n a n=ln(n)n converges and to what it converges?
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Asked 12 years, 6 months ago
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I need to "study the limit behavior" and compute the limit if it exists.
This is what I have done so far. In order to study the limit behavior I tried to first check the monotonicity and boundedness of the sequence. The sequence was found to not be monotonic for all n. Since that failed, I tried to attempt to prove that the sequence was Cauchy, and if it was, that would lead me to have completed the "study the limit behavior" part. Here is my attempt at the cauchy part:
For every ϵ>0 0 there exists an N N such that m,n>N m,n>N implies |a n−a m|<ϵ a n−a m|<ϵ for n≥m n≥m
|a n−a m|=∣∣∣ln(n)n−ln(m)m∣∣∣≤∣∣∣ln(n)n∣∣∣+∣∣∣ln(m)m∣∣∣⇔ln(n)n+ln(m)m<ϵ|a n−a m|=|ln(n)n−ln(m)m|≤|ln(n)n|+|ln(m)m|⇔ln(n)n+ln(m)m<ϵ
ln(n)n+ln(m)m<ϵ⇔ln(n)n<ϵ−ln(m)m ln(n)n+ln(m)m<ϵ⇔ln(n)n<ϵ−ln(m)m
(Now I get stuck. I dont know if what I did so far is even correct, and if it is I don't know where to go from here)
Even though I got stuck at the cauchy part I went on to compute the limit.
lim a n=lim ln(n)n=lim 1 n(ln(n))=lim ln(n)1 n lim a n=lim ln(n)n=lim 1 n(ln(n))=lim ln(n)1 n
So, lim e a n=lim e ln(n)1 n=lim n 1 n=1 lim e a n=lim e ln(n)1 n=lim n 1 n=1 ( I proved that using the epsilon definition)
Therefore because lim e a n=1 lim ln(e a n)=lim a n=ln(1)=0 lim e a n=1 lim ln(e a n)=lim a n=ln(1)=0
Any help (on the cauchy part especially)? Thanks in advance.
sequences-and-series
limits
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edited Mar 19, 2013 at 6:23
learner
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asked Mar 15, 2013 at 6:57
user66807user66807
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It was Dominic who edited it for you, not me. It not that you're supposed to edit it like this, but that's how L A T E X L A T E X is supposed to be used.Git Gud –Git Gud 2013-03-15 07:20:38 +00:00 Commented Mar 15, 2013 at 7:20
Oh okay, thanks for the info.user66807 –user66807 2013-03-15 07:23:09 +00:00 Commented Mar 15, 2013 at 7:23
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You made a mistake
∣∣∣log(m)m−log(n)n∣∣∣≤∣∣∣log(m)m∣∣∣−∣∣∣log(n)n∣∣∣|log(m)m−log(n)n|≤|log(m)m|−|log(n)n|
is wrong, the rhs could be smaller zero the lhs not.
The function is monotone for n n large enough (for n>e n>e), so just prove monoticity for n>3 n>3.
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edited Mar 15, 2013 at 7:31
answered Mar 15, 2013 at 7:02
Dominic MichaelisDominic Michaelis
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I tried that but doesnt the sequence have to be monotone for all n and because the sequence isn't monotone for n<3 I can't use the monotonicity of the sequence.user66807 –user66807 2013-03-15 07:03:32 +00:00 Commented Mar 15, 2013 at 7:03
as i said go for n>3 n>3 there it is monotone, at what happens below 3 3 doesn't matter for the limit at all Dominic Michaelis –Dominic Michaelis 2013-03-15 07:07:26 +00:00 Commented Mar 15, 2013 at 7:07
That is what I did first, but my book says that a sequence is decreasing if s n>=s n+1 f o r a l l n s n>=s n+1 f o r a l l n and a decreasing sequence will be called monotone. Also, what I did was the triangle inequality that says |a+b|<=|a|+|b||a+b|<=|a|+|b| Does that not work?user66807 –user66807 2013-03-15 07:13:21 +00:00 Commented Mar 15, 2013 at 7:13
the triangle inequality works but it gives you |a−b|≥|a|−|b||a−b|≥|a|−|b|, which is not helpful. You could use |a−b|≤|a|+|b||a−b|≤|a|+|b| with that one you can go. Yeah your original series is not monotone for all n n, but we are interested in the limit, so the first few values don't matter at all.Dominic Michaelis –Dominic Michaelis 2013-03-15 07:17:19 +00:00 Commented Mar 15, 2013 at 7:17
Oh okay, I understand what you are saying now, but since the problem says to study the behavior of the limit don't I have to know the sequence converges before I can talk about the limit? And for the inequality part I understand what you are saying, do I just edit the question?user66807 –user66807 2013-03-15 07:21:21 +00:00 Commented Mar 15, 2013 at 7:21
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Consider the continuous version:
f(x)=ln x x f(x)=lnx x
Computing the derivative:
f′(x)=1−ln x x 2 f′(x)=1−lnx x 2
When x>e x>e, this derivative is negative and so it is a decreasing function for x>e x>e. On the other hand, it's clear that this function is also bounded above 0 0 and so the sequence a n a n is bounded above 0 0 and is decreasing for n>2 n>2. Monotone convergence theorem says that a n a n must then converge.
To find what it converges to, we could use L`Hopital's rule:
lim x→∞f(x)=lim x→∞1 x=0 lim x→∞f(x)=lim x→∞1 x=0
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answered Mar 15, 2013 at 7:00
muzzlatormuzzlator
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Although I have learned L`Hopital's rule and derivatives, because we haven't gotten to it in this class I don'think I'm allowed to use them.user66807 –user66807 2013-03-15 07:25:29 +00:00 Commented Mar 15, 2013 at 7:25
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You can use the fact that n 1/n→1 n 1/n→1 as n→∞n→∞.
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edited Mar 15, 2013 at 7:19
answered Mar 15, 2013 at 7:02
Abhra Abir KunduAbhra Abir Kundu
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10674 | https://www.bbc.co.uk/bitesize/articles/zwwcmbk | Sequences - GCSE Maths Revision - BBC Bitesize
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Key points
Check your understanding
Term-to-term rules
𝒏th term rules
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Higher – Quiz
Key points about sequences
A number sequence is a list of ordered numbers that follow a pattern or a rule. A term-to-term rule explains how to find the next term close term A value in a sequence. The 3rd term is the 3rd value in the sequence. of a sequence.
The 𝑛th term of a sequence is a position-to-term rule that can be used to find out any term in a sequence.
The 𝑛th term of an arithmetic sequence close arithmetic sequence A sequence that increases or decreases by the same number each time, eg 4, 7, 10, 13. Sometimes called a linear sequence. (sometimes known as a linear sequence) is found by comparing the sequence to an appropriate times table.
Higher tier - The 𝑛th term of a quadratic close quadratic Describing an expression of the form 𝑎𝑥² + 𝑏𝑥 + 𝑐 where 𝑎, 𝑏 and 𝑐 are integers. sequence is found by considering the second differences close second difference Once the first difference between values of a sequence is calculated, the second difference is the difference between these values. between the terms and comparing the sequence to another that contains 𝑛².
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Check your understanding
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Term-to-term rules
A sequence is a list of values that follow a rule. Each value is called a term.
A rule that explains how to find the next term in a sequence is called a term-to-term rule.
The most common type of sequence is an arithmetic (or linear) sequence. The difference between each term is the same every time, and is known as the common difference.
Follow the working out below
GCSE exam-style questions
–20, –17, –14, –11 are the first four terms of a sequence. Find the term-to-term rule.
Show answer Hide answer
The term-to-term rule is ‘add 3’.
The negative numbers are getting closer to zero and the terms are increasing by 3 each time.
The term-to-term rule of a sequence is ‘multiply by 5, then subtract 3’. The first term is 2. Work out the 3rd term.
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The third term is 32
To work out the second term, multiply the first term by 5, then subtract 3.
2 × 5 – 3 = 7.
To work out the third term, multiply the second term by 5, then subtract 3.
7 × 5 – 3 = 32.
The term-to-term rule of a sequence is ‘add 3 then multiply by 2’. The third term is 46. Calculate the first term.
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The first term is 7.
To calculate the second term, work backwards from the third term. The term-to-term rule in reverse is ‘divide by 2, then subtract 3.’
46 ÷ 2 = 23. Then 23 – 3 = 20. The second term is 20.
To calculate the first term, work backwards from the second term. Use the same term-to-term rule in reverse.
20 ÷ 2 = 10. Then 10 – 3 = 7. The first term is 7.
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𝒏th term rules
Rather than finding the next term or next two terms of a sequence, it may be necessary to work out the 50th term, for example.
To do this without writing out all 50 terms, a general rule called the 𝒏th term is found.
To find an expression close expression A mathematical sentence expressed either numerically or symbolically made up of one or more terms, eg 8 + 2, or 6𝑥, or 5𝑥² + 3𝑦, or 3𝑎𝑏𝑐. for the 𝑛th term of an arithmetic sequence close arithmetic sequence A sequence that increases or decreases by the same number each time, eg 4, 7, 10, 13. Sometimes called a linear sequence., work out the common difference close common difference The difference between each term in an arithmetic linear sequence. between the terms and treat the sequence as a times table that has been shifted.
For example, 3, 7, 11, 14 has a common difference of 4, and is the 4 times table with 1 subtracted. The 𝑛th term is therefore 4𝑛 – 1.
To find the 50th term, substitute close substitute In algebra, to replace a letter with a number. the value of 50 into the 𝑛th term rule.
Follow the working out below
GCSE exam-style questions
A linear sequence starts 3, 8, 13, 18, 23. Work out an expression for the 𝑛th term.
Show answer Hide answer
5𝑛 – 2
Write the numbers for 𝑛 above the sequence.
The common difference between the terms is 5, so write the 5 times table under Click here to enter text. the values of 𝑛. Label the 5 times table as 5𝑛.
Work out how to get from the 5𝑛 row to the sequence. The sequence is the 5 times table subtract 2. The 𝑛th term is 5𝑛 – 2.
An arithmetic sequence starts 14, 11, 8, 5, 2. Work out an expression for the 𝑛th term.
Show answer Hide answer
–3𝒏 + 17
Write the numbers for 𝑛 above the sequence.
The common difference is –3, so write the –3 “times table” under the values of 𝑛. Label the row as –3𝑛.
Work out how to get from the –3𝑛 row to the sequence. The –3𝑛 row has had 17 added.
The 𝑛th term is –3𝑛 + 17.
(Another way to write the answer is 17 – 3𝑛).
Write down the first three terms of a sequence where the 𝑛th term is given by 𝑛² + 5.
Show answer Hide answer
6, 9, 14
1. The first term is when 𝑛 = 1. Substitute 1 into 𝑛² + 5 to give 12 + 5 = 6.
The second term is when 𝑛 = 2. Substitute 2 into 𝑛² + 5 to give 22 + 5 = 9.
3. The third term is when 𝑛 = 3. Substitute 3 into 𝑛² + 5 to give 32 + 5 = 14.
4. The first three terms are 6, 9, 14.
(Note: this sequence is not an arithmetic sequence as it does not go up by the same number each time.)
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Quiz – Sequences
Practise what you have learned about sequences with this quiz.
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Higher – 𝒏th term of a quadratic sequence
Quadratic sequences have an 𝑛th term rule that contains 𝑛².
Follow the working out below
Example 1
Example 2
The differences between the terms are not equal, but the second differences close second difference Once the first difference between values of a sequence is calculated, the second difference is the difference between these values. between the terms are equal.
To find the 𝑛th term, follow these steps:
Work out the first differences between the terms. The first differences are not the same. Work out the second differences.
The second differences will be the same. The coefficient close coefficient A number or symbol multiplied with a variable or an unknown quantity in an algebraic term. For example, 4 is the coefficient of 4n². (𝑎) of 𝑛² in the 𝑛th term rule is always half of the second difference.
Compare the numbers of the sequence 𝑎𝑛² with the original quadratic sequence. The difference between them will be a constant close constant A number that does not vary. Constants are different to variables such as 𝑥 and 𝑦 that can take many values., or should make an arithmetic sequence close arithmetic sequence A sequence that increases or decreases by the same number each time, eg 4, 7, 10, 13. Sometimes called a linear sequence..
Add the constant or 𝑛th term for the arithmetic sequence to 𝑎𝑛² to give the 𝑛th term for the quadratic sequence.
GCSE exam-style questions
A quadratic sequence has an 𝑛th term of 2𝑛² + 4𝑛 – 3. Find the first 3 terms.
Show answer Hide answer
3, 13, 27
Substitute 𝑛 = 1 into the expression. The first term is 2(1²) + 4(1) – 3 = 2(1) + 4 – 3 = 3
Substitute 𝑛 = 2 into the expression. The second term is 2(2²) + 4(2) – 3 = 2(4) + 8 – 3 = 13
Substitute 𝑛 = 3 into the expression. The third term is 2(3²) + 4(3) – 3 = 2(9) + 12 – 3 = 27
Work out the 𝑛th term of the sequence 6, 9, 14, 21, 30.
Show answer Hide answer
𝒏² + 5
The first differences are + 3, + 5, + 7, + 9.
The second differences are all +2. Half of 2 is 1, so the coefficient of 𝑛² is 1.
Write the values of 1𝑛² and compare it to the sequence. The constant value of 5 is always added to 𝑛² to make the sequence.
Write the final 𝑛th term rule as 𝑛² + 5.
Work out the 𝑛th term of the sequence 3, 9, 17, 27, 39.
Show answer Hide answer
𝑛² + 3𝑛 – 1
The first differences are + 6, + 8, + 10, + 12.
The second differences are all + 2. Half of 2 is 1, so the coefficient of 𝑛² is 1.
Write the values of 1𝑛² and compare it to the sequence. The differences 2, 5, 8, 11, 14 form an arithmetic sequence whose 𝑛th term is 3𝑛 –1 (the 3 times table subtract 1).
Add 3𝑛 – 1 to 𝑛² to give the final 𝑛th term rule
𝑛² + 3𝑛 – 1.
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Quadratic sequences – interactive activity
This interactive activity will help you to learn how to create quadratic sequences by selecting different coefficient values.
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Higher – Quiz – Sequences
Practise what you have learned about sequences with this quiz for Higher tier.
Now that you have revised sequences, why not try looking at geometric and special sequences?
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10675 | https://www.mmv.org/sites/default/files/uploads/docs/artemisinin/010b_IPCA_MALARIA_RESISTANCE_WHO_MODIFIED.pdf | Dr. Anil Pareek President – Medical Affairs & Clinical Research Ipca Laboratories Ltd.
DRUG RESISTANT MALARIA CURRENT STATUS DRUG RESISTANT MALARIA Ability of a parasite strain to survive and/or to multiply despite the administration and absorption of a drug given in doses equal or higher than those usually recommended, but within the limits of tolerance of the patients Main obstacle to malaria control Resistance to nearly all antimalarials in current use Curtails the lifespan of antimalarial drugs Increases malaria morbidity, mortality and treatment cost DRUG RESISTANT STRAINS OF MALARIA Predominant – P. falciparum Recent development – P. vivax Chloroquine resistant P. malariae has been described in Indonesia BURDEN OF DRUG RESISTANT MALARIA •Recurrent infections •More malaria – Work/school, productivity, anaemia, pregnant, birth weight •Epidemics •More deaths •Greater financial costs (Health service, community, individual) DISTRIBUTION OF DRUG RESISTANT MALARIA Chloroquine resistance Sulfadoxine + Pyrimethamine resistance Mefloquine resistance ETIOLOGY OF DRUG RESISTANT MALARIA Naturally occurring genetic mutations in the malaria parasite Inadequate treatment (subtherapeutic dose, suboptimal drug) of a high biomass infection – main selective pressure for resistance Resistant parasites are transmitted to other individuals by mosquitoes Drugs with long half lives (J Postgrad Med March 2004, Vol. 50, No. 1, p. 41) DRUG RESISTANT MALARIA Chloroquine resistance Chloroquine is ineffective in almost all malaria endemic countries In India chloroquine resistance was first detected in 1973 in Assam.
Severe in northeast and southeastern regions of India with high morbidity and mortality. DRUG RESISTANT MALARIA Sulfadoxine/pyrimethamine resistance Resistance to SP was first described from the Thai Cambodian border in 1960s Ineffective in South East Asia and the Amazon Basin for several years In Africa, SP resistance was detected in the late 1980s In India resistance to sulpha drugs has been reported from northeast states and Orissa Resistance in P. falciparum to sulphadoxine/ pyrimethamine combination was first detected in Delhi in 1987 (J Vect Borne Dis 41, September & December 2004, pp 45–53) DRUG RESISTANT MALARIA Quinine resistance The first case of quinine resistance was observed from Thai Cambodian border in mid 1960s. The clinical resistance to quinine therapy has been noticed sporadically in Southeast Asia and western Oceania It is less frequent in South America and Africa.
In India resistance has emerged against quinine in northeastern states and Kolar district in Karnataka.
(J Vect Borne Dis 41, September & December 2004, pp 45–53) DRUG RESISTANT MALARIA Mefloquine resistance Mefloquine resistance was first observed in late 1980s near the ThaiCambodian border It is frequent in some parts of Southeast Asia and has been reported in the Amazon region of South America and sporadically in Africa Resistance in P. falciparum to mefloquine in India was detected in Surat district in Gujarat state (J Vect Borne Dis 41, September & December 2004, pp 45–53) MECHANISMS OF ANTIMALARIAL DRUG RESISTANCE Chloroquine resistance Increased capacity for the parasite to expel chloroquine at a rate that does not allow chloroquine to reach levels required for inhibition of heamepolymerization This chloroquine efflux occurs at a rate 40 to 50 fold faster among resistant parasites than that in sensitive ones Mutations in pfmdr1 & 2 and pfcrt gene have also been associated with chloroquine resistance.
(J Vect Borne Dis 41, September & December 2004, pp 45–53) MECHANISMS OF ANTIMALARIAL DRUG RESISTANCE Sulphadoxine/pyrimethamine resistance Specific gene mutations encoding for resistance to dihydrofolate reductase and dihydropteroate synthetase have been identified.
The dehydrofolate reductase enzymes of resistant strains bind to pyrimethamine 400–800 fold less readily than to the enzymes of drug sensitive strains Quinine resistance pfmdr1 mutation associated with chloroquine resistance may also account for reduced susceptibility to quinine. The exact mechanism of resistance is not clear (J Vect Borne Dis 41, September & December 2004, pp 45–53) MECHANISMS OF ANTIMALARIAL DRUG RESISTANCE Mefloquine resistance Polymorphism of pfmdr1 gene is associated with mefloquine resistance.
(J Vect Borne Dis 41, September & December 2004, pp 45–53) • Recent development • Misperception that P. vivax is benign and easily treated • Gravity of the threat posed by vivax malaria to public health has been poorly appreciated (Clin Microbiol Rev. 2009 Jul;22(3):50834.) • Severe and fatal disease have been associated with P. vivax infection • Resistance in P. vivax is more serious as hypnozoites will cause relapse of resistant parasites (J Vect Borne Dis 41, September & December 2004, pp 45–53) CHLOROQUINE RESISTANT P. VIVAX • Reported in focal areas of India, Burma, Indonesia, Papua New Guinea, Brazil, Guyana, Colombia and Solomon Islands • In Papua, chloroquine resistant P. vivax constitutes a significant public health problem. (J Postgrad Med March 2004, Vol. 50, No. 1, p. 41) CHLOROQUINE RESISTANT P. VIVAX • Artemisinin resistance has been obtained in laboratory models • Genetically stable and transmissible artemisinin (ART) and artesunate (ATN)resistant malaria parasites has been selected in the rodent malaria parasite Plasmodium chabaudi (Antimicrob Agents Chemother. 2006 Feb;50(2):4809) • Decreased susceptibility to artesunate has been reported in Western Cambodia (N Engl J Med. 2009 Jul 30;361(5):45567) • Resistant parasites have mutations in PfATP6, a Ca++ ATPase and putative drug target ARTEMISININ RESISTANCE WHO CLASSIFICATION OF RESISTANCE Traditionally, response to treatment was categorised purely on parasitological ground as sensitive, RI, RII and RIII level of resistance RI: (low grade): recrudescence of the infection between 7 and 28 days of completing treatment following initial resolution of symptoms and parasite clearance RII: (high grade): Reduction of parasitaemia by >75% at 48 hours, but failure to clear parasites within 7 days RIII: Parasitaemia does not fall by >75% within 48 hours (Manson’s Tropical Diseases, 21st Ed., 2003, p. 1262) WHO CLASSIFICATION OF RESISTANCE Modified based on clinical, parasitological and fever assessment Early treatment failure (ETF) (If the patient develops one of the following during the first three days of followup) • Development of danger signs or severe malaria on Day 1, Day 2 or Day 3, in the presence of parasitemia; • Parasitaemia on Day 2 higher than Day 0 count irrespective of axillary temperature; • Parasitemia on Day 3 with axillary temperature > 37.5°C; • Parasitemia on Day 3 > 25 % of count on Day 0. WHO CLASSIFICATION OF RESISTANCE Late Clinical Failure (LCF) (If the patient develops one of the following during the followup period from day 4 to day 28) • Development of danger signs or severe malaria after Day 3 in the presence of parasitemia, without previously meeting any of the criteria of Early Treatment Failure • Presence of parasitemia and axillary temperature > 37.5 °C (or history of fever) on any day from Day 4 to Day 28, without previously meeting any of the criteria of Early Treatment Failure WHO CLASSIFICATION OF RESISTANCE Late Parasitological Failure (LPF) (If the patient develops one of the following during the followup period from day 7 to day 28) • Presence of parasitemia on any day from Day 7 to Day 28 and axillary temperature < 37.5 °C, without previously meeting any of the criteria of Early Treatment Failure or Late Clinical Failure Adequate Clinical Response (ACR) (if the patient shows one of the following during the followup period (Up to day 28) • Absence of parasitemia on Day 28 irrespective of axillary temperature without previously meeting any of the criteria of Early Treatment Failure or Late Clinical Failure or Late NEW APPROACHES TO TACKLE DRUG RESISTANCE Research into new compounds with novel mechanism of action Reversing resistance of existing drugs Combination Therapy (Artemisinin Combination Therapy) Approach taken from Tuberculosis LIMITATION OF ARTEMISININ MONOTHERAPY High recrudescence rates (1015%) is reported with artemisinin monotherapy Artemisinin compound clears most but not all parasites very rapidly 7 day dosage is required with monotherapy Postgrad. Med. J. 2005;81;7178 COMBINATION THERAPY IN MALARIA Combination therapy with antimalarial drugs is the simultaneous use of two or more blood schizonticidal drugs with independent modes of action and different biochemical targets in the parasite. The aim is to improve efficacy and to retard the development of resistance to the individual components of the combination. 0 1 2 week artemisinin level partner level parasitemia ARTEMISININ COMBINATION THERAPY WHO GUIDELINES (2006) WHO has endorsed ACT as firstline treatment for acute uncomplicated malaria, where the potentially life threatening parasite P. falciparum is the predominant infecting species. ACT: Artemisininbased combination therapy WHO INITIATIVE Artesunate + amodiaquine Artemether/lumefantrine Artesunate + SP Artesunate + mefloquine FDC MDT WHO Technical Consultation on “Antimalarial Combination Therapy” – April 2001 Clinicians should keep a watch on resistance Clinicians should not use Artemisinin and its derivatives as first line agent in malaria Artemisinin and its derivatives should not be used in vivax malaria Clinicians should use Artemisinin Combination Therapy in uncomplicated falciparum malaria ROLE OF CLINICIANS TO COMBAT DRUG RESISTANT MALARIA Development of synthetic trioxane in collaboration with CDRI Development of artesunate/curcumin copackage in collaboration with IISc and NIMR Stopping the manufacture of single ingredient oral artemisinin derivatives CME’s on ACT CME’s by expert Dr. Peter Weina from Walter Reed Institute IPCA’S INITIATIVE FOR RESTRICTION OF DRUG RESISTANT MALARIA The emergence and spread of drug resistant malaria represents a considerable challenge to controlling malaria. Very few new drugs are in pipeline It is essential to ensure rational deployment of the few remaining effective drugs, to maximize their useful therapeutic life WHO advocates Artemisinin combination therapy for uncomplicated falciparum malaria Clinician play an important role in restricting drug resistant malaria CONCLUSIONS |
10676 | https://byjus.com/maths/square-root-of-32/ | In Mathematics, the square root of 32 is a number, which when multiplied by itself results in the original number 32. The value of square root of 32 is an irrational number, as it cannot be expressed in the form of p/q. In this article, we are going to discuss two different methods such as the long division method and prime factorization method to find the square root of 32 in detail.
Table of Contents:
What is the Square Root of 32?
Square Root of 32 in Radical Form
Square Root of 32 by Prime Factorization Method
Square Root of 32 by Long Division Method
Examples
FAQs
What is the Square Root of 32?
The square of 32 is a number if it is multiplied by itself and gives the result as 32. The square root of 32 is symbolically expressed as √32.
Hence, √32 = √(Number × Number)
Thus, if we multiply the number 5.656 two times, we get the original value 32.
(i.e) √32 = √(5.656× 5.656)
√32 = √(5.656)2
Now, remove square and square root, we get
√32 = ± 5.656.
| Square Root of 32 in Decimal Form: 5.656. |
Square Root of 32 in Radical Form
The square root of 32 can also be expressed in the radical form. If we know the prime factorization of 32, we can write the simplest radical form of the square root of 32. Thus, the prime factorization of 32 is 2 × 2 × 2 × 2 × 2. If it is written in the radical form, then the simplest radical form of the square root of 32 is 4√2.
| Square Root of 32 in Radical Form: 4√2. |
Square Root of 32 by Prime Factorization Method
To find the square root of 32 using the prime factorization method, first, write the prime factorization of 32. Thus, the prime factorization of 32 is 2 × 2 × 2 × 2 × 2.
Thus, √32 = (√2)2. (√2)2. √2
√32 = (2×2)×√2.
√32 = 4√2
We know that,
√2 = 1.414
Now, substitute the values of √2 in the above equation.
√32 = 4×1.414
√32 =5.656 (approximately)
Hence, the square root of 32 in decimal form is approximately equal to 5.656.
Square Root of 32 by Long Division Method
The procedure to find the square root of 32 using the long division method is given as follows:
Step 1: Write the number 32 in decimal form. To find the exact value of the square root of 32, add 6 zeros after the decimal point. Hence, 32 in decimal form is 32.000000. Now, pair the number 32 from right to left by putting the bar on the top of the number.
Step 2: Now, divide the number 32 by a number, such that the product of the same number should be less than or equal to 32. Thus, 5×5 =25, which is less than 32. Thus, we obtained the quotient = 5 and remainder = 7.
Step 3: Double the quotient value, so we get 10, and assume that 100 is the new divisor. Now, bring down the value 00 for division operation. So, the new dividend obtained is 700. Now, find the number, such that (100 + new number) × new number should give the product value, that should be less than or equal to 700. Hence, (100+6) × 6 = 636, which is less than 700.
Step 4: Now subtract 636 from 700, and we get 64 as the new reminder, and 56 as a quotient.
Step 5: The new quotient obtained is 56, and double that. Hence, we get 112 and assume that 1120 is our new divisor. Now, bring down the two zeros and, we have 6400 as the new dividend.
Step 6: Find the number, such that (1120 + new number) × new number should give the product value, that should be less than or equal to 6400. Thus, (1120+5)× 5 = 5625, which is less than 6400.
Step 7: Subtract 5625 from 6400, and we get 775 as the new reminder.
Step 8: Continue this process until we get the approximate value of the square root of 32 up to three decimal places. (Note: keep the decimal point in the quotient value after bringing down all the values in the dividend).
Step 9: Thus, the approximate value of the square root of 32, √32 is 5.656.
| |
| Learn More on Square Root of a Number: Square Root of 30 Square Root of 343 Square Root of 1444 Square Root of 11 |
Video Lessons on Square Roots
Visualising square roots
Finding Square roots
Examples
Example 1:
Simplify 40 -√32.
Solution:
Given: 40 – √32.
We know that the square root of 32 is 5.656
Now, substitute the value in the expression, we get:
40 -√32 = 40 – 5.656
40 -√32 = 34. 344
Therefore, 40 -√32 is 34.344.
Example 2:
Simplify 4√32×√32.
Solution:
Given: 4√32×√32
We know that the square root of 32 in the simplest radical form is 4√2.
Now, substitute √32 = 4√2 in the given expression, we get
4√32×√32= 4(√32)2
4√32×√32 = 4(32)
4√32×√32 = 128 .
Hence, the simplified form of 4√32×√32 is 128.
Example 3:
Find the value of m, if m√32 +15 = 70.
Solution:
Given equation: m√32 +15 = 70…(1)
We know that √32 = 5.656.
Now, substitute the value in equation (1), we get
m(5.656) + 15 = 70
m(5.656) = 70-15
5.656m = 55
m = 55/5.656
m = 9.724
Therefore, the value of m is 9.724.
Frequently Asked Questions on Square Root of 32
Q1
What is the value of the square root of 32?
The value of the square root of 32 is approximately equal to 5.656.
Q2
What is the square root of 32 in radical form?
The square root of 32 in radical form is 4√2.
Q3
Is the square root of 32 a rational number?
No, the square root of 32 is not a rational number, as it cannot be expressed in the form of p/q.
Q4
What is the value of 5 plus square root of 32?
We know that √32 = 5.656.Hence, 5+√32 = 5+5.656 = 10.656.
Q5
What is the value of the square of square root of 32?
The square of square root of 32 is 32.
(i.e) (√32)2 = 32.
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10677 | https://chimpsnw.org/tag/brow-ridge/ | brow ridge Archives - Chimpanzee Sanctuary Northwest
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brow ridge
The Eyes Have It
January 21, 2017 by Diana
I took some photos of Burrito this morning that caused me to admire his face and think about the chimpanzee facial features that I love.
The top of my personal “awesome chimpanzee facial features” list is the brow ridge (also known as supraorbital torus, ridge, or arch).
Though much less pronounced in chimpanzees, humans have a supraorbital ridge too – our eyebrows normally sit near the bottom of our arches.
Each chimpanzee has a somewhat unique brow ridge – I wonder if you could identify a chimpanzee by a brow ridge “print” like a fingerprint.
I will never say that I have a favorite chimpanzee, but I will admit that I have a favorite brow ridge: Jody’s. I like how deeply arched it is with lots of wrinkles, giving it a decidedly heart-shaped appearance:
Here’s a look at everyone else’s supraorbital tori:
Missy:
Negra:
Foxie:
Jamie (that’s also her profile in the 2nd photo in this post):
Annie:
Burrito:
Filed Under: SanctuaryTagged With: arch, brow ridge, chimp, chimpanzee, close up, csnw, photo, profile, rescue, Sanctuary, shelter, supraorbital, torus
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10678 | https://www.reddit.com/r/calculus/comments/574xa5/could_someone_explain_what_yfafaxa_means/ | Could someone explain what y-f(a)=f'(a)(x-a) means.. : r/calculus
Skip to main contentCould someone explain what y-f(a)=f'(a)(x-a) means.. : r/calculus
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•9 yr. ago
Kirshy123
Could someone explain what y-f(a)=f'(a)(x-a) means..
I have this question asking me to find all the points on the graph of y=f(x) at which the tangent line contains the point (-20,2). f(x) = x/x-4. I'm pretty sure I have to use that equation in the title and I know it's the equation formula I'm just not sure what plugs in where and what the a's mean in the formula. If that makes sense :P
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10679 | https://theteacherstudio.com/comparison-problems-with-tape-diagrams-strip-diagrams/ | Skip to content
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Comparison Problems with Tape Diagrams/Strip Diagrams
Comparison Problems: Using Tape Diagrams to Represent Math Thinking
Word problems are a key part of rigorous standards everywhere, but teachers and students alike often dread tackling them. Even textbooks sometimes save them for the bottom few problems on a page or a separate lesson toward the end of a chapter. The thing is–word problems (at least good ones!) are the “real world” part of math. One type of word problems, comparison problems, can be particularly challenging for students. Let’s look at some teaching tips that might make them more accessible!
Critical Reading of Math Problems
As educators, we should always be striving to help our students understand that the skills we are teaching are them are FOREVER…not just to complete a math page or worksheet. One skill that we really want to make sure our students understand is the need to critically read math problems to figure out what is being asked, what information is given, and to make a plan for solving. So often we do the thinking and hard work for our students.
For example, just consider how many of our math books are organized. A lesson entitled “Solving Addition Stories” doesn’t leave much room for student thinking, does it? It seems pretty clear what operation students will need to choose! Providing students with a constantly spiraling variety of problems forces them to think for themselves, learn to look for key information in problems, and make solution decisions accordingly.
One idea? Use highlighters to find and notate important information. Underline or circle the question. One thing I do NOT recommend? Looking for key words like “fewer” or “total”. These words may seem like a quick fix for students…but they can lead them down the wrong path. How? How about this problem…
“Larry has 14 baseball cards. This is 25 fewer than his sister Kara has. How many cards does Kara have?”
By teaching “fewer” as a signal word that indicates subtraction, a student will certainly not think through this problem correctly!
Visualization and Modeling with Comparison Problems
One strategy that can really help students make sense of problems is to be able to visualize and draw models of different problem types. Comparison problems–sometimes represented with “tape” or “strip” diagrams are a GREAT way to help students visualize the math! I thought I’d share a few ways that these can be super helpful for students–whether used as whole class lessons or for intervention groups.
Using a blank “template” or outline of a tape diagram with manipulatives can really help students see the comparison that appears in a problem. The photo below showcases a visual representation of the following problem:
“Bucky has 13 candies. If Samantha has 5 more than Bucky, how many candies will Samantha have?”
Now…this is a VERY simple example and is perfect to use with our students who really need to back up and see the actual manipulation of the numbers. Often we make too many assumptions. Sometimes we move to the “abstract” too quickly and need to make things more concrete for students to truly “see” the math happening.
Next Steps with Comparison Problems
As students get more adept at these problems, you might see that a sketch with only numbers placed in the diagram is appropriate. Check out this lesson where we “filled” a diagram and then brainstormed a ton of different questions that could work with this problem. Again, part of making sense of problems is realizing that the QUESTION matters.. Students need to learn to be detectives as they problem solve. Finding the missing part is the first step!
Another strategy to get students really visualizing is to take a blank diagram and create different stories to go along with them. This is a fantastic way to help students get flexible with their thinking. I start with modeling some problems and then have students progress to writing and sharing their own. It truly helps them understand the problems at a deeper level.
Hope you found this information valuable! These problems don’t HAVE to be as challenging if we help students bridge the gap between the problem and a meaningful strategy to solve them!
Looking to try some of these on your own?
Click the image below to see more! |
10680 | https://tutco.com/conductive/ask-ian-temperature-scales | Journey Through Temperature Scales-fahrenheit, Celsius, Kelvin, Rankine?
Measure temperature and the use of temperature scales have played a pivotal role in human history, aiding in scientific discoveries, technological advancements, and everyday life. Over the centuries, various temperature scales have been developed, each with its unique origin and applications. In this article, we will explore four different temperature scales, Fahrenheit, Celsius, Kelvin, and Rankine, tracing their historical origins and delving into their modern day uses.
The Fahrenheit Scale (°F)
The Fahrenheit scale was created by Daniel Gabriel Fahrenheit in 1724. He was Polish, of German decent. At the age of 15 he moved to the Netherlands where he lived the rest of his life. His scale is one of the oldest temperature scales still in use today. Fahrenheit initially based his scale on the temperature of a brine of ice, water, and salt. He assigned 0°F to the lowest temperature he could achieve using this mixture, and 92°F or 96°F to the average human body temperature, depending on which historical source you choose. The scale was adjusted slightly so that there would be 180 degrees between the freezing and boiling points of water. That explains why we see them as such unusual numbers as 32°F and 212°F.
The Fahrenheit scale was the common temperature scale in most English speaking countries until the 1960s at which time it was replaced with the Celsius scale over a period of about a decade. The scale remains popular in the United States where it has been used for weather measurements for many years, as well as for cooking and baking measurements. It is preferred in the United States primarily for historical reasons and familiarity. It might be seen in the UK where degrees Fahrenheit might appear alongside degrees Celsius on the news or in newspapers, regarding weather temperatures. The only other countries that use the Fahrenheit scale on a regular basis are the Cayman Islands and Liberia. Some scientific research fields, such as chemistry and engineering, continue to use the Fahrenheit scale for specific applications.
The Celsius Scale (°C)
The Celsius scale was developed by Swedish astronomer Anders Celsius in 1742. Celsius originally defined 0°C as the boiling point of water and 100°C as the freezing point of water, which is backwards from today’s conversion. This scale was later inverted to its current form by either the Carl Linnaeus (Swedish, in 1745) or Jean-Pierre Christin (French, in 1777), depending on your source of history. With that inversion in place the scale was called the “forward Celsius scale” for a while before the ‘forward’ was dropped.
It should be noted that craftsman Pierre Casati, under the direction of Jean-Pierre Christin, built the first mercury filled thermometer. It provided much better accuracy and repeatability than the alcohol based thermometer that was in use until then.
Outside of Sweden, it was called the Centigrade scale (from the Latin centum, meaning 100, and gradus, meaning steps), but that had its own problems. In French, Centigrade was already a word that meant one hundredth of a gradient, which is an angular measurement. The name Centesimal degree was tried for a while but that caused problems in the French and Spanish languages, meaning one hundredth of a right angle. In 1948 the world changed the name to Celsius in honor of its inventor.
In 1972, however, Australia was still using Centigrade in their televised weather forecasts. It wasn’t until February of 1985 that it was mandated that all measurements must be given in degrees Celsius. The Celsius scale gained widespread adoption in Europe and most of the world. Its simplicity and alignment with the freezing and boiling points of water making it ideal for most scientific purposes. When it comes to standardization, Celsius has been through a heck of a run.
The Kelvin Scale (K)
The Kelvin scale, named after the Scottish physicist Lord Kelvin (William Thomson), was developed in the mid-19th century. Unlike the Fahrenheit and Celsius scales, the Kelvin scale is an absolute temperature scale, starting from absolute zero or 0 K (-273.15°C or -459.67°F). Absolute zero is the theoretical temperature at which all molecular movement stops or a system has zero thermal energy. There is no temperature lower than 0 K, meaning there are no negative numbers on the Kelvin scale.
The Kelvin scale revolutionized the study of heat. It provided a fundamental framework for understanding the behavior of gases and the laws of thermodynamics. The scale is primarily used in scientific and engineering disciplines, such as physics, chemistry, and engineering, where reference to an absolute zero value is critical. It is an essential scale for research involving extreme temperatures, such as in cryogenics and (strangely enough) high-temperature physics.
The Kelvin scale is one of the few temperature scales that isn’t marked in degrees. There are no degrees Kelvin or °K. It’s not 293 degrees Kelvin or 293 °K, but simply 293 K or 293 Kelvin. 293 Kelvins (the common plural) is acceptable too, though not preferred.
The Kelvin scale and Celsius scale are related. The size of a step in each scale is exactly the same. If it’s 20°C (293.15 K) outside, and the temperature goes down by 1°C, the temperature in Kelvin goes down by 1 Kelvin to 292.15 K. That makes conversion between the two temperature scales very easy. K = °C + 273.15 and °C = K-273.15.
Some interesting information about absolute zero and how absolute it really is….The average temperature of the universe (remembering that it’s mostly empty space) is 2.3 Kelvin, with areas like the Boomerang Nebula in the Centaurus constellation at a chillier 1 Kelvin. That’s about as cold is at gets.
But not quite. In August of 2021, the lowest temperature ever achieved in a laboratory was 38 picokelvins or 38 trillionths of a Kelvin (0.000000000038 K) using a cloud of rubidium atoms, a magnetic field, a vacuum, and putting the equipment into freefall to eliminate the effect of gravity. Now that’s COLD. The temperature was achieved for about 2 seconds.
The Rankine Scale (°R)
The Rankine scale is named after the Scottish engineer and physicist William John Macquorn Rankine. It is another absolute temperature scale developed in the 19th century though it is the rarest of the four scales discussed here. It is similar to the Fahrenheit scale but starts from absolute zero, meaning that a step in temperature in the Rankine Scale is the same size as a step in temperature in the Fahrenheit scale. Just as the Kelvin scale has no negative numbers, neither does the Rankine scale because its 0°R is absolute zero.
The Rankine scale is used in engineering applications, particularly in the United States and sometimes the UK. Much like Kelvin, it provides a convenient absolute temperature scale for calculations involving gases and thermodynamics. It is used where calculations are performed using imperial units instead of metric units (as with the Kelvin scale). Some specific engineering disciplines are thermodynamics, fluid mechanics, and refrigeration.
Since it is a temperature scale based on absolute zero, the National Institute of Standards and Technology (NIST) advises against using the degree symbol (°) when citing Rankine temperatures, though that is often not the case and degrees (°R) are used.
Here’s a bonus temperature scale to discuss; the Réaumur scale (°Ré)
The Réaumur temperature scale, denoted as °Ré, is a historical temperature scale named after René-Antoine Ferchault de Réaumur, a French scientist who introduced it in 1730. The Réaumur scale was widely used in Europe during the 18th and 19th centuries, particularly in France, where it was developed. Unlike modern temperature scales like Celsius and Fahrenheit, the Réaumur scale had specific applications. More on that later. His scale was based on the freezing and boiling points of water set at 0 °Ré and 80° Ré, respectively. The scale divided this range into 80 equal parts. Réaumur thermometer was alcohol based (look at the year, it was 1730!), so it was not as accurate as today’s thermometers.
The Réaumur scale found applications primarily in the cheese industry, used for the cheese-making processes. Cheese production is highly sensitive to temperature control, and the Réaumur scale provided a convenient reference for cheese makers at the time. Cheese-making involves several temperature-dependent steps such as milk pasteurization, curd formation, and fermentation. The Réaumur scale allowed cheese makers to monitor and control these processes accurately (at the time), and repeatedly. Specific cheeses like Camembert and Brie, for instance, require precise temperature control during maturation. Cheese makers could use the Réaumur thermometer to maintain the appropriate conditions. At the time, the Réaumur scale was revolutionary.
Its limitations became apparent, though, as scientific understanding and temperature measurement technology advanced. The Réaumur temperature scale did not account for temperature behavior at extreme values (being alcohol based), and over time it lacked universality, making it less suitable for modern, globalized industries.
As temperature measurement standards evolved, it was largely replaced by the Celsius scale, which offered greater precision and general acceptance. While the Réaumur scale is no longer the primary choice in cheese production, it is rumored to be found in use by a few cheesemakers in a few remote valleys in Switzerland and Italy. It remains a part of the historical heritage of temperature measurement.
While doing research for this document, the author discovered 73 different temperature scales that had been developed over the years, the vast majority of which are now obsolete. They were created for many different applications, and some were completely incorrect.
A scale called the Wedgwood scale (°W), for example, was used for temperatures above the boiling point of mercury (673°F). It’s purpose was in the fields of pottery, glass making and metallurgy where the scale would be used to measure the temperature of kilns. The only problem was, they had the temperature of the boiling point of mercury incorrect (by a lot!), so their scale was useless. It fell out of favor very quickly!
A most peculiar temperature scale that still exists today, but is mostly unused, is the Planck scale, named after the German physicists Max Planck. It ranges from 0 to 1 Plancks, and that’s it. There’s nothing outside that range. 0 is absolute zero and 1 is the hottest temperature ever achievable in the universe, believed to be ≈2.555 hundred thousand trillion trillion trillion °F.
Temperature scales have evolved over centuries, with each having their unique historical origins and applications. While Fahrenheit, Celsius, Kelvin, and Rankine serve different purposes and are used in various fields, they all contribute to our understanding of temperature and its impact on the world around us. Whether measuring the weather, conducting scientific research, or designing engineering systems, these temperature scales continue to play a crucial role in our lives.
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10681 | https://www.proprep.com/questions/what-does-the-unit-joule-per-coulomb-represent-in-the-context-of-electric-potential-difference-and-h | What does the unit joule per coulomb represent in the context of electric potential difference, and how does this quantity relate to the work done in moving a charge within an electric field?
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What does the unit joule per coulomb represent in the context of electric potential difference, and ...
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The unit joule per coulomb (J/C) represents the electric potential difference, also known as voltage. In the context of electricity and magnetism, voltage is a measure of the electric potential energy per unit charge. It quantifies the work that would be done on a charge by an electric field as the charge moves from one point to another.
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Intro To Potential
Exercise 4 - Electric Field of Infinite Plane
Exercise 1 - Electric Field of Infinite Wire
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Intro To Potential
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00:00 Hello. In this lesson,
00:01 we're going to be speaking about the electric potential.
00:04 A lot of people symbolize the electric potential with a capital V_E,
00:11 so that's the electric potential.
00:12 However, I am going to symbolize it with the Greek letter Phi,
00:17 and I'll explain why soon.
00:19 The electric potential is equal to negative the integral on our
00:26 E field.dr where dr is across our trajectory.
00:34 What this integral means is that I'm going along my trajectory and summing up all
00:40 of my tangential values for my E field across my trajectory.
00:47 Let's explain what that means.
00:49 Let's say I have some trajectory in 3D space, like so.
00:54 Let's say that over here,
00:57 I have my origin.
00:58 At any point from on my trajectory,
01:03 so let's say over here,
01:04 I'm going to have some vector r,
01:07 which goes from my origin and points to this point on my trajectory.
01:13 Imagine a moment later,
01:15 I move further along my trajectory and I'm now at this point over here.
01:20 Then my r vector is going to be from the origin until my new point,
01:26 so now this is my i vector a moment later.
01:30 Because the time difference between these 2 points is tiny,
01:35 the vector connecting these 2 points,
01:38 this blue arrow over here is going to be called the dr vector and that's this over here.
01:46 We can see that our dr is tangential to our trajectory.
01:51 What we'll see is that our trajectory is in fact made up of lots
01:56 of these dr vectors that are just going along,
02:01 and they build up our trajectory.
02:03 Then if I sum up all of my dr's,
02:06 I'll get the displacement of my trajectory.
02:10 Now let's see what our E field is.
02:13 Let's say that we're located at this point over here,
02:16 and let's say that at this point over here,
02:19 our E field is pointing in a direction like so.
02:22 What is my integral do?
02:24 It's going to sum on my component for my E field,
02:29 which is tangential to a trajectory.
02:32 How does it do that?
02:34 It uses the help of this dot product over here.
02:38 The dot product between any 2 vectors is going to
02:44 give the component of this vector in the direction of this vector,
02:51 so the projection of this vector on this vector.
02:55 This integral is going to sum up all of
02:59 the tangential components of our E field along our trajectory.
03:06 This is E, E,
03:07 E. Of course, when we're tangential,
03:11 so it's parallel, so the parallel component of our E field to our trajectory.
03:17 If our E field has no component which is parallel to our trajectory at that point,
03:23 so our E field is always perpendicular to our trajectory.
03:27 Then we're going to be summing up on 0.
03:30 We're going to explain why we're doing
03:33 this soon and the motivation for the electric potential.
03:36 But in the meantime, the last basic explanation is going to be that if our E field is,
03:44 or the parallel component of our E field is a uniform,
03:48 if it's not changing,
03:50 so we can write this integral of E.dr
03:54 with a minus simply as if we just have a parallel component
04:00 and it's uniform as the parallel component dr.
04:04 Then we can just simply write that as the E-field multiplied by L,
04:10 where L is the length of the trajectory,
04:14 and this is only if E is equal to a constant.
04:19 In our mechanics course,
04:21 we saw a similar integral to this,
04:24 where we worked out that the work that a system does is
04:28 equal to the integral of our force vector dot dr.
04:34 It's exactly the same just for our electric potential,
04:38 we have a minus and we're integrating along the E rather than F. Now we know
04:43 that work is simply the change in potential energy,
04:50 and of course, there's a minus over here.
04:52 We've already seen something similar to this integration,
04:56 and not just that,
04:58 what this integral does is exactly what our work integral does.
05:04 We already know from our first chapter on Coulomb's law that our force is
05:09 equal to the charge multiplied by our electric field.
05:15 Our electric field is simply a mathematical function
05:19 that tells me that if I place some kind of charge in some area of space,
05:25 then I can know what the force between that charge and my original charge will be.
05:30 Or we can say that our E field is the force per unit charge.
05:36 We can say that this equation for electric potential is the work done per unit charge.
05:44 As we saw in our mechanics chapter,
05:47 that our energy or our potential energy is equal to negative this integral, so F.dr.
05:56 Now we can see that our electric potential is exactly
06:01 like our potential energy per unit charge.
06:07 In that case, why do we even need electric potential?
06:11 Why do we calculate it?
06:12 We do that in order to find out what our potential energy is in the system.
06:19 Our potential energy of the system is going to be equal to
06:24 our charge q multiplied by our electric potential.
06:30 Just like when we have an electric field,
06:32 if we place some kind of point charge in the electric field,
06:36 we can know what the force is going to be between
06:38 our original charge and our newly placed point charge.
06:42 Similarly, with our electric potential,
06:45 we know we can have some kind of equation for
06:48 the potential and if we put a point charge or some kind of charge,
06:52 a test charge in this function for our electric potential,
06:57 we can know what the potential energy is going to be.
07:02 Our electric potential is almost the same as our potential energy,
07:07 just without knowing what kind of charge I have in my system.
07:13 Why do people sometimes write this out as V_E?
07:18 And why do people sometimes write it as Phi?
07:21 I prefer to symbolize the potential as PHI,
07:25 but people use this V_E.
07:27 V_E is for voltage,
07:31 so let's write volts, and our voltage
07:35 is the difference in potential, Delta is difference.
07:40 It's the difference in our potential,
07:42 so that is equal to our final potential minus our initial potential.
07:50 Our voltage is the difference,
07:53 or the potential difference.
07:54 Then I can say that if I have the potential at any point,
07:58 so the potential difference,
08:00 or our voltage, is going to be the potential at
08:03 that point minus the potential at point 0,
08:07 which is equal to 0.
08:08 This voltage that we know,
08:10 which is our potential difference,
08:12 we have that in our plugs.
08:15 Imagine that you have a plug that looks like so.
08:19 We know that here we have a positive and here
08:23 we have a negative and we have a voltage difference of,
08:28 let's say, 220 volts between our plus and minus.
08:34 Let's try and find out what this potential difference
08:37 or what this voltage in our plugs is equal to.
08:41 We already know that our work is equal
08:44 to negative our difference in our potential energy.
08:50 That means that it's equal to our potential energy at
08:53 the end minus our initial potential energy.
08:58 Then we can say that this is equal to negative.
09:01 Now, our potential energy,
09:03 we can see that it's q multiplied by our electric potential.
09:07 There'll be negative q and then we have our electric potential at the end,
09:14 minus our electric potential at the beginning.
09:18 Then our Phi_f minus Phi_i is what we have over here,
09:24 is equal to our voltage.
09:26 That's equal to negative q,
09:29 our charge, multiplied by our voltage.
09:33 What we can see is that voltage multiplied by
09:36 charge with a negative sign is going to give us work.
09:41 The voltage that I get is my work per unit charge.
09:47 What we have for our plug system is how much voltage we need
09:52 in order to move 1 coulomb of charge from here to here.
09:58 The most important equations to take from this lesson and first of all, all of them.
10:04 But the 3 most important are really these that our electric potential is
10:10 also sometimes known as our voltage and it's equal to the negative integral of E.dr.
10:16 Also, to remember that our voltage is defined as
10:20 our electric potential difference
10:23 and to remember that our potential energy is equal to q,
10:26 or some kind of charge multiplied by our electric potential.
10:30 Now, another important equation to know is what if we want to find the opposite?
10:35 Here, if we have our electric field,
10:37 we can find our potential.
10:38 But what if we have our potential and we want to find our electric field?
10:43 Our electric field is going to be equal to negative grad Phi.
10:52 Remember that Phi is some scalar quantity,
10:55 and we're taking the negative grad of this scalar quantity,
10:59 and then we find our electric field.
11:02 Write this in as well. This is
11:04 the fourth very important equation to take from this lesson.
11:08 That's the end of this lesson.
This lesson discusses the electric potential, symbolized by the Greek letter Phi, which is equal to the negative integral of the electric field (E) multiplied by the displacement vector (dr) along a trajectory. The electric potential is related to the potential energy of a system, which is equal to the charge (q) multiplied by the electric potential. The electric potential difference, or voltage, is the difference between the potential at a point and the potential at point 0, which is equal to 0. Voltage multiplied by charge with a negative sign gives work, and the voltage is the work per unit charge. Finally, the electric field can be found by taking the negative gradient of the scalar quantity Phi.
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10682 | https://byjus.com/maths/adjacent-angles-vertical/ | Adjacent angles and vertical angles are two different types of pairs of angles. These angles are the major concept of geometry, introduced to us in Class 4 and 5. The angles are the basic concept that is studied throughout academics. There are various types of angles we learn at school.
The measure of rotation of a ray, when it is rotated about its endpoint is known as the angle, formed by the ray between its initial and final position. An angle is formed by two rays, that are connected end-to-end.
At times, in geometry, a pair of angles are used. There are various kinds of pairs of angles, like – supplementary angles, complementary angles,adjacent angles, linear pair of angles, opposite angles, etc. In this article, we are going to discuss the definition of adjacent angles and vertical angles in detail.
What are Adjacent Angles?
The two angles are said to be adjacent angles when they share the common vertex and side. The endpoint of the rays, forming the sides of an angle, is called the vertex of an angle. Adjacent angles can be a complementary angle or supplementary angle when they share the common vertex and side.
If the sum of adjacent angles is equal to 90 degrees, then they are complementary to each other
If the sum of adjacent angles is equal to 180 degrees, then they are supplementary to each other
Adjacent Angle Examples
Consider a wall clock, The minute hand and second hand of the clock form one angle represented as ∠AOC and the hour hand forms another angle with the second hand represented as∠COB. Both these pairs of angles i.e.∠AOC and ∠COB lie next to each other and are known as adjacent angles.
∠AOC and ∠COB have a common vertex, a common arm and the uncommon arms lie on either side of the common arms. Such angles are known as adjacent angles.
Properties of Adjacent Angles
Some of the important properties of the adjacent angles are as follows:
Two angles are adjacent-angles, such that
They share the common vertex
They share the common arm
Adjacent angles do not overlap each other
It does not have a common interior-point
Adjacent angles can be complementary or supplementary angles when they share the common vertex.
There should be a non-common arm on both sides of the common arm
Adjacent Supplementary Angles
What is the sum of adjacent angles? The adjacent angles will have the common side and the common vertex. Two angles are said to be supplementary angles if the sum of both the angles is 180 degrees. If the two supplementary angles are adjacent to each other then they are called linear pair.
Sum of two adjacent supplementary angles = 180 o.
Here are some examples of Adjacent angles:
Adjacent angles – Linear Pair
Pair of adjacent angles whose measures add up to form a straight angle is known as a linear pair. The angles in a linear pair are supplementary.
Consider the following figure in which a ray O P→stand on the line segment A B―as shown:
The angles ∠POB and ∠POA are formed at O. ∠POB and ∠POA are adjacent angles and they are supplementary i.e. ∠POB + ∠POA = ∠AOB = 180°
∠POB and ∠POA are adjacent to each other and when the sum of adjacent angles is 180° then such angles form a linear pair of angles.
What are Vertical Angles?
When a pair of lines intersect, as shown in the fig. below, four angles are formed. ∠AOD and ∠COB are vertically opposite to each other and ∠AOC and ∠BOD are vertically opposite to each other. These angles are also known as vertical angles or vertically opposite angles.
Thus, when two lines intersect, two pairs of vertically opposite angles are formed i.e. ∠AOD, ∠COB and ∠AOC, ∠BOD.
According to the vertical angle theorem, in a pair of intersecting lines, the vertically opposite angles are equal.
Difference Between Adjacent and Vertical Angles
Adjacent anglesVertical angles
Two angles with a common arm and vertex are called adjacent angles When two lines intersect each other, then the pair of opposite angles formed at the vertex are called vertical angles
They share a common arm and common vertex They share a common vertex, but no common arm
Adjacent angles are not always equal in measure Vertically opposite angles are equal in measure
Solved Examples
Example 1:Find the value of x.
If m∠AOB = 110° , m∠COB = x and m∠AOC= 70°
Solution:
From figure:
m∠AOB = 110°
m∠AOC = 70°
m∠ COB = x
Now, from figure: ∠AOB = ∠AOC + ∠COB
m∠AOB = m∠AOC + m∠COB
110° = 70° + x
x = 110° – 70°
x = 30°
Example 2:In the given figure, is∠1 adjacent to∠2? Give an answer with justification.
Solution:
In the given figure,∠1 does not share the vertex of∠2.
As it does not obey the important property of adjacent angles, therefore,
∠1 is not adjacent to∠2.
Related Articles
Angles
Lines and Angles
Vertical Angles
Alternate Angles
Corresponding Angles
Linear Pair Of Angles
Complementary Angles and Supplementary Angles
To learn more about angles and other concepts download BYJU’S-The Learning App. Happy Learning!!!
Frequently Asked Questions on Adjacent and Vertical angles
Q1
What are Adjacent Angles?
Adjacent angles can be defined as two angles that have a common vertex and a common side. Two adjacent angles can be either complementary or supplementary based on their sum value.
Q2
What are Vertical Angles?
Vertical angles are defined as the angles opposite to each other when two lines cross (i.e. intersect). It is also known as vertically opposite angles. It should be noted that two vertical angles are always equal.
Q3
Can Vertical Angles be Adjacent?
No, vertical angles can never be adjacent.Adjacent angles are the ones next to each other while vertical angles are opposite from each other.
Q4
Are adjacent angles equal to 180 degrees?
Adjacent angles, that are supplementary to each other, always add up to 180 degrees. If the angles are not linear pairs, then the sum of the two angles is not 180 degrees
Q5
Can two adjacent angles be complementary?
Two adjacent angles that sum up to 90 degrees are complementary to each other.
Test your knowledge on Adjacent angles and Vertical angles
Q 5
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But then,limx→af(x)g(x)=limx→af(x)−f(a)g(x)−g(a)=limx→af(x)−f(a)x−ag(x)−g(a)x−a=limx→a(f(x)−f(a)x−a)limx→a(g(x)−g(a)x−a)=f′(a)g′(a). limx→af(x)g(x)====limx→af(x)−f(a)g(x)−g(a)limx→af(x)−f(a)x−ag(x)−g(a)x−alimx→a(f(x)−f(a)x−a)limx→a(g(x)−g(a)x−a)f′(a)g′(a). By definition, f′(a)=limx→af(x)−f(a)x−af′(a)=limx→af(x)−f(a)x−a and g′(a)=limx→ag(x)−g(a)x−ag′(a)=limx→ag(x)−g(a)x−a. Since f′f′ and g′g′ are assumed to be continuous, f′(a)/g′(a)f′(a)/g′(a) is also limx→af′(x)limx→ag′(x)=limx→af′(x)g′(x).QED limx→af′(x)limx→ag′(x)=limx→af′(x)g′(x).QED | To prove Macho L'Hospital's Rule we first need a lemma: | | | Souped Up Mean Value Theorem: If f(x)f(x) and g(x)g(x) are continuous on a closed interval [a,b][a,b] and differentiable on the open interval (a,b)(a,b), then there is a point cc, between aa and bb, where (f(b)−f(a))g′(c)=(g(b)−g(a))f′(c). (f(b)−f(a))g′(c)=(g(b)−g(a))f′(c). (When g(x)=xg(x)=x, this is the same as the usual MVT.) Proof: Consider the function h(x)=(f(x)−f(a))(g(b)−g(a))−(f(b)−f(a))(g(x)−g(a)). h(x)=(f(x)−f(a))(g(b)−g(a))−(f(b)−f(a))(g(x)−g(a)). This is continuous on [a,b][a,b] and differentiable on (a,b)(a,b), with h′(x)=f′(x)(g(b)−g(a))−g′(x)(f(b)−f(a)). h′(x)=f′(x)(g(b)−g(a))−g′(x)(f(b)−f(a)). Note that h(a)=0=h(b)h(a)=0=h(b). By Rolle's Theorem, there a spot cc where h′(c)=0h′(c)=0. But h′(c)=(f(b)−f(a))g′(c)−(g(b)−g(a))f′(c).h′(c)=(f(b)−f(a))g′(c)−(g(b)−g(a))f′(c). Since this is zero, (f(b)−f(a))g′(c)=(g(b)−g(a))f′(c)(f(b)−f(a))g′(c)=(g(b)−g(a))f′(c). QEDQED | | | | Proof of Macho L'Hospital's Rule: By assumption, ff and gg are differentiable to the right of aa, and the limits of ff and gg as x→a+x→a+ are zero. Define f(a)f(a) to be zero, and likewise define g(a)=0g(a)=0. Since these values agree with the limits, ff and gg are continuous on some half-open interval [a,b)[a,b) and differentiable on(a,b)(a,b). For any x∈(a,b)x∈(a,b), we have that ff and gg are differentiable on (a,x)(a,x) and continuous on [a,x][a,x]. By the Souped up MVT, there is apoint cc between aa and xx such that f′(c)g(x)=f′(x)g(c)f′(c)g(x)=f′(x)g(c). In other words, f′(c)/g′(c)=f(x)/g(x)f′(c)/g′(c)=f(x)/g(x). Also, as xx approaches aa, cc also approaches aa, since cc is somewhere between xx and aa. But then limx→a+f(x)g(x)=limx→a+f′(c)g′(c)=limc→a+f′(c)g′(c). limx→a+f(x)g(x)=limx→a+f′(c)g′(c)=limc→a+f′(c)g′(c). That last expression is the same as limx→a+f′(x)/g′(x)limx→a+f′(x)/g′(x). QEDQED | | | | Proof of the Extended L'Hospital's Rule: We're going to use a single trick, over and over again. Namely, we can always rewrite xx as 11/x11/x, f(x)f(x) as 11/f(x)11/f(x) and g(x)g(x) as 11/g(x)11/g(x). Suppose L=limx→af(x)g(x)L=limx→af(x)g(x), where both ff and gg go to ∞∞ (or −∞−∞) as x→a. Also suppose that L is neither 0 nor infinite. Then L=limx→af(x)g(x)=limx→a1/g(x)1/f(x). Since 1/g(x) and 1/f(x) go to zero as x→a, we can apply the (baby or macho) L'Hospital's rule to this limit: L=limx→a(1/g)′(1/f)′=limx→a−g′(x)/g(x)2−f′(x)/f(x)2=limx→af(x)2⋅g′(x)g(x)2⋅f′(x)=limx→af(x)2g(x)2⋅limx→ag′(x)f′(x)=L2limx→a(f′(x)/g′(x)). Since L=L2limx→a(f′(x)/g′(x)), L must equal limx→a(f′(x)/g′(x)), which is what we wanted to prove. This argument only works for finite and nonzero values of L. However, if L=0, we can apply the same argument to the limit of (f(x)+g(x))/g(x), which then does not equal zero. The upshot is that 1+limx→af(x)g(x)=limx→af(x)+g(x)g(x)=limx→af′(x)+g′(x)g′(x)=1+limx→af′(x)g′(x), hence that lim(f/g)=lim(f′/g′). Finally, if lim(f/g)=±∞, look instead at lim(g/f), which is then zero, so the previous reasoning applies. Since 0=lim(g/f)=lim(g′/f′), lim(f′/g′) must be infinite. By the Souped up MVT, f/g has the same sign as f′/g′, so we must have lim(f/g)=lim(f′/g′). Now that we have L'Hopital's Rule for limits as x→a (or x→a+ or x→a−), we consider what happens as x→∞. Define a new variable t=1/x, so that x→∞ is the same as t→0+.Then limx→∞f(x)g(x)=limt→0+f(1/t)g(1/t). But we know how to apply L'Hospital's Rule to limits as t→0, so this turns into limt→0+ddtf(1/t)ddtg(1/t)=limt→0+−f′(1/t)⋅1t2−g′(1/t)⋅1t2=limt→0+f′(1/t)g′(1/t). Converting back to x=1/t, we get limx→∞f′(x)g′(x), which is what we wanted. Computing a limit as x→−∞ is similar, only with t→0− instead of t→0+. That completes the proof of L'Hospital's Rule. QED | << Prev Next >> | |
10684 | https://www.scribd.com/doc/138686102/Ethers-and-Epoxides | Paula Yurkanis Bruice: Organic Chemistry 6 Edition | PDF
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Paula Yurkanis Bruice: Organic Chemistry 6 Edition
This document summarizes Chapter 10 from the 6th Edition of Organic Chemistry by Paula Yurkanis Bruice. The chapter covers reactions of alcohols, amines, ethers, epoxides, and sulfur-contain…
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Chapter 10
Reactions of Alcohols,Amines, Ethers, Epoxides, and Sulfur-Containing Compounds Organic Chemistry 6
th
Edition
Paula Yurkanis Bruice
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Nucleophilic Subs titution Reactions of Ethers
Ethers, like alcohols, can be activated by protonation:
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Ether cleavage: an S
N
1 reaction:Ether cleavage: an S
N
2 reaction:
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Reagents such as SOCl
2
and PCl
3
can activate alcohols but not ethers Ethers are frequently used as solvents because only they react with hydrogen halides
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10685 | https://www.omnicalculator.com/biology/punnett-square | Punnett Square Calculator
How does the inheritance of traits work? The Punnett square calculator provides you with an answer to that and many other questions. It comes as handy if you want to calculate the genotypic ratio, the phenotypic ratio, or if you're looking for a simple, ready-to-go, dominant and recessive traits chart. Moreover, our Punnet square maker allows you to calculate the probability that a rare, recessive genetic disease will be inherited.
Hey, perhaps you're looking for a more advanced dihybrid cross calculator (with 2 traits and 4 alleles), or an extreme, gigantic trihybrid cross calculator (a three trait punnett square)?
This Punnett square generator will teach you the basics of genetics, and will guide you, step-by-step, on how to create your own genetic square. Read on!
How to do a Punnett square? - examples
Making a simple 1 trait gene chart is extremely easy! You must remember that not all genes can be used to create a Punnett square. Here's a short list of rules to follow:
Given traits must be inherited independently (their genes can not be located close to each other in the genetic material);
External factors cannot influence the inheritance of a gene; and
A given trait must be defined only by the alleles we're going to use in the genetic square.
✅ The blood type inheritance makes a good example of a trait that is perfect to use in the Punnett square calculator.
❌ The height of a child cannot be predicted using the Punnett square method — there are too many variables and genes affecting this trait.
Traits are inherited through genes, the memory banks of the cell. Every gene has two versions, called alleles. We use capital letters for dominant alleles (A) and lowercase for recessive alleles (a). Dominant alleles are superior in terms of strength - if a dominant allele is present, the trait it carries will always be visible. Recessive alleles' features will only be visible if there are no dominant alleles.
If you already know your blood type... why don't you check who you could possibly donate it to?🩸Try using our Blood donor calculator.
Punnett square calculator in practice
Let's say we need to know the probability that our patients' baby will inherit a genetic disorder called cystic fibrosis.
Find out the manner of inheritance.
Autosomal recessive. (Autosomal inheritance means that described genes are located on regular chromosomes [1-22] and not sex chromosomes [X,Y])
2. Study the parents' genetics.
There are children with cystic fibrosis in both of families. Both parents are healthy, but they still may be carriers since the disorder is inherited in an autosomal recessive manner.
3. Fill in the square! We need two Punnett squares for this particular case.
A — Healthy, dominant allele
a — Recessive allele of Cystic Fibrosis
The first situation: both parents are carriers.
| ♂️\♀️ | A | a |
---
| A | AA | Aa |
| a | Aa | aa |
There is a 25% chance (1/4) of giving birth to a child with cystic fibrosis. On average, 75% of children born to these parents will be healthy: out of those 2/3 will be carriers, and 1/3 will inherit no cystic fibrosis alleles. (When the percentages get confusing — try the percentage tool.)
Second situation: only one parent is a carrier.
| ♂️\♀️ | A | a |
---
| A | AA | Aa |
| A | AA | Aa |
In this situation, 100% of babies will be born healthy. 50% of them will inherit one improper allele, making them carriers.
Go ahead, play with our Punnett square calculator and try all of the possible options!
Genotypic ratio and phenotypic ratio
Phenotype describes the appearance, that is, what's visible. Genotype describes hidden genetic properties of a trait. What's the difference? Why does it matter? Let's look at the genetic table below.
| ♂️\♀️ | A | a |
---
| A | AA | Aa |
| a | Aa | aa |
Now, let's calculate the genotypic and phenotypic ratios:
| Result | Genotype | Phenotype |
---
| AA | AA | A |
| Aa | Aa | A |
| aa | aa | a |
Genotypic ratio
AA : Aa : aa = 1 : 2 : 1
Phenotypic ratio:
A : a = 3 : 1
Because allele a is recessive, when it appears with a dominant allele, the trait it carries is not visible, but the allele is still there, ready to potentially be inherited in the future.
Autosomal alleles - homozygous or heterozygous?
Here are some basic definitions that may be crucial for the proper use of the genetic calculator:
Homozygous dominant — Where one set of alleles of one gene describes a particular trait. We can use this concept when both of those alleles are dominant (AA).
Homozygous recessive — We use it when both of described alleles are recessive (aa)
Heterozygous — We use it where one allele is recessive (a), and the other is dominant (A).
Mendelian inheritance
The basic rules of genetics were created by Gregor Mendel in 1865, thanks to his simple experiments conducted on garden peas. During that era, humanity had no microscopes, complex scientific technology, or the slightest concept of genes. With simple experiments and insightful observations, he was able to draw conclusions that are useful up to this day - it's no wonder he's called the Father of genetics.
Traits are unitary (red color vs. yellow color);
There are two versions of every gene (now we call them alleles);
There are types of alleles which are superior to the other types (dominant alleles);
Alleles are segregated in a random way;
The chance either allele will be inherited is equal; and
Genes are inherited independently.
A few centuries later, we can undoubtedly say that Mendel was not entirely right - some of the genes are inherited together because of their close proximity on the chromosome. Moreover, some of the genes are codominant: two different dominant alleles can coexist and be visible in the phenotype at the same time. Blood type inheritance is an excellent example of that since dominant alleles A and B cooperate in creating the AB blood type.
Types of Punnett squares
Our Punnett square maker works on autosomal alleles (chromosomes 1-22), but it can be used for other things.
Let's think about X-linked diseases — disorders that are inherited only via the female line of the family. Every woman has two different X chromosomes inherited from her parents. If one of them is faulty or sick, the second, healthy one may take its function. Every man, however, is equipped with only one X chromosome. This way, only one incorrect allele can cause diseases among men, but not among women.
Hemophilia is a rare genetic, X-linked disease. We want to know the chances that a male patient with hemophilia will have a baby with this disorder. His partner is healthy and has no traces of the disease in their family.
XD — Healthy X chromosome;
Xd — X chromosome with Hemophilia gene; and
Y — Y chromosome.
| ♂️\♀️ | XD | XD |
---
| Xd | XdXD | XdXD |
| Y | XDY | XDY |
We can clearly see that all of the patient's children will be healthy. However, all of his daughters will be carriers, and may transfer the disease to the next generation. All of his sons will be completely free of the disease.
FAQs
How to use a punnett square?
Find the genotypes of both parents. Consider if they are homozygous dominant, recessive, or heterozygous.
Fill the first column and row with the parent's alleles.
Mix each allele of one parent with the alleles of the other.
For example, if both parents are heterozygous, the Punnett square will look like this:
| ♂️\♀️ | A | a |
---
| A | AA | Aa |
| a | Aa | aa |
There's a 75% chance of carrying the dominant allele.
How to find genotype and phenotype from a Punnett square?
Look at the result of the Punnett square.
To find possible genotypes, locate different combinations of alleles - AA, Aa, or aa. You can determine the genotypic ratio by counting the number of occurrences of each genotype.
Based on the possible genotypes, you can assess the phenotypes. For example, if allele A is dominant and a is recessive:
Genotype AA will be expressed by phenotype A;
Genotype Aa will be expressed by phenotype A; and
Genotype aa will be expressed by phenotype a.
How to tell if a genotype is heterozygous or homozygous?
An organism with two different alleles at a gene locus (one dominant and one recessive — Aa) has a heterozygous genotype.
Homozygous genotype signifies the presence of two identical alleles (both normal or identically mutated — AA or aa).
What is the purpose of a Punnett square?
By using the Punnett square, we can find the probability of getting specific genotypes and phenotypes as a result of cross-breeding. While it's a good method to learn mendelian rules of inheritance, it's often not applicable to studying humans, as multiple genes often determine human traits. For example, more than ten genes influence eye color!
VPD Calculator (Vapor Pressure Deficit)
Sod Calculator
Learning genetics? Enter parent genotypes to quickly predict offspring traits with our Punnett Square calculator!
A - Dominant allelea - Recessive allele
Punnett Square
| ♂️\♀️ | A | A |
---
| A | AA | AA |
| A | AA | AA |
If you need 2 traits and 4 alleles, search for the larger, dihybrid cross Punnett Square calculator.
Genotype to phenotype
| Genotype | Phenotype |
--- |
| AA | A |
| Aa | A |
| aa | a |
Share result
Reload calculatorClear all changes
Did we solve your problem today?
Check out 5 similar genetics calculators 🧬
Allele frequency
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DNA copy number |
10686 | https://www.andrews.edu/~rwright/Precalculus-RLW/Text/10-05.html | Precalculus by Richard Wright
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My flesh and my heart may fail, but God is the strength of my heart and my portion forever. Psalms 73:26 NIV
10-05 Mathematical Induction
Summary: In this section, you will:
Write a proof for a sum formula using mathematical induction.
Prove other mathematical statements using mathematical induction.
SDA NAD Content Standards (2018): PC.7.2
Note: This assignment is long and should take two days.
The derivations for the sum of an arithmetic series and geometric series are given in lessons 10-03 and 10-04, but how do we know other sum formulas are correct? This lesson shows how to prove sum formulas and a few other mathematical statements.
Mathematical Induction
Mathematical induction is a method of proving mathematical statements by showing that the statement works for the first value and then works for the next value. As long as the next value works, all the values will work and the statement is proven.
Mathematical Induction
To prove a formula using mathematical induction
Show the formula works for n = 1.
Assume the formula works for n = k.
Show the formula works for n = k + 1.
Specifically for sum formulas
Show S1 = a1.
Assume Sk.
Show Sk+1 = Sk + ak+1.
Prove a Sum Formula
Prove 5 + 9 + 13 + 17 + ⋯ + (4n + 1) = n(2n + 3).
Solution
The left side of the equation is terms; the right side is the sum formula.
5a1+,9a2+,13a3+,17a4+,⋯+(4n+1)an==n(2n+3)Sn
Show the formula works for n = 1 by plugging in a 1 for n. In other words, show that S1 = a1.
S1 = 1(2(1) + 3) = 5 = a1
Assume the formula works for n = k. So, plug in a k for n in the sum formula.
Sk = k(2k + 3)
Show the formula works for Sk+1. For sum formulas that means to show that Sk+1 = Sk + ak+1. This means that the sum through the next term equals the sum through the current term plus the next term.
Sk+1 = Sk + ak+1
(k + 1)(2(k + 1) + 3) = k(2k + 3) + 4(k + 1) + 1
Simplify both sides and show that they are equal.
(k + 1)(2k + 2 + 3) = 2k2 + 3k + 4k + 4 + 1
(k + 1)(2k + 5) = 2k2 + 7k + 5
2k2 + 7k + 5 = 2k2 + 7k + 5
Since the formula works for the next term (k + 1), it will work for all the terms and the proof is complete.
Prove a Sum Formula
Prove 3+9+19+33+⋯+(2n2+1)=n(2n2+3n+4)3.
Solution
The left side of the equation is terms; the right side is the sum formula.
3a1+,9a2+,19a3+,33a4+,⋯+(2n2+1)an==n(2n2+3n+4)3Sn
Show the formula works for n = 1 by plugging in a 1 for n. In other words, show that S1 = a1.
S1=1(2(1)2+3(1)+4)3=3=a1
Assume the formula works for n = k. So, plug in a k for n in the sum formula.
Sk=k(2k2+3k+4)3
Show the formula works for Sk+1. For sum formulas that means to show that Sk+1 = Sk + ak+1. This means that the sum through the next term equals the sum through the current term plus the next term.
Sk+1 = Sk + ak+1
(k+1)(2(k+1)2+3(k+1)+4)3=k(2k2+3k+4)3+2(k+1)2+1
Simplify both sides and show that they are equal.
(k+1)(2(k2+2k+1)+3(k+1)+4)3=2k3+3k2+4k3+2(k2+2k+1)+1
(k+1)(2k2+4k+2+3k+3+4)3=2k3+3k2+4k3+2k2+4k+3
(k+1)(2k2+7k+9)3=2k3+3k2+4k3+6k2+12k+93
2k3+9k2+16k+93=2k3+9k2+16k+93
Since the formula works for the next term (k + 1), it will work for all the terms and the proof is complete.
Prove 2+6+12+20+⋯+n(n+1)=n(n2+3n+2)3.
Solution
Show your work. The final step should end with k3+6k2+11k+63.
Prove a Mathematical Statement
Prove n! ≥ 2n where n ≥ 4.
Solution
Start by checking n = 4 because that is the lowest value of n.
4! ≥ 24
24 ≥ 16
Assume the formula is valid for n = k.
k! ≥ 2k
Show it works when n = k + 1. Do this by plugging in k + 1. Then simplify both sides so that the 2nd step is in each side. Finally compare the two sides.
(k + 1)! ≥ 2k+1
(k + 1)k(k − 1)(k − 2)… ≥ 2k·21
(k + 1)k! ≥ 2k · 2
The blue part is the assumption and is true. The black part is always true when k ≥ 4. Thus, the proof is complete.
Prove n2 < 3n where n > 2.
Answer
22 < 32 → 4 < 9
Assume k2 < 3k
(k + 1)2 < 3k + 1
k2 + 2k + 1 < 3k · 31
Prove a Factor
Prove that 2 is a factor of 3n − 1.
Solution
Show that it is true when n = 1.
31 − 1 = 2
Assume 2 is a factor when n = k.
3k − 1
Show that it works when n = k + 1.
3k+1 − 1
The trick is to subtract and add 3k.
3k+1 − 3k + 3k − 1
(3k+1 − 3k) + (3k − 1)
(3k·31 − 3k) + (3k − 1)
3k(3 − 1) + (3k − 1)
2·3k + (3k − 1)
2 is a factor of the blue part and 2 is a factor of the red part from the assumption, thus 2 is a factor of the whole thing.
Prove that 3 is a factor of 4n − 1.
Answer
41 − 1 = 3
Assume 3 is a factor of 4k − 1
4k+1 − 1
4k+1} − 4k + 4k − 1
(4k+1 − 4k) + (4k − 1)
(4k·41 − 4k) + (4k − 1)
4k(4 − 1) + (4k − 1)
3·4k + (4k − 1)
3 is a factor of the blue part and 3 is a factor of the red part from the assumption, thus 3 is a factor of the whole thing.
Lesson Summary
Mathematical Induction
To prove a formula using mathematical induction
Show the formula works for n = 1.
Assume the formula works for n = k.
Show the formula works for n = k + 1.
Specifically for sum formulas
Show S1 = a1.
Assume Sk.
Show Sk+1 = Sk + ak+1.
Helpful videos about this lesson.
Mr. Wright Teaches the Lesson (
Practice Exercises
Substitute k+1 into the expression and simplify.
nn+3
n(n+1)4
Prove the sum formulas using mathematical induction.
2 + 4 + 6 + 8 + ⋯ + 2n = n(n + 1)
2+7+12+17+⋯+(5n−3)=n2(5n−1)
1 + 2 + 4 + 8 + ⋯ + 2n−1 = 2n − 1
1+2+3+4+⋯+n=n(n+1)2
∑i=1ni(i+1)=n(n+1)(n+2)3
Prove the inequality using mathematical induction.
2n ≥ 2n where n ≥ 2
n! > 2n where n ≥ 4
Prove the property using mathematical induction.
(ab)n = anbn
3 is a factor of n3 + 3n2 + 2n
5 is a factor of 6n − 1
Mixed Review
(10-04) Is 3(2)n−1 geometric, arithmetic, or neither?
(10-04) Write the rule for the nth term of 3, 9, 27, 81, …
(10-03) Evaluate 3 + 6 + 9 + 12 + 15 + ⋯ + 39.
Answers
k+1k+4
k2+3k+24
Show work and final step should have k2 + 3k + 2
Show work and final step should have 5k2+9k+42
Show work and final step should have 2k+1 − 1
Show work and final step should have k2+3k+22
Show work and final step should have k3+6k2+11k+63
Show work
Show work
Show work
Show work
Show work
geometric because it is exponential
an = 3n
273 |
10687 | https://t5k.org/glossary/xpage/Residue.html | residue
Suppose a and m are any two integers with m not zero. We say r is a residue of a modulo m if a = r (mod m). This is the same as m divides a − r (see congruence), or a = r + qm for some integer q. The division algorithm tells us that there is a unique residue r satisfying 0 < r < |m|, and this remainder r is called the least nonnegative residue of a modulo m.
A set of integers form a complete system of residues modulo m if every integer is congruent modulo m to exactly one integer in the set. So a complete system of residues includes exactly one element from each congruence class modulo m.
For example, if m is positive, then
{0, 1, 2, 3,..., m−1}
is a complete system of residues (called the least nonnegative residues modulo m). If m is positive and odd, then we sometimes use the system
{ − (m−1)/2, − (m−3)/2, ..., −1, 0, 1, ..., (m−3)/2, (m−1)/2}
There are infinitely many complete residue systems for each modulus m.
Printed from the PrimePages © Reginald McLean. |
10688 | https://www.statcrunch.com/help/view?example=11 | Analyzing data from a randomized block design
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Analyzing data from a randomized block design
This tutorial covers the steps for doing a randomized block design analysis using repeated measures ANOVA in StatCrunch. To begin, load the Granola comparison opens in new window data set, which will be used throughout this tutorial. Ten subjects in this fictional study were each asked to sample three kinds of granola cereal, labelled simply "A", "B", and "C", and to rate the granola's taste on a scale of 1 to 10. Each subject was given the three granola samples in random order. The goal is to determine whether the mean ratings for the three granola types differ significantly.
Analysis using two way ANOVA
This is a randomized block design, where each of the ten subjects is a "block". Choose Stat > ANOVA > Two Way. Select the Rating column for Responses in. Select the Subject column for Row factor in. Select the Granola column for Column factor in. Check the boxes for Display means table and for Fit additive model,then click Compute!The resulting ANOVA table shows the following mean ratings: mean A=6.1, mean B=6.2, and mean C=7.3. The P-value for Granola is shown as 0.0007, meaning that there is about a 7 in 10,000 chance that differences this large would occur by random chance.
Analysis using repeated measures ANOVA
To analyze the data using repeated measures ANOVA,choose Stat > ANOVA > Repeated Measures. Select the Rating column for Responses in. Select the Granola column for Treatments in. Select the Subject column for Blocks in and click Compute!The resulting ANOVA table shows exactly the same means and P-value for Granola as does two way ANOVA. In fact, two way ANOVA with Fit additive model specified will always give the same results as repeated measures ANOVA. Fit additive model indicates that there is no interaction between the two factors, and is required when there is only one observation for each combination of factor levels. (Repeated measures ANOVA always assumes an additive model, so it is not specified in that procedure.)
If a one way ANOVA had been performed instead, without the benefit of blocking, the mean for each Granola type would have been the same as with two way or repeated measures ANOVA, but the P-value would instead be 0.113, meaning that there is about a 1 in 9 chance of differences this large occurring by random chance. The two way ANOVA gives a far more sensitive test (that is, a much smaller P-value) in this case because it removes the variation due to subjects. Looking at the means table for two way ANOVA (above), it is clear that there is not much variation for the ratings of an individual subject across the three granola types. There is more variation in ratings across subjects.
Help topic 5 of 38
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10689 | https://nrich.maths.org/problems/tilted-squares | Problem-Solving Schools can now access the Hub!
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Tilted squares
It's easy to work out the areas of most squares that we meet, but what if they were tilted?
Age
11 to 14
Challenge level
Secondary
GEOMETRY
Angles, Polygons and Geometrical Proof
Exploring and noticing Working systematically Conjecturing and generalising Visualising and representing Reasoning, convincing and proving
Being curious Being resourceful Being resilient Being collaborative
Problem
Tilted Squares printable sheet
It's easy to work out the areas of squares drawn on a grid if they are oriented in the usual way:
Image
Can you find a quick and easy method to work out the areas of tilted squares? Here are some squares with a tilt of 1:
Image
See Getting started for suggested ways to calculate their areas.
Notice anything special about their areas?Can you predict the areas of other squares with a tilt of 1? What about squares with a tilt of 2? Or 3? Or 4? Or...?Notice anything interesting?Can you make any conjectures about the areas of tilted squares?Can you prove your conjectures?You might like to use the interactivity below to help you to draw tilted squares.Full Screen tablet/mobile friendly version
Getting Started
Two standard methods for working out the areas of shapes may help.
Boxing in the tilted squares, working out the area of the box and subtracting the areas of the unwanted right-angled triangles:
Image
Splitting the tilted squares into right-angled triangles and squares and adding the areas of the different parts:
Image
Student Solutions
Thank you to everyone who submitted a solution to this problem - we received a very large number!
Can you find a quick and easy method to work out the areas of tilted squares?
Phoebe from Long Field Spencer Academy, Harry B from St Johns College School and Meher, all in the UK, used counting methods. Phoebe and Harry counted the squares in the underlying grid. Harry B's work is shown below. Harry has labelled with '1' the pieces which add up to the 'first' square counted, then with '2' the pieces which add up to the 'second' square counted, and so on.
Image
Meher counted dots using the method shown below. Why does Meher's method work?
To get the area of a tilted square, you have to count how many dots there are inside a square. This will clearly tell you the area.
Leonardo from The British School of Milan and Cole suggested rotating the tilted square to find its area. Leonardo wrote:
A quick method to find their area is just to bring the tilt the square back to normal and the area will be the same as the tilted one.A way to prove them is easy. Draw a 2 by 2 square. Tilt it once. As you can see the square is still 2 by 2, so the area will still be 4.The answer will still be a square number.But for the tilted squares shown in the problem, the side length of the tilted square might not be a whole number...
Sophia from Kings Way School in New Zealand, Elaine and Jenna from New Zealand, Grace and Isla from Long Field Spencer Academy, Adam from Heckmondwike Grammar School in the UK, Holly-Hope and Annabelle from Jersey School for Girls in Jersey, Rafsan from Dulwich College in the UK, Adam M, Adam B, Anika and Lucia from King's School Dubai, Tejas, Ruhi, Valerie, Mithravinda, Rivaan, Kabir, Akshobhyan, Nithya, Deethya, Uday, Udit, Tarikha, Kashvi, Swathi, Devasena, Arnav, Samarth and Shubhangee from Ganit Kreeda in India, Riley and Mar all used a method that involved drawing a larger square and working out the areas of triangles. This is Annabelle's work:
Image
Dan and Freya from Long Field Spencer Academy used a similar method that involved using a smaller square inside the tilted square, and four triangles. This is Dan's work:
Image
Adam and Jack from Heckmondwike Grammar School, Lamech from Dulwich College in the UK and Monty J used Pythagoras' Theorem to find the areas of the tilted squares. This is Lamech's work:
Image
Notice anything special about their areas?Can you predict the areas of other squares with a tilt of 1?
Using the method of drawing a larger, horizontal/vertical square and subtracting the four triangles, Rafsan noticed something about the areas of the triangles:
0.5 is really and truly the star number here as when moving [between two] tilted squares, we add 0.5 [to the area of each little triangle].
Image
Freya noticed a pattern in the areas of the squares - and used Rafsan's idea:
I looked at the pattern of the previous areas of the squares within the squares. They were all square numbers in order, so I based my prediction on this. I figured out how to find the area of the triangles by looking back at what I did earlier. This followed a pattern of increasing by half a square cm per triangle.
Image
Harry B and the students from Ganit Kreeda recorded their results in a table to help spot patterns. This is Ganit Kreeda's work:
Image
Harry B noticed the same thing that Ganit Kreeda have in the "Observation" column:
along$^2$ + 1 = area$a^2+a =$areaConclusion: The area of a square that has been tilted up one is the horizontal base squared add 1.
Ci Hui from Queensland Academy of Science Mathematics and Technology in Australia noticed the same pattern. Ci Hui also made a table to show the areas of the squares. Ci Hui's method for finding the areas is also different to the methods shown above (although similar to Dan and Freya's method), and is explained in the table. Click to see Ci Hui's table and method.
Click on the image to open a larger version
Image
What about squares with a tilt of 2? Or 3? Or 4? Or...?
The students from Ganit Kreeda, Harry B and Ci Hui also made tables to record the areas of squares with other tilts. This is Ganit Kreeda's work:
Image
Harry B, Ci Hui and Shaunak from Ganit Manthan described the same pattern. Harry B began from the original counting squares method, and Ci Hui used the same method as before. Click to see Ci Hui's method applied to squares of tilt 2 and 3, and Ci Hui's table.
Click on the image to open a larger version
Image
Can you make any conjectures about the areas of tilted squares?Can you prove your conjectures?
Harry B had noticed that The area of a square that has been tilted up one is the horizontal base squared add 1. Harry B used the method described by Dan and Freya to prove this. Here is Harry B's proof:
Image
The students from Ganit Kreeda had noticed that Tilt area = square area + (tilt)$^2$. They used another method involving four triangles to prove their conjecture:
Image
From fig 1:Area of tilted square = Area of white partWe then cut blue 4 triangles and rearrange them to form 2 rectanges as shown in fig 2:So, area of the tilted square = area of the white part = $n^2 + t^2$
They also noticed some other things:
Area of the outer square (with side $n+t$) $= (n+t)^2$It is clear from fig 2:Area of outer square $= n^2 + t^2 + 2nt$
Therefore $(n+t)^2 = n^2 + t^2 + 2nt$
We proved and explained pictorially the identity $(a+b)^2=a^2+b^2+2ab$
Also, we proved Pythagoras theorem, sinceArea of tilted square $= n^2+t^2$Side of tilted square $\sqrt{n^2+t^2}$.
Adam M, Adam B, Anika and Lucia derived a formula for the area of a tilted square using the method of subtracting the areas of four triangles from the area of the larger square. This is their work, with two diagrams and a small correction added:
Image
Notice that they use different letters for the base and height of the larger square. If they used $h$ for both, then the final formula would be $h^2 - 2th + 2t^2$, which can be factorised to $(h-t)^2 + t^2.$ Can you see how this is related to the formulas proved by Harry B and the students from Ganit Kreeda?
The students from Ganit Kreeda and Harry B came across Pythagoras' Theorem when they explored and proved their conjectures. Some people worked the other way round - they used Pythagoras' Theorem to prove the formula. Umar from Eastcourt Independent School in the UK, Wenxuan from St George's International British School in Italy, Tamin from Canada, Muhammad, Adam from Heckmondwike Grammar School and Lamech used Pythagoras' Theorem to find the area of a tilted square. This is Tamin's work:
Image
This is Adam's work:
By using the Pythagorean theorem ($a^2 + b^2 = c^2$) you can calculate one side of the square which you can then use to find the area.
Image
This calculation can be further simplified into one equation.
Area $=\sqrt{a^2+b^2}\times\sqrt{a^2+b^2}=a^2+b^2$
This can be shown through a well-known diagram explaining the Pythagorean theorem.
Image
Ci Hui used the same method as before to prove the formula for the area of a tilted square. Ci Hui also made a conjecture about the areas of tilted rectangles. Click to see Ci Hui's work.
Click on the image to open a larger version
Image
Teachers' Resources
Why do this problem?
This problem offers an opportunity to spot patterns, make generalisations and eventually discover Pythagoras's Theorem, while giving students the chance to practise working out areas of squares and right-angled triangles.
Possible approach
This printable worksheet may be useful: Tilted Squares
Tilted Squares - Teaching Using Rich Tasks has a video of two members of the NRICH team teaching a 'Tilted Squares' lesson and sharing the teaching points they consider to be important.The video below shows Charlie and Alison from the NRICH team introducing the problem to a group of Year 9 students who came to visit the Maths Department at Cambridge. Video footage of the later stages of the lesson appear further down the page.
Video 1
Here is a description based on the approach used by a teacher with a Year 8 class, interspersed with some more video footage from the lesson in Cambridge:
I established that everyone could work out the area of squares when they were in the usual orientation:
"4 by 4?"
"16"
"5 by 5?"
"25"
etc.
"But what if we had to work out the area of a tilted square?"
I drew a point on the board, then moved 3 units to the right and 1 unit up and drew another point there.
I used this as the base of my square and then drew the other three sides.
The interactivity in this taskcould be used to familiarise students with drawing tilted squares.
"How might we work out the area of this square?"
One student suggested that the base was 3 cm long and that the area would be $\text{9 cm}^2$.
One student suggested we measure the length with a ruler and then square the result.
We discussed the problems with the two approaches and then I drew a 4 by 4 square around the tilted square and suggested that if we boxed the square in, worked out the area of the box (16 cm$^2$) and subtracted the area of the four unwanted triangles (6 cm$^2$) we would have the area of our tilted square (10 cm$^2$). See the hint for a diagram and an alternative method.
"It would be nice to be able to work out the areas of these tilted squares as fast as you worked out the areas of the original squares"
Students in one row were asked to draw a square with a base that went 4 along and 1 up.
Students in the next row were asked to draw a square with a base that went 5 along and 1 up.
Students in the next row were asked to draw a square with a base that went 6 along and 1 up. Students in the next row were asked to draw a square with a base that went 7 along and 1 up.
Here is a second video clip, showing the results being collected (approximately 5 minutes later) in the lesson in Cambridge:
Video 2
Students were all asked to work out the area of their squares and I then collected their results:
4 along and 1 up: 17
5 along and 1 up: 26
6 along and 1 up: 37
7 along and 1 up: 50
"Do you notice anything about the areas?"
"All 1 more than a square number"
"If you drew a square with a base 8 along and 1 up, what would you expect the area to be?"
"65"
"If you drew a square with a base x along and 1 up, what would you expect the area to be?"
"x$^2$ + 1"
"Great, it looks like we can now work out the areas of these tilted squares very easily."
But what would happen if they were more tilted, say, 3 along and 2 up, or 4 along and 2 up, or...?"
Someone suggested that the rule would be x$^2$ + 2
Here is the final video clip, showing the last part of the lesson in Cambridge, when students made the generalisation that leads to Pythagoras's Theorem:
Video 3
Again the class was split up to work out the areas of these tilted squares and we then collected their results:
3 along and 2 up: 13
4 along and 2 up: 20
5 along and 2 up: 29
6 along and 2 up: 40
"Was our conjecture (x$^2$ + 2) correct?"
"It's x$^2$ + 4"
"It's x$^2$ + 2$^2$"
"The first one should have been x$^2$ + 1$^2$"
At this point, it may be appropriate to work on some ways of justifying the $x^2 + 1^2$ and $x^2 + 2^2$ conjectures.
For groups who have met the idea of expanding $(x+1)^2$ this can be done algebraically.
Alternatively, a pictorial approach could be offered, and students could be asked to explain why this proves their conjecture:
Image
"What do you think will happen if the squares are even more tilted, say 3, or 4, or 5 up?"
"x$^2$ + 3$^2$"
"x$^2$ + 4$^2$"
"x$^2$ + 5$^2$"
Students were split into groups again in the following lesson to check these conjectures and report back.
Finally, algebraic or pictorial approaches used to justify earlier conjectures can be adapted to prove Pythagoras's Theorem:An 'x across, 1 up' tilted square can be enclosed by a square of side length x + 1. The areaof the tilted square is $(x + 1)^2$ minus the areas of the four unwanted triangles. This equals$(x^2 + 2x + 1) - 2x = x^2 + 1$, as predicted.An 'x across, 2 up' tilted square can be enclosed by a square of side length x + 2. The area of the tilted square is $(x + 2)^2$ minus the areas of the four unwanted triangles. This equals$(x^2 + 4x + 4) - 4x = x^2 + 4$, as predicted.An 'x across, y up' tilted square can be enclosed by a square of side length x + y. The area of the tilted square is $(x + y)^2$ minus the areas of the four unwanted triangles. This equals $(x^2 + 2xy + y^2) - 2xy = x^2 + y^2$, as predicted.Follow-up lessons could focus on working out the lengths of the sides of right-angled triangles when two lengths have been given.
Key questions
How could you work out the area of a tilted square?
If you drew a square with a base x along and 1 up, what would you expect the area to be?
If you drew a square with a base x along and 2 up, what would you expect the area to be?
If you drew a square with a base x along and y up, what would you expect the area to be?
Possible support
Start with Square It or Square Coordinates to help students become confident at drawing tilted squares.
Possible extension
It is possible to draw squares with areas of 1, 2, 4, 5, 8, 9... but not 3, 6, 7, 11, 12...
Students could explore some of the properties of numbers which are and are not possible areas of tilted squares.
Can they prove that numbers of the form 4n+3 are not possible areas of tilted squares?
Another possible follow-up task is Of All the Areas.
NOTES AND BACKGROUND Ken Nisbet, Mathematics teacher at Madras College in Fife, Scotland, has added:
"I used tilted squares as the basis of individual/group work with a top set (age 14). They were given time to explore this as an open ended question in groups in a brain storming session. Write ups were to be done individually, partly in class but completed at home. This is an important stage in the pupils' mathematical development where the idea of "proof" is coming to the fore. This excellent investigation allows algebra to come to the fore as the language of generalisation and the means of "proof" of patterns. At this stage algebra skills are limited but we have now used this investigation as a springboard to developing necessary algebra skills - e.g. double brackets, squares, expressions etc." |
10690 | http://bradburydunlop.pbworks.com/f/Factoring+%26+Place+Value.pdf | Divisibility Rules Triclcs that help us determine if a number is a multiple of another number are called divisibility rules. CIRCLE ALL THE CORRECT EXAMPLES 2 divides any number ending with an even digit (2,4,6,8,0). Examples) 32 1234 774 10106 3 divides any number when the sum ofthe digits is divisible by 3. Examples) 312 658 1130622 3232 4 divides any number where the last two digits form a number divisible by 4. Examples) 924 132 10234 7482 5 divides any number ending in 5 or 0. Examples) 105 10575 90637 72 m^m 6 divides any number divisible by 2 AND 3. Examples) 32 312 1130622 774 7 divides any number where the last digit, multiplied by 2 and subtracted from the remaining digits is divisible by 7. Examples) 287 331 455 875 8 divides any number where the last three digits are divisible by 8. Examples) 5008 5387 3244 331008 9 divides any number where the sum ofthe digits is divisible by 9. Examples) 5994 381921 504 775 ?w^.-%- • , , ^ -, • ' ' •'••'•• • " • • • ' -y^s . Divisibility Rules A number Is divisible by... 2 ,3 4 ,5 6 8 9 10 If... the last digit is even (0, 2, 4, 6, or 8) the sum of the digits is divisible by 3 the number formed hy the last two digits is divisible hy 2 at least twice the last digit is 0 or 5 the number is divisible hy both 2 and .^ the number is divisible by 2 at least three times the sum of the digits is divisihie hy 9 the last digit is 0 Numbers cannot be divided by 0. You can use the divisibility rules to find factors of a number. You can write fractions in lowest terms by dividing the numerator and the denominator by common factors until the only common factor is 1. CMmniintaMte tht Mtai 1. a) Why is a number that is divisible by 6 also divisible by 2 and 3? b) A number is divisible by 10. What other numbers is the number divisible by? How do you know? 2. a) Explain one method for determining the greatest common factor of 36 and 20. b) Share your answer with a partner. 3. Simone wrote - ^ in lowest terms as -j-r. i) Is she finished yet? Explain. 18 b) Show a method for writing ~ in lowest terms. 4. Explain what you know about divisibility by 0. Include an example in your explanation. f i 206 MHR. Chapter 6 1.2 More Patterns in Division Quick Review >- A number is divisible by 3 if the sum of its digits is divisible by 3. For example, 1035 is divisible by 3 because 1 + 0 + 3 + 5 = 9, and 9 is divisible by 3. 1036 is not divisible by 3 because 1 + 0 + 3 + 6=10, and 10 is not divisible by 3. >- A number is divisible by 6 if the number is divisible by 2 and by 3. For example, 1038 is divisible by 2 because the number is even. 1038 is divisible by 3 because 1 + 0 + 3 + 8=12, which is divisible by 3. So, 1038 is divisible by 6. V A number is divisible by 9 if the sum of its digits is divisible by 9. For example, 5418 is divisible by 9 because 5 + 4 + 1 + 8 = 18, and 18 is divisible by 9. 5428 is /tot divisible by 9 because 5 + 4 + 2 + 8= 19, and 19 is ttof divisible by 9. >" No number is divisible by 0. >- You can use a Carroll diagram to show numbers that are divisible by two numbers. This Carroll diagram shows divisibility by 6 and by 9. Divisible by 9 Not divisible by 9 Divisible by 6 18,36, 126, 162 6, 12,204,402 Not divisible by 6 27,45,963,711 10, 29, 325, 802 >- You can use divisibility rules to help list the factors of a number. To list the factors of 156: Try each rule in turn. Divide by 2: 156 + 2 = 78 Divide by 3: 156 + 3 = 52 Divide by 4: 156 + 4 = 39 156 is not divisible by 5. Divide by 6: 156 + 6 = 26 156 is not divisible by 7, by 8, by 9, or by 10. Use a calculator to check for divisibility by 11 and 12. 156 is not divisible by 11. Divide by 12: 156 + 12 = 13 Since the factors 12 and 13 are close in value, you have found all the factors. In order, the factors of 156 are: 1, 2, 3, 4, 6, 12, 13, 26, 39, 52, 78, 156 g Copyright © 2008 Pearson Education Canada. The right to reproduce this page is restricted to the purchasing school. Practice 1. Circle the numbers that are divisible by 2. 23 98 21 44 11 77 34 2. Circle the numbers that are divisible by 5. 55 10 7 59 105 775 1025 3. Circle the numbers that are divisible by 2 and by 5. 10 30 25 55 1000 52 4. Write each number in the correct place in the Venn diagram. 16, 20, 33, 64, 80, 95, 97, 105, 214, 216, 324, 405 5. Write four 3-digit numbers that are divisible by 10. Divisible by 4 Divisible by 5 6. Write three 4-digit numbers that are divisible by 8. 7. a) Write each number in the correct place in the Venn diagram. 115, 116, 120, 168, 450, 753, 800,928,1008,1110 Divisible by 8 b) Write 4 more numbers in the Venn diagram - one in each loop and one outside the loops. How do you know you placed each number correctly? Divisible by 10 Copyright © 2008 Pearson Education Canada. The right to 5 reproduce this page is restricted to the purchasing school. Practice 1. Match the niunber with the correct divisibility statement. Draw more than one line if it is needed. 54 Divisible by 10. 56 Divisible by 3. 50 Divisible by 9. 92 Divisible by 8. 75 Divisible by 5. 93 Divisible by 2. 30 Divisible by 6. 2. Cross out the numbers that are not divisible by 2. 12 79 98 134 227 2469 How do you know the numbers are not divisible by 2? 3. Circle the numbers that are divisible by 9. 91 331 333 153 99 12 321 How do you know you are correct? 4. Write four numbers that are divisible by 6: How did you choose those numbers? 5. Solve each riddle. a) I am divisible by 2 and by 3. I am between 21 and 29. Which number am I? 21 22 23 24 25 26 27 28 29 b) I am divisible by 5 and by 10. I am between 56 and 64. Which number am I? c) I am divisible by 2 and by 9. I am between 424 and 449. Which number am I? Copyright © 2008 Pearson Education Canada. The right to 7 reproduce this page is restricted to the purchasing school. 6, Which numbers below are divisible by 3? By 6? By 9? How do you know? a) 124 „ _ _ _ _ _ _ _ _ ^ _ _ _ _ _ „ „ b) 215 c) 330 d) 450 e) 150 7. Use your answers to question 6 to help you list the factors of each number. a) 124: b) 215: c) 150: 8- a) Sort these numbers in the Carroll diagram below. 16, 18,27,37, 120, 180,281,288,352,411,432,540 Divisible by 4 Not divisible by 4 Divisible by 9 Not divisible by 9 b) Write one more number in each part of the Carroll diagram. Explain how you knew where to place each number. 9. a) Sort these numbers in the Venn diagram. 12, 28, 36, 54, 72, 79, 135, 256, 270, 318, 371, 432 b) Which loop is empty? Explain why there is no number that belongs in that loop. Multiples of 4 Multiples of 6 8 Copyright © 2008 Pearson Education Canada. The right to reproduce this page is restricted to the purchasing school. Multiples of 9 Practise For help with #5 to #8, refer to Example 1 on page 202. 5. Which of the following numbers are divisible by 5? Explain how you know. 1010 554 605 902 900 325 6. Which of the following numbers are divisible by 4? Explain how you know. 124 330 3048 678 982 1432 7. a) Use a diagram or table to sort the numbers according to divisibility by 4 and 8. 312 330 148 164 264 13 824 b) If a number is divisible by 4 and 8, what is the smallest number other than I that it is also divisible by? How do you know? 8. a) Using a diagram or table, sort the numbers based on divisibility by 6 and 10. 5832 35 010 243 9810 31990 b) If a number is divisible by 6 and 10, what is the smallest number other than 1 that it is also divisible by? How do you know? For help with #9 to #14, refer to Example 2 on page 203. 9. Use the divisibility rules to list the factors of the following numbers. a) 36 b) 15 c) 28 10. What are the factors of these numbers? a) 18 b) 54 c) 72 11. Use the divisibility rules to determine the common factors for each pair of numbers. a) 3 and 6 b) 4 and 8 c) 6 and 12 12. What are the common factors for each pair of numbers? a) 5 and 10 b) 4 and 12 c) 24 and 15 13. a) Use the divisibility rules to determine the common factors of 16 and 20. Include a Venn diagram as part of your answer. b) What is the greatest common factor of 16 and 20? 14. a) What are the common factors of 10 and 30? Include a Venn diagram with your answer. b) Identify the greatest common factor of 10 and 30. For help with #15 and #16, refer to Example 3 on pages 204-205. 15. Write the following fractions in lowest terms. a) d) ii 20 _9 12 b) e) 6_ 18 4 10 c) f) 10 16 9_ 15 16. Write each fraction in lowest terms. 16 ' ' M 2 '' 20 14 .. 5 ,, 12 a) d) 24 e) 0 J I ip :ll •i': P. i l l 1 f I ' :i 1 8 1 i . t ii ' i ''• !' i ft 15 6.1 Divisibility. MHR 207 ' \ Nam,e 1.8 Factors and Divisibility MATHPOWERTW Seven, pp. 26-27 The factors of a number each divide the number evenly. 4S - 1 = 48 48 - 2 = 24 48 + 3 = 16 48 - 4 = 12 48 - 6 = 8 48 ~ 48 = 1 48 ^ 24 = 2 48 - 16 = 3 48 - 12 = 4 48 -- 8 --- 6 The factors of 48 are 1, 2, 3, 4, 6, 8, 12, 16, 24, and 48. The divisibility rules can help you find the factors of a number. A number is divisible by 2 if it ends in 0, 2, 4, 6, or 8. 6 if it is divisible by 2 and 3. 3 if the sum of the digits is divisible by 3. 8 if the last thr€>e digits are divisible by 8. 4 if the last two digits are divisible by 4. 9 if the sum of the digits is divisible by 9. 5 if it ends in 0 or 5. 10 if it ends in 0. 15. 65: 1, ,., 13, ^ 16. SO: _ , 2, , 5,.., H), _ , 20, _ _ _ , 80 State the mi L 5 X 3. 11 X 5. X 7. X Write tivo p 9. X 10. X 11. X 12. X 13. X Complete ea 14. 32: 1,2, ssing factor. _ = 4 5 := 77 7 - 4 9 4 = 28 2. 4. 6. 8. airs of factors for = 24; - 26; = 48; - 80; - 108 , zh Ust of factors. Q in 2 X 9 X each _ X _ X X X , y--X 3 = 21 X 6 = 54 - 16 - 3 6 number. - 24 --^ - 26 ^ 48 = 80 - 108 23. In questions 18-22, which numbc^rs a) are divisible bv 2? b) are divisible bv 3? c) are divisible bv 2 and 3? State the missing factors. Use factors greater than 1. 24. 2 X X - 180 25. X X =: 300 26. X 4 X = 240 27. 8 X X - 192 28. X X 2 = 168 29. X 9 X ^ 225 30. X X - 212 Find the smaUest number ivhose factors are 31. 3. 4. and 5. •VJ 1 A 3nf{ < ^ — 17 . 3, 4, 9, 12, 18, , l:i'-t aU the fictors nf eacn number. J 8. -2 19. ' " ^ -6 33. 11, 4, and 2. 34. 10, 2, and 6. Use the divisibdity rules to determine which of the f}nlovzing numbers are drvisibk tn/ 8 nnd zvhich are divisible bu 9. 21. 35. 729 __ 37. 14 112 36. 1520 38. Tnere are 24 desks in the classroom. In how manv wavs can the desks be arranged in equal groups? i Copyright X '19s 27 > 2.7 O O O O O O O rhe symbol > means "Is greater than." 2. Arrange each set of numbers from greatest to least. a) 1.8,2.8,1.9 3. Use the symbol > to show the numbers arranged from greatest to least. b) 365.7,358, 365.9 14 MHR • Chapter 2: Operations on Decimal Numbers a) 1.9, 2.4, 2 b) 5, 4.3, 0.7 • ^ I) Name 1.3 Place Value MATHPOWERTM Seven, pp. 8-9 A place value table can help you read and write numbers. 7 'S^l TS7 Read decim.al numbers by naming the ,'' Thousands .' Ones ^ ' ^ / c- /• ^ / S/ c / eSf / / $•'/ S 7 # / A'-^^/ >c 8 j 7 i 6 t 2 L l 5 9 6 3 place value of the final digit. 0.247 -^ ti^vo hundred fort\'-seven thousandths Read "and" for the decimal point, 29 375.56 -^ twenty-nine thousand three hundred seventy-five and fifty-six hundredths State the total value ofeach imderlined digit. 1. 23.45 2. 19 789 456 3. 457.3864 Write each nurnber in standard form. 4- eight thousand three hundred two 5. two million thirty-three thousand five hundred four 6. six and five tenths 7. one thciusand thirteen and eight hundred fortv-nine thousandths Write each number in ivords. 8. 2894 9. 687.95 10. 0.35: 11. 1976.089 Wnte each number in standard fomi. 12. 10 000 - 3000 + 8 13. 4 X 1 000 000 - 6 X in 000 X - 9 X 1000 6 X 100 -r 4 X 1 14. 3 X 10 + 5 X 1 + 4 X 0.1 + 2 X 0.01 15. 300 + 4 1 , 1 4- 2 f 0.1 + 0.08 + 0.007 Write in expanded form. 16. 203 17. 34.127 18. 276.13 19, 34 123 006 Insert >, =, or < to make each statement true 20. 34.56 n 30 -r 4 + 0.05 + 0.006 21. 12 309 • 10 000 + 2000 + 300 4 - 90 22. 35.7 Q 30 X 0 .
0 5 -^ 0 , 0 0 7 23. 5 .
0 1 D 5 X 0 .
0 1 24. The points awarded to the winners m the men's Olympic platform diving competition are show^n. Order them from, highest to lowest. Year 1972 Points 504.12 Year | Points 1984 T 710.91 1976 1980 600.51 1^88 ^^5.65 1 quo 638.61 677.31 CfspvritdU X 1996 McGraw-Hill Ryerson iJmiteci Name 1.9 Problem Solving: Make an Assumption MATHPOWER™ Seven, pp. 30-31 Understand the Problem Carry Out the Plan Look Back Determine a pattern and make an assumption. Then, list the next 3 terms. 1. 100, 98, 94, 88, 2. 15,21,27,33, 3. 0, 3, 7, 10, ^ ^ 4. The Kerels paid $98.84 for hydro m March this year. How much will they pay for hydro in one year? Wliat assumptions have you made? 5. Connor rode his bicycle 3 km in 4 min. How far could he ride in 2 h? WTiat assumptions have vou made? 6. The student council purchased 6 cases of cola, 2 cases of'ginger ale, and 2 ca.ses of orange soda for the dance. What assumptions did they make? 7. Each week, Mr. Blake purchases two 4-Lbags of milk for his family. This week, milk cost $3.29 for a bag. How much will he spend on milk in a year? WTiat a.ssumptions have vou made? For eadi pair of numbers, name a third number that could be next in the sequence. 9. 9,, 6, __ 11. 12, 24, 10. 3, 8, 12. 2, 8, 13. a) What assumptions did you make in questions 9-12 to determine each pattern? b) Suggest a different third number for each original pair of numbers in questions 9-12. What assumptions have you made this time? 14. A citv^ councillor conducttHl a suney. She asked every 10th household whether they were in favour of the city developing a community playground in the neighbourhood. The results showed that 65'?<! were in favour of the plan. a) What assumptions might the councillor make from. Iier surx'ev^ x fn her jib at the supermarket, [ei^nv :o unpack a case nf 48 cans of soup. She hc>ught she vvould stack them m 2 lnvin vith H row-^; of 3 cans in each iaver, Wha assumptions did she nia,ke? lad b) Do you think the sun-'ev is accurate for the u'hoie neighibourhotxl? Explain. Lovvn-sbl h^ McC.Xatva-iiJi Kyerson Lrmiiec |
10691 | https://micros23.classes.ryansafner.com/files/assignment/03-problem-set-answers-pdf.pdf | Problem Set 3 Solutions ECON 306 — Spring 2023 Note: Answers may be longer than I would deem sufficient on an exam. Some might vary slightly based on points of interest, examples, or personal experience. These suggested answers are designed to give you both the answer and a short explanation of why it is the answer.
Concepts and Critical Thinking 1. Describe, in your own words, what the marginal rate of technical substitution means. How is it different from the slope of the isocost line?
The marginal rate of technical substitution (MRST) is the tradeoff (or exchange rate) between two inputs for a firm (based on its technology or production function). The number literally means the amount of capital the firm would remove (rate of substitution) if they were to use 1 more unit of labor to produce the same amount of output.
The 𝑀𝑅𝑇𝑆= 𝑀𝑃𝑙 𝑀𝑃𝑘is the slope of the isoquant curves, which expresses all the combinations of 𝑙and 𝑘that produce the same output 𝑞.
The slope of the isocost line 𝑤 𝑟is the rate at which the market trades off (or the exchange rate) between 𝑙 and 𝑘, based on relative factor prices.
2. Describe, in your own words, what is true at the least-cost input combination (the optimum) for a firm. Why is it the optimum? What does the equality of the slope of the isoquant curve and the slope of the isocost line mean, in English?
At the optimum point, the producer minimizes their total cost (reaches the lowest possible isocost line) for a given level of output (a given isoquant curve) – thus their optimum is at a point where the two are tangent. At a tangency, the slopes between the budget constraint and the indifference curve are the same.
We have seen that at this point: 𝑀𝑃𝑙 𝑤 = 𝑀𝑃𝑘 𝑟 This means that at the optimum, the marginal product (output gained) for every dollar spent on either labor or capital is the same. That is, you can get no more output by spending a dollar more on labor, or by spending a dollar more on capital. This combination is the best that you can possibly do.
1 3. Explain the difference between the short run and the long run in production.
In the short run, at least one factor of production is fixed, meaning it is too costly to change. In the long run, all factors are variable, meaning they can be changed.
In our analysis, we often assume that capital (𝑘) is fixed in the short run - it is too difficult for a firm to change the number of locations or factories that it has (capital). It can only change labor in the short run.
𝑞𝑆𝑅= 𝑓( ̄ 𝑘, 𝑙) In the long run, firms can change the number of locations or factories that it has (capital), so all factors are variable.
𝑞𝐿𝑅= 𝑓(𝑘, 𝑙) 4. Describe, in your own words, what the law of diminishing marginal returns means. How can firms increase output?
The law of diminishing returns means that adding more of one input, holding all others fixed, the marginal product of that input will diminish.
For example, if we have a fixed amount of capital (one oven), and we keep adding chefs (labor), the more chefs we add, the smaller and smaller the marginal product of labor (we get fewer and fewer additional pizzas for every chef we add). It may be such that you ultimately get negative marginal product - if you keep adding enough labor, it might actually reduce total output since there are “too many cooks in the kitchen.” It’s crucial to understand that this relationship consists of adding more of just one factor, and holding constant all other factors. The key problem was that there was just a single oven that we kept adding chefs to. If we want to sustainably increase our output, we need to add more of both labor and capital.
This concept is largely attributed to David Ricardo, the classical economist, who famously said that if the law of diminishing returns was not true, we would be able to grow the entire world’s food supply in a single flower pot!
2 Problems Show all work for calculations. You may lose points, even if correct, for missing work. Be sure to label graphs fully, if appropriate.
5. Suppose a firm can hire labor at $5/hour and rent capital for $20 per hour.
a. Write an equation for the total cost of the firm.
If 𝑤= $5 and 𝑟= 20 𝐶= 𝑤𝐿+ 𝑟𝐾 𝐶= 5𝐿+ 20𝐾 b. Suppose the firm wants to spend exactly $100. With labor on the horizontal axis and capital on the vertical axis, find the equation of the isocost line (in a graphable form), and graph it.
100 = 5𝐿+ 20𝐾 100 −5𝐿= 20𝐾 5 −0.25𝐿= 𝐾 C=$100 C=$100 0 1 2 3 4 5 6 7 8 9 10 0 2 4 6 8 10 12 14 16 18 20 Labor (l) Capital (k) c. If the firm is completely automated (i.e. it uses only capital), how many units of capital can they employ for $100?
𝐶 𝑟= 100 20 = 5 d. If the firm uses only labor, how many units of labor can they employ for $100?
3 𝐶 𝑤= 100 5 = 20 e. What is the slope of the isocost line? What does it represent?
The slope is 0𝑤 𝑟= −5 20 = −1 4. It represents the tradeoff in the market between 𝑙and 𝑘based on their relative prices.
f. Suppose a tax on capital makes renting capital raises the price of capital to $25 per hour. What is the new (graphable) equation of the $100 isocost line? Graph the new isocost line on the same graph.
100 = 5𝐿+ 25𝐾 100 −5𝐿= 25𝐾 5 −0.2𝐿= 𝐾 An easier way is to think about the new endpoint for capital: 𝐶 𝑟′ = 100 25 = 4 C=$100 C=$100 C'=$100 C'=$100 0 1 2 3 4 5 6 7 8 9 10 0 2 4 6 8 10 12 14 16 18 20 Labor (l) Capital (k) 4 6. For each of the following production functions, identify whether the production process exhibits constant returns to scale, increasing returns to scale, or decreasing returns to scale. Be sure to show your work!
a. 𝑞= 2𝐿+ 4𝐾 Suppose for example, we have 2 L and 2 K: 𝑞= 2(2) + 4(2) = 4 + 8 = 12 Now we double inputs to 4L and 4K: 𝑞= 2(4) + 4(4) = 8 + 16 = 24 Output has doubled, from 12 to 24 units, when we have doubled inputs from 2 to 4 K & L, so we have constant returns to scale.
b. 𝑞= 6𝐿0.25𝐾0.75 Suppose for example, we have 2 L and 2 K: 𝑞= 6(2)0.25(2)0.75 = 12 Now we double inputs to 4L and 4K: 𝑞= 6(4)0.25(4)0.75 = 24 Output has doubled, from 12 to 24 units, when we have doubled inputs from 2 to 4 K & L, so we have constant returns to scale.
c. 𝑞= 2𝐿0.8𝐾0.4 Suppose for example, we have 2 L and 2 K: 𝑞= 2(2)0.8(2)0.4 = 4.59 Now we double inputs to 4L and 4K: 𝑞= 2(4)0.8(4)0.4 = 10.56 Output has more than doubled, from 4.59 to 10.56 units, when we have doubled inputs from 2 to 4 K & L, so we have increasing returns to scale.
5 d. 𝑞= 2𝐿0.25𝐾0.25 Suppose for example, we have 2 L and 2 K: 𝑞= 2(2)0.25(2)0.25 = 2.82 Now we double inputs to 4L and 4K: 𝑞= 2(4)0.25(4)0.25 = 4 Output has less than doubled, from 2.82 to 4 units, when we have doubled inputs from 2 to 4 K & L, so we have increasing returns to scale.
6 7. Jerry’s Berries is a small farm that has the following production function for strawberries using com-binations of labor (𝑙) and land (𝑡): 𝑞= 2 𝑙𝑡 The marginal products (of labor, 𝑙; and land, 𝑡) are: 𝑀𝑃𝑙= 2𝑡 𝑀𝑃𝑡= 2𝑙 Put labor, 𝑙on the horizontal axis and land, 𝑡on the vertical axis.
a. Write an equation for 𝑀𝑅𝑇𝑆𝑙,𝑡.
𝑀𝑅𝑇𝑆𝑙,𝑡= 𝑀𝑃𝑙 𝑀𝑃𝑡 = 2𝑡 2𝑙 = 𝑡 𝑙 b. Suppose the farm is currently using 4 units of labor and 1 unit of land. How much output (tons of strawberries) is the farm producing?
Plug this input combination into the production function: 𝑞= 2 𝑙𝑡 𝑞= 2(4)(1) 𝑞= 8 c. From its current production, how much more output would the farm get by utilizing 1 more unit of labor? What about 1 more unit of land (instead of labor)?
This is measuring the marginal product of 𝑙and the marginal product of 𝑡, evaluated at the firm’s current input combination in production of 𝑙= 4 and 𝑡= 1.
𝑀𝑃𝑙= 2𝑡 𝑀𝑃𝑙= 2(1) 𝑀𝑃𝑙= 2 Hiring one additional unit of labor, 𝑙, will increase output by 2 units of strawberries.
7 𝑀𝑃𝑡= 2𝑙 𝑀𝑃𝑡= 2(4) 𝑀𝑃𝑡= 8 Using one additional unit of land, 𝑡, increase output by 8 units of strawberries.
d. From its current production, how many units of land would the farm be willing to forgo in order to use one more unit of labor and still produce the same output as before? How many units of labor would the farm be willing to forgo in order to use one more unit of land and still produce the same output as before?
This is measuring the marginal rate of technical substitution (i.e. the slope of the isoquant curve) evaluated at the firm’s his current input combination of 𝑙= 4 and 𝑡= 1. From part A, we found the equation for the 𝑀𝑅𝑇𝑆𝑙,𝑡: 𝑀𝑅𝑇𝑆𝑙,𝑡= 𝑡 𝑙 𝑀𝑅𝑆4,1 = 1 4 At this current input combination, the firm would give up 1 4 units of land (𝑡) to hire one more unit of labor (𝑙) to produce the same amount of output. This is the slope of the isoquant curve at this point: to go one unit to the right, we go 1 4 units down.
To use one more unit of land, 𝑡, and produce the same amount of putput, the firm would give up 4 units of labor, 𝑙. This is the inverse of the isoquant curve slope at this point. Consider: to go up one unit, we go 4 units to the left.
e. Suppose the farm can choose between input combinations of 𝑎= (4, 1), 𝑏= (2, 2), 𝑐= (2, 1), and 𝑑= 𝑑(3, 2). What outputs does each combination yield?
Check the output each input combination yields.
𝑞= 2𝑙𝑡 𝑞= 2(4)(1) 𝑞= 8 Input combination 𝑎provides output of 8 units of strawberries.
8 𝑞= 2𝑙𝑡 𝑞= 2(2)(2) 𝑞= 8 Input combination 𝑏provides output of 8 units of strawberries.
𝑞= 2𝑙𝑡 𝑞= 2(2)(1) 𝑞= 4 Input combination 𝑐provides output of 4 units of strawberries.
𝑞= 2𝑙𝑡 𝑞= 2(3)(2) 𝑞= 12 Input combination 𝑑provides output of 12 units of strawberries.
f. Sketch a graph, plotting bundles 𝑎, 𝑏, 𝑐, and 𝑑. Indicate any isoquant curve(s) they are on, and how much output each provides.
9 q = 8 q = 8 q = 4 q = 4 q = 12 q = 12 a b c d 0 1 2 3 4 5 0 1 2 3 4 5 Labor (l) Land (t) 10 8. Dunder Mifflin paper company produces reams of paper each week according to the production function: 𝑞= 10𝑙0.5𝑘0.5 𝑀𝑃𝑙= 5𝑙−0.5𝑘0.5 𝑀𝑃𝑘= 5𝑙0.5𝑘−0.5 They have determined that they need to ship 1,000 reams of paper this week to Scranton, PA. Using capital costs $20, whereas labor costs $10.
a. What is the cost-minimizing combination of labor and capital that will yield 1,000 reams of paper?
Round each to the nearest whole number.
We know that at the optimum: 𝑀𝑃𝐿 𝑀𝑃𝐾 = 𝑤 𝑟 Definition of optimum 5𝐿−0.5𝐾0.5 5𝐿0.5𝐾−0.5 = 10 20 Plugging in known values 𝐿(−0.5−0.5)𝐾(0.5−(−0.5) = 0.5 Exponent rule for division 𝐿−1𝐾1 = 0.5 𝐾 𝐿= 0.5 Exponent rule for negative exponents 𝐾= 0.5𝐿 Multiplying both sides by 𝐿 To get exact quantities, plug this into the production function: 𝑞= 10 √ 𝐿𝐾 The production function 1000 = 10√𝐿(0.5𝐿) Plugging in our function of K and 𝑞∗= 1000 100 = √𝐿(0.5𝐿) Dividing both sides by 10 100 = √ 0.5𝐿2 Multiplying 10000 = 0.5𝐿2 Squaring both sides 20000 = 𝐿2 Dividing both sides by 0.5 141 ≈𝐿 Square rooting both sides Knowing 𝐿, we can find 𝐾: 𝐾= 0.5𝐿 𝐾= 0.5(141) 𝐾= 71 11 b. What is the total cost of using this combination of inputs?
𝑤𝐿+ 𝑟𝐾= 𝐶 $10(141) + $20(71) = 𝐶 $1410 + $1420 = $2830 c. Now suppose that they need to double their output this week, and need to produce 2,000 reams of paper. How does their optimal combination of inputs change?1 One quick hint is that recognizing the production function as a Cobb-Douglas production function and looking at the exponents on 𝐿and 𝐾(as they are square roots, the exponents are each 0.5).
𝑞= 2𝐿0.5𝐾0.5 1 = 0.5 + 0.5 The sum of the exponents is 1, so the production function experiences constant returns to scale: doubling inputs will double output. We know that output doubles, so all inputs must double from about 141 workers and 71 capital to about 282 workers and 141 capital. Since the MRTS is not changing (no marginal products changed), nor did any input prices, the ratio of capital to labor used is still 𝐾= 0.5𝐿.
Anyway, let’s check manually. Knowing 𝐾= 0.5𝐿still, we need to find the exact quantities used in production. Plug this into the production function, as before (with double 𝑞): 𝑞= 10 √ 𝐿𝐾 The production function 2000 = 10√𝐿(0.5𝐿) Plugging in our function of K and 𝑞∗= 2000 200 = √𝐿(0.5𝐿) Dividing both sides by 10 200 = √ 0.5𝐿2 Multiplying 40000 = 0.5𝐿2 Squaring both sides 80000 = 𝐿2 Dividing both sides by 0.5 283 ≈𝐿 Square rooting both sides Knowing 𝐿, we can find 𝐾: 𝐾= 0.5𝐿 𝐾= 0.5(283) 𝐾= 141 Which we anticipated (although with rounding error on 𝐿) before.
1Hint: neither the equation for MRTS nor any prices are changing!
12 e. What is the total cost of this new level of output?
𝑤𝐿+ 𝑟𝐾= 𝐶 $10(283) + $20(141) = 𝐶 $2830 + $2800 = $5630 It has just about doubled the cost of before, as should be intuitive (we’re producing twice as much as before, with twice as many inputs at the same prices).
A graph was not necessary for these problems, but it can help us visualize what we were solving for: f. Suppose management at Dunder Mifflin develops a new program that magically makes everyone at the firm more productive, such that the firm’s new production function becomes: 𝑞= 20𝑙0.5𝑘0.5 𝑀𝑃𝑙= 10𝑙−0.5𝑘0.5 𝑀𝑃𝑘= 10𝑙0.5𝑘−0.5 Still needing to supply 2,000 reams of paper this week at the same input prices, what is their new optimal combination of labor and capital?
The optimal ratio remains 𝐾= 0.5𝐿. To get the new quantities, plug this into the updated production function: 𝑞= 20 √ 𝐿𝐾 The new production function 2000 = 20√𝐿(0.5𝐿) Plugging in our function of K and 𝑞∗= 2000 100 = √𝐿(0.5𝐿) Dividing both sides by 10 100 = √ 0.5𝐿2 Multiplying 10000 = 0.5𝐿2 Squaring both sides 20000 = 𝐿2 Dividing both sides by 0.5 141 ≈𝐿 Square rooting both sides Knowing 𝐿, we can find 𝐾: 𝐾= 0.5𝐿 𝐾= 0.5(141) 𝐾= 71 This is the same optimal combination as when the firm produced 1,000 copies!
13 g. How much does this combination cost? What does this show you about technological improvement (or “total factor productivity”)?
With the same prices of labor and capital, this is the same total cost as in part b. Notice the doubling of “total factor productivity” in the production function (from 10 to 20) means the firm can use half the amount of inputs to produce the same amount as before!
q=1,000 q=1,000 q=2,000 q=2,000 C=$2,830 C=$2,830 C=$5,630 C=$5,630 0 50 100 150 200 250 300 350 400 450 500 0 50 100 150 200 250 300 350 400 450 500 Labor Capital q = 10 lk, w = $10, r = $20 14 |
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Affine Geometry
The document contains 17 problems involving affine geometry concepts such as transformations, ratios of line segments, parallelism, and properties of geometric shapes. Many of the problems c…
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Practice problems on affine geometry.
Proble m 1
An inscri bed ell ipse touches the paral lel sides
AB
and
C D
of a trapezoid
ABCD
at
P
and
Q
. If the ellipse touches the side
B C
at its midpoint, prove that
P B
=
QC
.
Proble m 2
An e llip se i s in scribed in a quadri late ral
ABCD
. Giv en that the ellip se touc hes side s
BC
and
DA
at their midpoints and that the other pair of sides
AB
and
CD
are parallel, prove that the quadrilateral is a parallelogram.
Proble m 3
An ellip se is inscri bed in a quadrilat eral so as to touc h all four side s at their mid point s. Prov e that the quadrilateral is a parallelogram.
Proble m 4
Let
ABCD
be a trapezoid with parallel sides
AB
and
CD
. Let
X
and
Y
be the mid-points of its diagonals.Let
f
be an affine transformation that transforms
A
to
A
′
= (0
,
0),
B
to
B
′
= (2
,
0), and
D
to
D
′
= (0
,
2). Suppose that
f
transforms
C
,
X
, and
Y
to
C
′
,
X
′
, and
Y
′
.(a) Write a general expression for the co-ordinates of
C
′
. (The expression should involve an unkown quantity
c
.)(b) Determine the co-ordinates of
X
′
and
Y
′
.(c) Prove that the line
X Y
is parallel to
AB
.(d) How is the length of
X Y
related to the lengths of
AB
and
C D
?
Proble m 5
Suppose that
ABC
is a triangle and that
D
,
E
, and
F
are trisection points that divide the sides
BC
,
C A
and
AB
in the ratio 1 : 2. (Thus
BD DC
=
1 2
, etc.).(i) Explain why there must be an affine transformation that maps all of the points
A,B,C,D,E,F
to points with int eger co-ordin ates. W rite down a possibl e set of integer co-ord inate s for their images. (F or examp le, you could choose to let
A
′
= (0
,
0). You must specify the images of the other points under your chosen affine transformation.)(ii) Suppose that
AD
meets
E F
at the point
X
. Determine the ratio
E X
:
X F
.
Proble m 6
Find an affine t ransfo rmati on tha t turn s the poin ts (1
,
1), (3
,
4), (4
,
5) into the points (1
,
0), (2
,
2),(3
,
1), respectively. (Give a formula
f
(x) =
M
x + b for the transformation.)
Proble m 7
Find a n affine transf ormat ion that t urns t he poin ts (
−
1
,
−
1), (1
,
2), (2
,
3) into the points (0
,
1),(1
,
3), (2
,
2), respectively. (Give a formula
f
(x) =
M
x+ b for the transf ormat ion.)
Proble m 8
Suppose that
O
is the center of an ellipse and that
X
and
Y
are points on the ellipse. Suppose that the tangents at
X
and
Y
(to the ellipse) meet at
P
. Let
Q
be the point of intersection of
X Y
with
OP
, and let
R
be the point of intersection of the segment
O P
with the ellipse.Prove that
OQ OR
=
OR OP
.
Proble m 9
Find an affine t ransfo rmati on tha t turn s the poin ts (1
,
0), (3
,
5), (2
,
3) into the points (0
,
1), (3
,
2),(4
,
5), respectively. (Give a formula
f
(x) =
M
x + b for the transformation.)
Proble m 10
Let
ABCD
be a parallelogram. Let
X
and
Y
be trisecti on points of the diago nal
AC
(where
X
is closer to
A
and
Y
is closer to
C
). Le t
R
=
B X
∩
AD
and
S
=
B Y
∩
CD
. Let
f
be an affine transformation that maps
B
to
B
′
= (0
,
0),
A
to
A
′
= (3
,
0), and
C
to
C
′
= (0
,
3).(a) Find the co-ordinates of the images
D
′
,
X
′
,
Y
′
,
R
′
and
S
′
of
D
,
X
,
Y
,
R
, and
S
under
f
. Give brief reasons for your answers.(b) Find the slopes of
A
′
C
′
and
R
′
S
′
.(c) What can you say about the type of the quadrilateral
ACSR
? (Is it a squar e, rect angle, parall elogra m, ...).
Proble m 11
Prove the foll owing theorem. (Hint: Use an affine transformation to an easier confi guration.)If a trapezoid is inscribed in an ellipse, then the line joining the midpoints of the two parallel sides of the trapezoid passes through the cen ter of the ellipse.
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Proble m 12
Whic h of the follow ing geom etric al config urati ons are affine co ncept s? Answ er with yes fo r any affine concept s and no for non-conce pts.(a) parallelogram(b) trapezoid(c) rectangle(d) isosceles trapezoid(e) conic(f) ellipse(g) ratio of perpendicular segments(h) ratio of parallel segments(i) incircle of a triangle(j) collinear triple of points, one on each of three sides of a triangle
Proble m 13
Let
AB C
be a triangle. Let
B D
be a median of triangle
AB C
. Let
X
and
Y
be two points on the segment
B C
such that
B X
=
X Y
=
Y C
. (So
X
and
Y
divide
B C
into 3 equal parts.) Let
AX
meet
B D
at
Z
. Use affine geometry to prove that
Z
is the midpoint of
B D
.
Proble m 14
Let
ABCD
be a parallelogram. Let
X
and
Y
be the midpoints of the sides
AB
and
B C
. Let
AY
meet
C X
at
Z
. Use affine geometry to prove that
Z
lies on the diagonal
B D
and to determine the ratio
DZ/ZB
.
Proble m 15
Let
AB C
be a triangle. Let
B D
be a median of triangle
AB C
. Let
X
,
Y
, and
Z
be three points on the segment
BC
such that
BX
=
X Y
=
Y Z
=
ZC
. (S o
X
,
Y
and
Z
divide
BC
int o 4 equal parts.) Let
AX
meet
B D
at
P
. Use affine geometry to prove that
AP/AX
= 4
/
5.
Proble m 16
Let
ABC
be a triangle and let
D
be a point on the side
BC
. Le t
X
and
Y
be the centroids of triangles
AB D
and
AC D
. Use affine geometry to prove that the line
X Y
is parallel to
B C
.
Proble m 17
Let
T
be a trapezoid. Prove that there is an affine transformation that transforms
T
to an isosceles trapezoid. (An isosceles trapezoid is a trapezoid with two equal base angles.)
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10693 | https://www.math.ucdavis.edu/~hunter/intro_analysis_pdf/ch13.pdf | Chapter 13 Metric, Normed, and Topological Spaces A metric space is a set X that has a notion of the distance d(x, y) between every pair of points x, y ∈X. A fundamental example is R with the absolute-value metric d(x, y) = |x −y|, and nearly all of the concepts we discuss below for metric spaces are natural generalizations of the corresponding concepts for R.
A special type of metric space that is particularly important in analysis is a normed space, which is a vector space whose metric is derived from a norm. On the other hand, every metric space is a special type of topological space, which is a set with the notion of an open set but not necessarily a distance.
The concepts of metric, normed, and topological spaces clarify our previous discussion of the analysis of real functions, and they provide the foundation for wide-ranging developments in analysis. The aim of this chapter is to introduce these spaces and give some examples, but their theory is too extensive to describe here in any detail.
13.1. Metric spaces A metric on a set is a function that satisfies the minimal properties we might expect of a distance.
Definition 13.1. A metric d on a set X is a function d : X × X →R such that for all x, y, z ∈X: (1) d(x, y) ≥0 and d(x, y) = 0 if and only if x = y (positivity); (2) d(x, y) = d(y, x) (symmetry); (3) d(x, y) ≤d(x, z) + d(z, y) (triangle inequality).
A metric space (X, d) is a set X with a metric d defined on X.
271 272 13. Metric, Normed, and Topological Spaces In general, many different metrics can be defined on the same set X, but if the metric on X is clear from the context, we refer to X as a metric space.
Subspaces of a metric space are subsets whose metric is obtained by restricting the metric on the whole space.
Definition 13.2. Let (X, d) be a metric space. A metric subspace (A, dA) of (X, d) consists of a subset A ⊂X whose metric dA : A × A →R is is the restriction of d to A; that is, dA(x, y) = d(x, y) for all x, y ∈A.
We can often formulate intrinsic properties of a subset A ⊂X of a metric space X in terms of properties of the corresponding metric subspace (A, dA).
When it is clear that we are discussing metric spaces, we refer to a metric subspace as a subspace, but metric subspaces should not be confused with other types of subspaces (for example, vector subspaces of a vector space).
13.1.1. Examples. In the following examples of metric spaces, the verification of the properties of a metric is mostly straightforward and is left as an exercise.
Example 13.3. A rather trivial example of a metric on any set X is the discrete metric d(x, y) = ( 0 if x = y, 1 if x ̸= y.
This metric is nevertheless useful in illustrating the definitions and providing counter-examples.
Example 13.4. Define d : R × R →R by d(x, y) = |x −y|.
Then d is a metric on R. The natural numbers N and the rational numbers Q with the absolute-value metric are metric subspaces of R, as is any other subset A ⊂R.
Example 13.5. Define d : R2 × R2 →R by d(x, y) = |x1 −y1| + |x2 −y2| x = (x1, x2), y = (y1, y2).
Then d is a metric on R2, called the ℓ1 metric. (Here, “ℓ1” is pronounced “ell-one.”) For example, writing z = (z1, z2), we have d(x, y) = |x1 −z1 + z1 −y1| + |x2 −z2 + z2 −y2| ≤|x1 −z1| + |z1 −y1| + |x2 −z2| + |z2 −y2| ≤d(x, z) + d(z, y), so d satisfies the triangle inequality. This metric is sometimes referred to infor-mally as the “taxicab” metric, since it’s the distance one would travel by taxi on a rectangular grid of streets.
Example 13.6. Define d : R2 × R2 →R by d(x, y) = p (x1 −y1)2 + (x2 −y2)2 x = (x1, x2), y = (y1, y2).
Then d is a metric on R2, called the Euclidean, or ℓ2, metric. It corresponds to the usual notion of distance between points in the plane. The triangle inequality is geometrically obvious but an analytical proof is non-trivial (see Theorem 13.26 below).
13.1. Metric spaces 273 0 0.2 0.4 0.6 0.8 1 −0.1 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 Figure 1. The graph of a function f ∈C([0, 1]) is in blue. A function whose distance from f with respect to the sup-norm is less than 0.1 has a graph that lies inside the dotted red lines y = f(x) ± 0.1 e.g., the green graph.
Example 13.7. Define d : R2 × R2 →R by d(x, y) = max (|x1 −y1| , |x2 −y2|) x = (x1, x2), y = (y1, y2).
Then d is a metric on R2, called the ℓ∞, or maximum, metric.
Example 13.8. Define d : R2 × R2 →R for x = (x1, x2), y = (y1, y2) as follows: if (x1, x2) ̸= k(y1, y2) for k ∈R, then d(x, y) = q x2 1 + x2 2 + q y2 1 + y2 2; and if (x1, x2) = k(y1, y2) for some k ∈R, then d(x, y) = q (x1 −y1)2 + (x2 −y2)2.
That is, d(x, y) is the sum of the Euclidean distances of x and y from the origin, unless x and y lie on the same line through the origin, in which case it is the Euclidean distance from x to y. Then d defines a metric on R2.
In England, d is sometimes called the “British Rail” metric, because all the train lines radiate from London (located at 0). To take a train from town x to town y, one has to take a train from x to 0 and then take a train from 0 to y, unless x and y are on the same line, when one can take a direct train.
Example 13.9. Let C(K) denote the set of continuous functions f : K →R, where K ⊂R is compact; for example, K = [a, b] is a closed, bounded interval. If f, g ∈C(K) define d(f, g) = sup x∈K |f(x) −g(x)| = ∥f −g∥∞, ∥f∥∞= sup x∈K |f(x)| .
274 13. Metric, Normed, and Topological Spaces −1.5 −1 −0.5 0 0.5 1 1.5 −1.5 −1 −0.5 0 0.5 1 1.5 x1 x2 Figure 2. The unit balls B1(0) on R2 for different metrics: they are the interior of a diamond (ℓ1-norm), a circle (ℓ2-norm), or a square (ℓ∞-norm).
The ℓ∞-ball of radius 1/2 is also indicated by the dashed line.
The function d : C(K) × C(K) →R is well-defined, since a continuous function on a compact set is bounded, and d is a metric on C(K). Two functions are close with respect to this metric if their values are close at every point x ∈K. (See Figure 1.) We refer to ∥f∥∞as the sup-norm of f. Section 13.6 has further discussion.
13.1.2. Open and closed balls. A ball in a metric space is analogous to an interval in R.
Definition 13.10. Let (X, d) be a metric space. The open ball Br(x) of radius r > 0 and center x ∈X is the set of points whose distance from x is less than r, Br(x) = {y ∈X : d(x, y) < r} .
The closed ball ¯ Br(x) of radius r > 0 and center x ∈X as the set of points whose distance from x is less than or equal to r, ¯ Br(x) = {y ∈X : d(x, y) ≤r} .
The term “ball” is used to denote a “solid ball,” rather than the “sphere” of points whose distance from the center x is equal to r.
Example 13.11. Consider R with its standard absolute-value metric, defined in Example 13.4. Then the open ball Br(x) = {y ∈R : |x −y| < r} is the open interval of radius r centered at x, and the closed ball ¯ Br(x) = {y ∈R : |x −y| ≤r} is the closed interval of radius r centered at x.
13.1. Metric spaces 275 Example 13.12. For R2 with the Euclidean metric defined in Example 13.6, the ball Br(x) is an open disc of radius r centered at x. For the ℓ1-metric in Exam-ple 13.5, the ball is a diamond of diameter 2r, and for the ℓ∞-metric in Exam-ple 13.7, it is a square of side 2r. The unit ball B1(0) for each of these metrics is illustrated in Figure 2.
Example 13.13. Consider the space C(K) of continuous functions f : K →R on a compact set K ⊂R with the sup-norm metric defined in Example 13.9. The ball Br(f) consists of all continuous functions g : K →R whose values are within r of the values of f at every x ∈K. For example, for the function f shown in Figure 1 with r = 0.1, the open ball Br(f) consists of all continuous functions g whose graphs lie between the red lines.
One has to be a little careful with the notion of balls in a general metric space, because they don’t always behave the way their name suggests.
Example 13.14. Let X be a set with the discrete metric given in Example 13.3.
Then Br(x) = {x} consists of a single point if 0 ≤r < 1 and Br(x) = X is the whole space if r ≥1. (See also Example 13.44.) An another example, what are the open balls for the metric in Example 13.8?
A set in a metric space is bounded if it is contained in a ball of finite radius.
Definition 13.15. Let (X, d) be a metric space. A set A ⊂X is bounded if there exist x ∈X and 0 ≤R < ∞such that d(x, y) ≤R for all y ∈A, meaning that A ⊂BR(x).
Unlike R, or a vector space, a general metric space has no distinguished origin, but the center point of the ball is not important in this definition of a bounded set.
The triangle inequality implies that d(y, z) < R + d(x, y) if d(x, z) < R, so BR(x) ⊂BR′(y) for R′ = R + d(x, y).
Thus, if Definition 13.15 holds for some x ∈X, then it holds for every x ∈X.
We can say equivalently that A ⊂X is bounded if the metric subspace (A, dA) is bounded.
Example 13.16. Let X be a set with the discrete metric given in Example 13.3.
Then X is bounded since X = Br(x) if r > 1 and x ∈X.
Example 13.17. A subset A ⊂R is bounded with respect to the absolute-value metric if A ⊂(−R, R) for some 0 < R < ∞.
Example 13.18. Let C(K) be the space of continuous functions f : K →R on a compact set defined in Example 13.9. The set F ⊂C(K) of all continuous functions f : K →R such that |f(x)| ≤1 for every x ∈K is bounded, since d(f, 0) = ∥f∥∞≤ 1 for all f ∈F. The set of constant functions {f : f(x) = c for all x ∈K} isn’t bounded, since ∥f∥∞= |c| may be arbitrarily large.
We define the diameter of a set in an analogous way to Definition 3.5 for subsets of R.
276 13. Metric, Normed, and Topological Spaces Definition 13.19. Let (X, d) be a metric space and A ⊂X. The diameter 0 ≤ diam A ≤∞of A is diam A = sup {d(x, y) : x, y ∈A} .
It follows from the definitions that A is bounded if and only if diam A < ∞.
The notions of an upper bound, lower bound, supremum, and infimum in R depend on its order properties. Unlike properties of R based on the absolute value, they do not generalize to an arbitrary metric space, which isn’t equipped with an order relation.
13.2. Normed spaces In general, there are no algebraic operations defined on a metric space, only a distance function. Most of the spaces that arise in analysis are vector, or linear, spaces, and the metrics on them are usually derived from a norm, which gives the “length” of a vector.
We assume that the reader is familiar with the basic theory of vector spaces, and we consider only real vector spaces.
Definition 13.20. A normed vector space (X, ∥· ∥) is a vector space X together with a function ∥· ∥: X →R, called a norm on X, such that for all x, y ∈X and k ∈R: (1) 0 ≤∥x∥< ∞and ∥x∥= 0 if and only if x = 0; (2) ∥kx∥= |k|∥x∥; (3) ∥x + y∥≤∥x∥+ ∥y∥.
The properties in Definition 13.20 are natural ones to require of a length. The length of x is 0 if and only if x is the 0-vector; multiplying a vector by a scalar k multiplies its length by |k|; and the length of the “hypoteneuse” x + y is less than or equal to the sum of the lengths of the “sides” x, y. Because of this last interpretation, property (3) is called the triangle inequality.
We also refer to a normed vector space as a normed space for short.
Proposition 13.21. If (X, ∥· ∥) is a normed vector space, then d : X × X →R defined by d(x, y) = ∥x −y∥is a metric on X.
Proof. The metric-properties of d follow directly from the properties of a norm in Definition 13.20. The positivity is immediate. Also, we have d(x, y) = ∥x −y∥= ∥−(x −y)∥= ∥y −x∥= d(y, x), d(x, y) = ∥x −z + z −y∥≤∥x −z∥+ ∥z −y∥= d(x, z) + d(y, z), which proves the symmetry of d and the triangle inequality.
□ If X is a normed vector space, then we always use the metric associated with its norm, unless stated specifically otherwise.
A metric associated with a norm has the additional properties that for all x, y, z ∈X and k ∈R d(x + z, y + z) = d(x, y), d(kx, ky) = |k|d(x, y), 13.2. Normed spaces 277 which are called translation invariance and homogeneity, respectively. These prop-erties imply that the open balls Br(x) in a normed space are rescaled, translated versions of the unit ball B1(0).
Example 13.22. The set of real numbers R with the absolute-value norm | · | is a one-dimensional normed vector space.
Example 13.23. The discrete metric in Example 13.3 on R, and the metric in Example 13.8 on R2 are not derived from a norm. (Why?) Example 13.24. The space R2 with any of the norms defined for x = (x1, x2) by ∥x∥1 = |x1| + |x2|, ∥x∥2 = q x2 1 + x2 2, ∥x∥∞= max (|x1|, |x2|) is a normed vector space. The corresponding metrics are the “taxicab” metric in Example 13.5, the Euclidean metric in Example 13.6, and the maximum metric in Example 13.7, respectively.
The norms in Example 13.24 are special cases of a fundamental family of ℓp-norms on Rn. All of the ℓp-norms reduce to the absolute-value norm if n = 1, but they are different if n ≥2.
Definition 13.25. For 1 ≤p < ∞, the ℓp-norm ∥· ∥p : Rn →R is defined for x = (x1, x2, . . . , xn) ∈Rn by ∥x∥p = (|x1|p + |x2|p + · · · + |xn|p)1/p .
The ℓ2-norm is called the Euclidean norm. For p = ∞, the ℓ∞-norm ∥·∥∞: Rn →R is defined by ∥x∥∞= max (|x1|, |x2|, . . . , |xn|) .
The notation for the ℓ∞-norm is explained by the fact that ∥x∥∞= lim p→∞∥x∥p.
Moreover, consistent with its name, the ℓp-norm is a norm.
Theorem 13.26. Let 1 ≤p ≤∞. The space Rn with the ℓp-norm is a normed vector space.
Proof. The space Rn is an n-dimensional vector space, so we just need to verify the properties of the norm.
The positivity and homogeneity of the ℓp-norm follow immediately from its definition. We verify the triangle inequality here only for the cases p = 1, ∞.
Let x = (x1, x2, . . . , xn) and y = (y1, y2, . . . , yn) be points in Rn. For p = 1, we have ∥x + y∥1 = |x1 + y1| + |x2 + y2| + · · · + |xn + yn| ≤|x1| + |y1| + |x2| + |y2| + · · · + |xn| + |yn| ≤∥x∥1 + ∥y∥1.
278 13. Metric, Normed, and Topological Spaces For p = ∞, we have ∥x + y∥∞= max (|x1 + y1|, |x2 + y2|, . . . , |xn + yn|) ≤max (|x1| + |y1|, |x2| + |y2|, . . . , |xn| + |yn|) ≤max (|x1|, |x2|, . . . , |xn|) + max (|y1|, |y2|, . . . , |yn|) ≤∥x∥∞+ ∥y∥∞.
The proof of the triangle inequality for 1 < p < ∞is more difficult and is given in Section 13.7.
□ We can use Definition 13.25 to define ∥x∥p for any 0 < p ≤∞. However, if 0 < p < 1, then ∥· ∥p doesn’t satisfy the triangle inequality, so it is not a norm.
This explains the restriction 1 ≤p ≤∞.
Although the ℓp-norms are numerically different for different values of p, they are equivalent in the following sense (see Corollary 13.29).
Definition 13.27. Let X be a vector space. Two norms ∥· ∥a, ∥· ∥b on X are equivalent if there exist strictly positive constants M ≥m > 0 such that m∥x∥a ≤∥x∥b ≤M∥x∥a for all x ∈X.
Geometrically, two norms are equivalent if and only if an open ball with re-spect to either one of the norms contains an open ball with respect to the other.
Equivalent norms define the same open sets, convergent sequences, and continuous functions, so there are no topological differences between them.
The next theorem shows that every ℓp-norm is equivalent to the ℓ∞-norm. (See Figure 2.) Theorem 13.28. Suppose that 1 ≤p < ∞. Then, for every x ∈Rn, ∥x∥∞≤∥x∥p ≤n1/p∥x∥∞.
Proof. Let x = (x1, x2, . . . , xn) ∈Rn. Then for each 1 ≤i ≤n, we have |xi| ≤(|x1|p + |x2|p + · · · + |xn|p)1/p = ∥x∥p, which implies that ∥x∥∞= max {|xi| : 1 ≤i ≤n} ≤∥x∥p.
On the other hand, since |xi| ≤∥x∥∞for every 1 ≤i ≤n, we have ∥x∥p ≤(n∥x∥p ∞)1/p = n1/p∥x∥∞, which proves the result □ As an immediate consequence, we get the equivalence of the ℓp-norms.
Corollary 13.29. The ℓp and ℓq norms on Rn are equivalent for every 1 ≤p, q ≤∞.
Proof. We have n−1/q∥x∥q ≤∥x∥∞≤∥x∥p ≤n1/p∥x∥∞≤n1/p∥x∥q.
□ With more work, one can prove that that ∥x∥q ≤∥x∥p for 1 ≤p ≤q ≤∞, meaning that the unit ball with respect to the ℓq-norm contains the unit ball with respect to the ℓp-norm.
13.3. Open and closed sets 279 13.3. Open and closed sets There are natural definitions of open and closed sets in a metric space, analogous to the definitions in R. Many of the properties of such sets in R carry over immediately to general metric spaces.
Definition 13.30. Let X be a metric space. A set G ⊂X is open if for every x ∈G there exists r > 0 such that Br(x) ⊂G.
A subset F ⊂X is closed if F c = X \ F is open.
We can rephrase this definition more geometrically in terms of neighborhoods.
Definition 13.31. Let X be a metric space. A set U ⊂X is a neighborhood of x ∈X if Br(x) ⊂U for some r > 0.
Definition 13.30 then states that a subset of a metric space is open if and only if every point in the set has a neighborhood that is contained in the set. In particular, a set is open if and only if it is a neighborhood of every point in the set.
Example 13.32. If d is the discrete metric on a set X in Example 13.3, then every subset A ⊂X is open, since for every x ∈A we have B1/2(x) = {x} ⊂A. Every set is also closed, since its complement is open.
Example 13.33. Consider R2 with the Euclidean norm (or any other ℓp-norm).
If f : R →R is a continuous function, then E = (x1, x2) ∈R2 : x2 < f(x1) is an open subset of R2. If f is discontinuous, then E needn’t be open. We leave the proofs as an exercise.
Example 13.34. If (X, d) is a metric space and A ⊂X, then B ⊂A is open in the metric subspace (A, dA) if and only if B = A ∩G where G is an open subset of X. This is consistent with our previous definition of relatively open sets in A ⊂R.
Open sets with respect to one metric on a set need not be open with respect to another metric. For example, every subset of R with the discrete metric is open, but this is not true of R with the absolute-value metric.
Consistent with our terminology, open balls are open and closed balls are closed.
Proposition 13.35. Let X be a metric space. If x ∈X and r > 0, then the open ball Br(x) is open and the closed ball ¯ Br(x) is closed.
Proof. Suppose that y ∈Br(x) where r > 0, and let ϵ = r −d(x, y) > 0. The triangle inequality implies that Bϵ(y) ⊂Br(x), which proves that Br(x) is open.
Similarly, if y ∈¯ Br(x)c and ϵ = d(x, y) −r > 0, then the triangle inequality implies that Bϵ(y) ⊂¯ Br(x)c, which proves that ¯ Br(x)c is open and ¯ Br(x) is closed.
□ The next theorem summarizes three basic properties of open sets.
Theorem 13.36. Let X be a metric space.
(1) The empty set ∅and the whole set X are open.
(2) An arbitrary union of open sets is open.
280 13. Metric, Normed, and Topological Spaces (3) A finite intersection of open sets is open.
Proof. Property (1) follows immediately from Definition 13.30. (The empty set satisfies the definition vacuously: since it has no points, every point has a neigh-borhood that is contained in the set.) To prove (2), let {Gi ⊂X : i ∈I} be an arbitrary collection of open sets. If x ∈ [ i∈I Gi, then x ∈Gi for some i ∈I.
Since Gi is open, there exists r > 0 such that Br(x) ⊂Gi, and then Br(x) ⊂ [ i∈I Gi.
Thus, the union S Gi is open.
The prove (3), let {G1, G2, . . . , Gn} be a finite collection of open sets. If x ∈ n \ i=1 Gi, then x ∈Gi for every 1 ≤i ≤n. Since Gi is open, there exists ri > 0 such that Bri(x) ⊂Gi. Let r = min(r1, r2, . . . , rn) > 0. Then Br(x) ⊂Bri(x) ⊂Gi for every 1 ≤i ≤n, which implies that Br(x) ⊂ n \ i=1 Gi.
Thus, the finite intersection T Gi is open.
□ The previous proof fails if we consider the intersection of infinitely many open sets {Gi : i ∈I} because we may have inf{ri : i ∈I} = 0 even though ri > 0 for every i ∈I.
The properties of closed sets follow by taking complements of the corresponding properties of open sets and using De Morgan’s laws, exactly as in the proof of Proposition 5.20.
Theorem 13.37. Let X be a metric space.
(1) The empty set ∅and the whole set X are closed.
(2) An arbitrary intersection of closed sets is closed.
(3) A finite union of closed sets is closed.
The following relationships of points to sets are entirely analogous to the ones in Definition 5.22 for R.
Definition 13.38. Let X be a metric space and A ⊂X.
(1) A point x ∈A is an interior point of A if Br(x) ⊂A for some r > 0.
(2) A point x ∈A is an isolated point of A if Br(x) ∩A = {x} for some r > 0, meaning that x is the only point of A that belongs to Br(x).
(3) A point x ∈X is a boundary point of A if, for every r > 0, the ball Br(x) contains a point in A and a point not in A.
13.3. Open and closed sets 281 (4) A point x ∈X is an accumulation point of A if, for every r > 0, the ball Br(x) contains a point y ∈A such that y ̸= x.
A set is open if and only if every point is an interior point and closed if and only if every accumulation point belongs to the set.
We define the interior, boundary, and closure of a set as follows.
Definition 13.39. Let A be a subset of a metric space. The interior A◦of A is the set of interior points of A. The boundary ∂A of A is the set of boundary points.
The closure of A is ¯ A = A ∪∂A.
It follows that x ∈¯ A if and only if the ball Br(x) contains some point in A for every r > 0. The next proposition gives equivalent topological definitions.
Proposition 13.40. Let X be a metric space and A ⊂X. The interior of A is the largest open set contained in A, A◦= [ {G ⊂A : G is open in X} , the closure of A is the smallest closed set that contains A, ¯ A = \ {F ⊃A : F is closed in X} , and the boundary of A is their set-theoretic difference, ∂A = ¯ A \ A◦.
Proof. Let A1 denote the set of interior points of A, as in Definition 13.38, and A2 = S {G ⊂A : G is open}.
If x ∈A1, then there is an open neighborhood G ⊂A of x, so G ⊂A2 and x ∈A2. It follows that A1 ⊂A2. To get the opposite inclusion, note that A2 is open by Theorem 13.36. Thus, if x ∈A2, then A2 ⊂A is a neighborhood of x, so x ∈A1 and A2 ⊂A1. Therefore A1 = A2, which proves the result for the interior.
Next, Definition 13.38 and the previous result imply that ( ¯ A)c = (Ac)◦= [ {G ⊂Ac : G is open} .
Using De Morgan’s laws, and writing Gc = F, we get that ¯ A = [ {G ⊂Ac : G is open}c = \ {F ⊃A : F is closed} , which proves the result for the closure.
Finally, if x ∈∂A, then x ∈ ¯ A = A ∪∂A, and no neighborhood of x is contained in A, so x / ∈A◦. It follows that x ∈¯ A\A◦and ∂A ⊂¯ A\A◦. Conversely, if x ∈¯ A \ A◦, then every neighborhood of x contains points in A, since x ∈¯ A, and every neighborhood contains points in Ac, since x / ∈A◦. It follows that x ∈∂A and ¯ A \ A◦⊂∂A, which completes the proof.
□ It follows from Theorem 13.36, and Theorem 13.37 that the interior A◦is open, the closure ¯ A is closed, and the boundary ∂A is closed. Furthermore, A is open if and only if A = A◦, and A is closed if and only if A = ¯ A.
Let us illustrate these definitions with some examples, whose verification we leave as an exercise.
282 13. Metric, Normed, and Topological Spaces Example 13.41. Consider R with the absolute-value metric. If I = (a, b) and J = [a, b], then I◦= J◦= (a, b), ¯ I = ¯ J = [a, b], and ∂I = ∂J = {a, b}. Note that I = I◦, meaning that I is open, and J = ¯ J, meaning that J is closed. If A = {1/n : n ∈N}, then A◦= ∅and ¯ A = ∂A = A ∪{0}. Thus, A is neither open (A ̸= A◦) nor closed (A ̸= ¯ A). If Q is the set of rational numbers, then Q◦= ∅ and ¯ Q = ∂Q = R. Thus, Q is neither open nor closed. Since ¯ Q = R, we say that Q is dense in R.
Example 13.42. Let A be the unit open ball in R2 with the Euclidean metric, A = (x, y) ∈R2 : x2 + y2 < 1 .
Then A◦= A, the closure of A is the closed unit ball ¯ A = (x, y) ∈R2 : x2 + y2 ≤1 , and the boundary of A is the unit circle ∂A = (x, y) ∈R2 : x2 + y2 = 1 , Example 13.43. Let A be the unit open ball with the x-axis deleted in R2 with the Euclidean metric, A = (x, y) ∈R2 : x2 + y2 < 1, y ̸= 0 .
Then A◦= A, the closure of A is the closed unit ball ¯ A = (x, y) ∈R2 : x2 + y2 ≤1 , and the boundary of A consists of the unit circle and the x-axis, ∂A = (x, y) ∈R2 : x2 + y2 = 1 ∪ (x, 0) ∈R2 : |x| ≤1 .
Example 13.44. Suppose that X is a set containing at least two elements with the discrete metric defined in Example 13.3. If x ∈X, then the unit open ball is B1(x) = {x}, and it is equal to its closure Br(x) = {x}. On the other hand, the closed unit ball is ¯ B1(x) = X. Thus, in a general metric space, the closure of an open ball of radius r > 0 need not be the closed ball of radius r.
13.4. Completeness, compactness, and continuity A sequence (xn) in a set X is a function f : N →X, where xn = f(n) is the nth term in the sequence.
Definition 13.45. Let (X, d) be a metric space. A sequence (xn) in X converges to x ∈X, written xn →x as n →∞or lim n→∞xn = x, if for every ϵ > 0 there exists N ∈N such that n > N implies that d(xn, x) < ϵ.
That is, xn →x in X if d(xn, x) →0 in R. Equivalently, xn →x as n →∞if for every neighborhood U of x there exists N ∈N such that xn ∈U for all n > N.
13.4. Completeness, compactness, and continuity 283 Example 13.46. If d is the discrete metric on a set X, then a sequence (xn) converges in (X, d) if and only if it is eventually constant. That is, there exists x ∈X and N ∈N such that xn = x for all n > N; and, in that case, the sequence converges to x.
Example 13.47. For R with its standard absolute-value metric, Definition 13.45 is the definition of the convergence of a real sequence.
As for subsets of R, we can give a sequential characterization of closed sets in a metric space.
Theorem 13.48. A subset F ⊂X of a metric space X is closed if and only if the limit of every convergent sequence in F belongs to F.
Proof. First suppose that F is closed, meaning that F c is open.
If (xn) be a sequence in F and x ∈F c, then there is a neighborhood U ⊂F c of x which contains no terms in the sequence, so (xn) cannot converge to x. Thus, the limit of every convergent sequence in F belongs to F.
Conversely, suppose that F is not closed. Then F c is not open, and there exists a point x ∈F c such that every neighborhood of x contains points in F. Choose xn ∈F such that xn ∈B1/n(x). Then (xn) is a sequence in F whose limit x does not belong to F, which proves the result.
□ We define the completeness of metric spaces in terms of Cauchy sequences.
Definition 13.49. Let (X, d) be a metric space. A sequence (xn) in X is a Cauchy sequence for every ϵ > 0 there exists N ∈N such that m, n > N implies that d(xm, xn) < ϵ.
Every convergent sequence is Cauchy: if xn →x then given ϵ > 0 there exists N such that d(xn, x) < ϵ/2 for all n > N, and then for all m, n > N we have d(xm, xn) ≤d(xm, x) + d(xn, x) < ϵ.
Complete spaces are ones in which the converse is also true.
Definition 13.50. A metric space is complete if every Cauchy sequence converges.
Example 13.51. If d is the discrete metric on a set X, then (X, d) is a complete metric space since every Cauchy sequence is eventually constant.
Example 13.52. The space (R, | · |) is complete, but the metric subspace (Q, | · |) is not complete.
In a complete space, we have the following simple criterion for the completeness of a subspace.
Proposition 13.53. A subspace (A, dA) of a complete metric space (X, d) is com-plete if and only if A is closed in X.
Proof. If A is a closed subspace of a complete space X and (xn) is a Cauchy sequence in A, then (xn) is a Cauchy sequence in X, so it converges to x ∈X.
Since A is closed, x ∈A, which shows that A is complete.
284 13. Metric, Normed, and Topological Spaces Conversely, if A is not closed, then by Proposition 13.48 there is a convergent sequence in A whose limit does not belong to A. Since it converges, the sequence is Cauchy, but it doesn’t have a limit in A, so A is not complete.
□ The most important complete metric spaces in analysis are the complete normed spaces, or Banach spaces.
Definition 13.54. A Banach space is a complete normed vector space.
For example, R with the absolute-value norm is a one-dimensional Banach space. Furthermore, it follows from the completeness of R that every finite-dimensional normed vector space over R is complete. We prove this for the ℓp-norms given in Definition 13.25.
Theorem 13.55. Let 1 ≤p ≤∞. The vector space Rn with the ℓp-norm is a Banach space.
Proof. Suppose that (xk)∞ k=1 is a sequence of points xk = (x1,k, x2,k, . . . , xn,k) in Rn that is Cauchy with respect to the ℓp-norm. From Theorem 13.28, |xi,j −xi,k| ≤∥xj −xk∥p , so each coordinate sequence (xi,k)∞ k=1 is Cauchy in R.
The completeness of R implies that xi,k →xi as k →∞for some xi ∈R. Let x = (x1, x2, . . . , xn). Then, from Theorem 13.28 again, ∥xk −x∥p ≤C max {|xi,k −xi| : i = 1, 2, . . . , n} , where C = n1/p if 1 ≤p < ∞or C = 1 if p = ∞. Given ϵ > 0, choose Ni ∈N such that |xi,k −xi| < ϵ/C for all k > Ni, and let N = max{N1, N2, . . . , Nn}. Then k > N implies that ∥xk −x∥p < ϵ, which proves that xk →x with respect to the ℓp-norm. Thus, (Rn, ∥· ∥p) is complete.
□ The Bolzano-Weierstrass property provides a sequential definition of compact-ness in a general metric space.
Definition 13.56. A subset K ⊂X of a metric space X is sequentially compact, or compact for short, if every sequence in K has a convergent subsequence whose limit belongs to K.
Explicitly, this condition means that if (xn) is a sequence of points xn ∈K then there is a subsequence (xnk) such that xnk →x as k →∞where x ∈K.
Compactness is an intrinsic property of a subset: K ⊂X is compact if and only if the metric subspace (K, dK) is compact.
Although this definition is similar to the one for compact sets in R, there is a significant difference between compact sets in a general metric space and in R.
Every compact subset of a metric space is closed and bounded, as in R, but it is not always true that a closed, bounded set is compact.
First, as the following example illustrates, a set must be complete, not just closed, to be compact. (A closed subset of R is complete because R is complete.) 13.4. Completeness, compactness, and continuity 285 Example 13.57. Consider the metric space Q with the absolute value norm. The set [0, 2] ∩Q is a closed, bounded subspace, but it is not compact since a sequence of rational numbers that converges in R to an irrational number such as √ 2 has no convergent subsequence in Q.
Second, completeness and boundedness is not enough, in general, to imply compactness.
Example 13.58. Consider N, or any other infinite set, with the discrete metric, d(m, n) = ( 0 if m = n, 1 if m ̸= n.
Then N is complete and bounded with respect to this metric. However, it is not compact since xn = n is a sequence with no convergent subsequence, as is clear from Example 13.46.
The correct generalization to an arbitrary metric space of the characterization of compact sets in R as closed and bounded replaces “closed” with “complete” and “bounded” with “totally bounded,” which is defined as follows.
Definition 13.59. Let (X, d) be a metric space. A subset A ⊂X is totally bounded if for every ϵ > 0 there exists a finite set {x1, x2, . . . , xn} of points in X such that A ⊂ n [ i=1 Bϵ(xi).
The proof of the following result is then completely analogous to the proof of the Bolzano-Weierstrass theorem in Theorem 3.57 for R.
Theorem 13.60. A subset K ⊂X of a metric space X is sequentially compact if and only if it is is complete and totally bounded.
The definition of the continuity of functions between metric spaces parallels the definitions for real functions.
Definition 13.61. Let (X, dX) and (Y, dY ) be metric spaces. A function f : X → Y is continuous at c ∈X if for every ϵ > 0 there exists δ > 0 such that dX(x, c) < δ implies that dY (f(x), f(c)) < ϵ.
The function is continuous on X if it is continuous at every point of X.
Example 13.62. A function f : R2 →R, where R2 is equipped with the Euclidean norm ∥· ∥and R with the absolute value norm | · |, is continuous at c ∈R2 if ∥x −c∥< δ implies that ∥f(x) −f(c)∥< ϵ Explicitly, if x = (x1, x2), c = (c1, c2) and f(x) = (f1(x1, x2), f2(x1, x2)) , this condition reads: p (x1 −c1)2 + (x2 −c2)2 < δ implies that |f(x1, x2) −f1(c1, c2)| < ϵ.
286 13. Metric, Normed, and Topological Spaces Example 13.63. A function f : R →R2, where R2 is equipped with the Euclidean norm ∥· ∥and R with the absolute value norm | · |, is continuous at c ∈R2 if |x −c| < δ implies that ∥f(x) −f(c)∥< ϵ Explicitly, if f(x) = (f1(x), f2(x)) , where f1, f2 : R →R, this condition reads: |x −c| < δ implies that q [f1(x) −f1(c)]2 + [f2(x) −f2(c)]2 < ϵ.
The previous examples generalize in a natural way to define the continuity of an m-component vector-valued function of n variables, f : Rn →Rm.
The definition looks complicated if it is written out explicitly, but it is much clearer if it is expressed in terms or metrics or norms.
Example 13.64. Define F : C([0, 1]) →R by F(f) = f(0), where C([0, 1]) is the space of continuous functions f : [0, 1] →R equipped with the sup-norm described in Example 13.9, and R has the absolute value norm. That is, F evaluates a function f(x) at x = 0. Thus F is a function acting on functions, and its values are scalars; such a function, which maps functions to scalars, is called a functional. Then F is continuous, since ∥f −g∥∞< ϵ implies that |f(0)−g(0)| < ϵ.
(That is, we take δ = ϵ).
We also have a sequential characterization of continuity in a metric space.
Theorem 13.65. Let X and Y be metric spaces.
A function f : X →Y is continuous at c ∈X if and only if f(xn) →f(c) as n →∞for every sequence (xn) in X such that xn →c as n →∞, We define uniform continuity similarly to the real case.
Definition 13.66. Let (X, dX) and (Y, dY ) be metric spaces. A function f : X → Y is uniformly continuous on X if for every ϵ > 0 there exists δ > 0 such that dX(x, y) < δ implies that dY (f(x), f(y)) < ϵ.
The proofs of the following theorems are identically to the proofs we gave for functions f : A ⊂R →R. First, a function on a metric space is continuous if and only if the inverse images of open sets are open.
Theorem 13.67. A function f : X →Y between metric spaces X and Y is continuous on X if and only if f −1(V ) is open in X for every open set V in Y .
Second, the continuous image of a compact set is compact.
Theorem 13.68. Let f : K →Y be a continuous function from a compact metric space K to a metric space Y . Then f(K) is a compact subspace of Y .
Third, a continuous function on a compact set is uniformly continuous.
Theorem 13.69. If f : K →Y is a continuous function on a compact set K, then f is uniformly continuous.
13.5. Topological spaces 287 13.5. Topological spaces A collection of subsets of a set X with the properties of the open sets in a metric space given in Theorem 13.36 is called a topology on X, and a set with such a collection of open sets is called a topological space.
Definition 13.70. Let X be a set. A collection T ⊂P(X) of subsets of X is a topology on X if it satisfies the following conditions.
(1) The empty set ∅and the whole set X belong to T .
(2) The union of an arbitrary collection of sets in T belongs to T .
(3) The intersection of a finite collection of sets in T belongs to T .
A set G ⊂X is open with respect to T if G ∈T , and a set F ⊂X is closed with respect to T if F c ∈T . A topological space (X, T ) is a set X together with a topology T on X.
We can put different topologies on a set with two or more elements. If the topology on X is clear from the context, then we simply refer to X as a topological space and we don’t specify the topology when we refer to open or closed sets.
Every metric space with the open sets in Definition 13.30 is a topological space; the resulting collection of open sets is called the metric topology of the metric space. There are, however, topological spaces whose topology is not derived from any metric on the space.
Example 13.71. Let X be any set. Then T = P(X) is a topology on X, called the discrete topology. In this topology, every set is both open and closed. This topology is the metric topology associated with the discrete metric on X in Example 13.3.
Example 13.72. Let X be any set. Then T = {∅, X} is a topology on X, called the trivial topology. The empty set and the whole set are both open and closed, and no other subsets of X are either open or closed.
If X has at least two elements, then this topology is different from the discrete topology in the previous example, and it is not derived from a metric. To see this, suppose that x, y ∈X and x ̸= y. If d : X × X →R is a metric on X, then d(x, y) = r > 0 and Br(x) is a nonempty open set in the metric topology that doesn’t contain y, so Br(x) / ∈T .
The previous example illustrates a separation property of metric topologies that need not be satisfied by non-metric topologies.
Definition 13.73. A topological space (X, T ) is Hausdorffif for every x, y ∈X with x ̸= y there exist open sets U, V ∈T such that x ∈U, y ∈V and U ∩V = ∅.
That is, a topological space is Hausdorffif distinct points have disjoint neigh-borhoods. In that case, we also say that the topology is Hausdorff. Nearly all topological spaces that arise in analysis are Hausdorff, including, in particular, metric spaces.
Proposition 13.74. Every metric topology is Hausdorff.
288 13. Metric, Normed, and Topological Spaces Proof. Let (X, d) be a metric space. If x, y ∈X and x ̸= y, then d(x, y) = r > 0, and Br/2(x), Br/2(y) are disjoint open neighborhoods of x, y.
□ Compact sets are defined topologically as sets with the Heine-Borel property.
Definition 13.75. Let X be a topological space. A set K ⊂X is compact if every open cover of K has a finite subcover. That is, if {Gi : i ∈I} is a collection of open sets such that K ⊂ [ i∈I Gi, then there is a finite subcollection {Gi1, Gi2, . . . , Gin} such that K ⊂ n [ k=1 Gik.
The Heine-Borel and Bolzano-Weierstrass properties are equivalent in every metric space.
Theorem 13.76. A metric space is compact if and only if it sequentially compact.
We won’t prove this result here, but we remark that it does not hold for general topological spaces.
Finally, we give the topological definitions of convergence, continuity, and con-nectedness which are essentially the same as the corresponding statements for R.
We also show that continuous maps preserve compactness and connectedness.
The definition of the convergence of a sequence is identical to the statement in Proposition 5.9 for R.
Definition 13.77. Let X be a topological space. A sequence (xn) in X converges to x ∈X if for every neighborhood U of x there exists N ∈N such that xn ∈U for every n > N.
The following definition of continuity in a topological space corresponds to Definition 7.2 for R (with the relative absolute-value topology on the domain A of f) and Theorem 7.31.
Definition 13.78. Let f : X →Y be a function between topological spaces X, Y .
Then f is continuous at x ∈X if for every neighborhood V ⊂Y of f(x), there exists a neighborhood U ⊂X of x such that f(U) ⊂V . The function f is continuous on X if f −1(V ) is open in X for every open set V ⊂Y .
These definitions are equivalent to the corresponding “ϵ-δ” definitions in a metric space, but they make sense in a general topological space because they refer only to neighborhoods and open sets. We illustrate them with two simple examples.
Example 13.79. If X is a set with the discrete topology in Example 13.71, then a sequence converges to x ∈X if an only if its terms are eventually equal to x, since {x} is a neighborhood of x. Every function f : X →Y is continuous with respect to the discrete topology on X, since every subset of X is open. On the other hand, if Y has the discrete topology, then f : X →Y is continuous if and only if f −1({y}) is open in X for every y ∈Y .
13.6. Function spaces 289 Example 13.80. Let X be a set with the trivial topology in Example 13.72. Then every sequence converges to every point x ∈X, since the only neighborhood of x is X itself. As this example illustrates, non-Hausdorfftopologies have the unpleasant feature that limits need not be unique, which is one reason why they rarely arise in analysis. If Y has the trivial topology, then every function X →Y is continuous, since f −1(∅) = ∅and f −1(Y ) = X are open in X. On the other hand, if X has the trivial topology and Y is Hausdorff, then the only continuous functions f : X →Y are the constant functions.
Our last definition of a connected topological space corresponds to Defini-tion 5.58 for connected sets of real numbers (with the relative topology).
Definition 13.81. A topological space X is disconnected if there exist nonempty, disjoint open sets U, V such that X = U ∪V . A topological space is connected if it is not disconnected.
The following proof that continuous functions map compact sets to compact sets and connected sets is the same as the proofs given in Theorem 7.35 and The-orem 7.32 for sets of real numbers. Note that a continuous function maps compact or connected sets in the opposite direction to open or closed sets, whose inverse image is open or closed.
Theorem 13.82. Suppose that f : X →Y is a continuous map between topologi-cal spaces X and Y . Then f(X) is compact if X is compact, and f(X) is connected if X is connected.
Proof. For the first part, suppose that X is compact. If {Vi : i ∈I} is an open cover of f(X), then since f is continuous {f −1(Vi) : i ∈I} is an open cover of X, and since X is compact there is a finite subcover f −1(Vi1), f −1(Vi2), . . . , f −1(Vin) .
It follows that {Vi1, Vi2, . . . , Vin} is a finite subcover of the original open cover of f(X), which proves that f(X) is compact.
For the second part, suppose that f(X) is disconnected.
Then there exist nonempty, disjoint open sets U, V in Y such that U ∪V ⊃f(X).
Since f is continuous, f −1(U), f −1(V ) are open, nonempty, disjoint sets such that X = f −1(U) ∪f −1(V ), so X is disconnected. It follows that f(X) is connected if X is connected.
□ 13.6. Function spaces There are many function spaces, and their study is a central topic in analysis.
We discuss only one important example here: the space of continuous functions on a compact set equipped with the sup norm.
We repeat its definition from Example 13.9.
290 13. Metric, Normed, and Topological Spaces Definition 13.83. Let K ⊂R be a compact set. The space C(K) consists of the continuous functions f : K →R. Addition and scalar multiplication of functions is defined pointwise in the usual way: if f, g ∈C(K) and k ∈R, then (f + g)(x) = f(x) + g(x), (kf)(x) = k (f(x)) .
The sup-norm of a function f ∈C(K) is defined by ∥f∥∞= sup x∈K |f(x)|.
Since a continuous function on a compact set attains its maximum and mini-mum value, for f ∈C(K) we can also write ∥f∥∞= max x∈K |f(x)|.
Thus, the sup-norm on C(K) is analogous to the ℓ∞-norm on Rn.
In fact, if K = {1, 2, . . . , n} is a finite set with the discrete topology, then it is identical to the ℓ∞-norm.
Our previous results on continuous functions on a compact set can be formu-lated concisely in terms of this space. The following characterization of uniform convergence in terms of the sup-norm is easily seen to be equivalent to Definition 9.8.
Definition 13.84. A sequence (fn) of functions fn : K →R converges uniformly on K to a function f : K →R if lim n→∞∥fn −f∥∞= 0.
Similarly, we can rephrase Definition 9.12 for a uniformly Cauchy sequence in terms of the sup-norm.
Definition 13.85. A sequence (fn) of functions fn : K →R is uniformly Cauchy on K if for every ϵ > 0 there exists N ∈N such that m, n > N implies that ∥fm −fn∥∞< ϵ.
Thus, the uniform convergence of a sequence of functions is defined in exactly the same way as the convergence of a sequence of real numbers with the absolute value | · | replaced by the sup-norm ∥· ∥. Moreover, like R, the space C(K) is complete.
Theorem 13.86. The space C(K) with the sup-norm ∥· ∥∞is a Banach space.
Proof. From Theorem 7.15, the sum of continuous functions and the scalar mul-tiple of a continuous function are continuous, so C(K) is closed under addition and scalar multiplication. The algebraic vector-space properties for C(K) follow immediately from those of R.
From Theorem 7.37, a continuous function on a compact set is bounded, so ∥·∥∞is well-defined on C(K). The sup-norm is clearly non-negative, and ∥f∥∞= 0 implies that f(x) = 0 for every x ∈K, meaning that f = 0 is the zero function.
13.6. Function spaces 291 We also have for all f, g ∈C(K) and k ∈R that ∥kf∥∞= sup x∈K |kf(x)| = |k| sup x∈K |f(x)| = |k| ∥f∥∞, ∥f + g∥∞= sup x∈K |f(x) + g(x)| ≤sup x∈K {|f(x)| + |g(x)|} ≤sup x∈K |f(x)| + sup x∈K |g(x)| ≤∥f∥∞+ ∥g∥∞, which verifies the properties of a norm.
Finally, Theorem 9.13 implies that a uniformly Cauchy sequence converges uniformly so C(K) is complete.
□ For comparison with the sup-norm, we consider a different norm on C([a, b]) called the one-norm, which is analogous to the ℓ1-norm on Rn.
Definition 13.87. If f : [a, b] →R is a Riemann integrable function, then the one-norm of f is ∥f∥1 = Z b a |f(x)| dx.
Theorem 13.88. The space C([a, b]) of continuous functions f : [a, b] →R with the one-norm ∥· ∥1 is a normed space.
Proof. As shown in Theorem 13.86, C([a, b]) is a vector space. Every continuous function is Riemann integrable on a compact interval, so ∥· ∥1 : C([a, b]) →R is well-defined, and we just have to verify that it satisfies the properties of a norm.
Since |f| ≥0, we have ∥f∥1 = R b a |f| ≥0. Furthermore, since f is continuous, Proposition 11.42 shows that ∥f∥1 = 0 implies that f = 0, which verifies the positivity. If k ∈R, then ∥kf∥1 = Z b a |kf| = |k| Z b a |f| = |k| ∥f∥1, which verifies the homogeneity. Finally, the triangle inequality is satisfied since ∥f + g∥1 = Z b a |f + g| ≤ Z b a |f| + |g| = Z b a |f| + Z b a |g| = ∥f∥1 + ∥g∥1.
□ Although C([a, b]) equipped with the one-norm ∥· ∥1 is a normed space, it is not complete, and therefore it is not a Banach space. The following example gives a non-convergent Cauchy sequence in this space.
Example 13.89. Define the continuous functions fn : [0, 1] →R by fn(x) = 0 if 0 ≤x ≤1/2, n(x −1/2) if 1/2 < x < 1/2 + 1/n, 1 if 1/2 + 1/n ≤x ≤1.
292 13. Metric, Normed, and Topological Spaces If n > m, we have ∥fn −fm∥1 = Z 1/2+1/m 1/2 |fn −fm| ≤1 m, since |fn −fn| ≤1. Thus, ∥fn −fm∥1 < ϵ for all m, n > 1/ϵ, so (fn) is a Cauchy sequence with respect to the one-norm.
We claim that if ∥f −fn∥1 →0 as n →∞where f ∈C([0, 1]), then f would have to be f(x) = ( 0 if 0 ≤x ≤1/2, 1 if 1/2 < x ≤1, which is discontinuous at 1/2, so (fn) does not have a limit in (C([0, 1]), ∥· ∥1).
To prove the claim, note that if ∥f −fn∥1 →0, then R 1/2 0 |f| = 0 since Z 1/2 0 |f| = Z 1/2 0 |f −fn| ≤ Z 1 0 |f −fn| →0, and Proposition 11.42 implies that f(x) = 0 for 0 ≤x ≤1/2. Similarly, for every 0 < ϵ < 1/2, we get that R 1 1/2+ϵ |f −1| = 0, so f(x) = 1 for 1/2 < x ≤1.
The sequence (fn) is not uniformly Cauchy since ∥fn −fm∥∞→1 as n → ∞for every m ∈N, so this example does not contradict the completeness of (C([0, 1]), ∥· ∥∞).
The ℓ∞-norm and the ℓ1-norm on the finite-dimensional space Rn are equiva-lent, but the sup-norm and the one-norm on C([a, b]) are not. In one direction, we have Z b a |f| ≤(b −a) · sup [a,b] |f|, so ∥f∥1 ≤(b −a)∥f∥∞, and ∥f −fn∥∞→0 implies that ∥f −fn∥1 →0. As the following example shows, the converse is not true. There is no constant M such that ∥f∥∞≤M∥f∥1 for all f ∈C([a, b]), and ∥f −fn∥1 →0 does not imply that ∥f −fn∥∞→0.
Example 13.90. For n ∈N, define the continuous function fn : [0, 1] →R by fn(x) = ( 1 −nx if 0 ≤x ≤1/n, 0 if 1/n < x ≤1.
Then ∥fn∥∞= 1 for every n ∈N, but ∥fn∥1 = Z 1/n 0 (1 −nx) dx = x −1 2nx2 1/n 0 = 1 2n, so ∥fn∥1 →0 as n →∞.
Thus, unlike the finite-dimensional vector space Rn, an infinite-dimensional vector space such as C([a, b]) has many inequivalent norms and many inequivalent notions of convergence.
The incompleteness of C([a, b]) with respect to the one-norm suggests that we use the larger space R([a, b]) of Riemann integrable functions on [a, b], which 13.7. The Minkowski inequality 293 includes some discontinuous functions. A slight complication arises from the fact that if f is Riemann integrable and R b a |f| = 0, then it does not follows that f = 0, so ∥f∥1 = 0 does not imply that f = 0. Thus, ∥·∥1 is not, strictly speaking, a norm on R([a, b]). We can, however, get a normed space of equivalence classes of Riemann integrable functions, by defining f, g ∈R([a, b]) to be equivalent if R b a |f −g| = 0.
For instance, the function in Example 11.14 is equivalent to the zero-function.
A much more fundamental defect of the space of (equivalence classes of) Rie-mann integrable functions with the one-norm is that it is still not complete. To get a space that is complete with respect to the one-norm, we have to use the space L1([a, b]) of (equivalence classes of) Lebesgue integrable functions on [a, b]. This is another reason for the superiority of the Lebesgue integral over the Riemann integral: it leads function spaces that are complete with respect to integral norms.
The inclusion of the smaller incomplete space C([a, b]) of continuous functions in the larger complete space L1([a, b]) of Lebesgue integrable functions is analogous to the inclusion of the incomplete space Q of rational numbers in the complete space R of real numbers.
13.7. The Minkowski inequality Inequalities are essential to analysis. Their proofs, however, may require consider-able ingenuity, and there are often many different ways to prove the same inequality.
In this section, we complete the proof that the ℓp-spaces are normed spaces by prov-ing the triangle inequality given in Definition 13.25. This inequality is called the Minkowski inequality, and it’s one of the most important inequalities in mathemat-ics.
The simplest case is for the Euclidean norm with p = 2. We begin by proving the following fundamental Cauchy-Schwartz inequality.
Theorem 13.91 (Cauchy-Schwartz inequality). If (x1, x2, . . . , xn), (y1, y2, . . . , yn) are points in Rn, then n X i=1 xiyi ≤ n X i=1 x2 i !1/2 n X i=1 y2 i !1/2 .
Proof. Since | P xiyi| ≤P |xi| |yi|, it is sufficient to prove the inequality for xi, yi ≥0. Furthermore, the inequality is obvious if x = 0 or y = 0, so we as-sume that at least one xi and one yi is nonzero.
For every α, β ∈R, we have 0 ≤ n X i=1 (αxi −βyi)2 .
Expanding the square on the right-hand side and rearranging the terms, we get that 2αβ n X i=1 xiyi ≤α2 n X i=1 x2 i + β2 n X i=1 y2 i .
294 13. Metric, Normed, and Topological Spaces We choose α, β > 0 to “balance” the terms on the right-hand side, α = n X i=1 y2 i !1/2 , β = n X i=1 x2 i !1/2 .
Then division of the resulting inequality by 2αβ proves the theorem.
□ The Minkowski inequality for p = 2 is an immediate consequence of the Cauchy-Schwartz inequality.
Corollary 13.92 (Minkowski inequality). If (x1, x2, . . . , xn) and (y1, y2, . . . , yn) are points in Rn, then " n X i=1 (xi + yi)2 #1/2 ≤ n X i=1 x2 i !1/2 + n X i=1 y2 i !1/2 .
Proof. Expanding the square in the following equation and using the Cauchy-Schwartz inequality, we get n X i=1 (xi + yi)2 = n X i=1 x2 i + 2 n X i=1 xiyi + n X i=1 y2 i ≤ n X i=1 x2 i + 2 n X i=1 x2 i !1/2 n X i=1 y2 i !1/2 + n X i=1 y2 i ≤ n X i=1 x2 i !1/2 + n X i=1 y2 i !1/2 2 .
Taking the square root of this inequality, we obtain the result.
□ To prove the Minkowski inequality for general 1 < p < ∞, we first define the H¨ older conjugate p′ of p and prove Young’s inequality.
Definition 13.93. If 1 < p < ∞, then the H¨ older conjugate 1 < p′ < ∞of p is the number such that 1 p + 1 p′ = 1.
If p = 1, then p′ = ∞; and if p = ∞then p′ = 1.
The H¨ older conjugate of 1 < p < ∞is given explicitly by p′ = p p −1.
Note that if 1 < p < 2, then 2 < p′ < ∞; and if 2 < p < ∞, then 1 < p′ < 2. The number 2 is its own H¨ older conjugate. Furthermore, if p′ is the H¨ older conjugate of p, then p is the H¨ older conjugate of p′.
Theorem 13.94 (Young’s inequality). Suppose that 1 < p < ∞and 1 < p′ < ∞ is its H¨ older conjugate. If a, b ≥0 are nonnegative real numbers, then ab ≤ap p + bp′ p′ .
13.7. The Minkowski inequality 295 Moreover, there is equality if and only if ap = bp′.
Proof. There are several ways to prove this inequality. We give a proof based on calculus.
The result is trivial if a = 0 or b = 0, so suppose that a, b > 0. We write ap p + bp′ p′ −ab = bp′ 1 p ap bp′ + 1 p′ − a bp′−1 .
The definition of p′ implies that p′/p = p′ −1, so that ap bp′ = a bp′/p p = a bp′−1 p Therefore, we have ap p + bp′ p′ −ab = bp′f(t), f(t) = tp p + 1 p′ −t, t = a bp′−1 .
The derivative of f is f ′(t) = tp−1 −1.
Thus, for p > 1, we have f ′(t) < 0 if 0 < t < 1, and Theorem 8.36 implies that f(t) is strictly decreasing; moreover, f ′(t) > 0 if 1 < t < ∞, so f(t) is strictly increasing. It follows that f has a strict global minimum on (0, ∞) at t = 1. Since f(1) = 1 p + 1 p′ −1 = 0, we conclude that f(t) ≥0 for all 0 < t < ∞, with equality if and only if t = 1.
Furthermore, t = 1 if and only if a = bp′−1 or ap = bp′. It follows that ap p + bp′ p′ −ab ≥0 for all a, b ≥0, with equality if and only ap = bp′, which proves the result.
□ For p = 2, Young’s inequality reduces to the more easily proved inequality in Proposition 2.8.
Before continuing, we give a scaling argument which explains the appearance of the H¨ older conjugate in Young’s inequality. Suppose we look for an inequality of the form ab ≤Map + Naq for all a, b ≥0 for some exponents p, q and some constants M, N. Any inequality that holds for all positive real numbers must remain true under rescalings. Rescaling a 7→λa, b 7→µb in the inequality (where λ, µ > 0) and dividing by λµ, we find that it becomes ab ≤λp−1 µ Map + µq−1 λ Nbq.
We take µ = λp−1 to make the first scaling factor equal to one, and then the inequality becomes ab ≤Map + λrNbq, r = (p −1)(q −1) −1.
If the exponent r of λ is non-zero, then we can violate the inequality by taking λ sufficiently small (if r > 0) or sufficiently large (if r < 0), since it is clearly 296 13. Metric, Normed, and Topological Spaces impossible to bound ab by ap for all b ∈R. Thus, the inequality can only hold if r = 0, which implies that q = p′.
This argument does not, of course, prove the inequality, but it shows that the only possible exponents for which an inequality of this form can hold must satisfy q = p′. Theorem 13.94 proves that such an inequality does in fact hold in that case provided 1 < p < ∞.
Next, we use Young’s inequality to deduce H´ older’s inequality, which is a gen-eralization of the Cauchy-Schwartz inequality for p ̸= 2.
Theorem 13.95 (H¨ older’s inequality). Suppose that 1 < p < ∞and 1 < p′ < ∞ is its H¨ older conjugate. If (x1, x2, . . . , xn) and (y1, y2, . . . , yn) are points in Rn, then n X i=1 xiyi ≤ n X i=1 |xi|p !1/p n X i=1 |yi|p′ !1/p′ .
Proof. We assume without loss of generality that xi, yi are nonnegative and x, y ̸= 0. Let α, β > 0. Then applying Young’s inequality in Theorem 13.94 with a = αxi, b = βyi and summing over i, we get αβ n X i=1 xiyi ≤αp p n X i=1 xp i + βp′ p′ n X i=1 yp′ i .
Then, choosing α = n X i=1 yp′ i !1/p , β = n X i=1 xp i !1/p′ to “balance” the terms on the right-hand side, dividing by αβ, and using the fact that 1/p + 1/p′ = 1, we get H¨ older’s inequality.
□ Minkowski’s inequality follows from H¨ older’s inequality.
Theorem 13.96 (Minkowski’s inequality). Suppose that 1 < p < ∞and 1 < p′ < ∞is its H¨ older conjugate. If (x1, x2, . . . , xn) and (y1, y2, . . . , yn) are points in Rn, then n X i=1 |xi + yi|p !1/p ≤ n X i=1 |xi|p !1/p + n X i=1 |yi|p !1/p .
Proof. We assume without loss of generality that xi, yi are nonnegative and x, y ̸= 0. We split the sum on the left-hand side as follows: n X i=1 |xi + yi|p = n X i=1 |xi + yi| |xi + yi|p−1 ≤ n X i=1 |xi| |xi + yi|p−1 + n X i=1 |yi| |xi + yi|p−1 .
By H¨ older’s inequality, we have n X i=1 |xi| |xi + yi|p−1 ≤ n X i=1 |xi|p !1/p n X i=1 |xi + yi|(p−1)p′ !1/p′ , 13.7. The Minkowski inequality 297 and using the fact that p′ = p/(p −1), we get n X i=1 |xi| |xi + yi|p−1 ≤ n X i=1 |xi|p !1/p n X i=1 |xi + yi|p !1−1/p .
Similarly, n X i=1 |yi| |xi + yi|p−1 ≤ n X i=1 |yi|p !1/p n X i=1 |xi + yi|p !1−1/p .
Combining these inequalities, we obtain n X i=1 |xi + yi|p ≤ n X i=1 |xi|p !1/p + n X i=1 |yi|p !1/p n X i=1 |xi + yi|p !1−1/p .
Fianlly, dividing this inequality by (P |xi + yi|p)1−1/p, we get the result.
□ |
10694 | https://www.aafp.org/pubs/afp/issues/2005/1215/p2501.html | Treatment of Irritable Bowel Syndrome
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SUSAN K. HADLEY, M.D., AND STEPHEN M. GAARDER, PH.D.
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Am Fam Physician. 2005;72(12):2501-2508
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A more recent article on irritable bowel syndrome is available.
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Author disclosure: Nothing to disclose.
Irritable bowel syndrome affects 10 to 15 percent of the U.S. population to some degree. This condition is defined as abdominal pain and discomfort with altered bowel habits in the absence of any other mechanical, inflammatory, or biochemical explanation for these symptoms. Irritable bowel syndrome is more likely to affect women than men and is most common in patients 30 to 50 years of age. Symptoms are improved equally by diets supplemented with fiber or hydrolyzed guar gum, but more patients prefer hydrolyzed guar gum. Antispasmodic agents may be used as needed, but anticholinergic and other side effects limit their use in some patients. Loperamide is an option for treatment of moderately severe diarrhea. Antidepressants have been shown to relieve pain and may be effective in low doses. Trials using alosetron showed a clinically significant, although modest, gain over placebo, but it is indicated only for women with severe diarrhea-predominant symptoms or for those in whom conventional treatment has failed. Tegaserod has an advantage over placebo in constipation-predominant irritable bowel syndrome; it is indicated for up to 12 weeks of treatment in women. However, postmarketing reports of severe diarrhea and ischemic colitis further limit its use. Herbal therapies such as peppermint oil also may be effective in the treatment of irritable bowel syndrome. Therapies should focus on specific gastrointestinal dysfunctions (e.g., constipation, diarrhea, pain), and medications only should be used when nonprescription remedies do not work or when symptoms are severe.
Irritable bowel syndrome (IBS) is defined as abdominal pain and discomfort with altered bowel habits that are not explained by any other mechanical, biochemical, or inflammatory cause. Approximately 10 to 15 percent of the U.S. population is affected by IBS, and women are more likely to have symptoms than are men.1 Diagnosis is based on clinical signs and symptoms that include abdominal pain, bloating, constipation, and diarrhea. The criteria in Table 1 were developed to aid in the diagnosis of IBS.2
SORT: KEY RECOMMENDATIONS FOR PRACTICE
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| Clinical recommendation | Evidence rating | References |
---
| Patients with alarm symptoms for malignancy, infection, or inflammatory bowel disease should undergo endoscopic evaluation. | C | 12 |
| Guar gum, fiber, exercise, episodic use of antispasmodics, peppermint oil, and adequate fluid intake are recommended as initial therapy for patients with constipation-predominant IBS. | B | 1, 12, 27 |
| Loperamide (Imodium), episodic use of antispasmodic agents, peppermint oil, and dietary manipulation are recommended as initial therapy for patients with diarrhea-predominant IBS. | B | 1, 12, 27 |
| Tricyclic antidepressants and psychotherapy should be considered for patients with pain-predominant IBS or for any patient with more severe symptoms. | B | 1, 10, 12, 16 |
| Use of newer agents such as alosetron (Lotronex) and tegaserod (Zelnorm) should be limited to selected patients with more severe disease because of adverse effects, high cost, and limited efficacy. | B | 21–23 |
IBS = irritable bowel syndrome.
A = consistent, good-quality patient-oriented evidence; B = inconsistent or limited-quality patient-oriented evidence; C = consensus, disease-oriented evidence, usual practice, expert opinion, or case series. For information about the SORT evidence rating system, see page 2416 or
TABLE 1
Diagnostic Criteria for IBS
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| | |
--- |
| Abdominal discomfort or pain, for at least 12 weeks (which need not be consecutive) in the preceding 12 months, with two of the following features: | |
| | Relief with defecation |
| | Onset associated with a change in stool frequency |
| | Onset associated with a change in form or appearance of stool |
| These additional symptoms cumulatively support the diagnosis of IBS: | |
| | Abnormal stool frequency (more than three times per day or less than three times per week) |
| | Abnormal stool form (loose and watery or lumpy and hard) |
| | Abnormal stool passage (urgency, frequency, feeling of incomplete evacuation) |
| Passage of mucus (white material) | |
| Bloating or sensation of abdominal distention | |
IBS = irritable bowel syndrome.
Information from reference 2.
Alarm factors are signs or symptoms requiring immediate attention and careful diagnostic evaluation to exclude diagnoses other than IBS. These factors, and other less urgent symptoms that may lead to a diagnosis other than IBS, are summarized in Table 2.3,4
TABLE 2
Signs and Symptoms Suggesting Alternative Diagnosis to Irritable Bowel Syndrome
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| Sign or symptoms | Suggested diagnosis |
--- |
| Alarm factors | |
| Anemia | Cancer, IBD |
| Chronic severe diarrhea | Cancer, infection, IBD |
| Family history of colon cancer | Cancer |
| Hematochezia, melena, or other signs of intestinal bleeding | Cancer, arteriovenous malformation, colonic polyps, IBD |
| Recurrent fever | Infection, IBD |
| Weight loss | Cancer, IBD |
| Other signs and symptoms | |
| Travel to areas with parasitic diseases | Infection |
| Family history of colon cancer, irritable bowel syndrome, celiac disease | Cancer, celiac disease |
| Signs or symptoms of malabsorption | Celiac disease |
| Nighttime symptoms (e.g., encopresis) | Infection, trauma |
| Onset after 50 years of age | Cancer |
| Arthritis | Arthritis |
| Thyroid dysfunction | Hypothyroidism, hyperthyroidism |
IBD = inflammatory bowel disease.
Information from references 3 and 4.
The severity of the symptoms and their effects on the patient's quality of life should guide the decision to investigate and treat IBS. Given the limited benefits of pharmacologic therapy and the psychosocial issues involved, effective treatment of IBS requires a comprehensive, multifaceted approach.
Pathophysiology
The pathophysiology of IBS is not clearly understood, but likely factors include altered gastrointestinal motility, increased gut sensitivity, and increased intestinal contractions. Proposed mechanisms include: (1) stress as an aggravating factor because of corticosporin releasing factor, gastric emptying delay, and accelerated colonic transit; (2) visceral hypersensitivity, with a decreased threshold after exposure; (3) abnormal brain activation; (4) altered colonic motility and disturbed motor function; (5) response to eating as a stimulus to colonic activity; (6) abnormal gas propulsion and expulsion; (7) dietary intolerance, most commonly to wheat and dairy products; and (8) inflammation, with production of prostaglandins, bradykinins, nerve growth factors, adenosine, and 5-hydroxytryptamine.5
Treatment
The evidence to support dietary, pharmacologic, behavioral, and herbal therapies for IBS is summarized and accompanied by a suggested management algorithm (Figure 11 ). Prescribing information for commonly used medications is listed in Table 3.
Figure 1.
Management of IBS
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TABLE 3
Medications for Treatment of Irritable Bowel Syndrome
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| Treatment | Initial dosage | Maintenance dosage | Cost (generic) | Comments |
--- ---
| Dicyclomine (Bentyl) | 20 mg fourtimes per day | 20 to 40 mg four times per day | $63 (22 to 82) | If not effective in two weeks, discontinue. |
| Hyoscyamine (Levsin) | 0.125 to 0.250 mg every four hours | Same | 145 (52 to 61) | Anticholinergic effects; maximum 1.5 gm per day |
| Loperamide (Imodium) | 4 mg | 4 to 8 mg per day | 49 (15 to 42) | — |
| Amitriptyline | 10 to 25 mg every night at bedtime | 10 to 100 mg every night at bedtime | 8 (2 to 10) | Large dosing range; start low, and titrate as needed. |
| Desipramine (Norpramin) | 10 to 50 mg every night at bedtime | 10 to 150 mg every night at bedtime | 21 (12 to 21) | Large dosing range; start low, and titrate as needed. |
| Alosetron (Lotronex) | 1 mg per day for four weeks | 1 mg once or twice per day | 216 | Use in women with diarrheapredominant irritable bowel syndrome; use with caution; available only through prescribing program; associated with ischemic colitis. |
| Tegaserod (Zelnorm) | 6 mg twice per day for four to 12 weeks | Same | 169 | Use in constipation with same caveats as alosetron; only indicated for 12 weeks of therapy. |
— Estimated 30-day cost to the pharmacist based on average wholesale prices (rounded to the nearest dollar) in Red Book. Montvale, N.J.: Medical Economics Data, 2005. Cost to the patient will be higher, depending on prescription filling fee.
DIETARY THERAPY
Reported dietary triggers of IBS include caffeine, citrus, corn, dairy lactose, wheat, and wheat gluten. Lactose and caffeine, in particular, may be associated with diarrhea-predominant IBS.1 Food diaries are recommended because they may help patients identify and avoid dietary triggers. Although one study showed a reduction of IBS symptoms in 48 percent of patients on an elimination diet,6 other specific diets have not been effective, and few studies have been done. A complicating factor is that patients may experience symptoms as a generalized effect to eating any foods.
Increasing dietary fiber has long been recommended as a treatment for IBS. The proposed mechanism of action is the enhancement of the stool's water-holding properties, gel formation to provide lubrication, bulking of the stool, and binding of agents such as bile.7 A systematic review8 of 13 randomized controlled trials (RCTs) found no convincing evidence that bulking agents relieve global symptoms of IBS. However, a second systematic review9 did find significant improvement in the ease of stool passage and in general satisfaction with bowel movements.
Because of its safety and low cost, a trial of fiber is reasonable, particularly in patients whose predominant symptom is constipation. There are many types of fiber, and not all have been studied. Synthetic fibers are more soluble than natural fibers but may cause more gas symptoms. Psyllium seed and linseed are natural fibers containing mucilages and are bulking agents with lubrication properties. Wheat bran fiber should be avoided in patients with gluten sensitivity.9 Patients with very slow colonic transit may benefit from scheduled use of osmotic laxatives such as magnesium salts, phosphate salts, and polyethylene glycol, although these agents have not been well studied.1
Partially hydrolyzed guar gum has been successful in softening and improving fecal output. One recent nonblinded RCT10 found that symptoms of IBS were improved equally by diets supplemented with fiber or guar gum, but more patients preferred guar gum. This was especially true of patients with IBS who could not tolerate fiber or reported a worsening of symptoms.10
ANTISPASMODIC AGENTS
Antispasmodic agents relax smooth muscle in the gut and reduce contractions. Dicyclomine (Bentyl) and hyoscyamine (Levsin) act through anticholinergic or antimuscarinic properties.8 One meta-analysis11 of 23 trials found an advantage over placebo in global improvement (56 versus 38 percent), pain (53 versus 41 percent), and abdominal distension (44 versus 35 percent), but no difference regarding constipation. However, the studies were generally of poor quality.11 The anticholinergic effects of antispasmodics limit their use, especially in the long term.
ANTIDIARRHEAL AGENTS
A systematic review9 of loperamide (Imodium) for the treatment of IBS found that it improved diarrhea symptoms; two of the four studies in the review also reported improved global symptoms. Loperamide slows intestinal transit, increases intestinal water absorption, and increases resting sphincter tone.12 Because loperamide does not cross the blood-brain barrier, side effects are less than other opioids, but it should be used with caution.
ANTIDEPRESSANT AND ANTIANXIETY MEDICATIONS
Antidepressants have been shown to relieve pain with low doses.13 The successful use of low doses supports a mechanism of action separate from the recognized psychiatric effects.14 It is thought that tricyclic antidepressants facilitate endogenous endorphin release and block norepinephrine reuptake, which leads to enhancement of descending inhibitory pathways blockage of the pain neuromodulator serotonin.15 Tricyclic antidepressants may slow intestinal transit time and aid in the treatment of diarrhea. Two recent meta-analyses9,16 reviewed RCTs of patients taking low-dose tricyclic antidepressants including amitriptyline, clomipramine (Anafranil), desipramine (Norpramin), doxepin (Sinequan), and trimipramine (Surmontil). These studies showed that tricyclic antidepressants improve global symptoms, abdominal pain, and diarrhea. On average, for every three patients treated with a tricyclic antidepressant, one experiences a significant benefit.16 Side effects may cause patients to discontinue use, particularly because tricyclic antidepressants may worsen constipation.
Selective serotonin reuptake inhibitors (SSRIs) are being examined for the treatment of IBS. The evidence is limited, but one RCT17 found that patients taking 10 to 40 mg of paroxetine (Paxil) per day were more likely than those taking placebo to have a clinically significant improvement in overall well-being (63 versus 26 percent, number needed to treat [NNT] = 2). This benefit also was present in the subset without depression. Given the limited evidence, SSRIs are not recommended as routine or first-line therapy for IBS except in patients who also have comorbid depression.
Although anxiolytics (e.g., benzodiazepines) may be beneficial in patients with comorbid anxiety disorders, they are not recommended routinely for treatment of IBS because of adverse effects, dependence potential, and interactions with alcohol and other medications.18
PSYCHOTHERAPY
A variety of psychotherapies, including cognitive behavior therapy, hypnosis, and stress management/relaxation therapy, reduce abdominal pain and diarrhea.18 For example, in one study19 patients were more likely to experience a significant benefit with cognitive behavior therapy than education (70 versus 37 percent). Psychotherapies should be considered for motivated patients who have more severe or disabling symptoms.
5-HT3 RECEPTOR ANTAGONISTS
Antagonism of serotonin receptor subtype 5-hydroxy-tryptamine-3 (5-HT3) reduces noxious stimuli perception, increases colonic compliance, and decreases gastrocolonic reflexes. Alosetron (Lotronex), the first IBS-specific medication approved by the U.S. Food and Drug Administration (FDA), is a highly selective central penetrating 5-HT3 antagonist. Alosetron did show a clinically significant, although modest, gain over placebo (41 versus 26 percent) in alleviating IBS symptoms such as bloating and pain.20 Initially, alosetron was removed from the market after being linked to ischemic colitis and the deaths of five women. It became available again in late 2002, but with strict prescribing regulations. Alosetron is indicated only for women with severe diarrhea-predominant symptoms and for whom conventional treatment has failed. Constipation may result from its use.
5-HT4 RECEPTOR AGONISTS
Stimulation of the serotonin receptor subtype 5-hydroxy-tryptamine-4 (5-HT4) increases colonic transit time and inhibits visceral sensitivity. Tegaserod (Zelnorm), a partial 5-HT4 receptor agonist, is an aminoguanidine indole similar to serotonin. Tegaserod stimulates the release of neurotransmitters and increases colonic motility; it is more effective than placebo in constipation-predominant IBS.21 However, a systematic review22 showed that global benefits are small, with an NNT of 17. Tegaserod is approved for up to 12 weeks of use for treatment of constipation-predominant IBS in women.23 The long-term safety of tegaserod has not been established, and post-marketing reports of tegaserod causingischemic colitis and diarrhea resulting in hypovolemia and syncope prompted an FDA advisory in April 2004.24 In summary, tegaserod improves global symptoms of IBS in women, but the benefits are small; further studies are needed to clarify any long-term adverse effects.22
OTHER AGENTS
There are a variety of other agents with reported advantages in treating IBS symptoms. Antibiotics may be recommended for the treatment of refractory diarrhea if bacterial infection is suspected.12 However, antibiotics should not be used routinely for treatment of IBS. Antibiotics are not indicated for long-term use because they may increase diarrhea through changes in the bowel flora.
Probiotics consist of a preparation containing a single- or mixed-culture of live microbes that exert beneficial health effects by altering the gastrointestinal flora.6 Probiotics are presumed to restore normal bowel flora. Studies with probiotics demonstrate a trend toward improvement of IBS symptoms and are promising enough to warrant further investigation.25 Cisapride (Propulsid), a promotility agent, has been studied for treatment of IBS, but its use was not supported in a recent systematic literature review.12 Cisapride was removed from the market and is only available for compassionate use in the United States. The alpha adrenoceptor antagonist clonidine (Catapres) has been shown in a single small study26 to provide overall relief at a dosage of 0.1 mg twice a day when compared with placebo (67 versus 46 percent).
COMPLEMENTARY THERAPIES
Given the absence of a cure and the adverse effects of medications, patients with IBS often turn to complementary therapies. Peppermint possesses antispasmodic properties and has long been associated with improvement of digestive function. Peppermint leaves contain oils that have mild anesthetic properties, relieve nausea, and relax smooth muscle spasticity caused by histamine and cholinergic stimulation.27 A systematic review27 identified five trials that showed that peppermint oil relieved IBS symptoms. Three of these trials showed statistically significant benefit of peppermint over placebo (P < .001). The placebo response ranged from 13 to 52 percent with a mean of 31 percent including all five trials.27 A randomized double-blind placebo-controlled study28 of enteric-coated peppermint oil involving 110 patients showed 79 percent with less pain, 83 percent with decreased stool frequency, and 79 percent with less flatulence. Peppermint is contraindicated in patients with gastroesophageal reflux disease.
The herb ginger also may play a role in IBS treatment. One component, gingerois, functions as a serotonin 5-HT antagonist and enhances motility.29 Aloe vera has been recommended for constipation-dominant IBS,30 and fennel has been recommended for IBS-related bloating. None of these agents has been studied in any clinical trials measuring patient-oriented outcomes.
Approach to the Patient
Given the variability of IBS, the most successful treatment will be comprehensive, involving multiple strategies (Figure 11). Patients should be allowed to participate actively in their care, and therapies should focus on particular types of gastrointestinal dysfunction.1 Initial treatment should include education, reassurance, stress management, and relaxation techniques. Further treatments are based on the type and severity of symptoms. Constipation-predominant IBS with mild symptoms may benefit from additional fluids, guar gum, exercise, and fiber. For constipation-predominant IBS with moderate symptoms, an antispasmodic, peppermint oil, or osmotic laxative may be appropriate. In severe cases, the aforementioned may be supplemented with tricyclic antidepressants, psychotherapy, and consideration of serotonin 5-HT4-agonist.
For diarrhea-predominant IBS, begin with dietary changes and add an antispasmodic, loperamide, or peppermint oil if symptoms are moderate. In severe diarrhea-predominant IBS, consider tricyclic antidepressants, therapy, and a serotonin 5-HT3 antagonist. In pain-predominant IBS, use an antispasmodic; a tricyclic antidepressant; and, if severe with diarrhea, consider a serotonin 5-HT3 agonist.
SUSAN K. HADLEY, M.D., is on staff at the Middlesex Hospital Family Practice Residency Program in Middletown, Conn., and is assistant professor at the University of Connecticut School of Medicine, Hartford. Dr. Hadley received her medical degree from the University of Arizona College of Medicine, Tucson, and completed a family practice residency at Montefiore Medical Center, Bronx, N.Y.
STEPHEN M. GAARDER, PH.D., is director of Applied Research Group in Middletown, Conn., which specializes in research and evaluation for health and social service providers. He received his doctorate degree from the University of Arizona.
Address correspondence to Susan K. Hadley, M.D., Middlesex Hospital Family Practice Residency Program, 90 S. Main St., Middletown, CT 06457. (e-mail: Susan_Hadley@midhosp.org). Reprints are not available from the authors.
Author disclosure: Nothing to disclose.
Mertz HR. Irritable bowel syndrome. N Engl J Med. 2003;349:2136-46.
Thompson WG, Longstreth GF, Drossman DA, Heaton KW, Irvine EJ, Müller-Lissner SA. Functional bowel disorders and functional abdominal pain. Gut. 1999;45:43-7.
Olden KW. Irritable bowel syndrome: an overview of diagnosis and pharmacologic treatment. Cleve Clin J Med. 2003;70(suppl 2):S3-7.
American College of Gastroenterology Functional Gastrointestinal Disorders Task Force. Evidence-based position statement on the management of irritable bowel syndrome in North America. Am J Gastroenterol. 2002;97(11 suppl):S1-5.
Talley NJ, Spiller R. Irritable bowel syndrome: a little understood organic bowel disease?. Lancet. 2002;360:555-64.
Nanda R, James R, Smith H, Dudley CR, Jewell DP. Food intolerance and the irritable bowel syndrome. Gut. 1989;30:1099-104.
Friedman G. Diet and the irritable bowel syndrome. Gastroenterol Clin North Am. 1991;20:313-24.
Brandt LJ, Bjorkman D, Fennerty MB, Locke GR, Olden K, Peterson W, et al. Systematic review on the management of irritable bowel syndrome in North America. Am J Gastroenterol. 2002;97(11 suppl):S7-26.
Jailwala J, Imperiale TF, Kroenke K. Pharmacologic treatment of the irritable bowel syndrome: a systematic review of randomized, controlled trials. Ann Intern Med. 2000;133:136-47.
Parisi GC, Zilli M, Miani MP, Carrara M, Bottona E, Verdianelli G, et al. High-fiber diet supplementation in patients with irritable bowel syndrome (IBS): a multicenter, randomized, open trial comparison between wheat bran diet and partially hydrolyzed guar gum (PHGG). Dig Dis Sci. 2002;47:1696-704.
Poynard T, Regimbeau C, Benhamou Y. Meta-analysis of smooth muscle relaxants in the treatment of irritable bowel syndrome. Aliment Pharmacol Ther. 2001;15:355-61.
Viera AJ, Hoag S, Shaughnessy J. Management of irritable bowel syndrome. Am Fam Physician. 2002;66:1867-74.
Spiller R. Pharmacotherapy: non-serotonergic mechanisms. Gut. 2002;51(suppl 1):i87-90.
Clouse RE. Antidepressants for irritable bowel syndrome. Gut. 2003;52:598-9.
Bueno L, Fioramonti J, Delvaux M, Frexinos J. Mediators and pharmacology of visceral sensitivity: from basic to clinical investigations. Gastroenterology. 1997;112:1714-43.
Jackson JL, O'Malley PG, Tomkins G, Balden E, Santoro J, Kroenke K. Treatment of functional gastrointestinal disorders with antidepressant medications: a meta-analysis. Am J Med. 2000;108:65-72.
Tabas G, Beaves M, Wang J, Friday P, Mardini H, Arnold G. Paroxetine to treat irritable bowel syndrome not responding to high-fiber diet: a double-blind, placebo-controlled trial. Am J Gastroenterol. 2004;99:914-20.
Drossman DA, Camilleri M, Mayer EA, Whitehead WE. AGA technical review on irritable bowel syndrome. Gastroenterology. 2002;123:2108-31.
Drossman DA, Toner BB, Whitehead WE, Diamant NE, Dalton CB, Duncan S, et al. Cognitive-behavioral therapy versus education and desipramine versus placebo for moderate to severe functional bowel disorders. Gastroenterology. 2003;125:19-31.
Camilleri M, Chey WY, Mayer EA, Northcut AR, Heath A, Dukes GE, et al. A randomized controlled trial of serotonin type 3 receptor antagonist alosetron in women with diarrhea-predominant irritable bowel syndrome. Arch Intern Med. 2001;161:1733-40.
Talley NJ. Serotoninergic neuroenteric modulators. Lancet. 2001;358:2061-8.
Evans BW, Clark WK, Moore DJ, Whorwell PJ. Tegaserod for the treatment of irritable bowel syndrome. Cochrane Database Syst Rev. 2004(1):CD003960.
Tegaserod maleate (Zelnorm) for IBS with constipation. Med Lett Drugs Ther. 2002;44:79-80.
2004 safety alerts for drugs, biologics, medical devices, and dietary supplements. Accessed online August 24, 2005, at:
Spanier JA, Howden CW, Jones MP. A systematic review of alternative therapies in the irritable bowel syndrome. Arch Intern Med. 2003;163:265-74.
Camilleri M, Kim DY, McKinzie S, Kim HJ, Thomforde GM, Burton DD, et al. A randomized, controlled exploratory study of clonidine in diarrhea-predominant irritable bowel syndrome. Clin Gastroenterol Hepatol. 2003;1:111-21.
Pittler MH, Ernst E. Peppermint oil for irritable bowel syndrome: a critical review and meta-analysis. Am J Gastroenterol. 1998;93:1131-5.
Liu JH, Chen GH, Ye HZ, Huang CK, Poon SK. Entericcoated peppermint-oil capsules in the treatment of irritable bowel syndrome: a prospective, randomized trial. J Gastroenterol. 1997;32:765-8.
Bensoussan A, Talley NJ, Hing M, Menzies R, Guo A, Ngu M. Treatment of irritable bowel syndrome with Chinese herbal medicine: a randomized controlled trial. JAMA. 1998;280:1585-9.
Hadley SK, Petry JJ. Medicinal herbs: a primer for primary care. Hosp Pract (Off Ed). 1999;34:105-6.
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10695 | https://emanuel.mit.edu/hurricane-potential-intensity-maps/ | Hurricane Potential Intensity Maps – Kerry Emanuel
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Kerry Emanuel
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Hurricane Potential Intensity Maps
Interpretation
Two maps are available at each analysis or forecast time: A map showing the minimum attainable central pressure of tropical cyclones (mb), and a map showing the maximum sustainable surface wind speed. The latter is the maximum gradient wind at low levels reduced by 20% to account for the reduction of surface winds by friction. The method used in calculating these quantities is described in a separate document – “Maximum Intensity Estimation”eqs. Experience using potential intensity estimates based on climatology indicate that few, if any, storms will ever exceed the potential intensity value; most storms will not achieve their potential intensity (see figure).new page
On the other hand, there has been almost no experience to date using potential intensity calculations from real-time data. Hurricanes cannot be expected to develop where the potential intensity estimate is small or zero; but storms moving rapidly from regions of high potential intensity to regions of low intensity can be expected to maintain strength well in excess of the local potential intensity estimate.
Where a storm already exists and its presence is reflected in the NCEP data, the potential intensity estimate may be in error at the location of the storm, because the calculation of the potential intensity assumes that the vertical sounding is an environmental sounding.
Atmospheric data used in these calculations reflects the latest data set available from NCEP; generally the 1 x 1 degree gridded data. Sea surface temperatures are updated weekly. The date on the maps is that of the atmospheric gridded data used.
Location of maps
The maps may be obtained by clicking here. I am most grateful to Jim Kinter, Paul Dirmeyer, Brian Doty and Jennifer Adams of the Center for Ocean-Land-Atmosphere Studies for preparing and maintaining these maps. Information about the graphics package used to make these charts as well as other weather information can be obtained directly from the COLA Home Page.
Comments
Please direct any comments about the format of the maps or the algorithm used to calculate potential intensity to me at [mylastname] @ mit.edu
Kerry Emanuel 1 July 1996
Massachusetts Institute of Technology
77 Massachusetts Avenue, Cambridge, MA, USA
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10696 | https://chem.libretexts.org/Courses/Minnesota_State_Community_and_Technical_College/CHEM_1112%3A_General_Inorganic_Chemistry_II_Lab_Manual-Online_Section/02%3A_Experiments/2.08%3A_Kinetics-_Iodine_Clock_using_Vitamin_C_and_the_Calculation_of_Initial_Rate-Home | Skip to main content
2.8: Kinetics- Iodine Clock using Vitamin C and the Calculation of Initial Rate-Home
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: Jun 27, 2022
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2.7: Volumetric Titration-Home
2.9: Determination of the Molar Mass by Freezing Point Depression (Experiment)-Home Version
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Learning Objectives
To observe and record the time needed for a reaction to happen
To develop skills to set-up a redox reaction
To complete the calculations to find out the initial rates
To graphically analyze the data to find the rate law for the reaction
Theory
Vitamin C (Ascorbic acid C6H8O6) is oxidized by free iodine molecule as follows. This reaction is often used in volumetry to quantify the amount of Vitamin C in a sample. This we will see in the titration lab experiment. Since iodine is involved, and these are redox reactions not neutralization reactions, these titrations are called iodometric titrations. However, in kinetics lab experiments, this reaction is often used as an example where the speed of a reaction is measured or rather calculated. The following equation shows the 1:1 reaction between Vitamin C (Ascorbic acid) and Iodine.
C6H8O6 + I2 à C6H6O6 + 2I- + 2H+
Iodine (I2) solution is a brown solution and dissolved Iodide (I-) ions are colorless. Free iodine in the presence of starch makes an I2-Starch complex that is dark blue in color. If there is vitamin C, it will react with I2. So, there will not be any free iodine to complex with starch to give the blue color. Once the vitamin C is oxidized completely, the free iodine will be available, and it will complex with starch to give the blue color. This is an indication of the completion of vitamin C oxidation. If you have more Vitamin C, it will take more time for the completion, and the appearance of blue colored is delayed. Therefore, the rate of the Vitamin C oxidation can be followed by the time taken for the appearance of the blue free iodine-starch complex. That is why this reaction is known as the iodine-clock reaction.
A set of four to five reactions will be set up with increasing amount of Vitamin C, keeping the amount all other reagents the same. Use a stopwatch (cellphone) to measure the time taken for the appearance of blue color in each case and are recorded. The mols of Vitamin C taken in each sample, and the corresponding time are used to calculate the initial rate of the reaction. Use a graph sheet or excel spread sheet to manipulate the data to find the rate law (zero, first, second, or?).
Watch the following video to see the set-up on iodine clock reaction at home using household items. The concentration calculations in molarity is more of an estimation as we are not using standard pure chemicals. We will be using a different amounts of reagents for our reactions. This video is meant only for a demonstration purpose.
Procedure
Materials Required
Vitamin C tablets (Nature Best brand) 500 or 1000 mg supplements, 2-3 #, Tincture iodine antiseptic (2%) 30ml, Hydrogen peroxide, 3% solution, 100 ml, Starch (corn or potato), 1-2 spoons, Beakers or clear bowls for mixing, graduated cylinder, electronic balance, spoons. No filtration is required.
Note
Wear gloves. Iodine can cause stains.
Part A: Calculating the Rate of Vitamin C Oxidation using Initial Rate Method
Even though we are doing five samples with different amounts of Vitamin C, we will do one at a time. That means prepare one set of A and B solution as shown in the video, mix and measure the time time taken for the appearance of the blue color using a stopwatch.
Crush 2-3 Vitamin C tablets.
Sample 1
Solution A: Dissolve 200 mg (0.2g) of Vitamin C powder in 25 ml of water in a beaker or a clear cup, and then add 6 ml 2% tincture iodine solution: Solution B: Dissolve a pinch of starch in 15 ml 3% Hydrogen peroxide, add another 16 ml water to this. Mix solutions A and B and start the stop-watch immediately to measure the time taken for the appearance of the signature blue color. The total volume of the solution is 62 ml.
Sample 2
Repeat the above procedure, but with 400 mg (0.4g) Vitamin C in Solution A. Keep the measurement of all other reagents the same as in sample 1. Prepare B with no change. Record the time take to see the blue color after mixing A and B.
Sample 3, 4, and 5. Repeat the above procedure with various amounts of Vitamin C in making solution A, 600 mg (0.6g) in the third sample, 800 mg (0.8 g) in the fourth, and 1000 mg (1.0 g) in the 5th. Remember to keep the amount of all other ingredients the same in all the repetitions for preparing solutions A and B.. Record the time taken to see the blue color in each experiment.
Part B-Graphical Analysis to Find the Rate Law (Graph Paper or Excel Spreadsheet)
Create a table with Concentration of Vitamin C (millimols) in each sample versus the time recorded
Create two other manipulated tables. Table 2, take the natural logarithm of the concentrations of Vitamin C vs the time. Table 3, take the inverse of all the Vitamin C concentrations versus the time.
Check which of the graphs give a straight line. If it is the first one, the rate law is zero order. If it is the second, it is a first order rate law. The thirs one would be second-order if it gives a straight line.
Report Sheet Part A: Calculating the Rate of Vitamin C Oxidation using Initial Rate Method
The vitamin C tables are not standardized are pure ascorbic acid available in a real laboratory. In order to find the concentration or mols of vitamin C in a particular amount of the tablet, it is titrated with a standardized base like NaOH. Since we do not have NaOH. We are going to calculate the millimols/mg of the tablet based on the titration values given in the video. This is why we need to use the same brand of Vitamin C tables as used in this reaction shown in the video. This can be used to find the mols of Vitamin C used in the following 5 samples.
M1V1=mols of NaOH. Since it is a 1:1 reaction between Vitamin C and NaOH, the number of mols of Vitamin C should be equal to the mols of NaOH used. Divide the mols of Vitamin C with the weight in milli grams of the tablet used we will get the mols per every mg of Vitamin C tablet. This calculations in the video is used to find the amount of Vitamin C is milli mol/mg of the tablet. We are going to borrow this to calculate the mols of Vitamin C in the report table for each sample.
It comes to 0.001milli mol Vitamin C per every milli gram of the Vitamin C tablet. For example, for sample 1
Molarity of Vitamin C would be
200mg Vitamin C tablet X 0.001mmol Vitamin C/mg tablet = 0.2mmol Vitamin C in sample 1
To calculate molarity M, divide the mols by volume of the solution (62 ml is the total volume of solutions in all the samples).
For sample 1, it would be Molarity, M=0.2mmol/63 ml = 0.003M. Calculate the molarity of vitamin C for all the other sample using the same method and enter in the table below.
Initial Rate calculation
| | | | | |
--- ---
| | | | | |
| Sample Vitamin C in grams | Mols of Vitamin C | Molarity, M of Vitamin C milli mol/milliliter | Time recorded Δt | Rate calculated = Rate= (-) Δ [Vitamin C]/Δt |
| Sample 1-200mg (0.2 g) Vitamin C | 200mg tablet X0.001mmol Vitamin C /mg tablet =0.2mmol Vitamin C | 0.2 mmol/62ml=0.003M | | (-)(0-0.003) M/(Δt) s =0.003M/(Δt)s = |
| Sample 1-400mg (0.4 g) Vitamin C | | | | |
| Sample 1-600mg (0.6 g)Vitamin C | | | | |
| Sample 1-800mg(0.8 g) Vitamin C | | | | |
| Sample 1-1000mg (1.0 g) Vitamin C | | | | |
Report Sheet-Graphical: Vitamin C Analysis to Find the Rate Law
Complete the following tables and plot the graphs of dependent variable vs. independent variable (t) in each case and check for the rate law. Watch the following video to review the concept.
Case 1
| | |
--- |
| Concentration = Molarity of Vitamin C | Time, s |
| Sample 1 =0.003M | |
| Sample 2 | |
| Sample 3 | |
| Sample 4 | |
| Sample 5 | |
Case 2
| | |
--- |
| Ln (concentration) | Time, s |
| Sample 1, Ln (0.003) = (-5.809) | |
| | |
| | |
| | |
| | |
Case 3
| | |
--- |
| 1/(Concentration of Vitamin C) 1/M | Time, s |
| Sample 1= 1/(0.003)= 333.3 | |
| | |
| | |
| | |
| | |
Data Analysis
Are the initial rates for all the samples the same?
Is the initial rate increasing or decreasing with increasing concentration of Vitamin C?
How would the temperature impact the rate? Will it increase or decrease the speed of the reaction?
The change in concentration is (final concentration - initial concentration) for each sample. What is the final concentration of Vitamin C for all the samples?
In PART B, which of the three cases resulted in a straight line plot? Include all the three graphs with the lab report.
What is the rate law for the reaction, zero, first, or second? Write down the rate law equation.
Calculate the rate constant, k from the slope of the straight-line plot. k=
Contributors and Attributions
Manjusha Saraswathiamma, Minnesota State University, Moorhead has developed this experiment to be performed at home with less hazardous and cost-effective lab supplies. Manjusha would like to acknowledge all the contributors of the YouTube videos embedded on this page.
2.7: Volumetric Titration-Home
2.9: Determination of the Molar Mass by Freezing Point Depression (Experiment)-Home Version |
10697 | https://books.google.com/books/about/Fundamentals_of_Physics_Chapters_1_11.html?id=yx2hLqTFKLwC | Fundamentals of Physics, Chapters 1-11 - David Halliday, Robert Resnick, Jearl Walker - Google Books
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Fundamentals of Physics, Chapters 1-11 ====================================== David Halliday, Robert Resnick, Jearl Walker John Wiley & Sons, Nov 23, 2009 - Science - 352 pages Measurement -- Motion along a straight line -- Vectors -- Motion in two and three dimensions -- Force and motion I -- Force and motion II -- Kinetic energy and work -- Potential energy and conservation of energy -- Center of mass and linear momentum -- Rotation -- Rolling, torque, and angular momentum. More » Preview this book »
Contents
Section 1 168
Section 2 173
Section 3 176
Section 4 166
Section 5 168
Section 6 173
Section 7 176
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8-8 CONSERVATIONAEmecAEthair dragangleapplied forceaxisballblock and floorblock slidesCalculationsCHAPTER 8 POTENTIALcoefficient of kineticCONSERVATION OF ENERGYconservation of mechanicalconservative force actscratedecreasedistanceEarth systemelastic potential energyENERGY AND CONSERVATIONenergy statementenergy transfersequationergyexternal forceforce Fforce F(xforce is conservativefricgravitational forcegravitational potential energygroundheight hhighest pointhorizontalincrease in thermalinitial speedisolated systemKEY IDEAkg blockkinetic energykinetic frictional forcelowest pointmagnitudemaximum heightmechanical energy Emecmg ymomentarily stopsmotionmoves from pointnegativenonconservative forcenormal forcepackageparticle movesparticle’spathplotpositionpotential energy functionpushreleased from restspring constantspring forcespring is compressedspring of springthermal energytional forcetomatototal energyturning pointvelocityverticalzero
Bibliographic information
Title Fundamentals of Physics, Chapters 1-11
Part 1 of Fundamentals of Physics
AuthorsDavid Halliday, Robert Resnick, Jearl Walker
Edition 9
Publisher John Wiley & Sons, 2009
ISBN 047054791X, 9780470547915
Length 352 pages
SubjectsScience
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10698 | https://en.wikipedia.org/wiki/Elementary_flow | Jump to content
Elementary flow
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From Wikipedia, the free encyclopedia
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| | This article includes a list of general references, but it lacks sufficient corresponding inline citations. Please help to improve this article by introducing more precise citations. (February 2018) (Learn how and when to remove this message) |
In the larger context of the Navier-Stokes equations (and especially in the context of potential theory), elementary flows are basic flows that can be combined, using various techniques, to construct more complex flows. In this article the term "flow" is used interchangeably with the term "solution" due to historical reasons.
The techniques involved to create more complex solutions can be
for example by superposition, by techniques such as topology or considering them as local solutions on a certain neighborhood, subdomain or boundary layer and to be patched together. Elementary flows can be considered the basic building blocks (fundamental solutions, local solutions and solitons) of the different types of equations derived from the Navier-Stokes equations. Some of the flows reflect specific constraints such as incompressible or irrotational flows, or both, as in the case of potential flow, and some of the flows may be limited to the case of two dimensions.
Due to the relationship between fluid dynamics and field theory, elementary flows are relevant not only to aerodynamics but to all field theory in general. To put it in perspective boundary layers can be interpreted as topological defects on generic manifolds, and considering fluid dynamics analogies and limit cases in electromagnetism, quantum mechanics and general relativity one can see how all these solutions are at the core of recent developments in theoretical physics such as the ads/cft duality, the SYK model, the physics of nematic liquids, strongly correlated systems and even to quark gluon plasmas.
Two-dimensional uniform flow
[edit]
For steady-state, spatially uniform flow of a fluid in the xy plane, the velocity vector is
where
: is the absolute magnitude of the velocity (i.e., );
: is the angle the velocity vector makes with the positive x axis ( is positive for angles measured in a counterclockwise sense from the positive x axis); and
: and are the unit basis vectors of the xy coordinate system.
Because this flow is incompressible (i.e., ) and two-dimensional, its velocity can be expressed in terms of a stream function, :
where
and is a constant.
In cylindrical coordinates:
and
This flow is irrotational (i.e., ) so its velocity can be expressed in terms of a potential function, :
where
and is a constant.
In cylindrical coordinates
Two-dimensional line source
[edit]
The case of a vertical line emitting at a fixed rate a constant quantity of fluid Q per unit length is a line source. The problem has a cylindrical symmetry and can be treated in two dimensions on the orthogonal plane.
Line sources and line sinks (below) are important elementary flows because they play the role of monopole for incompressible fluids (which can also be considered examples of solenoidal fields i.e. divergence free fields). Generic flow patterns can be also de-composed in terms of multipole expansions, in the same manner as for electric and magnetic fields where the monopole is essentially the first non-trivial (e.g. constant) term of the expansion.
This flow pattern is also both irrotational and incompressible.
This is characterized by a cylindrical symmetry:
Where the total outgoing flux is constant
Therefore,
This is derived from a stream function
or from a potential function
Two-dimensional line sink
[edit]
The case of a vertical line absorbing at a fixed rate a constant quantity of fluid Q per unit length is a line sink. Everything is the same as the case of a line source a part from the negative sign.
This is derived from a stream function
or from a potential function
Given that the two results are the same a part from a minus sign we can treat transparently both line sources and line sinks with the same stream and potential functions permitting Q to assume both positive and negative values and absorbing the minus sign into the definition of Q.
Two-dimensional doublet or dipole line source
[edit]
If we consider a line source and a line sink at a distance d we can reuse the results above and the stream function will be
The last approximation is to the first order in d.
Given
It remains
The velocity is then
And the potential instead
Two-dimensional vortex line
[edit]
This is the case of a vortex filament rotating at constant speed, there is a cylindrical symmetry and the problem can be solved in the orthogonal plane.
Dual to the case above of line sources, vortex lines play the role of monopoles for irrotational flows.
Also in this case the flow is also both irrotational and incompressible and therefore a case of potential flow.
This is characterized by a cylindrical symmetry:
Where the total circulation is constant for every closed line around the central vortex
and is zero for any line not including the vortex.
Therefore,
This is derived from a stream function
or from a potential function
Which is dual to the previous case of a line source
Generic two-dimensional potential flow
[edit]
Given an incompressible two-dimensional flow which is also irrotational we have:
Which is in cylindrical coordinates
We look for a solution with separated variables:
which gives
Given the left part depends only on r and the right parts depends only on , the two parts must be equal to a constant independent from r and . The constant shall be positive[clarification needed].
Therefore,
The solution to the second equation is a linear combination of and
In order to have a single-valued velocity (and also a single-valued stream function) m shall be a positive integer.
therefore the most generic solution is given by
The potential is instead given by
References
[edit]
Fitzpatrick, Richard (2017), Theoretical fluid dynamics, IOP science, ISBN 978-0-7503-1554-8
Faber, T.E. (1995), Fluid Dynamics for Physicists, Cambridge university press, ISBN 9780511806735
Specific
^ Oliver, David (2013-03-14). The Shaggy Steed of Physics: Mathematical Beauty in the Physical World. Springer Science & Business Media. ISBN 978-1-4757-4347-0.
^ Laplace operator
Further reading
[edit]
Batchelor, G.K. (1973), An introduction to fluid dynamics, Cambridge University Press, ISBN 978-0-521-09817-5
Chanson, H. (2009), Applied Hydrodynamics: An Introduction to Ideal and Real Fluid Flows, CRC Press, Taylor & Francis Group, Leiden, The Netherlands, 478 pages, ISBN 978-0-415-49271-3
Lamb, H. (1994) , Hydrodynamics (6th ed.), Cambridge University Press, ISBN 978-0-521-45868-9
Milne-Thomson, L.M. (1996) , Theoretical hydrodynamics (5th ed.), Dover, ISBN 978-0-486-68970-8
External links
[edit]
Richard Fitzpatrick University of Texas, Austin (2017). "Fluid Mechanics". University of Texas, Austin. Retrieved 2018-02-07.
(c) Aerospace, Mechanical & Mechatronic Engg. 2005 University of Sydney (2005). "Elements of Potential Flow". University of Sydney. Retrieved 2019-04-19.{{cite web}}: CS1 maint: numeric names: authors list (link)
Retrieved from "
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Fluid dynamics
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10699 | https://uk.farnell.com/i2r-loss-definition | Reduced prices
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I2R Loss
In an ideal circuit, all the power applied to the input terminals would reach the critical load with no energy wasted or dissipated in the wiring or components along the power path. In real circuits, however, these components always have some resistance, however small. This occurs with both AC and DC supplies, causing electrical losses which are dissipated as heat. These losses can be calculated as below:
Ohm’s Law: V = IR where V = voltage (Volts) across a component, R is the component’s resistance (in Ohms) and I is the current in Amps through it.
Power Law: W = VI where V and I are as above, and W = power dissipated in Watts.
From combining these, we can see that the loss W = (IR)I or I2R. This is known as copper loss.
Copper losses can be significant in AC circuits involving wound components like transformers. Such losses occur in their windings, so they are sometimes called winding losses. However, further losses will arise from induced currents flowing through resistance in the components’ iron core; these are called core losses.
Transformer or motor copper losses can be reduced by increasing the conductor’s cross-sectional area, improving the winding technique, and using materials with higher electrical conductivities.
With high-frequencies, the proximity effect and skin effect cause the current to be unevenly distributed across the conductor, increasing its effective resistance. This can be counteracted by using Litz wire, which forces current to be distributed evenly over its cross-section.
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